- Create an array of coefficients need to be replaced using syms function.
- Through an iteration substitute all the coefficients using subs function.
- Collect required expression through collect function.
Coefficients from a symbolic 'nested' function
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Unfortunatly, I'm stuck. In a previous question, I asked about a possibility to do a comparison of coefficients within a symbolic function. The tricky part seems to be (at least I think it is) that the function consist of a initial function as I have documented below. In the end, Im trying to receive a system of equations from every comparison of coefficients.
This is the code so far:
% coeffs a_k, b_k
coeffs = [sym('a0') ,sym('a', [1 2]), sym('b', [1 2])];
% Output: [a0, a1, a2, b1, b2]
% sin cos parts of function
trigs = [cos([1 2]*sym('tau')), sin([1 2]*sym('tau'))];
% Output: [cos(tau), cos(2*tau), sin(tau), sin(2*tau)]
% combine
initFC = symfun(sum([1 trigs] .* coeffs), [coeffs sym('tau')]);
% Output: initFC(a0, a1, a2, b1, b2, tau) = a0 + b2*sin(2*tau) + a1*cos(tau) + b1*sin(tau) + a2*cos(2*tau)
So that is my final inital function. Let me put that in any equation, e.g.
% Equation
F([coeffs sym('tau')]) = initFC.^2 == 0;
% Output: F(a0, a1, a2, b1, b2, tau) = (a0 + b1*sin(tau) + a1*cos(tau) + b2*sin(2*tau) + a2*cos(2*tau))^2 == 0
I can subs() any of the symbolic variables, superb! After expanding(F), combine(F, 'sincos') I receive a term that consists only of trigonometric terms:
F = expand(F);
F = combine(F, 'sincos')
% Ouput: (a1^2*cos(2*tau))/2 + (a2^2*cos(4*tau))/2 - ...
Now here is the problem: I need to substract the coefficients of this equation for e.g. cos(tau), cos(2*tau), ... .
I did like someone recommended before:
sym Z
% trigs(1) = cos(tau)
F = subs(F, trigs(1), Z)
% Output:
Error using symfun/subsindex (line 157)
Indexing values must be positive integers, logicals, or symbolic variables.
I also tried collect:
collect(F, Z)
% Output:
(2*a0*a1 + a1*a2 + b1*b2)*Z + (a1^2*cos(2*tau))/2 + ...
which is almost the solution which I'm seeking but there are still remaining parts of the equation within the output that I cant remove.
The final system of equations should look like this:
[2*a0*a1 + a1*a2 + b1*b2;
... next row from another comparison of coefficient;
...;
...]
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Respuestas (1)
Guru Mohanty
el 13 de Mzo. de 2020
Hi, I understand you are getting issues in removing all the coefficients from a given equation. You can do this by doing following adjustments to your code-
Here is a sample code for it.
clc;
clear all;
% coeffs a_k, b_k
syms a0
syms a [1,2]
syms b [1,2]
coeffs = [a0 ,a1, a2, b1, b2];
% sin cos parts of function
trigs = [cos([1 2]*sym('tau')), sin([1 2]*sym('tau'))];
% combine
initFC = symfun(sum([1 trigs] .* coeffs), [coeffs sym('tau')]);
% Equation
F([coeffs sym('tau')]) = initFC.^2 == 0;
F = expand(F);
F = combine(F, 'sincos');
syms Z [8,1]
syms tau
%Coefficients to be replaced
coff_eqn=[cos([1 2 3 4]*tau), sin([1 2 3 4]*tau)];
for i=1:length(coff_eqn)
F = subs(F, coff_eqn(i), Z(i));
end
expr=collect(F,[Z(1:7)]);
display(expr);
espr=children(expr); % Find LHS of the expression
final=children(espr{1});
for i=1:length(coff_eqn)
Out(i)=final{i};
Out(i)=subs(Out(i), Z(i), 1);
end
display(Out);
The output is
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