COMPLEX ANALYSIS: Mathematica 3.0 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 1998
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
3rd Edition,  1997,  ISBN: 0-7637-0270-6
Jones & Bartlett Publishers, Inc.
40 Tall Pine Drive, Sudbury, MA 01776
Tele.  (800) 832-0034, FAX:  (508)  443-8000
E-mail:  mkt@jbpub.com,     http://www.jbpub.com/
This free software is compliments of the authors.
mathews@fullerton.edu,  howell@westmont.edu

CHAPTER 4  SEQUENCES, SERIES, and JULIA and MANDELBROT SETS
Section 4.1  Definitions and Basic Theorems for Sequences and Series
Section 4.2  Power Series Functions
Section 4.3  Julia and Mandelbrot Sets
GoTo Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
GoTo Contents

CHAPTER 4  SEQUENCES, SERIES, and JULIA and MANDELBROT SETS

Clear[a, absZn, c, cx, cy, i, juliaC, k, lim, limX, limY, limZ, 
      mandelbrotC, n, Pn, Qn, R, s, Sn, set, sum, sumN, sumX, sumY, sumZ, x, 
      Xn, Yn, z, Z, Zn];

Section 4.1  Definitions and Basic Theorems for Sequences and Series

Informally, the sequence  {(Subscript[z, n])}  has  ζ  as a limit as  n  approaches infinity, provided the terms  (Subscript[z, n])   can be made as close as we please to  ζ  by making  n  large enough.  When this happens, we write   (lim_(n -> ∞))(Subscript[z, n])=ζ.

Theorem 4.1,  Page 97.  Let  (Subscript[z, n])=(Subscript[x, n])+i (Subscript[y, n])  and  ζ=u+i v.  Then 
(lim_(n -> ∞))(Subscript[z, n])=ζ    if and only if
(lim_(n -> ∞))(Subscript[x, n])=u    and     (lim_(n -> ∞))(Subscript[y, n])=v.

Theorem 4.2,  Page 98.  If   {(Subscript[z, n])}  is a Cauchy sequence,  {(Subscript[z, n])}  converges.

Theorem 4.3,  Page 100.  Let  (Subscript[z, n])=(Subscript[x, n])+i (Subscript[y, n])  and  S = U + i V.  Then 
S=(∑_(n = 1)^∞)(Subscript[z, n])=(∑_(n = 1)^∞)((Subscript[x, n])+i (Subscript[y, n]))    if and only if both
U=(∑_(n = 1)^∞)(Subscript[x, n])    and     V=(∑_(n = 1)^∞)(Subscript[y, n]).

Theorem 4.4,  Page 100.  If  (∑_(n = 1)^∞)(Subscript[z, n])  is a convergent complex series, then  (lim_(n -> ∞))(Subscript[z, n])=0.

Theorem 4.5,  Page 101.  Let  (∑_(n = 1)^∞)(Subscript[z, n])  and  (∑_(n = 1)^∞)(Subscript[w, n])  be convergent series, and let  c  be a complex number.  Then  
(∑_(n = 1)^∞)c (Subscript[z, n])=c(∑_(n = 1)^∞)(Subscript[z, n])     and   (∑_(n = 1)^∞)((Subscript[z, n])+(Subscript[w, n]))=  (∑_(n = 1)^∞)(Subscript[z, n]) + (∑_(n = 1)^∞)(Subscript[w, n]).

Theorem 4.6,  Page 101.  If the Cauchy product converges, then  (∑_(n = 1)^∞)(Subscript[c, n])=( (∑_(n = 1)^∞)(Subscript[a, n]))( (∑_(n = 1)^∞)(Subscript[b, n])),  where  (Subscript[c, n]) = (∑_(k = 1)^n)(Subscript[a, k]) (Subscript[b, n - k]).

Theorem 4.7,  Page 101.  (Comparison Test)  Let  (∑_(n = 1)^∞)(Subscript[M, n])  be a convergent series of real nonnegative terms.  If   {(Subscript[z, n])}  is a sequence of complex numbers and   |(Subscript[z, n])|≤(Subscript[M, n])   holds for all  n, then   (∑_(n = 1)^∞)(Subscript[z, n])=(∑_(n = 1)^∞)((Subscript[x, n])+i (Subscript[y, n]))   converges.

Theorem 4.8,  Page 103.  (Geometric Series)  If   |z| < 1,  the series   (∑_(n = 0)^∞)(z^n)   converges to   f (z)=(1/(1 - z)).

