Vanier College
Calculus 2 (201-NYB-05)
Section: 007, Semester: Winter 2011
§11.10 Taylor and Maclaurin Series (p734-p745)
1. Theorem 11.10.5:
Theorem 11.10.5

If f has a power series representation at a f ( x)   c n ( x  a) n
| x  a | R , then
n 0
f n  a 
.
cn 
n!
Proof
Suppose
f ( x)  c0  c1 ( x  a)  c 2 ( x  a) 2  c3 ( x  a) 3    c n ( x  a) n   | x  a | R
By Term-By-Term Differentiation Theorem, we have
f ( x)  c1  2c 2 ( x  a)  3c3 ( x  a) 2    nc n ( x  a) n 1   | x  a | R
f ( x)  2c 2  2  3c3 ( x  a)    n(n  1)c n ( x  a) n 2   | x  a | R
f ( x)  2  3c3  2  3  4c 4 ( x  a)    n(n  1)( n  2)c n ( x  a) n 3   | x  a | R

f n  ( x)  n!c n  a sum of terms with ( x  a) as a factor | x  a | R
Put x = a in each equation. We obtain,
f (a )  1!c1
f (a )  2!c 2
f (a )  3!c3

f n  (a )  n!c n
Solving the equation for the nth coefficient cn, we have
f n  a 
cn 
n!
f 0  a 
And also, c0 
, where 0! = 1 and f(0)(a) = f(a)
0!
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2. Definition of Taylor Series and Maclaurin Series:
Definition 11.10.6 (p735)
If f has derivatives of all orders at a, then we define the Taylor series of the
function at a (or about a or centered at a) to be

f n  a 
( x  a) n

n!
. n 0

f (a )
f (a )
 x  a 2  f ( a ) ( x  a ) 3  
 f (a) 
( x  a) 
1!
2!
3!
And the polynomial
n
f i  a 
Tn ( x)  
( x  a) i
i!
i 0
f (a)
f (a)
f n  (a)
2
x  a    
 f (a) 
( x  a) 
( x  a) n
1!
2!
n!
is called the nth-degree Taylor polynomial of f at a.
Definition 11.10.7 (p736)
For the special case a = 0 the Taylor series

f n  0 n
( x)

n!
n 0
f (0)
f (0) 2 f (0) 3
x  
 f (0) 
( x) 
( x)  
1!
2!
3!
is called the Maclaurin series of the function f.
And the polynomial
n
f i  0 i
( x)

i!
i 0
n 
f (0)
f (0) 2
x     f (0) ( x) n
( x) 
1!
2!
n!
is called the nth-degree Maclaurin polynomial of f at a.
 f (0) 
Definition of Remainder of the Taylor Series (p737)
Rn x   f ( x)  Tn x  is called the remainder of the nth-degree Taylor
polynomial of f(x)
Note: We define that 0! = 1 and f (0) = f.
Example 1
(p736)
x
Find the Maclaurin series for f ( x)  e and its radius of convergence.
Solution
If f ( x)  e x , then f n  ( x)  e x , so f n  (0)  e 0  1 for all n.
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Therefore, the Maclaurin series is


f n  0 n
xn
x x2 x3
x

1






n!
1! 2 ! 3 !
n 0
n  0 n!
Let an = xn /n!. Then
x
an1
x n1
n!

 n 
 0 1
an
(n  1)! x
n 1
By the Ratio Test, the series converges for all x and the radius of convergence is
R = ∞.
3. Two Theorems about Taylor Series
(p737)
When does a Taylor series converge to its generating function?
Theorem 11.10.8
f n  a 
on | x - a | < R
( x  a) n
n!
n 0
if and only if lim Rn ( x)  0 for | x - a | < R, where Rn x   f ( x)  Tn x  is the

