Re: domain for sum of geometric series
- To: mathgroup at smc.vnet.net
- Subject: [mg39124] Re: [mg39100] domain for sum of geometric series
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 30 Jan 2003 01:06:26 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
On further reflection, the p in your original question may actually be
a non-real complex number e.g.
Sum[1/(1+I)^n,{n,1,Infinity}]
-I
So in fact the correct domain in your problem is just Abs[1-p]<1 This
seems like another good reason why why Mathematica does not try to
tell you the domain of convergence. If you want to know it you should
determine it separately, using any appropriate assumptions you wish to
make (e.g. p is real).
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/
On Wednesday, January 29, 2003, at 08:06 PM, Andrzej Kozlowski wrote:
> The rather obvious answer why it doesn't state the domain of
> convergence is that nobody programmed it to do so. I can only
> speculate as to why, but the most likely explanation that comes to my
> mind is that it was not considered worth the necessary programming
> effort to do this. For a start, to do so in general would be pretty
> difficult and time consuming. Just take a simple modification of your
> case:
>
>
> Sum[(1/(p^4 - 3*p^3 + p^2 - 1))^i, {i, 1, Infinity}]
>
>
> 1/(-2 + p^2 - 3*p^3 + p^4)
>
> if Mathematica wanted to tell you the domain of convergence it would
> have to solve the inequalities:
>
> << Algebra`InequalitySolve`
>
>
> InequalitySolve[p^4 - 3*p^3 + p^2 - 1 > 1, p]
>
>
> p < 1 - Sqrt[3] || p > 1 + Sqrt[3]
>
>
> InequalitySolve[p^4-3p^3+p^2-1<-1,p]
>
>
> 3/2 - Sqrt[5]/2 < p < 3/2 + Sqrt[5]/2
>
> The ability to solve such inequalities appeared only in version 4,
> while Sum is a much older function. But in any case, it is easy to
> modify this further so that InequalitySolve won't be able to help, > e.g.
>
>
> Sum[(1/(p*E^p - p^p*Sin[p] + p^3))^i, {i, 1, Infinity}]
>
>
> -(1/(1 - E^p*p - p^3 + p^p*Sin[p]))
>
> or something even more complicated.
>
>
> A good principle in designing mathematical software is that if
> something can't be done in sufficient generality that includes at
> least a substantial number of non-trivial cases and not just the ones
> where you know the answer anyway, then it's better not to do it at
> all. Besides, Mathematica is not meant to replace mathematical
> knowledge and skill, only to provide tools to make it easier to apply
> such knowledge.
>
> As for your second (related) point: Sum does not return answers such
> as Infinity or -Infinity, it considers such series as divergent. For
> example:
>
> In[35]:=
> Sum[n, {n, 1, Infinity}]
>
> From In[35]:=
> Sum::div:Sum does not converge.
>
> Out[35]=
> Sum[n, {n, 1, Infinity}]
>
>
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
> On Wednesday, January 29, 2003, at 05:38 PM, Frank Buss wrote:
>
>> If I enter Sum[p^i, {i, 0, Infinity}] Mathematica says, it is
>> 1/(1-p), but
>> doesn't say something about the domain for p: 1/(1-p) is only valid
>> for
>> -1<p<1. How can I display the domain and why Mathematica doesn't say
>> me the
>> other result, infinity for p>=1 and p<=-1?
>>
>> PS: you can find a nice animation for the geometric series at
>> http://www.matheprisma.de/Module/Craps/summe.htm
>>
>> --
>> Frank Buß, fb at frank-buss.de
>> http://www.frank-buss.de, http://www.it4-systems.de
>>
>>
>>
>
>
>