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Re: Tough Limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg34252] Re: [mg34235] Tough Limit
  • From: Vladimir Bondarenko <vvb at mail.strace.net>
  • Date: Sun, 12 May 2002 03:25:50 -0400 (EDT)
  • Reply-to: Vladimir Bondarenko <vvb at mail.strace.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Dear James,

You may wish to consider Mathematica as the greatest mathematical
encyclopaedia ever created. You have just run across a typo ;-)

I am afraid, the package NLimit has been designed incorrectly.
If you will have a close look at its code (NLimit.m) you see that
it does not use the fundamental concept of interval.

That is why the answer you get from NSolve for your limit is
a howler.

In fact, the answer to your limit is  Interval[{-1/Pi, 1/Pi}] ,
that is approximately Interval[{-0.318308, 0.318308}].

The simplest way to see it might be to look at this sequence

Table[Plot[Binomial[-1/2, n]*Sqrt[n*Pi], {n, 10^k, 2 10^k}], {k, 1, 4}]


By the way, absence of interval concept makes the difference
between the Calculus`Limit` and NumericalMath`NLimit` packages.
The former uses the notion of interval (but, alas, not always can
apply it successfully, as in the following example)

Limit[Sin[z], z -> Infinity]

   Interval[{-1, 1}]

<< Calculus`Limit`
Limit[Sin[z], z -> Infinity]

   Indeterminate


Regards,

Vladimir Bondarenko

.......................................................................................

rjmilazzo at aol.com (RJMilazzo) writes:


Can anyone suggest how I can use Mathematica to get the following limit:

limit ( Binomial [-1/2  ,n ] * Sqrt[ n* Pi ] ) as n-> Infinity

I have tried both Calculus`Limit` and the standard Limit functions. I can
verify with NLimit that this limit equals  approximately 1. I don't think that
this is rigorous enough for a proof.

Thanks
James
rjmilazzo at aol.com




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