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

*ξ**n* =

*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.