Is there any discrete mathematics solver for symbolic equations ?

Hi, :

Excuse me, I have surveyed some, but no similar solution. So that may I ask some opinions from you.

That's if it's possible to solve the symbolic equations (simplify), which similar to the 'solve()', in continuous time questions.

For example:

(1). w[n] = x[n] + a*w[n-1]

(2). y[n] = w[n] + b*w[n-1]

The above (1), (2), equations represent a signal flow, the w[n] is intermediate node, what the users would want is y[n] = expressions in {x , y}.

From the handwrote deriven , the final is (3)

(3). y[n] = x[n] +b*x[n-1] + a*y[n-1]

is derived from,

from (1),

w[n] = x[n] +a*z^-1*w[n], such that, w[n] = x[n]*( 1 - a*z^-1 )^-1 --------(1-1)

from (2),

y[n] = w[n] + b*z^-1*w[n], such that, y[n] = w[n]*( 1 + b*z^-1 ) ---------(2-1)

from (1-1), (2-1),

y[n] = x[n]* ( 1 + b*z^-1) / ( 1 - a*z^-1)

then , y[n]*( 1 - a*z^-1 ) = x[n]*( 1 + b*z^-1 )

then , y[n] - a*y[n -1] = x[n] + b*x[n-1]

finally, y[n] = x[n] + b*x[n-1] + a*y[n-1] ------------(3)

Like the above, is there any method in Matlab to derive the (1), (2) to (3)

Thank you fro any ideas.

Best regards.

Answer 1

[Walter Roberson]

you will need to use evalin or feval into the symbolic engine, and use rec() to build the recurrence equation and then solve() to solve it. There is no exposed interface for this, and it is no longer documented in R2020a but you can find the MuPad description in older documentation

[Tamura Kentai]

Hi, Walter:

First, thank you for the reply. :)

I think you might misunderstood the purpose which I asked. I didn't try to input numerical data into a user-defined math function, specifically in integer (discrete). I didn't try finding same topics in MuPad yet, maybe there will be some there. I guess that will take more time, since I haven't the right 'keyword' to search.

I guess, possibly no explicity feature for this kind of question, no idea if there exists any workaround. I found 'Solve Difference Equations Using Z-Transform' may or may not help, it looks like the 'title' helps, but from the content it's not. It's about numerical data plot, etc.

Thank you very much.

[Walter Roberson]

https://books.google.ca/books?id=Ca0MBwAAQBAJ&pg=PA157&lpg=PA157&dq=recurrent+mupad&source=bl&ots=y0w-xUJeH0&sig=ACfU3U3zNucjN_BQGeZOHKhbHVeUXPIJOA&hl=en&sa=X&ved=2ahUKEwjGmNSLzLvqAhUiIDQIHf4rCF4Q6AEwAnoECAkQAQ#v=onepage&q=recurrent%20mupad&f=false

gives the MuPAD example

equation := rec(y(n+2) = y(n+1) + 2*y(n), y(n)):
solve(equation);

To use MATLAB to invoke that example you would use

feval(symengine, 'solve', 'rec(y(n+2) = y(n+1) + 2*y(n), y(n))')

getting back the result

(-1)^n*C3 + 2^n*C4

Here C3 and C4 are constants that depend upon the boundary conditions.

In case the google books URL goes bad, the example is the same as from the old official documentation, https://sites.calvin.edu/tmk5/research/mupad_tutorial.pdf see PDF page 202 (Mathworks numbering page 8-15)

At the MATLAB level, with a couple of helper functions:

%construct a MuPAD-level recurrence equation
%first input is a MATLAB type symbolic equation
%second input is the variable that would be solved for.
%This helper permits you to construct and store recurrence equations at the MATLAB level
MuRCN = @(A,B) feval(symengine, 'rec', A, B);

%solve MuPAD-level recurrence equations that have already been constructed.
%Note: the MATLAB level solve() will **not** work for these.
SolveMuRCN = @(A) feval(symengine, 'solve', A);

%example
syms y(n)
EQN = MuRCN(y(n+2) == y(n+1) + 2*y(n), y(n));
SolveMuRCN(EQN)

Unfortunately there is not much more documentation available (other than showing how to construct initial conditions). I can get at the source code:

feval(symengine,'expose', 'rec::solve')
feval(symengine,'expose', 'rec::Solve')   %called by rec::solve

but it is not easy to read.

I have not figured out yet how to solve multiple recurrence equations at the same time.

[Tamura Kentai]

Hi, Walter:

First, I appreciate your answers. I haven't understand those quite well, but I think it will take times.

I'm sorry, I don't know those advanced features behind rather than in 'Help'.

I mean those code, they are not included in 'Help'.

feval(symengine, 'solve', 'rec(y(n+2) = y(n+1) + 2*y(n), y(n))')

It would be impolite if I don't reply until I completely understand those, it's truely hard to read.

Thank you very much.

Best regards.

[Walter Roberson]

https://www.mathworks.com/help/symbolic/feval.html documents feval(symengine)

The short summary is that feval(symengine, 'SomeFunction', arg1, arg2, arg3, ...) evaluates arg1, arg2, arg3 using the usual MATLAB evaluation rules. It then takes the results and constructs a MuPAD call with SomeFunction as the name of the function to call, and the results of evaluating arg1, arg2, arg3 as the arguments. The call that is constructed is a character vector, and the character vector is

The effect is similar to evalin(symengine, 'SomeCommand') except that using feval(symengine) constructs arguments with variable names properly.

The programming language is MuPAD.

MuPAD is barely documented starting with R2020a, so you would need to look at a previous release for documentation.

The basic syntax is rec() with first parameter being an equation in = form (not == form) that sets out a recursion, and the second parameter indicates which expression you would want it to be solved for. The third parameter, not shown, is a set of initial conditions in the form of indexed_expression = value, such as {y(0)=1 y(1) = 7} . It takes a bit of extra work to construct a set at the MATLAB level: MATLAB's {} for cell arrays is not the same.

Answer 2

[madhan ravi]

You may have a look into collect(), isolate() which may be of use for you to obtain those results.

doc collect
doc isolate

[Tamura Kentai]

Hi, madhan:

Thank you for the reply.

But I don't think collect() & isolate() is about that.

I read the help of them, they are still about the continuous time equations, and arrange the expression style.

Originally, I shouldn't say that 'discrete mathematics', I should say about 'z-transform'.

So, it's something like doing the z-transform of equations, then re-arrange them, then inverse z-transform, I don't know what's the terminology, I just know it's from digital signal processing.

Thank you very much for the ideas.

Best regards.