1. ## existence of root

Prove that if an/(n+1) + an-1/n + ... + a1/2 + a0 = 0 (ak in R), then the equation anx^n + an-1x^(n-1) + ... + a1x + a0 = 0 has at least one real root between 0 and 1.
This is my work.
Let p(x) = (an/n+1)x^(n+1) + ..... a0x.
Then p'(x) = anx^n + ......... + a0 (precisely what we need to show has a root). This part I'm not sure about. If p(x) has two distinct real roots (call them a and b and WLOG b < d), then p(b) = p(d) = 0. Rolle's Theorem on the interval [b,d] tells us that there exists a number r such that p'(r) = 0, which is the root I'm looking for. Does this make sense? Then, all I'm unsure about is how to show p(x) has two distinct real roots and how the interval [0,1] plays in the problem.

2. Originally Posted by agentZERO
Prove that if an/(n+1) + an-1/n + ... + a1/2 + a0 = 0 (ak in R), then the equation anx^n + an-1x^(n-1) + ... + a1x + a0 = 0 has at least one real root between 0 and 1.
Consider $\displaystyle f(x) = a_nx^n+...+a_1x+a_0$ on $\displaystyle [0,1]$. But the Mean-Value theorem for integrals we have $\displaystyle \int_0^1 f(x) dx = (1-0)f(c)$ for some $\displaystyle c\in (0,1)$.
Thus,
$\displaystyle \frac{a_n}{n+1}+\frac{a_{n-1}}{n}+...+\frac{a_1}{2}+a_0 = 0 = f(c)$.

3. Thanks for helping, but I can't use that theorem (we haven't done anything with integrals yet). Is my idea hopeless?

4. Originally Posted by agentZERO
Thanks for helping, but I can't use that theorem (we haven't done anything with integrals yet). Is my idea hopeless?
Consider $\displaystyle f(x) = \frac{a_n}{n+1}x^n+\frac{a_{n-1}}{n}x^{n-1}+...+\frac{a_1}{2}x+a_0x$ on the interval $\displaystyle [0,1]$. This function is continous on the closed interval and differenciable on the open interval. Thus, by mean value theorem: $\displaystyle f'(c)(1-0) = f(1)-f(0)$ but $\displaystyle f(1)=f(0)=0$. Thus, $\displaystyle f'(c)=a_nc^n+...+a_0 = 0$ for some $\displaystyle c\in (0,1)$.

5. I realized after I wrote the second post that my p(0)=p(1)=0. Thanks for the help.