Semester 2 (1)

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CHEN 14010 Matlab II AJM & PC

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0) Introduction

This hand-out contains the material you need to cover for the mid-semester online

assessment that will take place on Tuesday 1st March between 2 5 pm. Detailed

arrangements about this test will be given later.

Note the second online assessment will take place on Tuesday 5 April.

As before, it is advisable to save your answers to the exercises in My Documents

and/or a pen-drive, as then you will be able to make use of these programs when

doing the online test.

In what follows, use format short e unless told otherwise. You will also find it

helpful to have the first semester notes with you.

Finally, before starting, please note that:

i) Ctrl C stops a program (very useful if you are in an infinite loop!)

ii) You can clear the screen by typing clc.This does not, however, wipe out the valuesand types of previously defined variables.

iii) To start completely afresh, type clear. This is like starting again from the

beginning. If you get strange error messages, one possible cure is to type clear and try

again!

1) Function M-files

Last semester, you wrote a program that would calculate an integral numerically using

the trapezium rule. It would have looked something like this (for Q25):

a = input( 'Value of a: ' );b = input( 'Value of b: ' );n = input( 'Value of n: ' );h=(b-a)/n;sum=0.5*(exp(a) + exp(b));for i=1:n-1

sum=sum+exp(a+i*h);end

answer=h*sum;disp('answer is: ')disp(answer);truth=exp(b)-exp(a);diff=abs(answer-truth);disp(truth);disp(diff);

This estimates b

a

dxx)exp( using n strips and the result is given in the variable answer.

Suppose, however, we wanted to integrate something like

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2x exp(x) 3sinx /x ln(1 x 2) dxa

b

We could easily edit the program to do this, replacing exp(x) by this lengthy

expression at all points in the program, but the end result would be rather messy with

long lines containing this lengthy integrand. There is also a danger that we would

make a mistake and forget to change all the lines that needed changing. Finally it

makes the program hard to follow.

A nice way round this is to define our own function in a separate procedure and then

let our main program call this. We introduce the idea offunction M-files.

Type

function y = f(x)

y = 2*x*exp(-x)+3*sin(x)/x-log(1+x^2);

and save it in your directory (e.g. My Documents) as f.m.

If, in the workspace, you now typef(2), you get

ans =

2.9585e-001

which is just the value of )1ln(/sin3)exp(2 2xxxxx evaluated atx = 2.

You have created your own function!

We can now edit the trapezium program so that it integrates f(x), where we have

definedf(x) as our very own function.

Thus type:

a = input( 'Value of a: ' );b = input( 'Value of b: ' );n = input( 'Value of n: ' );

h=(b-a)/n;sum=0.5*(f(a) + f(b));for i=1:n-1

sum=sum+f(a+i*h);endanswer=h*sum;disp('answer is: ')disp(answer);

and save this as trap.m in the same directory as f.m.

This is just the program for Q25 (except for the last few lines) but exp(x) is replaced

everywhere byf(x).

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If you now type trap in the command window and then set a = 1, b = 3 and n = 8, you

get

answer is:

6.3939e-001

The important point is that you can integrate any function you like with this program.

You simply have to edit the f.m file. The trap program need not be altered at all.

Ex. 1: Edit f.m so you can integrate :

a)

1

1 x x3between 0 and 1, using 16 strips

answer is:

6.3050e-001

b)4

1

)1ln(

x

x

between 2 and 3, using 12 strips.

answer is:

3.4371e-002

Important note

Let us now do the integral

1

c x1

2

dx

where we input the value ofc into the main program.

Thus you might think the main program might look like thissave it as trapc.m

a = input( 'Value of a: ' );b = input( 'Value of b: ' );

c = input( 'Value of c: ' );

n = input( 'Value of n: ' );h=(b-a)/n;sum=0.5*(f(a) + f(b));fori=1:n-1

sum=sum+f(a+i*h);endanswer=h*sum;disp('answer is: ')disp(answer);

and the f.m program would be

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function y = f(x)

y = 1/(c+x);

Make these changes and try running trapc with a = 1, b = 2, c = 1 and n=8.

It wont workyou get an error message!

The problem is that the f.m procedure does not know the value of c. The main

program (trapc) knows what c is, but the main program has not told the function

anything about it! All f.m knows is what is given in its argumentin this case,x.

