Function Definition Clarification in Matlab - matlab

I wrote some code that works just fine to evaluate theta on its own with some test input. However, I would like to take this code and turn it into a function that I can call within another matlab file. I keep getting the error message, "Function definitions are not permitted in this context."
I want to be able to define four vectors in another matlab file and call SP1 to evaluate theta for those inputs. I'm not sure where I'm going wrong, though. Please help!
Thanks so much.
clc
clear all
function theta = SP1(p,q1,w1,r)
% INPUT:
%function theta = SP1(p,q1,w1,r)
% p = [5; -7; 12];
% q1 = [17.3037; -3.1128; 2.48175];
% w1 = [1/sqrt(8); sqrt(3/8); 1/sqrt(2)];
% r = [1; 2; -3];
% Define vectors u and v as well as u' and v'.
u = p - r;
v = q1 - r;
w1_t = transpose(w1);
u_prime = u - w1 * w1_t * u;
v_prime = v - w1 * w1_t * v;
% Calculate theta if conditions are met for a solution to exist.
if (abs(norm(u_prime)-norm(v_prime))<0.01) & (abs((w1_t * u)-(w1_t * v))<0.01)
X = w1_t*cross(u_prime,v_prime);
Y = dot(u_prime,v_prime);
theta = atan2(X,Y)
else if (norm(u_prime) == 0 | norm(v_prime) == 0)
disp('Infinite Number of Solutions')
else
disp('Conditions not satisfied to find a solution')
end
end

I think you can just remove the top two lines,
clc
clear all
and save the rest of the code starting with function as SP1.m file.
Then you should be able to call this function as SP1 from other m files.

I think you're confused about how functions work. The first line of a function definition defines how many inputs and outputs MATLAB expects:
function theta = SP1(p,q1,w1,r)
This means that calling a function SP1 will require you to give four inputs, and will return one output. It doesn't mean that:
Your inputs need to be named p, q1 and so on
Your output will be called theta automatically
The function will automatically take in the input variables p, q1, etc if they exist in the workspace.
It also doesn't do any checking on the inputs; so if you require that inputs be of a certain type, size, etc. you need to write your own error checking at the start of the file. You might intend that those inputs be 3x1 vectors, but there's nothing in the function to tell MATLAB that. So, SP1(1,2,3,4) will work, to some extent - it will take those inputs and try to run them through the function, and if they don't cause an error it will give you an output. The output might be wrong, but the computer doesn't know that.
Once you have a function you can call it multiple ways from the command line or from within other functions or scripts. As previously mentioned you don't have to stick to the naming of variables within the function, as long as input variables exist when the function is called MATLAB will accept them:
theta = SP1(p8,q27,w35,not_r);
myoutput = SP1(any,variable,I,like);
I don't necessarily have to give an output (but then the first output will be routed to ans)
SP1(this,will,also,work);
If I have some variables stored in a *.mat file (the case you seem to be asking about), I can do it like this:
load('mydata.mat'); %this file contains stored variables p, q1, w1 and r
theta = SP1(p,q1,w1,r);

