(Fixed-Point Iteration). Consider the problem of finding the root of the function f(x) = 3.294a – cos(x?) in [0, 1]. Observe that cos(a?) 3.294x – cos(æ) = 0 + x = 3.294 on (0, 1), and thus the problem is reduced to a fixed-point problem for the function cos(22) g(x) 3.294 on [0, 1]. Find (for your own use) the first g' (æ) and the second derivative g" (x) of the function g(æ). (i) We claim that the iteration function g(x) takes the interval 0, 1| into itself. Indeed, the derivative g' (x) of the function g(x) is negative positive at every point x € (0, 1). In effect, the function g(x) is strictly increasing decreasing on [0, 1], and hence its minimum value on [0, 1] is equal to g(0) g(1) and its maximum value on [0, 1] is equal to g(0) g(1) We complete the argument by observing that (here and below, round your result to a 7-digit floating point number) g(0) = and g(1) = which demonstrates that 9([0, 1) C [0, 1], as imed (iii) Next, by analyzing the second derivative g" (a) on (0, 1), we see that for every x E (0,1), |g' (x)| < |g'(7), where %3D and |9'(2) =
(Fixed-Point Iteration). Consider the problem of finding the root of the function f(x) = 3.294a – cos(x?) in [0, 1]. Observe that cos(a?) 3.294x – cos(æ) = 0 + x = 3.294 on (0, 1), and thus the problem is reduced to a fixed-point problem for the function cos(22) g(x) 3.294 on [0, 1]. Find (for your own use) the first g' (æ) and the second derivative g" (x) of the function g(æ). (i) We claim that the iteration function g(x) takes the interval 0, 1| into itself. Indeed, the derivative g' (x) of the function g(x) is negative positive at every point x € (0, 1). In effect, the function g(x) is strictly increasing decreasing on [0, 1], and hence its minimum value on [0, 1] is equal to g(0) g(1) and its maximum value on [0, 1] is equal to g(0) g(1) We complete the argument by observing that (here and below, round your result to a 7-digit floating point number) g(0) = and g(1) = which demonstrates that 9([0, 1) C [0, 1], as imed (iii) Next, by analyzing the second derivative g" (a) on (0, 1), we see that for every x E (0,1), |g' (x)| < |g'(7), where %3D and |9'(2) =
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
Numerical analysis
4
![(Fixed-Point Iteration). Consider the problem of finding the root of the
function
f(x) = 3.294a – cos(x?)
in [0, 1]. Observe that
cos(2')
3.294x – cos(æ²) = 0 + x =
3.294
on 0, 1), and thus the problem is reduced to a fixed-point problem for
the function
cos (2)
9(x) =
3.294
on [0, 1]. Find (for your own use) the first g' (x) and the second
derivative g" (x) of the function g(x).
(i) We claim that the iteration function g(x) takes the interval (0, 1|
into itself. Indeed, the derivative g' (x) of the function g(x) is
negative
positive
at every point x € (0, 1). In effect, the function g(x) is strictly
increasing
decreasing
on [0, 1], and hence its minimum value on [0, 1] is equal to
g(0)
9(1)
and its maximum value on 0, 1 is equal to
g(0)
g(1)
We complete the argument by observing that (here and below, round
your result to a 7-digit floating point number)
g(0) =
and
g(1) =
which demonstrates that
g(10, 1) 드 [0, 1],
as claimed
(iii) Next, by analyzing the second derivative g" (r) on (0, 1), we see
that for every x E (0,1),
|g' (2)| < |g'(7)),
where
%3D
and
lg'(?D =
(iv) Accordingly, by (ii) and (iii), the corresponding fixed-point iteration
. (enter a correct word)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c909ed8-88db-4eee-8878-ac3c652129b1%2F2ca85962-cbc9-4e9c-be08-89f16f2a4e73%2Fn4tsfd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(Fixed-Point Iteration). Consider the problem of finding the root of the
function
f(x) = 3.294a – cos(x?)
in [0, 1]. Observe that
cos(2')
3.294x – cos(æ²) = 0 + x =
3.294
on 0, 1), and thus the problem is reduced to a fixed-point problem for
the function
cos (2)
9(x) =
3.294
on [0, 1]. Find (for your own use) the first g' (x) and the second
derivative g" (x) of the function g(x).
(i) We claim that the iteration function g(x) takes the interval (0, 1|
into itself. Indeed, the derivative g' (x) of the function g(x) is
negative
positive
at every point x € (0, 1). In effect, the function g(x) is strictly
increasing
decreasing
on [0, 1], and hence its minimum value on [0, 1] is equal to
g(0)
9(1)
and its maximum value on 0, 1 is equal to
g(0)
g(1)
We complete the argument by observing that (here and below, round
your result to a 7-digit floating point number)
g(0) =
and
g(1) =
which demonstrates that
g(10, 1) 드 [0, 1],
as claimed
(iii) Next, by analyzing the second derivative g" (r) on (0, 1), we see
that for every x E (0,1),
|g' (2)| < |g'(7)),
where
%3D
and
lg'(?D =
(iv) Accordingly, by (ii) and (iii), the corresponding fixed-point iteration
. (enter a correct word)
![towards the fixed point of the function g(x) in [0, 1].
(ii) Use the fixed-point iteration to find an approximation PN of the
fixed-point p of the function g(x) in [0, 1], the root of the function
f(x) in [0, 1], satisfying RE(pN PN-1) < 10-6 by taking
Po = 1 as the initial approximation. All calculations are to be carried
out in the FPA7.
(a) Enter the terms
P1, P2, . .., PN
you will generate in the process of obtaining the required
approximation in the corresponding input fields below; type the
asterisks * in the input fields for the terms Pk with k > N.
Pi =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
Ps =
Pg =
(b) The stopping criterion has been triggered at the N-th step, since
the corresponding relative error (round the value to a 7-digit floating-
point number)
RE(PN - PN-1) =
became less than the tolerance for the first time.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c909ed8-88db-4eee-8878-ac3c652129b1%2F2ca85962-cbc9-4e9c-be08-89f16f2a4e73%2Fncqlhjs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:towards the fixed point of the function g(x) in [0, 1].
(ii) Use the fixed-point iteration to find an approximation PN of the
fixed-point p of the function g(x) in [0, 1], the root of the function
f(x) in [0, 1], satisfying RE(pN PN-1) < 10-6 by taking
Po = 1 as the initial approximation. All calculations are to be carried
out in the FPA7.
(a) Enter the terms
P1, P2, . .., PN
you will generate in the process of obtaining the required
approximation in the corresponding input fields below; type the
asterisks * in the input fields for the terms Pk with k > N.
Pi =
P2 =
P3 =
P4 =
P5 =
P6 =
P7 =
Ps =
Pg =
(b) The stopping criterion has been triggered at the N-th step, since
the corresponding relative error (round the value to a 7-digit floating-
point number)
RE(PN - PN-1) =
became less than the tolerance for the first time.
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