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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.
>
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