Exam problem essiential singulairity or not complex function

[xdrgnh]

Homework Statement

1/(1-cos(z))

Homework Equations

taylor expansion at 0 for cos(z)=1-x^2/2+x^4/24 and so on.
1/(x^2/2+x^4/24...)

The Attempt at a Solution

Because all of the powers are negative wouldn't that make it a essential singularity and 0 + 2∏n. Also it just explodes at those points.

 

[xdrgnh]

Oh btw the exam was already graded this isn't a take home. For the exam I got this question wrong and I don;t understand.

 

[micromass]

You haven't computed the Laurent series at 0 yet. Right now you have something of the form

\frac{1}{a_0+a_1x+a_2x^2+...}

But the Laurent series is supposed to be something of the form

...+b_{-2}\frac{1}{x^2}+b_{-1}\frac{1}{x}+b_0+b_1x+b_2x^2+...

Notice the difference between the two??

In fact, it is easy to see that

\frac{z^2}{1-cos(z)}

has a removable singularity at 0 (why?). What does that say about the original function?

 

[xdrgnh]

But in this case the Laurant series of cos(z) is equivalent to it's Taylor Series and it has all negative powers. That one has a removable singularity at 0 because it's limit as z approaches 0 is zero. Wouldn't that say that the original function doesn't have a removable singularity at 0? Also its not a pole so wouldn't it be a essential singularity like I think it is?

 

[micromass]

But in this case the Laurant series of cos(z) is equivalent to it's Taylor Series and it has all negative powers.

Sure, the Laurent series of cos(z) is equal to its Taylor series, but I don't see how the Taylor series of cos(z) has all negative powers??

Let me give another example. Consider the function

\frac{1}{1-z}

What is its Laurent series at 0?? You seem to argue that 1-z is the Taylor series of 1-z and thus that

\frac{1}{1-z}

is its own Laurent series and that it apparently has a pole of order 1 there.
But actually, the Laurent series of that function is

1+z+z^2+z^3+...

and it has a removable singularity at 0.

That one has a removable singularity at 0 because it's limit as z approaches 0 is zero.

Certain that the limit is 0?

Wouldn't that say that the original function doesn't have a removable singularity at 0?
Also its not a pole so wouldn't it be a essential singularity like I think it is?

It indeed means that the original function does not have a removable singularity at 0. It means that it actually has a pole of order 2.

 

[xdrgnh]

cos(z) doesn't have negative powers. But for 1/(1-cos(z)) does because it equals as stated above 1/(z^2/2+z^4/24 and so on) You're right my bad it's 2 lol. I tried doing L'hospital rule in my head. I was thinking of 1/(1-z) and if you sub in Cosine it would be 1+cos(z)+coz(z)^2 and so on. Did I make my error by only taking the Laurent series of Cos(z)?

 

[micromass]

cos(z) doesn't have negative powers. But for 1/(1-cos(z)) does because it equals as stated above 1/(z^2/2+z^4/24 and so on)

Yes, but

\frac{1}{z^2/2 + z^4/24+...}

is not the Laurent series of 1/(1-cos(z)). Do you see that? Compare to my example of 1/(1-z).

 

[xdrgnh]

Ah I see where I went wrong. so 1+cos(z)+cos(z)^2+cos(z)^3+cos(z)^4 ...is the Laurent Series and it at z=0 it goes to infinity and this is not a essential singularity.


I'll be posting another question up also in a minute or two.

 

[micromass]

Ah I see where I went wrong. so 1+cos(z)+cos(z)^2+cos(z)^3+cos(z)^4 ...is the Laurent Series and it at z=0 it goes to infinity and this is not a essential singularity.

No, that is not the Laurent series either. The Laurent series is of the form

...+b_{-2}\frac{1}{z^2}+b_{-1}\frac{1}{z}+b_0+b_1 z + b_2 z^2 +...

Your series is not of that form.

 

[xdrgnh]

Whoops no wonder I got it wrong if I messed up the 2nd time. I thought you were typing a general form for all of them. And now I see why it has a removable singularity because z^2/z^2 is of course 1 so the limit would be whatever the b_2. Well I see how it is a pole or order 2 when you give me the general form of this series for 1/(1-cos(z)). But I don't see how you got that form.

 

[micromass]

To find the actual Laurent series you need to make the division

\frac{1}{z^2/2 + z^4/24+...}

We can write

\frac{1}{1-cos(z)}=\frac{1}{z^2} \frac{1}{1/2 + z^2/24 + ...}

The left factor is now defined at 0 so it has a Taylor series. So we can write

\frac{1}{1/2 + z^2/24 +...}=a_0+a_1z+a_2z^2+...

or equivalently

(1/2 + z^2/24 +...)(a_0+a_1z+a_2z^2+...)=1

When we multiply that out, you can find the coefficients you're looking for.

 

[xdrgnh]

Yah that was my biggest fear. I really didn't want to divide while taking the exam so I tried to work my way around that and it didn't work. But I see how you can use that 1/1-z to get it into that form and that the power are all not negative and the highest negative power is 2 and the it eventually becomes positive thus turning it into a pole of order 2.

Alright thanks.