compare Smythe and App G REVIEWED

Comparing Smythe's potential on cap with G.25 for the same thing PhL 12.24.15 A recent doc, has my new bowl picture, and the plot is now in Appendix G. 12.31.15 First my picture, Smythe says this for potential on the cap V(θ) = V0(2/π)sin-1[cos(α/2)/cos(θ/2)], (D.7) where θ = θ in my picture above, so θ runs from 0 to u0 on the cap α = u0 I claim below (D.7). Meanwhile, I get this fancy other result f2(θ) = (1/π) Σn=0∞{ sin(nθ0)/n + sin[(n+1)θ0] /(n+1) } Pn(cosθ) . (G.25) Here we have θ being the θ of (E.14), which is my angle η in the above picture, the angle measured from the bowl center. And θ0 is the lip measured the same way. So θ = η = π-θ runs from θ = θ0 to π on the cap. θ0 = π-u0 So if I want to compare these two expressions, I have to get them in terms of the same variable! So start with V(θ) = V0(2/π)sin-1[cos(u0/2)/cos(θ/2)] using my θ and u0 f2(θ) = (1/π) Σn=0∞{ sin(nθ0)/n + sin[(n+1)θ0] /(n+1) } Pn(cosθ) = (1/π) Σn=0∞{ sin(n[π-u0])/n + sin[(n+1)[π-u0]] /(n+1) } Pn(cos[π-θ]) When I plot these two babies, they match! Here with 30 terms So let's maybe write them the other way f2(θ) = (1/π) Σn=0∞{ sin(nθ0)/n + sin[(n+1)θ0] /(n+1) } Pn(cosθ) . (G.25) V(θ) = V0(2/π)sin-1[cos(α/2)/cos(θ/2)], α = u0 = π-θ0 θ = π-θ Then the Smythe form becomes V(θ) = (2/π)sin-1[cos(α/2)/cos(θ/2)], = (2/π)sin-1[cos([π-θ0]/2)/cos([π-θ]/2)], = (2/π)sin-1[cos(π/2-θ0/2)/cos(π/2-θ/2)], = (2/π)sin-1[sin(θ0/2)/sin(θ/2)], So I am now claiming this amazing sum (1/π) Σn=0∞{ sin(nθ0)/n + sin[(n+1)θ0] /(n+1) } Pn(cosθ) = (2/π)sin-1[sin(θ0/2)/sin(θ/2)] or Σn=0∞{ sin(nθ0)/n + sin[(n+1)θ0] /(n+1) } Pn(cosθ) = 2sin-1[sin(θ0/2)/sin(θ/2)] This must exist somewhere! I cannot find it anywhere! It is a Legendre transform of a certain function. I will add this at the end of Appendix. Start with