Difference between revisions of "Taylor series for error function"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+1}}{k!(2k+1)},$$ | $$\mathrm{erf}(z) = \dfrac{2}{\sqrt{\pi}} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+1}}{k!(2k+1)},$$ | ||
| − | where $\mathrm{erf}$ denotes the [[error function]] and $\pi$ denotes [[pi]]. | + | 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== | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=}}: $7.1.5$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
| + | [[Category:Justify]] | ||
Latest revision as of 04:16, 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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): $7.1.5$
