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MATLAB CODE NEWTON METHOD

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Adomas Bazinys
Adomas Bazinys on 5 Mar 2018
f = @(x) exp(x)-3*x;
fp = @(x) exp(x)-3;
x0 = 1;
N = 10;
tol = 1E-10;
x(1) = x0; % Set initial guess
n = 2;
nfinal = N + 1;
while (n <= N + 1)
fe = f(x(n - 1));
fpe = fp(x(n - 1));
x(n) = x(n - 1) - fe/fpe;
if (abs(fe) <= tol)
nfinal = n;
break;
end
n = n + 1;
end
plot(0:nfinal - 1,x(1:nfinal),'o-')
title('Solution:')
xlabel('Iterations')
ylabel('X')
I have a few questions about my code. What is happening with variable n during execution? And how could I make my code printing an iteration table? I can see iterations in the graphic but I need to print a table also.

Answers (5)

Geoff Hayes
Geoff Hayes on 5 Mar 2018
Adomas - your code is using n as an index into x. On each iteration of the loop, you increment n by one in preparation for the next iteration. n will be the length of your array x and so will tell you how many iterations have occurred until the tolerance has been satisfied (or until the maximum N has been reached).
As for displaying the data in a table, do you mean your x? Something like
table([1:length(x)]',x']
I'm using the ' so that the data is transposed (into a column).

Jan
Jan on 5 Mar 2018
What is happening with variable n during execution?
It is increased by 1 in each iteration. It counts the number of iterations.
how could I make my code printing an iteration table?
Insert a command to display the current values:
...
x(n) = x(n - 1) - fe/fpe;
fprintf('%3d: %20g %20g\n', n, x(n), abs(fe));
if (abs(fe) <= tol)
...

Mohamed Elsayed Abo Heiba
Mohamed Elsayed Abo Heiba on 18 Oct 2019
f = @(x) exp(x)-3*x;
fp = @(x) exp(x)-3;
x0 = 1;
N = 10;
tol = 1E-10;
x(1) = x0; % Set initial guess
n = 2;
nfinal = N + 1;
while (n <= N + 1)
fe = f(x(n - 1));
fpe = fp(x(n - 1));
x(n) = x(n - 1) - fe/fpe;
if (abs(fe) <= tol)
nfinal = n;
break;
end
n = n + 1;
end
plot(0:nfinal - 1,x(1:nfinal),'o-')
title('Solution:')
xlabel('Iterations')
ylabel('X')

Hamza saeed  khan
Hamza saeed khan on 24 Nov 2020
f = @(x) exp(x)-3*x;
fp = @(x) exp(x)-3;
x0 = 1;
N = 10;
tol = 1E-10;
x(1) = x0; % Set initial guess
n = 2;
nfinal = N + 1;
while (n <= N + 1)
fe = f(x(n - 1));
fpe = fp(x(n - 1));
x(n) = x(n - 1) - fe/fpe;
if (abs(fe) <= tol)
nfinal = n;
break;
end
n = n + 1;
end
plot(0:nfinal - 1,x(1:nfinal),'o-')
title('Solution:')
xlabel('Iterations')
ylabel('X')
table([1:length(x)]',x']
x(n) = x(n - 1) - fe/fpe;
fprintf('%3d: %20g %20g\n', n, x(n), abs(fe));
if (abs(fe) <= tol)

Aristi Christoforou
Aristi Christoforou on 10 Apr 2021
function[root]=newtonmethod(f,df,xo,tol,n)
xn=xo;
for k=1:n
xn1=xn-f(xn)/df(xn);
dx=abs(xn1-xn);
xn=xn1;
if dx<tol
dis("newton method has converged")
root=xn;
return
end
end
disp("no convergence after n interactions")

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