show me the steps of determine green and all information is here 2.8. Case 8: Pn = Xn, qn=Yn, rn=Yn, Sn = XnIn this case, system (1.8) is written as in the formXp+1 =f'(af (xg-1)+bf (Yn-2)), Yn+l =f"(af (yn-1)+bf (Xn-2)), (2.61)for n E No. Since f is "1 – 1", from (2.61)f (xn+1) = af (xn-1) +bf (yn-2), f (yn+1) = af (yn-1)+bf (xn-2),(2.62)for n E No. By using the change of variablesf (xn) == un, and f (yn) = Vn, n -2,(2.63)system (2.62) is transformed to the following oneUn+1 = aun-1+bvn-2, Vn+1 = avn-1+ bun-2, n E No.(2.64)By summing the equations in (2.64) we getUn+1+Vn+l = a(un-1+ Vn-1) +b(un-2+Vn-2), ne No,(2.65)whereas by subtracting the second one from the first, we haveUn+1 - Vn+1= a(un-1 – Vn-1) – b (un-2 – Vn-2), n E No.(2.66)From (2.5), we can write the solution of equation (2.65) asUn + Vn = (uo+ vo) Jn+1+(u-1+v-1) Jn+2+b (u_2+v–2) Jn,(2.67)for n -2. On the other hand, by taking a = 0, b = a, c= -b in (1.4) andS, = (-1)" Jn+1, for all n -2, which is called generalized Padovan sequence, in(1.5), from (2.66), we also have thatUn – Vn = (-1)" ( (u0 – vo) Jn+l – (u-1 – v-1) Jn+2+b (u-2 – v-2) Jn),(2.68)for n -2. From (2.68) we haveU2n - V2n=(4o – vo) J2n+1 - (u-1 – v–1) J2n+2+b (u-2 - v-2) J2n,(2.69)for n -1 andU2n+1 - V2n+1 =-(40 - vo) J2n+2+(u-1 - v-1)J2n+3-b (u_2-v-2) J2n+1 (2.70)for n -1. From (2.67)U2n+ V2n = (uo+ vo) J2n+1+(u_1+v-1)J2n+2+b (u_2+v_2)J2n(2.71)for n -1. By summing the equations (2.69) and (2.71)uzn = uoJ2n+1 +v-1J2n+2 + bu-2J2n, n -1.(2.72)By subtracting equation (2.69) from equation (2.71), we haveV2n =VOJ21+1+u-1J20+2+ bv-2J2n, n-1.(2.73)From (2.67)U2n+1+V2n+1(uo +vo)J2n+2+(u-1+v-1) J2n+3+b(u_2+v_2) J2n+1, (2.74)for n -1. By summing the equations (2.70) and (2.74) we getU2n+1 = VoJ2n+2 +u-1J21+3 +bv-2J2n+1, n -1.(2.75)By subtracting equation (2.70) from equation (2.74), we haveV2n+1 =uoJ2n+2+v-1J2n+3+ bu-2J2n+1, n -1.(2.76)From (2.63), (2.72), (2.73), (2.75), (2.76) and after some calculation, we obtainX2, = f(f (x0) J2n+1+f(y-1)J2n+2 +bf (x_2) J2n), n2-1,y2n = f(f(yo) J2n+1+f (x-1)J2n+2+bf (y-2) J2n), n2-1,f(f(vo)Jzn+2+f (x-1)J2n+3+bf (y_2)J2n+1), n –1,(2.77)(2.78)X2n+1 =(2.79)andy2n+1 = f (f (x0)Jzn+2+f(y-1)J2n+3+bf (x-2)J2n+1),n -1.(2.80) constant coefficientsXn+1 = axn+bxn-1+cxn-2, nE No,(1.4)which has actually the general solutionX = xoSn+x-1 (Sn+1 - aSn) +cx_2Sn-1, nE No,(1.5)where (S.) of equation (1L4) satisfving the initial values S 2=S-=0, So= L.The equationaxn-IXn-kXn+1 =nE No,(1.1)bxn-p±cxn-q'where the initial conditions are arbitrary positive real numbers, k, 1, p, q are non-negative integers and a, b, c are positive constants, is one of the difference equationswhose solutions are associated with