VOLUME 2 (25)
FASCICOL 2

1960
${}^{\mathrm{C}}\mathrm{L}^{\mathrm{U}}\mathrm{U}\mathrm{J}$

Vol. 2 (25), fasc. 2
1960

In Cluj,
W. A. ​​Markov, in his work [2] on the generalization of the famous AA Markov inequality, gave, as a preliminary lemma, the following theorem:

If the roots of two polynomials of degree$n$, having all their real roots, separate, the same is true for the roots of the derivatives of these polynomials.

In the second part of this work we will give a proof of this theorem. Our proof differs slightly from that of Wa Markov and also from that of P. Monteil [3] given almost 30 years ago in this same journal.

The demonstration we present is based on the continuity and monotonicity of the roots of the derivative of a polynomial with all its roots being real, with respect to the roots of the polynomial. In the first part of this work, we will analyze this property of monotonicity.

Finally, in the third part of this work, we will give a new theorem on polynomials having all their roots real, analogous to that of WA Markov, cited above.

We consider only polynomials of one variable whose roots are all real, and by the degree of a polynomial we always mean its actual degree, even if these properties are not explicitly stated. Accents denote derivatives. We can consider as equal two polynomials that differ only by a non-zero multiplicative constant.

1. 1.

   If a polynomial has all its roots real, its derivative also has all its roots real. There is an important, well-known property that separates the roots of the derivative from those of the polynomial. We will use this property in what follows.

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The roots of a polynomial whose highest coefficient is equal to 1 (i.e., of the form$x^{n}+$degree polynomial$<n$The coefficients of the polynomial are continuous functions with respect to the coefficients, and the coefficients are continuous functions (of the polynomials) with respect to the roots of the polynomial. Considering the relationships between the roots and coefficients of a polynomial, we deduce the continuity of the roots of the derivative with respect to the roots of the polynomial.
2. The monotonicity property of the roots of the derivative with respect to the roots of the polynomial can be stated as follows:

The roots of the derivative are non-decreasing functions of the roots of the polynomial.

This property is well known and has been used a lot, for example by Laguerre in his research on polynomials having all their roots real.

To better highlight the monotonicity property, we introduce the relationship$P\xrightarrow{c}Q$between two polynomials, which occurs if and only if:
$1^{0}$I polynomials$P,Q$are of the same degree$n\geqq 1$.
$2^{\circ}$The respective roots

of these polynomials satisfy the inequalities

This relationship is (reflexive and) transitive. It is unnecessary to consider the case$n=0$when the previous relationship makes no sense. If$n=1$It suffices to mention only the inequality (2) of definition and we see that in this case at least one of the relations$P\stackrel{{\scriptstyle\mathrm{c}}}Q,Q\stackrel{{\scriptstyle\mathrm{c}}}P$is still true. For all$n>1$We can construct polynomials$P,Q$such that none of the relationships$P\xrightarrow{\mathrm{c}}Q,Q\xrightarrow{\mathrm{c}}P$that it is not true.
$I_{1}a$The monotonicity property of the roots of the derivative with respect to those of the polynomial is then expressed by the

THEOREM 1. If$P$,$Q$are two polynomials of degree$n>1$, of$P\xrightarrow{\mathrm{c}}Q$it follows that$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$.

We also introduce the relationship$P\xrightarrow{cc}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$are of the same degree$n\geqq 1$and both have very simple roots.
$2^{\circ}$The respective roots
(1')

of these polynomials satisfy the inequalities

This relation, which is also transitive, is a special case of the previous relation and is obtained when everywhere in (1) and (2) the sign$\lesssim$is replaced by$<$.

The two relations considered are also linked by a kind of mixed transitivity, analogous to the corresponding property of inequality relations$<$And$\leqq$. If$P\xrightarrow{\mathrm{cc}}R,R\xrightarrow{\mathrm{c}}Q$and if$Q$to all its simple roots, we have$P\xrightarrow{\mathrm{cc}}Q$. Of$P\xrightarrow{\mathrm{cc}}R,R\xrightarrow{\mathrm{c}}S,S\xrightarrow{\mathrm{cc}}Q$it follows that$P\xrightarrow{\mathrm{cc}}Q$.

If the roots of$P,Q$are continuous functions of a parameter$\lambda$on an interval that contains the point$\lambda_{0}$and if the relationship$P\xrightarrow{cc}Q$is checked for$\lambda\neq\lambda_{0}$, We have$P\stackrel{{\scriptstyle c}}Q$but not in general$P\xrightarrow{cc}Q$, For$\lambda=\lambda_{0}$This property also holds true for pairs of relationships.$\xrightarrow[\rightarrow]{s},\xrightarrow{ss};\xrightarrow{m};\xrightarrow{mn},\xrightarrow{mc}$which we will introduce later.

