From: israel@math.ubc.ca (Robert Israel) Subject: Re: HypergeometricU Date: 17 May 2000 23:56:39 GMT Newsgroups: sci.math.research Summary: Borel summability In article <8fu91c$4q8$1@nnrp1.deja.com>, Dmitry Karp wrote: >In article <39217AA7.652Ec@physik.uni-wuerzburg.de>, > Michael Bechmann wrote: >> i got a question concerning a nonconvergent infinite series. in >> mathematica format the series in question reads >> Sum[(2 l)!/l! 1/x^(2 l+1),{l,0,Infinity}]. This series does not converge >> for any real z. nevertheless, mathematica converts this Sum into the >> expression >> Sqrt[Pi] /2/z/Exp[-z^2/4] ( Sqrt[-z^2/4] + Sqrt[z^1/4] >> Erfi[Sqrt[z^2/4]]) which can also be related to >> -z^2/4 HypergeometricU[1,3/2,-z^2/4]. >> for real z these expressions give complex values. how can that be?? how >> can the sum have an imaginary part although all parts of the sum are >> strictly real??? >> >> i'd be very gratefull for any usefull answer... >This seems to be a good question to Wolfram Research. Of course this >series diverges strongly and cannot be summed by any reasonable method, >since, if you apply Legedre's doubling formula to Gamma(2l+1) and thus >get rid of denominator you get roughly a series of the form >sum_l l!y^l, which is not so easy to give sense to. No, you can use Borel summability. For convenience, rewrite the series as S(y) = sum_{n=0}^infinity a_n y^n where a_n = (-1)^n (2n)!/n!. The Borel transform is h(y) = sum_{n=0}^infinity a_n/n! y^n which has radius of convergence 1/4, and in fact h(y) = 1/sqrt(1+4 y). Thus the Borel sum of the series is S(y) = int_0^infinity exp(-t) h(y t) dt which converges everywhere except on the negative real axis. In fact, according to Maple, / 1 \ exp(1/4 1/y) sqrt(Pi) |-1 + erf(1/2 -------)| \ sqrt(y) / S(y) = - 1/2 --------------------------------------------- sqrt(y) which is analytic except for a branch cut, which for the "standard branch" would be taken on the negative real axis, with a rather nasty branch point at y=0. Now what Michael wanted was 1/x S(-1/x^2), and Maple's result appears to agree with Mathematica's. The reason for the imaginary component is that branch cut: when y is on the negative axis, the two choices of branch of the square roots give you answers with an imaginary component, which are complex conjugates of each other. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2