I don't know an appropriate topic for this. Here is my question though.

How does one define change in something? For example, regarding Riemann sums one defines change in the width of the rectangles to be $$ \Delta x_i = x_i - x_{i-1}~~~,~~~i=1,2,\ldots, n$$

Would I be wrong in defining change as $$\Delta x_i = x_{i+1} - x_i~~?$$ In other words, if I have an interval $[x_0, x_1]$, would I write $\Delta x_0 = x_1 - x_0$ or $\Delta x_1 = x_1 - x_0$ and why?

  • 2
    $\begingroup$ math isn't an opinion, but notation sure is. $\endgroup$ – Exodd Jun 10 '15 at 16:01
  • $\begingroup$ Both are fine to use. $\endgroup$ – muaddib Jun 10 '15 at 16:04

As you mention these definitions for change of an interval appear to be very similar. For Reimann sums it seems that the issue is just a notational shift, but when averaging over multiple intervals we get different types of properties. You are free to define these intervals or grids however you want, but sometimes interesting properties emerge, especially in the context of finite differences and derivatives.

For example, a common definition for an approximation to the derivative could be $f'(x) = (f(x+h) - f(x))/h = (f(x_{i+1})-f(x_i))/h$. This is known as a 'forward difference' and has first order accuracy in the size of your interval $h$. So if you decrease the size of your $h$, the error in your derivative approximation will correspondingly decrease by that amount.

On the other hand if we define the approximation to the derivative be the 'centered difference' : $f'(x) = (f(x+h) - f(x-h))/(2h) = (f(x_{i+1})-f(x_{i-1}))/(2h)$ we can show that this results in $O(h^2)$ accuracy - i.e. if your $h$ is twice as small, the error in your derivative approximation will be 4 times less.

Thus in summary, the notational indexing for your grid or spacing might be just a choice of notation, but how you use that grid can make very big differences.

| cite | improve this answer | |
  • 1
    $\begingroup$ I think it's $2h$ in the central difference $\endgroup$ – Exodd Jun 10 '15 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.