1. 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?
[/quote]

2.

3. 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. Thatâ€™s how you eliminate s.

4. No , i mean where are the Xn, X(n+1)...?

5. 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>.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement