30 Jan
2013
30 Jan
'13
8:46 p.m.
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).
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