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T. Popoviciu, Sur les suites de polynômes, Mathematica, 5 (1931), pp. 36-48 (in French).
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This paper is the republishing of: T. Popoviciu, Sur les suites de polynômes, Buletinul Soc. de Ştiinţe din Cluj, 5 (1931), pp. 492-504 (in French)
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ON SEQUENCES OF POLYNOMIES
by
T. Popoviciu
Former student of the Ecole Normale Supérieure of Paris
Received on December 4, 1980.
MR Lagrange in a memoir published in "Acta Mathematica."^((1)){ }^{(1)}He studied sequences of numbers from an algebraic point of view. He applied these principles to certain sequences of polynomials. In this work, we will complement the "algebra of sequences of numbers" with an "algebra of sequences of polynomials." We will address the applications in other papers.
MR Lagrange in a memoir published in "Acta Mathematica."^((1)){ }^{(1)}He studied sequences of numbers from an algebraic point of view. He applied these principles to certain sequences of polynomials. In this work, we will complement the "algebra of sequences of numbers" with an "algebra of sequences of polynomials." We will address the applications in other papers.
We only provide definitions and results. The proofs often lead to somewhat lengthy calculations, but these present no difficulty.
- Definition of a sequence of polynomials. Consider a sequence of polynomials inxx
{:(1)P_(0)","P_(i)","dotsP_(n)","dots:}\begin{equation*} \mathrm{P}_{0}, \mathrm{P}_{\mathrm{i}}, \ldots \mathrm{P}_{n}, \ldots \tag{1} \end{equation*}
taken in a specific order. This order is characterized by the: numbernnwhich is the index or rank of the polynomialP_(n)\mathbb{P}_{n}. Let us designate byp(n)p(n)the degree of the polynomialP_(n)\mathrm{P}_{n}. IfP_(n)\mathrm{P}_{n}, is identically zero, we assume thatp(n)p(n)has a negative value as large as desired. Lat difference
{:(2)p(n)-n:}\begin{equation*} p(n)-n \tag{2} \end{equation*}
is the order of the polynomialP_(n)\mathrm{P}_{n}. Ifp(n) < 0p(n)<0So yes.P_(n)^(')\mathrm{P}_{n}^{\prime}is identical. null we say that it is of order -oo\infty.
If the orders of all the elements of a sequence are equal to-oo-\inftyWe say that this sequence is the zero sequence. For all other sequences, the order of the elements has an upper bound.mmFinite or infinite. Ifmmis finished, we say that the sequence is of finite ordermm. In;
(1) R. Lagrange. „Mémoire sur les suites de polynomes" Acla. Math. 51 (1928), p. 201..
this case the sequence (1) has at least one element of ordermmWe call the smallest value of the characteristic indexnnfor which
(1) R. Lagrange. „Mémoire sur les suites de polynomes" Acla. Math. 51 (1928), p. 201..
this case the sequence (1) has at least one element of ordermmWe call the smallest value of the characteristic indexnnfor which
p(n)-n=mp(n)-n=m
We say that a sequence is complete if all its elements have the same order. A complete sequence is necessarily of finite order and its characteristic index is 0.
Ifmmis infinite, the sequence is of infinite order. Such a sequence cannot be complete and does not admit a characteristic index. We call the class of the sequence (1) the numberkksuch as
p(0) < 0,quad p(1) < 0,dotsquad p(k-1) < 0,quad p(k) >= 0.p(0)<0, \quad p(1)<0, \ldots \quad p(k-1)<0, \quad p(k) \geqslant 0 .
A sequence of negative ordermmis at least class-m-mWe
ask
ask
((i)/(j))=(i(i-1)dots(i-j+1))/(1*2*3dots j);((i)/(j))=0,j > i\binom{i}{j}=\frac{i(i-1) \ldots(i-j+1)}{1 \cdot 2 \cdot 3 \ldots j} ;\binom{i}{j}=0, j>i
Since accents denote derivatives, we say that a class sequencekkand order-k-kis normal if
{:[(3)sum_(j=0)^(i)((i)/(j))P_(kj)^((j))!=0],[i=0","1","2","dots]:}\begin{gather*} \sum_{j=0}^{i}\binom{i}{j} \mathrm{P}_{kj}^{(j)} \neq 0 \tag{3}\\ i=0,1,2, \ldots \end{gather*}
We see that the first members are numbers.
We denote by[P][\mathrm{P}]the sequence (1). When considering several: sequences[P],[Q],dots[\mathrm{P}],[\mathrm{Q}], \ldotswe designate byp(n),q(n),dotsp(n), q(n), ...the degree of polynomialsP_(n),Q_(n),dots\mathrm{P}_{n}, Q_{n}, \ldots
We denote by[P][\mathrm{P}]the sequence (1). When considering several: sequences[P],[Q],dots[\mathrm{P}],[\mathrm{Q}], \ldotswe designate byp(n),q(n),dotsp(n), q(n), ...the degree of polynomialsP_(n),Q_(n),dots\mathrm{P}_{n}, Q_{n}, \ldots
All the relationships we write between several polynomials are verified identically with respect toxx2.
The algebra of polynomial sequences. 10. We denote by [0] the zero sequence.
20. The sequence
The algebra of polynomial sequences. 10. We denote by [0] the zero sequence.
20. The sequence
P_(0)=1;quadP_(n)=0,n > 0\mathrm{P}_{0}=1; \quad \mathrm{P}_{n}=0, n>0
-is the unit sequence this will be designated by [1]. It is normal and of class 0.
The sequel
P_(k)=1,quadP_(n)=0,quad n!=k\mathrm{P}_{k}=1, \quad \mathrm{P}_{n}=0, \quad n \neq k
is the unit sequence of classk,[l]_(k)k,[\mathrm{l}]_{k}30.
