Fourier Series on non-periodic functions

Fourier Series on non-periodic functions. PhL 10.20.09 If the function in question is defined over all argument and is non-periodic, we usually switch to the Fourier Transform. But suppose the function is only defined on part of a 2π interval, perhaps going out to some θmax. Let's say the function is not even defined beyond θmax for some problem we are considering. A. One approach is to say that f(θ) = 0 in the range (θmax, 2π). Then on a larger scale, we regard the function as periodic with the above shape in each 2π period. Since we then have a periodic function, we can apply our Fourier Series machinery. Example 1: Just as an example, suppose f(θ) = sinθ and θmax = π and we set it to 0 in the right half of the interval. Schaum p 131 then tells us that 2L = 2π and we have an = (1/π) !Syntax Error, Idθ f(θ) cos(nθ) = (1/π) !Syntax Error, I sin(θ) cos(nθ) This integral from Schaum p 96 is this !Syntax Error, I sin(θ) cos(nθ) = [ 0 if n = odd, else 2/(1-n2) if n = even ] Therefore an = 0 n odd = (2/π)/(1-n2) n even = 0,2,4.... And then we also have bn = (1/π) !Syntax Error, Idθ f(θ) sin (nθ) = (1/π) !Syntax Error, I sin(θ) sin(nθ) This integral from Schaum p 96 is this !Syntax Error, I sin(θ) sin(nθ) = [ π/2 if n = 1, else = 0 ] which tells us that bn = (1/π)(π/2) δn,1 = (1/2) δn,1 We can then examine the reconstructed function f(θ) = a0/2 + Σn=1∞ ( an cos(nθ) + bnsin(nθ) ) = (1/π) + (1/2) sin(θ) + (2/π) Σn=2,4...cos(nθ) /(1-n2) One might duly wonder how this is going to reproduce "0" on the right half of the interval. The series certainly looks convergent, off hand. Before attempting a theoretical understanding, let Maple lead the way, and here we are only putting n = 2 and 4 in the series! > g := 0: > for n from 2 by 2 while n < 5 do > g := g + cos(n*theta)/(1-n*n) > od: > g: > ff := (1/Pi) + 0.5*sin(theta) + (2/Pi)*g: > plot(ff,theta=0..2*Pi); So it must be true. It must just be a fact that (1/π) + (1/2) sin(θ) + (2/π) Σn=2,4...cos(nθ) /(1-n2) = sin(θ) H(π-θ) (2/π) Σn=2,4...cos(nθ) /(1-n2) = sin(θ) [H(π-θ) - 1/2] - (1/π) Σn=2,4...cos(nθ) /(1-n2) = (π/2) sin(θ) [H(π-θ) - 1/2] - (1/2) Σm=1,2...cos(2mθ) /(1-(2m)2) = (π/2) sin(θ) [H(π-θ) - 1/2] - (1/2) Σm=1,2...cos(2mθ) /((2m)2 - 1) = (π/2) sin(θ) [1/2 - H(π-θ)] + (1/2) The Fourier Series expansion generates an infinite number of weird expansion series like this one. It happens that this sum appears in GR p 39 (last on page) but says valid 0 ≤ θ ≤ π/2. But we know from a above that it is valid in fact for 0 ≤ θ ≤ π, and, as I have written it, for all of (0,2π). Bu the way, if you plot just the series (Maple g), things are perhaps less mysterious. If I use the first 50 terms, I find this plot; g = Σm=1,2...cos(2mθ) /((2m)2 - 1) The series just adds up to this function which is periodic with period π. The humps are sines and of course it we add this to a regular 2π sin the right side can cancel and that is