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Thread: characterristic equation

  1. #1 characterristic equation 
    Forum Junior ArezList's Avatar
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    there're some problems about sequence of number that are solve by using the characterristic equation, but i don't full understand the principle. for e.g : It is said that:
    . a sequence of number X(n+2)=C1X(n+1)+C2X(n)
    given r,s then X(n+2)-rX(n+1)=s[X(n+1)-rXn]
    so X(n+2)=(s+r)X(n+1)-srXn

    C1=s+r
    C2=-sr
    then eliminate s , then we got the so-called characterristic equation will be : r*r-C1*r-C2=0

    I just don't understand that how can the 'S' can be eliminated, since X(n+2), Xn and X(n+1) can' t be combined?
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  3. #2  
    Forum Ph.D.
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    C<sub>1</sub> = s + r
    C<sub>2</sub> = −sr

    Multiplying the first equaiton by r gives

    C<sub>1</sub>r = sr + r<sup>2</sup>

    Now add this to the second equation. That’s how you eliminate s.


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  4. #3  
    Forum Junior ArezList's Avatar
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    No , i mean where are the Xn, X(n+1)...?
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  5. #4  
    Forum Professor serpicojr's Avatar
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    I don't understand your argument, but I know what you're trying to solve. So suppose we have a recurrence relation:

    x<sub>n+1</sub> = c<sub>1</sub>x<sub>n</sub>+c<sub>2</sub>x<sub>n-1</sub>

    We can describe this by a matrix equation:



    Then the characteristic polynomial of the 2x2 matrix is t<sup>2</sup>-c<sub>1</sub>t-c<sub>2</sub>, and this helps us express x<sub>n</sub> as a sum of exponentials. Namely, let r<sub>i</sub>, i = 1,2, be the roots of the characteristic polynomial. Then it's not too hard to see that, if the roots are distinct, we have that:



    are the eigenvectors of the matrix, each with eigenvalue r<sub>i</sub>. (If the roots are not distinct, then there's a unique eigenvector, which looks the same as above, but this implies the matrix is not diagonalizable, which makes things a little murkier, so let's ignore this case.) So express:



    And this gives you:



    In particular, you have x<sub>n</sub> = a<sub>1</sub>r<sub>1</sub><sup>n</sup> + a<sub>2</sub>r<sub>2</sub><sup>n</sup>.
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