Consider the equation f (x) = 4e-*sinx – 1 = 0, a) Please justify if the bisection method can be used to determine the roots in the intervals [0, 1] nd [1, 2] respectively. If so, how many iterations are needed to satisfy the error bound of 10-6; b) Compute 2 iterations of the Secant method including the estimated error with x, = 0,x1 = 1. :) To solve the given problem, a Matlab code is generated. Please specify the method mplemented and the output for n=1.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question
3.
Consider the equation f (x) = 4e-*sinx – 1 = 0,
(a) Please justify if the bisection method can be used to determine the roots in the intervals [0, 1]
and [1, 2] respectively. If so, how many iterations are needed to satisfy the error bound of 106;
(b) Compute 2 iterations of the Secant method including the estimated error with x, = 0,x1 = 1.
%3|
(c) To solve the given problem, a Matlab code is generated. Please specify the method
implemented and the output for n=1.
clear all
syms x
f(x)=4*sin (x) *exp(-1*x)-1;
df (x) =diff (f (x));
x0=0;
et=10^-6;
nmax=20;
es=1;
n=0;
while es>et && n<nmax
x1=x0-eval (f(x0))/eva1(df (x0));
es=x1-x0;
х0-х1;
n=n+1;
end
disp('Root:')
disp (x1)
disp ('Number of the iterations')
disp (n)
Transcribed Image Text:3. Consider the equation f (x) = 4e-*sinx – 1 = 0, (a) Please justify if the bisection method can be used to determine the roots in the intervals [0, 1] and [1, 2] respectively. If so, how many iterations are needed to satisfy the error bound of 106; (b) Compute 2 iterations of the Secant method including the estimated error with x, = 0,x1 = 1. %3| (c) To solve the given problem, a Matlab code is generated. Please specify the method implemented and the output for n=1. clear all syms x f(x)=4*sin (x) *exp(-1*x)-1; df (x) =diff (f (x)); x0=0; et=10^-6; nmax=20; es=1; n=0; while es>et && n<nmax x1=x0-eval (f(x0))/eva1(df (x0)); es=x1-x0; х0-х1; n=n+1; end disp('Root:') disp (x1) disp ('Number of the iterations') disp (n)
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