Sum of Lagrange polynomials: $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $.

Given $\{x_0,...,x_n\}$ I am asked to show that $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $.

I already showed that $\sum_{i=0}^{n}L_i(x)x_i^{j} = x^j$ for $j=1,...,n$ and that $\sum_{i=0}^{n}L_i(x)=1$ but I don't know how to use this to solve my problem.

If I use the same method by which I solve the other problems then I am basically looking for a polynomial P of degree less than n such that $P(x_i) = x_i^{n+1}$ and $P(0) = (-1)^{n}x_0\cdot\cdots\cdot x_n$

Solution 1:

It can be proven with determinant form of Lagrange polynomial that interpolates $(x_0;y_0)$, $\dots$, $(x_n;y_n)$ $$ P(x) = (-1) \frac{ \det \begin{pmatrix} 0 & y_0 & y_1 & \cdots & y_n \\ x^n & x_0^n & x_1^n & \cdots & x_n^n \\ x^{n-1} & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & 1 & \cdots & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }, $$ presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation". Obviously, setting all $y_i = x_i^{n+1}$ and $x = 0$ we'll get the expression you need, because $$ P(x) = \sum_{i=0}^n\, y_i L_i(x) = \sum_{i=0}^n x_i^{n+1} L_i(x). $$ So I just plug them in $$ P(0) = (-1) \frac{ \det \begin{pmatrix} 0 & x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\ 0 & x_0^n & x_1^n & \cdots & x_n^n \\ 0 & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & 1 & \cdots & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} } $$ Now consider Laplace expansion along the first column $$ P(0) = (-1)^n \frac{ \det \begin{pmatrix} x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\ x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ x_0 & x_1 & \cdots & x_n \\ \end{pmatrix} }{ \det \begin{pmatrix} x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} } $$ and use properties of determinant $$ P(0) = (-1)^n x_0 x_1 \dots x_n \frac{ \det \begin{pmatrix} x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} x_0^n & x_1^n & \cdots & x_n^n \\ x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & 1 & \cdots & 1 \\ \end{pmatrix} }. $$ The latter finishes the proof $$ P(0) = (-1)^n x_0 x_1 \dots x_n $$

For more practical examples you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".

Solution 2:

Hint: $$P(x)=x^{n+1}-(x-x_0)(x-x_1)\cdot\cdot\cdot(x-x_n)$$ is a polynomial of degree $n$.