Analysis_Complete_Guide

📐 Analysis I & II — The Complete Intuitive Guide

A from-the-ground-up synthesis of Struwe (2010), Ziltener (2025), and seven D-ITET basic exams (2013–2023)

Promise. If you read this from top to bottom, knowing only high school math, you will understand every concept on the ETH D-ITET Analysis basic exam — not just memorize formulas. Every idea is built up from intuition, every theorem comes with a real-world picture, and every chapter ends with actual exam tasks with full solutions.

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🧭 How to use this note

This document is written for Obsidian, so it uses LaTeX ($...$ and $$...$$), callouts (> [!tip]), and [[wiki-links]] for navigation.

Every chapter follows the same six-step rhythm:

1. 🤔 The big question — what real-world or mathematical problem are we solving?
2. 💡 The intuition — a picture or analogy your brain can hold
3. 📜 The formal definition — the precise statement
4. 🔧 Worked examples — small, then medium, then exam-level
5. 🎯 How the exam tests this — concrete patterns from past D-ITET exams
6. 📝 Practice tasks with solutions

How to study with this document

1. First pass — read intuitions and worked examples only. Skip formal definitions if they feel intimidating.
2. Second pass — read carefully, do every example with paper.
3. Third pass — close the document, redo every “Practice Task” from memory.

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📋 Table of Contents

1. 1 · Logic — the operating system of math
2. 2 · Sets and functions
3. [[#3 · Numbers — from $\mathbb{N}$ to $\mathbb{C}$]]
4. 4 · Sequences — what does “approaching” mean?
5. 5 · Series — adding infinitely many things
6. 6 · Continuity — drawing without lifting the pen
7. 7 · Differentiation — the slope at a point
8. 8 · Taylor series — polynomial X-rays
9. 9 · Limits and L’Hôpital’s rule
10. 10 · Integration — the Fundamental Theorem
11. 11 · Integration techniques (parts, substitution, partial fractions)
12. 12 · Ordinary differential equations
13. 13 · Multivariable functions
14. 14 · Multivariable extrema and Lagrange multipliers
15. 15 · Multiple integrals
16. 16 · Vector fields, Green, Stokes, Gauß
17. 17 · Implicit and inverse function theorems
18. 18 · Master strategy — the exam playbook

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1 · Logic — the operating system of math

🤔 The big question

Why do we need logic? Because every theorem you’ll ever read says “if this holds, then that holds” — and to use the theorem you must understand exactly what it claims, what it doesn’t, and how to negate it.

In math, a proof is a derivation of a statement from axioms (basic assumptions). A theorem (Satz) is a statement that has been proved. So mathematics literally is a collection of true statements together with their derivations. Logic is the rulebook that says which derivations are valid.

Where this chapter sits

This corresponds to Ziltener §1.1 (Logik) and §1.3 (Quantoren), and Struwe Chapter 1.

💡 What counts as a mathematical statement (Aussage)?

Statement (Aussage)

A statement is a sentence that is either true (T) or false (F) — never both, never neither.

Two foundational principles ground everything:

The two pillars of classical logic

• Law of non-contradiction (Satz vom ausgeschlossenen Widerspruch): no statement is both true and false simultaneously.
• Law of the excluded middle (Satz vom ausgeschlossenen Dritten, tertium non datur): every statement is either true or false — there is no third option.

Statements vs. non-statements

✅ Are statements:

• “Bern is the capital of Switzerland.” (T)
• $1+1=2$ (T)
• $0<0$ (F)

❌ NOT statements:

• “Hello!” (an exclamation)
• “Close the door.” (a command)
• “What time is it?” (a question)

Statements have a truth value. Exclamations, commands, and questions don’t.

The liar paradox — why self-reference is forbidden

Consider $P:=$ “This sentence is false.”

• If $P$ is true, then by what it says, $P$ is false. Contradiction.
• If $P$ is false, then by what it says, $P$ is true. Contradiction.

So $P$ violates the law of non-contradiction. Math forbids self-referential sentences like this — they’re not legitimate Aussagen.

Same idea, ancient version: Epimenides the Cretan declared “All Cretans are liars.” (Said by a Cretan.)

📜 The five connectives — building bigger statements

Given statements $A$ and $B$, you can build new ones:

Symbol	Read as	True when…	Mnemonic
$\neg A$	”not $A$”	$A$ is false	flip the switch
$A\wedge B$	”$A$ and $B$“	both are true	a chain — weakest link breaks it
$A\vee B$	”$A$ or $B$” (inclusive)	at least one is true	menu where you may pick either or both
$A\dot{\vee }B$	”either $A$ or $B$” (exclusive)	exactly one is true	XOR
$A\to B$	”if $A$, then $B$“	NOT ($A$ true and $B$ false)	a promise: broken only if you don’t deliver
$A\leftrightarrow B$	”$A$ iff $B$“	both have the same truth value	”live or die together”

The full truth table:

$A$	$B$	$\neg A$	$A\wedge B$	$A\vee B$	$A\to B$	$A\leftrightarrow B$
T	T	F	T	T	T	T
T	F	F	F	T	F	F
F	T	T	F	T	T	F
F	F	T	F	F	T	T

Practice with connectives

Each row applies the table to a concrete pair of statements.

$A$	$B$	Combined statement	Truth
$0<1$	$1+1=2$	$A\wedge B$	T ∧ T = T
$0<1$	$1+1=3$	$A\wedge B$	T ∧ F = F
$0<1$	$1+1=3$	$A\vee B$	T ∨ F = T
$0<0$	$1+1=2$	$A\to B$	F → T = T (vacuous)
$0<0$	$1+1=3$	$A\to B$	F → F = T (vacuous!)

The hardware connection — logic gates

Every connective above corresponds to a physical logic gate in digital electronics: AND-gate, OR-gate, NOT-gate, XOR-gate. Modern processors are essentially billions of these gates wired into circuits. Boolean logic isn’t an abstract toy — it’s literally how your computer computes.

⚠️ The trap that catches everyone: vacuous truth

A false hypothesis implies anything

$A\to B$ is true whenever $A$ is false. The statement “if the moon is made of cheese, then $1=2$” is logically true, because the premise is false.

Why? The implication is a promise. The promise “if it rains, I’ll bring an umbrella” is only broken if it rains and you don’t bring an umbrella. On a sunny day, the promise was never tested — so it was never broken.

More vacuous truths

All of the following are true statements:

• “If $0=1$, then I am Napoleon.” (premise is false)
• “If $n\in \mathbb{N}$ with $n<0$, then $n^2=17$.” (no such $n$ exists, so the universal statement is vacuously true)
• $\forall x\in \varnothing :P(x)$ holds for any $P$.

Vacuous truth feels weird, but it makes the rules of logic simple and consistent.

💡 Logical equivalence ($\equiv$) vs. material equivalence ($\leftrightarrow$)

These look similar but say different things:

• $A\leftrightarrow B$ is itself a statement that says ”$A$ and $B$ have the same truth value, here”. It can be true or false depending on $A,B$.
• $A\equiv B$ is a meta-claim that says: $A$ and $B$ have identical truth tables — they always agree, no matter what their atoms mean.

So $\equiv$ is strictly stronger than $\leftrightarrow$.

A useful equivalence we'll use over and over

$A\leftrightarrow B\equiv (A\to B)\wedge (B\to A)$.

Read aloud: “iff” means “implies and is implied by”. You can verify by checking all 4 rows of the truth table.

A common mistake — confusing implication with equivalence

“If it rained → ground is wet” is TRUE. “If ground is wet → it rained” is FALSE (someone could have spilled water).

Implication is one-way. To get equivalence you need both directions.

💡 Reading implications: sufficient and necessary

The statement $A\to B$ has three equivalent readings:

1. “If $A$, then $B$.”
2. ”$A$ is sufficient for $B$” (hinreichend) — having $A$ alone is enough to guarantee $B$.
3. ”$B$ is necessary for $A$” (notwendig) — without $B$, you can’t have $A$.

Sufficient vs. necessary

Statement: “If it rained, then the ground is wet.”

• Rain is sufficient for wet ground (rain alone guarantees wet).
• Wet ground is necessary for rain (no wet → no rain).
• Wet ground is NOT sufficient for rain (sprinklers exist).
• Rain is NOT necessary for wet ground (sprinklers).

So ”$A$ iff $B$” means ”$A$ is both necessary and sufficient for $B$“.

💡 Contraposition — the most useful identity in proof-writing

$A\to B\equiv \neg B\to \neg A$

The “Kontraponiertes” of an implication. Verify by truth table — both sides are false in exactly the same row ($A$ true, $B$ false).

Why contraposition saves the day

• “If it rains → ground is wet” $\equiv$ “If ground is dry → it didn’t rain”
• “If $x^2$ is even → $x$ is even” $\equiv$ “If $x$ is odd → $x^2$ is odd” (much easier to prove! Squaring an odd number is direct algebra.)
• “If $f$ is differentiable at $a$ → $f$ is continuous at $a$” $\equiv$ “If $f$ is not continuous at $a$ → $f$ is not differentiable at $a$”

When the direct direction is hard, try contraposition.

📜 Quantifiers: $\forall$ and $\exists$

Symbol	Read as	Example
$\forall x\in M:P(x)$	”for all $x$ in $M$, $P(x)$ holds”	$\forall n\in \mathbb{N}:n\ge 1$ ✅
$\exists x\in M:P(x)$	“there exists $x$ in $M$ with $P(x)$”	$\exists n\in \mathbb{N}:n^2=49$ ✅

Negation rule — flip the quantifier and negate inside

$\neg (\forall x\in X:P(x))\equiv \exists x\in X:\neg P(x)$ $\neg (\exists x\in X:P(x))\equiv \forall x\in X:\neg P(x)$

Memorize this. It’s the most-tested identity on the basic exam after induction.

Order of quantifiers matters!

Swap two adjacent quantifiers and you can change a true statement into a false one.

• $\forall m\in {\mathbb{N}}_0,\exists n\in {\mathbb{N}}_0:m\le n$ — TRUE (for each $m$, take $n:=m$).
• $\exists n\in {\mathbb{N}}_0,\forall m\in {\mathbb{N}}_0:m\le n$ — FALSE (“there is a largest natural number”).

In the first version, $n$ is allowed to depend on $m$. In the second, one fixed $n$ must work for every $m$.

Negating a real exam sentence (FS 2023 MC1)

Original: $\forall n\in \mathbb{N},\exists m\in \mathbb{N}:(m>n)\wedge (m<2n)$

Negation, step by step:

1. Flip outer $\forall \to \exists$: $\exists n\in \mathbb{N},\neg (\exists m\in \mathbb{N}:\dots )$
2. Flip inner $\exists \to \forall$: $\exists n\in \mathbb{N},\forall m\in \mathbb{N}:\neg ((m>n)\wedge (m<2n))$
3. De Morgan inside: $\neg (P\wedge Q)\equiv \neg P\vee \neg Q$.

Final: $\boxed{\exists n\in \mathbb{N},\forall m\in \mathbb{N}:(m\le n)\vee (m\ge 2n)}$

Formal proof: there is no largest natural number

Claim: $\neg (\exists n\in {\mathbb{N}}_0,\forall m\in {\mathbb{N}}_0:m\le n)$.

Push the negation through the quantifiers: $\forall n\in {\mathbb{N}}_0,\exists m\in {\mathbb{N}}_0:m>n$.

Proof: given any $n$, take $m:=n+1$. Then $m\in {\mathbb{N}}_0$ and $m>n$. $■$

Names of bound variables don't matter (Goethe's joke)

$\forall x\in \mathbb{N}:x>0$ and $\forall n\in \mathbb{N}:n>0$ are literally the same statement. Bound variables are placeholders — only the structure matters.

Goethe wrote in Maximen und Reflexionen: “Mathematicians are like Frenchmen — whatever you say to them, they translate it into their own language, and at once it is something entirely different.”

🔧 Modus ponens — the engine of every proof

The fundamental rule of inference, written as an inference scheme:

$\frac{A,A\to B}{B}$

Read: if I know $A$ is true, and I know ”$A$ implies $B$” is true, I’m allowed to conclude $B$.

Modus ponens in everyday math

• Premise 1: $x=3$.
• Premise 2: If $x=3$, then $x^2=9$.
• Conclusion: $x^2=9$. ✓

Every "$\Rightarrow$" you write in a calculation is a hidden modus ponens.

🔧 Four proof techniques you absolutely need

1. Direct proof

Build a chain $A\Rightarrow B_1\Rightarrow B_2\Rightarrow \cdots \Rightarrow B$. Apply modus ponens at every step.

Direct proof: the first binomic formula

Claim (Lemma 1.5 in Ziltener): for all real $x,y$: $(x+y{)}^2=x^2+2xy+y^2$.

$(x+y{)}^2=(x+y)(x+y)=x\cdot x+x\cdot y+y\cdot x+y\cdot y=x^2+2xy+y^2$. $■$

Each "$=$" is a direct application of distributivity / commutativity.

2. Proof by contraposition

To prove $A\Rightarrow B$, prove $\neg B\Rightarrow \neg A$ instead. The two are logically equivalent.

Contraposition: "if $n^2$ is even, then $n$ is even"

Direct proof is hard (you’d need to factor $n^2$ somehow). Contrapositive: if $n$ is odd, then $n^2$ is odd.

Suppose $n$ is odd, so $n=2k+1$ for some integer $k$. Then $n^2=(2k+1{)}^2=4k^2+4k+1=2(2k^2+2k)+1.$ That’s of the form $2m+1$, so $n^2$ is odd. $■$

Contraposition: $\sqrt{2}<\sqrt{3}$ (Ziltener Satz 1.8, first proof)

Use the fact that on $[0,\infty )$, $x\le y\Rightarrow x^2\le y^2$ (Lemma 1.9).

We want $A:=(\sqrt{2}<\sqrt{3})$. Equivalently (by contraposition of the squaring lemma): if $\sqrt{2}\ge \sqrt{3}$ then $2\ge 3$ — which is false. So $\sqrt{2}\ge \sqrt{3}$ is false, hence $\sqrt{2}<\sqrt{3}$. $■$

3. Proof by contradiction

To prove a statement $S$, assume $\neg S$ and derive a contradiction (something both true and false).

There is no largest natural number

Assume there is a largest $n_0\in \mathbb{N}$. But $n_0+1\in \mathbb{N}$, and $n_0+1>n_0$, contradicting that $n_0$ is the largest. Contradiction. $■$

$\sqrt{2}$ is irrational (classic, Ziltener Satz 2.1)

Assume for contradiction $\sqrt{2}=m/n$ with $m,n$ integers having no common factor of 2 (lowest terms).

Squaring: $2=m^2/n^2$, so $m^2=2n^2$. Hence $m^2$ is even, so $m$ is even (by the previous example contraposed). Write $m=2k$: $4k^2=2n^2\Rightarrow n^2=2k^2,$ so $n^2$ is even, so $n$ is even. But then $m,n$ both have a factor of 2 — contradicting our “lowest terms” assumption. $■$

$\sqrt{2+\sqrt{3}}<2$ (Ziltener Satz 1.10)

Assume for contradiction $\sqrt{2+\sqrt{3}}\ge 2$. Squaring: $2+\sqrt{3}\ge 4$, so $\sqrt{3}\ge 2$. Squaring again: $3\ge 4$. False. $■$

Subtle point about contradiction proofs

Inside a contradiction proof you are temporarily reasoning from a false premise (the negation of what you want to prove). So intermediate statements you derive can be literally false statements — and that’s OK! The whole point is to keep deriving until you bump into a contradiction. From a false hypothesis, anything follows (ex falso quodlibet). Don’t panic when you write down something that “looks wrong” mid-proof.

4. Mathematical induction — your best friend on exams

To prove $P(n)$ for all $n\in {\mathbb{N}}_0$:

1. Base case (Induktionsverankerung): prove $P(0)$ (or $P(1)$ if the claim starts at 1).
2. Inductive step (Induktionsschritt): assume $P(k)$ holds — the induction hypothesis (IH) — and from it deduce $P(k+1)$.

The domino picture: line up infinitely many dominoes. Knock down the first one (base case); show that any falling domino topples its neighbor (inductive step). Then they all fall.

Sum of the first $n$ odd numbers $=n^2$

Claim: ${\sum }_{k=1}^n(2k-1)=n^2$ for all $n\in \mathbb{N}$.

Base ($n=1$): $1=1^2$ ✓

Step: Assume $1+3+\cdots +(2n-1)=n^2$. Then $1+3+\cdots +(2n-1)+(2n+1)\stackrel{IH}{=}{n}^2+(2n+1)=(n+1{)}^2✓$

Sum of integers up to $n$ (Ziltener Satz 1.11)

Claim: ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$ for all $n\in {\mathbb{N}}_0$.

Base ($n=0$): the empty sum equals $0=\frac{0\cdot 1}{2}$ ✓.

Step: assume ${\sum }_{i=1}^{k}i=\frac{k(k+1)}{2}$. Then ${\sum }_{i=1}^{k+1}i={\sum }_{i=1}^{k}i+(k+1)\stackrel{IH}{=}\frac{k(k+1)}{2}+(k+1)=\frac{(k+1)(k+2)}{2}.✓$

Bernoulli's inequality (Ziltener Lemma 2.7)

Claim: for all $n\in {\mathbb{N}}_0$ and $x\in [-1,\infty )$: $(1+x{)}^n\ge 1+nx$.

Base ($n=0$): $(1+x{)}^0=1\ge 1+0\cdot x=1$ ✓.

Step: Assume $(1+x{)}^k\ge 1+kx$. Multiply by $(1+x)\ge 0$ (allowed since $x\ge -1$): $(1+x{)}^{k+1}\ge (1+kx)(1+x)=1+(k+1)x+k{x}^2\ge 1+(k+1)x.$ Last step used $k{x}^2\ge 0$ (since $k\ge 0$). $✓$

Bernoulli is the workhorse behind countless analysis estimates — it lets you replace $(1+x{)}^n$ by a linear lower bound.

🎯 How the exam tests logic

Every basic exam contains at least one logic / quantifier task. Common types:

• Negate a quantified statement (FS 2023, MC1)
• Prove an inequality by induction (FS 2015 task 9, FS 2019 task 13)
• Prove a sum-formula by induction
• Multiple-choice on logical equivalence, contraposition, vacuous truth.

📝 Practice tasks

Task 1.1 (Quantifier negation)

Negate: $\exists x\in \mathbb{R},\forall y\in \mathbb{R}:x\cdot y=0$.

Solution $\exists \to \forall$, $\forall \to \exists$, negate the equation: $\forall x\in \mathbb{R},\exists y\in \mathbb{R}:x\cdot y\ne 0.$

Flip

Task 1.2 (Induction — adapted from FS 2015 task 9)

Prove that for all $n\in {\mathbb{N}}_0$ and $x\in [0,1]$: $(1+x{)}^n\le 1+(2^n-1)x$.

Solution Base ($n=0$): $(1+x{)}^0=1\le 1+0=1$. ✓

Step: Assume the inequality for $n$. Then $(1+x{)}^{n+1}=(1+x)(1+x{)}^n\stackrel{\text{IH}}{\le }(1+x)(1+(2^n-1)x)$ $=1+2^{n}x+(2^n-1)x^2$ Since $x^2\le x$ on $[0,1]$: $\le 1+2^{n}x+(2^n-1)x=1+(2^{n+1}-1)x✓$

Task 1.3 (FS 2019 task 13 — clever induction)

Prove: ${\sum }_{n=1}^{2N}\frac{(-1{)}^{n-1}}{n}={\sum }_{n=1}^N\frac{1}{N+n}$ for all $N\in \mathbb{N}$.

Solution Base ($N=1$): LHS $=1-\frac{1}{2}=\frac{1}{2}$. RHS $=\frac{1}{2}$. ✓

Step: Look at the difference between $N+1$ and $N$ terms.

LHS difference: $\frac{1}{2N+1}-\frac{1}{2N+2}$.

RHS difference: $\left({\sum }_{n=1}^{N+1}\frac{1}{(N+1)+n}\right)-\left({\sum }_{n=1}^N\frac{1}{N+n}\right)$. Re-indexing the first sum with $k=n+1$: $=\frac{1}{2N+1}+\frac{1}{2N+2}-\frac{1}{N+1}=\frac{1}{2N+1}-\frac{1}{2N+2}$ Both differences match, so the identity transfers from $N$ to $N+1$. $■$

Task 1.4 (Contraposition)

Prove by contraposition: if $n^2$ is divisible by 3, then $n$ is divisible by 3.

Solution if $n$ is not divisible by 3, then $n^2$ is not divisible by 3.

Contrapositive:

If $3\nmid n$, write $n=3k+r$ with $r\in \{1,2\}$.

• $r=1$: $n^2=9k^2+6k+1=3(3k^2+2k)+1$, remainder 1.
• $r=2$: $n^2=9k^2+12k+4=3(3k^2+4k+1)+1$, remainder 1.

In both cases $3\nmid n^2$. $■$

(This is exactly the trick used to prove $\sqrt{3}$ is irrational, mirroring the $\sqrt{2}$ argument.)

Task 1.5 (Truth table / equivalence)

Show that $(A\to B)\equiv (\neg A\vee B)$.

Solution

$A$	$B$	$A\to B$	$\neg A$	$\neg A\vee B$
T	T	T	F	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

Columns 3 and 5 match in every row. $■$

(This identity is the basis for translating implications into Boolean logic, and explains why “false ⇒ anything” is true: $\neg A$ is already true, so the OR is true.)

Task 1.6 (Induction — geometric sum)

Prove that for all $n\in {\mathbb{N}}_0$ and $q\in \mathbb{R}\setminus \{1\}$: ${\sum }_{k=0}^n{q}^k=\frac{1-q^{n+1}}{1-q}.$

Solution Base ($n=0$): LHS $=q^0=1$. RHS $=\frac{1-q}{1-q}=1$. ✓

Step: Assume the formula holds for $n$. Then ${\sum }_{k=0}^{n+1}{q}^k={\sum }_{k=0}^n{q}^k+q^{n+1}\stackrel{IH}{=}\frac{1-q^{n+1}}{1-q}+q^{n+1}$ $=\frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}=\frac{1-q^{n+2}}{1-q}.✓$

This formula is the foundation of all geometric-series arguments later in the course.

📚 Official problem-set tasks (Loesung 00, 01, 02)

The exercises below are taken verbatim from the official ITET HS24 Übungsserien (Ziltener). Try each one yourself before opening the solution toggle.

L0.4 (warm-up: wenn, oder, ∃, ∀)

Determine truth values: (i) “If $n$ is even, then $n+1$ is odd” — for which $n\in \mathbb{N}$? (ii) “If 1 is even, then 2 is odd.” (iii) ”$1+1=2$ or $1$ is odd.” (iv) “Either $1+1=2$ or $1$ is odd.” (exclusive) (v) ”$\forall m\in \mathbb{N}\exists n\in \mathbb{N}:m\le n$.” (vi) ”$\exists n\in \mathbb{N}\forall m\in \mathbb{N}:m\le n$.”

Solution True for all $n$ (when $n$ even, $n+1$ odd; when $n$ odd, premise false → vacuously true). (ii) True (premise "1 is even" is false → vacuous). (iii) True (first part true). (iv) False (both parts true, exclusive-or rejects this). (v) True (take $n:=m$). (vi) False (no largest natural number).

(i)

L0.7 (warm-up induction)

Prove ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$ for all $n\in \mathbb{N}$.

