Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $f$ is $n$ times differentiable on an interval $I$ and there are $n + 1$ points $x_0, x_1, \ldots, x_n \in I, x_0 < x_1 < \cdots < x_n$, such that $f(x_0) = f(x_1) = \cdots = f(x_n) = 0$. Prove that there exists a point $z \in I$ such that $f^{(n)}(z) = 0$.

I am trying to solve this, but other than using then using Rolle's Theorem, I am not sure how to proceed.

share|improve this question

2 Answers 2

Use the Mean Value Theorem $n$ times:

By the Mean Value Theorem, there are $n$ points $y_0\in(x_0,x_1)$, $y_1\in(x_1,x_2)$, $\dots$, $y_{n-1}\in(x_{n-1},x_n)$ so that $f'(y_0)=f'(y_1)=\dots=f'(y_{n-1})=0$.

By the Mean Value Theorem, there are $n-1$ points $z_0\in(y_0,y_1)$, $z_1\in(y_1,y_2)$, $\dots$, $z_{n-2}\in(y_{n-2},y_{n-1})$ so that $f''(z_0)=f''(z_1)=\dots=f''(z_{n-2})=0$.

Repeat until you have $1$ point $w_0$ so that $f^{(n)}(w_0)=0$.

share|improve this answer

Here’s a fairly broad hint:

You know from Rolle’s theorem that it’s true when $n=1$. Try it for $n=2$; Rolle’s theorem gives you points $y_0$ and $y_1$ such that $x_0<y_0<x_1<y_1<x_2$ and $f\;'(y_0)=f\;'(y_1)=0$. Can you now apply Rolle’s theorem to $f\;'$ on some interval to get something useful?

In order to generalize the hint from $2$ to $n$, you’ll need to use mathematical induction.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.