Erf Gaussian Error Function
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that occurs in probability, statistics, and partial differential equations describing diffusion. It is defined as:[1][2] erf ( x ) = 1 π ∫ − x x e − t erf error function ti-89 2 d t = 2 π ∫ 0 x e − t 2 excel error function erf d t . {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − gaussian error function table 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined as erfc ( x ) gaussian error function matlab = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle {\begin 2\operatorname 1 (x)&=1-\operatorname 0 (x)\\&={\frac Φ 9{\sqrt {\pi }}}\int _ Φ 8^{\infty }e^{-t^ Φ 7}\,\mathrm Φ 6 t\\&=e^{-x^ Φ 5}\operatorname Φ 4 (x),\end Φ 3}} which also defines
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erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). Another form of erfc ( x ) {\displaystyle \operatorname 2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 ) = 2 π ∫ 0 π / 2 exp ( − x 2 sin 2 θ ) d θ . {\displaystyle \operatorname 0 (x|x\geq 0)={\frac Φ 9{\pi }}\int _ Φ 8^{\pi /2}\exp \left(-{\frac Φ 7}{\sin ^ Φ 6\theta }}\right)d\theta \,.} The imaginary error function, denoted erfi, is defined as erfi ( x ) = − i erf ( i x ) = 2 π ∫ 0 x e t 2 d t = 2 π e x 2 D ( x ) , {\displaystyle {\begin Φ 0\operatorname − 9 (x)&=-i\operatorname − 8 (ix)\\&={\frac − 7{\sqrt {\pi }}}\int _ − 6^ − 5e^ − 4}\,\mathrm − 3 t\\&={\frac − 2{\sqrt {\pi }}}e^ − 1}D(x),\end − 0}} where D(x) is the Dawson function (which can be used instead of erfi to avoid arithmetic overflow[3]). Despite the name "imaginary error funct
that occurs in probability, statistics, and partial differential equations describing diffusion. It is defined as:[1][2] erf ( x ) = 1 π ∫ − x x e − t 2 d t = 2 π ∫ 0 x e − t 2 d t .
Gaussian Error Function Ti 84
{\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm erf q function − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, erf normal distribution is defined as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle https://en.wikipedia.org/wiki/Error_function {\begin 2\operatorname 1 (x)&=1-\operatorname 0 (x)\\&={\frac Φ 9{\sqrt {\pi }}}\int _ Φ 8^{\infty }e^{-t^ Φ 7}\,\mathrm Φ 6 t\\&=e^{-x^ Φ 5}\operatorname Φ 4 (x),\end Φ 3}} which also defines erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). Another form of erfc ( x ) {\displaystyle \operatorname 2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 https://en.wikipedia.org/wiki/Error_function ) = 2 π ∫ 0 π / 2 exp ( − x 2 sin 2 θ ) d θ . {\displaystyle \operatorname 0 (x|x\geq 0)={\frac Φ 9{\pi }}\int _ Φ 8^{\pi /2}\exp \left(-{\frac Φ 7}{\sin ^ Φ 6\theta }}\right)d\theta \,.} The imaginary error function, denoted erfi, is defined as erfi ( x ) = − i erf ( i x ) = 2 π ∫ 0 x e t 2 d t = 2 π e x 2 D ( x ) , {\displaystyle {\begin Φ 0\operatorname − 9 (x)&=-i\operatorname − 8 (ix)\\&={\frac − 7{\sqrt {\pi }}}\int _ − 6^ − 5e^ − 4}\,\mathrm − 3 t\\&={\frac − 2{\sqrt {\pi }}}e^ − 1}D(x),\end − 0}} where D(x) is the Dawson function (which can be used instead of erfi to avoid arithmetic overflow[3]). Despite the name "imaginary error function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usually discussed in scaled form as the Faddeeva function: w ( z ) = e − z 2 erfc ( − i z ) = erfcx ( − i z ) . {\displaystyle w(z)=e^{-z^ 6}\operatorname 5 (-iz)=\operatorname 4 (-iz).} Contents 1 The name "error function" 2 Properties 2.1 Taylor series 2.2 Derivative and integral 2.3 Bürmann series 2
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting http://math.stackexchange.com/questions/37889/why-is-the-error-function-defined-as-it-is ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Why is the error function defined as it is? up vote 35 down vote favorite 6 error function $\newcommand{\erf}{\operatorname{erf}}$ This may be a very naïve question, but here goes. The error function $\erf$ is defined by $$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt.$$ Of course, it is closely related to the normal cdf $$\Phi(x) = P(N < x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2}dt$$ (where $N \sim N(0,1)$ is a standard normal) by the expression $\erf(x) = 2\Phi(x \sqrt{2})-1$. My question is: Why is it natural or useful to define $\erf$ normalized in this way? I may be biased: as a probabilist, I gaussian error function think much more naturally in terms of $\Phi$. However, anytime I want to compute something, I find that my calculator or math library only provides $\erf$, and I have to go check a textbook or Wikipedia to remember where all the $1$s and $2$s go. Being charitable, I have to assume that $\erf$ was invented for some reason other than to cause me annoyance, so I would like to know what it is. If nothing else, it might help me remember the definition. Wikipedia says: The standard normal cdf is used more often in probability and statistics, and the error function is used more often in other branches of mathematics. So perhaps a practitioner of one of these mysterious "other branches of mathematics" would care to enlighten me. The most reasonable expression I've found is that $$P(|N| < x) = \erf(x/\sqrt{2}).$$ This at least gets rid of all but one of the apparently spurious constants, but still has a peculiar $\sqrt{2}$ floating around. probability statistics special-functions normal-distribution share|cite|improve this question asked May 8 '11 at 20:19 Nate Eldredge 49.1k356129 I had assumed it was because you can expand both $\erf(x)$ and $\erf^{-1}(x)$ in a Taylor series about $0$, while you can't with $\Phi^{-1}$. I'm not sure about the scaling with $\sqrt{2}$, though. –Mike Spivey May 8 '11 at 21:03 What about symmetry: $\text{erf}(x)$ is an odd function... –Fabian May 8 '11 at 21:34 @Fabian: But
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