- #1

LogarithmLuke

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- 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.

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##

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