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: Solution? by Richard)

Yes, the only solutions are exactly x=4.  There is no n such that fn(x)=2x for x not equal 4.

Here is the proof:

obviously only non-negative x need apply.

by induction fn(0)>0 for all n because
f0(0)=8,
if fn(0)>0 then fn+1(x)=sqrt(6fn(x)) which is also >0

by induction fn(x)>2x for 0<x<4 for all n because
f0(x)=8>2x
if fn(x)>2x then
  fn+1(x)>sqrt(x^2+12x)
             >x*sqrt(1+12/x)
             >x*sqrt(1+12/4)
             >2x

by induction fn(x)<2x for 4<x for all n because
f0(x)=8<2x
if fn(x)<2x then
  fn+1(x)<sqrt(x^2+12x)
             <x*sqrt(1+12/x)
             <x*sqrt(1+12/4)
             <2x

Posted by Joel on 2006-11-21 23:22:33