ON ALGEBRAIC EQUATIONS ALL HAVING THEIR REAL ROOTS

by
Tiberius Popoviciu
Former student of the Higher Normal School

Received on September 1, 1933.

CHAPTER I.

On equations where the first three coefficients are given.

1.

Consider the family of equations of degree $n$

f(x)=x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots+a_{n}=0

(1)

having all their real roots and for which $a_{1},a_{2}$ have: given values. It is easy to see that the roots, and therefore also the other coefficients, remain bounded.

We have

\Delta=(n-1)a_{1}^{2}-2na_{2}\geq 0,

Equality is only possible if all the roots of (1) are equal.
Equations (1), taking the values $a_{1},a_{2}$ and having at most two distinct roots are:

\begin{gathered}f_{k}(x)=\left(x+\frac{ka_{1}-\sqrt{k(n-k)\Delta}}{kn}\right)^{k}\left(x+\frac{(n-k)a_{1}+\sqrt{k(n-k)\Delta}}{n(n-k)}\right)^{n-k}=0\\ k=1,2,\ldots,n-1\end{gathered}

And $x_{i}$ is a root of equation (1) we have

(n-2)\left(a_{1}+x_{i}\right)^{2}-2(n-1)\left(a_{2}+a_{1}x_{i}+x_{i}^{2}\right)\geq 0

equality is only possible if all the roots, other than $x_{i}$ , are equal.

We therefore have the following property:
The upper limit of the roots of equations (1) is only reached for equation $f_{1}(x)=0$ and their lower limit only for the equation $f_{n-1}(x)=0$ .

We have as our only root $x_{i}$

-\frac{a_{1}+\sqrt{(n-1)\Delta}}{n}\leq x_{i}\leq-\frac{a_{1}-\sqrt{(n-1)\Delta}}{n}

For equation (1) to always have roots of the same sign, it is necessary and sufficient that

(n-1)a_{1}^{2}-2na_{2}\geq 0,\quad(n-2)a_{1}^{2}-2(n-1)x_{2}\leq 0

The roots are then always non-negative or non-positive depending on whether $a_{1}\leq 0$ or $a_{1}\geq 0$ 2.
Note that two consecutive coefficients of an equation with all its roots real cannot be zero at the same time. Therefore, by looking at the graph of the equation

f^{(n-3)}(x)=0

(2)

and taking into account a possible linear transformation, we deduce that
the coefficient $a_{3}$ reaches its minimum only for the equation $f_{1}(x)=0$ and its maximum only for $f_{n-1}(x)=0$ .

This property is none other than the condition for the reality of roots $x_{1}{}^{\prime\prime\prime}\leq x_{2}{}^{\prime\prime\prime}\leq x_{3}{}^{\prime\prime\prime}$ of equation (2) and is explicitly expressed by inequality
(3) $\quad(n-2)^{2}\Delta^{3}-(n-1)\left[(n-1)(n-2)a_{1}^{3}-3n(n-2)a_{1}a_{2}+3n^{2}a_{3}\right]^{2}\geq 0$ equality is only
possible for equations $f_{1}(x)=0,f_{n-1}(x)=0$ 3.
Let us propose to determine the maximum of the smallest interval containing the roots. This maximum is necessarily attained.

Suppose that equation (1) has at least four distinct roots
(4)

\begin{gathered}f(x)=g(x)\cdot\mathrm{P}(x)\\ g(x)=x^{n-4}+b_{1}x^{n-5}+\cdots+b_{n-4}\\ \mathrm{P}(x)=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)=x^{4}+c_{1}x^{3}+c_{2}x^{2}+c_{3}x+c_{4}\\ \delta<\gamma<\beta<\alpha\end{gathered}

⁰⁰footnotetext: (1) This property was found by Lauuerre. See, Works, vol. I, p. 93.

æt being the largest and ô the smallest root of (1). We have

	$\displaystyle c_{1}+b_{1}$	$\displaystyle=a_{4}$
	$\displaystyle c_{2}+c_{4}b_{1}+b_{2}$	$\displaystyle=a_{2}$
	$\displaystyle c_{4}+c_{3}\alpha+c_{2}\alpha^{2}+c_{1}\alpha^{3}+\alpha^{4}$	$\displaystyle=0$
	$\displaystyle c_{4}+c_{3}\delta+c_{2}\delta^{2}+c_{1}\delta^{3}+\delta^{4}$	$\displaystyle=0$

Let's leave the polynomial fixed $g(x)$ We then have a system of four linear equations in $c_{4},c_{2},c_{3},c_{4}$ whose determinant is not zero. Substitute $\alpha,\delta$ about $\alpha^{\prime},\delta^{\prime}$ respectively we have a new system to which the polynomial corresponds $\mathrm{P}_{1}(x)$ having zeros $\alpha^{\prime},\beta^{\prime},\gamma^{\prime},\delta^{\prime}$ And $\alpha^{\prime},\delta^{\prime}$ are quite similar to $\alpha,\delta$ respectively, $\beta^{\prime},\gamma^{\prime}$ are real and we have $\delta^{\prime}<\gamma^{\prime}<\beta^{\prime}<\alpha^{\prime}$ The polynomial $f^{*}(x)=g(x)$ . $\mathrm{P}_{1}(x)$ takes the given coefficients $a_{1},a_{2}$ Taking $\alpha^{\prime}>\alpha$ , $\delta>\delta^{\prime}$ We can see that the maximum in question cannot be reached for an equation having at least four distinct roots.

