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Series Summation Question

Started byclashton <clashton@gmail.com>
First post2012-01-25 22:46 -0800
Last post2012-02-02 12:21 +0100
Articles 9 — 4 participants

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  Series Summation Question clashton <clashton@gmail.com> - 2012-01-25 22:46 -0800
    Re: Series Summation Question Axel Vogt <&noreply@axelvogt.de> - 2012-01-26 19:41 +0100
    Re: Series Summation Question Joe Riel <joer@san.rr.com> - 2012-01-26 08:21 -0800
      Re: Series Summation Question clashton <clashton@gmail.com> - 2012-01-26 16:45 -0800
        Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-01-27 07:31 +0100
          Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-01-27 07:55 +0100
            Re: Series Summation Question clashton <clashton@gmail.com> - 2012-01-28 07:59 -0800
            Re: Series Summation Question Axel Vogt <&noreply@axelvogt.de> - 2012-01-28 20:21 +0100
              Re: Series Summation Question Peter Pein <petsie@dordos.net> - 2012-02-02 12:21 +0100

#330 — Series Summation Question

Fromclashton <clashton@gmail.com>
Date2012-01-25 22:46 -0800
SubjectSeries Summation Question
Message-ID<1fe7f908-df47-4b54-949f-630a50b0551e@j14g2000vba.googlegroups.com>
I am a little curious about whether Maple can evaluate a certain
infinite sum that fails with Mathematica (I do not have Maple so
cannot test it myself).

Here is the sum in Mathematica:
Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
This leads to and error,
but
(Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q -> 0.1)
                                 +
(Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
gives the correct answer, which I find a little strange.

How does Maple perform on the original sum?

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#331

FromAxel Vogt <&noreply@axelvogt.de>
Date2012-01-26 19:41 +0100
Message-ID<9odmcsFmigU1@mid.individual.net>
In reply to#330
On 26.01.2012 07:46, clashton wrote:
> I am a little curious about whether Maple can evaluate a certain
> infinite sum that fails with Mathematica (I do not have Maple so
> cannot test it myself).
>
> Here is the sum in Mathematica:
> Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q ->  0.1}
> This leads to and error,
> but
> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q ->  0.1)
>                                   +
> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q ->  0.1})
> gives the correct answer, which I find a little strange.
>
> How does Maple perform on the original sum?

Using Maple 15:

   Sum(q^(-6+4*n)/(1-q^(-5+4*n)),n = 0 .. infinity);
   eval(%, q=0.1);
   evalf(%);
                              -21.10120010

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#332

FromJoe Riel <joer@san.rr.com>
Date2012-01-26 08:21 -0800
Message-ID<87vcnywi1u.fsf@san.rr.com>
In reply to#330
clashton <clashton@gmail.com> writes:

> I am a little curious about whether Maple can evaluate a certain
> infinite sum that fails with Mathematica (I do not have Maple so
> cannot test it myself).
>
> Here is the sum in Mathematica:
> Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
> This leads to and error,
> but
> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q -> 0.1)
>                                  +
> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
> gives the correct answer, which I find a little strange.
>
> How does Maple perform on the original sum?

Summing numerically (and guessing at the meaning of the Mathematica):

(**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
                                  infinity
                                   -----      (-6 + 4 n)
                                    \        q
                             S :=    )     ---------------
                                    /           (-5 + 4 n)
                                   -----   1 - q
                                   n = 0

(**) evalf(eval(S,q=1/10));
                                     -21.10120010



-- 
Joe Riel

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#333

Fromclashton <clashton@gmail.com>
Date2012-01-26 16:45 -0800
Message-ID<39d1ff7b-301a-48ee-99a6-3e4d0df84b7a@o13g2000vbf.googlegroups.com>
In reply to#332
On Jan 26, 11:21 am, Joe Riel <j...@san.rr.com> wrote:
> clashton <clash...@gmail.com> writes:
> > I am a little curious about whether Maple can evaluate a certain
> > infinite sum that fails with Mathematica (I do not have Maple so
> > cannot test it myself).
>
> > Here is the sum in Mathematica:
> > Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
> > This leads to and error,
> > but
> > (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q -> 0.1)
> >                                  +
> > (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
> > gives the correct answer, which I find a little strange.
>
> > How does Maple perform on the original sum?
>
> Summing numerically (and guessing at the meaning of the Mathematica):
>
> (**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
>                                   infinity
>                                    -----      (-6 + 4 n)
>                                     \        q
>                              S :=    )     ---------------
>                                     /           (-5 + 4 n)
>                                    -----   1 - q
>                                    n = 0
>
> (**) evalf(eval(S,q=1/10));
>                                      -21.10120010
>
> --
> Joe Riel

Maple wins this round.

