# Lagrange's basis function and Interpolation

Let $x_0,...,x_n$ be distinct real numbers and $l_k(x)$ be the Lagrange's basis function. $δ_n = ∏^n _{k=0}(x-x_k)$. Prove that

a - $\sum^n_{k=0}x^j_kl_k(x) ≡ x^j$. for $j = 0,1,...,n$

b - Let $P_n(x)$ interpolate $f(x)$ at $x_0,...,x_n$. Then $$P_n(x) = \frac{\sum^n_{k=0}f(x_k)/[\delta_n'(x_k)(x-x_k)]}{\sum^n_{k=0}1/[\delta_n'(x_k)(x-x_k)]}$$.

a - From the fact that $\sum^n_{k=0}l_k(x) ≡ 1$ since that the interpolating polynomial $p$ for $f=1$ is the sum $S$ of the Lagrange polynomials. Since $f$ is a polynomial of degree $0$ and the interpolating polynomial is unique: $p = f$. As $p = s$, the sum of the Lagrange polynomials equals one. So from that since $l_k(x)\equiv 1$ then doesn't it follow from that fact it will be $\equiv x^j$? How can I show that using a proof?

b - No idea how to do.

For (a), you were on the right track. Suppose you interpolate $f(x) := x^j$ at the nodes $x_0, \ldots, x_n$. The interpolating polynomial is $$p(x) = \sum_{k=0}^n x_k^j \ell_k(x).$$ Since, as you correctly say, the interpolating polynomial is unique (because the nodes are all distinct) and because $f$ is itself an interpolating polynomial of degree $\leq n$, we must have $p=f$.
For (b), first establish that $\delta_n'(x_j) = \Pi_{k \neq j} (x_j - x_k)$. Then, note that the $j$-th Lagrange polynomial may be written $$\ell_j(x) = \frac{\delta_n(x)}{\delta_n'(x_j) \ (x - x_j)}.$$ So the interpolating polynomial may be written $$p(x) = \sum_{j=0}^n f_j \ \ell_j(x) = \delta_n(x) \sum_{j=0}^n \frac{f_j}{\delta_n'(x_j) \ (x - x_j)}.$$ Since $\sum_{j=0}^n \ell_j(x) = 1$, we have, using the expression of $\ell_j$ above $$1 = \sum_{j=0}^n \frac{\delta_n(x)}{\delta_n'(x_j) \ (x - x_j)} = \delta_n(x) \sum_{j=0}^n \frac{1}{\delta_n'(x_j) \ (x - x_j)},$$ and therefore $$\delta_n(x) = \left( \sum_{j=0}^n \frac{1}{\delta_n'(x_j) \ (x - x_j)} \right)^{-1}.$$
The resulting expression for $p$ is called the barycentric formula. It turns out to be the most useful in practice since it can be computed as cheaply as the Newton interpolant and also updated relatively cheaply if new nodes are added.
• Thank you Dominique! For a I have edited to make it more distinguishable. Also for b what do you mean in your very last line "implies using the expression of $l_j(x)$ above $\delta_n(x) = \sum_{j=0}^n \frac{1}{\delta_n'(x_j) \ (x - x_j)}.$?" – user60514 Feb 3 '13 at 4:22