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
, …
P
n , …
P
0
= 1 ,
P
1
, …
P
n , …
P_(0)=1,P_(1),dotsP_(n,dots) P_{0}=1, P_{1}, \ldots P_{n, \ldots} P 0 = 1 , P 1 , … P n , …
(2)
A
n
P
n
+
B
n
P
n − 1
+
C
n
P
n − 2
= 0
(2)
A
n
P
n
+
B
n
P
n − 1
+
C
n
P
n − 2
= 0
{:(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*} (2) A n P n + B n P n − 1 + C n P n − 2 = 0
A
n
,
B
n
,
C
n
A
n
,
B
n
,
C
n
A_(n),B_(n),C_(n) \boldsymbol{A}_{n}, \boldsymbol{B}_{n}, \boldsymbol{C}_{n} A n , B n , C n finding polynomials in x degrees
0 , 1 , 2
0 , 1 , 2 0,1,2 0,1,2 0 , 1 , 2 respectively.
In particular, we studied the case when
A
n
≠ 0
A
n
≠ 0 A_(n)!=0 A_{n} \neq 0 A n ≠ 0 for anything
n
n n n n and the case
A
n
= 0
A
n
= 0 A_(n)=0 \boldsymbol{A}_{n}=0 A n = 0 for anything
n
n n n n We now propose to study the third case reported in the above-mentioned article, when some of the
A
n
A
n
Year) \boldsymbol{A}_{n} A n are zero. We observe that it is sufficient to always assume
C
n
≠ 0
C
n
≠ 0 C_(n)!=0 C_{n} \neq 0 C n ≠ 0 , because if we had a con= ditic of the form
C
n
≡ 0
C
n
≡ 0 C_(n)-=0 \boldsymbol{C}_{n} \equiv 0 C n ≡ 0 , we have a particular case of the condition
A
n
= 0
A
n
= 0 A_(n)=0 \boldsymbol{A}_{n}=0 A n = 0 . It may happen that the relation (3) is not unique; we agree to say that
A
n
≠ 0
A
n
≠ 0 A_(n)!=0 \boldsymbol{A}_{n} \neq 0 A n ≠ 0 when there is no relationship (3) where
A
n
= 0
A
n
= 0 A_(n)=0 A_{n}=0 A n = 0 We 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
A
n + 1
= 0
A
n + 1
= 0 A_(n+1)=0 A_{n+1}=0 A n + 1 = 0 It immediately follows that we can take
A
n + 2
=
A
n + 3
= … =
A
n
= … = 1
A
n + 2
=
A
n + 3
= … =
A
n
= … = 1 A_(n+2)=A_(n+3)=dots=A_(n)=dots=1 A_{n+2}=A_{n+3}=\ldots=A_{n}=\ldots=1 A n + 2 = A n + 3 = … = A n = … = 1 without losing the generality of the problem, because if
A
m
= 0
A
m
= 0 A_(m)=0 \boldsymbol{A}_{m}=0 A m = 0 or we can
1
1
^(1) { }^{1} 1 ) Second note. function
_ _ _ _
_ _ _ _ ____ \_\_\_\_ _ _ _ _
The recurrence formulas established in No. 4 of the previous article give us,
λ
2
=
λ
m
m − 1 −
λ
m
( m − 2 )
λ
2
=
λ
m
m − 1 −
λ
m
( m − 2 )
lambda_(2)=(lambda_(m))/(m-1-lambda_(m)(m-2)) \lambda_{2}=\frac{\lambda_{m}}{m-1-\lambda_{m}(m-2)} λ 2 = λ m m − 1 − λ m ( m − 2 ) whatever it may be
