Re: Simplifying Bessel Functions

• To: mathgroup at smc.vnet.net
• Subject: [mg121421] Re: Simplifying Bessel Functions
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 14 Sep 2011 05:16:17 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Reply-to: hanlonr at cox.net

```a = BesselJ[1, BesselJZero[0, 1]]/BesselJ[2, BesselJZero[0, 1]];

b = BesselJZero[0, 1]/2;

{a - b, 1/a - 1/b} // FullSimplify

{0, 0}

sub = {a -> b, 1/a -> 1/b};

1/(2 rho) (6 - (3 BesselJ[2, BesselJZero[0, 1]] BesselJZero[0, 1])/
BesselJ[1, BesselJZero[0, 1]] + BesselJZero[0, 1]^2)  /. sub

BesselJZero[0, 1]^2/(2 rho)

Bob Hanlon

---- Sam Takoy <sam.takoy at yahoo.com> wrote:

=============
Hi,

As a follow up, when I try

1/(2 rho) (6 - (3 BesselJ[2, BesselJZero[0, 1]] BesselJZero[0, 1])/
BesselJ[1, BesselJZero[0, 1]] +
BesselJZero[0, 1]^2) // FullSimplify

I don't get

1/2 BesselJZero[0, 1]

which is what it is!

Thanks again,

Sam

```

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