matlabFunction forces me (unnecessarily) to include a dummy integration variable as an argument of the anonymous function it creates.

In the example below, I use matlabFunction to create the anonymous function intF{2}. The function to be created has five arguments, but one (tau) is a dummy variable of integration. Since tau is not going to be assigned a value, I define intF{2} with a set of 'vars' specified which does not include tau. When evaluated, however, intF{2} returns an error, saying that tau is undefined. When I type out the anonymous function directly (i.e., intF{1}), then it evaluates fine, without including tau as an argument. It's clear from the code sample below that intF{1} and intF{2} are identical. So why should the second version fail and the first one succeed? Is this a bug in matlabFunction?

function nothing
syms x0 t T a tau
f = @(tau,x0,t,a) log(x0*exp(a*(tau-t)));
intF{2} = matlabFunction(int(f(tau,x0,t,a),tau,t,T),'vars',[x0,t,T,a]);
intF{1} = @(x0,t,T,a)    int(log(x0.*exp(-a.*(t-tau))),tau,t,T);
intF{:}
x0      = 16; t = 1 ; T = 2 ; a = 0.5;
for ii=1:numel(intF);
    disp(['Evaluating intF{' num2str(ii) '}']);
    eval(intF{ii}(x0,t,T,a)) 
end;
keyboard;
~

[jgg]

I'm not 100% sure because I'm not at my PC to check, but I believe the issue is that f is an anonymous function given your syntax. I have no idea how matlabFunction handles this object type, especially when composed of syms. It's not in the documentation.

Try getting rid of the @(tau, x0, t,a) part of your code. I think it should work at that point.

[Leo Simon]

Thanks for the suggestions, jgg. I replaced the third and fourth lines with

f = log(x0*exp(a*(tau-t)));
intF{2} = matlabFunction(int(f,tau,t,T),'vars',[x0,t,T,a]);

which I believe is what you suggested, and got the same error,i.e.

Undefined function or variable 'tau'.
Error in symengine>makeFhandle/@(x0,t,T,a)int(log(x0.*exp(-a.*(t-tau))),tau,t,T)

[jgg]

I did the following as well: replacing lines 3-4 with

 f = log(x0*exp(a*(tau-t)));
 intF{2} = matlabFunction(int(f,tau,t,T),'vars',[x0,t,T,a]);

And it seems to work properly here. I get:

 @(x0,t,T,a)int(log(x0.*exp(-a.*(t-tau))),tau,t,T)

without any errors. Which version of Matlab are you using? It might be a bug.

[Leo Simon]

I get that too, but did you try to evaluate, say,

intF{2}(rand,rand,rand,rand)

Indeed I believe it's a bug. See my answer below.

Answer 1

[Leo Simon]

I talked to matlab support about this. The problem goes away if you set the IgnoreAnalyticConstraints flag to true.

syms x0 t T a tau
f = log(x0*exp(a*(tau-t)));
intF2 = matlabFunction(int(f,tau,t,T,'IgnoreAnalyticConstraints',true),'vars',[x0,t,T,a]);
x0 = 16; t = 1 ; T = 2 ; a = 0.5;
intF2(x0,t,T,a)

This now does not throw an error