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

$p'(c)=\dfrac{p(1)-p(0)}{1-0}=\dfrac{0-0}{1}=0$

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

$p'(x)=c_0+c_1x+c_2x^2+\dots+c_nx^n$

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 $[0,1]$.

$\Box$

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.