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Bernoulli number

From Wikipedia, the free encyclopedia
Rational number sequence

Bernoulli numbersB±
n
nfractiondecimal
01+1.000000000
1±⁠1/2⁠±0.500000000
2⁠1/6⁠+0.166666666
30+0.000000000
4−⁠1/30⁠−0.033333333
50+0.000000000
6⁠1/42⁠+0.023809523
70+0.000000000
8−⁠1/30⁠−0.033333333
90+0.000000000
10⁠5/66⁠+0.075757575
110+0.000000000
12−⁠691/2730⁠−0.253113553
130+0.000000000
14⁠7/6⁠+1.166666666
150+0.000000000
16−⁠3617/510⁠−7.092156862
170+0.000000000
18⁠43867/798⁠+54.97117794
190+0.000000000
20−⁠174611/330⁠−529.1242424

Inmathematics, theBernoulli numbersBn are asequence ofrational numbers which occur frequently inanalysis. The Bernoulli numbers appear in (and can be defined by) theTaylor series expansions of thetangent andhyperbolic tangent functions, inFaulhaber's formula for the sum ofm-th powers of the firstn positive integers, in theEuler–Maclaurin formula, and in expressions for certain values of theRiemann zeta function.

The values of the first 20 Bernoulli numbers are given in the adjacent table. Two conventions are used in the literature, denoted here byBn−{\displaystyle B_{n}^{-{}}} andBn+{\displaystyle B_{n}^{+{}}}; they differ only forn = 1, whereB1−=−1/2{\displaystyle B_{1}^{-{}}=-1/2} andB1+=+1/2{\displaystyle B_{1}^{+{}}=+1/2}. For every oddn > 1,Bn = 0. For every evenn > 0,Bn is negative ifn is divisible by 4 and positive otherwise. The Bernoulli numbers are special values of theBernoulli polynomialsBn(x){\displaystyle B_{n}(x)}, withBn−=Bn(0){\displaystyle B_{n}^{-{}}=B_{n}(0)} andBn+=Bn(1){\displaystyle B_{n}^{+}=B_{n}(1)}.[1]

The Bernoulli numbers were discovered around the same time by the Swiss mathematicianJacob Bernoulli, after whom they are named, and independently by Japanese mathematicianSeki Takakazu. Seki's discovery was posthumously published in 1712[2][3][4] in his workKatsuyō Sanpō; Bernoulli's, also posthumously, in hisArs Conjectandi of 1713.Ada Lovelace'snote G on theAnalytical Engine from 1842 describes analgorithm for generating Bernoulli numbers withBabbage's machine;[5] it is disputedwhether Lovelace or Babbage developed the algorithm. As a result, the Bernoulli numbers have the distinction of being the subject of the first published complexcomputer program.

Notation

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The superscript± used in this article distinguishes the two sign conventions for Bernoulli numbers. Only then = 1 term is affected:

In the formulas below, one can switch from one sign convention to the other with the relationBn+=(−1)nBn−{\displaystyle B_{n}^{+}=(-1)^{n}B_{n}^{-}}, or for integern = 2 or greater, simply ignore it.

SinceBn = 0 for all oddn > 1, and many formulas only involve even-index Bernoulli numbers, a few authors write "Bn" instead ofB2n . This article does not follow that notation.

History

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Early history

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The Bernoulli numbers are rooted in the early history of the computation of sums of integer powers, which have been of interest to mathematicians since antiquity.

A page fromSeki Takakazu'sKatsuyō Sanpō (1712), tabulating binomial coefficients and Bernoulli numbers

Methods to calculate the sum of the firstn positive integers, the sum of the squares and of the cubes of the firstn positive integers were known, but there were no real 'formulas', only descriptions given entirely in words. Among the great mathematicians of antiquity to consider this problem werePythagoras (c. 572–497 BCE, Greece),Archimedes (287–212 BCE, Italy),Aryabhata (b. 476, India),Al-Karaji (d. 1019, Persia) andIbn al-Haytham (965–1039, Iraq).

