How do I solve this recursive equation which involves square root?

Question

Please suggest a way to solve the following recursive equation: $$x=\sqrt{1+a^2_{n} e^2+2 a_{n} e f}$$ where $$a_{n}=\sqrt{1+a^2_{n-1} e^2+2 a_{n-1} e f}, a_{n-1}=\sqrt{1+a^2_{n-2} e^2+2 a_{n-2} e f}$$ and so on. Does this type of equations have a special name?

Answer 1

Such non-linear recurrences very seldom have closed form solutions.

What are $e$, $f$? Ranges of interest? For instance, for $e = f = 1$ it reduces to $a_{n + 1} = \lvert a_n + 1 \rvert$. What is $a_0$ (your starting point)? Any further information, i.e., is $a_n$ expected to be large or small with respect to $1$?

In what are you interested? Is it enough to e.g. get a limit for $a_n$ as $n \to \infty$? Is an asymptotic bound enough?

Answer 2

If you meant to have $a_{n+1}$ instead of $x$, these kinds of equations may be solved only asymptotically. I replace constants to avoid confusion of what you meant by $e$.

$$a_{n+1}^2=1+a_{n}^2d^2+2a_{n}dc$$

Let us write

$$a_{n}=g_{n}e^n$$

This leads to

$$g_{n+1}^2=\frac1{e^{2(n+1)}}+\frac{g_{n}^2e^{2n}d^2}{e^{2(n+1)}}+\frac{2g_{n}e^{n}dc}{e^{2(n+1)}}$$

$$g_{n+1}^2 \sim \frac{g_{n}^2d^2}{e^2}=g_{n}^2h$$

or

$$g_{n+1} \sim g_{n}\sqrt{h}$$

which has an obvious solution $g_{n+1} \sim kc^n$

Obviously the asymptotic solution, the solution at infinity, is

$$a_{n} \sim k_1e^{k_2n}$$

where you need to adjust $k_1$ and $k_2$ based on the constants involved and initial conditions.

Typically what you do next once you have asymptotic solution is to express $a_n$ using multiplication or addition with another function trying to get closer to the expression for smaller values of $n$. This typically leads to obtaining a solution in form of series. If you achieve a convergent series and there is some generic way of expressing the expansion terms, technically, you have found a solution.