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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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