Comparing two series to calculate pi

Many series are used to calculate the value of pi.  In this blog, we compare two series, one by Gregory and another by Ramanujan.

Here is a MATLAB program that does the comparison for you.  The MATLAB program can be downloaded as a Mfile (better to download it, as single quotes from the web-post do not translate correctly with the MATLAB editor).  The html file showing the mfile and the command window output is also available.

%% COMPARING TWO SERIES FOR VALUE OF PI
% Language : Matlab 2007a
% Authors : Autar Kaw
% Last Revised : October 30, 2008
% Abstract: This program compares results for the value of
% pi using a) Gregory series and b) Ramanajun series
clc
clear all
clf
format long
disp(‘This program compares results for the value of’)
disp(‘pi using a) Gregory series and b) Ramanajun series’)
disp(‘  ‘)
disp(‘Gregory series’)
disp(‘pi=sum over k from 0 to inf of (4*((-1)^k/(2*k+1))’)
disp(‘  ‘)
disp(‘Ramanajun Series’)
disp(‘1/pi=sum over k from 0 to infinity of 2*sqrt(2)/9801*((4k)!*(1103+26390k)/(k!)^4*396^(4*k))’)

%% INPUTS.
%If you want to experiment this the only parameter
% you should and can change.
% Maximum number of terms
n=30;

%% PROGRAM

%% GREGORY SERIES
pi_gregory=0;
for i=1:1:n
pi_gregory=pi_gregory+(-1)^(i+1)*4*(1/(2*i-1));
pi_gregory_array(i)=pi_gregory;
end

%% RAMANUJAN SERIES
pi_ram=0;
for i=0:1:n-1
pi_ram=pi_ram+2*sqrt(2)/9801.0*(factorial(4*i))*(1103.0+26390.0*i)/((factorial(i)^4)*(396)^(4*i));
pi_ram_array(i+1)=1/pi_ram;
end

%% THE OUTPUT
disp(‘ ‘)
fprintf(‘\nNumber of Terms = %g’,n)
fprintf(‘\nGregory Series Value = %g’,pi_gregory)
fprintf(‘\nRamanujan Series Value = %g’,1/pi_ram)
disp( ‘   ‘)

%% PLOTTING THE TWO SERIES AS A FUNCTION OF TERMS
x=1:1:n;
hold on
xlabel(‘Number of terms’)
ylabel(‘Value of pi’)
title(‘Comparing Gregory and Ramanujan series’)
plot(x,pi_gregory_array,’color’,’blue’,’LineWidth’,2)
hold on
plot(x,pi_ram_array,’color’,’black’,’LineWidth’,2)
legend(‘Gregory Series’,’Ramanajun Series’,1)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu.

An abridged (for low cost) book on Numerical Methods with Applications will be in print (includes problem sets, TOC, index) on December 10, 2008 and available at lulu storefront.

An automatic integrator using Trapezoidal rule

How would you know how many segments to use in a Trapezoidal rule of integration to get an accurate value of the integral?  This can be done by applying the Trapezoidal rule for 1 segment rule, then 2 segment rule, followed by 4 segment rule and so on.  As soon as the absolute relative approximate error (page 5-6) between the consecutive answers becomes less than the pre-specified tolerance chosen by the user, you would have your integral within the accuracy you desired.

Here is a MATLAB program that does that for you.  The MATLAB program that can be downloaded at http://numericalmethods.eng.usf.edu/blog/trapezoidal_rule_automatic.m (better to download it as single quotes from the web-post do not translate correctly with the MATLAB editor).  The html file showing the mfile and the command window output is here: http://numericalmethods.eng.usf.edu/blog/html/trapezoidal_rule_automatic.html

% Simulation : Using Trapezoidal rule as an automatic integrator

% Language : Matlab 2007a

% Authors : Autar Kaw, http://numericalmethods.eng.usf.edu

% Mfile available at
% http://numericalmethods.eng.usf.edu/blog/trapezoidal_rule_automatic.m

% Last Revised : October 12, 2008

% Abstract: This program uses multiple-segment Trapezoidal
% rule to integrate f(x) from x=a to x=b within a pre-specified tolerance

clc
clear all

disp(‘This program uses multiple-segment Trapezoidal rule as an automatic integrator’)
disp(‘to integrate f(x) from x=a to x=b within a pre-specified tolerance’)
disp(‘ ‘)
disp(‘Author: Autar K Kaw.’)
disp(‘https://autarkaw.wordpress.com&#8217;)
disp(‘http://numericalmethods.eng.usf.edu&#8217;)
disp(‘ ‘)

%INPUTS.  If you want to experiment, these are the only variables
% you should and can change.
% a = Lower limit of integration
% b = Upper limit of integration
% nmax = Maximum number of segments
% tolerance = pre-specified tolerance in percentage
% f = inline function as integrand
a=5.3;
b=10.7;
nmax=20000;
tolerance=0.005;
f=inline(‘exp(x)*sin(2*x)’);

