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Re: Simplify 0/0 to 1?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg77622] Re: Simplify 0/0 to 1?
*From*: "Dana DeLouis" <dana.del at gmail.com>
*Date*: Thu, 14 Jun 2007 05:18:10 -0400 (EDT)
Interesting subject. I got interested in a related subject.
When the numerator has a machine precision number, I was under the
impression the whole equation is evaluated at Machine precision
(Documentation). This equation returns 1 in a spreadsheet like Excel
because it works at machine precision only.
However, this returns "ComplexInfinity." (Note the use of 1.0). The
denominator appears to be evaluated at full precision. (small arbit.
Number)/0
(1. - Cos[2]^2 - Sin[2]^2)/(1 - Cos[2]^2 - Sin[2]^2)
Power::infy:Infinite expression 1/0. encountered. >>
ComplexInfinity
However, this appears to return 0 since the numerator is exactly zero.
0 / (small arbit. Number)
(1 - Cos[2]^2 - Sin[2]^2)/(1. - Cos[2]^2 - Sin[2]^2)
0.
I was expecting a value of 1.0 if just one of the numbers was a machine
precision number no matter where it was located.
I realize that this would return 1 only if the equation is complex enough to
generate a very small value other than 0 at machine precision. Which
apparently it does..
InputForm[1. - Cos[2]^2 - Sin[2]^2]
-1.1102230246251565*^-16
Thanks.
- - -
Dana
(Mathematica 6.0, and Help files 5.2)
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