欧拉-马歇罗尼常数

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欧拉-马歇罗尼常数是一个数学常数,定义为调和级数自然对数的差值:

\gamma = \lim_{n \rightarrow \infty } \left[ \left(
\sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right]=\int_1^\infty\left({1\over\lfloor x\rfloor} - {1\over x}\right)\,dx

它的近似值为\gamma \approx  0.57721 56649 01532 86060 65120 90082 40243 10421 59335[1]

欧拉-马歇罗尼常数主要应用于数论

历史[编辑]

该常数最先由瑞士数学家莱昂哈德·欧拉(Leonhard Euler)在1735年发表的文章De Progressionibus harmonicus observationes中定义。欧拉曾经使用C作为它的符号,并计算出了它的前6位小数。1761年他又将该值计算到了16位小数。1790年,意大利数学家马歇罗尼(Lorenzo Mascheroni)引入了γ作为这个常数的符号,并将该常数计算到小数点后32位。但后来的计算显示他在第20位的时候出现了错误。

目前尚不知道该常数是否为有理数,但是分析表明如果它是一个有理数,那么它的分母位数将超过10242080[2]

性质[编辑]

与伽玛函数的关系[编辑]

 \ -\gamma = \Gamma'(1) = \Psi(1)
 \gamma =   \lim_{x \to \infty} \left [ x - \Gamma \left ( \frac{1}{x} \right ) \right ]
 \gamma = \lim_{n \to \infty} \left [ \frac{ \Gamma(\frac{1}{n}) \Gamma(n+1)\, n^{1+\frac{1}{n}}}{\Gamma(2+n+\frac{1}{n})} - \frac{n^2}{n+1} \right ]

与ζ函数的关系[编辑]

\gamma = \sum_{m=2}^{\infty} \frac{(-1)^m\zeta(m)}{m}
=  \ln \left ( \frac{4}{\pi} \right ) + \sum_{m=1}^{\infty} \frac{(-1)^{m-1} \zeta(m+1)}{2^m (m+1)}
 \gamma = \frac{3}{2} - \ln 2 - \sum_{m=2}^\infty (-1)^m\,\frac{m-1}{m} [\zeta(m) - 1]
 = \lim_{n \to \infty} \left [ \frac{2\,n-1}{2\,n} - \ln\,n + \sum_{k=2}^n \left ( \frac{1}{k} - \frac{\zeta(1-k)}{n^k} \right ) \right ]
 = \lim_{n \to \infty} \left [ \frac{2^n}{e^{2^n}} \sum_{m=0}^\infty \frac{2^{m \,n}}{(m+1)!} \sum_{t=0}^m \frac{1}{t+1} - n\, \ln 2+ O \left ( \frac{1}{2^n\,e^{2^n}} \right ) \right ]
 \gamma = \lim_{s \to 1^+} \sum_{n=1}^\infty \left ( \frac{1}{n^s} - \frac{1}{s^n} \right )  = \lim_{s \to 1^+} \left ( \zeta(s) - \frac{1}{s - 1} \right )
 \gamma =   \lim_{x \to \infty} \left [ x - \Gamma \left ( \frac{1}{x} \right ) \right ]
 =   \lim_{n \to \infty} \frac{1}{n}\, \sum_{k=1}^n \left ( \left \lceil \frac{n}{k} \right \rceil - \frac{n}{k} \right )
\gamma = \sum_{k=1}^n \frac{1}{k} - \ln(n) -
\sum_{m=2}^\infty \frac{\zeta (m,n+1)}{m}

积分[编辑]

\gamma = - \int_0^\infty { e^{-x} \ln x }\,dx = \int_\infty^ 0 { e^{-x} \ln x }\,dx [證明 1]
 = - \int_0^1 { \ln\ln \frac{1}{x} }\,dx
 = \int_0^\infty {\left (\frac{1}{1 - e^{-x}} - \frac{1}{x} \right )e^{ - x}  }\,dx
 = \int_0^\infty { \frac{1}{x} \left ( \frac{1}{1+x} - e^{ - x} \right ) }\,dx
  \int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi}
 \int_0^\infty { e^{-x} \ln^2 x }\,dx  = \gamma^2 + \frac{\pi^2}{6}
 \gamma = \int_{0}^{1}\int_{0}^{1} \frac{x - 1}{(1 - x\,y)\ln(x\,y)} \, dx\,dy = \sum_{n=1}^\infty \left ( \frac{1}{n} - \ln\frac{n+1}{n} \right ).
 \sum_{n=1}^\infty \frac{N_1(n) + N_0(n)}{2n(2n+1)} = \gamma

级数展开式[编辑]

\gamma = \sum_{k=1}^\infty \left[ \frac{1}{k} - \ln \left( 1 + \frac{1}{k} \right) \right]

 \gamma = 1 - \sum_{k=2}^{\infty}(-1)^k\frac{\lfloor\log_2 k\rfloor}{k+1} .

