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