Two suites[P],[Q][\mathrm{P}],[\mathrm{Q}]are equal if, and only if
Two suites[P],[Q][\mathrm{P}],[\mathrm{Q}]are equal if, and only if
{:[P_(n)=Q_(n)],[n=0","1","2","dots]:}\begin{gathered}
P_{n}=Q_{n} \\
n=0,1,2, \ldots
\end{gathered}
We write to you
[P]=[Q][\mathrm{P}]=[\mathrm{Q}]
- The product of a sequel[P][\mathrm{P}]by a numberlambda\lambdais by definition a new sequel[Q][Q]given by
{:[Q_(n)=lambdaP_(n)],[n=0","1","2","dots],[lambda*∣P]=[lambda P]]:}\begin{gathered}
Q_{n}=\lambda P_{n} \\
n=0,1,2, \ldots \\
\lambda \cdot \mid P]=[\lambda P]
\end{gathered}
We write
P]\mathrm{P}]is the opposite of[P][\mathrm{P}]and is equal to-[P]-[\mathrm{P}]50.
The sum of two sequences[P],[Q][P],[Q]is a new sequel|R|∣|R| \middefined by
P]\mathrm{P}]is the opposite of[P][\mathrm{P}]and is equal to-[P]-[\mathrm{P}]50.
The sum of two sequences[P],[Q][P],[Q]is a new sequel|R|∣|R| \middefined by
{:[R_(n)=P_(n)+Q_(n)],[n=0","1","2","dots],[[P]+[Q]=[R].]:}\begin{gathered}
\mathrm{R}_{n}=\mathrm{P}_{n}+\mathrm{Q}_{n} \\
n=0,1,2, \ldots \\
{[\mathrm{P}]+[\mathrm{Q}]=[\mathrm{R}] .}
\end{gathered}
We write
{:[6^(0)". Le produit élémentaire de "[P]" par "[Q]" est une nouvelle suite- "],[" léfinie par les égalités "]:}\begin{aligned}
& 6^{0} \text {. Le produit élémentaire de }[\mathrm{P}] \text { par }[\mathrm{Q}] \text { est une nouvelle suite- } \\
& \text { léfinie par les égalités }
\end{aligned}ééééé
{:(4)R_(n)=sum_(i=0)^(oo)Q_(i)[sum_(j=0)^(n)((i)/(j))P_(n-j)^((i-j))]","quad n=0","1","2","dots:}\begin{equation*}
\mathrm{R}_{n}=\sum_{i=0}^{\infty} Q_{i}\left[\sum_{j=0}^{n}\binom{i}{j} P_{n-j}^{(i-j)}\right], \quad n=0,1,2, \ldots \tag{4}
\end{equation*}
It is easily verified that the second term contains a finite number of terms. For sequences of negative or zero order, we can write the condensed formulas
{:(5)R_(n)=sum_(i=0)^(n)Q[sum_(j=0)^(i)((i)/(j))P_(n rarr j)^((i-j))]","quad n=0","1","2","dots:}\begin{equation*}
\mathrm{R}_{n}=\sum_{i=0}^{n} \mathrm{Q}\left[\sum_{j=0}^{i}\binom{i}{j} \mathrm{P}_{n \rightarrow j}^{(i-j)}\right], \quad n=0,1,2, \ldots \tag{5}
\end{equation*}
We write
[Q]*[P]=[R].[\mathrm{Q}] \cdot[\mathrm{P}]=[\mathrm{R}] .
The definitions4^(0),5^(0)4^{0}, 5^{0}allow us to find the difference between two sequences.
The equality and addition of sequences enjoys all the properties of ordinary equality and addition.
Elementary multiplication is associative and distributive with respect to addition, but it is not generally commutative. (If...)
[Q]*[P]=[P]*[Q][\mathrm{Q}] \cdot[\mathrm{P}]=[\mathrm{P}] \cdot[\mathrm{Q}]
we say that the sequels[P],[Q][\mathrm{P}],[\mathrm{Q}]are interchangeable. The unit sequence is interchangeable with any sequence.
Multiplication by a non-zero number does not change the order, class, normality, or permutability of a sequence.
The order of a sum is at most equal to the highest order of the sums added, and its class is at least equal to the lowest class of the
sums added. In the formula
sums added. In the formula
[R]=[P]+[Q][\mathrm{R}]=[\mathrm{P}]+[\mathrm{Q}]
we have indeed
r(n) <= max[p(n),q(n)]r(n) \leq \max [p(n), q(n)]
Whatever happens next[P][\mathrm{P}]we have
[P]+[0]=[P][\mathrm{P}]+[0]=[\mathrm{P}]
The order of a product is at most equal to the sum of the orders of its factors, and its class is at least equal to the class of the sequence being multiplied. Indeed, we have
r(n) <= max[q(0)+p_(1)(n),q(1)+p_(1)(n)-1,q(2)+p_(1)(n)-2,dots q(p_(1)(n))]r(n) \leq \max \left[q(0)+p_{1}(n), q(1)+p_{1}(n)-1, q(2)+p_{1}(n)-2, \ldots q\left(p_{1}(n)\right)\right]Or
r(n) <= max[q(0)+p_(1)(n),q(1)+p_(1)(n)-1,q(2)+p_(1)(n)-2,dots q(p_(1)(n))]r(n) \leq \max \left[q(0)+p_{1}(n), q(1)+p_{1}(n)-1, q(2)+p_{1}(n)-2, \ldots q\left(p_{1}(n)\right)\right]Or
p_(1)(n)=max[p(n),p(n-1)+1,p(n-2)+2,dots,p(0)+n]p_{1}(n)=\max [p(n), p(n-1)+1, p(n-2)+2, \ldots, p(0)+n]
If one of the factors is [0], then the product is equal to [0].
