Conjugate of Symbolic Root doesn't change anything

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James Tursa
James Tursa on 21 Feb 2024
Edited: James Tursa on 26 Feb 2024
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?
  2 Comments
Walter Roberson
Walter Roberson on 21 Feb 2024
Interesting.
I could see conj(root(ex,z)) (without the root selector) being silently dropped, but it does seem wrong to drop it when there is a root selector.
James Tursa
James Tursa on 22 Feb 2024
Ran in:
It can also lead to strange results such as this:
syms z
ex = z^5 + z^4 + z^3 + z^2 + z^1 + 1 == 0;
s = root(ex,z,1)
s = 
vpa(s*conj(s))
ans = 
You would expect the result to be real (within numerical effects), but it isn't. Forces you to do the vpa( ) first:
vpa(s)*conj(vpa(s))
ans = 
1.0

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Answers (1)

Sufiyan
Sufiyan on 26 Feb 2024
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 Comment
James Tursa
James Tursa on 26 Feb 2024
Edited: James Tursa on 26 Feb 2024
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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