Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

How to find the machine number nearest to an exact number -- N[] fails

  • To: mathgroup at smc.vnet.net
  • Subject: [mg103869] How to find the machine number nearest to an exact number -- N[] fails
  • From: Alain Cochard <alain at geophysik.uni-muenchen.de>
  • Date: Sat, 10 Oct 2009 07:07:47 -0400 (EDT)

Hello.

  6.0 for Linux x86 (32-bit) (February 7, 2008)

I had always thought that N[exact number] gives the machine number
closest to that exact number.  But consider the following exact number

  ex=14539079527638557142952619724749887685325578333570\
     9865657255467349279813823794598008431302162071157\
    /                                                 \
     8000000000000000000000000000000000000000000000000\
    0000000000000000000000000000000000000000000000000; 

find its machine number representation:

  num=N[ex]; 

and determine the corresponding rounding error:  				      

  err=N[Abs[ex-  SetPrecision[num,Infinity]]  ] 

For me this gives 3.59393 10^-16, which is larger than Eps =
$MachineEpsilon = 2^-52 ~ 2.22 10^-16.  But, between 1 and 2, the
spacing between machine numbers is Eps.  So the rounding error should
be smaller than or equal to Eps/2 ~ 1.11*10^-16 if the rounding is to
the nearest machine number.

To investigate further, consider the binary expansions of ex:

  binex=RealDigits[ex,2,55]

which gives:

  Out[4]= {{1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 

  >     0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0,

  >     0, 1, 0, 1, 1, 0, 0, 1, 0}, 1}
                          ^ 
                          ^-53rd digit

Thus I would expect that the nearest binary number should end up
either like (rounding below):

  >     0, 1, 0, 1, 1, 0, 0}, 1}

or like (rounding above):

  >     0, 1, 0, 1, 1, 0, 1}, 1}

whereas the actual binary expansion is given by:

  binum=RealDigits[num,2]

and ends as

  >     0, 1, 0, 1, 0, 1, 1}, 1}

In order to check, let us artificially construct those 2 other
rounding numbers:

  binum2=binum;
  binum2[[1,51]]=1;
  binum2[[1,52]]=0;
  binum2[[1,53]]=0;

  binum3=binum2;
  binum3[[1,53]]=1;

and compute again the rounding errors:

  err2= N[ Abs[ ex-FromDigits[binum2,2] ]] 
  err3= N[ Abs[ ex-FromDigits[binum3,2] ]] 

This gives 1.37348 10^-16 for err2, which is this time less than Eps
and 8.46964 10^-17 for err3, which is less then Eps/2, as expected.

So, in this case, not only does N[] not give the nearest machine
number, but it "jumps" a machine number.

Is this a normal behavior?  Is there a flaw in my reasoning above? Is
there an easy way to automatically generate the nearest machine
number?

Many thanks in advance.
Alain

PS: all outputs are identical if num is evaluated as
SetPrecision[ex,MachinePrecision] instead of by N[ex].


  • Prev by Date: Re: generating submultisets with repeated elements
  • Next by Date: Re: Re: Re: generating submultisets with
  • Previous by thread: Re: Convert Graphics3D to Graphics2D. Is it possible?
  • Next by thread: Re: How to find the machine number nearest to an exact