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can anyone explain this result ?

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diadalina
diadalina on 23 Oct 2017
Commented: Walter Roberson on 23 Oct 2017
i have a polynomial p = [ 3 -5 2 ]
i have calculated his roots r = roots(p) r= 1.0000 0.6667
then i wanted to obtain the expression of my polynomial i get:
expression=poly(r)
expression= 1.0000 -1.6667 0.6667
which is not my first polynomial, can anyone explain to me?

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Answers (2)

Cam Salzberger
Cam Salzberger on 23 Oct 2017
Hello Diadalina,
According to the poly function's documenation, poly and roots are inverses, but allowing for roundoff error, ordering, and scaling. If you notice in your code, expression == p/3, which is just p scaled by its first element. It's still solves the same equation:
3*x^2 - 5*x + 2 = 0
or
x^2 - (5/3)*x + 2/3 = 0
-Cam

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Walter Roberson
Walter Roberson on 23 Oct 2017
Note that this means that given the roots of an polynomial, you can only recreate the equation to within a non-zero constant multiple.

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Andrei Bobrov
Andrei Bobrov on 23 Oct 2017
:)
>> expression= [1.0000 -1.6667 0.6667]
expression =
1.0000 -1.6667 0.6667
>> expression*3
ans =
3.0000 -5.0001 2.0001
>>

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Andrei Bobrov
Andrei Bobrov on 23 Oct 2017
Through two points on the axis 0x there passes an infinite set of polynomials. You need a third point for a single polynomial.
diadalina
diadalina on 23 Oct 2017
that's mean that the found roots are not exacts?
Cam Salzberger
Cam Salzberger on 23 Oct 2017
If you are asking that because there are multiple possible polynomials with the same roots, then no, they can be exact answers. If you multiple any polynomial by a constant, you will have a new polynomial. However, they will all have the same roots. So while you can't necessarily expect to get the exact same polynomial out of "poly" based solely on the original roots, you can expect to get a multiple of it (at least for vectors).
If you're asking that because the result of "expression*3" was -5.0001 instead of exactly -5, that's simply due to Andrei's example vector including -1.6667 instead of -5/3. Always bear in mind, however, that machine precision is a thing. There is no infinite accuracy, so you may run into precision errors sooner or later.

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