Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 $ be a continuous function defined on $ [0,\pi] $. Suppose that

$$ \int_{0}^{\pi}f(x)\sin {x} dx=0, \int_{0}^{\pi}f(x)\cos {x} dx=0 $$

Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

share|cite|improve this question
I have the feeling that the Fourier expansion $f(x) = a_0 + \sum_{n\geq 1} (a_n \sin 2n x + b_n \cos 2n x)$ might help, but can't see how exactly. – Lior B-S Nov 29 '12 at 17:20
@LiorB-S Since $f$ is "just@ continuous you can't be sure that Fourier series of $f$ converges to $f$. – Norbert Nov 29 '12 at 17:27
@Norbert: You are right, but I have no proof for smooth $f$'s either. Moreover it would be nice, I think, to have if a proof using Fourier expansion exists... – Lior B-S Nov 29 '12 at 17:32
It's worth remarking that the function $f(x)=\sin(3x)$ satisfies the conditions and has exactly 2 roots in $(0,\pi)$, so 2 is the best possible integer in this problem. – James Fennell Nov 29 '12 at 21:04

Here is one root: Let $F(x) = \int_{0}^x f(t) \sin t dt$. Then $F(0)=0$ and $F(\pi)=\int_{0}^\pi f(t)\sin tdt=0$. So by the intermediate value theorem, there exists $0<c<\pi$ such that $$ 0=F'(c) = f(c)\sin c. $$
But since $\sin c\neq 0$, we get that $f(c)=0$.

share|cite|improve this answer

If f has only one real root on $ (0,\pi)$, say $ a \in (0,\pi) $, then define $ g(x) = f(x) \sin(x-a) = f(x) (\sin(x)\cos(a) - \cos(x)\sin(a))$, then $ g(x) $ is either non-positive or non-negative, not identically zero, and has integral $ 0 $. Contradiction.

share|cite|improve this answer
Why is $g(x)$ either non-positive or non-negative? – TonyK Nov 29 '12 at 19:32
@Lev - I think you assume that $f$ changes sign at its root, which doesn't necessarily happen (the function could be tangent to the $x$-axis). However, if $f$ doesn't change sign and is not identically zero, the integral $\int_0^\pi f(x) \sin(x)dx$ would be strictly positive (or strictly negative, if you take $f\leq 0$). This is again a contradiction. – James Fennell Nov 29 '12 at 19:53
@TonyK: Note that $\sin (x-a)>0$ for $a<x<\pi$ and $\sin(x-a) <0$ for $0<x<a$. – Lior B-S Nov 29 '12 at 19:57
Yes, OK -- if $f$ doesn't change sign at $a$, then the $\sin$ integral can't be zero. So $f$ does change sign at $a$, and therefore $f(x)\sin(x-a)$ doesn't. – TonyK Nov 29 '12 at 20:01
@James Fennell: thank you, I missed the case when f does not change sign. – Lev Nov 29 '12 at 20:36

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.