20, Rigorously Proved Theorems about the global limits of certain second-order rational recurreces with arbitrary posiitive initial conditions By Shalosh B. Ekhad Theorem number, 1, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 3 x[n - 1] + 37 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 3 x[1] + 37 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 3 x[1] + 37 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+3*a[1]+37*a[2])], [(1+2*a[1]+38*a[2 ])/(1+3*a[1]+37*a[2]), (6*a[1]*a[2]+74*a[2]^2+79*a[1]+1483*a[2]+39)/(9*a[1]*a[2 ]+111*a[2]^2+77*a[1]+1446*a[2]+38)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (9*a[1]^2*a[2]^2+222*a[1]*a[2]^3+1369*a[2]^4-18 *a[1]^2*a[2]-438*a[1]*a[2]^2-2664*a[2]^3+10*a[1]^2+208*a[1]*a[2]+1223*a[2]^2+6* a[1]+72*a[2]+1)/(1+3*a[1]+37*a[2])^2, (162*a[1]^4*a[2]^2+5832*a[1]^3*a[2]^3+ 82332*a[1]^2*a[2]^4+585192*a[1]*a[2]^5+1886482*a[2]^6+1278*a[1]^4*a[2]+34410*a[ 1]^3*a[2]^2+114942*a[1]^2*a[2]^3-1688626*a[1]*a[2]^4-3326004*a[2]^5+5965*a[1]^4 +213604*a[1]^3*a[2]+1704101*a[1]^2*a[2]^2-3537034*a[1]*a[2]^3+3670964*a[2]^4+ 5912*a[1]^3+100882*a[1]^2*a[2]-174128*a[1]*a[2]^2+304294*a[2]^3+1481*a[1]^2-\ 1784*a[1]*a[2]+9584*a[2]^2+10*a[1]+148*a[2]+1)/(9*a[1]*a[2]+111*a[2]^2+77*a[1]+ 1446*a[2]+38)^2/(1+3*a[1]+37*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(72900*a[1]^6*a[2]^2+3596400*a[1]^5*a[2]^3+66606300*a[1]^4*a[2]^4+ 550648800*a[1]^3*a[2]^5+1753278300*a[1]^2*a[2]^6+547052400*a[1]*a[2]^7+ 1686744900*a[2]^8+1247400*a[1]^6*a[2]+69481800*a[1]^5*a[2]^2+1431669600*a[1]^4* a[2]^3+12981414000*a[1]^3*a[2]^4+44248307400*a[1]^2*a[2]^5+8584451400*a[1]*a[2] ^6+40664228400*a[2]^7+5336100*a[1]^6+330991800*a[1]^5*a[2]+7559719738*a[1]^4*a[ 2]^2+75021916968*a[1]^3*a[2]^3+268619151568*a[1]^2*a[2]^4-35447924592*a[1]*a[2] ^5+204885528418*a[2]^6-1848000*a[1]^5-268424878*a[1]^4*a[2]-9756701010*a[1]^3*a [2]^2-136902379542*a[1]^2*a[2]^3-702516906374*a[1]*a[2]^4-460532588196*a[2]^5-\ 2174965*a[1]^4-22322804*a[1]^3*a[2]+3317084299*a[1]^2*a[2]^2+82374593234*a[1]*a [2]^3+519181964436*a[2]^4+7695488*a[1]^3+487116318*a[1]^2*a[2]+9565421328*a[1]* a[2]^2+58773379906*a[2]^3+7899019*a[1]^2+281198584*a[1]*a[2]+2397524416*a[2]^2+ 2613390*a[1]+43046652*a[2]+288699)/(9*a[1]*a[2]+111*a[2]^2+77*a[1]+1446*a[2]+38 )^2/(1+3*a[1]+37*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 72900*a[1]^6*a[2]^2+3596400*a[1]^5*a[2]^3+66606300*a[1]^4*a[2]^4+550648800*a[1] ^3*a[2]^5+1753278300*a[1]^2*a[2]^6+547052400*a[1]*a[2]^7+1686744900*a[2]^8+ 1247400*a[1]^6*a[2]+69481800*a[1]^5*a[2]^2+1431669600*a[1]^4*a[2]^3+12981414000 *a[1]^3*a[2]^4+44248307400*a[1]^2*a[2]^5+8584451400*a[1]*a[2]^6+40664228400*a[2 ]^7+5336100*a[1]^6+330991800*a[1]^5*a[2]+7559719738*a[1]^4*a[2]^2+75021916968*a [1]^3*a[2]^3+268619151568*a[1]^2*a[2]^4-35447924592*a[1]*a[2]^5+204885528418*a[ 2]^6-1848000*a[1]^5-268424878*a[1]^4*a[2]-9756701010*a[1]^3*a[2]^2-136902379542 *a[1]^2*a[2]^3-702516906374*a[1]*a[2]^4-460532588196*a[2]^5-2174965*a[1]^4-\ 22322804*a[1]^3*a[2]+3317084299*a[1]^2*a[2]^2+82374593234*a[1]*a[2]^3+ 519181964436*a[2]^4+7695488*a[1]^3+487116318*a[1]^2*a[2]+9565421328*a[1]*a[2]^2 +58773379906*a[2]^3+7899019*a[1]^2+281198584*a[1]*a[2]+2397524416*a[2]^2+ 2613390*a[1]+43046652*a[2]+288699 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 2, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 4 x[n - 1] + 36 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 4 x[1] + 36 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 4 x[1] + 36 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+4*a[1]+36*a[2])], [(1+2*a[1]+38*a[2 ])/(1+4*a[1]+36*a[2]), (8*a[1]*a[2]+72*a[2]^2+80*a[1]+1482*a[2]+39)/(16*a[1]*a[ 2]+144*a[2]^2+76*a[1]+1408*a[2]+37)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (16*a[1]^2*a[2]^2+288*a[1]*a[2]^3+1296*a[2]^4-\ 32*a[1]^2*a[2]-568*a[1]*a[2]^2-2520*a[2]^3+20*a[1]^2+264*a[1]*a[2]+1157*a[2]^2+ 8*a[1]+70*a[2]+1)/(1+4*a[1]+36*a[2])^2, 4*(512*a[1]^4*a[2]^2+13312*a[1]^3*a[2]^ 3+136192*a[1]^2*a[2]^4+709632*a[1]*a[2]^5+1700352*a[2]^6+2176*a[1]^4*a[2]+50560 *a[1]^3*a[2]^2+87424*a[1]^2*a[2]^3-2050432*a[1]*a[2]^4-2953728*a[2]^5+5840*a[1] ^4+206912*a[1]^3*a[2]+1630096*a[1]^2*a[2]^2-3268768*a[1]*a[2]^3+3483520*a[2]^4+ 5720*a[1]^3+96672*a[1]^2*a[2]-156248*a[1]*a[2]^2+290816*a[2]^3+1421*a[1]^2-1430 *a[1]*a[2]+9290*a[2]^2+12*a[1]+146*a[2]+1)/(16*a[1]*a[2]+144*a[2]^2+76*a[1]+ 1408*a[2]+37)^2/(1+4*a[1]+36*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(102400*a[1]^6*a[2]^2+3686400*a[1]^5*a[2]^3+49868800*a[1]^4*a[2]^4+ 302284800*a[1]^3*a[2]^5+721612800*a[1]^2*a[2]^6+298598400*a[1]*a[2]^7+671846400 *a[2]^8+972800*a[1]^6*a[2]+44134400*a[1]^5*a[2]^2+717772800*a[1]^4*a[2]^3+ 5038489600*a[1]^3*a[2]^4+13203302400*a[1]^2*a[2]^5+3160166400*a[1]*a[2]^6+ 11831961600*a[2]^7+2310400*a[1]^6+126208000*a[1]^5*a[2]+2463442688*a[1]^4*a[2]^ 2+20121407488*a[1]^3*a[2]^3+56294027008*a[1]^2*a[2]^4-14503237632*a[1]*a[2]^5+ 40128162048*a[2]^6-1216000*a[1]^5-123534976*a[1]^4*a[2]-3552479360*a[1]^3*a[2]^ 2-39603357824*a[1]^2*a[2]^3-161535082368*a[1]*a[2]^4-96991817472*a[2]^5-961840* a[1]^4+3463488*a[1]^3*a[2]+1510663104*a[1]^2*a[2]^2+26038398368*a[1]*a[2]^3+ 116597772080*a[2]^4+3012680*a[1]^3+170609128*a[1]^2*a[2]+2800455448*a[1]*a[2]^2 +13359386984*a[2]^3+2695504*a[1]^2+81574430*a[1]*a[2]+552416335*a[2]^2+759138*a [1]+10054804*a[2]+68349)/(16*a[1]*a[2]+144*a[2]^2+76*a[1]+1408*a[2]+37)^2/(1+4* a[1]+36*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 102400*a[1]^6*a[2]^2+3686400*a[1]^5*a[2]^3+49868800*a[1]^4*a[2]^4+302284800*a[1 ]^3*a[2]^5+721612800*a[1]^2*a[2]^6+298598400*a[1]*a[2]^7+671846400*a[2]^8+ 972800*a[1]^6*a[2]+44134400*a[1]^5*a[2]^2+717772800*a[1]^4*a[2]^3+5038489600*a[ 1]^3*a[2]^4+13203302400*a[1]^2*a[2]^5+3160166400*a[1]*a[2]^6+11831961600*a[2]^7 +2310400*a[1]^6+126208000*a[1]^5*a[2]+2463442688*a[1]^4*a[2]^2+20121407488*a[1] ^3*a[2]^3+56294027008*a[1]^2*a[2]^4-14503237632*a[1]*a[2]^5+40128162048*a[2]^6-\ 1216000*a[1]^5-123534976*a[1]^4*a[2]-3552479360*a[1]^3*a[2]^2-39603357824*a[1]^ 2*a[2]^3-161535082368*a[1]*a[2]^4-96991817472*a[2]^5-961840*a[1]^4+3463488*a[1] ^3*a[2]+1510663104*a[1]^2*a[2]^2+26038398368*a[1]*a[2]^3+116597772080*a[2]^4+ 3012680*a[1]^3+170609128*a[1]^2*a[2]+2800455448*a[1]*a[2]^2+13359386984*a[2]^3+ 2695504*a[1]^2+81574430*a[1]*a[2]+552416335*a[2]^2+759138*a[1]+10054804*a[2]+ 68349 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 3, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 5 x[n - 1] + 35 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 5 x[1] + 35 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 5 x[1] + 35 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+5*a[1]+35*a[2])], [(1+2*a[1]+38*a[2 ])/(1+5*a[1]+35*a[2]), (10*a[1]*a[2]+70*a[2]^2+81*a[1]+1481*a[2]+39)/(25*a[1]*a [2]+175*a[2]^2+75*a[1]+1370*a[2]+36)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (25*a[1]^2*a[2]^2+350*a[1]*a[2]^3+1225*a[2]^4-\ 50*a[1]^2*a[2]-690*a[1]*a[2]^2-2380*a[2]^3+34*a[1]^2+312*a[1]*a[2]+1095*a[2]^2+ 10*a[1]+68*a[2]+1)/(1+5*a[1]+35*a[2])^2, 9*(1250*a[1]^4*a[2]^2+25000*a[1]^3*a[2 ]^3+197500*a[1]^2*a[2]^4+805000*a[1]*a[2]^5+1531250*a[2]^6+3250*a[1]^4*a[2]+ 67750*a[1]^3*a[2]^2+31250*a[1]^2*a[2]^3-2358750*a[1]*a[2]^4-2607500*a[2]^5+5725 *a[1]^4+200700*a[1]^3*a[2]+1559725*a[1]^2*a[2]^2-3009250*a[1]*a[2]^3+3300700*a[ 2]^4+5540*a[1]^3+92730*a[1]^2*a[2]-138940*a[1]*a[2]^2+277630*a[2]^3+1365*a[1]^2 -1084*a[1]*a[2]+9000*a[2]^2+14*a[1]+144*a[2]+1)/(25*a[1]*a[2]+175*a[2]^2+75*a[1 ]+1370*a[2]+36)^2/(1+5*a[1]+35*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(1562500*a[1]^6*a[2]^2+43750000*a[1]^5*a[2]^3+460937500*a[1]^4*a[2]^4+ 2187500000*a[1]^3*a[2]^5+4210937500*a[1]^2*a[2]^6+2143750000*a[1]*a[2]^7+ 3751562500*a[2]^8+9375000*a[1]^6*a[2]+365625000*a[1]^5*a[2]^2+4906250000*a[1]^4 *a[2]^3+27843750000*a[1]^3*a[2]^4+58734375000*a[1]^2*a[2]^5+16690625000*a[1]*a[ 2]^6+51450000000*a[2]^7+14062500*a[1]^6+700125000*a[1]^5*a[2]+12049176250*a[1]^ 4*a[2]^2+82741525000*a[1]^3*a[2]^3+181414410000*a[1]^2*a[2]^4-78642620000*a[1]* a[2]^5+123028281250*a[2]^6-9000000*a[1]^5-767429250*a[1]^4*a[2]-18857484750*a[1 ]^3*a[2]^2-174932406250*a[1]^2*a[2]^3-588416796250*a[1]*a[2]^4-324012307500*a[2 ]^5-6126525*a[1]^4+54293700*a[1]^3*a[2]+8915102475*a[1]^2*a[2]^2+122635038250*a [1]*a[2]^3+416968778700*a[2]^4+16645140*a[1]^3+876479430*a[1]^2*a[2]+ 12574642460*a[1]*a[2]^2+48405890330*a[2]^3+13621815*a[1]^2+364501356*a[1]*a[2]+ 2030456600*a[2]^2+3400074*a[1]+37481904*a[2]+258291)/(25*a[1]*a[2]+175*a[2]^2+ 75*a[1]+1370*a[2]+36)^2/(1+5*a[1]+35*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 1562500*a[1]^6*a[2]^2+43750000*a[1]^5*a[2]^3+460937500*a[1]^4*a[2]^4+2187500000 *a[1]^3*a[2]^5+4210937500*a[1]^2*a[2]^6+2143750000*a[1]*a[2]^7+3751562500*a[2]^ 8+9375000*a[1]^6*a[2]+365625000*a[1]^5*a[2]^2+4906250000*a[1]^4*a[2]^3+ 27843750000*a[1]^3*a[2]^4+58734375000*a[1]^2*a[2]^5+16690625000*a[1]*a[2]^6+ 51450000000*a[2]^7+14062500*a[1]^6+700125000*a[1]^5*a[2]+12049176250*a[1]^4*a[2 ]^2+82741525000*a[1]^3*a[2]^3+181414410000*a[1]^2*a[2]^4-78642620000*a[1]*a[2]^ 5+123028281250*a[2]^6-9000000*a[1]^5-767429250*a[1]^4*a[2]-18857484750*a[1]^3*a [2]^2-174932406250*a[1]^2*a[2]^3-588416796250*a[1]*a[2]^4-324012307500*a[2]^5-\ 6126525*a[1]^4+54293700*a[1]^3*a[2]+8915102475*a[1]^2*a[2]^2+122635038250*a[1]* a[2]^3+416968778700*a[2]^4+16645140*a[1]^3+876479430*a[1]^2*a[2]+12574642460*a[ 1]*a[2]^2+48405890330*a[2]^3+13621815*a[1]^2+364501356*a[1]*a[2]+2030456600*a[2 ]^2+3400074*a[1]+37481904*a[2]+258291 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 4, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 6 x[n - 1] + 34 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 6 x[1] + 34 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 6 x[1] + 34 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+6*a[1]+34*a[2])], [(1+2*a[1]+38*a[2 ])/(1+6*a[1]+34*a[2]), (12*a[1]*a[2]+68*a[2]^2+82*a[1]+1480*a[2]+39)/(36*a[1]*a [2]+204*a[2]^2+74*a[1]+1332*a[2]+35)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (36*a[1]^2*a[2]^2+408*a[1]*a[2]^3+1156*a[2]^4-\ 72*a[1]^2*a[2]-804*a[1]*a[2]^2-2244*a[2]^3+52*a[1]^2+352*a[1]*a[2]+1037*a[2]^2+ 12*a[1]+66*a[2]+1)/(1+6*a[1]+34*a[2])^2, 16*(2592*a[1]^4*a[2]^2+41472*a[1]^3*a[ 2]^3+263232*a[1]^2*a[2]^4+874752*a[1]*a[2]^5+1377952*a[2]^6+4464*a[1]^4*a[2]+ 85200*a[1]^3*a[2]^2-51024*a[1]^2*a[2]^3-2616208*a[1]*a[2]^4-2286432*a[2]^5+5620 *a[1]^4+194944*a[1]^3*a[2]+1493060*a[1]^2*a[2]^2-2758552*a[1]*a[2]^3+3122528*a[ 2]^4+5372*a[1]^3+89056*a[1]^2*a[2]-122204*a[1]*a[2]^2+264736*a[2]^3+1313*a[1]^2 -746*a[1]*a[2]+8714*a[2]^2+16*a[1]+142*a[2]+1)/(36*a[1]*a[2]+204*a[2]^2+74*a[1] +1332*a[2]+35)^2/(1+6*a[1]+34*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(1166400*a[1]^6*a[2]^2+26438400*a[1]^5*a[2]^3+225892800*a[1]^4*a[2]^4+ 875404800*a[1]^3*a[2]^5+1427428800*a[1]^2*a[2]^6+848966400*a[1]*a[2]^7+ 1202702400*a[2]^8+4795200*a[1]^6*a[2]+165888000*a[1]^5*a[2]^2+1885464000*a[1]^4 *a[2]^3+8890771200*a[1]^3*a[2]^4+15565118400*a[1]^2*a[2]^5+5121542400*a[1]*a[2] ^6+13371220800*a[2]^7+4928400*a[1]^6+227553600*a[1]^5*a[2]+3511502032*a[1]^4*a[ 2]^2+20455634112*a[1]^3*a[2]^3+34485591472*a[1]^2*a[2]^4-24267911808*a[1]*a[2]^ 5+22908142192*a[2]^6-3552000*a[1]^5-270266656*a[1]^4*a[2]-5888733600*a[1]^3*a[2 ]^2-46636631904*a[1]^2*a[2]^3-132519918368*a[1]*a[2]^4-66999713472*a[2]^5-\ 2229880*a[1]^4+25003024*a[1]^3*a[2]+2968940260*a[1]^2*a[2]^2+34492756208*a[1]*a [2]^3+92705125788*a[2]^4+5349512*a[1]^3+266501676*a[1]^2*a[2]+3423830016*a[1]*a [2]^2+10917572356*a[2]^3+4092973*a[1]^2+99009184*a[1]*a[2]+464929369*a[2]^2+ 926286*a[1]+8708382*a[2]+60846)/(36*a[1]*a[2]+204*a[2]^2+74*a[1]+1332*a[2]+35)^ 2/(1+6*a[1]+34*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 1166400*a[1]^6*a[2]^2+26438400*a[1]^5*a[2]^3+225892800*a[1]^4*a[2]^4+875404800* a[1]^3*a[2]^5+1427428800*a[1]^2*a[2]^6+848966400*a[1]*a[2]^7+1202702400*a[2]^8+ 4795200*a[1]^6*a[2]+165888000*a[1]^5*a[2]^2+1885464000*a[1]^4*a[2]^3+8890771200 *a[1]^3*a[2]^4+15565118400*a[1]^2*a[2]^5+5121542400*a[1]*a[2]^6+13371220800*a[2 ]^7+4928400*a[1]^6+227553600*a[1]^5*a[2]+3511502032*a[1]^4*a[2]^2+20455634112*a [1]^3*a[2]^3+34485591472*a[1]^2*a[2]^4-24267911808*a[1]*a[2]^5+22908142192*a[2] ^6-3552000*a[1]^5-270266656*a[1]^4*a[2]-5888733600*a[1]^3*a[2]^2-46636631904*a[ 1]^2*a[2]^3-132519918368*a[1]*a[2]^4-66999713472*a[2]^5-2229880*a[1]^4+25003024 *a[1]^3*a[2]+2968940260*a[1]^2*a[2]^2+34492756208*a[1]*a[2]^3+92705125788*a[2]^ 4+5349512*a[1]^3+266501676*a[1]^2*a[2]+3423830016*a[1]*a[2]^2+10917572356*a[2]^ 3+4092973*a[1]^2+99009184*a[1]*a[2]+464929369*a[2]^2+926286*a[1]+8708382*a[2]+ 60846 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 5, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 7 x[n - 1] + 33 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 7 x[1] + 33 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 7 x[1] + 33 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+7*a[1]+33*a[2])], [(1+2*a[1]+38*a[2 ])/(1+7*a[1]+33*a[2]), (14*a[1]*a[2]+66*a[2]^2+83*a[1]+1479*a[2]+39)/(49*a[1]*a [2]+231*a[2]^2+73*a[1]+1294*a[2]+34)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (49*a[1]^2*a[2]^2+462*a[1]*a[2]^3+1089*a[2]^4-\ 98*a[1]^2*a[2]-910*a[1]*a[2]^2-2112*a[2]^3+74*a[1]^2+384*a[1]*a[2]+983*a[2]^2+ 14*a[1]+64*a[2]+1)/(1+7*a[1]+33*a[2])^2, 25*(4802*a[1]^4*a[2]^2+63112*a[1]^3*a[ 2]^3+330652*a[1]^2*a[2]^4+922152*a[1]*a[2]^5+1239282*a[2]^6+5782*a[1]^4*a[2]+ 102130*a[1]^3*a[2]^2-156842*a[1]^2*a[2]^3-2825434*a[1]*a[2]^4-1989636*a[2]^5+ 5525*a[1]^4+189620*a[1]^3*a[2]+1430173*a[1]^2*a[2]^2-2516746*a[1]*a[2]^3+ 2949028*a[2]^4+5216*a[1]^3+85650*a[1]^2*a[2]-106040*a[1]*a[2]^2+252134*a[2]^3+ 1265*a[1]^2-416*a[1]*a[2]+8432*a[2]^2+18*a[1]+140*a[2]+1)/(49*a[1]*a[2]+231*a[2 ]^2+73*a[1]+1294*a[2]+34)^2/(1+7*a[1]+33*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/4*(470596*a[1]^6*a[2]^2+8874096*a[1]^5*a[2]^3+63223132*a[1]^4*a[2]^4+ 206096352*a[1]^3*a[2]^5+295193052*a[1]^2*a[2]^6+197222256*a[1]*a[2]^7+232440516 *a[2]^8+1402184*a[1]^6*a[2]+43879304*a[1]^5*a[2]^2+429625728*a[1]^4*a[2]^3+ 1714621552*a[1]^3*a[2]^4+2545189416*a[1]^2*a[2]^5+953728776*a[1]*a[2]^6+ 2153345040*a[2]^7+1044484*a[1]^6+45126424*a[1]^5*a[2]+629714138*a[1]^4*a[2]^2+ 3112642024*a[1]^3*a[2]^3+3813140848*a[1]^2*a[2]^4-4524449424*a[1]*a[2]^5+ 2623568706*a[2]^6-817600*a[1]^5-57108454*a[1]^4*a[2]-1126303802*a[1]^3*a[2]^2-\ 7734754958*a[1]^2*a[2]^3-18872467326*a[1]*a[2]^4-8771326740*a[2]^5-485917*a[1]^ 4+6067164*a[1]^3*a[2]+594622227*a[1]^2*a[2]^2+6012621746*a[1]*a[2]^3+ 13114023812*a[2]^4+1048760*a[1]^3+49911454*a[1]^2*a[2]+583012312*a[1]*a[2]^2+ 1568904578*a[2]^3+759427*a[1]^2+16845248*a[1]*a[2]+67890688*a[2]^2+158118*a[1]+ 1290916*a[2]+9147)/(49*a[1]*a[2]+231*a[2]^2+73*a[1]+1294*a[2]+34)^2/(1+7*a[1]+ 33*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 470596*a[1]^6*a[2]^2+8874096*a[1]^5*a[2]^3+63223132*a[1]^4*a[2]^4+206096352*a[1 ]^3*a[2]^5+295193052*a[1]^2*a[2]^6+197222256*a[1]*a[2]^7+232440516*a[2]^8+ 1402184*a[1]^6*a[2]+43879304*a[1]^5*a[2]^2+429625728*a[1]^4*a[2]^3+1714621552*a [1]^3*a[2]^4+2545189416*a[1]^2*a[2]^5+953728776*a[1]*a[2]^6+2153345040*a[2]^7+ 1044484*a[1]^6+45126424*a[1]^5*a[2]+629714138*a[1]^4*a[2]^2+3112642024*a[1]^3*a [2]^3+3813140848*a[1]^2*a[2]^4-4524449424*a[1]*a[2]^5+2623568706*a[2]^6-817600* a[1]^5-57108454*a[1]^4*a[2]-1126303802*a[1]^3*a[2]^2-7734754958*a[1]^2*a[2]^3-\ 18872467326*a[1]*a[2]^4-8771326740*a[2]^5-485917*a[1]^4+6067164*a[1]^3*a[2]+ 594622227*a[1]^2*a[2]^2+6012621746*a[1]*a[2]^3+13114023812*a[2]^4+1048760*a[1]^ 3+49911454*a[1]^2*a[2]+583012312*a[1]*a[2]^2+1568904578*a[2]^3+759427*a[1]^2+ 