Draw a closed curve passing through the points and : 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 . But polar graphs should work. (Aside: in the variable usually taken to be independent is listed first; in it is listed second. Is consistent notation too much to ask for? Yes, I know the answer…) Since must be -periodic, it is natural to interpolate by a trigonometric polynomial. The polynomial 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 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 (red) and the function that describes the ellipse in polar coordinates (blue):

I thought of pre-processing: apply some monotone function to the -values, interpolate, and then compose the interpolant with . The function should have a large derivative near , so that its inverse has a small derivative there, flattening the minima.

The first two candidates and did not improve the picture enough. But when I tried , the effect surprised me. Interpolation between yields , which after application of produces — **exactly the ellipse pictured above**.

Next, I tried a less symmetric example: curve through .

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

Finally, a curve with five-fold symmetry, with ranging from to .

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 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 and then raises the interpolant to power .

```
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, vs vs ; the last one loses convexity again.

And then to :

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.

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!

Exactly what I needed, thanks!