why its not removable singular point??

Nov 2008
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2
\(\displaystyle \frac{z-\sin z}{z^4}\)
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?
 

Jose27

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\(\displaystyle \frac{z-\sin z}{z^4}\)
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 \(\displaystyle \lim_{z\rightarrow 0} \frac{\cos (z)}{24z} =\infty\)
 
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Nov 2008
1,401
2
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
??
 

mr fantastic

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Dec 2007
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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.
 
Nov 2008
1,401
2
i want to solve it by the derivative way
not by developing into a series
 

mr fantastic

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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?