m ≥ 2
m ≥ 2 m >= 2 m \geq 2 m ≥ 2 . The equality immediately follows depending on
λ
n + 2
,
μ
n + 2
,
n
n + 2
,
β
n + 2
λ
n + 2
,
μ
n + 2
,
n
n + 2
,
β
n + 2
lambda_(n+2),mu_(n+2),nu_(n+2),beta_(n+2) \lambda_{n+2}, \mu_{n+2}, \nu_{n+2}, \beta_{n+2} λ n + 2 , μ n + 2 , n n + 2 , β n + 2 , which is reduced
1 +
λ
n + 2
+
A
n + 2
= 0
1 +
λ
n + 2
+
A
n + 2
= 0 1+lambda_(n+2)+a_(n+2)=0 1+\lambda_{n+2}+a_{n+2}=0 1 + λ n + 2 + A n + 2 = 0 immediately results in equality
λ
n + 2
n + 1 − n
λ
n + 2
=
λ
m
m − 1 − ( m − 2 )
λ
n
λ
n + 2
n + 1 − n
λ
n + 2
=
λ
m
m − 1 − ( m − 2 )
λ
n
(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}} λ n + 2 n + 1 − n λ n + 2 = λ m m − 1 − ( m − 2 ) λ n from which
(4)
λ
m
=
( m − 1 )
λ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
(4)
λ
m
=
( m − 1 )
λ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
{:(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*} (4) λ m = ( m − 1 ) λ n + 2 n + 1 + ( m − n − 2 ) λ n + 2 FROM
λ
m
λ ^
n + 2
=
μ
m
μ
n
+ 2
=
V
m
V
n + 2
λ
m
λ
^
n + 2
=
μ
m
μ
n
+ 2
=
V
m
V
n + 2
(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}} λ m λ ^ n + 2 = μ m μ n + 2 = V m V n + 2 deduce
(5)
μ
m
=
( m − 1 )
μ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
,
V
m
=
( m − 1 )
V
ε + 2
n + 1 + ( m − n − 2 )
λ
n + 2
(5)
μ
m
=
( m − 1 )
μ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
,
V
m
=
( m − 1 )
V
ε + 2
n + 1 + ( m − n − 2 )
λ
n + 2
{:(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*} (5) μ m = ( m − 1 ) μ n + 2 n + 1 + ( m − n − 2 ) λ n + 2 , V m = ( m − 1 ) V ε + 2 n + 1 + ( m − n − 2 ) λ n + 2 (6)
A
m
= −
n + 1 + ( 2 m − n − 3 )
λ
n
+ 2
n + 1 + ( m − n − 2 )
λ
n
+ 2
A
m
= −
n + 1 + ( 2 m − n − 3 )
λ
n
+ 2
n + 1 + ( m − n − 2 )
λ
n
+ 2
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} A m = − n + 1 + ( 2 m − n − 3 ) λ n + 2 n + 1 + ( m − n − 2 ) λ n + 2 We then have the recurrence relation:
β
m
[ ( n + 1 ) + ( m − n − 2 )
λ
n + 2
]
−
β
m − 1
| n + 1 + ( m − n − 3 )
λ
n + 2
|
+
μ
n + 2
= 0
β
m
( n + 1 ) + ( m − n − 2 )
λ
n + 2
−
β
m − 1
n + 1 + ( m − n − 3 )
λ
n + 2
+
μ
n + 2
= 0 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}=0 β m [ ( n + 1 ) + ( m − n − 2 ) λ n + 2 ] − β m − 1 | n + 1 + ( m − n − 3 ) λ n + 2 | + μ n + 2 = 0 from where
