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96
(iii)
Denote a formal power series
σ
=
(
s
0
,
s
1
,
s
2
,...,
s
n
,...)
by
σ
=
s
0
+
s
1
x
+
s
2
x
2
+···
.
Prove that if
σ
=
1
+
x
+
x
2
+···
, then
σ
=
1
/(
1
−
x
)
is in
R
[[
x
]]
.
Solution.
We have
1
+
x
+
x
2
+···=
1
+
x
(
1
+
x
+
x
2
+···
).
Hence, if
σ
=
1
+
x
+
x
2
+···
, then
σ
=
1
+
x
σ.
Solving for
σ
gives
σ
=
1
/(
1
−
x
)
.
3.40
If
σ
=
(
s
0
,
s
1
,
s
2
,...,
s
n
,...)
is a nonzero formal power series, de
f
ne
ord
(σ)
=
m
, where
m
is the smallest natural number for which
s
m
±=
0.
(i)
Prove that if
R
is a domain, then
R
[[
x
]]
is a domain.
Solution.
If
σ
=
(
s
0
,
s
1
,...)
and
τ
=
(
t
0
,
t
1
,...)
are nonzero
power series, then each has an order (
σ
±=
0 if and only if it has an
order); let ord
(σ)
=
p
and ord
(τ)
=
q
. Write
στ
=
(
c
0
,
c
1
,...).
For any
n
≥
0, we have
c
n
=
∑
i
+
j
=
n
s
i
t
j
. In particular, if
n
<
p
+
q
, then
i
<
p
and
s
i
=
0or
j
<
q
and
t
j
=
0; it follows
that
c
n
=
0 because each summand
s
i
t
j
=
0. The same analysis
shows that
c
p
+
q
=
s
p
t
q
, for all the other terms are 0. Since
R
is a
domain,
s
p
±=
0 and
t
q
±=
0 imply
s
p
t
q
±=
0. Therefore,
ord
(στ)
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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