Polynomial interpolation

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.^[1]

Given a set of n + 1 data points ${\displaystyle (x_{0},y_{0}),\ldots ,(x_{n},y_{n})}$, 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 ${\displaystyle j\in \{0,1,\dotsc ,n\}}$.

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

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.

For any ${\displaystyle n+1}$ bivariate data points ${\displaystyle (x_{0},y_{0}),\dotsc ,(x_{n},y_{n})\in \mathbb {R} ^{2}}$, 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. ${\displaystyle p(x_{0})=y_{0},\ldots ,p(x_{n})=y_{n}}$.^[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 ${\displaystyle (y_{0},\ldots ,y_{n})\in \mathbb {R} ^{n+1}}$ and the vector space ${\displaystyle P(n)}$ of real polynomials of degree at most n: $${\displaystyle L_{n}:\mathbb {R} ^{n+1}{\stackrel {\sim }{\longrightarrow }}\,P(n).}$$

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.

Consider the Lagrange basis functions ${\displaystyle L_{1}(x),\ldots ,L_{n}(x)}$ given by: $${\displaystyle L_{j}(x)=\prod _{i\neq j}{\frac {x-x_{i}}{x_{j}-x_{i}}}={\frac {(x-x_{0})\cdots (x-x_{j-1})(x-x_{j+1})\cdots (x-x_{n})}{(x_{j}-x_{0})\cdots (x_{j}-x_{j-1})(x_{j}-x_{j+1})\cdots (x_{j}-x_{n})}}.}$$

Notice that ${\displaystyle L_{j}(x)}$ is a polynomial of degree ${\displaystyle n}$, and we have ${\displaystyle L_{j}(x_{k})=0}$ for each ${\displaystyle j\neq k}$, 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 ${\displaystyle k=0,\dotsc ,n}$. 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,^[a] so ${\displaystyle p(x)-q(x)}$ must be the zero polynomial, i.e. ${\displaystyle p(x)=q(x)}$.^[3]

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: $${\displaystyle {\begin{bmatrix}x_{0}^{n}&x_{0}^{n-1}&x_{0}^{n-2}&\ldots &x_{0}&1\\x_{1}^{n}&x_{1}^{n-1}&x_{1}^{n-2}&\ldots &x_{1}&1\\\vdots &\vdots &\vdots &&\vdots &\vdots \\x_{n}^{n}&x_{n}^{n-1}&x_{n}^{n-2}&\ldots &x_{n}&1\end{bmatrix}}{\begin{bmatrix}a_{n}\\a_{n-1}\\\vdots \\a_{0}\end{bmatrix}}={\begin{bmatrix}y_{0}\\y_{1}\\\vdots \\y_{n}\end{bmatrix}}.}$$

An interpolant ${\displaystyle p(x)}$ corresponds to a solution ${\displaystyle A=(a_{n},\ldots ,a_{0})}$ 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 ${\displaystyle \textstyle \det(X)=\prod _{1\leq i<j\leq n}(x_{j}-x_{i}),}$ 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.

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.

We may write down the polynomial immediately in terms of Lagrange polynomials as: $${\displaystyle {\begin{aligned}p(x)&={\frac {(x-x_{1})(x-x_{2})\cdots (x-x_{n})}{(x_{0}-x_{1})(x_{0}-x_{2})\cdots (x_{0}-x_{n})}}y_{0}\\[4pt]&+{\frac {(x-x_{0})(x-x_{2})\cdots (x-x_{n})}{(x_{1}-x_{0})(x_{1}-x_{2})\cdots (x_{1}-x_{n})}}y_{1}\\[4pt]&+\cdots \\[4pt]&+{\frac {(x-x_{0})(x-x_{1})\cdots (x-x_{n-1})}{(x_{n}-x_{0})(x_{n}-x_{1})\cdots (x_{n}-x_{n-1})}}y_{n}\\[7pt]&=\sum _{i=0}^{n}{\Biggl (}\prod _{\stackrel {\!0\,\leq \,j\,\leq \,n}{j\,\neq \,i}}{\frac {x-x_{j}}{x_{i}-x_{j}}}{\Biggr )}y_{i}=\sum _{i=0}^{n}{\frac {p(x)}{p'(x_{i})(x-x_{i})}}\,y_{i}\end{aligned}}}$$For matrix arguments, this formula is called Sylvester's formula and the matrix-valued Lagrange polynomials are the Frobenius covariants.

