On polynomials with all roots real

Tiberiu Popoviciu
(original), paper, polynomials

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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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1952 b -190 -Popoviciu- Stud. Cerc. St., Cluj – Asupra polinoamelor cu toate radacinile reale

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Studii si Cercetari Matematice

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Academy of the Republic of S.R.

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https://ictp.acad.ro/scm/journal/article/view/246

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1952 b -190 -Popoviciu- Stud. Cerc. St., Cluj - On polynomials with all roots real
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ON POLYNOMIALS WITH ALL REAL ROOTS

OFTIBERIU POPOVICIUCorresponding member of the RPR AcademyCommunication presented at the meeting of October 12, 1951of the Cluj Branch of the RPR Academy

  1. -- Either P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)a polynomial of (effective) degree n. This polynomial has n n nnndistinct roots or not. Successive derivatives P ′ ( x ) , P ′ ′ ( x ) , … , P ( n − 1 ) ( x ) P ′ ( x ) , P ′ ′ ( x ) , … , P ( n − 1 ) ( x ) P^(')(x),P^('')(x),dots,P^((n-1))(x)\mathrm{P}^{\prime}(x), \mathrm{P}^{\prime \prime}(x), \ldots, \mathrm{P}^{(n-1)}(x)P′(x),P′′(x),…,P(n−1)(x)have respectively n − 1 , n − 2 , … n − 1 , n − 2 , … n-1,n-2,dotsn-1, n-2, \ldotsn−1,n−2,…, 1 distinct roots or not. Therefore, considering all the roots of the polynomial and its successive derivatives, we have
1 + 2 + … + n = n ( n + 1 ) 2 1 + 2 + … + n = n ( n + 1 ) 2 1+2+dots+n=(n(n+1))/(2)1+2+\ldots+n=\frac{n(n+1)}{2}1+2+…+n=n(n+1)2
such distinct roots or not.
The problem may arise of determining the number N N NNNof the distinct numbers between these n ( n + 1 ) 2 n ( n + 1 ) 2 (n(n+1))/(2)\frac{n(n+1)}{2}n(n+1)2numbers thus obtained and which represent the roots considered. The number N is the number of distinct roots of the polynomial P ( x ) P ′ ( x ) … P ( n − 1 ) ( x ) P ( x ) P ′ ( x ) … P ( n − 1 ) ( x ) P(x)P^(')(x)dotsP^((n-1))(x)\mathrm{P}(x) \mathrm{P}^{\prime}(x) \ldots \mathrm{P}^{(n-1)}(x)P(x)P′(x)…P(n−1)(x)obtained by taking the product of the given polynomial and its successive derivatives.
We can obviously assume n > 1 n > 1 n > 1n>1n>1It is clear
that if P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)has all the roots confused we have N = 1 N = 1 N=1\mathrm{N}=1N=1Otherwise
we have N > 1 N > 1 N > 1N>1N>1and the question arises how can this inequality be specified in this case?
Assuming that the roots of P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)are all real, we will prove that:
I. If P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)has at least two distinct roots we have N ≥ n + 1 N ≥ n + 1 N >= n+1\mathrm{N} \geq n+1N≥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 variable x x xxx, 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 ) ( ≠ 0 ) (!=0)(\neq 0)(≠0)his/her P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)It follows that in the following proofs one can always assume that the first coefficient of the polynomial (of x n x n x^(n)x^{n}xn) is equal to 1 and that if P ( x ) P ( x ) P(x)\mathrm{P}(x)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.
