Stoneham's series

[Albert Chan]

(06-04-2024 07:57 PM)Thomas Klemm Wrote:  In both cases we can use the infinite product representation of the sinc function:

\(
\begin{align}
\frac {\sin \pi z}{\pi z}=\prod _{n=1}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)
\end{align}
\)

Another way, identities derived from cos(x) infinite product.

Si(pi*z)    = (1-(z/1)²) * (1-(z/2)²) * (1-(z/3)²) * (1-(z/4)²) ...
Si(pi*z/2) = (1-(z/2)²) * (1-(z/4)²) * (1-(z/6)²) * (1-(z/8)²) ...

Si(pi*z/2) * cos(pi*z/2) = Si(pi*z)
cos(pi*z/2) = (1-(z/1)²) * (1-(z/3)²) * (1-(z/5)²) * (1-(z/7)²) ...

Set z=1/2, we have:

√2/2 = (1-¼/1²) * (1-¼/3²) * (1-¼/5²) * (1-¼/7²) ...

Set z=1, we have 0 = 0 * ?. We move RHS first term to the left, to find out what ? is.
(this is why pi/4 infinite product skipped over first odd number)

cos(pi*z/2) / (1-z²) = (1-(z/3)²) * (1-(z/5)²) * (1-(z/7)²) * (1-(z/9)²) ...

limit(LHS, z=1) = limit(-sin(pi*z/2)*(pi/2) / (-2z), z=1) = pi/4

pi/4 = (1-1/3²) * (1-1/5²) * (1-1/7²) * (1-1/9²) ...