Theorem 4.9,  Page 105.  (d'Alembert's Ratio Test)  If   (∑_(n = 1)^∞)(Subscript[ζ, n])  is a complex series with the property that  (lim_(n -> ∞))((|(Subscript[ζ, n + 1]) | 
                )/(|(Subscript[ζ, n]) | ))=L,  then the series converges absolutely if  L < 1  and diverges if  L > 1.  

Theorem 4.10,  Page 107.  (The Root Test)  Given the series  (∑_(n = 1)^∞)(Subscript[ζ, n]),  suppose  lim (sup_(n -> ∞))((|(Subscript[ζ, n]) | )^(1/n))=L.  Then the series is absolutely convergent if  L < 1  and divergent if  L > 1.

Example 4.1,  Page 98.  Find the limit of the sequence  (Subscript[z, n]) = (((Sqrt[n]) + i (n + 1))/n).

\(Clear[limX, limY, limZ, n, Xn, Yn, Zn]; 
Xn\  = \ \@n\/n; 

    Yn\  = \ \(n + 1\)\/n; 
Zn\  = \ Xn\  + \ I\ Yn; 

    limX\  = \ Limit[Xn, \ n -> ∞]; 

    limY\  = \ Limit[Yn, \ n -> ∞]; 

    limZ\  = \ Limit[Zn, \ n -> ∞]; 

    Print[\*""\!\(lim\+\(\\ n -> ∞\)\) "", Xn, " = ", 
      limX]; 
Print[\*""\!\(lim\+\(\\ n -> ∞\)\) "", Yn, 
      " = ", limY]; 

    Print[\*""\!\(lim\+\(\\ n -> ∞\)\) ("", Zn, ") = ", 
      limZ]; \)

Example 4.2,  Page 98.  Show that the sequence  (Subscript[z, n])  =  (((1  +  i))^n)   is divergent.

\(Clear[lim, limX, limY, limZ, n, Xn, Yn, Zn]; 

    Zn\  = \ \((1\  + \ I)\)\^n; 

    lim\  = \ Limit[Zn, \ n -> ∞]; 

    Print[\*""\!\(lim\+\(\\ n -> ∞\)\) "", Zn, " = ", 
      lim]; \)

\(Xn\  = \ \((\@2)\)\^n\ Cos[\(n\ π\)\/4]; 

    Yn\  = \(\((\@2)\)\^n\) Sin[\(n\ π\)\/4]; 

    Zn\  = \ Xn\  + \ I\ Yn; 

    limX\  = \ Limit[Xn, \ n -> ∞]; 

    limY\  = \ Limit[Yn, \ n -> ∞]; 

    limZ\  = \ limX\  + \ I\ limY; 

    Print[\*""\!\(lim\+\(\\ n -> ∞\)\) "", Xn, " = ", 
      limX]; 
Print[\*""\!\(lim\+\(\\ n -> ∞\)\) "", Yn, 
      " = ", limY]; 

    Print[\*""\!\(lim\+\(\\ n -> ∞\)\) ("", Zn, ") = ", 
      limZ]; \)

Therefore, the limit of the sequence does NOT exist.

Infinite Series

     Informally, the series   S =(∑_(n = 1)^∞)(Subscript[z, n])   converges if the sequence of partial sums
(Subscript[S, n]) = (∑_(k = 1)^n)(Subscript[z, k])  has  S  as its limit.

Example 4.3,  Page 100.  Show that the series  (∑_( n = 1)^∞)((1 + i n (((-1))^n))/((n^2)))  is convergent.

Clear[n, sumX, sumY, sumZ];
sumX =(∑_( n = 1)^∞)(1/((n^2)));
sumY = (∑_( n = 1)^∞)((n (((-1))^n))/((n^2)));
sumZ  =  sumX  +  I sumY;
\(Print[\*""          (∑_(n = 1)^∞)(1/(n^2)) = "", sumX]\);
\(Print[\*""     (∑_(n = 1)^∞)((n (((-1))^n))/(n^2)) = "", sumY]\);
\(Print[\*""(∑_(n = 1)^∞)\!\(\(\\ 1\\  + \\ I\\ n \((\(-1\))\)\^n\)\/n\^2\) = "", sumZ]\);
\(Print[\*""(∑_(n = 1)^∞)\!\(\(\\ 1\\  + \\ I\\ n \((\(-1\))\)\^n\)\/n\^2\) = "", N[sumZ]]\);

Therefore, the series converges.

