MATLAB code for the efficient automatic integrator

In the previous post, we discussed why doubling the number of segments in the automatic integrator based on multiple-segment trapezoidal rule is more efficient than increasing the number of segments one at a time. But this advantage involves having to store the individual function values from previous calculations and then having to retrieve them properly. This drawback was circumvented very efficiently by using the formula derived in another previous post where there is no need to store individual function values.

The matlab file for finding a definite integral by directly using the multiple segment trapezoidal rule from this post is given here (matlab file, html file), while the matlab file that uses the more efficient formula from this post is given here (matlab file, html file).  Here are the inputs to the programs.

% 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=200000;
tolerance=0.000005;
f=inline(‘exp(x)*sin(2*x)’)

We ran both the program on a PC and found that the more efficient algorithm (51 seconds) ran in half the time as the other one (82 seconds).  This is expected, as only n function evaluations are made for 2n-segments rule with the efficient formula, while 2n+1 functions evaluations are made for the original formula.

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://www.youtube.com/numericalmethodsguy.

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An efficient formula for an automatic integrator based on trapezoidal rule

In the previous post, we discussed why doubling the number of segments in the automatic integrator based on multiple-segment trapezoidal rule is more efficient than increasing the number of segments one at a time. But this advantage involves having to store the individual function values from previous calculations and then having to retrieve them properly. This drawback can be circumvented very efficiently as explained below. What you will see is that there is no need to store individual function values.

Automatic Integrator Formula
Automatic Integrator Formula

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://www.youtube.com/numericalmethodsguy.

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.

Why keep doubling the segments for an automatic integrator based on Trapezoidal rule?

Automatic Integrator
Automatic Integrator
Automatic Integrator
Automatic Integrator

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://www.youtube.com/numericalmethodsguy.  

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.

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’)
disp(‘http://numericalmethods.eng.usf.edu’)
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.

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.

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.

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.

Is it just a coincidence – true error in multiple segment Trapezoidal rule gets approximately quartered as the number of segments is doubled?

Look at the table below. This is a table that shows the approximate value of the integral

\int_{8}^{30} 2000 ln \frac{140000}{140000-2100t}-9.8t dt 

as a function of the number of segments used in the Trapezoidal rule and the corresponding true error.

n

Value

Et

1

11868

-807

2

11266

-205

3

11153

-91.4

4

11113

-51.5

5

11094

-33.0

6

11084

-22.9

7

11078

-16.8

8

11074

-12.9

The true error for n=1 is -807 and for n=2 is -205. As you can see the quarter of -807 is approximately -201.75 and close to the true error for n=2. Is this a coincidence?

Look at the true error for n=2 which is -205 and for n=4 is -51.5. As you can see the quarter of -205 is approximately -51.75 and close to the value of the true error for n=4. Is this a coincidence?

No. This is because the true error in a single segment trapezoidal rule is

\frac{(b-a)^3}{12} f^{\prime\prime} (c)

where c is some point not known but in the domain [a,b] of \int_{a}^{b} f(x) dx. It can be then shown (see page 14 of this pdf file for full proof) that for the multiple segment trapezoidal rule, the true error is

\frac{(b-a)^3}{12n^2} f^{\prime\prime}

where the f^{\prime\prime} is an average value of the second derivative of the function f(x) calculated at some point within each of the n segments. Since a and b are constant, and f^{\prime\prime} becomes almost a constant as n increases, the true error is approximately inversely proportional to the square of the number of segments.

Note to the reader: Develop a similar table as given above for an integral of your choice and see it for yourself if the true error gets approximately quartered as the number of segments is doubled.

_____________________________________________________

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

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.

Can I use Trapezoidal rule to calculate an improper integral?

For example, \int_{a}^{infinity} e^{-x} dx is an improper integral which can be calculated exactly as e^{-a}.

Can we solve this integral by multiple segment Trapezoidal rule when we already know that the upper limit is infinity?

Yes, we can solve it in spite of the upper limit being infinity. We first need to make a change of variables such as t=\frac{1}{x}, then dx=-\frac{1}{t^2}dt giving \int_{1/a}^{0} e^{-1/t} (-1/t^2) dt

But even for this integral we have issues as the integrand gives division by zero problems at t=0 (although the limit itself is zero at t=0).

The MATLAB program that can be downloaded at http://numericalmethods.eng.usf.edu/blog/trapezoidal_improper.m (better to download it as the single quotes do not translate well with matlab editor) shows that we get accurate results for transformed integral with 256 segments.
% 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.m
% Last Revised : July 15, 2008
% Abstract: This program shows use of multiple segment Trapezoidal
% rule to integrate exp(-x) from x=a to infinity, where a>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.
% a = Lower limit of integration
a=1/2;

fprintf(‘\nFinding the integral of exp(-x) from a to infinity where a=%g’,a)

% SIMULATION
syms x
% integrand exp(-x)
f=exp(-x);
valexact=double(int(f,x,a,inf));
fprintf(‘\n\nExact value of integral = %f’,valexact)
disp( ‘ ‘)

% Transformed integrand by change of variables
ft=inline(‘exp(-1/t)*(-1/t^2)’);
% Limits of integration
at=1/a;
b=0;

%finding value of the integral using 16, 32, 64, 128 and 256 segments
for k=4:1:8
n=2^k;
h=(b-at)/n;
sum=0;
for i=1:1:n-1
sum=sum+ft(at+i*h);
end
% See how f(b) is not added as f(b)=infinity. But
% using a different value of the function at one point or
% finite number of points does not change the value of the
% integral. We are substituting f(b)=0
sum=2*sum+ft(at)+0;
sum=(b-at)/(2*n)*sum;
fprintf(‘\nApproximate value of integral =%f with %g segments’,sum,n)
end
disp(‘ ‘)
This program shows the convergence of getting the value of
an improper integral using multiple segment Trapezoidal rule
Author: Autar K Kaw. autarkaw.wordpress.com

Finding the integral of exp(-x) from a to infinity where a=0.5

Exact value of integral = 0.606531

Approximate value of integral =0.605971 with 16 segments
Approximate value of integral =0.606496 with 32 segments
Approximate value of integral =0.606521 with 64 segments
Approximate value of integral =0.606528 with 128 segments
Approximate value of integral =0.606530 with 256 segments

Published with MATLAB® 7.3

Now someone may rightly point out that it works well because the limit of the integrand is zero at the lower limit of integration for the above mentioned integral.

So go ahead, modify the program for doing this improper integral \int_{0}^{4} \frac{1}{\sqrt {x}} dx and see how well it works. In this case the integrand is singular at x=0 and one would have to assume the function to be a number other than infinity at x=0 for applying the multiple segment Trapezoidal rule.

How many segments do you need to get an accurate answer for this integral?

_____________________________________________________

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

Subscribe to the blog via a reader or email to stay updated with this blog. Let the information follow you.