|November 18th, 2011, 01:23 AM||#2|
Joined: Dec 2006
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|
Joined: Nov 2006
From: UTC -5
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|
Joined: Dec 2006
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.
|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|