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>