Re: HP-48: extract coefficient of power series Message #4 Posted by Han on 21 Oct 2012, 11:42 p.m., in response to message #2 by Gilles Carpentier
If you simply look at how polynomials multiply, it will tell you a lot.
(1+ax)(1+bx) = 1 + ax + bx + abx^2
If you look at the coefficient of x, which is a+b, it basically "counts" how many x's there are. There's a contribution of "a" from 1+ax and a contribution of "b" from 1+bx.
In the original problem, we have z representing a single population. z^2 would represent when the population splits and now there are two of them. z^3 would mean that we have 3 bacteria -- but they could have arrived in several ways: x splits int x1 and x2, and then x2 splits again while x1 remains fixed would be one example. This would be represented by a product of z and z^2. Or we could also have x1 splitting while x2 remains fixed (this would be a product of z^2 and z).
So (1+2z+z^2)/4 = 1/4 z^0 + 1/2 z^1 + 1/4 z^2
1/4 z^0 = 1/4 probability 0 bacteria
1/2 z^1 = 1/2 probability 1 bacteria remains at 1
1/4 z^2 = 1/4 probability 1 bacteria splits into two (z^2).
[(1+2z+z^2)/4 ]^2 -- this would represent two cycles, and let's consider the coefficients of z's.
z^0 = 0 bacteria:
How do we obtain the coefficient of z^0 -- the only way is from 1/4z^0 * 1/4z^0 = 1/16z^0
z^1 = 1 bacteria:
How do we obtain the coefficient of z^1 -- the only way is from 1/4z^0 * 1/2z^1 OR 1/2z^1 * 1/4z^0 = 1/8z^1 + 1/8z^1 = 1/4z^1
(the 1/4z^0 from the first (1+2z+z^2)/4 multiplied with the 1/2z from the second (1+2z+z^2)/4, and vice versa).
z^2 = 2 bacteria:
How do we obtain the coefficient of z^2 -- the only way is from
1/4z^0 * 1/4z^2 = 1/16*z^2
OR
1/2z^1 * 1/2z^1 = 1/4*z^2
OR
1/4z^2 * 1/4z^0 = 1/16*z^2
Summing up: 3/8*z^2
etc.
So when you consider how polynomial multiplication works, it matches up precisely with the sum and/or product of probabilities based on possible outcomes.
Edited: 21 Oct 2012, 11:58 p.m.
|