Related

MATLAB function handles and parameters

When I type help gmres in MATLAB I get the following example:
n = 21; A = gallery('wilk',n); b = sum(A,2);
tol = 1e-12; maxit = 15;
x1 = gmres(#(x)afun(x,n),b,10,tol,maxit,#(x)mfun(x,n));
where the two functions are:
function y = afun(x,n)
y = [0; x(1:n-1)] + [((n-1)/2:-1:0)'; (1:(n-1)/2)'].*x+[x(2:n); 0];
end
and
function y = mfun(r,n)
y = r ./ [((n-1)/2:-1:1)'; 1; (1:(n-1)/2)'];
end
I tested it and it works great. My question is in both those functions what is the value for x since we never give it one?
Also shouldn't the call to gmres be written like this: (y in the #handle)
x1 = gmres(#(y)afun(x,n),b,10,tol,maxit,#(y)mfun(x,n));
Function handles are one way to parametrize functions in MATLAB. From the documentation page, we find the following example:
b = 2;
c = 3.5;
cubicpoly = #(x) x^3 + b*x + c;
x = fzero(cubicpoly,0)
which results in:
x =
-1.0945
So what's happening here? fzero is a so-called function function, that takes function handles as inputs, and performs operations on them -- in this case, finds the root of the given function. Practically, this means that fzero decides which values for the input argument x to cubicpoly to try in order to find the root. This means the user just provides a function - no need to give the inputs - and fzero will query the function with different values for x to eventually find the root.
The function you ask about, gmres, operates in a similar manner. What this means is that you merely need to provide a function that takes an appropriate number of input arguments, and gmres will take care of calling it with appropriate inputs to produce its output.
Finally, let's consider your suggestion of calling gmres as follows:
x1 = gmres(#(y)afun(x,n),b,10,tol,maxit,#(y)mfun(x,n));
This might work, or then again it might not -- it depends whether you have a variable called x in the workspace of the function eventually calling either afun or mfun. Notice that now the function handles take one input, y, but its value is nowhere used in the expression of the function defined. This means it will not have any effect on the output.
Consider the following example to illustrate what happens:
f = #(y)2*x+1; % define a function handle
f(1) % error! Undefined function or variable 'x'!
% the following this works, and g will now use x from the workspace
x = 42;
g = #(y)2*x+1; % define a function handle that knows about x
g(1)
g(2)
g(3) % ...but the result will be independent of y as it's not used.

Why can't I pass a function handle to fsolve, but I can write an equivalent anonymous function in fsolve?

I find this somewhat odd. I am currently writing a simple function to solve a system of equations using fsolve. Here is what I have:
%Variable Declarations
I0 = 10e-12;
n = 1;
Vt = 0.0259;
R = 10e3;
Vs = 3;
%Function 1 (Some may recognize that this is the Shockley Diode Equation, if anyone cares...)
i1 = #(v1)(I0) * (exp((v1)/(n*Vt))-1);
%Function 2
i2 = #(v1) ((Vs-v1)/R);
%This is what I originally tried
h = #(v1) i1(v1)-i2(v1);
fsolve(h(v1), 1)
%After running this, I receive "Undefined function or variable 'v1.'"
% However, if I write
fsolve(#(v1)i1(v1)-i2(v1),1)
%The function works. With the result, I plugged that value into h(v1), and it produces the expected result (very close to 0)
That said, why doesn't matlab allow me to pass a function handle to fsolve?
You want to pass a function handle, which is h, not h(v1). h(v1) itself can't be evaluated because v1 is not defined.
Try fsolve(h, 1)

How can I plot data to a “best fit” cos² graph in Matlab?

