Fix Fourier itransform with even number of points

[dlfivefifty]

julia> f = Fun(Fourier(), randn(5));

julia> f.(points(f)) ≈ values(f)
true

julia> f = Fun(Fourier(), randn(6));

julia> f.(points(f)) ≈ values(f)
false

[jishnub]

This is a bit tricky, as an even number of coefficients requires the highest-order term to be a sin, whereas in a real DFT, one requires all sin terms to be matched by a cos term, with potentially an extra Nyquist cos term without a matching sin. In this case, either we evaluate the inverse transform as f.(points(f)) if ncoefficients(f) is even, or we pad a dummy zero cos coefficient and require the number of points to be odd. In the latter case, we may use FFTs for the inverse transform, so this is the approach that I prefer.