Quote:
Originally Posted by EarlyMon
Um. f(x) is literally function of x  and does not mean the variable f is applied to x.
How about functions of the form 
where the last sum is simply the first sum rewritten using the definitions ξn = n/ T, and Δ ξ = ( n + 1)/ T − n/ T = 1/ T.No. In the above example, there is no variable f that is applied to variable x.
The f in f(x) is simply a placeholder  for the operations that are applied to x.
We've been trained to think that way for a reason. Commutation  it comes from commutation notation and means 2*3.
Honest.
So, by direct evaluation, the equation must become 6/2*3  as had been said before.

This post makes me realize how much I dislike math. (That is math, right?
)