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T. Popoviciu, Asupra polinoamelor cari formează un şir Appell (II), Bull. Mathematique de la Soc. Roumaine des Sciences, 33/34 (1932), pp. 22-27 (in Romanian) (second note).
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Bulletin mathématique de la Société Roumaine des Sciences
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Societatea de Științe Matematice din România
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ON POLYNOMIALS WHICH FORM AN APPELL SEQUENCE^(1){ }^{1}) by TIB. POPOVICIU
- In our first article published under this title we searched for the sequences of polynomials in x :
(1)
which verify the conditions
P_(0)=1,P_(1),dotsP_(n,dots)P_{0}=1, P_{1}, \ldots P_{n, \ldots}
{:(2)A_(n)P_(n)+B_(n)P_(n-1)+C_(n)P_(n-2)=0:}\begin{equation*} A_{n} P_{n}+B_{n} P_{n-1}+C_{n} P_{n-2}=0 \tag{2} \end{equation*}
A_(n),B_(n),C_(n)\boldsymbol{A}_{n}, \boldsymbol{B}_{n}, \boldsymbol{C}_{n}finding polynomials in x degrees0,1,20,1,2respectively.
In particular, we studied the case whenA_(n)!=0A_{n} \neq 0for anythingnnand the caseA_(n)=0\boldsymbol{A}_{n}=0for anythingnnWe now propose to study the third case reported in the above-mentioned article, when some of theYear)\boldsymbol{A}_{n}are zero. We observe that it is sufficient to always assumeC_(n)!=0C_{n} \neq 0, because if we had a con= ditic of the formC_(n)-=0\boldsymbol{C}_{n} \equiv 0, we have a particular case of the conditionA_(n)=0\boldsymbol{A}_{n}=0. It may happen that the relation (3) is not unique; we agree to say thatA_(n)!=0\boldsymbol{A}_{n} \neq 0when there is no relationship (3) whereA_(n)=0A_{n}=0We saw in No. 3 of the previous article when this is possible.
In particular, we studied the case whenA_(n)!=0A_{n} \neq 0for anythingnnand the caseA_(n)=0\boldsymbol{A}_{n}=0for anythingnnWe now propose to study the third case reported in the above-mentioned article, when some of theYear)\boldsymbol{A}_{n}are zero. We observe that it is sufficient to always assumeC_(n)!=0C_{n} \neq 0, because if we had a con= ditic of the formC_(n)-=0\boldsymbol{C}_{n} \equiv 0, we have a particular case of the conditionA_(n)=0\boldsymbol{A}_{n}=0. It may happen that the relation (3) is not unique; we agree to say thatA_(n)!=0\boldsymbol{A}_{n} \neq 0when there is no relationship (3) whereA_(n)=0A_{n}=0We saw in No. 3 of the previous article when this is possible.
We will use the notations from the previous article, to which the reader is asked to refer.
2. Suppose
2. Suppose
A_(n+1)=0A_{n+1}=0
It immediately follows that we can take
A_(n+2)=A_(n+3)=dots=A_(n)=dots=1A_{n+2}=A_{n+3}=\ldots=A_{n}=\ldots=1
without losing the generality of the problem, because ifA_(m)=0\boldsymbol{A}_{m}=0or we can
^(1){ }^{1}) Second note. function____\_\_\_\_
