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Re: Re: RE:Working Precision

  • To: mathgroup at
  • Subject: [mg24093] Re: [mg24030] Re: [mg23928] RE:Working Precision
  • From: Richard Fateman <fateman at>
  • Date: Tue, 27 Jun 2000 00:51:51 -0400 (EDT)
  • Organization: University of California, Berkeley
  • References: <8iuvto$>
  • Sender: owner-wri-mathgroup at

David Withoff wrote:
> > Here's an easily defended rule:  Do all arithmetic exactly. When
> > you must store the result into a finite sized memory location,
> > round it to the nearest representable number exactly representable
> > in that memory location. In case of a tie, round to even.
> The problem is that that rule by itself makes no allowance for
> keeping track of the accumulated effects of all that rounding.

That was merely the rule for finite-precision arithmetic. 

If you want to do blind "automatic" tracking of arithmetic, you can
try to do correct interval arithmetic.  This has mostly been debugged
in recent Mathematica versions (but see below). Better is to take a
more global view (see below.)

> > On a binary machine with a binary exponent, multiplication by 2 is
> > always exact.  Think about it:  it adds one to the exponent, leaving
> > the mantissa alone.  There is no error on such a machine operation.
> Actually, the absolute error is multipled by 2.  For example,
> multiplying 1 +/- 1 by 2 gives (or should give) 2 +/- 2.  It
> shouldn't matter what sort of bit-twiddling the machine is
> doing internally to arrive at this result.  The mathematics
> is the same regardless.
Is this meant to contradict the previous paragraph?  I'm not sure it
is at all related.  Here are examples (written in decimal) of binary
numbers that are exact.
2, 1, 0.5, 0.25, 0.125,  etc.  They are represented typically as though
they were
1 *2^1,  1*2^0, 1*2^-1, 1*2^-2, 1*2^-3   etc.

multiplying or for that matter dividing any of these by 2 provides the
exact result. A sequence of multiplying by 2 n times and then dividing
by 2 n times returns you to the same number.  (Assuming n is not so
as to cause overflow).  The absolute error is always zero.

If a person is truly concerned about absolute error in a calculation,
there are ways of doing this intelligently, instead of blindly. It does
matter how the computer is twiddling bits in error computations.  The
example is this:
  let x = be the interval [-1, 1].  Computing x*x  by  [-1,1]*[-1,1]
the interval from -1 to 1.   But computing it as [-1,1]^2 gives [0,1].
Error analysis that assumes related numbers are unrelated is unlikely to
be of much use.  Clearly x*x and x^2 should have the same errors. Not
so in Mathematica.

There are really much more compelling examples, where a bit of
will tell you that some function f[Interval[{low,high}]   should result
not in some clumsy local computation of error as is done in Mathematica,
but by computing the maximum of the function f on the interval
This can involve (say) computing the zeros of the derivative of f to
local max/min, as well as the endpoints.

By providing two models, one for "machine arithmetic" and another for
multiple-word arithmetic, Mathematica is inevitably harder to defend
that a system with just one model.  (My view is that the multiple-word
"precision" and "accuracy" model, even if it were the only model, would
still be difficult to defend...)

> Dave Withoff
> Wolfram Research

--Richard Fateman

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