How can I answer this Putnam question more rigorously?

Question

Given real numbers $a_0, a_1, ..., a_n$ such that $\dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1}=0,$ prove that $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n=0$ has at least one real solution.

My solution:

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Let $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$

$$\int f(x) = \dfrac {a_0}{1} x + \dfrac {a_1}{2}x^2 + \cdots + \dfrac {a_n}{n+1} x^{n+1} + C$$

$$\int_0^1 f(x) = \left[ \dfrac {a_0}{1} + \dfrac {a_1}{2} + \cdots + \dfrac {a_n}{n+1} \right]-0$$

$$\int_0^1 f(x) = 0$$

Since $f$ is continuous, by the area interpretation of integration, it must have at least one zero.

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My question is, is this rigorous enough? Do I need to prove the last statement, perhaps by contradiction using Riemann sums? Is this a theorem I can/should quote?

Answer 1

Why not write it the other way round?

The polynomial function $$F(x)=\sum_{k=0}^n\frac{a_k}{k+1}x^{k+1} $$ is a differentiable function $\Bbb R\to\Bbb R$ with derivative $$F'(x)=\sum_{k=0}^na_kx^k.$$ We are given that $F(1)=0$, and clearly $F(0)=0$. Hence by Rolle's theorem, there exists $x\in(0,1)$ such that $F'(x)=0$, as was to be shown.

Answer 2

Your proof looks fine. If you wanted to expand, you could add the following:

Suppose $f(x)>0$ for all $x>0$. Then we must have $$\int_0^1f(x)\ dx>0,$$ But we have already shown that $\int_0^1f(x)\ dx=0$, a contradiction.

If we assume $f(x)<0$ for all $x>0$, we arrive at a similar contradiction.

Answer 3

You can also prove it using the mean value theorem. You showed that $$\int_{0}^{1}f\left(x\right)dx=0 $$ now since $f$ is continuous by the mean value theorem for integrals we have that exists some $c\in\left(0,1\right) $ such that $$f\left(c\right)=\int_{0}^{1}f\left(x\right)dx $$ so $$f\left(c\right)=0$$ as wanted.