Does the Fundamental Theorem of Algebra apply to non-integer degree polynomials?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
14 replies · 3K views
DaTario
Messages
1,097
Reaction score
46
Hi All,

According to the fundamental theorem of algebra: "every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots".
My question is: what about polynomials with degree say 2.3 or 3.02, as in the polynomial:
## p(x) = x^{2.3} - 5x + 6 ?##
Do these polynomials take part in the FTA ?

Best wishes,

DaTario
 
Mathematics news on Phys.org
That is not a polynomial. By definition, polynomials only have integer powers of the variable.

Your function cannot even be defined as continuous function over the whole complex plane.
 
  • Like
Likes   Reactions: Logical Dog
mfb said:
Your function cannot even be defined as continuous function over the whole complex plane.
Further, the function isn't defined on the negative real numbers.
 
Mark44 said:
Further, the function isn't defined on the negative real numbers.
##\displaystyle x^{2.3}=e^{2.3 \ln(x)}## can easily be defined for negative real numbers, you just have to choose a branch, and you have to define where to make it discontinuous.
 
mfb said:
##\displaystyle x^{2.3}=e^{2.3 \ln(x)}## can easily be defined for negative real numbers, you just have to choose a branch, and you have to define where to make it discontinuous.
The context of my comment, which I didn't state, was polynomials with real variables. I thought those were what he was asking about in writing p(x) = ... instead of ##p(z) = z^{2.3} - 5z + 6##.

In any case, this is not a polynomial, as you have already said.
 
mfb said:
The fundamental theorem of algebra works with complex numbers and DaTario mentioned them in the first post as well.
I was so dumbfounded by the 2.3 exponent on what he called a polynomial that I didn't notice that he had mentioned complex coefficients.
 
Thank you all.
I am satisfied with the comments. I was curious because the graph of the function:
## f(x) = x^{2.3} - 5x + 2 ##
as shown below, suggested to me that we could also have some control over its roots.
Obs: The command to this plot was:
Plot[x^2.3 - 5x + 2, {x, -5, 5}].

power23div10.jpg
 
Last edited:
PeroK said:
Further, no equation can have 2.3 roots!

I must confess that, when I formulated this question, I was in the spirit of that child that asks if someone can multiply a number by itself 2.3 times.
 
DaTario said:
I must confess that, when I formulated this question, I was in the spirit of that child that asks if someone can multiply a number by itself 2.3 times.
And the answer is yes -- with conditions. As long as you can take the 10'th root of x, then (x1/10)23 can be considered and studied. So your question has some interesting aspects.

No matter how strange you think your question is, there is still a chance that someone has studied it seriously -- and maybe even applied it somewhere. There are also fractional derivatives, which I have a very hard time thinking about.
 
FactChecker said:
And the answer is yes -- with conditions. As long as you can take the 10'th root of x, then (x1/10)23 can be considered and studied..
But this gets us well beyond the concept of multiplication, depending as it does on being able to find the 10th root of a number.
 
  • Like
Likes   Reactions: FactChecker
Mark44 said:
But this gets us well beyond the concept of multiplication, depending as it does on being able to find the 10th root of a number.
It has some practical aspects - in a tempered scale, the ratio between two tones a half tone apart is [itex]\sqrt[12]{2}[/itex].
 
FactChecker said:
And the answer is yes -- with conditions. As long as you can take the 10'th root of x, then (x1/10)23 can be considered and studied. So your question has some interesting aspects.

No matter how strange you think your question is, there is still a chance that someone has studied it seriously -- and maybe even applied it somewhere. There are also fractional derivatives, which I have a very hard time thinking about.

I had contact with fractional derivatives once in my life and it was also "hard time".
 
  • Like
Likes   Reactions: FactChecker