On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k
In this paper a solution of the one difference equation was investigated.
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irk-123456789-1693782020-06-12T01:26:20Z On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k Simsek, D. Abdullayev, F.G. In this paper a solution of the one difference equation was investigated. 2017 Article On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k / D. Simsek, F.G. Abdullayev // Український математичний вісник. — 2017. — Т. 14, № 4. — С. 564-574. — Бібліогр.: 28 назв. — англ. 1810-3200 http://dspace.nbuv.gov.ua/handle/123456789/169378 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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In this paper a solution of the one difference equation was investigated. |
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Simsek, D. Abdullayev, F.G. |
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Simsek, D. Abdullayev, F.G. On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k Український математичний вісник |
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Simsek, D. Abdullayev, F.G. |
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Simsek, D. |
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On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
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On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
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On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
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On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
| title_full_unstemmed |
On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
| title_sort |
on the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k |
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Інститут прикладної математики і механіки НАН України |
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2017 |
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http://dspace.nbuv.gov.ua/handle/123456789/169378 |
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On the recursive sequence x(n+1)=xn-(k+1)/1+xnxn-1...xn-k / D. Simsek, F.G. Abdullayev // Український математичний вісник. — 2017. — Т. 14, № 4. — С. 564-574. — Бібліогр.: 28 назв. — англ. |
| series |
Український математичний вісник |
| work_keys_str_mv |
AT simsekd ontherecursivesequencexn1xnk11xnxn1xnk AT abdullayevfg ontherecursivesequencexn1xnk11xnxn1xnk |
| first_indexed |
2025-07-15T04:07:17Z |
| last_indexed |
2025-07-15T04:07:17Z |
| _version_ |
1837684425396060160 |
| fulltext |
Український математичний вiсник
Том 14 (2017), № 4, 564 – 574
On the recursive sequence xn+1 =
xn−(k+1)
1+xnxn−1...xn−k
Dağıstan Simsek, Fahreddin G. Abdullayev
Abstract. In this paper a solution of the following difference equation
was investigated
xn+1 =
xn−(k+1)
1 + xnxn−1...xn−k
, n = 0, 1, 2, . . .
where x−(k+1), x−k, . . . , x−1, x0 ∈ (0,∞) and k = 0, 1, 2, . . . .
Key words and phrases. Difference equation, period k + 2 solution.
1. Introduction
The study of difference equations has been growing continuously for
the last decade. This is largely due to the fact that difference equations
manifest themselves as mathematical models describing real life situations
in probability theory, queuing theory, statistical problems, stochastic time
series, combinatorial analysis, number theory, geometry, electrical net-
work, quanta in radiation, genetics in biology, economics, psychology,
sociology, etc. In fact, now it occupies a central position in applicable
analysis and will no doubt continue to play an important role in mathe-
matics as a whole. Recently there has been a lot of interest in studying
the periodic nature of nonlinear difference equations. For some recent
results, concerning among other problems, the periodic nature of scalar
nonlinear difference equations see, for example, [1–28].
Cinar [2, 3, 4] has studied the following problems with positive initial
values
xn+1 =
xn−1
1 + axnxn−1
,
xn+1 =
xn−1
−1 + axnxn−1
,
Received 19.12.2017
ISSN 1810 – 3200. c⃝ Iнститут прикладної математики i механiки НАН України
D. Simsek, F. G. Abdullayev 565
xn+1 =
axn−1
1 + bxnxn−1
for n = 0, 1, 2, ..., respectively.
In [18] Stevic solved the following problem
xn+1 =
xn−1
1 + xn
for n = 0, 1, 2, . . . ,
where x−1, x0 ∈ (0,∞). Also, this result was generalized to the equation
of the following form:
xn+1 =
xn−1
g(xn)
for n = 0, 1, 2, . . . ,
where x−1, x0 ∈ (0,∞).
Simsek et al. [19, 20, 21, 24], studied the following problems with
positive initial values
xn+1 =
xn−3
1 + xn−1
,
xn+1 =
xn−5
1 + xn−2
,
xn+1 =
xn−5
1 + xn−1xn−3
,
xn+1 =
xn−3
1 + xnxn−1xn−2
,
for n = 0, 1, 2, . . . , respectively.
In this paper we investigate the following nonlinear difference equation
xn+1 =
xn−(k+1)
1 + xnxn−1...xn−k
for n = 0, 1, 2, . . . (1.1)
where x−(k+1), x−k, ..., x−1, x0 ∈ (0,∞) and k = 0, 1, 2, . . . .
2. Main result
Theorem 1. Consider the difference equation (1.1). Then the following
statements are true.
a) The sequences (x(k+2)n−(k+1)), (x(k+2)n−k), . . . , (x(k+2)n) are de-
creasing and there exist a1, a2, . . . , ak+2 > 0 such that
lim
n→∞
x(k+2)n−(k+1) = a1, lim
n→∞
x(k+2)n−k = a2, . . . , lim
n→∞
x(k+2)n = ak+2.
