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From: Ludovicus on 31 May 2010 18:50
On May 30, 4:31 pm, master1729 <tommy1...(a)gmail.com> wrote:
> Let T(x) denote the number of squarefree integers between 1 and x.
>
> T(x) = 6 * pi^-2 * x + O( x^(1-ln(2)) )

> tommy1729

1 - ln(2) = .308
How can be the error O(x^.308)? This is too much.
Because for x = 1000000 T(x) = 607925
but 6*x/pi^2 = 607927. That is: Error = 2 = x^(.062)
The calcul shows that the exponent of x disminish
as x augment.
Ludovicus
From: master1729 on 1 Jun 2010 13:18
Rob Johnson wrote :

> In article
> <2126894270.251348.1275251507459.JavaMail.root(a)gallium
> .mathforum.org>,
> master1729 <tommy1729(a)gmail.com> wrote:
> >Let T(x) denote the number of squarefree integers
> between 1 and x.
> >
> >T(x) = 6 * pi^-2 * x + O( x^(1-ln(2)) )
> >
> >this is my teenage challange , tommy's jugendtraum
> if you like.
>
> The fact that T(x)/x ~ 6/pi^2 is easy, so your
> question is about the
> remainder: O(x^{-ln(2)}).
>
> Considering empirical data for small n, it looks as
> if the remainder
> very small:
>
> n
> --- 2 6n
> n > mu (k) ---- |diff|
> --- pi^2
> k=1
>
> 100 61 61 0
> 1000 608 608 0
> 10000 6083 6079 4
> 100000 60794 60793 1
> 1000000 607926 607927 1
> 10000000 6079291 6079271 20
> 100000000 60792694 60792710 16
>
> However,
> <http://en.wikipedia.org/wiki/Square-free_integer>
> says that The Riemann Hypothesis implies that the
> remainder is
> O(x^{-37/54+e}) for any e > 0. Unfortunately, -ln(2)
> < -37/54,
> so getting the remainder estimate you want may be
> difficult.
>
> Rob Johnson <rob(a)trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a
> monospaced font

huh ?

1 - ln(2) , not - ln(2)

and your link gives 17/54 and not -37/54.

where the **** did you get those numbers ?

especially the minus sign is weird !?


regards

tommy1729
From: master1729 on 1 Jun 2010 13:19
Luis A Rodriguez wrote :

> On May 30, 4:31 pm, master1729 <tommy1...(a)gmail.com>
> wrote:
> > Let T(x) denote the number of squarefree integers
> between 1 and x.
> >
> > T(x) = 6 * pi^-2 * x + O( x^(1-ln(2)) )
>
> > tommy1729
>
> 1 - ln(2) = .308
> How can be the error O(x^.308)? This is too much.
> Because for x = 1000000 T(x) = 607925
> but 6*x/pi^2 = 607927. That is: Error = 2 = x^(.062)
> The calcul shows that the exponent of x disminish
> as x augment.
> Ludovicus

fascinating ...
From: Rob Johnson on 1 Jun 2010 18:21
In article <1748062447.262303.1275427128672.JavaMail.root(a)gallium.mathforum.org>,
master1729 <tommy1729(a)gmail.com> wrote:
>Rob Johnson wrote :
>
>> In article
>> <2126894270.251348.1275251507459.JavaMail.root(a)gallium
>> .mathforum.org>,
>> master1729 <tommy1729(a)gmail.com> wrote:
>> >Let T(x) denote the number of squarefree integers
>> between 1 and x.
>> >
>> >T(x) = 6 * pi^-2 * x + O( x^(1-ln(2)) )
>> >
>> >this is my teenage challange , tommy's jugendtraum
>> if you like.
>>
>> The fact that T(x)/x ~ 6/pi^2 is easy, so your
>> question is about the
>> remainder: O(x^{-ln(2)}).
>>
>> Considering empirical data for small n, it looks as
>> if the remainder
>> very small:
>>
>> n
>> --- 2 6n
>> n > mu (k) ---- |diff|
>> --- pi^2
>> k=1
>>
>> 100 61 61 0
>> 1000 608 608 0
>> 10000 6083 6079 4
>> 100000 60794 60793 1
>> 1000000 607926 607927 1
>> 10000000 6079291 6079271 20
>> 100000000 60792694 60792710 16
>>
>> However,
>> <http://en.wikipedia.org/wiki/Square-free_integer>
>> says that The Riemann Hypothesis implies that the
>> remainder is
>> O(x^{-37/54+e}) for any e > 0. Unfortunately, -ln(2)
>> < -37/54,
>> so getting the remainder estimate you want may be
>> difficult.
>>
>> Rob Johnson <rob(a)trash.whim.org>
>> take out the trash before replying
>> to view any ASCII art, display article in a
>> monospaced font
>
>huh ?
>
>1 - ln(2) , not - ln(2)
>
>and your link gives 17/54 and not -37/54.
>
>where the **** did you get those numbers ?
>
>especially the minus sign is weird !?

