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MathGroup Archive 2005

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Re: Newbie Limit problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53473] Re: Newbie Limit problem
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Thu, 13 Jan 2005 03:12:22 -0500 (EST)
  • Organization: The University of Western Australia
  • References: <cs2olb$9ni$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <cs2olb$9ni$1 at smc.vnet.net>,
 Ken Tozier <kentozier at comcast.net> wrote:

> I'm trying to get a limit for a sum that I know converges as s->infinity
> 
> \!\(Limit[\[Sum]\+\(k = 0\)\%s\((\(d\^2 + 2\ s\^2 - 2\ s\ \((s\ 
> Cos[\(2\ \
> \[Pi]\)\/s] + d\ \((Cos[\(2\ k\ \[Pi]\)\/s] - Cos[\(2\ \((1 + k)\)\ \
> \[Pi]\)\/s])\))\)\)\/s\^2)\)\^0.5`, s \[Rule] \[Infinity]]\)
> 
> but all I'm getting for a result is the exact expression I plug into 
> the Limit function.
> 
> \!\(Limit[\[Sum]\+\(k = 0\)\%s\((\(d\^2 + 2\ s\^2 - 2\ s\ \((s\ 
> Cos[\(2\ \
> \[Pi]\)\/s] + d\ \((Cos[\(2\ k\ \[Pi]\)\/s] - Cos[\(2\ \((1 + k)\)\ \
> \[Pi]\)\/s])\))\)\)\/s\^2)\)\^0.5`, s \[Rule] \[Infinity]]\)

And you are suprised by this? Some observations:

[1] x^0.5 is not the same as x^(1/2) (or Sqrt[x])

[2] The factor 1/s^2 in the denominator of the Sqrt can be factored out.

[3] What sort of functional form do you expect as s->infinity? If you 
compute the sum for fixed s, say s=3, you will see that form of the 
expression is quite complicated

 FullSimplify[With[{s = 3}, 
  Sum[Sqrt[ (d^2 + 2 s^2 -  2 s (s Cos[(2 Pi)/s] + 
    d (Cos[(2 k Pi)/s] - Cos[(2 (1 + k) Pi)/s])))]/s, 
  {k, 0, s}]], d > 0]

Try this for s = 7. Then try and guess the form as s -> Infinity.

> I read the documentation which describes this scenario like so: "Limit 
> returns unevaluated when it encounters functions about which it has no 
> specific information. Limit therefore by default makes no explicit 
> assumptions about symbolic functions."
> 
> Next I tried to explicitly give it some assumptions like so:
> 
> \!\(Assuming[{s \[Element] Integers, d \[Element] Reals},
>      Limit[\[Sum]\+\(k = 0\)\%s\((\(1\/s\^2\) \((d\^2 + 2\ s\^2 - 2\ s\ 
> \((s\ \
> Cos[\(2\ \[Pi]\)\/s] + d\ \((Cos[\(2\ k\ \[Pi]\)\/s] - Cos[\(2\ \((1 + 
> k)\)\ \
> \[Pi]\)\/s])\))\))\))\)\^0.5`, s \[Rule] \[Infinity]]]\)
> 
> which yields the same result.

Which is not suprising. These assumptions do not help in reducing an 
intractable sum.
> 
> \!\(Limit[\[Sum]\+\(k = 0\)\%s\((\(d\^2 + 2\ s\^2 - 2\ s\ \((s\ 
> Cos[\(2\ \
> \[Pi]\)\/s] + d\ \((Cos[\(2\ k\ \[Pi]\)\/s] - Cos[\(2\ \((1 + k)\)\ \
> \[Pi]\)\/s])\))\)\)\/s\^2)\)\^0.5`, s \[Rule] \[Infinity]]\)
> 
> Am I using "Limit" wrong? Or is there some other way to write the 
> expression to get Mathematica to give me the limit?

Why do you think Mathematica should be able to compute the limit?

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
School of Physics, M013                         Fax: +61 8 6488 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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