Determining the Number of Solutions Using Generating Functions

  • Thread starter alec_tronn
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Homework Statement


Determine the number of solutions in nonegative intergers to the equation:
a + 2b + 4c = 10[tex]^{30}[/tex]


Homework Equations


The generating function I've found is f(x) = 1/[(1-x[tex]^{4}[/tex])(1-x[tex]^{2}[/tex])(1-x)]


The Attempt at a Solution



I'm pretty sure I need to get from here to an explicit formula, but I'm not sure how to start. Any hints to get me started on this one?
 

Answers and Replies

  • #2
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Alright, using the Apart function in Mathematica, I separated the generating function into:

[-1/(8(-1+x)^3)] + [1/(4(-1+x)^2)]- [9/(32(-1+x))]+ [1/(16(1+x)^2)]+ [5/(32(1+x))]+ [(1+x)/(8(1+x^2))]

Then, I turned those each into the followings infinite series:
[(-1/2)[tex]\Sigma[/tex](n+1)x^n][(1\2)[tex]\Sigma[/tex]x^n], so the coefficient for 10^30 is -1\4[((10^30)/2) +1],

(-1\2)[tex]\Sigma[/tex](n+1)x^n, so the coefficient is -1\2(10^30+1)

(9\30)[tex]\Sigma[/tex]x^n, so the coefficient is 9\32

(1\4)[tex]\Sigma[/tex](n+1)(-x)^n, so the coefficient is (1\4)(10^30+1)

(5\32)[tex]\Sigma[/tex](-x)^n, so the coefficient is 5\32

(1\8)[tex]\Sigma[/tex]x^(2n+1), so the coefficent is 1\8,

I add up all the coefficients is (-1\2)((10^30)+1)+(1\2)

Thats negative, so it can't be the answer. I'm awfully rusty in the Calc 2 skills of making functions like that into series... so if anybody could catch any of my mistakes I'd be forever grateful!!!
 

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