# Rolles Theorem

• Mar 31st 2011, 02:53 PM
worc3247
Rolles Theorem
Let $f(x)=(x^2-1)^n$. Using Rolle's theorem, prove that $f^{(n)}(x)$ is a polynomial of degree n, with distinct roots, all of which lie in the interval (-1,1).
Can anyone help to get me started with this question?
• Mar 31st 2011, 07:59 PM
kalyanram
If $f(a)=f(b)=0$ with $f$ satisfying the conditions of Rolle's Theorem we have $f(c)=0$ $c \in (a,b)$ now we observe that between two roots of $f$ there lies a root of $f^'$. Now observe that $f(x) = (x-1)^{n}(x+1)^{n} f^'$ has a root at $c \in (-1,1)$ in fact c=0. Now for $f^" \exists$ a roots between (-1,0) and (0,1). Now $f^{(n)}$ is indeed an polynomial of order n now use induction to prove that it has distinct roots between (-1,1).

Kalyan.
• Apr 1st 2011, 12:00 AM
chisigma
Quote:

Originally Posted by worc3247
Let $f(x)=(x^2-1)^n$. Using Rolle's theorem, prove that $f^{(n)}(x)$ is a polynomial of degree n, with distinct roots, all of which lie in the interval (-1,1).
Can anyone help to get me started with this question?

For any $0 \le k \le n$ is $f^{(k)} (-1)= f^{(k)} (1) =0$. The function $f(x)$ is 'even' , i.e. is $f(x)= f(-x)$, and that means that $f^{'} (x)$ is 'odd' , i.e. is $f^{'}(x)=-f^{'} (-x)$ , so that is $f^{'} (0)=0$ and that is also consequence of the Rolle's theorem. The $f^{'}(x)$ vanishes in $x=-1$, $x=0$ and $x=1$ and You can repeat the procedure for $f(x)$ in $-1\le x \le 1$ to $f^{'} (x)$ in $-1\le x \le 0$ and $0\le x \le 1$ demonstrating that $f^{(2)} (x)$ has a zero in each interval. Of course the same is for the derivatives of higher order...

Kind regards

$\chi$ $\sigma$