Length of a curve experiment

In a previous post, I mentioned that I have incorporated experiments in my Numerical Methods course. Here I will discuss the second experiment.

Length of the curve experimentIn this experiment, we find the length of two curves generated from the same points – one curve is a polynomial interpolant and another one is a spline interpolant.

Motivation behind the experiment: In 1901, Runge conducted a numerical experiment to show that higher order interpolation is a bad idea. It was shown that as you use higher order interpolants to approximate f(x)=1/(1+25x2) in [-1,1], the differences between the original function and the interpolants becomes worse. This concept also becomes the basis why we use splines rather than polynomial interpolation to find smooth paths to travel through several discrete points.

What do students do in the lab: A flexible curve (see Figure) of length 12″ made of lead-core construction with graduations in both millimeters and inches is provided. The student needs to draw a curve similar in shape to the Runge’s curve on the provided graphing paper as shown. It just needs to be similar in shape – the student can make the x-domain shorter and the maximum y-value larger or vice-versa. The student just needs to make sure that there is a one-to-one correspondence of values.

Assigned Exercises: Use MATLAB to solve problems (3 thru 6). Use comments, display commands and fprintf statements, sensible variable names and units to explain your work. Staple all the work in the following sequence.

  1. Signed typed affidavit sheet.
  2. Attach the plot you drew in the class. Choose several points (at least nine – do not need to be symmetric) along the curve, including the end points. Write out the co-ordinates on the graphing paper curve as shown in the figure.
  3. Find the polynomial interpolant that curve fits the data. Output the coefficients of the polynomial.
  4. Find the cubic spline interpolant that curve fits the data. Just show the work in the mfile.
  5. Illustrate and show the individual points, polynomial and cubic spline interpolants on a single plot.
  6. Find the length of the two interpolants – the polynomial and the spline interpolant. Calculate the relative difference between the length of each interpolant and the actual length of the flexible curve.
  7. In 100-200 words, type out your conclusions using a word processor. Any formulas should be shown using an equation editor. Any sketches need to be drawn using a drawing software such as Word Drawing. Any plots can be imported from MATLAB.

Where to buy the items for the experiment:

  1. Flexible curves – I bought these via internet at Art City. The brand name is Alvin Tru-Flex Graduated Flexible Curves. Prices range from $5 to $12. Shipping and handling is extra – approximately $6 plus 6% of the price. You may want to buy several 12″ and 16″ flexible curves. I had to send a query to the vendor when I did not receive them within a couple of weeks. Alternatively, call your local Art Store and see if they have them.
  2. Engineering Graph Paper – Staples or Office Depot. Costs about $12 for a pack for 100-sheet pad.
  3. Pencil – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore. $1 for a dozen.
  4. Scale – Anywhere – My favorite store is the 24-hour Wal-Mart Superstore. $1 per unit.

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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A legend used in the movie “The Happening”

Well M. Night Shyamalan may have made another disappointing movie – The Happening, but I somewhat liked it. I would give it a grade of B.

In the movie, John Leguzomo’s character, a math teacher, is distracting his fellow panicking passenger in the Jeep with a mathematical question. The question he asks her is if he gave her a penny on Day 1 of the month, two pennies on Day 2 of the month, four pennies on Day 3 of the month, and so on, how much would money would she have after a month. She shouts $300 or some odd number like that. But, do you know that the amount is actually more than a 10 million dollars (Thanks to a student who mentioned that it was a penny that John offered on the first day, not a dollar – sometimes I do feel generous).

This question is based on a story from India and it goes as follows.

King Shriham of India wanted to reward his grand minister Ben for inventing the game of chess. When asked what reward he wanted, Ben asked for 1 grain of rice on the first square of the board, 2 on the second square of the board, 4 on the third square of the board, 8 on the fourth square of the board, and so on till all the 64 squares were covered. That is, he was doubling the number of grains on each successive square of the board. Although Ben’s request looked less than modest, King Shriham quickly found that the amount of rice that Ben was asking for was humongous.

