Difference between revisions of "Taylor series for error function"

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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\mathrm{erf}(z) = \dfrac{2}{...")
 
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<strong>[[Taylor series for error function|Theorem]]:</strong> The following formula holds:
 
<strong>[[Taylor series for error function|Theorem]]:</strong> The following formula holds:
 
$$\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+1}}{k!(2k+1)},$$
 
$$\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+1}}{k!(2k+1)},$$
where $\mathrm{erf}$ denotes the [[error function]] and $\pi$ denotes [[pi]].
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where $\mathrm{erf}$ denotes the [[error function]] and $\pi$ denotes [[pi]], and $k!$ denotes the [[factorial]].
 
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<strong>Proof:</strong>  █  
 
<strong>Proof:</strong>  █  
 
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Revision as of 17:24, 23 May 2016

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: