Notice that
MatLab can work with complex numbers, as implied by
the ‘real’, ‘imag’, ‘conj’, ‘abs’, and ‘angle’
commands. For example:
>> x = 3 + 4i
Creates the complex number x with real part 3 and imaginary part 4.
If in addition, we define:
>> y = 5 + 6i
Then these
numbers can be manipulated. For example:
>> z = x*y
makes z =
-9 + 38i.
>> z = x/y
makes z =
0.6393 + 0.0328i
One more example. It is known that a right
triangle with opposite length 1 and adjacent length 1 has a 45 degree (π/4
radian) angle. So try this:
>> a = 1 + 1i
>> b = angle(a)*180/pi
As
expected, this gives a value of 45 degrees.
Note that the result must be multiplied by 180/pi to convert it to
degrees from radians.
Matlab is
even more powerful than this in that it can manipulate entire matrices. Let’s start with a simple array.
>> A = [1 2 3 4 5 6]
>> B = [2 3 4 5 6 7]
This
identifies two arrays (or vectors) A and B.
We can select any element of the array with B(3),
which would give us the third element of B, i.e. 4. We can “multiply” A and B with:
>> C=A.*B
Which
yields C=[2 6 12 20 30 42]. Why did I put “multiply” in quotes? Because this is neither a
cross product nor a dot product.
It is simply the element-by-element product of the two vectors. Notice that you will not be able to say:
>> C=A*B
This is
because A and B are both row vectors, and there is no
definition for the product of two row vectors.
However, there is a definition for the product of a row vector and a
column vector such as:
This is
just the dot product. To change B into a column vector, add an apostrophe after
it. Thus:
>> A*B’
Will give the result 40 as above.
Now
extrapolate this to matrices with multiple columns. For example, to make a 3 by 3 matrix, you can
say:
>> C=[[1 2 4]’ [5 7 9]’ [3 5 3]’]
(Note the
apostrophes). This will create the
matrix:
Which can then be manipulated. For example, define
C as above and take the determinant:
>> d=det(C)
You could
also define this same matrix with:
>> C = [[1 5 3]; [2 7 5]; [4 9 3]]
Notice that
the semicolon separates rows of the matrix.
How do I:
1.
Set
up a vector of 100 evenly spaced numbers?
For example, a vector starting at zero and going to 10 in steps of 0.2
can be set up like this:
>> z=[0:0.1:10];
Most of the time you will want to put the semicolon on the
end of this because it can get to be annoying when all 100 values are printed
out.
2.
Use
this to calculate a function at all of the values of this vector? Try squaring the vector:
>> zsq = z.*z
Or taking the exponential
>> zx = exp(z)
Or something like this:
>> zdv = exp(z)./z
Notice the
need for the ‘.’ in this
expression. This
expression exponentiates each component of the vector
and then divides by that same component.
For example, if z=[1 2 3], zdv
becomes [exp(1)/1 exp(2)/2 exp(3)/3]
m-files
Typing in
formulas gets old fast, so there should be a way of putting a formula into a
file for later use. This can be done
with the m-files. In an m-file you can
type your commands in just as when you’re on the command line of Matlab, and then when you type the name of the m file,
these commands are executed.
For
example, from the Matlab menu, call File | New |
m-file. Then type the following command:
z = x+y;
Then call
File | Save As
Under the
prompt type the file name “myadd.m”. Now go back to the main window of MatLab and type:
>> x=3;
>> y=21;
myadd
To find the
value of z, simply type
>> z
MatLab
should then respond with 24.
Of course,
if you did not put the semicolon after z=x+y in the
m-file, MatLab printed the value of z for you as soon
as you typed “myadd.”
To close the m-file, enter m-file window and choose File | Exit
Editor/Debugger.
Functions
Functions
have a similarity to m-files, but they are different in that the variables are
passed to them, as opposed to the file using whatever variables are set up on
the command line. If, for example, “myadd” were a function, it would have no idea what the
values of x and y were unless you told it.
