1 Introduction

Let \({{\mathbb {N}}}\), \({{\mathbb {Z}}}\), \({{\mathbb {R}}}\), \({{\mathbb {C}}}\) stand for the sets of natural, integer, real and complex numbers respectively, \({{\mathbb {N}}}_0={{\mathbb {N}}}\cup \{0\}\) and \({{\mathbb {R}}}_+=(0,\infty )\). Assume that \(s,t\in {{\mathbb {Z}}}\), then we use the notation \(j=\overline{s,t}\) instead of writing \(s\le j\le t\), \(j\in {{\mathbb {Z}}}\). By \(C^n_j\), \(n\in {{\mathbb {N}}}_0\), \(j=\overline{0,n},\) we denote the binomial coefficients.

There are many recursive relations by which square roots of numbers can be found. One of the most known ones is the following

$$\begin{aligned} x_{n+1}=\frac{1}{2}\left( x_n+\frac{a}{x_n}\right) ,\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$
(1)

where \(a\in {{\mathbb {R}}}_+\). It is well known that for any \(x_0\in {{\mathbb {R}}}_+\), such defined sequence \(x_n\) converges to \(\sqrt{a}\), when \(n\rightarrow +\infty \) (see, e.g., [5, 7, 8]).

Equation (1) can be obtained by choosing the function

$$\begin{aligned} f(x)=x^2-a\end{aligned}$$
(2)

in the Newton iteration process

$$\begin{aligned} x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(3)

(see, e.g., [8, 37]).

A square root of a number a can be also obtained by using the recursive relation

$$\begin{aligned} x_{n+2}=\frac{x_{n+1}x_n+a}{x_{n+1}+x_n},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(4)

It is interesting that (4) can be obtained by using the secant method

$$\begin{aligned} x_{n+2}=\frac{f(x_{n+1})x_n-f(x_n)x_{n+1}}{f(x_{n+1})-f(x_n)},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$
(5)

to the function in (2) (see, e.g., [39]).

Another recursive relation for finding a square root of a number a is the following

$$\begin{aligned} x_{n+1}=\frac{x_n^3+3ax_n}{3x_n^2+a},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(6)

This relation seems not so known as (1), but also frequently appears in the literature (see, e.g., [17, 23, 30]).

Since (6) is a recursive relation of the form

$$\begin{aligned} x_{n+1}=f(x_n),\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

where the function

$$\begin{aligned} f(t)=\frac{t^3+3at}{3t^2+a} \end{aligned}$$

for \(a\in {{\mathbb {R}}}_+\), maps \({{\mathbb {R}}}_+\) into itself, convergence of its positive solutions can be dealt with by using some standard arguments (see, e.g., [5, Problems 9.34, 9.35]) in various ways (see, e.g., the arguments in [23, 30]).

What is interesting related to recursive relations (1) and (4), is the fact that their solutions can be found in closed form (see, e.g., [7, 8, 19, 37, 39]). These are two examples of nonlinear recursive relations/difference equations solvable in closed form. Generally speaking, solvability of nonlinear difference eqations is a rare phenomena. This is one of the reasons why for nonlinear difference equations is also studied the existence of invariants. For some recent related results on solvability and invariants for nonlinear difference equations see, e.g., [3, 4, 25,26,27, 29, 31, 32, 34,35,36, 38, 40, 41] and the related references therein. Some old results on the topics can be found, e.g., in [1, 2, 6, 10, 11, 14, 20,21,22,23,24]. For some other related equations see also [13, 15, 16, 18, 28, 33, 42].

Here we show how recursive relation (6) can be obtained from an iteration process in the literature. Beside this, we show that the recursive relation is also solvable in closed form by presenting its general solution, and present a class of recursive relations for finding square roots which are also solvable in closed form, extending and unifying recursive relations (1) and (6).

2 Main results

This section presents our main results in this paper.

2.1 An iteration process forming (6)

It is a natural question if recursive relation (6) can be also obtained from a known iteration process. Here we deal with the problem.

The considerations in the previous section suggest that the requested iteration process should be related to the Newton one. To find the roots of the algebraic equation

$$\begin{aligned} f(x)=0 \end{aligned}$$

on an interval I, where f is a given function, is the same as to find the roots of the equation

$$\begin{aligned} f(x)g(x)=0 \end{aligned}$$

if the roots of the function g does not belong to the interval. Hence, in (3), instead of f we can use the function fg ( [9]), and obtain

$$\begin{aligned} x_{n+1}=x_n-\frac{f(x_n)g(x_n)}{f'(x_n)g(x_n)+f(x_n)g'(x_n)},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(7)

By suitable choice of function g we can obtain various iteration processes.

