Conjugate of Symbolic Root doesn't change anything

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James Tursa
James Tursa 2024-2-21
评论： Walter Roberson 2024-10-15
Ran in:
Reference question first asked here:
https://stackoverflow.com/questions/78010011/matlab-bug-with-root-and-conj
Basically, the Symbolic Toolbox is making a simplification that, while it can be considered correct in the big picture, is nevertheless annoying. Take this code:
syms z
ex = z^5 + z^4 + z^3 + z^2 + z^1 + 1 == 0;
s = root(ex,z,1)
s = 
c = conj(s)
c = 
vpa(s)
ans = 
vpa(c)
ans = 
You can see that the conj( ) had no effect, and seems to have been ignored since both vpa values are the same.
In the big picture, it seems the Symbolic Toolbox knows that roots come in complex conjugate pairs when the coefficients of a polynomial are real. And since there is no published preferred order of returning the roots, it has apparently implicitly changed the order in the background as reasoning for doing away with the conjugate. It is as if root(ex,z,1) is interpreted as "one of the roots of ex given a changeable background ordering depending on how you use it", and not "the first root of ex given some fixed order in the background". At least that is my take on what is happening.
If you ask for all the roots and convert them you get this:
vpa(root(ex,z))
ans = 
which of course gives you all the roots. Taking the conj( ) of this gives the expected results. It seems to me a description of this type of behavior in the doc is in order, but I see no mention of it in either the root( ) or conj( ) doc. Is there a mention of this somewhere that I missed? Is there a way to have conj( ) treat symbolic root( ) (and maybe other related functions) differently?
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Sufiyan
Sufiyan 2024-2-26
Ran in:
root() is used to represent roots of the polynomial. If you refer to this link root documentation, you can see that solve() is recommended to find the roots.
syms z
ex = z^5 + z^4 + z^3 + z^2 + z + 1;
s = (solve(ex, z));
% consider the second root
sroot = s(2);
product = vpa(sroot * conj(sroot))
product = 
1.0
I believe this would resolve your query.
  1 个评论
James Tursa
James Tursa 2024-2-26
编辑：James Tursa 2024-2-26
Ran in:
Except when it doesn't. E.g., when solve() can't find explicit roots, it punts to root():
syms z
ex = 3*z^5 + z^4 + z^3 + z^2 + z + 1;
s = (solve(ex, z))
s = 
vpa(s(2)*conj(s(2)))
ans = 
So, same problem.

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