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     In this work, we consider only algebraic equations with real coefficients. If the coefficient of the highest-degree term of such an equation is equal to 1, the roots are continuous functions with respect to the other coefficients. It follows, in particular, that if such an equation has exactly N roots in the neighborhood of a certain point, and if N-1 of these roots are real, then the Nth root is also necessarily real. We will also consider the fact that if a sequence of polynomials of degree$N$, having all their zeros real, tends towards a limiting polynomial, this limit is of degree$\leq N$and has all its zeros real. By taking the limit, some of the roots of the limiting equation may disappear at infinity, and then the degree is less than$N$.

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Given a polynomial$\mathrm{P}(x)$of degree$n(n\geq 1)$We propose to find a condition for the existence of another polynomial$Q(x)$such that the derivative of the equation$P(x) ⇔ Q(x) = 0$has all its real roots. More precisely, we will establish that if such a polynomial$Q(x)$If a polynomial exists, we can find another such that the derivative of the product has a certain limited number, a number that depends only on the degree of the polynomial.$\mathrm{P}(x)$, of distinct roots.

In the general case, the polynomial$\mathrm{P}(x)$can be in the form

Or,$q_{\nu}\geq 2,v=1,2,\ldots,r;\beta_{1},\beta_{2},\ldots,\beta_{r}$are real and distinct and$\mathrm{P}_{1}(x)$is a polynomial of degree$n-\sum_{\nu=1}^{r}q_{\nu}$whose real zeros are all distinct and different from$\beta_{\nu}$. Let us pose$k=n-\sum_{\nu=1}^{r}q_{\nu}+r$the number$k$then represents the sum of the number of distinct real zeros and the number of complex zeros of the polynomial$\mathrm{P}(x)$.

Suppose there exists a polynomial$Q(x)$of degree$m$verifying the property, therefore a polynomial

such as if$\mathrm{F}(x)=\mathrm{P}(x)\cdot\mathrm{Q}(x)$, the equation

has all its real roots.
We then have
${}^{T}{}^{\prime}(x)=\mathrm{C}\left(x-\beta_{1}\right)^{q_{1}-1}\left(x-\beta_{2}\right)^{q_{2}\cdot 1}\ldots\left(x-\beta_{r}\right)^{qr-1}\left(x-\alpha_{1}\right)^{p_{1}}\left(x-\alpha_{2}\right)^{p_{2}}\ldots\left(x-\alpha_{s}\right)^{p_{s}}$Or,$\quad p_{\mu}\geq 1,\mu=1,2,\ldots s,\sum_{\nu=1}^{r}q_{\nu}-r+\sum_{\mu=1}^{s}p_{\mu}=n+m-1$,
$\mathscr{H}_{1}<\alpha_{2}<\ldots<\alpha_{s}$and C is a constant.

2. - Let's assume, for the moment, that the zeros$\alpha_{\mu}$are all distinct from zeros$\beta_{\nu}$and suppose that$s\geq k$.

Let's first replace the polynomial$Q(x)$by a polynomial

whose coefficients$\xi_{1},\xi_{2},\ldots,\xi_{m}$are variables.
Let's assume$\Phi(x)=\mathrm{P}(x).\phi(x)$Let's write the conditions
(2)

$j_{\mu}=1,2,\ldots,p_{\mu}-1,\quad\mu=1,2,\ldots,s\quad$(none$j_{\mu}$if$p_{\mu}=1$)
$j_{\mu}=p_{\mu},\quad\mu=k+1,\quad k+2,\ldots,s\quad$(none if$s=k$).

Let us consider the systems of$m$equations obtained by successively adding to system (2) the equations, linear with respect to the unknowns$\xi_{\mathrm{j}}$,

Neither equation (2) nor (3) reduces identically to zero, since otherwise$\mathrm{P}(x)$should have at least one zero$\alpha_{\mu}$with a degree of multiplicity at least equal to 2. The total system formed by equations (2) and (3) is compatible since it is, by construction, satisfied by the values

:either the first one which, with (2), gives a system of$m$equations with a non-zero determinant. We can then, among equations (2) and some$\sigma-1$first equations (3), choose$m-1$that are linearly independent and that, with (5), give a system with a non-zero determinant. The neglected equations will be consequences of these$m-1$equations.

