Finite differences second derivative as successive application of the first derivative

Using these four relations:

$$f(x+\delta) = f(x) +\delta f'(x)+ \delta^2 \frac{1}{2!} f^{''}(x) + \delta^3 \frac{1}{3!} f^{(3)}(x) + \delta^4 \frac{1}{4!} f^{(4)}(x) + \delta^5 \frac{1}{5!} f^{(5)}(a) $$

$$f(x-\delta) = f(x) -\delta f'(x)+ \delta^2 \frac{1}{2!} f^{''}(x) - \delta^3 \frac{1}{3!} f^{(3)}(x) + \delta^4 \frac{1}{4!} f^{(4)}(x) - \delta^5 \frac{1}{5!} f^{(5)}(b) $$

$$f(x+2\delta) = f(x) +2\delta f'(x)+ 4\delta^2 \frac{1}{2!} f^{''}(x) + 8\delta^3 \frac{1}{3!} f^{(3)}(x) + 16 \delta^4 \frac{1}{4!} f^{(4)}(x) + 32 \delta^5 \frac{1}{5!} f^{(5)}(c) $$

$$f(x-2\delta) = f(x) -2\delta f'(x)+ 4\delta^2 \frac{1}{2!} f^{''}(x) - 8\delta^3 \frac{1}{3!} f^{(3)}(x) + 16 \delta^4 \frac{1}{4!} f^{(4)}(x) - 32 \delta^5 \frac{1}{5!} f^{(5)}(d) $$

we can show that:

$$\frac{- f(x- 2\delta) + 16f(x-\delta)-30 f(x) + 16 f(x+\delta) -f(x+2\delta) }{12\delta^2} = f^{''}(x) +O(\delta^4) $$

We note that:

$$\frac{- f(x- 2\delta) + 16f(x-\delta)-30 f(x) + 16 f(x+\delta) -f(x+2\delta) }{12\delta^2} = $$

$$ = \frac{- \frac{f(x+2\delta)-f(x+\delta)}{\delta} + \frac{ f(x-\delta) -f(x-2\delta)}{\delta} -15\frac{f(x)-f(x-\delta)}{\delta} +15\frac{f(x+\delta)-f(x)}{\delta} }{12\delta}$$