During the late sixteenth and early seventeenth centuries mathematicians made significant progress. In the WestThomas Harriot (1560–1621) of England,Johann Faulhaber (1580–1635) of Germany,Pierre de Fermat (1601–1665) and fellow French mathematicianBlaise Pascal (1623–1662) all played important roles.

Thomas Harriot seems to have been the first to derive and write formulas for sums of powers using symbolic notation, but even he calculated only up to the sum of the fourth powers. Johann Faulhaber gave formulas for sums of powers up to the 17th power in his 1631Academia Algebrae, far higher than anyone before him, but he did not give a general formula.

Blaise Pascal in 1654 provedPascal's identity relating(n+1)k+1 to the sums of thepth powers of the firstn positive integers forp = 0, 1, 2, ...,k.

The Swiss mathematician Jacob Bernoulli (1654–1705) was the first to realize the existence of a single sequence of constantsB0,B1,B2,... which provide a uniform formula for all sums of powers.[9]

The joy Bernoulli experienced when he hit upon the pattern needed to compute quickly and easily the coefficients of his formula for the sum of thecth powers for any positive integerc can be seen from his comment. He wrote:

"With the help of this table, it took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum 91,409,924,241,424,243,424,241,924,242,500."

Bernoulli's result was published posthumously inArs Conjectandi in 1713.Seki Takakazu independently discovered the Bernoulli numbers and his result was published a year earlier, also posthumously, in 1712.[2] However, Seki did not present his method as a formula based on a sequence of constants.

Bernoulli's formula for sums of powers is the most useful and generalizable formulation to date. The coefficients in Bernoulli's formula are now called Bernoulli numbers, following a suggestion ofAbraham de Moivre.

Bernoulli's formula is sometimes calledFaulhaber's formula after Johann Faulhaber who found remarkable ways to calculate sum of powers but never stated Bernoulli's formula. According to Knuth[9] a rigorous proof of Faulhaber's formula was first published byCarl Jacobi in 1834.[10] Knuth's in-depth study of Faulhaber's formula concludes (the nonstandard notation on the LHS is explained further on):

"Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a single sequence of constantsB0,B1,B2,... would provide a uniform
∑nm=1m+1(B0nm+1−(m+11)B1nm+(m+12)B2nm−1−⋯+(−1)m(m+1m)Bmn){\textstyle \sum n^{m}={\frac {1}{m+1}}\left(B_{0}n^{m+1}-{\binom {m+1}{1}}B_{1}n^{m}+{\binom {m+1}{2}}B_{2}n^{m-1}-\cdots +(-1)^{m}{\binom {m+1}{m}}B_{m}n\right)}
for all sums of powers. He never mentioned, for example, the fact that almost half of the coefficients turned out to be zero after he had converted his formulas forΣnmfrom polynomials inN to polynomials inn."[11]

In the above Knuth meantB1−{\displaystyle B_{1}^{-}}; instead usingB1+{\displaystyle B_{1}^{+}} the formula avoids subtraction:

∑nm=1m+1(B0nm+1+(m+11)B1+nm+(m+12)B2nm−1+⋯+(m+1m)Bmn).{\textstyle \sum n^{m}={\frac {1}{m+1}}\left(B_{0}n^{m+1}+{\binom {m+1}{1}}B_{1}^{+}n^{m}+{\binom {m+1}{2}}B_{2}n^{m-1}+\cdots +{\binom {m+1}{m}}B_{m}n\right).}

Reconstruction of "Summae Potestatum"

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Jacob Bernoulli's "Summae Potestatum", 1713[a]

