ON A SUFFICIENT CONDITION FOR A POLYNOMIAL TO BE POSITIVE by
Tiberiu Popoviciu in Cluj.
Received on May 5, 1935.
Consider a polynomial of degree
2 m
2 m 2m 2 m 2 m which we will write in the following form
f ( x ) =
has
0
c
0
+
has
1
c
1
x + , … , +
has
k
c
k
x
k
+ … , +
has
2 m
c
2 m
x
2 m
f ( x ) =
has
0
c
0
+
has
1
c
1
x + , … , +
has
k
c
k
x
k
+ … , +
has
2 m
c
2 m
x
2 m
f(x)=a_(0)c_(0)+a_(1)c_(1)x+,dots,+a_(k)c_(k)x^(k)+dots,+a_(2m)c_(2m)x^(2m) f(x)=a_{0} c_{0}+a_{1} c_{1} x+, \ldots,+a_{k} c_{k} x^{k}+\ldots,+a_{2 m} c_{2 m} x^{2 m} f ( x ) = has 0 c 0 + has 1 c 1 x + , … , + has k c k x k + … , + has 2 m c 2 m x 2 m Or
has
0
,
has
1
, … ,
has
k
, … ,
has
m
has
0
,
has
1
, … ,
has
k
, … ,
has
m
a_(0),a_(1),dots,a_(k),dots,a_(m) a_{0}, a_{1}, \ldots, a_{k}, \ldots, a_{m} has 0 , has 1 , … , has k , … , has m are positive constants and
c
L
c
L
c_(l) c_{l} c L real coefficients. Let's put this polynomial in the form
f ( x ) =
∑
i = 0
m − 1
[
α
i
c
2 i
x
2 i
( 1 +
β
i
c
2 i + 1
c
2 i
x )
2
+
γ
i
1
λ
2
c
2 i
c
2 i + 2
−
c
2 i + 1
2
c
2 i
x
2 i + 2
]
+
α
m
c
2 m
x
2 m
f ( x ) =
∑
i = 0
m − 1
α
i
c
2 i
x
2 i
1 +
β
i
c
2 i + 1
c
2 i
x
2
+
γ
i
1
λ
2
c
2 i
c
2 i + 2
−
c
2 i + 1
2
c
2 i
x
2 i + 2
+
α
m
c
2 m
x
2 m
f(x)=sum_(i=0)^(m-1)[alpha_(i)c_(2i)x^(2i)(1+beta_(i)(c_(2i+1))/(c_(2i))x)^(2)+gamma_( i)((1)/(lambda^(2))c_(2i)c_(2i+2)-c_(2i+1)^(2))/(c_(2i))x^(2i+2)]+alpha_(m)c_(2m)x^(2m) f(x)=\sum_{i=0}^{m-1}\left[\alpha_{i} c_{2 i} x^{2 i}\left(1+\beta_{i} \frac{c_{2 i+1}}{c_{2 i}} x\right)^{2}+\gamma_{i} \frac{\frac{1}{\lambda^{2}} c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}}{c_{2 i}} x^{2 i+2}\right]+\alpha_{m} c_{2 m} x^{2 m} f ( x ) = ∑ i = 0 m − 1 [ α i c 2 i x 2 i ( 1 + β i c 2 i + 1 c 2 i x ) 2 + γ i 1 λ 2 c 2 i c 2 i + 2 − c 2 i + 1 2 c 2 i x 2 i + 2 ] + α m c 2 m x 2 m
and let's determine the constants
α
i
,
β
i
,
γ
i
α
i
,
β
i
,
γ
i
alpha_(i),beta_(i),gamma_(i) \alpha_{i}, \beta_{i}, \gamma_{i} α i , β i , γ i by identification. We obtain
α
0
=
has
0
,
2
α
i
β
i
=
has
2 i + 1
,
β
i
2
α
i
=
γ
i
,
α
i + 1
+
γ
i
λ
2
=
has
2 i + 2
i = 0 , 1 , … , m − 1
α
0
=
has
0
,
2
α
i
β
i
=
has
2 i + 1
,
β
i
2
α
i
=
γ
i
,
α
i + 1
+
γ
i
λ
2
=
has
2 i + 2
i = 0 , 1 , … , m − 1
{:[alpha_(0)=a_(0)","quad2alpha_(i)beta_(i)=a_(2i+1)","quadbeta_(i)^(2)alpha_(i)=gamma_(i)","alpha_(i+1)+(gamma_(i))/(lambda^(2))=a_(2i+2)],[i=0","1","dots","m-1]:} \begin{gathered}
\alpha_{0}=a_{0}, \quad 2 \alpha_{i} \beta_{i}=a_{2 i+1}, \quad \beta_{i}^{2} \alpha_{i}=\gamma_{i}, \alpha_{i+1}+\frac{\gamma_{i}}{\lambda^{2}}=a_{2 i+2} \\
i=0,1, \ldots, m-1
\end{gathered} α 0 = has 0 , 2 α i β i = has 2 i + 1 , β i 2 α i = γ i , α i + 1 + γ i λ 2 = has 2 i + 2 i = 0 , 1 , … , m − 1 The constants
α
i
α
i
alpha_(i) \alpha_{i} α i are therefore determined by recurrence relations
α
0
=
a
0
,
α
i
+
1
+
a
2
i
+
1
2
4
λ
2
α
i
=
a
2
i
+
2
,
i
=
0
,
1
,
…
,
m
−
1
.
α
0
=
a
0
,
α
i
+
1
+
a
2
i
+
1
2
4
λ
2
α
i
=
a
2
i
+
2
,
i
=
0
,
1
,
…
,
m
−
1
.
alpha_(0)=a_(0),alpha_(i+1)+(a_(2i+1)^(2))/(4lambda^(2)alpha_(i))=a_(2i+2),quad i=0,1,dots,m-1. \alpha_{0}=a_{0}, \alpha_{i+1}+\frac{a_{2 i+1}^{2}}{4 \lambda^{2} \alpha_{i}}=a_{2 i+2}, \quad i=0,1, \ldots, m-1 . α 0 = has 0 , α i + 1 + has 2 i + 1 2 4 λ 2 α i = has 2 i + 2 , i = 0 , 1 , … , m − 1 . We can write
α
i
=
P
^
i
+
1
(
λ
)
λ
P
i
(
λ
)
α
i
=
P
^
i
+
1
(
λ
)
λ
P
i
(
λ
)
alpha_(i)=( hat(P)_(i+1)(lambda))/(lambdaP_(i)(lambda)) \alpha_{i}=\frac{\hat{P}_{i+1}(\lambda)}{\lambda P_{i}(\lambda)} α i = P ^ i + 1 ( λ ) λ P i ( λ ) and then
P
k
(
λ
)
P
k
(
λ
)
P_(k)(lambda) P_{k}(\lambda) P k ( λ ) is a polynomial of degree
k
k
k k k in
λ
λ
lambda \lambda λ
These polynomials satisfy the recurrence relations .
P
0
(
λ
)
=
1
,
P
1
(
λ
)
=
a
0
λ
,
P
i
+
2
(
λ
)
−
λ
a
2
i
+
2
P
i
+
1
(
λ
)
+
a
2
i
+
1
2
4
P
i
(
λ
)
=
0
P
0
(
λ
)
=
1
,
P
1
(
λ
)
=
a
0
λ
,
P
i
+
2
(
λ
)
−
λ
a
2
i
+
2
P
i
+
1
(
λ
)
+
a
2
i
+
1
2
4
P
i
(
λ
)
=
0
P_(0)(lambda)=1,quadP_(1)(lambda)=a_(0)lambda,quadP_(i+2)(lambda)-lambdaa_(2i+2)P_(i+1)(lambda)+(a_(2i+1)^(2))/(4)P_(i)(lambda)=0 \mathrm{P}_{0}(\lambda)=1, \quad \mathrm{P}_{1}(\lambda)=a_{0} \lambda, \quad \mathrm{P}_{i+2}(\lambda)-\lambda a_{2 i+2} \mathrm{P}_{i+1}(\lambda)+\frac{a_{2 i+1}^{2}}{4} \mathrm{P}_{i}(\lambda)=0 P 0 ( λ ) = 1 , P 1 ( λ ) = has 0 λ , P i + 2 ( λ ) − λ has 2 i + 2 P i + 1 ( λ ) + has 2 i + 1 2 4 P i ( λ ) = 0 Let's ask
Q
0
λ
)
=
P
0
(
λ
)
,
Q
i
(
λ
)
=
P
i
(
λ
)
a
0
a
2
,
…
,
a
2
i
−
2
Q
0
λ
=
P
0
(
λ
)
,
Q
i
(
λ
)
=
P
i
(
λ
)
a
0
a
2
,
…
,
a
2
i
−
2
{:Q_(0)lambda)=P_(0)(lambda),quadQ_(i)(lambda)=(P_(i)(lambda))/(a_(0)a_(2),dots,a_(2i-2)) \left.Q_{0} \lambda\right)=P_{0}(\lambda), \quad Q_{i}(\lambda)=\frac{P_{i}(\lambda)}{a_{0} a_{2}, \ldots, a_{2 i-2}} Q 0 λ ) = P 0 ( λ ) , Q i ( λ ) = P i ( λ ) has 0 has 2 , … , has 2 i − 2 we then
Q
J
(
λ
)
=
1
,
Q
I
(
λ
)
=
λ
,
Q
I
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
a
2
i
+
1
2
4
a
2
i
a
2
i
+
2
Q
i
(
λ
)
=
0
Q
J
(
λ
)
=
1
,
Q
I
(
λ
)
=
λ
,
Q
I
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
a
2
i
+
1
2
4
a
2
i
a
2
i
+
2
Q
i
(
λ
)
=
0
Q_(J)(lambda)=1,Q_(I)(lambda)=lambda,Q_(I+2)(lambda)-lambdaQ_(i+1)(lambda)+(a_(2i+1)^(2))/(4a_(2i)a_(2i+2))Q_(i)(lambda)=0 Q_{J}(\lambda)=1, Q_{I}(\lambda)=\lambda, Q_{I+2}(\lambda)-\lambda Q_{i+1}(\lambda)+\frac{a_{2 i+1}^{2}}{4 a_{2 i} a_{2 i+2}} Q_{i}(\lambda)=0 Q I ( λ ) = 1 , Q I ( λ ) = λ , Q I + 2 ( λ ) − λ Q i + 1 ( λ ) + has 2 i + 1 2 4 has 2 i has 2 i + 2 Q i ( λ ) = 0 The polynomial
Q
k
(
λ
)
Q
k
(
λ
)
Q_(k)(lambda) Q_{k}(\lambda) Q k ( λ ) has all its real zeros and the zeros of
Q
k
+
1
(
λ
)
Q
k
+
1
(
λ
)
Q_(k+1)(lambda) Q_{k+1}(\lambda) Q k + 1 ( λ ) are separated by those of
Q
k
(
λ
)
Q
k
(
λ
)
Q_(k)(lambda) Q_{k}(\lambda) Q k ( λ ) . So that the constants
α
i
α
i
alpha_(i) \alpha_{i} α i all things must be positive, it is necessary that
λ
λ
lambda \lambda λ be greater than the largest zero of
Q
m
+
1
(
λ
)
Q
m
+
1
(
λ
)
Q_(m+1)(lambda) Q_{m+1}(\lambda) Q m + 1 ( λ ) In this case, the constants
β
i
β
i
beta_(i) \beta_{i} β i And
γ
i
γ
i
gamma_(i) \gamma_{i} γ i are all positive as well. We can see, therefore, that if we have
c
0
>
0
,
1
λ
2
c
2
i
r
2
i
+
2
−
c
2
i
+
1
2
>
0
(
i
=
0
,
1
,
…
,
m
−
1
)
c
0
>
0
,
1
λ
2
c
2
i
r
2
i
+
2
−
c
2
i
+
1
2
>
0
(
i
=
0
,
1
,
…
,
m
−
1
)
{:[c_(0) > 0","quad(1)/(lambda^(2))c_(2i)r_(2i+2)-c_(2i+1)^(2) > 0],[(i=0","1","dots","m-1)]:} \begin{gathered}
c_{0}>0, \quad \frac{1}{\lambda^{2}} c_{2 i} r_{2 i+2}-c_{2 i+1}^{2}>0 \\
(i=0,1, \ldots, m-1)
\end{gathered} c 0 > 0 , 1 λ 2 c 2 i r 2 i + 2 − c 2 i + 1 2 > 0 ( i = 0 , 1 , … , m − 1 ) Or
λ
λ
lambda \lambda λ is at least equal to the largest root
λ
m
+
1
λ
m
+
1
lambda_(m+1) \lambda_{m+1} λ m + 1 of the equation
Q
m
+
1
(
λ
)
=
0
Q
m
+
1
(
λ
)
=
0
Q_(m+1)(lambda)=0 \mathrm{Q}_{m+1}(\lambda)=0 Q m + 1 ( λ ) = 0 the polynomial
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) is positive.
