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

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Leo Simon
Leo Simon 2016-1-1
回答： Leo Simon 2016-1-5
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;
~
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jgg
jgg 2016-1-4
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.

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Leo Simon
Leo Simon 2016-1-5
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

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