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===Taylor series=== The error function is an [[entire function]]; it has no singularities (except that at infinity) and its [[Taylor expansion]] always converges. For {{math|''x'' >> 1}}, however, cancellation of leading terms makes the Taylor expansion unpractical. The defining integral cannot be evaluated in [[Closed-form expression|closed form]] in terms of [[Elementary function (differential algebra)|elementary functions]] (see [[Liouville's theorem (differential algebra)|Liouville's theorem]]), but by expanding the [[integrand]] {{math|''e''<sup>β''z''<sup>2</sup></sup>}} into its [[Maclaurin series]] and integrating term by term, one obtains the error function's Maclaurin series as: <math display="block">\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}</math> which holds for every [[complex number]] {{mvar|z}}. The denominator terms are sequence [[oeis:A007680|A007680]] in the [[OEIS]]. For iterative calculation of the above series, the following alternative formulation may be useful: <math display="block">\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}</math> because {{math|{{sfrac|β(2''k'' β 1)''z''<sup>2</sup>|''k''(2''k'' + 1)}}}} expresses the multiplier to turn the {{mvar|k}}th term into the {{math|(''k'' + 1)}}th term (considering {{mvar|z}} as the first term). The imaginary error function has a very similar Maclaurin series, which is: <math display="block">\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}</math> which holds for every [[complex number]] {{mvar|z}}.
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