(7)
β
m
=
( n + 1 )
β
n + 2
− ( m − n − 2 )
μ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
(7)
β
m
=
( n + 1 )
β
n + 2
− ( m − n − 2 )
μ
n + 2
n + 1 + ( m − n − 2 )
λ
n + 2
{:(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*} (7) β m = ( n + 1 ) β n + 2 − ( m − n − 2 ) μ n + 2 n + 1 + ( m − n − 2 ) λ n + 2 To completely resolve the problem, it is necessary to determine
λ
m
,
μ
m
λ
m
,
μ
m
lambda_(m),mu_(m) \lambda_{m}, \mu_{m} λ m , μ m ,
V
m
,
p
m
,
α
m
V
m
,
p
m
,
α
m
v_(m),p_(m),alpha_(m) v_{m}, p_{m}, \alpha_{m} V m , p m , α m for
m ≤ n + 2
m ≤ n + 2 m <= n+2 m \leq n+2 m ≤ n + 2 . But I saw that
P
n
= (
x
+ A
)
k
(
x
+ b
)
n − k
P
n
= (
x
+ A
)
k
(
x
+ b
)
n − k
P_(n)=(x+a)^(k)(x+b)^(nk) P_{n}=(\mathrm{x}+a)^{k}(\mathrm{x}+b)^{n-k} P n = ( x + A ) k ( x + b ) n − k We can write for simplification
P
n
=
r
n
−
k
(
x
+
b
1
)
k
P
n
=
r
n
−
k
x
+
b
1
k
P_(n)=r^(n-k)(x+b_(1))^(k) P_{n}=\mathrm{r}^{n-k}\left(\mathrm{x}+b_{1}\right)^{k} P n = R n − k ( x + b 1 ) k and the general result will be obtained by substituting for
x
,
b
1
,
pc
x
+
b
,
a
−
b
x
,
b
1
,
pc
x
+
b
,
a
−
b
x,b_(1),pcx+b,a-b x, b_{1}, \mathrm{pc} x+b, a-b x , b 1 , pc x + b , A − b res= spectively. Nothing of generality is thus lost, for
d
d
(
x
+
b
)
=
d
d
x
d
d
(
x
+
b
)
=
d
d
x
(d)/(d(x+b))=(d)/(dx) \frac{d}{d(x+b)}=\frac{d}{d x} d d ( x + b ) = d d x the character of the polynomials is therefore maintained by this simplification
3. Let
k
=
0
k
=
0
k=0 k=0 k = 0 , so
P
n
=
x
n
P
n
=
x
n
P_(n)=x^(n) P_{n}=\mathrm{x}^{n} P n = x n polynomials
P
i
,
i
<
n
P
i
,
i
<
n
P_(i,)i < n \boldsymbol{P}_{i,} \boldsymbol{i}<\boldsymbol{n} P and , and < n , are determined, the constants
λ
m
,
μ
m
,
ν
m
,
α
m
,
β
m
cu
=
λ
m
,
μ
m
,
ν
m
,
α
m
,
β
m
cu
=
lambda_(m),mu_(m),nu_(m),alpha_(m),beta_(m)cu= \lambda_{m}, \mu_{m}, \nu_{m}, \alpha_{m}, \beta_{m} \mathrm{cu}= λ m , μ m , n m , α m , β m with = known for
m
≤
n
+
1
m
≤
n
+
1
m <= n+1 m \leq n+1 m ≤ n + 1 . Let's put
P
n
+
1
=
x
n
+
1
+
c
,
P
n
+
2
=
x
n
+
2
+
(
n
+
2
)
c
x
+
c
1
P
n
+
1
=
x
n
+
1
+
c
,
P
n
+
2
=
x
n
+
2
+
(
n
+
2
)
c
x
+
c
1
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} P n + 1 = x n + 1 + c , P n + 2 = x n + 2 + ( n + 2 ) c x + c 1 c.
