Difference between revisions of "Series for erf with exponential factored out"

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==Theorem==
<strong>[[Series for erf with exponential factored out|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}}e^{-z^2}\displaystyle\sum_{k=0}^{\infty} \dfrac{2^k}{1 \cdot 3 \cdot \ldots \cdot (2k+1)} z^{2k+1},$$
 
$$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}}e^{-z^2}\displaystyle\sum_{k=0}^{\infty} \dfrac{2^k}{1 \cdot 3 \cdot \ldots \cdot (2k+1)} z^{2k+1},$$
 
where $\mathrm{erf}$ denotes the [[error function]] and $\pi$ denotes [[pi]].
 
where $\mathrm{erf}$ denotes the [[error function]] and $\pi$ denotes [[pi]].
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<strong>Proof:</strong> █
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==Proof==
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 03:54, 3 October 2016

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