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 a complete solution | Comment 10 of 12 |
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (2)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information