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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  [Graphics:Images/TaylorSeriesMod_gr_1.gif]  is analytic at [Graphics:Images/TaylorSeriesMod_gr_2.gif], then the series

            [Graphics:Images/TaylorSeriesMod_gr_3.gif]
    
is called the Taylor series for f(z) centered at [Graphics:Images/TaylorSeriesMod_gr_4.gif].  When the center is  [Graphics:Images/TaylorSeriesMod_gr_5.gif], 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  [Graphics:Images/TaylorSeriesMod_gr_6.gif]  are complex numbers  with  [Graphics:Images/TaylorSeriesMod_gr_7.gif],  and  [Graphics:Images/TaylorSeriesMod_gr_8.gif],  then  

            [Graphics:Images/TaylorSeriesMod_gr_9.gif]  

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 [Graphics:Images/TaylorSeriesMod_gr_10.gif] is any disk contained in G.  Then the Taylor series for f(z) converges to f(z) for all z in [Graphics:Images/TaylorSeriesMod_gr_11.gif];  that is,

            [Graphics:Images/TaylorSeriesMod_gr_12.gif]   for all   [Graphics:Images/TaylorSeriesMod_gr_13.gif].  

Furthermore, for any r, 0<r<R, the convergence is uniform on the closed subdisk  [Graphics:Images/TaylorSeriesMod_gr_14.gif].  

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 [Graphics:Images/TaylorSeriesMod_gr_15.gif] 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  [Graphics:Images/TaylorSeriesMod_gr_16.gif].  Let  [Graphics:Images/TaylorSeriesMod_gr_17.gif]  be a nonremovable singular point of minimum distance to the point  [Graphics:Images/TaylorSeriesMod_gr_18.gif].  If  [Graphics:Images/TaylorSeriesMod_gr_19.gif],  then

    (i)  the Taylor series  [Graphics:Images/TaylorSeriesMod_gr_20.gif]  converges to [Graphics:Images/TaylorSeriesMod_gr_21.gif] on all of  [Graphics:Images/TaylorSeriesMod_gr_22.gif],  
and
    (ii)  if [Graphics:Images/TaylorSeriesMod_gr_23.gif],  the series  [Graphics:Images/TaylorSeriesMod_gr_24.gif]  does not converges to [Graphics:Images/TaylorSeriesMod_gr_25.gif].  

Proof.

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

 

Example 7.3.  Show that  [Graphics:Images/TaylorSeriesMod_gr_26.gif]  is valid for all  [Graphics:Images/TaylorSeriesMod_gr_27.gif].  

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  [Graphics:Images/TaylorSeriesMod_gr_28.gif],  then a standard induction argument (which we leave as an exercise) will show that  [Graphics:Images/TaylorSeriesMod_gr_29.gif]  for  [Graphics:Images/TaylorSeriesMod_gr_30.gif].  Thus  [Graphics:Images/TaylorSeriesMod_gr_31.gif],  and Taylor's theorem gives  

            [Graphics:Images/TaylorSeriesMod_gr_32.gif],  

and since f(z) is analytic in [Graphics:Images/TaylorSeriesMod_gr_33.gif], this series expansion is valid for all [Graphics:Images/TaylorSeriesMod_gr_34.gif].

                    [Graphics:Images/TaylorSeriesMod.0_gr_1.gif]          [Graphics:Images/TaylorSeriesMod.0_gr_2.gif]

                    [Graphics:Images/TaylorSeriesMod.0_gr_3.gif]          [Graphics:Images/TaylorSeriesMod.0_gr_4.gif]

               The disk    [Graphics:Images/TaylorSeriesMod.0_gr_5.gif]  and it's images under the mappings:
               [Graphics:Images/TaylorSeriesMod.0_gr_6.gif],    [Graphics:Images/TaylorSeriesMod.0_gr_7.gif],    and    [Graphics:Images/TaylorSeriesMod.0_gr_8.gif].  
               Remark. The accuracy of the image points for the approximation    [Graphics:Images/TaylorSeriesMod.0_gr_9.gif]    is  
               [Graphics:Images/TaylorSeriesMod.0_gr_10.gif].

Explore Solution 7.3.

 

Example 7.4.  Show that, for  [Graphics:Images/TaylorSeriesMod_gr_52.gif],   

(7-12)            (a) [Graphics:Images/TaylorSeriesMod_gr_53.gif]    and    (b) [Graphics:Images/TaylorSeriesMod_gr_54.gif].  

Solution.  For [Graphics:Images/TaylorSeriesMod_gr_59.gif],

(7-13)            [Graphics:Images/TaylorSeriesMod_gr_60.gif].

If we let [Graphics:Images/TaylorSeriesMod_gr_61.gif] take the role of z in (7-13), we get that  

            [Graphics:Images/TaylorSeriesMod_gr_62.gif],  

for  [Graphics:Images/TaylorSeriesMod_gr_63.gif].  But  [Graphics:Images/TaylorSeriesMod_gr_64.gif]  iff  [Graphics:Images/TaylorSeriesMod_gr_65.gif],  thus we have proven that  [Graphics:Images/TaylorSeriesMod_gr_66.gif]  for  [Graphics:Images/TaylorSeriesMod_gr_67.gif].

Next, let  [Graphics:Images/TaylorSeriesMod_gr_68.gif]  take the role of z in Equation (7-13), we get that  

            [Graphics:Images/TaylorSeriesMod_gr_69.gif]
            
gives the second part of Equations (7-12).

