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Re: with machine-precision input
- To: mathgroup at smc.vnet.net
- Subject: [mg128403] Re: with machine-precision input
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Tue, 16 Oct 2012 03:22:56 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
- Delivered-to: mathgroup-newsend@smc.vnet.net
On 10/14/12 at 11:40 PM, fateman at cs.berkeley.edu (Richard Fateman)
wrote:
>try this on Mathematica 8:
>q = 429515858585961022071539/6922263581864661506963;
>Note that q was chosen so that
>Rationalize[N[q,45],0] == q.
Not here:
In[1]:= q = 429515858585961022071539/6922263581864661506963;
Rationalize[N[q, 45], 0] == q
Out[2]= False
but
In[4]:= Rationalize[N[q, 45]] == q
Out[4]= True
>Peculiarly, then SetPrecision[N[q,45],100]-N[q,100] is not zero but
>about -3.9E-59
Why would you expect this to be zero? Per the documentation:
SetPrecision will first expose any hidden extra digits in the
internal binary representation of a number, and, only after
these are exhausted, add trailing zeros.
So unless q can be exactly expressed with 45 digits there will
be a non-zero difference. Here:
In[4]:= SetPrecision[N[q, 45], 100] - N[q, 100]
Out[4]= 3.3475286706596422492*10^-78
but note:
In[5]:= SetPrecision[N[q, 45], 100] - N[q, 45]
Out[5]= 0.*10^-44
as it should.
And here is:
In[6]:= $Version
Out[6]= 8.0 for Mac OS X x86 (64-bit) (October 5, 2011)
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