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So lets suppose that we have the following

$x_0=-1, y_0=2$

$x_1= 0, y_1=1$

$x_2=1, y_2=2$

$x_3=3, y_3=10$

and we know that all the above $x_i,y_i$ belong to $p_2(x)=x^2+1$ , and we want to add and extra $x_4=2$ where $y_4=-7$ and we want to find by using the Divided Differences method a new polynomial $p$ that includes $x_4$. My question is do i have to do all the calculations again from the beginning? Since it's only one extra $x_i$ i feel like i dont have to calculate all the divided differences again but i am not sure. Isn't there a smarter way to find the new polynomial?

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Indeed, the Newton-(interpolation)polynomial can be as $P_n = P_{n-1} + [y_0,...,y_n](x-x_0)...(x-x_{n-1})$. And the divided differences $[y_0,..,y_4]$ (in your case) can be expressed in recursion: $[y_0,...,y_4] = \frac{[y_0,...,y_3]-[y_1,...,y_4]}{x_0-x_4}$. Or more directly: $$[y_0,...,y_n] = \sum^n_{i=0}\frac{y_i}{\prod^n_{j=0,j\neq i}(x_i-x_j)}$$ to compute the polynomial. If you would continue adding points, use the recursive expression and make table of the 'divided differences' for convenience. (https://en.wikipedia.org/wiki/Divided_differences)

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