The following problem was posted recently in a math-forum:

Quote:

Every 10 minutes a bacteria has 3 possible actions. It will die, remain the same, or split into two. The probability of it dying is 1/4, staying the same is 1/4, and splitting is 1/2. All future bacteria share this probability. If you start with one bacteria, what is the probability that there will be exactly 6 bacteria after 30 min?

Among others the following solution was posted:

Quote:

There is a huge literature on this sort of random processes, going under the name of branching processes. For the current problem, define the function f (z)=1/4+1/4z+1/2z^{2}, and consider the third iterate f(f(f(z))), and especially the coefficient in front of z^{6}.

My first attempt to use WolframAlpha failed so I decided to give it a try using my HP-48GX:

f: \<< \-> z '(1+z+2*z^2)/4' \>>

g: \<< \-> z 'f(f(f(z)))' \>>

When I calculated g(z) I wondered how to extract the 6th coefficient. So I came up with this solution:

\<< \-> n

\<<

z g

z n TAYLR

1 n START

z \.d

NEXT

n ! /

\>>

\>>

Would there be an easier way to find that coefficient? And does somebody happen to know how to do this calculation with WolframAlpha?

Many thanks in advance

Thomas

*Edited: 20 Oct 2012, 8:39 a.m. *