What is polynomial interpolation

Polynomial interpolation

Under Polynomial interpolation one understands the solution to the problem of finding a polynomial that contains n + 1 given assignments and thus interpolates. For n + 1 given pairwise different points there is exactly one polynomial of the nth degree which satisfies them. That is what we call that Interpolation polynomial.
A polynomial of degree has n + 1 coefficients, i.e. the same number of degrees of freedom. The solution of the interpolation problem can therefore be determined by a linear system of equations. How this system of equations looks exactly depends on the representation or basis chosen. The uniqueness follows from the fact that a polynomial of the nth degree with real or complex coefficients, which is not the zero polynomial, has at most n zeros. The difference between two different solutions would be 0 at all n + 1 interpolation points.

Newton basis

The so-called Newton basis has proven itself here:
Nk (x) = ⎩⎪⎨⎪⎧ 1j = 0∏k − 1 (x − xj): k = 0: k≥1
The interpolation polynomial is given by
P (x) = k = 0∑n ck Nk (x)
The unknown coefficients ck can be calculated efficiently and stably by means of the Neville-Aitken scheme (also called "Aitken-Neville scheme" or "scheme of divided differences").
Furthermore, the evaluation of a polynomial in Newton representation by means of the Horner scheme is possible in a linear expenditure of time.

Lagrange base

A representation in the Lagrange basis is more favorable for theoretical considerations. Here the basis functions are called Lagrange polynomials:
ℓi (x) = j = 0, j = / i∏n xi −xj x − xj
The solution to the interpolation problem can then simply be given as
P (x) = i = 0∑n f (xi) ℓi (x)
This is often used to prove the existence of the solution to the interpolation problem.
The great advantage of the Newton basis is that new points can be inserted there very easily by simply adding a term to the end. With the Lagrange basis you have to completely recalculate all basis functions.



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