Author Русская версия


Discriminant

Denote by \mathbb A_{} any of the sets \mathbb Z_{}, \mathbb Q, \mathbb R or \mathbb C_{}.

Discriminant1) of the polynomial

f(x)=a_{0}x^n+a_1x^{n-1}+\dots+a_n, \ (n>1, a_0\ne 0)

coincides with the resultant of this polynomial and its derivative:

{\mathcal D}(f)=\frac{(-1)^{n(n-1)/2}}{a_0}{\mathcal R}(f(x),f^{\prime}(x)) \ .
Th

Theorem. If \lambda_{1},\dots,\lambda_n denotes the zeros of f_{}(x) counted with their multiplicities, then

{\mathcal D}(f) =(-1)^{n(n-1)/2} a_0^{n-2}\prod_{j=1}^n f^{\prime}(\lambda_j)= a_0^{2n-2} \prod_{1\le j < k \le n} (\lambda_k - \lambda_j)^2 \ .

=>

{\mathcal D}(f_{}) = 0 iff f_{}(x) possesses a multiple zero.

Ex

Example. For the quadratic polynomial one has

{\mathcal D}(a_0x^2+a_1x+a_2)=\frac{1}{a_0}\left|\begin{array}{ccc} a_0&a_1&a_2\\ 0&2a_0&a_1\\ 2a_0&a_1&0 \end{array}\right|=a_1^2-4\,a_0a_2 \ .

?

Prove the following:

a) \displaystyle {\mathcal D}(x^{3}+p\,x+q)=-108\left(\frac{q^2}{4}+\frac{p^3}{27}\right);

b) {\mathcal D}(a_{0}x^3+a_1x^2+a_2x+a_3)= a_1^2a_2^2-4a_1^3a_3-4\,a_0a_2^3+18\,a_0a_1a_2a_3-27\,a_0^2a_3^2;

c) \displaystyle {\mathcal D}(x^{4}+px+q)=6912\left(\frac{q^3}{27}-\frac{p^4}{256}\right).

Ex

Example.

{\mathcal D} (a_0x^4+a_1x^3+a_2x^2+a_3x+a_4)=4I_2^3-27I_3^2 \ .

Here

I_2=4a_0a_4-a_1a_3 +\frac{1}{3}a_2^2 \ ,
I_3=-a_0a_3^2-a_1^2a_4+\frac{8}{3}a_0a_2a_4+ \frac{1}{3}a_1a_2a_3-\frac{2}{27}a_2^3

stand for the invariants of the quartic polynomial.

Properties

1.

{\mathcal D}(A\cdot f(x)) = A^{2n-2} {\mathcal D}(f) ;

here A_{} stands for a constant and \deg f = n_{}>1.

2.

{\mathcal D}(f(x)\cdot g(x))={\mathcal D}(f){\mathcal D}(g)\left[{\mathcal R}(f, g) \right]^2 \ ;

here {\mathcal R}(f, g_{}) denotes the resultant of the polynomials f(x)_{} and g_{}(x); and it is assumed that \deg f> 1_{} and \deg g> 1_{}.

3.

{\mathcal D}(f(x)(x-a))={\mathcal D}(f)\left[f(a) \right]^2 \ .

4. If a_{0}\ne 0, a_n \ne 0 then

{\mathcal D}(a_0x^n+a_1x^{n-1}+\dots+a_n)={\mathcal D}(a_0+a_1x+\dots+a_{n-1}x^{n-1}+a_nx^n)

5.

{\mathcal D}(f(g(x))=\left[{\mathcal D}(f) \right]^m \prod_{j=1}^n {\mathcal D}(g(x)-\lambda_j) \ ;

here n=\deg f, m=\deg g, \lambda_{1},\dots,\lambda_n stand for the zeros of f_{}(x), and the leading coefficients of f_{}(x) and g_{}(x) are assumed to be equal 1_{}.

?

Express a) {\mathcal D}(f_{}(A\,x+B)); b) {\mathcal D}(f(x^{m})); c) {\mathcal D}( {\tilde f}_{}(x)) via {\mathcal D}(f)_{}. Here

{\tilde f}(x)\equiv (C\,x+D)^n f\left(\frac{A\,x+B}{C\,x+D} \right) \ .

Determinantal representation

According to the definition, the discriminant can be represented as the (2n-1)_{}-th odrer determinant:

{\mathcal D}(f)
=\frac{1}{a_0} \left|\begin{array}{cccccccccc} a_0&a_1&a_2&\ldots&\ldots&a_n&0&\dots &0 &0\\ 0&a_0&a_1&\ldots&\ldots&a_{n-1}&a_n&\dots&0 &0\\ \vdots&&\ddots&&&&&&\ddots\\ 0&0&\ldots&a_0&a_1&\ldots & & \ldots &a_{n-1} &a_n\\ 0&0&\ldots&&na_0&(n-1)a_1&\ldots& \ldots &2a_{n-2}&a_{n-1}\\ 0&0&\ldots&na_0&(n-1)a_1&\ldots &&\ldots &a_{n-1}&0\\ \vdots&&&\ldots&&&& &&\vdots\\ 0&na_0&\ldots&\ldots&&a_{n-1}&\ldots&&\ldots&0\\ na_0&\ldots&\ldots&&a_{n-1}&0&\ldots&&&0 \end{array}\right| \begin{array}{l} \left.\begin{array}{l} \\ \\ \\ \\ \end{array}\right\} n-1 \\ \left. \begin{array}{l} \\ \\ \\ \\ \\ \end{array}\right\} n \end{array}

