﻿ A Fundamental Relationship of Polynomials and Its Proof

Vol.08 No.06(2018), Article ID:85519,5 pages
10.4236/apm.2018.86033

A Fundamental Relationship of Polynomials and Its Proof

Serdar Beji

Faculty of Naval Architecture and Ocean Engineering, Istanbul Technical University, Istanbul, Turkey    Received: May 21, 2018; Accepted: June 23, 2018; Published: June 26, 2018

ABSTRACT

A fundamental algebraic relationship for a general polynomial of degree n is given and proven by mathematical induction. The stated relationship is based on the well-known property of polynomials that the nth-differences of the subsequent values of an nth-order polynomial are constant.

Keywords:

Polynomials of Degree n, nth-Order Finite-Differences, Recurrence Relationship for Polynomials 1. Introduction

The “Fundamental Theorem of Algebra” states that a polynomial of degree n has n roots. Its first assertion in a different form is attributed to Peter Rothe in 1606 and later Albert Girard in 1629. Euler gave a clear statement of the theorem in a letter to Gauss in 1742 and at different times Gauss gave four different proofs (see  , p. 292-306).

A nearly as important property of a polynomial is the constancy of the nth‑differences of its subsequent values. To clarify this point let us begin with some demonstrations. While it is customary to use polynomials with real coefficients, here a second-order polynomial with complex coefficients is considered first,

${P}_{2}\left(x\right)=\left(1+i\right){x}^{2}-3ix+2$ (1)

where $i=\sqrt{-1}$ is the imaginary unit. Taking a real starting point ${x}_{0}=-2$ and a real step value $s=1$ the following Table 1 of differences can be established for the subsequent values of the polynomial.

The first differences are computed by taking the differences of the subsequent values of the polynomial as in ${P}_{2}\left(-2\right)-{P}_{2}\left(-1\right)=\left(6+10i\right)-\left(3+4i\right)=3+6i$ .

Table 1. Second-order differences for a sample second-order polynomial.

The second differences are obtained similarly using the first difference values: $\left(3+6i\right)-\left(1+4i\right)=2+2i$ .

Expressing the first differences in terms of polynomial values ${P}_{2}\left(-2\right)-{P}_{2}\left(-1\right)=3+6i$ and ${P}_{2}\left(-1\right)-{P}_{2}\left(0\right)=1+4i$ , the first value of the second differences may be written as

$\begin{array}{l}\left[{P}_{2}\left(-2\right)-{P}_{2}\left(-1\right)\right]-\left[{P}_{2}\left(-1\right)-{P}_{2}\left(0\right)\right]\\ ={P}_{2}\left(-2\right)-2{P}_{2}\left(-1\right)+{P}_{2}\left(0\right)=2+2i\end{array}$ (2)

which is a particular form, $n=2$ , of the general theorem presented in Section 2. The constant value $2+2i$ of the second-differences can be calculated from the general expression ${\left(-1\right)}^{n}n!{a}_{0}{s}^{n}$ where n is the degree of the polynomial and ${a}_{0}$ the coefficient of the nth-order term. For this particular example $n=2$ and ${a}_{0}=1+i$ hence the constant becomes ${\left(-1\right)}^{2}2!\left(1+i\right){1}^{2}=2+2i$ as found above.

Another example is now given for a third-order polynomial with real coefficients

${P}_{3}\left(x\right)=2{x}^{3}-{x}^{2}-3x+5$ (3)

In this example a complex starting point ${x}_{0}=1-i$ and a complex step value $s=-3+2i$ are used so that Table 2 of differences is constructed, where the constant value of the third-differences can be calculated from the general formula ${\left(-1\right)}^{n}n!{a}_{0}{s}^{n}$ as ${\left(-1\right)}^{3}3!2{\left(-3+2i\right)}^{3}=-108-552i$ .

A direct connection with the finite-difference approximation of derivatives of a polynomial is of course possible. Finite-difference approximation of the third-derivative of a third-order polynomial is given by

${{P}^{‴}}_{3}\text{}\left(x\right)=\frac{{P}_{3}\left({x}_{0}+3s\right)-3{P}_{3}\left({x}_{0}+2s\right)+3{P}_{3}\left({x}_{0}+s\right)-{P}_{3}\left({x}_{0}\right)}{{s}^{3}}$ (4)

where s is the incremental step. If the third-order polynomial is defined as ${P}_{3}\left(x\right)={a}_{0}{x}^{3}+{a}_{1}{x}^{2}+{a}_{2}x+{a}_{3}$ its third-derivative is ${{P}^{‴}}_{3}\text{}\left(x\right)=6{a}_{0}$ . Now using this in (4) results in

${P}_{3}\left({x}_{0}\right)-3{P}_{3}\left({x}_{0}+s\right)+3{P}_{3}\left({x}_{0}+2s\right)-{P}_{3}\left({x}_{0}+3s\right)=-6{a}_{0}{s}^{3}$ (5)

which exactly corresponds to the tabulated constant of third-differences. A remarkable point is that while finite-difference approximations are typically formulated for real and relatively small incremental step sizes, for the general expression no such restrictions apply: the incremental step s may be complex and arbitrarily large while the result is always exact.