QUESTIONS:

Write a MATLAB (you can use any other programming language) program for the following using the for or while loop.

  1. Find out how many grains of rice Ben was asking for.
  2. If the mass of a grain of rice is 2 mg, and the world production of rice in recent years has been approximately 600,000,000 tons (1 ton=1000 kg), how many times the modern world production was Ben’s request?
  3. Do the inverse problem – find out how many squares are covered if the the number of grains on the chess board are given to you. For example, how many squares will be covered if the number of grains on the chess board are 16?

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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Shortest path for a robot

Imagine a robot that had to go from point to point consecutively (based on x values) on a two dimensional x-y plane. The shortest path in this case would simply be drawing linear splines thru consecutive data. What if the path is demanded to be smooth? Then what!

Well one may use polynomial or quadratic/cubic spline interpolation to develop the path. Which path would be shorter? To find out thru an anecdotal example, click here.

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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Do quadratic splines really use all the data points?

There are two reasons that linear splines are seldom used

  1. Each spline uses information only from two consecutive data points
  2. The slope of the splines is discontinuous at the interior data points

The answer to resolving the above concerns are to use higher order splines such as quadratic splines. Read the quadratic spline textbook notes before you go any further. You do want what I have to say to make sense to you.

In quadratic splines, a quadratic polynomial is assumed to go thru consecutive data points. So you cannot just find the three constants of each quadratic polynomial spline by using the information that the spline goes thru two consecutive points (that sets up two equations and three unknowns). Hence, we incorporate that the splines have a continuous slope at the interior points.

So does all this incorporation make the splines to depend on the values of all given data points. It does not seem so.

For example, in quadratic splines you have to assume that the first or last spline is linear. For argument sake, let that be the first spline. If the first spline is linear, then we can find the constants of the linear spline just by the knowledge of the value of the first two data points. So now we know that we can set up three equations for the three unknown constants of the second spline as follows

  1. the slope of the first spline at the 2nd data point and the slope of the second spline at the 2nd point are the same,
  2. the second spline goes thru the 2nd data point
  3. the second spline goes thru the 3rd data point

That itself is enough information to find the three constants of the second spline. We can keep using the same argument for all the other splines.

So the mth spline constants are dependent on data from the data points 1, 2, .., m, m+1 but not beyond that.

Can you now make the call on the kind of dependence or independence you have in the constants of the quadratic splines?

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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Extrapolation is inexact and may be dangerous

The NASDAQ was booming – people were dreaming of riches – early retirement and what not. The year was 1999 and NASDAQ was at an all time high of 4069 on the last day of 1999.

Yes, Prince was right, not just about the purple rain, but – “‘Cuz they say two thousand zero zero party over, Oops out of time, So tonight I’m gonna party like it’s 1999 party like 1999.”

But as we know the party did not last too long. The dot com bubble burst and as we know it today (June 2008), the NASDAQ is hovering around 2400.

Year ………………NASDAQ on December 31st

1994………………………… 751

1995 ……………………….1052

1996 ………………………..1291

1997 ………………………..1570

1998 ………………………..2192

1999 ………………………..4069

• End of Year NASDAQ Composite Data taken from www.bigcharts.com

So how about extrapolating the value of NASDAQ to not too far ahead – just to the end of 2000 and 2001. This is what you obtain from using a 5th order interpolant for approximation from the above six values.

End of Year …Actual Value …..5th Order Poly Extrapo……………Abs Rel True Error
2000 ……………..2471…………………. 9128 ………………………………….. 269%
2001………………1950……………….. 20720 ………………………………….. 962%

Do you know what would be the extrapolated value of NASDAQ on June 30, 2008 -a whopping 861391! On June 30, 2008, compare it with the actual value.