The recurrence formulas established in No. 4 of the previous article give us,
^(1){ }^{1}) Second note. function____\_\_\_\_
The recurrence formulas established in No. 4 of the previous article give us,
lambda_(2)=(lambda_(m))/(m-1-lambda_(m)(m-2))\lambda_{2}=\frac{\lambda_{m}}{m-1-\lambda_{m}(m-2)}
whatever it may bem >= 2m \geq 2. The equality immediately follows depending onlambda_(n+2),mu_(n+2),nu_(n+2),beta_(n+2)\lambda_{n+2}, \mu_{n+2}, \nu_{n+2}, \beta_{n+2}, which is reduced
1+lambda_(n+2)+a_(n+2)=01+\lambda_{n+2}+a_{n+2}=0
immediately results in equality
(lambda_(n+2))/(n+1-nlambda_(n+2))=(lambda_(m))/(m-1-(m-2)lambda_(n))\frac{\lambda_{n+2}}{n+1-n \lambda_{n+2}}=\frac{\lambda_{m}}{m-1-(m-2) \lambda_{n}}
from which
{:(4)lambda_(m)=((m-1)lambda_(n+2))/(n+1+(mn-2)lambda_(n+2)):}\begin{equation*} \lambda_{m}=\frac{(m-1) \lambda_{n+2}}{n+1+(mn-2) \lambda_{n+2}} \tag{4} \end{equation*}
FROM
(lambda_(m))/( hat(lambda)_(n+2))=(mu_(m))/(mu_(n)+2)=(v_(m))/(v_(n+2))\frac{\lambda_{m}}{\hat{\lambda}_{n+2}}=\frac{\mu_{m}}{\mu_{n}+2}=\frac{v_{m}}{v_{n+2}}
deduce
{:(5)mu_(m)=((m-1)mu_(n+2))/(n+1+(mn-2)lambda_(n+2))","v_(m)=((m-1)v_(epsilon+2))/(n+1+(mn-2)lambda_(n+2)):}\begin{equation*} \mu_{m}=\frac{(m-1) \mu_{n+2}}{n+1+(mn-2) \lambda_{n+2}}, v_{m}=\frac{(m-1) v_{\epsilon+2}}{n+1+(mn-2) \lambda_{n+2}} \tag{5} \end{equation*}
(6)
a_(m)=-(n+1+(2m-n-3)lambda_(n)+2)/(n+1+(mn-2)lambda_(n)+2)a_{m}=-\frac{n+1+(2 mn-3) \lambda_{n}+2}{n+1+(mn-2) \lambda_{n}+2}
We then have the recurrence relation:beta_(m)[(n+1)+(mn-2)lambda_(n+2)]-beta_(m-1)|n+1+(mn-3)lambda_(n+2)|+mu_(n+2)=0\beta_{m}\left[(n+1)+(mn-2) \lambda_{n+2}\right]-\beta_{m-1}\left|n+1+(mn-3) \lambda_{n+2}\right|+\mu_{n+2}=0from where
{:(7)beta_(m)=((n+1)beta_(n+2)-(mn-2)mu_(n+2))/(n+1+(mn-2)lambda_(n+2)):}\begin{equation*} \beta_{m}=\frac{(n+1) \beta_{n+2}-(mn-2) \mu_{n+2}}{n+1+(mn-2) \lambda_{n+2}} \tag{7} \end{equation*}
To completely resolve the problem, it is necessary to determinelambda_(m),mu_(m)\lambda_{m}, \mu_{m},v_(m),p_(m),alpha_(m)v_{m}, p_{m}, \alpha_{m}form <= n+2m \leq n+2.
But I saw that
P_(n)=(x+a)^(k)(x+b)^(nk)P_{n}=(\mathrm{x}+a)^{k}(\mathrm{x}+b)^{n-k}
We can write for simplification
P_(n)=r^(n-k)(x+b_(1))^(k)P_{n}=\mathrm{r}^{n-k}\left(\mathrm{x}+b_{1}\right)^{k}
and the general result will be obtained by substituting forx,b_(1),pcx+b,a-bx, b_{1}, \mathrm{pc} x+b, a-bres= spectively. Nothing of generality is thus lost, for
(d)/(d(x+b))=(d)/(dx)\frac{d}{d(x+b)}=\frac{d}{d x}
the character of the polynomials is therefore maintained by this simplification
3. Letk=0k=0, so
3. Letk=0k=0, so
P_(n)=x^(n)P_{n}=\mathrm{x}^{n}
polynomialsP_(i,)i < n\boldsymbol{P}_{i,} \boldsymbol{i}<\boldsymbol{n}, are determined, the constantslambda_(m),mu_(m),nu_(m),alpha_(m),beta_(m)cu=\lambda_{m}, \mu_{m}, \nu_{m}, \alpha_{m}, \beta_{m} \mathrm{cu}=known form <= n+1m \leq n+1.
