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T. Popoviciu, Sur une condition suffisante pour qu’un polynôme soit positif, Mathematica, 11 (1935), pp. 247-256 (in French).
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ON A SUFFICIENT CONDITION FOR A POLYNOMIAL TO BE POSITIVE
by
Received on May 5, 1935.
-
- Consider a polynomial of degree2m2 mwhich we will write in the following form
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}
Ora_(0),a_(1),dots,a_(k),dots,a_(m)a_{0}, a_{1}, \ldots, a_{k}, \ldots, a_{m}are positive constants andc_(l)c_{l}real coefficients.
Let's put this polynomial in the form
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}
and let's determine the constantsalpha_(i),beta_(i),gamma_(i)\alpha_{i}, \beta_{i}, \gamma_{i}by identification. We obtain
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}
and let's determine the constantsalpha_(i),beta_(i),gamma_(i)\alpha_{i}, \beta_{i}, \gamma_{i}by identification. We obtain
{:[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}
The constantsalpha_(i)\alpha_{i}are therefore determined by recurrence relations
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 .
We can write
alpha_(i)=( hat(P)_(i+1)(lambda))/(lambdaP_(i)(lambda))\alpha_{i}=\frac{\hat{P}_{i+1}(\lambda)}{\lambda P_{i}(\lambda)}
and thenP_(k)(lambda)P_{k}(\lambda)is a polynomial of degreekkinlambda\lambda
These polynomials satisfy the recurrence relations .
These polynomials satisfy the recurrence relations .
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
Let's ask
{: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}}
we then
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)=0Q_{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
The polynomialQ_(k)(lambda)Q_{k}(\lambda)has all its real zeros and the zeros ofQ_(k+1)(lambda)Q_{k+1}(\lambda)are separated by those ofQ_(k)(lambda)Q_{k}(\lambda).
So that the constantsalpha_(i)\alpha_{i}all things must be positive, it is necessary thatlambda\lambdabe greater than the largest zero ofQ_(m+1)(lambda)Q_{m+1}(\lambda)In this case, the constantsbeta_(i)\beta_{i}Andgamma_(i)\gamma_{i}are all positive as well.
We can see, therefore, that if we have
{:[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}
Orlambda\lambdais at least equal to the largest rootlambda_(m+1)\lambda_{m+1}of the equationQ_(m+1)(lambda)=0\mathrm{Q}_{m+1}(\lambda)=0the polynomialf(x)f(x)is positive.
2. - The limit foundlambda_(m+1)\lambda_{m+1}is the best possible. One can see it directly, or in the following way:
2. - The limit foundlambda_(m+1)\lambda_{m+1}is the best possible. One can see it directly, or in the following way:
Let's ask
{:[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}
and determinelambda\lambdain the manner that the equationf(x)=0f(x)=0can 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)+,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)Or
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)Or
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}}
lambda\lambdamust therefore be between the maximum and minimum of the right-hand side, in other words between the maximum and minimum of the quadratic formsum_(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}when the variables are linked by the relationsum_(i=0)^(m)a_(2i)x_(1)^(2)=1\sum_{i=0}^{m} a_{2 i} x_{1}^{2}=1.
The result is thatlambda\lambdais between the smallest and largest roots of the characteristic equation
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
It is immediately apparent that
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
It is immediately apparent that
{:[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}
The result is that
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)
The stated property follows from this identity.
3. - Let's examine some specific cases of this problem.
3. - Let's examine some specific cases of this problem.
Let us first assume that
a_(:)=a_(1)=dots=a_(2m)=1a_{:}=a_{1}=\ldots=a_{2 m}=1
The polynomialQ^(')(x)Q^{\prime}(x)is none other than the trigonometric polynomial
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}}}
and then we have
lambda_(m+1)=cos((pi)/(m+2))\lambda_{m+1}=\cos \frac{\pi}{m+2}
We therefore deduce the following property:
If the coefficientsc_(0),c_(1),c_(2),dots,c_(2m)c_{0}, c_{1}, c_{2}, \ldots, c_{2 m}of the polynomial
(1)
If the coefficientsc_(0),c_(1),c_(2),dots,c_(2m)c_{0}, c_{1}, c_{2}, \ldots, c_{2 m}of the polynomial
(1)
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}
check the inequalities
c_(0) > 0,quad muc_(2i)c_(2i+2)-c_(ii+1)^(2) > 0,quad1=0,1,dots,m-1c_{0}>0, \quad \mu c_{2 i} c_{2 i+2}-c_{i i+1}^{2}>0, \quad 1=0,1, \ldots, m-1
Ormu <= (1)/(cos^(2)((pi)/(m+2)))\mu \leq \frac{1}{\cos ^{2} \frac{\pi}{m+2}}, this polynomial is positive.
