Solution 9 (b).
Answer. Series
I.
for
,
Answer. Series
II.
for
.
Solution. Second
series.
Start by writing the function in the form
.
Series
II. Differentiate the geometric
series
which
is valid for
to
establish the identity
which
is valid for
.
Make the substitution
and
get:
Therefore,
is
valid for
.
We are done.
Aside. We can let Mathematica double check our work.
The second series.
We are really done.
Aside. As an
optional experiment, we can plot some of the partial sums of the
series
.
The mapping
is
not one-to-one in the punctured
disk
,
and so we will choose the small portion
.
![[Graphics:../Images/LaurentSeriesModHome_gr_378.gif]](../Images/LaurentSeriesModHome_gr_378.gif)
![[Graphics:../Images/LaurentSeriesModHome_gr_380.gif]](../Images/LaurentSeriesModHome_gr_380.gif)
The
image of
under
the mapping
for
.
![[Graphics:../Images/LaurentSeriesModHome_gr_385.gif]](../Images/LaurentSeriesModHome_gr_385.gif)
The
image of
under
the mapping
.
Recall Exercise
8. Differentiate the geometric
series
which
is valid for
to
establish the identity
which
is valid for
.
The details are similar to those given in Exercise 8.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell