[math-fun] error function: series development at infinity
dear math funners, does anyone know of a series development of erf(x) round x-> infinity, or equivalently series of erf (1/x) around x=0? Alternatively, how to get approximations to log(1-erf(x)) for x around, say, 50 to 1000? Googled for it in vain, Wouter.
One problem you face is the essential singularity, so getting an expansion valid _around_ x=0 of erf(1/x) will be difficult. However, there is a continued fraction in Abramowitz & Stegun, formula 7.1.4, which you might be able to work with. 2 e^z^2 integral( z to infinity, e^-t^2 dt) = 1/(z+ (1/2)/( z+ 1/( z + (3/2)/(z + 2/(z+ ... ))))) supposedly valid for Re(z)>0. After the first 1, 1/2, the numerators are just stepping by 1/2. This makes the convergence somewhat dawdling for small z (and eventually for any z), but it should work for your range of interest. Rich ------------------ Quoting wouter meeussen <wouter.meeussen@pandora.be>:
dear math funners,
does anyone know of a series development of erf(x) round x-> infinity, or equivalently series of erf (1/x) around x=0? Alternatively, how to get approximations to log(1-erf(x)) for x around, say, 50 to 1000? Googled for it in vain,
Wouter. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Not exactly my idea of fun, but here's how you do it: erf(x) = 1 - 2/sqrt(pi) int_x^inf exp(-t^2) dt Now integrate "by parts": int_x^inf exp(-t^2) dt = int_x^inf (d/dt exp(-t^2)) (- dt/(2t)) = exp(-t^2) (-1/(2t))|_x^inf - int_x^inf exp(-t^2) (dt/(2t^2)) = exp(-x^2) (1/(2x)) - int_x^inf exp(-t^2) (dt/(2t^2)) You can keep going, integrating by parts the resulting integral. Notice that each time you do this the integrand gets another factor of t^2 in the denominator. This allows you to bound the error term. Unfortunately, for fixed x you cannot get arbitrarily small errors by working out sufficiently many terms. That's because the numerical coefficients grow faster than a power (from repeated differentiation of 1/t^n with increasing n). That's no surprise because the function erf(x) has an essential singularity at x = inf. The successive integrations by parts give an asymptotic series, not a convergent series. As a result you are limited in what you can say about the error for x fixed. On the other hand, if you fix the number of terms, say keeping only those we calculated above, erf(x) = 1 - exp(-x^2)/(sqrt(pi) x) + ... then you can say a lot about how the error behaves as x --> inf. This should suffice for x ~ 50. Here's my idea of fun. Sum the series 1 - 1 + 2 - 6 + 24 - 120 + 720 - 5040 + ... Veit On Mar 19, 2009, at 11:22 PM, wouter meeussen wrote:
dear math funners,
does anyone know of a series development of erf(x) round x-> infinity, or equivalently series of erf (1/x) around x=0? Alternatively, how to get approximations to log(1-erf(x)) for x around, say, 50 to 1000? Googled for it in vain,
Wouter. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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