Example 4.4,  Page 100.  Show that the series   (∑_( n = 1)^∞)(((((-1))^n)   +  i)/n)  is divergent.

Clear[n, sumN, sumX, sumY, sumZ];
sumX = (∑_( n = 1)^∞)(((((-1))^n))/n);
sumY = (∑_( n = 1)^∞)(1/n);
sumZ  =  sumX  +  I sumY;
\(Print[\*""   (∑_(n = 1)^∞)((((-1))^n)/n) = "", sumX]\);
\(Print[\*""       (∑_(n = 1)^∞)(1/n) = "", sumY]\);
\(Print[\*""(∑_(n = 1)^∞)\!\(\(\((\(-1\))\)\^n + \\ I\)\/n\) = "", 
        sumZ]\);

It is known that the partial sums of the harmonic series grow slowly without bound.  For example, adding 10000 terms yields.

\(Print[\*""\!\(∑\+\(n = 1\)\%m\)\!\(1\/n\) = "", 
      ∑\+\(n = 1\)\%m 1\/n]; 

    For[i = 1, i <= 4, \(i++\), 
\ \ 
      Print["m = ", 10\^i, 
        \*""     \!\(∑\+\(n = 1\)\%m\)\!\(1\/n\) = "", 
        N[EulerGamma + PolyGamma[0, 1 + 10\^i]]]]; \)

Therefore, the series slowly diverges.

Example 4.5,  Page 100.  Show that the series  (∑_( n = 1)^∞)(((1  +  i))^n)  is divergent.

Clear[absZn, n, sum, Zn];
Zn  =  (((1  +  I ))^n);
absZn  =  Abs[1 + I]^n;
sum = (∑_( n=1)^∞)(((1 +I))^n);
\(Print[\*""   (∑_(n = 1)^∞)(1+I)^n = "", sum]\);
\(Print[\*""\!\(lim\+\(n -> ∞\)\) |"", Zn, "| = ", 
        Limit[absZn, n -> ∞]]\);

The terms do not go to zero in absolute value, therefore the series diverges.

Example 4.6,  Page 102.  Show that the series   (∑_( n = 1)^∞)(((((3  +  4 i))^n))/((5^n) (n^2)))  is convergent.

\(\(
Clear[n, sum, sumX, sumY, sumZ, x, Xn, y, Yn, z, Zn]; 

    Zn\  = \ \((3 + 4  I)\)\^n\/\(5\^n\ n\^2\); 

    Xn\  = \ ComplexExpand[Re[Zn]]; 
Yn\  = \ ComplexExpand[Im[Zn]]; 

    sumX\  = \ ∑\+\(n = 1\)\%∞ Xn; 

    sumY\  = \ ∑\+\(n = 1\)\%∞ Yn; 

    sumZ\  = \ 
      NSum[\((3 + 4  I)\)\^n\/\(5\^n\ n\^2\), 
        \ {n, 1, ∞}]; 
Print[\*""\!\(z\_n\) = "", Zn]; 

    Print[\*""\!\(z\_n\) = "", ComplexExpand[Zn]]; 

    Print[\*""\!\(x\_n\) = "", Xn]; 

    Print[\*""\!\(y\_n\) = "", Yn]; 

    Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(x\_n\) = "", 
      sumX]; 
Print["       ", Chop[N[sumX]]]; 

    Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(y\_n\) = "", 
      sumY]; 
Print["       ", Chop[N[sumY]]]; 

    Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(z\_n\) = "", 
      sumZ]; \)\)

Or we could use the comparison test to show that the series converges.

\(absZn\  = \ ComplexExpand[Abs[Zn]]; 

    sum\ \  = \ \ ∑\+\(n = 1\)\%∞ 1\/n\^2; 

    Print["|", Zn, "| = ", absZn]; 

    Print[\*""(∑_(n = 1)^∞)|\!\(\((3 + 4  I)\)\^n\/\(5\^n\\ n\^2\)\)| = (∑_(n = 1)^∞)(1/(n^2)) = "", sum]; \)

The series of absolute values converges, therefore the series converges.

Example 4.7,  Page 104.  Show that   (∑_( n = 0)^∞)(((((1 -  i))^n))/((2^n))) =  1 - i .

We can use the definition of convergence of a series and find the limit of the partial sums.