c
1
c
1
c_(1) \boldsymbol{c}_{1} c 1 being two constants. It is seen that to have an interesting case, irreducible to those studied so far, we must assume
c
≠
0
c
≠
0
c!=0 c \neq 0 c ≠ 0 . I have then
(
n
+
2
)
c
x
+
c
1
+
c
B
n
+
2
=
0
x
2
+
x
B
n
+
2
+
C
+
2
=
0
(
n
+
2
)
c
x
+
c
1
+
c
B
n
+
2
=
0
x
2
+
x
B
n
+
2
+
C
+
2
=
0
{:[(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} ( n + 2 ) c x + c 1 + c B n + 2 = 0 x 2 + x B n + 2 + C + 2 = 0 and it is removed immediately
α
n
+
2
=
−
(
n
+
2
)
λ
n
+
2
=
n
+
1
β
n
+
2
=
−
c
1
c
=
σ
μ
n
+
2
=
−
σ
v
n
+
2
=
0
α
n
+
2
=
−
(
n
+
2
)
λ
n
+
2
=
n
+
1
β
n
+
2
=
−
c
1
c
=
σ
μ
n
+
2
=
−
σ
v
n
+
2
=
0
{:[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} α n + 2 = − ( n + 2 ) λ n + 2 = n + 1 β n + 2 = − c 1 c = σ μ n + 2 = − σ V n + 2 = 0 Formulas (4), (5), (6), (7) now give us
pc
λ
m
,
μ
m
,
ν
m
,
α
m
,
β
m
pc
λ
m
,
μ
m
,
ν
m
,
α
m
,
β
m
pclambda_(m),mu_(m),nu_(m),alpha_(m),beta_(m) \mathrm{pc} \lambda_{m}, \mu_{m}, \nu_{m}, \alpha_{m}, \beta_{m} pc λ m , μ m , n m , α m , β m and all relations (3) are determined. We can write in the above case
P
m
=
x
m
+
Q
m
−
n
−
1
P
m
=
x
m
+
Q
m
−
n
−
1
P_(m)=x^(m)+Q_(m-n-1) \boldsymbol{P}_{m}=\mathbf{x}^{m}+Q_{m-n-1} P m = x m + Q m − n − 1 where the sequence of polynomials
(8)
Q
0
=
c
,
Q
1
,
Q
2
,
…
Q
s
…
(8)
Q
0
=
c
,
Q
1
,
Q
2
,
…
Q
s
…
{:(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*} (8) Q 0 = c , Q 1 , Q 2 , … Q S … check the relationships
(9)
d
Q
m
−
n
−
1
d
x
=
m
Q
m
−
n
−
2
Q
m
−
n
−
1
+
B
m
Q
m
−
n
−
2
+
C
m
Q
m
−
n
−
3
=
0
(9)
d
Q
m
−
n
−
1
d
x
=
m
Q
m
−
n
−
2
Q
m
−
n
−
1
+
B
m
Q
m
−
n
−
2
+
C
m
Q
m
−
n
−
3
=
0
{:[(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*} (9) d Q m − n − 1 d x = m Q m − n − 2 Q m − n − 1 + B m Q m − n − 2 + C m Q m − n − 3 = 0 So the sequence of polynomials:
(10)
R
0
=
c
,
R
1
,
…
R
s
,
…
(10)
R
0
=
c
,
R
1
,
…
R
s
,
…
{:(10)R_(0)=c","R_(1)","dotsR_(s)","dots:} \begin{equation*}
R_{0}=c, R_{1}, \ldots R_{s}, \ldots \tag{10}
\end{equation*} (10) R 0 = c , R 1 , … R S , … where
Q
s
=
(
n
+
s
+
1
s
)
R
s
Q
s
=
(
n
+
s
+
1
s
)
R
s
Q_(s)=((n+s+1)/(s))R_(s) Q_{s}=\binom{n+s+1}{s} R_{s} Q S = ( n + S + 1 S ) R S is an APPELL string checking the condition:
(11)
(
n
+
s
+
1
s
)
R
s
+
(
n
+
s
s
−
1
)
B
n
+
s
+
1
R
s
−
1
(
n
+
s
−
1
s
−
2
)
(
n
+
s
+
1
s
)
R
s
+
(
n
+
s
s
−
1
)
B
n
+
s
+
1
R
s
−
1
(
n
+
s
−
1
s
−
2
)
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} ( n + S + 1 S ) R S + ( n + S S − 1 ) B n + S + 1 R S − 1 ( n + S − 1 S − 2 )
C
n
+
s
+
1
R
s
−
2
=
0
C
n
+
s
+
1
R
s
−
2
=
0
C_(n+s+1)R_(s-2)=0 \boldsymbol{C}_{n+s+1} \boldsymbol{R}_{s-2}=0 C n + S + 1 R S − 2 = 0 And polynomials
R
s
R
s
R_(s) \boldsymbol{R}_{s} R S falls into the case studied in the previous chapter under No. 4.