We have 1st
THLOREME 2. If$P$,$Q$are two polynomials of degree$n>1$, of$P\rightarrow Q$it follows that$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$3.
We will first demonstrate that Theorem 1 follows from Theorem 2. Indeed, let$P\xrightarrow{c}Q$(1) the roots of the polynomials$P,Q,n>1$And

the respective roots of the polynomials$P^{\prime},Q^{\prime}$Let us consider polynomials$P_{\varepsilon},Q_{\varepsilon}$degree$n$, having respectively the roots$x_{i}+i\varepsilon,i=1,2,\ldots n$,$y_{i}+(i+1)\varepsilon,i=1,2,\ldots,n$Or$\varepsilon$is a positive number. Polynomials
$P_{\varepsilon},Q_{\varepsilon}$all have their simple roots, we have$P_{\varepsilon}\xrightarrow{\mathrm{cc}}Q_{e}$and if

are respectively the roots of the polynomials$P_{e}^{\prime},Q_{e}^{\prime}$, We have

If we assume that Theorem 2 is true, it follows that$P_{\varepsilon}^{\prime}\xrightarrow{cc}Q_{\varepsilon}^{\prime}$and from (4) we deduce, by$\varepsilon\rightarrow 0,P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$We have thus demonstrated that Theorem 1 is a continuation of Theorem 2.
4. It remains to prove Theorem 2. Let$P,Q$two polynomials of degree$n>1$such as$P\xrightarrow{cc}Q$and be$\left(1^{\prime}\right)$the respective roots of these polynomials. Let$P_{i}$a polynomial of degree$n$having as its roots$x_{1},x_{2},\ldots,x_{n-i},y_{n-i+1},y_{n-i+2},\ldots,y_{n}$, For$i=1,2,\ldots,n$The polynomial$P_{0}$is equal to$P$And$P_{n}$is equal to$Q$Your polynomials$P_{i}$all have their simple roots, but we generally only have$P_{i\rightarrow+}^{\mathrm{c}}P_{i+1},i=0,1$,$\ldots.n-1$If we demonstrate that

Therefore, due to the transitivity of the relation in question, it follows that$P^{\prime cc}\rightarrow Q^{\prime}$and theorem 2 is proven.

It remains to demonstrate relations (5). These relations result from
$I_{1}$em me 1. The roots of the derivative of a polynomial whose roots are all real and simple, are functions that are increasing with respect to each of the roots of the polynomial.

Either$g$a polynomial of degree$n(\geqq 1)$having all its roots$\alpha_{1}<\alpha_{2}<<\ldots<\alpha_{n}$real and simple. The property of Lemma 1 amounts to the fact that each of the roots of the derivative of the polynomial$f_{a}=(x-\alpha)g$is an increasing function of$\alpha$These roots$\beta_{1}<\beta_{2}<\ldots<\beta_{n}$are continuous functions of$\alpha$and remain distinct. We can first demonstrate that they are strictly monotonic functions of$\alpha$In effect, if, for example,$\beta_{k}$would not be a strictly monotonic function of$\alpha$we could find two different values$\alpha^{\prime},\alpha^{\prime\prime}$of$\alpha$for which the polynomials

have a common root equal to$\beta_{k}$This is impossible because any common root of the polynomials (6) would have to be a common root of the polynomials$g,g^{\prime}$which contradicts the hypothesis that$g$has only simple roots. The strict monotony of the roots$\beta_{i}$functions of$\alpha$, is thus demonstrated. It only remains to specify the meaning of this monotonicity. If we note that the roots of the derivative are separated by those of the polynomial and if we take into account

we find it good that the$\beta_{i}$are increasing functions of$\alpha$
Lemma 1 is therefore proven. Other
proofs of Lemma 1 can be given. In particular, proofs based on considerations similar to those used in the second and third parts of this work can be given. We will not dwell on these proofs here.

Note. If

are the roots of the polynomial$g^{\prime}$the roots$\beta_{i},i=1,2,\ldots,n$vary respectively within the intervals$\left(-\infty,\alpha_{1}^{\prime}\right],\left[\alpha_{i-1}^{\prime},\alpha_{i}^{\prime}\right],i=2,3,\ldots,n-1,\left[\alpha_{n-1}^{\prime},\infty\right)$, if$n>2$. If$n=1$the root$\beta_{1}$varies from$-\infty$has$\infty$and if$n=2$the roots$\beta_{1}$,$\beta_{2}$vary respectively in the intervals$\left(-\infty,\frac{\alpha_{1}+\alpha_{2}}{2}\right],\left[\frac{\alpha_{1}+\alpha_{2}}{2},\infty\right)$.

1. 5.

