The generalized Lambert W-function

[saltydog]

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:

<br /> f(x)=xe^x=y<br />

then:

f^{-1}(y)=x=W(y)

with W being the Lambert W-function for y&gt;-e^{-1}

I need help showing the following:

If:

g(x)=x^2e^x=y

then:

g^{-1}(y)=2W(\frac{\sqrt y}{2})

and in general if:

h(x)=x^ne^x=y

then:

h^{-1}(y)=nW(\frac{y^\frac{1}{n}}{n})<br />

Thanks,
Salty

 

[StatusX]

you need to get one side into the form:

f(x) e^{f(x)}

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

 

[dextercioby]

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

Daniel.

 

[saltydog]

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).

 

[saltydog]

you need to get one side into the form:

f(x) e^{f(x)}

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

 

[dextercioby]

It's straightforward.
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}

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

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

Daniel.

 

[saltydog]

It's straightforward.
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}

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

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