Using the elementary transformation of its rows one can reduce it to the (2n-2)_{}-th order determinant:

{\mathcal D}(f)
=\frac{1}{n^{n-2}} \left|\begin{array}{cccccccc} a_1&2a_2&3a_3&\ldots&na_n&0&\ldots&0\\ 0& a_1&2a_2&3a_3&\ldots&na_n&\ldots&0\\ & & &\ldots&\ldots& & & \\ 0&\ldots&0&a_1&2a_2&3a_3&\ldots&na_n\\ 0&\ldots&0&na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}\\ 0&\ldots&na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}&0\\ & & &\ldots&\ldots& & & \\ na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}&0 &\ldots&0 \end{array}\right|

The last determinant can be obtained from an alternative definition of the discriminant. Consider a homogeneous bivariate polynomial (form) in x_{} and y_{}:

F(x,y)= a_0x^n + a_1x^{n-1}y +a_2x^{n-2}y^2+\dots+a_{n-1}xy^{n-1}+a_ny^n .

Compute its partial derivatives

\phi(x,y)=\partial F / \partial x ,\ \psi(x,y)=\partial F / \partial y \ .

The discriminant of F(x,y)_{} is taken to be equal the resultant of \psi(x,1)_{} and \phi(x,1)_{}.

Discriminant as a function in the polynomial coefficients

By definition, the discriminant is a homogeneous polynomial over \mathbb Z_{} in the coefficients a_{0},\dots,a_n of the polynomial f_{}(x):

{\mathcal D}(a_0x^n+\dots+a_n)\equiv D(a_0,\dots,a_n) \in {\mathbb Z}[a_0,\dots,a_n] \ ;

one has \deg D= 2n-2_{} and this polynomial in a_{0},\dots,a_n contains the term (-1)^{n(n-1)/2}n^n a_{0}^{n-1}a_n^{n-1}.

Th

Theorem. If f(x)_{} possesses a unique multiple zero \lambda_{} and its multiplicity equals 2_{} then

1 : \lambda : \lambda^2 : \dots = \frac{\partial D}{\partial a_n} : \frac{\partial D}{\partial a_{n-1}} : \frac{\partial D}{\partial a_{n-2}} : \dots

Ex

Example. Deduce the general formula for the double zero for f(x)=a_{0}x^4+a_1x^3+a_2x^2+a_3x+a_4 under the assumption of its uniqueness.

Solution. Using the formula for the discriminant of the quartic polynomial ( v. ABOVE ), one gets:

\lambda = -\frac{2\,a_1I_2^2+9\,(-2\,a_0a_3+1/3\, a_1a_2)I_3}{8\,a_0I_2^2-9\,I_3(-a_1^2+8/3\,a_0a_2)} \ .

For f(x)= x^{4}-5x^3+4x^2+3x+9 one has I_{2}=169/3, I_3=-4394/27, {\mathcal D}(f)=0_{} and formula from the above theorem yields \lambda_{} = 3.

§

The discriminant is the invariant of the polynomial f(x)_{} (strictly speaking, the invariant of the homogeneous polynomial (form) y^{n}f(x/y)).

Subdiscriminants

The determinant obtained from

\left|\begin{array}{cccccccc} a_1&2a_2&3a_3&\ldots&na_n&0&\ldots&0\\ 0& a_1&2a_2&3a_3&\ldots&na_n&\ldots&0\\ & & &\ldots&\ldots& & & \\ 0&\ldots&0&a_1&2a_2&3a_3&\ldots&na_n\\ 0&\ldots&0&na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}\\ 0&\ldots&na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}&0\\ & & &\ldots&\ldots& & & \\ na_0&(n-1)a_1&(n-2)a_2&\ldots&a_{n-1}&0 &\ldots&0 \end{array}\right|

by deleting its k_{} first rows and its k_{} last rows, k_{} first columns and k_{} last columns we will call the k_{}-th subdiscriminant of the discriminant \mathcal D (f_{}) and will denote by {\mathcal D}_{k}. For the convenience of presentation of some results we will take the zero subdiscriminant to be equal the determinant itself, i.e.

{\mathcal D}_0=n^{n-2}{\mathcal D} (f) \quad and \quad {\mathcal D}_{n-1} = 1 \ .
§

In the classical and contemporary sourses I did not find the common name for this object. In [2] a similar determinant is called aperiodical or bigradient.

Th

Theorem. The polynomial f_{}(x) possesses exactly d_{} common zeros with its derivative (or, more strictly, \deg( \operatorname{GCD} (f,f^{\prime}))=d_{}) iff

\underbrace{{\mathcal D}_0=0, {\mathcal D}_1=0,\dots, {\mathcal D}_{d-1}=0}_d,{\mathcal D}_d\ne 0 \ .

=>

If f(x)_{} possesses a unique multiple zero \lambda_{} and its multiplicity equals 2_{} then this zero can be expessed as a rational function of the coefficients of the polynomial

\lambda=-\frac{\tilde {\mathcal D}_1}{{\mathcal D}_1} \ ,

here \tilde {\mathcal D}_{1} stands for the determinant obtained from {\mathcal D}_{0} by deleting its first and its last row, and its first and its last-but-one column (thus \tilde {\mathcal D}_{1} differs from {\mathcal D}_{1} only in its last column).

Ex

Example. Find all the values of the parameter \alpha_{} under which the polynomial f(x)=2\,x^5+3\,x^4+4\,x^3+x^{2}-\alpha possesses a unique multiple zero; compute this zero.

Solution. Compute the determinant {\mathcal D}_{0}:

{\mathcal D}_0=\left|\begin{array}{rrrrrrrr} 3 & 8 & 3 & 0 & -5 \alpha & 0 & 0 & 0 \\ 0 & 3 & 8 & 3 & 0 & -5 \alpha & 0 & 0 \\ 0 & 0 & 3 & 8 & 3 & 0 & -5 \alpha & 0 \\ 0 & 0 & 0 & 3 & 8 & 3 & 0 & -5 \alpha \\ 0 & 0 & 0 & 10 & 12 & 12 & 2 & 0 \\ 0 & 0 & 10 & 12 & 12 & 2 & 0 & 0 \\ 0 & 10 & 12 & 12 & 2 & 0 & 0 & 0 \\ 10 & 12 & 12 & 2 & 0 & 0 & 0 & 0 \end{array} \right| =50000\,\alpha^4-100608\,\alpha^3+51216\,\alpha^2-608\,\alpha \ .

This polynomial in \alpha_{} vanishes iff \alpha_{} \in \{0,1, 38/3125 \}. Deleting from {\mathcal D}_{0} its boundary rows and columns one gets

{\mathcal D}_1=\left|\begin{array}{rrrrrr} 3 & 8 & 3 & 0 & -5 \alpha & 0 \\ 0 & 3 & 8 & 3 & 0 & -5 \alpha \\ 0 & 0 & 3 & 8 & 3 & 0 \\ 0 & 0 & 10 & 12 & 12 & 2 \\ 0 & 10 & 12 & 12 & 2 & 0 \\ 10 & 12 & 12 & 2 & 0 & 0 \end{array} \right| = 110000\,\alpha^2-102400\,\alpha-7600 \ .

Substitution for \alpha_{} the values discovered above yields:

{\mathcal D}_1 \ne 0 \ for \ \alpha \in \{0, 38/3125 \}; \ {\mathcal D}_1 = 0 \ for \ \alpha =1 \ .

Consequently f_{}(x) possesses a unique double zero iff \alpha_{} \in \{0, 38/3125 \}, while for \alpha_{} =1 it has either several multiple zeros or a zero of multiplicity higher than 2_{}. For the evaluation of the multiple zero, compute the determinant \tilde {\mathcal D}_{1}:

\tilde {\mathcal D}_1=\left|\begin{array}{rrrrrr} 3 & 8 & 3 & 0 & -5 \alpha & 0 \\ 0 & 3 & 8 & 3 & 0 & 0 \\ 0 & 0 & 3 & 8 & 3 & -5 \alpha \\ 0 & 0 & 10 & 12 & 12 & 0 \\ 0 & 10 & 12 & 12 & 2 & 0 \\ 10 & 12 & 12 & 2 & 0 & 0 \end{array} \right| = 147000\alpha^2-147000\alpha \ ,

and substitute into the formula

\lambda=- \tilde {\mathcal D}_1 / {\mathcal D}_1

the obtained values for \alpha_{}:

\lambda = 0 \ \ for \ \alpha =0 ; \ \lambda = -1/5 \ \ for \ \alpha = 38/3125 \ .

Representation of the discriminant via the Hankel determinant

For the polynomial f_{}(x) its k_{}-th Newton sum is defined as the sum of k_{}-th powers of its zeros

s_k=\sum_{j=1}^n\lambda_j^k \ .

Newton sums can be expressed as rational functions of the coefficients of f_{}(x) with the aid of the following recursive Newton formulas:

s_0=n,\ s_1=-a_1/a_0,
s_k=\left\{\begin{array}{lr} -(a_1s_{k-1}+a_2s_{k-2}+\dots+a_{k-1}s_1+a_kk)/a_0, & if \ k\le n ;\\ -(a_1s_{k-1}+a_2s_{k-2}+\dots+a_ns_{k-n})/a_0, & if \ k > n \end{array} \right.

Explicit expressions for the Newton sums via a_{0}, \dots, a_n are given by the Waring formula.

