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T. Popoviciu, Remarques sur les polynômes binomiaux, Mathematica, 6 (1932), pp. 8-10 (in French).
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Mathematica
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published by the Romanian Academy Publishing House
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[JFM 58.0376.06]
https://zbmath.org/?q=an:58.0376.06
Note: this paper is republished from: T. Popoviciu, Remarques sur les polynômes binomiaux, Buletinul Soc. de Ştiinţe din Cluj, 6, pp.146-148 (in French)
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NOTES ON BINOMIAL POLYNOMIES
by
T. Popoviciu
Former student of the École Normale Supérieure.
Admitted on April 28, 1981.
We say that the sequence of polynomials inxx
(1)
Admitted on April 28, 1981.
We say that the sequence of polynomials inxx
(1)
P_(0),P_(1),dotsP_(n),dots\mathrm{P}_{0}, \mathrm{P}_{1}, \ldots \mathrm{P}_{n}, \ldots
the degree of each being equal to the corresponding index, is a binomial sequence if it satisfies the equations
{:(2)P_(n)(x+y)=sum_(i=0)^(n)P_(i)(x)P_(ni)(y):}\begin{equation*} P_{n}(x+y)=\sum_{i=0}^{n} P_{i}(x) P_{ni}(y) \tag{2} \end{equation*}
Forn=0,1,2,dotsn=0,1,2, \ldotsand similarly inxxAndyyWe then have
P_(0)=1,quadP_(n)(0)=0,quad n=1,2,dotsP_{0}=1, \quad P_{n}(0)=0, \quad n=1,2, \ldots
Binomial sequences are obtained through the following formal expansion
{:(3)(1+a_(1)z+a_(2)z^(2)+cdots)^(x)=e^(x(c_(1)z+c_(2)z^(2)+cdots))=sum_(n=0)^(oo)P_(n)(x)z^(pi):}\begin{equation*} \left(1+a_{1} z+a_{2} z^{2}+\cdots\right)^{x}=e^{x\left(c_{1} z+c_{2} z^{2}+\cdots\right)}=\sum_{n=0}^{\infty} P_{n}(x) z^{\pi} \tag{3} \end{equation*}
It is easy to find the coefficientsc_(n)c_{n}depending on theyear)year}and vice versa.
- If
c_(n) >= 0,quad n=1,2,3,dotsc_{n} \geq 0, \quad n=1,2,3, \ldots
we also have
and if
and if
a_(n) >= 0,quad n=1,2,3,dotsa_{n} \geq 0, \quad n=1,2,3, \ldots
we also have
{:[c_(n) >= 0",",n=1","2","3","dotsa_(1) > 0],[a_(n) > 0",",n=2","3","dots]:}\begin{array}{ll} c_{n} \geq 0, & n=1,2,3, \ldots a_{1}>0 \\ a_{n}>0, & n=2,3, \ldots \end{array}
We easily find that
a is a positive number so that we have
a is a positive number so that we have
P_(n)(x) >= 0P_{n}(x) \geq 0
n=0,1.9,dotsn=0,1.9, \ldots
in the interval (0, a) it is necessary and sufficient that
{:(5)c_(n) >= 0","quad n=1","2","dots:}\begin{equation*}
c_{n} \geq 0, \quad n=1,2, \ldots \tag{5}
\end{equation*}
It also follows from (2) that inequalities (4) are verified throughout the interval(0,oo)(0, \infty).
Eitherf(x)f(x)a uniform and continuous function in the interval Cermé(0,1)(0,1).
Given that inequalities (5) are verified and assuming thata_(1) > 0a_{1}>0so that the polynomial
mu_(n)=(sum_(nu=0)^(n)f((nu )/(n))P_(nu)(x)P_(n-nu)(1-x))/(a_(n))\mu_{n}=\frac{\sum_{\nu=0}^{n} f\left(\frac{\nu}{n}\right) P_{\nu}(x) P_{n-\nu}(1-x)}{a_{n}}
converges uniformly towardsf(x)f(x)throughout the interval(0,1)(0,1)It is necessary and sufficient that:
{:(6)lim_(n rarr oo)(sum_(nu=0)^(n)v^(k)P_(nu)(x)P_(n-nu)(1-x))/(n^(k)a_(n))=x_(k):}\begin{equation*}
\lim _{n \rightarrow \infty} \frac{\sum_{\nu=0}^{n} v^{k} P_{\nu}(x) P_{n-\nu}(1-x)}{n^{k} a_{n}}=x_{k} \tag{6}
\end{equation*}
whatever the positive integerkk2.
