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MathGroup Archive 1999

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Re: Zeta Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg20586] Re: [mg20556] Zeta Function
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Sat, 30 Oct 1999 14:54:58 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

The second question is easy to answer. You have to enter:

Plot[Evaluate[Normal[Series[Zeta[x], {x, 1, 2}]]], {x, 0, 2}]

This sort of thing is explained on this list over and over again so you
should just look at past postings if you don't understand why.

The other problem seems like s serious bug to me. Mathematica correctly
gives:

In[3]:=
Series[Zeta[z], {z, 1, 5}]
Out[3]=
  1
------ + EulerGamma - StieltjesGamma[1] (-1 + z) +
-1 + z
 
                            2                             3
  StieltjesGamma[2] (-1 + z)    StieltjesGamma[3] (-1 + z)             4
  --------------------------- - --------------------------- + O[-1 + z]
               2                             6
However, 



In[4]:=
Series[Zeta'[z], {z, 1, 4}]
Out[4]=
-StieltjesGamma[1] + StieltjesGamma[2] (-1 + z) -
 
                            2                             3
  StieltjesGamma[3] (-1 + z)    StieltjesGamma[4] (-1 + z)
  --------------------------- + --------------------------- -
               2                             6
 
                            4
  StieltjesGamma[5] (-1 + z)             5
  --------------------------- + O[-1 + z]
              24

Shows that something has gone badly wrong. Basically the term -1/(1-z)^2 is
missing at the beginning but it is not just "missing". Mathematic aclearly
consdiers this to be an expansion of an analytic function (compare the O
terms in the two expressions). Strangely enough Mathematica does now the
correct formula for Zeta'[z]:

In[7]:=
Sum[-Log[n]/n^z, {n, 1, Infinity}]
Out[7]=
Zeta'[z]

and evaluating N[Zeta'[a]] numerically for various a gives correct values
except in the crucial case:

In[9]:=
Zeta'[1]
Out[9]=
-StieltjesGamma[1]

In[10]:=
N[%]
Out[10]=
0.0728158

While, for example,
In[11]:=
Zeta'[0.999999999]
Out[11]=
551.472
. 


-- 
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp




> From: Andre Heinemann <A.Heinemann at ifw-dresden.de>
To: mathgroup at smc.vnet.net
> Date: Sat, 30 Oct 1999 00:14:06 -0400
> To: mathgroup at smc.vnet.net
> Subject: [mg20586] [mg20556] Zeta Function
> 
> In Mathematica 2.2 and 4.0 a friend
> pushed me to the following
> problem. And for me it is a bug
> in Mathematica because we solved it
> with Mapple in the right way.
> 
> At x== 1 the zeta function has a 1/(x -1)
> singular point as you can see in a plot.
> So the Series[D[Zeta[x],x],{x,1,2}]
> should have the leading term of -1/(x-1)^2.
> But it hasn't !
> 
> Only in the D[Series[Zeta[x], {x, 1, 2}], x]
> expression I can find this leading part.
> 
> I made a test with a other singular point,
> the x == -1 for Gamma[x].
> 
> The leading terms of
> 
> D[ Series[Gamma[x], {x, -1, 1}], x]
> 
> and 
> 
> Series[D[Gamma[x], x], {x, -1, 1}]
> 
> are both of the type 1/(1+x)^2 as one expect.
> 
> But what is wrong with the Zeta function ???
> 
> And by the way, Plot[Normal[Series[Zeta[x],{x,1,2}]],{x,0,2}]
> doesn't work and I can't see why not ?
> Can you help me ?
> 
> Regards,
> 
> Andre 
> -- 
> +-----------------------------------+-----------------------------+
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> 



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