But we have
B
n
+
s
+
1
=
−
n
+
2
s
s
×
+
σ
(
n
+
s
)
s
(
n
+
1
)
C
n
+
s
+
1
=
n
+
s
s
x
2
−
σ
(
n
+
s
)
s
(
n
+
1
)
×
B
n
+
s
+
1
=
−
n
+
2
s
s
×
+
σ
(
n
+
s
)
s
(
n
+
1
)
C
n
+
s
+
1
=
n
+
s
s
x
2
−
σ
(
n
+
s
)
s
(
n
+
1
)
×
{:[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} B n + S + 1 = − n + 2 S S × + σ ( n + S ) S ( n + 1 ) C n + S + 1 = n + S S x 2 − σ ( n + S ) S ( n + 1 ) ×
(12)
(
n
+
s
+
1
)
R
s
+
[
−
(
n
+
2
s
)
×
+
σ
(
n
+
s
)
n
+
1
]
R
s
−
1
+
(
s
−
1
)
(
x
2
−
σ
n
+
1
×
)
R
s
−
2
=
0
(12)
(
n
+
s
+
1
)
R
s
+
−
(
n
+
2
s
)
×
+
σ
(
n
+
s
)
n
+
1
R
s
−
1
+
(
s
−
1
)
x
2
−
σ
n
+
1
×
R
s
−
2
=
0
{:[(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*} (12) ( n + S + 1 ) R S + [ − ( n + 2 S ) × + σ ( n + S ) n + 1 ] R S − 1 + ( S − 1 ) ( x 2 − σ n + 1 × ) R S − 2 = 0 So we enter the general case by taking:
λ
=
1
n
+
3
,
μ
=
−
σ
(
n
+
1
)
(
n
+
3
)
,
ν
=
0
,
β
=
σ
(
n
+
2
)
(
n
+
1
)
(
n
+
3
)
λ
=
1
n
+
3
,
μ
=
−
σ
(
n
+
1
)
(
n
+
3
)
,
ν
=
0
,
β
=
σ
(
n
+
2
)
(
n
+
1
)
(
n
+
3
)
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)} λ = 1 n + 3 , μ = − σ ( n + 1 ) ( n + 3 ) , n = 0 , β = σ ( n + 2 ) ( n + 1 ) ( n + 3 ) It is then obtained
⋃
s
=
0
∞
7
s
s
!
R
s
=
c
⋅
e
α
x
G
.
(
n
+
1
,
n
+
2
,
−
σ
n
+
1
z
)
⋃
s
=
0
∞
7
s
s
!
R
s
=
c
⋅
e
α
x
G
.
n
+
1
,
n
+
2
,
−
σ
n
+
1
z
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) ⋃ S = 0 ∞ 7 S S ! R S = c ⋅ it is α x G . ( n + 1 , n + 2 , − σ n + 1 z ) for
R
0
=
c
R
0
=
c
R_(0)=c \boldsymbol{R}_{0}=\boldsymbol{c} R 0 = c ,
∑
s
=
0
∞
z
n
+
s
+
1
(
n
+
s
+
1
)
!
Q
s
=
z
n
+
1
(
n
+
1
)
!
∑
0
∞
z
s
s
!
R
s
∑
s
=
0
∞
z
n
+
s
+
1
(
n
+
s
+
1
)
!
Q
s
=
z
n
+
1
(
n
+
1
)
!
∑
0
∞
z
s
s
!
R
s
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} ∑ S = 0 ∞ z n + S + 1 ( n + S + 1 ) ! Q S = z n + 1 ( n + 1 ) ! ∑ 0 ∞ z S S ! R S and so the generating function is
∑
m
=
0
∞
z
m
m
!
P
m
=
e
2
x
[
1
+
c
z
n
+
1
(
n
+
1
)
!
G
(
n
+
1
,
n
+
2
,
−
σ
n
+
1
z
)
]
∑
m
=
0
∞
z
m
m
!
P
m
=
e
2
x
1
+
c
z
n
+
1
(
n
+
1
)
!
