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Show that the general solution to the recurrence equation $a_n+ba_{n-1} + ca_{n-2} = 0$ can be written as $a_n = Ar_1^n + Br_2^n$ if the roots $r_1$ and $r_2$ of the characteristic equation are unequal. Also account for the case where the roots are equal.

So my attempt is as follows:

Making the ansatz above, the homogenous equation becomes \begin{align*} aAr_1^n + aBr_2^n + bAr_1^{n-1} + bBr_2^{n-1} + cAr_1^{n-2} + cBr_2^{n-2} &= \\ Ar_1^{n-2}(ar_1^2 + br_1^{1} + cr_1) + r_2^{n-2}B(ar_2^2 + br_2^{1} + cr_2) &= 0 \end{align*} since we know that both of the paranthesis are zero, by the characteristic polynomial. I'm not quite sure how to do the latter part, and I'm not entirely sure about my proof either, since I never seem to use the fact that the roots are distinct.

Peatherfed
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  • there's an easy direct way to do this if you know what a Companion Matrix is. i.e. your kth term in the recurrence is given by $\mathbf e_1 C^k \mathbf v$ where $\mathbf v$ has your two initial conditions. So the zeroth term is $v_0$, first term is $v_1$ (k=1) and $v_2= \mathbf e_1 C^2 \mathbf v$ and so on. Then with distinct roots $C$ is diagonalizable so $v_k = \mathbf e_1 C^k \mathbf v = \big(\mathbf e_1S\big) D^k \big(S^{-1}\mathbf v\big) = \alpha r_1^k + \beta r_2^k$. Companion matrices are defective with repeated roots so use JCF or take limits as $r_2 \to r_1$ for second part. – user8675309 Jan 07 '20 at 19:12

1 Answers1

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If $r_1=r_2$, then you have obtained a solution with just $1$ arbitrary constant, since you can replace $A+B$ by a constant $C$.

In the equal roots case you must use $(An+B)(ar^{n+2}+br^{n+1}+cr^{n})$.

  • Yes, I know that, but why? – Peatherfed Jan 08 '20 at 09:08
  • If you let $a_k$ and $a_{k+1}$ satisfy these formulae then do the algebra (as you have done in your post) to show that $a_{k+2}$ also satisfies the formula. Therefore, by induction, you have the general solution. –  Jan 08 '20 at 09:19
  • No, since I don't know that I've got all solutions, which is what the question asks. – Peatherfed Jan 08 '20 at 09:36
  • Any sequence is completely determined by its initial two terms. Any formula which works as described above and which has two arbitrary constants can be equated to the first two terms (which then determines the two constants). Such a formula therefore is the general solution. –  Jan 08 '20 at 09:47
  • Why? That's just stating the question in a different form. – Peatherfed Jan 08 '20 at 09:58