[Numpy-discussion] Normalization of ifft

[Lutz Maibaum]

Hi Michael,

> this documentation is saying that the difference between the equations
> for the fft and ifft is a factor of 1/n (not the numpy implementations).
> if you do
>
>  output = numpy.ifft( numpy.fft( input ) )
>
> and you get output = input, then the normalizations are appropriately
> weighted.

what you say is of course correct, but I am wondering if there is a
mistake in the user guide (p. 180 of
http://numpy.scipy.org/numpybook.pdf): according to the expressions in
the user guide, both fft and ifft are not normalized. The
implementation if ifft, on the other hand, has the additional 1/n
factor, consistent with the online documentation.

  Lutz