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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


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    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
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