Intuitively why does the central difference method provide us a more accurate value of the derivative of a function than forward/backward difference? From what I have read recently,there are multiple methods to obtain the value of delta y to compute the derivative of a function, delta y/delta x. For as long as I have been learning about derivatives at school, I have only been exposed to the forward difference method, taught as the 'First Principle Method' which was then superseded by the direct usage of equations of common functions, such as trigonometric functions for solving questions
Recently, I learnt about how the central difference method is more accurate. After a lot of pondering of the equation for central difference, I can not quite understand precisely what makes it more accurate.
From what I understand of the forwards/backwards method, delta y is obtained by subtracting a function f(x) by another function f(z) where z = x incremented/decremented by a very small value. This is fairly intuitive because it is quite literally the very definition of delta y or change in a function.
But the central difference method doesn't make sense to me. It is defined as delta y = f(x + h/2) - f(x - h/2)
How is this equal to the change in a function? f(x + h/2) is not the same as f(x) by any means. From the few explanations I saw online, they claim this method is the 'average' of the forwards and backwards methods. I can't quite tell how it is the average of both the methods. Wouldn't the average be something similar to:
delta y = (f(x+h) - f(x-h))/2.
I have also seen another explanation which states that through the Taylor's Series, it is more accurate. For one, the explanation seems more like stating a fact than intuition to me, so I am still left confused. For another, I tried searching up as to what a Taylor's series is, as I have not heard of it before, and I am left with more questions than before.
Thus I am left with the following questions:
(i) I still can't quite tell how this is a change in a function
(ii) How is it the average of both the functions? Why is it more accurate simply because it is the average of both methods?
(iii) Intuitively why does it work?
 A: One way to explain the accuracy of these various finite-difference approximations is to use Taylor series to estimate the errors in the approximations.
Suppose that $f: \mathbb R \to \mathbb R$ is a smooth function and $x \in \mathbb R$. A Taylor series approximation tells us that
$$
\tag{1} f(x + h) \approx f(x) + f'(x)h + \frac12 f''(x) h^2 + \frac16 f'''(x) h^3
$$
which implies that
$$
\frac{f(x+h) - f(x)}{h} \approx f'(x) + \underbrace{\frac12 f''(x) h}_{\text{error term}}.
$$
(If $h$ is small then the higher order terms are negligible.)
So the error is proportional to $h$, to a good approximation.
Note also that
$$
\tag{2} f(x-h) \approx f(x) - f'(x) h + \frac12 f''(x) h^2 - \frac16 f'''(x) h^3.
$$
Subtracting (2) from (1) and then dividing both sides by $2h$, we obtain
$$
\frac{f(x + h) - f(x-h)}{2h} \approx f'(x) + \underbrace{\frac{1}{3!} f'''(x) h^2}_{\text{error term}}.
$$
So the error is proportional to $h^2$, to a good approximation.
If $h$ is small then $h^2$ is much smaller than $h$. This explains why the centered difference approximation of $f'(x)$ is more accurate than the forward-difference approximation (at least when $h$ is sufficiently small).
A: There's something called the mean value theorem.  The exact statement can be found at that Wikipedia link, but the basic gist of it is that if you draw a secant (a bit like a chord in a circle) across two points on a smooth curve, then somewhere in between those two points is a point where the tangent is parallel to that secant.  Like this:

The theorem doesn't tell you exactly where this blue point is, but it's there somewhere.  Intuitively, though, if the curve changes direction somewhat consistently, you can tell that the blue point will be close to halfway between the two black points, if not necessarily exactly halfway.
That's the intuition between the central difference estimate of the derivative.  Let's take the function $f(x) = x^2$, and try to estimate the derivative at $x = 1$, $f'(1)$.  Suppose we have three choices of how to do this:

*

*forward: $\frac{f(2)-f(1)}{2-1} = \frac{4-1}{1} = 3$

*backward: $\frac{f(1)-f(0)}{1-0} = \frac{1-0}{1} = 1$

*center: $\frac{f(3/2)-f(1/2)}{(3/2)-(1/2)} = \frac{(9/4)-(1/4)}{1} = 2$
Depicted in a graph, those look like this:

The forward and backward are shown with dotted black lines, while the center is shown with a solid black line.