Ifmu > (1)/(cos^(2)((pi)/(m+2)))\mu>\frac{1}{\cos ^{2} \frac{\pi}{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 takingmu=1\mu=1, we have the following theorem, due to MEB Van Vleck (^(1){ }^{1}).
If the coefficients ci satisfy the inequalities
c_(0) > 0,quadc_(2l)c_(2l+2)-c_(2l+1)^(2) > 0,quad i=0,1,dots,m-1c_{0}>0, \quad c_{2 l} c_{2 l+2}-c_{2 l+1}^{2}>0, \quad i=0,1, \ldots, m-1
Polynomial (1) is positive.
4. - By specifying the values ​​ofa_(0),a_(1),dots,a_(2m)a_{0}, a_{1}, \ldots, a_{2 m}We find: various statements.
4. - By specifying the values ​​ofa_(0),a_(1),dots,a_(2m)a_{0}, a_{1}, \ldots, a_{2 m}We find: various statements.
If we take
a_(0)=1,quada_(1)=a_(2)=dots=a_(2m)=2a_{0}=1, \quad a_{1}=a_{2}=\ldots=a_{2 m}=2
the polynomialQ_(I)(x)Q_{I}(x)becomes the trigonometric polynomial
Q_(t)(x)=(cos i(arccos x))/(2^(i-1))Q_{t}(x)=\frac{\cos i(\arccos x)}{2^{i-1}}
and then we have
lambda_(m+1)=cos((pi)/(2(m+1)))\lambda_{m+1}=\cos \frac{\pi}{2(m+1)}
which gives us the following statement:
If the coefficientsc_(0),c_(1),dots,c_(2m)duc_{0}, c_{1}, \ldots, c_{2 m} d upolynomial (1) satisfy the inequalities
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-1c_{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-1Ormu <= (1)/(cos^(2)((pi)/(2(m+1))))\mu \leq \frac{1}{\cos ^{2} \frac{\pi}{2(m+1)}}, this polynomial is positive.
If the coefficientsc_(0),c_(1),dots,c_(2m)duc_{0}, c_{1}, \ldots, c_{2 m} d upolynomial (1) satisfy the inequalities
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-1c_{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-1Ormu <= (1)/(cos^(2)((pi)/(2(m+1))))\mu \leq \frac{1}{\cos ^{2} \frac{\pi}{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)((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)}}(Form > 1)m>1)
Let's mention one more case. If
(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)}}(Form > 1)m>1)
Let's mention one more case. If
a_(i)=i+1,quad i=0,1,2,dots,2ma_{i}=i+1, \quad i=0,1,2, \ldots, 2 m
(1) EB Van Vleck "A sufficient condition for the maximum number of imaginary roots of an equation of thenn-the degree" Annals of Math. (2), t. 4 (1902-03) p. 191.
Qi (xx) is none other (up to a constant factor) than the Legendre polynomial of degreeii
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}
We therefore find a property that can be stated in the following form:
If the coefficientsc_(0),c_(1),dots,c_(2m)c_{0}, c_{1}, \ldots, c_{2 m}of the polynomial (1) satisfy the inequalities
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-1c_{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-1Orlambda\lambdais at least equal to the greatest zero of the Legendre polynomial. of degreem+1m+1, this polynomial is positive.
5. - Now let's consider the case where
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-1c_{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-1Orlambda\lambdais at least equal to the greatest zero of the Legendre polynomial. of degreem+1m+1, this polynomial is positive.
5. - Now let's consider the case where
a_(i)=((2m)/(i)),quad i=0,1,dots,2ma_{i}=\binom{2 m}{i}, \quad i=0,1, \ldots, 2 m
We know that if the equationf(x)=0f(x)=0to all its real roots. we have
c_(2i+1)^(2)-c_(2i)c_(2i+2) >= 0,quad i=0,1,dots,m-1c_{2 i+1}^{2}-c_{2 i} c_{2 i+2} \geq 0, \quad i=0,1, \ldots, 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))/((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
Therefore, applying the previous results yields the opposite property. In this caselambda_(m+1)\lambda_{m+1}is the largest root of the equation
{:((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}
We therefore have the following property:
If the coefficientsc_(0),c_(1),dots,c_(2n)c_{0}, c_{1}, \ldots, c_{2 n}dıc polynomial (1) verify the: inequalities
If the coefficientsc_(0),c_(1),dots,c_(2n)c_{0}, c_{1}, \ldots, c_{2 n}dıc polynomial (1) verify the: inequalities
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-1c_{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
Or,lambda\lambdais at least equal to the largest root of the equationS_(m+1)(x)=0\mathrm{S}_{m+1}(x)=0, this polynomial is positive.
