1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

N roots?

  1. Mar 10, 2010 #1

    Mentallic

    User Avatar
    Homework Helper

    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

    User Avatar

    Staff: Mentor

    According to wiki:

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

    Mentallic

    User Avatar
    Homework Helper

    Was that definition placed before or after this exception was noticed?
     
  5. Mar 10, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook