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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.

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    My answer here may be helpful: http://math.stackexchange.com/questions/523907/explanation-of-lagrange-interpolating-polynomial/523922#523922 – Spencer Apr 09 '14 at 14:50

2 Answers2

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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)

Robert Fan
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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.

Joel
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