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Empty
 one year ago
What are all possible functions on the reals that satisfy this functional equation?
$$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$$
source:
http://www.imo2015.org/solution.php?lang=en
Empty
 one year ago
What are all possible functions on the reals that satisfy this functional equation? $$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$$ source: http://www.imo2015.org/solution.php?lang=en

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jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Well... the identity function works for starters... :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Haha yeah, the trouble is showing this is the only possible function.

danica518
 one year ago
Best ResponseYou've already chosen the best response.0really thats the only one?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3In fact, f(x) = c does not appear to work for any constant....

Empty
 one year ago
Best ResponseYou've already chosen the best response.0No, that's just wishful thinking coming in, that or find all the other solutions whatever they may be.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea what the general solution is!

danica518
 one year ago
Best ResponseYou've already chosen the best response.0can a function be both multiplicative and also obey lienar Super position f(xy)=f(x)+f(y) and f(x)*F(y)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@danica518 the obvious solution \(f(x)=x\) is both

danica518
 one year ago
Best ResponseYou've already chosen the best response.0but the identity works? lemme see

danica518
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437170220728:dw

danica518
 one year ago
Best ResponseYou've already chosen the best response.0sorry i shouldnt have said Linear SSP i just mean a function that can do this f(xy)=f(x)+f(y) and f(x)*F(y)? which i dont think is possible

Empty
 one year ago
Best ResponseYou've already chosen the best response.0It looks like to what you're saying dan, f(x)=0 is the only solution, since you have: $f(x)f(y)=f(xy)=f(x)+f(y)$ which is basically just saying $ab=a+b$ so unless you put more conditions on it idk...?

danica518
 one year ago
Best ResponseYou've already chosen the best response.0i nothing something similar going there

danica518
 one year ago
Best ResponseYou've already chosen the best response.0like for the addition the x is added for multiplication the y is multiplied f(x+f(x+y))= x+f(x+y) and f(xy)=y*f(x)

danica518
 one year ago
Best ResponseYou've already chosen the best response.0suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely

danica518
 one year ago
Best ResponseYou've already chosen the best response.0f(x+f(x+y))= x+f(x+y) f(xy)=y*f(x) is there a solution to both of these equations, other than f(x)=0 and f(x)=x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(x=y=0\) so $$f(0+f(0))+f(0)=f(0)\\f(f(0))=0$$ now consider \(x=f(0),y=0\) so $$f(f(0)+f(f(0)+0))+f(0)=f(0)+f(f(0))\\f(f(0)+f(f(0)))=f(0)+f(f(0))\\f(f(0))=f(0)$$ implying \(f\) has a fixed point, \(f(0)\to f(0)\) now consider \(x=0,y=f(0)\) $$f(0+f(f(0)))+f(0)=f(f(0))+f(0)f(0)\\f(0)+f(0)=f(0)^2\\f(0)^22f(0)=0\\f(0)\left(f(0)2\right)=0$$ so either \(f(0)=0\) or \(f(0)=2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but if \(f(0)=2\) then \(f(2)=0\) which contradicts that \(f(0)\) is a fixed point (\(f(2)=2\)), so it must be that \(f(0)=0\) i think

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Potential typo in lines 6 and 7 of original calculation @oldrin.bataku

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now consider the general case \(x=0\) so $$f(f(y))+f(0)=f(y)+yf(0)$$using the fact \(f(0)=0\) this reduces to teh functional equation $$f(f(y))=f(y)$$ so all \(f(y)\) are fixed points, which suggests that the only functions that work are restrictions of \(f(x)=x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jtvatsim ? what are you talking about

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3The part where you plug in x = f(0) and y = 0 does not yield f(f(0)) = f(0) rather, it duplicates the previous substitution giving f(f(0)) = 0 again.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3However, I don't think it changes the main thrust of the proof. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which line is this? can you type it out?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Let x = f(0) and y = 0. Then, \[f(f(0) + f(f(0)+0)) + f(0) = f(0) + f(f(0))\] \[f(f(0) + f(f(0))) + f(0) = f(0) + f(f(0))\] \[f(f(0)) = 0\] rather than \[f(f(0)) = f(0) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, oops, so then \(f(0)\) need not be a fixed point, in which case \(f(0)=2\) might still work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but $$f(f(x))=x$$ still holds so \(f^2\) is still the identity, which means we want a function $f$ that is an involution but yields \(f(0)=2\). the most obvious one is \(f(x)=2x\), which seems to be a valid solution, but there might be more

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Is there some thing we can do with this \(x+f(x+y)\) term that appears twice inside and outside the function and do some sort of infinite recursive plugging in?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thought about it, but i'm not sure how to formalize it

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3This is International Math Olympiad...

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Here's my solution for integer n I made by combining both into one with a cute little discrete fourier transform... :P \[f(x)=1+(1)^n(x1)\]
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