If a polynomial has all its roots real, its derivative also has all its roots real. Let a the → , b the a the → , b the thing(a_(o)),b_(o)\overrightarrow{a_{o}}, b_{o}athe→,bthethe extreme roots, that is, the smallest and the largest root, of P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)and a 1 , b 1 a 1 , b 1 a_(1),b_(1)a_{1}, b_{1}a1,b1the analogous extreme roots of the derivative P ′ ( x ) P ′ ( x ) P^(')(x)\mathrm{P}^{\prime}(x)P′(x)If a the a the to the)to the}athe resp. b the b the b_(o)b_{o}btheis an at least double root of P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x), we have a 1 = a 0 a 1 = a 0 a_(1)=a_(0)a_{1}=a_{0}a1=a0resp. b 1 = b 0 b 1 = b 0 b_(1)=b_(0)b_{1}=b_{0}b1=b0If a 0 a 0 a_(0)a_{0}a0resp. b 0 b 0 b_(0)b_{0}b0If it is a simple root, we have a the < a 1 a the < a 1 a_(o) < a_(1)a_{o}<a_{1}athe<a1resp. b 1 < b the b 1 < b the b_(1) < b_(o)b_{1}<b_{o}b1<bthe. In fine a 1 a 1 a_(1)a_{1}a1resp. b 1 b 1 b_(1)b_{1}b1is a simple root of the derivative if and only if a 0 a 0 a_(0)a_{0}a0resp. b 0 b 0 b_(0)b_{0}b0is a simple or double root of the polynomial.
Let's assume that P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)does not have all the roots confused and be, in general, a i a i a_(i)a_{i}aithe smallest one and b i b i b_(i)b_{i}bithe largest root of the derivative P ( 1 ) ( x ) P ( 1 ) ( x ) P^((1))(x)\mathrm{P}^{(1)}(x)P(1)(x)of the order i , i = 0 , 1 … , n − 1 i , i = 0 , 1 … , n − 1 i,i=0,1dots,n-1i, i=0,1 \ldots, n-1i,i=0,1…,n−1We have obviously a n − 1 = b n − 1 a n − 1 = b n − 1 a_(n-1)=b_(n-1)a_{n-1}=b_{n-1}an−1=bn−1, but a n − 2 << b n − 2 a n − 2 << b n − 2 a_(n-2)<<b_(n-2)a_{n-2}< <b_{n-2}an−2<<bn−2This latter inequality results from the fact that if the derivative of a polynomial with all real roots has a root of order i > 1 i > 1 i > 1i>1i>1of multiplicity, the polynomial necessarily has this root of order i + 1 i + 1 i+1i+1i+1of multiplicity. In the hypothesis of the reality of all roots the equality a n − 2 = b n − 2 a n − 2 = b n − 2 a_(n-2)=b_(n-2)a_{n-2}=b_{n-2}an−2=bn−2is 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
(1) a i , b i , i = 0 , 1 , … ; n − − 1 (1) a i , b i , i = 0 , 1 , … ; n − − 1 {:(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*}(1)ai,bi,i=0,1,…;n−−1
From the above it follows that, if a the a the to the)to the}atheis a root of order l l lllof multiplicity and b the b the b_(o)b_{o}bthea root of the order k k kkkof multiplicity of the polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x), we have
a 0 = a 1 = ⋯ = a l − 1 < a l < a l + 1 < ⋯ < a n − 2 < a n − 1 = = b n − 1 < b n − 2 < ⋯ < b k < b k − 1 = b k − 2 = ⋯ = b 0 a 0 = a 1 = ⋯ = a l − 1 < a l < a l + 1 < ⋯ < a n − 2 < a n − 1 = = b n − 1 < b n − 2 < ⋯ < b k < b k − 1 = b k − 2 = ⋯ = b 0 {:[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}a0=a1=⋯=al−1<al<al+1<⋯<an−2<an−1==bn−1<bn−2<⋯<bk<bk−1=bk−2=⋯=b0
From this it follows that
E = 2 n − l − k + 1 . E = 2 n − l − k + 1 . E=2n-l-k+1.\mathrm{E}=2 n-l-k+1 .AND=2n−l−k+1.
But