The Bernoulli numbersOEIS: A164555(n)/OEIS: A027642(n) were introduced by Jacob Bernoulli in the bookArs Conjectandi published posthumously in 1713 page 97. The main formula can be seen in the second half of the corresponding facsimile. The constant coefficients denotedA,B,C andD by Bernoulli are mapped to the notation which is now prevalent asA =B2,B =B4,C =B6,D =B8. The expressionc·c−1·c−2·c−3 meansc·(c−1)·(c−2)·(c−3) – the small dots are used as grouping symbols. Using today's terminology these expressions arefalling factorial powersck. The factorial notationk! as a shortcut for1 × 2 × ... ×k was not introduced until 100 years later. The integral symbol on the left hand side goes back toGottfried Wilhelm Leibniz in 1675 who used it as a long letterS for "summa" (sum).[b] The lettern on the left hand side is not an index ofsummation but gives the upper limit of the range of summation which is to be understood as1, 2, ...,n. Putting things together, for positivec, today a mathematician is likely to write Bernoulli's formula as:

∑k=1nkc=nc+1c+1+12nc+∑k=2cBkk!ck−1_nc−k+1.{\displaystyle \sum _{k=1}^{n}k^{c}={\frac {n^{c+1}}{c+1}}+{\frac {1}{2}}n^{c}+\sum _{k=2}^{c}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}.}

This formula suggests settingB1 =⁠1/2⁠ when switching from the so-called 'archaic' enumeration which uses only the even indices 2, 4, 6... to the modern form (more on different conventions in the next paragraph). Most striking in this context is the fact that thefalling factorialck−1 has fork = 0 the value⁠1/c + 1⁠.[12] Thus Bernoulli's formula can be written

∑k=1nkc=∑k=0cBkk!ck−1_nc−k+1{\displaystyle \sum _{k=1}^{n}k^{c}=\sum _{k=0}^{c}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}}

ifB1 = 1/2, recapturing the value Bernoulli gave to the coefficient at that position.

The formula for∑k=1nk9{\displaystyle \textstyle \sum _{k=1}^{n}k^{9}} in the first half of the quotation by Bernoulli above contains an error at the last term; it should be−320n2{\displaystyle -{\tfrac {3}{20}}n^{2}} instead of−112n2{\displaystyle -{\tfrac {1}{12}}n^{2}}.

Definitions

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Many characterizations of the Bernoulli numbers have been found in the last 300 years, and each could be used to introduce these numbers. Here only four of the most useful ones are mentioned:

  • a recursive equation,
  • an explicit formula,
  • a generating function,
  • an integral expression.

For the proof of theequivalence of the four approaches, seeIreland & Rosen (1990) orConway & Guy (1996).

Recursive definition

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The Bernoulli numbers obey the sum formulas[1]

∑k=0m(m+1k)Bk−=δm,0∑k=0m(m+1k)Bk+=m+1{\displaystyle {\begin{aligned}\sum _{k=0}^{m}{\binom {m+1}{k}}B_{k}^{-{}}&=\delta _{m,0}\\\sum _{k=0}^{m}{\binom {m+1}{k}}B_{k}^{+{}}&=m+1\end{aligned}}}

wherem=0,1,2...{\displaystyle m=0,1,2...} andδ denotes theKronecker delta.

The first of these is sometimes written[13] as the formula (for m > 1)(B+1)m−Bm=0,{\displaystyle (B+1)^{m}-B_{m}=0,}where the power is expanded formally using the binomial theorem andBk{\displaystyle B^{k}} is replaced byBk{\displaystyle B_{k}}.

Solving forBm∓{\displaystyle B_{m}^{\mp {}}} gives the recursive formulas[14]

Bm−=δm,0−∑k=0m−1(mk)Bk−m−k+1Bm+=1−∑k=0m−1(mk)Bk+m−k+1.{\displaystyle {\begin{aligned}B_{m}^{-{}}&=\delta _{m,0}-\sum _{k=0}^{m-1}{\binom {m}{k}}{\frac {B_{k}^{-{}}}{m-k+1}}\\B_{m}^{+}&=1-\sum _{k=0}^{m-1}{\binom {m}{k}}{\frac {B_{k}^{+}}{m-k+1}}.\end{aligned}}}

