Why its not removable singular point?

[nhrock3]

why its not removable singular point??

$$\frac{z-\sin z}{z^4}$$

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?

 

[Office_Shredder]

If you do L'Hopital on it three times you don't get something you can do L'Hopital on again

Try writing out sine as a power series and see if you can get some info on what the pole should be from that

 

[nhrock3]

so in a complex function
what are the conditions for me doing lhotipal
is it just 0/0 inf/inf linke in real functions
or are the other conditions?

 

[Office_Shredder]

Same as for real functions. After doing L'Hopital three times the numerator is a cosine, that does NOT give you 0/0

 

[nhrock3]

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
??

 

[lanedance]

have you tried the power series suggested in post #2?

 

[vela]

It's a simple pole because the smallest positive integer n for which

$$\lim_{z \to 0}\,(z-0)^n\,\frac{z-\sin z}{z^4}$$

exists is n=1. This is probably what you had in mind when you decided to take limits.

From a practical standpoint, it's not a very efficient way of determining the order of a pole. It's better to find the Laurent series as the others have suggested.

 

[nhrock3]

the derivative way is a way from my textbook
do you know it?
if so can you tell me where did i got wrong
??
i tool g=1/f and i put x=a if every derivative till i get to the derivative for ach
it differs zero

where is my mistake
?

 

[hunt_mat]

A removable singular point is a point which if you evaluate the function you'll get 0/0 but when you expand the function via a taylor series then it vanishes. To give an example:
$$\frac{\sin z}{z}$$
When you evaluate at z=0, you get 0/0, but when you expand in a taylor series you arrive at:
$$\frac{\sin z}{z}=1-\frac{z^{2}}{3!}+\cdots$$
There isn't a problem on the RHS.

 

[lanedance]

ok so show us your derivatives...

as you won't try the power series, i'll show you the utility of it:
$$sin(x) = x - \frac{x^3}{3!}+O(x^5)$$

then
$$f(x) = \frac{z-sin(z)}{z^4} = \frac{z-(z-\frac{z^3}{3!}+O(z^5))}{z^4} = \frac{z^3}{3! z^4}+O(z) = \frac{1}{3! z}+O(z)$$

which shows its a 1st order pole

 

[vela]

0/0 is an indeterminate form. You need to use the hospital rule to calculate the limit.

 

[nhrock3]

but its not a limit
its a derivative

 

[nhrock3]

ok so show us your derivatives...

as you won't try the power series, i'll show you the utility of it:
$$sin(x) = x - \frac{x^3}{3!}+O(x^5)$$

then
$$f(x) = \frac{z-sin(z)}{z^4} = \frac{z-(z-\frac{z^3}{3!}+O(z^5))}{z^4} = \frac{z^3}{3! z^4}+O(z) = \frac{1}{3! z}+O(z)$$

which shows its a 1st order pole

ok suppose if i have a singularity at point a in this function
so i put (z-a) instead of z everywhere

 

[Office_Shredder]

but its not a limit
its a derivative

Whenever you have a function of the form 0/0 at a point and you want to know if there's a removable singularity there you can just calculate what the limit as you approach that point is to see if you get continuity

 

[i2c]

Am I the only one thinking, "Graph (z-sinz)/z^4 and you will see why it's the limit does not exist (i.e. not a single removable point) at z = 0 ?" (hint hint, it has the same shape as a 1/x graph)