Lagrange interpolation, syntax help I am told, the basic interpolation problem can be formulated as:
Given a set of nodes, $ \{x_i, i=0, ..., n\} $ and corresponding data values$\{y_i, i=0, ..., n\}$, find the polynomial $p(x)$ of degree less or equal to $n$ such that $p(x_i)=y_i$.
Which makes sense to me.  However, the explanation gets a little less wordy then and says:
Consider the family of functions:
$$ L_i^{(n)}(x)=\prod_{j=0,j\neq{k}}^n\frac{x-x_j}{x_k-x_j}, k=0,1,..., n\tag1 $$
We can see that they are polynomials of order $n$ and have the property (interpolatory condition):
$$ L_i^{(n)}(x_j)=\delta_{i,j}=\begin{cases}
1,  & i=j \\
0, & i\neq{j}  \\
\end{cases}\tag2 $$
Then if we define the polynomial by:
$$ p_n(x) = \sum_{k=0}^ny_kL_k^{(n)}(x)\tag3$$
then:
$$ p_n(x_i) = \sum_{k=0}^ny_kL_k^{(n)}(x_i)=y_i\tag4$$
Could someone please elaborate a little on (1-4) in words?  i.e, what does $L_{i}^{(n)}$ mean?
Thanks.
 A: Using the matrix notation for interpolation
$$\mathbf{y}=V\mathbf{a}$$
where $y$ output data vector and $a$ is the coefficients of the polynomial to be resolved. $V$ is a (n+1) by (n+1) Vandermonde matrix whose each row is of the form:
[1, x_i, x_i^2, ...x_i^n];
If you know how to solve the determinant of a Vandermonde matrix, you can see $L_i$ is just another form of the solution of the vector equation.(use Cramer's law)
A: The notation $L_i^{(n)}$ is a made up notation to tell you that each of these polynomials is a polynomial of degree $n$ and this is the $i$'th one considered.
This way when we take a sum of these polynomials, we know that the resulting polynomial is a polynomial of degree $n$.
The $L_i^{(n)}$ polynomials are rigged so that you can make interpolation easier. Notice that $L^{(n)}_i$ is zero for each $x_j$ where $j\neq i$ and is exactly $1$ when we input $x_i$. Thus $y_i L^{(n)}_i(x_i)=y_i$. The zeros for the other points ensures us that $L_i^{(n)}$ wont interfere with the value at the other points.
For instance take the interpolation problem $(1,2)$ and $(3,4)$.
$L_0^{(1)}(x) = (x-3)/(1-3)$ and $L_1^{(1)}(x) = (x-1)/(3-1)$.
$L_0^{(1)}(3) = 0 = L_1^{(1)}(1)$
Also $L_0^{(1)}(1) = 1 = L_1^{(1)}(3)$.
Now $p(x) = 2L_0^{(1)}(x) + 4L_1^{(1)}(x)$ satisfies $p(1)=2$ and $p(3)=4$ as desired.
