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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:

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.

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?

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    $\begingroup$ There's a standard lemma that if $f$ is a strictly positive (negative) function, then $\int f$ is strictly positive (negative). I suppose you could prove this easily with Riemann sums. Then apply the contrapositive to conclude that $f$ changes sign (or is identically zero), and invoke continuity. $\endgroup$ – user296602 Jul 28 '16 at 17:59
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    $\begingroup$ I just want to complement you on your beautiful solution. Did you find it yourself ? What is your mathematical level (undergrad, grad school, etc)? $\endgroup$ – Saikat Jul 29 '16 at 4:00
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    $\begingroup$ @user230452 Thank you very much. I did find it myself, but this problem was posed as a challenge in my calculus book in an integration chapter (Calculus by Larson/Edwards, $5^{th}$ ed., Page $346$ #$175$) so that was a giveaway. Once I saw that the integral had coefficients like the series, I remembered another problem I saw before somewhere else: "Find the sum of the coefficients of $f(x)=(x+3)^{30} (x+1)^{10}$". The answer is $4^{30} \cdot 2^{10}$; it is found by evaluating $f(1)$ (for any function). This led me to set one of the limits of integration to $1$, and $0$ was the first thing $\endgroup$ – Ovi Jul 29 '16 at 5:32
  • $\begingroup$ @user230452 I tried for the other limit. I am an undergrad who just finished discrete math, I am about to start the more exciting classes like analysis and number theory :) $\endgroup$ – Ovi Jul 29 '16 at 5:33
  • $\begingroup$ @Ovi Well done, Ovi ... Can you tell me which book you used for your discrete Math course ? Was it Kenneth Rosen or such ? I like discrete math a lot. Problems do become easier when you know their context, like you did here. Now that you've done it once, always look at a sum and ask yourself if the terms are the coefficients of a series of its derivative or integral. It's a common trick with sums and products. Keep up the hard work ! $\endgroup$ – Saikat Jul 29 '16 at 6:09
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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.

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    $\begingroup$ Indeed, this is the rigorous proof of the area-interpretation described in the OP. $\endgroup$ – Greg Martin Jul 28 '16 at 22:32
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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.

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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.

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