NOTES ON HIGHER ORDER CONVEX FUNCTIONS (I)

by
Tiberiu Popoviciu

Received on September 1, 1933.

CHAPTER I.

On equations whose 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. We can easily 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 being possible only 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(nk)\Delta}}{kn}\right)^{k}\left(x+\frac{(nk)a_{1}+\sqrt{k(nk)\Delta}}{n(nk)}\right)^{nk}=0\\ k=1,2,\ldots,n-1\end{gathered}

If $x_{i}$ 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 being possible only 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 reached only for the equation $f_{1}(x)=0$ and their lower limit only for the equation $f_{n-1}(x)=0$ .

We have for all 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 signs 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 having all its real roots cannot be zero at the same time. From there, looking at the curve representing 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 of reality of the roots $x_{1}{}^{\prime\prime\prime}\leq x_{2}{}^{\prime\prime\prime}\leq x_{3}{}^{\prime\prime\prime}$ of equation (2) and is expressed explicitly 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 being possible only for equations $f_{1}(x)=0,f_{n-1}(x)=0$ .
3. Let us determine the maximum of the smallest interval containing the roots. This maximum is necessarily reached.

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 discovered by Lauerre. 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 different from zero. Substitute $\alpha,\delta$ by $\alpha^{\prime},\delta^{\prime}$ respectively we have a new system to which corresponds the polynomial $\mathrm{P}_{1}(x)$ having zeros $\alpha^{\prime},\beta^{\prime},\gamma^{\prime},\delta^{\prime}$ . If $\alpha^{\prime},\delta^{\prime}$ 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 see that the maximum in question cannot be reached for an equation having at least four distinct roots.

The maximum can therefore 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 posing $a^{\prime}<\alpha_{p}$ $\delta<\delta^{\prime}$ 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.

We easily find that:
The smallest interval containing all the roots of equations (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 being reached only for the equation $\frac{f_{n}}{2}(x)=0$ if $n$ is even and for the equations $\frac{f_{\frac{n-1}{2}}(x)}{2}=0,\frac{f_{n+1}}{2}(x)=0$ . if $n$ 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. We show as above that for the determination of $\lambda$ it is enough to consider only the equations of the form (5). A simple discussion, which we do not reproduce, 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}

CHAPTER II.

On the equations whose 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 us 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 rule out 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 $f_{n-1}(x)$ .

Let us suppose that

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

(7)

If we have $\gamma=x_{2}{}^{\prime\prime\prime}$ 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 us pose for simplicity

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 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)

Let us also note that the second member of inequalities (9) and (10) increases with $n$ and tends towards $1+\sqrt{3}$ , so if e is outside (in the strict sense) of 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 whatever the degree of the polynomial $f(x)$ .
6. Any root $x_{i}$ of the equation must verify a certain inequality." This inequality is obtained from (3) by replacing $n,a_{1.,}a_{2},a_{3}$ by $n-1,\quad a_{1}+x_{i},\quad a_{2}+a_{1}x_{i}+x_{i}^{2},\quad a_{3}+a_{2}x_{i}+a_{1}x_{i}^{2}+x_{i}^{3}$ respectively. The maximum and minimum of $x_{i}$ will cancel the first member of this inequality. On the other hand, if $x_{i}$ cancels this expression the polynomial (1) is necessarily of the form (6). The maximum and minimum of the roots can therefore 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 into account (9) we easily check 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 the 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
included in 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 maximum for $\rho=1$ , so
if $x_{1}{}^{\prime\prime},x_{3}{}^{\prime\prime}$ are the smallest and largest root of the bone $(n-3)$ th 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 $f^{*}(x)=0$ . So we can 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 real roots 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 being possible only 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 be easily shown that this maximum, necessarily reached, is only reached by an equation of the form (5). It should be noted that the form of the maximizing equation is not invariable and changes according to the values of $\rho$ We do not insist on this point here.

One of the problems discussed above generalizes without difficulty as we will see in the next chapter.

CHAPTER III.

On the roots of the derived 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 greatest root of the kth derivative $f^{(k)}(x)=0$ .

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 root of order $k>1$ of multiplicity of the derived equation it is necessarily root of order $k+1$ of the given equation. We can easily deduce 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 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 $j=i+1$ , $i+2,\ldots,n-1$ .

The 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}

If $n_{i}=1$ all the different roots of $\beta_{j}$ remain fixed.
If $n_{i}>1$ there is a root of the derivative which 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 the $\beta_{j}$ remain fixed.

