COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 7  TAYLOR AND LAURENT SERIES

a = b = c = x = y = z = Z = 
                    0; 
Clear[a, A, b, c, f, f1, f2, g, h, i, lim1, lim2, L, L1, L2, L3, L4, 
        m, n, p, solset, S, SI, Sinf, S0, S7, S9, S11, S12, S13, S17, S21, 
        S22, S25, x, y, z, Z, Z1, Z2, Z3, Z4]; 
Needs["Graphics`ComplexMap`"]; 
Needs["Graphics`Colors`"];

Section 7.1	Uniform Convergence

Section 7.2  Taylor Series Representations

In Section 4.4 we saw that functions defined by power series have derivatives of all orders (Theorem 4.16). In Section 6.5 we saw 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.

To investigate when the preceding series converges we will need the following lemma.

Definition  7.2  (Taylor Series),  Page 269.  If  f (z)  is analytic at z = α, then the series 

	\(f \((α)\) + 
        f' \((α)\) \((z - α)\)\  + \(\(f\^\((2)\)\) \((α)\)\)\/\(2!\)\ \((z - α)\)\^2\)\(\(\(\(+\(\(\(f\^\((2)\)\) \((α)\)\)\/\(3!\)\)\)\ \((z - α)\)\^3\)\(\ \)\(+\)\) ... \)  
	
	\(\(\(=\)\(\ \)\(∑\+\(k = 0\)\%∞\(\(\( 
                      f\^\((k)\)\) \((α)\)\)\/\(k!\)\) \((z - α)\)\^k\)\)\).  

is called the Taylor series for f centered around α. When the center is  α = 0, the series is called the Maclaurin series for f.

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

Lemma 7.1,  Page 269.  If  z, (Subscript[z, 0])  and  α  are complex numbers  with  z ≠ (Subscript[z, 0]),  and  z ≠ α,  then  

	(1/(z  -  (Subscript[z, 0])))=\(\(1\/\(z\  - \ α\) + \(z\_0\  - \ α\)\/\((z\  - \ α)\)\^2 + \((z\_0\  - \ α)\)\^2\/\((z\  - \ α)\)\^3 +  ... \) + \((z\_0\  - \ α)\)\^n\/\((z\  - \ α)\)\^\(n + 1\)

          \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(+\ \(1\/\(z - 
                    z\_0\)\)\) \((z\_0\  - \ α)\)\^\(n + 1\)\/\((z\  - \ α)\)\^\(n + 1\)\)  

where n is a positive integer.

Proof of Lemma 7.1, see text Page 269.

Theorem 7.4 (Taylor's Theorem), Page 269. Suppose f is analytic in a domain G, and that \(\(D\_R\) \((α)\) = {z : \(\(|\)\(z - α\)\(|\)\(\(<\)\(R\)\)\)}\) is any disk contained in G. Then the Taylor series for f converges to f(z) in (Subscript[D, R]) (α);  that is, 

	\(f \((z)\) = ∑\+\(k = 0\)\%∞\(\(\( 
                      f\^\((k)\)\) \((α)\)\)\/\(k!\)\) \((z - α)\)\^k\) for all z ∈ (Subscript[D, R]) (α).  

Furthermore, the convergence is uniform on any closed subdisk  \(\(\(D\_r\)\&_\) \((α)\) = {z : \(\(|\)\(z - α\)\(|\)\(\(≤\)\(r\)\)\)}\ \ for\ \ 0\  < \ r\  < \ R\).

Proof of Theorem 7.4, see text Page 269.

Recall that 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 non-removable 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 f (z) = (1/(1  -  z)) has a non-removable 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 just a bit.

Corollary 7.3,  Page 272.  Suppose  f(z)  is analytic in the domain  G  that contains the point  z = α.  Let  (Subscript[z, 0])  be a singular point of minimum distance to  α.  If  \(\(\(|\)\(\(z\_0 - α\)\(|\)\)\) = \ R\),  then

	(i)  the Taylor series \(∑\+\(k = 0\)\%∞\(\(\( 
                    f\^\((k)\)\) \((α)\)\)\/\(k!\)\) \((z - α)\)\^k\) converges to f (z) on all of  (Subscript[D, R]) (α),  
and
	(ii)  if \(\(\(|\)\(z\_1 - α\)\(|\)\) = S > \ R\), the Taylor series \(∑\+\(k = 0\)\%∞\(\(\( 
                    f\^\((k)\)\) \((α)\)\)\/\(k!\)\) \((z\_1 - α)\)\^k\) does not converges to f ((Subscript[z, 1])).