f ( x)  
n 
remainder of the Taylor series
Proof
Suppose lim Rn ( x)  0 for | x - a | < R.
n 
Then lim Tn ( x)  lim [ f ( x)  Rn ( x)]  f ( x)  lim Rn ( x)  f ( x) for | x - a | < R.
n 
n 
n 
Conversely, suppose f ( x)  lim Tn ( x) for | x - a | < R, then we have
n 
lim Rn ( x)  lim [ f ( x)  Tn ( x)]  f ( x)  lim Tn ( x ) = f(x) – f(x) = 0 for | x - a | < R.
n 
n 
n 
How accurately do a function’s Taylor polynomials approximate the function on a given
interval?
Taylor’s Inequality Theorem 11.10.9
(p737)
If | f n1 ( x) | M for | x  a | d , then the remainder Rn(x) of the Taylor series
satisfies the inequality
M
| Rn ( x) |
| x  a | n 1 for | x  a | d
(n  1) !
That inequality is called the Taylor’s inequality.
Proof
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To prove this is true for n = 1, that is, prove that if | f ( x) | M
M
then | R1 ( x) | | x  a | 2 for | x  a | d
2!
3
for | x  a | d ,
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Given | f ( x) | M
(1)
for |x – a|  d
Assume f ( x)  M , for a  x  a + d, then
x
Thus

a
x
a
x
f t dt    f a   M t  a dt
a
a
Then f ( x)  f (a)  f a x  a  
f ( x)  f (a)  f a  x  a  
That is, R1  x  
(2)

x
f t dt   Mdt
a
By FTC2, we have  f t   Mt  ,
Then f x  f a  M x  a , and
f x  f a  M x  a
x
a
x
M
x  a 2 , and
2
M
 x  a 2
2
M
 x  a 2
2
Assume f ( x)   M , for a  x  a + d, then
x
x
a
a
 f t dt    Mdt
M
Similarly, we have R1  x     x  a 2
2
From (1) and (2), we have By FTC2, we have | R1 ( x) |
M
| x  a | 2 for a  x  a + d
2!
Using similar calculations, we can show this inequality is also true for a-d  x  a
M
Therefore, we have | R1 ( x) |
| x  a | 2 for |x – a|  d
2!
4. A Helpful Limit
(p738)
xn
0
n  n !
for every real number x
lim
(11.10.10)
Proof

xn
is converges for all x from Example 1.
n 0 n!
xn
0
By Theorem 11.2.6, we have lim
n  n !
Because the series 
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Example 2
(p738)
x
Prove that e is equal to the sum of its Maclaurin series.
Solution
Proof
Let f(x) = ex, then f(n+1)(x)= ex for all n.
If d is any positive number and | x |  d, then | f(n+1)(x) | = ex  ed.
By the Taylor’s Inequality, we have
ed
| Rn ( x) |
| x | n 1 for | x | d where a = 0, and M = ed.
(n  1) !
n 1
x
ed
lim
| x | n 1  e d lim
0
n  ( n  1) !
n  ( n  1)!
(Limit 11.10.10)
By the Squeeze Theorem, we have lim | Rn ( x) | 0 , and therefore
n 
lim Rn ( x)  0 for all values of x.
n 

xn
x x2 x3
By the Theorem 11.10.8, e  
 1 

  for all x.
1! 2 ! 3 !
n  0 n!
x
Formula 11.10.11

xn
x x2 x3
ex  
 1 


1! 2 ! 3 !
n  0 n!
for every real number x
Property 11.10.12
1 1 1
e  1   
1! 2 ! 3 !
(11.10.12)
Proof
If put x = 1 in Equation 11.10.11, we obtain the expression 11.10.12.
Example 3 (p739)
Find the Taylor series for f(x) = e x at a = 2
Solution
f (n)(2) = e2
2


f ( n ) 2
Then the Taylor series of e x is 
x  2n   e x  2n
n!
n 0
n 0 n!
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n 1
x2
a n 1
e 2 x  2
n!

