Derivative of a Polynomial with respect to a Point. A Theorem of Grace

By the derivative of a polynomial $f(z)$ of degree $n$ with respect to the point $\zeta$, denoted by $A_{\zeta }f(z)$, we mean the ordinary derivative if $\zeta =\infty$, and

if $\zeta$ is finite. $A_{\zeta }f(z)$ is a polynomial of degree $n-1$. (When using homogeneous coordinates this expression is called the first polar, or, by older authors, emanant; cf. Laguerre, Oeuvres, Vol. 1, p. 48. Paris: Gauthier-Villars, 1898.)

Let the $n+1$ coefficients of $f(z)$ (in the notation of §2 preceding 111) be $a_0, a_1, a_2, \ldots , a_n$. What are the $n$ coefficients of $[A_{\zeta }f(z)]/n$?

Let $\zeta$ be arbitrary. Show that the operation $A_{\zeta }f(z)$ is distributive: If we denote by $f_1(z)$ and $f_2(z)$ two polynomials of degree $n$, and by $c_1$ and $c_2$ arbitrary constants, then we have

Let $\zeta$ be arbitrary and $f(z)$ a polynomial of degree $n$, $f(z)=g(z)h(z)$ a factorization of $f(z)$ into the two factors $g(z)$ and $h(z)$ of degree $k$ and $l$ respectively, $k+l=n$. Show that

Let $f(z)$ be a polynomial of degree $n$, $\zeta_1$ and $\zeta_2$ two arbitrary constants. The operations $A_{\zeta_1}f(z)$ and $A_{\zeta_2}f(z)$ are commutative, that is to say

Show that $A_{\zeta }f(z)$ vanishes identically if and only if all the zeros of $f(z)$ coincide with $\zeta$.

By the derived system of the $n$ points $z_1, z_2, \ldots , z_n$ with respect to an $(n+1)$th point $\zeta$ we mean the $n-1$ zeros $z_1', z_2', \ldots , z_{n-1}'$ of $A_{\zeta }f(z)$, where $f(z)$ is the polynomial of degree $n$ whose zeros are $z_1, z_2, \ldots , z_n$, all the zeros being correctly counted (§2, preceding 111). The derived system is completely determined, except in the single exceptional case where the $n+1$ points $z_1, z_2, \ldots , z_n, \zeta$ all coincide (cf. 132).

If $\zeta$ is finite, then to the derived system with respect to $\zeta$ in general there belong four different kinds of points:

(1) The finite points distinct from $z_1, z_2, \ldots , z_n, \zeta$, with respect to which the center of gravity of $z_1, z_2, \ldots , z_n$ is precisely $\zeta$.

(2) The multiple zeros of $f(z)$, distinct from $\zeta$, each with multiplicity reduced by $1$.

(3) $\zeta$ itself only if it is a zero of $f(z)$ and in that case with the same multiplicity.

(4) The point $\infty$ if it is at least a double zero of $f(z)$; moreover also if it is not a zero of $f(z)$ but if $\zeta$ is the ordinary center of gravity of $f(z)$, i.e. the one constructed with respect to $\infty$.

A circular domain, whose boundary contains $\zeta$ as well as a point of the derived system with respect to $\zeta$, itself contains points of the original system.

If a circular domain contains the points $z_1, z_2, \ldots , z_n$ and does not contain the point $\zeta$, then it contains also the derived system of $z_1, z_2, \ldots , z_n$ with respect to $\zeta$.

The derived system of the points $z_1, z_2, \ldots , z_n$ lies in the smallest circular-arc polygon, circularly convex with respect to $\zeta$, that contains $z_1, z_2, \ldots , z_n$. (Generalization of Problem 31 of Part Three.)

If $f(z)$ is of degree $n$, then $(A_{\zeta })^{n-1}f(z)$ is of the first degree. Calculate the only zero of $(A_{\zeta })^{n-1}f(z)$!

Let

be an arbitrary polynomial of degree $n$ and let $\zeta_1, \zeta_2, \ldots , \zeta_n$ be arbitrary points in the complex plane. The expression

is a symmetric multilinear function of the variables $\zeta_1, \zeta_2, \ldots , \zeta_n$, which we can write as follows:

Here $\Sigma_0, \Sigma_1, \Sigma_2, \ldots , \Sigma_{n-1}, \Sigma_n$ are the elementary symmetric functions of $\zeta_1, \zeta_2, \ldots , \zeta_n$:

If the last $k$ of the numbers $\zeta_1, \zeta_2, \ldots , \zeta_n$ (and only these) lie at infinity, then we set $\Sigma_0=\Sigma_1=\cdots =\Sigma_{k-1}=0$, $\Sigma_k=1$, $\Sigma_{k+1}=\zeta_1+\zeta_2+\cdots +\zeta_{n-k}$, $\Sigma_{k+2}=\zeta_1\zeta_2+\zeta_1\zeta_3+\cdots +\zeta_{n-k-1}\zeta_{n-k}, \ldots , \Sigma_n=\zeta_1\zeta_2\ldots \zeta_{n-k}$.

