
Math Events Math Events, Competitions, Meetups  Local, Regional, State, National, International 
 LinkBack  Thread Tools  Display Modes 
September 18th, 2011, 02:31 AM  #1 
Joined: Sep 2011 Posts: 3 Thanks: 0  solve equation
determine all functions such that : , for each , hint please 
November 18th, 2011, 01:23 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 12,143 Thanks: 474 
Try giving x and y various values. For example, substituting x = f(0), y = f(0) into the equation gives f(f(0) + f(0)) = f(f(0) + f(0)) + (f(f(0)))². Hence f(f(0)) = 0. 
November 18th, 2011, 06:42 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,255 Thanks: 748 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic  Re: solve equation
I only get f(f(0)) = 0 from that. With x = y = 0 you can get two possibilities for f(0); eventually I was able to make one lead to contradiction leaving only f(0) = 0. Once you have that, take y = x to find f(2x) = f(x)  f(x)^2. If I had continuity I feel that this would give me a function for each choice of (say) f(1), but I don't have that. 
November 18th, 2011, 09:35 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 12,143 Thanks: 474 
Thanks. I've corrected my previous post. Now for a better attempt . . . For any real u, substituting x = u + f(u), y = f(u) gives f(u + 2f(u)) = (f(u + f(u)))² + f(u + 2f(u)), so f(u + f(u)) = 0. Also, substituting x = u, y = 0 gives f(u + f(u)) = (f(u))² + f(u), so f(u)(f(u) + 1) = 0, i.e. f(u) = 0 or 1. If f(0) = 1, substituting x = 0, y = u in the equation gives f(f(u)) = 1 + f(u), which implies f(u) = 1 and f(f(u)) = 0, but that gives both f(1) = 1 and f(1) = 0 (by putting u = 1 and u = 0, respectively), a contradiction. Hence f(0) = 0. Putting y = x gives f(x + f(0)) = (f(x))² + f(2x), i.e. f(x) = (f(x))² + f(2x), which implies f(x) cannot be 1. Hence f(x) = 0 for all x. 

Tags 
equation, solve 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Solve the Equation for X  Shamieh  Algebra  6  May 21st, 2013 09:46 PM 
solve equation  mmmmxxx  Algebra  3  September 4th, 2012 06:22 PM 
Solve the equation...  boomer029  Algebra  8  March 17th, 2012 05:09 PM 
Solve the equation indicated  ishoothigh  Algebra  1  December 14th, 2011 09:09 AM 
solve this equation for x  sagelady  Calculus  2  July 24th, 2011 08:35 PM 