Let $f$ be an entire function. If there exist a positive number $\rho$ and constants $A,B>0$ such that
$$|f(z)| \le Ae^{B|z|^\rho} \; \text{for all}\; z \in \mathbb{C},$$ then we say that $f$ has an order of growth $\le \rho$.
Consider $f(z)=\cos z^{1/2}$, which we define by
$$\cos z^{1/2} = \sum_{n=0}^\infty (-1)^n \frac{z^n}{(2n)!}.$$
Then $f$ is entire. The order of growth is given to be $1/2$.
From Taylor expansion, it is easy to see that $|\cos z^{1/2}| \le e^{|z|^{1/2}}$. However, how do we show that the order of growth is exactly $1/2$, i.e. if we had $\rho < 1/2, A,B>0$, then there is $z\in \mathbb{C}$ such that $|\cos z^{1/2}| > A e^{B|z|^\rho}$?
Best Answer
From $\cos(z) = \cosh(iz)$ one can see that the cosine growth maximally along the imaginary axis. That leads to the following argument:
For $z = -x $ with $x \in \Bbb R$ and $x> 0$ you have $$ \cos(z^{1/2}) = \frac{e^{x^{1/2}}+ e^{-x^{1/2}}}{2} \ge \frac 1 2 e^{x^{1/2}} = \frac 1 2 e^{|z|^{1/2}} $$ so that the growth order is at least $1/2$.