# Lagrange polynomial $x^n$ coefficient

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)