MAT A26
Lecture 46
1
Differentiating and Integrating Power Se
ries
proposition
1.1.
Let
∞
n
=0
a
n
x
n
be a real power series with radius of con
vergence
R
, defining a function
f
(
x
) for

x

< R
. Then
(a) the power series
∞
n
=0
na
n
x
n

1
obtained by differentiating the original series term by term also has
radius of convergence
R
and sums to
f
(
x
).
(b) the power series
∞
n
=0
a
n
x
n
+1
n
+ 1
obtained by integrating the original series term by term also has radius
of convergence
R
and sums to
x
0
f
(
x
)
dx
.
(This is also true for complex power series, although we have never yet
defined the derivative and integral for complex functions.
This is another
cliffhanger!! It will be covered in the course Complex Variables as well. )
example
1.2.
Integrating the equation
1
1

x
=
∞
n
=0
x
n
(for

x

<
1) yields
x
0
dx
1

x
=
∞
n
=0
x
n
+1
n
+ 1
=
∞
n
=1
x
n
n
Now
x
0
dx
1

x
=

ln(1

x
)

x
0
=

ln(1

x
), so
ln(1

x
) =

∞
n
=1
x
n
n
for

x

<
1
1
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remark
1.3.
It is tempting to “prove” 1

1
2
+
1
3

1
4
+
· · ·
= ln 2 by writing
ln 2 =
lim
x
→ 
1
ln(1

x
) =

lim
x
→ 
1
∞
n
=1
x
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 Fall '08
 HIMONAS
 Calculus, Power Series, Taylor Series, Mathematical analysis

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