Therefore, the maximum can only be reached for an equation of the form
(5)

\begin{gathered}f(x)=(x-\alpha)^{n_{1}}(x-\beta)^{n_{2}}(x-\gamma)^{n_{3}}=0\\ n_{1}+n_{2}+n_{3}=n,\alpha\geq\beta\geq\gamma.\end{gathered}

By a process similar to that used above, by posing $f(x)=g(x)\mathrm{P}(x)$ with $\mathrm{P}(x)=(x-\alpha)(x-\beta)(x-\gamma)$ It is easily shown that the maximum cannot be reached if $n_{1}+n_{3}>2$ A simple calculation then shows us that we must take $n_{1}=n_{3}=1,\beta=-\frac{a_{1}}{n}$ SO,

The smallest interval containing the roots of equation (1) is at most equal to

\sqrt{\frac{2}{n}\Delta}

•

The limit is only reached for the equation

\left(x^{2}+\frac{2}{n}a_{1}x+a_{2}-\frac{(n+1)(n-2)}{2n^{2}}a_{1}^{2}\right)\left(x+\frac{a_{4}}{n}\right)^{n-2}=0.

This property can also be stated as follows:
If $x_{1}{}^{\prime\prime}\leq x_{2}{}^{\prime\prime}$ are the roots of the $(n-2)^{\text{àme }}$ derivative $f^{(n-2)}(x)=0$ , Ibs roots ide of equation (1) are all in an interval of length at most equal to

\left(x_{2}^{\prime\prime}-x_{1}^{\prime\prime}\right)\sqrt{\frac{n(n-1)}{2}}

4.

We can also look for the minimum length of the smallest interval containing the recines. Let's put equation (1) in the form (4). Let's construct the polynomial again. $f^{*}(x)$ by asking $a^{\prime}<\alpha_{p}$ $\delta<\delta^{\prime}$ and if $\alpha,\delta$ are not simple zeros, let's repeat the same operation on $f^{*}(x)$ until we arrive at a polynomial whose extreme zeros are simple. We see in this way that the minimum can only be reached by an equation of the form (5). A simple discussion shows us that the minimum can only be reached if $\alpha=\beta$ or $\beta=\gamma$ so if the equation has at most two distinct roots.

It is easily found that:
The smallest interval containing all the roots of equation (1) is at least equal to

\frac{2}{n}\sqrt{\Delta}\text{ si }n\text{ est pair, }\frac{2}{\sqrt{n^{2}-1}}\sqrt{\Delta}\text{ si }n\text{ est impair }

the limit is only reached for the equation $\frac{f_{n}}{2}(x)=0$ and $n$ is even and for the equations $\frac{f_{\frac{n-1}{2}}(x)}{2}=0,\frac{f_{n+1}}{2}(x)=0$ and $n$ is odd.

These equations also arise if we seek to determine the smallest centered interval $-\frac{a_{1}}{n}$ always containing at least one root of equation (1). Indeed, $\left(-\lambda-\frac{a_{1}}{n},\lambda-\frac{a_{1}}{n}\right)$ this interval. It can be shown, as above, that for the determination of $\lambda$ It suffices to consider only equations of the form (5). A simple discussion, which we will not reproduce here, shows us that:

Equation (1) always has at least one root in the interval: (closed)

\begin{array}[]{ll}\left(-\frac{a_{1}}{n}-\frac{1}{n}\sqrt{\Delta},-\frac{a_{1}}{n}+\frac{1}{n}\sqrt{\Delta}\right)&n\text{ pair }\\ \left(-\frac{a_{1}}{n}-\frac{1}{n}\sqrt{\frac{n-1}{n+1}\Delta},-\frac{a_{1}}{n}+\frac{1}{n}\sqrt{\frac{n-1}{n+1}\Delta}\right)&n\text{ impair }\end{array}

GHAPITRE II.

On equations where the first four coefficients are given

•
1. 1.
  
  Let us now consider the set of equations (1) having all their roots real and for which the coefficients $a_{1},a_{2}$ , - $a_{3}$ are given.

Let's try to determine an equation of the form

(6)\quad(x-\alpha)(x-\beta)(x-\gamma)^{n-2}=0,\quad\alpha,\beta,\gamma\text{ réels }

taking the given coefficients.
We have at most three equations of this form with $\gamma$ real depending on whether this root is equal to one of the roots $x_{1}{}^{\prime\prime\prime}\leq x_{2}{}^{\prime\prime\prime}\leq x_{3}{}^{\prime\prime\prime}$ - of the equation $f^{(n-8)}(x)=0$ .

Let's set aside the cases $x_{1}{}^{\prime\prime\prime}=x_{2}^{\prime\prime\prime}\leq x_{3}{}^{\prime\prime\prime},x_{4}{}^{\prime\prime\prime}\leq x_{2}{}^{\prime\prime\prime}=x_{3}{}^{\prime\prime\prime}$ where the polynomial (1) is necessarily of the form $f_{1}(x)$ or $f_{n-1}(x)$ .

Assuming that

x_{1}^{\prime\prime\prime}<x_{2}^{\prime\prime\prime}<x_{3}^{\prime\prime\prime}.

