How to solve a symbolic script expression within a cell and assign the answer to a new variable

I have a 1x1 symbolic object (X) containing the following symbolic expression:

int(T - 2*t, t, 0, T)

I want to create a new variable (a0) that is simply the result of the above integration.

I can obtain the result by calling out the symbolic expression in the command window, copying and pasting it back into the command window and hitting enter (answer = 0).

How do I do this without copy and pasting?

Thanks

Answer 1

[Star Strider]

Assign it to a variable:

syms t T 

a0 = int(T - 2*t, t, 0, T)

produces:

a0 =

0

[Nate]

Understand-

But I don't want to have to search for and type "int(T - 2*t, t, 0, T)" - as it will be a part of a more complex loop.

I need something in the form of:

a0 = solve(X)

(Where X is a 1x1 symbolic object containing a single cell with the symbolic expression "int(T - 2*t, t, 0, T)" )

But that doesn't work, of course. I can't find the right function to call out the symbolic expression contained within X and "run" it as if hitting "enter" in command window after simply copy and pasting. I hope that makes sense.

Thanks again

[Star Strider]

I’m not certain that I understand, but symfun might come to the rescue again:

syms t T T1 T2

a0 = symfun(int(T - 2*t, t, T1, T2), [T1 T2]);
Va0 = a0(0,T)

producing:

Va0 =

0

with ‘a0’ now being the function, and in this instance, ‘Va0’ evaluating it with the desired limits of integration. The Symbolic Math Toolbox evaluates the symbolic expression in the process, so that internally:

a0(T1, T2) =

(T1 - T2)*(T1 - T + T2)

Since I’m not certain that I understand everything in your Comment, a bit on solve as well:

The solve function will give you solutions in one of two formats, depending on what you ask for. If you ask it for more than one solution for an expression but give it only one output, it will produce a structure ‘a0’ with fields for each solution, for example ‘C’ and ‘D’:

a0 = solve(X)

it will produce ‘a0.C’ and ‘a0.D’. You can then define ‘C’ and ‘D’ as:

C = a0.C
D = a0.D

If you solve for and ask it specifically for ‘C’ and ‘D’, it will provide them:

[C,D] = solve(X)

assuming of course that it can find a solution.

[Nate]

Thanks again - I will clarify:

X is a 1x1 symbolic object containing a symbolic expression "int(T - 2*t, t, 0, T)"

I am asking for a method to simply extract the contents of X and solve the symbolic expression (ie: do the integration). I want this automated such that I do not have to search for "int(T - 2*t, t, 0, T)" within X and manually copy it to a new variable.

Eg:

Step 1.

X returns

X = int(T - 2*t, t, 0, T)

Step 2.

int(T - 2*t, t, 0, T) returns

ans = 0

I want to skip having to do the second step such that:

unknownfunction(X) returns

ans = 0

What is the unknownfunction that will do this?

Thanks again

[Star Strider]

If you only want to evaluate ‘X’ as those limits and store it as a variable, my original Answer lets you do just that. With ‘a0’ as I wrote it, the Toolbox solves and evaluates it and stores the result in ‘a0’. Replace my ‘a0’ with ‘X’ in your last Comment and you have the same result.

I have no idea what ‘unknownfunction’ does, other than to be the default behaviour of the Symbolic Math Toolbox.

[Nate]

I will give example code, note this is a different function:

clear all

syms T t positive real

syms a0 A0

x(T,t)= (t-floor(t/T)*T);

a0 = int(x(T,t),t,0,T);

A0 = simplify(a0,'IgnoreAnalyticConstraints',true)

Output:

A0 = int(t, t, 0, T)

BUT: If I copy and paste "int(t, t, 0, T)" into command window and hit enter:

ans = T^2/2

What is the shortcut to creating a new variable whose value is the symbolic solution to the integral expression in the output of the symbolic 1x1 object A0 (ie. whose value in this case is simply "T^2/2" and not "int(T - 2*t, t, 0, T)")

[Nate]

Found it.

It's simply

eval(X)

Thanks

[Star Strider]

When I run your code, I get:

A0 =

T^2/2

I’m lost. It seems to be doing exactly what it should do.

You may have to ask it to simplify expressions (sometimes over several steps, using the 'steps' option in simplify) to get the answer. That’s just the nature of the MuPAD engine.

[Nate]

Strange - I have R2014a. A0 returns the long form with int() for me, whereas eval(A0) returns T^2/2. All good, thanks

[Star Strider]

My pleasure!

I have R2014a as well. I have no idea what the difference in output could be due to.