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Characteristic equation for recurrence equation

  1. Feb 14, 2016 #1
    An ODE of second order with constants coefficients, linear and homogeneous: [tex] Af''(x) + Bf'(x) +Cf(x) = 0 [/tex] has how caractherisc equation this equation here: [tex] Ax^2 + Bx +C = 0 [/tex] and has how solution this equation here: [tex] f(x) = a \exp(u x) + b \exp(v x)[/tex] 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: [tex] f(n) = a u^n + b v^n[/tex] Caractherisc equation: [tex] Ax^2 + Bx +C = 0 [/tex] 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!
     
  2. jcsd
  3. Feb 17, 2016 #2

    pasmith

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    [tex]Af(n+2) + Bf(n+1) + Cf(n) = 0[/tex]
     
  4. Feb 19, 2016 #3
    Thank you very much!!!

    ...

    Can be too this diference equation: [tex]Af(n-2) + Bf(n-1) + Cf(n) = 0[/tex] ???
     
  5. Feb 22, 2016 #4

    HallsofIvy

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    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 [itex]f(n)= a^m[/itex]. [itex]f(m+1)= a^{m+1}= a(a^m)[/itex] and [itex]f(m+2)= a^{m+2}= a^2(a^m)[/itex]. Putting those into the equation, it becomes [itex]A(a^m)+ Ba(a^m)+ Ca^2(a^m)= 0[/itex]. Dividing through by [itex]a^m[/itex] gives the "characteristic equation" [itex]Ca^2+ Ba+ A= 0[/itex], a quadratic equation which will, in general, have two real roots, [itex]a_1[/itex] and [itex]a_2[/itex]. In that case the general solution to the difference equation is [itex]f(n)= C_1a_1^m+ C_2a_2^m= C_1a_1^{n+2}+ C_2a^{-n-2}[/itex]. If the characteristic equation has a "double root" or two complex roots, the general solution is more complicated but similar.
     
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