Solutions to Exam III (April 4, 2005)
Math 12 Spring 2005
Tufts University
(1a)
. One computes:
1
x

2
=

1
2

x
=

1
2
·
1
1

x/
2
=

1
2
∞
X
n
=0
x
2
n
=
∞
X
n
=0

1
2
n
+1
x
n
.
This geometric series converges to the indicated value provided that
x
2
<
1; the radius of
convergence is
R
= 2.
(1b)
. Since
d
dx
1
1 +
x
=

1
(1 +
x
)
2
, and since
1
1 +
x
=
∞
X
n
=0
(

1)
n
x
n
whenever

x

<
1, one
finds that
1
(1 +
x
)
2
= (

1)
d
dx
1
1 +
x
=
d
dx
∞
X
n
=0
(

1)
n
+1
x
n
!
=
∞
X
n
=1
(

1)
n
+1
nx
n

1
whenever

x

<
1.
(2a)
. Recognize that
∞
X
n
=1
nx
n

1
=
d
dx
∞
X
n
=0
x
n
!
=
d
dx
1
1

x
=
1
(1

x
)
2
; the radius of
convergence is
R
= 1 since the equality
1
1

x
=
∞
X
n
=0
x
n
holds whenever

x

<
1, and since the
radius of convergence of a power series is unchanged upon differentiation,
(2b)
. We have
∞
X
n
=1
n
3
n
=
1
3
∞
X
n
=1
n
1
3
n

1
; since

1
/
3

< R
= 1, the series
∞
X
n
=1
n
1
3
n

1
is
the value at
x
= 1
/
3 of the series found in (2a) to represent
1
(1

x
)
2
; thus
∞
X
n
=1
n
3
n
=
1
3
∞
X
n
=1
n
1
3
n

1
=
1
3
1
(1

1
/
3)
2
=
1
3
·
1
4
/
9
=
1
3
·
9
4
=
3
4
.
(3)
. Since
d
dx
ln(1

x
) =

1
1

x
, one knows that
ln(1

x
) =
Z

1
1

x
dx
=

Z
∞
X
n
=0
x
n
!
dx
=
C

∞
X
n
=0
x
n
+1
n
+ 1
=
C

∞
X
n
=1
x
n
n
for some constant
C
. Since 0 = ln(1) = ln(1

0), and since the value of the RightHandSide
at
x
= 0 is
C
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 Spring '08
 GARANT
 Calculus, Geometric Series, Power Series, Mathematical Series, lim, Tufts University

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