Laurent Series Explained | How to Determine Laurent Series | Complex Analysis #9
Summary
TLDRThis educational video introduces and explores Laurent Series, starting with a theorem that if a function is analytic on an annulus domain, it equals its Laurent Series there. The instructor divides the series into an analytic part with positive powers and a principal part with negative powers. Through examples, the video demonstrates how to determine Laurent Series for different functions using geometric series and the significance of the annulus domain. It highlights the residue's importance in contour integrations and explains the convergence of the series' parts. The tutorial makes extensive use of visual aids to clarify concepts, making complex mathematical principles accessible and engaging.
Takeaways
- ๐ The video covers Laurent Series, starting with the theorem and proceeding to find Laurent Series for various functions.
- ๐ Laurent Series consist of two parts: an analytic part (positive powers of z minus z-naught) and a principal part (negative powers of z minus z-naught).
- ๐ The video emphasizes the significance of an annulus domain in complex analysis, illustrating it as a doughnut-shaped region between two circles.
- ๐ The coefficients in a Laurent Series can be determined using a specific formula involving a closed curve, but practical examples often use manipulation of known geometric series.
- ๐ Laurent Series are particularly useful for functions with isolated singularities, expanding on points where Taylor Series may not work.
- ๐ก The coefficient a-minus-one in the Laurent Series is termed the 'Residue' and is key in contour integrations later in the course.
- ๐ If a function is analytic across the whole disk, the Laurent Series reduces to a Taylor Series, as all coefficients for negative powers will be zero.
- ๐ The two parts of a Laurent Series converge in different regions: the analytic part within the larger circle and the principal part outside the smaller circle in an annulus domain.
- ๐ข The video uses geometric series to derive Laurent Series, adjusting coefficients to fit specific annulus domains for given functions.
- ๐งฎ Multiple examples illustrate how to determine Laurent Series for functions across various domains, including how to handle poles and apply partial fraction expansion.
Q & A
What is the Laurent Series?
-The Laurent Series is a representation of a complex function as a series of powers of (z - z_0), where (z_0) is a point in the complex plane. It consists of two parts: an analytic part with positive powers and a principal part with negative powers.
How does the Laurent Series differ from the Taylor Series?
-While both are series representations of functions, the Laurent Series includes both positive and negative powers of (z - z_0), allowing it to represent functions with isolated singularities. The Taylor Series includes only non-negative powers and applies to functions analytic at (z_0).
What is an annulus domain in the context of Laurent Series?
-An annulus domain refers to the region between two concentric circles in the complex plane, characterized by radii (R_1) and (R_2). It is the domain within which the Laurent Series of a function converges.
What is the significance of the residue in Laurent Series?
-The residue, denoted as the coefficient (a_{-1}), is crucial for calculating contour integrations around singularities. It specifically refers to the coefficient of the (z - z_0)^{-1} term in the Laurent Series.
How can the coefficients of the Laurent Series be determined?
-Although there's a formal formula involving contour integration around a closed curve in the annulus, in practice, coefficients are often determined by manipulating known geometric series or using partial fraction decomposition for specific functions.
What is meant by an isolated singularity in the context of Laurent Series?
-An isolated singularity refers to a point (z_0) where a function is not analytic (i.e., it cannot be represented by a Taylor Series), but the function is analytic at every other point in some neighborhood of (z_0).
How are Laurent Series used to determine functions' behavior around singularities?
-By representing a function as a Laurent Series around a singularity, one can analyze the function's behavior in the vicinity of the singularity, distinguishing between different types of singularities and understanding the function's analytic structure.
What role do geometric series play in deriving Laurent Series?
-Geometric series are used as a tool to express parts of the function in series form, especially when direct calculation of coefficients is complex. By fitting parts of a function into known geometric series forms, the Laurent Series can be constructed.
How is the principal part of the Laurent Series defined, and why is it important?
-The principal part of the Laurent Series consists of the series of terms with negative powers of (z - z_0). It is important because it characterizes the nature of singularities and is essential for calculating residues and performing contour integrations.
Can a Laurent Series reduce to a Taylor Series, and under what condition?
-Yes, a Laurent Series can reduce to a Taylor Series when the function is analytic at (z_0) and thus has no negative powers in its series expansion. This occurs when all the coefficients of the negative powers are zero.
