5
$\begingroup$

The following infinite sum of exponential terms gives a Dirac comb:
$$ \sum_{n=-\infty}^\infty e^{i n x} = 2 \pi \sum_{n=-\infty}^\infty \delta(x - 2 \pi n) $$ Of course the sum doesn't strictly converge, but only in the same sense in which the Dirac delta-function is defined.

What is the result of a semi-infinite sum of such terms?
$$ \sum_{n=1}^\infty e^{i n x} =~? $$

$\endgroup$
3
$\begingroup$

Hint:

(1) write the semi-infinite sum in the following way $$\sum_{n=1}^\infty e^{i n x} = \frac{1}{2} \sum_{n=-\infty}^\infty e^{i n x} + \frac{1}{2} \sum_{n=-\infty}^\infty \mathop{\rm sgn}(n) e^{i n x} - \frac{1}{2}.$$

The first summand is the Dirac-comb. So your question reduces to figuring out what the second term equals to...

(2) Lagrange's trigonometric identities might help there...

$\endgroup$
1
  • $\begingroup$ Got it - thanks! $\endgroup$
    – Joe
    Aug 5 '12 at 11:17

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.