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Topic: By Royal Decree: An Exact Calendar! (+ Estimating reals by 1/N's --
Egyptian style)

Replies: 5   Last Post: Jan 11, 2014 11:17 PM

 Messages: [ Previous | Next ]
 Rock Brentwood Posts: 116 Registered: 6/18/10
Estimating reals by 1/N's -- Egyptian style (& the "Brentwood
Calendar" reform)

Posted: Jan 11, 2014 2:34 PM

On Wednesday, January 8, 2014 7:44:36 PM UTC-6, federat...@netzero.com wrote:
> Gregorian Calendar... 365 + 1/4 - 1/100 + 1/400 - 1/3000 (or so)
> days per year.
> mean terrestrial year ... 1900 ... [365d 5h 48m 46s]
> going down about 1/2 second every century

> Taking advantage of this, I hereby issue the [Brentwood Calendar]
> [to be effective by 2200]:
> (1) Every 4th year shall continue to be a leap year as before
> (2) Every 32nd leap year, the extra day shall be removed.
> 365 + 1/4 - 1/128 days ... [365d 5h 48m 45s]

As mentioned before, these numbers don't come out of the blue, but by a process that converges at the same speed as Newton's Method and replicates the key point of the ancient Egyptian method for representing real numbers.

Every number can be estimated to the order e^{k 2^n} by an integer plus or minus the reciprocal of n integers.

Every number in the interval [0,1] can be accurately estimated as follows:
-- with 0 reciprocals to 1/2,
-- with 1 reciprocal to 1/12 = 1/2 (1/2 - 1/3),
-- with 2 reciprocals to 1/312 = 1/2 (1/12 - 1/13),
-- with 3 reciprocals to 1/195312 = 1/2 (1/312 - 1/313),
-- with 4 reciprocals to 1/76293945312 = 1/2 (1/195312 - 1/195313),
and so on.
[The 2nd number should be 312, I wrote 352 before out of haste].

For instance,
-- pi = 3 + 1/7 - 1/791 - 1/3748629 - 1/151648960887729 to 30 places,
-- 2^{1/2} = 1 + 1/2 - 1/12 - 1/408 - 1/470832 - 1/627013566048 to 25 places,
-- 3^{1/2} = 2 - 1/4 - 1/56 - 1/10864 - 1/408855276 to 19 places.
The "most remote" number between 0 and 1 is just:
Z def= 1 - 1/2 - 1/12 - 1/312 - 1/195312 - 1/76293945312 - ...
Z = 0.4134564184353240279837718402038750100819135...
I've never seen or heard of any reference to this number before.

The sequence
(a_0, a_1, a_2, a_3, ...) = (2, 12, 312, 195312, 76293945312, ...)
is just that given by a_n = (5^{2^n} - 1)/2. Thus
Z = 1 - 2/(5 - 1) - 2/(5^2 - 1) - 2/(5^4 - 1) - 2/(5^8 - 1) - ...

Date Subject Author
1/8/14 Rock Brentwood
1/8/14 fom
1/8/14 fom
1/9/14 David Bernier
1/11/14 Rock Brentwood
1/11/14 David Bernier