John - this converges much more quickly.
While awaiting your reply, I used Cliff's suggestion (which seems to
disagree with Leo's but I better understand it) which can be applied to work
with the series you suggest:
10^.-/%/"1(1 3,:1 1)&(+/ .*)^:(20+i.2)]1 0x NB. Fewer than 11 digits
_10.79625847621708
13j11":%/(10x^11)*(1 3,:1 1)&(+/ .*)^:20]1 0x NB. Last digit incorrect,
as expected
1.73205080758
9!:11]16
%:3
1.732050807568877
NB. Generalizing this
]pcsn=. 10^.-/%/"1 sqrts3=. (1 3,:1 1)&(+/ .*)^:(100+i.2)]1 0x
_56.55206227888699
]numdig=. <.-pcsn NB. Safe enough to round precision down?
56
((2+numdig)j. numdig)":%/1{sqrts3 NB. "1{" to be extra sure...
1.73205080756887729352744634150587236694280525381038062806
#'1.732050807568877293527446341' NB. From
http://www.research.att.com/~njas/sequences/A040001
29
So, it looks like "format" is smart enough to preserve precision according
to its left argument when given a rational fraction as its right argument?
That's helpful.
Using the method John sent:
]pcsn=. 10^.-/%/"1 sqrts3=. f^:(7 8) 1 1x
_72.66969546124619
]numdig=. <.-pcsn
72
((2+numdig)j. numdig)":%/1{sqrts3
1.732050807568877293527446341505872366942805253810380628055806979451933017
NB. "1{" is gilding the lily:
((2+numdig)j. numdig)":%/0{sqrts3
1.732050807568877293527446341505872366942805253810380628055806979451933017
((2+numdig)j. numdig)":%/0{sqrts3
1.7320508075688772935274463415058723669428052538103806280558069794519330171227462181522446767406659571
((2+numdig)j. numdig)":%/1{sqrts3
1.7320508075688772935274463415058723669428052538103806280558069794519330169088000370811461867572485757
On Fri, Mar 5, 2010 at 12:28 PM, John Randall <[email protected]
> wrote:
> Devon McCormick wrote:
> > Members of the Forum -
> >
> > If I'm approximating, e.g. the square root of 3, with a matrix method
> > which
> > returns an extended precision numerator and denominator, when I work out
> > the
> > decimal equivalent of this, at what point do I run out of significant
> > digits?
> >
>
> Devon:
>
> You can do much better with these Pell-type estimates, since you can
> generate estimates 2,4,8,16,... directly: If your last estimate was p/q,
> your next estimate is ((*:p)+3**:q)%2*p*q.
>
>
> f=:((1 3)&(+/ .*)@:*:),+:@(*/)
>
> f^:(i.5) 1 1x
> 1 1
> 4 2
> 28 16
> 1552 896
> 4817152 2781184
>
> (%:3)-%/f^:3 ] 1 1x
> _9.20496e_5
> (%:3)-%/f^:4 ] 1 1x
> _2.44585e_9
> (%:3)-%/f^:5 ] 1 1x
> 0
>
> Best wishes,
>
> John
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
