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 $$D_N(x)=\frac{\sin [(N+(1/2))t]}{\sin (t/2)}$$ be the Dirichlet kernel. Let $x(N)$ be the number in $0<x<\pi/N$ such that $D_N(x)=1$. Is $$\left|\int_{x(N)}^{\pi/N} D_N(t)\mathrm dt \right|=O\left(\frac1{N}\right)$$ true? This question arises from my attempt to give a rigorous proof of Gibbs phenomenon. In fact, I only need that the limit is 0 as $N\to\infty$.

share|cite|improve this question
up vote 4 down vote accepted

Looks true to me.

$\displaystyle x(N) = \dfrac{\pi}{N+1}$

Now $\displaystyle D_n(x) = 1 + 2\sum_{k=1}^{n} \cos(kx)$

and $\displaystyle |\sin x - \sin y| \leq |x-y|$ (easily seen using Mean Value Theorem)


$$\displaystyle \left|\int_{\pi/(N+1)}^{\pi/N} D_N(t) \ \text{dt}\right| = \left|\int_{\pi/(N+1)}^{\pi/N} 1 + 2\sum_{k=1}^{N} \cos(kx) \ \text{dt}\right|$$

$$\displaystyle = \frac{\pi}{N(N+1)} + 2 \sum_{k=1}^{N} \mathcal{O}(\frac{1}{N^2}) = \mathcal{O}(\frac{1}{N})$$

share|cite|improve this answer
Very nice. I did not see that $x(N)$ can be so simple!! – TCL Dec 5 '10 at 13:09

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.