Polynomial interpolation

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In numerical analysis, polynomial interpolation is the interpolation of a given bivariate data set by the polynomial of lowest possible degree that passes through the points of the dataset.

Given a set of n + 1 data points ${\displaystyle (x_0,y_0),\dots ,(x_n,y_n)}$, with no two ${\displaystyle x_j}$ the same, a polynomial function ${\displaystyle p(x)=a_0+a_{1}x+\cdots +a_n{x}^n}$ is said to interpolate the data if ${\displaystyle p(x_j)=y_j}$ for each ${\displaystyle j\in \{0,1,\dots ,n\}}$.

There is always a unique such polynomial, commonly given by two explicit formulas, the Lagrange polynomials and Newton polynomials.

Applications

The original use of interpolation polynomials was to approximate values of important transcendental functions such as natural logarithm and trigonometric functions. Starting with a few accurately computed data points, the corresponding interpolation polynomial will approximate the function at an arbitrary nearby point. Polynomial interpolation also forms the basis for algorithms in numerical quadrature (Simpson's rule) and numerical ordinary differential equations (multigrid methods).

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.

In computer science, polynomial interpolation also leads to algorithms for secure multi party computation and secret sharing.

Interpolation theorem

For any ${\displaystyle n+1}$ bivariate data points ${\displaystyle (x_0,y_0),\dots ,(x_n,y_n)\in {\mathbb{R}}^2}$, where no two ${\displaystyle x_j}$ are the same, there exists a unique polynomial ${\displaystyle p(x)}$ of degree at most ${\displaystyle n}$ that interpolates these points, i.e. ${\displaystyle p(x_0)=y_0,\dots ,p(x_n)=y_n}$.^[2]

Equivalently, for a fixed choice of interpolation nodes ${\displaystyle x_j}$, polynomial interpolation defines a linear bijection ${\displaystyle L_n}$ between the (n+1)-tuples of real-number values ${\displaystyle (y_0,\dots ,y_n)\in {\mathbb{R}}^{n+1}}$ and the vector space ${\displaystyle P(n)}$ of real polynomials of degree at most n:

$${\displaystyle L_n:{\mathbb{R}}^{n+1}\stackrel{\sim }{⟶}P(n).}$$

This is a type of unisolvence theorem. The theorem is also valid over any infinite field in place of the real numbers ${\displaystyle \mathbb{R}}$, for example the rational or complex numbers.

First proof

Consider the Lagrange basis functions ${\displaystyle L_1(x),\dots ,L_n(x)}$ given by:

$${\displaystyle L_j(x)=\prod _{i\ne j}\frac{x-x_i}{x_j-x_i}=\frac{(x-x_1)\cdots (x-x_{j-1})(x-x_{j+1})\cdots (x-x_n)}{(x_j-x_1)\cdots (x_j-x_{j-1})(x_j-x_{j+1})\cdots (x_j-x_n)}.}$$

Notice that ${\displaystyle L_j(x)}$ is a polynomial of degree ${\displaystyle n}$, and we have ${\displaystyle L_j(x_k)=0}$ for each ${\displaystyle j\ne k}$, while ${\displaystyle L_k(x_k)=1}$. It follows that the linear combination:

$${\displaystyle p(x)=\sum _{j=0}^n{y}_j{L}_j(x)}$$

has ${\displaystyle p(x_k)=\sum _j{y}_j{L}_j(x_k)=y_k}$, so ${\displaystyle p(x)}$ is an interpolating polynomial of degree ${\displaystyle n}$.

To prove uniqueness, assume that there exists another interpolating polynomial ${\displaystyle q(x)}$ of degree at most ${\displaystyle n}$, so that ${\displaystyle p(x_k)=q(x_k)}$ for all ${\displaystyle k=0,\dots ,n}$. Then ${\displaystyle p(x)-q(x)}$ is a polynomial of degree at most ${\displaystyle n}$ which has ${\displaystyle n+1}$ distinct zeros (the ${\displaystyle x_k}$). But a non-zero polynomial of degree at most ${\displaystyle n}$ can have at most ${\displaystyle n}$ zeros, so ${\displaystyle p(x)-q(x)}$ must be the zero polynomial, i.e. ${\displaystyle p(x)=q(x)}$.

