Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Bug with Limit, Series and ProductLog ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg61426] Re: Bug with Limit, Series and ProductLog ?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 19 Oct 2005 02:16:05 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <dj26kc$bc7$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

did wrote:
> With Mathematica 5.2 Windows I obtain
> 
> In[1]:=Limit[ ProductLog[Exp[a/x]/x]-a/x,x->0]
> Out[1]= -Log[a]
> 
> which seems correct. But, setting a=1, I get
> 
> In[2]:=Limit[ ProductLog[Exp[1/x]/x]-1/x,x->0]
> Out[2]=-8
> 
> which is inconsistent with the previous result
> (except if Log[1] is Infinity !).
> 
> Worse, with Series I get
> 
> In[3]:=Series[ ProductLog[Exp[a/x]/x]-a/x,{x,0,5}]
> 
> Out[3]=\!\(\*
>   InterpretationBox[
>     RowBox[{\(-\(a\/x\)\), "+", "Indeterminate", "+",
>       InterpretationBox[\(O[x]\^6\),
>         SeriesData[ x, 0, {}, -1, 6, 1],
>         Editable->False]}],
>     SeriesData[ x, 0, {
>       Times[ -1, a], Indeterminate}, -1, 6, 1],
>     Editable->False]\)
> 
> Setting a=1 in the Series gives a complex answer.
> 
> How can I workaround the problem and get the correct
> expansion for In[3]?
> Thanks,
> D.
> 
No bug here.

What result do you expect since the limit from the right is 
indeterminate of the form infinity - infinity and it is negative 
infinity from the left?

In[1]:=
expr = ProductLog[Exp[a/x]/x] - a/x

Out[1]=
-(a/x) + ProductLog[E^(a/x)/x]

In[2]:=
Limit[expr, x -> 0, Assumptions -> a > 0]

Out[2]=
-Infinity

In[3]:=
Limit[expr, x -> 0, Direction -> -1, Assumptions -> a > 0]

Out[3]=
-Infinity

In[4]:=
Limit[expr, x -> 0, Direction -> 1, Assumptions -> a > 0]

Out[4]=
Limit[-(a/x) + ProductLog[E^(a/x)/x], x -> 0, Direction -> 1, 
Assumptions -> a > 0]

In[5]:=
Limit[-a/x, x -> 0, Direction -> 1, Assumptions -> a > 0]

Out[5]=
Infinity

In[6]:=
Limit[ProductLog[Exp[a/x]/x], x -> 0, Direction -> 1, Assumptions -> a > 0]

Out[6]=
-Infinity

In[7]:=
Limit[-a/x, x -> 0, Direction -> -1, Assumptions -> a > 0]

Out[7]=
-Infinity

In[8]:=
Limit[ProductLog[Exp[a/x]/x], x -> 0, Direction -> -1, Assumptions -> a > 0]

Out[8]=
25/12 + a - Log[a]

Regards,
/J.M.


  • Prev by Date: Re: Stylesheets vs. DTDs or XML Schemas
  • Next by Date: Re: Re: Language vs. Library why it matters
  • Previous by thread: Re: Bug with Limit, Series and ProductLog ?
  • Next by thread: Generating unitary aliquot cycles