FS25 VMP

Single choice

1

Trick Question → If sup M = inf M, then it can only contain a single point → be constant.
Thus it converges automatically

C is wrong since sup or inf not existing means they aren’t $\pm \infty$.

5

B → wrong since it’s big O not small o

• $f(x)-f(x_0)=o(|x-x_0|)$ would be correct, although it forces the derivative to be $0$
• Only $f(x)-f(x_0)=O(|x-x_0|)$ means that the quotient is bounded, not that it converges to a value (can oscillate in there)

D → correct

• $f(x)-f(x_0)-K(x-x_0)=o(|x-x_0|)$ with $K\in \mathbb{R}$ for $x\to x_0$.
  • thus $\frac{f(x)-f(x_0)}{x-x_0}-K=0$ as $o$ forces it to converge to $0$
  • So $K=f^{\prime }(x_0)$.

10

Differentiability ⇒ stetigkeit ⇒ Riemann-Integrierbar

Aber für eine Funktion $V$ gilt: $V^{\prime }$ die Ableitung muss nicht Integrierbar sein → Volterra type functions.

11

$e^x=p(x)$ for $p(x)$ any polynomial. How many solutions?

For every $p(x)$ at most a finite amount of solutions →

• true for $x\in \mathbb{R}$, the polynomial would have to have an infinite amount of degrees
• false for $x\in \mathbb{C}$, then we can have $p(x)=1$ and circle around as often as we want with sin, cos

For every $N\in \mathbb{N}$ there exists a polynomial $p_N(x)$ s.t. $e^x=p_N(x)$ has at least $N$ solutions. → True, the taylorseries of $e^x$ again.

Proofs

18.a)

Wir leiten $f_n(x)$ ab und können dann $f_n^{\prime }(x)\le 1$ beschränken.
Proof: $f_n(x)$ bounded und dadurch Lipschitz (via MVT)

• for $x<y$ $\exists \xi \in (x,y)$ s.t. $f_n(x)-f_n(y)=f_n^{\prime }(\xi )(x-y)$ hence $|f_n(x)-f_n(y)|\le \sup |f_n^{\prime }|\cdot |x-y|$.
• And the supremum for all $x,y$ is $1$.

Dadurch ist $f_n$ Lipschitz stetig ⇒ glm. stetig.
Proof: Sei $|f_n^{\prime }(x)|\le L=1$

• Dann gilt Lipschitz-Stetigkeit $\forall x,y$ $|f_n(x)-f_n(y)|\le L|x-y|$
• Sei $ϵ>0$ fix. Dann $\exists \delta :=ϵ/L=ϵ$. s.t.
  • $\forall x,y$ $|x-y|<ϵ⟹|f_n(x)-f_n(y)|\le L|x-y|<Lϵ=ϵ$.
  • This gives us uniform continuity

18.b)

Für x = 0 trivial. Für $x\ne 0$, sei $x$ fixiert. Dann wächst der untere Term ins unendliche → geht zu 0.

18.c)

Wir leiten $f_n^{\prime }(x)$ ab und schauen nach kritischen Punkten. Da ${\lim }_{x\to \pm \infty }{f}_n(x)=0$ muss einer das maximum, der andere das Minimum sein.

Dann gilt $\sup |f_n(x)|=\max \left\{f_n\left(\frac{1}{\sqrt{n}}\right),f_n\left(-\frac{1}{\sqrt{n}}\right)\right\}$
Dan beide für ${\lim }_{n\to \infty }=0$, ist $f_n$ gleichmäßig gegen $f\equiv 0$ stetig.

19.a)

We use the MVT to show that $|sin(x)|<|x|$ for $x\ne 0$
Then we can just place into the inequality and get our result out.

19.b)

First we transform $x_n$ to $y_n$ with $y_n=|x_n|$.
Then $0\le |y_n|$ and monoton decreasing as $y_{n+1}=|f(y_n)|<y_n$ and therefore it converges.

Then by continuity of $f$ and stuff ${\lim }_{n\to \infty }{y}_n=L$. And since $f$ has a single fixedpoint $L=0$.
Thus ${\lim }_{n\to \infty }|x_n|=0$ $⟹$ ${\lim }_{n\to \infty }{x}_n=0$. as $|0|=0$!