Formal power series and non/homogeneous recurrence relations

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SUMMARY

The discussion focuses on solving recurrence relations using generating functions, specifically addressing both homogeneous and non-homogeneous types. The closed form derived is F(z) = z / (1 - z - z²), and the participant explores the implications of partial fractions and coefficient grouping for A(z). The conversation highlights the challenges in transitioning from closed forms to explicit forms, particularly in non-homogeneous cases, and suggests methods for simplifying complex recurrence relations.

PREREQUISITES
  • Understanding of generating functions in combinatorics
  • Familiarity with recurrence relations, both homogeneous and non-homogeneous
  • Knowledge of partial fraction decomposition
  • Experience with explicit formulas and their derivation from closed forms
NEXT STEPS
  • Study the method of generating functions for solving recurrence relations
  • Learn about partial fraction decomposition techniques in detail
  • Explore explicit formulas for Fibonacci-like sequences
  • Research advanced techniques for simplifying non-homogeneous recurrence relations
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Students and educators in mathematics, particularly those focusing on combinatorics and recurrence relations, as well as anyone seeking to deepen their understanding of generating functions and their applications in solving complex problems.

Ian_Brooks
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Homework Statement



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Homework Equations




We're using generating functions, and recurrence relations of homogeneous and non-homogeneous types

The mark allocation is 2, 3, 3 and 2


The Attempt at a Solution



I think I've done the first part correctly. The closed form is in terms of z, right? I get:

F(z) = z / (1 - z - z2)

but I'm unsure if I have to go into partial fractions for this in my answer. For part b, I get:

A(z)(1 - 2z + z2) = a0 + z(a1 - 2a0) ...
= F(z)
so I take my closed formula from part a and substitute F(z) for it so I get

A(z) = z / [ (1 - z - z2) (1-z)2) ]

I'm still unsure if this is right, since looking at previous examples, they express A(z) in terms of itself, that is: group together coefficients of an etc. and get a more complex remainder; They suggest to do this:

A(z)(1 - 2z + z2) = a0 + z (a1 - 2a0) + z2 (a2 -2a1 + a0) ...

and group together the coefficients of a1 or a2 etc. so there is a generating function within the brackets. I used F(z) in this case, but is that really in terms of itself?

For part (c), I want to use the bn and pn way for non homogeneous recurrence relations, but I got lost; we've only gone over the pn being a power of n (n2, n or even just integers) but the function here, even in the explicit form, is an integer to the power of n (I solved the Fibonacci sequence via the polynomial method). I used the partial fraction method, but it ended up very ugly, substituting alpha for (1 - root5 / 2) and beta for (1 - alpha), their product as negative 1, and their sum as positive 1. This is only a three mark question, and my working is over 3 pages, incomplete. Even if I finish it, I don't remember how to get from the closed form to explicit (that's in terms of n, right?). I can use a/(1-bz) = a times bn , but what about with multiple powers of (1-z) at the denominator? I only recall the generating function version of that (\sum (n+k)Ck z2 ) (bounds from n=0 to infinity), but not in it's explicit form.

Then for part (d), I'm unsure how to use part (c) to calculate the sum as I am completely lost for it. Any pointers?
 
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i'd agree with a) & b) and think you're essentially doing the same thing as the solution

for d) try writing out an in terms of the Fj's and see if you can spot a pattern
 
Last edited:
for c) you want to try and probably get rid of the F term so how about looking at
a_{n+1}-a_n{n}

after manipulation, this should leave you with an equation with a negative F term. Then adding that to the original, with correct choice of index, should leave you with a pure recurrence relation of aj that may be easier to solve...
 

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