## 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

1. jtvatsim

Well... the identity function works for starters... :)

2. Empty

Haha yeah, the trouble is showing this is the only possible function.

3. danica518

really thats the only one?

4. jtvatsim

In fact, f(x) = c does not appear to work for any constant....

5. Empty

No, that's just wishful thinking coming in, that or find all the other solutions whatever they may be.

6. Empty

I have no idea what the general solution is!

7. danica518

can a function be both multiplicative and also obey lienar Super position f(xy)=f(x)+f(y) and f(x)*F(y)?

8. danica518

ok i guess not

9. anonymous

@danica518 the obvious solution $$f(x)=x$$ is both

10. danica518

other than identity

11. danica518

but the identity works? lemme see

12. danica518

|dw:1437170220728:dw|

13. 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

14. 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...?

15. danica518

i nothing something similar going there

16. 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)

17. danica518

suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely

18. 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

19. 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 $$f(0)=2$$

20. 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

21. jtvatsim

Potential typo in lines 6 and 7 of original calculation @oldrin.bataku

22. jtvatsim

The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS.

23. 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 $$f(x)=x$$

24. anonymous

@jtvatsim ? what are you talking about

25. 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.

26. jtvatsim

However, I don't think it changes the main thrust of the proof. :)

27. anonymous

which line is this? can you type it out?

28. jtvatsim

Sure.

29. 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 $f(f(0)) = f(0)$

30. anonymous

oh, oops, so then $$f(0)$$ need not be a fixed point, in which case $$f(0)=2$$ might still work

31. 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

32. 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?

33. anonymous

thought about it, but i'm not sure how to formalize it

34. anonymous

this is hard

35. jtvatsim

Here's my solution for integer n I made by combining both into one with a cute little discrete fourier transform... :P $f(x)=1+(-1)^n(x-1)$