Symbolic dependency apparently sometimes gives wrong answers

5 views (last 30 days)
Hello,
I have a simple question but could not figour out how to tackle it by myself. Consider the following:
syms p1 y
f=[p1*y;p1];
hasSymType(f,'symfunDependingOn',y)
0
0
I am not happy with this since clearly I expected to get the answer [1,0]. I do not know why it behaves like this!!!
Please not that my actual problem are much more complicated and this is just to explain to you my question. So, please do not give me any answer which
just works for this simple example, rather I would like a generic sort of answer.
Any idea?
thanks in advance!
Babak

Accepted Answer

Walter Roberson
Walter Roberson on 28 Oct 2022
Not a bug. Your first element is a symbolic expression involving the desired variable, but symfundependingon is for unresolved symbolic functions that have the variable as a parameter.
syms f(x) g(x, y) h(y)
expressions involving f or g would be symfun depending on x, but h would not be.
  7 Comments

Sign in to comment.

More Answers (2)

John D'Errico
John D'Errico on 28 Oct 2022
Edited: John D'Errico on 28 Oct 2022
To be honest, this appears to be more of a bug report than anything else, since that option in hasSymType does not apprear to be working as per the documentation. So I would report it as such, but Answers is not the correct place to report bugs.
Could you find some other solution that would work? Even if someone could find something that works here, it does not get past the fact that hasSymType is the correct tool to solve your specidfic problem, and it should work as per the help.
  1 Comment
Mohammad Shojaei Arani
Mohammad Shojaei Arani on 28 Oct 2022
Hi John,
Thanks!
Of course, I have my alternative way to tackle this issue (I had to write a rather complicated function which
finds the functional dependencies. I first convert my symbolic expression to a character or string and then detect functional dependencies using character manipulations). But, I would be happier when MATLAB fixes this bug.
Thanks again,
Babak

Sign in to comment.


Paul
Paul on 28 Oct 2022
Hi Mohammad,
This works, at least for the examples cited in this comment. Haven't tested for anything but those examples.
syms x y z f(x,y)
F = @(expr,var) cellfun(@(s) any(has(s,var)),arrayfun(@(s) findSymType(s,'variable'),expr,'UniformOutput',false));
E = [x y f(x,y)] % then I expect to get the answer [1 0 1].
E = 
F(E,x)
ans = 1×3 logical array
1 0 1
E =[1 2 x^2;sin(x*y) x z] % then I expect to get the answer [0 0 1;1 1 0].
E = 
F(E,x)
ans = 2×3 logical array
0 0 1 1 1 0
  1 Comment
Mohammad Shojaei Arani
Mohammad Shojaei Arani on 29 Oct 2022
Hi Paul,
This is great!
I hope it works always and is not limitted to some case examples.
Thanks!
Babak

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by