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Re: on 2^O(lg n) being a polynomial

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On 2017-03-28, Ben Bacarisse <ben.usenet@bsb.me.uk> wrote:
> Daniel Bastos <dbastos@toledo.com> writes:
>
>> Here's an argument to prove that 2^O(lg n) is a polynomial.
>
> You are tripping up over your terminology.  2^O(lg n) is a set of
> functions ans do can't be polynomial.  You can't even say that the
> member function are polynomials.  What you can say (if it's true) is
> that the functions in it are asymptotically bounded above by a
> polynomial.  In other words, you are trying to prove that
>
>   if f(n) in 2^O(lg n) the there exists k such that f(n) in O(n^k).

That seems to condense to:

   ∃k:2^O(lg n) ⊂  O(n^k)

(Benefits from additional assertion that k is an integer > 1).

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Thread

on 2^O(lg n) being a polynomial Daniel Bastos <dbastos@toledo.com> - 2017-03-24 22:25 -0300
  Re: on 2^O(lg n) being a polynomial Kaz Kylheku <336-986-7787@kylheku.com> - 2017-03-25 02:39 +0000
  Re: on 2^O(lg n) being a polynomial Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-28 14:25 +0100
    Re: on 2^O(lg n) being a polynomial Kaz Kylheku <336-986-7787@kylheku.com> - 2017-03-28 17:45 +0000
    Re: on 2^O(lg n) being a polynomial Daniel Bastos <dbastos@toledo.com> - 2017-03-28 20:28 -0300
      Re: on 2^O(lg n) being a polynomial Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-29 14:12 +0100

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