Clear[lim, n, sum, Sn, z, Z];
Z[n_]  =  ((((1  -  I))^n)/(2^n));
\(Print[\*""\!\(z\_n\) = "", Z[n]]\);
Sn = (∑_( k=0)^n)((((1- I))^k)/(2^k));
sum = (∑_( k=0)^∞)((((1- I))^k)/(2^k));
lim = Limit[Sn,n->∞];
Print["(Subscript[S, n]) = (∑_(k = 0)^n)\!\(\((1 - \\ \*
StyleBox[\(I\),
FontSize->14])\)\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",Sn];
\(Print[\*""          \!\(lim\+\(n -> ∞\)\) "", 
        \*""\!\(S\_n\) = "", lim]\);
Print["     (∑_(k = 0)^∞)\!\(\((1 - \\ \*
StyleBox[\(I\),
FontSize->14])\)\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",sum];

Or we can see that this is a geometric series with ratio:

\(Print[\*"" \!\(z\_n\) = "", Z[n]]; 

    Print[\*""\!\(z\_\(n + 1\)\) = "", Z[n + 1]]; 

    z\  = \ Z[n + 1]\/Z[n]; 

    Print[\*""  z = \!\(z\_\(n + 1\)\/z\_n\) = "", z]; \)

The sum is given by

sum = (∑_( n = 0)^∞)((((1- I))^n)/((2^n)));
Print["          z = ", z];
Print["        |z| = ", Abs[z]];
Print["     Is |z| < 1 ?  ", Abs[z] <  1];
Print[" "];
\(Print[\*""      (1/(1 - z)) = \!\(1\/\(1\\  - \\ \((1\/2\\  - \(\\ I\)\/2)\)\)\)""]\);
\(Print[\*""      \!\(1\/\(1 - z\)\) = "", \ 
        1\/\(1\  - \ \((1\/2\  - \(\ I\)\/2)\)\)]\);
Print["(∑_(n = 0)^∞)\!\(\((1 - \\ \*
StyleBox[\(I\),
FontSize->14])\)\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",sum];

Example 4.8,  Page 104.  Evaluate  (∑_( n = 3)^∞)((i^n)/((2^n))).

We can use the definition of convergence of a series and find the limit of the partial sums.

Clear[a, k, lim, n, sum, Sn, z, Z];
Z[n_]  = (( (I^n))/(2^n));
Sn = (∑_( k=3)^n)((I^k)/(2^k));
sum = (∑_( k=3)^∞)((I^k)/(2^k));
lim = Limit[Sn,n->∞];
\(Print[\*""\!\(z\_n\) = "", Z[n]]\);
Print["(Subscript[S, n]) = (∑_(k = 3)^n)\!\(\*
StyleBox[\(I\),
FontSize->14]\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",Sn];
\(Print[\*""     \!\(lim\+\(n -> ∞\)\) "", 
        \*""\!\(S\_n\) = "", lim]\);
Print["     (∑_(k = 3)^∞)\!\(\*
StyleBox[\(I\),
FontSize->14]\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",sum];

Or we can see that this is a geometric series with ratio:

\(Print[\*"" \!\(z\_n\) = "", Z[n]]; 

    Print[\*""\!\(z\_\(n + 1\)\) = "", Z[n + 1]]; 

    z\  = \ Z[n + 1]\/Z[n]; 

    Print[\*""  z = \!\(z\_\(n + 1\)\/z\_n\) = "", z]; \)

The sum is given by

a  =  Z[3];
sum = (∑_( k=3)^∞)((I^k)/(2^k));
\(Print[\*""        a = ((I/2))^3 = "", a]\);
Print["        z = ", z];
Print["      |z| = ", Abs[z]];
Print["   Is |z| < 1 ?  ", Abs[z] <  1];
Print[" "];
\(Print[\*""    (a/(1 - z)) = \!\(\((I\/2)\)\^3\/\(1\\  - \(\\ I\)\/2\)\)""]\);
\(Print[\*""    \!\(a\/\(1 - z\)\) = "", a\/\(1 - z\)]\);
Print["    (∑_(k = 3)^∞)\!\(\*
StyleBox[\(I\),
FontSize->14]\^\*
StyleBox[\(k\),
FontSize->14]\/\*
StyleBox[\(2\^\*
StyleBox[\(k\),
FontSize->14]\),
FontSize->14]\) = ",sum];

Example 4.9,  Page 105.  Show that  (∑_( n = 0)^∞)(((((1  -  i))^n))/(n!))  converges.

