Using a Taylor expansion to prove equality

[LogarithmLuke]

Homework Statement

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 $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$

Relevant Equations

$$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$

Homework Statement: 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 $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
Homework Equations: $$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$

Mentor note:
Thread moved from Precalc section, as it is well beyond precalculus types of questions. @LogarithmLuke, please post your questions in the appropriate forum section.

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 $0$

 

[PeroK]

Homework Statement: 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 $$\partial^+\partial^-u(x) = \frac{u(x+h) - 2u(x) + u(x-h)}{h^2} \approx u''(x)$$
Homework Equations: $$u(x+h) = u(x) + hu'(x) \frac{h^2}{2!}u''(x) + ... + \frac{h^{k-1}}{(k-1)!}u^{(k-1)}(x) + \mathcal{O}(h^k)$$

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 $0$.

Hint: if you know how to expand $u(x + h)$ you must know how to expand $u(x -h)$ as well!