16845248*a[1]*a[2]+67890688*a[2]^2+158118*a[1]+1290916*a[2]+9147 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 6, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 8 x[n - 1] + 32 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 8 x[1] + 32 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 8 x[1] + 32 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+8*a[1]+32*a[2])], [(1+2*a[1]+38*a[2 ])/(1+8*a[1]+32*a[2]), (16*a[1]*a[2]+64*a[2]^2+84*a[1]+1478*a[2]+39)/(64*a[1]*a [2]+256*a[2]^2+72*a[1]+1256*a[2]+33)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (64*a[1]^2*a[2]^2+512*a[1]*a[2]^3+1024*a[2]^4-\ 128*a[1]^2*a[2]-1008*a[1]*a[2]^2-1984*a[2]^3+100*a[1]^2+408*a[1]*a[2]+933*a[2]^ 2+16*a[1]+62*a[2]+1)/(1+8*a[1]+32*a[2])^2, 36*(8192*a[1]^4*a[2]^2+90112*a[1]^3* a[2]^3+397312*a[1]^2*a[2]^4+950272*a[1]*a[2]^5+1114112*a[2]^6+7168*a[1]^4*a[2]+ 117760*a[1]^3*a[2]^2-283648*a[1]^2*a[2]^3-2989056*a[1]*a[2]^4-1716224*a[2]^5+ 5440*a[1]^4+184704*a[1]^3*a[2]+1371136*a[1]^2*a[2]^2-2283904*a[1]*a[2]^3+ 2780224*a[2]^4+5072*a[1]^3+82512*a[1]^2*a[2]-90448*a[1]*a[2]^2+239824*a[2]^3+ 1221*a[1]^2-94*a[1]*a[2]+8154*a[2]^2+20*a[1]+138*a[2]+1)/(64*a[1]*a[2]+256*a[2] ^2+72*a[1]+1256*a[2]+33)^2/(1+8*a[1]+32*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(6553600*a[1]^6*a[2]^2+104857600*a[1]^5*a[2]^3+635699200*a[1]^4*a[2]^4+ 1782579200*a[1]^3*a[2]^5+2306867200*a[1]^2*a[2]^6+1677721600*a[1]*a[2]^7+ 1677721600*a[2]^8+14745600*a[1]^6*a[2]+422707200*a[1]^5*a[2]^2+3606118400*a[1]^ 4*a[2]^3+12337152000*a[1]^3*a[2]^4+15767961600*a[1]^2*a[2]^5+6658457600*a[1]*a[ 2]^6+13212057600*a[2]^7+8294400*a[1]^6+336691200*a[1]^5*a[2]+4259663872*a[1]^4* a[2]^2+17715208192*a[1]^3*a[2]^3+13524897792*a[1]^2*a[2]^4-31660802048*a[1]*a[2 ]^5+11055726592*a[2]^6-6912000*a[1]^5-449958912*a[1]^4*a[2]-8130186240*a[1]^3*a [2]^2-48884589568*a[1]^2*a[2]^3-103561194496*a[1]*a[2]^4-44311670784*a[2]^5-\ 3936960*a[1]^4+52404864*a[1]^3*a[2]+4456253376*a[1]^2*a[2]^2+39956119936*a[1]*a [2]^3+71990563584*a[2]^4+7751952*a[1]^3+354648192*a[1]^2*a[2]+3806518832*a[1]*a [2]^2+8763680384*a[2]^3+5354136*a[1]^2+110003046*a[1]*a[2]+385731239*a[2]^2+ 1036170*a[1]+7449708*a[2]+53541)/(64*a[1]*a[2]+256*a[2]^2+72*a[1]+1256*a[2]+33) ^2/(1+8*a[1]+32*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 6553600*a[1]^6*a[2]^2+104857600*a[1]^5*a[2]^3+635699200*a[1]^4*a[2]^4+ 1782579200*a[1]^3*a[2]^5+2306867200*a[1]^2*a[2]^6+1677721600*a[1]*a[2]^7+ 1677721600*a[2]^8+14745600*a[1]^6*a[2]+422707200*a[1]^5*a[2]^2+3606118400*a[1]^ 4*a[2]^3+12337152000*a[1]^3*a[2]^4+15767961600*a[1]^2*a[2]^5+6658457600*a[1]*a[ 2]^6+13212057600*a[2]^7+8294400*a[1]^6+336691200*a[1]^5*a[2]+4259663872*a[1]^4* a[2]^2+17715208192*a[1]^3*a[2]^3+13524897792*a[1]^2*a[2]^4-31660802048*a[1]*a[2 ]^5+11055726592*a[2]^6-6912000*a[1]^5-449958912*a[1]^4*a[2]-8130186240*a[1]^3*a [2]^2-48884589568*a[1]^2*a[2]^3-103561194496*a[1]*a[2]^4-44311670784*a[2]^5-\ 3936960*a[1]^4+52404864*a[1]^3*a[2]+4456253376*a[1]^2*a[2]^2+39956119936*a[1]*a [2]^3+71990563584*a[2]^4+7751952*a[1]^3+354648192*a[1]^2*a[2]+3806518832*a[1]*a [2]^2+8763680384*a[2]^3+5354136*a[1]^2+110003046*a[1]*a[2]+385731239*a[2]^2+ 1036170*a[1]+7449708*a[2]+53541 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 7, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------ 1 + 9 x[n - 1] + 31 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 9 x[1] + 31 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 9 x[1] + 31 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+9*a[1]+31*a[2])], [(1+2*a[1]+38*a[2 ])/(1+9*a[1]+31*a[2]), (18*a[1]*a[2]+62*a[2]^2+85*a[1]+1477*a[2]+39)/(81*a[1]*a [2]+279*a[2]^2+71*a[1]+1218*a[2]+32)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (81*a[1]^2*a[2]^2+558*a[1]*a[2]^3+961*a[2]^4-\ 162*a[1]^2*a[2]-1098*a[1]*a[2]^2-1860*a[2]^3+130*a[1]^2+424*a[1]*a[2]+887*a[2]^ 2+18*a[1]+60*a[2]+1)/(1+9*a[1]+31*a[2])^2, 49*(13122*a[1]^4*a[2]^2+122472*a[1]^ 3*a[2]^3+461052*a[1]^2*a[2]^4+961992*a[1]*a[2]^5+1001362*a[2]^6+8586*a[1]^4*a[2 ]+131310*a[1]^3*a[2]^2-428886*a[1]^2*a[2]^3-3109702*a[1]*a[2]^4-1465308*a[2]^5+ 5365*a[1]^4+180172*a[1]^3*a[2]+1316021*a[1]^2*a[2]^2-2060098*a[1]*a[2]^3+ 2616140*a[2]^4+4940*a[1]^3+79642*a[1]^2*a[2]-75428*a[1]*a[2]^2+227806*a[2]^3+ 1181*a[1]^2+220*a[1]*a[2]+7880*a[2]^2+22*a[1]+136*a[2]+1)/(81*a[1]*a[2]+279*a[2 ]^2+71*a[1]+1218*a[2]+32)^2/(1+9*a[1]+31*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(53144100*a[1]^6*a[2]^2+732207600*a[1]^5*a[2]^3+3836216700*a[1]^4*a[2]^4+ 9419263200*a[1]^3*a[2]^5+11263592700*a[1]^2*a[2]^6+8687055600*a[1]*a[2]^7+ 7480520100*a[2]^8+93166200*a[1]^6*a[2]+2466498600*a[1]^5*a[2]^2+18475873200*a[1 ]^4*a[2]^3+54656060400*a[1]^3*a[2]^4+60809556600*a[1]^2*a[2]^5+28778682600*a[1] *a[2]^6+50835362400*a[2]^7+40832100*a[1]^6+1558593000*a[1]^5*a[2]+17834553322*a [1]^4*a[2]^2+61143960072*a[1]^3*a[2]^3+14738384752*a[1]^2*a[2]^4-137614132608*a [1]*a[2]^5+26916689362*a[2]^6-35784000*a[1]^5-2189651914*a[1]^4*a[2]-\ 36499764390*a[1]^3*a[2]^2-193219707186*a[1]^2*a[2]^3-357960016802*a[1]*a[2]^4-\ 141296067708*a[2]^5-19669885*a[1]^4+273813172*a[1]^3*a[2]+20694259771*a[1]^2*a[ 2]^2+166655614202*a[1]*a[2]^3+251057544540*a[2]^4+35654540*a[1]^3+1575206742*a[ 1]^2*a[2]+15659108772*a[1]*a[2]^2+31158453706*a[2]^3+23617831*a[1]^2+452907220* a[1]*a[2]+1396509880*a[2]^2+4281522*a[1]+27410136*a[2]+199851)/(81*a[1]*a[2]+ 279*a[2]^2+71*a[1]+1218*a[2]+32)^2/(1+9*a[1]+31*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 53144100*a[1]^6*a[2]^2+732207600*a[1]^5*a[2]^3+3836216700*a[1]^4*a[2]^4+ 9419263200*a[1]^3*a[2]^5+11263592700*a[1]^2*a[2]^6+8687055600*a[1]*a[2]^7+ 7480520100*a[2]^8+93166200*a[1]^6*a[2]+2466498600*a[1]^5*a[2]^2+18475873200*a[1 ]^4*a[2]^3+54656060400*a[1]^3*a[2]^4+60809556600*a[1]^2*a[2]^5+28778682600*a[1] *a[2]^6+50835362400*a[2]^7+40832100*a[1]^6+1558593000*a[1]^5*a[2]+17834553322*a [1]^4*a[2]^2+61143960072*a[1]^3*a[2]^3+14738384752*a[1]^2*a[2]^4-137614132608*a [1]*a[2]^5+26916689362*a[2]^6-35784000*a[1]^5-2189651914*a[1]^4*a[2]-\ 36499764390*a[1]^3*a[2]^2-193219707186*a[1]^2*a[2]^3-357960016802*a[1]*a[2]^4-\ 141296067708*a[2]^5-19669885*a[1]^4+273813172*a[1]^3*a[2]+20694259771*a[1]^2*a[ 2]^2+166655614202*a[1]*a[2]^3+251057544540*a[2]^4+35654540*a[1]^3+1575206742*a[ 1]^2*a[2]+15659108772*a[1]*a[2]^2+31158453706*a[2]^3+23617831*a[1]^2+452907220* a[1]*a[2]+1396509880*a[2]^2+4281522*a[1]+27410136*a[2]+199851 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 8, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 10 x[n - 1] + 30 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 10 x[1] + 30 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 10 x[1] + 30 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+10*a[1]+30*a[2])], [(1+2*a[1]+38*a[ 2])/(1+10*a[1]+30*a[2]), (20*a[1]*a[2]+60*a[2]^2+86*a[1]+1476*a[2]+39)/(100*a[1 ]*a[2]+300*a[2]^2+70*a[1]+1180*a[2]+31)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (100*a[1]^2*a[2]^2+600*a[1]*a[2]^3+900*a[2]^4-\ 200*a[1]^2*a[2]-1180*a[1]*a[2]^2-1740*a[2]^3+164*a[1]^2+432*a[1]*a[2]+845*a[2]^ 2+20*a[1]+58*a[2]+1)/(1+10*a[1]+30*a[2])^2, 64*(20000*a[1]^4*a[2]^2+160000*a[1] ^3*a[2]^3+520000*a[1]^2*a[2]^4+960000*a[1]*a[2]^5+900000*a[2]^6+10000*a[1]^4*a[ 2]+142000*a[1]^3*a[2]^2-590000*a[1]^2*a[2]^3-3190000*a[1]*a[2]^4-1236000*a[2]^5 +5300*a[1]^4+176000*a[1]^3*a[2]+1264900*a[1]^2*a[2]^2-1845400*a[1]*a[2]^3+ 2456800*a[2]^4+4820*a[1]^3+77040*a[1]^2*a[2]-60980*a[1]*a[2]^2+216080*a[2]^3+ 1145*a[1]^2+526*a[1]*a[2]+7610*a[2]^2+24*a[1]+134*a[2]+1)/(100*a[1]*a[2]+300*a[ 2]^2+70*a[1]+1180*a[2]+31)^2/(1+10*a[1]+30*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(25000000*a[1]^6*a[2]^2+300000000*a[1]^5*a[2]^3+1375000000*a[1]^4*a[2]^4+ 3000000000*a[1]^3*a[2]^5+3375000000*a[1]^2*a[2]^6+2700000000*a[1]*a[2]^7+ 2025000000*a[2]^8+35000000*a[1]^6*a[2]+860000000*a[1]^5*a[2]^2+5685000000*a[1]^ 