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.