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

3 thoughts on “Peanut allergy and Polar interpolation”

  1. Very interesting subject.. But what is special about r to the negative 2 power? I am curious about what would happen if we raise it to negative 3 and so on.

    In general, one can write r as an infinite sum of theta’s, and optimize over the coefficients to minimize the integral of absolute value of first or second derivative of r, with satisfying the interpolation points as equality constrains. This theoretically should give the smoothest picture.

    1. I see. I bet if we raise r to negative infinity it will converge to a diamond, which has the smallest length connecting the 4 points.

      I thought the argument is that because trigonometric interpolation produces sharp minima and smooth maxima, you want to apply an inverse-like function to map the minima to maxima and maxima to minima, then do interpolation, and map it back. Now you will have sharp maxima and smooth minima.

      Of all the functions log will preserve the extremums so it should not work. But the negative-power functions will reverse the extremums so all of them should produce sharp maxima and smooth minima. But it seems only satisfying this requirement is not sufficient for convexity.

      It seems your argument in the article is a little different. Can you expand on that? Thanks!

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