Parts of what Robert wrote don't use the specified domain and range. But he's right that the constant function f(n) == 2 satisfies all conditions. OK, let me add that f is *nonconstant*. I hope this makes the asymptotic expression unique. But I don't know that and haven't checked it out OK, forget the puzzle -- please treat it as a question. --Dan On 2013-01-30, at 7:45 PM, Robert Munafo wrote:
Well, since f(x) = {-1 if x is even; 1 if x is odd} works, I guess you could use
f(n) = -1 * cos(pi*n)
But that's needlessly complicated. The constant function f(n)=-1 works just fine for all n; or you could use f(n)=2 for all n.
Maybe you could explain what you mean by "asymptotic".
On 1/30/13, Dan Asimov <dasimov@earthlink.net> wrote:
While studying something apparently unrelated, I came upon a function
f: N_0 -> N_0
(where N_0 denotes {0,1,2,3,...})
with this curious property:
f(2n) + 2 = f(n)^2.
Puzzle: Find a closed-form asymptotic expression for f(n).
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
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