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  • Empty
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jtvatsim
  • jtvatsim
Well... the identity function works for starters... :)
Empty
  • Empty
Haha yeah, the trouble is showing this is the only possible function.
danica518
  • danica518
really thats the only one?

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jtvatsim
  • jtvatsim
In fact, f(x) = c does not appear to work for any constant....
Empty
  • Empty
No, that's just wishful thinking coming in, that or find all the other solutions whatever they may be.
Empty
  • Empty
I have no idea what the general solution is!
danica518
  • danica518
can a function be both multiplicative and also obey lienar Super position f(xy)=f(x)+f(y) and f(x)*F(y)?
danica518
  • danica518
ok i guess not
anonymous
  • anonymous
@danica518 the obvious solution \(f(x)=x\) is both
danica518
  • danica518
other than identity
danica518
  • danica518
but the identity works? lemme see
danica518
  • danica518
|dw:1437170220728:dw|
danica518
  • danica518
sorry 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
  • Empty
It 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
  • danica518
i nothing something similar going there
danica518
  • danica518
like 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
  • danica518
suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely
danica518
  • danica518
f(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
  • anonymous
consider \(x=y=0\) 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 \(x=0,y=f(0)\) $$f(0+f(f(0)))+f(0)=f(f(0))+f(0)f(0)\\f(0)+f(0)=f(0)^2\\f(0)^2-2f(0)=0\\f(0)\left(f(0)-2\right)=0$$ so either \(f(0)=0\) or \(f(0)=2\)
anonymous
  • anonymous
but 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
  • jtvatsim
Potential typo in lines 6 and 7 of original calculation @oldrin.bataku
jtvatsim
  • jtvatsim
The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS.
anonymous
  • anonymous
now 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
  • anonymous
@jtvatsim ? what are you talking about
jtvatsim
  • jtvatsim
The 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
  • jtvatsim
However, I don't think it changes the main thrust of the proof. :)
anonymous
  • anonymous
which line is this? can you type it out?
jtvatsim
  • jtvatsim
Sure.
jtvatsim
  • jtvatsim
Let 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
  • anonymous
oh, oops, so then \(f(0)\) need not be a fixed point, in which case \(f(0)=2\) might still work
anonymous
  • anonymous
but $$f(f(x))=x$$ 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 \(f(x)=2-x\), which seems to be a valid solution, but there might be more
Empty
  • Empty
Is 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
  • anonymous
thought about it, but i'm not sure how to formalize it
anonymous
  • anonymous
this is hard
jtvatsim
  • jtvatsim
This is International Math Olympiad...
Empty
  • Empty
Here's my solution for integer n I made by combining both into one with a cute little discrete fourier transform... :P \[f(x)=1+(-1)^n(x-1)\]

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