Transcript Ch.2 Limits and derivatives

Example
( x  3)n
Ex. For what values of x is the power series 
n
n 1
convergent?
Sol. By ratio test,



an 1
n( x  3)
lim
 lim
| x  3 |
n  a
n 
n 1
n
the power series absolutely converges when | x  3 | 1
and diverges when | x  3 | 1. When | x  3 | 1, we easily
see that it diverges when x=4 and converges when x=1. Thus
the power series converges for 2  x  4.
Example


Ex. Find the domain of the Bessel function defined by

(1) n x 2 n
J ( x)   2 n
(n !) 2
n 0 2
Sol. By ratio test,
an 1
x2
lim
 lim
0
2
n  a
n  4( n  1)
n
the power series absolutely converges for all x. In other words,
The domain of the Bessel function is (, ).
Characteristic of convergence

Theorem For a given power series  cn ( x  a)
n 0
there are only three possibilities:
(i) The series converges only when x=a.
(ii) The series converges for all x.
(iii) There is a positive number R such that the series
converges if | x  a | R and diverges if | x  a | R.
 The number R is called the radius of convergence of the
power series. By convention, in case (i) the radius of
convergence is R=0, and in case (ii) R  .

n
Characteristic of convergence
The interval of convergence of a power series is the
interval that consists of all x for which the series converges.
 To find the interval of convergence, we need to determine
whether the series converges or diverges at endpoints |x-a|=R.
 Ex. Find the radius of convergence and interval of

convergence of the series
(3)n ( x  2)n


n 1
n 0
an 1
 Sol lim |
| 3 | x  2 |
n 
an
3 | x  2 | 1 | x  2 | 1/ 3  R radius of convergence is 1/3.
At two endpoints: diverge at 5/3, converge at 7/3. (5/3,7/3]
Radius and interval of convergence
From the above example, we found that the ratio test or the
root test can be used to determine the radius of convergence.
cn 1
|  , then R  1/  .
 Generally, by ratio test, if lim |
n 
cn




By root test, if lim n | cn |   , then R  1/  .
n 
Ex. Find the interval of convergence of the series

n  1 n2 n
(
) x

n
n 1
Sol. lim n | cn |  e  R  1/ e when |x|=1/e, the
n 
general does not have limit zero, so diverge. (-1/e,1/e)
Representations of functions as power series

1
 We know that the power series  x converges to
1 x
n0
when –1<x<1. In other words, we can represent the function
as a power series

1
  x n  1  x  x 2  x3 
(| x | 1)
1  x n 0
n


Ex. Express 1/(1  x2 ) as the sum of a power series and
find the interval of convergence.
Sol. Replacing x by  x 2 in the last equation, we have

1
1
2 n
2
4
6
8


(

x
)

1

x

x

x

x


2
2
1 x
1  ( x ) n 0
Example


Ex. Find a power series representation for 1/( x  2).
1
1
Sol. 1  1  1
 
x  2 2 1  x / 2 2 1  ( x / 2)
1
1 
x n  (1)n n
  ( )   n1 x
x  2 2 n 0 2
n 0 2
The series converges when |-x/2|<1, that is |x|<2. So the
interval of convergence is (-2,2).
Question: find a power series representation for x3 /( x  2).
n
n


Sol. x3
1
(

1)
(

1)
 x3
 x3  n1 x n   n1 x n3
x2
x2
n 0 2
n 0 2
Differentiation and integration
Theorem If the power series cn ( x  a)n has radius of
convergence R>0, then the sum function


f ( x)   cn ( x  a)n  c0  c1 ( x  a)  c2 ( x  a)2 
n 0
is differentiable on the interval (a R, a  R) and
(i) f ( x)  c1  2c2 ( x  a)    ncn ( x  a) n1
n 1
( x  a) 2

(ii)  f ( x)dx  C  c0 ( x  a)  c1
2
( x  a)n1
 C   cn
n 1
n 0

The above two series have same radius of convergence R.
Example
The above formula are called term-by-term
differentiation and integration.
2
 Ex. Express 1/(1  x) as a power series and find the radius
of convergence.

1
 Sol. Differentiating
  x n gives
1  x n 0

1
n 1
2

nx

1

2
x

3
x


2
(1  x)
n 1
By the theorem, the radius of convergence is same as the
original series, namely, R=1.

Example


Ex. Find a power series representation of f ( x)  arctan x.
Sol.


1
2 n
n 2n
2
4
6

(

x
)

(

1)
x

1

x

x

x



2
1 x
n 0
n 0
2 n 1

1
x
n
 arctan x  
dx

C

(

1)

2
1 x
2n  1
n 0
2 n 1
x
arctan 0  0  C  0  arctan x   (1) n
2n  1
n 0

Example
Ex. Find a power series representation for f ( x)  ln(1  x)
and its radius of convergence.