(3) l + k ≦ n (3) l + k ≦ n {:(3)l+k <= n:}\begin{equation*} l+k \leqq n \tag{3} \end{equation*}(3)l+k≦n
(si) N ≧ E . (si) N ≧ E . {:(si)N >= E.:}\begin{equation*} \mathrm{N} \geqq \mathrm{E} . \tag{si} \end{equation*}(and)N≧AND.
But from (3) it follows that E ≧ n + 1 E ≧ n + 1 E >= n+1E \geqq n+1AND≧n+1, so N ≧ n + 1 N ≧ n + 1 N >= n+1N \geqq n+1N≧n+1and property I I IIIis demonstrated.
4. We now propose to determine all the polynomials P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)(with all real roots) of degree n n nnnfor which we have N = n + 1 N = n + 1 N=n+1\mathrm{N}=n+1N=n+1.
From (2), (3), (4) it follows that this can only happen if l + k = n l + k = n l+k=nl+k=nl+k=n, so only if the polynomial has exactly two distinct roots, therefore, apart from a constant factor ( ≠ 0 ≠ 0 !=0\neq 0≠0) and apart from a linear transformation of the variable, only for polynomials of the form
P ( x ) = x n − k ( x − 1 ) k , 2 k ≦ n . P ( x ) = x n − k ( x − 1 ) k , 2 k ≦ n . P(x)=x^(n-k)(x-1)^(k),2k <= n.\mathrm{P}(x)=x^{n-k}(x-1)^{k}, 2 k \leqq n .P(x)=xn−k(x−1)k,2k≦n.
In this case we have E = n + 1 E = n + 1 E=n+1\mathrm{E}=n+1AND=n+1and so that we have N = n + 1 N = n + 1 N=n+1N=n+1N=n+1it is necessary and sufficient that all the roots of any derivative P ( i ) ( x ) P ( i ) ( x ) P^((i))(x)\mathrm{P}^{(i)}(x)P(i)(x)not be different from the roots of (1) and this for i = 1 , 2 , … , n − 3 i = 1 , 2 , … , n − 3 i=1,2,dots,n-3i=1,2, \ldots, n-3i=1,2,…,n−3.
If n ≧ 2 , k = 1 n ≧ 2 , k = 1 n >= 2,k=1n \geqq 2, k=1n≧2,k=1there are no such roots and so we then have N = n + 1 N = n + 1 N=n+1N=n+1N=n+1In this way the cases n = 2 , 3 n = 2 , 3 n=2,3n=2,3n=2,3are exhausted.
If n ≧ 4 , 2 ≦ k ≦ n 2 , P ( n − 3 ) ( x ) n ≧ 4 , 2 ≦ k ≦ n 2 , P ( n − 3 ) ( x ) n >= 4,2 <= k <= (n)/(2),P(n-3)(x)n \geqq 4,2 \leqq k \leqq \frac{n}{2}, \mathrm{P}(n-3)(x)n≧4,2≦k≦n2,P(n−3)(x)has distinct roots and a root different from a n − 3 a n − 3 a_(n-3)a_{n-3}an−3and b n − 3 b n − 3 b_(n-3)b_{n-3}bn−3of this polynomial cannot coincide with a n − 1 a n − 1 a_(n-1)a_{n-1}an−1. In order to have N = n + 1 N = n + 1 N=n+1N=n+1N=n+1it is therefore necessary that the polynomials P ( n − 3 ) ( x ) , P ( n − 1 ) ( x ) P ( n − 3 ) ( x ) , P ( n − 1 ) ( x ) P^((n-3))(x),P^((n-1))(x)\mathrm{P}^{(n-3)}(x), \mathrm{P}^{(n-1)}(x)P(n−3)(x),P(n−1)(x)to have a common root. But a n − 1 = b n − 1 = k n a n − 1 = b n − 1 = k n a_(n-1)=b_(n-1)=(k)/(n)a_{n-1}=b_{n-1}=\frac{k}{n}an−1=bn−1=kn, that is, it is necessary to have
p ( n − 3 ) ( k n ) = 0 . p ( n − 3 ) k n = 0 . p^((n-3))((k)/(n))=0.p^{(n-3)}\left(\frac{k}{n}\right)=0 .p(n−3)(kn)=0.
But
P ( n − 3 ) ( x ) = ( n − 3 ) ! 3 ! [ n ( n − 1 ) ( n − 2 ) x 3 − 3 k ( n − 1 ) ( n − 2 ) x 2 + + 3 k ( k − 1 ) ( n − 2 ) x − k ( k − 1 ) ( k − 2 ) ] ducem P ( n − 3 ) ( x ) = ( n − 3 ) ! 3 ! n ( n − 1 ) ( n − 2 ) x 3 − 3 k ( n − 1 ) ( n − 2 ) x 2 + + 3 k ( k − 1 ) ( n − 2 ) x − k ( k − 1 ) ( k − 2 ) ]  ducem  {:[{:[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)(x)=(n−3)!3![n(n−1)(n−2)x3−3k(n−1)(n−2)x2++3k(k−1)(n−2)x−k(k−1)(k−2)] take 
P ( n − 3 ) ( k n ) = − k ( n − k ) ( n − 2 k ) 3 ⋅ ( n − 3 ) ! n 2 P ( n − 3 ) k n = − k ( n − k ) ( n − 2 k ) 3 ⋅ ( n − 3 ) ! n 2 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}}P(n−3)(kn)=−k(n−k)(n−2k)3⋅(n−3)!n2