2. - The limit found
λ
m
+
1
λ
m
+
1
lambda_(m+1) \lambda_{m+1} λ m + 1 is the best possible. One can see it directly, or in the following way: Let's ask
c
2
i
=
x
i
2
,
1
λ
2
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
=
0
,
c
2
i
+
1
=
−
1
λ
x
i
x
+
1
i
=
0
,
1
,
…
,
m
(
x
i
≠
0
)
i
=
0
,
1
,
…
,
m
−
1
c
2
i
=
x
i
2
,
1
λ
2
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
=
0
,
c
2
i
+
1
=
−
1
λ
x
i
x
+
1
i
=
0
,
1
,
…
,
m
x
i
≠
0
i
=
0
,
1
,
…
,
m
−
1
{:[c_(2i)=x_(i)^(2)","quad(1)/(lambda^(2))c_(2i)c_(2i+2)-c_(2i+1)^(2)=0","quadc_(2i+1)=-(1)/(lambda)x_(i)x_(+1)],[i=0","1","dots","m quad(x_(i)!=0)quad i=0","1","dots","m-1]:} \begin{gathered}
c_{2 i}=x_{i}^{2}, \quad \frac{1}{\lambda^{2}} c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}=0, \quad c_{2 i+1}=-\frac{1}{\lambda} x_{i} x_{+1} \\
i=0,1, \ldots, m \quad\left(x_{i} \neq 0\right) \quad i=0,1, \ldots, m-1
\end{gathered} c 2 i = x i 2 , 1 λ 2 c 2 i c 2 i + 2 − c 2 i + 1 2 = 0 , c 2 i + 1 = − 1 λ x i x + 1 i = 0 , 1 , … , m ( x i ≠ 0 ) i = 0 , 1 , … , m − 1 and determine
λ
λ
lambda \lambda λ in the manner that the equation
f
(
x
)
=
0
f
(
x
)
=
0
f(x)=0 f(x)=0 f ( x ) = 0 can have at least one real root. We can immediately see that it suffices to examine the cases where +1 is a root of this equation. We then have
a
0
x
0
2
+
a
2
x
1
2
+
,
…
,
+
a
2
m
x
m
2
=
1
λ
(
a
1
x
0
x
1
+
a
3
x
m
x
2
+
…
+
a
2
m
−
1
x
m
−
1
x
m
)
a
0
x
0
2
+
a
2
x
1
2
+
,
…
,
+
a
2
m
x
m
2
=
1
λ
a
1
x
0
x
1
+
a
3
x
m
x
2
+
…
+
a
2
m
−
1
x
m
−
1
x
m
a_(0)x_(0)^(2)+a_(2)x_(1)^(2)+,dots,+a_(2m)x_(m)^(2)=(1)/(lambda)(a_(1)x_(0)x_(1)+a_(3)x_(m)x_(2)+dots+a_(2m-1)x_(m-1)x_(m)) a_{0} x_{0}^{2}+a_{2} x_{1}^{2}+, \ldots,+a_{2 m} x_{m}^{2}=\frac{1}{\lambda}\left(a_{1} x_{0} x_{1}+a_{3} x_{m} x_{2}+\ldots+a_{2 m-1} x_{m-1} x_{m}\right) has 0 x 0 2 + has 2 x 1 2 + , … , + has 2 m x m 2 = 1 λ ( has 1 x 0 x 1 + has 3 x m x 2 + … + has 2 m − 1 x m − 1 x m ) Or
λ
=
a
1
x
0
x
1
+
a
3
x
1
x
2
+
,
…
+
a
2
m
−
1
x
m
−
1
x
m
a
0
x
0
2
+
a
2
x
1
2
+
…
+
+
a
2
m
x
m
2
λ
=
a
1
x
0
x
1
+
a
3
x
1
x
2
+
,
…
+
a
2
m
−
1
x
m
−
1
x
m
a
0
x
0
2
+
a
2
x
1
2
+
…
+
+
a
2
m
x
m
2
lambda=(a_(1)x_(0)x_(1)+a_(3)x_(1)x_(2)+,dots+a_(2m-1)x_(m-1)x_(m))/(a_(0)x_(0)^(2)+a_(2)x_(1)^(2)+dots++a_(2m)x_(m)^(2)) \lambda=\frac{a_{1} x_{0} x_{1}+a_{3} x_{1} x_{2}+, \ldots+a_{2 m-1} x_{m-1} x_{m}}{a_{0} x_{0}^{2}+a_{2} x_{1}^{2}+\ldots++a_{2 m} x_{m}^{2}} λ = has 1 x 0 x 1 + has 3 x 1 x 2 + , … + has 2 m − 1 x m − 1 x m has 0 x 0 2 + has 2 x 1 2 + … + + has 2 m x m 2
λ
λ
lambda \lambda λ must therefore be between the maximum and minimum of the right-hand side, in other words between the maximum and minimum of the quadratic form
∑
i
=
0
m
−
1
a
2
i
+
1
x
i
x
i
+
1
∑
i
=
0
m
−
1
a
2
i
+
1
x
i
x
i
+
1
sum_(i=0)^(m-1)a_(2i+1)x_(i)x_(i+1) \sum_{i=0}^{m-1} a_{2 i+1} x_{i} x_{i+1} ∑ i = 0 m − 1 has 2 i + 1 x i x i + 1 when the variables are linked by the relation
∑
i
=
0
m
a
2
i
x
1
2
=
1
∑
i
=
0
m
a
2
i
x
1
2
=
1
sum_(i=0)^(m)a_(2i)x_(1)^(2)=1 \sum_{i=0}^{m} a_{2 i} x_{1}^{2}=1 ∑ i = 0 m has 2 i x 1 2 = 1 . The result is that
λ
λ
lambda \lambda λ is between the smallest and largest roots of the characteristic equation
R
n
+
1
(
x
)
=
|
−
2
x
a
0
a
1
0
0
…
…
…
0
a
1
−
2
x
a
2
a
3
0
…
…
…
0
0
a
3
−
2
x
a
4
a
5
…
…
…
…
⋯
…
…
…
…
…
…
0
0
0
0
…
…
a
2
m
−
1
−
2
x
a
2
m
|
=
0
R
n
+
1
(
x
)
=
−
2
x
a
0
a
1
0
0
…
…
…
0
a
1
−
2
x
a
2
a
3
0
…
…
…
0
0
a
3
−
2
x
a
4
a
5
…
…
…
…
⋯
…
…
…
…
…
…
0
0
0
0
…
…
a
2
m
−
1
−
2
x
a
2
m
=
0
R_(n+1)(x)=|[-2xa_(0),a_(1),0,0,dots,dots,dots,0],[a_(1),-2xa_(2),a_(3),0,dots,dots,dots,0],[0,a_(3),-2xa_(4),a_(5),dots,dots,dots,dots],[cdots,dots,dots,dots,dots,dots,dots],[0,0,0,0,dots,dots,a_(2m-1),-2xa_(2m)]|=0 \mathbb{R}_{n+1}(x)=\left|\begin{array}{cccccccc}-2 x a_{0} & a_{1} & 0 & 0 & \ldots & \ldots & \ldots & 0 \\ a_{1} & -2 x a_{2} & a_{3} & 0 & \ldots & \ldots & \ldots & 0 \\ 0 & a_{3} & -2 x a_{4} & a_{5} & \ldots & \ldots & \ldots & \ldots \\ \cdots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & 0 & \ldots & \ldots & a_{2 m-1} & -2 x a_{2 m}\end{array}\right|=0 R n + 1 ( x ) = | − 2 x has 0 has 1 0 0 … … … 0 has 1 − 2 x has 2 has 3 0 … … … 0 0 has 3 − 2 x has 4 has 5 … … … … ⋯ … … … … … … 0 0 0 0 … … has 2 m − 1 − 2 x has 2 m | = 0
It is immediately apparent that
R
1
(
x
)
=
−
2
x
a
0
,
R
2
(
x
)
=
4
a
0
a
2
(
x
2
−
a
1
2
4
a
0
a
2
)
R
i
+
2
(
x
)
+
2
x
a
2
i
+
2
R
i
+
1
(
x
)
+
a
2
i
+
1
2
R
i
(
x
)
=
0
R
1
(
x
)
=
−
2
x
a
0
,
R
2
(
x
)
=
4
a
0
a
2
x
2
−
a
1
2
4
a
0
a
2
R
i
+
2
(
x
)
+
2
x
a
2
i
+
2
R
i
+
1
(
x
)
+
a
2
i
+
1
2
R
i
(
x
)
=
0
{:[R_(1)(x)=-2xa_(0)","quadR_(2)(x)=4a_(0)a_(2)(x^(2)-(a_(1)^(2))/(4a_(0)a_(2)))],[R_(i+2)(x)+2xa_(2i+2)R_(i+1)(x)+a_(2i+1)^(2)R_(i)(x)=0]:} \begin{gathered}
\mathrm{R}_{1}(x)=-2 x a_{0}, \quad \mathrm{R}_{2}(x)=4 a_{0} a_{2}\left(x^{2}-\frac{a_{1}^{2}}{4 a_{0} a_{2}}\right) \\
\mathrm{R}_{i+2}(x)+2 x a_{2 i+2} \mathrm{R}_{i+1}(x)+a_{2 i+1}^{2} \mathrm{R}_{i}(x)=0
\end{gathered} R 1 ( x ) = − 2 x has 0 , R 2 ( x ) = 4 has 0 has 2 ( x 2 − has 1 2 4 has 0 has 2 ) R i + 2 ( x ) + 2 x has 2 i + 2 R i + 1 ( x ) + has 2 i + 1 2 R i ( x ) = 0 The result is that
R
i
(
x
)
=
(
−
1
)
i
2
i
a
0
a
2
…
a
2
i
−
2
Q
i
(
x
)
R
i
(
x
)
=
(
−
1
)
i
2
i
a
0
a
2
…
a
2
i
−
2
Q
i
(
x
)
R_(i)(x)=(-1)^(i)2^(i)a_(0)a_(2)dotsa_(2i-2)Q_(i)(x) \mathrm{R}_{i}(x)=(-1)^{i} 2^{i} a_{0} a_{2} \ldots a_{2 i-2} Q_{i}(x) R i ( x ) = ( − 1 ) i 2 i has 0 has 2 … has 2 i − 2 Q i ( x ) The stated property follows from this identity.
