Selçuk J. Appl. Math. Selçuk Journal of Vol. 8. No.1. pp. 15-26 , 2007 Applied Mathematics
On The Recursive Sequence+1= 1+ −11
−1−3−5−7−9
Dagistan ¸Sim¸sek
Selcuk University, Engineering-Architecture Faculty, Department of Industrial Engi-neering, 42075 Campus / Konya, Turkey.
e-mail:dsim sek@ selcuk.edu.tr
Received : December 18, 2006
Summary. In this paper a solution of the following difference equation was investigated +1= −11 1 + −1−3−5−7−9 = 0 1 2 where −11 −10 −2 −1 0∈ (0 ∞)
Key words:Difference equation, period twelve solution. 1. Introduction
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 examples [1,2,4,5].
In [3] the following problem was posed. There is a solution of the following difference equation
+1=
−1
+
for = 0 1 2 where −1 0∈ (0 ∞), 0 such that → 0 as → ∞.
In [6] Stevic assumed that = 1 and solved the following problem
+1=
−1
1 +
for = 0 1 2
where −1 0∈ (0 ∞). Also, this result was generalized to the equation of the
+1=
−1
()
for = 0 1 2 where −1 0∈ (0 ∞)
In this paper we investigated the following nonlinear difference equation
(1) +1= −11 1 + −1−3−5−7−9 for = 0 1 2 where −11 −10 −2 −1 0∈ (0 ∞) 2. Main Result
Teorem 1Consider the difference equation (1). Then the following statements are true.
a)The sequences (12−11) (12−10) (12−9) (12−8) (12−7) (12−6)
(12−5) (12−4) (12−3) (12−2) (12−1) and (12) are decreasing
and there exist
1 2 3 4 5 6 7 8 9 10 11 12> 0 such that
lim
→∞12−11= 1 lim→∞12−10 = 2 lim→∞12−9= 3 lim→∞12−8= 4
lim
→∞12−7= 5 lim→∞12−6 = 6 lim→∞12−5= 7 lim→∞12−4= 8
lim
→∞12−3= 9 lim→∞12−2= 10 lim→∞12−1 = 11 lim→∞12 = 12
b) (1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12 ) is a solution of equation (1) of
period twelve.
c)1357911= 0, 24681012= 0
d) If there exists 0∈ such that
−1−3−5−7−9 ≥ +1−1−3−5−7
for all > 0, then
lim
→∞ = 0
e)The following formulas
12+1= −11(1− −1−3−5−7−9 1 + −1−3−5−7−9 X =0 6 Y =1 1 1 + 2−92−72−52−32−1 )
12+2 = −10(1 − 0−2−4−6−8 1 + 0−2−4−6−8 X =0 6 Y =1 1 1 + 2−82−62−42−22 ) 12+3= −9(1− −1−3−5−7−11 1 + −1−3−5−7−9 X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 ) 12+4= −8(1 − 0−2−4−6−10 1 + 0−2−4−6−8 X =0 6+1Y =1 1 1 + 2−82−62−42−22 ) 12+5= −7(1− −1−3−5−9−11 1 + −1−3−5−7−9 X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 ) 12+6= −6(1 − 0−2−4−8−10 1 + 0−2−4−6−8 X =0 6+2Y =1 1 1 + 2−82−62−42−22 ) 12+7= −5(1− −1−3−7−9−11 1 + −1−3−5−7−9 X =0 6+3Y =1 1 1 + 2−92−72−52−32−1 ) 12+8= −4(1 − 0−2−6−8−10 1 + 0−2−4−6−8 X =0 6+3Y =1 1 1 + 2−82−62−42−22 ) 12+9= −3(1− −1−5−7−9−11 1 + −1−3−5−7−9 X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 ) 12+10= −2(1 − 0−4−6−8−10 1 + 0−2−4−6−8 X =0 6+4Y =1 1 1 + 2−82−62−42−22 ) 12+11 = −1(1− −3−5−7−9−11 1 + −1−3−5−7−9 X =0 6+5Y =1 1 1 + 2−92−72−52−32−1 )
12+12 = 0(1 − −2−4−6−8−10 1 + 0−2−4−6−8 X =0 6+5Y =1 1 1 + 2−82−62−42−22 ) hold. f )If 12+1 → 16= 0 , 12+3→ 36= 0 , 12+5→ 56= 0 , 12+7→ 76= 0 and 12+9→ 96= 0 then 12+11→ 0 as → ∞. If 12+2 → 26= 0 , 12+4→ 4 6= 0 , 12+6→ 66= 0 , 12+8→ 8 6= 0 and 12+10→ 106= 0 then 12+12→ 0 as → ∞.