 2

 0  1 as n 
n
an
(n  1)!
n 1
e x  2
So the Taylor series is convergent for all x and the radius of convergence is R =
.
Using Ratio Test,
f(n+1)(x)= ex for all n.
If d is any positive number and | x |  d, then | f(n+1)(x) | = ex  ed.
By the Taylor’s Inequality, we have
ed
| Rn ( x) |
| x  2 | n 1 for | x | d where a = 2, and M = ed.
(n  1) !
n 1
x2
ed
(Limit 11.10.10)
lim
| x  2 | n 1  e d lim
0
n  ( n  1) !
n  ( n  1)!
By the Squeeze Theorem, we have lim | Rn ( x) | 0 , and therefore
n 
lim Rn ( x)  0 for all values of x.
n 

e2
x  2n for all x.

n 0 n!
Example 4
(p739)
Find the Maclaurin series for sin x and prove that it represents sin x for all x.
Solution
f (x) = sin x
f (0) = 0
By the Theorem 11.10.8, e x 
f '(x) = cos x
f '(0) = 1
f "(x) = -sin x
f "(0) = 0
f (3)(x) = -cos x
f (3)(0) = -1
f
(4)
(x) = sin x
f (4)(0) = 0
Since the derivatives repeat in a cycle of four, then we have the Maclaurin series:
f (0)
f (0) 2
f n  (0) n
x    
f (0) 
( x) 
( x)
1!
2!
n!

x3 x5 x7
x 2 n 1
n
 x


    (1)
3! 5! 7!
(2n  1)!
n 0
(n+1)
Since f
(x) is sin x or cos x, we know that | f (n+1) (x) |  1 for all x.
So we take M = 1 in Taylor’s Inequality:
M
| x | n 1
n 1
| Rn ( x) |
| x  0| 
(n  1) !
(n  1)!
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By the Limit 11.10.10 (Property 3), we have
| x | n 1
lim
0 .
n  ( n  1)!
So lim | Rn ( x) | 0 by the Squeeze Theorem
n 
It follows that lim Rn ( x)  0 by the theorem 11.1.6
n 
Therefore,
Formula 11.10.15
sin x  x 

x3 x5 x7
x 2 n1
for all x


    (1) n
3! 5! 7!
(2n  1)!
n 0
Example 5 (p740)
Find the Maclaurin series for cos x and prove that it represents sin x for all x.
Sol.
3
5


d
sin x   d  x  x  x  
cos x 
dx
dx 
3! 5!

Term by term
differentiation
2n

x2 x4 x6
n x


     1
2n !
2! 4! 6!
n 0
By the Term-by-term Differentiation Theorem, the differentiated series for cos x
is convergent for all x, since Maclaurin series fo sin x converges for all x.
Therefore,

1
Formula 11.10.16
cos x  1 
2n

x2 x4 x6
n x
for all x


     1
2n!
2! 4! 6!
n 0
Example 6 (p740)
Find the Maclaurin series for the function f(x) = x cos x.
Sol.
2n
2 n 1


n x
n x
x cos x  x  1
  1
2n! 
2n!
n 0
n 0
Example 7 (p740)
Represent f(x) = sin x as the sum of its Taylor series centered at π/3.
3
Sol.
f (x) = sin x
f (π/3) =
2
f '(x) = cos x
f '(π/3) = ½
3
f "(x) = -sin x
f "( π/3) = 
2
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f (3)(x) = -cos x
f (3)( π/3) = 1/2
This pattern repeats indefinitely. Therefore the Taylor series at π/3 is
 
 
 
f  
f  
f n   
2
3



3
3
3
 



f 
x  
x   
 x   
1! 
3
2! 
3
3! 
3
3
3
1 

1 

1 


x 
x   
 x   
2 2 1 ! 
3  2  2 !
3
2  3 !
3
(n+1)
(n+1)
Since f
(x) is sin x or cos x, we know that | f
(x) |  1 for all x.
So we take M = 1 in Taylor’s Inequality:
M
| x   / 3 | n 1
| Rn ( x) |
| x   / 3 | n 1 
(n  1) !
(n  1)!
By the Limit 11.10.10, we have
| x   / 3 | n 1
lim
0 .
n 
(n  1)!
So lim | Rn ( x) | 0 by the Squeeze Theorem
2
n

n 
It follows that lim Rn ( x)  0 by the theorem 11.1.6
n 
Therefore, this series represents sin x for all x.
Example 8
(p741)
Find the Maclaurin series for f ( x)  1  x k , where k is any real number
Sol
f ( x)  1  x 
k
f ( x)  k 1  x 
f (0) = 1
k 1
f ( x)  k (k  1)1  x 
f’(0) = k
k 2
f ( x)  k (k  1)( k  2)1  x 
…
f
n 
f’’(0) = k(k-1)
k 3
( x)  k (k  1)  k  n  11  x 
f’’’(0) = k(k-1)(k-2)
k n
f n  (0)  k (k  1) k  n  1
Therefore the Maclaurin series of f ( x)  1  x k is

k k  1 k  n  1 n
f n  0 n
x
x



n!
n!
n 0
n 0
Using Ratio Test,
a n 1
k (k  1)  (k  n  1) x n 1
n!


