MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Newbie Limit problem

  • To: mathgroup at
  • Subject: [mg53471] Re: [mg53439] Newbie Limit problem
  • From: Andrzej Kozlowski <akozlowski at>
  • Date: Thu, 13 Jan 2005 03:12:16 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

On 12 Jan 2005, at 09:41, Ken Tozier 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]]\)
> 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\ \
> 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.
> \!\(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?
> Thanks
> Ken

You are indeed doing a few things wrong but they are not responsible
for the lack of result. The first  very bad thing you are doing is
writing 0.5 for the power exponent 1/2. In Mathematica these two things

(0.5 and 1/2) are quite different and using the former in non-numerical

problems can cause all sorts of weird problems. Secondly, your
assumptions are obviously not very helpful. For a start, what is the
use of telling Mathematica that s is an integer if you also do not
include the assumption that k is also an integer? But in any case even

if you add that information it will make any difference; in fact
telling Mathematica that a parameter is an integer in most cases only
reduces the number of transformations it can perform and makes it less

rather than more likely it can solve the problem.

But actually, the main points seem to be:

1) Mathematica just can't do this for general d

2) While it is easy to prove that the sum is convergent (see below)  do

you have any special reason to expect that there is an explicit
"closed" formula for it? Such closed formulas are actually quite rare
so unless you are lucky neither Mathematica nor anyone else will find

However, there is quite a lot in this connection with this problem that

Mathematica can be helpful with. In fact we can get quite reasonable
bounds for the value of the limit. For convenience I shall assume that

  First of all, let's define a function of two variables:

p[s_,d_]:=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^2], {k, 0, s}]

Note that I am using InputForm, which makes it easy to paste
expressions into Mathematica in a readable way.

To start, observe that while Mathematica can't find the general limit
Limit[p[s,d],s->Infinity], it can find it for the special value d=0:

Limit[p[s, 0], s -> Infinity]


In general Mathematica can't find any closed form expression for the
limit Limit[p[s, d], s -> Infinity]. It is easy to see that the
difficulty is caused by the term d*(Cos[(2*k*Pi)/s] - Cos[(2*(1 +
k)*Pi)/s]). Fortunately it is rather easy to find bounds for this term:

First, d*(Cos[(2*k*Pi)/s] - Cos[(2*(1 + k)*Pi)/s]) can be re-written as

Simplify /@ TrigFactor[d*(Cos[(2*k*Pi)/s] - Cos[(2*(1 + k)*Pi)/s])]

2*d*Sin[Pi/s]*Sin[(Pi + 2*k*Pi)/s]

We see that this always between -2d Pi/s and 2d Pi/s . This means that

our limit will lie between

       s*(s*Cos[(2*Pi)/s]-2*Pi (d/s)))/s^2],{k,0,s}],s->Infinity],d>0]

d+2 =B9


       s*(s*Cos[(2*Pi)/s]+2*Pi (d/s)))/s^2],{k,0,s}],s->Infinity],d>0]

Abs[d-2 =B9]

Let's try checking this for some random values of  d, e.g. d=5 and d=

23. Let's use a large value of s, say


we obtain


The upper bound gives:

N[d+2 =B9]


while the lower bouund gives:

N[Abs[d-2 =B9]]


The average of the two is 6.2832, not that far off.

let's try




N[d+2 =B9]


N[Abs[d-2 =B9]]


This time the average is 23, which is quite good.

Andrzej Kozlowski
Chiba, Japan

  • Prev by Date: Re: Re: easy question about random numbers
  • Next by Date: Re: PlotLabel in two lines
  • Previous by thread: Re: Newbie Limit problem
  • Next by thread: Re: Re: Newbie Limit problem