       Re: Same Limit: OK in 5.2, fails in 6.0; Packages gone

• To: mathgroup at smc.vnet.net
• Subject: [mg77999] Re: [mg77959] Same Limit: OK in 5.2, fails in 6.0; Packages gone
• From: Carl Woll <carlw at wolfram.com>
• Date: Thu, 21 Jun 2007 05:40:20 -0400 (EDT)
• References: <200706200941.FAA10343@smc.vnet.net>

```jrc wrote:

>Why?
>
>I have,
>
>\$Assumptions = {a > 0, k1 > 0, k2 >0, x e(in) Reals}
>(repeats ok)
>
>
It's better to put in real Mathematica code, preferably in InputForm,
rather than pseudo code. With the pseudo code, we have to figure out
what you mean, and then we have to type it in. This extra effort reduces
the number of people who are willing to investigate your issue.

There is a problem with your \$Assumptions definition, since \$Assumptions
should always be a Boolean expression. Use

\$Assumptions = a>0 && k1>0 && k2>0 && Element[x, Reals];

>i1b = (1/a)* Integral from -a/2 to +a/2 of integrand:
>
>              exp(- i k2 x) exp(i k1 x) dx
>
>with the expected result,
>
>i1b = fraction with numerator = 2 Sin[(1/2)a(k1-k2)]
>                 and denominator = a(k1-k2)
>
>Now I want the limit of this result, as the parameter a goes to infinity:
>
>Limit[i1b, a -> Infinity]
>
>Mathematica 5.2 gives correct result, zero;
>   (note this is just lim(sin(x)/x, x -> inf), which is obviously zero)
>
>Mathematica 6.0 is unable to evaluate the limit.
>
>
With \$Assumptions corrected as above, we have:

In:= res =  1/a Integrate[Exp[-I k2 x] Exp[I k1 x], {x, -a/2, a/2}]

Out= (2 sin(1/2 a (k1-k2)))/(a (k1-k2))

and

In:= Limit[res, a -> Infinity]

Out= 0

Carl Woll
Wolfram Research

>Ruskeepaa's "Navigator", 2nd ed, (written for v5.2) claims there
>is a package, "Calculus`Limit`" that makes 'Limit' work better
>(p. 395 and p. 430). However, no such package seems to exist in
>v6.0.
>
>How many packages no longer exist in 6.0 ???? Is there a list ????
>
>Can anyone give me a reason for this obvious failure?
>
>jrc
>
>

```

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