Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Fourier transform of the product of two functions f(x) and g(x) is given as:

$\mathcal{F}[ f(x)g(x)] = \int_{-\infty}^{+\infty} F(\omega^\prime) G(\omega - \omega^\prime) d\omega^\prime \; = \; \mbox{convolution of} \; \; F(\omega^\prime )G(\omega^\prime)$

where $F(\omega^\prime)$ and $G(\omega^\prime)$ are the Fourier transforms of $f(x)$ and $g(x)$ respectively.

Although I understand the derivation of this formula, I've got difficulty making sense of two frequency terms $\omega$ and $\omega^\prime$. I'm fine with $\omega^\prime$ but I don't know what to make of $\omega$. Should I treat it as a constant, or should I set it to zero?

I'm really interested in the Fourier transform of the square of the second derivative of a function e.g. $\mathcal{F}[ f^{\prime\prime}(x)^2 ]$. Because this problem does not involve a shift, I don't know what to make of the shift term $\omega$.

share|improve this question
1  
Please ask on math.stackexchange.com instead. –  Deane Yang Sep 6 '10 at 12:52
add comment

closed as not a real question by Charles Matthews, Daniel Moskovich, Deane Yang, José Figueroa-O'Farrill, Gjergji Zaimi Sep 6 '10 at 13:27

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

The notation is not quite right, let F(w) and not F(w') be the Fourier transform of f(x) (resp G(w) for g(x)), so that w is the "frequency term", and w' (or better $\lambda$ to avoid confusion) be just a dummy variable for integration.

Then the convolution integral $\int^{\infty}_{-\infty}{F(\lambda)G(w-\lambda)d\lambda}$ see $\lambda$ as the variable of integration and after integration $\lambda$ will vanish and the result will be a function of w.

For the second question: If $F(w)$ is the Fourier transform of f(x), recall $\frac{d^nf(x)}{dx^n}$ ~ $(2 \pi iw)^n F(w)$.

Then $f''(x) f''(x)$ has Fourier Transform a convolution of $(2 \pi iw)^2 F(w)$ with itself, i.e $(2\pi )^4 \int^{\infty}_{-\infty}{\lambda^2 F(\lambda) (w-\lambda)^2 F(w-\lambda)d\lambda}$.

share|improve this answer
add comment

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