# Wallis product.

• Aug 16th 2010, 07:46 PM
roza
Wallis product.
• Aug 16th 2010, 08:53 PM
Vlasev
The first part is straightforward, just plug in x = 1. For the second part, what is meant by compare? Are you asked to show which one converges faster per the number of terms?

Btw, the Wallis product for pi is

$$\displaystyle \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}$$
• Aug 16th 2010, 10:24 PM
chisigma
The Leibnitz series is...

$\displaystyle \frac{\pi}{4} = (1-\frac{1}{3}) + (\frac{1}{5} - \frac{1}{7}) + ... + (\frac{1}{2n-1} - \frac{1}{2n+1}) + ...$ (1)

... so that after 2n iterations the last summed term is $\frac{1}{2n-1} - \frac{1}{2n+1} = \frac{2}{4n^{2}-1}$...

The Wallis product [in logarithmic version...] is...

$\displaystyle \ln \frac{\pi}{2} = \ln (1+\frac{1}{3}) + \ln (1 + \frac{1}{15}) + ... + \ln(1+\frac{1}{4n^{2}-1}) + ...$ (2)

... so that after 2n iterations the last summed term is $\ln (1+ \frac{1}{4 n^{2} -1}) \sim \frac{1}{4 n^{2}-1}$ ...

The two methods seem to be comparable in speed: both are very slow (Worried) ...

Kind regards

$\chi$ $\sigma$