1
Sol. ln(1  x)   
dx    (1  x  x 2  x 3 
1 x

x2
xn
C  x  C 
2
n 1 n

xn
ln(1  0)  0  C  0  ln(1  x)   .
n 1 n

1
R  1.
 ln 2.

n
n 1 n 2
)dx
Taylor series

Theorem If f has a power series representation (expansion)
at a, that is, if

f ( x)   cn ( x  a)n | x  a | R
n 0
Then its coefficients are given by the formula
f ( n ) (a)
cn 
n!
 This is called the Taylor series of f at a (or about a)
f ( n ) (a)
f (a)
f (a)
n
f ( x)  
( x  a)  f ( a) 
( x  a) 
( x  a) 2 
n!
1!
2!
n 0

Maclaurin series


The Taylor series of f at a=0 is called Maclaurin series

f ( n ) (0) n
f (0)
f (0) 2
f ( x)  
x  f (0) 
x
x 
n!
1!
2!
n 0
Ex. Find the Maclaurin series of the function f ( x)  ex and
its radius of convergence.

n
2
3
x
x
x
x
ex  
 1 


1! 2! 3!
n 0 n !
  x  
Maclaurin series


Ex. Find the Maclaurin series for sinx.
Sol. f ( n ) ( x)  sin( x  n )  f ( n ) (0)  sin n
2
2
So the Maclaurin series is
x3 x5
x  
3! 5!
(1)n x 2 n1

n 0 (2n  1)!

Important Maclaurin series

Important Maclaurin series and their convergence interval

1
  x n  1  x  x 2  x3 
(1,1)
1  x n 0
n
2
3

x
x
x
x
ex    1    
(, )
1! 2! 3!
n 0 n !
(1)n x 2 n1
x3 x5 x 7
sin x  
 x   
3! 5! 7!
n 0 (2n  1)!

(1)n x 2 n
x2 x4 x6
cos x  
 1   
(2n)!
2! 4! 6!
n 0

(, )
(, )
Example


Ex. Find the Maclaurin series of f ( x)  1  x.
Sol.
1
1/ 2

f ( x)  ( x  1)  f ( x)  ( x  1) 1/ 2 ,
2
1
3
3/ 2
f ( x)   ( x  1) , f ( x)  ( x  1) 5/ 2 ,
4
8
f
(n)
( x)  (1)
n 1
35
 (2n  3)
 (2 n 1) / 2
(
x

1)
2n
(2n  3)!!
n 1 (2 n  3)!!
f (0)  (1)
, an  ( 1)
n
2
2n n !
x 
n 1 (2n  3)!! n
1  x  1    (1)
x .
n
2 n2
2 n!
(n)
n 1
Multiplication of power series

Ex. Find the first 3 terms in the Maclaurin series for e x sin x
Sol I. Find f (0), f (0), f (0) and the Maclaurin
series is found.
x
 Sol II. Multiplying the Maclaurin series of e and sinx
collecting terms:
2
3
3
x
x
x
e x sin x  (1  x    )( x   )
2
6
6
3
x
 x  x2  
3

Division of power series



Ex. Find the first 3 terms in the Maclaurin series for tan x
Sol I. Find f (0), f (0), f (0) and the Maclaurin
series is found.
Sol II. Use long division
x3 x5
x 

6 120
tan x 
x2 x4
1 

2 24
x3 2 x5
 x 

3 15
Application of power series
2
n
 n 1
n
s   (1)
.
n
2
n 0


Ex. Find


2
n
Sol. Let S ( x)   (n  n  1) x , then s=S(-1/2).
n 0
To find S(x), we rewrite it as


S ( x)  x n(n  1) x n 1   x n  xT ( x)  1/(1  x),
 
n 1
x
x
0
0


n 0
T ( x)dx dx   x n1  x 2 /(1  x)
n 1
 x 2 
2
T ( x)  


3
1

x
(1

x
)


1 2
2 10
s
 
.
3
2 (3 / 2) 3 27
Exercise
n2  n  1
s
.
n
3
n 0

Ex. Find

Sol.


S ( x)   (n2  n  1) x n ,
n 0


S ( x)  x 2  n(n  1) x n2   x n  x 2T ( x)  1/(1  x),
n2
 
x
x
0
0
n 0


T ( x)dx dx   x n  x 2 /(1  x)
n2
9
s .
4
Application of power series


ex  x 1
Ex. Find lim
by the Maclaurin series expansion.
2
x 0
x
x 2 x3
(1  x    )  x  1
Sol.
ex  x 1
2! 3!
lim

lim
x 0
x 0
x2
x2
x 2 x3
 
2

 1
1
x
x
2!
3!
 lim
 lim       ,
2
x 0
x 0 2
x
3! 4!

 2
where in the last limit, we have used the fact that power
series are continuous functions.
Homework 25

Section 11.8: 10, 17, 24, 35

Section 11.9: 12, 18, 25, 38, 39

Section 11.10: 41, 47, 48