Inverse Error Function and Inverse Complementary Error Function Calculator

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y (dimensionless) erf⁻¹ needs -1 < y < 1; erfc⁻¹ needs 0 < y < 2

Display digits 6 10 14 18 22 26 30 34 38 42 46 50

Formula

Show calculation steps (1)

1. Inverse Complementary Error Function

   

   Uses the identity erfc^-1(y) = erf^-1(1 - y); returns x such that erfc(x) = y.

Results

Inverse Error Function erf^-1(y)

0.2724628197

value x where erf(x) = y

erfc-1(y) = erf-1(1 - y)	0.7328691494
Input y	0.3

What is the Inverse Error Function Calculator?

This tool computes the inverse error function, \(\operatorname{erf}^{-1}(y)\), and the inverse complementary error function, \(\operatorname{erfc}^{-1}(y)\), for a given dimensionless argument \(y\). The error function \(\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}}\, dt\) arises throughout probability, diffusion, and signal-processing problems. Its inverse answers the reverse question: given a value \(y\), what \(x\) produces \(\operatorname{erf}(x) = y\)?

How to use it

Enter a value \(y\) and choose how many display digits you want. For \(\operatorname{erf}^{-1}\) the valid domain is \(-1 < y < 1\); for \(\operatorname{erfc}^{-1}\) the valid domain is \(0 < y < 2\). The two outputs are linked by the identity \(\operatorname{erfc}^{-1}(y) = \operatorname{erf}^{-1}(1 - y)\), because \(\operatorname{erfc}(x) = 1 - \operatorname{erf}(x)\). At the boundaries the results diverge to \(\pm\infty\) (for example \(\operatorname{erf}^{-1}(1) = +\infty\)).

The formula explained

There is no elementary closed form for \(\operatorname{erf}^{-1}\). We start from the Giles (2010) rational approximation seed, accurate to about 1e-7, then refine with Newton's method: $$x_{n+1} = x_{n} - \frac{\operatorname{erf}(x_{n}) - y}{\frac{2}{\sqrt{\pi}} e^{-x_{n}^{2}}}.$$ The derivative of erf is \(\frac{2}{\sqrt{\pi}} e^{-x^{2}}\). Iterating until the residual \(|\operatorname{erf}(x_{n}) - y|\) is below 1e-15 yields full double precision.

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erf^-1 is defined on (-1,1); erfc^-1 is defined on (0,2) and equals erf^-1(1-y).

The inverse error function finds the x whose erf equals a given y.

Worked example

For \(y = 0.3\): \(\operatorname{erf}^{-1}(0.3) \approx 0.2724627\) because \(\operatorname{erf}(0.2724627) \approx 0.3\). Then $$\operatorname{erfc}^{-1}(0.3) = \operatorname{erf}^{-1}(1 - 0.3) = \operatorname{erf}^{-1}(0.7) \approx 0.7328691,$$ so that \(\operatorname{erfc}(0.7328691) = 1 - 0.7 = 0.3\).

FAQ

Why are \(\operatorname{erf}^{-1}\) and \(\operatorname{erfc}^{-1}\) equal at \(y = 0.5\)? Because \(\operatorname{erfc}^{-1}(0.5) = \operatorname{erf}^{-1}(1 - 0.5) = \operatorname{erf}^{-1}(0.5)\); the arguments coincide only at \(y = 0.5\).

What happens at the domain edges? \(\operatorname{erf}^{-1}(\pm 1) = \pm\infty\) and \(\operatorname{erfc}^{-1}(0) = +\infty\), \(\operatorname{erfc}^{-1}(2) = -\infty\). Out-of-domain inputs return an error.

Is \(\operatorname{erf}^{-1}\) odd? Yes: \(\operatorname{erf}^{-1}(-y) = -\operatorname{erf}^{-1}(y)\).