My Math Forum function equation

 Algebra Pre-Algebra and Basic Algebra Math Forum

 March 30th, 2014, 09:12 AM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 function equation Find all $f:R -> R$ so that $f(xy - f(y) )= (1 - f(x)) f(y)$
March 30th, 2014, 07:39 PM   #2
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,693
Thanks: 2677

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by gelatine1 Find all $f:R -> R$ so that $f(xy - f(y) )= (1 - f(x)) f(y)$
$f(xy - f(y) )= (1 - f(x)) f(y)$
or
$f(xy) - f(y)= (1 - f(x)) f(y)$
?

The latter makes more sense to me on the face of it.

 March 30th, 2014, 07:54 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 For the equation originally posted, f(x) ≡ 0 and f(x) ≡ -x work. I suspect they are the only solutions, but I don't have a proof of that.
 March 31st, 2014, 10:39 AM #4 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Yes, that is correct. Although I have a proof (in Dutch) I don't completely understand it. But at least your solutions are correct
 March 31st, 2014, 09:56 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra Well, I can see those two solutions: If we assume that there exists a value of y for which f(y) = 0, then we have that f(xy) = 0 for all x. This gives us two possibilities:f(x) = 0 for all x; or y = 0, which means that f(0) = 0. The first is clearly degenerate, so let's take f(0) = 0. Note that we got this far on the basis that there exists at least one value for which f(y) = 0. Moving on, if we let x = 0 in the functional equation, we get $f(-f(y))= f(y)$ because f(0) = 0. Now, if we assume that there exists an inverse function g(x) such that g(f(x)) = x, we can apply this function to both sides of our equation to get $-f(y)= y$ so $f(y)= -y$ for all y. Thus, if there exists a value of x for which f(x) = 0, we can say that $f(x) = \begin{cases} 0 \; \forall x \; \text{or}\\ -x \; \forall x \\ \end{cases}$ I can't make anything work if we assume that there is no value for which f(x) = 0. That is, I can neither see such a function nor prove that no such function exists.

 Tags equation, function

### signum function equation

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Lucida Algebra 8 May 23rd, 2012 07:11 PM unlimited Algebra 2 August 13th, 2011 05:38 AM felipe21 Differential Equations 6 April 16th, 2009 12:38 PM FutureMD2Be24 Calculus 1 April 15th, 2009 07:49 AM Simonp2 Differential Equations 0 February 13th, 2009 04:35 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top