How to call element of matrix of symbolic variables

I have two symbolic variables

syms th1(t) Lc1

and generated a matrix in terms of th(t) and Lc1

tc1(t) = 
[ cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t))]
[ sin(th1(t)),  cos(th1(t)), 0, Lc1*sin(th1(t))]
[           0,            0, 1,               0]
[           0,            0, 0,               1]

now when I try to call the first three elements of the last column

pc1 = tc1(1:3,4);

I got this error:

Error using symfun/subsref (line 169)
Symbolic function expected 1 inputs and received 2.

Error in Fk (line 10)
pc1 = tc1(1:3,4);

thanks

Answer 1

[Walter Roberson]

It is not possible to index the entries of a symbolic function that computes a matrix.

The closest you can get is to call the function passing in variables for the parameters, get an array output, index into the array, and use the resulting expression or turn it into a function if needed.

[Hazim Nasir]

thank you Walter I have to generate a matrix with symbolic variables many times so as to differentiate them. I am beginner to find a solution. could you please attach an example

[Walter Roberson]

tc1mat = tc1(t);

then you can index tc1mat, each element of which will be a symbolic expression (not a function.)

[Hazim Nasir]

It is working many thanks

[Faizan ul Haq]

try using:

A=formula (tc1);

It will remove the function arguments and only save the equations in the A matrix.

then you can call functions like:

A(1)

A(1,1)

A(1:2,1,:)... etc

[Mr.Gh]

It is working, thanks.

[Niklas Kurz]

Wow, good job Matlab.

[Fenglei Gu]

hope it can be improve for the next version. lol

[ermiyas lakew]

How can you get back to the function argument after using "formula()"

[Walter Roberson]

@Fenglei Gu

Using () indexing to access the components of a symbolic function that defines an array, is something that is not going to be permitted. The () indexing method for symbolic function is already overloaded to evaluate the function, and there is no way to say "when I write these () brackets I mean indexing instead of evaluation".

There is already formula to retrieve the array expression in a way that can be later indexed.

[Walter Roberson]

@ermiyas lakew I think you might be looking for argnames

Answer 2

[Nathan Hardenberg]

Since the code answer is a bit hidden in the comments I'm posting the solution as full code here:

syms th1(t) Lc1

tc1(t) = [
     cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t));
     sin(th1(t)),  cos(th1(t)), 0, Lc1*sin(th1(t));
               0,            0, 1,               0;
               0,            0, 0,               1;]

pc1 = tc1(t);  % call the function with t first

pc1 = pc1(1:3,4)  % now you can index

[Osvaldo]

Very usefull! Tks

[Stu]

Wow, thanks for that. It's beautiful.

Now I can pass a vector (which itself contains symbolic variables) into another symbolic function which requires that vector input.

Answer 3

[João Vitor Fraga]

Here's a simple solution that doesn't turn your symfun into a sym: simply left-multiply and right-multiply your symfun by appropriate matrices whose entries are either zero or one, depending on which elements you wish to extract. Your symfun will remain a symfun. In your case:

syms th1(t) Lc1

tc1(t) = [ cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t))
           sin(th1(t)),  cos(th1(t)), 0, Lc1*sin(th1(t))
           0,            0,           1,               0
           0,            0,           0,               1];

Lmatrix = [1 0 0 0
           0 1 0 0
           0 0 1 0];

Rmatrix = [0
           0
           0
           1];

pc1 = Lmatrix*tc1*Rmatrix

pc1(t) = 

class(pc1)

ans = 'symfun'

This can easily be condensed into a simple function you could call whenever needed.

[Walter Roberson]

Cute ;-)

Answer 4

[Peyman Amiri]

Even we can make the brilliant soultion of @Nathan Hardenberg better by the following improvement (this solution does not change anything about the initial tc1 matrix):

syms th1(t) Lc1
tc1(t) = [ cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t));
          sin(th1(t)),  cos(th1(t)), 0, Lc1*sin(th1(t));
          0,            0, 1,               0;
          0,            0, 0,               1];
tc1=tc1(t);
pc1 = tc1(1:3,4)

pc1 =