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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

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)

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