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N roots?

  1. Mar 10, 2010 #1

    Mentallic

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    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)?
     
  2. jcsd
  3. Mar 10, 2010 #2

    Borek

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    Staff: Mentor

    According to wiki:

    so perhaps that's where the problem lies.
     
  4. Mar 10, 2010 #3

    Mentallic

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    Was that definition placed before or after this exception was noticed?
     
  5. Mar 10, 2010 #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.
     
  6. Mar 10, 2010 #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.
     
  7. Mar 10, 2010 #6
    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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