Erfc

Author

Eric W. Weisstein
January 19, 2006

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/Erfc.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/Erfc.html.

©2006 Wolfram Research, Inc. except for portions noted otherwise

Plots

<<Utilities`Plot`

Real Axis

PlotDiscontinuous[Erfc[x],{x,-5,5},
AxesLabel->TraditionalForm/@{x,Erfc[x]},
HorizontalAsymptotes->{0,2}
]

Complex Plane

ComplexPlot3D[Erfc[z],{z,-2,2},PlotPoints->100]//Timing

ComplexContourPlot[Erfc[z],{z,-2,2}]//Timing

Representations

FunctionExpand[Erfc[z]]

FullSimplify[1-Erf[x]]

(2/(Sqrt[Pi]))Integrate[Exp[-t^2],{t,x,∞}]

FullSimplify[Gamma[1/2,(z^2)]/Sqrt[π],z<0]

FullSimplify[Gamma[1/2,(z^2)]/Sqrt[π],z>0]

Special Values

Erfc/@{-∞,0,∞}

Identities

FullSimplify[Subtract@@{Erfc[-z],2-Erfc[z]}]

Derivative

D[Erfc[z],z]

Indefinite Integrals

∫erfc(z)dz==z erfc(z)-((e^(-(z^2)))/(Sqrt[π]))

Integrate[Erfc[z],z]

Definite Integrals

(∫_0^∞)((erfc(z))^n)dz

Integrate[Erfc[z]^n,{z,0,∞}]

Integrate[#,{z,0,∞}]&/@Table[Erfc[z]^n,{n,4}]//FullSimplify

NIntegrate[Erfc[z]^3,{z,0,∞},WorkingPrecision->20]

(∫_0^∞)sin((z^2)) erfc(z)dz

Integrate[Sin[(z^2)]Erfc[z],{z,0,∞}]

Repeated Integrals

FullSimplify[-2/Sqrt[Pi]Integrate[(t-z)^n/n! Exp[-t^2],{t,z,∞}],n∈Integers&&n≥0]

FullSimplify[-2/Sqrt[Pi]Integrate[(t-z)^n/n! Exp[-t^2],{t,z,∞}],n∈Integers&&n≥0&&Re[z]<0]

FullSimplify[-2/Sqrt[Pi]Integrate[(t-z)^n/n! Exp[-t^2],{t,z,∞}],n∈Integers&&n≥0&&Re[z]>0]

These should all give the same answer...

v1=Rest[NestList[Integrate[#,{z,z,Infinity}]&,Erfc[z],4]]//FullSimplify

((E^(-(z^2)))/((Sqrt[π]) n!)) (Γ(((n+1)/2)) (Subscript[, 1])(Subscript[F, 1])(((n+1)/2);(1/2);(z^2))-n z Γ((n/2)) (Subscript[, 1])(Subscript[F, 1])((n/2)+1;(3/2);(z^2)))

v2=Table[-(1/((Sqrt[π]) n!))((E^(-(z^2))) (n z Gamma[(n/2)] Hypergeometric1F1[1+(n/2),(3/2),(z^2)]-Gamma[((1+n)/2)] Hypergeometric1F1[((1+n)/2),(1/2),(z^2)])),{n,4}]//FullSimplify

(2^(-n)) (E^(-(z^2))) ((((Subscript[, 1])(Subscript[F, 1])(((n+1)/2);(1/2);(z^2)))/(Γ((n/2)+1)))-((2 z (Subscript[, 1])(Subscript[F, 1])((n/2)+1;(3/2);(z^2)))/(Γ(((n+1)/2)))))

FullSimplify[(2^(-n)) (E^(-(z^2))) (-((2 z Hypergeometric1F1[1+(n/2),(3/2),(z^2)])/(Gamma[((1+n)/2)]))+((Hypergeometric1F1[((1+n)/2),(1/2),(z^2)])/(Gamma[1+(n/2)]))),n∈Inetegrs&&n>0]

v3=Table[(2^(-n)) (E^(-(z^2))) (-((2 z Hypergeometric1F1[1+(n/2),(3/2),(z^2)])/(Gamma[((1+n)/2)]))+((Hypergeometric1F1[((1+n)/2),(1/2),(z^2)])/(Gamma[1+(n/2)]))),{n,4}]//FullSimplify

Equal@@@Transpose[{v1,v2,v3}]

erfc[n_,z_]:=Nest[Integrate[#,{z,z,Infinity}]&,Erfc[z],n]

Integrate[Erfc[t],{t,z,Infinity}]

Integrate[Erfc[t],t]

<<Graphics`Colors`

Plot[Evaluate[Table[erfc[n,z],{n,0,5}]],{z,-3,2},PlotStyle->{Black,Red,Yellow,Green,Blue,Violet},AxesLabel->TraditionalForm/@{x,"(Subscript[erfc, n])(x)"}]

FullSimplify[Table[erfc[n,z],{n,0,4}]]//ColumnForm//TraditionalForm

Inequality

Integrate[Exp[-(t^2)],{t,x,∞}]

l={(2/(Sqrt[π]))((Exp[-(x^2)])/(x+(Sqrt[(x^2)+2]))),Erfc[x],(2/(Sqrt[π]))((Exp[-(x^2)])/(x+(Sqrt[(x^2)+(4/π)])))};

<<Utilities`Typesetting`

colors={Red,Green,Blue};
Plot[Evaluate[l],{x,-1,3},PlotStyle->colors,
AxesLabel->TraditionalForm/@{x,MapIndexed[StyleForm[#1,FontColor->colors[[#2[[1]]]]]&,l]}]

And@@LessEqual@@@Table[l,{x,0,5,.01}]//Timing

Series[l,{x,Infinity,3}]

ser=Normal[%]

Simplify[LessEqual@@ser,x>0]