I want to find the order of the entire function $f(z)=\sin(z)$. I have this result
Let
$$f(z)=\sum_{n=0}^\infty a_n z^n$$
be an entire function, non-constant and with finite order. Then, the order is given by
$$\lambda=\limsup_{n\to \infty} \frac{n\log n}{-\log|a_n|}.$$
I've tried to apply this previous theorem to get (using Stirling's approximation)
$$\lambda=\limsup_{n\to \infty} \frac{n\log n}{-\log|a_n|}=
\limsup_{n\to \infty} \frac{n\log n}{(2n+1)\log(2n+1)-2n-1}=\frac{1}{2}.$$
However, the result should be $1$. I suspect that I cannot use this result because $a_{2n}=0$ is this case, and this will give problems in the expression.
How should I proceed?
Best Answer
First of all, the order of $\sin(z)$ is the same as the order of $\frac{\sin(z)}{z}.$ This, in turn, can be written as
$$ \frac{\sin(z)}{z} = g(z^2),$$ for some entire $g$. Thus, the order of $\sin(z)$ is twice the order of this $g$. But, from the Taylor coefficients of the sine function, we know that this $g$ can be written as $$ g(z) = \sum_{n=0}^{\infty} b_n z^n,$$ where $b_n = \frac{(-1)^n}{(2n+1)!}.$ Now you are allowed to use your result, together with Stirling's formula, to conclude.
P.S.: Notice also that the way you applied this result is slightly inaccurate, due to the fact that the given approximation holds only for $a_{2n}$, which implies there should be a $(2n)\log(2n)$ in the numerator instead of $n \log n.$