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T. Popoviciu, Sur un théorème de Laguerre, Buletinul Soc. de Ştiinţe din Cluj, 8 (1934), pp. 1-4 (in Romanian).
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Note: this paper has been republished in T. Popoviciu, Sur un théorème de Laguerre, Mathematica, 10 (1935), pp. 128-131 (in French)
[JFM 60.1042.05]. https://zbmath.org/?q=an:60.1042.05
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ON A LAGUERRE THEOREM
by
Tiberiu Popoviciu,
former student of the École Normale Supérieure
Received on January 31, 1934.
-
- Let us consider a polynomialf(x)f(x)whose degree is at least: equal ton-1n-1.
EitherU(x_(1),x_(2),dots,x_(n);f(x))\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n}; f(x)\right)the determinant whose general line is
1x_(i)x_(i)^(2)dotsx_(i)^(n-2)f(x_(i))1 x_{i} x_{i}^{2} \ldots x_{i}^{n-2} f\left(x_{i}\right)
and let's askV(x_(1),x_(2),dots,x_(n))=U(x_(1),x_(2),dots,x_(n);x^(n-1))\mathrm{V}\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n} ; x^{n-1}\right)which is the Van der Monde determinant of quantitiesx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, ..., x_{n}.
We have identity
(1)U(x_(1),x_(2),dots,x_(n);f(x))=sum_(i=1)^(i=n)(-1)^(ni)V(x_(1),x_(2),dots,x_(i-1),x_(i+1),dots,x_(n))f(x_(i))\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n} ; f(x)\right)=\sum_{i=1}^{i=n}(-1)^{ni} V\left(x_{1}, x_{2}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\right) f\left(x_{i}\right)which we will need later.
(1)U(x_(1),x_(2),dots,x_(n);f(x))=sum_(i=1)^(i=n)(-1)^(ni)V(x_(1),x_(2),dots,x_(i-1),x_(i+1),dots,x_(n))f(x_(i))\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n} ; f(x)\right)=\sum_{i=1}^{i=n}(-1)^{ni} V\left(x_{1}, x_{2}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\right) f\left(x_{i}\right)which we will need later.
Let us consider the quotient
[x_(1),x_(2),dots,x_(n);f(x)]=(U)/((x_(1),x_(2),dots,x_(n);f(x))).\left[x_{1}, x_{2}, \ldots, x_{n}; f(x)\right]=\frac{\mathrm{U}}{\left(x_{1}, x_{2}, \ldots, x_{n} ; f(x)\right)} .
Let us now recall that we say that the polynomialf(x)f(x)is convex of order (n-2n-2) (^(1){ }^{1}) in the interval (a,ba, b) if for any group of:nndistinct pointsx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, ..., x_{n}from this interval we have the inequality
{:(2)[x_(1),x_(2),dots,x_(n);f(x)] > 0:}\begin{equation*} \left[x_{1}, x_{2}, \ldots, x_{n}; f(x)\right]>0 \tag{2} \end{equation*}
So that the polynomialf(x)f(x)either convex of order (n-2n-2) In (a,ba, b) it is necessary and sufficient thatf^((n-1))(x) >= 0f^{(n-1)}(x) \geq 0in this interval. Inequality (2) will hold for any system ofnnpoint not all together.
2. - So nowP(x)=(1-x_(1)x)(1-x_(2)x)dots,(1-x_(n)x)\mathrm{P}(x)=\left(1-x_{1} x\right)\left(1-x_{2} x\right) \ldots,\left(1-x_{n} x\right)a polynomial having all its zeros real and consider the expansion
(3)
(3)
We know that
(1)/(P(x))=S_(0)+S_(1)x+S_(2)x^(2)+cdots\frac{1}{\mathrm{P}(x)}=\mathrm{S}_{0}+\mathrm{S}_{1} x+\mathrm{S}_{2} x^{2}+\cdots
S_(m)=sumx_(1)^(i_(1))x_(2)^(i_(2))dotsx_(n)^(i_(n))\mathrm{S}_{m}=\sum x_{1}^{i_{1}} x_{2}^{i_{2}} \ldots x_{n}^{i_{n}}
where the summation is extended for all non-negative integer values ​​ofi_(1),i_(2),dots,i_(n)i_{1}, i_{2}, \ldots, i_{n}such asi_(1)+i_(2)+cdots+i_(n)=mi_{1}+i_{2}+\cdots+i_{n}=m.
