- #1
rbzima
- 84
- 0
Hey all, I'm having a difficult time seeing why the approximation of the Wallis Product equals the square root of 2pi.
I know for a fact that:
[tex]\pi\approx\left(\frac{2*4*6*8*\ldots*\left(2n\right)}{1*3*5*7*\ldots*\left(2n-1\right)}\right)\frac{1}{n}[/tex]
How then does this produce:
[tex]\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n\right)!}=\sqrt{2\pi}[/tex]
I know for a fact that:
[tex]\pi\approx\left(\frac{2*4*6*8*\ldots*\left(2n\right)}{1*3*5*7*\ldots*\left(2n-1\right)}\right)\frac{1}{n}[/tex]
How then does this produce:
[tex]\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n\right)!}=\sqrt{2\pi}[/tex]