Lets say we have an arbitrary periodic function $u(x, y, z)$ where $x, y, z \in [0, 2 \pi]$ and represented on a computational mesh $\text{Ni} \times \text{Nj} \times \text{Nk}$. If we take the Discrete 3D-Fourier Transform on $u$ to obtain $\hat{u}(\hat{x}, \hat{y}, \hat{z})$, how would we find what the domains for $\hat{x}, \hat{y}, \hat{z}$ would be?
Conceptually, it seems that the "longest" wave that could fit within this mesh would be the length of the domain, and the "shortest" wave that could fit within this mesh would be twice the interval between grid points. Wouldn't that mean that the domain would be: $$\hat{x} \in \left[\frac{2 \pi}{2 \pi}, \frac{2 \pi}{2 \pi /\text{Ni}} \right] = [1, \text{Ni}]$$
How would that work with the fact that the Fourier Transform would also output both pairs of the complex conjugate?
Note: I originally posted this on the Computational Physics SE, but deleted it as I felt it would be more relevant here.
@HyeongmukLIM, let's say I have a function $y(x) = \sin(x)$. I can compute this discretely over [-π, π] and Ni = 10. Y(ξ) will now be:
[-1.2246468e-16+0.00000000e+00j -1.2246468e-16+5.00000000e+00j
-1.2246468e-16+3.41691513e-16j -1.2246468e-16+0.00000000e+00j
-1.2246468e-16-8.06624243e-17j -1.2246468e-16-2.22044605e-16j
-1.2246468e-16+8.06624243e-17j -1.2246468e-16+2.22044605e-16j
-1.2246468e-16-3.41691513e-16j -1.2246468e-16-5.00000000e+00j]
Discounting random floating point errors, you can quickly see that there are:
-1.2246468e-16+5.00000000e+00j
-1.2246468e-16-5.00000000e+00j