Analysis für InformatikA chapter-by-chapter guide to the lecture notes
One page per chapter of the lecture script by Silke Rolles. Each page explains the chapter as one connected argument, works the matching exercise sheets into the text, and ends with the exam questions that chapter has produced in past exams, plus a short self-test.
Source labels on every box: Skript 2.3 = the lecture script · Blatt 2 · 2.2 = exercise sheet · Klausur WiSe20 = old exam · Extra = my own addition (intuition, a missing proof, an extra example).
→How the course fits together
The course is one long chain. Each chapter uses the tools of the previous ones, so it pays to read them in order:
- Foundations (1–3). ℝ has no holes (supremum axiom) → so monotone bounded sequences converge → so we can add up infinitely many numbers (series) and define \(e^x\).
- Functions (4–5). Continuity is defined through sequences. It gives the intermediate value theorem and the existence of max/min. Then ln and general powers come from inverting exp, and sin/cos are defined through \(e^{ix}\).
- Differential calculus (6–7). The derivative is the best linear approximation (with Landau symbols). The mean value theorem turns derivative information into statements about the function: monotonicity, extrema, l’Hospital.
- Integral calculus (8–9). The integral is a limit of sums; the fundamental theorem links it to derivatives. Then improper integrals and the comparison between sums and integrals.
- Power series (10). Taylor polynomials approximate functions to higher order; power series are "infinite polynomials".
- Several variables and ODEs (11–12). Gradient and Hessian generalise f′ and f″; differential equations use everything from integration.
§Chapters
!What the exams ask
I sorted every task in the nine uploaded exams (WiSe18 to WiSe24) by topic. A dot means the topic appears in that exam. This shows what has been asked before; it does not predict the next exam.
Reading the table: series convergence, integration and computing limits appear in practically every exam. Gradient/Hessian extrema and ODEs appear in every recent Rolles/Gantert exam. Sup/inf of a set is a common warm-up task.
Basics: sets, maps, fieldsGrundlagen: Mengen, Abbildungen, Körper
The script starts with the sentence "\((\mathbb R,+,\cdot)\) ist ein kommutativer Körper" and moves on. This page explains everything that sentence and the first chapters take for granted: the language of sets, the basic proof techniques, maps (injective, surjective, bijective), the algebraic structures group, ring and field, ordered fields, and the algebra of powers and roots with the domains each rule needs. Only a handful of exam items ask about this material directly (they are listed at the end of B.3); everything else in the course uses it silently. The last section is a German–English list for reading the exam paper itself.
Source labels: Skript §1.1 where the script uses the idea · Extra background the script assumes (everything on this page is background, so most boxes carry this label).
0Why this page, and how to use it
Analysis builds numbers up from a handful of rules. To follow that, you need to know what a "rule for numbers" is — that is what fields and ordered fields are — and how mathematicians argue with such rules — that is what proofs and quantifiers are. If you already know these topics from Diskrete Strukturen or Lineare Algebra, skim the summary at the end and do the self-test. If not, read the page top to bottom once; it takes about an hour.
| Topic on this page | Where the course needs it |
|---|---|
| Sets, \(\in,\subseteq,\cup,\cap,\times\), set-builder notation | every page — e.g. \(\{x\in\mathbb R: x^2>1\}\) in the sup/inf exam tasks |
| Implication, contrapositive, contradiction, induction | every proof; contradiction for \(\sqrt2\notin\mathbb Q\), Cantor, Archimedes (Ch. 1) |
| Injective / surjective / bijective, inverse map | countability (Ch. 1), inverse functions ln, arcsin (Ch. 5, 6) |
| Field axioms and what follows from them | "ℝ is a field" (Ch. 1); every algebraic manipulation you ever do |
| Ordered field and the rules for inequalities | "ℂ cannot be ordered" (Ch. 1); every ε-estimate from Ch. 2 on |
| Powers and roots: the five rules, \(\sqrt{x^2}=|x|\), fractional and negative exponents | root limits and the conjugate trick (Ch. 2); \(x^a:=e^{a\ln x}\) (Ch. 5); differentiating \(x^x\) (Ch. 6, 7); every integral of \(x^a\) (Ch. 8) |
B.1Sets and notation [Mengen]
Intuition. A set is a collection of objects, called its elements. Order and repetition do not matter: \(\{1,2,3\}=\{3,1,2\}=\{1,1,2,3\}\). A set is completely determined by which elements it contains.
| symbol | read as | example |
|---|---|---|
| \(a\in A\), \(a\notin A\) | \(a\) is (not) an element of \(A\) | \(2\in\mathbb N\), \(-1\notin\mathbb N\) |
| \(A\subseteq B\) | \(A\) is a subset of \(B\): every element of \(A\) is in \(B\) | \(\mathbb N\subseteq\mathbb Z\) |
| \(A\subsetneq B\) | proper subset: \(A\subseteq B\) and \(A\neq B\) | \(\mathbb Q\subsetneq\mathbb R\) |
| \(\emptyset\) | the empty set [leere Menge] | \(\{x\in\mathbb R:x^2<0\}=\emptyset\) |
| \(A\cup B\) | union [Vereinigung]: in \(A\) or in \(B\) (or both) | \(\{1,2\}\cup\{2,3\}=\{1,2,3\}\) |
| \(A\cap B\) | intersection [Schnitt]: in \(A\) and in \(B\) | \(\{1,2\}\cap\{2,3\}=\{2\}\) |
| \(A\setminus B\) | difference: in \(A\) but not in \(B\) | \(\mathbb R\setminus\{0\}\) = all nonzero reals |
| \(A\times B\) | Cartesian product: all pairs \((a,b)\) with \(a\in A,\ b\in B\) | \(\mathbb R\times\mathbb R=\mathbb R^2\) (the plane) |
| \(\{x\in A:\ \text{condition}\}\) | "all \(x\) in \(A\) such that …" (set-builder notation) | \(\{x\in\mathbb R:|x-1|<1\}=(0,2)\) |
| \(\{f(n):n\in\mathbb N\}\) | the set of all values \(f(1),f(2),\dots\) | \(\{\frac1n:n\in\mathbb N\}=\{1,\frac12,\frac13,\dots\}\) |
Four things in these examples get their proper definition in Chapter 1 and are used here only as familiar notation: the number sets \(\mathbb N=\{1,2,3,\dots\}\) — in this course without 0 — and \(\mathbb Z,\mathbb Q,\mathbb R\) (§1.1); intervals such as \((0,2)=\{x:0<x<2\}\) (§1.2); the absolute value \(|x|\), the distance from \(x\) to 0 (§1.3); and countability, which is what maps are used for there.
Show both inclusions: \(A\subseteq B\) (take any \(x\in A\), show \(x\in B\)) and \(B\subseteq A\). This "⊆ and ⊇" pattern appears in the script, e.g. in Lemma 12.6(c).
Example: \(\{x\in\mathbb R:|x-1|<1\}=(0,2)\). "⊆": if \(|x-1|<1\), then \(-1<x-1<1\), so \(0<x<2\). "⊇": if \(0<x<2\), then \(-1<x-1<1\), so \(|x-1|<1\). This exact set is an exam question (SoSe19 A1c, worked in Chapter 1).
Reading sums and products: \(\sum\) and \(\prod\) Extra
\[\sum_{k=m}^{n}a_k=a_m+a_{m+1}+\dots+a_n,\qquad \prod_{k=m}^{n}a_k=a_m\cdot a_{m+1}\cdots a_n.\] \(k\) is a dummy index: its name carries no meaning, \(\sum_{k=1}^3a_k=\sum_{i=1}^3a_i\). Conventions: an empty sum (\(n<m\)) is \(0\), an empty product is \(1\) — this is why \(0!=\prod_{k=1}^{0}k=1\).
Examples: \(\sum_{k=1}^nk=\frac{n(n+1)}2\); \(n!=\prod_{k=1}^nk\); \(\binom nk=\frac{n!}{k!(n-k)!}\); the series of Chapter 3 are sums with \(n\to\infty\).
- Linearity: \(\sum_k(\alpha a_k+\beta b_k)=\alpha\sum_ka_k+\beta\sum_kb_k\). Constants come out, sums split — but only term by term.
- Splitting off a term: \(\sum_{k=0}^{n}a_k=a_0+\sum_{k=1}^{n}a_k=\sum_{k=0}^{n-1}a_k+a_n\). Used whenever a formula starts at a different index than your sum (Chapter 3 does this constantly).
- Index shift [Indexverschiebung]: substituting \(j=k-1\) shifts both limits, \(\sum_{k=1}^{n}a_k=\sum_{j=0}^{n-1}a_{j+1}\). It is the discrete version of a substitution in an integral (§3.1, and in every power-series computation in Chapter 10).
- Telescoping: if \(a_k=b_k-b_{k+1}\), almost everything cancels: \(\sum_{k=m}^{n}a_k=b_m-b_{n+1}\) (§3.1).
Warning: \(\sum_k a_kb_k\ne\left(\sum_ka_k\right)\left(\sum_kb_k\right)\). Multiplying two sums produces all pairs \(a_ib_j\); sorting them by \(i+j\) is exactly the Cauchy product of §3.3.
B.2Logic and proof techniques [Beweistechniken]
- \(A\Rightarrow B\) ("\(A\) implies \(B\)", "if \(A\) then \(B\)") is false only when \(A\) is true and \(B\) false. It says nothing when \(A\) is false.
- \(A\Leftrightarrow B\) ("\(A\) if and only if \(B\)", "iff", [genau dann, wenn]): both \(A\Rightarrow B\) and \(B\Rightarrow A\).
- \(B\) is necessary for \(A\) [notwendig] if \(A\Rightarrow B\); \(A\) is sufficient for \(B\) [hinreichend]. Example (Ch. 3): "\(a_k\to0\)" is necessary for \(\sum a_k\) to converge, but not sufficient.
\(A\Rightarrow B\) does not give \(B\Rightarrow A\). "Differentiable ⇒ continuous" is true (Ch. 6); "continuous ⇒ differentiable" is false (\(|x|\) at 0). What is equivalent to \(A\Rightarrow B\) is the contrapositive [Kontraposition] \(\neg B\Rightarrow\neg A\): "not continuous ⇒ not differentiable".
Four proof techniques cover almost every proof in the course:
| technique | to prove \(A\Rightarrow B\) you … | example in the course |
|---|---|---|
| direct | assume \(A\), derive \(B\) step by step | triangle inequality (Satz 1.12) |
| contrapositive | assume "not \(B\)", derive "not \(A\)" | "if \(a_k\not\to0\), the series diverges" (Lemma 3.4) |
| contradiction [Widerspruch] | assume \(A\) and "not \(B\)", derive something impossible | \(\sqrt2\notin\mathbb Q\), Cantor, Archimedes (Ch. 1) |
| induction [Induktion] | for statements about all \(n\in\mathbb N\): base case + step \(n\to n+1\) | Bernoulli's inequality (Satz 2.18) — recipe in Ch. 2 |
- Direct. Claim: if \(a\) and \(b\) are even, so is \(a+b\). Proof: even means \(a=2m\), \(b=2k\) with \(m,k\in\mathbb Z\). Then \(a+b=2(m+k)\), and \(m+k\in\mathbb Z\), so \(a+b\) is even. ∎
- Contrapositive. Claim: if \(n^2\) is even, then \(n\) is even. A direct attack would have to take a square root; the contrapositive is easy. Proof: assume \(n\) is odd, \(n=2m+1\). Then \(n^2=4m^2+4m+1=2(2m^2+2m)+1\) is odd. That is "not \(B\) ⇒ not \(A\)", which is the same statement. ∎ (This is the step the script needs inside the \(\sqrt2\notin\mathbb Q\) proof, §1.1.)
- Contradiction. Claim: there is no smallest positive real number. Proof: suppose \(x>0\) were the smallest. Then \(\frac x2\) is again positive and \(\frac x2<x\) — a positive number smaller than the smallest one. Contradiction, so no such \(x\) exists. ∎
Notice the shape: a proof starts by writing down what the assumption means (a definition unpacked into a formula) and ends with what the claim means. Most exam proofs in this course are three to six such lines.
Quantifiers (\(\forall\), \(\exists\)) and how to negate them are explained at the start of Chapter 1 (§1.1 · Quantifiers); induction is explained with Bernoulli's inequality in Chapter 2. "o.B.d.A." [ohne Beschränkung der Allgemeinheit] = "without loss of generality": the other case works the same way.
B.3Maps [Abbildungen, Funktionen]
Intuition. A map \(f:A\to B\) is a rule that assigns to every element \(a\) of \(A\) exactly one element \(f(a)\) of \(B\). Picture arrows: every element of \(A\) sends out exactly one arrow into \(B\).
- \(A\) is the domain [Definitionsbereich], \(B\) the codomain [Zielmenge], and \(f(A)=\{f(a):a\in A\}\) the image [Bild].
- \(f\) is injective [injektiv] if different inputs give different outputs: \(f(a_1)=f(a_2)\Rightarrow a_1=a_2\). No two arrows land on the same point.
- \(f\) is surjective [surjektiv] if every \(b\in B\) is hit: \(\forall b\in B\ \exists a\in A: f(a)=b\). Every point of \(B\) receives at least one arrow.
- \(f\) is bijective [bijektiv] if it is both. Every point of \(B\) receives exactly one arrow. Then the arrows can be reversed: the inverse map \(f^{-1}:B\to A\) sends \(b\) to the unique \(a\) with \(f(a)=b\).
- The composition [Verkettung] of \(g:A\to B\) and \(f:B\to C\) is \(f\circ g:A\to C\), \(x\mapsto f(g(x))\) — "first \(g\), then \(f\)". The identity \(\operatorname{id}_A(x)=x\) satisfies \(f^{-1}\circ f=\operatorname{id}_A\) and \(f\circ f^{-1}=\operatorname{id}_B\) for bijective \(f\) — this is Satz 5.3 in the script.
- The preimage [Urbild] of a set \(C\subseteq B\) is \(f^{-1}(C)=\{a\in A:f(a)\in C\}\). It exists for every map — \(f\) need not be bijective, and \(f^{-1}(C)\) may be empty or large. Do not confuse this set with the inverse map \(f^{-1}\), which only exists when \(f\) is bijective. Example: for \(f(x)=x^2\), \(f^{-1}(\{4\})=\{-2,2\}\).
- The restriction [Einschränkung] \(f|_{A_0}\) is the same rule with the smaller domain \(A_0\subseteq A\). This is how Chapter 5 makes \(\sin\) invertible: \(\sin\) on all of \(\mathbb R\) is not injective, but \(\sin|_{[-\pi/2,\pi/2]}\) is, and its inverse is \(\arcsin\) (§5.3).
- \(f:\mathbb R\to\mathbb R,\ x\mapsto x^2\): not injective (\(f(-1)=f(1)\)), not surjective (\(-1\) is never hit).
- \(f:[0,\infty)\to\mathbb R,\ x\mapsto x^2\): injective, still not surjective.
- \(f:[0,\infty)\to[0,\infty),\ x\mapsto x^2\): bijective, inverse \(y\mapsto\sqrt y\).
Same formula, three different maps: whether a map is injective/surjective depends on the chosen domain and codomain, not just on the formula. For finite sets you can read off: injective needs \(|A|\le|B|\), surjective needs \(|A|\ge|B|\), bijective needs \(|A|=|B|\) — try it in the widget.
- WiSe23 6a: is \(f:[0,\infty)\to\mathbb N_0,\ x\mapsto\lfloor x\rfloor\) injective, surjective, bijective? Answer each with a reason: one counterexample kills injectivity (\(f(6.2)=f(6.3)=6\)), and one preimage per target proves surjectivity (\(f(n)=n\)). Worked in Chapter 5.
- WiSe19 1i: draw the graph of the inverse function — mirror at the line \(y=x\) (§5.1).
- SoSe19 A1c: describing a set given by an inequality, then reading off sup/inf (Chapter 1).
- The field and order rules themselves are never asked on their own, with one exception: "ℂ cannot be ordered" (Lemma 1.4) is a standard short item — see Chapter 1.
B.4Groups, rings and fields [Gruppen, Ringe, Körper]
Why this matters. When you simplify \(\frac{2x+6}{2}=x+3\) or move a term to the other side of an equation, you use rules like "you may divide by any nonzero number". A field is exactly a set with \(+\) and \(\cdot\) in which all the familiar rules of school algebra hold. Saying "ℝ is a field" is the script's way of saying "all of that works in ℝ". Rings and groups are the same idea with fewer rules — useful to see which rule fails where.
All three are built from binary operations [Verknüpfungen]: rules that combine two elements of a set \(M\) into a third element of \(M\), like \(+\) on \(\mathbb Z\).
Group [Gruppe]
A set \(M\) with one operation \(\ast\) is a group if:
- (G1) associative [assoziativ]: \((a\ast b)\ast c=a\ast(b\ast c)\) — brackets don't matter;
- (G2) neutral element [neutrales Element]: there is an \(e\) with \(e\ast a=a\ast e=a\) for all \(a\);
- (G3) inverses [inverses Element]: for every \(a\) there is an \(a'\) with \(a\ast a'=a'\ast a=e\).
If also (G4) commutative [kommutativ]: \(a\ast b=b\ast a\), the group is abelian [abelsch].
Examples: \((\mathbb Z,+)\) with \(e=0\), inverse \(-a\). \((\mathbb Q\setminus\{0\},\cdot)\) with \(e=1\), inverse \(\frac1a\). Not groups: \((\mathbb N,+)\) — no \(0\), no negatives; \((\mathbb Z,\cdot)\) — \(2\) has no inverse in \(\mathbb Z\) since \(\frac12\notin\mathbb Z\).
Ring [Ring]
A set \(R\) with two operations \(+\) and \(\cdot\) is a ring if
- \((R,+)\) is an abelian group (neutral element called \(0\), inverse of \(a\) called \(-a\));
- \(\cdot\) is associative, and there is a \(1\) with \(1\cdot a=a\cdot1=a\);
- distributive laws [Distributivgesetze]: \(a\cdot(b+c)=a\cdot b+a\cdot c\) and \((a+b)\cdot c=a\cdot c+b\cdot c\).
If \(a\cdot b=b\cdot a\) always, it is a commutative ring. In a ring you can add, subtract and multiply — but not necessarily divide.
Examples: \(\mathbb Z\) (the model example); polynomials \(\mathbb R[x]\); \(\mathbb Z/n\) (arithmetic "mod \(n\)", see the widget); \(2\times2\) matrices — a ring that is not commutative, since in general \(AB\neq BA\).
Field [Körper]
A field is a commutative ring \(K\) with \(0\neq1\) in which every nonzero element has a multiplicative inverse: for every \(a\neq0\) there is \(a^{-1}\) with \(a\cdot a^{-1}=1\). Equivalently: \((K,+)\) and \((K\setminus\{0\},\cdot)\) are abelian groups, linked by the distributive law.
In words: a field is a set where you can add, subtract, multiply, and divide by anything except 0, with all the usual rules. Division is defined as \(\frac ab=a\cdot b^{-1}\).
| \((+)\) group | \(\cdot\) commutative | inverses for \(\cdot\) | structure | |
|---|---|---|---|---|
| \(\mathbb N\) | ✗ (no 0, no \(-a\)) | ✓ | ✗ | none of the three |
| \(\mathbb Z\) | ✓ | ✓ | ✗ (\(\frac12\notin\mathbb Z\)) | commutative ring |
| \(\mathbb Q,\ \mathbb R,\ \mathbb C\) | ✓ | ✓ | ✓ | fields |
| \(\mathbb Z/p\), \(p\) prime | ✓ | ✓ | ✓ | field (finite!) |
| \(\mathbb Z/n\), \(n\) not prime | ✓ | ✓ | ✗ (in \(\mathbb Z/6\): no \(2^{-1}\), because \(2\cdot3=0\)) | commutative ring |
| \(2\times2\) matrices | ✓ | ✗ | ✗ | non-commutative ring |
| polynomials \(\mathbb R[x]\) | ✓ | ✓ | ✗ (\(\frac1x\) is no polynomial) | commutative ring |
What follows from the field axioms
Everything you do in algebra follows from the axioms. Here are the rules the course uses silently, each with its two-line proof — they show how "obvious" rules are actually consequences of the axioms.
- \(a\cdot0=0\).
- \((-a)\cdot b=-(a\cdot b)\), and in particular \((-1)\cdot(-1)=1\).
- No zero divisors [nullteilerfrei]: \(a\cdot b=0\Rightarrow a=0\) or \(b=0\).
- \(a\cdot x=b\) with \(a\neq0\) has exactly one solution, \(x=a^{-1}b\).
Proofs
- \(a\cdot0=a\cdot(0+0)=a\cdot0+a\cdot0\) (neutral element, then distributivity). Add \(-(a\cdot0)\) to both sides: \(0=a\cdot0\).
- \(a\cdot b+(-a)\cdot b=(a+(-a))\cdot b=0\cdot b=0\) by rule 1. So \((-a)b\) is the additive inverse of \(ab\): \((-a)b=-(ab)\). With \(a=1,\ b=-1\): \((-1)(-1)=-(1\cdot(-1))=-(-1)=1\).
- Suppose \(ab=0\) and \(a\neq0\). Multiply by \(a^{-1}\): \(b=a^{-1}ab=a^{-1}\cdot0=0\).
- \(x=a^{-1}b\) works: \(a\cdot a^{-1}b=b\). If \(x\) and \(y\) both work, \(a(x-y)=0\), so \(x-y=0\) by rule 3.
Rule 3 is what you use every time you solve \((x-2)(x+3)=0\) by "one of the factors must be 0". In \(\mathbb Z/6\) it fails (\(2\cdot3=0\)), and so does that way of solving equations — see the widget.
B.5Ordered fields [angeordnete Körper]
Intuition. A field says how to calculate. An ordering adds the notion of "positive", and with it \(<\). Analysis lives on inequalities (\(|a_n-a|<\varepsilon\)), so this is the second half of what makes ℝ suitable.
A field \(K\) is ordered if there is a notion "\(a>0\)" (positive) with
- for every \(a\): exactly one of \(a=0\), \(a>0\), \(-a>0\);
- \(a>0,\ b>0\ \Rightarrow\ a+b>0\) and \(ab>0\).
Then \(a>b\) is defined as \(a-b>0\).
- Adding keeps the direction: \(a<b\Rightarrow a+c<b+c\).
- Multiplying by a positive keeps it: \(a<b,\ c>0\Rightarrow ac<bc\).
- Multiplying by a negative flips it: \(a<b,\ c<0\Rightarrow ac>bc\).
- Squares: \(a^2>0\) for \(a\neq0\); in particular \(1=1^2>0\).
- Reciprocals flip: \(0<a<b\Rightarrow\frac1b<\frac1a\).
- Transitivity: \(a<b,\ b<c\Rightarrow a<c\).
Proofs
- \((b+c)-(a+c)=b-a>0\).
- \(bc-ac=(b-a)c\), a product of two positives, so \(>0\) by (b).
- \(-c>0\), so by 2: \(a(-c)<b(-c)\), i.e. \(-ac<-bc\); add \(ac+bc\) to both sides (rule 1): \(bc<ac\).
- If \(a>0\), \(a\cdot a>0\) by (b). If \(-a>0\), then \(a^2=(-a)(-a)>0\) (using \((-1)(-1)=1\) from B.4).
- \(ab>0\), so \(\frac1{ab}>0\) (its product with \(ab\) is \(1>0\); a negative would give a negative). Multiply \(a<b\) by \(\frac1{ab}\) (rule 2): \(\frac1b<\frac1a\).
- \(c-a=(c-b)+(b-a)\), a sum of two positives.
Rule 3 is the one people get wrong: from \(-2x<6\) follows \(x>-3\), not \(x<-3\).
- \(\mathbb Q\) and \(\mathbb R\): yes, with the usual "positive".
- \(\mathbb C\): no (Lemma 1.4): rule 4 would force \(i^2=-1>0\), but \(1>0\) gives \(-1<0\).
- \(\mathbb Z/p\): no: \(1>0\) and rule (b) give \(1+1+\dots+1>0\) for any number of ones — but \(p\) ones add up to \(0\) in \(\mathbb Z/p\).
\(\mathbb Q\) and \(\mathbb R\) are both ordered fields. The difference is completeness: in ℝ every nonempty set that is bounded above has a least upper bound (the supremum axiom), in ℚ not. So ℝ is a complete ordered field — this is the precise answer to "what are the real numbers?", and it is the starting point of Chapter 1.
B.6Powers and roots [Potenzen und Wurzeln]
Intuition. \(a^n\) starts as a shorthand: "multiply \(a\) by itself \(n\) times". Everything else — \(a^0\), \(a^{-3}\), \(a^{1/2}\), \(a^{\sqrt2}\) — is a later extension, and each extension was chosen for one reason: so that the five rules in the second box below keep holding. That is worth knowing, because it explains every restriction you are about to see. The rules are easy; what costs points is the domains — which rule needs \(a\ge0\), which needs \(a>0\), and why \(\sqrt{x^2}\) is not \(x\).
The course makes all of this rigorous much later: the \(n\)-th root exists by Anwendung 4.13 (intermediate value theorem), and the general power \(x^a:=e^{a\ln x}\) is Definition 5.11. This section is the algebra you are assumed to have from school, stated with the domains the course silently relies on.
| expression | means | needs |
|---|---|---|
| \(a^n\), \(n\in\mathbb N\) | \(\underbrace{a\cdot a\cdots a}_{n\text{ factors}}\) | nothing — any \(a\) in any field |
| \(a^0\) | \(1\) | \(a\neq0\) |
| \(a^{-n}\), \(n\in\mathbb N\) | \(\frac1{a^n}\) | \(a\neq0\) |
| \(a^{1/n}=\sqrt[n]a\) | the one \(y>0\) with \(y^n=a\) | \(a>0\) (Skript 4.13; \(\sqrt[n]0=0\) is added by hand) |
| \(a^{m/n}\) | \(\left(\sqrt[n]a\right)^m=\sqrt[n]{a^m}\) | \(a>0\) |
| \(a^x\), \(x\in\mathbb R\) | \(e^{x\ln a}\) | \(a>0\) (Skript 5.11) |
Why the middle rows look the way they do: rule 1 below forces them. From \(a^n\cdot a^0=a^{n+0}=a^n\) you get \(a^0=1\); from \(a^n\cdot a^{-n}=a^0=1\) you get \(a^{-n}=\frac1{a^n}\); from \(\left(a^{1/n}\right)^n=a^{\frac1n\cdot n}=a\) you get that \(a^{1/n}\) must be an \(n\)-th root. Nothing was invented — the rules were kept and the definitions followed.
\(0^0\) is left undefined. Both \(a^0=1\) and \(0^x=0\) would apply and they disagree. In the binomial theorem (Satz 2.19) and in power series (Ch. 10) the convention \(0^0=1\) is used silently, because there the exponent is a counting index, not a limit.
For \(a,b>0\) and all real \(x,y\):
- \(a^x\,a^y=a^{x+y}\) — multiplying powers adds exponents (this one is Lemma 5.12)
- \(\dfrac{a^x}{a^y}=a^{x-y}\)
- \(\left(a^x\right)^y=a^{xy}\) — a power of a power multiplies exponents
- \((ab)^x=a^x\,b^x\)
- \(\left(\dfrac ab\right)^x=\dfrac{a^x}{b^x}\)
Read with \(x=\frac1n\), the same five rules are the root rules:
\[\sqrt[n]{ab}=\sqrt[n]a\,\sqrt[n]b,\qquad \sqrt[n]{\frac ab}=\frac{\sqrt[n]a}{\sqrt[n]b},\] \[\sqrt[m]{\sqrt[n]a}=\sqrt[mn]a,\qquad \left(\sqrt[n]a\right)^m=\sqrt[n]{a^m}=a^{m/n}.\]For integer exponents the five rules hold for every \(a,b\neq0\) — negatives included — because they are then just bookkeeping of factors. The restriction \(a>0\) is needed exactly when the exponent stops being an integer.
Proof of all five in one line each Extra
With \(a^x=e^{x\ln a}\) (Definition 5.11) everything reduces to two facts about \(\exp\) and \(\ln\): \(e^{u+v}=e^ue^v\) and \(\ln(ab)=\ln a+\ln b\).
- \(a^xa^y=e^{x\ln a}e^{y\ln a}=e^{(x+y)\ln a}=a^{x+y}\).
- Same with \(e^{-y\ln a}=\frac1{e^{y\ln a}}\).
- \(\left(a^x\right)^y=e^{y\ln(a^x)}=e^{yx\ln a}=a^{xy}\), using \(\ln(a^x)=\ln\!\left(e^{x\ln a}\right)=x\ln a\).
- \((ab)^x=e^{x\ln(ab)}=e^{x\ln a+x\ln b}=a^xb^x\).
- Rule 4 with \(b\) replaced by \(\frac1b\), since \(\ln\frac1b=-\ln b\).
The script proves only rule 1 (Lemma 5.12) and the case \(y=n\in\mathbb N\) of rule 3; the rest it uses without stating. This is why \(a>0\) appears everywhere: \(\ln a\) has to exist. ∎
- \((a+b)^n\neq a^n+b^n\). For \(n=2\) the truth is \(a^2+2ab+b^2\); in general the binomial theorem (Satz 2.19). There is no rule for a power of a sum — only for products and quotients.
- \(\sqrt{a+b}\neq\sqrt a+\sqrt b\). Test \(a=b=1\): \(\sqrt2\approx1.41\), not \(2\). What is true is \(\sqrt{a+b}\le\sqrt a+\sqrt b\) (square both sides: \(a+b\le a+2\sqrt{ab}+b\)).
- \(a^x+a^y\neq a^{x+y}\). Exponents add when powers are multiplied. Adding powers does nothing at all; \(2^3+2^4=24\), while \(2^7=128\).
- \(\left(a^x\right)^y\) and \(a^{x^y}\) are different. \(\left(2^2\right)^3=4^3=64\), but \(2^{2^3}=2^8=256\). A tower without brackets is read top-down.
- \(\frac1{a+b}\neq\frac1a+\frac1b\) — the same mistake as the first two, with exponent \(-1\).
All four wrong lines have the same shape: a rule for products being applied to a sum. If you remember only one thing from this box: powers and roots commute with \(\cdot\) and \(/\), never with \(+\) and \(-\).
- \(\sqrt{x^2}=|x|\), not \(x\). The symbol \(\sqrt{\ }\) returns the nonnegative root by definition, so for \(x=-3\): \(\sqrt{(-3)^2}=\sqrt9=3=|-3|\). Generally \(\sqrt[n]{x^n}=|x|\) for even \(n\) and \(=x\) for odd \(n\). This is the identity behind \(|x|\) (§1.3) and behind the Euclidean norm \(\|x\|=\sqrt{x_1^2+\dots+x_n^2}\) of Chapter 11.
- Even powers destroy the sign, odd powers keep it. \((-2)^4=16\), \((-2)^3=-8\). So \(x^n=c\) with \(c>0\) has two real solutions \(x=\pm\sqrt[n]c\) when \(n\) is even and one when \(n\) is odd. The \(\pm\) comes from solving the equation, never from the root symbol.
- \(\sqrt a\sqrt b=\sqrt{ab}\) needs \(a,b\ge0\). Ignore that and you get the classic fallacy \(-1=i\cdot i=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt1=1\). Rule 4 was applied outside its domain; nothing else went wrong.
- Fractional powers of negative numbers are not defined here. \((-8)^{1/3}\) has no meaning in this course, even though \(y^3=-8\) does have the solution \(y=-2\). The reason is rule 3: if \((-8)^{1/3}\) were \(-2\), then \(-2=(-8)^{1/3}=\left((-8)^2\right)^{1/6}=64^{1/6}=2\). Either a rule or the definition has to go, and the course keeps the rules by demanding \(a>0\).
- \(a^x=e^{x\ln a}\) needs \(a>0\) simply because \(\ln a\) must exist. That is why the exam functions \(x^x\), \(x^{1/x}\), \(x^a\) are always studied on \((0,\infty)\) (Ch. 6, Ch. 7) — not because someone chose a convenient domain, but because there is no other one.
- \(a^{1/n}\) is \(\sqrt[n]a\): the number \(y>0\) with \(y^n=a\). That such a \(y\) exists is not obvious — it needs completeness of ℝ (Anwendung 4.13 via the IVT). In ℚ it is false: \(\sqrt2\notin\mathbb Q\).
- \(a^{m/n}=\left(\sqrt[n]a\right)^m=\sqrt[n]{a^m}\) — both orders give the same number. Take the root first if you want small numbers: \(8^{2/3}=\left(\sqrt[3]8\right)^2=2^2=4\) beats \(\sqrt[3]{64}\).
- A minus sign in the exponent means "reciprocal", never "negative result": \(4^{-3/2}=\frac1{4^{3/2}}=\frac1{\left(\sqrt4\right)^3}=\frac18\), and \(\left(\frac14\right)^{-1/2}=4^{1/2}=2\).
- \(\sqrt{\ }\) is a function, which is why it picks the nonnegative root: \(x^2=4\) has the solutions \(\pm2\), but \(\sqrt4=2\) alone.
- Cancelling in the exponent is safe for \(a>0\) (\(a^{2/4}=a^{1/2}\)) but not in the root notation, where the radicand may be negative: \(\sqrt[4]{a^2}=\sqrt{|a|}\) is defined for every real \(a\), while \(\sqrt a\) is not.
Sanity checks (cover the right-hand sides): \(27^{-1/3}=\frac13\); \(\sqrt{9x^2}=3|x|\); \(\sqrt[4]{16x^8}=2x^2\) (no bars needed — \(x^2\ge0\) already); \(\sqrt{x^2-2x+1}=|x-1|\); \(\left(\frac{a^3b^{-1}}{a\,b^2}\right)^{-1}=\frac{b^3}{a^2}\).
- Pull the dominant power out of the root. \(\sqrt{n^2+n}=n\sqrt{1+\frac1n}\) for \(n>0\) (rule 4, backwards), and \(\sqrt{1+\frac1n}\to1\). This is how almost every root limit in Chapter 2 is done.
- Conjugate trick [Erweitern mit dem Konjugierten]. A difference of two roots that both grow is "\(\infty-\infty\)", which says nothing. Multiply by \(\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}\) and use \((u-v)(u+v)=u^2-v^2\): \[\sqrt{n+1}-\sqrt n=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt{n+1}+\sqrt n}\to0.\] Appears on Blatt 2.2 and in the exams (§2.2).
- Anything to the power of anything goes to base \(e\): \(f(x)^{g(x)}=e^{g(x)\ln f(x)}\), valid where \(f>0\). This is the only way the course differentiates \(x^x\) (Ch. 6) or evaluates \(\lim n^{1/n}\) — a power with the variable in the exponent is not a polynomial and has no other derivative rule.
- Swap base and exponent past a logarithm: \(a^{\log_bn}=n^{\log_ba}\), since both equal \(e^{\frac{\ln a\,\ln n}{\ln b}}\) (Lemma 5.13). This turns a strange-looking growth comparison into a comparison of exponents — it is exactly the step in the exam item \(3^{\log_bn}\) vs \(n^3\) (Ch. 6, SoSe23 1g).
- Compare sizes by monotonicity, not by calculator. On \([0,\infty)\) the map \(x\mapsto x^n\) is strictly increasing, so for \(0\le a<b\): \(a^n<b^n\) and \(\sqrt[n]a<\sqrt[n]b\). Raising an inequality between nonnegative numbers to a power, or taking roots, is allowed and changes nothing — unlike multiplying by a negative, which flips it (rule 3 of B.5).
One more pattern that surprises people: for \(0<a<1\) a bigger exponent gives a smaller number (\(\left(\frac12\right)^3<\left(\frac12\right)^2\)), so \(\sqrt a>a\) there, while for \(a>1\) everything is the other way round. Both follow from rule 5 of B.5 read through \(x\mapsto a^x=e^{x\ln a}\): the sign of \(\ln a\) decides the direction.
B.7Reading the exam in German [Prüfungssprache]
This guide is in English, but the exam paper is in German. Most of the mathematical vocabulary appears in brackets throughout the chapters; what is easy to overlook are the instruction words and the sentences that tell you how much you have to write. Those decide your points as much as the mathematics does.
| German | English | what is expected |
|---|---|---|
| Zeigen Sie / Beweisen Sie | show / prove | a complete argument, every step justified by a named theorem or definition |
| Begründen Sie | justify, give reasons | shorter: name the theorem or give the counterexample, no full proof |
| Berechnen Sie / Bestimmen Sie | compute / determine | a result plus the computation |
| Geben Sie … an | state … | the answer alone is enough (often "ohne Begründung", without reasons) |
| Untersuchen Sie … auf Konvergenz | investigate … for convergence | decide convergent/divergent and say which criterion you used |
| Entscheiden Sie, ob … | decide whether … | yes/no with a reason; a counterexample counts as a reason |
| Skizzieren Sie / Zeichnen Sie | sketch / draw | a labelled drawing; axes, intercepts, asymptotes |
| Kreuzen Sie an | tick | multiple choice; usually no reason needed |
| Ermitteln / Ableiten / Auflösen nach | find out / differentiate / solve for | as it says |
- "Bei dieser Aufgabe wird nur das Ergebnis bewertet." — only the result is graded. Write the answer; working earns nothing (but may still help you).
- "Bei dieser Aufgabe sind alle Zwischenschritte zu begründen." — every intermediate step must be justified. Name the theorem you use ("nach dem Mittelwertsatz", "mit dem Quotientenkriterium"); a correct result without reasons loses most of the points.
- "Es genügt, … anzugeben." — it is enough to state …; "ohne Beweis" — no proof required; "Hinweis" — hint, and it is nearly always meant to be used.
- "Aussage i)* ist schwierig und wird als Bonus gewertet." — that item is a bonus; do it last (this exact sentence is in WiSe20 A2).
- "(bitte wenden)" — please turn over; "Platz für den Rechenweg" — space for your working; "Zusatzblatt" — extra sheet.
| German | English | German | English |
|---|---|---|---|
| Sei / Seien … gegeben | let … be given | Behauptung | claim |
| für alle / es existiert (es gibt) | for all / there exists | Voraussetzung | hypothesis, assumption |
| genau dann, wenn | if and only if | Widerspruch | contradiction |
| notwendig / hinreichend | necessary / sufficient | Gegenbeispiel | counterexample |
| es gilt | we have, it holds that | o.B.d.A. | without loss of generality |
| höchstens / mindestens / genau | at most / at least / exactly | paarweise verschieden | pairwise distinct |
| beliebig (beliebig oft) | arbitrary (arbitrarily often) | jeweils | in each case |
| Ansatz | trial form you insert (Ch. 12) | Probe | check by substituting back |
| Satz / Lemma / Korollar | theorem / lemma / corollary | Folgerung, Bemerkung | consequence, remark |
| Definitionsbereich / Wertebereich | domain / range | ausgelassen | skipped (the script marks material it left out) |
Subject vocabulary (Folge, Reihe, Grenzwert, Häufungspunkt, Schranke, Stammfunktion, …) is given in brackets next to each notion in the chapters, so it is not repeated here. If a word in an old exam is not in this list, the chapter that owns the topic has it.
ΣSummary
- Sets: read \(\in,\subseteq,\cup,\cap,\setminus,\times\) and \(\{x\in A:\dots\}\); prove \(A=B\) by showing both inclusions.
- Proofs: \(A\Rightarrow B\) is not \(B\Rightarrow A\); the contrapositive \(\neg B\Rightarrow\neg A\) is the same statement; contradiction and induction are the other two standard tools.
- Maps: injective = no collisions, surjective = everything hit, bijective = both = invertible. Domain and codomain are part of the map.
- Structures: group (one operation, invertible) → ring (+, ·, no division) → field (division by nonzero), each adding rules to the previous one. ℤ is a ring, ℚ, ℝ, ℂ are fields.
- Consequences of the field axioms: \(a\cdot0=0\), \((-1)(-1)=1\), \(ab=0\Rightarrow a=0\) or \(b=0\).
- Order: "positive" with rules (a), (b) ⇒ all inequality rules; multiplying by a negative flips. ℚ, ℝ ordered; ℂ, \(\mathbb Z/p\) not. ℝ = the complete ordered field.
- Sums: \(\sum\) is linear, splits, shifts (\(j=k-1\)) and telescopes; empty sum \(=0\), empty product \(=1\). A product of two sums is not the sum of the products.
- Powers and roots: the five rules (\(a^xa^y=a^{x+y}\), \(\frac{a^x}{a^y}=a^{x-y}\), \((a^x)^y=a^{xy}\), \((ab)^x=a^xb^x\), \(\left(\frac ab\right)^x=\frac{a^x}{b^x}\)) hold for \(a,b>0\) and all real exponents, and for \(a,b\neq0\) with integer exponents. They never apply to a sum: \(\sqrt{a+b}\neq\sqrt a+\sqrt b\). \(\sqrt{x^2}=|x|\); fractional powers need a positive base; \(f^g=e^{g\ln f}\) is the way past a variable exponent.
- Exam German: zeigen/beweisen = full proof, begründen = give the reason, angeben = answer only; watch for "nur das Ergebnis wird bewertet" and "alle Zwischenschritte sind zu begründen".
?Self-test
Real NumbersReelle Zahlen
What the real numbers are, why analysis needs them, and the three tools this chapter gives you for the rest of the course: supremum and infimum, the Archimedean property, and inequalities. Exercise Sheet 1 and the recurring exam question type are worked into the text at the point where they belong.
Source labels: Skript 1.5 = stated in your lecture script · Blatt 1 · 1.2 = exercise sheet · Klausur WiSe20 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§1.2 = Section 1.2) rather than a numbered result (1.2 = Definition 1.2).
0The big picture
This chapter assumes you can read set notation, know what injective/surjective means, and know what a field is. If any of that is shaky, read the Basics page first (about an hour, with its own self-test).
Every new number set in this chapter exists because the previous one had a problem. Reading the chain left to right is the storyline of Section 1.1:
The step that matters most for analysis is ℚ → ℝ. Analysis is about limits, and you can only take limits if they actually land somewhere. In ℚ, the numbers 1, 1.4, 1.41, 1.414, … creep towards a point that is not there, namely √2. The real numbers fill all such holes. The precise statement of "no holes" is the supremum axiom in 1.2, and it is the foundation for everything in Chapters 2–4.
So the chapter has three parts:
- 1.1 Which number sets exist, how they differ (countable vs. uncountable, ordered vs. not ordered).
- 1.2 What makes ℝ special: sup/inf exist, no infinitely large or small numbers, fractions are everywhere.
- 1.3 The inequalities you will use to estimate things in every later proof: triangle inequality and Cauchy–Schwarz.
1.1Number sets at a glance [Zahlenmengen im Überblick]
Quantifiers [Quantoren]
Before any numbers, the script introduces two symbols you will read thousands of times:
- \(\forall x\): "for all \(x\)" [für alle]
- \(\exists x\): "there exists an \(x\)" [es existiert]
The single most useful skill with them is negation, because every proof by contradiction starts by negating a statement. The rule: flip every \(\forall\) into \(\exists\), every \(\exists\) into \(\forall\), and negate the statement at the end.
Statement: "every real number is smaller than some natural number":
\[\forall a\in\mathbb R\ \exists n\in\mathbb N:\ a
The order of quantifiers matters. "\(\forall\) people \(\exists\) a mother" is true; "\(\exists\) a mother \(\forall\) people" (one mother of everyone) is false. You will see this again in the definition of convergence, where \(n_0\) is allowed to depend on \(\varepsilon\).
The number sets
- \(\mathbb N=\{1,2,3,\dots\}\) natural numbers [natürliche Zahlen]. Careful: in this course 0 ∉ ℕ. With 0: \(\mathbb N_0=\mathbb N\cup\{0\}\).
- \(\mathbb Z=\{\dots,-2,-1,0,1,2,\dots\}\) integers [ganze Zahlen]
- \(\mathbb Q=\{\tfrac ab: a\in\mathbb Z,\ b\in\mathbb N\}\) rational numbers [rationale Zahlen]
- \(\mathbb R\) real numbers [reelle Zahlen]. \((\mathbb R,+,\cdot)\) is a commutative field [kommutativer Körper].
"Field" just means: you can add, subtract, multiply and divide (except by 0), and all the usual rules hold (commutative, associative, distributive, 0 and 1 behave as expected). ℚ is also a field; ℤ is not, since \(1/2\notin\mathbb Z\). The precise axioms, what follows from them (like \((-1)(-1)=1\) and "\(ab=0\Rightarrow a=0\) or \(b=0\)"), and how rings and groups differ are on the Basics page.
√2 is not rational
This is the proof that ℚ has holes, so it is the reason ℝ is needed at all.
\(\sqrt2\notin\mathbb Q\).
Proof, step by step
Proof by contradiction [Widerspruchsbeweis]: assume \(\sqrt2\in\mathbb Q\).
- Then \(\sqrt2=\frac ab\) with \(a\in\mathbb Z,\ b\in\mathbb N\), and we may cancel the fraction completely, so \(a\) and \(b\) are coprime [teilerfremd] (no common prime factor).
- Squaring: \(2=\frac{a^2}{b^2}\). Squaring does not create new prime factors, so \(a^2\) and \(b^2\) are still coprime. The fraction \(\frac{a^2}{b^2}\) is therefore fully cancelled.
- A fully cancelled fraction equal to the integer 2 must have denominator 1, so \(b^2=1\) and \(a^2=2\).
- No integer squares to 2 (\(1^2=1,\ 2^2=4\)). Contradiction. ∎
The script compresses steps 3–4 into "Das ist ein Widerspruch". The version you may know from school (\(a^2=2b^2\Rightarrow a\) even \(\Rightarrow b\) even, contradiction) is equally valid.
Countable and uncountable [abzählbar, überabzählbar]
Intuition. A set is countable if you can write its elements in one infinite list \[f(1),\ f(2),\ f(3),\ \dots\] so that every element of the set turns up somewhere in the list — maybe late, but at a definite, finite position. "Countable" means "can be numbered 1st, 2nd, 3rd, …, and nobody is left out".
A set \(A\) is countable if there is a surjective map \(f:\mathbb N\to A\), i.e. for every \(a\in A\) there is an \(n\in\mathbb N\) with \(f(n)=a\).
Surjective [surjektiv] means "every element of \(A\) is hit at least once". The number \(n\) with \(f(n)=a\) is the position at which \(a\) appears in the list. Repetitions are allowed: the same element may appear at several positions. A set that is not countable is called uncountable [überabzählbar].
- ℕ itself: \(f(n)=n\). Trivially countable.
- The even numbers \(\{2,4,6,\dots\}\): \(f(n)=2n\). There are "as many" even numbers as natural numbers — infinite sets behave strangely.
- ℤ: the list \(0,\ 1,\ -1,\ 2,\ -2,\ 3,\ -3,\dots\) reaches every integer. The integer \(k>0\) sits at position \(2k\), and \(-k\) at position \(2k+1\). The trick: don't list all positive numbers first (you would never get to the negative ones); alternate instead.
That last trick is the whole idea for ℚ, one dimension up: arrange the list so that you never get stuck in an infinite part before reaching the rest.
\(\mathbb Q\) is countable. \(\mathbb R\) is not countable.
The script proves only the second half and leaves the first as an exercise ("Übung"). Both halves are worked out below: first how to list ℚ, then why no list can ever contain all of ℝ.
Part 1 — Why ℚ is countable: Cantor's diagonal walk Extra (the script's "Übung")
The problem. Between any two fractions there are infinitely many others, so you cannot list fractions "in order of size" — there is no next fraction after 0. A naive list like \(\frac11,\frac12,\frac13,\frac14,\dots\) never ends, so it never reaches \(\frac21\). We need a cleverer order.
The idea: put the fractions in a grid. Every positive fraction is \(\frac pq\) with \(p,q\in\mathbb N\) (numerator \(p\), denominator \(q\)). Write \(\frac pq\) in column \(p\), row \(q\):
| p = 1 | p = 2 | p = 3 | p = 4 | … | |
|---|---|---|---|---|---|
| q = 1 | 1/1 | 2/1 | 3/1 | 4/1 | … |
| q = 2 | 1/2 | 2/2 | 3/2 | 4/2 | … |
| q = 3 | 1/3 | 2/3 | 3/3 | 4/3 | … |
| q = 4 | 1/4 | 2/4 | 3/4 | 4/4 | … |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋱ |
Every positive rational number appears in this grid (usually several times: \(\frac12=\frac24=\frac36\)). Walking along a row or a column fails, since each is infinite. Instead, walk along the short diagonals:
- Group the grid by diagonals. Diagonal number \(d\) contains all fractions with \(p+q=d\). Diagonal 2 is just \(\frac11\); diagonal 3 is \(\frac12,\frac21\); diagonal 4 is \(\frac13,\frac22,\frac31\); in general diagonal \(d\) has exactly \(d-1\) entries. Each diagonal is finite — that is the whole point.
- List the diagonals one after the other, each from its left end to its right end: \[\underbrace{\tfrac11}_{d=2},\ \underbrace{\tfrac12,\ \tfrac21}_{d=3},\ \underbrace{\tfrac13,\ \tfrac22,\ \tfrac31}_{d=4},\ \underbrace{\tfrac14,\ \tfrac23,\ \tfrac32,\ \tfrac41}_{d=5},\ \dots\] Because every diagonal is finite, we finish it and move on to the next — we never get stuck.
- Every fraction is reached after finitely many steps. Take any \(\frac pq\). It lies on diagonal \(d=p+q\). Before that diagonal come the diagonals \(2,3,\dots,d-1\), with \(1+2+\dots+(d-2)=\frac{(d-2)(d-1)}2\) entries in total, and inside its own diagonal \(\frac pq\) is entry number \(p\). So it appears at position \[n(p,q)=\frac{(p+q-2)(p+q-1)}2+p,\] a definite natural number. Example: \(\frac23\): \(d=5\), so \(n=\frac{3\cdot4}2+2=8\). Check against the list above: \(\frac11,\frac12,\frac21,\frac13,\frac22,\frac31,\frac14,\frac23\) — eighth. ✓
- Add 0 and the negatives with the ℤ trick: put 0 first and each negative right after its positive twin: \[0,\ \tfrac11,\ -\tfrac11,\ \tfrac12,\ -\tfrac12,\ \tfrac21,\ -\tfrac21,\ \tfrac13,\ -\tfrac13,\ \dots\] If \(r>0\) is at position \(n\) in the positive list, then \(r\) is at position \(2n\) and \(-r\) at position \(2n+1\) here. Every rational number is hit, so this is a surjection \(f:\mathbb N\to\mathbb Q\). ∎
What about the repeats? \(\frac22\) is the same number as \(\frac11\), so 1 appears more than once. The script's definition only asks for a surjective map, so repeats are completely fine. (If you want each number exactly once — a bijection — just skip every fraction that is not fully cancelled, like \(\frac22\) or \(\frac24\). The widget can do both.)
The diagonal walk works because every fraction is described by two whole numbers \(p,q\) — finitely much information — so the fractions fit in a grid with finite diagonals. A real number like \(\sqrt2=1.41421356\ldots\) needs infinitely many digits to describe. There is no grid of reals with finite diagonals, and Cantor's argument below shows that no clever order at all can work: every attempted list of real numbers misses some number.
Part 2 — Why ℝ is not countable: Cantor's diagonal argument
Strategy (proof by contradiction): assume someone hands us a list that supposedly contains every number in \([0,1]\). We then construct, from the list itself, a number in \([0,1]\) that is guaranteed not to be in it. So no list can be complete.
- Assume ℝ is countable. Then \([0,1]\) is countable as well (drop from the list every entry outside \([0,1]\)), so there is a surjection \(f:\mathbb N\to[0,1]\): a list that contains every number in \([0,1]\).
- Write every entry of the list as a decimal: \(f(i)=0.d_{i1}d_{i2}d_{i3}\dots\), where \(d_{ij}\) is the \(j\)-th digit of the \(i\)-th number. Arranged in rows, these digits form an infinite table; the digits \(d_{11},d_{22},d_{33},\dots\) run down its diagonal.
- Build a new number \(x=0.d_1d_2d_3\dots\) digit by digit: \(d_i=2\) if \(d_{ii}=1\), otherwise \(d_i=1\). In words: the \(i\)-th digit of \(x\) is chosen to be different from the \(i\)-th digit of the \(i\)-th number.
- Then \(x\neq f(1)\) (they differ in digit 1), \(x\neq f(2)\) (they differ in digit 2), and in general \(x\neq f(i)\) because they differ in digit \(i\). So \(x\in[0,1]\) is missing from the list.
- That contradicts surjectivity. ∎
Extra Why only the digits 1 and 2? Some numbers have two decimal expansions (\(0.1999\ldots = 0.2000\ldots\)), so two different digit strings can be the same number. Using only 1 and 2 means \(x\) has no 0s or 9s, so it has exactly one expansion, and "different digit" really means "different number".
A computer-science application from the script:
The set \(B\) of computable numbers [berechenbare Zahlen] (real numbers for which some program outputs the first \(n\) decimal digits for every \(n\)) is countable, so \(B\subsetneq\mathbb R\). A number drawn uniformly from \([0,1]\) is not computable with probability 1.
Reason: a program is a finite string over a finite alphabet, and there are only countably many finite strings. Countably many programs can only compute countably many numbers, but ℝ is uncountable. So almost all real numbers can never be printed by any program.
Ordering [Anordnung]
ℝ is not just a field, it is an ordered field. The script defines "ordered" through a single predicate "\(a>0\)" (being positive) with two rules:
- For every \(a\in\mathbb R\), exactly one of the following holds: \(a=0\), \(a>0\), or \(-a>0\).
- If \(a>0\) and \(b>0\), then \(a+b>0\) and \(a\cdot b>0\).
Everything else is notation: \(a>b\) means \(a-b>0\); \(a<b\) means \(b-a>0\); \(a\ge b\) means \(a>b\) or \(a=b\).
From these two rules alone you can prove that every nonzero square is positive: if \(a>0\) then \(a\cdot a>0\) by (b); if \(-a>0\) then \(a^2=(-a)(-a)>0\), again by (b). That little fact is the key to the next result.
Complex numbers [komplexe Zahlen]
Intuition: a complex number is a point in the plane. We just add a rule for multiplying points.
\(\mathbb C=\{z=(x,y):x,y\in\mathbb R\}=\mathbb R^2\). With the imaginary unit \(i=(0,1)\), every \(z=(x,y)\) can be written as \(z=x+iy\). Here \(x=\operatorname{Re}z\) is the real part [Realteil] and \(y=\operatorname{Im}z\) the imaginary part [Imaginärteil].
Addition: \((x+iy)+(x'+iy')=(x+x')+i(y+y')\)
Multiplication: \(i^2=-1\) and \((x+iy)(x'+iy')=(xx'-yy')+i(xy'+x'y)\)
You don't need to memorise the multiplication formula. Just expand the brackets as usual and replace \(i^2\) by \(-1\):
Check with the formula: \(xx'-yy'=2\cdot1-3\cdot(-1)=5\) and \(xy'+x'y=2\cdot(-1)+1\cdot3=1\). ✓
Exercise Sheet 1 adds three tools you will use constantly, especially in Chapter 5 (Euler's formula):
The complex conjugate [komplex Konjugierte] of \(z=a+ib\) is \(\bar z=a-ib\): the point mirrored at the real axis. Then:
\[\text{(a)}\ \operatorname{Re}z=\tfrac12(z+\bar z),\qquad \text{(b)}\ \operatorname{Im}z=\tfrac1{2i}(z-\bar z),\qquad \text{(c)}\ |z|=\sqrt{z\bar z}.\]Solution
- \(z+\bar z=(a+ib)+(a-ib)=2a\), so \(\tfrac12(z+\bar z)=a=\operatorname{Re}z\).
- \(z-\bar z=(a+ib)-(a-ib)=2ib\), so \(\tfrac1{2i}(z-\bar z)=b=\operatorname{Im}z\).
- \(z\bar z=(a+ib)(a-ib)=a^2-i^2b^2+(iab-iab)=a^2+b^2=|z|^2\). Taking the root gives the claim.
Why (c) matters: \(|z|=\sqrt{a^2+b^2}\) is the distance of the point \((a,b)\) from 0 (Pythagoras). Example: \(z=3+4i\) gives \(z\bar z=9+16=25\), so \(|z|=5\). And \(z\bar z\) is always a real number \(\ge0\), which is why the trick "multiply by the conjugate" gets rid of \(i\) in a denominator.
\(\mathbb C\) cannot be ordered. (There is no way to define "positive" on ℂ that satisfies rules (a) and (b).)
Proof
- Assume ℂ were ordered. As shown above, rules (a) and (b) force \(a^2>0\) for every \(a\neq0\).
- In particular \(1=1\cdot1>0\), so \(-1<0\).
- But \(-1=i^2\), and \(i\neq0\), so \(i^2>0\). Contradiction. ∎
Never write \(z<w\) for complex numbers. Inequalities in ℂ only make sense for real quantities like \(|z|\), \(\operatorname{Re}z\), \(\operatorname{Im}z\). That is why all later statements for complex sequences are phrased with \(|a_n-a|\).
1.2Properties of the real numbers [Eigenschaften der reellen Zahlen]
For \(a\le b\): \((a,b)=\{x: a<x<b\}\) open [offen], \([a,b]=\{x:a\le x\le b\}\) closed [abgeschlossen], \((a,b]\) and \([a,b)\) half-open [halboffen]. A round bracket means "endpoint excluded", a square bracket "endpoint included".
Upper bounds, supremum and infimum
Intuition: an upper bound is a ceiling for a set, a number that nothing in the set exceeds. A set has many ceilings. The supremum is the lowest ceiling, the one that touches the set.
A set \(M\subseteq\mathbb R\) is bounded above [nach oben beschränkt] if there is an \(s_0\in\mathbb R\) with \[a\le s_0\quad\text{for all }a\in M.\] Such an \(s_0\) is called an upper bound [obere Schranke] of \(M\). "Bounded below" and "lower bound" [untere Schranke] are defined the same way with \(\ge\).
- \(M=(0,1)\): every \(s_0\ge1\) is an upper bound. The smallest one is \(1\), but \(1\notin M\): no upper bound lies inside \(M\).
- \(M=[0,1]\): again the smallest upper bound is 1, and this time \(1\in M\).
This pair of examples is the whole difference between supremum and maximum.
Every nonempty subset \(M\subseteq\mathbb R\) that is bounded above has a least upper bound \(\sup M\in\mathbb R\), the supremum of \(M\).
It follows: every nonempty set that is bounded below has a greatest lower bound \(\inf M\in\mathbb R\), the infimum.
Why the infimum version follows (the script says only "Daraus folgt")
Extra Mirror the set at 0: let \(-M=\{-a:a\in M\}\). Multiplying by \(-1\) turns every inequality around, so
- \(c\) is a lower bound of \(M\) \(\iff\) \(c\le a\) for all \(a\in M\) \(\iff\) \(-a\le-c\) for all \(a\in M\) \(\iff\) \(-c\) is an upper bound of \(-M\).
- \(M\) is nonempty and bounded below, so \(-M\) is nonempty and bounded above, and the supremum axiom gives \(s=\sup(-M)\).
- By step 1, "the smallest upper bound of \(-M\)" becomes "the largest lower bound of \(M\)" after flipping the sign. So \(\inf M=-\sup(-M)\) exists. ∎
Example: \(M=\{1,2,3\}\), \(-M=\{-1,-2,-3\}\), \(\sup(-M)=-1\), so \(\inf M=1\). ✓
If \(\sup M\in M\), it is also called the maximum [Maximum]. If \(\inf M\in M\), it is the minimum [Minimum].
The script lists some conventions for the cases the axiom doesn't cover: \(\sup M=\infty\) if \(M\) is not bounded above, \(\inf M=-\infty\) if \(M\) is not bounded below, \(\sup\emptyset=-\infty\), and \(-\infty<a<\infty\) for every real \(a\).
- The supremum always exists (for nonempty bounded sets); the maximum may not. The max exists ⇔ sup ∈ M, and then max = sup.
- \(\sup M\) does not have to be an element of \(M\). In \(\{-\frac1n:n\in\mathbb N\}\) the supremum is 0, but no \(-\frac1n\) equals 0 (Skript 1.7).
- "∞ is the supremum" is only a convention. If \(\sup M=\infty\), then the maximum does not exist.
Why ℚ is not enough
ℚ does not satisfy the supremum axiom. The set \(M=\{a\in\mathbb Q: a^2\le2\}\) is nonempty and bounded above (e.g. by 2), but it has no least upper bound in ℚ. In ℝ, \(\sup M=\sqrt2\notin\mathbb Q\).
This is the precise meaning of "ℝ has no holes": whenever a set pushes upward against a ceiling, the lowest ceiling actually exists as a real number. In ℚ, the lowest ceiling of \(M\) would be √2, and that spot is empty.
How to prove that a number is the supremum
This is what exams ask for, so here is a reusable recipe. It follows directly from the definition ("least upper bound" = an upper bound, and nothing smaller is one). The script uses exactly this reasoning inside its proofs of Satz 1.9 and Satz 2.17.
- \(s\) is an upper bound: show \(a\le s\) for all \(a\in M\).
- No smaller number is an upper bound: for every \(\varepsilon>0\), find an element \(a\in M\) with \(a>s-\varepsilon\). (So \(s-\varepsilon\) is "pierced" by the set.)
- Max or not? Check whether \(s\in M\).
For the infimum, mirror everything: \(s\le a\) for all \(a\in M\), and for every \(\varepsilon>0\) some \(a\in M\) with \(a<s+\varepsilon\).
A useful shortcut for sets given by a sequence \(M=\{x_n:n\in\mathbb N\}\): if \(x_n\) is increasing, then \(x_1\) is the minimum, and the supremum is the limit — or \(\infty\) if the sequence is unbounded. If it is decreasing, \(x_1\) is the maximum and the infimum is the limit (or \(-\infty\)). (Chapter 2, Satz 2.17, turns "the limit is the sup" into a theorem.)
For nonempty bounded sets \(A,B\subseteq\mathbb R\) and \(c\in\mathbb R\), write \(A+B=\{a+b:a\in A,\ b\in B\}\) and \(cA=\{ca:a\in A\}\). Then
- \(\sup(-A)=-\inf A\) and \(\inf(-A)=-\sup A\) — mirroring at 0 swaps the two (this is the proof of the infimum version of the axiom above);
- \(\sup(A\cup B)=\max\{\sup A,\sup B\}\);
- \(\sup(A+B)=\sup A+\sup B\);
- \(\sup(cA)=c\sup A\) for \(c>0\), but \(\sup(cA)=c\inf A\) for \(c<0\).
