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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: 7,664 Thanks: 2644 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 $\pi$: $$\begin{matrix} n & S_n & \text{1st Richardson} & \text{2nd Richardson} & \text{3rd Richardson} & \text{4th Richardson} \\ 1 & 4.000000 & 3.200000 & 3.136508 & 3.140445 & 3.141557 \\ 3 & 3.466667 & 3.149206 & 3.139882 & 3.141434 & 3.141600 \\ 5 & 3.339683 & 3.143878 & 3.140917 & 3.141555 & 3.141596 \\ 7 & 3.283738 & 3.142562 & 3.141265 & 3.141580 & 3.141594 \end{matrix}$$ Tags acceleration, day, series, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post atari Calculus 4 March 6th, 2016 08:59 AM jiasyuen Applied Math 5 May 12th, 2015 06:31 PM Mike7remblay Physics 3 February 1st, 2012 06:48 PM imcutenfresa Calculus 5 October 7th, 2009 02:51 PM imcutenfresa Calculus 2 September 15th, 2009 04:11 PM

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