newtons method using function

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Catherine Chen
Catherine Chen 2020-2-29
回答： David Hill 2020-2-29
I have been testing the code but it does not seem right. could you please tell me where the issue occurs and how to correct it?
tnx.
function [x1] = tutorial1(x0,nMax,tol)
% Calculate the root of the function f(x) = x^3 - 3^x + 1
% using the Newton Method of root-finding.
% Inputs:
% - x0 initial guess
% - nMax number of iterations
% - tol solution accuracy tolerance
% Output:
% - x1 converged root of the equation
x0 = 1.5;
nMax = 15;
tol = 1e-4;
for i = 1 : nMax
fx= (x0)^3-3^(x0)+1;
df= 3*(x0)^2-3*(x0)*log(3);
x1 = x0 - fx/df;
fx1 = (x1)^3-3^(x1)+1;
err = abs(fx1-fx);
if err < tol
break
end
end
% Sample output code for monitoring (this should be included in your loop structure.
fprintf('Iteration = %d, x0 = %.4f, x1 = %.4f, fx1 = %.4f\n',i,x0,x1,fx1);
return

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回答（1 个）

David Hill
David Hill 2020-2-29
function x = tutorial1(x,nMax,tol)%it would be easier if everything was just x
f=@(x)x^3-3^x+1;%should place functions outside your loop
df=@(x)3*x^2-3;%not sure why your equation had a log(3)
for i = 1 : nMax
x = x - f(x)/df(x);
fprintf('Iteration = %d, x1 = %.4f, f(x) = %.4f\n',i,x,f(x));
if abs(f(x)) < tol%I assume you are measuring error with respect to the zero root
return;
end
end

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