All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Another function problem (Posted on 2006-11-19) Difficulty: 3 of 5
Define a sequence of functions f0, f1, f2, ......, by

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

Solve the equation

fn(x)= 2x

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re(2): Solution? Comment 12 of 12 |
(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

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information