Another BBP formula for ln(2)
Alexander R.Povolotsky
ln(2) =
230166911/331873080-(1/35917)*(Sum((1/2)^k*(11/k+10/(k+1)+9/(k+2)+8/(k
+3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9)-9/(k+10)-10/(k+11)),k=
1 .. infinity))
Alexander R. Povolotsky
clicl...@freenet.de
"Alexander R.Povolotsky" schrieb:
Simplification on Derive 6.10 confirms that this reduces to ln(2). I
have no experience with BBP formulae, but I am inclined to prefer
something as simple as
sum((1/2)^k (1/k), k, 1, inf)
as an exact representation of ln(2).
Martin.
Axel Vogt
I think it is though for computing specific digits of ln(2)
without computing the whole value up to that position.
clicl...@freenet.de
Axel Vogt schrieb:
> clicl...@freenet.de wrote:
> > "Alexander R.Povolotsky" schrieb:
> >> FYI - I came up with the following exact BBP formula for ln(2)
> >>
> >> ln(2) =
> >> 230166911/331873080-(1/35917)*(Sum((1/2)^k*(11/k+10/(k+1)+9/(k+2)+8/(k
> >> +3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9)-9/(k+10)-10/(k+11)),k=
> >> 1 .. infinity))
> >>
> >
> > Simplification on Derive 6.10 confirms that this reduces to ln(2). I
> > have no experience with BBP formulae, but I am inclined to prefer
> > something as simple as
> >
> > sum((1/2)^k (1/k), k, 1, inf)
> >
> > as an exact representation of ln(2).
>
> without computing the whole value up to that position.
I see, the constants from 1 to 11 in the "BBP" formula then probably
simplify the computation of high-order digits of ln(2) in the decimal
system, although I don't immediately see how. Brute-force numerical
approximation with Derive gives
9019820183 9254629180 5310193462 2212786078 0179157540 ...
for the 100000th and subsequent digits of ln(2). How can the "BBP"
formula be used to do this more efficiently?
Martin.
Axel Vogt
Bailey, A Compendium of BBP-Type Formulas for Mathematical Constants
at http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf
Any C = Sum( p(k)/q(k) * 1/b^k, k = 0 ... infinity) allows to compute
directly the n-th digit in base b, if deg(p) < deg(q) and the rational
fraction is nonsingular through that series.
The posted identity is for base 2, I have not tried the algorithms.
BBP for Pi is for hexadecimal and can be used in usual double precision
Axel
ralph
> Axel Vogt schrieb:
>
> - Show quoted text -
It is a pretty formula, P(1,2,12,(11,10,9,8,7,6,0,-6,-7,-8,-9,-10)).
In Maple 12 I just do
Oevalf[100050](log(2));
0.6931471805599453094172321214581765680755001343602552541206800094933936219696
Snip … Bad idea to print 33 pages, took 18 seconds and 4.5Meg
35920220546948769685927379019820183925462918053101934622212786078017915754
The difference is you can check the 43112609000 BINARY digit of log
(2).
C
Cheers,
Ralph