In computer graphics, polynomials can be used to approximate complicated plane curves given a few specified points, for example the shapes of letters in typography. This is usually done with Bézier curves, which are a simple generalization of interpolation polynomials (having specified tangents as well as specified points).
In numerical analysis, polynomial interpolation is essential to perform sub-quadratic multiplication and squaring, such as Karatsuba multiplication and Toom–Cook multiplication, where interpolation through points on a product polynomial yields the specific product required. For example, given a = f(x) = a0x0 + a1x1 + ··· and b = g(x) = b0x0 + b1x1 + ···, the product ab is a specific value of W(x) = f(x)g(x). One may easily find points along W(x) at small values of x, and interpolation based on those points will yield the terms of W(x) and the specific product ab. As fomulated in Karatsuba multiplication, this technique is substantially faster than quadratic multiplication, even for modest-sized inputs, especially on parallel hardware.
For any bivariate data points , where no two are the same, there exists a unique polynomial of degree at most that interpolates these points, i.e. .
Equivalently, for a fixed choice of interpolation nodes , polynomial interpolation defines a linear bijection between the (n+1)-tuples of real-number values and the vector space of real polynomials of degree at most n:
This is a type of unisolvence theorem. The theorem is also valid over any infinite field in place of the real numbers , for example the rational or complex numbers.
Notice that is a polynomial of degree , and we have for each , while . It follows that the linear combination:
has , so is an interpolating polynomial of degree .
To prove uniqueness, assume that there exists another interpolating polynomial of degree at most , so that for all . Then is a polynomial of degree at most which has distinct zeros (the ). But a non-zero polynomial of degree at most can have at most zeros,[a] so must be the zero polynomial, i.e. .
An interpolant corresponds to a solution of the above matrix equation . The matrix X on the left is a Vandermonde matrix, whose determinant is known to be which is non-zero since the nodes are all distinct. This ensures that the matrix is invertible and the equation has the unique solution ; that is, exists and is unique.
For a polynomial of degree less than or equal to n, that interpolates at the nodes where . Let be the polynomial of degree less than or equal to n+1 that interpolates at the nodes where . Then is given by:
where and .
This can be shown for the case where :
and when :
By the uniqueness of interpolated polynomials of degree less than , is the required polynomial interpolation. The function can thus be expressed as:
Lozenge diagram is a diagram that is used to describe different interpolation formulas that can be constructed for a given data set. A line starting on the left edge and tracing across the diagram to the right can be used to represent an interpolation formula if the following rules are followed:
Left to right steps indicate addition whereas right to left steps indicate subtraction
If slope of step is positive, the term to be used is product of the difference and the factor immediately below it. If slope of step is negative, the term to be used is product of the difference and the factor immediately above it.
If step is horizontal and passes through a factor, use the product of the factor and average of two terms immediately above and below it. If step is horizontal and passes through a difference, use the product of the difference and average of two terms immediately above and below it.
If a path goes from to , it can connect through three intermediate steps, (a) through , (b) through or (c) through . Proving the equivalence of these three two-step paths should prove that all (n-step) paths can be morphed with the same starting and ending, all of which represents the same formula.
Subtracting contributions from path a and b:
Thus, the contribution of either path (a) or path (b) is the same. Since path (c) is the average of path (a) and (b), it also contributes identical function to the polynomial. Hence the equivalence of paths with same starting and ending points is shown. To check if the paths can be shifted to different values in the leftmost corner, taking only two step paths is sufficient: (a) to through or (b) factor between and , to through or (c) starting from .
Since , substituting in the above equations shows that all the above terms reduce to and are hence equivalent. Hence these paths can be morphed to start from the leftmost corner and end in a common point.
The Vandermonde matrix in the second proof above may have large condition number, causing large errors when computing the coefficients ai if the system of equations is solved using Gaussian elimination.
Several authors have therefore proposed algorithms which exploit the structure of the Vandermonde matrix to compute numerically stable solutions in O(n2) operations instead of the O(n3) required by Gaussian elimination. These methods rely on constructing first a Newton interpolation of the polynomial and then converting it to the monomial form above.
To find the interpolation polynomial p(x) in the vector space P(n) of polynomials of degree n, we may use the usual monomial basis for P(n) and invert the Vandermonde matrix by Gaussian elimination, giving a computational cost of O(n3) operations. To improve this algorithm, a more convenient basis for P(n) can simplify the calculation of the coefficients, which must then be translated back in terms of the monomial basis.
One method is to write the interpolation polynomial in the Newton form and use the method of divided differences to construct the coefficients, e.g. Neville's algorithm. The cost is O(n2) operations. Furthermore, you only need to do O(n) extra work if an extra point is added to the data set, while for the other methods, you have to redo the whole computation.
Another method is preferred when the aim is not to compute the coefficients of p(x), but only a single valuep(a) at a point x = a not in the original data set. The Lagrange form computes the value p(a) with complexity O(n2).