Proof of 3, and why the exam's function version only gives "≤"
"\(\le\)": every element of \(A+B\) is some \(a+b\le\sup A+\sup B\), so the right side is an upper bound, and \(\sup(A+B)\) is the least one.
"\(\ge\)": let \(\varepsilon>0\). By step 2 of the recipe there are \(a\in A\) with \(a>\sup A-\frac\varepsilon2\) and \(b\in B\) with \(b>\sup B-\frac\varepsilon2\). Then \(a+b\in A+B\) and \(a+b>\sup A+\sup B-\varepsilon\). Since this works for every \(\varepsilon>0\), no number below \(\sup A+\sup B\) is an upper bound. ∎
The catch: here \(a\) and \(b\) may be chosen independently. For functions, \(\sup_x(f(x)+g(x))\) has to use the same \(x\) in both terms, so the two suprema may be attained at different points and only "\(\le\)" survives — that is exactly Klausur WiSe18 A5 (worked in the exam section below).
"Determine infimum, supremum, minimum and maximum (if they exist) of \(M=\left\{\frac{n+1}{n}: n\in\mathbb N\right\}\), with a short justification."
Solution
Look first. \(x_n=\frac{n+1}{n}=1+\frac1n\): the elements are \(2,\ \frac32,\ \frac43,\ \frac54,\dots\), falling towards 1.
Monotone: \(x_{n+1}=1+\frac1{n+1}<1+\frac1n=x_n\), so the sequence is decreasing. The exam solution checks this via \(\frac{x_{n+1}}{x_n}=\frac{n^2+2n}{n^2+2n+1}<1\).
Supremum and maximum. Decreasing, so every \(x_n\le x_1=2\): 2 is an upper bound, and it is an element of \(M\). So \(\sup M=\max M=2\).
Infimum. Every \(x_n=1+\frac1n>1\), so 1 is a lower bound (step 1). No larger number is a lower bound (step 2): given \(x'>1\), we need some \(n\) with \(1+\frac1n<x'\), i.e. \(n>\frac1{x'-1}\). Such an \(n\) exists because ℝ is Archimedean (Satz 1.9, below). So \(\inf M=1\).
Minimum? \(1+\frac1n=1\) has no solution, so \(1\notin M\): no minimum.
The Archimedean property [archimedisch]
ℝ is Archimedean: for every \(a\in\mathbb R\) there is an \(n\in\mathbb N\) with \(a<n\). In particular, there are no infinitely large numbers in ℝ.
Proof, step by step
- Assume the statement is false. Its negation (see quantifiers) is: \(\exists a\in\mathbb R\ \forall n\in\mathbb N:\ a\ge n\). So \(a\) is an upper bound of ℕ, and ℕ is bounded above.
- ℕ is nonempty and bounded above, so by the supremum axiom \(s=\sup\mathbb N\) exists.
- \(s\) is the least upper bound, so \(s-1\) is not an upper bound: there is an \(n\in\mathbb N\) with \(s-1<n\).
- Then \(s<n+1\), and \(n+1\in\mathbb N\). So \(s\) is not an upper bound of ℕ. Contradiction. ∎
Notice step 3: it is exactly step 2 of the supremum recipe, with \(\varepsilon=1\).
For every \(a>0\) there is an \(n\in\mathbb N\) with \(\frac1n<a\). So there are no infinitely small positive numbers either.
Why: \(\frac1a\) is a real number, so by Archimedes some \(n>\frac1a\), and flipping gives \(\frac1n<a\). This is the most-used line in Chapter 2. It is exactly how the script proves \(\lim\frac1n=0\) (Beispiel 2.4): given \(\varepsilon\), choose \(n_0>\frac1\varepsilon\).
\(a=0.003\): we need \(n>\frac1{0.003}\approx333.3\), so \(n=334\) works: \(\frac1{334}\approx0.00299<0.003\).
ℚ is dense [dicht] in ℝ: for all \(a<b\) in ℝ there is an \(r\in\mathbb Q\) with \(a<r<b\).
Consequence: every real number can be approximated arbitrarily well by fractions. For every \(n\) there is \(r\in\mathbb Q\) with \(a-10^{-n}<r<a\).
Proof (the script states it without proof, so I supply one)
Extra Idea: zoom in by a factor \(n\) until the gap between \(a\) and \(b\) is longer than 1. Then an integer must fit inside it.
- By Archimedes choose \(n\in\mathbb N\) with \(n>\frac1{b-a}\), so \(nb-na>1\).
- Let \(m\) be the smallest integer with \(m>na\). Such an \(m\) exists: the set \(S=\{m\in\mathbb Z:m>na\}\) is nonempty by Archimedes, and it is bounded below, because Archimedes applied to \(-na\) gives a \(k\in\mathbb N\) with \(-k<na\), so every element of \(S\) is \(>-k\). A nonempty set of integers that is bounded below has a smallest element. Minimality means \(m-1\le na\).
- Then \(na<m\le na+1<nb\).
- Divide by \(n\): \(a<\frac mn<b\), and \(r=\frac mn\in\mathbb Q\). ∎
Example: \(a=0.31,\ b=0.32\). The gap is 0.01, so take \(n=101\). Then \(na=31.31\), \(m=32\), \(r=\frac{32}{101}\approx0.3168\). ✓
Decimal representation [Dezimaldarstellung]
Every real number has a decimal expansion \(\pm d_0.d_1d_2d_3\ldots\) with \(d_0\in\mathbb N_0\) and digits \(d_i\in\{0,\dots,9\}\). The script connects decimals to the supremum in both directions:
- Decimal → real number: \(d_0.d_1d_2\ldots\) means \(\sup\{d_0.d_1\ldots d_n: n\in\mathbb N\}\). Every finite truncation is rational (\(1.37=\frac{137}{100}\)), and the real number is the supremum of these truncations.
- Real number → decimal: for \(a\ge0\) set \(p_n=\max\{p\in\mathbb N_0: p\le10^na\}\), and \(\frac{p_n}{10^n}\) gives the first \(n\) digits.
\(a=\pi\): \(p_0=3,\ p_1=\max\{p\le31.4\ldots\}=31,\ p_2=314,\ p_3=3141\). So \(\frac{p_3}{10^3}=3.141\).
This is the supremum axiom at work: the truncations 3, 3.1, 3.14, 3.141, … form a bounded set of rationals, and π is its supremum. In ℚ that supremum would not exist.
The script says the second procedure gives a bijection between ℝ and decimal expansions. That is only true because the procedure never produces an expansion ending in endless 9s: for \(a=1\) it gives \(1.000\ldots\), never \(0.999\ldots\). Both strings describe the same number — by the first bullet, \(0.999\ldots=\sup\{0.9,\ 0.99,\ 0.999,\dots\}=1\), because 1 is an upper bound and for every \(\varepsilon>0\) some \(0.99\ldots9=1-10^{-n}\) exceeds \(1-\varepsilon\) (Archimedes). This is exactly why Cantor's argument above avoids the digits 0 and 9.
What you must know for 1.2 (script: "Minimale Inhalte"): supremum, infimum, minimum and maximum: know them and be able to determine them.
1.3Important inequalities [Wichtige Ungleichungen]
Why inequalities? From Chapter 2 on, almost every proof reads "make \(|a_n-a|\) smaller than \(\varepsilon\)". You rarely compute \(|a_n-a|\) exactly. You estimate it, and these inequalities are the estimating tools.
The absolute value [Betrag]: \(|x|=x\) if \(x\ge0\), and \(|x|=-x\) if \(x<0\).
Picture: \(|x|\) is the distance of \(x\) from 0, and \(|x-y|\) is the distance between \(x\) and \(y\) on the number line. Read \(|a_n-a|<\varepsilon\) as "\(a_n\) is closer than \(\varepsilon\) to \(a\)".
Triangle inequality [Dreiecksungleichung]
For all \(x,y\in\mathbb R\): \[|x+y|\le|x|+|y|,\] with equality if and only if \(xy\ge0\) (same sign). More generally \(\left|\sum_{i=1}^nx_i\right|\le\sum_{i=1}^n|x_i|\), with equality iff all \(x_i\) have the same sign.
Proof (not in the script, so I supply one)
Extra Use the fact \(|c|\le d \iff -d\le c\le d\).
- Always \(-|x|\le x\le|x|\) and \(-|y|\le y\le|y|\).
- Add the two chains: \(-(|x|+|y|)\le x+y\le|x|+|y|\).
- By the fact above, this says \(|x+y|\le|x|+|y|\). ∎
When is it an equality? If \(x,y\ge0\): \(|x+y|=x+y=|x|+|y|\). If \(x,y\le0\): \(|x+y|=-(x+y)=|x|+|y|\). If they have opposite signs, say \(x>0>y\), then \(x+y\) lies strictly between \(y\) and \(x\), so \(|x+y|<\max(|x|,|y|)\le|x|+|y|\) — strict. So equality holds exactly when \(xy\ge0\).
The version with \(n\) terms follows by induction [vollständige Induktion]: it holds for \(n=2\) (just proved), and if \(\left|\sum_{i=1}^nx_i\right|\le\sum_{i=1}^n|x_i|\), then treating \(\sum_{i=1}^nx_i\) as a single number and applying the two-term case gives \(\left|\sum_{i=1}^{n+1}x_i\right|\le\left|\sum_{i=1}^nx_i\right|+|x_{n+1}|\le\sum_{i=1}^{n+1}|x_i|\). (Induction as a method is explained in Chapter 2, where the course first leans on it.)
Intuition: if you walk \(x\) steps and then \(y\) steps, you end up at most \(|x|+|y|\) away from the start. You land exactly that far only if you never turned around.
Reverse triangle inequality [umgekehrte Dreiecksungleichung]: for all \(x,y\in\mathbb R\) \[|x+y|\ge\big||x|-|y|\big|,\] with equality if and only if \(xy\le0\) (opposite signs).
Intuition: walking forward and then back leaves you at least the difference of the two distances away.
Proof: the script gives none. It is Tutoriumsaufgabe 1.1a below (replace \(y\) by \(-y\) there, using \(|-y|=|y|\)). Equality case: if \(x,y\) have opposite signs, \(x+y\) is the difference of two distances, e.g. \(3+(-5)=-2\) and \(\big|3-5\big|=2\). If they have the same sign, \(|x+y|=|x|+|y|\), which is bigger than \(\big||x|-|y|\big|\) unless one of them is 0.
Prove the reverse triangle inequality in the form \[\big||x|-|y|\big|\le|x-y|\qquad\forall x,y\in\mathbb R.\] (Same statement as Satz 1.13, with \(y\) replaced by \(-y\), since \(|-y|=|y|\).)
Solution, with the reasoning spelled out
Strategy: \(|c|\le d\) means both \(c\le d\) and \(-c\le d\). So we show \(|x|-|y|\le|x-y|\) and \(|y|-|x|\le|x-y|\) separately, each time using the normal triangle inequality with a clever split.
- Write \(x=(x-y)+y\). Triangle inequality: \(|x|\le|x-y|+|y|\), so \(|x|-|y|\le|x-y|\).
- Write \(y=(y-x)+x\). Triangle inequality: \(|y|\le|y-x|+|x|=|x-y|+|x|\), so \(|y|-|x|\le|x-y|\).
- Both \(c=|x|-|y|\) and \(-c\) are \(\le|x-y|\), so \(\big||x|-|y|\big|\le|x-y|\). ∎
The official solution does the same with \(z=-y\), starting from \(|a+b|-|b|\le|a|\).
Reusable trick: "add and subtract" (\(x=(x-y)+y\)). You will use it constantly, e.g. in the proof of the squeeze theorem (Skript 2.14): \(|b_n-\alpha|\le|b_n-a_n|+|a_n-\alpha|\).
What it says in words: if \(x\) and \(y\) are close, then \(|x|\) and \(|y|\) are close. Later this shows: \(a_n\to a\) implies \(|a_n|\to|a|\).
Young's ε-inequality
For all \(x,y\in\mathbb R\) and all \(\varepsilon>0\): \[xy\le\frac{x^2}{2\varepsilon}+\frac{\varepsilon y^2}{2}.\]
Solution
Idea: squares are never negative. Pick a square that contains \(xy\) when you expand it.
\[0\le\left(\frac{x}{\sqrt\varepsilon}-\sqrt\varepsilon\,y\right)^2=\frac{x^2}{\varepsilon}-2xy+\varepsilon y^2.\]Add \(2xy\) and divide by 2: \(xy\le\frac{x^2}{2\varepsilon}+\frac{\varepsilon y^2}{2}\). ∎
Special case \(\varepsilon=1\): \(xy\le\frac{x^2+y^2}{2}\), which is just \(0\le(x-y)^2\). Keep this one in mind: it is exactly the key step (1.3) in the proof of Cauchy–Schwarz below.
Why the ε? It lets you split a product \(xy\) into an \(x\)-part and a \(y\)-part, and make one of them as small as you like ("rob Peter to pay Paul"). With \(\varepsilon=0.01\): \(xy\le 50x^2+0.005y^2\).
Vectors in ℝⁿ
\(\mathbb R^n=\{x=(x_1,\dots,x_n): x_i\in\mathbb R\}\). The length [Länge / Norm] and the dot product [Skalarprodukt] are \[\|x\|=\sqrt{\sum_{i=1}^nx_i^2},\qquad \langle x,y\rangle=\sum_{i=1}^nx_iy_i,\qquad\text{so }\|x\|^2=\langle x,x\rangle.\]
Example: \(x=(3,4)\): \(\|x\|=\sqrt{9+16}=5\). This is Pythagoras in \(n\) dimensions.
Cauchy–Schwarz inequality
For all \(x,y\in\mathbb R^n\): \[|\langle x,y\rangle|\le\|x\|\cdot\|y\|,\] with equality iff \(x=0\), \(y=0\), or \(x=\lambda y\) for some \(\lambda\in\mathbb R\) (the vectors are parallel).
Intuition Extra: from linear algebra, \(\langle x,y\rangle=\|x\|\|y\|\cos\theta\), where \(\theta\) is the angle between the vectors. Since \(|\cos\theta|\le1\), the inequality is no surprise, and equality means \(\cos\theta=\pm1\): parallel vectors. But this course has not defined angles yet, so the script proves it using only algebra.
Proof, step by step
- Trivial case: if \(\|x\|=0\), then \(x=0\) and both sides are 0. Same for \(y\). From now on \(\|x\|,\|y\|>0\).
- Remove signs: by the triangle inequality, \(|\langle x,y\rangle|=\left|\sum x_iy_i\right|\le\sum|x_iy_i|\). It suffices to show \(\sum|x_i||y_i|\le\|x\|\|y\|\).
- Normalise: set \(a_i=\frac{|x_i|}{\|x\|}\) and \(b_i=\frac{|y_i|}{\|y\|}\). These are the entries of \(x\) and \(y\) rescaled to length 1: \(\sum a_i^2=\sum b_i^2=1\).
- The key step (Young with ε = 1): \(0\le(a_i-b_i)^2\), so \(a_ib_i\le\frac12(a_i^2+b_i^2)\).
- Sum over i: \(\sum a_ib_i\le\frac12\left(\sum a_i^2+\sum b_i^2\right)=\frac12(1+1)=1\).
- Undo the normalisation: \(\sum\frac{|x_i|}{\|x\|}\frac{|y_i|}{\|y\|}\le1\), i.e. \(\sum|x_iy_i|\le\|x\|\|y\|\). Together with step 2, done. ∎
Equality needs equality in both step 2 (all \(x_iy_i\) have the same sign) and step 4 (\(a_i=b_i\) for all \(i\), i.e. \(|x_i|=\frac{\|x\|}{\|y\|}|y_i|\)). Together this is exactly \(x_i=\lambda y_i\) for all \(i\).
Why normalise? Without it, step 5 would give \(\sum|x_i||y_i|\le\frac12(\|x\|^2+\|y\|^2)\). That is true but too weak. Rescaling both vectors to length 1 makes the right side exactly 1.
\(x=(1,2),\ y=(3,4)\): \(\langle x,y\rangle=3+8=11\) and \(\|x\|\|y\|=\sqrt5\cdot5\approx11.18\). Indeed \(11\le11.18\). They are almost equal because the vectors point in almost the same direction.
\(x=(1,2),\ y=(2,4)=2x\): \(\langle x,y\rangle=10=\sqrt5\cdot\sqrt{20}\). Equality, as the theorem predicts for parallel vectors.
Triangle inequality in ℝⁿ: \(\|x+y\|\le\|x\|+\|y\|\) for all \(x,y\in\mathbb R^n\). Geometrically: one side of a triangle is never longer than the other two together.
The script only says "the proof uses Cauchy–Schwarz". Exercise 1.2(b) carries it out: see the next box, property (N2).
Norms: different ways to measure length
A norm [Norm] on ℝⁿ is a map \(\|\cdot\|:\mathbb R^n\to\mathbb R\) with, for all \(x,y\in\mathbb R^n\), \(\lambda\in\mathbb R\):
- (N1) Definiteness [Definitheit]: \(\|x\|=0\iff x=0\)
- (N2) Triangle inequality: \(\|x+y\|\le\|x\|+\|y\|\)
- (N3) Absolute homogeneity [absolute Homogenität]: \(\|\lambda x\|=|\lambda|\,\|x\|\)
These are the three properties anything must have to behave like "length".
(a) A norm is never negative
Hint from the sheet: look at \(\|x-x\|\).
\[0\overset{(N1)}=\|0\|=\|x+(-x)\|\overset{(N2)}\le\|x\|+\|-x\|\overset{(N3)}=\|x\|+|-1|\,\|x\|=2\|x\|.\]So \(\|x\|\ge0\). Nice: non-negativity doesn't need to be an axiom, it follows from (N1)–(N3).
(b) The Euclidean norm \(\|x\|_2=\sqrt{\langle x,x\rangle}\) is a norm
(N1) \(\|x\|_2=0\iff\sum x_j^2=0\iff\) every \(x_j=0\) (a sum of non-negative squares is 0 only if each square is 0) \(\iff x=0\).
(N2) Square the left side and expand: \[\|x+y\|_2^2=\sum(x_j+y_j)^2=\|x\|_2^2+\|y\|_2^2+2\langle x,y\rangle\le\|x\|_2^2+\|y\|_2^2+2\|x\|_2\|y\|_2=(\|x\|_2+\|y\|_2)^2,\] using Cauchy–Schwarz for the middle term. Taking square roots (both sides \(\ge0\)) gives the triangle inequality. This is the proof of Satz 1.14.
(N3) \(\|\lambda x\|_2=\sqrt{\sum\lambda^2x_j^2}=\sqrt{\lambda^2}\sqrt{\sum x_j^2}=|\lambda|\,\|x\|_2\). Note \(\sqrt{\lambda^2}=|\lambda|\), not \(\lambda\).
(c) \(\|x\|_1=\sum|x_i|\) is a norm, but \(\sum x_i\) is not
(N1) \(\sum|x_j|=0\iff\) all \(|x_j|=0\iff x=0\). (N2) \(\sum|x_j+y_j|\le\sum(|x_j|+|y_j|)=\|x\|_1+\|y\|_1\), using the real triangle inequality (Satz 1.12) in each coordinate. (N3) \(\sum|\lambda x_j|=|\lambda|\sum|x_j|\).
\(\sum x_i\) is not a norm: counterexample \(x=(1,-1)\neq0\), but \(1+(-1)=0\). (N1) fails. (It can even be negative, e.g. \(x=(-1,0)\).)
(d) \(\|x\|_\infty=\max\{|x_1|,\dots,|x_n|\}\) is a norm
(N1) The largest \(|x_j|\) is 0 \(\iff\) all are 0. (N2) \(\max|x_j+y_j|\le\max(|x_j|+|y_j|)\le\max|x_j|+\max|y_j|\). The second step holds because each \(|x_j|\le\max_k|x_k|\) and each \(|y_j|\le\max_k|y_k|\). (N3) \(\max|\lambda x_j|=|\lambda|\max|x_j|\).
(e) Unit "circles". The sets \(\{x\in\mathbb R^2:\|x\|=1\}\) look completely different for the three norms: a circle for \(\|\cdot\|_2\), a diamond (rotated square) for \(\|\cdot\|_1\), and an axis-parallel square for \(\|\cdot\|_\infty\). Drag the point below to see them.
What you must know for 1.3 (script: "Minimale Inhalte"): be able to apply the different variants of the triangle inequality and the Cauchy–Schwarz inequality.
!What the exam asks from Chapter 1
I went through all nine uploaded exams (WiSe18 – WiSe24). Chapter 1 appears in five of them, and almost always in the same form. Here is the complete list:
| Exam | Task | Type |
|---|---|---|
| SoSe19 A1 (10 P) | sup, inf, max, min of three sets: \(\{n\in\mathbb N:n^2=(2k-1)n\text{ for some }k\}\), \(\{x:x^2-1>0\}\), \(\{x:|x-1|-1<0\}\) | sup/inf |
| WiSe19 1a | sup and min of \(\{\frac n{n+2}:n\in\mathbb N\}\) | sup/inf |
| WiSe20 1b | sup, inf, max, min of \(\{\frac{n+1}n:n\in\mathbb N\}\) | sup/inf |
| SoSe23 1d | \(\inf\) and \(\sup\) of \(a_n=e^{1/n}+e^{-1/n}\) | sup/inf |
| WiSe18 A5 (8 P) | \(\sup(f+g)\le\sup f+\sup g\), with an example of strict inequality | sup proof |
| WiSe19 1f | determine \(\{z\in\mathbb C:z=|z|^2\}\) | complex numbers |
Conclusion: the one skill you must have cold is determining and justifying sup, inf, max and min. Countability, Cantor, Archimedes and the inequalities are not asked as standalone tasks — but Archimedes and the triangle inequality are used inside nearly every ε-proof of Chapter 2, so you need them anyway. The four tasks not yet worked above follow (the other exam sets are worked in the explorer).
Determine sup, inf, max and min of \(A=\{n\in\mathbb N:\ n^2=(2k-1)n\text{ for some }k\in\mathbb N\}\).
Solution — and a warning about the official one
- Understand the set first. For \(n\neq0\) divide \(n^2=(2k-1)n\) by \(n\): \(n=2k-1\). So \(n\) belongs to \(A\) exactly when \(n\) is odd. With \(\mathbb N=\{1,2,3,\dots\}\) (the convention of your script): \(A=\{1,3,5,7,\dots\}\).
- Infimum and minimum. Every element is \(\ge1\) and \(1\in A\) (take \(k=1\)). So \(\inf A=\min A=1\).
- Supremum. \(A\) contains \(2k-1\) for every \(k\), and these grow beyond every bound (Archimedes). So \(A\) is not bounded above: \(\sup A=\infty\), no maximum.
Warning. The official SoSe19 solution answers \(\inf A=\min A=0\), because that exam counted \(0\) as a natural number (\(0^2=(2k-1)\cdot0\) holds for every \(k\)). In your course \(0\notin\mathbb N\) (Skript §1.1), so the answer is 1. In an exam, state the convention you use in one line — then either answer is gradeable.
Let \(a_n=e^{1/n}+e^{-1/n}\), \(n\in\mathbb N\). Determine \(\inf_{n}a_n\) and \(\sup_na_n\).
Solution
- Look first. \(a_1=e+\frac1e\approx3.086\), \(a_2=e^{0.5}+e^{-0.5}\approx2.255\), \(a_{10}\approx2.010\). The terms fall towards 2.
- The sequence is decreasing. Put \(g(t)=e^t+e^{-t}\). For \(0<s<t\): \[g(t)-g(s)=(e^t-e^s)+(e^{-t}-e^{-s})=(e^t-e^s)\left(1-\tfrac1{e^{s+t}}\right)>0,\] since \(e^t>e^s\) and \(e^{s+t}>1\). (Check the factorisation by multiplying out: \(e^{-t}-e^{-s}=-\frac{e^t-e^s}{e^{s+t}}\).) So \(g\) is increasing on \((0,\infty)\). As \(n\) grows, \(t=\frac1n\) shrinks, so \(a_n=g(\frac1n)\) decreases.
- Supremum = maximum = the first term: \(\sup a_n=\max a_n=a_1=e+e^{-1}\).
- Infimum. Each \(a_n>2\): by the inequality \(\frac{u+v}2\ge\sqrt{uv}\) (equivalent to \((\sqrt u-\sqrt v)^2\ge0\)) with \(u=e^{1/n}\), \(v=e^{-1/n}\), \(uv=1\): \(a_n\ge2\), and "=" only if \(u=v\), i.e. \(\frac1n=0\), impossible. And \(a_n\to e^0+e^0=2\) (continuity of exp, Chapter 4 — or simply: the terms get as close to 2 as you like). So \(\inf a_n=2\), not attained: no minimum.
Let \(f,g:D\to\mathbb R\) with \(D\subseteq\mathbb R\). (a) Show: if \(\sup_{x\in D}|f(x)-g(x)|=0\), then \(\sup(f+g)=\sup f+\sup g\). (b) Give an example with \(\sup(f+g)<\sup f+\sup g\). (c) Show that always \(\sup_{x\in D}(f(x)+g(x))\le\sup_{x\in D}f(x)+\sup_{x\in D}g(x)\).
Solution
Notation: \(\sup_{x\in D}f(x)\) means the supremum of the set of values \(\{f(x):x\in D\}\).
(c) first, because it is the heart. Take any \(x\in D\). Since \(\sup f\) is an upper bound of all values of \(f\), \(f(x)\le\sup f\); likewise \(g(x)\le\sup g\). Adding: \(f(x)+g(x)\le\sup f+\sup g\). This holds for every \(x\), so \(\sup f+\sup g\) is an upper bound of \(\{f(x)+g(x):x\in D\}\). The supremum is the least upper bound, hence \(\sup(f+g)\le\sup f+\sup g\). ∎
(b) \(D=[0,1]\), \(f(x)=x\), \(g(x)=-x\). Then \(f+g\equiv0\), so \(\sup(f+g)=0\), but \(\sup f+\sup g=1+0=1\). The two suprema are reached at different points (\(f\) at \(x=1\), \(g\) at \(x=0\)); the sum cannot use both.
(a) If \(f=g\), then \(\sup(f+g)=\sup(2f)=2\sup f=\sup f+\sup g\) (multiplying all values by 2 multiplies every upper bound, and hence the least one, by 2).
Where (a) starts: \(\sup_{x\in D}|f(x)-g(x)|=0\) means \(|f(x)-g(x)|\le0\) for every \(x\), and an absolute value is never negative, so \(f(x)=g(x)\) for all \(x\) — the exam's way of writing "\(f=g\)". Then the computation above applies.
The reusable pattern from (c): to show \(\sup S\le c\), show that \(c\) is an upper bound of \(S\). The exam does not assume \(f\) and \(g\) bounded; with the conventions \(\sup=\infty\) for an unbounded set the statements stay true, and in an exam it is worth one line to say which case you are in.
Determine \(A=\{z\in\mathbb C:\ z=|z|^2\}\).
Solution
- \(|z|^2\) is a real number \(\ge0\). So if \(z=|z|^2\), then \(z\) itself is real and \(\ge0\): \(z=x\) with \(x\ge0\), and then \(|z|=x\).
- The equation becomes \(x=x^2\), i.e. \(x(x-1)=0\), so \(x=0\) or \(x=1\).
- Check: \(0=|0|^2\) ✓, \(1=|1|^2\) ✓. So \(A=\{0,1\}\).
Alternative: write \(z=a+ib\); then \(a+ib=a^2+b^2\). Comparing imaginary parts gives \(b=0\), real parts \(a=a^2\).
Drill: sup, inf, max, min
Ten sets of the kind the exam uses. Decide the four values before you click — and, for each bound, which of the two steps of the recipe you would write down.
- For the exam: determine sup, inf, max, min of a set given by a formula, an inequality, or a sequence — with justification (upper bound + ε-step, or monotonicity + limit). Solve simple equations in ℂ by comparing real and imaginary parts.
- For Chapter 2 you will use constantly: (1) negating statements with quantifiers; (2) Archimedes in the form "for every \(\varepsilon>0\) there is \(n\) with \(\frac1n<\varepsilon\)"; (3) the triangle inequality and its reverse to estimate \(|a_n-a|\); (4) the supremum axiom (it is what makes monotone bounded sequences converge, Satz 2.17).
- For Chapters 3 and 5: computing with complex numbers, \(|z|\), \(\bar z\).
ΣSummary sheet
The whole chapter as one chain of reasoning:
- ℚ is a perfectly good field, but it has holes: √2 ∉ ℚ (Lemma 1.1), and \(\{a\in\mathbb Q:a^2\le2\}\) has no supremum in ℚ (Remark 1.8).
- ℝ closes these holes. That is the supremum axiom: nonempty + bounded above ⇒ sup exists.
- Filling the holes makes ℝ much bigger: ℚ is countable, ℝ is uncountable (Cantor, Lemma 1.3). So almost no real number is computable (Lemma 1.11).
- From the supremum axiom: ℝ is Archimedean (Satz 1.9) ⇒ for every \(a>0\) some \(\frac1n<a\) ⇒ ℚ is dense (Satz 1.10) ⇒ every real is approximated by fractions / decimals.
- ℝ is ordered; ℂ = ℝ² gains \(i^2=-1\) but loses the order (Lemma 1.4). In ℂ only \(|z|\) can be compared.
- To estimate distances, use the triangle inequality (1.12), its reverse (1.13), Cauchy–Schwarz (1.15) and the triangle inequality in ℝⁿ (1.14). These are the tools for all ε-proofs from Chapter 2 on.
| Result | Statement | Typical use |
|---|---|---|
| Supremum axiom | nonempty, bounded above ⇒ sup M ∈ ℝ exists | existence of limits (Ch. 2: monotone sequences) |
| sup recipe | (1) upper bound, (2) ∀ε ∃a∈M: a > s−ε | exam: find sup/inf/max/min |
| Satz 1.9 | ∀a ∃n∈ℕ: a < n; so 1/n < a for a > 0 | choosing n₀ in ε-proofs |
| Satz 1.10 | between any a < b lies a rational | approximation by fractions |
| Satz 1.12 | |x+y| ≤ |x|+|y|, = iff xy ≥ 0 | splitting |aₙ − a| into pieces |
| Satz 1.13 / Blatt 1.1a | | |x|−|y| | ≤ |x−y| | lower bounds; aₙ→a ⇒ |aₙ|→|a| |
| Blatt 1.1b | xy ≤ x²/(2ε) + εy²/2 | splitting products |
| Satz 1.15 | |⟨x,y⟩| ≤ ‖x‖‖y‖, = iff parallel | proving the triangle inequality in ℝⁿ |
| Blatt 1.2 | N1–N3; ‖·‖₁, ‖·‖₂, ‖·‖∞ are norms | checking whether something is a norm |
| Blatt 1.3 | Re z = (z+z̄)/2, Im z = (z−z̄)/(2i), |z|² = z z̄ | computing with ℂ (Ch. 5: sin, cos) |
Exam focus
Sup/inf/max/min of a set is the Chapter 1 exam task (five of nine exams, see "What the exam asks" above). Every exam set is in the explorer and fully worked on this page.
- Write out the first few elements, or rewrite the set as an interval (solve the inequality, e.g. \(|x-1|<1\iff 0<x<2\)).
- Monotone sequence? Then the first element is max (decreasing) or min (increasing), and the limit is the other bound.
- Justify the "not attained" bound with the ε-step (usually via Archimedes).
- Decide max/min: is the bound an element of the set?
- Unbounded ⇒ sup = ∞ / inf = −∞, and max/min don't exist.
?Self-test
Click an answer. You get an explanation either way.
SequencesFolgen
A sequence is an infinite list of numbers. This chapter makes precise what it means for such a list to approach a number, gives you the rules for computing limits, and proves the most important existence theorem of the course: monotone and bounded ⇒ convergent. With it we meet Euler's number \(e\).
Source labels: Skript 2.3 = stated in your lecture script · Blatt 2 · 2.2 = exercise sheet · Klausur WiSe22 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§2.2 = Section 2.2) rather than a numbered result (2.2 = Beispiel 2.2).
0The big picture
Chapter 1 gave us ℝ without holes. Chapter 2 is where that pays off. The storyline:
- Definition of convergence with ε and \(n_0\): "eventually the terms stay as close to \(a\) as you like". Everything else rests on this.
- Limit rules (sum, product, quotient) and the squeeze theorem: with these you almost never need ε again when computing limits.
- Improper convergence (to ±∞) and asymptotic equality (\(a_n\simeq b_n\)): the language for "how fast things grow".
- Monotone sequences: a monotone sequence converges exactly when it is bounded, and its limit is its sup (or inf). This is the supremum axiom of Chapter 1 at work. It gives us \(e\) and shows that the harmonic numbers \(H_n\) grow to ∞.
2.1Convergence [Konvergenz]
A sequence [Folge] \((a_n)_{n\in\mathbb N_0}\) of complex numbers is a map \(\mathbb N_0\to\mathbb C,\ n\mapsto a_n\). We also write \(a_0,a_1,a_2,\dots\) Sometimes the index starts at 1 instead of 0.
Sequences often come from recursions [Rekursionen], where each term is computed from the previous one, the way a loop computes its next value. Example (Skript 2.2): \(a_0=1,\ a_{n+1}=q\,a_n\) gives the explicit formula \(a_n=q^n\). What happens for large \(n\) depends on \(q\): for \(q=2\) the terms grow to ∞, for \(q=\frac12\) they shrink to 0, and for \(q=-1\) they jump between \(-1\) and \(1\) forever.
The definition of convergence
Intuition. "\(a_n\to a\)" means: pick any tolerance ε, however tiny. From some index on, all terms lie within ε of \(a\). A finite number of early terms may do anything; only the "tail" of the sequence matters.
A sequence \((a_n)\) converges to \(a\in\mathbb C\) if for every accuracy \(\varepsilon>0\) there is an \(n_0\in\mathbb N\) such that \(|a_n-a|<\varepsilon\) for all \(n\ge n_0\):
\[\forall\varepsilon>0\ \exists n_0\in\mathbb N\ \forall n\ge n_0:\ |a_n-a|<\varepsilon.\]We write \(\lim_{n\to\infty}a_n=a\) or \(a_n\xrightarrow{n\to\infty}a\); \(a\) is the limit [Grenzwert]. A sequence that does not converge is divergent [divergent].
Read the quantifiers as a game. Your opponent picks ε. You answer with an \(n_0\), which may depend on ε. You win if every term from \(n_0\) on is within ε of \(a\). The sequence converges if you can win against every ε.
Claim: \(\lim_{n\to\infty}\frac1n=0\).
How do you find the proof? Work backwards. You want \(|a_n-0|=\frac1n<\varepsilon\), which is the same as \(n>\frac1\varepsilon\). So choose \(n_0>\frac1\varepsilon\).
Proof, written forwards: Let \(\varepsilon>0\). Since ℝ is Archimedean (Satz 1.9) there is \(n_0\in\mathbb N\) with \(n_0>\frac1\varepsilon\), i.e. \(\frac1{n_0}<\varepsilon\). For all \(n\ge n_0\): \(|a_n-0|=\frac1n\le\frac1{n_0}<\varepsilon\). ∎
Concretely: for \(\varepsilon=0.01\) you need \(n_0>100\), so \(n_0=101\) works.
- Scratch work: simplify \(|a_n-a|\) and bound it from above by something simple that clearly goes to 0, e.g. \(|a_n-a|\le\frac{C}{n}\).
- Solve "simple bound \(<\varepsilon\)" for \(n\). That gives the condition on \(n_0\).
- Write it up: "Let \(\varepsilon>0\). Choose \(n_0\in\mathbb N\) with … (exists by Archimedes). Then for all \(n\ge n_0\): \(|a_n-a|\le\dots<\varepsilon\)."
You may estimate generously. You do not need the smallest possible \(n_0\), just one that works.
Show with the definition that \(a_n=\frac1{\sqrt n}\to0\).
Solution
Let \(\varepsilon>0\). \(|a_n-0|=\frac1{\sqrt n}\le\varepsilon\iff\sqrt n\ge\frac1\varepsilon\iff n\ge\varepsilon^{-2}\). So \(n_0:=\lceil\varepsilon^{-2}\rceil\) works: for all \(n\ge n_0\), \(\frac1{\sqrt n}\le\varepsilon\). ∎
(Here the sheet uses "\(\le\varepsilon\)" instead of "\(<\varepsilon\)". That is allowed; see 2.1b below.)
\(a_n=(-1)^n\) does not converge. To show divergence, negate the definition (flip all quantifiers, as in Chapter 1):
\[\forall a\in\mathbb R\ \exists\varepsilon>0\ \forall n_0\in\mathbb N\ \exists n\ge n_0:\ |a_n-a|\ge\varepsilon.\]Every real \(a\) has distance \(\ge\frac12\) from 1 or from \(-1\) (they are 2 apart). The sequence hits both values infinitely often. So with \(\varepsilon=\frac12\), beyond any \(n_0\) there is a term at distance \(\ge\frac12\) from \(a\). ∎
The script defines sequences and convergence for complex numbers; \(|a_n-a|\) is then the distance in the plane (Chapter 1). A useful criterion: \(a_n\to a\) in ℂ if and only if \(\operatorname{Re}a_n\to\operatorname{Re}a\) and \(\operatorname{Im}a_n\to\operatorname{Im}a\).
Why: \(|\operatorname{Re}z|\le|z|\) and \(|\operatorname{Im}z|\le|z|\) give "⇒"; \(|z|\le|\operatorname{Re}z|+|\operatorname{Im}z|\) (triangle inequality for \(z=x+iy\)) gives "⇐".
Does \(a_n=i^n\) converge?
Solution
The powers of \(i\) repeat with period 4: \(i,\ -1,\ -i,\ 1,\ i,\ -1,\dots\) Their real parts are \(0,-1,0,1,0,-1,\dots\), which do not converge (same argument as for \((-1)^n\): the values \(1\) and \(-1\) keep recurring). By the criterion above, \((i^n)\) diverges. (Only the result was graded; the official remark is "the real parts diverge".)
(a) The limit is unique. A sequence cannot converge to two different numbers.
Solution (a)
Picture: two disjoint bands around \(a\) and \(b\); the tail cannot lie in both. Formally: suppose \(a_n\to a\) and \(a_n\to b\) with \(a\neq b\). Choose any \(\varepsilon<\frac{|a-b|}2\) and \(N_1,N_2\) from the two convergences. For \(n\ge\max(N_1,N_2)\) both \(|a_n-a|<\varepsilon\) and \(|a_n-b|<\varepsilon\), so by the triangle inequality \[|a-b|\le|a-a_n|+|a_n-b|<2\varepsilon<|a-b|,\] a contradiction. ∎ (Trick from Chapter 1: add and subtract \(a_n\).)
(b) Constants don't matter. The definition is equivalent to the versions with "\(|a_n-a|\le\varepsilon\)" and with "\(|a_n-a|<17\varepsilon\)".
Solution (b)
One direction is trivial each time (\(<\varepsilon\) implies \(\le\varepsilon\) and \(<17\varepsilon\)). For the other: given ε, apply the weaker version with a smaller number. For "≤": use \(\varepsilon_0<\varepsilon\); then \(|a_n-a|\le\varepsilon_0<\varepsilon\). For "17ε": use \(\varepsilon_0=\frac\varepsilon{17}\); then \(|a_n-a|<17\varepsilon_0=\varepsilon\).
Why this matters: in proofs you may end up with \(|a_n-a|<2\varepsilon\) or \(<3\varepsilon\). That is fine. (The script's squeeze proof uses \(\frac\varepsilon3\) to land exactly on ε, but either way works.)
(c) Null sequence × bounded = null sequence. If \(b_n\to0\) and \(|c_n|\le C\) for all \(n\), then \(b_nc_n\to0\).
Solution (c)
Let \(\varepsilon>0\). Since \(b_n\to0\), there is \(n_0\) with \(|b_n|<\frac\varepsilon C\) for \(n\ge n_0\). Then \(|b_nc_n|=|b_n||c_n|\le C|b_n|<\varepsilon\). ∎
Very useful in practice: \(\frac{\sin n}{n}\to0\) and \(\frac{(-1)^n}{\sqrt n}\to0\) immediately, because \(\sin n\) and \((-1)^n\) are bounded. Note: the limit rule for products (Satz 2.6c) does not apply here, since \(\sin n\) does not converge.
Rules for limits [Rechenregeln für Grenzwerte]
If \(\lim a_n=a\in\mathbb R\) and \(\lim b_n=b\in\mathbb R\), then
- \(\lim(a_n+b_n)=a+b\)
- \(\lim c\,a_n=c\,a\) for all \(c\in\mathbb R\)
- \(\lim a_nb_n=ab\)
- \(\lim\frac{a_n}{b_n}=\frac ab\), provided \(b\neq0\)
Special case of (d) with \(a_n=1\): \(\lim\frac1{b_n}=\frac1b\) if \(b\neq0\).
Proofs (the script states them without proof)
Extra All four use the same two tools: the triangle inequality from Chapter 1, and "a convergent sequence is bounded" (Satz 2.12 below).
(a) Sum. Given \(\varepsilon>0\), choose \(n_1\) with \(|a_n-a|<\frac\varepsilon2\) for \(n\ge n_1\) and \(n_2\) with \(|b_n-b|<\frac\varepsilon2\) for \(n\ge n_2\). For \(n\ge\max(n_1,n_2)\): \[|(a_n+b_n)-(a+b)|\le|a_n-a|+|b_n-b|<\tfrac\varepsilon2+\tfrac\varepsilon2=\varepsilon.\]
(b) is (c) with the constant sequence \(b_n=c\).
(c) Product. Add and subtract \(a_nb\): \[a_nb_n-ab=a_n(b_n-b)+b(a_n-a),\qquad\text{so}\qquad|a_nb_n-ab|\le|a_n|\,|b_n-b|+|b|\,|a_n-a|.\] \((a_n)\) is bounded, \(|a_n|\le C\). Both terms on the right are "bounded × something that tends to 0", hence tend to 0 (Blatt 2.1c), and so does their sum by (a).
(d) Quotient. By (c) it suffices to show \(\frac1{b_n}\to\frac1b\). Step 1: \(b_n\) stays away from 0. Take \(\varepsilon=\frac{|b|}2\): for large \(n\), \(|b_n-b|<\frac{|b|}2\), and the reverse triangle inequality (Satz 1.13) gives \(|b_n|>|b|-\frac{|b|}2=\frac{|b|}2\). (Same idea as the exam task SoSe23 A5 below.) Step 2: \[\left|\frac1{b_n}-\frac1b\right|=\frac{|b-b_n|}{|b_n|\,|b|}\le\frac{2}{|b|^2}\,|b_n-b|\to0.\ \ ∎\]
The pattern "add and subtract, then bound each piece" is the single most important technique in ε-proofs. You will see it again in the squeeze theorem, in Chapter 3 (Cauchy product) and Chapter 4 (continuity of products).
The rules need both limits to exist as real numbers. You may not split \(\lim\frac{4n^2+3n}{2n^2+1}\) into \(\frac{\lim(4n^2+3n)}{\lim(2n^2+1)}\), because numerator and denominator both go to ∞ and "\(\frac\infty\infty\)" is not a number. First rewrite, then apply the rules.
For large \(n\), the numerator of \(\frac{4n^2+3n-7}{2n^2+1}\) is about \(4n^2\) and the denominator about \(2n^2\), so the quotient is about 2. Formally, divide numerator and denominator by the highest power \(n^2\):
\[\lim_{n\to\infty}\frac{4n^2+3n-7}{2n^2+1}=\lim_{n\to\infty}\frac{4+\frac3n-\frac7{n^2}}{2+\frac1{n^2}}=\frac{4+0-0}{2+0}=2.\]Each piece is justified by a rule: \(\frac1n\to0\) (Ex. 2.4), so \(\frac3n\to0\) by (b), \(\frac1{n^2}=\frac1n\cdot\frac1n\to0\) by (c), and the quotient by (d) since \(2\neq0\). Similarly \(\frac{n^3+4n^2+2}{8n^5-1}\to0\) (divide by \(n^5\)), Skript 2.8.
Investigate convergence: (a) \(\frac{4n^2+3n-6}{(n+2)^2}\), (b) \(\frac{3n^3+5n}{7n^2+4n+4}\), (c) \(\frac{2n+3}{n^2+3n}\), (e) \(\sqrt{4n+1}-\sqrt{4n-3}\).
Solutions
(a) \((n+2)^2=n^2+4n+4\). Divide by \(n^2\): \(\frac{4+\frac3n-\frac6{n^2}}{1+\frac4n+\frac4{n^2}}\to\frac41=4\).
(b) Diverges to \(+\infty\). Estimate: for \(n\ge1\), \(4n\le4n^2\) and \(4\le4n^2\), so \(7n^2+4n+4\le15n^2\), hence \[a_n\ge\frac{3n^3}{15n^2}=\frac n5\to\infty.\] Comparison trick: to show \(a_n\to\infty\), bound \(a_n\) from below by something that goes to ∞ (make the numerator smaller, the denominator larger).
(c) Divide by \(n^2\): \(\frac{\frac2n+\frac3{n^2}}{1+\frac3n}\to\frac01=0\).
(e) "∞ − ∞" is undetermined, so use the conjugate trick [Erweitern]: multiply by \(\frac{\sqrt{4n+1}+\sqrt{4n-3}}{\sqrt{4n+1}+\sqrt{4n-3}}\) and use \((x-y)(x+y)=x^2-y^2\): \[a_n=\frac{(4n+1)-(4n-3)}{\sqrt{4n+1}+\sqrt{4n-3}}=\frac4{\sqrt{4n+1}+\sqrt{4n-3}}\le\frac4{2\sqrt n}=\frac2{\sqrt n},\] because both roots are \(\ge\sqrt n\) (as \(4n-3\ge n\) for \(n\ge1\)). Since \(0\le a_n\le\frac2{\sqrt n}\to0\) (by (d)), the squeeze theorem gives \(a_n\to0\).
General rule for fractions of polynomials Extra (the sheet asks you to guess it): compare the degrees \(p\) (numerator) and \(q\) (denominator). \(p<q\): limit 0. \(p=q\): limit = ratio of leading coefficients. \(p>q\): ±∞ (sign of the ratio of leading coefficients).
Compute \(\lim_{n\to\infty}\frac{3n^2+6n+3}{4n^2+6n+2}\) with justification.
Solution
Divide by \(n^2\): \(\frac{3+\frac6n+\frac3{n^2}}{4+\frac6n+\frac2{n^2}}\). Numerator → 3, denominator → 4 ≠ 0 (sum rule), so by the quotient rule the limit is \(\frac34\). The official solution explicitly names the rules used. Do the same: that is where the points are.
Improper convergence [uneigentliche Konvergenz]
\((a_n)\) diverges to \(+\infty\) if \(\forall K>0\ \exists n_0\in\mathbb N\ \forall n\ge n_0:\ a_n\ge K\). Similarly to \(-\infty\) with \(a_n\le-K\). One also says the sequence converges improperly. Examples: \(n^2\), \(n^3-n\), \(2^n\) all go to \(+\infty\).
Same game as before, but now the opponent names a height \(K\) and you must show that the tail stays above it. A real sequence therefore has three possibilities: it converges to some \(a\in\mathbb R\), it diverges to \(\pm\infty\), or neither (like \((-1)^n\), or \((-1)^nn^2\)).
Show using the definition that \(a_n=n^3\) diverges to \(+\infty\). State the definition explicitly.
Solution
Definition: \((a_n)\) diverges to \(+\infty\) if \(\forall K>0\ \exists n_0\in\mathbb N\ \forall n\ge n_0:\ a_n\ge K\).
Proof: Let \(K>0\). Choose \(n_0=\lceil\sqrt[3]K\rceil\in\mathbb N\). Then for all \(n\ge n_0\): \(a_n=n^3\ge n_0^3\ge K\). ∎
Points were given for (1) stating the definition, (2) the choice of \(n_0\), (3) the chain of inequalities.
Determine the behaviour of (a) \((-1)^nn^2\), (b) \(\frac{2^n}{n^3}\), (c) \(q^n\), (d) \(nq^n\) for \(|q|<1\), (e) \(\sqrt{b_n}\) when \(b_n\to b\ge0\).
(a) \((-1)^nn^2\): divergent, and not even to ±∞
\(|a_n|=n^2\ge K\) for \(n\ge\lceil\sqrt K\rceil\), so the sequence is unbounded, hence divergent (convergent sequences are bounded, Satz 2.12 below). It doesn't go to \(+\infty\) either: the odd terms are negative, so the tail never stays above \(K=1\). Likewise not to \(-\infty\).
(b) \(\frac{2^n}{n^3}\to\infty\) (exponential beats polynomial)
Binomial theorem: \(2^n=(1+1)^n=\sum_{k=0}^n\binom nk\ge\binom n4\) for \(n\ge4\) (all terms are positive, keep just one). So \[\frac{2^n}{n^3}\ge\frac{n(n-1)(n-2)(n-3)}{4!\,n^3}\ge\frac{n-6}{4!},\] using \((n-1)(n-2)(n-3)=n^3-6n^2+11n-6\ge n^3-6n^2\) for all \(n\in\mathbb N\). And \(\frac{n-6}{24}\ge K\) for \(n\ge24K+6\). ∎
Idea to remember: to beat \(n^3\), pick the binomial term with \(n^4\) in it.
(c) \(q^n\): full classification
- \(q=1\): constant 1, converges to 1.
- \(q=-1\): alternates, divergent (Skript 2.5).
- \(q>1\): write \(q=1+(q-1)\). Bernoulli (Satz 2.18 below): \(q^n\ge1+n(q-1)\to\infty\). Diverges to ∞.
- \(q<-1\): \(|q^n|\to\infty\), unbounded with alternating sign, divergent.
- \(|q|<1\): \(q^n\to0\). The sheet takes logs: \(|q|^n<\varepsilon\iff n>\frac{\log\varepsilon}{\log|q|}\) (the inequality flips because \(\log|q|<0\)). The script proves the same with Bernoulli in Lemma 3.2.
(d) \(nq^n\to0\) for \(|q|<1\)
For \(q=0\) trivial. Otherwise write \(|q|=\frac1{1+p}\) with \(p>0\). Binomial theorem, keep only the \(p^2\) term: \[|q|^n=\frac1{(1+p)^n}\le\frac1{\binom n2p^2}=\frac{2}{p^2\,n(n-1)}\quad\Rightarrow\quad n|q|^n\le\frac2{p^2}\cdot\frac1{n-1}\to0.\] Linear growth \(n\) loses against exponential decay \(q^n\). (The same holds for \(n^mq^n\) with any fixed \(m\); it follows from "exp beats every power", Satz 5.6 in Chapter 5.)
(e) \(\sqrt{b_n}\to\sqrt b\)
Key inequality: \(|\sqrt b-\sqrt{b_n}|\le\sqrt{|b-b_n|}\). (Check for \(b_n\le b\): square both sides; it reduces to \(b_n\le\sqrt{b_nb}\), true.) Given ε, choose \(N\) with \(|b_n-b|\le\varepsilon^2\) for \(n\ge N\). Then \(|\sqrt{b_n}-\sqrt b|\le\varepsilon\). ∎ (Later, in Chapter 4, this is just "√ is continuous".)
Asymptotic equality [asymptotisch gleich]
Sequences \((a_n)\), \((b_n)\) of nonzero numbers are asymptotically equal, \(a_n\simeq b_n\) for \(n\to\infty\), if \[\lim_{n\to\infty}\frac{a_n}{b_n}=1\quad(\text{equivalently }\lim\tfrac{b_n}{a_n}=1).\]
Intuition: the relative error goes to 0. So \(n+1\simeq n\), \(n^2+7n+13\simeq n^2\), and any polynomial is \(\simeq\) its leading term (Skript 2.11). Only the large terms matter; finitely many terms play no role.
- If \(a_n\simeq b_n\) and \(a_n\to a\neq0\), then \(b_n\to a\). Proof: \(b_n=\frac{b_n}{a_n}\cdot a_n\to1\cdot a\).
- \(a_n\simeq b_n\) does not imply \(a_n-b_n\to0\). Counterexample: \(a_n=n+1,\ b_n=n\). Ratio → 1, difference = 1. (Asymptotic equality is about ratios, not differences.)
- If \(a_n\simeq c_n\) and \(b_n\simeq d_n\), then \(\frac{a_n}{b_n}\simeq\frac{c_n}{d_n}\). Proof: \(\frac{a_n/b_n}{c_n/d_n}=\frac{a_n}{c_n}\cdot\frac{d_n}{b_n}\to1\cdot1\).
- If \(|a_n|\to\infty\), then eventually \(a_n\neq0\) and \(\frac1{a_n}\to0\). Proof: given ε, take \(K=\frac1\varepsilon\): eventually \(|a_n|\ge\frac1\varepsilon\), i.e. \(\left|\frac1{a_n}\right|\le\varepsilon\).
How this is used (sheet example): replace numerator and denominator by their leading terms. \[\frac{9n^8+28n^3+17n^6+544}{-\pi^2n^8+42n^{15/2}-98n^{5/4}+9}\simeq\frac{9n^8}{-\pi^2n^8}=-\frac9{\pi^2}\quad\Rightarrow\quad\lim=-\frac9{\pi^2}\text{ by (a), (c).}\] Careful: \(n^{15/2}\) is smaller than \(n^8\), so the leading term below is \(-\pi^2n^8\).
Boundedness, inequalities, squeeze theorem
Every convergent real sequence is bounded [beschränkt]: \(\exists C>0\ \forall n:\ |a_n|\le C\). The converse is false: \((-1)^n\) is bounded but divergent.
Proof (the script omits it, so I supply one)
Extra Take \(\varepsilon=1\) in the definition: there is \(n_0\) with \(|a_n-a|<1\), hence \(|a_n|\le|a|+1\), for all \(n\ge n_0\). Only finitely many terms \(a_1,\dots,a_{n_0-1}\) are left, and a finite set has a maximum. So \(C=\max\{|a_1|,\dots,|a_{n_0-1}|,|a|+1\}\) works. ∎
This "tail + finitely many" pattern is exactly the exam task SoSe23 A5 below.
If \(a_n\to a\), \(b_n\to b\), and \(a_n\le b_n\) for all but finitely many \(n\), then \(a\le b\).
Careful: strict inequality is not preserved. \(\frac1n>0\) for all \(n\), but the limit is \(0\), not \(>0\).
Proof (the script omits it)
Extra By contradiction: suppose \(a>b\). Take \(\varepsilon=\frac{a-b}2>0\), the half-distance. For large \(n\), \(a_n>a-\varepsilon=\frac{a+b}2\) and \(b_n<b+\varepsilon=\frac{a+b}2\), so \(b_n<\frac{a+b}2<a_n\) — for all large \(n\), contradicting \(a_n\le b_n\) for all but finitely many \(n\). ∎
Picture: two disjoint bands around \(b\) (lower) and \(a\) (upper); eventually \(b_n\) sits in the lower band and \(a_n\) in the upper one.
If \(a_n\le b_n\le c_n\) for all but finitely many \(n\), and \(\lim a_n=\alpha=\lim c_n\), then \(\lim b_n=\alpha\).
Proof
Let \(\varepsilon>0\). Choose \(n_1,n_2\) with \(|a_n-\alpha|<\frac\varepsilon3\) for \(n\ge n_1\) and \(|c_n-\alpha|<\frac\varepsilon3\) for \(n\ge n_2\). For \(n\ge\max(n_1,n_2)\): \[|b_n-\alpha|\le|b_n-a_n|+|a_n-\alpha|,\qquad |b_n-a_n|=b_n-a_n\le c_n-a_n\le|c_n-\alpha|+|\alpha-a_n|<\tfrac{2\varepsilon}3.\] Together \(|b_n-\alpha|<\varepsilon\). ∎ (Uses the "add and subtract" trick from Chapter 1 twice.)
Example (Skript 2.15): \(0\le\frac1{n+\sqrt n+5}\le\frac1n\), and both bounds → 0, so the middle → 0.
Let \(a_n>0\) for all \(n\) with \(\lim a_n=a>0\). Show: \(\exists c>0\ \forall n\in\mathbb N:\ a_n\ge c\).
Solution
- Take \(\varepsilon=\frac a2\) in the definition: there is \(n_0\) with \(|a_n-a|<\frac a2\) for \(n\ge n_0\), so \(a_n>a-\frac a2=\frac a2\) for \(n\ge n_0\).
- The finitely many remaining terms \(a_1,\dots,a_{n_0-1}\) are all \(>0\).
- Set \(c:=\min\{a_1,\dots,a_{n_0-1},\frac a2\}\). A minimum of finitely many positive numbers is positive, and \(a_n\ge c\) for all \(n\). ∎
Why ε = a/2? You need the band around \(a\) to stay away from 0; any \(\varepsilon<a\) works, \(\frac a2\) is the standard choice.
2.3 Products. If \(a_n\to0\) and \(b_n\) diverges, \(a_nb_n\) can do anything:
- converge: \(a_n=\frac1{n^2},\ b_n=n\Rightarrow a_nb_n=\frac1n\to0\); or \(a_n=\frac1n,\ b_n=n\Rightarrow a_nb_n=1\to1\).
- diverge: \(a_n=\frac1n,\ b_n=n^2\Rightarrow a_nb_n=n\to\infty\).
So "0 · ∞" is undetermined, and 2.1c needs \(b_n\) bounded.
2.4 Averages (Cesàro means). Let \(s_n=\frac1n(a_1+\dots+a_n)\). (a) If \(a_n\to a\), then \(s_n\to a\). (b) The converse fails: for \(a_n=(-1)^n\), \(s_{2n}=0\) and \(s_{2n+1}=-\frac1{2n+1}\), so \(s_n\to0\) although \(a_n\) diverges.
Proof of (a)
Idea: split the average into the finitely many early terms (their contribution is divided by \(n\), so it dies) and the tail (where every term is close to \(a\)). Given ε, choose \(N\) with \(|a_n-a|<\frac\varepsilon2\) for \(n\ge N\). For \(k>N\): \[|s_k-a|\le\frac1k\sum_{n=1}^N|a_n-a|+\frac1k\sum_{n=N+1}^k|a_n-a|\le\frac{A}{k}+\frac{\varepsilon}2,\quad A:=\sum_{n=1}^N|a_n-a|.\] For \(k>\frac{2A}\varepsilon\) the first term is \(<\frac\varepsilon2\) too. ∎
What you must know for 2.1 (script: "Minimale Inhalte"): the definition of convergence and improper convergence with quantifiers; asymptotic equality; computing limits; the squeeze theorem.
2.2Monotone sequences [Monotone Folgen]
\((a_n)\) is monotonically increasing [monoton wachsend] if \(a_n\le a_{n+1}\) for all \(n\), and monotonically decreasing [monoton fallend] if \(a_n\ge a_{n+1}\). With strict inequalities: strictly increasing/decreasing [streng monoton].
How to check monotonicity in practice: look at the sign of \(a_{n+1}-a_n\), or (for positive terms) compare \(\frac{a_{n+1}}{a_n}\) with 1.
For every increasing sequence, \(\lim a_n=\sup_{n}a_n\). For every decreasing sequence, \(\lim a_n=\inf_na_n\).
In particular: every bounded monotone sequence converges, and every unbounded monotone sequence diverges to \(\pm\infty\).
Proof for increasing sequences
Case 1: bounded. By the supremum axiom \(s=\sup a_n\in\mathbb R\) exists. Let \(\varepsilon>0\). Since \(s\) is the least upper bound, \(s-\varepsilon\) is not an upper bound, so some \(a_{n_0}>s-\varepsilon\) (this is step 2 of the sup recipe from Chapter 1). By monotonicity, for all \(n\ge n_0\): \[s-\varepsilon<a_{n_0}\le a_n\le s\quad\Rightarrow\quad|a_n-s|<\varepsilon.\]
Case 2: unbounded. Being increasing, it can only be unbounded above: for every \(K\) some \(a_{n_0}>K\), and then \(a_n\ge a_{n_0}>K\) for all \(n\ge n_0\). So \(a_n\to\infty=\sup a_n\). ∎
Why this theorem is so important: it proves a limit exists without knowing its value in advance. Most limits in this course (\(e\), series with positive terms, Heron's method) are found this way. It is also the reason ℚ would not work: in ℚ, the increasing bounded sequence 1, 1.4, 1.41, … has no limit.
Let \(0<a_n<1\) for all \(n\). Show that \(x_0=a_0,\ x_n=x_{n-1}a_n\) converges.
Solution
\(x_n=a_0a_1\cdots a_n>0\), so the sequence is bounded below by 0. It is decreasing: \(x_n=x_{n-1}a_n<x_{n-1}\) because \(a_n<1\) and \(x_{n-1}>0\). Bounded + monotone ⇒ convergent (Satz 2.17). ∎
Note: we don't know the limit (it may or may not be 0), and we don't need it.
\(a_0=1\), \(a_n=\frac12\left(a_{n-1}+\frac2{a_{n-1}}\right)\). Show that \((a_n)\) converges and find the limit.
Idea: if \(a\) is too big, \(\frac2a\) is too small, and their average is a better guess for \(\sqrt2\). This is the Babylonian square-root algorithm, and it is what a computer effectively does.
Solution in four steps
- \(a_n>0\) by induction: \(a_0=1>0\); if \(a_{n-1}>0\), then \(a_n\) is half a sum of positives.
- \(a_n^2\ge2\) for \(n\ge1\): \[a_n^2-2=\tfrac14\left(a_{n-1}+\tfrac2{a_{n-1}}\right)^2-2=\tfrac14\left(a_{n-1}-\tfrac2{a_{n-1}}\right)^2\ge0.\] (Expand: \((x+y)^2-4xy=(x-y)^2\) with \(xy=2\).)
- Decreasing from \(n=1\) on: \(a_n-a_{n+1}=a_n-\frac12\left(a_n+\frac2{a_n}\right)=\frac{a_n^2-2}{2a_n}\ge0\). Bounded below by 0 ⇒ convergent (Satz 2.17), limit \(a\ge0\). In fact \(a\ge\sqrt2>0\) because \(a_n\ge\sqrt2\) for all \(n\ge1\) (Satz 2.13).