G
n
+
1
,
n
+
2
,
−
σ
n
+
1
z
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] ∑ m = 0 ∞ z m m ! P m = it is 2 x [ 1 + c z n + 1 ( n + 1 ) ! G ( n + 1 , n + 2 , − σ n + 1 z ) ] But it can be said for the general case
P
n
=
(
x
+
b
)
n
P
n
=
(
x
+
b
)
n
P_(n)=(x+b)^(n) P_{n}=(\mathrm{x}+b)^{n} P n = ( 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
,
θ
z
)
]
e
ℓ
(
x
+
b
)
1
+
c
z
n
+
1
(
n
+
1
)
!
G
(
n
+
1
,
n
+
2
,
θ
z
)
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] it is ℓ ( x + b ) [ 1 + c z n + 1 ( n + 1 ) ! G ( n + 1 , n + 2 , θ z ) ]
b
,
c
,
d
b
,
c
,
d
b,c,d b, c, d b , c , d being arbitrary constants.
The previous result was obtained under the assumption
σ
≠
0
σ
≠
0
sigma!=0 \sigma \neq 0 σ ≠ 0 If
σ
=
0
σ
=
0
sigma=0 \sigma=0 σ = 0 The 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
)
1
+
c
z
n
+
1
(
n
+
1
)
!
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] it is z ( x + b ) [ 1 + c z n + 1 ( n + 1 ) ! ]
Let's move on to the case.
k
=
1
k
=
1
k=1 k=1 k = 1 , which is treated in the same way.
P
n
=
x
n
−
19
(
x
+
b
1
)
P
n
=
x
n
−
19
x
+
b
1
P_(n)=x^(n-19)(x+b_(1)) P_{n}=\mathrm{x}^{n-19}\left(\mathrm{x}+b_{1}\right) P n = x n − 19 ( x + b 1 )
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
)
c
x
+
c
1
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
)
c
x
+
c
1
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} 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 ) c x + c 1 and
c
≠
0
c
≠
0
c!=0 \boldsymbol{c} \neq 0 c ≠ 0 . It is found that
α
n
+
2
=
−
(
n
+
2
)
β
n
+
2
=
−
c
1
c
=
−
n
b
1
α
n
+
2
=
−
(
n
+
2
)
β
n
+
2
=
−
c
1
c
=
−
n
b
1
{:[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} α n + 2 = − ( n + 2 ) β n + 2 = − c 1 c = − n b 1 We can also write here
P
m
=
x
m
−
1
(
x
+
m
n
b
1
)
+
Q
m
−
n
−
1
P
m
=
x
m
−
1
x
+
m
n
b
1
+
Q
m
−
n
−
1
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} P m = x m − 1 ( x + m n b 1 ) + Q m − n − 1 and polynomials
Q
s
Q
s
Q_(s) Q_{s} Q S verify the relation (9). The polynomials are deduced
R
R
R \boldsymbol{R} R , with relation (11), which is written
(
n
+
s
+
1
)
R
s
+
[
−
(
n
+
2
s
)
×
+
(
1
−
n
−
s
)
b
1
]
R
s
−
1
+
(
s
−
1
)
(
x
2
+
b
x
)
R
s
−
2
=
0
(
n
+
s
+
1
)
R
s
+
−
(
n
+
2
s
)
×
+
(
1
−
n
−
s
)
b
1
R
s
−
1
+
(
s
−
1
)
x
2
+
b
x
R
s
−
2
=
0
{:[(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} ( n + S + 1 ) R S + [ − ( n + 2 S ) × + ( 1 − n − S ) b 1 ] R S − 1 + ( S − 1 ) ( x 2 + b x ) R S − 2 = 0 and it can be taken
λ
=
1
n
+
3
,
μ
=
b
1
n
+
3
,
ν
=
0
,
β
=
−
n
+
1
n
+
3
b
1
λ
=
1
n
+
3
,
μ
=
b
1
n
+
3
,
ν
=
0
,
β
=
−
n
+
1
n
+
3
b
1
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} λ = 1 n + 3 , μ = b 1 n + 3 , n = 0 , β = − n + 1 n + 3 b 1 and the generating function is:
∑
0
∞
z
s
s
!
R
s
=
c
⋅
e
2
x
G
(
n
,
n
+
2
,
b
1
z
)
∑
0
∞
z
s
s
!