\(Clear[L, n, sum, r, z, Z]; 

    Z[n_]\  = \ \(\((1\  - \ I)\)\^n\ \)\/\(n!\); 

    Print[\*""\!\(z\_n\) = "", Z[n]]; \)

The ratio of consecutive terms is:

\(Print[\*""     \!\(z\_\(n + 1\)\) = "", Z[n + 1]]\);
\(Print[\*""      \!\(z\_n\) = "", Z[n]]\);
r  =  Z[n + 1]/Z[n];
\(Print[\*""    \!\(z\_\(n + 1\)\/z\_n\) = "", r]\);
r = FullSimplify[r];
\(Print[\*""    \!\(z\_\(n + 1\)\/z\_n\) = "", r]\);
L  =  Limit[ r ,  n -> ∞];
\(Print[\*""(lim_(n -> ∞)) ((Subscript[z, n + 1])/(Subscript[z, n])) = L = "", L]\);

L < 1, thus the series converges.   To what ?

\(sum\ \  = \ \ 
      ∑\+\(n = 0\)\%∞\((1 - I)\)\^n\/\(n!\); 

    Print[\*""        (Subscript[z, n]) = "", Z[n]]; 

    Print[\*""(∑_(n = 0)^∞)((((1 - I))^n)/(n!)) = "", sum]; \)

Example 4.10,  Page 106.  Show that  (∑_( n = 0)^∞)(((((z  -  i))^n))/((2^n))) converges 
for all  z  in the disk  |z - i| < 2.

\(Clear[L, n, sum, r, z, Z]; 

    Z[n_]\  = \ \((z\  - \ I)\)\^n\/2\^n; 

    Print[\*""\!\(z\_n\) = "", Z[n]]; \)

The ratio of consecutive terms is:

\(Print[\*""    \!\(z\_\(n + 1\)\) = "", Z[n + 1]]\);
\(Print[\*""      \!\(z\_n\) = "", Z[n]]\);
r  =  Z[n + 1]/Z[n];
\(Print[\*""    \!\(z\_\(n + 1\)\/z\_n\) = "", r]\);
r = FullSimplify[r];
\(Print[\*""    \!\(z\_\(n + 1\)\/z\_n\) = "", r]\);
L  =  Limit[ r ,  n -> ∞];
\(Print[\*""(lim_(n -> ∞)) ((Subscript[z, n + 1])/(Subscript[z, n])) = L = "", L]\);

Solve  L  =  |((z  -  i)/2)|  <  1  and obtain the disk  |z - i| < 2.

So L < 1, thus the series converges.   To what ?

sum = (∑_( n = 0)^∞)((((z -I))^n)/((2^n)));
\(Print[\*""        \!\(z\_n\) = "", Z[n]]\);
\(Print[\*""(∑_(n = 0)^∞)((((z - I))^n)/(2^n)) = "", Simplify[sum]]\);

Example 4.11,  Page 106.  The limit supremum of the sequence {(Subscript[t, n])} = {4.1, 5.1, 4.01, 5.01, 4.001, 5.001, ⋯}
is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 5.

\(Clear[limsup, n, set]; 

    limsup = Max[Limit[4 + 1\/10\^n, n -> ∞], 
        Limit[5 + 1\/10\^n, n -> ∞]]; 

    Print[\*""{(Subscript[t, n])} = "", 
      Flatten[Table[{4 + 1.\/10\^n, 5 + 1.\/10\^n}, {n, 1, 5}]]]; 

    Print[\*""\!\(\(lim\\ sup\)\+\(n -> ∞\)\) (Subscript[t, n]) = "", limsup]; \)

Example 4.12,  Page 106.  The limit supremum of the sequence {(Subscript[t, n])} = {1, 2, 3, 1, 2, 3, 1, 2, 3, ⋯}
is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 3.

\(Print[\*""{(Subscript[t, n])}={1,2,3,1,2,3,1,2,3,⋯}""]; 
    
Print[\*""\!\(\(lim\\ sup\)\+\(n -> ∞\)\) (Subscript[t, n]) = "", Max[1, 2, 3]]; \)

Example 4.13,  Page 107.  The limit supremum of the Fibonacci sequence {(Subscript[t, n])} = {1, 1, 2, 3, 5, 8, 13, 21, 34  ⋯}
is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = ∞.

