就讓我們從白努利多項式的定義︰

$B_n(x) = \sum \limits_{k=0}^{n} \binom {n}k B_k x^{n-k}$ ，計算 $B_n (0)$ 開始︰ $B_n(0) = \sum \limits_{k=0}^{n} \binom {n}k B_k 0^{n-k} = B_n \cdot 0^0 = B_n$ 。

輔之以『對稱公式』

$B_n (1-x) = {(-1)}^n B_n (x), \ n \ge 0$ ，那麼

$B_{2k} (1) = B_{2k} (0) = B_{2k} , \ B_{2k+1} (1) = - B_{2k+1} (0) = - B_{2k+1}$ 。

因為白努利奇數除 $B_1 = - \frac{1}{2} = B_1(0)$ 外，都是零 $B_{2k+1} = 0$ 。且 $B_0 (x) = 1$ 是個『常數函數』，故而 $B_0 (1) = B_0 (0) = 1$ ，所以

$\therefore B_n (1) = B_n(0) , \ n \ge 0 , \ n \neq 1$ 。

假設 $n \neq 1$ ，如果我們用白努利多項式的定義直接計算 $B_n (1)$

$= \sum \limits_{k=0}^{n} \binom {n}k B_k 1^{n-k} = \sum \limits_{k=0}^{n} \binom {n}k B_k = B_n (0) = B_n$

$= \sum \limits_{k=0}^{n-1} \binom {n}k B_k + B_n =B_n$

$\therefore \sum \limits_{k=0}^{n-1} \binom {n}k B_k = 0$ ，因此

$\therefore \sum \limits_{k=0}^{n-2} \binom {n}k B_k + \binom {n}{n-1} B_{n-1} = 0$ 。可以 $m = n -1$ 將之改寫成『遞迴關係式』

$B_{m} = -\frac{1}{m+1} \sum \limits_{k=0}^{m-1} \binom {m+1}k B_k, \ m \ge 1$ ，如是 $n \neq 1$ ，意味著 $m \neq 0$ ，妙哉！ $B_1 = -\frac{1}{1+1} B_0 = - \frac{1}{2}$ 也！！

不過用此『對稱公式』計算 $B_n (\frac{1}{2})$ ，對偶次白努利多項式無攸利吧！ $B_{2k} (\frac{1}{2}) = B_{2k} (\frac{1}{2})$ 。但能知其為奇次白努利多項式之『根』哩！ $B_{2k+1} (\frac{1}{2}) = - B_{2k+1} (\frac{1}{2})$ 。

即使從白努利多項式的『微分公式』

$B_n^{'} (x) = n \cdot B_{n-1} (x) , \ n \ge 1$ ，

能知相鄰奇偶白努利多項式之『根』和『極值』間的『關係』，

加上已知的『推步公式』

$B_n (x+1) = B_n (x) +n x^{n-1}, \ n \ge 1$ ，

皆無助於求得 $B_{2k} ( \frac{1}{2} )$ 的『數值』正負大小或其與 $B_n (0)$ 以及 $B_n (1)$ 之『關係』乎？此所以『復製公式』

$B_k (x) + B_k (x+\frac{1}{2}) = 2^{1-k} B_k (2x), \ k \ge 0$ 、

『乘法定理』發現之濫觴耶？？

Multiplication theorem

In mathematics, the multiplication theorem is a certain type of identity obeyed by many special functions related to the gamma function. For the explicit case of the gamma function, the identity is a product of values; thus the name. The various relations all stem from the same underlying principle; that is, the relation for one special function can be derived from that for the others, and is simply a manifestation of the same identity in different guises.

Finite characteristic

The multiplication theorem takes two common forms. In the first case, a finite number of terms are added or multiplied to give the relation. In the second case, an infinite number of terms are added or multiplied. The finite form typically occurs only for the gamma and related functions, for which the identity follows from a p-adic relation over a finite field. For example, the multiplication theorem for the gamma function follows from the Chowla–Selberg formula, which follows from the theory of complex multiplication. The infinite sums are much more common, and follow from characteristic zero relations on the hypergeometric series.

The following tabulates the various appearances of the multiplication theorem for finite characteristic; the characteristic zero relations are given further down. In all cases, n and k are non-negative integers. For the special case of n = 2, the theorem is commonly referred to as the duplication formula.

Bernoulli polynomials

For the Bernoulli polynomials, the multiplication theorems were given by Joseph Ludwig Raabe in 1851:

k^{{1-m}}B_{m}(kx)=\sum _{{n=0}}^{{k-1}}B_{m}\left(x+{\frac {n}{k}}\right)

and for the Euler polynomials,

k^{{-m}}E_{m}(kx)=\sum _{{n=0}}^{{k-1}}(-1)^{n}E_{m}\left(x+{\frac {n}{k}}\right)\quad {\mbox{ for }}k=1,3,\dots

and

k^{{-m}}E_{m}(kx)={\frac {-2}{m+1}}\sum _{{n=0}}^{{k-1}}(-1)^{n}B_{{m+1}}\left(x+{\frac {n}{k}}\right)\quad {\mbox{ for }}k=2,4,\dots .

The Bernoulli polynomials may be obtained as a special case of the Hurwitz zeta function, and thus the identities follow from there.

※ 註︰恰是幾何級數等比數列關係

$\sum \limits_{k=0}^{\infty} \left[ B_k (x) + B_k (x+\frac{1}{m}) + B_k (x+\frac{2}{m}) + \cdots + B_k (x+\frac{m-1}{m}) \right] \frac{t^k}{k!}$

$= \frac{t e^{xt}}{e^t - 1} + \frac{t e^{(x+\frac{1}{m})t}}{e^t - 1} + \frac{t e^{(x+\frac{2}{m})t}}{e^t - 1} + \cdots + \frac{t e^{(x+\frac{m-1}{m})t}}{e^t - 1}$