For
example, from the MatLab main menu, choose File | Open | m-file
Then type
the following
function b=doadd(p,q)
y=p+q;
b=y;
Save this with File
| Save (give the file the name “doadd.m”).
Now, back
in the main MatLab window type:
>> z=doadd(x,y)
You should
get the same answer as you did before.
The difference is that you have to tell “doadd”
what you want to add, and it now returns that value to a variable of your
choice (in this case ‘z’). Notice that
the variable names in the file doadd (p and q) do not
need to match those that you specify in the main MatLab
window. The function knows what you want
added from the argument list (i.e., (x,y)).
Newton
Now apply
this to a more complicated example.
Let’s say you wish to be able to calculate the value of a polynomial of
arbitrary size. You could define a function
“poly” which takes as its argument the coefficients of your polynomial and the
value at which you want it evaluated. For example, if your polynomial looked
like:
y = 12x3
+ 3x2 + 5x + 8
The
function might look like this:
function p=dopoly(x,a1,a2,a3,a4)
p = a1*x*x*x + a2*x*x + a3*x + a4;
You can then call this function with:
>> y = dopoly(x,12,3,5,8)
to
evaluate the polynomial.
But what if
you want a more general function that will calculate the value of any
polynomial? Consider the following:
function p=dopoly(x,A)
n=length(A);
xpow=1;
sum = 0;
for i=1:n
sum = sum + A(i)*xpow;
xpow
= xpow*x;
end
p = sum;
If you call
this with
>> A
= [1 2 4 8 3];
>> x
= 37;
>> y
= dopoly(x,A);
you will
get the value of
y = 3x4 + 8x3
+ 4x2 + 2x + 1
In fact,
this will work regardless of the order of the polynomial. You need only set the length of A correctly.
Newton
Now let’s say you want to find the
roots of the polynomial. There is a
numerical method for doing this called the Newton Raphson
Iteration. Consider the figure below,
which shows a polynomial curve with three roots (three points of intersection
with the x axis). Let’s say that we
guess a value of the root (Initial Guess).
It’s probably not right, but we can still evaluate the value of the
polynomial. We can then calculate the
slope of the polynomial at that point (we need another function for this, but
it’s easy to take the derivative of a polynomial) and draw a tangent line and follow
it until it hits the x axis. This is
also not correct, but it’s closer than our initial guess. If we repeat the process several times, and
if our function behaves itself, we will eventually converge on the correct
answer. This all hinges on our being
able to determine the intersection point of the tangent line and the
x-axis. This is not so hard. The tangent line has the equation:
(y-y0) = m(x-x0)
You will
recognize this as the point-slope formula for a straight line. If we knew only one point on the line (x0,y0) and we knew the slope m, we would have an
equation for the line. But we have both
of these. If we evaluate the function
at our guess (call the guess x0), then we would get y0 (call
this f(x0)), so we would have a point on the line (x0,y0). The slope of the line is just the first
derivative of f(x) evaluated at x0.
We are interested in where the line crosses the line y=0, so if we set
y=0 in the point-slope formula, we get:
-y0
= m(x-x0)
x = -y0/m + x0
In terms
of the function we are interested, this can be written as:
x = -f(x0)/f’(x0) + x0
Now it is
time to see if you understand all of this:
Exercise:
1.
Write a MatLab function to calculate an
arbitrary length polynomial, where the input is the value of x and an array of
the coefficients. Test it with the
polynomial
y = 3x4 - 8x3 +
4x2 - 2x + 1
at the points x = 1, 2, and 3 to make
sure it works (use your calculator to make sure the function works).
2.
Write a matlab function to evaluate the first
derivative of the given polynomial. The
input is again the value of x and the coefficients of the polynomial (not the
coefficients of the derivative!). Again,
check to make sure it provides the correct answers.
3. Find two roots of the polynomial
given above using a Newton Raphson function. The input to the Newton-Raphson
function will be just the coefficients of the polynomial.
Steven A. Jones
Biomedical Systems,