Let

$$\begin{aligned} g(t)=t^b,\quad b\in {{\mathbb {R}}}. \end{aligned}$$

Then, by some simple calculations (7) becomes

$$\begin{aligned} x_{n+1}=x_n-\frac{f(x_n)x_n}{f'(x_n)x_n+bf(x_n)}=x_n\frac{f'(x_n)x_n+(b-1)f(x_n)}{f'(x_n)x_n+bf(x_n)},\quad n\in {{\mathbb {N}}}_0.\qquad \end{aligned}$$
(8)

Since we want to find a root of a number a it is natural to choose the function f as in (2). By using it in (8) we obtain

$$\begin{aligned} x_{n+1}=x_n\frac{2x_n^2+(b-1)(x_n^2-a)}{2x_n^2+b(x_n^2-a)}=x_n\frac{(b+1)x_n^2-a(b-1)}{(b+2)x_n^2-ab},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(9)

To get recursive relation (6) from (9) is only possible if

$$\begin{aligned} \frac{b+1}{b+2}=\frac{1}{3}, \end{aligned}$$

from which it follows that \(b=-1/2.\) By using such obtained b in (9) we really obtain (6).

The above consideration shows that recursive relation (6) is obtained from the iteration process in (7) with

$$\begin{aligned} g(t)=t^{-\frac{1}{2}}.\end{aligned}$$
(10)

Note that function (10) does not have a zero on \({{\mathbb {R}}}_+\), so the functions

$$\begin{aligned} t^2-a\quad \text{ and } \quad (t^2-a)t^{-\frac{1}{2}} \end{aligned}$$

have the same set of zeros on \({{\mathbb {R}}}_+.\)

2.2 Solvability of recursive relation (6)

The fact that recursive relations (1) and (4) are solvable in closed form naturally suggests investigation of solvability of recursive relation (6). Here we deal with the problem.

It should be pointed out that it is a quite rare situation when nonlinear recursive relation is solvable in closed form. This is why majority of authors use other methods for dealing with the nonlinear relations, which was also the case in [23] and [30], where was considered relation (6).

Let \(x^*\) be an equilibrium of (6). Then we have

$$\begin{aligned} x^*=x^*\frac{(x^*)^2+3a}{3(x^*)^2+a}, \end{aligned}$$

from which it follows that

$$\begin{aligned} \frac{2x^*((x^*)^2-a)}{3(x^*)^2+a}=0. \end{aligned}$$

Hence

$$\begin{aligned} x^*\in \{0,\sqrt{a},-\sqrt{a}\}. \end{aligned}$$

Since we calculate square root of a number different from zero, the equilibrium \(x^*=0\) is of no interest in the investigation.

What is interesting, is to consider the sequence

$$\begin{aligned} e_n=x_n-\sqrt{a},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

The difference is usually considered when is studied convergence of a sequence.

Simple calculation shows that

$$\begin{aligned} x_{n+1}-\sqrt{a}=\frac{(x_n-\sqrt{a})^3}{3x_n^2+a},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(11)

Further, similarly is obtained

$$\begin{aligned} x_{n+1}+\sqrt{a}=\frac{(x_n+\sqrt{a})^3}{3x_n^2+a},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(12)

From (11) and (12) we have

$$\begin{aligned} \frac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left( \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}\right) ^3,\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

from which it easily follows that

$$\begin{aligned} \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(13)

Finally, from (13) we obtain

$$\begin{aligned} x_n=\sqrt{a}\, \frac{1+\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n}}{1-\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n}},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(14)

Let

$$\begin{aligned} y_n:=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Then, (14) can be written as follows

$$\begin{aligned} x_n=\sqrt{a}\, \frac{1+y_n}{1-y_n},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Since

$$\begin{aligned} \frac{x_n^3+3ax_n}{3x_n^2+a}=&\frac{\left( \sqrt{a}\, \frac{1+y_n}{1-y_n}\right) ^3+3a\sqrt{a}\, \frac{1+y_n}{1-y_n}}{3(\sqrt{a}\, \frac{1+y_n}{1-y_n})^2+a}\\ =&\sqrt{a}\,\frac{(1+y_n)^3+3(1+y_n)(1-y_n)^2}{(1-y_n)(3(1+y_n)^2+(1-y_n)^2)}\\ =&\sqrt{a}\,\frac{1+y_n^3}{1-y_n^3}=\sqrt{a}\, \frac{1+\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^{n+1}}}{1-\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^{n+1}}}=x_{n+1} \end{aligned}$$

we have that the sequence defined in (14) is a solution to Eq. (6). In fact, (14) is the general solution to the equation.

