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I would like to know if it is possible that two functions $f(n), g(n)$ can exist such that both of the following conditions are met:

  1. $g(n) = o(1)$

  2. $f(n-g(n)) \neq \Theta (f(n))$

I though I found $1/n$ and $1/n^2$ but I think I got it all wrong.

Is it even possible?!

Raphael
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wannabe programmer
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1 Answers1

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Let $g(n) = 1/n$ and define $$ f(n) = \begin{cases} n & \text{if $n$ is an integer}, \\ \log \lceil n \rceil & \text{otherwise}. \end{cases} $$ For an integer $n>1$, $f(n) = n$ while $f(n-g(n)) = \log n$.

Yuval Filmus
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  • And again, the question arises: what about continuous functions? (As far as I can tell, the definition of continuity mandates that $f(n - g(n)) = f(n)$ holds in the limit for all $g$ that tend to zero. – Raphael Nov 13 '14 at 17:19
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    @Raphael Since it is enough to demand that $f(n) = n$ and $f(n-1/n) = \log n$ for all integers $n>1$, it is probably possible to construct a $C^\infty$ function $f$ going through all these points. – Yuval Filmus Nov 13 '14 at 18:41
  • I imagine the OP was assuming f was monotonically increasing. – Simd Nov 25 '14 at 21:57
  • @lembik I don't, since they considered the example $f(n)=1/n$. – Yuval Filmus Nov 25 '14 at 22:10
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    It looks to me that the OP meant $g(n) = o(1)$ and $f(n)$ monotonically increasing but I could be wrong. – Simd Nov 26 '14 at 06:19