(Note: \(a_0=1<a_1=1.5\), so the sequence is monotone only from index 1 on. That's fine, since finitely many terms don't matter.) - Find the limit: let \(n\to\infty\) on both sides of the recursion. \(a_n\to a\) and \(a_{n-1}\to a\), and since \(a\neq0\), the limit rules give \(a=\frac12\left(a+\frac2a\right)\), i.e. \(a^2=2\). With \(a>0\): \(a=\sqrt2\).
General method for recursions: first show the limit exists (monotone + bounded), then pass to the limit in the recursion. Doing step 4 without steps 1–3 is a classic mistake: \(a_{n+1}=2a_n\), \(a_0=1\) "gives" \(a=2a\Rightarrow a=0\), but the sequence goes to ∞.
Three tools: Bernoulli, binomial theorem, geometric sum
The script proves Bernoulli's inequality below "mit vollständiger Induktion" without explaining the method, and induction is used again in Chapters 5, 6, 8 and 10. Induction [vollständige Induktion] proves a statement \(A(n)\) for every \(n\in\mathbb N\) in two steps:
- Base case [Induktionsanfang]: show \(A(n_0)\) for the smallest \(n_0\) (often 0 or 1).
- Induction step [Induktionsschritt]: show that if \(A(n)\) holds (the induction hypothesis [Induktionsvoraussetzung]), then \(A(n+1)\) holds.
Picture: a row of dominoes. Step 1 knocks over the first one; step 2 says each domino knocks over the next; therefore all of them fall. Formally you get \(A(n_0)\Rightarrow A(n_0+1)\Rightarrow A(n_0+2)\Rightarrow\dots\), reaching every \(n\) in finitely many moves.
Tiny example: \(1+2+\dots+n=\frac{n(n+1)}2\). Base \(n=1\): \(1=\frac{1\cdot2}2\) ✓. Step: \(1+\dots+n+(n+1)=\frac{n(n+1)}2+(n+1)=\frac{(n+1)(n+2)}2\) ✓. The step is where the hypothesis gets used — always say out loud where you used it.
For all \(n\in\mathbb N_0\) and \(x>-1\): \((1+x)^n\ge1+nx\).
Extra When is it an equality? Exactly when \(x=0\), or when \(n\in\{0,1\}\) (then both sides are literally the same). For \(n\ge2\) and \(x\neq0\) the inequality is strict, and the binomial theorem says by how much: the terms you dropped, starting with \(\binom n2x^2>0\).
Proof by induction (the script only says "by induction")
Extra \(n=0\): \(1\ge1\). Step \(n\to n+1\): since \(1+x>0\), multiplying the induction hypothesis by \(1+x\) keeps the direction: \[(1+x)^{n+1}\ge(1+nx)(1+x)=1+(n+1)x+nx^2\ge1+(n+1)x. \ \ ∎\]
Use: it turns a power into something linear, which is easy to compare. E.g. \(q^n\to\infty\) for \(q>1\) (Blatt 3.1c) and \(q^n\to0\) for \(|q|<1\) (Lemma 3.2).
Why Extra: multiply out \((a+b)(a+b)\cdots(a+b)\) (\(n\) brackets). Each product term picks \(a\) from some brackets and \(b\) from the rest. The term \(a^kb^{n-k}\) arises once for every way of choosing which \(k\) of the \(n\) brackets give an \(a\), and there are \(\binom nk\) such choices ("\(n\) choose \(k\)"). Example: \((a+b)^3=a^3+3a^2b+3ab^2+b^3\), with \(\binom30,\binom31,\binom32,\binom33=1,3,3,1\).
Use in estimates: if all terms are \(\ge0\), dropping terms makes the sum smaller. That gave \(2^n\ge\binom n4\) and \((1+p)^n\ge\binom n2p^2\) above.
For \(q\in\mathbb C\), \(q\neq1\), \(n\in\mathbb N_0\): \[\sum_{j=0}^nq^j=\frac{1-q^{n+1}}{1-q}.\]
Proof (telescoping sum)
\((1-q)\sum_{j=0}^nq^j=\sum_{j=0}^n(q^j-q^{j+1})=(1-q)+(q-q^2)+\dots+(q^n-q^{n+1})=1-q^{n+1}\). Everything in between cancels. Divide by \(1-q\neq0\). ∎
Example: \(1+\frac12+\frac14+\dots+\frac1{2^n}=\frac{1-(1/2)^{n+1}}{1/2}=2-\frac1{2^n}<2\).
Euler's number \(e\)
The sequence \(a_n=\left(1+\frac1n\right)^n\) converges. Its limit \[e:=\lim_{n\to\infty}\left(1+\frac1n\right)^n=\sup_n\left(1+\frac1n\right)^n\approx2.718\] is Euler's number [Eulersche Zahl].
Intuition: 100% interest per year, credited \(n\) times a year at \(\frac1n\) each. As \(n\to\infty\) (continuous compounding) the money grows by the factor \(e\). With interest rate \(x\) instead of 1 the same computation gives \(\left(1+\frac xn\right)^n\to e^x\) — that is Lemma 6.24, proved in Chapter 6 once the logarithm is available.
Proof: increasing and bounded by 3
Increasing. For \(n\ge2\), show \(\frac{a_n}{a_{n-1}}\ge1\). Rewrite: \[\frac{a_n}{a_{n-1}}=\left(\frac{1+\frac1n}{1+\frac1{n-1}}\right)^n\left(1+\frac1{n-1}\right),\qquad \frac{1+\frac1n}{1+\frac1{n-1}}=\frac{(n+1)(n-1)}{n^2}=1-\frac1{n^2}.\] Bernoulli with \(x=-\frac1{n^2}>-1\) gives \(\left(1-\frac1{n^2}\right)^n\ge1-\frac1n\), so \[\frac{a_n}{a_{n-1}}\ge\left(1-\frac1n\right)\frac n{n-1}=\frac{n-1}n\cdot\frac n{n-1}=1.\]
Bounded. Binomial theorem, then estimate each term: \[a_n=\sum_{k=0}^n\binom nk\frac1{n^k},\qquad \binom nk\frac1{n^k}=\frac1{k!}\cdot\underbrace{\frac{n(n-1)\cdots(n-k+1)}{n^k}}_{\le1}\le\frac1{k!}\le\frac1{2^{k-1}}\ (k\ge1),\] since \(k!=1\cdot2\cdot3\cdots k\ge2^{k-1}\). With the geometric sum: \[a_n\le1+\sum_{i=0}^{n-1}\frac1{2^i}\le1+\frac1{1-\frac12}=3.\] Increasing + bounded ⇒ convergent (Satz 2.17). ∎
Numerically \(a_{100}=2.70481\), and \(a_{100}\le a_n\le3\) for \(n\ge100\), so \(e\in[2.7,3]\) by Satz 2.13.
Harmonic numbers and Quicksort
\(H_n=1+\frac12+\frac13+\dots+\frac1n\). The average number of comparisons of Quicksort on \(n\) random numbers is \(C_n=2(n+1)H_n-2n\), so the growth of \(H_n\) decides how fast Quicksort is.
Claim: \(H_n\to\infty\). Trick: group the terms in blocks of length 1, 2, 4, 8, … and bound each term in a block below by the smallest one: \[H_{2^k}=1+\tfrac12+\underbrace{\left(\tfrac13+\tfrac14\right)}_{\ge2\cdot\frac14=\frac12}+\underbrace{\left(\tfrac15+\dots+\tfrac18\right)}_{\ge4\cdot\frac18=\frac12}+\dots\ge1+\frac k2.\] \((H_n)\) is increasing and unbounded, so \(H_n\to\infty\) by Satz 2.17. More precisely \(H_n\ge1+\frac12\lfloor\log_2n\rfloor\).
The divergence is extremely slow: \(H_n\) first reaches 100 at \(n\approx1.5\cdot10^{43}\). This grouping trick reappears in Chapter 3 (harmonic series) and the growth \(H_n\approx\ln n\) in Chapter 9.
What you must know for 2.2: Satz 2.17 (monotone convergence), the binomial theorem and the geometric sum formula, and the definition of \(e\).
Mixed exercises from the script [Kapitel 13: Verschiedene Aufgaben]
Chapter 13 of the script is a pool of exam-style exercises with full solutions. Two of them are limits of the kind this chapter is about; they are worth doing now, because exams love exactly this format.
Does \(\lim_{x\to\infty}\sqrt x\left(\sqrt{x+1}-\sqrt x\right)\) exist?
Here \(x\) runs through the real numbers, a limit defined only in Chapter 4. Nothing changes for us: read it as \(n\to\infty\), and the computation is the same — in fact "\(x\to\infty\)" follows from the sequence version, since \(f(x_n)\to L\) for every sequence \(x_n\to\infty\).
Solution
Directly this is "\(\infty\cdot0\)", which says nothing. The standard trick for a difference of roots: multiply and divide by the conjugate \(\sqrt{x+1}+\sqrt x\), using \((u-v)(u+v)=u^2-v^2\): \[\sqrt x\left(\sqrt{x+1}-\sqrt x\right)=\sqrt x\cdot\frac{(x+1)-x}{\sqrt{x+1}+\sqrt x}=\frac{\sqrt x}{\sqrt{x+1}+\sqrt x}.\] Now divide numerator and denominator by \(\sqrt x\): \[=\frac1{\sqrt{1+\frac1x}+1}\xrightarrow[x\to\infty]{}\frac1{1+1}=\frac12.\] The method to reuse: a difference of two square roots that both blow up ⇒ multiply by the conjugate.
\[G=\lim_{n\to\infty}\frac{5n^8+23(n+2)^{-9/2}+\pi(\ln n)^5n^7+n^9e^{-n^2}\arctan n}{5\sqrt{7n^{10}+28n^{-3/2}+14n}+3n^8+2n^7(\ln n)^3}.\]
Solution
Step 1: find the leading term on each side (this is the whole task; the rest is bookkeeping).
- Numerator: \(23(n+2)^{-9/2}\to0\) and \(n^9e^{-n^2}\arctan n\to0\) (exp beats every power, Satz 5.6; \(\arctan\) is bounded by \(\frac\pi2\)). \((\ln n)^5n^7\) is smaller than \(n^8\), because \(\frac{(\ln n)^5}{n}\to0\) (Satz 5.8). So the numerator behaves like \(5n^8\).
- Denominator: \(\sqrt{7n^{10}+\dots}\approx\sqrt7\,n^5\), which is far below \(n^8\); \(n^7(\ln n)^3\) is below \(n^8\) for the same reason as above. So the denominator behaves like \(3n^8\).
Step 2: make it precise by dividing numerator and denominator by \(n^8\). Every term listed as negligible turns into something with limit 0, and \[G=\frac{5+0+0+0}{0+3+0}=\frac53.\] The method to reuse: divide by the largest power that occurs, then use the growth hierarchy \((\ln n)^a\ll n^b\ll e^{cn}\) (Chapter 5).
!What the exam asks from Chapter 2
Across the nine exams, Chapter 2 appears in five (SoSe19, WiSe19, WiSe22, SoSe23, WiSe23), always as one of five task types. Every one is worked on this page:
| Type | Exam tasks | What earns the points |
|---|---|---|
| Compute a limit | WiSe23 1a (\(\frac{3n^2+6n+3}{4n^2+6n+2}\)), WiSe19 1c (\(n^7\sin 3^{-n}\)) | divide by the highest power; name the rules you use (sum, quotient, squeeze) |
| ε- or K-definition proof | WiSe22 A3 (\(n^3\to\infty\)) | write the definition with quantifiers; choose \(n_0\) explicitly; clean inequality chain |
| "tail + finitely many" proof | SoSe23 A5 (\(a_n\to a>0\Rightarrow a_n\ge c>0\)) | ε = a/2 for the tail, minimum over the finitely many rest |
| monotone + bounded | SoSe19 A2 (\(x_n=x_{n-1}a_n\)) | show both properties, then cite Satz 2.17 |
| complex sequence | WiSe22 1e (\(i^n\)) | look at real/imaginary parts |
- Fraction of polynomials (or roots of them)? Divide by the highest power; compare degrees.
- Difference of two roots that both grow? Conjugate trick, then squeeze.
- Something oscillating (\(\sin n\), \((-1)^n\)) times something → 0? "Null × bounded" (Blatt 2.1c) or squeeze.
- Power \(q^n\), or \(n^k\) against \(q^n\)? Table in the summary; exponential beats polynomial.
- Defined by a recursion? Monotone + bounded first, then pass to the limit in the recursion.
- Asked "with the definition"? Write the quantifier statement, do the scratch work backwards, write it up forwards.
Drill: which tool, and what is the limit?
Ten items in exam format. Decide the answer and the rule you would name in the exam.
- For the exam: the five task types above.
- For Chapter 3 (series), which is built entirely on this chapter: a series is the sequence of its partial sums, so every convergence question about series is a sequence question. You need: the geometric sum formula (Satz 2.20) and \(q^n\to0\) for \(|q|<1\) — together they give the geometric series; monotone + bounded ⇒ convergent — it is the reason series with positive terms converge exactly when their partial sums are bounded (Satz 3.6); asymptotic equality — it becomes a convergence test (Satz 3.10); the harmonic numbers \(H_n\to\infty\) — the harmonic series diverges.
- For Chapter 4: the definition of continuity uses sequences: "\(x_n\to x\Rightarrow f(x_n)\to f(x)\)".
ΣSummary sheet
- \(a_n\to a\iff\forall\varepsilon>0\ \exists n_0\ \forall n\ge n_0:\ |a_n-a|<\varepsilon\). Limits are unique; constants in front of ε don't matter; only the tail counts.
- Compute limits with the rules (only when both limits exist and are finite, denominator ≠ 0), by dividing by the highest power, with the conjugate trick for √-differences, and with the squeeze theorem.
- \(a_n\to\infty\iff\forall K\ \exists n_0\ \forall n\ge n_0:\ a_n\ge K\). To show it, bound \(a_n\) from below by something that goes to ∞.
- \(a_n\simeq b_n\iff a_n/b_n\to1\): replace sums by their largest term. It says nothing about \(a_n-b_n\).
- Convergent ⇒ bounded. Null sequence × bounded ⇒ null sequence.
- Monotone + bounded ⇒ convergent, limit = sup (increasing) or inf (decreasing). For recursions: first existence, then pass to the limit in the recursion.
- \(e=\lim(1+\frac1n)^n\in[2.7,3]\); \(H_n\to\infty\) (slowly, about like \(\log n\)).
| Sequence | Behaviour | Why |
|---|---|---|
| \(q^n\) | → 0 for |q| < 1; → 1 for q = 1; → ∞ for q > 1; divergent for q ≤ −1 | Bernoulli (Blatt 3.1c) |
| \(n\,q^n\), |q| < 1 | → 0 | binomial, keep the \(p^2\) term (Blatt 3.1d) |
| \(2^n/n^3\) | → ∞ | binomial, keep \(\binom n4\) (Blatt 3.1b) |
| \(1/\sqrt n\), \(\sqrt{4n+1}-\sqrt{4n-3}\) | → 0 | definition; conjugate + squeeze (Blatt 2.2) |
| polynomial / polynomial | 0, ratio of leading coefficients, or ±∞ | compare degrees |
| \((1+1/n)^n\) | ↗ e | monotone + bounded (Satz 2.21) |
| \(H_n\) | ↗ ∞ | block grouping |
Exam focus
See "What the exam asks" above for the full list; the one task not worked elsewhere on the page follows.
Compute \(\lim_{n\to\infty}n^7\sin(3^{-n})\). (Only the result counted: 0.)
Why 0
Use \(|\sin x|\le|x|\) (it follows from the mean value theorem in Chapter 7, since \(|\cos|\le1\)): \(0\le|n^7\sin(3^{-n})|\le\frac{n^7}{3^n}\). And \(\frac{n^7}{3^n}=n^7q^n\) with \(q=\frac13\) goes to 0, because exponential decay beats every power of \(n\) (Blatt 3.1d idea; in general it follows from Satz 5.6 in Chapter 5). Squeeze theorem ⇒ 0.
?Self-test
SeriesReihen
A series is what you get when you keep adding the terms of a sequence: \(a_1+a_2+a_3+\dots\). This chapter explains when such an infinite sum has a value, gives you a toolbox of convergence tests (the most frequent exam topic of the whole course), warns you about rearranging infinite sums, and uses all of it to define the exponential function.
Source labels: Skript 3.2 = stated in your lecture script · Blatt 4 · 4.2 = exercise sheet · Klausur WiSe20 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§3.4 = Section 3.4) rather than a numbered result (3.4 = Lemma 3.4).
0The big picture
There is nothing mysterious about an infinite sum. It is simply the limit of a sequence, namely the sequence of partial sums \(s_n=a_1+\dots+a_n\). So everything from Chapter 2 applies. The chapter in four steps:
- 3.1 Definition and the three model series you must know by heart: geometric (converges iff \(|q|<1\)), harmonic \(\sum\frac1k\) (diverges), telescoping (everything cancels).
- 3.2 Tests that decide convergence without computing the sum: necessary condition \(a_k\to0\), comparison (majorant/minorant), asymptotic comparison, ratio test, Leibniz test for alternating series.
- 3.3 Rules: sums of series behave well; rearranging terms is only safe for absolutely convergent series; products via the Cauchy product.
- 3.4 The exponential series \(\exp(z)=\sum\frac{z^k}{k!}\) and its functional equation \(e^{w+z}=e^we^z\).
This chapter is Chapter 2 applied to one special kind of sequence. You will need, from Chapter 2:
- the definition of convergence and the limit rules (ε–n₀, rules);
- monotone + bounded ⇒ convergent (Satz 2.17, §2.2) — it is behind almost every test below;
- the geometric sum \(1+q+\dots+q^n=\frac{1-q^{n+1}}{1-q}\) (Satz 2.20) and Bernoulli's inequality (tools);
- asymptotic equality \(a_n\simeq b_n\) (Def. 2.10) and the harmonic numbers \(H_n\to\infty\) (§2.2);
- for complex series: \(|z|\) and the triangle inequality (Chapter 1).
3.1Definition and examples [Definition und Beispiele]
\(\displaystyle\sum_{k=1}^n a_k\) is shorthand for \(a_1+a_2+\dots+a_n\): let \(k\) run from the lower number to the upper one and add up. The letter \(k\) is a placeholder [Summationsindex]; \(\sum_{k=1}^3k^2=\sum_{j=1}^3j^2=1+4+9\).
Index shift [Indexverschiebung]: substituting \(j=k-1\) moves both limits: \(\sum_{k=1}^{n}a_k=\sum_{j=0}^{n-1}a_{j+1}\). The same list of numbers is added, only the counter is renamed. The script uses this constantly (e.g. in the proof of Satz 2.21 and in 3.23).
Why an infinite sum needs a definition at all. You cannot actually perform infinitely many additions. Picture walking towards a wall 1 m away, each step covering half of the remaining distance: \(\frac12+\frac14+\frac18+\dots\). After \(n\) steps you have walked \(s_n=1-\frac1{2^n}\) metres. You never "finish", but the numbers \(s_n\) approach 1, so it is natural to say the infinite sum is 1. That is exactly the definition: add finitely many terms, then take the limit.
Let \((a_n)\) be a sequence of complex numbers and \(s_n=\sum_{k=1}^na_k\) the \(n\)-th partial sum [Partialsumme]. The sequence \((s_n)\) is called an infinite series [unendliche Reihe], written \(\sum_{k=1}^\infty a_k\).
If \((s_n)\) converges, the series is convergent and its limit \(s=\lim s_n\) is the value or sum of the series: \(s=\sum_{k=1}^\infty a_k\). Otherwise the series is divergent. For real terms with \(s_n\to\pm\infty\) one writes \(\sum a_k=\pm\infty\).
Keep apart the terms \(a_k\) and the partial sums \(s_n\). "The series converges" is a statement about \(s_n\). Example: \(a_k=\frac1k\to0\) (terms converge), but \(s_n=H_n\to\infty\) (series diverges).
The harmonic series [harmonische Reihe]
Its partial sums are the harmonic numbers \(H_n\), and in Chapter 2 we showed \(H_{2^k}\ge1+\frac k2\) by grouping terms into blocks of length 1, 2, 4, 8, … This is the standard divergent series: terms go to 0, and yet the sum is infinite.
The geometric series [geometrische Reihe]
For \(z\in\mathbb C\) with \(|z|<1\): \[\sum_{k=0}^\infty z^k=\frac1{1-z}.\] For real \(q\ge1\): \(\sum_{k=0}^\infty q^k=\infty\). For \(q=-1\) the series diverges.
Proof
For \(z\neq1\) the partial sums are known from Satz 2.20: \(s_n=\frac{1-z^{n+1}}{1-z}\). So everything depends on \(z^{n+1}\).
\(|z|<1\): show \(z^{n+1}\to0\). If \(z=0\), clear. Otherwise \(\frac1{|z|}>1\), write \(\frac1{|z|}=1+x\) with \(x>0\). Bernoulli: \(\frac1{|z|^{n+1}}=(1+x)^{n+1}\ge1+(n+1)x\ge nx\), hence \[0\le|z|^{n+1}\le\frac1x\cdot\frac1n\to0.\] Squeeze ⇒ \(|z^{n+1}|\to0\) ⇒ \(z^{n+1}\to0\) ⇒ \(s_n\to\frac1{1-z}\).
\(q\ge1\): every term \(q^k\ge1\), so \(s_n\ge n+1\to\infty\).
\(q=-1\): \(s_n=1,0,1,0,\dots\) diverges. ∎
| q | 999/1000 | 99/100 | 1/2 | 1/100 | −1/100 |
|---|---|---|---|---|---|
| \(\sum_{k\ge0}q^k=\frac1{1-q}\) | 1000 | 100 | 2 | 100/99 | 100/101 |
Table from Skript 3.2: the closer \(q\) is to 1, the larger the sum. Watch the starting index: \(\sum_{k=0}^\infty\frac1{2^k}=2\), but \(\sum_{k=1}^\infty\frac1{2^k}=1\). In general \(\sum_{k=m}^\infty q^k=\frac{q^m}{1-q}\) ("first term over one minus ratio"), because \(q^m+q^{m+1}+\dots=q^m(1+q+q^2+\dots)\) Extra.
A telescoping series [Teleskopreihe]
Proof
Partial fractions: \(\frac1{k(k+1)}=\frac1k-\frac1{k+1}\) (check: \(\frac{(k+1)-k}{k(k+1)}\)). Then almost everything cancels: \[s_n=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\dots+\left(\tfrac1n-\tfrac1{n+1}\right)=1-\frac1{n+1}\to1.\ \ ∎\]
Convergent or divergent? Compute the value if possible: \(\displaystyle\sum_{n=2}^\infty\frac{n-1}{n(n^2-1)}\).
Solution
Factor \(n^2-1=(n-1)(n+1)\) and cancel: \(\frac{n-1}{n(n-1)(n+1)}=\frac1{n(n+1)}\). This is the telescoping series, but it starts at \(n=2\): \[\sum_{n=2}^N\left(\frac1n-\frac1{n+1}\right)=\frac12-\frac1{N+1}\to\frac12.\] (Convergence alone would also follow from the majorant \(\frac1{n(n+1)}<\frac1{n^2}\).) Value: \(\frac12\).
"\(\sum_{n\ge0}(-1)^n=1+\sum_{n\ge1}(-1)^n=1-\sum_{n\ge0}(-1)^n\), so \(2\sum(-1)^n=1\), so \(\sum_{n\ge0}(-1)^n=\frac12\)."
Where it goes wrong
\(\sum(-1)^n\) diverges. It has no value, so you can't treat it as a number \(S\) and solve "\(S=1-S\)". With partial sums (which are real numbers) the same manipulation reads \(\sum_{k=0}^n(-1)^k=1-\sum_{k=0}^{n-1}(-1)^k\). The two sums on the right and left are different partial sums (one is 0, the other 1), so nothing can be combined.
Lesson: calculate with \(\sum a_k\) like with a number only after you know it converges.
What you must know for 3.1: the definition of convergent/divergent series; the harmonic and geometric series and their behaviour; the value of the geometric series.
3.2Convergence tests [Konvergenzkriterien]
Only the tail matters. Leaving out finitely many terms changes neither convergence of a sequence nor its limit. For a series, dropping terms does not change whether it converges, but it does change the value: \(\sum_{k=0}^\infty2^{-k}=2\) but \(\sum_{k=1}^\infty2^{-k}=1\). So in convergence questions you may always say "for \(k\ge k_0\)".
Extra Why: for \(n\ge k_0\), \(s_n=\underbrace{(a_1+\dots+a_{k_0-1})}_{\text{a fixed number }C}+\sum_{k=k_0}^na_k\). A sequence plus a constant converges exactly when the sequence does (sum rule), and the limit is shifted by \(C\).
The necessary condition
If \(\sum a_k\) converges, then \(a_k\to0\).
Proof
\(a_k=s_k-s_{k-1}\), and both \(s_k\) and \(s_{k-1}\) tend to the same \(s\). So \(a_k\to s-s=0\). ∎
Contrapositive (how you use it): if \(a_k\not\to0\), the series diverges. This is the fastest test: always check it first.
Not sufficient (Skript 3.5): \(\frac1k\to0\), but \(\sum\frac1k=\infty\).
- \(\sum\frac k{2k+1}\): terms → \(\frac12\neq0\) ⇒ divergent. Blatt 4.2a
- \(\sum_{k\ge1}\exp\left(-\frac1{k^2}\right)\): terms → \(e^0=1\) ⇒ divergent. WiSe22 1b
- \(\sum k\sin\frac2k\): terms → 2 (since \(\frac{\sin x}x\to1\), Chapter 5) ⇒ divergent. SoSe23 1b
- \(\sum(-1)^n\frac{n^n}{n!}\): \(|a_n|=\frac{n^n}{n!}=\prod_{k=1}^n\frac nk\ge1\), since every factor \(\frac nk\ge1\). So \(a_n\not\to0\) ⇒ divergent. WiSe20 1a(i)
(a) If \(\sum x_k\) converges with \(x_k>0\), then \(\sum\frac1{x_k}\) diverges. (b) The converse is false.
Solution
(a) By Lemma 3.4, \(x_k\to0\), so \(\frac1{x_k}\to\infty\) (for every \(M\), eventually \(x_k<\frac1M\), i.e. \(\frac1{x_k}>M\)). The terms of \(\sum\frac1{x_k}\) don't tend to 0 ⇒ divergent.
(b) \(x_k=\sqrt k\): \(\sum\sqrt k\) diverges (terms ↛ 0) and \(\sum\frac1{\sqrt k}\) diverges too (minorant \(\frac1k\), see below).
Let \(x_n>0\) with \(\sum x_n\) convergent. Show (a) there is \(n_0\) with \(x_n<1\) for \(n\ge n_0\); (b) \(\sum x_n^p\) converges for every \(p>1\).
Solution
(a) \(x_n\to0\) (Lemma 3.4). Take \(\varepsilon=1\) in the definition of convergence: \(x_n=|x_n-0|<1\) for \(n\ge n_0\).
(b) For \(n\ge n_0\): \(0<x_n<1\) and \(p>1\) give \(x_n^p\le x_n\). So \(\sum_{n\ge n_0}x_n\) is a convergent majorant, and the finitely many terms before \(n_0\) don't matter. ∎
Comparison: majorant and minorant
If \(a_k\ge0\) for all \(k\): \(\sum a_k\) converges \(\iff\) the partial sums are bounded.
Proof: with \(a_k\ge0\) the partial sums increase, and Satz 2.17 says an increasing sequence converges iff it is bounded. So for nonnegative series there are only two options: a finite value, or \(+\infty\).
Let \(a_k\in\mathbb C\), \(b_k\in\mathbb R\) with \(|a_k|\le b_k\) for all \(k\). \(\sum b_k\) is called a majorant [Majorante] of \(\sum a_k\).
- Majorant test: if \(\sum b_k\) converges, then \(\sum a_k\) converges and \(\left|\sum a_k\right|\le\sum|a_k|\le\sum b_k\).
- Minorant test: if \(\sum|a_k|\) diverges, then \(\sum b_k\) diverges.
In the same situation \(\sum|a_k|\) is called a minorant [Minorante] of \(\sum b_k\): majorant = the bigger one, minorant = the smaller one.
In words: smaller than something convergent ⇒ convergent; larger than something divergent ⇒ divergent. (The inequality only needs to hold for \(k\ge k_0\).) Picture two non-negative series as two staircases of partial sums, one always above the other: if the upper one stays below a ceiling, so does the lower one; if the lower one climbs to ∞, so does the upper one.
Proof Extra — the script proves only part (b)
(a) for real \(a_k\ge0\): here \(|a_k|=a_k\le b_k\), so the partial sums satisfy \(\sum_{k=1}^na_k\le\sum_{k=1}^nb_k\le\sum_{k=1}^\infty b_k=:B\) (the last step because a series with non-negative terms increases towards its value). So the partial sums of \(\sum a_k\) are increasing and bounded by \(B\), hence convergent by Satz 3.6, and the limit is \(\le B\).
(a) in general: apply the previous paragraph to \(|a_k|\) (which satisfies \(0\le|a_k|\le b_k\)): \(\sum|a_k|\) converges. Now split \(a_k\) into non-negative pieces. For real \(a_k\) put \(a_k^+=\max(a_k,0)\) and \(a_k^-=\max(-a_k,0)\); both are \(\ge0\) and \(\le|a_k|\), so \(\sum a_k^+\) and \(\sum a_k^-\) converge by the non-negative case, and \(\sum a_k=\sum a_k^+-\sum a_k^-\) converges by Satz 3.20. For complex \(a_k\) apply this to real and imaginary parts, using \(|\operatorname{Re}a_k|\le|a_k|\) and \(|\operatorname{Im}a_k|\le|a_k|\). (This is also the proof of the statement in Bemerkung 3.9 that every absolutely convergent series converges.) The estimate \(|\sum a_k|\le\sum|a_k|\) is the triangle inequality for each partial sum (Satz 1.12; Satz 1.14 for complex numbers), carried to the limit with Satz 2.13.
(b) is (a) read backwards: if \(\sum b_k\) converged, (a) would make \(\sum|a_k|\) converge, contradicting the assumption. ∎
Comparing in the wrong direction proves nothing. "\(\frac1{k^2+1}\le\frac1k\) and \(\sum\frac1k\) diverges" says nothing: being smaller than a divergent series is no information. Before comparing, decide what you expect (converge or diverge) and estimate in the matching direction.
\(\sum a_k\) converges absolutely [absolut konvergent] if \(\sum|a_k|\) converges.
By the majorant test (with \(b_k=|a_k|\)), absolutely convergent ⇒ convergent. The converse is false (Skript 3.9): the Leibniz series \(\sum\frac{(-1)^k}{2k+1}\) converges, but \(\sum\frac1{2k+1}=\infty\).
If \(a_n,b_n>0\) and \(a_n\simeq b_n\) (i.e. \(\frac{a_n}{b_n}\to1\)), then \(\sum a_n\) and \(\sum b_n\) both converge or both diverge.
Proof
With \(\varepsilon=\frac12\): for \(n\ge n_0\), \(\left|\frac{a_n}{b_n}-1\right|<\frac12\), i.e. \(\frac12b_n<a_n<\frac32b_n\). If \(\sum b_n\) converges, \(\frac32b_n\) is a convergent majorant of \(a_n\). If \(\sum b_n\) diverges, \(\frac12b_n\) is a divergent minorant. ∎
This is the most practical test for "rational-looking" terms: replace the term by its leading behaviour and compare with \(\sum\frac1{k^\alpha}\).
- \(\sum\frac1{k^2}\) converges: \(\frac1{k^2}\simeq\frac1{k(k+1)}\) (ratio \(1+\frac1k\to1\)), and the telescoping series converges. Or directly: \(\frac1{k^2}\le\frac1{(k-1)k}\) for \(k\ge2\).
- \(\sum\frac1{\sqrt{k(k+1)}}=\infty\): \(\frac1{\sqrt{k(k+1)}}\simeq\frac1k\). Or: \(\frac1{\sqrt{k(k+1)}}\ge\frac1{\sqrt{2k\cdot2k}}=\frac1{2k}\), a divergent minorant.
- \(\sum\frac1{\sqrt k}=\infty\): \(\sqrt k\le k\Rightarrow\frac1{\sqrt k}\ge\frac1k\).
Extra Rule of thumb (proved in Chapter 9 with the integral test): \(\sum\frac1{k^\alpha}\) converges iff \(\alpha>1\). The comparisons above are the cases α = 2 and α = ½.
(b) \(a_k=\frac{k^2+k}{k^4+k^3-k}\): converges
Upper estimate: numerator \(\le2k^2\), denominator \(\ge k^4\) (as \(k^3\ge k\)). So \(|a_k|\le\frac{2k^2}{k^4}=\frac2{k^2}\), a convergent majorant.
(c) \(a_k=\frac{(k^3+k+1)^2}{k^7}\): diverges
Lower estimate: \((k^3+k+1)^2\ge k^6\), so \(a_k\ge\frac{k^6}{k^7}=\frac1k\), a divergent minorant.
(f) \(a_k=(-1)^k\frac{k+7}{k^5+k^2-2k}\): converges (absolutely)
For \(k\ge2\): \(k+7\le7k\) (since \(6k\ge7\)) and \(k^5+k^2-2k\ge k^5\) (since \(k^2\ge2k\)). So \(|a_k|\le\frac{7k}{k^5}=\frac7{k^4}\le\frac7{k^2}\). Majorant test.
(g) \(a_k=\sqrt{k+1}-\sqrt k\): diverges
Conjugate trick: \(a_k=\frac1{\sqrt{k+1}+\sqrt k}\ge\frac1{2\sqrt{k+1}}\ge\frac1{2(k+1)}\), a divergent minorant. Extra Even simpler: it telescopes, \(s_n=\sqrt{n+1}-1\to\infty\).
(a) convergent + convergent = convergent; (b) convergent + divergent = divergent. Proof of (b): if \(\sum(x_k+y_k)\) converged, then \(\sum y_k=\sum(x_k+y_k)-\sum x_k\) would converge by (a). Application:
- \(\sum\left(\frac1k+\frac{(-1)^k}{k^2}\right)\): divergent (harmonic) + convergent ⇒ divergent.
- \(\sum\left(\frac{(-1)^k}k+\frac1{k^2}\right)\): convergent (Leibniz) + convergent ⇒ convergent.
(Divergent + divergent can be anything: \(\sum1+\sum(-1)=\sum0\).)
WiSe19 1d: for which \(a\in\mathbb R\) does \(\sum\frac{3\sin(na)+5\cos(n^2a)}{n^2}\) converge absolutely?
For every \(a\): \(|3\sin(na)+5\cos(n^2a)|\le3+5=8\), so the absolute terms are \(\le\frac8{n^2}\), a convergent majorant. Answer \(M=\mathbb R\). Same idea: \(\sum\frac{\sin(2k)}{k^2}\) converges (SoSe23 1b).
SoSe23 1c: for which \(x>0\) does \(\sum\frac1{k+x^k}\) converge?
\(x>1\): \(\frac1{k+x^k}\le\frac1{x^k}=\left(\frac1x\right)^k\), convergent geometric majorant. \(0<x\le1\): \(x^k\le1\), so \(\frac1{k+x^k}\ge\frac1{k+1}\), divergent minorant. Answer \(M=(1,\infty)\).
WiSe19 A6: \(a_n,b_n\ge0\), \(b_n\to b>0\). Show \(\sum a_nb_n\) converges \(\iff\sum a_n\) converges.
This is the proof idea of Satz 3.10. With \(\varepsilon=\frac b2\): for \(n\ge n_0\), \(\frac b2\le b_n\le\frac{3b}2\), hence \(\frac b2a_n\le a_nb_n\le\frac{3b}2a_n\). Right inequality: \(\sum a_n\) conv ⇒ \(\sum a_nb_n\) conv (majorant). Left inequality: \(\sum a_nb_n\) conv ⇒ \(\sum a_n\) conv (majorant \(\frac2ba_nb_n\)). ∎
The ratio test [Quotientenkriterium]
Let \(a_k\neq0\) for all but finitely many \(k\), and suppose \(q=\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|\) exists. Then:
- \(q<1\) ⇒ \(\sum a_k\) converges (even absolutely),
- \(q>1\) ⇒ \(\sum a_k\) diverges,
- \(q=1\) ⇒ no conclusion.
Proof for q < 1
Idea: eventually each term is at most \((q+\varepsilon)\) times the previous one, so the terms are dominated by a geometric series.
Choose \(\varepsilon>0\) with \(q+\varepsilon<1\). For \(k\ge k_0\): \(\left|\frac{a_{k+1}}{a_k}\right|<q+\varepsilon\). Multiply the ratios in a chain: for \(n>k_0\) \[|a_n|=\left|\frac{a_n}{a_{n-1}}\right|\cdots\left|\frac{a_{k_0+1}}{a_{k_0}}\right||a_{k_0}|\le|a_{k_0}|(q+\varepsilon)^{n-k_0}=\underbrace{\frac{|a_{k_0}|}{(q+\varepsilon)^{k_0}}}_{\text{const}}(q+\varepsilon)^n.\] The geometric series \(\sum(q+\varepsilon)^n\) converges, so the majorant test finishes the proof. Because the majorant bounds \(|a_n|\), we even get absolute convergence.
Proof for q > 1 Extra — the script proves only q < 1: take \(\varepsilon=q-1>0\). For \(k\ge k_0\): \(\left|\frac{a_{k+1}}{a_k}\right|>q-\varepsilon=1\), i.e. \(|a_{k+1}|>|a_k|\). So from \(k_0\) on the absolute values increase: \(|a_k|\ge|a_{k_0}|>0\). The terms cannot tend to 0, and Lemma 3.4 gives divergence. ∎
\(\sum\frac1k\) (divergent): \(\left|\frac{a_{k+1}}{a_k}\right|=\frac k{k+1}\to1\). \(\sum\frac1{k(k+1)}\) (convergent): \(\frac k{k+2}\to1\). Both have ratios \(<1\) for every \(k\) and limit 1. So also "\(\left|\frac{a_{k+1}}{a_k}\right|<1\) for all \(k\)" is not enough; you need the limit \(q<1\). Consequence: for rational terms like \(\frac{k+1}{k^3}\) the ratio test always gives 1. Use comparison instead. The ratio test is made for factorials and powers (\(k!\), \(2^k\), \(k^k\)); for a term that is itself a \(k\)-th power, see the root test below.
(d) \(\sum\frac{k^2}{2^k}\): converges
\(\frac{a_{k+1}}{a_k}=\frac{(k+1)^2}{2^{k+1}}\cdot\frac{2^k}{k^2}=\frac12\left(\frac{k+1}k\right)^2\to\frac12<1\).
(h*) \(\sum\frac{k!}{k^k}\): converges
\(\frac{a_{k+1}}{a_k}=\frac{(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{k!}=\frac{(k+1)k^k}{(k+1)^{k+1}}=\left(\frac k{k+1}\right)^k=\frac1{\left(1+\frac1k\right)^k}\to\frac1e<1\) (definition of \(e\), Satz 2.21).
Does \(\sum\frac{3^n(n!)^2}{(2n)!}\) converge? Name the test.
Solution
Ratio test. Most things cancel; keep track of \((2n+2)!=(2n+2)(2n+1)(2n)!\): \[\frac{b_{n+1}}{b_n}=\frac{3^{n+1}((n+1)!)^2}{(2n+2)!}\cdot\frac{(2n)!}{3^n(n!)^2}=\frac{3(n+1)^2}{(2n+2)(2n+1)}=\frac{3n^2+6n+3}{4n^2+6n+2}\to\frac34<1.\] So the series converges (absolutely). (Part (a) of the same exam was exactly this limit, see Chapter 2.)
\(\sum_{k=0}^\infty\frac{z^k}{k!}\) converges for every \(z\in\mathbb C\): \(\left|\frac{a_{k+1}}{a_k}\right|=\frac{|z|^{k+1}/(k+1)!}{|z|^k/k!}=\frac{|z|}{k+1}\to0<1\). (The factorial eventually beats any power.)
Many students know a second test of the same kind. If \(q=\lim_{k\to\infty}\sqrt[k]{|a_k|}\) exists, then \(q<1\) ⇒ convergent (absolutely), \(q>1\) ⇒ divergent, \(q=1\) ⇒ no conclusion. The proof is the one above with \(|a_k|\le(q+\varepsilon)^k\) straight away, so again the majorant is a geometric series.
It is the better tool when the term is a \(k\)-th power: \(\sum\left(\frac{k}{2k+1}\right)^k\) has \(\sqrt[k]{a_k}=\frac k{2k+1}\to\frac12<1\), while the ratio test here is a mess. Where both give a limit, they give the same one.
In the exam, use a test from the script (comparison, ratio, Leibniz) — those are what the solutions expect. The root idea does appear in the script, but only in Chapter 10: the radius of convergence formula of Cauchy–Hadamard, \(r=1/\limsup_k\sqrt[k]{|a_k|}\) (Satz 10.7), is the root test applied to \(\sum a_kx^k\).
Alternating series: the Leibniz test [Leibnizkriterium]
Let \((a_n)_{n\ge0}\) be monotonically decreasing with \(a_n\to0\). Then the alternating series \(s=\sum_{k=0}^\infty(-1)^ka_k\) converges, and the error of the partial sum \(s_n\) is at most the first omitted term: \[|s-s_n|\le a_{n+1}.\]
Proof (nested intervals)
Picture: step forward \(a_0\), back \(a_1\), forward \(a_2\), … Each step is no longer than the previous one, so the position swings around a point with shrinking amplitude.
- \(s_{n+1}-s_{n-1}=(-1)^n(a_n-a_{n+1})\), and \(a_n-a_{n+1}\ge0\). So the odd partial sums increase and the even ones decrease: \(s_1\le s_3\le s_5\le\dots\) and \(\dots\le s_4\le s_2\le s_0\).
- \(a_n\ge0\) (a decreasing sequence with limit 0 converges to its infimum, which is 0). Hence \(s_{2n+1}=s_{2n}-a_{2n+1}\le s_{2n}\), and all odd sums lie below all even sums: \[s_1\le s_3\le\dots\le s_{2n+1}\le s_{2n}\le\dots\le s_2\le s_0.\]
- \((s_{2n+1})\) is increasing and bounded by \(s_0\); \((s_{2n})\) is decreasing and bounded by \(s_1\). Both converge (Satz 2.17), and to the same limit, because \(s_{2n}-s_{2n+1}=a_{2n+1}\to0\).
- The limit \(\tilde s\) always lies between \(s_n\) and \(s_{n+1}\), so \(|\tilde s-s_n|\le|s_{n+1}-s_n|=a_{n+1}\to0\). ∎
The script states Leibniz with the sum starting at \(k=0\) and the first term positive. Other starts or signs, like \(\sum_{k\ge1}\frac{(-1)^{k+1}}k\), are covered too: shifting the index or multiplying the whole series by \(-1\) does not change convergence Extra.
Alternating sign, \(a_n\to0\), and \(a_n\) decreasing. Monotonicity is often forgotten. The WiSe24 exam solution explicitly deducted points for applying Leibniz where it doesn't belong, and warned: "alternating sequences can converge too", so don't confuse the sequence \((-1)^na_n\) with the series.
- Leibniz series \(\sum\frac{(-1)^k}{2k+1}=1-\frac13+\frac15-\dots\) converges (\(\frac1{2k+1}\) decreases to 0). The nesting (3.7) gives concrete bounds: \(0.78514\ldots=s_{1001}\le s\le s_{1000}=0.78564\ldots\), so \(s=0.785\ldots\) Skript 3.16. Extra Its value is \(\frac\pi4\). The error bound shows how slow it is: for \(m\) correct digits you need \(a_{n+1}=\frac1{2n+3}\le10^{-m}\), i.e. \(n\approx\frac12\cdot10^m\). For 10 digits that is about \(5\cdot10^9\) terms.
- \(\sum\frac{(-1)^{k+1}}{\sqrt k}\) and \(\sum\frac{(-1)^k}{\sqrt k}\) (Blatt 4.2e) converge: \(\frac1{\sqrt k}\) decreases to 0. Not absolutely, since \(\sum\frac1{\sqrt k}=\infty\).
- \(\sum\frac{(-1)^k}{k+\sqrt k}\) converges (WiSe22 1b): \(k+\sqrt k\) increases, so its reciprocal decreases to 0.
Careful: taking only the even or only the odd terms of the Leibniz series gives divergent series (\(\sum\frac1{4k+1}=\infty\), \(-\sum\frac1{4k+3}=-\infty\)). The convergence lives entirely on the cancellation.
- Do the terms go to 0? If not ⇒ divergent. Done.
- Is it a known series (geometric, harmonic, \(\frac1{k^2}\), telescoping, exponential)? Possibly after rewriting (partial fractions, factoring).
- Positive rational-looking terms (polynomials, roots)? Find the leading behaviour \(a_k\simeq\frac c{k^\alpha}\) and compare: majorant with \(\frac1{k^2}\) or minorant with \(\frac1k\).
- Factorials, \(q^k\), \(k^k\)? Ratio test.
- Alternating sign? Leibniz (check decreasing!). Also check absolute convergence if asked.
- Bounded oscillating factor like \(\sin\), \(\cos\), \((-1)^k\) times something summable? Take absolute values and use a majorant.
In the exam, always name the test you use. That is typically worth a point on its own.
\(\sum_{k\ge1}\frac{3^{-k}}{1-3^{-k}}\) converges. (1) Asymptotic: \(\frac{3^{-k}}{1-3^{-k}}\simeq3^{-k}\), a convergent geometric series. (2) Majorant: \(1-3^{-k}\ge\frac23\), so the term is \(\le\frac32\cdot3^{-k}\). (3) Ratio test: the ratio tends to \(\frac13<1\). Pick whichever you find fastest.
What you must know for 3.2: \(a_k\to0\) is necessary but not sufficient; majorant, minorant and ratio tests; the exponential series, the Leibniz series and \(\sum\frac1{k^2}\) and how they behave; deciding convergence of series.
3.3Rules for series [Rechenregeln]
If \(\sum a_k\) and \(\sum b_k\) converge, so does \(\sum(a_k+b_k)\), and \(\sum(a_k+b_k)=\sum a_k+\sum b_k\).
Proof: this is the sum rule for limits applied to the partial sums.
Extra In the same way (rule 2.6(b)): \(\sum ca_k=c\sum a_k\) for any constant \(c\). That is why "\(\sum\frac6{k^2}\) converges" needs no extra argument.
Rearranging terms [Umordnung]
\(\sum a_k\) converges absolutely \(\iff\) for every bijection \(\sigma:\mathbb N\to\mathbb N\) the rearranged series \(\sum a_{\sigma(k)}\) converges to the same value.
Riemann (1866): a convergent but not absolutely convergent series can be rearranged to converge to any real number you like.
\(s=1-\frac12+\frac13-\frac14+\dots\) converges by Leibniz, and \(|s-s_1|\le\frac13\) gives \(s\ge\frac16>0\). Rearrange: one positive term, then two negative ones: \[r_n=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2(2k-1)}-\frac1{4k}\right).\] Since \(\frac1{2k-1}-\frac1{2(2k-1)}=\frac12\cdot\frac1{2k-1}\), we get \(r_n=\frac12\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)=\frac12s_{2n-1}\to\frac s2\neq s\). The same terms, a different sum. Extra (\(s=\ln2\), Chapter 9.)
Multiplying series: the Cauchy product
Multiplying two finite sums produces all products \(a_lb_m\). Group them by index sum \(l+m=k\) (like collecting powers of \(x\) when multiplying polynomials):
If \(\sum_{k\ge0}a_k\) and \(\sum_{k\ge0}b_k\) converge absolutely, then so does \(\sum c_k\) with \(c_k=\sum_{l=0}^ka_lb_{k-l}\), and \[\left(\sum_{k=0}^\infty a_k\right)\left(\sum_{k=0}^\infty b_k\right)=\sum_{k=0}^\infty\sum_{l=0}^ka_lb_{k-l}.\] This is the Cauchy product [Cauchy-Produkt].
Square the geometric series (\(|x|<1\)): \(a_l=b_l=x^l\), so \(c_k=\sum_{l=0}^kx^lx^{k-l}=(k+1)x^k\). Hence \[\frac1{(1-x)^2}=\sum_{k=0}^\infty(k+1)x^k=1+2x+3x^2+\dots\] The script derives the same formula in Chapter 10 by differentiating. It is behind WiSe19 1e: \(\sum_{n\ge1}n(3x)^n=\frac{3x}{(1-3x)^2}\) for \(|x|<\frac13\).
\(\sum_{k\ge0}\frac{(-1)^k}{\sqrt{k+1}}\) converges (Leibniz) but not absolutely. Its Cauchy product with itself has \(c_k=(-1)^k\sum_{l=0}^k\frac1{\sqrt{(l+1)(k-l+1)}}\). Each of the \(k+1\) terms is \(\ge\frac2{k+2}\), because \(\sqrt{xy}\le\frac{x+y}2\) gives \(\sqrt{(l+1)(k-l+1)}\le\frac{k+2}2\). So \(|c_k|\ge\frac{2(k+1)}{k+2}\ge1\): the terms do not tend to 0 and the product series diverges.
What you must know for 3.3: only absolutely convergent series may be rearranged arbitrarily; the alternating series; the Cauchy product and how to use it.
3.4The exponential function [Die Exponentialfunktion]
It converges absolutely for every \(z\) (ratio test, Skript 3.14). This is the definition of exp in this course. Everything else is derived from it.
For all \(w,z\in\mathbb C\): \(\exp(w+z)=\exp(w)\exp(z)\).
Proof: Cauchy product + binomial theorem
Both series converge absolutely, so Satz 3.23 applies: \[\exp(w)\exp(z)=\sum_{k=0}^\infty\sum_{l=0}^k\frac{z^l}{l!}\frac{w^{k-l}}{(k-l)!}=\sum_{k=0}^\infty\frac1{k!}\sum_{l=0}^k\binom klz^lw^{k-l}=\sum_{k=0}^\infty\frac{(w+z)^k}{k!}=\exp(w+z).\] The middle step multiplies and divides by \(k!\) to create \(\binom kl=\frac{k!}{l!(k-l)!}\); the next one is the binomial theorem. ∎
- \(\exp(0)=1\)
- \(\exp(z)\neq0\) for all \(z\in\mathbb C\)
- \(\exp(-z)=\frac1{\exp(z)}\)
- \(\exp(x)>0\) for all real \(x\)
- \(\exp:\mathbb R\to\mathbb R\) is strictly increasing
- \(|\exp(z)|\le\exp(|z|)\)
Proofs
(a) Only the \(k=0\) term survives: \(\frac{0^0}{0!}=1\).
(b), (c) \(1=\exp(0)=\exp(-z+z)=\exp(-z)\exp(z)\). A product equal to 1 has no zero factor, and \(\exp(-z)=\frac1{\exp z}\).
(d) For \(x>0\) every term is positive, so \(\exp(x)>\frac{x^0}{0!}=1>0\). For \(x<0\): \(\exp(x)=\frac1{\exp(-x)}>0\).
(e) For \(x<y\): \(y-x>0\), so \(1<\exp(y-x)=\frac{\exp y}{\exp x}\). Since \(\exp x>0\), \(\exp x<\exp y\).
(f) Triangle inequality for the partial sums: \(\left|\sum_{k\le n}\frac{z^k}{k!}\right|\le\sum_{k\le n}\frac{|z|^k}{k!}\); then let \(n\to\infty\) (using \(|s_n|\to|s|\), reverse triangle inequality + squeeze). ∎
One can show that \(\exp(1)=e=\lim(1+\frac1n)^n\) (Chapter 6 proves the more general \(\lim(1+\frac xn)^n=e^x\)). With the functional equation, \(\exp(k)=e^k\) for all integers \(k\), e.g. \(\exp(2)=e\cdot e\) and \(\exp(-1)=\frac1e\). That is why we write \(e^z\) for \(\exp(z)\), and the functional equation reads \(e^{w+z}=e^we^z\).
For all \(m\in\mathbb N_0\): (a) \(\lim_{x\to\infty}e^x=\infty\), (b) \(\lim_{x\to-\infty}e^x=0\), (c) \(\lim_{x\to\infty}\frac{e^x}{x^m}=\infty\), (d) \(\lim_{x\to-\infty}x^me^x=0\).
Solution (the key is: drop terms of the series)
(For "\(\lim_{x\to\infty}\)" read: for every sequence \(x_n\to\infty\), Chapter 4.)
(a) For \(x\ge0\) all terms are \(\ge0\), so \(e^x\ge1+x\to\infty\).
(b) \(e^{x}=\frac1{e^{-x}}\), and \(-x\to\infty\), so \(e^{-x}\to\infty\) and its reciprocal → 0 (Blatt 3.2d).
(c) Keep only the term with \(k=m+1\): \(e^x\ge\frac{x^{m+1}}{(m+1)!}\) for \(x\ge0\), so \(\frac{e^x}{x^m}\ge\frac{x}{(m+1)!}\to\infty\).
(d) With \(y=-x\to\infty\): \(|x^me^x|=\frac{y^m}{e^y}\le\frac{y^m(m+1)!}{y^{m+1}}=\frac{(m+1)!}y\to0\); squeeze.
Take-away: the exponential function beats every polynomial. This is used constantly (Chapter 5, Satz 5.6).
What you must know for 3.4: the functional equation; \(\exp(1)=e\).
Mixed exercises from the script [Kapitel 13]
Three convergence questions from the script’s exercise pool — exactly the format of the "does it converge?" exam task.
Does \(\displaystyle\sum_{k=1}^\infty\frac{k^4}{e^{k^2}}\) converge?
Solution (two ways)
Comparison. Intuition: \(e^{k^2}\) is enormously bigger than \(k^4\). To make that precise, keep one term of the exponential series: \(e^{k^2}=\sum_{j\ge0}\frac{(k^2)^j}{j!}\ge\frac{(k^2)^3}{3!}=\frac{k^6}6\). Hence \[0\le\frac{k^4}{e^{k^2}}\le\frac{6k^4}{k^6}=\frac6{k^2},\] and \(\sum\frac6{k^2}\) converges (Anwendung 3.11), so the series converges by the majorant test.
Ratio test. \(\left|\frac{a_{k+1}}{a_k}\right|=\left(1+\frac1k\right)^4e^{-2k-1}\to0<1\). Converges.
The trick "\(e^x\ge\frac{x^n}{n!}\) for \(x>0\)" (drop all other non-negative terms of the series) is the cheapest way to beat an exponential against a power, and it is used again in Satz 5.6.
Does \(\displaystyle\sum_{k=1}^\infty(-1)^{k+1}\sqrt[k]{k}\) converge?
Solution
Check the necessary condition first (Lemma 3.4). \(\sqrt[k]k=k^{1/k}=e^{\frac{\ln k}k}\to e^0=1\), so the terms \((-1)^{k+1}\sqrt[k]k\) oscillate between values near \(+1\) and \(-1\) and do not tend to 0. Therefore the series diverges.
Warning: "alternating" is not enough. Leibniz needs the size of the terms to decrease to 0; here they increase to 1.
Does \(\displaystyle\sum_{k=2}^\infty\frac k{(\ln k)^k}\) converge?
Solution
Ratio test with \(a_k=\frac k{(\ln k)^k}\): \[0\le\frac{a_{k+1}}{a_k}=\frac{k+1}k\cdot\left(\frac{\ln k}{\ln(k+1)}\right)^k\cdot\frac1{\ln(k+1)}\le\left(1+\frac1k\right)\cdot\frac1{\ln(k+1)}\to0,\] because \(\frac{\ln k}{\ln(k+1)}\le1\) makes the middle factor \(\le1\). By the squeeze theorem the quotient tends to \(0<1\), so the series converges.
The point: bounding an awkward factor by 1 is allowed as long as the factor really is \(\le1\) — say why.
!What the exam asks from Chapter 3
Series appear in every one of the nine exams — usually two or three short "converges?" items in Task 1 (often answer-only, 1–2 points each) plus sometimes a proof task. The task types:
| Type | Exam tasks | What earns the points |
|---|---|---|
| Converges or not? | WiSe20 1a, WiSe22 1b, SoSe23 1b | first check \(a_k\to0\); then Leibniz or a majorant with a bounded numerator over \(k^2\) |
| "Name the test" | WiSe23 1b (\(\frac{3^n(n!)^2}{(2n)!}\)) | ratio test, clean cancellation of factorials, the limit, "\(<1\)" |
| For which x? + value | WiSe22 1c, 1d · WiSe19 1d, 1e · SoSe23 1c | recognise a geometric or exponential series; treat the cases of \(x\) separately; check the boundary |
| Telescoping value | WiSe20 1a(ii) | partial fractions, write \(s_N\), watch the start index |
| Proofs about an abstract series | WiSe18 A3, WiSe19 A6, WiSe20 A2a, SoSe21 A2 | \(a_k\to0\) (Lemma 3.4), then a majorant/minorant for the tail |
| Later tools on series | SoSe19 A3 (uniform), WiSe24 6d, SoSe23 A4e (integral test) | the majorant idea again (Ch. 9); integral test in Chapter 9 |
Determine \(M_1=\{x\in\mathbb R:\ S_1(x)=\sum_{k=0}^\infty4^{2k}x^k\text{ converges}\}\) and the value \(S_1(x)\).
Solution
Recognise the pattern. \(4^{2k}x^k=(4^2)^kx^k=(16x)^k\). So \(S_1(x)=\sum_{k\ge0}z^k\) with \(z=16x\): a geometric series.
Cases. \(|16x|<1\): converges to \(\frac1{1-16x}\) (Lemma 3.2). \(|16x|\ge1\): \(|(16x)^k|\ge1\), the terms do not tend to 0, so it diverges (Lemma 3.4). Answer: \(M_1=\left(-\frac1{16},\frac1{16}\right)\), \(S_1(x)=\frac1{1-16x}\).
Determine \(M_2=\{x\in\mathbb R:\ S_2(x)=\sum_{k=1}^\infty\frac{x^{2k}}{5^kk!}\text{ converges}\}\) and the value.
Solution
\(\frac{x^{2k}}{5^kk!}=\frac{(x^2/5)^k}{k!}\): the exponential series at \(z=\frac{x^2}5\), which converges for every \(z\) (Skript 3.14). But it starts at \(k=1\), so the \(k=0\) term (which is 1) is missing: \[M_2=\mathbb R,\qquad S_2(x)=\exp\left(\tfrac{x^2}5\right)-1.\]
Determine \(M=\{x\in\mathbb R:\ S(x)=\sum_{n=1}^\infty n3^nx^n\text{ converges}\}\) and \(S(x)\).
Solution
Put \(y=3x\), so the terms are \(ny^n\). Ratio test (for \(y\neq0\)): \(\left|\frac{(n+1)y^{n+1}}{ny^n}\right|=\frac{n+1}n|y|\to|y|\). So convergence for \(|y|<1\), divergence for \(|y|>1\). For \(|y|=1\): \(|ny^n|=n\not\to0\), divergent. Hence \(M=\left(-\frac13,\frac13\right)\).
Value. The Cauchy product example in §3.3 gave \(\sum_{k\ge0}(k+1)y^k=\frac1{(1-y)^2}\). Shift the index (\(n=k+1\)) and multiply by \(y\): \(\sum_{n\ge1}ny^n=y\sum_{k\ge0}(k+1)y^k=\frac y{(1-y)^2}\). With \(y=3x\): \(S(x)=\frac{3x}{(1-3x)^2}\). (Chapter 10 gets the same by differentiating the geometric series.)
\(a_n>0\) and \(\sum a_n\) converges. True or false? (i)* \(\lim na_n=0\) (bonus) (ii) \(\lim na_n=\infty\) (iii) \(\liminf na_n=0\).
\(\liminf\) ("limes inferior") is the smallest accumulation point of a sequence [Häufungspunkt], defined in Chapter 4. Here you only need: "\(\liminf na_n>0\)" would mean that from some \(N\) on, \(na_n\ge c\) for a fixed \(c>0\).
Solution
(ii) false: \(a_n=\frac1{n^2}\) gives a convergent series and \(na_n=\frac1n\to0\), not ∞.
(iii) true (proof by contradiction): if \(na_n\ge c>0\) for all \(n\ge N\), then \(a_n\ge\frac cn\), and \(\sum\frac cn\) is a divergent minorant. So \(\sum a_n\) would diverge — contradiction.
(i)* false — the official counterexample: \(a_n=\frac1{k^2}\) if \(n=2^k\) is a power of 2 (\(k\ge1\)), and \(a_n=\frac1{n^2}\) otherwise. The sum is at most \(\sum_k\frac1{k^2}+\sum_n\frac1{n^2}<\infty\), but at \(n=2^k\): \(na_n=\frac{2^k}{k^2}\to\infty\). So \(na_n\) does not tend to 0 (it only has some small values — which is exactly (iii)).
Prove that \(\sum_{k=1}^\infty\frac1{k^2+1+\cos(kx)}\) converges uniformly [gleichmäßig] in \(x\in\mathbb R\).
Solution
Since \(\cos(kx)\ge-1\), the denominator is \(\ge k^2\), so \(0<a_k(x)\le\frac1{k^2}=:b_k\) for all \(x\) and \(k\), and \(\sum b_k\) converges. For each \(x\) this is the majorant test. "Uniformly" means that one majorant works for all \(x\) at once — which is exactly what the official solution checks. (The theorem behind it, Satz 9.10, is in Chapter 9.)
\(a_n>0\), \(S=\sum a_n<\infty\), and \(b_n=\frac{a_n}{\sum_{k\ge n}a_k}\) (term divided by the remaining tail). Find examples with (a) \(b_n\to0\), (b) \(\liminf b_n>0\).
Solution
(a) Telescoping \(a_n=\frac1n-\frac1{n+1}=\frac1{n(n+1)}\): the tail is \(\sum_{k\ge n}a_k=\frac1n\), so \(b_n=\frac{1/(n(n+1))}{1/n}=\frac1{n+1}\to0\).
(b) Geometric \(a_n=q^n\), \(0<q<1\): the tail is \(\frac{q^n}{1-q}\), so \(b_n=1-q\) for all \(n\), and \(\liminf b_n=1-q>0\).
Interpretation: for slowly decaying terms, each term is a vanishing fraction of what is left; for geometric terms, a fixed fraction.
Drill: which test decides it?
The most common exam task, as a quick drill. For each series, pick the argument that settles it; the explanation gives the full answer you would write.
