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-1 n-1 n − 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 ( x 1 , x 2 , … , x n ; 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) 1 x i x i 2 … x i n − 2 f ( x i ) 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 ( x 1 , x 2 , … , x n ) = U ( x 1 , x 2 , … , x n ; x n − 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} x 1 , x 2 , … , x n . 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 ( 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 ) 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)} . [ x 1 , x 2 , … , x n ; f ( x ) ] = U ( x 1 , x 2 , … , x n ; 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-2 n-2 n − 2 ) (
1
1
^(1) { }^{1} 1 ) in the interval (
has , b
has , b a,b a, b has , b ) if for any group of:
n
n n n n distinct 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} x 1 , x 2 , … , x n from 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) [ x 1 , x 2 , … , x n ; 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-2 n-2 n − 2 ) In (
has , b
has , b a,b a, b has , b ) it is necessary and sufficient that
f
( n − 1 )
( x ) ≥ 0
f
( n − 1 )
( x ) ≥ 0 f^((n-1))(x) >= 0 f^{(n-1)}(x) \geq 0 f ( n − 1 ) ( x ) ≥ 0 in this interval. Inequality (2) will hold for any system of
n
n n n n point 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 − x 1 x ) ( 1 − x 2 x ) … , ( 1 − x n x ) 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}+\cdots 1 P ( x ) = S 0 + S 1 x + S 2 x 2 + ⋯
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}} S m = ∑ x 1 i 1 x 2 i 2 … x n i n 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} i 1 , i 2 , … , i n such as
i
1
+
i
2
+ ⋯ +
i
n
= m
i
1
+
i
2
+ ⋯ +
i
n
= m i_(1)+i_(2)+cdots+i_(n)=m i_{1}+i_{2}+\cdots+i_{n}=m i 1 + i 2 + ⋯ + i n = m . Especially if
x
1
=
x
2
= … =
x
n
= 1
x
1
=
x
2
= … =
x
n
= 1 x_(1)=x_(2)=dots=x_(n)=1 x_{1}=x_{2}=\ldots=x_{n}=1 x 1 = x 2 = … = x n = 1 we 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} S m = ( n + n − 1 m ) = ( m + n − 1 n − 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)} 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 ) 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] S m = [ x 1 , x 2 , … , x n ; x m + n − 1 ]
Let us consider the equation of degree
2 p
2 p 2p 2 p 2 p
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}>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 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 ) = x n + k − 1 ( has 0 + has 1 x z + has 2 ( x z ) 2 + ⋯ + has 2 p ( x z ) 2 p ) 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 ) = [ x 1 , x 2 , … , x n ; 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 ) ! 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 ]
We can state the following properties:
If
k
k
k k k is 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 = 0 i = 2 p ( k + n + i − 1 n − 1 ) has i z i 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 = 0 i = 2 p has i S k + i z i 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} x 1 , x 2 , … , x n are not all equal. If
k
k
k k k is 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} x 1 , x 2 , … , x n are 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)=0 P ( x ) = 0 to 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 = 0 2 p S i z i 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} Q has , 0 = ∑ i = 1 2 p ( α + i α ) z i is positive for
α
α
alpha \alpha α entire
>
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 . Q has , β = ∑ i = 0 2 p ( α + 3 + i α ) z i , 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} Q has , β = Q has , β − 1 + Q has − 1 , β and the fact that
Q
0
,
β
Q
0
,
β
Q_(0,beta) Q_{0, \beta} Q 0 , β is 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 ( Q 0 , β = 1 + z + z 2 + ⋯ + z 2 p ) show us that the polynomial
Q
α
,
β
Q
α
,
β
Q_(alpha,beta) Q_{\alpha, \beta} Q α , β is 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)=1 a_{1}=a_{2}=\cdots=a_{2 p}=1 has 1 = has 2 = ⋯ = has 2 p = 1 In (5) we deduce the following properties: If
k
k
k k k is 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 = 0 2 p S k + i z i 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} x 1 , x 2 , … , x n are not all equal. If
k
k
k k k is odd, the polynomial (7) is certainly positive as long as the roots
x
i
x
i
x_(i) x_{i} x i are 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}} σ m = S m ( m + n − 1 n − 1 ) , M m = σ m m (1) See Lagumrre “Works” t. J. p. 111.
M
m
M
m
M_(m) \mathrm{M}_{m} M m is 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=0,1,2 p=1, \quad a_{i}=\frac{\binom{2}{i}}{\binom{k+n+i-1}{n-1}}, \quad 1=0,1,2 p = 1 , has i = ( 2 i ) ( k + n + i − 1 n − 1 ) , 1 = 0 , 1 , 2 My polynomial (5) is non-negative, therefore assuming the
x
l
x
l
x_(l) x_{l} x L non-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 0 σ k + 2 σ k + 1 x + σ k + 2 x 2 ≥ 0 "regardless of
k
k
k k k and equality is only possible if all the
x
i
x
i
x_(i) x_{i} x i are 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} σ k + 1 2 ≤ σ k , σ k + 2 σ 0 = 1 , k = 0 , 1 , 2 , … aet we can then state the following property:
If the
x
l
x
l
x_(l) x_{l} x L are 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) <= dots M_{1} \leq M_{2} \leq \ldots \leq M_{m} \leq \ldots M 1 ≤ M 2 ≤ … ≤ M m ≤ … equality between two consecutive terms can only occur if the
x
1
x
1
x_(1) x_{1} x 1 are all identical.