R
s
=
c
⋅
e
2
x
G
n
,
n
+
2
,
b
1
z
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) ∑ 0 ∞ z S S ! R S = c ⋅ it is 2 x G ( n , n + 2 , b 1 z ) For the general case we have immediately:
∑
0
∞
z
m
m
!
P
m
=
e
z
(
x
+
b
)
[
1
+
2
n
(
a
−
b
)
z
+
c
G
(
n
,
n
+
2
(
a
−
b
)
z
)
]
∑
0
∞
z
m
m
!
P
m
=
e
z
(
x
+
b
)
1
+
2
n
(
a
−
b
)
z
+
c
G
(
n
,
n
+
2
(
a
−
b
)
z
)
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] ∑ 0 ∞ z m m ! P m = it is z ( x + b ) [ 1 + 2 n ( A − b ) z + c G ( n , n + 2 ( A − b ) z ) ]
case
k
>
1
k
>
1
k > 1 k>1 k > 1 , that is
P
n
=
x
n
−
k
(
x
+
b
1
)
k
P
n
=
x
n
−
k
x
+
b
1
k
P_(n)=x^(n-k)(x+b_(1))^(k) P_{n}=\mathrm{x}^{n-k}\left(\mathrm{x}+b_{1}\right)^{k} P n = x n − k ( x + b 1 ) k
n
−
k
≥
k
,
k
>
1
,
n
−
k
≥
k
,
k
>
1
,
n-k >= k,quad k > 1, n-k \geq k, \quad k>1, n − k ≥ k , k > 1 , It can be attached to the general case by taking
λ
=
−
1
n
−
1
λ
=
−
1
n
−
1
lambda=-(1)/(n-1) \lambda=-\frac{1}{n-1} λ = − 1 n − 1 as we saw in the previous article.
We found:
λ
=
−
1
n
−
1
,
α
=
−
n
−
2
n
−
1
,
μ
=
−
b
1
n
−
1
,
ν
=
0
,
β
=
−
(
k
−
1
)
b
1
n
−
1
λ
=
−
1
n
−
1
,
α
=
−
n
−
2
n
−
1
,
μ
=
−
b
1
n
−
1
,
ν
=
0
,
β
=
−
(
k
−
1
)
b
1
n
−
1
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} λ = − 1 n − 1 , α = − n − 2 n − 1 , μ = − b 1 n − 1 , n = 0 , β = − ( k − 1 ) b 1 n − 1 The equation that gives the generating function is written:
we put
z
F
′
′
−
[
z
b
1
+
n
]
F
′
+
k
b
1
F
=
0
z
F
′
′
−
z
b
1
+
n
F
′
+
k
b
1
F
=
0
zF^('')-[zb_(1)+n]F^(')+kb_(1)F=0 z F^{\prime \prime}-\left[z b_{1}+n\right] F^{\prime}+k b_{1} F=0 z F ′ ′ − [ z b 1 + n ] F ′ + k b 1 F = 0
t
F
′
′
t
2
−
[
t
+
n
]
F
′
1
+
k
F
=
0
t
F
′
′
t
2
−
[
t
+
n
]
F
′
1
+
k
F
=
0
tF^('')_(t^(2))-[t+n]F^(')_(1)+kF=0 t F^{\prime \prime}{ }_{t^{2}}-[t+n] F^{\prime}{ }_{1}+k F=0 t F ′ ′ t 2 − [ t + n ] F ′ 1 + k F = 0 All integrals of this equation are holomorphic about the origin. Int'a= indeed we have a polynomial
∑
i
=
0
k
k
(
k
−
1
)
…
k
−
i
+
1
)
i
!
n
(
n
−
1
)
…
(
n
−
i
+
1
)
t
i
=
P
1
(
t
)
∑
i
=
0
k
k
(
k
−
1
)
…
k
−
i
+
1
i
!
