has_root.m

   1 function has_root = has_root(fa, fb)
   2   ## Use the intermediate value theorem to determine whether or not some
   3   ## function has an odd number of roots on an interval. If the function
   4   ## in question has an even number of roots, the result will be
   5   ## incorrect.
   6   ##
   7   ## Call the function whose roots we're concerned with 'f'. The two
   8   ## parameters `fa` and `fb` should correspond to f(a) and f(b).
   9   ##
  10   ##
  11   ## INPUTS:
  12   ##
  13   ##   * ``fa`` - The value of `f` at one end of the interval.
  14   ##
  15   ##   * ``fb`` - The value of `f` at the other end of the interval.
  16   ##
  17   ## OUTPUTS:
  18   ##
  19   ##  * ``has_root`` - True if we can use the I.V.T. to conclude that
  20   ##    there is a root on [a,b], false otherwise.
  21   ##
  22
  23   ## If either f(a) or f(b) is zero, the product of their signs will be
  24   ## zero and either a or b is a root. If the product of their signs is
  25   ## negative, then f(a) and f(b) are non-zero and have opposite sign,
  26   ## so there must be a root on (a,b). The only case we don't want is
  27   ## when f(a) and f(b) have the same sign; in this case, the product of
  28   ## their signs would be one.
  29   if (sign(fa) * sign(fb) != 1)
  30     has_root = true;
  31   else
  32     has_root = false;
  33   end
  34 end