- For the exam: decide convergence of a given series and name the test; find the set of \(x\) for which a geometric or exponential series converges and its value; short proofs using \(a_k\to0\) plus a majorant/minorant.
- For Chapters 4–6: \(\exp\) is defined by its series. Its continuity (Lemma 4.4), its growth \(\frac{e^x}{x^m}\to\infty\) (Satz 5.6, "drop all but one term") and its derivative (Beispiel 6.13) are all proved by estimating the series. \(\sin\) and \(\cos\) are defined from \(e^{ix}\) (§5.3).
- For Chapter 9: the integral test decides \(\sum\frac1{k^\alpha}\) and series like \(\sum\frac1{n\ln n}\); \(\ln 2=1-\frac12+\frac13-\dots\) is proved with the Leibniz error bound.
- For Chapter 10: a power series \(\sum a_kx^k\) is a series for each \(x\). The geometric series is the model case (radius 1); the ratio test finds the radius; the Cauchy product becomes the product of power series (Satz 10.10).
ΣSummary sheet
| Series | Behaviour | Reason |
|---|---|---|
| \(\sum_{k\ge0}q^k\) | \(=\frac1{1-q}\) for |q| < 1, else divergent | geometric sum formula (3.2) |
| \(\sum\frac1k\) | \(=\infty\) | block grouping |
| \(\sum\frac1{k^2}\) | converges | \(\le\frac1{(k-1)k}\), telescoping |
| \(\sum\frac1{k^\alpha}\) | converges iff α > 1 | comparison; proof in Ch. 9 |
| \(\sum\frac1{k(k+1)}\) | \(=1\) | telescoping (3.3) |
| \(\sum\frac{(-1)^k}{2k+1}\), \(\sum\frac{(-1)^{k+1}}k\) | converge, not absolutely | Leibniz (3.15) |
| \(\sum\frac{z^k}{k!}\) | \(=e^z\), all z | ratio test |
- A series is the sequence of its partial sums. Finitely many terms change the value, never the convergence.
- Convergent ⇒ terms → 0 (not conversely). Absolutely convergent ⇒ convergent (not conversely).
- Comparison: \(|a_k|\le b_k\), \(\sum b_k<\infty\) ⇒ convergent; \(a_k\ge b_k\ge0\), \(\sum b_k=\infty\) ⇒ divergent. Asymptotic version: \(a_k\simeq b_k\) (positive) ⇒ same behaviour.
- Ratio test: limit \(q<1\) conv., \(q>1\) div., \(q=1\) no information.
- Leibniz: alternating + decreasing + → 0 ⇒ converges; error ≤ first omitted term.
- Rearranging and multiplying (Cauchy product) are safe for absolutely convergent series.
- \(e^{w+z}=e^we^z\); \(e^x>0\), strictly increasing; beats every polynomial.
?Self-test
ContinuityStetigkeit
A function is continuous if small changes in the input cause small changes in the output: no jumps, no wild oscillation. The script defines this with sequences, so everything from Chapter 2 carries over. The payoff is three big theorems you will use until the end of the course: the intermediate value theorem, Bolzano–Weierstrass, and the existence of maxima and minima on compact sets.
Source labels: Skript 4.2 = stated in your lecture script · Blatt 5 · 5.1 = exercise sheet · Klausur SoSe19 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§4.3 = Section 4.3) rather than a numbered result (4.3 = Beispiel 4.3).
0The big picture
- 4.1 \(f\) is continuous at \(x\) if \(x_n\to x\) always implies \(f(x_n)\to f(x)\): "\(f\) commutes with limits". Sums, products, quotients and compositions of continuous functions are continuous, so almost everything built from polynomials and exp is continuous.
- 4.2 Intermediate value theorem: a continuous function on an interval cannot skip values. It gives zeros of equations and the existence of \(\sqrt[n]y\).
- 4.3 Subsequences and accumulation points. Bolzano–Weierstrass: every bounded sequence has a convergent subsequence (found by repeated halving).
- 4.4 Compact = closed + bounded. A continuous function on a compact set attains its maximum and minimum. This is why optimisation problems on \([a,b]\) always have a solution.
Everything here is built on Chapter 2: the definition of a convergent sequence, the limit rules, the squeeze theorem and "limits keep ≤" (Satz 2.13), and monotone + bounded ⇒ convergent (Satz 2.17). Two proofs (IVT, Lemma 4.26) use the supremum from §1.2. The exponential function comes from its series (§3.4).
Some examples use \(\sin\), \(\cos\) and \(\ln\), which the script only defines in Chapter 5. Treat them as the familiar school functions for now; nothing here depends on their precise definition.
4.1Definition and simple properties [Definition und einfache Eigenschaften]
The idea. A function is continuous at \(x\) if nearby inputs give nearby outputs: if you approach \(x\), the values approach \(f(x)\). "Approach" is exactly what sequences are good for, so the script defines continuity by testing \(f\) with every sequence that runs into \(x\). Informally: you can draw the graph near \(x\) without lifting the pen.
A sequence \((x_n)\) in \(\mathbb R^d\) converges iff every component converges. (Example in \(\mathbb R^2\): \(x_n=(\frac1n,\,2+\frac{(-1)^n}{n})\to(0,2)\), because \(\frac1n\to0\) and \(2+\frac{(-1)^n}n\to2\).)
A function \(f:D\subseteq\mathbb R^d\to\mathbb R^q\) with domain [Definitionsbereich] \(D\) is continuous at the point \(x\in D\) if for every sequence \((x_n)\) in \(D\) with \(x_n\to x\): \[\lim_{n\to\infty}f(x_n)=f(x).\] One also writes \(\lim_{y\to x}f(y)=f(x)\). \(f\) is continuous [stetig] if it is continuous at every \(x\in D\).
Two things are packed into this: the limit of \(f(x_n)\) must exist (for every approaching sequence, and always be the same), and it must equal the value \(f(x)\).
To show \(f\) is not continuous at \(x\), one sequence suffices: find \(x_n\to x\) with \(f(x_n)\not\to f(x)\), either because \(f(x_n)\) has no limit, or because it has the wrong one. Alternatively find two sequences with different limits of \(f(x_n)\).
\(\lim_{x\downarrow x_0}f(x)=c\) (right-hand limit [rechtsseitiger Grenzwert]) means \(f(x_n)\to c\) for every sequence \(x_n\to x_0\) with \(x_n>x_0\). Left-hand limit \(\lim_{x\uparrow x_0}\) analogously with \(x_n<x_0\).
The sheet's own examples — \(\lim_{x\downarrow0}\ln x=-\infty\), \(\lim_{x\downarrow0}x\ln x=0\) and the behaviour of \(x^a\) at both ends — need the logarithm, so they are worked in §5.2 and the growth section of Chapter 5.
For a point \(x_0\) inside an interval: \(f\) is continuous at \(x_0\) ⇔ both one-sided limits exist and equal \(f(x_0)\). This is the practical test for piecewise functions: compute the limit from the left, the limit from the right, and compare with the value. (Why: every sequence \(x_n\to x_0\) splits into the terms left of \(x_0\), those right of it, and those equal to it; if all three groups give \(f(x_0)\) in the limit, so does the whole sequence.)
Let \(f:[0,1]\to\mathbb R\) with \(f(\frac1n)=(-1)^n\), linearly interpolated in between. It zig-zags between \(-1\) and \(1\) ever faster as \(x\to0\). It is continuous on \((0,1]\), but for \(x_n=\frac1n\to0\) the values \(f(x_n)=(-1)^n\) do not converge. So \(f\) is discontinuous at 0 whatever value \(f(0)\) has.
\(f:[0,1]\to\mathbb R,\ f(x)=2x^2+5\) is continuous; \(d(x)=0\) for \(x\neq0\), \(d(0)=1\) is not continuous at 0.
(a) With sequences (Definition 4.2)
Let \(x_n\to x_0\). By the limit rules \(f(x_n)=2x_n^2+5\to2x_0^2+5=f(x_0)\). Done.
(b) With the ε–δ criterion
The sheet uses the equivalent ε–δ criterion [ε-δ-Kriterium] (not stated in the script, but equivalent to Def. 4.2): \[\forall\varepsilon>0\ \exists\delta>0\ \forall x\in D:\ |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon.\] "Output within ε whenever input within δ." For \(f(x)=2x^2+5\) on \([0,1]\): \(|f(x)-f(x_0)|=2|x-x_0|\,|x+x_0|\le4|x-x_0|\) because \(x+x_0\le2\). So \(\delta=\frac\varepsilon4\) works.
(c), (d) The jump function is discontinuous at 0
(c) \(x_n=\frac1n\to0\), but \(d(x_n)=0\to0\neq1=d(0)\). (d) With \(\varepsilon=\frac12\): for every \(\delta\), any \(x\neq0\) with \(|x|<\delta\) has \(|d(x)-d(0)|=1>\varepsilon\).
Why the two definitions say the same thing Extra
ε–δ ⇒ sequences: let \(x_n\to x_0\) and \(\varepsilon>0\). Take the δ from ε–δ. Since \(x_n\to x_0\), \(|x_n-x_0|<\delta\) for \(n\ge n_0\), hence \(|f(x_n)-f(x_0)|<\varepsilon\) for \(n\ge n_0\). That is \(f(x_n)\to f(x_0)\).
sequences ⇒ ε–δ (by contraposition): if ε–δ fails, there is an ε such that no δ works; for \(\delta=\frac1n\) pick \(x_n\) with \(|x_n-x_0|<\frac1n\) but \(|f(x_n)-f(x_0)|\ge\varepsilon\). Then \(x_n\to x_0\) while \(f(x_n)\not\to f(x_0)\): the sequence definition fails too. ∎
The exponential function is continuous on ℂ.
Proof
At 0: for \(|z|\le1\), estimate the series without its first term: \[|e^z-1|=\left|\sum_{k\ge1}\frac{z^k}{k!}\right|\le|z|\sum_{k\ge1}\frac{|z|^{k-1}}{(k-1)!}=|z|e^{|z|}\le e|z|.\] If \(z_n\to0\), squeeze gives \(e^{z_n}\to1=e^0\).
At any z: functional equation: \(|e^{z_n}-e^z|=|e^{z_n-z}-1|\,|e^z|\to0\), since \(z_n-z\to0\) and continuity at 0. ∎
The composition \(f\circ g:x\mapsto f(g(x))\) of continuous functions is continuous. Proof: \(x_n\to x\Rightarrow g(x_n)\to g(x)\Rightarrow f(g(x_n))\to f(g(x))\). (The composition must be defined: \(g\) must map into the domain of \(f\).)
- Continuous: constants, \(|x|\), \(\sqrt x\) on \([0,\infty)\), \((x,y)\mapsto x+y\), \(xy\), \(\frac xy\) (for \(y\neq0\)). So sums, products and quotients of continuous functions are continuous.
- Polynomials are continuous on ℂ (built from constants and \(z\mapsto z\) by sums and products). Rational functions \(\frac{p(z)}{q(z)}\) are continuous on their domain.
- \(f:\mathbb R^d\to\mathbb R^q\) is continuous iff every component \(f_i\) is. E.g. \((x,y)\mapsto(3x+7,\ y^{25}-4xy+x^3,\ 3-x^2y^2)\).
In exams, "continuous as a composition of continuous functions" is a complete justification away from problem points (division by 0, gluing points of piecewise definitions). All the work goes into the problem points.
Find the maximal domain and check continuity.
- \(\frac1x\): \(D=\mathbb R\setminus\{0\}\), rational ⇒ continuous on \(D\).
- \(\frac{7x^5+13x^4+8x+\pi}{x^2+1}\): denominator never 0 ⇒ \(D=\mathbb R\), continuous.
- \(\sqrt x\): \(D=[0,\infty)\); continuous since \(x_n\to x\Rightarrow\sqrt{x_n}\to\sqrt x\) (Blatt 3.1e).
- \(\exp(\sqrt{x^2-9})\): need \(x^2\ge9\), so \(D=(-\infty,-3]\cup[3,\infty)\); composition ⇒ continuous.
- \(f(x)=\lim_{n\to\infty}x^n\): exists only for \(x\in(-1,1]\), with \(f=0\) on \((-1,1)\) and \(f(1)=1\). Not continuous at 1: \(x_n=\frac{n-1}n\to1\) but \(f(x_n)=0\). Moral: a pointwise limit of continuous functions (\(x^n\)) need not be continuous.
WiSe24 2a: Is \(f(x)=\frac{\sin x}x\) (\(x\neq0\)), \(f(0)=1\) continuous everywhere? WiSe23 A2: For which \(a\) is \(f_a(x)=\frac{x\sin x}{1-\cos x}\) (\(x\neq0\)), \(f_a(0)=a\) continuous on \((-\pi,\pi)\)?
Solution pattern
- Away from the special point: quotient/composition of continuous functions with nonzero denominator ⇒ continuous. (On \((-\pi,\pi)\setminus\{0\}\), \(\cos x\neq1\).)
- At the special point: compute \(L=\lim_{x\to0}f(x)\). \(f\) is continuous there iff \(f(0)=L\).
WiSe24: \(\lim_{x\to0}\frac{\sin x}x=1=f(0)\) (Lemma 5.19) ⇒ yes, continuous everywhere.
WiSe23: split with the two standard limits from Lemma 5.19: \[\frac{x\sin x}{1-\cos x}=\frac{\sin x}{x}\cdot\frac{x^2}{1-\cos x}=\frac{\sin x}x\cdot2\cdot\frac{\frac12x^2}{1-\cos x}\to1\cdot2\cdot1=2.\] So \(f_a\) is continuous iff \(a=2\); for \(a\neq2\) the only discontinuity is \(x=0\). (The official solution used l’Hospital twice, Chapter 7; the route above only needs Chapter 5.)
\(f\) is Lipschitz continuous on \([a,b]\) if \(|f(x)-f(y)|\le L|x-y|\) for some constant \(L\). Show: Lipschitz ⇒ continuous.
Solution
Let \(x_n\to x\) in \([a,b]\). Then \(0\le|f(x_n)-f(x)|\le L|x_n-x|\to0\); squeeze ⇒ \(f(x_n)\to f(x)\). (Part b, "\(f'\) continuous ⇒ Lipschitz", takes \(L=\max_{[a,b]}|f'|\) — this maximum exists by the max–min theorem 4.29 below, because \(|f'|\) is continuous on the compact \([a,b]\) — and then uses the mean value theorem, Chapter 7.)
What you must know for 4.1: the notion of continuity (at a point); deciding whether a given function is continuous; knowing that exp, polynomials, rational functions and compositions are continuous on their domain.
4.2The intermediate value theorem [Zwischenwertsatz]
Let \(f:[a,b]\to\mathbb R\) be continuous and \(y\) a number strictly between \(f(a)\) and \(f(b)\). Then there is \(x\in(a,b)\) with \(f(x)=y\).
In words: a continuous function on an interval takes every value between \(f(a)\) and \(f(b)\). Picture: you cannot draw the graph from below the line \(y\) to above it without crossing it. A jump (Figure 3 in the script) is exactly what breaks this.
Proof (via the supremum)
WLOG \(f(a)<y<f(b)\). Let \(M=\{t\in[a,b]: f(t)\le y\}\) ("the points still at or below \(y\)"). \(a\in M\), and \(M\) is bounded, so \(x:=\sup M\in[a,b]\) exists (supremum axiom). Claim: \(f(x)=y\).
- \(f(x)\le y\): choose \(x_n\in M\) with \(x_n\to x\). This is possible because \(x-\frac1n\) is not an upper bound of \(M\) (\(x\) is the smallest one), so some \(x_n\in M\) has \(x-\frac1n<x_n\le x\); squeeze gives \(x_n\to x\). \(f(x_n)\le y\) and continuity give \(f(x)=\lim f(x_n)\le y\) (Satz 2.13).
- \(f(x)\ge y\): since \(f(x)\le y<f(b)\), \(x<b\), so \(x+\frac1n\in[a,b]\) for large \(n\). These points are beyond the sup, so not in \(M\): \(f(x+\frac1n)>y\). Continuity: \(f(x)=\lim f(x+\frac1n)\ge y\). ∎
- Bring everything to one side: \(g(x)=0\), e.g. \(\cos x=x\iff g(x)=\cos x-x=0\).
- Argue that \(g\) is continuous on \([a,b]\) (composition of continuous functions).
- Find \(a,b\) with \(g(a)<0<g(b)\) (or the other way round). Evaluate and justify the signs.
- IVT ⇒ some \(x\in(a,b)\) with \(g(x)=0\). For uniqueness additionally show \(g\) is strictly monotone.
Does \(\cos x=x\) have a solution in \([0,\frac\pi2]\)? \(g(x)=\cos x-x\) is continuous with \(g(0)=1>0\) and \(g(\frac\pi2)=-\frac\pi2<0\). IVT ⇒ a zero in \((0,\frac\pi2)\). \(\cos\) and \(-x\) are strictly decreasing there, so \(g\) is too ⇒ exactly one solution (≈ 0.739).
Show with the IVT that there is \(x\in(\frac\pi2,\pi)\) with \(\ln\left(\frac12+\cos^2x\right)e^{-\sin(\ln x)}=0\).
Solution
\(f(x)=\ln(\frac12+\cos^2x)\,e^{-\sin(\ln x)}\) is continuous on \([\frac\pi2,\pi]\) (products and compositions of continuous functions; \(\frac12+\cos^2x>0\), \(x>0\)). The exponential factor is always \(>0\), so the sign is that of the log: \(f(\frac\pi2)=\ln(\frac12)\cdot(\text{pos.})<0\), \(f(\pi)=\ln(\frac12+1)\cdot(\text{pos.})=\ln\frac32\cdot(\text{pos.})>0\). IVT ⇒ a zero in \((\frac\pi2,\pi)\).
Tip: don't compute the ugly factor, just its sign.
\(f:[a,b]\to\mathbb R\) continuous, \(\int_a^bf(x)\,dx<0\), and \(f(x_0)>0\) for some \(x_0\in(a,b)\). Show \(f(c)=0\) for some \(c\in(a,b)\).
Solution
If \(f\ge0\) on \((a,b)\), then \(\int_a^bf\ge0\) (positivity of the integral, Chapter 8), contradiction. So there is \(y_0\in(a,b)\) with \(f(y_0)<0\). IVT on the interval between \(x_0\) and \(y_0\) gives a zero \(c\) there, in particular \(c\in(a,b)\).
(a) Fixed points: every continuous \(f:[a,b]\to[a,b]\) has \(f(x)=x\) for some \(x\)
\(g(x)=x-f(x)\) is continuous. Since \(f\) maps into \([a,b]\): \(g(a)=a-f(a)\le0\) and \(g(b)=b-f(b)\ge0\). If one is 0, done (\(x=a\) or \(x=b\)); otherwise IVT gives a zero of \(g\), i.e. \(f(x)=x\). Graphically: the graph must cross the diagonal.
(b) Every real polynomial of odd degree has a real zero
WLOG leading coefficient \(>0\). Odd degree: \(P(x)\to+\infty\) for \(x\to\infty\) and \(\to-\infty\) for \(x\to-\infty\). So there are \(x_1<0<x_2\) with \(P(x_1)<0<P(x_2)\); IVT on \([x_1,x_2]\). (For even degree this fails: \(x^2+1\).)
(c*) For continuous \(f\) on an interval: injective ⇔ strictly monotone
"⇐" is clear (and needs no continuity): \(x<y\) gives \(f(x)<f(y)\) or \(f(x)>f(y)\), in any case \(f(x)\neq f(y)\).
"⇒": let \(f\) be continuous and injective on the interval \(I\). Fix two points \(a<b\) in \(I\); by injectivity \(f(a)\neq f(b)\), say \(f(a)<f(b)\) (otherwise apply the argument to \(-f\)). We show \(f(x)<f(y)\) for all \(x<y\) in \(I\), i.e. \(f\) is strictly increasing.
- Slide the pair \((x,y)\) to \((a,b)\). For \(t\in[0,1]\) put \(x_t=(1-t)x+ta\) and \(y_t=(1-t)y+tb\). Both stay in \(I\) (an interval contains the segment between two of its points), and \(y_t-x_t=(1-t)(y-x)+t(b-a)>0\), so \(x_t<y_t\) for every \(t\).
- The difference never vanishes. \(h(t)=f(y_t)-f(x_t)\) is continuous on \([0,1]\) (composition of continuous maps), and \(h(t)\neq0\) because \(x_t\neq y_t\) and \(f\) is injective.
- IVT. A continuous function on \([0,1]\) without zeros cannot change sign. \(h(1)=f(b)-f(a)>0\), so also \(h(0)=f(y)-f(x)>0\). ∎
This is what makes inverse functions (Chapter 5) work: a continuous injective function on an interval is automatically strictly monotone, so Satz 5.3 applies to it. Without continuity it fails: \(f(x)=x\) on \([0,1)\), \(f(x)=3-x\) on \([1,2]\) is injective but not monotone.
Application: n-th roots
\(\lim_{x\to\infty}f(x)=\infty\) means \(f(x_n)\to\infty\) for every sequence \(x_n\to\infty\) (similarly for other combinations of \(\pm\infty\)). \(f\) is (strictly) increasing if \(x\le y\Rightarrow f(x)\le f(y)\) (\(x<y\Rightarrow f(x)<f(y)\)); decreasing analogously.
For every \(n\in\mathbb N\) and \(y>0\) there is exactly one \(x>0\) with \(x^n=y\). We write \(x=\sqrt[n]y=y^{1/n}\).
Proof: \(f(x)=x^n\) is continuous and strictly increasing on \([0,\infty)\) (so at most one solution) with \(f(x)\to\infty\). Choose \(b\) with \(f(0)=0<y<f(b)\); IVT on \([0,b]\). This finally proves that \(\sqrt2\) exists as a real number, and \(x=\sup\{t\ge0:t^n\le y\}\), exactly the sup from Chapter 1.
What you must know for 4.2: the intermediate value theorem and how to apply it.
4.3Accumulation points [Häufungspunkte]
Let \(n_1<n_2<n_3<\dots\) be a strictly increasing sequence of indices. Then \((a_{n_k})_{k\in\mathbb N}\) is a subsequence [Teilfolge] of \((a_n)\). A point \(a^*\) is an accumulation point [Häufungspunkt] of \((a_n)\) if some subsequence converges to \(a^*\).
Examples of subsequences: \((a_{n^2})\), \((a_{2n})\), \((a_{2^n})\): you pick infinitely many terms, keeping their order.
\(a\) is an accumulation point of \((a_n)\) \(\iff\) for every \(\varepsilon>0\) there are infinitely many indices \(n\) with \(|a_n-a|<\varepsilon\).
Proof
"⇒": a subsequence \(a_{n_k}\to a\) eventually stays within ε, and that is infinitely many indices \(n_k\).
"⇐": build the subsequence step by step. Choose \(n_1\) with \(|a_{n_1}-a|<1\), then \(n_2>n_1\) with \(|a_{n_2}-a|<\frac12\) (possible since there are infinitely many candidates), …, \(n_k>n_{k-1}\) with \(|a_{n_k}-a|<\frac1k\). Then \(a_{n_k}\to a\). ∎
Compare with convergence: convergent to a = all but finitely many terms within ε; accumulation point a = infinitely many terms within ε.
- A convergent sequence has exactly one accumulation point, its limit. (Every subsequence has the same limit: if all but finitely many \(a_n\) lie within ε of \(a\), the same holds for the terms \(a_{n_k}\), because \(n_k\ge k\to\infty\).)
- \((-1)^n\) has the accumulation points \(1\) (even indices) and \(-1\) (odd indices).
- Not every sequence has a convergent subsequence: \(a_n=n\). Every subsequence is unbounded.
The converse of the first fact, for bounded sequences Extra
Claim: a bounded sequence with exactly one accumulation point \(a\) converges to \(a\). (This is what the exam tasks use when they conclude "convergent" or "not convergent" from the accumulation points.)
Proof by contradiction. If \(a_n\not\to a\), there is an \(\varepsilon>0\) with \(|a_n-a|\ge\varepsilon\) for infinitely many \(n\). These terms form a bounded subsequence, so by Bolzano–Weierstrass (below) it has a convergent sub-subsequence. Its limit \(b\) is an accumulation point of \((a_n)\) with \(|b-a|\ge\varepsilon\) (Satz 2.13), so \(b\neq a\) — a second accumulation point. Contradiction. ∎
Without boundedness it fails: \(1,1,1,2,1,3,1,4,\dots\) has only the accumulation point 1 but diverges (Blatt 7.3b below).
- Infinitely many accumulation points: \(1;\ 1,2;\ 1,2,3;\ 1,2,3,4;\dots\) Every \(n\in\mathbb N\) appears infinitely often, so every \(n\) is an accumulation point.
- Exactly one accumulation point, but divergent: \(a_{2k-1}=1,\ a_{2k}=k\). The only accumulation point is 1 (the even terms run off to ∞). So "one accumulation point ⇒ convergent" needs boundedness.
WiSe24 1a: accumulation points of \(a_n=\frac{n^3(-1)^n}{3+n^3}\)? Convergent?
Divide by \(n^3\): \(a_n=\frac{(-1)^n}{1+3/n^3}\). Even subsequence → 1, odd subsequence → −1. Every index is even or odd, so these are the only accumulation points. Two accumulation points ⇒ not convergent. The official solution warns: citing Leibniz here costs points (the Leibniz criterion is about series!).
SoSe21 1a: accumulation points of \(c_n=e^{i\pi n/2}(1-e^{i\pi n})\)?
With \(e^{i\pi/2}=i\) and \(e^{i\pi}=-1\) (Euler, Chapter 5): \(c_n=i^n(1-(-1)^n)=i^n-(-i)^n\). The sequence is periodic with period 4: \(c_{4k-3}=2i,\ c_{4k-2}=0,\ c_{4k-1}=-2i,\ c_{4k}=0\). Accumulation points: \(2i,0,-2i\).
For a bounded real sequence, \(\limsup a_n\) is the largest and \(\liminf a_n\) the smallest accumulation point. They exist: by Bolzano–Weierstrass (below) there is at least one accumulation point, the set of them is bounded, and one can check with the criterion of Blatt 6.3 that its supremum and infimum are again accumulation points. The sequence converges iff \(\liminf=\limsup\). Example: \(\limsup(-1)^n=1\), \(\liminf(-1)^n=-1\).
Bolzano–Weierstrass
Every bounded sequence in ℝ has at least one convergent subsequence, hence an accumulation point.
Proof (the construction you must know)
Let \(|a_n|\le c\). Start with \(I_1=[b_1,c_1]=[-c,c]\), which contains all terms, and \(n_1=1\).
- Halve \(I_k\) into two halves. At least one half contains infinitely many terms of the remaining sequence (two halves with finitely many each would give finitely many overall). Call it \(I_{k+1}=[b_{k+1},c_{k+1}]\).
- Pick an index \(n_{k+1}>n_k\) with \(a_{n_{k+1}}\in I_{k+1}\) (possible: infinitely many candidates).
- Repeat. Now \(b_k\le a_{n_k}\le c_k\) and the length \(c_k-b_k=\frac{2c}{2^{k-1}}\to0\).
- \((b_k)\) is increasing and bounded ⇒ converges to some \(b^*\); \((c_k)\) is decreasing and bounded ⇒ converges to \(c^*\) (Satz 2.17). \(c^*-b^*=\lim(c_k-b_k)=0\).
- Squeeze: \(a_{n_k}\to b^*=c^*\). ∎
In \(\mathbb R^d\) (Skript 4.20): a bounded sequence (\(\|x_n\|\le M\)) has bounded components; take a convergent subsequence for component 1, from it a sub-subsequence for component 2, and so on.
What you must know for 4.3: the procedure for finding a convergent subsequence from the proof of Bolzano–Weierstrass.
4.4Existence of maxima and minima
For \(f:D\to\mathbb R\), a point \(x\in D\) is a minimum point [Minimumstelle] and \(f(x)\) the minimum if \(f(x)\le f(y)\) for all \(y\in D\). Maximum analogously.
- \(x\mapsto x\) on ℝ: neither. On \((a,b)\): neither (inf \(=a\), sup \(=b\) not attained). On \([a,b]\): min at \(a\), max at \(b\).
- \(\frac1x\) on \((0,\infty)\): neither. On \((\varepsilon,M)\): neither. On \([\varepsilon,M]\): max at ε, min at \(M\).
What goes wrong: an open end (the extreme value is approached but not attained) or an unbounded domain. Hence the next definition.
\(A\subseteq\mathbb R^d\) is closed [abgeschlossen] if limits of convergent sequences in \(A\) stay in \(A\): \(x_n\in A,\ x_n\to x\Rightarrow x\in A\). \(K\) is compact [kompakt] if it is closed and bounded.
Intuition: closed = "contains its boundary points": you cannot walk inside \(A\) towards a point and find that the point is missing. \((0,1]\) fails at 0. Bounded = fits in some ball. Compact = both. In \(\mathbb R\), the typical compact sets are closed intervals \([a,b]\) and finite unions of them; in \(\mathbb R^d\), closed boxes and closed balls.
Skript 4.25: \([0,1]\) compact; \([0,\infty)\) closed, not bounded, not compact; \((0,1]\) and \((0,1)\) not closed (\(\frac1n\to0\notin\) set), not compact.
- 4.26 A nonempty compact \(K\subseteq\mathbb R\) has a max and a min. (Proof: \(\inf K\) exists since \(K\) is bounded below; \(\inf K+\frac1n\) is not a lower bound, so there is \(x_n\in K\) with \(\inf K\le x_n<\inf K+\frac1n\), hence \(x_n\to\inf K\) by squeeze; closed ⇒ \(\inf K\in K\), so it is the minimum. The maximum works the same way with \(\sup K\).)
- 4.27 \(K\) compact \(\iff\) every sequence in \(K\) has a subsequence converging to a point of \(K\). (⇒: Bolzano–Weierstrass gives a convergent subsequence, closedness puts its limit in \(K\). ⇐: closed: if \(x_n\in K\), \(x_n\to x\), a subsequence converges to some \(x^*\in K\); but subsequences of a convergent sequence have the same limit, so \(x=x^*\in K\). Bounded: if not, pick \(x_n\in K\) with \(\|x_n\|\ge n\); every subsequence is unbounded, so none converges — contradiction.)
- 4.28 Continuous images of compact sets are compact: \(f\) continuous, \(K\) compact ⇒ \(f(K)\) compact.
Proof of 4.28
Use 4.27. Take a sequence \(y_n=f(x_n)\) in \(f(K)\). \(K\) compact ⇒ a subsequence \(x_{n_j}\to x\in K\). \(f\) continuous ⇒ \(y_{n_j}=f(x_{n_j})\to f(x)\in f(K)\). ∎
Let \(\emptyset\neq K\subseteq\mathbb R^d\) be compact. Every continuous \(f:K\to\mathbb R\) attains its maximum and minimum on \(K\): there are \(\underline x,\overline x\in K\) with \(f(\underline x)\le f(x)\le f(\overline x)\) for all \(x\in K\).
Proof: \(f(K)\subseteq\mathbb R\) is compact (4.28) and has a max and a min (4.26). The points need not be unique (constant \(f\)). Notation: \(\underline x=\operatorname{argmin}_{x\in K}f(x)\), \(\overline x=\operatorname{argmax}_{x\in K}f(x)\).
Find an example or explain why none exists. (These show that each assumption of the max–min theorem is needed.)
- \(D=(0,1)\), continuous, max but no min: \(f(x)=-(\frac12-x)^2\): max 0 at \(\frac12\); inf \(-\frac14\) is approached at the open ends, not attained.
- \(D=(0,1)\), continuous, unbounded above and below: \(f(x)=\frac1x+\frac1{x-1}\) (→ +∞ at 0, → −∞ at 1).
- \(D=[0,1]\), bounded, no max, no min: must be discontinuous (otherwise 4.29 applies): \(f(x)=x\) on \((0,1)\), \(f(0)=f(1)=\frac12\).
- Strictly increasing on \([0,\infty)\) with infinitely many discontinuities: \(f(x)=x+n\) on \([n,n+1)\): a jump at every integer.
- Continuous bijection \([0,1]\to(0,1)\)? Impossible: the image of the compact \([0,1]\) would be compact (4.28), but \((0,1)\) is not.
- Countable non-compact / compact sets: \(\{\frac1n\}\) is not closed (0 missing); \(\{\frac1n\}\cup\{0\}\) is bounded and closed, so compact.
What you must know for 4.4: minimum, maximum, minimum and maximum points; a continuous function on a compact set attains its max and min; which subsets of \(\mathbb R^d\) are compact.
!What the exam asks from Chapter 4
Chapter 4 appears in six of the nine exams, as one of four task types. All of them are worked on this page:
| Type | Exam tasks | What earns the points |
|---|---|---|
| Continuous? For which parameter? | WiSe23 A2, WiSe24 2a | "composition of continuous functions" away from the special point; at the point: limit = value |
| IVT existence proof | SoSe19 A5, SoSe23 A6 | name \(g\), say why it is continuous on a closed interval, show the sign change, cite the IVT |
| Accumulation points | WiSe24 1a, SoSe21 1a | split into subsequences that cover all indices, compute each limit, conclude (one point + bounded ⇔ convergent) |
| Short proofs | WiSe18 A4a (Lipschitz), WiSe18 A4b uses the max–min theorem | sequence definition + squeeze; cite 4.29 when you need a maximum |
- Where is \(f\) built from continuous pieces with no problem (no zero denominator, no gluing)? There: "continuous as a composition of continuous functions" — done.
- List the problem points (gluing points of a piecewise definition, zeros of denominators).
- At each: compute the limit (from both sides if the formula differs). Useful limits: \(\frac{\sin x}x\to1\), \(\frac{1-\cos x}{x^2/2}\to1\) (Lemma 5.19), \(\frac{e^x-1}x\to1\).
- Continuous there ⇔ the limit exists and equals the value. For "which \(a\)": set the value equal to the limit.
- To show discontinuity: give one explicit sequence \(x_n\to x_0\) with \(f(x_n)\not\to f(x_0)\).
Drill: continuity, IVT, accumulation points, compactness
Ten short items covering the four exam task types. Decide first, then read the explanation — it is the sentence you would write.
- For the exam: the four task types above.
- For Chapter 5: inverse functions. A continuous strictly monotone function on an interval is a bijection onto its image, and the inverse is continuous (Satz 5.3). This is how \(\ln\), \(\arcsin\), \(\arctan\) and \(\sqrt[n]{\ }\) are obtained; the IVT is what guarantees that no value is skipped.
- For Chapter 7: Rolle's theorem and the mean value theorem start with "a continuous function on \([a,b]\) attains its max and min" (Satz 4.29).
- For Chapter 8: continuous functions are integrable (Satz 8.3); the mean value theorem of integration uses max–min + IVT.
- For Chapters 10–11: \(\limsup\) in the radius-of-convergence formula (Satz 10.7) is the largest accumulation point; compact sets in \(\mathbb R^d\) guarantee extrema of functions of several variables.
ΣSummary sheet
- Continuity at \(x\): \(x_n\to x\Rightarrow f(x_n)\to f(x)\) for every sequence. Discontinuity: one bad sequence suffices. Equivalent: ε–δ.
- Polynomials, rational functions, exp, √, |·| and all their sums, products, quotients and compositions are continuous on their domains. Exam work happens only at gluing points / zeros of denominators: continuous there iff value = limit.
- IVT: continuous on \([a,b]\) + sign change ⇒ zero. Add strict monotonicity for uniqueness. Consequences: fixed points, odd-degree polynomials have roots, \(\sqrt[n]y\) exists.
- Accumulation point: limit of a subsequence ⇔ infinitely many terms in every ε-neighbourhood. Convergent ⇒ exactly one.
- Bolzano–Weierstrass: bounded ⇒ convergent subsequence (halve, keep the half with infinitely many terms).
- Compact = closed + bounded. Continuous on compact ⇒ max and min are attained. Open ends or unbounded domains or discontinuities can destroy this.
?Self-test
Important FunctionsWichtige Funktionen
With continuity and the exponential series in hand, we can now build the rest of the standard functions: inverse functions in general, the logarithm as the inverse of exp, general powers \(x^a\) and \(b^x\), and sine and cosine via Euler's formula \(e^{ix}=\cos x+i\sin x\). Along the way you get the growth hierarchy log ≪ power ≪ exponential that exam questions test constantly.
Source labels: Skript 5.5 = stated in your lecture script · Blatt 7 · 7.1 = exercise sheet · Klausur WiSe23 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§5.2 = Section 5.2) rather than a numbered result (5.2 = Beispiel 5.2).
0The big picture
- 5.1 A continuous, strictly increasing function on an interval can be undone. Its inverse is again continuous and increasing, and its graph is the mirror image at the diagonal \(y=x\).
- 5.2 \(\ln\) = inverse of \(\exp\). The functional equation \(e^{a+b}=e^ae^b\) turns into \(\ln(xy)=\ln x+\ln y\). Then \(x^a:=e^{a\ln x}\) defines powers with any real exponent. Growth: \(e^x\) beats every power, every power beats every power of \(\ln\).
- 5.3 Plug imaginary numbers into exp: \(e^{ix}\) lies on the unit circle, and its coordinates define \(\cos x\) and \(\sin x\). Trig identities follow from \(e^{i(x+y)}=e^{ix}e^{iy}\). Key limits: \(\frac{\sin x}x\to1\), \(\frac{1-\cos x}{x^2/2}\to1\).
- injective, surjective, bijective and the inverse map (Basics);
- the exponential function: its series, \(e^{w+z}=e^we^z\), \(e^x>0\), strictly increasing (§3.4);
- continuity and the intermediate value theorem (§4.2) — they are why inverse functions exist;
- limits of functions: \(\lim_{x\to\infty}f(x)\) via sequences (Def. 4.11) and one-sided limits (§4.1);
- complex numbers, \(|z|\) and the conjugate (§1.1).
5.1Inverse functions [Umkehrfunktion]
\(f:D\to B\) is bijective [bijektiv] if for every \(y\in B\) there is exactly one \(x\in D\) with \(f(x)=y\). The inverse function is \(f^{-1}:B\to D,\ y\mapsto x\).
Bijective = injective (at most one \(x\) per \(y\), "no two inputs collide") + surjective (at least one, "every \(y\) is hit").
\(f:[0,\infty)\to[0,\infty),\ x\mapsto x^2\) is bijective with \(f^{-1}(y)=\sqrt y\) (solve \(x^2=y\) for \(x\ge0\)). The domain and target matter:
- \(x^2:\mathbb R\to[0,\infty)\) is not injective: \(f(-1)=f(1)\).
- \(x^2:[0,\infty)\to\mathbb R\) is not surjective: nothing maps to \(-1\).
Let \(I\) be an interval and \(f:I\to\mathbb R\) continuous and strictly increasing. Then \(f:I\to f(I)\) is bijective, and \(f^{-1}:f(I)\to I\) is also continuous and strictly increasing, with \[f^{-1}(f(x))=x\ \ \forall x\in I,\qquad f(f^{-1}(y))=y\ \ \forall y\in f(I).\]
Why: strictly increasing ⇒ injective; onto \(f(I)\) by definition; and the IVT makes \(f(I)\) an interval (no gaps). Graphically \(f(x)=y\iff x=f^{-1}(y)\), so the graph of \(f^{-1}\) is the graph of \(f\) mirrored at the diagonal [erste Winkelhalbierende] (Skript 5.4).
Full proof Extra — the script states 5.3 without proof
Bijective: \(x<x'\Rightarrow f(x)<f(x')\), so different inputs give different outputs (injective); onto \(f(I)\) by definition of the image.
\(f(I)\) is an interval: if \(y_1=f(x_1)<y<y_2=f(x_2)\), the IVT on the interval between \(x_1\) and \(x_2\) gives \(x\) with \(f(x)=y\). So \(f(I)\) has no gaps.
\(f^{-1}\) strictly increasing: let \(y_1<y_2\) and \(x_i=f^{-1}(y_i)\). If \(x_1\ge x_2\), then \(y_1=f(x_1)\ge f(x_2)=y_2\) — contradiction.
\(f^{-1}\) continuous: let \(y_n\to y\) in \(f(I)\), \(x_n=f^{-1}(y_n)\), \(x=f^{-1}(y)\). Suppose \(x_n\not\to x\). Then for some ε>0 infinitely many \(x_n\) satisfy \(x_n\ge x+\varepsilon\) (or infinitely many \(x_n\le x-\varepsilon\); that case is symmetric). Then \(x+\varepsilon\) lies between \(x\) and \(x_n\), so in \(I\), and for those \(n\): \(y_n=f(x_n)\ge f(x+\varepsilon)>f(x)=y\). The gap \(f(x+\varepsilon)-y>0\) is fixed, contradicting \(y_n\to y\). ∎
Extra — worth noticing The script states 5.3 only for strictly increasing \(f\). You also need the decreasing case (for \(\arccos\), and for \(\cos\) on \([0,\pi]\) below). It follows at once: if \(f\) is continuous and strictly decreasing, then \(-f\) is continuous and strictly increasing, so \((-f)^{-1}\) exists, is continuous and increasing, and \(f^{-1}(y)=(-f)^{-1}(-y)\) — continuous and strictly decreasing. In an exam you may simply say "5.3 analogously for strictly decreasing \(f\)".
- Choose the domain \(D\) so that \(f\) is injective there (e.g. strictly monotone), and determine the image \(B=f(D)\).
- Write \(y=f(x)\) and solve for \(x\). The result is \(x=f^{-1}(y)\), defined for \(y\in B\).
- Check both compositions: \(f(f^{-1}(y))=y\) and \(f^{-1}(f(x))=x\).
Example: \(f:\mathbb R\to(1,\infty),\ f(x)=e^{2x}+1\). Strictly increasing (composition of increasing functions); as \(x\to-\infty\) the values approach 1, as \(x\to\infty\) they go to ∞, so the image is \((1,\infty)\). Solve: \(y=e^{2x}+1\iff e^{2x}=y-1\iff 2x=\ln(y-1)\). So \(f^{-1}:(1,\infty)\to\mathbb R,\ f^{-1}(y)=\frac12\ln(y-1)\).
Example: \(g(x)=\frac{2x+1}{x-3}\) on \(\mathbb R\setminus\{3\}\). \(y(x-3)=2x+1\iff x(y-2)=3y+1\iff x=\frac{3y+1}{y-2}\) for \(y\neq2\). So \(g:\mathbb R\setminus\{3\}\to\mathbb R\setminus\{2\}\) is bijective with \(g^{-1}(y)=\frac{3y+1}{y-2}\) (the value 2 is never hit: it would need \(0=7\)).
WiSe23 6a: \(f:[0,\infty)\to\mathbb N_0,\ f(x)=\lfloor x\rfloor\) (round down). Injective, surjective, bijective?
Solution
Not injective: \(f(6.2)=6=f(6.3)\) but \(6.2\neq6.3\). Surjective: every \(n\in\mathbb N_0\) is hit, e.g. by \(x=n\). Not bijective (not injective). To disprove a property, one concrete counterexample is enough; to prove surjectivity, give a preimage for every target.
WiSe19 1i: "Draw the graph of the inverse function of the function shown." Mirror the graph at \(y=x\): every point \((a,b)\) becomes \((b,a)\). The widget above does exactly this.
What you must know for 5.1: the definition of the inverse function; determining the inverse in concrete cases.
5.2The logarithm [Logarithmusfunktion]
\(\exp:\mathbb R\to(0,\infty)\) is continuous and strictly increasing (Chapter 3/4). Its inverse is the natural logarithm [natürlicher Logarithmus] \[\ln:(0,\infty)\to\mathbb R,\qquad e^{\ln x}=x\ (x>0),\qquad\ln e^x=x\ (x\in\mathbb R).\]
The working rule for all proofs: \(\ln y=x\iff y=e^x\) (equation (5.3) in the script). Why exp is a bijection \(\mathbb R\to(0,\infty)\): continuous and strictly increasing (Lemma 4.4, Satz 3.25), \(e^x\to\infty\) for \(x\to\infty\) and \(e^x\to0\) for \(x\to-\infty\) (Blatt 5.3), so by the IVT every positive number is a value. And therefore, by Satz 5.3 applied to \(\exp\), \(\ln\) is itself continuous and strictly increasing — that is the fact every later proof in this section quietly uses (for example "\(\ln\) is increasing, hence \(\ln x_n>K\)" in Satz 5.7).
- \(\ln1=0,\quad\ln e=1\)
- \(\ln(xy)=\ln x+\ln y\) for \(x,y>0\)
- \(\ln\frac xy=\ln x-\ln y\)
- \(\ln(x^k)=k\ln x\) for \(k\in\mathbb Z\) (later for all real \(k\))
Proofs (each is the matching exp rule, read backwards)
(a) \(e^0=1\Rightarrow\ln1=0\); \(e^1=e\Rightarrow\ln e=1\).
(b) \(e^{\ln x+\ln y}=e^{\ln x}e^{\ln y}=xy\), so \(\ln(xy)=\ln x+\ln y\).
(c) \(0=\ln1=\ln(y\cdot\frac1y)=\ln y+\ln\frac1y\Rightarrow\ln\frac1y=-\ln y\); then (b).
(d) Induction for \(k\ge0\): \(\ln(x^{k+1})=\ln(x\cdot x^k)=\ln x+k\ln x\). For negative \(k\): \(\ln(x^k)=-\ln(x^{-k})=-(-k)\ln x\). ∎
Extra — redeeming the "later" Rule (d) holds for every real exponent \(a\), and once \(x^a\) is defined below it is no longer an induction but a one-liner: \[\ln(x^a)=\ln\!\left(e^{a\ln x}\right)=a\ln x\qquad(x>0,\ a\in\mathbb R),\] because \(\ln e^t=t\) for every real \(t\). So \(\ln(x^{\sqrt2})=\sqrt2\,\ln x\). This is what lets you pull any exponent out of a logarithm in the growth and Landau tasks.
Growth rates
- 5.6 \(\displaystyle\lim_{x\to\infty}\frac{e^x}{x^m}=\infty\) for every \(m\in\mathbb N\): exp grows faster than every polynomial.
- 5.7 \(\lim_{x\to\infty}\ln x=\infty\).
- 5.8 \(\displaystyle\lim_{x\to\infty}\frac x{(\ln x)^m}=\infty\) for every \(m\): \(x\) grows faster than every power of \(\ln\).
- 5.9 \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{\ln\ln x}=\infty\).
- 5.10 For \(a>b>0\): \(\frac{a^n}{b^n}\to\infty\).
Proofs (5.6 by dropping terms, 5.8 by substituting)
5.6 For \(x>0\) all terms of the series are positive; keep only one: \(e^x\ge\frac{x^{m+1}}{(m+1)!}\), so \(\frac{e^x}{x^m}\ge\frac{x}{(m+1)!}\to\infty\).
5.7 \(\ln(e^n)=n\to\infty\), and ln is increasing.
5.8 Let \(x_n\to\infty\) and \(y_n=\ln x_n\to\infty\). Then \(\frac{x_n}{(\ln x_n)^m}=\frac{e^{y_n}}{y_n^m}\to\infty\) by 5.6. Trick: substitute \(x=e^y\) to turn a log-statement into an exp-statement.
5.9 Same substitution in 5.8 with \(m=1\). 5.10 \(\frac{a^n}{b^n}=e^{n(\ln a-\ln b)}\) and \(\ln a-\ln b>0\). ∎
For \(x\to\infty\), each rung is negligible against the next (any positive exponents \(a,b\), any base \(q>1\)): \[\ln\ln x\ \ll\ (\ln x)^a\ \ll\ x^b\ \ll\ q^x\ \ll\ e^{x^2}.\] Recipe for "\(f\simeq\,?\)" questions: in a sum, find the term highest on the ladder (compare the \(x\)-powers first, then the log-powers). That term alone is asymptotically equal to the whole sum. Its coefficient comes along.
Why the ladder holds for real exponents Extra — the script proves only the integer cases 5.6 and 5.8
Satz 5.6 is stated for \(\frac{e^x}{x^m}\) with \(m\in\mathbb N\), and 5.8 for \(\frac x{(\ln x)^m}\) with \(m\in\mathbb N\). Exam tasks use exponents like \(x^{4}\), \((\ln x)^{3}\), \(x^\pi\), \(1.1^x\). Both extensions are one step:
(i) \(x^b=o(q^x)\) for every \(b>0\), \(q>1\). Pick an integer \(m\ge b\). For \(x\ge1\) we have \(x^b\le x^m\) (bigger exponent, base \(\ge1\)). Substitute \(u=x\ln q\to\infty\) (note \(\ln q>0\)): \[0\le\frac{x^b}{q^x}\le\frac{x^m}{e^{x\ln q}}=\frac1{(\ln q)^m}\cdot\frac{u^m}{e^{u}}\longrightarrow0\] by Satz 5.6. Squeeze. (With \(q=e\) this is the plain "exp beats every power".)
(ii) \((\ln x)^a=o(x^b)\) for all \(a,b>0\). Pick an integer \(m\ge a\). For \(x\ge e\) we have \(\ln x\ge1\), so \((\ln x)^a\le(\ln x)^m\). Substitute \(x=e^y\), i.e. \(y=\ln x\to\infty\): \[0\le\frac{(\ln x)^a}{x^b}\le\frac{y^m}{e^{by}}=\frac1{b^m}\cdot\frac{(by)^m}{e^{by}}\longrightarrow0\] by Satz 5.6 again. ∎
The same two tricks — round the exponent up to an integer, and substitute so that the exponent becomes the variable — also give \(\ln\ln x=o\!\left((\ln x)^a\right)\): apply (ii) with \(x\) replaced by \(\ln x\).
Both exams prescribe the shape of the answer: "find a function \(g\) of the form \(g(x)=ax^b(\ln x)^c\) or \(g(x)=ax^b\ln(\ln x)\) with suitable constants \(a,b,c\in\mathbb R\) such that \(f(x)\simeq g(x)\) for \(x\to\infty\)". So your answer is always one monomial in \(x\) and \(\ln x\) (with its coefficient) — never a sum, and never a limit value.
WiSe19 1h: \(f(x)=(2x^2-3x+7)\ln(\ln x^2)+(3x-2)^4\ln x-(53x+2)(2x^2+4)(\ln x)^2\simeq\ ?\)
Leading power of each term: \(2x^2\ln\ln(x^2)\) (power 2), \((3x)^4\ln x=81x^4\ln x\) (power 4), \(-106x^3(\ln x)^2\) (power 3). The highest \(x\)-power wins no matter what the logs do: \(f(x)\simeq81x^4\ln x\).
WiSe22 1f: \(f(x)=17x^4(\ln x)^3+5x^3(\ln x)^7+23x^4(\ln|\ln x|)^9\simeq\ ?\)
Two terms have \(x^4\); compare their log parts: \((\ln x)^3\) beats \((\ln\ln x)^9\) (5.9 ladder). The \(x^3\) term loses despite the 7th power of ln. Answer: \(17x^4(\ln x)^3\). Check: \(\frac f{17x^4(\ln x)^3}=1+\frac{5(\ln x)^4}{17x}+\frac{23(\ln\ln x)^9}{17(\ln x)^3}\to1\).
General powers and logarithms to base b
For \(x>0\), \(a\in\mathbb R\): \(\displaystyle x^a:=e^{a\ln x}\).
- Consistent with the old meaning: \(x^n=e^{n\ln x}=x\cdots x\), and \(x^{1/n}=\sqrt[n]x\) since \((e^{\frac1n\ln x})^n=e^{\ln x}=x\).
- \(x^{a+b}=x^ax^b\) (Lemma 5.12) — proof: \(x^{a+b}=e^{(a+b)\ln x}=e^{a\ln x}e^{b\ln x}\), the functional equation of exp.
- Extra \((x^a)^b=x^{ab}\) — proof: \((x^a)^b=e^{b\ln(x^a)}=e^{ba\ln x}=x^{ab}\), using \(\ln(x^a)=a\ln x\) from above. (The script proves only the special case \(b=n\in\mathbb N\) for the \(n\)-th root.)
- For \(b>1\), \(x\mapsto b^x=e^{x\ln b}\) is strictly increasing on all of \(\mathbb R\), with values in \((0,\infty)\); its inverse is \(\log_b:(0,\infty)\to\mathbb R\), and \(\displaystyle\log_b(a)=\frac{\ln a}{\ln b}\).
Note The script writes "\(f:(0,\infty)\to\mathbb R,\ f(x)=b^x\)" here. That looks like a slip: \(b^x=e^{x\ln b}\) makes sense for every real \(x\), and it is \(\log_b\) whose domain is \((0,\infty)\). Use \(b^x:\mathbb R\to(0,\infty)\) and \(\log_b:(0,\infty)\to\mathbb R\); everything the script proves is unaffected.
Proof of \(\log_b a=\frac{\ln a}{\ln b}\) (Lemma 5.13)
Put \(u=\log_ba\), i.e. \(b^u=a\) (definition of the inverse). Apply ln: \(\ln a=\ln(b^u)=\ln(e^{u\ln b})=u\ln b\). Since \(b>1\), \(\ln b>0\), and dividing gives \(u=\frac{\ln a}{\ln b}\). ∎
Why rewrite \(x^a\) as \(e^{a\ln x}\)? It turns every power question into an exp/ln question, where the rules above apply. Example: \(\log_2(1024)=\frac{\ln1024}{\ln2}=\frac{10\ln2}{\ln2}=10\). And the sorting bound \(\log_2(n!)=\ln(n!)\log_2e\) (Chapter 9).
(a) \(\lim_{x\downarrow0}\ln x=-\infty\) and \(\lim_{x\downarrow0}x\ln x=0\). (b) For \(x\downarrow0\): \(x^a\to0\) (\(a>0\)), \(1\) (\(a=0\)), \(+\infty\) (\(a<0\)). For \(x\to\infty\): \(x^a\to\infty,1,0\) respectively. (c) \(\lim_{x\downarrow0}x^x=1\).
Notation (right-hand limit [rechtsseitiger Grenzwert]): \(\lim_{x\downarrow x_0}f(x)=c\) means \(f(x_n)\to c\) for every sequence \(x_n\to x_0\) with \(x_n>x_0\). Left-hand limits \(x\uparrow x_0\) analogously. The sheet states this definition itself, in the preamble of 7.1 — in the script it appears only much later, as Definition 9.3 (§9.1, improper integrals). So if you look for it around Chapter 5 you will not find it; that is not an oversight on your part.
Solutions
(a) Substitute \(y=\frac1x\to\infty\): \(\ln x=-\ln y\to-\infty\), and \(x\ln x=-\frac{\ln y}y\to0\) by 5.8. (The log loses against the power \(x\) also near 0.)
(b) \(x^a=e^{a\ln x}\). For \(x\downarrow0\), \(\ln x\to-\infty\), so \(a\ln x\to-\infty\) if \(a>0\) and \(e^{a\ln x}\to0\) (Blatt 5.3b). For \(a<0\) use \(x^a=\frac1{x^{-a}}\). For \(x\to\infty\) analogously.
(c) \(x^x=e^{x\ln x}\to e^0=1\) by (a) and continuity of exp. Other parts: \(\lim_{x\downarrow2}\frac{x^2-2x}{x^2-4x+4}=\lim\frac{x(x-2)}{(x-2)^2}=\lim\frac x{x-2}=+\infty\); \(\lim_{x\downarrow0}x^\pi=0\); \(\lim_{x\downarrow0}x^{-\sqrt3}=+\infty\).
SoSe23 4a: maximal domains of \(f(x)=\ln\ln x\) and \(g(x)=x\ln x\)
ln needs a positive argument. \(g\): \(x>0\), so \(D_g=(0,\infty)\). \(f\): need \(x>0\) and \(\ln x>0\iff x>1\), so \(D_f=(1,\infty)\).
WiSe22 1a: \(\lim\frac{\sqrt x-\frac1{\sqrt x}}{\sqrt x+\frac1{\sqrt x}}\) for \(x\to\infty\) and \(x\downarrow0\)
Multiply by \(\frac{\sqrt x}{\sqrt x}\): \(\frac{x-1}{x+1}\). For \(x\to\infty\): \(\frac{1-\frac1x}{1+\frac1x}\to1\). For \(x\downarrow0\): \(\frac{0-1}{0+1}=-1\).
WiSe20 3b: \(\lim_{x\to0}\ln(x^{-1})\,(e^x-1)\) (only the result counted: 0)
First, which limit is meant? The exam writes \(x\to0\), but \(\ln(x^{-1})\) is defined only for \(x^{-1}>0\), i.e. \(x>0\). So the expression only exists to the right of 0 and "\(x\to0\)" here can only mean the right-hand limit \(x\downarrow0\). Saying this in one line costs nothing and shows you checked the domain.
Rewrite: \(\ln(x^{-1})(e^x-1)=-(x\ln x)\cdot\frac{e^x-1}x\), using \(\ln(x^{-1})=-\ln x\). As \(x\downarrow0\), \(x\ln x\to0\) (Blatt 7.1a) and \(\frac{e^x-1}x\to1\) (from the series: \(\frac{e^x-1}x=1+\frac x2+\frac{x^2}6+\dots\to1\); Skript 6.13 proves this carefully, it is the derivative of exp at 0). Product → 0.
WiSe19 1b: \(\lim_{x\to0}\frac{\ln(1+7x^2)}{3x^2}\) (only the result counted: \(\frac73\))
Step 1, a key limit: \(\lim_{u\to0}\frac{\ln(1+u)}u=1\). With Chapter 5 tools: put \(v=\ln(1+u)\), i.e. \(u=e^v-1\) (a bijection, and \(u\to0\iff v\to0\) by continuity of ln and exp). Then \(\frac{\ln(1+u)}u=\frac v{e^v-1}\to1\), because \(\frac{e^v-1}v=1+\frac v2+\frac{v^2}6+\dots\to1\) (series; Skript 6.13).
Step 2: with \(u=7x^2\to0\): \(\frac{\ln(1+7x^2)}{3x^2}=\frac73\cdot\frac{\ln(1+u)}u\to\frac73\). Same pattern as \(\frac{\sin 3x}{x}\): create exactly the standard quotient. (Later: l’Hospital, Chapter 7, or \(\ln(1+u)=u+o(u)\), Chapter 6.)
SoSe23 1a: \(\lim_{x\to\infty}(x+1)^4e^{-x}\) and \(\lim_{x\to-\infty}(x+1)^4e^{-x}\)
\(x\to\infty\): \(\frac{(x+1)^4}{e^x}=\frac{(x+1)^4}{x^4}\cdot\frac{x^4}{e^x}\to1\cdot0=0\) (Satz 5.6: exp beats every power). \(x\to-\infty\): no competition at all — \((x+1)^4\to\infty\) and \(e^{-x}\to\infty\), so the product → ∞. Check the direction before quoting "exp wins".
What you must know for 5.2: the rules for the logarithm; the growth behaviour of log and exp; the functions \(x^a\), \(b^x\) and \(\log_bx\).
5.3Trigonometric functions [Trigonometrische Funktionen]
\(\overline{x+iy}=x-iy\), \(\ \overline{\bar z}=z,\ \overline{z+z'}=\bar z+\bar z',\ \overline{zz'}=\bar z\,\bar z'\), and \(|x+iy|=\sqrt{x^2+y^2}=\sqrt{z\bar z}\) (Blatt 1.3).
For real \(x\), conjugating the series term by term gives \[\overline{e^{ix}}=\sum_{k=0}^\infty\frac{\overline{(ix)^k}}{k!}=\sum_{k=0}^\infty\frac{(-ix)^k}{k!}=e^{-ix},\] because \(\overline{i}=-i\) and conjugation respects products and sums (and limits, so it may be taken inside the series). Hence \[|e^{ix}|^2=e^{ix}\,\overline{e^{ix}}=e^{ix}e^{-ix}=e^0=1.\] So \(e^{ix}\) always lies on the unit circle.
Equivalently, and this is how the script defines sine and cosine: \[\cos x=\frac12\left(e^{ix}+e^{-ix}\right),\qquad\sin x=\frac1{2i}\left(e^{ix}-e^{-ix}\right).\]
Picture: walk along the unit circle an arc length \(x\) from \((1,0)\). The \(x\)-coordinate is \(\cos x\), the \(y\)-coordinate \(\sin x\). The formulas come from adding/subtracting \(e^{ix}=\cos x+i\sin x\) and \(e^{-ix}=\cos x-i\sin x\) (since cos is even and sin odd). Periodicity: \(e^{ix}=e^{i(x+2\pi)}\).
Extra The script uses π and the geometric picture from school without constructing them. For the record: one can show from the series that \(\cos\) has a smallest positive zero, and define \(\pi\) as twice that zero; then \(e^{i\pi/2}=i\), \(e^{i\pi}=-1\), \(e^{2\pi i}=1\). You are not expected to reproduce this.
\(e^{i(x+y)}=e^{ix}e^{iy}\). With Euler on both sides: \[\cos(x+y)+i\sin(x+y)=(\cos x+i\sin x)(\cos y+i\sin y)=\cos x\cos y-\sin x\sin y+i(\cos x\sin y+\sin x\cos y).\] Real and imaginary parts must agree: \[\cos(x+y)=\cos x\cos y-\sin x\sin y,\qquad\sin(x+y)=\sin x\cos y+\cos x\sin y.\] Extra With \(y=x\): \(\sin(2x)=2\sin x\cos x\), \(\cos(2x)=\cos^2x-\sin^2x\). From \(|e^{ix}|=1\): \(\cos^2x+\sin^2x=1\).
Find three different \(z_0,z_1,z_2\in\mathbb C\) with \(z^3=i\), in exponential form.
Solution
\(|z|^3=|i|=1\Rightarrow|z|=1\), so \(z=e^{i\varphi}\). \(z^3=e^{3i\varphi}\) (functional equation) and \(i=e^{i\pi/2}\). Two points on the unit circle agree iff their angles differ by a multiple of \(2\pi\): \[3\varphi=\tfrac\pi2+2\pi k\ \Rightarrow\ \varphi=\tfrac\pi6+\tfrac{2\pi k}3.\] \(k=0,1,2\): \(z_0=e^{i\pi/6},\ z_1=e^{i5\pi/6},\ z_2=e^{i9\pi/6}=e^{i3\pi/2}=-i\). (Check: \((-i)^3=-i^3=i\) ✓.) The roots form an equilateral triangle; see the widget above.
Related (WiSe19 1f): \(\{z\in\mathbb C: z=|z|^2\}\). The right side is real and \(\ge0\), so \(z=r\ge0\) with \(r=r^2\): \(A=\{0,1\}\).