\(Clear[n, t]; 

    t[n_] = \(-\@5\)\/5\ 2\^\(\(-1\) - n\)\ \((1 - \@5)\)\^\(n + 1\)\  + 
        \((2\/\(1 + \@5\))\)\^\(\(-1\) - n\)\/\@5; 

    Print[\*""{(Subscript[t, n])} = "", Table[Simplify[t[n]], {n, 0, 12}]]; 

    Print[\*""\!\(\(lim\\ sup\)\+\(n -> ∞\)\) (Subscript[t, n]) = "", Limit[t[n], n -> ∞]]; \)

Example 4.14,  Page 107.  The sequence 
 {(Subscript[t, n])} = {2, (3/2), (4/3), (5/4), (6/5), (7/6), (8/7), (9/8), (10/9), 
          (11/10), (12/11), (13/12), ⋯}
has  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 1.

\(Clear[n, t]; 
t[n_] = 1 + 1\/n; 

    Print[\*""{(Subscript[t, n])} = "", Table[Simplify[t[n]], {n, 1, 12}]]; 

    Print[\*""\!\(\(lim\\ sup\)\+\(n -> ∞\)\) (Subscript[t, n]) = "", Limit[t[n], n -> ∞]]; \)

GoTo Chapter 4

Section 4.2  Power Series Functions

The function  f(z)  = (∑_( n = 0)^∞)(Subscript[a, n])(z^n)  is called a power series.

Example 4.15,  Page 111.  Find the radius of convergence of  
f(z) = (∑_( n = 0)^∞)(((((n + 2)/(3 n + 1))))^n)(((z-4))^n).

\(Clear[c, L, n, Pn, R]; 
c[n_]\  = \ \((\(n + 2\)\/\(3  n + 1\))\)\^n; 
    
Pn\  = \ c[n]\^\(1/n\); 
Print[\*""    \!\(c\_n\) = "", c[n]]; 

    Print[\*""|\!\(c\_n\)\!\(\( | \^\(1/n\)\)\) = "", Pn]; 

    Pn\  = \ PowerExpand[Pn]; 

    Print[\*""|\!\(c\_n\)\!\(\( | \^\(1/n\)\)\) = "", Pn]; \)

Use the root test and find the limit and then the radius of convergence R.

\(L\  = \ Limit[\ Pn, \ n -> ∞]; 
R\  = \(\ 1\)\/L; 

    Print[\*""  L = (lim_(n -> ∞)) |(Subscript[c, n])\!\(\( | \^\(1/n\)\)\) = "", L]; 
Print["The limit  is  L = ", L]; 

    Print[\*""The radius is  R = (1/L) = "", R]; \)

Example 4.16,  Page 111.  The series 
f(z)=(∑_( n = 0)^∞)(Subscript[c, n])(z^n)=1+4 z+(5^2)(z^2)+(4^3) (z^3)+(5^4)(z^4)+(4^5)(z^5)+⋯
has radius of convergence  by the Cauchy-Hadamard formula because
(Underscript[(lim sup), n -> ∞]) 
          (((Subscript[( ∣ c), n]) ∣ )^(1/n)) = 5.

\(Clear[n, ρ, R]; 

    ρ = Max[Limit[\@\(4\^n\)\%n, n -> ∞], 
        Limit[\@\(5\^n\)\%n, n -> ∞]]; 

    Print[\*""\!\(\(lim\\ sup\)\+\(n -> ∞\)\) (( ∣ (Subscript[c, n]) ∣ )^(1/n))  = "", ρ]; 

    Print[\*""R = (1/ρ) = "", 1\/ρ]; \)

Example 4.17,  Page 111.  Find the radius of convergence of  f(z) = (∑_( n = 0)^∞)(((z^n))/(n!)).

\(Clear[c, L, n, Qn, R, z]; 
c[n_]\  = \ 1\/\(n!\); 

    Qn\  = \ c[n + 1]\/c[n]; 
Print[\*""    \!\(c\_n\) = "", c[n]]; 

    Print[\*""|\!\(c\_\(n + 1\)\/c\_n\)| = "", Qn]; 

    Qn\  = \ FullSimplify[Qn]; 

    Print[\*""|\!\(c\_\(n + 1\)\/c\_n\)| = "", Qn]; \)

Use the ratio test and find the limit and then the radius of convergence R.

