Error Function Value Infinity
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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 value of infinity gauntlet comics d t = 2 π ∫ 0 x e − t 2 d what is the value of infinity divided by infinity t . {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − complementary error function 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 ) error function calculator = 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 erfcx,
Error Function Table
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 function", erfi
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Inverse Error Function
Created, developed, and nurturedbyEricWeisstein at WolframResearch Calculus and Analysis>Special Functions>Erf> Calculus error function matlab and Analysis>Complex Analysis>Entire Functions> Interactive Entries>webMathematica Examples> More... History and Terminology>Wolfram Language Commands> MathWorld Contributors>D'Orsogna> Less... Erf is error function python 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) Note that https://en.wikipedia.org/wiki/Error_function 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 of the first http://mathworld.wolfram.com/Erf.html 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 integrals include (31) (M.R.D'Orsogna, pers. comm., D
// additional overloads for integral types Compute error function Returns the error function value for x. Header
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/* erf error function example */ #include
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 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 Error Function limit up vote 5 down vote favorite 1 $$\prod_{n=1}^{\infty}{\frac{2}{\sqrt{\pi}}\int_0^n e^{-x^{2}} \mathrm{d}x} \approx 0.83874 $$ Is it a known constant? I couldn't find anything about it. Do you know ways to calculate the value efficiently? integration limits convergence special-functions infinite-product share|cite|improve this question edited Apr 9 '13 at 18:16 1015 34.8k34199 asked Nov 6 '12 at 15:49 Gunnar 9710 2 An interesting(?) but unfortunately totally irrelevant observation: Googling 0.83874 yields about 12600 hits. –Harald Hanche-Olsen Nov 6 '12 at 16:04 add a comment| 1 Answer 1 active oldest votes up vote 4 down vote I don't know the constant, but when it comes to calculating the product, we might note that it can be written $$\prod_{n=1}^\infty\Bigl(1-\frac{2}{\sqrt{\pi}}\int_n^\infty e^{-x^2}\,dx\Bigr).$$ When $n$ grows large, partial integration yields $$\int_n^\infty e^{-x^2}\,dx=\frac{e^{-n^2}}{2n}+\frac12\int_n^\infty\frac{e^{-x^2}}{x^2}\,dx,$$ which gives you a reasonable handle on the integral on the left. Taking the log of the infinite product, you get a sum in which this will give you a somewhat decent approximation for the tails of the sum. There are many details to fill in, but off the top of my head this outlines how I would go about computing the product. share|cite|improve this answer answered Nov 6 '12 at 16:19 Harald Hanche-Olsen 24k23047 Could you tell me, how you did this? $$\int_n^\infty e^{-x^2}\,dx=\frac{e^{-n^2}}{2n}+\frac12\int_n^\infty\frac{e^{-x^2}}{x^2}\,dx$$ –Gunnar Nov 6 '12 at 16:29 @Gunnar: Write the integrand as $xe^{-x^2}\cdot x^{-1}$ and do a partial integration, integrating $xe^{-x^2}$ to get $-\frac12e^{-x^2}$ and differentiatiing $x^{-1}$ to get $-x^{-2}$. –Haral