Editing Entire function (section)

=={{anchor|order of an entire function}} Order and type ==
The '''order''' (at infinity) of an entire function <math>f(z)</math> is defined using the [[limit superior]] as:

<math display="block">\rho = \limsup_{r\to\infty}\frac{\ln \left (\ln\| f \|_{\infty, B_r} \right ) }{\ln r},</math>

where <math>B_r</math> is the disk of radius <math>r</math> and <math>\|f \|_{\infty, B_r}</math> denotes the [[supremum norm]] of <math>f(z)</math> on <math>B_r</math>. The order is a non-negative [[real number]] or infinity (except when <math>f(z) = 0</math> for all <math>z</math>). In other words, the order of <math>f(z)</math> is the [[infimum]] of all <math>m</math> such that:

<math display="block">f(z) = O \left (\exp \left (|z|^m \right ) \right ), \quad \text{as } z \to \infty.</math>

The example of <math>f(z) = \exp(2z^2)</math> shows that this does not mean <math>f(z)=O(\exp(|z|^m))</math> if 
<math>f(z)</math> is of order <math>m</math>.

If <math>0<\rho < \infty,</math> one can also define the '''''type''''':

<math display="block">\sigma=\limsup_{r\to\infty}\frac{\ln  \| f\|_{\infty, B_r}} {r^\rho}.</math>

If the order is 1 and the type is <math>\sigma</math>, the function is said to be "of [[exponential type]] <math>\sigma</math>". If it is of order less than 1 it is said to be of exponential type 0.

If <math display="block"> f(z)=\sum_{n=0}^\infty a_n z^n,</math> then the order and type can be found by the formulas
<math display="block">\begin{align}
\rho &=\limsup_{n\to\infty} \frac{n\ln n}{-\ln|a_n|} \\[6pt]
(e\rho\sigma)^{\frac{1}{\rho}} &= \limsup_{n\to\infty} n^{\frac{1}{\rho}} |a_n|^{\frac{1}{n}}
\end{align}</math>

Let <math>f^{(n)}</math> denote the <math>n</math>-th derivative of <math>f</math>. Then we may restate these formulas in terms of the derivatives at any arbitrary point <math>z_0</math>:

<math display="block">\begin{align}
\rho &=\limsup_{n\to\infty}\frac{n\ln n}{n\ln n-\ln|f^{(n)}(z_0)|}=\left(1-\limsup_{n\to\infty}\frac{\ln|f^{(n)}(z_0)|}{n\ln n}\right)^{-1} \\[6pt]
(\rho\sigma)^{\frac{1}{\rho}} &=e^{1-\frac{1}{\rho}} \limsup_{n\to\infty}\frac{|f^{(n)}(z_0)|^{\frac{1}{n}}}{n^{1-\frac{1}{\rho}}}
\end{align}</math>

The type may be infinite, as in the case of the [[reciprocal gamma function]], or zero (see example below under {{slink||Order 1}}).

Another way to find out the order and type is [[Matsaev's theorem]].

===Examples===
Here are some examples of functions of various orders:

====Order ''ρ''====
For arbitrary positive numbers <math>\rho</math> and <math>\sigma</math> one can construct an example of an entire function of order <math>\rho</math> and type <math>\sigma</math> using:

<math display="block">f(z)=\sum_{n=1}^\infty \left (\frac{e\rho\sigma}{n} \right )^{\frac{n}{\rho}} z^n</math>

====Order 0====
* Non-zero polynomials
*<math>\sum_{n=0}^\infty 2^{-n^2} z^n</math>

====Order 1/4====
<math display="block">f(\sqrt[4]z)</math> 
where <math display="block">f(u)=\cos(u)+\cosh(u)</math>

====Order 1/3====
<math display="block">f(\sqrt[3]z)</math>
where
<math display="block">f(u)=e^u+e^{\omega u}+e^{\omega^2 u} = e^u+2e^{-\frac{u}{2}}\cos \left (\frac{\sqrt 3u}{2} \right ), \quad \text{with } \omega \text{ a complex cube root of 1}.</math>

====Order 1/2====
<math display="block">\cos \left (a\sqrt z \right )</math> with <math>a\neq 0</math> (for which the type is given by <math>\sigma=|a|</math>)

====Order 1====
*<math>\exp(az)</math> with <math>a\neq 0</math> (<math>\sigma=|a|</math>)
*<math>\sin(z)</math>
*<math>\cosh(z)</math>
*the [[Bessel function]]s <math>J_n(z)</math> and spherical Bessel functions <math>j_n(z)</math> for integer values of <math>n</math><ref>See asymptotic expansion in Abramowitz and Stegun, [https://personal.math.ubc.ca/~cbm/aands/page_377.htm p. 377, 9.7.1].</ref>
*the [[reciprocal gamma function]] <math>1/\Gamma(z)</math> (<math>\sigma</math> is infinite)
*<math>\sum_{n=2}^\infty \frac{z^n}{(n\ln n)^n}. \quad (\sigma=0)</math>

====Order 3/2====
* [[Airy function]] <math>Ai(z)</math>

====Order 2====
*<math>\exp(az^2)</math> with <math>a\neq 0</math> (<math>\sigma=|a|</math>)
*The [[Barnes G-function]] (<math>\sigma</math> is infinite).

====Order infinity====
*<math>\exp(\exp(z))</math>