عدد برنولي

أعداد برنولي B±n

n	الكسر	العشري
0	1	+1.000000000
1	±1/2	±0.500000000
2	1/6	+0.166666666
3	0	+0.000000000
4	−1/30	−0.033333333
5	0	+0.000000000
6	1/42	+0.023809523
7	0	+0.000000000
8	−1/30	−0.033333333
9	0	+0.000000000
10	5/66	+0.075757575
11	0	+0.000000000
12	−691/2730	−0.253113553
13	0	+0.000000000
14	7/6	+1.166666666
15	0	+0.000000000
16	−3617/510	−7.092156862
17	0	+0.000000000
18	43867/798	+54.97117794
19	0	+0.000000000
20	−174611/330	−529.1242424

في الرياضيات ، أعداد برنولي Bernoulli numbers ، هي متتالية من الأعداد الكسرية ذات علاقة وثيقة مع نظرية الأعداد، ويرتبطون بقيم اقتران ريمان الزائي للأرقام الصحيحة السالبة.

وفي اوروبا، كان أول من درس تلك الأعداد جاكوب برنولي، مما حدا بعالم الرياضيات ابراهام ده مواڤر، أن يسميهم "أعداد برنولي". وفي اليابان، ربما قبل اوروبا، فقد اكتشفهم سـِكي كـُوا. ويظهرون في فرد متسلسلة تايلور لدوال الظل (ظا) والظل الزائدي، في صيغة اويلر–مكلورين، وفي التعبيرات عن قيم معينة لاقتران ريمان الزائي.

وفي ملحوظة G في ملاحظات ادا لڤليس على الآلة التحليلية التي كتبتها سنة 1842، تصف لڤليس خوارزمية لتوليد أعداد برنولي بواسطة ماكينة بابدج . ونتيجة لذلك، فأعداد برنولي نالهم شرف أنهم كانوا موضوع أول برنامج حاسوب.

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 مقدمة

أعداد برنولي B_n، أحياناً يـُرمز لها b_n. تم اكتشافهم لأول مرة فيما يرتبط مع الأشكال المغلقة لعمليات الجمع

$\sum _{k=0}^{m-1}k^{n}=0^{n}+1^{n}+2^{n}+\cdots +{(m-1)}^{n}$

لمختلف القيم المحددة للمتغير n. الأشكال المغلقة هم دوماً متعددات حدود في m من الدرجة n + 1. معاملات عديدات الحدود تلك يرتبطن بدرجة وثيقة بأعداد برنولي، التي على صلة بصيغة فاولهابر:

$\sum _{k=0}^{m-1}k^{n}={1 \over {n+1}}\sum _{k=0}^{n}{n+1 \choose {k}}B_{k}m^{n+1-k}.$

فعلى سبيل المثال، لو أخذنا n لتساوي 1،

$0+1+2+\cdots +(m-1)={\frac {1}{2}}\left(B_{0}m^{2}+2B_{1}m^{1}\right)={\frac {1}{2}}\left(m^{2}-m\right).$

 الدوال المولـِّدة

${\frac {x}{e^{x}-1}}=\sum _{n=0}^{\infty }B_{n}{\frac {x^{n}}{n!}}$

$x=\left(\sum _{j=1}^{\infty }{\frac {x^{j}}{j!}}\right)\left(\sum _{k=0}^{\infty }{\frac {B_{k}x^{k}}{k!}}\right).$

وبفردها كحاضل ضرب كوشي وإعادة ترتيب قليلاً، نحصل على

$x=\sum _{m=0}^{\infty }\left(\sum _{j=0}^{m}{m+1 \choose j}B_{j}\right){\frac {x^{m+1}}{(m+1)!}}.$

 أشكال ورموز أخرى

$\sum _{k=0}^{m-1}k^{n}={\frac {B_{n+1}(m)-B_{n+1}(0)}{n+1}},$

$\sum _{j=0}^{m}{m+1 \choose {j}}B_{j}=0$

فلكل m > 0، و B₀ = 1.