(7)

Are you him and $\gamma=x_{2}{}^{\prime\prime\prime}$ the polynomial (6) must necessarily be identical to

	$\displaystyle f^{*}(x)=\left[\left(xx_{2}^{\prime\prime\prime}\right)^{2}+\frac{n}{3}\left(2x_{2}^{\prime\prime\prime}-x_{4}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)\left(x-x_{2}^{\prime\prime\prime}\right)+\right.$
		$\displaystyle\left.\quad+\frac{n(n-1)}{6}\left(x_{2}^{\prime\prime\prime}-x_{1}^{\prime\prime\prime}\right)\left(x_{2}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)\right]\cdot\left(x-x_{2}^{\prime\prime\prime}\right)^{n-2}$

and the equation $f^{*}(x)=0$ a, by virtue of (7), all its real roots.
If the form (6) is peeled with $\gamma=x_{1}{}^{\prime\prime\prime}{}^{(2)}$ it is necessarily -identical to

		$\displaystyle{\left[\left(x-x_{1}^{\prime\prime\prime}\right)^{2}+\frac{n}{3}\left(2x_{1}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)\left(x-x_{1}^{\prime\prime\prime}\right)+\right.}$
		$\displaystyle\left.\quad+\frac{x(n-1)}{6}\left(x_{1}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime\prime}\right)\left(x_{4}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}\right)\right]\cdot\left(x-x_{1}^{\prime\prime\prime}\right)^{n-2}$

hence the condition of reality
(8) $\quad\frac{n^{2}}{9}\left(2x_{1}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)^{2}-\frac{2n(n-1)}{3}\left(x_{1}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)\left(x_{1}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}\right)\geq 0$

Let's simplify.

p=\frac{x_{3}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}}{x_{2}^{\prime\prime\prime}-x_{1}^{\prime\prime\prime}}

⁽²⁾ We say, for simplicity, that the form (6) is real if $\alpha$ And $\beta$ are real.
which is a positive number. Inequality (8) becomes

\rho\geq\frac{n-3+\sqrt{3(n-1)(n-3)}}{n}.

(9)

Similarly, we find that the form (6) is real for $\gamma=\boldsymbol{x}_{\mathbf{3}}{}^{\text{308- }}$ only if

\frac{1}{\rho}\geq\frac{n-3+\sqrt{3(n-1)(n-3)}}{n}

(10)

Note also that the second member of inequalities (9) and (10) grows with $n$ and tends towards $1+\sqrt{3}$ , therefore if e is outside (in the strict sense) the interval

\left(\frac{\sqrt{3}-1}{2},1+\sqrt{3}\right)\quad\text{ ou }\quad(0,366\ldots,2,732\ldots)

one of the forms (6) with $\gamma=x_{1}$ or $\gamma=x_{3}$ "' is real regardless of the degree of the polynomial $f(x)$ 6.
Every root $x_{i}$ of the equation must satisfy a certain inequality." This inequality is obtained from (3) by substituting $n,a_{1.,}a_{2},a_{3}$ about $n-1,\quad a_{1}+x_{i},\quad a_{2}+a_{1}x_{i}+x_{i}^{2},\quad a_{3}+a_{2}x_{i}+a_{1}x_{i}^{2}+x_{i}^{3}$ respectively. The maximum and minimum of $x_{i}$ We will cancel the first member of this inequality. On the other hand, if $x_{i}$ If this expression is nullified, then polynomial (1) is necessarily of the form (6). Therefore, the maximum and minimum roots can only be reached by an equation of the form (6). This statement is justified by the fact that there is always at least one real form (6).

Taking (9) into account, it is easily verified that if (6) is real. with $\gamma=x_{1}{}^{\prime\prime}$ "the roots of the equation
$\left(x-x_{1}^{\prime\prime\prime}\right)^{2}+\frac{n}{3}\left(2x_{1}^{\prime\prime\prime}-x_{2}^{\prime\prime}-x_{3}^{\prime\prime\prime}\right)\left(x-x_{1}^{\prime\prime\prime}\right)+\frac{n(n-1)}{6}\left(x_{1}^{\prime\prime\prime}-x_{3}^{\prime\prime\prime}\right)\left(x_{1}^{\prime\prime\prime}-x_{2}^{\prime\prime\prime}\right)=0$
are always included between the roots of the equation
$\left(x-x_{2}{}^{\prime\prime\prime}\right)^{2}+\frac{n}{3}\left(2x_{2}{}^{\prime\prime}-x_{1}{}^{\prime\prime\prime}-x_{3}{}^{\prime\prime\prime}\right)\left(x-x_{2}{}^{\prime\prime}\right)+\frac{n(n-1)}{6}\left(x_{2}{}^{\prime\prime}-x_{1}{}^{\prime\prime\prime}\right)\left(x_{2}{}^{\prime\prime\prime}-x_{3}{}^{\prime\prime\prime}\right)=0$ We
have the same property for form (6) with $\gamma=x_{3}$ "'if it is real.

We can therefore state the following property:
The upper and lower limits of the roots are reached only by the equation $f^{*}=0$ We
also see that the roots of equation (1) are always
contained within an interval of length at most equal to

\frac{x_{3}^{\prime\prime\prime}-x_{1}^{\prime\prime\prime}}{3}\sqrt{n^{2}+2n(n-3)\frac{\rho}{(1+\rho)^{2}}}

The radical is at its maximum for $\rho=1$ , therefore
If $x_{1}{}^{\prime\prime},x_{3}{}^{\prime\prime}$ are the smallest and largest roots of los $(n-3)$ nth derivative $f^{(n-3)}(x)=0$ , the roots of the given equation are all in an interval of length at most equal to

\left(x_{3}^{\prime\prime\prime}-x_{1}^{\prime\prime\prime}\right)\sqrt{\frac{n(n-1)}{6}}.

7.

