> help bisect

  bisection method looks for a zero of function f on the interval
  [a,b].  It stops when the interval length < 2 tol. 
 
  function root = bisect(a,b,tol)

% first we find the roots of x^2-1.  These are +/- 1.  We use the 
% bisection method.  It takes 17 steps to reach the tolerance when
% starting with the interval [-1/2,3/2]
> root = bisect(-1/2,3/2,10^(-5));
i =
    17
> root
root =
    1.0000

> format long e, format compact
% we see that the approximate root is 7.6 10^(-6) from the true root.
> root
root =
    1.000007629394531e+000

% if we ask for a smaller tolerance then it takes 34 iterations:
> root = bisect(-1/2,3/2,10^(-10));
i =
    34
% we see that the approximate root is 5.8 10^(-11) from the true root.
> root
root =
    1.000000000058208e+000

% change the interval
> root = bisect(-1/2,3,10^(-10));
i =
    35
% it takes longer, but the approximate root is still close 2.2
% 10^(-11). this isn't surprising since we know _exactly_ when the
% bisection method will stop: (b-a)*(1/2)^n determines it all
> root
root =
    9.999999999781721e-001

% go after the negative root:
> root = bisect(-10,3,10^(-10));
i =
    36
> root
root =
   -1.000000000036380e+000

> help newton
  Newton's method looks for a zero of function f using an
  iterative method, with first guess x=a.  It stops when |f(x_n)| <
  tol. it returns the root and the iterates (x).
 
  function [root,x] = newton(a,tol)

% try newtons method:
> [root,x] = newton(-2,10^(-14));
> format short e
> root
root =
    -1
% it meets the tolerance in 6 steps:
> x
x =
 -2.0000e+00  -1.2500e+00  -1.0250e+00  -1.0003e+00  -1.0000e+00  -1.0000e+00

% starting further away, it takes 7 steps:
> [root,x] = newton(-3,10^(-14));
> root
root =
    -1
% the iterates are converging to the root x=-1
> x
  Columns 1 through 6 
 -3.0000e+00  -1.6667e+00  -1.1333e+00  -1.0078e+00  -1.0000e+00  -1.0000e+00
  Column 7 
 -1.0000e+00
% the error is going to zero quadratically once we get close enough
% (the third iterate)
> x + 1
  Columns 1 through 6 
 -2.0000e+00  -6.6667e-01  -1.3333e-01  -7.8431e-03  -3.0518e-05  -4.6566e-10
  Column 7 
           0
% let's look at the significant digits:
> format long e
% we see that we get 1, then 3, then 5, then 10 digits of accuracy.
% Then we run into round-off error.
> x
  Columns 1 through 3 
   -3.000000000000000e+00    -1.666666666666667e+00    -1.133333333333333e+00
  Columns 4 through 6 
   -1.007843137254902e+00    -1.000030518043793e+00    -1.000000000465661e+00
  Column 7 
    -1.000000000000000e+00

% now we change the function.  Now we'll look for the zeros of f(x) =
% x^2.
> [root,x] = newton(1,10^(-14));
% it takes 25 iterates to meet the tolerance that we met in 7 steps
% when zeroing x^2-1.  Plus, we know the root is x=0.  The final
% iterate is 6.0 10^(-8) which isn't that close.  We do check that
% f is small then --- (6.0 10^(-8))^2 = 3.6 10^(-16) which is smaller
% than the tolerance.
> x
x =
 Columns 1 through 6 
  1.0000e+00   5.0000e-01   2.5000e-01   1.2500e-01   6.2500e-02   3.1250e-02
 Columns 7 through 12 
  1.5625e-02   7.8125e-03   3.9062e-03   1.9531e-03   9.7656e-04   4.8828e-04
 Columns 13 through 18 
  2.4414e-04   1.2207e-04   6.1035e-05   3.0518e-05   1.5259e-05   7.6294e-06
 Columns 19 through 24 
  3.8147e-06   1.9073e-06   9.5367e-07   4.7684e-07   2.3842e-07   1.1921e-07
 Column 25 
  5.9605e-08
> root^2
ans =
  3.5527e-15

> diary off