It remains to examine the variation of a root $\beta_{1}$ .
Let us suppose $j\leq i-1$ to fix the ideas and then pose

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

We have

\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

adoù $\phi^{\prime}\left(\beta_{/}\right)\neq 0$ 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)

from where

\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 $\alpha_{j}$ we have

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

And, $\alpha^{\prime}{}_{i}$ being close enough to $\alpha_{i},\mathrm{\penalty 10000\ F}^{\prime}(x)=0$ has only one root in the interval $\left(\alpha_{j},\alpha_{j+1}\right)$ [if $j=i-1\quad a_{i}$ 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 given the demonstration for the 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 we take care to number the roots of the derived equation beforehand and not to change this numbering even if these roots pass through each other.
11. The property previously demonstrated 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 at
(3) We have as usual sig. $z=1,0,-1$ following that $z>,=,<0$ .
more equal to $\alpha$ , we have

\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 see that

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

We will in fact approach this minimum indefinitely by tending towards $-\infty$ the zeros of $h(x)$ . We can also avoid infinities by a simple transformation. We can assume $\alpha>0$ without restricting the generality. It is then sufficient to make the transformation $x\left\lvert\,\frac{1}{x}\right.$ on the equation $f(x)=0$ and apply the results of 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.

If $u>3$ 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 $c$ , we have
$\frac{2(a+b+c)+(n-3)(a+b)+\sqrt{2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]+(n-3)(n+1)(n-b)^{2}}}{2n}\geq$

\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 of the previous No. 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)

ot be $\phi(x)$ 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 is no root of the derived 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 propose to determine this number which is obviously smaller 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 ( $-1,1$ ) ; be $\alpha,\beta$ the largest and smallest root 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 $\alpha^{\prime}-\beta^{\prime}$ .

If $\alpha=1,\beta=-1$ We have $\alpha^{\prime}-\beta^{\prime}=2$ .
If $\alpha=1,\beta>-1$ Or $\alpha<1,\beta=-1$ from 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 different from zero by applying a reasoning analogous to that of No. 3 we show that the polynomial $f(x)$ can be replaced by another
⁽⁴⁾ This result is also deduced 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. Then taking

\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}

—

we have $\beta<\gamma<\alpha$ 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 come across a non-zero determinant or else by passing to 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 derived 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 ò

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

We can also see that we can state the following property more generally:

If the roots of the kth derivative $f^{(k)}(x)=0$ are all in 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 reported in Chap. I and II.

⁰⁰footnotetext: (5) I have just read the memorandum 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 MI Schur's inequality*

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

If $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)$ 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 also a triple minus root.

If $\mathrm{R}(f)>\mathrm{R}\left(f^{\prime}\right)$ 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 different from zero we can,… as a result of continuity, determine a polynomial $P_{1}(x)=\left(x-\alpha_{1}\right)\left(x-\beta_{1}\right)$ . such 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 increases 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 that

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

We 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 the system (12) is different from zero. The maximum is therefore only reached for the equation $(x-1)(x+2)^{2}=0$ .

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

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

We will demonstrate this by induction. We have seen that the property is true for degrees $3,4,\ldots,n-1$ and demonstrate it for the degree $n$ . Choosing the roots properly $\alpha$ And $\beta$ we see that the property is demonstrated by recurrence.

This property is due to MI Schur ( ⁶ ) who stated it in the following way:

If equation (1) has all its roots real 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 - outside the trivial case $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)==\mathrm{R}\left(f^{\prime\prime}\right)$ - that 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 again, as above, that the maximum of $\mathrm{R}\left(f^{(i+2)}\right)$ can only be achieved 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 of them, one could transform one into the other by a simple linear transformation and one would come across 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 „Zwei Sätze über algebrai she Gleichungen mit lauter reelslen Wurzeln ${}^{\text{" }}$ , 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 $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)>\mathrm{R}\left(f^{\prime\prime}\right)$ we have

\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 being possible only 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 then equation (14) is completely determined. The same is true if $\mathrm{R}(f)=\mathrm{R}\left(f^{\prime}\right)$ .

Let us suppose 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 $f^{\prime\prime}(x)$ got 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 deleting the case $p=0$ which leads to $\xi=-1$ , a case that 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}

has at least three variations. It follows that 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)

Let's 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. Afterwards it increases to a maximum and then decreases towards $b=-\infty$ .

It follows 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 exists a number $\mid\xi_{1}<-1$ such as for $\xi$ included in $\left(-1,\xi_{1}\right)$ equation (16) has three real roots, two of which are $<-(n-1)$ . If $\xi=\xi_{4}$ 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 root different from the multiple root.

Examining the ratio (18) shows us the existence of a number $\xi_{2}\leq\xi_{1}$ such 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 actually have $\xi_{2}<\xi_{1}$ . In any case the discriminant cannot be cancelled 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 proposition we had in mind.

An equation of the form (15) with $a\geq b\geq c$ is completely determined by knowledge of 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, etc. derivative.

In the following we will show the extremal properties of the 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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NOTES ON HIGHER ORDER CONVEX FUNCTIONS (I)

CHAPTER I.

On equations whose first three coefficients are given.

CHAPTER II.

On the equations whose first four coefficients are given

CHAPTER III.

On the roots of the derived equation

CHAPTER IV.

On MI Schur's inequality*

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