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From: Archimedes Plutonium on 10 Sep 2009 01:30
Chip Eastham wrote:
> On Sep 9, 10:47 pm, "plutonium.archime...(a)gmail.com"
> <plutonium.archime...(a)gmail.com> wrote:
>
> > I seem to be rusty on permutations, but rusty on
> all computations.
> > I am more of a logician than a calculator. Anyone,
> I was wonder
> > where the Factorial takes over from the Exponent:
> >
> > 10^10 is greater than 10!
> >
> > but somewhere down that line the Factorial such as
> >
> > 94! is larger than 10^94
> >
> > So if Dik is reading, where is that boundary line?
>
> Interested readers may wish to consult the Wikipedia
> article on Stirling's approximation to the factorial
> (and its refinements):
>
> [Stirling's approximation -- Wikipedia]
> http://en.wikipedia.org/wiki/Stirling%27s_approximatio
> n
>
> A refinement due to Gosper is:
>
> n! ~ (n/e)^n * SQRT(2pi*n + 1/3)
>
> See these Mathworld pages for further details:
>
> [Stirling's Series -- MathWorld/A Wolfram Web
> Resource]
> http://mathworld.wolfram.com/StirlingsSeries.html
>
> [Stirling's Approximation -- MathWorld/A Wolfram Web
> Resource]
> http://mathworld.wolfram.com/StirlingsApproximation.ht
> ml
>
> In terms of the common logarithm (log) this says:
>
> log(n!) ~ n*(log n - log e) + 0,5*log(2pi*n + 1/3)
>
> For what n will n! roughly equal 10^n ? That is:
>
> n ~ n*(log n - log e) + 0.5*log(2pi*n + 1/3)
>
> A crude estimate is that log n - log e ~ 1 when n is
> a little more than 27. Indeed log(27!) is just
> slightly more than 28, so 27! exceeds 10^28.
>
> Checking back from there:
>
> 26! ~ 10^26.6 > 10^26
>
> 25! ~ 10^25.2 > 10^25
>
> 24! ~ 10^23.8 < 10^24
>
> So the crossover where 10^n and n! are equal lies
> between n = 24 and 25.
>
> > Also, what are these factorials approx equal to in
> terms of exponents?
> >
> > 94! = approx 10^?
> >
> > 190! = approx 10^?
> >
> > 231! = approx 10^?
> >
> > And what is 10^500 equal approx to in factorial ??
> >
> > So, anyone feel free to jump in, if Dik is still
> angry sulking away.
>
> A decent scientific calculator can find this:
>
> log(94!) ~ 146.04
>
> and by using Gosper's refinement of Stirling's
> approximation,
> I got these:
>
> log(190!) ~ 351.986
>
> log(231!) ~ 447.253
>
> Conversely 10^500 is approximately equal to 253!
> times 2.73.
>
>
> regards, chip
>
>

Thank you very much for your high quality post that
answered all my questions.

Now I need to try to convince the physics community
that they have one important, and the most important
Planck Unit missing. The total number of Coulomb Interactions within a
single atom. So the Element of
109 goes up to 266 nucleons and the Element 99 goes
up to 254 nucleons, so my original guess of about 10^500 is not that
far off the mark as to the highest number that has Physics meaning.

So if the Universe has or had one atom of Element
109, means that there was a Physics need of the
number 266! and that is much larger than 10^500,
but my early guess-estimate was not far off.

Now I have to convince the Physics community that
they need a Planck Unit of the total Coulomb Interactions at any
instant of time reaches as far as
266!

Thanks again Chip for all that information.