I’m currently a Physics student and for several weeks have been compiling data related to ‘Quantum Entanglement’. I’ve now got to a point where I have to plot my data (which should resemble a cos² graph - and does) to a sort of “best fit” cos² graph. The lab script says the following:
A more precise determination of the visibility V (this is basically how 'clean' the data is) follows from the best fit to the measured data using the function:
f(b) = A/2[1-Vsin(b-b(center)/P)]
Granted this probably doesn’t mean much out of context, but essentially A is the amplitude, b is an angle and P is the periodicity. Hence this is also a “wave” like the experimental data I have found.
From this I understand, as previously mentioned, I am making a “best fit” curve. However, I have been told that this isn’t possible with Excel and that the best approach is Matlab.
I know intermediate JavaScript but do not know Matlab and was hoping for some direction.
Is there a tutorial I can read for this? Is it possible for someone to go through it with me? I really have no idea what it entails, so any feed back would be greatly appreciated.
Thanks a lot!
Initial steps
I guess we should begin by getting a representation in Matlab of the function that you're trying to model. A direct translation of your formula looks like this:
function y = targetfunction(A,V,P,bc,b)
y = (A/2) * (1 - V * sin((b-bc) / P));
end
Getting hold of the data
My next step is going to be to generate some data to work with (you'll use your own data, naturally). So here's a function that generates some noisy data. Notice that I've supplied some values for the parameters.
function [y b] = generateData(npoints,noise)
A = 2;
V = 1;
P = 0.7;
bc = 0;
b = 2 * pi * rand(npoints,1);
y = targetfunction(A,V,P,bc,b) + noise * randn(npoints,1);
end
The function rand generates random points on the interval [0,1], and I multiplied those by 2*pi to get points randomly on the interval [0, 2*pi]. I then applied the target function at those points, and added a bit of noise (the function randn generates normally distributed random variables).
Fitting parameters
The most complicated function is the one that fits a model to your data. For this I use the function fminunc, which does unconstrained minimization. The routine looks like this:
function [A V P bc] = bestfit(y,b)
x0(1) = 1; %# A
x0(2) = 1; %# V
x0(3) = 0.5; %# P
x0(4) = 0; %# bc
f = #(x) norm(y - targetfunction(x(1),x(2),x(3),x(4),b));
x = fminunc(f,x0);
A = x(1);
V = x(2);
P = x(3);
bc = x(4);
end
Let's go through line by line. First, I define the function f that I want to minimize. This isn't too hard. To minimize a function in Matlab, it needs to take a single vector as a parameter. Therefore we have to pack our four parameters into a vector, which I do in the first four lines. I used values that are close, but not the same, as the ones that I used to generate the data.
Then I define the function I want to minimize. It takes a single argument x, which it unpacks and feeds to the targetfunction, along with the points b in our dataset. Hopefully these are close to y. We measure how far they are from y by subtracting from y and applying the function norm, which squares every component, adds them up and takes the square root (i.e. it computes the root mean square error).
Then I call fminunc with our function to be minimized, and the initial guess for the parameters. This uses an internal routine to find the closest match for each of the parameters, and returns them in the vector x.
Finally, I unpack the parameters from the vector x.
Putting it all together
We now have all the components we need, so we just want one final function to tie them together. Here it is:
function master
%# Generate some data (you should read in your own data here)
[f b] = generateData(1000,1);
%# Find the best fitting parameters
[A V P bc] = bestfit(f,b);
%# Print them to the screen
fprintf('A = %f\n',A)
fprintf('V = %f\n',V)
fprintf('P = %f\n',P)
fprintf('bc = %f\n',bc)
%# Make plots of the data and the function we have fitted
plot(b,f,'.');
hold on
plot(sort(b),targetfunction(A,V,P,bc,sort(b)),'r','LineWidth',2)
end
If I run this function, I see this being printed to the screen:
>> master
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the function tolerance.
A = 1.991727
V = 0.979819
P = 0.695265
bc = 0.067431
And the following plot appears:
That fit looks good enough to me. Let me know if you have any questions about anything I've done here.
I am a bit surprised as you mention f(a) and your function does not contain an a, but in general, suppose you want to plot f(x) = cos(x)^2
First determine for which values of x you want to make a plot, for example
xmin = 0;
stepsize = 1/100;
xmax = 6.5;
x = xmin:stepsize:xmax;
y = cos(x).^2;
plot(x,y)
However, note that this approach works just as well in excel, you just have to do some work to get your x values and function in the right cells.

MATLAB Function (Solving an Error)

