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How can I construct an entire function whose growth rate at infinity satisfies

$$ \lim_{r \to \infty} \frac{\log M(r)}{\sqrt {r}} =1$$

where $M(r) = \max_{|z|=r} |f(z)|$?

Based on the above limit, I think that I can say $M(r) \to e^{\sqrt{r}}$, as r grows to infinity. So, this shows that $f(z)$ has exponential growth like $e^\sqrt{r}$.

Where can I go from here? Can I manipulate the known series for $e^z$ and claim that this series

$$\sum \frac{(\sqrt{z})^n}{n!}$$ is the entire function that we want?

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    $\begingroup$ Well, that won't give you an entire function. $(\sqrt{z})^n$ poses a problem for odd $n$. What happens if you drop the problem-makers? $\endgroup$ – Daniel Fischer Oct 10 '15 at 9:36
  • $\begingroup$ That's such a cool hint, @DanielFischer :-) I'll try again now - thanks so much... $\endgroup$ – User001 Oct 10 '15 at 9:43
  • $\begingroup$ Hi @DanielFischer, if I drop the odd power terms and keep the even power summands, I now have the series for $cosh(\sqrt{z})$, which is entire and grows just like M(r). What do you think? Also, you mentioned that the odd powers of $\sqrt{z}$ are problematic; I think you are referring to the fact that we must choose a branch of the complex logarithm. But, wouldn't the even power terms also require branch cuts -- and also be problematic? Thanks, $\endgroup$ – User001 Oct 10 '15 at 10:14
  • $\begingroup$ Oh ... the summands will become z^n / (2n)! so the square root is gone and there's no need to choose a branch of log anymore @DanielFischer. Thanks again :-) $\endgroup$ – User001 Oct 10 '15 at 10:30
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    $\begingroup$ Since $\cosh$ is an even function, the function $\cosh \sqrt{z}$ is well-defined (ditto $\cos \sqrt{z}$ or $\dfrac{\sin \sqrt{z}}{\sqrt{z}}$ and so on), the ambiguity of the choice of $\sqrt{z}$ is annihilated by the evenness of the function one applies to it. $\endgroup$ – Daniel Fischer Oct 10 '15 at 11:18
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Your idea goes in the right direction, but doesn't quite work out. If we choose a branch of $\sqrt{z}$ on a domain where one exists, and expand $e^{\sqrt{z}}$, we get

$$\sum_{n = 0}^\infty \frac{(\sqrt{z})^n}{n!} = \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k}}{(2k)!} + \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k+1}}{(2k+1)!} = \sum_{k = 0}^\infty \frac{z^k}{(2k)!} + \sqrt{z}\sum_{k = 0}^\infty \frac{z^k}{(2k+1)!}.$$

Both series clearly yield entire functions, but the $\sqrt{z}$ factor on the second sum makes it impossible for the whole thing to be an entire function. So what happens if we take only part of it? If we take the first series, we get

$$\sum_{k = 0}^\infty \frac{z^k}{(2k)!} = \cosh \sqrt{z} = \frac{1}{2}\bigl(e^{\sqrt{z}} + e^{-\sqrt{z}}\bigr).$$

That is an entire function (the ambiguity of $\sqrt{z}$ is annihilated by the evenness of $\cosh$/the fact that $\sqrt{z}$ and $-\sqrt{z}$ are both used in the same way), and the exponential representation strongly suggests - or makes it evident - that this function has the right growth behaviour. If we had taken the other series, the resulting function $\dfrac{\sinh \sqrt{z}}{\sqrt{z}}$ would also have the right growth, the $\sqrt{z}$ in the denominator is insignificant in comparison with the exponential involved in $\sinh$.

We can extend this approach systematically to obtain functions such that

$$\lim_{r\to \infty} \frac{\log M(r)}{r^\alpha} = 1$$

for rational $\alpha > 0$. Basically, we want something containing $e^{z^{\alpha}}$, but for $\alpha \notin \mathbb{N}$ we need a modification to obtain an entire function. If $\alpha = \frac{m}{k}$, we retain only the terms of

$$e^{z^\alpha} = \sum_{n = 0}^\infty \frac{(z^\alpha)^n}{n!}$$

for which the exponent $n \alpha$ is an integer, so $k \mid n$. That gives the function

$$g_{\alpha}(z) = \sum_{n = 0}^\infty \frac{(z^\alpha)^{kn}}{(kn)!} = \sum_{n = 0}^\infty \frac{z^{mn}}{(kn)!}.$$

