## 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. When the sines don’t get in the way

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: Periodic extension of a continuous function may be discontinuous

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 ${N=4}$.

## Peanut allergy and Polar interpolation

Draw a closed curve passing through the points ${(\pm 3,0)}$ and ${(0,\pm 1)}$: more specifically, from (3,0) to (0,1) to (-3,0) to (0,-1) and back to (3,0).

I’ll wait.

… Done?

Okay, you may read further.

This is an interpolation problem. The geometry of the problem does not allow for curves of the form ${y=f(x)}$. But polar graphs ${\rho=f(\theta)}$ should work. (Aside: in ${(x,y)}$ the variable usually taken to be independent is listed first; in ${(\rho,\theta)}$ it is listed second. Is consistent notation too much to ask for? Yes, I know the answer…) Since ${f}$ must be ${2\pi}$-periodic, it is natural to interpolate ${(3,0),(1,\pi/2),(3,\pi),(1,3\pi/2)}$ by a trigonometric polynomial. The polynomial ${f(\theta)=2+\cos 2\theta}$ does the job. But does it do it well?

This is not what I would draw. My picture would be more like the ellipse with the given points as vertices:

The unnecessary concavity of the peanut graph comes from the second derivative ${f''}$ being too large at the points of minimum (and too small at the points of maximum). We need a function with flat minima and sharp maxima. Here is the comparison between ${f(\theta)=2+\cos 2\theta}$ (red) and the function that describes the ellipse in polar coordinates (blue):

I thought of pre-processing: apply some monotone function ${\varphi\colon (0,\infty)\rightarrow \mathbb R}$ to the ${\rho}$-values, interpolate, and then compose the interpolant with ${\varphi^{-1}}$. The function ${\varphi}$ should have a large derivative near ${0}$, so that its inverse has a small derivative there, flattening the minima.

The first two candidates ${\varphi(r)=\log r}$ and ${\varphi(r)=1/r}$ did not improve the picture enough. But when I tried ${\varphi(r)=1/r^2}$, the effect surprised me. Interpolation between ${(1/9,0),(1,\pi/2),(1/9,\pi),(1,3\pi/2)}$ yields ${(5-4\cos 2\theta)/9}$, which after application of ${\varphi^{-1}}$ produces ${\rho = 3/\sqrt{5-4\cos 2\theta}}$exactly the ellipse pictured above.

Next, I tried a less symmetric example: curve through ${(5,0);(0,2);(-3,0);(0,-2)}$.

Again, the second approach produces a more natural looking curve.

Finally, a curve with five-fold symmetry, with ${\rho}$ ranging from ${2}$ to ${5}$.

The interpolation and plotting were done in Scilab, by invoking the functions given below with commands polar1([3,1,3,1]) or polar2([2,5,2,5,2,5,2,5,2,5]), etc. Because I considered equally spaced interpolation nodes, the coefficients of interpolated polynomials are nothing but the (inverse) discrete Fourier transform of the given values. First function interpolates the radius ${\rho}$ itself.

function polar1(R)
clf()
p = fft(R,1)
t = 0:.01:2*%pi
rho = zeros(t)
for i = 1:length(p)
rho = rho + real(p(i)*exp(-%i*(i-1)*t))
end
polarplot(t,rho,style=5)
t = resize_matrix(linspace(0,2*%pi,length(R)+1),1,length(R))
plot(R.*cos(t), R.*sin(t), 'o')
endfunction

The second function includes preprocessing: it interpolates ${1/\rho^2}$ and then raises the interpolant to power ${-1/2}$.

function polar2(R)
clf()
p = fft(R^(-2),1)
t = 0:.01:2*%pi
rho = zeros(t)
for i = 1:length(p)
rho = rho + real(p(i)*exp(-%i*(i-1)*t))
end
rho = max(rho,0.01*ones(rho))
polarplot(t,rho^(-1/2),style=5)
t = resize_matrix(linspace(0,2*%pi,length(R)+1),1,length(R))
plot(R.*cos(t), R.*sin(t), 'o')
endfunction

Added: one more comparison, ${\rho^{1}}$ vs ${\rho^{-2}}$ vs ${\rho^{-3}}$; the last one loses convexity again. Raising to -3 is going too far.

And then to ${\rho^{-7}}$: