Asupra polinoamelor cari formează un şir Appell (I)

Tiberiu Popoviciu
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1932 a -Popoviciu- Bull. Math. Soc. Roum. Sci. – Asupra polinoamelor cari formeaza un sir Appell (I)

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T. Popoviciu, Asupra polinoamelor cari formează un şir Appell (I), Bull. Mathematique de la Soc. Roumaine des Sciences, 33/34 (1932), pp. 11-21 (in Romanian)
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1932 a -Popoviciu- Bull. Math. Soc. Roum. Sci. - Asupra polinoamelor cari formeaza un sir Appell (I)

ASUPRA POLINOAMELOR CARI FORMEAZA UN ŞIR APPELL de TIBERIU POPOVICIU

  1. Še zice că un şir de polinoame în x x xxx :
(1) P 0 , P 1 , … … P n , … … (1) P 0 , P 1 , … … P n , … … {:(1)P_(0)","P_(1)","dots dotsP_(n)","dots dots:}\begin{equation*} P_{0}, P_{1}, \ldots \ldots P_{n}, \ldots \ldots \tag{1} \end{equation*}(1)P0,P1,……Pn,……
unde pentru simplificare se poate presupune:
P 0 = 1 , P 0 = 1 , P_(0)=1,P_{0}=1,P0=1,
fără a restrânge generalifatea, formează un şir Appell, dacă
(2) d P n d x = n P n − 1 (2) d P n d x = n P n − 1 {:(2)(dP_(n))/(dx)=nP_(n-1):}\begin{equation*} \frac{d P_{n}}{d x}=n P_{n-1} \tag{2} \end{equation*}(2)dPndx=nPn−1
E uşor de văzut că F n F n F_(n)F_{n}Fn este de forma
P n = x n + n a 1 x n − 1 + n ( n − 1 ) a 2 x n − 2 + … + n ! a n , P n = x n + n a 1 x n − 1 + n ( n − 1 ) a 2 x n − 2 + … + n ! a n , P_(n)=x^(n)+na_(1)x^(n-1)+n(n-1)a_(2)x^(n-2)+dots+n!a_(n),P_{n}=x^{n}+n a_{1} x^{n-1}+n(n-1) a_{2} x^{n-2}+\ldots+n!a_{n},Pn=xn+na1xn−1+n(n−1)a2xn−2+…+n!an,
unde a 1 , a 2 , … . n u a 1 , a 2 , … . n u a_(1),a_(2),dots.nua_{1}, a_{2}, \ldots . n ua1,a2,….nu depind de n n nnn. Şirul (1) depinde deci de succesiu= nca de numere
a ˙ 1 , a 2 , … … a m , … 1 ) a ˙ 1 , a 2 , … … a m , … 1 {:a^(˙)_(1),a_(2),dots dotsa_(m),dots^(1))\left.\dot{a}_{1}, a_{2}, \ldots \ldots a_{m}, \ldots{ }^{1}\right)a˙1,a2,……am,…1)
Polinoamele considerate intervin în multe chesliuni importante d e a = d e a = dea=\mathrm{d} e \mathrm{a}=dea= nalizǎ. Polinoamele lui Bernoulli şi Euler formează fiecare un şir Appell. Ele intervin in calculul difercnțelor finite. Polinoamele R v ( x , k ) R v ( x , k ) Rv(x,k)R v(x, k)Rv(x,k) studiate de DI. R. Racliş 2 2 ^(2){ }^{2}2 ) verifică şi cle conditia (2). Polinoamele lui Hermite, putin transformate,
2 n H n = e − x 2 d n d x n e x 2 2 n H n = e − x 2 d n d x n e x 2 2^(n)H_(n)=e^(-x^(2))(d^(n))/(dx^(n))e^(x^(2))2^{n} H_{n}=e^{-x^{2}} \frac{d^{n}}{d x^{n}} e^{x^{2}}2nHn=e−x2dndxnex2
cari se întâlnesc în teoria ecuaţiilor cu derivate partiale şi în multe alte chestiuni, formează şi ele un şir Appell.
2. In lucrarea de faţă ne propunem a găsi toate şirurile (1) cari sax tisfac conditia (2), astfel ca între trei polinoame consecutive P n , P n − 1 P n , P n − 1 P_(n),P_(n-1)P_{n}, P_{n-1}Pn,Pn−1, P n − 2 P n − 2 P_(n-2)P_{n-2}Pn−2 să avem o relaţie identică de forma:
(3)
A n P n + B n P n − 1 + C n P n − 2 = 0 , A n P n + B n P n − 1 + C n P n − 2 = 0 , A_(n)P_(n)+B_(n)P_(n-1)+C_(n)P_(n-2=0,)A_{n} P_{n}+B_{n} P_{n-1}+C_{n} P_{n-2=0,}AnPn+BnPn−1+CnPn−2=0,
unde A n , B n , C n A n , B n , C n A_(n),B_(n),C_(n)A_{n}, B_{n}, C_{n}An,Bn,Cn sunt polinoame în x x xxx de grade 0 , 1 , 2 0 , 1 , 2 0,1,20,1,20,1,2, şi egalitatea are loc oricare ar fi n n nnn. De exemplu polinoamele sus-amintite a lui Hermite safisfac o astfel de conditie.
3. Să presupunem întâi că A n A n A_(n)A_{n}An e nul oricare ar fì n n nnn. Relaţia (3) devine
(4) B n P n − 1 + C n P n − 2 = 0 (4) B n P n − 1 + C n P n − 2 = 0 {:(4)B_(n)P_(n-1)+C_(n)P_(n-2)=0:}\begin{equation*} B_{n} P_{n-1}+C_{n} P_{n-2}=0 \tag{4} \end{equation*}(4)BnPn−1+CnPn−2=0
Această relatie ne arată că P n − 1 , P n − 2 P n − 1 , P n − 2 P_(n-1),P_(n-2)P_{n-1}, P_{n-2}Pn−1,Pn−2, au cel puțin n − 3 n − 3 n-3n-3n−3 zerori comune şi cum P n − 2 P n − 2 P_(n-2)\mathrm{P}_{n-2}Pn−2 diferă numai printr'un factor constant de derivata lui P n − 1 P n − 1 P_(n-1)\mathrm{P}_{n-1}Pn−1, urmează că P n − 1 P n − 1 P_(n-1)\mathrm{P}_{n-1}Pn−1 este de forma:
P n − 1 = ( x + a ) n − 2 ( x + b n − 1 ) P n − 1 = ( x + a ) n − 2 x + b n − 1 P_(n-1)=(x+a)^(n-2)(x+b_(n-1))P_{n-1}=(x+a)^{n-2}\left(x+b_{n-1}\right)Pn−1=(x+a)n−2(x+bn−1)
Relaţia fundamentală (2) ne dă imediat
b n n = b n − 1 n − 1 − a n ( n − 1 ) = b n − 1 n − 1 − a ( 1 n − 1 − 1 n ) , b n n = b n − 1 n − 1 − a n ( n − 1 ) = b n − 1 n − 1 − a 1 n − 1 − 1 n , (b_(n))/(n)=(b_(n-1))/(n-1)-(a)/(n(n-1))=(b_(n-1))/(n-1)-a((1)/(n-1)-(1)/(n)),\frac{b_{n}}{n}=\frac{b_{n-1}}{n-1}-\frac{a}{n(n-1)}=\frac{b_{n-1}}{n-1}-a\left(\frac{1}{n-1}-\frac{1}{n}\right),bnn=bn−1n−1−an(n−1)=bn−1n−1−a(1n−1−1n),
din care se are
b n = n ( b 2 − a ) + 2 a 2 b n = n b 2 − a + 2 a 2 b_(n)=(n(b_(2)-a)+2a)/(2)b_{n}=\frac{n\left(b_{2}-a\right)+2 a}{2}bn=n(b2−a)+2a2
Se poate aşa dal scrie
P 0 = 1 , P 1 = x + a + b 2 , P 2 = ( x + a ) ( x + b ) (I) P n = ( x + a ) n − 1 ( x + n ( b − a ) + 2 a 2 ) P 0 = 1 , P 1 = x + a + b 2 , P 2 = ( x + a ) ( x + b ) (I) P n = ( x + a ) n − 1 x + n ( b − a ) + 2 a 2 {:[P_(0)=1","P_(1)=x+(a+b)/(2)","P_(2)=(x+a)(x+b)],[(I)P_(n)=(x+a)^(n-1)(x+(n(b-a)+2a)/(2))]:}\begin{gather*} \mathrm{P}_{0}=1, \mathrm{P}_{1}=x+\frac{a+b}{2}, \mathrm{P}_{2}=(x+a)(x+b) \\ \mathrm{P}_{n}=(x+a)^{n-1}\left(x+\frac{n(b-a)+2 a}{2}\right) \tag{I} \end{gather*}P0=1,P1=x+a+b2,P2=(x+a)(x+b)(I)Pn=(x+a)n−1(x+n(b−a)+2a2)
Se vede apoi că
B n = x + ( n − 2 ) ( b − a ) + 2 a 2 C n = ( x + a ) ( x + n ( b − a ) + 2 a 2 B n = x + ( n − 2 ) ( b − a ) + 2 a 2 C n = ( x + a ) x + n ( b − a ) + 2 a 2 {:[B_(n)=x+((n-2)(b-a)+2a)/(2)],[C_(n)=(x+a)(x+(n(b-a)+2a)/(2):}]:}\begin{aligned} & B_{n}=x+\frac{(n-2)(b-a)+2 a}{2} \\ & C_{n}=(x+a)\left(x+\frac{n(b-a)+2 a}{2}\right. \end{aligned}Bn=x+(n−2)(b−a)+2a2Cn=(x+a)(x+n(b−a)+2a2
Se ştie că un şir de polinoame Appel are o funcţie generatoare de forma.
F ( z ) e z x = ∑ n = 0 ∞ z n n ! P n , F ( z ) e z x = ∑ n = 0 ∞   z n n ! P n , F(z)e^(zx)=sum_(n=0)^(oo)(z^(n))/(n!)P_(n),F(z) e^{z x}=\sum_{n=0}^{\infty} \frac{z^{n}}{n!} P_{n},F(z)ezx=∑n=0∞znn!Pn,
F ( z ) F ( z ) F(z)F(z)F(z) find funcţie numai de z z zzz. Să căutăm funcţia generatoare a po= linoamelor (I).
∑ 0 ∞ z n n ! P n = ∑ 0 ∞ z n n ! ( x + a ) n − 1 ( x + a + n ( b − a ) 2 ) = ∑ 0 ∞ z n n ! ( x + a ) n + b − − a 2 ∑ n = 1 ∞ n z n n ! ( x + a ) n − 1 = e z ( x + a ) + b − a 2 z e z ( x + a ) = ( 1 + b − a 2 z ) e z a e z x ∑ 0 ∞   z n n ! P n = ∑ 0 ∞   z n n ! ( x + a ) n − 1 x + a + n ( b − a ) 2 = ∑ 0 ∞   z n n ! ( x + a ) n + b − − a 2 ∑ n = 1 ∞   n z n n ! ( x + a ) n − 1 = e z ( x + a ) + b − a 2 z e z ( x + a ) = 1 + b − a 2 z e z a e z x {:[sum_(0)^(oo)(z^(n))/(n!)P_(n)=sum_(0)^(oo)(z^(n))/(n!)(x+a)^(n-1)(x+a+(n(b-a))/(2))],[=sum_(0)^(oo)(z^(n))/(n!)(x+a)^(n)+(b--a)/(2)sum_(n=1)^(oo)n(z^(n))/(n!)(x+a)^(n-1)],[=e^(z(x+a))+(b-a)/(2)ze^(z(x+a))],[=(1+(b-a)/(2)z)e^(za)e^(zx)]:}\begin{aligned} \sum_{0}^{\infty} \frac{z^{n}}{n!} P_{n} & =\sum_{0}^{\infty} \frac{z^{n}}{n!}(x+a)^{n-1}\left(x+a+\frac{n(b-a)}{2}\right) \\ & =\sum_{0}^{\infty} \frac{z^{n}}{n!}(x+a)^{n}+\frac{b--a}{2} \sum_{n=1}^{\infty} n \frac{z^{n}}{n!}(x+a)^{n-1} \\ & =e^{z(x+a)}+\frac{b-a}{2} z e^{z(x+a)} \\ & =\left(1+\frac{b-a}{2} z\right) e^{z a} e^{z x} \end{aligned}∑0∞znn!Pn=∑0∞znn!(x+a)n−1(x+a+n(b−a)2)=∑0∞znn!(x+a)n+b−−a2∑n=1∞nznn!(x+a)n−1=ez(x+a)+b−a2zez(x+a)=(1+b−a2z)ezaezx
deci
F ( z ) = ( 1 + b − a 2 z ) e z a F ( z ) = 1 + b − a 2 z e z a F(z)=(1+(b-a)/(2)z)e^(za)F(z)=\left(1+\frac{b-a}{2} z\right) e^{z a}F(z)=(1+b−a2z)eza
  1. Să presupunem acum că A n A n A_(n)A_{n}An este diferit de zero oricare ar fi n = 2 , 3 , … n = 2 , 3 , … n=2,3,dotsn=2,3, \ldotsn=2,3,… In acest caz din cauza omogeneitătii formulei (3) putem presupune
    (5)
Punem acum
A n = 1 , n = 2 , 3 , … A n = 1 , n = 2 , 3 , … A_(n)=1,n=2,3,dotsA_{n}=1, n=2,3, \ldotsAn=1,n=2,3,…
B n = α n x + β n B n = α n x + β n B_(n)=alpha_(n)x+beta_(n)B_{n}=\alpha_{n} x+\beta_{n}Bn=αnx+βn
C n ′ = λ n x 2 + μ n x + v n C n ′ = λ n x 2 + μ n x + v n C_(n)^(')=lambda_(n)x^(2)+mu_(n)x+v_(n)C_{n}^{\prime}=\lambda_{n} x^{2}+\mu_{n} x+v_{n}Cn′=λnx2+μnx+vn
μ n , β n , λ n , μ n , v n μ n , β n , λ n , μ n , v n mu_(n),beta_(n),lambda_(n),mu_(n),v_(n)\mu_{n}, \beta_{n}, \lambda_{n}, \mu_{n}, v_{n}μn,βn,λn,μn,vn depinzând numai de n n nnn
Derivând relația (3), avem
(6) ( n + B n ′ ) P n − 1 + ( B n ( n − 1 ) + C n ) P n − 2 + C n ( n − 2 ) P n − 3 = 0 n + B n ′ P n − 1 + B n ( n − 1 ) + C n P n − 2 + C n ( n − 2 ) P n − 3 = 0 (n+B_(n)^('))P_(n-1)+(B_(n)(n-1)+C_(n))P_(n-2)+C_(n)(n-2)P_(n-3)=0\left(n+B_{n}^{\prime}\right) P_{n-1}+\left(B_{n}(n-1)+C_{n}\right) P_{n-2}+C_{n}(n-2) P_{n-3}=0(n+Bn′)Pn−1+(Bn(n−1)+Cn)Pn−2+Cn(n−2)Pn−3=0 dar avem şi
(7)
P n − 1 + B n − 1 P n − 2 + C n − 1 P n − 3 = 0 P n − 1 + B n − 1 P n − 2 + C n − 1 P n − 3 = 0 P_(n-1)+B_(n-1)P_(n-2)+C_(n-1)P_(n-3)=0P_{n-1}+B_{n-1} \mathrm{P}_{n-2}+C_{n-1} \mathrm{P}_{n-3}=0Pn−1+Bn−1Pn−2+Cn−1Pn−3=0
Dacă relatiile (6), (7) sunt distincte se poate deduce, prin eliminarea lui P n − 1 P n − 1 P_(n-1)P_{n-1}Pn−1 o relaţie de forma (4) și cădem peste cazul studiat mai înainte. Să presupunem deci că relaţiile (6), (7) sunt echivalente, trebue atunci ca:
(8) n + B n ′ = B ( n − 1 ) + C n ′ B n − 1 = C n ( n − 2 ) C n − 1 (8) n + B n ′ = B ( n − 1 ) + C n ′ B n − 1 = C n ( n − 2 ) C n − 1 {:(8)n+B_(n)^(')=(B(n-1)+C_(n)^('))/(B_(n-1))=(C_(n)(n-2))/(C_(n-1)):}\begin{equation*} n+B_{n}^{\prime}=\frac{B(n-1)+C_{n}^{\prime}}{B_{n-1}}=\frac{C_{n}(n-2)}{C_{n-1}} \tag{8} \end{equation*}(8)n+Bn′=B(n−1)+Cn′Bn−1=Cn(n−2)Cn−1
E clar că ambele relațiuni trebue să fie satisfăcute, căci dacă ar fi satisfăcută numai una am deduce o relaţie de forma
K . P n = 0 , K . P n = 0 , K.P_(n)=0,K . P_{n}=0,K.Pn=0,
care nu pot fi identic satisfăcute decât dacă K ≡ 0 K ≡ 0 K-=0K \equiv 0K≡0.
Relaţiile (8) sunt nişte identităti în x x xxx, trebue deci ca :
(9)
n + α n = α n ( n − 1 ) + 2 λ n α n − 1 = β n ( n − 1 ) + μ n β n − 1 = λ n λ n − 1 ( n − 2 ) = μ n μ n − 1 ( n − 2 ) = v n v n ( n − 2 ) n + α n = α n ( n − 1 ) + 2 λ n α n − 1 = β n ( n − 1 ) + μ n β n − 1 = λ n λ n − 1 ( n − 2 ) = μ n μ n − 1 ( n − 2 ) = v n v n ( n − 2 ) {:[n+alpha_(n)=(alpha^(n)(n-1)+2lambda_(n))/(alpha_(n-1))=(beta_(n)(n-1)+mu_(n))/(beta_(n-1))=(lambda_(n))/(lambda_(n-1))(n-2)=],[(mu_(n))/(mu_(n-1))(n-2)=(v_(n))/(v_(n))(n-2)]:}\begin{gathered} n+\alpha_{n}=\frac{\alpha^{n}(n-1)+2 \lambda_{n}}{\alpha_{n-1}}=\frac{\beta_{n}(n-1)+\mu_{n}}{\beta_{n-1}}=\frac{\lambda_{n}}{\lambda_{n-1}}(n-2)= \\ \frac{\mu_{n}}{\mu_{n-1}}(n-2)=\frac{v_{n}}{v_{n}}(n-2) \end{gathered}n+αn=αn(n−1)+2λnαn−1=βn(n−1)+μnβn−1=λnλn−1(n−2)=μnμn−1(n−2)=vnvn(n−2)
  • ABERIU POPOVICIU
    (10)
A n = 1 / сӑ : A n = 1 /  сӑ  : A_(n)=1//" сӑ ":A_{n}=1 / \text { сӑ }:An=1/ сӑ :
ceeace se obtine anulând coeficientul lui x n x n x^(n)x^{n}xn. Se deduce imediat egalitatea
n − 1 − λ n = λ n λ n − 1 ( n − 2 ) n − 1 − λ n = λ n λ n − 1 ( n − 2 ) n-1-lambda_(n)=(lambda_(n))/(lambda_(n-1))(n-2)n-1-\lambda_{n}=\frac{\lambda_{n}}{\lambda_{n-1}}(n-2)n−1−λn=λnλn−1(n−2)
sau
n − 1 λ n = n − 2 λ n − 1 + 1 n − 1 λ n = n − 2 λ n − 1 + 1 (n-1)/(lambda_(n))=(n-2)/(lambda_(n)-1)+1\frac{n-1}{\lambda_{n}}=\frac{n-2}{\lambda_{n}-1}+1n−1λn=n−2λn−1+1
din care
(11)
λ n = ( n − 1 ) λ 2 1 + λ 2 ( n − 2 ) λ n = ( n − 1 ) λ 2 1 + λ 2 ( n − 2 ) lambda_(n)=((n-1)lambda_(2))/(1+lambda_(2)(n-2))\lambda_{n}=\frac{(n-1) \lambda_{2}}{1+\lambda_{2}(n-2)}λn=(n−1)λ21+λ2(n−2)
apoi se deduce
(12) μ n = ( n − 1 ) μ 2 1 + λ 2 ( n − 2 ) , ν 2 = ( n − 1 ) v 2 1 + λ 2 ( n − 2 ) (12) μ n = ( n − 1 ) μ 2 1 + λ 2 ( n − 2 ) , ν 2 = ( n − 1 ) v 2 1 + λ 2 ( n − 2 ) {:(12)mu_(n)=((n-1)mu_(2))/(1+lambda_(2)(n-2))","nu_(2)=((n-1)v_(2))/(1+lambda_(2)(n-2)):}\begin{equation*} \mu_{n}=\frac{(n-1) \mu_{2}}{1+\lambda_{2}(n-2)}, \nu_{2}=\frac{(n-1) v_{2}}{1+\lambda_{2}(n-2)} \tag{12} \end{equation*}(12)μn=(n−1)μ21+λ2(n−2),ν2=(n−1)v21+λ2(n−2)
Din ( 10 ) : Din ( 10 ) : Din(10):\operatorname{Din}(10):Din(10):
(13) α n = − 1 − ( n − 1 ) λ 2 1 + λ 2 ( n − 2 ) = − 1 + λ 2 ( 2 n − 3 ) 1 + λ 2 ( n − 2 ) (13) α n = − 1 − ( n − 1 ) λ 2 1 + λ 2 ( n − 2 ) = − 1 + λ 2 ( 2 n − 3 ) 1 + λ 2 ( n − 2 ) {:(13)alpha_(n)=-1-((n-1)lambda_(2))/(1+lambda_(2)(n-2))=-(1+lambda_(2)(2n-3))/(1+lambda_(2)(n-2)):}\begin{equation*} \alpha_{n}=-1-\frac{(\mathrm{n}-1) \lambda_{2}}{1+\lambda_{2}(\mathrm{n}-2)}=-\frac{1+\lambda_{2}(2 n-3)}{1+\lambda_{2}(n-2)} \tag{13} \end{equation*}(13)αn=−1−(n−1)λ21+λ2(n−2)=−1+λ2(2n−3)1+λ2(n−2)
In fine din raportul al treilea şi al patrulea al relatiei (9) se deduce.
β n [ 1 + λ 2 ( n − 2 ) ∣ − β n − 1 [ 1 + λ 2 ( n − 3 ) ] + μ 2 = 0 β n 1 + λ 2 ( n − 2 ) ∣ − β n − 1 1 + λ 2 ( n − 3 ) + μ 2 = 0 beta_(n)[1+lambda_(2)(n-2)∣-beta_(n-1)[1+lambda_(2)(n-3)]+mu_(2)=0:}\beta_{n}\left[1+\lambda_{2}(n-2) \mid-\beta_{n-1}\left[1+\lambda_{2}(n-3)\right]+\mu_{2}=0\right.βn[1+λ2(n−2)∣−βn−1[1+λ2(n−3)]+μ2=0
din care:
(14) β n = β 2 − ( n − 2 ) μ 2 1 + λ 2 ( n − 2 ) (14) β n = β 2 − ( n − 2 ) μ 2 1 + λ 2 ( n − 2 ) {:(14)beta_(n)=(beta_(2)-(n-2)mu_(2))/(1+lambda_(2)(n-2)):}\begin{equation*} \beta_{n}=\frac{\beta_{2}-(n-2) \mu_{2}}{1+\lambda_{2}(n-2)} \tag{14} \end{equation*}(14)βn=β2−(n−2)μ21+λ2(n−2)
Formulele (11), (12), (13), (14) rczolvă problema şi se vede că po= linoamele verifică relația:
(15) [ 1 + λ ( n − 2 ) ] P n + [ − x [ 1 + λ ( 2 n − 3 ) ] + β − ( n − 2 ) μ ] P n − 1 + ( n − 1 ) ( λ x 2 + μ x + v ) P n − 2 = 0 (15) [ 1 + λ ( n − 2 ) ] P n + − x 1 + λ 2 n − 3 + β − ( n − 2 ) μ P n − 1 + ( n − 1 ) λ x 2 + μ x + v P n − 2 = 0 {:[(15){:[1+lambda(n-2)]P_(n)+[-x[1+lambda(2_(n)-3)]+beta-(n-2)mu]P_(n-1):}],[+(n-1)(lambdax^(2)+mu x+v)P_(n-2)=0]:}\begin{gather*} {[1+\lambda(n-2)] P_{n}+\left[-x\left[1+\lambda\left(2_{n}-3\right)\right]+\beta-(n-2) \mu\right] P_{n-1}} \tag{15}\\ +(n-1)\left(\lambda x^{2}+\mu x+v\right) P_{n-2}=0 \end{gather*}(15)[1+λ(n−2)]Pn+[−x[1+λ(2n−3)]+β−(n−2)μ]Pn−1+(n−1)(λx2+μx+v)Pn−2=0
şi depind de patru constante arbitrare λ , μ , ν , β λ , μ , ν , β lambda,mu,nu,beta\lambda, \mu, \nu, \betaλ,μ,ν,β, unde am suprimat indi= ccle 2 pentru simpliticare.
Se poate calcula acum functia generatoare. Pentru aceasta se înmulteşte relatia (15) cu z n − 2 ( n − 2 ) ! cu z n − 2 ( n − 2 ) ! cu(z^(n-2))/((n-2)!)\mathrm{cu} \frac{z^{n-2}}{(n-2)!}cuzn−2(n−2)! se face n = 2 , 3 , … n = 2 , 3 , … n=2,3,dotsn=2,3, \ldotsn=2,3,… și se admă. Găsim astfel
d z 3 Z z + d 2 Z d z 2 [ 1 − z ( 2 λ x + μ ) ] + d Z d [ β − x − λ x + (16) + Z ( λ x 2 + μ x + v ) ] + Z ( λ x 2 + μ x + v ) = 0 Z = ∑ n = 0 ∞ z ′ ′ n ! P n d z 3 Z z + d 2 Z d z 2 [ 1 − z ( 2 λ x + μ ) ] + d Z d [ β − x − λ x + (16) + Z λ x 2 + μ x + v + Z λ x 2 + μ x + v = 0 Z = ∑ n = 0 ∞   z ′ ′ n ! P n {:[dz^(3)Zz+(d^(2)Z)/(dz^(2))[1-z(2lambda x+mu)]+(dZ)/(d)[beta-x-lambda x+],[(16){:+Z(lambdax^(2)+mu x+v)]+Z(lambdax^(2)+mu x+v)=0],[Z=sum_(n=0)^(oo)(z^(''))/(n!)P_(n)]:}\begin{align*} d z^{3} Z z+ & \frac{d^{2} Z}{d z^{2}}[1-z(2 \lambda x+\mu)]+\frac{d Z}{d}[\beta-x-\lambda x+ \\ & \left.+Z\left(\lambda x^{2}+\mu x+v\right)\right]+Z\left(\lambda x^{2}+\mu x+v\right)=0 \tag{16}\\ Z & =\sum_{n=0}^{\infty} \frac{z^{\prime \prime}}{n!} P_{n} \end{align*}dz3Zz+d2Zdz2[1−z(2λx+μ)]+dZd[β−x−λx+(16)+Z(λx2+μx+v)]+Z(λx2+μx+v)=0Z=∑n=0∞z′′n!Pn
dacă facem
Z = e z x F ( z ) Z = e z x F ( z ) Z=e^(zx)F(z)Z=e^{z x} F(z)Z=ezxF(z)
F ( z ) F ( z ) F(z)F(z)F(z) e functie de z z zzz nu şi de x x xxx; ecuaţia devine :
unde :
d Φ d z + x Φ = 0 d Φ d z + x Φ = 0 (d Phi)/(dz)+x Phi=0\frac{d \Phi}{d z}+x \Phi=0dΦdz+xΦ=0
(17) Φ = λ z F ′ ′ + ( 1 − λ − z μ ) F ′ + ( z v + β + μ ) F = 0 (17) Φ = λ z F ′ ′ + ( 1 − λ − z μ ) F ′ + ( z v + β + μ ) F = 0 {:(17)Phi=lambda zF^('')+(1-lambda-z mu)F^(')+(zv+beta+mu)F=0:}\begin{equation*} \Phi=\lambda z F^{\prime \prime}+(1-\lambda-z \mu) F^{\prime}+(z v+\beta+\mu) F=0 \tag{17} \end{equation*}(17)Φ=λzF′′+(1−λ−zμ)F′+(zv+β+μ)F=0
Funcţia F F FFF este prin urmare dată de ecuația (17).
5. Să presupunem λ = μ = 0 λ = μ = 0 lambda=mu=0\lambda=\mu=0λ=μ=0, ecuația (17) se scrie
F ′ + ( z v + β ) F = 0 F ′ + ( z v + β ) F = 0 F^(')+(zv+beta)F=0F^{\prime}+(z v+\beta) F=0F′+(zv+β)F=0
şi avem imediat
F ( z ) = e − v 2 z 2 − β z F ( z ) = e − v 2 z 2 − β z F(z)=e^(-(v)/(2)z^(2)-beta z)F(z)=e^{-\frac{v}{2} z^{2}-\beta z}F(z)=e−v2z2−βz
căci P 0 = 1 P 0 = 1 P_(0)=1P_{0}=1P0=1. Polinoamele satisfac relatia
( 15 ) ′ ( 15 ) ′ (15)^(')(15)^{\prime}(15)′
P n − ( x − β ) P l − 1 + ( n − 1 ) ∨ P n − 2 = 0 P n − ( x − β ) P l − 1 + ( n − 1 ) ∨ P n − 2 = 0 P_(n)-(x-beta)P_(l-1)+(n-1)vvP_(n-2)=0P_{n}-(x-\beta) P_{l-1}+(n-1) \vee P_{n-2}=0Pn−(x−β)Pl−1+(n−1)∨Pn−2=0
Aceste polinoame se reduc la polinoamele lui Hermite. Intr'adevăr functia generatoare a polinoamelor H n H n H_(n)H_{n}Hn este
e z 2 4 + z x = ∑ 0 ∞ z n n ! H n ( x ) e z 2 4 + z x = ∑ 0 ∞   z n n ! H n ( x ) e^((z^(2))/(4)+zx)=sum_(0)^(oo)(z^(n))/(n!)H_(n)(x)e^{\frac{z^{2}}{4}+z x}=\sum_{0}^{\infty} \frac{z^{n}}{n!} H_{n}(x)ez24+zx=∑0∞znn!Hn(x)
iar functia generatoare a polinoamelor noastre este :
e − v 2 z 2 − β z + z x = e ( − 2 v z ) 2 4 + 2 ( x − β ) e − v 2 z 2 − β z + z x = e ( − 2 v z ) 2 4 + 2 ( x − β ) e^(-(v)/(2)z^(2)-beta z+zx)=e^(((sqrt(-2vz))^(2))/(4)+2(x-beta))e^{-\frac{v}{2} z^{2}-\beta z+z x}=e^{\frac{(\sqrt{-2 v z})^{2}}{4}+2(x-\beta)}e−v2z2−βz+zx=e(−2vz)24+2(x−β)
deci
(15) P n ( x ) = ( − 2 v ) n H n ( x − β − 2 v ) (15) P n ( x ) = ( − 2 v ) n H n x − β − 2 v {:(15)P_(n)(x)=(sqrt(-2v))^(n)H_(n)((x-beta)/(sqrt(-2v))):}\begin{equation*} P_{n}(x)=(\sqrt{-2 v})^{n} H_{n}\left(\frac{x-\beta}{\sqrt{-2 v}}\right) \tag{15} \end{equation*}(15)Pn(x)=(−2v)nHn(x−β−2v)
Să presupunem acum că λ λ lambda\lambdaλ, u nu sunt ambele nule și fie ω ω omega\omegaω una din rădăcinile ecuațici
(18) λ z 2 − μ z + v = 0 (18) λ z 2 − μ z + v = 0 {:(18)lambdaz^(2)-mu z+v=0:}\begin{equation*} \lambda z^{2}-\mu z+v=0 \tag{18} \end{equation*}(18)λz2−μz+v=0
Să puncm
F = e w z F 1 F = e w z F 1 F=e^(wz)F_(1)F=e^{w z} F_{1}F=ewzF1
ecuatia (17) devine
(19) λ z F ′ ′ 1 + ( 2 ω λ z − z μ + 1 − λ ) F ′ 1 + [ ω ( 1 − λ ) + β + μ ] F = 0 λ z F ′ ′ 1 + ( 2 ω λ z − z μ + 1 − λ ) F ′ 1 + [ ω ( 1 − λ ) + β + μ ] F = 0 lambda zF^('')_(1)+(2omega lambda z-z mu+1-lambda)F^(')_(1)+[omega(1-lambda)+beta+mu]F=0\lambda z F^{\prime \prime}{ }_{1}+(2 \omega \lambda z-z \mu+1-\lambda) F^{\prime}{ }_{1}+[\omega(1-\lambda)+\beta+\mu] F=0λzF′′1+(2ωλz−zμ+1−λ)F′1+[ω(1−λ)+β+μ]F=0 și după definitia ce am dat funcției generatoare trebue să căutăm soluția holomorfă în jurul originei şi care pentru z = 0 z = 0 z=0z=0z=0 este egală cu 1. Ori accasta este totdeauna posibil în general.
Inainte de a merge mai departe să reamintim câteva rezultate : Ecuațiile
(20) x y ′ ′ + ( γ − x ) y ′ − α y = 0 x y ′ ′ + ( γ − x ) y ′ − α y = 0 xy^('')+(gamma-x)y^(')-alpha y=0x y^{\prime \prime}+(\gamma-x) y^{\prime}-\alpha y=0xy′′+(γ−x)y′−αy=0 (Ecuatia lui Kummer)
admit solutiunile
x y ′ ′ + γ y ′ − y = 0 (Ecuaţia lui Bessel) x y ′ ′ + γ y ′ − y = 0  (Ecuaţia lui Bessel)  xy^('')+gammay^(')-y=0quad" (Ecuaţia lui Bessel) "x y^{\prime \prime}+\gamma y^{\prime}-y=0 \quad \text { (Ecuaţia lui Bessel) }xy′′+γy′−y=0 (Ecuaţia lui Bessel) 
G ( α , γ , x ) = ∑ n = 0 ∞ α ( α + 1 ) … ( α + n − 1 ) γ ( γ + 1 ) … ( γ + n − 1 ) ⋅ x n n ! T ( γ , x ) = ∑ 0 ∞ 1 γ ( γ + 1 ) … ( γ + n − 1 ) ⋅ x n n ! G ( α , γ , x ) = ∑ n = 0 ∞   α ( α + 1 ) … ( α + n − 1 ) γ ( γ + 1 ) … ( γ + n − 1 ) ⋅ x n n ! T ( γ , x ) = ∑ 0 ∞   1 γ ( γ + 1 ) … ( γ + n − 1 ) ⋅ x n n ! {:[G(alpha","gamma","x)=sum_(n=0)^(oo)(alpha(alpha+1)dots(alpha+n-1))/(gamma(gamma+1)dots(gamma+n-1))*(x^(n))/(n!)],[T(gamma","x)=sum_(0)^(oo)(1)/(gamma(gamma+1)dots(gamma+n-1))*(x^(n))/(n!)]:}\begin{gathered} G(\alpha, \gamma, x)=\sum_{n=0}^{\infty} \frac{\alpha(\alpha+1) \ldots(\alpha+n-1)}{\gamma(\gamma+1) \ldots(\gamma+n-1)} \cdot \frac{x^{n}}{n!} \\ T(\gamma, x)=\sum_{0}^{\infty} \frac{1}{\gamma(\gamma+1) \ldots(\gamma+n-1)} \cdot \frac{x^{n}}{n!} \end{gathered}G(α,γ,x)=∑n=0∞α(α+1)…(α+n−1)γ(γ+1)…(γ+n−1)⋅xnn!T(γ,x)=∑0∞1γ(γ+1)…(γ+n−1)⋅xnn!
cari în general sunt nişte functiuni întregi
Facem acum mai multe ipoteze :
a) λ ≠ 0 λ ≠ 0 lambda!=0\lambda \neq 0λ≠0, ecuaţia (18) având rădăcinile distincte.
