Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (xn , yn−k ), yn+1 = f (yn ,xn−k )" potx

Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (xn , yn−k ), yn+1 = f (yn ,xn−k )" potx

Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (xn , yn−k ), yn+1 = f (yn ,xn−k )" potx

... hat M j = M, P k+ j = g(M ), for j ∈{ 1,2 , ,2 k +1 }, (2.15) from which it follows that lim n→∞ x l n − j = M,forj ∈{ 0,1 , ,2 k +1 }, lim n→∞ y l n − j = g(M ), for j ∈{k + 1, ,3 k +1}. (2.16) In view (2.16 ), ... 2005 We study the global asymptotic behavior of the positive solutions of the system of rational difference...

Ngày tải lên: 22/06/2014, 22:20

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Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (yn−q ,xn−s ), yn+1 = g(xn−t , yn− p )" ppt

Báo cáo hóa học: "ON THE SYSTEM OF RATIONAL DIFFERENCE EQUATIONS xn+1 = f (yn−q ,xn−s ), yn+1 = g(xn−t , yn− p )" ppt

... system of r ational difference equations x n+1 = f (y n−q ,x n−s ), y n+1 = g(x n−t , y n−p ), n = 0,1 , 2, ,wherep,q,s, t ∈ { 0,1 , 2, } with s ≥ t and p ≥ q, the initial values x −s ,x −s+1 , ,x 0 , ... E,wehave x = f (y,x) > ;f( y +1,x) ≥ a, y = g(x, y) >g(x +1,y) ≥ b. (2.2) Claim 1. g(A,b) < y<Band f (B,a)...

Ngày tải lên: 22/06/2014, 22:20

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Báo cáo hóa học: " On the behavior of solutions of the system of rational difference equations" pptx

Báo cáo hóa học: " On the behavior of solutions of the system of rational difference equations" pptx

... of the system of difference equations z n+1 = t n z n−1 + a t n + z n −1 , t n+1 = z n t n−1 + a z n + t n −1 In [11 ], Irićanin and Stević studied the positive solutions of the system of difference equations x (1) n+1 = 1+x (2) n x (3) n−1 , ... stability of a system of difference equations. Appl Anal. 87(6 ), :689–699 (2008). do...

Ngày tải lên: 20/06/2014, 22:20

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báo cáo hóa học: "On the identification of sensory information from mixed nerves by using single-channel cuff electrodes" pot

báo cáo hóa học: "On the identification of sensory information from mixed nerves by using single-channel cuff electrodes" pot

... However, the main goal of these studies was to identify the onset (and offset) of a specific neural activ- ity, with the aim of triggering stimulation. The aim of our work was to investigate the ... amount of samples for post-process- ing rule [3 0,3 1] ), the identification of these parameters was analyzed first. Therefore, different observation win- dow length...

Ngày tải lên: 19/06/2014, 08:20

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Báo cáo hóa học: " On the epidemiology of influenza" pdf

Báo cáo hóa học: " On the epidemiology of influenza" pdf

... of influenza, one heavily dependent on a profound, even controlling, effect of solar radiation. Furthermore, while agreeing the sick could infect the well, Hope-Simpson's principal hypothesis ... early in their illness, "A few of the donors were in the first day of the disease. Others were in the second or third day of the disease." "Then we proceed...

Ngày tải lên: 20/06/2014, 01:20

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Báo cáo hóa học: " On the solvability of discrete nonlinear Neumann problems involving the p(x)-Laplacian" docx

Báo cáo hóa học: " On the solvability of discrete nonlinear Neumann problems involving the p(x)-Laplacian" docx

... solutions of (1.1). The idea of the proof is to transfer the problem of the existence of solutions for (1.1) into the problem of the existence of a minimizer for some associated energy functional. ... + T  k=1 1 p(k) |u(k)| p(k) − T  k=1 F + (k, u(k)) + T  k=1 F − (k, u(k) ) ≥ T+1  k =1 A(k − 1, u(k − 1)) + T  k =1 1 p(k) |u(k)| p(k) − T  k =1...

Ngày tải lên: 20/06/2014, 22:20

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Báo cáo hóa học: " On the solvability of a boundary value problem on the real line" pdf

Báo cáo hóa học: " On the solvability of a boundary value problem on the real line" pdf

... assumptions of Theorem 2.3 are satisfied, with the exception of (2.2 ), and with (2.5) replaced by lim y→+∞ θ(y) y =0 . (2:17) Then, the assertion of Theorem 2.3 follows. Proof. The proof is quite ... conse- quence of Theorem 2. 5, by the same proof of Theorem 3.1. Theorem 3.3. Consider F( y )=| y| p-2 y, p >1,and let all the assumptions of Theorem 3.1...

Ngày tải lên: 20/06/2014, 22:20

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Báo cáo hóa học: "On the feasibility of a channel-dependent scheduling for the SC-FDMA in 3GPP-LTE (mobile environment) based on a prioritizedbifacet Hungarian method" doc

Báo cáo hóa học: "On the feasibility of a channel-dependent scheduling for the SC-FDMA in 3GPP-LTE (mobile environment) based on a prioritizedbifacet Hungarian method" doc

... of specific UE speeds (1 5, 3 0, 6 0, and 120 km/h, respec- tively, constant ), and after the transmission of 1 0,0 00 SC-FDMA symbols (i.e ., per E b N o ), the system sperfor- mance for both the fixed ... to Make them Work? (Kluwer Academic Publishers, 2002 ), pp. 33–50 3. H Holma, A Toskala, LTE for UMTS OFDMA and SC-FDMA Based Radio Access (Wiley,...

Ngày tải lên: 21/06/2014, 00:20

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báo cáo hóa học: " On the regularity of the solution for the second initial boundary value problem for hyperbolic systems in domains with conical points" pptx

báo cáo hóa học: " On the regularity of the solution for the second initial boundary value problem for hyperbolic systems in domains with conical points" pptx

... L(x, t, D ), N j (x, t, D) in the form L = L(x, t, ∂ x )=  | p |≤2m a p (x, t) D p N j = N j (x, t, D )=  | p |≤2m− j b jp (x, t) D p , j =1 , , m . Let L 0 (x, t, D ), N 0j (x, t, D ), be the principal ... assertion of the theorem by induction on h. Let us consider first the case h = 1. We rewrite (2.6 ), (2.7) in the form L (...

Ngày tải lên: 21/06/2014, 00:20

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Báo cáo hóa học: " On the stability of an AQCQ-functional equation in random normed spaces" potx

Báo cáo hóa học: " On the stability of an AQCQ-functional equation in random normed spaces" potx

... 680-74 9, Republic of Korea 3 Department of Mathematics, Daejin University, Kyeonggi 487- 71 1, Republic of Korea 4 Department of Mathematics, University of Seoul, Seoul 130-74 3, Republic of Korea Authors’ ... follows from (2.6) and (2.7) that μ f (4x)−1 0f (2x)+1 6f (x) (t ) = μ ( 4f (3x)−1 6f( 2x)+2 0f( x))+ (f (4x)− 4f (3x)+ 6f( 2x)− 4f (x)) (t ) ≥ T  μ...

Ngày tải lên: 21/06/2014, 01:20

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