More on Laurent Series

Note to readers: I reiterate once more that this set of blog posts isn’t intended to be comprehensive nor should they be seen as replacing the hard work of diligently working problems on your own using available text books. They are merely intended to spur on curiosity and perhaps offer a few different perspectives from what one might have seen already.

Included here are the coefficients in typical Laurent series which are generally obtained by other means (i.e. than appealing directly to integral representations). Some examples I have included below might suffice to expose this.

 

Ex. (1):  You have the Maclaurin series:

 

exp (z) = å^¥ _{n = 0}   (z) ⁿ  / n!  = 1 + z + z² / 2!  + z³ / 3!  +…..

 

 

Now, replace z by 1/z in the expansion:

 

 

exp( 1/z) = å^¥ _{n = 0}   1  / n!   z ⁿ  = 1 + 1/ 1! z   + 1/ 2!  z² 

 +  1/ 3! z³   +…..

 

for: 0   <   ÷ z ÷    <     ¥ 

 

Which then becomes a Laurent series expansion.

 

To establish this on inspection note that no positive powers of z appear, only negative, i.e. then 1/z is 

z ^{- 1}   

 So, in effect we can say that the coefficients of the positive powers are zero.  It’s also important to note here that the coefficient of : 

 

1/ 1! z     = 1/ z

 

is unity, so according to Laurent’s theorem we designate that the coefficient:

 

c _n  =  1/2 pi  ò_C  exp( 1/z)  dz

 

where C is any positively oriented simple closed contour around the origin.  Now, since from our prior  Dec. 12, Introducing Laurent Series) examination c _n  =  1 then:
 

1  =1/2 pi  ò_C  exp( 1/z)  dz   and:  ò_C  exp( 1/z)  dz    =  2 pi 

Ex. (2):  Consider the function:

f(z) = 1/ (z – i) ²

 

This is already in the form of a Laurent series , where z ₀  =   i.  That is,

 

å^¥ _{n = -¥}   c _n (z – i) ⁿ     (For:  0   <   ÷ z - i ÷    <     ¥  )

Where in this case,  c _{- 2}  =  1  and all other coefficients are zero.  Then from the previous theorems, relations we’ve explored (blog post of December 12, ‘Introduction to Laurent Series’, sub-header: ‘More intricacies and singularities’):

 

c _n  =  1/2 pi  ò_C   dz/ ( z – i) ^{n +3}      n = 0, +1, +2, +3 …..

 

 

where C denotes any positively oriented circle ÷ z - i ÷   = R  about the point: z ₀  =   i

 

 

Then it follows from this:

 

 

a)    ò_C   dz/ ( z – i) ^{n +3}      =  0   (when n ¹   2)

 

b) ò_C   dz/ ( z – i) ^{n +3}      =  1    (when n=    2)

 

Problems for the Math Maven:

 

1)     The function: f(z) =  -1/ (z – 1) (z – 2)
 

a) Rewrite it using the partial fraction form

 

b) Identify the two singular points.

 

 

c) If  the function is analytic in the domains:

 

÷ z ÷    <   1;   1     < ÷ z ÷     <      2   and:   2    <  ÷ z ÷    <     ¥,  

 

 

Draw a sketch showing the different domains.

 

 

2) Find the Laurent series that represents the function:

 

 

f(z) =  z²  sin (1 / z² )

 

In the domain:  : 0   <  ÷ z ÷   <     ¥ 

 

 

Hint:  Recall the series for sin(z) = z -  z³ / 3!  +  z⁵ / 5!  …..-  

   (-1) ^{n- 1} z ^{2n -1}/  (2n -1)!  +   ……   (For ÷ z ÷   <    ¥  )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Looking At Complex Power Series

With complex power series we move one more step toward Laurent series and then calculus of residues.  Terms of a complex series depend no the complex variable z. Most such series are in the form of power series, such as:

å^¥ _{n = 0}   z _n  =  1 + z + z² + z ³ + …..

