Can't get a proper solution with modified secant method

10 vues (au cours des 30 derniers jours)
Ishmum Monjur Nilock
Ishmum Monjur Nilock le 16 Août 2020
Hi all, I have built a code for modified secant method to find the root of the function f(x)= which is given below. The output of the code should give an answer of about 0.5671 but instead it is giving 56.71.... something answer. HOw to solve this problem? Code is mentioned below:
clear all
close all
clc
n=100;
f=@(x) exp(-x)-x;
dx=0.01;
x0=1;
x(1)=x0;
i=0;
err=0.0048;
for i=1:n
x(i+1)=x(i)-((f(x(i)*dx)/(f(x(i)+dx)-f(x(i)))));
if abs((x(i+1)-x(i))/x(i+1))*100<err
roots=x(i);
break;
end
end
disp(roots)

Réponse acceptée

Walter Roberson
Walter Roberson le 16 Août 2020
x(i+1)=x(i)-((f(x(i)*dx)/(f(x(i)+dx)-f(x(i)))));
Notice that you have f(x(i)*dx) and dx is 0.01 . When you find the root of f(x(i)/100) == 0, then x(i) would be 100 times the root of f(x)
Meanwhile in the denominator you have f(x(i)+dx) with +dx not *dx . It seems to me you need to be consistent

Plus de réponses (1)

Peter
Peter le 9 Mar 2023
clear all
close all
clc
n=100;
f=@(x) exp(-x)-x;
dx=0.01;
x0=1;
x(1)=x0;
i=0;
err=0.0048;
for i=1:n
x(i+1)=x(i)-((f(x(i)*dx)/(f(x(i)+dx)-f(x(i)))));
if abs((x(i+1)-x(i))/x(i+1))*100<err
roots=x(i);
break;
end
end
disp(roots)

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