NOTES ON BINOMIAL POLYNOMIES by
T. Popoviciu Former student of the École Normale Supérieure.
Admitted on April 28, 1981.
We say that the sequence of polynomials in
x
x x x x
(1)
P
0
,
P
1
, …
P
n
, …
P
0
,
P
1
, …
P
n
, … P_(0),P_(1),dotsP_(n),dots \mathrm{P}_{0}, \mathrm{P}_{1}, \ldots \mathrm{P}_{n}, \ldots P 0 , P 1 , … P n , … the degree of each being equal to the corresponding index, is a binomial sequence if it satisfies the equations
(2)
P
n
( x + y ) =
∑
i = 0
n
P
i
( x )
P
n − i
( y )
(2)
P
n
( x + y ) =
∑
i = 0
n
P
i
( x )
P
n − i
( y )
{:(2)P_(n)(x+y)=sum_(i=0)^(n)P_(i)(x)P_(ni)(y):} \begin{equation*} P_{n}(x+y)=\sum_{i=0}^{n} P_{i}(x) P_{ni}(y) \tag{2} \end{equation*} (2) P n ( x + y ) = ∑ i = 0 n P i ( x ) P n − i ( y ) For
n = 0 , 1 , 2 , …
n = 0 , 1 , 2 , … n=0,1,2,dots n=0,1,2, \ldots n = 0 , 1 , 2 , … and similarly in
x
x x x x And
y
y y y y We then have
P
0
= 1 ,
P
n
( 0 ) = 0 ,
n = 1 , 2 , …
P
0
= 1 ,
P
n
( 0 ) = 0 ,
n = 1 , 2 , … P_(0)=1,quadP_(n)(0)=0,quad n=1,2,dots P_{0}=1, \quad P_{n}(0)=0, \quad n=1,2, \ldots P 0 = 1 , P n ( 0 ) = 0 , n = 1 , 2 , … Binomial sequences are obtained through the following formal expansion
(3)
( 1 +
has
1
z +
has
2
z
2
+ ⋯ )
x
=
e
x
(
c
1
z +
c
2
z
2
+ ⋯ )
=
∑
n = 0
∞
P
n
( x )
z
π
(3)
1 +
has
1
z +
has
2
z
2
+ ⋯
x
=
e
x
c
1
z +
c
2
z
2
+ ⋯
=
∑
n = 0
∞
P
n
( x )
z
π
{:(3)(1+a_(1)z+a_(2)z^(2)+cdots)^(x)=e^(x(c_(1)z+c_(2)z^(2)+cdots))=sum_(n=0)^(oo)P_(n)(x)z^(pi):} \begin{equation*} \left(1+a_{1} z+a_{2} z^{2}+\cdots\right)^{x}=e^{x\left(c_{1} z+c_{2} z^{2}+\cdots\right)}=\sum_{n=0}^{\infty} P_{n}(x) z^{\pi} \tag{3} \end{equation*} (3) ( 1 + has 1 z + has 2 z 2 + ⋯ ) x = e x ( c 1 z + c 2 z 2 + ⋯ ) = ∑ n = 0 ∞ P n ( x ) z π It is easy to find the coefficients
c
n
c
n
c_(n) c_{n} c n depending on the
has
n
has
n
year) year} has n and vice versa.
