Question about the Fundamental Theorem of Algebra

[DaTario]

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

 

[mfb]

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.

 

[Mark44]

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.

 

[PeroK]

Further, no equation can have 2.3 roots!

 

[mfb]

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.

 

[Mark44]

$\displaystyle x^{2.3}=e^{2.3 \ln(x)}$ 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]

The fundamental theorem of algebra works with complex numbers and DaTario mentioned them in the first post as well.

 

[Mark44]

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.

 

[DaTario]

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}].

 

[Svein]

The "superellipse" can have non-integer powers (https://en.wikipedia.org/wiki/Superellipse).

 

[DaTario]

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.

 

[FactChecker]

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 (x^1/10)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.

 

[Mark44]

And the answer is yes -- with conditions. As long as you can take the 10'th root of x, then (x^1/10)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.

 

[Svein]

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 \sqrt[12]{2}.

 

[DaTario]

And the answer is yes -- with conditions. As long as you can take the 10'th root of x, then (x^1/10)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".