How to find the coefficient matrix of a set of nonlinear equations?

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I want to find the coefficient matrix (A) of the following equations:
f1 = (a+2*b+c+5*d)^3;
f2 = (3*a+b+4*c+d)^3;
f3 = (a+b+5*c+d)^3;
f4 = (a+2*b+c+6*d)^3;
The above equations can be written in the following form:
f=A(a,b,c,d)X
Which X= Transpose{a b c d}
and A is a matrix which its element are functions of a,b,c,d.
How to find such matrices in Matlab?
  6 Comments
Mehdi
Mehdi on 8 Mar 2022
Edited: Mehdi on 8 Mar 2022
No does not matter, just rational elements. My equations are something like below:
f1 = (a+2*b+c+5*d+...)^3;
f2 = (3*a+b+4*c+d+...)^3;
f3 = (a+b+5*c+d+...)^3;
f4 = (a+2*b+c+6*d+...)^3;
.
.
fn = (a+7*b+c+3*d+...)^3;
No restrictions on powers.

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Answers (2)

Torsten
Torsten on 10 Mar 2022
Edited: Torsten on 11 Mar 2022
There might be more elegant solutions, but this is one way you could generalize.
I did it in octave ; maybe you will have to access the elements of subfexp by subfexp{i} instead of subfexp(i).
You will have to apply the code to all 4 functions involved - I did it for one only.
If n variables are involved, you will have to write the code for an array of coefficients (A) instead of addressing each coefficient individually (a,b,c,d).
fa, fb, fc and fd give the matrix entries in row i if f = fi.
syms a b c d
f = (a+2*b+c+5*d)^3;
fexp = expand(f);
f1 = collect(fexp,a)
subfexp = children(fexp);
subfexpa = [];
subfexpb = [];
subfexpc = [];
subfexpd = [];
for i=1:length(subfexp)
if (has(subfexp(i),a))
subfexpa = [subfexpa,subfexp(i)/a];
continue;
end
if (has(subfexp(i),b))
subfexpb = [subfexpb,subfexp(i)/b];
continue;
end
if (has(subfexp(i),c))
subfexpc = [subfexpc,subfexp(i)/c];
continue;
end
if (has(subfexp(i),d))
subfexpd = [subfexpd,subfexp(i)/d];
continue;
end
end
fa = sum( subfexpa);
fb = sum( subfexpb);
fc = sum( subfexpc);
fd = sum( subfexpd);
res = simplify(a*fa + b*fb + c*fc + d*fd - fexp);
  2 Comments
Torsten
Torsten on 11 Mar 2022
Edited: Torsten on 11 Mar 2022
Here is a much simpler solution (I wonder why I didn't come up with it until now):
f =
(a+2*b+c+5*d)^3 =
(a+2*b+c+5*d)*(a+2*b+c+5*d)^2 =
a*(a+2*b+c+5*d)^2 + 2*b*(a+2*b+c+5*d)^2 + c*(a+2*b+c+5*d)^2 + 5*d*(a+2*b+c+5*d)^2
So choose
fa = (a+2*b+c+5*d)^2
fb = 2*(a+2*b+c+5*d)^2
fc = (a+2*b+c+5*d)^2
fd = 5*(a+2*b+c+5*d)^2
This is easy to implement and generalize.

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John D'Errico
John D'Errico on 11 Mar 2022
Edited: John D'Errico on 11 Mar 2022
This is a linear problem, IF you recognize that each of the right hand sides are linear in a,b,c,d except for the cube around them.
f1 = (a+2*b+c+5*d)^3;
f2 = (3*a+b+4*c+d)^3;
f3 = (a+b+5*c+d)^3;
f4 = (a+2*b+c+6*d)^3;
If a solution exists, then we can simply choose the real cube roots of each of f1,f2,f3,f4.
So those equations reduce to:
nthroot(f1,3) = a+2*b+c+5*d;
nthroot(f2,3) = 3*a+b+4*c+d;
nthroot(f3,3) = a+b+5*c+d;
nthroot(f4,3) = a+2*b+c+6*d;
Note that the cube roots of any real number always have exactly one real solution. nthroot returns it, so this is unambiguous for the problem as stated. If f is negative, then nthroot will return the negative cube root. If f is positive, then nthroot returns the positive cube root.
The right hand sides are now entirely and purely linear in the variables a,b,c,d.
So we can solve the problem trivially as
A = [1 2 1 5;3 1 4 1;1 1 5 1;1 2 1 6];
abcd = A\nthroot([f1;f2;f3;f4],3)
This yields a vector of the UNIQUE solutions to the problem. No iterative schemes are needed. Nothing symbolic even.
As long as the matrix A is not singular, a unique solution exists. And what is the rank of A?
rank(A)
ans =
4
So A is full rank.
I'm not sure I see the difficulty here. Perhaps if those were other powers than 3 on the right hand sides. For example, had the problem been one where you were raising things to the 4th or 6th power, then you would have some ambiguity in the solution. But even then, you would merely solve for all cases, so the solution would be non-unique. And with other powers, like the 2nd or 5th or 7th powers, again, only one real root can eist. Still no substantial problem in any case.
  2 Comments
John D'Errico
John D'Errico on 11 Mar 2022
I don't see how you would conclude that from the question, or even from the comments. Anything could be the case from such a confusing question.

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