6. - Let us examine in particular the casesm=2,3,4m=2,3,4.
6. - Let us examine in particular the casesm=2,3,4m=2,3,4.
Eitherm=2m=2If the equation
{:(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*}
to all its real roots we have
8c_(0)c_(2)-3c_(1)^(2) <= 0,quad8c_(2)c_(4)-3c_(3)^(2) <= 08 c_{0} c_{2}-3 c_{1}^{2} \leq 0, \quad 8 c_{2} c_{4}-3 c_{3}^{2} \leq 0
In this caseS_(3)(x)=-3(3x^(3)-4x)\mathrm{S}_{3}(x)=-3\left(3 x^{3}-4 x\right)Equation (2) will therefore have all its roots imaginary if
c_(0) > 0,quad8muc_(0)c_(2)-3c_(1)^(2) > 0,quad8muc_(2)c_(4)-3c_(3)^(2) > 0c_{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
Ormu <= (3)/(4)=0,75\mu \leq \frac{3}{4}=0,75.
Eitherm=3m=3If equation
(3)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
to all its real roots we have
Eitherm=3m=3If equation
(3)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
to all its real roots we have
12c_(0)c_(2)-0^(ˇ)c_(1)^(2) <= 0,16 cc_(4)-9c_(3)^(2) <= 0,12c_(4)c_(6)-5c_(5)^(2) <= 012 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
In this caseS_(4)(x)=225x^(4)-370x^(2)+8L\mathrm{S}_{4}(x)=225 x^{4}-370 x^{2}+8 \mathrm{~L}Equation (3) therefore has all its roots imaginary when
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) > 0c_{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
Ormu <= (5(37-8sqrt10))/(81)=0,7,23 dots\mu \leq \frac{5(37-8 \sqrt{10})}{81}=0,7,23 \ldots
So finallym=4m=4If equation
(4)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
to all its real roots we have
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) <= 016 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.
In this caseS_(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)Equation (4) therefore has all its roots imaginary if
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) > 0c_{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
Ormu <= (5(37-8sqrt10))/(81)=0,7,23 dots\mu \leq \frac{5(37-8 \sqrt{10})}{81}=0,7,23 \ldots
So finallym=4m=4If equation
(4)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
to all its real roots we have
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) <= 016 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.
In this caseS_(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)Equation (4) therefore has all its roots imaginary if
{:[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}
- Ormu <= (35)/(69)=0,5072 dots\mu \leq \frac{35}{69}=0,5072 \ldots
-
- Consider a polynomial of degree2m+12 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}
We have
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}
Taking the derivative, we see that if
{:(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*}
the polynomial is positiveF(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 coefficientsc_(0),c_(1),dots,c_(2n)c_{0}, c_{1}, \ldots, c_{2 n}of the equation
{:(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*}
verify the inequalities
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-1c_{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
Orlambda\lambdais at least equal to the maximum zero of the Legendre polynomial of degreem+1m+1, the equation has at most one real root.
In particular, we deduce from this the second Van Vleck SEM theorem (^(2){ }^{2}).
If the coefficientsc_(0),c_(1),dots,c_(2m)c_{0}, c_{1}, \ldots, c_{2 m}of equation (6) satisfy the inequalities
c_(0) > 0,quadc_(2i)c_(2i+2)-c_(2i+1)^(2) > 0,quad i=0,1,dots,m-1c_{0}>0, \quad c_{2 i} c_{2 i+2}-c_{2 i+1}^{2}>0, \quad i=0,1, \ldots, m-1
This equation has at most one real root.
8. - Consider the equation of degree n.
(7)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.
8. - Consider the equation of degree n.
(7)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.
Letj,rj, rtwo positive integers such thatj >= 0,j+2r <= nj \geq 0, j+2 r \leq n, and let's take the derivative of orderjjofg(x)g(x)
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]
Let us consider the transformation into(1)/(x)\frac{1}{x}(leaving aside one factor-
(^(2))\left({ }^{2}\right)loe cit(^(1))\left({ }^{1}\right).
constant)
(^(2))\left({ }^{2}\right)loe cit(^(1))\left({ }^{1}\right).
constant)
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}
and let's take the derivative of ordern-j-2rn-j-2 rof this polynomial
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]
So finally
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}
Suppose that equation (7) hasppreal roots. The equationg^((1))(x)=0g^{(1)}(x)=0, therefore alsog_(1)(x)=0g_{1}(x)=0, then at leastp-jp-jreal roots.
