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MathGroup Archive 2004

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Re: FindRoot for an oscillating function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50936] Re: FindRoot for an oscillating function
  • From: drbob at bigfoot.com (Bobby R. Treat)
  • Date: Tue, 28 Sep 2004 00:59:05 -0400 (EDT)
  • References: <cj86qq$78r$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here's an approach that takes advantage of the Plot itself. It finds
consecutive data points that bracket roots, averages the x-values,
uses those as guesses in FindRoot, and finally graphs the original
function with roots superimposed. It will only find roots internal to
the plotted interval, so I reduced the lower limit to get the root at
zero.

Needs["Graphics`"]
p = 1.234;
q = .7654;
gr[x_] = Sin[p x]/p + Sin[q x]/q;
plot = Plot[gr@x, {x, -1, 25}, DisplayFunction -> Identity];
points = First@Cases[plot, Line[a_] -> a, Infinity];
guesses = Mean /@ Extract[Partition[points[[All, 1]], 2, 1], 
        Position[Partition[points[[All, -1]], 2, 
    1], _?(Times @@ # <= 0 &), {1}]]
roots = x /. FindRoot[gr@x, {x, #}] & /@ guesses
rootPts = {#, gr@#} & /@ roots
DisplayTogether[plot, Graphics at {PointSize[0.02], 
        Red, Point /@ rootPts}, DisplayFunction -> $DisplayFunction];

Bobby

mathma18 at hotmail.com (Narasimham G.L.) wrote in message news:<cj86qq$78r$1 at smc.vnet.net>...
> How to find all real/complex roots by sweeping through the domain
> {x,0,25} using Mathematica capability?   p = 1.234; q = .7654; gr = Sin[p x]/p
> + Sin[q x]/q ; Plot[gr,{x, 0, 25}]; FindRoot[gr == 0, {x, 0, 25}]


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