Order of an entire function

Remarks

Equivalently, $\rho$ is the infimum of the numbers $\lambda \ge 0$ such that $M(r) < \exp(r^\lambda)$ for all sufficiently large $r$. Example: for $f(z) = e^{z^2}$, $M(r) = e^{r^2}$, so $\log \log M(r)/\log r = 2$ and the order is $2$. Non-example for a bound: $e^{z^2}$ does not satisfy $M(r) < \exp(r^{\lambda})$ for large $r$ when $\lambda < 2$, so its order is not at most $1$. For a nonconstant polynomial, $\log M(r) \sim d \log r$ with $d$ the degree, so the order is $0$. In terms of the Taylor coefficients of $f(z) = \sum_n a_n z^n$, the order is $\limsup_{n \to \infty} \dfrac{n \log n}{\log(1/\abs{a_n})}$, where terms with $a_n = 0$ are omitted. If $0 < \rho < \infty$, the type of $f$ is $\sigma = \limsup_{r \to \infty} r^{-\rho} \log M(r)$; functions of order $1$ and finite type, or of order less than $1$, are those of exponential type.