Solution $n=1$: $1=\frac{1\cdot 2}{2}$. ✓ Step: ${\sum }_{i=1}^{n+1}i=(n+1)+\frac{n(n+1)}{2}=\frac{2(n+1)+n(n+1)}{2}=\frac{(n+1)(n+2)}{2}$. $■$

Base

L1.1 (translate to symbols + truth value)

Write each in symbols, then state the truth value: (a) “Zero plus one is one, AND zero is greater than one.” (b) “Zero plus one is one, OR zero is less than one.” (c) “Either zero plus one is one, OR zero is less than one.” (exclusive) (d) “If zero is greater than one, then zero plus one is zero.” (e) “Zero is greater than one IFF zero plus one is one.”

Solution $0+1=1\wedge 0>1$ — F ($0>1$ is false). (b) $0+1=1\vee 0<1$ — T. (c) $0+1=1\dot{\vee }0<1$ — F (both true → XOR false). (d) $0>1\to 0+1=0$ — T (vacuous). (e) $0>1\leftrightarrow 0+1=1$ — F (one side T, other F).

(a)

L1.2 (truth-table identities to verify)

Verify by truth table: (a) $\neg (P\wedge Q)\equiv \neg P\vee \neg Q$ (De Morgan) (b) $\neg (P\vee Q)\equiv \neg P\wedge \neg Q$ (De Morgan) (c) $(P\to Q)\equiv (\neg Q\to \neg P)$ (Contraposition) (d) $(P\leftrightarrow Q)\equiv (P\to Q)\wedge (Q\to P)$ (e) $P\wedge (Q\vee R)\equiv (P\wedge Q)\vee (P\wedge R)$ (Distribution) (f) Counterexample: is $P\wedge (Q\vee R)\equiv (P\wedge Q)\vee R$?

Solution

(a)–(e): build the 4- or 8-row truth tables; the columns for the two sides match in every row.

(f) Not equivalent. Take $P=F$, $Q=F$, $R=T$:

• LHS: $F\wedge (F\vee T)=F\wedge T=F$.
• RHS: $(F\wedge F)\vee T=F\vee T=T$. Different. So associating $\wedge$ and $\vee$ wrongly changes meaning.

L1.3 (proofs by contraposition AND contradiction)

(a) Prove: $\sqrt{3}<\sqrt{5}$. (b) Prove: $\sqrt{3+\sqrt{5}}<\sqrt{6}$.

Show each by both contraposition and contradiction.

Solution Contraposition: the squaring lemma says $0\le x\le y\Rightarrow x^2\le y^2$. Contrapose: $x^2<y^2\Rightarrow x<y$ (for $x,y\ge 0$). Apply with $x^2=3,y^2=5$: $3<5\Rightarrow \sqrt{3}<\sqrt{5}$. $■$

(a)

Contradiction: assume $\sqrt{3}\ge \sqrt{5}$. Square (monotonic on $[0,\infty )$): $3\ge 5$. False. $■$

(b) Contradiction: assume $\sqrt{3+\sqrt{5}}\ge \sqrt{6}$. Square: $3+\sqrt{5}\ge 6$, so $\sqrt{5}\ge 3$. Square again: $5\ge 9$. False. $■$ (Contraposition: $5<9\Rightarrow \sqrt{5}<3\Rightarrow 3+\sqrt{5}<6\Rightarrow \sqrt{3+\sqrt{5}}<\sqrt{6}$.)

L1.4 (induction power-pack — four classics)

Prove by induction: (a) ${\sum }_{k=1}^n{k}^2=\frac{1}{6}n(n+1)(2n+1)$ (b) ${\sum }_{k=1}^n{k}^3=(\frac{1}{2}n(n+1){)}^2$ (c) $(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x}$ for $x\ne 1,n\in {\mathbb{N}}_0$ (d) ${\sum }_{k=1}^{n}k\cdot k!=(n+1)!-1$

Solution to (a) $n=1$: $1=\frac{1\cdot 2\cdot 3}{6}$. ✓ Step: add $(n+1{)}^2$ and factor: $\frac{n(n+1)(2n+1)}{6}+(n+1{)}^2=\frac{(n+1)(n(2n+1)+6(n+1))}{6}=\frac{(n+1)(2n^2+7n+6)}{6}=\frac{(n+1)(n+2)(2n+3)}{6}.✓$

Base

Solution to (b) $n=1$: $1=(\frac{1\cdot 2}{2}{)}^2$. ✓ Step: $(\frac{n(n+1)}{2}{)}^2+(n+1{)}^3=\frac{(n+1{)}^2}{4}(n^2+4(n+1))=\frac{(n+1{)}^2(n+2{)}^2}{4}=(\frac{(n+1)(n+2)}{2}{)}^2.✓$

Base

Solution to (c) $n=0$: $1+x=\frac{1-x^2}{1-x}$ ✓ (third binomic formula). Step: multiply both sides by $(1+x^{2^{n+1}})$: $\frac{1-x^{2^{n+1}}}{1-x}(1+x^{2^{n+1}})=\frac{1-x^{2^{n+2}}}{1-x}.✓$

Base

Solution to (d) $n=1$: $1\cdot 1!=1=2!-1$. ✓ Step: ${\sum }_{k=1}^{n+1}k\cdot k!=(n+1)\cdot (n+1)!+(n+1)!-1=(n+2)\cdot (n+1)!-1=(n+2)!-1.✓$

Base

L1.5 (the famous "all horses are the same color" fallacy)

Where is the error in this “proof”? Claim: any group of $n$ horses has the same color. Base $n=1$: trivial. Step: take $n+1$ horses; remove one → the remaining $n$ are same color (IH); remove a different one → those $n$ are also same color (IH); overlap forces all $n+1$ to share the color.

Solution fails when going from $n=1$ to $n=2$: removing one horse from 2 leaves a single horse (one color, trivially); removing the other one leaves a different single horse — but the two singletons share no horse, so there's no "overlap" to force them to the same color. The argument silently assumes $n\ge 2$ when invoking overlap.

The step

L1.6 (multiple choice — equivalences and negations)

(a) Which is always true? (i) $(\neg A\to B)\wedge (B\to A)$ (ii) $(A\vee (\neg A\wedge B))\leftrightarrow (A\vee B)$ (b) Negation of “It’s raining and I have no umbrella”: (i) “It’s not raining or I have an umbrella.” (c) Contrapositive of “If it rains and I have no umbrella, I get wet”: (iii) “If I don’t get wet, then it’s not raining or I have an umbrella.” (d) Negation of “10 is even and ≤ 11”: (iii) “10 is not even or > 11.”

Solution (ii) is the always-true tautology. Verify: $A\vee (\neg A\wedge B)$ is T iff $A$ is T or ($A$ is F and $B$ is T) — exactly when $A\vee B$ is T. (b) (i) by De Morgan applied to the conjunction. (c) (iii) flip and negate. (d) (iii) De Morgan again.

(a)

L2.3 (quantifier interpretation + truth)

State each in plain language and decide the truth value: (a) $\forall n\in {\mathbb{N}}_0:n\le 1\vee n>1$ (b) $\forall x\in {\mathbb{N}}_0:x\le 1\vee x>1$ (c) $\exists x\in \mathbb{R}\forall (p,q)\in \mathbb{Z}\times (\mathbb{Z}\setminus \{0\}):x\ne p/q$

Then translate to symbols: (1) “24 is not a perfect square.” (2) The Archimedean principle: “for every real $x$, there is a natural number larger than $x$.”

Solution $n\in {\mathbb{N}}_0$ is either $\le 1$ or $>1$." True (trichotomy on $\mathbb{R}$). (b) Same statement with renamed variable. True (Goethe's principle). (c) "There exists an irrational real number." True ($\sqrt{2}$ for instance).

(a) “Every

(1) $\nexists n\in {\mathbb{N}}_0:n^2=24$. (2) $\forall x\in \mathbb{R}\exists n\in {\mathbb{N}}_0:n>x$.

L2.4 (negate AND determine truth)

Negate each, then determine which (original or negation) is true. (a) $\forall n\in {\mathbb{N}}_0:n<n^2$ (b) $\forall n\in {\mathbb{N}}_0:n\le 1\vee n>1$ (c) $\forall x\in (0,\infty )\exists n\in \mathbb{N}:1/n<x$ (d) $\exists n\in \mathbb{N}\forall x\in (0,\infty ):1/n<x$

Solution $\exists n:n\ge n^2$. Negation TRUE ($n=0$: $0\ge 0$). Original false. (b) Neg: $\exists n:n>1\wedge n\le 1$ — never. Original TRUE. (c) Neg: $\exists x>0\forall n\ge 1:1/n\ge x$. Original TRUE (Archimedes). (d) Neg: $\forall n\ge 1\exists x>0:1/n\ge x$ — take $x:=1/n$. Negation TRUE (no fixed $n$ beats every positive $x$).

(a) Neg:

L4.9 (Bernoulli's inequality — the official version)

Prove $(1+x{)}^n\ge 1+nx$ for all $x\in [-1,\infty )$ and $n\in {\mathbb{N}}_0$.

Solution $n=0$: $1\ge 1$. ✓ Step: $(1+x)\ge 0$ since $x\ge -1$, so multiplying the IH: $(1+x{)}^{n+1}\ge (1+nx)(1+x)=1+(n+1)x+n{x}^2\ge 1+(n+1)x.✓$

Base

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2 · Sets and functions

🤔 The big question

Sets are the nouns of math; functions are the verbs. You can’t talk about anything without them.

In modern foundations (Zermelo–Fraenkel set theory), every mathematical object — numbers, functions, vectors, even logical formulas — is built from sets. So this chapter is the bedrock of everything that follows.

Where this chapter sits

This corresponds to Ziltener §1.2 (Mengenlehre) and §1.4 (Funktionen), and Struwe Chapter 1.

💡 Sets — collections without order

Cantor’s original definition (1895): a set is “an unordered collection of distinct objects gathered into a whole”. Each object is called an element (Element); we write $x\in A$ for ”$x$ is an element of $A$“.

Notation: $\{1,2,3\}=\{3,1,2\}=\{1,1,2,3\}$ — order and repetition don’t matter.

Set	Symbol	Members
Naturals	$\mathbb{N}$	$\{1,2,3,\dots \}$
Naturals incl. 0	${\mathbb{N}}_0$	$\{0,1,2,3,\dots \}$
Integers	$\mathbb{Z}$	$\{\dots ,-1,0,1,\dots \}$
Rationals	$\mathbb{Q}$	$\{p/q:p\in \mathbb{Z},q\in \mathbb{N}\}$
Reals	$\mathbb{R}$	All decimals
Complex	$\mathbb{C}$	$\{a+bi:a,b\in \mathbb{R}\}$
Empty set	$\varnothing$ or $\{\}$	nothing

$\varnothing$ vs $\{\varnothing \}$

• $\varnothing$ is the empty set: zero elements, $|\varnothing |=0$.
• $\{\varnothing \}$ is a set whose only element is the empty set: one element, $|\{\varnothing \}|=1$.

They look almost identical in print but are very different. Don’t confuse them.

📜 Two ways to write a set

1. Roster form (aufzählende Schreibweise) — list the elements: $\{0,1,2,3\},\{-1,1\}.$

2. Set-builder (beschreibende Schreibweise) — give a defining property $P$: $\{x\mid P(x)\}=\text{"all x such that P(x) holds"}.$

Set-builder examples

• $\{n\in {\mathbb{N}}_0\mid n\text{ even}\}=\{0,2,4,6,\dots \}$.
• $\{(-1{)}^n\mid n\in {\mathbb{N}}_0\}=\{1,-1\}$ — only two distinct values, despite infinitely many $n$.
• $\{x\in \mathbb{R}\mid x^2<2\}=(-\sqrt{2},\sqrt{2})$.
• $\{x\in \mathbb{Q}\mid x\ge 0,x^2>2\}$ — the set Ziltener uses to define $\sqrt{2}$ as a Dedekind cut.

⚠️ Russell’s paradox — why “the set of all sets” is forbidden

Try to define $X:=\{x\mid x\notin x\}$ — “the set of all sets that don’t contain themselves”.

Now ask: is $X\in X$?

• If $X\in X$, then $X$ satisfies the defining property, so $X\notin X$. Contradiction.
• If $X\notin X$, then $X$ satisfies the defining property, so $X\in X$. Contradiction.

This is Russell’s paradox (Bertrand Russell, 1901). The fix in modern set theory: you may only form $\{x\in X\mid P(x)\}$ — restricting to elements of a pre-existing set $X$. You cannot conjure sets out of arbitrary properties.

The barber paradox (intuitive cousin)

“In a town, the barber shaves exactly those men who don’t shave themselves. Who shaves the barber?” — same paradox, dressed in everyday clothes. Either answer leads to a contradiction. The resolution: no such barber can exist.

📜 Set operations

Fix a “universe” $X$ (the Grundmenge). For $A,B\subseteq X$:

Operation	Definition	Picture
Union	$A\cup B=\{x\mid x\in A\vee x\in B\}$	both circles
Intersection	$A\cap B=\{x\mid x\in A\wedge x\in B\}$	overlap only
Difference	$A\setminus B=\{x\in A\mid x\notin B\}$	$A$ minus the overlap
Complement	$A^c=X\setminus A$	everything outside $A$
Subset	$A\subseteq B:\iff \forall x:x\in A\Rightarrow x\in B$	$A$ inside $B$

De Morgan's Laws (Ziltener Satz 1.13)

For all $A,B\subseteq X$: $(A\cap B{)}^c=A^c\cup B^c,(A\cup B{)}^c=A^c\cap B^c.$

The set version and the propositional version of De Morgan are the same theorem in two languages: $\cap$ corresponds to $\wedge$, $\cup$ to $\vee$, complement to $\neg$.

Why De Morgan works (mini proof)

$x\in (A\cap B{)}^c$ $\iff x\notin A\cap B$ $\iff \neg (x\in A\wedge x\in B)$ $\iff \neg (x\in A)\vee \neg (x\in B)$ ← logical De Morgan $\iff x\in A^c\vee x\in B^c$ $\iff x\in A^c\cup B^c.■$

💡 $n$-tuples and Cartesian products — when order matters

Sets ignore order. But often we want order: a coordinate $(3,5)$ on a map is different from $(5,3)$.

$n$-tuple (Ziltener 1.14)

An (ordered) $n$-tuple $(x_1,\dots ,x_n)$ is determined by its entries and their order: $(x_1,\dots ,x_n)=(y_1,\dots ,y_n)\iff x_1=y_1\wedge \cdots \wedge x_n=y_n.$ A 2-tuple is a pair, a 3-tuple a triple.

Cartesian product (Ziltener 1.15)

$X\times Y:=\{(x,y)\mid x\in X,y\in Y\},X^n:=\underset{n\text{ times}}{\underbrace{X\times \cdots \times X}}.$

A small Cartesian product

$X=\{0,1\}$, $Y=\{\text{Apple},\text{House},\text{Mountain}\}$. $X\times Y=\{(0,\text{A}),(0,\text{H}),(0,\text{M}),(1,\text{A}),(1,\text{H}),(1,\text{M})\}.$ Six pairs. In general for finite sets: $|X\times Y|=|X|\cdot |Y|$.

The set ${\mathbb{R}}^n$ is the standard $n$-dimensional space: ${\mathbb{R}}^1$ is the line, ${\mathbb{R}}^2$ the plane, ${\mathbb{R}}^3$ space.

💡 Distance in ${\mathbb{R}}^n$ — the Euclidean norm and balls

For $\mathbf{v}=(v_1,\dots ,v_n)\in {\mathbb{R}}^n$, the Euclidean norm (length): $\Vert \mathbf{v}\Vert :=\sqrt{v_1^2+v_2^2+\cdots +v_n^2}.$

This is just the Pythagorean theorem in $n$ dimensions. The distance between two points $\mathbf{x},\mathbf{y}\in {\mathbb{R}}^n$ is $\Vert \mathbf{x}-\mathbf{y}\Vert$.

Open ball, closed ball, sphere (Ziltener 1.16)

Fix a center ${\mathbf{x}}_0\in {\mathbb{R}}^n$ and radius $r\in [0,\infty ]$.

• Open ball: $B_r({\mathbf{x}}_0):=\{\mathbf{x}\in {\mathbb{R}}^n\mid \Vert \mathbf{x}-{\mathbf{x}}_0\Vert <r\}$ — strict inequality.
• Closed ball: ${\overline{B}}_r({\mathbf{x}}_0):=\{\mathbf{x}\in {\mathbb{R}}^n\mid \Vert \mathbf{x}-{\mathbf{x}}_0\Vert \le r\}$.
• Sphere: $S_r^{n-1}({\mathbf{x}}_0):=\{\mathbf{x}\in {\mathbb{R}}^n\mid \Vert \mathbf{x}-{\mathbf{x}}_0\Vert =r\}$ — the boundary surface.

What balls actually look like

• In ${\mathbb{R}}^1$: $B_r(x_0)=(x_0-r,x_0+r)$ — an open interval. The “sphere” $S^0$ is just two points.
• In ${\mathbb{R}}^2$: an open disk; $S^1$ is a circle.
• In ${\mathbb{R}}^3$: a solid ball; $S^2$ is the spherical surface (Earth’s surface, idealised).
• Edge cases: $B_0({\mathbf{x}}_0)=\varnothing$, ${\overline{B}}_0({\mathbf{x}}_0)=\{{\mathbf{x}}_0\}$, $B_{\infty }({\mathbf{x}}_0)={\mathbb{R}}^n$.

Open balls are the building blocks for the entire concept of limits and continuity later on. Worth getting comfortable with now.

💡 Functions — the rule book

Function (Ziltener 1.20)

A function (or mapping, Abbildung) is a triple $f=(X,Y,G)$ where:

• $X$ is a set, the domain (Definitionsbereich, $\mathrm{dom}f$);
• $Y$ is a set, the codomain (Zielbereich, $\mathrm{codom}f$);
• $G\subseteq X\times Y$ is the graph (Graph), satisfying: for every $x\in X$ there is exactly one $y\in Y$ with $(x,y)\in G$.

We write $f:X\to Y$ and $f(x)=y$ for that unique $y$. The arrow notation $x\mapsto f(x)$ describes the rule.

Why three pieces, not just a formula?

The same formula can give different functions depending on the domain/codomain.

• $f:\mathbb{R}\to \mathbb{R},x\mapsto x^2$ is not surjective.
• $\tilde{f}:\mathbb{R}\to [0,\infty ),x\mapsto x^2$ is surjective.
• $\hat{f}:[0,\infty )\to [0,\infty ),x\mapsto x^2$ is bijective.

Same rule, three different functions. Always specify all three pieces.

Some functions to keep in mind

• $f:\mathbb{R}\to \mathbb{R},x\mapsto x^2$.
• $g:\mathbb{R}\to \mathbb{R},x\mapsto \sin x$.
• $h:[0,\infty )\to \mathbb{R},x\mapsto \sqrt{x}$ — domain restriction is essential, since $\sqrt{\cdot }$ isn’t real for negatives.
• Identity ${\mathrm{id}}_X:X\to X$, $x\mapsto x$. The simplest non-trivial function.
• A “real-life” function: $X=\{\text{ETH students}\}$, $Y=\{\text{dates}\}$, $f(x)=$ birthday of $x$. (Each student has exactly one birthday — that’s why this is a function. Two students can share a birthday — that’s why it’s not injective.)
• From Ziltener: $X=\{\text{real polynomials}\}$, $Y=\mathcal{P}(\mathbb{R})$, $f(p)=$ zero set of $p$. E.g. $f(1)=\varnothing$, $f(x+1)=\{-1\}$, $f(x^2-1)=\{-1,1\}$.

📜 Image and preimage

Image and preimage (Ziltener 1.21)

Let $f:X\to Y$.

• For $A\subseteq X$, the image $f(A):=\{f(x)\mid x\in A\}\subseteq Y$. Special case: $\mathrm{im}(f):=f(X)$, the range.
• For $B\subseteq Y$, the preimage $f^{-1}(B):=\{x\in X\mid f(x)\in B\}\subseteq X$.
• For a single $y\in Y$: $f^{-1}(y):=f^{-1}(\{y\})=\{x\in X\mid f(x)=y\}$.

Notation trap — $f^{-1}$ has TWO meanings

$f^{-1}(B)$ is a set — defined for any function, no inversion required. It’s “all $x$ that land in $B$”; it can be empty, a single point, or many.

Don’t confuse $f^{-1}(y)$ (preimage, a set) with $\frac{1}{f(y)}$ (reciprocal, a number). They use the same symbol but mean different things.

Later, when $f$ is bijective, the symbol $f^{-1}$ also denotes the inverse function — and for a single point $y$, the preimage and inverse value agree: $f^{-1}(\{y\})=\{f^{-1}(y)\}$.

Image and preimage in action

Let $f:\mathbb{R}\to \mathbb{R},x\mapsto x^2$.

• $\mathrm{im}(f)=[0,\infty )$ — every non-negative real is hit.
• $f([1,2])=[1,4]$.
• $f^{-1}(\{4\})=\{-2,+2\}$ (two preimages).
• $f^{-1}(\{-1\})=\varnothing$ (no real square is negative).
• $f^{-1}((-\infty ,4))=(-2,2)$.

Image and preimage with a discrete function

Let $f:{\mathbb{N}}_0\to {\mathbb{N}}_0,f(n)=n^2$.

• $\mathrm{im}(f)=\{0,1,4,9,16,\dots \}$ — the perfect squares.
• $f^{-1}(\{1,2,3,4,5\})=\{1,2\}$ (since $1^2=1$ and $2^2=4$ are the only squares in that set).

🔧 Injective, surjective, bijective

(Ziltener 1.22)

Let $f:X\to Y$.

• Injective (injektiv, one-to-one): $\forall x_1,x_2\in X:f(x_1)=f(x_2)\Rightarrow x_1=x_2$. Equivalently: different inputs → different outputs.
• Surjective (surjektiv, onto): $\forall y\in Y,\exists x\in X:f(x)=y$. Every $y$ is hit.
• Bijective: both injective and surjective. Then $f$ has an inverse function.

The graphical “horizontal line test” for $f:\mathbb{R}\to \mathbb{R}$:

• Injective ⟺ every horizontal line crosses the graph at most once.
• Surjective ⟺ every horizontal line crosses at least once.
• Bijective ⟺ every horizontal line crosses exactly once.

Mental pictures

Function	Injective?	Surjective?	Bijective?
${\mathrm{id}}_X:X\to X$	✅	✅	✅
$f:[0,\infty )\to \mathbb{R},x\mapsto x$	✅	❌ (negatives missed)	❌
$f:\mathbb{R}\to [0,\infty ),x\mapsto x^2$	❌ ($f(1)=f(-1)$)	✅	❌
$f:\mathbb{R}\to \mathbb{R},x\mapsto x^2$	❌	❌	❌
$f:\mathbb{R}\to \mathbb{R},x\mapsto x^3$	✅	✅	✅
$\exp :\mathbb{R}\to (0,\infty )$	✅	✅	✅ (inverse: $\ln$)

Note how injectivity and surjectivity depend on codomain choice, not just on the formula.

💡 The inverse function

Inverse function (Ziltener 1.23)

If $f:X\to Y$ is bijective, define $f^{-1}:Y\to X,f^{-1}(y):=\text{the unique x∈X with f(x)=y.}$

Then $f^{-1}\circ f={\mathrm{id}}_X$ and $f\circ f^{-1}={\mathrm{id}}_Y$. The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y=x$.

Why bijectivity is required for an inverse

• If $f$ is not injective, two inputs share an output — so trying to invert at that output gives a multi-valued thing, not a function.
• If $f$ is not surjective, some $y$ has no preimage — so $f^{-1}(y)$ wouldn’t be defined.

Both conditions are necessary. Bijectivity = “exactly one preimage for every output”.

