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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...
> 




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