Compute the Newton sums s_{0},s_1,\dots,s_{2n-2} for f_{}(x) and compose the Hankel matrix

S=\left[s_{j+k} \right]_{j,k=0}^{n-1} = \left[\begin{array}{llllll} s_0 &s_1&s_2&\dots&s_{n-2}& s_{n-1}\\ s_1 &s_2&s_3&\dots&s_{n-1}& s_{n}\\ s_2 &s_3&s_4&\dots&s_{n}& s_{n+1}\\ \dots& & &&& \dots\\ s_{n-1} &s_n&s_{n+1}&\dots &s_{2n-3}&s_{2n-2} \end{array}\right]_{n\times n} \ .

Denote by S_{1},\dots, S_n its leading principal minors.

Th

Theorem. The followng formula connects the minors of the matrix S_{} with the subdiscriminants of the polynomial f(x)_{}:

{\mathcal D}_{k}=n^{n-k-2}a_0^{2(n-k-1)}S_{n-k} \ .

In particular,

{\mathcal D}(f)=a_0^{2n-2}\det S \ .

Proof follows from the repesentation for \mathcal R(f_{},f^{\prime}) in the Kronecker form.

Discriminant ifluence on the polynomial zeros

Closeness of zeros

!

Discriminant is responsible for the closeness of zeros of f_{}(x): the more they are congested the smaller it is and vice versa. The discriminant value can be used for the estimation of the distance between the zeros.

Th

Theorem. The following estimations are valid

\frac{\sqrt{\left|{\mathcal D}(f) \right|}}{(2\rho)^{n(n-1)/2-1}|a_0|^{n-1}} \le \min_{j,k\in \{1,\dots,n \} \atop j\ne k} \left|\lambda_j - \lambda_k \right| \le \frac{\left|{\mathcal D}(f)\right|^{1/[n(n-1)]}}{|a_0|^{2/n}} \quad for \quad \rho = \max_{j \in \{1,\dots,n \}} |a_j| \ .

Solution of algebraic equations in radicals

In the middle school course in Algebra the following formula is known for the expression of the zeros of a quadric polynomial f(x)=a_{0}x^2+a_1x+a_2=0 as functions of its coefficients:

\lambda_{1,2}=\frac{-a_1\pm\sqrt{{\mathcal D}(f)}}{2a_0} \ .

Here {\mathcal D}(f)=a_{1}^2-4a_0a_2 is the discriminant of the quadric polynomial. While initially introduced for the quadric with real coefficients and for the case {\mathcal D}(f) \ge 0_{}, the formula remains also valid for the case of imaginary coefficients.

For the cubic equation f(x)=a_{0}x^3+a_1x^2+a_2x+a_3=0, there also exists a formula for the explicit representation of polynomial zeros via coefficients — namely, Cardano's formula. In a particular case of a polynomial f(x)=x^{3}+px+q, this formula is

\lambda = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+ \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} \ .

There is a special agreement [3] for combining the values of the cubic roots in the above sum (generally the radicans are imaginary numbers even for the case of real p_{} and q_{}). One may notice that the radicand of the square root coincides with the discriminant:

\frac{q^2}{4}+\frac{p^3}{27} = -108 {\mathcal D}(x^3+px+q) \ .

Reality of zeros

Polynomial f(x)=a_0x^{n}+a_1x^{n-1}+\dots+a_{n} with real coefficients a_{0}\ne 0, a_1,\dots a_n may possess both real and non-real (imaginary) zeros \lambda_{1},\dots,\lambda_n. Although these zeros cannot be expressed in «good» functions from the polynomial coefficients, the conditions for the existence of the prescribed number of real zeros can be expressed in terms of polynomial inequalities imposed on a_{0}, a_1,\dots a_n. The «most essential» from these inequalities is the one imposed on the sign of the discriminant {\mathcal D}(f_{}).

Ex

Example. The necessary and sufficient condition for the reality of all the zeros of the polynomial

a) f(x)=a_{0}x^2+a_1x+a_{2} is {\mathcal D}(f)=a_1^2-4a_0a_{2} \ge 0;

b) f(x)=x^{3}+p\,x+q_{} is {\mathcal D}(f) = -4\,p^3-27\,q^{2} \ge 0 \ \iff \ \frac{q^2}{4}+\frac{p^3} {27} \le 0.

For the polynomial of the degree n \ge 4_{}, the nonnegativity of discriminant is not the necessary and sufficient condition for the reality of all the polynomial zeros.

Th

Theorem. For the reality of all the zeros of f(x)_{} it is necessary that {\mathcal D}(f) \ge 0_{}.

Proof evidently follows from the representation of \mathcal D (f)_{} via the zeros of f(x)_{}.

The more general result connects the number of distinct zeros of f(x)_{} with the signs of subdiscriminants.

Th

Theorem. Let {\mathcal D}(f) \ne 0_{}. If the sequence of subdiscriminants

{\mathcal D}_0,{\mathcal D}_{1},\dots,{\mathcal D}_{n-1}=1

does not contain two consecutive zeros, then all the zeros of the polynomial f(x)_{} are distinct and the number of real zeros equals

{\mathcal P}(1,{\mathcal D}_{n-1},\dots,{\mathcal D}_0) - {\mathcal V}(1,{\mathcal D}_{n-1},\dots,{\mathcal D}_0) \ .

