Re: Zeta Function

• To: mathgroup at smc.vnet.net
• Subject: [mg20582] Re: [mg20556] Zeta Function
• From: BobHanlon at aol.com
• Date: Sat, 30 Oct 1999 14:54:55 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Andre,

Plot[Zeta[x], {x, 0, 2}];

To Plot your series you must use Evaluate.

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

I will leave it to mathematicians to discuss the utility of having
expectations for operations at singularities.

Bob Hanlon

In a message dated 10/30/1999 2:53:57 AM, A.Heinemann at ifw-dresden.de writes:

>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].
>
>
>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 ?
>

```

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