Solution of the Rational Difference Equation

Difference equations appear naturally as discrete analogs and as numerical solutions of differential and delay differential equations, having applications in biology, ecology, physics.

Recently, a high attention to studying the periodic nature of nonlinear difference equations has been attracted. For some recent results concerning the periodic nature of scalar nonlinear difference equations, among other problems, see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33].

Amleh [1], studied the global stability, the boundedness character, and the periodic nature of the positive solutions of the difference equation: $x_{n + 1} = α + \frac{x_{n - 1}}{x_{n}}$ {x_{n + 1}} = \alpha + {{{x_{n - 1}}} \over {{x_n}}} where α ∈ [0, ∞) and where the initial conditions x₋₁ and x₀ are arbitrary positive real numbers.

De Vault [8], studied the following problems $x_{n + 1} = \frac{A}{x_{n}} + \frac{1}{x_{n - 2}}, n = 0, 1, \dots,$ {x_{n + 1}} = {A \over {{x_n}}} + {1 \over {{x_{n - 2}}}},\quad n = 0,1, \ldots, and showed every positive solution of the equation where A ∈ (0, ∞).

Elsayed [15], studied the global result, boundedness, and periodicity of solutions of the difference equation $x_{n + 1} = a + \frac{{bx}_{n - 1} + {cx}_{n - k}}{{dx}_{n - 1} + {ex}_{n - k}}, n = 0, 1, \dots,$ {x_{n + 1}} = a + {{{bx_{n - 1}} + {cx_{n - k}}} \over {{dx_{n - 1}} + {ex_{n - k}}}},\quad n = 0,1, \ldots, where the parameters a, b, c, d and e are positive real numbers and the initial conditions x_−t,x_−t+1,...,x₀ are positive real numbers where t = max{l,k}, l ≠ k.

Ibrahim [18], studied the solutions of non-linear difference equation $x_{n + 1} = \frac{x_{n} x_{n - 2}}{x_{n - 1} (a + {bx}_{n} x_{n - 2})}, n = 0, 1, \dots,$ {x_{n + 1}} = {{{x_n}{x_{n - 2}}} \over {{x_{n - 1}}(a + {bx_n}{x_{n - 2}})}},\quad n = 0,1, \ldots, where the initial values x₀,x₋₁ and x₋₂ non-negative real numbers with bx₀x₋₂ ≠ −a and x₋₁ ≠ 0. He investigated some properties for this difference equation such as the local stability and the boundedness for the solutions.

Simsek et. al. [25,26,27] and [30], studied the following problems with positive initial values $\begin{matrix} x_{n + 1} = \frac{x_{n - 3}}{1 + x_{n - 1}}, n = 0, 1, \dots, \\ x_{n + 1} = \frac{x_{n - 5}}{1 + x_{n - 2}}, n = 0, 1, \dots, \\ x_{n + 1} = \frac{x_{n - 5}}{1 + x_{n - 1} x_{n - 3}}, n = 0, 1, \dots, \\ x_{n + 1} = \frac{x_{n - 3}}{1 + x_{n} x_{n - 1} x_{n - 2}}, n = 0, 1, \dots, \end{matrix}$ \matrix{ {{x_{n + 1}} = {{{x_{n - 3}}} \over {1 + {x_{n - 1}}}},\quad n = 0,1, \ldots,} \cr {{x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 2}}}},\quad n = 0,1, \ldots,} \cr {{x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 1}}{x_{n - 3}}}},\quad n = 0,1, \ldots,} \cr {{x_{n + 1}} = {{{x_{n - 3}}} \over {1 + {x_n}{x_{n - 1}}{x_{n - 2}}}},\quad n = 0,1, \ldots,}} respectively.

In this work, the following non-linear difference equation is studied (1) $x_{n + 1} = \frac{x_{n - 13}}{1 + x_{n - 1} x_{n - 3} x_{n - 5} x_{n - 7} x_{n - 9} x_{n - 11}}, n = 0, 1, \dots,$ {x_{n + 1}} = {{{x_{n - 13}}} \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}},\quad n = 0,1, \ldots, where x₋₁₃,x₋₁₂,...,x₋₁,x₀ ∈ (0, ∞) is investigated.

