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