COMPLEX ANALYSIS: Maple Worksheets,
2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu
Complimentary software to accompany the textbook:
COMPLEX ANALYSIS: for Mathematics &
Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and 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/
CHAPTER 4 SEQUENCES, JULIA and
MANDELBROT SETS, and Power Series
Section 4.3 Geometric Series and
Convergence Theorems
We begin this section by studying
series of the form
, which is called a
geometric series
, one of the most important series in
mathematics.
Theorem 4.11, (Geometric
series) If
, the series
converges to
.
That is, if
, then
=
=
.
If
, then the series
diverges.
Corollary 4.2
If
, the series
converges to
.
That is, if
, then
=
=
.
If
, then the series
diverges.
Corollary 4.3
If
, then for all
.
Example 4.13, Page 146.
Show that
,
> n:='n':
Zn:='Zn': z:='z':
Z := n -> (1-I)^n/2^n:
z[n] = Z(n);
This is a geometric series with
ratio:
> R :=
Z(n+1)/Z(n):
`R = `, z[n+1]/z[n] = R;
R := simplify(R):
`R = `, z[n+1]/z[n] = R;
`|R| = `, abs(z[n+1]/z[n]) = abs(R);
`|R| = `, abs(z[n+1]/z[n]) =
evalf(abs(R));
![]()
![]()
![]()
Since
, the sum is given by the
calculation:
> S:='S':
S := 1/(1-R):
`R = `, z[n+1]/z[n] = R;
`S = 1/(1 - R) ` = S;
![]()
Or we can use Maple to sum the
series directly.
> n:='n': S:='S':
Zn:='Zn': z:='z':
Zn := (1 - I)^n/2^n:
z[n] = Zn;
S := sum(Zn,n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
Error, (in convert/hypergeom) cannot evaluate boolean: 1/2*2^(1/2) < 1
![Sum(z[n],n = 1 .. infinity) = Sum((1-I)^n/(2^n),n =...](images/C04-334.gif)
![Sum(z[n],n = 1 .. infinity) = S](images/C04-335.gif)
Example 4.14, Page 146.
Evaluate
.
> n:='n': Z:='Z':
Zn:='Zn': z:='z':
Z := n -> I^n/2^n:
z[n] = Z(n);
This is a geometric series with
ratio:
> R :=
Z(n+1)/Z(n):
`R = `, z[n+1]/z[n] = R;
R := simplify(R):
`R = `, z[n+1]/z[n] = R;
`|R| = `, abs(z[n+1]/z[n]) = abs(R);
`|R| = `, abs(z[n+1]/z[n]) =
evalf(abs(R));
![]()
![]()
![]()
Since |R| < 1, the sum is given
by the calculation:
> S:='S':
S := 1/(1-R) - 1 - R - R^2:
`R = `, z[n+1]/z[n] = R;
`S = 1/(1 - R) - 1 - R - R^2 ` = S;
![]()
Or we can use Maple to sum the
series directly.
> n:='n': S:='S':
Zn:='Zn': z:='z':
Zn := I^n/2^n:
z[n] = Zn;
S := sum(Zn, n=3..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
![]()
![Sum(z[n],n = 1 .. infinity) = Sum(I^n/(2^n),n = 1 ....](images/C04-346.gif)
![Sum(z[n],n = 1 .. infinity) = 1/20-1/10*I](images/C04-347.gif)
The geometric series is used in the proof of the following theorem, known as the ratio test . It is one of the most commonly used tests for determining the convergence or divergence of series. The proof is similar to the one used for real series, and is left for the reader to establish.
Theorem 4.12 (d'Alembert's ratio test)
If
is a complex series with the
property that
,
then the series is absolutely
convergent if
, and divergent if
.
Example 4.15, Page 147.
Show that
converges.
> L:='L': n:='n':
R:='R': Z:='Z': Zn:='Zn': z:='z':
Z := n -> (1 - I)^n / n!:
z[n] = Z(n);
R := Z(n+1)/Z(n):
z[n+1]/z[n] = R;
R := simplify(R):
z[n+1]/z[n] = R;
L := limit(Z(n+1)/Z(n) , n=infinity):
limit(z[n+1]/z[n], n=infinity) = L;
![z[n+1]/z[n] = (1-I)^(n+1)/(n+1)!/((1-I)^n)*n!](images/C04-354.gif)
![]()
![]()
Since
, the series converges. Let us use
Maple to find the sum.
> n:='n': S:='S':
Zn:='Zn': z:='z':
Zn := (1-I)^n/n!:
z[n] = Zn;
S := sum(Zn, n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
Sum(z[n], n=1..infinity) = evalc(S);
![]()
![Sum(z[n],n = 1 .. infinity) = Sum((1-I)^n/n!,n = 1 ...](images/C04-360.gif)
![Sum(z[n],n = 1 .. infinity) = exp(1-I)](images/C04-361.gif)
![Sum(z[n],n = 1 .. infinity) = exp(1)*cos(1)-I*exp(1...](images/C04-362.gif)
Example 4.16, Page 147.
Show that
converges for all
in the disk
.
> L:='L': n:='n':
R:='R': Z:='Z': Zn:='Zn': z:='z':
Z := n -> (z-I)^n/2^n:
z[n] = Z(n);
R := Z(n+1)/Z(n):
z[n+1]/z[n] = R;
R := simplify(R):
z[n+1]/z[n] = R;
L := limit(Z(n+1)/Z(n) , n=infinity):
`L = `, limit(z[n+1]/z[n], n=infinity) = L;
`|L| ` = abs(L);
![z[n+1]/z[n] = (z-I)^(n+1)/(2^(n+1))/((z-I)^n)*2^n](images/C04-367.gif)
![]()
![]()
![]()
Hence
, now solve
and obtain
.
Therefore, the series converges the disk
.
Let us use Maple to find the
sum.
> n:='n': S:='S':
Zn:='Zn': z:='z':
Zn := (z-I)^n/2^n:
z[n] = Zn;
S := sum(Zn, n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
![]()
![Sum(z[n],n = 1 .. infinity) = Sum((z-I)^n/(2^n),n =...](images/C04-377.gif)
![Sum(z[n],n = 1 .. infinity) = 1/(1+1/2*I-1/2*z)](images/C04-378.gif)
End of Section 4.3.