Section 11.9 Lecture Slides

1. Chapter 11: Sequences, Series, and Power Series
Section 11.9: Representation of Functions as Power Series
Alea Wittig
SUNY Albany

2. Outline
Representations of Functions as Power Series
Differentiation and Integration of Power Series

3. Representations of Functions as Power Series

4. Representations of Functions as Power Series
Recall
f (x) =
1
1 − x
=
∞
X
n=0
xn
for |x| < 1 (1)
▶ We say that
∞
X
n=0
xn
, is the power series representation of
f (x) =
1
1 − x
on the interval (−1, 1).
▶ We can use this power series representation to find the power
series representation of other functions as well.

5. Example 1
1/2
Use (1) to express the function
1
1 + x2
as the sum of a power series and find the interval of convergence.

6. Example 1
2/2
▶ We notice that 1
1+x2 is just the function f (x) = 1
1−x evaluated
at −x2:
f (−x2
) =
1
1 − (−x2)
=
1
1 + x2
▶ Since
1
1 − x
=
∞
X
n=0
xn
for |x| < 1, by plugging in −x2 to both
sides, we have that
1
1 + x2
=
∞
X
n=0
(−x2
)n
for | − x2
| < 1
=
∞
X
n=0
(−1)n
x2n
for |x| < 1
since (−x2
)n
= (−1)n
(x2
)n
= (−1)n
x2n
and
| − x2
| < 1 ⇐⇒ |x2
| < 1 ⇐⇒ |x| < 1

7. Example 2
1/3
Find a power series representation for
1
x + 2

8. Example 2
2/3
▶ Again we use equation (1) by noticing that 1
x+2 is a
transformation of f (x) = 1
1−x :
1
x + 2
=
1
2 + x
=
1
2(1 + (x
2 ))
=
1
2(1 − (−x
2 ))
ie,
1
x + 2
=
1
2
f
−
x
2
.
▶ So we plug in −x
2 to both sides of (1) and get
1
1 − (−x
2 )
=
∞
X
n=0
−
x
2
n
for

11. −
x
2

14. 1
=
∞
X
n=0
(−1)nxn
2n
for |x| 2

15. Example 2
3/3
▶ Now multiplying both sides of the above by 1
2 we get
1
2(1 − (−x
2 ))
=
1
2
∞
X
n=0
(−1)nxn
2n
for |x| 2
=
∞
X
n=0
(−1)nxn
2n+1
for |x| 2
=⇒
1
x + 2
=
∞
X
n=0
(−1)nxn
2n+1
for |x| 2

16. Example 3
1/2
Find a power series representation of
x3
x + 2

17. Example 3
2/2
▶ Now since we found in example 2 that
1
x + 2
=
∞
X
n=0
(−1)nxn
2n+1
for |x| 2
we have that
x3
x + 2
= x3
∞
X
n=0
(−1)nxn
2n+1
for |x| 2
=
∞
X
n=0
(−1)nx3xn
2n+1
for |x| 2
=
∞
X
n=0
(−1)nxn+3
2n+1
for |x| 2

18. Differentiation and Integration of Power Series

19. Differentiation and Integration of Power Series
▶ Theorem: If
P
cn(x − a)n has a radius of convergence R 0,
then the function f defined by
f (x) =
∞
X
n=0
cn(x−a)n
= c0+c1(x−a)+c2(x−a)2
+c3(x−a)3
+. . .
is differentiable on the interval (a − R, a + R) and
(i) f ′
(x) = c1 + 2c2(x − a) + 3c3(x − a)2
+ . . .
=
∞
X
n=1
ncn(x − a)n−1
(ii)
Z
f (x)dx = C + c0(x − a) + c1
(x − a)2
2
+ c2
(x − a)3
3
+ . . .
= C +
∞
X
n=0
cn
(x − a)n+1
n + 1
The radii of convergence in (i) and (ii) are both R.

20. Remark
▶ Note that the intervals of convergence for the derivative and
integral may be different (even though the radii are the same.)

21. Example 4
1/2
Express
1
(1 − x)2
as a power series by differentiating equation (1).

22. Example 4
2/2
f (x) =
1
1 − x
=
∞
X
n=0
xn
=⇒
f ′
(x) =
1
(1 − x)2
=
∞
X
n=0
nxn−1
▶ The radius of convergence of 1
(1−x)2 =
P∞
n=0 nxn−1 is R = 1
▶ The series is centered at 0 so we just need to test the
endpoints of the interval (−1, 1) to find the interval of
convergence.
▶ x = 1 :
∞
X
n=0
nxn−1
=
∞
X
n=0
n; diverges by test for divergence.
▶ x = −1 :
∞
X
n=0
nxn−1
=
∞
X
n=0
(−1)n
n; diverges by test for
divergence.
▶ So the interval of convergence is I = (−1, 1).

23. Example 5
1/3
Find the power series representation and its radius of convergence
for the function
ln(1 + x)

24. Example 5
2/3
▶ We start by noticing the fact that
d
dx
ln (1 + x) =
1
1 + x
▶ We can find the power series of 1
1+x using (1):
1
1 + x
=
1
1 − (−x)
=
∞
X
n=0
(−x)n
for |x| 1
=
∞
X
n=0
(−1)n
xn
for |x| 1

25. Example 5
3/3
▶ Now we can integrate the power series of 1
1+x according to the
theorem to get back the power series of ln (1 + x).
ln (1 + x) =
Z
d
dx
ln (1 + x)dx [by FTC]
=
Z
1
1 + x
dx
= C +
∞
X
n=0
(−1)n xn+1
n + 1
▶ To find the constant of integration we plug in x = 0 on both
sides and get C = ln 1 = 0 so that
ln (1 + x) =
∞
X
n=0
(−1)n xn+1
n + 1
for |x| 1

26. Example 6
1/2
Find the power series representation and radius of convergence for
f (x) = tan−1
(x)

27. Example 6
2/2
▶ Recall that
d
dx
tan−1
(x) =
1
1 + x2
▶ In example 1 we found that
1
1 + x2
=
∞
X
n=0
(−1)n
x2n
for |x| 1
▶ So taking the integral of both sides we have
tan−1
x =
Z
d
dx
tan−1
(x)dx [by FTC]
=
Z
1
1 + x2
dx
= C +
∞
X
n=0
(−1)n x2n+1
2n + 1
for |x| 1