Here {\mathcal P} and {\mathcal V} stand for correspondingly the number of permanences and the number of variations of signs in the considered sequence.

The previous theorem is just a reformulation of the following result based on the representaion of the discriminant as the determinant of the Hankel matrix S_{}.

Th

Theorem [Jacobi]. The number of distinct zeros of a polynomial f(x)_{} equals the rank, while the number of distinct real zeros of f(x)_{} equals the signature of the matrix S_{}.

The constructive computation of the rank and the signature of a symmetric matrix S_{} is possible via evaluation of the signs of its leading principal minors S_{1},\dots,S_n.

=>

Let

S_n=0,\dots,S_{{\mathfrak r}+1}=0,S_{\mathfrak r}\ne 0, \dots, S_1 \ne 0 \ .

Then \operatorname{rank} (S)={\mathfrak r}_{} and the number of distinct real zeros of f(x)_{} equals

{\mathcal P}(1,S_1,\dots,S_{\mathfrak r}) -{\mathcal V}(1,S_1,\dots,S_{\mathfrak r}) \ .

=>

For the reality of all the zeros of f(x) \in \mathbb R[x]_{} it is necessary and sufficient that all the leading principal minors of the matrix S_{} be positive:

S_1>0,\dots,S_n > 0 \ .

Ex

Example. Find the number of real zeros of x^{5}-3\,x^3-x-1.

Solution. Newton sums:

\{s_j \}_{j=0}^8=\{5,\, 0,\, 6,\, 0,\, 22,\, 5,\, 72,\, 21,\,238 \} \ .

Compose the Hankel matrix:

S= \left[ \begin{array}{rrrrr} 5 & 0 & 6 & 0 & 22 \\ 0 & 6 & 0 & 22 & 5 \\ 6 & 0 & 22 & 5 & 72 \\ 0 & 22 & 5 & 72 & 21 \\ 22 & 5 & 72 & 21 & 238 \end{array} \right]

and compute its leading principal minors:

S_1=5,\, S_2=30,\, S_3=444,\, S_4=-4598,\, S_5=-56\,123 \ .

Since S_{5}\ne 0, all the zeros of f(x)_{} are distinct.

{\mathcal P}(1,\,5,\,30,\,444,\,-4598,\,-56\,123)=4,\ {\mathcal V}(1,\,5,\,30,\,444,\,-4598,\,-56\,123)=1 \ .

Answer. Three real zeros.

Ex

Example. Find the number of real zeros of the polynomial 2\,x^{5}+3\,x^4+4\,x^3+x^{2}-\alpha in dependency of the parameter \alpha_{} \in \mathbb R.

Solution. The discriminant of the polynomial and its first subdiscriminant have been already computed above:

{\mathcal D}_0= 16\alpha(3125\,\alpha-38)(\alpha-1)^2,\ {\mathcal D}_1=400\,(275\,\alpha+19)(\alpha-1) \ .
{\mathcal D}_2=\left|\begin{array}{rrrr} 3 & 8 & 3 & 0 \\ 0 & 3 & 8 & 3 \\ 0 & 10 & 12 & 12 \\ 10 & 12 & 12 & 2 \end{array} \right| = -2940 \ , {\mathcal D}_3=\left|\begin{array}{rr} 3 & 8 \\ 10 & 12 \end{array} \right|=-44 \ .

Analyze the signs of {\mathcal D}_{0} and {\mathcal D}_{1} under the variation of parameter \alpha_{}; the critical values for the latter are those annihilating at least one of subdiscriminants:

\begin{array}{c|c|c|c} & {\mathcal D}_0 & {\mathcal D}_1 & \begin{array}{l} {\mathcal P}(1,1,{\mathcal D}_3,{\mathcal D}_2, {\mathcal D}_1,{\mathcal D}_0) - \\ {\mathcal V}(1,1,{\mathcal D}_3,{\mathcal D}_2, {\mathcal D}_1,{\mathcal D}_0) \end{array} \\ \hline \alpha>1 & >0 & <0 & 3-2 \\ \hline \frac{38}{3125}<\alpha<1 & >0 & <0 & 3-2 \\ \hline 0<\alpha<\frac{38}{3125} & <0 & <0 & 4-1 \\ \hline -\frac{19}{275} < \alpha < 0 & >0 & <0 & 3-2 \\ \hline \alpha < -\frac{19}{275} & >0 & >0 & 3-2 \end{array}

Answer. Polynomial possesses one real zero if \alpha<0_{} and if \alpha_{} > 38/3125; polynomial possesses three real zero if \alpha_{} \in [0,38/3125].

§

Note, that the conditions in the answer to the previous example have the following structure: the end point for the intervals for the parameter values, providing the prescribed number for the real zeros for the polynomial f(x,\alpha), turn out to be the zeros of the discriminant \mathcal D_{x} (f(x,\alpha)). This condition is a demonstration of a more general principle: among all the inequalities imposed on the subdiscriminants, the most crucial one is that on the sign of the discriminant itself.