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#334

FromPeter Pein <petsie@dordos.net>
Date2012-01-27 07:31 +0100
Message-ID<jftgbn$bt$1@online.de>
In reply to#333
Am 27.01.2012 01:45, schrieb clashton:
> On Jan 26, 11:21 am, Joe Riel<j...@san.rr.com>  wrote:
>> clashton<clash...@gmail.com>  writes:
>>> I am a little curious about whether Maple can evaluate a certain
>>> infinite sum that fails with Mathematica (I do not have Maple so
>>> cannot test it myself).
>>
>>> Here is the sum in Mathematica:
>>> Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q ->  0.1}
>>> This leads to and error,
>>> but
>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q ->  0.1)
>>>                                   +
>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q ->  0.1})
>>> gives the correct answer, which I find a little strange.
>>
>>> How does Maple perform on the original sum?
>>
>> Summing numerically (and guessing at the meaning of the Mathematica):
>>
>> (**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
>>                                    infinity
>>                                     -----      (-6 + 4 n)
>>                                      \        q
>>                               S :=    )     ---------------
>>                                      /           (-5 + 4 n)
>>                                     -----   1 - q
>>                                     n = 0
>>
>> (**) evalf(eval(S,q=1/10));
>>                                       -21.10120010
>>
>> --
>> Joe Riel
>
> Maple wins this round.

Well, if you are interested in the approximation of the sum's value for 
q=1/10, you should have used NSum in Mathematica:

NSum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)) /. q->1/10, {n, 0, Infinity}, 
WorkingPrecision->20]

-21.101200102001001100

or - using your workaround - get

6/(5*(-1 + q)) + (1 + 2*q + 3*q^2 + 4*q^3)/(5*(1 + q + q^2 + q^3 + q^4)) 
+ (-2*Log[q^4] + Log[1 - q^4] + QPolyGamma[0, 2 - Log[q^5]/Log[q^4], 
q^4])/(q*Log[q^4])

as (hopefully) exact value.

Cheers, Peter

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#335

FromPeter Pein <petsie@dordos.net>
Date2012-01-27 07:55 +0100
Message-ID<jfthpq$18d$1@online.de>
In reply to#334
Am 27.01.2012 07:31, schrieb Peter Pein:
> Am 27.01.2012 01:45, schrieb clashton:
>> On Jan 26, 11:21 am, Joe Riel<j...@san.rr.com> wrote:
>>> clashton<clash...@gmail.com> writes:
>>>> I am a little curious about whether Maple can evaluate a certain
>>>> infinite sum that fails with Mathematica (I do not have Maple so
>>>> cannot test it myself).
>>>
>>>> Here is the sum in Mathematica:
>>>> Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
>>>> This leads to and error,
>>>> but
>>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q -> 0.1)
>>>> +
>>>> (Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
>>>> gives the correct answer, which I find a little strange.
>>>
>>>> How does Maple perform on the original sum?
>>>
>>> Summing numerically (and guessing at the meaning of the Mathematica):
>>>
>>> (**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
>>> infinity
>>> ----- (-6 + 4 n)
>>> \ q
>>> S := ) ---------------
>>> / (-5 + 4 n)
>>> ----- 1 - q
>>> n = 0
>>>
>>> (**) evalf(eval(S,q=1/10));
>>> -21.10120010
>>>
>>> --
>>> Joe Riel
>>
>> Maple wins this round.
>
> Well, if you are interested in the approximation of the sum's value for
> q=1/10, you should have used NSum in Mathematica:
>
> NSum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)) /. q->1/10, {n, 0, Infinity},
> WorkingPrecision->20]
>
> -21.101200102001001100
>
> or - using your workaround - get
>
> 6/(5*(-1 + q)) + (1 + 2*q + 3*q^2 + 4*q^3)/(5*(1 + q + q^2 + q^3 + q^4))
> + (-2*Log[q^4] + Log[1 - q^4] + QPolyGamma[0, 2 - Log[q^5]/Log[q^4],
> q^4])/(q*Log[q^4])
>
> as (hopefully) exact value.
>
> Cheers, Peter

Sorry for posting too fast. One gets the result in a more simple form by 
doing sth. more complicated:

In[1]:= f[q_] = q^(-6 + 4*n)/(1 - q^(-5 + 4*n));
assum = SumConvergence[f[q], n]
s[q_] = Together[Subtract @@ (Limit[Sum[f[q], n], n -> #1, Assumptions 
-> assum] & ) /@ {Infinity, 0}]
N[s[1/10]]


Out[2]= q != 0 && Abs[q]^4 < 1
Out[3]= (Log[1 - q^4] + QPolyGamma[0, -(Log[q^5]/Log[q^4]), 
q^4])/(q*Log[q^4])
Out[4]= -21.1012

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#336

Fromclashton <clashton@gmail.com>
Date2012-01-28 07:59 -0800
Message-ID<57c5484a-a9c3-4a41-8b3b-097e3faf92bb@l14g2000vbe.googlegroups.com>
In reply to#335
Thanks to everyone.

I am happy to see that there was a way of doing it in Mathematica, as
I do not have Maple.