To verify that this has the right growth behaviour, let $\zeta_k = \exp \frac{2\pi i}{k}$ and consider the function

$$h_k(z) = \frac{1}{k}\sum_{s = 0}^{k-1} e^{\zeta_k^s\cdot z} = \sum_{n = 0}^\infty \Biggl(\frac{1}{k}\sum_{s = 0}^{k-1} \zeta_k^{n\cdot s}\Biggr)\frac{z^n}{n!} = \sum_{n = 0}^\infty \frac{z^{kn}}{(kn)!}.$$

Then we see that $g_{\alpha}(z) = h_k({z^{\alpha}})$ has the right growth:

Since $\lvert e^{\zeta_k^s\cdot z}\rvert \leqslant e^{\lvert z\rvert}$, we have $\lvert h_k(z)\rvert \leqslant e^{\lvert z\rvert}$, and for $x > 0$ we have $\operatorname{Re} (\zeta_k^s x) = \bigl(\cos \frac{2\pi s}{k}\bigr)x \leqslant \bigl(\cos \frac{2\pi}{k}\bigr)x$ for $0 < s < k$, so

$$\lvert h_k(x)\rvert \geqslant \frac{1}{k} e^{x} - \frac{k-1}{k} e^{x\cos \frac{2\pi}{k}} = \frac{e^x}{k}\Bigl(1 - (k-1)e^{-(1-\cos \frac{2\pi}{k})x}\Bigr) \sim \frac{e^x}{k}.$$

Altogether it follows that $\log M_{h_k}(r) \sim r$ and therefore $\log M_{g_\alpha}(r) \sim r^\alpha$.

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  • $\begingroup$ Hi @DanielFischer, thanks so much for this awesome answer - and for the generalization of the technique. Can I ask you a quick follow-up question? Can you elaborate just a bit more about how the ambiguity of $\sqrt{z}$ is annihilated by the evenness of cosh / the fact that $\sqrt{z}$ and -$\sqrt{z}$ are both used in the same way? $\endgroup$ – User001 Oct 11 '15 at 1:12
  • $\begingroup$ What do you mean, when you say this? When I look at the manipulated series of $e^z$, and arrive at the series for cosh($\sqrt{z}$), I am happy with it, primarily because the summands do not have a square root anymore. But when I look the exponential formula for cosh($\sqrt{z}$), there are obviously two exponential terms involving the (complex) square root, both terms divided by two. Looking at this formula would make me think that I'd have to choose a branch of logarithm for each term, and so this makes cosh($\sqrt{z}$)... not entire? Hmm...what do you think? Thanks, @DanielFischer $\endgroup$ – User001 Oct 11 '15 at 1:13
  • $\begingroup$ Oh..do you mean that the multivaluedness cancels each other out, @DanielFischer? I expanded out the logarithms of the exponential formula for cosh and notice that when the argument jumps by 2pi, the other term will jump by -2pi, resulting in a net effect of 0 jump in the argument. So the function is not multivalued anymore. Am I thinking correctly about this now? Thanks, $\endgroup$ – User001 Oct 11 '15 at 1:24
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    $\begingroup$ Yes, the multivaluedness cancels out. We have an expression of the form $G(z) = F(\sqrt{z}) + F(-\sqrt{z})$. For $z\neq 0$, on a neighbourhood $U$ of $z$ we have two branches, call them $a$ and $b$ of the square root. Then we have $a(z) = -b(z)$ for all $z\in U$ and $G(z) = F(a(z)) + F(-a(z)) = F(a(z)) + F(b(z)) = F(-b(z)) + F(b(z))$ is the same, whichever branch we choose on $U$. Hence $G$ is well-defined and analytic on $\mathbb{C}\setminus \{0\}$. At $0$, Riemann's removable singularity theorem gives us analyticity. $\endgroup$ – Daniel Fischer Oct 11 '15 at 10:26
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    $\begingroup$ It's the same for the more general case, the possible choices of $z^{\alpha}$ differ by factors of $\zeta_k^r$, and since we sum over $F(\zeta_k^s z)$, the ambiguity cancels out. $\endgroup$ – Daniel Fischer Oct 11 '15 at 10:28

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