Table 2. Third-order differences for a sample third-order polynomial.

Finally, a possible application of (5) or its general form for an nth-order polynomial, is its use as a recurrence formula for evaluating a given polynomial at equal intervals once the polynomial is evaluated at n distinct points. For instance for a third-order polynomial it is sufficient to know ${P}_{3}\left({x}_{0}\right)$ , ${P}_{3}\left({x}_{0}+s\right)$ , and ${P}_{3}\left({x}_{0}+2s\right)$ to obtain ${P}_{3}\left({x}_{0}+3s\right)$ from (5). Then, by setting ${x}_{0}$ to ${x}_{0}+s$ in (5), ${P}_{3}\left({x}_{0}+4s\right)$ can be obtained from the same recurrence relationship and continuing in this manner gives ${P}_{3}\left({x}_{0}+5s\right)$ , ${P}_{3}\left({x}_{0}+6s\right)$ , etc. with considerably less arithmetic operations compared to straightforward evaluation of polynomial.

2. Main Theorem and Proof

The main theorem which expresses the constancy of nth-order differences for an nth-order polynomial is stated first and then proven by the method of induction.

Theorem 1

For an nth-order polynomial ${P}_{n}\left(x\right)={a}_{0}{x}^{n}+{a}_{1}{x}^{n-1}\cdots +{a}_{n-1}x+{a}_{n}$ with ${a}_{0}\ne 0$ the following relationship holds

$\sum _{m=0}^{n}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left(x+ms\right)={\left(-1\right)}^{n}n!{a}_{0}{s}^{n}$ (6)

where $n\ge 1$ and ${a}_{j}$ 's, $x$ , $s\in ℝ$ or $ℂ$ .

Proof of Theorem 1. The base case: Setting $n=1$ in (6) results in

$\left(\begin{array}{c}1\\ 0\end{array}\right){P}_{1}\left(x\right)-\left(\begin{array}{c}1\\ 1\end{array}\right){P}_{1}\left(x+s\right)={\left(-1\right)}^{1}1!{a}_{0}{s}^{1}$ (7)

Substituting ${P}_{1}\left(x\right)={a}_{0}x+{a}_{1}$ and ${P}_{1}\left(x+s\right)={a}_{0}\left(x+s\right)+{a}_{1}$ gives

$\left({a}_{0}x+{a}_{1}\right)-\left[{a}_{0}\left(x+s\right)+{a}_{1}\right]=-{a}_{0}s$ (8)

which is correct.

The inductive step: Assuming that the statement (6) holds true for any integer n it is now proven that it also holds true for $\left(n+1\right)$ :

$\sum _{m=0}^{n+1}{\left(-1\right)}^{m}\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n+1}\left(x+ms\right)={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}$ (9)

${P}_{n+1}\left(\overline{x}\right)$ can be expressed in terms of ${P}_{n}\left(\overline{x}\right)$ as

${P}_{n+1}\left(\overline{x}\right)=\overline{x}{P}_{n}\left(\overline{x}\right)+{a}_{n+1}$ (10)

Letting $\overline{x}=x+ms$ in (10) and using it in (9) result in

$\begin{array}{l}\sum _{m=0}^{n+1}{\left(-1\right)}^{m}\left(\begin{array}{c}n+1\\ m\end{array}\right)\left[\left(x+ms\right){P}_{n}\left(x+ms\right)+{a}_{n+1}\right]\\ ={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}\end{array}$ (11)

Since ${\sum }_{m=0}^{n+1}{\left(-1\right)}^{m}\left(\begin{array}{c}n+1\\ m\end{array}\right)=0$ for any n (odd or even) the summation proportional to the constant ${a}_{n+1}$ vanishes, reducing (11) to

$\begin{array}{l}x\sum _{m=0}^{n+1}{\left(-1\right)}^{m}\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n}\left(x+ms\right)+s\sum _{m=0}^{n+1}{\left(-1\right)}^{m}m\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n}\left(x+ms\right)\\ ={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}\end{array}$ (12)

