The Real And Complex Number Systems

Proof: Consider
f t+1 (x) − f t (x) =
so as n = 1, 2, ..., we have
∞∑
n=0
[P n (t + 1) − P n (t)] xn
n!
= xe xt by f t (x) = xe xt / (e x − 1)
∞∑
= (n + 1) t n xn+1
(n + 1)! ,
n=0
P n (t + 1) − P n (t) = nt n−1 .
by (b)
(f) P n (1 − t) = (−1) n P n (t)
Proof: Note that
so we have
n=0
Hence, P n (1 − t) = (−1) n P n (t) .
(g) B 2n+1 = 0 if n = 1, 2, ...
f t (−x) = f 1−t (x) ,
∞∑
∞
(−1) n P n (t) xn
n! = ∑
P n (1 − t) xn
n! .
Proof: With help of (e) and (f), let t = 0 and n = 2k + 1, then it is clear
that B 2k+1 = 0 if k = 1, 2, ...
n=0
(h) 1 n + 2 n + ... + (k − 1) n = P n+1(k)−P n+1 (0)
n+1
(n = 2, 3, ...)
Proof: With help of (e), we know that
P n+1 (t + 1) − P n+1 (t)
n + 1
= t n
which implies that
1 n + 2 n + ... + (k − 1) n = P n+1 (k) − P n+1 (0)
n + 1
(n = 2, 3, ...)
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