CHAPTER 02.02: DIFFERENTIATION OF CONTINUOUS FUNCTIONS: Accuracy of Divided Difference Formulas: Part 2 of 2   So in order to be able to do that, I will do as follows, I will say f of x plus delta x is equal to f of x, plus f prime of x times delta x, plus f double-prime of x times delta x squared by 2, plus order of delta x cubed terms, okay, because those are the terms which are left off now. Now, what I can do is, I can choose a point delta x to the left of x, so, or you can algebraically look at it as that I'm substituting for delta x, I'm substituting minus delta x.  So I get f of x, plus f prime of x times minus delta x, plus f double-prime of x times delta x squared, divided by . . . or minus delta x, divided by factorial 2, this should be factorial 2, plus the order of delta x cubed terms here.  If I subtract the two, what I'm going to get is, subtract the two, I'm going to get f of x plus delta x, minus f of x minus delta x, and what's going to happen is these two are going to cancel, these two are going to add up, I get 2 times f prime of x, delta x.  This term is going to cancel because this is minus delta x squared, minus delta x squared is going to become plus delta x squared, so it's going to cancel with this term here, so there will be nothing left over, plus the order of delta x cubed terms.  So I'll have this term and this term here give me another term which is of the order of delta x cubed.  So from here, if I divided by delta x throughout, I get f of x plus delta x, minus f of x minus delta x, divided by 2 times delta x will be equal to f prime of x, but since I'm dividing by 2 times delta x, this means that the order of the terms which are left over here will be of the order of delta x squared terms.  So, that gives me f prime of x is equal to f of x plus delta x, minus f of x minus delta x, divided by 2 times delta x, plus the order of delta x squared, just because you're taking it to the left-hand side, it does not . . . does not matter what happens to the sign, because these are simply terms of the order of delta x squared.  And what you are finding out here is that this particular formula here is nothing but your central divided difference formula, that's your central divided difference formula, and the central divided difference formula has the remaining, or the error term, is proportional to delta x squared, not delta x anymore, as opposed to when we were talking about forward divided difference scheme.  So that's why what happens is that as delta x is halved, your error gets approximately quartered.  So, that's why the central divided difference scheme gives you a much more accurate answer than your forward or backward divided difference scheme, because the central divided difference scheme, the error is proportional to the order of delta x squared, while in forward and backward divided difference scheme, this term is not squared, but just delta x.  And that's the end of this segment.