Good evening,
I am trying to compute a matrix in Matlab whose elements are given by:
K(n,m) = int_{-inf}^{+inf} dx int_{-inf}^{+inf} dy f(n,m,x,y),
with f being the elements of an (x,y)-dependent matrix whose form I already know.
I have tried different methods for this:
Computing directly the matrix F with elements f and then two univariate integrals (function integral) with the corresponding 'ArrayValued' options.
Computing directly the matrix F with elements f and then the double integral (function integral2).
Trying to compute the elements of K with two consecutive univariate integrals (again, with integral).
Computing the elements of K with a double integral (again, with integral2).
None of this seems to be working, so I was wondering if anyone could provide a different approach.
Cheers
Related
I have a Matlab function G(x,y,z). At each given (x,y,z), G(x,y,z) is a scalar. x=(x1,x2,...,xK) is a Kx1 vector.
Let us fix y,z at some given values. I would like your help to understand how to compute the derivative of G with respect to xk evaluated at a certain x.
For example, suppose K=3
function f= G(x1,x2,x3,y,z)
f=3*x1*sin(z)*cos(y)+3*x2*sin(z)*cos(y)+3*x3*sin(z)*cos(y);
end
How do I compute the derivative of G(x1,x2,x3,4,3) wrto x2 and then evaluate it at x=(1,2,6)?
You're looking for the partial derivative of dG/dx2
So the first thing would be getting rid of your fixed variables
G2 = #(x2) G(1,x2,6,4,3);
The numerical derivatives are finite differences, you need to choose an step h for your finite difference, and an appropriate method
The simplest one is
(G2(x2+h)-G2(x2))/h
You can make h as small as your numeric precision allows you to. At the limit h -> 0 the finite difference is the partial derivative
I am working on something and I haven't found any solution, maybe I didn't know how to correctly search for it...
I have two arrays of experimental data (x and y). x is a list of certain energies (512 values from 0 to 100 kev) and I want to fit them to a function which returns a vector of values of y for every x in the list (the energies are always the same, 512 certain values). This is because my function model contains several matrix and other functions.
So, I can't evaluate my function as f(x,a,b,c...) (with a,b,c the parameters to fit) and expect a single scalar, but I have to evaluate f(a,b,c...), and it returns a vector of y(x1),y(x2)...
Now, I want to fit my data to my model. But lsqcurvefit needs a function of the form f(x), I suppose that it evaluates every f(x). I could write my function so that every time it is called it evaluates the vector result, and then returns y for the given x, but it would be quite inefficient... And I'm sure there must be another way.
Any idea?
Maybe you can do an fminsearch on the sum of square errors? It is best to put all fitting parameters into one vector. Here I call it p.
f = #(x,p) (p(3)+p(1)*x.^p(2)).^(1/p(4); %example function with four free parameters
sqerr = #(x,y,p) sum((y-f(x,p)).^2); %sum of squared errors
p = [1,1,1,1]; %four starting conditions
p = fminsearch(#(p) sqerr(x,y,p),p); %fit
Then you can find your y(x_i) values by calling the function with the fitted paramters
f(x,p)
I want to create a D by D matrix, where D is an even positive integer. I then want to fill it with values depending on a lambda and calculate the eigenvalues of the matrix. I'm interested in the smallest eigenvalue as a function of D and lambda.
Is this doable? I absolutely cannot find any way of doing this, and the help only mentions creating matrices of symbolic variables with known dimension.
So I have these three vectors:
And I have to find out for what value of k these three vectors are linearly dependent. I have tried using rref and linsolve with syms for this but that did not work out. I'm relatively new to MatLab and matrices so please keep that in mind.
I know in order to check if vectors are linearly dependent that c1...cn have to be non-zero.
I also want to know how you can use variables in general when solving these types of equations in MatLab.
