伯努利数
一.伯努利公式
伯努利数是一个用于解决 \(n\) 次方和的数列。
它的递归定义公式如下:
\[\sum_{i=0}^n \binom {n+1} i B_i=[n=0] ~~~~~~~~ (1.1)
\]
通过这个定义可以得到伯努利数的前几项:\(1,-\frac{1}{2},\frac{1}{6},0...\)
令 \(S_m(n)=\sum_{i=0}^{n-1} i^m\) ,伯努利通过找规律发现了伯努利公式:
\[S_m(n)=\frac{1}{m+1} \sum_{i=0}^m \binom {m+1} i B_i n^{m+1-i} ~~~~~~~~ (1.2)
\]
《具体数学》上给出的证明如下:
\[\begin{aligned}
S_{m+1}(n)+n^{m+1}&=\sum_{i=0}^ni^{m+1}\\
&=\sum_{i=0}^{n-1} (i+1)^{m+1} \\
&=\sum_{i=0}^{n-1}\sum_{j=0}^{m+1} \binom {m+1}{j} i^j \\
&=\sum_{j=0}^{m+1}\binom {m+1}{j} \sum_{i=0}^{n-1} i^j \\
&=\sum_{j=0}^{m+1}\binom {m+1}{j} S_j(n) & (1.3)
\end{aligned}\]
由\((1.3)\)两边同时减去 \(n^{m+1}\) 得,
\[S_{m+1}(n)=\sum_{j=0}^{m}\binom {m+1}{j} S_j(n) ~~~~~~~~ (1.4)
\]
设 \(S'_ m(n)\) 为\((1.2)\) 右式 , \(\Delta=S_m(n)-S'_m(n)\)。
归纳证明 \(S_m(n)=S'_m(n)\):
1.当 \(m=0\) 时成立。
2.设对于 \(\forall i\in[0,m),S_i(n)=S'_ i(n)\), 由 \((3.4)\) 得:
\[\begin{aligned}
n^{m+1}&=\sum_{j=0}^{m} \binom{m+1}{j} S'_ {j}(n)+ \binom{m+1}{m} \Delta & \text{(只有j=m时有差异)} \\
&=\sum_{j=0}^{m} \binom{m+1}{j} \frac{1}{j+1} \sum_{k=0}^j \binom {j+1} k B_k n^{j+1-k} + (m+1) \Delta & \text{(化简代入)} \\
&=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} k \frac{B_k}{j+1} n^{j+1-k} + (m+1) \Delta \\
&=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {j-k} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & \text{(将k换为j-k)} \\
&=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & (\binom {n}{m} = \binom {n}{n-m}) \\
&=\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta \\
&=\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} * \frac{j+1}{k+1} * \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta & (\binom{n+1}{m+1}=\binom{n}{m} \times \frac{n+1}{m+1}) \\
&=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} B_{j-k} + (m+1) \Delta \\
&=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=k}^{m} \binom{m+1-k}{j-k} B_{j-k} + (m+1) \Delta \\
&=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=0}^{m-k} \binom{m+1-k}{j} B_{j} + (m+1) \Delta \\
&=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} [m-k=0] + (m+1) \Delta & (\text{将(1.1)带入)} \\
&=\frac{n^{m+1}}{m+1} \binom{m+1}{m} + (m+1) \Delta \\
&=n^{m+1} + (m+1) \Delta \\
\end{aligned}\]
所以 \((m+1)\Delta=0\) , 又 \(m \ge 1\) , 所以 \(\Delta=0\) , 证毕。