From the previous detailed consideration we see that the following theorem on solvability of relation (6) holds.

Theorem 1

Assume that \(a\in {{\mathbb {R}}}_+\) and \(x_0\in {{\mathbb {R}}}_+\). Then, recursive relation (6) is solvable in closed form and its general solution is given by the formula

$$\begin{aligned} x_n=\sqrt{a}\, \frac{(x_0+\sqrt{a})^{3^n}+(x_0-\sqrt{a})^{3^n}}{(x_0+\sqrt{a})^{3^n}-(x_0-\sqrt{a})^{3^n}},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$
(15)

Remark 1

Note that in the above consideration we have not used the assumption \(x_0\in {{\mathbb {R}}}_+\). This means that formula (15) also holds if \(x_0\in {{\mathbb {R}}}\) or \(x_0\in {{\mathbb {C}}}.\) Beside this, the consideration also holds for any \(a\in {{\mathbb {C}}}\setminus \{0\}.\) Hence, formula (15) also holds in this case.

From (15) we also see that a solution to (6) is well defined if and only if

$$\begin{aligned} \left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n}\ne 1\end{aligned}$$
(16)

for every \(n\in {{\mathbb {N}}}.\) Note that if \(x_0, a\in {{\mathbb {R}}}_+\), then the relation in (16) holds for every \(n\in {{\mathbb {N}}}.\)

Remark 2

If \(a=0\), then (6) becomes

$$\begin{aligned} x_{n+1}=\frac{x_n^3}{3x_n^2},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(17)

from which it easily follows that

$$\begin{aligned} x_n=\frac{x_0}{3^n},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(18)

when \(x_0\ne 0.\) This means that recursive relation (6) is also solvable in this case and that the general solution in this case is given by formula (18). If \(x_0=0\), then the solution to (17) is not well-defined.

Remark 3

Note that from (11) we have

$$\begin{aligned} |x_{n+1}-\sqrt{a}|\le \frac{|x_n-\sqrt{a}|^3}{a},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

from which it follows that for \(a>1\) the sequence \(x_n\) converges to \(\sqrt{a}\) very quickly. Namely, according to the standard terminology, the rate of the convergence has the third order [12].

Remark 4

By using formula (15) the convergence result in [23, 30] immediately follows. Namely, if \(a>0\), then every positive solution to (6) is convergent. Indeed, if \(x_0>0\), then we have

$$\begin{aligned} \left| \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right| <1. \end{aligned}$$

Hence

$$\begin{aligned} \lim _{n\rightarrow +\infty }\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{3^n}=0. \end{aligned}$$

By letting \(n\rightarrow +\infty \) in formula (15) and using the last relation, the result easily follows.

2.3 Two other recursive relations for calculating square roots

It is a natural problem to find some other recursive relations for calculating square roots which are solvable in closed form.

Here we present two examples of such recursive relations, which are related to relations (1) and (6), and give some comments with respect to the obtained formulas for their solutions.

Consider the recursive relation

$$\begin{aligned} x_{n+1}=\frac{x_n^4+6ax_n^2+a^2}{4x_n^3+4ax_n},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(19)

Let \(x^*\) be an equilibrium of (19). Then we have

$$\begin{aligned} x^*=\frac{(x^*)^4+6a(x^*)^2+a^2}{4(x^*)^3+4ax^*}, \end{aligned}$$

from which it follows that

$$\begin{aligned} \frac{(3(x^*)^2+a)((x^*)^2-a)}{4(x^*)^3+4ax^*}=0. \end{aligned}$$

Hence

$$\begin{aligned} x^*\in \{\sqrt{a},-\sqrt{a}\}. \end{aligned}$$

By some calculation we have

$$\begin{aligned} x_{n+1}-\sqrt{a}=\frac{(x_n-\sqrt{a})^4}{4x_n^3+4ax_n},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(20)

and

$$\begin{aligned} x_{n+1}+\sqrt{a}=\frac{(x_n+\sqrt{a})^4}{4x_n^3+4ax_n},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(21)

From (20) and (21) we have

$$\begin{aligned} \frac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left( \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}\right) ^4,\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Hence

$$\begin{aligned} \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{4^n},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

and consequently

$$\begin{aligned} x_n=\sqrt{a}\, \frac{1+\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{4^n}}{1-\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{4^n}},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

As is the case of relation (6) it is verified that the last expression is a solution to relation (19).

