Error in line 12. I cant use 'inline' command. How to fix it? There are 3 iterations. Equation is f(x)= e^-x-x and the derivative is -e^-x-1. Using Newton Raphson Method

7 次查看（过去 30 天）
Christian
Christian 2022-11-15
编辑： Stephen23 2022-11-16
%%%% MATLAB CODE of Newton Raphson Method
%%% Method for finding the ROOT of
%%%% equation f(x)= exp(1)^-x-x
format short
clear
clc
syms x
% Write the function here
% f= @ (x) exp(1)^-x-x;
f = @ (x) exp(1)^-x-x;
df = diff (f,x);
dfx = inline(df); %derivative of funtion df
x0 = 0; %initial guess
n = 4; % number of decimal places
Variables={'Iter','x','f_x0','Error'};
iter = 1;
err = abs (f(x0));
epsilon = 5*10^(-n-1);
itermax = 70;
HG = [];
if dfx(x0)<10^(-9)
disp('Wrong choice of initial Guess');
else
while (iter<=itermax && err>epsilon)
x1 = x0-f(x0)/dfx(x0);
err = abs (f(x0));
HG = [iter x0 f(x0) err];
iter = iter+1;
x0 = x1;
end
end
disp ('======================================')
disp ('Output Table with Iteration wise')
Result= array2table(HG);
Result.Properties.VariableNames(1:size(HG,2)) = Variables
x0 = x0-rem(x0,10^-n);
fprintf('Converged solution after %d iterations \n',iter);
fprintf('Root is %1.5f \n',x0)

请先登录，再回答此问题。

回答（1 个）

Askic V
Askic V 2022-11-15
编辑：Askic V 2022-11-15
Ran in:
Try this:
format short
clear
clc
syms x
% Write the function here
% f= @ (x) exp(1)^-x-x;
f = @ (x) exp(1)^-x-x;
df = diff (f,x);
dfx = @(x) eval(df);
%inline(df); %derivative of funtion df
% inline will be removed in future releases
x0 = 0; %initial guess
n = 4; % number of decimal places
Variables={'Iter','x','f_x0','Error'};
iter = 1;
err = abs (f(x0));
epsilon = 5*10^(-n-1);
itermax = 70;
HG = [];
% Use abs function here!
if abs ( dfx(x0) ) < 10^(-9)
disp('Wrong choice of initial Guess');
else
while (iter<=itermax && err>epsilon)
x1 = x0-f(x0)/dfx(x0);
err = abs (f(x0));
HG = [iter x0 f(x0) err];
iter = iter+1;
x0 = x1;
end
end
disp ('======================================')
======================================
disp ('Output Table with Iteration wise')
Output Table with Iteration wise
Result= array2table(HG);
Result.Properties.VariableNames(1:size(HG,2)) = Variables
Result = 1×4 table
Iter x f_x0 Error ____ _______ __________ __________ 4 0.56714 1.9648e-07 1.9648e-07
x0 = x0-rem(x0,10^-n);
fprintf('Converged solution after %d iterations \n',iter);
Converged solution after 5 iterations
fprintf('Root is %1.5f \n',x0)
Root is 0.56710
  2 个评论
Askic V
Askic V 2022-11-15
The number of iterations generally depends of initial guess and the epsilon.
In your while loop the stop criteria is maximum iteration reached or error is below predifed epsilon. If you want 3 iterations, the easiest way (for the same initial guess 0) is to make epsilon larger, such as:
epsilon = 5*10^(-n+1);
This would produce the following result:
Iter x f_x0 Error
____ _______ _________ _________
3 0.56631 0.0013045 0.0013045

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Number Theory 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by