Which Laurent Series for $\frac{1}{z(z+2)}$ in the Region $1 < |z-1| < 3$?

[ognik]

Please help me with this Laurent series example for $\frac{1}{z(z+2)}$ in the region 1 < |z-1| < 3

Let w = z-1, then $ f(z) = \frac{1}{(w+1)(w+3)}=\frac{1}{2} \left[ \frac{1}{w+1}-\frac{1}{w+3} \right]$

I get $ \frac{1}{1-(-w)} = \sum_{n=0}^{\infty}(-1)^n w^n, \:for\: |w|<1;$
$ = - \sum_{n=1}^{\infty}(-1)^n \left(\frac{1}{w}\right)^{\!{n}}, \:for\:|w|>1 $

AND \(\displaystyle \frac{1}{3}.\frac{1}{1-\left( -\frac{w}{3}\right)} = \sum_{n=0}^{\infty}(-1)^n\frac{w^n}{3^{n+1}}, \:for\:|w|<3\)

\(\displaystyle = \sum_{n=1}^{\infty}(-1)^n\frac{3^{n-1} }{w^n}, \:for\:|w|>3 \)
So far, so good.

My problem is understanding which of the above 4 series we need for the region 1 < |z-1| < 3:

My logic tells me that the Taylor-type part for |w| < 3 minus the Taylor-type for |w| < 1 gives us the series in the region.

However this example I am looking at says to instead use the series' for |w| > 1 and |w| < 3

Could someone please explain why I am wrong and they are right?

 

[alyafey22]

My logic tells me that the Taylor-type part for |w| < 3 minus the Taylor-type for |w| < 1 gives us the series in the region.

What do you mean by minus? you are finding the Laurent expansion of the product, right?

 

[ognik]

Hi - so the question is to find the expansion of the (product, yes ) entire function between 1 and 3, and I arrive at 4 series which are each on a different region. So why can't I take the series greater than 1 and subtract the series less than 3?

The book says to instead use the series' for |w| > 1 and |w| < 3 - but to me the series for |w| > 1 includes stuff greater than 3 and the series for |w| < 3 includes stuff less than 1 ?

 

[alyafey22]

The intersection of the regions $|w|>1$ and $|w|<3$ gives a the region $1<|w|<3$.

 

[ognik]

By intersection I understand the 'and' of sets - which makes sense here, but adding those 2 regions is not the same as 'and'-ing them; isn't adding them is a union?

Perhaps someone could explain what region I would be covering with my version to help me see where I'm understanding something wrong?

 

[alyafey22]

Suppose that

$$F(w) = g(w)+h(w)$$

and we need the Laurent expansion for $F$ in the region $a<|w|<b$, then if we have

$$g(w) = \sum^{\infty}_{n=-\infty} a_n w^n \,\,\,\, |w|>a$$

and

$$h(w) = \sum^{\infty}_{n=-\infty} b_n w^n \,\,\,\, |w|<b$$

Then we have

$$F(w) = \sum^{\infty}_{n=-\infty} (a_n+b_n) w^n \,\,\,\, a<|w|<b$$

For $F$ this expansion is valid because it lies in the intersection between the two regions.

 

[ognik]

But doesn't the series for g(w) extend also greater than b?

 

[alyafey22]

But doesn't the series for g(w) extend also greater than b?

Yes, but since the other series does not extend greater than $b$ the whole series does not. Remember always that the solution must be valid (in the area of convergence) for both. Hence, we must always take the intersection of the two regions. Suppose that you are given a value $w$ such that $|w|>b$ then our expansion is not valid since the other function $h$ is expanded under the assumption that $|w|<b$.

 

[ognik]

Ah, sorry if this is obvious just making sure I have it right; you are saying that because w will only be chosen between a and b, then all series that include the region between a and b are valid, it is the choice w itself that defines the intersection?