It follows that if n ≧ 4 , 2 ≦ k < n 2 n ≧ 4 , 2 ≦ k < n 2 n >= 4,2 <= k < (n)/(2)n \geqq 4,2 \leqq k<\frac{n}{2}n≧4,2≦k<n2we certainly have N > n + 1 N > n + 1 N > n+1\mathrm{N}>n+1N>n+1It
remains to examine the case n = 2 k n = 2 k n=2kn=2 kn=2kIf k = 2 k = 2 k=2k=2k=2, based on the above it can be seen that we have N = n + 1 N = n + 1 N=n+1\mathrm{N}=n+1N=n+1If k > 2 k > 2 k > 2k>2k>2, his roots P ( n − 4 ) ( x ) P ( n − 4 ) ( x ) P^((n-4))(x)\mathrm{P}^{(n-4)}(x)P(n−4)(x)are all distinct and symmetrically placed opposite each other a − 1 a − 1 a_(-1)a_{-1}a−1. From this it immediately follows that none of the roots distinct from the extreme roots of P ( n − 4 ) ( x ) P ( n − 4 ) ( x ) P^((n-4))(x)\mathrm{P}^{(n-4)}(x)P(n−4)(x)does not coincide with a n − 1 a n − 1 a_(n-1)a_{n-1}an−1and that in order to have N = n + 1 N = n + 1 N=n+1\mathrm{N}=n+1N=n+1it is necessary that these roots coincide with a n − 2 , b n − 2 a n − 2 , b n − 2 a_(n-2),b_(n-2)a_{n-2}, b_{n-2}an−2,bn−2respectively, so that P ( n − 4 ) ( x ) P ( n − 4 ) ( x ) P^((n-4))(x)\mathrm{P}^{(n-4)}(x)P(n−4)(x)to be divisible by P ( n − 2 ) ( x ) P ( n − 2 ) ( x ) P^((n-2))(x)\mathrm{P}^{(n-2)}(x)P(n−2)(x)For simplification we can now take
P ( x ) = ( x 2 − 1 ) k P ( x ) = x 2 − 1 k P(x)=(x^(2)-1)^(k)\mathrm{P}(x)=\left(x^{2}-1\right)^{k}P(x)=(x2−1)k
We have snow.
P ( n − 4 ) ( x ) = ( 2 k − 4 ) ! k ( k − 1 ) 3 ! [ ( 2 k − 1 ) ( 2 k − 3 ) x 4 − 6 ( 2 k − 3 ) x 2 + 3 ] P ( n − 2 ) ( x ) = ( 2 k − 2 ) ! k [ ( 2 k − 1 ) x 2 − 1 ] P ( n − 4 ) ( x ) = ( 2 k − 4 ) ! k ( k − 1 ) 3 ! ( 2 k − 1 ) ( 2 k − 3 ) x 4 − 6 ( 2 k − 3 ) x 2 + 3 P ( n − 2 ) ( x ) = ( 2 k − 2 ) ! k ( 2 k − 1 ) x 2 − 1 {:[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}P(n−4)(x)=(2k−4)!k(k−1)3![(2k−1)(2k−3)x4−6(2k−3)x2+3]P(n−2)(x)=(2k−2)!k[(2k−1)x2−1]
Whence, doing the calculations, it turns out that P ( n − 4 ) ( x ) P ( n − 4 ) ( x ) P^((n-4))(x)\mathrm{P}^{(n-4)}(x)P(n−4)(x)will divide by P ( n − 2 ) ( x ) P ( n − 2 ) ( x ) P^((n-2))(x)\mathrm{P}^{(n-2)}(x)P(n−2)(x)only if k = 3 k = 3 k=3k=3k=3.
If so n = 2 k , k > 3 n = 2 k , k > 3 n=2k,k > 3n=2 k, k>3n=2k,k>3, we have safety egg N > n + 1 N > n + 1 N > n+1N>n+1N>n+1For
​ n = 6 , k = 3 n = 6 , k = 3 n=6,k=3n=6, k=3n=6,k=3, it is directly verified that N := n + 1 N := n + 1 N:=n+1\mathbb{N}:=n+1N:=n+1.
Finally, we have the following property
II. - If the polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)of the degree n n nnn(with all real roots) has at least two distinct roots, the equality N = n + 1 N = n + 1 N=n+1\mathrm{N}=n+1N=n+1occurs if and only if this polynomial has a multiple root of order n − 1 n − 1 n-1n-1n−1of multiplicity, as well as only in the following two cases:
1 0 1 0 1^(0)1^{0}10If n = 4 n = 4 n=4n=4n=4and the polynomial has two double roots.
20. If n = 6 n = 6 n=6n=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 ​​of n n nnn.
10. For n = 2 n = 2 n=2n=2n=2, the property is already proven.
