       Re: Re: Newbie Limit problem

```Unfortunately (as often happens) in my posting there was a certian
disharmony between the words and the deeds ;-) It was not all "input
form". In particular

> d+2 =B9

should have been d + 2*Pi
and

> Abs[d-2 =B9]

should have been:

Abs[b-2*Pi]

Andrzej

On 13 Jan 2005, at 09:12, Andrzej Kozlowski wrote:

> 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
>>
>> 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.
>>
>> \!\(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
> one.
>
> 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
>
> d>0.
>   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]
>
> 2*Pi
>
> 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
>
>
> Simplify[Limit[Sum[Sqrt[(d^2+2*s^2-2*
>        s*(s*Cos[(2*Pi)/s]-2*Pi (d/s)))/s^2],{k,0,s}],s->Infinity],d>0]
>
> d+2 =B9
>
> and
>
>
> Simplify[Limit[Sum[Sqrt[(d^2+2*s^2-2*
>        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
>
> s=10000;
> putting
> d=5;
>
> we obtain
> N[p[s,d]]
>
> 7.32662
>
> The upper bound gives:
>
> N[d+2 =B9]
>
> 11.2832
>
> while the lower bouund gives:
>
> N[Abs[d-2 =B9]]
>
>
> 1.28319
>
> The average of the two is 6.2832, not that far off.
>
> let's try
>
> d=23;
>
>
> N[p[s,d]]
>
>
> 23.4335
>
>
> N[d+2 =B9]
>
>
> 29.2832
>
>
> N[Abs[d-2 =B9]]
>
>
> 16.7168
>
> This time the average is 23, which is quite good.
>
>
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/~andrzej/
> http://www.mimuw.edu.pl/~akoz/
>

```

• Prev by Date: Re: Mathematica Graphics output in an ASP.NET WebApplication
• Next by Date: Problem with DSolve
• Previous by thread: Re: Newbie Limit problem
• Next by thread: Re: Newbie Limit problem