Re: Limit of Error function Erf
- To: mathgroup at smc.vnet.net
- Subject: [mg73565] Re: Limit of Error function Erf
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Thu, 22 Feb 2007 04:27:07 -0500 (EST)
- Organization: The Open University, Milton Keynes, UK
- References: <ergpqn$hbq$1@smc.vnet.net>
ben wrote:
> Dear group
>
> I dont understand the following behaviour of Mathematica,
> I would say the first result is simply wrong
>
> 1.)
>
> Limit[Erf[1\/Sqrt[\[Alpha]\ t]], t -> \[Infinity],
> Assumptions -> {\[Alpha] > 0}]
>
> gives infinity
>
> 2.)
>
> Limit[Erf[1\/Sqrt[\[Alpha]\ t]], t -> \[Infinity],
> Assumptions -> {\[Alpha] == 1}]
>
> gives zero
>
> Any suggestions? Is this a bug?
> Bye
> Ben
>
>
Same behavior with Mathematica 5.2 for Microsoft Windows (June 20, 2005).
It seems that the issue is related to the combination parameter and
fractional powers.
In[1]:=
Limit[Erf[1/Sqrt[α*t]], t -> Infinity,
Assumptions -> {α > 0}]
Out[1]=
Infinity
(* It is usually better to use a clause With to temporarily set the
value of a parameter rather than using Assumption -> param == value. *)
In[2]:=
With[{α = 1}, Limit[Erf[1/Sqrt[α*t]],
t -> Infinity]]
Out[2]=
0
In[3]:=
Limit[Erf[1/Sqrt[t]], t -> Infinity]
Out[3]=
0
In[4]:=
Limit[Erf[1/(α*t)], t -> Infinity,
Assumptions -> {α > 0}]
Out[4]=
0
In[5]:=
Limit[Erf[1/(α*t)^(1/2)], t -> Infinity,
Assumptions -> {α > 0}]
Out[5]=
Infinity
In[6]:=
Limit[Erf[1/(α*t)^(1/4)], t -> Infinity,
Assumptions -> {α > 0}]
Out[6]=
Infinity
In[7]:=
Limit[Erf[(α*t)^(-(1/2))], t -> Infinity,
Assumptions -> {α > 0}]
Out[7]=
Infinity
Regards,
Jean-Marc