Outlines
๐ Introduction to Laurent Series
This segment provides a comprehensive introduction to Laurent Series, beginning with an explanation of the theorem that states if a function is analytic on some annulus domain, it will be equal to its Laurent Series within this domain. The Laurent Series is described as consisting of two parts: the analytic part, containing positive powers of z minus z-naught, and the principal part, containing negative powers. The segment explains the process of determining the coefficients of the Laurent Series without relying on the formula directly, by manipulating known geometric series, similar to the approach used for Taylor series. It also introduces the concept of an annulus domain, illustrating it with examples of different radii and explaining its significance in capturing isolated singularities, which allows the Laurent Series to be applicable in scenarios where Taylor series might not. Furthermore, the importance of the residue, the coefficient of the (a_{-1}) term, is highlighted for its utility in contour integrations. The segment concludes by distinguishing between the convergence zones of the analytic and principal parts of the Laurent Series and emphasizing that the entire series is valid only within the annulus domain defined by two circles.
๐งฉ Applying Laurent Series to Functions
This part delves into the application of Laurent Series to specific functions, illustrating the process through examples. It begins by emphasizing the flexibility in generating Laurent Series by choosing constants smartly in geometric series. The instructor then demonstrates how to determine Laurent Series for a function with a pole at z equals two, using geometric series and adjustments to fit the series' requirements, such as ensuring the series converges within a specified domain. Different domains around the pole lead to different Laurent Series. Through detailed step-by-step examples, the segment shows how to manipulate functions to fit into the forms required for the geometric series, adjusting constants as needed for convergence. The approach is applied to multiple domains around poles, highlighting the method's utility in handling functions with singularities at different points and in different expansion domains. The examples provided illustrate the practical steps in breaking down functions into parts that can be expressed as geometric series, thereby constructing the Laurent Series for each domain.
๐ Advanced Laurent Series Examples
The final section addresses more complex scenarios involving multiple singularities and domains, guiding through the construction of Laurent Series for each situation. It suggests using partial fraction expansion as a strategy to simplify the expression of a function, making it easier to determine its Laurent Series. The instructor works through examples with functions having multiple poles, showing how to create Laurent Series that are valid in different regions: inside a smaller circle, between two circles, and outside a larger circle. Each case involves identifying the principal and analytic parts of the Laurent Series and combining them appropriately. The examples demonstrate the versatility of the Laurent Series in handling complex functions with several singularities, by dividing the complex plane into regions and determining the series for each. The segment underscores the practicality of Laurent Series in advanced mathematical problems, like contour integration, and its superiority in capturing the nuances of functions across different domains.
Mindmap
Keywords
๐กLaurent Series
๐กAnnulus Domain
๐กAnalytic Part
๐กPrincipal Part
๐กResidue
๐กGeometric Series
๐กSingularity
๐กConvergence
๐กPole
๐กPartial Fraction Expansion
Highlights
Introduction to Laurent Series with a focus on its theorem, stating that if a function is analytic on an annulus domain, it equals its Laurent Series on this domain.
Explanation of the Laurent Series, consisting of an analytic part with positive powers and a principal part with negative powers.
Method to determine Laurent Series coefficients using a closed curve within the annulus, highlighting manipulation of geometric series as a practical approach.
Description of an annulus domain as the area between two circles, relevant for defining the domain of Laurent Series convergence.
Introduction of the concept of isolated singularity and its importance for the applicability of Laurent Series.
The residue at a singularity, termed as the coefficient a minus one, and its significance in contour integrations.
Clarification that a Laurent Series becomes a Taylor Series if the function is analytic across the whole disk, highlighting the principal part's role.
Differentiation of convergence zones for the analytic and principal parts of a Laurent Series within the complex plane.
Application of geometric series for deriving Laurent Series for specific functions, demonstrating through examples.
Illustration of domain-specific Laurent Series development by factoring and applying geometric series to functions with poles.
Explanation on how to handle functions with multiple poles across different domains using partial fraction expansion.
Demonstration of Laurent Series derivation for functions across various domains, emphasizing the analytic and principal parts.
Technique to rewrite functions to align with geometric series formulations for Laurent Series construction.
Example-based approach to explaining the creation of Laurent Series for complex functions with isolated singularities.
Conclusion emphasizing the utility of Laurent Series in complex analysis, especially in handling isolated singularities and contour integrations.
Transcripts
- [Instructor] Hello, and welcome.
In this video we're going to talk a bit about
Laurent Series.
We'll start with the theorem, and then we will continue
by determining all possible Laurent Series
for each function here below.
So let's get going.
