Re: domain for sum of geometric series
- To: mathgroup at smc.vnet.net
- Subject: [mg39120] Re: [mg39100] domain for sum of geometric series
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 30 Jan 2003 01:06:06 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
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
>
>
>