Upper Bounds and Lower Bounds on the members of the orbit of the generalized Lynnes Equation x[n+1]=(p+x[n])/x[n-1] By Shalosh B. Ekhad Let a,b,p be a positive real numbers, and consider the terms of the non-line\ ar recurrence p + x[n] x[n + 1] = -------- x[n - 1] and in Maple notation p + x[n] x[n + 1] = -------- x[n - 1] Subject to the initial conditions x[1] = a, x[2] = b Then x[n] is always between m and M where m and M are the middle two roots o\ f quartic equation in X 2 2 4 3 2 2 3 3 2 2 2 3 a b X + (-2 a b - 2 a b - 2 a b - 4 a b - 2 a b p - 2 a b 2 2 2 3 4 2 3 3 2 4 - 2 a b p - 2 a b - 2 a b - 2 a b p) X + (a b + 2 a b + a b 4 3 2 3 2 3 2 2 4 3 + 2 a b + 2 a b + 2 a b p + 2 a b + 2 a b p + 2 a b + 2 a b p 4 3 3 2 2 2 3 2 2 + a + 2 a b + 2 a p + 4 a b p + a p + 2 a b + 4 a b p + 2 a b p 4 3 2 2 3 2 2 2 2 + b + 2 b p + b p + 2 a + 2 a b + 4 a p + 2 a b + 4 a b p + 2 a p 3 2 2 2 2 2 2 + 2 b + 4 b p + 2 b p + a + 2 a b + 2 a p + b + 2 b p + p ) X + ( 3 2 2 3 3 2 3 2 3 2 2 -2 a b p - 2 a b p - 2 a b - 2 a b p - 2 a b - 8 a b p 2 2 3 2 2 3 2 2 2 3 - 2 a b p - 2 a b p - 2 a b p - 2 a b - 4 a b - 4 a b p - 2 a b 2 2 2 2 2 2 2 - 4 a b p - 2 a b p - 2 a b - 2 a b - 2 a b p) X + a b p 2 2 2 2 - 2 a b p + a b = 0 and in Maple notation a^2*b^2*X^4+(-2*a^3*b^2-2*a^2*b^3-2*a^3*b-4*a^2*b^2-2*a^2*b*p-2*a*b^3-2*a*b^2*p -2*a^2*b-2*a*b^2-2*a*b*p)*X^3+(a^4*b^2+2*a^3*b^3+a^2*b^4+2*a^4*b+2*a^3*b^2+2*a^ 3*b*p+2*a^2*b^3+2*a^2*b^2*p+2*a*b^4+2*a*b^3*p+a^4+2*a^3*b+2*a^3*p+4*a^2*b*p+a^2 *p^2+2*a*b^3+4*a*b^2*p+2*a*b*p^2+b^4+2*b^3*p+b^2*p^2+2*a^3+2*a^2*b+4*a^2*p+2*a* b^2+4*a*b*p+2*a*p^2+2*b^3+4*b^2*p+2*b*p^2+a^2+2*a*b+2*a*p+b^2+2*b*p+p^2)*X^2+(-\ 2*a^3*b^2*p-2*a^2*b^3*p-2*a^3*b^2-2*a^3*b*p-2*a^2*b^3-8*a^2*b^2*p-2*a^2*b*p^2-2 *a*b^3*p-2*a*b^2*p^2-2*a^3*b-4*a^2*b^2-4*a^2*b*p-2*a*b^3-4*a*b^2*p-2*a*b*p^2-2* a^2*b-2*a*b^2-2*a*b*p)*X+a^2*b^2*p^2-2*a^2*b^2*p+a^2*b^2 = 0 Proof: It is readily seen, by pluging in the recurrence that for any two consecutiv\ e terms of the orbit,x[n-1],x[n] we have the following invariant (p + x[n - 1] + x[n]) (1 + x[n - 1]) (1 + x[n]) ----------------------------------------------- x[n - 1] x[n] and in Maple notation (p+x[n-1]+x[n])*(1+x[n-1])*(1+x[n])/x[n-1]/x[n] Plugging in we n=2, we have that x[n-1]=X, and x[n]=Y must obey (p + X + Y) (1 + X) (1 + Y) (p + a + b) (1 + a) (1 + b) --------------------------- - --------------------------- = 0 X Y a b simplifying, we see that 2 2 2 2 2 2 2 X Y a b + X Y a b - X Y a b - X Y a b + X a b - X Y a - X Y a p - X Y b 2 - X Y b p + X a b p + Y a b + Y a b p - X Y a - X Y b - X Y p + X a b + Y a b + a b p = 0 and in Maple notation X^2*Y*a*b+X*Y^2*a*b-X*Y*a^2*b-X*Y*a*b^2+X^2*a*b-X*Y*a^2-X*Y*a*p-X*Y*b^2-X*Y*b*p +X*a*b*p+Y^2*a*b+Y*a*b*p-X*Y*a-X*Y*b-X*Y*p+X*a*b+Y*a*b+a*b*p = 0 Both lower bounds and upper bounds must be roots of the discriminant with re\ spect to Y of the above quadratic in Y, that is the above-mentioned quar\ tic equation in X, namely 2 2 4 3 2 2 3 3 2 2 2 3 a b X + (-2 a b - 2 a b - 2 a b - 4 a b - 2 a b p - 2 a b 2 2 2 3 4 2 3 3 2 4 - 2 a b p - 2 a b - 2 a b - 2 a b p) X + (a b + 2 a b + a b 4 3 2 3 2 3 2 2 4 3 + 2 a b + 2 a b + 2 a b p + 2 a b + 2 a b p + 2 a b + 2 a b p 4 3 3 2 2 2 3 2 2 + a + 2 a b + 2 a p + 4 a b p + a p + 2 a b + 4 a b p + 2 a b p 4 3 2 2 3 2 2 2 2 + b + 2 b p + b p + 2 a + 2 a b + 4 a p + 2 a b + 4 a b p + 2 a p 3 2 2 2 2 2 2 + 2 b + 4 b p + 2 b p + a + 2 a b + 2 a p + b + 2 b p + p ) X + ( 3 2 2 3 3 2 3 2 3 2 2 -2 a b p - 2 a b p - 2 a b - 2 a b p - 2 a b - 8 a b p 2 2 3 2 2 3 2 2 2 3 - 2 a b p - 2 a b p - 2 a b p - 2 a b - 4 a b - 4 a b p - 2 a b 2 2 2 2 2 2 2 - 4 a b p - 2 a b p - 2 a b - 2 a b - 2 a b p) X + a b p 2 2 2 2 - 2 a b p + a b = 0 Let's illustrate the theorem with p=5, a=3,b=7 The above lower and upper bounds are [1.117899093, 7.076831620] and the minimum and maximum in the first 20000 terms are [1.117899093, 7.076831574] The difference is [0., .46e-7] Not Bad! ------------------------------------- ------------------------------ This took, 2.179, seconds.