BC 3 More on Taylor
(1)
Name:
Write out series for the following:
cos  x 3 
x   x   x 
= 1
3 2
3 4
3 6
2!
4!
6!
6
12
18
x
x
x
= 1 


2! 4! 6!

3
5
7
2x  2x  2x

sin(2 x) 1 
 2 x 



x
x 
3!
5!
7!
23 x 2 25 x 4 27 x 6
=2



3!
5!
7!

x 2 x3
xe x  x 1  x   
2! 3!

x3 x 4
2
 xx   
2! 3!






3x
2
3
4

 3x 1  2 x   2 x    2 x    2 x  


1  2x
 3x  3  2 x 2  3  22 x3  3  23 x 4  3  24 x5 
x2
e
2 x 1
IMSA BC 3
2
3
4
2x  2x  2x

x 2 2 x x 2 
 e  1  2 x 



e
e 
2!
3!
4!
x 2 2 x 3 2 2 x 4 23 x 5 2 4 x 6
 




e
e
e  2! e  3! e  4!
Taylor Rev p.1



Fall 03
(2)
Write out Taylor’s formula with remainder for ƒ (x) = sinx, with a = 3/4, and
n = 3.
2
 3 
f  
 4  2
 3
f 
 4
f ( x)  sin x 
f ( x)  cos x 
f ( x)   sin x 
f ( x)   cos x 
So,
2
 3 
f     
2
 4 
 3
f  
 4
f (4) ( x)  sin x
2


2

2


 2
 3 
f   
2
3
3 
3 
3 
4 
 3 
 3 
 3 



sin x  f    f   x   
 x    f   x    R3 . Then
4 
2! 
4 
4 
 4 
 4 
 4 
2
2
2
3
4
2
2
3  2 
3 
3  sin c 
3 

2
sin x 

x

x
 
x
 
x
 , where c is some
2
2 
4  2! 
4 
3! 
4 
4! 
4 
3
number between x and
.
4
(3)
If sinx is to be approximated by x - x3/3! , with |x| < .4, find an upper bound for the error.
f (4) (c)
( x  0)4 where c is some number between 0 and x, and f (4) (c)  sin c .
4!
sin c 4
1
( x) 
(.4) 4  .001067
So error =
4!
24
sin x  P3  R3 
(4)
If terms through n = 8 are used to approximate e2, find the error.
f (9) (c)
(2  0)4 where c is some number between 0 and 2 and f (4) (c)  ec .
9!
ec
e2 9 9
4
9
So error = (2)  (2)  (2)9 
 .013
9!
9!
9!
315
e2  P8  R8 
IMSA BC 3
Taylor Rev p.2
Fall 03
(5)
What values of x can be used if cos(2x) is to be approximated by two non-zero terms with
an error less than .005?
24 x 2 23 sin(2c) 3

x where c is some number between 0 and x. We want
2!
3!
8sin(2c) 3
8 3
x  .005 . Since sin(2c)  1 , we need
x  .005  x  .155 or  .155  x  .155
3!
3!
cos(2 x)  1 
(6)
Find the value of each series by recognizing the function and the point at which it is
evaluated.
(.2)2 (.2) 4
1

  cos(.2)
2!
4!
1 (1/ 2) 2 (1/ 2)3
1 


2
2!
3!
1  .4 
(7)
(.4) 2 (.4)3


2
6

 e.4
0.9
2
Use series to approximate  0 cos(x ) dx with an error less than .005.
0.9
0.9
0
1
 e2  e
cos( x ) dx 
2




0
 x 4 x8 x12
1  2!  4!  6! 

x5
x9
x13
 x



5  2! 9  4! 13  6!
 .9 

 dx

.9
0
(.9)5 (.9)9 (.9)13



5  2! 9  4! 13  6!
Since this is an alternating series with an decreasing, positive, and tending toward zero. We need
to find the first term less that .005 and add all the terms previous to that term.
(.9)5
(.9)9
0.9
 .841 , with error less than .005.
Since,
 .0018  .005 ,  0 cos( x 2 ) dx  .9 
5  2!
9  4!
IMSA BC 3
Taylor Rev p.3
Fall 03
(8)
ln(1  x)
. Then use l’Hôpital’s Rule to confirm your answer.
x 0 sin(2x)
Use series to find lim
Series:

x 2 x3 x 4
x   

ln(1  x)
2 3 4
lim
 lim 
3
5
7
x 0 sin(2 x )
x 0 
2x  2x  2x




 2x 
3!
5!
7!

L’Hôpital’s Rule :
1
1


 1 x 
ln(1  x)
1
lim
 lim 
 1 0 

x 0 sin(2 x)
x 0 2cos(2 x)

 2cos(0) 2



(9)
If ƒ(x) =


x x 2 x3
1

  


2 3 4
  lim 
23 x 2 25 x 4 27 x 6
 x 0 
2






3!
5!
7!


 1

 2

 x / 2 , find the interval of convergence for ƒ, for ƒ’, and for F.
n
n0

n
x
x
 x
f ( x)     is a geometric series with  r   1  2  x  2 .
2
2
n 0  2 

n x
f ( x)    
n 0 2  2 
n 1
has the same radius of convergence as f. We need only check the endpoints.

n  2 
x  2  f ( x)    
n 0 2  2 

n2
x  2  f ( x)    
n 0 2  2 

x
f ( x)   n  
n 0  2 
n 1
diverges by the nth term test.
n 1
diverges by the nth term test. So the interval of convergence for
n 1
is (  2, 2).

2 x
F ( x)  
 
n0 n  1  2 
n 1
has the same radius of convergence as f. We need only check the endpoints.

2  2 
 
n 0 (n  1  2 
x  2  F ( x)  

2 2
x  2  F ( x)  
 
n 0 n  1  2 
n 1
converges by AST.
n 1
diverges by the p  test.

2 x
So the interval of convergence for F ( x)  
 
n 0 n  1  2 
IMSA BC 3
Taylor Rev p.4
n 1
is [  2, 2).
Fall 03
(10)
Let ƒ(2) = 1, ƒ'(2) = -2, ƒ''(2) = 4, and | ƒ(n) (x) | < 6 for all x.
Find the second degree Taylor Polynomial for ƒ at a = 2.
f (2)
( x  2) 2 .
2!
P2 ( x)  f (2)  f (2)( x  2) 
If these three terms are used to approximate ƒ(2.3), find an upper bound for the error.
f (2.3)  P2 (2.3)  R3 (2.3) 
f (c)
6
(2.3  2)3  (.3)3  .027
3!
6

(11)
n2
(FDWK P. 496) We can show that the series  n converges by the Ratio Test, but
n0 2
what does it equal? To find out, follow the subsequent steps.
Express as a series:
1
 1  x  x 2  x3  x 4 
1 x
Differentiate both sides.
1
 1  2 x  3x 2  4 x3  5 x 4 
2
1  x 
Multiply both sides by x.
x
 x  2 x 2  3x3  4 x 4  5 x5 
2
1  x 
Differentiate both sides (again).
(1  x)
 1  4 x  9 x 2  16 x3  25 x 4 
3
1  x 
Multiply by x (again).
 x  (1  x)
 x  4 x 2  9 x3  16 x 4  25 x 5 
3
1  x 
Let x = 1/2.
1  1
  1  
2
3
4
5
2  2 1
1
1
1
1
  4    9    16    25   
3
2
2
2
2
2
 1
1  
 2
2
What do you get?
IMSA BC 3
6
3
4
5
1
1
1
1
1
 4    9    16    25   
2
2
2
2
2
Taylor Rev p.5
Fall 03