Discrete derivatives with finite-differences

[mnb96]

Hello,
I have a function in discrete domain f:\mathbb{Z}\rightarrow \mathbb{R}, and I assume that f is an approximation of another differentiable function g:\mathbb{R}\rightarrow \mathbb{R}.
In other words f(n)=g(n), n\in \mathbb{Z}.

When one wants to approximate the first derivative of g, it is possible to use the forward difference or backward difference operators, which are respecively:

\Delta f(n)=f(n+1)-f(n)
\nabla f(n)=f(n)-f(n-1)

My question is: is it common or allowed to use a mixture of these two operators in the following way:

g'(x) \approx \Delta f(x) for x\geq 0
g'(x) \approx \nabla f(x) for x<0

 

[HallsofIvy]

No, it is not "common" and I can see no reason to do that. Why would you use one for positive x and the other for negative x? It is the change that is positive or negative in the two differences and whether x is positive or negative has nothing to do with that.

 

[mnb96]

I would like to use different operators for positive/negative x's in order to obtain a symmetry I need. In fact, in my case the continuous function g has the property:

g'(x)=-g'(-x), for x\in \mathbb{R}

If f(n)=g(n), n\in \mathbb{Z}, you get \Delta f(n) \neq -\Delta f(-n).
Instead, given g I was able to obtain:

\Delta f(n) = -\nabla f(-n), n\in \mathbb{Z}^+ - \{ 0 \}

EDIT: in simpler words g is an even function and its first derivative is an odd function. The discretized f is even too, but \Delta f is not odd.

 

[Martin Rattigan]

You could presumably use \frac{1}{2}(\Delta f+\nabla f)=(f(n+1)-f(n-1))/2, but that may be too much arithmetic (the shift operation is usually quite quick if you're doing it by computer).

 

[mnb96]

Thanks Martin,
yours is probably a better idea. In fact, the central-difference operator keeps the property "f is even \Rightarrow f' is odd".