3. - Let's examine some specific cases of this problem.
Let us first assume that
a
:
=
a
1
=
…
=
a
2
m
=
1
a
:
=
a
1
=
…
=
a
2
m
=
1
a_(:)=a_(1)=dots=a_(2m)=1 a_{:}=a_{1}=\ldots=a_{2 m}=1 has : = has 1 = … = has 2 m = 1 The polynomial
Q
′
(
x
)
Q
′
(
x
)
Q^(')(x) Q^{\prime}(x) Q ′ ( x ) is none other than the trigonometric polynomial
Q
i
(
x
)
=
sin
(
i
+
1
)
arccos
x
2
′
1
−
x
2
Q
i
(
x
)
=
sin
(
i
+
1
)
arccos
x
2
′
1
−
x
2
Q_(i)(x)=(sin(i+1)arccos x)/(2^(')sqrt(1-x^(2))) Q_{i}(x)=\frac{\sin (i+1) \arccos x}{2^{\prime} \sqrt{1-x^{2}}} Q i ( x ) = si ( i + 1 ) arccos x 2 ′ 1 − x 2 and then we have
λ
m
+
1
=
cos
π
m
+
2
λ
m
+
1
=
cos
π
m
+
2
lambda_(m+1)=cos((pi)/(m+2)) \lambda_{m+1}=\cos \frac{\pi}{m+2} λ m + 1 = cos π m + 2 We therefore deduce the following property:
If the coefficients
c
0
,
c
1
,
c
2
,
…
,
c
2
m
c
0
,
c
1
,
c
2
,
…
,
c
2
m
c_(0),c_(1),c_(2),dots,c_(2m) c_{0}, c_{1}, c_{2}, \ldots, c_{2 m} c 0 , c 1 , c 2 , … , c 2 m of the polynomial
(1)
c
0
+
c
1
x
+
c
2
x
2
+
…
+
c
2
m
x
2
m
c
0
+
c
1
x
+
c
2
x
2
+
…
+
c
2
m
x
2
m
c_(0)+c_(1)x+c_(2)x^(2)+dots+c_(2m)x^(2m) c_{0}+c_{1} x+c_{2} x^{2}+\ldots+c_{2 m} x^{2 m} c 0 + c 1 x + c 2 x 2 + … + c 2 m x 2 m check the inequalities
c
0
>
0
,
μ
c
2
i
c
2
i
+
2
−
c
i
i
+
1
2
>
0
,
1
=
0
,
1
,
…
,
m
−
1
c
0
>
0
,
μ
c
2
i
c
2
i
+
2
−
c
i
i
+
1
2
>
0
,
1
=
0
,
1
,
…
,
m
−
1
c_(0) > 0,quad muc_(2i)c_(2i+2)-c_(ii+1)^(2) > 0,quad1=0,1,dots,m-1 c_{0}>0, \quad \mu c_{2 i} c_{2 i+2}-c_{i i+1}^{2}>0, \quad 1=0,1, \ldots, m-1 c 0 > 0 , μ c 2 i c 2 i + 2 − c i i + 1 2 > 0 , 1 = 0 , 1 , … , m − 1 Or
μ
≤
1
cos
2
π
m
+
2
μ
≤
1
cos
2
π
m
+
2
mu <= (1)/(cos^(2)((pi)/(m+2))) \mu \leq \frac{1}{\cos ^{2} \frac{\pi}{m+2}} μ ≤ 1 cos 2 π m + 2 , this polynomial is positive. If
μ
>
1
cos
2
π
m
+
2
μ
>
1
cos
2
π
m
+
2
mu > (1)/(cos^(2)((pi)/(m+2))) \mu>\frac{1}{\cos ^{2} \frac{\pi}{m+2}} μ > 1 cos 2 π m + 2 The property is no longer true. Moreover, it can be shown that in this case the polynomial, while still satisfying the written inequalities, can change its sign. In particular, by taking
μ
=
1
μ
=
1
mu=1 \mu=1 μ = 1 , we have the following theorem, due to MEB Van Vleck (
1
1
^(1) { }^{1} 1 ). If the coefficients ci satisfy the inequalities
c
0
>
0
,
c
2
l
c
2
l
+
2
−
c
2
l
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c
0
>
0
,
c
2
l
c
2
l
+
2
−
c
2
l
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(0) > 0,quadc_(2l)c_(2l+2)-c_(2l+1)^(2) > 0,quad i=0,1,dots,m-1 c_{0}>0, \quad c_{2 l} c_{2 l+2}-c_{2 l+1}^{2}>0, \quad i=0,1, \ldots, m-1 c 0 > 0 , c 2 L c 2 L + 2 − c 2 L + 1 2 > 0 , i = 0 , 1 , … , m − 1 Polynomial (1) is positive.
4. - By specifying the values of
a
0
,
a
1
,
…
,
a
2
m
a
0
,
a
1
,
…
,
a
2
m
a_(0),a_(1),dots,a_(2m) a_{0}, a_{1}, \ldots, a_{2 m} has 0 , has 1 , … , has 2 m We find: various statements. If we take
a
0
=
1
,
a
1
=
a
2
=
…
=
a
2
m
=
2
a
0
=
1
,
a
1
=
a
2
=
…
=
a
2
m
=
2
a_(0)=1,quada_(1)=a_(2)=dots=a_(2m)=2 a_{0}=1, \quad a_{1}=a_{2}=\ldots=a_{2 m}=2 has 0 = 1 , has 1 = has 2 = … = has 2 m = 2 the polynomial
Q
I
(
x
)
Q
I
(
x
)
Q_(I)(x) Q_{I}(x) Q I ( x ) becomes the trigonometric polynomial
Q
t
(
x
)
=
cos
i
(
arccos
x
)
2
i
−
1
Q
t
(
x
)
=
cos
i
(
arccos
x
)
2
i
−
1
Q_(t)(x)=(cos i(arccos x))/(2^(i-1)) Q_{t}(x)=\frac{\cos i(\arccos x)}{2^{i-1}} Q t ( x ) = cos i ( arccos x ) 2 i − 1 and then we have
λ
m
+
1
=
cos
π
2
(
m
+
1
)
λ
m
+
1
=
cos
π
2
(
m
+
1
)
lambda_(m+1)=cos((pi)/(2(m+1))) \lambda_{m+1}=\cos \frac{\pi}{2(m+1)} λ m + 1 = cos π 2 ( m + 1 ) which gives us the following statement:
If the coefficients
c
0
,
c
1
,
…
,
c
2
m
d
u
c
0
,
c
1
,
…
,
c
2
m
d
u
c_(0),c_(1),dots,c_(2m)du c_{0}, c_{1}, \ldots, c_{2 m} d u c 0 , c 1 , … , c 2 m d u polynomial (1) satisfy the inequalities
c
0
>
0
,
2
μ
c
0
c
2
−
c
1
2
>
0
,
μ
c
2
i
c
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
1
,
2
,
…
,
m
−
1
c
0
>
0
,
2
μ
c
0
c
2
−
c
1
2
>
0
,
μ
c
2
i
c
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
1
,
2
,
…
,
m
−
1
c_(0) > 0,quad2muc_(0)c_(2)-c_(1)^(2) > 0,quad muc_(2i)c_(i+2)-c_(2i+1)^(2) > 0,quad i=1,2,dots,m-1 c_{0}>0, \quad 2 \mu c_{0} c_{2}-c_{1}^{2}>0, \quad \mu c_{2 i} c_{i+2}-c_{2 i+1}^{2}>0, \quad i=1,2, \ldots, m-1 c 0 > 0 , 2 μ c 0 c 2 − c 1 2 > 0 , μ c 2 i c i + 2 − c 2 i + 1 2 > 0 , i = 1 , 2 , … , m − 1 Or
μ
≤
1
cos
2
π
2
(
m
+
1
)
μ
≤
1
cos
2
π
2
(
m
+
1
)
mu <= (1)/(cos^(2)((pi)/(2(m+1)))) \mu \leq \frac{1}{\cos ^{2} \frac{\pi}{2(m+1)}} μ ≤ 1 cos 2 π 2 ( m + 1 ) , this polynomial is positive. It should be noted that this criterion is distinct from the previous one. Indeed, we have
1
cos
2
π
m
+
2
>
1
cos
2
π
2
(
m
+
1
)
,
1
cos
2
π
m
+
2
<
2
cos
2
π
2
(
m
+
1
)
(
1
cos
2
π
m
+
2
>
1
cos
2
π
2
(
m
+
1
)
,
1
cos
2
π
m
+
2
<
2
cos
2
π
2
(
m
+
1
)
(
(1)/(cos^(2)((pi)/(m+2))) > (1)/(cos^(2)((pi)/(2(m+1)))),(1)/(cos^(2)((pi)/(m+2))) < (2)/(cos^(2)((pi)/(2(m+1))))( \frac{1}{\cos ^{2} \frac{\pi}{m+2}}>\frac{1}{\cos ^{2} \frac{\pi}{2(m+1)}}, \frac{1}{\cos ^{2} \frac{\pi}{m+2}}<\frac{2}{\cos ^{2} \frac{\pi}{2(m+1)}}( 1 cos 2 π m + 2 > 1 cos 2 π 2 ( m + 1 ) , 1 cos 2 π m + 2 < 2 cos 2 π 2 ( m + 1 ) ( For
m
>
1
)
m
>
1
)
m > 1) m>1) m > 1 )
Let's mention one more case. If
a
i
=
i
+
1
,
i
=
0
,
1
,
2
,
…
,
2
m
a
i
=
i
+
1
,
i
=
0
,
1
,
2
,
…
,
2
m
a_(i)=i+1,quad i=0,1,2,dots,2m a_{i}=i+1, \quad i=0,1,2, \ldots, 2 m has i = i + 1 , i = 0 , 1 , 2 , … , 2 m (1) EB Van Vleck "A sufficient condition for the maximum number of imaginary roots of an equation of the
n
n
n n n -the degree" Annals of Math. (2), t. 4 (1902-03) p. 191. Qi (
x
x
x x x ) is none other (up to a constant factor) than the Legendre polynomial of degree
i
i
i i i
Q
i
(
x
)
=
(
2
i
)
!
