# why its not removable singular point??

• Jul 12th 2010, 10:12 AM
transgalactic
why its not removable singular point??
$\displaystyle \frac{z-\sin z}{z^4}$
my singular point is 0
if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
which is singular point because the result is constant

so why its a first order pole?
• Jul 12th 2010, 10:21 AM
Jose27
Quote:

Originally Posted by transgalactic
$\displaystyle \frac{z-\sin z}{z^4}$
my singular point is 0
if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
which is singular point because the result is constant

so why its a first order pole?

Expand $\displaystyle z-\sin (z)$ in its Taylor series and you'll find it has an order 3 zero at 0

Your reasoning is wrong since applying L'hopital gives you $\displaystyle \lim_{z\rightarrow 0} \frac{\cos (z)}{24z} =\infty$
• Jul 12th 2010, 10:19 PM
transgalactic
ok now i tried to determine the power of the pole
i did
g(x)=1/f(x)
where x=0
and i got that g'(0)=0/0 which means that its not a first order pole
actualyy i should get zero but i got 0/0
??
• Jul 13th 2010, 03:31 AM
mr fantastic
Quote:

Originally Posted by transgalactic
ok now i tried to determine the power of the pole
i did
g(x)=1/f(x)
where x=0
and i got that g'(0)=0/0 which means that its not a first order pole
actualyy i should get zero but i got 0/0
??

Post #2 has given you a big hint.

The singularity of $\displaystyle \displaystyle f(z) = \frac{z - \sin z}{z^4}$ at z = 0 is a simple pole. Do what post #2 said to see why.
• Jul 13th 2010, 04:12 AM
transgalactic
i want to solve it by the derivative way
not by developing into a series
• Jul 13th 2010, 04:57 AM
mr fantastic
Quote:

Originally Posted by transgalactic
i want to solve it by the derivative way
not by developing into a series

Well then, where is your trouble in using l'Hopital's Rule a couple of times to find that $\displaystyle \displaystyle {\lim_{z \to 0} \frac{z(z - \sin z)}{z^4} = {\lim_{z \to 0} \frac{z - \sin z}{z^3}}$ exists and is finite?