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        riddles >> putnam exam (pure math) >> f(x +2f(y)) = f(x) + f(y) + y
        
(Message started by: THUDandBLUNDER on Dec 20^th, 2004, 10:39am)

Title: f(x +2f(y)) = f(x) + f(y) + y
Post by THUDandBLUNDER on Dec 20^th, 2004, 10:39am

Find all functions f: [bbr] [mapsto] [bbr] such that
f(x +2f(y)) = f(x) + f(y) + y

Title: Re: f(x +2f(y)) = f(x) + f(y) + y
Post by John_Gaughan on Dec 20^th, 2004, 2:01pm

I found one: [hide]f(x) = x[/hide] :)

Title: Re: f(x +2f(y)) = f(x) + f(y) + y
Post by BNC on Dec 20^th, 2004, 2:58pm

Not sure if it's enough for proof, but::
[hide]
if x=y, we get
f(x +2f(x)) = f(x) + f(x) + x = x + 2f(x)
z=x+2f(x)

f(z) = z

[/hide]

Title: Re: f(x +2f(y)) = f(x) + f(y) + y
Post by Icarus on Dec 20^th, 2004, 5:07pm

It is not sufficient for a proof. There is at least (and probably at most, though I haven't proved it yet) one other solution: [hide]f(x) = -x/2[/hide].

Title: Re: f(x +2f(y)) = f(x) + f(y) + y
Post by BNC on Dec 20^th, 2004, 10:33pm

FWIW, it's easy to prove that if f(x)=0, it must happen only at x=0.

Title: Re: f(x +2f(y)) = f(x) + f(y) + y
Post by Eigenray on Dec 21^st, 2004, 12:28am

It's also easy to show injectivity, and that for any r in the image of f (which is unbounded),
f(x+2r)+f(x-2r) = 2f(x),
which hints at linearity.