Finite differences on scalar? Matrix?

[divB]

Hi,

In a paper I have

v_{n,k} = \Delta^K ( (-1)^n n^k y_n )

with n = K, \dots , N-1, k = 0, \dots, K and N = 2K

where \Delta^K is the Kth finite difference operator.

As you can see, all v_{n,k} consistute an (N-K) \times (K+1) matrix.

So without the \Delta's, each v_{n,k} would be a scalar. I do not see how to calculate the finite difference of a scalar?!

Well, probably it is not a finite difference. But can anybody tell me what could be meant with that?

Regards,
divB

 

[HallsofIvy]

I don't understand. You say that \Delta^K is the "Kth finite difference operator" and ask what would v_{n, k} be without the \Delta? It would no longer be a finite difference!

(The finite difference of a scalar function, at some n, would be f(n+1)- f(n).)

 

[divB]

Hi,

Yes, this is exactly my question! I do not understand how it is meant!

Maybe you can take a look at http://biblion.epfl.ch/EPFL/theses/2001/2369/EPFL_TH2369.pdf, page 55-56.

I again try to explain: Suppose you have a sequence y_n consisting of N values. Now consider the expression

(-1)^n P(n) y_n where P(n) is a polynomial of order K in the variable n. The authors argue that this term vanishes if K finite differences are applied. This yields equation 3.14-3.16 in the link above and can be written as matrix equation:

<br /> \Delta^K ((-1)^n P(n) y_n) = \sum_{k=0}^K p_k \underbrace{\Delta^K ((-1)^n n^k y_n)}_{v_{n,k}} = \mathbf{V} \cdot \mathbf{p} = 0<br />

My question is how to calculate the matrix V. The algorithm 3.1 in the link above tells me exactly what I have asked in the original post but I do not know how to actually calculate it.

It would be easier to understand if there would be just one dimension.

But this brings me to an idea: The finite difference is the difference dependent on n, isn't it?

So for the first column I set k=0 and have the sequence (-1)^n n^0 y_n and the column are just the discrete differences of the sequence?

E.g. if (-1)^n n^0 y_n = \left\{1, 4, 9, 8, 10\right\} then the first column would be \left[3 , 5 , -1 , 2\right]^T?

And the same for columns 1,...K?

Maybe this is the solution?

Regards, divB