Use Taylor expansion to show that for $u \in C^4([0,1]) $ $$ max |\partial^+\partial^-u(x) - u''(x)| = \mathcal{O}(h^2)$$ For $x \in [0,1]$ and where the second order derivative $u''$ can be approximated by the central difference operator defined by $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
I know what the Taylor expansion of $u$ around $x+h$ looks like, but I don't know how to evaluate $u''(x)$ other than setting it equal to the approximation $\partial^+\partial^-u(x)$ which makes the left side equal $0$.