Re: Mod Bessel function bug ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg23019] Re: [mg22988] Mod Bessel function bug ?*From*: BobHanlon at aol.com*Date*: Tue, 11 Apr 2000 23:18:36 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Plot[Evaluate[Table[BesselK[n, x], {n, 0, 2, 1/3}]], {x, 0.1, 1.5}]; Series[BesselK[n, x], {x, 0, 5}] == Series[BesselK[-n, x], {x, 0, 5}] // Normal // Simplify True FullSimplify[BesselK[n, x] == BesselK[-n, x]] True Abramowitz & Stegun 10.2.4 argues against your statement that "the I function with negative, noninteger order is the same as the K function" Sqrt[1/2*Pi/z]*BesselK[n + 1/2, z] == Pi/2 * (-1)^(n + 1)* Sqrt[1/2*Pi/z]*(BesselI[n + 1/2, z] - BesselI[-n - 1/2, z]); Simplify[FunctionExpand[%], Element[n, Integers]] Sqrt[Pi/2]*Sqrt[1/z]*(-BesselK[-(1/2) - n, z] + BesselK[1/2 + n, z]) == 0 FullSimplify[% /. n -> (m - 1/2)] True Bob Hanlon In a message dated 4/10/2000 10:16:07 AM, jrchaff at nwlink.com writes: >Well, something is very strange. > >Thank both of you for your replies. I am using Mathematica >Student Version 4.0; supposedly same as full version >capabilities, or at least so advertised. > >My plots show there is a difference near zero; however, both >functions come together (and become large) for large argument, >precisely opposite to what bessel theory says. Supposedly, >the I function with negative, noninteger order is the same as >the K function (not with Mathematica), and the K function >goes to zero exponentially with large argument. > >Something is definitely wrong. >