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

To prove the statement, we can use the definition of the Lagrange basis polynomials and the properties of interpolation.<br /><br />The Lagrange basis polynomials are defined as follows:<br />$L_{i}(x) = \prod_{j=0, j \neq i}^{N} \frac{x - x_{j}}{x_{i} - x_{j}}, \quad i = 0, 1, \ldots, N$<br /><br />Now, let's consider the sum $\sum_{i=0}^{N} L_{i}(x)$. We can rewrite this sum as:<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 />Notice that each term in the sum is a product of terms of the form $\frac{x - x_{j}}{x_{i} - x_{j}}$, where $i \neq j$. When we expand this product, each term will cancel out with another term in the sum, except for the terms where $i = j$. Therefore, we have:<br />$\sum_{i=0}^{N} L_{i}(x) = \sum_{i=0}^{N} \prod_{j=0, j \neq i}^{N} \frac{x - x_{j}}{x_{i} - x_{j}} = 1$<br /><br />Thus, we have proved that $\sum_{i=0}^{N} L_{i}(x) = 1$.