Difference between revisions of "Error function is odd"
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==Proof== | ==Proof== | ||
| + | From the definition, | ||
| + | $$\mathrm{erf}(x)=\dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^x e^{-\tau^2} \mathrm{d}\tau.$$ | ||
| + | So, | ||
| + | $$\begin{array}{ll} | ||
| + | \mathrm{erf}(-x) &= \dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^{-x} e^{-\tau^2} \mathrm{d}\tau \\ | ||
| + | &\stackrel{u=-\tau}{=} \dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^{x} e^{-u^2} \mathrm{d}u, | ||
| + | \end{array}$$ | ||
| + | proving that $\mathrm{erf}$ is odd. $\blacksquare$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
==References== | ==References== | ||
Revision as of 03:41, 28 March 2017
Theorem
The following formula holds: $$\mathrm{erf}(-z)=-\mathrm{erf}(z),$$ where $\mathrm{erf}$ denotes the error function (i.e. $\mathrm{erf}$ is an odd function).
Proof
From the definition, $$\mathrm{erf}(x)=\dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^x e^{-\tau^2} \mathrm{d}\tau.$$ So, $$\begin{array}{ll} \mathrm{erf}(-x) &= \dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^{-x} e^{-\tau^2} \mathrm{d}\tau \\ &\stackrel{u=-\tau}{=} \dfrac{2}{\sqrt{\pi}} \displaystyle\int_0^{x} e^{-u^2} \mathrm{d}u, \end{array}$$ proving that $\mathrm{erf}$ is odd. $\blacksquare$
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 7.1.9
