# Rolles Theorem

• Mar 31st 2011, 01:53 PM
worc3247
Rolles Theorem
Let $\displaystyle f(x)=(x^2-1)^n$. Using Rolle's theorem, prove that $\displaystyle f^{(n)}(x)$ 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, 06:59 PM
kalyanram
If $\displaystyle f(a)=f(b)=0$ with $\displaystyle f$ satisfying the conditions of Rolle's Theorem we have $\displaystyle f(c)=0$ $\displaystyle c \in (a,b)$ now we observe that between two roots of $\displaystyle f$ there lies a root of $\displaystyle f^'$. Now observe that $\displaystyle f(x) = (x-1)^{n}(x+1)^{n} f^'$ has a root at $\displaystyle c \in (-1,1)$ in fact c=0. Now for $\displaystyle f^" \exists$ a roots between (-1,0) and (0,1). Now $\displaystyle f^{(n)}$ is indeed an polynomial of order n now use induction to prove that it has distinct roots between (-1,1).

Kalyan.
• Mar 31st 2011, 11:00 PM
chisigma
Quote:

Originally Posted by worc3247
Let $\displaystyle f(x)=(x^2-1)^n$. Using Rolle's theorem, prove that $\displaystyle f^{(n)}(x)$ 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 $\displaystyle 0 \le k \le n$ is $\displaystyle f^{(k)} (-1)= f^{(k)} (1) =0$. The function $\displaystyle f(x)$ is 'even' , i.e. is $\displaystyle f(x)= f(-x)$, and that means that $\displaystyle f^{'} (x)$ is 'odd' , i.e. is $\displaystyle f^{'}(x)=-f^{'} (-x)$ , so that is $\displaystyle f^{'} (0)=0$ and that is also consequence of the Rolle's theorem. The $\displaystyle f^{'}(x)$ vanishes in $\displaystyle x=-1$, $\displaystyle x=0$ and $\displaystyle x=1$ and You can repeat the procedure for $\displaystyle f(x)$ in $\displaystyle -1\le x \le 1$ to $\displaystyle f^{'} (x)$ in $\displaystyle -1\le x \le 0$ and $\displaystyle 0\le x \le 1$ demonstrating that $\displaystyle f^{(2)} (x)$ has a zero in each interval. Of course the same is for the derivatives of higher order...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$