Th

Theorem. In the n_{}-dimensional parameter space (a_{1},\dots, a_n), the domains corresponding to the polynomials f(x)=x^n+a_1x^{n-1}+\dots+a_{n} with equal number of real zeros, are separated by the discriminant manifold, i.e. the surface defined by

D(1,a_1,\dots,a_n) = \mathcal D_x(x^n+a_1x^{n-1}+\dots + a_n)=0 \ .

Ex

Example. For the polynomial x^{3}+px+q the discriminant manifold becomes a curve in the (p,q)-plane: 4\,p^3+27\,q^2 =0; it separates the domain for the parameter values corresponding to polynomials with three real zeros (обозначена in yellow) frome the domain of values defining the polynomials with precisely one real zero (in blue).

§

I failed to establish the authorship for the previous theorem. In [3] at p. 252 a reference is presented to the work by Brill of 1877, as well as to the Kronecker works on Charakteristik Theorie; this object was referred to as Diskriminantenfläche.

Applications

Critical values of a polynomial

Problem. For the polynomial f(x)=a_0x^{n}+a_1x^{n-1}+\dots+a_n with real coefficients a_0\ne 0, a_{1},\dots a_n, evaluate its critical values. In particular, for an even n_{} and for a_{0}<0, find the absolute maximum

\max_{x\in {\mathbb R}}f(x) \ .
Th

Theorem. The critical values of the polynomial f_{}(x) are the real zeros of the polynomial

{\mathcal F}(z)={\mathcal D}_x(f(x)-z) \ .

Here the discriminant is treated with respect to the variable x_{}, while z_{} is tackled as a numerical parameter.

=>

For an even n_{} and a_{0}<0, the maximal value of f_{}(x) coincides with the maximal real zero of the polynomial {\mathcal F}(z_{}) provided that this zero is not a multipe one.

Ex

Example. For f(x)=x^{4}+p\,x+q one has

{\mathcal F}(z)=256\ z^3-768q\ z^2+768q^2\ z+(27 p^4-256 q^3) \ .

Ex

Example. Find the maximum of the polynomial f(x)=-x^{6}+12\,x^2+12\,x+2.

Solution. One has

\mathcal F(z)= 46656(-z^5+10\,z^4+472\,z^3+16208\,z^2-16272\,z-32800)\ ,

maximal real zero of the last polynomial \approx 35.6321_{} coincides with the value of f_{}(x) at a zero of its derivative: \mu_{} \approx 1.51851.

Resume. The standard algorithm for finding the maximal value of a polynomial f_{}(x) consists in first finding all the real zeros of its derivative f{'}(x), then their substitution into f_{}(x) and ordering the obtained values to find the maximal one. Instead of this approach, we suggest to evaluate only one zero of the constructed polynomail \mathcal F_{}(z), namely the maximal one. The last problem, i.e. the search for a particular zero of a polynomial, is oftenly easier in solving. Thus, for instance, one may try to find the maximal zero for \mathcal F_{}(z) via the Bernoulli method. If its procedure converges to a positive value, then this value is2) the maximum of f_{}(x).

?

Construct the polynomial {\mathcal F}_{}(z) for

(a) f(x)=-x^{4}-4x^3+2x^2+12x,

(b) f(x)=-x^{4}+4x^3-4x^2,

( c) f(x)=-x^{6}-10x^3+12 x

and establish that \max f_{}(x) is attained at two stationary points of f_{}(x).

The importance of the simplicity condition for the maximal zero of {\mathcal F}(z) is clarified by the following example:

Ex

Example. [5]. For f(x)=-x^{6}-135x^2-324x one gets:

{\mathcal F}(z)= 46656(z^3+1080\,z^2+1603800\,z-354294000)(z-540)^2

possesses the maximal zero z_{}=540; however, the latter corresponds to the nonreal zeros \mu_{1,2}=(-3\pm \mathbf i\sqrt {15})/2_{} of the derivative f{'}(x)=-6(x^{5}+45\,x+54). Maximum for f_{}(x) is attained at the zero \mu_{3}=1-\sqrt[3]{10} and it equals 90(-4+5 \sqrt[3]{10}-\sqrt[3]{100}_{}) \approx 191.7526154.

Critical values of an implicit function

Let us generalize the problem from the previous section:

Problem. Find critical values of the function y=f_{}(x) defined implicitly by the algebraic equation \Phi_{}(x,y)=0. Here \Phi_{}(x,y) is a polynomial in x_{} and y_{} with real coefficients.

Th

Theorem. Critical values of the implicit function are among the real zeros of the polynomial

{\mathcal F}(y)={\mathcal D}_x(\Phi(x,y)) \ .

Here the discriminant is considered with respect to x_{}, while y_{} is assumed to be a numerical parameter.

Proof. The necessary condition for the existence of a stationary point for y=f_{}(x) at x=x_{0} consists in vanishing the derivative f^{\prime}(x_{}). Differentiating the identity \Phi(x,f(x)) \equiv 0 as to x_{}

\frac{\partial \Phi}{\partial x} + \frac{\partial \Phi}{\partial y} f^{\prime}(x) \equiv 0

we conclude that at x=x_0, y_0=f(x_0) the conditions have to be fulfilled:

\Phi(x,y)=0,\ \partial \Phi / \partial x = 0 \ .

