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 $f\colon\mathbb R\to\mathbb R$ defined by

$$f(x)=a_1\sin x+a_2\sin2x+a_3\sin3x+\cdots+a_n\sin nx,$$ for some values $a_1,a_2,a_3,\cdots,a_n\in\mathbb R.$ Prove that $$ |f(x)|\le|\sin x|\quad \forall x\in \mathbb{R}$$ implies: $$|a_1+2a_2+3a_3+\cdots+na_n|\le1$$

share|cite|improve this question
Should that be $3a_3$ in your last displayed equation? – Dilip Sarwate Dec 20 '11 at 18:01
up vote 6 down vote accepted

For $x$ such that $\sin x\neq 0$, we have $\left|\sum_{j=1}^na_j\frac{\sin(jx)}{\sin x}\right|\leq 1$ so $$\left|\sum_{j=1}^nja_j\frac{\sin(jx)}{jx}\frac x{\sin x}\right|\leq 1,$$ and taking the limit $x\to 0$ we get $$\left|\sum_{j=1}^nja_j\right|\leq 1,$$ which is the wanted result.

share|cite|improve this answer

Here's another perspective on Davide's answer:

$f$ is a finite sum of differentiable functions and therefore differentiable, and $|f(x)|\le|\sin x|$ everywhere implies $|f'(0)|\le |\sin'(0)|=1$. Now the sought inequality follows simply by evaluating $f'(0)$.

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.