\(\tan x=\frac{\sin x}{\cos x}\) (\(x\neq\frac\pi2+k\pi\)), \(\cot x=\frac{\cos x}{\sin x}\) (\(x\neq k\pi\)). Since exp is continuous, sin, cos, tan, cot are continuous on their domains.
Proof
\(\cos x=\frac12\sum_k\frac{(ix)^k+(-ix)^k}{k!}\), and \((ix)^k+(-ix)^k=(ix)^k(1+(-1)^k)\) is 0 for odd \(k\), \(2(ix)^k\) for even \(k\). Keep \(k=2j\): \((ix)^{2j}=(i^2)^jx^{2j}=(-1)^jx^{2j}\). Sine: the same with the odd terms. ∎
(\(\lim_{y\to x}f(y)=a\) means \(f(y_n)\to a\) for every sequence \(y_n\to x\), Skript 5.18.)
Proof
From the sine series, for \(x\neq0\): \(\frac{\sin x}x=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\dots=1+c(x)x^2\) with \(c(x)=\sum_{j\ge0}(-1)^{j+1}\frac{x^{2j}}{(2j+3)!}\) (index shift \(k=j+1\)).
Bound the correction: \((2j+3)!\ge j!\), so each term is at most \(\frac{(x^2)^j}{j!}\) in absolute value and \(|c(x)|\le\sum_j\frac{(x^2)^j}{j!}=e^{x^2}\le e\) for \(|x|\le1\). Hence \(0\le|c(x)x^2|\le ex^2\to0\), and squeeze gives \(\frac{\sin x}x\to1\).
Second limit Extra — the script says "analogously": \(1-\cos x=\frac{x^2}2-\frac{x^4}{4!}+\dots\), so \(\frac{1-\cos x}{x^2/2}=1-\frac{2x^2}{4!}+\frac{2x^4}{6!}-\dots=1+d(x)x^2\) with \(|d(x)|\le2e\) for \(|x|\le1\) by the same estimate. ∎
Meaning: near 0, \(\sin x\approx x\) and \(\cos x\approx1-\frac{x^2}2\). These two approximations solve most "limit at 0" exam tasks.
How to use the second one safely. Write \(1-\cos x=\frac{x^2}2\,r(x)\) with \(r(x)=\frac{1-\cos x}{x^2/2}\to1\). Now \(r\) can be carried through products, powers and quotients by the limit rules — no hand-waving needed. That is the whole trick in WiSe24 1c below.
WiSe24 1c: does \(\lim_{x\to0}\frac{x^4}{(1-\cos x)^2}\) exist, and what is it?
With \(1-\cos x=\frac{x^2}2r(x)\), \(r(x)\to1\) (Lemma 5.19): \[\frac{x^4}{(1-\cos x)^2}=\frac{x^4}{\frac{x^4}4\,r(x)^2}=\frac4{r(x)^2}\longrightarrow\frac4{1^2}=4.\] The limit exists and equals 4. Note \(1-\cos x\neq0\) for \(0<|x|<2\pi\), so the quotient is defined near 0 (but not at 0 — which is exactly why it is a limit question).
The official solution uses Landau symbols (\(1-\cos x=\frac{x^2}2(1+o(1))\), Chapter 6) and l'Hospital is also possible but needs four rounds. The version above uses nothing beyond this chapter. Same task discussed again in §7.4.
WiSe24 2a ("is \(\frac{\sin x}x\), patched by \(f(0)=1\), continuous everywhere?" — yes) and WiSe23 A2 (for which \(a\) is \(\frac{x\sin x}{1-\cos x}\), patched by \(f(0)=a\), continuous? — \(a=2\)) are continuity questions whose whole content is one of these two limits. They are worked in §4.1. Note the pattern: a key limit rarely appears as "compute the limit"; it appears as "is this patched-together function continuous?", and the patch value must be the limit.
Compute with Chapter 5 methods only: \(\lim_{x\to\infty}\frac x{(3x+1)\ln x}\), \(\lim_{x\to0}\frac{\sin3x}x\), \(\lim_{x\to0}\frac{\tan2x}{\tan3x}\), \(\lim_{x\to1}\frac{\sin(x-1)}{x^2-1}\), \(\lim_{x\to\infty}x^\pi\), \(\lim_{x\to\infty}x^{-\sqrt3}\).
Solutions (substitution + the key limits)
- \(\frac x{(3x+1)\ln x}=\frac1{(3+\frac1x)\ln x}\to0\).
- \(\frac{\sin3x}x=3\cdot\frac{\sin(3x)}{3x}\). With \(y=3x\to0\): \(\to3\cdot1=3\).
- \(\frac{\tan2x}{\tan3x}=\frac{\sin2x}{2x}\cdot\frac{3x}{\sin3x}\cdot\frac{\cos3x}{\cos2x}\cdot\frac23\to1\cdot1\cdot1\cdot\frac23=\frac23\).
- \(y=x-1\to0\), \(x^2-1=y(y+2)\): \(\frac{\sin y}{y}\cdot\frac1{y+2}\to\frac12\).
- \(\lim_{x\to\infty}x^\pi=+\infty\) and \(\lim_{x\to\infty}x^{-\sqrt3}=0\): Blatt 7.1(b) with \(a=\pi>0\) and \(a=-\sqrt3<0\).
Pattern: create exactly \(\frac{\sin(\square)}{\square}\) with the same expression \(\square\to0\) above and below. The sheet remarks that such substitutions are allowed because \(x\mapsto3x\) etc. are bijective.
- \(\sin:[-\frac\pi2,\frac\pi2]\to[-1,1]\) is continuous and strictly increasing ⇒ inverse \(\arcsin:[-1,1]\to[-\frac\pi2,\frac\pi2]\).
- \(\cos:[0,\pi]\to[-1,1]\) strictly decreasing ⇒ \(\arccos:[-1,1]\to[0,\pi]\), continuous and strictly decreasing (the decreasing version of 5.3, see the note under Satz 5.3).
- \(\tan:(-\frac\pi2,\frac\pi2)\to\mathbb R\) strictly increasing ⇒ \(\arctan:\mathbb R\to(-\frac\pi2,\frac\pi2)\).
sin is not injective on ℝ, so we restrict it to a piece where it is (Satz 5.3). The chosen piece decides the range of arcsin. Note how each target set is exactly the image of the restricted function — that is what Satz 5.3 gives you (\(f:I\to f(I)\)), and it is where marks are lost: writing \(\arcsin:[-1,1]\to\mathbb R\) or \(\arctan:\mathbb R\to\mathbb R\) is not the inverse of anything.
What you must know for 5.3: Euler's formula and deriving trig identities from it; the definitions of sin, cos, tan, cot and their inverses; the asymptotic behaviour of \(\sin x\) and \(1-\cos x\) for \(x\to0\).
!What the exam asks from Chapter 5
Chapter 5 appears in six of the nine exams — WiSe19, WiSe20, WiSe22, WiSe23, SoSe23, WiSe24 — mostly as short answer-only items in Task 1:
| Type | Exam tasks | What earns the points |
|---|---|---|
| "\(f\simeq\,?\)" with logs | WiSe19 1h, WiSe22 1f | the growth ladder: highest \(x\)-power first, then compare the log factors; keep the coefficient |
| Limits | WiSe19 1b, WiSe20 3b, WiSe22 1a, SoSe23 1a, WiSe24 1c | create a standard quotient (\(\frac{\sin u}u\), \(\frac{\ln(1+u)}u\), \(\frac{e^u-1}u\), \(\frac{1-\cos u}{u^2/2}\)) or use exp ≫ power ≫ log; check which one-sided limit the domain allows |
| A key limit inside a continuity question | WiSe24 2a, WiSe23 A2 | continuity away from the point is "quotient of continuous functions"; at the point, the patch value must equal the limit (worked in §4.1) |
| Domains | SoSe23 4a | ln needs argument \(>0\), √ needs \(\ge0\), denominators \(\neq0\) — apply from the inside out |
| Complex roots / sets | WiSe23 6c, WiSe19 1f | write \(z=re^{i\varphi}\); compare absolute values and angles (angles up to \(2\pi k\)) |
| Injective / surjective / inverse | WiSe23 6a, WiSe19 1i | concrete counterexample to disprove; a preimage for every target to prove surjectivity; mirror at \(y=x\) |
And as a first step inside later tasks. Chapter 5 is also what makes several non-Chapter-5 exam tasks start: SoSe21 4b (extrema of \(x^x\) on \((0,\infty)\)) only becomes differentiable-looking after writing \(x^x=e^{x\ln x}\); WiSe24 4a differentiates \(\log\frac1{1+e^x}\); SoSe21 1a asks for the accumulation points of \(e^{i\pi n/2}(1-e^{i\pi n})\), i.e. Euler's formula on the unit circle; WiSe19 1c (\(n^7\sin 3^{-n}\to0\)) needs \(|\sin u|\le|u|\) plus "exponential beats power". If a task looks unfamiliar, check first whether a Chapter 5 rewrite turns it into something standard.
Drill: which term dominates?
Answer-only "≃" and limit items are quick points once the growth ladder is automatic. Pick the answer; the explanation shows the reasoning to write down.
- For the exam: the six task types above.
- For Chapter 6: Landau symbols \(o\), \(O\) are the growth ladder in formal dress; the derivatives of exp, sin, cos are computed from the key limits of this chapter; \((\ln x)'=\frac1x\) comes from the inverse-function rule, and \(x^a=e^{a\ln x}\) gives \((x^a)'=ax^{a-1}\).
- For Chapter 7: l’Hospital gives a second route to the limits here; curve sketches of \(x^2e^{-x}\), \(x^x\) etc. need the growth rates.
- For Chapter 8: antiderivatives \(\ln|x|\), \(\arctan x\), \(\arcsin x\).
- For Chapters 10 and 12: the series of sin, cos, ln(1+x) are Taylor series; oscillations \(e^{-\frac a2x}(c_1\cos\omega x+c_2\sin\omega x)\) come from \(e^{\lambda x}\) with complex \(\lambda\) and Euler's formula.
ΣSummary sheet
| Object | Definition / rule | Use |
|---|---|---|
| \(f^{-1}\) | continuous + strictly monotone on an interval ⇒ inverse exists on the image, same properties; mirror at \(y=x\) | ln, √, arcsin, arctan |
| \(\ln\) | \(\ln y=x\iff y=e^x\); \(\ln(xy)=\ln x+\ln y\), \(\ln x^a=a\ln x\) (every real \(a\)) | turn products/powers into sums |
| \(x^a\) | \(e^{a\ln x}\); \(x^{a+b}=x^ax^b\), \((x^a)^b=x^{ab}\) | every power question → exp/ln |
| \(\log_b a\) | \(\frac{\ln a}{\ln b}\) | base change |
| growth | \(\ln\ll x^b\ll q^x\) (\(b>0,q>1\)); also \(x\ln x\to0\), \(x^x\to1\) as \(x\downarrow0\) | \(\simeq\), o/O questions |
| Euler | \(e^{ix}=\cos x+i\sin x\), \(|e^{ix}|=1\) | trig identities, roots of unity |
| key limits | \(\frac{\sin x}x\to1\), \(\frac{1-\cos x}{x^2/2}\to1\) as \(x\to0\) (Lemma 5.19); worth adding: \(\frac{e^x-1}x\to1\) and \(\frac{\ln(1+x)}x\to1\) (from the series / Skript 6.13, both used by exam tasks in this chapter) | limits at 0, continuity questions |
?Self-test
DifferentiabilityDifferenzierbarkeit
The derivative answers: which straight line approximates \(f\) best near \(x_0\)? To say "best" precisely, the chapter first introduces the Landau symbols \(o\) and \(O\), the language of error terms that computer scientists know from running times. Then comes the definition of the derivative, the derivatives of exp, sin, cos from their series, and the rules (product, quotient, chain, inverse function) that let you differentiate everything else.
Source labels: Skript 6.7 = stated in your lecture script · Blatt 8 · 8.2 = exercise sheet · Klausur WiSe24 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§6.2 = Section 6.2) rather than a numbered result (6.2 = Beispiel 6.2).
0The big picture
- 6.1 \(f=O(g)\): \(f\) is at most a constant times \(g\) (near \(x_0\)). \(f=o(g)\): \(f\) is negligible compared to \(g\) (\(\frac fg\to0\)).
- 6.2 \(f\) is differentiable at \(x_0\) if \(f(x)=f(x_0)+f'(x_0)(x-x_0)+o(|x-x_0|)\): the tangent line has an error that is small even compared to the distance \(|x-x_0|\). Equivalently, the difference quotient converges. Differentiable ⇒ continuous, not conversely (\(|x|\)).
- 6.3 Rules: sum, product, quotient, chain rule, derivative of the inverse. With them, \((x^n)'=nx^{n-1}\), \((a^x)'=a^x\ln a\), \((\ln x)'=\frac1x\), \((\tan)'=1+\tan^2\), \((\arctan)'=\frac1{1+x^2}\), …
- Limits of functions \(\lim_{x\to x_0}f(x)\): defined through sequences (Skript 5.18, Def. 4.11), including one-sided limits (§4.1). Every derivative is such a limit.
- The key limits \(\frac{\sin x}x\to1\) and \(\frac{1-\cos x}{x^2/2}\to1\) (Lemma 5.19, §5.3) and the growth ladder log ≪ power ≪ exp (§5.2) — the Landau questions are the ladder in new notation. The third standard limit \(\frac{e^x-1}x\to1\) is not in Chapter 5: it is Beispiel 6.13 of this chapter (it is the derivative of exp at 0), even though the Chapter-5 exam solutions already lean on it.
- Composition \(f\circ g\) and inverse functions (Basics, §5.1) — for the chain rule and the inverse rule.
- \(\|x\|\) = length of a vector in \(\mathbb R^d\) (§1.3), used when \(x_0\in\mathbb R^d\).
6.1Landau symbols [Landau-Symbole]
Let \(f,g\) be functions and \(x_0\in\mathbb R^d\) (or \(x_0=\pm\infty\)).
- \(f(x)=O(g(x))\) for \(x\to x_0\) if there are \(\varepsilon>0\), \(C>0\) such that \(|f(x)|\le C|g(x)|\) for all \(x\) with \(\|x-x_0\|<\varepsilon\). For \(x_0=\infty\): for all \(x>M\) (some \(M\)). "\(f\) is bounded by \(g\) up to a constant."
- \(f(x)=o(g(x))\) for \(x\to x_0\) if \(\displaystyle\lim_{x\to x_0}\frac{f(x)}{g(x)}=0\). "\(f\) is negligible compared to \(g\)."
\(o(h(x))\) inside a formula stands for "some function that is \(o(h(x))\)". So \[f(x)=g(x)+o(h(x))\quad(x\to x_0)\qquad\text{means}\qquad r(x):=f(x)-g(x)\ \text{satisfies}\ \lim_{x\to x_0}\frac{r(x)}{h(x)}=0.\] Read it as "\(f\) equals \(g\) up to an error that is negligible compared to \(h\)". Example: \(e^x=1+x+o(x)\) for \(x\to0\), because \(\frac{e^x-1-x}x=\frac{e^x-1}x-1\to0\). Same for \(O\): \(\sin x=x+O(x^3)\) for \(x\to0\).
The same pair can behave oppositely at 0 and at ∞: \(x^3=o(x^2)\) for \(x\to0\), but \(x^2=o(x^3)\) for \(x\to\infty\). Near 0, higher powers are smaller; near ∞, lower powers are smaller. Also, "\(=\)" here is a one-way statement: \(x^2=O(x)\) for \(x\to0\) does not mean \(O(x)=x^2\).
- \(f=O(1)\) for \(x\to x_0\): \(f\) is bounded near \(x_0\). \(f=o(1)\): \(f(x)\to0\). And \(o(1)\) implies \(O(1)\).
- \(f(h)=o(h)\) for \(h\to0\) implies \(f(h)=o(1)\), since \(f(h)=\frac{f(h)}h\cdot h\to0\cdot0\). But \(\sqrt h=o(1)\) and \(\sqrt h\neq o(h)\), since \(\frac{\sqrt h}h=\frac1{\sqrt h}\to\infty\).
- \(f(x)=5x^7+2x^3-x^2\): for \(x\to0\), \(f=O(x^2)\): for \(|x|\le1\), \(|f(x)|\le5x^2+2x^2+x^2=8x^2\). For \(x\to\infty\), \(f=O(x^7)\): for \(x\ge1\), \(|f(x)|\le8x^7\).
If \(\displaystyle\lim_{x\to x_0}\left|\frac{f(x)}{g(x)}\right|=c\) exists (finite), then \(f=O(g)\).
Proof: near \(x_0\), \(\left|\frac fg\right|\le c+1\). So in practice: compute \(\lim\frac fg\). Limit 0 ⇒ \(o(g)\) (and \(O(g)\)). Finite limit ⇒ \(O(g)\). Limit ∞ ⇒ neither.
Find simple \(g,h\) with \(f=o(g)\) and \(f=O(h)\).
| f | x₀ | O(h) | o(g) | why |
|---|---|---|---|---|
| \(5x^6+4x^5+2x^2\) | 0 | \(x^2\) | \(x\) | \(f/x^2\to2\); lowest power dominates near 0 |
| same | ∞ | \(x^6\) | \(x^7\) | \(f/x^6\to5\); highest power dominates near ∞ |
| \(x^3+3x^2-24x+28\) | 2 | \((x-2)^2\) | \(x-2\) | factor: \(f=(x-2)^2(x+7)\), \(f/(x-2)^2\to9\) |
| same | −7 | \(x+7\) | \(1\) | \(f/(x+7)\to81\) |
| \(e^x-1\) | 0 | \(x\) | \(1\) | \(\frac{e^x-1}x\to1\) |
| \(e^x\) | −∞ | \(e^x\) | \(1\) | \(e^x\to0\) |
Lesson from (c), the third row: near a zero \(x_0\) of a polynomial, factor out \((x-x_0)\) as often as possible. The multiplicity of the zero gives the order.
- Products are fine: \(f_j=o(g_j)\Rightarrow f_1f_2=o(g_1g_2)\); \(f_j=O(g_j)\Rightarrow f_1f_2=O(g_1g_2)\) (multiply the limits / the constants).
- Sums need care: \(f_1+f_2=O(g_1+g_2)\) can fail because \(g_1+g_2\) may cancel: for \(x\to0\), \(x^2-x=O(-x)\) and \(x^2+x=O(x)\), but the sum \(2x^2\neq O(0)\). What is true: \(f_1+f_2=O(|g_1|+|g_2|)\) and similarly for \(o\).
WiSe20 3a: all \(p\) with \(\frac{3x^2\cos x}{x+\sin(x)/x}=o(x^p)\) for \(x\to0\)
Denominator: \(x+\frac{\sin x}x\to0+1=1\). Numerator \(\simeq3x^2\). So \(\frac{f(x)}{x^p}\simeq3x^{2-p}\to0\iff2-p>0\iff p<2\). (Check \(p=2\): the ratio → 3, which is \(O\) but not \(o\).) Note: the uploaded Musterlösung prints “\(p>2\)”. That contradicts the computation (for \(p>2\) the ratio \(3x^{2-p}\) blows up), so I believe it is a typo. Worth asking your tutor if you want certainty.
WiSe19 A3: all \(a\) with \((2x^7-x^2)e^x+x^5e^{-x}+e^{2x}=o(x^ae^{2x})\) for \(x\to\infty\)
Divide by \(x^ae^{2x}\): \((2x^{7-a}-x^{2-a})e^{-x}+x^{5-a}e^{-3x}+x^{-a}\). The first two terms → 0 for every \(a\) (exp beats every power). The last, \(x^{-a}\to0\iff a>0\). Answer: \(a\in(0,\infty)\).
WiSe24 3a: \(a\) with \(2e^x-a=o(1)\) for \(x\to0\); 3b: smallest \(a\) with \(200x^2+10^{-5}x^3+\sin x=O(x^a)\) for \(x\to\infty\)
3a: \(o(1)\) means "→ 0". \(2e^x\to2\), so \(a=2\).
3b: \(|\dots|\le Cx^3\) for large \(x\) (the tiny coefficient does not matter!), so \(a=3\) works. For \(a<3\), \(\frac{10^{-5}x^3}{x^a}\to\infty\), so it fails. Answer \(a=3\).
SoSe23 1g: \(n^2+2n\log n+24=o(n^2\log n)\)? \(3^{\log n}=O(n^3)\)?
First: divide, \(\frac1{\log n}+\frac2{n}+\frac{24}{n^2\log n}\to0\): true. Second: \(3^{\ln n}=e^{\ln3\cdot\ln n}=n^{\ln3}\approx n^{1.0986}\le n^3\): true. (Trick: \(a^{\ln b}=b^{\ln a}\).)
Extra — which log? The exam writes \(\log\) without a base, so the answer should not depend on one. It does not: in general \(3^{\log_b n}=n^{\log_b3}\) by the same rewriting, and \(\log_b3\le\log_23\approx1.585<3\) for every base \(b\ge2\), so \(3^{\log_b n}=O(n^3)\) either way. In the first statement the base cancels out of \(\frac1{\log n}\to0\) as well.
What you must know for 6.1: working with o and O.
6.2Differentiability [Differenzierbarkeit]
Goal: approximate \(f\) near \(x_0\) by a straight line that describes how \(f\) changes there.
\(f:I\to\mathbb R\) on an open interval \(I\) is differentiable at \(x_0\in I\) if there is a number \(f'(x_0)\) such that the linearisation holds: \[f(x)=f(x_0)+f'(x_0)(x-x_0)+o(|x-x_0|)\qquad(x\to x_0).\] The line \(x\mapsto f(x_0)+f'(x_0)(x-x_0)\) is the tangent [Tangente], its slope \(f'(x_0)\) the derivative [Ableitung].
Divide the linearisation by \(x-x_0\) (for \(x\neq x_0\)): "the error \(f(x)-f(x_0)-f'(x_0)(x-x_0)\) is \(o(|x-x_0|)\)" means \(\frac{f(x)-f(x_0)-f'(x_0)(x-x_0)}{x-x_0}\to0\), i.e. \(\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\to0\). (Dividing by \(|x-x_0|\) or by \(x-x_0\) makes no difference for "→ 0".) Now set \(x=x_0+h\). The definition is equivalent to \[f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}.\] \(\frac{f(x_0+h)-f(x_0)}h\) is the difference quotient [Differenzenquotient]: the slope of the secant through \((x_0,f(x_0))\) and \((x_0+h,f(x_0+h))\). The tangent is the limit of secants.
Definition 6.10 Skript 6.10: \(f\) is differentiable if it is differentiable at every point of its domain. Notation: \(f'(x_0)=\frac{df}{dx}(x_0)=Df(x_0)\).
6.8 The tangent approximates \(f\) better than any other line: another line \(a(x-x_0)+b\) has error \(O(1)\) if \(b\neq f(x_0)\), and error \(O(x-x_0)\) if \(b=f(x_0)\) but \(a\neq f'(x_0)\). Only the tangent achieves \(o(|x-x_0|)\).
Why: \(f(x)-[a(x-x_0)+b]=\big(f(x_0)-b\big)+\big(f'(x_0)-a\big)(x-x_0)+o(|x-x_0|)\). If \(b\neq f(x_0)\), the error tends to the nonzero constant \(f(x_0)-b\). If \(b=f(x_0)\) but \(a\neq f'(x_0)\), dividing by \(x-x_0\) gives \(f'(x_0)-a+o(1)\not\to0\). Extra That last line says slightly more than the script's \(O(x-x_0)\): the order is exact, because the quotient tends to the nonzero number \(f'(x_0)-a\), so the error is not \(o(|x-x_0|)\).
6.9 Differentiable at \(x_0\) ⇒ continuous at \(x_0\): let \(x\to x_0\) in the linearisation; everything after \(f(x_0)\) tends to 0.
Not conversely: \(|x|\) is continuous but not differentiable at 0 — the script states this without an argument. Extra Here is the argument: the difference quotient \(\frac{|h|}h\) is \(+1\) for \(h>0\) and \(-1\) for \(h<0\), so it has no limit (a corner has no single tangent).
- 6.11 \((ax+b)'=a\): \(\frac{a(x_0+h)+b-ax_0-b}h=a\).
- 6.12 \((x^2)'=2x\): \(\frac{(x_0+h)^2-x_0^2}h=\frac{2x_0h+h^2}h=2x_0+h\to2x_0\).
- 6.13 \((e^x)'=e^x\): at 0, the series gives \(\frac{e^h-1}h=\frac1h\sum_{k\ge1}\frac{h^k}{k!}=1+h\,c(h)\) with \(c(h)=\sum_{j\ge0}\frac{h^j}{(j+2)!}\), and \(|c(h)|\le\sum_j\frac{|h|^j}{j!}=e^{|h|}\) because \((j+2)!\ge j!\). So \(|h\,c(h)|\le|h|e^{|h|}\to0\) and the quotient → 1. At any \(x\): \(\frac{e^{x+h}-e^x}h=e^x\cdot\frac{e^h-1}h\to e^x\) (functional equation).
- 6.14 \((\sin)'=\cos\): addition theorem, \(\frac{\sin(x+h)-\sin x}h=\sin x\,\frac{\cos h-1}h+\cos x\,\frac{\sin h}h\to\sin x\cdot0+\cos x\cdot1\). Here \(\frac{\cos h-1}h=-\frac h2\cdot\frac{1-\cos h}{h^2/2}\to0\) (Lemma 5.19). Extra For cosine the script only says "analogously"; the computation is \(\frac{\cos(x+h)-\cos x}h=\cos x\,\frac{\cos h-1}h-\sin x\,\frac{\sin h}h\to-\sin x\), using \(\cos(x+h)=\cos x\cos h-\sin x\sin h\).
\(f:\mathbb R\to\mathbb R\) with (i) \(f(x+y)=f(x)f(y)\), (ii) \(f(0)=1\), (iii) \(f\) differentiable at 0 with \(f'(0)=1\). (a) Show with the definition that \(f\) is differentiable everywhere with \(f'=f\). (b)–(d) Show \(g(x)=f(x)e^{-x}\) is constant \(=1\), so \(f=\exp\).
Solution
(a) Exactly the argument for \(e^x\) above, using only (i)–(iii): \[\frac{f(x+h)-f(x)}h\overset{(i)}=f(x)\,\frac{f(h)-1}h\overset{(ii)}=f(x)\,\frac{f(h)-f(0)}h\to f(x)f'(0)=f(x).\] (b) Product rule: \(g'=f'e^{-x}-fe^{-x}=0\), so \(g\) is constant (Satz 7.13, Chapter 7). (c) \(g(0)=f(0)e^0=1\). (d) \(f(x)e^{-x}=1\iff f(x)=e^x\).
Moral: exp is the only differentiable function turning sums into products with \(f'(0)=1\).
The sheet asks about two functions: (a) \(f(x)=2x^3\sin\frac1x\) for \(x\neq0\), \(f(0)=0\), and (b) \(f(x)=x^2\sin\frac1x\) for \(x\neq0\), \(f(0)=0\). Differentiable on ℝ? Is \(f'\) continuous?
Solution
Away from 0 use the rules. At 0 you must use the definition (the rules don't apply at the gluing point). Part (b) is the interesting one, so take it first:
(b) \(x^2\sin\frac1x\): \(\frac{f(h)-f(0)}h=h\sin\frac1h\to0\) (bounded × null sequence). So \(f'(0)=0\). For \(x\neq0\): \(f'(x)=2x\sin\frac1x-\cos\frac1x\). As \(x\to0\), the first term → 0, but \(\cos\frac1x\) has no limit (it is 0 at \(x=\frac1{\pi/2+\pi n}\), 1 at \(x=\frac1{2\pi n}\)). So \(f\) is differentiable everywhere, but \(f'\) is not continuous at 0.
(a) \(2x^3\sin\frac1x\): \(f'(0)=\lim2h^2\sin\frac1h=0\), and \(f'(x)=6x^2\sin\frac1x-2x\cos\frac1x\to0=f'(0)\): here \(f'\) is continuous. One power of \(x\) more is enough to tame the oscillation.
SoSe21 A6 asked the same about \(x^2\sin\frac1x\): "(i) \(f'\) continuous on ℝ" false; "(ii) \(f\) differentiable on ℝ" true; "(iii) \(f\) continuously differentiable on \(\mathbb R\setminus\{0\}\)" true.
What you must know for 6.2: the definitions of derivative and tangent; the derivatives of exp, sin and cos.
6.3Rules of differentiation [Ableitungsregeln]
If \(f,g\) are differentiable at \(x\):
- \((cf)'=cf'\)
- \((f+g)'=f'+g'\) [Summenregel]
- \((fg)'=f'g+fg'\) [Produktregel]
- \(\left(\frac fg\right)'=\frac{gf'-fg'}{g^2}\) where \(g(x)\neq0\) [Quotientenregel]
Proof of the product rule
Add and subtract \(f(x)g(x+h)\) (the Chapter-1 trick once more): \[\frac{f(x+h)g(x+h)-f(x)g(x)}h=\frac{f(x+h)-f(x)}h\,g(x+h)+f(x)\,\frac{g(x+h)-g(x)}h\to f'(x)g(x)+f(x)g'(x).\] We used \(g(x+h)\to g(x)\), i.e. that \(g\) is continuous, which holds because \(g\) is differentiable (6.9). ∎
Proof of the quotient rule Extra — the script proves only the product rule
First \(\frac1g\): \(\frac1h\left(\frac1{g(x+h)}-\frac1{g(x)}\right)=-\frac{g(x+h)-g(x)}h\cdot\frac1{g(x+h)g(x)}\to-\frac{g'(x)}{g(x)^2}\) (again \(g(x+h)\to g(x)\neq0\), and \(g(x+h)\neq0\) for small \(h\) by continuity). Then the product rule for \(f\cdot\frac1g\): \[\Big(\frac fg\Big)'=f'\cdot\frac1g+f\cdot\Big(-\frac{g'}{g^2}\Big)=\frac{f'g-fg'}{g^2}.\ \ ∎\] Sum rule and constant factor: directly from the limit rules for the difference quotient.
\((x^n)'=nx^{n-1}\) (induction: base \(n=1\): \((x)'=1=1\cdot x^0\); step: \((x^{n+1})'=(x\cdot x^n)'=1\cdot x^n+x\cdot nx^{n-1}=(n+1)x^n\) by the product rule and the induction hypothesis). Hence \(\left(\sum a_kx^k\right)'=\sum ka_kx^{k-1}\). E.g. \((4x^{12}-7x^4+3x^3+7)'=48x^{11}-28x^3+9x^2\).
\((\tan)'=\frac{\cos\cdot\cos-\sin\cdot(-\sin)}{\cos^2}=\frac1{\cos^2x}=1+\tan^2x\) (quotient rule and \(\sin^2+\cos^2=1\)).
If \(g\) is differentiable at \(x_0\) and \(f\) at \(g(x_0)\), then \[(f\circ g)'(x_0)=f'(g(x_0))\cdot g'(x_0).\] "Outer derivative, evaluated at the inner function, times inner derivative."
Definition 6.19 (needed for the precise statement): \(x_0\) is an interior point [innerer Punkt] of \(A\subseteq\mathbb R\) if some interval \((x_0-\varepsilon,x_0+\varepsilon)\) lies completely in \(A\). Satz 6.20 requires \(x_0\) interior in the domain of \(g\) and \(g(x_0)\) interior in the domain of \(f\), so that both difference quotients can be formed from both sides. Example: 0 is an interior point of \((-1,1)\), but not of \([0,1]\).
Proof idea
Where \(g(x)\neq g(x_0)\), expand the quotient: \[\frac{f(g(x))-f(g(x_0))}{x-x_0}=\underbrace{\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)}}_{\to f'(g(x_0))}\cdot\underbrace{\frac{g(x)-g(x_0)}{x-x_0}}_{\to g'(x_0)}.\] (The first factor converges because \(g(x)\to g(x_0)\), by continuity of \(g\).) Where \(g(x)=g(x_0)\), the quotient is 0; if this happens for \(x\) arbitrarily close to \(x_0\), then \(g'(x_0)=0\), so the claimed value \(f'(g(x_0))\,g'(x_0)\) is 0 as well. ∎
- \(a^x=e^{x\ln a}\) for \(a>0\) (the script says so; \(\ln a\) needs it): outer \(e^u\), inner \(x\ln a\). \((a^x)'=e^{x\ln a}\ln a=a^x\ln a\), for every \(x\in\mathbb R\).
- \(x^a=e^{a\ln x}\) on \((0,\infty)\), with \(a\neq0\) in Lemma 6.25 (for \(a=0\) the function is the constant 1 and the formula \(ax^{a-1}\) reads 0, which is still right): \((x^a)'=e^{a\ln x}\cdot\frac ax=ax^{a-1}\). Special cases \((\sqrt x)'=\frac1{2\sqrt x}\), \(\left(\frac1x\right)'=-\frac1{x^2}\), \(\left(\frac1{x^2}\right)'=-\frac2{x^3}\).
If \(f\) is bijective, differentiable at \(y_0\) with \(f'(y_0)\neq0\), then \(f^{-1}\) is differentiable at \(x_0=f(y_0)\) with \[(f^{-1})'(x_0)=\frac1{f'(f^{-1}(x_0))}.\]
Where it comes from: differentiate \(f^{-1}(f(y))=y\) with the chain rule: \((f^{-1})'(f(y))\cdot f'(y)=1\). Extra — what the script's proof does and does not do The script's proof assumes that \(f^{-1}\) is differentiable and only computes its value. To close that gap one writes the difference quotient of \(f^{-1}\) at \(x_0\) as the reciprocal of the difference quotient of \(f\) at \(y_0\) — legitimate because \(f\) is injective, so the denominator is nonzero — and then needs \(f^{-1}(x)\to f^{-1}(x_0)\) as \(x\to x_0\), i.e. continuity of \(f^{-1}\). Satz 5.3 supplies exactly that, but only under its own hypotheses (\(f\) continuous and strictly monotone on an interval), which is the situation of all the applications below. Geometrically: mirroring at \(y=x\) turns slope \(m\) into slope \(\frac1m\) — and a horizontal tangent (\(m=0\)) becomes vertical, which is why \(f'(y_0)\neq0\) is needed (e.g. \(\sqrt[3]x\) at 0).
- \((\ln)'(x)=\frac1{\exp(\ln x)}=\frac1x\).
- \((\arcsin)'(x)=\frac1{\cos(\arcsin x)}=\frac1{\sqrt{1-\sin^2(\arcsin x)}}=\frac1{\sqrt{1-x^2}}\) on \((-1,1)\) (cos ≥ 0 on \([-\frac\pi2,\frac\pi2]\), so \(\cos y=+\sqrt{1-\sin^2y}\)). For arccos: \((\arccos)'(x)=\frac1{-\sin(\arccos x)}=-\frac1{\sqrt{1-x^2}}\), since \(\sin\ge0\) on \([0,\pi]\).
- \((\arctan)'(x)=\frac1{1+\tan^2(\arctan x)}=\frac1{1+x^2}\).
\(\displaystyle\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\) for all \(x\in\mathbb R\) (for \(x=1\): this is how \(e\) was defined in Chapter 2).
Proof: take logs (ln is continuous): \(n\ln(1+\frac xn)=x\cdot\frac{\ln(1+\frac xn)-\ln1}{\frac xn}\to x\cdot(\ln)'(1)=x\), a difference quotient of ln at 1.
- Simplify first (log rules, cancel, rewrite roots as powers). E.g. \(\log\frac1{1+e^x}=-\log(1+e^x)\).
- Identify the outermost operation: sum → sum rule; product → product rule; quotient → quotient rule; function of something → chain rule.
- Variable in the base and in the exponent (\(x^x\), \(x^{2x}\), \((1+x^3)^{\cos x}\))? Rewrite as \(e^{(\text{exponent})\cdot\ln(\text{base})}\) first.
- Check with a simple value or special case if you can.
Blatt 8.3 (a)–(e), in the sheet's order
- (a) \((8x^5+8x^3+12x^2+5)'=40x^4+24x^2+24x\) (Lemma 6.16 term by term).
- (b) \((e^{2x^2+7x+3})'=(4x+7)e^{2x^2+7x+3}\) (chain rule).
- (c) \(\left(\frac1x\right)'=\frac{x\cdot0-1\cdot1}{x^2}=-\frac1{x^2}\) on \(\mathbb R\setminus\{0\}\) (quotient rule).
- (d) On \(\mathbb R\setminus\{-3\}\), using \(x^2+6x+9=(x+3)^2\): \[\left(\frac{2x^4+5x^3+x+2}{x^2+6x+9}\right)'=\frac{(x+3)^2(8x^3+15x^2+1)-2(x+3)(2x^4+5x^3+x+2)}{(x+3)^4}=\frac{4x^4+29x^3+45x^2-x-1}{(x+3)^3}.\] Cancel one factor \((x+3)\) before expanding, or the algebra gets unpleasant.
- (e) \((x^5e^{-2x^2+3x+1})'=\big(5x^4+x^5(-4x+3)\big)e^{-2x^2+3x+1}=(-4x^6+3x^5+5x^4)e^{-2x^2+3x+1}\) (product + chain rule).
Blatt 9.2 (a)–(i), in the sheet's order — only (a)–(d) were to be handed in
- (a) \((4\sin^2x+4\cos^2x)'=8\sin x\cos x-8\cos x\sin x=0\). Of course: the function is the constant 4.
- (b) \((\tan x)'=\frac1{\cos^2x}\) on \((-\frac\pi2,\frac\pi2)\) — this is Lemma 6.18 above.
- (c) \(\left(\sqrt{\frac{x+1}{x-1}}\right)'=\frac12\left(\frac{x+1}{x-1}\right)^{-1/2}\cdot\frac{-2}{(x-1)^2}=-\frac1{(x-1)^{3/2}\sqrt{x+1}}\) on \((1,\infty)\).
- (d) \((x^2\sin(3x+2))'=2x\sin(3x+2)+3x^2\cos(3x+2)\).
- (e) \(\left(xe^{1/x}\right)'=e^{1/x}+x\,e^{1/x}\left(-\frac1{x^2}\right)=e^{1/x}\left(1-\frac1x\right)\).
- (f) \((\sin(e^{\cos x}))'=\cos(e^{\cos x})\cdot e^{\cos x}\cdot(-\sin x)\) (chain rule twice).
- (g) \((\arccos)'(x)=\frac1{-\sin(\arccos x)}=-\frac1{\sqrt{1-x^2}}\) on \((-1,1)\) — the inverse-function rule, as in Beispiel 6.26 above.
- (h) Variable exponent, so rewrite: \(x^{2x}=e^{2x\ln x}\), hence \((x^{2x})'=x^{2x}(2\ln x+2)=2x^{2x}(\ln x+1)\) on \((0,\infty)\).
- (i) \((1+x^3)^{\cos x}=e^{\cos x\ln(1+x^3)}\), derivative \((1+x^3)^{\cos x}\left(-\sin x\ln(1+x^3)+\frac{3x^2\cos x}{1+x^3}\right)\) on \((-1,\infty)\).
(h) and (i) are the case the method box above warns about: a variable in the base and in the exponent must be rewritten with \(e\) and \(\ln\) before any rule applies.
Blatt 9.1: even and odd functions
\(f\) even: \(f(-x)=f(x)\); odd: \(f(-x)=-f(x)\). A polynomial with only even (odd) powers is even (odd). \(x^4-3\), \(\cos\), \(\sin(x^2)\) are even; \(\sin\), \(\tan\) are odd; \(x^3-3\) is neither. Derivative swaps parity: differentiate \(f(-x)=f(x)\) with the chain rule: \(-f'(-x)=f'(x)\), so even \(f\) ⇒ odd \(f'\), and odd \(f\) ⇒ even \(f'\).
WiSe24 4a: \(f(x)=\log\frac1{1+e^x}\). Simplify: \(f=-\log(1+e^x)\), so \(f'(x)=-\frac{e^x}{1+e^x}\).
SoSe23 4b: \(f(x)=\ln\ln x\): \(f'(x)=\frac1{\ln x}\cdot\frac1x=\frac1{x\ln x}\).
What you must know for 6.3: the sum, product, quotient and chain rules and the rule for the inverse function; the derivatives of important functions, in particular \(x^n,a^x,x^a,\tan,\arctan\).
Mixed exercises from the script [Kapitel 13]
Three true/false statements about \(o\) and \(O\) from the script’s exercise pool. The method is always the same: translate the symbol into a limit, then compute the limit.
True or false: \(\sin x\cos x\ln(1+x)=o(x)\) for \(x\to0\)?
Solution
Translate: the claim says \(\lim_{x\to0}\frac{\sin x\cos x\ln(1+x)}x=0\). Split the quotient so that the known limit \(\frac{\sin x}x\to1\) (Lemma 5.19) appears: \[\frac{\sin x\cos x\ln(1+x)}x=\underbrace{\frac{\sin x}x}_{\to1}\cdot\underbrace{\cos x}_{\to1}\cdot\underbrace{\ln(1+x)}_{\to\ln1=0}\to1\cdot1\cdot0=0.\] True.
True or false: \(x^xe^{-x^2}=O(e^{-x})\) for \(x\to\infty\)?
Solution
Translate: the claim says \(\frac{x^xe^{-x^2}}{e^{-x}}=x^xe^{-x^2+x}\) stays bounded for large \(x\). Write everything as one exponential — the standard move when a variable sits in the exponent (Chapter 5): \(x^x=e^{x\ln x}\), so \[x^xe^{-x^2}e^{x}=e^{\,x\ln x-x^2+x}.\] In the exponent, \(-x^2\) dominates \(x\ln x\) and \(x\) (again \(\frac{\ln x}x\to0\)), so the exponent tends to \(-\infty\) and the whole expression tends to 0. A function with limit 0 is certainly bounded near \(\infty\): true.
Note the direction of the definitions: \(o\) (limit 0) is stronger than \(O\) (bounded quotient). Having shown the limit is 0 we have even proved \(x^xe^{-x^2}=o(e^{-x})\).
True or false: \(x^5(\ln x)^{12}=o(x^4)\) for \(x\downarrow0\)?
Solution
Translate: \(\lim_{x\downarrow0}\frac{x^5(\ln x)^{12}}{x^4}=\lim_{x\downarrow0}x(\ln x)^{12}\stackrel?=0\). Substitute \(y=\frac1x\), so \(y\to\infty\) and \(\ln x=-\ln y\), hence \((\ln x)^{12}=(\ln y)^{12}\) (even power): \[x(\ln x)^{12}=\frac{(\ln y)^{12}}y\to0\] by Satz 5.8 (\(y\) beats every power of the logarithm). True.
\(x\downarrow0\) means the one-sided limit from the right, which is the only side where \(\ln x\) is defined.
!What the exam asks from Chapter 6
Chapter 6 appears in six of the nine exams. Landau items are answer-only and quick; derivatives show up everywhere later (curve sketching, Taylor, ODEs), so fluency matters more than any single task.
| Type | Exam tasks | What earns the points |
|---|---|---|
| Landau: find all p / smallest a / true–false | WiSe20 3a, WiSe19 A3, WiSe24 3a–b, SoSe23 1g | translate into \(\lim\frac fg\); growth ladder; note the direction \(x\to0\) vs \(x\to\infty\) |
| Compute a derivative | WiSe24 4a, SoSe23 4b | simplify first (log rules), then chain/product rule |
| Differentiability from the definition | SoSe19 A4, SoSe21 A6 | at the special point use the difference quotient; elsewhere the rules |
WiSe20 3a: the Musterlösung prints \(p>2\); the computation gives \(p<2\) (see the solution above).
Extra Beyond these three types, a chapter-6 step is buried inside many later tasks, and that is where most of the marks for this chapter actually are: every curve sketch needs \(f'\) and \(f''\) (WiSe23 A3, WiSe18 A1, WiSe20 A4, §7.7); SoSe21 4b needs the rewrite \(x^x=e^{x\ln x}\) before it can be differentiated at all; l’Hospital (WiSe18 1a, WiSe24 1c) differentiates numerator and denominator separately; WiSe22 4b differentiates an integral with the fundamental theorem; and the several-variable tasks (WiSe19 1g, WiSe24 5a, SoSe23 A3) are these same rules applied one variable at a time.
Drill: derivatives and Landau
Ten quick items in exam style. Do them on paper first, then check.
- For the exam: the three task types above, and the derivative table in the summary by heart.
- For Chapter 7: \(f'(x_0)=0\) at interior extrema, the mean value theorem, monotonicity from the sign of \(f'\), l’Hospital (which needs \(f'\), \(g'\)).
- For Chapter 8: antiderivatives are derivatives read backwards; substitution is the chain rule backwards, integration by parts the product rule backwards.
- For Chapter 10: Taylor's formula \(f(a+h)=\sum\frac{f^{(k)}(a)}{k!}h^k+o(h^n)\) extends the linearisation \(f(a)+f'(a)h+o(h)\) — the \(o\)-notation of §6.1 is used throughout.
- For Chapters 11–12: partial derivatives are ordinary derivatives in one variable; total differentiability \(f(x+h)=f(x)+Df(x)h+o(\|h\|)\) is the same definition in \(\mathbb R^n\); checking an ODE solution means differentiating it.
ΣSummary sheet
| f(x) | f′(x) | f(x) | f′(x) |
|---|---|---|---|
| \(x^a\) | \(ax^{a-1}\) | \(\sin x\) | \(\cos x\) |
| \(e^x\) | \(e^x\) | \(\cos x\) | \(-\sin x\) |
| \(a^x\) | \(a^x\ln a\) | \(\tan x\) | \(1+\tan^2x=\frac1{\cos^2x}\) |
| \(\ln x\) | \(\frac1x\) | \(\arcsin x\) | \(\frac1{\sqrt{1-x^2}}\) |
| \(\ln|x|\) Extra | \(\frac1x\) (\(x\neq0\)) | \(\arctan x\) | \(\frac1{1+x^2}\) |
Domains: \(x^a\) and \(\ln x\) on \((0,\infty)\), \(a^x\) for \(a>0\) on all of \(\mathbb R\), \(\arcsin\) on \((-1,1)\), \(\tan\) away from \(\frac\pi2+k\pi\). The \(\ln|x|\) row is not in Chapter 6 — the script only does \(\ln\) on \((0,\infty)\) (Beispiel 6.23) — but it is the antiderivative you need in Chapter 8, and it follows from the chain rule: for \(x<0\), \((\ln(-x))'=\frac{-1}{-x}=\frac1x\).
- \(f=O(g)\): \(|f|\le C|g|\) near \(x_0\). \(f=o(g)\): \(f/g\to0\). Check via \(\lim|f/g|\). Near 0 the lowest power dominates, near ∞ the highest.
- \(f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}h\iff f(x)=f(x_0)+f'(x_0)(x-x_0)+o(|x-x_0|)\).
- Differentiable ⇒ continuous. \(|x|\) at 0: continuous, not differentiable. \(x^2\sin\frac1x\): differentiable, \(f'\) not continuous.
- At gluing points of piecewise functions, use the definition (difference quotient), not the rules.
- Rules: \((fg)'=f'g+fg'\), \(\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}\), \((f\circ g)'=f'(g)\,g'\), \((f^{-1})'=\frac1{f'(f^{-1})}\).
?Self-test
Applications of the DerivativeAnwendungen der Ableitung
This chapter turns derivatives into information about functions: where the extrema are, where the function increases, how it curves, and how to compute tricky limits (l’Hospital). The engine behind all of it is one theorem, the mean value theorem. The chapter ends with the full recipe for a curve sketch [Kurvendiskussion], a guaranteed exam task.
Source labels: Skript 7.4 = stated in your lecture script · Blatt 10 · 10.2 = exercise sheet · Klausur WiSe23 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§7.7 = Section 7.7, curve sketching) rather than a numbered result (7.7 = Definition 7.7).
0The big picture
- 7.1 At an interior local extremum, \(f'=0\) (necessary, not sufficient).
- 7.2 Mean value theorem: the average slope \(\frac{f(b)-f(a)}{b-a}\) is attained as a derivative \(f'(\xi)\) somewhere in between. This is the bridge from "derivative info" to "function info".
- 7.3 Consequences: \(f'>0\) ⇒ strictly increasing; \(f'=0\) ⇒ constant; sign change of \(f'\) ⇒ extremum; inequalities like \(1+x\le e^x\).
- 7.4 l’Hospital: for "\(\frac00\)" or "\(\frac\infty\infty\)", replace \(\frac fg\) by \(\frac{f'}{g'}\).
- 7.5–7.6 Higher derivatives; \(f''\ge0\) ⇔ convex; \(f'(x_0)=0\) and \(f''(x_0)>0\) ⇒ local minimum.
- 7.7 Curve sketching puts it all together.
- Derivatives and all rules (§6.3) — every section here computes \(f'\) or \(f''\).
- "A continuous function on \([a,b]\) attains max and min" (Satz 4.29, §4.4) — the first line of the proof of Rolle's theorem.
- "Limits keep ≤" (Satz 2.13) and one-sided limits (§4.1) — for the necessary condition \(f'=0\).
- Growth rates and the key limits (§5.2, §5.3) — boundary behaviour in curve sketches, and an alternative to l’Hospital.
7.1Extrema [Extrema]
\(f:[a,b]\to\mathbb R\) has at \(x_0\) a
- global maximum [globales Maximum] if \(f(x_0)\ge f(x)\) for all \(x\in[a,b]\);
- local maximum [lokales Maximum] if there is \(\varepsilon>0\) with \(f(x_0)\ge f(x)\) for all \(x\in(x_0-\varepsilon,x_0+\varepsilon)\cap[a,b]\).
Minima analogously. Maxima and minima are extrema; with strict inequalities (for \(x\neq x_0\)) they are strict.
In words: global = best value on the whole domain; local = best value in some small window around \(x_0\). Every global extremum is also a local one; a function can have many local maxima but only one global maximum value. (Skript Figure 6: several local maxima, one global.)
If \(f\) is differentiable on \((a,b)\) and has a local extremum at an interior point \(x_0\in(a,b)\), then \(f'(x_0)=0\).
Proof
Say local max: \(f(x_0+h)-f(x_0)\le0\) for small \(|h|\). Divide by \(h\): the difference quotient is \(\le0\) for \(h>0\) and \(\ge0\) for \(h<0\). Its limit \(f'(x_0)\) is therefore both \(\le0\) and \(\ge0\) (Satz 2.13), so \(=0\). ∎
- Not sufficient: \(f(x)=x^3\) has \(f'(0)=0\) but no extremum at 0 (it increases through 0). Skript 7.3
- Boundary points are not covered: \(x^2\) on \([-1,1]\) has maxima at \(\pm1\) with \(f'(\pm1)=\pm2\neq0\). Always check the endpoints separately.
- Find the zeros of \(f'\) in \((a,b)\) (the critical points).
- Throw out the zeros that are not extrema (sign of \(f'\) does not change, or \(f''\) test).
- Check the endpoints \(a,b\): does \(f\) have a local max or min there?
- The largest local max is the global max, the smallest local min the global min.
Extra — the two cases the script's recipe does not mention The script states the recipe for a closed interval \([a,b]\). If the domain is open or unbounded, step 3 has no endpoint values to read off: compute the limits at the open ends instead. And if such a limit is \(+\infty\) (resp. \(-\infty\)), then there is no global maximum (resp. minimum) on that side, however large the local ones are — that is exactly what settles WiSe23 A3 and SoSe21 4b below.
What you must know for 7.1: local and global min/max; \(f'(x)=0\) is necessary for a local extremum of a differentiable function; finding local and global extrema.
7.2The mean value theorem [Mittelwertsatz]
Let \(f:[a,b]\to\mathbb R\) be continuous on \([a,b]\) and differentiable on \((a,b)\). Then there is \(\xi\in(a,b)\) with \[\frac{f(b)-f(a)}{b-a}=f'(\xi).\]
Picture: some tangent is parallel to the secant. Car version: if you drove 100 km in 1 hour, at some moment your speedometer showed exactly 100 km/h. Order in the script: Rolle (7.5) and the generalised MVT (7.6) are proved first; the MVT is the special case \(g(x)=x\).
If additionally \(f(a)=f(b)\), there is \(\xi\in(a,b)\) with \(f'(\xi)=0\).
Proof
If \(f\) is constant, any \(\xi\) works (\(f'=0\) everywhere). Otherwise \(f\) attains max and min on the compact \([a,b]\) (Satz 4.29), and they differ. Since \(f(a)=f(b)\), at least one of them is attained at an interior point \(\xi\). There \(f'(\xi)=0\) (Lemma 7.2). ∎
Let \(f,g\) be continuous on \([a,b]\), differentiable on \((a,b)\), with \(g'\neq0\) on \((a,b)\). Then \(g(a)\neq g(b)\) and there is \(\xi\in(a,b)\) with \[\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}.\]
Proof (reduce to Rolle)
\(g(a)\neq g(b)\): otherwise Rolle gives \(g'(\xi)=0\). Now tilt \(f\) so that both ends are level: \[F(x)=f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}\,(g(x)-g(a)).\] Then \(F(a)=F(b)=f(a)\). Rolle: \(0=F'(\xi)=f'(\xi)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(\xi)\). Divide by \(g'(\xi)\neq0\). The ordinary MVT is the case \(g(x)=x\). ∎
WiSe19 2b: "State the mean value theorem." Learn the statement with both hypotheses (continuous on \([a,b]\), differentiable on \((a,b)\)) and \(\xi\in(a,b)\).
WiSe22 4a: \(f\) continuous on \([a,b]\), differentiable on \((a,b)\), \(f(a)<f(b)\). Show \(f'(\xi)>0\) for some \(\xi\).
MVT: \(f'(\xi)=\frac{f(b)-f(a)}{b-a}\), numerator and denominator \(>0\). Done in one line.
WiSe18 A4b: \(f\in C^1(\mathbb R)\) ⇒ \(f\) is Lipschitz on \([a,b]\)
For \(x,y\in[a,b]\), MVT: \(f(x)-f(y)=f'(z)(x-y)\) with \(z\) between them. \(f'\) is continuous on the compact \([a,b]\), so \(L:=\max_{[a,b]}|f'|\) exists (Satz 4.29). Hence \(|f(x)-f(y)|\le L|x-y|\).
SoSe23 A4: MVT for \(f(x)=\ln\ln x\) on \([n,n+1]\) ⇒ \(\sum_{n\ge2}\frac1{n\ln n}\) diverges
\(f'(x)=\frac1{x\ln x}\). MVT: \(\ln\ln(n+1)-\ln\ln n=\frac1{\xi\ln\xi}\) for some \(\xi\in(n,n+1)\). Since \(g(x)=x\ln x\) is increasing for \(x>\frac1e\) (\(g'=\ln x+1\)), \(\frac1{\xi\ln\xi}<\frac1{n\ln n}\). So \[\sum_{n=2}^k\frac1{n\ln n}>\sum_{n=2}^k\left(\ln\ln(n+1)-\ln\ln n\right)=\ln\ln(k+1)-\ln\ln2\to\infty\] (telescoping!). A slick divergence proof; Chapter 9 gets it again with the integral test.
WiSe20 A6 / SoSe21 A5: fixed-point iteration \(x_{n+1}=f(x_n)\) with \(\alpha=\sup|f'|\)
MVT: \(|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|=|f'(z)||x_n-x_{n-1}|\le\alpha|x_n-x_{n-1}|\). If \(\alpha<1\), the steps shrink geometrically and the sequence converges (Extra: then \(|x_{n+1}-x_n|\le\alpha^n|x_1-x_0|\), so \(\sum(x_{k+1}-x_k)\) converges absolutely by comparison with the geometric series, and \(x_n=x_0+\sum_{k<n}(x_{k+1}-x_k)\) converges.) But \(\alpha<1\) is not necessary: \(f(x)=x^2\), \(|x_0|<1\) gives \(x_n=x_0^{2^n}\to0\) although \(f'\) is unbounded. (WiSe20 answer: (ii).) In SoSe21 A5, \(f(x)=\frac x4+\frac{\cos x}3\): \(|f'(x)|=\left|\frac14-\frac{\sin x}3\right|\le\frac14+\frac13=\frac7{12}<1\), so the sequence converges for every \(x_0\), always to the same limit, the unique fixed point (answer (i)).
What you must know for 7.2: the mean value theorem, the generalised MVT and Rolle's theorem, and how to apply them.
7.3Monotonicity [Monotonie]
\(f\) is (monotonically) increasing if \(x\le y\Rightarrow f(x)\le f(y)\), strictly increasing if \(x<y\Rightarrow f(x)<f(y)\); decreasing analogously (Def. 7.7 = Def. 4.12). Examples (7.8): \(x\) strictly increasing; a constant is both increasing and decreasing; \(x^2\) strictly increasing on \([0,\infty)\), strictly decreasing on \((-\infty,0]\).
Let \(f\) be continuous on \([a,b]\), differentiable on \((a,b)\).
- \(f'>0\) on \((a,b)\) ⇒ \(f\) strictly increasing on \([a,b]\).
- \(f'<0\) on \((a,b)\) ⇒ strictly decreasing.
- \(f'\ge0\) on \((a,b)\) \(\iff\) increasing.
- \(f'\le0\) on \((a,b)\) \(\iff\) decreasing.
Proof of (a)
For \(x<y\): MVT gives \(\frac{f(y)-f(x)}{y-x}=f'(\xi)>0\), and \(y-x>0\), so \(f(y)>f(x)\).
(b)–(d): "⇒" in (c) is the same argument with \(f'(\xi)\ge0\). "⇐" in (c): if \(f\) is increasing, then for \(h>0\) the difference quotient \(\frac{f(x+h)-f(x)}h\ge0\), and its limit \(f'(x)\ge0\) (Satz 2.13). (b) and (d) follow by applying (a) and (c) to \(-f\). ∎
(a) has no converse: \(x^3\) is strictly increasing on \([-1,1]\), but \(f'(0)=0\). Example 7.10: \(\ln\) is strictly increasing because \((\ln)'(x)=\frac1x>0\) (we knew it already as the inverse of exp).
each with equality only at \(x=0\).
Proof (the standard technique for proving inequalities)
Let \(f(x)=e^x-1-x\), \(f'(x)=e^x-1\): negative for \(x<0\), positive for \(x>0\). So \(f\) decreases then increases: global minimum at 0, where \(f(0)=0\). Hence \(f\ge0\), with equality only at 0. Apply the increasing function ln to \(1+x\le e^x\) to get the second inequality. ∎
Method: to prove \(g\le h\), study \(f=h-g\): find its minimum with \(f'\) and show it is \(\ge0\).
\(f\) differentiable on \((a,b)\), \(f'(x_0)=0\). If \(f'\ge0\) left of \(x_0\) and \(f'\le0\) right of it, then \(x_0\) is a (global on \((a,b)\)) maximum; for the reverse sign pattern, a minimum. Restricted to \((x_0-\varepsilon,x_0+\varepsilon)\): local extremum. In short: a sign change of \(f'\) from + to − means a max, from − to + a min.
Proof
Take the maximum case. On \((a,x_0)\) we have \(f'\ge0\), so \(f\) is increasing there by the monotonicity criterion (Satz 7.9(c)); on \((x_0,b)\) we have \(f'\le0\), so \(f\) is decreasing. Increasing up to \(x_0\) gives \(f(x)\le f(x_0)\) for every \(x<x_0\); decreasing after \(x_0\) gives \(f(x)\le f(x_0)\) for every \(x>x_0\). Together \(f(x)\le f(x_0)\) on all of \((a,b)\), which is the definition of a maximum there. (Both halves also use that \(f\) is continuous at \(x_0\) itself, which holds because \(f\) is differentiable, Lemma 6.9.) For the minimum, apply this to \(-f\). ∎
Note what the criterion does not need: no second derivative. That is why it still works at points where \(f''(x_0)=0\) and the \(f''\)-test below says nothing.
- Factor \(f'\) as far as possible and drop factors that are always positive (\(e^{\dots}\), \(x^2+1\), squares away from their zero).
- Mark the zeros (and the points where \(f'\) is undefined) on the number line. Between two consecutive marks the sign of a continuous \(f'\) cannot change (IVT!).
- Determine the sign on each piece: from the signs of the factors, or by plugging in one test point.
- Read off: + means increasing, − decreasing; + → − is a max, − → + a min, no change means no extremum.
Example \(f'(x)=(3-x)x^2e^{-x}\) (WiSe23 A3): \(e^{-x}>0\) always; \(x^2>0\) except at 0; \(3-x\) is + left of 3, − right of 3.
| \((-\infty,0)\) | 0 | \((0,3)\) | 3 | \((3,\infty)\) | |
|---|---|---|---|---|---|
| \(x^2\) | + | 0 | + | + | + |
| \(3-x\) | + | + | + | 0 | − |
| \(f'\) | + | 0 | + | 0 | − |
| \(f\) | ↗ | no extremum | ↗ | max | ↘ |
On an interval: \(f'=0\iff f\) is constant. (MVT: \(f(y)-f(x)=f'(\xi)(y-x)=0\).) Hence if \(f'=g'\), then \(f=g+c\).
The interval matters Extra: \(f(x)=\text{sign}(x)\) on \(\mathbb R\setminus\{0\}\) has \(f'=0\) but is not constant. This is why antiderivatives are unique only "up to a constant" on an interval (Chapter 8).
(a) \(f'=0\) on an interval ⇒ \(f\) constant (the MVT proof above). (b) If \(f'=cf\) on an open interval, then \(f(x)=be^{cx}\).
Solution (b)
Consider \(g(x)=f(x)e^{-cx}\). Product rule: \(g'=f'e^{-cx}-cfe^{-cx}=(f'-cf)e^{-cx}=0\). By (a), \(g=b\) is constant, so \(f=be^{cx}\). This is the first differential equation of the course (Chapter 12), the same trick as SoSe19 A4.
WiSe22 4b: where is \(f(x)=\int_0^x(t-1)^9e^{-t}\,dt\) (on \((0,\infty)\)) strictly increasing, where strictly decreasing?
Do not compute the integral. The fundamental theorem (Satz 8.6, §8.3) gives the derivative directly: \[f'(x)=(x-1)^9e^{-x}.\] Now the sign chart: \(e^{-x}>0\) always, and \((x-1)^9\) has the sign of \(x-1\) (an odd power keeps the sign). So \(f'<0\) on \((0,1)\) and \(f'>0\) on \((1,\infty)\), and the monotonicity criterion 7.9(a),(b) gives: \(f\) is strictly decreasing on \((0,1]\), strictly increasing on \([1,\infty)\), with a global minimum at \(x=1\).
