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Topic: f(x +2f(y)) = f(x) + f(y) + y (Read 601 times) |
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ThudnBlunder
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f(x +2f(y)) = f(x) + f(y) + y
« on: Dec 20th, 2004, 10:39am » |
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Find all functions f: [bbr] [mapsto] [bbr] such that
f(x +2f(y)) = f(x) + f(y) + y
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| « Last Edit: Dec 20th, 2004, 11:08am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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John_Gaughan
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Re: f(x +2f(y)) = f(x) + f(y) + y
« Reply #1 on: Dec 20th, 2004, 2:01pm » |
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I found one: f(x) = x 
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x = (0x2B | ~0x2B)
x == the_question
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BNC
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Re: f(x +2f(y)) = f(x) + f(y) + y
« Reply #2 on: Dec 20th, 2004, 2:58pm » |
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Not sure if it's enough for proof, but::
if x=y, we get
f(x +2f(x)) = f(x) + f(x) + x = x + 2f(x)
z=x+2f(x)
f(z) = z
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Icarus
wu::riddles Moderator
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Re: f(x +2f(y)) = f(x) + f(y) + y
« Reply #3 on: Dec 20th, 2004, 5:07pm » |
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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: f(x) = -x/2.
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BNC
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Re: f(x +2f(y)) = f(x) + f(y) + y
« Reply #4 on: Dec 20th, 2004, 10:33pm » |
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FWIW, it's easy to prove that if f(x)=0, it must happen only at x=0.
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Eigenray
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Re: f(x +2f(y)) = f(x) + f(y) + y
« Reply #5 on: Dec 21st, 2004, 12:28am » |
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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.
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| « Last Edit: Dec 21st, 2004, 12:42am by Eigenray » |
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