Complex Analysis - Maths KU

Important
f(z) =
☞
6.2 Taylor Series 315
be carried out term by term:
�
�∞
ak(z − z0) k ∞�
dz =
k=0
=
k=0
∞�
k=0
ak
�
(z − z0) k dz
ak
k +1 (z − z0) k+1 + constant.
The ratio test given in Theorem 6.4 can be used to be prove that both
∞�
ak(z − z0) k
and
∞� ak
k+1
(z − z0)
k +1
k=0
k=0
have the same circle of convergence | z − z0| = R.
Taylor Series Suppose a power series represents a function f within
| z − z0| = R, that is,
∞�
k=0
ak(z − z0) k = a0 + a1(z − z0)+a2(z − z0) 2 + a3(z − z0) 3 + ··· . (1)
It follows from Theorem 6.7 that the derivatives of f are the series
f ′ ∞�
(z) =
f ′′ (z) =
f ′′′ (z) =
k=1
akk(z − z0) k−1 = a1 +2a2(z − z0)+3a3(z − z0) 2 + ··· , (2)
∞�
akk(k − 1)(z − z0) k−2 =2· 1a2 +3· 2a3(z − z0)+··· , (3)
k=2
∞�
akk(k − 1) (k − 2) (z − z0) k−3 =3· 2 · 1a3 + ··· , (4)
k=3
and so on.Since the power series (1) represents a differentiable function f
within its circle of convergence | z − z0| = R, where R is either a positive
number or infinity, we conclude that a power series represents an analytic
function within its circle of convergence.
There is a relationship between the coefficients ak in (1) and the derivatives
of f.Evaluating (1), (2), (3), and (4) at z = z0 gives
f(z0) =a0, f ′ (z0) =1!a1, f ′′ (z0) =2!a2, and f ′′′ (z0) =3!a3,
respectively.In general, f (n) (z0) =n! an, or
an = f (n) (z0)
,n≥ 0. (5)
n!
When n = 0 in (5), we interpret the zero-order derivative as f(z0) and 0! = 1
so that the formula gives a0 = f(z0). Substituting (5) into (1) yields
f(z) =
∞�
k=0
f (k) (z0)
k!
(z − z0) k . (6)