comp.lang.idl-pvwave archive - RDF feed
type: channel
description: Messages from Usenet group comp.lang.idl-pvwave, compiled by Paulo Penteado
items: Inaccuracies
items: Re: Inaccuracies

Inaccuracies
type: item
description: Ok, I am sure this has been discussed before, but let me start this thread again. I wish to create a 15-element vector which contains the numbers -1.4 to 1.4 by an increment of 0.2 I also wish the sum of these elements to be zero (No, this isn't the new math). Here is what I tried... TRIAL #1 ======== IDL> a = findgen(15)*.2 - 1.4 IDL> print, total(a) 7.15256e-07 Hmmm! Not so good. TRIAL #2 ======== IDL> a = dindgen(15)*(.2D)-1.4D IDL> print, total(a, /double) 4.4408921e-15 Ok, this is better but not correct. And what are the values of a? IDL> print, a -1.4000000 -1.2000000 -1.0000000 -0.80000000 -0.60000000 -0.40000000 -0.20000000 2.2204460e-16 0.20000000 0.40000000 0.60000000 0.80000000 1.0000000 1.2000000 1.4000000 I seem to have lost a zero somewhere, and for me this matters!!! TRIAL #3 ======== What if I only needed 13 numbers between -1.2 and 1.2. IDL> a = findgen(13)*.2 - 1.2 IDL> print, total(a) 0.00000 Now how can I get this to work for 15 numbers? Maybe I am missing something here, but this kind of behavior makes IDL a bit problematical for scientific use. With only 15 numbers and double precision arithmetic, I can't believe this would fail in FORTRAN or C! -- Andrew F. Loughe (afl@cdc.noaa.gov) University of Colorado, CIRES * Campus Box 449 * Boulder, CO 80309 phone: (303) 492-0707 fax: (303) 497-7013
creator: Andy Loughe
date: 1995-11-13T08:00:00-00:00

Re: Inaccuracies
type: item
description: In article <30A7BC4D.7018@cdc.noaa.gov>, Andy Loughe <afl@cdc.noaa.gov> wrote: > IDL> a = findgen(15)*.2 - 1.4 > IDL> print, total(a) > 7.15256e-07 ~7 significant figures is what you expect from single precision arithmetic. > IDL> a = dindgen(15)*(.2D)-1.4D > IDL> print, total(a, /double) > 4.4408921e-15 ~15 is what you expect from double precision. You can do this: IDL> i = LINDGEN(15)*2L - 14L IDL> print, i -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 IDL> c = DOUBLE(i)/DOUBLE(10) IDL> print, c -1.4000000 -1.2000000 -1.0000000 -0.80000000 -0.60000000 -0.40000000 -0.20000000 0.0000000 0.20000000 0.40000000 0.60000000 0.80000000 1.0000000 1.2000000 1.4000000 which gives you an exact 0 (division of 0 by anything should(!) be exactly 0), but the other terms are not necessarily exact, and you still get IDL> print, total(c) 2.2204460e-16 which is the best you can hope for. That's floating point arithemtic ... Regards, Ken Bowman -- Kenneth P. Bowman, Assoc. Prof. 409-862-4060 Department of Meteorology 409-862-4132 fax Texas A&M University bowman@csrp.tamu.edu College Station, TX 77843-3150
creator: bowman
date: 1995-11-13T08:00:00-00:00