i
!
d
i
d
x
i
(
x
2
−
1
)
i
Q
i
(
x
)
=
(
2
i
)
!
i
!
d
i
d
x
i
x
2
−
1
i
Q_(i)(x)=((2i)!)/(i!)(d^(i))/(dx^(i))(x^(2)-1)^(i) Q_{i}(x)=\frac{(2 i)!}{i!} \frac{d^{i}}{d x^{i}}\left(x^{2}-1\right)^{i} Q i ( x ) = ( 2 i ) ! i ! d i d x i ( x 2 − 1 ) i We therefore find a property that can be stated in the following form:
If the coefficients
c
0
,
c
1
,
…
,
c
2
m
c
0
,
c
1
,
…
,
c
2
m
c_(0),c_(1),dots,c_(2m) c_{0}, c_{1}, \ldots, c_{2 m} c 0 , c 1 , … , c 2 m of the polynomial (1) satisfy the inequalities
c
0
>
0
,
1
λ
2
⋅
4
(
i
+
1
)
2
(
2
i
+
1
)
(
2
i
+
3
)
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c
0
>
0
,
1
λ
2
⋅
4
(
i
+
1
)
2
(
2
i
+
1
)
(
2
i
+
3
)
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(0) > 0,quad(1)/(lambda^(2))*(4(i+1)^(2))/((2i+1)(2i+3))c_(2i)c_(2i+2)-c_(2i+1)^(2) > 0,quad i=0,1,dots,m-1 c_{0}>0, \quad \frac{1}{\lambda^{2}} \cdot \frac{4(i+1)^{2}}{(2 i+1)(2 i+3)} c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}>0, \quad i=0,1, \ldots, m-1 c 0 > 0 , 1 λ 2 ⋅ 4 ( i + 1 ) 2 ( 2 i + 1 ) ( 2 i + 3 ) c 2 i c 2 i + 2 − c 2 i + 1 2 > 0 , i = 0 , 1 , … , m − 1 Or
λ
λ
lambda \lambda λ is at least equal to the greatest zero of the Legendre polynomial. of degree
m
+
1
m
+
1
m+1 m+1 m + 1 , this polynomial is positive.
5. - Now let's consider the case where
a
i
=
(
2
m
i
)
,
i
=
0
,
1
,
…
,
2
m
a
i
=
(
2
m
i
)
,
i
=
0
,
1
,
…
,
2
m
a_(i)=((2m)/(i)),quad i=0,1,dots,2m a_{i}=\binom{2 m}{i}, \quad i=0,1, \ldots, 2 m has i = ( 2 m i ) , i = 0 , 1 , … , 2 m We know that if the equation
f
(
x
)
=
0
f
(
x
)
=
0
f(x)=0 f(x)=0 f ( x ) = 0 to all its real roots. we have
c
2
i
+
1
2
−
c
2
i
c
2
i
+
2
≥
0
,
i
=
0
,
1
,
…
,
m
−
1
c
2
i
+
1
2
−
c
2
i
c
2
i
+
2
≥
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(2i+1)^(2)-c_(2i)c_(2i+2) >= 0,quad i=0,1,dots,m-1 c_{2 i+1}^{2}-c_{2 i} c_{2 i+2} \geq 0, \quad i=0,1, \ldots, m-1 c 2 i + 1 2 − c 2 i c 2 i + 2 ≥ 0 , i = 0 , 1 , … , m − 1 If we consider the polynomial in the form (1), these inequalities can be written in the following form:
4
(
i
+
1
)
(
m
−
i
)
(
2
i
+
1
)
(
2
m
−
2
ı
−
1
)
c
2
t
c
2
i
+
2
−
c
2
i
+
1
2
≤
0
,
i
=
0
,
1
,
…
,
m
−
1
4
(
i
+
1
)
(
m
−
i
)
(
2
i
+
1
)
(
2
m
−
2
ı
−
1
)
c
2
t
c
2
i
+
2
−
c
2
i
+
1
2
≤
0
,
i
=
0
,
1
,
…
,
m
−
1
(4(i+1)(m-i))/((2i+1)(2m-2ı-1))c_(2t)c_(2i+2)-c_(2i+1)^(2) <= 0,quad i=0,1,dots,m-1 \frac{4(i+1)(m-i)}{(2 i+1)(2 m-2 \imath-1)} c_{2 t} c_{2 i+2}-c_{2 i+1}^{2} \leq 0, \quad i=0,1, \ldots, m-1 4 ( i + 1 ) ( m − i ) ( 2 i + 1 ) ( 2 m − 2 ı − 1 ) c 2 t c 2 i + 2 − c 2 i + 1 2 ≤ 0 , i = 0 , 1 , … , m − 1 Therefore, applying the previous results yields the opposite property. In this case
λ
m
+
1
λ
m
+
1
lambda_(m+1) \lambda_{m+1} λ m + 1 is the largest root of the equation
(
b
i
=
(
2
i
+
1
)
(
2
m
−
2
i
+
1
)
γ
i
=
i
(
2
i
+
1
)
)
(
b
i
=
(
2
i
+
1
)
(
2
m
−
2
i
+
1
)
γ
i
=
i
(
2
i
+
1
)
)
{:((b_(i)=(2i+1)(2m-2i+1))/(gamma_(i)=i(2i+1))):} \begin{aligned}
& \binom{b_{i}=(2 i+1)(2 m-2 i+1)}{\gamma_{i}=i(2 i+1)}
\end{aligned} ( b i = ( 2 i + 1 ) ( 2 m − 2 i + 1 ) γ i = i ( 2 i + 1 ) ) We therefore have the following property:
If the coefficients
c
0
,
c
1
,
…
,
c
2
n
c
0
,
c
1
,
…
,
c
2
n
c_(0),c_(1),dots,c_(2n) c_{0}, c_{1}, \ldots, c_{2 n} c 0 , c 1 , … , c 2 n dıc polynomial (1) verify the: inequalities
c
0
>
0
,
1
λ
2
4
(
i
+
1
)
(
m
−
i
)
(
2
i
+
1
)
(
2
m
−
2
ι
−
1
)
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c
0
>
0
,
1
λ
2
4
(
i
+
1
)
(
m
−
i
)
(
2
i
+
1
)
(
2
m
−
2
ι
−
1
)
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(0) > 0,(1)/(lambda^(2))(4(i+1)(m-i))/((2i+1)(2m-2iota-1))c_(2i)c_(2i+2)-c_(2i+1)^(2) > 0,quad i=0,1,dots,m-1 c_{0}>0, \frac{1}{\lambda^{2}} \frac{4(i+1)(m-i)}{(2 i+1)(2 m-2 \iota-1)} c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}>0, \quad i=0,1, \ldots, m-1 c 0 > 0 , 1 λ 2 4 ( i + 1 ) ( m − i ) ( 2 i + 1 ) ( 2 m − 2 ι − 1 ) c 2 i c 2 i + 2 − c 2 i + 1 2 > 0 , i = 0 , 1 , … , m − 1 Or,
λ
λ
lambda \lambda λ is at least equal to the largest root of the equation
S
m
+
1
(
x
)
=
0
S
m
+
1
(
x
)
=
0
S_(m+1)(x)=0 \mathrm{S}_{m+1}(x)=0 S m + 1 ( x ) = 0 , this polynomial is positive.
6. - Let us examine in particular the cases
m
=
2
,
3
,
4
m
=
2
,
3
,
4
m=2,3,4 m=2,3,4 m = 2 , 3 , 4 . Either
m
=
2
m
=
2
m=2 m=2 m = 2 If the equation
(2)
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
=
0
(2)
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
=
0
{:(2)c_(0)+c_(1)x+c_(2)x^(2)+c_(3)x^(3)+c_(4)x^(4)=0:} \begin{equation*}
c_{0}+c_{1} x+c_{2} x^{2}+c_{3} x^{3}+c_{4} x^{4}=0 \tag{2}
\end{equation*} (2) c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 = 0 to all its real roots we have
8
c
0
c
2
−
3
c
1
2
≤
0
,
8
c
2
c
4
−
3
c
3
2
≤
0
8
c
0
c
2
−
3
c
1
2
≤
0
,
8
c
2
c
4
−
3
c
3
2
≤
0
8c_(0)c_(2)-3c_(1)^(2) <= 0,quad8c_(2)c_(4)-3c_(3)^(2) <= 0 8 c_{0} c_{2}-3 c_{1}^{2} \leq 0, \quad 8 c_{2} c_{4}-3 c_{3}^{2} \leq 0 8 c 0 c 2 − 3 c 1 2 ≤ 0 , 8 c 2 c 4 − 3 c 3 2 ≤ 0 In this case
S
3
(
x
)
=
−
3
(
3
x
3
−
4
x
)
S
3
(
x
)
=
−
3
3
x
3
−
4
x
S_(3)(x)=-3(3x^(3)-4x) \mathrm{S}_{3}(x)=-3\left(3 x^{3}-4 x\right) S 3 ( x ) = − 3 ( 3 x 3 − 4 x ) Equation (2) will therefore have all its roots imaginary if
c
0
>
0
,
8
μ
c
0
c
2
−
3
c
1
2
>
0
,
8
μ
c
2
c
4
−
3
c
3
2
>
0
c
0
>
0
,
8
μ
c
0
c
2
−
3
c
1
2
>
0
,
8
μ
c
2
c
4
−
3
c
3
2
>
0
c_(0) > 0,quad8muc_(0)c_(2)-3c_(1)^(2) > 0,quad8muc_(2)c_(4)-3c_(3)^(2) > 0 c_{0}>0, \quad 8 \mu c_{0} c_{2}-3 c_{1}^{2}>0, \quad 8 \mu c_{2} c_{4}-3 c_{3}^{2}>0 c 0 > 0 , 8 μ c 0 c 2 − 3 c 1 2 > 0 , 8 μ c 2 c 4 − 3 c 3 2 > 0 Or
μ
≤
3
4
=
0
,
75
μ
≤
3
4
=
0
,
75
mu <= (3)/(4)=0,75 \mu \leq \frac{3}{4}=0,75 μ ≤ 3 4 = 0 , 75 .