Let us consider this last system of$m$equations and substitute the following for equation (5):

Or,$\varepsilon$is a positive number. The determinant of the system thus obtained will then be different from zero, at least for sufficiently small values ​​of$\varepsilon$Solving this system, we find for the$\xi_{\mathrm{j}}$of _S values ​​that are continuous functions of$\varepsilon$For$\varepsilon$small enough and reduce to the initial values ​​(4) for$\varepsilon=0$.

It follows that we can determine the polynomial$\phi(x)$in such a way
that$\Phi^{\prime}(x)=0$has the roots indicated, with their order of multiplicities, by the equalities (2) and in addition$k$other roots$\alpha_{1}^{\prime},\alpha_{2}^{\prime},\ldots,\alpha^{\prime}{}_{k}$Here we have$\alpha_{1}^{\prime}=\alpha_{1},\alpha_{2}^{\prime}=\alpha_{2},\ldots,\alpha_{\sigma-1}^{\prime}=\alpha_{\sigma-1},\alpha_{\sigma}^{\prime}=\alpha_{\sigma}+\varepsilon$And$\alpha_{\sigma+1}^{\prime}$,$\alpha^{\prime}{}_{\sigma+2},\ldots,\alpha^{\prime}{}_{k}$are continuous functions of$\varepsilon$and are reduced$\alpha_{\sigma+1},\alpha_{\sigma+2},\ldots,\alpha_{k}$For$\varepsilon=0$. For$\varepsilon$These roots are small enough, they are real, and we have

Let's grow$\varepsilon$Starting from the initial value 0, two cases are then possible a priori.
$1^{0}$For a first value$\varepsilon=\varepsilon_{1}>0$There are coincidences between the roots (7). Let's give to$\varepsilon$this value and either$Q_{4}(x)$the polynomial$\phi(x)$correspond. The product$\mathrm{F}_{1}(x)=\mathrm{P}(x).\mathrm{Q}_{1}(x)$then enjoys the property that the derivative equation$\mathrm{F}^{\prime}{}_{1}(x)=0$still has all its real roots. The$p_{1}+p_{2}+\cdots+p_{\sigma}-1$The first roots of this equation coincide with the$p_{1}+p_{2}+\ldots+p_{\sigma}-1$first roots of equation (1) but its$\left(p_{1}+p_{2}+\ldots+p_{\sigma}\right)^{\text{ome }}$root is greater than ($\left.p_{1}+p_{2}+\ldots+p_{\sigma}\right)^{\text{èmu }}$square root of (1).
20. There is a first value$0<\varepsilon_{2}<\varepsilon_{1}$such as if$\varepsilon\rightarrow\varepsilon_{2}$some of the coefficients$\xi_{\mathrm{j}}$become infinite. In this case, by making tend$\varepsilon$towards$\varepsilon_{2}$Inequalities (7) persist, but some are being eradicated$\alpha^{\prime}\mu(\mu>\sigma)$tend towards infinity. By taking the limit, we find a polynomial$Q^{*}(x)$of a smaller degree than$m$such as if$\mathrm{F}^{*}(x)=\mathrm{P}(x)\cdot\mathrm{Q}^{*}(x)$the equation$\mathrm{F}^{*\prime}(x)=0$has all its real roots

Note that we are still in this situation if$\varepsilon_{1}$does not exist, then$\varepsilon_{2}$may be equal to$+\infty$4.
- If for the polynomial$\mathrm{F}_{1}(x)$we still have$s\geq k$We repeat the procedure used on this polynomial. We should eventually arrive at the following property

If we have$s\geq k$The following three cases may occur:
10. We encounter a polynomial of the type$\mathrm{F}(x)$for which$\mathrm{s}<k_{\text{a }}$2.
We arrive at a polynomial of the type$\mathrm{F}^{*}(x)=\mathrm{P}(x)\cdot\mathrm{Q}^{*}(x)$oì$\mathrm{Q}^{*}(x)$is of degree < m.
$3^{\circ}$We arrive at a polynomial of the type$\mathrm{F}_{1}(x)$for which the smallest root of the derivative equation$\mathrm{F}_{1}^{\prime}(x)=0$is larger than$\alpha_{1}$.

