( § 11.7 and § 11.8)

Summary of important concepts
§ 11.7
[1] Power Series
A power series centered at a is given by

 n

0 n
(
 a ) n  c
0

(
 a )

2
(
 a )
2   c x
 a ) n 
...
We call a the center of the series, and the coefficients are constants.
Think of a power series as a polynomial function in ( x
 a ) . But this polynomial function is in the form of a series. Remember there is always a variable such as x in a power series.
Why do we study power series?
It can be used to study the behavior of a much more complicated function in the neighborhood of a point of interest, and that point is the center of the power series. Power series is a powerful tool for solving some difficult differential equations, which many of you will take in the following year.
Since a power series is still a series, it can be divergent or convergent. However, what makes power series
different from other series is that its convergence/divergence depends on the x -values. So, our main task here is to figure out how to test a power series, find the radius of convergence and the interval of convergence. Here are the steps:
(i) Use the Ratio test or nth Root test (but need to take the absolute value of the terms, why?), to find the interval where the series converges absolutely. Let R represent the radius of convergence, then x a R and the interval for absolute convergence is ( a

,

R )
(ii) It is possible that the series converges absolutely or conditionally at the end points of the interval.
Therefore, we need to test them here. Plug x a R in the original series, then use previously learned testing methods, such as Integral test, Comparison test, or Alternating series test.
Special cases for the radius of convergence: (a) R
 
means the series converges for all x -values; (b)
R

0 implies that the power series only converges when x
 a .
Example: Find the radius and interval of convergence. For what values of x does the series converge
(a) absolutely (b) conditionally?

 n

0
 x

2) n
4 n
Notice that the series is not in the standard form; we need to factor out 3 from (3 x

2) n
as follows:

n



0
 x

2) n
4 n
 n



0
 x

2 / 3)) n
4 n
 n



0
 x

2 / 3) n
4 n
, also recognize that the center of the power series is a
 
2
3
. Now, we proceed to use the Ratio test (actually, nth Root test works too). Pay attention here that we use the absolute value: n lim
 a n

1 a n
 lim n

 n

1
3 n

1
( x

2 / 3) n

1
4 n

1
 x

2 / 3) n
4 n
 lim n

 x

2 / 3)
4
 lim n

3( x

2 / 3)

3( x

2 / 3)
4 4

1
Now, solve the absolute-value inequality:
3( x

2 / 3)
4

1,
3
4 x

2 / 3

1, x

2 / 3

4
3
, hence the radius of convergence is R

4
3
. Next, we solve for x from interval for absolute convergence is


2,
2
3
 x

2 / 3

4
,
4
3 3
2 / 3

4
3
, 2 x
2
3
. Therefore the
. Now, let’s test power series at the end points of the interval. At x
 
2 , n



0
At x

2
3
,

 n

0
 x

2) n
4 n
4 n

 x

2) n n



0


 n

0
  
2) n
4 n
 n
4 n
 n



0


( 1) n


0 n
  n
4 n
 n



0
  n
, diverges, why?
4 n



1 , diverges, why? n

0
Finally, the answer to (a) is x
2
3
, and the answer to (b) is none.
Example: n



1
( x

1) n n !
with the same questions as the previous example.
The center of the power series is a

1 . Use Ratio test first to the power series: lim n
 a n

1 a n
 lim n

( x

1) n

1
( n

1)!
( x

1) n
 lim n

( x

1) n

1 n !
( x

1) ( n

1)!
 lim n
 x n

1

1 why?

0 1 n !
This implies that the radius of convergence is R
 
because no matter what x is, lim n
 a n

1 a n to the Ratio test, the power series converges absolutely, and the interval of convergence is ( , ) .
, according
Now, try

 n

1 n n
( x

1) n
We use the nth-Root test this time, lim n
 n a n
 lim n
 n n n
( x

1) n  lim n ( n
 n x

1) n  lim n
 n x


0
 if if x x


1
1
The radius of convergence is zero because the power series converges only if x

1 convergence is only a point x

1.
. The interval of

Example:

 n

1
( x

2) n n ln n
Use Ratio test: lim n
 a n

1 a n
 lim n

( x

2) n

1
( n

1) ln( n

1)
( x

2) n n ln n
 n lim

( x

2) n

1 n ln n
( x

2) ( n

1) ln( n

1)
 n lim
 x

2 lim n
 n n

1 lim n
 ln n ln( n

1)
 n lim
 x

2 x 2 1
Radius of convergence is R

1 . Solve the absolute-value inequality to get 1 absolute convergence is (1,3) . Now, test the end points of the interval. At x

1
3 . Therefore, interval of
, the series becomes n



1
( x

2) n n ln n


 n

1 n ln n n


 n

1 n n ln n
. This is an alternating series, use the Alternating Series test…your assignment. This alternating series converges after using Alternating Series test. Next, we test the power series at x

3 :

 n

1
( x

2) n n ln n
 n



1
 n n ln n
 n



1
1 n ln n
. This series can be tested with the Integral test…your assignment again. This series turns out to be divergent. Now, the interval of convergence has to be updated as [1,3).
Caveat : remember to test the end points of your interval!
§ 11.8
[1] Maclaurin series (a special case of Taylor series)
Why study Taylor series?
Polynomial functions can be used to approximate functions that are infinitely differentiable. Instead of studying complicated functions, one may choose to study the associated power series (consists of polynomial functions) of the function. Taylor series are useful in engineering sciences.
We begin with the easier version of the Taylor series, called Maclaurin series.
The Maclaurin series of ( ) is given by k