Proof of Corollary 7.3, see text Page 272.

Example 7.3,  Page 272.  Show that (1/(((1  -  z))^2)) = (∑_(n = 0)^∞)(n + 
                1) (z^n) is valid for all |z|<1.

Solution 7.3.

Enter the function  f[z] = (1/(((1  -  z))^2)) and find the first few terms of the Maclaurin series.

Remove[z]; 
Clear[f, n, S17, S∞]; 
f[z_]  =  (1/(((1 - z))^2)); 
S17[z_]  =  Normal[Series[f[z], {z, 0, 17}]]; 
Print[" f[z] = ", f[z]]; 
Print[""]; 
\(\(Print[\*""\!\(S\_17\)[z] = "", 
        S17[z]];\)\ \)
Print[""]; 
Print[" f[z] = ", 
        S17[z], " + ..."];

Sum up the terms to verify that we have things right.

S17[
          z_]    =  (∑_(n = 0)^17)(n + 1) (z^n); 
S∞[
          z_]  =  (∑_(n = 0)^∞)(n + 1) 
            (z^n); 
\(\(Print[\*""\!\(S\_17\)[z] = "", 
        S17[z]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[S, ∞])[z] =  (∑_(n = 0)^∞)(n+1)(z^n) = "", 
        S∞[z]];\)\ \)

We can use Mathematica to plot some partial sums for  f[z] = (1/(((1  -  z))^2)).  Since convergence will be more rapid in a smaller disk  |z|< r  we will use the disk  |z|< 0.6.

Clear[f, j, m, n, s, 
        z]; 
f[z_] = (1/(((1 - z))^2)); 
For [j = 2, 
        j ≤ 4, j++, 
m = 
          4 j;  
s[
            z_] = (∑_(n = 0)^m)(n + 1) 
              (z^n);  
PolarMap[
          s, {0, 0.6, 0.05}, {0, 2  π, (π/16)}, 
          PlotRange -> {{0, 4}, {-2, 2}}, AspectRatio -> 1, 
          Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Blue, 
              Cyan}];   
Print["w = s[z,", 
          m, "] = ", s[z]]; ]; 
PolarMap[f, {0, 0.6, 0.05}, {0, 2  π, (π/16)}, 
        PlotRange -> {{0, 4}, {-2, 2}}, AspectRatio -> 1, 
        Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}];  
Print["  The mapping  w = ", 
        f[z]];

Example 7.4,  Page 273.  (a)  Show that (1/(1  -  (z^2))) = (∑_(n = 0)^∞) (z^(2  n)) and  (b) \(1\/\(1\  + \ z\^2\) = ∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) z\^\(2  n\)\) are valid for all |z|<1.

Solution 7.4 (a).

(a)  Show that (1/(1  -  (z^2))) = (∑_(n = 0)^∞) (z^(2  n))  is valid for all |z|<1.  

Enter the function  f[z] = (1/(1  -  (z^2))) and find the first few terms of the Maclaurin series.

Remove[z]; 
Clear[f, n, S50, S∞]; 
f[z_]  =  (1/(1 - (z^2))); 
S50[z_]  =  Normal[Series[f[z], {z, 0, 50}]]; 
Print[" f[z] = ", f[z]]; 
Print[""]; 
\(\(Print[\*""\!\(S\_50\)[z] = "", 
        S50[z]];\)\ \)
Print[""]; 
Print[" f[z] = ", 
        S50[z], " + ..."];

Sum up infinitely many terms to verify that we have things right.