Proof 1 a)Firstly, we consider equation (1). From this equation, we obtain +1(1 + −1−3−5−7−9) = −11
If −1 −3 −5 −7 −9 ∈ (0 +∞), then (1+−1−3−5−7−9) ∈
(1 +∞). Since +1 −11 ∈ , we obtain that there exist
lim
→∞12−11= 1 lim→∞12−10= 2 lim→∞12−9= 3 lim→∞12−8= 4
lim
→∞12−7= 5 lim→∞12−6 = 6 lim→∞12−5= 7 lim→∞12−4= 8
lim
→∞12−3= 9 lim→∞12−2 = 10 lim→∞12−1= 11 lim→∞12= 12
b) (1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12 ) is a solution of equation (1) of
period twelve.
c) In view of equation (1), we obtain 12+1=
12−11
1 + 12−112−312−512−712−9
Take the limits on both sides of the above equality
1= 1 1 + 119753 ⇒ 1 +1197531= 1⇒ 1357911= 0 Also, we obtain 12+2 = 12−10 1 + 1212−212−412−612−8 Take the limits on both sides of the above equality
2=
2
1 + 1210864 ⇒ 2
+12108642= 2⇒ 24681012= 0
d) If there exists 0∈ such that
for all > 0, then 1≤ 3 ≤ 5≤ 7 ≤ 9 ≤ 11 ≤ 1 and 2 ≤ 4 ≤ 6 ≤
8≤ 10≤ 12≤ 2
Since 1357911= 0 and 24681012= 0 we obtain the result.
e)Subracting −11from the left and right-hand sides of equation (1) we obtain
+1− −11=
1
1 + −1−3−5−7−9
(−1− −13)
and the following formula (2) > 2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 2−3− 2−15= (1− −11) Y−2 =1 1 1+2−92−72−52−32−1 2−2− 2−14= (2− −10) Y−2 =1 1 1+2−82−62−42−22
holds. Replacing by 6 in (2) and summing from = 0 to = , we obtain (3) 12+1−−11= (1−−11) X =0 6 Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 ) and (4) 12+2−−10= (2−−10) X =0 6 Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
Also, replacing by 6 + 1 in (2) and summing from = 0 to = , we obtain (5) 12+3−−9= (1−−11) X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 ) and (6) 12+4−−8 = (2−−10) X =0 6+1Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
Also, replacing by 6 + 2 in (2) and summing from = 0 to = , we obtain (7) 12+5−−7= (1−−11) X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 )
and (8) 12+6−−6 = (2−−10) X =0 6+2Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
Also, replacing by 6 + 3 in (2) and summing from = 0 to = , we obtain (9) 12+7−−5= (1−−11) X =0 6+3Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 ) and (10) 12+8−−4 = (2−−10) X =0 6+3Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
Also, replacing by 6 + 4 in (2) and summing from = 0 to = , we obtain (11) 12+9−−3= (1−−11) X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 ) and (12) 12+10−−2 = (2−−10) X =0 6+4Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
