Let $x_0,…,x_n$ be distinct real numbers and $l_k(x)$ be the
Lagrange's basis function. $\delta_n = \prod^n_{k=0}(x-x_k)$.Prove that:
a. – $\sum^n_{k=0}(x_k-x)^jl_k(x)\equiv 0$, for $j = 0,1,…,n$
b. – Let $p(x)$ be any polynomial of degree $n+1$ with its highest
degree coefficient $a_{n+1} =1$. Then, $p(x) –
\sum^n_{k=0}p(x_k)l_k(x) = \delta_n(x)$
a. – I know that Lagrange functions $l_k(x)$ there are k terms in the product where each term contains an $x$ but then how does multiplying it by $(x_k-x)^j$ make it equivalent to $0$?
b. – I do not know how to do the proof. I tried googling it and found "Lagrange's fundamental interpolating polynomials" which seems similar to it.
Best Answer
(The Lagrange basis functions are $$ l_k(x):=\frac{\prod_{0\le i\le n,\ i\ne k} (x - x_i)}{\prod_{0\le i\le n, \ i\ne k} (x_k - x_i)}, \qquad k=0,\dots,n.) $$
(a) This is not true for $j=0$, but it works for $j=1$, $\dots$, $n$.
For any polynomial $f(x)$, let $${\overline{f(x)}}:=\sum_{0\le k\le n} f(x_k) l_k(x).$$ Then $\deg {\overline{f(x)}}\le n$ and $f(x)$ and ${\overline{f(x)}}$ agree at $x=x_0$, $\dots$, $x=x_n$. Since a nonzero polynomial of degree $n$ or less can have at most $n$ zeroes, it follows that, if $\deg f(x)\le n$, ${\overline{f(x)}}=f(x)$.
Now, in the sum in (a), replace the first $x$ by $y$ to give \begin{eqnarray*} \sum_{0\le k\le n} (x_k-y)^j l_k(x). \end{eqnarray*} Treat this as a polynomial in $x$ with coefficients which are functions of $y$. Then, it is ${\overline{(x-y)^j}}$. But $j\le n$, so by the last paragraph, this equals $(x-y)^j$. Substituting the $x$ back in gives $(x-x)^j=0$.
(b) This can be rewritten as $$ p(x)-{\overline{p(x)}} = \delta_n(x).\ \ \ \ (*) $$ By the above, since $\deg (p(x)-\delta_n(x))\le n$, $$ p(x)-\delta_n(x) = {\overline{p(x)- \delta_n(x)}}. \ \ \ \ (**) $$ But since $\delta_n(x_0)=\dots=\delta_n(x_n)=0$, ${\overline{p(x)-\delta_n(x)}}=\overline{p(x)}$. Substituting this into (**) gives (*).