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Re: how can I solve a function Erfc

  • To: mathgroup at
  • Subject: [mg48447] Re: how can I solve a function Erfc
  • From: Daniel Lichtblau <danl at>
  • Date: Sun, 30 May 2004 06:12:08 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

David W. Cantrell wrote:
> montgermont.aude at (aude) wrote:
>>I have to solve this function:
>>Erfc[L/(4*(Dg*t)^(1/2))]= 0.9
>>Dg is constant.
>>I want to plot t as a function of L.
> If plotting t as a function of L is all you want to do, you can just use
> ImplicitPlot. But trying to use Mathematica to solve an equation of the
> form
>  Erfc[L/(4*(Dg*t)^(1/2))] == k
> for t in terms of L, Dg, and k raises some interesting points, one of which
> I suspect is a bug.
> First, note that Mathematica has a built-in function InverseErfc. With that
> in mind, you might think that one of the following would work, but alas,
> no.
> In[1]:= Solve[Erfc[L/(4*(Dg*t)^(1/2))] == k, t]
> ...
> Out[1]= {}
> In[2]:= Reduce[Erfc[L/(4*(Dg*t)^(1/2))] == k, t]
> ...
> Out[2]= Reduce[Erfc[L/(4*Sqrt[Dg*t])] == k, t]
> But it seems that we should be able, trivially, to assist Mathematica by
> rewriting the equation using InverseErfc. (Note that I've now used your
> specific k value, so that there is no doubt that InverseErfc is defined
> there.)
> In[3]:= Solve[L/(4*Sqrt[Dg*t]) == InverseErfc[9/10], t]
> Out[3]= {}
> Surely this is a bug. The solution for t should have been
>  L^2/(16*Dg*InverseErf[Infinity, -(9/10)]^2)
> which, BTW, is obtainable with Reduce. I can't imagine why Solve didn't
> work there!
> David Cantrell

The Solve verification process did not like that value (no doubt due to 
presence of the Infinity). That may indicate a bug, I'm not sure.

In[6]:= InputForm[Solve[Erfc[L/(4*(Dg*t)^(1/2))] == k, t, 

Solve::ifun: Inverse functions are being used by Solve, so some 
solutions may
      not be found; use Reduce for complete solution information.

Out[6]//InputForm= {{t -> L^2/(16*Dg*InverseErf[Infinity, -k]^2)}}

Daniel Lichtblau
Wolfram Research

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