I have a question from a past exam paper, with no solutions provided, that I'm stuck on.
Consider the set of interpolation nodes:
$$\begin{array}{|l|cr}
x & -1 & 1 & 2\\
\hline
y & -2 & 0 & 4
\end{array}$$
(a) Use Lagrange interpolation to calculate an estimate of $y$ at $x=0$.
(b) Given that $\lvert f^n(x)\rvert \lt 10^\frac {n}{3}$, for $-4\le x\le 4$ and $n=1,2,…,5$, what is an upper bound on error in your approximation?
So for part (a) I got that $y=-2$ but I'm really having a hard time with part (b). My notes explain that the error formula is $E=\frac {1}{(n+1)!}f^{(n+1)}(\xi)(x-x_0)(x-x_1)…(x-x_n)$ and I have no idea how to work out what $f^{(6)}(\xi )$ is in this case is. I've seen examples in places like page two of here: https://math.okstate.edu/people/binegar/4513-F98/4513-l16.pdf where it's a sine or cosine function, or $\frac 1x$, and that makes sense to me. If anyone could help me out that'd be great. Thanks in advance.
Best Answer
Since their formula $$\lvert f^{(n)}(x)\rvert \lt 10^\frac {n}{3}$$ is a general formula for $n=1,2,3,4,5$, the question is not asking you to estimate the error for all different $n$'s. In your case, $n=2$.
So the error bound will be $$\frac{1}{3!}f^{(3)}(\xi)(x+1)(x-1)(x-2)$$
where $x=0$ and $|f^{(3)}(\xi)|<10^{2/3}$. This gives $$\frac{1}{3}10^{2/3}$$