Let's put
P_(n+1)=x^(n+1)+c,P_(n+2)=x^(n+2)+(n+2)cx+c_(1)P_{n+1}=\mathrm{x}^{n+1}+c, P_{n+2}=\mathrm{x}^{n+2}+(n+2) c \mathrm{x}+c_{1}
c.c_(1)\boldsymbol{c}_{1}being two constants. It is seen that to have an interesting case, irreducible to those studied so far, we must assumec!=0c \neq 0.
I have then
{:[(n+2)cx+c_(1)+cB_(n+2)=0],[x^(2)+xB_(n+2)+C+2=0]:}\begin{gathered}
(n+2) c \mathrm{x}+c_{1}+c B_{n+2}=0 \\
\mathrm{x}^{2}+\mathrm{x} B_{n+2}+C+2=0
\end{gathered}
and it is removed immediately
{:[alpha_(n+2)=-(n+2),lambda_(n+2)=n+1],[beta_(n+2)=-(c_(1))/(c)=sigma,mu_(n+2)=-sigma],[v_(n+2)=0]:}\begin{array}{ll}
\alpha_{n+2}=-(n+2) & \lambda_{n+2}=n+1 \\
\beta_{n+2}=-\frac{c_{1}}{c}=\sigma & \mu_{n+2}=-\sigma \\
v_{n+2}=0
\end{array}
Formulas (4), (5), (6), (7) now give uspclambda_(m),mu_(m),nu_(m),alpha_(m),beta_(m)\mathrm{pc} \lambda_{m}, \mu_{m}, \nu_{m}, \alpha_{m}, \beta_{m}and all relations (3) are determined.
We can write in the above case
P_(m)=x^(m)+Q_(m-n-1)\boldsymbol{P}_{m}=\mathbf{x}^{m}+Q_{m-n-1}
where the sequence of polynomials
{:(8)Q_(0)=c","Q_(1)","Q_(2)","dotsQ_(s)dots:}\begin{equation*}
Q_{0}=c, Q_{1}, Q_{2}, \ldots Q_{s} \ldots \tag{8}
\end{equation*}
check the relationships
{:[(9)(dQ_(m-n-1))/(dx)=mQ_(m-n-2)],[Q_(m-n-1)+B_(m)Q_(m-n-2)+C_(m)Q_(m-n-3)=0]:}\begin{gather*}
\frac{d Q_{m-n-1}}{d \mathrm{x}}=m Q_{m-n-2} \tag{9}\\
Q_{m-n-1}+B_{m} Q_{m-n-2}+C_{m} Q_{m-n-3}=0
\end{gather*}
So the sequence of polynomials:
{:(10)R_(0)=c","R_(1)","dotsR_(s)","dots:}\begin{equation*}
R_{0}=c, R_{1}, \ldots R_{s}, \ldots \tag{10}
\end{equation*}
where
Q_(s)=((n+s+1)/(s))R_(s)Q_{s}=\binom{n+s+1}{s} R_{s}
is an APPELL string checking the condition:
(11)quad((n+s+1)/(s))R_(s)+((n+s)/(s-1))B_(n+s+1)R_(s-1)((n+s-1)/(s-2))\quad\binom{n+s+1}{s} R_{s}+\binom{n+s}{s-1} B_{n+s+1} R_{s-1}\binom{n+s-1}{s-2}
(11)quad((n+s+1)/(s))R_(s)+((n+s)/(s-1))B_(n+s+1)R_(s-1)((n+s-1)/(s-2))\quad\binom{n+s+1}{s} R_{s}+\binom{n+s}{s-1} B_{n+s+1} R_{s-1}\binom{n+s-1}{s-2}
C_(n+s+1)R_(s-2)=0\boldsymbol{C}_{n+s+1} \boldsymbol{R}_{s-2}=0
And polynomialsR_(s)\boldsymbol{R}_{s}falls into the case studied in the previous chapter under No. 4.
