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MathGroup Archive 2004

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Re: how can I solve a function Erfc

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48431] Re: how can I solve a function Erfc
  • From: "Curt Fischer" <crf3 at po.cwru.edu>
  • Date: Sat, 29 May 2004 03:06:53 -0400 (EDT)
  • References: <c96h2m$idk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Aude,

Check the help files!  This is very easy to do in Mathematica.

In[1]:=
?Erfc

"Erfc[z] gives the complementary error function erfc(z)."

In[2]:=
dg=5*^5

Out[2]=
500000

In[3]:=
root[l_?NumericQ]:=t/.FindRoot[Erfc[l /(4(dg t)^(1/2))] == 0.9, { t,
l^2/dg }]

In[4]:=
root[.2]

Out[4]=
\!\(6.332811503848235`*^-7\)

I'll leave it up to you to figure out how to plot this function.

Curt



aude wrote:
> Hi,
>
> I have to solve this function:
>
> Erfc[L/(4*(Dg*t)^(1/2))]= 0.9
>
> Dg is constant.
>
> Dg=5*10^5
>
> I want to plot t as a function of L.
>
> thanks.
>
> aude


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