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Find $\alpha, \beta, \gamma$ for recursive equation:

$$ \alpha a_{n+3}-3a_{n+1}+\beta a_n = 18n$$ $$a_0=0,a_1=\gamma, a_2=3 $$

$$\lim_{n\rightarrow\infty}\frac{3a_n+(-2)^{n}}{n^3}=3$$

Hey guys,

normally I know my way around recursive equations, but when it comes to exercise with limits like this one I don't know how to use the information from the limit condition. I looked in to answers and found out, that first you have to figure out that $a_n$ is formed like $n^3-\frac{1}{3}(-2)^n+ ...$

How can I come to this and move on from there?

My attempt was something like this: $$\frac{3a_n+(-2)^{n}}{n^3}=3/*n^3$$ $$3a_n+(-2)^{n}=3n^3/-(-2)^{n}$$ $$3a_n=3n^3-(-2)^{n}/*\frac{1}{3}$$ $$a_n=n^3-\frac{1}{3}(-2)^{n}$$

but I don't know if this is correct and which parts are missing in $+...$ Please help

Matthew
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1 Answers1

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First, convert the recursion to a linear recursion:

$$\alpha~a_{n+3}-3~a_{n+1}+\beta~ a_{n} = 18~n \tag{1}$$ $$\alpha~a_{n+4}-3~a_{n+2}+\beta~ a_{n+1} = 18~n +18 \tag{2}$$ $$\alpha~a_{n+5}-3~a_{n+3}+\beta~ a_{n+2} = 18~n + 36\tag{3}$$

Eliminate the nonlinear terms with linear algebra:

$$a_{n+5} = 2~a_{n+4} + \left(\frac{3}{\alpha} - 1\right)~a_{n+3} + \left(-\frac{\beta}{\alpha} - \frac{6}{\alpha}\right)~a_{n+2} + \left(\frac{2\beta}{\alpha} + \frac{3}{\alpha}\right)~a_{n+1} - \left(\frac{\beta}{\alpha} \right )~a_{n} \tag{4}$$

This is a linear recursion represented by the matrix:

$$\begin{bmatrix} a_{n+4} \\ a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ a_{n+0} \end{bmatrix} = \underbrace{\begin{bmatrix} 2 & \frac{3}{\alpha} - 1 & -\frac{\beta}{\alpha} - \frac{6}{\alpha} & \frac{2\beta}{\alpha} + \frac{3}{\alpha} & -\frac{\beta}{\alpha} \\ 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ \end{bmatrix}^n}_M \begin{bmatrix}a_{4} \\ a_{3} \\ a_{2} \\ a_{1} \\ a_{0}\end{bmatrix} \tag{5}$$

For the recurrence to have a cubic $k^nn^3$ and an exponential $r^n$ term, then Jordan form of the matrix must contain subblocks:

$$\begin{bmatrix} k & 1 & 0 & 0 \\ 0 & k & 1 & 0 \\ 0 & 0 & k & 1 \\ 0 & 0 & 0 & k \\ \end{bmatrix} \text{ and } \begin{bmatrix} r \end{bmatrix}$$

Since the matrix is 5 by 5, with $k = 1$ and $r = -2$ the Jordan form can only be:

$${\rm Jordan}(M) = \begin{bmatrix} 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -2 \end{bmatrix}$$

So $M$ has an eigenvalue $1$ with multiplicity $4$ and eigenvalue $-2$ with multiplicity $1$. Expanding $\det (\lambda I - M) = 0$ you get:

$$\begin{cases}\lambda^5 - 2~\lambda^4 - \left(\frac 3{\alpha} - 1\right)~\lambda^3 + \frac{\beta+6}{~\alpha}~\lambda^2 - \frac{2~\beta+3}{\alpha}\lambda + \frac \beta{\alpha} = 0 \\ (\lambda - 1)^4(\lambda + 2) = 0 \end{cases}$$

Solving gives $~\alpha=1$, $\beta=2$.

DanielV
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  • Thanks, but is there any other way? – Matthew Mar 28 '15 at 16:43
  • There's probably a way to rewrite the limit as $$G(j, k) : \lim_{n \to \infty} k~a_{n+j} - \frac{1}{3} (-2)^{n+j} - (n+j)^3 + f(n) = 0$$ where $f(n)$ is some function asymptotically less than $n^3$. Add up $G(3, \alpha) + G(1, -3) + G(0, \beta)$ and substitute the original recursion to eliminate the $a$ terms and equate coefficients of the $n$ terms to zero to make the limit exist and it might work. Maybe. Maybe not. I think the approach in the answer is easier and more confident. – DanielV Mar 28 '15 at 23:06
  • One thing you have to somehow realize is that because the lineared equation is only 6 terms, it can't contain an exponential term, and cubic, and another function of $n$. There just aren't enough eigenvalues to make it happen. I don't think there is an "ansatz and check" way to epiphany that. – DanielV Mar 28 '15 at 23:10