Derivative Of Error Function
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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, developed, and nurturedbyEricWeisstein at WolframResearch Calculus and Analysis>Special Functions>Erf> Calculus and Analysis>Complex Analysis>Entire Functions> derivative of error function complement Interactive Entries>webMathematica Examples> More... History and Terminology>Wolfram Language Commands> MathWorld Contributors>D'Orsogna> Less... Erf is
Differentiation Error Function
the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function). It is an entire
Derivative Q Function
function defined by (1) 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
Derivative Gamma Function
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 , (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 , derivative normal distribution (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 the continued fraction (32) (33) (Wall 1948, p.357), first stated by Laplace in 1805 and Legendre in 1826 (Olds 1963, p.139), proved by Jacobi, and rediscovered by Ramanujan (Watson 1928; Hardy 1999, pp.8-9). Definite integrals involving include Definite integrals involving include (34) (35) (36) (37) (38) The first two of these appear in Prudnikov et al. (1990, p.123, eqns. 2.8.19.8 and 2.8.19.11), with , . A complex generalization of is defined as (39) (40) Integral representations valid only in the upper half-plane are given by (41)
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the derivative gaussian workings and policies of this site About Us Learn more about error function values Stack Overflow the company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags erf(2) 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 http://mathworld.wolfram.com/Erf.html 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 derivative of error function up vote 0 down vote favorite How can I calculate the derivatives $$\frac{\partial \mbox{erf}\left(\frac{\ln(t)-\mu}{\sqrt{2}\sigma}\right)}{\partial \mu}$$ and $$\frac{\partial \mbox{erf}\left(\frac{\ln(t)-\mu}{\sqrt{2}\sigma}\right)}{\partial \sigma}$$ where $\mbox{erf}$ denotes the http://math.stackexchange.com/questions/1755149/derivative-of-error-function error function can be given by $$\mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}\exp(-t^2)\,dt$$ I have tried it using WA derivative calculator but I am not able to understand the steps. derivatives error-function share|cite|improve this question edited Apr 23 at 9:02 kamil09875 4,3592729 asked Apr 23 at 7:44 Rakesh 11 The error function erf($x$) is just $\frac{2}{\sqrt\pi}\int_0^xe^{-t^2}\ dt$, so its derivative is just $\frac{2}{\sqrt\pi}e^{-x^2}$. All you have to do for your examples is use the chain rule. –almagest Apr 23 at 7:58 add a comment| 1 Answer 1 active oldest votes up vote 0 down vote You have error in your definition of error function :-). The definition of error function is $$\operatorname{erf}(x) = \frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\,\mathrm dt = \int_0^x \frac{2}{\sqrt\pi}e^{-t^2}\,\mathrm dt.$$ Derivative of this integral with variable is it's integrand applied to upper boundary and multiplicated by boundary's derivative. ($\frac{\partial x}{\partial x}=1$) $$\frac{\partial \operatorname{erf}(x) }{\partial x}=1\cdot\frac{2}{\sqrt\pi}e^{-x^2}$$ The next step is calculating derivative of a composite function. I hope you can do it yourself. ==Added== You should treat $t$ and $\mu$ as a parameters. For example: $$\frac{\partial \frac{\
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company http://math.stackexchange.com/questions/364112/how-to-prove-that-integration-of-exp-x2-is-error-function Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags https://www.youtube.com/watch?v=CcFUQhorgdc Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top error function How to prove that integration of exp(-x^2) is error function? [closed] up vote -1 down vote favorite Prove that $$\int e^{-x^2} dx=\frac{\sqrt{\pi}}{2}\rm erf (x).$$ integration share|cite|improve this question edited Apr 17 '13 at 5:13 Mhenni Benghorbal 39.7k52965 asked Apr 17 '13 at 5:10 Litun John 399215 closed as unclear what you're asking by Micah, Adriano, Dominic Michaelis, Danny Cheuk, draks ... Jul 24 '13 at 5:30 Please clarify your specific problem or add derivative of error additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question. 4 isn't the error function defined over the integral ? –Dominic Michaelis Apr 17 '13 at 5:12 2 By definition, $$\text{erf}(x)=\frac2{\sqrt\pi}\int_0^x\exp(-w^2)\,dw.$$ Is this a different definition than you've been given? Also, since your integral is indefinite, don't forget your constant of integration. –Cameron Buie Apr 17 '13 at 5:17 add a comment| 1 Answer 1 active oldest votes up vote 3 down vote accepted There's nothing here to prove, the definition of the error function is that $$ \text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx $$ I suppose that you might be interested in why the $\frac{2}{\sqrt{\pi}}$. Basically, it works like this: let $$ I = \int_0^\infty e^{-x^2}dx $$ Now, we can state that $$\begin{align} I^2 &= \int_0^\infty e^{-x^2}dx\int_0^\infty e^{-y^2}dy\\ &=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy\\ &=\int_0^{\frac{\pi}2}\int_0^\infty e^{-r^2}rdrd\theta \end{align}$$ The last line, there, is because we can convert from cartesian to radial coordinates, for which $dxdy = rdrd\theta$ and $r^2=x^2+y^2$. Now, as the inner integral doesn't depend on $\theta$, we may let $r^2=s$ (and so, $rdr=\frac{ds}{2}$) to get $$\begin{align} I^2 &= \frac{\pi}{2}\int_0^\infty e^{-s}\frac{ds}2\\ &= \frac{\pi}{4}\left[-e^{-s}\right]_0^\infty\\ &=
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