$\sum _{j=0}^{m}{m+1 \choose {j}}B_{j}=m+1.$

$\sum _{j=0}^{m}{m \choose {j-1}}{\frac {B_{j}}{j}}=1$

${\frac {xe^{x}}{e^{x}-1}}.$

 قيم أعداد برنولي

B_n = 0 لكل قيم n الفردية عدا 1. B₁ = 1/2 أو −1/2 حسب الاصطلاح المستخدم (انظر لاحقاً). أول بضعة أداد برنولي غير صفرية (المتتاليات OEIS: A027641 و OEIS: A027642 في OEIS) وبعض الأعداد الأكبر مذكورين أدناه.

n	N	D	Bn = N / D
0	1	1	+1.00000000000
1	-1	2	-0.50000000000
2	1	6	+0.16666666667
4	-1	30	-0.03333333333
6	1	42	+0.02380952381
8	-1	30	-0.03333333333
10	5	66	+0.07575757576
12	-691	2730	-0.25311355311
14	7	6	+1.16666666667
16	-3617	510	-7.09215686275
18	43867	798	+54.9711779448

 الحساب الفعال لأعداد برنولي

في بعض التطبيقات، من المفيد أن تستطيع حساب أعداد برنولي B₀ إلى B_p − 3 modulo p، حيث p هو a prime; for example to test whether Vandiver's conjecture holds for p, or even just to determine whether p is an irregular prime. It is not feasible to carry out such a computation using the above recursive formulae, since at least (a constant multiple of) p² arithmetic operations would be required. Fortunately, faster methods have been developed (Buhler et al. 2001) which require only O(p (log p)²) operations (see big-O notation).

وصف ديڤد هارڤي (Harvey 2008) خوارزمية لحساب أعداد برنولي بحساب B_n modulo p for للعديد من الأعداد الأولية الصغيرة p, ثم يعيد بناء B_n باستخدام Chinese Remainder Theorem. Harvey writes that the asymptotic time complexity of this algorithm is O(n² log(n)^2+eps) and claims that this implementation is significantly faster than implementations based on other methods. Harvey's implementation is included in Sage since version 3.1. Using this implementation Harvey computed B_n for n = 10⁸, which is a new record (October 2008). Prior to that Bernd Kellner (Kellner 2002) computed B_n to full precision for n = 10⁶ on December 2002 واُلكسندر پاڤليك (Pavlyk 2008) for n = 10⁷ with 'Mathematica' on April 2008.

تاريخ حساب أعداد برنولي

الحاسب	السنة	n	الأرقام
ي. برنولي	~1689	10	1
ل. اويلر	1748	30	8
J.C. Adams	1878	62	36
د. إ. كنوث، ت.ج. بكهولتس	1967	360	478
G. Fee, S. Plouffe	1996	10000	27677
G. Fee, S. Plouffe	1996	100000	376755
B.C. Kellner	2002	1000000	4767529
اُلكسندر پاڤليك	2008	10000000	57675260
د. هارڤي	2008	100000000	676752569

 إعادة بناء 'Summae Potestatum'

ياكوب برنولي: Summae Potestatum، سنة 1713

$\sum _{0<k\leq n}k^{c}=\sum _{k\geq 0}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}\ .$

 وجهات نظر مختلفة

• أرقام برنولى هى كيان رياضى قائم بذاته .
  إرتباط تسلسلى: 1/6, −1/30, 1/42, −1/30,...
  هذه هي وجهة نظر ياكوب بيرنولي. (أنظر الى النبذة من Ars Conjectandi, الطبعة الأولى, 1713). أعداد برنولي تفهم كأرقام , تكرارية في طبيعتها,أبتكرت لحل مسائل أو معضلات رياضية معينة, جمع الأسس (الجمع من أس), الذى يعنىparadigmatic application لبرنولي. أنه لمضلل أن نسمى وحهة النظر تلك 'قديمة'. فمثلا چين -بيير سيريه ويستخدم في تقريره الكتاب الرائع فصل في دراسة الرياضيات وهو من الكتب الأساسية المستخدمة في كثير من الجامعات اليوم .
• أعداد بيرنولي أشياء تجميعية .
  تسلسل إرتباطى: 1, +1/2, 1/6, 0,....
  هذا الرأي ، ويركز على العلاقة بين ستيرلنغ وأرقام وأعداد بيرنولي من الطبيعي أن تنشأ في حساب التفاضل والتكامل المحدد الإختلافات. في أكثر صوره عمومية و إدماجا من تلك العلاقة نلخصها بالتعريف أن ستيرلنج عديدة الحدود σ_n(x), formula (6.52) في Concrete Mathematics by Graham, Knuth and Patashnik.