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

Finding the length of curve using MATLAB

As per integral calculus, the length of a continuous and differentiable curve f(x) from x=a to x=b is given by

S=\int_a^b \sqrt{(1+(dy/dx)^2} dx

Now how do we find the length of a curve in MATLAB.

Let us do this via an example. Assume one asked you to find the length of x^2*sin(x) from Π to 2Π. In the book, How People Learn, the authors mention that learning a concept in multiple contexts prolongs retention. Although it may not be the context that the authors of the book are talking about, let us find the length of the curve multiple ways within MATLAB. Try the program for functions and limits of your own choice to evaluate the difference.

METHOD 1: Use the formula S=\int_a^b \sqrt{(1+(dy/dx)^2} dx by using the diff and int function of MATLAB

METHOD 2: Generate several points between a and b, and join straight lines between consecutive data points. Add the length of these straight lines to find the length of the curve.

METHOD 3. Find the derivative dy/dx numerically using forward divided difference scheme, and then use trapezoidal rule (trapz command in MATLAB) for discrete data with unequal segments to find the length of the curve.

QUESTIONS TO BE ANSWERED:

  1. Why does METHOD 3 giving inaccurate results? Can you make them better by using better approximations of derivative like central divided difference scheme?
  2. Redo the problem with f(x)=x^{\frac{3}{2}}   with a=1 and b=4 as the exact length can be found for such a function.

% Simulation : Find length of a given Curve
% Language : Matlab 2007a
% Authors : Autar Kaw
% Last Revised : June 14 2008
% Abstract: We are finding the length of the curve by three different ways
% 1. Using the formula from calculus
% 2. Breaking the curve into bunch of small straight lines
% 3. Finding dy/dx of the formula numerically to use discrete function
% integration
clc
clear all

disp(‘We are finding the length of the curve by three different ways’)
disp(‘1. Using the formula from calculus’)
disp(‘2. Breaking the curve into bunch of small straight lines’)
disp(‘3. Finding dy/dx of the formula numerically to use discrete function integration’)

%INPUTS – this is where you will change input data if you are doing
% a different problem
syms x;
% Define the function
curve=x^2*sin(x)
% lower limit
a=pi
% b=upper limit
b=2*pi
% n = number of straight lines used to approximate f(x) for METHOD 2
n=100
%p = number of discrete data points where dy/dx is calculated for METHOD 3
p=100

% OUTPUTS
% METHOD 1. Using the calculus formula
% S=int(sqrt(1+dy/dx^2),a,b)
% finding dy/dx
poly_dif=diff(curve,x,1);
% applying the formula
integrand=sqrt(1+poly_dif^2);
leng_exact=double(int(integrand,x,a,b));
fprintf (‘\nExact length =%g’,leng_exact)
%***********************************************************************

% METHOD 2. Breaking the curve as if it is made of small length
% straight lines
% Generating n x-points from a to b

xi= a:(b-a)/n:b;
% generating the y-values of the function
yi=subs(curve,x,xi);
% assuming that between consecutive data points, the
% curve can be approximated by linear splines.
leng_straight=0;
m=length(xi);
% there are m-1 splines for m points
for i=1:1:m-1
dx=xi(i+1)-xi(i);
dy= yi(i+1)-yi(i);
leneach=sqrt(dx^2+dy^2);
leng_straight=leng_straight+leneach;
end
fprintf (‘\n\nBreaking the line into short lengths =%g’,leng_straight)

% METHOD 3. Same as METHOD1, but calculating dy/dx
% numerically and integrating using trapz
xi=a:(b-a)/p:b;
% generating the dy/dx-values
m=length(xi);
for i=1:1:m-1
numer=yi(i+1)-yi(i);
den=xi(i+1)-xi(i);
dydxv(i)=numer/den;
end
% derivative at last point using Backward divided difference formula
% is same as Forward divided difference formula
dydxv(m)=dydxv(m-1);
integrandi=sqrt(1+dydxv.*dydxv);
length_fdd=trapz(xi,integrandi);
disp(‘ ‘)
disp(‘ ‘)
disp (‘Using numerical value of dy/dx coupled’)
disp (‘with discrete integration’)
fprintf (‘ =%g’,length_fdd)

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

A simple MATLAB program to show that High order interpolation is a bad idea

In a previous post, we talked about that higher order interpolation is a bad idea.