For normal sequences of class 0, the converse is true, but in the general case, the product of two non-zero sequences can be zero; for example, for the sequences
For normal sequences of class 0, the converse is true, but in the general case, the product of two non-zero sequences can be zero; for example, for the sequences
{:[[P],x","0","0","dots0","dots],[[Q],1","-x","(x^(2))/(2);dots((-1)^(n)x^(n))/(n!)","dots]:}\begin{array}{ll}
{[\mathrm{P}]} & x, 0,0, \ldots 0, \ldots \\
{[\mathrm{Q}]} & 1,-x, \frac{x^{2}}{2} ; \ldots \frac{(-1)^{n} x^{n}}{n!}, \ldots
\end{array}
we have
[Q]*[P]=[0][Q] \cdot[P]=[0]
Knowing the product of two sequences, we can calculate any positive integer power. We will denote by[P]^(m)[\mathrm{P}]^{m}, Or[mP][\mathrm{mP}]therem^("èm ")m^{\text {èm }}è. power of the suite[P][\mathrm{P}]We posit by definition.
[P_(1)^(0)=1_(0)P]=[1].\left[P_{1}^{0}=1{ }_{0} P\right]=[1] .
We then have form,km, kpositive integers or zero
[P]^(m)*[P]^(k)=[P]^(m+k);{[P]^(m)}^(k)=[P]^(mk).[\mathrm{P}]^{m} \cdot[\mathrm{P}]^{k}=[\mathrm{P}]^{m+k} ;\left\{[\mathrm{P}]^{m}\right\}^{k}=[\mathrm{P}]^{m k} .
We still need to say a few more things about the elementary division of sequences.
We say that a soot[P][\mathrm{P}]is divisible on the left thereafter[Q][Q]if there is a sequel[R][R]unique and well-defined such as
(6)
(6)
Equation (6) is not always possible in [R] and if it is possible the solution is not always unique.
If[R][\mathrm{R}]is the sequence [1] we say that [P] has a left inverse. Let us denote by[P]_(g)^(-1)[\mathrm{P}]_{g}^{-1}This is the opposite. We have
[P]*[P]_(g)^(-1)=[I][\mathrm{P}] \cdot[\mathrm{P}]_{\mathrm{g}}^{-1}=[\mathrm{I}]
from which we deduce that the following[P][\mathrm{P}]is divisible from the left by any sequence that has a left inverse.
Similarly, we define right division and right division.[Pi_(d)^(-1):}\left[\mathrm{Pi}_{d}^{-1}\right.So that[P]_(d)^(-1)[\mathrm{P}]_{d}^{-1}exists, the sequel must follow[P][\mathrm{P}]either of class 0.
If a sequel[P][\mathrm{P}]has a left inverse and a right inverse and if
[P]_(g)^(-1)=[P]_(d)^(-1)[\mathrm{P}]_{g}^{-1}=[\mathrm{P}]_{d}^{-1}
We say that it is reversible. We then designate by[P]^(-1)[\mathrm{P}]^{-1}the inverse of [P]. All integer powers of such a sequence are determined.
3. We will point out some properties of normal sequences.
3. We will point out some properties of normal sequences.
The product of two normal sequences is a normal sequence of class equal to the sum of the classes of the factors
We have been posing
[Q]*[P]=[R][\mathrm{Q}] \cdot[\mathrm{P}]=[\mathrm{R}]
{:[sum_(j=0)^(i)((i)/(j))R_(k+k^(')+j)^((i))={sum_(j=0)^(i)((i)/(j))Q_(k^(')+j)^((j)\\{sum_(j=0)^(k^(')+i)((k^(')+i)/(j))P_(k+i)^((j))}:}],[i=0","1","2","dots]:}\begin{aligned}
\sum_{j=0}^{i}\binom{i}{j} \mathrm{R}_{k+k^{\prime}+j}^{(i)}= & \left\{\sum_{j=0}^{i}\binom{i}{j} Q_{k^{\prime}+j}^{(j} \backslash\left\{\sum_{j=0}^{k^{\prime}+i}\binom{k^{\prime}+i}{j} \mathrm{P}_{k+i}^{(j)}\right\}\right. \\
& i=0,1,2, \ldots
\end{aligned}
k,k^(')\boldsymbol{k}, \boldsymbol{k}^{\prime}being the classes of[P][\mathrm{P}]And[Q][\mathrm{Q}]It
follows that the product of two normal sequences is always different from [0].
follows that the product of two normal sequences is always different from [0].
Every normal sequence of class 0 is invertible and its inverse is also a normal sequence of class 0.
Let [__1 P] be the inverse sequence; we have
{:[|sum_(j=0)^(i)((i)/(j))-1P_(j)^((j))|{sum_(j=0)^(i)((i)/(j))P_(j)^((j))}=1],[i=0","1","2","dots]:}\begin{gathered}
\left|\sum_{j=0}^{i}\binom{i}{j}-1 P_{j}^{(j)}\right|\left\{\sum_{j=0}^{i}\binom{i}{j} P_{j}^{(j)}\right\}=1 \\
i=0,1,2, \ldots
\end{gathered}
The normal sequences of class zero form a group. This group is not permutable, but it contains permutable subgroups.
We already know any integer power of a normal sequence of class 0.
