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From: Ted Ersek on 9 Apr 2010 03:35
What does Mathematica version 6 give when the following is evaluated?
Here (x) is 'zero' with 80 digits of accuracy.

x = SetAccuracy[0, 80];
{ Abs[x] < 10^-900, Positive[x], Negative[x], NonPositive[x],
NonNegative[x] }//InputForm


{Sign[x], SetPrecision[x, 20], Sign[0], KroneckerDelta[x] }


UnitStep[{0, 0.0, x}]


Unitize[{0, 0.0, x}]


----------------------
I have version 7, but not version 6. I am finishing an update to my
RootSearch package. Knowing what version 6 returns above will help me make
one version of this package that is optimized for Mathematica versions 6 and
7.


Thanks,
Ted Ersek


From: Albert Retey on 10 Apr 2010 06:54
Am 09.04.2010 09:35, schrieb Ted Ersek:
> x = SetAccuracy[0, 80];
> { Abs[x] < 10^-900, Positive[x], Negative[x], NonPositive[x],
> NonNegative[x] }//InputForm
>
>
> {Sign[x], SetPrecision[x, 20], Sign[0], KroneckerDelta[x] }
>
>
> UnitStep[{0, 0.0, x}]
>
>
> Unitize[{0, 0.0, x}]
>

This is on Windows XP with the 32bit version of Mathematica 6.0.2:

In[1]:= x = SetAccuracy[0, 80];
{Abs[x] < 10^-900, Positive[x], Negative[x], NonPositive[x],
NonNegative[x]} // InputForm

Out[1]//InputForm=
{False, Positive[0``80.], Negative[0``80.], NonPositive[0``80.],
NonNegative[0``80.]}

In[2]:= {Sign[x], SetPrecision[x, 20], Sign[0], KroneckerDelta[x]}

Out[2]= {0, 0, 0, 1}

In[3]:= UnitStep[{0, 0.0, x}]

Out[3]= {1, 1, 1}

In[4]:= Unitize[{0, 0.0, x}]

Out[4]= {0, 0, 0}

hth,

albert

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