# Fourier transform in Mathematica [closed]

R and Mathematica software differ when computing fft(c(1,1)) and Fourier[{1,1}],

2+0i 0+0i

and

{1.41421+ 0i, 0} respectively. How can this be????

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## closed as off topic by Gerald Edgar, Yemon Choi, Torsten Ekedahl, Felipe Voloch, Andres CaicedoJul 30 '11 at 22:21

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off topic, see the FAQ –  Gerald Edgar Jul 30 '11 at 18:44
Obviously, they're using different definitions. The notation and conventions associated with the Fourier transform differ between different authors, although it's usually easy to figure out the differences and adjust your results accordingly. A look at the documentation for the R and Mathematica functions should help you figure this out. –  Brian Borchers Jul 30 '11 at 18:51

Normalizing factor.

It looks like R defines the Discrete Fourier Transform matrix as $F = [1$ $1; 1$ $-1]$ while Mathematica defines it as $F = \frac{1}{\sqrt{2}}[1$ $1; 1$ $-1]$.

If you do inverse fft - R would define it to be $F^{-1} = \frac{1}{2}[1$ $1; 1$ $-1]$ while Mathematica would define it as $F^{-1} = F^{H} = \frac{1}{\sqrt{2}}[1$ $1; 1$ $-1]$ where $H$ is Hermitian transpose.

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