Re: question
- To: mathgroup at smc.vnet.net
- Subject: [mg78019] Re: [mg77935] question
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 21 Jun 2007 05:50:44 -0400 (EDT)
- References: <200706200928.FAA09665@smc.vnet.net>
On 20 Jun 2007, at 18:28, dimitris wrote:
> Hi.
>
> Say
>
> o=(16-x^2)^(1/2)-(4-x)^(1/2)*(4+x)^(1/2);
>
> In another CAS one must use assumptions
> in order to simplify
>
>> o:=sqrt(16-x^2)-sqrt(4-x)*sqrt(4+x);
>> combine(u) assuming x>-4,x<4 ;
>> simplify(%);
>
>
> 2 1/2 1/2 1/2
> o := (16 - x ) - (4 - x) (4 + x)
>
> 2 1/2 1/2
> (16 - x ) - ((4 - x) (4 + x))
>
>
> 0
>
>
> In Mathematica one simply gets
>
> In[10]:=
> Simplify[o]
>
> Out[10]=
> 0
>
>
> I would like to hear your comments on this issue.
>
>
> Dimitris
>
>
"Bad performance" by "the other CAS" ;-)
Note that Mathematica gives, generally:
Simplify[Sqrt[a - x]*Sqrt[a + x], a > 0]
Sqrt[a^2 - x^2]
It is not very hard to prove directly by hand but for a fixed value
of a, e.g. a=4 one can give quite convincing numerical verification
with Mathematica:
ls = RandomComplex[{-10 - 10 I, 10 + 10 I}, 10^5];
Max[Abs[Map[Function[{x}, Sqrt[16 - x^2] - Sqrt[4 - x] Sqrt[x + 4]],
ls]]] // Chop
0
Andrzej Kozlowski
- References:
- question
- From: dimitris <dimmechan@yahoo.com>
- question