 

[alyafey22]

Ah, sorry if this is obvious just making sure I have it right; you are saying that because w will only be chosen between a and b, then all series that include the region between a and b are valid, it is the choice w itself that defines the intersection?

It is actually the opposite. I'll give you a simpler example, Look at the following

$$F(x) = \sqrt{2-x}+\sqrt{x-1}$$

The domain of the function is the intersection of the domains of the two functions which is $1\leq x\leq 2$. Similarly, in our example since the function $F$ has a representation using two functions it is region of convergence is defined as the intersection of the two regions of convergence of both functions.

 

[ognik]

Thanks for your patience Zaid.

In your example, doesn't the first function have a domain < 2 AND > 2 up to infinity?
And the 2nd function has a domain < 1 AND > 1 also up to infinity?

Is it perhaps by definition that we choose the annulus defined by both functions?

 

[alyafey22]

Thanks for your patience Zaid.

In your example, doesn't the first function have a domain < 2 AND > 2 up to infinity?
And the 2nd function has a domain < 1 AND > 1 also up to infinity?

Is it perhaps by definition that we choose the annulus defined by both functions?

No, the domain of the first function is $x\leq 2$ and the domain of the second function is $x\geq 1$ since we are assuming that the function is real valued so we cannot have the root of a negative value.

 

[ognik]

But going back to my original problem, the 1st part of the function has a laurent series for w < 1 and w > 1, and the 2nd part a laurent series for w < 3 and w > 3.

1) I don't see anything that restricts this to a real solution, w/z are complex?
2) So my question remains - there is an intersection for both parts for w > 3 and w < 1, why are we only taking the intersection between 1 and 3?

 

[alyafey22]

But going back to my original problem, the 1st part of the function has a laurent series for w < 1 and w > 1, and the 2nd part a laurent series for w < 3 and w > 3.

1) I don't see anything that restricts this to a real solution, w/z are complex?
2) So my question remains - there is an intersection for both parts for w > 3 and w < 1, why are we only taking the intersection between 1 and 3?

Ok, I think I got your question now. The question asked for a Laurent expansion valid on the region $1<|w|<3$. So the only way to find such an expansion is to take the series for $|w|>1$ and the series for $|w|<3$ because their intersection gives the required region.

 

[ognik]

Ok, I think I got your question now. The question asked for a Laurent expansion valid on the region $1<|w|<3$. So the only way to find such an expansion is to take the series for $|w|>1$ and the series for $|w|<3$ because their intersection gives the required region.

Thanks Zaid.
...just checking - and if the question hadn't specified a domain, then there would also be intersections for w < 1 and w > 3?

 

[alyafey22]

Thanks Zaid.
...just checking - and if the question hadn't specified a domain, then there would also be intersections for w < 1 and w > 3?

Well, usually the question has to specify in which region to expand the function otherwise the question is vague!

 

[ognik]

Well, usually the question has to specify in which region to expand the function otherwise the question is vague!

But they could have chosen from the other 2 regions (w < 1 and w > 3) couldn't they, those are still intersections?

 

[alyafey22]

But they could have chosen from the other 2 regions (w < 1 and w > 3) couldn't they, those are still intersections?

I don't quite understand what you meant by $w<1$ because we are speaking about complex numbers!

 

[ognik]

sorry, |w| for both of them

 

[alyafey22]

But they could have chosen from the other 2 regions (w < 1 and w > 3) couldn't they, those are still intersections?

There is no intersection between these two regions. $|w|<1$ represents a ball of radius 1. And $|w|>3$ is the complement of the ball of radius 3.

 

[ognik]

Sorry, still don't get it. Copying from a previous post:

"But going back to my original problem, the 1st part of the function has a laurent series for w < 1 and w > 1, and the 2nd part a laurent series for w < 3 and w > 3." I assume this is correct.