This is the cleanest illustration of why §7.3 and Chapter 8 belong together: all the monotonicity information sits in \(f'\), and the fundamental theorem hands you \(f'\) for free.
SoSe23 A4c: monotonicity of \(g(x)=x\ln x\) on \((0,\infty)\)
\(g'(x)=\ln x+1\), which is \(<0\) for \(x<\frac1e\) and \(>0\) for \(x>\frac1e\). So \(g\) is strictly decreasing on \((0,\frac1e]\) and strictly increasing on \([\frac1e,\infty)\), with minimum \(g(\frac1e)=-\frac1e\).
The rest of SoSe23 A4 uses exactly this: because \(x\ln x\) is increasing beyond \(\frac1e\), \(\frac1{\xi\ln\xi}<\frac1{n\ln n}\) for \(\xi>n\ge2\) — the step the divergence proof above needs.
What you must know for 7.3: the definition of monotonicity and the monotonicity criterion; determining monotonicity and extrema of a given function.
7.4Computing limits: l’Hospital [Regel von l’Hospital]
Let \(f,g\) be differentiable on \((a,b)\) (\(a,b\) may be \(\pm\infty\)), \(g'(x)\neq0\), and \(x_0\in\{a,b\}\). Suppose
- \(\lim_{x\to x_0}f=0=\lim_{x\to x_0}g\) ("\(\frac00\)"), or
- \(\lim f=\infty=\lim g\) ("\(\frac\infty\infty\)").
If \(\displaystyle\lim_{x\to x_0}\frac{f'(x)}{g'(x)}\) exists, then \(\displaystyle\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}\).
Proof of case (a) at a finite endpoint
Extend \(f,g\) continuously by \(f(a)=g(a)=0\). For \(x\in(a,b)\), the generalised MVT gives \(\xi\in(a,x)\) with \[\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\xi)}{g'(\xi)}.\] As \(x\to a\), also \(\xi\to a\) (squeeze), so the right side tends to \(\lim\frac{f'}{g'}\). ∎
The script proves only this case; the "\(\frac\infty\infty\)" case and \(x_0=\pm\infty\) are used without proof (the latter reduces to the former by substituting \(x=\frac1t\)).
- Only for "\(\frac00\)" or "\(\frac\infty\infty\)". \(\lim_{x\to0}\frac{\cos x}{x+1}=1\), but "l’Hospital" would wrongly give \(\frac{-\sin0}1=0\).
- Differentiate numerator and denominator separately. This is not the quotient rule.
- Other forms (\(0\cdot\infty\), \(\infty-\infty\), \(1^\infty\), \(0^0\)) must first be rewritten as a quotient (common denominator; or take ln).
- 7.16 \(\lim_{x\to0}\frac{\sin x}x=\lim\frac{\cos x}1=1\).
- 7.17 \(\lim_{x\to0}\left(\frac1{\sin x}-\frac1x\right)=\lim\frac{x-\sin x}{x\sin x}\overset{\frac00}=\lim\frac{1-\cos x}{\sin x+x\cos x}\overset{\frac00}=\lim\frac{\sin x}{2\cos x-x\sin x}=\frac02=0\) (twice; first a common denominator).
- 7.18 \(\lim_{x\to\infty}\frac{\ln x}{x^\alpha}=\lim\frac{1/x}{\alpha x^{\alpha-1}}=\lim\frac1{\alpha x^\alpha}=0\) for \(\alpha>0\). Consequence: \(\frac{\ln\ln n}{(\ln n)^\alpha}\to0\).
(a) \(\lim_{x\to\infty}\frac{5^x}{x^2}\), (b) \(\lim_{x\to0}\frac{\sin x-x\cos x}{x\sin x}\), (c) \(\lim_{x\to\infty}x\ln(1+\frac1x)\), (d*) \(\lim_{x\downarrow0}(\sin x)^{1/\ln x}\)
(a) "\(\frac\infty\infty\)" twice: \(\frac{5^x\ln5}{2x}\to\frac{5^x(\ln5)^2}2\to\infty\).
(b) "\(\frac00\)": \(\frac{x\sin x}{x\cos x+\sin x}\), again "\(\frac00\)": \(\frac{x\cos x+\sin x}{2\cos x-x\sin x}\to\frac02=0\).
(c) "\(\infty\cdot0\)" → rewrite \(\frac{\ln(1+1/x)}{1/x}\) ("\(\frac00\)"): \(\frac{\frac1{1+1/x}\cdot(-\frac1{x^2})}{-\frac1{x^2}}=\frac x{x+1}\to1\).
(d*) "\(0^0\)": take ln: \(\frac{\ln\sin x}{\ln x}\) ("\(\frac{-\infty}{-\infty}\)") → \(\frac{\cos x/\sin x}{1/x}=\cos x\cdot\frac x{\sin x}\to1\). So the limit is \(e^1=e\).
WiSe24 1c: \(\lim_{x\to0}\frac{x^4}{(1-\cos x)^2}\)
Official solution with Landau: \(1-\cos x=\frac{x^2}2(1+o(1))\), so \((1-\cos x)^2=\frac{x^4}4(1+o(1))\) and the quotient → 4. (l’Hospital also works but needs four rounds; the \(o\)-expansion is much faster. Cf. Lemma 5.19.)
WiSe18 1a: \(\lim_{x\to\infty}x^2e^{-x}\) and \(\lim_{x\to-\infty}x^2e^{-x}\)
\(x\to\infty\): \(\frac{x^2}{e^x}\) "\(\frac\infty\infty\)", l’Hospital twice → \(\frac2{e^x}\to0\). \(x\to-\infty\): both factors → ∞, no quotient issue: \(\infty\).
WiSe20 3b: \(\lim_{x\to0}\ln(x^{-1})\,(e^x-1)\) (only the result counted: 0)
Which limit is meant? \(\ln(x^{-1})\) needs \(x>0\), so "\(x\to0\)" can only be the right-hand limit \(x\downarrow0\).
The shape is "\(\infty\cdot0\)", so it must be rewritten before l’Hospital applies. Two routes:
- Factor into two known limits (no l’Hospital at all): \(\ln(x^{-1})=-\ln x\), so the product is \(-(x\ln x)\cdot\frac{e^x-1}x\). Here \(x\ln x\to0\) (Blatt 7.1a) and \(\frac{e^x-1}x\to1\) (Beispiel 6.13), so the limit is \(-0\cdot1=0\).
- Or make it a quotient: \(\frac{-\ln x}{1/(e^x-1)}\) is "\(\frac\infty\infty\)" — workable but messy. The first route is much cleaner, which is the general lesson: before reaching for l’Hospital, check whether the expression splits into standard limits.
WiSe23 A2: for which \(a\) is \(f_a(x)=\frac{x\sin x}{1-\cos x}\) (\(x\neq0\)), \(f_a(0)=a\), continuous on \((-\pi,\pi)\)?
Away from 0 the function is a quotient of continuous functions with \(1-\cos x\neq0\), hence continuous. So everything hangs on \(L=\lim_{x\to0}f_a(x)\), and \(f_a\) is continuous at 0 exactly when \(a=L\).
With l’Hospital (the official route), twice, since numerator and denominator both → 0: \[\frac{x\sin x}{1-\cos x}\overset{\frac00}=\frac{\sin x+x\cos x}{\sin x}\overset{\frac00}=\frac{2\cos x-x\sin x}{\cos x}\longrightarrow\frac21=2.\] Without it (Chapter 5 only): \(\frac{x\sin x}{1-\cos x}=\frac{\sin x}x\cdot2\cdot\frac{x^2/2}{1-\cos x}\to1\cdot2\cdot1=2\) by Lemma 5.19.
So \(f_a\) is continuous on \((-\pi,\pi)\) iff \(a=2\); for every other \(a\) the only point of discontinuity is \(x=0\). (The same task is worked from the continuity side in §4.1.)
What you must know for 7.4: the rule of l’Hospital and how to apply it.
7.5Higher derivatives [Höhere Ableitungen]
\(f^{(0)}=f,\ f^{(1)}=f',\ f^{(2)}=f'',\ f^{(n+1)}=(f^{(n)})'\) if it exists (\(f^{(n)}\) must be defined on a neighbourhood). \(f\in C^n(a,b)\): \(f',\dots,f^{(n)}\) exist on \((a,b)\) and are continuous. \(f\in C^\infty\): all derivatives exist.
Polynomials (Lemma 7.22) and exp (7.23: \((e^x)^{(n)}=e^x\)) are \(C^\infty\). For \(f(x)=\sum_{k=0}^na_kx^k\): \(\displaystyle a_l=\frac{f^{(l)}(0)}{l!}\): the coefficients are read off from the derivatives at 0. (Each derivative lowers the degree by one; \(f^{(l)}(0)\) keeps only the constant term \(l!\,a_l\).) Example: \(f=x^3-2x^2+7x+4\): \(f'=3x^2-4x+7\), \(f''=6x-4\), \(f'''=6\), \(f^{(n)}=0\) for \(n\ge4\). (Beispiel 7.21.) This is the seed of Taylor's formula (Chapter 10).
Warning (script): to define \(f^{(n+1)}(x)\), the function \(f^{(n)}\) must exist on a whole neighbourhood \((x-\varepsilon,x+\varepsilon)\), not only at \(x\).
Extra A consequence of Definition 7.20 worth stating separately: "\(n\) times differentiable" is weaker than \(C^n\), because \(C^n\) also demands that the derivatives be continuous. The standard example is \(x^2\sin\frac1x\) (with value 0 at 0) from Chapter 6: differentiable everywhere, but \(f'\) is not continuous at 0, so it is not \(C^1\).
This distinction is an exam item in its own right. For \(f(x)=x^2\sin\frac1x\) (\(x\neq0\)), \(f(0)=0\), decide: (i) \(f'\) is continuous on \(\mathbb R\); (ii) \(f\) is differentiable on \(\mathbb R\); (iii) \(f\) is continuously differentiable on \(\mathbb R\setminus\{0\}\).
Answer: (i) false, (ii) true, (iii) true
(ii) holds: away from 0 by the rules, at 0 because \(\frac{f(h)-f(0)}h=h\sin\frac1h\to0\), so \(f'(0)=0\). (iii) holds: on \(\mathbb R\setminus\{0\}\), \(f'(x)=2x\sin\frac1x-\cos\frac1x\) is built from continuous functions. (i) fails at the single point 0: \(\cos\frac1x\) has no limit there, so \(f'\) is not continuous at 0 — even though \(f'(0)\) exists. Worked in full in §6.2.
The trap is reading (iii) as if it implied (i). It does not: (iii) says nothing about the point 0, which is exactly where \(f'\) misbehaves.
What you must know for 7.5: computing higher derivatives.
7.6Convexity [konvex / konkav] — the script calls the section Krümmungsverhalten, curvature behaviour
\(f\) is convex [konvex] if for all \(x_0\neq x_1\) and \(\lambda\in(0,1)\): \[f((1-\lambda)x_0+\lambda x_1)\le(1-\lambda)f(x_0)+\lambda f(x_1).\] Strictly convex with \(<\). \(f\) is concave if \(-f\) is convex.
Picture: the graph lies below every chord. Walking in the \(+x\) direction, the graph turns left. The left side of the inequality is the function at a point between \(x_0,x_1\); the right side is the chord at the same point.
For twice differentiable \(f\) on \((a,b)\): \(f\) convex \(\iff f''\ge0\); \(f''>0\Rightarrow\) strictly convex. Concave: \(f''\le0\); \(f''<0\Rightarrow\) strictly concave.
The converse of "\(f''>0\Rightarrow\) strictly convex" fails: \(x^4\) is strictly convex but \(f''(0)=0\) (Skript 7.26). Examples (Skript 7.27–7.29): \(x^2\) (\(f''=2\)) and \(e^x\) (\(f''=e^x\)) strictly convex; \(\ln x\) strictly concave (\(f''=-\frac1{x^2}\)).
Proof of "\(f''\ge0\Rightarrow\) convex" Extra — the script states 7.25 without proof
Take \(x_0<x_1\), \(\lambda\in(0,1)\), \(x=(1-\lambda)x_0+\lambda x_1\). Convexity at \(x\) is equivalent to "slope of the left chord ≤ slope of the right chord": \[\frac{f(x)-f(x_0)}{x-x_0}\le\frac{f(x_1)-f(x)}{x_1-x}.\] (Multiply out: \(x-x_0=\lambda(x_1-x_0)\), \(x_1-x=(1-\lambda)(x_1-x_0)\).) By the MVT the left side is \(f'(\xi_1)\) with \(\xi_1\in(x_0,x)\), the right side \(f'(\xi_2)\) with \(\xi_2\in(x,x_1)\). Since \(f''\ge0\), \(f'\) is increasing (7.9c) and \(\xi_1<\xi_2\), so \(f'(\xi_1)\le f'(\xi_2)\). With \(f''>0\) everything is strict. ∎
"⇐": if \(f\) is convex, the chord slopes increase, and letting the chord shrink shows \(f'\) is increasing, so \(f''\ge0\).
\(f\in C^2(a,b)\), \(f'(x_0)=0\). Then \(f''(x_0)>0\Rightarrow\) strict local min; \(f''(x_0)<0\Rightarrow\) strict local max.
Proof of the min case
\(f''\) is continuous and \(f''(x_0)>0\), so \(f''>0\) on some \((x_0-\varepsilon,x_0+\varepsilon)\). Then \(f'\) is strictly increasing there (7.9(a) applied to \(f'\)), and since \(f'(x_0)=0\): \(f'<0\) left of \(x_0\), \(f'>0\) right of it. Now 7.9(b) makes \(f\) strictly decreasing on \((x_0-\varepsilon,x_0]\) and 7.9(a) makes it strictly increasing on \([x_0,x_0+\varepsilon)\), so \(f(x)>f(x_0)\) for every \(x\neq x_0\) in that window: a strict local minimum. ∎
Extra — why not just cite 7.12? Satz 7.12 needs only \(f'\ge0\) / \(f'\le0\) and therefore concludes only a (non-strict) extremum. Strictness comes from the strict inequalities \(f'<0\), \(f'>0\) that \(f''(x_0)>0\) hands you, via 7.9(a),(b). That is the step the script's own proof of 7.30 takes.
If \(f'(x_0)=f''(x_0)=0\), no conclusion: \(x^3\) (no extremum), \(x^4\) (min), \(-x^4\) (max) all have \(f'(0)=f''(0)=0\). Then use the sign of \(f'\) around \(x_0\). (Exactly the trap in WiSe23 A3.)
\(f\) strictly convex, \(\alpha<0\), \(x\neq y\). Which holds? (i) \(\alpha f(x)+(1-\alpha)f(y)<f(\alpha x+(1-\alpha)y)\), (ii) the reverse, (iii) not enough information.
Solution: (i)
With \(\alpha<0\), the point \(z=\alpha x+(1-\alpha)y\) lies outside the segment, beyond \(y\). So \(y\) lies between \(x\) and \(z\): \(y=\frac{-\alpha}{1-\alpha}x+\frac1{1-\alpha}z\) (weights in \((0,1)\), summing to 1). Convexity at \(y\): \(f(y)<\frac{-\alpha}{1-\alpha}f(x)+\frac1{1-\alpha}f(z)\). Multiply by \(1-\alpha>0\): \(\alpha f(x)+(1-\alpha)f(y)<f(z)\). That is (i). (Outside the segment, the graph lies above the extended chord.) The official solution checks it with \(f=x^2\), \(x=0\).
What you must know for 7.6: the sufficient criterion for local min/max; determining the curvature behaviour of a function.
7.7Curve sketching [Kurvendiskussion]
- Domain and behaviour at the boundary (limits at \(\pm\infty\) and at gaps).
- Derivatives \(f',f''\) (state differentiability, e.g. "as a composition of \(C^\infty\) functions").
- Critical points: solve \(f'(x)=0\) (only inside the domain!).
- Classify: sign of \(f''\) there, or sign change of \(f'\) (needed when \(f''=0\)).
- Monotonicity from the sign of \(f'\) on the intervals between critical points.
- Global extrema: compare local extrema with boundary values/limits.
- Curvature from the sign of \(f''\) (convex where \(f''\ge0\), concave where \(f''\le0\)).
- Sketch.
Extra — inflection points The word is not in this chapter of the script (it appears once, undefined, in Chapter 10), and the script's own six-step recipe stops at "Krümmung". But it is the natural thing to record while reading the sign of \(f''\), so: an inflection point [Wendepunkt] is a point where \(f''\) changes sign, i.e. where the curvature flips between convex and concave. Example: \(x^3\) at 0. Since the script does not ask for them, an exam will not either — treat them as a free extra line in your sketch, not as something you must find.
Script example: \(f(x)=x+\frac1x\) on \((0,\infty)\): → ∞ at both ends; \(f'=1-\frac1{x^2}=0\iff x=1\) (\(-1\) is not in the domain!); \(f''=\frac2{x^3}>0\), so min at 1 with \(f(1)=2\), global because of the boundary behaviour; decreasing on \((0,1)\), increasing on \((1,\infty)\); strictly convex.
\(f(x)=x^3e^{-x}\) on ℝ. (a) \(f'\) and its zeros. (b) Local extrema with classification; name critical points that are not extrema. (c) Global extrema.
Solution
(a) Product rule: \(f'=3x^2e^{-x}-x^3e^{-x}=(3-x)x^2e^{-x}\). Since \(e^{-x}>0\): zeros \(x=0\) (double) and \(x=3\).
(b) Sign of \(f'\): positive on \((-\infty,0)\) and \((0,3)\), negative on \((3,\infty)\). So \(x=3\) is a local max (+ to −). At \(x=0\) there is no sign change: not an extremum (a "plateau point"). The \(f''\)-test alone fails at 0: \(f''=(x^3-6x^2+6x)e^{-x}\), \(f''(0)=0\); at 3: \(f''(3)=-9e^{-3}<0\) ✓.
(c) \(f\) increases on \((-\infty,3]\), decreases on \([3,\infty)\) ⇒ \(x=3\) is the global max, \(f(3)=27e^{-3}\). No global min, since \(f(x)\to-\infty\) as \(x\to-\infty\).
WiSe18 A1 (= Blatt 10.2d): \(f(x)=x^2e^{-x}\)
\(f'=(2x-x^2)e^{-x}\), zeros 0 and 2. Signs: − on \((-\infty,0)\), + on \((0,2)\), − on \((2,\infty)\): local min at 0, local max at 2. Limits: \(\infty\) at \(-\infty\), 0 at \(+\infty\). Since \(f\ge0=f(0)\): global min at 0; no global max. (Second-derivative route: \(f''=(x^2-4x+2)e^{-x}\), \(f''(0)=2>0\), \(f''(2)=-2e^{-2}<0\).)
WiSe20 A4: \(f(x)=10+\int_{-3}^x(3t^2-12)\,dt\) on \([-3,\infty)\)
By the fundamental theorem (Chapter 8), \(f'(x)=3x^2-12=3(x-2)(x+2)\), and \(f(x)=x^3-12x+1\). Candidates: \(x=\pm2\) and the boundary \(x=-3\). Values \(f(-3)=10\), \(f(-2)=17\), \(f(2)=-15\), \(f\to\infty\). Signs of \(f'\): + on \([-3,-2)\), − on \((-2,2)\), + beyond. So: local min 10 at the boundary \(-3\), local max 17 at \(-2\), global min \(-15\) at 2, no global max.
SoSe21 4b: extrema of \(x^x\) on \((0,\infty)\)
\(x^x=e^{x\ln x}\), \(f'=x^x(\ln x+1)\). \(x^x>0\), so \(f'=0\iff\ln x=-1\iff x=\frac1e\). Sign − before, + after ⇒ global min at \(\frac1e\), value \(e^{-1/e}\).
10.1 The perfect tin can: open cylinder with volume 60 cm³, minimise the sheet metal
Volume \(\pi r^2h=60\Rightarrow h=\frac{60}{\pi r^2}\). Area (bottom + side) \(f(r)=\pi r^2+2\pi rh=\pi r^2+\frac{120}r\) on \((0,\infty)\). \(f'(r)=2\pi r-\frac{120}{r^2}=0\iff r^3=\frac{60}\pi\iff r=\sqrt[3]{60/\pi}\). \(f\to\infty\) at both ends, so this unique critical point is the global min: \(f\approx67.34\) cm².
Recipe: write the target as a function of one variable using the constraint, fix the domain, then find the global extremum.
10.3 Among rectangles with perimeter U, the square has maximal area
\(y=\frac U2-x\), area \(f(x)=\frac U2x-x^2\) on \((0,\frac U2)\). \(f'=\frac U2-2x=0\iff x=\frac U4\); \(f'\) changes from + to − ⇒ global max; then \(y=\frac U4\) too.
10.2 Extrema on restricted domains (boundary points!)
- \(2x^2-16x+36\) on ℝ: global min at 4. On \([0,\infty)\): additionally a local max at the boundary 0 (\(f\) decreases from there), not global.
- \(-4x^3+24x^2+12x+6\): \(f'=-12(x^2-4x-1)=0\iff x=2\pm\sqrt5\). Local min at \(2-\sqrt5\), local max at \(2+\sqrt5\); no global extrema (limits \(\pm\infty\)).
- \(x^2-4x\) on \([-2,5]\): local min at 2 (\(-4\), global); local maxima at both ends, \(f(-2)=12\) global, \(f(5)=5\).
- \(\frac{x-1}{x+1}\) on \((-1,4]\): \(f'=\frac2{(x+1)^2}>0\): global max at 4 (\(\frac35\)); no min (\(\to-\infty\) at \(-1\)).
- \(x^2e^x\): \(f'=x(x+2)e^x\): local max at \(-2\) (\(4e^{-2}\)), global min at 0; no global max.
What you must know for 7.7: for a given function, determining boundary behaviour, derivatives, extrema, monotonicity and curvature.
!What the exam asks from Chapter 7
Chapter 7 is in every exam, usually with the biggest single task (a curve sketch worth 8–11 points):
| Type | Exam tasks | What earns the points |
|---|---|---|
| Curve sketch / all extrema | WiSe18 A1, WiSe20 A4, SoSe21 4b, WiSe23 A3 | factorised \(f'\), sign chart, classification (sign change when \(f''=0\)), boundary values and limits for global extrema |
| MVT proofs | WiSe19 2b, WiSe22 4a, SoSe23 A4, WiSe18 A4b, WiSe20 A6, SoSe21 A5 | state the MVT with both hypotheses; apply it to the right function and interval |
| Limits with l’Hospital | WiSe18 1a, WiSe24 1c, WiSe20 3b, WiSe23 A2 | check "0/0" or "∞/∞" first; rewrite the other forms into a quotient; often a standard limit or an \(o\)-expansion is faster |
| Monotonicity from \(f'\) | WiSe22 4b, SoSe23 A4c | get \(f'\) (in WiSe22 4b from the fundamental theorem, without computing the integral), factor it, read the sign, cite 7.9 |
| \(C^n\) vs \(n\)-times differentiable | SoSe21 A6 | the derivative may exist everywhere and still be discontinuous; decide each statement separately |
| Convexity | SoSe21 A3 | definition with λ, or \(f''\ge0\) |
| Derivative estimates | WiSe20 A2b | work with \((\ln f)'=\frac{f'}f\) and the MVT |
\(f\) continuous on \([0,\infty)\), differentiable on \((0,\infty)\), \(f(0)=0\), \(f(x)>0\) for \(x>0\). Which hold? (i) \(\liminf_{x\downarrow0}\frac{f'(x)}{f(x)}=-\infty\), (ii) \(\lim_{x\downarrow0}\frac{f'(x)}{f(x)}=-\infty\), (iii) \(\liminf_{x\downarrow0}\frac{f'(x)}{f(x)}=\infty\).
Solution: official answer (iii), but see the flag below
(i), (ii) false: \(f(x)=\sqrt x\) gives \(\frac{f'}f=\frac1{2x}\to+\infty\).
What the argument proves. Put \(L(x)=\ln f(x)\) on \((0,\infty)\); then \(L'=\frac{f'}f\) (chain rule), and \(L(x)\to-\infty\) as \(x\downarrow0\) because \(f(x)\to f(0)=0\) (continuity). Suppose \(\limsup_{x\downarrow0}\frac{f'(x)}{f(x)}<\infty\): then there are \(M\) and \(t>0\) with \(L'(\xi)<M\) for all \(\xi\in(0,t)\). For \(0<x<t\), the MVT gives \(L(t)-L(x)=L'(\xi)(t-x)<M(t-x)\le|M|t\), so \(L(x)>L(t)-|M|t\) stays bounded below — contradicting \(L(x)\to-\infty\). Hence \(\limsup_{x\downarrow0}\frac{f'(x)}{f(x)}=\infty\). (The official solution integrates \(L'\) instead of using the MVT; same idea.)
Flag: the official solution uses exactly this argument for (iii), which is about the liminf. But "\(\liminf<\infty\)" only gives points arbitrarily close to 0 with \(L'<M\), not a whole interval \((0,t)\). Counterexample: \(f(x)=x\big(2+\sin\frac1x\big)\), \(f(0)=0\), satisfies all hypotheses, and \(\frac{f'(x)}{f(x)}=\frac1x-\frac{\cos(1/x)}{x^2(2+\sin(1/x))}\) equals \(2\pi n-2\pi^2n^2\to-\infty\) at \(x=\frac1{2\pi n}\). So for this \(f\) the liminf is \(-\infty\) and (iii) fails. As stated, none of (i)–(iii) holds for every such \(f\); what always holds is \(\limsup=\infty\). Worth asking your tutor.
Drill: classify points and compute limits
- For the exam: a complete curve sketch without gaps; MVT short proofs; l’Hospital with its conditions.
- For Chapter 8: "\(F'=G'\) on an interval ⇒ \(F=G+c\)" (7.14) is why antiderivatives are unique up to a constant; the fundamental theorem is proved with the mean value theorem of integration.
- For Chapter 9: monotone functions allow comparing sums with integrals (integral test).
- For Chapter 10: Lemma 7.22 (\(a_l=\frac{f^{(l)}(0)}{l!}\)) becomes Taylor's formula; the Lagrange remainder is a generalised MVT.
- For Chapter 11: the second-derivative test becomes the Hessian test (eigenvalues instead of the sign of \(f''\)).
ΣSummary sheet
- Interior local extremum ⇒ \(f'=0\). Not conversely (\(x^3\)); endpoints separately.
- MVT: \(\frac{f(b)-f(a)}{b-a}=f'(\xi)\). Rolle: equal ends ⇒ \(f'(\xi)=0\). Generalised: \(\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}\).
- \(f'>0\Rightarrow\) strictly increasing; \(f'\ge0\iff\) increasing; \(f'=0\iff\) constant (on intervals).
- Extremum tests: sign change of \(f'\) (always works), or \(f''(x_0)\neq0\) (fails when \(f''(x_0)=0\)).
- \(1+x\le e^x\), \(\ln(1+x)\le x\). Prove inequalities by minimising the difference.
- l’Hospital only for \(\frac00,\frac\infty\infty\); rewrite other forms first; often an \(o\)-expansion is faster.
- \(f''\ge0\iff\) convex (graph below chords).
?Self-test
IntegrationIntegration
The integral measures the (signed) area under a graph, defined as the limit of rectangle sums. The fundamental theorem of calculus then reveals that integrating undoes differentiating, so areas can be computed with antiderivatives. Two techniques, integration by parts and substitution, are the reverse of the product and chain rules. Every exam contains at least one integral to compute.
Source labels: Skript 8.6 = stated in your lecture script · Blatt 11 · 11.1 = exercise sheet · Klausur WiSe23 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§8.1 = Section 8.1) rather than a numbered result (8.1 = Definition 8.1).
0The big picture
- 8.1 Chop \([a,b]\) into small pieces, add up rectangles \(f(\xi_j)\cdot\text{width}\) (Riemann sums), let the pieces shrink. The limit is \(\int_a^bf(x)\,dx\).
- 8.2 The integral is linear, monotone, additive over intervals; every continuous function is integrable; mean value theorem of integration.
- 8.3 Fundamental theorem: \(F(x)=\int_a^xf(t)\,dt\) has \(F'=f\). Consequence: \(\int_a^bf=F(b)-F(a)\) for any antiderivative \(F\).
- 8.4 Rules: \(\int fg'=[fg]-\int f'g\) (by parts); \(\int f(g(t))g'(t)\,dt=\int f(x)\,dx\) (substitution).
- Σ notation and limits of sequences (§3.1, §2.1) — the integral is a limit of sums.
- Max–min theorem and IVT (§4.4, §4.2) — together they give the mean value theorem of integration, which proves the fundamental theorem.
- The derivative table and the product and chain rules (§6.3) — integration by parts and substitution are these rules read backwards.
- "\(F'=G'\) on an interval ⇒ \(F=G+c\)" (Lemma 7.14, §7.3).
8.1Definition of the integral
For a constant function \(f=c\) on \([a,b]\), the area is \(c(b-a)\). For other functions, approximate by rectangles:
A partition [Zerlegung] \(Z\): \(a=x_0<x_1<\dots<x_n=b\), with mesh [Feinheit] \(|Z|=\max_j|x_j-x_{j-1}|\). Choose intermediate points \(\xi_j\in[x_{j-1},x_j]\). The Riemann sum is \[S_Z=\sum_{j=1}^nf(\xi_j)(x_j-x_{j-1}).\] \(f\) is (Riemann) integrable if for every sequence of partitions with \(|Z_n|\to0\) and every choice of intermediate points, the Riemann sums converge to the same limit \(I(f)=\int_a^bf(x)\,dx\).
Why "every partition sequence and every choice of \(\xi_j\)": for a fixed partition, the Riemann sum depends on where in each strip you measure the height. Integrable means that this freedom stops mattering as the strips get thin: all choices lead to the same number. Extra A function where it never stops mattering: \(f(x)=1\) for rational \(x\), \(0\) for irrational \(x\) on \([0,1]\). Every strip contains both kinds of points (rationals by Satz 1.10; irrationals too, e.g. \(r+\frac{\sqrt2}n\) for rational \(r\) and large \(n\)), so choosing rational \(\xi_j\) gives sums 1, irrational \(\xi_j\) gives sums 0 — not integrable.
Conventions: \(\int_a^af=0\), \(\int_b^af=-\int_a^bf\). The name of the variable does not matter: \(\int_a^bf(x)\,dx=\int_a^bf(t)\,dt\). Equidistant partition: \(x_j=a+j\frac{b-a}n\), mesh \(\frac{b-a}n\to0\).
Try "smallest value" and "largest value" in the widget: they aim at the lower and upper sum, the smallest and largest Riemann sum a given partition allows. (The widget finds each strip's smallest and largest value by sampling 25 points in it, so for a wiggly \(f\) it is very close rather than exact.) For a continuous \(f\) the two squeeze together as \(n\) grows — and every other choice of \(\xi_j\) lies between them, which is why all choices share the same limit.
What you must know for 8.1: the definition of the integral.
8.2Important properties
For integrable \(f,g\) on \([a,b]\):
- Normalisation: \(\int_a^b1\,dx=b-a\).
- Positivity: \(f\ge0\Rightarrow\int_a^bf\ge0\).
- Monotonicity: \(f\ge g\Rightarrow\int_a^bf\ge\int_a^bg\).
- Linearity: \(\int(\lambda f+\mu g)=\lambda\int f+\mu\int g\).
- Additivity: \(\int_a^bf=\int_a^cf+\int_c^bf\) (for any \(a,b,c\) where defined).
- Triangle inequality: \(\left|\int_a^bf\right|\le\int_a^b|f|\).
8.3 Every continuous \(f:[a,b]\to\mathbb R\) is integrable.
Why the properties hold Extra — the script says they "follow from the definition"
Each property already holds for every Riemann sum and survives the limit. (a) For \(f=1\): \(S_Z=\sum(x_j-x_{j-1})=b-a\) (telescoping). (b), (c): \(f\ge g\) gives \(S_Z(f)\ge S_Z(g)\), and limits keep ≥ (Satz 2.13). (d): \(S_Z(\lambda f+\mu g)=\lambda S_Z(f)+\mu S_Z(g)\), then the limit rules. (e): use partitions that contain \(c\) as a point. (f): \(|S_Z(f)|\le S_Z(|f|)\) by the triangle inequality for sums (Satz 1.12).
Satz 8.3 is stated without proof in the script; its proof needs "uniform continuity" (beyond this course). You only need to apply it.
"Signed" area: parts below the \(x\)-axis count negatively. \(\int_0^{2\pi}\cos x\,dx=[\sin x]_0^{2\pi}=0\) (Skript 8.14): the area above and below cancel exactly.
\(f\) continuous, \(p\ge0\) integrable (a weight). Then there is \(\xi\in[a,b]\) with \(\int_a^bf(x)p(x)\,dx=f(\xi)\int_a^bp(x)\,dx\). With \(p=1\): \(\int_a^bf=f(\xi)(b-a)\): the integral equals the area of a rectangle whose height is some value of \(f\), the average value.
Proof
\(f\) has min \(\underline m\) and max \(\overline m\) on \([a,b]\) (Satz 4.29). Since \(p\ge0\): \(\underline mp\le fp\le\overline mp\), and monotonicity of the integral gives \(\underline m\int p\le\int fp\le\overline m\int p\), so \(\int fp=\mu\int p\) for some \(\mu\in[\underline m,\overline m]\). By the IVT, \(\mu=f(\xi)\) for some \(\xi\). ∎
Extra — the case the script skips Dividing by \(\int p\) to get \(\mu\) needs \(\int p>0\); the script writes "Folglich existiert \(\mu\)" without saying so. The degenerate case is harmless: if \(\int p=0\), then \(\underline m\int p\le\int fp\le\overline m\int p\) forces \(\int fp=0\) as well, so both sides of the claim are 0 and any \(\xi\in[a,b]\) works. Worth one line if you are asked to prove 8.4.
For integrable \(f\) on \([-a,a]\): \(f\) even ⇒ \(\int_{-a}^af=2\int_0^af\); \(f\) odd ⇒ \(\int_{-a}^af=0\).
Blatt 12.3 states this for continuous \(f\), but the two exams deliberately weaken that: WiSe18 A2 says only "integrable on every bounded interval" and WiSe20 5a "\(f:[-a,a]\to\mathbb R\) integrable and odd". The proof below never uses continuity, so it gives the stronger version — do not add a hypothesis the task did not give you.
Proof (substitution y = −x)
Split \(\int_{-a}^a=\int_{-a}^0+\int_0^a\). In the first, substitute \(x=-y\), \(dx=-dy\): \(\int_{-a}^0f(x)\,dx=\int_a^0f(-y)(-1)\,dy=\int_0^af(-y)\,dy\). If \(f\) is even this is \(\int_0^af\) (total \(2\int_0^af\)); if odd it is \(-\int_0^af\) (total 0).
Exam use (SoSe21 1b): \(\int_{-N}^N\sum_ka_k\sin(b_kx)\,dx=0\) immediately: linearity + every \(\sin(b_kx)\) is odd. No computation needed.
\(f:[a,b]\to\mathbb R\) continuous with \(\int_a^bf(x)\,dx<0\), and \(f(x_0)>0\) for some \(x_0\in(a,b)\). Show that \(f(c)=0\) for some \(c\in(a,b)\).
Solution — two properties, no computation
- \(f\) must go negative somewhere. Suppose \(f\ge0\) on \((a,b)\). Then positivity of the integral (Satz 8.2(b)) gives \(\int_a^bf\ge0\), contradicting \(\int_a^bf<0\). So there is \(y_0\in(a,b)\) with \(f(y_0)<0\).
- Now the intermediate value theorem. \(f\) is continuous, \(f(x_0)>0>f(y_0)\), so on the closed interval between \(x_0\) and \(y_0\) — which lies inside \((a,b)\) — the IVT (Satz 4.10) gives a \(c\) with \(f(c)=0\). ∎
The point of the task: you are not meant to compute anything. It tests whether you can turn "the integral is negative" into "the function is negative somewhere" — which is exactly property (b) read backwards — and then reach for the IVT. The same task is worked from the continuity side in §4.2.
What you must know for 8.2: applying the properties of the integral.
8.3Antiderivatives [Stammfunktionen]
Let \(f:[a,b]\to\mathbb R\) be continuous. Then \(F(x)=\int_a^xf(t)\,dt\) is differentiable on \((a,b)\) with \[F'(x)=f(x).\] In this sense, integration is the inverse of differentiation.
Proof
\(\frac{F(x+h)-F(x)}h=\frac1h\int_x^{x+h}f(t)\,dt=f(\xi_h)\) for some \(\xi_h\in[x,x+h]\) (MVT of integration). As \(h\to0\), \(\xi_h\to x\), and \(f\) is continuous, so \(f(\xi_h)\to f(x)\). ∎
In detail: \(F(x+h)-F(x)=\int_a^{x+h}f-\int_a^xf=\int_x^{x+h}f\) by additivity (8.2e); the MVT of integration with \(p=1\) turns this into \(f(\xi_h)\cdot h\); and \(x\le\xi_h\le x+h\) (or the reverse for \(h<0\)) gives \(\xi_h\to x\) by squeeze.
Picture: if you move the right edge by \(h\), the area grows by a thin strip of height ≈ \(f(x)\) and width \(h\). So the rate of change of the area is the height \(f(x)\).
\(F\) is an antiderivative [Stammfunktion] of \(f\) on \((a,b)\) if \(F'=f\) there. Written \(\int f(x)\,dx\) (indefinite integral [unbestimmtes Integral]). Antiderivatives are unique up to an additive constant: \(F+c\) is one too, and two antiderivatives differ by a constant (Lemma 7.14).
If \(f\) is continuous on \([a,b]\) and \(F\) is any antiderivative, then \[\int_a^bf(x)\,dx=F(b)-F(a)=\big[F(x)\big]_a^b.\]
Proof: \(G(x)=\int_a^xf\) is an antiderivative (8.6), so \(G=F+c\), and \(\int_a^bf=G(b)-G(a)=F(b)-F(a)\); the constant cancels.
| f(x) | ∫f(x)dx | f(x) | ∫f(x)dx |
|---|---|---|---|
| \(x^a,\ a\neq-1\) | \(\frac{x^{a+1}}{a+1}\) | \(\frac1{\sqrt{1-x^2}}\) | \(\arcsin x\) |
| \(\frac1x\) (on \(x\neq0\)) | \(\ln|x|\) | \(-\frac1{\sqrt{1-x^2}}\) | \(\arccos x\) |
| \(e^x\) | \(e^x\) | \(\frac1{1+x^2}\) | \(\arctan x\) |
| \(\sin x\) | \(-\cos x\) | \(a^x\) (\(a>0,\ a\neq1\)) Blatt 11.2d | \(\frac{a^x}{\ln a}\) |
| \(\cos x\) | \(\sin x\) | \(\ln x\) (\(x>0\)) Bsp. 8.17 | \(x\ln x-x\) |
The first eight entries are the script's own table (after Satz 8.9). The last two are added: \(\int\ln x\,dx\) is Beispiel 8.17 further down this page, and \(\int a^x\,dx=\frac{a^x}{\ln a}\) comes from the solution to Blatt 11.2(d) — both are standard exam fare, which is why they sit here rather than only in the text.
Skript 8.13: the simplest use of the table — term by term (linearity), then \(F(1)-F(0)\): \[\int_0^1(x^7+5x^3+2)\,dx=\left[\frac{x^8}8+\frac54x^4+2x\right]_0^1=\frac18+\frac54+2=\frac{27}8.\]
Why \(\ln|x|\): for \(x<0\), \((\ln(-x))'=\frac{-1}{-x}=\frac1x\). Always check an antiderivative by differentiating it. That takes seconds.
To integrate \(\frac1{\text{product of distinct linear factors}}\), split it into simple fractions:
- Ansatz: \(\frac1{(x-1)(x-2)(x+1)}=\frac a{x-1}+\frac b{x-2}+\frac c{x+1}\).
- Multiply out: \(1=a(x-2)(x+1)+b(x-1)(x+1)+c(x-1)(x-2)\), compare coefficients of \(x^2,x,1\) (or plug in \(x=1,2,-1\): each kills two terms!). Result \(a=-\frac12,b=\frac13,c=\frac16\).
- Integrate each: \(-\frac12\ln|x-1|+\frac13\ln|x-2|+\frac16\ln|x+1|\).
Simpler case (Skript 8.10): \(\frac1{x^2-4}=\frac14\left(\frac1{x-2}-\frac1{x+2}\right)\Rightarrow\int=\frac14\ln\left|\frac{x-2}{x+2}\right|\). And (Skript 8.11) \(\int\frac{dx}{x^2+4}=\frac14\int\frac{dx}{(x/2)^2+1}=\frac12\arctan\frac x2\) (no real factorisation: use arctan; check by differentiating).
Why plugging in works Extra: the equation \(1=a(x-2)(x+1)+\dots\) holds for all \(x\) except the three zeros; both sides are polynomials, so it holds for those too (continuity). At \(x=1\): \(1=a(1-2)(1+1)=-2a\), so \(a=-\frac12\). Before the ansatz: if the numerator's degree is ≥ the denominator's, divide first (as in Blatt 11.2: \(\frac{x^2+x+1}x=x+1+\frac1x\)).
SoSe19 A6: \(f\) continuous and \(>0\) on \([a,b]\). Prove \(\int_a^bf>0\) using the FTC.
\(F(x)=\int_a^xf\) is differentiable on \((a,b)\) with \(F'=f>0\), so \(F\) is strictly increasing (monotonicity criterion; \(F\) is continuous on \([a,b]\)). Hence \(\int_a^bf=F(b)-F(a)>0\).
WiSe22 4b: where is \(f(x)=\int_0^x(t-1)^9e^{-t}\,dt\) (\(x>0\)) increasing/decreasing?
FTC: \(f'(x)=(x-1)^9e^{-x}\). \(e^{-x}>0\) and \((x-1)^9\) has the sign of \(x-1\). So \(f\) is strictly decreasing on \((0,1)\), strictly increasing on \((1,\infty)\). No need to compute the integral!
WiSe20 A4 (\(f(x)=10+\int_{-3}^x(3t^2-12)dt\)) is solved in Chapter 7.
What you must know for 8.3: finding an antiderivative; the fundamental theorem (important!); computing integrals with an antiderivative; partial fractions.
8.4Integration rules [Integrationsregeln]
Integration by parts [Partielle Integration]
For continuously differentiable \(f,g\) on \([a,b]\): \[\int_a^bf(x)g'(x)\,dx=\big[f(x)g(x)\big]_a^b-\int_a^bf'(x)g(x)\,dx.\]
Proof: integrate the product rule \((fg)'=f'g+fg'\) from \(a\) to \(b\).
- f = the factor that gets simpler when differentiated: polynomials (\(x\to1\)), \(\ln x\to\frac1x\), \(\arctan x\to\frac1{1+x^2}\), \(\arcsin\).
- g′ = the factor you can integrate easily: \(e^x\), \(\sin\), \(\cos\), \(x^n\).
- A lone \(\ln x\) or \(\arctan x\): write it as \(1\cdot\ln x\) with \(g'=1\).
- The integral comes back — then solve for it. \(e^x\sin x\) needs parts twice before \(I\) reappears on the right. \(\sin^2x\) needs it only once: after one round you are left with \(\int\cos^2x\), and \(\cos^2=1-\sin^2\) turns that back into \(I\). Either way you end with an equation in \(I\), not an antiderivative — isolate \(I\).
Write the four pieces in a small table (\(f\), \(g'\), \(f'\), \(g\)) like the script does. It prevents sign errors.
\(\int_0^2xe^x\,dx\)
\(f=x,\ g'=e^x,\ f'=1,\ g=e^x\): \([xe^x]_0^2-\int_0^2e^x=2e^2-(e^2-1)=e^2+1\).
\(\int\ln x\,dx\)
\(f=\ln x,\ g'=1,\ f'=\frac1x,\ g=x\): \(\int_1^x\ln t\,dt=[t\ln t]_1^x-\int_1^x1\,dt=x\ln x-x+1\). So \(x\ln x-x\) is an antiderivative (check: \(\ln x+1-1\) ✓).
\(\int_0^\pi\sin^2x\,dx=\frac\pi2\) (Skript 8.18)
\(f=\sin x,\ g'=\sin x,\ f'=\cos x,\ g=-\cos x\): \(I=[-\sin x\cos x]_0^\pi+\int_0^\pi\cos^2x=0+\int_0^\pi(1-\sin^2x)=\pi-I\). So \(2I=\pi\).
SoSe21 4a asks the same up to \(\frac{n\pi}2\): \(\int_0^{n\pi/2}\sin^2x\,dx=\frac{n\pi}4\) (the boundary term \(\sin x\cos x\) vanishes at multiples of \(\frac\pi2\)).
\(\int_0^\pi e^x\sin x\,dx=\frac12(e^\pi+1)\) (also Klausur WiSe23 A4, 7 points)
First: \(f=\sin x,\ g'=e^x\): \(I=[e^x\sin x]_0^\pi-\int_0^\pi e^x\cos x=-\int_0^\pi e^x\cos x\). Second: \(f=-\cos x,\ g'=e^x\): \(I=[-e^x\cos x]_0^\pi-\int_0^\pi e^x\sin x=e^\pi+1-I\). So \(I=\frac12(e^\pi+1)\). The exam asked to name the technique (partial integration, twice) and evaluate \(\sin,\cos\) numerically.
11.1(II), all four parts: \(\int\sin x\cos x\), \(\int x^2\sqrt{1-x}\), \(\int x^2\cos x\), \(\int e^{-x}\cos5x\)
- (a) \(\int\sin x\cos x=\sin^2x-\int\cos x\sin x\Rightarrow\int\sin x\cos x=\frac12\sin^2x+C\). (The integral comes back after one round — solve for it.)
- (b) \(\int x^2\sqrt{1-x}\,dx\): take \(f=x^2\), \(g'=(1-x)^{1/2}\), so \(g=-\frac23(1-x)^{3/2}\). Two rounds of parts, each lowering the power of \(x\) by one and raising the root's exponent by 1: \[\int x^2\sqrt{1-x}\,dx=-\tfrac23x^2(1-x)^{3/2}-\tfrac8{15}x(1-x)^{5/2}-\tfrac{16}{105}(1-x)^{7/2}+C.\] Check by differentiating, or compare with the definite version \(\int_0^1x^2\sqrt{1-x}\,dx=\frac{16}{105}\), which Blatt 12.2 does by substitution \(y=\sqrt{1-x}\) — the same integrand, two different techniques, and both are worth knowing.
- (c) \(\int x^2\cos x=x^2\sin x-2\int x\sin x=x^2\sin x+2x\cos x-2\sin x+C\) (by parts twice, each time lowering the power).
- (d) \(\int e^{-x}\cos5x=-e^{-x}\cos5x-5\int e^{-x}\sin5x=\dots-25\int e^{-x}\cos5x\) ⇒ \(\int e^{-x}\cos5x=\frac{e^{-x}(5\sin5x-\cos5x)}{26}+C\).
11.1(I) \(\int_0^1x^p(1-x)^q\,dx=\frac{p!\,q!}{(p+q+1)!}\) (induction on q)
\(q=0\): \(\int_0^1x^p=\frac1{p+1}=\frac{p!0!}{(p+1)!}\) ✓. Step: by parts with \(g'=x^p\), \(f=(1-x)^{q+1}\); the boundary term vanishes at 0 and 1: \[\int_0^1x^p(1-x)^{q+1}=\frac{q+1}{p+1}\int_0^1x^{p+1}(1-x)^q=\frac{q+1}{p+1}\cdot\frac{(p+1)!\,q!}{(p+q+2)!}=\frac{p!\,(q+1)!}{(p+q+2)!}.\]
11.2 definite integrals
- \[\int_0^2(3x^4+5x^2+6x+2)\,dx=\left[\frac35x^5+\frac53x^3+3x^2+2x\right]_0^2=\frac{96}5+\frac{40}3+16=48+\frac8{15}.\]
- \(\int_0^3(x^2+4x+3)e^x=16e^3-1\) (by parts twice).
- Divide first! \(\displaystyle\int_2^4\frac{x^2+x+1}x\,dx=\int_2^4\left(x+1+\frac1x\right)dx=8+\ln2\).
- \(\int_0^12^{x+1}=2\left[\frac{2^x}{\ln2}\right]_0^1=\frac2{\ln2}\).
- \(\int_0^12x(1-e^x)=[x^2]_0^1-2\int_0^1xe^x=1-2=-1\).
- \(\int_0^{5\pi/2}\sin\frac x5\,dx=5\int_0^{\pi/2}\sin y\,dy=5\) (substitution \(y=\frac x5\)).
WiSe19 4a: \(\int_0^4te^{-5t}\,dt\)
\(u=t,\ v'=e^{-5t},\ u'=1,\ v=-\frac15e^{-5t}\): \(\left[-\frac t5e^{-5t}\right]_0^4+\frac15\int_0^4e^{-5t}=-\frac45e^{-20}+\frac1{25}(1-e^{-20})=\frac1{25}-\frac{21}{25}e^{-20}\).
WiSe22 2a: the antiderivative \(F\) of \(x\cos x\) with \(F(0)=0\)
\(\int x\cos x=x\sin x-\int\sin x=x\sin x+\cos x+c\). \(F(0)=1+c=0\Rightarrow c=-1\): \(F(x)=x\sin x+\cos x-1\). (Variant with \(F(0)=3\): \(c=2\).)
SoSe23 2b: all antiderivatives of \(\cos(x)e^x\)
Twice by parts and solve: \(F(x)=\frac12(\sin x+\cos x)e^x+c\). Check: \(F'=\frac12(\cos x-\sin x+\sin x+\cos x)e^x=\cos x\,e^x\) ✓.
WiSe20 3c: all antiderivatives of \(\frac{\ln\sqrt x}2\)
Simplify first: \(\frac{\ln\sqrt x}2=\frac{\ln x}4\). With \(\int\ln x=x\ln x-x\): \(F(x)=\frac14(x\ln x-x)+c\).
Substitution [Substitutionsregel]
Let \(g:[a,b]\to\mathbb R\) be continuously differentiable and \(f\) continuous on \(g([a,b])\). Then \[\int_a^bf(g(t))\,g'(t)\,dt=\int_{g(a)}^{g(b)}f(x)\,dx.\]
Proof: if \(F'=f\), then \((F\circ g)'=f(g)g'\) (chain rule), so both sides equal \(F(g(b))-F(g(a))\).
About "\(dx=g'(t)\,dt\)": this is a memory aid, not a new rule: it reminds you which factor must be present for the chain rule to run backwards (\(\frac{dx}{dt}=g'(t)\)). For an indefinite integral: \(\int f(g(t))g'(t)\,dt=F(g(t))+C\).
- Spot an inner function \(g(t)\) whose derivative also appears (up to a constant).
- Write \(x=g(t)\), \(dx=g'(t)\,dt\).
- Change the limits: \(t=a\Rightarrow x=g(a)\), \(t=b\Rightarrow x=g(b)\). (Or, for indefinite integrals, substitute back at the end.)
- Integrate in \(x\).
Three cases worth naming: shift \(\int_a^bf(t+c)\,dt=\int_{a+c}^{b+c}f\) (Lemma 8.21; for \(f\) periodic with period \(c\) this gives \(\int_a^b=\int_{a+c}^{b+c}\)); scaling \(\int_a^bf(ct)\,dt=\frac1c\int_{ac}^{bc}f\) (Lemma 8.22); and the log-derivative pattern \(\int_a^b\frac{g'}g=\ln\frac{g(b)}{g(a)}\) for \(g>0\), which the script gives as a worked example rather than a named rule (Beispiel 8.23) — e.g. \(\int\tan t\,dt=-\ln|\cos t|\). Only the first two are Lemmas; quote 8.23 as an example, not as a theorem.
Definite: \(\int_0^{\ln2}e^x\sqrt{e^x-1}\), \(\int_0^1\frac{2x-1}{1+x^2}\), \(\int_0^1x^2\sqrt{1-x}\)
- \(y=e^x-1\), \(dy=e^xdx\), limits \(0\to1\): \(\int_0^1\sqrt y\,dy=\frac23\).
- Split: \(\int_0^1\frac{2x}{1+x^2}-\int_0^1\frac1{1+x^2}\). First with \(y=1+x^2\): \(\ln2\). Second: \(\arctan1=\frac\pi4\). Result \(\ln2-\frac\pi4\).
- \(y=\sqrt{1-x}\), \(x=1-y^2\), \(dx=-2y\,dy\): \(2\int_0^1(1-y^2)^2y^2\,dy=2\left(\frac17-\frac25+\frac13\right)=\frac{16}{105}\). (Substitution "backwards": \(x\) expressed through \(y\).)
Indefinite: \(\int2x\sin x^2\), \(\int\frac{4x\arcsin x^2}{\sqrt{1-x^4}}\), \(\int\sin(\ln x)\), \(\int\arcsin x\)
- \(y=x^2\): \(\int\sin y\,dy=-\cos x^2\).
- \(y=\arcsin x^2\), \(dy=\frac{2x}{\sqrt{1-x^4}}dx\): \(2\int y\,dy=(\arcsin x^2)^2\).
- \(y=\ln x\), \(x=e^y\), \(dx=e^ydy\): \(\int e^y\sin y\,dy=\frac12e^y(\sin y-\cos y)\) (Ex. 8.19) \(=\frac x2(\sin\ln x-\cos\ln x)\).
- By parts with \(1\cdot\arcsin x\): \(x\arcsin x-\int\frac x{\sqrt{1-x^2}}=x\arcsin x+\sqrt{1-x^2}\).
SoSe23 2a: \(\int_0^{\pi/2}\sin x\,e^{\cos x}\,dx\)
\(y=\cos x\), \(dy=-\sin x\,dx\), limits \(1\to0\): \(-\int_1^0e^y\,dy=\int_0^1e^y\,dy=e-1\). (Variant \(\int_0^\pi\): \(\int_{-1}^1e^y=e-\frac1e\).)
WiSe24 6a: \(\int_0^1x^{2024}e^{-x^{2025}}\,dx\)
The derivative of the exponent, \(-2025x^{2024}\), is there up to a constant. Antiderivative \(-\frac1{2025}e^{-x^{2025}}\) (check by the chain rule). Value \(\frac{1-e^{-1}}{2025}\). Don't be scared by big numbers: look for \(g'\) next to \(g\).
What you must know for 8.4: integration by parts and the substitution rule.
Mixed exercises from the script [Kapitel 13]
Three integrals from the script’s exercise pool, one per technique.
\(\displaystyle I=\int_0^{\pi/2}\frac{\sin(2x)}{\cos x+\cos^2x}\,dx\).
Solution
The denominator vanishes at \(x=\frac\pi2\) (since \(\cos\frac\pi2=0\)), which looks like an improper integral. But use \(\sin2x=2\sin x\cos x\) and cancel: \[\frac{\sin2x}{\cos x+\cos^2x}=\frac{2\sin x\cos x}{\cos x(1+\cos x)}=\frac{2\sin x}{1+\cos x},\] which is perfectly well defined on all of \([0,\frac\pi2]\) (the denominator is \(\ge1\)). So the integral is an ordinary one. Substitute \(u=\cos x\), \(du=-\sin x\,dx\); the limits become \(u(0)=1\) and \(u(\frac\pi2)=0\): \[I=-2\int_1^0\frac{du}{1+u}=2\int_0^1\frac{du}{1+u}=2\big[\ln(1+u)\big]_0^1=2\ln2.\]
Lesson: simplify first. A zero in the denominator that cancels is not a singularity.
\(\displaystyle F=\int x\sin(x^2)\cos^2(x^2)\,dx\).
Solution
First \(u=x^2\), \(du=2x\,dx\) (the lone factor \(x\) is the signal): \(F=\frac12\int\sin u\cos^2u\,du\). Then \(z=\cos u\), \(dz=-\sin u\,du\): \[F=-\frac12\int z^2\,dz=-\frac16z^3=-\frac16\cos^3(x^2)+C.\] Check by differentiating (chain rule twice): \(\frac{d}{dx}\left(-\frac16\cos^3(x^2)\right)=-\frac12\cos^2(x^2)\cdot(-\sin(x^2))\cdot2x=x\sin(x^2)\cos^2(x^2)\) ✓.
\(\displaystyle\int\arctan x\,dx\).
Solution
There is only one factor, so write the second one as 1 and integrate by parts with \(f=\arctan x\), \(g'=1\), \(f'=\frac1{1+x^2}\), \(g=x\): \[\int\arctan x\,dx=x\arctan x-\int\frac x{1+x^2}\,dx=x\arctan x-\frac12\ln(1+x^2)+C,\] the last integral being of the form \(\frac12\int\frac{g'}g\) with \(g=1+x^2>0\).
Same trick as for \(\int\ln x\,dx=x\ln x-x\) (Skript 8.17): for a lone \(\ln\), \(\arctan\), \(\arcsin\), take \(g'=1\).
!What the exam asks from Chapter 8
Every exam has at least one integral, often a proof about integrals as well:
| Type | Exam tasks | What earns the points |
|---|---|---|
| By parts | WiSe19 4a, WiSe22 2a, SoSe23 2b, WiSe23 A4, SoSe21 4a, WiSe20 3c | table \(f,g',f',g\); for \(e^x\sin x\) twice + solve; state the technique |
| Substitution | SoSe23 2a, WiSe24 6a | spot \(g\) and \(g'\); change the limits |
| Symmetry | WiSe18 A2, WiSe20 5a, SoSe21 1b | substitution \(y=-x\) on \([-a,0]\) plus additivity |
| Fundamental theorem | SoSe19 A6, WiSe22 4b, WiSe20 A4 | \(F'=f\) without computing the integral; then monotonicity/extrema (Ch. 7) |
| Antiderivative with a condition | WiSe22 2a (\(F(0)=0\)) | find \(F+c\), then fix \(c\) |
| Properties of the integral, no computation | SoSe23 A6 | positivity (8.2b) to locate a sign, then the IVT; resist the urge to integrate |
Drill: which technique?
Deciding the technique is half the task. For each integral, pick the method (and check the answer in the explanation).
- For the exam: the five task types above, and the antiderivative table by heart.
- For Chapter 9: improper integrals \(\int_1^\infty\frac{dx}{x^\alpha}\) are limits of ordinary ones; the integral test compares sums with integrals; parameter integrals differentiate under the integral sign.
- For Chapter 10: Taylor's remainder in integral form is proved by repeated integration by parts.
- For Chapter 12: separation of variables is substitution; variation of constants needs \(\int f(t)e^{A(t)}dt\), usually by parts.
ΣSummary sheet
- \(\int_a^bf=\lim\) of Riemann sums \(\sum f(\xi_j)\Delta x_j\); continuous ⇒ integrable; signed area.
- Linear, monotone, additive; \(|\int f|\le\int|f|\); \(\int_a^bf=f(\xi)(b-a)\) for continuous \(f\).
- FTC: \(\frac{d}{dx}\int_a^xf(t)\,dt=f(x)\), and \(\int_a^bf=F(b)-F(a)\). Use it to differentiate integral-defined functions without computing them.
- By parts: \(\int fg'=[fg]-\int f'g\); choose \(f\) to simplify; twice + solve for cyclic cases.
- Substitution: \(x=g(t)\), \(dx=g'(t)dt\), change the limits. Look for "function and its derivative".
- Partial fractions for rational functions; odd functions on symmetric intervals integrate to 0.
- Always check antiderivatives by differentiating.
?Self-test
More on IntegralsMehr über Integrale
Four extensions of the integral. Improper integrals over unbounded intervals or up to singularities; integrals that depend on a parameter; when you may swap sum and integral; and, most useful for computer science, estimating sums by integrals: \(H_n\approx\ln n\), \(\ln(n!)\approx n\ln n\) (the comparison count of sorting), and the integral test for series.
Source labels: Skript 9.6 = stated in your lecture script · Blatt 12 · 12.4 = exercise sheet · Klausur WiSe24 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§9.1 = Section 9.1) rather than a numbered result (9.1 = Definition 9.1).
0The big picture
- 9.1 \(\int_a^\infty f=\lim_{M\to\infty}\int_a^Mf\): an improper integral is a limit of ordinary integrals, just like a series is a limit of partial sums. Benchmark: \(\int_1^\infty x^{-\alpha}\) converges iff \(\alpha>1\); \(\int_0^1x^{-\alpha}\) iff \(\alpha<1\).
- 9.2 For continuous integrands you may swap integration order (Fubini) and differentiate under the integral sign.
- 9.3 \(\int\sum=\sum\int\) if there is a convergent majorant. Gives \(\ln(1+x)=\sum(-1)^k\frac{x^{k+1}}{k+1}\) and \(\ln2=1-\frac12+\frac13-\dots\)
- 9.4 For monotone \(f\), \(\sum f(k)\) is squeezed between integrals. Integral test: \(\sum f(k)\) and \(\int f\) converge together.
- The fundamental theorem, antiderivatives and substitution (§8.3, §8.4) — every improper integral is computed with them first.
- Series: majorant/minorant and "non-negative terms: convergent ⇔ partial sums bounded" (§3.2, Satz 3.6). §9.1 and §9.4 copy these ideas for integrals.
- Monotone + bounded ⇒ convergent (Satz 2.17), one-sided limits (§4.1), and the growth ladder (§5.2).
9.1Improper integrals [Uneigentliche Integrale]
If \(f:[a,b)\to\mathbb R\) is integrable on every \([a,\beta]\), \(\beta<b\), the improper integral is \[\int_a^bf(x)\,dx:=\lim_{\beta\to b}\int_a^\beta f(x)\,dx,\] if the limit exists (\(b=\infty\) allowed). Same at the lower limit. If both ends are critical, split at some \(c\in(a,b)\) and require both limits to exist separately: \[\int_a^bf=\lim_{\alpha\to a}\int_\alpha^cf+\lim_{\beta\to b}\int_c^\beta f.\]
One-sided limits (Skript 9.3): \(\lim_{x\downarrow x_0}f(x)=c\) means \(f(x_n)\to c\) for all \(x_n\to x_0\) with \(x_n>x_0\); \(\lim_{x\uparrow x_0}\) analogously from the left.
- An integration limit is \(\pm\infty\), or
- the integrand blows up at a point of \([a,b]\) (a singularity), possibly inside the interval.
