finding root using false position method
120 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Good evening\morning
I try to write a code that calculate the root of a nonlinear function using False Position Method, but I get an infinite loop. I use the same loop for the Bisection Method and it's work.
clc
x0 = input('enter the value of x0 = ');
x1 = input('enter the value of x1 = ');
tolerance=input('inter the tolerance = ');
f =@(x) sin(2*pi*x)+ exp(1.2*x) + x - 2.5;
for i=0:inf
x2= x1 - (f(x1)* (x1-x0)/(f(x1)-f(x0)))
c = f(x2)
absolute_c= abs(c);
if absolute_c < tolerance
break
end
if f(x0)*c <0
x1=x2;
continue
else
x0=x2;
continue
end
end
i
1 commentaire
Samanta
le 7 Fév 2024
1.Use the False Position Method to find the root of the equation e^x−3x=0. Start with the initial guesses x_0=0 and x_1 =1
Réponse acceptée
Fangjun Jiang
le 22 Nov 2011
Do a plot to find out the curve. If you put the right initial value, it could solve the problem.
ezplot(f)
x0=-6
x1=6
Tolerance=0.001
It reached the end at i==590
A better approach is to check whether f(x0)*f(x1)<0 right after the input().
2 commentaires
Fangjun Jiang
le 22 Nov 2011
A better approach is to check whether f(x0)*f(x1)<0 right after the input().
Plus de réponses (4)
Polash Roy
le 10 Avr 2021
Modifié(e) : Walter Roberson
le 7 Fév 2024
clc
clear all
close all
f=@(r) exp((-5e-3)*r)*cos((sqrt(2000-.01*r^2)*.05))-.01;
a=100;
b=550;
for i=1:10
x0=a;
x1=b;
fprintf('\n Hence root lies between (%.4f,%.0f)',a,b)
x2(i)=x0-(x1-x0)/(f(x1)-f(x0))*f(x0);
if f(x2(i))*f(x0)>0
b=x2(i);
else
a=x2(i);
end
fprintf('\n Therefore, x2=%.4f \n Here, f(x20=%.4f',x2(i),f(x2(i)))
p=x2(i);
end
for i=1:10
eror(i)=p-x2(i);
end
Answer=p
plot(eror)
grid on;
title('Plot of error')
xlabel('iterations')
ylabel('Error')
0 commentaires
Aman Pratap Singh
le 3 Déc 2021
clc
% Setting x as symbolic variable
syms x;
% Input Section
y = input('Enter non-linear equations: ');
a = input('Enter first guess: ');
b = input('Enter second guess: ');
e = input('Tolerable error: ');
% Finding Functional Value
fa = eval(subs(y,x,a));
fb = eval(subs(y,x,b));
% Implementing Bisection Method
if fa*fb > 0
disp('Given initial values do not bracket the root.');
else
c = a - (a-b) * fa/(fa-fb);
fc = eval(subs(y,x,c));
fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');
while abs(fc)>e
fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);
if fa*fc< 0
b =c;
fb = eval(subs(y,x,b));
else
a =c;
fa = eval(subs(y,x,a));
end
c = a - (a-b) * fa/(fa-fb);
fc = eval(subs(y,x,c));
end
fprintf('\nRoot is: %f\n', c);
end
1 commentaire
Walter Roberson
le 3 Déc 2021
Modifié(e) : Walter Roberson
le 3 Déc 2021
You should never eval() a symbolic expression. Symblic expressions are not character vectors containing MATLAB code. Sometimes they look like MATLAB code, but they contain parts that are not MATLAB, and some of the functions have a different parameter order than MATLAB uses.
If you have fully substituted for all numeric variables, then
fa = double(subs(y,x,a));
fb = double(subs(y,x,b));
Samanta
le 7 Fév 2024
Use the False Position Method to find the root of the equation e^x−3x=0. Start with the initial guesses x_0=0 and x_1 =1
1 commentaire
Walter Roberson
le 7 Fév 2024
This solution does not explain how to use False Position Method, so it is not clear how it answers the question?
Voir également
Catégories
En savoir plus sur Symbolic Math Toolbox dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!