Normal Error Integral
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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 2 d t = 2 π
Integral Of Error Function
∫ 0 x e − t 2 d t . {\displaystyle {\begin − 6\operatorname error function calculator − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _
Error Function Table
9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2 π inverse error function ∫ 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 erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). error function matlab 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 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
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Erf(inf)
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Error Function Python
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 https://en.wikipedia.org/wiki/Error_function 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 $\newcommand{\erf}{\operatorname{erf}}$ 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 $$\Phi(x) = P(N < x) http://math.stackexchange.com/questions/37889/why-is-the-error-function-defined-as-it-is = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2}dt$$ (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 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 const
Tour Start 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 http://stats.stackexchange.com/questions/187828/how-are-the-error-function-and-standard-normal-distribution-function-related the company Business Learn more about hiring developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. 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 error function to the top How are the Error Function and Standard Normal distribution function related? up vote 3 down vote favorite If the Standard Normal PDF is $$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$ and the CDF is $$F(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-x^2/2}\mathrm{d}x\,,$$ how does this turn into an error function of $z$? normal-distribution cdf share|improve this question edited Dec 22 '15 at 14:59 whuber♦ 145k18284544 asked Dec 21 '15 at 22:44 TH4454 1019 johndcook.com/erf_and_normal_cdf.pdf normal error integral –Mark L. Stone Dec 21 '15 at 22:54 I saw this, but it starts with ERF already defined. –TH4454 Dec 21 '15 at 22:57 Well, there's a definition of erf and a definition of the Normal CDF.. The relations, derivable by some routine calculations, are shown as to how to convert between them, and how to convert between their inverses. –Mark L. Stone Dec 21 '15 at 23:43 Sorry, I don't see many of the details. For example, the CDF is from -Inf to x. So how does the ERF go from 0 to x? –TH4454 Dec 22 '15 at 0:13 Are you familiar with the calculus technique of change of variable? If not, learn how to do it. –Mark L. Stone Dec 22 '15 at 0:19 add a comment| 1 Answer 1 active oldest votes up vote 3 down vote accepted Because this comes up often in some systems (for instance, Mathematica insists on expressing the Normal CDF in terms of $\text{Erf}$), it's good to have a thread like this that documents the relationship. By definition, the Error Function is $$\text{Erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} \mathrm{d}t.$$ Writing $t^2 = z^2/2$ implies $t = z / \sqrt{2}$ (because $t$ is not negative), whence $\mathrm{d}t = \mathrm{d}z/\sqrt{2}$. The endpoints $t=0$ and $t=x$ be