I just want to evaluate an integral as follows:
enter image description here
The function rho_delta is defined as follows:
p = 3;
delta = 0.1;
A_p = 1/integral(#(s) exp(-s.^2),-p,p);
rho_delta = #(t) A_p*exp(-t.^2/delta^2)/delta;
I try to use int function to evaluate the symbolic expression of the integrand function, and than convert it to anonymous function. Finally, calculate it using the integral function, but it don't works.
The following code can calculate this integral:
I = quadgk(#(t1) arrayfun(#(t1) quadgk(#(s) rho_delta(t1-s),1,1.1).*(E(t1)-quadgk(#(s) rho_delta(t1-s).*E(s),0,1)),t1),0.9,1);
Related
I would like to solve the following polynomial numerically for r:
I am trying to use fzero() as follows:
r = (5/(r^2*9))- ((2)/(9*(6-r)^2))
x0 = 10; % some initial point
x = fzero(r,x0)
How can this be done with fzero()?
Input variable and function name should be different
Just change the function name to pol
To use fzero the function pol have to be a function handle
defined via #
pol =#(r) (5/(r^2*9))- ((2)/(9*(6-r)^2))
x0 = 10; % some initial point
x = fzero(pol,x0)
solution
x = 3.6754
It should be noted that, the first argument in fzero() should be "a function handle, inline function, or string containing the name of the function to evaluate", but yours is just an expression, which is not valid.
Besides the approach by #Adam (using function handle), another way is to use anonymous function, i.e.,
x = fzero(#(r) (5/(r^2*9))- ((2)/(9*(6-r)^2)) ,x0)
where
#(r) (5/(r^2*9))- ((2)/(9*(6-r)^2))
is the anonymous function with respect to argument r.
You will get the same result that x = 3.6754.
I have some code here which illustrates the nested nature of some integrals that I want to perform in matlab. When I run the following code I get the error
Error using .*
Matrix dimensions must agree.
Error in fun1/integrand (line 7)
f = x.^2.*t;
it seems that the way I have set it up now does not allow for vectorized output of the integral function. What can I do to make my code run? I have commented below what I want the functions to do.
My code is as follows:
%Main script:
z = linspace(0,10,10);
I = zeros(10,1);
for i = 1:10
I(i) = fun2(z(i));
end
%Function 1 in separate file:
function I = fun1(x)
I = integral(#integrand,-1,1);
function f = integrand(t)
f = x.^2.*t;
end
end
%Function 2 in separate file:
function I = fun2(z)
I = integral2(#integrand,-1,1,-1,1);
function f = integrand(x,y)
f = z.*fun1(x).*y; // here fun1 should return the value for all x that is swept through by integral2 during its call.
end
end
The problem with your code is you are using integral2 and integral1 in a combination. This way integral2 generates a grid for x and y and integral1 generates the t values. This way the t and x values don't match in size, and even if they would they are part of a different grid.
As already discussed, using ingetral3 is the right choice. Alternatively you could simply "fix" this introducing a loop, but this results in much slower code:
function I = fun1(x)
I=nan(size(x));
for ix=1:numel(x)
I(ix) = integral(#(t)integrand(x(ix),t),-1,1);
end
function f = integrand(x,t)
f = x.^2.*t;
end
end
I want to optimize an unconstrained multivariable problem using fminunc function in MATLAB. Here is an example:
Minimize the function f(w)=x'Ax
Create a file myfun.m:
function f = myfun(x)
f = x'*A*x + b'x
Then call fminunc to find a minimum of myfun near x0:
[x,fval] = fminunc(#myfun,x0).
My problem is that in my algorithm, the matrix A and vector b in myfun.m are not fixed, but can be changed over loops, so I cannot type them by hand. How can I pass values to A and b?
There are a few options for passing additional arguments to an objective function. For a simple one like yours, you could just make an anonymous function, which will save the values of A and b when it was created:
A = myMatA();
b = myVecb();
myfun = #(x) x.'*A*x + b.'*x;
[x,fval] = fminunc(myfun,x0); % use no # with an anonymous function
The other two options are global variables (yuck!) and nested functions. A nested function version looks like this:
function [x,fval] = myopt(A,B,x0)
[x,fval] = fminunc(#myfunnested,x0);
function y = myfunnested(x)
y = x.'*A*x + b.'*x;
end
end
But I think you would not use fminunc to solve minimization of x'Ax + b'x...
I a struggling writing a fully parametrised fitness function for the global optimisation toolbox in matlab.
Approach:
[x fvall,exitflag,output]=ga(fitnessfcn,nvars,A,b,Aeq,beq,lb,ub)
I have a fitness function that I call with
fitnessfcn=#fitnessTest;
hence, the function is stated in a separate file.
Problem:
My issue is now that my optimisation is a simple but super long sum like
cost=f1*x1+f2x2+...fnxn
n should be parameterised (384 at the moment). In all matlab help files, the objective function is always short and neat like
y = 100 * (x(1)^2 - x(2)) ^2 + (1 - x(1))^2;
I have tried several approaches "writing" the objective function intelligently, but then, I cannot call the function correctly:
if I write the fitness function manually (for fi=1)
function y = simple_fitness(x)
y = x(1)+ x(2)+ x(3)+ x(4)+ x(5)+ x(6)+ x(7)+ x(8);
the global optimisation works
But if I use my automated approach:
n = 8; %# number of function handles
parameters = 1:1:n;
store = cell(2,3);
for i=1:n
store{1,i} = sprintf('x(%i)',parameters(i));
store{2,i} = '+'; %# operator
end
%# combine such that we get
%# sin(t)+sin(t/2)+sin(t/4)
funStr = [store{1:end-1}];%# ignore last operator
endFunction=';';
%functionHandle = str2func(funStr)
y=strcat(funStr,endFunction)
matlab does not recognise the function properly:
error:
Subscripted assignment dimension mismatch.
Error in fcnvectorizer (line 14)
y(i,:) = feval(fun,(pop(i,:)));
thanks! I cannot write the objective function by hand as I will have several hundred variables.
You can use the sum(x) directly using function handle, instead of writing all indexes, do: fitnessfcn = #(x) sum(x).
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.