Find the Root of the Equations Newton Method

% Find the roots of the equation
%Use initial guesses of x=1. and x=-1. What happens if you give an initial guess of x=0? Explain.
% Find the root of the equation
%starting from the initial condition x=0.
%How many iterations are required to obtain accuracy of 6 decimal places?
% Below is what I have so far

x = -2;
Tol = 0.0000001;
count = 0;
dx=1;   %this is a fake value so that the while loop will execute
f=-13;    % because f(-2)=-13
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 = 3*x^2 + 3;
    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^3 + 3*x + 1;       % compute the new value of f(x)
    fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
end

% This produces the following output:

step      x           dx           f(x)
----  -----------  ---------    ----------
  0  -2.00000000   1.00000000 -13.00000000
  1  -1.13333333   0.86666667  -3.85570370
  2  -0.57073065   0.56260268  -0.89809804
  3  -0.34491904   0.22581161  -0.07579185
  4  -0.32234116   0.02257788  -0.00051597
  5  -0.32218536   0.00015580  -0.00000002
  6  -0.32218535   0.00000001   0.00000000

[John D'Errico]

And so what is your question? It seems like you have made some progress in that task. As a hint, is it converging from a start point of x=-2?

format long g
roots([1 0 3 1])
ans =
          0.161092677313043 +      1.75438095978372i
          0.161092677313043 -      1.75438095978372i
         -0.322185354626086 +                     0i

Now you need to answer the remainder of the questions. But that is merely a question of re-running your code with a new start point, and perhaps thinking about what happened, in terms of what you understand about Newton's method. But that is probably the point of why you were assigned this specific problem, to think about what you see. In fact I might argue f(x)=x^3+3*x+1 is quite an innocuous problem to solve using Newton's method. Hey, you are lucky in that respect. Were I your instructor, I would have been more "nasty" in my choice of problem to solve. But maybe that will be your next assignment. ;-)

[SAPAN NAYAK]

Hello Sir, My question is how to get rhe complex roots of this equation like =4.1185e-07+i4.0939e-06

[Walter Roberson]

I have marked the only real difference. The other changes do not affect the computation, and were only made because fprintf() normally ignores the imaginary portion of values.

% Find the roots of the equation
%Use initial guesses of x=1. and x=-1. What happens if you give an initial guess of x=0? Explain.
% Find the root of the equation
%starting from the initial condition x=0.
%How many iterations are required to obtain accuracy of 6 decimal places?
% Below is what I have so far

x = -2 + pi*1j;       %<---- THIS IS THE ONLY REAL DIFFERENCE
Tol = 0.0000001;
count = 0;
dx=1;   %this is a fake value so that the while loop will execute
f=-13;    % because f(-2)=-13
fprintf('step      x           dx           f(x)\n')

step      x           dx           f(x)

fprintf('----  -----------  ---------    ----------\n')

----  -----------  ---------    ----------

fprintf('%3i %45s %12.8f %45s\n', count, mat2str(x), dx, mat2str(f))

  0                          -2+3.14159265358979i   1.00000000                                           -13

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 = 3*x^2 + 3;
    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^3 + 3*x + 1;       % compute the new value of f(x)
    fprintf('%3i %45s %12.8f %45s\n', count, mat2str(x), dx, mat2str(f))
end

  1           -2.11618125003165+3.44140687381899i   0.32153794             60.362313822926+15.8008341471295i
  2           -1.30567192182889+2.41418111503511i   1.30847932             17.686529205077+5.51897989055528i
  3          -0.699361801093534+1.77941870614676i   0.87780139            5.20308460398037+2.31500568396276i
  4            -0.1632106840138+1.44388453515715i   0.63248810             1.5268065373346+1.43682485125081i
  5            0.406353487669999+1.6428115912685i   0.60330367           -1.00387802350351+1.30856439976018i
  6              0.141387836351897+1.69655460025i   0.27036107          0.206121152824967+0.308219943874544i
  7           0.162669423670935+1.75733557103833i   0.06439901       -0.0147682355158745-0.0155423767087743i
  8           0.161099322766261+1.75438738854152i   0.00334021   -5.18091918002561e-05-2.83047357569899e-05i
  9           0.161092677383075+1.75438095979388i   0.00000925   -4.48330705893341e-10+5.62003776849451e-11i
 10           0.161092677313043+1.75438095978372i   0.00000000                         -2.22044604925031e-16