21 Power Series - Handout

Power Series

Math 55 - Elementary Anlaysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Power Series

Recall

R Alternating Series Test

If a given alternating series

n=1

(1)nbn orn=1

(1)n1bnsatisfies

(i) limn bn = 0;

(ii) bn+1 bn for all n, i.e., the sequence {bn} is decreasingthen it is convergent.


Recall

R Ratio Test

Suppose

n=1

an is a series with nonzero terms.

(i) If limn

an+1an = L < 1, then the series is absolutely

convergent, and is therefore convergent.

(ii) If limn

an+1an = L > 1 or limn

an+1an =, then the series

is divergent.

R Root Test

Given a seriesn=1

an,

(i) If limn |an|

1n = L < 1, then the series is absolutely

convergent, and is therefore convergent.(ii) If lim

n |an|1n = L > 1 or lim

n |an|1n =, then the series is

divergent.


Power Series

Definition

A series of the form

n=0

cn(x a)n = c0 + c1(x a) + c2(x a)2 +

where x is a variable, a is a fixed real number and the cns areconstants, is called a power series centered at a or powerseries about a

Remark. We adopt the convention that (x a)0 = 1 evenwhen x = a.


Power Series

Examples

1

n=0

xn = 1 +x+x2 +x3 + . . . is a power series centered at 0.

2

n=0

(x 1)nn!

= 1 + (x 1) + (x 1)2

2!+

(x 1)33!

+ . . . is a

power series centered at 1.

When does a power series converge?

R When x = a, all of the terms of the power series are 0 andso the power series converges for x = a.


When does a Power Series Converge?

Theorem

Given a power seriesn=0

cn(x a)n, there are only threepossibilities:

(i) The series converges only at x = a.

(ii) The series converges for all real values of x.

(iii) There is a positive number R such that the series convergesif |x a| < R and diverges if |x a| > R.

Remarks.

R Such number R is called the radius of convergence ofthe power series.

R The set of values of x for which the series converges iscalled the interval of convergence of the power series.


Radius and Interval of Convergence

Example

Determine the radius of convergence and interval of convergence

of the power series

n=0

n! (x 1)n.

Solution. Let an = n! (x 1)n. By Ratio Test,

limn

an+1an = limn

(n + 1)! (x 1)n+1n! (x 1)n

= limn(n + 1)|x 1|

= if x 6= 1

Hence, the series diverges when x 6= 1 so the radius ofconvergence is 0 and the interval of convergence is {1}.



Example


of the power series

n=0

xn

n!.

Solution. Let an =xn

n!and by Ratio Test,

limn

an+1an = limn

xn+1(n + 1)! n!xn

= limn

|x|n + 1

= 0 for all x

Hence, the series converges for all x so the radius of convergenceis and the interval of convergence is (,) = R.



Example


of the power series

n=0

xn

n.

Solution. Let an =xn

n .

limn

an+1an = limn

xn+1n + 1 nxn = limn

(n

n + 1

)|x| = |x|

By Ratio Test, the series converges when |x| < 1, i.e., on theinterval (1, 1) so R = 1. How about at the endpoints?R If x = 1, the series becomes

n=1

(1)nn

which converges.

R If x = 1, the series becomesn=1

1

nwhich diverges.

Therefore, the interval of convergence is [1, 1).Math 55 Power Series

Power Series Representations

The sum of a power series

n=0

cn(x a)n is a function f(x) suchthat

f(x) = c0 + c1(x a) + c2(x a)2 + c3(x a)3 +

whose domain is the interval of convergence of the power series.

That means we can represent certain types of functions aspower series. This is useful for integrating functions that donthave elementary antiderivatives, for solving differentialequations, and for approximating functions by polynomials.



We begin with a familiar series:

1

1 x = 1 + x + x2 + . . . =

n=0

xn |x| < 1

We now regard the series

n=0

xn as a power series representation

for the function f(x) =1

1 x .

Using this representation, we can obtain power seriesrepresentation for certain type of functions.