Looking at the curve $f^{(n-4)}(x)=0$ we see that

The minimum of a $a_{4}$ is only achieved for the equation $f^{*}(x)=0$
Consider the symmetric function of the roots ( 11
)

\Sigma\left(x_{i}-x_{j}\right)^{2}\left(x_{j}-x_{k}\right)^{2}\left(x_{k}-x_{l}\right)^{2}

of equation (1).
This expression being of the form $\mathrm{A}-2\Delta a_{4}$ , where A depends only on $a_{1},a_{2},a_{3}$ will be maximum for the equation $f^{*}(x)=0$ We can therefore say that if $a_{4},a_{2}$ are given (11) is maximum only for an equation of the form (5) in which $n_{1}=n_{3}=1$ , $n_{2}=n-2$ A simple calculation then shows us that this maximum is reached only if $2\beta=\alpha+\gamma$ .

We therefore deduce the following property:
If equation (1) has all its roots real, then we have

\Sigma\left(x_{i}-x_{j}\right)^{2}\left(x_{i}-x_{k}\right)^{2}\left(x_{k}-x_{i}\right)^{2}\leq\frac{n-2}{2n^{3}}\Delta^{3}

equality is only possible for the equation

\left(x^{2}+\frac{2}{n}a_{1}x+a_{2}-\frac{(n+1)(n-2)}{2n^{2}}a_{1}^{2}\right)\left(x+\frac{a_{1}}{n}\right)^{n-2}=0.

8.

We can also look for the maximum of the coefficient $a_{4}$ It can easily be shown that this maximum, which is necessarily attained, is only reached by an equation of the form (5). It should be noted that the form of the maximizing equation is not invariant and changes according to the values of $\rho$ We will not dwell on this point here.

One of the problems discussed above can be generalized without difficulty, as we will see in the next chapter.

CHAPTER III.

On the roots of the derivative equation

9.

Let us designate by $\mathrm{R}(f)$ the largest root (or one of them if there are several) of equation (1). We will therefore denote by $\mathrm{R}\left(f^{(k)}\right)$ the largest root of the derivative chemistry $f^{(k)}(x)=0$ .

According to a classical theorem, we have

\mathrm{R}\left(f^{\prime}\right)\geq\mathrm{R}\left(f^{\prime}\right)\geq\ldots\geq\mathrm{R}\left(f^{(n-1)}\right)

Note that if $x_{0}$ is a root of order $k>1$ of multiplicity of the derivative equation it is necessarily a root of order $k+1$ of the given equation. It is easily deduced that the only possible general arrangement is the following:

\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)=\ldots=\mathrm{R}\left(f^{(i)}\right)>\mathrm{R}\left(f^{(i+1)}\right)>\ldots>\mathrm{R}\left(f^{(i-1)}\right)

and then $\mathrm{R}(f)$ is a root of order $i+1$ of multiplicity for equation (1) and $\mathrm{R}\left(f^{(j)}\right)$ is a simple root of the equation $f^{(j)}(x)=0$ For $j=i+1$ , $i+2,\ldots,n-1$ .

Polynomial (1) is generally of the following form:

\begin{gathered}f(x)=\left(x-\alpha_{1},n_{1}\left(x-\alpha_{2}\right)^{n_{2}}\ldots\left(x-\alpha_{k}\right)^{n_{k}}\right.\\ n_{1}+n_{2}+\ldots+n_{k}=n,\quad\alpha_{1}>\alpha_{2}>\ldots>\alpha_{k}\end{gathered}

The derivative equation $f(x)=0$ has two kinds of roots. First $k-1$ roots $\beta_{1},\beta_{2},\ldots,\beta_{k-1}$ distinct from $\alpha_{i}$ and separated by the latter

\alpha_{1}>\beta_{1}>\alpha_{2}>\beta_{2}>\ldots>\alpha_{k-1}>\beta_{k-1}>\alpha_{k}

and then $n-k$ roots coinciding with a $\alpha_{i}$ .

And $n_{i}=1$ all the different roots of $\beta_{j}$ remain fixed.
If $n_{i}>1$ there is a root of the derivative that detaches from $\boldsymbol{\alpha}_{i}$ but it obviously varies in the same direction as $\alpha_{l}$ This root is therefore an increasing function of the varied root of the given equation. All other distinct roots of $\beta_{j}$ remain fixed.

It remains to examine the variation of a root $\beta_{1}$ Suppose
$j\leq i-1$ to clarify things, let's then ask ourselves...

\phi(x)=\frac{f(x)}{x-\alpha_{i}}

On a

\phi\left(\beta_{j}\right)\neq 0\text{ et sig. }\phi(\beta i)=\text{ sig. }(-1)^{n_{1}+n_{2}+\ldots n_{i}(3)}

But we also have

\phi^{\prime}\left(\beta_{j}\right)\left(\beta_{j}-\alpha_{i}\right)+\phi\left(\beta_{j}\right)==0

ads $\phi^{\prime}\left(\beta_{/}\right)\neq 0$ and as $\beta_{j}-\alpha_{i}>0$ we find

\operatorname{sig}.\phi^{\prime}\left(\beta_{j}\right)=\operatorname{sig}.(-1)^{n_{1}+n_{2}+\ldots+n_{j}-1}

Let's now ask

F(x)=\phi(x)\left(x-\alpha_{i}^{\prime}\right)

$\alpha_{i}^{*}$ being in the vicinity of $\alpha_{i}$ We find

\mathrm{F}^{\prime}\left(\beta_{j}\right)=\left(\alpha_{i}-\alpha_{j}^{\prime}\right)\phi^{\prime}\left(\beta_{j}\right)

hence

\text{ sig. }\mathrm{F}^{\prime}\left(\beta_{l}\right)=\operatorname{sig}\cdot(-1)^{n_{i}+n_{2}+\cdots+n_{j}-1}\times\operatorname{sig}\cdot\left(\alpha_{i}-\alpha_{i}^{\prime}\right)

