The definition of order I'm using is the following:
An entire function $f$ is said to be of finite order if there is a number $t > 0$ such that $$|f(z)| \leq e^{|z|^{t}}$$
for all sufficiently large $z$. The infimum of all such $t$ is called the order of $f$.
In another problem I had to prove that $cos(z^{\frac{1}{2}})$ was of order $\frac{1}{2}$. So, for this problem I'm wondering whether it would be possible to show that $cos(z^{\frac{1}{4}})$ was of order $\frac{1}{4}$.
From Taylor Series expansions:
$$ |cos(z^{\frac{1}{4}})|= |\Sigma^{\infty}_{n=0} (-1)^n \frac{z^{\frac{n}{4}}}{(2n)!} |$$
$$ e^{|z|^{\frac{1}{4}}} = \Sigma^{\infty}_{n=0} \frac{|z|^{\frac{1}{4}}}{n!} $$
So, $|cos(z^{\frac{1}{4}})| \leq e^{|z|^{\frac{1}{4}}} $ for sufficiently large $z$.
But,
$$ cos(z^{\frac{1}{4}} ) = \frac{e^{\frac{x}{4}}+ e^{\frac{-x}{4}} }{2} \geq \frac{e^{\frac{x}{4}}}{2} = \frac{e^{|z|^{\frac{1}{4}}}}{2} $$
Hence, $\frac{1}{4}$ is the infimum and the order of the function is $\frac{1}{4}$. Is my proof correct?