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 March 14th, 2016, 07:04 AM #1 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 Acceleration of a sum of a series for Pi Day . For the series $\displaystyle \ \ \ 4\bigg(1 \ - \ \frac{1}{3} \ + \ \frac{1}{5} \ - \ \frac{1}{7} \ + \ \ ... \ \bigg), \ \$ 628 terms are needed to equal to pi correct to two decimal places for the first time for an even number of terms. (An even number of terms always gives a sum that is less than pi.) When 628 terms are used, the sum equals 3.1400002979... Source: WolframAlpha - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - I "discovered" a tweak for a related sum that allows for only eight terms to sum to two correct decimal places for pi. $\displaystyle \ \ \ 4\bigg[1 \ - \ \frac{1}{3} \ + \ \frac{1}{5} \ - \ \frac{1}{7} \ + \ \frac{1}{9} \ - \ \frac{1}{11} \ + \ \frac{3}{4}\bigg(\frac{1}{13}\bigg) \ - \ \frac{1}{4}\bigg(\frac{1}{15}\bigg) \ \bigg] \ \approx \ \$3.1401487 The coefficients on the last two terms can be used on many other convergent alternating series to accelerate their sums. Thanks from 123qwerty
 March 14th, 2016, 09:29 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra Here's a different approach Richardson Extrapolation. Let $S = \lim \limits_{n \to \infty} S_n$ where $S_n = 4\sum \limits_{k=1}^n {(-1)^{n+1} \over 2n+1}$ Then we assume that for odd $n$ (to get a monotonic sequence) $S_n = S + {a_1 \over n} + {a_2 \over n^2} + \ldots$. Now, if $n$ is very large (as we would like for our estimate of $\pi$) then $\frac1n$ is very small and so we can write \begin{aligned}S_n &\approx S + {a_1 \over n} \\ S_{n+2} &\approx S + {a_1 \over n+2} \end{aligned} And now we have two (approximate) equations in two unknowns ($S$ and $a_1$) so we can eliminate $a_1$ to get $$(n+2)S_{n+1} - (n+2)S \approx nS_n - nS \implies S \approx \frac12\big((n+2)S_{n+2}-nS_n\big)$$ This is the first Richardson extrapolation for the series. We can improve the approximation by taking the first three terms of the series for $S_n$, $S_{n+2}$ and $S_{n+4}$ and eliminating $a_1$ and $a_2$. This is the second Richardson extrapolation $$S \approx \frac14 \cdot \frac12 \big( (n+4)^2S_{n+4} - 2(n+2)^2S_{n+2} + n^2S_n \big)$$ We can then get the following estimates for $\pi$: $$\begin{matrix} n & S_n & \text{1st Richardson} & \text{2nd Richardson} & \text{3rd Richardson} & \text{4th Richardson} \\ 1 & 4.000000 & 3.200000 & 3.136508 & 3.140445 & 3.141557 \\ 3 & 3.466667 & 3.149206 & 3.139882 & 3.141434 & 3.141600 \\ 5 & 3.339683 & 3.143878 & 3.140917 & 3.141555 & 3.141596 \\ 7 & 3.283738 & 3.142562 & 3.141265 & 3.141580 & 3.141594 \end{matrix}$$

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