Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(x)=\sum_{k=0}^n a_k x^k$, where $a_k$'s satisfy $\sum_{k=0}^n \frac{a_k}{k+1}=0$.Show that there exists a root of $f(x)=0$ in the interval $(0,1)$.

share|cite|improve this question
I am stuck at first stage. So, I give no work that I have done. – A.D Jan 1 '13 at 10:30
up vote 7 down vote accepted

Hint: Consider $g(x)=\sum_{k=0}^n\frac{a_k}{k+1}x^{k+1}$. Show that $g(0)=g(1)$. Use mean value theorem to infer that $g'(x)=0$ for some $x\in(0,1)$. Now, what is the relation between $g'$ and $f$?

share|cite|improve this answer
Is $g'=f$ the relationship? – A.D Jan 1 '13 at 10:45
@A.D Just differentiate $g$ and you'll see. – user1551 Jan 1 '13 at 11:07

Note that $$\int_0^1 f(x) dx=\sum_{k=0}^n\frac{a_k}{k+1}=0$$ So none of $f(x)>0$ or $f(x)<0$ in whole of $(0,1)$ is possible, so $f$ must change its sign and hence by Intemediate Value Property it must have a root in $(0,1).$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.