At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Well... the identity function works for starters... :)
Haha yeah, the trouble is showing this is the only possible function.
really thats the only one?
In fact, f(x) = c does not appear to work for any constant....
No, that's just wishful thinking coming in, that or find all the other solutions whatever they may be.
I have no idea what the general solution is!
can a function be both multiplicative and also obey lienar Super position f(xy)=f(x)+f(y) and f(x)*F(y)?
ok i guess not
@danica518 the obvious solution \(f(x)=x\) is both
other than identity
but the identity works? lemme see
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
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...?
i nothing something similar going there
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)
suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely
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
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 \(f(0)=2\)
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
Potential typo in lines 6 and 7 of original calculation @oldrin.bataku
The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS.
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 \(f(x)=x\)
@jtvatsim ? what are you talking about
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.
However, I don't think it changes the main thrust of the proof. :)
which line is this? can you type it out?
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 \[f(f(0)) = f(0) \]
oh, oops, so then \(f(0)\) need not be a fixed point, in which case \(f(0)=2\) might still work
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
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?
thought about it, but i'm not sure how to formalize it
this is hard
This is International Math Olympiad...
Here's my solution for integer n I made by combining both into one with a cute little discrete fourier transform... :P \[f(x)=1+(-1)^n(x-1)\]