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futurebird

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## Homework Statement

Prove that if f(z) is entire and [tex]|f(z)|<|z|^{n}, \forall |z|> R[/tex] then f(z) must be a polynomial.

## Homework 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]

## 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?

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