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Re: Re: error with Sum and Infinity
*To*: mathgroup at smc.vnet.net
*Subject*: [mg102457] Re: [mg102437] Re: error with Sum and Infinity
*From*: George Woodrow III <georgevw3 at mac.com>
*Date*: Sun, 9 Aug 2009 18:22:01 -0400 (EDT)
*References*: <h5bk64$hlm$1@smc.vnet.net> <200908070932.FAA15211@smc.vnet.net>
Also including a private response from Richard:
[me]
> You can get the performance you want by using Piecewise[]. That way,
> there is only one function definition.
>
[Richard]
Yes, one can often get the predetermined right answer by asking a
subtlely different question. Still, I think you agree the answer to
the original question was wrong...
[my comment]
The point to this is that yes, the original 'naive' attempt at a sum
gave the wrong answer. The wrong answer should have prompted a journey
to find out why.
The reason why the first attempt gave the wrong answer is that the
question was ill-posed for Mathematica to "always" do the right thing.
My solution (using Piecewise) and another poster's solution using
DiracDelta posed the question in a way that when Mathematica switched
from brute force (below SymbolicSumThreshold) to symbolic, it would do
the right thing.
You get this type of thing all the time. What would you want
Mathematica to do with a finite sum of {1, -1, 1, -1, ..... 1, -1}? It
should give 0 in this case. However, for an infinite sum, you'd want
it to fail, as it does.
(forgive the quick and dirty way to get a sequence of 1,-1.)
t[x_] := Cos[Pi x]
In[21]:= u = Table[t[a], {a, 0, 100}]
Out[21]= {1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1,
-1, 1, -1, 1, -1, \
1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1,
1, -1, \
1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1,
1, -1, \
1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1,
1, -1, \
1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1}
In[22]:= Sum[t[a], {a, 100}]
Out[22]= 0
In[18]:= Sum[t[a], {a, 10000000}]
Out[18]= 0
In[23]:= Sum[t[a], {a, 10000001}]
Out[23]= -1
In[24]:= Sum[t[a],{a,Infinity}]
During evaluation of In[24]:= Sum::div: Sum does not converge. >>
Out[24]= \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(a\), \(\[Infinity]\)]\(Cos[a\ \[Pi]]
\)\)
The point of this is to show that you need to know some maths, and at
least a little bit about the conventions of Mathematica. An infinite
sum cannot be treated the same as a finite sum, and care needs to be
taken to be sure that Mathematica knows what the sequence really is.
In the case of a sequence of all 0, except for a single 1, then the
function that defines the sequence needs to make the exception clear
in the definition.
Could Mathematica be designed to always give the right answer, not
matter what the input? Perhaps. Most of the time, it would be a waste
of time.
The proper lesson to learn is that if you are going to be evaluating
infinite sums, then it would be worth finding out how Mathematica does
this and what needs to be done to make the code fool-proof.
In this case, one could learn that Mathematica will sometimes ignore
exceptional cases (separate definitions of the function at special
values) when computing sums, and if you want these cases to be
included in an infinite sum, you need to define the function so that
these cases are handled symbolically.
You can be upset that mathematica does not do exactly what you expect
all the time, but it does no good to rant about it. The fact is, there
are well documented ways to deal with this problem so that mathematica
*does* do the right thing, not just in this toy example, but in more
complex problems. Every language has its way of interpreting code.
Mathematica is special in that there are a lot more ways to get the
code 'almost' right and usually a lot of different ways to get the
code exactly right. A similar coding error in c would simply give a
compiler error or crash.
As for your gratuitous comment about Andrzej's post, I think that
anyone familiar with any type of computer algebra system would be able
to guess, just in terms of practicality, that there is a threshold
from brute force to symbolic summation. Otherwise, Mathematica could
never compute an infinite sum. I have been using Mathematica for 20
years, and I did not know the name for the threshold, let alone its
value, but for my purposes, I did not need to know. However, I did
know (and it was in the documentation) that there is a point where one
type of computation ends and another begins.
The function SystemOptions[] is defined and documented in version 7
(It was new in version 6).
If you execute SystemOptions["*"], you get an impressive list of all
the options and their current values. Most appear to be self-
explanatory, but a few look like more explanation is needed. It also
appears that if your code depends on the values for these options, the
code may be fragile at a minimum.
Note that the original problem had a resolution that did not involve
special knowledge of these variables. I'd say that knowing that these
options exist belongs to advanced users of Mathematica. It might be
nice to have more complete documentation for some of these options --
and Tech Support has provided this information when I have needed it,
but this knowledge would not have solved the original question.
george
On Aug 9, 2009, at 6:06 AM, Richard Fateman wrote:
> Andrzej Kozlowski wrote:
>> On 8 Aug 2009, at 02:16, Richard Fateman wrote:
>>
>> <attempt at humor snipped>
>>
>> he or she can
>> always evaluate
>>
>> "SymbolicSumThreshold" /. SystemOptions[]
>> 1000000
>>
>> which, I think, deals with the rest of your post.
>>
>> Andrzej
>
> I assumed there was some name for that magic number, and now you
> tell me
> this is the name. Thanks.
>
> 1. In the Alphabetical List of functions and other objects in
> Mathematica, available as guide/AlphabeticalListing in the online
> manual
> for version 6.0, there is no appearance of SymbolicSumThreshold.
> Options[Sum] returns {} On the plus side, searching via ? *Sum*
> reveals the presence of Global`SymbolicSumThreshold but with no
> information on it.
>
> 2. I assume you are implicitly agreeing with me that it is
> unsatisfactory to require any user to guess at the existence, name,
> and
> effect of an undocumented global flag. But you can't bring yourself to
> say so.
>
>
> RJF
>
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