I am reading following Lemma. I am confused about determining the end points $a$ and $b$ as suggested in the lemma. If possible any numerical examples which uses this lemma will be enough to clarify my doubt.
Thank you for your kind help.
Help to understand result on divided differnece
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numerical-methods
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0It may help you to see this as a generalization of the mean value theorem. when $n=0$, you have $a=\min(x,x_0)$, $b=\max(x,x_0)$. And the statement says there exists an $\xi$ between $a$ and $b$ ( ie. between $x$ and $x_0$ ) such that $f'(\xi)=\frac{f(x)-f(x_0)}{x-x_0}$ – 2017-01-10
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0Let me remark that this lemma allows us to define $f[x,\dots,x]=\frac{f^{(n)}(x)}{n!}$ ($x$ is taken $n+1$ times), which is useful in writing the Newton form of an interpolating polynomial with multiple nodes. – 2017-01-10
1 Answers
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It will suffice to demonstrate that the divided differences become derivatives at some point. Applying Rolle's theorem you can conclude the following statement:
Let $\{x_0,\cdots,x_n\} \in [a,b] : \exists \zeta \in(a,b) \Rightarrow f[x_0\cdots x_n]n! = f^{(n)}(\zeta)$.
Taking the $nth$ derivative of the $nth$ degree interpolation polynomial:
$$P_n(x) = f[x_0,\cdots,x_n] + \cdots + f[x_0,x_1]$$
With $f[x_0,\cdots,x_n]$ as the leading coefficient of $P_n(x)$. Then take the $nth$ derivative of $P_n$ that's why the factorial appears.
$$P_n^{(n)}(x) = f[x_0,\cdots,x_n]n!$$
Now as we initially know that:
$$f^{(n)}(\zeta) = P_n^{(n)}(x) = f[x_0,\cdots,x_n]n!$$
$$\frac{f^{(n)}(\zeta)}{n!} = f[x_0,\cdots,x_n]$$