n  1!
an
k ( k  1)  k  n  1x n

k
k n
n

x 
x  x
as n  
1
n 1
1
n
So the series converges if |x| < 1, and diverges if |x| > 1.
1
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Then prove that f ( x)  1  x k is equal to the sum of its Maclaurin series.

k k  1 k  n  1 n
x , then
Let g ( x)  
n!
n 0
Term by term
Differentiation 
k k  1 k  n  1 n 1
nx
n!
n 1


1  x g ( x)   k k  1k  n  1 nx n1   k k  1k  n  1 nx n
n!
n!
n 1
n 1

n  m 1  k k  1 k  ( m  1)  1n
k k  1 k  n  1 n
 
(m  1) x m  
nx
m  1!
n!
m 0
n 0

m  n  k k  1 k  n  1k  n 
k k  1 k  n  1 n
 
xn  
nx
n!
n!
n 0
n 0

k k  1 k  n  1
k  n   nx n

n
!
n 0

k k  1 k  n  1 n
 k
x
n!
n 0
 kg( x)
kg( x)
So we have g ( x) 
for |x| < 1
1 x
g ( x)


To prove g ( x)  1  x  , we need to prove g ( x)1  x 
k
k
1
Let h( x)  g ( x)1  x 
k
h( x)  k 1  x 
 k 1
g ( x)  (1  x)  k g ( x)
kg( x)
 k 1
 k 1  x  g ( x)  (1  x)  k
1 x
0
k
By the Theorem 4.2.5 (p284), h( x)  g ( x)1  x  is a constant on (-1, 1)
Then h(x) = h(0) = g(0) =1

k k  1 k  n  1 n
k
x
Therefore, 1  x   
n!
n 0
Then
Note
k k  1 k  n  1 n
x is called the binomial series, and
n!
n 0
k (k  1)  k  n  1
is called the binomial coefficient, its traditional notation is
n!
 k  k (k  1)  k  n  1
  
n!
n

The series
c34.doc

9
April 29, 2011
Binomial Series 11.10.17
(p742)
If k is any real number and |x| < 1, then

1  x k   k k  1k  n  1 x n  1  kx  k (k  1) x 2  k (k  1)(k  2) x 3  
n!
2!
3!
n 0
Example 9
(p742)
Find the Maclaurin series for the function f ( x) 
1
4 x
and its radius of
convergence..
Sol.
1
4 x


1
x

 1  
x 2 4
x

41   2 1 
4
 4
1
1

1
2
n
1
1     x 

 2 
2 n 0  n  4 


2
 12  32  52   12  n  1  x  n 
1   1  x   12  32   x 
 1       
  
    
2   2  4 
2!
n!
 4
 4

1 1
1 3 2 1 3  5 3
1  3  5 2n  1 n

1 x 
x 
x 
x  
2
3
n

2 8
2!8
3!8
n!8

From 11.10.17, this series converges when |-x/4| < 1, that is |x| < 4, so the radius of
convergence is R = 4.