 \gamma = \sum_{k=2}^\infty (-1)^k \frac{ \left \lfloor \log_2 k \right \rfloor}{k}
  = \tfrac12-\tfrac13
  + 2\left(\tfrac14 - \tfrac15 + \tfrac16 - \tfrac17\right)
  + 3\left(\tfrac18 - \dots - \tfrac1{15}\right) + \dots

 \gamma + \zeta(2) = \sum_{k=1}^{\infty} \frac1{k\lfloor\sqrt{k}\rfloor^2}
  = 1 + \tfrac12 + \tfrac13 + \tfrac14\left(\tfrac14 + \dots + \tfrac18\right)
    + \tfrac19\left(\tfrac19 + \dots + \tfrac1{15}\right) + \dots

 \gamma = \sum_{k=2}^{\infty} \frac{k - \lfloor\sqrt{k}\rfloor^2}{k^2\lfloor\sqrt{k}\rfloor^2}
 =  \tfrac1{2^2} + \tfrac2{3^2}
  + \tfrac1{2^2}\left(\tfrac1{5^2} + \tfrac2{6^2} + \tfrac3{7^2} + \tfrac4{8^2}\right)
  + \tfrac1{3^2}\left(\tfrac1{10^2} + \dots + \tfrac6{15^2}\right) + \dots

 \gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx

 \gamma 连分数展开式为:

 \gamma = [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...]\, OEIS中的数列A002852).

渐近展开式[编辑]

\gamma  \sim H_n  - \ln \left( n \right) - \frac{1}{{2n}} + \frac{1}{{12n^2 }} - \frac{1}{{120n^4 }} + .
\gamma  \sim H_n  - \ln \left( {n + \frac{1}{2} + \frac{1}{{24n}} - \frac{1}{{48n^3 }} + ...} \right)
\gamma  \sim H_n  - \frac{{\ln \left( n \right) + \ln \left( {n + 1} \right)}}{2} - \frac{1}{{6n\left( {n + 1} \right)}} + \frac{1}{{30n^2 \left( {n + 1} \right)^2 }} - .

已知位数[编辑]

\boldsymbol{\gamma}的已知位数
日期 位数 计算者
1734年 5 莱昂哈德·欧拉
1736年 15 莱昂哈德·欧拉
1790年 19 Lorenzo Mascheroni
1809年 24 Johann G. von Soldner
1812年 40 F.B.G. Nicolai
1861年 41 Oettinger
1869年 59 William Shanks
1871年 110 William Shanks
1878年 263 约翰·柯西·亚当斯
1962年 1,271 高德纳
1962年 3,566 D.W. Sweeney
1977年 20,700 Richard P. Brent
1980年 30,100 Richard P. Brent埃德温·麦克米伦
1993年 172,000 Jonathan Borwein
1997年 1,000,000 Thomas Papanikolaou
1998年12月 7,286,255 Xavier Gourdon
1999年10月 108,000,000 Xavier Gourdon和Patrick Demichel
2006年7月16日 2,000,000,000 Shigeru Kondo和Steve Pagliarulo
2006年12月8日 116,580,041 Alexander J. Yee
2007年7月15日 5,000,000,000 Shigeru Kondo和Steve Pagliarulo
2008年1月1日 1,001,262,777 Richard B. Kreckel
2008年1月3日 131,151,000 Nicholas D. Farrer
2008年6月30日 10,000,000,000 Shigeru Kondo和Steve Pagliarulo
2009年1月18日 14,922,244,771 Alexander J. Yee和Raymond Chan
2009年3月13日 29,844,489,545 Alexander J. Yee和Raymond Chan

參見[编辑]

相关证明[编辑]