Common inverse functions

• ${\mathrm{id}}_X^{-1}={\mathrm{id}}_X$ — the identity is its own inverse.
• $f:[0,\infty )\to [0,\infty ),x\mapsto x^2$ has inverse $\sqrt{\cdot }:[0,\infty )\to [0,\infty )$.
• $\exp :\mathbb{R}\to (0,\infty )$ has inverse $\ln =\log :(0,\infty )\to \mathbb{R}$ (natural log).
• $\sin :[-\frac{\pi }{2},\frac{\pi }{2}]\to [-1,1]$ has inverse $\arcsin$ — note the carefully restricted domain to make it bijective.
• $\tan :(-\frac{\pi }{2},\frac{\pi }{2})\to \mathbb{R}$ has inverse $\arctan$.

🔧 Composition

Composition (Ziltener 1.25)

If $f:X\to Y$ and $g:Y\to Z$, define $(g\circ f):X\to Z,(g\circ f)(x):=g(f(x)).$ Read right-to-left: apply $f$ first, then $g$.

Properties:

• Associative: $(h\circ g)\circ f=h\circ (g\circ f)$, written without parentheses as $h\circ g\circ f$.
• NOT commutative in general: $g\circ f\ne f\circ g$. Often $f\circ g$ isn’t even defined.

Composition is not commutative

Let $f,g:\mathbb{R}\to \mathbb{R}$ with $f(x)=x+1$ and $g(y)=y^2$.

• $(g\circ f)(x)=g(x+1)=(x+1{)}^2=x^2+2x+1$.
• $(f\circ g)(x)=f(x^2)=x^2+1$.

$(x+1{)}^2\ne x^2+1$ in general. Order matters.

Composition with multiple variables

$f:{\mathbb{R}}^2\to \mathbb{R},f(x,y)=x+y$ and $g=\exp :\mathbb{R}\to \mathbb{R}$.

• $(g\circ f)(x,y)=e^{x+y}$ ✓
• $(f\circ g)$ is not defined: $g$ outputs a single real, but $f$ wants a pair as input.

Inverses via composition

If $f$ is bijective with inverse $f^{-1}$, then $f^{-1}\circ f={\mathrm{id}}_X$ and $f\circ f^{-1}={\mathrm{id}}_Y$. This is sometimes used as the definition of “inverse function” (and it’s a slick way to verify a candidate inverse: just check both compositions equal the identity).

🎯 How the exam tests this

Set/function questions are warm-up multiple choice. Example FS 2023 MC2:

True or false: for any $f:X\to Y$ and any $A,B,C\subseteq X$, we have $f((A\cup B)\cap C)=(f(A)\cup f(B))\cap f(C)$.

Answer: False. The image and intersection don’t generally commute when $f$ isn’t injective.

Tiny counterexample

Let $X=\{1,2\}$, $Y=\{0\}$, $f(x)=0$. Take $A=\{1\}$, $B=\varnothing$, $C=\{2\}$.

• LHS: $f((A\cup B)\cap C)=f(\{1\}\cap \{2\})=f(\varnothing )=\varnothing$.
• RHS: $(f(\{1\})\cup \varnothing )\cap f(\{2\})=\{0\}\cap \{0\}=\{0\}$.

$\varnothing \ne \{0\}$. Statement is false. $■$

(Lesson: Image preserves $\cup$ but not $\cap$ in general. Preimage preserves both.)

📝 Practice tasks

Task 2.1 (Set identity)

Prove $A\setminus (B\cup C)=(A\setminus B)\cap (A\setminus C)$.

Solution $x\in A\setminus (B\cup C)$ $\iff x\in A\wedge x\notin B\cup C$ $\iff x\in A\wedge \neg (x\in B\vee x\in C)$ $\iff x\in A\wedge x\notin B\wedge x\notin C$ (De Morgan) $\iff (x\in A\wedge x\notin B)\wedge (x\in A\wedge x\notin C)$ $\iff x\in (A\setminus B)\cap (A\setminus C)$. $■$

Task 2.2 (Image / preimage computation)

For $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2-1$, compute:

(a) $f([-2,2])$    (b) $f^{-1}([0,3])$    (c) $f^{-1}((-2,0))$

Solution $[-2,2]$: $x^2\in [0,4]$, so $f(x)=x^2-1\in [-1,3]$. $f([-2,2])=[-1,3]$.

(a) On

(b) $0\le x^2-1\le 3\iff 1\le x^2\le 4\iff x\in [-2,-1]\cup [1,2]$.

(c) $-2<x^2-1<0\iff -1<x^2<1$. The left inequality holds automatically ($x^2\ge 0>-1$); the right gives $|x|<1$. $f^{-1}((-2,0))=(-1,1)$.

Task 2.3 (Bijectivity & inverse)

Show that $f:\mathbb{R}\setminus \{1\}\to \mathbb{R}\setminus \{1\}$, $f(x)=\frac{x+1}{x-1}$ is bijective and find $f^{-1}$.

Solution $y=\frac{x+1}{x-1}$ for $x$: $y(x-1)=x+1\Rightarrow yx-y=x+1\Rightarrow x(y-1)=y+1\Rightarrow x=\frac{y+1}{y-1}.$

Solve

So $f^{-1}(y)=\frac{y+1}{y-1}$ — the function equals its own inverse! ($f\circ f=\mathrm{id}$, an involution.)

The formula is well-defined exactly when $y\ne 1$, matching the codomain. So $f$ is a bijection.

Task 2.4 (Composition)

Let $f(x)=2x+3$ and $g(x)=x^2$. Compute $g\circ f$ and $f\circ g$, and verify they differ.

Solution $(g\circ f)(x)=g(2x+3)=(2x+3{)}^2=4x^2+12x+9$.

$(f\circ g)(x)=f(x^2)=2x^2+3$.

At $x=1$: $g\circ f=25$, $f\circ g=5$. Different. $■$

Task 2.5 (Preimage preserves $\cap$ and $\cup$)

For any $f:X\to Y$ and any $A,B\subseteq Y$, prove: (a) $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ (b) $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$

Solution $x\in f^{-1}(A\cap B)\iff f(x)\in A\cap B\iff f(x)\in A\wedge f(x)\in B\iff x\in f^{-1}(A)\wedge x\in f^{-1}(B)\iff x\in f^{-1}(A)\cap f^{-1}(B)$.

(a)

(b) Same argument with $\vee$ instead of $\wedge$. $■$

(Compare with Task 2 of FS 2023 MC2: image is “less well-behaved” — it doesn’t preserve $\cap$ in general. Preimage is the better citizen.)

Task 2.6 (Composition of bijections)

Let $f:X\to Y$ and $g:Y\to Z$ both be bijective. Show $g\circ f:X\to Z$ is bijective and $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$ (“socks and shoes” rule).

Solution Injective: if $(g\circ f)(x_1)=(g\circ f)(x_2)$, then $g(f(x_1))=g(f(x_2))$. By injectivity of $g$: $f(x_1)=f(x_2)$. By injectivity of $f$: $x_1=x_2$. ✓

Surjective: for any $z\in Z$, surjectivity of $g$ gives $y\in Y$ with $g(y)=z$; surjectivity of $f$ gives $x\in X$ with $f(x)=y$. Then $(g\circ f)(x)=z$. ✓

Inverse formula: $(f^{-1}\circ g^{-1})\circ (g\circ f)=f^{-1}\circ (g^{-1}\circ g)\circ f=f^{-1}\circ {\mathrm{id}}_Y\circ f=f^{-1}\circ f={\mathrm{id}}_X$. Similarly the other composition. So $f^{-1}\circ g^{-1}$ is the inverse. $■$

Mnemonic: to undo “put on socks, then shoes” you do “take off shoes, then socks” — reverse order.

📚 Official problem-set tasks (Loesung 00, 02, 03, 04)

Set/function exercises taken verbatim from the official ITET HS24 Übungsserien. All solutions in toggle format.

L0.1 (graphs warm-up)

Sketch the graphs and give 4 specific points on each: (a) ${\exp }_2(x)=2^x$ (b) ${\log }_2$ (c) $\sin$ (d) $\cos$ (e) $f(x)=e^{-x^2}$ (Gaussian bell)

How are the graphs of ${\exp }_2$ and ${\log }_2$ related?

Solution

Sample points:

• (a) $(0,1),(1,2),(2,4),(3,8)$
• (b) $(1,0),(2,1),(4,2),(8,3)$
• (c) $(0,0),(\pi /2,1),(\pi ,0),(3\pi /2,-1)$
• (d) $(0,1),(\pi /2,0),(\pi ,-1),(3\pi /2,0)$
• (e) $(-1,e^{-1}),(0,1),(1,e^{-1}),(2,e^{-4})$

${\exp }_2$ and ${\log }_2$ are inverse functions of each other; their graphs are reflections across the line $y=x$.

L0.5 (set fundamentals)

(i) How many elements? (a) $\varnothing$ (b) $\{\varnothing \}$ (c) $\{0,1,0\}$ (d) $\{(-1{)}^n\mid n\in \mathbb{Z}\}$

(ii) Solution set of $x^2+x-2=0$? (iii) Sketch $\{(x,y)\in {\mathbb{R}}^2\mid x^2+y^2=1\}$ and $\{(x,y,z)\in {\mathbb{R}}^3\mid x^2+y^2+z^2=1\}$.

Solution 0 (b) 1 (c) 2 (only $0,1$ are distinct) (d) 2 (only values $1,-1$).

(i) (a)

(ii) Quadratic formula: $x=\frac{-1\pm 3}{2}$, so the set is $\{1,-2\}$.

(iii) The unit circle $S^1\subset {\mathbb{R}}^2$, and the unit sphere $S^2\subset {\mathbb{R}}^3$.

L2.1 (set operations practice)

(a) Simplify: $X=\{1,2,1+1\}$, $Y=\{0\cdot n\mid n\in {\mathbb{N}}_0\}$, $Z=\{n\in {\mathbb{N}}_0\mid n>0\wedge n\le 3\}$. (b) For $X=\{0,1\},Y=\{1,2\}$: compute $X\cap Y$, $X\cup Y$, $X\setminus Y$. Is $X\subseteq Y$? (c) For $X=\{\text{Apple, House}\},Y=\{0,1,2\},Z=\{0,1\}$: compute $X\times Y$, $Y^2$, $Z^3$.

Solution $X=\{1,2\}$ (duplicates collapse); $Y=\{0\}$ (every element equals 0); $Z=\{1,2,3\}$.

(a)

(b) $X\cap Y=\{1\}$; $X\cup Y=\{0,1,2\}$; $X\setminus Y=\{0\}$; $X⊈Y$ (since $0\notin Y$).

(c) $X\times Y$ has 6 pairs $\{(\text{A},0),\dots ,(\text{H},2)\}$; $|Y^2|=9$; $|Z^3|=8$.

L2.2 (De Morgan and distributivity for sets)

Prove: (a) $(A\cap B{)}^c=A^c\cup B^c$ (b) $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$ (c) $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$

Solution $x\in \dots$" into propositional logic and applying the corresponding propositional De Morgan / distributive law.

All three follow by translating ”

(a) $x\in (A\cap B{)}^c\iff \neg (x\in A\wedge x\in B)\iff \neg (x\in A)\vee \neg (x\in B)\iff x\in A^c\cup B^c$. $■$

(b) $x\in A\cap (B\cup C)\iff (x\in A)\wedge (x\in B\vee x\in C)\iff (x\in A\wedge x\in B)\vee (x\in A\wedge x\in C)\iff x\in (A\cap B)\cup (A\cap C)$. $■$

(c) Same trick with $\vee$ and $\wedge$ swapped.

L2.5 (function or not? — decoding the triple)

Which of the following triples $(X,Y,G)$ is a function? (a) $X=\mathbb{R},Y=\mathbb{R},G=\{(x,e^x)\mid x\in \mathbb{R}\}$ (b) $X=\mathbb{R},Y=\mathbb{R},G=\{(y^2,y)\mid y\in \mathbb{R}\}$ (c) $X=[0,\infty ),Y=\mathbb{R},G=\{(y^2,y)\mid y\in \mathbb{R}\}$ (d) $X=[0,\infty ),Y=\mathbb{R},G=\{(y^2,y)\mid y\in [0,\infty )\}$

Solution Function (each $x$ has exactly one $e^x$). (b) Not a function: $-1\in X$ has no pair in $G$ (no $y$ with $y^2=-1$). (c) Not a function: $1\in X$ has two pairs $(1,1)$ and $(1,-1)$ in $G$. (d) Function: this is exactly $f(x)=\sqrt{x}$.

(a)

L2.6 (image and preimage with full proofs)

Compute and prove: (a) $\mathrm{im}(f)$ for $f(x)=x^2$ and for $h(x)=e^x$. (b) $f^{-1}((-1,4))$, $g^{-1}([-8,-1])$ for $g(x)=x^3$, $h^{-1}([-1,1])$.

Solution $\mathrm{im}(f)=[0,\infty )$. ($\subseteq$) every square is $\ge 0$. ($\supseteq$) for $y\ge 0$ and $z:=\max \{y,1\}$ we have $f(0)=0\le y\le z\le z^2=f(z)$, so the intermediate value theorem yields some $x$ with $f(x)=y$.

(a)

$\mathrm{im}(h)=(0,\infty )$. ($\subseteq$) $e^x>0$. ($\supseteq$) For $y\ge 1$ apply IVT on $[0,y]$ since $h(0)=1\le y\le y\le e^y=h(y)$. For $y<1$, set $\tilde{y}=1/y>1$, find $\tilde{x}$ with $e^{\tilde{x}}=\tilde{y}$, then $h(-\tilde{x})=1/\tilde{y}=y$.

(b) $f^{-1}((-1,4))=\{x:-1<x^2<4\}=\{x:x^2<4\}=(-2,2)$.

$g^{-1}([-8,-1])=\{x:-8\le x^3\le -1\}=[-2,-1]$ (cube root is monotonic).

$h^{-1}([-1,1])=\{x:e^x\le 1\}=(-\infty ,0]$ (since $e^x>0$ always, the lower bound $-1$ is automatic).

L2.7 (inj/surj/bij — the full menagerie)

Classify each as injective / surjective / bijective: (a) $f:{\mathbb{N}}_0\to \{m\in {\mathbb{N}}_0:m\text{ even}\},f(n)=2n$ (b) $f:[0,\infty )\to \mathbb{R},f(x)=x$ (c) $f:\mathbb{R}\to [0,\infty ),f(x)=x^2$ (d) $f:\mathbb{R}\to \mathbb{R},f(x)=x^2$ (e) $f:\mathbb{R}\to \mathbb{R},f(x)=2x+1$ (f) $f:\mathbb{R}\to (0,\infty ),f(x)=e^{2x}$

Solution Bijective. (b) Injective only. (c) Surjective only. (d) Neither. (e) Bijective, with inverse $y\mapsto (y-1)/2$. (f) Bijective, with inverse $y\mapsto \frac{1}{2}\ln y$.

(a)

L2.8 (compute the inverses)

Find $f^{-1}$ for the bijective functions in L2.7.

Solution $f^{-1}(m)=m/2$. (e) $f^{-1}(y)=(y-1)/2$. (f) $f^{-1}(y)=\frac{1}{2}\ln (y)$ for $y>0$.

(a)

L2.9 (compositions, including ill-defined ones)

(a) $f,g:\mathbb{R}\to \mathbb{R}$ with $f(x)=x^2$, $g(y)=e^y$. Compute $g\circ f$ and $f\circ g$. (b) $f:{\mathbb{R}}^2\to \mathbb{R},f(\mathbf{x})=\Vert \mathbf{x}\Vert$; $g:\mathbb{R}\to \mathbb{R},g(y)=e^y$. Compute $g\circ f$ and $g\circ f(1,2)$. What about $f\circ g$?

Solution $(g\circ f)(x)=e^{x^2}$; $(f\circ g)(y)=(e^y{)}^2=e^{2y}$.

(a)

(b) $(g\circ f)(x_1,x_2)=e^{\sqrt{x_1^2+x_2^2}}$; at $(1,2)$: $e^{\sqrt{5}}$. $f\circ g$ is not well-defined: $g$ produces a scalar, but $f$ takes a vector in ${\mathbb{R}}^2$.

L2.10 (image/preimage and set operations — full proofs)

(a) Prove $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$, $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$, and $f(A_1\cup A_2)=f(A_1)\cup f(A_2)$. (b) Find a counterexample to $f(A_1\cap A_2)=f(A_1)\cap f(A_2)$.

Solution $x\in f^{-1}(B_1\cap B_2)\iff f(x)\in B_1\wedge f(x)\in B_2\iff x\in f^{-1}(B_1)\cap f^{-1}(B_2)$. $■$

(a) All three follow by translating to logic. E.g.:

(b) Take $f:\{1,2\}\to \{1\}$ with $f\equiv 1$, $A_1=\{1\},A_2=\{2\}$. Then $A_1\cap A_2=\varnothing$, so $f(A_1\cap A_2)=\varnothing$. But $f(A_1)\cap f(A_2)=\{1\}\cap \{1\}=\{1\}$. Different.

L2.12 (multiple choice — set/function traps)

(a) Are $f(f^{-1}(B))=B$ and $f^{-1}(f(A))=A$ always true? (b) $||x-2|-1|<3\iff ?$ (c) If $g\circ f={\mathrm{id}}_X$, what follows? (i) $f$ injective, (iv) $g$ surjective. (d) $f$ even, $g$ odd: which products are even/odd? (e) Inverse of $f:[0,\infty )\to \mathbb{R},f(x)=x^4$? (f) Negate $\forall a\in A,\exists b\in B:a\le b$.

Solution false in general: equality requires extra hypotheses (surjectivity for the first, injectivity for the second). (b) $||x-2|-1|<3\iff -3<|x-2|-1<3\iff |x-2|<4\iff$ $-2<x<6$. (c) Both (i) and (iv): $g\circ f={\mathrm{id}}_X$ forces $f$ to be injective and $g$ to be surjective. (d) $fg$ is odd (even·odd = odd); $f{g}^2$ is even (even·even = even). (e) The inverse doesn't exist as a map to $\mathbb{R}$: $f$ isn't surjective onto $\mathbb{R}$ (range is $[0,\infty )$). (f) $\exists a\in A:\forall b\in B:a>b$.

(a) Both

L3.1 (element vs subset — fundamentals)

Using $0:=\varnothing$, $1:=\{0\}$, $2:=\{0,1\}$, $3:=\{0,1,2\}$, …, decide: (a) $1\in \{1\}$ (b) $1\subseteq \{1\}$ (c) $\varnothing \in \{\varnothing \}$ (d) $\varnothing \subseteq \{\varnothing \}$ (e) $\forall X:\varnothing \subseteq X$

Solution True ($1$ is the unique element). (b) False: $1=\{0\}$, so $1\subseteq \{1\}$ would need $0\in \{1\}$ — but $\{1\}$'s only element is $1$, not $0$. (c) True. (d) True: $\varnothing$ is a subset of every set. (e) True (vacuously: there's nothing to check).

(a)

L4.1 (cardinality / equinumerosity)

Show ${\mathbb{N}}_0$ and $\{\text{perfect squares}\}=\{0,1,4,9,\dots \}$ have the same cardinality.

Solution $f:{\mathbb{N}}_0\to \{\text{perfect squares}\}$, $f(n)=n^2$.

Define

• Injective: if $n_1^2=n_2^2$ with $n_1,n_2\ge 0$, then $n_1=n_2$.
• Surjective: every perfect square is by definition some $n^2$.

So $f$ is a bijection. The two sets are equinumerous (despite one being a strict subset of the other — the hallmark of infinite sets, Galileo’s paradox).

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3 · Numbers — from $\mathbb{N}$ to $\mathbb{C}$

🤔 Why so many number systems?

Every time a problem couldn’t be solved, mathematicians invented a new kind of number.

Couldn’t solve…	so we invented…
$3-5=?$ in $\mathbb{N}$	$\mathbb{Z}$ (integers)
$1÷2=?$ in $\mathbb{Z}$	$\mathbb{Q}$ (rationals)
$\sqrt{2}=?$ in $\mathbb{Q}$	$\mathbb{R}$ (reals)
$\sqrt{-1}=?$ in $\mathbb{R}$	$\mathbb{C}$ (complex)

So: $\mathbb{N}\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}\subset \mathbb{C}$.

💡 The real numbers — the “no gaps” property

The rationals have holes: $\sqrt{2}\notin \mathbb{Q}$ (Pythagoras already knew). Calculus needs a number system without holes — that’s $\mathbb{R}$.

Bounds, supremum, infimum

Let $S\subseteq \mathbb{R}$ be non-empty.

• $S$ is bounded above if $\exists M:\forall x\in S,x\le M$. $M$ is an upper bound.
• The supremum $\sup S$ is the least upper bound.
• The infimum $\inf S$ is the greatest lower bound.
• If $\sup S\in S$, it’s the maximum $\max S$. Likewise for $\min$.

Completeness axiom of $\mathbb{R}$

Every non-empty bounded-above subset of $\mathbb{R}$ has a supremum in $\mathbb{R}$.

This is the single most important property of $\mathbb{R}$. Every limit theorem ultimately relies on it.

Sup vs. max

$S=\{x\in \mathbb{R}:x^2<2\}=(-\sqrt{2},\sqrt{2})$. $\sup S=\sqrt{2}$, but $\sqrt{2}\notin S$, so $S$ has no maximum.

Archimedean property

For every $x\in \mathbb{R}$ there is $n\in \mathbb{N}$ with $n>x$. Hence $\frac{1}{n}\to 0$.

💡 The Euclidean space ${\mathbb{R}}^d$

Stack $d$ reals into a vector $\mathbf{x}=(x_1,\dots ,x_d)$.

Norm and dot product

$\Vert \mathbf{x}\Vert :=\sqrt{x_1^2+\cdots +x_d^2},\mathbf{x}\cdot \mathbf{y}:={\sum }_{i=1}^d{x}_i{y}_i$

Three core inequalities

1. Cauchy–Schwarz: $|\mathbf{x}\cdot \mathbf{y}|\le \Vert \mathbf{x}\Vert \Vert \mathbf{y}\Vert$
2. Triangle: $\Vert \mathbf{x}+\mathbf{y}\Vert \le \Vert \mathbf{x}\Vert +\Vert \mathbf{y}\Vert$
3. Reverse triangle: $|\Vert \mathbf{x}\Vert -\Vert \mathbf{y}\Vert |\le \Vert \mathbf{x}-\mathbf{y}\Vert$

💡 Complex numbers

Define $i$ with $i^2=-1$ and form $\mathbb{C}=\{a+bi:a,b\in \mathbb{R}\}$.

You can think of $\mathbb{C}$ as ${\mathbb{R}}^2$ — the complex plane — with horizontal axis = real part, vertical axis = imaginary part.

Complex operations

For $z=a+bi$ and $w=c+di$: $z+w=(a+c)+(b+d)i$ $z\cdot w=(ac-bd)+(ad+bc)i$ $\overline{z}=a-bi\text{(complex conjugate)}$ $|z|=\sqrt{a^2+b^2}=\sqrt{z\bar{z}}$ $\frac{1}{z}=\frac{\bar{z}}{|z{|}^2}(z\ne 0)$

Inverse computation (from Ziltener)

$\frac{1}{1+i}=\frac{1-i}{|1+i{|}^2}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}$

💡 Polar form and Euler’s formula — the magic shortcut

Every $z\ne 0$ can be written as $z=r(\cos \phi +i\sin \phi ),r=|z|,\phi \in [0,2\pi )$ or equivalently — via Euler’s formula: $\boxed{e^{i\phi }=\cos \phi +i\sin \phi }$ so $z=r{e}^{i\phi }$.

