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Complex Analysis - Maths KU

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6.2 Taylor Series 313<br />

45. Consider the power series f(z) = �∞ k=0 akz k where<br />

⎧<br />

⎨ 2<br />

ak =<br />

⎩<br />

k , k =0, 2, 4,...<br />

1<br />

k =1, 3, 5,... .<br />

7k (a) Show that neither lim<br />

n→∞ |an+1/an| nor limn→∞ |an|1/n exist.<br />

(b) Find the radius of convergence of each power series:<br />

∞�<br />

f1(z) = 2 2k z 2k<br />

and<br />

∞�<br />

f2(z) =<br />

k=0<br />

k=0<br />

1<br />

7 2k+1 z2k+1 .<br />

(c) Verify that f(z) =f1(z) +f2(z). Discuss: How can the radius of convergence<br />

R for the original power series be found from the foregoing observation?<br />

What is R?<br />

46. Proceed as in Problem 45 to find the radius of convergence R for the power<br />

series<br />

1+3z + z 2 +27z 3 + z 4 + 243z 5 + z 6 + ··· .<br />

47. In this problem you are guided through the proof of the proposition:<br />

If the series �∞ k=1 zk converges absolutely, then the series converges.<br />

6.2Taylor Series<br />

Proof We begin with the hypothesis that � ∞<br />

then � ∞<br />

k=1 | ak| ≤ � ∞<br />

k=1<br />

� a 2 k + b 2 k = � ∞<br />

k=1 |zk|.<br />

k=1 |zk| converges. If zk = ak+ibk,<br />

(a) First, explain why the foregoing inequality is true. Second, explain why<br />

this inequality shows that the series �∞ k=1 | ak| converges.<br />

(b) Explain how your reasoning in part (a) also shows that �∞ k=1 | bk|<br />

converges.<br />

(c) Explain how parts (a) and (b) show that � ∞<br />

k=1 zk converges.<br />

The correspondence 6.2 between a complex number z within the circle of convergence and<br />

the number to which the series � ∞<br />

k=0 ak(z − z0) k converges is single-valued.In this sense,<br />

a power series defines or represents a function f; for a specified z within the circle of<br />

convergence, the number L to which the power series converges is defined to be the value of<br />

f at z, that is, f(z) =L.In this section we present some important facts about the nature<br />

of this function f.<br />

In the preceding section we saw that every power series has a radius of convergence<br />

R.Throughout the discussion in this section we will assume that a power series<br />

� ∞<br />

k=0 ak(z − z0) k has either a positive or an infinite radius R of convergence.<br />

Differentiation and Integration of Power Series The three<br />

theorems that follow indicate a function f that is defined by a power series is<br />

continuous, differentiable, and integrable within its circle of convergence.

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