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Co-efficient of an Algebraic Term


Date: 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/   
    
Associated Topics:
High School Permutations and Combinations
High School Sequences, Series

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