How can we show using Lagrange interpolation polynomial that $$\sum_{i=0}^n y_i \prod_{j=0, j\ne i}^n \frac{1}{x_i-x_j}$$ is the coefficient of $x^{n}$?

I know that $f[x_0,x_1, .. x_n]$ is the coefficient of $x^{n}$ in Newton polynomial $$\sum_{i=0}^n f[x_0,x_1, .. x_i] \prod_{j=0, }^{i-1} (x-x_i)$$ So $$\bigg[f[x_0,x_1,..x_n]\bigg] * x^n $$should be equal (?) to $$\bigg[\sum_{i=0}^n y_i \prod_{j=0, j\ne i}^n \frac{1}{x_i-x_j}\bigg] * x^n $$ There is possible to show that using above facts?


The (interpolation)polynomial $P_n$ can be written as $\sum_{i=0}^nC_ix^i$ (Monomial basis),

and $C_n$ can be expressed as: $C_n = \sum^n_{i=0}y_i\prod^n_{j=0,j\neq i}\frac{1}{x_i-x_j}$

Proof from Langrange-interpolation:

$L_n = \sum^n_{i=0}y_i\prod^n_{j=0,j\neq i}\frac{x-x_j}{x_i-x_j}$

If we expand the product $(x-x_0)...(x-x_{n_1})$ to $x^n + ...x^{n-1} + etc.$

And we only look at the $x^n$ component, $C_n$ follows.

The same holds up for Newton-interpolation:

$N_n = \sum_{i=0}^nf[x_0,...,x_i](x-x_0)...(x-x_{i-1})$

Then we get: $C_n = f[x_0,...,x_n]$.

Thus, written differently:

$f[x_0,...,x_n] = \sum^n_{i=0}\frac{y_i}{\prod^n_{j=0,j\neq i}(x_i-x_j)}$ (https://en.wikipedia.org/wiki/Divided_differences)


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