The Lagrange Interpolation Formula

Let the polynomial of $n$th degree

have only distinct zeros $x_1, x_2, \ldots , x_n$. Then we have

Furthermore $\sigma_k$ is independent of $a_n$ for $k = n, n+1, \ldots , 2n-2$ and is linear in $a_n$ with coefficient $-a_0^{-2}$ for $k = 2n-1$.

(Continuation of problem 67.) We have

(Continuation of problem 67.) Let $x_1, x_2, \ldots , x_n$ be different from $0$ and $-1$. Prove that

Let $x_0, x_1, x_2, \ldots , x_n$ be arbitrary integers, $x_0 < x_1 < x_2 < \cdots < x_n$. Every polynomial of $n$th degree of the form

assumes at the points $x_0, x_1, x_2, \ldots , x_n$ values, at least one of which is of absolute value $\geqq n!/2^n$.

If one chooses the zeros of $T_n(x)$ (problem 1), i.e.the numbers $x_\nu = \cos (2\nu -1)\pi /(2n)$, $\nu = 1, 2, \ldots , n$, as interpolation points, then the basis polynomials are

Hence every polynomial $P(x)$ of degree $n-1$ may be represented as follows:

The square roots are to be taken everywhere as positive.

If one chooses the zeros of $U_n(x)$ (problem 1), that is the numbers $x_\nu = \cos \nu \pi /(n+1)$, $\nu =1,2,\ldots ,n$, as interpolation points, then the basis polynomials are

Hence every polynomial of degree $n-1$ may be represented as follows:

If one chooses the zeros of $U_{n-1}(x)(x^2-1)$, that is the numbers $x_\nu = \cos \nu \pi /n$, $\nu =1,2,\ldots ,n-1$, and in addition $x_0=1$, $x_n=-1$ as interpolation points, then the basis polynomials $f_0(x), f_1(x), f_2(x), \ldots , f_n(x)$ are the following:

Hence every polynomial $P(x)$ of degree $n$ may be represented in the following form:

If one chooses the $n$th roots of unity $\varepsilon_\nu = e^{2\pi i \nu /n}$, $\nu = 1, 2, \ldots , n$, as interpolation points, then the basis polynomials are

Hence for an arbitrary polynomial $P(x)$ of degree $n-1$ we have the representation

Let the polynomials $P(x)$ and $Q(x)$ be both of degree $n$, and let $x_0, x_1, \ldots , x_n$ be any $n+1$ distinct or in part coinciding real or complex numbers. From the $n+1$ equations

it follows that we have identically

It is possible, and in fact in a unique manner, to determine, corresponding to $n+1$ arbitrary given numbers $c_0, c_1, c_2, \ldots , c_n$, a polynomial $P(x)$ of degree $\leqq n$ such that

(cf.problem 75). This polynomial $P(x)$ may be represented in the form

where $A_0(x), A_1(x), \ldots , A_n(x)$ are arithmetically determined polynomials, independent of the arbitrarily assigned numbers $c_0 = P(0)$, $c_1 = P'(1)$, , $c_n = P^{(n)}(n)$. (They may be thought of as the ``basis polynomials'' of this interpolation, which is essentially different from the Lagrange interpolation.)