  $\left({\frac {ze^{z}}{e^{z}-1}}\right)^{x}=x\sum _{n\geq 0}\sigma _{n}(x)z^{n}$

  In consequence B_n = n! σ_n(1) for n ≥ 0.
• Bernoulli numbers as values of a sequence of certain polynomials.
  Assuming the Bernoulli polynomials as already introduced the Bernoulli numbers can be defined in two different ways:
  B_n = B_n(0). Associated sequence: 1, −1/2, 1/6, 0,....
  B_n = B_n(1). Associated sequence: 1, +1/2, 1/6, 0,....
  The two definitions differ only in the sign of B₁. The choice B_n = B_n(0) is the convention used in the Handbook of Mathematical Functions.
• Bernoulli numbers as values of the Riemann zeta function.
  Associated sequence: 1, +1/2, 1/6, 0,....
  
  

  The Bernoulli numbers as given by the Riemann zeta function.

  This convention agrees with the convention B_n = B_n(1) (for example J. Neukirch and M. Kaneko). The sign '+' for B₁ matches the representation of the Bernoulli numbers by the Riemann zeta function. In fact the identity nζ(1−n) = (−1)ⁿ⁺¹B_n valid for all n > 0 is then replaced by the simpler nζ(1−n) = -B_n. (See the paper of S. C. Woon.)
  (Note that in the foregoing equation for n = 0 and n = 1 the expression −nζ(1 − n) is to be understood as lim_x → n −xζ(1 − x).)

$\ B_{n}=n!\sigma _{n}(1)=B_{n}(1)=-n\zeta (1-n)\quad (n\geq 0)\$

 تطبيقات أعداد برنولي

Arguably the most important application of the Bernoulli number in mathematics is their use in the Euler-MacLaurin formula. Assuming that f is a sufficiently often differentiable function the Euler-MacLaurin formula can be written as

$\sum \limits _{a\leq k<b}f(k)=\int _{a}^{b}f(x)dx\ +\sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

This formulation assumes the convention B₁ = -1/2. However, if one sets B₁ = 1/2 then this formula can also be written as

$\sum \limits _{a<k\leq b}f(k)=\int _{a}^{b}f(x)dx\ \ +\ \sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

Here f⁽⁰⁾ = f which is a commonly used notation identifying the zero-th derivative of f with f. Moreover, let f^(-1) denote an antiderivative of f. By the fundamental theorem of calculus $\int _{a}^{b}f(x)dx\ =f^{(-1)}(b)-f^{(-1)}(a)$. Thus the last formula can be further simplified to the following succinct form of the Euler-Maclaurin formula

$\sum \limits _{a<k\leq b}f(k)=\sum \limits _{k=0}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

This form is for example the source for the important Euler-MacLaurin expansion of the zeta function

${\begin{aligned}\zeta (s)&=\sum _{k=0}^{m}{\frac {{B}_{k}}{k!}}s^{\overline {k-1}}+R(s,m)\\&={\frac {{B}_{0}}{0!}}s^{\overline {-1}}+{\frac {{B}_{1}}{1!}}s^{\overline {0}}+{\frac {{B}_{2}}{2!}}s^{\overline {1}}+\cdots +R(s,m)\\&={\frac {1}{s-1}}+{\frac {1}{2}}+{\frac {1}{12}}s+\cdots +R(s,m)\ .\\\end{aligned}}$

Here $s^{\overline {k}}$ denotes the rising factorial power.

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 تعاريف إندماجية

 Connection with the Worpitzky number

$W_{n,k}=\sum _{v=0}^{k}(-1)^{v+k}\left(v+1\right)^{n}{\frac {k!}{v!(k-v)!}}\ .$

$W_{n,k}=k!\left\{{\begin{matrix}n+1\\k+1\end{matrix}}\right\}.$

A Bernoulli number is then introduced as an inclusion-exclusion sum of Worpitzky numbers weighted by the sequence 1, 1/2, 1/3,...

$B_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {W_{n,k}}{k+1}}\ .$

Worpitzky's representation of the Bernoulli number
B0	=	1/1
B1	=	1/1 − 1/2
B2	=	1/1 − 3/2 + 2/3
B3	=	1/1 − 7/2 + 12/3 − 6/4
B4	=	1/1 − 15/2 + 50/3 − 60/4 + 24/5
B5	=	1/1 − 31/2 + 180/3 − 390/4 + 360/5 − 120/6
B6	=	1/1 − 63/2 + 602/3 − 2100/4 + 3360/5 − 2520/6 + 720/7

This representation has B₁ = 1/2.

