This project graphs a finite truncation of the infinite product: ...(2x-4)/-3 * (2x-2)/-1 * (2x)/1 * (2x+2)/3 * (2x+4)/5 ... for sin(πx). I derived this product by integrating my infinite sum for cot(x) : ... 1/(x-2π)+1/(x-π)+1/x+1/(x+π)+1/(x+2π) ... to give an infinite product for ln(sin(x))+ a constant , then raising e^this and dividing by the resulting constant factor to give the following infinite product for sin(x) : ...(2x-4π)/-3π * (2x-2π)/-π *(2x)/π * (2x+2π)/3π * (2x+4π)/5π ... and substituted x with πx to give this product.
As differentiating sin(πx) at 0 gives π, the x coefficient of this product when expanded should equal π (as any higher power differentiates to 0) and it is easy to show by rearranging the terms that this is in fact 2/1*2/3*4/3*4/5..., the Wallace infinite product for π ! Edit: I recently discovered a mathologer video about euler basically doing this in reverse and discovering all these pi series, albeit in a different way `\(._.)/`