0 f k !
(0) x k
. Compare to power series, c k
 f k !
(0)
and the center of the series is 0. Students will need to memorize this formula. The difficulty of the Maclaurin series is to find the pattern in the coefficients formula for c k c k
 f (0)
. Students will find the first few terms of c k in order to find a k !
. Let’s familiarize ourselves with this concept via the following two examples:
Example: Find the Maclaurin series of ( )
 e

2 x
.
Obviously, 0!=1, f (0) 1 (note: f
(0)  f ). Now, we need to find f (0) :

f x
'( )
 
2 e

2 x f ''( )
  2 e

2 x f '''( )
  3 e

2 x f
(4) x ( 2)
4 e

2 x f
(5) x ( 2)
5 e

2 x f (0) f '(0)
 
2 f ''(0)
  2 f '''(0)
  3 f
(4)
(0)
  4
Based on pattern from the above few trials, we get f (0) ( 2) k
(notice that I did not rewrite
This way it will help us see the pattern. The Maclaurin series of ( )
 e

2 x
is k



0 f k !
(0) x k
Let’s try another example.
Example: Find the Maclaurin series of ( )
 sin 3 x .
 k



0
 k x k
. k !
Notice that f (0)

0.
We build a similar table: f
(5)
(0)
  5
 2
as 4, etc. f x
'( )

3cos3 x f x
 
3 sin 3 x f x
 
3 cos3 x f
(4) x

3 sin 3 x f
(5) x

3 cos 3 x f f f
(0)
'(0)

3
''(0)

0 f '''(0)
 
3
3 f
(4)
(0)

0 f
(5)
(0)

3
5
So, what is the pattern of f (0) ? You may cross out the 0’s. Let’s see, f '(0)

3 , f '''(0)
 
3
3
, f
(5)
(0)

3
5
.
Only the odd-order of derivatives are none-zero. Remember that odd numbers can be written as 2 k

1 (even numbers are 2 k ). Therefore, we make our first attempt f (0)

0 and f
(2 k

1)
(0)
  k 2 k

1
, k

1, 2,3...
Let’s double check (students will do the same when they solve similar problems): k

1, f '(0)
   
3 , woops, the sign is wrong, try another one, k

2, f
(3)
(0) ( 1) 3

3
3
, the sign is wrong again. So, let’s go back and fix it, f
(2 k

1)
(0)
  k

3 k

1
, k

1, 2,3...
Now, k

1, f '(0)
 
3

3 , k

2, f
(3)
(0) ( 1) 3
 
3
3
, and match! We are ready to find the Maclaurin series: k



0 f k !
(0) x k  k



1 f
(2 k

1)
(0)
(2 k

1)!
x
2 k

1 because f all the terms with even index should be dropped
 k



1 k

3, f
(5)
(0) ( 1) 3
 k

3 k

1
(2 k

1)!
x
2 k

1
.

3
5
, perfect
[2] Taylor series
The Taylor series can be used to expand a function at any point x
 a , i.e. the center of the power series is not zero. The Taylor series of an infinitely differentiable function is given by

k



0 f ( )
( x
 a ) k  k !


1!
( x
 a )
 f

( )
( x
 a )
2 
2!
f
(3)
( )
( x
 a )
3 
.....
3!
We use the same strategy developed for the Maclaurin series for the Taylor series.
Example: Find the Taylor series of

1 x

1
at x

1 . (Note: we will be using ( x – 1) k )
Here is the table f x

 x
1

1

1
( x

1)
2 f ''( )

(1)(2)
( x

1)
3 f '''( )


(1)(2)(3)
( x

1)
4 f (1) f f (1)

1
'(1)
2

1

2
2 f ''(1)

2!
2
3 f '''(1)


3!
2
4 f
(4)
( )

(1)(2)(3)(4)
( x

1)
5 f
(5)
( )


(1)(2)(3)(4)(5)
( x

1)
6 f
(4)
(1)

4!
2
5 f
(5)
(1)


5!
2
6
We conjecture that f (1)


2 k k

1 k !
, k

0,1, 2,3,...
This can be verified by plugging in different k-values and compare to the table above. Now, the Taylor series of k



0 f k !
(1)
( x

1) k  k



0

1 x

1
at x

1 is found to be
 k k !
2 k

1
( x

1) k  k !
k



0
2 k

1 k
( x

1) k
.
[3] Taylor polynomial
A Taylor polynomial can be treated as a partial sum of the Taylor series. It is formally defined as the following: n
 k n


0 f ( )
( x
 a ) k k !
, this is called a Taylor polynomial of ORDER n (an nth degree polynomial).
Example: Find the Taylor polynomial of orders 1, 2, and 4 generated by of
We already found the Taylor series, by definition of a Taylor polynomial,
( )
1 1
2 2
2
Homework:
( x

1)
,
( )
1 1
2 2
2
( x
 
1
2
3
( x

1)
2
, and
( ) n

 k
1 x

1
at x

1 . n


0
2 k

1 k
( x

1) k
. Hence,
1 1
2 2
2
( x
 
1
2
3
( x

1)
2 
1
2
4
( x

1)
3 
1
2
5
( x

1)
4
§ 11.7: 1, 3, 4, 5, 7, 9, 11, 13, 15, 17, 18, 19, 21, 25, 29, 31.
§ 11.8: 1, 3, 5, 7, 9, 11, 13, 19, 21, 23, 25, 27, 33, 35, 37.

( § 11.7 and § 11.8)
NAME: _______________________________ (25-points total)
(1-10 pts) Find the series’ radius and interval of convergence, show details of work.

 n

0 n x

2) n n
3 
1

(2-5pts) Find the Taylor polynomial of order 4 generated by of ( ) x sin x at x


2
.

(3-10pts) Find the Maclaurin series at x

0 for the function ( )
 cos(2 ) .