Error Function From 0 To Infinity
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Random Entry New in MathWorld MathWorld Classroom About MathWorld Contribute to MathWorld Send a Message to the Team MathWorld Book Wolfram Web Resources» 13,594 entries Last updated: Tue Sep 27 integral of delta function from 0 to infinity 2016 Created, developed, and nurturedbyEricWeisstein at WolframResearch Calculus and Analysis>Special Functions>Erf>
Error Function Values
Calculus and Analysis>Complex Analysis>Entire Functions> Interactive Entries>webMathematica Examples> More... History and Terminology>Wolfram Language Commands> MathWorld Contributors>D'Orsogna> Less... integral of error function Erf is the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function). It is an entire function defined by (1) erf function calculator Note that some authors (e.g., Whittaker and Watson 1990, p.341) define without the leading factor of . Erf is implemented in the Wolfram Language as Erf[z]. A two-argument form giving is also implemented as Erf[z0, z1]. Erf satisfies the identities (2) (3) (4) where is erfc, the complementary error function, and is a confluent hypergeometric function
Error Function Table
of the first kind. For , (5) where is the incomplete gamma function. Erf can also be defined as a Maclaurin series (6) (7) (OEIS A007680). Similarly, (8) (OEIS A103979 and A103980). For , may be computed from (9) (10) (OEIS A000079 and A001147; Acton 1990). For , (11) (12) Using integration by parts gives (13) (14) (15) (16) so (17) and continuing the procedure gives the asymptotic series (18) (19) (20) (OEIS A001147 and A000079). Erf has the values (21) (22) It is an odd function (23) and satisfies (24) Erf may be expressed in terms of a confluent hypergeometric function of the first kind as (25) (26) Its derivative is (27) where is a Hermite polynomial. The first derivative is (28) and the integral is (29) Min Max Re Im Erf can also be extended to the complex plane, as illustrated above. A simple integral involving erf that Wolfram Language cannot do is given by (30) (M.R.D'Orsogna, pers. comm., May 9, 2004). More complicated inte
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 inverse error function = 2 π ∫ 0 x e − t 2 d t . error function matlab {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac −
Error Function Excel
0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined as erfc ( x ) = 1 − erf http://mathworld.wolfram.com/Erf.html ( 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 erfcx, the scaled complementary error function[3] (which can https://en.wikipedia.org/wiki/Error_function 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 function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x i
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 ads with us http://math.stackexchange.com/questions/869150/a-function-fx-that-increases-from-0-to-1-when-x-increases-from-0-to-infinity 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 A function f(x) that increases from 0 to 1 when x increases from 0 to infinity. up vote 21 down vote error function favorite 4 I am looking for a function f(x) with a value range of [0,1]. f(x) should increase from 0 to 1 while its parameter x increases from 0 to +infinity. f(x) increases very fast when x is small, and then very slow and eventually approach 1 when x is infinity. Here is a figure. The green curve is what I am looking for: Thanks. It would be great if I can adjust the slope of the increase. Although this is not a compulsory function from 0 requirement. functions share|cite|improve this question asked Jul 16 '14 at 17:18 Changwang Zhang 23329 2 Do you want it to actually take on the values 0 and 1? –Mike Miller Jul 16 '14 at 17:20 @MikeMiller value 0 yes. value 1 no. f(x)=0 when x=0, =1 when x=infinity, and = a value between 0 and 1 otherwise. –Changwang Zhang Jul 16 '14 at 17:22 4 Consider $f(x)=1-\frac1{x+1}$ –abiessu Jul 16 '14 at 17:22 @Leo do you want this $f$ to be $\mathcal{C}^0$? or something more strict perhaps? –DanZimm Jul 16 '14 at 19:20 1 It is always going to be relative fast when $x$ is small, a simple consequence of the fact that you require it to be bounded and monotone. –Gina Jul 16 '14 at 19:21 | show 2 more comments 8 Answers 8 active oldest votes up vote 42 down vote accepted I think this should work well for your purposes: $$ f(x) = \frac{x}{x + a} $$ Where $a$ can be any number bigger than $0$. The smaller $a$ is, the sharper the increase will be. ADDENDUM: if you want to extend this to an odd (and continuously differentiable) function, simply take $$ f(x) = \frac{x}{|x| + a} $$ share|cite|improve this answer edited Jul 17 '14 at 1:48 answered Jul 16 '14 at 17:25 Omnomnomnom 79.6k550102 1 I love this function so much. So simple, elegant, and nice.... –Changwang Zhang Jul 16 '14 at 17:30 2 A