When we interpolate a sequence, we want to "fill the gaps" between samples of a sequence in such a way that

The first method to do this is called polynomial interpolation. Given a set of $$N$$ points, we can always define a polynomial of degree $$N-1$$ that passes through these N points. Beyond the simple naive approach and without loss of generality, we can define a set of $$2N+1$$ Lagrange polynomials of degree 2N

$$L_n^{(N)} (t) = \prod_{k=-N, k\neq n}^N \frac{t-k}{k-n}, n = -N, …, N$$

that have the property that

$$L_n^{(N)} (t) = \begin{cases} 1 \text{ for }t=n \\ 0 \text{ at integer multiples.} \end{cases}$$

From this set, the interpolation formula is

$$x(t) = \sum_{n=-N}^N x[n] L_n^{(N)} (t)$$

Lagrange interpolation is maximally smooth but has the drawback that the interpolation kernel depends on $$N$$.

To avoid this drawback, we can define another interpolation method called local interpolation. This method uses only a neighborhood of $$2N+1$$ points and an interpolation kernel $$i_c(t)$$ defined over these 2N+1 points that is equal to 1 at zero and 0 at nonzero integer points. Then, the interpolation formula is given by the weighted sum of these kernels by the sample value

$$x(t) = \sum_{n=-N}^{N} x[n]i_c(t-n)$$

In particular, for $$c=0$$, the interpolation kernel is a rect function and we obtain a so-called piecewise-constant interpolation. For $$c=1$$, we draw a line between consecutive samples. For $$c \rightarrow \infty$$, the interpolation kernel is a sinc function. We have now an interpolation kernel that does not depends on $$N$$ but we loose some of the smoothness in the result.

Finally, it can be shown that when N goes to infinity, the Lagrange interpolation corresponds also to a sinc interpolation. In the limit, global and local interpolation are equivalent.