Abstract
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T. Popoviciu
Institutul de Calcul
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T. Popoviciu, Asupra polinoamelor cu toate rădăcinile reale. Studii și cercetări științifice (Cluj), tom. III, nr. 1-2, pag. 7-10 (1952).
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Studii si Cercetari Matematice
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Academy of the Republic of S.R.
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1220-269X
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ON POLYNOMIALS WITH ALL REAL ROOTS
- -- EitherP(x)\mathrm{P}(x)a polynomial of (effective) degree n. This polynomial hasnndistinct roots or not. Successive derivativesP^(')(x),P^('')(x),dots,P^((n-1))(x)\mathrm{P}^{\prime}(x), \mathrm{P}^{\prime \prime}(x), \ldots, \mathrm{P}^{(n-1)}(x)have respectivelyn-1,n-2,dotsn-1, n-2, \ldots, 1 distinct roots or not. Therefore, considering all the roots of the polynomial and its successive derivatives, we have
1+2+dots+n=(n(n+1))/(2)1+2+\ldots+n=\frac{n(n+1)}{2}
such distinct roots or not.
The problem may arise of determining the numberNNof the distinct numbers between these(n(n+1))/(2)\frac{n(n+1)}{2}numbers thus obtained and which represent the roots considered. The number N is the number of distinct roots of the polynomialP(x)P^(')(x)dotsP^((n-1))(x)\mathrm{P}(x) \mathrm{P}^{\prime}(x) \ldots \mathrm{P}^{(n-1)}(x)obtained by taking the product of the given polynomial and its successive derivatives.
The problem may arise of determining the numberNNof the distinct numbers between these(n(n+1))/(2)\frac{n(n+1)}{2}numbers thus obtained and which represent the roots considered. The number N is the number of distinct roots of the polynomialP(x)P^(')(x)dotsP^((n-1))(x)\mathrm{P}(x) \mathrm{P}^{\prime}(x) \ldots \mathrm{P}^{(n-1)}(x)obtained by taking the product of the given polynomial and its successive derivatives.
We can obviously assumen > 1n>1It is clear
that ifP(x)\mathrm{P}(x)has all the roots confused we haveN=1\mathrm{N}=1Otherwise
we haveN > 1N>1and the question arises how can this inequality be specified in this case?
that ifP(x)\mathrm{P}(x)has all the roots confused we haveN=1\mathrm{N}=1Otherwise
we haveN > 1N>1and the question arises how can this inequality be specified in this case?
Assuming that the roots ofP(x)\mathrm{P}(x)are all real, we will prove that:
I. IfP(x)\mathrm{P}(x)has at least two distinct roots we haveN >= n+1\mathrm{N} \geq n+1.
I. IfP(x)\mathrm{P}(x)has at least two distinct roots we haveN >= n+1\mathrm{N} \geq n+1.
This property is very likely true even when the restriction on the reality of the roots is lifted, i.e. for any polynomial of a complex variable.
Property I does not depend on a linear transformation of the variablexx, so it will be true if the roots of the polynomial are located on a line in the complex plane. Also, the property does not depend, obviously, on a constant factor(!=0)(\neq 0)his/herP(x)\mathrm{P}(x)It follows that in the following proofs one can always assume that the first coefficient of the polynomial (ofx^(n)x^{n}) is equal to 1 and that ifP(x)\mathrm{P}(x)has at least two distinct roots, one of which is equal, e.g., to 0, and one to 1.
2. The proof of property I in the case of real roots is based on several properties that are consequences of Rolle's theorem.
2. The proof of property I in the case of real roots is based on several properties that are consequences of Rolle's theorem.
