Attachment 18620

Printable View

- Aug 16th 2010, 06:46 PMrozaWallis product.
- Aug 16th 2010, 07:53 PMVlasev
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

[Math]\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}[/tex] - Aug 16th 2010, 09:24 PMchisigma
The Leibnitz series is...

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

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

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

$\displaystyle \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 $\displaystyle \ln (1+ \frac{1}{4 n^{2} -1}) \sim \frac{1}{4 n^{2}-1}$ ...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$