Re: some question
- To: mathgroup at smc.vnet.net
- Subject: [mg21005] Re: [mg20937] some question
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Thu, 2 Dec 1999 21:41:18 -0500 (EST)
- References: <199912010650.BAA07569@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
ahn wrote:
>
> Dear,
>
> I have some question about the mathematica commands.
> I know that \sqrt{3+2 \sqrt{3}} = 3^{1/4}*(1+\sqrt{3})/\sqrt{2}.
> Then how one can get this result from Mathematica? Which command I
> should use in order to
> get the right hand side in the above expression?
>
> Regards, Changhyun Ahn(ahn at kyungpook.ac.kr)
It's not a pretty sight, but here is something that works provided you
know in advance that the result can be expressed in terms of the square
root of two and the fourth root of three.
ee = Sqrt[3+2*Sqrt[3]];
algnum = RootReduce[ee];
poly = First[algnum][x];
fpoly = Factor[poly, Extension->{2^(1/2),3^(1/4)}];
rts = x /. Solve[fpoly==0, x];
rt = Select[rts, N[#]==N[algnum]&];
In[15]:= InputForm[eenew = Factor[rt]]
Out[15]//InputForm= {(3^(1/4)*(1 + Sqrt[3]))/Sqrt[2]}
Maybe someone has a simpler or more automatic method?
Daniel Lichtblau
Wolfram Research
- References:
- some question
- From: ahn <ahn@kyungpook.ac.kr>
- some question