If a polynomial has all its roots real, its derivative also has all its roots real. Letthing(a_(o)),b_(o)\overrightarrow{a_{o}}, b_{o}the extreme roots, that is, the smallest and the largest root, ofP(x)\mathrm{P}(x)anda_(1),b_(1)a_{1}, b_{1}the analogous extreme roots of the derivativeP^(')(x)\mathrm{P}^{\prime}(x)Ifto the)to the} resp. b_(o)b_{o}is an at least double root ofP(x)\mathrm{P}(x), we havea_(1)=a_(0)a_{1}=a_{0}resp.b_(1)=b_(0)b_{1}=b_{0}Ifa_(0)a_{0}resp.b_(0)b_{0}If it is a simple root, we havea_(o) < a_(1)a_{o}<a_{1}resp.b_(1) < b_(o)b_{1}<b_{o}. In fine a_(1)a_{1}resp.b_(1)b_{1}is a simple root of the derivative if and only ifa_(0)a_{0}resp.b_(0)b_{0}is a simple or double root of the polynomial.
Let's assume thatP(x)\mathrm{P}(x)does not have all the roots confused and be, in general,a_(i)a_{i}the smallest one andb_(i)b_{i}the largest root of the derivativeP^((1))(x)\mathrm{P}^{(1)}(x)of the orderi,i=0,1dots,n-1i, i=0,1 \ldots, n-1We have obviouslya_(n-1)=b_(n-1)a_{n-1}=b_{n-1}, buta_(n-2)<<b_(n-2)a_{n-2}< <b_{n-2}This latter inequality results from the fact that if the derivative of a polynomial with all real roots has a root of orderi > 1i>1of multiplicity, the polynomial necessarily has this root of orderi+1i+1of multiplicity. In the hypothesis of the reality of all roots the equalitya_(n-2)=b_(n-2)a_{n-2}=b_{n-2}is therefore not compatible with the hypothesis that the roots of the polynomial are not all equal.
3. Let us now denote by E the number of distinct numbers among the numbers
3. Let us now denote by E the number of distinct numbers among the numbers
{:(1)a_(i)","b_(i)","i=0","1","dots;n_(--1):}\begin{equation*}
a_{i}, b_{i}, i=0,1, \ldots ; n_{--1} \tag{1}
\end{equation*}
From the above it follows that, ifto the)to the}is a root of orderllof multiplicity andb_(o)b_{o}a root of the orderkkof multiplicity of the polynomialP(x)\mathrm{P}(x), we have
{:[a_(0)=a_(1)=cdots=a_(l-1) < a_(l) < a_(l+1) < cdots < a_(n-2) < a_(n-1)=],[=b_(n-1) < b_(n-2) < cdots < b_(k) < b_(k-1)=b_(k-2)=cdots=b_(0)]:}\begin{gathered}
a_{0}=a_{1}=\cdots=a_{l-1}<a_{l}<a_{l+1}<\cdots<a_{n-2}<a_{n-1}= \\
=b_{n-1}<b_{n-2}<\cdots<b_{k}<b_{k-1}=b_{k-2}=\cdots=b_{0}
\end{gathered}
From this it follows that
E=2n-l-k+1.\mathrm{E}=2 n-l-k+1 .
But
{:(3)l+k <= n:}\begin{equation*}
l+k \leqq n \tag{3}
\end{equation*}
{:(si)N >= E.:}\begin{equation*}
\mathrm{N} \geqq \mathrm{E} . \tag{si}
\end{equation*}
But from (3) it follows thatE >= n+1E \geqq n+1, soN >= n+1N \geqq n+1and propertyIIis demonstrated.
4. We now propose to determine all the polynomialsP(x)\mathrm{P}(x)(with all real roots) of degreennfor which we haveN=n+1\mathrm{N}=n+1.
4. We now propose to determine all the polynomialsP(x)\mathrm{P}(x)(with all real roots) of degreennfor which we haveN=n+1\mathrm{N}=n+1.
From (2), (3), (4) it follows that this can only happen ifl+k=nl+k=n, so only if the polynomial has exactly two distinct roots, therefore, apart from a constant factor (!=0\neq 0) and apart from a linear transformation of the variable, only for polynomials of the form
P(x)=x^(n-k)(x-1)^(k),2k <= n.\mathrm{P}(x)=x^{n-k}(x-1)^{k}, 2 k \leqq n .
