RE: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
Fascinating! I had a different objective in mind when I wrote the problem. My goal was to reach the fairly difficult pi^2/6 result using series manipulation, starting from a definition of pi based on 1 - 1/3 + 1/5 - ... or the like. Ideally, one would avoid calculus and complex integrations and the theory of dilogarithms, or Euler's trick with the power series for sin x. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Robert Baillie Sent: Wednesday, February 22, 2006 8:39 AM To: math-fun Subject: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8 HAKMEM 239 has this interesting question: "119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?" Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8? Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 --++ ... Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + ... Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan's "Advanced Calculus". The series for Pi/Sqrt(8) comes from the Fourier series for f1(x) = 0 for -Pi <= x < 0, f(x) = 1 for 0 <= x <= Pi. The series is f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}] f[Pi/4] gives the series for Pi/Sqrt[8]. The series for Pi^2/8 comes from the Fourier series for g1(x)= Abs[x], -Pi <= x <= Pi. g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}] g[0] gives the series for Pi^2/8. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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. ----- Schroeppel, Richard wrote:
Fascinating!
I had a different objective in mind when I wrote the problem. My goal was to reach the fairly difficult pi^2/6 result using series manipulation, starting from a definition of pi based on 1 - 1/3 + 1/5 - ... or the like. Ideally, one would avoid calculus and complex integrations and the theory of dilogarithms, or Euler's trick with the power series for sin x.
Rich
--- 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
participants (3)
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Eugene Salamin -
Robert Baillie -
Schroeppel, Richard