Revision as of 17:25, 23 May 2016

The error function $\mathrm{erf}$ is defined by $$\mathrm{erf}(x)=\dfrac{2}{\sqrt{\pi}}\displaystyle\int_0^x e^{-\tau^2} d\tau.$$

Properties

Theorem

The following formula holds: $$\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+1}}{k!(2k+1)},$$ where $\mathrm{erf}$ denotes the error function and $\pi$ denotes pi, and $k!$ denotes the factorial.

Proof

From the Taylor series of the exponential function, $$e^{-\tau^2}=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-\tau^2)^k}{k!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k}}{k!}.$$ So, integrating term by term (justify this), $$\displaystyle\int_0^x e^{-\tau^2} \mathrm{d}\tau = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)}.$$ Therefore multiplying by $\dfrac{2}{\sqrt{\pi}}$ and comparing to the definition of the error function, we get $$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)},$$ as was to be shown.

References

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): $7.1.5$

Theorem

The following formula holds: $$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}}e^{-z^2}\displaystyle\sum_{k=0}^{\infty} \dfrac{2^k}{1 \cdot 3 \cdot \ldots \cdot (2k+1)} z^{2k+1},$$ where $\mathrm{erf}$ denotes the error function and $\pi$ denotes pi.

Proof

References

Videos

The Laplace transform of the error function $\mathrm{erf}(t)$

References

Relating $\phi$ and erf

Error function

Complementary $\mathrm{erf}$

Erfi

Inverse $\mathrm{erf}$