N roots?

  • Thread starter Mentallic
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  • #1
Mentallic
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Main Question or Discussion Point

The function [tex]y=e^x[/tex] can be expanded using the power series, thus [tex]y=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...[/tex]
This is a polynomial of infinite degree, and the theorem that says a polynomial must have at least one root in the complex field, and thus this extends to a polynomial of nth degree having n roots (not necessarily distinct).

However, for [itex]e^x=0[/itex] there are zero roots. Why is this possible when, clearly by the power series for [itex]e^x[/itex] it should be an infinite degree polynomial with infinite roots (in the complex plane)?
 

Answers and Replies

  • #2
Borek
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According to wiki:

In mathematics, a polynomial is an expression of finite length
so perhaps that's where the problem lies.
 
  • #3
Mentallic
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Was that definition placed before or after this exception was noticed?
 
  • #4
Borek
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No idea.

Actually, I have checked other source - Wolfram Mathworld - and their definition of polynomial doesn't state it has to be finite.
 
  • #5
HallsofIvy
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In any case, the statement "a polynomial of degree n has n zeros (counting multiplicity) over the complex field" is only true for n finite.
 
  • #6
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The phrase "polynomial of infinite degree" is never never used by real mathematicians. In particular [tex]e^z[/tex] is not (repeat not) a polynomial in [tex]z[/tex].
 

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