formation of a function like in Mathematica

Hello, I am new user in Matlab, My question is that how can i form a function in editor file like we can make in Mathematica like

F[n_] := ComplexExpand[Im[etp[n]/(et[n] - (x + y I))]]

Also Is there any equivalent command for

 ComplexExpand

in Matlab? Thanks

[hparacha777]

Hi, you can write the function in Matlab Editor as:

 function [out1, out2, ...] = myfun(in1, in2, ...)

Just save the file with the name of the function. Hope it helps.

Answer 1

[Walter Roberson]

f = @(n) ComplexExpand(imag(etp(n)/(et(n) - (x + y * i))))

however, ComplexExpand, etp, and et are not defined for MATLAB.

Mathematica's ComplexExpand is a non-trivial symbolic process so to get it right you would need to be working with the Symbolic Toolbox. At the moment I do not see an equivalent routine in the Symbolic Toolbox. In the programming language Maple that the Symbolic Toolbox has many similarities to, the routine would be named evalc(), but MATLAB'S evalc() has nothing to do with that.

[Kashif]

Ok thanks for reply, I am still facing problem regarding ComplexExpand, Also please tell How can i find partial derivative of above function with respect to both variables in Matlab

[Walter Roberson]

After you have found the complex expansion (somehow), use gradient()

As you are taking the imaginary part of the function anyhow, you can probably substitute

syms etp etn x y real
syms f(n)
f(n) = simplify(imag(etp(n)/(et(n) - (x + y * i))));
gradient(f(n), [x, y])

[Kashif]

Thanks for the reply, in my problem etp and etn are known values and these are not variables,

        %  Actually In Mathematica
% With the following function
% F[n_] := ComplexExpand[Im[etp[n]/(et[n] - (x + y I))]]
% n1 = n/8;
% n2 = n + n/8;
% et(n1)=8.33663 + 6.8417*1i;
% etp(n1)=-4.67008 + 10.1188*1i;
% et(n2)=0.641945 - 0.526831*1i;
% etp(n2)=-1.00014 - 0.25351*1i;
% I am  getting like following values of F(n1) and F(n2)
% F[n1]=116.308/((8.33663 - x)^2 + (6.8417 - y)^2) - (
%  10.1188 x)/((8.33663 - x)^2 + (6.8417 - y)^2) - (
%  4.67008 y)/((8.33663 - x)^2 + (6.8417 - y)^2);
% And
%  F[n2]=-(0.689644/((0.641945 - x)^2 + (-0.526831 - y)^2)) + (
%  0.25351 x)/((0.641945 - x)^2 + (-0.526831 - y)^2) - (
%  1.00014 y)/((0.641945 - x)^2 + (-0.526831 - y)^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % I am looking for the same function in Matlab, which may give me like 
% same values of f(n1) and f(n2) in Matlab, But with following
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
syms  x y real
syms f(n)
f(n) = simplify(imag(etp(n))/((et(n)) - (x + y * 1i)));
gradient(f(n), [x, y])
% with your command i am not getting the same values as above

[Walter Roberson]

I am not positive this will work but try

syms et(n)
n1 = n/8;
n2 = n + n/8;
feval(symengine,'_assign', et(n1), 8.33663 + 6.8417*1i);
feval(symengine,'_assign', et(n2), 0.641945 - 0.526831*1i);

then see what

et(n1), et(n2), et(n)

return

If not then.... ummm, maybe

syms et n
evalin(symengine, 'et := N -> piecewise([N=n/8, 8.33663 + 6.8417*1i], [N=n+n/8, 0.641945 - 0.526831*1i], [Otherwise, procname(N)]);');

A bit tricky here is the unassigned behaviour for et at values other than n/8 and n+n/8.

You asked for the partial derivative with respect to both variables. I interpreted that as gradient. Perhaps you wanted

diff(f(n),x,y)

[Kashif]

Thanks, But my problem is not with et or etp, as these are known in my program, but only problem that i need function defining and ComplexExpand like command in matlab to get the same expressions as above.

[Walter Roberson]

You are not likely to get those any time soon. MATLAB has a limited interface subset to the Symbolic Toolbox ('MuPAD') which is more powerful.

I was assisting you under the assumption that you wanted to make something work. The primary barrier to making something work is the definition of et and etp as objects that need to recognize two specific inputs and return specific arbitrary outputs for those two inputs while being meaningfully defined symbolically for the formula with indefinite n.

[Kashif]

ok Sir , Thanks a lot for your time, i will try it