# How to write Newton method with exact number of iterations and see all digit ? for example if method is required to take exact 5 iterations of newton's method

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Malcolm Kevin on 10 Jan 2021
Edited: Mischa Kim on 10 Jan 2021
Here is my code so far, I haven't set up the N yet
function c = newton(x0, delta)
format long e
x0 = 5;
delta = 10^-8;
c = x0;
fc = f(x0);
fprintf('initial guess: c=%d, fc=%d\n',c,fc)
if abs(fc) <= delta
return;
end;
while abs(fc) > delta,
fpc = fprime(c);
if fpc==0,
error('fprime is 0')
end;
c = c - fc/fpc;
fc = f(c);
fprintf(' c=%d, fc=%d\n',c,fc)
end;
function fx = f(x)
fx = sinx;
return;
function fprimex = fprime(x)
fprimex = cosx;
return

Mischa Kim on 10 Jan 2021
Edited: Mischa Kim on 10 Jan 2021
Add a loop index, e.g. ii. Also, you set your code as a function. This means you can call it, e.g., from the command window using
c = newton(1, 1e-3)
and thus it does not make sense to define x0 and delta in the function.
function c = newton(x0, delta)
format long e
% x0 = 5;
% delta = 10^-8;
c = x0;
fc = f(x0);
fprintf('Initial guess: c = %d, fc = %d\n',c,fc)
if abs(fc) <= delta
return;
end
ii = 0;
while abs(fc) > delta
ii = ii + 1;
fpc = fprime(c);
if (fpc==0)
error('fprime is 0')
end
c = c - fc/fpc;
fc = f(c);
fprintf('ii = %d: c = %10.7e, fc = %10.7e\n',ii,c,fc)
end
end
function fx = f(x)
fx = sin(x);
end
function fprimex = fprime(x)
fprimex = cos(x);
end
##### 2 CommentsShowHide 1 older comment
Mischa Kim on 10 Jan 2021
Well, with Newton's method you do not know the number of iteration steps. It typically depends on the tolerance, the delta you set. The smaller the delta the more iteration steps it will take to find the solution. Of course, provided that the initial guess is close enough to the solution and that the algorithm converges.
If you just want to do a certain number of iterations and then quit the algorithm you can add a condition like
while (abs(fc) > delta) && (ii < 6)

R2018a

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