Interpolations as linear combinations of values
Given a set of (position, value) data points where no two positions are the same, the interpolating polynomial may be considered as a linear combination of the values , using coefficients which are polynomials in depending on the . For example the interpolation polynomial in the Lagrange form is the linear combination
with each coefficient given by the corresponding Lagrange basis polynomial on the given positions :
Since the coefficients depend only on the positions , not the values , we can use the same coefficients to find the interpolating polynomial for a second set of data points at the same positions:
Furthermore, the coefficients only depend on the relative spaces between the positions. Thus, given a third set of data whose points are given by the new variable (an affine transformation of , inverted by ):
we can use a transformed version of the previous coefficient polynomials:
and write the interpolation polynomial as:
Data points often have equally spaced positions, which may be normalized by an affine transformation to . For example, consider the data points
The case of equally spaced points can also be treated by the method of finite differences. The first difference of a sequence of values is the sequence defined by . Iterating this operation gives the nth difference operation , defined explicitly by:
Furthermore, there is a Lagrange remainder form of the error, for a function f which is n + 1 times continuously differentiable on a closed interval , and a polynomial of degree at most n that interpolates f at n + 1 distinct points . For each there exists such that
This error bound suggests choosing the interpolation points xi to minimize the product , which is achieved by the Chebyshev nodes.
Set the error term as , and define an auxiliary function:
But since is a polynomial of degree at most n, we have , and:
Now, since xi are roots of and , we have , which means Y has at least n + 2 roots. From Rolle's theorem, has at least n + 1 roots, and iteratively has at least one root ξ in the interval I. Thus:
This parallels the resoning behind the Lagrange remainder term in the Taylor theorem; in fact, the Taylor remainder is a special case of interpolation error when all interpolation nodes xi are identical. Note that the error will be zero when for any i. Thus, the maximum error will occur at some point in the interval between two successive nodes.
We fix the interpolation nodes x0, ..., xn and an interval [a, b] containing all the interpolation nodes. The process of interpolation maps the function f to a polynomial p. This defines a mapping X from the space C([a, b]) of all continuous functions on [a, b] to itself. The map X is linear and it is a projection on the subspace of polynomials of degree n or less.
In other words, the interpolation polynomial is at most a factor (L + 1) worse than the best possible approximation. This suggests that we look for a set of interpolation nodes that makes L small. In particular, we have for Chebyshev nodes:
We conclude again that Chebyshev nodes are a very good choice for polynomial interpolation, as the growth in n is exponential for equidistant nodes. However, those nodes are not optimal.
It is natural to ask, for which classes of functions and for which interpolation nodes the sequence of interpolating polynomials converges to the interpolated function as n → ∞? Convergence may be understood in different ways, e.g. pointwise, uniform or in some integral norm.
The situation is rather bad for equidistant nodes, in that uniform convergence is not even guaranteed for infinitely differentiable functions. One classical example, due to Carl Runge, is the function f(x) = 1 / (1 + x2) on the interval [−5, 5]. The interpolation error || f − pn||∞ grows without bound as n → ∞. Another example is the function f(x) = |x| on the interval [−1, 1], for which the interpolating polynomials do not even converge pointwise except at the three points x = ±1, 0.
One might think that better convergence properties may be obtained by choosing different interpolation nodes. The following result seems to give a rather encouraging answer:
Theorem — For any function f(x) continuous on an interval [a,b] there exists a table of nodes for which the sequence of interpolating polynomials converges to f(x) uniformly on [a,b].
It is clear that the sequence of polynomials of best approximation converges to f(x) uniformly (due to the Weierstrass approximation theorem). Now we have only to show that each may be obtained by means of interpolation on certain nodes. But this is true due to a special property of polynomials of best approximation known from the equioscillation theorem. Specifically, we know that such polynomials should intersect f(x) at least n + 1 times. Choosing the points of intersection as interpolation nodes we obtain the interpolating polynomial coinciding with the best approximation polynomial.
The defect of this method, however, is that interpolation nodes should be calculated anew for each new function f(x), but the algorithm is hard to be implemented numerically. Does there exist a single table of nodes for which the sequence of interpolating polynomials converge to any continuous function f(x)? The answer is unfortunately negative:
Theorem — For any table of nodes there is a continuous function f(x) on an interval [a, b] for which the sequence of interpolating polynomials diverges on [a,b].
The proof essentially uses the lower bound estimation of the Lebesgue constant, which we defined above to be the operator norm of Xn (where Xn is the projection operator on Πn). Now we seek a table of nodes for which
Due to the Banach–Steinhaus theorem, this is only possible when norms of Xn are uniformly bounded, which cannot be true since we know that
For example, if equidistant points are chosen as interpolation nodes, the function from Runge's phenomenon demonstrates divergence of such interpolation. Note that this function is not only continuous but even infinitely differentiable on [−1, 1]. For better Chebyshev nodes, however, such an example is much harder to find due to the following result:
Theorem — For every absolutely continuous function on [−1, 1] the sequence of interpolating polynomials constructed on Chebyshev nodes converges to f(x) uniformly.
Runge's phenomenon shows that for high values of n, the interpolation polynomial may oscillate wildly between the data points. This problem is commonly resolved by the use of spline interpolation. Here, the interpolant is not a polynomial but a spline: a chain of several polynomials of a lower degree.
Hermite interpolation problems are those where not only the values of the polynomial p at the nodes are given, but also all derivatives up to a given order. This turns out to be equivalent to a system of simultaneous polynomial congruences, and may be solved by means of the Chinese remainder theorem for polynomials. Birkhoff interpolation is a further generalization where only derivatives of some orders are prescribed, not necessarily all orders from 0 to a k.
Collocation methods for the solution of differential and integral equations are based on polynomial interpolation.
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