Let x_(0)lt x_(1)lt ... lt x_(N) be N+1 distinct interpolation points. Prove that sum _(i=0)^NL_(i)(x)=1 where L_(i)(x) for i=0,1,... ,N are the Lagrange basis polynomials.

To prove that $\sum_{i=0}^{N}L_{i}(x)=1$, we can use the properties of the Lagrange basis polynomials.<br /><br />The Lagrange basis polynomials are defined as follows:<br />$L_{i}(x) = \prod_{j=0, j \neq i}^{N} \frac{x-x_j}{x_i-x_j}$<br /><br />Now, let's consider the sum $\sum_{i=0}^{N}L_{i}(x)$:<br />$\sum_{i=0}^{N}L_{i}(x) = \sum_{i=0}^{N} \prod_{j=0, j \neq i}^{N} \frac{x-x_j}{x_i-x_j}$<br /><br />We can rewrite this sum as:<br />$\sum_{i=0}^{N}L_{i}(x) = \prod_{j=0}^{N} \frac{x-x_j}{x_i-x_j} \sum_{i=0}^{N} \frac{1}{x_i-x_j}$<br /><br />Notice that the sum $\sum_{i=0}^{N} \frac{1}{x_i-x_j}$ is equal to 1, since it is a telescoping sum.<br /><br />Therefore, we have:<br />$\sum_{i=0}^{N}L_{i}(x) = \prod_{j=0}^{N} \frac{x-x_j}{x_i-x_j} \cdot 1 = 1$<br /><br />Hence, we have proved that $\sum_{i=0}^{N}L_{i}(x)=1$.