Prin transformarea
z = λ t μ − 2 ω λ z = λ t μ − 2 ω λ z=(lambda t)/(mu-2omega lambda)z=\frac{\lambda t}{\mu-2 \omega \lambda}z=λtμ−2ωλ
ecuatia (19) devine
deci
t d 2 F 1 d t 2 + ( 1 − λ λ − t ) d F 1 λ t − ω ( 1 − λ ) + β + μ 2 ω λ − μ F 1 = 0 t d 2 F 1 d t 2 + 1 − λ λ − t d F 1 λ t − ω ( 1 − λ ) + β + μ 2 ω λ − μ F 1 = 0 t(d^(2)F_(1))/(dt^(2))+((1-lambda)/(lambda)-t)(dF_(1))/(lambda t)-(omega(1-lambda)+beta+mu)/(2omega lambda-mu)F_(1)=0t \frac{d^{2} F_{1}}{d t^{2}}+\left(\frac{1-\lambda}{\lambda}-t\right) \frac{d F_{1}}{\lambda t}-\frac{\omega(1-\lambda)+\beta+\mu}{2 \omega \lambda-\mu} F_{1}=0td2F1dt2+(1−λλ−t)dF1λt−ω(1−λ)+β+μ2ωλ−μF1=0
F 1 = G ( ω ( 1 − λ ) + β + μ 2 ω λ − μ , 1 − λ λ , ( μ − 2 ω λ ) z λ ) F 1 = G ω ( 1 − λ ) + β + μ 2 ω λ − μ , 1 − λ λ , ( μ − 2 ω λ ) z λ F_(1)=G((omega(1-lambda)+beta+mu)/(2omega lambda-mu),(1-lambda)/(lambda),((mu-2omega lambda)z)/(lambda))F_{1}=G\left(\frac{\omega(1-\lambda)+\beta+\mu}{2 \omega \lambda-\mu}, \frac{1-\lambda}{\lambda}, \frac{(\mu-2 \omega \lambda) z}{\lambda}\right)F1=G(ω(1−λ)+β+μ2ωλ−μ,1−λλ,(μ−2ωλ)zλ)
b) λ ≠ 0 λ ≠ 0 lambda!=0\lambda \neq 0λ≠0, ecuația (18) având rădăcinile cgale, deci
Facem
μ = 2 ω λ μ = 2 ω λ mu=2omega lambda\mu=2 \omega \lambdaμ=2ωλ
şi avem
z = − λ ˙ t ω ( 1 + λ ) + β z = − λ ˙ t ω ( 1 + λ ) + β z=-((lambda^(˙))t)/(omega(1+lambda)+beta)z=-\frac{\dot{\lambda} t}{\omega(1+\lambda)+\beta}z=−λ˙tω(1+λ)+β
t d 2 F 1 d t 2 + 1 − λ λ d F 1 d t − F 1 = 0 t d 2 F 1 d t 2 + 1 − λ λ d F 1 d t − F 1 = 0 t(d^(2)F_(1))/(dt^(2))+(1-lambda)/(lambda)(dF_(1))/(dt)-F_(1)=0t \frac{d^{2} F_{1}}{d t^{2}}+\frac{1-\lambda}{\lambda} \frac{d F_{1}}{d t}-F_{1}=0td2F1dt2+1−λλdF1dt−F1=0
şi deci
F 1 = T ( 1 − λ λ , − ω ( 1 + λ ) + β λ z ) F 1 = T 1 − λ λ , − ω ( 1 + λ ) + β λ z F_(1)=T((1-lambda)/(lambda),-(omega(1+lambda)+beta)/(lambda)z)F_{1}=T\left(\frac{1-\lambda}{\lambda},-\frac{\omega(1+\lambda)+\beta}{\lambda} z\right)F1=T(1−λλ,−ω(1+λ)+βλz)
c) λ = c λ = c lambda=c\lambda=\mathrm{c}λ=c, averm
w = v μ w = v μ w=(v)/( mu)w=\frac{v}{\mu}w=vμ
şi ecur."a devine
( 1 − z μ ) F 1 ′ + ( v μ + β + μ ) F 1 = 0 ( 1 − z μ ) F 1 ′ + v μ + β + μ F 1 = 0 (1-z mu)F_(1)^(')+((v)/( mu)+beta+mu)F_(1)=0(1-z \mu) F_{1}^{\prime}+\left(\frac{v}{\mu}+\beta+\mu\right) F_{1}=0(1−zμ)F1′+(vμ+β+μ)F1=0
deci
F 1 = ( 1 − μ z ) ν μ + β + μ μ F 1 = ( 1 − μ z ) ν μ + β + μ μ F_(1)=(1-mu z)^(((nu )/(mu)+beta+mu)/(mu))F_{1}=(1-\mu z)^{\frac{\frac{\nu}{\mu}+\beta+\mu}{\mu}}F1=(1−μz)νμ+β+μμ
Functia generatoare în acest caz este
e v μ z + z x ( 1 − μ z ) v μ + β + μ μ e v μ z + z x ( 1 − μ z ) v μ + β + μ μ e^((v)/( mu)z+zx)(1-mu z)((v)/( mu)+beta+mu)/(mu)e^{\frac{v}{\mu} z+z x}(1-\mu z) \frac{\frac{v}{\mu}+\beta+\mu}{\mu}evμz+zx(1−μz)vμ+β+μμ
Cazul studiat în § 3 intră aci. Dacă v=o avem nişte polinoame P n P n P_(n)P_{n}Pn care sunt tocmai polinoamele Q n ( x , λ ) Q n ( x , λ ) Q_(n)(x,lambda)Q_{n}(\mathrm{x}, \lambda)Qn(x,λ) studiate de noi într'o lucrare ante= rioară 2 2 ^(2){ }^{2}2 )
Nu mai insist asupra cazurilor particulare ce pot proveni de exemplu din faptul că
1 − λ λ 1 − λ λ (1-lambda)/(lambda)\frac{1-\lambda}{\lambda}1−λλ
este un număr întreg negativ. Acest caz se deduce din teoria ecuaților lui Kummer și Bessel.
6. Polinoamele P n P n P_(n)P_{n}Pn studiate de noi se pot ataşa funcţiei hipergeome= trice și a confluentelor ei. Vom da resultatele fără a insista asupra detaliilor calculelor. Dīı formulele (2), (15) deducem că polinomul P n P n P_(n)P_{n}Pn satisface ecua= ţia diferentială lincară.
(21)
( λ x 2 + μ x + ν ) y ′ ′ + [ − x ( 1 + λ ( 2 n − 3 ) ) + β − ( n − 2 ) μ ∣ y ′ + + n [ 1 + λ ( n − 2 ) ] y = 0 λ x 2 + μ x + ν y ′ ′ + − x ( 1 + λ ( 2 n − 3 ) ) + β − ( n − 2 ) μ ∣ y ′ + + n [ 1 + λ ( n − 2 ) ] y = 0 {:[(lambdax^(2)+mu x+nu)y^('')+[-x(1+lambda(2n-3))+beta-(n-2)mu∣y^(')+:}],[+n[1+lambda(n-2)]y=0]:}\begin{gathered} \left(\lambda x^{2}+\mu x+\nu\right) y^{\prime \prime}+\left[-x(1+\lambda(2 n-3))+\beta-(n-2) \mu \mid y^{\prime}+\right. \\ +n[1+\lambda(n-2)] y=0 \end{gathered}(λx2+μx+ν)y′′+[−x(1+λ(2n−3))+β−(n−2)μ∣y′++n[1+λ(n−2)]y=0
Deosebim cazurile :
a) λ = 0 , μ ≠ 0 λ = 0 , μ ≠ 0 lambda=0,mu!=0\lambda=0, \mu \neq 0λ=0,μ≠0, prin transformarea
t = μ x + v μ x t = μ x + v μ x t=(mu x+v)/(mu^(x))t=\frac{\mu x+v}{\mu^{x}}t=μx+vμx
ecuatia (21) devine
t y ′ ′ + [ − t + v μ ′ + β μ − ( n − 2 ) ] y ′ + n y = 0 , ε = v μ 2 + β μ t y ′ ′ + − t + v μ ′ + β μ − ( n − 2 ) y ′ + n y = 0 , ε = v μ 2 + β μ ty^('')+[-t+(v)/(mu^('))+(beta )/(mu)-(n-2)]y^(')+ny=0,quad epsi=(v)/(mu^(2))+(beta )/(mu)t y^{\prime \prime}+\left[-t+\frac{v}{\mu^{\prime}}+\frac{\beta}{\mu}-(n-2)\right] y^{\prime}+n y=0, \quad \varepsilon=\frac{v}{\mu^{2}}+\frac{\beta}{\mu}ty′′+[−t+vμ′+βμ−(n−2)]y′+ny=0,ε=vμ2+βμ
deci
P n = ( − 1 ) n μ n ( ξ + 1 ) ξ ( ξ − 1 ) … ( ξ − n + 2 ) n ! G ( − n , ξ − n + 2 , μ x + ν μ 2 ) P n = ( − 1 ) n μ n ( ξ + 1 ) ξ ( ξ − 1 ) … ( ξ − n + 2 ) n ! G − n , ξ − n + 2 , μ x + ν μ 2 P_(n)=((-1)^(n)mu^(n)(xi+1)xi(xi-1)dots(xi-n+2))/(n!)G(-n,xi-n+2,(mu x+nu)/(mu^(2)))P_{n}=\frac{(-1)^{n} \mu^{n}(\xi+1) \xi(\xi-1) \ldots(\xi-n+2)}{n!} G\left(-n, \xi-n+2, \frac{\mu x+\nu}{\mu^{2}}\right)Pn=(−1)nμn(ξ+1)ξ(ξ−1)…(ξ−n+2)n!G(−n,ξ−n+2,μx+νμ2)
observând că coeficientul lui x n x n x^(n)x^{n}xn este 1 ,
b) λ = 0 , μ = 0 λ = 0 , μ = 0 lambda=0,mu=0\lambda=0, \mu=0λ=0,μ=0. Acesta c cazul polinoamelor lui Hermite.
