# Roots of higher derivatives

1. Jun 12, 2010

### rainwyz0706

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!

2. Jun 12, 2010

### Hurkyl

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

3. Jun 13, 2010

### rainwyz0706

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.

4. Jun 13, 2010

### HallsofIvy

Staff Emeritus
$f(x)= (x-1)^n(x+ 1)^n$ 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.

5. Jun 13, 2010

### rainwyz0706

Could you please be a bit more specific about your second line? The k here means the k-th derivatives. The power of (x-1)(x+1) is a fix n, and I don't think I'm supposed to do an induction on that. f^(r)(x) has to be a pretty messy function, is there a clear way to take derivative out of that?

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