Second proof

Write out the interpolation polynomial in the form

$${\displaystyle p(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_2{x}^2+a_{1}x+a_0.}$$

(1)

Substituting this into the interpolation equations ${\displaystyle p(x_j)=y_j}$, we get a system of linear equations in the coefficients ${\displaystyle a_j}$, which reads in matrix-vector form as the following multiplication:

$${\displaystyle \left[\begin{array}{cccccc}{x}_0^n& x_0^{n-1}& x_0^{n-2}& \dots & x_0& 1\\ x_1^n& x_1^{n-1}& x_1^{n-2}& \dots & x_1& 1\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ x_n^n& x_n^{n-1}& x_n^{n-2}& \dots & x_n& 1\end{array}\right]\left[\begin{array}{c}{a}_n\\ a_{n-1}\\ ⋮\\ a_0\end{array}\right]=\left[\begin{array}{c}{y}_0\\ y_1\\ ⋮\\ y_n\end{array}\right].}$$

An interpolant ${\displaystyle p(x)}$ corresponds to a solution ${\displaystyle A=(a_n,\dots ,a_0)}$ of the above matrix equation ${\displaystyle X\cdot A=Y}$. The matrix X on the left is a Vandermonde matrix, whose determinant is known to be ${\displaystyle det(X)=\prod _{1\le i<j\le n}(x_j-x_i),}$ which is non-zero since the nodes ${\displaystyle x_j}$ are all distinct. This ensures that the matrix is invertible and the equation has the unique solution ${\displaystyle A=X^{-1}\cdot Y}$; that is, ${\displaystyle p(x)}$ exists and is unique.

Corollary

If ${\displaystyle f(x)}$ is a polynomial of degree at most ${\displaystyle n}$, then the interpolating polynomial of ${\displaystyle f(x)}$ at ${\displaystyle n+1}$ distinct points is ${\displaystyle f(x)}$ itself.

Constructing the interpolation polynomial

Main article: Lagrange polynomial

The red dots denote the data points (x_k, y_k), while the blue curve shows the interpolation polynomial.

The Vandermonde matrix in the second proof above may have large condition number, causing large errors when computing the coefficients a_i 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.

Alternatively, we may write down the polynomial immediately in terms of Lagrange polynomials:

$${\displaystyle \begin{array}{rl}p(x)& =\frac{(x-x_1)(x-x_2)\cdots (x-x_n)}{(x_0-x_1)(x_0-x_2)\cdots (x_0-x_n)}{y}_0\\ & +\frac{(x-x_0)(x-x_2)\cdots (x-x_n)}{(x_1-x_0)(x_1-x_2)\cdots (x_1-x_n)}{y}_1\\ & +\cdots \\ & +\frac{(x-x_0)(x-x_1)\cdots (x-x_{n-1})}{(x_n-x_0)(x_n-x_1)\cdots (x_n-x_{n-1})}{y}_n\\ & =\sum _{i=0}^n(\prod _{\stackrel{0\le j\le n}{j\ne i}}\frac{x-x_j}{x_i-x_j})y_i=\sum _{i=0}^n\frac{p(x)}{p^{\prime }(x_i)(x-x_i)}{y}_i\end{array}}$$

For matrix arguments, this formula is called Sylvester's formula and the matrix-valued Lagrange polynomials are the Frobenius covariants.

Non-Vandermonde algorithms

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 value p(a) at a point x = a not in the original data set. The Lagrange form computes the value p(a) with complexity O(n²).

The Bernstein form was used in a constructive proof of the Weierstrass approximation theorem by Bernstein and has gained great importance in computer graphics in the form of Bézier curves.