 العلاقة بعدد فئة سترلنگ

A similar combinatorial representation derives from

$B_{n}=\sum _{k=0}^{n}(-1)^{k+n}{\frac {k!}{k+1}}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}\ .$

$B_{0}=+\left\vert \ \left\{\varnothing \right\}\ \right\vert$

$B_{1}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{a\right\}\right\vert$

$B_{2}=+{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert -{\frac {1}{2}}\left\vert \left\{aa\right\}\right\vert +{\frac {1}{3}}\left\vert \left\{ab,ba\right\}\right\vert$

$B_{3}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{aaa\right\}\right\vert -{\frac {1}{3}}\left\vert \left\{aab,aba,baa,abb,bab,bba\right\}\right\vert +{\frac {1}{4}}\left\vert \left\{abc,acb,bac,bca,cab,cba\right\}\right\vert$

 العلاقة بعدد دورة سترلنگ

Let $\left[{\begin{matrix}n\\m\end{matrix}}\right]$ denote the signless Stirling cycle number. The two main formulas relating these number to the Bernoulli number (B₁ = 1/2) are

${\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{\begin{matrix}m+1\\k+1\end{matrix}}\right]B_{k}={\frac {1}{m+1}}\ ,$

and the inversion of this sum (for n ≥ 0, m ≥ 0)

${\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{\begin{matrix}m+1\\k+1\end{matrix}}\right]B_{n+k}=A_{n,m}\ .$

Here the number A_n,m are the rational Akiyama-Tanigawa number, the first few of which are displayed in the following table.

Akiyama-Tanigawa number
n \ m	0	1	2	3	4
0	1	1/2	1/3	1/4	1/5
1	1/2	1/3	1/4	1/5	...
2	1/6	1/6	3/20	...	...
3	0	1/30	...	...	...
4	-1/30	...	...	...	...

These relations lead to a simple algorithm to compute the Bernoulli number. The input is the first row, A_0,m = 1/(m + 1) and the output are the Bernoulli number in the first column A_n,0 = B_n . This transformation is shown in pseudo-code below.

Akiyama-Tanigawa algorithm for Bn

Enter integer n. For m from 0 by 1 to n do     A[m] ← 1/(m + 1)     For j from m by −1 to 1 do         A[j − 1] ← j × (A[j − 1] − A[j]) Return A[0]  (which is Bn).

 العلاقة بعدد اويلر

$\sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }=2^{n+1}(2^{n+1}-1){\frac {B_{n+1}}{n+1}}\ ,$

$\sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }{\binom {n}{m}}^{-1}=(n+1)B_{n}\ .$

Both formulas are valid for n ≥ 0 if B₁ is set to 1/2. If B₁ is set to −1/2 they are valid only for n ≥ 1 and n ≥ 2 respectively.

 تقريب مقارب

$B_{2n}=(-1)^{n+1}{\frac {2(2n)!}{(2\pi )^{2n}}}\left[1+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots \;\right].$

It then follows from the Stirling formula that, as n goes to infinity,

$|B_{2n}|\sim 4{\sqrt {\pi n}}\left({\frac {n}{\pi e}}\right)^{2n}.$

 Integral representation and continuation

The integral

$b(s)=2e^{si\pi /2}\int _{0}^{\infty }{\frac {st^{s}}{1-e^{2\pi t}}}{\frac {dt}{t}}$

has as special values b(2n) = B_2n for n > 0. The integral might be considered as a continuation of the Bernoulli numbers to the complex plane and this was indeed suggested by Peter Luschny in 2004.^{[بحاجة لمصدر]}

For example b(3) = (3/2)ζ(3)Π⁻³Ι and b(5) = −(15/2) ζ(5) Π ⁻⁵Ι. Here ζ(n) denotes the Riemann zeta function and Ι the imaginary unit. It is remarkable that already Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

$p={\frac {3}{2\pi ^{3}}}\left(1+{\frac {1}{2^{3}}}+{\frac {1}{3^{3}}}+{\text{etc.}}\ \right)=0.0581522\ldots \ ,$