Also, replacing by 6 + 5 in (2) and summing from = 0 to = , we obtain (13) 12+11−−1 = (1−−11) X =0 6+5Y =1 1 1 + 2−92−72−52−32−1 ( = 0 1 2 ) and (14) 12+12−0= (2−−10) X =0 6+5Y =1 1 1 + 2−82−62−42−22 ( = 0 1 2 )
12+1= −11(1− −1−3−5−7−9 1 + −1−3−5−7−9 X =0 6 Y =1 1 1 + 2−92−72−52−32−1 ) 12+2 = −10(1 − 0−2−4−6−8 1 + 0−2−4−6−8 X =0 6 Y =1 1 1 + 2−82−62−42−22 ) 12+3= −9(1− −1−3−5−7−11 1 + −1−3−5−7−9 X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 ) 12+4= −8(1 − 0−2−4−6−10 1 + 0−2−4−6−8 X =0 6+1Y =1 1 1 + 2−82−62−42−22 ) 12+5= −7(1− −1−3−5−9−11 1 + −1−3−5−7−9 X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 ) 12+6= −6(1 − 0−2−4−8−10 1 + 0−2−4−6−8 X =0 6+2Y =1 1 1 + 2−82−62−42−22 ) 12+7= −5(1− −1−3−7−9−11 1 + −1−3−5−7−9 X =0 6+3Y =1 1 1 + 2−92−72−52−32−1 ) 12+8= −4(1 − 0−2−6−8−10 1 + 0−2−4−6−8 X =0 6+3Y =1 1 1 + 2−82−62−42−22 ) 12+9= −3(1− −1−5−7−9−11 1 + −1−3−5−7−9 X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 ) 12+10= −2(1 − 0−4−6−8−10 1 + 0−2−4−6−8 X =0 6+4Y =1 1 1 + 2−82−62−42−22 )
12+11 = −1(1− −3−5−7−9−11 1 + −1−3−5−7−9 X =0 6+5Y =1 1 1 + 2−92−72−52−32−1 ) 12+12 = 0(1 − −2−4−6−8−10 1 + 0−2−4−6−8 X =0 6+5Y =1 1 1 + 2−82−62−42−22 )
f) Suppose that 1= 3= 5= 7= 9= 11= 0. By (e), we have
lim →∞12+1 = lim→∞−11(1 − −1−3−5−7−9 1 + −1−3−5−7−9 X =0 6 Y =1 1 1 + 2−92−72−52−32−1 ) 1= −11(1 − −1−3−5−7−9 1 + −1−3−5−7−9 ∞ X =0 6 Y =1 1 1 + 2−92−72−52−32−1 ) (15) 1= 0 ⇒ 1 + −1−3−5−7−9 −1−3−5−7−9 = ∞ X =0 6 Y =1 1 1 + 2−92−72−52−32−1 and, 3= −9(1 − −1−3−5−7−11 1 + −1−3−5−7−9 ∞ X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 ) (16) 3= 0 ⇒ 1 + −1−3−5−7−9 −1−3−5−7−11 = ∞ X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 Similarly, we write 5= −7(1 − −1−3−5−9−11 1 + −1−3−5−7−9 ∞ X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 ) (17) 5= 0 ⇒ 1 + −1−3−5−7−9 −1−3−5−9−11 = ∞ X =0 6+2Y =1 1 1 + 2−92−72−52−32−1
Then, 7= −5(1 − −1−3−7−9−11 1 + −1−3−5−7−9 ∞ X =0 6+3 Y =1 1 1 + 2−92−72−52−32−1 ) (18) 7= 0 ⇒ 1 + −1−3−5−7−9 −1−3−7−9−11 = ∞ X =0 6+3Y =1 1 1 + 2−92−72−52−32−1
In the same manner, we obrain
9= −3(1 − −1−5−7−9−11 1 + −1−3−5−7−9 ∞ X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 ) (19) 9= 0 ⇒ 1 + −1−3−5−7−9 −1−5−7−9−11 = ∞ X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 Finally, 11= −1(1 − −3−5−7−9−11 1 + −1−3−5−7−9 ∞ X =0 6+5Y =1 1 1 + 2−92−72−52−32−1 ) (20) 11= 0 ⇒ 1 + −1−3−5−7−9 −3−5−7−9−11 = ∞ X =0 6+5 Y =1 1 1 + 2−92−72−52−32−1
