Friday, January 11, 2013

Complex Analysis: #6 Power Series

  • Complex Analysis: #6 Power Series

Definition 4
A power series is a sum of the form

Complex Analysis: #6 Power Series equation pic 1
where (an)n∈ℕ0 is some arbitrary sequence of complex numbers and z0 is a given complex number. (ℕ0 is the set of non-negative integers)

So the question is, for which z does the power series converge? Well it obviously converges for z = z0. But more generally, we can say the following.


Theorem 8
Let the power series ∑an(z − z0)n, (with n = 0, . . .,∞) be given. Then there exists 0 ≤ R ≤ ∞, the radius of convergence, such that
  1. The series is absolutely convergent for |z − z0| < R, and uniformly convergent for |z − z0| ≤ ρ, for 0 ≤ ρ < R fixed. 
  2. It diverges for |z − z0| > R.
  3. The radius of convergence is given by 1/R = limn→∞ sup n|an|
  4. The function given by f(z) = ∑an(z − z0)n, (with n = 0, . . .,∞) is analytic in the region |z − z0| < R. For each such z, the derivative is given by the series f '(z) = ∑nan(z − z0)n-1, (with n = 1, . . .,∞), and the radius of convergence of this derivative series is also R. 

Proof
Parts 1 and 2 are proved in the analysis lecture. For 3, let |z − z0| < ρ < R with 1/R = limn→∞ sup n|an|. Thus there exists some N0 ∈ ℕ with √|an| < 1/R for all n ≥ N0. That is, |an| < 1/Rn. Therefore

Complex Analysis: #6 Power Series equation pic 2

with ρ/R < 1. This is a geometric series which, as is well known, converges. On the other hand, if |z − z0| ≥ ρ > R then there exist arbitrarily large n with n|an|  > 1/R. That is,  |an| > 1/Rn or

Complex Analysis: #6 Power Series equation pic 3

So the series cannot possibly converge, since the terms of the series do not converge to zero.

As far as part 4 is concerned, it is clear that limn→∞ sup n|nan| = limn→∞ sup n|an| since limn→∞ sup nn = 1. So let f1(z) = ∑nan(z − z0)n-1, (with n = 1, . . .,∞), be the function which is defined in the region |z − z0| < R. We must show that f is analytic here, with f ' = f1. To simplify the notation, let us assume from now on that z0 = 0. Choose some complex number w with |w| < R. We must show that the derivative of f exists at w, and it equals f1(w).

To begin with, we write

Complex Analysis: #6 Power Series equation pic 4

So take some ρ with |w| < ρ < R and we restrict ourselves to examining complex numbers z with |z| < ρ. Furthermore, choose ∈ > 0. We must show that there exists a δ > 0 such that if 0 < |z −w| < δ then

Complex Analysis: #6 Power Series equation pic 5

(Remember that the series is absolutely and uniformly convergent in the closed disc with radius ρ.) Thus for some N1 ∈ ℕ, the “tail” of the series beyond N1 sums to something less than ∈ /3. Similarly, the series defining f1 is absolutely and uniformly convergent in this disc. Therefore take N2 to be sufficiently large that

Complex Analysis: #6 Power Series equation pic 6

for all n ≥ N2. Let N be the larger of N1 and N2. Finally we must determine the number δ. For this, we note that since Sn is just a polynomial, and thus analytic, we have a δ > 0 such that

Complex Analysis: #6 Power Series equation pic 7

for all z with |z − w| < δ. In particular, if necessary, we can choose a smaller δ to ensure that such z are in our disc of radius ρ. The fact that
Complex Analysis: #6 Power Series equation pic 8



Theorem 9
Let f : G → ℂ be an analytic function defined in a region G, and let z0 ∈ G be given. Then there exists a unique power series ∑an(z − z0)n, (with n = 0, . . .,∞) whose radius of convergence is greater than zero, and which converges to f(z) in a neighborhood of z0.

Proof
Let r > 0 be sufficiently small that B(z0, r) = {z ∈ ℂ : |z − z0| < r} ⊂ G. In fact, we will also assume the r is sufficiently small that z ∈ G for all z with |z − z0| = r. Once again, in order to simplify the notation, we will assume that z0 = 0. That is to say, we will imagine that we are dealing with the function f(z − z0) rather than the function f(z). But obviously if the theorem is true for this simplified function, then it is also true for the original function. According to theorem 7, for |z| < r we then have
Complex Analysis: #6 Power Series equation pic 9


Here are a few points to think about in this proof.
  • The third equation is true since |z/ζ| < 1, and thus the sum is absolutely convergent. 
  • The fifth equation is true since the partial sums are uniformly convergent, thus the sum and integral operations can be exchanged. 
  • Although the function f(ζ)/ζn+1 is not differentiable at zero, it is defined and continuous on the (compact) circle |ζ| = r. Thus, although cn is not always zero, still it is always a well defined complex number, for all n. 
  • It looks like cn might vary with r. But this is not the case. Theorem 8 implies that f(n)(0) = n!cn, for all n, and this is certainly independent of r. 
  • The power series converges to f(z) at all points of B(z0, r).

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