|
|
| Line 1: |
Line 1: |
| In [[number theory]], the '''von Staudt–Clausen theorem''' is a result determining the [[fractional part]] of [[Bernoulli number]]s, found independently by
| | <br><br>I woke up another day and noticed - I've been single for a little while now and following much bullying from buddies I today find myself opted for web dating. They guaranteed me that there are a lot of pleasant, regular and [http://enjoyablepeople.org/ enjoyable people] to meet, so here goes the pitch!<br>I make an effort [http://www.hotelsedinburgh.org vip tickets to luke bryan] to stay as toned as [http://Www.adobe.com/cfusion/search/index.cfm?term=&potential+staying&loc=en_us&siteSection=home potential staying] at the fitness center several times per week. I appreciate my sports and make an effort to perform luke bryan about ([http://www.museodecarruajes.org www.museodecarruajes.org]) or watch while many a possible. Being wintertime I will often at Hawthorn suits. Note: I've noticed the carnage of wrestling fits at stocktake revenue, In case that you really contemplated purchasing an activity I really don't brain.<br>My buddies and household are awe-inspiring and spending time with them at tavern gigs or dinners is obviously essential. I haven't ever been in to nightclubs as I discover that one can not get a nice dialogue together with the sound. I likewise have [http://www.senatorwonderling.com luke bryan dates] 2 unquestionably [http://www.banburycrossonline.com luke bryan luke bryan] cheeky and quite cunning canines that are almost always excited to meet up fresh people.<br><br>my webpage; [http://lukebryantickets.pyhgy.com luke bryan concert tours] |
| {{harvs|txt|authorlink=Karl von Staudt|first=Karl |last=von Staudt|year=1840}} and {{harvs|txt|authorlink=Thomas Clausen (mathematician)|first=Thomas|last= Clausen|year= 1840}}.
| |
| | |
| Specifically, if ''n'' is a positive integer and we add 1/''p'' to the Bernoulli number ''B''<sub>2''n''</sub> for every [[prime number|prime]] ''p'' such that ''p'' − 1 divides 2''n'', we obtain an integer.
| |
| | |
| This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers ''B''<sub>2''n''</sub> as the product of all primes ''p'' such that ''p'' − 1 divides 2''n''; consequently the denominators are [[Square-free integer|square-free]] and divisible by 6.
| |
| | |
| These denominators are
| |
| : 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, ... {{OEIS|A002445}}
| |
| | |
| == Proof ==
| |
| A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:
| |
| :<math> B_{2n}=\sum_{j=0}^{2n}{\frac{1}{j+1}}\sum_{m=0}^{j}{(-1)^{m}{j\choose m}m^{2n}} \!</math>
| |
| and as a corollary:
| |
| :<math> B_{2n}=\sum_{j=0}^{2n}{\frac{j!}{j+1}}(-1)^jS(2n,j) \!</math>
| |
| where <math> S(n,j) \!</math> are the [[Stirling numbers of the second kind]].
| |
| | |
| Furthermore the following lemmas are needed:<br>
| |
| Let p be a prime number then,<br>
| |
| '''1'''. If '''p-1 divides 2n''' then,
| |
| :<math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv{-1}\pmod p \!</math> | |
| '''2'''. If '''p-1 does not divide 2n''' then,
| |
| :<math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv0\pmod p \!</math>
| |
| '''Proof of (1) and (2)''': One has from [[Fermat's little theorem]],
| |
| :<math> m^{p-1}\equiv 1\pmod p \!</math>
| |
| for <math> m=1,2,...,p-1 \!</math>.<br>
| |
| If '''p-1 divides 2n''' then one has,
| |
| :<math> m^{2n}\equiv 1\pmod p \!</math> | |
| for <math> m=1,2,...,p-1 \!</math>.<br>
| |
| Thereafter one has,
| |
| :<math> \sum_{m=1}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv \sum_{m=1}^{p-1}{(-1)^m{p-1\choose m}}\pmod p\!</math>
| |
| from which '''(1)''' follows immediately.<br>
| |
| If '''p-1 does not divide 2n''' then after Fermat's theorem one has,
| |
| :<math> m^{2n}\equiv m^{2n-(p-1)}\pmod p \!</math>
| |
| If one lets <math> \wp=[\frac{2n}{p-1}] \!</math> ([[Floor and ceiling functions|Greatest integer function]]) then after iteration one has,
| |
| :<math> m^{2n}\equiv m^{2n-\wp(p-1)}\pmod p \!</math> | |
| for <math> m=1,2,...,p-1 \!</math> and <math> 0<2n-\wp(p-1)<p-1 \!</math>.<br>
| |
| Thereafter one has,
| |
| :<math> \sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n}}\equiv\sum_{m=0}^{p-1}{(-1)^m{p-1\choose m} m^{2n-\wp(p-1)}}\pmod p\!</math>
| |
| Lemma '''(2)''' now follows from the above and the fact that ''S''(''n'',''j'')=0 for ''j''>''n''.<br>
| |
| '''(3)'''. It is easy to deduce that for '''a>2 and b>2, ab divides (ab-1)!'''.<br>
| |
| '''(4). Stirling numbers of second kind are integers'''.