\(\int_0^1\frac1x\,dx=\lim_{\varepsilon\downarrow0}[\ln x]_\varepsilon^1=\lim(-\ln\varepsilon)=\infty\). And \(\int_{-1}^1\frac1x\,dx\) is not defined: it splits into \(-\infty+\infty\). Writing \([\ln|x|]_{-1}^1=0\) is wrong because the FTC needs continuity on the whole interval.
But (Skript 13.3): \(\int_0^{\pi/2}\frac{\sin2x}{\cos x+\cos^2x}\) only looks improper. After cancelling \(\cos x\), the integrand \(\frac{2\sin x}{1+\cos x}\) is continuous on the whole interval.
Proof
For \(\alpha\neq1\): \(\int_1^M x^{-\alpha}=\left[\frac{x^{1-\alpha}}{1-\alpha}\right]_1^M=\frac{M^{1-\alpha}-1}{1-\alpha}\). As \(M\to\infty\), \(M^{1-\alpha}\to0\) iff \(\alpha>1\) (limit \(\frac1{\alpha-1}\)), else → ∞. For \(\alpha=1\): \(\ln M\to\infty\). Near 0 it is the other way round: \(\varepsilon^{1-\alpha}\to0\) iff \(\alpha<1\).
Memory hook: at ∞ the function must decay faster than \(\frac1x\); at 0 it may blow up only slower than \(\frac1x\). \(\frac1x\) itself fails at both ends.
- Write it as a limit: \(\lim_{M\to\infty}\int_a^M\) or \(\lim_{\varepsilon\downarrow0}\int_\varepsilon^b\) (both, if needed, at a split point).
- Compute the proper integral with an antiderivative (substitution/by parts).
- Take the limit. Finite ⇒ converges, value = limit. Otherwise diverges.
- If no antiderivative is available: compare with a benchmark \(\frac1{x^\alpha}\) (smaller than convergent ⇒ convergent; larger than divergent ⇒ divergent, for nonnegative integrands). Same logic as majorant/minorant for series.
Let \(0\le f(x)\le g(x)\) on \([a,\infty)\), both integrable on every \([a,M]\). If \(\int_a^\infty g\) converges, so does \(\int_a^\infty f\) (and \(\int f\le\int g\)). If \(\int_a^\infty f\) diverges, so does \(\int_a^\infty g\). The same holds at a singular endpoint.
Proof (the Satz 3.6 argument for integrals)
\(I(M)=\int_a^Mf\) is increasing in \(M\) (since \(f\ge0\), additivity) and bounded by \(\int_a^Mg\le\int_a^\infty g\) (monotonicity). An increasing bounded function has a limit as \(M\to\infty\) (the sup — proof as in Satz 2.17). The second statement is the first read backwards. ∎
Also as in Chapter 3: if \(f,g>0\) and \(\frac{f(x)}{g(x)}\to c\in(0,\infty)\), the two integrals converge or diverge together (asymptotic comparison).
Blatt 12.4 — "check whether the improper integral exists and compute its value if it does". Parts (a)–(b) were the ones handed in; (c) is marked voluntary on the sheet.
- (a) \(\int_{-\infty}^\infty\frac{dx}{1+x^2}\): split at 0, as the sheet's hint says. \(\int_0^\beta=\arctan\beta\to\frac\pi2\), and \(\int_\alpha^0\to\frac\pi2\) as \(\alpha\to-\infty\). Total \(\pi\).
- (b) \(\int_3^\infty\frac{4x}{x^2-4}dx\): substitution \(y=x^2-4\): \(2\ln(\beta^2-4)-2\ln5\to\infty\). Diverges (integrand \(\simeq\frac4x\)).
- (c) \(\int_0^e\ln x\,dx\) (singular at 0): \([x\ln x-x]_\varepsilon^e=0-(\varepsilon\ln\varepsilon-\varepsilon)\to0\), using \(\varepsilon\ln\varepsilon\to0\). Value 0.
Blatt 13.2 (a) — a convergence question with no value asked for:
- \(\int_1^\infty\frac{dx}{x^4(1+2\ln x)}\): for \(x\ge1\), \(1+2\ln x\ge1\), so the integrand is \(\le\frac1{x^4}\). Comparison with \(\int_1^\infty x^{-4}=\frac13\) ⇒ converges. (Part (b) of the same task is a series estimate and sits in §9.4.)
\(\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt\) extends the factorial (\(\Gamma(n)=(n-1)!\)). For \(x\in(0,1)\) it is improper at both ends, and the task is to show the integral is well defined [wohldefiniert] — in four steps, one per part of the sheet.
Solution, part by part
(a) At 0. The sheet writes \(A(s)=\int_s^1t^{x-1}e^{-t}dt\) and asks for: bounded above, and decreasing in \(s\). On \((0,1)\), \(e^{-t}<1\), so \(A(s)\le\int_s^1t^{x-1}dt=\frac{1-s^x}x\le\frac1x\) — bounded. Shrinking \(s\) adds a piece of a non-negative integrand, so \(A\) decreases as \(s\) grows, i.e. increases as \(s\downarrow0\).
(b) At ∞. \(B(s)=\int_1^st^{x-1}e^{-t}dt\) is increasing in \(s\) (again non-negative integrand) and bounded: on \((1,\infty)\) we have \(t^{x-1}<1\) for \(x<1\), so \(B(s)\le\int_1^se^{-t}dt\le e^{-1}\).
(c) Conclusion. This is the part that answers the question. Both one-sided limits exist by "monotone + bounded ⇒ convergent" (Satz 2.17 transported to functions), so \(\Gamma(x)=\lim_{s\downarrow0}A(s)+\lim_{s\to\infty}B(s)\) is a finite number for every \(x\in(0,1)\) — the Gamma function is well defined there. Note the split at 1 is what makes each piece monotone; one single limit would not be monotone in anything.
(d) \(\Gamma(\frac12)\): substitute \(t=u^2\), \(dt=2u\,du\): \(\int_0^\infty\frac1ue^{-u^2}2u\,du=2\int_0^\infty e^{-u^2}du=2\cdot\frac{\sqrt\pi}2=\sqrt\pi\), using \(\int_0^\infty e^{-u^2}du=\frac{\sqrt\pi}2\), which the sheet lets you quote without proof.
WiSe19 4b: \(\int_0^\infty\min\left(x^2,\frac1{x^2}\right)dx\)
\(\min=x^2\) on \([0,1]\), \(=\frac1{x^2}\) on \((1,\infty)\) (continuous). \(\int_0^1x^2=\frac13\), \(\int_1^\infty x^{-2}=\lim_\alpha\left(1-\frac1\alpha\right)=1\). Total \(\frac43\).
WiSe20 5b: for which \(a\) does \(\int_0^\infty\frac x{(1+x^2)^a}dx\) converge?
Substitute \(y=1+x^2\), \(dy=2x\,dx\): \(\frac12\int_1^\infty y^{-a}\,dy\). By the benchmark: converges iff \(a>1\), value \(\frac1{2(a-1)}\). (\(a=1\): \(\frac12\ln M\to\infty\).)
WiSe22 2b: \(\int_0^\infty x^4e^{-x^5}dx\)
\(y=x^5\), \(dy=5x^4dx\): \(\lim_M\frac15\int_0^{M^5}e^{-y}dy=\lim\frac15(1-e^{-M^5})=\frac15\). (Variant with \(x^3e^{-x^4}\): \(\frac14\).)
WiSe24 6c: does \(\int_0^\infty|\sin x|^{\cos x}\,xe^{-x}\,dx\) exist? (hard)
No. Look near \(x=\pi\): there \(\cos x\approx-1\), so \(|\sin x|^{\cos x}\approx\frac1{|\sin x|}\approx\frac1{|x-\pi|}\), and the integrand behaves like \(\frac{\pi e^{-\pi}}{|x-\pi|}\), which is not integrable (\(\int\frac{dh}h\) diverges at 0).
The common wrong solution (partial credit only): "\(|\sin x|^{\cos x}\le1\), so the integral is \(\le\int xe^{-x}=1\)". The bound is false where \(\cos x<0\): a number in \((0,1)\) raised to a negative power is \(>1\). Always check where the integrand can blow up.
What you must know for 9.1: recognising improper integrals; for which α the integrals \(\int_0^1x^{-\alpha}\) and \(\int_1^\infty x^{-\alpha}\) converge or diverge.
9.2Parameter-dependent integrals [Parameterabhängige Integrale]
For \(f(x,y)\) and fixed \(x\), the derivative of \(y\mapsto f(x,y)\) is the partial derivative \(\frac{\partial f}{\partial y}=\partial_yf=\partial_2f\) (differentiate in \(y\), treat \(x\) as a constant; Chapter 11 in detail).
Let \(f:[a,b]\times[c,d]\to\mathbb R\) be continuous.
- \(F(y)=\int_a^bf(x,y)\,dx\) is continuous.
- Fubini: \(\int_c^d\int_a^bf(x,y)\,dx\,dy=\int_a^b\int_c^df(x,y)\,dy\,dx\). The order does not matter.
- If \(\partial_yf\) exists and is continuous, then \(F\) is \(C^1\) and \(\displaystyle\frac{d}{dy}\int_a^bf(x,y)\,dx=\int_a^b\frac{\partial f}{\partial y}(x,y)\,dx\).
\(\varphi(t)=\int_a^be^{tx}g(x)\,dx\) with continuous \(g\). Differentiating under the integral: \(\varphi^{(k)}(t)=\int_a^bx^ke^{tx}g(x)\,dx\). If \(g\) is a probability density of \(X\), then \(\varphi(t)=E[e^{tX}]\) and \(E[X^k]=\varphi^{(k)}(0)\): all moments come from one function (used in probability theory).
What you must know for 9.2: computing partial derivatives; Satz 9.8.
9.3Swapping summation and integration
Let \(f_k\) be integrable on \([a,b]\) with \(|f_k(x)|\le a_k\) for all \(x\) and \(\sum a_k<\infty\) (a convergent majorant independent of \(x\)). Then \(f=\sum f_k\) converges and \[\int_a^b\sum_{k=1}^\infty f_k(x)\,dx=\sum_{k=1}^\infty\int_a^bf_k(x)\,dx.\]
Why the majorant makes it work Extra — the script gives no proof
For each \(x\), \(\sum f_k(x)\) converges absolutely by the majorant test (Satz 3.7). Split off the first \(n\) terms: \(f=\sum_{k\le n}f_k+R_n\) with the tail \(|R_n(x)|\le\sum_{k>n}a_k=:\varepsilon_n\) — the same bound for every \(x\), and \(\varepsilon_n\to0\). For finitely many terms, integral and sum may be swapped (linearity), so \[\Big|\int_a^bf-\sum_{k\le n}\int_a^bf_k\Big|=\Big|\int_a^bR_n\Big|\le(b-a)\,\varepsilon_n\to0.\] (That \(f\) itself is integrable is part of the theorem; you may take it as given.) The key is that the tail bound does not depend on \(x\) — exactly what fails in 9.14. ∎
For \(|t|<1\): \(\frac1{1+t}=\sum_{k\ge0}(-t)^k\) (geometric). For \(|t|\le|x|<1\), \(|(-t)^k|\le|x|^k\), a convergent majorant. So integrate term by term: \[\ln(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1}=x-\frac{x^2}2+\frac{x^3}3-\dots\qquad(|x|<1).\] This is a power series [Potenzreihe] with radius of convergence [Konvergenzradius] \(r=1\) (Def. 9.12; at \(x=-1\) the series is \(-\sum\frac1{k+1}=-\infty\)).
Lemma 9.13: the formula even holds at \(x=1\): \[\ln2=1-\frac12+\frac13-\frac14+\dots\] Proof idea: for \(0<x<1\) the Leibniz error bound gives \(\left|\ln(1+x)-s_n(x)\right|\le\frac{x^{n+2}}{n+2}\); let \(x\uparrow1\) (ln is continuous), then \(n\to\infty\).
The script's second power series (stated right after Lemma 9.13, and named in the Minimale Inhalte) is the geometric series itself: \[\sum_{k=0}^\infty x^k=\frac1{1-x}\qquad(|x|<1),\quad\text{radius of convergence }r=1.\] It is the same series that started this box, with \(t\) replaced by \(-x\); both are the templates Chapter 10 builds on. You are expected to be able to write both down from memory.
The script builds functions \(f_k\) whose partial sums are spikes of height \(n\) on \((0,\frac1n)\) (the \(f_k\) themselves are steps taking the values \(1\) and \(-(k-1)\); see the details below) with \(\sum_k\int_0^1f_k=1\) but \(\int_0^1\sum_kf_k=0\). The same phenomenon appears in WiSe24 6b: \(f_n(x)=\frac n{(1+x)^n}\) on \([0,1]\). For each \(x\in(0,1]\), \(f_n(x)\to0\) (exponential beats linear), yet \[\int_0^1f_n=\frac n{n-1}\left(1-2^{1-n}\right)\to1\neq0.\] The mass concentrates in an ever thinner spike near 0. Moral: "limit of integrals = integral of the limit" needs a justification (such as a majorant).
The script's counterexample in detail (Bemerkung 9.14)
Let \(f_1=\mathbb 1_{(0,1)}\) and \(f_k=k\,\mathbb 1_{(0,\frac1k)}-(k-1)\,\mathbb 1_{(0,\frac1{k-1})}\) for \(k\ge2\) (\(\mathbb 1_A\) = 1 on \(A\), 0 elsewhere). Then \(\int_0^1f_k=k\cdot\frac1k-(k-1)\frac1{k-1}=0\) for \(k\ge2\) and \(\int f_1=1\), so \(\sum_k\int f_k=1\). But the partial sums telescope: \(\sum_{k\le n}f_k=n\,\mathbb 1_{(0,\frac1n)}\), a spike of height \(n\) and width \(\frac1n\). For every fixed \(x\) it is 0 as soon as \(\frac1n\le x\), so \(\sum_kf_k(x)=0\) and \(\int\sum=0\). No majorant exists: \(\sup_x|f_k|=k-1\), and \(\sum(k-1)=\infty\).
A sequence of functions \(f_n:D\to\mathbb R\) converges pointwise [punktweise] to \(f\) if \(f_n(x)\to f(x)\) for every single \(x\in D\). It converges uniformly [gleichmäßig] if the worst-case distance goes to 0: \[\sup_{x\in D}|f_n(x)-f(x)|\xrightarrow{n\to\infty}0.\] Uniform means one \(n_0\) works for all \(x\) at once; pointwise allows \(n_0\) to depend on \(x\). Uniform ⇒ pointwise, not conversely. To disprove uniform convergence, find points \(x_n\) (depending on \(n\)) with \(|f_n(x_n)-f(x_n)|\not\to0\).
Link to Satz 9.10: a convergent majorant \(|f_k(x)|\le a_k\) makes the partial sums converge uniformly (the tail bound \(\varepsilon_n\) above does not depend on \(x\)). That is the "Weierstrass test" behind SoSe19 A3 (Chapter 3).
\(f_n:[0,1]\to\mathbb R\), \(f_n(x)=\frac{nx}{(1+x)^n}\). Pointwise and/or uniform convergence?
Solution: pointwise to 0, not uniformly
Pointwise: \(f_n(0)=0\). For \(x\in(0,1]\), \((1+x)^n\) grows exponentially and \(nx\) only linearly, so \(f_n(x)\to0\) (e.g. \((1+x)^n\ge\binom n2x^2\) by the binomial theorem, so \(0\le f_n(x)\le\frac{2}{(n-1)x}\to0\)).
Not uniform: take \(x_n=\frac1n\). Then \(f_n(x_n)=\frac1{(1+\frac1n)^n}\to\frac1e\neq0\) (Satz 2.21), so \(\sup_x|f_n(x)-0|\ge f_n(\frac1n)\not\to0\).
Flagged: the uploaded Musterlösung writes "\(f_n(x_n)\to\infty\)". With \(x_n=\frac1n\) the value tends to \(\frac1e\), not ∞; the conclusion "not uniform" is still right, because \(\frac1e\neq0\).
What you must know for 9.3: the sufficient criterion for swapping sum and integral; the power series of \(\ln(1+x)\) and \(\frac1{1-x}\).
9.4Estimating sums and series [Abschätzungen von Summen und Reihen]
Let \(a<b\) be integers and \(f:[a,b]\to\mathbb R\) continuous and monotone.
- \(f\) increasing: \(\displaystyle\sum_{k=a}^{b-1}f(k)\le\int_a^bf(x)\,dx\le\sum_{k=a+1}^bf(k)\).
- \(f\) decreasing: \(\displaystyle\sum_{k=a+1}^bf(k)\le\int_a^bf(x)\,dx\le\sum_{k=a}^{b-1}f(k)\).
Proof: on each \([k,k+1]\), \(f(k)\le f(x)\le f(k+1)\) (increasing case); integrate over the unit interval and sum up. Picture: the sum is a staircase of unit-width rectangles, lying below or above the graph (see the widget).
Note on numbering Blatt 12.1 tells you to "Betrachten Sie Satz 9.18 (b) aus dem Skript" — that is this result under an older numbering of the script. In the version you have it is Satz 9.15, and 9.18 is the lemma \(\ln(n!)\simeq n\ln n\) below. Quote the number from your script.
- Power sums (\(\alpha>0\), \(f=x^\alpha\) increasing): \(\frac{n^{\alpha+1}}{\alpha+1}\le\sum_{k=1}^nk^\alpha\le\frac{(n+1)^{\alpha+1}-1}{\alpha+1}\), hence \(\sum_{k=1}^nk^\alpha\simeq\frac{n^{\alpha+1}}{\alpha+1}\) (e.g. \(\sum k\simeq\frac{n^2}2\)).
- Harmonic numbers (Blatt 12.1): (a) with \(f=\frac1x\) decreasing, \(\ln(n+1)\le H_n\le1+\ln n\) (the sheet's hint is \(\int_1^y\frac{dx}x=\ln y\)); (b) divide by \(\ln n\) and squeeze to get \(\frac{H_n}{\ln n}\to1\), written \(H_n\simeq\ln n\). Feeding this into the formula for the average number of comparisons of Quicksort, \(C_n=2(n+1)H_n-2n\) (stated in the script's Chapter 2 motivation, not here), gives \(C_n\simeq2n\ln n\).
- How many comparisons does sorting need? Skript 9.17 Sorting \(n\) elements takes up to \(\log_2(n!)\) comparisons — that is the script's wording — and \(\log_2(n!)=\ln(n!)\log_2e\), so the growth is decided by \(\ln(n!)=\sum_{k=2}^n\ln k\). With \(f=\ln x\) increasing:
\[n\ln n-n+1\le\ln(n!)\le(n+1)\ln(n+1)-n.\]
Hence \(\ln(n!)\simeq n\ln n\) (Lemma 9.18) and even \(\ln(n!)=n\ln n-n+O(\ln n)\) (Lemma 9.19). Either way \(\log_2(n!)\) grows like \(n\log_2n\).
Extra — which direction is the famous one? In computer science \(\log_2(n!)\) is normally quoted as a lower bound: a comparison-based sorting algorithm is a decision tree that must distinguish \(n!\) orderings, so some branch has depth \(\ge\log_2(n!)\), and no such algorithm can beat \(\approx n\log_2n\) comparisons in the worst case. Mergesort attains that order, so \(n\log n\) is both a lower and an upper bound (\(\Theta(n\log n)\)). Your script writes "bis zu" (up to) and proves nothing about the direction — so take the estimate \(\log_2(n!)\simeq n\log_2n\) from Beispiel 9.17 and leave the word "lower bound" out of an exam answer unless you prove it.
Proofs of 9.18 and 9.19
9.18: divide the bounds by \(n\ln n\): the lower bound gives \(1-\frac1{\ln n}+\frac1{n\ln n}\to1\); the upper gives \(\frac{n+1}n\cdot\frac{\ln(n+1)}{\ln n}-\frac1{\ln n}\to1\), because \(\frac{\ln(n+1)}{\ln n}=1+\frac{\ln(1+\frac1n)}{\ln n}\to1\). Squeeze.
9.19: subtract \(n\ln n-n\): \(1\le\ln(n!)-n\ln n+n\le(n+1)\ln(n+1)-n\ln n=\ln n+r_n\) with \(r_n=\frac{\ln(1+\frac1n)}{1/n}+\ln(1+\frac1n)\to1\) (using \(\frac{\ln(1+u)}u\to1\), Chapter 5). So the difference is at most \(\ln n+\) a bounded amount, i.e. \(O(\ln n)\).
Let \(f:[1,\infty)\to[0,\infty)\) be decreasing. Then \[0\le\sum_{k=1}^nf(k)-\int_1^{n+1}f(x)\,dx\le f(1),\] and \(\sum_{k\ge1}f(k)\) converges iff \(\int_1^\infty f(x)\,dx\) converges. In that case \(0\le\sum_{k=1}^\infty f(k)-\int_1^\infty f\le f(1)\).
Starting at another index \(m\) works the same (Blatt 13.2 remark): \(0\le\sum_{k=m}^nf(k)-\int_m^{n+1}f\le f(m)\). Corollary: \(\sum\frac1{k^s}\) converges iff \(s>1\) (compare \(\int_1^\infty x^{-s}\)), which proves the rule of thumb from Chapter 3.
Proof (it is Satz 9.15 with \(a=1\), \(b=n+1\))
\(f\) is decreasing, so part (b) of Satz 9.15 with \(a=1\), \(b=n+1\) gives \[\sum_{k=2}^{n+1}f(k)\le\int_1^{n+1}f(x)\,dx\le\sum_{k=1}^nf(k).\] The right inequality is \(\sum_{k=1}^nf(k)-\int_1^{n+1}f\ge0\). For the other bound, add \(f(1)-f(n+1)\) to the left sum: \(\sum_{k=2}^{n+1}f(k)=\sum_{k=1}^nf(k)-f(1)+f(n+1)\), so \[\sum_{k=1}^nf(k)-\int_1^{n+1}f\le f(1)-f(n+1)\le f(1),\] using \(f\ge0\) in the last step. That is the displayed double inequality.
Convergence now follows in both directions, because both sides are increasing in \(n\) (terms and integrand are \(\ge0\)): if \(\int_1^\infty f<\infty\), the partial sums are bounded by \(\int_1^\infty f+f(1)\), hence the series converges (Satz 3.6). If the series converges, the integrals \(\int_1^{n+1}f\) are bounded by \(\sum_{k\ge1}f(k)\), and being increasing in \(n\) they converge as well. Letting \(n\to\infty\) in the double inequality gives the final estimate. ∎
Picture (same as in the widget above): \(\sum_{k=1}^nf(k)\) is a staircase of unit-width rectangles whose left ends sit on the graph. Since \(f\) falls, each rectangle covers its piece of the area under the graph, and the total overshoot is at most the first rectangle, \(f(1)\).
\(\zeta(s)=\sum_{k\ge1}k^{-s}\) converges for \(s>1\), and the integral test gives \(\frac1{s-1}\le\zeta(s)\le\frac1{s-1}+1\). For \(s\to\infty\), \(\zeta(s)=1+O(2^{-s})\): the first term dominates.
\(\sum\frac{\ln k}{k^2}\) converges, with bounds; \(\sum\frac{\ln k}k\) diverges
\(f(x)=\frac{\ln x}{x^2}\): \(f'=\frac{1-2\ln x}{x^3}<0\) for \(x>\sqrt e\), so decreasing on \([2,\infty)\) (the \(k=1\) term is 0). With \(y=\ln x\), \(x=e^y\): \(\int_2^{n+1}\frac{\ln x}{x^2}dx=\int_{\ln2}^{\ln(n+1)}ye^{-y}dy\to\frac{1+\ln2}2\). Integral test from \(m=2\): \[\frac{1+\ln2}2\le\sum_{k\ge2}\frac{\ln k}{k^2}\le\frac{1+\ln2}2+\frac{\ln2}4.\] \(\frac{\ln x}x\) (decreasing for \(x>e\)): \(\int_3^{n+1}\frac{\ln x}x=\frac12\left((\ln(n+1))^2-(\ln3)^2\right)\to\infty\): diverges. (Faster: \(\frac{\ln k}k\ge\frac1k\) for \(k\ge3\), harmonic minorant.)
Does \(\displaystyle\sum_{n\ge3}\frac1{n\ln n\,(\ln\ln n)^2}\) converge?
Solution: yes, integral test
The function \(\frac1{x\ln x(\ln\ln x)^2}\) is positive and decreasing for \(x\ge3\). Substitute \(y=\ln\ln x\), \(dy=\frac{dx}{x\ln x}\): \[\int_3^\infty\frac{dx}{x\ln x(\ln\ln x)^2}=\int_{\ln\ln3}^\infty\frac{dy}{y^2}<\infty\] (since \(\ln\ln3>0\)). So the series converges. Compare: the neighbouring series \(\sum_{n\ge2}\frac1{n\ln n}\) diverges, because the same substitution leaves \(\int\frac{dy}y\). That divergence is its own exam question — SoSe23 A4e, where you are asked to derive it from the mean value theorem instead of the integral test. One extra power of \(\ln\ln\) is the whole difference between the two.
\(\displaystyle\sum_{n\ge2}\frac1{n\ln n}\) diverges — and this exam does not let you use the integral test. Parts (a)–(c) set up \(f(x)=\ln\ln x\) and \(g(x)=x\ln x\), (d) says "apply the mean value theorem to \(f\) on \([n,n+1]\)", and (e) asks for the conclusion.
The whole argument is worked in §7.2, where the MVT lives. In one line: the MVT turns \(\ln\ln(n+1)-\ln\ln n\) into \(\frac1{\xi\ln\xi}\) with \(\xi\in(n,n+1)\); monotonicity of \(x\ln x\) replaces \(\xi\) by \(n\); and the resulting sum telescopes to \(\ln\ln(N+1)-\ln\ln2\to\infty\).
Extra — read the wording Here the integral test would finish in one line (\(y=\ln x\) turns \(\int\frac{dx}{x\ln x}\) into \(\int\frac{dy}y=\infty\)), but "Wenden Sie den Mittelwertsatz an" removes the choice. When a task prescribes a tool, the prescribed route is the answer; a different correct proof earns little. Conversely, WiSe24 6d prescribes nothing, so use the integral test there.
What you must know for 9.4: estimating sums with integrals; the integral test for series.
Mixed exercises from the script [Kapitel 13]
\(\displaystyle I=\int_0^\infty xe^{-x^2/2}\,dx\).
Solution
Write it as a limit, then substitute \(u=\frac{x^2}2\), \(du=x\,dx\) (the factor \(x\) is exactly what the substitution needs); the limits \(0\) and \(M\) become \(0\) and \(\frac{M^2}2\): \[I=\lim_{M\to\infty}\int_0^Mxe^{-x^2/2}\,dx=\lim_{M\to\infty}\int_0^{M^2/2}e^{-u}\,du=\lim_{M\to\infty}\left(1-e^{-M^2/2}\right)=1.\]
\(\displaystyle\int_{-\infty}^{\infty}xe^{-x^2/2}\,dx\).
Solution
The integrand \(f(x)=xe^{-x^2/2}\) is odd: \(f(-x)=-f(x)\). Splitting at 0 and substituting \(u=-x\) in the left half turns it into the right half with a minus sign: \[\int_{-\infty}^0f(x)\,dx=-\int_0^\infty f(u)\,du=-1,\] using the previous exercise. Hence \(\int_{-\infty}^\infty f=-1+1=0\).
Important: this argument is only allowed because both halves converge separately (each equals \(\pm1\)). For \(\int_{-1}^1\frac{dx}x\) the same symmetry is useless, because each half is infinite and "\(-\infty+\infty\)" is undefined (Bemerkung 9.4). Up to the factor \(\frac1{\sqrt{2\pi}}\) this integral is the expected value of a standard normal random variable — 0, as symmetry demands.
!What the exam asks from Chapter 9
Chapter 9 appears in six of the nine exams, most heavily in WiSe24 (four subtasks):
| Type | Exam tasks | What earns the points |
|---|---|---|
| Compute an improper integral | WiSe19 4b, WiSe22 2b, Skript 13.6 | write the limit explicitly; substitution; evaluate the limit |
| Not improper at all | WiSe24 6a | \(\int_0^1x^{2024}e^{-x^{2025}}dx=\frac{1-e^{-1}}{2025}\) — an ordinary Chapter-8 substitution sitting in the middle of the improper-integral block. Check the interval and the integrand before reaching for a limit |
| For which parameter? | WiSe20 5b | reduce to the benchmark \(\int y^{-a}\) |
| Does it exist? (trap) | WiSe24 6c | find every point where the integrand can blow up |
| Integral test | WiSe24 6d; companion: SoSe23 A4e | check positive + decreasing; substitute \(y=\ln x\) or \(\ln\ln x\). SoSe23 A4e is the neighbouring series \(\sum\frac1{n\ln n}\), which diverges, and the exam has you show it with the mean value theorem instead |
| Limits and integrals | WiSe24 6b, WiSe24 2b | pointwise limit first, then a moving point \(x_n\) to kill uniformity or to show \(\int f_n\not\to\int f\) |
| Prove uniform convergence | SoSe19 A3 | the opposite direction: produce a constant majorant \(b_k\) with \(\sum b_k<\infty\) (here \(b_k=\frac1{k^2}\)) — the Weierstrass test, no moving point involved |
Drill: does it converge?
- For the exam: the five task types above; the benchmarks \(\int_1^\infty x^{-\alpha}\), \(\int_0^1x^{-\alpha}\) and \(\sum k^{-s}\) by heart.
- For Chapter 10: \(\ln(1+x)=\sum(-1)^k\frac{x^{k+1}}{k+1}\) and \(\frac1{1-x}=\sum x^k\) are the model power series; term-by-term integration (Satz 10.10c) is Satz 9.10 again; "radius of convergence" was introduced here (Def. 9.12).
- For Chapter 11: partial derivatives \(\partial_yf\) (§9.2) become the main tool.
ΣSummary sheet
- Improper integral = limit of proper integrals; split at an interior point if both ends are critical; each part must converge separately.
- \(\int_1^\infty x^{-\alpha}<\infty\iff\alpha>1\); \(\int_0^1x^{-\alpha}<\infty\iff\alpha<1\). Compare other integrands with these.
- Continuous \(f(x,y)\): Fubini; \(\frac d{dy}\int f\,dx=\int\partial_yf\,dx\).
- \(\int\sum=\sum\int\) with a convergent majorant; otherwise it can fail (spikes). \(\ln(1+x)=\sum(-1)^k\frac{x^{k+1}}{k+1}\) on \((-1,1]\).
- Monotone \(f\): \(\sum f(k)\) is sandwiched by integrals. \(H_n\simeq\ln n\), \(\sum k^\alpha\simeq\frac{n^{\alpha+1}}{\alpha+1}\), \(\ln n!\simeq n\ln n\).
- Integral test: for decreasing \(f\ge0\), \(\sum f(k)\) and \(\int_1^\infty f\) converge together.
?Self-test
Power SeriesPotenzreihen
The tangent line (Chapter 6) approximates \(f\) to first order. Taylor polynomials go further: they approximate \(f\) near a point by a polynomial of degree \(n\), with an explicit error term. Letting \(n\to\infty\) gives power series, "infinite polynomials" that converge inside a radius of convergence and can be differentiated and integrated term by term.
Source labels: Skript 10.1 = stated in your lecture script · Blatt 14 · 14.2 = exercise sheet · Klausur SoSe23 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§10.1 = Section 10.1) rather than a numbered result (10.1 = Satz 10.1).
0The big picture
- Known power series so far: \(e^x=\sum\frac{x^k}{k!}\) (all \(x\)) and \(\ln(1+x)=\sum(-1)^k\frac{x^{k+1}}{k+1}\) — these two are the script's own list at the head of the chapter — plus the geometric series \(\frac1{1-x}=\sum x^k\) (\(|x|<1\)) from Chapter 3.
- Why they matter: truncating gives polynomial approximations (numerics); conversely, a closed formula for \(\sum a_kx^k\) (a generating function) tells you things about the \(a_k\) (combinatorics, probability).
- 10.1 Taylor's formula: \(f(a+h)=\sum_{k\le n}\frac{f^{(k)}(a)}{k!}h^k+\text{remainder}\). Taylor series = limit, when the remainder → 0.
- 10.2 Every power series has a radius \(r\): absolutely convergent for \(|x|<r\), divergent for \(|x|>r\). Inside, differentiate/integrate term by term.
- Higher derivatives and Lemma 7.22 (\(a_l=\frac{f^{(l)}(0)}{l!}\) for polynomials) (§7.5); l’Hospital (§7.4); \(o(h^n)\) notation (§6.1).
- Integration by parts and the mean value theorem of integration (§8.4, §8.2) — they produce the two remainder formulas.
- Series: geometric series, ratio test, Cauchy product (§3.1, §3.2, §3.3); \(\limsup\) = largest accumulation point (§4.3).
- Power series are not new here. The script already defined them, together with the words Potenzreihe and Konvergenzradius, in Definition 9.12, and proved \(\ln(1+x)=\sum(-1)^k\frac{x^{k+1}}{k+1}\) on \((-1,1]\) in Lemma 9.13. What Chapter 10 adds is (i) where the coefficients come from — Taylor — and (ii) that the radius always exists, with a formula for it (Satz 10.7).
10.1Taylor's formula [Taylorsche Formel]
Where the coefficients come from. Suppose \(f(a+h)=f(a)+f'(a)h+a_2h^2+\dots+a_nh^n+o(h^n)\). Then each \(a_k\) is a limit of the form \(\frac{f(a+h)-(\dots)}{h^k}\), which is "\(\frac00\)". Applying l’Hospital \(k\) times gives \(a_k=\frac{f^{(k)}(a)}{k!}\). This matches Lemma 7.22 for polynomials: the coefficients are the derivatives at the centre, divided by factorials.
A Taylor polynomial is built around a centre [Entwicklungspunkt] \(a\). The script writes the input either as \(a+h\) (\(h\) = distance from the centre) or as \(x\) (then \(h=x-a\)); both mean the same. Degree by degree: \(n=0\) gives the constant \(f(a)\), \(n=1\) the tangent \(f(a)+f'(a)h\) (Chapter 6), \(n=2\) the parabola that also matches the curvature \(f''(a)\), and so on — each new term matches one more derivative at \(a\).
If \(f\) is \(n\) times continuously differentiable near \(a\), then \[f(a+h)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}h^k+o(h^n)\qquad(h\to0).\] \(T_{n,a}f(h)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}h^k\) is the \(n\)-th Taylor polynomial [Taylorpolynom] of \(f\) at \(a\). (For \(n=1\): the tangent.)
Where the coefficients come from, and why the error is \(o(h^n)\)
Finding the coefficients. Suppose such an approximation \(f(a+h)=f(a)+f'(a)h+a_2h^2+\dots+a_nh^n+o(h^n)\) exists. Solve for \(a_2\) and let \(h\to0\): \[a_2=\lim_{h\to0}\frac{f(a+h)-f(a)-f'(a)h}{h^2}\overset{\text{l'Hospital}}{=}\lim_{h\to0}\frac{f'(a+h)-f'(a)}{2h}=\frac{f''(a)}2,\] where l'Hospital (Satz 7.15) applies because numerator and denominator both tend to 0, and the last step is the difference quotient of \(f'\) at \(a\). Repeating this \(k\) times gives \(a_k=\frac{f^{(k)}(a)}{k!}\) — the factor \(k!\) appears because differentiating \(h^k\) that often produces \(k!\). Sanity check: for a polynomial this is exactly Lemma 7.22, \(a_k=\frac1{k!}f^{(k)}(0)\), and there the "approximation" is exact.
Why the error really is \(o(h^n)\). This follows from the version with remainder below: the Lagrange form gives \(R_{n+1}=\frac{f^{(n+1)}(\xi)}{(n+1)!}h^{n+1}\), and \(h^{n+1}=o(h^n)\). (Satz 10.1 assumes only \(n\) continuous derivatives, one less than Satz 10.2; in that case one applies the same argument to \(T_{n-1}\) and uses continuity of \(f^{(n)}\) at \(a\).)
If \(f\) is \((n+1)\) times continuously differentiable on an open interval \(I\), then for all \(x,a\in I\): \[f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+R_{n+1}f(a,x),\] \[R_{n+1}f(a,x)=\frac1{n!}\int_a^x(x-t)^nf^{(n+1)}(t)\,dt\ \ \text{(Cauchy)}\ =\ \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\ \ \text{(Lagrange)}\] for some \(\xi\) between \(a\) and \(x\).
Proof
Cauchy form, induction on \(n\). \(n=0\): \(f(x)=f(a)+\int_a^xf'(t)\,dt\) (FTC). Step: integrate the remainder by parts with \(u=f^{(n)}(t)\), \(v'=(x-t)^{n-1}\), \(v=-\frac1n(x-t)^n\): \[\frac1{(n-1)!}\int_a^x(x-t)^{n-1}f^{(n)}(t)\,dt=\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac1{n!}\int_a^x(x-t)^nf^{(n+1)}(t)\,dt.\] The first term is the next Taylor term; the second is the new remainder.
Lagrange form. \((x-t)^n\) has a fixed sign for \(t\) between \(a\) and \(x\), so the MVT of integration (Satz 8.4) with weight \(p(t)=(x-t)^n\) pulls out \(f^{(n+1)}(\xi)\); and \(\int_a^x(x-t)^n\,dt=\frac{(x-a)^{n+1}}{n+1}\) (substitute \(u=x-t\)). ∎
Extra — the sign, carefully Satz 8.4 is stated for a weight \(p\ge0\). For \(a<x\) that is fine, since \(x-t\ge0\). For \(x<a\) and odd \(n\) the weight is \(\le0\), so apply the theorem to \(-p\) instead: the same \(\xi\) comes out, because multiplying both sides by \(-1\) changes nothing about which \(\xi\) exists. This is why the statement only needs "\(\xi\) between \(a\) and \(x\)" and never says which of the two is larger.
How to use Lagrange: the unknown \(\xi\) is harmless if you bound \(|f^{(n+1)}|\) on the whole interval: \(|R_{n+1}|\le\frac{\max|f^{(n+1)}|}{(n+1)!}|x-a|^{n+1}\).
All derivatives of \(e^x\) are \(e^x\), all equal to 1 at 0. Lagrange: \(e^x=\sum_{k=0}^n\frac{x^k}{k!}+\frac{e^\xi}{(n+1)!}x^{n+1}\). At \(x=1\), \(\xi\in[0,1]\) and \(1\le e^\xi\le3\): \[\sum_{k=0}^n\frac1{k!}+\frac1{(n+1)!}\le e\le\sum_{k=0}^n\frac1{k!}+\frac3{(n+1)!}.\] With \(n=10\) the uncertainty is below \(10^{-7}\).
\(f(x)=e^{x^2}\). (a) \(T_{5,0}f\). (b) A constant \(C\) with \(|f(x)-T_{5,0}f(x)|\le C\) on \([-1,1]\).
Solution (two ways)
(a) Fast way: substitute \(x^2\) into the exp series: \(e^{x^2}=\sum\frac{x^{2k}}{k!}=1+x^2+\frac{x^4}2+\frac{x^6}6+\dots\) Keep the terms of degree ≤ 5: \(T_{5,0}f=1+x^2+\frac12x^4\). (By uniqueness of power series coefficients, Satz 10.10b, this is the Taylor polynomial.) The direct way computes \(f'(0)=0,f''(0)=2,f'''(0)=0,f^{(4)}(0)=12,f^{(5)}(0)=0\) and gets the same.
(b) With \(y=x^2\in[0,1]\) and the Lagrange remainder of \(e^y\) of order 3: \(e^y=1+y+\frac{y^2}2+\frac{e^\xi}{6}y^3\), so \(|f(x)-T_5(x)|=\frac{e^\xi}6x^6\le\frac36=\frac12\). (The direct route via \(f^{(6)}\) gives the weaker \(C=\frac{173}{30}\). Both are valid answers.)
If \(f\in C^\infty\) and \(R_nf(a,x)\to0\), then \(f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k\), the Taylor series [Taylorreihe].
The condition \(R_n\to0\) cannot be dropped (Skript 10.5): \(f(x)=e^{-1/x^2}\) (\(f(0)=0\)) is \(C^\infty\) with \(f^{(n)}(0)=0\) for all \(n\). Its Taylor series at 0 is the zero series, but \(f(x)>0\) for \(x\neq0\). The function is too flat at 0 for any polynomial to "see" it. (The script computes \(f'(0)=f''(0)=0\) by the difference quotient: e.g. \(f'(0)=\lim_{h\to0}\frac1he^{-1/h^2}=\lim_{y\to\infty}\frac{\pm y}{e^{y^2}}=0\) with \(y=\frac1h\); the higher derivatives work the same way, because every derivative is a polynomial in \(\frac1x\) times \(e^{-1/x^2}\), and the exponential wins.)
For \(a\in\mathbb R\), \(f(x)=(1+x)^a\) has \(\frac{f^{(k)}(0)}{k!}=\frac{a(a-1)\cdots(a-k+1)}{k!}=:\binom ak\), and for \(|x|<1\): \[(1+x)^a=\sum_{k=0}^\infty\binom akx^k.\] For \(a=n\in\mathbb N\), \(\binom nk=0\) for \(k>n\): the ordinary binomial theorem. Example: \(\sqrt{1+x}=1+\frac12x-\frac18x^2+\dots\) (\(\binom{1/2}2=\frac{\frac12\cdot(-\frac12)}2=-\frac18\)).
That the remainder tends to 0 for \(|x|<1\) is stated in the script without proof ("man kann zeigen"). You only need the formula and its range.
WiSe19 2a: Taylor series of \(e^x\) at 0: \(e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\), \(x\in\mathbb R\).
SoSe23 1e: \(T_{20,0}f\) for \(f(x)=\ln(1+x^2)\)
Substitute \(x^2\) into \(\ln(1+y)=\sum_{k\ge1}(-1)^{k+1}\frac{y^k}k\): \(\ln(1+x^2)=\sum_{k\ge1}(-1)^{k+1}\frac{x^{2k}}k\). Degree ≤ 20 means \(k\le10\): \[T_{20,0}f(x)=\sum_{k=1}^{10}(-1)^{k+1}\frac{x^{2k}}k=x^2-\frac{x^4}2+\frac{x^6}3-\dots-\frac{x^{20}}{10}.\] Never differentiate 20 times; substitute into a known series.
SoSe23 1f: value of \(\sum_{k\ge0}2\binom{1/3}k(-8)^{-k}x^k\)
Binomial series with \(a=\frac13\) and \(-\frac x8\) in place of \(x\): \(2\left(1-\frac x8\right)^{1/3}=2\cdot\frac{\sqrt[3]{8-x}}{\sqrt[3]8}=\sqrt[3]{8-x}\), valid for \(|x|<8\).
WiSe23 6b: Taylor series of \(f(x)=\frac13x^3+x+\frac23\) at \(x_0=1\), written out
\(f'=x^2+1\), \(f''=2x\), \(f'''=2\), higher = 0. At 1: \(f=2,f'=2,f''=2,f'''=2\). So \[T_{x_0}f(x)=2+2(x-1)+\frac22(x-1)^2+\frac26(x-1)^3.\] Expanding gives back \(\frac13x^3+x+\frac23\): the Taylor series of a polynomial is the polynomial, just re-centred.
Watch the notation The exam writes \(T_{x_0}f\) for the whole Taylor series about \(x_0\), with only the centre in the subscript. This guide (and the script, Satz 10.1) writes \(T_{n,a}f\) with the degree first, so the same object here is \(T_{3,1}f\) — everything beyond degree 3 vanishes. When a task uses a subscript, check from its own definition whether it means a degree or a centre before you start.
What you must know for 10.1: Taylor's formula including the Cauchy and Lagrange remainders; when a function can be expanded into its Taylor series; the binomial series.
10.2Radius of convergence and analytic functions
For every power series \(\sum a_kz^k\) there is \(r\in[0,\infty]\), the radius of convergence, such that the series converges absolutely for \(|z|<r\) and diverges for \(|z|>r\). It is \[r=\frac1{\limsup_{k\to\infty}\sqrt[k]{|a_k|}}\qquad\left(\tfrac10=\infty,\ \tfrac1\infty=0\right).\] Nothing is said about \(|z|=r\): the boundary points must be checked separately.
\(\limsup_k\sqrt[k]{|a_k|}\) is the largest accumulation point of the sequence \(\sqrt[k]{|a_k|}\) (Chapter 4); if the sequence converges, it is simply the limit.
What is actually new The name "radius of convergence" was already given in Definition 9.12, as the largest bound up to which a power-series expansion is valid. Satz 10.7 is what makes that definition safe: it says such a radius exists for every power series — the set of \(x\) where it converges is always an interval around 0, never something stranger — and it hands you a formula to compute it.
Why a radius exists — the idea Extra — the script states 10.7 without proof
Suppose \(\sqrt[k]{|a_k|}\to L\) (the common case). Then \(\sqrt[k]{|a_kz^k|}=\sqrt[k]{|a_k|}\,|z|\to L|z|\). If \(L|z|<1\), pick \(q\) with \(L|z|<q<1\); eventually \(|a_kz^k|\le q^k\), a convergent geometric majorant (Satz 3.7) — absolute convergence. If \(L|z|>1\), eventually \(|a_kz^k|\ge1\), so the terms do not tend to 0 (Lemma 3.4) — divergence. The border is \(|z|=\frac1L=r\). With lim sup instead of lim the same argument works for the largest accumulation point.
Ratio shortcut: if \(\left|\frac{a_{k+1}}{a_k}\right|\to q\), the ratio test (Satz 3.13) applied to \(\sum a_kz^k\) gives convergence for \(q|z|<1\) and divergence for \(q|z|>1\), so \(r=\frac1q\) (with \(\frac10=\infty\)).
- Ratio test on the terms (often easiest): \(\left|\frac{a_{k+1}x^{k+1}}{a_kx^k}\right|\to q|x|\). Converges for \(q|x|<1\), so \(r=\frac1q\). (Or root test / Cauchy–Hadamard with \(\sqrt[k]{|a_k|}\); useful fact: \(\sqrt[k]k\to1\).)
- Substitution if the series is geometric in disguise: \(\sum(3x)^k\) is geometric in \(y=3x\).
- Check both endpoints \(x=\pm r\) by hand (Leibniz, harmonic, terms ↛ 0, …).
(a) For each series, give the radius of convergence \(r\) and the full convergence set \(D\) — so the endpoints have to be decided separately. (Part (b) of the same task asks for the value of \(\sum_{k\ge1}\frac k{5^k}\) and is done below, with the differentiated geometric series.)
| series | r | D | boundary |
|---|---|---|---|
| \(\sum_{k\ge1}\frac{(-x)^k}k\) | 1 | \((-1,1]\) | \(x=1\): \(\sum\frac{(-1)^k}k\) Leibniz ✓; \(x=-1\): harmonic ✗ |
| \(\sum_{k\ge0}(3x)^k\) | \(\frac13\) | \((-\frac13,\frac13)\) | terms \((\pm1)^k\not\to0\) ✗ |
| \(\sum_{k\ge1}\frac{x^{2k}}{3^k}\) | \(\sqrt3\) | \((-\sqrt3,\sqrt3)\) | terms \(=1\) ✗ (ratio \(\frac{x^2}3<1\)) |
| \(\sum_{k\ge0}k!\,x^k\) | 0 | \(\{0\}\) | ratio \((k+1)|x|\to\infty\) for \(x\neq0\) |
The geometric series \(\sum x^k\) has \(a_k=1\), so \(\sqrt[k]{|a_k|}=1\) and \(r=1\).
\(f\) is analytic at 0 [analytisch] if \(f(x)=\sum a_kx^k\) for \(|x|<r\) with some \(r>0\). (\(e^{-1/x^2}\) is \(C^\infty\) but not analytic at 0.)
Let \(f(x)=\sum a_kx^k\) (\(|x|<r_f\)), \(g(x)=\sum b_kx^k\) (\(|x|<r_g\)).
- Differentiate term by term: \(f'(x)=\sum_{k\ge1}ka_kx^{k-1}\) for \(|x|<r_f\).
- Coefficients are unique: \(a_k=\frac{f^{(k)}(0)}{k!}\). (So any way of finding a power series, e.g. substitution, gives the Taylor series.)
- Integrate term by term: \(\int_0^xf=\sum\frac{a_k}{k+1}x^{k+1}\).
- \(f+g\) and \(fg\) are analytic, with \(\sum(a_k+b_k)x^k\) and the Cauchy product \(\sum_k\left(\sum_{j\le k}a_jb_{k-j}\right)x^k\), for \(|x|<\min(r_f,r_g)\).
Stated without proof in the script. Useful facts behind it: the differentiated and integrated series have the same radius (since \(\sqrt[k]k\to1\)); (b) follows from (a) by differentiating \(k\) times and setting \(x=0\) — exactly as in Lemma 7.22; (c) is Satz 9.10 with the majorant \(|a_k|\rho^k\) for \(|x|\le\rho<r\); (d) is the Cauchy product, Satz 3.23.
For \(|x|<1\): \(\frac1{(1-x)^2}=\left(\frac1{1-x}\right)'=\sum_{k\ge1}kx^{k-1}\), \(\ \frac2{(1-x)^3}=\sum_{k\ge2}k(k-1)x^{k-2}\).
- Blatt 14.2b: multiply by \(x\): \(\sum_{k\ge1}kx^k=\frac x{(1-x)^2}\). At \(x=\frac15\): \(\sum\frac k{5^k}=\frac{1/5}{(4/5)^2}=\frac5{16}\).
- WiSe19 1e: \(\sum_{n\ge1}n3^nx^n=\sum n(3x)^n=\frac{3x}{(1-3x)^2}\), converging iff \(|3x|<1\): \(M=(-\frac13,\frac13)\).
- Geometric distribution Skript, Anwendung after 10.11 — marked ausgelassen: \(P(X=k)=p(1-p)^{k-1}\) (first success at trial \(k\)). \(E[X]=\sum kp(1-p)^{k-1}=\frac p{(1-(1-p))^2}=\frac1p\). On average \(\frac1p\) trials until the first success.
Note That last bullet is the one place in this chapter where the script writes "ausgelassen" in the margin — the lecture skipped it, and it is not in the Minimale Inhalte. Keep it as a sanity check that you can differentiate a power series and as motivation (it is the same computation you will meet in Diskrete Wahrscheinlichkeitstheorie), but do not expect it in the exam. The two bullets above it are examinable.
WiSe24 1b: radius of convergence of \(\sum_{k\ge1}k!\,e^{-k^2}x^k\)
Ratio: \(\frac{(k+1)!e^{-(k+1)^2}}{k!e^{-k^2}}|x|=(k+1)e^{-2k-1}|x|\to0\) for every \(x\) (exp beats polynomial). So \(r=\infty\).
WiSe22 1c/1d: determine \(M_1,M_2\) and the value of the series on them
Both parts read "Bestimmen Sie die Menge \(M\)" of all \(x\) for which the series converges, and its value there. Recognising the series is the whole task — no criterion is needed.
1c: \(S_1(x)=\sum_{k\ge0}4^{2k}x^k=\sum_{k\ge0}(16x)^k\), since \(4^{2k}=16^k\). Geometric, so it converges iff \(|16x|<1\): \[M_1=\left(-\tfrac1{16},\tfrac1{16}\right),\qquad S_1(x)=\frac1{1-16x}.\]
1d: \(S_2(x)=\sum_{k\ge1}\frac{x^{2k}}{5^kk!}=\sum_{k\ge1}\frac{\left(\frac{x^2}5\right)^k}{k!}\) — the exponential series in the variable \(y=\frac{x^2}5\), with the \(k=0\) term (which is 1) missing: \[M_2=\mathbb R,\qquad S_2(x)=e^{x^2/5}-1.\] The exponential series converges for every real argument, and \(\frac{x^2}5\) is a real number for every \(x\), so there is no restriction at all.
Extra — the trick both parts share Each series is a known series in a substituted variable: \(16x\) in 1c, \(\frac{x^2}5\) in 1d. So the method is: force the series into the shape \(\sum u^k\) or \(\sum\frac{u^k}{k!}\), read off the value from the known formula, and translate the condition on \(u\) back into a condition on \(x\). Watch the starting index — here \(k\ge1\), which costs exactly the \(k=0\) term.
What you must know for 10.2: what the radius of convergence is and how to determine it; the derivative, an antiderivative and the coefficients of a power series.
!What the exam asks from Chapter 10
Chapter 10 appears in five of the nine exams, almost always as short answer-only items:
| Type | Exam tasks | What earns the points |
|---|---|---|
| Taylor polynomial of a composite | SoSe23 1e, Blatt 13.3 | substitute into a known series; cut at the right degree |
| Taylor series written out / re-centred | WiSe19 2a, WiSe23 6b | derivatives at the centre, divide by \(k!\) |
| Value of a power series | SoSe23 1f, WiSe19 1e, WiSe22 1c/1d | recognise a known series in a substituted variable — \(16x\), \(\frac{x^2}5\), \(3x\), \(-\frac x8\) — possibly differentiated; then translate the condition back to \(x\), and mind the starting index |
| Radius of convergence | WiSe24 1b, Blatt 14.2 | ratio of consecutive coefficients; boundary by hand |
| Error estimate | Blatt 13.3b, Skript 10.3 | Lagrange remainder with a bound for \(|f^{(n+1)}|\) |
Drill: Taylor polynomials, values, radii
- For the exam: the five task types above, and the table of standard series in the summary by heart.
- For Chapter 11: the second-order Taylor expansion in several variables, \(f(x+h)\approx f(x)+\nabla f(x)^th+\frac12h^tH_f(x)h\), is why the Hessian decides extrema — the same idea as \(f''(x_0)\) in one variable.
- For Chapter 12: \(e^{\lambda x}\) with complex \(\lambda\) (Euler's formula) produces the oscillating solutions of second-order ODEs.
ΣSummary sheet
| function | series at 0 | valid for |
|---|---|---|
| \(e^x\) | \(\sum\frac{x^k}{k!}\) | ℝ |
| \(\sin x\), \(\cos x\) | \(\sum(-1)^k\frac{x^{2k+1}}{(2k+1)!}\), \(\sum(-1)^k\frac{x^{2k}}{(2k)!}\) | ℝ |
| \(\frac1{1-x}\) | \(\sum x^k\) | \(|x|<1\) |
| \(\frac1{(1-x)^2}\) | \(\sum_{k\ge1}kx^{k-1}\) | \(|x|<1\) |
| \(\ln(1+x)\) | \(\sum_{k\ge1}(-1)^{k+1}\frac{x^k}k\) | \((-1,1]\) — that \(x=1\) is included is Lemma 9.13 |
| \((1+x)^a\) | \(\sum\binom akx^k\) | \(|x|<1\) |
- \(T_{n,a}f(x)=\sum_{k\le n}\frac{f^{(k)}(a)}{k!}(x-a)^k\); error \(o((x-a)^n)\), or exactly \(\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\) (Lagrange).
- To get Taylor polynomials of composite functions, substitute into known series (uniqueness makes it legitimate).
- Radius via ratio or root test; boundary points by hand.
- Inside the radius: differentiate/integrate term by term.
?Self-test
Several VariablesDifferentialrechnung mehrerer Veränderlicher
Functions of several inputs, like \(f(x,y)\). Differentiate in one variable at a time (partial derivatives), collect them in the gradient, the second derivatives in the Hessian matrix, and use both to find and classify extrema. This extrema task appears in most recent exams.
Source labels: Skript 11.6 = stated in your lecture script · Blatt 14 · 14.3 = exercise sheet · Klausur SoSe23 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§11.3 = Section 11.3) rather than a numbered result (11.3 = Definition 11.3).
0The big picture
Everything is the one-variable theory with vectors and matrices in place of numbers:
| one variable | several variables |
|---|---|
| \(f'(x)\) | gradient \(\nabla f(x)\) (vector of partial derivatives) |
| tangent line | tangent plane \(f(x_0)+Df(x_0)(x-x_0)\) |
| \(f''(x)\) | Hessian \(Hf(x)\) (matrix of second partials) |
| \(f'(x_0)=0\) | \(\nabla f(x_0)=0\) (critical point) |
| \(f''(x_0)>0\) ⇒ min | \(Hf(x_0)\) positive definite (all eigenvalues \(>0\)) ⇒ min |
| — | saddle point: eigenvalues of both signs |
Roadmap. §11.1 treats functions out of one variable into \(\mathbb R^n\) (curves): differentiate each component. §11.2 treats functions from \(\mathbb R^n\) into \(\mathbb R\): differentiate in one variable at a time, and Lemma 11.7 says these partial derivatives together give the best linear approximation. §11.3 goes to second order (Hessian) — which is exactly what decides extrema, just as \(f''\) did in Chapter 7 — and ends with maps \(\mathbb R^n\to\mathbb R^m\) (Jacobian), where everything is combined.
- All one-variable derivative rules — product, chain, exp/ln rewriting (§6.3) — and \(o(\cdot)\) notation (§6.1). Partial derivatives are nothing but these rules applied with the other variables frozen.
- Local extrema in one variable: \(f'(x_0)=0\) is necessary, \(f''(x_0)\gtrless0\) sufficient (§7.1, §7.5); the mean value theorem (§7.2), used in the proof of Lemma 11.7.
- The second-order Taylor formula (§10.1): its several-variable version is why the Hessian decides.
- Length of a vector \(\|x\|=\sqrt{x_1^2+\dots+x_n^2}\) and the scalar product (norms, Chapter 1).
- Matrices, eigenvalues and definiteness — not in the script's earlier chapters, so they are explained from scratch right below.
The linear algebra you need here Extra
The script assumes matrices, eigenvalues and definiteness are known. Here is everything this chapter actually uses, from scratch. If it is familiar, skip ahead.
A vector \(x\in\mathbb R^n\) is a list of \(n\) numbers; we write it as a column. The transpose [Transponierte] \(x^t\) is the same list written as a row. A matrix \(A\in\mathbb R^{n\times n}\) is a square table of numbers \(A_{ij}\) (row \(i\), column \(j\)); \(A^t\) is the table mirrored along its diagonal, and \(A\) is symmetric [symmetrisch] if \(A=A^t\), i.e. \(A_{ij}=A_{ji}\).
Matrix times vector: \((Ax)_i=\sum_jA_{ij}x_j\) — row \(i\) of \(A\), multiplied entrywise with \(x\) and summed. The number \(x^tAx=\sum_{i,j}x_iA_{ij}x_j\) is a quadratic form [quadratische Form]. For \(n=2\), with \(A=\begin{pmatrix}a&b\\b&c\end{pmatrix}\): \[\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\b&c\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+2bxy+cy^2.\] So a quadratic form is just a homogeneous quadratic polynomial, and \(x^tAx\) is the several-variable version of \(ax^2\). That is why the Hessian plays the role that \(f''\) plays in one variable.
\(\lambda\in\mathbb R\) is an eigenvalue [Eigenwert] of \(A\) if there is a vector \(v\neq0\) (an eigenvector) with \[Av=\lambda v.\] In words: \(v\) is a direction that \(A\) only stretches (by the factor \(\lambda\)) instead of turning. Eigenvalues are the roots of the characteristic polynomial \(\det(A-\lambda I)=0\). For \(2\times2\) this is \[\lambda^2-\underbrace{(a+c)}_{\text{trace}}\lambda+\underbrace{(ac-b^2)}_{\det}=0,\qquad \lambda_{1,2}=\frac{a+c}2\pm\sqrt{\left(\frac{a+c}2\right)^2-(ac-b^2)}.\] A symmetric matrix always has real eigenvalues, and one can choose eigenvectors that are perpendicular to each other. Note \(\lambda_1\lambda_2=\det A\) and \(\lambda_1+\lambda_2=a+c\) — this is what makes the shortcut below work.
Example: \(A=\begin{pmatrix}2&0\\0&4\end{pmatrix}\) has eigenvalues 2 and 4 (the coordinate directions are stretched by 2 and 4). \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\): \(\lambda^2-4\lambda+3=0\), so \(\lambda=1,3\).
A symmetric \(A\) is called
- positive definite if \(x^tAx>0\) for every \(x\neq0\) (a bowl opening upwards) ⇔ all eigenvalues \(>0\);
- negative definite if \(x^tAx<0\) for every \(x\neq0\) ⇔ all eigenvalues \(<0\);
- indefinite if \(x^tAx\) takes both signs ⇔ eigenvalues of both signs (a saddle);
- semidefinite if \(\ge0\) (resp. \(\le0\)) everywhere, with 0 occurring ⇔ all eigenvalues \(\ge0\) (resp. \(\le0\)) and some eigenvalue is 0.
Why the eigenvalues decide: write \(x\) in a perpendicular eigenvector basis, \(x=t_1v_1+t_2v_2\) with \(\|v_i\|=1\). Then \(Ax=\lambda_1t_1v_1+\lambda_2t_2v_2\) and \[x^tAx=\lambda_1t_1^2+\lambda_2t_2^2.\] The signs of \(\lambda_1,\lambda_2\) are exactly the signs this expression can take. With \(\lambda_1,\lambda_2>0\) it is a sum of positive terms; with opposite signs you get \(+\) along \(v_1\) and \(-\) along \(v_2\) — a saddle.
For \(H=\begin{pmatrix}a&b\\b&c\end{pmatrix}\) compute \(\det H=ac-b^2\) and look at \(a\):
| \(\det H\) | \(a\) | conclusion at a critical point |
|---|---|---|
| \(>0\) | \(>0\) | positive definite ⇒ strict local minimum |
| \(>0\) | \(<0\) | negative definite ⇒ strict local maximum |
| \(<0\) | any | indefinite ⇒ saddle point |
| \(=0\) | any | an eigenvalue is 0 ⇒ test says nothing, look at \(f\) directly |
Why: \(\det H=\lambda_1\lambda_2\), so \(\det H>0\) means equal signs and \(\det H<0\) opposite signs; with equal signs, the sign of \(a=e_1^tHe_1\) tells you which. You may always compute the eigenvalues instead — the exam solutions usually do — but the determinant test is faster and harder to get wrong.
Three or more variables (e.g. WiSe24 5a has \(x_1,x_2,x_3\)): there is no one-line determinant test in this course; compute the eigenvalues. When the Hessian is diagonal — common in exams, whenever \(f\) is a sum of one-variable functions like \(x^3+y^4\) — the eigenvalues are simply the diagonal entries.
