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 Post subject: deriving the formula for linear recursionsPosted: Tue, 17 Feb 2009 23:34:38 UTC
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Joined: Fri, 21 Sep 2007 17:15:28 UTC
Posts: 47
Hello!

I know there is a formula related to recursions of the type

where are known. I'm now interested in deriving this formula using the propriety

So if we denote

then it follows that

If we assume is diagonizable then we can write

After I multiply to obtain the equality for I get some ugly multiplications, so I'm wondering what did I do wrong or, eventually, how to simplify the obtained equation to get the correct formula for linear recursions.

Thanks

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 Post subject: Posted: Wed, 18 Feb 2009 06:19:54 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Tue, 20 Nov 2007 04:36:12 UTC
Posts: 826
Location: Las Cruces
You should tell us what "this formula" is.

Suppose we take the matrix approach.If is diagonal and

Then

Since the last expression is easy to compute, some people would call that "the formula" for the solution.

On the other hand, there is an approach to solving the equation based on the "lag operator", as given in the pdf file:

It would interesting to understand how these two approaches are related.

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