| |
| | |
| <big>'''Proof of the theorem'''</big>: Now we are ready to prove Von-Staudt Clausen theorem,<br>
| |
| If '''j+1 is composite''' and '''j>3''' then from (3), j+1 divides j!.<br>
| |
| For j=3,
| |
| :<math> \sum_{m=0}^{3}{(-1)^m{3\choose m}m^{2n}}=3 \cdot 2^{2n}-3^{2n}-3\equiv0 \pmod 4 \!</math>
| |
| If '''j+1 is prime then we use (1) and (2)''' and if '''j+1 is composite then we use (3) and (4)''' to deduce:
| |
| :<math> B_{2n}=I_n-\sum_{(p-1)|2n}{\frac{1}{p}} \!</math>
| |
| where <math> I_n \!</math> is an integer, which is the Von-Staudt Clausen theorem.<ref>H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.</ref><ref>T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 197.</ref>
| |
| | |
| ==See also==
| |
| | |
| *[[Kummer's congruence]]
| |
| | |
| == References ==
| |
| <references /> | |
| *{{Citation | last1=Clausen | first1=Thomas | title=Theorem | doi=10.1002/asna.18400172204 | year=1840 | journal=[[Astronomische Nachrichten]] | volume=17 | issue=22 | pages=351–352}}
| |
| *{{Citation | last1=Rado | first1=R. | title=A New Proof of a Theorem of V. Staudt | doi=10.1112/jlms/s1-9.2.85 | year=1934 | journal=J. London Math. Soc. | volume=9 | issue=2 | pages=85–88}}
| |
| *{{Citation | last1=von Staudt | first1=Ch. | title=Beweis eines Lehrsatzes, die Bernoullischen Zahlen betreffend | url=http://resolver.sub.uni-goettingen.de/purl?GDZPPN002142562 | id={{ERAM|021.0672cj}} | year=1840 | journal=Journal für Reine und Angewandte Mathematik | issn=0075-4102 | volume=21 | pages=372–374}}
| |
| | |
| ==External links==
| |
| * {{MathWorld |urlname=vonStaudt-ClausenTheorem |title=von Staudt-Clausen Theorem}}
| |
| | |
| {{DEFAULTSORT:Von Staudt Clausen Theorem}}
| |
| [[Category:Theorems in number theory]]
| |
I woke up another day and noticed - I've been single for a little while now and following much bullying from buddies I today find myself opted for web dating. They guaranteed me that there are a lot of pleasant, regular and enjoyable people to meet, so here goes the pitch!
I make an effort vip tickets to luke bryan to stay as toned as potential staying at the fitness center several times per week. I appreciate my sports and make an effort to perform luke bryan about (www.museodecarruajes.org) or watch while many a possible. Being wintertime I will often at Hawthorn suits. Note: I've noticed the carnage of wrestling fits at stocktake revenue, In case that you really contemplated purchasing an activity I really don't brain.
My buddies and household are awe-inspiring and spending time with them at tavern gigs or dinners is obviously essential. I haven't ever been in to nightclubs as I discover that one can not get a nice dialogue together with the sound. I likewise have luke bryan dates 2 unquestionably luke bryan luke bryan cheeky and quite cunning canines that are almost always excited to meet up fresh people.
my webpage; luke bryan concert tours