Module

for

Taylor Series Representations

7.2  Taylor Series Representations

In Section 4.4 we showed that functions defined by power series have derivatives of all orders (Theorem 4.16).  In Section 6.5 we demonstrated that analytic functions also have derivatives of all orders (Corollary 6.2).  It seems natural, therefore, that there would be some connection between analytic functions and power series.  As you might guess, the connection exists via the Taylor and Maclaurin series  of analytic functions.

Definition  7.2  (Taylor Series).  If    is analytic at , then the series

is called the Taylor series for f(z) centered at .  When the center is  , the series is called the Maclaurin series for f(z).

To investigate when these series converge we will need the following lemma.

Lemma 7.1.  If    are complex numbers  with  ,  and  ,  then

where n is a positive integer.

Proof.

Proof of Lemma 7.1 is in the book.
Complex Analysis for Mathematics and Engineering

We are now ready for the main result of this section.

Theorem 7.4 (Taylor's Theorem). Suppose f(z) is analytic in a domain G, and that is any disk contained in G.  Then the Taylor series for f(z) converges to f(z) for all z in ;  that is,

for all   .

Furthermore, for any r, 0<r<R, the convergence is uniform on the closed subdisk  .

Proof.

Proof of Theorem 7.4 is in the book.
Complex Analysis for Mathematics and Engineering

A singular point of a function is a point at which the function fails to be analytic.  You will see in Section 7.4 that singular points of a function can be classified according to how badly the function behaves at those points.  Loosely speaking, a nonremovable singular point of a function has the property that it is impossible to redefine the value of the function at that point so as to make it analytic there.  For example, the function has a nonremovable singularity at z=1.  We give a formal definition of this concept in Section 7.4, but with this language we can nuance Taylor's theorem a bit.

Corollary 7.3.  Suppose that f(z)  is analytic in the domain G that contains the point  .  Let    be a nonremovable singular point of minimum distance to the point  .  If  ,  then

(i)  the Taylor series    converges to on all of  ,
and
(ii)  if ,  the series    does not converges to .

Proof.

Proof of Corollary 7.3 is in the book.
Complex Analysis for Mathematics and Engineering

Example 7.3.  Show that    is valid for all  .

Solution.  In Example 4.24 (see Section 4.4) we established this identity with the use of Theorem 4.17.  We now do so via Theorem 7.4.  If  ,  then a standard induction argument (which we leave as an exercise) will show that    for  .  Thus  ,  and Taylor's theorem gives

,

and since f(z) is analytic in , this series expansion is valid for all .

The disk      and it's images under the mappings:
,    ,    and    .
Remark. The accuracy of the image points for the approximation        is
.

Explore Solution 7.3.

Example 7.4.  Show that, for  ,

(7-12)            (a)     and    (b) .

Solution.  For ,

(7-13)            .

If we let take the role of z in (7-13), we get that

,

for  .  But    iff  ,  thus we have proven that    for  .

Next, let    take the role of z in Equation (7-13), we get that

gives the second part of Equations (7-12).

The disk      and it's images under the mappings:
,    ,   and    .

Remark 1. The accuracy of the image points for the approximation      is
.

Remark 2. The images of     under the mappings:   ,
,   and       will appear like those shown above,
because    rotates the plane about the origin and   .
Also, the accuracy of the image points for the approximation      will be

.

Explore Real Solution 7.4 (a).

Explore Complex Solution 7.4 (a).

Explore Real Solution 7.4 (b).

Explore Complex Solution 7.4 (b).

Remark 7.1  Corollary 7.3 clears up what often seems to be a mystery when series are first introduced in calculus.  The calculus analog of Equations (7-12) is

(7-14)                and       for   .

For many students, it makes sense that the first series in Equations (7-14) converges only on the interval    because is undefined at the points  .  It seems unclear as to why this should also be the case for the series representing , since the real-valued function is defined everywhere.  The explanation, of course, comes from the complex domain.  The complex function is not defined everywhere.  In fact, the singularities of    are at the points  ,  and the distance between them and the point    equals 1. According to Corollary 7.3, therefore, Equations (7-14) are valid only for , and thus Equations (7-14) are valid only for the real numbers  .

Alas, there is a potential fly in this ointment: Corollary 7.3 applies to Taylor series.  To form the Taylor series of a function, we must compute its derivatives.  We didn't get the series in Equations (7-12) by computing derivatives, so how do we know that they are indeed the Taylor series centered at  ?  Perhaps the Taylor series would give completely different expressions from those given by Equations (7-12).  Fortunately, Theorem 7.5 removes this possibility.

Theorem 7.5  (Uniqueness of Power Series).  Suppose that in some disk we have

.

Then  .

Proof.

Example 7.5.  Find the Maclaurin series for .

Solution.  Computing derivatives for f(z) would be an onerous task. Fortunately, we can make use of the trigonometric identity

.

Recall that the series for sin z (valid for all z) is  .  Using the identity for  ,  we obtain

By the uniqueness of power series, this last expression is the Maclaurin series for  .

Explore Solution 7.5.

In the preceding argument we used some obvious results of power series representations that we haven't yet formally stated.  The requisite results are part of Theorem 7.6.

Theorem 7.6.  Let f(z) and g(z) have the power series representations

,
and
.

If    is any complex constant, then

(7-15)            ,

(7-16)            ,  and

(7-17)            ,   where

(7-18)            .

Identity (7-17) is known as the Cauchy product of the series for f(z) and g(z).

Proof.

Proof of Theorem 7.6 is in the book.
Complex Analysis for Mathematics and Engineering

Example 7.6.  Use the Cauchy product of series to show that

for   .

Solution.  We let  ,  for .   In terms of Theorem 7.6, we have  ,  for all n, and thus Equation (7-17) gives

Explore Solution 7.6.

Extra Example 1.  Use the Cauchy product of series to show that

for   .

Explore Solution for Extra Example 1.

Extra Example 2.  Show that     for   .

Solution.  Use the result of Example 7.6 and (7-16) and obtain

Explore Solution for Extra Example 2.

The Next Module is
Laurent Series Representations

(c) 2012 John H. Mathews, Russell W. Howell