What we need to do now is to edit the programs so f.m is told the value ofc. To do

this we write

function y = g(x,c)

y = 1/(c+x);

and save this as g.m. You can give your own name to functions (within certain

restrictions)it does not always have to be f.m!

trapc.m is then given by

a = input( 'Value of a: ' );b = input( 'Value of b: ' );

c = input( 'Value of c: ' );

n = input( 'Value of n: ' );h=(b-a)/n;sum=0.5*(g(a,c) + g(b,c));for i=1:n-1

sum=sum+g(a+i*h,c);endanswer=h*sum;disp('answer is: ')disp(answer);

Now try running trapc with a = 1, b = 2, c = 1 and n=8.

answer is:

4.0565e-001

Now the program worksg.m now knows what c is.

Ex. 2:

(a) Integrate

f(x) exp(cx) between 0 and 1 for n = 8, reading in the value ofc = 2 into the main program.

answer is:4.3458e-001

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(b) Repeat the above for

f(x) 1

c2 x2, between 0 and 1, n = 8, reading in c = 2.

answer is:

2.3172e-001

(c)Repeat the above for

f(x) 1c2 dx x2

, between 0 and 1, n = 8, reading in

c = 2 and d= 3.

answer is:

1.7857e-001

The message is that f.m is in a watertight bubblethe only information it has is what

you put in its argument. It has no knowledge of any other variable that might happen

to exist in the calling program.

Let us now go a stage further with our trapezium rule program and create a function

trap.m. We return to integrating a functionf(x) contained in f.m viz.

function y = trap(a, b, n)

h=(b-a)/n;sum=0.5*(f(a) + f(b));for i=1:n-1

sum=sum+f(a+i*h);endy=h*sum;

This is very like our previous main program, but is now itself a function. You tell it

the limits of the integral, a and b, and the number of strips, n, and it goes away and

uses the trapezium rule to estimate the integral off(x). It then returns the value to the

main program.

Note we assign the final answer in the last line to the variable y. The important thing

is that the variable at the beginning and the end should be the same.

Edit the f.m program sof(x) = 1/x.

The main program might be:

a = input( 'Value of a: ' );b = input( 'Value of b: ' );n = input( 'Value of n: ' );answer = trap(a,b,n);disp(answer);

Save this as quad.m, in the same directory as trap.m and f.m.

If you run quad in the command window, then quad will call trap and trap will call

f, ending up integratingf(x) between a and b using n strips.

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Check this by running quad and then setting a = 1, b = 2 and n = 8.

Value of n: 8

6.9412e-001

Why do this?

One reason is, in practice, you do not know the true answer to your integral and thus

you do not know what value ofn to choose in order to get the right answer.

Now it is very easy to edit quad.m so it prints out the trapezium rule results for a

range ofn-values.

Ex. 3: Set

f(x) 1

x, and let a = 1 and b = 2. Calculate the value of this integral

analyticallyi.e. add in lines to calculate truth in the program above.

Now use the trapezium rule to calculate this integral using n = 4, 8 and 16 strips and

check you are steadily approaching the right answer.

e.g.

Value of a: 1

Value of b: 2

Value of n: 16

1.6000e+001 6.9339e-001 6.9315e-001

Ex. 4: Modify the program by introducing a for loop into the main program, so that

the program prints out a table, showing how the trapezium result varies for n = 4, 6, 8,

10, up to n = 20.

To print out the numbers a, b &c on one line, type disp([a, b, c]).

Ex. 5: Further modify the program so that for each value of n, you get a print out of

the error, i.e. the absolute value of the difference between the trapezium value and

truth. The MatLab function abs(x) returns the absolute value ofx.

i.e. error = abs(truth-answer);

Ex. 6: Use MatLab to plot the error against

1 n 2 . Do you get a straight line for

reasonably large values ofn?

To do this, it is helpful to have two arrays, x and y. You will need to think of

providing a counter, step for example, so that step is incremented by one each time

you go round the loop. Then x(step) will contain the value of

1 n 2 for that loop and

y(step) will contain the corresponding error.

At the end of the loop, you have: plot(x,y).

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Ex. 7: Now modify the program so that at each iteration, n is doubled. Thus you

should get a print out for n = 4, 8, 16, 32 up to n = 1024.