4*a[2]^3+14620000000*a[1]^3*a[2]^4+14265000000*a[1]^2*a[2]^5+7560000000*a[1]*a[ 2]^6+12015000000*a[2]^7+12250000*a[1]^6+439000000*a[1]^5*a[2]+4508930000*a[1]^4 *a[2]^2+12224440000*a[1]^3*a[2]^3-5579570000*a[1]^2*a[2]^4-36507360000*a[1]*a[2 ]^5+3277350000*a[2]^6-11200000*a[1]^5-647260000*a[1]^4*a[2]-9987572000*a[1]^3*a [2]^2-46650860000*a[1]^2*a[2]^3-75871860000*a[1]*a[2]^4-27715824000*a[2]^5-\ 5969800*a[1]^4+85974000*a[1]^3*a[2]+5853254100*a[1]^2*a[2]^2+42742156400*a[1]*a [2]^3+54259348700*a[2]^4+10009880*a[1]^3+428603860*a[1]^2*a[2]+3971477680*a[1]* a[2]^2+6881410220*a[2]^3+6385705*a[1]^2+115010984*a[1]*a[2]+314460265*a[2]^2+ 1091166*a[1]+6276406*a[2]+46434)/(100*a[1]*a[2]+300*a[2]^2+70*a[1]+1180*a[2]+31 )^2/(1+10*a[1]+30*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 25000000*a[1]^6*a[2]^2+300000000*a[1]^5*a[2]^3+1375000000*a[1]^4*a[2]^4+ 3000000000*a[1]^3*a[2]^5+3375000000*a[1]^2*a[2]^6+2700000000*a[1]*a[2]^7+ 2025000000*a[2]^8+35000000*a[1]^6*a[2]+860000000*a[1]^5*a[2]^2+5685000000*a[1]^ 4*a[2]^3+14620000000*a[1]^3*a[2]^4+14265000000*a[1]^2*a[2]^5+7560000000*a[1]*a[ 2]^6+12015000000*a[2]^7+12250000*a[1]^6+439000000*a[1]^5*a[2]+4508930000*a[1]^4 *a[2]^2+12224440000*a[1]^3*a[2]^3-5579570000*a[1]^2*a[2]^4-36507360000*a[1]*a[2 ]^5+3277350000*a[2]^6-11200000*a[1]^5-647260000*a[1]^4*a[2]-9987572000*a[1]^3*a [2]^2-46650860000*a[1]^2*a[2]^3-75871860000*a[1]*a[2]^4-27715824000*a[2]^5-\ 5969800*a[1]^4+85974000*a[1]^3*a[2]+5853254100*a[1]^2*a[2]^2+42742156400*a[1]*a [2]^3+54259348700*a[2]^4+10009880*a[1]^3+428603860*a[1]^2*a[2]+3971477680*a[1]* a[2]^2+6881410220*a[2]^3+6385705*a[1]^2+115010984*a[1]*a[2]+314460265*a[2]^2+ 1091166*a[1]+6276406*a[2]+46434 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 9, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 11 x[n - 1] + 29 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 11 x[1] + 29 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 11 x[1] + 29 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+11*a[1]+29*a[2])], [(1+2*a[1]+38*a[ 2])/(1+11*a[1]+29*a[2]), (22*a[1]*a[2]+58*a[2]^2+87*a[1]+1475*a[2]+39)/(121*a[1 ]*a[2]+319*a[2]^2+69*a[1]+1142*a[2]+30)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (121*a[1]^2*a[2]^2+638*a[1]*a[2]^3+841*a[2]^4-\ 242*a[1]^2*a[2]-1254*a[1]*a[2]^2-1624*a[2]^3+202*a[1]^2+432*a[1]*a[2]+807*a[2]^ 2+22*a[1]+56*a[2]+1)/(1+11*a[1]+29*a[2])^2, 81*(29282*a[1]^4*a[2]^2+202312*a[1] ^3*a[2]^3+572572*a[1]^2*a[2]^4+946792*a[1]*a[2]^5+809042*a[2]^6+11374*a[1]^4*a[ 2]+149050*a[1]^3*a[2]^2-764434*a[1]^2*a[2]^3-3232578*a[1]*a[2]^4-1027412*a[2]^5 +5245*a[1]^4+172164*a[1]^3*a[2]+1217845*a[1]^2*a[2]^2-1639882*a[1]*a[2]^3+ 2302228*a[2]^4+4712*a[1]^3+74706*a[1]^2*a[2]-47104*a[1]*a[2]^2+204646*a[2]^3+ 1113*a[1]^2+824*a[1]*a[2]+7344*a[2]^2+26*a[1]+132*a[2]+1)/(121*a[1]*a[2]+319*a[ 2]^2+69*a[1]+1142*a[2]+30)^2/(1+11*a[1]+29*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(177156100*a[1]^6*a[2]^2+1868191600*a[1]^5*a[2]^3+7565004700*a[1]^4*a[2]^ 4+14852895200*a[1]^3*a[2]^5+15945948700*a[1]^2*a[2]^6+12984703600*a[1]*a[2]^7+ 8558100100*a[2]^8+202045800*a[1]^6*a[2]+4619901000*a[1]^5*a[2]^2+27027044000*a[ 1]^4*a[2]^3+60562642800*a[1]^3*a[2]^4+52035726600*a[1]^2*a[2]^5+31014902600*a[1 ]*a[2]^6+44748937200*a[2]^7+57608100*a[1]^6+1931153400*a[1]^5*a[2]+17553199258* a[1]^4*a[2]^2+34396754728*a[1]^3*a[2]^3-56681669232*a[1]^2*a[2]^4-151947538352* a[1]*a[2]^5+2293052098*a[2]^6-54648000*a[1]^5-2989474494*a[1]^4*a[2]-\ 42728357250*a[1]^3*a[2]^2-175905847846*a[1]^2*a[2]^3-251305944982*a[1]*a[2]^4-\ 85405711428*a[2]^5-28355445*a[1]^4+419159916*a[1]^3*a[2]+25920600555*a[1]^2*a[2 ]^2+172925091442*a[1]*a[2]^3+185801598132*a[2]^4+44033328*a[1]^3+1833012414*a[1 ]^2*a[2]+15918000224*a[1]*a[2]^2+24148099874*a[2]^3+27144747*a[1]^2+461650056*a [1]*a[2]+1126737536*a[2]^2+4395294*a[1]+22884108*a[2]+171819)/(121*a[1]*a[2]+ 319*a[2]^2+69*a[1]+1142*a[2]+30)^2/(1+11*a[1]+29*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 177156100*a[1]^6*a[2]^2+1868191600*a[1]^5*a[2]^3+7565004700*a[1]^4*a[2]^4+ 14852895200*a[1]^3*a[2]^5+15945948700*a[1]^2*a[2]^6+12984703600*a[1]*a[2]^7+ 8558100100*a[2]^8+202045800*a[1]^6*a[2]+4619901000*a[1]^5*a[2]^2+27027044000*a[ 1]^4*a[2]^3+60562642800*a[1]^3*a[2]^4+52035726600*a[1]^2*a[2]^5+31014902600*a[1 ]*a[2]^6+44748937200*a[2]^7+57608100*a[1]^6+1931153400*a[1]^5*a[2]+17553199258* a[1]^4*a[2]^2+34396754728*a[1]^3*a[2]^3-56681669232*a[1]^2*a[2]^4-151947538352* a[1]*a[2]^5+2293052098*a[2]^6-54648000*a[1]^5-2989474494*a[1]^4*a[2]-\ 42728357250*a[1]^3*a[2]^2-175905847846*a[1]^2*a[2]^3-251305944982*a[1]*a[2]^4-\ 85405711428*a[2]^5-28355445*a[1]^4+419159916*a[1]^3*a[2]+25920600555*a[1]^2*a[2 ]^2+172925091442*a[1]*a[2]^3+185801598132*a[2]^4+44033328*a[1]^3+1833012414*a[1 ]^2*a[2]+15918000224*a[1]*a[2]^2+24148099874*a[2]^3+27144747*a[1]^2+461650056*a [1]*a[2]+1126737536*a[2]^2+4395294*a[1]+22884108*a[2]+171819 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 10, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 12 x[n - 1] + 28 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 12 x[1] + 28 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 12 x[1] + 28 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+12*a[1]+28*a[2])], [(1+2*a[1]+38*a[ 2])/(1+12*a[1]+28*a[2]), (24*a[1]*a[2]+56*a[2]^2+88*a[1]+1474*a[2]+39)/(144*a[1 ]*a[2]+336*a[2]^2+68*a[1]+1104*a[2]+29)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (144*a[1]^2*a[2]^2+672*a[1]*a[2]^3+784*a[2]^4-\ 288*a[1]^2*a[2]-1320*a[1]*a[2]^2-1512*a[2]^3+244*a[1]^2+424*a[1]*a[2]+773*a[2]^ 2+24*a[1]+54*a[2]+1)/(1+12*a[1]+28*a[2])^2, 100*(41472*a[1]^4*a[2]^2+248832*a[1 ]^3*a[2]^3+617472*a[1]^2*a[2]^4+924672*a[1]*a[2]^5+727552*a[2]^6+12672*a[1]^4*a [2]+151680*a[1]^3*a[2]^2-949632*a[1]^2*a[2]^3-3240064*a[1]*a[2]^4-838656*a[2]^5 +5200*a[1]^4+168640*a[1]^3*a[2]+1174928*a[1]^2*a[2]^2-1443616*a[1]*a[2]^3+ 2152448*a[2]^4+4616*a[1]^3+72640*a[1]^2*a[2]-33800*a[1]*a[2]^2+193504*a[2]^3+ 1085*a[1]^2+1114*a[1]*a[2]+7082*a[2]^2+28*a[1]+130*a[2]+1)/(144*a[1]*a[2]+336*a [2]^2+68*a[1]+1104*a[2]+29)^2/(1+12*a[1]+28*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is (2985984*a[1]^6*a[2]^2+27869184*a[1]^5*a[2]^3+100528128*a[1]^4*a[2]^4+179601408 *a[1]^3*a[2]^5+186052608*a[1]^2*a[2]^6+151732224*a[1]*a[2]^7+88510464*a[2]^8+ 2820096*a[1]^6*a[2]+60051456*a[1]^5*a[2]^2+311150592*a[1]^4*a[2]^3+606978048*a[ 1]^3*a[2]^4+459454464*a[1]^2*a[2]^5+311291904*a[1]*a[2]^6+410941440*a[2]^7+ 665856*a[1]^6+20760576*a[1]^5*a[2]+163057408*a[1]^4*a[2]^2+179546112*a[1]^3*a[2 ]^3-880132352*a[1]^2*a[2]^4-1554739200*a[1]*a[2]^5-59897600*a[2]^6-652800*a[1]^ 5-33819520*a[1]^4*a[2]-446905728*a[1]^3*a[2]^2-1613689728*a[1]^2*a[2]^3-\ 2018431616*a[1]*a[2]^4-644060928*a[2]^5-330928*a[1]^4+4985536*a[1]^3*a[2]+ 281559232*a[1]^2*a[2]^2+1726884896*a[1]*a[2]^3+1572488112*a[2]^4+474968*a[1]^3+ 19277976*a[1]^2*a[2]+157669128*a[1]*a[2]^2+210176344*a[2]^3+283768*a[1]^2+ 4579534*a[1]*a[2]+10030183*a[2]^2+43746*a[1]+207444*a[2]+1581)/(144*a[1]*a[2]+ 336*a[2]^2+68*a[1]+1104*a[2]+29)^2/(1+12*a[1]+28*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 2985984*a[1]^6*a[2]^2+27869184*a[1]^5*a[2]^3+100528128*a[1]^4*a[2]^4+179601408* a[1]^3*a[2]^5+186052608*a[1]^2*a[2]^6+151732224*a[1]*a[2]^7+88510464*a[2]^8+ 2820096*a[1]^6*a[2]+60051456*a[1]^5*a[2]^2+311150592*a[1]^4*a[2]^3+606978048*a[ 1]^3*a[2]^4+459454464*a[1]^2*a[2]^5+311291904*a[1]*a[2]^6+410941440*a[2]^7+ 665856*a[1]^6+20760576*a[1]^5*a[2]+163057408*a[1]^4*a[2]^2+179546112*a[1]^3*a[2 ]^3-880132352*a[1]^2*a[2]^4-1554739200*a[1]*a[2]^5-59897600*a[2]^6-652800*a[1]^ 5-33819520*a[1]^4*a[2]-446905728*a[1]^3*a[2]^2-1613689728*a[1]^2*a[2]^3-\ 2018431616*a[1]*a[2]^4-644060928*a[2]^5-330928*a[1]^4+4985536*a[1]^3*a[2]+ 281559232*a[1]^2*a[2]^2+1726884896*a[1]*a[2]^3+1572488112*a[2]^4+474968*a[1]^3+ 