Complex Power Series JL

Convergence and Absolute Convergence
•
∑*∞Cn CONVerges iff

lim
N → ∞ 
 N
∑
* 
 Cn
exists (finite).
•
If Cn = xn + i yn, then ∑*∞Cn CONVerges iff ∑*∞xn CONVerges and ∑*∞yn CONVerges.
•
∑*∞Cn CONVerges ABSsolutely iff ∑*∞|Cn| converges or

lim
N → ∞ 
 N
∑
* 
 |Cn|
exists (finite).
•
If Cn = xn + i yn, then ∑*∞Cn CONVerges ABSolutely iff ∑*∞xn and ∑*∞yn CONVerge ABSolutely.
•
∑*∞|Cn| converges iff the sequence ∑*N |Cn| is bounded.


Comparison Test for Absolute Convergence
•
If
|Cn| ≤ Bn,
then
0 ≤ ∞
∑
* 
 |Cn| ≤ ∞
∑
* 
 Bn.
So that if the series ∑*∞Bn CONVerges, the ∑*∞An CONVerges ABSolutely. Moreover there is the obvious error estimate
⎢
⎢ 		∞
∑
N+1 
 Cn⎢
⎢
≤ ∞
∑
N+1 
 |Cn|
≤ ∞
∑
N+1 
 Bn

The Geometric Series
We consider the geometric series
∞
∑
0 
 zn.
A bare hands calculation shows that
(1 − z) N
∑
0 
 zn
=(1 − z) (1 + z + …+ zN)
= 1 − zN+1.
so that
∞
∑
0 
 zn     ⎧
⎪
⎨
⎪
⎩
CONVerges ABSolutely to 1
1−z
 for |z| < 1,
DIVerges for |z| ≥ 1.
For |z| < r < 1, there is the error estimate
⎢
⎢ 		∞
∑
N+1 
 zn⎢
⎢ 		≤ ∞
∑
N+1 
 |z|n
= |z|N+1
1 − |z|
≤ ⎢
⎢ 		z
r
⎢
⎢ 		N+1

 
 rN+1
1 − r
≤ rN+1
1 − r
 .
Ratio Test for ABSolute CONVergence
For the series ∑*∞CN, suppose that

lim
n → ∞ 
 ⎢
⎢ 		Cn+1
Cn
⎢
⎢ 		= L.
•
If 0 ≤ L < 1, the series ∑*∞CN CONVerges ABSolutely.
Why? Choose an r, L < r < 1. For n sufficiently large, |Cn+1| < r |Cn| and for N large enough,
∞
∑
N 
 |Cn|
≤ ∞
∑
N 
 |CN|rn − N
= |CN| 1
1−r
 .
•
If 1 < L ≤ ∞, the series ∑*∞CN DIVerges.
•
If L = 1, we are not sure - additional information is needed to decide DIVergence or CONVergence and/or ABS0lute CONVergence.
Power Series, Radius of Convergence, and Circle/Disk of Convergence
•
If the power series ∑n=0∞ an zn, converges for a nonzero z = z0, then for all z, |z| < |z0|, the power series CONVerges ABSolutely.
Why? We have that limn → ∞ |an z0n| = 0. Then we can compare
∞
∑
N+1 
 |an zn|
= ∞
∑
N+1 
 |an z0n|⎢
⎢ 		z
z0
⎢
⎢ 		n

 
≤ ⎛
⎝
max
n ≥ N 
 |an z0n|⎞
⎠ 		∞
∑
N+1 
 ⎢
⎢ 		z
z0
⎢
⎢ 		n

 
= o(1) θN+1
1 − θ
 ,
where o(1) means limN → ∞ o(1) = 0, and θ = |[z/(z0)]|.
Thus the convergence of the series at a nonzero z0 forces the absolute convergence of the series in the entire open disk centered at 0 with radius |z0|.
•
For a power series ∑n=0∞ an zn, there is a number R, 0 ≤ R ≤ ∞ for which
∞
∑
n=0 
 an zn⎧
⎪
⎨
⎪
⎩
CONVerges ABSolutely for |z| < R,
DIVerges for |z| > R.
The number R is called the radius of convergence of the power series. R can often be determined by the Ratio Test.
•
If f(z) is represented by a convergent power series for |z| < R, then f(z) is an analytic function in the region |z| < R and its derivative is represented by the convergent series ∑n=1∞ n an zn−1, |z| < R.
Thus the power series for f′ has radius of convergence at least R, and the formally differentiated series converges to the analytic function f′(z). Within the open disk of convergence, it follows that function represented by a power series has derivatives of all orders which are represented by the series differentiated term by term.
•
If
f(z) = ∞
∑
n=0 
 an zn, |z| < R,
then
f′(z) = ∞
∑
n=1 
 n an zn−1 = ∞
∑
n=0 
 (n+1) an+1 xn, |x| < R,
and1
⌠
⌡ 		z

0 
 f(ζ)  dζ = ∞
∑
n=0 
 an
n+1
 zn+1 = ∞
∑
n=1 
 an−1
n
 zn, |z| < R

Footnotes:

1The integral must be very carefully defined. For now interpret the integral as being on the straight line path from 0 to the point z.

File translated from TEX by TTH, version 4.03.
On 27 Mar 2012, 18:01.