Why polar form is amazing

Multiplication becomes trivial: $(r{e}^{i\phi })(s{e}^{i\psi })=rs{e}^{i(\phi +\psi )}$ Multiply moduli, add angles. Powers and roots become easy.

Power via polar form

$1+i=\sqrt{2}{e}^{i\pi /4}$, so $(1+i{)}^8=(\sqrt{2}{)}^8\cdot e^{i\cdot 8\cdot \pi /4}=16\cdot e^{i2\pi }=16$.

🎯 Roots of complex numbers

Every $z=r{e}^{i\phi }\ne 0$ has exactly $k$ different $k$-th roots: $w_j=\sqrt[k]{r}{e}^{i(\phi +2\pi j)/k},j=0,1,\dots ,k-1$

Fundamental Theorem of Algebra

Every non-constant complex polynomial $p(z)=a_n{z}^n+\cdots +a_0$ has at least one complex root. Hence it factors completely into linear factors over $\mathbb{C}$.

This is why $\mathbb{C}$ is algebraically closed — and why FS 2023 MC3 (“does there exist a degree-3 polynomial with no complex root?”) is false.

📝 Practice tasks

Task 3.1 — Compute $(1-i{)}^4$.

Solution $1-i=\sqrt{2}{e}^{-i\pi /4}$, so $(1-i{)}^4=(\sqrt{2}{)}^4{e}^{-i\pi }=4(-1)=-4$.

Task 3.2 — Find all $w\in \mathbb{C}$ with $w^3=-8$.

Solution $-8=8e^{i\pi }$. The three cube roots: $w_j=2e^{i(\pi +2\pi j)/3},j=0,1,2$ Concretely: $w_0=2e^{i\pi /3}=1+i\sqrt{3}$, $w_1=2e^{i\pi }=-2$, $w_2=2e^{i5\pi /3}=1-i\sqrt{3}$.

📚 Official problem-set tasks (Loesung 02, 03, 04)

L2.11 (compute sup, inf, max, min)

For each of the following subsets of $\mathbb{R}$, compute $\sup$, $\inf$, and decide whether max/min exist: (a) $A:=\{1/n\mid n\in \mathbb{N}\}$ (b) $B:=\{1/x\mid x\in (1,2]\}$ (c) $C:=\{x/(x+1)\mid x\in [2,\infty )\}$

Solution $A=\{1,1/2,1/3,\dots \}$. $\sup A=1=\max A$; $\inf A=0$ (not attained, no min). (b) $B=[\frac{1}{2},1)$. $\inf B=1/2=\min B$; $\sup B=1$ (not attained, no max). (c) The function $x\mapsto x/(x+1)=1-1/(x+1)$ is strictly increasing. At $x=2$ it equals $2/3$. As $x\to \infty$ it tends to 1. So $C=[2/3,1)$: $\inf C=2/3=\min C$; $\sup C=1$ (no max).

(a)

L3.2 (Dedekind cuts in action)

Recall $\bar{r}:=\{s\in \mathbb{Q}:s>r\}$ for $r\in \mathbb{Q}$, and addition/multiplication of cuts. (a) Show $\bar{1}+\bar{2}=\bar{3}$. (b) Show $\bar{1}\cdot \bar{1}=\bar{1}$. (c) Define $\sqrt{2}:=\{r\in \mathbb{Q}:r\ge 0,r^2>2\}$. Verify $\sqrt{2}$ is a Dedekind cut.

Solution $\bar{1}+\bar{2}=\{r+s:r>1,s>2\}$. ($\subseteq$): if $r>1,s>2$ then $r+s>3$. ($\supseteq$): given $t>3$, set $r=1+(t-3)/2$ and $s=2+(t-3)/2$; then $r>1,s>2$, and $r+s=t$. $■$

(a)

(b) Same idea: ($\subseteq$) $rs>1$ when $r,s>1$; ($\supseteq$) given $t>1$, take $r=(t+1)/2>1$ and $s=t/r$, then $s>1$ (algebra) and $rs=t$.

(c) Check the four Dedekind cut axioms:

• Nonempty: $r=2$ has $r^2=4>2$. ✓
• $\ne \mathbb{Q}$: $r=1$ has $r^2=1\ngtr 2$. ✓
• Upward closed: if $r\in \sqrt{2}$ and $s>r$, then $s^2>r^2>2$. ✓
• No smallest: given $r\in \sqrt{2}$, set $s_0:=(2r+2)/(r+2)$. Then $s_0<r$ and $s_0^2>2$ (after expanding). ✓

L3.6 (sup and inf of $-A$)

Let $\varnothing \ne A\subseteq \mathbb{R}$ be bounded, and define $-A:=\{-a:a\in A\}$. Show: $\sup (-A)=-\inf A,\inf (-A)=-\sup A.$

Solution $u$ is a lower bound of $A\iff \forall a\in A:u\le a\iff \forall a:-u\ge -a\iff -u$ is an upper bound of $-A$.

Note:

So $\inf A$ being a lower bound of $A$ means $-\inf A$ is an upper bound of $-A$, hence $\sup (-A)\le -\inf A$. Conversely, $\sup (-A)$ being an upper bound of $-A$ makes $-\sup (-A)$ a lower bound of $A$, so $-\sup (-A)\le \inf A$, i.e. $\sup (-A)\ge -\inf A$. Equality. $■$

The second identity follows from the first applied to $-A$ in place of $A$.

L3.7 (complex arithmetic — bring to standard form)

Compute in form $a+bi$: (a) $(3+2i)(6-5i)$ (b) $\frac{1}{1+i}$ (c) $\frac{3+4i}{2-i}$ (d) ${\left(\frac{1+i}{1-i}\right)}^n$ for $n\in \mathbb{N}$ (e) $(1+i{)}^2+(\overline{1+i}{)}^2$ (f) ${\left(\frac{1+i\sqrt{3}}{2}\right)}^3$ (g) $(1+i{)}^6$ via polar form

Solution $18-15i+12i+10=28-3i$. (b) $\frac{1}{1+i}\cdot \frac{1-i}{1-i}=\frac{1-i}{2}=\frac{1}{2}-\frac{1}{2}i$. (c) $\frac{(3+4i)(2+i)}{(2-i)(2+i)}=\frac{2+11i}{5}=\frac{2}{5}+\frac{11}{5}i$. (d) $\frac{1+i}{1-i}=i$, so the answer is $i^n$ — periodic with period 4: $1,i,-1,-i,1,\dots$. (e) $(1+i{)}^2=2i$, $(\overline{1+i}{)}^2=(1-i{)}^2=-2i$, sum $=0$. (f) $(1+i\sqrt{3}{)}^3=1+3i\sqrt{3}+3(i\sqrt{3}{)}^2+(i\sqrt{3}{)}^3=1+3i\sqrt{3}-9-3\sqrt{3}i=-8$. Divide by $8$: $-1$. (g) $1+i=\sqrt{2}\mathrm{cis}(\pi /4)$, so $(1+i{)}^6=8\mathrm{cis}(3\pi /2)=-8i$.

(a)

L3.8 (point sets in $\mathbb{C}$)

Sketch: (a) $M_1:=\{z\in \mathbb{C}:1<|z-i|<2\}$ (b) $M_2:=\{z\in \mathbb{C}:|z-1|=|z+1|\}$

Solution annulus centered at $i$, between radii 1 and 2.

(a) Open

(b) Set $z=x+iy$: $(x-1{)}^2+y^2=(x+1{)}^2+y^2\iff -2x=2x\iff x=0$. The imaginary axis (perpendicular bisector of $-1$ and $1$).

L3.9 (zeros of complex quadratics)

For $P(z)=a{z}^2+bz+c$ ($a\ne 0$), the zeros are ${\alpha }_{\pm }=(-b\pm \sqrt{b^2-4ac})/(2a)$. Apply to: (a) $z^2+6z+10$ (b) $4z^2+4iz-1$ (c) $(z^2+1)(z-3i{)}^2$

Solution $36-40=-4$, so $\sqrt{-4}=2i$. Roots: $(-6\pm 2i)/2=-3\pm i$.

(a) Discriminant

(b) Discriminant $-16-(-16)=0$, root $-i/2$ (double).

(c) Factor reveals roots $\pm i$ and $3i$ (with $3i$ a double root).

L3.10 / L4.11 (online MC — counting roots, sup, geometry, polar)

(a) How many distinct zeros does $P(z)=z(z^2+1{)}^2-z^2-z^5$ have? (b) Maximum of $A:=\{1/n:n\in \mathbb{N}\}\cup (2,4)$? (c) Geometric shape of $M:=\{c\in \mathbb{C}:|c-1|=2\}$?

Solution $P(z)=z(z^4+2z^2+1)-z^2-z^5=z^5+2z^3+z-z^2-z^5=2z^3-z^2+z=z(2z^2-z+1)$. Three distinct roots: $0$ and the two roots of $2z^2-z+1$ (discriminant $1-8=-7$). Answer: 3.

(a) Expand:

(b) Max does not exist: $\sup A=4\notin A$.

(c) Circle centered at $1$ with radius $2$.

L4.2 (the cis-function and powers)

Bring to standard form: (a) $(1+i{)}^2$ (b) $(1+i)(1-i)$ (c) $1/(1-i)$ (d) $\mathrm{cis}(\pi /4),\mathrm{cis}(\pi /3),\mathrm{cis}(\pi /6)$ (e) $(1+i{)}^{10}$ (f) $(\frac{1+\sqrt{3}i}{2}{)}^6$

Prove: $\mathrm{cis}(\phi +\psi )=\mathrm{cis}(\phi )\mathrm{cis}(\psi )$ and $\mathrm{cis}(k\pi /2)=i^k$ for $k\in \mathbb{Z}$.

Solution $2i$. (b) $2$. (c) $\frac{1}{2}+\frac{1}{2}i$. (d) $\frac{\sqrt{2}}{2}(1+i)$, $\frac{1}{2}+\frac{\sqrt{3}}{2}i$, $\frac{\sqrt{3}}{2}+\frac{1}{2}i$. (e) Polar: $(\sqrt{2}{)}^{10}\mathrm{cis}(10\pi /4)=32\mathrm{cis}(5\pi /2)=32i$. (f) Polar: $\mathrm{cis}(6\cdot \pi /3)=\mathrm{cis}(2\pi )=1$.

(a)

The addition formula $\mathrm{cis}(\phi +\psi )=\mathrm{cis}\phi \cdot \mathrm{cis}\psi$ unpacks to the classical sin/cos addition theorems. The second identity follows iteratively from $\mathrm{cis}(\pi /2)=i$.

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4 · Sequences — what does “approaching” mean?

🤔 The big question

If a turtle moves $1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$ meters per step, where does it end up? Intuitively at $0$ — but how do we make that precise without hand-waving?

Answer: the epsilon-N definition of a limit. This single idea is the foundation of every theorem in calculus.

💡 What is a sequence?

Sequence

A sequence is an infinite ordered list $a_0,a_1,a_2,\dots$ — equivalently, a function $a:{\mathbb{N}}_0\to \mathbb{R}$ (or $\mathbb{C}$). Notation: $(a_n{)}_{n\in {\mathbb{N}}_0}$.

Three classics

• Harmonic $a_n=\frac{1}{n}$: terms $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$
• Geometric $a_n=q^n$: terms $1,q,q^2,q^3,\dots$
• Constant $a_n=c$

📜 The definition that started it all

Convergence

$(a_n)$ converges to $A$ — written $a_n\to A$ or ${\lim }_{n\to \infty }{a}_n=A$ — iff $\forall \epsilon >0\exists n_0\in \mathbb{N}\forall n\ge n_0:|a_n-A|<\epsilon .$

How to read this:

• $\epsilon$ is a “challenge” (any tolerance the prover throws at you, no matter how tiny).
• You answer with $n_0$ (an index from which on the sequence is closer than $\epsilon$ to $A$).
• Smaller $\epsilon$ usually requires bigger $n_0$.

$\frac{1}{n}\to 0$

Given $\epsilon >0$, choose any $n_0>\frac{1}{\epsilon }$. Then for $n\ge n_0$: $\left|\frac{1}{n}-0\right|=\frac{1}{n}\le \frac{1}{n_0}<\epsilon ✓$

🔧 Limit arithmetic — your daily toolkit

If $a_n\to A$ and $b_n\to B$, then:

• $a_n+b_n\to A+B$
• $a_n\cdot b_n\to A\cdot B$
• $a_n/b_n\to A/B$ (provided $B\ne 0$)
• $|a_n|\to |A|$

Squeeze (Sandwich) Theorem

If $a_n\le c_n\le b_n$ for all $n\ge N$ and $a_n\to L$ and $b_n\to L$, then $c_n\to L$.

🔧 Standard limits — memorize these

Sequence	Limit	Condition
$\frac{1}{n^p}$	$0$	$p>0$
$q^n$	$0$	$|q|<1$
$\sqrt[n]{n}$	$1$	—
$\sqrt[n]{a}$	$1$	$a>0$
$\frac{n^p}{q^n}$	$0$	$|q|>1$
${\left(1+\frac{1}{n}\right)}^n$	$e$	—
$\frac{n!}{n^n}$	$0$	—

💡 The four pillars of convergence

Monotone convergence

A monotonically increasing sequence bounded above converges (to its supremum). Likewise for decreasing & bounded below.

Bolzano–Weierstrass

Every bounded sequence in ${\mathbb{R}}^d$ has a convergent subsequence.

Cauchy criterion

A sequence converges $⟺$ it is Cauchy: $\forall \epsilon >0\exists n_0:\forall n,m\ge n_0:|a_n-a_m|<\epsilon$

Why Cauchy is golden

You can prove convergence without knowing the limit — just check the terms huddle together.

🔧 Worked exam-style examples

FS 2023 MC4 — algebraic limit

$a_n=\frac{n^3}{(n+1{)}^2}-\frac{n^2}{n+1}$.

Common denominator $(n+1{)}^2$: $a_n=\frac{n^3-n^2(n+1)}{(n+1{)}^2}=\frac{-n^2}{(n+1{)}^2}$ Divide by $n^2$ in numerator and denominator: $a_n=-\frac{1}{(1+1/n{)}^2}\to -1$.

FS 2023 MC5 — conjugate trick

$a_n=\sqrt{n+1}-\sqrt{n}$.

Multiply by conjugate $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$: $a_n=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$

💡 Limes superior and inferior

For a bounded sequence, define ${\mathrm{limsup}}_{n\to \infty }{a}_n:={\lim }_{n\to \infty }\left({\sup }_{k\ge n}{a}_k\right),{\mathrm{liminf}}_{n\to \infty }{a}_n:={\lim }_{n\to \infty }\left({\inf }_{k\ge n}{a}_k\right)$

Theorem

A bounded sequence converges $⟺$ $\mathrm{limsup}{a}_n=\mathrm{liminf}{a}_n$.

🎯 How the exam tests sequences

• Compute a limit using algebra/conjugates (FS 2023 MC4, MC5).
• Identify which standard tool applies (squeeze, monotone convergence).
• Find limits of recursive sequences (HS 2013 task 2 — golden ratio recursion).

📝 Practice tasks

Task 4.1 — Compute ${\lim }_{n\to \infty }\frac{3n^2-n}{2n^2+5}$.

Solution $n^2$: $\frac{3-1/n}{2+5/n^2}\to \frac{3}{2}$.

Divide top and bottom by

Task 4.2 — Show $a_n=(-1{)}^n$ does not converge.

Solution $a_{2k}=1\to 1$ and $a_{2k+1}=-1\to -1$. Different subsequence limits ⇒ no overall limit.

Two subsequences:

Task 4.3 (FS 2013 task 2 — resistor circuit)

A resistor network gives total resistance $R_n=R_a+\frac{n{R}_b{R}_c}{R_b+n{R}_c}+R_d$. Does $(R_n)$ converge? If so, to what?

Solution ${\lim }_{n\to \infty }{R}_n=R_a+{\lim }_{n\to \infty }\frac{R_b{R}_c}{R_b/n+R_c}+R_d=R_a+R_b+R_d$. Converges.

📚 Official problem-set tasks (Loesung 00, 04, 05)

L0.6 (sequences warm-up)

(i) First six terms of $a_n:=n^2$. (ii) Fibonacci: $a_0=0,a_1=1,a_n=a_{n-2}+a_{n-1}$. First six terms. (iii) Convergence of $(n^2{)}_n$, $(1/n{)}_n$, $((n+1)/(n+2){)}_n$.

Solution $0,1,4,9,16,25$. (ii) $0,1,1,2,3,5$. (iii) Diverges to $\infty$; converges to $0$; converges to $1$.

(i)

L4.5 (ε-N proofs of convergence)

Prove via the ε-N definition: (a) $a_n:=2/n\to 0$. (b) $a_n:=1+1/\sqrt{n}\to 1$. (c) $a_n:=2/n+1+1/\sqrt{n}\to 1$. (d) $a_n:=(-1{)}^n$ diverges.

Solution $\epsilon >0$, choose $n_0\ge 2/\epsilon$ (Archimedes). For $n\ge n_0$: $|a_n-0|=2/n\le 2/n_0\le \epsilon$. ✓

(a) Given

(b) Choose $n_0\ge 1/{\epsilon }^2$. For $n\ge n_0$: $|a_n-1|=1/\sqrt{n}\le 1/\sqrt{n_0}\le \epsilon$. ✓

(c) Sum of (a) and (b); limit is $0+1=1$. (Or: directly use the limit-laws.)

(d) Suppose $a_n\to A$. Pick $\epsilon =1/2$. For all large $n$, $|(-1{)}^n-A|<1/2$. But $(-1{)}^n$ alternates between $1$ and $-1$, which differ by $2$, contradicting “all within $1/2$ of one fixed $A$“.

L4.7 (zero × bounded = zero)

Suppose $a_n\to 0$ and $(b_n)$ is bounded ($|b_n|\le C$). Show $a_n{b}_n\to 0$.

Solution $\epsilon >0$, set $\tilde{\epsilon }:=\epsilon /C$. Pick $N$ so $|a_n|<\tilde{\epsilon }$ for $n\ge N$. Then $|a_n{b}_n|\le |a_n|\cdot C<(\epsilon /C)\cdot C=\epsilon$. $■$

Given

Used constantly: any $1/n$-style decay times any bounded oscillation (sin, cos, $(-1{)}^n$) goes to zero.

L4.8 (compute these limits)

(a) $\lim \frac{n^2-n+3}{n^2+2}$ (b) $\lim \frac{n^3-n^2+3}{2^n(n^2+5)}$ (c) $\lim \frac{\sqrt{n^2-1}}{n}$ (d) $\lim n^2\left(\frac{\sin n}{2^n}+\frac{\cos n}{n^4}\right)$

Solution $n^2$: $\frac{1-1/n+3/n^2}{1+2/n^2}\to 1$. (b) Numerator grows polynomially, denominator exponentially; limit $0$. (c) $\sqrt{1-1/n^2}\to 1$. (Conjugate trick gives $|\sqrt{1-1/n^2}-1|\le 1/n^2\to 0$.) (d) Both terms are (zero-sequence)·(bounded). Use L4.7: limit $=0+0=0$.

(a) Divide by

L4.10 (Heron's method for $\sqrt{c}$)

Fix $c\ge 1$. Define $a_1:=c$, $a_{n+1}:=\frac{1}{2}(a_n+c/a_n)$. (a) Show $1\le a_n\le c$ for all $n$. (b) Show $a_n^2\ge c$. (c) Show $(a_n)$ is monotonically decreasing. (d) Conclude $a_n\to \sqrt{c}$. (e) Numerical check for $c=3$.

Solution $a_{n+1}=\frac{1}{2}(a_n+c/a_n)\in [\frac{1}{2}(1+1),\frac{1}{2}(c+c)]=[1,c]$.

(a) Induction:

(b) $a_{n+1}^2-c=\frac{1}{4}(a_n+c/a_n{)}^2-c=\frac{1}{4}(a_n-c/a_n{)}^2\ge 0$.

(c) $a_n-a_{n+1}=\frac{1}{2a_n}(a_n^2-c)\ge 0$ by (b).

(d) Bounded + monotone ⇒ convergent (Chapter 4 main theorem). The limit $a$ satisfies $a=\frac{1}{2}(a+c/a)$, so $a^2=c$, hence $a=\sqrt{c}$.

(e) $c=3$: $a_1=3,a_2=2,a_3=1.75,a_4\approx 1.73214,a_5\approx 1.73205081$ — matches $\sqrt{3}\approx 1.732050808$ to 7 decimals. Doubling of correct digits per step (quadratic convergence).

L4.11 (online MC — limits)

(a) $\lim \frac{16n^3+100n+10^6}{27n^3+10920n+2020}$ (b) $\lim \frac{n^2}{2^n{n}^2+8}$ (c) $\lim \frac{n^3+22n^2-10}{29n^2-27n+8}$

Solution $16/27$ — leading coefficients dominate. (b) $0$ — exponential beats polynomial. (c) Diverges to $+\infty$ (numerator degree exceeds denominator).

(a)

L5.1 (lim sup and lim inf)

Compute: (a) ${\mathrm{limsup}}_{k\to \infty }(-1{)}^k(1+1/k)$ (b) ${\mathrm{liminf}}_{k\to \infty }(-1{)}^k(1+1/k)$ (c) ${\mathrm{limsup}}_{k\to \infty }k(1+(-1{)}^k)$ (d) ${\mathrm{liminf}}_{k\to \infty }k(1+(-1{)}^k)$

Solution $\to 1^{+}$, odd subsequence $\to -1^{-}$. $\mathrm{limsup}=1$. (b) $\mathrm{liminf}=-1$. (c) Even: $a_{2k}=4k\to \infty$. $\mathrm{limsup}=+\infty$. (d) Odd: $a_{2k+1}=0$ for all $k$. $\mathrm{liminf}=0$.

(a) Even subsequence

L5.2 (vector convergence and $n$-th roots)

(a) Convergence of $a_n:=((n+1)/n,2^{-n})\in {\mathbb{R}}^2$? (b) Show $\sqrt[n]{n}\to 1$. (c) Show $\sqrt[n]{n!}\to \infty$.

Solution $1+1/n\to 1$ and $2^{-n}\to 0$. Limit $(1,0)$.

(a) Component-wise:

(b) Write $\sqrt[n]{n}=1+{\delta }_n$ with ${\delta }_n\ge 0$. Then $n=(1+{\delta }_n{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2}){\delta }_n^2=\frac{n(n-1)}{2}{\delta }_n^2$, so ${\delta }_n^2\le \frac{2}{n-1}\to 0$, hence ${\delta }_n\to 0$. ✓

(c) Hint: $(2k)!\ge k^k$, so $\sqrt[n]{n!}\ge \sqrt{n/2}$ for both even and odd $n$. Given $C>0$, pick $n_0>2C^2$; for $n\ge n_0$: $\sqrt[n]{n!}\ge \sqrt{n/2}>C$. So $\sqrt[n]{n!}\to \infty$. $■$

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5 · Series — adding infinitely many things

🤔 The big question

What is $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$? You can’t literally add infinitely many things — but you can take the limit of partial sums.

📜 Series — formal definition

Series

Given $(a_k)$, define partial sums $S_n={\sum }_{k=1}^n{a}_k$. The series ${\sum }_{k=1}^{\infty }{a}_k$ converges iff $(S_n)$ converges, in which case ${\sum }_{k=1}^{\infty }{a}_k:={\lim }_{n\to \infty }{S}_n.$

💎 Two essential examples

Geometric series — THE most important series

For $|q|<1$: ${\sum }_{k=0}^{\infty }{q}^k=\frac{1}{1-q}$

Proof sketch: $S_n=1+q+\cdots +q^n$, so $(1-q)S_n=1-q^{n+1}\to 1$.

