ok so we have
let T(k)=x and T(k-1)=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)^2-8y^2)]/(4y)
x=[y^2+2+-sqrt(y^4+4y^2+4-8y^2)]/(4y)
x=[y^2+2+-sqrt(y^4-4y^2+4)]/(4y)
x=[y^2+2+-sqrt((y^2-2)^2)]/(4y)
x=[y^2+2+-(y^2-2)]/(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(k-1)/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
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Posted by Daniel
on 2009-09-20 15:27:47 |