2
$\begingroup$

The error function is define like this: $$\operatorname{erf}(x):=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$

If i take the derivative i get $$\operatorname{erf'}(x)=\frac{2}{\sqrt\pi}e^{-x^2}$$because say $$\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt=\frac{2}{\sqrt{\pi}}\left(G(x)-G(0)\right)$$and $G(0)$ is constant.


Because $\forall x,\frac{2}{\sqrt\pi}e^{-x^2}>0$ we know that $\operatorname{erf}(x)$ is strictly increasing, so it suppose to have an inverse function.

I can understand few of things about $\operatorname{erf^{-1}}(x)$:

  • the domain is $\operatorname{erf}((-\infty,\infty))=(-1,1)$

  • the codomain is $(-\infty,\infty)$

  • 3 points:

    • $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1\implies\lim\limits_{x\to1}\operatorname{erf^{-1}}(x)=\infty$
    • $\operatorname{erf}(0)=0\implies \operatorname{erf^{-1}}(0)=0$
    • $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1\implies\lim\limits_{x\to-1}\operatorname{erf^{-1}}(x)=-\infty$
  • it is strictly increasing


But apart from those 4 things I can't think about any other information.

So my question is the following: Is there a way to define inverse error function using combination of known functions and transforms? And if not, what more can we know about the inverse function?

$\endgroup$
  • $\begingroup$ You might look at L. Carlitz, The inverse of the error function, Pacific J. Math. Volume 13, Number 2 (1963), 459-470, for some properties of the Maclaurin series $$\text{erf}^{-1}(x) = \frac{\sqrt{\pi} x}{2} + \frac{\pi^{3/2} x^3}{24} + \frac{7 \pi^{5/2} x^5}{960} + \ldots$$ $\endgroup$ – Robert Israel Dec 20 '17 at 22:44
0
$\begingroup$

Recall that $$\frac{d}{dx}f^{-1}(x)=\frac{1}{(f'\circ f^{-1})(x)}$$ and so we have $$\frac{d}{dx}\text{erf}^{-1}(x)=\frac{1}{(\text{erf}'\circ \text{erf}^{-1} )(x)}$$ or $$\frac{d}{dx}\text{erf}^{-1}(x)=\frac{\sqrt \pi}{2}e^{(\text{erf}^{-1}(x))^2}$$ and so the inverse error function can be defined by the differential equation $$y'=\frac{\sqrt\pi}{2}\exp (y^2)$$ with initial condition $y(0)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.