From equation (15) and (16), we obtain
1 + −1−3−5−7−9 −1−3−5−7−9 = ∞ X =0 6 Y =1 1 1 + 2−92−72−52−32−1 (21) 1 + −1−3−5−7−9 −1−3−5−7−11 = ∞ X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 thus, −11 −9
1 + −1−3−5−7−9 −1−3−5−7−11 = ∞ X =0 6+1Y =1 1 1 + 2−92−72−52−32−1 (22) 1 + −1−3−5−7−9 −1−3−5−9−11 = ∞ X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 thus, −9 −7
From equation (17) and (18), we obtain
1 + −1−3−5−7−9 −1−3−5−9−11 = ∞ X =0 6+2Y =1 1 1 + 2−92−72−52−32−1 (23) 1 + −1−3−5−7−9 −1−3−7−9−11 = ∞ X =0 6+3Y =1 1 1 + 2−92−72−52−32−1 thus, −7 −5
From equation (18) and (19), we obtain
1 + −1−3−5−7−9 −1−3−7−9−11 = ∞ X =0 6+3Y =1 1 1 + 2−92−72−52−32−1 (24) 1 + −1−3−5−7−9 −1−5−7−9−11 = ∞ X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 thus, −5 −3
From equation (19) and (20), we obtain
1 + −1−3−5−7−9 −1−5−7−9−11 = ∞ X =0 6+4Y =1 1 1 + 2−92−72−52−32−1 (25) 1 + −1−3−5−7−9 −3−5−7−9−11 = ∞ X =0 6+5Y =1 1 1 + 2−92−72−52−32−1 thus, −3 −1
We, therefore, here −11 −9 −7 −5 −3 −1We arrive at a conradiction.
Suppose that 2= 4= 6= 8 = 10= 12= 0 . From that equation (26) in
(e) follows. It is noted that since proof of equation (21) is similar, it is omitted here.
1 + 0−2−4−6−8 0−2−4−6−8 = ∞ X =0 6 Y =1 1 1 + 2−82−62−42−22 (26) 1 + 0−2−4−6−8 0−2−4−6−10 = ∞ X =0 6+1Y =1 1 1 + 2−82−62−42−22 thus, −10 −8
From that equation (27) in (e) follows. Proof of equation (22) is similar and will be omitted. 1 + 0−2−4−6−8 0−2−4−6−10 = ∞ X =0 6+1Y =1 1 1 + 2−82−62−42−22 (27) 1 + 0−2−4−6−8 0−2−4−8−10 = ∞ X =0 6+2Y =1 1 1 + 2−82−62−42−22 thus, −8 −6
From that equation (28) in (e) follows. Proof of equation (23) is similar and will be omitted. 1 + 0−2−4−6−8 0−2−4−8−10 = ∞ X =0 6+2Y =1 1 1 + 2−82−62−42−22 (28) 1 + 0−2−4−6−8 0−2−6−8−10 = ∞ X =0 6+3Y =1 1 1 + 2−82−62−42−22 thus, −6 −4
From that equation (29) in (e) follows. Proof of equation (24) is similar and will be omitted. 1 + 0−2−4−6−8 0−2−6−8−10 = ∞ X =0 6+3Y =1 1 1 + 2−82−62−42−22 (29) 1 + 0−2−4−6−8 0−4−6−8−10 = ∞ X =0 6+4Y =1 1 1 + 2−82−62−42−22 thus, −4 −2
From that equation (30) in (e) follows. Proof of equation (25) is similar and will be omitted.
1 + 0−2−4−6−8 0−4−6−8−10 = ∞ X =0 6+4Y =1 1 1 + 2−82−62−42−22 (30) 1 + 0−2−4−6−8 −2−4−6−8−10 = ∞ X =0 6+5Y =1 1 1 + 2−82−62−42−22 thus, −2 0
We obtain here −10 −8 −6 −4 −2 0We arrive at a
conradic-tion which completes the proof of the theorem. References
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