$q={\frac {15}{2\pi ^{5}}}\left(1+{\frac {1}{2^{5}}}+{\frac {1}{3^{5}}}+{\text{etc.}}\ \right)=0.0254132\ldots \ .$

 العلاقة بأعداد اويلر و ط π

The Euler numbers are a sequence of integers intimately connected with the Bernoulli numbers. Comparing the asymptotic expansions of the Bernoulli and the Euler numbers shows that the Euler numbers E_2n are in magnitude approximately (2/π)(4²ⁿ − 2²ⁿ) times larger than the Bernoulli numbers B_2n. In consequence:

$\pi \ \sim \ 2\left(2^{2n}-4^{2n}\right){\frac {B_{2n}}{E_{2n}}}\ .$

This asymptotic equation reveals that π lies in the common root of both the Bernoulli and the Euler numbers. In fact π could be computed from these rational approximations.

Bernoulli numbers can be expressed through the Euler numbers and vice versa. Since for n odd B_n = E_n = 0 (with the exception B₁), it suffices to regard the case when n is even.

$B_{n}=\sum _{k=0}^{n-1}{\binom {n-1}{k}}{\frac {n}{4^{n}-2^{n}}}E_{k}\quad (n=2,4,6,\ldots )$

$E_{n}=\sum _{k=1}^{n}{\binom {n}{k-1}}{\frac {2^{k}-4^{k}}{k}}B_{k}\quad (n=2,4,6,\ldots ).$

These conversion formulas express an inverse relation between the Bernoulli and the Euler numbers. But more important, there is a deep arithmetic root common to both kinds of numbers, which can be expressed through a more fundamental sequence of numbers, also closely tied to π. These numbers are defined for n > 1 as

$S_{n}=2\left({\frac {2}{\pi }}\right)^{n}\sum _{k=-\infty }^{\infty }\left(4k+1\right)^{-n}\quad (k=0,-1,1,-2,2,\ldots )$

and S₁ = 1 by convention. The magic of these numbers lies in the fact that they turn out to be rational numbers. This was first proved by Leonhard Euler 1734 in a landmark paper `De summis serierum reciprocarum' (On the sums of series of reciprocals) and fascinated mathematicians ever since. The first few of these numbers are

$1,1,{\frac {1}{2}},{\frac {1}{3}},{\frac {5}{24}},{\frac {2}{15}},{\frac {61}{720}},{\frac {17}{315}},{\frac {277}{8064}},{\frac {62}{2835}},\ldots$

The Bernoulli numbers and Euler numbers are best understood as special views of these numbers, selected from the sequence S_n and scaled for use in special applications.

$B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n\ \operatorname {even} \right]{\frac {n!}{2^{n}-4^{n}}}\,S_{n}\ ,\quad (n=2,3,\ldots )\ ,$

$E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n\ \operatorname {even} \right]n!\,S_{n+1}\quad \qquad (n=0,1,\ldots )\ .$

The expression [n even] has the value 1 if n is even and 0 otherwise (Iverson bracket).

These identities show that the quotient of Bernoulli and Euler numbers at the beginning of this section is just the special case of R_n = 2 S_n / S_n+1 when n is even. The R_n are rational approximations to π and two successive terms always enclose the true value of π. Beginning with n = 1 the sequence starts

$2,4,3,{\frac {16}{5}},{\frac {25}{8}},{\frac {192}{61}},{\frac {427}{136}},{\frac {4352}{1385}},\ldots \quad \longrightarrow \pi \ .$

These rational numbers also appear in the last paragraph of Euler's paper cited above. But it was only in September 2007 that this classical sequence found its way into the Encyclopedia of Integer Sequences (A132049).

 نظرة خوارزمية: مثلث سايدل

The sequence S_n has another unexpected yet important property: The denominators of S_n divide the factorial (n − 1)!. In other words: the numbers T_n = S_n(n − 1)! are integers.

$T_{n}=1,1,1,2,5,16,61,272,1385,7936,50521,353792,\ldots \quad (n=1,2,\ldots )$

Thus the above representations of the Bernoulli and Euler numbers can be rewritten in terms of this sequence as

$B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n{\text{ even}}\right]{\frac {n}{2^{n}-4^{n}}}\,T_{n}\ ,\quad (n=2,3,\ldots )\ ,$

$E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n{\text{ even}}\right]T_{n+1}\quad \quad \qquad (n=0,1,\ldots )\ .$

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers E_n are given immediately by T_2n + 1 and the Bernoulli numbers B_2n are obtained from T_2n by some easy shifting, avoiding rational arithmetic.

