2016-11-10 12:47:47 +0100 received badge ● Famous Question (source) 2013-11-19 12:03:05 +0100 received badge ● Notable Question (source) 2012-09-19 12:00:34 +0100 received badge ● Popular Question (source) 2012-05-17 17:18:38 +0100 received badge ● Nice Question (source) 2011-05-19 00:22:31 +0100 commented answer evaluating inverse erf Thanks for the answer (I have something to work with now) and for creating the ticket! 2011-05-19 00:21:42 +0100 answered a question evaluating inverse erf Thanks for the answer (I have something to work with now) and creating the ticket! 2011-05-18 14:57:12 +0100 received badge ● Student (source) 2011-05-18 10:29:53 +0100 received badge ● Supporter (source) 2011-05-18 10:22:39 +0100 asked a question evaluating inverse erf Hi - The code below results in an expression, y, involving the inverse erf function i.e. exp(sqrt(2) inverse_erf(1/2)) which is approximately 1.96 according to another mathematical package. I tried to get a numerical value for y using y.n() but that crashes. Can anyone please advise how to evaluate such symbolic expressions? var('z,t') PDF = exp(-x^2 /2)/sqrt(2*pi) integralExpr = integrate(PDF,x,z,oo).subs(z==log(t)) y = solve(integralExpr==z,t)[0].rhs().subs(z==1/4) y