number sequences. Positive solutions of concreteMotivated by this line of investigations, here we show that the systems of differ-ence equationsXn+l =f'(af (Pn-1)+bf(qn-2)), Yntl =f"(af (ra-1)+bf (Sn-2)), (1.8)for n E No, where the sequences Pn, qn, Tn and Sn are some of the sequences x, andYn, f: Df R is a "1– 1" continuous function on its domain Dr C R, the initialvalues x-j, y-j, je {0,1,2} are arbitrary real numbers and the parameters and a, b2.6. Case 6: Pn = Yn, qn = yn, rn = Xn, Sn = XnIn this case, we obtain the systemXn+1 =5"(af (Yn-1)+bf (Yn-2)), Yn+1 = '(af (xn-1)+bf (xn-2)), (2.29)for n e No. Since f is "1– 1", from (2.29)f (Xn+1) = af (Yn-1)+bf (yn-2), f(yn+l) = af (Xn-1) +bf (xn-2),(2.30)for n E No. By using the change of variablesf (xn) == Un,and f(yn) = Vn, n-2,(2.31)system (2.30) is transformed to the following oneUn+l = avn-1+bvn-2, Vn+1 = aun-1+ bun-2, nE No.(2.32)By summing the equations in (2.32) we getUn+l +Vn+l = a (un-1+ Vn-1) +b(un-2+Vn-2), nE No,(2.33)whereas by subtracting the second one from the first, we haveUn+l - Vn+l = -a(un-1 - Vn-1) -6(un-2- Vn-2), ne No.(2.34)|In this section, we consider the eight special cases of systems (1.8), where thesequences Pn, qn, ľn, Sn are some of the sequences xn and yn, for n-2, and initialvalues x-j, y-j, j e {0,1,2}, are arbitrary real numbers.2.1Case 1: Pn =xn, qn = Xn, ľn = yn, Sn == ynIn this case, system (1.8) is expressed asXn+1 =f (af (xn-1)+bf (xn-2)), Yn+1 =f(af (Vn-1)+bf (Yn-2)), (2.1)for n E No. Since f is "1– 1", from (2.1)f (Xn+1) = af (x,-1)+bf (xn-2), f (Yn+1) = af (yn-1)+bf (yn–2),(2.2)for n E No. By using the change of variablesf (xn)= Un, and f(yn) == Vn, n -2,(2.3)system (2.2) is transformed to the following oneUn+1 = au,-1+bun-2, Vn+1 = avn-1+bvn-2,(2.4)for n E No. By taking a = 0, b= a, c = b in (1.4) and S = Jn+1, for all n -2,which is called generalized Padovan sequence, in (1.5), the solutions to equations in(2.4) are given byUn = ugJn+1+u-1Jn+2+ bu_2Jn,(2.5)Vn = voJn+1+v_1Jn+2+bv_2Jn,(2.6)for n E No. From (2.3), (2.5) and (2.6), it follows that the general solution to system(2.2) is given byXn = f(f (x0) Jn+1+f (x-1) Jn+2+bf (x-2)Jn), n -2,Yn =f(f (vo) Jn+1+f(y-1)Ja+2+bf (y-2) J.), n -2.(2.7)(2.8)2.2. Case 2: Pn = Xn, qn = Xn, ľn = Xn, Sn XnIn this case, system (1.8) becomesXp+1 =f"(af (x,-1)+bf (xp-2)), Yn+1=f"(af (xn-1)+bf (xn-2)), (2.9)for n e No. It should be first note that from the equations in (2.9) it immediatelyfollows that x, = yn, for all n E N. From (2.7), the general solution to system (2.9) is%3DX, = Ya = f'(f (xo) Jn+1+f (x=1) Jn+2+bf (x_2)Jn), n E N.(2.10)

Question