Main Result

Let $\bar{x}$ \bar x be the unique positive equilibrium of the equation 1, then clearly, $\bar{x} = \frac{\bar{x}}{1 + \bar{x} . \bar{xxxxx}} \Rightarrow \bar{x} + {\bar{x}}^{7} = \bar{x} \Rightarrow {\bar{x}}^{7} = 0 \Rightarrow \bar{x} 0,$ \bar x = {{\bar x} \over {1 + \bar x.\overline {xxxxx} }} \Rightarrow \bar x + {\bar x^7} = \bar x \Rightarrow {\bar x^7} = 0 \Rightarrow \bar x0, so, $\bar{x} = 0$ \bar x = 0 can be obtained. For any k ≥ 0 and m > k, notation $i = \bar{k, m}$ i = \overline {k,m} means i = k,k + 1,...,m.

Theorem 1

Consider the difference equation 1. Then the following statements are true: a)

The sequences (x_14n−13), (x_14n−12),...(x_14n−1), (x_14n) are decreased and a₁,..., a₁₄ ≥ 0 is existed in such that: $lim_{n \to \infty} x_{14 n - 13 + k} = a_{1 + k}, k = \bar{0, 13} .$ \lim\limits_{n \to \infty } {x_{14n - 13 + k}} = {a_{1 + k}},\quad k = \overline {0,13}.

(a₁,a₂,a₃,a₄,a₅,a₆,a₇,a₈,a₉,a₁₀,a₁₁,a₁₂,a₁₃,a₁₄,...) is a solution of 1 having period fourteen.

$\prod_{k = 0}^{6} lim_{n \to \infty} x_{14 n - j + 2 k} = 0, j = \bar{0, 1};$ \prod\limits_{k = 0}^6 \lim\limits_{n \to \infty } {x_{14n - j + 2k}} = 0,\quad j = \overline {0,1}; or $\prod_{k = 0}^{6} a_{2 k + i} = 0, i = \bar{0, 1} .$ \prod\limits_{k = 0}^6 {a_{2k + i}} = 0,\quad i = \overline {0,1}.

If there exist n₀ ∈ ℕ such that x_n+1 ≤ x_n−11 for all n ≥ n₀, then, $lim_{n \to \infty} x_{n} = 0 .$ \lim\limits_{n \to \infty } {x_n} = 0.

The following formulas below: $\begin{matrix} x_{14 n + k + 1} = x_{- 13 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 1}} = {x_{ - 13 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 3} = x_{- 11 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 3}} = {x_{ - 11 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 5} = x_{- 9 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 5}} = {x_{ - 9 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 7} = x_{- 7 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 7}} = {x_{ - 7 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 9} = x_{- 5 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 9}} = {x_{ - 5 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 11} = x_{- 3 + k} (1 - \frac{x_{- 1 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 11}} = {x_{ - 3 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 13} = x_{- 1 + k} (1 - \frac{x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 6} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 13}} = {x_{ - 1 + k}}(1 - {{{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} { \times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 6} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $k = \bar{0, 1}$ k = \overline {0,1} , holds.

If x_14n+1+k → a_k+1 ≠ 0, x_14n+3+k → a_k+3 ≠ 0, x_14n+5+k → a_k+5 ≠ 0, x_14n+7+k → a_k+7 ≠ 0, x_14n+9+k → a_k+9 ≠ 0, x_14n+11+k → a_k+11 ≠ 0 then x_14n+13+k → 0 as n → ∞, $k = \bar{0, 1}$ k = \overline {0,1} .

Proof

Firstly, from 1, $x_{n + 1} (1 + x_{n - 1} x_{n - 3} x_{n - 5} x_{n - 7} x_{n - 9} x_{n - 11}) = x_{n - 13},$ {x_{n + 1}}(1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}) = {x_{n - 13}}, is obtained. If x_n−1x_n−3x_n−5x_n−7x_n−9x_n−11 ∈ (0, +∞), then 1 + x_n−1x_n−3x_n−5x_n−7x_n−9x_n−11 ∈ (1, +∞).

