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!

Liouville's theorem (extended) proof

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    f is an entire function with pos. constants A and m such that |f(z)| ≤ A|z|[itex]^{m}[/itex] for all z: |z|≥R[itex]_{0}[/itex]

    Show that f is polynomial of degree m or less

    2. Relevant equations

    Cauchy estimates need to be used here

    |f[itex]^{n}[/itex](z[itex]_{0}[/itex])|≤[itex]\frac{n!}{r^{n}}[/itex]max[itex]_{z-z_{0}=r}[/itex]|f(z)| , n=0,1,2,3,....

    3. The attempt at a solution
    I was thinking that the right side of the inequality would just be a constant and could be treated as such.. meaning that taking the derivative of the left side would eventually result in a constant and then 0... but i'm not sure how to show that you have to take the derivative m+1 times...
     
  2. jcsd
  3. Mar 18, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If f(z) is entire then it has a power series expansion at z=0. f(z)=a0+a1*z+a2*z^2+... [itex]f^n(0)[/itex] is related to an. Use your Cauchy estimate on that.
     
  4. Mar 19, 2012 #3
    ok,thats what I did and I got the answer, Thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Liouville's theorem (extended) proof
  1. Liouville's Theorem (Replies: 8)

Loading...