   We will now deal with the aforementioned proof of WA Markov's theorem.

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We introduce the relationship$P\xrightarrow{s}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$have the same degree$n\geqq 1$.
$2^{\circ}$The respective roots (1) of these polynomials satisfy the inequalities

If$P\xrightarrow{s}Q$Or$Q\xrightarrow{s}P$we can say that the roots of the polynomials$P,Q$separate. Generally, from$P\xrightarrow{s}Q$It follows that$P\xrightarrow{c}Q$and, for$n=1$relationships$P\xrightarrow{s}Q,P\xrightarrow{\mathrm{c}}Q$are equivalent. Taking into account a previous remark, we can find, for all$n>1$, two polynomials$P,Q$degree$n$such that none of the relationships$P\xrightarrow{S}Q$,$Q\xrightarrow{s}P$not to be verified.

10. W. A. ​​Markov's theorem can be stated as follows:
THEOREM 3. If$P,Q$are two polynomials of degree$n>1$, of$P\xrightarrow{\mathrm{s}}Q$it follows that$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$.

If$n=2$Theorem 3 follows from Theorem 1. Indeed, in this case, of$P\xrightarrow{\mathrm{s}}Q$it follows that$P\xrightarrow{\mathrm{c}}Q$Therefore, by virtue of Theorem 1, it follows that$P^{\prime}\xrightarrow{\mathrm{c}}Q^{\prime}$But this relationship is equivalent to$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$and ownership is demonstrated.

We also introduce the relationship$P\xrightarrow{ss}Q$between two polynomials, which occurs if and only if:
$1^{\circ}$Polynomials$P,Q$have the same degree$n\geq 1$and both have their very simple roots.

$2^{\circ}$The respective roots$\left(1^{\prime}\right)$of these polynomials satisfy the inequalities

Of$P\xrightarrow{\mathrm{ss}}Q$it follows that$P\xrightarrow{\mathrm{cc}}Q$and for$u=1$These relationships are equivalent.

We have the following special case of W. A. ​​Markov's theorem:
THEOREM 4. If$P,Q$are two polynomials of degree$n>1$, of$P\xrightarrow{\mathrm{ss}}Q$it follows that$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$.

It is demonstrated, as above, that for$n=2$Theorem 4 follows from Theorem 2.

6. It suffices to prove Theorem 4, since Theorem 3 then follows from it. To see this, we proceed as in No. 3, where we showed that Theorem 1 follows from Theorem 2.
If we have$P\xrightarrow{\mathrm{s}}Q$and if we now consider polynomials$P_{\varepsilon},Q_{\varepsilon}$having respectively as roots$x_{i}+(2i-1)\varepsilon,i=1,2,\ldots,n$,$y_{i}+2i\varepsilon,i=1,2,\ldots,n$, Or$\varepsilon$is a positive number, we have$P_{\varepsilon}\xrightarrow{\mathrm{ss}}Q_{\varepsilon}$If we assume that Theorem 4 is true, it follows that$P_{\varepsilon}^{\prime}\xrightarrow{\mathrm{ss}}Q_{\varepsilon}^{\prime}$If we do$\varepsilon\rightarrow 0$, the roots of$P_{\varepsilon}^{\prime},Q_{\varepsilon}^{\prime}$tend towards the respective roots of$P^{\prime},Q^{\prime}$and we deduce that$P^{\prime}\xrightarrow{\mathrm{s}}Q^{\prime}$Theorem 3 is therefore proven.

7. We can prove Theorem 4, based on Theorem 2, Lemma 1 and the continuity of the roots of the derivative.

If$P\xrightarrow{\mathrm{ss}}Q$it follows that$P\xrightarrow{\mathrm{cc}}Q$so also$P^{\prime}\xrightarrow{\mathrm{cc}}Q^{\prime}$To demonstrate that we have more$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$It suffices to show that the derivatives of the polynomials$P,Q$(which all have simple roots) cannot have common roots. Indeed, it is easy to see that then the relationship$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$maintains itself while the roots of$P$grow towards the respective roots of$Q$.

But, if$(x_{i}^{\prime})$are the roots of the polynomials$P,Q$degree$n$the relationship$P\xrightarrow{\mathrm{ss}}Q$is equivalent to equality

Or$a$is a non-zero constant and$a_{1},a_{2},\ldots,a_{n}$constants other than zero and of the same sign. Moreover, the product$aa_{i}$is of opposite sign with the highest coefficient of$P$, therefore with$P$For$x$very large.

By derivation of (8) it follows that
(9)

We can see that if$P^{\prime},Q^{\prime}$had a common root, this root must also make the polynomial zero$P$which is impossible since, by hypothesis,$P$has only simple roots.

