It is true that if a function $f$ has fourier transform $\hat{f}$ then, $(\hat{f})^n = FT(f^{n})$ (where $FT$ denotes the fourier transform). If so, why?
Fourier transform of $f^{n}$
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fourier-analysis
fourier-transform
1 Answers
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In general, no. The Fourier Transform of a product of functions is the convolution of their Fourier Transforms, e.g.
$$\widehat{(fg)}(x) = (\hat{f}* \hat{g})(x) = \int_{-\infty}^{\infty} \hat{f}(x-y) \hat{g}(y) \, \mathrm{d} y. $$
Showing this is a play with the definition of the Fourier Transform, and a good exercise to do to convince yourself.
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2$\widehat{f \ast g} = \hat{f} \hat{g}$ the opposite direction needs the Fourier inversion theorem – 2017-02-18