Section 11.6 – Taylor’s Formula with Remainder
The Lagrange Remainder of a Taylor Polynomial
Rn x  
 z  n 1
x
 n  1 !
f
n 1
where z is some number between x and c
The Error of a Taylor Polynomial
Rn x  
M
 n  1 !
bc
n 1
where M is the maximum value of f
on the interval [b, c] or [c, b]
 n  1
x 
Let f be a function that has derivatives of all orders on the
Interval (-1, 1). Assume f(0) = 1, f ‘ (0) = ½, f ”(0) = -1/4,
f ’’’(0) = 3/8 and f  4   x   6 for all x in the interval (0, 1).
a. Find the third-degree Taylor polynomial about x = 0 for f.
p3  x   1 
p3  x   1 
1
2
1
2
x 
1/ 4
x
1
2!
8
x 
2
x 
2
3/8
1
x
3
3!
x
3
16
b. Use your answer to part a to estimate the value of f(0.5)
p 3  0 .5   1 
p 3  0 .5  
1
2
157
128
1
 0 .5    0 .5 
8
 1 .2 2 7
2

1
16
 0 .5 
3
Let f be a function that has derivatives of all orders on the
Interval (-1, 1). Assume f(0) = 1, f ‘ (0) = ½, f ”(0) = -1/4,
f ’’’(0) = 3/8 and f  4   x   6 for all x in the interval (0, 1).
c. What is the maximum possible error for the approximation
made in part b?
p 3  0 .5  
R3 x  
R3 x  
R3 x  
157
128
 1 .2 2 7
M
 3  1 !
4
f x 
 3  1 !
6
 4 !
bc
bc
0 .5  0
4
3 1
3 1

1
64
 0 .0 1 6
Estimate the error that results when arctan x is replaced by
x
x
3
3
if x  0 .2 a n d f
4
x 
R3 x  
R3 x  
R3 x  
 4
M
 3  1 !
4
 3  1 !
1
6
0 .2
4
bc
3 1
0 .2  0
3 1
 0 .0 0 0 2 6 7
Estimate the error that results when ln(x + 1) is replaced by
x
1
2
x if x  0 .1
2
f  x   ln  x  1
f 'x  
f "x  
f "'  x  
1
R2 x  
f '''   0.1
 2  1 !
R 2  x   0.000457
x 1
1
 x  1
2
2
 x  1
3
F ‘’’ (x) has a maximum value at x = -0.1
0  0.1
2 1
Find an approximation of ln 1.1 that is accurate to three decimal
places.
We just determined that the error using the second degree
expansion is 0.000457.
p2 x   x 
1
x
2
2
p 2  0 .1   0 .1 
p 2  0.1  0.095
1
2
 0 .1
2
Use a Taylor Polynomial to estimate cos(0.2) to 3 decimal places
co s x  1 
x
2
2!

x
4
4!

x
6
6!
 ...
If x = 0.2, Alternating Series Test works for convergence
 0.2 
 2n  !
2n
 0.005
cos x  1 
0 .2
2!
2
 0 .9 8
1
Use a Taylor Polynomial to estimate
place accuracy.
1

0
3
5

1
x
x

...  dx
x 
x
3!
5! 

sin x
0
dx with three decimal
x
1
 2n  1  2n  1 !
 0 .0 0 5
2
4


x
x
  1  3 !  5 ! ...  dx

0 
1
3
5
7

 1
x
x
x


 ...  |o
x 
3  3! 5  5! 7  7!


1
1
1
1


 ...
3  3! 5  5! 7  7!
Satisfies Alternating Series Test
1
1
3  3!

1
5  5!
 0 .9 4 6
Suppose the function f is defined so that
f  1 
1
2
, f '  1 
1
2

2 3x  1
, f " x  
x
2
2
1


3
a. Write a second degree Taylor polynomial for f about x = 1
p2 x  
p2 x  
1
2
1
2


1
2
1
2
 x  1 
1/ 2
 x  1 
1
2!
4
 x  1
 x  1
2
2
b. Use the result from (a) to approximate f(1.5)
p 2  1 .5  
1
2

1
2
 1 .5  1 
1
4
 1 .5  1
2
 0 .3 1 2 5
Suppose the function f is defined so that
f  1 
c. If f "  x  
1
1
2
, f '  1 
1
2
, f " x  

2 3x  1
x
R2 x  
2
M
 n  1 !
M
 2  1 !
R2 x   M
2
1


3
for all x in [1, 1.5], find an upper bound for the
approximation error in part b if f "'  x  
Rn x  
2
0 .5
bc

24 x 1  x
x
2

4
0 .5
3
1
2

n 1
1 .5  1
2 1
3
6
R 2  x   0 .4 2 0 9
6
 0 .0 0 8 7 7
The first four derivatives of f  x  
f 'x  
1
1 x
are
1
2 1  x 
f "x  
f "'  x  
f ""  x  
a. Find the third-degree Taylor
approximation to f at x = 0
3/2
3
4 1  x 
5/2
b. Use your answer in (a) to find
an approximation of f(0.5)
15
8 1  x 
7/2
105
1 6 1  x 
9/2
c. Estimate the error involved in the
approximation in (b). Show your
reasoning.
1
The first four derivatives of f  x  
f 'x  
1
2 1  x 
f "x  
f "'  x  
f ""  x  
1 x
a. Find the third-degree Taylor
approximation to f at x = 0
3/2
f 0   1
3
4 1  x 
f ' 0  
5/2
p3  x   1 
15
8 1  x 
are
1
2
1
f " 0  
2
x
3/4
x
3
2!
x 
2
3
4
15 / 8
15
f "'  0  
x
8
3
3!
7/2
p3  x   1 
105
1 6 1  x 
9/2
1
2
8
x 
2
5
x
3
16
b. Use your answer in (a) to find
an approximation of f(0.5)
p 3  0 .5   1 
p 3  0 .5  
1
2
103
128
3
 0 .5    0 .5 
8
2

5
16
 0 .5 
3
The first four derivatives of f  x  
f 'x  
1
2 1  x 
f "x  
f "'  x  
f ""  x  
1 x
3
5/2
R3 x  
15
8 1  x 
7/2
105
1 6 1  x 
are
c. Estimate the error involved in the
approximation in (b). Show your
reasoning.
3/2
4 1  x 
1
9/2
M
 3  1 !
R3 x  
105 / 16
R3 x  
105
 3  1 !
384
0 .5
0 .5  0
0 .5  0
4

3 1
3 1
35
2048
 0 .0 1 7