The connection between the functions of Riemann zeta and Bernoulli Number
碩士 === 國立臺中教育大學 === 數學教育學系 === 96 === This research hung over from the extended functions for the sum of powers of consecutive integers, we colleted the literatures of the related research about the functions of Riemann zeta and Bernoulli Number, both newly interpreted and predigested the properti...
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ndltd-TW-096NTCTC4800122015-10-13T11:31:59Z http://ndltd.ncl.edu.tw/handle/17154599310613619902 The connection between the functions of Riemann zeta and Bernoulli Number 黎曼-傑塔函數與伯努力數的關聯 Chih -Shiuan Liu 劉志璿 碩士 國立臺中教育大學 數學教育學系 96 This research hung over from the extended functions for the sum of powers of consecutive integers, we colleted the literatures of the related research about the functions of Riemann zeta and Bernoulli Number, both newly interpreted and predigested the properties of the functions of Riemann zeta and Bernoulli Number. Thus we built the connection between the functions of Riemann zeta and Bernoulli Number, according to \zeta(2 k)=(-1)^{k-1} 2^{2k-1} \frac{B_{2k} \pi^{2k}}{(2k)!}, \ k \in \mathbb{N},and S_{2k}^{\prime}(-1)=\frac{(-1)^{k-1} (2k)!}{2^{2k-1} (\pi)^{2k}}\zeta(2k), S_{2k+1}^{\prime}(-1)=0,Take the function of Riemann zeta as bridge, we find that S_{2k}^{\prime}(-1)=B_{2k},B_{2k}=\frac{1}{2k+1} \left \{ C_{2k}^{2k+1} S_{1}^{\prime}(-1)+ \sum_{i=1}^{k} C_{2i+1}^{2k+1} S_{2k-2i}^{\prime}(-1) \right \},where $S_k^{\prime}(x)$ denotes the first derivative of $S_k(x)$ for each positive integer $k$. Tao-Ming Wang Feng-Rung Hu 王道明 胡豐榮 2008 學位論文 ; thesis 84 zh-TW |
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碩士 === 國立臺中教育大學 === 數學教育學系 === 96 === This research hung over from the extended functions for the sum of powers of consecutive integers,
we colleted the literatures of the related research about the functions of Riemann zeta and Bernoulli Number,
both newly interpreted and predigested the properties of the functions of Riemann zeta and Bernoulli Number.
Thus we built the connection between the functions of Riemann zeta and Bernoulli Number, according to \zeta(2 k)=(-1)^{k-1} 2^{2k-1} \frac{B_{2k} \pi^{2k}}{(2k)!}, \ k \in \mathbb{N},and S_{2k}^{\prime}(-1)=\frac{(-1)^{k-1} (2k)!}{2^{2k-1} (\pi)^{2k}}\zeta(2k), S_{2k+1}^{\prime}(-1)=0,Take the function of Riemann zeta as bridge, we find that S_{2k}^{\prime}(-1)=B_{2k},B_{2k}=\frac{1}{2k+1}
\left \{
C_{2k}^{2k+1} S_{1}^{\prime}(-1)+
\sum_{i=1}^{k}
C_{2i+1}^{2k+1} S_{2k-2i}^{\prime}(-1)
\right \},where $S_k^{\prime}(x)$ denotes the first derivative of $S_k(x)$ for
each positive integer $k$.
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Tao-Ming Wang |
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Tao-Ming Wang Chih -Shiuan Liu 劉志璿 |
author |
Chih -Shiuan Liu 劉志璿 |
spellingShingle |
Chih -Shiuan Liu 劉志璿 The connection between the functions of Riemann zeta and Bernoulli Number |
author_sort |
Chih -Shiuan Liu |
title |
The connection between the functions of Riemann zeta and Bernoulli Number |
title_short |
The connection between the functions of Riemann zeta and Bernoulli Number |
title_full |
The connection between the functions of Riemann zeta and Bernoulli Number |
title_fullStr |
The connection between the functions of Riemann zeta and Bernoulli Number |
title_full_unstemmed |
The connection between the functions of Riemann zeta and Bernoulli Number |
title_sort |
connection between the functions of riemann zeta and bernoulli number |
publishDate |
2008 |
url |
http://ndltd.ncl.edu.tw/handle/17154599310613619902 |
work_keys_str_mv |
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