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

In summary, the function does not have a proper inverse as y(x)=y(-x-c) and the problem may be a differential equation that can be solved numerically with methods such as the midpoint method or Runge-Kutta.
  • #1
jforeman83
4
0
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.
 
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  • #2
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.
 
  • #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.
 
  • #4
quasar987 said:
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.
 
  • #5
lzkelley said:
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?
 
  • #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.
 
  • #7
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.
 
  • #8
If you only want to evaluate it numerically, then why don't you just use Newton-Rhapson to determine when y(x) = K ?
 
  • #9
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 =]
 
  • #10
daudaudaudau said:
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!
 

1. How do I find the inverse of a 1-to-1 function?

To find the inverse of a 1-to-1 function, you need to switch the x and y variables and solve for y. This will give you the inverse function.

2. What is a 1-to-1 function?

A 1-to-1 function is a function in which each input value has only one unique output value. In other words, each x-value has only one corresponding y-value.

3. Can all functions have an inverse?

No, not all functions have an inverse. For a function to have an inverse, it must be 1-to-1, meaning each input value has only one unique output value.

4. How do I know if a function is 1-to-1?

To determine if a function is 1-to-1, you can use the horizontal line test. If a horizontal line can intersect the graph of the function at more than one point, then the function is not 1-to-1.

5. Why is it important to find the inverse of a 1-to-1 function?

Finding the inverse of a 1-to-1 function allows you to "undo" the function and find the original input value given an output value. This can be useful in many real-world applications, such as solving equations and finding inverse relationships between variables.

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