n
(
n
−
1
)
…
(
n
−
i
+
1
)
t
i
=
P
1
(
t
)
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) ∑ and = 0 k k ( k − 1 ) … k − and + 1 ) and ! n ( n − 1 ) … ( n − and + 1 ) t and = P 1 ( t ) which satisfies the equation. Putting
F
=
e
λ
φ
F
=
e
λ
φ
F=e^(lambda)varphi F=e^{\lambda} \varphi F = it is λ φ HAVE
t
φ
′
′
−
(
n
−
t
)
φ
′
−
(
n
−
k
)
φ
=
0
t
φ
′
′
−
(
n
−
t
)
φ
′
−
(
n
−
k
)
φ
=
0
tvarphi^('')-(n-t)varphi^(')-(n-k)varphi=0 t \varphi^{\prime \prime}-(n-t) \varphi^{\prime}-(n-k) \varphi=0 t φ ′ ′ − ( n − t ) φ ′ − ( n − k ) φ = 0 or changing to
t
t
t t t in -
t
t
t t t :
t
φ
′
′
−
(
n
+
t
)
φ
′
+
(
n
−
k
)
φ
=
0
t
φ
′
′
−
(
n
+
t
)
φ
′
+
(
n
−
k
)
φ
=
0
tvarphi^('')-(n+t)varphi^(')+(n-k)varphi=0 t \varphi^{\prime \prime}-(n+t) \varphi^{\prime}+(n-k) \varphi=0 t φ ′ ′ − ( n + t ) φ ′ + ( n − k ) φ = 0 So we have the solution distinct from the previous one:
e
t
∑
i
=
0
n
−
k
(
−
1
)
i
(
n
−
k
)
(
n
−
k
−
1
)
…
(
n
−
k
−
i
+
1
)
i
!
n
(
n
−
1
)
…
(
n
−
i
+
1
)
i
i
=
e
t
P
2
(
t
)
e
t
∑
i
=
0
n
−
k
(
−
1
)
i
(
n
−
k
)
(
n
−
k
−
1
)
…
(
n
−
k
−
i
+
1
)
i
!
n
(
n
−
1
)
…
(
n
−
i
+
1
)
i
i
=
e
t
P
2
(
t
)
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) it is t ∑ and = 0 n − k ( − 1 ) and ( n − k ) ( n − k − 1 ) … ( n − k − and + 1 ) and ! n ( n − 1 ) … ( n − and + 1 ) and and = it is t P 2 ( t )
So there is
∑
n
=
0
∞
z
n
n
!
P
n
=
e
ι
x
[
c
P
1
(
b
1
z
)
+
c
1
e
z
b
1
P
2
(
b
2
z
)
]
∑
n
=
0
∞
z
n
n
!
P
n
=
e
ι
x
c
P
1
b
1
z
+
c
1
e
z
b
1
P
2
b
2
z
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] ∑ n = 0 ∞ z n n ! P n = it is I x [ c P 1 ( b 1 z ) + c 1 it is z b 1 P 2 ( b 2 z ) ] constants
c
,
c
1
c
,
c
1
c,c_(1) \boldsymbol{c}, \boldsymbol{c}_{1} c , c 1 being linked by the relationship
c
+
c
1
=
1
c
+
c
1
=
1
c+c_(1)=1 c+c_{1}=1 c + c 1 = 1 In general we will have the polynomials
∑
n
=
0
∞
z
n
n
!
P
n
=
e
2
[
x
+
b
]
[
c
P
1
[
(
a
−
b
)
z
]
+
c
1
e
[
a
−
b
]
z
P
2
[
(
a
−
b
)
z
)
]
∑
n
=
0
∞
z
n
n
!
P
n
=
e
2
[
x
+
b
]
c
P
1
[
(
a
−
b
)
z
]
+
c
1
e
[
a
−
b
]
z
P
2
[
(
a
−
b
)
z
)
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] ∑ n = 0 ∞ z n n ! P n = it is 2 [ x + b ] [ c P 1 [ ( A − b ) z ] + c 1 it is [ A − b ] z P 2 [ ( A − b ) z ) ]
6. If a
B
n
B
n
B_(n) B_{n} B n it is identical we must have
λ
=
−
1
2
n
−
3
λ
=
−
1
2
n
−
3
lambda=-(1)/(2n-3) \lambda=-\frac{1}{2 n-3} λ = − 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.
