(In reply to
re(5): Solution? by Richard)
I'll try to state my original idea a little more clearly, and add that x = 4 also should work. If you set f_1(x) equal to 2x, you get:
2x = sqrt(x^2 + 48)
4x^2 = x^2 + 48
3x^2 = 48
x^2 = 16
x = 4 or 4
Plugging x = 4 back into the equation for f_1(x):
f_1(x) = sqrt(16+48) = 8 (or choose 8, if you plug in x = 4)
But then f_0(x) = f_1(x) = 8, so solving f_2(x) will be exactly the same as f_1(x), and so on for f_3, f_4, etc. It will always ultimately reduce to:
2x = sqrt(64)
No matter what n you choose x=4 and x=4 will be the only solutions.
If you want to try to solve them nonrecursively, it quickly gets cumbersome (at least if you do it by hand):
f_2: x^4  4x^2  192 = 0
f_3: x^8  8x^6 + 16x^4  576x^2  27648 = 0
etc...

Posted by tomarken
on 20061119 19:45:48 