Interpolations as linear combinations of values

Given a set of (position, value) data points ${\displaystyle (x_0,y_0),\dots ,(x_j,y_j),\dots ,(x_n,y_n)}$ where no two positions ${\displaystyle x_j}$ are the same, the interpolating polynomial ${\displaystyle y(x)}$ may be considered as a linear combination of the values ${\displaystyle y_j}$, using coefficients which are polynomials in ${\displaystyle x}$ depending on the ${\displaystyle x_j}$. For example the interpolation polynomial in the Lagrange form is the linear combination

$${\displaystyle y(x):=\sum _{j=0}^k{y}_j{c}_j(x)}$$

with each coefficient ${\displaystyle c_j(x)}$ given by the corresponding Lagrange basis polynomial on the given positions ${\displaystyle x_j}$:

$${\displaystyle c_j(x)=L_j(x_0,\dots ,x_n;x)=\prod _{\genfrac{}{}{0ex}{}{0\le i\le n}{i\ne j}}\frac{x-x_i}{x_j-x_i}=\frac{(x-x_0)}{(x_j-x_0)}\cdots \frac{(x-x_{j-1})}{(x_j-x_{j-1})}\frac{(x-x_{j+1})}{(x_j-x_{j+1})}\cdots \frac{(x-x_n)}{(x_j-x_n)}.}$$

Since the coefficients depend only on the positions ${\displaystyle x_j}$, not the values ${\displaystyle y_j}$, we can use the same coefficients to find the interpolating polynomial for a second set of data points ${\displaystyle (x_0,v_0),\dots ,(x_n,v_n)}$ at the same positions:

$${\displaystyle v(x):=\sum _{j=0}^k{v}_j{c}_j(x).}$$

Furthermore, the coefficients ${\displaystyle c_j(x)}$ only depend on the relative spaces ${\displaystyle x_i-x_j}$ between the positions. Thus, given a third set of data whose points are given by the new variable ${\displaystyle t=ax+b}$ (an affine transformation of ${\displaystyle x}$, inverted by ${\displaystyle x=\frac{t-b}{a}}$):

$${\displaystyle (t_0,w_0),\dots ,(t_j,w_j)\dots ,(t_n,w_n)\text{with}{t}_j=a{x}_j+b,}$$

we can use a transformed version of the previous coefficient polynomials:

${\displaystyle {\tilde{c}}_j(t):=c_j(\frac{t-b}{a})=c_j(x),}$

and write the interpolation polynomial as:

$w(t):=\sum _{j=0}^k{w}_j{\tilde{c}}_j(t).$

Data points ${\displaystyle (x_j,y_j)}$ often have equally spaced positions, which may be normalized by an affine transformation to ${\displaystyle x_j=j}$. For example, consider the data points

${\displaystyle (0,y_0),(1,y_1),(2,y_2)}$.

The interpolation polynomial in the Lagrange form is the linear combination

$${\displaystyle \begin{array}{rl}y(x):=\sum _{j=0}^2{y}_j{c}_j(x)& =y_0\frac{(x-1)(x-2)}{(0-1)(0-2)}+y_1\frac{(x-0)(x-2)}{(1-0)(1-2)}+y_2\frac{(x-0)(x-1)}{(2-0)(2-1)}\\ & =\frac{1}{2}{y}_0(x-1)(x-2)-y_1(x-0)(x-2)+\frac{1}{2}{y}_2(x-0)(x-1).\end{array}}$$

For example, ${\displaystyle y(3)=y_3=y_0-3y_1+3y_2}$ and ${\displaystyle y(1.5)=y_{1.5}=\frac{1}{8}(-y_0+6y_1+3y_2)}$.

