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Is there a way to solve this function?

  • #1
296
0

Homework Statement



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

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

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.
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14
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...

[tex]W(x) e^{W(x)} = x[/tex]
 
  • #3
296
0
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:confused:

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

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

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:

[tex] exp\left(\frac{-x}{a}\right) \approx 1-\frac{x}{a} [/tex]

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

[tex] \frac{b}{x^{2}}=1 [/tex]

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

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

Is that correct? I feel like I'm missing something because I almost felt I had it.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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 [itex]x^2e^{-x/a}= a^2y^2e^{y}= b[/itex],
[tex]y^2e^{y}= \frac{b}{a^2}[/tex]
and so
[tex]y= BM\left(\frac{b}{a^2}\right)[/tex]
and, finally,
[tex]x= -ay= -aBM\left(\frac{b}{a^2}\right)[/itex]
is the negative root to the equation.
 
  • #5
Char. Limit
Gold Member
1,204
14
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 [itex]x^2e^{-x/a}= a^2y^2e^{y}= b[/itex],
[tex]y^2e^{y}= \frac{b}{a^2}[/tex]
and so
[tex]y= BM\left(\frac{b}{a^2}\right)[/tex]
and, finally,
[tex]x= -ay= -aBM\left(\frac{b}{a^2}\right)[/itex]
is the negative root to the equation.
I must humbly beg to differ. As you can see, Wolfram-Alpha gives a solution for x that involves only the Lambert W function and the square root function.
 

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