The case$3^{0}$means that we arrive at$\sigma=1$Indeed, if none of the cases$1^{0},2^{0},3^{0}$doesn't happen for$\mathrm{F}_{1}(x)$we repeat the process by deducing from$\mathrm{F}_{1}(x)$A$\mathrm{F}_{2}(x)$, of$\mathrm{F}_{2}(x)$A$\mathrm{F}_{3}(x)$and so on. If we cannot prove the property after a finite number of such operations, we consider the infinite sequence$\mathrm{F}_{1}(x),\mathrm{F}_{2}(x),\ldots,\mathrm{F}_{\nu}(x),\ldots$

Doing$\nu\rightarrow\infty$we arrive at a limiting equation$\mathrm{F}_{\omega}(x)=0$of the same ^type . If$\mathrm{F}_{\omega}(x)$If the property does not hold, we repeat the process on this polynomial and so on indefinitely. We thus form a set of polynomials of the type$F(x)$such that none of them satisfies the property. There is a corresponding set formed by the first roots of the derived equations, another set formed by the second roots of the derived equations, and so on. One of these sets is unbounded (above all) by the very definition of the set of polynomials of the type$\mathrm{F}(x)$However, this contradicts the assumption that we do not arrive at the case$2^{0}$.

The property is therefore proven.
5. - Using these observations, we can now prove the following property:

If there exists a polynomial$Q(x)$(of degree$m$) such that the derivative of the equation$\mathrm{P}(x),\mathrm{Q}(x)=0$Given all its real roots, there certainly exists a polynomial$\mathrm{R}(x)$(of degree$\leq m$) such that the derivative of the equation$\mathrm{P}(x)\cdot\mathrm{R}(x)=0$has all its real roots, of which at most$k+r-1$distinct.

In the case$1^{0}$The property is demonstrated.
In the case$3^{0}$We continue the process. Reasoning as above, we see that we must arrive at$1^{0}$Or$9^{0}$.

In the case$2^{0}$we lowered the degree of the polynomial$Q(x)$and we then combine the demonstration with a complete inductive argument by noting that if$m=0$we are still in the case$1^{0}$since then$s\leq\sum_{\mu=1}^{s}p_{\mu}=n-\sum_{\nu=1}^{r}q_{\nu}+r-1=k-1$.

The property is therefore completely proven.
6. - Let's not forget that we assumed that none of the roots$\alpha_{\mu}$does not coincide with a root$\beta_{\nu}$Otherwise, _γ, we modify the polynomials .$\mathrm{P}(x)$And$Q(x)$For example, if$\alpha_{\mu}=\beta_{\nu}$,$\mathrm{F}(x)$is divisible by$\left(x-\beta_{\nu}\right)^{q_{\nu}+p_{\mu}}$SO$Q(x)$by$\left(x-\beta_{\nu}\right)^{p_{\mu}}$We write$\left(x-\beta_{\nu}\right)^{p_{\mu}}\mathrm{P}(x)$instead of$\mathrm{P}(x)$and we take for$\mathrm{Q}(x)$its quotient by$\left(x-\beta_{\nu}\right)^{p_{\mu}}$We make this change for each couple.$\alpha_{\mu},\beta_{\nu}$coinciding and every time this happens. Note that: the numbers$k$And$r$do not change with this modification. We can therefore deduce that the property is completely general.

Note also that, without specifying the nature of the zeros of the polynomial$\mathrm{P}(x)$We can state the following property,

If there exists a polynomial$\mathrm{Q}(x)$such that the derivative of the equation$\mathrm{P}(x)\cdot\mathrm{Q}(x)=0$Given all its real roots, there certainly exists a polynomial$\mathrm{R}(x)$such that the derivative of the equation$\mathrm{P}(x).\mathrm{R}(x)=0$has all its real roots, of which at most$n-1$distinct.

%. - Consider a polynomial of the form

$Q(x)$being another polynomial,$c,d\neq 0$real constants and$p$a positive integer.

