- To: mathgroup at smc.vnet.net
- Subject: [mg57077] Re: ChineseRemainder
- From: "Dana DeLouis" <delouis at bellsouth.net>
- Date: Sat, 14 May 2005 04:58:19 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Thank you everyone. I see now that when the interval of list2 falls within
the interval of list1, it is possible to get a returned value 'R' such that
Mod[R, list2] will "not" match list1. I'm still doing some study on my own
with my "number theory" book in hand. Thanks again. :>)
"Dana DeLouis" <delouis at bellsouth.net> wrote in message
news:d5ski6$nvj$1 at smc.vnet.net...
> Hello. I have a question on the function "ChineseRemainder." Could
> offer an explanation on the following behavior? Thanks.
Here's the package...
> However, it does not correctly return list1. (According to help)
> This happens often in my program, and I'm having a tough time trusting the
> solution. Does anyone familiar with this have any insight? Thanks.
> I note that Mathematica's built-in ChineseRemainder function returns the
> same answer, so I'm sure Mathematica is doing it correctly. I guess I
> understand why Mod[r,list2] will often not return list1 (According to
> Reduce`RChineseRemainder[list1, list2]
> Thank you for any insight.
> "5.1 for Microsoft Windows (January 27, 2005)"
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