Code to estimate pi

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John Jamison
John Jamison 2017-2-1
评论： Rena Berman 2017-2-1
see above image.
Please help for both calculations.
thanks
  2 个评论
Image Analyst
Image Analyst 2017-2-1
John, somehow the formula image got deleted. Can you put it back?
Rena Berman
Rena Berman 2017-2-1
(Answers dev) Restored edit

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采纳的回答

Star Strider
Star Strider 2017-2-1
Use the MATLAB vector capabilities:
N = 1E+8; % Number Of Terms
pie1 = sqrt(sum(6./[1:N].^2))
pie2 = sqrt(8 + sum(16./((2*[1:N] - 1).^2 .* (2*[1:N] + 1).^2)))
pie1 =
3.1415926440655
pie2 =
3.14159265358975

更多回答（1 个）

Image Analyst
Image Analyst 2017-2-1
Here's code that gets to pi even faster. Dubbed the world's ugliest formula by mathematicians, it's so accurate after one iteration that MATLAB can't even detect a difference after just one term! Very interesting article if you want to read it. If someone can show me how to get the true difference, which is less than 1E-16, I'd appreciate it. By the way, most of the mathematicians agreed on which equations were beautiful and which were ugly, with Euler's identity, 1+eiπ=0, consistently rated the most attractive equation in the lot.
% Computes Ramanujan's formula for pi.
% http://www.scientificamerican.com/article/equations-are-art-inside-a-mathematicians-brain/
% On the bottom of the heap of beautiful formulas, mathematicians consistently rated
% Srinivasa Ramanujan's infinite series for estimating pi as the most ugly.
% Problem: convergence it so fast that double precision is not
% precise enough to show the convergence curve.
clc; % Clear the command window.
close all; % Close all figures.
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 24;
% Compute the sum for a bunch of different number of terms
% in the series to see the shape of the error curve
% as it converges towards the true value of pi.
for kMax = 1 : 10
thisSum = 0;
for k = 0 : kMax
numerator = factorial(4 * k) * (1103 + 26390*k);
denominator = factorial(k)^4 * 396^(4*k);
thisTerm = numerator / denominator;
thisSum = thisSum + thisTerm;
end
theConstant = 2 * sqrt(2) / 9801;
oneOverPi = theConstant * thisSum;
estimatedPi(kMax) = 1 / oneOverPi
theError(kMax) = estimatedPi(kMax) - pi
end
% Plot the value of estimated pi.
subplot(2, 1, 1);
plot(estimatedPi, 'b*-', 'LineWidth', 2);
grid on;
title('Estimated \pi', 'FontSize', fontSize);
xlabel('Number of Terms in the Sum', 'FontSize', fontSize);
% Plot the value of the error (difference between true pi and estimated pi).
subplot(2, 1, 2);
plot(theError, 'b*-', 'LineWidth', 2);
grid on;
title('Error', 'FontSize', fontSize);
xlabel('Number of Terms in the Sum', 'FontSize', fontSize);
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Estimation of pi', 'NumberTitle', 'Off')

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