Prove Lagrange form of interpolation for $x^j$

Question

To interpolate a polynomial of degree $n$ using the Lagrange form,

$$p(x)=\sum_{i=0}^ny_iL_i(x)$$ with $$L_i(x)=\frac{\prod_{i\not=j}(x-x_j)}{\prod_{i\not=j}(x_i-x_j)}$$

How can I show that for $y_i=x^j_i$ for $j=0, 1, ..., n$ $$\sum_{i=0}^nx_i^jL_i(x)=x^j$$

I read through this question which is the same identity, but the answers were unclear. I tried proving by expanding, so $$p(x)=\sum_{i=0}^n x_i^j \frac{\prod_{i\ne j} (x - x_j)}{\prod_{i \ne j} (x_i - x_j)}=\sum_{i=0}^n x_i^j \frac{(x - x_0)(x - x_1)...(x - x_{i-1})(x - x_{i+1})...(x - x_n)} {(x_i - x_0)(x_i - x_1)...(x_i - x_{i-1})(x_i - x_{i+1})...(x_i - x_n)}$$ but since the top product has $x$ where the denominator product has $x_i$, nothing seems to cancel out... but if I let $x=x_k$ as suggested in the link would that mean $x_i=x_k$ as well, leaving $\sum x_k^j=x_k^j$?

Answer 1

So we are speaking of "over-interpolation", meaning that we are interpolating $(n+1)$ points laying on a polynomial curve of degree $q$ ($q \le n$) with a polynomial of higher degree, which then will reduce to having degree $q$.

In fact, in general, interpolating $n+1$ points will produce a polynomial of degree at most $n$. $$ p_{\,n} (x) = \sum\limits_{0 \le \,\,i\, \le \,n} {y_{\,i} \frac{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x - x_{\,j} } \right)} }}{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x_{\,i} - x_{\,j} } \right)} }}} \quad \Rightarrow \quad \deg \left( {p_{\,n} (x)} \right) \le n $$ and if the points are actually laying on a polynomial of degree <= $n$, we will get exactly that polynomial. $$ y_{\,i} = p_{\,q} (x_{\,i} )\quad \left| \begin{array}{l} \;0 \le q \le n \\ \;\forall i\;:\;0 \le i \le n \\ \end{array} \right.\quad \Rightarrow \quad p_{\,n} (x) = \sum\limits_{0 \le \,\,i\, \le \,n} {p_{\,q} (x_{\,i} )\frac{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x - x_{\,j} } \right)} }}{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x_{\,i} - x_{\,j} } \right)} }}} \quad = p_{\,q} (x) $$ In particular
$$ x^{\,q} \quad \left| {\;0 \le q \le n\;} \right.\quad = \sum\limits_{0 \le \,\,i\, \le \,n} {x_{\,i} ^{\,q} \frac{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x - x_{\,j} } \right)} }}{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {x_{\,i} - x_{\,j} } \right)} }}} $$

The algebric demonstration is not straightforward, it requires to expand the product in ${\left( {x - x_{\,j} } \right)}$, recurring to Vieta's formulas and then collect the terms of the summation in $x^0$, $x^1$, and/or recurring to Divided Differences, …
But we can take a blick on what is going on, if we take the simple case of interpolating the points $\left\{ {\left( {0,0} \right), \ldots ,\left( {k,k^{\,q} } \right), \ldots ,\left( {n,n^{\,q} } \right)} \right\}$ and limit our attention to the coefficient of $x^n$ $$ \begin{array}{l} \left[ {x^{\,n} } \right]p_{\,n} (x) = \sum\limits_{0 \le \,\,i\, \le \,n} {\frac{{i^{\,q} }}{{\prod\limits_{j\;:\;0 \le \,j\, \ne \,i\, \le \,n} {\left( {i - j} \right)} }}} = \\ = \sum\limits_{0 \le \,\,i\, \le \,n} {\frac{{i^{\,q} }}{{\prod\limits_{0 \le \,j\,\, \le \,i - 1} {\left( {i - j} \right)} \prod\limits_{i + 1 \le \,j\,\, \le \,n} {\left( {i - j} \right)} }}} = \\ = \sum\limits_{0 \le \,\,i\, \le \,n} {\frac{{i^{\,q} }}{{i!\prod\limits_{1 \le \,k\, \le \,n - i} {\left( { - k} \right)} }}} = \sum\limits_{0 \le \,\,i\, \le \,n} {\left( { - 1} \right)^{\,n - i} \frac{{i^{\,q} }}{{i!\left( {n - i} \right)!}}} = \\ = \frac{1}{{n!}}\sum\limits_{0 \le \,\,i\, \le \,n} {\left( { - 1} \right)^{\,n - i} \left( \begin{array}{c} n \\ i \\ \end{array} \right)i^{\,q} = } \frac{1}{{n!}}\Delta _{\,x} ^n \,x^{\,q} \left| {_{x\, = \,0} } \right. = \\ = n{\rm th}\,{\rm coefficient}\,{\rm in}\,{\rm Newton}\,{\rm expansion}\,{\rm of}\,x^{\,q} = \\ = \left\{ \begin{array}{l} 0\;\;q < n \\ 1\;\;q = n \\ \end{array} \right. \\ \end{array} $$