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Re: Re: computing residues
Maxim wrote:
> On Fri, 4 Mar 2005 10:29:00 +0000 (UTC), Andrzej Kozlowski
> <akoz at mimuw.edu.pl> wrote:
>
>
>>I have to admit Mathematica is smarter than I had thought and in fact:
>>
>>
>>Residue[1/Sin[x],{x,Root[8*#1^3-6*#1-1&,3]-Cos[Pi/9]}]
>>
>>1
>>
>>I made a mistake by using Root[8*#1^3-6*#1-1&,1] instead of
>>Root[8*#1^3-6*#1-1&,3] in the first part of my example below. In fact
>>Residue deals with this case impressively well. This certainly seems to
>>reduce the strength of my argument, though I still would prefer to get
>>an unevaluated input in the non-numerical case.
>>
>>Andrzej
>>
>
>
> This is simply a case where Mathematica assumes sufficiently close values
> to be equal:
>
> In[1]:=
> Residue[Csc[x],
> {x, Root[8*#^3 - 6*# - 1&, 3] - Cos[Pi/9] + 10^-75}]
>
> Out[1]=
> 1
>
> which is incorrect. Series and Limit make the same 'error of the second
> kind'.
>
> Maxim Rytin
> m.r at inbox.ru
To clarify a bit, I will point out that all the above go through common
code for zero testing.
In[6]:= Developer`ZeroQ[Root[8*#^3 - 6*# - 1&, 3] - Cos[Pi/9] + 10^-75]
Out[6]= True
In our development kernel this can be influenced by a system option.
In[9]:= Developer`SetSystemOptions[ZeroTestNumericalPrecision->200];
In[10]:= Developer`ZeroQ[Root[8*#^3 - 6*# - 1&, 3] - Cos[Pi/9] + 10^-75]
Out[10]= False
Daniel Lichtblau
Wolfram Research
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