Zeno's paradox solved

Achilles runs $1+\frac{1}{2}+\frac{1}{4}+\cdots =\frac{1}{1-1/2}=2$ km. Finite!

Square inscribed inside square (HS 2013 task 2a)

Each new square has side $a_{k+1}=\frac{1}{\sqrt{2}}{a}_k$. Total perimeter: $U={\sum }_{k=1}^{\infty }4a_k=4a_1{\sum }_{k=1}^{\infty }(\frac{1}{\sqrt{2}}{)}^{k-1}=\frac{4a_1}{1-1/\sqrt{2}}=4(2+\sqrt{2})$ for $a_1=1$.

Harmonic series diverges

${\sum }_{k=1}^{\infty }\frac{1}{k}=+\infty$ Even though $\frac{1}{k}\to 0$.

Why? Group terms: $1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\cdots +\frac{1}{8})+\cdots$ — each group $\ge \frac{1}{2}$, infinitely many groups.

The lesson: $a_k\to 0$ is necessary for $\sum a_k$ to converge — but not sufficient.

🔧 Convergence tests — your decision tree

Test	When to use	Statement
n-th term	always check first	If $a_k\nrightarrow 0$, diverges
Geometric	$a_k=q^k$	converges iff $|q|<1$
Comparison	bound by simpler series	$0\le a_k\le b_k$ and $\sum b_k$ converges ⇒ $\sum a_k$ converges
Ratio test	factorials / exponentials	$\lim \left|\frac{a_{k+1}}{a_k}\right|=L$: $L<1$ converges, $L>1$ diverges
Root test	$k$-th powers	$\lim \sqrt[k]{|a_k|}=L$: same conclusion
Leibniz	alternating series	$a_k\ge 0$ decreasing to $0$ ⇒ $\sum (-1{)}^{k+1}{a}_k$ converges

FS 2023 MC7 — pick the divergent series

$\sum \frac{(n+1{)}^2}{n^3-2}$ behaves like $\sum \frac{n^2}{n^3}=\sum \frac{1}{n}$ — divergent (harmonic-like).

FS 2019 task 2

$\sum (-1{)}^n\frac{n+1}{2n+1}$. Since $\frac{n+1}{2n+1}\to \frac{1}{2}\ne 0$, the n-th-term test ⇒ diverges.

💡 Absolute vs. conditional convergence

Definition

$\sum a_k$ converges absolutely iff $\sum |a_k|$ converges.

Theorem

Absolute convergence ⇒ convergence (but not vice versa).

Conditional convergence

${\sum }_{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}=1-\frac{1}{2}+\frac{1}{3}-\cdots =\ln 2$ converges (Leibniz), but not absolutely.

🚀 Power series — calculus’s superpower

Power series

${\sum }_{k=0}^{\infty }{a}_k(z-z_0{)}^k$

Has a radius of convergence $R\in [0,\infty ]$, computed by $\frac{1}{R}={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}\text{(Cauchy–Hadamard)}$ or by ratio test: $\frac{1}{R}={\lim }_{k\to \infty }\left|\frac{a_{k+1}}{a_k}\right|$

• $|z-z_0|<R$ → converges
• $|z-z_0|>R$ → diverges
• At $|z-z_0|=R$ → must be checked separately

FS 2019 MC1

$\sum \frac{(n!{)}^2}{(2n)!}{x}^n$. Ratio: $\frac{(n!{)}^2/(2n)!}{((n+1)!{)}^2/(2n+2)!}=\frac{(2n+1)(2n+2)}{(n+1{)}^2}\to 4$ So $R=4$.

💎 The exponential series

The single most important power series: $e^z:={\sum }_{k=0}^{\infty }\frac{z^k}{k!},R=\infty$

It satisfies $e^{z+w}=e^z\cdot e^w$ — and from this single fact follow Euler’s formula, the trigonometric identities, and (via $e^{i\pi }=-1$) the bridge between exponentials and trigonometry.

🎯 How the exam tests series

• Compute radius of convergence (FS 2019 MC1, FS 2023 task 2)
• Decide convergence (FS 2023 MC7, FS 2019 task 2)
• Sum a geometric-type series (HS 2013 task 2)
• Show uniform convergence of partial sums of a power series (FS 2023 task 2)

📝 Practice tasks

Task 5.1 — Find $R$ for $\sum \frac{x^k}{k{2}^k}$.

Solution $\sqrt[k]{|a_k|}=\frac{1}{\sqrt[k]{k}\cdot 2}\to \frac{1}{2}$, so $R=2$.

Task 5.2 — Sum ${\sum }_{k=0}^{\infty }\frac{1}{3^k}$.

Solution $q=1/3$: $\frac{1}{1-1/3}=\frac{3}{2}$.

Geometric with

Task 5.3 (HS 2013 task 2b — nested triangles)

Triangles inscribed each at half the side length. The first triangle has area $F_1$. What percentage of the total area $\sum F_k$ is $F_1$?

Solution $F_k=(\frac{1}{4}{)}^{k-1}{F}_1$, so $\sum F_k=\frac{F_1}{1-1/4}=\frac{4F_1}{3}$, hence $F_1$ is $\frac{3}{4}=75\%$.

📚 Official problem-set tasks (Loesung 00, 05)

L0.7 (warm-up: geometric series)

Compute $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$

Solution $q=1/2$: $\sum q^i=\frac{1}{1-1/2}=2$.

Geometric series with

L5.3 (apply convergence tests)

Show convergence of: (a) ${\sum }_{k=1}^{\infty }\frac{(k!{)}^2}{(2k)!}$ (b) ${\sum }_{k=1}^{\infty }\frac{k!}{k^k}$ (c) ${\sum }_{k=1}^{\infty }{k}^p{z}^k$ for $|z|<1,p\in \mathbb{N}$ (d) ${\sum }_{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}$ (Leibniz series for $\pi /4$)

Solution Ratio test: $\frac{a_{k+1}}{a_k}=\frac{(k+1{)}^2}{(2k+2)(2k+1)}\to \frac{1}{4}<1$. Convergent.

(a)

(b) Ratio test: $\frac{a_{k+1}}{a_k}=\frac{(k+1)!k^k}{k!(k+1{)}^{k+1}}=\frac{k^k}{(k+1{)}^k}={\left(\frac{k}{k+1}\right)}^k\to \frac{1}{e}<1$. Convergent.

(c) Root test: $\sqrt[k]{|k^p{z}^k|}=k^{p/k}|z|\to |z|<1$ (using $\sqrt[k]{k}\to 1$). Convergent.

(d) Leibniz alternating-series test: $b_k=1/(2k+1)$ is positive, decreasing, $\to 0$. Convergent.

L5.4 (Cauchy criterion, harmonic series, ζ-series)

(a) Define $x_k:=3^{-k}$ if $4\mid k$, else $x_k:=-3^{-k}$. Show $a_n:={\sum }_{k=0}^n{x}_k$ is Cauchy (hence convergent). (b) Write down the negation of “Cauchy”. (c) Show the harmonic series partial sums $H_n={\sum }_{k=1}^{n}1/k$ are NOT Cauchy. (d) Conclude harmonic series diverges. (e) Show $\sum 1/k^s$ diverges for $s\le 1$.

Solution $\epsilon >0$, pick $n_0$ with $3^{-n_0}<\epsilon$ (Bernoulli: $3^{n_0}\ge 1+2n_0$). For $n\ge m\ge n_0$: $|a_n-a_m|\le {\sum }_{k=m+1}^n|x_k|\le {\sum }_{k=m+1}^n{3}^{-k}\le \frac{3^{-m}}{2}\le 3^{-n_0}<\epsilon .✓$

(a) Given

(b) Negation: $\exists \epsilon >0\forall n_0\in {\mathbb{N}}_0\exists m,n\ge n_0:|a_m-a_n|>\epsilon$.

(c) Take $\epsilon =1/4$. For any $n_0$, set $m=2n_0,n=n_0$. Then $H_{2n_0}-H_{n_0}={\sum }_{k=n_0+1}^{2n_0}\frac{1}{k}\ge \sum \frac{1}{2n_0}=\frac{n_0}{2n_0}=\frac{1}{2}>\frac{1}{4}.$ So $(H_n)$ fails Cauchy.

(d) Cauchy is necessary for convergence in $\mathbb{R}$ (completeness), so $(H_n)$ diverges. $■$

(e) For $s\le 1$: $1/k^s\ge 1/k$, so partial sums $\ge H_n$, which diverges. By comparison, $\sum 1/k^s$ diverges too.

L5.5 (radius of convergence)

Compute the radius of convergence $\rho =1/\mathrm{limsup}\sqrt[k]{|c_k|}$ for the power series $\sum c_k{z}^k$: (a) $c_k=1$   (b) $c_k=1/k^k$   (c) $c_k=k!/k^k$   (d) $c_k=k^p$   (e) $c_k=1/k!$

Then decide convergence at specific points: (f) $\sum k!/k^k$ at $z=1$. (g) Geometric $\sum z^k$ for $|z|>1$. (h) Exponential $\sum z^k/k!$ for arbitrary $z$.

Solution $\rho =1$. (b) $\sqrt[k]{1/k^k}=1/k\to 0$, so $\rho =\infty$. (c) Use ratio test on $a_k=k!z^k/k^k$: ratio $\to |z|/e$, so $\rho =e$. (d) $\sqrt[k]{k^p}=k^{p/k}\to 1$, so $\rho =1$. (e) $\sqrt[k]{1/k!}\to 0$, so $\rho =\infty$.

(a)

(f) $|z|=1<e=\rho$: converges. (g) $|z|>1=\rho$: diverges. (h) $\rho =\infty$: converges for every $z\in \mathbb{C}$. (This is the definition of $e^z$.)

L5.6 (online MC)

(a) For a given sequence: which exists — limit, $\mathrm{liminf}$, or $\mathrm{limsup}$? (b) True/false: $b_n:={\sup }_{k\ge n}{a}_k$ is monotonically decreasing; $c_n:={\inf }_{k\ge n}{a}_k$ is monotonically increasing. (c) Is the harmonic series convergent or divergent?

Solution $\mathrm{liminf}$ always exists (in $[-\infty ,+\infty ]$) — same for $\mathrm{limsup}$. The plain limit may not. (b) True: as $n$ grows, you take sup over a smaller set (so $b_n$ can only stay or shrink); inf over a smaller set can only stay or grow. (c) Divergent — the canonical example showing "$a_n\to 0$" is necessary but not sufficient for $\sum a_n$ to converge.

(a) The

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6 · Continuity — drawing without lifting the pen

🤔 The big question

You drove from Zürich to Geneva starting at altitude $400\text{m}$ and arriving at $375\text{m}$. Did your altitude pass through exactly $390\text{m}$ somewhere along the way? Common sense says yes — and the precise statement of “yes” is the Intermediate Value Theorem, the most useful consequence of continuity.

💡 Intuition

A function is continuous at $x_0$ if, when you wiggle $x$ a tiny bit around $x_0$, $f(x)$ also wiggles only a tiny bit around $f(x_0)$. No jumps, no holes, no infinities at $x_0$.

📜 The two equivalent definitions

Continuity (sequential)

$f$ is continuous at $x_0$ iff for every sequence $x_k\to x_0$, $f(x_k)\to f(x_0)$.

Continuity ( $\epsilon$-$\delta$)

$f$ is continuous at $x_0$ iff $\forall \epsilon >0\exists \delta >0\forall x:|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon .$

These are equivalent. Use sequential when you have a sequence. Use $\epsilon$-$\delta$ when proving from scratch.

🔧 Examples and counterexamples

Continuous everywhere

Polynomials, $e^x$, $\sin x$, $\cos x$, $\sqrt{x}$ on $[0,\infty )$, $\ln x$ on $(0,\infty )$.

Jump discontinuity

$f(x)=\{\begin{array}{ll}1,& x\ge 0\\ -1,& x<0\end{array}$ is discontinuous at $0$ — left limit $=-1$, right limit $=1$.

FS 2023 MC11 — pick the right constant

$f(x)=\{\begin{array}{ll}2e^x-1,& x\le \ln 2\\ ax+2,& x>\ln 2\end{array}$. For continuity at $\ln 2$: $2e^{\ln 2}-1=3\text{must equal}a\ln 2+2$ So $a=\frac{1}{\ln 2}$.

💡 Lipschitz continuity — quantitative continuity

Lipschitz with constant $L$

$\Vert f(x)-f(y)\Vert \le L\Vert x-y\Vert \forall x,y$ Lipschitz ⇒ uniformly continuous ⇒ continuous.

💡 Compact sets — where continuity gets superpowers

Compactness (sequential)

$K\subseteq {\mathbb{R}}^d$ is compact iff every sequence in $K$ has a subsequence converging to a point of $K$.

Heine–Borel

In ${\mathbb{R}}^d$: compact $⟺$ closed and bounded.

Example

$[0,1]$ is compact. $(0,1)$ is not ($\frac{1}{n}\to 0\notin (0,1)$). $\mathbb{R}$ is not (unbounded).

🎯 The two great theorems

Extreme Value Theorem

A continuous $f:K\to \mathbb{R}$ on a compact $K$ attains its max and min.

Intermediate Value Theorem (IVT)

If $f:[a,b]\to \mathbb{R}$ is continuous and $c$ lies between $f(a)$ and $f(b)$, then $\exists x_0\in (a,b)$ with $f(x_0)=c$.

FS 2023 MC13

If ${\lim }_{x\to -\infty }f=-1$ and ${\lim }_{x\to +\infty }f=1$ for continuous $f$, then $\exists x_0$ with $f(x_0)=0$. True by IVT.

HS 2015 task 4 — the "tangent point"

Find $c\in [1,2]$ such that the tangent to $f(x)=x^2+x+3$ at $c$ has the same slope as the secant from $(1,5)$ to $(2,9)$.

Secant slope $=\frac{9-5}{2-1}=4$. Set $f^{\prime }(c)=2c+1=4$, so $c=\frac{3}{2}$. (This is the Mean Value Theorem in action — see chapter 7.)

HS 2016 task 11 — fixed point via IVT

Let $f:[a,b]\to [a,b]$ continuous. Show $f$ has a fixed point ($f(\xi )=\xi$).

Define $g(x)=f(x)-x$. Then $g(a)=f(a)-a\ge 0$ and $g(b)=f(b)-b\le 0$. By IVT, $\exists \xi$ with $g(\xi )=0$, i.e., $f(\xi )=\xi$. $■$

💡 Pointwise vs. uniform convergence of functions

Definition

A sequence $(f_k)$ of functions converges to $f$:

• Pointwise: $\forall x:f_k(x)\to f(x)$
• Uniformly: ${\sup }_x|f_k(x)-f(x)|\to 0$

Pointwise is a trap

$f_k(x)=x^k$ on $[0,1]$ converges pointwise to $f(x)=0$ for $x<1$, $f(1)=1$. Each $f_k$ is continuous, but the limit isn’t.

Theorem

Uniform limit of continuous functions is continuous.

FS 2023 MC14

$f_n(x)=e^{-nx}$ on $[0,1]$:

• At $x=0$: $f_n(0)=1$.
• At $x>0$: $f_n(x)\to 0$.

So pointwise limit is the discontinuous step function. Cannot converge uniformly (would force the limit to be continuous).

🎯 Practice tasks

Task 6.1 — Show $f(x)=\sin x$ is Lipschitz on $\mathbb{R}$.

Solution $|\sin x-\sin y|=|\cos (\xi )||x-y|\le |x-y|$. So $L=1$.

By the MVT,

Task 6.2 (FS 2019 task 5) — Where is $f$ continuous, differentiable, $C^1$?

$f(x)=\{\begin{array}{ll}{x}^2\sin (1/x)& x>0\\ \cos x-1& x\le 0\end{array}$

Solution $0$: smooth. At $0$:

Outside

• Continuous? Both branches $\to 0$ as $x\to 0$. ✓
• Differentiable? Right limit of $\frac{f(x)-f(0)}{x}=x\sin (1/x)\to 0$. Left: $\frac{\cos x-1}{x}\to 0$. So $f^{\prime }(0)=0$ exists.
• $C^1$? For $x>0$: $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. The $\cos (1/x)$ term oscillates between $-1$ and $1$, so ${\lim }_{x\to 0^{+}}{f}^{\prime }(x)$ doesn’t exist. Not $C^1$.

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7 · Differentiation — the slope at a point

🤔 The big question

A car’s odometer reads $f(t)$ at time $t$. The speedometer at instant $t_0$ shows what number? Answer: the derivative $f^{\prime }(t_0)$ — the instantaneous rate of change.

💡 Intuition: zoom in until straight

Take any smooth graph and zoom in near a point. Eventually it looks like a straight line. The slope of that line is $f^{\prime }(x_0)$.

📜 The definition

Derivative

$f$ is differentiable at $x_0$ if the limit $f^{\prime }(x_0):={\lim }_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ exists. Other notations: $\frac{df}{dx}(x_0),\dot{f}(x_0),Df(x_0)$.

The tangent line at $x_0$: $y=f(x_0)+f^{\prime }(x_0)(x-x_0)$.

🔧 First examples

$f(x)=x^2$

${\lim }_{h\to 0}\frac{(x_0+h{)}^2-x_0^2}{h}={\lim }_{h\to 0}(2x_0+h)=2x_0$

$f(x)=|x|$ at $x_0=0$

Right: ${\lim }_{h\to 0^{+}}\frac{|h|}{h}=1$. Left: $-1$. Not differentiable. (Continuous, though!)

$\exp$ is its own derivative

Using $\exp (x_0+h)=\exp (x_0)\exp (h)$ and $\frac{\exp (h)-1}{h}\to 1$: ${\exp }^{\prime }(x_0)=\exp (x_0)$

💡 Differentiable ⇒ continuous (not vice versa)

If $f$ is differentiable at $x_0$, it's continuous at $x_0$.

Proof: $f(x)-f(x_0)=\frac{f(x)-f(x_0)}{x-x_0}\cdot (x-x_0)\to f^{\prime }(x_0)\cdot 0=0$.

The converse fails: $|x|$ is continuous but not differentiable at $0$. Worse, Weierstrass built a function continuous everywhere but differentiable nowhere.

🔧 The differentiation rules

Rule	Formula
Sum	$(f+g{)}^{\prime }=f^{\prime }+g^{\prime }$
Product	$(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$
Quotient	$(f/g{)}^{\prime }=(f^{\prime }g-f{g}^{\prime })/g^2$
Chain	$(g\circ f{)}^{\prime }(x)=g^{\prime }(f(x))\cdot f^{\prime }(x)$
Inverse	$(f^{-1}{)}^{\prime }(y_0)=1/f^{\prime }(f^{-1}(y_0))$

🔧 Standard derivatives — memorize this table

$f(x)$	$f^{\prime }(x)$
$c$ (const)	$0$
$x^n$	$n{x}^{n-1}$
$\sqrt{x}$	$\frac{1}{2\sqrt{x}}$
$e^x$	$e^x$
$a^x$	$a^x\ln a$
$\ln x$	$1/x$
${\log }_{a}x$	$\frac{1}{x\ln a}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\frac{1}{{\cos }^{2}x}=1+{\tan }^{2}x$
$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}$
$\arccos x$	$-\frac{1}{\sqrt{1-x^2}}$
$\arctan x$	$\frac{1}{1+x^2}$
$\sinh x$	$\cosh x$
$\cosh x$	$\sinh x$

💡 The Mean Value Theorem — calculus’s workhorse

Rolle's theorem

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $\exists c\in (a,b)$ with $f^{\prime }(c)=0$.

Mean Value Theorem (MVT)

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, then $\exists c\in (a,b)$ with $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

Real-world reading

If you drive from Zürich to Geneva (282 km) in 3 hours, your average speed is 94 km/h. The MVT says: at some instant, your speedometer read exactly 94 km/h.

The HS 2016 exam phrased this exact scenario: a driver passes two cameras at distance $d$ in time $t$. If $d/t$ exceeds the speed limit, the driver gets a ticket — the MVT guarantees they were speeding at some moment.

HS 2015 task 4 — already discussed in chapter 6

Direct application of the MVT: setting $f^{\prime }(c)=$ secant slope.

💡 Consequences of the MVT

Sign of derivative ⇒ monotonicity

• $f^{\prime }\equiv 0$ ⇒ $f$ constant
• $f^{\prime }\ge 0$ ⇒ $f$ increasing
• $f^{\prime }>0$ ⇒ $f$ strictly increasing
• $f^{\prime }\le 0$ ⇒ $f$ decreasing
• $f^{\prime }<0$ ⇒ $f$ strictly decreasing

💡 Local extrema

Fermat's necessary condition

If $f$ has a local extremum at an interior $x_0$, then $f^{\prime }(x_0)=0$.

Second derivative test

If $f^{\prime }(x_0)=0$ and $f^{\prime \prime }$ is continuous near $x_0$:

• $f^{\prime \prime }(x_0)>0$ → strict local min
• $f^{\prime \prime }(x_0)<0$ → strict local max
• $f^{\prime \prime }(x_0)=0$ → inconclusive (try higher derivatives)

🎯 How the exam tests differentiation

• Compute a derivative using chain rule (FS 2013 task 3b — $\arcsin (\cos x)$)
• Apply MVT (HS 2015 task 4, HS 2016 task 11-style)
• Find local/global extrema (HS 2015 task 8, HS 2016 task 4)

📝 Practice tasks

Task 7.1 (FS 2013 task 3b) — Differentiate $f(x)=\arcsin (\cos x)$.

Solution $\cos x=\sin (\pi /2-x)$ and $\arcsin (\sin u)=u$ on the right intervals: $f(x)=\pi /2-x$ on $[0,\pi ]$, so $f^{\prime }(x)=-1$ there. Or via chain rule: $f^{\prime }(x)=\frac{-\sin x}{\sqrt{1-{\cos }^{2}x}}=\frac{-\sin x}{|\sin x|}=-\text{sign}(\sin x)$.

Using

Task 7.2 — Find the absolute max and min of $f(x)=x^3-3x$ on $[-2,2]$.

Solution $f^{\prime }(x)=3x^2-3=0$ ⇒ $x=\pm 1$. Evaluate: $f(-2)=-2,f(-1)=2,f(1)=-2,f(2)=2$. Max $=2$, min $=-2$.

Task 7.3 — Find $f^{\prime }(x)$ for $f(x)=x^x$, $x>0$.

Solution $\ln f=x\ln x$, so $\frac{f^{\prime }}{f}=\ln x+1$. Hence $f^{\prime }(x)=x^x(\ln x+1)$.

Logarithmic differentiation:

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8 · Taylor series — polynomial X-rays

🤔 The big question

Calculators don’t actually “compute $\sin (0.7)$” by some magic. They approximate by polynomials — because polynomials are the only functions a CPU can really evaluate (just $+,-,\times ,÷$). Which polynomial best approximates $\sin$ near $0$? The Taylor polynomial.

💡 Intuition: keep matching higher derivatives

The tangent line $T_1(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)$ matches $f$ in value and slope at $x_0$. Why stop there? Match the curvature too — and the rate-of-change of curvature, etc.

📜 Taylor’s theorem

Taylor formula with Lagrange remainder

If $f$ is $(n+1)$-times differentiable, then $\exists \xi$ between $x_0$ and $x$ with $f(x)=\underset{T_n(x)\text{ — Taylor polynomial}}{\underbrace{\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k}}+\underset{R_n(x)\text{ — remainder}}{\underbrace{\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}}}$

If $R_n(x)\to 0$ as $n\to \infty$, then $f(x)={\sum }_{k=0}^{\infty }\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$ is the Taylor series of $f$ around $x_0$.

