1 Introduction
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:
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x n + 1 = α + 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 x 0 are arbitrary positive real numbers.
De Vault [8 ], studied the following problems
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x n + 1 = A x n + 1 x n − 2 , n = 0 , 1 , … , {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
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x n + 1 = a + bx n − 1 + cx n − k dx n − 1 + ex n − k , n = 0 , 1 , … , {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 ,...,x 0 are positive real numbers where t = max {l ,k }, l ≠ k .
Ibrahim [18 ], studied the solutions of non-linear difference equation
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x n + 1 = x n x n − 2 x n − 1 ( a + bx n x n − 2 ) , n = 0 , 1 , … , {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 x 0 ,x −1 and x −2 non-negative real numbers with bx 0 x −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
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x n + 1 = x n − 3 1 + x n − 1 , n = 0 , 1 , … , x n + 1 = x n − 5 1 + x n − 2 , n = 0 , 1 , … , x n + 1 = x n − 5 1 + x n − 1 x n − 3 , n = 0 , 1 , … , x n + 1 = x n − 3 1 + x n x n − 1 x n − 2 , n = 0 , 1 , … , \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 = 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 , … , {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 ,x 0 ∈ (0, ∞) is investigated.
2 Main Result
Let
x ¯ \bar x
be the unique positive equilibrium of the equation 1 , then clearly,
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x ¯ = x ¯ 1 + x ¯ . xxxxx ¯ ⇒ x ¯ + x ¯ 7 = x ¯ ⇒ x ¯ 7 = 0 ⇒ 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,
x ¯ = 0 \bar x = 0
can be obtained. For any k ≥ 0 and m > k , notation
i = 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 a 1 ,..., a 14 ≥ 0 is existed in such that:
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lim n → ∞ x 14 n − 13 + k = a 1 + k , k = 0 , 13 ¯ . \lim\limits_{n \to \infty } {x_{14n - 13 + k}} = {a_{1 + k}},\quad k = \overline {0,13}.
b) (a 1 ,a 2 ,a 3 ,a 4 ,a 5 ,a 6 ,a 7 ,a 8 ,a 9 ,a 10 ,a 11 ,a 12 ,a 13 ,a 14 ,...) is a solution of 1 having period fourteen.
c) ![]()
∏ k = 0 6 lim n → ∞ x 14 n − j + 2 k = 0 , j = 0 , 1 ¯ ; \prod\limits_{k = 0}^6 \lim\limits_{n \to \infty } {x_{14n - j + 2k}} = 0,\quad j = \overline {0,1};
or
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∏ k = 0 6 a 2 k + i = 0 , i = 0 , 1 ¯ . \prod\limits_{k = 0}^6 {a_{2k + i}} = 0,\quad i = \overline {0,1}.
d) If there exist n 0 ∈ ℕ such that x n +1 ≤ x n −11 for all n ≥ n 0 , then,
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lim n → ∞ x n = 0 . \lim\limits_{n \to \infty } {x_n} = 0.
e) The following formulas below:
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x 14 n + k + 1 = x − 13 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j 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 ) , \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}}}}),}} ![]()
x 14 n + k + 3 = x − 11 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 1 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 ) , \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}}}}),}} ![]()
x 14 n + k + 5 = x − 9 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 2 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 ) , \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}}}}),}} ![]()
x 14 n + k + 7 = x − 7 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 3 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 ) , \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}}}}),}} ![]()
x 14 n + k + 9 = x − 5 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 4 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 ) , \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}}}}),}} ![]()
x 14 n + k + 11 = x − 3 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 5 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 ) , \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}}}}),}} ![]()
x 14 n + k + 13 = x − 1 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 6 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 ) , \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 = 0 , 1 ¯ k = \overline {0,1}
, holds.
f) 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 = 0 , 1 ¯ k = \overline {0,1}
.
Proof
a) Firstly, from 1 ,
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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
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x n + 1 < x n − 13 , {x_{n + 1}} < {x_{n - 13}}, n ∈ ℕ
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lim n → ∞ x 14 n − 13 + k = a 1 + k , for k = 0 , 13 ¯ \lim\limits_{n \to \infty } {x_{14n - 13 + k}} = {a_{1 + k}},\quad for\;\;k = \overline {0,13}
existed formulas are obtained.
b) (a 1 ,a 2 ,a 3 ,a 4 ,a 5 ,a 6 ,a 7 ,a 8 ,a 9 ,a 10 ,a 11 ,a 12 ,a 13 ,a 14 ,...) is a solution of 1 having period fourteen.
c) In view of the 1 ,
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n = 14 n ⇒ x 14 n + 1 = x 14 n − 13 1 + ∏ k = 0 5 lim n → ∞ 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,
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∏ k = 0 6 lim n → ∞ x 14 n − 13 + 2 k = 0 or ∏ 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,
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n = 14 n + 1 ⇒ x 14 n + 2 = x 14 n − 12 1 + ∏ k = 0 5 lim n → ∞ 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,
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∏ k = 0 6 lim n → ∞ x 14 n − 12 + 2 k = 0 or ∏ 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.
d) If there exist n 0 ∈ ℕ such that x n +1 ≤ x n −11 for all n ≥ n 0 , then, a 1 ≤ a 3 ≤ a 5 ≤ a 7 ≤ a 9 ≤ a 11 ≤ a 13 ≤ a 1 , a 2 ≤ a 4 ≤ a 6 ≤ a 8 ≤ a 10 ≤ a 12 ≤ a 14 ≤ a 2 . Using (c) we get
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∏ k = 0 6 a 2 k + i = 0 , i = 1 , 2 ¯ . \prod\limits_{k = 0}^6 {a_{2k + i}} = 0,\quad i = \overline {1,2}.
