Question

(1) When using an iterative method such as Newton's Method one can be pretty sure that pn is accurate to k decimal places if the digits in the first k decimal places of pn agree with those in Pn-1. Use Newton's method to approximate ♡2 to ten decimal places. How many iterations did this require? What value of po did you use?

Answer 1

*Matlab code for finding Root using Newton method clear all close all fun=@(x) x.*3-2; fprintf('Function for which root have to find\n') disp(fun) x0=0.5; Initial guess tol-10-10; {maximum iteration [root, iter ]=newton_method(fun, x0, tol); fprintf('Root using Newton method for initial guess &f is $2.15f with iteration count %d.\n\n',x0,root, iter); $Matlab function for Newton Method function (root,iter)=newton_method (fun, x0, tol) syms x gl (x) =diff(fun,x); $1st Derivative of this function xx=x0; Sinitial guess] Loop for all intial guesses n=tol; $error limit for close itteration maxit=1000; for i=1:maxit x2=double(xx-(fun(xx). /g1(xx))); Newton Raphson Formula cc=abs(fun(x2)); SError cc=abs(xx-x2); err(i)=CC; xx=x2; if cc<=n break end end root=xx; iter=i; end 888888888888%88%% End of Code $$$$$$$$$$$$$$$$ Function for which root have to find @(x)x. 3-2 Root using Newton method for initial guess 0.500000 is 1.259921049894873 with iteration count 9. Published with MATLABIO R2018a

%Matlab code for finding Root using Newton method
clear all
close all
fun=@(x) x.^3-2;
fprintf('Function for which root have to find\n')
disp(fun)
x0=0.5; %Initial guess
tol=10-10; %maximum iteration
[root,iter]=newton_method(fun,x0,tol);
fprintf('Root using Newton method for initial guess %f is %2.15f with iteration count %d.\n\n',x0,root,iter);

%Matlab function for Newton Method
function [root,iter]=newton_method(fun,x0,tol)
syms x
g1(x) =diff(fun,x);   %1st Derivative of this function
xx=x0;            %initial guess]
%Loop for all intial guesses
    n=tol; %error limit for close itteration
    maxit=1000;
    for i=1:maxit
        x2=double(xx-(fun(xx)./g1(xx))); %Newton Raphson Formula
        %cc=abs(fun(x2));                 %Error
        cc=abs(xx-x2);
        err(i)=cc;
        xx=x2;
        if cc<=n
            break
        end
      
    end
    root=xx;
    iter=i;
end

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