Does this equation have any solution?

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Discussion Overview

The discussion centers around the equation \( x = e^x \) and whether it has any solutions, specifically exploring the existence of complex solutions as opposed to real solutions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asserts that there are no real solutions to the equation \( x = e^x \) but questions the existence of complex solutions.
  • Another participant claims that there are two distinct complex solutions, specifically \( x_{1} = -W(-1) \) and \( x_{2} = -W(-1, -1) \), where \( W \) denotes the Lambert function.
  • A participant expresses confusion regarding the nature of the solution \( x_{1} = -W(-1) \) and asks if it is a complex number.
  • The responding participant confirms that \( x_{1} = -W(-1) \) does indeed have a nonzero imaginary part, reiterating that the equation does not have real solutions.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus; while one participant claims the existence of complex solutions, another expresses uncertainty about their nature.

Contextual Notes

There are unresolved aspects regarding the interpretation of the Lambert function and the specific characteristics of the solutions, particularly in relation to their complex nature.

Petrushka
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[tex]x\multsp =\multsp {{e}^x}[/tex]

I'm aware there's no real solution, but does any complex solution exist?
 
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Yes,it has 2 distinct solutions:

[tex]x_{1}=-W(-1)[/tex]

[tex]x_{2}=-W(-1,-1)[/tex]

,where W is the Lambert function.

Daniel.
 
I see. Having read a bit about the Lambert function on Http://www.mathworld.com it doesn't make much sense to me.

Is the following value:

[tex]x_{1}=-W(-1)[/tex]

...a complex number, or something entirely different?
 
You mean a # of nonzero imaginary part...?Yes it is.As u said and as a graph and elementary calculus would show,for real #-s the equation does not have solution.

Daniel.
 

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