Either
m
=
3
m
=
3
m=3 m=3 m = 3 If equation
(3)
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
c
6
x
6
=
0
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
c
6
x
6
=
0
quadc_(0)+c_(1)x+c_(2)x^(2)+c_(3)x^(3)+c_(4)x^(4)+c_(5)x^(5)+c_(6)x^(6)=0 \quad c_{0}+c_{1} x+c_{2} x^{2}+c_{3} x^{3}+c_{4} x^{4}+c_{5} x^{5}+c_{6} x^{6}=0 c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 + c 6 x 6 = 0
to all its real roots we have
12
c
0
c
2
−
0
ˇ
c
1
2
≤
0
,
16
c
c
4
−
9
c
3
2
≤
0
,
12
c
4
c
6
−
5
c
5
2
≤
0
12
c
0
c
2
−
0
ˇ
c
1
2
≤
0
,
16
c
c
4
−
9
c
3
2
≤
0
,
12
c
4
c
6
−
5
c
5
2
≤
0
12c_(0)c_(2)-0^(ˇ)c_(1)^(2) <= 0,16 cc_(4)-9c_(3)^(2) <= 0,12c_(4)c_(6)-5c_(5)^(2) <= 0 12 c_{0} c_{2}-\check{0} c_{1}^{2} \leq 0,16 c c_{4}-9 c_{3}^{2} \leq 0,12 c_{4} c_{6}-5 c_{5}^{2} \leq 0 12 c 0 c 2 − 0 ˇ c 1 2 ≤ 0 , 16 c c 4 − 9 c 3 2 ≤ 0 , 12 c 4 c 6 − 5 c 5 2 ≤ 0 In this case
S
4
(
x
)
=
225
x
4
−
370
x
2
+
8
L
S
4
(
x
)
=
225
x
4
−
370
x
2
+
8
L
S_(4)(x)=225x^(4)-370x^(2)+8L \mathrm{S}_{4}(x)=225 x^{4}-370 x^{2}+8 \mathrm{~L} S 4 ( x ) = 225 x 4 − 370 x 2 + 8 L Equation (3) therefore has all its roots imaginary when
c
0
>
0
,
12
μ
c
0
c
2
−
5
c
1
2
>
0
,
16
μ
c
2
c
4
−
9
c
3
2
>
0
,
12
μ
c
4
c
6
−
5
c
5
2
>
0
c
0
>
0
,
12
μ
c
0
c
2
−
5
c
1
2
>
0
,
16
μ
c
2
c
4
−
9
c
3
2
>
0
,
12
μ
c
4
c
6
−
5
c
5
2
>
0
c_(0) > 0,quad12 muc_(0)c_(2)-5c_(1)^(2) > 0,16 muc_(2)c_(4)-9c_(3)^(2) > 0,12 muc_(4)c_(6)-5c_(5)^(2) > 0 c_{0}>0, \quad 12 \mu c_{0} c_{2}-5 c_{1}^{2}>0,16 \mu c_{2} c_{4}-9 c_{3}^{2}>0,12 \mu c_{4} c_{6}-5 c_{5}^{2}>0 c 0 > 0 , 12 μ c 0 c 2 − 5 c 1 2 > 0 , 16 μ c 2 c 4 − 9 c 3 2 > 0 , 12 μ c 4 c 6 − 5 c 5 2 > 0
Or
μ
≤
5
(
37
−
8
10
)
81
=
0
,
7
,
23
…
μ
≤
5
(
37
−
8
10
)
81
=
0
,
7
,
23
…
mu <= (5(37-8sqrt10))/(81)=0,7,23 dots \mu \leq \frac{5(37-8 \sqrt{10})}{81}=0,7,23 \ldots μ ≤ 5 ( 37 − 8 10 ) 81 = 0 , 7 , 23 …
So finally
m
=
4
m
=
4
m=4 m=4 m = 4 If equation
(4)
c
0
+
c
4
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
c
6
x
6
+
c
7
x
7
+
c
8
x
8
=
0
c
0
+
c
4
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
c
6
x
6
+
c
7
x
7
+
c
8
x
8
=
0
quadc_(0)+c_(4)x+c_(2)x^(2)+c_(3)x^(3)+c_(4)x^(4)+c_(5)x^(5)+c_(6)x^(6)+c_(7)x^(7)+c_(8)x^(8)=0 \quad c_{0}+c_{4} x+c_{2} x^{2}+c_{3} x^{3}+c_{4} x^{4}+c_{5} x^{5}+c_{6} x^{6}+c_{7} x^{7}+c_{8} x^{8}=0 c 0 + c 4 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 + c 6 x 6 + c 7 x 7 + c 8 x 8 = 0
to all its real roots we have
16
c
0
c
2
−
7
c
1
2
≤
0
,
8
c
2
c
4
−
5
c
3
2
≤
0
,
8
c
4
c
6
−
5
c
5
2
≤
0
,
16
c
6
c
8
−
7
c
7
2
≤
0
16
c
0
c
2
−
7
c
1
2
≤
0
,
8
c
2
c
4
−
5
c
3
2
≤
0
,
8
c
4
c
6
−
5
c
5
2
≤
0
,
16
c
6
c
8
−
7
c
7
2
≤
0
16c_(0)c_(2)-7c_(1)^(2) <= 0,quad8c_(2)c_(4)-5c_(3)^(2) <= 0,quad8c_(4)c_(6)-5c_(5)^(2) <= 0,quad16c_(6)c_(8)-7c_(7)^(2) <= 0 16 c_{0} c_{2}-7 c_{1}^{2} \leq 0, \quad 8 c_{2} c_{4}-5 c_{3}^{2} \leq 0, \quad 8 c_{4} c_{6}-5 c_{5}^{2} \leq 0, \quad 16 c_{6} c_{8}-7 c_{7}^{2} \leq 0 16 c 0 c 2 − 7 c 1 2 ≤ 0 , 8 c 2 c 4 − 5 c 3 2 ≤ 0 , 8 c 4 c 6 − 5 c 5 2 ≤ 0 , 16 c 6 c 8 − 7 c 7 2 ≤ 0 .
In this case
S
5
(
x
)
=
45
x
(
7
x
2
−
4
)
(
69
−
35
x
2
)
S
5
(
x
)
=
45
x
7
x
2
−
4
69
−
35
x
2
S_(5)(x)=45 x(7x^(2)-4)(69-35x^(2)) \mathrm{S}_{5}(x)=45 x\left(7 x^{2}-4\right)\left(69-35 x^{2}\right) S 5 ( x ) = 45 x ( 7 x 2 − 4 ) ( 69 − 35 x 2 ) Equation (4) therefore has all its roots imaginary if
c
0
>
0
,
16
μ
c
0
c
2
−
7
c
1
2
>
0
,
8
μ
c
2
c
4
−
o
ˇ
c
3
2
>
0
8
μ
c
4
c
6
−
5
c
5
2
>
0
,
16
μ
c
6
c
8
−
7
c
7
2
>
0
c
0
>
0
,
16
μ
c
0
c
2
−
7
c
1
2
>
0
,
8
μ
c
2
c
4
−
o
ˇ
c
3
2
>
0
8
μ
c
4
c
6
−
5
c
5
2
>
0
,
16
μ
c
6
c
8
−
7
c
7
2
>
0
{:[c_(0) > 0","16 muc_(0)c_(2)-7c_(1)^(2) > 0","8muc_(2)c_(4)-o^(ˇ)c_(3)^(2) > 0],[8muc_(4)c_(6)-5c_(5)^(2) > 0","quad16 muc_(6)c_(8)-7c_(7)^(2) > 0]:} \begin{gathered}
c_{0}>0,16 \mu c_{0} c_{2}-7 c_{1}^{2}>0,8 \mu c_{2} c_{4}-\check{o} c_{3}^{2}>0 \\
8 \mu c_{4} c_{6}-5 c_{5}^{2}>0, \quad 16 \mu c_{6} c_{8}-7 c_{7}^{2}>0
\end{gathered} c 0 > 0 , 16 μ c 0 c 2 − 7 c 1 2 > 0 , 8 μ c 2 c 4 − o ˇ c 3 2 > 0 8 μ c 4 c 6 − 5 c 5 2 > 0 , 16 μ c 6 c 8 − 7 c 7 2 > 0
Or
μ
≤
35
69
=
0
,
5072
…
μ
≤
35
69
=
0
,
5072
…
mu <= (35)/(69)=0,5072 dots \mu \leq \frac{35}{69}=0,5072 \ldots μ ≤ 35 69 = 0 , 5072 …
Consider a polynomial of degree
2
m
+
1
2
m
+
1
2m+1 2 m+1 2 m + 1
F
(
x
)
=
c
0
+
c
1
x
+
…
+
c
2
m
+
1
x
2
m
+
1
F
(
x
)
=
c
0
+
c
1
x
+
…
+
c
2
m
+
1
x
2
m
+
1
F(x)=c_(0)+c_(1)x+dots+c_(2m+1)x^(2m+1) \mathrm{F}(x)=c_{0}+c_{1} x+\ldots+c_{2 m+1} x^{2 m+1} F ( x ) = c 0 + c 1 x + … + c 2 m + 1 x 2 m + 1 We have
x
2
m
+
1
F
(
1
x
)
=
c
0
x
2
m
+
1
+
…
+
c
2
m
+
1
x
2
m
+
1
F
1
