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Re: Re: simplifying inside sum, Mathematica 5.1
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53780] Re: [mg53749] Re: simplifying inside sum, Mathematica 5.1
*From*: Richard Fateman <fateman at cs.berkeley.edu>
*Date*: Thu, 27 Jan 2005 05:41:18 -0500 (EST)
*Organization*: University of California, Berkeley
*References*: <ct4h70$av2$1@smc.vnet.net> <ct56p1$eca$1@smc.vnet.net> <200501260936.EAA00194@smc.vnet.net> <opsk7uawdtiz9bcq@monster.ma.dl.cox.net>
*Sender*: owner-wri-mathgroup at wolfram.com
DrBob wrote:
> Richard,
>
> If the sum doesn't simplify because Mathematica is missing an "arguably
> correct" rule, that's unfortunate ("arguably"). But it doesn't reach the
> level of "bug" until it simplifies to something WRONG, and I don't mean
> "arguably" wrong.
>
> If a[1]=Infinity, (or the same for ANY i>0) it's hard to see why
> Sum[a[i]x^i,{i,0,Infinity}]/.x->0 should resolve to a[0] -- since one of
> the terms is Indeterminate.
Your same argument holds for the expression a[1]-a[1]. Mathematica should
not simplify that to 0 just in case a[1] is Indeterminate, or a[1]
is Interval[{-1,1}]. Nevertheless, Mathematica treats a[1]-a[1] by
simplifying it to zero. So your point, that Mathematica is being
smart, and is being cautious about simplifications just doesn't
hold water. Mathematica is not so cautious elsewhere.
>
> Allowing the possibility that unknown terms might be Infinity,
> ComplexInfinity, or Indeterminate can be inconvenient, and it may not be
> a good reason Sum shouldn't simplify in this case. It probably isn't the
> reason Mathematica doesn't simplify it, in fact.
I agree here. You are back on track.
>
> Still, if it did simplify, and if one of the a[i] WERE one of those
> values, the simplification would be wrong, and we'd rightly call it a
> bug. As things are, your "arguably correct" expectation is simply a rule
> Mathematica doesn't know, among infinitely many other rules no CAS
> happens to know and apply.
>
You are off track again.
> Anyway, a simple way to get the result you're seeking is the statement:
>
Now in the material below, you veer way off track. My initial question
is not "can you write a complicated program to produce a particular result?"
It is shorthand for "I think this result is wrong. Maybe Mathematica should
be able to produce the right result. Am I somehow using Mathematica
the wrong way? Is there something like "PrincipalValue->True" or "Assumptions->{i>=0}"
etc that will make this work? Or is it a bug? If so, can it be fixed?"
Your "solutions" below are presumably intended to be silly.
> a[0]
>
> or, if you prefer,
>
> First[Sum[a[i]*x^i, {i, 0, Infinity}]] /. i -> 0
>
> A more complete method is:
>
> firstTerm[s_Sum] :=
> Block[{i = s[[2,1]]}, If[s[[2,2]] > 0, 0, s[[1]] /. i -> 0]]
> firstTerm[Sum[a[i]*x^i, {i, 0, Infinity}]]
>
> a[0]
>
> firstTerm[Sum[a[i]*x^i, {i, 1, Infinity}]]
>
> 0
>
> This kind of control would be harder if Mathematica had already applied
> a rule that, in a given situation, turned out to be wrong. Also note
> that all three methods avoid the 0^0 quandary -- as of course we should,
> in mathematics or Mathematica.
As for avoiding 0^0, you might read about it on the internet. Making it 1
is often just the right thing to do.
See
http://db.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html
>
> Still, I agree that we often want the simple answer, without working for
> it.
Um, back on track??
>
> Bobby
>.....<snip>
As for Andrzej's comment, that this does the job...
Block[{Power,Infinity},
0^(i_) := KroneckerDelta[i, 0]; Sum[a[i]*x^i, {i, 0, Infinity}]/. x
-> 0]
Here are some comments:
1. There is no need for Infinity to be bound inside the Block.
2. It is neat that one can Unprotect and temporarily redefine Power
by a Block binding. I was not aware of this feature. I wonder if
it is new?
3. The simplification of the Sum form takes into account the
presence of KroneckerDelta, which is nice to know because
then it could also use a similar technique for 0^i within Sum.
4. Your solution gives the wrong answer for
Sum[a[i]*x^i, {i, -1, Infinity}]
It also doesn't work for
Sum[a[i]*x^(i^2), {i, -1, Infinity}]
This latter problem suggests an inadequacy in the treatment of the
simplification of Sum[KroneckerDelta[...]....]
RJF
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