- Find the order of growth of the following entire functions:
\(p ( z )\) where p is a polynomial.
\(e ^ { b z ^ { n } }\) for \(b \neq 0\)
\(e ^ { e ^ { z } }\)
Solution
Recall that the order of an entire function \(f\) is the infimum of the positive numbers \(\rho\) for which \[ |f(z)|\le A e^{B|z|^\rho} \] for suitable \(A,B>0\).
For (a), every polynomial has order \(0\). Indeed, if \(p\) has degree \(m\), then \(|p(z)|=O(|z|^m)\), and for every \(\rho>0\) one has \[ |z|^m\le C_\rho e^{|z|^\rho} \] for all \(z\). Thus the infimum of admissible positive \(\rho\) is \(0\).
For (b), \[ |e^{bz^n}|=e^{\Re(bz^n)}\le e^{|b||z|^n}, \] so the order is at most \(n\). Choose \(\theta\) so that \[ b e^{in\theta}=|b|. \] Then along \(z=re^{i\theta}\), \[ |e^{bz^n}|=e^{|b|r^n}. \] No bound \(Ae^{Br^\rho}\) with \(\rho<n\) can dominate this as \(r\to\infty\). Hence the order is exactly \[ \boxed n. \]
For (c), along the positive real axis, \[ |e^{e^x}|=e^{e^x}. \] If this function had finite order \(\rho\), then for suitable \(A,B>0\) we would have \[ e^{e^x}\le A e^{Bx^\rho} \] for all large \(x\), equivalently \[ e^x\le \log A+Bx^\rho, \] which is impossible. Therefore \(e^{e^z}\) has infinite order.