From this we obtain the following theorem.

Theorem 2

Assume that \(a\in {{\mathbb {C}}}\setminus \{0\}\) and \(x_0\in {{\mathbb {C}}}\). Then, recursive relation (19) is solvable in closed form and its general solution is given by the formula

$$\begin{aligned} x_n=\sqrt{a}\, \frac{(x_0+\sqrt{a})^{4^n}+(x_0-\sqrt{a})^{4^n}}{(x_0+\sqrt{a})^{4^n}-(x_0-\sqrt{a})^{4^n}},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(22)

Remark 5

If \(a=0\), then (19) becomes

$$\begin{aligned} x_{n+1}=\frac{x_n^4}{4x_n^3},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(23)

from which it follows that

$$\begin{aligned} x_n=\frac{x_0}{4^n},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(24)

when \(x_0\ne 0.\) Hence, recursive relation (19) is also solvable in this case and its general solution is given by formula (24). If \(x_0=0\), then the solution to (23) is not well-defined.

Remark 6

By using formula (22), we easily see that for \(a>0\) every positive solution to (19) converges to \(\sqrt{a}.\)

Remark 7

Recursive relation (19) could be also obtained by a known iteration process or by a modification of a known iteration process, but, at the moment, we do not see from which one.

Now we consider the recursive relation

$$\begin{aligned} x_{n+1}=\frac{x_n^5+10ax_n^3+5a^2x_n}{5x_n^4+10ax_n^2+a^2},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(25)

Let \(x^*\) be an equilibrium of (25). Then we have

$$\begin{aligned} x^*=\frac{(x^*)^5+10a(x^*)^3+5a^2x^*}{5(x^*)^4+10a(x^*)^2+a^2}, \end{aligned}$$

from which it follows that

$$\begin{aligned} \frac{4x^*((x^*)^2+a)((x^*)^2-a)}{5(x^*)^4+10a(x^*)^2+a^2}=0. \end{aligned}$$

Hence

$$\begin{aligned} x^*\in \{0,\sqrt{a},-\sqrt{a}\}. \end{aligned}$$

The equilibrium \(x^*=0\) is of no special interest for calculating square roots.

We have

$$\begin{aligned} x_{n+1}-\sqrt{a}=\frac{(x_n-\sqrt{a})^5}{5x_n^4+10ax_n^2+a^2},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(26)

and

$$\begin{aligned} x_{n+1}+\sqrt{a}=\frac{(x_n+\sqrt{a})^5}{5x_n^4+10ax_n^2+a^2},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(27)

From (26) and (27) we have

$$\begin{aligned} \frac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left( \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}\right) ^5,\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Hence

$$\begin{aligned} \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{5^n},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

and consequently

$$\begin{aligned} x_n=\sqrt{a}\, \frac{1+\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{5^n}}{1-\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{5^n}},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

As is the case of relation (6) it is verified that the last expression is a solution to relation (25).

So, we get the following theorem.

Theorem 3

Assume that \(a\in {{\mathbb {C}}}\setminus \{0\}\) and \(x_0\in {{\mathbb {C}}}\). Then, recursive relation (25) is solvable in closed form and its general solution is given by the formula

$$\begin{aligned} x_n=\sqrt{a}\, \frac{(x_0+\sqrt{a})^{5^n}+(x_0-\sqrt{a})^{5^n}}{(x_0+\sqrt{a})^{5^n}-(x_0-\sqrt{a})^{5^n}},\end{aligned}$$
(28)

for \(n\in {{\mathbb {N}}}_0.\)

Remark 8

If \(a=0\), then (25) becomes

$$\begin{aligned} x_{n+1}=\frac{x_n^5}{5x_n^4},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(29)

from which it follows that

$$\begin{aligned} x_n=\frac{x_0}{5^n},\quad n\in {{\mathbb {N}}}_0,\end{aligned}$$
(30)

when \(x_0\ne 0.\) So, recursive relation (25) is also solvable in this case and its general solution is given by formula (30). If \(x_0=0\), then the solution to (29) is not well-defined.

