Catégorie:Quadrangle de Sinus Sabaeus
v · m Quadrangles de Mars | |||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| |||||||||||||||||||||||||||||||||||||||||||||||||
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...