19277976*a[1]^2*a[2]+157669128*a[1]*a[2]^2+210176344*a[2]^3+283768*a[1]^2+ 4579534*a[1]*a[2]+10030183*a[2]^2+43746*a[1]+207444*a[2]+1581 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 11, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 13 x[n - 1] + 27 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 13 x[1] + 27 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 13 x[1] + 27 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+13*a[1]+27*a[2])], [(1+2*a[1]+38*a[ 2])/(1+13*a[1]+27*a[2]), (26*a[1]*a[2]+54*a[2]^2+89*a[1]+1473*a[2]+39)/(169*a[1 ]*a[2]+351*a[2]^2+67*a[1]+1066*a[2]+28)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (169*a[1]^2*a[2]^2+702*a[1]*a[2]^3+729*a[2]^4-\ 338*a[1]^2*a[2]-1378*a[1]*a[2]^2-1404*a[2]^3+290*a[1]^2+408*a[1]*a[2]+743*a[2]^ 2+26*a[1]+52*a[2]+1)/(1+13*a[1]+27*a[2])^2, 121*(57122*a[1]^4*a[2]^2+298792*a[1 ]^3*a[2]^3+653692*a[1]^2*a[2]^4+895752*a[1]*a[2]^5+654642*a[2]^6+13858*a[1]^4*a [2]+149110*a[1]^3*a[2]^2-1143038*a[1]^2*a[2]^3-3215086*a[1]*a[2]^4-668844*a[2]^ 5+5165*a[1]^4+165404*a[1]^3*a[2]+1136221*a[1]^2*a[2]^2-1256674*a[1]*a[2]^3+ 2007484*a[2]^4+4532*a[1]^3+70842*a[1]^2*a[2]-21068*a[1]*a[2]^2+182654*a[2]^3+ 1061*a[1]^2+1396*a[1]*a[2]+6824*a[2]^2+30*a[1]+128*a[2]+1)/(169*a[1]*a[2]+351*a [2]^2+67*a[1]+1066*a[2]+28)^2/(1+13*a[1]+27*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(482680900*a[1]^6*a[2]^2+4009964400*a[1]^5*a[2]^3+12975262300*a[1]^4*a[2] ^4+21307384800*a[1]^3*a[2]^5+21473934300*a[1]^2*a[2]^6+17297420400*a[1]*a[2]^7+ 8981352900*a[2]^8+382717400*a[1]^6*a[2]+7582725800*a[1]^5*a[2]^2+34753227600*a[ 1]^4*a[2]^3+58732102000*a[1]^3*a[2]^4+38994485400*a[1]^2*a[2]^5+30631559400*a[1 ]*a[2]^6+37255982400*a[2]^7+75864100*a[1]^6+2182577800*a[1]^5*a[2]+14221634138* a[1]^4*a[2]^2-78903032*a[1]^3*a[2]^3-116026726832*a[1]^2*a[2]^4-156755263392*a[ 1]*a[2]^5-12148462782*a[2]^6-76648000*a[1]^5-3756918418*a[1]^4*a[2]-45694246910 *a[1]^3*a[2]^2-143260587002*a[1]^2*a[2]^3-155475327594*a[1]*a[2]^4-47372720076* a[2]^5-38099965*a[1]^4+580682916*a[1]^3*a[2]+30057487659*a[1]^2*a[2]^2+ 170424199754*a[1]*a[2]^3+131276553836*a[2]^4+50281028*a[1]^3+1995577318*a[1]^2* a[2]+15444200428*a[1]*a[2]^2+18126035066*a[2]^3+29192119*a[1]^2+449181884*a[1]* a[2]+886502296*a[2]^2+4303770*a[1]+18685312*a[2]+144579)/(169*a[1]*a[2]+351*a[2 ]^2+67*a[1]+1066*a[2]+28)^2/(1+13*a[1]+27*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 482680900*a[1]^6*a[2]^2+4009964400*a[1]^5*a[2]^3+12975262300*a[1]^4*a[2]^4+ 21307384800*a[1]^3*a[2]^5+21473934300*a[1]^2*a[2]^6+17297420400*a[1]*a[2]^7+ 8981352900*a[2]^8+382717400*a[1]^6*a[2]+7582725800*a[1]^5*a[2]^2+34753227600*a[ 1]^4*a[2]^3+58732102000*a[1]^3*a[2]^4+38994485400*a[1]^2*a[2]^5+30631559400*a[1 ]*a[2]^6+37255982400*a[2]^7+75864100*a[1]^6+2182577800*a[1]^5*a[2]+14221634138* a[1]^4*a[2]^2-78903032*a[1]^3*a[2]^3-116026726832*a[1]^2*a[2]^4-156755263392*a[ 1]*a[2]^5-12148462782*a[2]^6-76648000*a[1]^5-3756918418*a[1]^4*a[2]-45694246910 *a[1]^3*a[2]^2-143260587002*a[1]^2*a[2]^3-155475327594*a[1]*a[2]^4-47372720076* a[2]^5-38099965*a[1]^4+580682916*a[1]^3*a[2]+30057487659*a[1]^2*a[2]^2+ 170424199754*a[1]*a[2]^3+131276553836*a[2]^4+50281028*a[1]^3+1995577318*a[1]^2* a[2]+15444200428*a[1]*a[2]^2+18126035066*a[2]^3+29192119*a[1]^2+449181884*a[1]* a[2]+886502296*a[2]^2+4303770*a[1]+18685312*a[2]+144579 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 12, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 14 x[n - 1] + 26 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 14 x[1] + 26 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 14 x[1] + 26 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+14*a[1]+26*a[2])], [(1+2*a[1]+38*a[ 2])/(1+14*a[1]+26*a[2]), (28*a[1]*a[2]+52*a[2]^2+90*a[1]+1472*a[2]+39)/(196*a[1 ]*a[2]+364*a[2]^2+66*a[1]+1028*a[2]+27)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (196*a[1]^2*a[2]^2+728*a[1]*a[2]^3+676*a[2]^4-\ 392*a[1]^2*a[2]-1428*a[1]*a[2]^2-1300*a[2]^3+340*a[1]^2+384*a[1]*a[2]+717*a[2]^ 2+28*a[1]+50*a[2]+1)/(1+14*a[1]+26*a[2])^2, 144*(76832*a[1]^4*a[2]^2+351232*a[1 ]^3*a[2]^3+680512*a[1]^2*a[2]^4+861952*a[1]*a[2]^5+589472*a[2]^6+14896*a[1]^4*a [2]+140560*a[1]^3*a[2]^2-1342096*a[1]^2*a[2]^3-3160272*a[1]*a[2]^4-517088*a[2]^ 5+5140*a[1]^4+162432*a[1]^3*a[2]+1101796*a[1]^2*a[2]^2-1079128*a[1]*a[2]^3+ 1867360*a[2]^4+4460*a[1]^3+69312*a[1]^2*a[2]-8908*a[1]*a[2]^2+172096*a[2]^3+ 1041*a[1]^2+1670*a[1]*a[2]+6570*a[2]^2+32*a[1]+126*a[2]+1)/(196*a[1]*a[2]+364*a [2]^2+66*a[1]+1028*a[2]+27)^2/(1+14*a[1]+26*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(188238400*a[1]^6*a[2]^2+1398342400*a[1]^5*a[2]^3+4083620800*a[1]^4*a[2]^4 +6221196800*a[1]^3*a[2]^5+6134564800*a[1]^2*a[2]^6+4822854400*a[1]*a[2]^7+ 2239182400*a[2]^8+126772800*a[1]^6*a[2]+2331302400*a[1]^5*a[2]^2+9416388800*a[1 ]^4*a[2]^3+13641465600*a[1]^3*a[2]^4+7846238400*a[1]^2*a[2]^5+7397062400*a[1]*a [2]^6+8341569600*a[2]^7+21344400*a[1]^6+560616000*a[1]^5*a[2]+2810135248*a[1]^4 *a[2]^2-4830239552*a[1]^3*a[2]^3-35122577232*a[1]^2*a[2]^4-38993382272*a[1]*a[2 ]^5-4136776592*a[2]^6-22176000*a[1]^5-1026049056*a[1]^4*a[2]-11399388960*a[1]^3 *a[2]^2-30468542944*a[1]^2*a[2]^3-28131791008*a[1]*a[2]^4-8459860032*a[2]^5-\ 10850040*a[1]^4+165952848*a[1]^3*a[2]+7890847044*a[1]^2*a[2]^2+41574876208*a[1] *a[2]^3+26945302140*a[2]^4+13056840*a[1]^3+508291068*a[1]^2*a[2]+3740357888*a[1 ]*a[2]^2+3866269844*a[2]^3+7385949*a[1]^2+108890880*a[1]*a[2]+194252105*a[2]^2+ 1045998*a[1]+4176414*a[2]+32814)/(196*a[1]*a[2]+364*a[2]^2+66*a[1]+1028*a[2]+27 )^2/(1+14*a[1]+26*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 188238400*a[1]^6*a[2]^2+1398342400*a[1]^5*a[2]^3+4083620800*a[1]^4*a[2]^4+ 6221196800*a[1]^3*a[2]^5+6134564800*a[1]^2*a[2]^6+4822854400*a[1]*a[2]^7+ 2239182400*a[2]^8+126772800*a[1]^6*a[2]+2331302400*a[1]^5*a[2]^2+9416388800*a[1 ]^4*a[2]^3+13641465600*a[1]^3*a[2]^4+7846238400*a[1]^2*a[2]^5+7397062400*a[1]*a [2]^6+8341569600*a[2]^7+21344400*a[1]^6+560616000*a[1]^5*a[2]+2810135248*a[1]^4 *a[2]^2-4830239552*a[1]^3*a[2]^3-35122577232*a[1]^2*a[2]^4-38993382272*a[1]*a[2 ]^5-4136776592*a[2]^6-22176000*a[1]^5-1026049056*a[1]^4*a[2]-11399388960*a[1]^3 *a[2]^2-30468542944*a[1]^2*a[2]^3-28131791008*a[1]*a[2]^4-8459860032*a[2]^5-\ 10850040*a[1]^4+165952848*a[1]^3*a[2]+7890847044*a[1]^2*a[2]^2+41574876208*a[1] *a[2]^3+26945302140*a[2]^4+13056840*a[1]^3+508291068*a[1]^2*a[2]+3740357888*a[1 ]*a[2]^2+3866269844*a[2]^3+7385949*a[1]^2+108890880*a[1]*a[2]+194252105*a[2]^2+ 1045998*a[1]+4176414*a[2]+32814 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 13, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 15 x[n - 1] + 25 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 15 x[1] + 25 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 15 x[1] + 25 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+15*a[1]+25*a[2])], [(1+2*a[1]+38*a[ 2])/(1+15*a[1]+25*a[2]), (30*a[1]*a[2]+50*a[2]^2+91*a[1]+1471*a[2]+39)/(225*a[1 ]*a[2]+375*a[2]^2+65*a[1]+990*a[2]+26)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (225*a[1]^2*a[2]^2+750*a[1]*a[2]^3+625*a[2]^4-\ 450*a[1]^2*a[2]-1470*a[1]*a[2]^2-1200*a[2]^3+394*a[1]^2+352*a[1]*a[2]+695*a[2]^ 2+30*a[1]+48*a[2]+1)/(1+15*a[1]+25*a[2])^2, 169*(101250*a[1]^4*a[2]^2+405000*a[ 1]^3*a[2]^3+697500*a[1]^2*a[2]^4+825000*a[1]*a[2]^5+531250*a[2]^6+15750*a[1]^4* a[2]+125250*a[1]^3*a[2]^2-1544250*a[1]^2*a[2]^3-3078250*a[1]*a[2]^4-382500*a[2] ^5+5125*a[1]^4+159700*a[1]^3*a[2]+1071725*a[1]^2*a[2]^2-911050*a[1]*a[2]^3+ 1732100*a[2]^4+4400*a[1]^3+68050*a[1]^2*a[2]+2680*a[1]*a[2]^2+161830*a[2]^3+ 1025*a[1]^2+1936*a[1]*a[2]+6320*a[2]^2+34*a[1]+124*a[2]+1)/(225*a[1]*a[2]+375*a [2]^2+65*a[1]+990*a[2]+26)^2/(1+15*a[1]+25*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(1139062500*a[1]^6*a[2]^2+7593750000*a[1]^5*a[2]^3+20123437500*a[1]^4*a[2 ]^4+28687500000*a[1]^3*a[2]^5+27773437500*a[1]^2*a[2]^6+21093750000*a[1]*a[2]^7 +8789062500*a[2]^8+658125000*a[1]^6*a[2]+11188125000*a[1]^5*a[2]^2+39555000000* a[1]^4*a[2]^3+48153750000*a[1]^3*a[2]^4+23315625000*a[1]^2*a[2]^5+28078125000*a [1]*a[2]^6+29531250000*a[2]^7+95062500*a[1]^6+2247375000*a[1]^5*a[2]+7315326250 *a[1]^4*a[2]^2-39358695000*a[1]^3*a[2]^3-161228690000*a[1]^2*a[2]^4-\ 153340050000*a[1]*a[2]^5-19506968750*a[2]^6-101400000*a[1]^5-4412911750*a[1]^4* a[2]-44259692250*a[1]^3*a[2]^2-97528896750*a[1]^2*a[2]^3-73271750750*a[1]*a[2]^ 4-23365732500*a[2]^5-49031125*a[1]^4+745860700*a[1]^3*a[2]+32611308475*a[1]^2*a [2]^2+160483592450*a[1]*a[2]^3+86658020100*a[2]^4+53167400*a[1]^3+2037377550*a[ 1]^2*a[2]+14330191080*a[1]*a[2]^2+13026531730*a[2]^3+29384875*a[1]^2+417326416* a[1]*a[2]+674355520*a[2]^2+4016454*a[1]+14804244*a[2]+118131)/(225*a[1]*a[2]+ 375*a[2]^2+65*a[1]+990*a[2]+26)^2/(1+15*a[1]+25*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 1139062500*a[1]^6*a[2]^2+7593750000*a[1]^5*a[2]^3+20123437500*a[1]^4*a[2]^4+ 28687500000*a[1]^3*a[2]^5+27773437500*a[1]^2*a[2]^6+21093750000*a[1]*a[2]^7+ 8789062500*a[2]^8+658125000*a[1]^6*a[2]+11188125000*a[1]^5*a[2]^2+39555000000*a [1]^4*a[2]^3+48153750000*a[1]^3*a[2]^4+23315625000*a[1]^2*a[2]^5+28078125000*a[ 1]*a[2]^6+29531250000*a[2]^7+95062500*a[1]^6+2247375000*a[1]^5*a[2]+7315326250* a[1]^4*a[2]^2-39358695000*a[1]^3*a[2]^3-161228690000*a[1]^2*a[2]^4-153340050000 *a[1]*a[2]^5-19506968750*a[2]^6-101400000*a[1]^5-4412911750*a[1]^4*a[2]-\ 44259692250*a[1]^3*a[2]^2-97528896750*a[1]^2*a[2]^3-73271750750*a[1]*a[2]^4-\ 23365732500*a[2]^5-49031125*a[1]^4+745860700*a[1]^3*a[2]+32611308475*a[1]^2*a[2 ]^2+160483592450*a[1]*a[2]^3+86658020100*a[2]^4+53167400*a[1]^3+2037377550*a[1] ^2*a[2]+14330191080*a[1]*a[2]^2+13026531730*a[2]^3+29384875*a[1]^2+417326416*a[ 1]*a[2]+674355520*a[2]^2+4016454*a[1]+14804244*a[2]+118131 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 14, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 16 x[n - 1] + 24 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 16 x[1] + 24 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 16 x[1] + 24 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+16*a[1]+24*a[2])], [(1+2*a[1]+38*a[ 2])/(1+16*a[1]+24*a[2]), (32*a[1]*a[2]+48*a[2]^2+92*a[1]+1470*a[2]+39)/(256*a[1 ]*a[2]+384*a[2]^2+64*a[1]+952*a[2]+25)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (256*a[1]^2*a[2]^2+768*a[1]*a[2]^3+576*a[2]^4-\ 512*a[1]^2*a[2]-1504*a[1]*a[2]^2-1104*a[2]^3+452*a[1]^2+312*a[1]*a[2]+677*a[2]^ 2+32*a[1]+46*a[2]+1)/(1+16*a[1]+24*a[2])^2, 196*(131072*a[1]^4*a[2]^2+458752*a[ 1]^3*a[2]^3+704512*a[1]^2*a[2]^4+786432*a[1]*a[2]^5+479232*a[2]^6+16384*a[1]^4* a[2]+102400*a[1]^3*a[2]^2-1746944*a[1]^2*a[2]^3-2971648*a[1]*a[2]^4-264192*a[2] ^5+5120*a[1]^4+157184*a[1]^3*a[2]+1046080*a[1]^2*a[2]^2-752512*a[1]*a[2]^3+ 1601728*a[2]^4+4352*a[1]^3+67056*a[1]^2*a[2]+13696*a[1]*a[2]^2+151856*a[2]^3+ 1013*a[1]^2+2194*a[1]*a[2]+6074*a[2]^2+36*a[1]+122*a[2]+1)/(256*a[1]*a[2]+384*a [2]^2+64*a[1]+952*a[2]+25)^2/(1+16*a[1]+24*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(419430400*a[1]^6*a[2]^2+2516582400*a[1]^5*a[2]^3+6081740800*a[1]^4*a[2]^4 +8178892800*a[1]^3*a[2]^5+7785676800*a[1]^2*a[2]^6+5662310400*a[1]*a[2]^7+ 2123366400*a[2]^8+209715200*a[1]^6*a[2]+3276800000*a[1]^5*a[2]^2+10027008000*a[ 1]^4*a[2]^3+9876275200*a[1]^3*a[2]^4+3745382400*a[1]^2*a[2]^5+6547046400*a[1]*a [2]^6+6458572800*a[2]^7+26214400*a[1]^6+547225600*a[1]^5*a[2]+603471872*a[1]^4* a[2]^2-14938062848*a[1]^3*a[2]^3-44531314688*a[1]^2*a[2]^4-37272403968*a[1]*a[2 ]^5-5327290368*a[2]^6-28672000*a[1]^5-1167855616*a[1]^4*a[2]-10385254400*a[1]^3 *a[2]^2-17642399744*a[1]^2*a[2]^3-9482600448*a[1]*a[2]^4-3884193792*a[2]^5-\ 13761280*a[1]^4+206102784*a[1]^3*a[2]+8284630080*a[1]^2*a[2]^2+38286889088*a[1] *a[2]^3+16950273728*a[2]^4+13230752*a[1]^3+501260656*a[1]^2*a[2]+3390520096*a[1 ]*a[2]^2+2700535856*a[2]^3+7167088*a[1]^2+98674694*a[1]*a[2]+144590599*a[2]^2+ 950586*a[1]+3244972*a[2]+26301)/(256*a[1]*a[2]+384*a[2]^2+64*a[1]+952*a[2]+25)^ 2/(1+16*a[1]+24*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 419430400*a[1]^6*a[2]^2+2516582400*a[1]^5*a[2]^3+6081740800*a[1]^4*a[2]^4+ 8178892800*a[1]^3*a[2]^5+7785676800*a[1]^2*a[2]^6+5662310400*a[1]*a[2]^7+ 2123366400*a[2]^8+209715200*a[1]^6*a[2]+3276800000*a[1]^5*a[2]^2+10027008000*a[ 1]^4*a[2]^3+9876275200*a[1]^3*a[2]^4+3745382400*a[1]^2*a[2]^5+6547046400*a[1]*a [2]^6+6458572800*a[2]^7+26214400*a[1]^6+547225600*a[1]^5*a[2]+603471872*a[1]^4* a[2]^2-14938062848*a[1]^3*a[2]^3-44531314688*a[1]^2*a[2]^4-37272403968*a[1]*a[2 ]^5-5327290368*a[2]^6-28672000*a[1]^5-1167855616*a[1]^4*a[2]-10385254400*a[1]^3 *a[2]^2-17642399744*a[1]^2*a[2]^3-9482600448*a[1]*a[2]^4-3884193792*a[2]^5-\ 13761280*a[1]^4+206102784*a[1]^3*a[2]+8284630080*a[1]^2*a[2]^2+38286889088*a[1] *a[2]^3+16950273728*a[2]^4+13230752*a[1]^3+501260656*a[1]^2*a[2]+3390520096*a[1 ]*a[2]^2+2700535856*a[2]^3+7167088*a[1]^2+98674694*a[1]*a[2]+144590599*a[2]^2+ 950586*a[1]+3244972*a[2]+26301 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 15, : For any positive initial conditions x[1], x[2], the recurrence 1 + 2 x[n - 1] + 38 x[n] x[n + 1] = ------------------------- 1 + 17 x[n - 1] + 23 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 2 x[1] + 38 x[2] [x[1],x[2]] goes to, [x[2], ---------------------] 1 + 17 x[1] + 23 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 2 x[1] + 38 x[2] [x[2], ---------------------], has the Global Attractor, [1, 1], 1 + 17 x[1] + 23 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+2*a[1]+38*a[2])/(1+17*a[1]+23*a[2])], [(1+2*a[1]+38*a[ 2])/(1+17*a[1]+23*a[2]), (34*a[1]*a[2]+46*a[2]^2+93*a[1]+1469*a[2]+39)/(289*a[1 ]*a[2]+391*a[2]^2+63*a[1]+914*a[2]+24)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (289*a[1]^2*a[2]^2+782*a[1]*a[2]^3+529*a[2]^4-\ 578*a[1]^2*a[2]-1530*a[1]*a[2]^2-1012*a[2]^3+514*a[1]^2+264*a[1]*a[2]+663*a[2]^ 2+34*a[1]+44*a[2]+1)/(1+17*a[1]+23*a[2])^2, 225*(167042*a[1]^4*a[2]^2+510952*a[ 1]^3*a[2]^3+701692*a[1]^2*a[2]^4+747592*a[1]*a[2]^5+432722*a[2]^6+16762*a[1]^4* a[2]+71230*a[1]^3*a[2]^2-1947622*a[1]^2*a[2]^3-2843094*a[1]*a[2]^4-161276*a[2]^ 5+5125*a[1]^4+154860*a[1]^3*a[2]+1024933*a[1]^2*a[2]^2-603586*a[1]*a[2]^3+ 1476268*a[2]^4+4316*a[1]^3+66330*a[1]^2*a[2]+24140*a[1]*a[2]^2+142174*a[2]^3+ 1005*a[1]^2+2444*a[1]*a[2]+5832*a[2]^2+38*a[1]+120*a[2]+1)/(289*a[1]*a[2]+391*a [2]^2+63*a[1]+914*a[2]+24)^2/(1+17*a[1]+23*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/4*(96550276*a[1]^6*a[2]^2+522507376*a[1]^5*a[2]^3+1156932892*a[1]^4*a[2]^4+ 1478930912*a[1]^3*a[2]^5+1383878812*a[1]^2*a[2]^6+956423536*a[1]*a[2]^7+ 323496196*a[2]^8+42094584*a[1]^6*a[2]+599818344*a[1]^5*a[2]^2+1559985008*a[1]^4 *a[2]^3+1147358832*a[1]^3*a[2]^4+262930296*a[1]^2*a[2]^5+960380456*a[1]*a[2]^6+ 893544480*a[2]^7+4588164*a[1]^6+82343784*a[1]^5*a[2]-139160678*a[1]^4*a[2]^2-\ 3201511352*a[1]^3*a[2]^3-7644969168*a[1]^2*a[2]^4-5738982080*a[1]*a[2]^5-\ 887889470*a[2]^6-5140800*a[1]^5-194689770*a[1]^4*a[2]-1492128582*a[1]^3*a[2]^2-\ 1654670482*a[1]^2*a[2]^3-266901634*a[1]*a[2]^4-398540412*a[2]^5-2460357*a[1]^4+ 35868084*a[1]^3*a[2]+1323275283*a[1]^2*a[2]^2+5778575914*a[1]*a[2]^3+2044136028 *a[2]^4+2052252*a[1]^3+77319654*a[1]^2*a[2]+506748692*a[1]*a[2]^2+351346106*a[2 ]^3+1093887*a[1]^2+14716356*a[1]*a[2]+19553912*a[2]^2+141714*a[1]+449256*a[2]+ 3699)/(289*a[1]*a[2]+391*a[2]^2+63*a[1]+914*a[2]+24)^2/(1+17*a[1]+23*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 96550276*a[1]^6*a[2]^2+522507376*a[1]^5*a[2]^3+1156932892*a[1]^4*a[2]^4+ 1478930912*a[1]^3*a[2]^5+1383878812*a[1]^2*a[2]^6+956423536*a[1]*a[2]^7+ 323496196*a[2]^8+42094584*a[1]^6*a[2]+599818344*a[1]^5*a[2]^2+1559985008*a[1]^4 *a[2]^3+1147358832*a[1]^3*a[2]^4+262930296*a[1]^2*a[2]^5+960380456*a[1]*a[2]^6+ 