But, in the left neighborhood of $\alpha_{j}$ on a

\text{ sig. }\mathrm{F}^{\prime}(x)=\text{ sig. }(-1)^{n_{1}+n_{2}+\ldots+n_{j}-1}

And, $\alpha^{\prime}{}_{i}$ being close enough $\alpha_{i},\mathrm{\penalty 10000\ F}^{\prime}(x)=0$ has a single root in the interval $\left(\alpha_{j},\alpha_{j+1}\right)$ [and $j=i-1\quad a_{i}$ is replaced here by $\alpha^{\prime}{}_{i}$ which is precisely the root $\beta_{j}$ varied; or $\varepsilon_{j}^{\prime}$ It follows that

\operatorname{sig}.\left(\beta_{1}-\beta_{j}^{\prime}\right)=\operatorname{sig}.\left(\alpha_{i}-\alpha_{i}^{\prime}\right)

done $\beta_{j}$ is an increasing function of the root $\alpha_{i}$ varied.
We obtain the same property and demonstrate it in the same way if $j\geq i$ .

We have only provided the demonstration for variations of $\boldsymbol{\alpha}_{i}$ around its initial position. It is easy to see that the property remains true for any variation of $\alpha_{i}$ if care is taken to number the roots of the derivative equation beforehand and not to change this numbering even if these roots pass through each other.
11. The property demonstrated above has some interesting consequences.

For example, we can see that if

f(x)=g(x)(x-\alpha)h(x)

Or $g(x)$ is a fixed polynomial of degree $k(k<n-1)$ whose zeros are at least equal to $\alpha$ And $h(x)$ a polynomial whose zeros are in
(3) We set as usual sig. $z=1,0,-1$ depending on $z>,=,<0$ .
more equal to $\alpha$ , on a

\mathrm{R}\left(f^{\prime}\right)\leq\mathrm{R}\left(\left(g(x)(x-\alpha)^{n-k}\right)^{\prime}\right)

equality is only possible if $h(x)=(x-\alpha)^{n-k-1}$ By
always leaving the root fixed $\alpha$ the polynomial $g(x)$ we can see that

\min.\mathrm{R}\left(f^{\prime}\right)=\mathrm{R}\left(\left(g(x)(x-\alpha)^{\prime}\right)\right.

We will indeed approach this minimum indefinitely by making it tend towards $-\infty$ the zeros of $h(x)$ We can also avoid infinities with a simple transformation. We can assume $\alpha>0$ without restricting the generality. It is then sufficient to perform the transformation $x\left\lvert\,\frac{1}{x}\right.$ on the equation $f(x)=0$ and apply the results from the previous No. to the smallest positive root of this equation.

a+\frac{b-a}{n}\geq R\left(f^{\prime}\right)>\frac{a+b}{2}

equality can only occur for $g(x)=(x-b)^{n-2}$ and the lower limit cannot be replaced by any other smaller number.

And $u>3$ And $f(x)=(x-a)(x-b)(x-c)g(x),a>a\geq c$ , where the zeros: of $g(x)$ are at most equal to $c$ , on a
$\frac{2(a+b+c)+(n-3)(a+b)+\sqrt{2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]+(n-3)(n+1)(n-b)^{2}}}{2n}\geq$

\geq\mathrm{R}\left(f^{\prime}\right)>\frac{2(a+b+c)+\sqrt{2\left[(a-b)^{2}+(b-c)^{2}-(c-a)^{2}\right]}}{6}

equality being possible only for. $g(x)=(x-c)^{n-3}$ and the lower limit cannot be replaced by any smaller number.

The results from the previous issue also apply to the roots of the equations. $f^{\prime\prime}(x)=0,f^{\prime\prime\prime}(x)=0,\ldots$ etc. $\mathrm{C}\theta$ that we said about the limitation of $\mathrm{R}\left(f^{\prime}\right)$ can easily be extended to the roots $\mathrm{R}\left(f^{\prime\prime}\right)$ , $\mathrm{R}\left(f^{\prime\prime\prime}\right),\ldots$ etc. We can therefore obtain various inequalities for these roots as before. Suppose, for example, that

\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)=\ldots=\mathrm{R}\left(f^{(k-1)}\right)>\mathrm{R}\left(f^{(k)}\right)

from soit $\phi(x)$ a non-constant factor of the polynomial $f(x)$ . We have

\mathrm{R}\left(f^{(i)}\right)>\mathrm{R}\left(\phi^{(i)}\right),\quad i\geq k

Or $i+1$ must not exceed the degree of the polynomial $\phi(x)$ Laguerre
demonstrated that if $\alpha<\beta$ are two consecutive roots of equation (1); there are no roots of the derivative equation in the intervals $\left(\alpha,\alpha+\frac{\beta-\alpha}{n}\right),\left(\beta-\frac{\beta-\alpha}{n},\beta\right)$ We can see more precisely if there is $k$ roots to the left of $\alpha$ or confuse with $\alpha$ There is no root of the derivative in the intervals $\left(\alpha,\alpha+\frac{\beta-\alpha}{n-k}\right),\left(\beta-\frac{\beta-\alpha}{k+2},\beta\right)\left({}^{4}\right)$
12. Either $\lambda$ the length of the smallest interval containing the roots of equation (1). The roots of the derived equation are all: in an interval of length at least equal to $A.\lambda$ .

We aim to determine this number, which is obviously less than 1. Without restricting the generality, we can take...

f(x)=\left(x^{2}-1\right)g(x)

where the roots of $g(x)=0$ are all within the interval ( $-1,1$ ); are $\alpha,\beta$ the largest and smallest roots of this equation and $\alpha^{\prime},\beta^{\prime}$ the largest and smallest root of $f^{\prime}(x)=0$ The goal is to determine the minimum of $\alpha^{\prime}-\beta^{\prime}$ .

And $\alpha=1,\beta=-1$ We have $\alpha^{\prime}-\beta^{\prime}=2$ And
$\alpha=1,\beta>-1$ or $\alpha<1,\beta=-1$ According to the results of No. 10, we obtain the smallest value of $\alpha^{\prime}-\beta^{\prime}$ For $(x-1)^{n-1}(x+1)=0$ , $(x-1)(x+1)^{n-1}=0$ respectively; hence

\alpha^{\prime}-\beta^{\prime}\geq 2\cdot\frac{n-1}{n}

Let's now suppose that $\alpha<1,\beta>-1$ We can then write

f(x)=\psi(x)(x-\alpha)(x-\beta)

By writing $f^{\prime}\left(\alpha^{\prime}\right)=0,f^{\prime}\left(\beta^{\prime}\right)=0$ we have a system of two linear equations in $\alpha+\beta,\alpha\beta$ If the determinant of this system is not zero, applying reasoning analogous to that of No. 3, we show that the polynomial $f(x)$ can be replaced by another
⁽⁴⁾ This result also follows from the generalization given to Laguerre's theorem by MJ v. Sz. Nagy "Ueber algebraische Gleichungen mit lauter reellen Wurzeln" Jahreshericht der Deutschen Math. Ver. 27 (1918) p. 37-43.
for which $\alpha^{\prime}-\beta^{\prime}$ smaller solt. We take into account here that $\alpha^{\prime}$ , $\beta$ , are simple roots.

If the determinant is zero, one of the equations is a consequence of the other. Taking then

\begin{gathered}F(x)-\psi(x)(x-\gamma)^{2}\\ \alpha^{\prime}-\gamma=\frac{2\left(\alpha^{\prime}-\alpha\right)\left(\alpha^{\prime}-\beta\right)}{2\alpha^{\prime}-\alpha-\beta}\end{gathered}

•

on a $\beta<\gamma<\alpha$ And $\alpha^{\prime},\beta^{\prime}$ are the largest and smallest root - of $\mathrm{F}^{\prime}(x)=0$ We then repeat the same operation indefinitely. We see that either we arrive at a non-zero determinant, or by taking the limit we find an equation of the form $\left(x^{2}-1\right)(x-\lambda)^{n-2}=0$ for which $\alpha^{\prime},\beta^{\prime}$ are still the largest and - and the smallest root of its derivative. In any case, to find the minimum of $\alpha^{\prime}-\beta^{\prime}$ We simply need to examine the equations. $\left(x^{2}-1\right)(x-\lambda)^{n-2}=0$ .

The minimum value is $2\sqrt{\frac{n-2}{n}}$ and is obtained for $\lambda=0$ We
therefore have the following property:
If the roots of the derivative equation are in an interval of length $\lambda$ the roots of the given equation are all in an interval of length at most equal to 0

\lambda\sqrt{\frac{n}{n-2}}

Moreover, we can see that we can state the following property more generally:

If the roots of the derived kth $f^{(k)}(x)=0$ are all within a length interval $\lambda$ the roots of the given equation are all in an interval of length at most equal to

\lambda\sqrt{\frac{n(n-1)}{(n-k)(n-k-1)}}^{(5)}

This is the generalization of cases $k=n-2,k=n-3$ already mentioned in Chapters I and II.

⁰⁰footnotetext: (5) I have just become aware of the memoir of MJ v. Sz. Nagy, * loc. cit. ( ⁴ ), unfortunately after having made the corrections. These results are due to MJ v. Sz. Nagy.

CHAPTER IV.

On the inequality of MI Schur*

13.

Consider the family of equations (1) for which $\mathrm{R}(f)$ law and $\mathrm{R}\left(f^{\prime}\right)$ have given values. Let us propose to determine the maximum of $\mathrm{R}\left(f^{\prime\prime}\right)$ .

And $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)$ we obviously have max. $\mathrm{R}\left(f^{\prime\prime}\right)=\mathrm{R}\left(f^{\prime}\right)$ and this maximum is reached by any equation for which $\mathrm{B}(f)$ is a root of at least three.

And $\mathrm{R}(f)>\mathrm{R}\left(f^{\prime}\right)$ we can take, without restricting the generality, $\mathrm{R}(f)=1,\mathrm{R}\left(f^{\prime}\right)=0$ .

Suppose that equation (1) has at least two distinct roots $\approx$ of $\mathrm{R}(f)=1$ We can write the decomposition (4) with

\begin{gathered}g(x)=\text{ polynome de degré }n-1\\ \mathrm{P}(x)=(x-\alpha)(x-\beta)=x^{2}+c_{1}x+c_{2},\quad 1>\alpha>\beta.\end{gathered}

We have the system

	$\displaystyle c_{1}g^{\prime}(0)+c_{2}g(0)=0$
	$\displaystyle{\left[x^{2}g(x)\right]_{x=\xi}^{\prime\prime}+c_{1}[xg(x)]_{x=-\xi}^{\prime\prime}+c_{2}[g(x)]_{x=\xi}^{\prime\prime}=0,\quad\xi=\mathrm{R}\left(f^{\prime\prime}\right)}$		(12)

of two linear equations in $c_{1},c_{2}$
If the determinant of this system is not zero, we can, by virtue of continuity, determine a polynomial . $P_{1}(x)=\left(x-\alpha_{1}\right)\left(x-\beta_{1}\right)$ . tel that if $\mathrm{F}(x)==(x).\mathrm{P}_{1}(x)$ we have

