Re: Changed order of multiplication in Times[a, b, c] in version

*To*: mathgroup at smc.vnet.net*Subject*: [mg77621] Re: Changed order of multiplication in Times[a, b, c] in version*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Thu, 14 Jun 2007 05:17:39 -0400 (EDT)*References*: <f4okvs$6sk$1@smc.vnet.net>

Hi, with machine precission numbers you will always get wrong aswers .. some aswers where wronger than others but wrong is wrong -- anyway. Use the exact arithmetic in Mathematica and you will get the right aswer. Regards Jens Andrew Moylan wrote: > I think the order in which the pairwise multiplication in Times[a, b, > c] has changed between version 5.2 and version 6! Can anyone comment > on this? > > Example: > > Here are three exact numbers whose product is Sqrt[-1]: > > Times[(1 + I)*10^2000, (1 + I)*1/2*10^-2000, 1] > > Here are those numbers in approximate form, labelled a, b, and c: > > {a, b, c} = > { > 1.`15.954589770191005*^2000*(1 + I), > 5.`15.954589770191005*^-2001*(1 + I), > 1. + 0.*I > }; > > Here are their various pairwise products: > > Inputting: > { > a*b, > b*c, > a*c > } > yields the output: > { > 0``15.653559774527023 + 1.`15.653559774527023*I, > 0. + 0.*I, > 1.`15.954589770191005*^2000 + 1.`15.954589770191005*^2000*I > } > > So far, all these results are identical between versions 5.2 and 6. > > Now consider a*b*c. Here is what is given version 5.2: > > 0``15.653559774526984 + 1.`15.653559774526984*I > > This happens to be the right answer, because it has done the > multiplication in the order (a*b)*c. > > Here is what is given for a*b*c in version 6: > > 0. + 0.*I > > This happens to be the wrong answer, because it has done the > multiplication in the order a*(b*c). > >