If
c
n
≥ 0 ,
n = 1 , 2 , 3 , …
c
n
≥ 0 ,
n = 1 , 2 , 3 , … c_(n) >= 0,quad n=1,2,3,dots c_{n} \geq 0, \quad n=1,2,3, \ldots c n ≥ 0 , n = 1 , 2 , 3 , … we also have
and if
has
n
≥ 0 ,
n = 1 , 2 , 3 , …
has
n
≥ 0 ,
n = 1 , 2 , 3 , … a_(n) >= 0,quad n=1,2,3,dots a_{n} \geq 0, \quad n=1,2,3, \ldots has n ≥ 0 , n = 1 , 2 , 3 , … we also have
c
n
≥ 0 ,
n = 1 , 2 , 3 , …
has
1
> 0
has
n
> 0 ,
n = 2 , 3 , …
c
n
≥ 0 ,
n = 1 , 2 , 3 , …
has
1
> 0
has
n
> 0 ,
n = 2 , 3 , …
{:[c_(n) >= 0",",n=1","2","3","dotsa_(1) > 0],[a_(n) > 0",",n=2","3","dots]:} \begin{array}{ll} c_{n} \geq 0, & n=1,2,3, \ldots a_{1}>0 \\ a_{n}>0, & n=2,3, \ldots \end{array} c n ≥ 0 , n = 1 , 2 , 3 , … has 1 > 0 has n > 0 , n = 2 , 3 , … We easily find that
a is a positive number so that we have
P
n
( x ) ≥ 0
P
n
( x ) ≥ 0 P_(n)(x) >= 0 P_{n}(x) \geq 0 P n ( x ) ≥ 0
n = 0 , 1.9 , …
n = 0 , 1.9 , … n=0,1.9,dots n=0,1.9, \ldots n = 0 , 1.9 , … in the interval (0, a) it is necessary and sufficient that
(5)
c
n
≥ 0 ,
n = 1 , 2 , …
(5)
c
n
≥ 0 ,
n = 1 , 2 , …
{:(5)c_(n) >= 0","quad n=1","2","dots:} \begin{equation*}
c_{n} \geq 0, \quad n=1,2, \ldots \tag{5}
\end{equation*} (5) c n ≥ 0 , n = 1 , 2 , … It also follows from (2) that inequalities (4) are verified throughout the interval
(
0
,
∞
)
(
0
,
∞
)
(0,oo) (0, \infty) ( 0 , ∞ ) . Either
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) a uniform and continuous function in the interval Cermé
(
0
,
1
)
(
0
,
1
)
(0,1) (0,1) ( 0 , 1 ) . Given that inequalities (5) are verified and assuming that
a
1
>
0
a
1
>
0
a_(1) > 0 a_{1}>0 has 1 > 0 so that the polynomial
μ
n
=
∑
ν
=
0
n
f
(
ν
n
)
P
ν
(
x
)
P
n
−
ν
(
1
−
x
)
a
n
μ
n
=
∑
ν
=
0
n
f
ν
n
P
ν
(
x
)
P
n
−
ν
(
1
−
x
)
a
n
mu_(n)=(sum_(nu=0)^(n)f((nu )/(n))P_(nu)(x)P_(n-nu)(1-x))/(a_(n)) \mu_{n}=\frac{\sum_{\nu=0}^{n} f\left(\frac{\nu}{n}\right) P_{\nu}(x) P_{n-\nu}(1-x)}{a_{n}} μ n = ∑ ν = 0 n f ( ν n ) P ν ( x ) P n − ν ( 1 − x ) has n converges uniformly towards
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) throughout the interval
(
0
,
1
)
(
0
,
1
)
(0,1) (0,1) ( 0 , 1 ) It is necessary and sufficient that:
(6)
lim
n
→
∞
∑
ν
=
0
n
v
k
P
ν
(
x
)
P
n
−
ν
(
1
−
x
)
n
k
a
n
=
x
k
(6)
lim
n
→
∞
∑
ν
=
0
n
v
k
P
ν
(
x
)
P
n
−
ν
(
1
−
x
)
n
k
a
n
=
x
k
{:(6)lim_(n rarr oo)(sum_(nu=0)^(n)v^(k)P_(nu)(x)P_(n-nu)(1-x))/(n^(k)a_(n))=x_(k):} \begin{equation*}
\lim _{n \rightarrow \infty} \frac{\sum_{\nu=0}^{n} v^{k} P_{\nu}(x) P_{n-\nu}(1-x)}{n^{k} a_{n}}=x_{k} \tag{6}
\end{equation*} (6) limit n → ∞ ∑ ν = 0 n v k P ν ( x ) P n − ν ( 1 − x ) n k has n = x k whatever the positive integer
k
k
k k k 2.
Let us consider in particular the polynomials (1) arising from the expansion
e
x
z
1
−
z
=
Σ
L
n
′
(
x
)
⋅
z
n
e
x
z
1
−
z
=
Σ
L
n
′
(
x
)
⋅
z
n
e^((xz)/(1-z))=SigmaL_(n)^(')(x)*z^(n) e^{\frac{x z}{1-z}}=\Sigma L_{n}^{\prime}(x) \cdot z^{n} e x z 1 − z = Σ L n ′ ( x ) ⋅ z n If
L
∗
n
(
x
)
L
∗
n
(
x
)
L^(**)_(n)(x) L^{*}{ }_{n}(x) L * n ( x ) is the Laguerre polynomial of degree
n
n
n n n we have
L
n
(
x
)
=
L
n
∗
(
−
x
)
−
L
n
−
1
∗
(
−
x
)
L
n
(
x
)
=
L
n
∗
(
−
x
)
−
L
n
−
1
∗
(
−
x
)
L_(n)(x)=L_(n)^(**)(-x)-L_(n-1)^(**)(-x) \mathrm{L}_{n}(x)=\mathrm{L}_{n}^{*}(-x)-\mathrm{L}_{n-1}^{*}(-x) L n ( x ) = L n * ( − x ) − L n − 1 * ( − x ) We obtain easily
L
n
(
x
)
=
e
−
x
x
n
!
