FFTW: Logical array size in r2r transform? I'm taking a multidimensional discrete cosine transform (FFTW_REDFT11) in FFTW, and am unsure how to compute the scaling factor.  According to the documentation, taking the forward and reverse transforms results in the original image multiplied by a scaling factor:


  [C]omputing a transform followed by its inverse yields the original
  array scaled by N, where N is the logical DFT size. For REDFT00,
  N=2(n-1); for RODFT00, N=2(n+1); otherwise, N=2n.


The documentation also defines logical arrays for 1D transforms:


  [I]f you specify a size-5 REDFT00 (DCT-I) of the data abcde, it
  corresponds to the DFT of the logical even array abcdedcb of size 8. A
  size-4 REDFT10 (DCT-II) of the data abcd corresponds to the size-8
  logical DFT of the even array abcddcba, shifted by half a sample.


I'm not sure what n is in a multidimensional transform.  Say I have a 10x10 matrix.  Is n twice the total number of pixels (200)?  Or is it the number of pixels in the image if it were reflected in all dimensions (400)?
You're on the right track.  The distinction must be made between the physical size of your array, n (generally $n_0 × n_1 × n_2 × … × n_{d-1}$ except for in-place R2C transforms), and the logical size, N.  For a 10x10 matrix, the physical size is n=10*10=100, so the logical size for this particular transform is N=2n=200,  the scaling factor for the unnormalized transform-then-inverse result.

They talk a little more about that here in the Real even/odd FFTW3 docs, although they don't give much more info than the link you provided.

It's always a good idea to test with a problem that you already know the answer to,