In this case we haveE=n+1\mathrm{E}=n+1and so that we haveN=n+1N=n+1it is necessary and sufficient that all the roots of any derivativeP^((i))(x)\mathrm{P}^{(i)}(x)not be different from the roots of (1) and this fori=1,2,dots,n-3i=1,2, \ldots, n-3.
Ifn >= 2,k=1n \geqq 2, k=1there are no such roots and so we then haveN=n+1N=n+1In this way the casesn=2,3n=2,3are exhausted.
Ifn >= 4,2 <= k <= (n)/(2),P(n-3)(x)n \geqq 4,2 \leqq k \leqq \frac{n}{2}, \mathrm{P}(n-3)(x)has distinct roots and a root different froma_(n-3)a_{n-3}andb_(n-3)b_{n-3}of this polynomial cannot coincide witha_(n-1)a_{n-1}. In order to haveN=n+1N=n+1it is therefore necessary that the polynomialsP^((n-3))(x),P^((n-1))(x)\mathrm{P}^{(n-3)}(x), \mathrm{P}^{(n-1)}(x)to have a common root. Buta_(n-1)=b_(n-1)=(k)/(n)a_{n-1}=b_{n-1}=\frac{k}{n}, that is, it is necessary to have
p^((n-3))((k)/(n))=0.p^{(n-3)}\left(\frac{k}{n}\right)=0 .
But
{:[{:[P^((n-3))(x)=((n-3)!)/(3!)[n(n-1)(n-2)x^(3)-3k(n-1)(n-2)x^(2)+:}],[+3k(k-1)(n-2)x-k(k-1)(k-2)]]:}],[" ducem "quad]:}\begin{aligned}
& \begin{aligned}
\mathrm{P}^{(n-3)}(x)=\frac{(n-3)!}{3!} & {\left[n(n-1)(n-2) x^{3}-3 k(n-1)(n-2) x^{2}+\right.} \\
& +3 k(k-1)(n-2) x-k(k-1)(k-2)]
\end{aligned} \\
& \text { ducem } \quad
\end{aligned}
P^((n-3))((k)/(n))=-(k(n-k)(n-2k))/(3)*((n-3)!)/(n^(2))\mathrm{P}^{(n-3)}\left(\frac{k}{n}\right)=-\frac{k(n-k)(n-2 k)}{3} \cdot \frac{(n-3)!}{n^{2}}
It follows that ifn >= 4,2 <= k < (n)/(2)n \geqq 4,2 \leqq k<\frac{n}{2}we certainly haveN > n+1\mathrm{N}>n+1It
remains to examine the casen=2kn=2 kIfk=2k=2, based on the above it can be seen that we haveN=n+1\mathrm{N}=n+1Ifk > 2k>2, his rootsP^((n-4))(x)\mathrm{P}^{(n-4)}(x)are all distinct and symmetrically placed opposite each othera_(-1)a_{-1}. From this it immediately follows that none of the roots distinct from the extreme roots ofP^((n-4))(x)\mathrm{P}^{(n-4)}(x)does not coincide witha_(n-1)a_{n-1}and that in order to haveN=n+1\mathrm{N}=n+1it is necessary that these roots coincide witha_(n-2),b_(n-2)a_{n-2}, b_{n-2}respectively, so thatP^((n-4))(x)\mathrm{P}^{(n-4)}(x)to be divisible byP^((n-2))(x)\mathrm{P}^{(n-2)}(x)For simplification we can now take
remains to examine the casen=2kn=2 kIfk=2k=2, based on the above it can be seen that we haveN=n+1\mathrm{N}=n+1Ifk > 2k>2, his rootsP^((n-4))(x)\mathrm{P}^{(n-4)}(x)are all distinct and symmetrically placed opposite each othera_(-1)a_{-1}. From this it immediately follows that none of the roots distinct from the extreme roots ofP^((n-4))(x)\mathrm{P}^{(n-4)}(x)does not coincide witha_(n-1)a_{n-1}and that in order to haveN=n+1\mathrm{N}=n+1it is necessary that these roots coincide witha_(n-2),b_(n-2)a_{n-2}, b_{n-2}respectively, so thatP^((n-4))(x)\mathrm{P}^{(n-4)}(x)to be divisible byP^((n-2))(x)\mathrm{P}^{(n-2)}(x)For simplification we can now take
P(x)=(x^(2)-1)^(k)\mathrm{P}(x)=\left(x^{2}-1\right)^{k}
We have snow.
