Complex Error Function Integral
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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 d t = 2 π error function integral table ∫ 0 x e − t 2 d t . {\displaystyle {\begin − 5\operatorname
Error Function Integral Calculation
− 4 (x)&={\frac − 3{\sqrt {\pi }}}\int _{-x}^ − 2e^{-t^ − 1}\,\mathrm − 0 t\\&={\frac 9{\sqrt {\pi }}}\int _
Integral Complementary Error Function
8^ 7e^{-t^ 6}\,\mathrm 5 t.\end 4}} The complementary error function, denoted erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2
Integral Of Error Function With Gaussian Density Function
π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle {\begin 1\operatorname 0 (x)&=1-\operatorname Φ 9 (x)\\&={\frac Φ 8{\sqrt {\pi }}}\int _ Φ 7^{\infty }e^{-t^ Φ 6}\,\mathrm Φ 5 t\\&=e^{-x^ Φ 4}\operatorname Φ 3 (x),\end Φ 2}} which also defines erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic complex error function matlab underflow[3][4]). Another form of erfc ( x ) {\displaystyle \operatorname 1 (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 Φ 9 (x|x\geq 0)={\frac Φ 8{\pi }}\int _ Φ 7^{\pi /2}\exp \left(-{\frac Φ 6}{\sin ^ Φ 5\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 − 9\operatorname − 8 (x)&=-i\operatorname − 7 (ix)\\&={\frac − 6{\sqrt {\pi }}}\int _ − 5^ − 4e^ − 3}\,\mathrm − 2 t\\&={\frac − 1{\sqrt {\pi }}}e^ − 0}D(x),\end − 9}} 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 7 (x)} is real when x is real. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error fun
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 gamma function integral hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question error function values _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; normal distribution integral 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 erf(a+ib) error function separate into real and imaginary https://en.wikipedia.org/wiki/Error_function part up vote 5 down vote favorite 3 Is there an easy way to separate erf(a+ib) into real and imaginary part? calculus integration complex-analysis contour-integration share|cite|improve this question edited Mar 14 '14 at 22:49 Ron Gordon 109k12130221 asked Mar 14 '14 at 19:04 Sleepyhead 1385 add a comment| 3 Answers 3 active oldest votes up vote 6 down vote I'm not sure if you are interested in an analytical answer or a computational answer; these are two http://math.stackexchange.com/questions/712434/erfaib-error-function-separate-into-real-and-imaginary-part different things. The analytical answer is...not really, unless you consider GEdgar's answer useful. (And one might.) The computational answer is a resounding yes. A result found in Abramowitz & Stegun claims the following: $$\operatorname*{erf}(x+i y) = \operatorname*{erf}{x} + \frac{e^{-x^2}}{2 \pi x} [(1-\cos{2 x y})+i \sin{2 x y}]\\ + \frac{2}{\pi} e^{-x^2} \sum_{k=1}^{\infty} \frac{e^{-k^2/4}}{k^2+4 x^2}[f_k(x,y)+i g_k(x,y)] + \epsilon(x,y) $$ where $$f_k(x,y) = 2 x (1-\cos{2 x y} \cosh{ k y}) + k\sin{2 x y} \sinh{k y}$$ $$g_k(x,y) = 2 x \sin{2 x y} \cosh{k y} + k\cos{2 x y} \sinh{k y}$$ Then $$\left |\epsilon(x,y) \right | \le 10^{-16} |\operatorname*{erf}{(x+i y)}| $$ This accuracy is valid for all $x$ and $y$, i.e., the complex plane. I will present a derivation of this result to show you where the error term comes from. Consider the definition of the error function in the complex plane: $$\operatorname*{erf}{z} = \frac{2}{\sqrt{\pi}} \int_{\Gamma} d\zeta \, e^{-\zeta^2}$$ where $\Gamma$ is any path in the complex plane from $\zeta = 0$ to $\zeta=z$. Consider, then, the special case where $\Gamma$ is the path that runs from $0$ to $x$ along the real axis, then from $x$ to $z=x+i y$ parallel to the imaginary axis. Seen this way, the error function of a complex number is equal to $$\operatorname*{erf}{(x+i y)} = \operatorname*{erf}{x} + i \frac{2}{\sqrt{\pi}} e^{-x^2} \int_0^y du \, e^{u^2} \cos{2 x u} \\ + \frac{2}{\sqrt{\pi}} e^{-x^2} \int_0^y du \, e^{u^2} \sin{
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 http://stackoverflow.com/questions/6805164/complex-error-function-in-mathematica Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Complex Error Function in Mathematica up error function vote 6 down vote favorite The complex error function w(z) is defined as e^(-x^2) erfc(-ix). The problem with using w(z) as defined above is that the erfc tends to explode out for larger x (complemented by the exponential going to 0 so everything stays small), so that Mathematica reverts to arbitrary precision calculations that make life VERY slow. The function is used in implementing the voigt profile error function integral - a line shape commonly used in spectroscopy and other related areas. Right now I'm reverting to calculating the lineshape once and using an interpolation to speed things up, however this doesn't let me alter the parameters of the lineshape (or fit to them) easily. scipy has a nice and fast implementation of w(z) as scipy.special.wofz, and I was wondering if there is an equivalent in Mathematica. wolfram-mathematica share|improve this question edited Jul 25 '11 at 1:12 asked Jul 24 '11 at 5:47 crasic 613522 2 Shouldn't it be the other way around in that the error function blows up and the exponential decays for x large (and real)? –Heike Jul 24 '11 at 8:52 I tried playing with SetSystemOptions["CatchMachineUnderflow"->False], which, however, results in getting 0. +0. I for large arguments. Then I tried defining the function as Exp[Log[-z^2+Log@Erfc[-I*z]]], but this turns out to not be any faster than with automatic switching. So, it seems hard to speed this up, except as @Daniel Lichtblau suggests –acl Jul 24 '11 at 23:46 @Heike you are right, a bit of a slip on my part. –crasic Jul 25 '11 at 1:11 add a co
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