First what if we go backwards?
Using my function f(x) = sqrt((x+2)+(x2)f(x+1))
f(0) = sqrt(2+2*f(1)) = sqrt(2+2) = 0
f(1) = sqrt(1+3*f(0)) = sqrt(1+0) = 1
which is promising but
f(2) = sqrt(0+4*f(1)) = sqrt(4) = 2i
which is less so
f(3) =sqrt(1+5*f(2)) = sqrt(110i) = 2.125192.35052i
and they seem to continue as irrational complex numbers without too much pattern.
Second the recursive proof that f(x)=x for x a positive integer:
We now know f(3)=3
assume f(x)=x to prove f(x+1)=x+1
f(x+1) = (f(x)^2  (x+2)) / (x2)
= ((x)^2  x  2) / (x2)
=(x+1)(x2) / (x2)
=x+1
[I was very close to this in my other post but it requires x>2 which we didn't have at the time.]
Finally to show this formula works for any real number > 0
We prove f(x+a) = x+a where x is a positive integer and a is a positive real.
x+a = f(x)+a
so we have f(x+a)=f(x)+a and we are done.

Posted by Jer
on 20100206 15:20:25 