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95
Solution.
Assume that
x
−
a

f
(
x
)
.I
f
(
x
−
a
)
2

f
(
x
)
, then
f
(
x
)
=
(
x
−
a
)
2
h
(
x
)
, and so
f
0
(
x
)
=
2
(
x
−
a
)
h
(
x
)
+
(
x
−
a
)
2
h
0
(
x
)
. Hence,
x
−
a

f
0
. (The assumption
(
x
−
a
)

f
(
x
)
is not needed in this direction.)
Conversely, assume that
x
−
a

f
(
x
)
and
x
−
a

f
0
; hence
f
(
x
)
=
(
x
−
a
)
g
(
x
)
and
f
0
(
x
)
=
(
x
−
a
)
k
(
x
)
.Now
(
x
−
a
)
k
(
x
)
=
f
0
(
x
)
=
((
x
−
a
)
g
)
0
=
g
+
(
x
−
a
)
g
0
,
and so
x
−
a

g
; that is,
g
(
x
)
=
(
x
−
a
)`(
x
)
. Hence,
f
(
x
)
=
(
x
−
a
)
g
(
x
)
=
(
x
−
a
)
2
`(
x
)
.
3.38
(i)
If
f
(
x
)
=
ax
2
p
+
bx
p
+
c
∈
F
p
[
x
]
, prove that
f
0
(
x
)
=
0.
Solution.
If
f
(
x
)
=
ax
2
p
+
bx
p
+
c
∈
F
p
[
x
]
, then
f
(
x
)
=
2
pax
2
p
−
1
+
pbx
p
−
1
=
0
.
(ii)
State and prove a necessary and suf
f
cient condition that a polyno
mial
f
(
x
)
∈
F
p
[
x
]
have
f
0
(
x
)
=
0.
Solution.
The condition is that there should be a polynomial
g
(
x
)
=
∑
b
n
x
n
with
f
(
x
)
=
g
(
x
p
)
; that is,
f
(
x
)
=
∑
a
n
x
np
and
b
p
n
=
a
n
for all
n
. As in part (i), if
f
(
x
)
has this form, then
f
0
=
0. Conversely, if
f
(
x
)
=
∑
a
n
x
n
has
a
n
±=
0 for some
n
not divisible by
p
, then
f
0
has a nonzero coef
f
cient
na
n
and so
f
0
±=
0.
3.39
If
R
is a commutative ring, de
f
ne
R
[[
x
]]
to be the set of all formal power
series over
R
.
(i)
Show that the formulas de
f
ning addition and multiplication on
R
[
x
]
make sense for
R
[[
x
]]
, and prove that
R
[[
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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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