S50[
          z_]    =  (∑_(n = 0)^25) (z^(2  n)); 
S∞[
          z_]  =   (∑_(n = 0)^∞) (z^(2  n)); 
\(\(Print[\*""\!\(S\_50\)[z] = "", 
        S50[z]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)(z^(2  n)) = "", 
        S∞[z]];\)\ \)

We can use Mathematica to plot some partial sums for  f[z] = (1/(1  -  (z^2))).  Since convergence will be more rapid in a smaller disk  |z|< r  we will use the half disk  Im[z] > 0  and   | z | < 0.9.

Clear[f, j, m, n, s, 
        z]; 
f[z_] =  (1/(1 - (z^2))); 
\(\(For\ [j = 3, 
        j ≤ 5, \(j++\), 
m = 
          4\ j; \ 
s[
            z_] = ∑\+\(n = 0\)\%m z\^\(2  n\); \ 
PolarMap[s, {0, 0.9, 0.05}, {0, π, π\/24}, 
          PlotRange -> {{0, 4}, {\(-2\), 2}}, AspectRatio -> 1, 
          Ticks -> {Range[0, 4, 1], Range[\(-2\), 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {DarkGreen, 
              Green}]; \ \ 
Print["w = s[z,", 
          2  m, "] = ", s[z]];\ ];\)\ \)
PolarMap[f, {0, 0.9, 0.05}, {0, π, (π/24)}, 
        PlotRange -> {{0, 4}, {-2, 2}}, AspectRatio -> 1, 
        Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["  The mapping  w = ", 
        f[z]];

Solution 7.4 (b).

(b)  Show that \(1\/\(1\  + \ z\^2\) = ∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) z\^\(2  n\)\) is valid for all |z|<1.  

Enter the function  f[z] = (1/(1  +  (z^2))) and find the first few terms of the Maclaurin series.

Remove[z]; 
Clear[f, n, S50, S∞]; 
f[z_]  =  (1/(1 + (z^2))); 
S50[z_]  =  Normal[Series[f[z], {z, 0, 50}]]; 
Print[" f[z] = ", f[z]]; 
Print[""]; 
\(\(Print[\*""\!\(S\_50\)[z] = "", 
        S50[z]];\)\ \)
Print[""]; 
Print[" f[z] = ", 
        S50[z], " + ..."];

Sum up infinitely many terms to verify that we have things right.

S50[
          z_]    =  (∑_(n = 0)^25)(((-1))^n) 
            (z^(2  n)); 
S∞[
          z_]  =   (∑_(n = 0)^∞)(((-1))^n) 
            (z^(2  n)); 
\(\(Print[\*""\!\(S\_25\)[z] = "", 
        S50[z]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)(((-1))^n)(z^(2  n)) =  "", 
        S∞[z]];\)\ \)

We can use Mathematica to plot some partial sums for  f[z] = (1/(1  +  (z^2))).  Since convergence will be more rapid in a smaller disk  |z|< r  we will use the half disk  Im[z] > 0  and   | z | < 0.9.

Clear[f, j, m, n, s, 
        z]; 
f[z_] =  (1/(1 + (z^2))); 
For [j = 3, 
        j ≤ 5, j++, 
m = 
          4 j;  
s[
            z_] = (∑_(n = 0)^m)(((-1))^n) 
              (z^(2  n));  
PolarMap[
          s, {0, 0.9, 0.05}, {0, π, (π/24)}, 
          PlotRange -> {{0, 4}, {-2, 2}}, AspectRatio -> 1, 
          Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Blue, 
              Cyan}];   
Print["w = s[z,", 
          2  m, "] = ", s[z]]; ]; 
PolarMap[f, {0, 0.9, 0.05}, {0, π, (π/24)}, 
        PlotRange -> {{0, 4}, {-2, 2}}, AspectRatio -> 1, 
        Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["  The mapping  w = ", 
        f[z]];

Question.  Why do the graph of  w = (Subscript[f, 1])[
            z] = (1/(1  -  (z^2)))   and   w = (Subscript[f, 2])[z] = 
              (1/(1  +  (z^2))) appear to be the same ?

Theorem 7.5  (Uniqueness of Power Series),  Page 274.  Suppose that in some disk (Subscript[D, r]) (α) we have  

	\(f \((z)\) = \(∑\+\(n = 0\)\%∞\( 
                  a\_n\) \((z - α)\)\^n = ∑\+\(n = 0\)\%∞\( b\_n\) \((z - α)\)\^n\)\).  

