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MathGroup Archive 1999

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Re: Zeta Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg20587] Re: [mg20556] Zeta Function
  • From: David Withoff <withoff at wolfram.com>
  • Date: Sat, 30 Oct 1999 14:54:59 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

> 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.
> 
> But what is wrong with the Zeta function ???

The definition of Zeta by default excludes the singular term:

In[1]:= Options[Zeta]

Out[1]= {IncludeSingularTerm -> False}

As your example shows, the Series function sometimes includes the
singular term anyway.  Both results are correct, but they are
correct in different situations.  I will report this inconsistency.

> 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 ?

This is unrelated.  This is intended behavior.  The Plot function
inserts numerical values for x, and series expansion with a number
as the variable doesn't work.  There are a variety of ways to
address this, one of which is to force evaluation of the argument
of Plot:

In[3]:= Plot[Evaluate[Normal[Series[Zeta[x],{x,1,2}]]], {x,0,2}]

Out[3]= -Graphics-

Dave Withoff
Wolfram Research


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