Re: Substitute values in functions!!!
- To: mathgroup at smc.vnet.net
- Subject: [mg52255] Re: Substitute values in functions!!!
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Thu, 18 Nov 2004 01:44:55 -0500 (EST)
- Organization: The University of Western Australia
- References: <cneu7f$q8j$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <cneu7f$q8j$1 at smc.vnet.net>,
davidx at x-mail.net (david Lebonvieux) wrote:
> How can substitute some values(the zeros of BesselJ function) in this
> Integrals?
> j0[n_] := x /. FindRoot[BesselJ[0, x] == 0, {x, n 2.5}]
Change this to
j0[n_] := FindRoot[BesselJ[0, x] == 0, {x, n*2.5}]
so that the result is a replacement rule instead of a value.
> \!\(s\_l = Array[j0, \ 10]\)
>
> Integrate[BesselJ[0,s0*r/R] BesselJ[0,s2*r/R] BesselJ[0,s3*r/R],{r,0,R}]
By a change of variables r -> R r, this integral becomes
R Integrate[BesselJ[0,s0 r] BesselJ[0,s2 r] BesselJ[0,s3 r],{r,0,1}]
> where s0, s1,s2,the Zero-Poles of BesselJ.
Substitute the first three zeros into the integrand:
Times @@ (BesselJ[0, x r] /. Array[j0, 3])
Compute the integral numerically:
R NIntegrate[%, {r, 0, 1}]
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
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