# Revision history [back]

### 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

 2 No.2 Revision kcrisman 11972 ●37 ●124 ●241

### 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')

var('z,t')
PDF = exp(-x^2 /2)/sqrt(2*pi) /2)/sqrt(2*pi)
integralExpr = integrate(PDF,x,z,oo).subs(z==log(t)) integrate(PDF,x,z,oo).subs(z==log(t))
y = solve(integralExpr==z,t)[0].rhs().subs(z==1/4)solve(integralExpr==z,t)[0].rhs().subs(z==1/4)      yy