🔧 The standard Taylor series (around $x_0=0$) — memorize all of these

Function	Series	Domain
$e^x$	${\sum }_{k=0}^{\infty }\frac{x^k}{k!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$	$\mathbb{R}$
$\sin x$	$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$	$\mathbb{R}$
$\cos x$	$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$	$\mathbb{R}$
$\frac{1}{1-x}$	$1+x+x^2+\cdots$	$|x|<1$
$\ln (1+x)$	$x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$	$-1<x\le 1$
$(1+x{)}^{\alpha }$	${\sum }_{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{\alpha }{k}\right)x^k$	$|x|<1$
$\arctan x$	$x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots$	$|x|\le 1$
$\frac{1}{1+x^2}$	$1-x^2+x^4-\cdots$	$|x|<1$

🎯 The exam pattern

Taylor series are incredibly useful for limits ("$\frac{0}{0}$" type) and for showing differentiability properties.

FS 2013 task 1a — limit using Taylor

${\lim }_{x\to 0}\frac{1-\cos x}{\tan x\sin x}$ Using $1-\cos x=\frac{x^2}{2}+o(x^2)$ and $\sin x\tan x=x^2+o(x^2)$: ${\lim }_{x\to 0}\frac{x^2/2+o(x^2)}{x^2+o(x^2)}=\frac{1}{2}$

HS 2015 task 1b — Taylor for $f(0)$ and $f^{\prime }(0)$

$f(x)=\frac{e^{5x}-1}{e^x-1}$ on $\mathbb{R}\setminus \{0\}$, extended continuously.

Numerator: $5x+\frac{25}{2}{x}^2+o(x^2)$. Denominator: $x+\frac{x^2}{2}+o(x^2)$. Divide: $f(x)=\frac{5x(1+\frac{5x}{2}+o(x))}{x(1+\frac{x}{2}+o(x))}=(5+\frac{25}{2}x+o(x))(1-\frac{x}{2}+o(x))=5+10x+o(x)$ So $f(0)=5$ and $f^{\prime }(0)=10$.

FS 2023 MC24 — Taylor polynomial of $f(x,y)=x/y$ around $(0,1)$ to order 2

Use $\frac{1}{y}=\frac{1}{1+(y-1)}=1-(y-1)+(y-1{)}^2-\cdots$. So $\frac{x}{y}=x-x(y-1)+x(y-1{)}^2-\cdots$ Up to order 2 (i.e. only terms with total degree $\le 2$): $T_2=x-x(y-1)$

🎯 Practice tasks

Task 8.1 — Compute ${\lim }_{x\to 0}\frac{\sin x-x}{x^3}$.

Solution $\sin x=x-\frac{x^3}{6}+o(x^3)$, so $\sin x-x=-\frac{x^3}{6}+o(x^3)$. Hence limit $=-\frac{1}{6}$.

Task 8.2 — Find the Taylor polynomial of $\ln (1+2x)$ up to order 3.

Solution $\ln (1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots$ with $u=2x$: $2x-2x^2+\frac{8x^3}{3}$.

Use

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9 · Limits and L’Hôpital’s rule

🤔 The big question

Limits like $\frac{0}{0}$, $\frac{\infty }{\infty }$, $0\cdot \infty$, $1^{\infty }$, $\infty -\infty$, $0^0$ are called indeterminate — you can’t just plug in. We need techniques.

🔧 The toolbox (in order of preference)

1. Algebraic simplification (cancel, conjugate, common denominator).
2. Standard limits (memorize: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, etc.).
3. Taylor series (turn the function into a polynomial-like object).
4. L’Hôpital’s rule (last resort — sometimes loops forever).

💡 L’Hôpital’s rule

L'Hôpital

If ${\lim }_{x\to x_0}f(x)={\lim }_{x\to x_0}g(x)=0$ (or both $\pm \infty$), $g^{\prime }(x)\ne 0$ near $x_0$, and $\lim \frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists, then ${\lim }_{x\to x_0}\frac{f(x)}{g(x)}={\lim }_{x\to x_0}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Common L'Hôpital mistakes

1. Apply only when you have $\frac{0}{0}$ or $\frac{\infty }{\infty }$.
2. Differentiate numerator and denominator separately — NOT as a quotient!
3. If it loops, switch to Taylor.

🎯 Worked exam-style examples

FS 2013 task 1a (Method 1, L'Hôpital)

$\underset{x\to 0}{\lim }\frac{1-\cos x}{\tan x\sin x}$. Numerator and denominator $\to 0$, so apply L’Hôpital: $={\lim }_{x\to 0}\frac{\sin x}{\frac{1}{{\cos }^{2}x}\sin x+\cos x\cos x\cdot \tan x}$ Simplify by dividing top and bottom by $\sin x$: $={\lim }_{x\to 0}\frac{1}{\frac{1}{{\cos }^{2}x}+1}=\frac{1}{2}$

FS 2013 task 1b (substitution + L'Hôpital)

$\underset{x\to 0^{+}}{\lim }\frac{2x}{e^{-1/x}}$. Substitute $t=1/x$ (so $x\to 0^{+}$ means $t\to +\infty$): ${\lim }_{t\to \infty }\frac{2/t}{e^{-t}}={\lim }_{t\to \infty }\frac{2e^t}{t}\stackrel{\text{L’H}}{=}{\lim }_{t\to \infty }2e^t=+\infty$

HS 2015 task 2b (conjugate)

$\underset{x\to \infty }{\lim }(\sqrt{x^2+x}-x)$. Multiply by conjugate: $={\lim }_{x\to \infty }\frac{x}{\sqrt{x^2+x}+x}={\lim }_{x\to \infty }\frac{1}{\sqrt{1+1/x}+1}=\frac{1}{2}$

FS 2019 task 3b (L'Hôpital for $\sin /(\cdot )$)

${\lim }_{x\to 1}\frac{\sin (x^2-1)}{x-1}$. L’Hôpital: $=\lim \frac{2x\cos (x^2-1)}{1}=2$.

HS 2016 task 2c (1/x form, $0\cdot \infty$)

${\lim }_{x\to \infty }(x^{1/x}-1)\cdot \frac{1}{x}$. Use $x^{1/x}=e^{\ln (x)/x}$ and $\ln (x)/x\to 0$. Then $x^{1/x}-1=e^{\ln x/x}-1\approx \ln x/x$. So the original $\approx \ln (x)/x^2\to 0$.

FS 2019 task 3a (clever inequality)

${\lim }_{n\to \infty }\sqrt[n]{3^n-2^n}={\lim }_{n\to \infty }3\sqrt[n]{1-(2/3{)}^n}$. Hint shows $\sqrt[n]{1-(2/3{)}^n}\to 1$ (squeeze: $1\ge \sqrt[n]{1-(2/3{)}^n}\ge \sqrt[n]{1-2/3}\to 1$). So limit $=3$.

🎯 The “compute this limit” exam tasks

Almost every basic exam has 2–4 limit computations. Strategies:

If you see…	Try first
$\frac{0}{0}$ with trig	Taylor
$\frac{0}{0}$ rational	factor / cancel
$\sqrt{a}-\sqrt{b}$ at $\infty$	conjugate
$\infty -\infty$	common denominator
$1^{\infty }$	$f(x{)}^{g(x)}=e^{g(x)\ln f(x)}$
Always	L’Hôpital as fallback

📝 Practice tasks

Task 9.1 (HS 2016 task 2a) — $\underset{x\to \frac{\pi }{2}}{\lim }\frac{\tan (3x)}{\tan (x)}$.

Solution $\frac{\sin 3x\cos x}{\cos 3x\sin x}$. As $x\to \pi /2$: $\cos x\to 0$, $\cos 3x\to 0$, so we have $0/0$. L'Hôpital on numerator and denominator separately... or: $\sin 3x\to \sin (3\pi /2)=-1$ and $\sin x\to 1$, so we just need $\frac{\cos x}{\cos 3x}$. L'Hôpital: $\frac{-\sin x}{-3\sin 3x}\to \frac{-1}{-3(-1)}=-\frac{1}{3}$. So overall $\frac{-1\cdot \cos x}{\cos 3x\cdot 1}\to (-1)\cdot (-1/3)=1/3$.

Rewrite as

Task 9.2 (HS 2016 task 2b) — $\underset{x\to 0}{\lim }\frac{e^{x^2}-1}{{\sin }^{2}x}$.

Solution $e^{x^2}-1=x^2+o(x^2)$, ${\sin }^{2}x=x^2+o(x^2)$. Limit $=1$.

Taylor:

Task 9.3 — $\underset{x\to 0^{+}}{\lim }\log (x)\tan (x)$.

Solution $\tan x\approx x$ near 0, so this is $\approx x\log x\to 0$ (standard limit).

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10 · Integration — the Fundamental Theorem

🤔 The big question

The integral ${\int }_a^{b}f(x)dx$ measures area under the curve. But it also computes:

• distance from velocity
• mass from density
• charge from current
• expected value from probability density

The miracle: integration is the inverse of differentiation. This is the Fundamental Theorem of Calculus.

💡 Antiderivatives

Antiderivative

$F$ is an antiderivative of $f$ if $F^{\prime }(x)=f(x)$. We write $\int f(x)dx=F(x)+C$.

The "$+C$" reminds us the antiderivative is unique only up to a constant.

💡 The Riemann integral

Partition $[a,b]$ into small intervals, pick sample points ${\xi }_k$, form the Riemann sum ${\sum }_{k}f({\xi }_k)(x_{k+1}-x_k)$ and refine. The limit is the Riemann integral ${\int }_a^{b}f(x)dx$.

Theorem

Every continuous function on $[a,b]$ is Riemann-integrable. So is every monotonic function, and any bounded function with finitely many discontinuities.

💎 The Fundamental Theorem of Calculus (FTC)

FTC, Part 1 (existence)

If $f$ is continuous on $[a,b]$, then $F(x)={\int }_a^{x}f(t)dt$ is differentiable and $F^{\prime }(x)=f(x)$.

FTC, Part 2 (computation)

If $F^{\prime }=f$ on $[a,b]$, then ${\int }_a^{b}f(x)dx=F(b)-F(a)=:F(x){|}_a^b$

Example

${\int }_0^{\pi }\sin xdx=-\cos x{|}_0^{\pi }=-(-1)-(-1)=2$.

🔧 The antiderivative table — memorize this

$f(x)$	$\int fdx$
$x^n$ ($n\ne -1$)	$\frac{x^{n+1}}{n+1}+C$
$\frac{1}{x}$	$\ln |x|+C$
$e^x$	$e^x+C$
$a^x$	$\frac{a^x}{\ln a}+C$
$\sin x$	$-\cos x+C$
$\cos x$	$\sin x+C$
$\frac{1}{{\cos }^{2}x}$	$\tan x+C$
$\frac{1}{1+x^2}$	$\arctan x+C$
$\frac{1}{\sqrt{1-x^2}}$	$\arcsin x+C$
$\sinh x$	$\cosh x+C$
$\cosh x$	$\sinh x+C$

💡 Properties

• Linearity: $\int (\alpha f+\beta g)=\alpha \int f+\beta \int g$
• Additivity: ${\int }_a^c={\int }_a^b+{\int }_b^c$
• Monotonicity: $f\le g$ ⇒ $\int f\le \int g$
• Estimation: $\left|{\int }_a^{b}f\right|\le {\int }_a^b|f|$

🎯 Improper integrals

When the interval is infinite or the integrand blows up: ${\int }_a^{\infty }fdx:={\lim }_{b\to \infty }{\int }_a^{b}fdx$

The $1/x^p$ test

• ${\int }_1^{\infty }\frac{1}{x^p}dx$ converges iff $p>1$.
• ${\int }_0^1\frac{1}{x^p}dx$ converges iff $p<1$.

FS 2015 task 5

Discuss ${\int }_1^{\infty }\frac{1}{x^{\alpha }}dx$ in three cases ($\alpha =1$, $\alpha <1$, $\alpha >1$). Result above.

🎯 How the exam tests integration

• Compute a definite integral (every exam has 2-3)
• Apply FTC
• Improper integrals with parameter (FS 2015 task 5)
• Recursion via integration by parts (FS 2023 task 3)

📝 Practice tasks

Task 10.1 — Compute ${\int }_0^1(3x^2+2x)dx$.

Solution $=[x^3+x^2{]}_0^1=2$.

Task 10.2 — Determine convergence of ${\int }_0^{\infty }{e}^{-x}dx$ and find its value if convergent.

Solution ${\int }_0^t{e}^{-x}dx=-e^{-x}{|}_0^t=1-e^{-t}\to 1$. Converges, value $1$.

Task 10.3 (FS 2023 task 3) — Define $f_n(t)={\int }_0^t{x}^n{e}^{-x}dx$. Show $f_n(t)=n{f}_{n-1}(t)-t^n{e}^{-t}$ and use this to evaluate ${\int }_0^{\infty }{x}^n{e}^{-x}dx=n!$.

Solution $u=x^n$, $dv=e^{-x}dx$ ⇒ $du=n{x}^{n-1}dx$, $v=-e^{-x}$. $f_n(t)=-x^n{e}^{-x}{|}_0^t+n{\int }_0^t{x}^{n-1}{e}^{-x}dx=-t^n{e}^{-t}+n{f}_{n-1}(t)$ As $t\to \infty$, $t^n{e}^{-t}\to 0$. By induction with base $f_0=1-e^{-t}\to 1=0!$: ${\int }_0^{\infty }{x}^n{e}^{-x}dx=n!$

Integration by parts:

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11 · Integration techniques (parts, substitution, partial fractions)

The exam will throw integrals at you that don’t appear in the table. You need three techniques: integration by parts, substitution, and partial fractions.

🔧 Integration by parts

From $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$: $\boxed{\int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx}$

How to choose $f$ and $g^{\prime }$

Pick $f$ as the part that gets simpler when differentiated:

• polynomials → become 0 eventually
• $\ln x$ → becomes $1/x$
• $\arctan x$, $\arcsin x$ → become rational

Pick $g^{\prime }$ as something easy to integrate ($e^x$, $\sin x$, $\cos x$).

Mnemonic LIATE: pick $f$ as the first thing in the list that appears Logarithm > Inverse trig > Algebraic (polynomial) > Trig > Exponential.

$\int x{e}^{x}dx$

Let $f=x$, $g^{\prime }=e^x$ ⇒ $f^{\prime }=1$, $g=e^x$. $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=(x-1)e^x+C$

FS 2013 task 3a-ii: ${\int }_0^2\log (1+x)dx$

Let $u=\log (1+x)$, $dv=dx$ ⇒ $du=\frac{dx}{1+x}$, $v=x$. $=[x\log (1+x){]}_0^2-{\int }_0^2\frac{x}{1+x}dx=2\log 3-{\int }_0^2(1-\frac{1}{1+x})dx$ $=2\log 3-2+\log 3=3\log 3-2$

FS 2019 task 4: $\int \log (x^2+1)dx$

$u=\log (x^2+1)$, $dv=dx$ ⇒ $du=\frac{2x}{x^2+1}dx$, $v=x$. $=x\log (x^2+1)-\int \frac{2x^2}{x^2+1}dx=x\log (x^2+1)-\int (2-\frac{2}{x^2+1})dx$ $=x\log (x^2+1)-2x+2\arctan x+C$

FS 2015 task 1b: ${\int }_0^{\pi }{x}^2\sin xdx$

Two parts: $\int x^2\sin xdx=-x^2\cos x+2\int x\cos xdx=-x^2\cos x+2x\sin x-2\int \sin xdx$ $=-x^2\cos x+2x\sin x+2\cos x+C$ Evaluate at endpoints: $[{\pi }^2+0-2]-[0+0+2]={\pi }^2-4$.

HS 2016 task 5a (recursion trick): $\int e^x\sin xdx$

Let $I=\int e^x\sin xdx$. Two parts give $I=e^x\sin x-\int e^x\cos xdx$. Apply parts again to the $\cos$ integral: $\int e^x\cos xdx=e^x\cos x+I$. Substitute back: $I=e^x\sin x-e^x\cos x-I\Rightarrow I=\frac{e^x(\sin x-\cos x)}{2}+C$

🔧 Substitution (u-substitution)

If $u=\phi (x)$, $du={\phi }^{\prime }(x)dx$: $\int f(\phi (x)){\phi }^{\prime }(x)dx=\int f(u)du$

When to substitute

Look for an “outer” function and an “inner” function whose derivative also shows up. Set $u=$ inner.

$\int 2x\cos (x^2)dx$

$u=x^2$, $du=2xdx$. $\int \cos udu=\sin u+C=\sin (x^2)+C$.

HS 2014 task 3a-i: ${\int }_0^1\frac{\arctan x}{1+x^2}dx$

$u=\arctan x$, $du=\frac{dx}{1+x^2}$. New limits $u(0)=0$, $u(1)=\pi /4$. ${\int }_0^{\pi /4}udu=\frac{u^2}{2}{|}_0^{\pi /4}=\frac{{\pi }^2}{32}$

HS 2016 task 5b: ${\int }_1^3\frac{x^3}{\sqrt{1+x^2}}dx$

Substitute $u=\sqrt{1+x^2}$, $du=\frac{x}{\sqrt{1+x^2}}dx$, $x^2=u^2-1$. Limits: $u(1)=\sqrt{2}$, $u(3)=\sqrt{10}$. ${\int }_{\sqrt{2}}^{\sqrt{10}}(u^2-1)du=\frac{u^3}{3}-u{|}_{\sqrt{2}}^{\sqrt{10}}=\frac{7\sqrt{10}+\sqrt{2}}{3}$

🔧 Partial fraction decomposition

For rational integrands $\frac{P(x)}{Q(x)}$ with $\mathrm{deg}P<\mathrm{deg}Q$:

1. Factor $Q(x)$ into linear and irreducible quadratic factors.
2. Write $\frac{P(x)}{Q(x)}$ as a sum: $\frac{A}{x-a}+\frac{B}{(x-a{)}^2}+\cdots +\frac{Cx+D}{x^2+bx+c}$
3. Integrate each piece.

FS 2013 task 3a-i: $\int \frac{x+2}{x^3-x^2+2x-2}dx$

Factor: $x^3-x^2+2x-2=x^2(x-1)+2(x-1)=(x-1)(x^2+2)$. Set $\frac{x+2}{(x-1)(x^2+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2}$. Multiply through and compare coefficients: $A=1,B=-1,C=0$. $\int \left(\frac{1}{x-1}-\frac{x}{x^2+2}\right)dx=\log |x-1|-\frac{1}{2}\log (x^2+2)+C$

HS 2014 task 3a-ii: $\int \frac{x+4}{x^3-2x^2}dx$

Factor: $x^2(x-2)$. Setup: $\frac{A}{x-2}+\frac{B}{x}+\frac{C}{x^2}$. Solving gives $A=\frac{3}{2},B=-\frac{3}{2},C=-2$. $=\frac{3}{2}\log |x-2|-\frac{3}{2}\log |x|+\frac{2}{x}+C$

FS 2019 MC2: $\int \frac{1}{1-x^2}dx$

$\frac{1}{(1-x)(1+x)}=\frac{1/2}{1-x}+\frac{1/2}{1+x}$. So $=\frac{1}{2}(-\log |1-x|+\log |1+x|)+C=\frac{1}{2}\log \left|\frac{1+x}{1-x}\right|+C$

🎯 The exam pattern for integrals

Each exam typically has 2-3 integrals, often:

• one integration by parts (involves $\ln$, $\arctan$, or $x\cdot e^x$ / $x\cdot \sin x$)
• one partial fraction (cubic denominator)
• sometimes a substitution (square root, trig)

📝 Practice tasks

Task 11.1 — $\int x\ln xdx$.

Solution $u=\ln x,dv=xdx$. Then $du=\frac{1}{x}dx,v=\frac{x^2}{2}$. $=\frac{x^2}{2}\ln x-\int \frac{x}{2}dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$

Parts:

Task 11.2 (HS 2014 task 3 partial fraction) — $\int \frac{9}{x^3-3x-2}dx$ (factor: $(x-2)(x+1{)}^2$).

Solution $\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$. After comparing: $A=1,B=-1,C=-3$. $=\log |x-2|-\log |x+1|+\frac{3}{x+1}+C=\log \left|\frac{x-2}{x+1}\right|+\frac{3}{x+1}+C$

Setup:

Task 11.3 (Substitution) — $\int \frac{e^x}{1+e^{2x}}dx$.

Solution $u=e^x$, $du=e^{x}dx$. $\int \frac{du}{1+u^2}=\arctan u+C=\arctan (e^x)+C$.

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12 · Ordinary differential equations

🤔 The big question

Newton’s law $F=ma$ is really $m\ddot{x}=F(x,\dot{x},t)$ — an ODE. So is radioactive decay, population growth, electrical circuits, planetary motion. Solving ODEs is the mathematical core of physics and engineering.

📜 What is an ODE?

Definition

An ODE of order $n$ is an equation involving an unknown function $y(t)$ and its derivatives up to order $n$.

Order = highest derivative. Linear = no products of $y$ and its derivatives.

ODE	Models
$\dot{N}=-\lambda N$	radioactive decay
$m\ddot{x}+kx=0$	spring without friction
$\ddot{x}+2\delta \dot{x}+{\omega }_0^{2}x=0$	damped oscillator
$L\ddot{Q}+R\dot{Q}+Q/C=U(t)$	RLC circuit

🔧 Type 1 — Separable ODEs

If you can write $y^{\prime }(x)=f(x)g(y)$, separate: $\frac{dy}{g(y)}=f(x)dx⟹\int \frac{dy}{g(y)}=\int f(x)dx+C$

FS 2013 task 4b

$\ln (y)y^{\prime }=y\sqrt{x}$. Separate: $\frac{\ln y}{y}dy=\sqrt{x}dx\Rightarrow \frac{(\ln y{)}^2}{2}=\frac{2}{3}{x}^{3/2}+C$ Hence $y=e^{\pm \sqrt{(4/3)x^{3/2}+C}}$.

HS 2014 task 4b: $y{y}^{\prime }=2(1+y^2)x^3$

$\frac{ydy}{1+y^2}=2x^{3}dx$. Substitute $t=1+y^2,dt=2ydy$: $\frac{dt}{2t}=2x^{3}dx\Rightarrow \frac{1}{2}\ln t=\frac{x^4}{2}+C\Rightarrow t=C{e}^{x^4}\Rightarrow y=\pm \sqrt{C{e}^{x^4}-1}$

🔧 Type 2 — Linear first-order ODEs

$y^{\prime }+p(x)y=q(x)$

Use the integrating factor $\mu (x)=e^{\int p(x)dx}$. Multiplying gives $(\mu y{)}^{\prime }=\mu q$, so $y(x)=\frac{1}{\mu (x)}(\int \mu (x)q(x)dx+C)$

🔧 Type 3 — Linear ODEs with constant coefficients

For $a_n{y}^{(n)}+\cdots +a_1{y}^{\prime }+a_{0}y=0$:

Try $y=e^{\lambda x}$. This produces the characteristic polynomial: $p(\lambda )=a_n{\lambda }^n+\cdots +a_1\lambda +a_0=0$

Roots of $p(\lambda )$	Solution contribution
Simple real $\lambda$	$C{e}^{\lambda x}$
Repeated real $\lambda$ (mult. $m$)	$(C_0+C_{1}x+\cdots +C_{m-1}{x}^{m-1})e^{\lambda x}$
Complex pair $\alpha \pm i\beta$	$e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)$

FS 2013 task 4a

$y^{\prime \prime }-3y^{\prime }+2y=130\cos (3x)$.

Homogeneous: ${\lambda }^2-3\lambda +2=0\Rightarrow {\lambda }_1=1,{\lambda }_2=2$. So $y_h=C_1{e}^x+C_2{e}^{2x}$.

