ok so we have
let T(k)=x and T(k1)=y then we have
y+2/y=2x+1/x since neither x,y=0
y^2x+2x=2yx^2+y
2yx^2(y^2+2)x+y=0
x=[y^2+2+sqrt((y^2+2)^28y^2)]/(4y)
x=[y^2+2+sqrt(y^4+4y^2+48y^2)]/(4y)
x=[y^2+2+sqrt(y^44y^2+4)]/(4y)
x=[y^2+2+sqrt((y^22)^2)]/(4y)
x=[y^2+2+(y^22)]/(4y)
so we have either
x=y/2 or x=1/y
now lets say we start with T0=x and we use Tk=T(k1)/2 for each k from 1 to 1994 and then T1995=1/T1994 then we end up with
T0=x=2^1994/x and thus
x^2=2^1994
x=2^997
Thus it would appear that 2^997 is the maximum value for T0

Posted by Daniel
on 20090920 15:27:47 