Re: What we get from (0.0*x), (0.0^x) and similar stuff
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- Subject: [mg58757] Re: What we get from (0.0*x), (0.0^x) and similar stuff
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Sun, 17 Jul 2005 03:04:06 -0400 (EDT)
- References: <db2g2e$dh7$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
ted.ersek at tqci.net wrote:
> Many users will see nothing wrong with what we get from the following:
>
> In[1]:= Clear[x,y];
> {x^0.0, x^0, 0*x*y}
>
> Out[2]= {1., 1, 0}
In a sense, there is something wrong. These simplifications are what David
Stoutemyer called "misdemeanors" in his article "Crimes and Misdemeanors in
the Computer Algebra Trade", _Notices of the American Mathematical Society_
38:7 (1991) 778-785.
In case anyone doesn't see why simplifying, say, x^0 to 1 is a misdemeanor:
In Mathematica, 0^0 is regarded as being Indeterminate. Thus, without
knowing that x is nonzero, Mathematica should not bluntly simplify x^0 to
1. Rather, it would seem that, in Mathematica, x^0 should simplify to, say,
If[x == 0, Indeterminate, 1]. But such a result is normally considered to
be too cumbersome to be practical; the "misdemeanor" simplification is
considered preferable. All CASs known to me make such simplifications.
Perhaps it should be noted that many mathematicians (including me) consider
x^0 = 1 always (in particular, even when x is zero). For them, simplifying
x^0 to 1 is perfectly correct. [But I would still say that Mathematica's
simplifying x^0 to 1 is a misdemeanor because Mathematica considers 0^0 to
be Indeterminate.]
IMO, simplifying x^0.0 to 1.0 is a substantially worse misdemeanor. Since
the exponent is only close to 0, the result need not be close to 1 at all.
> But the kernel does nothing to simplify the next input. I suppose this
> is because the result is Indeterminate in the case of (0.0*Infinity), or
> (0^0). So then why didn't the kernel account for that with the previous
> example?
>
> In[3]:= {0.0*x*y,0^x,0.0^x}
>
> Out[3]= {0. x y, 0^x, 0.^x}
I'm glad that 0.0*x*y was not simplified to 0.0. Since one of the factors
is only close to 0, the product need not be close to 0 at all. (This is
analogous to the case x^0.0 considered above. Why Mathematica chooses to
simplify x^0.0 to 1.0, but not 0.0*x*y to 0.0, I don't know. That behavior
seems inconsistent to me.)
But 0^x and 0.0^x are very different from the previously considered cases.
For example, 0^x is 0 if x is a positive real, but ComplexInfinity if x is
a negative real. I would consider simplifying 0^x to 0 to be a "crime", not
a mere "misdemeanor", if x might potentially be negative.
> We can use the definitions below to ensure the input above returns {0. ,
> 1, 0.} and some will consider this a nice improvement.
You meant to say {0., 0, 0.}, but I would not consider that an improvement.
Indeed, I consider the latter two simplifications to be dangerous.
Regards,
David Cantrell
> In[4]:= Unprotect[Times,Power];
> Times[0.0,__?(#=!=Indeterminate&&Head[#]=!=DirectedInfinity&)]=0.0;
> Power[0,
> _?((!NumericQ[#]||Positive[Re[#]])&&Head[#]=!=DirectedInfinity&)]=0;
> Power[0.0,
> _?((!NumericQ[#]||Positive[Re[#]])&&Head[#]=!=DirectedInfinity&)]=0.0;
>
> In[9]:= {0.0*x*y, 0^x, 0.0^x}
>
> Out[9]= {0., 0, 0.}
>
> After making the above definitions we still get Indeterminate or
> ComplexInfinity whenever we should. See the next input as an example.
>
> In[10]:= {0*x*¥, 0*x*Indeterminate, 0^0, 0^Indeterminate, 0^(I-3)}
>
> Out[10]= {Indeterminate, Indeterminate, Indeterminate, Indeterminate,
> ComplexInfinity}
>
> However some users might want to always acount for the possibity that we
> might have (0*Infinity) or (0^0) and so prefer that the list below would
> return itself. However, I think it isn't possible to do that because
> since Mathematica Version 3 Times and Power use built-in definitions
> before user definitions. Am I wrong? Can anyone change the outcome
> below?
>
> In[11]:= Clear[Times,Power];
> {x^0.0,x^0,0*x*y}
>
> Out[11]= {1., 1, 0}
>
> --------
> Regards,
> Ted Ersek