What remains is to find a convenient way to compute the numbers T_n. However, already in 1877 Philipp Ludwig von Seidel published an ingenious algorithm which makes it extremely simple to calculate T_n.

Seidel's algorithm for Tn

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 نظرة توافيقية: تباديل متغايرة

•  مقالة مفصلة: تباديل متغايرة

$\tan x=1{\frac {x}{1!}}+2{\frac {x^{3}}{3!}}+16{\frac {x^{5}}{5!}}+272{\frac {x^{7}}{7!}}+7936{\frac {x^{9}}{9!}}+\cdots$

$\sec x=1+1{\frac {x^{2}}{2!}}+5{\frac {x^{4}}{4!}}+61{\frac {x^{6}}{6!}}+1385{\frac {x^{8}}{8!}}+50521{\frac {x^{10}}{10!}}+\cdots$

The coefficients are the Euler numbers of odd and even index, respectively. In consequence the ordinary expansion of tan x + sec x has as coefficients the rational numbers S_n.

$\tan x+\sec x=1+1x+{\frac {1}{2}}x^{2}+{\frac {1}{3}}x^{3}+{\frac {5}{24}}x^{4}+{\frac {2}{15}}x^{5}+{\frac {61}{720}}x^{6}+\cdots$

 الخصائص الحسابية لأعداد برنولي

 مبرهنة فون شتاوت-كلاوسن

Clausen's algorithm for the denominator of Bn
Clausen: function(integer n)	|	n = 88
S = divisors(n);	|	{1, 2, 4, 8, 11, 22, 44, 88}
S = map(k → k + 1, S);	|	{2, 3, 5, 9, 12, 23, 45, 89}
S = select(isprime, S);	|	{2, 3, 5, 23, 89}
return product(S);	|	61410

Von Staudt–Clausen representation of Bn
B0	=	1
B1	=	1 − 1/2
B2	=	1 − 1/2 − 1/3
B4	=	1 − 1/2 − 1/3 − 1/5
B6	=	1 − 1/2 − 1/3 − 1/7
B8	=	1 − 1/2 − 1/3 − 1/5
B10	=	1 − 1/2 − 1/3 − 1/11

The theorem affirms the existence of an integer I_n such that

$B_{n}={\frac {I_{n}}{2}}-\sum _{(p-1)|n}{\frac {1}{p}}\quad (n\geq 0)\ .$

The sum is over the primes p for which p − 1 divides n. These are the same primes which are employed in the Clausen algorithm. The proposition holds true for all n ≥ 0, not only for even n. I₁ = 2 and for odd n > 1, I_n = 1.

Consequences of the von Staudt–Clausen theorem are: the denominators of the Bernoulli numbers are square-free and for n ≥ 2 divisible by 6.

 لماذا تختفي أعداد برنولي الفردية؟

The sum

$\varphi _{k}(n)=\sum _{i=0}^{n}i^{k}-{\frac {n^{k}}{2}}$

can be evaluated for negative values of the index n. Doing so will show that it is an odd function for even values of k, which implies that the sum has only terms of odd index. This and the formula for the Bernoulli sum imply that B_2k+1−m is 0 for m odd and greater than 1; and that the term for B₁ is cancelled by the subtraction. The von Staudt-Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for n > 1).

From the von Staudt-Clausen theorem it is known that for odd n > 1 the number 2B_n is an integer. This seems trivial if one knows beforehand that in this case B_n = 0. However, by applying Worpitzky's representation one gets

$2B_{n}=\sum _{m=0}^{n}\left(-1\right)^{m}{\frac {2}{m+1}}m!\left\{{\begin{matrix}n+1\\m+1\end{matrix}}\right\}=0\quad \left(n>1\ {\text{odd}}\right)$

as a sum of integers, which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let S_n,m be the number of surjective maps from {1, 2, ..., n} to {1, 2, ..., m}, then $S_{n,m}=m!\left\{{\begin{matrix}n\\m\end{matrix}}\right\}$. The last equation can only hold if

