I am trying to solve this recurrence inequality for $ a_n $, hopefully in terms of $ a_0,a_1,c $ and $ x $,

$$(1-x)a_{n+1} \le xa_n + (1+x)a_{n-1} + c\quad\forall\ n\ge1$$

What I tried: RSolve. But RSolve expects an equality, and not an inequality. Any tips on how to proceed? $a_i \in \mathbb{R}$, $x\in\mathbb{Q}$, $c > 0$, $0 < x < 1$ and $a_i$ is not constant for all $i$.

Edit: I have some control over $x$ so I chose $x=1/2$ and used an equality instead of an inequality, then applied RSolve.

Simplify[RSolve[(1/2) a[n] == (1/2) a[n - 1] + (3/2)*a[-2 + n] + c, a[n], n]]

I got the result $$a(n)\to c_1 \left(\frac{1}{2} \left(1-\sqrt{13}\right)\right)^n+c_2 \left(\frac{1}{2} \left(\sqrt{13}+1\right)\right)^n-\frac{2 c}{3}.$$ However what I need is the inequality, and in terms of $x$. Do you know any other method in Mathematica available to do this?

  • $\begingroup$ As the error message says, RSolve is expecting an equation. You gave it an inequality. Change <= to == and it will return a result. $\endgroup$
    – Carl Woll
    Oct 29 '18 at 2:37
  • $\begingroup$ Hi, I am open to other ways to solve the recurrence without RSolve. I have searched but I can't find the functions I need. $\endgroup$
    – Cogicero
    Oct 29 '18 at 2:38
  • $\begingroup$ Thanks Bill, that works, but my problem is that $a$ is definitely not constant for all $n$ in my case. $\endgroup$
    – Cogicero
    Oct 29 '18 at 3:14
  • $\begingroup$ Thanks. This is one step in some numerical analysis. I know that $a$ depends on $n$ i.e. $a_i$ is not constant for all $i$. I know that $x > 0$ and $c > 0$ while $n >= 1$. $x \in \mathbb{Q}$ while $a_i \in \mathbb{R}$ and $c \in \mathbb{R}$. $\endgroup$
    – Cogicero
    Oct 29 '18 at 3:25
  • 1
    $\begingroup$ I don't understand the problem yet. Let's take a simple special case, say a0=0 and a1=1. What kind of answer would you expect to see? $\endgroup$
    – bill s
    Oct 29 '18 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.