Re: Is it a bug?
- To: mathgroup at smc.vnet.net
- Subject: [mg17640] Re: Is it a bug?
- From: "F. Mittermayr" <mitterma at linz.vai.co.at>
- Date: Fri, 21 May 1999 03:37:33 -0400
- Organization: IBM Global Services - Remote Access Mail & News Services
- References: <37393190.0@webpc1.vai.co.at>
- Sender: owner-wri-mathgroup at wolfram.com
Thanks to all who have told me that it is not a strange result.
But now look at the online-help under "Cases - Further Examples". You can
find the following lines:
This finds all elements in the list that are integers.
Cases[{.....}, _Integer]
And now we test it:
x = Sqrt[I]-1/Sqrt[2](1+I);
Cases[{x, x//FullSimplify},_Integer]
Out[]= {0}
It cannot be that a number becomes an integer by doing a "FullSimplify".
Shouldn't the text in the online-help be changed to the following:
This finds s o m e elements in the list that are integers and when you are
in doubt look at the FullForm!!
The statement "Cases[{...},_Integer]" is quite useless because often I want
to know if elements are integers but seldom I want to know if the FullForm
of an element is integer.
Why doesn't Mathematica do any calculation on the expression x =
Sqrt[I]-1/Sqrt[2](1+I); ??
If you have Cases[ { I I }, _Integer ] the expression is calculated to -1.
Is the expression to complicated ??
Regards
Franz Mittermayr
F. Mittermayr wrote in message <37393190.0 at webpc1.vai.co.at>...
>A = {1 + Sqrt[2] I , 1 + 2 * 3 I};
>Cases[A, _Complex]
>
>Out[1] = {1+6 I}
>
>Isn't it a strange result?
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