Ch7p275debug2

Math Problem PhL 7.10.11 The problem is to start with this g(r,φ|r0,π) = (1/π2) !Syntax Error, Idγ Kiγ(kr) Kiγ(kr0) { ch[γ(|φ-π|-π)] – ch[γ(φ-π)] / ch(πγ) } and transform it into this g(r,φ|r0,π) = (1/π) Σm=1∞ sin(mφ/2)sin(mπ/2) Im/2(kr<) Km/2(kr>) = (1/π) Σm=1,3,5 ∞ sin(mφ/2)sin(mπ/2) Im/2(kr<) Km/2(kr>) This is a completely self-contained problem so we can forget about everything else on Stak p 275. We do have these basic facts concerning I and K for general complex α , I-α - Iα = (2/π) sin(πα) Kα K-α = Kα We can clearly see that the big factor {.... } is analytic in γ except for the poles of 1/ch(γπ). I think Kν(x) is analytic in ν everywhere, but perhaps that needs another look. Let's start with this series We can write (z/2)ν = eν ln(z/2) = eaν which function is analytic everywhere in ν !! The gamma function has some poles in ν, but they just become zeros in I. My conclusion is this: Iν(z) is analytic in ν everywhere Now consider, We conclude at once that Kν(z) has no cuts in ν, but could have poles in ν at ν = integers, though the numerator might cancel these poles. We do know that In = I-n so indeed the poles are in fact cancelled since the formula above says K = 0/0 at these poles. I then conclude that Kν(z) is analytic in ν everywhere I cannot find confirmation of these claims on the web or in my books or PDF's, but I have just proven that it is true. Plan A. Can I write one of the K functions as a contour integral of an I function? That would at least cause a contour integral to appear in the problem. Did not pan out. Status: I am starting with this integral g(r,φ|r0,π) = (1/π2) !Syntax Error, Idγ Kiγ(kr) Kiγ(kr0) { ch[γ(|φ-π|-π)] – ch[γ(φ-π)] / ch(πγ) } and I know that the entire integrand is analytic in γ except for the poles due to ch(πγ). Plan B. Write the above in this way g(r,φ|r0,π) = (1/2π2) !Syntax Error, Idγ Kiγ(kr) Kiγ(kr0) f(γ) where f(γ) = ch[γ(|φ-π|-π)] – ch[γ(φ-π)] / ch(πγ) = a function even in γ. Of course KK is also even in γ, that is why we can alter the integrand. NOW, the next step is to replace one of the K's like this, using the AS fact quoted above Kiγ(kr) = (π/2) [I-iγ(kr) - Iiγ(kr)] / sin(iγπ) = (π/2i) [I-iγ(kr) - Iiγ(kr)] / sh(γπ) = (iπ/2) [Iiγ(kr) - I-iγ(kr)] / sh(γπ) We than have g(r,φ|r0,π) = (1/2π2) !Syntax Error, Idγ {(iπ/2) [Iiγ(kr) - I-iγ(kr)] / sh(γπ)} Kiγ(kr0) f(γ) = (i/4π) !Syntax Error, Idγ { [Iiγ(kr) - I-iγ(kr)] / sh(γπ)} Kiγ(kr0) f(γ) = (i/4π) !Syntax Error, Idγ [Iiγ(kr) - I-iγ(kr)] Kiγ(kr0) f(γ)/ sh(γπ) We know write this as the sum of two integrals so that g = g1+g2, where we use K-iγ = Kiγ , g1 = (i/4π) !Syntax Error, Idγ Iiγ(kr) Kiγ(kr0) f(γ)/ sh(γπ) g2 = - (i/4π) !Syntax Error, Idγ I-iγ(kr) K-iγ(kr0) f(γ)/ sh(γπ) Now in the second integral, replace γ → -γ, use the fact that f(γ) is even in γ, and we get g2 = (i/4π) !Syntax Error, Idγ Iiγ(kr) Kiγ(kr0) f(γ)/ sh(γπ) = g1 So these two integrals are exactly the same !!!! So that is an easy way to "change a K into an I". We then have shown that either of the following is true: g(r,φ|r0,π) = (i/2π) !Syntax Error, Idγ Iiγ(kr) Kiγ(kr0) f(γ)/ sh(γπ) g(r,φ|r0,π) = -(i/2π) !Syntax Error, Idγ I-iγ(kr) K-iγ(kr0) f(γ)/ sh(γπ) and we have taken finally a giant leap forward. Note: At the very start, I could have written Kiγ(kr) Kiγ(kr0) = Kiγ(kr<) Kiγ(kr>) Then I would have ended up with these results which are what we really will need below, g(r,φ|r0,π) = (i/2π) !Syntax Error, Idγ Iiγ(kr<) Kiγ(kr>) f(γ)/ sh(γπ) g(r,φ|r0,π) = -(i/2π) !Syntax Error, Idγ I-iγ(kr<) K-iγ(kr>) f(γ)/ sh(γπ) NOW maybe we can close top or bottom! First, if we go high up or down in the γ plane, f(γ) becomes oscillatory with some poles, so does not help convergence. The same is true with our sh(γπ) factor. So it all hinges on the IK product. I requote earlier results