Chebyshev spectral differentiation via FFT

Question

I am using the Chebyshev spectral differentiation technique that is described concisely under "details" here. The idea is to take the initial data $v_0,v_1\,...,v_N$ and store it in union with itself as the vector $$V = [v_0\,v_1\,...\,v_{N-1}\,v_N\,v_{N-1}\,v_{N-2}\,...\,v_1]^\top$$

From there, the Fourier transform of this vector $V$ is taken. However, for the Fourier transform to provide a good interpolation of the data in $V$, $V$ should be smooth and periodic. Although $V$ is continous and periodic, there is (generally) a discontinuity in its first derivative (around the entries $v_{N-1},v_N,v_{N-1}$). Why, then, is this method of differentiation still so effective?

Answer 1

Ah, I have realized the answer to my own question:

It is important to recognize that the initial data $v_0,...,v_N$ is not stored on a uniform grid, but rather at the Chebyshev points $$ x_j = \cos\frac{\pi j}{N},\qquad j=0,...,N.$$

Now as long as the initial data has a decent polynomial interpolation, then \begin{align} v_j = p(x_j) &= a_0 + a_1x_j + \cdots + a_Nx_j^N \\ &=a_0 + a_1\cos\frac{\pi j}{N} + \cdots + a_N\cos^N\frac{\pi j}{N} \\ &=a_0 + a_1\cos\theta_j + \cdots + a_N\cos^N\theta_j = f(\theta_j) \end{align} where $\theta_j = \pi j/N\in[0,\pi]$ is a uniform grid. Therefore, on the new uniform grid the data is an even function (hence the powers of cosine), and in particular $df/d\theta|_{\theta=0} = 0$. Thus the function can easily be extended to $[-\pi,\pi]$, giving a smooth, even, periodic function with data at uniformly-spaced gridpoints: ripe for the Fourier transform.