Of course what we really want to do is to have a loop in quad.m that keeps on going

until we have calculated the integral to the accuracy we want, rather than a forend

loop that just carries on for a set number of iterations. We return to this problem later,but first here are some more examples to practice functions.

Ex. 8: Consider a sphere, radius r. Write two functionsarea.m and volume.mthat

return values of the spheres area and volume respectively, given the value of r. Write

a main program - sphere.m - to which you input the r-value and which then calls

area(r) and volume(r) and prints out the values. Remember pi gives 3.14159 in

MatLab. What are the values for r= 2?

(area = 5.0265e+001; volume = 3.3510e+001)

Ex. 9: Now consider a cylinder, height h and radius r. Repeat the above, this timewriting a main program cylinder.m to which you input h and rand which then

calls area(h, r) and volume(h, r). What are they for h = 2 and r= 3?

(area = 9.4248e+001; volume = 5.6549e+001)

2) Making decisions with if

Often a program branches, depending on whether two numbers are the same, or

whether one exceeds the other. These tests use relational operators. Here are some of

those used in MATLAB:

< less than

greater than

> = greater than or equal

Type the following on the command line

x = 2;if x < 0disp( 'neg' )else

disp ( 'non-neg' )end

Did you get the non-neg message? Repeat the exercise, this time withx =1.

This is the basic ifelse end construction. You must remember to put in end, or

MATLAB will wait forever!

We can also write programs using this facility.

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Try the following:

r = rand % MATLAB random number between 0 and 1if r < 0.5

disp ( 'move left' )else

disp ( 'move right')end

This is the basis for doing a random walk along a line.

Suppose, however, you wanted to do a random walk on a square lattice. An efficient

way to do it makes use of the elseifconstruction.

Try this:

r = rand % Matlab random number between 0 and 1if r < 0.25

disp ( 'move left' )elseif r < 0.5

disp ( 'move up')elseif r < 0.75

disp ( 'move right')else

disp ( 'move down')end

Run this a few times to check you see how it works. Note that if, for example, r= 0.1,

sor< 0.25, then you only get the move left message, even though

r< 0.5, etc. aswell!

Ex. 10: Modify this program so the particle does a random walk on a cubic lattice

i.e. each step can be along one of three perpendicular directions, so there are a total of

six possible steps.

Ex. 11: Write a program to solve the quadratic equation 02 cbxax , where thevalues ofa, b and c are read into the program.

In this program calculate

disc = b^2 - 4*a*c;

Then use an if construction, so that if disc < 0, the program prints out No real

roots, but otherwise calculates the two roots and prints them out.

Ex. 12: Strictly you should also allow for the possibility disc = 0 and there is only one

root. Use an if elseif end construction to allow for this.

Ex. 13: One way to solve non-linear equations is to use a numerical method called the

bisection method (you will do this later in your mathematics unit). A key element of

the routine is to call a function,f(x), at two pointsx1 andx1and decide whether theproductf(x1)*f(x1) is positive or negative.

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Firstly write a function programfun.mto calculate fun(x) = exp(-x)x.

Then write a main program in which you enter x1 and x2. You then use an if

construction , so that if fun(x1)*fun(x2) > 0, you write positive product, else you

write negative product.

Test this on (a) x1 = 0 x2=1, (b) x1 = 1, x2=2, (c) x1=-1, x2 = 0.

3) While loops

Sometimes you do not know how many times you will need to repeat some lines of

program, so a forend loop is not well-suited for the task.

For example, suppose you had 1000 in an investment and the compound interest on

it was 10%. How many years must you wait for the investment to double?

Here is a MATLAB program to do this.

a = 1000; % initial investmentr = 0.1; % rate of interestbal = a;year = 0;disp( 'year balance' )

while bal < 2*abal = bal + r * bal;year = year + 1;

disp( [year bal] )end

The program keeps on advancing the years and adding on the interest until the

statement bal < 2*a is false, whereupon the loop ends.

Let us do another example.

Consider the series 3

1

2

11

1

1n nUse the while-end construction to find the

minimum number of terms you need in this series for the sum to be greater than 3.

sum = 0.0; % Initialise the sum to zeron = 0; % Initialise the counter in the sum to be zerodisp('n sum')

while sum < 3n = n+1; % Increment the countersum = sum + 1/n; %Accumulate the sumdisp( [n sum])

end

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Again the loop keeps on going until the condition sum < 3 is false. The last value

printed gives the sum and the value of n when the sum first exceeds 3.

