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From: Ludovicus on 2 Jun 2010 11:48
>
> >>>                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
>
>

>             6n
>     T(n) = ---- + O(sqrt(n))                                 [2]
>            pi^2
>
> simply by taking the absolute value of the summands in [1].
>
If your estimate of the error is O(sqrt(n))
how you explain the little errors in your table?
Ludovicus
From: Rob Johnson on 2 Jun 2010 13:05
In article <93c3ded9-a8d5-4cdc-a848-0c6611fb75f8(a)a16g2000vbr.googlegroups.com>,
Ludovicus <luiroto(a)yahoo.com> wrote:
>>
>> >>> 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
>>
>>
>
>> 6n
>> T(n) = ---- + O(sqrt(n)) [2]
>> pi^2
>>
>> simply by taking the absolute value of the summands in [1].
>>
>If your estimate of the error is O(sqrt(n))
>how you explain the little errors in your table?

Even if the differences were 0, the error would still be O(sqrt(n)),
would it not? I never claimed that O(sqrt(n)) is the best, just that
it follows easily from formula [1]:

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

Furthermore, the constant could be very small for O(sqrt(n)).

Since mu(d) is positive and negative, there is cancellation. For n
equal to 100000000, there is evidently quite a bit of cancellation.
However, the best that has been proven so far, even assuming the
Riemann Hypothesis, is O(x^(17/54+e)) for any e > 0. This gives an
exponent less than .314815, a bit better than the .5 gotten above.

Rob Johnson <rob(a)trash.whim.org>
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