Let's ask
U_(i)^('//))=sum_(s=0)^(j)((i)/(s))P_(s+i)^((s))\mathrm{U}_{i}^{\prime /)}=\sum_{s=0}^{j}\binom{i}{s} \mathrm{P}_{s+i}^{(s)}
and let us introduce the following notation
[a_(1),a_(2),dotsa_(k)]=|[1,a_(1),a_(1)^(2),dots,a_(1)^(k-2)],[1,a_(2),a_(2)^(2),dots,a_(2)^(k-2)],[-,0,0,cdots,a_(2)^(m)],[1,a_(k),a_(k)^(2),dots,a_(k)^(k-2)],[a_(k)^(m),dots,a_(k)^(m)]|:|[1,a_(1),a_(1)^(2),dots,a_(1)^(k-1)],[1,a_(2),a_(2)^(2),dots,a_(2)^(k-1)],[1,-,,,.],[1,a_(k),a_(k)^(2),dots,a_(k)^(k-1)]|\left[a_{1}, a_{2}, \ldots a_{k}\right]=\left|\begin{array}{ccccc}
1 & a_{1} & a_{1}^{2} & \ldots & a_{1}^{k-2} \\
1 & a_{2} & a_{2}^{2} & \ldots & a_{2}^{k-2} \\
- & 0 & 0 & \cdots & a_{2}^{m} \\
1 & a_{k} & a_{k}^{2} & \ldots & a_{k}^{k-2} \\
a_{k}^{m} & \ldots & a_{k}^{m}
\end{array}\right|:\left|\begin{array}{ccccc}
1 & a_{1} & a_{1}^{2} & \ldots & a_{1}^{k-1} \\
1 & a_{2} & a_{2}^{2} & \ldots & a_{2}^{k-1} \\
1 & - & & & . \\
1 & a_{k} & a_{k}^{2} & \ldots & a_{k}^{k-1}
\end{array}\right|
Let us now consider the following[mP][m \mathrm{P}]defined by the relationships
{:[_(m)P_(n)=sum_(n_(1),n_(2),dots,n_(k))U_(n_(1))^((0))*U_(n_(2))^((n_(1)))*U_(n_(3))^((n_(1)+n_(2)))dotsU_(n_(k))^((n_(1)+n_(2)+cdots+n_(k-1)))],[*[U_(0)^((0)),U_(0)^((n_(1))),U_(0)^((n_(1)+n_(2))),dotsU_(0)^((n_(1)+n_(2)+cdots+n_(k)))]],[n=0","1","2","dots]:}\begin{gathered}
{ }_{m} \mathrm{P}_{n}=\sum_{n_{1}, n_{2}, \ldots, n_{k}} \mathrm{U}_{n_{1}}^{(0)} \cdot \mathrm{U}_{n_{2}}^{\left(n_{1}\right)} \cdot \mathrm{U}_{n_{3}}^{\left(n_{1}+n_{2}\right)} \ldots \mathrm{U}_{n_{k}}^{\left(n_{1}+n_{2}+\cdots+n_{k-1}\right)} \\
\cdot\left[\mathrm{U}_{0}^{(0)}, \mathrm{U}_{0}^{\left(n_{1}\right)}, \mathrm{U}_{0}^{\left(n_{1}+n_{2}\right)}, \ldots \mathrm{U}_{0}^{\left(n_{1}+n_{2}+\cdots+n_{k}\right)}\right] \\
n=0,1,2, \ldots
\end{gathered}
the summation being extended to positive values ​​ofn_(1),n_(2),dotsn_(k)n_{1}, n_{2}, \ldots n_{k}verifying equality
n_(1)+n_(2)+cdots+n_(k)=nn_{1}+n_{2}+\cdots+n_{k}=n
k\boldsymbol{k}taking all possible values.
It can be shown that ifmmis whole
It can be shown that ifmmis whole
{:(7)[_(m)P]=[P]^(m).:}\begin{equation*}
\left[{ }_{m} \mathrm{P}\right]=[\mathrm{P}]^{m} . \tag{7}
\end{equation*}
It is then demonstrated that
{:[[m^(P)]*[m^(')P]=[m^(')P]*[m^(P)P]=[m+m^(')P]],[[m^(')[mP]]=[m^(')[m^(')P]]=[mm^(')P]]:}\begin{gathered}
{\left[m^{\mathrm{P}}\right] \cdot\left[m^{\prime} \mathrm{P}\right]=\left[m^{\prime} \mathrm{P}\right] \cdot\left[m^{\mathrm{P}} \mathrm{P}\right]=\left[m+m^{\prime} \mathrm{P}\right]} \\
{\left[m^{\prime}[m \mathrm{P}]\right]=\left[m^{\prime}\left[m^{\prime} \mathrm{P}\right]\right]=\left[m m^{\prime} \mathrm{P}\right]}
\end{gathered}
we can therefore keep equality (7) as defining any power of the sequence[P][\mathrm{P}], normal and class 0.
We have
sum_(j=0)^(i)((i)/(j))_(m)P_(j)^((j))=(sum_(j=0)^(i)((i)/(j))P_(j)^((j)))^(m)\sum_{j=0}^{i}\binom{i}{j}_{m} \mathrm{P}_{j}^{(j)}=\left(\sum_{j=0}^{i}\binom{i}{j} \mathrm{P}_{j}^{(j)}\right)^{m}
For example, for the normal binomial sequence of class 0,
we have
we have
{:[P_(0)","P_(1)","0","0","dots0","dots],[_(m)P_(n)=P^(n)*[U_(0)^((0)),U_(0)^((1)),U_(0)^((2)),dotsU_(0)^((n))]]:}\begin{gathered}
\mathrm{P}_{0}, \mathrm{P}_{1}, 0,0, \ldots 0, \ldots \\
{ }_{m} \mathrm{P}_{n}=\mathrm{P}^{n} \cdot\left[\mathrm{U}_{0}^{(0)}, \mathrm{U}_{0}^{(1)}, \mathrm{U}_{0}^{(2)}, \ldots \mathrm{U}_{0}^{(n)}\right]
\end{gathered}
The number[U_(0)^((0)),U_(0)^((1)),dotsU_(0)^((n))]\left[\mathrm{U}_{0}^{(0)}, \mathrm{U}_{0}^{(1)}, \ldots \mathrm{U}_{0}^{(n)}\right]generalize the number((m)/(n))\binom{m}{n}and is reduced to the latter forP^(')_(1)=0\mathrm{P}^{\prime}{ }_{1}=0.