Ex. 14: A population of bacteria doubles its size every hour. Starting from 1

bacterium, how many hours are needed for the population to first exceed 1,000,000 ?use a while end loop to do this.

(21 hours. Population is 1048576)

Ex. 15: Consider the series

22

1

2321

n

n Use the while-end construction to

find the minimum number of terms you need in this series for the sum to be greater

than 100,000.

(n = 67 sum = 102510)

Now that we have understood the while end construction, we can return to the

trapezium rule programs and answer the question about how to calculate an unknown

integral to a desired level of accuracy.

We return to the program you wrote for ex. 7 and we are integrating f(x) = 1/x. We

now use the while construction, so that the computer keeps on doubling n until the

value of the error gets less than a user defined tolerance, tol. What value of n is

needed, for example, to get the error less that 1.0e-3, 1.0e-6 or 1.0e-8 ?

Ex. 16: Write a program to integrate 1/x between a = 1 and b = 2, doubling the

number of strips (starting from n = 4) until tol < 1.0e-8.

(n = 4.0960e+003 answer = 6.9315e-001 error = 3.7253e-009)

You need to alter the main program so it has a structure like

User provides value of tol.

error = 100*tol % make sure error>tol before you start!

truth = ??? % provide the true answer for the integral

n = 2;

Provide values of a and bwhile error > tol

n = 2*n; % double the value of n. The first value is 4.

get trapezium value of integral

calculate the error = abs(answertruth)

print out n and the error

end;

print out n, error

Please note you cannot type in the above and expect it to work. Youhave to translate the English phrases intoMatLab!

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Make sure you understand how the while loop works. The computer will carry on

doubling n until the condition error > tol is false, whereupon it will leave the loop.

The value ofn finally printed out is the value you need. Along with the first value of

the error that is less than the tolerance.

Ex. 17: A problem with while loops is that if something goes wrong, the conditionthat stops the loop might never be met and the loop will never stop. If you run a

program and it never stops, type ctrl c to end it!

It is good practice to prevent while loops lasting for ever. To do this you can have a

counter in the loop and ensure that the counter never exceeds a certain value.

Modify your quadrature program so the while loop cannot do more than nmax

iterations.

You will need some extra lines like:

User provides nmax

steps = 0;

while error > tol & (steps < nmax)

steps = steps + 1;

Etc.

end;

if (steps >= nmax)

Print message saying nmax steps exceeded.

else

Print out n and error

end if

Usually you do not use numerical methods to work out an integral you already know!

If you do not know the value oftruth, then you usually estimate the accuracy of your

estimate by comparing the difference between two successive iterations.

Ex. 18: Modify your program so that it keeps on doubling n until the difference

between the estimates from two successive iterations (diff) is less than tol = 1.0e-6.

To do this, you could save in an array the trapezium estimate for each value ofn used,

but this is a bit wasteful in terms of storage and is not very elegant.

Better is just to use two variables and continually update them. Try something along

these lines:

Provide normal input data including tol;

diff = 100*tol; % we need to ensure diff > tol at the start

n=4;

old = trap (a, b, n); % this contains the first estimate

while diff > tol and (steps < nmax)

n = 2*n; %double n

new = value of trap with new value of n;diff = absolute difference between old and new;

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old = new;

% we copy the last estimate, new, into old

End;

Make sure you understand how this old/new business works! Each time, when you go

into an iteration, old contains the previous value of the estimate and new contains thenew one. The while loop keeps on going until diff becomes less than tol (or you

exceed the maximum number of iterations if something has gone wrong!).

Test your program by working out how many iterations are needed to get

diff< 1.0e-3, 1.0e-6 and 1.0e-8.

You are now in a position to calculate the integral of whatever function you like to

your specified accuracy!

Value of a: 1Value of b: 2

Value of tol: 1.0e-6

n = 5.1200e+002 old = 6.9315e-001 new = 6.9315e-001 diff = 3.7253e-009

Ex. 19: Calculate the following integrals to a tolerance of 1.0e-6. Note the value of

the integral as given in the final value for new and the number of strips used to

calculate this value.

a) dxx

1

0

2exp


n = 2.0480e+003 old = 1.4627e+000 new = 1.4627e+000 diff = 3.2404e-007

b) dxx 5

1

21ln



c) dxx3

1sin



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