Re: Same Limit in v5.2 and v6.0 - conclusion, I hope...

*To*: mathgroup at smc.vnet.net*Subject*: [mg78070] Re: Same Limit in v5.2 and v6.0 - conclusion, I hope...*From*: David Bailey <dave at Remove_Thisdbailey.co.uk>*Date*: Fri, 22 Jun 2007 06:40:21 -0400 (EDT)*References*: <f5di2q$quv$1@smc.vnet.net>

jrc wrote: > This may be of use to others, so I furnish the conclusion, > prompted by yet another offline communication. Here is the > code that works in v6.0: (I have been advised to post 'actual > mathematica code' - so some lines may have to be text...) > > In[1]= $Assumptions = a > 0 && k1 > 0 && k2 > 0 && Element[x, Reals] > > Out[1]= a > 0 && k1 > 0 && k2 > 0 && x e(in) Reals > > In[2]= i1b = (1/a)*Integrate[Exp[-i*k2*x]*Exp[i*k1*x],x] > > 2*Sin[(1/2)*a*(k1-k2)] > Out[2]= ------------------------ > a*(k1-k2) > > In[3]= Limit[i1b, a -> Infinity] > > Out[3]= 0 > ---------------- > This is the obviously correct result, since it is the same > limit as sin(x)/x at infinity. > > However, the use of $Assumptions is in fact different in v6.0, > and I can find no instance in the 'new' 6.0 documentation that > shows this, including the little 'new in 6' links. > > In v5.2, the assumptions statement, shown in a number of > books, > > $Assumptions = {a > 0, k1 > 0, k2 > 0, x e(in) Reals} > > works perfectly fine. In fact, the use of a list enclosed in > brackets is shown in one of the examples in the 5.2 Help Browser > rather clearly under '$Assumptions'. But this form will not > work in 6.0 until three modifications are made: > > 1) The list should be joined with Boolean 'and': "&&" operators; > > 2) Apparently one must use the 'Element' function instead of the > funny 'e' symbol (not sure this is necessary); > > But finally, strangely enough and this *IS* a change from 5.2, > enclosing the list in brackets does not work in 6.0 - at least for > the 'Limit' function. > > Once again, I would like to see a clear statement of differences > in 6.0 for the oft-used functions. "$Assumptions" falls into this > category. > > jrc > I can't understand your point 2 because e<esc>elem<esc>Reals is the same expression as Element[e,Reals], as can be seen using FullForm! David Bailey http://www.dbaileyconsultancy.co.uk