For a polynomial ${\displaystyle p_{n}}$ of degree less than or equal to n, that interpolates ${\displaystyle f}$ at the nodes ${\displaystyle x_{i}}$ where ${\displaystyle i=0,1,2,3,\cdots ,n}$. Let ${\displaystyle p_{n+1}}$ be the polynomial of degree less than or equal to n+1 that interpolates ${\displaystyle f}$ at the nodes ${\displaystyle x_{i}}$ where ${\displaystyle i=0,1,2,3,\cdots ,n,n+1}$. Then ${\displaystyle p_{n+1}}$ is given by:$${\displaystyle p_{n+1}(x)=p_{n}(x)+a_{n+1}w_{n}(x)}$$where ${\textstyle w_{n}(x):=\prod _{i=0}^{n}(x-x_{i})}$ also known as Newton basis and ${\textstyle a_{n+1}:={f(x_{n+1})-p_{n}(x_{n+1}) \over w_{n}(x_{n+1})}}$.

Proof:

This can be shown for the case where ${\displaystyle i=0,1,2,3,\cdots ,n}$:$${\displaystyle p_{n+1}(x_{i})=p_{n}(x_{i})+a_{n+1}\prod _{j=0}^{n}(x_{i}-x_{j})=p_{n}(x_{i})}$$and when ${\displaystyle i=n+1}$:$${\displaystyle p_{n+1}(x_{n+1})=p_{n}(x_{n+1})+{f(x_{n+1})-p_{n}(x_{n+1}) \over w_{n}(x_{n+1})}w_{n}(x_{n+1})=f(x_{n+1})}$$By the uniqueness of interpolated polynomials of degree less than ${\displaystyle n+1}$, ${\textstyle p_{n+1}(x)=p_{n}(x)+a_{n+1}w_{n}(x)}$ is the required polynomial interpolation. The function can thus be expressed as:

${\textstyle p_{n}(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})(x-x_{1})+\cdots +a_{n}(x-x_{0})\cdots (x-x_{n-1}).}$

To find ${\displaystyle a_{i}}$, we have to solve the lower triangular matrix formed by arranging ${\textstyle p_{n}(x_{i})=f(x_{i})=y_{i}}$ from above equation in matrix form:

${\displaystyle {\begin{bmatrix}1&&\ldots &&0\\1&x_{1}-x_{0}&&&\\1&x_{2}-x_{0}&(x_{2}-x_{0})(x_{2}-x_{1})&&\vdots \\\vdots &\vdots &&\ddots &\\1&x_{k}-x_{0}&\ldots &\ldots &\prod _{j=0}^{n-1}(x_{n}-x_{j})\end{bmatrix}}{\begin{bmatrix}a_{0}\\\\\vdots \\\\a_{n}\end{bmatrix}}={\begin{bmatrix}y_{0}\\\\\vdots \\\\y_{n}\end{bmatrix}}}$

The coefficients are derived as

${\displaystyle a_{j}:=[y_{0},\ldots ,y_{j}]}$

where

${\displaystyle [y_{0},\ldots ,y_{j}]}$

is the notation for divided differences. Thus, Newton polynomials are used to provide a polynomial interpolation formula of n points.^[3]

