How can I calculate the residue at 0 of the next function:
$\displaystyle f(z)=\frac{1}{z^2 \sin z}$ ??
Hope you'll be able to help me
Thanks a lot!
By calculating the residue "at infinity" of $\displaystyle \frac{1}{f(z)}= z^2 sin(z)$.
That is, write sin(z) as a Taylor's series, then multiply by $\displaystyle z^2$. That, of course, will be a power series with only positive powers. Now invert the powers getting a Laurent series with only negative powers. The residue is the coefficient of the $\displaystyle z^{-1}$ term.
Here is my try:
$\displaystyle z^2 sin(z)=z^2 (z-\frac{z^3}{3!} + \frac{z^5}{5!} -...)=$
$\displaystyle z^3 - \frac{z^5}{3!} +\frac{z^7}{5!} -... $
Hence:
$\displaystyle \frac{1}{z^2 sin(z) } = \frac{1}{z^3 - \frac{z^5}{3!} +\frac{z^7}{5!} -...} =$
$\displaystyle = \frac{1}{\sum_{n=3}^{\infty}\frac{z^n}{(n-2)!} } $
and this is where I got stuck in my calculation... We can't just right it with negative powers because all the sum is in negative power...
Hope you'll be able to correct my misunderstanding and help me continue...
Thanks a lot !
Let's start with the series expansion of the function...
$\displaystyle f(z)= \frac{z}{\sin z} = a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots$ (1)
... from the well know result...
$\displaystyle \frac{\sin z}{z} = \sum_{n=0}^{\infty} (-1)^{n}\cdot \frac{z^{2n}}{(2n+1)!} = 1 - \frac{z^{2}}{6} + \frac{z^{4}}{120} - \dots$ (2)
The most 'spontaneous' approach is to impose...
$\displaystyle (a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots) \cdot (1 - \frac{z^{2}}{6} + \frac{z^{4}}{120} - \dots)= 1$ (3)
... and that permits us to compute the $\displaystyle a_{n}$ in recursive way...
$\displaystyle a_{0}=1$
$\displaystyle a_{1} =0$
$\displaystyle a_{2} = \frac{a_{0}}{6} = \frac{1}{6}$
$\displaystyle a_{3} =0$
$\displaystyle a_{4} = \frac{1}{36} - \frac{1}{120} = \frac{7}{360}$
... and so one...
Now we can write...
$\displaystyle \frac{z}{\sin z} = 1 + \frac{z^{2}}{6} + \frac{7}{360}\cdot z^{4} + \dots $ (4)
... from which it follows...
$\displaystyle \frac{1}{z^{2}\cdot \sin z} = z^{-3} + \frac{z^{-1}}{6} + \frac{7}{360}\cdot z + \dots$ (5)
... so that the residue of $\displaystyle \frac{1}{z^{2}\cdot \sin z}$ in $\displaystyle z=0$ is $\displaystyle \frac{1}{6}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Actually I didn't understand your last result:
$\displaystyle
\frac{z}{\sin z} = 1 + \frac{z^{2}}{6} + \frac{7}{360}\cdot z^{4} + \dots
$
It means that:
$\displaystyle
\frac{\frac{1}{z^2}}{\sin z} = 1 + \frac{(\frac{1}{z^2})^{2}}{6} + \frac{(\frac{1}{z^2})^4}{360}\cdot + \dots
$
Which means the Residue is excatly 0, isn't it?
Thanks!
If You start from $\displaystyle f(z) = \frac{z}{\sin z}$ , then is...
$\displaystyle \frac{1}{z^{2}\cdot \sin z} = \frac{f(z)}{z^{3}}$ (1)
Your Laurent expression is very interesting however, but the correct expression is...
$\displaystyle \frac{1}{z^{2}\cdot \sin \frac{1}{z^{2}}} = 1 + \frac{1}{6}\cdot z^{-4} + \frac{7}{360}\cdot z^{-8} + \dots$ (2)
The function represented in (2) has in $\displaystyle z=0$ an essential singularity, that is not the case for $\displaystyle \frac{1}{z^{2}\cdot \sin z}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$