Posts by Tiberiu Popoviciu

Tiberiu Popoviciu

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T. Popoviciu

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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)

On a theorem of Laguerre

[JFM 60.1042.05]. https://zbmath.org/?q=an:60.1042.05

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1935 k -Popoviciu- Mathematica - On a theorem of Laguerre
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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 polynomial f ( x ) f ( x ) f(x)f(x)f(x)whose degree is at least: equal to n − 1 n − 1 n-1n-1n−1.
Either U ( x 1 , x 2 , … , x n ; f ( x ) ) U x 1 , x 2 , … , x n ; f ( x ) U(x_(1),x_(2),dots,x_(n);f(x))\mathrm{U}\left(x_{1}, x_{2}, \ldots, x_{n}; f(x)\right)U(x1,x2,…,xn;f(x))the determinant whose general line is
1 x i x i 2 … x i n − 2 f ( x i ) 1 x i x i 2 … x i n − 2 f x i 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)1xixi2…xin−2f(xi)
and let's ask V ( x 1 , x 2 , … , x n ) = U ( x 1 , x 2 , … , x n ; x n − 1 ) V x 1 , x 2 , … , x n = U x 1 , x 2 , … , x n ; x n − 1 V(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)V(x1,x2,…,xn)=U(x1,x2,…,xn;xn−1)which is the Van der Monde determinant of quantities x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, ..., x_{n}x1,x2,…,xn.
We have identity
(1) U ( x 1 , x 2 , … , x n ; f ( x ) ) = ∑ i = 1 i = n ( − 1 ) n − i V ( x 1 , x 2 , … , x i − 1 , x i + 1 , … , x n ) f ( x i ) U x 1 , x 2 , … , x n ; f ( x ) = ∑ i = 1 i = n   ( − 1 ) n − i V x 1 , x 2 , … , x i − 1 , x i + 1 , … , x n f x i 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)U(x1,x2,…,xn;f(x))=∑i=1i=n(−1)n−iV(x1,x2,…,xi−1,xi+1,…,xn)f(xi)which we will need later.
Let us consider the quotient
[ x 1 , x 2 , … , x n ; f ( x ) ] = U ( x 1 , x 2 , … , x n ; f ( x ) ) . x 1 , x 2 , … , x n ; f ( x ) = U x 1 , x 2 , … , x n ; f ( x ) . [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)} .[x1,x2,…,xn;f(x)]=U(x1,x2,…,xn;f(x)).
Let us now recall that we say that the polynomial f ( x ) f ( x ) f(x)f(x)f(x)is convex of order ( n − 2 n − 2 n-2n-2n−2) ( 1 1 ^(1){ }^{1}1) in the interval ( has , b has , b a,ba, bhas,b) if for any group of: n n nnndistinct points x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, ..., x_{n}x1,x2,…,xnfrom this interval we have the inequality
(2) [ x 1 , x 2 , … , x n ; f ( x ) ] > 0 (2) x 1 , x 2 , … , x n ; f ( x ) > 0 {:(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*}(2)[x1,x2,…,xn;f(x)]>0
So that the polynomial f ( x ) f ( x ) f(x)f(x)f(x)either convex of order ( n − 2 n − 2 n-2n-2n−2) In ( has , b has , b a,ba, bhas,b) it is necessary and sufficient that f ( n − 1 ) ( x ) ≥ 0 f ( n − 1 ) ( x ) ≥ 0 f^((n-1))(x) >= 0f^{(n-1)}(x) \geq 0f(n−1)(x)≥0in this interval. Inequality (2) will hold for any system of n n nnnpoint not all together.
2. - So now P ( x ) = ( 1 − x 1 x ) ( 1 − x 2 x ) … , ( 1 − x n x ) P ( x ) = 1 − x 1 x 1 − x 2 x … , 1 − x n x P(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)P(x)=(1−x1x)(1−x2x)…,(1−xnx)a polynomial having all its zeros real and consider the expansion
(3)
We know that
1 P ( x ) = S 0 + S 1 x + S 2 x 2 + ⋯ 1 P ( x ) = S 0 + S 1 x + S 2 x 2 + ⋯ (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}+\cdots1P(x)=S0+S1x+S2x2+⋯
S m = ∑ x 1 i 1 x 2 i 2 … x n i n S m = ∑ x 1 i 1 x 2 i 2 … x n i n 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}}Sm=∑x1i1x2i2…xnin
where the summation is extended for all non-negative integer values ​​of i 1 , i 2 , … , i n i 1 , i 2 , … , i n i_(1),i_(2),dots,i_(n)i_{1}, i_{2}, \ldots, i_{n}i1,i2,…,insuch as i 1 + i 2 + ⋯ + i n = m i 1 + i 2 + ⋯ + i n = m i_(1)+i_(2)+cdots+i_(n)=mi_{1}+i_{2}+\cdots+i_{n}=mi1+i2+⋯+in=m.
Especially if x 1 = x 2 = … = x n = 1 x 1 = x 2 = … = x n = 1 x_(1)=x_(2)=dots=x_(n)=1x_{1}=x_{2}=\ldots=x_{n}=1x1=x2=…=xn=1we have
S m = ( n + n − 1 m ) = ( m + n − 1 n − 1 ) S m = ( n + n − 1 m ) = ( m + n − 1 n − 1 ) 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}Sm=(n+n−1m)=(m+n−1n−1)
Decomposing the left-hand side of (3) into simple fractions, we have
1 P ( x ) = − ∑ i = 1 i = n ( − 1 ) n − i V ( x 1 , x 2 , … , x i − 1 , x i + 1 , … , x n ) x i n − 1 V ( x 1 , x 2 , … , x n ) ( 1 − x i x ) 1 P ( x ) = − ∑ i = 1 i = n   ( − 1 ) n − i V x 1 , x 2 , … , x i − 1 , x i + 1 , … , x n x i n − 1 V x 1 , x 2 , … , x n 1 − x i x (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)}1P(x)=−∑i=1i=n(−1)n−i V(x1,x2,…,xi−1,xi+1,…,xn)xin−1 V(x1,x2,…,xn)(1−xix)
and taking into account (1) we find
(4)
S m = [ x 1 , x 2 , … , x n ; x m + n − 1 ] S m = x 1 , x 2 , … , x n ; x m + n − 1 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]Sm=[x1,x2,…,xn;xm+n−1]
    • Let us consider the equation of degree 2 p 2 p 2p2 p2p
F ( z ) ≡ has 0 S k + has 1 S k + 1 z … has 2 p S 2 p + k z 2 p = 0 , has 2 p > 0 F ( z ) ≡ has 0 S k + has 1 S k + 1 z … has 2 p S 2 p + k z 2 p = 0 , has 2 p > 0 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}>0F(z)≡has0 Sk+has1 Sk+1z…has2p S2p+kz2p=0,has2p>0
If we pose
φ ( x ) = x n + k − 1 ( has 0 + has 1 x z + has 2 ( x z ) 2 + ⋯ + has 2 p ( x z ) 2 p ) φ ( x ) = x n + k − 1 has 0 + has 1 x z + has 2 ( x z ) 2 + ⋯ + has 2 p ( x z ) 2 p 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)φ(x)=xn+k−1(has0+has1xz+has2(xz)2+⋯+has2p(xz)2p)
we have according to (4)
F ( z ) = [ x 1 , x 2 , … , x n ; φ ( x ) ] F ( z ) = x 1 , x 2 , … , x n ; φ ( x ) 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]F(z)=[x1,x2,…,xn;φ(x)]
If we notice that
φ ( n − 1 ) ( x ) = ( n − 1 ) ! x k [ ( k + n − 1 n − 1 ) has 0 + ( k + n n − 1 ) has 1 x z + ⋯ + ( k + n + 2 p − 1 n − 1 ) has 2 p ( x z ) 2 p ] φ ( n − 1 ) ( x ) = ( n − 1 ) ! x k ( k + n − 1 n − 1 ) has 0 + ( k + n n − 1 ) has 1 x z + ⋯ + ( k + n + 2 p − 1 n − 1 ) has 2 p ( x z ) 2 p 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]φ(n−1)(x)=(n−1)!xk[(k+n−1n−1)has0+(k+nn−1)has1xz+⋯+(k+n+2p−1n−1)has2p(xz)2p]
We can state the following properties:
If k k kkkis even and if the polynomial
(5) ∑ i = 0 i = 2 p ( k + n + i − 1 n − 1 ) a i z i (5) ∑ i = 0 i = 2 p   ( k + n + i − 1 n − 1 ) a i z i {:(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*}(5)∑i=0i=2p(k+n+i−1n−1)hasizi
is non-negative, the polynomial
(6) ∑ i = 0 i = 2 p a i S k + i z i (6) ∑ i = 0 i = 2 p   a i S k + i z i {:(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*}(6)∑i=0i=2phasi Sk+izi
is positive as long as the roots x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,…,xnare not all equal.
If k k kkkis odd and if the polynomial (5) is non-negative the polynomial (6) is positive as long as the roots x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,…,xnare non-negative and not all equal.
4. - Laguerre demonstrated the following proposition:
If the equation P ( x ) = 0 P ( x ) = 0 P(x)=0\mathrm{P}(x)=0P(x)=0to all its real roots the polynomial
∑ i = 0 2 p S i z i ∑ i = 0 2 p   S i z i sum_(i=0)^(2p)S_(i)z^(i)\sum_{i=0}^{2 p} \mathrm{~S}_{i} z^{i}∑i=02p Sizi