Proof

The first few coefficients can be calculated using the system of equations. The form of n-th coefficient is assumed for proof by mathematical induction. ${\displaystyle {\begin{aligned}a_{0}&=y_{0}=[y_{0}]\\a_{1}&={y_{1}-y_{0} \over x_{1}-x_{0}}=[y_{0},y_{1}]\\\vdots \\a_{n}&=[y_{0},\cdots ,y_{n}]\quad {\text{(let)}}\\\end{aligned}}}$ Let Q be polynomial interpolation of points ${\displaystyle (x_{1},y_{1}),\ldots ,(x_{n},y_{n})}$. Adding ${\displaystyle (x_{0},y_{0})}$ to the polynomial Q: ${\displaystyle Q(x)+a'_{n}(x-x_{1})\cdot \ldots \cdot (x-x_{n})=P_{n}(x),}$ where ${\textstyle a'_{n}(x_{0}-x_{1})\ldots (x_{0}-x_{n})=y_{0}-Q(x_{0})}$. By uniqueness of the interpolating polynomial of the points ${\displaystyle (x_{0},y_{0}),\ldots ,(x_{n},y_{n})}$, equating the coefficients of ${\displaystyle x^{n-1}}$ we get, ${\textstyle a'_{n}=[y_{0},\ldots ,y_{n}]}$. Hence the polynomial can be expressed as:${\displaystyle P_{n}(x)=Q(x)+[y_{0},\ldots ,y_{n}](x-x_{1})\cdot \ldots \cdot (x-x_{n}).}$ Adding ${\displaystyle (x_{n+1},y_{n+1})}$ to the polynomial Q, it has to satisfiy: ${\textstyle [y_{1},\ldots ,y_{n+1}](x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n})=y_{n+1}-Q(x_{n+1})}$ where the formula for ${\textstyle a_{n}}$ and interpolating polynomial are used. The ${\textstyle a_{n+1}}$ term for the polynomial ${\textstyle P_{n+1}}$ can be found by calculating:$${\displaystyle {\begin{aligned}&[y_{0},\ldots ,y_{n+1}](x_{n+1}-x_{0})\cdot \ldots \cdot (x_{n+1}-x_{n})\\&={\frac {[y_{1},\ldots ,y_{n+1}]-[y_{0},\ldots ,y_{n}]}{x_{n+1}-x_{0}}}(x_{n+1}-x_{0})\cdot \ldots \cdot (x_{n+1}-x_{n})\\&=\left([y_{1},\ldots ,y_{n+1}]-[y_{0},\ldots ,y_{n}]\right)(x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n})\\&=[y_{1},\ldots ,y_{n+1}](x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n})-[y_{0},\ldots ,y_{n}](x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n})\\&=(y_{n+1}-Q(x_{n+1}))-[y_{0},\ldots ,y_{n}](x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n})\\&=y_{n+1}-(Q(x_{n+1})+[y_{0},\ldots ,y_{n}](x_{n+1}-x_{1})\cdot \ldots \cdot (x_{n+1}-x_{n}))\\&=y_{n+1}-P(x_{n+1}).\end{aligned}}}$$which implies that ${\displaystyle a_{n+1}={y_{n+1}-P_{n}(x_{n+1}) \over w_{n}(x_{n+1})}=[y_{0},\ldots ,y_{n+1}]}$. Hence it is proved by principle of mathematical induction.

The Newton polynomial can be expressed in a simplified form when ${\displaystyle x_{0},x_{1},\dots ,x_{k}}$ are arranged consecutively with equal spacing.

If ${\displaystyle x_{0},x_{1},\dots ,x_{k}}$ are consecutively arranged and equally spaced with ${\displaystyle {x}_{i}={x}_{0}+ih}$ for i = 0, 1, ..., k and some variable x is expressed as ${\displaystyle {x}={x}_{0}+sh}$, then the difference ${\displaystyle x-x_{i}}$ can be written as ${\displaystyle (s-i)h}$. So the Newton polynomial becomes

${\displaystyle {\begin{aligned}N(x)&=[y_{0}]+[y_{0},y_{1}]sh+\cdots +[y_{0},\ldots ,y_{k}]s(s-1)\cdots (s-k+1){h}^{k}\\&=\sum _{i=0}^{k}s(s-1)\cdots (s-i+1){h}^{i}[y_{0},\ldots ,y_{i}]\\&=\sum _{i=0}^{k}{s \choose i}i!{h}^{i}[y_{0},\ldots ,y_{i}].\end{aligned}}}$

Since the relationship between divided differences and forward differences is given as:^[4]$${\displaystyle [y_{j},y_{j+1},\ldots ,y_{j+n}]={\frac {1}{n!h^{n}}}\Delta ^{(n)}y_{j},}$$Taking ${\displaystyle y_{i}=f(x_{i})}$, if the representation of x in the previous sections was instead taken to be ${\displaystyle x=x_{j}+sh}$, the Newton forward interpolation formula is expressed as:$${\displaystyle f(x)\approx N(x)=N(x_{j}+sh)=\sum _{i=0}^{k}{s \choose i}\Delta ^{(i)}f(x_{j})}$$which is the interpolation of all points after ${\displaystyle x_{j}}$. It is expanded as:$${\displaystyle f(x_{j}+sh)=f(x_{j})+{\frac {s}{1!}}\Delta f(x_{j})+{\frac {s(s-1)}{2!}}\Delta ^{2}f(x_{j})+{\frac {s(s-1)(s-2)}{3!}}\Delta ^{3}f(x_{j})+{\frac {s(s-1)(s-2)(s-3)}{4!}}\Delta ^{4}f(x_{j})+\cdots }$$