% SIMULATION
disp(‘INPUTS’)
func=[‘     The integrand is =’ char(f)];
disp(func)
fprintf(‘     Lower limit of integration, a= %g’,a)
fprintf(‘\n     Upper limit of integration, b= %g’,b)
fprintf(‘\n     Maximum number of segments, nmax = %g’,nmax)
fprintf(‘\n     Pre-specified percentage tolerance, eps = %g’,tolerance)
disp(‘  ‘)
disp(‘  ‘)

% Doing the automatic integration
% Calculating the integral using 1-segment rule
previous_integral=(b-a)/2*(f(a)+f(b));
% Initializing ea as greater than pre-specified tolerance for loop to work
ea=2*tolerance;
% Starting with 2-segments inside the while loop
n=2;
while (ea>tolerance) & (n<=nmax)
h=(b-a)/n;
% Keeping track of used number of segments
nused=n;
current_integral=0;
for i=1:1:n-1
current_integral=current_integral+f(a+i*h);
end
current_integral=2*current_integral+f(a)+f(b);
current_integral=(b-a)/(2*n)*current_integral;
% Calculating the absolute relative approximate error
ea = abs((current_integral-previous_integral)/current_integral)*100;
previous_integral=current_integral;
% Doubling the number of segments for next estimate of the integral
n=n*2;
end

disp(‘OUTPUTS’)
fprintf(‘      Number of segments used  =%g’, nused)
fprintf(‘\n      Approximate value of integral is =%g’,current_integral)
fprintf(‘\n      Absolute percentage relative approximate error =%g’, ea)
if (ea>tolerance)
disp(‘  ‘)
disp(‘  ‘)
disp(‘     NOTE: The value of integral is not within the pre-specified tolerance’)
end
disp(‘  ‘)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu.

An abridged (for low cost) book on Numerical Methods with Applications will be in print (includes problem sets, TOC, index) on December 10, 2008 and available at lulu storefront.

Another improper integral solved using trapezoidal rule

In a previous post, I showed how Trapezoidal rule can be used to solve improper integrals.  The example used in the post was an improper integral with an infinite interval of integration.

In an example in this post, we use Trapezoidal rule to solve an improper integral where the integrand becomes infinite.  The integral is $\int_{0}^{b} 1/sqrt{x} dx$.  The integrand becomes infinite at x=0.  Since x=0 would be one of the points where the integrand will be sought by the multiple-segment Trapezoidal rule, we choose the value of the integrand at x=0 to be zero (any other value would do too – a better assumption would be f(h), where h is the segment width in the multiple-segment Trapezoidal rule).

Here is a MATLAB program that shows you the exact value of the integral and then compares it with the multiple-segment Trapezoidal rule.  The convergence is slow but you can integrate improper integrals using Trapezoidal rule.

The MATLAB program that can be downloaded at http://numericalmethods.eng.usf.edu/blog/trapezoidal_improper_sqrtx.m (better to download it as single quotes from the web-post do not translate correctly with the MATLAB editor).  The html file showing the mfile and the command window output is here: http://numericalmethods.eng.usf.edu/blog/html/trapezoidal_improper_sqrtx.html

% Simulation : Can I use Trapezoidal rule for an improper integral?

% Language : Matlab 2007a

% Authors : Autar Kaw, http://numericalmethods.eng.usf.edu

% Mfile available at
% http://numericalmethods.eng.usf.edu/blog/trapezoidal_improper_sqrt.m

% Last Revised : October 8, 2008

% Abstract: This program shows use of multiple segment Trapezoidal
% rule to integrate 1/sqrt(x) from x=0 to b, b>0.

clc
clear all

disp(‘This program shows the convergence of getting the value of ‘)
disp(‘an improper integral using multiple segment Trapezoidal rule’)
disp(‘Author: Autar K Kaw.  autarkaw.wordpress.com’)

%INPUTS.  If you want to experiment, these are the only variables
% you should and can change.
% b  = Upper limit of integration
% m = Maximum number of segments is 2^m
b=9;
m=14;

% SIMULATION
fprintf(‘\nFinding the integral of 1/sqrt(x) with limits of integration as x=0 to x=%g’,b)

% EXACT VALUE OF INTEGRAL
% integrand 1/sqrt(x)
syms x
f=1/sqrt(x);
a=0;
valexact=double(int(f,x,a,b));
fprintf(‘\n\nExact value of integral = %f’,valexact)
disp( ‘  ‘)

f=inline(‘1/sqrt(x)’);
%finding value of the integral using 16,…2^m segments
for k=4:1:m
n=2^k;
h=(b-a)/n;
sum=0;
for i=1:1:n-1
sum=sum+f(a+i*h);
end
% See below how f(a) is not added as f(a)=infinity.  Instead we
% use a value of f(a)=0.  How can we do that? Because as per integral calculus,
% using a different value of the function at one point or
% at finite number of points does not change the value of the
% integral.
sum=2*sum+0+f(b);
sum=(b-a)/(2*n)*sum;
fprintf(‘\nApproximate value of integral =%f with %g segments’,sum,n)
end
disp(‘  ‘)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu.

An abridged (for low cost) book on Numerical Methods with Applications will be in print (includes problem sets, TOC, index) on December 10, 2008 and available at lulu storefront.