Define a sequence of functions f₀, f₁, f₂, ......, by

f₀(x)= 8, for all real x, and
f_n+1(x) = sqrt(x² + 6f_n(x)); for all real x and all non-negative integers n.

Solve the equation

f_n(x)= 2x

(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 non-recursively, 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 2006-11-19 19:45:48