  1. ^ \gamma = - \int_0^\infty { e^{-x} \ln x }\,dx 的证明:
    首先根据放缩法(\int_k^{k+1} \frac 1x \,dx < \frac 1k < \int_{k-1}^k \frac 1x \,dx)容易知道, \int_k^{k-1} \frac 1x \,dx - \frac 1k < \frac 1{k(k-1)},以及\ln n < \sum_{k=1}^n \frac 1k < 1 + \ln n。因此\gamma存在并有限。
    \sum_{k=1}^n \frac{1}{k}
    = \sum_{k=1}^n \int_0^1 t^{k-1} \,dt
    = \int_0^1 \sum_{k=1}^n t^{k-1} \,dt
    = \int_0^1 \frac {1 - t^n}{1 - t} \,dt
    = \int_n^0 \frac {1 - (1-x/n)^n}{1 - (1-x/n)} \,d(1-x/n)
    = \int_n^0 \frac {1 - (1-x/n)^n}{x/n} (- \frac 1n) \,dx
    = \int_0^n \frac {1 - (1-x/n)^n}{x} \,dx,
    \ln n = \int_1^n \frac 1x \,dx,
    所以\gamma = \lim_{n \to \infty} [\sum_{k=1}^n \frac{1}{k} - \ln n]
    = \lim_{n \to \infty} \left[ \int_0^n \frac {1 - (1-x/n)^n}{x} \,dx - \int_1^n \frac 1x \,dx \right]
    = \lim_{n \to \infty} \left[ \int_0^1 \frac {1 - (1-x/n)^n}{x} \,dx - \int_1^n \frac {(1-x/n)^n}{x} \right]
    = \int_0^1 \frac {1 - \lim_{n \to \infty}(1-x/n)^n}{x} \,dx - \int_1^{\infty} \frac {\lim_{n \to \infty}(1-x/n)^n}{x} (单调收敛定理)
    = \int_0^1 \frac {1 - e^{-x}}{x} \,dx - \int_1^{\infty} \frac {e^{-x}}{x}
    = \left. (1 - e^{-x}) \ln x \right|_0^1 - \int_0^1 \ln x \,d(1 - e^{-x}) - \left. e^{-x} \ln x \right|_1^{\infty} + \int_1^{\infty} \ln x \,de^{-x}
    = - \int_0^{\infty} e^{-x} \ln x \,dx.

參考文獻[编辑]

  1. ^ A001620 oeis.org [2014-7-17]
  2. ^ Havil 2003 p 97.
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  2. Gourdon, Xavier, and Sebah, P. (2002) "Collection of formulas for Euler's constant, γ."
  3. Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ."
  4. Donald Knuth (1997) The Art of Computer Programming, Vol. 1, 3rd ed. Addison-Wesley. ISBN 0-201-89683-4
  5. Krämer, Stefan (2005) Die Eulersche Konstante γ und verwandte Zahlen. Diplomarbeit, Universität Göttingen.
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  7. Sondow, Jonathan (2002) "A hypergeometric approach, via linear forms involving logarithms, to irrationality criteria for Euler's constant." With an Appendix by Sergey Zlobin, Mathematica Slovaca 59: 307-314.
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  11. Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π."
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  13. G. Vacca (1926), "Nuova serie per la costante di Eulero, C = 0,577…". Rendiconti, Accademia Nazionale dei Lincei, Roma, Classe di Scienze Fisiche, Matematiche e Naturali (6) 3, 19–20.
  14. James Whitbread Lee Glaisher (1872), "On the history of Euler's constant". Messenger of Mathematics. New Series, vol.1, p. 25-30, JFM 03.0130.01
  15. Carl Anton Bretschneider (1837). "Theoriae logarithmi integralis lineamenta nova". Crelle Journal, vol.17, p. 257-285 (submitted 1835)
  16. Lorenzo Mascheroni (1790). "Adnotationes ad calculum integralem Euleri, in quibus nonnulla problemata ab Eulero proposita resolvuntur". Galeati, Ticini.
  17. Lorenzo Mascheroni (1792). "Adnotationes ad calculum integralem Euleri. In quibus nonnullae formulae ab Eulero propositae evolvuntur". Galeati, Ticini. Both online at: http://books.google.de/books?id=XkgDAAAAQAAJ
  18. Havil, Julian. Gamma: Exploring Euler's Constant. Princeton University Press. 2003. ISBN 0-691-09983-9. 
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外部連結[编辑]