## Example to show how numerical ODE solutions can be used to find integrals

In a previous post, I enumerated how we can use numerical ODE techniques like Euler and Runge-Kutta methods to find approximate value of definite integrals. Here is an example. Be sure to do the exercises at the end of the post to appreciate the procedure.  _____________________________________________________

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## Comparing Runge-Kutta 2nd order methods

Many a times, students ask me

Which of the Runge-Kutta 2nd order methods gives the most accurate answer to solving a first order ODE?

dy/dx=f(x,y), y(0)=y0

There is no direct answer, although Ralston’s method gives a minimum bound for the truncation error (Ralston, A., Runge-Kutta Methods with Minimum Error Bounds, Match. Compu., Vol 16, page 431, 1962).

They also ask me if using the first three terms of the Taylor series would give a more accurate answer if we calculate $f^{\prime}(x,y)$ symbolically.

The equations for the four methods are given below Here is the comparison graph for

dy/dx=sin(5*x)-0.4*y, y(0)=5

with

step size of h=1.1 and

step size of h=0.55 This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu

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## Can I use numerical solution of ODE techniques to do numerical integration?

Yes.

If you are finding the value of the $y=\int_{a}^{b} f(x) dx$, then we can solve the integral as an ordinary differential equation as

dy/dx=f(x), y(a)=0

We can now use any of the numerical techniques such as Euler’s methods and Runge-Kutta methods to find the value of y(b) which would be the approximate value of the integral. Use exact techniques of solving linear ODEs with fixed coefficients such as Laplace transforms, and you can have the exact value of the integral.

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## Time of death – a classic ODE problem

One of the classical applied problems in ordinary differential equations is that of finding the time of death of a homicide victim.

The estimation of time of death is generally based on the temperature of the body at two times – 1) when the victim is found and 2) then a few hours later. Assuming the ambient temperature stays the same and the body is treated as a lumped system, one can use a simple linear ODE to solve the problem.

It is somewhat an inverse problem as we are trying to find the value of the independent variable – the time of death. Here is a solved problem (a pdf version also available).   _____________________________________________________

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## 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.

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## 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?

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## A metric for measuring wildness of a college football season

A Metric to Quantify the Topsy-Turvyness (Wildness) of a College Football Season

The 2008 college football season is almost here, and news media, sports commentators, and bloggers will hope for something to hype about. Luckily for them, the 2007 season did give them something to talk about; you would be hard pressed to recall a more topsy-turvy season. Ranked teams regularly lost to low-ranked and unranked teams.

In just Week#1 of the 2007 season, Associated Press (AP) No. 5 team University of Michigan lost to an unranked Division II team – Appalachian State. The Associated Press wasted no time in booting out Michigan out of the Top AP 25. Two weeks later, No. 11 UCLA lost to unranked Utah by a wide margin of 44-6. UCLA also met the same fate as Michigan; UCLA was dropped from the AP Top 25.

The topsy-turvyness continued in the season, especially for No. 2 ranked teams. The University of South Florida, where I work, was ranked No. 2 when they lost to unranked Rutgers 30-27 in Week#8. This was the same week when three other teams (South Carolina, Kentucky, and California) ranked in the Top 10 of the AP poll also lost their games.

To top off the season, for the first time in history of the Bowl Championship Series (BCS), the title bowl game had a team (Louisiana State University (LSU)) with two regular season losses, and LSU ended up winning the national championship.

Although many ranted and raved about the anecdotal evidence of a topsy-turvy season, is it possible that the media and fans over exaggerated the topsy-turvyness of the 2007 college football season. Were there other seasons that were more topsy-turvy than 2007?

To answer this question scientifically, this article proposes an algorithm to quantify the topsy-turvyness of the college football season. The author does not know of any previous literature that has attempted to develop a metric that quantifies the topsy-turvyness of any sport, which is ranked regularly in the season.

The TT factor

The Topsy-Turvy Factor (TT factor) is a metric that quantifies the topsy-turvyness of a college football season. Two different TT factors are calculated: one for each of week of the season, referred to as the Week TT factor, and one for the cumulative topsy-turvyness at the end of each week of the season, referred to as the Season TT factor.

The method to find the Week TT Factor is based on comparing the AP Top 25 poll rankings of schools from the current week to that of the previous week. The difference in the rankings of each school in the AP Top 25 from the current week to the previous week is squared. How do we account for teams that fall out of the rankings? A team that gets unranked from the previous week is assumed as having become the No. 26 team in the current week. All the squares of the differences in the rankings are then added together and normalized on a scale of 0-100.

The other TT factor, the Season TT factor is also calculated for the end of each week to gage how topsy-turvy the season has been so far. The Season TT factor is calculated using weighted averages of the Week TT factors. As the season progresses, the Week TT factors are given more weight in the calculation of the Season TT factor because toward the end of the season, an upset of a ranked team is more topsy-turvy than an upset in the beginning of the season when the strength of a ranked team is less established.

Season End of Season TT factor
2007 ……………..60
2006 ……………..46
2005 ……………..48
2004 ……………..40
2003 ……………. 55
2002 ……………. 48

Table 1: End of Season TT factors for 2002-2007 Seasons

Table 1 shows the end of Season TT factor of the last six football seasons. It is clear that 2007 was the most topsy-turvy season in recent history, with the 2003 season not too far behind. In contrast, the 2004 season was the least topsy-turvy.

Read the complete paper including formulas and detailed analysis at

http://www.eng.usf.edu/~kaw/TT_factor_paper_media.pdf

Can you write a program in MATLAB or any other language to find the Week TT factors and the Season TT factors?

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