If the nodes are reordered as ${\displaystyle {x}_{k},{x}_{k-1},\dots ,{x}_{0}}$, the Newton polynomial becomes

${\displaystyle N(x)=[y_{k}]+[{y}_{k},{y}_{k-1}](x-{x}_{k})+\cdots +[{y}_{k},\ldots ,{y}_{0}](x-{x}_{k})(x-{x}_{k-1})\cdots (x-{x}_{1}).}$

If ${\displaystyle {x}_{k},\;{x}_{k-1},\;\dots ,\;{x}_{0}}$ are equally spaced with ${\displaystyle {x}_{i}={x}_{k}-(k-i)h}$ for i = 0, 1, ..., k and ${\displaystyle {x}={x}_{k}+sh}$, then,

${\displaystyle {\begin{aligned}N(x)&=[{y}_{k}]+[{y}_{k},{y}_{k-1}]sh+\cdots +[{y}_{k},\ldots ,{y}_{0}]s(s+1)\cdots (s+k-1){h}^{k}\\&=\sum _{i=0}^{k}{(-1)}^{i}{-s \choose i}i!{h}^{i}[{y}_{k},\ldots ,{y}_{k-i}].\end{aligned}}}$

Since the relationship between divided differences and backward differences is given as:^{[citation needed]}$${\displaystyle [{y}_{j},y_{j-1},\ldots ,{y}_{j-n}]={\frac {1}{n!h^{n}}}\nabla ^{(n)}y_{j},}$$taking ${\displaystyle y_{i}=f(x_{i})}$, if the representation of x in the previous sections was instead taken to be ${\displaystyle x=x_{j}+sh}$, the Newton backward interpolation formula is expressed as:$${\displaystyle f(x)\approx N(x)=N(x_{j}+sh)=\sum _{i=0}^{k}{(-1)}^{i}{-s \choose i}\nabla ^{(i)}f(x_{j}).}$$which is the interpolation of all points before ${\displaystyle x_{j}}$. It is expanded as:$${\displaystyle f(x_{j}+sh)=f(x_{j})+{\frac {s}{1!}}\nabla f(x_{j})+{\frac {s(s+1)}{2!}}\nabla ^{2}f(x_{j})+{\frac {s(s+1)(s+2)}{3!}}\nabla ^{3}f(x_{j})+{\frac {s(s+1)(s+2)(s+3)}{4!}}\nabla ^{4}f(x_{j})+\cdots }$$

A 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:^[5]

1. Left to right steps indicate addition whereas right to left steps indicate subtraction
2. If the slope of a step is positive, the term to be used is the product of the difference and the factor immediately below it. If the slope of a step is negative, the term to be used is the product of the difference and the factor immediately above it.
3. If a step is horizontal and passes through a factor, use the product of the factor and the average of the two terms immediately above and below it. If a step is horizontal and passes through a difference, use the product of the difference and the average of the two terms immediately above and below it.

The factors are expressed using the formula:$${\displaystyle C(u+k,n)={\frac {(u+k)(u+k-1)\cdots (u+k-n+1)}{n!}}}$$

If a path goes from ${\displaystyle \Delta ^{n-1}y_{s}}$ to ${\displaystyle \Delta ^{n+1}y_{s-1}}$, it can connect through three intermediate steps, (a) through ${\displaystyle \Delta ^{n}y_{s-1}}$, (b) through ${\textstyle C(u-s,n)}$ or (c) through ${\displaystyle \Delta ^{n}y_{s}}$. 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.