IfP^(')_(1)!=0\mathrm{P}^{\prime}{ }_{1} \neq 0the series
sum_(n=1)^(oo)[U_(0)^((0)),U_(0)^((1)),dotsU_(0)^((n))]\sum_{n=1}^{\infty}\left[\mathrm{U}_{0}^{(0)}, \mathrm{U}_{0}^{(1)}, \ldots \mathrm{U}_{0}^{(n)}\right]
converges absolutely regardless ofmmIt can indeed be easily demonstrated
that
that
|[U_(0)^((0)),U_(0)^((1)),dotsU_(0)^((n))]| < (((n)/(n^('))))/(n!(P_(1)^('))^(n))sum_(i=0)^(n)|U_(0)^((i))|^(m)\left|\left[\mathrm{U}_{0}^{(0)}, \mathrm{U}_{0}^{(1)}, \ldots \mathrm{U}_{0}^{(n)}\right]\right|<\frac{\binom{n}{n^{\prime}}}{n!\left(\mathrm{P}_{1}^{\prime}\right)^{n}} \sum_{i=0}^{n}\left|\mathrm{U}_{0}^{(i)}\right|^{m}
nnbeing equal to(n)/(2)\frac{n}{2}Or(n-1)/(2)\frac{n-1}{2}following thatnnis even or odd.
4. On some particular sequences. Let us designate by[a],[b],dots[a],[b], \ldotsthe: sequences of numbers, therefore the sequence (1) such thatp(n) <= 0n=0,1,2,dotsp(n) \leq 0 n=0,1,2, \ldotsA sequence of numbers is always normal. Its order is equal to its class with the sign reversed.
4. On some particular sequences. Let us designate by[a],[b],dots[a],[b], \ldotsthe: sequences of numbers, therefore the sequence (1) such thatp(n) <= 0n=0,1,2,dotsp(n) \leq 0 n=0,1,2, \ldotsA sequence of numbers is always normal. Its order is equal to its class with the sign reversed.
Number sequences of class 0 form a permutable subgroup of the group of normal sequences of class 0.
A normal sequence of class 0 that is permutable with a sequence of numbers of class 0 is not necessarily a sequence of numbers. If the sequence of numbers has all its elements non-zero, then any sequence permutable with it is a sequence of numbers.
Let us consider sequences of the form
{:(8)P_(n)=a_(n)(x^(n))/(n!)","quad n=0","1","2","dots:}\begin{equation*}
\mathrm{P}_{n}=a_{n} \frac{x^{n}}{n!}, \quad n=0,1,2, \ldots \tag{8}
\end{equation*}
These sequences are characterized by the sequence of numbers [aaWe will refer to them as [P;[a]\mathrm{P} ;[a]For such a sequence to be normal, it must belong to class 0. The conditions for normality are then
{:[sum_(j=0)^(i)((i)/(j))a_(j)!=0],[i=0","1","2","dots]:}\begin{gathered}
\sum_{j=0}^{i}\binom{i}{j} a_{j} \neq 0 \\
i=0,1,2, \ldots
\end{gathered}
Normal sequences of the form (8) form a permutable subgroup of the group of normal sequences of class 0.
The reverse sequence
[P;[a]]^(-1)=[_(-1)P;[_(-1)a]][\mathrm{P} ;[a]]^{-1}=\left[{ }_{-1} \mathrm{P} ;\left[{ }_{-1} a\right]\right]
is determined by the equations
{sum_(i=0)^(n)((n)/(i))-1a_(i)}{sum_(i=0)^(n)((n)/(i))a_(i)}=1\left\{\sum_{i=0}^{n}\binom{n}{i}-1 a_{i}\right\}\left\{\sum_{i=0}^{n}\binom{n}{i} a_{i}\right\}=1
The product
n=0l,2,dotsn=0 l, 2, \ldots
[Q;[b]]*[P;[a]=[R;[c]][Q ;[b]] \cdot[\mathrm{P} ;[a]=[\mathrm{R} ;[c]]
is determined by the equations
{:(9)sum_(i=0)^(n)((n)/(i))c_(i)=(sum_(i=0)^(n)((n)/(i))a_(i))(sum_(i=0)^(n)((n)/(i))b_(i)).:}\begin{equation*}
\sum_{i=0}^{n}\binom{n}{i} c_{i}=\left(\sum_{i=0}^{n}\binom{n}{i} a_{i}\right)\left(\sum_{i=0}^{n}\binom{n}{i} b_{i}\right) . \tag{9}
\end{equation*}
The power[P;[a]]^(m)=[_(m)P;[_(m)a]∣:}[\mathrm{P} ;[a]]^{m}=\left[{ }_{m} \mathrm{P} ;\left[{ }_{m} a\right] \mid\right.is given by
{:[sum_(i=0)^(n)((n)/(i))_(m)a_(i)=(sum_(i=0)^(n)((n)/(i))a)^(m)],[n=0","1","2","dots]:}\begin{gathered}
\sum_{i=0}^{n}\binom{n}{i}_{m} a_{i}=\left(\sum_{i=0}^{n}\binom{n}{i} a\right)^{m} \\
n=0,1,2, \ldots
\end{gathered}
From equations (9) we easily derive
c_(n)=sum_(i=0)^(n)((n)/(i))b_(i)(sum_(r=0)^(i)((i)/(r))a_(n-i+r))=sum_(i=0)^(n)((n)/(i))a_(i)(sum_(r=0)^(i)((i)/(r))b_(n-i+))c_{n}=\sum_{i=0}^{n}\binom{n}{i} b_{i}\left(\sum_{r=0}^{i}\binom{i}{r} a_{n-i+r}\right)=\sum_{i=0}^{n}\binom{n}{i} a_{i}\left(\sum_{r=0}^{i}\binom{i}{r} b_{n-i+}\right)
If the series
suma_(n)z^(n),quad sumb_(n)z^(n)\sum a_{n} z^{n}, \quad \sum b_{n} z^{n}
converge inside circles of radius respectively equal toR_(a),\mathrm{R}_{a},.R_(b)\mathrm{R}_{b}, the series
certainly converges inside the circle of radius
certainly converges inside the circle of radius
R_(c)=(1)/((1+(1)/(R_(a)))(1+(1)/(R_(b)))-1)\mathrm{R}_{c}=\frac{1}{\left(1+\frac{1}{\mathrm{R}_{a}}\right)\left(1+\frac{1}{\mathrm{R}_{b}}\right)-1}
but it may eventually converge outside this circle.
