I doubt erfinv(z) is entire, since as x -> +-oo, erf'(x) -> 0. In fact erfinv'(x) -> oo as |x| -> 1/2. This means its power series about 0 can't have a radius of convergence > 1/2. --Dan P.S. This is related to a simple -- some would say obvious -- insight I hadn't thought of before today: If you want to pick a real number at random from a distribution given by a certain pdf equal to f(x), and you can invert its cdf F(x) = Integral_{-oo,x} f(t) dt to get Finv(x), then you just need to pick a number u at random from U[0,1]. Then Finv(u) is your random pick from the distribution in question. (Of course you can also just start with the cdf F(x).) On Jul 8, 2014, at 10:31 PM, Mike Stay <metaweta@gmail.com> wrote:
(sqrt(pi) x)/2 + 1/24 pi^(3/2) x^3 + 7/960 pi^(5/2) x^5 + O(x^7)
http://www.wolframalpha.com/input/?i=taylor+series+of+InverseErf
On Tue, Jul 8, 2014 at 10:24 PM, Warren D Smith <warren.wds@gmail.com> wrote:
If x=erf(y), then say y=erfinv(x).
SMP jocks: What is the Maclaurin series of erfinv(x)?
Obviously the erfinv function is useful for statistics. It also occurs to me: This series should have infinite radius of convergence, i.e. erfinv(x) should be a very well behaved "entire" function. Because: the derivative of erf is never zero anywhere in the complex plane, and the value of erf is always finite everywhere in the plane, so its inverse function should exist everywhere.
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