Co-efficient of an Algebraic TermDate: 07/18/99 at 19:58:12 From: Elizabeth Subject: Co-efficient of an Algebraic Term Find [z^40] (1 + z + z^2 + ... + z^9)^100 My generating function is: (1 - x^(9+1))/(1 - x). So, you can break that down to 1 - x^10 * (1/(1 - x)) = [z^40] (1/(1 - x)) - [z^??] 1/(1 - x) But what is the second z coefficient? How do I figure it out? Date: 07/19/99 at 07:55:44 From: Doctor Anthony Subject: Re: Co-efficient of an Algebraic Term I assume that you want the COEFFICIENT of z^40 in the expansion of (1 + z + z^2 + z^3 + ...... + z^9)^100 First, express the series as the sum of a GP. (1 - z^10)^100 ----------------------- = (1 - z^10)^100 x (1 - z)^(-100) (1 - z)^100 = (1 - z^10)^100 [ 1 + C(100, 1)z + C(101, 2)z^2 + C(102, 3)z^3 + ....] The first bracket can be expanded with powers of z^10 = (1 - C(100, 1)z^10 + C(100, 2)z^20 - C(100, 3)z^30 + C(100, 4)z^40 - And now associate each of these terms with a term from the second bracket such that we have a term in z^40. 1 x C(139, 40) - C(100, 1) x C(129, 30) + C(100, 2) x C(119, 20) - C(100, 3) x C(109, 10) + C(100, 4) x 1 = 1.262595 x 10^35 This will be the coefficient of z^40 in the generating function. It represents the number of ways that 40 objects could be chosen from 100 urns, each urn containing 9 objects, indistinguishable from each other, but different from objects in the other urns. For example, the first urn could contain all blue beads, the second urn could contain all red beads, the third urn all white beads and so on. With 100 such urns each containing beads of one colour for that urn, the number of ways of selecting 40 beads is the answer given above. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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