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How would you solve for the closed form solution of a(n) given the general form of the third order linear homogenous recurrence relation with real constant coefficients.

$a_n=Pa_{n-1}+Qa_{n−2}+Ra_{n−3}$

with the initial terms of a_1, a_2, and a_3

given that the roots of the characteristic equations have

1)two repeated roots and a real root

2)three repeated roots

(can you give answers for both cases please)

When I search the web I get these results

1)$a_n = nAx_1^n + Bx_1^n + Cx_2^n$,for the case when there are two repeated roots

and

2)$a_n = n^2Ax^n + nBx^n + Cx^n$, for the case when there are three repeated roots

Can anyone help derive the closed form of each case in order to get such results?

Please help

I'm new to the system so i didn't quite know how to get the symbols right (sorry) if you're uncertain about anything please ask

Shuchang
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Brian
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    A standard technique would be to use a generating function, split into partial fractions, and use the binomial expansion. – Mark Bennet Aug 29 '13 at 09:35
  • Could you explain more please? I just don't know how to get to the final answer. I've searched online nearly everywhere. If you could give me a detailed explanation that would be great or even a scan of the procedure? I don't have an awful lot of time please help – Brian Aug 29 '13 at 09:43
  • Already you got the solutions for both cases! You only need to use the initial conditions to find the constants $A,B$, and $C$. – Mhenni Benghorbal Aug 29 '13 at 10:05
  • Yes I understand that I have the closed forms however it was the steps I have to take to get there that was the problem – Brian Sep 01 '13 at 13:12

2 Answers2

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Let the generating function $$g(x)=\sum_{n=0}^\infty a_nx^n$$ with the recurrence $a_n-pa_{n-1}-qa_{n-2}-ra_{n-3}=0$

Now consider $$(1-px-qx^2-rx^3)g(x)=a_0+(a_1-pa_0)x+(a_2-pa_1-qa_2)x^2=A(x)$$ Note that all the other terms go to zero because of the recurrence. We then have $$g(x)=\frac{A(x)}{(1-px-qx^2-rx^3)}$$

$A(x)$ is quadratic (and we have an explicit expression for it). If the denominator factors as $(1-sx)^2(1-tx)$ we have the partial fraction decomposition$$g(x)=\frac B{(1-sx)^2}+\frac C{1-sx}+ \frac D{1-tx}$$ [$B,C,D$ are constant]

We expand using the binomial theorem and equate coefficients - the coefficients involving $n$ come from the quadratic factor in the denominator.

If the denominator factors as $(1-ex)^3$ we have the partial fraction decomposition$$g(x)=\frac B{(1-sx)^3}+\frac C{(1-sx)^2}+ \frac D{1-sx}$$ The cubic factor gives the $n^2$ factor which appears as a coefficient in the expression for $a_n$.

Note: this was adjusted in the light of Brian's comment, which highlighted a careless basic error in the original text.

Mark Bennet
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  • This can be used to prove the general form. Once the form is known it can be applied each time without proof. – Mark Bennet Aug 29 '13 at 09:51
  • Note minor adjustments are necessary if you want to begin with $a_1$ - alternatively $a_0$ can be defined by running the recursion backwards. – Mark Bennet Aug 29 '13 at 09:53
  • Is it not possible to just replace a0 for a1? sorry I don't have so much depth in my knowledge in this field (I'm only a highschooler that ended up having to do this) I understand how you got to divide the fraction but I'm not sure of the procedure afterward. Could you provide a little more detail please? – Brian Aug 29 '13 at 09:56
  • @Brian You can use $g(x)=\sum_{n=1}^\infty a_nx^{n-1}$ which just shifts everything along (add $1$ to every suffix) - you just need to make sure you equate the right coefficients, because $a_n$ goes with $x^{n-1}$ [Equivalently let $b_r=a_{r+1}$ so that $b_0=a_1$ and you can calculate the $b_r$] – Mark Bennet Aug 29 '13 at 10:00
  • sorry but in this proof there was an error. I multiplied g(x) by the characteristic equation which then lead me to the partial fraction however that was wrong because then the terms after x^2 do not cancel out. In fact it should have been (1-Px-Qx^2-Rx^3) that was supposed to be multiplied. Then the method of dividing into partial factions cannot apply. What am I supposed to do? – Brian Sep 01 '13 at 08:09
  • @brian - I carelessly got my coefficients the wrong way round - I want factors $(1-sx)$ to apply binomial - I've adjusted everything to correct it. Note that $s$ and $t$ are still the roots of the characteristic equation. – Mark Bennet Sep 01 '13 at 11:12
  • Thank you for the clarification. So just to check, if you factorize (1-Px-Qx^2-Rx^3) the answer will still be (1-sx)^2(1-tx)? Is there any way of showing that this is right? rather than just stating? – Brian Sep 01 '13 at 13:07
  • @Brian: Suppose $s,t$ are roots of $(x-s)^2(x-t)=x^3-Px^2-Qx-R=0$. Now set $y=1/x$ and we have $(1-sy)^2(1-ty)=1-Py-Qy^2-Ry^3=0$ – Mark Bennet Sep 01 '13 at 13:29
  • Thank you so much. It really helped and I finally got the solution I was struggling to find for 3 months now. I'm so glad you helped. Thanks again :) – Brian Sep 01 '13 at 13:43
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There are many such methods as linear algebra and generating function to linear homogeneous recurrence relations with constant coefficients You can refer to https://en.wikipedia.org/wiki/Recurrence_relation for different methods. For detailed proof, please see here http://www.cs.bsu.edu/homepages/fischer/math215/recurrence.pdf

Shuchang
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