On the number of extremum

Question

Let $f$ be infinitely differentiable on $[-1,1]$, $f^{(n)}(-1)=f^{(n)}(1)=0,n=0,1,2,\cdots,$ and $f>0$ in $(-1,1)$. Prove that there is a positive integer $k$ such that $\dfrac{f(x)}{(1-x^2)^k}$ has at least three extreme points in $(- 1,1)$. I tried to use Taylor and $$f(x)=f(1)+f'(1)(x-1)+\dfrac{f''(1)}{2}(x-1)^2+\cdots$$ $$f(x)=f(-1)+f'(-1)(x+1)+\dfrac{f''(1)}{2}(x+1)^2+\cdots$$ But this leads to an expression of $0$. I'm a little confused...

Answer 1

Since $f^{(0)}$ is the function $f$ itself one has $f(-1)=f(1)=0$

Put $F(x)=\dfrac{f(x)}{(1-x^2)^k}$. The derivative of $F$ should have at least three roots for some integer $k\gt0$.

Choose $k=2$. We have $$\frac{dF}{dx}=\frac{f'(x)(1-x^2)-2xf(x)}{(1-x^2)^{3}}=0$$ so $x=0\Rightarrow f'(0)=0$ and,the function $F$ has a minimum at zero. On the other hand, since $f(-1)=0$ we have $F(-1)$ has an indeterminated form but $$\lim_{x\to{-1}}F(x)=\frac{f'(-1)}{2x^3-2x}\\\lim_{x\to{-1}}F(x)=\frac{f''(-1)}{6x^2-2}=0$$Now by prolongation by continuity we can do $F(-1)=0$ and because of $F$ is continuous on the compact $[-1,0]$, by the extreme value theorem, has a maximum at the interior of $[-1,0]$ i.e. on the open interval $(-1,0)$.

Similarly happens on the interval $[0,1]$ with which we have in total three extremums at least as required.