Maybe this result can be trusted for the IK product and we would then have IνKν → (1/2) ν-1 Iiγ(kr) Kiγ(kr0) → (1/2) (iγ)-1 Then as we to up or down, we are looking at an integrand which is roughly 1/γ * oscillatory, and maybe this causes the GC to vanish. It at least has a chance. It seems that the chance is the same up or down, so let's close "up" and then we get the right sense on our pole residues. Now write f(γ) = ch[γ(|φ-π|-π)] – ch[γ(φ-π)] / ch(πγ) = a function even in γ. = fE(γ) + fv(γ) where fv(γ) = – ch[γ(φ-π)] / ch(πγ) and this is the part that arises from our v contribution to g = g + E. We are then interested in v(r,φ|r0,π) = (i/2π) !Syntax Error, Idγ Iiγ(kr<) Kiγ(kr>) fv(γ)/ sh(γπ) = (i/2π) !Syntax Error, Idγ Iiγ(kr<) Kiγ(kr>) {– ch[γ(φ-π)] / ch(πγ)}/ sh(γπ) = -(i/2π) !Syntax Error, Idγ Iiγ(kr<) Kiγ(kr>) ch[γ(φ-π)] / ch(πγ)}/ sh(γπ) = -(i/π) !Syntax Error, Idγ Iiγ(kr<) Kiγ(kr>) ch[γ(φ-π)] / sh(2πγ) = (1/iπ) ∫up_ccw dγ Iiγ(kr<) Kiγ(kr>) ch[γ(φ-π)] / sh(2πγ) // I moved the i downstairs I now want to take "the other form" for the I and K subscripts. Change them and put an overall minus sign, then we have v(r,φ|r0,π) = - (1/iπ) ∫up_ccw dγ I-iγ(kr<) K-iγ(kr>)ch[γ(φ-π)] / sh(2πγ) Now back in my raw2 notes I already dealt with the poles here and I showed there that sh(2πγ) ≈ 2π (-1)n (γ - γn) for γ near the pole position γn = ni/2, -iγn = n/2 There is the same issue with how much of the pole at φ = 0 we "get". But looking now at the full f(γ), we can see that when γ = 0, we have f(0) = 0 and I think things cancel between the two terms. So let's just ignore the pole at 0 for both the v and E terms (I will later remove it from the known E sum). Then we get v(r,φ|r0,π) = -(1/iπ) (2πi) 1/(2πi) ∫up_ccw dγ I-iγ(kr<) K-iγ(kr>) ch[γ(φ-π)] / sh(2πγ) = - 2 Σn=1,2,3.. In/2(kr<) Kn/2(kr>) ch[γn(φ-π)] / [ 2π(-1)n ] = - (1/π) Σn=1,2,3.. (-1)n In/2(kr<) Kn/2(kr>) ch[γn(φ-π)] and we can write ch[γn(φ-π)] = ch[ni(φ-π)/2] = cos[n(φ-π)/2] so our final v result is then v(r,φ|r0,π) = - (1/π) Σn=1,2,3.. (-1)n In/2(kr<) Kn/2(kr>) cos[n(φ-π)/2] This is similar to -- but not the same as -- a result I obtained in the raw2 notes, because here the sum is only over positive integers. Now lets quote our E term sum, but exclude the n=0 term as noted above, E(r,φ|ro,π) = (1/π) Σn=1∞ (-1)n cos(nφ) In(kr<)Kn(kr>) In order to attempt addition of these, change to m = 2n or n = m/2 in this last sum E(r,φ|ro,π) = (1/π) Σm=2,4,6..∞ (-1)m/2 cos(mφ/2) Im/2(kr<)Km/2(kr>) Then we have these results prior to addition v(r,φ|r0,π) = - (1/π) Σn=1,2,3.. ∞ (-1)n In/2(kr<) Kn/2(kr>) cos[n(φ-π)/2] E(r,φ|ro,π) = (1/π) Σn=2,4,6..∞ (-1)n/2 In/2(kr<) Kn/2(kr>) cos(nφ/2) But we are not yet ready to add. First note that cos[n(φ-π)/2] = cos(nφ/2)cos(nπ/2) + sin(nφ/2)sin(nπ/2) = cos(nφ/2)cos(nπ/2) when n = even integers = sin(nφ/2) sin(nπ/2) when n = odd integers Let's now separate the v series into two series, odds and evens, so we then want to add three series v(r,φ|r0,π)o = + (1/π) Σn=1,3,5.. ∞ In/2(kr<) Kn/2(kr>) sin(nφ/2) sin(nπ/2) v(r,φ|r0,π)e = - (1/π) Σn=2,4,6.. ∞ In/2(kr<) Kn/2(kr>) cos(nφ/2)cos(nπ/2) E(r,φ|ro,π) = (1/π) Σn=2,4,6..∞ (-1)n/2 In/2(kr<) Kn/2(kr>) cos(nφ/2) Now observe that for n even cos(nπ/2) = (-1)n/2 so we then have v(r,φ|r0,π)o = + (1/π) Σn=1,3,5.. ∞ In/2(kr<) Kn/2(kr>) sin(nφ/2) sin(nπ/2) v(r,φ|r0,π)e = - (1/π) Σn=2,4,6.. ∞ In/2(kr<) Kn/2(kr>) cos(nφ/2) cos(nπ/2) E(r,φ|ro,π) = (1/π) Σn=2,4,6..∞ In/2(kr<) Kn/2(kr>) cos(nφ/2) cos(nπ/2) Finally after 2-3 days, we now get cancellation of the two even series! We are then left with g(r,φ|r0,π) = (1/π) Σn=1,3,5.. ∞ In/2(kr<) Kn/2(kr>) sin(nφ/2) sin(nπ/2) We can then add the even indices since sin(nπ/2) = 0 and we have g(r,φ|r0,π) = (1/π) Σn=1 ∞ In/2(kr<) Kn/2(kr>) sin(nφ/2) sin(nπ/2) which finally agrees with (7.183) !!!