show me the steps of determine green and all information is here 2.8. Case 8: Pn = Xn, qn=Yn, rn=Yn, Sn = XnIn this case, system (1.8) is written as in the formXp+1 =f'(af (xg-1)+bf (Yn-2)), Yn+l =f&#34;(af (yn-1)+bf (Xn-2)), (2.61)for n E No. Since f is &#34;1 – 1&#34;, from (2.61)f (xn+1) = af (xn-1) +bf (yn-2), f (yn+1) = af (yn-1)+bf (xn-2),(2.62)for n E No. By using the change of variablesf (xn) == un, and f (yn) = Vn, n> -2,(2.63)system (2.62) is transformed to the following oneUn+1 = aun-1+bvn-2, Vn+1 = avn-1+ bun-2, n E No.(2.64)By summing the equations in (2.64) we getUn+1+Vn+l = a(un-1+ Vn-1) +b(un-2+Vn-2), ne No,(2.65)whereas by subtracting the second one from the first, we haveUn+1 - Vn+1= a(un-1 – Vn-1) – b (un-2 – Vn-2), n E No.(2.66)From (2.5), we can write the solution of equation (2.65) asUn + Vn = (uo+ vo) Jn+1+(u-1+v-1) Jn+2+b (u_2+v–2) Jn,(2.67)for n > -2. On the other hand, by taking a = 0, b = a, c= -b in (1.4) andS, = (-1)&#34; Jn+1, for all n> -2, which is called generalized Padovan sequence, in(1.5), from (2.66), we also have thatUn – Vn = (-1)&#34; ( (u0 – vo) Jn+l – (u-1 – v-1) Jn+2+b (u-2 – v-2) Jn),(2.68)for n > -2. From (2.68) we haveU2n - V2n=(4o – vo) J2n+1 - (u-1 – v–1) J2n+2+b (u-2 - v-2) J2n,(2.69)for n >-1 andU2n+1 - V2n+1 =-(40 - vo) J2n+2+(u-1 - v-1)J2n+3-b (u_2-v-2) J2n+1> (2.70)for n > -1. From (2.67)U2n+ V2n = (uo+ vo) J2n+1+(u_1+v-1)J2n+2+b (u_2+v_2)J2n(2.71)for n> -1. By summing the equations (2.69) and (2.71)uzn = uoJ2n+1 +v-1J2n+2 + bu-2J2n, n > -1.(2.72)By subtracting equation (2.69) from equation (2.71), we haveV2n =VOJ21+1+u-1J20+2+ bv-2J2n, n>-1.(2.73)From (2.67)U2n+1+V2n+1(uo +vo)J2n+2+(u-1+v-1) J2n+3+b(u_2+v_2) J2n+1, (2.74)for n >-1. By summing the equations (2.70) and (2.74) we getU2n+1 = VoJ2n+2 +u-1J21+3 +bv-2J2n+1, n> -1.(2.75)By subtracting equation (2.70) from equation (2.74), we haveV2n+1 =uoJ2n+2+v-1J2n+3+ bu-2J2n+1, n > -1.(2.76)From (2.63), (2.72), (2.73), (2.75), (2.76) and after some calculation, we obtainX2, = f(f (x0) J2n+1+f(y-1)J2n+2 +bf (x_2) J2n), n2-1,y2n = f(f(yo) J2n+1+f (x-1)J2n+2+bf (y-2) J2n), n2-1,f(f(vo)Jzn+2+f (x-1)J2n+3+bf (y_2)J2n+1), n> –1,(2.77)(2.78)X2n+1 =(2.79)andy2n+1 = f (f (x0)Jzn+2+f(y-1)J2n+3+bf (x-2)J2n+1),n > -1.(2.80) constant coefficientsXn+1 = axn+bxn-1+cxn-2, nE No,(1.4)which has actually the general solutionX = xoSn+x-1 (Sn+1 - aSn) +cx_2Sn-1, nE No,(1.5)where (S.) of equation (1L4) satisfving the initial values S 2=S-=0, So= L.The equationaxn-IXn-kXn+1 =nE No,(1.1)bxn-p±cxn-q'where the initial conditions are arbitrary positive real numbers, k, 1, p, q are non-negative integers and a, b, c are positive constants, is one of the difference equationswhose solutions are associated with number sequences. Positive