Let's just look at the region |w| > 3.
the 1st part of the function has a laurent series ... for |w| > 1 - which must extend into |w| > 3 and therefore intersects with the 2nd function which has a laurent series for w < 3 and w > 3 - in the region |w| > 3?

 

[alyafey22]

Sorry, still don't get it. Copying from a previous post:

"But going back to my original problem, the 1st part of the function has a laurent series for w < 1 and w > 1, and the 2nd part a laurent series for w < 3 and w > 3." I assume this is correct.

Let's just look at the region |w| > 3.
the 1st part of the function has a laurent series ... for |w| > 1 - which must extend into |w| > 3 and therefore intersects with the 2nd function which has a laurent series for w < 3 and w > 3 - in the region |w| > 3?

You are misunderstanding the meaning of $|w|>3$. First of all $w$ is a complex number so you can NOT say $w<3$ and $w>3$. Hence if we want to draw $|w|<1$ and $|w|>3$ on the same plain we get

View attachment 5125

As you can see no intersection of the two regions.

 

[ognik]

I get $ \frac{1}{1-(-w)} = \sum_{n=0}^{\infty}(-1)^n w^n, \:for\: |w|<1;$
$ = - \sum_{n=1}^{\infty}(-1)^n \left(\frac{1}{w}\right)^{\!{n}}, \:for\:|w|>1 $

AND \(\displaystyle \frac{1}{3}.\frac{1}{1-\left( -\frac{w}{3}\right)} = \sum_{n=0}^{\infty}(-1)^n\frac{w^n}{3^{n+1}}, \:for\:|w|<3\)

\(\displaystyle = \sum_{n=1}^{\infty}(-1)^n\frac{3^{n-1} }{w^n}, \:for\:|w|>3 \)

I have that same understanding of the meaning of |w| and would draw the same diagram.

But looking at the 4 series I found above, there are 2 'taylor like' series, function 1 for |w| < 1 and function 2 for |w| < 3
Then there are 2 'laurant extension' type series, function1 for |w| > 1 and function 2 for |w| > 3

I get the intersection as you explained for the range 1 < |w| < 3 , what I am asking is that there are 2 other areas, (not asked for in this problem, so just to confirm my understanding) where the 2 functions also intersect, namely |w| < 1 and |w| > 3?

So for example if the problem had asked for |z-1| < 1, the intersection would be the series for function1 for |w| < 1 plus the series for function 2 for |w| < 3, would that be right?

 

[alyafey22]

I get the intersection as you explained for the range 1 < |w| < 3 , what I am asking is that there are 2 other areas, (not asked for in this problem, so just to confirm my understanding) where the 2 functions also intersect, namely |w| < 1 and |w| > 3?

There is no intersection between these two regions as the diagram above illustrated.

 

[ognik]

Sorry, I am really lost with this. I have $ \frac{1}{1-(-w)} = \sum_{n=0}^{\infty}(-1)^n w^n, \:for\: |w|<1;$ and

\(\displaystyle \frac{1}{3}.\frac{1}{1-\left( -\frac{w}{3}\right)} = \sum_{n=0}^{\infty}(-1)^n\frac{w^n}{3^{n+1}}, \:for\:|w|<3\) , why do these two not intersect please?

 

[alyafey22]

Sorry, I am really lost with this. I have $ \frac{1}{1-(-w)} = \sum_{n=0}^{\infty}(-1)^n w^n, \:for\: |w|<1;$ and

\(\displaystyle \frac{1}{3}.\frac{1}{1-\left( -\frac{w}{3}\right)} = \sum_{n=0}^{\infty}(-1)^n\frac{w^n}{3^{n+1}}, \:for\:|w|<3\) , why do these two not intersect please?

In your previous post you said $|w|<1$ and $|w|>3$. On the other hand the intersection of $|w|<3$ and $|w|<1$ is clearly $|w|<1$.

 

[ognik]

My apologies, too focused on one thing I mis-typed. Thanks for your patience, it is all clear now :-)