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).
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
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 :