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.

---

🧭 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.

---

📋 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

---

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.

💡 Intuition: statements are switches

A mathematical statement is a sentence that is either true or false — like a light switch with only “on” and “off”. No “kinda”, no “maybe”, no “depends on the weather”.

Example

• $4>2$ — TRUE ✅
• $5<3$ — FALSE ❌
• “This sentence is false” — not allowed (paradox: if it’s true it’s false, if it’s false it’s true). Math forbids self-reference like this.

📜 The five connectives

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

Symbol	Read as	Meaning	Mnemonic
$\neg A$	”not $A$“	Flips the truth value	think: “the opposite"
$A\wedge B$	"$A$ and $B$“	True only if both are true	think: a chain — weakest link breaks it
$A\vee B$	”$A$ or $B$“	True if at least one is true	inclusive or — like a menu where you can pick either or both
$A\to B$	”if $A$, then $B$“	Only false when $A$ true, $B$ false	a promise: only broken if you don’t deliver
$A\leftrightarrow B$	”$A$ iff $B$“	Same truth value	”they live or die together”

⚠️ 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.

💡 Contraposition — the most useful identity in proof-writing

$A\to B⟺\neg B\to \neg A$

Example

• “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” (easier to prove!)

📜 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 the statement

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

Negating a real exam-style sentence (FS 2023 task 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: $\neg (P\wedge Q)\equiv \neg P\vee \neg Q$

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

🔧 Three proof techniques you absolutely need

Direct proof

Build a chain $A\Rightarrow B_1\Rightarrow B_2\Rightarrow \cdots \Rightarrow S$.

Proof by contradiction

To prove $A\Rightarrow B$, assume the opposite ($A\wedge \neg B$) and derive nonsense.

There is no largest natural number

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

Mathematical induction — your best friend on exams

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

1. Base case: prove $P(1)$ (or $P(0)$).
2. Inductive step: assume $P(n)$ holds — the induction hypothesis (IH) — and prove $P(n+1)$.

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

Claim: ${\sum }_{k=1}^n(2k-1)=n^2$.

Base ($n=1$): $1=1^2$ ✓

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

🎯 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

📝 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 $N+1$. $■$

---

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.

💡 Sets — collections without order

Set

A set is an unordered collection of distinct objects, called elements.

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

📜 Set operations and De Morgan

Operation	Definition	Picture
Union	$A\cup B=\{x:x\in A\vee x\in B\}$	both circles
Intersection	$A\cap B=\{x:x\in A\wedge x\in B\}$	overlap only
Difference	$A\setminus B=\{x\in A:x\notin B\}$	$A$ minus the overlap
Complement	$A^c=X\setminus A$	everything outside $A$
Cartesian product	$A\times B=\{(a,b):a\in A,b\in B\}$	grid of pairs

De Morgan's Laws

$(A\cap B{)}^c=A^c\cup B^c,(A\cup B{)}^c=A^c\cap B^c$

💡 Functions — the rule book

Function

A function $f:X\to Y$ assigns to every $x\in X$ exactly one $f(x)\in Y$.

• $X$ — domain
• $Y$ — codomain
• Notation: $x\mapsto f(x)$

Example

• $f:\mathbb{R}\to \mathbb{R},x\mapsto x^2$
• $g:\mathbb{R}\to \mathbb{R},x\mapsto \sin x$
• $h:[0,\infty )\to \mathbb{R},x\mapsto \sqrt{x}$ — only on non-negatives!

🔧 Injective, surjective, bijective

Definition

• Injective (one-to-one): different inputs → different outputs. $f(x_1)=f(x_2)\Rightarrow x_1=x_2$.
• Surjective (onto): every $y\in Y$ is hit. $\forall y\in Y,\exists x:f(x)=y$.
• Bijective: both. Then an inverse $f^{-1}:Y\to X$ exists.

Mental picture

• $f(x)=x^2$ on $\mathbb{R}$: NOT injective ($f(1)=f(-1)$), NOT surjective onto $\mathbb{R}$ (negative numbers aren’t hit).
• $f(x)=e^x:\mathbb{R}\to (0,\infty )$: bijective, inverse is $\ln$.

🔧 Composition

If $f:X\to Y$ and $g:Y\to Z$: $(g\circ f)(x)=g(f(x))$ Apply $f$ first, then $g$. Associative: $(h\circ g)\circ f=h\circ (g\circ f)$.

🎯 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.

---

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: $w_0=2e^{i\pi /3}=1+i\sqrt{3}$, $w_1=2e^{i\pi }=-2$, $w_2=2e^{i5\pi /3}=1-i\sqrt{3}$.

---

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.

---

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 $\frac{3}{4}=75\%$.

---

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 $C^1$.

---

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:

---

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

---

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).

---

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:

---

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

---

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:

---

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

---

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 $(1,1)$.

---

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

---

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 $\frac{2\pi }{5}$.

---

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

---

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.

---

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.