Explicit definition

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In 1893Louis Saalschütz listed a total of 38 explicit formulas for the Bernoulli numbers,[15] usually giving some reference in the older literature. One of them is (form≥1{\displaystyle m\geq 1}):

Bm−=∑k=0m1k+1∑j=0k(kj)(−1)jjmBm+=∑k=0m1k+1∑j=0k(kj)(−1)j(j+1)m.{\displaystyle {\begin{aligned}B_{m}^{-}&=\sum _{k=0}^{m}{\frac {1}{k+1}}\sum _{j=0}^{k}{\binom {k}{j}}(-1)^{j}j^{m}\\B_{m}^{+}&=\sum _{k=0}^{m}{\frac {1}{k+1}}\sum _{j=0}^{k}{\binom {k}{j}}(-1)^{j}(j+1)^{m}.\end{aligned}}}

Generating function

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The exponentialgenerating functions are

tet−1=t2(coth⁡t2−1)=∑m=0∞Bm−tmm!tetet−1=t1−e−t=t2(coth⁡t2+1)=∑m=0∞Bm+tmm!.{\displaystyle {\begin{alignedat}{3}{\frac {t}{e^{t}-1}}&={\frac {t}{2}}\left(\operatorname {coth} {\frac {t}{2}}-1\right)&&=\sum _{m=0}^{\infty }{\frac {B_{m}^{-{}}t^{m}}{m!}}\\{\frac {te^{t}}{e^{t}-1}}={\frac {t}{1-e^{-t}}}&={\frac {t}{2}}\left(\operatorname {coth} {\frac {t}{2}}+1\right)&&=\sum _{m=0}^{\infty }{\frac {B_{m}^{+}t^{m}}{m!}}.\end{alignedat}}}

where the substitution ist→−t{\displaystyle t\to -t}. The two generating functions only differ byt.

Proof

If we letF(t)=∑i=1∞fiti{\displaystyle F(t)=\sum _{i=1}^{\infty }f_{i}t^{i}} andG(t)=1/(1+F(t))=∑i=0∞giti{\displaystyle G(t)=1/(1+F(t))=\sum _{i=0}^{\infty }g_{i}t^{i}} then

G(t)=1−F(t)G(t).{\displaystyle G(t)=1-F(t)G(t).}

Theng0=1{\displaystyle g_{0}=1} and form>0{\displaystyle m>0} the mth term in the series forG(t){\displaystyle G(t)} is:

gmtm=−∑j=0m−1fm−jgjtm{\displaystyle g_{m}t^{m}=-\sum _{j=0}^{m-1}f_{m-j}g_{j}t^{m}}

If

F(t)=et−1t−1=∑i=1∞ti(i+1)!{\displaystyle F(t)={\frac {e^{t}-1}{t}}-1=\sum _{i=1}^{\infty }{\frac {t^{i}}{(i+1)!}}}

then we find that

G(t)=t/(et−1){\displaystyle G(t)=t/(e^{t}-1)}
m!gm=−∑j=0m−1m!j!j!gj(m−j+1)!=−1m+1∑j=0m−1(m+1j)j!gj{\displaystyle {\begin{aligned}m!g_{m}&=-\sum _{j=0}^{m-1}{\frac {m!}{j!}}{\frac {j!g_{j}}{(m-j+1)!}}\\&=-{\frac {1}{m+1}}\sum _{j=0}^{m-1}{\binom {m+1}{j}}j!g_{j}\\\end{aligned}}}

showing that the values ofi!gi{\displaystyle i!g_{i}} obey the recursive formula for the Bernoulli numbersBi−{\displaystyle B_{i}^{-}}.

The (ordinary) generating function

z−1ψ1(z−1)=∑m=0∞Bm+zm{\displaystyle z^{-1}\psi _{1}(z^{-1})=\sum _{m=0}^{\infty }B_{m}^{+}z^{m}}

is anasymptotic series. It contains thetrigamma functionψ1.

