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. >