Trigonometric approximation and the Clenshaw-Curtis quadrature

Trying to approximate a generic continuous function on {[-1,1]} with the Fourier trigonometric series of the form {\sum_n (A_n\cos \pi nx+B_n\sin \pi nx)} is in general not very fruitful. Here’s such an approximation to {f(x)=\exp(x)}, with the sum over {n\le 4}:

Poor approximation to exp(x)
Poor approximation to exp(x)

It’s better to make a linear change of variable: consider {f(2x-1)} on the interval {[0,1]}, and use the formula for the cosine series. This results in {\exp(2x-1)}, which is approximated by the partial sum {\sum_{n=0}^4 A_n\cos \pi nx} of its cosine Fourier series as follows.

Better approximation after a change of variable
Better approximation after a change of variable

But one can do much better with a different, nonlinear change of variable. Consider {f(\cos x)} on the interval {[0,\pi]}, and again use the formula for the cosine series. This results in {\exp(\cos x)}, which is approximated by the partial sum {\sum_{n=0}^4 A_n\cos nx} of its cosine Fourier series as follows.

Excellent approximation after nonlinear change of variable
Excellent approximation after nonlinear change of variable

Yes, I can’t see any difference either: the error is less than {10^{-3}}.

The composition with cosine improves approximation because {f(\cos t)} is naturally a periodic function, with no jumps or corners in its graph. Fourier series, which are periodic by nature, follow such functions more easily.


A practical implication of this approximation of {f(\cos t)} is the Clenshaw-Curtis integration method. It can be expressed in one line:

\displaystyle    \int_{-1}^1 f(x)\,dx = \int_0^\pi f(\cos t)\sin t\,dt \approx \int_0^\pi \sum_{n=0}^N a_n \cos nt \sin t\,dt   = \sum_{n=0}^N \frac{(1+(-1)^n) a_n}{1-n^2}

The integral {\int_{-1}^1 f(x)\,dx} is approximated by summing {2a_{2k}/(1-4k^2)}, where {a_{2k}} are even-numbered cosine coefficients of {f(\cos t)}. In the example with {f(x)=\exp(x)} using just three coefficients yields

\displaystyle    \frac{2a_0}{1}+\frac{2a_2}{-3}+\frac{2a_4}{-15} \approx 2.350405

while the exact integral is {\approx 2.350402}.


At first this doesn’t look practical at all: after all, the Fourier coefficients are themselves found by integration. But since {f(\cos t)} is so close to a trigonometric polynomial, one can sample it at equally spaced points and apply the Fast Fourier transform to the result, quickly obtaining many coefficients at once. This is what the Clenshaw-Curtis quadrature does (at least in principle, the practical implementation may streamline these steps.)

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