From the Taylor series expansion, we can write f(x₁) in terms of f(x₀) and its derivatives as follows:

We can also write f(x_-1) in terms of f(x₀) and its derivatives as follows:

Now, let b º (x₁ - x₀)/(x_-1 - x₀); multiply f(x_-1) by b; and subtract it from f(x₁).

(2!)-1(x1 - x0)2(1 - 1/b) [ ¶x2f ]0	=	[ f(x1) - f(x0) ] - b [ f(x-1) - f(x0) ]
		- { (3!)-1(x1 - x0)3 (1 - 1/b2) [ ¶x3 f ]0 + ... }
[ ¶x2f ]0	»	[ f(x1) - (1 - b)f(x0) - bf(x-1)] 2 (x1 - x0)-2(1 - 1/b)-1
		- { (3)-1(x1 - x0) (1 - 1/b2)(1 - 1/b)-1 [ ¶x3 f ]0 + ... }

With this expression, then, we have a FD expression for [ ¶_x²f ]₀ that is accurate to 1^st-order in Dx for a discrete grid with arbitrary grid spacing.

Now, notice what happens when the discrete grid has uniform spacing, i.e., when b = -1 . The coefficient of the third-derivative term goes to zero, and we obtain,

Hence, we are (accidently) able to achieve 2^nd-order accuracy from just a 3-point FD stencil.