Liouville's theorem (extended) proof

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SUMMARY

Liouville's theorem states that if an entire function f satisfies the growth condition |f(z)| ≤ A|z|^{m} for |z| ≥ R_{0}, where A and m are positive constants, then f must be a polynomial of degree m or less. The proof utilizes Cauchy estimates, specifically the inequality |f^{n}(z_{0})| ≤ (n!/r^{n}) max_{|z-z_{0}|=r} |f(z)|, for n = 0, 1, 2, 3, etc. The solution involves recognizing that the derivatives of f must be evaluated m+1 times to demonstrate that they converge to zero, confirming that f is indeed a polynomial.

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Homework Statement



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

Homework 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,...

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...
 
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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.
 
ok,thats what I did and I got the answer, Thank you
 

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