2

Let $L_i(x)$ be the Lagrange polynomials for pairwise different support abscissas $x_0, \ldots, x_n$. Show that

$$\sum_{i=0}^n L_i(0)x_i^j = (-1)^nx_0x_1\cdots x_n \text{ for } j = n+1$$

I have previously shown that $\sum_{i=0}^n L_i(x) = 1$ and $\sum_{i=0}^n L_i(x)x_i^j = x^j$ for $j = 0, 1, \ldots, n$, but I am unsure of how to extend this to the $j = n+1$ case. Any help would be appreciated!

Michael Hardy
  • 1
Walter
  • 21

1 Answers1

2

If you interpolate $f(x)=x^{n+1}$ with the Lagrange polynomial $P(x) = \sum_{i=0}^{n} L_i(x)x_i^{n+1}$ you have that the error is $x^{n+1} - \sum_{i=0}^{n} L_i(x)x_i^{n+1} = \frac{1}{(n+1)!}f^{(n+1)}(\theta_x)\Pi_{i=0}^n(x-x_i)$

Now evaluate this equation in zero and you get what you are trying to prove.

McNuggets666
  • 1,583