Does this look right?

xpack · #1 $Old$ October 19th, 2009, 07:49 PM

Use Newton's method to find all roots of the equation correct to the six decimal place.
2cos(x) = 2 - x

This is the newton code our teacher gave us, its all explained here

Code:

function r = newton(f, fp, xg, mx, tol, vu)
% NEWTON: Finds zero of a function via Newton's method
% Syntax: newton(f, fp, xg, mx, tol, vu)
%   f   - function for which a zero [f(x)=0] is sought;
%           use function handles: @f, @fp, etc.
%   fp  - function representing derivative f'(x)
%   xg  - initial guess for zero
%   mx  - maximum number of iterates
%   tol - tolerance (how close last two iterates
%            must be to declare convergence)
%   vu  - 1=view iterates, 0=suppress viewing
fprintf('Newton''s Method\n\n')
xn = xg; ae = 2*tol;
fprintf('initial guess: x = %+13.9f\n', xn)
for k = 1:mx
    xo = xn;
    fo = feval(f, xo);
    fpo = feval(fp, xo);
    xn = xo - fo/fpo;
    ae = abs(xn - xo);
    if vu
        fprintf('x = %+13.9f\n', xn)
    end
    if ae < tol
        break
    end
end
if ae < tol
    disp('Converged!')
    fprintf('solution: %+13.9f, #iterations: %i\n\n', xn, k)
else
    disp('Failed to converge!')
    fprintf('last iterate: %+13.9f, #iterations: %i\n\n', xn, mx)
end
r = xn;

This is what I got

Code:

delete s232x17.txt; diary s232x17.txt
clear; clc; close all; echo on
%
% Stewart 232/17
%
f=@(x) 2*cos(x)+x-2
fp=@(x) 1-2*sin(x)
x=[0:.01:2];
y=f(x);
plot(x,y)
hold on
plot(x,0)
format long
newton(f,fp,1.5,10,1e-6,1)

%answer 1.109144

%
echo off; diary off

Can someone tell me if this is correct and if not, what am I doing wrong. Thank you

CaptainBlack · #2 $Old$ October 20th, 2009, 01:49 AM

Quote:

Originally Posted by xpack $View Post$

Use Newton's method to find all roots of the equation correct to the six decimal place.
2cos(x) = 2 - x

This is the newton code our teacher gave us, its all explained here

Code:

function r = newton(f, fp, xg, mx, tol, vu)
% NEWTON: Finds zero of a function via Newton's method
% Syntax: newton(f, fp, xg, mx, tol, vu)
%   f   - function for which a zero [f(x)=0] is sought;
%           use function handles: @f, @fp, etc.
%   fp  - function representing derivative f'(x)
%   xg  - initial guess for zero
%   mx  - maximum number of iterates
%   tol - tolerance (how close last two iterates
%            must be to declare convergence)
%   vu  - 1=view iterates, 0=suppress viewing
fprintf('Newton''s Method\n\n')
xn = xg; ae = 2*tol;
fprintf('initial guess: x = %+13.9f\n', xn)
for k = 1:mx
    xo = xn;
    fo = feval(f, xo);
    fpo = feval(fp, xo);
    xn = xo - fo/fpo;
    ae = abs(xn - xo);
    if vu
        fprintf('x = %+13.9f\n', xn)
    end
    if ae < tol
        break
    end
end
if ae < tol
    disp('Converged!')
    fprintf('solution: %+13.9f, #iterations: %i\n\n', xn, k)
else
    disp('Failed to converge!')
    fprintf('last iterate: %+13.9f, #iterations: %i\n\n', xn, mx)
end
r = xn;

This is what I got

Code:

delete s232x17.txt; diary s232x17.txt
clear; clc; close all; echo on
%
% Stewart 232/17
%
f=@(x) 2*cos(x)+x-2
fp=@(x) 1-2*sin(x)
x=[0:.01:2];
y=f(x);
plot(x,y)
hold on
plot(x,0)
format long
newton(f,fp,1.5,10,1e-6,1)
 
%answer 1.109144
 
%
echo off; diary off

Can someone tell me if this is correct and if not, what am I doing wrong. Thank you

You can check this by substituting the final solution back into $f(x)$ . In this case:

$f(1.109144)=2*\cos(1.109144)+1.109144-2\approx 1.4\times 10^{-7}$

which looks pretty good.

(Also the code looks good as well)

The only problem is that the question asks for all of the roots, there appear to be three near $x=0,\ 1.1,\ 3.7$ (you can find these approximatly by plotting $f(x)$

CB

xpack · #3 $Old$ November 1st, 2009, 03:18 PM

Oh okay, I got it, thank you very much!