Closed Form Formula for Nonlinear Recurrence $a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$

Question

I'm trying to find a closed form solution to the sequence

$a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$

I tried using a generating function approach in the following way:

Let $$f(x) = \sum_{n=1}^\infty a_n x^n$$ Then multiplying the original equation with $x^n$ and summing over all $n$, we get:

$$\sum_{n=1}^\infty a_{n+1}x^n= \sum_{n=1}^\infty \frac{a_{n}}{2} x^n + \sum_{n=1}^\infty\frac{5}{a_{n}} x^n$$ $$= \frac{f(x) - a_1}{x} = \frac{f(x)}{2} + 5 \sum_{n=1}^\infty \frac{x^n}{a_n}$$ But now I don't know how to simplify the last term and I get stuck. I put it on Wolframalpha and found the general solution to the recurrence relation as: $$a_n = -i \sqrt{10} \cot\left(c_1 2^n\right)$$

Answer 1

Here is an intuition: Rewrite the recurrence relation s

$$ a_{n+1} = \frac{1}{2}\left(a_n + \frac{\ell^2}{a_n}\right), \qquad \ell = \sqrt{10}. $$

This is precisely what we get when we apply the Newton-Raphson method to $x^2 - \ell^2$. So,

• for any positive initial value, $(a_n)$ converges to $\ell$, and
• for any negative initial value, $(a_n)$ converges to $-\ell$.

This naturally leads us to study the behavior of the sequences $(x_n^+)$ and $(x_n^-)$ defined by

$$ x_n^{\pm} = a_n \pm \ell. $$

Let us find the recurrence relation for these sequences:

$$ x_{n+1}^{\pm} = a_{n+1} \pm \ell = \frac{1}{2}\left(a_n + \frac{\ell^2}{a_n}\right) \pm \ell = \frac{(a_n \pm \ell)^2}{2a_n} = \frac{(x_n^{\pm})^2}{2a_n}. $$

This gives an intuitive explanation to @NN2's wonderful trick. Indeed, taking ratio of the two sequences,

$$ \frac{x_{n+1}^+}{x_{n+1}^-} = \left( \frac{x_{n}^+}{x_{n}^-} \right)^2, \qquad\text{i.e.,}\qquad \frac{a_{n+1} + \ell}{a_{n+1} - \ell} = \left( \frac{a_n + \ell}{a_n - \ell} \right)^2. $$

Iterating this, we get

$$ \frac{a_n + \ell}{a_n - \ell} = \left( \frac{a_0 + \ell}{a_0 - \ell} \right)^{2^n}. $$

Now solving this for $a_n$ gives the desired recurrence relation for $(a_n)$.

Answer 2

Let's make a change of variable $x_n = \frac{a_n}{\sqrt{10}}$, then $$x_{n+1} = \frac{x_n^2 +1}{2x_n}$$

We have $$x_{n+1} -1 = \frac{(x_n -1)^2}{2x_n}$$ $$x_{n+1} + 1 = \frac{(x_n +1)^2}{2x_n}$$ Then

$$\frac{x_{n+1} -1}{x_{n+1} + 1} = \left(\frac{x_n -1}{x_n +1}\right)^2$$

By reccurence, we obtain $$\frac{x_{n} -1}{x_{n} + 1} = \left(\frac{x_0 -1}{x_0 +1}\right)^{2^n}$$

and we can deduce the closed form solution of $x_n$ and then $a_n$

Answer 3

Based on the Wolfram Alpha answer, this is how one would get a closed form for $a_n$:

First, the substitution $a_n = b_n\sqrt{10}$ yield the equation

$$b_{n+1}=\frac{b_n+b_n^{-1}}{2}$$

Now take $b_n = \coth c_n$ to get

$$\coth c_{n+1} = \frac{\coth c_n + \tanh c_n}{2} = \coth (2c_n)$$

From this, $c_{n+1} = 2 c_n$. This means that $c_n = k\, 2^n$ for some $k$.

So $a_n = \sqrt{10} \coth(k\, 2^n)$.

It's not hard to check that this satisfies the recurrence relation.