Thread: 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.

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 on . But the Mean-Value theorem for integrals we have for some .
Thus,
.

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

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 on the interval . This function is continous on the closed interval and differenciable on the open interval. Thus, by mean value theorem: but . Thus, for some .

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