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 May 27th, 2013, 09:50 AM #1 Newbie   Joined: May 2013 Posts: 6 Thanks: 0 Functional equation Find all functions f:R->R such that $f(x+f(x)f(y))=f(x)+xf(y)$
 May 29th, 2013, 08:53 AM #2 Newbie   Joined: May 2013 Posts: 6 Thanks: 0 Re: Functional equation A couple of simple observation: f(x)=0 iff x=0 $f(-1)^2=-1$ ( so we must face both cases). f(x)=0 is a solution.
 May 29th, 2013, 07:28 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: Functional equation f(x) = x
 May 29th, 2013, 09:46 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, 05:15 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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