complementry error fucntion

Status
Not open for further replies.

pdcom

Newbie level 6
Joined
Jul 17, 2008
Messages
14
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,369
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)
 

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?
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…