Path (a):

${\displaystyle C(u-s,n)\Delta ^{n}y_{s-1}+C(u-s+1,n+1)\Delta ^{n+1}y_{s-1}}$

Path (b):

${\displaystyle C(u-s,n)\Delta ^{n}y_{s}+C(u-s,n+1)\Delta ^{n+1}y_{s-1}}$

Path (c):

${\displaystyle C(u-s,n){\frac {\Delta ^{n}y_{s-1}+\Delta ^{n}y_{s}}{2}}\quad +{\frac {C(u-s+1,n+1)+C(u-s,n+1)}{2}}\Delta ^{n+1}y_{s-1}}$

Subtracting contributions from path a and b:

${\displaystyle {\begin{aligned}{\text{Path a - Path b}}=&C(u-s,n)(\Delta ^{n}y_{s-1}-\Delta ^{n}y_{s})+(C(u-s+1,n+1)-C(u-s,n-1))\Delta ^{n+1}y_{s-1}\\=&-C(u-s,n)\Delta ^{n+1}y_{s-1}+C(u-s,n){\frac {(u-s+1)-(u-s-n)}{n+1}}\Delta ^{n+1}y_{s-1}\\=&C(u-s,n)(-\Delta ^{n+1}y_{s-1}+\Delta ^{n+1}y_{s-1})=0\\\end{aligned}}}$

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) ${\displaystyle y_{s+1}}$ to ${\displaystyle y_{s}}$ through ${\displaystyle \Delta y_{s}}$ or (b) factor between ${\displaystyle y_{s+1}}$ and ${\displaystyle y_{s}}$, to ${\displaystyle y_{s}}$ through ${\displaystyle \Delta y_{s}}$ or (c) starting from ${\displaystyle y_{s}}$.

Path (a)

${\displaystyle y_{s+1}+C(u-s-1,1)\Delta y_{s}-C(u-s,1)\Delta y_{s}}$

Path (b)

${\displaystyle {\frac {y_{s+1}+y_{s}}{2}}+{\frac {C(u-s-1,1)+C(u-s,1)}{2}}\Delta y_{s}-C(u-s,1)\Delta y_{s}}$

Path (c)

${\displaystyle y_{s}}$

Since ${\displaystyle \Delta y_{s}=y_{s+1}-y_{s}}$, substituting in the above equations shows that all the above terms reduce to ${\displaystyle y_{s}}$ and are hence equivalent. Hence these paths can be morphed to start from the leftmost corner and end in a common point.^[5]

Taking negative slope transversal from ${\displaystyle y_{0}}$ to ${\displaystyle \Delta ^{n}y_{0}}$ gives the interpolation formula of all the ${\displaystyle n+1}$ consecutively arranged points, equivalent to Newton's forward interpolation formula:

${\displaystyle {\begin{aligned}y(s)&=y_{0}+C(s,1)\Delta y_{0}+C(s,2)\Delta ^{2}y_{0}+C(s,3)\Delta ^{3}y_{0}+\cdots \\&=y_{0}+s\Delta y_{0}+{\frac {s(s-1)}{2}}\Delta ^{2}y_{0}+{\frac {s(s-1)(s-2)}{3!}}\Delta ^{3}y_{0}+{\frac {s(s-1)(s-2)(s-3)}{4!}}\Delta ^{4}y_{0}+\cdots \end{aligned}}}$

whereas, taking positive slope transversal from ${\displaystyle y_{n}}$ to ${\displaystyle \nabla ^{n}y_{n}=\Delta ^{n}y_{0}}$, gives the interpolation formula of all the ${\displaystyle n+1}$ consecutively arranged points, equivalent to Newton's backward interpolation formula:

${\displaystyle {\begin{aligned}y(u)&=y_{k}+C(u-k,1)\Delta y_{k-1}+C(u-k+1,2)\Delta ^{2}y_{k-2}+C(u-k+2,3)\Delta ^{3}y_{k-3}+\cdots \\&=y_{k}+(u-k)\Delta y_{k-1}+{\frac {(u-k+1)(u-k)}{2}}\Delta ^{2}y_{k-2}+{\frac {(u-k+2)(u-k+1)(u-k)}{3!}}\Delta ^{3}y_{k-3}+\cdots \\y(k+s)&=y_{k}+(s)\nabla y_{k}+{\frac {(s+1)s}{2}}\nabla ^{2}y_{k}+{\frac {(s+2)(s+1)s}{3!}}\nabla ^{3}y_{k}+{\frac {(s+3)(s+2)(s+1)s}{4!}}\nabla ^{4}y_{k}+\cdots \\\end{aligned}}}$

where ${\displaystyle s=u-k}$ is the number corresponding to that introduced in Newton interpolation.

