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Let the sequence $a_n$ be defined as such: $$\begin{cases}a_0=1\\ a_1=x\\a_n=xa_{n-1}-a_{n-2},\ n \in \mathbb{N}\setminus\{1,2\} \end{cases}$$ I am looking for the $x$ for which the sequence is always positive.

For $x\ge 2$, the sequence is always positive, since it is the sequence of natural numbers for $x=2$.Then, a sequence for $x>2$ is always greater or equal to the sequence for which $x=2$. I tried finding the limit $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}$, and since $x>1$ ($a_2=x^2-1$), this limit only exists for $x\ge 2$, as $L=\frac{x\pm \sqrt(x^2-4)}{2}$, so for $2>x>1$ the sequence can't converge. (Since this limit is always 1 for positive sequences that converge.)

Also, if for $2>x>1$ $\ \ \ a_n\le a_{n-1}\Leftrightarrow a_{n+1}\le(x-1)a_{n-1}\Rightarrow a_{n+1}<a_{n-1}$ for positive $a_n$. Meaning that the sequence has to be stricly increasing or at some point it will be negative, because it can't converge since the limit above (of the ratio of $a_{n+1}$ and $a_n$) doesn't exist. But now I'm stuck. How can I show that for all $1<x<2$ the sequence isn't strictly increasing? Or can you find some value for which it is? (I tested experimentally and it seems that for all values under 2 the sequence decreases at some point, but one can never be sure.)

P.S. I found a way to write the general expression of the sequence, but it didn't help me much. $a_n = x^n - {n-1 \choose 1}x^{n-2}+{n-2 \choose 2}x^{n-4}-{n-3 \choose 3}x^{n-6}+\cdots $

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  • $\begingroup$ If you want explicit expressions, you could also write $a_n=U_n(x/2)$ (Chebyshev polynomials of second kind). So for $|x|<2$, they're certainly not all positive. $\endgroup$ Commented Nov 28, 2020 at 20:19

2 Answers 2

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The characteristic equation has two roots: $r=\frac{x+\sqrt{x^2-4}}{2}, s=\frac{x-\sqrt{x^2-4}}{2}$.

Note that $a_n-(r+s)a_{n-1}+rsa_{n-2}=0$

Case 1: $x>2$, $r, s$ are real and $r>s>0$.

$$ a_{n+1} - s a_n = r(a_n-sa_{n-1}) \Rightarrow a_{n+1} - s a_n = r^n(a_1-sa_0) =r^n (x-s) =r^{n+1}\tag1\\ $$ $$a_{n+1} - r a_n = s(a_n-ra_{n-1}) \Rightarrow a_{n+1} - r a_n = s^n(a_1-ra_0)= s^n (x-r) = s^{n+1}\tag2 $$

(1)-(2), $$a_n=\frac{r^{n+1}-s^{n+1} }{r-s} >0 \tag 3$$

(Note that $a_n=\frac{r^{n+1}-s^{n+1} }{r-s}$ as long as $r\neq s$, even if they are complex).

Case 2: $x=2$.

$$a_n-2a_{n-1}+a_{n-2}=0 \implies a_n-a_{n-1}=a_{n-1}-a_{n-2} \\ \implies a_n-a_{n-1}=a_1-a_0=1 \implies a_n=n$$

So $a_n>0, \forall n$.

Case 3: $x<2$, $r$ and $s$ are conjugate complex numbers.

Let $r=\exp\{i\theta\}, s=\exp\{i(-\theta)\}$ (since $rs=1$ we know $|r|=|s|=1$).

Now $$a_n=\frac{r^{n+1}-s^{n+1} }{r-s}=\frac{2\sin((n+1)\theta)}{2 \sin \theta} = \frac{\sin((n+1)\theta)}{\sin \theta}$$

Note that $\theta=\tan^{-1}\frac{4-x^2}{x} \in (0,\frac{\pi}{2}) \implies \lambda :=\frac{\theta}{\pi} \in (0,\frac 12)$.

If $a_n>0, \forall n$, then $\lfloor n \lambda \rfloor$ is always even. Let $k$ be the smallest integer such that $k\lambda \ge 2$, then $(k-1)\lambda < 1, \lambda = k\lambda - (k-1)\lambda > 2-1=1$, a contradiction.

So for some $n$, $a_n \leq 0$.

Conclusion: The sequence is always positive if and only if $x\geqslant 2$.

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$a_n=xa_{n-1}-a_{n-2};\;a_0=1,a_1=x$

has characteristic equation

$\lambda^2-x\lambda+1=0$

$$\lambda=\frac{x\pm\sqrt{x^2-4}}{2}$$

therefore the solution of the recurrence are

$$a_n=\alpha\left(\frac{x+\sqrt{x^2-4}}{2}\right)^n+\beta\left(\frac{x-\sqrt{x^2-4}}{2}\right)^n$$ setting the initial values $$a_0=\alpha+\beta=1;\;a_1=\alpha\,\frac{x+\sqrt{x^2-4}}{2}+\beta\,\frac{x-\sqrt{x^2-4}}{2}=x$$ $$\alpha=\frac{1}{2}+\frac{x}{2 \sqrt{x^2-4}};\;\beta=\frac{1}{2}-\frac{x}{2 \sqrt{x^2-4}}$$ Finally the solution $$a_n=\left(\frac{1}{2}+\frac{x}{2 \sqrt{x^2-4}}\right)\left(\frac{x+\sqrt{x^2-4}}{2}\right)^n+\left(\frac{1}{2}-\frac{x}{2 \sqrt{x^2-4}}\right)\left(\frac{x-\sqrt{x^2-4}}{2}\right)^n$$

$a_n>0$ for all $n$ if $x-\sqrt{x^2-4}>0\to x>2$

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