Difference between revisions of "Error function"

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(Properties)
(Properties)
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<strong>Proof:</strong>  █  
 
<strong>Proof:</strong>  █  
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<strong>Theorem:</strong> The following formula holds:
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$$\dfrac{1}{2} \left( 1 + \mathrm{erf} \left( \dfrac{x-\mu}{\sqrt{2}\sigma} \right) \right)=\dfrac{1}{\sigma \sqrt{2 \pi}} \displaystyle\int_{-\infty}^x \exp \left( -\dfrac{(t-\mu)^2}{2\sigma^2} \right)dt.$$
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<strong>Proof:</strong> █
 
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Revision as of 04:26, 7 May 2015

$$\mathrm{erf}(x)=\dfrac{2}{\sqrt{\pi}}\displaystyle\int_0^x e^{-\tau^2} d\tau$$

500px

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 $\mathrm{erf}(t)$

References

Relating $\phi$ and erf