Characteristic equation for recurrence equation

[Bruno Tolentino]

An ODE of second order with constants coefficients, linear and homogeneous: Af''(x) + Bf'(x) +Cf(x) = 0 has how caractherisc equation this equation here: Ax^2 + Bx +C = 0 and has how solution this equation here: f(x) = a \exp(u x) + b \exp(v x) where u and v are the solutions (roots) of the characteristic equation and a and b are arbitrary constants.

Very well, until here, no problems!

But, in domain of discrete math, exist an analog equation for each equation above.

Solution equation: f(n) = a u^n + b v^n Caractherisc equation: Ax^2 + Bx +C = 0 Differential equation: ?

I don't know what's the "differential" equation in discrete domain whose solution and characteristic equation are the two equations above. This my question!

Thanks!

 

[pasmith]

Af(n+2) + Bf(n+1) + Cf(n) = 0

 

[Bruno Tolentino]

Af(n+2) + Bf(n+1) + Cf(n) = 0

Thank you very much!

...

Can be too this diference equation: Af(n-2) + Bf(n-1) + Cf(n) = 0 ?

 

[HallsofIvy]

Essentially the same thing. Let m= n- 2 and the difference equation becomes Af(m)+ Bf(m+1)+ Cf(m+ 2)= 0. Now 'look for' a solution of the form f(n)= a^m. f(m+1)= a^{m+1}= a(a^m) and f(m+2)= a^{m+2}= a^2(a^m). Putting those into the equation, it becomes A(a^m)+ Ba(a^m)+ Ca^2(a^m)= 0. Dividing through by a^m gives the "characteristic equation" Ca^2+ Ba+ A= 0, a quadratic equation which will, in general, have two real roots, a_1 and a_2. In that case the general solution to the difference equation is f(n)= C_1a_1^m+ C_2a_2^m= C_1a_1^{n+2}+ C_2a^{-n-2}. If the characteristic equation has a "double root" or two complex roots, the general solution is more complicated but similar.