| Inaccuracies [message #5313] |
Mon, 13 November 1995 00:00 |
Andy Loughe
Messages: 174
Registered: November 1995
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Senior Member |
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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
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| Re: Inaccuracies [message #5314 is a reply to message #5313] |
Mon, 13 November 1995 00:00  |
bowman
Messages: 121
Registered: September 1991
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Senior Member |
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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
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