The theorem tell us that if a function is analytic
on some annulus domain, then we know that the function
is going to be equal to its Laurent Series on this domain,
and the Laurent Series is in fact made up of
two separate series:
The first one, which include all the positive powers
of z minus z-naught is called an analytic part,
since this part will always be analytic,
while the second one contains all the negative powers
of z minus z-naught, and is called a principal part.
And the coefficients in the Laurent Series can be determined
by using this formula here, and here c is a closed curve
inside the annulus surrounding z-naught, but you don't
really have to remember this formula,
since we never really use it.
Instead we manipulate known geometric series
to determine the coefficients, just like we did
in the case with Taylor series.
But what is an annulus domain?
An annulus domain looks something like this.
It's all the z values which lies between two circles.
The inner circle is determined by the radius R-one.
While the second circle is determined by radius R-two.
And this is a typical annulus domain.
Looks pretty much like a doughnut, right?
But our annulus domain doesn't need to look like this,
we can have one case where R-one is equal to zero,
then the domain contains all the z values inside the
circle R-two except z-naught.
But we can also have that R-two is equal to infinity.
Then the domain would be all the z-values outside
the circle R-one, but we can also have these two cases
at the same time.
And then the domain would look something like this.
Which means that the domain would include
all the z values on the whole plane
except z-naught.
And by using this annulus domain we're able to
capture, isolate the single lattice, and that is why
Laurent series, unlike Taylor series, still works
if the point you are expanding about, z-naught
is an isolated singularity.
And some other good things to know, is that
the coefficient a minus one is called the Residue
of the function f at z is equal to z-naught.
And this residue will be really useful
later on in the course when we're starting
to do contour integrations.
And note that if you find that a function is
analytic on the whole disk, then all the coefficients
for the negative powers of z minus z-naught
will be equal to zero, and that means that the
Laurent series is going to be reduced to a Taylor series,
and this comes from a fact that the only thing that
separates a Taylor series and a Laurent series
is that a Laurent series also contains this principal part.
And the last thing I would like to touch on is that
these two parts of a Laurent series
converges on different places in the complex plane.
The analytic part converges for all z values
inside the big circle R-two.
While the principal part only converges for z values
outside the smaller circle, R-one.
And that is why the sum of these two series,
which makes up our Laurent series will only be
valid for z values in between these two circles.
Let's continue by doing some examples.
So here I would like us to determine all possible
Laurent series for each function here below.
And we can do that by using geometric series,
because we know that the Laurent series for the function
f around some point z-naught should look
something like this.
And we can get something that looks pretty much exactly
like that by using these two geometric series here above.
And here I can also note that the second of these
two geometric series is in fact only a special case
of a first series, since if you insert one divided
by W as the W in the first series, you get the second one.
So if we know for some constant b, insert b divided
by z minus z-naught as our one divided by W,
in the second geometric series, we get the following.
And this thing will only be valid if the absolute value
of b is smaller than the absolute value of z minus z-naught.
And if we now for some constant c insert
c times z minus z-naught as our W in the first
geometric series, we get the following.
And this thing will only be valid if the absolute value
of the z minus z-naught is smaller than one
divided by the absolute value of c.
And here we can see that if we take the sum
of these two series, we get something that looks
pretty much exactly the same as our Laurent series.
And the sum will only be valid inside some annulus
determined by our coefficients b and c.
So in short I'm saying that if we use these two
geometric series and choose our constants b and c smart,
we will be able to create whatever Laurent series
we would like.
So let's continue with our first example.
And the easiest way to determine all possible Laurent
series is to mark all the important points
on a graph, so we know that we are expanding about
the point z is equal to zero, and we also know
that the function has a pole at z is equal to two.
And that means that we can create two separate domains,
one inside the circle and one outside of it.
And each of these two domains will have
their own Laurent series.
For our first domain, we would like to use
the first geometric series, since the series
must be valid inside the circle.
So we can start by factoring out the minus one,
to switch places on the numbers on the denominator.
And then we can continue by factoring out the half
to create a one in the denominator.
And now we can use our first geometric series
with W is equal to z, and c is equal to a half,
which will give us the following.
And by moving one divided by two from outside
to the inside of the sum, we get our final expression.
And this thing will only be valid if the absolute
value of z is less than one divided by one half,
which is the same as saying that
the absolute value of z must be less than two,
which is exactly what we wanted.
