make you a deal, expand as a summation series and give me a closed form expression for the 3rd term (coefficient for x^3) and I'll reveal all ...
It can be written as
f(x) = A \prod_{m=1}^\oo (1 - \frac{4z^2}{(4m-1)^2})
where z = x/\pi - 1/2, and
A = \prod_{m=1}^\oo \frac{(4m-1)^2}{4m(4m-2)}
That is an even function of z. Therefore, the the coefficient of (x-\pi/2)^3 in the Taylor series around \pi/2 is zero.
Is that close enough?
Jorge: thanks for this, it really is an interesting approach. It shifts the curve by pi/2, and normalises (by A) creating a normalised even function.
You might be interested to know the following:
A = f(pi/2) = ([Gamma(1/4)]^2)/(2.pi.sqrt(pi)) = [3.3.7.7.11.11.15.15.....]/[2.4.6.8.10.12.14.16] ; where Gamma() is the gamma function
and
f(-pi/2) = (4.sqrt(pi))/([Gamma(1/4)]^2) = [1.5.5.9.9.13.13.....]/[2.4.6.8.10.12.14.....]
and
f(pi/2).f(-pi/2) = [1.3.3.5.5.7.7.9.9.11.11.15.15....]/[2.2.4.4.6.6.8.8.10.10.12.12.14.14.16....] = 2/pi
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