defining inverse error function

Question

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.

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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

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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?

Answer 1

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 $y(0)=0$.

Answer 2

As a Special Case $\def \erf{\operatorname{erf}}$

Let’s use a more general inverse function implemented called Inverse Gamma Regularized which existed before $2000$ in Mathematica, so it is a well established function. It finds the quantile of a cdf based on Gamma Regularized $Q(a,z)$. There also exists Gamma P Inverse for the inverse with respect to $z$ for $\text P(a,z)=1-Q(a,z)$. The inverse functions are quantile functions for Incomplete Gamma function based distributions and $\erf ^{-1}$ is the quantile function for the Normal Distribution:

$$Q(a,z)=\frac{\int_x^\infty t^{a-1}e^{-t}dt}{\int_0^\infty t^{a-1}e^{-t}dt}=y\mathop\implies^{a,z\ge0} z=Q^{-1}(a,y)$$

$$\text P(a,z)=\frac{\int_0^x t^{a-1}e^{-t}dt}{\int_0^\infty t^{a-1}e^{-t}dt}=y\mathop\implies^{a,z\ge0} z=\text{gammapinv}(a,y)=Q^{-1}(a,1-x)\ \ \ \text {”=“}\ \ \ \text P^{-1}(a,x)$$

Notice that:

$$1-Q\left(\frac12,y^2\right)=\text P\left(\frac12,y^2\right)=\text{erf}(y)=x\implies y=\boxed{\sqrt{Q^{-1}\left(\frac12,1-x\right)}=\sqrt{\text{gammapinv}\left(\frac12,x\right)}\mathop=^{x\ge0}\erf^{-1}(x)}$$

Here is a graph:

These identities give another special case of $Q^{-1}(a,z)$. Please correct me and give me feedback!

Other Differential Equation:

Here is a simpler differential equation based off of @Fraklin Pezzuti Dyer’s solution:

$$y=\erf^{-1}(x):y’= \frac{\sqrt\pi}2e^{y^2} \implies y’’= \underbrace{\frac{\sqrt\pi}2e^{y^2}}_{y’}yy’\implies \boxed{y’’=2y’^2y\iff y=\text{erf}^{-1}(c_0x+c_1)}$$

Integral Representation:

The following is experimental, but

$$\erf^{-1}(x)=\frac{\lim\limits_{m\to\infty}}2\int_{-1}^1\erf^{-1}(t)\csc\left(\frac\pi2(t-x)\right)\sin\left(\pi m(t-x)\right)dt=\frac{\lim\limits_{m\to\infty}}{\sqrt\pi}\int_{-\infty}^\infty te^{-t^2}\csc\left(\frac\pi2(\erf(t)-x)\right)\sin\left(\pi m(\erf(t)-x)\right)dt,|x|<1 $$

shown here