Calculus
:
Math 21C, Fall 2010
Summary of basic results for Midterm 2.
1. Power series
Convergence of power series.
A power series centered at
x
=
a
is a series
of the form
∞
∑
n
=0
c
n
(
x
−
a
)
n
.
(1)
Every power series has a radius of convergence 0
≤
R
≤ ∞
such that the
series converges absolutely if

x
−
a

< R
and diverges if

x
−
a

> R
. The
series may or may not converge at the endpoints of the interval of convergence
where

x
−
a

=
R
.
Ratio test.
The power series (1) converges absolutely for all
x
such that
lim
n
→∞
±
±
±
±
c
n
+1
(
x
−
a
)
n
+1
c
n
(
x
−
a
)
n
±
±
±
±
<
1
if the limit exists. The radius of convergence
R
is given by
1
R
= lim
n
→∞
±
±
±
±
c
n
+1
c
n
±
±
±
±
if the limit exists (with the natural conventions for
R
= 0 and
R
=
∞
).
Diﬀerentiation and integration of power series.
If a function
f
(
x
) =
∞
∑
n
=0
c
n
(
x
−
a
)
n
is the sum of a power series with nonzero radius of convergence
R >
0, then
f
(
x
) has derivatives of all orders inside the interval of convergence

x
−
a

< R
.
Its derivative
f
′
is given by diﬀerentiating the power series of
f
termbyterm,
f
′
(
x
) =
∞
∑
n
=0
nc
n
(
x
−
a
)
n
−
1
,
and this power series has the same radius of convergence as the power series
for
f
. Power series for higherorder derivatives of
f
are obtained by the
1
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View Full Documentrepeated application of this result. Similarly, the integral of
f
is given by
termbyterm integration of the power series of
f
∫
f
(
x
)
dx
=
∞
∑
n
=0
c
n
n
+ 1
(
x
−
a
)
n
+1
+
C,
and this power series has the same radius of convergence as the power series
for
f
.
Multiplication of power series.
If
f
(
x
),
g
(
x
) have power series expansions
f
(
x
) =
∞
∑
n
=0
a
n
(
x
−
a
)
n
,
g
(
x
) =
∞
∑
n
=0
b
n
(
x
−
a
)
n
,
which both converge in

x
−
a

< R
, then
h
(
x
) =
f
(
x
)
g
(
x
) has the power
series expansion
h
(
x
) =
∞
∑
n
=0
c
n
(
x
−
a
)
n
,
c
n
=
n
∑
k
=0
a
k
b
n
−
k
,
which converges in

x
−
a

< R
. Writing out the ﬁrst few terms explicitly, we
have
[
a
0
+
a
1
(
x
−
a
) +
a
2
(
x
−
a
)
2
+
a
3
(
x
−
a
)
3
+
. . .
]
·
[
b
0
+
b
1
(
x
−
a
) +
b
2
(
x
−
a
)
2
+
b
3
(
x
−
a
)
3
+
. . .
]
=
a
0
b
0
+ (
a
0
b
1
+
a
1
b
0
) (
x
−
a
) + (
a
0
b
2
+
a
1
b
1
+
a
2
b
0
) (
x
−
a
)
2
+ (
a
0
b
3
+
a
1
b
2
+
a
2
b
1
+
a
3
b
0
) (
x
−
a
)
3
+
. . . .
Taylor series.
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