BC-3
Taylor Review Key
1. Use Maclaurin series to write power series expanded about x0  0 (in simplified -notation) for
each of the following functions.
a.
3x
2  5x
3x 1

2 1  5x
2
2
3

3x   5 x   5 x   5 x 

1           L L 
2   2   2   2 

f  x 



3  5n1
n 1
2
n
xn
b.
f  x 
x
3
e2 x 1
 x3  e2 x 1
 e x e
3
2 x
2
3
4

2 x   2 x   2 x 

 e  x  1   2 x  


L
2!
3!
4!

3



n 0
2.



e  (2) n3
x
n!
n
Write the 3rd degree Taylor polynomial for f  x   sin  x  , expanded about x0 
f ( x)  sin( x), f   x   cos  x  , f   x    sin  x  , f   x    cos  x 
2
2
2
2
, f   x0   
, f ( x0 )  
, f   x0  
.
2
2
2
2
2
3
2
2
3 
2 
3 
2 
3 

x


x


x

Then, P3 ( x) 






2
2 
4  2  2! 
4  2  3! 
4 
 f ( x0 ) 
3
.
4
BC-3
3.
Taylor Review Key
Find an upper bound for the error made in approximating the value of e
Maclaurin series.
e =1 
3
 3
 3   3

2
3
2!
3!
 3
L 
4!
3
using the 6th degree
6
 Rn , Where R6 
f (7) (c)
( 3)7 , where 0  c  3 .
7!
Thus,
error 
4.
f (7) (c)
ec
e 3
9
( 3)7 
( 3)7 
( 3)7  ( 3)7  .0835
7!
7!
7!
7!
When approximating ln  0.8 , how many nonzero terms of its Maclaurin series do you have to use
in order to have an error that is less than 0.0001?
1 2 1 3 1 4
x  x  x L
2
3
4
1
2 1
3 1
4
So, ln(0.8)  ln(1  .2)  .2  .2   .2   .2  L
2
3
4
ln(1  x)  x 
Then, ln(0.8)  Sn  Rn 
f ( x)  ln(1  x)  f ( x) 
L L f n ( x) 
f
( n 1)
(c )
.2 n1 , where 0.2  c  0
 n  1!
1
1
2!
 f ( x) 
 f ( x) 

2
(1  x)
(1  x)
(1  x)3
(1)n (n  1)!
(1  x)n
n!
Then, ln(0.8)  Sn  Rn 
When n = 5,
1
 n  1 (.8)
n 1
1
(1  c) n1
n 1
.2
.
.2 n1 
n 1  
 n  1!
 n  1 (.8)
.2 n1  .00004069  .0001 ,
So we use n = 5. That is we use the 5th degree MacLaurin Polynomial or 5 non-zero terms.
BC-3
5.
Taylor Review Key
What are the possible values for x if we want the error in approximating cos  2x  using the first
four nonzero terms of its Maclaurin series to be less than 0.01?
  2 x 2  2 x 4  2 x 6   2 x 8

cos  2 x   1 


, since this is an alternating series.

2!
4!
6! 
8!


So,
6.
 2 x 8
8!
 .01  28  x8  403.2  x8  1.575  1.058  x  1.058
When approximating tan 1  0.75  , how many nonzero terms of its Maclaurin series do you have to
use in order to have an error that is less than 0.0001?
1
1
1
tan 1 ( x)  x  x3  x5  x 7 L .
3
5
7
So,
1
1
1
.753  .755  .757 L
3
5
7
Since this is an Alternating Series

 1n1 .75 2n1   1 .75 2n1
1
3 1
5 1
7
tan 1 (.75)   .75   .75   .75   .75   L 
  
 

3
5
7
2n  1
2n  1


1
.752n1  .0001  n  11.
2n  1
So we need 11 non-zero terms
tan 1 (.75)  .75  
BC-3
7.
Taylor Review Key
n
Suppose f  2   1, f   2   2, f   2   4, f   2   18, and f    x   6 for all x and for n > 3.
a.
Write the 3rd degree Taylor polynmial for  expanded about x0  2 .
f (2)
f (2)
( x  2) 2 
( x  2)3
2!
3!
4
18
 1  2( x  2)  ( x  2) 2  ( x  2)3
2!
3!
2
 1  2( x  2)  2( x  2)  3( x  2)3
P3 ( x)  f (2)  f (2)( x  2) 
b.
Find an upper bound for the error made in approximating the value of f  2.3 using the first
four nonzero terms of Taylor series for f expanded about x0  2 from part (a).
f (2.3)  P3 (2.3)  R3 (2.3)

f (4) (c)
(2.3  2) 4
4!
,
c is between 2 and 2.3.
6
(.3) 4
24
 .002025

8.
Use Maclaurin series to evaluate lim
x 0
ln 1  x 
.
sin  3 x 
1 2 1 3 1 4
1 2 1 3


 1

x  x  x  x L 
x 1  x  x  x  L 

ln 1  x 
2
3
4
3
4
  lim  2
1
lim
 lim 
3
5
7
3
2
5
4
7
6
x 0 sin  3 x 
x 0 
 3x    3x    3x   x0 x  3  3 x  3 x  3 x  3
 3x 

3!
5!
7! 

3!
5!
7! 



BC-3
9.
Taylor Review Key
Use power series to approximate the value of

0.9
0

0.9
  dx  
sin x
0
2
0.9
0
     
 
sin x 2 dx with error less than 0.0001.

2 3
2 5
2 7
x
x
x
 2
 x  3!  5!  7!  L



 dx

7
11
15
 x3
.9 
x
x
x
 


L 
0 
 3 7  3! 11 5! 15  7!
 (.9)3 (.9)7 (.9)11 (.9)15





L 
7  3! 11  5! 15  7!
 3

Since this is an alternating series we look for the first term that is less than .0001.
(.9)15
Since,
is the first term less than .0001,
15  7!

0.9
0
 
sin x
2
(.9)3 (.9)7 (.9)11
dx 


 .2318 with error less than 0.0001
3
7  3! 11 5!
BC-3
10.
Taylor Review Key
Suppose that f is a function such that f 1  1, f  1  2, f   x  
a.

1
1 x
3

, for all x > –1.
Estimate the value of f 1.5 using a quadratic Taylor polynomial.
P2 ( x)  f (1)  f (1)( x  1) 
f (1)
( x  1) 2
1!
1
3
 1  2( x  1)  1  1 ( x  1) 2
2!
1
 1  2( x  1)  ( x  1) 2 
4
2
 1  1  1  33
P2 (1.5)  1  2      
 2  4  2  16
b.
Determine an upper bound for the error made in your approximating in part (a).
f (1.5) 
33
 R3
16

f (c)
(1.5  1)3 ,
3!
where 1  c  1.5
3c 2
(1  c3 ) 2 1


6
8
3
(1  13 ) 2

,
48
1

64
since for 1  c  1.5,
3c 2
is largest at c  1.
(1  c3 ) 2