Its known that newton’s interpolation and lagrange interpolation gives the same value All i need is to prove it

Well I came across a problem to find a generalized version ($n+1$ nodes) of first and second order derivatives for Lagrange interpolation polynomial. In some former post, I found an expression for deriving $L_j(x)$, where $L_j$ stands for Lagrange basis polynomial. The expression is as follows: \begin{align} L_j'(x) = \sum_{l\not = j} \frac{1}{x_j-x_l}\prod_{m\not = (j,l)} […]

A foreign book mentioned that “when the Lagrange’s interpolation formula fails (for example with large sample due to Runge’s phenomenon), you should use approximation methods such as Least-squares-method.” I am confused because I have always thought that interpolation/extrapolations are approximations. My confusion lies in the fact that the book used the three terms as disjoint […]

Using Lagrange interpolation (I think identity $\sum \limits_{k=0}^{n}k^{m}\prod \limits_{\substack{i=0\\i\neq k}}^{n}\frac{x-i}{k-i}=x^m$) shows that $$\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}=0 \text{ if }\ 0≤m<n$$ and $$\sum \limits_{k=0}^{n}(-1)^{k}k^{n}\binom{n}{k}=(-1)^{n}n!$$ (same sum with $m=n$). EDIT: I found it quite straightforward to prove first equality (just put x=0 into identity), the second part is definitely harder.

Question: Is there an easier way to solve this problem? Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$. My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manually plugged in each point to get the following system:$$\begin{align*} & 27a+9b+3c+d=2\\ & 64a+16b+4c+d=4\\ […]

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