I need help please !! Newton method

Hi , I know there's a lot of subjects on this but I didn't find what I'm looking for... This is my problem: I have create this code and it works perfectly well :

fonc=inline('fonction');
dfonc=inline('derivée de la fonction');
x=0.8;
;xn=0;
eps=1e-6;
k=0;
fprintf('   k        x_k          f(x_k)   \n');
fprintf('%4d   %12.4e  %12.4e  \n',k, x,fonc(x));
k=k+1;
while abs(x-xn) > eps & k < 20
    xn=x;
    x=x-fonc(x)/dfonc(x);
    fprintf('%4d   %12.4e  %12.4e  \n',k, x,fonc(x));
    k=k+1;
end

In this case fonction = x^4-1.3*x^3-2.97*x^2+5.929*x-2.662 and dérivée de la fonction is 4*x^3-3.9*x^2-5.94*x+5.929

   k        x_k          f(x_k)   
       0     8.0000e-01   -7.5600e-02  
       1     9.0370e-01   -2.1963e-02  
       2     9.7064e-01   -6.4300e-03  
       3     1.0144e+00   -1.8908e-03  
       4     1.0432e+00   -5.5753e-04  
       5     1.0623e+00   -1.6467e-04  
       6     1.0749e+00   -4.8691e-05  
       7     1.0833e+00   -1.4407e-05  
       8     1.0889e+00   -4.2649e-06  
       9     1.0926e+00   -1.2629e-06  
      10     1.0951e+00   -3.7405e-07  
  ...
      29     1.1000e+00    0.0000e+00

My problem is that I have to add a new thing to the table that would be create by this fonction:

 abs((xn2-xn)/(xn-x))

and I don't know how to do it... I really need help. It should be looking like this

     k        x_k          f(x_k)     |En+1/En| 
     0     8.0000e-01   -7.5600e-02  0,645515911
     1     9.0370e-01   -2.1963e-02  0,653719749
     2     9.7064e-01   -6.4300e-03  0,658135283
     3     1.0144e+00   -1.8908e-03  0,663194444
     4     1.0432e+00   -5.5753e-04  0,659685864
     5     1.0623e+00   -1.6467e-04  ...
     6     1.0749e+00   -4.8691e-05  ...
     7     1.0833e+00   -1.4407e-05  ...
     8     1.0889e+00   -4.2649e-06  ...
     9     1.0926e+00   -1.2629e-06  ...
    10     1.0951e+00   -3.7405e-07  ...
    11     1.0967e+00   -1.1080e-07  …

Thank you. ( by the way , english is not my native language so I'm sorry for all the mistakes I may have made. )

[Geoff Hayes]

Thierry - in your equation

 abs((xn2-xn)/(xn-x))

does xn2 really mean xn^2 or does it mean something else?

[Thierry Gelinas]

Hi , xn2 represent the second point found with the newton method. For example :

       x    0     8.0000e-01    
       xn   1     9.0370e-01    
      xn2   2     9.7064e-01

So the first iteration would be ( with abs((xn2-xn)/(xn-x)) )

abs((9.7064e-01- 9.0370e-01 )/( 9.0370e-01 - 8.0000e-01  )

Is it more clear now ?

Answer 1

[Geoff Hayes]

Thierry - in order to calculate this new value, you will need to save the data from the previous iterations. So outside of the while loop, create a variable for the historical data as

 maxIters    = 20;
 histData    = zeros(maxIters+1,1);
 histData(1) = x;

then inside the while loop handle the case for when you have enough data to do the calculation (i.e. at least three historical points)

 while abs(x-xn) > eps && k < maxIters
    xn = x;
    x  = x-fonc(x)/dfonc(x);

    histData(k+1) = x;

    fprintf('%4d   %12.4e  %12.4e  ',k, x,fonc(x));

    if k==1
        fprintf('\n');
    else
        fprintf('%12.4e\n',abs((histData(k+1)-histData(k))/(histData(k)-histData(k-1))));
    end

    k = k + 1;
 end

Try the above and see what happens!

[Thierry Gelinas]

Thank you ! You are a MVP. Have a nice day !