Show that $$f(z)=\frac{\sin\sqrt z}{\sqrt z}$$ is an entire function of finite order $\rho$ and determine $\rho$.

I observed that the two determinations of the square root differ only for the signum. Since $\sin(-z)=-\sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing. For the order i use the Taylor expansion $$\sin z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$ which for $z=\sqrt z$ gives $$\sin\sqrt z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{n}\sqrt z}{(2n+1)!}$$ Thus $$\frac{\sin\sqrt z}{\sqrt z}=\sum_{n=0}^{\infty}(-1)^n\frac{z^{n}}{(2n+1)!}$$ Then we have $$(2n+1)!\geq2^n n!$$ hence $$\bigg|\frac{\sin\sqrt z}{\sqrt z}\bigg|\leq\large\sum_{n=0}^{\infty}\left(\frac{|z|}{2}\right)^{n}\cdot\frac{1}{n!}=e^{|z|/2}$$

This (if correct) shows that $\rho\leq\frac{1}{2}$. How can be shown the identity?

  • $\begingroup$ There are a few mistakes here. The series for $\sin z$ is not quite right. You mean: $$\sin z = \sum_{n=0}^{\infty} \frac{(-1)^nz^{n+1}}{(2n+1)!} \, . $$ When you change from the series for $\sin z$ to the series for $\sin \sqrt{z}$, you seem to change $z^{2n}$ to $z^n\sqrt{z}$. This isn't correct. $(\sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$\sum_{n=0}^{\infty} \frac{(-1)^nz^n\sqrt{z}}{(2n+1)!} = \frac{\sqrt{z}\sinh \sqrt{-z}}{\sqrt{-z}} \, . $$ $\endgroup$ Jan 9 '13 at 20:08
  • $\begingroup$ sorry, there was an error in the $sin z$ series, i've edited. thank you $\endgroup$ Jan 9 '13 at 20:09
  • $\begingroup$ Remember: You can use \sin \cos \tan \sec \cot and \csc for the trig' functions in LaTeX, i.e. $\sin z$ instead of $sin z$. $\endgroup$ Jan 9 '13 at 20:14

To show $\rho \geq 1/2$, write

$$ f(z) = \frac{\sin \sqrt{z}}{\sqrt{z}} = \frac{e^{i\sqrt{z}}-e^{-i\sqrt{z}}}{2i\sqrt{z}} $$

and show that

$$ f(-x) \sim \frac{e^{\sqrt{x}}}{2\sqrt{x}} $$

as $x \to \infty$ with $x > 0$.

Clarification: If $f$ is of order $\rho'$ then for any $\rho>\rho'$ there is $C$ such that $$|f(z)| \leq C \exp(|z|^\rho)$$ so that $|f(z)| \exp(-|z|^\rho)$ is bounded. In particular, $$|f(-x)| \exp(-x^\rho)$$ is bounded as $x \to \infty$. Now suppose $\rho'< 1/2$, pick $\rho\in (\rho',1/2)$ and use the asymptotic estimate above.

  • $\begingroup$ $f(-x)=\frac{e^{-\sqrt x} -e^{\sqrt x}}{-2\sqrt x}$, dividing this by $-\frac{e^{\sqrt x}}{2\sqrt x}$ i get something which tends to 1 as x goes to $\infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$? $\endgroup$ Jan 9 '13 at 20:57
  • $\begingroup$ @FedericaMaggioni, If $f$ is of order $\rho$ then $$|f(z)| \leq C \exp(|z|^\rho)$$ for $|z|$ large enough, so that $$|f(z)| \exp(-|z|^\rho)$$ is bounded. In particular, $$|f(-x)| \exp(-x^\rho)$$ is bounded as $x \to \infty$. Now suppose $\rho < 1/2$ and use the asymptotic estimate. $\endgroup$ Jan 9 '13 at 21:10
  • $\begingroup$ substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness. $\endgroup$ Jan 9 '13 at 21:17
  • $\begingroup$ @FedericaMaggioni exactly. $\endgroup$ Jan 9 '13 at 21:17
  • 1
    $\begingroup$ But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review. $\endgroup$
    – user147263
    Jun 27 '15 at 3:52

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