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I have the following:

Suppose $p(x) =(x^{2}-1)^{m}$ for some $m \geq2$.

Applying Rolle's Theorem on each of the intervals $[-1,c]$ and $[c,1]$, show that $p''(x)$ has at least two roots in $(-1,1)$.

By induction, prove that for $k=1,2,\ldots,m$ the $k$-th derivative $p^{(k)}(x)$ has at least $k$ distict real roots lying in the interval $(-1,1)$.

I have first of all shown that for $p'(x)$ that there is a root in $(-1,1)$, but struggling with the other two parts, any help would be most appreciated.

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2 Answers 2

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We have that $$p(x) = (x^2-1)^m$$ Note we need $m >2$.

Taking derivatives we have

$$p'(x)=2xm(x^2-1)^{m-1}$$

$$p''(x)=4x^2m(m-1)(x^2-1)^{m-2}+2m(x^2-1)^{m-1}$$

Rolle's Theorem asserts that if $f(x)$ is continuous in $I=[a,b]$ and $f(a)=f(b)$, then there is a $c \in I$ such that

$$f'(c)=0$$

We have that $f(-1) = f(1)$, so there is a $c$ such that $f'(c)=0$. We can immediatelly see such $c$ is $c=0$.

Moreover, we see that $f'(-1) = f'(1)=0$, but since $f'(0)=0$, we can assert that there exists $c_1 \in (-1,0)$ and $c_2 \in (0,1)$ such that $f''(c_1)=f''(c_2)=0$. Note that if $m=1,2$ the test would fail, since the first (third) derivative is a line.

You want to prove now that for $k=1,2,3,\dots ,m$, it is the case $p^{(k)}(c_i)=0$ for $i=1,2,\dots, m \in (-1,1) $. The induction hypothesis would then be:

$\mathcal H.$ It is true that $p^{(k)}(c_i)=0$ for $i=1,2,\dots,k \in (-1,1) $

This means that $p^{(k)}(c_1) = p^{(k)}(c_2) = \cdots = p^{(k)}(c_i)=0$. Since we also have two more roots, namely $1$ and $-1$, then we can assert by Rolle's Theorem, that $p^{(k+1)}(x)$ will have $n=k+2-1=k+1$ roots. Since $k=2$ is true, this completes the induction phase.

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You know that $p'(x)$ has a root somewhere in $(-1,1)$. Call that "somewhere" $c$.

Now, $p'(x) = m(x^2-1)^{m-1}(2x)$. Since $m-1\gt 0$, it is still the case that $p'(-1)=p'(1)=0$. But you also know that $p'(c)=0$. So set $f(x)=p'(x)$, and note that we have $f(-1)=f(c)=0$, and you can use Rolle's Theorem to conclude that $f'(x) = p''(x)$ has a root in $(-1,c)$; likewise, since $f(c)=f(1)=0$, it follows that $f'(x)=p''(x)$ has a root in $(c,1)$.

Now assume that you know that $p^{(r)}(x)$ has $r$ roots on $(-1,1)$; call them $-1\lt c_1\lt c_2\lt\cdots \lt c_r\lt 1$, and that $p^{(r+1)}$ still is zero at $-1$ and at $1$ (this requires $r\lt m$, as is not hard to check). Now you know that $p^{(r)}(-1)=p^{(r)}(c_1)=\cdots=p^{(r)}(c_r) = p^{(r)}(1) = 0$. You can apply Rolle's Theorem to $p^{(r)}$ on $[-1,c_1]$, on $[c_1,c_2]$, on $[c_2,c_3]$, and so on.

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