How To Solve 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 ∫ integral of error function 0 x e − t 2 d t . {\displaystyle {\begin − 6\operatorname −
Erf Function Calculator
5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac − 0{\sqrt {\pi }}}\int _ 9^ erf function table 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x error function properties ∞ 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, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). Another form of
Inverse Error Function
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 ( 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 th
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Error Function Excel
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 https://en.wikipedia.org/wiki/Error_function the top How can I solve the integral in the error function $\mbox{erf}(x)$? up vote 1 down vote favorite To get from this To this series I can't seem find the step-by-step solution anywhere. integration taylor-expansion error-function share|cite|improve this question edited Apr 3 '15 at 1:22 asked Apr 3 '15 at 0:28 Agustina 62 Your calculus textbook should have information on integrating a power series term-by-term. –GEdgar Apr 3 '15 http://math.stackexchange.com/questions/1218007/how-can-i-solve-the-integral-in-the-error-function-mboxerfx at 0:30 Integrate the series term by term to get a series for your answer. Then, if you need a "practical" answer, approximate the resulting series with as much precision as you need. –TravisJ Apr 3 '15 at 0:31 1 Is the question how to evaluate $\int \mbox{erf}(t)\,dt$, or is the question how to evaluate $\mbox{erf}(x)$ itself, which is a function defined by an integral? –David K Apr 3 '15 at 0:36 @DavidK the error function itself –Agustina Apr 3 '15 at 0:47 @Agustina then your question title is misleading. –Kaster Apr 3 '15 at 0:50 | show 2 more comments 1 Answer 1 active oldest votes up vote 3 down vote Note that \begin{align*} \int e^{-x^2}\,dx &= \int\left\{ \sum_{k=0}^\infty \frac{1}{k!}\left(-x^2\right)^{k} \right\}\,dx \\ &= \int \left\{ \sum_{k=0}^\infty \frac{1}{k!}(-1)^kx^{2k} \right\}\,dx \\ &= \sum_{k=0}^\infty(-1)^k\frac{1}{k!} \left\{ \int x^{2k}\,dx \right\} \\ &= \sum_{k=0}^\infty (-1)^k\frac{1}{k!}\frac{1}{2k+1}x^{2k+1} \end{align*} share|cite|improve this answer edited Apr 3 '15 at 16:39 answered Apr 3 '15 at 1:05 Brian Fitzpatrick 18.7k42450 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you
Search All Support Resources Support Documentation MathWorks Search MathWorks.com MathWorks Documentation Support Documentation Toggle navigation Trial Software Product Updates Documentation Home Symbolic Math Toolbox Examples Functions and Other https://www.mathworks.com/help/symbolic/erf.html Reference Release Notes PDF Documentation Mathematics Mathematical Functions Symbolic Math Toolbox Functions erf On this page Syntax Description Examples Error Function for Floating-Point and Symbolic Numbers Error Function for Variables and Expressions Error Function for Vectors and Matrices Special Values of Error Function Handling Expressions That Contain Error Function Plot Error Function Input Arguments X More About Error Function Tips Algorithms References error function See Also This is machine translation Translated by Mouse over text to see original. Click the button below to return to the English verison of the page. Back to English × Translate This Page Select Language Bulgarian Catalan Chinese Simplified Chinese Traditional Czech Danish Dutch English Estonian Finnish French German Greek Haitian Creole Hindi Hmong Daw Hungarian Indonesian Italian Japanese Korean Latvian how to solve Lithuanian Malay Maltese Norwegian Polish Portuguese Romanian Russian Slovak Slovenian Spanish Swedish Thai Turkish Ukrainian Vietnamese Welsh MathWorks Machine Translation The automated translation of this page is provided by a general purpose third party translator tool. MathWorks does not warrant, and disclaims all liability for, the accuracy, suitability, or fitness for purpose of the translation. Translate erfError functioncollapse all in page Syntaxerf(X) exampleDescriptionexampleerf(X
) represents the error function of X. If X is a vector or a matrix, erf(X) computes the error function of each element of X.ExamplesError Function for Floating-Point and Symbolic Numbers Depending on its arguments, erf can return floating-point or exact symbolic results. Compute the error function for these numbers. Because these numbers are not symbolic objects, you get the floating-point results:A = [erf(1/2), erf(1.41), erf(sqrt(2))]A = 0.5205 0.9539 0.9545Compute the error function for the same numbers converted to symbolic objects. For most symbolic (exact) numbers, erf returns unresolved symbolic calls:symA = [erf(sym(1/2)), erf(sym(1.41)), erf(sqrt(sym(2)))]symA = [ erf(1/2), erf(141/100), erf(2^(1/2))]Use vpa to approximate symbolic results with the required number of digits:d = digits(10); vpa(symA) digits(d)ans = [ 0.5204998778, 0.9538524394, 0.95449