       Re: Approximate Zero Times A Symbol

• To: mathgroup at smc.vnet.net
• Subject: [mg127076] Re: Approximate Zero Times A Symbol
• From: "djmpark" <djmpark at comcast.net>
• Date: Thu, 28 Jun 2012 04:04:06 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201206270811.EAA18698@smc.vnet.net> <19783652.3140.1340794328091.JavaMail.root@m06>

```Thanks Christoph. That does sound like a reasonable argument, yet I would
like to dissent from it.

I don't see why a non-numeric symbol should be considered to have a
precision at all.  Don't we have fields of mathematics, such as algebra or
multilinear algebra, where we have undefined objects or elements to which we
apply various operations? These objects never have numerical values,
although they might have numerical weights. It doesn't seem proper to me to
treat them as numerical quantities so as to combine them with other
numerical factors. If x and y are such objects I can see some justification
for dropping terms such as 0 x, but not 0.0 x and certainly not
automatically converting it to 0.0. The symbol x has a meaning, as one of
the basic elements of the algebra, that the scalar 0.0 does not have. Using
0.0 x -> 0.0 changes the mathematical meaning of the expression.

For example, in Grassmann algebra one can have general Grassmann numbers
that are multi-graded and have a scalar term + vector terms + bivector terms
etc. The above transformation would incorrectly change a pure vector
expression into a multi-graded expression that has an approximate zero
scalar term (instead of an approximately zero x vector term).

David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/index.html

From: Christoph Lhotka [mailto:christoph.lhotka at fundp.ac.be]

Hello,

my argumentation would be as follows:

0 x = 0  if (0,x) have infinite precision, while 0. x = 0. since the
result has a precision at most of 0., therefore there is no need to keep the
x, which has infinite precision:
for the expression we only
need x up to precision of 0., therefore 0. * approximate x = 0., which is
true up to the precision of 0.

If I understand well, what you would like to have and if it only concerns
the format it is printed, I would define:

{8,8}]&/@{x,y})))

which would give

dis/@{0.x+1.y,0.34324324324324324324324x+1.000000000004y}
{0.00000000 x+1.00000000 y,0.34324324 x+1.00000000 y}

Hope that helps,

Christoph

On 06/27/2012 10:11 AM, djmpark wrote:
> What is the justification for the following?
>
>
>
> 0. x + 1. y
>
>
>
> 0. + 1. y
>
>
>
> I want to display a dynamic weighted sum of x and y and sometimes one
> of the coefficients becomes zero. I would like to keep both terms (for
> display) and format with NumberForm.  If Mathematica is going to drop
> the x, why doesn't it at least also drop the approximate zero?
>
>
>
> If I use SetPrecision we obtain:
>
>
>
> SetPrecision[0. x + 1. y, 10]
>
>
>
> 1.000000000 y
>
>
>
> which is at least more consistent, but not what I want either.
>
>
>
> David Park
>
> djmpark at comcast.net
>
> http://home.comcast.net/~djmpark/index.html
>

```

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