Especially ifx_(1)=x_(2)=dots=x_(n)=1x_{1}=x_{2}=\ldots=x_{n}=1we have
S_(m)=((n+n-1)/(m))=((m+n-1)/(n-1))\mathrm{S}_{m}=\binom{n+n-1}{m}=\binom{m+n-1}{n-1}
Decomposing the left-hand side of (3) into simple fractions, we have
(1)/(P(x))=-sum_(i=1)^(i=n)((-1)^(ni)(V)(x_(1),x_(2),dots,x_(i-1),x_(i+1),dots,x_(n))x_(i)^(n-1))/((V)(x_(1),x_(2),dots,x_(n))(1-x_(i)x))\frac{1}{\mathrm{P}(x)}=-\sum_{i=1}^{i=n} \frac{(-1)^{ni} \mathrm{~V}\left(x_{1}, x_{2}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\right) x_{i}^{n-1}}{\mathrm{~V}\left(x_{1}, x_{2}, \ldots, x_{n}\right)\left(1-x_{i} x\right)}
and taking into account (1) we find
(4)
(4)
S_(m)=[x_(1),x_(2),dots,x_(n);x^(m+n-1)]\mathrm{S}_{m}=\left[x_{1}, x_{2}, \ldots, x_{n}; x^{m+n-1}\right]
-
- Let us consider the equation of degree2p2 p
F(z)-=a_(0)S_(k)+a_(1)S_(k+1)z dotsa_(2p)S_(2p+k)z^(2p)=0,quada_(2p) > 0\mathrm{F}(z) \equiv a_{0} \mathrm{~S}_{k}+a_{1} \mathrm{~S}_{k+1} z \ldots a_{2 p} \mathrm{~S}_{2 p+k} z^{2 p}=0, \quad a_{2 p}>0
If we pose
varphi(x)=x^(n+k-1)(a_(0)+a_(1)xz+a_(2)(xz)^(2)+cdots+a_(2p)(xz)^(2p))\varphi(x)=x^{n+k-1}\left(a_{0}+a_{1} x z+a_{2}(xz)^{2}+\cdots+a_{2 p}(xz)^{2 p}\right)
we have according to (4)
F(z)=[x_(1),x_(2),dots,x_(n);varphi(x)]\mathrm{F}(z)=\left[x_{1}, x_{2}, \ldots, x_{n}; \varphi(x)\right]
If we notice that
varphi^((n-1))(x)=(n-1)!x^(k)[((k+n-1)/(n-1))a_(0)+((k+n)/(n-1))a_(1)xz+cdots+((k+n+2p-1)/(n-1))a_(2p)(xz)^(2p)]\varphi^{(n-1)}(x)=(n-1)!x^{k}\left[\binom{k+n-1}{n-1} a_{0}+\binom{k+n}{n-1} a_{1} x z+\cdots+\binom{k+n+2 p-1}{n-1} a_{2 p}(x z)^{2 p}\right]
We can state the following properties:
Ifkkis even and if the polynomial
varphi^((n-1))(x)=(n-1)!x^(k)[((k+n-1)/(n-1))a_(0)+((k+n)/(n-1))a_(1)xz+cdots+((k+n+2p-1)/(n-1))a_(2p)(xz)^(2p)]\varphi^{(n-1)}(x)=(n-1)!x^{k}\left[\binom{k+n-1}{n-1} a_{0}+\binom{k+n}{n-1} a_{1} x z+\cdots+\binom{k+n+2 p-1}{n-1} a_{2 p}(x z)^{2 p}\right]
We can state the following properties:
Ifkkis even and if the polynomial
{:(5)sum_(i=0)^(i=2p)((k+n+i-1)/(n-1))a_(i)z^(i):}\begin{equation*}
\sum_{i=0}^{i=2 p}\binom{k+n+i-1}{n-1} a_{i} z^{i} \tag{5}
\end{equation*}
is non-negative, the polynomial
{:(6)sum_(i=0)^(i=2p)a_(i)S_(k+i)z^(i):}\begin{equation*}
\sum_{i=0}^{i=2 p} a_{i} \mathrm{~S}_{k+i} z^{i} \tag{6}
\end{equation*}
is positive as long as the rootsx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}are not all equal.
Ifkkis odd and if the polynomial (5) is non-negative the polynomial (6) is positive as long as the rootsx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}are non-negative and not all equal.