But we have
But we have
{:[B_(n+s+1)=-(n+2s)/(s)xx+(sigma(n+s))/(s(n+1))],[C_(n+s+1)=(n+s)/(s)x^(2)-(sigma(n+s))/(s(n+1))xx]:}\begin{gathered}
B_{n+s+1}=-\frac{n+2 s}{s} \times+\frac{\sigma(n+s)}{s(n+1)} \\
C_{n+s+1}=\frac{n+s}{s} x^{2}-\frac{\sigma(n+s)}{s(n+1)} \times
\end{gathered}
{:[(12)(n+s+1)R_(s)+[-(n+2s)xx+(sigma(n+s))/(n+1)]],[R_(s-1)+(s-1)(x^(2)-(sigma)/(n+1)xx)R_(s-2)=0]:}\begin{align*}
& (n+s+1) R_{s}+\left[-(n+2 s) \times+\frac{\sigma(n+s)}{n+1}\right] \tag{12}\\
& R_{s-1}+(s-1)\left(\mathrm{x}^{2}-\frac{\sigma}{n+1} \times\right) R_{s-2}=0
\end{align*}
So we enter the general case by taking:
lambda=(1)/(n+3),mu=-(sigma)/((n+1)(n+3)),nu=0,beta=(sigma(n+2))/((n+1)(n+3))\lambda=\frac{1}{n+3}, \mu=-\frac{\sigma}{(n+1)(n+3)}, \nu=0, \beta=\frac{\sigma(n+2)}{(n+1)(n+3)}
It is then obtained
uuu_(s=0)^(oo)(7^(s))/(s!)R_(s)=c*e^(alpha x)quad G.(n+1,n+2,-(sigma)/(n+1)z)\bigcup_{s=0}^{\infty} \frac{\boldsymbol{7}^{s}}{\boldsymbol{s}!} \boldsymbol{R}_{s}=\boldsymbol{c} \cdot \boldsymbol{e}^{\alpha x} \quad G .\left(n+1, n+2,-\frac{\sigma}{n+1} \mathrm{z}\right)
forR_(0)=c\boldsymbol{R}_{0}=\boldsymbol{c},
sum_(s=0)^(oo)(z^(n+s+1))/((n+s+1)!)Q_(s)=(z^(n+1))/((n+1)!)sum_(0)^(oo)(z^(s))/(s!)R_(s)\sum_{s=0}^{\infty} \frac{z^{n+s+1}}{(n+s+1)!} Q_{s}=\frac{z^{n+1}}{(n+1)!} \sum_{0}^{\infty} \frac{z^{s}}{s!} R_{s}
and so the generating function is
sum_(m=0)^(oo)(z^(m))/(m!)P_(m)=e^(2x)[1+c(z^(n+1))/((n+1)!)G(n+1,n+2,-(sigma)/(n+1)z)]\sum_{m=0}^{\infty} \frac{\mathbf{z}^{m}}{m!} \boldsymbol{P}_{m}=e^{2 x}\left[1+c \frac{\mathbf{z}^{n+1}}{(n+1)!} G\left(n+1, n+2,-\frac{\sigma}{n+1} z\right)\right]
But it can be said for the general case
P_(n)=(x+b)^(n)P_{n}=(\mathrm{x}+b)^{n}
that polynomials are generated by the function
e^(ℓ(x+b))[1+c(z^(n+1))/((n+1)!)G(n+1,n+2,theta z)]e^{\ell(x+b)}\left[1+c \frac{z^{n+1}}{(n+1)!} G(n+1, n+2, \theta z)\right]
b,c,db, c, dbeing arbitrary constants.
The previous result was obtained under the assumptionsigma!=0\sigma \neq 0If
​sigma=0\sigma=0The procedure we applied in the previous paper does not make sense, but it is easy to see that the formula remains applicable. We obtain the generating function
The previous result was obtained under the assumptionsigma!=0\sigma \neq 0If
​sigma=0\sigma=0The procedure we applied in the previous paper does not make sense, but it is easy to see that the formula remains applicable. We obtain the generating function
e^(z(x+b))[1+c(z^(n+1))/((n+1)!)]e^{z(x+b)}\left[1+c \frac{z^{n+1}}{(n+1)!}\right]
- Let's move on to the case.k=1k=1, which is treated in the same way.
