The order of an entire function $f$ is defined to be $\lim_{R\to \infty}\sup_{\theta \in [0,2\pi)}\frac{\ln\ln |f(Re^{i\theta})|}{\ln R}$, consider $f=a_nz^n+\cdots +a_0$ a polynomial, then \begin{align}\operatorname{ord}(f) &= \lim_{R\to \infty}\sup_{\theta \in [0,2\pi)}\frac{\ln\ln |f(Re^{i\theta}|}{\ln R} \\&\leqslant \lim_{R\to \infty}\frac{\ln\ln\left(|a_n|R^n+\ldots+|a_0|\right)}{\ln R} \\&\leqslant \lim_{R\to \infty}\frac{\ln\ln(|a_n|+\ldots+|a_0|)R^n}{\ln R}\end{align} when $R$ large enough.

$$\lim_{R\to \infty}\frac{\ln\ln(|a_n|+\ldots+|a_0|)R^n}{\ln R}= \lim_{R\to \infty}\frac{\ln(\ln(|a_n|+\ldots+|a_0|)+n\ln R)}{\ln R}$$

How should I calculate the $ \lim_{R\to \infty}\frac{\ln(\ln(|a_n|+\ldots+|a_0|)+n\ln R)}{\ln R}$ ?

  • $\begingroup$ You need absolute value signs in places; please edit. $\endgroup$ – zhw. Nov 29 '16 at 22:10
  • $\begingroup$ sorry about that. $\endgroup$ – joseph Nov 30 '16 at 2:18

If $\ln(R) = t$, and $b = \ln(|a_n| + \ldots + |a_0|)$, this is $$ \lim_{t \to \infty} \dfrac{\ln(b+nt)}{t}$$ which is $0$, e.g. by l'Hôpital.


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