{:[P(n-4)(x)=((2k-4)!k(k-1))/(3!)[(2k-1)(2k-3)x^(4)-6(2k-3)x^(2)+3]],[P(n-2)(x)=(2k-2)!k[(2k-1)x^(2)-1]]:}\begin{aligned}
& P(n-4)(x)=\frac{(2 k-4)!k(k-1)}{3!}\left[(2 k-1)(2 k-3) x^{4}-6(2 k-3) x^{2}+3\right] \\
& P(n-2)(x)=(2 k-2)!k\left[(2 k-1) x^{2}-1\right]
\end{aligned}
Whence, doing the calculations, it turns out thatP^((n-4))(x)\mathrm{P}^{(n-4)}(x)will divide byP^((n-2))(x)\mathrm{P}^{(n-2)}(x)only ifk=3k=3.
If son=2k,k > 3n=2 k, k>3, we have safety eggN > n+1N>n+1For
​n=6,k=3n=6, k=3, it is directly verified thatN:=n+1\mathbb{N}:=n+1.
Finally, we have the following property
II. - If the polynomialP(x)\mathrm{P}(x)of the degreenn(with all real roots) has at least two distinct roots, the equalityN=n+1\mathrm{N}=n+1occurs if and only if this polynomial has a multiple root of ordern-1n-1of multiplicity, as well as only in the following two cases:
1^(0)1^{0}Ifn=4n=4and the polynomial has two double roots.
20. Ifn=6n=6and the polynomial has two triple roots.
5. -- As for the accuracy of property I in the case when the restriction on the reality of the roots is not made, it can be easily established for some of the first values ​​ofnn.
10. Forn=2n=2, the property is already proven.
20. Ifn=3n=3and if its rootsP(x)\mathrm{P}(x)are distinct, derivedP^(')(x)\mathrm{P}^{\prime}(x)will have distinct roots or not, but both different from hisP(x)\mathrm{P}(x). We therefore have, in this caseN >= 4\mathrm{N} \geq 4IfP(x)\mathrm{P}(x)does not have all distinct roots, through a linear transformation we return to the case of real roots.
3^(@)3^{\circ}Ifn=4n=4and if its rootsP(x)\mathrm{P}(x)are distinct, it is seen, as above, that we haveN >= 5\mathrm{N} \geq 5IfP(x)\mathrm{P}(x)has three distinct roots, it remains to examine the case when we cannot return to the case of all real roots by a linear transformation. If then the roots ofP^(')(x)\mathrm{P}^{\prime}(x)are distinct, two of them are different from the roots ofP(x)\mathrm{P}(x), so we haveN >= 5N \geqq 5. If finallyP^(')(x)P^{\prime}(x)does not have distinct roots, then one (simple) coincides with the double root ofP(x)\mathrm{P}(x), and the other distinct from this is a double root different from the roots ofP(x)P(x). By a linear transformation it can be done thatP^(')(x)\mathrm{P}^{\prime}(x)to have all real roots and thenP^(')(x)P^('')(x)P^(''')(x)\mathrm{P}^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \mathrm{P}^{\prime \prime \prime}(x)has at least 4 distinct real roots, andP(x)\mathrm{P}(x)at least one unreal root. So we have everythingN >= 5\mathrm{N} \geqq 5. If finallyP(x)\mathrm{P}(x)has only two distinct roots, through a linear transformation we can make all the roots real.