x
=
c
0
x
2
m
+
1
+
…
+
c
2
m
+
1
x^(2m+1)F((1)/(x))=c_(0)x^(2m+1)+dots+c_(2m+1) x^{2 m+1} \mathrm{~F}\left(\frac{1}{x}\right)=c_{0} x^{2 m+1}+\ldots+c_{2 m+1} x 2 m + 1 F ( 1 x ) = c 0 x 2 m + 1 + … + c 2 m + 1 Taking the derivative, we see that if
(5)
(
2
m
+
1
)
c
0
+
2
m
c
1
x
+
…
+
c
2
m
x
2
m
(5)
(
2
m
+
1
)
c
0
+
2
m
c
1
x
+
…
+
c
2
m
x
2
m
{:(5)(2m+1)c_(0)+2mc_(1)x+dots+c_(2m)x^(2m):} \begin{equation*}
(2 m+1) c_{0}+2 m c_{1} x+\ldots+c_{2 m} x^{2 m} \tag{5}
\end{equation*} (5) ( 2 m + 1 ) c 0 + 2 m c 1 x + … + c 2 m x 2 m the polynomial is positive
F
(
x
)
F
(
x
)
F(x) F(x) F ( x ) It changes sign at most once. More precisely, it only admits one real zero. Since polynomial (5) is positive, we can apply the result from No. 4 and we have the following property:
If the coefficients
c
0
,
c
1
,
…
,
c
2
n
c
0
,
c
1
,
…
,
c
2
n
c_(0),c_(1),dots,c_(2n) c_{0}, c_{1}, \ldots, c_{2 n} c 0 , c 1 , … , c 2 n of the equation
(6)
c
0
+
c
1
x
+
…
+
c
2
m
+
1
x
2
m
+
1
=
0
(6)
c
0
+
c
1
x
+
…
+
c
2
m
+
1
x
2
m
+
1
=
0
{:(6)c_(0)+c_(1)x+dots+c_(2m+1)x^(2m+1)=0:} \begin{equation*}
c_{0}+c_{1} x+\ldots+c_{2 m+1} x^{2 m+1}=0 \tag{6}
\end{equation*} (6) c 0 + c 1 x + … + c 2 m + 1 x 2 m + 1 = 0 verify the inequalities
c
2
m
>
0
,
(
c
0
>
0
)
,
1
λ
2
c
2
i
c
2
i
+
2
−
c
2
i
+
1
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c
2
m
>
0
,
c
0
>
0
,
1
λ
2
c
2
i
c
2
i
+
2
−
c
2
i
+
1
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(2m) > 0,(c_(0) > 0),(1)/(lambda^(2))c_(2i)c_(2i+2)-c_(2^(i+1)) > 0,quad i=0,1,dots,m-1 c_{2 m}>0,\left(c_{0}>0\right), \frac{1}{\lambda^{2}} c_{2 i} c_{2 i+2}-c_{2^{i+1}}>0, \quad i=0,1, \ldots, m-1 c 2 m > 0 , ( c 0 > 0 ) , 1 λ 2 c 2 i c 2 i + 2 − c 2 i + 1 > 0 , i = 0 , 1 , … , m − 1 Or
λ
λ
lambda \lambda λ is at least equal to the maximum zero of the Legendre polynomial of degree
m
+
1
m
+
1
m+1 m+1 m + 1 , the equation has at most one real root. In particular, we deduce from this the second Van Vleck SEM theorem (
2
2
^(2) { }^{2} 2 ). If the coefficients
c
0
,
c
1
,
…
,
c
2
m
c
0
,
c
1
,
…
,
c
2
m
c_(0),c_(1),dots,c_(2m) c_{0}, c_{1}, \ldots, c_{2 m} c 0 , c 1 , … , c 2 m of equation (6) satisfy the inequalities
c
0
>
0
,
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c
0
>
0
,
c
2
i
c
2
i
+
2
−
c
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
m
−
1
c_(0) > 0,quadc_(2i)c_(2i+2)-c_(2i+1)^(2) > 0,quad i=0,1,dots,m-1 c_{0}>0, \quad c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}>0, \quad i=0,1, \ldots, m-1 c 0 > 0 , c 2 i c 2 i + 2 − c 2 i + 1 2 > 0 , i = 0 , 1 , … , m − 1 This equation has at most one real root.
8. - Consider the equation of degree n.
(7)
g
(
x
)
=
c
0
+
(
n
1
)
c
1
x
+
,
…
,
+
(
n
k
)
c
k
x
k
+
…
,
+
c
n
x
n
=
0
g
(
x
)
=
c
0
+
(
n
1
)
c
1
x
+
,
…
,
+
(
n
k
)
c
k
x
k
+
…
,
+
c
n
x
n
=
0
quad g(x)=c_(0)+((n)/(1))c_(1)x+,dots,+((n)/(k))c_(k)x^(k)+dots,+c_(n)x^(n)=0 \quad g(x)=c_{0}+\binom{n}{1} c_{1} x+, \ldots,+\binom{n}{k} c_{k} x^{k}+\ldots,+c_{n} x^{n}=0 g ( x ) = c 0 + ( n 1 ) c 1 x + , … , + ( n k ) c k x k + … , + c n x n = 0 . Let
j
,
r
j
,
r
j,r j, r I , r two positive integers such that
j
≥
0
,
j
+
2
r
≤
n
j
≥
0
,
j
+
2
r
≤
n
j >= 0,j+2r <= n j \geq 0, j+2 r \leq n I ≥ 0 , I + 2 r ≤ n , and let's take the derivative of order
j
j
j j I of
g
(
x
)
g
(
x
)
g(x) g(x) g ( x )
g
(
j
)
(
x
)
=
n
!
(
n
−
j
)
!
[
c
j
+
(
n
−
j
1
)
c
j
+
i
x
+
,
…
,
+
c
n
x
n
−
j
]
g
(
j
)
(
x
)
=
n
!
(
n
−
j
)
!
c
j
+
(
n
−
j
1
)
c
j
+
i
x
+
,
…
,
+
c
n
x
n
−
j
g^((j))(x)=(n!)/((n-j)!)[c_(j)+((n-j)/(1))c_(j+i)x+,dots,+c_(n)x^(n-j)] g^{(j)}(x)=\frac{n!}{(n-j)!}\left[c_{j}+\binom{n-j}{1} c_{j+i} x+, \ldots,+c_{n} x^{n-j}\right] g ( I ) ( x ) = n ! ( n − I ) ! [ c I + ( n − I 1 ) c I + i x + , … , + c n x n − I ] Let us consider the transformation into
1
x
1
x
(1)/(x) \frac{1}{x} 1 x (leaving aside one factor-
(
2
)
2
(^(2)) \left({ }^{2}\right) ( 2 ) loe cit
(
1
)
1
(^(1)) \left({ }^{1}\right) ( 1 ) .
constant)
g
1
(
x
)
=
c
j
x
n
−
1
+
(
n
−
j
1
)
c
i
+
1
x
n
−
1
−
1
+
…
,
+
c
n
g
1
(
x
)
=
c
j
x
n
−
1
+
(
n
−
j
1
)
c
i
+
1
x
n
−
1
−
1
+
…
,
+
c
n
g_(1)(x)=c_(j)x^(n-1)+((n-j)/(1))c_(i+1)x^(n-1-1)+dots,+c_(n) g_{1}(x)=c_{j} x^{n-1}+\binom{n-j}{1} c_{i+1} x^{n-1-1}+\ldots,+c_{n} g 1 ( x ) = c I x n − 1 + ( n − I 1 ) c i + 1 x n − 1 − 1 + … , + c n and let's take the derivative of order
n
−
j
−
2
r
n
−
j
−
2
r
n-j-2r n-j-2 r n − I − 2 r of this polynomial
g
1
(
n
−
j
−
2
r
)
(
x
)
=
(
n
−
j
)
!
(
2
r
)
!
[
c
j
x
2
r
+
(
2
r
1
)
c
j
+
1
x
2
r
−
1
+
…
,
+
c
j
+
2
r
]
g
1
(
n
−
j
−
2
r
)
(
x
)
=
(
n
−
j
)
!
(
2
r
)
!
