Shanks transformation: Example 2

To illustrate the Shanks transformation we will use the following series:

$\displaystyle \ln(x) = (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\dots$

$\displaystyle \ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-1)^n}{n}$

This series is converging very slowly. To illustate this we will compute ln(2):

$\displaystyle \ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots$

$\displaystyle \ln(2) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}$

The corresponding partial sum of ln(2) is:

$\displaystyle S_N = \sum_{n=1}^{N} (-1)^{n+1} \frac{1}{n}$

Let calculate some terms of this partial sum:

$\displaystyle S_1 = 1 \quad S_{10} = 0.6456$

$\displaystyle S_2 = 0.5 \quad S_{400} = 0.6919$

$\displaystyle S_3 = 0.8333 \quad \ln(2) = 0.6931$

Perform the Shanks transform (using S_N = S₄):

$\displaystyle S_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{12}{12} - \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{7}{12}$

$\displaystyle S_4^2 = \frac{49}{144}$

$\displaystyle S_5 = \frac{7}{12} + \frac{1}{5} = \frac{35}{60} + \frac{12}{60} = \frac{47}{60}$

$\displaystyle S_3 = 1 -\frac{1}{2} + \frac{1}{3} = \frac{6}{6} - \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

$\displaystyle \mathcal{S} = \frac{S_N^2 - S_{N+1}S_{N-1}}{2S_N -S_{N+1}- S_{N-1}}$

$\displaystyle \mathcal{S} = \frac{\frac{49}{144} - (\frac{47}{60} \frac{5}{6})}{\frac{14}{12} - \frac{47}{60} -\frac{5}{6}}$

$\displaystyle \mathcal{S} = \frac{\frac{49}{144} - \frac{235}{360}}{\frac{70}{60} - \frac{47}{60} -\frac{50}{60}}$

$\displaystyle \mathcal{S} = \frac{-225/720}{-27/60}$

$\displaystyle \mathcal{S} = 0.6944$

Since $S_{358} = 0.6918$ and $\ln(2) = 0.6931$ It is necessary to calculate up to 358 terms of the partial sum S_N to obtain a relative error equal to that of the Shanks transformation $\mathcal{S}$ based on the first 4 terms.

Shanks transformation: Example 2

Comments

Leave a comment Cancel reply

More posts