I have one file with the following code:
function fx=ff(x)
fx=x;
I have another file with the following code:
function g = LaplaceTransform(s,N)
g = ff(x)*exp(-s*x);
a=0;
b=1;
If=0;
h=(b-a)/N;
If=If+g(a)*h/2+g(b)*h/2;
for i=1:(N-1)
If=If+g(a+h*i)*h;
end;
If
Whenever I run the second file, I get the following error:
Undefined function or variable 'x'.
What I am trying to do is integrate the function g between 0 and 1 using trapezoidal approximations. However, I am unsure how to deal with x and that is clearly causing problems as can be seen with the error.
Any help would be great. Thanks.
Looks like what you're trying to do is create a function in the variable g. That is, you want the first line to mean,
"Let g(x) be a function that is calculated like this: ff(x)*exp(-s*x)",
rather than
"calculate the value of ff(x)*exp(-s*x) and put the result in g".
Solution
You can create a subfunction for this
function result = g(x)
result = ff(x) * exp(-s * x);
end
Or you can create an anonymous function
g = #(x) ff(x) * exp(-s * x);
Then you can use g(a), g(b), etc to calculate what you want.
You can also use the TRAPZ function to perform trapezoidal numerical integration. Here is an example:
%# parameters
a = 0; b = 1;
N = 100; s = 1;
f = #(x) x;
%# integration
X = linspace(a,b,N);
Y = f(X).*exp(-s*X);
If = trapz(X,Y) %# value returned: 0.26423
%# plot
area(X,Y, 'FaceColor',[.5 .8 .9], 'EdgeColor','b', 'LineWidth',2)
grid on, set(gca, 'Layer','top', 'XLim',[a-0.5 b+0.5])
title('$\int_0^1 f(x) e^{-sx} \,dx$', 'Interpreter','latex', 'FontSize',14)
The error message here is about as self-explanatory as it gets. You aren't defining a variable called x, so when you reference it on the first line of your function, MATLAB doesn't know what to use. You need to either define it in the function before referencing it, pass it into the function, or define it somewhere further up the stack so that it will be accessible when you call LaplaceTransform.
Since you're trying to numerically integrate with respect to x, I'm guessing you want x to take on values evenly spaced on your domain [0,1]. You could accomplish this using e.g.
x = linspace(a,b,N);
EDIT: There are a couple of other problems here: first, when you define g, you need to use .* instead of * to multiply the elements in the arrays (by default MATLAB interprets multiplication as matrix multiplication). Second, your calls g(a) and g(b) are treating g as a function instead of as an array of function values. This is something that takes some getting used to in MATLAB; instead of g(a), you really want the first element of the vector g, which is given by g(1). Similarly, instead of g(b), you want the last element of g, which is given by g(length(g)) or g(end). If this doesn't make sense, I'd suggest looking at a basic MATLAB tutorial to get a handle on how vectors and functions are used.

Plotting graph error (values not showign up)

How do I plot the value of Approximation - Answer as s varies in the code below? If you look at my code below, you can see the method I used (I put it in a separate file).
However, it does not show me a graph from 1 to 1000. Instead the graph is from 999 to 1001 and does not have any points on it.
for s = 1:1000
error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);
title('Accuracy of Approximation');
xlabel('s');
ylabel('Approximation - Exact Answer');
The functions used:
function g = LaplaceTransform(s,N);
% define function parameters
a=0;
b=1;
h=(b-a)/N;
x = 0:h:1;
% define function
g = ff(x).*exp(-s*x);
% compute the exact answer of the integral
exact_answer=antiderivative(b,s)-antiderivative(a,s)
% compute the composite trapezoid sum
If=0;
for i=1:(N-1)
If=If+g(i).*h;
end;
If=If+g(1).*h/2+g(N).*h/2;
If
with
function fx=ff(x)
fx=x;
and
function fx=antiderivative(x,s);
fx= (-exp(-s*x)*(s*x+1))/(s^2);
Any help would be appreciated. Thanks.
The following
for s = 1:1000
error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);
already has several issues. The two main ones are that error is getting overwritten at each iteration, as #Amro has pointed out, and that s, your loop variable, is a scalar.
Thus, you need to write
difference = zeros(1000,1); %# preassignment is good for you
for s = 1:1000
difference(s) = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(1:1000,difference);
There is another error in the LaplaceTransform function
function g = LaplaceTransform(s,N);
[...]
g = ff(x).*exp(-s*x); %# g is an array
[...]
If %# If is calculated, but not returned.
I assume you want to write
function If = LaplaceTransform(s,N);
instead, because otherwise, you try to assign the array g to the scalar difference(s).