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MathGroup Archive 2004

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Re: SetPrecision - What does in find?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48456] Re: SetPrecision - What does in find?
  • From: "Drago Ganic" <drago.ganic at in2.hr>
  • Date: Mon, 31 May 2004 00:13:32 -0400 (EDT)
  • References: <c8kcs0$mpv$1@smc.vnet.net> <c8mumk$a4t$1@smc.vnet.net> <c944pq$r43$1@smc.vnet.net> <c96i11$ijt$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Does that mean that this holds for every real number:

SetPrecision[x, Infinity] ==
FromDigits[RealDigits[x, 2, Round[MachinePrecision*Log[10]/Log[2]]], 2]

SetPrecision[x,Infinity] should be documented !

Greeting from Croatia,
Drago Ganic

P.S.:  Round[MachinePrecision*Log[10]/Log[2]] stands (curently) for 53.

"Peltio" <peltio at twilight.zone> wrote in message
news:c96i11$ijt$1 at smc.vnet.net...
> "Kazimir" wrote
>
> >> If you compute with infinite precision the sum of the (negative) powers
> >> of 2 corresponding to that expansion, you should get
> >>     1261007895663739 / 9007199254740992
>
> >However, I would not agree that one can say that it is an infinite
> >precision of 0.14  in usual sens.
>
> Right, in fact that result is the infinite precision version of the
> _truncated binary expansion_ of 0.14. Truncated at the 53rd bit of
mantissa.
> I was referrring to that with "that expansion".
> My sentence was not clear, sorry.
>
> cheers,
> Peltio
> Invalid address in reply-to. Crafty demunging required to mail me.
>



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