Since $x_{n + 1} < x_{n - 13},$ {x_{n + 1}} < {x_{n - 13}},n ∈ ℕ $lim_{n \to \infty} x_{14 n - 13 + k} = a_{1 + k}, for k = \bar{0, 13}$ \lim\limits_{n \to \infty } {x_{14n - 13 + k}} = {a_{1 + k}},\quad for\;\;k = \overline {0,13} existed formulas are obtained.

(a₁,a₂,a₃,a₄,a₅,a₆,a₇,a₈,a₉,a₁₀,a₁₁,a₁₂,a₁₃,a₁₄,...) is a solution of 1 having period fourteen.

In view of the 1, $n = 14 n \Rightarrow x_{14 n + 1} = \frac{x_{14 n - 13}}{1 + \prod_{k = 0}^{5} lim_{n \to \infty} x_{14 n - 11 + 2 k}},$ n = 14n \Rightarrow {x_{14n + 1}} = {{{x_{14n - 13}}} \over {1 + \prod_{k = 0}^5 \lim\limits_{n \to \infty } {x_{14n - 11 + 2k}}}}, is obtained. If the limits are put on both sides of the above equality, $\prod_{k = 0}^{6} lim_{n \to \infty} x_{14 n - 13 + 2 k} = 0 or \prod_{k = 0}^{6} a_{2 k + 1} = 0,$ \prod\limits_{k = 0}^6 \lim\limits_{n \to \infty } {x_{14n - 13 + 2k}} = 0\quad or\quad \prod\limits_{k = 0}^6 {a_{2k + 1}} = 0, is obtained. Similarly for n = 14n + 1, $n = 14 n + 1 \Rightarrow x_{14 n + 2} = \frac{x_{14 n - 12}}{1 + \prod_{k = 0}^{5} lim_{n \to \infty} x_{14 n - 10 + 2 k}},$ n = 14n + 1 \Rightarrow {x_{14n + 2}} = {{{x_{14n - 12}}} \over {1 + \prod_{k = 0}^5 \lim\limits_{n \to \infty } {x_{14n - 10 + 2k}}}}, is obtained. If the limits are put on both sides of the above equality, $\prod_{k = 0}^{6} lim_{n \to \infty} x_{14 n - 12 + 2 k} = 0 or \prod_{k = 0}^{6} a_{2 k + 2} = 0,$ \prod\limits_{k = 0}^6 \lim\limits_{n \to \infty } {x_{14n - 12 + 2k}} = 0\quad or\quad \prod\limits_{k = 0}^6 {a_{2k + 2}} = 0, is obtained.

If there exist n₀ ∈ ℕ such that x_n+1 ≤ x_n−11 for all n ≥ n₀, then, a₁ ≤ a₃ ≤ a₅ ≤ a₇ ≤ a₉ ≤ a₁₁ ≤ a₁₃ ≤ a₁, a₂ ≤ a₄ ≤ a₆ ≤ a₈ ≤ a₁₀ ≤ a₁₂ ≤ a₁₄ ≤ a₂. Using (c) we get $\prod_{k = 0}^{6} a_{2 k + i} = 0, i = \bar{1, 2} .$ \prod\limits_{k = 0}^6 {a_{2k + i}} = 0,\quad i = \overline {1,2}. Then we see that, $lim_{n \to \infty} x_{n} = 0 .$ \lim\limits_{n \to \infty } {x_n} = 0. Hence the proof of (d) completed.

