Roots of a power series in an interval

Question

Let $ a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} = 0 $

Prove that $ a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n = 0 $ has real roots into the interval $ (0,1) $

I found this problem in a real analysis course notes, but I don't even know how to attack the problem. I tried to affirm that all coefficients are zero, but that is cleary not true, we have many cases when the result is 0 but $ a_i \ne 0$ for some $i$. I have tried derive/integrate, isolate and substitute some coefficients ($ a_0 $ and $a_n $ where my favorite candidates). Work with factorials (and derivatives and factorials) but could not find a way to prove. I have many pages of useless scratches.

Any tips are welcome.

The condition on ${a_k}$ is the $\displaystyle\int\limits_0^1 p(x),dx=0$ where $p(x)$ is the given function. The rest follows from e.g. mean value theorem, i.e. if no roots, the integral can't be $0$. — Alexey Burdin, Jun 05 '15 at 05:56
The idea behind if no roots integral can't be 0 escaped me. Thank you for the tip. — Lin, Jun 05 '15 at 06:09

score 6 · Accepted Answer · answered Jun 05 '15 at 05:59

Let $f(x)=a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$, then $f(x)$ is a continuous function in the interval $(0,1)$. Consider \begin{align*} \int_0^1 f(x) \, dx & = \int_0^1 a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \, dx\\ & = a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + \cdots + \frac{a_nx^n}{n+1}\Bigg|_0^1\\ & = a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1}\\ & =0. \end{align*} Since the integral vanishes on the interval $(0,1)$, by continuity either $f$ must be both positive and negative, or identically $0$ on the interval. In either case it must be $0$ at some point in $(0,1)$.

Thank you. I completelly was not able to use/think about the Rolle's theorem. Also I was doing indefinite integrals. Thank you for the good insight. — Lin, Jun 05 '15 at 06:14

score 3 · Answer 2 · answered Jun 05 '15 at 05:59

3

Consider the function $f(x)=a_0x+ \frac{a_1}{2}x^2+\cdots + \frac{a_n}{n+1}x^{n+1}$. Then $f(1)=0$ is given, and $f(0)=0$ is clear. Rolle's theorem now shows there is $x \in (0,1)$ such that $f'(x)=0$.

answered Jun 05 '15 at 05:59

Noah Olander

2,514

score 2 · Answer 3 · edited Jun 05 '15 at 09:19

As Alexey Burdin points out, if you integrate $f(x) = a_0+a_1x+a_2x^2+\cdots+a_nx^n$ from $0$ to $1$, you get the sum

$$ \int_{x=0}^1 f(x) \, dx = \sum_{k=0}^n \frac{a_k}{k+1} = 0 $$

Since $f(0) = a_0$ and $f(x)$ is continuous over the reals, either $f(x) = 0$ over the interval $(0, 1)$, or $f(x)$ must be negative and positive over some portions of the interval $(0, 1)$; either way, there exist real roots in that interval as well.

Roots of a power series in an interval

3 Answers3