The inverse function of x exp(-1/x^2)

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Dear All,

Is it possible to have an analytical inverse function of
[tex]
y=x e^{-\frac{1}{x^2}}.
[/tex]

Since y is monotonously increasing, its inverse function exists. But is it possible to get a close form? Thanks a lot!

Phonic
 

Answers and Replies

benorin
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Well, you can expand the function into an infinite series and then use series reversion.
 
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It looks like you found a case for the ProductLog function, which is a common special function defined as the principal solution for y in [tex] x= y e^y [/tex].

In other words there is no answer in terms of elementary functions.
 
arildno
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Well, you can expand the function into an infinite series and then use series reversion.
Just a comment added to this:
You have an essential singularity at x=0, so since you haven't got any power series representation of the function at that point, you'll certainly not be able to revert that series in that particular neighbourhood.

However, it should be perfectly possible to use series reversion about any point in the function's domain where it is analytic.
 
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I am myself looking for a similar answer.
I came to this question while looking for the fourier transform of causal impulse
responses. In what context did you come to yours?

I may have an answer for you: the Lambert W funtion may have the solution
you are looking for...
 
disregardthat
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By some manipulation without care of domain or codomain we get [tex]y = \sqrt{\frac{2}{W(2x^{-2})}}[/tex]. I believe it is a proper inverse for x > 0.
 
Last edited:
arildno
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I am myself looking for a similar answer.
I came to this question while looking for the fourier transform of causal impulse
responses. In what context did you come to yours?

I may have an answer for you: the Lambert W funtion may have the solution
you are looking for...
Since OP haven't logged in since december 2007, I'm not sure he is going to appreciate your answer. :smile:
 
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I find these so interesting as well as beautiful but no one else seems to and I don't know why:

[tex]y^2=x^2e^{-2/x^2}[/tex]

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

[tex]\frac{2}{y^2}=\frac{2}{x^2} e^{2/x^2}[/tex]

[tex]\frac{2}{x^2}=W\left(2/y^2\right)[/tex]

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

where the root symbol implies it's multi-valued version. That last expression is doubly-infinitely valued for all (finite and complex) y except 0 and represents a beautifully intricate geometric object in the complex plane. You guys wouldn't get off so easy if this was my class. :)

And also, he asked for an "analytical" inverse and I think I could argue the expression for x above is perfectly analytical except y=0 and when W(2/y^2)=0.
 
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