Re: Recursive function

*To*: mathgroup at smc.vnet.net*Subject*: [mg115937] Re: Recursive function*From*: Dana DeLouis <dana.del at gmail.com>*Date*: Wed, 26 Jan 2011 05:06:52 -0500 (EST)

On Jan 25, 4:20 am, StatsMath <stats.ma... at gmail.com> wrote: > I am trying to compute the following function: > > x[t] = a * x[t-1] + b * x[t-2] > x[0] = 1 > x[1] = 1 > > For different values of a & b. > > Would like to use Manipulate[] so that I can change the values of a & b <snip> Hi. I believe Table could be another option x[t_,a_,b_]:=a*x[t-1,a,b]+b*x[t-2,a,b]; x[0,a_,b_]=1; x[1,a_,b_]=1; Manipulate[Table[x[t,a,b],{t,0,10}],{a,-1,1,1/2},{b,-1,1,1/2}] Perhaps if you wanted to change the number of returned values: x[t_,a_,b_]:=a*x[t-1,a,b]+b*x[t-2,a,b]; x[0,a_,b_]=1; x[1,a_,b_]=1; Manipulate[Table[x[t,a,b],{t,0,terms}],{terms,0,20,1},{a,-1,1,1/2},{b,-1,1,1/2}] As a side note, what you started with looked like you were going the RSolve route: Not as elegant a solution as you have, but... equ={ x[t]=a*x[t-1]+b*x[t-2], x[0]=1, x[1]=1 }; x[t]/.RSolve[equ,x[t],t]//FullSimplify//First << output here >> Replace Sqrt[a^2+4 b] with 'k since Sqrt[a^2+4 b] occurs often. << a little better >> Since k is in the denominator, combinations of a & b could make k = 0 (Divide by zero error). Hence... Limit[ %, k->0] 2^-t a^(-1+t) (a+2 t-a t) Therefore... f[t_,a_,b_]:=Module[{k=Sqrt[a^2+4 b]}, If[Or[t=0,t=1],1, If[k=0, 2^-t a^(-1+t) (a+2 t-a t), (1/k)2^(-1-t) ((a-k)^t (-2+a+k)+(2-a+k) (a+k)^t)]//N//Chop//Rationalize]] Manipulate[Table[f[t,a,b],{t,0,terms}],{terms,0,20,1},{a,-1,1,1/2},{b,-1,1,1/2}] = = = = = = = = = = HTH : >) Dana DeLouis To understand recursion, one must first understand recursion. :>o