Can the Function x^2 * exp(-x/a) = b Be Solved for x Using Lambert W?

[Beer-monster]

Homework Statement

It could be the late night, but a problem I've been doing results in a function of the form

x^{2}exp\left(\frac{-x}{a}\right) = b

Where a and b are constants. I'm not sure how to go about solving for x.

Homework Equations

The Attempt at a Solution

I've trying taking natural logarithms of both sides but that doesn't seem to help me too much.

Any help would be appreciated. I get the funny feeling I've seen this before.

 

[Char. Limit]

You require the Lambert W function, defined as the inverse of the function x e^x. That is, if W(x) is the Lambert W function...

W(x) e^{W(x)} = x

 

[Beer-monster]

Thanks for the reply.

I looked up the Lambert W function, but not quite sure how to apply it in my circumstances. Most of the stuff I find online on the topic seems a little opaque to me

I did try another method which is to rearrange the equation into:

exp\left(\frac{-x}{a}\right) = \frac{b}{x^{2}}

Now, since I'm only interested in a specific value of x (x=X) corresponding to a saddle point (the above is the result of differentiating and setting to zero) I thought I could use the Taylor series of the exponential function and take the first term. So:

exp\left(\frac{-x}{a}\right) \approx 1-\frac{x}{a}

This means I can take the first term (1) and then rearrange to solve for X.

\frac{b}{x^{2}}=1

However, I realize now that this only applies for x = X = 0. So, as far as I can tell that means I'm back to:

exp\left(\frac{-X}{a}\right) = \frac{b}{X^{2}}

Is that correct? I feel like I'm missing something because I almost felt I had it.

 

[HallsofIvy]

Unfortunately, with that extra "x" in there, you cannot solve this equation with the Lambert W function, for the same reasons you cannot solve [math]xe^x= b[/math] with the logarithm function. Of course, we could define the "Beer-Monster", BM(x), to be the "inverse function to [math]f(x)= x^2e^{x}[/math] (well,for x> 0 we can- this function is not one-to-one for all x and so does not have a true "inverse").

If y= -x/a, then x= -ay so x^2e^{-x/a}= a^2y^2e^{y}= b,
y^2e^{y}= \frac{b}{a^2}
and so
y= BM\left(\frac{b}{a^2}\right)
and, finally,
x= -ay= -aBM\left(\frac{b}{a^2}\right)[/itex]<br /> is the negative root to the equation.

 

[Char. Limit]

Unfortunately, with that extra "x" in there, you cannot solve this equation with the Lambert W function, for the same reasons you cannot solve [math]xe^x= b[/math] with the logarithm function. Of course, we could define the "Beer-Monster", BM(x), to be the "inverse function to [math]f(x)= x^2e^{x}[/math] (well,for x> 0 we can- this function is not one-to-one for all x and so does not have a true "inverse").

If y= -x/a, then x= -ay so x^2e^{-x/a}= a^2y^2e^{y}= b,
y^2e^{y}= \frac{b}{a^2}
and so
y= BM\left(\frac{b}{a^2}\right)
and, finally,
x= -ay= -aBM\left(\frac{b}{a^2}\right)[/itex]<br /> is the negative root to the equation.

<br /> <br /> I must humbly beg to differ. <a href="http://www.wolframalpha.com/input/?i=x^2+e^(-x/a)+%3D+b" target="_blank" class="link link--external" rel="nofollow ugc noopener">As you can see</a>, Wolfram-Alpha gives a solution for x that involves only the Lambert W function and the square root function.