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from Rumen, MEXICO, petition

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  from Rumen, MEXICO, petition Rumen Ivanov <rumen5252@yahoo.com.mx> - 2011-04-09 11:12 +0000
    Re: from Rumen, MEXICO, petition Peter <petsie@dordos.net> - 2011-04-11 11:07 +0000
    Re: from Rumen, MEXICO, petition Noqsi <noqsiaerospace@gmail.com> - 2011-04-11 11:07 +0000
    Re: from Rumen, MEXICO, petition "Stephen Luttrell" <steve@_removemefirst_stephenluttrell.com> - 2011-04-11 11:05 +0000

#1553 — from Rumen, MEXICO, petition

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@yahoo.com.mx 
rumen@fisica.uaz.edu.mx

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#1593

Am 09.04.2011 13:12, schrieb Rumen Ivanov:
> 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@yahoo.com.mx
> rumen@fisica.uaz.edu.mx

Hello Rumen,

as has been said, there is no real solution, but to make things easier 
for Mathematica, I took the Log of the Exp-sum and set it equal to 
Log[-2.02912]. Then FindRoot gets an approx. solution:

In[1]:= expr=11*Exp[-Es/135]+10.5*Exp[-Es/425]+4.899*Exp[-Es/1754]+2.02912;

eqn = Log[expr - 2.02912] == Log[-2.02912];
{Es, expr} /. First @ FindRoot[eqn, {Es, 0}, MaxIterations->Infinity]
Out[3]= {298.34 -1494.67 I, 1.77636*10^-15 + 0. I}

Regards,
Peter

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#1594

On Apr 9, 5:12 am, Rumen Ivanov <rumen5...@yahoo.com.mx> wrote:
> 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.

As others have pointed out, it has no real solutions. But it appears
to have many complex solutions:

f=11*Exp[-Es/135]+10.5*Exp[-Es/425]+4.899*Exp[-Es/1754]+2.02912

FindRoot[f==0,{Es,500I}]
{Es->-55.0724+546.446 I}

FindRoot[f==0,{Es,2000I}]
{Es->81.4202+2009.63 I}

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#1602

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@yahoo.com.mx> wrote in message 
news:inpeua$8ul$1@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@yahoo.com.mx
> rumen@fisica.uaz.edu.mx 

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