Să presupunem ν = 1 2 , β = 0 ν = 1 2 , β = 0 nu=(1)/(2),beta=0\nu=\frac{1}{2}, \beta=0ν=12,β=0, ecuatia devine
y ′ ′ − 2 x y ′ + − 2 n y = 0 . y ′ ′ − 2 x y ′ + − 2 n y = 0 . y^('')-2xy^(')+-2ny=0.y^{\prime \prime}-2 x y^{\prime}+-2 n y=0 .y′′−2xy′+−2ny=0.
Prin transformarea
x 2 = t x 2 = t x^(2)=tx^{2}=tx2=t
se obtine
t y ′ ′ + ( 1 2 − t ) y ′ + n 2 y = 0 t y ′ ′ + 1 2 − t y ′ + n 2 y = 0 ty^('')+((1)/(2)-t)y^(')+(n)/(2)y=0t y^{\prime \prime}+\left(\frac{1}{2}-t\right) y^{\prime}+\frac{n}{2} y=0ty′′+(12−t)y′+n2y=0
care ne dă pentru n n nnn par, solutia
H n ( x ) = ( − 1 ) n 2 ⋅ 1 ⋅ 3 ⋅ 5 … ( n − 1 ) 2 n 2 G ( − n 2 , 1 2 , x 2 ) H n ( x ) = ( − 1 ) n 2 ⋅ 1 ⋅ 3 ⋅ 5 … ( n − 1 ) 2 n 2 G − n 2 , 1 2 , x 2 H_(n)(x)=((-1)^((n)/(2))*1*3*5dots(n-1))/(2^((n)/(2)))G(-(n)/(2),(1)/(2),x^(2))H_{n}(x)=\frac{(-1)^{\frac{n}{2}} \cdot 1 \cdot 3 \cdot 5 \ldots(n-1)}{2^{\frac{n}{2}}} G\left(-\frac{n}{2}, \frac{1}{2}, x^{2}\right)Hn(x)=(−1)n2⋅1⋅3⋅5…(n−1)2n2G(−n2,12,x2)
Pentru n n nnn impar luăm întâi ca nouă funcție
y = x z y = x z y=xzy=x zy=xz
şi obtinem, urmând acelaş procedcu:
H n ( x ) = ( − 1 ) n − 1 2 ⋅ 1 ⋅ 3 ⋅ 5 ⋯ n 2 n − 1 2 × G ( − n − 1 2 , 3 2 , x 2 ) H n ( x ) = ( − 1 ) n − 1 2 ⋅ 1 ⋅ 3 ⋅ 5 ⋯ n 2 n − 1 2 × G − n − 1 2 , 3 2 , x 2 H_(n)(x)=((-1)^((n-1)/(2))*1*3*5cdots n)/(2^((n-1)/(2)))xx G(-(n-1)/(2),(3)/(2),x^(2))H_{n}(x)=\frac{(-1)^{\frac{n-1}{2}} \cdot 1 \cdot 3 \cdot 5 \cdots n}{2^{\frac{n-1}{2}}} \times G\left(-\frac{n-1}{2}, \frac{3}{2}, x^{2}\right)Hn(x)=(−1)n−12⋅1⋅3⋅5⋯n2n−12×G(−n−12,32,x2)
Pentru cazul general vom avea deci, după ( 15 ) ′ ′ ( 15 ) ′ ′ (15)^('')(15)^{\prime \prime}(15)′′ :
P n ( x ) = { ( − 1 ) n 2 , 1.3 .5 … ( n − 1 ) ( − v ) n G ( − n 2 , 1 2 , − ( x − β ) 2 2 v ) ( − 1 ) n − 1 2 , 1.3 .5 . … n . ( − v ) n − 1 ( x − β ) G ( − n − 1 2 , 3 2 , − ( x − β ) 2 2 v ) P n ( x ) = ( − 1 ) n 2 , 1.3 .5 … ( n − 1 ) ( − v ) n G − n 2 , 1 2 , − ( x − β ) 2 2 v ( − 1 ) n − 1 2 , 1.3 .5 . … n . ( − v ) n − 1 ( x − β ) G − n − 1 2 , 3 2 , − ( x − β ) 2 2 v P_(n)(x)={[(-1)^((n)/(2))","1.3.5 dots(n-1)(sqrt(-v))^(n)G(-(n)/(2),(1)/(2),-((x-beta)^(2))/(2v))],[(-1)^((n-1)/(2))","1.3.5.dots n.(sqrt()-v)^(n-1)(x-beta)G(-(n-1)/(2),(3)/(2),-((x-beta)^(2))/(2v))]:}P_{n}(x)=\left\{\begin{array}{l}(-1)^{\frac{n}{2}}, 1.3 .5 \ldots(n-1)(\sqrt{-v})^{n} G\left(-\frac{n}{2}, \frac{1}{2},-\frac{(x-\beta)^{2}}{2 v}\right) \\ (-1)^{\frac{n-1}{2}}, 1.3 .5 . \ldots n .(\sqrt{ }-v)^{n-1}(x-\beta) G\left(-\frac{n-1}{2}, \frac{3}{2},-\frac{(x-\beta)^{2}}{2 v}\right)\end{array}\right.Pn(x)={(−1)n2,1.3.5…(n−1)(−v)nG(−n2,12,−(x−β)22v)(−1)n−12,1.3.5.…n.(−v)n−1(x−β)G(−n−12,32,−(x−β)22v)
după cum n n nnn este par sau impar.
c) λ ≠ 0 λ ≠ 0 lambda!=0\lambda \neq 0λ≠0 şi ecuatia (18) are două rădăcini distincte w 1 , w 2 w 1 , w 2 w_(1),w_(2)w_{1}, w_{2}w1,w2.
Făcând transformarea
x = ( w 1 − w 2 ) t − w 1 x = w 1 − w 2 t − w 1 x=(w_(1)-w_(2))t-w_(1)x=\left(w_{1}-w_{2}\right) t-w_{1}x=(w1−w2)t−w1
avem ecuatia in t t ttt :
t ( 1 − t ) y ′ ′ + [ ρ + 1 + λ ( 2 n − 3 ) λ t ] y ′ − n λ [ 1 + λ ( n − 2 ) ] y = 0 ρ = ω 1 [ 1 + λ ( 2 n − 3 ) ] + β − ( n − 2 ) μ λ ( ω 2 − ω 1 ) t ( 1 − t ) y ′ ′ + ρ + 1 + λ ( 2 n − 3 ) λ t y ′ − n λ [ 1 + λ ( n − 2 ) ] y = 0 ρ = ω 1 [ 1 + λ ( 2 n − 3 ) ] + β − ( n − 2 ) μ λ ω 2 − ω 1 {:[t(1-t)y^('')+[rho+(1+lambda(2n-3))/(lambda)t]y^(')-(n)/( lambda)[1+lambda(n-2)]y=0],[rho=(omega_(1)[1+lambda(2n-3)]+beta-(n-2)mu)/(lambda(omega_(2)-omega_(1)))]:}\begin{gathered} t(1-\mathrm{t}) y^{\prime \prime}+\left[\rho+\frac{1+\lambda(2 n-3)}{\lambda} t\right] y^{\prime}-\frac{n}{\lambda}[1+\lambda(n-2)] y=0 \\ \rho=\frac{\omega_{1}[1+\lambda(2 n-3)]+\beta-(n-2) \mu}{\lambda\left(\omega_{2}-\omega_{1}\right)} \end{gathered}t(1−t)y′′+[ρ+1+λ(2n−3)λt]y′−nλ[1+λ(n−2)]y=0ρ=ω1[1+λ(2n−3)]+β−(n−2)μλ(ω2−ω1)
care e de forma hipergeometrică :
t ( 1 − t ) y ′ ′ + [ ζ − ( ξ + η + 1 ) t ] y ′ − ξ η y = 0 t ( 1 − t ) y ′ ′ + [ ζ − ( ξ + η + 1 ) t ] y ′ − ξ η y = 0 t(1-t)y^('')+[zeta-(xi+eta+1)t]y^(')-xi eta y=0t(1-t) y^{\prime \prime}+[\zeta-(\xi+\eta+1) t] y^{\prime}-\xi \eta y=0t(1−t)y′′+[ζ−(ξ+η+1)t]y′−ξηy=0
cu solutia
Φ ( ξ , η , ζ , t ) = ∑ n = 0 ∞ ξ ( ξ + 1 ) … ( ξ + n − 1 ) η ( η + 1 ) … ( η + n − 1 ) τ ( ζ + 1 ) … ( ζ + n − 1 ) f n n ! Φ ( ξ , η , ζ , t ) = ∑ n = 0 ∞   ξ ( ξ + 1 ) … ( ξ + n − 1 ) η ( η + 1 ) … ( η + n − 1 ) τ ( ζ + 1 ) … ( ζ + n − 1 ) f n n ! Phi(xi,eta,zeta,t)=sum_(n=0)^(oo)(xi(xi+1)dots(xi+n-1)eta(eta+1)dots(eta+n-1))/(tau(zeta+1)dots(zeta+n-1))(f^(n))/(n!)\Phi(\xi, \eta, \zeta, t)=\sum_{n=0}^{\infty} \frac{\xi(\xi+1) \ldots(\xi+n-1) \eta(\eta+1) \ldots(\eta+n-1)}{\tau(\zeta+1) \ldots(\zeta+n-1)} \frac{f^{n}}{n!}Φ(ξ,η,ζ,t)=∑n=0∞ξ(ξ+1)…(ξ+n−1)η(η+1)…(η+n−1)τ(ζ+1)…(ζ+n−1)fnn!
