Does e^z - z^2 = 0 Have Infinite Solutions?

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Discussion Overview

The discussion revolves around the equation ez - z2 = 0, specifically exploring whether it has infinitely many solutions. The scope includes mathematical reasoning and potential applications of special functions.

Discussion Character

  • Exploratory, Mathematical reasoning

Main Points Raised

  • One participant inquires about demonstrating that the equation has infinitely many solutions.
  • Another participant suggests working within a fixed branch of log(z) and reformulating the equation as z2 = e2 log z, indicating that finding one solution leads to infinitely many due to the periodicity of ez.
  • A subsequent reply elaborates that this leads to the equation e{z - 2 log z} = 1, implying that solutions correspond to z - 2 log z = 2πn for n ∈ ℤ.
  • Another participant proposes that the Lambert W function might be useful in finding solutions to the equation derived from the previous discussions.

Areas of Agreement / Disagreement

Participants present various approaches and insights, but there is no consensus on a definitive method or conclusion regarding the existence of infinitely many solutions.

Contextual Notes

The discussion relies on assumptions related to the branches of the logarithm and the periodic nature of the exponential function, which may affect the interpretation of solutions.

Likemath2014
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How can we show that the following equation has infinitely many solutions
[tex]e^z-z^2=0[/tex].
Thanks
 
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Work within a fixed branch of logz and write : ## z^2=e^{2logz} ##

Once you find a solution, you have infinitely-many, by periodicity of ##e^z##.
 
Continuing WWGD's post: you get ##e^z=e^{2\log z}## and thus the equation ##e^{z-2log z}=1##. In other words you are looking at the solutions of each of the equations ##z-2\log z=2\pi n## for ##n\in\mathbb{Z}##.
 
It seems that something like Lambert's W function may be helpful in finding solutions to platetheduke's equation.
 

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