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.