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Re: simplification

• To: mathgroup at smc.vnet.net
• Subject: [mg121922] Re: simplification
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Thu, 6 Oct 2011 04:26:16 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j6e693$kef$1@smc.vnet.net> <j6h3d2$738$1@smc.vnet.net>

"Peter Pein" <petsie at dordos.net> schrieb im Newsbeitrag 
news:j6h3d2$738$1 at smc.vnet.net...
> Am 04.10.2011 07:40, schrieb dimitris:
>> Hello.
>>
>> Let
>>
>> o1 = 1 + Sqrt[15 + 2*Sqrt[35] + 2*Sqrt[6*(6 + Sqrt[35])]];
>> o2 = 1 + Sqrt[3] + Sqrt[5] + Sqrt[7];
>>
>> o1 is equal to o2.
>>
>> o1 == o2 // FullSimplify
>> True
>>
>> The question is how to make Mathematica to simplify o1 to o2.
>>
>> Thanks
>> Dimitris
>>
>
> With a lot of luck:
>
> In[1]:= o1 = 1 + Sqrt[15 + 2*Sqrt[35] + 2*Sqrt[6*(6 + Sqrt[35])]];
>  ext = Block[{x, poly = RootReduce[o1][[1]]},
>   Sqrt[Cases[Union @@ Divisors[Abs[CoefficientList[poly[x], x]]],
>     1 | _?PrimeQ, 1]]]
>  o2 = ((Rest[#1] / First[#1]) . ext & )[
>        FindIntegerNullVector[Prepend[ext, -o1]]]
>
> Out[3]= {1, Sqrt[2], Sqrt[3], Sqrt[5], Sqrt[7], Sqrt[19], Sqrt[31]}
>
> Out[4]= 1 + Sqrt[3] + Sqrt[5] + Sqrt[7]
>
> :-)
>
We could watch a very interesting paper at the virtual conference on 
how well Mathematica is able to analyse expressions before working on 
them. Unfortunately, this seems to go with integrations and other 
things but not with these (somewhat academic?) examples.

Wolfgang