\mathrm{R}(\mathrm{\penalty 10000\ F})=1,\mathrm{R}\left(\mathrm{\penalty 10000\ F}^{\prime}\right)=0,\mathrm{R}\left(\mathrm{\penalty 10000\ F}^{\prime\prime}\right)>\mathrm{R}\left(f^{\prime\prime}\right)

If the determinant is zero, the second equation (12) is a consequence of the first. In this case, when $\beta$ decreases towards $-\infty$ a grows towards a limit which is determined by the equation $\alpha g(0)-g^{\prime}(0)=0$ … We have $g(0)\neq 0$ [so also $g^{\prime}(0)\neq 0$ that is to say

\lim.\alpha=\frac{g^{\prime}(0)}{g(0)}

We can then see that if

G(x)=g(x)\left[x-\frac{g^{r}(0)}{g(0)}\right]

\mathrm{R}(\dot{G})=\mathrm{R}(f),\mathrm{R}\left(G^{\prime}\right)=\mathrm{R}\left(f^{\prime}\right),\mathrm{R}\left(G^{\prime\prime}\right)=\mathrm{R}\left(f^{\prime\prime}\right)

14.

We can now determine the maximum of $R\left(f^{\prime\prime}\right)$ b

Note that an equation of the form

f(x)=(x-a)(x-b)^{m}=0

is completely determined by the conditions $\mathrm{R}(f)=1,\mathrm{R}\left(f^{\prime}\right)=0$ We then have $a=1,b=-\mathrm{m}$ And $\mathrm{R}\left(f^{\prime\prime}\right)=-1$ regardless of $m$ .

Either $n=3$ . We have $g(x)=x-1$ and the determinant of system (12) is not zero. Therefore, the maximum is only reached for the equation $(x-1)(x+2)^{2}=0$ .

We will demonstrate the following property:
The maximum of $\mathrm{R}\left(f^{\prime\prime}\right)$ can only be achieved $p$ on equations of the form (13).

From the property demonstrated in the previous No., we see that the maximum is reached, or rather for the equation $(x-1)(x+n-1)^{n-1}=0$ or for an equation of degree $<n$ .

We will prove it by induction. We have seen that the property is true for degrees $3,4,\ldots,n-1$ and let's demonstrate it to the degree $n$ Choosing the roots properly $\alpha$ And $\beta$ We see that the property is proven by induction.

This property is due to M.I. Schur ( ⁶ ) who stated it as follows:

If equation (1) has all its roots real, then we have the inequality

\mathrm{R}(f)-\mathrm{R}\left(f^{\prime}\right)\leq\mathrm{R}\left(f^{\prime}\right)-\mathrm{R}\left(f^{\prime\prime}\right)

equality not being possible - except in the trivial case $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)==\mathrm{R}\left(f^{\prime\prime}\right)$ - than for equations of the form

(x-a)(x-b)^{m}=0

15.

We can extend the previous result a little in case $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)=\ldots=\mathrm{R}\left(f^{(i)}\right)>\mathrm{R}\left(f^{(i+1)}\right)$ are given. We show oncore, as above, that the maximum of $\mathrm{R}\left(f^{(i+2)}\right)$ can only be reached by an equation of the form

[x-\mathrm{R}(f)]^{4+1}(x-b)^{n-i-1}=0,\quad\mathrm{R}(f)>b

Such an equation is completely determined by the given values. Indeed, if there were two, one could be transformed into the other by a simple linear transformation and we would encounter a contradiction with the growth property demonstrated in No. 10.

It is clear that the minimum of the report

\frac{\mathrm{R}\left(f^{(i)}\right)-\mathrm{R}\left(f^{(i+2)}\right)}{\mathrm{R}\left(f^{(i)}\right)-\mathrm{R}\left(f^{(i+1)}\right)}

(6) I Schur “Two theorems on algebraic equations with purely real roots ${}^{\text{" }}$ , Journal für Math. B. 144 (4, 4) pp. 75-88.
is obtained by calculating its value for the equation $(x-1)^{i+1}x^{n-i-1}=0$ For example.

In particular, for $i=1$ We obtain the following property:
If equation (1) has all $s$ : s real roots and if $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)>\mathrm{R}\left(f^{\prime\prime}\right)$ on a

\frac{\mathrm{R}\left(f^{\prime}\right)-\mathrm{R}\left(f^{\prime\prime\prime}\right)}{\mathrm{R}\left(f^{\prime}\right)-\mathrm{R}\left(f^{\prime\prime}\right)}\geq\frac{3\sqrt{n-1}-\sqrt{3(n-3)}}{2\sqrt{n-1}-\sqrt{2(n-2)}}

equality is only possible for equations of the form

(x-a)^{2}(x-b)^{n-2}=0.\quad a>b

16.

Let us propose to determine an equation of the form

f(x)=(x-a)(x-b)(x-c)^{n-2}=0

taking the given values $\mathrm{R}(f),\mathrm{R}\left(f^{\prime}\right),\mathrm{R}\left(f^{\prime\prime}\right)$
The case $\mathrm{R}\left(f^{\prime\prime}\right)=2\mathrm{R}\left(f^{\prime}\right),-\mathrm{R}(f)$ has already been demonstrated, and we know that equation (14) is then completely determined. The same is true if $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)$ .

