fprintf("\nApproximate solution of %s is xn= %11.9f on %dst iterations \n\n',y, p, i); elseif i == 2 fprintf('\nApproximate solution of %s is xn= %11.9f on %dnd iterations \n\n',y, p, i); else fprintf('\nApproximate solution of %s is xn= %11.9f on %dth iterations \n\n',y, p, i); end break; else i = i+1; pO = p1; fO = f1; p1 = p; f1 = fp; end end Run the program and enter the following. Enter the given function: cos(x) Enter 1st approximation, p0: 0 Enter 2nd approximation, p1: pi Enter no. of iterations, n: 20 Enter tolerance, tol: 0.0001 Write the output below:
fprintf("\nApproximate solution of %s is xn= %11.9f on %dst iterations \n\n',y, p, i); elseif i == 2 fprintf('\nApproximate solution of %s is xn= %11.9f on %dnd iterations \n\n',y, p, i); else fprintf('\nApproximate solution of %s is xn= %11.9f on %dth iterations \n\n',y, p, i); end break; else i = i+1; pO = p1; fO = f1; p1 = p; f1 = fp; end end Run the program and enter the following. Enter the given function: cos(x) Enter 1st approximation, p0: 0 Enter 2nd approximation, p1: pi Enter no. of iterations, n: 20 Enter tolerance, tol: 0.0001 Write the output below:
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
Related questions
Question
Transcribed Image Text:fprintf("\nApproximate solution of %s is xn=
%11.9f on %dst iterations \n\n',y, p, i);
elseif i
== 2
fprintf("\nApproximate solution of %s is xn=
%11.9f on %dnd iterations \n\n',y, p, i);
else
fprintf("\nApproximate solution of %s is xn=
%11.9f on %dth iterations \n\n',y, p, i);
end
break;
else
i = i+1;
p0 = p1;
f0 = f1;
p1 = p;
f1 = fp;
%3D
%3D
end
end
Run the program and enter the following.
Enter the given function: cos(x)
Enter 1st approximation, p0: 0
Enter 2nd approximation, p1: pi
Enter no. of iterations, n: 20
Enter tolerance, tol: 0.0001
Write the output below:
Transcribed Image Text:SECANT METHOD
EXAMPLE 2: Find the root of the function f(x) = ex
-2x +1 using SM
e-
Input the following codes to your MATLAB edit window
% Secant Algorithm
syms x
%Input section
y = input('Enter the given function: ');
p0 = input('Enter 1st approximation, p0: ');
p1 = input('Enter 2nd approximation, p1: ');
n = input('Enter no. of iterations, n: ');
tol = input('Enter tolerance, tol: ');
i = 2;
fO = eval(subs(y,x,p0));
f1 = eval(subs(y,x,p1));
while i <= n
p = p1-(f1* (p1-p0))/(f1-f0);
fp = eval(subs(y,x,p));
if abs(p-p1) < tol
if i == 3
ニニ
fprintf('\nApproximate solution of %s is xn=
%11.9f on %drd iterations \n\n',y, p, i);
elseif i
==
fprintf('\nApproximate solution of %s is xn=
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