P_(n)=x^(n-19)(x+b_(1))P_{n}=\mathrm{x}^{n-19}\left(\mathrm{x}+b_{1}\right)
P_(n+1)=x^(n)(x+(n+1)/(n)b_(1))+c,P_(n+2)=x^(n+1)(x+(n+2)/(n)b_(1))+(n+2)cx+c_(1)P_{n+1}=\mathrm{x}^{n}\left(\mathrm{x}+\frac{n+1}{n} b_{1}\right)+c, P_{n+2}=\mathrm{x}^{n+1}\left(\mathrm{x}+\frac{n+2}{n} b_{1}\right)+(n+2) c \mathrm{x}+c_{1}andc!=0\boldsymbol{c} \neq 0.
It is found that
{:[alpha_(n+2)=-(n+2)],[beta_(n+2)=-(c_(1))/(c)=-nb_(1)]:}\begin{aligned}
& \alpha_{n+2}=-(n+2) \\
& \beta_{n+2}=-\frac{c_{1}}{c}=-n b_{1}
\end{aligned}
We can also write here
P_(m)=x^(m-1)(x+(m)/(n)b_(1))+Q_(m-n-1)\boldsymbol{P}_{m}=\mathrm{x}^{m-1}\left(\mathrm{x}+\frac{m}{n} b_{1}\right)+Q_{m-n-1}
and polynomialsQ_(s)Q_{s}verify the relation (9). The polynomials are deducedR\boldsymbol{R}, with relation (11), which is written
{:[(n+s+1)R_(s)+[-(n+2s)xx+(1-n-s)b_(1)]],[R_(s-1)+(s-1)(x^(2)+b_(x))R_(s-2)=0]:}\begin{gathered}
(n+s+1) R_{s}+\left[-(n+2 s) \times+(1-n-s) b_{1}\right] \\
R_{s-1}+(s-1)\left(\mathrm{x}^{2}+b_{x}\right) R_{s-2}=0
\end{gathered}
and it can be taken
lambda=(1)/(n+3),mu=(b_(1))/(n+3),nu=0,beta=-(n+1)/(n+3)b_(1)\lambda=\frac{1}{n+3}, \mu=\frac{b_{1}}{n+3}, \nu=0, \beta=-\frac{n+1}{n+3} b_{1}
and the generating function is:
sum_(0)^(oo)(z^(s))/(s!)R_(s)=c*e^(2x)G(n,n+2,b_(1)z)\sum_{0}^{\infty} \frac{\mathbf{z}^{s}}{\boldsymbol{s}!} \boldsymbol{R}_{s}=\boldsymbol{c} \cdot \boldsymbol{e}^{2 x} \boldsymbol{G}\left(\boldsymbol{n}, \boldsymbol{n}+2, b_{1} \mathbf{z}\right)
For the general case we have immediately:
sum_(0)^(oo)(z^(m))/(m!)P_(m)=e^(z(x+b))[1+(2)/(n)(a-b)z+cG(n,n+2(a-b)z)]\sum_{0}^{\infty} \frac{\mathrm{z}^{m}}{m!} P_{m}=e^{z(\mathrm{x}+b)}\left[1+\frac{2}{n}(a-b) \mathrm{z}+c G(n, n+2(a-b) \mathrm{z})\right]
- casek > 1k>1, that is
P_(n)=x^(n-k)(x+b_(1))^(k)P_{n}=\mathrm{x}^{n-k}\left(\mathrm{x}+b_{1}\right)^{k}
n-k >= k,quad k > 1,n-k \geq k, \quad k>1,
It can be attached to the general case by taking
lambda=-(1)/(n-1)\lambda=-\frac{1}{n-1}
as we saw in the previous article.
