Question

Study the roots of the nonlinear equation f(x) = cos(x) + (1 /(1 + e^2x)) both theoretically and numerically. (a) Plot f(x) on the interval x ∈ [−15, 15] and describe the overall behaviour of the function as well as the number and location of its roots. Use the “zoom” feature of Matlab’s plotting window (or change the axis limits) in order to ensure that you are identifying all roots – you may have to increase your plotting point density in order to see sufficient detail!

Please provide MATLAB code as well.

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) cos(x)+(1./(1+exp(2*x)));
fprintf('function for which root have to find f(x)=')
disp(fun)
xx=linspace(-15,15,100001);
yy=fun(xx);

plot(xx,yy)
xlabel('x')
ylabel('f(x)')
title('x vs. f(x) plot')
grid on
x0=-10;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=-5;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=2;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=5;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=8;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=10;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

x0=13;
[root]=newton_method(fun,x0,1000);
hold on
plot(root,fun(root),'r*')

%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 for initial guess %f\n',x0)
%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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