l = 
Operations and differentiation with symmatrix in MATLAB... possible bug?
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Hi
I have found matlab to behave unexpectedly when using the symmatrix. I dont understand it, maybe i misunderstood the way operators work, or maybe its a bug?
Any help would be greatly appreciated! Thank you!
I declare some symbolic objects using symmatrix
clear all
n = 2;
A = symmatrix('A', n);
lambda = symmatrix('lambda', [n,1]);
g = symmatrix('g', [n,1]);
phi = symmatrix('phi', [n,1]);
l = symmatrix('l', [n,1]);
item = symmatrix('l', [n,1]);
I declare 2 almost identical equations
arga = (l.'*(((lambda).*(A * g) + (phi.'*lambda) * (A * g))));
argb = (l.'*(((lambda).*(item) + (phi.'*lambda) * (A * g))));
arga and argb should be the interchangable, as both item and A*g are 2x1 parameter vectors.
I differenciate wrt lambda:
sola = diff(arga, lambda)
solb = diff(argb, lambda)
I get
sola =
l.'*(symmatrix(eye(2)) .* A*g + kron(phi.', A*g))
solb =
l.'*(kron(phi.', A*g) + symmatrix(eye(2)) .* l)
ProbIem 1: I cannot evaluate the sum in the parenthesis of sola due to dimensionality issues. the probelm doesnt appear for solb.
Problem 2: Running the numerical example below, sola and solb give two results different whereas this should not be the case
l=[1.2;1.1] ;
g=[1.9876;1.88] ;
phi=[1.0987;1.5192]
A=[132 123;1222 124] ;
item=A * g;
% sola, having removed only the symmatrix comand
l.'*((eye(2)) .* A*g + kron(phi.', A*g))
% solb, having removed only the symmatrix comand
l.'*(kron(phi.', A*g) + (eye(2)) .* l)
I get
ans =
1.0e+03 *
4.4392 5.9196
ans =
1.0e+03 *
3.8694 5.3495
Answers (1)
Hi dominik,
I'm not sure what Problem 1 is refering to.
I think Problem 2 can be explained as follows:
clear all
n = 2;
A = symmatrix('A', n);
lambda = symmatrix('lambda', [n,1]);
g = symmatrix('g', [n,1]);
phi = symmatrix('phi', [n,1]);
l = symmatrix('l', [n,1])
item = symmatrix('l', [n,1])
We see that the Matlab variables l and item both reference the same symbolic variable.
arga = (l.'*(((lambda).*(A * g) + (phi.'*lambda) * (A * g))))
argb = (l.'*(((lambda).*(item) + (phi.'*lambda) * (A * g))))
In argb, we see that the symbolic l shows up in the expression on the right.
sola = diff(arga, lambda)
solb = diff(argb, lambda)
l=[1.2;1.1] ;
g=[1.9876;1.88] ;
phi=[1.0987;1.5192] ;
A=[132 123;1222 124] ;
Here, item is set to A*g, but item is not a variable in the expression for solb. So setting item here has no affect on the results and because l ~= A*g we get different results for sola and solb.
%item=A * g;
% sola, having removed only the symmatrix comand
Also, I think A*g must be enclosed in parentheses.
l.'*((eye(2)) .* (A*g) + kron(phi.', A*g))
% solb, having removed only the symmatrix comand
l.'*(kron(phi.', A*g) + (eye(2)) .* l)
Maybe the intent was for item to be its own variable
%clear all
n = 2;
A = symmatrix('A', n);
lambda = symmatrix('lambda', [n,1]);
g = symmatrix('g', [n,1]);
phi = symmatrix('phi', [n,1]);
l = symmatrix('l', [n,1]);
item = symmatrix('Q', [n,1])
arga = (l.'*(((lambda).*(A * g) + (phi.'*lambda) * (A * g))));
argb = (l.'*(((lambda).*(item) + (phi.'*lambda) * (A * g))));
sola = diff(arga, lambda)
solb = diff(argb, lambda)
l=[1.2;1.1] ;
g=[1.9876;1.88] ;
phi=[1.0987;1.5192] ;
A=[132 123;1222 124] ;
%item=A * g;
Q = A * g;
% sola, having removed only the symmatrix comand
l.'*((eye(2)) .* (A*g) + kron(phi.', A*g))
% solb, having removed only the symmatrix comand
l.'*(kron(phi.', A*g) + (eye(2)) .* Q)
8 Comments
Dear Paul,
thank you for your extremely swift and comprehensive response.
I indeed had a typo in my code, but you figured that out yourself already... Sorry, and thank you!
The second half of your answer regards the problem which I did mean to post. And indeed, if you enclose A*g in parentheses as you suggest, the problem is solved.
However, why do I have to manually add parentheses? Why are those parentheses not in the symbolic expression that matlab displays? Here is what I get in 2023a, as a screenshot:
The parentheses are missing (marked in red).
Are symbolic matrix operators (their order of execution) not consistent with numeric matrix operators? (I only used symbolic scalars before and i cant recall having noted the same issue).
And if this is the case, is there a way to translate the symbolic expression into the equivalent numeric expression?
Problem 1, which i hadn't explained well enough, is bascially the same as problem 2. If I excecute the highlighted expression in the comand window, right after having execuded the code in thescreeshot, i get an error of matching dimensions. I.e. matlab doesnt accept its own symbolic output (without brackets) as symbolic input.
Thank you very much!
Best
Dominik
PS: How do you get this beautiful math font output?
