Re: difference equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg24361] Re: difference equation*From*: Leszek Sczaniecki <lsczan at home.com>*Date*: Sun, 9 Jul 2000 04:53:03 -0400 (EDT)*Organization*: @Home Network*References*: <8k3l8a$3af@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Wilfried Stock wrote: > Hi > I would like to built a difference equatition like: > Yt=A+b*Yt-1-c*Yt-2 > with Y=f(X) > I don't find an easy solution. > Can anyone provide a solution? > > Thanks in advance, > Wilfried In old days this type of equations used to be often solved in the following way (subtle mathematical details omitted) y[n+2] == a + b y[n+1] - c y[n] y[1] == y1 , y(0) == y0 There is time for some tricks now. We multiply both sides of the equation by x^n/n! y(n+2) x^n/n! == a x^n/n! + b y(n+1)x^n/n - c y(n)x^n/n (d/dx)^2 y[n+2] x^(n+2)/(n+1)! == a x^n/n! + b (d/dx)y(n+1)x^(n+1)/(n+1!) - c y(n)x^n/n! We define a function ( called the generating function) f(x) == Sum[y(n) x^(n)/(n+1)!, {n, 0, Infinity}] The original equation has been transformed to a second order differential equation with constant coefficients (and known solution :-)) (d/dx)^2 f[x] - b (d/dx) f[x] + c f[x] == a Exp[x] with initial conditions ((d/dx) f)[0] == y1 and f[0] == y0 . y[n] is the n-th derivative of the generating function f with respect to x taken at x==0. These days happy owners of Mathematica have another option. In[1]:= << "DiscreteMath`RSolve`" In[2]:= sol = RSolve[{y[n + 2] == a + b*y[n + 1] - c*y[n], y[1] == y1, y[0] == y0}, y[n], n] // FullSimplify Out[2]= {{y[n] -> 1/(Sqrt[b^2 - 4*c]*(-1 + b - c))* 2^(-1 - n)*(Sqrt[b^2 - 4*c]*(a*(-2^(1 + n) + (b - Sqrt[b^2 - 4*c])^n + (b + Sqrt[b^2 - 4*c])^n) + ((b - Sqrt[b^2 - 4*c])^n + (b + Sqrt[b^2 - 4*c])^ n)*(-1 + b - c)*y0) + (b - Sqrt[b^2 - 4*c])^n*(a*(-2 + b) + (-1 + b - c)*(b*y0 - 2*y1)) - (b + Sqrt[b^2 - 4*c])^n* (a*(-2 + b) + (-1 + b - c)*(b*y0 - 2*y1)))}} In[4]:= y[n_] = y[n] /. First[sol]; In[5]:= FullSimplify[y[0]] Out[5]= y0 In[6]:= FullSimplify[y[1]] Out[6]= y1 It looks like this is a correct solution (at least for 0 <= n <=10). In[7]:= Table[FullSimplify[y[n + 2] - (a + b*y[n + 1] - c*y[n])], {n, 1, 10}] Out[7]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0} Regards, -- Leszek Sczaniecki