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May 27th, 2013, 08:50 AM  #1 
Newbie Joined: May 2013 Posts: 6 Thanks: 0  Functional equation
Find all functions f:R>R such that 
May 29th, 2013, 07:53 AM  #2 
Newbie Joined: May 2013 Posts: 6 Thanks: 0  Re: Functional equation
A couple of simple observation: f(x)=0 iff x=0 ( so we must face both cases). f(x)=0 is a solution. 
May 29th, 2013, 06:28 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  Re: Functional equation
f(x) = x

May 29th, 2013, 08:46 PM  #4 
Newbie Joined: May 2013 Posts: 6 Thanks: 0  Re: Functional equation
Well also f(x)=x is a solution (one per each case). The difficult is to prove that f is surjective, once done you just need to put x=1 and it's done.

June 6th, 2013, 04:15 AM  #5 
Member Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0  Re: Functional equation
First, we will prove that f is injective. We have, f(x+f(x)f(y))=f(x)+xf(y). Let y=f(x), we obtain f(x+f(x)f(y))=y+xf(f(x)) x=0, f(f(0)f(y))=y, then f(f(y))=y/f(0). If f(y1)=f(y2), so f(f(y1))=f(f(y2)) imply that y1=y2. Hence, f is injective. since, f(f(0)f(y))=y=f(x). Then, f(0)f(y)=x f(y)=x/f(0), f(x)=y. Let y=1, multiply f(x)f(1)=x/f(0) f(x)=x/f(0)f(1). Therefore, f(x)=ax, a is a real number. But this is not satisfy with x=0. 

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