The property means that the error function is an
odd function. This directly results from the fact that the integrand is an
even function (the antiderivative of an even function which is zero at the origin is an odd function and vice versa). Since the error function is an
entire function which maps real numbers to real numbers, for any
complex number : \operatorname{erf}(\overline{z}) = \overline{\operatorname{erf}(z)} where \overline{z} denotes the
complex conjugate of z. The integrand and are shown in the complex -plane in the figures at right with
domain coloring. The error function at is exactly 1 (see
Gaussian integral). At the real axis, approaches unity at and −1 at . At the imaginary axis, it tends to .
Taylor series The error function is an
entire function; it has no singularities (except that at infinity) and its
Taylor expansion always converges. For , however, cancellation of leading terms makes the Taylor expansion impractical. The defining integral cannot be evaluated in
closed form in terms of
elementary functions (see
Liouville's theorem), but by expanding the
integrand into its
Maclaurin series, integrating term by term, and using the fact that \operatorname{erf}(0)=0, one obtains the error function's Maclaurin series as: \begin{align} \operatorname{erf}(z) &= \frac{2}{\sqrt\pi}\sum_{n=0}^\infty\frac{(-1)^n z^{2n+1}}{n! (2n+1)} \\[6pt] &= \frac{2}{\sqrt\pi} \left(z-\frac{z^3}{3}+\frac{z^5}{10}-\frac{z^7}{42}+\frac{z^9}{216}-\cdots\right) \end{align} which holds for every
complex number . The denominator terms are sequence
A007680 in the
OEIS. It is a special case of
Kummer's function: \operatorname{erf}(z) = \frac{2z}{\sqrt\pi}\,{}_1F_1(1/2;3/2;-z^2). For iterative calculation of the above series, the following alternative formulation may be useful: \begin{align} \operatorname{erf}(z) &= \frac{2}{\sqrt\pi}\sum_{n=0}^\infty\left(z \prod_{k=1}^n {\frac{-(2k-1) z^2}{k (2k+1)}}\right) \\[6pt] &= \frac{2}{\sqrt\pi} \sum_{n=0}^\infty \frac{z}{2n+1} \prod_{k=1}^n \frac{-z^2}{k} \end{align} because expresses the multiplier to turn the th term into the th term (considering as the first term). The imaginary error function has a very similar Maclaurin series, which is: \begin{align} \operatorname{erfi}(z) &= \frac{2}{\sqrt\pi}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)} \\[6pt] &=\frac{2}{\sqrt\pi} \left(z+\frac{z^3}{3}+\frac{z^5}{10}+\frac{z^7}{42}+\frac{z^9}{216}+\cdots\right) \end{align} which holds for every
complex number .
Derivative and integral The derivative of the error function follows immediately from its definition: \frac{d}{dz}\operatorname{erf}(z) =\frac{2}{\sqrt\pi} e^{-z^2}. From this, the derivative of the imaginary error function is also immediate: \frac{d}{dz}\operatorname{erfi}(z) =\frac{2}{\sqrt\pi} e^{z^2}.Higher order derivatives are given by \operatorname{erf}^{(k)}(z) = \frac{2 (-1)^{k-1}}{\sqrt\pi} \mathit{H}_{k-1}(z) e^{-z^2} = \frac{2}{\sqrt\pi} \frac{d^{k-1}}{dz^{k-1}} \left(e^{-z^2}\right),\qquad k=1, 2, \dots where are the physicists'
Hermite polynomials. An
antiderivative of the error function, obtainable by
integration by parts, is \int \operatorname{erf}(z) dz = z\operatorname{erf}(z) + \frac{e^{-z^2}}{\sqrt\pi}+C. An antiderivative of the imaginary error function, also obtainable by integration by parts, is \int \operatorname{erfi}(z) dz = z\operatorname{erfi}(z) - \frac{e^{z^2}}{\sqrt\pi}+C.
Bürmann series An expansion which converges more rapidly for all real values of than a Taylor expansion is obtained by using
Hans Heinrich Bürmann's theorem: \begin{align} \operatorname{erf}(x) &= \frac{2}{\sqrt\pi} \sgn(x) \cdot \sqrt{1-e^{-x^2}} \left( 1-\frac{1}{12} \left (1-e^{-x^2} \right ) -\frac{7}{480} \left (1-e^{-x^2} \right )^2 -\frac{5}{896} \left (1-e^{-x^2} \right )^3-\frac{787}{276 480} \left (1-e^{-x^2} \right )^4 - \cdots \right) \\[10pt] &= \frac{2}{\sqrt\pi} \sgn(x) \cdot \sqrt{1-e^{-x^2}} \left(\frac{\sqrt\pi}{2} + \sum_{k=1}^\infty c_k e^{-kx^2} \right) \end{align} where is the
sign function. By keeping only the first two coefficients and choosing and , the resulting approximation shows its largest
relative error at , where it is less than 0.0034361: \operatorname{erf}(x) \approx \frac{2}{\sqrt\pi}\sgn(x) \cdot \sqrt{1-e^{-x^2}} \left(\frac{\sqrt{\pi}}{2} + \frac{31}{200}e^{-x^2}-\frac{341}{8000} e^{-2x^2}\right).