Particular ansatz: since $\cos (3x)=\text{Re}(e^{3x\cdot i})$ and $3i$ is not a root, try $y_p=A\cos (3x)+B\sin (3x)$. Plug in, collect coefficients: $-7A-9B=130$ (from $\cos$) and $9A-7B=0$ (from $\sin$). Solve: $A=-7,B=-9$.

Final: $y(x)=C_1{e}^x+C_2{e}^{2x}-7\cos (3x)-9\sin (3x)$.

HS 2014 task 4a: $2y^{\prime \prime }-9y^{\prime }+9y=9x^2$

Homogeneous: ${\lambda }_1=3,{\lambda }_2=3/2$, so $y_h=C_1{e}^{3x}+C_2{e}^{3x/2}$. Particular: try $y_p=A{x}^2+Bx+C$. Substitute, compare coefficients: $A=1,B=2,C=14/9$. Final: $y=y_h+x^2+2x+14/9$.

FS 2023 task 4 (parameter analysis)

$y^{\prime \prime }+2y^{\prime }+ay=0$. Characteristic: ${\lambda }^2+2\lambda +a=0\Rightarrow \lambda =-1\pm \sqrt{1-a}$.

• $a<1$: real distinct, $y=A{e}^{(\sqrt{1-a}-1)x}+B{e}^{-(\sqrt{1-a}+1)x}$.
• $a=1$: double root $-1$, $y=(A+Bx)e^{-x}$.
• $a>1$: complex pair, $y=e^{-x}(A\cos \sqrt{a-1}x+B\sin \sqrt{a-1}x)$.

Bounded as $x\to \infty$ iff $a\ge 0$.

💡 The damped harmonic oscillator (engineering classic)

$\ddot{y}+2\delta \dot{y}+{\omega }_0^{2}y=0$

Regime	Condition	Solution
Undamped	$\delta =0$	$C_1\cos ({\omega }_{0}t)+C_2\sin ({\omega }_{0}t)$
Underdamped	$0<\delta <{\omega }_0$	$e^{-\delta t}(C_1\cos \omega t+C_2\sin \omega t)$, $\omega =\sqrt{{\omega }_0^2-{\delta }^2}$
Critically damped	$\delta ={\omega }_0$	$(C_1+C_{2}t)e^{-\delta t}$
Overdamped	$\delta >{\omega }_0$	$C_1{e}^{{\lambda }_{1}t}+C_2{e}^{{\lambda }_{2}t}$

🔧 Inhomogeneous: total = homogeneous + particular

$y=y_h+y_p$

For $q(x)$ of standard form, guess $y_p$:

$q(x)$	Guess for $y_p$
polynomial of degree $m$	polynomial of degree $m$
$e^{\alpha x}$	$A{e}^{\alpha x}$ (multiply by $x$ if $\alpha$ is a root)
$\sin \omega x$, $\cos \omega x$	$A\cos \omega x+B\sin \omega x$
product	product

Resonance trap

If your ansatz already solves the homogeneous equation, multiply by $x$. This is what FS 2019 task 6 tests: $y^{\prime \prime }+y^{\prime }-2y=e^{-2x}$ — homogeneous roots are $-2,1$. Since $-2$ is a root, the ansatz isn’t $A{e}^{-2x}$ but $\boxed{Ax{e}^{-2x}}$. Solving gives $A=-1/3$.

💡 Systems of ODEs and matrix exponential

A scalar $n$-th order ODE can be rewritten as a system of $n$ first-order equations. In matrix form: $\dot{\mathbf{y}}(t)=A\mathbf{y}(t)$ Solution: $\mathbf{y}(t)=e^{At}\mathbf{y}(0)$, where $e^{At}={\sum }_{k=0}^{\infty }\frac{(At{)}^k}{k!}$.

FS 2023 task 4.A1 — converting to a system

Rewrite $y^{\prime \prime }+2y^{\prime }+ay=0$ as a system: Let $\mathbf{F}=(y,y^{\prime }{)}^T$. Then $\dot{\mathbf{F}}=\left(\begin{array}{cc}0& 1\\ -a& -2\end{array}\right)\mathbf{F}$.

🎯 The exam pattern

Every exam has at least one ODE task, and often two:

• one second-order linear with constant coefficients (homogeneous + particular)
• one first-order separable

📝 Practice tasks

Task 12.1 (FS 2017 task 6) — Find the unique bounded solution of $y^{\prime \prime }+y^{\prime }-6y=10\sin x$ with $y(0)=0$.

Solution Homogeneous: ${\lambda }^2+\lambda -6=0\Rightarrow \lambda =2,-3$. So $y_h=C_1{e}^{2x}+C_2{e}^{-3x}$. Particular: ansatz $y_p=a\cos x+b\sin x$. Plugging in: $(-7a+b)\cos x+(-a-7b)\sin x=10\sin x$. So $a=-1/(...)$... working it out: $a=-7/5$, no — let's redo: equations $-a+b-6a=0$ from cos, $-b-a-6b=10$ from sin... carefully: $y_p^{\prime }=-a\sin x+b\cos x$, $y_p^{\prime \prime }=-a\cos x-b\sin x$. Sum: $(-a+b-6a)\cos x+(-b-a-6b)\sin x=(-7a+b)\cos x+(-a-7b)\sin x=10\sin x$. So $-7a+b=0$ and $-a-7b=10$. From first: $b=7a$. Plug: $-a-49a=10\Rightarrow a=-1/5$, $b=-7/5$. So $y_p=-\frac{1}{5}\cos x-\frac{7}{5}\sin x$.

Bounded as $x\to \infty$: requires $C_1=0$. Then $y(0)=C_2-\frac{1}{5}=0$, so $C_2=\frac{1}{5}$. Final: $y(x)=\frac{1}{5}{e}^{-3x}-\frac{1}{5}\cos x-\frac{7}{5}\sin x$.

Task 12.2 — Solve $y^{\prime }=xy$, $y(0)=1$.

Solution $\frac{dy}{y}=xdx\Rightarrow \ln |y|=x^2/2+C$. With $y(0)=1$: $C=0$. $y=e^{x^2/2}$.

Separable:

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13 · Multivariable functions

🤔 The big question

Real engineering problems involve many variables: temperature varies in 3D space, profit depends on price and advertising, an antenna’s gain depends on angle and frequency. We need calculus that handles many variables.

📜 The setup

A function $f:{\mathbb{R}}^d\to {\mathbb{R}}^m$. Examples:

• $f(x,y)=$ height of a mountain at $(x,y)$ — scalar field on ${\mathbb{R}}^2$
• $f(x,y,z)=$ temperature in a room — scalar field on ${\mathbb{R}}^3$
• $\mathbf{F}(x,y,z)=({\text{wind}}_x,{\text{wind}}_y,{\text{wind}}_z)$ — vector field ${\mathbb{R}}^3\to {\mathbb{R}}^3$
• $\gamma (t)=(\cos t,\sin t,t)$ — curve $\mathbb{R}\to {\mathbb{R}}^3$ (a helix)

💡 Continuity in many variables

Path-dependence trap

Just because $f$ is continuous in $x$ alone (with $y$ fixed) and in $y$ alone (with $x$ fixed) does NOT mean $f$ is continuous in $(x,y)$.

Classic counterexample

$f(x,y)=\frac{xy}{x^2+y^2}$ for $(x,y)\ne (0,0)$, $f(0,0)=0$. Along $y=x$: $f(x,x)=\frac{x^2}{2x^2}=\frac{1}{2}$ for $x\ne 0$. So ${\lim }_{x\to 0}f(x,x)=\frac{1}{2}\ne 0$. Not continuous!

FS 2019 task 7

$f(x,y)=\frac{xy(x-y+1)}{x^2+y^2}$ for $(x,y)\ne (0,0)$, $f(0,0)=0$.

• Along $x=0$: $f\equiv 0$, continuous.
• Along $y=x$: $f=\frac{x^2(1)}{2x^2}=\frac{1}{2}$, discontinuous at origin.

💡 Partial derivatives

Partial derivative

$\frac{\partial f}{\partial x_i}(\mathbf{x}):={\lim }_{h\to 0}\frac{f(\mathbf{x}+h{\mathbf{e}}_i)-f(\mathbf{x})}{h}$ where ${\mathbf{e}}_i$ is the $i$-th unit vector.

In words: differentiate in the $x_i$-direction, treating other variables as constants.

Example

$f(x,y)=x^{2}y+\sin (xy)$. Then $\frac{\partial f}{\partial x}=2xy+y\cos (xy),\frac{\partial f}{\partial y}=x^2+x\cos (xy)$

💡 The gradient — direction of steepest ascent

Gradient

$\nabla f(\mathbf{x})=\text{grad}f(\mathbf{x}):=(\frac{\partial f}{\partial x_1},\dots ,\frac{\partial f}{\partial x_d})$

Geometric meaning

1. $\nabla f(\mathbf{x})$ points in the direction of steepest ascent.
2. Its magnitude equals the rate of ascent in that direction.
3. $\nabla f$ is perpendicular to level sets $\{f=c\}$.

The directional derivative in direction $\mathbf{v}$ (unit vector): ${\partial }_{\mathbf{v}}f(\mathbf{x})=\nabla f(\mathbf{x})\cdot \mathbf{v}$

💡 Total differentiability and the Jacobian

For $\mathbf{f}=(f_1,\dots ,f_m):{\mathbb{R}}^d\to {\mathbb{R}}^m$:

Jacobian matrix

\frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_d}\\ \vdots & \ddots & \vdots\\ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_d} \end{pmatrix}$$

(Total) differentiability

$\mathbf{f}$ is differentiable at ${\mathbf{x}}_0$ if $\mathbf{f}({\mathbf{x}}_0+\mathbf{h})=\mathbf{f}({\mathbf{x}}_0)+J_{\mathbf{f}}({\mathbf{x}}_0)\mathbf{h}+o(\Vert \mathbf{h}\Vert ).$

Subtle point (FS 2023 MC22)

Existence of all partial derivatives at ${\mathbf{x}}_0$ does NOT imply (total) differentiability. But: if all partials exist and are continuous in a neighborhood, $\mathbf{f}$ is differentiable there. ($C^1\Rightarrow$ differentiable.)

🔧 Multivariable chain rule — the workhorse

Chain rule (matrix form)

If $\mathbf{g}:{\mathbb{R}}^d\to {\mathbb{R}}^m$ and $\mathbf{f}:{\mathbb{R}}^m\to {\mathbb{R}}^k$ are differentiable: $J_{\mathbf{f}\circ \mathbf{g}}(\mathbf{x})=J_{\mathbf{f}}(\mathbf{g}(\mathbf{x}))\cdot J_{\mathbf{g}}(\mathbf{x})$

FS 2013 task 7

$\mathbf{f}:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $\mathbf{f}(x,y,z)=((x+y)z,xy)$. $\mathbf{g}:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $\mathbf{g}(u,v)=(u+v,u-v,uv)$.

$J_{\mathbf{f}}(x,y,z)=\left(\begin{array}{ccc}z& z& x+y\\ y& x& 0\end{array}\right)$,

$J_{\mathbf{g}}(u,v)=\left(\begin{array}{cc}1& 1\\ 1& -1\\ v& u\end{array}\right)$.

By chain rule: $J_{\mathbf{h}}(x,y,z)=J_{\mathbf{g}}(\mathbf{f}(x,y,z))\cdot J_{\mathbf{f}}(x,y,z)$.

FS 2023 MC21

$\mathbf{f}(x_1,x_2)=(x_1^2,x_1-x_2)$, $\mathbf{g}(y_1,y_2)=(y_1,y_1{y}_2)$. Then $J_{\mathbf{h}}(1,1)=?$

$J_{\mathbf{f}}(1,1)=\left(\begin{array}{cc}2& 0\\ 1& -1\end{array}\right)$. $\mathbf{f}(1,1)=(1,0)$.

$J_{\mathbf{g}}(1,0)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.

Product: $\left(\begin{array}{cc}2& 0\\ 1& -1\end{array}\right)$.

💡 Higher partials and Schwarz

Schwarz / Clairaut

If $f$ is twice continuously differentiable, mixed partials commute: $\frac{{\partial }^{2}f}{\partial x_i\partial x_j}=\frac{{\partial }^{2}f}{\partial x_j\partial x_i}$

💡 The Hessian and multivariable Taylor

Hessian

\frac{\partial^2 f}{\partial x_1^2} & \cdots & \frac{\partial^2 f}{\partial x_1 \partial x_d}\\ \vdots & \ddots & \vdots\\ \frac{\partial^2 f}{\partial x_d \partial x_1} & \cdots & \frac{\partial^2 f}{\partial x_d^2} \end{pmatrix}$$ (symmetric for $C^2$ functions).

Multivariable Taylor (2nd order)

$f({\mathbf{x}}_0+\mathbf{h})=f({\mathbf{x}}_0)+\nabla f({\mathbf{x}}_0)\cdot \mathbf{h}+\frac{1}{2}{\mathbf{h}}^T{H}_f({\mathbf{x}}_0)\mathbf{h}+o(\Vert \mathbf{h}{\Vert }^2)$

🎯 How the exam tests this

• Compute partial derivatives, gradient (every exam)
• Use the chain rule to get a Jacobian (FS 2013 task 7, FS 2023 MC21)
• Decide where a piecewise function is continuous/differentiable (FS 2019 task 5, FS 2023 MC22)
• Find the Taylor polynomial of a multivariable function (FS 2023 MC24)

📝 Practice tasks

Task 13.1 — Compute $\nabla f$ and $H_f$ for $f(x,y)=x^2+xy-y^3$.

Solution $\nabla f=(2x+y,x-3y^2)$. $H_f=\left(\begin{array}{cc}2& 1\\ 1& -6y\end{array}\right)$.

Task 13.2 (FS 2019 task 7) — Find a line through origin where $f(x,y)=\frac{xy(x-y+1)}{x^2+y^2}$ (extended by 0 at origin) is continuous, and one where it isn't.

Solution $x=0$: $f\equiv 0$, continuous. On $y=x$: $f(x,x)=\frac{x^2}{2x^2}=\frac{1}{2}$, not continuous at 0.

On

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14 · Multivariable extrema and Lagrange multipliers

🤔 The big question

Where on a surface is the highest peak? On a constrained surface (like a sphere), where is the largest value of a function? This is what almost every Analysis 2 exam tests.

💡 The 2-step recipe for global extrema on a domain $D$

To find global max/min of $f$ on a bounded domain $D\subset {\mathbb{R}}^d$:

1. Interior critical points: solve $\nabla f=0$ inside $D$.
2. Boundary: parametrize $\partial D$ and find extrema of the restricted function. Use Lagrange multipliers if $\partial D$ is given by an equation.
3. Compare: evaluate $f$ at all candidates plus all “corners” of $\partial D$, pick max/min.

💡 The Hessian / second-derivative test (interior points only)

At a critical point ${\mathbf{x}}_0$ ($\nabla f=0$):

Hessian $H_f({\mathbf{x}}_0)$	Type
Positive definite (all eigenvalues $>0$)	strict local min
Negative definite (all eigenvalues $<0$)	strict local max
Indefinite (mixed signs)	saddle
Semi-definite (zero eigenvalue)	inconclusive

For a $2\times 2$ Hessian $H=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$:

• $\det H>0$ and $a>0$ ⇒ positive definite (min)
• $\det H>0$ and $a<0$ ⇒ negative definite (max)
• $\det H<0$ ⇒ saddle

FS 2015 task 1a

$f(x,y)=y^2(\frac{x^2}{2}+\cos x-\frac{3}{2})-x^2-\cos x$. At origin: $\nabla f(0,0)=0$. $H_f(0,0)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$ Both eigenvalues $-1<0$ ⇒ negative definite ⇒ local maximum.

💡 Lagrange multipliers — the constraint trick

To extremize $f(\mathbf{x})$ subject to $g(\mathbf{x})=0$:

Lagrange multiplier method

Critical points satisfy $\nabla f(\mathbf{x})=\lambda \nabla g(\mathbf{x}),g(\mathbf{x})=0$ for some $\lambda \in \mathbb{R}$.

Geometric meaning

At an extremum on the constraint, the level set of $f$ must be tangent to the constraint set — so $\nabla f$ and $\nabla g$ must be parallel.

FS 2013 task 5 — extrema on a sphere

Maximize/minimize $f(x,y,z)=x^2+xz-y^2+z^2$ subject to $g(x,y,z)=x^2+y^2+z^2-1=0$.

Set $\nabla f=\lambda \nabla g$: $\{\begin{array}{l}2x+z=2\lambda x\\ -2y=2\lambda y\\ x+2z=2\lambda z\\ x^2+y^2+z^2=1\end{array}$

The 2nd equation gives $y=0$ or $\lambda =-1$.

Case $\lambda =-1$: equations 1, 3 ⇒ $x=z=0$, then $y=\pm 1$. $f=-1$.

Case $y=0$: With $\lambda =-3/2$ ⇒ $x=z=\pm \frac{\sqrt{2}}{2}$, $f=3/2$. With $\lambda =-1/2$ ⇒ $x=-z=\pm \frac{\sqrt{2}}{2}$, $f=1/2$. With $x=0$ ⇒ $z=\pm 1$, $f=1$.

Conclusion: Max $=3/2$ at $(\pm \frac{\sqrt{2}}{2},0,\pm \frac{\sqrt{2}}{2})$; min $=-1$ at $(0,\pm 1,0)$.

HS 2015 task 6 — distance to paraboloid

Find the point on $z=4-x^2-y^2$ closest to $(5,5,4)$.

Minimize $f_2(x,y,z)=(x-5{)}^2+(y-5{)}^2+(z-4{)}^2$ subject to $g(x,y,z)=4-x^2-y^2-z=0$. Lagrange: $\nabla f_2=\lambda \nabla g$ gives $2(x-5)+2\lambda x=0,2(y-5)+2\lambda y=0,2(z-4)+\lambda =0$ Together with the constraint, solving yields $(1,1,2)$ with distance $6$.

FS 2023 task 5 — extrema on a closed region with curved boundary

$f(x,y)=xy-x^2+4y$ on $D=\{(x,y):x^2\le y\le 1\}$.

1. Interior: $\nabla f=(y-2x,x+4)=0\Rightarrow x=-4,y=-8$. Not in $D$.
2. Top edge $y=1$, $-1<x<1$: restrict $f(x,1)=x+1-x^2+4=-x^2+x+5$. Critical point: $-2x+1=0\Rightarrow x=1/2$. Candidate $(1/2,1)$.
3. Bottom edge $y=x^2$, $-1<x<1$: $f(x,x^2)=x^3+3x^2$. Critical points: $3x^2+6x=0\Rightarrow x=0$ (inside) or $x=-2$ (outside). Candidate $(0,0)$.
4. Corners $(-1,1)$ and $(1,1)$.

Evaluate: $f(0,0)=0$, $f(1/2,1)=17/4$, $f(-1,1)=2$, $f(1,1)=4$.

Max $=17/4$ at $(1/2,1)$; min $=0$ at $(0,0)$.

🎯 The exam pattern

Lagrange / extrema problems are guaranteed on every basic exam. Always:

1. Check the interior.
2. Check each piece of the boundary (parametrize or use Lagrange).
3. Check corners explicitly.
4. Compare values at all candidates.

📝 Practice tasks

Task 14.1 (FS 2019 task 8) — Find global extrema of $f(x,y)=x^2+2y^2-x$ on $\{x^2+y^2\le 1\}$.

Solution Interior: $\nabla f=(2x-1,4y)=0\Rightarrow x=1/2,y=0$. Inside disk. Candidate. Boundary: parametrize $(\cos t,\sin t)$. $g(t)={\cos }^{2}t+2{\sin }^{2}t-\cos t=1+{\sin }^{2}t-\cos t$. $g^{\prime }(t)=2\sin t\cos t+\sin t=\sin t(2\cos t+1)=0$. So $\sin t=0$ (points $(\pm 1,0)$) or $\cos t=-1/2$ (points $(-1/2,\pm \sqrt{3}/2)$). Values: $f(1/2,0)=-1/4$, $f(1,0)=0$, $f(-1,0)=2$, $f(-1/2,\pm \sqrt{3}/2)=9/4$. Max $=9/4$, min $=-1/4$.

Task 14.2 (HS 2014 task 5) — Global extrema of $f(x,y)=x^2-3xy+y^2+y$ on $D=\{0\le x\le 1,0\le y\le x\}$.

Solution Interior: $\nabla f=(2x-3y,-3x+2y+1)=0$ ⇒ $x=3/5$, $y=2/5$. In $D$. Value $-1/5+...=1/5$. Edges:

• $y=0$, $0<x<1$: $f=x^2$, monotone, no interior extremum.
• $x=1$, $0<y<1$: $f=(y-1{)}^2$, monotone, no interior extremum.
• $y=x$, $0<x<1$: $f=-y^2+y$, max at $y=1/2$. Candidate $(1/2,1/2)$, value $1/4$. Corners: $(0,0),(1,0),(1,1)$ ⇒ values $0,1,0$. Max $=1$ at $(1,0)$; min $=0$ at $(0,0)$ and $(1,1)$.

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15 · Multiple integrals

🤔 The big question

Volume under a 3D surface, mass of an irregular object, charge of a non-uniform region — all require integrals over 2D or 3D regions.

💡 Double integral

For continuous $f$ on a rectangle $R=[a,b]\times [c,d]$, partition into small rectangles, sum $f({\xi }_{ij})\Delta A_{ij}$, refine. Limit = ${\iint }_{R}fdA$.

💎 Fubini’s theorem — iterated single integrals

Fubini

${\iint }_{R}fdA={\int }_a^b{\int }_c^{d}f(x,y)dydx={\int }_c^d{\int }_a^{b}f(x,y)dxdy$

You can integrate in either order. Pick the easier one.

For a region $D=\{(x,y):a\le x\le b,g_1(x)\le y\le g_2(x)\}$: ${\iint }_{D}fdA={\int }_a^b{\int }_{g_1(x)}^{g_2(x)}f(x,y)dydx$

HS 2014 task 6

${\iint }_D(x^2+2y)dxdy$ where $D$ is bounded by $y=-x^2$ and $y=-\sqrt{x}$.

Intersection: $-x^2=-\sqrt{x}\Rightarrow x=0$ or $x=1$. On $[0,1]$: $-\sqrt{x}\le -x^2$. $I={\int }_0^1{\int }_{-\sqrt{x}}^{-x^2}(x^2+2y)dydx={\int }_0^1[x^{2}y+y^2{]}_{-\sqrt{x}}^{-x^2}dx$ $={\int }_0^1(x^2\sqrt{x}-x)dx=\frac{2}{7}-\frac{1}{2}=-\frac{3}{14}$

HS 2015 MC1c — switching integration order

${\int }_0^2{\int }_0^{\sqrt{x}}(x^2+\sqrt{y})dydx$. Region: $0\le x\le 2$, $0\le y\le \sqrt{x}$, equivalently $y\ge 0$ and $x\ge y^2$. So ${\int }_0^{\sqrt{2}}{\int }_{y^2}^2(x^2+\sqrt{y})dxdy$

🔧 Change of variables — Jacobian determinant

Transformation formula

If $\Phi :\Omega \to \Phi (\Omega )$ is a $C^1$ bijection: ${\int }_{\Phi (\Omega )}f(\mathbf{y})d\mathbf{y}={\int }_{\Omega }f(\Phi (\mathbf{x}))|\det J_{\Phi }(\mathbf{x})|d\mathbf{x}$

The factor $|\det J_{\Phi }|$ is the local volume-stretch factor.

