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Math Help - Rolles Theorem

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    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?
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  2. #2
    Member kalyanram's Avatar
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    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.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by worc3247 View Post
    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
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