893544480*a[2]^7+4588164*a[1]^6+82343784*a[1]^5*a[2]-139160678*a[1]^4*a[2]^2-\ 3201511352*a[1]^3*a[2]^3-7644969168*a[1]^2*a[2]^4-5738982080*a[1]*a[2]^5-\ 887889470*a[2]^6-5140800*a[1]^5-194689770*a[1]^4*a[2]-1492128582*a[1]^3*a[2]^2-\ 1654670482*a[1]^2*a[2]^3-266901634*a[1]*a[2]^4-398540412*a[2]^5-2460357*a[1]^4+ 35868084*a[1]^3*a[2]+1323275283*a[1]^2*a[2]^2+5778575914*a[1]*a[2]^3+2044136028 *a[2]^4+2052252*a[1]^3+77319654*a[1]^2*a[2]+506748692*a[1]*a[2]^2+351346106*a[2 ]^3+1093887*a[1]^2+14716356*a[1]*a[2]+19553912*a[2]^2+141714*a[1]+449256*a[2]+ 3699 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 16, : For any positive initial conditions x[1], x[2], the recurrence x[n + 1] = 1 converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 [x[1],x[2]] goes to, [x[2], 1] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , [x[2], 1], 2 has the Global Attractor, [1, 1], for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], 1], [1, 1]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (-1+a[2])^2, 0] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is a[1]^2+a[2]^2-2*a[1]-2*a[2]+2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial a[1]^2+a[2]^2-2*a[1]-2*a[2]+2 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 17, : For any positive initial conditions x[1], x[2], the recurrence 1 + 3 x[n - 1] + 37 x[n] x[n + 1] = ------------------------ 1 + 4 x[n - 1] + 36 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 3 x[1] + 37 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 4 x[1] + 36 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 3 x[1] + 37 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 4 x[1] + 36 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+3*a[1]+37*a[2])/(1+4*a[1]+36*a[2])], [(1+3*a[1]+37*a[2 ])/(1+4*a[1]+36*a[2]), (12*a[1]*a[2]+108*a[2]^2+115*a[1]+1408*a[2]+38)/(16*a[1] *a[2]+144*a[2]^2+112*a[1]+1372*a[2]+37)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (16*a[1]^2*a[2]^2+288*a[1]*a[2]^3+1296*a[2]^4-\ 32*a[1]^2*a[2]-568*a[1]*a[2]^2-2520*a[2]^3+17*a[1]^2+270*a[1]*a[2]+1154*a[2]^2+ 8*a[1]+70*a[2]+1)/(1+4*a[1]+36*a[2])^2, (512*a[1]^4*a[2]^2+13312*a[1]^3*a[2]^3+ 136192*a[1]^2*a[2]^4+709632*a[1]*a[2]^5+1700352*a[2]^6+3200*a[1]^4*a[2]+54144*a [1]^3*a[2]^2+32128*a[1]^2*a[2]^3-2082688*a[1]*a[2]^4-2870784*a[2]^5+12688*a[1]^ 4+289152*a[1]^3*a[2]+1373968*a[1]^2*a[2]^2-2912224*a[1]*a[2]^3+3294016*a[2]^4+ 8456*a[1]^3+90120*a[1]^2*a[2]-141992*a[1]*a[2]^2+280376*a[2]^3+1442*a[1]^2-1234 *a[1]*a[2]+9073*a[2]^2+14*a[1]+144*a[2]+1)/(16*a[1]*a[2]+144*a[2]^2+112*a[1]+ 1372*a[2]+37)^2/(1+4*a[1]+36*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(409600*a[1]^6*a[2]^2+14745600*a[1]^5*a[2]^3+199475200*a[1]^4*a[2]^4+ 1209139200*a[1]^3*a[2]^5+2886451200*a[1]^2*a[2]^6+1194393600*a[1]*a[2]^7+ 2687385600*a[2]^8+5734400*a[1]^6*a[2]+224460800*a[1]^5*a[2]^2+3271065600*a[1]^4 *a[2]^3+21097676800*a[1]^3*a[2]^4+51867648000*a[1]^2*a[2]^5+13536460800*a[1]*a[ 2]^6+45984153600*a[2]^7+20070400*a[1]^6+846284800*a[1]^5*a[2]+13185279488*a[1]^ 4*a[2]^2+89211685888*a[1]^3*a[2]^3+213004561408*a[1]^2*a[2]^4-47633144832*a[1]* a[2]^5+150670222848*a[2]^6-16844800*a[1]^5-1005878400*a[1]^4*a[2]-20758594944*a [1]^3*a[2]^2-183834105728*a[1]^2*a[2]^3-636553475712*a[1]*a[2]^4-366534844416*a [2]^5+2342512*a[1]^4+373637248*a[1]^3*a[2]+11261794032*a[1]^2*a[2]^2+ 122102774624*a[1]*a[2]^3+443480245184*a[2]^4+27511544*a[1]^3+1045656280*a[1]^2* a[2]+12881391592*a[1]*a[2]^2+51451096424*a[2]^3+16293658*a[1]^2+378606234*a[1]* a[2]+2154657327*a[2]^2+3572786*a[1]+39730856*a[2]+273699)/(16*a[1]*a[2]+144*a[2 ]^2+112*a[1]+1372*a[2]+37)^2/(1+4*a[1]+36*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 409600*a[1]^6*a[2]^2+14745600*a[1]^5*a[2]^3+199475200*a[1]^4*a[2]^4+1209139200* a[1]^3*a[2]^5+2886451200*a[1]^2*a[2]^6+1194393600*a[1]*a[2]^7+2687385600*a[2]^8 +5734400*a[1]^6*a[2]+224460800*a[1]^5*a[2]^2+3271065600*a[1]^4*a[2]^3+ 21097676800*a[1]^3*a[2]^4+51867648000*a[1]^2*a[2]^5+13536460800*a[1]*a[2]^6+ 45984153600*a[2]^7+20070400*a[1]^6+846284800*a[1]^5*a[2]+13185279488*a[1]^4*a[2 ]^2+89211685888*a[1]^3*a[2]^3+213004561408*a[1]^2*a[2]^4-47633144832*a[1]*a[2]^ 5+150670222848*a[2]^6-16844800*a[1]^5-1005878400*a[1]^4*a[2]-20758594944*a[1]^3 *a[2]^2-183834105728*a[1]^2*a[2]^3-636553475712*a[1]*a[2]^4-366534844416*a[2]^5 +2342512*a[1]^4+373637248*a[1]^3*a[2]+11261794032*a[1]^2*a[2]^2+122102774624*a[ 1]*a[2]^3+443480245184*a[2]^4+27511544*a[1]^3+1045656280*a[1]^2*a[2]+ 12881391592*a[1]*a[2]^2+51451096424*a[2]^3+16293658*a[1]^2+378606234*a[1]*a[2]+ 2154657327*a[2]^2+3572786*a[1]+39730856*a[2]+273699 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 18, : For any positive initial conditions x[1], x[2], the recurrence 1 + 3 x[n - 1] + 37 x[n] x[n + 1] = ------------------------ 1 + 5 x[n - 1] + 35 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 3 x[1] + 37 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 5 x[1] + 35 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 3 x[1] + 37 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 5 x[1] + 35 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+3*a[1]+37*a[2])/(1+5*a[1]+35*a[2])], [(1+3*a[1]+37*a[2 ])/(1+5*a[1]+35*a[2]), (15*a[1]*a[2]+105*a[2]^2+116*a[1]+1407*a[2]+38)/(25*a[1] *a[2]+175*a[2]^2+110*a[1]+1335*a[2]+36)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (25*a[1]^2*a[2]^2+350*a[1]*a[2]^3+1225*a[2]^4-\ 50*a[1]^2*a[2]-690*a[1]*a[2]^2-2380*a[2]^3+29*a[1]^2+322*a[1]*a[2]+1090*a[2]^2+ 10*a[1]+68*a[2]+1)/(1+5*a[1]+35*a[2])^2, 4*(1250*a[1]^4*a[2]^2+25000*a[1]^3*a[2 ]^3+197500*a[1]^2*a[2]^4+805000*a[1]*a[2]^5+1531250*a[2]^6+4750*a[1]^4*a[2]+ 69750*a[1]^3*a[2]^2-31750*a[1]^2*a[2]^3-2372750*a[1]*a[2]^4-2534000*a[2]^5+ 12325*a[1]^4+279300*a[1]^3*a[2]+1321925*a[1]^2*a[2]^2-2677450*a[1]*a[2]^3+ 3121500*a[2]^4+8160*a[1]^3+86840*a[1]^2*a[2]-125660*a[1]*a[2]^2+267620*a[2]^3+ 1390*a[1]^2-896*a[1]*a[2]+8787*a[2]^2+16*a[1]+142*a[2]+1)/(25*a[1]*a[2]+175*a[2 ]^2+110*a[1]+1335*a[2]+36)^2/(1+5*a[1]+35*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(390625*a[1]^6*a[2]^2+10937500*a[1]^5*a[2]^3+115234375*a[1]^4*a[2]^4+ 546875000*a[1]^3*a[2]^5+1052734375*a[1]^2*a[2]^6+535937500*a[1]*a[2]^7+ 937890625*a[2]^8+3437500*a[1]^6*a[2]+113281250*a[1]^5*a[2]^2+1365468750*a[1]^4* a[2]^3+7197187500*a[1]^3*a[2]^4+14446250000*a[1]^2*a[2]^5+4387031250*a[1]*a[2]^ 6+12487343750*a[2]^7+7562500*a[1]^6+285062500*a[1]^5*a[2]+3887123750*a[1]^4*a[2 ]^2+22369412500*a[1]^3*a[2]^3+42669865000*a[1]^2*a[2]^4-17349430000*a[1]*a[2]^5 +28779843750*a[2]^6-7150000*a[1]^5-368042250*a[1]^4*a[2]-6530569750*a[1]^3*a[2] ^2-49245618250*a[1]^2*a[2]^3-144720102250*a[1]*a[2]^4-76491453500*a[2]^5+ 1022675*a[1]^4+147208200*a[1]^3*a[2]+3796911200*a[1]^2*a[2]^2+34141564950*a[1]* a[2]^3+99314836625*a[2]^4+9462840*a[1]^3+325605160*a[1]^2*a[2]+3498562410*a[1]* a[2]^2+11664700130*a[2]^3+5033010*a[1]^2+102939496*a[1]*a[2]+495344163*a[2]^2+ 977584*a[1]+9262858*a[2]+64699)/(25*a[1]*a[2]+175*a[2]^2+110*a[1]+1335*a[2]+36) ^2/(1+5*a[1]+35*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 390625*a[1]^6*a[2]^2+10937500*a[1]^5*a[2]^3+115234375*a[1]^4*a[2]^4+546875000*a [1]^3*a[2]^5+1052734375*a[1]^2*a[2]^6+535937500*a[1]*a[2]^7+937890625*a[2]^8+ 3437500*a[1]^6*a[2]+113281250*a[1]^5*a[2]^2+1365468750*a[1]^4*a[2]^3+7197187500 *a[1]^3*a[2]^4+14446250000*a[1]^2*a[2]^5+4387031250*a[1]*a[2]^6+12487343750*a[2 ]^7+7562500*a[1]^6+285062500*a[1]^5*a[2]+3887123750*a[1]^4*a[2]^2+22369412500*a [1]^3*a[2]^3+42669865000*a[1]^2*a[2]^4-17349430000*a[1]*a[2]^5+28779843750*a[2] ^6-7150000*a[1]^5-368042250*a[1]^4*a[2]-6530569750*a[1]^3*a[2]^2-49245618250*a[ 1]^2*a[2]^3-144720102250*a[1]*a[2]^4-76491453500*a[2]^5+1022675*a[1]^4+ 147208200*a[1]^3*a[2]+3796911200*a[1]^2*a[2]^2+34141564950*a[1]*a[2]^3+ 