# Estimating linearly independent solutions to third order recurrence relation

I'm trying to prove something about two linearly independent solutions; $a_n$, $b_n$, to a recurrence relation I have - specifically that $\left| \frac{a_n}{b_n} \right|$ is eventually monotonically increasing using an estimate of the linearly independent solutions. The recurrence relation in question is: \begin{align*} n^3u_n&=-3\left(27(n-1)^3-8(n-1)-4\right)u_{n-1}\\&\quad {}-3^4\left(27(n-2)^3-8(n-2)+4\right)u_{n-2}-3^9\left(n-3\right)^3u_{n-3} \end{align*}

and the first few terms of the solutions $a_n$, $b_n$ are: \begin{align*} (a_n)_{n\geq0} &= \{0, 1/12, 3/32, -133/24, \ldots\} \\ (b_n)_{n\geq0} &= \{1/12, 1, -9, 127, \ldots\} \end{align*}

Unfortunately I've only ever seen recurrence relations of order up to 2, and this isn't really my area of knowledge at all. Would anyone happen to have any pointers? I've tried to use Mathematica's RSolve function, but it didn't yield anything helpful. Any advice (though not an explicit solution) would be very much appreciated! Thanks.

• Where does this come from? Nonlinear recurrences are a bear (worse than their counterpart differential equations, in fact). Perhaps more details of the problem suggest a way around solving the equation? – vonbrand Mar 20 '14 at 19:25
• @vonbrand they arise as coefficients of a series expansion of an overconvergent modular function. I personally cannot see a way of ascertaining that the ratio of the solutions is monotonically increasing in absolute value by using this fact - the specific recurrence relation they satisfy seems to be the only way. Note that an explicit solution isn't required, just some kind of estimate. I'm not sure if this is something that can be done by hand though; do you know if Sage would be able to deal with this problem? – ah11950 Mar 20 '14 at 20:16
• I don't get $-133/24$. I get $-59/6$. Has a typo' been introduced in editing? – Eric Towers Mar 23 '14 at 5:15
• You are indeed correct, I get -59/6 also... (the $b_n$ appear to be correct though). I'll go back and check my derivation of the first few $a_n$ - that is where my mistake will lie! Thanks for noticing! – ah11950 Mar 23 '14 at 11:38
• If you set $t_n = n^3 u_n$, then neglect the non-dominant term, you end with $t_n + 3^4 t_{n-1} + 3^7 t_{n-2} + 3^9 t_{n-3}=0$. Now the growth of the $t_n$ is easy to estimate, the only real root of $y^3+3^4 y^2+3^7 y+3^9$ is around $y\sim -9.75$. – Jack D'Aurizio Mar 29 '14 at 10:41

If you set $t_n = n^3 u_n$, you can approximate your recurrence with a linear one: $$t_n + 3^4 t_{n+1} + 3^7 t_{n-2} + 3^9 t_{n-3} \sim 0, \tag{1}$$ then study the roots of the characteristic polynomial: $$p(\lambda) = \lambda^3 + 3^4 \lambda^2 + 3^7 \lambda + 3^9 = (\lambda+27)^3, \tag{2}$$ in order to have that $t_n$ behaves like $(-27)^n$, so $u_n$ behaves like $(-1)^n\cdot\frac{27^n}{n^3}.$
• If I'm not wrong, $\frac{a_n}{b_n}$ is a convergent of $\zeta_3(3)$, so the "error terms" $\left|\zeta_3(3)-a_n/b_n\right|$ give a decreasing sequence. Moreover, the sequences of even-indexed and odd-indexed convergents are both monotonic, so you only need to prove that the index of $\frac{a_n}{b_n}$ as a convergent of $\zeta(3)$ is always even/odd for any $n\geq 2$, or just study the sign of $\zeta(3)-\frac{a_n}{b_n}$. – Jack D'Aurizio Mar 29 '14 at 12:06
• Indeed, the ratio converges to $\zeta_3(3)$. What I'm trying to show is that that we don't actually have $a_n/b_n = \zeta_3(3)$ for all large enough $n$. Yep, I can see that the even and odd-indexed convergents are both monotonic (this just follows by the estimate of the $a_n$ and $b_n$ growth and that they are linearly independent). What do you mean by "prove the index of $\frac{a_n}{b_n}$ as a convergent of $\zeta(3)$ is always even/odd" though? I'm afraid I don't follow! – ah11950 Mar 29 '14 at 12:13
• I meant that $\frac{a_n}{b_n}$ always has an even/odd position in the list of convergents of $\zeta(3)$. Anyway, you know that $b_n$ is alternating in sign from a certain point on. Moreover, $c_n = b_n\zeta(3) - a_n$ is a linear combination of $a_n$ and $b_n$. Given that $n^3 a_n$ and $n^3 b_n$ satisfy an approximate linear recurrence, $n^3 c_n$ satisfy the same approximate linear recurrence, hence it is alternating in sign from a point on. This gives that $\frac{a_n}{b_n}$ is monotonic from a point on. – Jack D'Aurizio Mar 29 '14 at 12:22
• Since $\left\{\Delta_n = \zeta(3)-\frac{a_n}{b_n}\right\}_{n\geq N}$ is a sequence of positive numbers converging to zero, just to be clear. You only need to prove that for any $N\geq 2$ the neglected terms in the original recurrence are not so big to affect the sign of $u_N$: this is straightforward. – Jack D'Aurizio Mar 29 '14 at 12:27