Re: from Rumen, MEXICO, petition
- To: mathgroup at smc.vnet.net
- Subject: [mg118059] Re: from Rumen, MEXICO, petition
- From: "Stephen Luttrell" <steve at _removemefirst_stephenluttrell.com>
- Date: Mon, 11 Apr 2011 07:05:18 -0400 (EDT)
- References: <inpeua$8ul$1@smc.vnet.net>
The roots are complex-valued. Here is a way of finding them:
Define the function whose roots you want.
f[z_] := 11 Exp[-(z/135)] + 10.5` Exp[-(z/425)] + 4.899` Exp[-(z/1754)] +
2.02912`;
Contour plot the absolute value over a large enough region to get a feel for
what the function looks like.
ContourPlot[Abs[f[x + I y]], {x, -200, 300}, {y, -3000, 3000}, Contours ->
50]
The zeros all lie at around Re(z) = -50, and they repeat periodically in
Im(z) as you would expect from the exponential dependence of f(z) on z.
Use your mouse to grab the approximate position of one of the zeros (I got
{-58.01, 478.8}), and find the minimum.
FindMinimum[Abs[f[x + I y]], {{x, -58.01}, {y, 478.8}}]
which gives
{2.04123*10^-7,{x->-55.0724,y->546.446}}
--
Stephen Luttrell
West Malvern, UK
"Rumen Ivanov" <rumen5252 at yahoo.com.mx> wrote in message
news:inpeua$8ul$1 at smc.vnet.net...
> Hello,
>
> 11*Exp[-Es/135]+10.5*Exp[-Es/425]+
> 4.899*Exp[-Es/1754]+ 2.02912 = 0
> Please tell me how Mathematica can solve this equation
> for Es? With the derivatives of "Solve" it is impossible.
>
> Best
>
> Regards,
>
> Rumen
>
> Prof. Dr. Rumen Ivanov Tsonchev
> Facultad de F=EDsica
> Universidad Aut=F3noma de Zacatecas
> e-mail=B4s:
> rumen5252 at yahoo.com.mx
> rumen at fisica.uaz.edu.mx