Making use of ${\sum }_{m=0}^{n+1}\left(\begin{array}{c}n+1\\ m\end{array}\right)={\sum }_{m=0}^{n}\left(\begin{array}{c}n\\ m\end{array}\right)+{\sum }_{m=1}^{n+1}\left(\begin{array}{c}n\\ m-1\end{array}\right)$ (  , p. 882) in the first summation above results in

$\begin{array}{l}x\left[\sum _{m=0}^{n}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left(x+ms\right)+\sum _{m=1}^{n+1}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m-1\end{array}\right){P}_{n}\left(x+ms\right)\right]\\ +s\sum _{m=0}^{n+1}{\left(-1\right)}^{m}m\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n}\left(x+ms\right)={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}\end{array}$ (13)

By re-defining the running index in the second summation (13) becomes

$\begin{array}{l}x\left[\sum _{m=0}^{n}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left(x+ms\right)+\sum _{m=0}^{n}{\left(-1\right)}^{m+1}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left[x+\left(m+1\right)s\right]\right]\\ +s\sum _{m=0}^{n+1}{\left(-1\right)}^{m}m\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n}\left(x+ms\right)={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}\end{array}$ (14)

Since x may be assigned to any value, substituting x + s in place of x in the base case (6) reveals that ${\sum }_{m=0}^{n}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left[x+\left(m+1\right)s\right]$ is too equal to the same quantity: ${\left(-\text{1}\right)}^{n}n!{a}_{0}{s}^{n}$ . Noting in the second summation in (14) that ${\left(-\text{1}\right)}^{m+\text{1}}=-{\left(-\text{1}\right)}^{m}$ renders the terms in square brackets zero. Thus, to complete the proof the remaining equality in the second line of (14) must be proven:

$s\sum _{m=0}^{n+1}{\left(-1\right)}^{m}m\left(\begin{array}{c}n+1\\ m\end{array}\right){P}_{n}\left(x+ms\right)={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}$ (15)

For m = 0 the first term of the summation in (15) is zero hence bringing no contribution. Therefore, we can start the summation from m = 1 without any error. Then, the summation may be expressed as

$\begin{array}{l}\sum _{m=1}^{n+1}{\left(-1\right)}^{m}m\left(\begin{array}{c}n+1\\ m\end{array}\right)\\ =\sum _{m=1}^{n+1}{\left(-1\right)}^{m}m\frac{\left(n+1\right)!}{\left(n+1-m\right)!m!}\\ =\sum _{m=1}^{n+1}{\left(-1\right)}^{m}\frac{\left(n+1\right)!}{\left(n+1-m\right)!\left(m-1\right)!}\end{array}$

$\begin{array}{l}=\sum _{m=1}^{n+1}{\left(-1\right)}^{m}\left(n+1\right)\frac{n!}{\left(n+1-m\right)!\left(m-1\right)!}\\ =\sum _{p=0}^{n}{\left(-1\right)}^{p+1}\left(n+1\right)\frac{n!}{\left(n-p\right)!p!}\\ =\left(n+1\right)\sum _{p=0}^{n}{\left(-1\right)}^{p+1}\left(np\right)\end{array}$

where an obvious change of running index m = p + 1 has been implemented in the final stage. Employing the last expression obtained above after changing p to m for the summation of (15) results in

$s\left(n+1\right)\sum _{m=0}^{n}{\left(-1\right)}^{m+1}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left[x+\left(m+1\right)s\right]={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}$ (16)

As indicated above, the main theorem may also be stated as ${\sum }_{m=0}^{n}{\left(-1\right)}^{m}\left(\begin{array}{c}n\\ m\end{array}\right){P}_{n}\left[x+\left(m+1\right)s\right]={\left(-1\right)}^{n}n!{a}_{0}{s}^{n}$ . Using this in (16) yields

$s\left(n+1\right)\left(-1\right){\left(-1\right)}^{n}n!{a}_{0}{s}^{n}={\left(-1\right)}^{n+1}\left(n+1\right)!{a}_{0}{s}^{n+1}$ (17)

which proves that the proposition holds true for (n + 1) as well. ☐

Cite this paper

Beji, S. (2018) A Fundamental Relationship of Polynomials and Its Proof. Advances in Pure Mathematics, 8, 559-563. https://doi.org/10.4236/apm.2018.86033

References

1. 1. Smith, D.E. (1959) A Source Book in Mathematics. Dover Publications, New York.

2. 2. Abramowitz, M. and Stegun, I.A. (1972) Handbook of Mathematical Functions. Dover Publications, New York.