Then (Subscript[a, n]) = (Subscript[b, n])  for  n = 0, 1, 2,  ... .

Proof of Theorem 7.5, see text Page 274.

Example 7.5,  Page 274.  Find the Maclaurin series for f (z) = (((sin (z)))^3).

Solution 7.5.

Enter the function  f[z] = Sin[z]^3  and find the first few terms of the Maclaurin series.

Remove[z]; 
Clear[f, f1, f2, n, S27]; 
f[z_]  =  Sin[z]^3; 
S27[z_]  =  Normal[Series[f[z], {z, 0, 27}]]; 
Print[" f[z] = ", f[z]]; 
Print[""]; 
\(\(Print[\*""\!\(S\_27\)[z] = "", 
        S27[z]];\)\ \)
Print[""]; 
Print[" f[z] = ", 
        S27[z], " + ..."];

Aside.  The Maclaurin polynomial  (Subscript[S, 27])[z] of degree n=27  is NOT equal to the cube of  (Subscript[S, 9])[z].

S9[z_]  =  
      Normal[Series[f[z], {z, 0, 9}]]
\(\(Print[\*""   \!\(S\_9\)[z] = "", 
        S9[z]];\)\ \)
\(\(Print[\*""(\!\(S\_9\)[z]\!\(\()\^3\)\) = "", 
        S9[z]\^3];\)\ \)
\(\(Print[\*""(\!\(S\_9\)[z]\!\(\()\^3\)\) = "", 
        Expand[S9[z]\^3]];\)\ \)
Print[""]; 
\(\(Print[\*""Warning ((Subscript[S, 9])[z])^3 ≠ (Subscript[S, 27])[z].""];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(S\_27\)[z] = "", 
        S27[z]];\)\ \)

Sum up infinitely many terms to verify that we have things right.

f1[z_]  = 
        PowerExpand[(∑_(n = 0)^∞)(((((-1))^n) 3)/(4 (2  n + 1)!)) (z^(2  n + 1))]; 
f2[z_]  =  
        PowerExpand[(∑_(n = 0)^∞)((- (((-1))^n) 3 (9^n))/(4 (2  n + 1)!)) (z^(2  n + 1))]; 
f[z_]  =  
        Factor[f1[z]  +  f2[z],  Trig -> True]; 
\(\(Print[\*""(Subscript[f, 1])[z] = (∑_(n = 0)^∞)\!\(\(\((\(-1\))\)\^n\\ 3\)\/\(4 \(\((2  n + 1)\)!\)\)\) (z^(2  n + 1)) =  "", f1[z]];\)\ \)
\(\(Print[\*""(Subscript[f, 2])[z] = (∑_(n = 0)^∞)\!\(\(\(-\\ \((\(-1\))\)\^n\)\\ 3\\ 9\^n\)\/\(4 \(\((2  n + 1)\)!\)\)\) (z^(2  n + 1)) =  "", f2[z]];\)\ \)
\(\(Print[\*"" f[z] = \!\(f\_1\)[z] + \!\(f\_2\)[z] = "", \ 
        f1[z]\  + \ f2[z]];\)\ \)
Print[" f[z] = ", f[z]];

We can use Mathematica to plot some partial sums for  f[z] =  Sin[z]^3.  For illustration we use the portion of the  disk  |z|< 1.2  that lies in the sector 0 < Arg[z] < ((23  π)/36).

Clear[f, j, m, n, s, 
        z]; 
f[z_]  =  Sin[z]^3; 
For [j = 1, 
        j ≤ 3, j++, 
m = 
          3 j;  
s[z_] = 
          Normal[Series[f[z], {z, 0, m}]];  
PolarMap[
          s, {0, 1.2, 0.1}, {0, ((23  π)/36), (π/36)}, 
          PlotRange -> {{-3, 3}, {-3.5, 2.5}}, 
          AspectRatio -> 1, 
          Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {DarkGreen, 
              Green}];   
Print["w = s[z,", 
          m, "] = ", s[z]]; ]; 
PolarMap[f, {0, 1.2, 0.1}, {0, ((23  π)/36), (π/36)}, 
        PlotRange -> {{-3, 3}, {-3.5, 2.5}}, 
        AspectRatio -> 1, 
        Ticks -> {Range[0, 4, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["  The mapping  w = ", 
        f[z]];

