# Finite differences on scalar? Matrix?

1. Sep 10, 2010

### 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

2. Sep 11, 2010

### 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).)

Last edited by a moderator: Sep 12, 2010
3. Sep 11, 2010

### 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:

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

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

Last edited by a moderator: Apr 25, 2017