Integral Expression

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From the generating functions above, one can obtain the following integral formula for the even Bernoulli numbers:

B2n=4n(−1)n+1∫0∞t2n−1e2πt−1dt{\displaystyle B_{2n}=4n(-1)^{n+1}\int _{0}^{\infty }{\frac {t^{2n-1}}{e^{2\pi t}-1}}\mathrm {d} t}

Bernoulli numbers and the Riemann zeta function

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Bernoulli numbers, using 1/2 for B1, related to the Riemann zeta function of negative real numbers.
Absolute value of even-index Bernoulli numbers and relation to the Riemann zeta function

The Bernoulli numbers can be expressed in terms of theRiemann zeta function:

Bn+=−nζ(1−n){\displaystyle B_{n}^{+}=-n\,\zeta (1-n)\quad } forn ≥ 1 .

Here the argument of the zeta function is0or negative. Asζ(k){\displaystyle \zeta (k)} is zero for negative even integers (thetrivial zeroes), ifn>1 is odd,ζ(1−n){\displaystyle \zeta (1-n)} is zero.

By means of the zetafunctional equation and the gammareflection formula the following relation can be obtained:[16]

B2n=(−1)n+12(2n)!(2π)2nζ(2n){\displaystyle B_{2n}={\frac {(-1)^{n+1}2(2n)!}{(2\pi )^{2n}}}\zeta (2n)\quad } forn ≥ 1 .

Now the argument of the zeta function is positive.

It then follows fromζ → 1 (n → ∞) andStirling's formula that

|B2n|∼4πn(nπe)2n{\displaystyle |B_{2n}|\sim 4{\sqrt {\pi n}}\left({\frac {n}{\pi e}}\right)^{2n}\quad } forn → ∞ .

Efficient computation of Bernoulli numbers

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In some applications it is useful to be able to compute the Bernoulli numbersB0 throughBp − 3 modulop, wherep is a prime; for example to test whetherVandiver's conjecture holds forp, or even just to determine whetherp is anirregular 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[17] which require onlyO(p (logp)2) operations (seebigO notation).

David Harvey[18] describes an algorithm for computing Bernoulli numbers by computingBn modulop for many small primesp, and then reconstructingBn via theChinese remainder theorem. Harvey writes that theasymptotictime complexity of this algorithm isO(n2 log(n)2 +ε) and claims that thisimplementation is significantly faster than implementations based on other methods. Using this implementation Harvey computedBn forn = 108. Harvey's implementation has been included inSageMath since version 3.1. Prior to that, Bernd Kellner[19] computedBn to full precision forn = 106 in December 2002 and Oleksandr Pavlyk[20] forn = 107 withMathematica in April 2008.

ComputerYearnDigits*
J. Bernoulli~1689101
L. Euler1748308
J. C. Adams18786236
D. E. Knuth, T. J. Buckholtz196716723330
G. Fee,S. Plouffe19961000027677
G. Fee, S. Plouffe1996100000376755
B. C. Kellner200210000004767529
O. Pavlyk20081000000057675260
D. Harvey2008100000000676752569
*Digits is to be understood as the exponent of 10 whenBn is written as a real number in normalizedscientific notation.

Applications of the Bernoulli numbers

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Asymptotic analysis

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Arguably the most important application of the Bernoulli numbers in mathematics is their use in theEuler–Maclaurin formula. Assuming thatf is a sufficiently often differentiable function the Euler–Maclaurin formula can be written as[21]

∑k=ab−1f(k)=∫abf(x)dx+∑k=1mBk−k!(f(k−1)(b)−f(k−1)(a))+R−(f,m).{\displaystyle \sum _{k=a}^{b-1}f(k)=\int _{a}^{b}f(x)\,dx+\sum _{k=1}^{m}{\frac {B_{k}^{-}}{k!}}(f^{(k-1)}(b)-f^{(k-1)}(a))+R_{-}(f,m).}