Let us consider in particular the polynomials (1) arising from the expansion
Let us consider in particular the polynomials (1) arising from the expansion
e^((xz)/(1-z))=SigmaL_(n)^(')(x)*z^(n)e^{\frac{x z}{1-z}}=\Sigma L_{n}^{\prime}(x) \cdot z^{n}
IfL^(**)_(n)(x)L^{*}{ }_{n}(x)is the Laguerre polynomial of degreennwe have
L_(n)(x)=L_(n)^(**)(-x)-L_(n-1)^(**)(-x)\mathrm{L}_{n}(x)=\mathrm{L}_{n}^{*}(-x)-\mathrm{L}_{n-1}^{*}(-x)
We obtain easily
L_(n)(x)=(e^(-x)x)/(n!)*(d^(n))/(dx^(n))x^(n-1)e^(x)=sum_(nu=1)^(n)((n-1)/(v-1))(x^(nu))/(v!)L_{n}(x)=\frac{e^{-x} x}{n!} \cdot \frac{d^{n}}{d x^{n}} x^{n-1} e^{x}=\sum_{\nu=1}^{n}\binom{n-1}{v-1} \frac{x^{\nu}}{v!}
Let us now return to the general case and suppose that the series
sum_(n=1)^(oo)c_(n)z^(n)\sum_{n=1}^{\infty} c_{n} z^{n}
converges inside a circle of radiusr^(')r^{\prime}and from the center of origin and let's taker < r^(')r<r^{\prime}According to a theorem of Cauchy, there exists a number
M such as
|c_(n)| < (M)/(r^(n))quad n=1,2,dots\left|c_{n}\right|<\frac{M}{r^{n}} \quad n=1,2, \ldots
And
sum_(n=1)^(oo)c_(n)z^(n) <= sum_(n=1)^(oo)(Mz^(n))/(r^(n))=(M(z)/(r))/(1-(z)/(r))quad(|z| < r)\sum_{n=1}^{\infty} c_{n} z^{n} \leq \sum_{n=1}^{\infty} \frac{\mathrm{M} z^{n}}{r^{n}}=\frac{\mathrm{M} \frac{z}{r}}{1-\frac{z}{r}} \quad(|z|<r)
We can deduce that forx >= 0x \geq 0
|P_(n)(x)| < (1)/(r^(n))L_(n)(Mx)\left|\mathrm{P}_{n}(x)\right|<\frac{1}{r^{n}} \mathrm{~L}_{n}(\mathrm{M} x)
It is easily demonstrated that ifx > 0x>0
(L_(2)(x))/(L_(1)(x)) > (L_(3)(x))/(L_(2)(x)) > cdots > (L_(n)(x))/(L_(n-1)(x)) > cdots\frac{L_{2}(x)}{L_{1}(x)}>\frac{L_{3}(x)}{L_{2}(x)}>\cdots>\frac{L_{n}(x)}{L_{n-1}(x)}>\cdots
the report(L_(n))/(L_(n-1))\frac{L_{n}}{L_{n-1}}Therefore, it has a finite limit. The recurrence relation
xL_(n)=(n+1)L_(n+1)-2nL_(n)+(n-1)L_(n-1)x L_{n}=(n+1) L_{n+1}-2 n L_{n}+(n-1) L_{n-1}
then shows us that
lim_(n rarr oo)(L_(n))/(L_(n-1))=1quad x > 0\lim _{n \rightarrow \infty} \frac{L_{n}}{L_{n-1}}=1 \quad x>0
We deduce from this
bar(lim)_(n rarr oo)root(n)(|P_(n)(x)|) <= (1)/(r).\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|\mathrm{P}_{n}(x)\right|} \leq \frac{1}{r} .
Let us now consider a sequence of numbers
lambda_(0),lambda_(1),dotslambda_(n),dots\lambda_{0}, \lambda_{1}, \ldots \lambda_{n}, \ldots
ot
bar(lim)_(n rarr oo)root(n)(|lambda_(n)|)=lambda\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|\lambda_{n}\right|}=\lambda
As a result, the series
lambda_(0)P_(0)(x)+lambda_(1)P_(1)(x)+cdots+lambda_(n)P_(n)(x)+cdots\lambda_{0} \mathrm{P}_{0}(x)+\lambda_{1} \mathrm{P}_{1}(x)+\cdots+\lambda_{n} \mathrm{P}_{n}(x)+\cdots
converges absolutely regardless oflambda < r^(')\lambda<r^{\prime}Andx >= 0x \geq 0.