Re: Output Precision Exploration
- To: mathgroup at smc.vnet.net
- Subject: [mg118449] Re: Output Precision Exploration
- From: Joseph Gwinn <joegwinn at comcast.net>
- Date: Thu, 28 Apr 2011 06:36:54 -0400 (EDT)
- References: <ip8o8c$piq$1@smc.vnet.net>
In article <ip8o8c$piq$1 at smc.vnet.net>, Rafael Dunn <worthless.trash.junk at gmail.com> wrote: > Mathematica 8.0.1.0, Mac OSX x86 > > In:= > Log[173.5/173.5] > > Out:= > -1.11022*10^-16 > > I expect an output of exactly 0. Although 10^-16 is small, it turned > out to be the largest factor in a chemical equation I was attempting to > compute. > > I discovered this is because Mathematica does not actually evaluate > 173.5/173.5 = 1. The output is actually some number 0.9999999999... > > However, for most decimal constants x/x produces an exact output of 1. > By entering a few decimals off the top of my head I also found 1733.5, > 26.44, and 27.44 do not produce an output of 1 when divided by > themselves. > > Why? I understand Mathematica's algorithms for working with decimals > must make approximations, but why is there so much variance among > decimal calculations? 173.49/173.49 = 1, while 173.5/173.5 != 1. > Furthermore, I find: > x=173.49999999999999 > x/x = 173.5/173.5, with infinite precision. If you add or remove a > single 9 to the end of x, this ceases to be true. > > Furthermore, this looks like a contradiction to me: > > In:= > 173.5/173.5 = 1 > Log[1] = 0 > Log[173.5/173.5] = 0 > > Out:= > True > True > False > > I have learned a lot about Mathematica's precision and approximation through > the help documentation, but I still can not explain this or see how I can > expect Log[x/x] = 0 for the sake of calculations on the 10^-16 scale. The computations are performed using 64-bit double precision floating point numbers, as defined in IEEE Std 754. This is by definition a finite-precision computation, and errors of order 10^-16 are to be expected, and cannot be removed unless one goes to Mathematica's arbitrary precision arithmetic, which is orders of magnitude slower to compute. You may wish to reformulate your problem. With an explanation of what you are trying to solve and why, people will be able to suggest alternatives. Joe Gwinn