# 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 ${n\le 4}$:

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.

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.

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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