Take the 2-minute tour ×
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.

This question already has an answer here:

Consider a function $f(x)$ and its Fourier Transform $\tilde{f}(k)$ given by $$ \tilde{f}(k) = \int_\mathbb{R}\!\!\!dx\; e^{-ikx}f(x). $$ Now, lets have the coordinate transform $\xi = \tau(x)$ and, thus, we have the Fourier Transform $\tilde{f}(\kappa)$ of $f(\xi)=f(\tau(x))$ with a new coordinate $\kappa$.

Is there a way to compute $\tilde{f}(\kappa)$ from a given $\tilde{f}(k)$ and coordinate transform $\xi = \tau(x)$?

Does the coordinate transform $k \rightarrow \kappa$ exist at all?

Thanks and regards.

share|improve this question

marked as duplicate by Dirk, Davide Giraudo, Chris Eagle, rschwieb, Asaf Karagila Mar 11 '13 at 11:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Yes, I found that, too. But it does not help me. Its clear that I can just plug the composition in and rearrange a bit. But I want to know if I can compute $\tilde{f}(\kappa)$ without doing another Fourier Transform. –  André Bergner Mar 11 '13 at 9:56

1 Answer 1

There is no such coordinate transform in general: If $\tau(x) = x-x_0$ then the Fourier transform of $f\circ \tau$ is the Fourier transform of $f$ multiplied by a complex exponential.

In the case of dilation $\tau(x) = Ax$ (with invertible $A$) there is indeed a coordinate transform but this seems to be a rare exception.

share|improve this answer

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