Re: [math-fun] Arithmetic-geometric mean analog calculator?
John Aspinall <j@jkmfamily.org> wrote:
I'm familiar with those. Did you have any particular kind of nomogram in mind? I don't think it would work with the classic nomogram, three parallel lines. Since AGM(a,b) = AGM(b,a), the two outer lines would presumably be the same, due to a symmetry argument. Then the middle line would also have to be the same, since AGM(a,a) = a. So it would fail for the exact same reason as the original single straight line failed. With a single curve, with agm(a,b) found at the point where the perpendicular at the midpoint of the line segment connecting a and b intersects the curve, at least you have fewer degrees of freedom. At each point (or rather infinitesimal interval) there's a single curvature and a single scale factor. Perhaps such a curve could be found by a hill-climbing algorithm, which starts with a straight line and a linear scale, and keeps making tiny changes to get the errors as small as possible. It would be very CPU intensive, but so what? Since there's always a skew in the same direction as a and b get further apart, I think the single curve would be a spiral. Assuming, of course, that a curve with the desired property even exists. I'm sure someone has worked on this before.
participants (1)
-
Keith F. Lynch