Suppose then that $\mathrm{R}\left(f^{\prime\prime}\right)<2\mathrm{R}\left(f^{\prime}\right)-\mathrm{R}(f).\mathrm{R}(f)>\mathrm{R}\left(f^{\prime}\right)$ Without
restricting the generality, we can assume that $\mathrm{R}(f)=1$ , $R\left(f^{\prime}\right)=0$ and then $R\left(f^{\prime\prime}\right)=\xi<-1$ Equation (14) becomes

f(x)=(x-1)(x-\lambda)\left(x+\frac{(n-2)\lambda}{1+\lambda}\right)^{n-2}=0.

(15)

If we write that $f^{\prime\prime}(x)$ rid of the postman $\left(x+\frac{(n-2)\lambda}{1+\lambda}\right)^{n-4}$ 5'cancels for $x=\xi$ we obtain an equation of the form

p\lambda^{3}-1-q\lambda^{2}+r\lambda+s=0

(16)

to determine $\lambda$ We have thus neglected the value -1 of $\lambda$ when (15) tends towards a second-degree equation.

Doing the calculations, we find

		$\displaystyle p=-(n-1)(n-2)(\xi+1)$
		$\displaystyle q=n(n-1)\xi^{2}+(n-1)(n-2)\xi-2(n-2)$
		$\displaystyle r=2n(n-1)\xi^{2}-(n-1)(n-2)\xi-(n-1)(n-2)$
		$\displaystyle s=(n-1)\xi[n\xi-(n-2)].$

The discriminant of equation (17) is of the form
(17)

\lambda_{0}\xi^{6}+\lambda_{1}\xi^{5}+\lambda_{2}\xi^{4}+\lambda_{3}\xi^{3}+\lambda_{4}\xi^{2}+\lambda_{5}\xi+\lambda_{6}

by removing the case $p=0$ which leads to $\xi=-1$ , a case we have already studied.

We have

		$\displaystyle\lambda_{0}=-8n^{2}(n-1)^{3}(n-2)^{3}$
		$\displaystyle\lambda_{4}=4n(n-1)(n-2)^{3}\left(5n^{3}-29n^{2}+60n-44\right)$
		$\displaystyle\lambda_{5}=-8n(n-1)(n-2)^{5}(n+3)$
		$\displaystyle\lambda_{6}=4n(n-1)^{2}(n-2)^{5}$

which shows that the following

\lambda_{0},\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}

presents at least three variations. It follows that the polynomial (17) has at most three negative zeros.

We first check that (17) vanishes for $\xi=-1$ Consider
equation (14) and the ratio

\frac{\mathrm{R}\left(f^{\prime}\right)-\mathrm{R}\left(f^{\prime\prime}\right)}{\mathrm{R}(f)-\mathrm{R}\left(f^{\prime}\right)}.

(18)

Suppose $a$ And $c<a$ fixed and let's vary $b$ of $a$ up to: $-\infty$ The ratio (18) decreases by $+\infty$ up to the value 1 for $b=c$ which is a minimum. Then it increases to a maximum and then decreases towards... $b=-\infty$ .

The result is that $\mathrm{R}(f),\mathrm{R}\left(f^{\prime}\right),\mathrm{R}\left(f^{\prime\prime}\right)$ given there always exists at least one equation of the form (14) with $a\geq b\geq c$ taking these values.

It also follows that there is a number $\mid\xi_{1}<-1$ such as for $\xi$ including some $\left(-1,\xi_{1}\right)$ equation (16) has three real roots, two of which are $<-(n-1)$ And $\xi=\xi_{4}$ equation (16) has a double root smaller than - ( $n-1$ ).

Note that (35) also gives the values of $\lambda$ for which $\xi$ is no longer the largest root of the second derivative but the other different root of the multiple root.

Examination of report (18) shows us the existence of a number $\xi_{2}\leq\xi_{1}$ tel that if $\xi<\xi_{2}$ equation (16) still has three real roots, two of which give equations for which $\xi$ is not the largest root of the second derivative. For $\xi=\xi_{2}$ equation (16) has a double root which enjoys the same property.

We have thus highlighted the three negative zeros - 1, $\xi_{1},\xi_{2}$ of the discriminant. We can show that we have indeed $\xi_{2}<\xi_{1}$ In any case, the discriminant cannot cancel out between $\xi_{1}$ And $\xi_{2}$ and changes sign when passing through these points. It follows that for $\xi$ included in the interval ( $\xi_{1},\xi_{2}$ ) equation (16) has only one real root.

We can now state the proposal we had in mind.

An equation of the form (15) with $a\geq b\geq c$ is completely determined by knowledge of the values $\mathrm{R}\left(f^{\prime}\right),\mathrm{R}\left(f^{\prime}\right),\mathrm{R}\left(f^{\prime\prime}\right)$ .

It can easily be seen from the previous property that if we consider equation (14) with $a\geq b\geq c$ in which $\mathrm{R}(f),\mathrm{R}\left(f^{\prime}\right)$ are given, the root $\mathrm{R}\left(f^{\prime\prime}\right)$ is a decreasing function of $b$ and increasing function of $c$ .

The previous results can be extended to the case where, instead of the largest root of the second derivative, we take the largest root of the third, fourth, ... derivative.

In what follows we will show the extremal properties of equations of the form (14) and we will determine in particular the maximum of $\mathrm{R}\left(f^{\prime\prime\prime}\right)$ When $\mathrm{R}(f),\mathrm{R}\left(f^{\prime}\right)$ And $\mathrm{R}\left(f^{\prime\prime}\right)$ are given.

On algebraic equations having all their roots real

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