### 5.9.6 Complementary error function: erfc

erfc takes as argument a number a.
erfc returns the value of the complementary error function at x=a, this function is defined by :

erfc(x)=

2

√

π

∫

+∞
x

e−t2dt=1−erf(x)

Hence erfc(0)=1, since :

∫

+∞
0

e−t2dt=

√

π

2

Input :

erfc(1)

Output :

0.15729920705

Input :

1- erfc(1/(sqrt(2)))*1/2

Output :

0.841344746069

Remark
The relation between erfc and normal_cdf is :

normal_cdf(x)=1−

1
2

erfc (

x

√

2

)

Check :
normal_cdf(1)=0.841344746069