Re: bug in Legendre polinomials

*To*: mathgroup at smc.vnet.net*Subject*: [mg20455] Re: bug in Legendre polinomials*From*: paul <paul at physics.uwa.edu.au>*Date*: Wed, 27 Oct 1999 02:04:33 -0400*Organization*: University of Western Australia*References*: <7u8vsu$qcn@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Peter Pollner wrote: > I have found a misterious bug (Version 4.0.1.0): > > I have checked the identity: > Sum[Binomial[ktmp, i]^2 x^i, {i, 0, ktmp}] = (1 - x)^ktmp LegendreP[ktmp, > (x + 1)/(1 - x)] > > using: > > Simplify[Sum[ > Binomial[ktmp, i]^2 x^i, {i, 0, ktmp}] - (1 - x)^ktmp LegendreP[ > ktmp, (x + 1)/(1 - x)]] > > which should be zero for arbitrary ktmp integers. > Mathematica gives only for ktmp<36 the correct result > for ktmp>=36 it gives a nonvanishing polinom. You will find that, for rational functions, Together is faster than Simplify and gives the desired result. > I am interested also to force Mathematica to give the result of the series > Sum[Binomial[ktmp, i]^2 x^i, {i, 0, ktmp}] > in terms of Legendre polinomials and not as terms of Hypergeometric > functions. In general, this is not so simple. There are many alternative forms in terms of orthogonal polynomials. However, you can define a pattern rule that will do this, e.g. Simplify[Sum[Binomial[k, i]^2*((y - 1)/(y + 1))^i, {i, 0, k}] /. Hypergeometric2F1[k_, k_, 1, x_] :> LegendreP[-k, (x + 1)/(1 - x)]/(1 - x)^k] Cheers, Paul ____________________________________________________________________ Paul Abbott Phone: +61-8-9380-2734 Department of Physics Fax: +61-8-9380-1014 The University of Western Australia Nedlands WA 6907 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul God IS a weakly left-handed dice player ____________________________________________________________________