Polinomul căutat e deci de forma :
P n = ( − 1 ) n λ n ρ ( ρ + 1 ) … ( ρ + n − 1 ) ( ω 1 − ω 2 ) n n ! [ λ ( n − 2 ) + 1 ] [ λ ( n − 1 ) + 1 ] … [ λ ( 2 ω − 3 ) + 1 ] Φ ( − n , − 1 + λ ( n − 2 ) λ , ρ , x 1 + ω 1 ω 1 − ω 2 ) P n = ( − 1 ) n λ n ρ ( ρ + 1 ) … ( ρ + n − 1 ) ω 1 − ω 2 n n ! [ λ ( n − 2 ) + 1 ] [ λ ( n − 1 ) + 1 ] … [ λ ( 2 ω − 3 ) + 1 ] Φ − n , − 1 + λ ( n − 2 ) λ , ρ , x 1 + ω 1 ω 1 − ω 2 {:[P_(n)=((-1)^(n)lambda^(n)rho(rho+1)dots(rho+n-1)(omega_(1)-omega_(2))^(n))/(n![lambda(n-2)+1][lambda(n-1)+1]dots[lambda(2omega-3)+1])],[Phi(-n,-(1+lambda(n-2))/(lambda),rho,(x_(1)+omega_(1))/(omega_(1)-omega_(2)))]:}\begin{gathered} P_{n}=\frac{(-1)^{n} \lambda^{n} \rho(\rho+1) \ldots(\rho+n-1)\left(\omega_{1}-\omega_{2}\right)^{n}}{n![\lambda(n-2)+1][\lambda(n-1)+1] \ldots[\lambda(2 \omega-3)+1]} \\ \Phi\left(-n,-\frac{1+\lambda(n-2)}{\lambda}, \rho, \frac{x_{1}+\omega_{1}}{\omega_{1}-\omega_{2}}\right) \end{gathered}Pn=(−1)nλnρ(ρ+1)…(ρ+n−1)(ω1−ω2)nn![λ(n−2)+1][λ(n−1)+1]…[λ(2ω−3)+1]Φ(−n,−1+λ(n−2)λ,ρ,x1+ω1ω1−ω2)
d) λ ≠ 0 λ ≠ 0 lambda!=0\lambda \neq 0λ≠0, şi ecuatia (18) are o rădăcină dublă w = μ 2 λ w = μ 2 λ w=(mu)/(2lambda)w=\frac{\mu}{2 \lambda}w=μ2λ.
Facem
t = − λ ( x + ω ) ω ( 1 + λ ) + β t = − λ ( x + ω ) ω ( 1 + λ ) + β t=-(lambda(x+omega))/(omega(1+lambda)+beta)t=-\frac{\lambda(x+\omega)}{\omega(1+\lambda)+\beta}t=−λ(x+ω)ω(1+λ)+β
acuatia devine :
t 2 y ′ ′ − [ 1 + 1 + λ ( 2 n − 3 ) λ t ] y ′ + n λ [ 1 + λ ( n − 2 ) ] y = 0 t 2 y ′ ′ − 1 + 1 + λ ( 2 n − 3 ) λ t y ′ + n λ [ 1 + λ ( n − 2 ) ] y = 0 t^(2)y^('')-[1+(1+lambda(2n-3))/(lambda)t]y^(')+(n)/( lambda)[1+lambda(n-2)]y=0t^{2} y^{\prime \prime}-\left[1+\frac{1+\lambda(2 n-3)}{\lambda} t\right] y^{\prime}+\frac{n}{\lambda}[1+\lambda(n-2)] y=0t2y′′−[1+1+λ(2n−3)λt]y′+nλ[1+λ(n−2)]y=0
Această ecuațic e de forma:
t 2 y ′ ′ − [ 1 − ( ξ + η + 1 ) t ] y ′ + ξ η y = 0 t 2 y ′ ′ − [ 1 − ( ξ + η + 1 ) t ] y ′ + ξ η y = 0 t^(2)y^('')-[1-(xi+eta+1)t]y^(')+xi eta y=0t^{2} y^{\prime \prime}-[1-(\xi+\eta+1) t] y^{\prime}+\xi \eta y=0t2y′′−[1−(ξ+η+1)t]y′+ξηy=0
căreia am ataşat seria divergentă 1 1 ^(1){ }^{1}1 ).
T ( ξ , η , l ) = ∑ ω = 0 ∞ ξ ( ξ + 1 ) … ( ξ + n − 1 ) η ( η + 1 ) … ( η + n − 1 ) n ! p n T ( ξ , η , l ) = ∑ ω = 0 ∞   ξ ( ξ + 1 ) … ( ξ + n − 1 ) η ( η + 1 ) … ( η + n − 1 ) n ! p n T(xi,eta,l)=sum_(omega=0)^(oo)(xi(xi+1)dots(xi+n-1)eta(eta+1)dots(eta+n-1))/(n!)p^(n)T(\xi, \eta, l)=\sum_{\omega=0}^{\infty} \frac{\xi(\xi+1) \ldots(\xi+n-1) \eta(\eta+1) \ldots(\eta+n-1)}{n!} p^{n}T(ξ,η,l)=∑ω=0∞ξ(ξ+1)…(ξ+n−1)η(η+1)…(η+n−1)n!pn
Rezultă dar, că polinomul e de forma
P n = ( − 1 ) n [ μ ( 1 + λ ) + 2 λ β ) n 2 n λ n [ λ ( n − 2 ) + 1 ] [ λ ( n − 3 ) + 1 ] … ( λ + 1 ) ( 1 − λ ) T ( − n , − 1 + λ ( n − 2 ) λ , 2 λ 2 x + λ μ μ ( 1 + λ ) + 2 λ β ) P n = ( − 1 ) n [ μ ( 1 + λ ) + 2 λ β ) n 2 n λ n [ λ ( n − 2 ) + 1 ] [ λ ( n − 3 ) + 1 ] … ( λ + 1 ) ( 1 − λ ) T − n , − 1 + λ ( n − 2 ) λ , 2 λ 2 x + λ μ μ ( 1 + λ ) + 2 λ β {:[P_(n)=((-1)^(n)[mu(1+lambda)+2lambda beta)^(n))/(2^(n)lambda^(n)[lambda(n-2)+1][lambda(n-3)+1]dots(lambda+1)(1-lambda))],[T(-n,-(1+lambda(n-2))/(lambda),(2lambda^(2)x+lambda mu)/(mu(1+lambda)+2lambda beta))]:}\begin{aligned} P_{n}= & \frac{(-1)^{n}[\mu(1+\lambda)+2 \lambda \beta)^{n}}{2^{n} \lambda^{n}[\lambda(n-2)+1][\lambda(n-3)+1] \ldots(\lambda+1)(1-\lambda)} \\ & T\left(-n,-\frac{1+\lambda(n-2)}{\lambda}, \frac{2 \lambda^{2} x+\lambda \mu}{\mu(1+\lambda)+2 \lambda \beta}\right) \end{aligned}Pn=(−1)n[μ(1+λ)+2λβ)n2nλn[λ(n−2)+1][λ(n−3)+1]…(λ+1)(1−λ)T(−n,−1+λ(n−2)λ,2λ2x+λμμ(1+λ)+2λβ)
  1. In fine pentru ca studiul de fată să fie complet, revenind la pros blema initială, ar trebui să examinăm cazul când unii din A n A n A_(n)A_{n}An sunt nuli. Pentru a nu lungi mult acest articol, vom da numai indicatiuni asupra acestui caz urmând ca să publicăm ulterior un studiu eomplet.
Să demonstrăm întâi lema:
Dacă polinomul f ( x ) f ( x ) f(x)f(\mathrm{x})f(x) este de forma
(22) f ( x ) = ( x + a ) n ( x + b ) m n > 1 , m > 1 , a ≠ b f ( x ) = ( x + a ) n ( x + b ) m n > 1 , m > 1 , a ≠ b f(x)=(x+a)^(n)(x+b)^(m)quad n > 1,m > 1,a!=bf(x)=(x+a)^{n}(x+b)^{m} \quad n>1, m>1, a \neq bf(x)=(x+a)n(x+b)mn>1,m>1,a≠b, nici una din derivatele f ′ , f ′ ′ , … f ( m + n − 3 ) f ′ , f ′ ′ , … f ( m + n − 3 ) f^('),f^(''),dotsf^((m+n-3))f^{\prime}, f^{\prime \prime}, \ldots f^{(m+n-3)}f′,f′′,…f(m+n−3) nu poate f i f i fif ifi de aceeaş formă. Cu alte cuvinte : ecuatia
  1. Vezi : T. Popoviciu loc. cit. p. 6.
    (23)
f ( k ) ( x ) = 0 k = 1 , 2 , … m + n − 3 f ( k ) ( x ) = 0 k = 1 , 2 , … m + n − 3 f^((k))(x)=0quad k=1,2,dots m+n-3f^{(k)}(x)=0 \quad k=1,2, \ldots m+n-3f(k)(x)=0k=1,2,…m+n−3
are cel putin trei rădăcini distincte.
