Re: A basic question about RecurrenceTable - warning messages

*To*: mathgroup at smc.vnet.net*Subject*: [mg122321] Re: A basic question about RecurrenceTable - warning messages*From*: victorphy <vbapst at gmail.com>*Date*: Tue, 25 Oct 2011 06:16:49 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j80qk9$ag5$1@smc.vnet.net>

On 23 oct, 12:34, Dana DeLouis <dan... at me.com> wrote: > 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. :>( > Hello Dana, thank you for your detailled answer. As a matter of fact this was just a silly example to test things and explain my problem ! But thank you for the advices you gave me anyway ! Best regards, victor