e^x
Badges: 15
Rep:
?
#1
Report Thread starter 3 years ago
#1
Can some1 check if the solution to Q2b is correct? And if it is correct can you explain why?
Attached files
0
reply
poorform
Badges: 10
Rep:
?
#2
Report 3 years ago
#2
(Original post by e^x)
Can some1 check if the solution to Q2b is correct? And if it is correct can you explain why?
Didn't work through it myself explicitly but it seems to be correct.

You basically want to manipulate whatever you have and get it into a form where you can exploit the properties of a geometric series.

Since you are talking about a Laurent series centred at 0 then the 1/z part is in fact the Laurent series for 1/z centred at 0 so you have to do no work there.

The other two parts are just about thinking about what annulus you want your function to converge on and then just manipulating the respective part to do that.

So for example for the second part of the partial fraction on 0<|z|<1 we want to find a Laurent series expansion for -1/(z-1) valid on 0<|z|<1.

One way to do this is to note that \displaystyle -1/(z-1)=1/(1-z)=\sum_{n=0}^{\infty} z^n by geometric series and this is valid for |z|<1 so it is certainly valid on 0<|z|<1 which is what we wanted. The rest of the question follows the same line the only thing that changes is how you manipulate the fractions to make sure it is valid for the region you want.

You will want something like 1/(1-z) for Laurent series expansions (for things like a<|z|) and something like 1/(1-(1/z)) for things regions like |z|<b.

It turns out that the Laurent series of functions are equal to the Taylor series of functions in some cases. This is the case when you are looking for series' that are defined on an annulus inside the singularity, like above we wanted to find out the laurent series for -1/z-1 inside |z|<1 so it gives in fact the Taylor series at the point (i.e. all the negative coeffiecents are zero)

No idea if that helped at all I may have just been rambling though.

There are some good youtube videos out there and you will need to practise problems to get familiar with it.
0
reply
e^x
Badges: 15
Rep:
?
#3
Report Thread starter 3 years ago
#3
(Original post by poorform)
Didn't work through it myself explicitly but it seems to be correct.

You basically want to manipulate whatever you have and get it into a form where you can exploit the properties of a geometric series.

Since you are talking about a Laurent series centred at 0 then the 1/z part is in fact the Laurent series for 1/z centred at 0 so you have to do no work there.

The other two parts are just about thinking about what annulus you want your function to converge on and then just manipulating the respective part to do that.

So for example for the second part of the partial fraction on 0<|z|<1 we want to find a Laurent series expansion for -1/(z-1) valid on 0<|z|<1.

One way to do this is to note that \displaystyle -1/(z-1)=1/(1-z)=\sum_{n=0}^{\infty} z^n by geometric series and this is valid for |z|<1 so it is certainly valid on 0<|z|<1 which is what we wanted. The rest of the question follows the same line the only thing that changes is how you manipulate the fractions to make sure it is valid for the region you want.

You will want something like 1/(1-z) for Laurent series expansions (for things like a<|z|) and something like 1/(1-(1/z)) for things regions like |z|<b.

It turns out that the Laurent series of functions are equal to the Taylor series of functions in some cases. This is the case when you are looking for series' that are defined on an annulus inside the singularity, like above we wanted to find out the laurent series for -1/z-1 inside |z|<1 so it gives in fact the Taylor series at the point (i.e. all the negative coeffiecents are zero)

No idea if that helped at all I may have just been rambling though.

There are some good youtube videos out there and you will need to practise problems to get familiar with it.
For the example you did that's what I got, but in the solutions I don't get why they have included the other parts as well
0
reply
poorform
Badges: 10
Rep:
?
#4
Report 3 years ago
#4
(Original post by e^x)
For the example you did that's what I got, but in the solutions I don't get why they have included the other parts as well
The parts for 1<|z|<2 and |z|>2?

Well you can define a Laurent series that converges on these regions so why would it be surprising for it to ask you to find them?

The question doesn't specify where you want the series to converge so they just calculated ones that converge on the different regions I suppose.

If you are talking about the other parts of the partial fractions then of course they need to be included; you want to find a Laurent series for f(z) on some annulus and f(z) is equal to the sum of the three parts of the partial fractions so you would need to make sure you are finding the Laurent series for these parts as well as otherwise you would only be finding the Laurent series for something that is certainly not f(z).
0
reply
e^x
Badges: 15
Rep:
?
#5
Report Thread starter 3 years ago
#5
(Original post by poorform)
The parts for 1<|z|<2 and |z|>2?

Well you can define a Laurent series that converges on these regions so why would it be surprising for it to ask you to find them?

The question doesn't specify where you want the series to converge so they just calculated ones that converge on the different regions I suppose.

If you are talking about the other parts of the partial fractions then of course they need to be included; you want to find a Laurent series for f(z) on some annulus and f(z) is equal to the sum of the three parts of the partial fractions so you would need to make sure you are finding the Laurent series for these parts as well as otherwise you would only be finding the Laurent series for something that is certainly not f(z).
that is what i was getting confused about. Thanks
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Suffolk
    Undergraduate Open Day - Ipswich Main Campus Undergraduate
    Mon, 9 Dec '19
  • University of Hertfordshire
    All Subjects Undergraduate
    Wed, 11 Dec '19
  • University of Lincoln
    Undergraduate Open Day Undergraduate
    Wed, 11 Dec '19

Which party will you be voting for in the General Election?

Conservatives (182)
18.69%
Labour (499)
51.23%
Liberal Democrats (133)
13.66%
Green Party (55)
5.65%
Brexit Party (15)
1.54%
Independent Group for Change (Change UK) (4)
0.41%
SNP (15)
1.54%
Plaid Cymru (4)
0.41%
Democratic Unionist Party (DUP) (0)
0%
Sinn Fein (7)
0.72%
SDLP (1)
0.1%
Ulster Unionist (4)
0.41%
UKIP (10)
1.03%
Other (10)
1.03%
None (35)
3.59%

Watched Threads

View All