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Re: on reading ``o(1)''

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Daniel Bastos <dbastos@toledo.com> wrote:
> Suppose f in o(1).  By definition, there's a natural number n(0) such
> that for any c > 0
> 
>   0 <= f < c * 1 = c
> 
> for any n >= n(0).  (AFAIK, c is a real number.)  That seems to imply
> that f = 0 because that's the only thing that could be less than any 
> c > 0. 

I believe there are two problems.  The first is that you are writing
`o(1)', but the paper is using `O(1)'. The two are rather different,
and I think you want to be talking about O(1).

The second is that your definition seems to have misplaced a
quantifier. The definition I know is that:

    f:R -> R is in O(1) if there exists a real number c and a real
    number N such that, for all n > N, |f(n)| < |c * 1| = |c|.

So f could be quite different from the zero function.  The classic
example is to let f(x) = sin(x).  Then you could take N = 0, c =
1.1, or perhaps N = 999, c = 999, etc.  However, f could not be
something like f(x) = x, or even f(x) = log(log(log(x))).

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on reading ``o(1)'' Daniel Bastos <dbastos@toledo.com> - 2017-03-24 21:29 -0300
  Re: on reading ``o(1)'' Kaz Kylheku <336-986-7787@kylheku.com> - 2017-03-25 02:40 +0000
  Re: on reading ``o(1)'' Michael Forsythe <forsythe@example.com> - 2017-03-25 05:51 +0000
    Re: on reading ``o(1)'' Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-26 22:14 +0100
      Re: on reading ``o(1)'' Michael Forsythe <forsythe@example.com> - 2017-03-28 10:32 +0000
      Re: on reading ``o(1)'' Michael Forsythe <forsythe@example.com> - 2017-03-28 10:33 +0000
  Re: on reading ``o(1)'' Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-26 22:11 +0100
    Re: on reading ``o(1)'' Daniel Bastos <dbastos@toledo.com> - 2017-03-28 21:05 -0300
      Re: on reading ``o(1)'' Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-29 14:23 +0100
        Re: on reading ``o(1)'' Daniel Bastos <dbastos@toledo.com> - 2017-04-03 09:40 -0300

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