A set of vectors (at least if you have n vectors in n dimensions) is linearly dependent if the matrix constructed from them is singular, i.e. if its determinant is 0. If you have the Symbolic Math Toolbox, you can construct a symbolic matrix:
syms k;
M = [1 k 0; -1 1 2; 0 0 3];
det(M)
This will tell you that det(M)==3*k+3, which you can solve by hand. But generally, you can ask matlab to solve it:
solve(det(M)==0,k);
which will tell you the answer is -1. So unless k==-1, these vectors are linearly independent (i.e. they comprise a basis of the Euclidean space R^3).
Update: If you don't have the Symbolic Math Toolbox, you could still try to find a numerical solution. First define a function
detfun=#(k) det([1 k 0; -1 1 2; 0 0 3]);
that for any value of k will give you the determinant of your matrix, for instance detfun(3) gives 12. Then you can use fsolve to find a numerical solution to the equation detfun(k)==0, by calling
fsolve(detfun,0)
in which the second argument, 0, refers to the starting point of the search performed by fsolve. This will tell you that the answer is k==-1, but a single call to fsolve will only give you a single solution. If your function has multiple roots, you have to play around with the starting points to find more of them. In this case, you can know that your function (i.e. det(M(k)) is linear in k, so it has a unique root.
I have the following equation:
I want to do a exponential curve fitting using MATLAB for the above equation, where y = f(u,a). y is my output while (u,a) are my inputs. I want to find the coefficients A,B for a set of provided data.
I know how to do this for simple polynomials by defining states. As an example, if states= (ones(size(u)), u u.^2), this will give me L+Mu+Nu^2, with L, M and N being regression coefficients.
However, this is not the case for the above equation. How could I do this in MATLAB?
Building on what #eigenchris said, simply take the natural logarithm (log in MATLAB) of both sides of the equation. If we do this, we would in fact be linearizing the equation in log space. In other words, given your original equation:
We get:
However, this isn't exactly polynomial regression. This is more of a least squares fitting of your points. Specifically, what you would do is given a set of y and set pair of (u,a) points, you would build a system of equations and solve for this system via least squares. In other words, given the set y = (y_0, y_1, y_2,...y_N), and (u,a) = ((u_0, a_0), (u_1, a_1), ..., (u_N, a_N)), where N is the number of points that you have, you would build your system of equations like so:
This can be written in matrix form:
To solve for A and B, you simply need to find the least-squares solution. You can see that it's in the form of:
Y = AX
To solve for X, we use what is called the pseudoinverse. As such:
X = A^{*} * Y
A^{*} is the pseudoinverse. This can eloquently be done in MATLAB using the \ or mldivide operator. All you have to do is build a vector of y values with the log taken, as well as building the matrix of u and a values. Therefore, if your points (u,a) are stored in U and A respectively, as well as the values of y stored in Y, you would simply do this:
x = [u.^2 a.^3] \ log(y);
x(1) will contain the coefficient for A, while x(2) will contain the coefficient for B. As A. Donda has noted in his answer (which I embarrassingly forgot about), the values of A and B are obtained assuming that the errors with respect to the exact curve you are trying to fit to are normally (Gaussian) distributed with a constant variance. The errors also need to be additive. If this is not the case, then your parameters achieved may not represent the best fit possible.
See this Wikipedia page for more details on what assumptions least-squares fitting takes:
http://en.wikipedia.org/wiki/Least_squares#Least_squares.2C_regression_analysis_and_statistics
One approach is to use a linear regression of log(y) with respect to u² and a³:
Assuming that u, a, and y are column vectors of the same length:
AB = [u .^ 2, a .^ 3] \ log(y)
After this, AB(1) is the fit value for A and AB(2) is the fit value for B. The computation uses Matlab's mldivide operator; an alternative would be to use the pseudo-inverse.
The fit values found this way are Maximum Likelihood estimates of the parameters under the assumption that deviations from the exact equation are constant-variance normally distributed errors additive to A u² + B a³. If the actual source of deviations differs from this, these estimates may not be optimal.