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

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

Answer 1

[Mischa Kim]

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

[Malcolm Kevin]

yes you're right. The question does not give tolerence "delta" value, but has the initial guess x0. My question is how to print out exact 5 iterations. I was just stuck in here not knowing how to set up this loop with certain variable like Maxiteration < 6 ?

[Mischa Kim]

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)