Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Here wikipedia it is said that the Dirac delta could be thought of as $$ \delta(x) = \left\{ \begin{array}{ll} \infty &, x = 0 \\ 0 &, x \ne 0 \end{array}\right. $$ and here that the Fourier-Transform of $\cos(2\pi k_0 x)$ is $$ \frac{1}{2} \left( \delta(k - k_0) + \delta(k + k_0) \right). $$ From computer programs I know that the spectrum gives the intensity of every cos/sin wave in the waveform, but where is the intensive (i.e. one) of the simple cosine waveform represented when i thought of $\delta(k-k_0)$ as being "stretched till infinity" at the point $k_0$?

share|cite|improve this question

You are right in a sense. In a Fourier series, you have a single coefficient that tells you the "intensity" of each frequency: $$f = c +\sum_{n=1}^\infty a_n sin(nx) + b_n cos(nx)$$ you can just read off the $a_n$ and $b_n$ to find the contribution of any sinusoidal component.

The Fourier transform on the other hand, may yield a continuous frequency. If $\hat{f}$ is the Fourier transform of $f$, then $\int_a^b \hat{f}(\xi)d\xi$ gives you the contribution of the "range" of frequencies from $a$ to $b$. Here, the contributing frequencies are $k_0$ and $-k_0$, and since $\hat{f}$ is 0 elsewhere, you know these are the only values that matter. The integral over neighborhoods of these points will tell you the contributions. Don't think of the function $\delta(x)$ has having a value at 0, think of it only as having a nonzero integral over an infinitesimal domain.

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.