Re: Newbie Limit problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg53471] Re: [mg53439] Newbie Limit problem*From*: Andrzej Kozlowski <akozlowski at gmail.com>*Date*: Thu, 13 Jan 2005 03:12:16 -0500 (EST)*References*: <200501120841.DAA09647@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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\ > \((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/

**Follow-Ups**:**Re: Re: Newbie Limit problem***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**Newbie Limit problem***From:*Ken Tozier <kentozier@comcast.net>

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