                    [Graphics:Images/TaylorSeriesMod.0_gr_11.gif]          [Graphics:Images/TaylorSeriesMod.0_gr_12.gif]

                    [Graphics:Images/TaylorSeriesMod.0_gr_13.gif]          [Graphics:Images/TaylorSeriesMod.0_gr_14.gif]

               The disk    [Graphics:Images/TaylorSeriesMod.0_gr_15.gif]  and it's images under the mappings:
               [Graphics:Images/TaylorSeriesMod.0_gr_16.gif],    [Graphics:Images/TaylorSeriesMod.0_gr_17.gif],   and    [Graphics:Images/TaylorSeriesMod.0_gr_18.gif].  
               
               Remark 1. The accuracy of the image points for the approximation   [Graphics:Images/TaylorSeriesMod.0_gr_19.gif]   is
               [Graphics:Images/TaylorSeriesMod.0_gr_20.gif].
               
               Remark 2. The images of   [Graphics:Images/TaylorSeriesMod.0_gr_21.gif]  under the mappings:   [Graphics:Images/TaylorSeriesMod.0_gr_22.gif],    
               [Graphics:Images/TaylorSeriesMod.0_gr_23.gif],   and    [Graphics:Images/TaylorSeriesMod.0_gr_24.gif]   will appear like those shown above,
               because  [Graphics:Images/TaylorSeriesMod.0_gr_25.gif]  rotates the plane about the origin and   [Graphics:Images/TaylorSeriesMod.0_gr_26.gif].  
               Also, the accuracy of the image points for the approximation   [Graphics:Images/TaylorSeriesMod.0_gr_27.gif]   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)            [Graphics:Images/TaylorSeriesMod_gr_171.gif]    and    [Graphics:Images/TaylorSeriesMod_gr_172.gif]   for   [Graphics:Images/TaylorSeriesMod_gr_173.gif].  

For many students, it makes sense that the first series in Equations (7-14) converges only on the interval  [Graphics:Images/TaylorSeriesMod_gr_174.gif]  because [Graphics:Images/TaylorSeriesMod_gr_175.gif] is undefined at the points  [Graphics:Images/TaylorSeriesMod_gr_176.gif].  It seems unclear as to why this should also be the case for the series representing [Graphics:Images/TaylorSeriesMod_gr_177.gif], since the real-valued function [Graphics:Images/TaylorSeriesMod_gr_178.gif] is defined everywhere.  The explanation, of course, comes from the complex domain.  The complex function [Graphics:Images/TaylorSeriesMod_gr_179.gif] is not defined everywhere.  In fact, the singularities of  [Graphics:Images/TaylorSeriesMod_gr_180.gif]  are at the points  [Graphics:Images/TaylorSeriesMod_gr_181.gif],  and the distance between them and the point  [Graphics:Images/TaylorSeriesMod_gr_182.gif]  equals 1. According to Corollary 7.3, therefore, Equations (7-14) are valid only for [Graphics:Images/TaylorSeriesMod_gr_183.gif], and thus Equations (7-14) are valid only for the real numbers  [Graphics:Images/TaylorSeriesMod_gr_184.gif].  

 

    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  [Graphics:Images/TaylorSeriesMod_gr_185.gif]?  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 [Graphics:Images/TaylorSeriesMod_gr_186.gif] we have  

            [Graphics:Images/TaylorSeriesMod_gr_187.gif].  

Then  [Graphics:Images/TaylorSeriesMod_gr_188.gif].  

Proof.

 

Example 7.5.  Find the Maclaurin series for [Graphics:Images/TaylorSeriesMod_gr_189.gif].  

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

            [Graphics:Images/TaylorSeriesMod_gr_190.gif].  

Recall that the series for sin z (valid for all z) is  [Graphics:Images/TaylorSeriesMod_gr_191.gif].  Using the identity for  [Graphics:Images/TaylorSeriesMod_gr_192.gif],  we obtain  

            [Graphics:Images/TaylorSeriesMod_gr_193.gif]

By the uniqueness of power series, this last expression is the Maclaurin series for  [Graphics:Images/TaylorSeriesMod_gr_194.gif].

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  

            [Graphics:Images/TaylorSeriesMod_gr_225.gif],
and
            [Graphics:Images/TaylorSeriesMod_gr_226.gif].  

If  [Graphics:Images/TaylorSeriesMod_gr_227.gif]  is any complex constant, then

(7-15)            [Graphics:Images/TaylorSeriesMod_gr_228.gif],  

(7-16)            [Graphics:Images/TaylorSeriesMod_gr_229.gif],  and  

(7-17)            [Graphics:Images/TaylorSeriesMod_gr_230.gif],   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

            [Graphics:Images/TaylorSeriesMod_gr_232.gif]   for   [Graphics:Images/TaylorSeriesMod_gr_233.gif].  

Solution.  We let  [Graphics:Images/TaylorSeriesMod_gr_234.gif],  for [Graphics:Images/TaylorSeriesMod_gr_235.gif].   In terms of Theorem 7.6, we have  [Graphics:Images/TaylorSeriesMod_gr_236.gif],  for all n, and thus Equation (7-17) gives  

            [Graphics:Images/TaylorSeriesMod_gr_237.gif]

Explore Solution 7.6.

 

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

            [Graphics:Images/TaylorSeriesMod_gr_242.gif]   for   [Graphics:Images/TaylorSeriesMod_gr_243.gif].  

Explore Solution for Extra Example 1.

 

Extra Example 2.  Show that  [Graphics:Images/TaylorSeriesMod_gr_249.gif]   for   [Graphics:Images/TaylorSeriesMod_gr_250.gif].  

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

            [Graphics:Images/TaylorSeriesMod_gr_251.gif]  

Explore Solution for Extra Example 2.

 

Exercises for Section 7.2.  Taylor Series Representations

 

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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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