temp2

can throw this out Cosine Transform Derivation for a particular function PhL 12.18.15 I want to derive these equations, 1/ = (1/π) Σn=0∞ εn Qn-1/2(a/b) cos(nx) // expansion !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) // projection (10.1.8) I may have done this before somewhere, perhaps worth a search. I do a partial derivation here where I use the general four properties of a cosine transform and the integral of Q, so I can fall back on that if I want. But let's attempt a direct derivation. One integral representation I show in the above doc is this Qn-1/2(z) = (1/) !Syntax Error, Idx cos(nx) (z-cos(x))-1/2 I then apply this to my situation and I find that !Syntax Error, Idx cos(nx)/ = 2 Qn-1/2(a/b) I guess I would rewrite this as !Syntax Error, Idx cos(nx)/ = 2 Qn-1/2(a/b) then I would say the integrand was even in x so I conclude that !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) So that is a quick proof of the second equation. What happens if I try to use the same integral representation in the other direction? Set b = 1 without LOG to get 1/ = (/π) Σn=0∞ εn Qn-1/2(a) cos(nx) // expansion Then RHS = (/π) Σn=0∞ εn { (1/) !Syntax Error, Idy cos(ny) (a-cos(y))-1/2 } cos(nx) = (1/π) !Syntax Error, Idy Σn=0∞ εn (a-cosy)-1/2 [ Σn=0∞ εn cos(nx)cos(ny) ] At this point I have to use my repaired completeness result Σn=0∞ εn cos(nx)cos(ny) = πδ(x-y) And then we have RHS = (1/π) !Syntax Error, Idy Σn=0∞ εn (a-cosy)-1/2 π δ(x-y) = (a-cosx)-1/2 QED So I cannot avoid that completeness relation. M&F p 1274: The use the phrase "the Neumann factor" for εn and define it as I do. But but they give no explanation of why Neumann is connected with this factor. WW don't mention "neumann factor" or "neumann's factor" . Most people trace the use of this term to his M&F usage!!! Volume 1