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).

How hard could it be?
How hard could it be?

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?

I did not order peanuts
I did not order peanuts

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

Ellipse
Ellipse

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):

We want blue, not red.
We want blue, not red.

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)}.

Egg without preprocessing
Egg without preprocessing
Preprocessed egg
Preprocessed egg

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

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

Star without preprocessing
Star without preprocessing
Star with preprocessing
Star with preprocessing

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.
Raising to -3 is going too far.

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

Power -7
Power -7