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Consider the sequence $\{a_n\}_{n=0}^\infty$ with $a_0 = 0, a_1 = 1, a_{n+2} = 6a_{n+1} - 9a_{n}$. Using the characteristic polynomial prove $a_{n} = n3^{n-1}$.

So I really wasn't sure where to start, as I don't recall doing this in lectures. I searched online, rearranged the recurrence relation, and got the characteristic polynomial as

$$x^2 - 6x + 9$$

Then I tried to solve the relation as you would a differential equation, I found it was a repeated root, $c1 = c2 = 3$. Giving one solution $e^{3t}$. I substituted this into the original equation $a_n = v(t)e^{3t}$, then differentiated to get $a_{n+1}$ and again for $a_{n+2}$.

Substituting these new values in for the original equations I got: $$e^{3t}(v''(t)) = 0$$ hence $v''(t) = 0$, hence $v'(t) = c_{1}$ and $v(t) = c_{1}t + c_{2}$ where $c_{1}$ and $c_{2}$ are arbitrary constants.

However solving for $c_{1}$ and $c_{2}$ I get nothing like the solution I was looking for, I think that I must have been on completely the wrong track, but can't find anything online, I good guide of what to do would be really appreciated.

JennySmart
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3 Answers3

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The procedure for handling the roots is similar but not identical to that for differential equations. The repeated root of $3$ gives you basic solutions $x_n=3^n$ and $y_n=n3^n$, and every possible solution to the recurrence is then a linear combination of these. Thus, there are constants $A$ and $B$ such that $a_n=A3^n+Bn3^n$ for each $n\ge 0$. Now you use the initial conditions $a_0=0$ and $a_1=1$ to solve for $A$ and $B$.

More generally, a root $r$ of multiplicity $m$ gives you basic solutions $n^kr^n$ for $k=0,\ldots,m-1$.

Brian M. Scott
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See page 3 of: http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf. It is basically a walk through to your problem. You should not end up with a differential equation. If you do, it means that there is no closed relation for your coefficient or that you did not use the proper path.

The reason is that such recurrence relation arise from the series solution of differential equations: if you end up with a differential equation, you only managed to map your recurence relation back to the differential equation.

As of latex formatting, adding '$' signs on both sides of every equation should do the trick. For single line equation you can use \begin{equation} a+b+c \end{equation}

jgyou
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Another method wich yields the correct solution slightly modify the original problem: Replace $9$ by $9 - \epsilon^{2}$. The purpose is to have two different roots of the characteristic polynomial: $x_{\pm} = 3 \pm \epsilon$. At the end of the calculation we´ll take the limit $\epsilon \to 0$. The general solution becomes $a_{n} = A_{-}\,x_{-}^{n} + A_{+}\,x_{+}^{n}$. Initial conditions yields:

$$ 0 = a_{0} = A_{-} + A_{+}\,, \qquad 1 = a_{1} = A_{-}\,x_{-} + A_{+}\,x_{+} $$ Then, $$ A_{-} = -A_{+} = {1 \over x_{-} - x_{+}} \quad\Longrightarrow\quad \color{#ff0000}{\large a_{n}} = \lim_{\epsilon \to 0}{x_{-}^{n} - x_{+}^{n} \over x_{-} - x_{+}} = \lim_{\epsilon \to 0}{-nx_{-}^{n - 1} - nx_{+}^{n - 1} \over -2} = \color{#ff0000}{\large n\,3^{n - 1}} $$

Felix Marin
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