Ex

Example. Find the mimimal value of the implicit function given by

-x^4-1/2 y^4+4\,x^2+3\,xy+4\,y=0 for \ y < 0, x\in ]-2,2[ \ .

Solution. We ignore here the questions of existence and representations for this function; but apply the theorem as it is:

{\mathcal F}(y)={\mathcal D}_x(-x^4+4\,x^2+3\,xy-1/2 y^4+4\,y) =
=32\,y^{12}-768\,y^9-512\,y^8+3552\,y^6+8192\,y^5-139\,y^4+4352\,y^3-30464\,y^2-16384\,y \ .

The real zeros of this polynomial are:

y_1 \approx -1.795011, \ y_2 \approx -0.490598, \ y_3= 0,\ y_4 \approx 2.741399 \ .

The minimal zero is y_{1}; with its value, one can restore the corresponding x_{1} — as a multiple zero for \Phi (x,y_1). Its value x_1 \approx -1.674506 lies within the demanded interval ]-2,2[, and one can additionally verify that for x = \pm 2 the real zeros of \Phi(x_{},y) are greater than y_{1}.

Answer. \min \approx -1.795011.

Distance evaluation

Th

Theorem[6,7]. The square of the distance to the quadric {\mathbb R}^{n} given by the equation

X^{\top}AX+2\,B^{\top}X-1=0 \ , (A=A^{\top})

from the point X_{0} \in {\mathbb R}^n not lying in the quadric (i.e. X_{0}^{\top}AX_0+2 B^{\top}X_0-1\ne 0) equals the minimal positive zero of the distance equation

{\mathcal F}(z)={\mathcal D}_{\mu} (\Phi(\mu,z))=0 \ ,

provided that the mentioned zero is not a multiple one. Here

\Phi(\mu,z)=\det \left( \left[ \begin{array}{cc} A & B \\ B^{\top} & -1 \end{array} \right] + \mu \left[ \begin{array}{cc} -I & X_0 \\ X_0^{\top} & z-X_0^{\top}X_0 \end{array} \right] \right)

while {\mathcal D}_{} stands for the discriminant of the polynomial (treated w.r.t. \mu), and I_{} is the identity matrix of the order n_{}.

=>

The square of the distance from {\mathbb O} \in {\mathbb R}^{n} to the quadric in {\mathbb R}^{n} given by the equation

X^{\top}AX+2B^{\top}X-1=0 \ ,

equals the minimal positive zero of the distance equation

{\mathcal F}(z)={\mathcal D}_{\mu} \left( f(\mu)(\mu z-1)-B^{\top}\operatorname{adj}(A,\mu)B \right)=0\ ,

provided that this zero is simple. Here f(\mu)=\det (A-\mu I) is the characteristic polynomial of the matrix A_{} while \operatorname{adj}(A,\mu) is the adjoint matrix for A-\mu I_{}.

=>

For the particular case B={\mathbb O}_{} (i.e. the quadric centered to the origin), one has:

{\mathcal F}(z)=\left[z^nf(1/z) \right]^2{\mathcal D}_{\mu}(f(\mu)) \ ,

and the distance from the origin to the quadric equals 1_{}/\sqrt{\lambda_{\max}}, where \lambda_{\max}^{} stands for the maximal eigenvalue of the matrix A_{}.

§

These and other applications of the discriminant to the problems of distance evaluation HERE

Equidistant curve

Consider a planar smooth curve \mathbf K_{}. In every its point A_{} raise a normal and take the points in it lying at the distance h_{} from A_{}. These points constitute two curves with each of them called equidistant curve for \mathbf K_{}; we will denote them {\mathbf K}_{+h}^{} and {\mathbf K}_{-h}^{}.

Th

Theorem. Equidistant curve for y=f_{}(x) where f_{}(x) is a polynomial with real coefficients, is given by the equation

\Phi(x,y)=0 \ with \ \Phi(x,y)= {\mathcal D}_{X}\left(\left[X-x \right]^2 + \left[f(X)-y \right]^2-h^2 \right) \ .

Here the discriminant is taken w.r.t. X_{}, while other variables are treated as parameters.

Ex

Example. Find equidistant curve for the parabola y=x^{2}.

Solution. Compute the discriminant, skip a numerical factor, and order the resulting polynomial in powers of h_{}:

\begin{array}{rcl} \Phi(x,y)&=&{\mathcal D}_{X}\left(X^4+(1-2y)X^2-2\ xX+x^2+y^2-h^2\right)= \\ &=&(16 y^2+16 x^2-8 y+1)(y-x^2)^2 + \\ &+&\left[8(-4y^2-8yx^2-y+1-8 x^4)(y-x^2)- (4 x^2+1)^3 \right]h^2+ \\ &+&8(2y^2+4 y+6 x^2-1)h^4-16 h^6 \ . \end{array}

Equation \Phi(x,y)=0 provides )implicitly) the equidistant curves {\mathbf K}_{+h}^{} and {\mathbf K}_{-h}^{} for the parabola y=x^{2}. In the figure, they ar shown for the choice h=1_{}

Envelope

Consider now th efamily of planar curves \left\{ \mathbf K(\lambda) \right\}_{} depending on the parameter \lambda_{}; the latter takes values from the interval [a,b] \in \mathbb R_{}. If there exists such a curve \mathbf L_{} which is tangent in any its point to some curve of the given family but does not coincide with any of these curves for some its segment, then this curve \mathbf L_{} is called the envelope for the family \left\{ \mathbf K(\lambda) \right\}_{}.

