old app B save

Old Appendix B Save PhL 12.20.15 this is as things were in my 11/30/11 release of bowl doc, it is archived there Appendix B: Comments on computing the capacitance of a toroid and its degenerate limit. Recall the capacitance formula (10.2.6) given above, C(z) = (2R/π) Σn=0∞ εn [Qn-1/2(z)/ Pn-1/2(z)] . z = chξ = ρc/R . (B.1) It is convenient to remove the leading factor and talk about this function T(z) ≡ Σn=0∞ εn [Qn-1/2(z) / Pn-1/2(z)] so C(z) = (2R/π)T(z) . (B.2) Basically this is a "special function" that has no name. We can call it the "toroidal capacitance function". It is defined for our purposes on the range z in (1,∞). One can see from this fact , Qν(z)/ Pν(z) ≈ π { Γ(1+ν)2/ [Γ(ν+3/2) Γ(ν+1/2)] } (2z)-2ν-1 , replace ? (B.3) that the series converges for z > 1. It is a simple matter to have Maple compute T(z) for a variable number of terms to make sure enough terms are included to give a stabilized result. As z→1, the number of required terms increases (Qν(z) has a log singularity at z=1). The magic question is to find the limit T(z=1), it if exists. This is the limit which determines the capacitance of a toroid for which the hole has just gone out of existence. This physical interpretation suggests that the limit must exist. We use appropriate definitions of the P and Q functions as follows, and do some cursory checks: After this warmup exercise, we turned to the T(z) series (B.2) and computed it the same way, always with a stable number of terms. Our results were as follows: This different versions of this table appears in toroidal cap function.doc page 11 and page 16 z T(z) terms used -log10(z-1) 1.1 2.8465 15 1 1.01 2.7987 140 2 1.001 2.7414 300 3 1.0001 2.7359 600 4 1.00001 2.7353 1200 5 1.000001 2.7353 4000 6 (B.5) suggesting that we are approaching a limit close to 2.7353 for the value of T(1). We can plot the above points as follows: where the x axis is -log10(z-1). The point is that we are coming into the limit T(1) with zero slope. 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) and we have then a second way to compute T(1). This series is more stable because it avoids the ln(z-1) singularity of the Q functions. Using this series Maple yielded the following results z = 1.1 above = 2.898517346 z = 1.01 above = 2.751570937 z = 1.001 above = 2.736973437 z = 1.0001 above = 2.735058209 z = 1.00001 above = 2.735362078 z = 1.000001 above = 2.735354499 10,000 terms z = 1.0000001 above = 2.735353885 40,000 terms (B.8) We verified most of these results by jamming strings like the following into the Wolfram Alpha website computation window ( the argument 3 indicates off-the-cut Legendre functions in Mathematica) sqrt(1.0000001^2-1)*(2*sum(1/LegendreP(n-1/2,0,3,1.0000001)^2,n=1 to 39999) + 1/LegendreP(-1/2,0,3,1.0000001) ^2) Though it timed out on this case, it did all the other cases listed, confirming Maple's results. Our conclusion is the following strange fact T(1) ≈ 2.735353 (B.9) and therefore we find that the capacitance of a degenerate toroid is given by C(ρc= R) = (2R/π)T(1) ≈ R 1.741380 . // *4πε0 for SI units (B.10) It is interesting that T(1) is slightly larger than e ≈ 2.718282, but for sure T(1) ≠ e. We do not know how to obtain this limit in any other way. We know of no integral representations of inverse Legendre functions, nor of any published series involving them, though PBM has very many series involving Legendre functions and products of same combined with other functions. Perhaps a variation for z ≥1 of Bateman HT I p 149 (11) could be used. We sometimes read about the interchange of limits giving different results in situations where uniform convergence is lacking in both limits, or where one of the limits does not exist. We have here exactly a case of this second situation: limz→1 limN→∞ { Σn=0N εn /[Pn-1/2(z)]2 } = T(1) = 2.735353 whereas limN→∞ limz→1 { Σn=0N εn /[Pn-1/2(z)]2 } = Σn=0∞ εn limz→1 { } since limz→1Pν(z) = 1 = ∞ * 0 The situation is even worse with the P/Q series since limz→1Qν(z)/Pν(z) = ∞/1 = ∞.