566 On the recursive sequence . . .
b) (a1, a2, . . . , ak+2, a1, a2, . . . , ak+2, . . .) is a solution of equation (1.1)
of period k+2.
c) a1 × a2 × . . .× ak+2 = 0.
d) If there exists n0 ∈ N such that xn−k ≥ xn+1 for all n > n0, then
lim
n→∞
xn = 0.
e) The following formulas
x(k+2)n+1 = x−(k+1)
(
1− x0x−1...x−k
1+x0x−1...x−k
n∑
j=0
(k+2)j∏
i=1
1
1+xixi−1...xi−k
)
,
.
.
.
x(k+2)n+k+2 = x0
(
1− x−1x−2...x−(k+1)
1+x0x−1...x−k
n∑
j=0
(k+2)j+(k+1)∏
i=1
1
1+xixi−1...xi−k
)
holds.
f) If x(k+2)n+1 → a1 ̸= 0, x(k+2)n+2 → a2 ̸= 0,..., x(k+2)n+k+1 →
ak+1 ̸= 0 then x(k+2)n+k+2 → 0 as n→ ∞.
Proof. a) Firstly, from the equation (1.1), we obtain
xn+1(1 + xnxn−1...xn−k) = xn−(k+1).
If xn, xn−1, ..., xn−k ∈ (0,+∞), then (1 + xnxn−1...xn−k) ∈ (1,+∞).
Since xn+1 < xn−(k+1), n ∈ N , we obtain that lim
n→∞
x(k+2)n−(k+1) =
a1, lim
n→∞
x(k+2)n−(k) = a2, ..., lim
n→∞
x(k+2)n = ak+2.
b) (a1, a2, ..., ak+2, a1, a2, ..., ak+2, ...) is a solution of equation (1.1)
of period k + 2.
c) In view of the equation (1.1), we obtain:
x(k+2)n+1 =
x(k+2)n−(k+1)
1 + x(k+2)n...x(k+2)n−k
.
Taking limit as n→ ∞ on both sides of the above equality, we get:
lim
n→∞
x(k+2)n+1 = lim
n→∞
x(k+2)n−(k+1)
1 + x(k+2)n...x(k+2)n−k
,
D. Simsek, F. G. Abdullayev 567(
lim
n→∞
x(k+2)n−(k+1)
)
×
(
lim
n→∞
x(k+2)n−(k)
)
× ...×
(
lim
n→∞
x(k+2)n
)
= 0.
Then a1 × a2 × ...× ak+2 = 0.
d) If there exists n0 ∈ N such that xn−k ≥ xn+1 for all n > n0, then
a1 ≤ a2 ≤ ... ≤ ak+2 ≤ a1. Since a1 × a2 × ...× ak+2 = 0, we obtain the
result.
e) Subracting xn−(k+1) from the left and right-hand sides of equation
(1.1), we obtain:
xn+1 − xn−(k+1) =
1
1 + xnx−1...xn−k
(xn − xn−(k+2))
and for n > 1 the following formula
xn − xn−(k+2) = (x1 − x−(k+1))
n−1∏
i=1
1
1 + xixi−1...xi−k
(2.1)
holds. Replacing n by (k+2)j in (2.1) and summing from j = 0 to j = n,
we obtain:
x(k+2)n+1 − x−(k+1) = (x1 − x−(k+1))
n∑
j=0
(k+2)j∏
i=1
1
1 + xixi−1...xi−k
,
(n = 0, 1, 2, ...),
. (2.2)
.
.
Also, replacing n by (k + 2)j + (k + 1) in (2.1) and summing from
j = 0 to j = n, we obtain:
x(k+2)n+k+2 − x0 = (xk+1 − x0)
n∑
j=0
(k+2)j+(k+1)∏
i=1
1
1 + xixi−1...xi−k
,
(n = 0, 1, 2, ...).
From the formulas above, we obtain:
568 On the recursive sequence . . .
x(k+2)n+1 = x−(k+1)
(
1− x0x−1...x−k
1+x0x−1...x−k
n∑
j=0
(k+2)j∏
i=1
1
1+xixi−1...xi−k
)
,
.
.
.
x(k+2)n+k+2 = x0(1−
x−1x−2...x−(k+1)
1+x0x−1...x−k
n∑
j=0
(k+2)j+(k+1)∏
i=1
1
1+xixi−1...xi−k
) . . .