The case of equally spaced points can also be treated by the method of finite differences. The first difference of a sequence of values ${\displaystyle v=\{v_j{\}}_{j=0}^{\mathrm{\infty }}}$ is the sequence ${\displaystyle \mathrm{\Delta }v=u=\{u_j{\}}_{j=0}^{\mathrm{\infty }}}$ defined by ${\displaystyle u_j=v_{j+1}-v_j}$. Iterating this operation gives the n^th difference operation ${\displaystyle {\mathrm{\Delta }}^{n}v=u}$, defined explicitly by:

$${\displaystyle u_j=\sum _{k=0}^n(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{n}{k})v_{j+k},}$$

where the coefficients form a signed version of Pascal's triangle, the triangle of binomial transform coefficients:

								1									Row n = 0
							1		−1								Row n = 1 or d = 0
						1		−2		1							Row n = 2 or d = 1
					1		−3		3		−1						Row n = 3 or d = 2
				1		−4		6		−4		1					Row n = 4 or d = 3
			1		−5		10		−10		5		−1				Row n = 5 or d = 4
		1		−6		15		−20		15		−6		1			Row n = 6 or d = 5
	1		−7		21		−35		35		−21		7		−1		Row n = 7 or d = 6

A polynoimial ${\displaystyle y(x)}$ of degree d defines a sequence of values at positive integer points, ${\displaystyle y_j=y(j)}$, and the ${\displaystyle (d+1{)}^{\text{th}}}$ difference of this sequence is identically zero:

${\displaystyle {\mathrm{\Delta }}^{d+1}y=0}$.

Thus, given values ${\displaystyle y_0,\dots ,y_n}$ at equally spaced points, where ${\displaystyle n=d+1}$, we have:

$${\displaystyle (-1{)}^n{y}_0+(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{n}{1})y_1+\cdots -(\genfrac{}{}{0ex}{}{n}{n-1})y_{n-1}+y_n=0.}$$

For example, 4 equally spaced data points ${\displaystyle y_0,y_1,y_2,y_3}$ of a quadratic ${\displaystyle y(x)}$ obey ${\displaystyle 0=-y_0+3y_1-3y_2+y_3}$, and solving for ${\displaystyle y_3}$ gives the same interpolation equation obtained above using the Lagrange method.

Interpolation error: Lagrange remainder formula

When interpolating a given function f by a polynomial ${\displaystyle p_n}$ of degree n at the nodes x₀,…, x_n we get the error

$${\displaystyle f(x)-p_n(x)=f[x_0,\dots ,x_n,x]\prod _{i=0}^n(x-x_i)}$$

where $f[x_0,\dots ,x_n,x]$ is the (n+1)^st divided difference of the data points

${\displaystyle (x_0,f(x_0)),\dots ,(x_n,f(x_n)),(x,f(x))}$.

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 ${\displaystyle I}$, and a polynomial ${\displaystyle p_n(x)}$ of degree at most n that interpolates f at n + 1 distinct points ${\displaystyle x_0,\dots ,x_n\in I}$. For each ${\displaystyle x\in I}$ there exists ${\displaystyle \xi \in I}$ such that

$${\displaystyle f(x)-p_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}\prod _{i=0}^n(x-x_i).}$$

This error bound suggests choosing the interpolation points x_i to minimize the product $\left|\prod (x-x_i)\right|$, which is achieved by the Chebyshev nodes.

Proof of Lagrange remainder

Set the error term as $R_n(x)=f(x)-p_n(x)$, and define an auxiliary function:

$${\displaystyle Y(t)=R_n(t)-\frac{R_n(x)}{W(x)}W(t)\text{where}W(t)=\prod _{i=0}^n(t-x_i).}$$

