FS20

1

c)

ii) MVT gives us that the value is assumed at least once.
$f^{\prime }(c)=f(1)-f(0)/1=f(1)-f(0)<0$. Since $f$ is twice differentiable, $f^{\prime }$ is differentiable on $[-1,1]$ and thus continuous → if not twice diff, then $f^{\prime }$ not continuous

Then set $ϵ:=-\frac{1}{2}{f}^{\prime }(c)>0$. By continuity of $f^{\prime }$ at $c$ there’s $\delta >0$ s.t.

$$|x-c|<\delta ⟹|f^{\prime }(x)-f^{\prime }(c)|<ϵ$$

hence for such $x$, $f^{\prime }(x)<f^{\prime }(c)+ϵ=\frac{1}{2}{f}^{\prime }(x)<0$.
To choose $a,b$ we can take $\delta <\min \{c,1-c\}$ and thus $(c-\delta ,c+\delta )\subseteq (0,1)$.

iii) Sei $f:[-1,1]\to \mathbb{R}$ eine gerade zweimal differenzierbare Funktion. Für alle $x\in ]0,1[$ gibt es $\eta \in ]0,x[$ so dass $f(x)-f(0)=\frac{f^{\prime \prime }(\eta )x^2}{2}$. → WAHR
Proof:

• $f(x)=f(0)+\frac{f^{\prime }(0)x^1}{1}+\frac{f^{\prime \prime }(\eta )x^2}{2}$ from Taylor expansions (differentiable up to 2 times) around $0$.
• As $f$ is even, $f^{\prime }$ is odd
  • $f(-x)=f(x)⟹$ $-f^{\prime }(-x)=f^{\prime }(x)⟹f^{\prime }(-x)=-f^{\prime }(x)$.
• Therefore $f^{\prime }(0)=-f^{\prime }(-0)⟹f^{\prime }(0)+f^{\prime }(-0)=0$
  • $2f^{\prime }(0)=0⟹f^{\prime }(0)=0$.
  • intuitiv: f(x) even muss kritischen punkt bei $0$ haben.
• Daher gilt $f(x)=f(0)+\frac{f^{\prime \prime }(\eta )x^2}{2}$
• $f(x)-f(0)=\frac{f^{\prime \prime }(\eta )x^2}{2}$.

Use Taylor

For $n$ times differnetiable function, we can expand it around $a$ using Taylor, up to term $n$, which then gets the $f^{(n)}(\eta )x^n/n!$.

Note: das braucht keine Konvergenz → ist ja keine Taylorreihe! Das ist die Taylorformel mit Lagrange-Expansion.

e)

iii) We have $f,g:[0,\infty )\to \mathbb{R}$ non-negative differentiable functions. If ${\int }_0^{\infty }f(x)dx$ converges and ${\int }_0^{\infty }g(x)dx$ converges, does ${\int }_0^{\infty }g(f(x))dx$ converge? → NO
Intuitiv: $f(x)\to 0$ deswegen sampled $f$ dann $g$ nahe null → wenn $g(0)$ z.B. 10 ist, summieren wir sehr oft 10…
Proof: We construct a counterexample:

• let $f,g=e^{-x}$
  • then both converge to $0$
• but $g(f(x))=e^{-(e^{-x})}$ converges to $e^{-0}=1$.

2

ii)

Calculate the limit of ${\int }_{-\infty }^{0}x{e}^{3x}dx$.
We expand using partial derivation to ${\left[\frac{x}{3}{e}^{3x}\right]}_{-\infty }^0-\frac{1}{3}{\int }_{-\infty }^0{e}^{3x}$.

1. ${\left[\frac{x}{3}{e}^{3x}\right]}_{-\infty }^0$ gives us:
   1. $0/3e^0=0$
   2. ${\lim }_{x\to -\infty }x/3e^{3x}=0$
      1. Prove using variable sub. and l’Hopital:
         1. $t=-x$ then we get ${\lim }_{t\to \infty }-t/e^{3t}$ gives -infty/infty
         2. with Hopital we get ${\lim }_{t\to \infty }-\frac{3}{e^{3t}}=0$.
   3. Thus we get $0$
2. $-\frac{1}{3}{\int }_{-\infty }^0{e}^{3x}$ gives $-\frac{1}{3}{\left[\frac{1}{3}{e}^{3x}\right]}_{-\infty }^0$
   1. gives $-1/3\cdot 0$
   2. ${\lim }_{x\to -\infty }-\frac{1}{3}{e}^{3x}=-\frac{1}{9}\cdot 1$.
      In the end we get $-1/9$.

3 (Proof)

e^x \ge x + 1

We need to justify $e^x\ge x+1$ for a fully correct proof and always at least state it!

Claim: $\forall x\in \mathbb{R}$, $e^x\ge 1+x$ with equality iff. $x=0$.
Proof: Define $g:\mathbb{R}\to \mathbb{R}$ with $g(x)=e^x-1-x$.

• Then $g$ differentiable (composition of differentiable) and gives $g^{\prime }(x)=e^x-1$.
• $g^{\prime }(x)<0$ for $x<0$ and $g^{\prime }(x)>0$ for $x>0$.
  • Since $\exp$ is strictly increasing and $e^0=1⟹e^0-1=0$ (i.e. critical point at $0$).
• $g$ attains a global minimum at $x=0$. That minimum is $g(0)=e^0-1-0=0$.
• Therefore $g(x)\ge g(0)=0$ forall $x$ with equality exactly at $0$.

continuity of e^x for the limit proof

We know that $a_k$ converges. Call the limit $l:={\lim }_{k\to \infty }{a}_k$.
Since $a_{k+1}$ is just shifted, it converges to the same $l$.

Now we use the recursion:

$$\underset{k\to \infty }{\lim }{e}^{a_k}-1=e^{{\lim }_{h\to \infty }{a}_k}-1$$

This is allowed because $e$ is continuous! This means we can commute the limit!

Thus we get $l={\lim }_{k\to \infty }{a}_k={\lim }_{k\to \infty }{e}^{a_k}-1=e^l-1$. Because $e^l-1=l$ has a unique solution $l=0$, we can conclude $l=0$ (fixed point).

4

5

We can use the monotonicity of the riemann integral and the min-max satz to construct a valid scenario for the ZWS → conclude.