2020-2021:teams:alchemist:hardict:powersum
| 两侧同时换到之前的修订记录 前一修订版 后一修订版 | 前一修订版 | ||
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2020-2021:teams:alchemist:hardict:powersum [2020/05/09 10:41] hardict [伯努利数以及生成函数] |
2020-2021:teams:alchemist:hardict:powersum [2020/05/09 10:56] (当前版本) hardict [伯努利数以及生成函数] |
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| 行 67: | 行 67: | ||
| $f_{k}(n-1) | $f_{k}(n-1) | ||
| =\sum_{i=0}^{n-1}i^{k} | =\sum_{i=0}^{n-1}i^{k} | ||
| - | =\frac{k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_{i}n^{k+1-i}$ | + | =\frac{1}{k+1}\sum_{i=0}^{k}\binom{k+1}{i}B_{i}n^{k+1-i}$ |
| + | |||
| + | $ | ||
| + | 考虑F(x)=\sum_{k=0}^{\infty}(\sum_{i=0}^{n-1}i^{k})\frac{x^{k}}{k!}\\\\ | ||
| + | F(x)=\sum_{i=0}^{n-1}\sum_{k=0}^{\infty}i^{k}\frac{x^{k}}{k!}=\sum_{i=0}^{n-1}e^{ix}=\frac{e^{nx}-1}{e^{x}-1}\\\\ | ||
| + | 注意到C(x)=\frac{x}{e^{x}-1},F(x)=C(x)\frac{e^{nx}-1}{x}, | ||
| + | \frac{e^{nx}-1}{x}=\sum_{i=0}^{\infty}\frac{n^{i+1}x^{i}}{(i+1)!}\\\\ | ||
| + | F(x)中x^{k}系数\frac{f_{k}(n-1)}{k!}=\sum_{i+j=k}\frac{B_{i}}{i!}\frac{n^{j+1}}{(j+1)!}即可得到上述公式 | ||
| + | $ | ||
2020-2021/teams/alchemist/hardict/powersum.1588992091.txt.gz · 最后更改: 2020/05/09 10:41 由 hardict
