Re: simulating limited precision
- To: mathgroup at smc.vnet.net
- Subject: [mg27820] Re: simulating limited precision
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Mon, 19 Mar 2001 01:29:11 -0500 (EST)
- References: <98vic1$91p@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Murray, SetAttributes[sp, HoldAll] sp[x_]:= Unevaluated[x]/.r_Real:> ToExpression[ToString[r]<>"`3"] $Pre=sp; 1.00- 1.00*^3 \!\(\(-999.`2.9991\)\) Which displays as -1.0 10^3, so maybe it is not performing as you want. $Pre=. 1.00- 1.00*^3 -999. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Murray Eisenberg" <murray at math.umass.edu> wrote in message news:98vic1$91p at smc.vnet.net... > I would like to explain creation of roundoff-error in addition by > working with a precision of 3 decimal digits. The following is an > example: > > 1.00`3 - 1.00`3*^3 > > which gives the "incorrect" result of -1.0 10^3. > > What I'd REALLY like to do is the same thing but without specifying > explicitly the precision of each number -- by using the Standard AddOn > package NumericalMath`ComputerArithmetic`. > > I tried the following: > > <<NumericalMath`ComputerArithmetic` > > SetArithmetic[3]; > > 1.00 - 1.00*^3 > -999. > > Unfortunately, Mathematica here reverts to full machine precision. > > How do I use ComputerNumber to get the "wrong" answer -1.00 10^3 ? >