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Let $C_0,C_1,\ldots,C_n$ are real constants. It is given that $$C_0 + \frac{C_1}{2} + \ldots + \frac{C_{n-1}}{n} + \frac{C_n}{n+1}= 0$$ We need to prove that the equation $C_0 + C_1 x + \ldots + C_{n-1} x^{n-1} + C_{n}x^{n} = 0$ has at least one real root between 0 and 1. This is taken from Rudin pg-114. I Need some help in the last part of the following proof:

Partial proof:

Let $f(x)= C_0 + C_1 x + \ldots + C_{n-1} x^{n-1} + C_{n}x^{n}$. We have: $$f'(x)= C_1+ 2C_2 x + 3C_3 x^2 + \ldots +nC_n x^{n-1}$$ $$f'(0)=C_1, \ f'(1)= C_1 + 2C_2 + 3C_3 + \ldots +(n-1)C_{n-1}+ nC_{n}$$ I would probably want to show that $f'(0)$ and $f'(1)$ are opposite in sign so that we can claim that there exists at least one $x \in (0,1)$ such that $f'(x)=0$. This is possible because $f(x)$ is a polynomial and hence continuous, differentiable function. Going forward, we use: $$C_0 + \frac{C_1}{2} + \ldots +\frac{C_{n-1}}{n} + \frac{C_n}{n+1}= 0$$ to get: $$C_1 = -2\left(C_0 + \frac{C_2}{3}+ \ldots +\frac{C_n}{n+1}\right)$$ Using this we get the following: $$f'(1)=-2C_0 + \sum_{k=2}^{n} \frac{(k+2)(k-1)}{k+1} C_{k}$$ $$f'(0)=-2C_0 -2 \sum_{k=2}^{n} \frac{1}{k+1} C_{k}$$ But how do we conclude that $f'(1)$ is opposite in sign of $f'(0)$ ?

Nathaniel Bubis
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Pradipta
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2 Answers2

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Consider the polynomial

$$ Q(x) = \int_0^x f(t)dt$$

We find that

$$ Q(x) = C_{0}x + C_{1}\frac{x^{2}}{2} + \cdots + C_{n}\frac{x^{n+1}}{n+1} $$

Then with the given condition you find that $Q(1) = 0$ and obviously, $Q(0) = 0$.

Now, use Rolle's Theorem.

Norbert
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Vishal Gupta
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The original question cries out for integration. Recall $\int x^n\sim x^{n+1}/(n+1)$. What can you do with this? (What does the integral mean visually?)

not all wrong
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