Question

why can't we use bisection methods or newton's method for nonconvex functions? x^4+x^3-2x^2-2x especially for this function?

Answer 1

%%Matlab code for finding root using newton secant bisection and false
clear all
close all

%function for which root have to find
fun=@(x) x.^4+x.^3-2.*x.^2-2.*x;
fprintf('For the function f(x)=')
disp(fun)

a=1;b=2;
xx=linspace(1,2);
yy=fun(xx);

plot(xx,yy)
xlabel('x')
ylabel('f(x)')
title('x vs. f(x) plot')

[root]=bisection_method(fun,a,b,1000);
fprintf('\tRoot using Bisection method is %f.\n',root)

[root]=newton_method(fun,a,1000);
fprintf('\tRoot using Newton method is %f.\n',root)

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%Matlab function for Bisection Method
function [root]=bisection_method(fun,x0,x1,maxit)
if fun(x0)<=0
    t=x0;
    x0=x1;
    x1=t;
end
fprintf('\nRoot using Bisection method\n')
%f(x1) should be positive
%f(x0) should be negative
k=10; count=0;
while k>5*10^-10
    count=count+1;
    xx(count)=(x0+x1)/2;
    mm=double(fun(xx(count)));
    if mm>=0
        x0=xx(count);
    else
        x1=xx(count);
    end
    err(count)=abs(fun(x1));
    k=abs(fun(x1));
    if count>=maxit
        break
      
    end
    %
end
fprintf('\tAfter %d iteration root using Bisection method is %f\n',count,xx(count))
root=xx(end);
end

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%Matlab function for Newton Method
function [root]=newton_method(fun,x0,maxit)
syms x
g1(x) =diff(fun,x);   %1st Derivative of this function
xx=x0;            %initial guess]
fprintf('\nRoot using Newton method\n')
%Loop for all intial guesses
    n=5*10^-15; %error limit for close itteration
    for i=1:maxit
        x2=double(xx-(fun(xx)./g1(xx))); %Newton Raphson Formula
        cc=abs(double(fun(x2)));                 %Error
        err(i)=cc;
        xx=x2;
        if cc<=n
            break
        end
      
    end
    fprintf('\tAfter %d iteration root using Newton method is %f\n',i,xx)
    root=xx;
end

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