Taking a zigzag line towards the right starting from ${\displaystyle y_{0}}$ with negative slope, we get Gauss forward formula:

${\displaystyle y(u)=y_{0}+u\Delta y_{0}+{\frac {u(u-1)}{2}}\Delta ^{2}y_{-1}+{\frac {(u+1)u\left(u-1\right)}{3!}}\Delta ^{3}y_{-1}+{\frac {(u+1)u\left(u-1\right)(u-2)}{4!}}\Delta ^{4}y_{-2}+\cdots }$

whereas starting from ${\displaystyle y_{0}}$ with positive slope, we get Gauss backward formula:

${\displaystyle y(u)=y_{0}+u\Delta y_{-1}+{\frac {(u+1)u}{2}}\Delta ^{2}y_{-1}+{\frac {(u+1)u\left(u-1\right)}{3!}}\Delta ^{3}y_{-2}+{\frac {(u+2)(u+1)u\left(u-1\right)}{4!}}\Delta ^{4}y_{-2}+\cdots }$

By taking a horizontal path towards the right starting from ${\displaystyle y_{0}}$, we get Stirling formula:

${\displaystyle {\begin{aligned}y(u)&=y_{0}+u{\frac {\Delta y_{0}+\Delta y_{-1}}{2}}+{\frac {C(u+1,2)+C(u,2)}{2}}\Delta ^{2}y_{-1}+C(u+1,3){\frac {\Delta ^{3}y_{-2}+\Delta ^{3}y_{-1}}{2}}+\cdots \\&=y_{0}+u{\frac {\Delta y_{0}+\Delta y_{-1}}{2}}+{\frac {u^{2}}{2}}\Delta ^{2}y_{-1}+{\frac {u(u^{2}-1)}{3!}}{\frac {\Delta ^{3}y_{-2}+\Delta ^{3}y_{-1}}{2}}+{\frac {u^{2}(u^{2}-1)}{4!}}\Delta ^{4}y_{-2}+\cdots \end{aligned}}}$

Stirling formula is the average of Gauss forward and Gauss backward formulas.

By taking a horizontal path towards the right starting from factor between ${\displaystyle y_{0}}$ and ${\displaystyle y_{1}}$, we get Stirling formula:

${\displaystyle {\begin{aligned}y(u)&=1{\frac {y_{0}+y_{1}}{2}}+{\frac {C(u,1)+C(u-1,1)}{2}}\Delta y_{0}+C(u,2){\frac {\Delta ^{2}y_{-1}+\Delta ^{2}y_{0}}{2}}+\cdots \\&={\frac {y_{0}+y_{1}}{2}}+\left(u-{\frac {1}{2}}\right)\Delta y_{0}+{\frac {u(u-1)}{2}}{\frac {\Delta ^{2}y_{-1}+\Delta ^{2}y_{0}}{2}}+{\frac {\left(u-{\frac {1}{2}}\right)u\left(u-1\right)}{3!}}\Delta ^{3}y_{0}+{\frac {(u+1)u(u-1)(u-2)}{4!}}{\frac {\Delta ^{4}y_{-1}+\Delta ^{4}y_{-2}}{2}}+\cdots \\\end{aligned}}}$

The Vandermonde matrix in the second proof above may have large condition number,^[6] 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.^[7][8][9] These methods rely on constructing first a Newton interpolation of the polynomial and then converting it to a monomial form.

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 (i.e. using Newton basis) 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).^[10]

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.