c
j
x
2
r
+
(
2
r
1
)
c
j
+
1
x
2
r
−
1
+
…
,
+
c
j
+
2
r
g_(1)^((n-j-2r))(x)=((n-j)!)/((2r)!)[c_(j)x^(2r)+((2r)/(1))c_(j+1)x^(2r-1)+dots,+c_(j+2r)] g_{1}^{(n-j-2 r)}(x)=\frac{(n-j)!}{(2 r)!}\left[c_{j} x^{2 r}+\binom{2 r}{1} c_{j+1} x^{2 r-1}+\ldots,+c_{j+2 r}\right] g 1 ( n − I − 2 r ) ( x ) = ( n − I ) ! ( 2 r ) ! [ c I x 2 r + ( 2 r 1 ) c I + 1 x 2 r − 1 + … , + c I + 2 r ] So finally
g
2
(
x
)
=
c
j
+
(
2
r
1
)
c
j
+
1
x
+
,
…
+
(
2
r
k
)
c
j
+
k
x
k
+
…
,
+
c
j
+
2
r
x
2
r
g
2
(
x
)
=
c
j
+
(
2
r
1
)
c
j
+
1
x
+
,
…
+
(
2
r
k
)
c
j
+
k
x
k
+
…
,
+
c
j
+
2
r
x
2
r
g_(2)(x)=c_(j)+((2r)/(1))c_(j+1)x+,dots+((2r)/(k))c_(j+k)x^(k)+dots,+c_(j+2r)x^(2r) g_{2}(x)=c_{j}+\binom{2 r}{1} c_{j+1} x+, \ldots+\binom{2 r}{k} c_{j+k} x^{k}+\ldots,+c_{j+2 r} x^{2 r} g 2 ( x ) = c I + ( 2 r 1 ) c I + 1 x + , … + ( 2 r k ) c I + k x k + … , + c I + 2 r x 2 r Suppose that equation (7) has
p
p
p p p real roots. The equation
g
(
1
)
(
x
)
=
0
g
(
1
)
(
x
)
=
0
g^((1))(x)=0 g^{(1)}(x)=0 g ( 1 ) ( x ) = 0 , therefore also
g
1
(
x
)
=
0
g
1
(
x
)
=
0
g_(1)(x)=0 g_{1}(x)=0 g 1 ( x ) = 0 , then at least
p
−
j
p
−
j
p-j p-j p − I real roots. We can deduce that the equation
g
2
(
x
)
=
0
g
2
(
x
)
=
0
g_(2)(x)=0 g_{2}(x)=0 g 2 ( x ) = 0 has at least
p
−
n
+
2
r
p
−
n
+
2
r
p-n+2r p-n+2 r p − n + 2 r real roots. It follows that if the polynomial
g
2
(
x
)
g
2
(
x
)
g_(2)(x) g_{2}(x) g 2 ( x ) equation (7) has all its imaginary roots and at most
n
−
2
r
n
−
2
r
n-2r n-2 r n − 2 r real roots. The results of No. 5 therefore allow us to state the following general property:
If
j
=
0
,
j
+
2
r
≤
n
j
=
0
,
j
+
2
r
≤
n
j=0,j+2r <= n j=0, j+2 r \leq n I = 0 , I + 2 r ≤ n and if the coefficients ci of equation (7) satisfy the inequalities
c
j
>
0
,
1
λ
2
c
j
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
r
−
1
c
j
>
0
,
1
λ
2
c
j
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
1
2
>
0
,
i
=
0
,
1
,
…
,
r
−
1
c_(j) > 0,(1)/(lambda2)c_(j+2i)c_(j+2i+2)-c_(j+2i+1)^(2) > 0,quad i=0,1,dots,r-1 c_{j}>0, \frac{1}{\lambda 2} c_{j+2 i} c_{j+2 i+2}-c_{j+2 i+1}^{2}>0, \quad i=0,1, \ldots, r-1 c I > 0 , 1 λ 2 c I + 2 i c I + 2 i + 2 − c I + 2 i + 1 2 > 0 , i = 0 , 1 , … , r − 1 Or,
λ
λ
lambda \lambda λ is at least equal to the largest root of the equation
S
r
+
1
(
x
)
=
0
S
r
+
1
(
x
)
=
0
S_(r+1)(x)=0 \mathrm{S}_{r+1}(x)=0 S r + 1 ( x ) = 0 , this equation (7) has at most
n
−
2
r
n
−
2
r
n-2r n-2 r n − 2 r real roots. We can also say that in this case equation (7) has at least
r
r
r r r pairs of imaginary conjugated roots. For
r
=
1
r
=
1
r=1 r=1 r = 1 we have this property, obvious a priori, that if we can find a
j
j
j j I such as
c
j
>
0
,
c
j
c
j
+
2
−
c
j
+
1
2
>
0
c
j
>
0
,
c
j
c
j
+
2
−
c
j
+
1
2
>
0
c_(j) > 0,quadc_(j)c_(j+2)-c_(j+1)^(2) > 0 c_{j}>0, \quad c_{j} c_{j+2}-c_{\mathrm{j}+1}^{2}>0 c I > 0 , c I c I + 2 − c I + 1 2 > 0 equation (7) has at least one pair of imaginary conjugate roots.
For
r
=
2
r
=
2
r=2 r=2 r = 2 , we find that if
c
j
>
0
,
μ
c
i
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
1
2
>
0
,
i
=
0
,
1
μ
≤
3
4
=
0
,
75
,
(
j
≥
0
,
j
+
4
≤
n
)
c
j
>
0
,
μ
c
i
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
1
2
>
0
,
i
=
0
,
1
μ
≤
3
4
=
0
,
75
,
(
j
≥
0
,
j
+
4
≤
n
)
{:[c_(j) > 0","quad muc_(i+2i)c_(j+2i+2)-c_(j+2i+1)^(2) > 0","quad i=0","1],[mu <= (3)/(4)=0","75","quad(j >= 0","quad j+4 <= n)]:} \begin{gathered}
c_{j}>0, \quad \mu c_{i+2 i} c_{j+2 i+2}-c_{j+2 i+1}^{2}>0, \quad i=0,1 \\
\mu \leq \frac{3}{4}=0,75, \quad(j \geq 0, \quad j+4 \leq n)
\end{gathered} c I > 0 , μ c i + 2 i c I + 2 i + 2 − c I + 2 i + 1 2 > 0 , i = 0 , 1 μ ≤ 3 4 = 0 , 75 , ( I ≥ 0 , I + 4 ≤ n ) equation (7) has at least two pairs of imaginary conjugate roots.
For
r
=
3
,
4
r
=
3
,
4
r=3,4 r=3,4 r = 3 , 4 We similarly find analogous sufficient conditions for the existence of three or four pairs of imaginary
conjugate roots. The values of the coefficients
μ
μ
mu \mu μ which are included in the respective inequalities are those specified in No. 6.9
. - Finally, if we write equation (7) in the form
c
0
+
c
1
x
+
c
2
x
2
+
⋯
+
c
n
x
n
=
0
c
0
+
c
1
x
+
c
2
x
2
+
⋯
+
c
n
x
n
=
0
c_(0)+c_(1)x+c_(2)x^(2)+cdots+c_(n)x^(n)=0 c_{0}+c_{1} x+c_{2} x^{2}+\cdots+c_{n} x^{n}=0 c 0 + c 1 x + c 2 x 2 + ⋯ + c n x n = 0 He polynomial
g
2
(
x
)
g
2
(
x
)
g_(2)(x) g_{2}(x) g 2 ( x ) will be written in the form
1
(
n
j
)
c
j
+
(
2
r
1
)
(
n
j
+
1
)
c
j
+
1
x
+
⋯
+
(
2
r
k
)
(
n
j
+
k
)
c
j
+
k
x
k
+
⋯
+
1
(
n
j
+
2
r
)
c
j
+
2
r
x
2
r
.
1
(
n
j
)
c
j
+
(
2
r
1
)
(
n
j
+
1
)
c
j
+
1
x
+
⋯
+
(
2
r
k
)
(
n
j
+
k
)
c
j
+
k
x
k
+
⋯
+
1
(
n
j
+
2
r
)
c
j
+
2
r
x
2
r
.
(1)/(((n)/(j)))c_(j)+(((2r)/(1)))/(((n)/(j+1)))c_(j+1)x+cdots+(((2r)/(k)))/(((n)/(j+k)))c_(j+k)x^(k)+cdots+(1)/(((n)/(j+2r)))c_(j+2r)x^(2r). \frac{1}{\binom{n}{j}} c_{j}+\frac{\binom{2 r}{1}}{\binom{n}{j+1}} c_{j+1} x+\cdots+\frac{\binom{2 r}{k}}{\binom{n}{j+k}} c_{j+k} x^{k}+\cdots+\frac{1}{\binom{n}{j+2 r}} c_{j+2 r} x^{2 r} . 1 ( n I ) c I + ( 2 r 1 ) ( n I + 1 ) c I + 1 x + ⋯ + ( 2 r k ) ( n I + k ) c I + k x k + ⋯ + 1 ( n I + 2 r ) c I + 2 r x 2 r . Let us turn our attention to the case where
j
=
0
j
=
0
j=0 j=0 I = 0 The preceding polynomial can then be written in the following form:
(
2
r
+
1
)
(
2
r
+
2
)
…
n
c
j
+
2
r
(
2
r
+
1
)
…
(
n
−
1
)
c
j
+
1
x
+
⋯
+
(
2
r
+
1
)
(
2
r
+
2
)
…
n
c
j
+
2
r
(
2
r
+
1
)
…
(
n
−
1
)
c
j
+
1
x
+
⋯
+
(2r+1)(2r+2)dots nc_(j)+2r(2r+1)dots(n-1)c_(j+1)x+cdots+ (2 r+1)(2 r+2) \ldots n c_{j}+2 r(2 r+1) \ldots(n-1) c_{j+1} x+\cdots+ ( 2 r + 1 ) ( 2 r + 2 ) … n c I + 2 r ( 2 r + 1 ) … ( n − 1 ) c I + 1 x + ⋯ +
⋯
+
(
2
r
−
k
+
1
)
(
2
r
−
k
+
2
)
…
(
n
−
k
)
c
j
+
k
x
k
+
⋯
+
1
,
2
…
(
n
−
2
r
)
c
j
+
2
r
x
2
r
⋯
+
(
2
r
−
k
+
1
)
(
2
r
−
k
+
2
)
…
(
n
−
k
)
c
j
+
k
x
k
+
⋯
+
1
,
2
…
(
n
−
2
r
)
c
j
+
2
r
x
2
r
cdots+(2r-k+1)(2r-k+2)dots(n-k)c_(j+k)x^(k)+cdots+1,2dots(n-2r)c_(j+2r)x^(2r) \cdots+(2 r-k+1)(2 r-k+2) \ldots(n-k) c_{j+k} x^{k}+\cdots+1,2 \ldots(n-2 r) c_{j+2 r} x^{2 r} ⋯ + ( 2 r − k + 1 ) ( 2 r − k + 2 ) … ( n − k ) c I + k x k + ⋯ + 1 , 2 … ( n − 2 r ) c I + 2 r x 2 r .
