Let $f:(-1,1) \rightarrow \mathbb{R}$ be smooth on $(-1,1)$, $f(0) = 1$. Find $f(x)$. Let $f:(-1,1) \rightarrow \mathbb{R}$ be infinitely differentiable on $(-1,1)$, $f(0)=1$ and 
$|f^{(n)}(x)| \le n!$ for every $x\in(-1,1)$ and for every $n\in \mathbb{N} $
$f'(\frac{1}{m+1})=0$ for every $m \in \mathbb{N}$
Determine the value of $f(x)$ for every $x\in(-1,1)$.
I guess $f(x)=1$ for every $x\in(-1,1)$.
 A: We show that $f(x) = 1$ for all $x$ on $(-1, 1)$ in a few steps. First, we show that $f^{(n)}(0) = 0$ for all $n \geq 1$. Then we use Taylor's Theorem and estimate $f(x)$. Finally, we take a limit and show that the Taylor series converges to $f(x)$.

Showing $f^{(n)}(0) = 0$
As $f'$ is continuous, we have immediately from $f'(1/(m+1)) = 0$ that $f'(0) = 0$. More importantly, between any two zeroes of $f'(x)$, there is a zero of $f''(x)$ (by the Mean Value Theorem). So there is a sequence $x_m$ where $\frac{1}{m+1} \leq x_m \leq \frac{1}{m}$ such that $f''(x_m) = 0$, and with $x_m \to 0$. Thus $f''(0) = 0$.
Inductively, one shows that $f^{(n)}(0) = 0$ for all $n \geq 1$. [I leave out the induction details --- the argument is the exact same].

Taylor's Theorem
We now represent $f(x)$ using its first $k$ terms of its Taylor series as
$$ f(x) = \sum_{n = 0}^k f^{(n)}(0) \frac{x^n}{n!} + f^{(k+1)}(y) \frac{x^{k+1}}{(k+1)!}.$$
for some $y$ in $[0, x]$. It is possible that you know a different statement of Taylor's Theorem, but I use this one as it's very basic and one of the weaker statements of the theorem.
As $f^{(n)}(0) = 0$ for all $n \geq 1$, we can write this is
$$ f(x) = 1 + f^{(k+1)}(y) \frac{x^{k+1}}{(k+1)!}.$$
By the given bound on the derivative, $f^{(k+1)}(y) \leq (k+1)!$, so that
$$ \lvert f(x) - 1 \rvert \leq \lvert x \rvert ^{k+1}. \tag{1}$$

Conclusion
As $(1)$ is true for all $k \geq 1$, we take the limit as $k \to \infty$. Since $-1 < x < 1$, we have that $\lvert x \rvert^{k+1} \to 0$. This shows that $f(x) - 1 = 0$ for all $\lvert x \rvert < 1$, and thus $f(x) \equiv 1$.
(Note that we also showed that $f$ is exactly equal to its Taylor series by showing the remainder goes to $0$. Of course, the Taylor series is identically 1, as all other terms vanish.) $\diamondsuit$
