Re: A basic question about RecurrenceTable - warning messages

*To*: mathgroup at smc.vnet.net*Subject*: [mg122290] Re: A basic question about RecurrenceTable - warning messages*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 23 Oct 2011 06:26:11 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

Hi. I see you found a solution. f[x_,a_]:=-(x-a)^3+(x-a) soltest[a_?NumericQ,b_?NumericQ]:=x/.FindRoot[f[x,a],{x,b}] eqns = {a[n+1]==soltest[n/10,a[n]], a[0]==-1}; RecurrenceTable[eqns, a, {n,10}] {-1,-1.,-0.9,-0.8,-0.7,-0.6,-0.5,-0.4,-0.3,-0.2,-0.1} I may be wrong, but I noticed you have two -1's in the beginning. In the Recurrence table, n goes from 1 to 10. However, there is no a[n-1] to refer to your initial a[0] = -1. Again, I may be wrong, but I -think- it should be a[1]== -1. If we did that, then... eqns = {a[n+1]==soltest[n/10,a[n]], a[1]==-1}; RecurrenceTable[eqns,a,{n,10}] {-1,-0.9,-0.8,-0.7,-0.6,-0.5,-0.4,-0.3,-0.2,-0.1} If we take your initial equation, and substitute z for (x-a), then we get: -(x-a)^3+(x-a) //.(x-a)->z z-z^3 We Factor to get the 3 solution for the zero function: Factor[%] -(-1+z) z (1+z) I believe you are using FindRoot to solve for the 3rd equation (1+z). If we substitute back in: (1+z)/.z -> x-a 1-a+x And solve for x. Basically we get that x is a - 1. Solve[%==0,x] {{x-> a - 1}} So, at 0, we have -1, and then as your numbers increase linearly... Range[0,9]/10. - 1 {-1.,-0.9,-0.8,-0.7,-0.6,-0.5,-0.4,-0.3,-0.2,-0.1} Again, I may be wrong. :>( = = = = = = = = = Dana DeLouis $Version "8.0 for Mac OS X x86 (64-bit) (February 23, 2011)" - To understand recurrence, one must first understand recurrence. = = = = = = = On Oct 21, 7:30 am, victorphy <vba... at gmail.com> wrote: > Hello, > > I have a small and basic question about RecurrenceTable. > > I would like to solve f(x,a)==0 where x is the unknown and a is a > parameter. As f has several zeros I want to follow one zero when > varying a. > > Thus I write something like that : > > f[x_,a_] := -(x - a)^3 + (x - a) > > soltest[a_, b_] := x /. FindRoot[f[x, a], {x, b}] > > RecurrenceTable[{a[n + 1] == soltest[n/10, a[n]], a[0] == -1}, a, {n, > 10}] > > I get the correct output but a lot of warning messages : > > FindRoot::srect: "Value a[n] in search specification {x,a[n]} is not a > number or array of numbers > ReplaceAll::reps: "{FindRoot[f[x,n/10],{x,a[n]}]} is neither a list of > replacement rules nor a valid dispatch table, and so cannot be used > for replacing > > Is there something that I am doing wrong ? Or is it just normal ? > > Thank you very much for your answers. > > Victor