Find OGF for Recurrence: a_n = 6a_{n-1} + a_{n-2}, a_0=2, a_1=1

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Homework Help Overview

The discussion revolves around finding the ordinary generating function (OGF) for a recurrence relation defined as a_{n} = 6a_{n-1} + a_{n-2} with initial conditions a_{0} = 2 and a_{1} = 1. Participants are exploring the setup and derivation of the generating function.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants describe their attempts to define the generating function A and derive expressions based on the recurrence relation. There are questions about the correctness of the definitions and the resulting expressions for A. Some participants express confusion about the different forms of A obtained and the implications of starting the generating function at different indices.

Discussion Status

The discussion is active, with participants sharing their derivations and questioning the validity of their approaches. There is an acknowledgment of differing results and a suggestion to clarify the definitions used in the generating function setup. Some participants have referenced external tools like Maple to verify their findings, but no consensus has been reached on the correct form of the generating function.

Contextual Notes

Participants are considering the implications of starting the generating function at different indices, which may affect the resulting expressions. There is also mention of potential discrepancies in the expected outcome compared to what has been derived, indicating a need for careful examination of the assumptions made in the problem setup.

Punkyc7
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Find the OGF for the recurrence

a[itex]_{n}[/itex]= 6 * a[itex]_{n-1}[/itex]+ a[itex]_{n-2}[/itex] a[itex]_{0}[/itex]=2, a[itex]_{1}[/itex]=1
So here is what I did

I said let A = [itex]\sum[/itex][itex]_{2>=n}[/itex]a[itex]_{n}[/itex]x[itex]^{n}[/itex]then I got

A = 6x (A+x) + x[itex]^{2}[/itex](A +x+2)

which gets me

A= [itex]\frac{6x^2+x^{3} +2x}{1-6x - x^2}[/itex]
ButI should get [itex]\frac{2-x}{1-6x - x^2}[/itex]Can anyone tell me what I am doing wrong ?
 
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Punkyc7 said:
Find the OGF for the recurrence

a[itex]_{n}[/itex]= 6 * a[itex]_{n-1}[/itex]+ a[itex]_{n-2}[/itex] a[itex]_{0}[/itex]=2, a[itex]_{1}[/itex]=1

So here is what I did

I said let A = [itex]\sum[/itex][itex]_{2>=n}[/itex]a[itex]_{n}[/itex]x[itex]^{n}[/itex]

then I got

A = 6x (A+x) + x[itex]^{2}[/itex](A +x+2)  = A(6x) + 6x2 + A(x2) + x3 +2x2
This gets you a different A.
which gets me

A= [itex]\frac{6x^2+x^{3} +2x}{1-6x - x^2}[/itex]

ButI should get [itex]\frac{2-x}{1-6x - x^2}[/itex]

Can anyone tell me what I am doing wrong ?

Congrats on your 400th post ! :smile:
 
Thank you on the congratulations. But I do not see how that get me a different A. Did I define what A is right. Usually the generating function start at zero but with this problem that gives you bad subscripts.
 
Punkyc7 said:
Find the OGF for the recurrence

a[itex]_{n}[/itex]= 6 * a[itex]_{n-1}[/itex]+ a[itex]_{n-2}[/itex] a[itex]_{0}[/itex]=2, a[itex]_{1}[/itex]=1



So here is what I did

I said let A = [itex]\sum[/itex][itex]_{2>=n}[/itex]a[itex]_{n}[/itex]x[itex]^{n}[/itex]


then I got

A = 6x (A+x) + x[itex]^{2}[/itex](A +x+2)

which gets me

A= [itex]\frac{6x^2+x^{3} +2x}{1-6x - x^2}[/itex]



ButI should get [itex]\frac{2-x}{1-6x - x^2}[/itex]


Can anyone tell me what I am doing wrong ?

I usually find it safer, to write things out a bit more:
[tex]A = a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots = (6a_1 + a_0)x^2 <br /> + (6a_2 + a_1)x^3 + (6a_3 + a_2)x^4 + \cdots\\<br /> = (6+2)x^2 + x^3 + 6x A + x^2 A = x^3 + 8x^2 + (x^2 + 6x)A.[/tex]
This will give you a different A from what you obtained.

Besides that difference, maybe the answer was for a GF starting at a_0*x^0, or at a_1*x^1. However, I checked that, and could not get the posted answer for any three variants of the problem. Just to be sure, I used Maple to get a solution, and got an answer in agreement with a_0 + a_1 x + A. Here is the Maple command, but I won't give the output, since that would deprive you of the fun of getting it yourself.
> rsolve({a(n)=6*a(n-1)+a(n-2),a(1)=1,a(0)=2},a,'genfunc'(x));

RGV
 

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