Finding the inverse of a nasty 1-to-1 function.

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Hey, everybody.

I have a function:
[tex]
\int\limits_{x}^{x+c} exp(-t^2) dt = y
[/tex]

c is a known constant here.

I am beating my head against the wall trying to find a good way to numerically evaluate the inverse here, i.e. I have y and c and I want to know x. I know that erf^-1 is readily available in mathematica and maple and the like but the limits of integration here make this a bit nastier. Any ideas? I don't need a perfect evaluation, just a moderately good approximation will work.
 

Answers and Replies

quasar987
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If you're interested only in values of x smaller than 1, then you can always Taylor expand exp(-t²) and integrate term by term and keep only the first 2 terms. I get y=c-xc+c²/2.
 
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i think the rigorous method would be to use a fourier inversion expansion; can't remember the details of how to do that - but with a smooth gaussian function i think its not that bad.
 
If you're interested only in values of x smaller than 1, then you can always Taylor expand exp(-t²) and integrate term by term and keep only the first 2 terms. I get y=c-xc+c²/2.
I wish I could say with certainty that this was the case, because you're right, that would be a good idea. Unfortunetly, I think I'll need something that covers a bit more values.
 
i think the rigorous method would be to use a fourier inversion expansion; can't remember the details of how to do that - but with a smooth gaussian function i think its not that bad.
Do you have any recommendations on sources where I might read up on this?
 
This function does not have a proper inverse as y(x)=y(-x-c)

I believe that the inverse may exist if you restrict x to be greater than zero.
 
Gib Z
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I think we can view this problem as a differential equation, and use some method of numerically solving ODE's such as the midpoint method or Runge-Kutta? Not 100% sure how it'll work out though.
 
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If you only want to evaluate it numerically, then why don't you just use Newton-Rhapson to determine when y(x) = K ?
 
Gib Z
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That involves evaluating the integral, which can't be done analytically though I do know there are many tables of data for that particular function (The error function). So Yes, I guess that method can do it numerically, good idea =]
 
If you only want to evaluate it numerically, then why don't you just use Newton-Rhapson to determine when y(x) = K ?
This is a fantastic idea. Thanks!
 

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