99314836625*a[2]^4+9462840*a[1]^3+325605160*a[1]^2*a[2]+3498562410*a[1]*a[2]^2+ 11664700130*a[2]^3+5033010*a[1]^2+102939496*a[1]*a[2]+495344163*a[2]^2+977584*a [1]+9262858*a[2]+64699 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 19, : For any positive initial conditions x[1], x[2], the recurrence 1 + 3 x[n - 1] + 37 x[n] x[n + 1] = ------------------------ 1 + 6 x[n - 1] + 34 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 3 x[1] + 37 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 6 x[1] + 34 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 3 x[1] + 37 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 6 x[1] + 34 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+3*a[1]+37*a[2])/(1+6*a[1]+34*a[2])], [(1+3*a[1]+37*a[2 ])/(1+6*a[1]+34*a[2]), (18*a[1]*a[2]+102*a[2]^2+117*a[1]+1406*a[2]+38)/(36*a[1] *a[2]+204*a[2]^2+108*a[1]+1298*a[2]+35)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (36*a[1]^2*a[2]^2+408*a[1]*a[2]^3+1156*a[2]^4-\ 72*a[1]^2*a[2]-804*a[1]*a[2]^2-2244*a[2]^3+45*a[1]^2+366*a[1]*a[2]+1030*a[2]^2+ 12*a[1]+66*a[2]+1)/(1+6*a[1]+34*a[2])^2, 9*(2592*a[1]^4*a[2]^2+41472*a[1]^3*a[2 ]^3+263232*a[1]^2*a[2]^4+874752*a[1]*a[2]^5+1377952*a[2]^6+6480*a[1]^4*a[2]+ 84816*a[1]^3*a[2]^2-119568*a[1]^2*a[2]^3-2614032*a[1]*a[2]^4-2221696*a[2]^5+ 11988*a[1]^4+270144*a[1]^3*a[2]+1272772*a[1]^2*a[2]^2-2450648*a[1]*a[2]^3+ 2953344*a[2]^4+7884*a[1]^3+83804*a[1]^2*a[2]-109876*a[1]*a[2]^2+255148*a[2]^3+ 1342*a[1]^2-566*a[1]*a[2]+8505*a[2]^2+18*a[1]+140*a[2]+1)/(36*a[1]*a[2]+204*a[2 ]^2+108*a[1]+1298*a[2]+35)^2/(1+6*a[1]+34*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/100*(4665600*a[1]^6*a[2]^2+105753600*a[1]^5*a[2]^3+903571200*a[1]^4*a[2]^4+ 3501619200*a[1]^3*a[2]^5+5709715200*a[1]^2*a[2]^6+3395865600*a[1]*a[2]^7+ 4810809600*a[2]^8+27993600*a[1]^6*a[2]+804556800*a[1]^5*a[2]^2+8249817600*a[1]^ 4*a[2]^3+36458726400*a[1]^3*a[2]^4+61356019200*a[1]^2*a[2]^5+21240806400*a[1]*a [2]^6+51881280000*a[2]^7+41990400*a[1]^6+1447632000*a[1]^5*a[2]+17590123872*a[1 ]^4*a[2]^2+86760913152*a[1]^3*a[2]^3+128609259712*a[1]^2*a[2]^4-88720944768*a[1 ]*a[2]^5+85382705632*a[2]^6-42768000*a[1]^5-1986019920*a[1]^4*a[2]-31307875344* a[1]^3*a[2]^2-205483754288*a[1]^2*a[2]^3-520990273712*a[1]*a[2]^4-252903420736* a[2]^5+5306508*a[1]^4+807755904*a[1]^3*a[2]+19012073052*a[1]^2*a[2]^2+ 148324597432*a[1]*a[2]^3+354063371904*a[2]^4+48188244*a[1]^3+1544728564*a[1]^2* a[2]+14880583284*a[1]*a[2]^2+42147044868*a[2]^3+23747422*a[1]^2+438648094*a[1]* a[2]+1816292255*a[2]^2+4190638*a[1]+34459740*a[2]+244091)/(36*a[1]*a[2]+204*a[2 ]^2+108*a[1]+1298*a[2]+35)^2/(1+6*a[1]+34*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 4665600*a[1]^6*a[2]^2+105753600*a[1]^5*a[2]^3+903571200*a[1]^4*a[2]^4+ 3501619200*a[1]^3*a[2]^5+5709715200*a[1]^2*a[2]^6+3395865600*a[1]*a[2]^7+ 4810809600*a[2]^8+27993600*a[1]^6*a[2]+804556800*a[1]^5*a[2]^2+8249817600*a[1]^ 4*a[2]^3+36458726400*a[1]^3*a[2]^4+61356019200*a[1]^2*a[2]^5+21240806400*a[1]*a [2]^6+51881280000*a[2]^7+41990400*a[1]^6+1447632000*a[1]^5*a[2]+17590123872*a[1 ]^4*a[2]^2+86760913152*a[1]^3*a[2]^3+128609259712*a[1]^2*a[2]^4-88720944768*a[1 ]*a[2]^5+85382705632*a[2]^6-42768000*a[1]^5-1986019920*a[1]^4*a[2]-31307875344* a[1]^3*a[2]^2-205483754288*a[1]^2*a[2]^3-520990273712*a[1]*a[2]^4-252903420736* a[2]^5+5306508*a[1]^4+807755904*a[1]^3*a[2]+19012073052*a[1]^2*a[2]^2+ 148324597432*a[1]*a[2]^3+354063371904*a[2]^4+48188244*a[1]^3+1544728564*a[1]^2* a[2]+14880583284*a[1]*a[2]^2+42147044868*a[2]^3+23747422*a[1]^2+438648094*a[1]* a[2]+1816292255*a[2]^2+4190638*a[1]+34459740*a[2]+244091 The minimum is indeed, 0, and this takes place at, [1, 1] Theorem number, 20, : For any positive initial conditions x[1], x[2], the recurrence 1 + 3 x[n - 1] + 37 x[n] x[n + 1] = ------------------------ 1 + 7 x[n - 1] + 33 x[n] converges to, 1 Proof: Introducing the corresponding transformation from [0,infinity]^2 into\ [0,infinity]^2 1 + 3 x[1] + 37 x[2] [x[1],x[2]] goes to, [x[2], --------------------] 1 + 7 x[1] + 33 x[2] We have to prove that its orbit always converges to, [1, 1], no matter where you start. So let's prove the following Proof that the transformation, [x[1], x[2]], goes to , 1 + 3 x[1] + 37 x[2] [x[2], --------------------], has the Global Attractor, [1, 1], 1 + 7 x[1] + 33 x[2] 2 for the positive orthant in, R By Shalosh B. Ekhad Starting at any point in the positive orthant, [a[1], a[2]], the first, 3, terms of the orbit are [[a[1], a[2]], [a[2], (1+3*a[1]+37*a[2])/(1+7*a[1]+33*a[2])], [(1+3*a[1]+37*a[2 ])/(1+7*a[1]+33*a[2]), (21*a[1]*a[2]+99*a[2]^2+118*a[1]+1405*a[2]+38)/(49*a[1]* a[2]+231*a[2]^2+106*a[1]+1261*a[2]+34)]] Let L be the distance-squared of the above members of the orbit to the fixed\ point, [1, 1], Then L is [a[1]^2+a[2]^2-2*a[1]-2*a[2]+2, (49*a[1]^2*a[2]^2+462*a[1]*a[2]^3+1089*a[2]^4-\ 98*a[1]^2*a[2]-910*a[1]*a[2]^2-2112*a[2]^3+65*a[1]^2+402*a[1]*a[2]+974*a[2]^2+ 14*a[1]+64*a[2]+1)/(1+7*a[1]+33*a[2])^2, 16*(4802*a[1]^4*a[2]^2+63112*a[1]^3*a[ 2]^3+330652*a[1]^2*a[2]^4+922152*a[1]*a[2]^5+1239282*a[2]^6+8330*a[1]^4*a[2]+ 98658*a[1]^3*a[2]^2-228914*a[1]^2*a[2]^3-2809066*a[1]*a[2]^4-1933008*a[2]^5+ 11677*a[1]^4+261660*a[1]^3*a[2]+1226581*a[1]^2*a[2]^2-2231890*a[1]*a[2]^3+ 2789572*a[2]^4+7628*a[1]^3+81012*a[1]^2*a[2]-94640*a[1]*a[2]^2+242960*a[2]^3+ 1298*a[1]^2-244*a[1]*a[2]+8227*a[2]^2+20*a[1]+138*a[2]+1)/(49*a[1]*a[2]+231*a[2 ]^2+106*a[1]+1261*a[2]+34)^2/(1+7*a[1]+33*a[2])^2] We claim that the first entry minus 101/100 times the , 3, th entry is always non-negative The difference between the distance-square to the point of the originating p\ oint and the distance to, 3, -th point is, times 101/100 is 1/25*(2941225*a[1]^6*a[2]^2+55463100*a[1]^5*a[2]^3+395144575*a[1]^4*a[2]^4+ 1288102200*a[1]^3*a[2]^5+1844956575*a[1]^2*a[2]^6+1232639100*a[1]*a[2]^7+ 1452753225*a[2]^8+12725300*a[1]^6*a[2]+326313050*a[1]^5*a[2]^2+2897230350*a[1]^ 4*a[2]^3+10919387500*a[1]^3*a[2]^4+15700469400*a[1]^2*a[2]^5+6111740250*a[1]*a[ 2]^6+13043334150*a[2]^7+13764100*a[1]^6+439524400*a[1]^5*a[2]+4804667492*a[1]^4 *a[2]^2+20288715052*a[1]^3*a[2]^3+21841237492*a[1]^2*a[2]^4-26364672708*a[1]*a[ 2]^5+15188444172*a[2]^6-14765800*a[1]^5-633567720*a[1]^4*a[2]-9046682832*a[1]^3 *a[2]^2-52368658544*a[1]^2*a[2]^3-115873747536*a[1]*a[2]^4-51685289568*a[2]^5+ 1505692*a[1]^4+257087560*a[1]^3*a[2]+5672584401*a[1]^2*a[2]^2+39377869160*a[1]* a[2]^3+78451349237*a[2]^4+14588288*a[1]^3+442486252*a[1]^2*a[2]+3887721910*a[1] *a[2]^2+9477034610*a[2]^3+6774508*a[1]^2+114873576*a[1]*a[2]+414807042*a[2]^2+ 1103720*a[1]+7988648*a[2]+57396)/(49*a[1]*a[2]+231*a[2]^2+106*a[1]+1261*a[2]+34 )^2/(1+7*a[1]+33*a[2])^2 The denominator is obviously positive We have to prove that the numerator is non-negative in the positive orthant The numerator is the polynomial 2941225*a[1]^6*a[2]^2+55463100*a[1]^5*a[2]^3+395144575*a[1]^4*a[2]^4+1288102200 *a[1]^3*a[2]^5+1844956575*a[1]^2*a[2]^6+1232639100*a[1]*a[2]^7+1452753225*a[2]^ 8+12725300*a[1]^6*a[2]+326313050*a[1]^5*a[2]^2+2897230350*a[1]^4*a[2]^3+ 10919387500*a[1]^3*a[2]^4+15700469400*a[1]^2*a[2]^5+6111740250*a[1]*a[2]^6+ 13043334150*a[2]^7+13764100*a[1]^6+439524400*a[1]^5*a[2]+4804667492*a[1]^4*a[2] ^2+20288715052*a[1]^3*a[2]^3+21841237492*a[1]^2*a[2]^4-26364672708*a[1]*a[2]^5+ 15188444172*a[2]^6-14765800*a[1]^5-633567720*a[1]^4*a[2]-9046682832*a[1]^3*a[2] ^2-52368658544*a[1]^2*a[2]^3-115873747536*a[1]*a[2]^4-51685289568*a[2]^5+ 1505692*a[1]^4+257087560*a[1]^3*a[2]+5672584401*a[1]^2*a[2]^2+39377869160*a[1]* a[2]^3+78451349237*a[2]^4+14588288*a[1]^3+442486252*a[1]^2*a[2]+3887721910*a[1] *a[2]^2+9477034610*a[2]^3+6774508*a[1]^2+114873576*a[1]*a[2]+414807042*a[2]^2+ 1103720*a[1]+7988648*a[2]+57396 The minimum is indeed, 0, and this takes place at, [1, 1] ------------------------------ This took, 315.853, seconds.