Then we see that,
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lim n → ∞ x n = 0 . \lim\limits_{n \to \infty } {x_n} = 0.
Hence the proof of (d) completed.
e) Subracting x n −13 from the left and right-hand sides in 1 ,
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x n + 1 − x n − 13 = 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) ![]()
x 2 n − 3 − x 2 n − 17 = ( x 1 − x − 13 ) ∏ i = 1 n − 2 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 ) ∏ i = 1 n − 2 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 , \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 = 0 , 1 ¯ k = \overline {0,1}
,
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x 14 n + 1 + k − x − 13 + k = ( x 1 + k − x − 13 + k ) ∑ j = 0 n ∏ i = 1 7 j 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 = 0 , 1 ¯ k = \overline {0,1}
,
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x 14 n + 3 + k − x − 11 + k = ( x 3 + k − x − 11 + k ) ∑ j = 0 n ∏ i = 1 7 j + 1 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 = 0 , 1 ¯ k = \overline {0,1}
,
![]()
x 14 n + 5 + k − x − 9 + k = ( x 5 + k − x − 9 + k ) ∑ j = 0 n ∏ i = 1 7 j + 2 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 = 0 , 1 ¯ k = \overline {0,1}
,
![]()
x 14 n + 7 + k − x − 7 + k = ( x 7 + k − x − 7 + k ) ∑ j = 0 n ∏ i = 1 7 j + 3 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 = 0 , 1 ¯ k = \overline {0,1}
,
![]()
x 14 n + 9 + k − x − 5 + k = ( x 9 + k − x − 5 + k ) ∑ j = 0 n ∏ i = 1 7 j + 4 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 = 0 , 1 ¯ k = \overline {0,1}
,
![]()
x 14 n + 11 + k − x − 3 + k = ( x 11 + k − x − 3 + k ) ∑ j = 0 n ∏ i = 1 7 j + 5 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 = 0 , 1 ¯ k = \overline {0,1}
,
![]()
x 14 n + 13 + k − x − 1 + k = ( x 13 + k − x − 1 + k ) ∑ j = 0 n ∏ i = 1 7 j + 6 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:
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x 14 n + k + 1 = x − 13 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j 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 ) , \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}}}}),}} ![]()
x 14 n + k + 3 = x − 11 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 1 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 ) , \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}}}}),}} ![]()
x 14 n + k + 5 = x − 9 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 2 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 ) , \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}}}}),}} ![]()
x 14 n + k + 7 = x − 7 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 3 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 ) , \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}}}}),}} ![]()
x 14 n + k + 9 = x − 5 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 4 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 ) , \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}}}}),}} ![]()
x 14 n + k + 11 = x − 3 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 5 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 ) , \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}}}}),}} ![]()
x 14 n + k + 13 = x − 1 + k ( 1 − 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 × ∑ j = 0 n ∏ i = 1 7 j + 6 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 ) . \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}}}}).}}
f) Suppose that a 1 = a 3 = a 5 = a 7 = a 9 = a 11 = a 13 = 0. By (e), we have
(3) ![]()
lim n → ∞ x 14 n + 1 = lim n → ∞ x − 13 ( 1 − 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 ∑ j = 0 n ∏ i = 1 7 j 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 − 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 ∑ j = 0 n ∏ i = 1 7 j 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j 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 . \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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 1 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 2 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 3 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 4 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 5 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 ⇒ 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 = ∑ j = 0 n ∏ i = 1 7 j + 6 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 ,
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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 = ∑ j = 0 n ∏ i = 1 7 j 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 1 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 , \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 ![]()
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 = ∑ j = 0 n ∏ i = 1 7 j + 1 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 2 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 , \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 ,
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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 = ∑ j = 0 n ∏ i = 1 7 j + 2 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 3 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 , \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 ,
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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 = ∑ j = 0 n ∏ i = 1 7 j + 3 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 4 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 \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 ,
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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 = ∑ j = 0 n ∏ i = 1 7 j + 4 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 5 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 \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 ,
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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 = ∑ j = 0 n ∏ i = 1 7 j + 5 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 > 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 = ∑ j = 0 n ∏ i = 1 7 j + 6 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 , \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 > x 0 . We arrive at a contradiction which completes the proof of theorem.
Example 2
If the initial conditions are selected as follows:
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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 . \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, xn = { 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,...}
Fig. 1 xn graph of the solution.