7.3 The Symmetric Approximation: f '(x) ~ (f(x+d) - f(x-d)) / (2d)

Using this formula for the "d-approximation" to the derivative is much more efficient than using the naïve formula

Why is it better?

The answer is that the "symmetric formula" is exactly right if f is a quadratic function, which means that the error made by it is proportional to d2 or less as d decreases. The naïve formula is wrong for quadratics and makes an error that is proportional to d.

How come?

Suppose f is a quadratic: f(x) = ax2 + bx + c.

Then we get

f(x + d) = a(x + d)2 + b(x + d) + x


On the other hand, we get

This means that the symmetric approximation is exact for any value of d for any quadratic; no need to make d small; and this is not true for the asymmetric formula.

In general, if our function being differentiated, f(x + d), can be expanded in a power series in d, the first error in our symmetric formula comes from cubic terms, and will be proportional to d2.

The reason this happens is that the d2 term in f(x + d) - f(x - d) cancels itself out, being the same in both terms. The same things happens for all even power terms, by the way; the errors in this approximation to the derivative all come from odd power terms in the power series expansion of f about x.

Thus, if we replace d by , the error in the symmetric approximation will decline by a factor of 4, while the asymmetric formula has error which declines only by a factor of 2 when we divide d by 2.

And so, the symmetric formula approaches the true answer for the derivative much faster than the naïve asymmetric one does, as we decrease d.

Now we ask: can we get even faster convergence?