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
From: David Bernier on 10 Sep 2009 03:04
Archimedes Plutonium wrote:
> Chip Eastham wrote:
>> On Sep 9, 10:47 pm, "plutonium.archime...(a)gmail.com"
>> <plutonium.archime...(a)gmail.com> wrote:
>>
>>> I seem to be rusty on permutations, but rusty on
>> all computations.
>>> I am more of a logician than a calculator. Anyone,
>> I was wonder
>>> where the Factorial takes over from the Exponent:
>>>
>>> 10^10 is greater than 10!
>>>
>>> but somewhere down that line the Factorial such as
>>>
>>> 94! is larger than 10^94
>>>
>>> So if Dik is reading, where is that boundary line?
>> Interested readers may wish to consult the Wikipedia
>> article on Stirling's approximation to the factorial
>> (and its refinements):
>>
>> [Stirling's approximation -- Wikipedia]
>> http://en.wikipedia.org/wiki/Stirling%27s_approximatio
>> n
>>
>> A refinement due to Gosper is:
>>
>> n! ~ (n/e)^n * SQRT(2pi*n + 1/3)
>>
>> See these Mathworld pages for further details:
>>
>> [Stirling's Series -- MathWorld/A Wolfram Web
>> Resource]
>> http://mathworld.wolfram.com/StirlingsSeries.html
>>
>> [Stirling's Approximation -- MathWorld/A Wolfram Web
>> Resource]
>> http://mathworld.wolfram.com/StirlingsApproximation.ht
>> ml
>>
>> In terms of the common logarithm (log) this says:
>>
>> log(n!) ~ n*(log n - log e) + 0,5*log(2pi*n + 1/3)
>>
>> For what n will n! roughly equal 10^n ? That is:
>>
>> n ~ n*(log n - log e) + 0.5*log(2pi*n + 1/3)
>>
>> A crude estimate is that log n - log e ~ 1 when n is
>> a little more than 27. Indeed log(27!) is just
>> slightly more than 28, so 27! exceeds 10^28.
>>
>> Checking back from there:
>>
>> 26! ~ 10^26.6 > 10^26
>>
>> 25! ~ 10^25.2 > 10^25
>>
>> 24! ~ 10^23.8 < 10^24
>>
>> So the crossover where 10^n and n! are equal lies
>> between n = 24 and 25.
>>
>>> Also, what are these factorials approx equal to in
>> terms of exponents?
>>> 94! = approx 10^?
>>>
>>> 190! = approx 10^?
>>>
>>> 231! = approx 10^?
>>>
>>> And what is 10^500 equal approx to in factorial ??
>>>
>>> So, anyone feel free to jump in, if Dik is still
>> angry sulking away.
>>
>> A decent scientific calculator can find this:
>>
>> log(94!) ~ 146.04
>>
>> and by using Gosper's refinement of Stirling's
>> approximation,
>> I got these:
>>
>> log(190!) ~ 351.986
>>
>> log(231!) ~ 447.253
>>
>> Conversely 10^500 is approximately equal to 253!
>> times 2.73.
>>
>>
>> regards, chip
>>
>>
>
> Thank you very much for your high quality post that
> answered all my questions.
>
> Now I need to try to convince the physics community
> that they have one important, and the most important
> Planck Unit missing. The total number of Coulomb Interactions within a
> single atom. So the Element of
> 109 goes up to 266 nucleons and the Element 99 goes
> up to 254 nucleons, so my original guess of about 10^500 is not that
> far off the mark as to the highest number that has Physics meaning.
>
> So if the Universe has or had one atom of Element
> 109, means that there was a Physics need of the
> number 266! and that is much larger than 10^500,
> but my early guess-estimate was not far off.
>
> Now I have to convince the Physics community that
> they need a Planck Unit of the total Coulomb Interactions at any
> instant of time reaches as far as
> 266!

Many of the nucleons in an atom of element 109 are neutral neutrons,
but let's assume 109 protons and 157 neutrons for a total
of 266 nucleons in a nucleus of element 109.

A Coulomb interaction occurs between pairs of particles.
So with 109 positive charge nucleons and 157 neutral
nucleons, the first member can be chosen in 266 ways
and the second in 265 ways. Then,
(nucleon X, nucleon Y) will be counted as well as
(nucleon Y, nucleon X) : a double count is the result.

The number of unordered pairs of distinct nucleons is
then 266x265/2 = 35,245.

This is the kind of formula that can be verified for small
numbers of nucleons by manually drawing of graph with
a line between points showing one pairing.

With 4 nucleons: vertices of a square (4 in number),
2 horizontal line, 2 vertical lines, 2 diagonals;
total: 6 lines. Check: 4x3/2 = 12/2 = 6 (check).

I don't follow why you want to use the factorial
function to count things here ...

David Bernier
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