$\sum _{m=1,3,5,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}=\sum _{m=2,4,6,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}\quad \left(n>2{\text{ even}}\right).\$

This equation can be proved by induction. The first two examples of this equation are

n = 4 :  2 + 8 = 7 + 3,  n = 6:  2 + 120 + 144 = 31 + 195 + 40.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

 p-adic continuity

An especially important congruence property of the Bernoulli numbers can be characterized as a p-adic continuity property. If b, m and n are positive integers such that m and n are not divisible by p − 1 and $m\equiv n\,{\bmod {\,}}p^{b-1}(p-1)$, then

$(1-p^{m-1}){B_{m} \over m}\equiv (1-p^{n-1}){B_{n} \over n}\,{\bmod {\,}}p^{b}.$

Since $B_{n}=-n\zeta (1-n)$, this can also be written

$(1-p^{-u})\zeta (u)\equiv (1-p^{-v})\zeta (v)\,{\bmod {\,}}p^{b}\,,$

where u = 1 − m and v = 1 − n, so that u and v are nonpositive and not congruent to 1 mod p − 1. This tells us that the Riemann zeta function, with 1 − p^−s taken out of the Euler product formula, is continuous in the p-adic numbers on odd negative integers congruent mod p − 1 to a particular $a\not \equiv 1\,{\bmod {\,}}p-1$, and so can be extended to a continuous function $\zeta _{p}(s)$ for all p-adic integers ${\mathbb {Z}}_{p},\,$ the p-adic Zeta function.

 أعداد برنولي وفرضية ريمان

For every ε > 1/4 there exists a constant C_ε > 0 (depending on ε) such that |R(x)| < C_ε x^ε as x → ∞. Here R(x) is the Riesz function

$R(x)=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\left(B_{2k}/(2k)\right)}}=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\beta _{2k}}}.\$

$n^{\overline {k}}$ denotes the rising factorial power in the notation of D. E. Knuth. The number β_n = B_n/n occur frequently in the study of the zeta function and are significant because β_n is a p-integer for primes p where p − 1 does not divide n. The β_n are called divided Bernoulli number.

 Ramanujan's congruences

The following relations, due to Ramanujan, provide a more efficient method for calculating Bernoulli numbers:

$m\equiv 0\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{m/6}{m+3 \choose {m-6j}}B_{m-6j}$

$m\equiv 2\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{(m-2)/6}{m+3 \choose {m-6j}}B_{m-6j}$

$m\equiv 4\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}=-{{m+3} \over 6}-\sum _{j=1}^{(m-4)/6}{m+3 \choose {m-6j}}B_{m-6j}.$

 استخدام أعداد برنولي في الطوبولجيا

The Kervaire-Milnor formula for the order of the cyclic group of diffeomorphism classes of exotic (4n − 1)-spheres which bound parallelizable manifolds for $n\geq 2$ involves Bernoulli numbers; if B_(n) is the numerator of B_4n/n, then

$2^{2n-2}(1-2^{2n-1})B_{(n)}$

is the number of such exotic spheres. (The formula in the topological literature differs because topologists use a different convention for naming Bernoulli numbers; this article uses the number theorists' convention.)

 هويات متنوعة

• Let n be non-negative and even

  $\zeta (n)={\frac {\left(-1\right)^{{\frac {n}{2}}-1}B_{n}\left(2\pi \right)^{n}}{2n!}}$

• The nth cumulant of the uniform probability distribution on the interval [−1, 0] is B_n/n.
• Let n¡ = 1/n! and n ≥ 1. Then B_n is n! times the determinant of the following matrix.

  Bn =

  1 ─── n¡

  2¡ 1¡ 0 0 0 ... 0 3¡ 2¡ 1¡ 0 0 ... 0 4¡ 3¡ 2¡ 1¡ 0 ... 0 ... ... ... ... ... ... ... (n − 2)¡ ... ... 3¡ 2¡ 1¡ 0 (n − 1)¡ (n − 2)¡ ... ... 3¡ 2¡ 1¡ n¡ (n − 1)¡ (n − 2)¡ ... ... 3¡ 2¡

  Thus the determinant is σ_n(1), the Stirling polynomial at x = 1.

• Let n ≥ 1.