Hi dominik,
How do you get this beautiful math font output?
Use LiveScript, which I think is also what's being used here on Answers.
Are symbolic matrix operators (their order of execution) not consistent with numeric matrix operators?
Not that I know of, and I would be absolutely shocked if that were the case.
Let's start with the code
n = 2;
A = symmatrix('A', n);
lambda = symmatrix('lambda', [n,1]);
g = symmatrix('g', [n,1]);
phi = symmatrix('phi', [n,1]);
l = symmatrix('l', [n,1]);
item = symmatrix('Q', [n,1]);
arga = (l.'*(((lambda).*(A * g) + (phi.'*lambda) * (A * g))));
argb = (l.'*(((lambda).*(item) + (phi.'*lambda) * (A * g))));
sola = diff(arga, lambda)
solb = diff(argb, lambda)
BTW, could the display of the Kronecker term also be ambiguous? It wouldn't be in Matlab because kron is a function, not an operator. But in math, which the display is trying to show, is it always the the case that (regular) product has precedence over the tensor product? Or are they both considered multiplication with precedence from left to right in the usual way?
However, why do I have to manually add parentheses? Why are those parentheses not in the symbolic expression that matlab displays?
No idea. Clearly, these two expressions are different
e1 = symmatrix(eye(2)).*(A*g);
e2 = symmatrix(eye(2)).*A*g;
size(e1)
size(e2)
Interestingly, even though the LiveScript displays of e1 and e2 are different, both expressions are the same using the normal, Matlab operator precedence rules (assuming that bull's eye symbol means .* , though I doubt that's standard math notation, is .* a formal math operator?)
e1,e2
And they are displayed exactly the same using pretty (I don't know if there is a way to get the normal "screen output" here on Answers.
pretty(e1)
pretty(e2)
is there a way to translate the symbolic expression into the equivalent numeric expression?
There are a couple of options.
One option is to use subs to substitute numeric values for symbolic variables if you just want the answer
lnum = [1.2;1.1] ;
gnum = [1.9876;1.88] ;
phinum = [1.0987;1.5192] ;
Anum = [132 123;1222 124] ;
subs(sola,{A l g phi},{Anum, lnum,gnum,phinum})
vpa(ans)
double(ans)
If don't care about overwriting variables, then
l = [1.2;1.1] ;
g = [1.9876;1.88] ;
phi = [1.0987;1.5192] ;
A = [132 123;1222 124] ;
double(subs(sola)) % no other arguments needed
Another option is to convert sola to a matlabFunction (doesn't work with symmatrix, so need to convert to sym)
ha = matlabFunction(symmatrix2sym(sola))
ha(Anum(1,1),Anum(1,2),Anum(2,1),Anum(2,2),gnum(1),gnum(2),lnum(1),lnum(2),phinum(1),phinum(2))
It appears that the actual processing of sola is correct based on comparison to solb. Are you able to verify that the expression for sola (with the parentheses on A*g) is, in fact, the correct derivative for your expression? I have no reason to doubt that it is.
Perhaps the only problem is how sola is displayed, though I can see that being a pretty serious problem if you want to copy/paste the expression for sola from the command window.
dominik
on 8 Apr 2024
Perhaps it would be worthwhile to file a bug report with a simple example to illustrate the problem with the display to the screen. Maybe use the e1 and e2 example from above. If you do file a bug report, please add a comment on this thread, or to the original question, with their response.
In the near term, all might not be lost. Starting with the code:
n = 2;
A = symmatrix('A', n);
lambda = symmatrix('lambda', [n,1]);
g = symmatrix('g', [n,1]);
phi = symmatrix('phi', [n,1]);
l = symmatrix('l', [n,1]);
item = symmatrix('Q', [n,1]);
arga = (l.'*(((lambda).*(A * g) + (phi.'*lambda) * (A * g))));
argb = (l.'*(((lambda).*(item) + (phi.'*lambda) * (A * g))));
sola = diff(arga, lambda);
funcsola = matlabFunction(symmatrix2sym(sola),'Vars',{symmatrix2sym(A), symmatrix2sym(g), symmatrix2sym(phi), symmatrix2sym(l)})
lnum = [1.2;1.1] ;
gnum = [1.9876;1.88] ;
phinum = [1.0987;1.5192] ;
Anum = [132 123;1222 124] ;
funcsola(Anum,gnum,phinum,lnum)
Unfortunately, the signature of funcsola is opaque in terms of its variable names. There might be a way to modify that, I'm not sure.
If more flexibility is needed to generate funcsola, like if you don't know its variables a priori, there may be other options, but I'm not quite sure what you're looking for.
Paul
on 9 Apr 2024
To make sure I understand correctly ....
sola is derived as symmatrix with n = 2 with the idea that the expression for sola is vaiid for any other value of n, and then we want to define a Matlab expression for sola that can be evaluated for numerical values of the variables using standard matrix operations with the variables being of dimension for any value of n?
dominik
on 11 Apr 2024
AFAIK there's no way to do that (but will keep thinking about it), maybe someone else will chime in with workable approach.
It would be cool to be able to use symmatrix with symbolic dimensions, and be able to matlabFucntion the same, something like
syms n m integer
A = symmatrix('A',[n n]);
B = symmatrix('B',[n m]);
C = A*B;
cfunc = matlabFunction(C);
Maybe you should consider submitting an enhancement request. They do improve functionality, e.g., diff on symmatrix arrived in r2023b.
Did you get a reponse on your bug report?
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