Posing
m
=
r
,
a
i
=
(
i
+
1
)
(
i
+
2
)
…
(
n
−
2
r
+
i
)
,
i
=
0
,
1
,
…
,
2
r
m
=
r
,
a
i
=
(
i
+
1
)
(
i
+
2
)
…
(
n
−
2
r
+
i
)
,
i
=
0
,
1
,
…
,
2
r
m=r,a_(i)=(i+1)(i+2)dots(n-2r+i),quad i=0,1,dots,2r m=r, a_{i}=(i+1)(i+2) \ldots(n-2 r+i), \quad i=0,1, \ldots, 2 r m = r , has i = ( i + 1 ) ( i + 2 ) … ( n − 2 r + i ) , i = 0 , 1 , … , 2 r the polynomial
Q
i
(
λ
)
Q
i
(
λ
)
Q_(i)(lambda) Q_{i}(\lambda) Q i ( λ ) the corresponding is given by the recurrence relations (8)
Q
0
(
λ
)
=
1
,
Q
1
(
λ
)
=
λ
,
Q
i
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
(
i
+
1
)
(
n
−
2
r
+
2
i
+
1
)
2
(
2
i
+
1
)
(
n
−
2
r
+
2
i
+
2
)
Q
1
(
λ
)
=
0
Q
0
(
λ
)
=
1
,
Q
1
(
λ
)
=
λ
,
Q
i
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
(
i
+
1
)
(
n
−
2
r
+
2
i
+
1
)
2
(
2
i
+
1
)
(
n
−
2
r
+
2
i
+
2
)
Q
1
(
λ
)
=
0
Q_(0)(lambda)=1,Q_(1)(lambda)=lambda,Q_(i+2)(lambda)-lambdaQ_(i+1)(lambda)+((i+1)(n-2r+2i+1))/(2(2i+1)(n-2r+2i+2))Q_(1)(lambda)=0 Q_{0}(\lambda)=1, Q_{1}(\lambda)=\lambda, Q_{i+2}(\lambda)-\lambda Q_{i+1}(\lambda)+\frac{(i+1)(n-2 r+2 i+1)}{2(2 i+1)(n-2 r+2 i+2)} Q_{1}(\lambda)=0 Q 0 ( λ ) = 1 , Q 1 ( λ ) = λ , Q i + 2 ( λ ) − λ Q i + 1 ( λ ) + ( i + 1 ) ( n − 2 r + 2 i + 1 ) 2 ( 2 i + 1 ) ( n − 2 r + 2 i + 2 ) Q 1 ( λ ) = 0 . We therefore have the following property:
If
j
≥
0
,
j
+
2
r
≤
n
j
≥
0
,
j
+
2
r
≤
n
j >= 0,j+2r <= n j \geq 0, j+2 r \leq n I ≥ 0 , I + 2 r ≤ n and if the coefficients
c
i
c
i
c_(i) c_{\mathrm{i}} c i of the equation
c
0
+
c
1
x
+
…
+
c
n
x
n
=
0
c
0
+
c
1
x
+
…
+
c
n
x
n
=
0
c_(0)+c_(1)x+dots+c_(n)x^(n)=0 c_{0}+c_{1} x+\ldots+c_{n} x^{n}=0 c 0 + c 1 x + … + c n x n = 0 verify the inequalities
c
j
>
0
,
1
λ
2
c
j
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
i
2
>
0
,
i
=
0
,
1
,
…
,
r
c
j
>
0
,
1
λ
2
c
j
+
2
i
c
j
+
2
i
+
2
−
c
j
+
2
i
+
i
2
>
0
,
i
=
0
,
1
,
…
,
r
c_(j) > 0,(1)/(lambda^(2))c_(j+2i)c_(j+2i+2)-c_(j+2i+i)^(2) > 0,quad i=0,1,dots,r c_{\mathrm{j}}>0, \frac{1}{\lambda^{2}} c_{\mathrm{j}+2 \mathrm{i}} c_{\mathrm{j}+2 \mathrm{i}+2}-c_{\mathrm{j}+2 \mathrm{i}+\mathrm{i}}^{2}>0, \quad i=0,1, \ldots, r c I > 0 , 1 λ 2 c I + 2 i c I + 2 i + 2 − c I + 2 i + i 2 > 0 , i = 0 , 1 , … , r Or,
λ
λ
lambda \lambda λ is at least equal to the greatest zero of the polynomial
Q
r
+
1
(
λ
)
=
0
Q
r
+
1
(
λ
)
=
0
Q_(r+1)(lambda)=0 Q_{r+1}(\lambda)=0 Q r + 1 ( λ ) = 0 defined by relations (8), this equation has at most
n
n
n n n -er real roots. The largest root of the equation
Q
r
+
1
(
λ
)
=
0
Q
r
+
1
(
λ
)
=
0
Q_(r+1)(lambda)=0 Q_{r+1}(\lambda)=0 Q r + 1 ( λ ) = 0 no, does not depend on
j
j
j j I and is bounded above. It follows from this, in effect, by a known theorem.
(
3
)
(
3
)
^((3)) { }^{(3)} ( 3 ) that the zeros of
Q
+
1
(
λ
)
Q
+
1
(
λ
)
Q_(+1)(lambda) Q_{+1}(\lambda) Q + 1 ( λ ) are in modulo at most equal to
max
.
(
b
2
+
b
3
,
b
3
+
b
4
,
…
,
b
r
+
b
r
+
1
―
)
max
.
b
2
+
b
3
,
b
3
+
b
4
,
…
,
b
r
+
b
r
+
1
¯
max.(sqrt(b_(2))+sqrt(b_(3)),sqrt(b_(3))+sqrt(b_(4)),dots,sqrt(b_(r))+sqrt( bar(b_(r+1)))) \operatorname{max.}\left(\sqrt{b_{2}}+\sqrt{b_{3}}, \sqrt{b_{3}}+\sqrt{b_{4}}, \ldots, \sqrt{b_{r}}+\sqrt{\overline{b_{r+1}}}\right) max . ( b 2 + b 3 , b 3 + b 4 , … , b r + b r + 1 ― ) (3) See e.g. Wolgang Hahn „Bericht über die Nullstellen der Laguerreschen und der Hermiteschen Polynome” Jahresbericht d. DMV Bd. 44 p. 215-236, sp. p. 227.
where
b
i
+
2
=
(
i
+
1
)
(
2
i
+
n
−
2
r
+
1
)
2
(
2
i
+
1
)
(
2
i
+
n
−
2
r
+
2
)
,
i
=
0
,
1
,
…
,
r
−
1
b
i
+
2
=
(
i
+
1
)
(
2
i
+
n
−
2
r
+
1
)
2
(
2
i
+
1
)
(
2
i
+
n
−
2
r
+
2
)
,
i
=
0
,
1
,
…
,
r
−
1
b_(i+2)=((i+1)(2i+n-2r+1))/(2(2i+1)(2i+n-2r+2)),quad i=0,1,dots,r-1 b_{\mathrm{i}+2}=\frac{(i+1)(2 i+n-2 r+1)}{2(2 i+1)(2 i+n-2 r+2)}, \quad i=0,1, \ldots, r-1 b i + 2 = ( i + 1 ) ( 2 i + n − 2 r + 1 ) 2 ( 2 i + 1 ) ( 2 i + n − 2 r + 2 ) , i = 0 , 1 , … , r − 1 It is immediately apparent that (
n
>
2
r
n
>
2
r
n > 2r n>2 r n > 2 r )
b
2
>
b
3
>
…
>
b
r
+
1
b
2
>
b
3
>
…
>
b
r
+
1
b_(2) > b_(3) > dots > b_(r+1) b_{2}>b_{3}>\ldots>b_{r+1} b 2 > b 3 > … > b r + 1 The largest root of the equation
Q
r
+
1
(
λ
)
=
0
Q
r
+
1
(
λ
)
=
0
Q_(r+1)(lambda)=0 Q_{r+1}(\lambda)=0 Q r + 1 ( λ ) = 0 is therefore at most equal to
b
2
+
b
3
=
α
+
1
2
(
α
+
2
)
+
α
+
3
3
(
α
+
4
)
<
1
2
+
1
3
=
1
,
24
…
b
2
+
b
3
=
α
+
1
2
(
α
+
2
)
+
α
+
3
3
(
α
+
4
)
<
1
2
+
1
3
=
1
,
24
…
sqrt(b_(2))+sqrt(b_(3))=sqrt((alpha+1)/(2(alpha+2)))+sqrt((alpha+3)/(3(alpha+4))) < sqrt((1)/(2))+sqrt((1)/(3))=1,24 dots \sqrt{b_{2}}+\sqrt{b_{3}}=\sqrt{\frac{\alpha+1}{2(\alpha+2)}}+\sqrt{\frac{\alpha+3}{3(\alpha+4)}}<\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{3}}=1,24 \ldots b 2 + b 3 = α + 1 2 ( α + 2 ) + α + 3 3 ( α + 4 ) < 1 2 + 1 3 = 1 , 24 … It can also be demonstrated that the true upper limit in question is
>
1
>
1
> 1 >1 > 1 Indeed, if we make it tender
n
n
n n n towards
∞
∞
oo \infty ∞ we obtain the polynomials
Q
0
(
λ
)
=
1
,
Q
1
(
λ
)
=
(
λ
)
,
Q
i
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
i
+
1
2
(
2
i
+
1
)
Q
i
(
λ
)
=
0
Q
0
(
λ
)
=
1
,
Q
1
(
λ
)
=
(
λ
)
,
Q
i
+
2
(
λ
)
−
λ
Q
i
+
1
(
λ
)
+
i
+
1
2
(
2
i
+
1
)
Q
i
(
λ
)
=
0
Q_(0)(lambda)=1,quadQ_(1)(lambda)=(lambda),quadQ_(i+2)(lambda)-lambdaQ_(i+1)(lambda)+(i+1)/(2(2i+1))Q_(i)(lambda)=0 Q_{0}(\lambda)=1, \quad Q_{1}(\lambda)=(\lambda), \quad Q_{i+2}(\lambda)-\lambda Q_{i+1}(\lambda)+\frac{i+1}{2(2 i+1)} Q_{i}(\lambda)=0 Q 0 ( λ ) = 1 , Q 1 ( λ ) = ( λ ) , Q i + 2 ( λ ) − λ Q i + 1 ( λ ) + i + 1 2 ( 2 i + 1 ) Q i ( λ ) = 0 and we deduce from this
Q
0
(
1
)
=
1
,
Q
1
(
1
)
=
1
,
Q
2
(
1
)
=
1
2
,
Q
3
(
1
)
=
1
2
−
1
3
=
1
6
Q
4
(
1
)
=
1
6
−
3
10
⋅
1
2
=
1
60
,
Q
5
(
1
)
=
1
60
−
2
7
⋅
1
6
=
−
13
420
<
0
Q
0
(
1
)
=
1
,
Q
1
(
1
)
=
1
,
Q
2
(
1
)
=
1
2
,
Q
3
(
1
)
=
1
2
−
1
3
=
1
6
Q
4
(
1
)
=
1
6
−
3
10
⋅
1
2
=
1
60
,
Q
5
(
1
)
=
1
60
−
2
7
⋅
1
6
=
−
13
420
<
0
{:[Q_(0)(1)=1","quadQ_(1)(1)=1","quadQ_(2)(1)=(1)/(2)","quadQ_(3)(1)=(1)/(2)-(1)/(3)=(1)/(6)],[Q_(4)(1)=(1)/(6)-(3)/(10)*(1)/(2)=(1)/(60)","quadQ_(5)(1)=(1)/(60)-(2)/(7)*(1)/(6)=-(13)/(420) < 0]:} \begin{gathered}
Q_{0}(1)=1, \quad Q_{1}(1)=1, \quad Q_{2}(1)=\frac{1}{2}, \quad Q_{3}(1)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \\
Q_{4}(1)=\frac{1}{6}-\frac{3}{10} \cdot \frac{1}{2}=\frac{1}{60}, \quad Q_{5}(1)=\frac{1}{60}-\frac{2}{7} \cdot \frac{1}{6}=-\frac{13}{420}<0
\end{gathered} Q 0 ( 1 ) = 1 , Q 1 ( 1 ) = 1 , Q 2 ( 1 ) = 1 2 , Q 3 ( 1 ) = 1 2 − 1 3 = 1 6 Q 4 ( 1 ) = 1 6 − 3 10 ⋅ 1 2 = 1 60 , Q 5 ( 1 ) = 1 60 − 2 7 ⋅ 1 6 = − 13 420 < 0 which proves our claim.