Subracting x_n−13 from the left and right-hand sides in 1, $x_{n + 1} - x_{n - 13} = \frac{1}{1 + x_{n - 1} x_{n - 3} x_{n - 5} x_{n - 7} x_{n - 9} x_{n - 11}} (x_{n - 1} - x_{n - 15}),$ {x_{n + 1}} - {x_{n - 13}} = {1 \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}}({x_{n - 1}} - {x_{n - 15}}), is obtained and the following formula is produced below, for n ≥ 2 (2) $\begin{array}{l} x_{2 n - 3} - x_{2 n - 17} = (x_{1} - x_{- 13}) \prod_{i = 1}^{n - 2} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}, \\ x_{2 n - 2} - x_{2 n - 16} = (x_{1} - x_{- 12}) \prod_{i = 1}^{n - 2} \frac{1}{1 + x_{2 i} x_{2 i - 2} x_{2 i - 4} x_{2 i - 6} x_{2 i - 8} x_{2 i - 10}}, \end{array}$ \matrix{ {{x_{2n - 3}} - {x_{2n - 17}} = ({x_1} - {x_{ - 13}})\prod\limits_{i = 1}^{n - 2} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}},} \cr {{x_{2n - 2}} - {x_{2n - 16}} = ({x_1} - {x_{ - 12}})\prod\limits_{i = 1}^{n - 2} {1 \over {1 + {x_{2i}}{x_{2i - 2}}{x_{2i - 4}}{x_{2i - 6}}{x_{2i - 8}}{x_{2i - 10}}}},}\,\,\,\,} 7 j inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 1 + k} - x_{- 13 + k} = (x_{1 + k} - x_{- 13 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 1 + k}} - {x_{ - 13 + k}} = ({x_{1 + k}} - {x_{ - 13 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 1 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 3 + k} - x_{- 11 + k} = (x_{3 + k} - x_{- 11 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 3 + k}} - {x_{ - 11 + k}} = ({x_{3 + k}} - {x_{ - 11 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 2 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 5 + k} - x_{- 9 + k} = (x_{5 + k} - x_{- 9 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 5 + k}} - {x_{ - 9 + k}} = ({x_{5 + k}} - {x_{ - 9 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 3 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 7 + k} - x_{- 7 + k} = (x_{7 + k} - x_{- 7 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 7 + k}} - {x_{ - 7 + k}} = ({x_{7 + k}} - {x_{ - 7 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 4 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 9 + k} - x_{- 5 + k} = (x_{9 + k} - x_{- 5 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 9 + k}} - {x_{ - 5 + k}} = ({x_{9 + k}} - {x_{ - 5 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 5 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 11 + k} - x_{- 3 + k} = (x_{11 + k} - x_{- 3 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 11 + k}} - {x_{ - 3 + k}} = ({x_{11 + k}} - {x_{ - 3 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Also, 7 j + 6 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k = \bar{0, 1}$ k = \overline {0,1} , $x_{14 n + 13 + k} - x_{- 1 + k} = (x_{13 + k} - x_{- 1 + k}) \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 6} \frac{1}{1 + x_{2 i - 1 + k} x_{2 i - 3 + k} x_{2 i - 5 + k} x_{2 i - 7 + k} x_{2 i - 9 + k} x_{2 i - 11 + k}} .$ {x_{14n + 13 + k}} - {x_{ - 1 + k}} = ({x_{13 + k}} - {x_{ - 1 + k}})\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 6} {1 \over {1 + {x_{2i - 1 + k}}{x_{2i - 3 + k}}{x_{2i - 5 + k}}{x_{2i - 7 + k}}{x_{2i - 9 + k}}{x_{2i - 11 + k}}}}. Now we obtained of the above formulas: $\begin{matrix} x_{14 n + k + 1} = x_{- 13 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 1}} = {x_{ - 13 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 3} = x_{- 11 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 3}} = {x_{ - 11 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} \begin{array}{l} x_{14 n + k + 5} = x_{- 9 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{array} \end{matrix}$ \eqalign{& {{x_{14n + k + 5}} = {x_{ - 9 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 7} = x_{- 7 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 7}} = {x_{ - 7 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 9} = x_{- 5 + k} (1 - \frac{x_{- 1 + k} x_{- 3 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 9}} = {x_{ - 5 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 11} = x_{- 3 + k} (1 - \frac{x_{- 1 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}), \end{matrix}$ \eqalign{& {{x_{14n + k + 11}} = {x_{ - 3 + k}}(1 - {{{x_{ - 1 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}),}} $\begin{matrix} x_{14 n + k + 13} = x_{- 1 + k} (1 - \frac{x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k} x_{- 13 + k}}{1 + x_{- 1 + k} x_{- 3 + k} x_{- 5 + k} x_{- 7 + k} x_{- 9 + k} x_{- 11 + k}} \\ \times \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 6} \frac{1}{1 + x_{2 i - 11 + k} x_{2 i - 9 + k} x_{2 i - 7 + k} x_{2 i - 5 + k} x_{2 i - 3 + k} x_{2 i - 1 + k}}) . \end{matrix}$ \eqalign{& {{x_{14n + k + 13}} = {x_{ - 1 + k}}(1 - {{{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}{x_{ - 13 + k}}} \over {1 + {x_{ - 1 + k}}{x_{ - 3 + k}}{x_{ - 5 + k}}{x_{ - 7 + k}}{x_{ - 9 + k}}{x_{ - 11 + k}}}}} \cr & {\kern 20pt}{\kern 50pt}{\kern 50pt}{\kern 50pt} {\times \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 6} {1 \over {1 + {x_{2i - 11 + k}}{x_{2i - 9 + k}}{x_{2i - 7 + k}}{x_{2i - 5 + k}}{x_{2i - 3 + k}}{x_{2i - 1 + k}}}}).}}

