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Re: Re: Troubles with HarmonicNumber, empty sums, and Zeta
*To*: mathgroup at smc.vnet.net
*Subject*: [mg86119] Re: [mg85934] Re: Troubles with HarmonicNumber, empty sums, and Zeta
*From*: Devendra Kapadia <dkapadia at wolfram.com>
*Date*: Sun, 2 Mar 2008 13:57:08 -0500 (EST)
*References*: <200802130915.EAA21588@smc.vnet.net> <fouma4$qu3$1@smc.vnet.net>
On Thu, 28 Feb 2008, David W.Cantrell wrote:
> "David W.Cantrell" <DWCantrell at sigmaxi.net> wrote:
> [snip]
>> Here's another bug, which might be unrelated to the other troubles I had
>> mentioned:
>>
>> In[8]:= Limit[Zeta[z, 9/10] - Zeta[z, 9/10 + z], z -> 1]
>>
>> Out[8]= Infinity
>>
>> But it's easy to guess the correct result:
>>
>> In[16]:= N[Zeta[z, 9/10] - Zeta[z, 9/10 + z] /. z -> 9999/10000, 12]
>> Out[16]= 1.11103060226
>> In[17]:= N[Zeta[z, 9/10] - Zeta[z, 9/10 + z] /. z -> 10001/10000, 12]
>> Out[17]= 1.11119161012
>> In[18]:= (% + %%)/2
>> Out[18]= 1.11111110619
>>
>> and so our guess would be 10/9.
>>
>> Furthermore, we can get the correct exact result by "converting" to
>> HarmonicNumber by hand (and then there is no need of Limit):
>>
>> In[19]:= -HarmonicNumber[-1/10, z] + HarmonicNumber[-1/10 + z, z] /. z ->
>> 1 Out[19]= -HarmonicNumber[-1/10] + HarmonicNumber[9/10]
>> In[20]:= FullSimplify[%]
>> Out[20]= 10/9
>>
>> Finally, I wonder about things like ZetaClassical and
>> ZetaClassicalRegularized. They are used at the Wolfram Functions site.
>> For example, the third and fourth entries at
>> <http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta2/02/>
>> concern ZetaClassical and ZetaClassicalRegularized, resp. And in the
>> output from one of the bugs mentioned in my original post in this thread,
>> there was mention of ZetaClassical in a system dump. So are ZetaClassical
>> and ZetaClassicalRegularized things that users of Mathematica have access
>> to, or are they strictly for internal work, to be hidden from the user?
>
> Having just now updated to version 6.0.2, I'm sorry to note that bugs
> mentioned earlier in this thread still exist. And here's another bug
> related to Zeta (but probably not directly related to bugs previously
> mentioned in this thread):
>
> In[27]:= Sum[i*(1/(i - 1/10)^2 - 1/(i + 9/10)^2), {i, 1, Infinity}]
>
> Out[27]= EulerGamma - (2*Sqrt[5/8 + Sqrt[5]/8]*Pi)/(-1 + Sqrt[5]) -
> (100/361)*HypergeometricPFQ[{19/10, 19/10, 2}, {29/10, 29/10}, 1] +
> 2*Log[4] + Log[5] + (1/4)*Log[5/8 - Sqrt[5]/8] - (1/4)*Sqrt[5]*Log[5/8 -
> Sqrt[5]/8] + (1/4)*Log[5/8 + Sqrt[5]/8] + (1/4)*Sqrt[5]*Log[5/8 +
> Sqrt[5]/8] - (1/2)*Log[-1 + Sqrt[5]] - (1/2)*Sqrt[5]*Log[-1 + Sqrt[5]] -
> (1/2)*Log[1 + Sqrt[5]] + (1/2)*Sqrt[5]*Log[1 + Sqrt[5]] +
> (1/10)*PolyGamma[1, 9/10]
>
> In[28]:= N[%]
>
> Out[28]= ComplexInfinity
>
> In[29]:= FullSimplify[%%]
>
> Out[29]= Indeterminate
>
> Out[27] is incorrect. (BTW, it seems strange to me that N[Out[27]] is
> ComplexInfinity while FullSimplify[Out[27]] is Indeterminate. But that is
> not my point.) A correct result from In[27] would have been simply
> Zeta[2, 9/10], and that's why I had said this bug is related to Zeta.
>
> In[37]:= N[Zeta[2, 9/10]]
>
> Out[37]= 1.92254
>
> In[38]:= NSum[i*(1/(i - 1/10)^2 - 1/(i + 9/10)^2), {i, 1, Infinity}]
>
> Out[38]= 1.92254
>
> David W. Cantrell
>
Hello David,
Thank you for informing us about the incorrect result from the above
infinite sum. As noted by you, this example should return a simple
answer equivalent to Zeta[2, 9/10].
A partial workaround for the problem is to expand the summand and then
evaluate the series, as shown below.
==========================
In[2]:= InputForm[res = Sum[Expand[i*(1/(i - 1/10)^2 - 1/(i + 9/10)^2)],
{i, 1, Infinity}]]
Out[2]//InputForm= PolyGamma[1, 9/10]
In[3]:= FullSimplify[res - Zeta[2, 9/10]]
Out[3]= 0
In[4]:= N[res]
Out[4]= 1.92254
In[5]:= NSum[i*(1/(i - 1/10)^2 - 1/(i + 9/10)^2), {i, 1, Infinity}]
Out[5]= 1.92254
=============================
I apologize for the confusion caused by this problem.
Sincerely,
Devendra Kapadia.
Wolfram Research, Inc.
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