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2011-05-19 00:22:31 +0200	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 +0200	answered a question	evaluating inverse erf Thanks for the answer (I have something to work with now) and creating the ticket!
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2011-05-18 10:22:39 +0200	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`