Question

Find root of the equation cos (x) = xe^{x} using
Bisection method. Make calculation for 4 iterations. Choose
x_{l}= 0 and x_{u}= 1. Determine the approximate
error in each iteration. Give the final answer in a tabular
form.

Answer #1

Let f(x) = x^3 + x - 4
a. Show that f(x) has a root on the interval [1,4]
b. Find the first three iterations of the bisection method on f
on this interval
c. Find a bound for the number of iterations needed of bisection
to approximate the root to within 10^-4

Find n for which the nth iteration by the fixed point method is
guaranteed to approximate the root of f(x) = x − cos x on [0, π/3]
with an accuracy within 10−8 using x0 = π/4
Answer: n = 127 iterations or n = 125 iterations.
Please show work to get to answer

Solve the following problem using the MATLAB environment
Write a function [approx_root, num_its] = bisection(f,a,b,tol)
that implements the bisection method. You function should take as
input 4 arguments with the last argument being optional, i.e, if
the user does not provide the accuracy tol use a default of 1.0e-6
(use varargin to attain this). Your function should output the
approximate root, approx_root and the number of iterations it took
to attain the root, num_its. However, if the user calls the...

For the following function, determine the highest real root
of
f(x) = 2x3 – 11.7x2 + 17.7x - 5
by using (a) graphical methods, (b) fixed point iteration (three
iterations, x0 = 3) (Hint: Be certain that you develop a solution
that converges on the root), and (c) Newton-Raphson method (three
iterations, x0 = 3).
Perform an error check on each of your final root approximations
(e.g. for the last of the three iterations).

for f=(x^4)-(6.4*x^3)+(6.45*x^2)+(20.538*x)- 31.752;
find the roots using bisection for five iterations

Find the root of the function: f(x)=2x+sin(x)-e^x,
using Newton Method and initial value of 0. Calculate the
approximate error in each step. Use maximum 4 steps (in case you do
not observe a convergence).

Consider the function
g (x) = 12x + 4 - cos x. Given
g (x) = 0 has a unique solution
x = b in the interval (−1/2, 0), and you can use this
without justification.
(a) Show that Newton's method of starting point
x0
= 0 gives a number sequence with
b <··· <xn+1
<xn
<··· <x1
<x0
= 0
(The word "curvature" should be included in the argument!)
(b) Calculate
x1
and x2.
Use theorem 2 in section...

Consider the function g (x) = 12x + 4 - cos x.
Given g (x) = 0 has a unique solution x =
b in the interval (−1/2, 0), and you can use this without
justification.
(a) Show that Newton's method of starting point
x0 = 0 gives a number sequence with
b <··· <xn+1 <xn <···
<x1 <x0 = 0
(The word "curvature" should be included in the argument!)
(b) Calculate x1 and x2. Use
theorem 2 in section...

Consider the function g (x) = 12x + 4 - cos x.
Given g (x) = 0 has a unique solution x =
b in the interval (−1/2, 0), and you can use this without
justification.
(a) Show that Newton's method of starting point
x0 = 0 gives a number sequence with
b <··· <xn+1 <xn <···
<x1 <x0 = 0
(The word "curvature" should be included in the argument!)
(b) Calculate x1 and x2. Use
theorem 2 in section...

Newton's method: For a function ?(?)=ln?+?2−3f(x)=lnx+x2−3
a. Find the root of function ?(?)f(x) starting with
?0=1.0x0=1.0.
b. Compute the ratio |??−?|/|??−1−?|2|xn−r|/|xn−1−r|2, for
iterations 2, 3, 4 given ?=1.592142937058094r=1.592142937058094.
Show that this ratio's value approaches
|?″(?)/2?′(?)||f″(x)/2f′(x)| (i.e., the iteration converges
quadratically). In error computation, keep as many digits as you
can.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 8 minutes ago

asked 8 minutes ago

asked 10 minutes ago

asked 13 minutes ago

asked 14 minutes ago

asked 17 minutes ago

asked 23 minutes ago

asked 27 minutes ago

asked 32 minutes ago

asked 33 minutes ago

asked 38 minutes ago