Given a set of (position, value) data points ${\displaystyle (x_{0},y_{0}),\ldots ,(x_{j},y_{j}),\ldots ,(x_{n},y_{n})}$ 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 ${\displaystyle x_{j}}$: $${\displaystyle c_{j}(x)=L_{j}(x_{0},\ldots ,x_{n};x)=\prod _{0\leq i\leq n \atop i\neq j}{\frac {x-x_{i}}{x_{j}-x_{i}}}={\frac {(x-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x-x_{j-1})}{(x_{j}-x_{j-1})}}{\frac {(x-x_{j+1})}{(x_{j}-x_{j+1})}}\cdots {\frac {(x-x_{n})}{(x_{j}-x_{n})}}.}$$

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 ${\displaystyle (x_{0},v_{0}),\ldots ,(x_{n},v_{n})}$ 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 ${\displaystyle x={\tfrac {t-b}{a}}}$): $${\displaystyle (t_{0},w_{0}),\ldots ,(t_{j},w_{j})\ldots ,(t_{n},w_{n})\qquad {\text{with}}\qquad t_{j}=ax_{j}+b,}$$

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

${\displaystyle {\tilde {c}}_{j}(t):=c_{j}({\tfrac {t-b}{a}})=c_{j}(x),}$

and write the interpolation polynomial as:

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

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

$${\displaystyle {\begin{aligned}y(x):=\sum _{j=0}^{2}y_{j}c_{j}(x)&=y_{0}{\frac {(x-1)(x-2)}{(0-1)(0-2)}}+y_{1}{\frac {(x-0)(x-2)}{(1-0)(1-2)}}+y_{2}{\frac {(x-0)(x-1)}{(2-0)(2-1)}}\\&={\tfrac {1}{2}}y_{0}(x-1)(x-2)-y_{1}(x-0)(x-2)+{\tfrac {1}{2}}y_{2}(x-0)(x-1).\end{aligned}}}$$

For example, ${\displaystyle y(3)=y_{3}=y_{0}-3y_{1}+3y_{2}}$ and ${\displaystyle y(1.5)=y_{1.5}={\tfrac {1}{8}}(-y_{0}+6y_{1}+3y_{2})}$.

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

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

Thus, given values ${\displaystyle y_{0},\ldots ,y_{n}}$ at equally spaced points, where ${\displaystyle n=d+1}$, we have: $${\displaystyle (-1)^{n}y_{0}+(-1)^{n-1}{\binom {n}{1}}y_{1}+\cdots -{\binom {n}{n-1}}y_{n-1}+y_{n}=0.}$$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.

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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 $${\displaystyle f(x)-p_{n}(x)=f[x_{0},\ldots ,x_{n},x]\prod _{i=0}^{n}(x-x_{i})}$$

where ${\textstyle f[x_{0},\ldots ,x_{n},x]}$ is the (n+1)^st divided difference of the data points

${\displaystyle (x_{0},f(x_{0})),\ldots ,(x_{n},f(x_{n})),(x,f(x))}$.

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 ${\displaystyle x_{0},\ldots ,x_{n}\in I}$. 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 ${\textstyle \left|\prod (x-x_{i})\right|}$, which is achieved by the Chebyshev nodes.

Set the error term as ${\textstyle R_{n}(x)=f(x)-p_{n}(x)}$, and define an auxiliary function:$${\displaystyle Y(t)=R_{n}(t)-{\frac {R_{n}(x)}{W(x)}}W(t)\qquad {\text{where}}\qquad W(t)=\prod _{i=0}^{n}(t-x_{i}).}$$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 ${\textstyle R_{n}^{(n+1)}(t)=f^{(n+1)}(t)}$, 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, ${\displaystyle Y^{\prime }(t)}$ 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 reasoning 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.^[11] 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.

In the case of equally spaced interpolation nodes where ${\displaystyle x_{i}=a+ih}$, for ${\displaystyle i=0,1,\ldots ,n,}$ and where ${\displaystyle h=(b-a)/n,}$ the product term in the interpolation error formula can be bound as^[12] $${\displaystyle \left|\prod _{i=0}^{n}(x-x_{i})\right|=\prod _{i=0}^{n}\left|x-x_{i}\right|\leq {\frac {n!}{4}}h^{n+1}.}$$

Thus the error bound can be given as $${\displaystyle \left|R_{n}(x)\right|\leq {\frac {h^{n+1}}{4(n+1)}}\max _{\xi \in [a,b]}\left|f^{(n+1)}(\xi )\right|}$$

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.

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): $${\displaystyle \left\|f-X(f)\right\|\leq (L+1)\left\|f-p^{*}\right\|.}$$

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

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.

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.^[13]

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:

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:

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

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

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

$${\displaystyle \|X_{n}\|\geq {\tfrac {2}{\pi }}\log(n+1)+C.}$$

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:

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.

• Newton series
• Polynomial regression