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