11.1Curves [Kurven]
A parametrised curve is a map \(\gamma:[a,b]\to\mathbb R^n,\ t\mapsto(\gamma_1(t),\dots,\gamma_n(t))\) with continuous components; think of \(\gamma(t)\) as the position of a moving point at time \(t\). The set \(\gamma([a,b])\) is its trace [Spur].
If differentiable: \(\gamma'(t)=(\gamma_1'(t),\dots,\gamma_n'(t))\) is the tangent / velocity vector, \(\|\gamma'(t)\|\) the speed, and \(T_\gamma(t)=\frac{\gamma'(t)}{\|\gamma'(t)\|}\) (if \(\gamma'(t)\neq0\)) the unit tangent vector.
- \(\alpha(t)=(\cos t,\sin t)\) and \(\beta(t)=(\cos t,-\sin t)\), \(t\in[0,2\pi]\): same trace (unit circle), different curves (opposite direction). \(\alpha'(t)=(-\sin t,\cos t)\) has length 1 and is perpendicular to \(\alpha(t)\) (\(\langle\alpha,\alpha'\rangle=0\)).
- Graph of \(f\) as a curve: \(\gamma(t)=(t,f(t))\), \(T_\gamma(t)=\frac{(1,f'(t))}{\sqrt{1+f'(t)^2}}\).
- Extra A computed example: the helix \(\gamma(t)=(\cos t,\sin t,t)\) has \(\gamma'(t)=(-\sin t,\cos t,1)\), speed \(\|\gamma'(t)\|=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2\), and \(T_\gamma(t)=\frac1{\sqrt2}(-\sin t,\cos t,1)\).
- Caution: \(\gamma(t_1)=\gamma(t_2)\) does not imply \(\gamma'(t_1)=\gamma'(t_2)\) (a curve can pass through a point twice in different directions).
What you must know for 11.1: what a curve is; determining its tangent vector.
11.2Partial derivatives [Partielle Ableitungen]
For \(f:\mathbb R^n\to\mathbb R\), the partial derivative \(\frac{\partial f}{\partial x_k}(x)=\partial_kf(x)\) is the ordinary derivative of \(\xi\mapsto f(x_1,\dots,x_{k-1},\xi,x_{k+1},\dots,x_n)\) at \(\xi=x_k\): differentiate in \(x_k\), treat all other variables as constants. The gradient is the column vector \(\nabla f(x)=\operatorname{grad}f(x)=(\partial_1f(x),\dots,\partial_nf(x))^t\); \(Df(x)=\nabla f(x)^t\) is the same as a row.
Extra Tiny example: \(f(x,y)=x^2y\). For \(\partial_xf\) pretend \(y\) is a number like 5: \((5x^2)'=10x\), so \(\partial_xf=2xy\). For \(\partial_yf\) pretend \(x\) is a number: \(\partial_yf=x^2\). So \(\nabla f(x,y)=(2xy,\ x^2)^t\), and at \((1,3)\): \(\nabla f=(6,1)^t\).
If \(f\) is continuously partially differentiable, then \[f(x+h)=f(x)+Df(x)\,h+o(\|h\|)\qquad(h\to0\text{ in }\mathbb R^n),\] exactly like the one-dimensional linearisation. The tangent plane to the graph at \((x_0,f(x_0))\) is \(x\mapsto f(x_0)+Df(x_0)(x-x_0)\).
What it says. The partial derivatives only look along the coordinate axes. The lemma says that, if they are continuous, they already determine the behaviour in every direction: the error of the linear approximation is small compared with \(\|h\|\), however \(h\) approaches 0. A function with this property is called (totally) differentiable [(total) differenzierbar] (Definition 11.18 below).
Proof for two variables (the script states the lemma without proof) Extra
Let \(x=(x_1,x_2)\), \(h=(h_1,h_2)\). Go from \(x\) to \(x+h\) in two axis-parallel steps: \[f(x+h)-f(x)=\underbrace{f(x_1+h_1,x_2+h_2)-f(x_1,x_2+h_2)}_{\text{only }x_1\text{ changes}}+\underbrace{f(x_1,x_2+h_2)-f(x_1,x_2)}_{\text{only }x_2\text{ changes}}.\] Each bracket is a difference of a one-variable function, so the mean value theorem (Satz 7.4) gives points \(\xi_1\) between \(x_1\) and \(x_1+h_1\), \(\xi_2\) between \(x_2\) and \(x_2+h_2\) with \[f(x+h)-f(x)=\partial_1f(\xi_1,x_2+h_2)\,h_1+\partial_2f(x_1,\xi_2)\,h_2.\] Subtract \(Df(x)h=\partial_1f(x)h_1+\partial_2f(x)h_2\): \[r(h):=f(x+h)-f(x)-Df(x)h=\big(\partial_1f(\xi_1,x_2+h_2)-\partial_1f(x)\big)h_1+\big(\partial_2f(x_1,\xi_2)-\partial_2f(x)\big)h_2.\] Since \(|h_i|\le\|h\|\), \(\frac{|r(h)|}{\|h\|}\le\big|\partial_1f(\xi_1,x_2+h_2)-\partial_1f(x)\big|+\big|\partial_2f(x_1,\xi_2)-\partial_2f(x)\big|\). As \(h\to0\) the points \((\xi_1,x_2+h_2)\) and \((x_1,\xi_2)\) tend to \(x\), so by continuity of the partial derivatives both terms tend to 0. Hence \(r(h)=o(\|h\|)\). For \(n\) variables, change one coordinate at a time in \(n\) steps; the argument is the same.
Why "continuous" matters — a counterexample Extra
\(f(x,y)=\frac{xy}{x^2+y^2}\) for \((x,y)\neq0\), \(f(0,0)=0\). On both axes \(f=0\), so \(\partial_xf(0,0)=\partial_yf(0,0)=0\) exist. But on the diagonal \(f(t,t)=\frac12\) for every \(t\neq0\): \(f\) is not even continuous at 0, so certainly \(f(h)\neq f(0)+0\cdot h+o(\|h\|)\). The partial derivatives exist but are not continuous at 0, and the lemma does not apply. Existence of partial derivatives alone says little about the other directions.
- Linear \(f(x)=\langle a,x\rangle=\sum a_kx_k\): \(\partial_kf=a_k\), \(\nabla f=a\) (like \((ax)'=a\)); the linearisation is exact.
- Quadratic \(f(x)=x^tAx\) with symmetric \(A\): \(\nabla f(x)=2Ax\) (like \((ax^2)'=2ax\)), and \(f(x+h)=f(x)+2x^tAh+h^tAh\) with \(h^tAh=o(\|h\|)\).
Where \(2x^tAh\) and \(o(\|h\|)\) come from Extra
Multiply out: \((x+h)^tA(x+h)=x^tAx+x^tAh+h^tAx+h^tAh\). The number \(h^tAx\) equals its own transpose \(x^tA^th=x^tAh\) because \(A\) is symmetric, so the middle terms give \(2x^tAh=(2Ax)^th\). This is the linear part, so \(Df(x)=2x^tA\) and \(\nabla f(x)=2Ax\).
The rest is small: \(|h^tAh|=\big|\sum_{i,j}h_iA_{ij}h_j\big|\le\sum_{i,j}|A_{ij}|\,|h_i|\,|h_j|\le C\|h\|^2\) with \(C=\sum_{i,j}|A_{ij}|\), using \(|h_i|\le\|h\|\). Hence \(\frac{|h^tAh|}{\|h\|}\le C\|h\|\to0\): it is \(o(\|h\|)\), even \(O(\|h\|^2)\). Check with \(n=1\): \((x+h)^2a=ax^2+2axh+ah^2\).
- \(f(x,y)=3x^4+xy^2-2\sin(e^y)\): \(\partial_xf=12x^3+y^2\), \(\partial_yf=2xy-2\cos(e^y)e^y\), \(\partial_x\partial_yf=2y\).
- \(f(x,y)=(x^2+5)^y=e^{y\ln(x^2+5)}\): \(\nabla f=\left(\frac{2xy}{x^2+5}(x^2+5)^y,\ \ln(x^2+5)(x^2+5)^y\right)^t\). Variable exponent → rewrite with exp/ln first (Chapter 6).
WiSe19 1g: \(f(x,y)=\sin(3x^2y)\): \(\partial_xf=6xy\cos(3x^2y)\); \(\partial_yf=3x^2\cos(3x^2y)\), so \(\partial_y\partial_yf=-9x^4\sin(3x^2y)\).
WiSe24 5a: \(f=\sin(x_1+x_2^2)e^{x_1+x_3}\): \(\partial_1f=\cos(x_1+x_2^2)e^{x_1+x_3}+\sin(x_1+x_2^2)e^{x_1+x_3}\) (product rule), \(\partial_2f=2x_2\cos(x_1+x_2^2)e^{x_1+x_3}\) (chain rule), \(\partial_3f=\sin(x_1+x_2^2)e^{x_1+x_3}\).
What you must know for 11.2: for \(f:\mathbb R^n\to\mathbb R\), the partial derivatives, grad f, Df and tangent planes; the notion of total differentiability.
11.3Higher derivatives and extrema
Second partial derivatives: \(\partial_j\partial_if=\frac{\partial^2f}{\partial x_j\partial x_i}\). The Hessian matrix [Hesse-Matrix] is \(Hf(x)=\left(\partial_j\partial_if(x)\right)_{i,j}\).
If all \(\partial_i\partial_jf\) are continuous, then \(\partial_i\partial_jf=\partial_j\partial_if\): the order of differentiation does not matter, and the Hessian is symmetric. Example 11.11: \(f=\cos(xy)+\ln x-\ln y\) has \(\partial_x\partial_yf=\partial_y\partial_xf=-\sin(xy)-xy\cos(xy)\).
For a vector field \(v=(v_1,v_2,v_3):\mathbb R^3\to\mathbb R^3\) the rotation [Rotation] is \(\operatorname{rot}(v)=(\partial_yv_3-\partial_zv_2,\ \partial_zv_1-\partial_xv_3,\ \partial_xv_2-\partial_yv_1)^t\). The exam defines it on the spot; you only need to apply the definition.
(a) For \(f\in C^\infty(\mathbb R^3)\) show \(\operatorname{rot}(\nabla f)=0\)
With \(v=\nabla f=(\partial_xf,\partial_yf,\partial_zf)\): \[\operatorname{rot}(\nabla f)=\begin{pmatrix}\partial_y\partial_zf-\partial_z\partial_yf\\\partial_z\partial_xf-\partial_x\partial_zf\\\partial_x\partial_yf-\partial_y\partial_xf\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},\] because \(f\) is at least twice continuously differentiable, so by Schwarz the order of the two partial derivatives may be swapped. Name the theorem and its hypothesis — that is where the points are.
(b) Give \(v_1,v_2,v_3\), polynomials of degree \(\ge2\) in each of \(x,y,z\), with \(\operatorname{rot}(v)=0\)
Use (a) backwards: any gradient field works. Take \(f=x^3y^3z^3\) and \(v=\nabla f=(3x^2y^3z^3,\ 3x^3y^2z^3,\ 3x^3y^3z^2)\). Every component has degree \(\ge2\) in each variable, and \(\operatorname{rot}(v)=0\) by (a) — no need to verify by hand (though e.g. \(\partial_yv_3=9x^3y^2z^2=\partial_zv_2\)).
(c) \(v=\big(\tfrac13z^3+\sin z-2y,\ \tfrac13x^3+\sin x-2z,\ \tfrac13y^3+\cos y-2x\big)\): find \(f\) with \(\nabla f=v\), or show none exists
Compute \(\operatorname{rot}(v)\): \(\partial_yv_3-\partial_zv_2=(y^2-\sin y)-(-2)=y^2-\sin y+2\), and similarly the others give \(z^2+\cos z+2\) and \(x^2+\cos x+2\). Since \(y^2\ge0\) and \(|\sin y|\le1\), the first component is \(\ge1>0\) everywhere. So \(\operatorname{rot}(v)\neq0\). If \(v=\nabla f\) with \(f\in C^\infty\), (a) would give \(\operatorname{rot}(v)=0\) — contradiction. No such \(f\) exists.
Extra The pattern: "rot \(\neq0\) ⇒ not a gradient" is always valid. The converse ("rot \(=0\) ⇒ gradient") is true on all of \(\mathbb R^3\) but is not part of this course.
\(B(x,\varepsilon)=\{y:\|x-y\|<\varepsilon\}\) is the ball of radius ε. \(U\subseteq\mathbb R^n\) is open if around every point of \(U\) some ball lies entirely in \(U\) (no boundary points belong to \(U\)). \((0,1)^2\) is open; \([0,1)^2\) and \([0,1]^2\) are not.
Let \(U\) be open and \(f:U\to\mathbb R\) twice continuously differentiable.
- Necessary: local extremum at \(x\) ⇒ \(\nabla f(x)=0\). At a local min, \(Hf(x)\) is positive semidefinite (all eigenvalues \(\ge0\)); at a local max, negative semidefinite.
- Sufficient: \(\nabla f(x)=0\) and \(Hf(x)\) positive definite (all eigenvalues \(>0\)) ⇒ strict local min. \(\nabla f(x)=0\) and negative definite (all \(<0\)) ⇒ strict local max.
Why this is true (the script omits the proof) Extra
Necessary condition. Fix a direction \(v\neq0\) and walk along the line through \(x\): \(g(t)=f(x+tv)\), a function of one variable, defined for small \(|t|\) because \(U\) is open. If \(f\) has a local extremum at \(x\), then \(g\) has one at \(t=0\), so \(g'(0)=0\) by Chapter 7 (Lemma 7.2). By the chain rule \(g'(0)=\nabla f(x)^tv\). This vanishes for every \(v\), and a vector perpendicular to everything is 0, so \(\nabla f(x)=0\). Differentiating once more, \(g''(0)=v^tHf(x)\,v\); at a local minimum \(g''(0)\ge0\) for every \(v\), which is exactly "positive semidefinite".
Sufficient condition. Taylor in several variables (Lemma 11.7 applied twice) gives, with \(\nabla f(x)=0\), \[f(x+h)=f(x)+\tfrac12h^tHf(x)h+o(\|h\|^2).\] Let \(\lambda_{\min}>0\) be the smallest eigenvalue of \(Hf(x)\). Writing \(h\) in the eigenvector basis as in the primer above, \(h^tHf(x)h\ge\lambda_{\min}\|h\|^2\). Hence \[f(x+h)-f(x)\ge\tfrac12\lambda_{\min}\|h\|^2+o(\|h\|^2)=\|h\|^2\left(\tfrac12\lambda_{\min}+\tfrac{o(\|h\|^2)}{\|h\|^2}\right).\] The bracket tends to \(\tfrac12\lambda_{\min}>0\) as \(h\to0\), so it is positive for all small \(h\neq0\): \(f(x+h)>f(x)\), a strict local minimum. For the maximum apply this to \(-f\).
Why "definite" and not just "semidefinite" is needed. With a zero eigenvalue the quadratic term vanishes in that direction and the \(o(\|h\|^2)\) decides — and it can go either way: \(f=x^4+y^2\) has a minimum at 0, \(f=-x^4+y^2\) a saddle, and both have \(Hf(0)=\operatorname{diag}(0,2)\).
- Find all critical points: solve \(\nabla f(x)=0\) (a system of equations).
- Compute \(Hf\) at each critical point.
- Look at the eigenvalues \(\lambda_i\):
- all \(>0\) ⇒ strict local min; all \(<0\) ⇒ strict local max;
- positive and negative ⇒ neither: saddle point [Sattelpunkt];
- some \(>0\) and some \(=0\) ⇒ no local max (but could be a min); some \(<0\) and some \(=0\) ⇒ no local min.
- Global extrema: check the behaviour as \(\|x\|\to\infty\) or at the boundary (e.g. \(f(x,y)\to\pm\infty\) along a line ⇒ no global max/min).
Extra For a symmetric \(2\times2\) matrix \(\begin{pmatrix}a&b\\b&c\end{pmatrix}\) you don't need to compute eigenvalues: \(\det=\lambda_1\lambda_2\), \(\operatorname{tr}=a+c=\lambda_1+\lambda_2\). \(\det>0,\ a>0\) ⇒ positive definite; \(\det>0,\ a<0\) ⇒ negative definite; \(\det<0\) ⇒ indefinite (saddle). For diagonal matrices the eigenvalues are the diagonal entries.
- \(f=x^2+2y^2\): \(\nabla f=(2x,4y)=0\iff(0,0)\); \(Hf=\operatorname{diag}(2,4)\), positive definite ⇒ strict local and global min.
- \(f=x^2-2y^2\): same critical point; \(Hf=\operatorname{diag}(2,-4)\) indefinite ⇒ saddle point.
SoSe23 A3: all local and global extrema of \(f=x^3+y^4-12x+4y\)
\(\nabla f=(3x^2-12,\ 4y^3+4)=0\iff x=\pm2,\ y=-1\). Two critical points. \(Hf=\operatorname{diag}(6x,12y^2)\).
- \((2,-1)\): \(Hf=\operatorname{diag}(12,12)\), both eigenvalues \(>0\) ⇒ strict local min.
- \((-2,-1)\): \(\operatorname{diag}(-12,12)\) ⇒ saddle point.
No global extrema: for fixed \(y\), \(f(x,y)\to\pm\infty\) as \(x\to\pm\infty\) (the \(x^3\) term).
WiSe24 5b: local extrema of \(f=x_1^2+x_1+x_2+e^{-x_1-x_2}\)
\(\nabla f=(2x_1+1-e^{-x_1-x_2},\ 1-e^{-x_1-x_2})\). Second component 0 ⇒ \(x_1+x_2=0\) ⇒ \(e^{0}=1\), then the first gives \(2x_1=0\). Only critical point \((0,0)\).
\(Hf=\begin{pmatrix}2+e^{-x_1-x_2}&e^{-x_1-x_2}\\e^{-x_1-x_2}&e^{-x_1-x_2}\end{pmatrix}\), at \((0,0)\): \(\begin{pmatrix}3&1\\1&1\end{pmatrix}\), \(\det=2>0\), top-left \(>0\) ⇒ positive definite ⇒ local min. (The official solution writes the top-left entry as 2, which is a slip; its determinant 1 leads to the same conclusion.)
Blatt 14.3c: extrema of \(f=y^2\,3^x\)
\(\nabla f=(\ln3\,y^23^x,\ 2y3^x)=0\iff y=0\) (since \(3^x>0\)): a whole line of critical points. Here the Hessian test is inconclusive, but directly: \(f\ge0\) everywhere and \(f(x,0)=0\), so every \((x,0)\) is a local and global minimum. Sometimes looking at the function beats the Hessian.
WiSe22 A5: sufficient condition; critical points of \(f=x\ln(x^2y^2)-y\); a Hessian from a given gradient
(a) Strict local max at \(x\): \(\nabla g(x)=0\) and \(Hg(x)\) negative definite. (The printed solution says "positive definite", which would be a minimum. Watch out for this in your own answers.)
(b) Rewrite \(f=2x(\ln x+\ln y)-y\). \(\nabla f=(2\ln(xy)+2,\ \frac{2x}y-1)=0\iff xy=e^{-1},\ y=2x\iff x=\frac1{\sqrt{2e}},\ y=\frac2{\sqrt{2e}}=\sqrt{\tfrac2e}\) (only the positive root, since \(x,y>0\)). Exactly one critical point.
(c) \(\nabla g(x,y)=\left(\frac{3y}x-1,\ \ln(x^3y^3)+3\right)^t\); find \(Hg\). You never need \(g\) itself: row \(i\) of the Hessian is the gradient of the \(i\)-th component of \(\nabla g\). Rewrite \(\ln(x^3y^3)=3\ln x+3\ln y\) (valid for \(x,y>0\)). Then \[Hg(x,y)=\begin{pmatrix}\partial_x(\frac{3y}x-1)&\partial_y(\frac{3y}x-1)\\\partial_x(3\ln x+3\ln y+3)&\partial_y(3\ln x+3\ln y+3)\end{pmatrix}=\begin{pmatrix}-\frac{3y}{x^2}&\frac3x\\\frac3x&\frac3y\end{pmatrix}.\] The two off-diagonal entries agree, as Schwarz predicts — a free check on your computation.
Vector-valued functions: the Jacobian matrix
\(f:U\subseteq\mathbb R^n\to\mathbb R^m\) is (totally) differentiable at \(x\) if \(f(x+h)=f(x)+Ah+o(\|h\|)\) for a linear map / \(m\times n\) matrix \(A\). Then \(a_{ij}=\frac{\partial f_i}{\partial x_j}(x)\): row \(i\) is the gradient (as a row) of component \(f_i\). \(A=Df(x)=J_f(x)\) is the Jacobian matrix [Jacobi-Matrix]. \(f\) is differentiable iff all components are.
Example: \(f(x,y)=(x^2y,\ \sin x+y)\): \(J_f=\begin{pmatrix}2xy&x^2\\\cos x&1\end{pmatrix}\).
Special cases you already know: for \(m=1\) the Jacobian is the row \(Df=\nabla f^t\); for \(n=1\) (a curve) it is the column \(\gamma'(t)\); and the Hessian of \(f\) is the Jacobian of \(\nabla f\).
Why \(a_{ij}=\partial f_i/\partial x_j\) (Lemma 11.19, the script's proof spelled out) Skript 11.19
Put \(h=te_j\) (move only in coordinate \(j\), by \(t\)). Then \(Ah=t\cdot(\text{column }j\text{ of }A)\), and \(\|h\|=|t|\). Row \(i\) of \(f(x+h)=f(x)+Ah+o(\|h\|)\) reads \[f_i(x+te_j)=f_i(x)+t\,a_{ij}+o(|t|).\] Divide by \(t\) and let \(t\to0\): \(\frac{f_i(x+te_j)-f_i(x)}t\to a_{ij}\). The left side is by definition \(\partial_jf_i(x)\). So the matrix \(A\) is forced: if a linear approximation exists at all, it must be the matrix of partial derivatives. The same argument, read row by row, shows that \(f\) is differentiable iff each component \(f_i\) is.
What you must know for 11.3: higher partial derivatives and in particular the Hessian; \(\partial_i\partial_jf=\partial_j\partial_if\) when continuous; necessary and sufficient conditions for local extrema; total differentiability; the Jacobian matrix.
!What the exam asks from Chapter 11
Chapter 11 appears in five of the nine exams provided (WiSe19, WiSe22, SoSe23, WiSe23, WiSe24; the others end before this chapter), usually as a full task of 4–11 points:
| Type | Exam tasks | What earns the points |
|---|---|---|
| Compute partial derivatives / gradient | WiSe19 1g, WiSe24 5a, Blatt 14.3a/b | freeze the other variables; product and chain rule; variable exponents via \(e^{\ln}\) |
| Critical points | SoSe23 A3, WiSe22 A5b, WiSe24 5b | solve \(\nabla f=0\) completely; respect the domain (e.g. \(x,y>0\)) |
| Classify with the Hessian | SoSe23 A3, WiSe24 5b, Blatt 14.3c | \(Hf\) at each point, eigenvalues or det test, the right conclusion (min/max/saddle/inconclusive) |
| Global extrema | SoSe23 A3, Blatt 14.3c | a line along which \(f\to\pm\infty\), or a direct bound like \(f\ge0\) |
| State the sufficient condition | WiSe22 A5a | \(\nabla g=0\) and \(Hg\) negative definite (max) / positive definite (min) |
| Hessian from a given gradient | WiSe22 A5c | differentiate each gradient component again; check symmetry |
| Schwarz arguments | WiSe23 A5 | quote Schwarz and its hypothesis; "rot \(\neq0\) ⇒ not a gradient" |
WiSe22 A5a says "positive definite" for a strict local maximum (it must be negative definite). WiSe24 5b writes the top-left Hessian entry as 2 instead of 3. Neither changes what you should write, but do not copy them.
Drill: partial derivatives, critical points, classification
- For the exam: the seven task types above; the procedure of §11.3 written out with every step justified; the \(2\times2\) determinant test and the eigenvalue criterion; the statement of Schwarz with its hypothesis.
- Common point losses: forgetting critical points (solve the system fully, e.g. both \(x=2\) and \(x=-2\)); calling a point with a zero eigenvalue a minimum; claiming a global extremum without checking behaviour at infinity.
- For Chapter 12: the condition "\(\partial f/\partial y\) continuous" in Picard–Lindelöf is a partial derivative, and a curve \(\gamma(t)\) with \(\gamma'(t)=F(\gamma(t))\) is exactly what a system of ODEs describes.
ΣSummary sheet
- Partial derivative = ordinary derivative in one variable, others frozen. \(\nabla f\) = column of partials.
- \(f(x+h)=f(x)+\nabla f(x)^th+o(\|h\|)\); tangent plane.
- Hessian \(Hf=(\partial_i\partial_jf)\), symmetric (Schwarz).
- Extrema: \(\nabla f=0\) (critical points) → eigenvalues of \(Hf\): all \(>0\) min, all \(<0\) max, mixed saddle, zero eigenvalues ⇒ look closer.
- Global: check behaviour at infinity / along lines.
- Jacobian: row \(i\) = gradient of component \(i\).
?Self-test
Differential EquationsDifferentialgleichungen
An equation where the unknown is a function \(y\), and the equation links \(y\) with its derivatives. The course teaches four solution recipes: integrate directly, separation of variables, variation of constants for linear 1st order equations, and the characteristic polynomial + ansatz for linear 2nd order equations with constant coefficients. Each recipe has appeared in the exams, usually as a task worth 3–8 points.
Source labels: Skript 12.5 = stated in your lecture script · Blatt 14 · 14.1 = exercise sheet · Klausur WiSe19 = old exam · Extra = my own addition (intuition, a proof the script leaves out, or an extra example). A § marks a whole section of the script (§12.4 = Section 12.4) rather than a numbered result (12.4 = Beispiel 12.4). "Aufgabe 13.x" refers to the solved problems in Chapter 13 of the script.
0The big picture
With an ordinary equation like \(x^2=4\) you look for a number. With a differential equation like \(y'=y\) you look for a function: which function equals its own derivative? (Answer: \(y=ce^x\).) Differential equations describe change: "the population grows proportionally to its size", "the spring pulls back proportionally to the stretch".
Before computing anything, recognise the type. The type tells you which recipe to use:
| the equation looks like … | name | recipe | section |
|---|---|---|---|
| \(y'=g(x)\) | pure integration | \(y=y_0+\int_{x_0}^xg\) | 12.1 |
| \(y'=f(x)\,g(y)\) | separable [getrennte Variablen] | \(\int\frac{dy}{g(y)}=\int f(x)\,dx\), solve for \(y\) | 12.2 |
| \(y'+a(x)y=f(x)\) | linear, 1st order | homogeneous \(ce^{-A}\), then variation of constants | 12.3 |
| \(y''+ay'+by=f(x)\), \(a,b\) constants | linear, 2nd order, constant coefficients | \(\lambda^2+a\lambda+b=0\), then ansatz for \(f\) | 12.4 |
| \(y'=f(x,y)\) in general | — | no recipe; Picard–Lindelöf says a unique solution exists | 12.5 |
And one habit for every task: check your answer ("Probe"). Differentiate your \(y\), plug it in, check the initial value. The script does this after every example, and in an exam it catches almost every sign error.
- Integration is the engine of this chapter: antiderivatives and the fundamental theorem (§8.3), substitution and integration by parts (§8.4). Separation of variables is a substitution; variation of constants ends with an integral.
- Derivative rules, especially the chain rule (§6.3); "\(f'=0\) on an interval ⇒ \(f\) constant" (§7.3, Satz 7.13) — every uniqueness proof below uses it.
- \(\exp\), \(\ln\), \(\arctan\), \(\tan\) and their domains (§5.2, §5.3): solving \(G(y)=F(x)\) for \(y\) means applying an inverse function, and its domain gives the maximal domain of the solution.
- Complex numbers and Euler's formula \(e^{i\omega x}=\cos\omega x+i\sin\omega x\) (Chapter 1, §5.3) for the oscillating case of 12.4; a \(2\times2\) determinant for the Wronskian.
12.1Initial value problems [Anfangswertprobleme]
Motivation Skript 12.1. A population of size \(N(t)\) with birth rate \(\beta\) and death rate \(\delta\) changes by \(N'(t)=(\beta-\delta)N(t)\). All solutions are \(N(t)=ce^{(\beta-\delta)t}\) (check: \(N'=c(\beta-\delta)e^{(\beta-\delta)t}=(\beta-\delta)N\)). There is one solution for every \(c\). Fixing the starting size \(N(0)=N_0\) picks one: \(c=N_0\).
An equation \(y'(x)=f(x,y(x))\) for an unknown function \(y\) of one real variable \(x\) is an ordinary differential equation [gewöhnliche Differentialgleichung] of 1st order [1. Ordnung] (only \(y'\) occurs). "Ordinary" means there is only one variable \(x\), so there are no partial derivatives. The order is the highest derivative that occurs.
An initial value problem (IVP) additionally prescribes \(y(x_0)=y_0\): for an interval \(I\ni x_0\) and a set \(\Omega_0\ni y_0\) of allowed values, find \(y:I\to\Omega_0\) with \[y'(x)=f(x,y(x))\ \ \forall x\in I,\qquad y(x_0)=y_0.\]
Why "initial value"? Think of \(x\) as time: the equation says how the state changes, the initial value says where you start. Without it you get a whole family of solutions, and with it (usually) exactly one.
- \(y'=g(x)\), \(y(x_0)=y_0\): \(y\) is an antiderivative of \(g\). By the fundamental theorem (Chapter 8), \(y(x)-y_0=\int_{x_0}^xg(t)\,dt\), so \(y(x)=y_0+\int_{x_0}^xg(t)\,dt\), and this solution is unique. E.g. \(y'=a\), \(y(0)=y_0\) gives \(y=y_0+ax\).
- \(y'=y\), \(y(0)=1\) (characterisation of \(\exp\)): since \(y(0)=1>0\) and \(y\) is continuous, \(y>0\) near 0, so we may divide: \(\frac{y'}{y}=1\). Integrate from 0 to \(x\): \([\ln y(t)]_0^x=x\), i.e. \(\ln y(x)=x\) and \(y(x)=e^x\). With \(y(x_0)=y_0>0\) the same computation gives \(y=y_0e^{x-x_0}\).
If \(f'=cf\) on an interval \(I\), then \(f(x)=be^{cx}\) for some \(b\in\mathbb R\). The computation above divided by \(y\), which only works where \(y\neq0\). This proof avoids that problem:
Proof (Blatt 10.4)
Put \(g(x)=f(x)e^{-cx}\). By the product rule, \(g'(x)=f'(x)e^{-cx}-cf(x)e^{-cx}=(cf(x)-cf(x))e^{-cx}=0\). A function with derivative 0 on an interval is constant (mean value theorem, Blatt 10.4a): \(g\equiv b\). Hence \(f(x)=be^{cx}\).
SoSe19 A4 uses the same trick. From \(f(x+y)=f(x)f(y)\), \(f(0)=1\), \(f'(0)=1\), part a) shows \(f'=f\) via the difference quotient \(\frac{f(x)f(h)-f(x)}h=f(x)\frac{f(h)-f(0)}h\to f(x)\). Then b) \(g=fe^{-x}\) has \(g'=0\), c) \(g(0)=1\), so d) \(f=e^x\).
What you must know for 12.1: what a differential equation and an initial value problem are; solving \(y'=g(x)\).
12.2Separation of variables [Trennung der Variablen]
The right-hand side splits into a factor depending only on \(x\) times a factor depending only on \(y\). Then you can put all \(y\)-terms on one side and all \(x\)-terms on the other, and integrate each side.
IVP: \(y'(x)=f(x)\,g(y(x))\), \(y(x_0)=y_0\), with \(f,g\) continuous and \(g(y_0)\neq0\).
- \(g\) and \(y\) are continuous and \(g(y_0)\ne0\), so \(g(y(x))\neq0\) for \(x\) near \(x_0\). Only there may we divide: \(\frac{y'(x)}{g(y(x))}=f(x)\).
- Integrate from \(x_0\) to \(x\): \(\int_{x_0}^x\frac{y'(t)}{g(y(t))}dt=\int_{x_0}^xf(t)\,dt\).
- Substitute \(s=y(t)\), \(ds=y'(t)\,dt\) on the left (Chapter 8): \(\displaystyle\int_{y_0}^{y(x)}\frac{ds}{g(s)}=\int_{x_0}^xf(t)\,dt.\)
- With \(G(y)=\int_{y_0}^y\frac{ds}{g(s)}\), \(F(x)=\int_{x_0}^xf\): \(G(y(x))=F(x)\), so \(y(x)=G^{-1}(F(x))\) if \(G\) is invertible.
Write \(y'=\frac{dy}{dx}\) and "multiply" by \(dx\) (a shorthand for steps 1–3): \[\frac{dy}{dx}=f(x)g(y)\ \Longrightarrow\ \frac{dy}{g(y)}=f(x)\,dx\ \Longrightarrow\ \int\frac{dy}{g(y)}=\int f(x)\,dx+c.\] Then: (1) solve for \(y\); (2) determine \(c\) from \(y(x_0)=y_0\); (3) state where the solution is defined (the exams often ask for the maximal domain [maximaler Definitionsbereich]); (4) check.
Extra If \(g(y_0)=0\), the constant function \(y\equiv y_0\) is a solution (both sides are 0), so there is nothing to separate. Example: \(y'=y^2\), \(y(0)=0\) has the solution \(y\equiv0\).
\(y'=1+y^2\), \(y(0)=0\). Since \(1+y^2>0\), divide: \(\frac{y'}{1+y^2}=1\). Integrate and substitute: \(x=\int_0^{y(x)}\frac{ds}{1+s^2}=\arctan y(x)\). So \(y(x)=\tan x\), but only on \(\left(-\frac\pi2,\frac\pi2\right)\), because \(\arctan\) takes values only there. Check: \(y'=1+\tan^2x=1+y^2\). The right-hand side \(1+y^2\) is harmless, yet the solution reaches \(\pm\infty\) at \(x=\pm\frac\pi2\).
(a) \(y'=e^{2x-y}\), \(y(0)=1\). The key observation is \(e^{2x-y}=e^{2x}e^{-y}\), so the equation is separable: \[\int e^y\,dy=\int e^{2x}dx\ \Rightarrow\ e^y=\tfrac12e^{2x}+c\ \Rightarrow\ y=\ln\!\left(\tfrac12e^{2x}+c\right).\] \(1=\ln(\tfrac12+c)\) gives \(c=e-\tfrac12\), so \(y(x)=\ln\!\left(\tfrac12e^{2x}+e-\tfrac12\right)\). The argument is \(>0\) because \(e-\frac12>0\), so \(y\) is defined on all of \(\mathbb R\).
(b) \(y'=(x+y)^2\) is not separable. Substitute \(z=x+y\): \(z'=1+y'=1+z^2\), which is separable (it is Skript 12.5 again). \(\arctan z-\arctan z_0=x-x_0\), so \(z=\tan(\arctan z_0+x-x_0)\) and \[y(x)=\tan\big(\arctan(x_0+y_0)+x-x_0\big)-x.\] With \(y(0)=0\): \(y=\tan x-x\) on \(\left(-\frac\pi2,\frac\pi2\right)\).
WiSe24 4b (3 P): \(y'=\frac{2x}{3y^2}\), \(y(0)=1\). Separate: \(\int_1^y3s^2\,ds=\int_0^x2t\,dt\), so \(y^3-1=x^2\) and \(y(x)=\sqrt[3]{x^2+1}\). Check: \(y'=\frac{2x}{3(x^2+1)^{2/3}}=\frac{2x}{3y^2}\) ✓.
SoSe23 A7 (5 P): \(y'=\frac{4xe^{-y}}{(1+x^2)^3}\), \(y(0)=2\), maximal domain?
Solution
- Separable: \(\int e^y\,dy=\int\frac{4x}{(1+x^2)^3}dx\).
- Right side: substitute \(u=1+x^2\), \(du=2x\,dx\): \(\int\frac{2}{u^3}du=-\frac1{u^2}=-\frac1{(1+x^2)^2}\). So \(e^y=-\frac1{(1+x^2)^2}+c\).
- Initial value: \(e^2=-1+c\), so \(c=e^2+1\) and \(y(x)=\ln\!\left(e^2+1-\frac1{(1+x^2)^2}\right)\).
- Domain: \(\frac1{(1+x^2)^2}\le1\), so the argument is \(\ge e^2>0\) for every \(x\). The maximal domain is \(\mathbb R\).
A solution of an IVP is a differentiable function on an interval containing \(x_0\). The maximal domain is the largest such interval on which your formula makes sense and satisfies the equation. Find it by listing where the formula breaks — a zero denominator, \(\ln\) of something \(\le0\), a root of something negative, \(\tan\) at \(\pm\frac\pi2\), or leaving the allowed region of \(y\) — and taking the piece around \(x_0\).
Example: \(y'=2xy^2\), \(y(0)=1\). Separate: \(-\frac1y=x^2+c\), \(c=-1\), so \(y=\frac1{1-x^2}\). The formula is defined on \(\mathbb R\setminus\{\pm1\}\), but the solution is only the piece through \(x_0=0\): maximal domain \((-1,1)\). The branches on \((1,\infty)\) and \((-\infty,-1)\) are separate solutions of the same equation with different initial values; they are not part of this solution, because a solution cannot jump across a pole.
Notice that the maximal domain depends on the initial value: with \(y(0)=y_0>0\) you get \(y=\frac1{1/y_0-x^2}\) on \(|x|<1/\sqrt{y_0}\), and with \(y_0\le0\) on all of \(\mathbb R\). Try it below.
What you must know for 12.2: recognise when separation of variables applies, and use it to solve the equation.
12.3Linear equations of 1st order [Lineare DGL 1. Ordnung]
With \(a,f:\mathbb R\to\mathbb R\) continuous, \[y'(x)+a(x)y(x)=f(x)\] is an inhomogeneous linear equation of 1st order. If \(f\equiv0\), it is homogeneous [homogen]. "Linear" means that \(y\) and \(y'\) appear only to the first power and are never multiplied together: no \(y^2\), no \(e^y\), no \(yy'\).
Let \(H\) = solutions of the homogeneous equation and \(I\) = solutions of the inhomogeneous one. (a) \(y_i\in I,\ y_h\in H\Rightarrow y_i+y_h\in I\). (b) \(y_{i1},y_{i2}\in I\Rightarrow y_{i1}-y_{i2}\in H\). (c) For any single \(y_i\in I\): \(I=y_i+H\).
In words: general solution = one particular solution + all homogeneous solutions. This is the same structure as for linear systems of equations \(Ax=b\) in linear algebra.
Proof
(a) \((y_i+y_h)'+a(y_i+y_h)=(y_i'+ay_i)+(y_h'+ay_h)=f+0=f\). (b) \((y_{i1}-y_{i2})'+a(y_{i1}-y_{i2})=f-f=0\). (c) "⊇" is (a). "⊆": for \(y\in I\), \(y-y_i\in H\) by (b), and \(y=y_i+(y-y_i)\).
All solutions of \(y'+a(x)y=0\) are \(y(x)=ce^{-A(x)}\), \(c\in\mathbb R\), where \(A\) is an antiderivative of \(a\) (\(A'=a\)).
Proof
They are solutions: \((ce^{-A})'=-A'ce^{-A}=-a\,ce^{-A}\). They are all solutions (the same trick as in Blatt 10.4): \(\big(y\,e^{A}\big)'=(y'+ay)e^{A}=0\), so \(ye^A=c\) is constant.
Example 12.8: \(y'=y\) means \(a=-1\), \(A=-x\), so \(y=ce^x\).
Replace the constant \(c\) by a function: ansatz \(y(x)=c(x)e^{-A(x)}\). Insert it: \[c'e^{-A}\underbrace{-c\,a\,e^{-A}+a\,c\,e^{-A}}_{=0}=f\ \Longrightarrow\ c'(x)=f(x)e^{A(x)}\ \Longrightarrow\ c(x)=\int_{x_0}^xf(t)e^{A(t)}dt.\] The two middle terms cancel because \(A'=a\). This cancellation happens every time, so it is a good check on your computation.
Satz 12.10: all solutions are \(\displaystyle y(x)=e^{-A(x)}\Big(c+\int_{x_0}^xf(t)e^{A(t)}dt\Big)\), \(c\in\mathbb R\). If \(A(x_0)=0\), the IVP solution has \(c=y_0\).
\(y'+y=\sin x\), \(y(0)=1\). Here \(a=1\), \(A=x\), \(A(0)=0\), so \(y=e^{-x}\big(1+\int_0^xe^t\sin t\,dt\big)\). Integrating by parts twice (the "loop" trick from Chapter 8) gives \(I(x)=\int_0^xe^t\sin t\,dt=\tfrac12e^x(\sin x-\cos x)+\tfrac12\). Hence \[y(x)=\tfrac32e^{-x}+\tfrac12(\sin x-\cos x).\] Check: \(y'=-\frac32e^{-x}+\frac12(\cos x+\sin x)\), so \(y'+y=\sin x\) ✓, and \(y(0)=\frac32-\frac12=1\) ✓. For large \(x\) the \(e^{-x}\) part dies out and \(y\) follows the oscillation. Choose this equation in the slope field above to see it.
\(y'+y=\frac1{1+e^x}\). The homogeneous solutions \(ce^{-x}\) are given. Find all solutions by variation of constants.
Solution
Ansatz \(y=c(x)e^{-x}\): \(c'e^{-x}-ce^{-x}+ce^{-x}=\frac1{1+e^x}\), so \(c'(x)=\frac{e^x}{1+e^x}\). This is of the form \(\frac{g'}{g}\) with \(g=1+e^x\), so \(c(x)=\ln(1+e^x)+c_1\). All solutions: \[y(x)=e^{-x}\big(\ln(1+e^x)+c_1\big),\quad c_1\in\mathbb R.\]
What you must know for 12.3: recognise homogeneous and inhomogeneous linear equations of 1st order; get the homogeneous solutions by separation of variables; find a particular solution by variation of constants; general solution = particular + homogeneous.
12.4Linear equations of 2nd order [Lineare DGL 2. Ordnung]
\(y''+a(x)y'+b(x)y=f(x)\). The structure from 12.3 carries over unchanged: all solutions = \(y_i+H\). There is no general recipe, but for constant coefficients \(a,b\in\mathbb R\) there is one, and that is the case the exams ask about.
How many homogeneous solutions? Linear independence
\(y_1,y_2:I\to\mathbb R\) are linearly independent [linear unabhängig] if \(\alpha_1y_1(x)+\alpha_2y_2(x)=0\) for all \(x\in I\) forces \(\alpha_1=\alpha_2=0\). In other words, neither function is a constant multiple of the other. The Wronskian [Wronski-Determinante] is \[W(x)=\det\begin{pmatrix}y_1(x)&y_2(x)\\y_1'(x)&y_2'(x)\end{pmatrix}.\]
Where it comes from: if \(\alpha_1y_1+\alpha_2y_2\equiv0\), differentiating gives \(\alpha_1y_1'+\alpha_2y_2'\equiv0\). Together this is a \(2\times2\) linear system for \((\alpha_1,\alpha_2)\). If \(W(x)\neq0\) at some point, its only solution is \(\alpha_1=\alpha_2=0\).
The script marks Definition 12.13 and Satz 12.14 as ausgelassen (skipped in the lecture). You need the idea "two independent solutions give all solutions", and the Wronskian is the tool that checks independence.
(a) Linearly dependent ⇒ \(W\equiv0\). (b) For two solutions of the homogeneous equation: independent ⇔ \(W(x)\ne0\) for some \(x\).
All solutions of the homogeneous equation are \(y=c_1y_1+c_2y_2\) with two linearly independent solutions \(y_1,y_2\), called fundamental solutions [Fundamentallösungen / Fundamentalsystem]. The solution space is 2-dimensional: one free constant per order.
Example: \(e^{2x}\) and \(e^{3x}\) give \(W=\det\begin{pmatrix}e^{2x}&e^{3x}\\2e^{2x}&3e^{3x}\end{pmatrix}=e^{5x}\neq0\), so they are independent.
Proof of Satz 12.14 (the script's short argument, with the missing step for (b) "⇒" filled in) Extra
Recall from linear algebra that the \(2\times2\) system \(M\alpha=0\) has only the solution \(\alpha=0\) exactly when \(\det M\neq0\). Apply this to \(M=\begin{pmatrix}y_1(x)&y_2(x)\\y_1'(x)&y_2'(x)\end{pmatrix}\), whose determinant is \(W(x)\).
(a) If \(y_1,y_2\) are linearly dependent, there is \((\alpha_1,\alpha_2)\neq(0,0)\) with \(\alpha_1y_1+\alpha_2y_2\equiv0\); differentiating, \(\alpha_1y_1'+\alpha_2y_2'\equiv0\). So for every \(x\) the system \(M\alpha=0\) has a non-zero solution, which forces \(W(x)=0\) for all \(x\).
(b) "⇐" If \(W(x_1)\ne0\) at some \(x_1\), then \(M(x_1)\alpha=0\) forces \(\alpha=0\). So any relation \(\alpha_1y_1+\alpha_2y_2\equiv0\) has \(\alpha_1=\alpha_2=0\): independent. (This direction works for any differentiable functions.)
(b) "⇒", shown in the contrapositive form "\(W(x_1)=0\) at some \(x_1\) ⇒ dependent". Here we need that \(y_1,y_2\) solve the same equation. Since \(\det M(x_1)=0\), there is \(\alpha\ne(0,0)\) with \(M(x_1)\alpha=0\). The function \(y=\alpha_1y_1+\alpha_2y_2\) solves the homogeneous equation (a sum of solutions) and has \(y(x_1)=0\), \(y'(x_1)=0\). The zero function solves the same problem. A second-order linear equation has only one solution with given \(y(x_1)\), \(y'(x_1)\) (for constant coefficients this follows from the proof of Satz 12.15 below; in general from Picard–Lindelöf applied to the pair \((y,y')\)). So \(y\equiv0\), a non-trivial relation: dependent. ∎
Why the script's short argument is not enough on its own: for arbitrary functions, "\(W\ne0\) somewhere" is sufficient but not necessary for independence. \(y_1=x^3\), \(y_2=|x|^3\) are independent on \(\mathbb R\), yet \(W\equiv0\). They are not solutions of a common equation of the form (12.4), which is exactly the hypothesis (b) needs.
Consequence worth remembering: for solutions of the same equation, \(W\) is either identically 0 or nowhere 0 — so a single test point decides.
Homogeneous case: the characteristic polynomial
Idea: \(y\), \(y'\), \(y''\) must cancel, so they should be "the same kind of function" up to constant factors. Exponentials have this property. Ansatz \(y=e^{\lambda x}\) gives \(e^{\lambda x}(\lambda^2+a\lambda+b)=0\), and since \(e^{\lambda x}\ne0\): \[\lambda^2+a\lambda+b=0,\qquad \lambda_{1,2}=-\frac a2\pm\sqrt D,\quad D=\frac{a^2}4-b.\]
| case | roots | general solution | behaviour |
|---|---|---|---|
| \(D>0\) | \(\lambda_1\ne\lambda_2\) real | \(c_1e^{\lambda_1x}+c_2e^{\lambda_2x}\) | no oscillation |
| \(D=0\) | \(\lambda=-\frac a2\) (double) | \(c_1e^{-\frac a2x}+c_2\,x\,e^{-\frac a2x}\) | critical case |
| \(D<0\) | \(-\frac a2\pm i\omega\), \(\omega=\sqrt{-D}\) | \(e^{-\frac a2x}\big(c_1\cos\omega x+c_2\sin\omega x\big)\) | oscillation; damped if \(a>0\) |
Why the extra \(x\) when \(D=0\), and why cos/sin when \(D<0\)
\(D=0\): the ansatz only produces one function \(e^{-\frac a2x}\), but we need two. The script checks directly that \(y=xe^{-\frac a2x}\) also works: \(y'=(1-\frac a2x)e^{-\frac a2x}\), \(y''=(\frac{a^2}4x-a)e^{-\frac a2x}\), and inserting gives \((b-\frac{a^2}4)xe^{-\frac a2x}=0\) because \(D=0\).
\(D<0\): the roots are complex, \(e^{\lambda_1x}=e^{-\frac a2x}(\cos\omega x+i\sin\omega x)\) (Euler, Chapter 5). The real and imaginary parts of a complex solution are real solutions (the equation has real coefficients), which gives \(e^{-\frac a2x}\cos\omega x\) and \(e^{-\frac a2x}\sin\omega x\).
Why these are all solutions (Satz 12.15; the script gives no proof) Extra
Factor \(\lambda^2+a\lambda+b=(\lambda-\lambda_1)(\lambda-\lambda_2)\), so \(a=-(\lambda_1+\lambda_2)\), \(b=\lambda_1\lambda_2\). Let \(y\) be any solution and put \(z=y'-\lambda_1y\). Then \[z'-\lambda_2z=y''-(\lambda_1+\lambda_2)y'+\lambda_1\lambda_2y=y''+ay'+by=0.\] So \(z\) solves a first-order homogeneous equation, and by Lemma 12.7 \(z=c\,e^{\lambda_2x}\). Now \(y'-\lambda_1y=c\,e^{\lambda_2x}\) is a first-order linear equation, and Satz 12.10 (\(A=-\lambda_1x\)) gives all its solutions: \[y=e^{\lambda_1x}\Big(c_1+c\int_0^xe^{(\lambda_2-\lambda_1)t}\,dt\Big).\] If \(\lambda_1\ne\lambda_2\), the integral is \(\frac{e^{(\lambda_2-\lambda_1)x}-1}{\lambda_2-\lambda_1}\), and \(y\) is a combination of \(e^{\lambda_1x}\) and \(e^{\lambda_2x}\). If \(\lambda_1=\lambda_2\), the integrand is 1, the integral is \(x\), and \(y=(c_1+cx)e^{\lambda_1x}\) — this is where the extra factor \(x\) comes from. (For complex roots the same computation works with complex numbers; taking real parts gives the cos/sin form.)
The proof also shows uniqueness for the IVP: \(y(x_0)\) and \(y'(x_0)\) fix \(z(x_0)\) and hence both constants. So there are exactly two free constants — the solution space is 2-dimensional.
\(y''-2y'-3y=0\) (WiSe19 A5a) with \(y(0)=2\), \(y'(0)=2\). General solution \(y=c_1e^{3x}+c_2e^{-x}\), so \(y'=3c_1e^{3x}-c_2e^{-x}\). At \(x=0\): \(c_1+c_2=2\) and \(3c_1-c_2=2\). Adding gives \(4c_1=4\), so \(c_1=1\), \(c_2=1\): \(y=e^{3x}+e^{-x}\). A second-order IVP needs two initial conditions, one per constant. For an inhomogeneous equation, fit the constants after adding the particular solution.
Inhomogeneous case: guess a particular solution
| right-hand side \(f(x)\) | ansatz \(y_i(x)\) |
|---|---|
| polynomial of degree \(n\) | general polynomial of degree \(n\): \(\sum_{k=0}^nb_kx^k\) |
| \(e^{\alpha x}(a_1\cos\omega x+a_2\sin\omega x)\) | \(e^{\alpha x}(b_1\cos\omega x+b_2\sin\omega x)\), with both cos and sin even if \(f\) has only one |
| \(e^{\alpha x}\) (special case \(\omega=0\)) | \(b\,e^{\alpha x}\) |
Insert, compare coefficients [Koeffizientenvergleich] of \(x^k\) (or of cos and sin), solve the linear system. Then general solution = \(y_i\) + homogeneous.
Why comparing coefficients is legitimate: if two polynomials agree for all \(x\), their coefficients agree, because \(a_k=\frac{p^{(k)}(0)}{k!}\) is determined by the function (Lemma 7.22). The same holds for \(b_1\cos\omega x+b_2\sin\omega x\), since \(\cos\) and \(\sin\) are linearly independent — set \(x=0\) to read off \(b_1\), then \(x=\frac\pi{2\omega}\) for \(b_2\).
Extra Resonance: the ansatz fails when it is already a homogeneous solution. In Skript 12.16 this is the case \(|\omega_0|=|\omega|\): the coefficient equations become \(0=1\). The fix (not needed in this course's exams) is to multiply the ansatz by \(x\). Method 2, variation of constants with \(c_1(x)y_1+c_2(x)y_2\), is mentioned in the script but not worked out.
- 12.16: \(y''+\omega^2y=\sin(\omega_0x)\), \(|\omega_0|\ne|\omega|\). Ansatz \(b_1\cos\omega_0x+b_2\sin\omega_0x\) has \(y''=-\omega_0^2y\), so \((\omega^2-\omega_0^2)(b_1\cos+b_2\sin)=\sin\) gives \(b_1=0\), \(b_2=\frac1{\omega^2-\omega_0^2}\). Homogeneous: \(\lambda=\pm i\omega\). All solutions: \(\frac{\sin\omega_0x}{\omega^2-\omega_0^2}+c_1\cos\omega x+c_2\sin\omega x\).
- 13.1: \(y''-3y'+2y=x^2+1\). \(\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2)\). Ansatz \(ax^2+bx+c\): \(2a-3(2ax+b)+2(ax^2+bx+c)=x^2+1\) gives \(2a=1\), \(-6a+2b=0\), \(2a-3b+2c=1\), so \(a=\frac12,b=\frac32,c=\frac94\). All solutions: \(\frac12x^2+\frac32x+\frac94+c_1e^x+c_2e^{2x}\).
- 13.2: \(y''-6y'+8y=e^{3x}\). \(\lambda=2,4\). Ansatz \(be^{3x}\): \((9-18+8)b=1\), so \(b=-1\). All solutions: \(-e^{3x}+c_1e^{2x}+c_2e^{4x}\).
\(y''+9y=5\cos2x\). Homogeneous: \(\lambda^2+9=0\), \(\lambda=\pm3i\), fundamental system \(\cos3x,\sin3x\). Ansatz \(b_1\cos2x+b_2\sin2x\): \(y_i''=-4y_i\), so \(5b_1\cos2x+5b_2\sin2x=5\cos2x\) and \(b_1=1,b_2=0\). \[y(x)=\cos2x+c_1\cos3x+c_2\sin3x.\]
WiSe19 A5a (4 P): \(y''-2y'-3y=0\). \(\lambda^2-2\lambda-3=(\lambda-3)(\lambda+1)\), so \(y=c_1e^{3x}+c_2e^{-x}\).
WiSe22 A6a: \(y''-7y'+12y=0\). \((\lambda-3)(\lambda-4)=0\), so \(y=c_1e^{3x}+c_2e^{4x}\). (Variant: \(y''-5y'+6y=0\) gives \(c_1e^{2x}+c_2e^{3x}\).)
WiSe22 A6b: one solution of \(y''-7y'+12y=12x^2+10x\).
Solution
Quadratic right side, so the ansatz is \(y=ax^2+bx+c\), \(y'=2ax+b\), \(y''=2a\): \[2a-7(2ax+b)+12(ax^2+bx+c)=12x^2+10x.\] Compare coefficients: \(x^2\): \(12a=12\); \(x\): \(-14a+12b=10\); \(1\): \(2a-7b+12c=0\). So \(a=1\), \(b=2\), \(c=1\): \(y=x^2+2x+1=(x+1)^2\). (Variant \(y''-5y'+6y=12x^2+4x+2\) gives \(2x^2+4x+3\).) The task asks for a solution; adding \(c_1e^{3x}+c_2e^{4x}\) gives all of them.
What you must know for 12.4: recognise homogeneous and inhomogeneous linear equations of 2nd order; find the solution space of the homogeneous equation; find a particular solution of the inhomogeneous equation when the right-hand side is a polynomial.
12.5General theory: existence and uniqueness
Let \(\Omega=(\underline x,\overline x)\times\Omega_0\), \(f:\Omega\to\mathbb R\) continuous with \(\frac{\partial f}{\partial y}\) continuous, \(x_0\in(\underline x,\overline x)\), \(y_0\in\Omega_0\). Then the IVP \(y'=f(x,y)\), \(y(x_0)=y_0\) has a solution continued in both directions (\(x<x_0\) and \(x>x_0\)), and it is unique.
Why the assumption on \(\partial f/\partial y\) matters Extra: \(y'=2\sqrt{|y|}\), \(y(0)=0\) has the two solutions \(y\equiv0\) and \(y=x^2\) (for \(x\ge0\)). Here \(f=2\sqrt{|y|}\) is continuous, but \(\partial f/\partial y\) does not exist at \(y=0\), so uniqueness fails. Geometrically, solution curves in the slope field cannot cross under the assumptions of the theorem, but here they can.
The theorem does not say that the solution lives on the whole interval. There are three possibilities:
| behaviour | meaning | example |
|---|---|---|
| existence to the end [Existenz bis zum Schluss] | solution exists on all of \((\underline x,\overline x)\) | every linear equation \(y'=a(x)y+b(x)\): global solution on \(\mathbb R\) |
| blow-up | \(y(x)\to\pm\infty\) as \(x\to x_+<\overline x\) | 12.18: \(y'=y^2\), \(y(0)=1\): \(y=\frac1{1-x}\), \(y\to\infty\) as \(x\uparrow1\) |
| premature collapse [vorzeitiger Kollaps] | \(y(x)\) reaches the boundary of \(\Omega_0\) at \(x_+<\overline x\) | 12.19: \(y'=-\frac1{\sqrt y}\), \(y>0\), \(y(0)=1\): \(y=(1-\frac32x)^{2/3}\to0\) as \(x\to\frac23\) |
- 12.18: \(f(x,y)=y^2\) and \(\partial_yf=2y\) are continuous, so Picard–Lindelöf applies near 0. Separate: \(x=\int_0^x\frac{y'}{y^2}=\big[-\frac1y\big]_0^x=1-\frac1{y(x)}\), so \(y=\frac1{1-x}\) on \((-\infty,1)\). The right-hand side \(y^2\) is defined everywhere, but it grows so fast that the solution runs off to infinity in finite time.
- 12.19: \(\sqrt y\,y'=-1\), so \(\int_1^{y(x)}\sqrt s\,ds=-x\), \(\frac23(y^{3/2}-1)=-x\), \(y=(1-\frac32x)^{2/3}\). At \(x=\frac23\) the solution hits \(y=0\), where the equation is no longer defined.
Both are among the equations in the slope-field widget in 12.1.
This section is marked as partly omitted [ausgelassen] in the script. Know the statement of Picard–Lindelöf and the three possible behaviours with their examples.
ΣSummary sheet
- Identify the type (table in section 0). Try in this order: \(y'=g(x)\)? separable \(f(x)g(y)\)? linear \(y'+ay=f\)? 2nd order with constant coefficients?
- Separable: \(\int\frac{dy}{g(y)}=\int f\,dx+c\) → solve for \(y\) → constant from the IVP → domain. Look for hidden products such as \(e^{2x-y}=e^{2x}e^{-y}\), or substitute \(z=x+y\).
- Linear 1st order: \(y_h=ce^{-A}\), ansatz \(c(x)e^{-A}\) → \(c'=fe^{A}\) → \(y=e^{-A}(c+\int fe^A)\).
- Linear 2nd order, constant coefficients: \(\lambda^2+a\lambda+b=0\); \(D>0\): \(e^{\lambda_1x},e^{\lambda_2x}\); \(D=0\): \(e^{\lambda x},xe^{\lambda x}\); \(D<0\): \(e^{-\frac a2x}\cos\omega x,\ e^{-\frac a2x}\sin\omega x\).
- Inhomogeneous: ansatz of the same type as \(f\) (polynomial of the same degree; \(b_1\cos+b_2\sin\) with the same frequency), compare coefficients. Result: \(y_i+y_h\).
- Theory: Picard–Lindelöf (\(f\), \(\partial_yf\) continuous ⇒ unique local solution); blow-up (\(y'=y^2\)); collapse (\(y'=-1/\sqrt y\)).
- Always check the solution by inserting it.
!What the exam asks from Chapter 12
Differential equations appear in four of the nine exams (WiSe19, WiSe22, SoSe23, WiSe24), always as a computation of 3–8 points; the \(f'=cf\) uniqueness argument appears once more as SoSe19 A4. Nothing from §12.5 (Picard–Lindelöf, blow-up) has been asked so far, and the script gives no "minimal contents" for it; knowing the statement and the examples is enough.
| Type | Exam tasks | What earns the points |
|---|---|---|
| Separable IVP | WiSe24 4b, SoSe23 A7, Blatt 14.1a | separate correctly, integrate (often a substitution), constant from \(y(x_0)\), solve for \(y\) |
| Maximal domain | SoSe23 A7 | where the formula is defined, the interval around \(x_0\), with a reason |
| Linear 1st order, variation of constants | WiSe19 A5b, Skript 12.11 | ansatz \(c(x)e^{-A}\), the cancellation, \(c'=fe^{A}\), integrate, "+ homogeneous" |
| 2nd order homogeneous | WiSe19 A5a, WiSe22 A6a | ansatz \(e^{\lambda x}\), characteristic polynomial, correct case, both constants |
| Particular solution, polynomial right side | WiSe22 A6b, Aufgabe 13.1 | ansatz of the same degree, comparison of coefficients, clean linear system |
| \(f'=cf\Rightarrow f=be^{cx}\) | SoSe19 A4, Blatt 10.4 | differentiate \(fe^{-cx}\), derivative 0 ⇒ constant (mean value theorem) |
- Forgetting "+ all homogeneous solutions" when the task asks for all solutions (WiSe19 A5b), or adding them when it asks for one (WiSe22 A6b — harmless, but read the question).
- Sign errors in \(e^{-A}\) (the homogeneous solution of \(y'+ay=0\) is \(ce^{-A}\), with a minus).
- Dropping the constant of integration and then fitting the initial value to nothing.
- Giving a domain that is not an interval around \(x_0\), or not giving one when asked.
- Skipping the check. Inserting the solution takes a minute and finds almost every mistake.
Drill: recognise, solve, check
- Recognise the type of a given equation in seconds (table in section 0) — this decides the method.
- Solve separable IVPs including the maximal domain; linear first-order equations by variation of constants; second-order equations with constant coefficients in all three cases, with a polynomial, exponential or cos/sin right-hand side.
- Explain why general solution = particular + homogeneous (Lemma 12.6), and state Picard–Lindelöf with the blow-up and collapse examples.
- Across the course: this chapter uses almost everything before it — integration (Ch. 8), the chain rule (Ch. 6), exp/ln/trig (Ch. 5), complex numbers (Ch. 1). Weak spots there show up here first; the drill above is a good final check.