I'm aware of the fminsearch function, but it only seems to be able to solve for one variable.
If my function looks like f(x,y,z) and I want to find the values of x,y,z that gives the lowest result, how would I do this in MatLab? The complexity of f would make it unreasonably difficult to calculate the partial derivatives.
Any help would be appreciated, thanks!
fminsearch is multivariable, for example:
a = fminsearch(#(x)((x(1)-1)^2+(x(2)-2)^2), [0,0]);
the parameter passed to the objective function can be a vector, just be sure to specify the x0 parameter (the second parameter in fminsearch) to the right size.
Related
I'm using scipy.optimize.minimize with the Newton-CG (Newton Conjugate Gradient) method since I have an objective function for which I know the analytical Jacobian and Hessian. However, I need to add a regularization term R=exp(max(s)) based on the maximum value inside the array parameter "s" that being fit. It isn't entirely obvious to me how to implement derivatives for R. Letting the minimization algorithm do numeric derivatives for the whole objective function isn't an option, by the way, because it is far too complex. Any thoughts, oh wise people of the web?
I am now doing an optimization problem. Currently, I have codes in optimization min f(x_1,x_2,...,x_n). It may give x_1,x_2,...,x_n to be different in values after optimization. However, suppose I want to make that x_1=x_2=...=x_n and do the optimization again, what I expect is to find y such that f(y,y,...,y) is minimized. Setting x_1,x_2,...,x_n to be all the same as the initial input, but still, it may produce different values of x_1,x_2,...,x_n. Are there any good ways to solve the problem without rewriting the codes? Any existing functions/techniques can help me to do it? If possible, you can treat the function is not known (the codes of functions is not accessible, what I know is that for the function, when input n parameters, it gives a value).
As pointed out by Erwin Kalvelagen, the most general approach is to define it as equality constraints, but if it is your goal to simplify your problem, you can define a new function which accepts a single input value and forwards it to every input of your function f. Assuming you are using fmincon, the solution is:
x = fmincon(#(x) f(x, x, x, ..., x), x0, ...)
let us suppose that we have following task:Find the optimal value of weights
so that minimize following equation
where var-means variance of given x1 variable, also we have constraint that sum of these weights should be equal to 1
i have initialized anonymous function and weights for initial points
w=[0.5; 0.5];
>> f=#(x1,x2) (w(1)*w(1)*var(x1)+w(2)*w(2)*var(x2))
f =
#(x1,x2)(w(1)*w(1)*var(x1)+w(2)*w(2)*var(x2))
i think i should use function fmincon,
i have created A matrix
A=[1;1];
and b column
b=[1];
then i tried following fun
weighs=fmincon(f(x1,x2),w,A,b)
but it gives me error
Error using optimfcnchk (line 287)
FUN must be a function, a valid string expression, or an inline function
object.
could you help me please what is wrong? thanks in advance
You need to specify the function in fmincon as a function handle or anonymous function; f(x1,x2) evaluates to a scalar double, not a function handle. fmincon will want to evaluate this function with current values of w to check for the quality of the solution, so it needs a way to feed w as an input.
Thus, you need to
Change the function definition to f(w,x1,x2), i.e.
f=#(w,x1,x2) (w(1)*w(1)*var(x1)+w(2)*w(2)*var(x2))
Write the fmincon call as fmincon(#(u)f(u,x1,x2),...)
However, I would suggest to substitute 1-w(2) for w(1) (or vice versa) in your problem to reformulate it as an unconstrained optimization of one variable (unless w is a real weight, and has to remain between 0 and 1, in which case you still need a constraint).
Lets say, I have a function 'x' and a function '2sin(x)'
How do I output the intersects, i.e. the roots in MATLAB? I can easily plot the two functions and find them that way but surely there must exist an absolute way of doing this.
If you have two analytical (by which I mean symbolic) functions, you can define their difference and use fzero to find a zero, i.e. the root:
f = #(x) x; %defines a function f(x)
g = #(x) 2*sin(x); %defines a function g(x)
%solve f==g
xroot = fzero(#(x)f(x)-g(x),0.5); %starts search from x==0.5
For tricky functions you might have to set a good starting point, and it will only find one solution even if there are multiple ones.
The constructs seen above #(x) something-with-x are called anonymous functions, and they can be extended to multivariate cases as well, like #(x,y) 3*x.*y+c assuming that c is a variable that has been assigned a value earlier.
When writing the comments, I thought that
syms x; solve(x==2*sin(x))
would return the expected result. At least in Matlab 2013b solve fails to find a analytic solution for this problem, falling back to a numeric solver only returning one solution, 0.
An alternative is
s = feval(symengine,'numeric::solve',2*sin(x)==x,x,'AllRealRoots')
which is taken from this answer to a similar question. Besides using AllRealRoots you could use a numeric solver, manually setting starting points which roughly match the values you have read from the graph. This wa you get precise results:
[fzero(#(x)f(x)-g(x),-2),fzero(#(x)f(x)-g(x),0),fzero(#(x)f(x)-g(x),2)]
For a higher precision you could switch from fzero to vpasolve, but fzero is probably sufficient and faster.
I would like to know if anybody knows how I can plot an integral calculated using quad/quadl, or if this is possible.
I read that I can set the trace parameter to be non-zero, and this results in the information of each iteration being provided, but I'm not sure how and if I can use the information to plot an integral.
Thanks.
quad and quadl do not compute an integral function anyway, i.e., an integral as a function of the parameter. And since tools like this work iteratively, refining their estimate until it satisfies a tolerance on the global value, they are not easily made to produce the plot you desire.
You can do what you desire by using a differential equation solver to generate the solution, ode45 for example.