Example

Express1

1 + x2as a power series and indicate the interval of

convergence.

Solution. Note that1

1 + x2=

1

1 (x2) is the sum of ageometric series with r = x2. Hence,

1

1 (x2) =n=0

(x2)n

1

1 + x2=

n=0

(1)nx2n

The geometric series converges when |r| = | x2| = x2 < 1, andhence, the interval of convergence is (1, 1).



Example

Find a power series representation for1

2 x .

Solution. We can manipulate the given expression so that itlooks like the sum of a geometric series:

1

2 x =1

2(

1 x2

)=

1

2

n=0

(x2

)n x2

< 1=

n=0

xn

2n+1|x| < 2



Example

Obtain a power series representation forx3

x + 3.

Solution. Again, manipulating the given expression,

x3

x + 3=

x3

3

1[1

(x

3

)]=

x3

3

n=0

(x

3

)n x3

< 1=

n=0

(1)nxn+3

3n+1|x| < 3


Differentiation of Power Series

Theorem

Ifn=0

cn(x a)n has radius of convergence R > 0, then thefunction f defined by

f(x) =

n=0

cn(x a)n = c0 + c1(x a) + c2(x a)2 +

is differentiable (hence continuous) on (aR, a + R) andf (x) =

n=1

ncn(x a)n1 = c1 + 2c2(x a) + 3c3(x a)2 +

The radius of convergence of the power series for f (x) is R.

In short,

d

dx

[ n=0

cn(x a)n]

=

n=0

d

dx[cn(x a)n]



Example

Find a power series representation for1

(1 x)2 .

Solution. Since

1

1 x =n=0

xn |x| < 1,

differentiating each side gives

d

dx

[1

1 x]

=n=0

d

dx(xn) |x| < 1

1

(1 x)2 =n=1

nxn1 |x| < 1



Example

Use the previous example to find the sum

n=1

n

2n.

Solution. Note that1

(1 x)2 =n=1

nxn1 |x| < 1.

If x = 12 , the series converges andn=1

nxn1 =n=1

n

2n1= 2

n=1

n

2n.

Hence,2

n=1

n

2n=

1(1 12

)22

n=1

n

2n= 4

n=1

n

2n= 2


Integration of Power Series

Theorem

Ifn=0

cn(x a)n has radius of convergence R > 0, then thefunction f defined by

f(x) =

n=0

cn(x a)n = c0 + c1(x a) + c2(x a)2 +

is differentiable (hence continuous) on (aR, a + R) andf(x) dx =

n=0

cn(x a)n+1n + 1

+C = c0(xa)+c1 (x a)2

2+ +C

The radius of convergence of the latter is R.

In short, ( n=0

cn(x a)n)dx =

n=0

cn(x a)n dx



Example

Obtain a power series representation for ln(1 + x).

Solution. Notice that

1

1 + xdx = ln(1 + x) + c. Hence,

ln(1 + x) =

1

1 (x)dx =n=0

(1)n(x)ndx | x| < 1

=

n=0

(1)n(x)n+1n + 1

+ C |x| < 1

To find C, let x = 0. Then C = ln 1 = 0. Hence,

ln(1 + x) =

n=0

(1)n(x)n+1n + 1

|x| < 1



Example

Use the previous example to show that

n=1

(1)n1n

= ln 2.

Solution.Note that ln(1 + x) =n=0

(1)n(x)n+1n + 1

|x| < 1.

If x = 1, the series becomes

n=0

(1)nn + 1

, which converges (by the

Alternating Series Test). Hence,

ln(1 + 1) =

n=1

(1)nn + 1

ln 2 = 1 12

+1

3 1

4+

=

n=1

(1)n1n


Exercises

1 Find the radius of convergence and interval of convergenceof the following power series

a.

n=0

(x 2)nn2 + 1

b.

n=1

(x 1)nn 3n

2 Find a power series representation for

a.1

2 + 3xb. tan1 x

3 Find a power series representation for1

(1 x2)2 and use it

to find the sum

n=1

n

9n.


References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/


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21 Power Series - Handout