Proof: f(z) is Entire & |f(z)|<|z|^{n} $\Rightarrow$ f(z) is a Polynomial

In summary, if f is analytic inside and on a circle C centered at z_{0}, and if M is the maximum value of |f(z)| over the contour C, then for any large R, the coefficient of the (n+1)th term in f(z) is zero.
  • #1
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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  • #2
Anyone?
 
  • #3
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.
 

1. What does it mean for a function to be entire?

Being entire means that the function is analytic (differentiable) at every point in the complex plane. This also implies that the function is continuous everywhere in the complex plane.

2. What is the significance of |f(z)|<|z|^n in the given statement?

This inequality tells us that the function f(z) is bounded by a polynomial function of z with degree n. This is an important condition because it allows us to prove that f(z) is a polynomial.

3. How does the condition |f(z)|<|z|^n prove that f(z) is a polynomial?

By using Cauchy's integral formula, we can show that if a function is analytic and bounded by a polynomial, then it must be a polynomial itself. This is essentially what the given statement is saying.

4. Can you provide an example of a function that satisfies the given conditions?

One example would be f(z) = z^2. This function is entire and satisfies the inequality |f(z)|<|z|^3 for all z in the complex plane.

5. Is the given statement a necessary and sufficient condition for a function to be a polynomial?

No, it is not. The statement only provides a sufficient condition, meaning that if a function satisfies the conditions, then it must be a polynomial. However, there are other conditions and properties that a function must satisfy in order to be considered a polynomial.

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