I originally asked this question yesterday and found the answer myself; however, I used the clear all command in Matlab and now the function throws an error Undefined function or variable 'y'.
I used the code from my answer
function [s1] = L_Analytic3(eqn,t0,h,numstep,y0)
%Differential Equation solver for specific inputs
% eqn is the differential equation
% t0 is start of evaluation interval
% h is stepize
% numstep is the number of steps
% y0 is the initial condition
syms y(x)
cond = y(0) == y0;
A = dsolve(eqn, cond);
s1 = A;
S1 = s1;
for x = t0 : h : h*(numstep)
subs(x);
if x == t0
S1 = subs(s1,x);
else
S1 = [subs(S1), subs(s1,vpa(x))];
end
end
end
and also put L_Analytic3(diff(y) == y,0,0.1,5,1) into the Command Window after entering clear all. I have to run a seperate code
syms y(x)
cond = y(0) == 1;
A = dsolve(diff(y) == y, cond);
before using my function in order for the function to work. Is this just because A,ans,cond,x, and y, are already defined by the 3 line code before using the function? If so, is there a way that I can use the function without having to use that 3 line code first?
When you do L_Analytic3(diff(y) == ...); you do not have variable y defined, so MATLAB complains - it has no way of knowing y is a symbol that will be defined in the function you are calling. You do not require all 3 lines of code. syms y(x) should be enough to define y and lets you use the function call you wanted.
Now, there are 2 easy ways to fix this that I see:
A script (or another function) that has syms y(x), followed by the call to L_Analytic3 the way you are doing it (which now does not need syms y(x), it has been defined already).
Give anonymous equation as the input instead, say #(x) diff(x)==x, and change a line of L_Analytic3 slightly to A = dsolve(eqn(y), cond);
Both ways work fine for this, no idea if 2nd one breaks in more complex cases. I would likely pick 1st version if you are doing symbolic stuff, and 2nd if you would like to have same function call to both numeric and symbolic functions.
Related
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.
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);
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)
I want to solve two nonlinear equations in MATLAB so i did the following:
part of my script
c=[A\u;A\v];
% parts of code are omitted.
x0=[1;1];
sol= fsolve(#myfunc,x0);
the myfunc function is as follows
function F = myfunc(x)
F=[ x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj];
end
i have two unknowns x(1) and x(2)
my question is How to pass a values(c,ii,jj) to myfunc in every time i call it?
or how to overcome this error Undefined function or method 'c' for input arguments of type 'double'.
thanks
Edit: The previous answer was bogus and not contributing at all. Hence has been deleted. Here is the right way.
In your main code create a vector of the coefficients c,ii,jj and a dummy function handle f_d
coeffs = [c,ii,jj];
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
sol = fsolve(f_d,x0);
Make your function myfunc capable of taking in 2 variables, x0 and coeffs
function F = myfunc(x, coeffs)
c = coeffs(1:end-2);
ii = coeffs(end-1);
jj = coeffs(end);
F(1) = x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
F(2) = x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj;
I think that should solve for x0(1) and x0(2).
Edit: Thank you Eitan_T. Changes have been made above.
There is an alternative option, that I prefer, if a function handle is not what you are looking for.
Say I have this function:
function y = i_have_a_root( x, a )
y = a*x^2;
end
You can pass in your initial guess for x and a value for a by just calling fsolve like so:
a = 5;
x0 = 0;
root = fsolve('i_have_a_root',x0,[],a);
Note: The [] is reserved for fsolve options, which you probably want to use. See the second call to fsolve in the documentation here for information on the options argument.
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.