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| ArezList |
 Posted: Mon Jun 23, 2008 7:59 pm Post subject: characterristic equation |
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Forum Freshman
Joined: 22 Mar 2008
Posts: 64
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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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| JaneBennet |
| Posted: Mon Jun 23, 2008 8:18 pm Post subject: |
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Forum Ph.D.
Joined: 06 Apr 2008
Posts: 830
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C1 = s + r
C2 = −sr
Multiplying the first equaiton by r gives
C1r = sr + r2
Now add this to the second equation. That’s how you eliminate s.
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| ArezList |
| Posted: Mon Jun 23, 2008 8:38 pm Post subject: |
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Forum Freshman
Joined: 22 Mar 2008
Posts: 64
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| No , i mean where are the Xn, X(n+1)...? |
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| serpicojr |
| Posted: Mon Jun 23, 2008 10:23 pm Post subject: |
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Forum Professor
Joined: 17 Jul 2007
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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:
xn+1 = c1xn+c2xn-1
We can describe this by a matrix equation:
Then the characteristic polynomial of the 2x2 matrix is t2-c1t-c2, and this helps us express xn as a sum of exponentials. Namely, let ri, 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 ri. (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 xn = a1r1n + a2r2n. |
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