--- Robert Baillie <rjbaillie@frii.com> wrote:
Here's another series with a similar weird property:
a1 + a2 + a3 + ... = a1^2 + a2^2 + a3^2 + ...
Specifically: Sum[ Sin[n]/n , {n, 1, Infinity}] = Sum[ (Sin[n]/n)^2 , {n, 1, Infinity}] = (Pi - 1)/2
The Fourier series for this function: f[x_] := -(Pi + x)/2 /; -Pi < x < -1; f[x_] := x(Pi - 1)/2 /; -1 < x < 1; f[x_] := (Pi - x)/2 /; 1 < x < Pi; is: Sum[ Sin[n]/(n^2) * Sin[n x] , {n, 1, Infinity}]. At x = 1, you get the sum of (Sin[n]/n)^2
The function g[x_] := (Pi-x)/2 has the Fourier series Sum[ Sin[n x]/n , {n, 1, Infinity}]. At x = 1, you get the sum of Sin[n]/n.
This was problem 6241 in the Monthly about 25 years ago. -----
Let f(x) have Fourier transform F(v), so that f(x) = int(F(v) exp(2 pi i v x), v=-inf..inf). Consider the relation between the integral int(f(x), x=-inf..inf) = F(0) and its approximation h sum(f(nh+y), n=-inf..inf). The sum equals h sum(int(F(v) exp(2 pi i v (nh + y)), v=-inf..inf), n=-inf..inf) = h int(F(v) exp(2 pi i v y) sum(exp(2 pi i n h v))) = h int(F(v) exp(2 pi i v y) sum(delta(h v - k), k=-inf..inf)) = int(F(v) exp(2 pi i v y) sum(delta(v - k/h)) = sum(F(k/h) exp(2 pi i k y/h), k=-inf..inf). Now, if f(x) is band limited, so that F(v) = 0 for v > v0, then the discrete sum equals the exact integral whenever 1/h < v0. Now let f(x) = sinc(x) = sin(pi x)/(pi x), F(v) = rect(v) = 1 for |x| < 1/2, 0 elsewhere. For any h < 2, h sum(sinc(nh + y), n=-inf..inf) = rect(0) = 1. And let g(x) = sinc(x)^2, G(v) = tri(v) = 1 - |x| for |x| < 1, 0 elsewhere. For any h < 1, h sum(sinc(n h + y), n=-inf..inf) = tri(0) = 1. Your example is the special case, h = 1/pi, y = 0. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com