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

Theorem 7.6,  Page 275.  Let f(z) and g(z) have the power series representations  

	\(f \((z)\) = ∑\+\(n = 0\)\%∞\( 
                  a\_n\) \((z - α)\)\^n\ \ \ for\ \ z ∈ \(D\_\(r\_1\)\) \((α)\)\), 
and 
	\(g \((z)\) = ∑\+\(n = 0\)\%∞\( 
                  b\_n\) \((z - α)\)\^n\ \ \ for\ \ z ∈ \(D\_\(r\_2\)\) \((α)\)\).  

If  r = min {(Subscript[r, 1]), (Subscript[r, 2])}  and  β  is any complex constant, then

	(i)  \(βf \((z)\) = ∑\+\(n = 0\)\%∞\( βa\_n\) \((z - α)\)\^n\ \ \ for\ \ z ∈ \(D\_\(r\_1\)\) \((α)\)\),  

	(ii)  \(f \((z)\) + 
          g \((z)\) = ∑\+\(n = 0\)\%∞\((a\_n + 
                  b\_n)\) \((z - α)\)\^n\ \ \ for\ \ z ∈ \(D\_r\) \((α)\)\), and  

	(iii)  \(f \((z)\)*
          g \((z)\) = ∑\+\(n = 0\)\%∞\( 
                  c\_n\) \((z - α)\)\^n\ \ \ for\ \ z ∈ \(D\_r\) \((α)\)\), 
	where (Subscript[c, n]) = (∑_(k = o)^n) (Subscript[a, n]) (Subscript[b, n - k]).  

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

Proof of Theorem 7.6, see text Page 275.

Example 7.6,  Page 276.  Use the Cauchy product of series to show that 

			(z/(((1  -  z))^2)) = (∑_(n = 0)^∞)(n + 
                1) (z^n)   for  z ϵ (Subscript[D, 1]) (0).

Solution 7.6.

Use the following fact that \(f \((z)\) = \(g \((z)\) = \(1\/\(1\  - \ z\) = ∑\+\(n = 0\)\%∞ z\^n\ \ \ for\ \ z\ ϵ\ \(D\_1\) \((0)\)\)\)\). 
Enter the coefficients (Subscript[a, n]) = 1   and   (Subscript[b, n]) = 1  and let Mathematica carry out the computations.

Remove[a, b, c, z]; 
Clear[f, g, h, k, n];  
(Subscript[a, n])_ = 1;  
(Subscript[b, n])_ = 1;  
(Subscript[c, n])_ = 
      FullSimplify[(∑_(k = 0)^n) (Subscript[a, k]) (Subscript[b, n - k])];  
f[z_] = 
      FullSimplify[(∑_(n = 0)^∞) (Subscript[a, n]) 
            (z^n)];  
g[z_] = 
      FullSimplify[(∑_(n = 0)^∞) (Subscript[b, n]) 
            (z^n)];  
h[z_] = 
      FullSimplify[(∑_(n = 0)^∞) (Subscript[c, n]) 
            (z^n)];  
\(Print[\*""\!\(a\_n\) = "", a\_n]; \ \)
\(Print[\*""\!\(b\_n\) = "", b\_n]; \ \)
Print[""]; 
\(\(Print[\*""f[z] = (∑_(n = 0)^∞)(Subscript[a, n])(z^n) = (∑_(n = 0)^∞)("", a\_k, \*"")(z^n) = "", 
        f[z]];\)\ \)
Print[""]; 
\(\(Print[\*""g[z] = (∑_(n = 0)^∞)(Subscript[b, n])(z^n) = (∑_(n = 0)^∞)("", b\_\(n - k\), \*"")(z^n) = "", 
        g[z]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[c, n]) = (∑_(k = 0)^n)(Subscript[a, k])(Subscript[b, n - k])""];\)\ \)
\(\(Print[\*""(Subscript[c, n]) = (∑_(k = 0)^n)("", 
        a\_k, ")(", 
        b\_\(n - k\), ") = ", \*""(∑_(k = 0)^n)("", \(a\_k\) b\_\(n - k\), ")"];\)\ \)
\(\(Print[\*""\!\(∑\+\(k = 0\)\%n\)("", \(a\_k\) 
          b\_\(n - k\), ") = ", c\_n];\)\ \)
\(\(Print[\*""\!\(c\_n\) = "", c\_n];\)\ \)
Print[""];  
\(\(Print[\*""h[z] = (∑_(n = 0)^∞)(Subscript[c, n])(z^n) = (∑_(n = 0)^∞)("", c\_n, \*"")(z^n) = "", 
        h[z]];\)\ \)

