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

  1. Apr 21, 2008 #1
    Hey, everybody.

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

    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.
  2. jcsd
  3. Apr 21, 2008 #2


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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.
  4. Apr 21, 2008 #3
    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.
  5. Apr 21, 2008 #4
    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.
  6. Apr 21, 2008 #5
    Do you have any recommendations on sources where I might read up on this?
  7. Apr 22, 2008 #6
    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.
  8. Apr 22, 2008 #7

    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.
  9. Apr 22, 2008 #8
    If you only want to evaluate it numerically, then why don't you just use Newton-Rhapson to determine when y(x) = K ?
  10. Apr 22, 2008 #9

    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 =]
  11. Apr 22, 2008 #10
    This is a fantastic idea. Thanks!
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