Mallaig
Cette page d’homonymie répertorie les différents sujets et articles partageant un même nom.
Mallaig est une ville canadienne de l'Alberta
Mallaig est un port écossais des Highlands.
Order of growth of the entire function $sin(sqrt{z})/sqrt{z}$
6
4
$begingroup$
Show that $$f(z)=frac{sinsqrt z}{sqrt z}$$ 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 $$sin z=sum_{n=0}^{infty}(-1)^nfrac{z^{2n+1}}{(2n+1)!}$$ which for $z=sqrt z$ gives $$sinsqrt z=sum_{n=0}^{infty}(-1)^nfrac{z^{n}sqrt z}{(2n+1)!}$$ Thus $$frac{sinsqrt z}{sqrt z}=sum_{n=0}^{infty}(-1)^nfrac{z^{n}}{(2n+1)!}$$ Then we have $$(2n+1)!geq2^n n!$$ hence $$bigg|frac{sinsqrt z}{sqrt z}bigg|leqlargesum_{n=0}^{infty}left(frac{|z|}{2}right)^{n}cdotfrac{1}{n!}=e^{|z|/2}$$ This (if correct...