  ${\frac {1}{n}}\sum _{k=1}^{n}{\binom {n}{k}}B_{k}B_{n-k}+B_{n-1}=-B_{n}\quad {\text{(L. Euler)}}$

• Let n ≥ 1. Then (von Ettingshausen 1827)

  $\sum _{k=0}^{n}{\binom {n+1}{k}}(n+k+1)B_{n+k}=0$

• Let n ≥ 1 and m ≥ 1. Then (Carlitz 1968)

  $(-1)^{m}\sum _{r=0}^{m}{\binom {m}{r}}B_{n+r}=(-1)^{n}\sum _{s=0}^{n}{\binom {n}{s}}B_{m+s}$

• Let n ≥ 4 and

  $H_{n}=\sum _{1\leq k\leq n}k^{-1}$

  the harmonic number. Then

  ${\frac {n}{2}}\sum _{k=2}^{n-2}{\frac {B_{n-k}}{n-k}}{\frac {B_{k}}{k}}-\sum _{k=2}^{n-2}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}=H_{n}B_{n}\qquad {\text{(H. Miki, 1978)}}$

• Let n ≥ 4. Yuri Matiyasevich found (1997)

  $(n+2)\sum _{k=2}^{n-2}B_{k}B_{n-k}-2\sum _{l=2}^{n-2}{\binom {n+2}{l}}B_{l}B_{n-l}=n(n+1)B_{n}$

• Let n ≥ 1

  ${\frac {n}{2}}\left(B_{n-1}(x)+\sum _{k=1}^{n-1}{\frac {B_{k}(x)}{k}}{\frac {B_{n-k}(x)}{n-k}}\right)-\sum _{k=0}^{n-1}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}(x)=H_{n-1}B_{n}(x)$

  This is an identity by Faber-Pandharipande-Zagier-Gessel. Choose x = 0 or x = 1 to get a Bernoulli number identity according to your favourite convention.
• The next formula is true for n ≥ 0 if B₁ = B₁(1) = 1/2, but only for n ≥ 1 if B₁ = B₁(0) = -1/2.

  $\sum _{k=0}^{n}{\binom {n}{k}}{\frac {B_{k}}{n-k+2}}={\frac {B_{n+1}}{n+1}}$

• Let n ≥ 0 and [b] = 1 if b is true, 0 otherwise.

  $-1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(1)=2^{n}$

  and

  $-1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(0)=[n=0]$

 فك لغز أرقام برنولي

محمد التميمي يشرح تفسير أعداد برنولي

استطاع محمد التميمي 16 سنة ، وهو فتى عراقي يعيش في سويسرا منذ ستة أعوام، وفي غضون أربعة أشهر فقط وضع صيغة لتفنسير وتبسيط أرقام برنولي. ولم يقتنع معلمي التميمي في البداية بعمله، مما دعاه إلى الاتصال بأساتذة فيجامعة ابسالا، وهي إحدى أعرق الجامعات السويدية ليطلب منهم التدقيق في عمله.وأضافت الوكالة أن الأساتذة وجدوا أن حله صحيح وعرضوا عليه الانضمام إلى جامعة ابسالا. التميمي أشار إلى أنه يريد التركيز حاليا على دراسته المدرسية وينوي تلقي دروس مكثفة في الرياضيات والفيزياء خلال الصيف، لافتا إلى رغبته في أن يكون باحثا في الفيزياء أو الرياضيات.

 انظر أيضا

• برنولي
• poly-Bernoulli number
• q-Bernoulli number
• Bernoulli polynomials
• Riemann zeta function
• Hurwitz zeta function
• Euler number
• Euler summation
• محمد التميمي

 المصادر

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• قالب:Neukirch ANT
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 وصلات خارجية

• The first 498 Bernoulli Numbers from Project Gutenberg
• The first 10,000 Bernoulli numbers
• A multimodular algorithm for computing Bernoulli numbers
• The Bernoulli Number Page
• Bernoulli number programs at LiteratePrograms
• Dilcher, K.; Skula, L.; Slavutskii, I. Sh. (1991), "Bernoulli numbers. Bibliography (1713-1990)", Queen's Papers in Pure and Applied Mathematics (Kingston, Ontario) (87), http://www.mscs.dal.ca/~dilcher/bernoulli.html 
• Eric W. Weisstein, Bernoulli Number at MathWorld.
• The Computation of Irregular Primes
• Online Bernoulli Numbers Generator
• Some properties,sums of Bernoulli-and related numbers