Erfc[z]
gives the complementary error function
.
Erfc
Erfc[z]
gives the complementary error function
.
Details
- Mathematical function, suitable for both symbolic and numerical manipulation.
- Erfc[z] is given by
. - For certain special arguments, Erfc automatically evaluates to exact values.
- Erfc can be evaluated to arbitrary numerical precision.
- Erfc automatically threads over lists.
- Erfc can be used with Interval and CenteredInterval objects. »
Examples
open all close allBasic Examples (5)
Erfc[0.95]Plot over a subset of the reals:
Plot[Erfc[x], {x, -5, 5}]Plot over a subset of the complexes:
ComplexPlot3D[Erfc[z], {z, -2 - 2I, 2 + 2I}, PlotLegends -> Automatic]Series expansion at the origin:
Series[Erfc[x], {x, 0, 10}]Series expansion at Infinity:
Series[Erfc[x], {x, ∞, 5}]//NormalScope (40)
Numerical Evaluation (6)
Erfc[1.5]N[Erfc[3 / 2], 50]The precision of the output tracks the precision of the input:
Erfc[1.5000000000000000000000000000]Evaluate for complex arguments:
Erfc[1.5 - I]Evaluate Erf efficiently at high precision:
Erfc[1.2`500]//TimingErfc[1.2`5000];//TimingCompute worst-case guaranteed intervals using Interval and CenteredInterval objects:
Erfc[Interval[{.2, .3}]]Erfc[CenteredInterval[1, 1 / 100]]Or compute average-case statistical intervals using Around:
Erfc[ Around[2, 0.01]]Compute the elementwise values of an array:
Erfc[{{0 , I π / 2}, {π , 0}}]//FunctionExpandOr compute the matrix Erfc function using MatrixFunction:
MatrixFunction[Erfc, {{0, I π / 2}, {π, 0}}]//FunctionExpandSpecific Values (3)
Simple exact values are generated automatically:
Erfc[0]Erfc[{Infinity, -Infinity}]Find the inflection point as the root of
:
xinfl = Solve[D[Erfc[x], {x, 2}] == 0 && -1 < x < 1, x][[1, 1, 2]]Plot[Erfc[x], {x, -4, 4}, Epilog -> Style[Point[{xinfl, Erfc[xinfl]}], PointSize[Large], Red]]Visualization (2)
Plot the Erfc function:
Plot[Erfc[x], {x, -3, 3}]ComplexContourPlot[Re[Erfc[z]], {z, -2 - 2 I, 2 + 2 I}, Contours -> 20]ComplexContourPlot[Im[Erfc[z]], {z, -2 - 2 I, 2 + 2 I}, Contours -> 20]Function Properties (9)
Erfc is defined for all real and complex values:
FunctionDomain[Erfc[x], x]FunctionDomain[Erfc[z], z, Complexes]Erfc takes all real values between 0 and 2:
FunctionRange[Erfc[x], x, y]Erfc has the mirror property
:
FullSimplify[Erfc[Conjugate[z]] == Conjugate[Erfc[z]]]Erfc is an analytic function of x:
FunctionAnalytic[Erfc[x], x]It has no singularities or discontinuities:
FunctionSingularities[Erfc[x], x]FunctionDiscontinuities[Erfc[x], x]Erfc is nonincreasing:
FunctionMonotonicity[Erfc[x], x]Erfc is injective:
FunctionInjective[Erfc[x], x]Plot[{Erfc[x], .5}, {x, -5, 5}]Erfc is not surjective:
FunctionSurjective[Erfc[x], x]Plot[{Erfc[x], -2}, {x, -10, 10}]Erfc is non-negative:
FunctionSign[Erfc[x], x]Erfc is neither convex nor concave:
FunctionConvexity[Erfc[x], x]Differentiation (3)
D[Erfc[x], x]Table[D[Erfc[x], {x, n}], {n, 1, 4}]Plot[Evaluate[%], {x, -4, 4}, PlotLegends -> {"First Derivative", "Second Derivative", "Third Derivative", "Fourth Derivative"}]D[Erfc[x], {x, n}]Integration (3)
Indefinite integral of Erfc:
Integrate[Erfc[x], x]Definite integral Erfc:
Integrate[Erfc[x], {x, -3, 3}]Integrate[z^αErfc[z^β], z]Integrate[z Exp[z]Erfc[z], z]Integrate[Erfc[z]^2, {z, 0, Infinity}]Series Expansions (4)
Taylor expansion for Erfc:
Series[Erfc[x], {x, 0, 7}]Plot the first three approximations for Erfc around
:
terms = Normal@Table[Series[Erfc[x], {x, 0, m}], {m, 1, 5, 2}];
Plot[{Erfc[x], terms}, {x, -2, 2}]General term in the series expansion of Erfc:
SeriesCoefficient[Erfc[x], {x, 0, n}]Asymptotic expansion of Erfc:
Series[Erfc[x], {x, Infinity, 1}]//NormalErfc can be applied to a power series:
Erfc[Sin[x] + O[x] ^ 10]Integral Transforms (3)
Compute the Fourier transform of Erfc using FourierTransform:
FourierTransform[Erfc[t], t, ω]LaplaceTransform[Erfc[t], t, s]MellinTransform[Erfc[x], x, s]Function Identities and Simplifications (3)
