[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.
--- Robert Baillie <rjbaillie@frii.com> wrote:
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.
Yes. Fourier series exchanges pointwise multiplication with convolution. The square wave convolved with itself gives the triangle wave, so the square of the Fourier coefficient of the former equals the Fourier coefficient of the latter. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
* Robert Baillie <rjbaillie@frii.com> [Feb 22. 2006 16:39]:
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 + ... ?"
[...]
Somewhat similar vein (but finite sums): sum(k)^2 == sum(k^3) which generalizes to (sum_{d\N}{nu(d)})^2 == sum_{d\N}{nu(d)^3} nu(x) counts divisors of x. N a perfect power gives former. (p.567 of my scribblings) I also like this one (specialized Clausen's product formula): F([1/4,1/4];[1];z)^2==F([1/2,1/2,1/2];[1,1],z) z a sixth power (e.g. z=1) The relation is an identity between the square of a sum of squares and a sum of cubes. (p.502) -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
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Robert Baillie