0. Tricks & Exam

Einfach Quadratische Gleichung 2. Grades Lösen

Für $x^2-ax+b=0$ gilt. $\alpha ,\beta$ sind Lösungen wenn $\alpha \cdot \beta =a$ und $\alpha +\beta =b$ halten.

Beispiel: Für $x^2-8x+15=0$ erfüllen $\alpha :=-3$, $\beta :=-5$ die Konditionen. Also sind $-3,-5$ die Lösungen.

Ableitbarkeit beweisen

Case 1: Komposition ableitbarer

If our function is a composition / product of other differentiable functions, we can argue via sum/product/quotient/chain rule.

On $\mathbb{R}\setminus \{0\}$, $f$ is a composition/product of differentiable functions, hence differentiable (using the chain rule, product rule) and $f^{\prime }(x)=...$

Case 2: Limit Punkte

There might be points where:

• a denominator becomes $0$
• $|\cdot |$
• $\sqrt[3]{\cdot }$
• or anything where the formula changes

Here we go back to the definition and we show that $f^{\prime }(a)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ exists.

If $f$ is continuous at $a$, differentiable on a punctured neighbourhood, and ${\lim }_{x\to a}{f}^{\prime }(x)=L$ exists, then $f^{\prime }(a)=L$ (by the MVT).

→ See Definition 5.2 in Differentialrechnung für warum MVT.

Trap: Make sure to actually compute the limit! $f^{\prime }(a)$ can exist without the limit existing!

Checklist:

1. continuity at $a$
2. compute both one-sided limits if they are different (piece wise function ex)

Tipp: squeeze theorem is usually very useful here: $\frac{|f(x+h)-f(x)}{h}|\le |h|\to 0$ usually works quick

Case 2 - Carathéodory

We can also use Carathéodory which asserts that if there is a $\varphi$ with $f(x)-f(a)=\varphi (x)(x-a)$ continuous, then $f$ is differentiable at $a$ → continuity check instead of limit.

Disproving

Show that:

• two one-sided limits disagree
• there is a discontinuity at some point

Prove Pointwise vs. Uniform Convergence

Step 1 - Pointwise limit

Fix $x\in D$. Compute $f(x):={\lim }_{n\to \infty }{f}_n(x)$.
Pay attention to case splits.

Step 2 - Uniform

Compute

$$M_n:=\underset{x\in D}{\sup }|f_n(x)-f(x)|$$

then $f_n\to f$ uniformly on $D$ $⟺$ $M_n\to 0$!

To find $M_n$:

1. estimate it away:
   1. find $|f_n(x)-f(x)|\le a_n$ with $a_n\to 0$ and $a_n$ does not depend on $x$.
2. Maximise:
   1. Differentiate the difference $g(x)=|f_n(x)-f(x)|$ and find the critical point
   2. plug it back in.

Disprove Uniform

We can find a witness sequence: $x_n\in D$ and $c>0$ s.t. $|f_n(x_n)-f(x_n)|\ge c$ for all large $n$.
→ this works for cases with a bump!

• ex: $f_n(x)$ approaches $0$ everywhere, except at $0$ where it’s $1$ (ex: $e^{-nx}$ for example).

Partial fractions

Factor in denominator	Term in partial fraction decomposition
$ax+b$	$\frac{A}{ax+b}$
$(ax+b{)}^k$	$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b{)}^2}+\cdots +\frac{A_k}{(ax+b{)}^k},k=1,2,3,\dots$
$a{x}^2+bx+c$	$\frac{Ax+B}{a{x}^2+bx+c}$
$(a{x}^2+bx+c{)}^k$	$\frac{A_{1}x+B_1}{a{x}^2+bx+c}+\frac{A_{2}x+B_2}{(a{x}^2+bx+c{)}^2}+\cdots +\frac{A_{k}x+B_k}{(a{x}^2+bx+c{)}^k},k=1,2,3,\dots$

Note: total number of coefficients always = $\mathrm{deg}Q$.

• linear factor of multiplicity $k$ contributes $k$ unknowns
• an irreducible quadratic of multiplicity $k$ contributes $2k$
  → sum = $\mathrm{deg}Q$.

To determine the coefficients, evaluate the linear system and find the solutions.

• group by $x^i$.