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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. Tags equation, functional Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pell's fish Real Analysis 2 February 3rd, 2014 03:08 AM nukem4111 Real Analysis 1 September 16th, 2013 03:44 AM Nobody1111 Real Analysis 7 March 1st, 2012 10:01 AM antros Applied Math 4 October 12th, 2008 11:12 AM desum Elementary Math 0 December 31st, 1969 04:00 PM

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