Thus:

$${\displaystyle Y^{(n+1)}(t)=R_n^{(n+1)}(t)-\frac{R_n(x)}{W(x)}(n+1)!}$$

But since ${\displaystyle p_n(x)}$ is a polynomial of degree at most n, we have $R_n^{(n+1)}(t)=f^{(n+1)}(t)$, and:

$${\displaystyle Y^{(n+1)}(t)=f^{(n+1)}(t)-\frac{R_n(x)}{W(x)}(n+1)!}$$

Now, since x_i are roots of ${\displaystyle R_n(t)}$ and ${\displaystyle W(t)}$, we have ${\displaystyle Y(x)=Y(x_j)=0}$, which means Y has at least n + 2 roots. From Rolle's theorem, ${\displaystyle Y^{\mathrm{\prime }}(t)}$ has at least n + 1 roots, and iteratively ${\displaystyle Y^{(n+1)}(t)}$ has at least one root ξ in the interval I. Thus:

$${\displaystyle Y^{(n+1)}(\xi )=f^{(n+1)}(\xi )-\frac{R_n(x)}{W(x)}(n+1)!=0}$$

and:

$${\displaystyle R_n(x)=f(x)-p_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}\prod _{i=0}^n(x-x_i).}$$

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 x_i are identical. Note that the error will be zero when ${\displaystyle x=x_i}$ for any i. Thus, the maximum error will occur at some point in the interval between two successive nodes.

Equally spaced intervals

In the case of equally spaced interpolation nodes where ${\displaystyle x_i=a+ih}$, for ${\displaystyle i=0,1,\dots ,n,}$ and where ${\displaystyle h=(b-a)/n,}$ the product term in the interpolation error formula can be bound as

$${\displaystyle \left|\prod _{i=0}^n(x-x_i)\right|=\prod _{i=0}^n\left|x-x_i\right|\le \frac{n!}{4}{h}^{n+1}.}$$

Thus the error bound can be given as

$${\displaystyle \left|R_n(x)\right|\le \frac{h^{n+1}}{4(n+1)}\underset{\xi \in [a,b]}{max}\left|f^{(n+1)}(\xi )\right|}$$

However, this assumes that ${\displaystyle f^{(n+1)}(\xi )}$ is dominated by ${\displaystyle h^{n+1}}$, i.e. ${\displaystyle f^{(n+1)}(\xi )h^{n+1}\ll 1}$. In several cases, this is not true and the error actually increases as n → ∞ (see Runge's phenomenon). That question is treated in the section Convergence properties.

Lebesgue constants

Main article: Lebesgue constant (interpolation)

We fix the interpolation nodes x₀, ..., x_n 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 ${\displaystyle P(n)}$ of polynomials of degree n or less.

The Lebesgue constant L is defined as the operator norm of X. One has (a special case of Lebesgue's lemma):

$${\displaystyle \Vert f-X(f)\Vert \le (L+1)\Vert f-p^{\ast }\Vert .}$$

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:

$${\displaystyle L\le \frac{2}{\pi }\log (n+1)+1.}$$

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.

Convergence properties

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 − p_n||_∞ 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 ${\displaystyle p_n(x)}$ converges to f(x) uniformly on [a,b].

Proof

It is clear that the sequence of polynomials of best approximation ${\displaystyle p_n^{\ast }(x)}$ converges to f(x) uniformly (due to the Weierstrass approximation theorem). Now we have only to show that each ${\displaystyle p_n^{\ast }(x)}$ 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 X_n (where X_n is the projection operator on Π_n). Now we seek a table of nodes for which

$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{X}_{n}f=f,\text{ for every }f\in C([a,b]).}$$

Due to the Banach–Steinhaus theorem, this is only possible when norms of X_n are uniformly bounded, which cannot be true since we know that

$${\displaystyle \Vert X_n\Vert \ge \frac{2}{\pi }\log (n+1)+C.}$$

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.

Related concepts

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.

Interpolation of periodic functions by harmonic functions is accomplished by Fourier transform. This can be seen as a form of polynomial interpolation with harmonic base functions, see trigonometric interpolation and trigonometric polynomial.

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.

The technique of rational function modeling is a generalization that considers ratios of polynomial functions.

At last, multivariate interpolation for higher dimensions.