This formulation assumes the conventionB−
1 = −⁠1/2⁠. Using the conventionB+
1 = +⁠1/2⁠ the formula becomes

∑k=a+1bf(k)=∫abf(x)dx+∑k=1mBk+k!(f(k−1)(b)−f(k−1)(a))+R+(f,m).{\displaystyle \sum _{k=a+1}^{b}f(k)=\int _{a}^{b}f(x)\,dx+\sum _{k=1}^{m}{\frac {B_{k}^{+}}{k!}}(f^{(k-1)}(b)-f^{(k-1)}(a))+R_{+}(f,m).}

Heref(0)=f{\displaystyle f^{(0)}=f} (i.e. the zeroth-order derivative off{\displaystyle f} is justf{\displaystyle f}). Moreover, letf(−1){\displaystyle f^{(-1)}} denote anantiderivative off{\displaystyle f}. By thefundamental theorem of calculus,

∫abf(x)dx=f(−1)(b)−f(−1)(a).{\displaystyle \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

∑k=a+1bf(k)=∑k=0mBkk!(f(k−1)(b)−f(k−1)(a))+R(f,m).{\displaystyle \sum _{k=a+1}^{b}f(k)=\sum _{k=0}^{m}{\frac {B_{k}}{k!}}(f^{(k-1)}(b)-f^{(k-1)}(a))+R(f,m).}

This form is for example the source for the important Euler–Maclaurin expansion of the zeta function

ζ(s)=∑k=0mBk+k!sk−1¯+R(s,m)=B00!s−1¯+B1+1!s0¯+B22!s1¯+⋯+R(s,m)=1s−1+12+112s+⋯+R(s,m).{\displaystyle {\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}}}

Heresk denotes therising factorial power.[22]

Bernoulli numbers are also frequently used in other kinds ofasymptotic expansions. The following example is the classical Poincaré-type asymptotic expansion of thedigamma functionψ.

ψ(z)∼ln⁡z−∑k=1∞Bk+kzk{\displaystyle \psi (z)\sim \ln z-\sum _{k=1}^{\infty }{\frac {B_{k}^{+}}{kz^{k}}}}

Sum of powers

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Main article:Faulhaber's formula

Bernoulli numbers feature prominently in theclosed form expression of the sum of themth powers of the firstn positive integers. Form,n ≥ 0 define

Sm(n)=∑k=1nkm=1m+2m+⋯+nm.{\displaystyle S_{m}(n)=\sum _{k=1}^{n}k^{m}=1^{m}+2^{m}+\cdots +n^{m}.}

This expression can always be rewritten as apolynomial inn of degreem + 1. Thecoefficients of these polynomials are related to the Bernoulli numbers byBernoulli's formula:

Sm(n)=1m+1∑k=0m(m+1k)Bk+nm+1−k=m!∑k=0mBk+nm+1−kk!(m+1−k)!,{\displaystyle S_{m}(n)={\frac {1}{m+1}}\sum _{k=0}^{m}{\binom {m+1}{k}}B_{k}^{+}n^{m+1-k}=m!\sum _{k=0}^{m}{\frac {B_{k}^{+}n^{m+1-k}}{k!(m+1-k)!}},}

where(m + 1
k) denotes thebinomial coefficient.

For example, takingm to be 1 gives thetriangular numbers0, 1, 3, 6, ...OEIS: A000217.

1+2+⋯+n=12(B0n2+2B1+n1)=12(n2+n).{\displaystyle 1+2+\cdots +n={\frac {1}{2}}(B_{0}n^{2}+2B_{1}^{+}n^{1})={\tfrac {1}{2}}(n^{2}+n).}

Takingm to be 2 gives thesquare pyramidal numbers0, 1, 5, 14, ...OEIS: A000330.