Use FunctionExpand to convert to other functions:
FunctionExpand[Erfc[x]]FullSimplify[%]Integral definition of Erfc:
(2/Sqrt[π])Integrate[Exp[-t^2], {t, x, Infinity}]Argument involving basic arithmetic operations:
Erfc[-z] == 2 - Erfc[z]//FullSimplifyErfc[Sqrt[z^2]]//FunctionExpandFunction Representations (4)
Erfc[z] == Erf[z, Infinity]//FullSimplifyErfc[z] == 1 - Erf[z]//FullSimplifyErfc can be represented as a DifferentialRoot:
DifferentialRootReduce[Erfc[x], x]Erfc can be represented in terms of MeijerG:
MeijerGReduce[Erfc[x], x]Activate[%]//FullSimplifyTraditionalForm formatting:
TraditionalForm[Erfc[x]]Applications (5)
The CDF of NormalDistribution can be expressed in terms of the complementary error function:
CDF[NormalDistribution[0, σ], x]The probability that a random value is greater than
:
Probability[x > n σ, xNormalDistribution[0, σ], Assumptions -> n > 0]Table[%, {n, 1, 3}]//NThe solution of the heat equation for a piecewise‐constant initial condition:
temp[x_, t_] := With[{η = (1/2 Sqrt[t])}, (Erfc[(x - 1)η] - Erfc[(x + 1)η]) / 2]A check that the solution fulfills the heat equation:
D[temp[x, t], t] == D[temp[x, t], x, x]// SimplifyThe plot of the solution for different times:
Plot[Evaluate[{temp[x, 0.01], temp[x, 0.2], temp[x, 0.6], temp[x, 1.5]}], {x, -4, 4}]Define the scaled complementary error function using HermiteH:
FullSimplify[2HermiteH[-1, x] / Sqrt[Pi]]Plot[%, {x, 0, 10}]Interference pattern at the edge of a shadow:
Plot[(1/2)Abs[Exp[-(π I x^2/2)]Erfc[((1 - I/2))Sqrt[π]x]]^2, {x, -8, 8}]The lifetime of a device follows a Birnbaum–Saunders distribution. Find the reliability of the device:
r = SurvivalFunction[BirnbaumSaundersDistribution[α, λ], t]The hazard function has the horizontal asymptote
:
Limit[Refine[HazardFunction[BirnbaumSaundersDistribution[α, 1], t], t > 0], t -> ∞, Assumptions -> α > 0]Limit[Refine[HazardFunction[BirnbaumSaundersDistribution[1, λ], t], t > 0], t -> ∞, Assumptions -> λ > 0]Find the reliability of two such devices in series:
rs = r r//PiecewiseExpandFind the reliability of two such devices in parallel:
rp = 1 - (1 - r)(1 - r)//PiecewiseExpandCompare the reliability of both systems for
and
:
Block[{α = 0.8, λ = 1.2}, Plot[{rs, rp}, {t, 0, 5}, PlotLegends -> {"in series", "in parallel"}]]Properties & Relations (3)
Use FunctionExpand to convert to other functions:
FunctionExpand[Erfc[x]]FullSimplify[%]Compose with inverse functions:
{Erfc[InverseErfc[z]], InverseErfc[Erfc[z]]}PowerExpand[%]Solve a transcendental equation:
Solve[Erfc[x] == 1 / 2, x]Possible Issues (3)
For large arguments, intermediate values may underflow:
Erfc[-2 10 ^ 8 - 10. ^ 3 I]The error function for large negative real-part arguments can be very close to 2:
Erfc[-30`16 + 10 ^ -1 I]Very large arguments can give unevaluated results:
Erfc[10. ^ 100 I]Neat Examples (1)
A continued fraction whose partial numerators are consecutive integers:
With[{n = 20}, 1 / (1 + ContinuedFractionK[HoldForm[k], 1, {k, n}])]Its limit can be expressed in terms of Erfc:
1 / (1 + ContinuedFractionK[k, 1, {k, ∞}])Tech Notes
Related Guides
Related Links
History
Introduced in 1991 (2.0) | Updated in 2021 (13.0) ▪ 2022 (13.1)
Text
Wolfram Research (1991), Erfc, Wolfram Language function, https://reference.wolfram.com/language/ref/Erfc.html (updated 2022).
CMS
Wolfram Language. 1991. "Erfc." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2022. https://reference.wolfram.com/language/ref/Erfc.html.
APA
Wolfram Language. (1991). Erfc. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Erfc.html
BibTeX
@misc{reference.wolfram_2026_erfc, author="Wolfram Research", title="{Erfc}", year="2022", howpublished="\url{https://reference.wolfram.com/language/ref/Erfc.html}", note=[Accessed: 12-June-2026]}
BibLaTeX
@online{reference.wolfram_2026_erfc, organization={Wolfram Research}, title={Erfc}, year={2022}, url={https://reference.wolfram.com/language/ref/Erfc.html}, note=[Accessed: 12-June-2026]}