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|f(z)| < z^n

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that if f(z) is entire and [tex]|f(z)|<|z|^{n}, \forall |z|> R[/tex] then f(z) must be a polynomial.


    2. Relevant equations

    Taylor Series

    [tex]f(z) =\displaystyle\sum_{j=0}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}[/tex]

    Cauchy's Estimate
    If f is analytic interior to and on a circle C centered at [tex]z_{0}[/tex], radius [tex]R = |z-z_{0}|[/tex]. Where M is the maximum value of |f(z)|.

    [tex]\left| f^{j}(z_{0}) \right| \leq \frac{j!M}{R^{n}}[/tex]

    3. The attempt at a solution

    f(z) is entire so it has Taylor series.

    [tex]f(z) =\displaystyle\sum_{j=0}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}[/tex]

    I need to show that this series has a finite number of terms.

    [tex]=\displaystyle\sum_{j=o}^{n}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}+\displaystyle\sum_{j=n+1}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}[/tex]

    The first sum is the polynomial part of the series, now to show the other part is zero. Consider the coefficient of the (n+1)th term:

    [tex]a_{n+1}=\frac{f^{(n+1)}(z_{0})}{(n+1)!}[/tex]

    Because,

    [tex]|f^{(n+1)}(z_{0})|\leq \frac{M(n+1)!}{R^{n+1}}[/tex]

    R is the radious of the circle around z0, and M is the max value of f(z).

    [tex]|a_{n+1}|\leq MR^{-n-1}[/tex]

    Now I'm lost... Is M the same thing as [tex]|z|^{n}[/tex]?
    Is this even making any sense?

    If M is at most [tex]|z|^{n}[/tex] then can I say that for R large this term is zero?
     
    Last edited: Mar 6, 2008
  2. jcsd
  3. Mar 6, 2008 #2
    Anyone?
     
  4. Mar 6, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It making sense. Set z0 to be 0. M is the maximum of |f(z)| over the contour. Since the contour is a circle of radius R, |z|=R. |f(z)|<=|z|^n=R^n. So you have shown |f^(n+1)(0)|<=(n+1)!/R for any R. So f^(n+1)(0)=0. Now just change the argument at little to show that f^(n+j)=0 for j>=1.
     
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