5. The continuation[A][\mathrm{A}]is a harmonic sequence if
5. The continuation[A][\mathrm{A}]is a harmonic sequence if
A_(0)=c^(te),A_(n)^(')=A_(n-1),quad n=1,2,dots\mathrm{A}_{0}=c^{t e}, \mathrm{~A}_{n}^{\prime}=\mathrm{A}_{n-1}, \quad n=1,2, \ldots
Such a sequence is characterized by a sequence of numbers[alpha][\alpha]and we have
A_(n)=(alpha_(0))/(n!)x^(n)+(alpha_(1))/((n-1)!)x^(n-1)+cdots+alpha_(n)\mathrm{A}_{n}=\frac{\alpha_{0}}{n!} x^{n}+\frac{\alpha_{1}}{(n-1)!} x^{n-1}+\cdots+\alpha_{n}
Every harmonic sequence is normal and of class 0. Let us consider the more general sequences of the form
P_(n)=a^(n)A_(n),quad n=0,1,2,dots\mathrm{P}_{n}=a^{n} \mathrm{~A}_{n}, \quad n=0,1,2, \ldots
[A] being a harmonic sequence. Let us denote these sequences by[P;[alpha],a][P ;[\alpha], a]Sequences of this form are normal and of class 0 ifa!=-1a \neq-1They form a group. The reverse sequence,[-1P;[-1alpha],-1a][-1 \mathrm{P} ;[-1 \alpha],-1 a]is such that:
-1a=-(a)/(1+a)-1 a=-\frac{a}{1+a}
and the rest[-1alpha][-1 \alpha]is such that the following
-1alpha_(0),--1x_(1),-1x_(2),dots(-1)^(n)_(-1)alpha_(n),dots-1 \alpha_{0},--1 x_{1},-1 x_{2}, \ldots(-1)^{n}{ }_{-1} \alpha_{n}, \ldots
is the opposite of[alpha][\alpha]
The product
[R;[gamma],c]=[Q;[beta],b]*[P;[alpha],a][\mathrm{R} ;[\gamma], c]=[\mathrm{Q} ;[\beta], b] \cdot[\mathrm{P} ;[\alpha], a]
is obtained using the formulas
{:[c=a+b+ab],[c^(n)gamma_(n)=sum_(i=0)^(n)b^(i)(a+1)^(i)a^(n-i)alpha_(n-i)beta_(i)],[n=0","1","2","dots]:}\begin{gathered}
c=a+b+a b \\
c^{n} \gamma_{n}=\sum_{i=0}^{n} b^{i}(a+1)^{i} a^{n-i} \alpha_{n-i} \beta_{i} \\
n=0,1,2, \ldots
\end{gathered}
Regarding power,
we generally have
we generally have
{:[[_(m)P;[_(m)alpha],_(m)a]=[P;[alpha]","a]^(m)],[_(m)a=(a+1)^(n)-1]:}\begin{gathered}
{\left[{ }_{m} \mathrm{P} ;\left[{ }_{m} \alpha\right],{ }_{m} a\right]=[\mathrm{P} ;[\alpha], a]^{m}} \\
{ }_{m} a=(a+1)^{n}-1
\end{gathered}
We see that this group is not permutable.
6. Transformation of a sequence with respect to a fundamental sequence. A sequence is said to be fundamental if it is normal and of class 1. Let
[G] be a fundamental sequence.
6. Transformation of a sequence with respect to a fundamental sequence. A sequence is said to be fundamental if it is normal and of class 1. Let
[G] be a fundamental sequence.
G_(0),G_(1),dotsG_(n),dots\mathrm{G}_{0}, \mathrm{G}_{1}, \ldots \mathrm{G}_{n}, \ldots
OrG_(0)=0,G_(1)=c^("te ")!=0\mathrm{G}_{0}=0, \mathrm{G}_{1}=c^{\text {te }} \neq 0We consider positive integer powers.{:_([m)G]\left.{ }_{[m} \mathrm{G}\right]of[G][\mathrm{G}]The sequel[mG][m \mathrm{G}]is normal and classmmThe sequence [G] will also be designated by[iG][\mathrm{i} \mathrm{G}]The sequel[_(0)G]\left[{ }_{0} \mathrm{G}\right]is the unit sequence.
We call the transformed sequence[P][P]in relation to the fundamental sequence[G][G]the new sequel
{:(Q])Q_(0)","Q_(1)","dotsQ_(n)","dots:}\begin{equation*}
Q_{0}, Q_{1}, \ldots Q_{n}, \ldots \tag{Q]}
\end{equation*}
obtained using the equations
{:[(10)P=sum_(i=0)^(n)Q_(i.i)G_(n)],[n=0","1","2","dots]:}\begin{gather*}
\mathrm{P}=\sum_{i=0}^{n} \mathrm{Q}_{i . i} \mathrm{G}_{n} \tag{10}\\
n=0,1,2, \ldots
\end{gather*}
The sequence [Q] is completely determined by these equations since
{:[_(n)G_(n)=prod_(i=0)^(n-1)(sum_(s=0)^(i)((i)/(s))G_(s+1)^((s)))!=0],[n=1","2","3","dots]:}\begin{gathered}
{ }_{n} \mathrm{G}_{n}=\prod_{i=0}^{n-1}\left(\sum_{s=0}^{i}\binom{i}{s} \mathrm{G}_{s+1}^{(s)}\right) \neq 0 \\
n=1,2,3, \ldots
\end{gathered}
precisely express the normality of the fundamental sequence. We will consider the transformed sequence as being taken with respect to the sequence[G][G]and we designate herpar[Q∣G]\operatorname{par}[Q \mid G].
For sequences considered in relation to a fundamental sequence, we can also establish an algebra. This algebra will be characterized by multiplication.