where a zero right half can come from. So not too mysterious. Let's do another example but "on the sphere" Example 2: Suppose we have f(θ,φ) which is defined only on half of a sphere. We can decide to set it to zero on the other half. Then what happens? f(θ,φ) = Σmn Ynm(θ,φ)fnm fnm = (1/Nnm ) ∫dΩ f(θ,φ) Ynm(θ,φ)* Here I am using the Stakgold Appendix A normalization and my own notation for fnm putting the n index first, which for some reason he does not do. Suppose in particular that f(θ,φ) =f(θ). Then fnm = (1/Nnm ) ∫dΩ f(θ,φ) Ynm(θ,φ)* = (1/Nnm ) ∫dΩ f(θ) Ynm(θ,φ)* = (1/Nnm ) ∫d(cosθ) f(θ)Pnm(cosθ) ∫dφ e-imφ = ( 2π δm,0/Nn0 ) ∫d(cosθ) Pn(cosθ) f(θ) Now suppose we take f(θ) = cosθ on the upper half sphere and f = 0 on the lower. Then we have fnm = ( 2π δm,0/Nn0 )!Syntax Error, Id(cosθ) Pn(cosθ) cosθ = ( 2π δm,0/Nn0 ) !Syntax Error, Idx Pn(x) x The integral here, not being full range, does not equal k δn,1 . Rather, it is just some set of numbers I will here call pn that are easy to compute in maple as ratios of integers. If n is odd and ≥ 3, we know that the full interval integral is 0, and can be written as twice our integral, so our integral = 0 for such cases. It n is even, we get the numbers pn. I don't think there is any closed form. Well, we could try GR p 795 7.121 with φ = π/2 so cosφ= 0 and sinφ = 1. It says then that !Syntax Error, Idx Pn(x) x = 1/[ (n-1)(n+2)] * 1 Pn(0) = Pn(0) / [(n-1)(n+2)] But Maple disagrees with the sign of this result. So let's instead use AS p 786 top which says !Syntax Error, Idx P2m(x) x = (-1)m Γ(m-1/2) Γ(1) / [ 2 Γ(-1/2) Γ(m+2) ] = (-1)m+1 Γ(m-1/2) / [ 4 Γ(m+2) ] m = 0,1,2,3 Maple agrees with this formula. So its a bit of a mess !Syntax Error, Idx Pn(x) x = (-1)n/2+1 Γ(n/2-1/2) / [ 4 Γ(n/2+2) ] = pn n = even only so fine, there is a closed form if you want one. Remember that if n is odd and ≥ 3, pn = 0. We need then one extra value by hand: p1 = 1/3. So we have found for our example with f(θ) = cosθ on the upper hemisphere and 0 on the lower that fnm = δm,0 ( 2π pn /Nn0 ) Now let's reconstruct the function. f(θ,φ) = Σmn Ynm(θ,φ)fnm = 2π Σn=1 (pn /Nn0 )Pn(cosθ) Again, you might wonder how this is going to come out being 0 on the lower hemisphere, so let's plot it in Maple. We need this extra fact Nn0 = 4π/(2n+1) So we want to compute and plot f(θ) = Σn=1 (n+1/2) pn Pn(cosθ) pn = (-1)n/2+1 Γ(n/2-1/2) / [ 4 Γ(n/2+2) ] n = 0,2,4,6.... pn = 0 n = 3,5,7... pn = 1/3 n = 1 which we can rewrite as f(θ) = (3/2) (1/3) cosθ + Σn=0,2,4.. (n+1/2) pn Pn(cosθ) = (1/2) cosθ + Σn=0,2,4.. (n+1/2) pn Pn(cosθ) Now we have Maple plot this for the sum having just a few terms: > restart; > with(orthopoly): > g := 0: > for n from 0 by 2 while n < 7 do > g := g + P(n,cos(theta))* (n+1/2)* (-1)^(n/2+1) * GAMMA(n/2-1/2)/ ( 4 * sqrt(Pi)*GAMMA(n/2+2)) > od: > > g: > plot(g + (1/2)*cos(theta),theta = 0..Pi); and there is the expected result, just as in the simpler Fourier series example.