We propose to demonstrate that,
The derivative equation$\mathrm{F}^{\prime}(x)=0$cannot have all its real roots.

First, note that the property is independent of the values ​​of the constants.$c$And$d$since a real and linear transformation does not affect the reality of the roots of the derivative.

To demonstrate the property, let's assume the opposite. There then exists a polynomial$Q(x)$of degree$m$

such that the derivative of the polynomial$F(x)=x^{p}\left(x^{2}+1\right)\cdot Q(x)$either of the form

Or,$p_{\mu}\geq 1,\mu=1,2,\ldots,s,p_{1}+p_{2}+\ldots+p_{s}=m+2$And$\alpha_{1},\alpha_{2},\ldots$
$\alpha_{s}$are real, distinct, and nonzero. We can assume that$\alpha_{1}$is positive and that the others$\alpha_{\mu}$are all negative or positive and greater than$\alpha_{1}$We fix two of these roots$\alpha_{\gamma}$And$\alpha_{\delta}$. If$p_{1}>1$we take$\gamma=1,\delta=2$and if$p_{1}=1$we take$\gamma=3$,$\delta=2$8.
- Suppose$s>1$Let's introduce instead$Q(x)$the polynomial

with variable coefficients and let$\Phi(x)=(x-\varepsilon)^{p}\left[(x-\varepsilon)^{2}+1\right]\phi(x),\varepsilon$being a positive number.

Let's determine the polynomial$\phi(x)$by conditions
(8)$\quad\Phi^{\left(j_{\mu}\right)}\left(\alpha_{\mu}\right)=0,\quad j_{\mu}=1,2,\ldots,p_{\mu}-1,\mu=\gamma,\delta\quad$(none if$p_{\mu}=1$)$j_{\mu}=1,2,\ldots,p_{\mu}$for all$\mu$different$\gamma$And$\delta$, which is a linear system with respect to$m$unknowns$\xi_{1},\xi_{2},\ldots,\xi_{m}$. Let us designate by$D(\varepsilon)$the determinant of this system.
9. - Let us first suppose$D(0)=0(1)$System (8) is indeterminate. Therefore, there exist two distinct polynomials.$\phi_{1}(x),\phi_{2}(x)$such as if$\Phi_{1}(x)=x^{p}\left(x^{2}+1\right)\phi_{1}(x),\Phi_{2}(x)=x^{n}\left(x^{2}+1\right)\phi_{2}(x)$, polynomials$\Phi_{1}^{\prime}(x),\Phi_{2}^{\prime}(x)$both be divisible by$\frac{\mathrm{F}^{\prime}(x)}{\left(x-\alpha_{\gamma}\right)\left(x-\alpha_{\delta}\right)}$If we form the difference$\Phi_{3}(x)=x^{p}\left(x^{2}+1\right)$. [$\phi_{1}(x)-\phi_{2}(x)$we see that the polynomial$\Phi^{\prime}{}_{3}(x)$of degree$<m+p+1$is divisible by the polynomial$\frac{\mathrm{F}^{\prime}(x)}{\left(x-\alpha_{\gamma}\right)\left(x-\alpha_{\delta}\right)}$of degree$m+p-1$. It follows that$\Phi_{3}(x)$has all its real zeros.

The problem is thus reduced to the case where the degree m of the polynomial$\mathrm{Q}(x)$has become smaller.
10. - Now suppose that$D(0)\neq 0$. SO$\xi_{1},\xi_{2},\ldots\xi_{m}$will be continuous functions of$\varepsilon$in the vicinity of$\varepsilon=0$and are respectively reduced to$a_{1},a_{2},\ldots,a_{m}$For$\varepsilon=0$. The equation$\Phi^{\prime}(x)=0$will have the roots$\alpha_{1},\alpha_{2},\ldots,\alpha_{s}$with the order of multiplicity indicated by (8) and in addition two roots$\alpha_{\gamma}^{\prime},\alpha_{\delta}^{\prime}$which are continuous in$\varepsilon$for sufficiently small values ​​of$\varepsilon$and are reduced to$\alpha_{\gamma}$And$\alpha_{\delta}$For$\varepsilon=0$.