We understand that in many cases such series will only converge if z is confined to a certain region. In the case of the series above, it converges (according to the ratio test), provided that ÷ z÷ < 1. 

In effect, the above power series converges for all points inside a circle of radius R = 1. This is called the radius of convergence.  This concept of radius of convergence can be applied to every power series. Thus, if a power series is convergent on a circle of some radius r then it is absolutely convergent everywhere inside this circle.  The key point is that R must be finite in order for us to assess a convergence. Else, we say the series diverges.


New Example: 

Consider the complex series:  å^¥ _{n = 1}   (z- a) ⁿ  / n 

Using the ratio test we look at:

lim _{n ® ¥}      [(z- a) ^{n +1}   / n + 1 /   (z- a) ⁿ  / n ]

= (z – a) lim _{n ® ¥}   [n/  n + 1]  =  ÷ z – a ÷

We thereby have convergence for all z such that: 
 ÷ z – a ÷ < 1

And we have divergence for all z such that:  ÷ z – a ÷ >  1

 We have ambiguous or no conclusive test result for: ÷ z – a ÷ = 1

Example (3):    å^¥ _{n = 1}   a ⁿ  (z- a) ⁿ 

Apply the ratio test to write: lim _{n ® ¥}      [a ⁿ⁺¹ (z- a) ^{n +1}    /  a ⁿ  (z- a) ⁿ ]

=  ÷ z – a ÷  lim _{n ® ¥}   [a ⁿ⁺¹ /  a ⁿ ] =   ÷ z – a ÷ × L

We have convergence for all z such that:

 ÷ z – a ÷ × L < 1 or ÷ z – a ÷  < 1/ L

We have divergence for all z such that:

÷ z – a ÷ × L >  1 or ÷ z – a ÷  >  1/ L

No test result for:  ÷ z – a ÷ =   1/ L

Using Taylor’s Theorem in the complex context:

Recall Taylor’s Theorem:

 Let f be a function that is continuous together with its first n + 1 derivatives on an interval containing a and x. Then the value of the function at x is given by:

f(x) = f(a) + f’(a) × (x – a) + f” (a)/ 2! (x – a)² + f”” (a) / 3! (x – a)³ +.......

R_n (x,a)

Where  R_n (x,a)  denotes the remainder.

This can also be applied to the case of complex functions. If f(z) is analytic inside a circle C with center at a (see diagram) then for all z inside C:

f(z) = f(a) + f’(a) × (z – a) + f” (a)/ 2! (z – a)² + f”” (a) / 3! (z – a)³ +......

R_n (z,a)

By reference consider the diagram shown.  Here, we let z be any point inside the circle C. We next construct a circle C1 with center at a, and which encloses z. Then by Cauchy’s integral formula:

f(z) =  1/2 pi  ò_C1  f(w) dw /(w – z)

 One can also show, using appropriate substitutions, that the remainder:

R_n (z,a) =  1/ 2 pi ò_C1  (z- a / w – a)ⁿ f(w) dw /(w – z)

And: lim _{n ® ¥}   R_n = 0. Interested readers are invited to show this by taking:

1 / w – z  = 1/ {w – a) – (z – a) = 1/ (w – a)[ 1/ 1 –(z –a)/(w – a)]

then expanding into terms of a complex series and substituting into Cauchy’s integral formula.

Problems for Math Mavens:

1) Confirm that for the power series: å^¥ _{n = 0}   z _n 

The radius of convergence is R= 1. (I.e. apply the ratio test to show this)

2) Consider the complex power series:

å^¥ _{n = 0}   (z) ⁿ  / n! 

Show that the radius of convergence R = ¥

3) In the Maclaurin series we use the Taylor series but with a = 0. Use the Maclaurin series to expand the function f(z) = exp(-z) to show that the radius of convergence R = ¥.