⋅
d
n
d
x
n
x
n
−
1
e
x
=
∑
ν
=
1
n
(
n
−
1
v
−
1
)
x
ν
v
!
L
n
(
x
)
=
e
−
x
x
n
!
⋅
d
n
d
x
n
x
n
−
1
e
x
=
∑
ν
=
1
n
(
n
−
1
v
−
1
)
x
ν
v
!
L_(n)(x)=(e^(-x)x)/(n!)*(d^(n))/(dx^(n))x^(n-1)e^(x)=sum_(nu=1)^(n)((n-1)/(v-1))(x^(nu))/(v!) L_{n}(x)=\frac{e^{-x} x}{n!} \cdot \frac{d^{n}}{d x^{n}} x^{n-1} e^{x}=\sum_{\nu=1}^{n}\binom{n-1}{v-1} \frac{x^{\nu}}{v!} L n ( x ) = e − x x n ! ⋅ d n d x n x n − 1 e x = ∑ ν = 1 n ( n − 1 v − 1 ) x ν v ! Let us now return to the general case and suppose that the series
∑
n
=
1
∞
c
n
z
n
∑
n
=
1
∞
c
n
z
n
sum_(n=1)^(oo)c_(n)z^(n) \sum_{n=1}^{\infty} c_{n} z^{n} ∑ n = 1 ∞ c n z n converges inside a circle of radius
r
′
r
′
r^(') r^{\prime} r ′ and from the center of origin and let's take
r
<
r
′
r
<
r
′
r < r^(') r<r^{\prime} r < r ′ According to a theorem of Cauchy, there exists a number M such as
|
c
n
|
<
M
r
n
n
=
1
,
2
,
…
c
n
<
M
r
n
n
=
1
,
2
,
…
|c_(n)| < (M)/(r^(n))quad n=1,2,dots \left|c_{n}\right|<\frac{M}{r^{n}} \quad n=1,2, \ldots | c n | < M r n n = 1 , 2 , … And
∑
n
=
1
∞
c
n
z
n
≤
∑
n
=
1
∞
M
z
n
r
n
=
M
z
r
1
−
z
r
(
|
z
|
<
r
)
∑
n
=
1
∞
c
n
z
n
≤
∑
n
=
1
∞
M
z
n
r
n
=
M
z
r
1
−
z
r
(
|
z
|
<
r
)
sum_(n=1)^(oo)c_(n)z^(n) <= sum_(n=1)^(oo)(Mz^(n))/(r^(n))=(M(z)/(r))/(1-(z)/(r))quad(|z| < r) \sum_{n=1}^{\infty} c_{n} z^{n} \leq \sum_{n=1}^{\infty} \frac{\mathrm{M} z^{n}}{r^{n}}=\frac{\mathrm{M} \frac{z}{r}}{1-\frac{z}{r}} \quad(|z|<r) ∑ n = 1 ∞ c n z n ≤ ∑ n = 1 ∞ M z n r n = M z r 1 − z r ( | z | < r ) We can deduce that for
x
≥
0
x
≥
0
x >= 0 x \geq 0 x ≥ 0
|
P
n
(
x
)
|
<
1
r
n
L
n
(
M
x
)
P
n
(
x
)
<
1
r
n
L
n
(
M
x
)
|P_(n)(x)| < (1)/(r^(n))L_(n)(Mx) \left|\mathrm{P}_{n}(x)\right|<\frac{1}{r^{n}} \mathrm{~L}_{n}(\mathrm{M} x) | P n ( x ) | < 1 r n L n ( M x ) It is easily demonstrated that if
x
>
0
x
>
0
x > 0 x>0 x > 0
L
2
(
x
)
L
1
(
x
)
>
L
3
(
x
)
L
2
(
x
)
>
⋯
>
L
n
(
x
)
L
n
−
1
(
x
)
>
⋯
L
2
(
x
)
L
1
(
x
)
>
L
3
(
x
)
L
2
(
x
)
>
⋯
>
L
n
(
x
)
L
n
−
1
(
x
)
>
⋯