I was talking about the mean density of square-free integers, T(x)/x.
As I mentioned above, this can easily be shown to tend to 6/pi^2.
The result on the web page talks about T(x), so I converted the
remainder for T(x)/x.

Since RH implies that T(x) ~ 6x/pi^2 + O(x^{17/54+e}), it also
implies that T(x)/x ~ 6/pi^2 + O(x^{-37/54+e}). Similarly, your
jugendtraum would require T(x)/x ~ 6/pi^2 + O(x^{-ln(2)}).

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Rob Johnson on 1 Jun 2010 20:34
In article <20100601.150656(a)whim.org>,
Rob Johnson <rob(a)trash.whim.org> wrote:
>In article <1748062447.262303.1275427128672.JavaMail.root(a)gallium.mathforum.org>,
>master1729 <tommy1729(a)gmail.com> wrote:
>>Rob Johnson wrote :
>>
>>> In article
>>> <2126894270.251348.1275251507459.JavaMail.root(a)gallium
>>> .mathforum.org>,
>>> master1729 <tommy1729(a)gmail.com> wrote:
>>> >Let T(x) denote the number of squarefree integers
>>> between 1 and x.
>>> >
>>> >T(x) = 6 * pi^-2 * x + O( x^(1-ln(2)) )
>>> >
>>> >this is my teenage challange , tommy's jugendtraum
>>> if you like.
>>>
>>> The fact that T(x)/x ~ 6/pi^2 is easy, so your
>>> question is about the
>>> remainder: O(x^{-ln(2)}).
>>>
>>> Considering empirical data for small n, it looks as
>>> if the remainder
>>> very small:
>>>
>>> n
>>> --- 2 6n
>>> n > mu (k) ---- |diff|
>>> --- pi^2
>>> k=1
>>>
>>> 100 61 61 0
>>> 1000 608 608 0
>>> 10000 6083 6079 4
>>> 100000 60794 60793 1
>>> 1000000 607926 607927 1
>>> 10000000 6079291 6079271 20
>>> 100000000 60792694 60792710 16
>>>
>>> However,
>>> <http://en.wikipedia.org/wiki/Square-free_integer>
>>> says that The Riemann Hypothesis implies that the
>>> remainder is
>>> O(x^{-37/54+e}) for any e > 0. Unfortunately, -ln(2)
>>> < -37/54,
>>> so getting the remainder estimate you want may be
>>> difficult.
>>>
>>> Rob Johnson <rob(a)trash.whim.org>
>>> take out the trash before replying
>>> to view any ASCII art, display article in a
>>> monospaced font
>>
>>huh ?
>>
>>1 - ln(2) , not - ln(2)
>>
>>and your link gives 17/54 and not -37/54.
>>
>>where the **** did you get those numbers ?
>>
>>especially the minus sign is weird !?
>
>I was talking about the mean density of square-free integers, T(x)/x.
>As I mentioned above, this can easily be shown to tend to 6/pi^2.
>The result on the web page talks about T(x), so I converted the
>remainder for T(x)/x.
>
>Since RH implies that T(x) ~ 6x/pi^2 + O(x^{17/54+e}), it also
>implies that T(x)/x ~ 6/pi^2 + O(x^{-37/54+e}). Similarly, your
>jugendtraum would require T(x)/x ~ 6/pi^2 + O(x^{-ln(2)}).

I imagine you may know this, but just in case you haven't come
across it, here is the exact formula for T(n):

oo
6n --- n
T(n) = ---- - > mu(d) { --- } [1]
pi^2 --- d^2
d=1

where {x} is the fractional part of x, and mu(x) is the Moebius mu
function. This formula immediately gives the asymptotic estimate

6n
T(n) = ---- + O(sqrt(n)) [2]
pi^2

simply by taking the absolute value of the summands in [1].

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
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