# MUIC Math

Mathematics at MUIC

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seminar:at_bernoulli

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 seminar:at_bernoulli [2017/05/17 08:23]aram [Abstract] seminar:at_bernoulli [2017/05/17 08:23] (current)aram [Abstract] Both sides previous revision Previous revision 2017/05/17 08:23 aram [Abstract] 2017/05/17 08:23 aram [Abstract] 2017/05/17 08:22 aram [Abstract] 2017/05/16 22:42 aram created 2017/05/17 08:23 aram [Abstract] 2017/05/17 08:23 aram [Abstract] 2017/05/17 08:22 aram [Abstract] 2017/05/16 22:42 aram created Line 11: Line 11: ==== Abstract ==== ==== Abstract ==== - The Bernoulli numbers $B_n$ appear as coefficients in a Maclaurin series expansion of the function $x\mapsto x/(e^x-1).$ Some of their number-theoretic properties have a connection with the Euler-Maclaurin-sum formula:For $k\ge 0$ and $n\ge 2,$ + The Bernoulli numbers $B_n$ appear as coefficients in a Maclaurin series expansion of the function $x\mapsto x/(e^x-1).$ Some of their number-theoretic properties have a connection with the Euler-Maclaurin-sum formula: For $k\ge 0$ and $n\ge 2,$ $$1^k + 2^k + \cdots + (n-1)^k = \sum_{r=0}^k \frac{1}{k+1-r}{k\choose r}n^{k+1-r}B_r.$$ Their fractional parts are realized by a celebrated theorem of von Staudt and Clausen (1840) which states the following: For $k\ge 1,$ $$1^k + 2^k + \cdots + (n-1)^k = \sum_{r=0}^k \frac{1}{k+1-r}{k\choose r}n^{k+1-r}B_r.$$ Their fractional parts are realized by a celebrated theorem of von Staudt and Clausen (1840) which states the following: For $k\ge 1,$ $$B_{2k}+\sum_{\substack{p\, \mathrm{prime} \\ p-1 \mid 2k}}\frac{1}{p} \in \mathbb{Z}.$$  In this talk, we will prove the above two results and discuss some recent developments of arithmetic properties of the Bernoulli numbers. $$B_{2k}+\sum_{\substack{p\, \mathrm{prime} \\ p-1 \mid 2k}}\frac{1}{p} \in \mathbb{Z}.$$  In this talk, we will prove the above two results and discuss some recent developments of arithmetic properties of the Bernoulli numbers.