Let

be an arbitrary polynomial of degree $n$ with zeros $z_1, z_2, \ldots , z_n$ and

an arbitrary polynomial of degree $n$ with zeros $\zeta_1, \zeta_2, \ldots , \zeta_n$. Calculate $A(\zeta_1, \zeta_2, \ldots , \zeta_n)f(z)$ and $A(z_1, z_2, \ldots , z_n)g(z)$.

If the two polynomials are related in such a manner that

or, equivalently [solution 139], $A(z_1, z_2, \ldots , z_n)g(z)=0$, then the two polynomials $f(z)$ and $g(z)$ are termed apolar. The term suggests the vanishing of the $n$th polar $A_{\zeta_1}A_{\zeta_2}\ldots A_{\zeta_n}f(z)$. The condition of apolarity is

[139.] The two systems $z_1, z_2, \ldots , z_n$ and $\zeta_1, \zeta_2, \ldots , \zeta_n$ are also termed mutually apolar.

What is the geometrical significance of the apolarity of $z_1, \zeta_1$, and that of $z_1, z_2$ and $\zeta_1, \zeta_2$?

Which are the systems $\zeta_1, \zeta_2, \ldots , \zeta_n$, that are apolar to the system consisting of the roots of the equation $z^n-1=0$?

Let $z_1, z_2, \ldots , z_n$ be an arbitrary system of points. There are $n$ systems $\zeta ,\zeta ,\ldots ,\zeta$ of coincident points that are apolar to the given system. Which are they?

The polynomial of degree $n$

is apolar to any system of numbers whose sum is $n$ and product is $0$.

Let $f(z)$ be an arbitrary polynomial of degree $n$, all of whose zeros lie in the circular domain $C$. If $\zeta_1, \zeta_2, \ldots , \zeta_k$, $k<n$, are arbitrary points lying outside $C$, then all the zeros of the polynomial of degree $n-k$, $A_{\zeta_1}A_{\zeta_2}\ldots A_{\zeta_k}f(z)$ lie in $C$.

Let the two polynomials $f(z)$ and $g(z)$ of degree $n$ be apolar to each other. Every circular domain that contains all the zeros of one polynomial also contains at least one zero of the other.

Let the two polynomials $f(z)$ and $g(z)$ of degree $n$ be apolar to each other. The two smallest convex polygons that contain all the zeros of $f(z)$ and $g(z)$ respectively have at least one common point. More generally, the two smallest circular-arc polygons that are circularly convex with respect to the same point and that contain all the zeros of $f(z)$ and $g(z)$ respectively, also have at least one common point.

The polynomial

always has one zero in the circle $\left|z\right| \leq 2$.

The polynomial

always has one zero in the circle $\left|z-1\right| \leq 1$.

The polynomial of $k+1$ terms

always has one zero in the circle

and hence in the circle

(Generalization of 147.)

If a polynomial of degree $n$ assumes the same value at two points $a$ and $b$, $a \neq b$, then its derivative has at least one zero in the circular disc with center at the midpoint of the line segment $ab$ and radius $(\left|a-b\right|/2)\cot (\pi /n)$. (Analogue of Rolle's theorem in the complex domain.)

Let

be a polynomial of degree $n$ all of whose zeros lie in a circular domain $C$. Also let

be a polynomial of degree $n$ with zeros $\beta_1, \beta_2, \ldots , \beta_n$. Every zero $\gamma$ of the polynomial

formed by ``composition'' of $f(z)$ and $g(z)$ then has the form

where $\nu$ denotes a suitably chosen index and $k$ a suitably chosen point in $C$. (Here we must set $\infty \cdot \infty = \infty$ and $0 \cdot \infty =$ indeterminate.)

Let the zeros of the polynomials $f(z)$ and $g(z)$ of degree $n$ all lie in the unit circle $\left|z\right| \leq 1$, and the zeros of at least one of the two polynomials in fact in $\left|z\right|<1$. Then also all the zeros of the polynomial of degree $n$ formed from $f(z)$ and $g(z)$ by composition [151] lie in the interior of the unit circle $\left|z\right|<1$.

Let the polynomial $f(z)$ of degree $n$ have all its zeros in a convex domain $\mathfrak {C}$ that contains the origin. Let the polynomial $g(z)$ of the same degree have only real zeros which lie in the interval $-1, 0$. The polynomial $h(z)$ of degree $n$ derived from the two polynomials by composition [151] then also has all its zeros in $\mathfrak {C}$.

If all the zeros of the polynomial

of degree $n$ lie in the real interval $-a, a$, and those of the polynomial

of degree $n$ in the interval $-b, 0$ (or $0, b$) then the zeros of the polynomial

of degree $n$ lie in the interval $-ab, ab$. ($a, b$ are positive numbers.)

Let all the zeros of the polynomial

of degree $n$ be real and those of the polynomial

of degree $n$ be real and of the same sign. Then all the zeros of the polynomial

are also real. ($z=\infty$ counts as a real zero.)

Under the assumptions of 155, all the zeros of the polynomial

are real.