12+22+⋯+n2=13(B0n3+3B1+n2+3B2n1)=13(n3+32n2+12n).{\displaystyle 1^{2}+2^{2}+\cdots +n^{2}={\frac {1}{3}}(B_{0}n^{3}+3B_{1}^{+}n^{2}+3B_{2}n^{1})={\tfrac {1}{3}}\left(n^{3}+{\tfrac {3}{2}}n^{2}+{\tfrac {1}{2}}n\right).}

Some authors use the alternate convention for Bernoulli numbers and state Bernoulli's formula in this way:

Sm(n)=1m+1∑k=0m(−1)k(m+1k)Bk−nm+1−k.{\displaystyle S_{m}(n)={\frac {1}{m+1}}\sum _{k=0}^{m}(-1)^{k}{\binom {m+1}{k}}B_{k}^{-{}}n^{m+1-k}.}

Bernoulli's formula is sometimes calledFaulhaber's formula afterJohann Faulhaber who also found remarkable ways to calculatesums of powers.

Faulhaber's formula was generalized by V. Guo and J. Zeng to aq-analog.[23]

Taylor series

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The Bernoulli numbers appear in theTaylor series expansion of manytrigonometric functions andhyperbolic functions.

tan⁡x=1x∑n=1∞(−1)n−122n(22n−1)B2n(2n)!x2n−1,|x|<π2.cot⁡x=1x∑n=0∞(−1)nB2n(2x)2n(2n)!,0<|x|<π.tanh⁡x=1x∑n=1∞22n(22n−1)B2n(2n)!x2n−1,|x|<π2.coth⁡x=1x∑n=0∞B2n(2x)2n(2n)!,0<|x|<π.{\displaystyle {\begin{aligned}\tan x&={\hphantom {1 \over x}}\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}}\;x^{2n-1},&&\left|x\right|<{\frac {\pi }{2}}.\\\cot x&={1 \over x}\sum _{n=0}^{\infty }{\frac {(-1)^{n}B_{2n}(2x)^{2n}}{(2n)!}},&0<&|x|<\pi .\\\tanh x&={\hphantom {1 \over x}}\sum _{n=1}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}}\;x^{2n-1},&&|x|<{\frac {\pi }{2}}.\\\coth x&={1 \over x}\sum _{n=0}^{\infty }{\frac {B_{2n}(2x)^{2n}}{(2n)!}},&0<&|x|<\pi .\end{aligned}}}

Laurent series

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The Bernoulli numbers appear in the followingLaurent series:[24]

Digamma function:ψ(z)=ln⁡z−∑k=1∞Bk+kzk{\displaystyle \psi (z)=\ln z-\sum _{k=1}^{\infty }{\frac {B_{k}^{+{}}}{kz^{k}}}}

Use in topology

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TheKervaire–Milnor formula for the order of the cyclic group of diffeomorphism classes ofexotic(4n − 1)-spheres which boundparallelizable manifolds involves Bernoulli numbers. LetESn be the number of such exotic spheres forn ≥ 2, then

ESn=(22n−2−24n−3)Numerator⁡(B4n4n).{\displaystyle {\textit {ES}}_{n}=(2^{2n-2}-2^{4n-3})\operatorname {Numerator} \left({\frac {B_{4n}}{4n}}\right).}

TheHirzebruch signature theorem for theL genus of asmoothorientedclosed manifold ofdimension 4n also involves Bernoulli numbers.

Connections with combinatorial numbers

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The connection of the Bernoulli number to various kinds of combinatorial numbers is based on the classical theory of finite differences and on the combinatorial interpretation of the Bernoulli numbers as an instance of a fundamental combinatorial principle, theinclusion–exclusion principle.

