<pre>
If p is the probability that a single amoeba's descendants will die
out eventually, the probability that N amoebas' descendents will all
die out eventually must be p^N, since each amoeba is independent of
every other amoeba.  Also, the probability that a single amoeba's
descendants will die out must be independent of time when averaged
over all the possibilities.  At t<code>0, the probability is p, at t</code>1 the
probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be
equal.  Extinction probability p is a root of f(p)=p.  In this case,
 p = sqrt(2)-1.

The generating function for the sequence P(n,i), which gives the
probability of i amoebas after n minutes, is f^n(x), where f^n(x) <code></code>
f^(n-1) ( f(x) ), f^0(x) <code></code> x .  That is, f^n is the nth composition
of f with itself.

Then f^n(0) gives the probability of 0 amoebas after n minutes, since
f^n(0) = P(n,0). We then note that:

	f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4

so that if f^(n+1)(0) -> f^n(0) we can solve the equation.

The generating function also gives an expression for the expectation
value of the number of amoebas after n minutes. This is d/dx(f^n(x))
evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x))
and since f'(1) <code> 1.5  and f(1) </code> 1, we see that the result is just
1.5^n, as might be expected.
</pre>