Inverse functions Given a complex number , there is not a
unique complex number satisfying , so a true inverse function would be multivalued. However, for , there is a unique
real number denoted satisfying \operatorname{erf}\left(\operatorname{erf}^{-1}(x)\right) = x. The
inverse error function is usually defined with domain , and it is restricted to this domain in many
computer algebra systems. However, it can be extended to the disk of the complex plane, using the Maclaurin series \operatorname{erf}^{-1}(z)=\sum_{k=0}^\infty\frac{c_k}{2k+1}\left (\frac{\sqrt\pi}{2}z\right )^{2k+1}, where and \begin{align} c_k & =\sum_{m=0}^{k-1}\frac{c_m c_{k-1-m}}{(m+1)(2m+1)} \\[1ex] &= \left\{1,1,\frac{7}{6},\frac{127}{90},\frac{4369}{2520},\frac{34807}{16200},\ldots\right\}. \end{align} So we have the series expansion (common factors have been canceled from numerators and denominators): \operatorname{erf}^{-1}(z) = \frac{\sqrt{\pi}}{2} \left (z + \frac{\pi}{12}z^3 + \frac{7\pi^2}{480}z^5 + \frac{127\pi^3}{40320}z^7 + \frac{4369\pi^4}{5806080} z^9 + \frac{34807\pi^5}{182476800}z^{11} + \cdots\right ). (After cancellation the numerator and denominator values in and respectively; without cancellation the numerator terms are values in .) The error function's value at is equal to . For , we have . The
inverse complementary error function is defined as \operatorname{erfc}^{-1}(1-z) = \operatorname{erf}^{-1}(z). For real , there is a unique
real number satisfying . The
inverse imaginary error function is defined as . For any real
x,
Newton's method can be used to compute , and for , the following Maclaurin series converges: \operatorname{erfi}^{-1}(z) =\sum_{k=0}^\infty\frac{(-1)^k c_k}{2k+1} \left( \frac{\sqrt\pi}{2} z \right)^{2k+1}, where is defined as above.
Asymptotic expansion A useful
asymptotic expansion of the complementary error function (and therefore also of the error function) for large real is \begin{align} \operatorname{erfc}(x) &= \frac{e^{-x^2}}{x\sqrt{\pi}}\left(1 + \sum_{n=1}^\infty (-1)^n \frac{1\cdot3\cdot5\cdots(2n - 1)}{\left(2x^2\right)^n}\right) \\[6pt] &= \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{\left(2x^2\right)^n}, \end{align} where is the
double factorial of , which is the product of all odd numbers up to . This series diverges for every finite , and its meaning as asymptotic expansion is that for any
integer one has \operatorname{erfc}(x) = \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^{N-1} (-1)^n \frac{(2n - 1)!!}{\left(2x^2\right)^n} + R_N(x) where the remainder is R_N(x) := \frac{(-1)^N \, (2 N - 1)!!}{\sqrt{\pi} \cdot 2^{N - 1}} \int_x^\infty t^{-2N}e^{-t^2}\, dt, which follows easily by induction, writing e^{-t^2} = -\frac{1}{2 t} \, \frac{d}{dt} e^{-t^2} and integrating by parts. The asymptotic behavior of the remainder term, in
Landau notation, is R_N(x) = O\left(x^{- (1 + 2N)} e^{-x^2}\right) as . This can be found by R_N(x) \propto \int_x^\infty t^{-2N}e^{-t^2}\, dt = e^{-x^2} \int_0^\infty (t+x)^{-2N}e^{-t^2-2tx}\,dt\leq e^{-x^2} \int_0^\infty x^{-2N} e^{-2tx}\,dt \propto x^{-(1+2N)}e^{-x^2}. For large enough values of , only the first few terms of this asymptotic expansion are needed to obtain a good approximation of (while for not too large values of , the above Taylor expansion at 0 provides a very fast convergence).
Continued fraction expansion A
continued fraction expansion of the complementary error function was found by
Laplace: \operatorname{erfc}(z) = \frac{z}{\sqrt\pi}e^{-z^2} \cfrac{1}{z^2+ \cfrac{a_1}{1+\cfrac{a_2}{z^2+ \cfrac{a_3}{1+\dotsb}}}},\qquad a_m = \frac{m}{2}.
Factorial series The inverse
factorial series: \begin{align} \operatorname{erfc}(z) &= \frac{e^{-z^2}}{\sqrt{\pi}\,z} \sum_{n=0}^\infty \frac{\left(-1\right)^n Q_n}{{\left(z^2+1\right)}^{\bar{n}}} \\[1ex] &= \frac{e^{-z^2}}{\sqrt{\pi}\,z} \left[1 -\frac{1}{2}\frac{1}{(z^2+1)} + \frac{1}{4}\frac{1}{\left(z^2+1\right) \left(z^2+2\right)} - \cdots \right] \end{align} converges for . Here \begin{align} Q_n &\overset{\text{def}}{{}={}} \frac{1}{\Gamma{\left(\frac{1}{2}\right)}} \int_0^\infty \tau(\tau-1)\cdots(\tau-n+1)\tau^{-\frac{1}{2}} e^{-\tau} \,d\tau \\[1ex] &= \sum_{k=0}^n \frac{s(n,k)}{2^{\bar{k}}}, \end{align} denotes the
rising factorial, and denotes a signed
Stirling number of the first kind. The Taylor series can be written in terms of the
double factorial: \operatorname{erf}(z) = \frac{2}{\sqrt\pi} \sum_{n=0}^\infty \frac{(-2)^n(2n-1)!!}{(2n+1)!}z^{2n+1} == Bounds and numerical approximations ==