# Down with sines!

Suppose you have a reasonable continuous function ${f}$ on some interval, say ${f(x)=e^x}$ on ${[-1,1]}$, and you want to approximate it by a trigonometric polynomial. A straightforward approach is to write

$\displaystyle f(x) \approx \frac12 A_0+\sum_{n=1}^N (A_n\cos n \pi x +B_n \sin n \pi x)$

where ${A_n}$ and ${B_n}$ are the Fourier coefficients:

$\displaystyle A_n= \int_{-1}^1 e^x \cos \pi n x \,dx = (-1)^n \frac{e-e^{-1}}{1+ \pi^2 n^2 }$

$\displaystyle B_n= \int_{-1}^1 e^x \sin \pi n x \,dx = (-1)^{n-1} \frac{\pi n (e-e^{-1})}{1+ \pi^2 n^2 }$

(Integration can be done with the standard Calculus torture device). With ${N=4}$, we get

$\displaystyle e^{x} \approx 1.175 -0.216\cos\pi x +0.679 \sin \pi x +0.058\cos 2\pi x -0.365 \sin 2\pi x -0.026\cos 3\pi x +0.247 \sin 3\pi x +0.015\cos 4\pi x -0.186 \sin 4\pi x$

which, frankly, is not a very good deal for the price.

Still using the standard Fourier expansion formulas, one can improve approximation by shifting the function to ${[0,2]}$ and expanding it into the cosine Fourier series.

$\displaystyle f(x-1) \approx \frac12 A_0+\sum_{n=1}^N A_n\cos \frac{n \pi x}{2}$

where

$\displaystyle A_n= \int_{0}^2 e^{x-1} \cos \frac{\pi n x}{2} \,dx = \begin{cases} \dfrac{e-e^{-1}}{1+ \pi^2 k^2 } \quad & n=2k \\ {} \\ (-1)^k \dfrac{4(e+e^{-1})}{4+ \pi^2 (2k-1)^2 } \quad & n = 2k-1 \end{cases}$

Then replace ${x}$ with ${x+1}$ to shift the interval back. With ${N=4}$, the partial sum is

$\displaystyle e^x \approx 1.175 -0.890\cos \frac{\pi(x+1)}{2} +0.216\cos \pi(x+1)-0.133\cos \frac{3\pi(x+1)}{2} +0.058\cos 2\pi (x+1)$

which gives a much better approximation with fewer coefficients to calculate.

To see what is going on, one has to look beyond the interval on which ${f}$ is defined. The first series actually approximates the periodic extension of ${f}$, which is discontinuous because the endpoint values are not equal:

Cosines, being even, approximate the symmetric periodic extension of ${f}$, which is continuous whenever ${f}$ is.

Discontinuities hurt the quality of Fourier approximation more than the lack of smoothness does.

Just for laughs I included the pure sine approximation, also with ${N=4}$.

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