why the equation $f^{(n)}(x)=0$ has at least $n-1$ distinct roots in $(-1,1)$ Let $f \in C^{(n)} ( (-1,1) )$ and $\sup_{-1<x<1}|f(x)|\leq 1$.
show  that:
There exists a number $\alpha_n$ depending only on $n$ such that if $|f'(0)|\geq \alpha_n$,then the equation $f^{(n)}(x)=0$ has at least $n-1$ distinct roots in $(-1,1)$
The problem is from Zorich analysis; the book hint: prove by induction that there exists a sequence $x_{k_{1}}<x_{k_{2}}<\cdots<x_{k_{k}}$ of pointsof the open inteval $(-1,1)$ such that $f^{(k)}(x_{k_{i}})\cdot f^{(k)}(x_{k_{i+1}})<0$ for $1\le i\le k-1$
By Now, I still can't use this hint to solve it
 A: We will need ''small'' values of the derivatives on certain intervals.
Lemma: For every positive integer $k$ and every interval $[a,b]\subset(-1,1)$, there exists some $\xi\in(a,b)$ such that $\big|f^{(k)}(\xi)\big|<\frac{k^k\cdot (k+1)!}{(b-a)^k}.$
Proof: Let $x_i=a+\frac{b-a}{k}i$ for $i=0,1,\ldots,k$. By the mean value theorem of the divided differences, there is some $\xi\in(a,b)$ such that
$$
\frac{f^{(k)}(\xi)}{k!} = f[x_0,x_1,\ldots,x_k] = \sum_{i=0}^k \frac{f(x_i)}{\prod\limits_{j\ne i} (x_i-x_i)}.
$$
Since $|f|\le 1$,
$$
\big|f^{(k)}(\xi)\big| \le k!\cdot (k+1)\cdot \frac{1}{\big((b-a)/k\big)^k} =
\frac{k^k\cdot (k+1)!}{(b-a)^k}. \qquad\qquad\Box
$$
(Using Chebyshev polynomials and more precise calculation, the upper bound can be improved to $\frac{2^{2k-1}\cdot k!}{(b-a)^k}$.)
Let $K=n^{2n}\cdot (n+1)!$. By the Lemma, for $k=1,2,\ldots,n$ there are some numbers 
$u_k\in\left(\frac{k-1}{2n},\frac{k}{2n}\right)$,
$v_k\in\left(-\frac{k}{2n},-\frac{k-1}{2n}\right)$, such that 
$\big|f^{(k)}(u_k)\big|<K$ and $\big|f^{(k)}(v_k)\big|<K$. Call a value ''small'' if its absolute value is less than $K$. Hence, we have 
$$ -\frac12 < v_{n}<\ldots<v_1<0<u_1<\ldots<u_{n}<\frac12; $$
the numbers $f^{(k)}(u_k)$ and $f^{(k)}(v_k)$ are all small, but $f'(0)\ge \alpha_n$ is ``big''.
Suppose that $f'(0)>2^n\cdot K$. From this start, we can prove by induction that for every $k=1,\ldots,n$ there are some numbers $x_{k,1},\ldots,x_{k,k}$ such that
$\qquad$ $v_k<x_{k,1}<\ldots<x_{k,k}<u_k$;
$\qquad$ $f^{(k)}(x_{k,i}) > 2^{n+1-k} K$ if $i$ is odd;
$\qquad$ $f^{(k)}(x_{k,i}) < -2^{n+1-k} K$ if $i$ is even.
For $k=1$, $x_{1,1}=0$ is an appropriate choice.
Suppose that $x_{k,1}<\ldots<x_{k,k}$ are already found, and add two elements, $x_{k,0}=v_k$ and $x_{k,k+1}=u_k$ at the two ends. For every $i=1,\ldots,k+1$, choose $x_{k+1,i+1}$ in such a way that
$$
f^{(k+1)}(x_{k+1,i+1}) = \frac{f^{(k)}(x_{k,i})-f^{(k)}(x_{k,i-1})}{x_{k,i}-x_{k,i-1}}.
$$
If $i$ is odd then
$f^{(k+1)}(x_{k+1,i}) > \frac{2^{n+1-k}K - K}{1} > 2^{n-k}K$.
If $i$ is even then 
$f^{(k+1)}(x_{k+1,i}) < \frac{-2^{n+1-k}K + K}{1} < -2^{n-k}K$.
We also have $v_{k+1}=x_{k,0}<x_{k+1,1}<x_{k,1}<\ldots<x_{k+1,k+1}<x_{k,k+1}=u_k$, so we have the prescribed ordering.
At the end, we constructed some numbers $\frac12<x_{n-1,1}<\ldots<x_{n-1,n}<\frac12$ such that the values $f^{(n)}(x_{n,k})$ have alternating signs.
Remark. It is easy to see that $(\arctan x)^{(n)}$ has precisely $n-1$ zeros. Choosing $f(x)=\frac2\pi \arctan (Kx)$, there are at most $n-1$ zeros but $f^{(n)}(0)$ can be arbitrarily big.
