Re: solution
- To: mathgroup at smc.vnet.net
- Subject: [mg118286] Re: solution
- From: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
- Date: Wed, 20 Apr 2011 04:28:01 -0400 (EDT)
Hi,
you have a sum of two Bessel-functions where the last part is just
weighted by k. The roots of BesselJ[1,k] and k*BesselJ[1,k] are
therefore the same (for k!=0).
The structure of your function suggests, that a root of it lies between
the corresponding roots of BesselJ[0,k] and BesselJ[1,k].
If you want the first say 20 roots of your function, why don't you use
that fact to create your start points for FindRoot?
term = BesselJ[0, k] + k BesselJ[1, k];
start = Mean[{BesselJZero[0, #], BesselJZero[1, #]}] & /@ N[Range[20]];
zeroes = k /.
FindRoot[BesselJ[0, k] + k BesselJ[1, k] == 0, {k, #}] & /@ start;
Plot[term, {k, 0, Max[zeroes] + 1},
Epilog :> {Black, PointSize[0.02], Point[{#, 0}] & /@ zeroes}]
Please note that I'm not that deep into Bessel-functions. It was just an
idea and it seems to work for the first 20 zeroes.
But I have a follow-up question to somebody who is more skilled in math
than I'm:
The Bourgets hypothesis tells that BesselJ[n,k] have all different roots
for different n. But if we have the roots BesselJ[0,k] and BesselJ[1,k]
in table, is there a way to compute the zeros of the sum of these two
functions?
Cheers
Patrick
On Tue, 2011-04-19 at 06:56 -0400, amelia Jackson wrote:
> Dear MathGroup,
>
> I have a problem. I want to find solution:
> r := Table[
> k /. FindRoot[BesselJ[0, k] + k BesselJ[1, k] == 0, {k, n}], {n, 1, 100}]
>
> but I get about 30 roots. I need about 100 or more.
> I think that "step" "n" tend to Pi
>
> Please for help...
>