Runge's function

In this post I am showing you a MATLAB program that will allow you to experiment by changing the number of data points you choose, that is, the value of n (see the input highlighted in red in the code – this is the only line you want to change) and see for yourself why high order interpolation is a bad idea. Just, cut and paste the code below (or download it from http://www.eng.usf.edu/~kaw/download/runge.m) in the MATLAB editor and run it.

% Simulation : Higher Order Interpolation is a Bad Idea

% Language : Matlab r12
% Authors : Autar Kaw
% Last Revised : June 10 2008
% Abstract: In 1901, Carl Runge published his work on dangers of high order
% interpolation. He took a simple looking function f(x)=1/(1+25x^2) on
% the interval [-1,1]. He took points equidistantly spaced in [-1,1]
% and interpolated the points with polynomials. He found that as he
% took more points, the polynomials and the original curve differed
% even more considerably. Try n=5 and n=25
clc
clear all
clf

disp(‘In 1901, Carl Runge published his work on dangers of high order’)
disp(‘interpolation. He took a simple looking function f(x)=1/(1+25x^2) on’)
disp(‘the interval [-1,1]. He took points equidistantly spaced in [-1,1]’)
disp(‘and interpolated the points with a polynomial. He found that as he’)
disp(‘took more points, the polynomials and the original curve differed’)
disp(‘even more considerably. Try n=5 and n=15’)

%
% INPUT:
% Enter the following
% n= number of equidisant x points from -1 to +1
n=15;

% SOLUTION
disp(‘ ‘)
disp(‘SOLUTION’)
disp(‘Check out the plots to appreciate: High order interpolation is a bad idea’)
fprintf(‘\nNumber of data points used =%g’,n)
% h = equidisant spacing between points
h=2.0/(n-1);
syms xx
% generating n data points equally spaced along the x-axis
% First data point
x(1)=-1;
y(1)=subs(1/(1+25*xx^2),xx,-1);
% Other data points
for i=2:1:n
x(i)=x(i-1)+h;
y(i)=subs(1/(1+25*xx^2),xx,x(i));
end

% Generating the (n-1)th order polynomial from the n data points
p=polyfit(x,y,n-1);

% Generating the points on the polynomial for plotting
xpoly=-1:0.01:1;
ypoly=polyval(p,xpoly);

% Generating the points on the function itself for plotting
xfun=-1:0.01:1;
yfun=subs(1/(1+25*xx^2),xx,xfun);

% The classic plot
% Plotting the points
plot(x,y,’o’,’MarkerSize’,10)
hold on
% Plotting the polynomial curve
plot(xpoly,ypoly,’LineWidth’,3,’Color’,’Blue’)
hold on
% Plotting the origianl function
plot(xfun,yfun,’LineWidth’,3,’Color’,’Red’)
hold off
xlabel(‘x’)
ylabel(‘y’)
title(‘Runges Phenomena Revisited’)
legend(‘Data points’,’Polynomial Interpolant’,’Original Function’)
%***********************************************************************
disp(‘ ‘)
disp(‘ ‘)
disp(‘What you will find is that the polynomials diverge for’)
disp(‘0.726<|x|<1. If you started to choose same number of points ‘)
disp(‘but more of them close to -1 and +1, you would avoid such divergence. ‘)
disp(‘ ‘)
disp(‘However, there is no general rule to pick points for a general ‘)
disp(‘function so that this divergence is avoided; but some rules do exist for ‘)
disp(‘certian types of functions.’)

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