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

The reason the only solution is x=4 for all values of n is as follows:

If f_n(x)=2x for all n then f_n+1(x)=2x since n+1 is just another value.

This means that f_n+1(x)=sqrt(x^2+6f_n(x)) can be reduced to 2x=sqrt(x^2+6*2x). This is solved below:

2x=sqrt(x^2+6*2x)
4x^2=x^2+12x
3x^2-12x=0
x(x-4)=0
Therefore x = 0 or 4
However x=0 is not a solution since for n=0 f_0(x)=2*0=0 but we have been given that f_0(x)=8
x=4 does of course work for n=0 and is therefore the only solution for all n.

Posted by Paul on 2006-11-20 10:07:25