Appendix B 12_11_15

this is obsolete, was never used Appendix B Recall the toroid capacitance expression from (10.2.6) and the drawing (10.2.4) C = (2/π) Σn=0∞ εn , εn = 2-δn,0 . (10.2.6) (10.2.4) We wish to hold ρc constant and increase R to R = ρc when the hole in the toroid just disappears, so the toroid becomes "degenerate", or a "horn torus". Our question of interest is this: what is the capacitance at this point? We set ρc = 1 and write C(R) = (2/π) Σn=0∞ εn , εn = 2-δn,0 . (10.2.6) T(R) = Σn=0∞ εn , εn = 2-δn,0 . (10.2.6) The two functions shown differ by the simple factor shown, and we shall deal with both to be compatible with an earlier version of this appendix. One cannot simply set R = 1 because Qν(z) has a singularity at z = 1. Rather, we have to carefully approach the limit. Here is an illustration of what happens as one approaches this limit : C(R) seems to have a value around 1.74 at R = .999, but as we get closer to 1, the sum fades off. The remedy is to include more terms in the sum. If we increase from 100 to 1000 terms, the result is instead this, This suggests that the limit is around C = 1.7414 . To obtain more accuracy, instead of using the Maple plot routine, we make our own curve of only 8 points so we can watch the approach to R = 1, while printing out the values. The code is this, where we start at R = .99999 and approach R→1 using 4000 terms. Here is the output of this code, This are a little rocky, but the limit now seems around 1.74137. We have found that a different series is more well-behaved. Using the Wronskian of P and Q, one can show that this slope is given by T'(z) = { T(z) – Σn=0∞ εn /[Pn-1/2(z)]2 } / (z2-1) . (B.6) If we conjecture that T'(1) is not infinite, which the picture certainly suggests, we may conclude that T(1) = limz→1{ Σn=0∞ εn /[Pn-1/2(z)]2 } (B.7) Pause : I would like to get T(z) back into the picture, but I am wearing down from the 3AM start. So let's do that manana.