20. If n = 3 n = 3 n=3n=3n=3and if its roots P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)are distinct, derived P ′ ( x ) P ′ ( x ) P^(')(x)\mathrm{P}^{\prime}(x)P′(x)will have distinct roots or not, but both different from his P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x). We therefore have, in this case N ≥ 4 N ≥ 4 N >= 4\mathrm{N} \geq 4N≥4If P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)does not have all distinct roots, through a linear transformation we return to the case of real roots.
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘If n = 4 n = 4 n=4n=4n=4and if its roots P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)are distinct, it is seen, as above, that we have N ≥ 5 N ≥ 5 N >= 5\mathrm{N} \geq 5N≥5If P ( x ) P ( x ) P(x)\mathrm{P}(x)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 of P ′ ( x ) P ′ ( x ) P^(')(x)\mathrm{P}^{\prime}(x)P′(x)are distinct, two of them are different from the roots of P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x), so we have N ≧ 5 N ≧ 5 N >= 5N \geqq 5N≧5. If finally P ′ ( x ) P ′ ( x ) P^(')(x)P^{\prime}(x)P′(x)does not have distinct roots, then one (simple) coincides with the double root of P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x), and the other distinct from this is a double root different from the roots of P ( x ) P ( x ) P(x)P(x)P(x). By a linear transformation it can be done that P ′ ( x ) P ′ ( x ) P^(')(x)\mathrm{P}^{\prime}(x)P′(x)to have all real roots and then P ′ ( x ) P ′ ′ ( x ) P ′ ′ ′ ( x ) P ′ ( x ) P ′ ′ ( x ) P ′ ′ ′ ( x ) P^(')(x)P^('')(x)P^(''')(x)\mathrm{P}^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \mathrm{P}^{\prime \prime \prime}(x)P′(x)P′′(x)P′′′(x)has at least 4 distinct real roots, and P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)at least one unreal root. So we have everything N ≧ 5 N ≧ 5 N >= 5\mathrm{N} \geqq 5N≧5. If finally P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)has only two distinct roots, through a linear transformation we can make all the roots real.
For n = 2 , 3 , 4 n = 2 , 3 , 4 n=2,3,4n=2,3,4n=2,3,4property I is therefore true in general.
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 polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)degrees has all real roots, the result P ( x ) P ′ ( x ) P ′ ′ ( x ) … P ( n − 1 ) P ( x ) P ′ ( x ) P ′ ′ ( x ) … P ( n − 1 ) P(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)}P(x)P′(x)P′′(x)…P(n−1)has 1 or at least n + 1 n + 1 n+1n+1n+1different roots. All polynomials are also determined P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)for which the limit n + 1 n + 1 n+1n+1n+1achieved.

RESUME

On polynomials having all real roots

about

TIBERIU POPOVICIU

In this work the author shows that if the polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)of degree n n nnnhas all real roots, the product P ( x ) P ′ ( x ) P ′ ′ ( x ) … P ( n − 1 ) P ( x ) P ′ ( x ) P ′ ′ ( x ) … P ( n − 1 ) P(x)P^(')(x)P^('')(x)dotsP(n-1)\mathrm{P}(x) P^{\prime}(x) \mathrm{P}^{\prime \prime}(x) \ldots \mathrm{P}(n-1)P(x)P′(x)P′′(x)…P(n−1)a 1 or at least n + 1 n + 1 n+1n+1n+1distinct roots. We also determine all the polynomials P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)for which the limit n + 1 n + 1 n+1n+1n+1Hest taken hit.
1952

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