Difference between revisions of "Error function"
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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
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| + | <strong>Theorem:</strong> The following formula holds: | ||
| + | $$\dfrac{1}{2} \left( 1 + \mathrm{erf} \left( \dfrac{x-\mu}{\sqrt{2}\sigma} \right) \right)=\dfrac{1}{\sigma \sqrt{2 \pi}} \displaystyle\int_{-\infty}^x \exp \left( -\dfrac{(t-\mu)^2}{2\sigma^2} \right)dt.$$ | ||
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| + | <strong>Proof:</strong> █ | ||
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Revision as of 04:26, 7 May 2015
$$\mathrm{erf}(x)=\dfrac{2}{\sqrt{\pi}}\displaystyle\int_0^x e^{-\tau^2} d\tau$$
Properties
Theorem: $\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2n+1}}{n!(2n+1)}$
Proof: █
Theorem: $\mathrm{erf}(-z)=-\mathrm{erf}(z)$
Proof: █
Theorem: $\mathrm{erf}(\overline{z}) = \overline{\mathrm{erf}}(z)$
Proof: █
Theorem: The following formula holds: $$\dfrac{1}{2} \left( 1 + \mathrm{erf} \left( \dfrac{x-\mu}{\sqrt{2}\sigma} \right) \right)=\dfrac{1}{\sigma \sqrt{2 \pi}} \displaystyle\int_{-\infty}^x \exp \left( -\dfrac{(t-\mu)^2}{2\sigma^2} \right)dt.$$
Proof: █
Videos
The Laplace transform of the error function $\mathrm{erf}(t)$
