1
$\begingroup$

Statement: Given the Fejer Kernel $F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2$. Show that $F_n(x)$ is unbounded for $x=0$ as $n\rightarrow \infty$

$\endgroup$
  • $\begingroup$ why don't you just compute the limit as x -> 0? Is that the right expression for the fejer kernel? wikipedia gives something else... $\endgroup$ – Tyler Mar 27 '14 at 3:47
  • 1
    $\begingroup$ That's not the Fejer kernel. You're missing some squares. $\endgroup$ – Pedro Tamaroff Mar 27 '14 at 3:54
3
$\begingroup$

There is a big square missing: $$ F_n(x) = \frac{1}{2n\pi}\left( \frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2\sim_0 \frac{1}{2n\pi}\left( \frac{\frac{nx}{2}}{\frac{x}{2}}\right)^2=\frac n{2\pi} $$

$\endgroup$
  • $\begingroup$ How did you get those approximation? Also could have I used the fact that $F_n(x) = \frac{1}{n} \frac{1-\cos(nx)}{1-\cos(x)}$ and applied L'Hopital's rule twice to see that the limit at 0 is divergent as $n \rightarrow 0$? $\endgroup$ – nonameswereavailable Mar 27 '14 at 4:33
  • 1
    $\begingroup$ I get them using the fact that $\sin$ has a derivative $=1$ on $x=0$. $\endgroup$ – mookid Mar 27 '14 at 4:41
1
$\begingroup$

Hint

When $a$ goes to $0$, $\sin(a) \simeq a$. Then replacing each sine by its argument, for $x$ going to $0$, you arrive to $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n$$ If you want to go further, you could use a Taylor expansion built at $x=0$ and obtain $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n-\frac{1}{12} \left (n-1)n(n+1\right) x^2+O\left(x^3\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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