Let the family \left\{ \mathbf K(\lambda) \right\}_{} is given implicitly by the equation

\Psi(x,y,\lambda)=0 \ ,

where \Psi(x,y,\lambda)_{} is a continuously differentiable function in its variables. The loci of the points satisfying the conditions

\Psi(x,y,\lambda)=0,\ \frac{\partial \Psi(x,y,\lambda)}{\partial \lambda} = 0

is called the discriminant curve of the family \left\{ \mathbf K(\lambda) \right\}_{}.

Th

Theorem. The discriminant curve of the family contains the envelope of the family and, probably, the set of critical points, i.e. the points which are reals solutions for the system

\frac{\partial \Psi(x,y,\lambda)}{\partial x} = 0,\ \frac{\partial \Psi(x,y,\lambda)}{\partial y} = 0 \ .

Ex

Example. Find the envelope for the family of ellipses

\frac{x^2}{a^2}+ \frac{y^2}{(1-a)^2}=1 for \ a \in ]0,1[ \ .

Solution. Here the equation for the discriminant curve is obtained as a result

of elimination of the parameter a_{} from the system

\frac{x^2}{a^2}+ \frac{y^2}{(1-a)^2}=1,\ \frac{x^2}{a^3}-\frac{y^2}{(1-a)^3} = 0 \ .

Resolve the second equation w.r.t. a_{}:

a=\frac{x^{2/3}}{x^{2/3}+y^{2/3}} \ ,

(here the restriction 0< a_{} < 1 is essential) and substitute the result into the first equation:

x^{2/3}+y^{2/3}=1 \ .

The obtained curve is known as the astroid.

Th

Theorem. If \Psi(x,y,\lambda)_ is a polynomial in \lambda_{} then the discriminant curve is given by the equation

{\mathcal D}_{\lambda} (\Psi(x,y,\lambda)) = 0 \ .

Here the discriminant is taken w.r.t. \lambda_{}, while other variables are treated as parameters.

Ex

Example. On rewriting the equation for the family of ellipses from the previous example in the form

(1-a)^2x^2+a^2y^2-a^2(1-a)^2=0 \quad \iff \quad -a^4+2\,a^3+(x^2+y^2-1)a^2-2\,x^2a+x^2 =0 \ ,

one can obtain a represenetation for the discriminant curve with the aid of the theorem:

-16\,x^2y^2((x^2+y^2-1)^3+27\,x^2y^2)=0 \ .

The cases x_{}=0 or y_{}=0 correspond to the values of the parameter a_{} lying at the endpoints of the given interval. The remaind factor from the last equality defined an astroid; this fact can be confirmed by the substitution u_{}=x^{2/3}, v_{}=y^{2/3}.

Ex

Example. It can be easily verified that the equidistant curves for the curve \mathbf K_{} intriduced in the previous section, are the envelopes for the family of circumferences of the radius h_{} centered at the given curve. One gets for the parabola Y=X^{2}:

\Psi(x,y,X)\equiv (x-X)^2+(y-X^2)^2 - h^2

with X_{} treated as the parameter generating the family:

References

[1]. Kalinina E.A., Uteshev A.Yu. Elimination Theory (in Russian) SPb, Nii khimii, 2002

[2]. Jury E.I. Inners and Stability of Dynamic Systems. J.Wiley & Sons, New York, NY, 1974.

[3]. Uspensky J.V. Theory of Equations. New York. McGraw-Hill. 1948

[4]. Encyklopädie der Mathematischen Wissenschaften mit Einschluss ihrer Anwendungen. Bd. I. Arithmetik und Algebra. Ed. Meyer W.F. 1898-1904. Leipzig, Teubner

[5]. Uteshev A.Yu., Cherkasov T.M. The search for the maximum of a polynomial. J. Symbolic Computation. 1998. Vol. 25, № 5. P. 587-618. Text in djvu HERE

[6]. Uteshev A.Yu., Yashina M.V. Distance Computation from an Ellipsoid to a Linear or a Quadric Surface in {\mathbb R}^n. Lect.Notes Comput. Sci. 2007. V.4770. P.392-401. Text in djvu HERE

[7]. Uteshev A.Yu., Yashina M.V. Metric Problems for Quadrics in Multidimensional Space. J.Symbolic Computation, 2015, Vol. 68, Part I, P. 287-315.

1) Discriminant (Lat.) — differentiating, separating, distinguishing; the word discrimination is originated from discriminatio — distinction.
2) Almoust certain, i.e. with probability 1_{}

2016/01/10 21:12 редактировал au