*Important Maclaurin series and their radius of convergence:

1
2
 1 x  x     xn
R=1
1 x
n 0

ex  
n 0
xn
x x 2 x3
 1 


n!
1! 2 ! 3 !
R=
(11.9.1)
(11.10.11)

x3 x5 x7
x 2 n1
R=
(11.10.15)


    (1) n
3! 5! 7!
(2n  1)!
n 0
2n

x2 x4 x6
n x
R=
(11.10.16)
cos x  1 


     1
2n!
2! 4! 6!
n 0

x3 x5 x7
x 2 n1
R=1
(11.9, p732)
tan 1 x  x 


    (1) n
3
5
7
(2n  1)
n 0

1  x k   k k  1k  n  1 x n  1  kx  k (k  1) x 2  k (k  1)(k  2) x 3  
n!
2!
3!
n 0
R=1
(11.10.17)
sin x  x 
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10
April 29, 2011
Example 10
(p744)
(1)
Evaluate  e  x dx as an infinite series.
(2)
Evaluate  e  x dx correct to within an error of 0.001.
2
1
2
0
Sol.
(1)
Replacing x by –x2 in the series for ex
e
 x2


n 0
 x 
2 n
n!
 1
x2 x4 x6



1! 2 ! 3 !
Then
e
 x2
2n

x2 x4 x6
n x

dx    1 


    1

1! 2! 3!
n!

Term by term
Integration

Cx

dx

x3
x5
x7
x 2 n 1
n


    1

2n  1  n!
3  1! 5  2! 7  3!
This series converges for all x because the original series for e  x converges for
all x.
2
(2)
1


x3
x5
x7
x9
0 e dx   x  3  1!  5  2!  7  3!  9  4!  
0
1
1
1
1
 1




3  1! 5  2! 7  3! 9  4!
This is the alternating series.
Using Alternating Series Estimation Theorem, notice that
1
1
b5 

 0.001
11  5! 1320
1
2
1 1
1
1

 0.7475
Then  e  x dx  s 4  1   
3 10 42 216
0
Example 11
(p744)
x
e 1 x
Evaluate lim
x 0
x2
Sol.
Using the Maclaurin series for ex, we have


x x2 x3

1

 1!  2!  3!    1  x
x
e 1 x

lim
 lim 
2
2
x 0
x

0
x
x
2
1 x x
 1
 lim   
  
x 0 2
3! 4!

 2
1
c34.doc
 x2
FTC 2
11
April 29, 2011
3. Multiplication and Division of Power Series (p745)
Some algebraic techniques to find power series are sometimes useful.
Theorem for Multiplication of Power Series:


Two power series A( x)   a n x n and B( x)   bn x n can be multiplied to
n 0
n 0

obtain A( x)  B( x)  C ( x)   c n x n , where cn  a0 bn  a1bn1    an b0 the result
n 0
being valid for each x within the common interval of convergence.
See Murray’s Advanced Calculus, ISBN 0-07-060229-8, p230
Theorem for Division of Power Series:

The power series A( x)   a n x n is divided by the power
n 0

series B( x)   bn x n where b0  0, the quotient can be written as a power series
n 0
which converges for sufficiently small |x|.
See Murray’s Advanced Calculus, ISBN 0-07-060229-8, p230
Example 12 (p769)
Find the first three nonzero terms in the Maclaurin series for
(a) ex sin x,
and (b) tan x.
Sol



x x2 x3
x3 x5
(a)
e x sin x   1  

   x 

  
1! 2 ! 3 !
3! 5!



Multiply these expressions, and collect like terms as for polynomials
( Factor 1)
1  x  12 x 2  16 x 3  
3
5
1

( Factor 2)
x
 16 x
 120
x
 x  x2
 12 x 3
 16 x 4

 x
 x

1
6
x  x2
c34.doc
3
 13 x 3
1
6
4

(Product)
12
April 29, 2011
Thus e x sin x  x  x 2 
(b)
1 3
x 
3!
sin x 
x3 x5
tan x 
 x


cos x 
3! 5!

  x2 x4
  1 

 
2! 4!
 

Using long division,
(Divisor) 1  12 x 2

1
24
x4

x  13 x 3
 152 x 5
x  16 x 3
1
 120
x5
x  12 x 3

1
3
1
3
1
24
x5
x
3
 301 x
x
3
5
1
6
 x
5
2
15
5
x

(Quotient)
  (Dividend)




1
2
Thus tan x  x  x 3  x 5  
3
15
Exercises 11.10
1,
3,
31,
35,
c34.doc
5,
37,
(p746)
7,
9,
43,
45,
15,
53,
13
17,
57,
19,
63,
21,
65,
25,
67
27,
April 29, 2011