# Thread: Second Order Differentiation of Discrete Data

1. ## Second Order Differentiation of Discrete Data

Hi,
It's been a couple years since I studied calculus but I wondered if it was possible to calculate the double differential of a set of discrete data points, and if so, how do I go about doing this. What I mean is:

If I have a sequence of six data points: [1,2,3,4,5,6], can I process these in some way so that I find out what the second order differential is in the middle of this sequence (i.e. between 3 and 4). Obviously for this data it's 0 but I wondered if there was a standard method that could be applied to any sequence?

Something in my head says it's to do with the Taylor series or finite difference, but I am struggling to follow them and I might be way off.
Thanks so much.

2. Originally Posted by CA90210
Hi,
It's been a couple years since I studied calculus but I wondered if it was possible to calculate the double differential of a set of discrete data points, and if so, how do I go about doing this. What I mean is:

If I have a sequence of six data points: [1,2,3,4,5,6], can I process these in some way so that I find out what the second order differential is in the middle of this sequence (i.e. between 3 and 4). Obviously for this data it's 0 but I wondered if there was a standard method that could be applied to any sequence?

Something in my head says it's to do with the Taylor series or finite difference, but I am struggling to follow them and I might be way off.
Thanks so much.
first derivate from step function is impulse function... you can't derivate impulse function $\displaystyle \delta (t)$
because of definition of impulse function it self

your point's are like if you wrote $\displaystyle \big{\{}\delta [k] , 2\delta [k], 3 \delta [k], 4 \delta [k], 5\delta [k], 6\delta [k] \big{\}}$

Edit: sorry i totally have seen there something else sorry.....

3. Originally Posted by yeKciM
first derivate from step function is impulse function... you can't derivate impulse function $\displaystyle \delta (t)$
because of definition of impulse function it self

your point's are like if you wrote $\displaystyle \big{\{}\delta [k] , 2\delta [k], 3 \delta [k], 4 \delta [k], 5\delta [k], 6\delta [k] \big{\}}$
Thanks for the very quick reply.

Sorry, I might be completely missunderstanding but I thought that a step function was only for single dimensional data? My understanding was that as my data is a sequence it could be almost be considered as a set of six two dimensional points.

i.e. if I take the sequence [a,b,c,d,e] then the points would be: [(1,a),(2,b),(3,c),(4,d),(5,e)].

Does that make any sense?

4. There are a number of different methods you might try. Derivatives are not defined at a single point, so you'd have to "connect the dots" in some way in order to get a continuous function (which is what you were talking about anyway: you seem to want the second derivative in-between data points). You have to decide how you want to do that. If you need the second derivative to be continuous, then you might try cubic splines to interpolate your data. If you go that route, pay attention to how you want the splines to look at the endpoints of your intervals.

5. Originally Posted by Ackbeet
There are a number of different methods you might try. Derivatives are not defined at a single point, so you'd have to "connect the dots" in some way in order to get a continuous function (which is what you were talking about anyway: you seem to want the second derivative in-between data points). You have to decide how you want to do that. If you need the second derivative to be continuous, then you might try cubic splines to interpolate your data. If you go that route, pay attention to how you want the splines to look at the endpoints of your intervals.
Ah, thank you.

Just to check my understanding. I would fit a cubic spline to my six data points (with some boundary condition), and then I could calculate the second derivative at any point i like using the standard method for a polynomial equation?

6. You got it.

7. Sweet, thanks very much for your help guys.

8. You're welcome. Have a good one!