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

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- jtvatsim

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

- Empty

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

- danica518

really thats the only one?

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## More answers

- jtvatsim

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

- Empty

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

- Empty

I have no idea what the general solution is!

- 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

ok i guess not

- anonymous

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

- danica518

other than identity

- danica518

but the identity works? lemme see

- danica518

|dw:1437170220728:dw|

- 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

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

i nothing something similar going there

- 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

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

- 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

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

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

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

- jtvatsim

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

- 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

@jtvatsim ? what are you talking about

- 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

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

- anonymous

which line is this? can you type it out?

- jtvatsim

Sure.

- 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

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

- 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

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

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

- anonymous

this is hard

- jtvatsim

This is International Math Olympiad...

- 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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