On Jan 27, 1:55 am, Peter Pein <pet...@dordos.net> wrote:
> Am 27.01.2012 07:31, schrieb Peter Pein:
>
>
> >>> Summing numerically (and guessing at the meaning of the Mathematica):
>
> >>> (**) S := Sum(q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), n=0..infinity);
> >>> infinity
> >>> ----- (-6 + 4 n)
> >>> \ q
> >>> S := ) ---------------
> >>> / (-5 + 4 n)
> >>> ----- 1 - q
> >>> n = 0
>
> >>> (**) evalf(eval(S,q=1/10));
> >>> -21.10120010
>
> >>> --
> >>> Joe Riel
>
> >> Maple wins this round.
>
> > Well, if you are interested in the approximation of the sum's value for
> > q=1/10, you should have used NSum in Mathematica:
>
> > NSum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)) /. q->1/10, {n, 0, Infinity},
> > WorkingPrecision->20]
>
> > -21.101200102001001100
>
> > or - using your workaround - get
>
> > 6/(5*(-1 + q)) + (1 + 2*q + 3*q^2 + 4*q^3)/(5*(1 + q + q^2 + q^3 + q^4))
> > + (-2*Log[q^4] + Log[1 - q^4] + QPolyGamma[0, 2 - Log[q^5]/Log[q^4],
> > q^4])/(q*Log[q^4])
>
> > as (hopefully) exact value.
>
> > Cheers, Peter
>
> Sorry for posting too fast. One gets the result in a more simple form by
> doing sth. more complicated:
>
> In[1]:= f[q_] = q^(-6 + 4*n)/(1 - q^(-5 + 4*n));
> assum = SumConvergence[f[q], n]
> s[q_] = Together[Subtract @@ (Limit[Sum[f[q], n], n -> #1, Assumptions
> -> assum] & ) /@ {Infinity, 0}]
> N[s[1/10]]
>
> Out[2]= q != 0 && Abs[q]^4 < 1
> Out[3]= (Log[1 - q^4] + QPolyGamma[0, -(Log[q^5]/Log[q^4]),
> q^4])/(q*Log[q^4])
> Out[4]= -21.1012

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#337

FromAxel Vogt <&noreply@axelvogt.de>
Date2012-01-28 20:21 +0100
Message-ID<9oj066Fr04U1@mid.individual.net>
In reply to#335
On 27.01.2012 07:55, Peter Pein wrote:
...
>> Well, if you are interested in the approximation of the sum's value for
>> q=1/10, you should have used NSum in Mathematica:
>>
>> NSum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)) /. q->1/10, {n, 0, Infinity},
>> WorkingPrecision->20]
>>
>> -21.101200102001001100
>>
>> or - using your workaround - get
>>
>> 6/(5*(-1 + q)) + (1 + 2*q + 3*q^2 + 4*q^3)/(5*(1 + q + q^2 + q^3 + q^4))
>> + (-2*Log[q^4] + Log[1 - q^4] + QPolyGamma[0, 2 - Log[q^5]/Log[q^4],
>> q^4])/(q*Log[q^4])
>>
>> as (hopefully) exact value.
>>
>> Cheers, Peter
>
> Sorry for posting too fast. One gets the result in a more simple form by doing sth. more complicated:
>
> In[1]:= f[q_] = q^(-6 + 4*n)/(1 - q^(-5 + 4*n));
> assum = SumConvergence[f[q], n]
> s[q_] = Together[Subtract @@ (Limit[Sum[f[q], n], n -> #1, Assumptions -> assum] & ) /@ {Infinity, 0}]
> N[s[1/10]]
>
>
> Out[2]= q != 0 && Abs[q]^4 < 1
> Out[3]= (Log[1 - q^4] + QPolyGamma[0, -(Log[q^5]/Log[q^4]), q^4])/(q*Log[q^4])
> Out[4]= -21.1012

I am not aware that Maple has a command to find a
condition for convergence.

And it does not provide a symbolic solution (where
I guess the above cryptic command does just that)

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#342

FromPeter Pein <petsie@dordos.net>
Date2012-02-02 12:21 +0100
Message-ID<jgdrk9$m77$1@online.de>
In reply to#337
Am 28.01.2012 20:21, schrieb Axel Vogt:
> On 27.01.2012 07:55, Peter Pein wrote:
> ....
>>
>> Sorry for posting too fast. One gets the result in a more simple form
>> by doing sth. more complicated:
>>
>> In[1]:= f[q_] = q^(-6 + 4*n)/(1 - q^(-5 + 4*n));
>> assum = SumConvergence[f[q], n]
>> s[q_] = Together[Subtract @@ (Limit[Sum[f[q], n], n -> #1, Assumptions
>> -> assum] & ) /@ {Infinity, 0}]
>> N[s[1/10]]
>>
>>
>> Out[2]= q != 0 && Abs[q]^4 < 1
>> Out[3]= (Log[1 - q^4] + QPolyGamma[0, -(Log[q^5]/Log[q^4]),
>> q^4])/(q*Log[q^4])
>> Out[4]= -21.1012
>
> I am not aware that Maple has a command to find a
> condition for convergence.
>
> And it does not provide a symbolic solution (where
> I guess the above cryptic command does just that)

yes, see: http://reference.wolfram.com/mathematica/ref/QPolyGamma.html
and/or
http://mathworld.wolfram.com/q-PolygammaFunction.html

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