4. - Laguerre demonstrated the following proposition:
4. - Laguerre demonstrated the following proposition:
If the equationP(x)=0\mathrm{P}(x)=0to all its real roots the polynomial
sum_(i=0)^(2p)S_(i)z^(i)\sum_{i=0}^{2 p} \mathrm{~S}_{i} z^{i}
is positive.
Let us always assume that Laguerre's theorem is true when all the zeros of the polynomialP(x)\mathrm{P}(x)are equal. In other words, the polynomial
Let us always assume that Laguerre's theorem is true when all the zeros of the polynomialP(x)\mathrm{P}(x)are equal. In other words, the polynomial
Q_(a,0)=sum_(i=1)^(2p)((alpha+i)/(alpha))z^(i)Q_{a, 0}=\sum_{i=1}^{2 p}\binom{\alpha+i}{\alpha} z^{i}
is positive foralpha\alphaentire> 0>0
Let us consider the more general polynomial
Let us consider the more general polynomial
Q_(a,beta)=sum_(i=0)^(2p)((alpha+3+i)/(alpha))z^(i),quad alpha,beta" entiers " >= 0.Q_{a, \beta}=\sum_{i=0}^{2 p}\binom{\alpha+3+i}{\alpha} z^{i}, \quad \alpha, \beta \text { entiers } \geq 0 .
The recurrence relation
Q_(a,beta)=Q_(a,beta-1)+Q_(a-1,beta)Q_{a, \beta}=Q_{a, \beta-1}+Q_{a-1, \beta}
and the fact thatQ_(0,beta)Q_{0, \beta}is independent ofbeta(Q_(0,beta)=1+z+z^(2)+cdots+z^(2p))\beta\left(Q_{0, \beta}=1+z+z^{2}+\cdots+z^{2 p}\right)show us that the polynomialQ_(alpha,beta)Q_{\alpha, \beta}is positive.
So doinga_(1)=a_(2)=cdots=a_(2p)=1a_{1}=a_{2}=\cdots=a_{2 p}=1In (5) we deduce the following properties:
Ifkkis even the polynomial
(7)
(7)
sum_(i=0)^(2p)S_(k+i)z^(i)\sum_{i=0}^{2 p} \mathrm{~S}_{k+i} z^{i}
is certainly positive, as long as the rootsx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}are not all equal.
Ifkkis odd, the polynomial (7) is certainly positive as long as the rootsx_(i)x_{i}are non-negative and not all equal.
5. - Let
5. - Let
sigma_(m)=(S^(m))/(((m+n-1)/(n-1))),quadM_(m)=root(m)(sigma_(m))\sigma_{m}=\frac{\mathrm{S}^{m}}{\binom{m+n-1}{n-1}}, \quad \mathrm{M}_{m}=\sqrt[m]{\sigma_{m}}
(1) See Lagumrre “Works” t. J. p. 111.
M_(m)\mathrm{M}_{m}is therefore an average value.
If
M_(m)\mathrm{M}_{m}is therefore an average value.
If
p=1,quada_(i)=(((2)/(i)))/(((k+n+i-1)/(n-1))),quad1=0,1,2p=1, \quad a_{i}=\frac{\binom{2}{i}}{\binom{k+n+i-1}{n-1}}, \quad 1=0,1,2
My polynomial (5) is non-negative, therefore assuming thex_(l)x_{l}non-negative soft we also have
sigma_(k)+2sigma_(k+1)x+sigma_(k+2)x^(2) >= 0\sigma_{k}+2 \sigma_{k+1} x+\sigma_{k+2} x^{2} \geq 0
"regardless ofkkand equality is only possible if all thex_(i)x_{i}are identical.
So we have
{:[sigma_(k+1)^(2) <= sigma_(k)","sigma_(k+2)],[sigma_(0)=1","k=0","1","2","dots]:}\begin{gathered}
\sigma_{k+1}^{2} \leq \sigma_{k}, \sigma_{k+2} \\
\sigma_{0}=1, k=0,1,2, \ldots
\end{gathered}
aet we can then state the following property:
If thex_(l)x_{l}are non-negative, we have the sequence of inequalities
If thex_(l)x_{l}are non-negative, we have the sequence of inequalities
M_(1) <= M_(2) <= dots <= M_(m) <= dotsM_{1} \leq M_{2} \leq \ldots \leq M_{m} \leq \ldots
equality between two consecutive terms can only occur if thex_(1)x_{1}are all identical.
- ^((1)){ }^{(1)}See: T. Popoviciu „On some properties of functions of one or two real variables" Thesis Paris 1933.