For our next domain it will act to use the second
geometric series, since the series must be
valid outside of the circle,
and to do that we have to make our function
look a little bit more like a function in the formula.
If we start by factoring out one divided by z
we get the following.
And now we can actually use the second geometric series
with W is equal to z and b is equal to two.
And this expression can be simplified to the following
by factoring in one divided by z.
But if you remember the principal part which
this thing corresponds to, should start at
N is equal to one and not N is equal to zero.
And we can easily do that by using this neat trick.
Let's say that we change the variables.
Let's say that k is equal to N plus one.
Then the sum would go from k is equal to one, to infinity,
and the expression inside would be two raised
to the power of k minus one, divided by z raised
to the power of k.
And now the trick is that k is only a name
for a variable.
And we can use whatever name we would like,
so I can rename it now to N and we get
exactly what we like to have.
And this sum will only be valid if
the absolute value of z is bigger than two,
which is exactly what we wanted.
In our next problem we will have a same function
but you are expanding about a different point.
This time we're expanding about z is equal to one,
which means that we will get two different domains
when we mark the important points.
And for first domain we like to use
the first geometric series again.
But this time we like to have that W is equal to
z minus one, since we're expanding about
point z is equal to one this time.
And we could do that by breaking up R minus two
in the denominator to minus one, minus one,
and then we can just factor out the minus one
to switch places on the terms in the denominator.
And now we can once again use our first
geometric series with W is equal to z minus one.
And the constant c is equal to one.
This gives us a series which is valid
inside the right domain.
And for our next domain we would like to write
the expression in the form of a second geometric series,
but since we still would like our W to be equal
to z minus one, we can start exactly the same
as before, and by now factoring out one divided
by z minus one, we can once again use
our geometric series.
And in this case we have that W is equal to
z minus one, and our constant b is going to be
equal to one, so that means that we get
the following expression.
And this expression can be simplified as the following.
And to get it to start with N is equal to one instead
of zero, we just have to decrease all the N
inside the sum by one since we are raising our
starting point by one.
And this sum will be valid outside of the circle,
like we wanted.
For our last example, we start by marking the important
points, which are the point of expansion,
z is equal to zero, and the poles for our function
which are one and two, and here we can see
that we can create three different regions.
One inside the smaller circle,
one between the two circles,
and one outside of a big circle.
So in all cases where you don't really see
how you can use a geometric series by the way,
I recommend you to always start by rewriting the
expression with the help of partial fraction expansion
so that you can determine the Laurent series
by observing each of these two terms,
one at a time.
For our first domain, we already determined earlier
that one divided by z minus two is equal to the following.
That since this is valid for z values inside the circle
with radius two, we know that it will also work
in our domain here.
And for our second term, one divided by z minus one,
we can rewrite it with the help of the first
geometric series by factoring out minus one
and then by using the first geometric series
we get the following expression.
And if we now add these two terms together again,
we get that the function's Laurent series
is equal to the following.
And since the sums start at the same points,
we can just add them together
and get the following results which will be valid if
the absolute value of z is less than one
since this is the only place in the complex plane
where both of these two series are valid
at the same time.
So, until now, the Laurent series we have determined
have only contained either the analytic part
or the principal part, but in this second region here
we will get them both at the same time.
And the trick is to let one of these two terms
up here be equal to the principal part
while letting the other term be equal to
the analytic part, and which one to choose to what
comes from experience.
But in most cases you will see that one way
is absolutely more easy than another.
Like in this case here we can still use that
one divided by z minus two is equal to the following.
So this term can correspond to the analytic part
we are searching for.
And if we now try to rewrite our second term,
so that corresponds to our principle part
we are all done.
So one divided by z minus one, can be rewritten
as the following, by factoring out one divided by z,
and if we now use our second geometric series
we get the following.
And by simplifying this expression a bit, we see
that they have managed to find the principal part
for a Laurent series, so if we just add these two parts
together, we get something that is valid inside
these two circles, which is exactly what we wanted.
For the last domain, we know that one divided
by z minus two from earlier is equal to the following.
And we can still use this expression for this bit here.
So the only thing we really have to do
is to determine the Laurent series for one divided
by z minus one.
That is valid in this domain, but we have actually
already done that, we did that in the last example.
So we already know that this thing
is going to be equal to the following.
And if we just add these two results together
we get that the Laurent series for the function f
is going to be equal to this thing here,
which only is valid if the absolute value of z
is bigger than two.
Thanks for watching.
(gentle acoustic folk music)
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