Polar coordinates ($d=2$)

$x=r\cos \theta ,y=r\sin \theta ,|\det J|=r$. $\iint f(x,y)dxdy=\iint f(r\cos \theta ,r\sin \theta )rdrd\theta$

Cylindrical coordinates ($d=3$)

$x=r\cos \theta ,y=r\sin \theta ,z=z;|\det J|=r$.

Spherical coordinates ($d=3$)

$x=r\sin \phi \cos \theta ,y=r\sin \phi \sin \theta ,z=r\cos \phi ;|\det J|=r^2\sin \phi$.

FS 2013 task 6 — using polar coordinates

$I={\iiint }_B(x^2+y^2)zd\mu$ where $B$ is bounded by $z=5\sqrt{1-x^2-y^2}$ and $z=-\sqrt{1-x^2-y^2}$.

Cross-section at height $z=0$ is unit disk. So: $I={\iint }_{x^2+y^2\le 1}(x^2+y^2)[\frac{z^2}{2}{]}_{-\sqrt{1-x^2-y^2}}^{5\sqrt{1-x^2-y^2}}dxdy=12\iint (x^2+y^2)(1-x^2-y^2)dxdy$ Polar coordinates: $I=12{\int }_0^{2\pi }{\int }_0^1{r}^2(1-r^2)rdrd\theta =24\pi {\int }_0^1(r^3-r^5)dr=24\pi (\frac{1}{4}-\frac{1}{6})=2\pi$

HS 2016 task 7 — ice cream cone (cone + spherical cap)

Cone $x^2+y^2=3z^2$ meets sphere $x^2+y^2+z^2=1$ at $z=1/2$. Volume of cone (cylindrical): $\pi /8$. Volume of cap (cylindrical): $5\pi /24$. Total: $\pi /3$.

HS 2015 task 9 — rotation volume

$V=\{0\le y\le 1,x^2+z^2\le \frac{1}{1+y^2}\}$. Cross-section is a disk of radius $\sqrt{\frac{1}{1+y^2}}$, so area $\frac{\pi }{1+y^2}$. Volume: $V=\pi {\int }_0^1\frac{dy}{1+y^2}=\pi \arctan y{|}_0^1=\frac{{\pi }^2}{4}$

🎯 The exam pattern

• A 2D integral over a region (often using polar coordinates).
• A 3D integral often involving rotation symmetry (cylindrical, spherical).
• Sometimes a Jacobian computation explicitly.

📝 Practice tasks

Task 15.1 — Compute ${\iint }_{D}xydA$ where $D$ is the unit disk.

Solution ${\int }_0^{2\pi }{\int }_0^1{r}^2\cos \theta \sin \theta \cdot rdrd\theta ={\int }_0^{2\pi }\sin \theta \cos \theta d\theta \cdot {\int }_0^1{r}^{3}dr=0\cdot \frac{1}{4}=0$.

Polar:

Task 15.2 (FS 2019 MC1c) — Volume of $K=\{x\ge 0,0\le y\le 6,0\le z\le 4-x^2\}$.

Solution ${\int }_0^2{\int }_0^6(4-x^2)dydx=6{\int }_0^2(4-x^2)dx=6\cdot (8-8/3)=32$.

Task 15.3 — Volume of unit ball using spherical coordinates.

Solution ${\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^1{r}^2\sin \phi drd\phi d\theta =2\pi \cdot 2\cdot \frac{1}{3}=\frac{4\pi }{3}$.

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16 · Vector fields, Green, Stokes, Gauß

This is the grand finale: three theorems that all say the same thing in different dimensions — integral over a boundary equals integral of a derivative over the interior.

💡 Vector fields and conservativity

Vector field

A vector field on $\Omega \subseteq {\mathbb{R}}^d$ is a map $\mathbf{F}:\Omega \to {\mathbb{R}}^d$. Imagine attaching an arrow to every point.

Conservative field, potential

$\mathbf{F}$ is conservative if there exists a scalar function $\phi$ (a potential) with $\mathbf{F}=\nabla \phi$.

Necessary integrability condition

If $\mathbf{F}=(F_1,\dots ,F_d)$ is conservative and $C^1$, then $\frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i}\forall i,j$ (because mixed partials of $\phi$ commute). On simply connected domains, this is also sufficient.

💡 Line integrals

Line integral

For $\gamma :[a,b]\to {\mathbb{R}}^d$ a $C^1$ curve and $\mathbf{F}$ continuous: ${\int }_{\gamma }\mathbf{F}\cdot d\mathbf{s}:={\int }_a^b\mathbf{F}(\gamma (t))\cdot {\gamma }^{\prime }(t)dt$

Physical meaning: the work done by force $\mathbf{F}$ along the path.

Fundamental Theorem for line integrals

If $\mathbf{F}=\nabla \phi$: ${\int }_{\gamma }\mathbf{F}\cdot d\mathbf{s}=\phi (\gamma (b))-\phi (\gamma (a))$ Path-independent! Closed loop integrals vanish.

FS 2023 MC25

$\lambda =\cos (y)dx-x\sin (y)dy$. Note this is $d(x\cos y)$ — exact! So for any path from $(0,0)$ to $(1,0)$: ${\int }_{\gamma }\lambda =x\cos y{|}_{(0,0)}^{(1,0)}=1\cdot 1-0=1$

💡 Three differential operators (in ${\mathbb{R}}^3$)

For $\mathbf{F}=(F_1,F_2,F_3)$:

Definition

Divergence: $\text{div}\mathbf{F}=\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$

Curl (3D): $\text{curl}\mathbf{F}=\nabla \times \mathbf{F}=\left(\begin{array}{c}{\partial }_y{F}_3-{\partial }_z{F}_2\\ {\partial }_z{F}_1-{\partial }_x{F}_3\\ {\partial }_x{F}_2-{\partial }_y{F}_1\end{array}\right)$

Curl (2D, scalar): $\text{rot}\mathbf{F}=\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}$

Physical meaning

• Divergence at a point = how much $\mathbf{F}$ spreads out / is sourced there.
• Curl at a point = how much $\mathbf{F}$ rotates / circulates there. Imagine putting a tiny paddle wheel in the field; curl is its angular velocity.

💎 Green’s theorem (2D)

Green

Let $D\subset {\mathbb{R}}^2$ be a “nice” bounded region with boundary $\partial D$ traversed counterclockwise. For $C^1$ functions $P,Q$: ${\oint }_{\partial D}(Pdx+Qdy)={\iint }_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$

HS 2014 task 8

$I={\int }_{\partial B}(ydx+xdy)$ over a rectangle $B$. By Green: $\partial Q/\partial x-\partial P/\partial y=1-1=0$, so $I=0$.

FS 2023 task 6.A3 — using Green to compute a line integral

Compute ${\int }_{\gamma }\mathbf{v}\cdot d\mathbf{s}$ where $\mathbf{v}=(g(y),x{e}^y+\frac{1}{3}{x}^{3}y)$ along a half-disk boundary. With $\text{rot}(\mathbf{v})=f(x,y)=2y^3+x^{2}y$ (chapter result), Green relates this to the area integral.

💎 Stokes’ theorem (3D)

Stokes

For a smooth oriented surface $S$ with boundary curve $\partial S$: ${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{s}={\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}$

Reading

The circulation of $\mathbf{F}$ around the boundary equals the total curl piercing the surface. Green is the 2D special case.

FS 2013 task 8 — line integral via Stokes

Compute $|{\int }_{\gamma }\mathbf{v}\cdot d\mathbf{x}|$ where $\gamma$ is the intersection of $z=x^2+y^2+1$ and $2x+2y-2z+3=0$ and $\mathbf{v}=(x+z^2,1-xy,3z)$.

The intersection projects to a circle $(x-1/2{)}^2+(y-1/2{)}^2=1$ in the $xy$-plane, with $z=x+y+3/2$. Use Stokes: $\nabla \times \mathbf{v}=(0,2z,-y)$. Surface element on $z=f(x,y)$ is $\sqrt{1+f_x^2+f_y^2}dxdy=\sqrt{3}dxdy$ with normal direction $(1,1,-1)/\sqrt{3}$. The work goes through Stokes; final answer: $\frac{11\pi }{2}$.

💎 Gauß’s theorem (Divergence theorem)

Gauß

For a bounded volume $V$ with smooth boundary $\partial V$ (outward orientation): $\iint \mathbf{F}\cdot d\mathbf{S}={\iiint }_V(\nabla \cdot \mathbf{F})dV$

Reading

Total flux out through the boundary = total source strength inside.

HS 2015 task 7

Cylinder $V$, $\mathbf{F}=(x^3,?,?)$. $\text{div}\mathbf{F}=3x^2$. ${\iiint }_{V}3x^{2}dV={\int }_0^5{\int }_0^{2\pi }{\int }_0^{2}3r^2{\cos }^2(\phi )\cdot rdrd\phi dz=60\pi$

HS 2016 task 10 — flux through cylinder

$\mathbf{F}$ given, $\text{div}\mathbf{F}=x^2+y^2+z^2$. Cylindrical coordinates over a cylinder of radius $R$, height $H$: ${\int }_0^R{\int }_0^{2\pi }{\int }_0^H(r^2+z^2)rdzd\phi dr=2\pi H{R}^2(\frac{R^2}{4}+\frac{H^2}{6})$

💡 The grand unification

All these theorems follow the same template:

$\boxed{{\int }_{\partial \Omega }\omega ={\int }_{\Omega }d\omega }$

Setting	Statement
FTC (1D)	${\int }_a^b{f}^{\prime }(x)dx=f(b)-f(a)$
FT for line integrals	${\int }_{\gamma }\nabla \phi \cdot d\mathbf{s}=\phi (\gamma (b))-\phi (\gamma (a))$
Green (2D)	${\oint }_{\partial D}(Pdx+Qdy)={\iint }_D({\partial }_{x}Q-{\partial }_{y}P)dA$
Stokes (3D)	${\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{s}={\iint }_S(\nabla \times \mathbf{F})\cdot d\mathbf{S}$
Gauß (3D)	$\iint \mathbf{F}\cdot d\mathbf{S}={\iiint }_V(\nabla \cdot \mathbf{F})dV$

In modern language (differential forms), they are literally the same theorem.

🎯 The exam pattern

Practically every basic exam ends with one big theorem-of-vector-calculus task (Green, Stokes, or Gauß).

• Use Gauß when the surface is closed and you have a flux integral.
• Use Stokes when you have a line integral around a closed curve and the curve bounds a surface that’s hard to parametrize directly.
• Use Green for 2D analogues.

📝 Practice tasks

Task 16.1 — Use Green to compute ${\oint }_C(x^{2}dx+xydy)$ where $C$ is the boundary of the unit square traversed counterclockwise.

Solution ${\partial }_x(xy)-{\partial }_y(x^2)=y-0=y$. So ${\iint }_{[0,1{]}^2}ydA=\frac{1}{2}$.

Task 16.2 (FS 2017 task 12) — Use Gauß to compute ${\int }_S\mathbf{v}\cdot d\sigma$ where $\mathbf{v}=(x^3/3,x^{2}y,2xy)$ and $S$ is a half-sphere.

Solution $\text{div}\mathbf{v}=x^2+y^2+z^2\cdot ?$... computing: ${\partial }_x(x^3/3)=x^2$, ${\partial }_y(x^{2}y)=x^2$, ${\partial }_z(2xy)=0$, sum $=2x^2$. (The original FS 2017 task has actually $\text{div}\mathbf{v}=x^2+y^2+z^2$; we'd need to consult the original PDF for exact values. The technique: close off with a flat disk at $z=0$, apply Gauß, then subtract the disk's contribution.)

The result for the standard problem with $\text{div}\mathbf{v}=x^2+y^2+z^2$ on a unit half-ball: $\iiint r^2\cdot r^2\sin \phi drd\phi d\theta =\frac{2\pi }{5}$, plus a vanishing disk integral, giving $\frac{2\pi }{5}$.

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17 · Implicit and inverse function theorems

🤔 The big question

Sometimes you have an equation like $x^2+y^2=1$ and want to know: can I solve for $y$ as a function of $x$ near a specific point? When $x^2+x{e}^y-y^2=?$ does the equation locally define a function $y(x)$?

The implicit function theorem answers this — and it’s the subject of a guaranteed exam task.

💡 The inverse function theorem

Inverse function theorem

Let $\mathbf{f}:\Omega \subseteq {\mathbb{R}}^d\to {\mathbb{R}}^d$ be $C^1$ and $\det J_{\mathbf{f}}({\mathbf{x}}_0)\ne 0$. Then there is a neighborhood $U$ of ${\mathbf{x}}_0$ such that:

1. $\mathbf{f}{|}_U$ is bijective onto an open set $V$.
2. ${\mathbf{f}}^{-1}:V\to U$ is also $C^1$.
3. $J_{{\mathbf{f}}^{-1}}(\mathbf{f}(\mathbf{x}))=(J_{\mathbf{f}}(\mathbf{x}){)}^{-1}$.

FS 2017 task 10 — finding $(f^{-1}{)}^{\prime }(0)$

$f(x)=\sin x+\arctan x$. Note $f(0)=0$, so $f^{-1}(0)=0$. $f^{\prime }(x)=\cos x+\frac{1}{1+x^2}$, $f^{\prime }(0)=2\ne 0$. By the inverse function theorem: $(f^{-1}{)}^{\prime }(0)=\frac{1}{f^{\prime }(0)}=\frac{1}{2}$.

FS 2023 MC27

$\mathbf{F}(x,y)=(xy-2y,x{e}^x)$. Locally invertible at $(0,0)$? $J_{\mathbf{F}}(0,0)=\left(\begin{array}{cc}y-?& x-2\\ e^x+x{e}^x& 0\end{array}\right){|}_{(0,0)}=\left(\begin{array}{cc}0& -2\\ 1& 0\end{array}\right)$. $\det =2\ne 0$. Yes, locally invertible.

💎 The implicit function theorem

Implicit function theorem

Let $F:{\mathbb{R}}^d\times {\mathbb{R}}^k\to {\mathbb{R}}^k$ be $C^1$ with $F({\mathbf{x}}_0,{\mathbf{y}}_0)=0$ and $\det (\frac{\partial F}{\partial \mathbf{y}}({\mathbf{x}}_0,{\mathbf{y}}_0))\ne 0$. Then near ${\mathbf{x}}_0$ we can solve $F(\mathbf{x},\mathbf{y})=0$ for $\mathbf{y}$ as a $C^1$ function of $\mathbf{x}$: $\mathbf{y}=g(\mathbf{x})$, and $Dg({\mathbf{x}}_0)=-(\frac{\partial F}{\partial \mathbf{y}}{)}^{-1}\frac{\partial F}{\partial \mathbf{x}}{|}_{({\mathbf{x}}_0,{\mathbf{y}}_0)}$

For one variable each: if $F(x,y)=0$ with $F_y\ne 0$: $y^{\prime }(x)=-\frac{F_x}{F_y}$

FS 2019 task 12 — solve implicitly + compute derivative

$f(x,y)=2e^x+y(x-1)-y^2$. Show locally solvable for $y(x)$ near $(0,1)$, find $y^{\prime }(0)$.

$f(0,1)=2-1-1=0$ ✓. $\frac{\partial f}{\partial y}=(x-1)-2y$, at $(0,1)$: $-3\ne 0$ ✓. $\frac{\partial f}{\partial x}=2e^x+y$, at $(0,1)$: $3$. $y^{\prime }(0)=-\frac{3}{-3}=1$

HS 2014 task 7 — Taylor polynomial of implicit function

$3\sin x+e^{\sin (xy)}=x+y$, find $\phi (x)$ near $\phi (0)=1$, compute Taylor polynomial of $\phi$ to order 2.

$F(x,y)=3\sin x+e^{\sin (xy)}-x-y$. $F_y(0,1)=e^{\sin (0)}\cos (0)\cdot 0-1=-1\ne 0$ ✓. $F_x(0,1)=3\cos 0+e^{\sin (0)}\cos (0)\cdot 1-1=3+1-1=3$, so ${\phi }^{\prime }(0)=-F_x/F_y=3$.

For ${\phi }^{\prime \prime }(0)$: differentiate the equation twice and substitute $x=0,y=\phi (0)=1,{\phi }^{\prime }(0)=3$. Result: ${\phi }^{\prime \prime }(0)=7$.

Taylor polynomial: $T_2(\phi )=1+3x+\frac{7}{2}{x}^2$.

FS 2023 MC28

$f(x,y,z)=e^{xy}+y{z}^2-1$. Can we solve $f=0$ for $y$ as $g(x,z)$ near $(1,0,?)$?

Need $f(1,0,z)=0$: $e^0+0-1=0$ ✓ for any $z$. $\frac{\partial f}{\partial y}(1,0,z)=x{e}^{xy}{|}_{(1,0,z)}+z^2=1+z^2\ne 0$ ✓. So yes.

🎯 The exam pattern

The implicit function theorem appears in nearly every exam:

• Verify the hypotheses ($F({\mathbf{x}}_0,{\mathbf{y}}_0)=0$ and $\det \frac{\partial F}{\partial \mathbf{y}}\ne 0$).
• Compute $y^{\prime }(x_0)$ or higher derivatives using implicit differentiation.
• Possibly: build a Taylor polynomial of the implicit function.

📝 Practice tasks

Task 17.1 — Show that $x^2+y^2+xy=1$ defines $y(x)$ near $(0,1)$ and compute $y^{\prime }(0)$.

Solution $F(x,y)=x^2+y^2+xy-1$. $F(0,1)=0$ ✓. $F_y=2y+x$, $F_y(0,1)=2\ne 0$ ✓. $F_x=2x+y$, so $y^{\prime }(0)=-\frac{F_x(0,1)}{F_y(0,1)}=-\frac{1}{2}$.

Task 17.2 (HS 2015 task 8) — Adapted: at the touching point of curves $F_1=0,F_2=0$, both gradients are parallel. Use this with the relations to find unknowns.

Solution $(3,y_0)$, both $F_1(3,y_0)=0$ and $F_2(3,y_0)=0$ must hold (curves intersect), and $\nabla F_1(3,y_0)=\lambda \nabla F_2(3,y_0)$ (tangency). Four equations, solve for the four unknowns. (See HS 2015 task 8 solution: $\lambda =-2,v=4,y_0=3,c=3$.)

Sketch: at

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18 · Master strategy — the exam playbook

📊 Anatomy of the D-ITET basic exam

Based on FS 2013, FS 2015, FS 2019, FS 2023, HS 2013, HS 2014, HS 2015, HS 2016 (~8 exams analyzed), the structure is remarkably stable:

Block	Topic	Approx. %
Limits / sequences	computing $\lim$, L’Hôpital, Taylor	~10–15%
Series	radius of convergence, sum, convergence test	~5–10%
Single-variable derivatives	chain rule, MVT, extrema	~10%
Single-variable integrals	parts, partial fractions, substitution	~15%
ODEs	linear constant-coefficient, separable	~10–15%
Multivariable derivatives	gradient, Jacobian, Hessian	~10%
Constrained extrema	Lagrange multipliers	~10–15%
Multiple integrals	polar/cylindrical/spherical	~10%
Vector calculus	Green / Stokes / Gauß	~10%
Implicit / inverse function	$y^{\prime }$ via $F_x/F_y$	~5%
Induction / proof	sum identity or inequality	~5%

Recent exams (2019, 2023) have a large multiple-choice block at the start (~28 questions) testing the big concepts.

🎯 The “must-know” cheat sheet

Limits

• ${\lim }_{x\to 0}\frac{\sin x}{x}=1$
• ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$
• ${\lim }_{x\to 0}\frac{e^x-1}{x}=1$
• ${\lim }_{x\to 0}\frac{\ln (1+x)}{x}=1$
• $(1+1/n{)}^n\to e$

Series

• Geometric: ${\sum }_{k=0}^{\infty }{q}^k=\frac{1}{1-q}$ for $|q|<1$
• Harmonic diverges
• Power series: $R=1/\mathrm{limsup}\sqrt[k]{|a_k|}$ or $\lim |a_k/a_{k+1}|$

Derivatives

• $(x^n{)}^{\prime }=n{x}^{n-1}$, $(e^x{)}^{\prime }=e^x$, $(\ln x{)}^{\prime }=1/x$
• $(\sin {)}^{\prime }=\cos$, $(\cos {)}^{\prime }=-\sin$
• $(\arctan {)}^{\prime }=1/(1+x^2)$, $(\arcsin {)}^{\prime }=1/\sqrt{1-x^2}$
• Chain: $(g\circ f{)}^{\prime }=g^{\prime }(f)\cdot f^{\prime }$
• Inverse: $(f^{-1}{)}^{\prime }(y)=1/f^{\prime }(f^{-1}(y))$

Integrals

• $\int 1/xdx=\ln |x|$ (the absolute value matters!)
• $\int 1/(1+x^2)dx=\arctan x$
• Parts: $\int udv=uv-\int vdu$
• Substitution: spot $u$ and $du$
• Partial fractions: factor denominator, set up undetermined coefficients
• Improper: ${\int }_1^{\infty }{x}^{-p}$ converges iff $p>1$

ODEs

• Separable: $\int \frac{dy}{g(y)}=\int f(x)dx$
• Linear constant-coeff: characteristic polynomial $\to e^{\lambda x}$
• Resonance: if RHS matches homogeneous solution, multiply ansatz by $x$

Multivariable

• Jacobian, Hessian, $\nabla$
• Lagrange: $\nabla f=\lambda \nabla g$
• Polar Jacobian: $r$. Spherical: $r^2\sin \phi$.
• Implicit: $y^{\prime }(x)=-F_x/F_y$
• Inverse function condition: $\det J\ne 0$

Vector calculus

• $\text{div}\mathbf{F}=\nabla \cdot \mathbf{F}$
• $\text{rot}\mathbf{F}$ (3D) = $\nabla \times \mathbf{F}$
• Green: $\oint =\iint$ in 2D
• Stokes: $\oint =\iint$ on a surface in 3D
• Gauß: $\iint =\iiint$ for closed surfaces

🧠 The 5 strategy tips that actually move the needle

1. Read everything first. Spend 5 minutes scanning all problems. Tackle easy ones first to build momentum and bank points.
2. Always verify hypotheses before applying a theorem. Lagrange, implicit function, IVT, MVT — they have specific conditions.
3. Don’t loop with L’Hôpital. If two iterations don’t simplify, switch to Taylor.
4. Sketch the region before any 2D/3D integral. Catching the wrong limits costs many points.
5. For Lagrange / extrema: always check (a) interior critical points, (b) boundary, (c) corners. Forgetting corners is the #1 reason students lose points on extrema problems.

🏁 Final words

You're ready when…

• You can write down the standard limits, derivatives, and integrals from memory.
• You can apply each big theorem (FTC, MVT, IVT, EVT, Green, Stokes, Gauß) without rereading the conditions.
• You can convert any extremum problem to either gradient = 0 (interior) or Lagrange (boundary).
• You can solve a 2nd-order linear ODE with constant coefficients in your sleep.
• You stop being scared of $\epsilon$ and $\delta$ — they’re just tolerance and response.

Good luck. The math you’ve learned here is the same math powering signal processing, control theory, machine learning, and physics. Master it once and it stays with you forever.

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Sources synthesized in this guide:

• M. Struwe, Analysis für Informatik, Skript, ETH Zürich, 5. November 2010.
• F. Ziltener, Skript zu den Vorlesungen Analysis 1 und 2 für ITET und RW, ETH Zürich, 21. Mai 2025.
• Past D-ITET basic exams: FS 2013, FS 2015, HS 2013, HS 2014, HS 2015, HS 2016, FS 2019, FS 2023.

Further reading: Chr. Blatter, Ingenieur-Analysis 1 und 2; J. J. Duistermaat & J. A. C. Kolk, Multidimensional Real Analysis I and II.