The Real And Complex Number Systems

and
f 0 (x) =
n=0
∞∑
n=0
P n (0) xn
n! = x
e x − 1 .
So, we have the identity
∞∑
P n (t) xn
n! = ∑ ∞ etx P n (0) xn
n! .
n=0
Use the identity with e tx = ∑ ∞ t n
n=0 n! xn , then we obtain
n∑
which implies that
P n (t)
n!
=
k=0
= 1 n!
P n (t) =
t n−k P k (0)
(n − k)! k!
n∑
( n k) P k (0) t n−k
k=0
n∑
( n k) P k (0) t n−k .
k=0
This shows that each function P n is a polynomial. There are the Bernoulli
polynomials. The numbers B n = P n (0) (n = 0, 1, 2, ...) are called the
Bernoulli numbers. Derive the following further properties:
(c) B 0 = 1, B 1 = − 1 2 , ∑ n−1
k=0 (n k ) B k = 0, if n = 2, 3, ...
Proof: Since 1 = p(x) , where p (x) := ∑ ∞ x n
p(x) n=0
, and 1 := ∑ ∞
(n+1)! p(x) n=0 P n (0) xn . n!
So,
where
1
1 = p (x)
p (x)
∞∑ x n
=
(n + 1)!
n=0
∞∑
= C n x n
C n =
n=0
1
(n + 1)!
n∑
k=0
36
∞∑
n=0
P n (0) xn
n!
( n+1
)
k Pk (0) .