Remark 9

From (26) we have

$$\begin{aligned} |x_{n+1}-\sqrt{a}|\le \frac{|x_n-\sqrt{a}|^5}{a^2},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

from which it follows that when \(a>1\) the sequence \(x_n\) converges to \(\sqrt{a}\) very quickly (the rate of convergence has the fifth order).

Remark 10

By using formula (28), we easily see that for \(a>0\) every positive solution to (25) converges to \(\sqrt{a}.\)

2.4 Two families of recursive relations generalizing the ones in (19) and (25)

Unlike relations (1) and (6), the recursive relations in (19) and (25) could be new (as far as we know they are new).

It is a natural question if the recursive relations are special cases of some classes of solvable nonlinear recursive relations. Here we deal with the problem. We present two families of recursive relations which naturally generalize the ones in (1), (6), (19) and (25). We show their solvability by presenting closed form formulas for their general solutions.

First, we consider a family of recursive relations, which generalizes relations (1) and (19).

Theorem 4

Let \(k\in {{\mathbb {N}}}\) and \(a\in {{\mathbb {C}}}\setminus \{0\}\). Consider the recursive relation

$$\begin{aligned} x_{n+1}=\frac{x_n^{2k}+C^{2k}_2ax_n^{2k-2}+C^{2k}_4a^2x_n^{2k-4}+\cdots +C^{2k}_{2k-2}a^{k-1}x_n^2+a^k}{C^{2k}_1x_n^{2k-1}+C^{2k}_3ax_n^{2k-3}+\cdots +C^{2k}_{2k-1}a^{k-1}x_n},\end{aligned}$$
(31)

for \(n\in {{\mathbb {N}}}_0,\) where \(x_0\in {{\mathbb {C}}}\). Then, relation (31) is solvable in closed form and its general solution is given by the formula

$$\begin{aligned} x_n=\sqrt{a}\, \frac{(x_0+\sqrt{a})^{(2k)^n}+(x_0-\sqrt{a})^{(2k)^n}}{(x_0+\sqrt{a})^{(2k)^n}-(x_0-\sqrt{a})^{(2k)^n}},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(32)

Proof

We have

$$\begin{aligned} x_{n+1}-\sqrt{a}=&\frac{x_n^{2k}+C^{2k}_2ax_n^{2k-2}+C^{2k}_4a^2x_n^{2k-4}+\cdots +C^{2k}_{2k-2}a^{k-1}x_n^2+a^k}{C^{2k}_1x_n^{2k-1}+C^{2k}_3ax_n^{2k-3}+\cdots +C^{2k}_{2k-1}a^{k-1}x_n}-\sqrt{a}\nonumber \\ =&\frac{(x_n-\sqrt{a})^{2k}}{C^{2k}_1x_n^{2k-1}+C^{2k}_3ax_n^{2k-3}+\cdots +C^{2k}_{2k-1}a^{k-1}x_n}, \end{aligned}$$
(33)

and

$$\begin{aligned} x_{n+1}+\sqrt{a}=&\frac{x_n^{2k}+C^{2k}_2ax_n^{2k-2}+C^{2k}_4a^2x_n^{2k-4}+\cdots +C^{2k}_{2k-2}a^{k-1}x_n^2+a^k}{C^{2k}_1x_n^{2k-1}+C^{2k}_3ax_n^{2k-3}+\cdots +C^{2k}_{2k-1}a^{k-1}x_n}+\sqrt{a}\nonumber \\ =&\frac{(x_n+\sqrt{a})^{2k}}{C^{2k}_1x_n^{2k-1}+C^{2k}_3ax_n^{2k-3}+\cdots +C^{2k}_{2k-1}a^{k-1}x_n}, \end{aligned}$$
(34)

for \(n\in {{\mathbb {N}}}_0.\)

From (33) and (34) we have

$$\begin{aligned} \frac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left( \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}\right) ^{2k},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Hence

$$\begin{aligned} \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{(2k)^n},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

from which formula (32) easily follows. \(\square \)

From Theorem 4 the following corollary easily follows.