We found:
We found:
lambda=-(1)/(n-1),alpha=-(n-2)/(n-1),mu=-(b_(1))/(n-1),nu=0,beta=-((k-1)b_(1))/(n-1)\lambda=-\frac{1}{n-1}, \alpha=-\frac{n-2}{n-1}, \mu=-\frac{b_{1}}{n-1}, \nu=0, \beta=-\frac{(k-1) b_{1}}{n-1}
The equation that gives the generating function is written:
we put
we put
zF^('')-[zb_(1)+n]F^(')+kb_(1)F=0z F^{\prime \prime}-\left[z b_{1}+n\right] F^{\prime}+k b_{1} F=0
tF^('')_(t^(2))-[t+n]F^(')_(1)+kF=0t F^{\prime \prime}{ }_{t^{2}}-[t+n] F^{\prime}{ }_{1}+k F=0
All integrals of this equation are holomorphic about the origin. Int'a= indeed we have a polynomial
sum_(i=0)^(k)(k(k-1)dots^(k)-i+1))/(i!n(n-1)dots(n-i+1))t^(i)=P_(1)(t)\sum_{i=0}^{k} \frac{\left.k(k-1) \ldots{ }^{k}-i+1\right)}{i!n(n-1) \ldots(n-i+1)} t^{i}=P_{1}(t)
which satisfies the equation. Putting
F=e^(lambda)varphiF=e^{\lambda} \varphi
HAVE
tvarphi^('')-(n-t)varphi^(')-(n-k)varphi=0t \varphi^{\prime \prime}-(n-t) \varphi^{\prime}-(n-k) \varphi=0
or changing tottin -tt:
tvarphi^('')-(n+t)varphi^(')+(n-k)varphi=0t \varphi^{\prime \prime}-(n+t) \varphi^{\prime}+(n-k) \varphi=0
So we have the solution distinct from the previous one:
e^(t)sum_(i=0)^(n-k)(-1)^(i)((n-k)(n-k-1)dots(n-k-i+1))/(i!n(n-1)dots(n-i+1))i^(i)=e^(t)P_(2)(t)e^{t} \sum_{i=0}^{n-k}(-1)^{i} \frac{(n-k)(n-k-1) \ldots(n-k-i+1)}{i!n(n-1) \ldots(n-i+1)} i^{i}=e^{t} P_{2}(t)
So there is
e^(t)sum_(i=0)^(n-k)(-1)^(i)((n-k)(n-k-1)dots(n-k-i+1))/(i!n(n-1)dots(n-i+1))i^(i)=e^(t)P_(2)(t)e^{t} \sum_{i=0}^{n-k}(-1)^{i} \frac{(n-k)(n-k-1) \ldots(n-k-i+1)}{i!n(n-1) \ldots(n-i+1)} i^{i}=e^{t} P_{2}(t)
So there is
sum_(n=0)^(oo)(z^(n))/(n!)P_(n)=e^(iota x)[cP_(1)(b_(1)z)+c_(1)e^(z)b_(1)P_(2)(b_(2)z)]\sum_{n=0}^{\infty} \frac{\mathrm{z}^{n}}{n!} P_{n}=e^{\iota x}\left[c P_{1}\left(b_{1} \mathrm{z}\right)+c_{1} e^{z} b_{1} P_{2}\left(b_{2} \mathrm{z}\right)\right]
constantsc,c_(1)\boldsymbol{c}, \boldsymbol{c}_{1}being linked by the relationship
c+c_(1)=1c+c_{1}=1
In general we will have the polynomials
sum_(n=0)^(oo)(z^(n))/(n!)P_(n)=e^(2[x+b])[cP_(1)[(a-b)z]+c_(1)e^([a-b]z)P_(2)[(a-b)z)]\sum_{n=0}^{\infty} \frac{\mathbf{z}^{n}}{n!} P_{n}=e^{2[x+b]}\left[c P_{1}[(a-b) \mathbf{z}]+c_{1} e^{[a-b] z} P_{2}[(a-b) \mathbf{z})\right]
6. If aB_(n)B_{n}it is identical we must have
sum_(n=0)^(oo)(z^(n))/(n!)P_(n)=e^(2[x+b])[cP_(1)[(a-b)z]+c_(1)e^([a-b]z)P_(2)[(a-b)z)]\sum_{n=0}^{\infty} \frac{\mathbf{z}^{n}}{n!} P_{n}=e^{2[x+b]}\left[c P_{1}[(a-b) \mathbf{z}]+c_{1} e^{[a-b] z} P_{2}[(a-b) \mathbf{z})\right]
6. If aB_(n)B_{n}it is identical we must have
lambda=-(1)/(2n-3)\lambda=-\frac{1}{2 n-3}
and we are in the previous case. This therefore does not present a case distinct from those studied.
Paris, November 8, 1928.