Error Function Evaluated 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 2016 Created, complementary error function developed, and nurturedbyEricWeisstein at WolframResearch Calculus and Analysis>Special Functions>Erf> Calculus and error function calculator Analysis>Complex Analysis>Entire Functions> Interactive Entries>webMathematica Examples> More... History and Terminology>Wolfram Language Commands> MathWorld Contributors>D'Orsogna> Less... Erf is the error function table "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) Note that some
Inverse Error Function
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 of the first kind. For error function matlab , (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 integrals include (31) (M.R.D'Orsogna, pers. comm., Dec.15, 2005). Erf has th
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Erf(1)
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Error Function Excel
Math Toolbox MuPAD Functions erf On this page Syntax Description Environment Interactions Examples Example 1 Example 2 Example 3 Parameters Return Values Algorithms See Also More About http://mathworld.wolfram.com/Erf.html This is machine translation Translated by Mouse over text to see original. Click the button below to return to the English verison of the page. Back to English × Translate This Page Select Language Bulgarian Catalan Chinese Simplified Chinese Traditional Czech Danish Dutch English Estonian Finnish French German Greek Haitian Creole Hindi Hmong Daw https://www.mathworks.com/help/symbolic/mupad_ref/erf.html Hungarian Indonesian Italian Japanese Korean Latvian Lithuanian Malay Maltese Norwegian Polish Portuguese Romanian Russian Slovak Slovenian Spanish Swedish Thai Turkish Ukrainian Vietnamese Welsh MathWorks Machine Translation The automated translation of this page is provided by a general purpose third party translator tool. MathWorks does not warrant, and disclaims all liability for, the accuracy, suitability, or fitness for purpose of the translation. Translate erfError functionexpand all in page MuPAD notebooks are not recommended. Use MATLAB live scripts instead.MATLAB live scripts support most MuPAD functionality, though there are some differences. For more information, see Convert MuPAD Notebooks to MATLAB Live Scripts.Syntaxerf(x) Descriptionerf(x) represents the error function 2π∫0xe−t2dt.This function is defined for all complex arguments x. For floating-point arguments, erf returns floating-point results. The implemented exact values are: erf(0) = 0, erf(∞) = 1, erf(-∞) = -1, erf(i ∞) = i ∞, and erf(-i ∞) = -i ∞. For all other arguments, the error function returns symbolic function calls.For the function call erf(x) = 1 - erfc(x) w
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies http://math.stackexchange.com/questions/108109/steps-in-evaluating-the-integral-of-complementary-error-function of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting 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 error function 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 Steps in evaluating the integral of complementary error function? up vote 5 down vote favorite 2 Could you please check the below and show me any errors? $$ \int_ x^ \infty {\rm erfc} ~(t) ~dt ~=\int_ x^ error function evaluated \infty \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ dt $$ If I let dv=dt and u equal the term inside the bracket, and do integration by parts, $$ \int u ~dv ~=uv - \int v~ du $$ v=t and du becomes $$ -\frac{2}{\sqrt\pi} e^{-t^2} $$ This was obtained from using the Leibniz rule below, $$ \frac {d} {dt} \left[ \int_ a^ b f(u)du \right]\ = \int_ a^ b \frac {d} {dt} f(u) du + f \frac {db} {dt} - f \frac {da} {dt} $$ Then, $$ \frac {d} {dt} \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ = \frac{2}{\sqrt\pi} \left[ \int_ t^ \infty \frac {d} {dt} \left( e^{-u^2} \right) du + e^{-\infty ^2} * 0 - e^{-t^2}*1 \right]= \frac{2}{\sqrt\pi} \left[0~+~0~- e^{-t^2} \right]$$ Is the first and second term going to zero correct? The upper limit b=infinity, and is db/dt=0 in the second term correct? The integral becomes $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty + \int_ x^ \infty t \left[\frac{2}{\sqrt\pi} e^{-t^2} \right]\ dt =$$ $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty - \left[\frac{1}{\sqrt\pi} e^{-t^2}