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 Δξ = (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.