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Re: 2 approximations and 1 exact formula for pi
Posted:
Jan 22, 2011 2:36 PM
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On Jan 22, 11:57 am, Jacob <a...@hsu.edu> wrote:
> On Fri, 21 Jan 2011 08:15:47 -0800 (PST), plouffe wrote:
>
> >Hello everybody, (this message needs FIXED font).
>
> > these days I am working on exponential sums and I
> >have found something of interest,
> >like
> > infinity
> > ----- 3
> > \ n
> > ) ---------------
> > / 2 Pi n
> > ----- exp(------) - 1
> > n = 1 13
>
> >= 119.00000000000000000000000000000009593745851025547335588584913...
>
> >the precision is 31 digits.
>
> > Another one is
> > infinity
> > ----- 3
> > \ n
> > ) ---------------
> > / 2 Pi n
> > ----- exp(------) - 1
> > n = 1 7
>
> > =
> >10.0000000000000001901617678886626755843593058554453334802548978434061099438\
> >The precision here is 15 digits.
>
> >For an argument of 2*Pi*n/163, the precision is 435 digits! I
> >don't know why I took 163... (of course).
>
> >in general, these sums will be near an integer if the argument
> >is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ?
> >These are the simplest I could find,
> >if the exponent is of the form 4m-1 then the sum is often an
> >integer with the same conditions. This, I believe extends
> >a little bit the known formula of Ramanujan/Berndt/etc.
>
> > A good question is : does someone has a simple explanation
> >of this ? I don't. I am preparing a paper on these results.
>
> >Because, I do have another one which is EXACT, namely for pi:
> >In general these sums with a fractional exponent are very
> >close to integers or pi, I have a couple of series for 1/pi, 1/pi^2,
> >etc.
>
> >here is the EXACT formula with fractional arguments mixed with
> >integer exponents.
>
> > /infinity \ /infinity \
> > | ----- | | ----- |
> > | \ 1 | | \ 1 |
> >10 | ) -----------------| - 40 | ) -------------------|
> > | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)|
> > | ----- | | ----- |
> > \ n = 1 / \ n = 1 /
>
> > /infinity \ /
> >infinity \
> > | ----- | |
> >----- |
> > | \ 1 | | \
> >1 |
> > + 10 | ) -------------------| - 10 | )
> >-----------------|
> > | / n (exp(4 Pi n) - 1)| | / / Pi
> >n \|
> > | ----- | | ----- n |exp(----) -
> >1||
> > \ n = 1 / \ n = 1 \
> >5 //
>
> > /infinity \ /
> >infinity \
> > | ----- | |
> >----- |
> > | \ 1 | | \
> >1 |
> > + 40 | ) -------------------| - 10 | )
> >-------------------|
> > | / / 2 Pi n \| | / / 4 Pi
> >n \|
> > | ----- n |exp(------) - 1|| | ----- n |exp(------)
> >- 1||
> > \ n = 1 \ 5 // \ n = 1 \
> >5 //
>
> >> evalf(%);
> >3.14159265358979323846264338327950288419716939937510582097494459230781640628\
> > 62089986280348253421170679821480865132823066470938442...
> >well, I verified up to 1000 digits and it holds.
>
> >This is trivial, ? I do not see how.
>
> >If someone has a piece of information on why this exist, I would
> >be glad to ear from it, references, known results, etc.
> >Me, I never saw these kind of formulas before,
>
> > have a good day,
> > Simon Plouffe
>
> Interesting that a relatively short expression yields a number close
> to an integer.
>
> But I think the title of the thread is misleading.
>
> Doesn't one need to know the exact value for Pi as input to the right
> hand side of the formula?
>
> If so, in what sense is Pi being approximated?
Yes, I think Simon did not mean to say that first two formulas have to
do with Pi being approximated - I think Simon presented those two as a
very impressive examples of obtaining "near integers" (thus hint to
Ramanujan via mentioning 163 ;-))
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