Polynomial/K/Interpolation/Section

A piecewise linear and

a polynomial interpolation.

The following theorem is called theorem about polynomial interpolation and describes the interpolation of given function values by a polynomial. If just one function value at one point is given, then this determines a constant polynomial, two values at two points determine a linear polynomial (the graph is a line), three values at three points determine a quadratic polynomial, etc.

Theorem

Let ${}K$ be a field, and let ${}n$ different elements ${}a_{1},\ldots ,a_{n}\in K$, and ${}n$ elements ${}b_{1},\ldots ,b_{n}\in K$ be given. Then there exists a unique polynomial ${}P\in K[X]$ of degree ${}\leq n-1$, such that ${}P{\left(a_{i}\right)}=b_{i}$

holds for all ${}i$.

Proof

We prove the existence and consider first the situation where ${}b_{j}=0$ for all ${}j\neq i$ for some fixed ${}i$. Then

$(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$

is a polynomial of degree ${}n-1$, which at the points ${}a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}$ has value ${}0$. The polynomial

${\frac {b_{i}}{(a_{i}-a_{1})\cdots (a_{i}-a_{i-1})(a_{i}-a_{i+1})\cdots (a_{i}-a_{n})}}(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$

has at these points still a zero, but additionally at ${}a_{i}$, its value is ${}b_{i}$. We denote this polynomial by ${}P_{i}$. Then

${}P=P_{1}+P_{2}+\cdots +P_{n}\,$

is the polynomial looked for, because for the point ${}a_{i}$, we have

${}P_{j}(a_{i})=0\,$

for ${}j\neq i$ and ${}P_{i}(a_{i})=b_{i}$.

The uniqueness follows from fact.

$\Box$

Remark

If the data ${}a_{1},\ldots ,a_{n}$ and ${}b_{1},\ldots ,b_{n}$ are given, then one can find the interpolating polynomial ${}P$ of degree ${}\leq n-1$, which exists by fact, in the following way: We write

${}P=c_{0}+c_{1}X+c_{2}X^{2}+\cdots +c_{n-2}X^{n-2}+c_{n-1}X^{n-1}\,$

with unknown coefficients ${}c_{0},\ldots ,c_{n-1}$, and determine then these coefficients. Each interpolation point ${}(a_{i},b_{i})$ yields a linear equation

${}c_{0}+c_{1}a_{i}+c_{2}a_{i}^{2}+\cdots +c_{n-2}a_{i}^{n-2}+c_{n-1}a_{i}^{n-1}=b_{i}\,$

over ${}K$. The resulting system of linear equations has exactly one solution ${}(c_{0},\ldots ,c_{n-1})$, which gives the polynomial.