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Recursive solution of determinants

  1. Aug 9, 2008 #1
    Hi,

    I'm reading a paper where the determinant of the following matrix is solved for using some kind of recurisve method.

    The matrix is given by [tex]M_{ij} = A \delta_{i,j} - B \delta_{i,j-1} - C \delta_{i,j+1}[/tex], with [tex]i,j = 1...N[/tex] and are NOT cyclic.

    The author sets [tex]D_N = \texttt{det}\[M_{(N)}\][/tex] and writes the equation

    [tex]D_N = A D_{N-1} - B C D_{N-2}[/tex],

    assumes a solution of the form [tex]D_N = \lambda ^N[/tex] and finds two solutions,

    [tex]\lambda_{\pm} = (A \pm \sqrt{A^2 - 4 B C}) / 2[/tex].


    He then notes the initial conditions of [tex]D_1 = A[/tex] and [tex]D_2 = A^2 - B C[/tex] and says that the answer is therefore

    [tex]D_N = \frac{\lambda_{+}^{N+1} - \lambda_{-}^{N+1}}{\lambda_{+} - \lambda_{-}}[/tex].


    It's the very last step I don't understand, how did he find [tex]D_N[/tex]?

    Thanks
     
  2. jcsd
  3. Aug 9, 2008 #2
    After a bit of search I see that the method to solve this kind of recurrence relations is to assume a solution of the form [tex]A \lambda_1^N + B \lambda_2^N[/tex] and find A and B from the initial conditions. However this is not exactly the form of the solution here... how come?
     
  4. Aug 9, 2008 #3

    HallsofIvy

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    Why isn't it?
    [tex]D_N= \frac{\lambda_+}{\lambda_+-\lambda_-}\lambda_+^n+ \frac{-1}{\lambda_-(\lambda_+-\lambda_-)}\lambda_-^n[/tex]
    is exactly of the form you give.
     
  5. Aug 9, 2008 #4
    Yes that would be correct, a bad case of dullness. Thanks.
     
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