Rudin: Ch. 5 #4

Problem Statement: If c_0+\dfrac{c_1}{2}+\dfrac{c_2}{3}+\dots+\dfrac{c_{n-1}}{n}+\dfrac{c_n}{n+1}=0 where c_0,\dots,c_n are real constants, prove that c_0+c_1x+c_2x^2+\dots+c_nx^n=0 has at least one real root between 0 and 1.

Proof: Consider the polynomial p(x)=c_0x+\dfrac{c_1}{2}x^2+\dfrac{c_2}{3}x^3+\dots+\dfrac{c_n}{n+1}x^{n+1} on [0,1]. Then it follows that p(0)=0 and p(1)=c_0+\dfrac{c_1}{2}+\dots+\dfrac{c_n}{n+1} which equals 0 by our assumption. Furthermore, since p(x) is a polynomial with real coefficients it is differentiable on [0,1] and so we may apply the Mean Value Theorem to p on the interval [0,1]. Applying MVT we see that there exists at least one point c\in(0,1) such that


So there is some c\in(0,1) such that p'(c)=0. But we may compute p'(x) since we know p(x).


Thus, we have shown that there is at least one real root to c_0+c_1x+c_2x^2+\dots+c_nx^n=0 on the interval [0,1].


Reflection: What a nice, sweet, and simple proof. I have to admit, when I first saw this problem I had no clue what to do. The trick came in remembering that I was in the chapter on differentiation and MVT, but that won’t happen on the qual…After completing the proof I understand a bit better how I would “see” the proof in the future. After realizing that the index of the numerator was 1 plus the value of the denominator I started to see that differentiation could be helpful here. I think the hardest part of this proof is realizing that this is one of those times where you want a “helpful function”. After writing out p(x) and recognizing that p(1)=0=p(0) the rest fell out.

This entry was posted in Analysis, Differentiable, Math, MVT. Bookmark the permalink.

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