(2.3)
f) Suppose that a1 = a2 = ... = ak+2 = 0. By (e) we have:
lim
n→∞
x(k+2)n+1 = lim
n→∞
x−(k+1)
(
1− x0x−1...x−k
1+x0x−1...x−k
n∑
j=0
(k+2)j∏
i=1
1
1+xixi−1...xi−k
)
,
a1 = x−(k+1)
1− x0x−1...x−k
1 + x0x−1...x−k
∞∑
j=0
(k+2)j∏
i=1
1
1 + xixi−1...xi−k
,
a1 = 0 ⇒ 1 + x0x−1...x−k
x0x−1...x−k
=
∞∑
j=0
(k+2)j∏
i=1
1
1 + xixi−1...xi−k
. (2.4)
Similarly,
lim
n→∞
x(k+2)n+2 = lim
n→∞
x−(k)(1−
x0x−1...x−(k+1)
1+x0x−1...x−k
n∑
j=0
(k+2)j+1∏
i=1
1
1+xixi−1...xi−k
),
a2 = x−(k)(1−
x0x−1...x−(k+1)
1 + x0x−1...x−k
∞∑
j=0
(k+2)j+1∏
i=1
1
1 + xixi−1...xi−k
),
a2 = 0 ⇒ 1+x0x−1...x−k
x0x−1...x−(k+1)
=
∞∑
j=0
(k+2)j+1∏
i=1
1
1+xixi−1...xi−k
,
.
.
.
(2.5)
D. Simsek, F. G. Abdullayev 569
Similarly,
lim
n→∞
x(k+2)n+k+1= lim
n→∞
x−1
(
1− x0x−2x−3...x−(k+1)
1+x0x−1...x−k
n∑
j=0
(k+2)j+k∏
i=1
1
1+xixi−1...xi−k
)
,
ak+1 = x−1
1−
x0x−2x−3...x−(k+1)
1 + x0x−1...x−k
∞∑
j=0
(k+2)j+k∏
i=1
1
1 + xixi−1...xi−k
,
ak+1 = 0 ⇒ 1 + x0x−1...x−k
x−2x−3...x−(k+1)
=
∞∑
j=0
(k+2)j+k∏
i=1
1
1 + xixi−1...xi−k
. (2.6)
lim
n→∞
x(k+2)n+k+2= lim
n→∞
x0
(
1− x−1x−2...x−(k+1)
1+x0x−1...x−k
n∑
j=0
(k+2)j+(k+1)∏
i=1
1
1+xixi−1...xi−k
)
,
ak+2 = x0
1−
x−1x−2...x−(k+1)
1 + x0x−1...x−k
∞∑
j=0
(k+2)j+(k+1)∏
i=1
1
1 + xixi−1...xi−k
,
ak+2 = 0 ⇒ 1 + x0x−1...x−k
x−1x−2...x−(k+1)
=
∞∑
j=0
(k+2)j+(k+1)∏
i=1
1
1 + xixi−1...xi−k
. (2.7)
From (2.4) and (2.5), we get:
1 + x0x−1...x−k
x0x−1...x−k
>
1 + x0x−1...x−k
x0x−1...x−(k+1)
,
thus x−(k+1) > x−k, and
.
.
.
Therefore, from (2.6) and (2.7), we have:
1 + x0x−1...x−k
x0x−2...x−(k+1)
>
1 + x0x−1...x−k
x−1x−2...x−(k+2)
,
thus, x−1 > x0.
From here we obtain x−(k+1) > x−k > ... > x−1 > x0. We arrive at a
contradiction which completes the proof of theorem.
570 On the recursive sequence . . .
3. Examples
Consider the following equation xn+1 = xn−3
1+xnxn−1xn−2
which is a
special case of (1.1) for k = 2.
Example 1. If the initial conditions are selected as follows: x[−3] =
0.99999; x[−2] = 0.99998; x[−1] = 0.99997; x[0] = 0.99996. The follow-
ing solutions are obtained:
x(n) = {0.500018, 0.666661, 0.74998, 0.799968, 0.357163, 0.549016, 0.648287,
0.709744, 0.285135, 0.485341, 0.590307, 0.656143, 0.240015, 0.44406,
0.551724, 0.619703, 0.208378, 0.414527, 0.523691, 0.592883, 0.184617,
0.392054, 0.502143, 0.572091, 0.165929, 0.374216, 0.484917, 0.555368,
0.150738, 0.359617, 0.470745, 0.541549, 0.138079, 0.347389, 0.458826,
0.529887, 0.127325, 0.336958, 0.448627, 0.519881, 0.118048, 0.327929,
0.439777, 0.511178, 0.109943, 0.32002, 0.432007, 0.503525, 0.102788,
0.313021, 0.42512, 0.49673, 0.0964144, 0.306775, 0.418964, 0.49065,
0.090695, 0.30116, 0.413424, 0.485172, 0.0855285, 0.296081, 0.408406,
0.480205, 0.0808346, 0.291461, 0.403837, 0.475679, 0.0765488, 0.287237,
0.399657, 0.471536, 0.0726179, 0.283359, 0.395817, 0.467726, 0.0689983,
0.279785, 0.392275, 0.464211, 0.0656534, 0.27648, 0.388997, 0.460956,
0.0625523, 0.273413, 0.385954, 0.457933, 0.0596689, 0.27056, 0.383122,
0.455118, 0.0569808, 0.267898, 0.380478, 0.45249, 0.0544686, 0.265409,
0.378006, 0.450031, 0.0521156, 0.263077, 0.375688, 0.447725, 0.0499071,
0.260887, 0.37351, 0.445558, 0.0478305, 0.258826, 0.371461, 0.443519,
0.0458743, 0.256885, 0.36953, 0.441596, 0.0440287, 0.255052, 0.367707,
0.43978, 0.0422847, 0.25332, 0.365983, 0.438062, 0.0406344, 0.251681,
0.36435, 0.436436, 0.0390707, 0.250127, 0.362803, 0.434894,
0.0375873, 0.248652, 0.361334, 0.43343, . . .}.
