The generalized Lambert W-function

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Discussion Overview

The discussion revolves around deriving the generalized Lambert W-function, particularly in the context of solving ordinary differential equations (ODEs). Participants explore the mathematical formulation and properties of the function, as well as its application in specific cases.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Salty presents an initial definition of the Lambert W-function and seeks help in deriving the generalized form for functions of the type g(x) = x^n e^x.
  • Some participants suggest transforming the equation into the form f(x)e^{f(x)} to facilitate the application of the W-function.
  • Daniel questions the introduction of the factor "n" in front of the W-function and seeks clarification on its derivation.
  • Daniel provides a step-by-step explanation of how to derive the inverse function using the Lambert W-function, asserting that it is straightforward.
  • Salty expresses appreciation for Daniel's explanation and indicates a plan to report back on the homework section regarding this problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivation process, as there are differing views on the necessity and origin of the factor "n" in the generalized form of the Lambert W-function. The discussion remains unresolved regarding the clarity of the derivation steps.

Contextual Notes

The discussion includes assumptions about the applicability of the Lambert W-function and its generalizations, but these assumptions are not explicitly stated or agreed upon by all participants.

saltydog
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Can anyone here help me derive the generalized Lambert function? I'm working on a solution for an ODE from the homework group which involves this function. This is what I have so far:

The W-function is defined as the inverse of the following:

[tex] f(x)=xe^x=y[/tex]

then:

[tex]f^{-1}(y)=x=W(y)[/tex]

with W being the Lambert W-function for [itex]y>-e^{-1}[/itex]

I need help showing the following:

If:

[tex]g(x)=x^2e^x=y[/tex]

then:

[tex]g^{-1}(y)=2W(\frac{\sqrt y}{2})[/tex]

and in general if:

[tex]h(x)=x^ne^x=y[/tex]

then:

[tex]h^{-1}(y)=nW(\frac{y^\frac{1}{n}}{n})[/tex]

Thanks,
Salty
 
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you need to get one side into the form:

[tex]f(x) e^{f(x)}[/tex]

then take the W of both sides, which will leave f(x) here.
 
Where did you come up with that factor "n" in front of the W function...?

Daniel.
 
dextercioby said:
Where did you come up with that factor "n" in front of the W function...?

Daniel.

Mathematica reports the results as I stated but I'd like to understand how it's arriving at it. I've checked it with real numbers (I don't have a support contract and they don't like me bothering them).
 
StatusX said:
you need to get one side into the form:

[tex]f(x) e^{f(x)}[/tex]

then take the W of both sides, which will leave f(x) here.

Thanks StatusX. I don't see that but will try and work with that logic in the morning.

Salty
 
It's straightforward.
[tex]x^{n}e^{x}=y \Rightarrow xe^{\frac{x}{n}}=y^{\frac{1}{n}}\Rightarrow \frac{x}{n}e^{\frac{x}{n}}=\frac{y^{\frac{1}{n}}}{n}[/tex]

Apply the W (Lambert function) on the last equality and u'll get
[tex]\frac{x}{n}=W(\frac{y^{\frac{1}{n}}}{n}) \Rightarrow x=n W(\frac{y^{\frac{1}{n}}}{n})[/tex]

which is the inverse function of the one you started with.

Daniel.
 
dextercioby said:
It's straightforward.
[tex]x^{n}e^{x}=y \Rightarrow xe^{\frac{x}{n}}=y^{\frac{1}{n}}\Rightarrow \frac{x}{n}e^{\frac{x}{n}}=\frac{y^{\frac{1}{n}}}{n}[/tex]

Apply the W (Lambert function) on the last equality and u'll get
[tex]\frac{x}{n}=W(\frac{y^{\frac{1}{n}}}{n}) \Rightarrow x=n W(\frac{y^{\frac{1}{n}}}{n})[/tex]

which is the inverse function of the one you started with.

Daniel.

Yep, would not have figured that on my own. I mean, it took me a while to even see what you were doing. Thanks Daniel.
I'll follow-up with a report (and plot) in the homework section for this problem. I know they're long-gone to other things but I tell you what, they missin' out (and they wouldn't like me as their teacher because I'd make them do this extra stuf for at least some of the problems).

Salty
 

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