1. We want to use a standard fixed point method to compute the real solution of f(x) = 3x³- 5 = 0. (Recall that two complex solutions also exist). A simple plot of f for x = [1.1, 1.3] shows that the real zero exists in this interval. 1.5 1.0 0.5 -0.5 -1.0 1.15 1.20 1.25 1.30 by solving 3 The standard fixed point method manipulates f(x): 0 into g(x): = =x so that the iterative scheme becomes xn+1 = g(xn). The iterative scheme will converge to the required solution if the root is in the interval defined by g'(x)| < 1. = (i) We begin by adding a to both sides of equation (1) to form x 3x3 3x3x 5. Will the root be found? - 5x such that g(x) = = == (ii) We now try adding -kx, k > 0 to both sides of equation (1) to form -kx 3x3 - 5 - kx such that g(x) (3x³ – kx – 5). What values of k will ensure the root is in the interval of convergence? = -1 k (iii) With a tolerance of 10-6 and letting k = 6.5, 10, 18, and 26, write a MATLAB livescript to com- pute the number of steps required for convergence. Plot the output for a sample of seed points, xo, from -5 to 5 in steps of 0.001. Compare and comment on the output.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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1. We want to use a standard fixed point method to compute the real solution of
f(x) = 3x³- 5 = 0.
(Recall that two complex solutions also exist).
A simple plot of f for x = [1.1, 1.3] shows that the real zero exists in this interval.
1.5
1.0
0.5
-0.5
-1.0
1.15
1.20
1.25
1.30
by solving
3
The standard fixed point method manipulates f(x): 0 into g(x):
=
=x so that the iterative scheme
becomes xn+1 = g(xn). The iterative scheme will converge to the required solution if the root is in the
interval defined by g'(x)| < 1.
=
(i) We begin by adding a to both sides of equation (1) to form x 3x3
3x3x 5. Will the root be found?
- 5x such that g(x) =
=
==
(ii) We now try adding -kx, k > 0 to both sides of equation (1) to form -kx 3x3 - 5 - kx
such that g(x) (3x³ – kx – 5). What values of k will ensure the root is in the interval of
convergence?
=
-1
k
(iii) With a tolerance of 10-6 and letting k = 6.5, 10, 18, and 26, write a MATLAB livescript to com-
pute the number of steps required for convergence. Plot the output for a sample of seed points,
xo, from -5 to 5 in steps of 0.001. Compare and comment on the output.
Transcribed Image Text:1. We want to use a standard fixed point method to compute the real solution of f(x) = 3x³- 5 = 0. (Recall that two complex solutions also exist). A simple plot of f for x = [1.1, 1.3] shows that the real zero exists in this interval. 1.5 1.0 0.5 -0.5 -1.0 1.15 1.20 1.25 1.30 by solving 3 The standard fixed point method manipulates f(x): 0 into g(x): = =x so that the iterative scheme becomes xn+1 = g(xn). The iterative scheme will converge to the required solution if the root is in the interval defined by g'(x)| < 1. = (i) We begin by adding a to both sides of equation (1) to form x 3x3 3x3x 5. Will the root be found? - 5x such that g(x) = = == (ii) We now try adding -kx, k > 0 to both sides of equation (1) to form -kx 3x3 - 5 - kx such that g(x) (3x³ – kx – 5). What values of k will ensure the root is in the interval of convergence? = -1 k (iii) With a tolerance of 10-6 and letting k = 6.5, 10, 18, and 26, write a MATLAB livescript to com- pute the number of steps required for convergence. Plot the output for a sample of seed points, xo, from -5 to 5 in steps of 0.001. Compare and comment on the output.
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