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MathGroup Archive 2007

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Re: Re: Simplification with Integers assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74996] Re: [mg74897] Re: Simplification with Integers assumption
  • From: Devendra Kapadia <dkapadia at wolfram.com>
  • Date: Fri, 13 Apr 2007 02:15:15 -0400 (EDT)
  • References: <eunqc2$7ic$1@smc.vnet.net><euqnn4$89o$1@smc.vnet.net>

On Tue, 10 Apr 2007, did wrote:

> On Apr 2, 12:59 pm, "David W.Cantrell" <DWCantr... at sigmaxi.net> wrote:
>> But your post did lead me to find something surprising. Suppose we fix n as
>> a positive integer -- say, 3 for example:
>>
>> In[6]:= Sum[3/((2p)!(3 - 2p)!), {p, 1, Infinity}]
>>
>> results in nothing but warning messages and gibberish. OTOH, if we replace
>> Infinity with a large integer -- say, 100 or 1000 -- Mathematica has no
>> trouble giving the correct answer.
>>
>> In[7]:= Sum[3/((2p)!(3 - 2p)!), {p, 1, 1000}]
>>
>> Out[7]= 3/2
>>
>> Of course, that answer is _also_ the correct one for the infinite sum at
>> which Mathematica balked. Why couldn't Mathematica do the infinite sum? Is
>> this problem fixed in the development version perhaps?
>
> Hi,
>
> Thanks for the reply.
> Actually it's not only infinite sums that have problems.
> Try, for example,
> Sum[3/((2p)!(3 - 2p)!), {p, 1, 10000000000000000000000}]
> and you'll get some nonsensical answers too.
>
> Did
>
>
>
Hello Didier,

In the example given by you, the terms in the series become
0 after p = 1, because we are dividing by (3-2p)! and this
is ComplexInfinity for p > 1.

The sum from 1 to 1000 mentioned in David Cantrell's message
is done without difficulty because it uses the procedural
approach but the symbolic algorithms in Sum (which evaluate
the remaining examples) have difficulty in handling the hidden
infinity. The incorrect behavior in these examples is 
fixed in the development version.

Thanks to you and David for these examples. I apologize for the
confusion caused by the wrong behavior.

Sincerely,

Devendra Kapadia,
Wolfram Research, Inc.



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