Extra Example,  Page 276.  Use the Cauchy product of series to show that 

			(z/(((z  -  1))^4)) = (∑_(n = 0)^∞)(( 
                  n (1 + n) (2 + n))/6) 
            (z^n)   for  z ϵ (Subscript[D, 1]) (0).

Solution for Extra Example, Page 276.

Use the following facts (z/(((z  -  1))^2)) = (∑_(n = 0)^∞) 
            n (z^n)   for  z ϵ (Subscript[D, 1]) (0) and (1/(((z  -  1))^2)) = (∑_(n = 0)^∞)(n + 
                1) (z^n)   for  z ϵ (Subscript[D, 1]) (0). 
Enter the coefficients (Subscript[a, n]) = n   and   (Subscript[b, n]) = n + 1  and let Mathematica carry out the computations.

Remove[a, b, c, z]; 
Clear[f, g, h, k, n]; 
(Subscript[a, n])_ = n; 
(Subscript[b, n])_ = n + 1; 
(Subscript[c, n])_ = 
        FullSimplify[(∑_(k = 0)^n) (Subscript[a, k]) 
              (Subscript[b, n - k])]; 
f[z_] = 
        FullSimplify[(∑_(n = 0)^∞) (Subscript[a, n]) 
              (z^n)]; 
g[z_] = 
        FullSimplify[(∑_(n = 0)^∞) (Subscript[b, n]) 
              (z^n)]; 
h[z_] = 
        FullSimplify[(∑_(n = 0)^∞) (Subscript[c, n]) 
              (z^n)]; 
\(\(Print[\*""\!\(a\_n\) = "", a\_n];\)\ \)
\(\(Print[\*""\!\(b\_n\) = "", b\_n];\)\ \)
Print[""]; 
\(\(Print[\*""f[z] = (∑_(n = 0)^∞)(Subscript[a, n])(z^n) = (∑_(n = 0)^∞)("", a\_k, \*"")(z^n) = "", 
        f[z]];\)\ \)
Print[""]; 
\(\(Print[\*""g[z] = (∑_(n = 0)^∞)(Subscript[b, n])(z^n) = (∑_(n = 0)^∞)("", b\_\(n - k\), \*"")(z^n) = "", 
        g[z]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[c, n]) = (∑_(k = 0)^n)(Subscript[a, k])(Subscript[b, n - k])""];\)\ \)
\(\(Print[\*""(Subscript[c, n]) = (∑_(k = 0)^n)("", 
        a\_k, ")(", 
        b\_\(n - k\), ") = ", \*""(∑_(k = 0)^n)("", \(a\_k\) b\_\(n - k\), ")"];\)\ \)
\(\(Print[\*""\!\(∑\+\(k = 0\)\%n\)("", \(a\_k\) 
          b\_\(n - k\), ") = ", c\_n];\)\ \)
\(\(Print[\*""\!\(c\_n\) = "", c\_n];\)\ \)
Print[""]; 
\(\(Print[\*""h[z] = (∑_(n = 0)^∞)(Subscript[c, n])(z^n) = (∑_(n = 0)^∞)("", c\_n, \*"")(z^n) = "", 
        h[z]];\)\ \)

Section 7.2  Exercises for Taylor Series Representations
See textbook page 276.

Section 7.3	Laurent Series Representations

Section 7.4	Singularities, Zeros and Poles

Section 7.5	Applications of Taylor and Laurent Series

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