Shanks transformation: Example 1

Let’s transform the geometric series:

$\displaystyle 1+x+x^2+x^3+x^4+\dots$

using the corresponding partial sum S₂:

$\displaystyle 1+x+x^2$

$\displaystyle S_2^2 = (1+x+x^2)^2 = 1 + 2x + 3x^2 + 2x^3 + x^4$

$\displaystyle S_3 S_1 = (1+x+x^2+x^3)(1+x) = 1 + 2x + 2x^2 + 2x^3 + x^4$

$\displaystyle 2S_2 = 2 + 2x + 2x^2$

$\displaystyle S_3 = 1+x+x^2+x^3$

$\displaystyle S_1 = 1 + x$

$\displaystyle \mathcal{S} = \frac{(1+2x+3x^2+2x^3+x^4) - (1+2x+2x^2+2x^3+x^4)}{(2+2x +2x^2) - (1+x+x^2+x^3) - (1+x)}$

$\displaystyle \mathcal{S} = \frac{x^2}{x^2-x^3}$

$\displaystyle \mathcal{S} = \frac{1}{1-x}$

Therefore the Shanks transformation of the geometric series is correctly $\frac{1}{1-x}$ . A similar result can be obtain for the series:

$\displaystyle 1-x+x^2-x^3+\dots$

The corresponding Shanks transformation is $\frac{1}{1+x}$ .

Remarkably, the Shanks transformation yields the exact sum of the geometric series using only a few terms, since the error model $S_N = S + \alpha \beta^N$ perfectly describes such a series. The Gregory series for $\pi$ is notoriously slow, requiring five hundred terms just to calculate two correct decimal places. However, by applying the Shanks transformation, we can accelerate this convergence dramatically, extracting high-precision results from only a few initial partial sums. This demonstration highlights how a simple mathematical shift can transform an inefficient infinite series into a powerful computational tool.

Shanks transformation: Example 1

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