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).rhs().subs(z==1/4) y