​n=6,k=3n=6, k=3, it is directly verified thatN:=n+1\mathbb{N}:=n+1.
Finally, we have the following property
II. - If the polynomialP(x)\mathrm{P}(x)of the degreenn(with all real roots) has at least two distinct roots, the equalityN=n+1\mathrm{N}=n+1occurs if and only if this polynomial has a multiple root of ordern-1n-1of multiplicity, as well as only in the following two cases:
1^(0)1^{0}Ifn=4n=4and the polynomial has two double roots.
20. Ifn=6n=6and the polynomial has two triple roots.
5. -- As for the accuracy of property I in the case when the restriction on the reality of the roots is not made, it can be easily established for some of the first values ​​ofnn.
10. Forn=2n=2, the property is already proven.
20. Ifn=3n=3and if its rootsP(x)\mathrm{P}(x)are distinct, derivedP^(')(x)\mathrm{P}^{\prime}(x)will have distinct roots or not, but both different from hisP(x)\mathrm{P}(x). We therefore have, in this caseN >= 4\mathrm{N} \geq 4IfP(x)\mathrm{P}(x)does not have all distinct roots, through a linear transformation we return to the case of real roots.
3^(@)3^{\circ}Ifn=4n=4and if its rootsP(x)\mathrm{P}(x)are distinct, it is seen, as above, that we haveN >= 5\mathrm{N} \geq 5IfP(x)\mathrm{P}(x)has three distinct roots, it remains to examine the case when we cannot return to the case of all real roots by a linear transformation. If then the roots ofP^(')(x)\mathrm{P}^{\prime}(x)are distinct, two of them are different from the roots ofP(x)\mathrm{P}(x), so we haveN >= 5N \geqq 5. If finallyP^(')(x)P^{\prime}(x)does not have distinct roots, then one (simple) coincides with the double root ofP(x)\mathrm{P}(x), and the other distinct from this is a double root different from the roots ofP(x)P(x). By a linear transformation it can be done thatP^(')(x)\mathrm{P}^{\prime}(x)to have all real roots and thenP^(')(x)P^('')(x)P^(''')(x)\mathrm{P}^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \mathrm{P}^{\prime \prime \prime}(x)has at least 4 distinct real roots, andP(x)\mathrm{P}(x)at least one unreal root. So we have everythingN >= 5\mathrm{N} \geqq 5. If finallyP(x)\mathrm{P}(x)has only two distinct roots, through a linear transformation we can make all the roots real.
Forn=2,3,4n=2,3,4property I is therefore true in general.
Mathematics Department
of the Cluj Branch of the RPR Academy
Mathematics Department
of the Cluj Branch of the RPR Academy
SUMMARY
On polynomials with all real roots
TIBERIA NOSHOVICHA
In this paper it is proved that if a polynomialP(x)\mathrm{P}(x)degrees has all real roots, the resultP(x)P^(')(x)P^('')(x)dotsP^((n-1))\mathrm{P}(x) \mathrm{P}^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \ldots \mathrm{P}^{(n-1)}has 1 or at leastn+1n+1different roots. All polynomials are also determinedP(x)\mathrm{P}(x)for which the limitn+1n+1achieved.
RESUME
On polynomials having all real roots
about
TIBERIU POPOVICIU
In this work the author shows that if the polynomialP(x)\mathrm{P}(x)of degreennhas all real roots, the productP(x)P^(')(x)P^('')(x)dotsP(n-1)\mathrm{P}(x) P^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \ldots \mathrm{P}(n-1)a 1 or at leastn+1n+1distinct roots. We also determine all the polynomialsP(x)\mathrm{P}(x)for which the limitn+1n+1Hest taken hit.