Re: sorting list of roots af a transcendental function

*To*: mathgroup at smc.vnet.net*Subject*: [mg65276] Re: [mg65260] sorting list of roots af a transcendental function*From*: "Carl K. Woll" <carlw at wolfram.com>*Date*: Thu, 23 Mar 2006 06:58:42 -0500 (EST)*References*: <200603221113.GAA10232@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Dule wrote: > Dear group, > > for calculating a model i need values for x which are given by the > transcendental function Cot[x] == x/a - a/(4*x). a is a parameter 0<a<200. > i obtained the roots with Table and FindRoot: > Table[FindRoot[Cot[x] == x/a - a/(4*x), {x, i}], {i, 1, 50}]] > > I have two questions: > 1. Is there a better way to do this? > 2. How can i construct a list, where the values for x, which appear > multiple are dropped? > > Thanks! One method to find subintervals of an interval guaranteed to contain all of the roots of a function f is to use the package IntervalRoots. With this package, f must be able to accept an Interval object as an argument, which is satisfied with your example. So, load the package: Needs["NumericalMath`IntervalRoots`"] We'll use IntervalBisection because your function is discontinuous. With IntervalNewton, the roots are found more quickly, but the discontinuities may cause some roots to be missed. The default value for MaxRecursion is a bit low, so we'll increase it. In[10]:= With[{a = 10}, IntervalBisection[Cot[x] - x/a + a/(4*x), x, Interval[{0, 50}], 0.1, MaxRecursion -> 30]] Out[10]= Interval[{2.24609, 2.34375}, {4.6875, 4.78516}, {7.42188, 7.51953}, {10.2539, 10.3516}, {13.2812, 13.3789}, {16.2109, 16.3086}, {19.3359, 19.4336}, {22.3633, 22.4609}, {25.4883, 25.5859}, {28.6133, 28.7109}, {31.6406, 31.7383}, {34.7656, 34.8633}, {37.8906, 37.9883}, {41.0156, 41.1133}, {44.1406, 44.2383}, {47.2656, 47.3633}] You can refine the search for roots by choosing a smaller eps than 0.1, or you can use FindRoot with these smaller intervals. Here we repeat with an eps of 10^-6: In[11]:= With[{a = 10}, IntervalBisection[Cot[x] - x/a + a/(4*x), x, Interval[{0, 50}], 10^-6, MaxRecursion -> 30]] Out[11]= Interval[{2.28445, 2.28445}, {4.76129, 4.76129}, {7.46368, 7.46368}, {10.3266, 10.3266}, {13.2862, 13.2862}, {16.3031, 16.3031}, {19.3552, 19.3552}, {22.4298, 22.4298}, {25.5197, 25.5197}, {28.6202, 28.6202}, {31.7285, 31.7285}, {34.8426, 34.8426}, {37.961, 37.961}, {41.0829, 41.0829}, {44.2075, 44.2075}, {47.3344, 47.3344}] Let me repeat the advantage of IntervalBisection. All of the roots to your equation between 0 and 50 are guaranteed to lie in one of the above intervals. Another nice thing with IntervalBisection is that for functions with a finite number of roots, your starting interval can be Interval[{-Infinity,Infinity}]. The disadvantage, of course, is that most special functions cannot accept Interval objects as input. Carl Woll Wolfram Research

**References**:**sorting list of roots af a transcendental function***From:*Dule <dule23@gmx.de>