\(L\  = \ Limit[\ Qn, \ n -> ∞]; 

    R\  = \ Limit[\(\ 1\)\/Qn, \ n -> ∞]; 

    Print[\*""  L = (lim_(n -> ∞)) |((Subscript[c, n + 1])/(Subscript[c, n]))| = "", L]; 
Print["The limit  is  L = ", L]; 

    Print["The radius is  R = ", R]; \)

This is a well known function.

sum = (∑_( n = 0)^∞)(((z^n))/(n!));
\(Print[\*""f(z) = (∑_(n = 0)^∞)((z^n)/(n!)) = "", sum]\);

Example 4.18,  Page 114.  Show that    f(z) = (∑_( n = 0)^∞)(n + 1)(z^n)   =  (1/((((1  -  z))^2))).

Solution:  Use the fact that   s(z)  = (∑_( n = 0)^∞)  (z^n)  =   (1/(1  -  z))   and   s ' (z)  =  f(z).

Clear[n, s, sum, z]sum   =   (∑_(n = 0)^∞) 1*(z^n);
s[z_]  =  sum;
\(Print[\*"" s[z] = (∑_(n = 0)^∞)(z^n) = "", s[z]]\);
\(Print[\*""s'[z] = (∑_(n = 0)^∞)(n+1)(z^n) = "", \(s'\)[z]]\);

Or do it directly.

sum = (∑_( n = 0)^∞)(n + 1)*(z^n) ;
\(Print[\*""f[z] = (∑_(n = 0)^∞)(n+1)(z^n) = "", sum]\);

Example 4.19,  Page 114.  Termwise differentiate the Bessel function  (Subscript[J, 0])(z)  
and  get  - (Subscript[J, 1])(z).

\(J0[z_]\  = \ Series[BesselJ[0, z], \ {z, 0, 10}]; 

    J1[z_]\  = \ Series[BesselJ[1, z], \ {z, 0, 9}]; 

    Print[\*"" \!\(J\_0\)[z] = "", J0[z]]\)\(Print[\*""\!\(J\_0\)'[z] = "", \(J0'\)[z]]\)\(Print[\*"" \!\(J\_1\)[z] = "", J1[z]]\)

GoTo Chapter 4

Section 4.3  Julia and Mandelbrot Sets

Julia Set

     The Julia set associated to a complex number c is found by iterating the 
map  z = (z^2) + c.  The set of points that do not escape to infinity is the Julia set.

Here is the subroutine from the Mathematica  Journal.

\(\(juliaC\  = 

      Compile[{x, y, lim, cx, cy}, 
        Module[{z, ct = 0}, z = x + I\ y; 
          While[Abs[z] < 2.0\  && \ ct ≤ lim, 
            z = z\^2 + \((cx + I\ cy)\); \(++ct\)]; ct]]; \)\)

Here is the Julia set generated by the value  c = -1.25.  This is a fuzzy picture of 
Color Plate 4.  Increase the number of PlotPoints to get a sharper picture.  Also, 
it will increase the computing time. It takes quite a bit of time with  PlotPoints->100
Things start to look better with  PlotPoints->300

DensityPlot[-juliaC[x, y, 50, -1.25, 0], {x, -1.5, 1.5}, 
       {y, -1.5, 1.5}, PlotPoints -> 100, Mesh -> False, 
      ColorFunction -> Hue];

Mandelbrot Set

     The Mandelbrot set is the set of points c that do not escape to infinity under 
iteration of the map  c = (c^2) + c.  Points in the Mandelbrot set have connected 
Julia sets.  This modification of the  juliaC  code can be used to plot the Mandelbrot set.
Here is the subroutine from the Mathematica  Journal.

mandelbrotC = 
      Compile[{x, y, lim}, 
        Module[{z, ct = 0}, z = x + I y; 
          While[Abs[z] < 2.0  &&  ct ≤ lim, 
            z = (z^2) + (x + I y); ++ct]; ct]];

This is a fuzzy picture of Color Plate 6. Increase the number of PlotPoints to get a 
sharper picture.  Also, it will increase the computing time.  It takes quite a bit of time 
with  PlotPoints->100.   Things start to look better with  PlotPoints->300

DensityPlot[-mandelbrotC[x, y, 50], {x, -2.0, 0.55}, 
       {y, -1.15, 1.15}, PlotPoints -> 100,  Mesh -> False, 
      Frame -> False, ColorFunction -> Hue];

GoTo Chapter 4