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Re: Puzzled by IntegerPart
*To*: mathgroup at smc.vnet.net
*Subject*: [mg114641] Re: Puzzled by IntegerPart
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Sat, 11 Dec 2010 01:54:41 -0500 (EST)
On 12/10/10 at 2:29 AM, sebhofer at gmail.com (Sebastian) wrote:
>On Dec 9, 12:00 pm, Themis Matsoukas <tmatsou... at me.com> wrote:
>>IntegerPart[100*1.15]
>>114
>>???
>I was going to say: Mathematica seems to floor numeric quantities
>(which would kindof make sense to me), e.g. In[86]:=
>In[86]:= IntegerPart[100*1.13]
>IntegerPart[100*(1.13 + $MachineEpsilon)]
>Out[86]= 112
>Out[87]= 113
>In[88]:= Floor[100*1.13]
>Floor[100*(1.13 + $MachineEpsilon)]
>Out[88]= 112
>Out[89]= 113
>On the other hand...
>In[90]:= IntegerPart[100*1.12]
>IntegerPart[100*(1.12 + $MachineEpsilon)]
>
>Out[90]= 112
>Out[91]= 112
>In[92]:= Floor[100*1.12]
>Floor[100*(1.12 + $MachineEpsilon)]
>Out[92]= 112
>Out[93]= 112 ... so I'm also puzzled.
The puzzle is explained by looking at how these values are
represented and understanding what the functions act on.
That is
In[1]:= RealDigits[1.12] == RealDigits[1.12 + $MachineEpsilon]
Out[1]= True
So, any function working on the decimal digits clearly gets the
same result for both 1.12 and 1.12+$MachineEpsilon
But note,
In[2]:= RealDigits[1.12, 2] == RealDigits[1.12 +
$MachineEpsilon, 2]
Out[2]= False
That is adding $MachineEpsilon does change the binary
representation but does so at bits that not used in the decimal
representation for this particular value. However
In[3]:= RealDigits[1.13] == RealDigits[1.13 + $MachineEpsilon]
Out[3]= False
So, functions working on the decimal representation for 1.13
will yield a different result when $MachineEpsilon is added.
The behavior depends on the precise value of the nearest machine
number to whatever decimal input you make.
There are a several tutorials in the documentation that expound
on this further. If you want to research this further a starting
point would be:
tutorial/MachinePrecisionNumbers
ComputerArithmetic/tutorial/ComputerArithmetic
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