I am currently trying to explain why it is the case that when given a Lagrange polynomial with coefficients of all $1$
$P_3(x)$ $=$ $L_0(x)$ + $L_1(x)$ + $L_2(x)$ + $L_3(x)$ $=$ $1$, for all $x$.
What I think I know is that when we are given n+1 data, then there exists a polynomial that passes through all of these data points, and this is simply just $y = 1$?
I'm having a really hard time intuitively understanding what exactly this means. Any guidance in the right direction is appreciated.
Lagrange polynomial for point $x_i$ in interpolating $f(x)$ at $\{x_0,x_1,\dots,x_n\}$, doesn't depend on values of $f$ at these points. the respective interpolated polynomial would be $p(x) = \sum_{i} f(x_i) L_i(x)$. so for any set of points $\{x_0,x_1,\dots,x_n\}$ you can always interpolate function $f(x)=1$ which is indeed a polynomial (of degree 0) and its interpolation is equal to itself. in this case the interpolated polynomial becomes: $$ 1= f(x) = p(x) = \sum_i f(x_i) L_i(x) = \sum_i L_i(x). \square $$
you can actually prove a lot of relations for lagrange polynomials with these kind of technique, e.g: $$ \sum_i \frac{d}{dx}L_i(x) = 0 \\ \sum_i x_i^k L_i(x) = x^k \qquad k\leq n $$