8. Relationships$P\xrightarrow{\mathrm{s}}Q$And$P\xrightarrow{\mathrm{ss}}Q$can be extended to the case where the polynomial$P$is of degree$n$and the polynomial$Q$degree$n-1$. If$n>1$and if

are respectively the roots of$P$and$Q$the relationship$P\xrightarrow{\mathrm{s}}Q$takes place if and only if

We can also say that the roots of polynomials$P,Q$They separate.
The relationship$P\xrightarrow{\mathrm{ss}}Q$takes place if and only if, in addition, the roots of the polynomials$P,Q$are all simple and if instead of (11) we have the inequalities

We then have
Consequence 1. If$P$is a polynomial of degree$n$And$Q$a polynomial of degree$n-1,n>1$, of$P\xrightarrow{s}Q$it follows that$P^{\prime}\xrightarrow{s}Q^{\prime}$.

The property follows from Theorem 3 by taking the limit. To see this, let (10) be the roots of$P$And$Q$which verify the relationship$P\xrightarrow{s}Q$Consider the polynomial$R=\left(x-y_{n}\right)Q$degree$n$. If$x_{n}\leqq y_{n}$no we have$P\xrightarrow{\mathrm{~s}}R$, hence, taking into account the theorem$3,P^{\prime}\xrightarrow{\mathrm{s}}R^{\prime}$If we do$y_{11}\rightarrow\infty$, one of the roots of$R^{\prime}$(the largest) tends towards$\infty$and the others towards the respective roots of$Q^{\prime}$Considering the continuity of the roots of the derivative with respect to the roots of the polynomial, we see that if we do$y_{n}\rightarrow\infty$, of$P^{\prime}\rightarrow R^{\prime}$it follows that$P^{\prime}\rightarrow Q^{\prime}$.

We also have
Consequence 2. If$P$is a polynomial of degree$n$And$Q$a polynomial of degree$n-1,n>1$, of$P\xrightarrow{\mathrm{ss}}Q$it follows that$P^{\prime}\xrightarrow{\mathrm{ss}}Q^{\prime}$.

This property follows from Theorem 4 in the same way as Consequence 1 of Theorem 3. We construct, as above, the polynomial$R$. If$P\xrightarrow{ss}Q$and if$x_{n}<y_{n}$, We have$P\xrightarrow{ss}R$, therefore, as a consequence of the theorem$4,P^{\prime}\xrightarrow{s}R^{\prime}$From there, if we do$j^{\prime}n\rightarrow\infty$it follows that$P^{\prime}\rightarrow Q^{\prime}$To demonstrate that we even$P^{\prime}\rightarrow Q^{\prime}$it suffices to demonstrate that if$P\xrightarrow{ss}Q$, polynomials$P^{\prime},Q^{\prime}$cannot have common roots.

If$P\xrightarrow{\mathrm{ss}}Q$we have formula (8), where$a=0$And$a_{i},i=1,2,\ldots,n$are$u$constants other than zero and of the same sign. Formula (9) shows us that$P^{\prime},Q^{\prime}$cannot have common roots.
$L_{\text{a conséquence }}2$is demonstrated.
9. As a function, consider a sequence of orthogonal polynomials

We know that the roots of$\Pi_{n}$are all real, simple, and the roots of$\Pi_{n-1}$are separated in the strict sense by the roots of$\Pi_{n}$So we have$\Pi_{n}\xrightarrow{\mathrm{ss}}\Pi_{n-1}$For$n>1$From consequence 2, it therefore follows that

Consequence 3. If$\Pi_{n-1},\Pi_{n}$are two consecutive terms of a sequence of orthogonal polynomials ($n>1$), We have$\Pi_{n}^{\prime}\xrightarrow{ss}\Pi_{n-1}^{\prime}$.

1. 10.

   We will deal with a theorem analogous to WA Markov's theorem.

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We introduce the relationship$P\xrightarrow{m}Q$between two polynomials, which occurs if and only if:
$10^{\circ}$Polynomials$P,Q$are of the same degree$n\geqq 1$.
$2^{\circ}$The roots (1) of these polynomials satisfy the inequalities
(12)$\quad x_{1}+x_{2}+\ldots+x_{i}\leqq y_{1}+y_{2}+\ldots+y_{i},\quad i=1,2,\ldots,n-1$
and equality

If$n=1$, we only have equality (13). If$n=1$the relationship$P\xrightarrow{m}Q$means that$P,Q$have the same root, therefore, according to the meaning adopted at the beginning of this work, they are equal. It is easy to see that for all$n>1$we can find two polynomials$P,Q$degree$n$such that none of the relationships$P\xrightarrow{\mathrm{~m}}Q,Q\xrightarrow{\mathrm{~m}}P$not to be verified.