The product
The product
[Q∣G]*[P∣G]==[R∣G][Q \mid G] \cdot[P \mid G]==[R \mid G]
is defined by the equalities
{:[(11)sum_(i=0)^(n)R_(i,i)G_(n)=sum_(i=0)^(oo)(sum_(r=0)^(i)Q_(r,r)G_(i))[sum_(j=0)^(n)((i)/(j))(sum_(s=0)^(n-j)P_(s,s)G_(n-j))^((i-j))]],[n=0","1","2","dots]:}\begin{gather*}
\sum_{i=0}^{n} \mathrm{R}_{i, i} \mathrm{G}_{n}=\sum_{i=0}^{\infty}\left(\sum_{r=0}^{i} \mathrm{Q}_{r, r} \mathrm{G}_{i}\right)\left[\sum_{j=0}^{n}\binom{i}{j}\left(\sum_{s=0}^{n-j} \mathrm{P}_{s, s} \mathrm{G}_{n-j}\right)^{(i-j)}\right] \tag{11}\\
n=0,1,2, \ldots
\end{gather*}
In this way, the product of the transforms of two sequences is equal^(-){ }^{-}to the transformed form of their elementary product.
We see that elementary multiplication corresponds to sequences taken with respect to the tondamental sequence.
0,1,0,0,dots0,dots0,1,0,0, \ldots 0, \ldots
The fundamental sequence[G][G]was taken herself in relation to this sequel.
Let[G],[H][G],[H]two fundamental sequences. In general, the transform[P∣H][P \mid H]compared to the following[G][G]is a sequel[Q∣G][Q \mid G]; defined by the equalities.
{:[sum_(i=0)^(n)P_(i.i)H_(n)=sum_(i=0)^(n)Q_(i.i)G_(n)],[n=0","1","2","dots]:}\begin{gathered}
\sum_{i=0}^{n} \mathrm{P}_{i . i} \mathrm{H}_{n}=\sum_{i=0}^{n} Q_{i . i} \mathrm{G}_{n} \\
n=0,1,2, \ldots
\end{gathered}
Let us designatebar(par)[G∣G]\overline{\operatorname{par}}[G \mid G]the transform of the fundamental sequence
(12)
(12)
0,1,0,0,dots0,dots0,1,0,0, \ldots 0, \ldots
compared to[G][G]This sequence is given by the equalities
sum_(i=0)^(n) bar(G)_(i,i)G_(n)={[1,n=1],[0,n!=1]:}\sum_{i=0}^{n} \overline{\mathrm{G}}_{i, i} \mathrm{G}_{n}= \begin{cases}1 & n=1 \\ 0 & n \neq 1\end{cases}
The sequelbar([G)∣G]\overline{[\mathrm{G}} \mid \mathrm{G}]is the inverse of[A][A]We can calculate them^(éme)m^{e ́ m e}positive whole power[m bar(G)∣G][m \bar{G} \mid \mathrm{G}]of[ bar(G)∣G][\overline{\mathrm{G}} \mid \mathrm{G}]We find that this power is given by the equations
{:[m bar(G)_(0)=m bar(G)_(1)=cdots=_(m) bar(G)_(m-1)=0],[sum_(i=m)^(n) bar(G)_(i,i)G_(n)={[1,n=m],[0,n > m]:}]:}\begin{gathered}
m \overline{\mathrm{G}}_{0}=m \overline{\mathrm{G}}_{1}=\cdots={ }_{m} \overline{\mathrm{G}}_{m-1}=0 \\
\sum_{i=m}^{n} \overline{\mathrm{G}}_{i, i} \mathrm{G}_{n}= \begin{cases}1 & n=m \\
0 & n>m\end{cases}
\end{gathered}
We also
{:[_(m)G_(0)=_(m)G_(1)=cdots=_(m)G_(m-1)=0],[sum_(i=m)^(n)_(m)G_(i,i) bar(G)_(n)={[1,n=m],[0,n > m]:}]:}\begin{gathered}
{ }_{m} \mathrm{G}_{0}={ }_{m} \mathrm{G}_{1}=\cdots={ }_{m} \mathrm{G}_{m-1}=0 \\
\sum_{i=m}^{n}{ }_{m} \mathrm{G}_{i, i} \overline{\mathrm{G}}_{n}= \begin{cases}1 & n=m \\
0 & n>m\end{cases}
\end{gathered}
Let us also mention the formulas
{:[sum_(i=0)^(n)m-i bar(G)_(m*m-n+i)G_(m)=0],[n=1","2","dots]:}\begin{gathered}
\sum_{i=0}^{n} m-i \overline{\mathrm{G}}_{m \cdot m-n+i} \mathrm{G}_{m}=0 \\
n=1,2, \ldots
\end{gathered}
If we consider the inverse sequence with respect to sequence (12), it is also a fundamental sequence. Using the inverse sequence, your formulas (10) and (11) can be written
{:[Q_(n)=sum_(i=0)^(n)P_(i*i) bar(G)_(n)],[n=0","1","2","dots],[R_(n)=sum_(i=0)^(n)i bar(G)_(n){sum_(alpha=0)^(oo)(sum_(r=0)^(alpha)Q_(r*r)G_(a))[sum_(i=0)^(i)((alpha )/(j))(sum_(s=0)^(i-1)P_(s*s)G_(i*j))^((alpha-j))]:}],[n=0","1","2","dots]:}\begin{gathered}
\mathrm{Q}_{n}=\sum_{i=0}^{n} \mathrm{P}_{i \cdot i} \overline{\mathrm{G}}_{n} \\
n=0,1,2, \ldots \\
\mathrm{R}_{n}=\sum_{i=0}^{n} i \overline{\mathrm{G}}_{n}\left\{\sum_{\alpha=0}^{\infty}\left(\sum_{r=0}^{\alpha} \mathrm{Q}_{r \cdot r} \mathrm{G}_{a}\right)\left[\sum_{i=0}^{i}\binom{\alpha}{j}\left(\sum_{s=0}^{i-1} \mathrm{P}_{s \cdot s} \mathrm{G}_{i \cdot j}\right)^{(\alpha-j)}\right]\right. \\
n=0,1,2, \ldots
\end{gathered}
Suppose in particular that these are sequences of negative or zero order; we can write
R=sum_(i=0)^(n)Q_(i)(sum_(i=0)^(n-j)i+iG_(n)*B_(i+1,i)),quad n=0,1,2,dots\mathrm{R}=\sum_{i=0}^{n} \mathrm{Q}_{i}\left(\sum_{i=0}^{n-j} i+i \mathrm{G}_{n} \cdot \mathrm{~B}_{i+1, i}\right), \quad n=0,1,2, \ldots
Or -
{:[B_(i,j)=sum_(r=0)^(i-j)G_(r+j)","A_(i,r+j)],[A_(i,j)=sum_(r=0)^(j)((j)/(r))(sum_(s=0)^(i-r)P_(s,s)G_(i-r))^((j-r))]:}\begin{gathered}
\mathrm{B}_{i, j}=\sum_{r=0}^{i-j} \mathrm{G}_{r+j}, \mathrm{~A}_{i, r+j} \\
\mathrm{~A}_{i, j}=\sum_{r=0}^{j}\binom{j}{r}\left(\sum_{s=0}^{i-r} \mathrm{P}_{s, s} \mathrm{G}_{i-r}\right)^{(j-r)}
\end{gathered}
- The class and order of a sequence are invariant under a transformation. The characteristic index is also independent of a transformation. Permutability is a property that is invariant under a transformation.:.\thereforeto a transformation.