We can deduce that the equationg_(2)(x)=0g_{2}(x)=0has at leastp-n+2rp-n+2 rreal roots. It follows that if the polynomialg_(2)(x)g_{2}(x)equation (7) has all its imaginary roots and at mostn-2rn-2 rreal roots.
The results of No. 5 therefore allow us to state the following general property:
Ifj=0,j+2r <= nj=0, j+2 r \leq nand if the coefficients ci of equation (7) satisfy the inequalities
c_(j) > 0,(1)/(lambda2)c_(j+2i)c_(j+2i+2)-c_(j+2i+1)^(2) > 0,quad i=0,1,dots,r-1c_{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
Or,lambda\lambdais at least equal to the largest root of the equationS_(r+1)(x)=0\mathrm{S}_{r+1}(x)=0, this equation (7) has at mostn-2rn-2 rreal roots. We can also say that in this case equation (7) has at leastrrpairs of imaginary conjugated roots.
Forr=1r=1we have this property, obvious a priori, that if we can find ajjsuch as
c_(j) > 0,quadc_(j)c_(j+2)-c_(j+1)^(2) > 0c_{j}>0, \quad c_{j} c_{j+2}-c_{\mathrm{j}+1}^{2}>0
equation (7) has at least one pair of imaginary conjugate roots.
Forr=2r=2, we find that if
Forr=2r=2, we find that if
{:[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}
equation (7) has at least two pairs of imaginary conjugate roots.
Forr=3,4r=3,4We similarly find analogous sufficient conditions for the existence of three or four pairs of imaginary
conjugate roots. The values ​​of the coefficientsmu\muwhich are included in the respective inequalities are those specified in No. 6.9
. - Finally, if we write equation (7) in the form
conjugate roots. The values ​​of the coefficientsmu\muwhich 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)+cdots+c_(n)x^(n)=0c_{0}+c_{1} x+c_{2} x^{2}+\cdots+c_{n} x^{n}=0
He polynomialg_(2)(x)g_{2}(x)will be written in the form
(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} .
Let us turn our attention to the case wherej=0j=0The preceding polynomial can then be written in the following form:
(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+
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}.
Posing
m=r,a_(i)=(i+1)(i+2)dots(n-2r+i),quad i=0,1,dots,2rm=r, a_{i}=(i+1)(i+2) \ldots(n-2 r+i), \quad i=0,1, \ldots, 2 r
the polynomialQ_(i)(lambda)Q_{i}(\lambda)the corresponding is given by the recurrence relations (8)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)=0Q_{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.
We therefore have the following property:
Ifj >= 0,j+2r <= nj \geq 0, j+2 r \leq nand if the coefficientsc_(i)c_{\mathrm{i}}of the equation
Ifj >= 0,j+2r <= nj \geq 0, j+2 r \leq nand if the coefficientsc_(i)c_{\mathrm{i}}of the equation
c_(0)+c_(1)x+dots+c_(n)x^(n)=0c_{0}+c_{1} x+\ldots+c_{n} x^{n}=0
verify the inequalities
c_(j) > 0,(1)/(lambda^(2))c_(j+2i)c_(j+2i+2)-c_(j+2i+i)^(2) > 0,quad i=0,1,dots,rc_{\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
Or,lambda\lambdais at least equal to the greatest zero of the polynomialQ_(r+1)(lambda)=0Q_{r+1}(\lambda)=0defined by relations (8), this equation has at mostnn-er real roots.
The largest root of the equationQ_(r+1)(lambda)=0Q_{r+1}(\lambda)=0no, does not depend onjjand is bounded above. It follows from this, in effect, by a known theorem.^((3)){ }^{(3)}that the zeros ofQ_(+1)(lambda)Q_{+1}(\lambda)are in modulo at most equal to
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)
(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
where
b_(i+2)=((i+1)(2i+n-2r+1))/(2(2i+1)(2i+n-2r+2)),quad i=0,1,dots,r-1b_{\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
It is immediately apparent that (n > 2rn>2 r)
b_(2) > b_(3) > dots > b_(r+1)b_{2}>b_{3}>\ldots>b_{r+1}
The largest root of the equationQ_(r+1)(lambda)=0Q_{r+1}(\lambda)=0is therefore at most equal to
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
It can also be demonstrated that the true upper limit in question is> 1>1Indeed, if we make it tendernntowardsoo\inftywe obtain the polynomials
Q_(0)(lambda)=1,quadQ_(1)(lambda)=(lambda),quadQ_(i+2)(lambda)-lambdaQ_(i+1)(lambda)+(i+1)/(2(2i+1))Q_(i)(lambda)=0Q_{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
and we deduce from this
{:[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}
which proves our claim.