Corollary 1

Assume that \(k\in {{\mathbb {N}}}\), \(a\in {{\mathbb {R}}}_+\). Then every positive solution to (31) converges to \(\sqrt{a}.\)

Now we consider a family of recursive relations, which generalizes relations (6) and (25).

Theorem 5

Let \(k\in {{\mathbb {N}}}\) and \(a\in {{\mathbb {C}}}\setminus \{0\}\). Consider the recursive relation

$$\begin{aligned}&x_{n+1}\nonumber \\&\quad =\frac{x_n^{2k+1}+C^{2k+1}_2ax_n^{2k-1}+C^{2k+1}_4a^2x_n^{2k-3}+\cdots +C^{2k+1}_{2k-2}a^{k-1}x_n^3+C^{2k+1}_{2k}a^kx_n}{C^{2k+1}_1x_n^{2k}+C^{2k+1}_3ax_n^{2k-2}+\cdots +C^{2k+1}_{2k-1}a^{k-1}x_n^2+a^k},\nonumber \\ \end{aligned}$$
(35)

for \(n\in {{\mathbb {N}}}_0,\) where \(x_0\in {{\mathbb {C}}}\). Then, relation (35) is solvable in closed form and its general solution is given by the formula

$$\begin{aligned} x_n=\sqrt{a}\, \frac{(x_0+\sqrt{a})^{(2k+1)^n}+(x_0-\sqrt{a})^{(2k+1)^n}}{(x_0+\sqrt{a})^{(2k+1)^n}-(x_0-\sqrt{a})^{(2k+1)^n}},\quad n\in {{\mathbb {N}}}_0.\end{aligned}$$
(36)

Proof

We have

$$\begin{aligned}&x_{n+1}-\sqrt{a}\nonumber \\&\quad =\frac{x_n^{2k+1}+C^{2k+1}_2ax_n^{2k-1}+C^{2k+1}_4a^2x_n^{2k-3}+\cdots +C^{2k+1}_{2k-2}a^{k-1}x_n^3+C^{2k+1}_{2k}a^kx_n}{C^{2k+1}_1x_n^{2k}+C^{2k+1}_3ax_n^{2k-2}+\cdots +C^{2k+1}_{2k-1}a^{k-1}x_n^2+a^k}-\sqrt{a}\nonumber \\&\quad =\frac{(x_n-\sqrt{a})^{2k+1}}{C^{2k+1}_1x_n^{2k}+C^{2k+1}_3ax_n^{2k-2}+\cdots +C^{2k+1}_{2k-1}a^{k-1}x_n^2+a^k}, \end{aligned}$$
(37)

and

$$\begin{aligned}&x_{n+1}+\sqrt{a}\nonumber \\&\quad = \frac{x_n^{2k+1}+C^{2k+1}_2ax_n^{2k-1}+C^{2k+1}_4a^2x_n^{2k-3}+\cdots +C^{2k+1}_{2k-2}a^{k-1}x_n^3+C^{2k+1}_{2k}a^kx_n}{C^{2k+1}_1x_n^{2k}+C^{2k+1}_3ax_n^{2k-2}+\cdots +C^{2k+1}_{2k-1}a^{k-1}x_n^2+a^k}+\sqrt{a}\nonumber \\&\quad = \frac{(x_n+\sqrt{a})^{2k+1}}{C^{2k+1}_1x_n^{2k}+C^{2k+1}_3ax_n^{2k-2}+\cdots +C^{2k+1}_{2k-1}a^{k-1}x_n^2+a^k}, \end{aligned}$$
(38)

for \(n\in {{\mathbb {N}}}_0.\)

From (37) and (38) we have

$$\begin{aligned} \frac{x_{n+1}-\sqrt{a}}{x_{n+1}+\sqrt{a}}=\left( \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}\right) ^{2k+1},\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

Hence

$$\begin{aligned} \frac{x_n-\sqrt{a}}{x_n+\sqrt{a}}=\left( \frac{x_0-\sqrt{a}}{x_0+\sqrt{a}}\right) ^{(2k+1)^n},\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

from which formula (36) follows. \(\square \)

From Theorem 5 the following corollary easily follows.

Corollary 2

Assume that \(k\in {{\mathbb {N}}}\), \(a\in {{\mathbb {R}}}_+\). Then every positive solution to (35) converges to \(\sqrt{a}.\)