Question about the 1D string Green's function

Question about the 1D string Green's function PhL 5.3.10 Stak always talks about a string on (0,l), both in Chapter 1 and Chapter 4. But I want to talk about a string on (-l/2,l/2). So make these replacements in his results: x'= x - l/2 so when his x = l, my x' = l/2. So replace x = x' + l/2 then rename x' to be x. Look then at his Green's function on page 261 g(x|ξ) = Σn=1∞ φn(x)φn(ξ)/λn = Σn=1∞ (2/l) sin(nπx/l) sin(nπξ/l) / [nπ/l]2 The lowest mode n = 1 has sin(nπx/l) has a peak at x = l/2 which will get moved now to 0, so it is the n = odd modes which peak in the center of the string. So I transcribe this into g(x|ξ) = Σn=1∞ φn(x)φn(ξ)/λn = Σn=1∞ (2/l) sin(nπ[x+l/2]/l) sin(nπ[ξ+l/2]/l) / [nπ/l]2 where φn(x) = (2/l)1/2 sin(nπ[x+l/2]/l) Notice now that φn(0) = (2/l)1/2 sin(nπ[l/2]/l) = (2/l)1/2 sin(nπ/2) = (2/l)1/2 (-1)n n = odd // peaks mentioned above = 0 n = even // no surprise here, central node So we can say g(x|0) = Σn=1∞ φn(x)φn(ξ)/λn = Σn=odd (2/l) (-1)n sin(nπ[x+l/2]/l) / [nπ/l]2 = (l2/π2) (2/l) Σn=odd (-1)n sin(nπ[x+l/2]/l)/n2 I conjecture that sin(nπ[x+l/2]/l) = sin [ nπx/l - nπ/2] = (-1)n cos(nπx/l) for n odd so we would then have g(x|0) =(l2/π2) (2/l) Σn=odd cos(nπx/l)/n2 = ( 2l/π2) Σn=odd cos(nπx/l)/n2 Now consider this infinite series S ≡ Σn=1,3,5 cos(ny)/n2 Let n = 2m-1 so that m = 1,2,3.. then we have S ≡ Σm=1,2,3 cos([2m-1]y)/(2m-1)2 We voyage over to GR7 page 47 to get So we conclude that S = (π/4)( π/2 - |α| ) Therefore with α = πx/l we get g(x|0) = ( 2l/π2) * (π/4)( π/2 - | πx/l | ) = ( 2l) * (1/4)( 1/2 - | x/l | ) = (1/2)( l /2 - | x | ) = - (1/2)| x | + l/4 This describes a down-facing V centered at x = 0 and having slope -1/2 on the right and +1/2 on the left. As we increase l, the height of g in the center keeps increasing. We can compare this to the 1D "fundamental solution" shown Stak p 51. This is normally taken to be simple E(x) = - (1/2)| x | where the constant is set to 0. My point is that we have obtained this result from the general form of the Green's function on the string (-l/2,l/2) where g(x|ξ) = Σn=1∞ φn(x)φn(ξ)/λn = Σn=1∞ (2/l) sin(nπ[x+l/2]/l) sin(nπ[ξ+l/2]/l) / [nπ/l]2 = Σn=1∞ (2/l)(l/π)2 sin(nπ[x+l/2]/l) sin(nπ[ξ+l/2]/l) /n2 = (2l/π2) Σn=1∞ sin(nπ[x+l/2]/l) sin(nπ[ξ+l/2]/l) /n2 Now on scratch I have shown that sin(nπ[x+l/2]/l) = (-1)n sin(nπx/l) n even = (-1)n+1 cos(nπx/l) n odd so we can write our Green's function as g(x|ξ) = (2l/π2) Σn=even sin(nπx/l) sin(nπξ/l) /n2 - (2l/π2) Σn=odd cos(nπx/l) cos(nπξ/l) /n2 I could go on to "do" all the series here to obtain Stak page 9 for the g, shifted by l/2 , g(x|ξ) = (1/l) (x< + l/2)(l - [x>+ l/2]) = - (1/l) (x< + l/2) (x< - l/2) For x < ξ this becomes - (1/l) (x + l/2) (ξ - l/2) = - (x/l + 1/2) (ξ - l/2) = -xξ/l + x/2 - ζ/2 + l/4 = - (1/2)(ξ-x) -xξ/l + l/4 and the result for x > ξ is obtained by doing x ↔ ξ, which is like taking l → -l which you see is a symmetry of the form shown above. Thus, the result is g(x|ξ) = - (1/2)|x-ξ| -xξ/l + l/4 and as l → ∞, this time we have a term xξ/l which vanishes, and we get the same l/4 constant which we ignore, and again we recover the fundy solution which is g(x|ξ) =- (1/2)|x-ξ|. So you can ask: what happens to the Green's Function as l → ∞ . The answer is that the non-constant part stays fixed at - (1/2)| x-ξ |. I conjecture that in 3D, something like this happens if we start with a sphere, do eigenfunctions, and take the sphere radius it to infinity -- the non-constant part will by 1/ |r-ξ| .