The limits of solutions approaching zero due to the fact that the initial
condition is selected. The graph of the solutions is given below:
Example 2. If the initial conditions are selected as follows: x[−3] =
0.1; x[−2] = 0.09; x[−1] = 0.08; x[0] = 0.07; The following solutions
D. Simsek, F. G. Abdullayev 571
Figure 3.1. x(n) graph of the solutions.
are obtained:
x(n) = {0.0999496, 0.0899497, 0.0799497, 0.0699497, 0.0998994, 0.0898994,
0.0798995, 0.0698996, 0.0998492, 0.0898493, 0.0798494, 0.0698495, 0.0997992,
0.0897993, 0.0797995, 0.0697996, 0.0997493, 0.0897495, 0.0797496, 0.0697498,
0.0996996, 0.0896997, 0.0796999, 0.0697001, 0.0996499, 0.0896501, 0.0796503,
0.0696506, 0.0996004, 0.0896006, 0.0796008, 0.0696011, 0.099551, 0.0895512,
0.0795515, 0.0695518, 0.0995017, 0.0895019, 0.0795022, 0.0695026, 0.0994525,
0.0894528, 0.0794531, 0.0694535, 0.0994034, 0.0894037, 0.0794041, 0.0694045,
0.0993544, 0.0893548, 0.0793552, 0.0693557, 0.0993056, 0.089306, 0.0793064,
0.0693069, 0.0992569, 0.0892573, 0.0792578, 0.0692583, 0.0992083, 0.0892087,
0.0792092, 0.0692098, 0.0991598, 0.0891602, 0.0791608, 0.0691614, 0.0991114,
0.0891119, 0.0791124, 0.0691131, 0.0990631, 0.0890637, 0.0790642, 0.0690649,
0.099015, 0.0890155, 0.0790161, 0.0690168, 0.0989669, 0.0889675, 0.0789681,
0.0689689, 0.098919, 0.0889196, 0.0789203, 0.068921, 0.0988712, 0.0888718,
0.0788725, 0.0688733, 0.0988235, 0.0888241, 0.0788248, 0.0688257, 0.0987759,
0.0887766, 0.0787773, 0.0687782, 0.0987284, 0.0887291, 0.0787299, 0.0687308,
0.098681, 0.0886817, 0.0786825, 0.0686835, 0.0986337, 0.0886345, 0.0786353,
0.0686363, 0.0985866, 0.0885874, 0.0785882, 0.0685892, 0.0985395, 0.0885403,
0.0785412, 0.0685422, 0.0984926, 0.0884934, 0.0784943, 0.0684954, 0.0984457,
0.0884466, 0.0784475, 0.0684486, 0.098399, 0.0883999, 0.0784009, 0.068402,
0.0983524, 0.0883533, 0.0783543, 0.0683554, . . .}.
In this case the limits of solutions are not approaching zero, since the
initial condition are hasn’t satisfied. The graph of the solutions is given
below:
572 On the recursive sequence . . .
Figure 3.2. x(n) graph of the solutions.
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yn−1
yn
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xn−1
1+axnxn−1
//
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// Appl. Math. Comp., 158 (2004), No. 3, 793–797.
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Contact information
Dağıstan Simsek Kyrgyz–Turkish Manas University,
Bishkek, Kyrgyzstan,
Selcuk University,
Konya, Turkey
E-Mail: dagistan.simsek@manas.edu.kg
dsimsek@selcuk.edu.tr
Fahreddin G.
Abdullayev
Kyrgyz–Turkish Manas University,
Bishkek, Kyrgyzstan,
Mersin University,
Mersin, Turkey
E-Mail: fabdul@mersin.edu.tr
fahreddin.abdullayev@manas.edu.kg
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