MATLAB: Newton’s method to determine roots / I’m not sure if what I got so far is correct

newton’s method to determine roots

    >> x = 3;
    Tol = 0.001;
    count = 0;
    dx=1;   %this is a fake value so that the while loop will execute
    f=-11;    % because f(3)=-11
    fprintf('step      x           dx           f(x)\n')
    fprintf('----  -----------  ---------    ----------\n')
    fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
    xVec=x;fVec=f;
    while (dx > Tol || abs(f)>Tol)  %note that dx and f need to be defined for this statement to proceed
        count = count + 1;
        fprime = 5*x^4-20*x^3+6*x^2+28*x;
        xnew = x - (f/fprime);   % compute the new value of x
        dx=abs(x-xnew);          % compute how much x has changed since last step
        x = xnew;
        f = x^5-5*x^4+2*x^3+14*x^2-29;       % compute the new value of f(x)
        fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
    end
    step      x           dx           f(x)
    ----  -----------  ---------    ----------
      0   3.00000000   1.00000000 -11.00000000
      1   6.66666667   3.66666667 4477.99588477
      2   5.64985329   1.01681337 1440.74956648
      3   4.86575495   0.78409835 457.60831319
      4   4.27678772   0.58896722 141.56939561
      5   3.85764079   0.41914694  41.17150694
      6   3.59446346   0.26317733  10.13724563
      7   3.47377942   0.12068404   1.53513578
      8   3.44792517   0.02585424   0.06059241
      9   3.44681794   0.00110723   0.00010774
     10   3.44681597   0.00000198   0.00000000
    >>

Best Answer

You can compare your results with the result obtained from the ROOTS command. If they agree, it will be a good sign that you've done it right.
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