Plot Error Function Mathematica
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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 ∫ 0 x e − t 2 d t . https://en.wikipedia.org/wiki/Error_function {\displaystyle {\begin − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted http://mathematica.stackexchange.com/questions/64635/inverse-error-function erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , error function {\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, 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 ≥ plot error function 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 ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usually discussed in scaled form as the Faddeeva function: w ( z ) = e − z 2 erfc ( − i z ) = erfcx ( − i z ) . {\displaystyle w(z)=e^{-z^ 6}\operatorname 5 (-iz)=\operatorname 4 (-iz).} Contents 1 The name "error function" 2 Properties 2.1 Taylor series 2.2 Derivative and integral 2
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 Mathematica Questions Tags Users Badges Unanswered Ask Question _ Mathematica Stack Exchange is a question and answer site for users of Mathematica. 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 Inverse error function up vote 6 down vote favorite I solved some equation in Mathematica and I obtained something like $$y(t)=\exp \left\lbrace \left[ \text{erf}^{-1} (\text{i}t) \right]^2\right\rbrace, (1)$$ where $\text{i}$ is imaginary unit and $\text{erf}^{-1}(x)$ is the inverse error function (it is not equal to $\frac{1}{\text{erf}(x)}$ !!), which is defined for $x \in [-1,1]$. The problem is that the $t$ is real and the function has to be also real, but I can't plot this function since $\text{erf}^{-1}$ accepts only real arguments in Mathematica. Is there any way how to plot the solution or convert it to some other expression, which can be plotted? I tried to use some approximations of inverse error functions, such as $$ \text{erf}^{-1}(x) = \sum_{k=0}^{N} \frac{c_k}{2k+1}\left(\frac{\sqrt \pi}{2}x\right)^{2k+1}, (2)$$ to finite $N$ (from http://en.wikipedia.org/wiki/Error_function#Inverse_functions) which holds if $x \in [-1,1]$ and then I just simply put $t \rightarrow \text{i}t$ in approximated version of (1) and obtained only real part (imaginary part was zero), but I'm not sure wheather it is correct. special-functions share|improve this question edited Aug 23 '15 at 4:58 J. M.♦ 68.2k8208336 asked Nov 1 '14 at 19:16 George 536 Can you show the Mathematica code? Be