0

Using Newton's difference form $f''(x_0) \approx 2f[x_{-1} x_0 x_1]$ as an approximation, I need to show that this approximation is first order accurate.

So WTS $f''(x_0) - P''_n(x_0)$ is first order accurate. Using the error formula, we have $f(x_0) - P_n(x_0)= \frac{f^{n+1}(\xi(x_0))}{(n+1)!} \prod_{k=-l}^u(x_0-x_k)$. However, I am having a hard time differentiating this twice. Is this the correct method to take, or is there an easier route?

Lutterbach
  • 627

1 Answers1

1

You know that $f[x_{-1}x_0x_1]=\frac12f''(\mu)$ for some $\mu\in(x_{-1},x_1)$. For a sufficiently smooth $f$, the difference of $f(\mu)$ and $f(x_0)$ is $O(\mu-x_0)$, from which it follows that it is first order accurate, as $|\mu-x_0|\le\max(|x_1-x_0|,|x_{-1}-x_0|)$.

Lutz Lehmann
  • 126,666