Suppose that a₁ = a₃ = a₅ = a₇ = a₉ = a₁₁ = a₁₃ = 0. By (e), we have (3) $\begin{matrix} lim n \to \infty x_{14 n + 1} = lim_{n \to \infty} x_{- 13} (1 - \frac{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}} \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}), \\ a_{1} = x_{- 13} (1 - \frac{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}} \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}) \\ a_{1} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} . \end{matrix}$ \matrix{ {\lim n \to \infty {x_{14n + 1}} = \lim\limits_{n \to \infty } {x_{ - 13}}\left( {1 - {{{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}}}\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}} \right),} \cr {{a_1} = {x_{ - 13}}\left( {1 - {{{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}}}\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}} \right)} \cr {{a_1} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}.}} Similarly, (4) $a_{3} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_3} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. Similarly, (5) $a_{5} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_5} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. Similarly, (6) $a_{7} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_7} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. Similarly, (7) $a_{9} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_9} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. Similarly, (8) $a_{11} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 5} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_{11}} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. Similarly, (9) $a_{13} = 0 \Rightarrow \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 6} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} .$ {a_{13}} = 0 \Rightarrow {{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 6} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}. From 3 and 4, $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}, \end{array}$ \matrix{ {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}}} =} & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} >} \hfill \cr & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}},}} thus, x₋₁₃ > x₋₁₁. From 4 and 5 $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 13}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 1} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}, \end{array}$ \matrix{ {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 13}}}} = } & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 1} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} > } \hfill \cr & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}},}} thus, x₋₁₁ > x₋₉. From 5 and 6, $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 11} x_{- 13}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 2} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}, \end{array}$ \matrix{{{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 11}}{x_{ - 13}}}} = } & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 2} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} > } \hfill \cr {} \hfill & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}},}} thus, x₋₉ > x₋₇. From 6 and 7, $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 5} x_{- 9} x_{- 11} x_{- 13}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 3} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} \end{array}$ \matrix{{{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = } \hfill & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 3} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} > } \hfill \cr {} \hfill & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}}} thus, x₋₇ > x₋₅. From 7 and 8, $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 3} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 4} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 5} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} \end{array}$ \matrix{{{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 3}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = } \hfill & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 4} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} > } \hfill \cr {} \hfill & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}}}} thus, x₋₅ > x₋₃. From 8 and 9, $\begin{array}{l} \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 1} x_{- 5} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = & \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 5} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}} > \\ \frac{1 + x_{- 1} x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11}}{x_{- 3} x_{- 5} x_{- 7} x_{- 9} x_{- 11} x_{- 13}} = \sum_{j = 0}^{n} \prod_{i = 1}^{7 j + 6} \frac{1}{1 + x_{2 i - 1} x_{2 i - 3} x_{2 i - 5} x_{2 i - 7} x_{2 i - 9} x_{2 i - 11}}, \end{array}$ \matrix{{{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 1}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = } \hfill & {\sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 5} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}} > } \hfill \cr {} \hfill & {{{1 + {x_{ - 1}}{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}} \over {{x_{ - 3}}{x_{ - 5}}{x_{ - 7}}{x_{ - 9}}{x_{ - 11}}{x_{ - 13}}}} = \sum\limits_{j = 0}^n \prod\limits_{i = 1}^{7j + 6} {1 \over {1 + {x_{2i - 1}}{x_{2i - 3}}{x_{2i - 5}}{x_{2i - 7}}{x_{2i - 9}}{x_{2i - 11}}}},}} thus, x₋₃ > x₋₁. From here we obtain x₋₁₃ > x₋₁₁ > x₋₉ > x₋₇ > x₋₅ > x₋₃ > x₋₁. Similarly, we can obtain x₋₁₂ > x₋₁₀ > x₋₈ > x₋₆ > x₋₄ > x₋₂ > x₀. We arrive at a contradiction which completes the proof of theorem.