If the following[P∣G][\mathrm{P} \mid \mathrm{G}]is taken with respect to the fundamental sequence [G] we say that it is normal if it is of classkk, of order -kkand if
{:[sum_(j=0)^(k)_(j+m) bar(G)_(k+m)quadB_(j+m)","!=0],[m=0","1","2","dots]:}\begin{gathered}
\sum_{j=0}^{k}{ }_{j+m} \overline{\mathrm{G}}_{k+m} \quad \mathrm{~B}_{j+m}, \neq 0 \\
m=0,1,2, \ldots
\end{gathered}
The quantities on the left-hand side are numbers. Indeed, in this case we have
A_(i,j)=0quad" si "quad i < j+k\mathrm{A}_{i, j}=0 \quad \text { si } \quad i<j+k
SO
sum_(j=0)^(k)j+m bar(G)_(k+m)*B_(j+m,m)=k+m bar(G)_(k+m*m)G_(m)*A_(k+m;m)=\sum_{j=0}^{k} j+m \overline{\mathrm{G}}_{k+m} \cdot \mathrm{~B}_{j+m, m}=k+m \overline{\mathrm{G}}_{k+m \cdot m} \mathrm{G}_{m} \cdot \mathrm{~A}_{k+m ; m}=
=_(k+m) bar(G)_(k+m*m)G_(m)sum_(r=0)^(m)((m)/(r))(sum_(s=0)^(m-r)((m-r)/(s))P_(k+sk+s)^((s))G_(k+m-r)^((m-r-s)))={ }_{k+m} \overline{\mathrm{G}}_{k+m \cdot m} \mathrm{G}_{m} \sum_{r=0}^{m}\binom{m}{r}\left(\sum_{s=0}^{m-r}\binom{m-r}{s} \mathrm{P}_{k+s k+s}^{(s)} \mathrm{G}_{k+m-r}^{(m-r-s)}\right)
We can therefore see that a normal sequence transforms into a normal sequence.
It follows that the transform with respect to [G] of a normal sequence of class 0 has an inverse which is equal to the transform of the inverse.
The multiplication of sequences of numbers taken with respect to a fundamental sequence[G][G]is done according to the ordinary rule
c_(n)=sum_(i=0)^(n)a_(n-i)b_(n)c_{n}=\sum_{i=0}^{n} a_{n-i} b_{n}
Therefore:
Sequences of class numbers0,[a∣G]0,[a \mid G]form a permutable subgroup of the group of normal sequences of class zero.
Sequences of class numbers0,[a∣G]0,[a \mid G]form a permutable subgroup of the group of normal sequences of class zero.
Let [H] be a normal sequence of class 1. The sequences [P] of the form
{:[P_(n)=sum_(i=0)^(n)a_(i*i)H_(n)],[n=0","i","2","dots]:}\begin{aligned}
\mathrm{P}_{n} & =\sum_{i=0}^{n} a_{i \cdot i} \mathrm{H}_{n} \\
n & =0, \mathrm{i}, 2, \ldots
\end{aligned}
- a_(i)a_{i}being constants, they form a permutable group.
- We could study various other questions relating to sequences of polynomials. The study of sequences with particular properties leads to interesting identities, as Mr. Lagrange did for sequences of numbers.(^((1)):}\left({ }^{(1)}\right.We further note that Mr. Lagrange's conception is as follows: A series of powers is associated with a sequence of numbers [a].
a_(0)+a_(1)z+a_(2)z^(2)+cdots+a_(n)z^(n)+cdotsa_{0}+a_{1} z+a_{2} z^{2}+\cdots+a_{n} z^{n}+\cdots
(') loc. cit.
The reader will easily notice that we associate with the following[P][P]The functional operation:
P_(0)+P_(1)D+P_(2)D^(2)+cdots+P_(n)D^(n)+cdotsquadD=(d)/(dx)\mathrm{P}_{0}+\mathrm{P}_{1} \mathrm{D}+\mathrm{P}_{2} \mathrm{D}^{2}+\cdots+\mathrm{P}_{n} \mathrm{D}^{n}+\cdots \quad \mathrm{D}=\frac{d}{d x}
This new perspective has allowed us to generalize Mr. Lagrange's theory. We have already presented it in a previous paper where we gave some applications. In particular, we gave interesting functional properties for binomial sequences, which Mr. Lagrange also studied under the name of interpolation sequences.(^(1))\left({ }^{1}\right).
- (^(1){ }^{1}) See T. Popovicu "Asupra unor polinoame remarcabile". The note at the end of the memoir. (Autographed 1927).