My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Thanks Tree1Thanks
  • 1 Post By Math Message Board tutor
Reply
 
LinkBack Thread Tools Display Modes
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
Math Message Board tutor is offline  
 
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 $\pi$:
$$\begin{matrix} n & S_n & \text{1st Richardson} & \text{2nd Richardson} & \text{3rd Richardson} & \text{4th Richardson} \\
1 & 4.000000 & 3.200000 & 3.136508 & 3.140445 & 3.141557 \\
3 & 3.466667 & 3.149206 & 3.139882 & 3.141434 & 3.141600 \\
5 & 3.339683 & 3.143878 & 3.140917 & 3.141555 & 3.141596 \\
7 & 3.283738 & 3.142562 & 3.141265 & 3.141580 & 3.141594 \end{matrix}$$
v8archie is offline  
Reply

  My Math Forum > Math Forums > Math

Tags
acceleration, day, series, sum



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Acceleration atari Calculus 4 March 6th, 2016 08:59 AM
Constant Acceleration jiasyuen Applied Math 5 May 12th, 2015 06:31 PM
acceleration Mike7remblay Physics 3 February 1st, 2012 06:48 PM
acceleration? imcutenfresa Calculus 5 October 7th, 2009 02:51 PM
Velocity and acceleration imcutenfresa Calculus 2 September 15th, 2009 04:11 PM





Copyright © 2019 My Math Forum. All rights reserved.