Transcript Section 11.6 - Taylor`s Formula with Remainder

Section 11.6 – Taylor’s Formula with Remainder
The Lagrange Remainder of a Taylor Polynomial
f n1  z  n1
Rn  x  
x
n  1!
where z is some number between x and c
The Error of a Taylor Polynomial
M
n1
Rn  x  
bc
n  1!

f
where M is the maximum value of
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.
1
1/ 4 2 3 / 8 3
p3  x   1  x 
x 
x
2
2!
3!
1
1 2 1 3
p3  x   1  x  x 
x
2
8
16
b. Use your answer to part a to estimate the value of f(0.5)
2
3
1
1
1
p3  0.5   1   0.5    0.5  
0.5 
2
8
16
157
p3  0.5  
 1.227
128
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?
157
p3  0.5  
 1.227
128
M
3 1
R3  x  
bc
3  1!
f
4
x
3 1
R3  x  
bc
 3  1!
R3  x  
6
1
4
0.5  0 
 0.016
64
 4 !
Estimate the error that results when arctan x is replaced by
x3
4
x
if x  0.2 and f    x   4
3
M
3 1
R3  x  
bc
3  1!
4
31
R3  x  
0.2  0
3  1!
R3  x  
1
4
0.2  0.000267
6
Estimate the error that results when ln(x + 1) is replaced by
1
x  x 2 if x  0.1
2
f '''  0.1
21
f  x   ln  x  1
R2  x  
0  0.1
 2  1!
1
f 'x  
R2  x   0.000457
x 1
1
f " x 
2
 x  1
f "'  x  
2
 x  1
3
F ‘’’ (x) has a maximum value at x = -0.1
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.
1
p2  x   x  x 2
2
2
1
p2  0.1   0.1   0.1
2
p2  0.1  0.095
Use a Taylor Polynomial to estimate cos(0.2) to 3 decimal places
x2 x4 x6
cos x  1 


 ...
2! 4! 6!
If x = 0.2, Alternating Series Test works for convergence
 0.2
 2n!
2n
 0.005
0.22
cos x  1 
 0.98
2!
1
sin x
Use a Taylor Polynomial to estimate 
dx with three decimal
x
0
place accuracy.
1
1
1
x3 x5 
 0.005
x


...
dx
0 x  3! 5! 
 2n  1 2n  1!

x2 x4 
0  1  3!  5! ...  dx
1

1
x3
x5
x7
 x  3  3!  5  5!  7  7!  ...  |o


1
1
1
1


 ...
3  3! 5  5! 7  7!
Satisfies Alternating Series Test
1
1
1

 0.946
3  3! 5  5!
Suppose the function f is defined so that
2
2
3x
1
1
1
f 1  , f ' 1  , f "  x  
3
2
2
2
x 1




a. Write a second degree Taylor polynomial for f about x = 1
2
1 1
1/ 2
p2  x     x  1 
x

1
 
2 2
2!
2
1 1
1
p2  x     x  1   x  1
2 2
4
b. Use the result from (a) to approximate f(1.5)
2
1 1
1
p2 1.5    1.5  1  1.5  1  0.3125
2 2
4
Suppose the function f is defined so that
2
2
3x
1
1
1
f 1  , f ' 1  , f "  x  
3
2
2
2
x 1
1
c. If f "  x  
for all x in [1, 1.5], find an upper bound for the
2
24x 1  x 2
approximation error in part b if f "'  x  
4
2
x 1
M
n1
Rn  x  
bc
n  1!
M
21
R2  x  
1.5  1
 2  1!




0.53
R2  x   M
6




0.53
R 2  x   0.4209
 0.00877
6
The first four derivatives of f  x  
f 'x  
1
2 1  x 
f " x  
f "'  x  
f ""  x  
1
1 x
are
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
16 1  x 
9/2
c. Estimate the error involved in the
approximation in (b). Show your
reasoning.
The first four derivatives of f  x  
f 'x  
1
2 1  x 
f " x  
f "'  x  
f ""  x  
3/2
3
4 1  x 
5/2
15
8 1  x 
7/2
105
16 1  x 
9/2
1
1 x
are
a. Find the third-degree Taylor
approximation to f at x = 0
1
3
15
f 0   1 f ' 0  
f " 0  
f "' 0  
2
4
8
1
3 / 4 2 15 / 8 3
p3  x   1  x 
x 
x
2
2!
3!
1
3 2 5 3
p3  x   1  x  x 
x
2
8
16
b. Use your answer in (a) to find
an approximation of f(0.5)
2
3
1
3
5
p3  0.5   1   0.5    0.5    0.5 
2
8
16
103
p3  0.5  
128
The first four derivatives of f  x  
f 'x  
1
2 1  x 
f " x  
f "'  x  
f ""  x  
3/2
3
4 1  x 
5/2
15
8 1  x 
7/2
105
16 1  x 
9/2
1
1 x
are
c. Estimate the error involved in the
approximation in (b). Show your
reasoning.
M
3 1
R3  x  
0.5  0
3  1!
105 /16
31
R3  x  
0.5  0
3  1!
105
35
4
R3  x  
0.5 
 0.017
384
2048