Error Function Evaluation

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

Integral Error Function

2 d t = 2 π ∫ 0 x e − t erf(0) 2 d t . {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm erf(2) − 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 ⁡ (

Erfc

x ) = 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

Erf(n)

also defines erfcx, 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 t

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Function Evaluation Worksheet

9 Documentation APIs and reference Dev centers Retired content Samples We’re sorry. The content function notation evaluation you requested has been removed. You’ll be auto redirected in 1 second. Visual Basic Reference Visual Basic Language Reference Error Messages Error Messages Function https://en.wikipedia.org/wiki/Error_function evaluation is disabled because a previous function evaluation timed out Function evaluation is disabled because a previous function evaluation timed out Function evaluation is disabled because a previous function evaluation timed out '#ElseIf' must be preceded by a matching '#If' or '#ElseIf' '#Region' and '#End Region' statements https://msdn.microsoft.com/en-us/library/ms234762.aspx are not valid within method bodies/multiline lambdas '' cannot be applied because the format of the GUID '' is not correct '' is not CLS-compliant because the interface '' it implements is not CLS-compliant '' is obsolete (Visual Basic Warning) '' is an event, and cannot be called directly '' cannot be used as a type constraint '' is not declared (Smart Device/Visual Basic Compiler Error) '.' is already implemented by the base class ''. Re-implementation of assumed '' is valid only within an instance method '' cannot expose type '' outside the project through '' '' is ambiguous across the inherited interfaces '' and '' This error could also be due to mixing a file reference with a project reference to assembly '' '' has multiple definitions with identical signatures '' is ambiguous in the namespace '

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 http://math.stackexchange.com/questions/108109/steps-in-evaluating-the-integral-of-complementary-error-function 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 function evaluation 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? $$ \int_ x^ \infty {\rm erfc} ~(t) ~dt ~=\int_ x^ \infty \left[\frac{2}{\sqrt\pi} \int_ t^ \infty error function evaluation e^{-u^2} du \right]\ dt $$ 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 $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty + \int_ x^ \infty t \left[\frac{2}{\sqrt\pi} e^{-t^2} \right]\ dt =$$ $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty - \left[\frac{1}{\sqrt\pi} e^{-t^2} \right] _{x}^\infty =$$ $$ \left[ 0 - ~ x~ \frac{2}{

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