complementry error fucntion

[pdcom]

what is square of erfc (complementry error fucntion), i.e (erfc(x))².

or what will be 1-(1-erfc(x))²

or why 1-(1-erfc(x))²≈2erfc(x)

 

[jcy]

well,

1-(1-erfc)^2 = 1 - [1-2erfc + erfc^2] =2erfc-erfc^2

The erfc is 2 for x->-∞, 1 for x=0, and 0 for x->∞.
For x greater than about 1 , I would agree that

1-(1-erfc)^2 = 1 - [1-2erfc + erfc^2] ≈ 2erfc

but for all x, I do not think the relationship holds. Are you sure there isn't some other restriction in the problem?