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Let $f:(-1,1)rightarrowmathbb R$ be infinitely differentiable on $(-1,1)$ with $f(0)=1$ and has the following properties:

(1) $vert f^{(n)}(x)vert le n!$ for every $xin (-1, 1)$ and for every $nin mathbb N$

(2) $f'(frac{1}{m+1})=0$ for every $min mathbb N$

I need to:

Find the value of $f^{(n)}(0)$ for each $nin mathbb N$

Determine the value of $f(x)$ for every $xin(-1,1)$

I claim that $f^{(n)}(0)=0$ for all $nin mathbb N$ and I wanted to prove this by induction. When $n=1$, we note that by (2), $f'(0)=0$ by the sequential criterion for limits with $mrightarrow infty$. However, I am not sure on how to prove the inductive step.

I note that from (1), we have $f(x)lefrac{1}{1-x}$ by Taylor's Theorem.

Any help is greatly appreciated!

Rolle's theorem implies that $f''$ has a zero in the interval $(frac{1}{m+2}, frac{1}{m+1})$
for each $m in Bbb N$, therefore $f''(0) = 0$.

The same argument can be repeated to show that each derivative $f^{(k)}$ has
a sequence of zeros converging to $0$.

(The estimate $vert f^{(n)}(x)vert le n!$ is not needed for this
conclusion.)

Hint. Let us consider the case $n=2$ (try by your own the case $n>2$)

For any $minmathbb{N}$, by using the Taylor expansion of $f'$ at $0$, we have that there exists $t_min (-1,1)$ such that
$$f'left(frac{1}{m+1}right)=f'(0)+f''(0)cdot frac{1}{m+1}+frac{f'''(t_m)}{2}cdot frac{1}{(m+1)^2}.$$
Since $f'left(frac{1}{m+1}right)=0$ and we already know that $f'(0)=0$, it follows
$$f''(0)=-frac{f'''(t_m)}{2}cdot frac{1}{(m+1)}$$
which implies that as $mto+infty$,
$$|f''(0)|=frac{|f'''(t_m)|}{2}cdot frac{1} {(m+1)}leqfrac{M_3} {2(m+1)}to 0.$$
where $M_3=max_{xin[-1/2,1/2]}| f^{(3)}(x)|$ (the assumption $vert f^{(n)}(x)vert le n!$ in $(-1,1)$ is not needed).