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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].
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