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Let

$f_K(x)=\frac{1}{\sqrt{K}}\sum_{k=0}^{K-1}\exp\{i\pi kx\}=\exp\{i\pi(K-1)x/2\}\frac{\sin(\pi Kx/2)}{\sqrt{K}\sin(\pi x/2)}$

where $i=\sqrt{-1}$.

I want to know whether the limit of $f_K(x)$ is a Dirac delta function when $K\to\infty$?

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Well, $$ \int_{-\pi}^\pi f_K(x)\,dx=\frac{2\,\pi}{\sqrt K} $$ converges to $0$ as $K\to\infty$, so the answer is no.

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  • $\begingroup$ Why the range of integration is set as $[-\pi,\pi]$ rather than $(-\infty,\infty)$? $\endgroup$ – nuse_li Feb 16 '17 at 2:05
  • $\begingroup$ In addition, if $f_K(x)$ is scaled with a factor $2\pi/\sqrt{K}$, will the new function tend to Dirac delta when $K\to\infty$? $\endgroup$ – nuse_li Feb 16 '17 at 2:07
  • $\begingroup$ Also, how are the integration results obtained? $\endgroup$ – nuse_li Feb 16 '17 at 3:15
  • $\begingroup$ We can chose to integrate on any interval we wish. I have chosen $[-\pi,\pi]$ because the sum looks a partial sum of a Fourier series on that interval. The integrals of the terms in the sum are all $0$ except the one corresponding to $k=0$. $\endgroup$ – Julián Aguirre Feb 16 '17 at 10:52
  • $\begingroup$ Is it possible to construct a function which tends to the Dirac delta function using the exponential sum mentioned above? $\endgroup$ – nuse_li Feb 16 '17 at 14:26

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