solutions of concreteMotivated by this line of investigations, here we show that the systems of differ-ence equationsXn+l =f'(af (Pn-1)+bf(qn-2)), Yntl =f&#34;(af (ra-1)+bf (Sn-2)), (1.8)for n E No, where the sequences Pn, qn, Tn and Sn are some of the sequences x, andYn, f: Df R is a &#34;1– 1&#34; continuous function on its domain Dr C R, the initialvalues x-j, y-j, je {0,1,2} are arbitrary real numbers and the parameters and a, b2.6. Case 6: Pn = Yn, qn = yn, rn = Xn, Sn = XnIn this case, we obtain the systemXn+1 =5&#34;(af (Yn-1)+bf (Yn-2)), Yn+1 = '(af (xn-1)+bf (xn-2)), (2.29)for n e No. Since f is &#34;1– 1&#34;, from (2.29)f (Xn+1) = af (Yn-1)+bf (yn-2), f(yn+l) = af (Xn-1) +bf (xn-2),(2.30)for n E No. By using the change of variablesf (xn) == Un,and f(yn) = Vn, n>-2,(2.31)system (2.30) is transformed to the following oneUn+l = avn-1+bvn-2, Vn+1 = aun-1+ bun-2, nE No.(2.32)By summing the equations in (2.32) we getUn+l +Vn+l = a (un-1+ Vn-1) +b(un-2+Vn-2), nE No,(2.33)whereas by subtracting the second one from the first, we haveUn+l - Vn+l = -a(un-1 - Vn-1) -6(un-2- Vn-2), ne No.(2.34)|In this section, we consider the eight special cases of systems (1.8), where thesequences Pn, qn, ľn, Sn are some of the sequences xn and yn, for n>-2, and initialvalues x-j, y-j, j e {0,1,2}, are arbitrary real numbers.2.1Case 1: Pn =xn, qn = Xn, ľn = yn, Sn == ynIn this case, system (1.8) is expressed asXn+1 =f (af (xn-1)+bf (xn-2)), Yn+1 =f(af (Vn-1)+bf (Yn-2)), (2.1)for n E No. Since f is &#34;1– 1&#34;, from (2.1)f (Xn+1) = af (x,-1)+bf (xn-2), f (Yn+1) = af (yn-1)+bf (yn–2),(2.2)for n E No. By using the change of variablesf (xn)= Un, and f(yn) == Vn, n> -2,(2.3)system (2.2) is transformed to the following oneUn+1 = au,-1+bun-2, Vn+1 = avn-1+bvn-2,(2.4)for n E No. By taking a = 0, b= a, c = b in (1.4) and S = Jn+1, for all n > -2,which is called generalized Padovan sequence, in (1.5), the solutions to equations in(2.4) are given byUn = ugJn+1+u-1Jn+2+ bu_2Jn,(2.5)Vn = voJn+1+v_1Jn+2+bv_2Jn,(2.6)for n E No. From (2.3), (2.5) and (2.6), it follows that the general solution to system(2.2) is given byXn = f(f (x0) Jn+1+f (x-1) Jn+2+bf (x-2)Jn), n> -2,Yn =f(f (vo) Jn+1+f(y-1)Ja+2+bf (y-2) J.), n> -2.(2.7)(2.8)2.2. Case 2: Pn = Xn, qn = Xn, ľn = Xn, Sn XnIn this case, system (1.8) becomesXp+1 =f&#34;(af (x,-1)+bf (xp-2)), Yn+1=f&#34;(af (xn-1)+bf (xn-2)), (2.9)for n e No. It should be first note that from the equations in (2.9) it immediatelyfollows that x, = yn, for all n E N. From (2.7), the general solution to system (2.9) is%3DX, = Ya = f'(f (xo) Jn+1+f (x=1) Jn+2+bf (x_2)Jn), n E N.(2.10)

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