Example 2

If the initial conditions are selected as follows: $\begin{array}{l} x_{- 13} = 0.98, x_{- 12} = 0.97, x_{- 11} = 0.96, x_{- 10} = 0.95, x_{- 9} = 0.94, x_{- 8} = 0.93, x_{- 7} = 0.92, \\ x_{- 6} = 0.91, x_{- 5} = 0.9, x_{- 4} = 0.89, x_{- 3} = 0.88, x_{- 2} = 0.87, x_{- 1} = 0.86, x_{0} = 0.85 . \end{array}$ \eqalign{&{{x_{ - 13}} = 0.98,\;\;{x_{ - 12}} = 0.97,\;\;{x_{ - 11}} = 0.96,\;\;{x_{ - 10}} = 0.95,\;\;{x_{ - 9}} = 0.94,\;\;{x_{ - 8}} = 0.93,\;\;{x_{ - 7}} = 0.92,} \cr & {{x_{ - 6}} = 0.91,\;\;{x_{ - 5}} = 0.9,\;\;{x_{ - 4}} = 0.89,\;\;{x_{ - 3}} = 0.88,\;\;{x_{ - 2}} = 0.87,\;\;{x_{ - 1}} = 0.86,\;\;{x_0} = 0.85.}}

The graph of the solution is given below, x_n = {0.62601, 0.634341, 0.701375, 0.701974, 0.737178, 0.734195, 0.753815, 0.748866, 0.759718, 0.753567, 0.759006, 0.752062, 0.753929, 0.746432, 0.535307, 0.545309, 0.621059, 0.622964, 0.664753, 0.66286, 0.687738, 0.683736, 0.698929, 0.693618, 0.702724, 0.696539, 0.701557, 0.694753, 0.487602, 0.498104, 0.576757, 0.579084, 0.623219, 0.621705, 0.64857, 0.644917, 0.661839, 0.656857, 0.667494,...}

eISSN:: 2444-8656
Idioma:: Inglés

Calendario de la edición:: Volume Open
Temas de la revista:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

RSS Feed de revista

Solution of the Rational Difference Equation xn+1=xn−131+xn−1xn−3xn−5xn−7xn−9xn−11{x_{n + 1}} = {{{x_{n - 13}}} \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}}

Publicado en línea: 10 abr 2020

Páginas: 485 - 494

Recibido: 21 may 2019

Aceptado: 29 jul 2019

DOI: https://doi.org/10.2478/amns.2020.1.00047

Palabras claveDifference equations, rational difference equations

© 2019 Dagistan Simsek et al., published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Fig. 1

Solution of the Rational Difference Equation $x_{n + 1} = \frac{x_{n - 13}}{1 + x_{n - 1} x_{n - 3} x_{n - 5} x_{n - 7} x_{n - 9} x_{n - 11}}$ {x_{n + 1}} = {{{x_{n - 13}}} \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}}

Palabras clave
Difference equations, rational difference equations