# Roots of higher derivatives

Let f (x) = (x^2 − 1)^n . Prove (by induction on r) that for r = 0, 1, 2, · · · , n,
f^ (r) (x)(the r-th derivative of f(x)) is a polynomial whose value is 0 at no fewer than r distinct points of (−1, 1).

I'm thinking about expanding f(x) as the sum of the (n+1) terms, then it's easier to take derivatives. But I don't know how to get the roots from there then. Could anyone please give me some hints? Thanks!

Hurkyl
Staff Emeritus
Gold Member
Well, have you figured out any special cases? Like maybe n small, or maybe r=0 and r=n?

I tried to prove by induction on r. But I'm not sure how to express the k-th derivative of f(x). r=0 or r=n are special cases, they clearly holds. My problem is how to generalize it.

HallsofIvy
$f(x)= (x-1)^n(x+ 1)^n$ and all derivatives are done by repeated use of the product rule.
As for using induction, $(x-1)^{k+1}(x-1)^{k+1}= (x^2- 1)^k (x-1)(x+ 1)$. Use the product rule on that.