Jul 18, 2008 #1 P 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)
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)
Jul 25, 2008 #2 J jcy Junior Member level 3 Joined Jul 21, 2006 Messages 26 Helped 3 Reputation 6 Reaction score 0 Trophy points 1,281 Activity points 1,479 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?
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?