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MathGroup Archive 2000

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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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