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

Home > Just Math
Positively Real (Posted on 2012-09-09) Difficulty: 3 of 5
R+denotes the set of positive real numbers.

Determine all possible functions f:R+ R+ such that:

f(f(x)) = 12x - f(x) for all x > 0.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Explanation of my solution. | Comment 3 of 4 |
Steve Herman showed that by letting f(x) = ax, then applying the above condition one arrives at a quadratic equation.  Note if the form was f(x) = ax + c, then one can easily prove c = 0, and that a = 3 or -4.  Steve correctly stated only a=3 satisfies the problem, since a = -4 doesn't map from R+ to R+.

Next, assume f(x) = ax^2 + bx + c.
Then, f(f(x)) = a(ax^2 + bx + c)^2 + b*(ax^2 + bx + c) + c.
From examination of the above expression, there exists one x^4 term in the expression.  This term is a^3x^4.  Since f(x) doesn't possess an x^4 term however, the condition f(f(x)) = 12x - f(x), implies that x^4 term of ff(x) must be 0, and thus a = 0.  If a = 0, then we find ourselves considering the equation of the form f(x) = ax + c, which Steve Herman already solved.

By continuing this reasoning, one can show through induction that f(x) cannot be a polynomial of degree 2 or more.

Thus any expression that can be represented by an infinite Taylor expansion, or any polynomial of degree greater than 2 cannot satisfy the expression.

Edited on September 10, 2012, 6:03 am
  Posted by Chris, PhD on 2012-09-10 06:02:57

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

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information