is positive.
Let us always assume that Laguerre's theorem is true when all the zeros of the polynomial P ( x ) P ( x ) P(x)\mathrm{P}(x)P(x)are equal. In other words, the polynomial
Q a , 0 = ∑ i = 1 2 p ( α + i α ) z i Q a , 0 = ∑ i = 1 2 p   ( α + i α ) z i 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}Qhas,0=∑i=12p(α+iα)zi
is positive for α α alpha\alphaαentire > 0 > 0 > 0>0>0
Let us consider the more general polynomial
Q a , β = ∑ i = 0 2 p ( α + 3 + i α ) z i , α , β entiers ≥ 0 . Q a , β = ∑ i = 0 2 p   ( α + 3 + i α ) z i , α , β  entiers  ≥ 0 . 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 .Qhas,β=∑i=02p(α+3+iα)zi,α,β whole ≥0.
The recurrence relation
Q a , β = Q a , β − 1 + Q a − 1 , β Q a , β = Q a , β − 1 + Q a − 1 , β Q_(a,beta)=Q_(a,beta-1)+Q_(a-1,beta)Q_{a, \beta}=Q_{a, \beta-1}+Q_{a-1, \beta}Qhas,β=Qhas,β−1+Qhas−1,β
and the fact that Q 0 , β Q 0 , β Q_(0,beta)Q_{0, \beta}Q0,βis independent of β ( Q 0 , β = 1 + z + z 2 + ⋯ + z 2 p ) β Q 0 , β = 1 + z + z 2 + ⋯ + z 2 p beta(Q_(0,beta)=1+z+z^(2)+cdots+z^(2p))\beta\left(Q_{0, \beta}=1+z+z^{2}+\cdots+z^{2 p}\right)β(Q0,β=1+z+z2+⋯+z2p)show us that the polynomial Q α , β Q α , β Q_(alpha,beta)Q_{\alpha, \beta}Qα,βis positive.
So doing a 1 = a 2 = ⋯ = a 2 p = 1 a 1 = a 2 = ⋯ = a 2 p = 1 a_(1)=a_(2)=cdots=a_(2p)=1a_{1}=a_{2}=\cdots=a_{2 p}=1has1=has2=⋯=has2p=1In (5) we deduce the following properties:
If k k kkkis even the polynomial
(7)
∑ i = 0 2 p S k + i z i ∑ i = 0 2 p   S k + i z i sum_(i=0)^(2p)S_(k+i)z^(i)\sum_{i=0}^{2 p} \mathrm{~S}_{k+i} z^{i}∑i=02p Sk+izi
is certainly positive, as long as the roots x 1 , x 2 , … , x n x 1 , x 2 , … , x n x_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}x1,x2,…,xnare not all equal.
If k k kkkis odd, the polynomial (7) is certainly positive as long as the roots x i x i x_(i)x_{i}xiare non-negative and not all equal.
5. - Let
σ m = S m ( m + n − 1 n − 1 ) , M m = σ m m σ m = S m ( m + n − 1 n − 1 ) , M m = σ m m 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}}σm=Sm(m+n−1n−1),Mm=σmm
(1) See Lagumrre “Works” t. J. p. 111.
M m M m M_(m)\mathrm{M}_{m}Mmis therefore an average value.
If
p = 1 , a i = ( 2 i ) ( k + n + i − 1 n − 1 ) , 1 = 0 , 1 , 2 p = 1 , a i = ( 2 i ) ( k + n + i − 1 n − 1 ) , 1 = 0 , 1 , 2 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=0,1,2p=1,hasi=(2i)(k+n+i−1n−1),1=0,1,2
My polynomial (5) is non-negative, therefore assuming the x l x l x_(l)x_{l}xLnon-negative soft we also have
σ k + 2 σ k + 1 x + σ k + 2 x 2 ≥ 0 σ k + 2 σ k + 1 x + σ k + 2 x 2 ≥ 0 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 0σk+2σk+1x+σk+2x2≥0
"regardless of k k kkkand equality is only possible if all the x i x i x_(i)x_{i}xiare identical.
So we have
σ k + 1 2 ≤ σ k , σ k + 2 σ 0 = 1 , k = 0 , 1 , 2 , … σ k + 1 2 ≤ σ k , σ k + 2 σ 0 = 1 , k = 0 , 1 , 2 , … {:[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}σk+12≤σk,σk+2σ0=1,k=0,1,2,…
aet we can then state the following property:
If the x l x l x_(l)x_{l}xLare non-negative, we have the sequence of inequalities
M 1 ≤ M 2 ≤ … ≤ M m ≤ … M 1 ≤ M 2 ≤ … ≤ M m ≤ … M_(1) <= M_(2) <= dots <= M_(m) <= dotsM_{1} \leq M_{2} \leq \ldots \leq M_{m} \leq \ldotsM1≤M2≤…≤Mm≤…
equality between two consecutive terms can only occur if the x 1 x 1 x_(1)x_{1}x1are all identical.
  1. ( 1 ) ( 1 ) ^((1)){ }^{(1)}(1)See: T. Popoviciu „On some properties of functions of one or two real variables" Thesis Paris 1933.

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