Approximating function derivatives

[Pekis]

I've found a more iterative way, and it seems interesting:

f'(x0) ~= a * f(x0-2h) + b * f(x0-h) + c * f(x0+h) + d * f(x0+2h):
=> a = 1/(12h), b = -8/(12h), c = 8/(12h), d = -1/(12h)

But after all, if the first derivative f'(x0) ~= a * f(x0-2h) + b * f(x0-h) + c * f(x0+h) + d * f(x0+2h) (already very good precision), then the 2nd derivative f"(x0) ~= a * f'(x0-2h) + b * f'(x0-h) + c * f'(x0+h) + d * f'(x0+2h) will have good precision, and so on, with the same a, b, c, d coefficients from above:

So, if you want it, f"(x0) ~= a * f'(x0-2h) + b * f'(x0-h) + c * f'(x0+h) + d * f'(x0+2h):
the f'(x0-2h) term has to be calculated as a * f(x0-4h) + b * f(x0-3h) + c * f(x0-h) + d * f(x0) (since f'(x0) ~= a * f(x0-2h) + b * f(x0-h) + c * f(x0+h) + d * f(x0+2h))
the f'(x0-h) term has to be calculated as a * f(x0-3h) + b * f(x0-2h) + c * f(x0) + d * f(x0+h)
the f'(x0+h) term has to be calculated as a * f(x0-h) + b * f(x0) + c * f(x0+2h) + d * f(x0+3h)
the f'(x0+2h) term has to be calculated as a * f(x0) + b * f(x0+h) + c * f(x0+3h) + d * f(x0+4h)
We can reuse previous calculations but also have to add new ones:
f(x0-4h), f(x0-3h), f(x0+3h), f(x0+4h).

And if you want it, f3(x0) ~= a * f"(x0-2h) + b * f"(x0-h) + c * f"(x0+h) + d * f"(x0+2h):
the f"(x0-2h) term has to be calculated as a * f'(x0-4h) + b * f'(x0-3h) + c * f'(x0-h) + d * f'(x0)
the f"(x0-h) term has to be calculated as a * f'(x0-3h) + b * f'(x0-2h) + c * f'(x0) + d * f'(x0+h)
the f"(x0+h) term has to be calculated as a * f'(x0-h) + b * f'(x0) + c * f'(x0+2h) + d * f'(x0+3h)
the f"(x0+2h) term has to be calculated as a * f'(x0) + b * f'(x0+h) + c * f'(x0+3h) + d * f('x0+4h)
So, we can reuse previous calculations but also have to add new ones:
f'(x0-4h) = a * f(x0-6h) + b * f(x0-5h) + c * f(x0-3h) + d * f(x0-2h)
f'(x0-3h) = a * f(x0-5h) + b * f(x0-4h) + c * f(x0-2h) + d * f(x0-h)
f'(x0+3h) = a * f(x0+h) + b * f(x0+2h) + c * f(x0+4h) + d * f(x0+5h)
f'(x0+4h) = a * f(x0+2h) + b * f(x0+3h) + c * f(x0+5h) + d * f(x0+6h)
We can reuse previous calculations but also have to add new ones:
f(x0-6h), f(x0-5h), f(x0+5h), f(x0+6h).

The precision seems good, as shown in the Excel attachment here, which also shows the building tree:

   

What do you think of it ?