(L_(2)(x))/(L_(1)(x)) > (L_(3)(x))/(L_(2)(x)) > cdots > (L_(n)(x))/(L_(n-1)(x)) > cdots \frac{L_{2}(x)}{L_{1}(x)}>\frac{L_{3}(x)}{L_{2}(x)}>\cdots>\frac{L_{n}(x)}{L_{n-1}(x)}>\cdots L 2 ( x ) L 1 ( x ) > L 3 ( x ) L 2 ( x ) > ⋯ > L n ( x ) L n − 1 ( x ) > ⋯ the report
L
n
L
n
−
1
L
n
L
n
−
1
(L_(n))/(L_(n-1)) \frac{L_{n}}{L_{n-1}} L n L n − 1 Therefore, it has a finite limit. The recurrence relation
x
L
n
=
(
n
+
1
)
L
n
+
1
−
2
n
L
n
+
(
n
−
1
)
L
n
−
1
x
L
n
=
(
n
+
1
)
L
n
+
1
−
2
n
L
n
+
(
n
−
1
)
L
n
−
1
xL_(n)=(n+1)L_(n+1)-2nL_(n)+(n-1)L_(n-1) x L_{n}=(n+1) L_{n+1}-2 n L_{n}+(n-1) L_{n-1} x L n = ( n + 1 ) L n + 1 − 2 n L n + ( n − 1 ) L n − 1 then shows us that
lim
n
→
∞
L
n
L
n
−
1
=
1
x
>
0
lim
n
→
∞
L
n
L
n
−
1
=
1
x
>
0
lim_(n rarr oo)(L_(n))/(L_(n-1))=1quad x > 0 \lim _{n \rightarrow \infty} \frac{L_{n}}{L_{n-1}}=1 \quad x>0 limit n → ∞ L n L n − 1 = 1 x > 0 We deduce from this
lim
―
n
→
∞
|
P
n
(
x
)
|
n
≤
1
r
.
lim
¯
n
→
∞
P
n
(
x
)
n
≤
1
r
.
bar(lim)_(n rarr oo)root(n)(|P_(n)(x)|) <= (1)/(r). \varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|\mathrm{P}_{n}(x)\right|} \leq \frac{1}{r} . limit — n → ∞ | P n ( x ) | n ≤ 1 r . Let us now consider a sequence of numbers
λ
0
,
λ
1
,
…
λ
n
,
…
λ
0
,
λ
1
,
…
λ
n
,
…
lambda_(0),lambda_(1),dotslambda_(n),dots \lambda_{0}, \lambda_{1}, \ldots \lambda_{n}, \ldots λ 0 , λ 1 , … λ n , … ot
lim
―
n
→
∞
|
λ
n
|
n
=
λ
lim
¯
n
→
∞
λ
n
n
=
λ
bar(lim)_(n rarr oo)root(n)(|lambda_(n)|)=lambda \varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|\lambda_{n}\right|}=\lambda limit — n → ∞ | λ n | n = λ As a result, the series
λ
0
P
0
(
x
)
+
λ
1
P
1
(
x
)
+
⋯
+
λ
n
P
n
(
x
)
+
⋯
λ
0
P
0
(
x
)
+
λ
1
P
1
(
x
)
+
⋯
+
λ
n
P
n
(
x
)
+
⋯
lambda_(0)P_(0)(x)+lambda_(1)P_(1)(x)+cdots+lambda_(n)P_(n)(x)+cdots \lambda_{0} \mathrm{P}_{0}(x)+\lambda_{1} \mathrm{P}_{1}(x)+\cdots+\lambda_{n} \mathrm{P}_{n}(x)+\cdots λ 0 P 0 ( x ) + λ 1 P 1 ( x ) + ⋯ + λ n P n ( x ) + ⋯ converges absolutely regardless of
λ
<
r
′
λ
<
r
′
lambda < r^(') \lambda<r^{\prime} λ < r ′ And
x
≥
0
x
≥
0
x >= 0 x \geq 0 x ≥ 0 .
