Question 2: Use the Maclaurin polynomials in example 6.12 part (c) to solve the following. (You can see example 6.14 which is similar). (a) Use the sixth Maclaurin polynomial for cos x to approximate cos() and bound the error. (b) For what values of x does the sixth Maclaurin polynomial approximate cos x to within 0.0001.
Question 2: Use the Maclaurin polynomials in example 6.12 part (c) to solve the following. (You can see example 6.14 which is similar). (a) Use the sixth Maclaurin polynomial for cos x to approximate cos() and bound the error. (b) For what values of x does the sixth Maclaurin polynomial approximate cos x to within 0.0001.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:Question 2:
Use the Maclaurin polynomials in example 6.12 part (c) to solve the following. (You
can see example 6.14 which is similar).
(a) Use the sixth Maclaurin polynomial for cos x to approximate cos() and bound
the error.
(b) For what values of x does the sixth Maclaurin polynomial approximate cos x to
within 0.0001.
![c. For f(x) = cos x, the values of the function and its first four derivatives at x = 0 are given as
follows:
and for n ≥ 0,
f (x)
ƒ' (x)
f" (x)
f" (x)
f(4) (x)
EXAMPLE 6.14
=
=
=
=
COS X
-sin x
-COS X
sin x
=x-
= COS X
P2m (x) = P2m+1 (x)
Solution
a. The fifth Maclaurin polynomial is
Since the fourth derivative is cos x, the pattern repeats. In other words, f(2m) (0) = (-1) and
f(2m+1) = 0 for m > 0. Therefore,
m
= Σ (−1)k (2k)! *
k=0
Po (x) = 1,
P₁ (x) = 1 + 0 = 1,
P₂ (x) = 1 + 0x² = 1 – 2,
P3 (x) = 1 + 0 − 1⁄x² + 0 = 1 − x,
P4 (x) = 1 + 0x² +0+
x¹ = 1 −
Ps (x) = 1 + 0x² +0+
x¹ + 0 = 1−
x4
3
2.
W
1
-1 0 1
-2+
14
R6
N
-1+
-3+
Graphs of the function and the Maclaurin polynomials appear in Figure 6.8.
ƒ (0)
ƒ' (0)
f" (0)
f" (0)
ƒ(4) (0)
= 1 − + ---- + (−1)" (2m)!
Approximating sin x Using Maclaurin Polynomials
From Example 6.12b., the Maclaurin polynomials for sin x are given by
P2m+1 (x) = P2m+2 (x)
[Show/Hide Solution]
p5 (x) = x −
Using this polynomial, we can estimate as follows:
sin () ≈ P5 (8)
3!
P4(x)
=
=
\ P₂(x)
5!
1/17 - ( 18
7!
b. We need to find the values of x such that
Figure 6.8 The function y = cos x and the Maclaurin polynomials Po, P2 and P4 are
plotted on this graph.
3 + 5 − + ··· + (−1)″,
= 0
1
0
= 1.
-1
f(¹) (c)
7
( ² ) = ( ² ) ²
18
7! 18
for m = 0, 1, 2, ....
a. Use the fifth Maclaurin polynomial for sinx to approximate sin () and bound the error.
b. For what values of x does the fifth Maclaurin polynomial approximate sin x to within 0.0001?
7x|7 ≤ 0.0001.
= 秀 - (六) + (六)
≈ 0.173648.
≤9.8 × 10-10.
+ ,
3 4 5
f(x) = cos(x)
+ 4,
Po(x)
+2m+1
(2m+1)!
To estimate the error, use the fact that the sixth Maclaurin polynomial is p6 (x) = p5 (x) and
calculate a bound on R6 (). By Uniqueness of Taylor Series, the remainder is
6x
for some c between 0 and. Using the fact that |ƒ(7) (x)| ≤ 1 for all x, we find that the
magnitude of the error is at most
Solving this inequality for x, we have that the fifth Maclaurin polynomial gives an estimate to
within 0.0001 as long as |x| < 0.907.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F293ec77a-bb0f-4498-a317-90c98547c742%2Fa100bd5d-2c6a-454d-9c52-1253a72089c9%2Ft6rkkv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:c. For f(x) = cos x, the values of the function and its first four derivatives at x = 0 are given as
follows:
and for n ≥ 0,
f (x)
ƒ' (x)
f" (x)
f" (x)
f(4) (x)
EXAMPLE 6.14
=
=
=
=
COS X
-sin x
-COS X
sin x
=x-
= COS X
P2m (x) = P2m+1 (x)
Solution
a. The fifth Maclaurin polynomial is
Since the fourth derivative is cos x, the pattern repeats. In other words, f(2m) (0) = (-1) and
f(2m+1) = 0 for m > 0. Therefore,
m
= Σ (−1)k (2k)! *
k=0
Po (x) = 1,
P₁ (x) = 1 + 0 = 1,
P₂ (x) = 1 + 0x² = 1 – 2,
P3 (x) = 1 + 0 − 1⁄x² + 0 = 1 − x,
P4 (x) = 1 + 0x² +0+
x¹ = 1 −
Ps (x) = 1 + 0x² +0+
x¹ + 0 = 1−
x4
3
2.
W
1
-1 0 1
-2+
14
R6
N
-1+
-3+
Graphs of the function and the Maclaurin polynomials appear in Figure 6.8.
ƒ (0)
ƒ' (0)
f" (0)
f" (0)
ƒ(4) (0)
= 1 − + ---- + (−1)" (2m)!
Approximating sin x Using Maclaurin Polynomials
From Example 6.12b., the Maclaurin polynomials for sin x are given by
P2m+1 (x) = P2m+2 (x)
[Show/Hide Solution]
p5 (x) = x −
Using this polynomial, we can estimate as follows:
sin () ≈ P5 (8)
3!
P4(x)
=
=
\ P₂(x)
5!
1/17 - ( 18
7!
b. We need to find the values of x such that
Figure 6.8 The function y = cos x and the Maclaurin polynomials Po, P2 and P4 are
plotted on this graph.
3 + 5 − + ··· + (−1)″,
= 0
1
0
= 1.
-1
f(¹) (c)
7
( ² ) = ( ² ) ²
18
7! 18
for m = 0, 1, 2, ....
a. Use the fifth Maclaurin polynomial for sinx to approximate sin () and bound the error.
b. For what values of x does the fifth Maclaurin polynomial approximate sin x to within 0.0001?
7x|7 ≤ 0.0001.
= 秀 - (六) + (六)
≈ 0.173648.
≤9.8 × 10-10.
+ ,
3 4 5
f(x) = cos(x)
+ 4,
Po(x)
+2m+1
(2m+1)!
To estimate the error, use the fact that the sixth Maclaurin polynomial is p6 (x) = p5 (x) and
calculate a bound on R6 (). By Uniqueness of Taylor Series, the remainder is
6x
for some c between 0 and. Using the fact that |ƒ(7) (x)| ≤ 1 for all x, we find that the
magnitude of the error is at most
Solving this inequality for x, we have that the fifth Maclaurin polynomial gives an estimate to
within 0.0001 as long as |x| < 0.907.
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