Find Taylor series approximations using n=1, 2, and 4 to the function f(x) = 1 + eª at x = 0.6, using the expansion point x = 1. For each n, compute the approximation, find an upper bound on the error using Taylor's theorem, and compare it to the actual error (see chapter 4, example 15).
Find Taylor series approximations using n=1, 2, and 4 to the function f(x) = 1 + eª at x = 0.6, using the expansion point x = 1. For each n, compute the approximation, find an upper bound on the error using Taylor's theorem, and compare it to the actual error (see chapter 4, example 15).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:4. Find Taylor series approximations using n=1, 2, and 4 to the function f(x) = 1 + eª at
X = 0.6, using the expansion point xo 1. For each n, compute the approximation, find an
upper bound on the error using Taylor's theorem, and compare it to the actual error (see
chapter 4, example 15).
=
![Example 16. Use Taylor's theorem with n = 2 to approximate sin(x) at x = 1.1, using
the expansion point xo = 1. Find a bound on the error using Taylor's theorem and com-
pare the bound to the actual error.
First, we calculate:
f(1) = sin(1),
f'(1) = cos(1), f" (1) =
sin(1).
Thus we can approximate f by: f(x) = sin(1) + cos(1)(x − 1) – sin(1)(x - 1)² with error
term cos(c)(x - 1)³, where c is not known precisely, but is known to live in the interval
[1, 1.1]. Hence at x = 1.1, we get the approximation
f(1.1)
sin(1) + (0.1) cos(1) - (0.01)-
The error term was given by - cos(c)(x - 1)³, with c unknown but in the interval [1, 1.1].
We know that | cos(x)| ≤ 1, so we can upper bound the error by
cos(c)
6
error:
(1.11)
1-1)³| ≤ (0.001):
sin(1)
2
Using MATLAB, we calculate this approximation, the actual answer, and the actual
| co
cos(c)
6
This is now quite close to the actual error.
(0.001) 1.6667e - 4.
(1.1-1)³| ≤ 3
<
=
=
>> approx-sin (1) + (0.1) *cos (1)-0.01/2*sin (1);
>> exact=sin (1.1);
.55
>> error approx-exact;
>> disp([approx, exact, error]);
0.891293860470671 0.891207360061435 0.000086500409236
0.891293860470671.
Note the actual error was smaller than the estimated error, but the estimated er-
ror was a guarantee of what the maximum error can be. Note that cos(1.1) ≈ 0.45 ≤
cos(c) ≤ 0.55= cos(1), so we can improve the error bound by using | cos(c)| ≤ 0.55,
which gives the improved upper bound on the error estimate:
6
-(0.001) 9.1667e - 5.
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Transcribed Image Text:Example 16. Use Taylor's theorem with n = 2 to approximate sin(x) at x = 1.1, using
the expansion point xo = 1. Find a bound on the error using Taylor's theorem and com-
pare the bound to the actual error.
First, we calculate:
f(1) = sin(1),
f'(1) = cos(1), f" (1) =
sin(1).
Thus we can approximate f by: f(x) = sin(1) + cos(1)(x − 1) – sin(1)(x - 1)² with error
term cos(c)(x - 1)³, where c is not known precisely, but is known to live in the interval
[1, 1.1]. Hence at x = 1.1, we get the approximation
f(1.1)
sin(1) + (0.1) cos(1) - (0.01)-
The error term was given by - cos(c)(x - 1)³, with c unknown but in the interval [1, 1.1].
We know that | cos(x)| ≤ 1, so we can upper bound the error by
cos(c)
6
error:
(1.11)
1-1)³| ≤ (0.001):
sin(1)
2
Using MATLAB, we calculate this approximation, the actual answer, and the actual
| co
cos(c)
6
This is now quite close to the actual error.
(0.001) 1.6667e - 4.
(1.1-1)³| ≤ 3
<
=
=
>> approx-sin (1) + (0.1) *cos (1)-0.01/2*sin (1);
>> exact=sin (1.1);
.55
>> error approx-exact;
>> disp([approx, exact, error]);
0.891293860470671 0.891207360061435 0.000086500409236
0.891293860470671.
Note the actual error was smaller than the estimated error, but the estimated er-
ror was a guarantee of what the maximum error can be. Note that cos(1.1) ≈ 0.45 ≤
cos(c) ≤ 0.55= cos(1), so we can improve the error bound by using | cos(c)| ≤ 0.55,
which gives the improved upper bound on the error estimate:
6
-(0.001) 9.1667e - 5.
=
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