Difference between revisions of "Taylor series for error function"

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==Proof==
 
==Proof==
 +
From the [[Taylor series of the exponential function]],
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$$e^{-\tau^2}=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-\tau^2)^k}{k!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k}}{k!}.$$
 +
So, integrating term by term (justify this),
 +
$$\displaystyle\int_0^x e^{-\tau^2} \mathrm{d}\tau = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)}.$$
 +
Therefore multiplying by $\dfrac{2}{\sqrt{\pi}}$ and comparing to the definition of the error function, we get
 +
$$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)},$$
 +
as was to be shown.
  
 
==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=}}: 7.1.5
+
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=}}: $7.1.5$
 +
 
 +
[[Category:Theorem]]
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[[Category:Proven]]

Revision as of 03:59, 3 October 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

From the Taylor series of the exponential function, $$e^{-\tau^2}=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-\tau^2)^k}{k!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k}}{k!}.$$ So, integrating term by term (justify this), $$\displaystyle\int_0^x e^{-\tau^2} \mathrm{d}\tau = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)}.$$ Therefore multiplying by $\dfrac{2}{\sqrt{\pi}}$ and comparing to the definition of the error function, we get $$\mathrm{erf}(z)=\dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k \tau^{2k+1}}{k! (2k+1)},$$ as was to be shown.

References