Dacă a , b a , b a,ba, ba,b, sunt reali lema rezultă din teorema lui Rolle. Ori în cazul general se poate face o transformare
y = u x + v y = u x + v y=ux+vy=u x+vy=ux+v
u , v u , v u,vu, vu,v fiind reali sau complexi, astfel ca rădăcinile ecuației transformate să fie reale. Cum o astfel de transformare nu schimbă caracterul ecuației (23) rezultă că lema este adevărată oricare ar fi a , b ; a ≠ b a , b ; a ≠ b a,b;a!=ba, b ; a \neq ba,b;a≠b.
Dacă acum A n = 0 A n = 0 A_(n)=0A_{n}=0An=0, din ecuațiile (2), (3) scoatem
(24) P n − 1 = ( x + a ) k ( x + b ) n − 1 − k (24) P n − 1 = ( x + a ) k ( x + b ) n − 1 − k {:(24)P_(n-1)=(x+a)^(k)(x+b)^(n-1-k):}\begin{equation*} P_{n-1}=(x+a)^{k}(x+b)^{n-1-k} \tag{24} \end{equation*}(24)Pn−1=(x+a)k(x+b)n−1−k
şi dacă încă A n ′ = 0 , n ′ > n A n ′ = 0 , n ′ > n A_(n^('))=0,n^(') > nA_{n^{\prime}}=0, n^{\prime}>nAn′=0,n′>n
P n ′ − 1 = ( x + a ′ ) k ′ ( x + b ′ ) n ′ − 1 − k ′ P n ′ − 1 = x + a ′ k ′ x + b ′ n ′ − 1 − k ′ P_(n^(')-1)=(x+a^('))^(k^('))(x+b^('))^(n^(')-1-k^('))P_{n^{\prime}-1}=\left(\mathrm{x}+\mathrm{a}^{\prime}\right)^{k^{\prime}}\left(\mathrm{x}+b^{\prime}\right)^{n^{\prime}-1-k^{\prime}}Pn′−1=(x+a′)k′(x+b′)n′−1−k′
Dar P n − 1 P n − 1 P_(n-1)P_{n-1}Pn−1 este o derivată a lui P n ′ − 1 P n ′ − 1 P_(n^(')-1)P_{n^{\prime}-1}Pn′−1, afară de un factor constant; deci relația precedentă e compatibilă cu (24), numai dacă k = 0 , k = 1 k = 0 , k = 1 k=0,k=1k=0, k=1k=0,k=1.
Să presupunem dar că
(25) P n = ( x + a ) k ( x + b ) n − k (25) P n = ( x + a ) k ( x + b ) n − k {:(25)P_(n)=(x+a)^(k)(x+b)^(n-k):}\begin{equation*} P_{n}=(\mathrm{x}+a)^{k}(\mathrm{x}+b)^{n-k} \tag{25} \end{equation*}(25)Pn=(x+a)k(x+b)n−k
se vede că dacă mai mulți A n A n A_(n)\mathrm{A}_{n}An sunt nuli există un n n nnn, cel puţin, pentru care p n p n p_(n)p_{n}pn e de forma (25) unde k = 0 k = 0 k=0k=0k=0 sau 1 .
Pe ecuația (15) se vede că se poate întâmpla ca un singur A n A n A_(n)A_{n}An să fie nul. Dacă
λ = − 1 n − 1 λ = − 1 n − 1 lambda=-(1)/(n-1)\lambda=-\frac{1}{n-1}λ=−1n−1
P n P n P_(n)P_{n}Pn verifică ecuatia diferentială:
( − x 2 n − 1 + μ x + v ) Y ′ + ( n n − 1 x + β − ( n − 1 ) μ ) Y = 0 − x 2 n − 1 + μ x + v Y ′ + n n − 1 x + β − ( n − 1 ) μ Y = 0 (-(x^(2))/(n-1)+mu x+v)Y^(')+((n)/(n-1)x+beta-(n-1)mu)Y=0\left(-\frac{x^{2}}{n-1}+\mu x+v\right) Y^{\prime}+\left(\frac{n}{n-1} x+\beta-(n-1) \mu\right) Y=0(−x2n−1+μx+v)Y′+(nn−1x+β−(n−1)μ)Y=0
Dar P n P n P_(n)P_{n}Pn fiind luat sub forma (25), avem
P n ′ = ( x + a ) k − 1 ( x + b ) n − 1 − k ( n x + k ( b − a ) + n a ) P n ′ = ( x + a ) k − 1 ( x + b ) n − 1 − k ( n x + k ( b − a ) + n a ) P_(n^('))=(x+a)^(k-1)(x+b)^(n-1-k)(nx+k(b-a)+na)P_{n^{\prime}}=(\mathrm{x}+a)^{k-1}(\mathrm{x}+b)^{n-1-k}(n \mathrm{x}+k(b-a)+n a)Pn′=(x+a)k−1(x+b)n−1−k(nx+k(b−a)+na)
unde presupunem k > 1 , n − k ≥ k k > 1 , n − k ≥ k k > 1,n-k >= kk>1, n-k \geq kk>1,n−k≥k.
Se vede numai decât că
μ = − a + b n − 1 , v = − a b n − 1 , β = − ( a + b ) + k ( b − a ) + n a n − 1 μ = − a + b n − 1 , v = − a b n − 1 , β = − ( a + b ) + k ( b − a ) + n a n − 1 mu=-(a+b)/(n-1),v=-(ab)/(n-1),beta=-(a+b)+(k(b-a)+na)/(n-1)\mu=-\frac{a+b}{n-1}, v=-\frac{a b}{n-1}, \beta=-(a+b)+\frac{k(b-a)+n a}{n-1}μ=−a+bn−1,v=−abn−1,β=−(a+b)+k(b−a)+nan−1
In ce priveşte funcţia generatoare suntem în cazul a) dela Nr. 5. Putem lua ω = b ω = b omega=b\omega=bω=b, şi ecuaţia în t t ttt a lui F 1 F 1 F_(1)F_{1}F1 este
(26) t d 2 F 1 d t 2 − ( n + t ) d F 1 d t + k F 1 = 0 (26) t d 2 F 1 d t 2 − ( n + t ) d F 1 d t + k F 1 = 0 {:(26)t(d^(2)F_(1))/(dt^(2))-(n+t)(dF_(1))/(dt)+kF_(1)=0:}\begin{equation*} t \frac{d^{2} F_{1}}{d t^{2}}-(n+t) \frac{d F_{1}}{d t}+k F_{1}=0 \tag{26} \end{equation*}(26)td2F1dt2−(n+t)dF1dt+kF1=0
t == ( a − b ) z t == ( a − b ) z t==(a-b)zt==(a-b) zt==(a−b)z
Ecuaţia (26) are soluţia
C ∑ i = 0 k k ( k − 1 ) … ( k − i + 1 ) n ( n − 1 ) … ( n − i + 1 ) ⋅ t i i ! C ∑ i = 0 k   k ( k − 1 ) … ( k − i + 1 ) n ( n − 1 ) … ( n − i + 1 ) ⋅ t i i ! Csum_(i=0)^(k)(k(k-1)dots(k-i+1))/(n(n-1)dots(n-i+1))*(t^(i))/(i!)C \sum_{i=0}^{k} \frac{k(k-1) \ldots(k-i+1)}{n(n-1) \ldots(n-i+1)} \cdot \frac{t^{i}}{i!}C∑i=0kk(k−1)…(k−i+1)n(n−1)…(n−i+1)⋅tii!
Tinand deci seamă de forma pe care trebue s'o aibe functia genem ratoare rezultă că
F 1 = ∑ i = 0 k k ( k − 1 ) … ( k − i + 1 ) n ( n − 1 ) … ( n − i + 1 ) ⋅ ( a − b ) 2 z i i ! F 1 = ∑ i = 0 k   k ( k − 1 ) … ( k − i + 1 ) n ( n − 1 ) … ( n − i + 1 ) ⋅ ( a − b ) 2 z i i ! F_(1)=sum_(i=0)^(k)(k(k-1)dots(k-i+1))/(n(n-1)dots(n-i+1))*((a-b)^(2)z^(i))/(i!)F_{1}=\sum_{i=0}^{k} \frac{k(k-1) \ldots(k-i+1)}{n(n-1) \ldots(n-i+1)} \cdot \frac{(a-b)^{2} z^{i}}{i!}F1=∑i=0kk(k−1)…(k−i+1)n(n−1)…(n−i+1)⋅(a−b)2zii!
Paris, 22 Octombrie 1928.
    1. Pentru legălura ce există între aceşti coeficienti și șirul (1) se poate vedea „ A A AAA. supra unor polinoame remarcabile" de T. Popoviciu (Arad 1927).
    2. Vezi Revista Universitară Matematică 1 (1), pag. 1-15,
    1. T. Popoviciu "Asupra unor polinoame remarcabile" pag. 14.

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