Connection with Worpitzky numbers

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The definition to proceed with was developed by Julius Worpitzky in 1883. Besides elementary arithmetic only the factorial functionn! and the power functionkm is employed. The signless Worpitzky numbers are defined as

Wn,k=∑v=0k(−1)v+k(v+1)n(kv) .{\displaystyle W_{n,k}=\sum _{v=0}^{k}(-1)^{v+k}(v+1)^{n}{k \choose v}\ .}

They can also be expressed through theStirling numbers of the second kind

Wn,k=k!{n+1k+1}.{\displaystyle W_{n,k}=k!\left\{{n+1 \atop k+1}\right\}.}

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

Bn=∑k=0n(−1)kWn,kk+1 = ∑k=0n1k+1∑v=0k(−1)v(v+1)n(kv) .{\displaystyle B_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {W_{n,k}}{k+1}}\ =\ \sum _{k=0}^{n}{\frac {1}{k+1}}\sum _{v=0}^{k}(-1)^{v}(v+1)^{n}{k \choose v}\ .}
B0 = 1
B1 = 1 −⁠1/2⁠
B2 = 1 −⁠3/2⁠ +⁠2/3⁠
B3 = 1 −⁠7/2⁠ +⁠12/3⁠ −⁠6/4⁠
B4 = 1 −⁠15/2⁠ +⁠50/3⁠ −⁠60/4⁠ +⁠24/5⁠
B5 = 1 −⁠31/2⁠ +⁠180/3⁠ −⁠390/4⁠ +⁠360/5⁠ −⁠120/6⁠
B6 = 1 −⁠63/2⁠ +⁠602/3⁠ −⁠2100/4⁠ +⁠3360/5⁠ −⁠2520/6⁠ +⁠720/7⁠

This representation hasB+
1 = +⁠1/2⁠.

Consider the sequencesn,n ≥ 0. From Worpitzky's numbersOEIS: A028246,OEIS: A163626 applied tos0,s0,s1,s0,s1,s2,s0,s1,s2,s3, ... is identical to the Akiyama–Tanigawa transform applied tosn (seeConnection with Stirling numbers of the first kind). This can be seen via the table:

Identity of
Worpitzky's representation and Akiyama–Tanigawa transform
101001000100001
1−102−2003−30004−4
1−3204−106009−2112
1−712−608−3854−24
1−1550−6024

The first row representss0,s1,s2,s3,s4.

Hence for the second fractional Euler numbersOEIS: A198631 (n) /OEIS: A006519 (n + 1):

E0 = 1
E1 = 1 −⁠1/2⁠
E2 = 1 −⁠3/2⁠ +⁠2/4⁠
E3 = 1 −⁠7/2⁠ +⁠12/4⁠ −⁠6/8⁠
E4 = 1 −⁠15/2⁠ +⁠50/4⁠ −⁠60/8⁠ +⁠24/16⁠
E5 = 1 −⁠31/2⁠ +⁠180/4⁠ −⁠390/8⁠ +⁠360/16⁠ −⁠120/32⁠
E6 = 1 −⁠63/2⁠ +⁠602/4⁠ −⁠2100/8⁠ +⁠3360/16⁠ −⁠2520/32⁠ +⁠720/64⁠

A second formula representing the Bernoulli numbers by the Worpitzky numbers is forn ≥ 1

Bn=n2n+1−2∑k=0n−1(−2)−kWn−1,k.{\displaystyle B_{n}={\frac {n}{2^{n+1}-2}}\sum _{k=0}^{n-1}(-2)^{-k}\,W_{n-1,k}.}

The simplified second Worpitzky's representation of the second Bernoulli numbers is:

OEIS: A164555 (n + 1) /OEIS: A027642(n + 1) =⁠n + 1/2n + 2 − 2⁠ ×OEIS: A198631(n) /OEIS: A006519(n + 1)

which links the second Bernoulli numbers to the second fractional Euler numbers. The beginning is:

⁠1/2⁠,⁠1/6⁠, 0, −⁠1/30⁠, 0,⁠1/42⁠, ... = (⁠1/2⁠,⁠1/3⁠,⁠3/14⁠,⁠2/15⁠,⁠5/62⁠,⁠1/21⁠, ...) × (1,⁠1/2⁠, 0, −⁠1/4⁠, 0,⁠1/2⁠, ...)

The numerators of the first parentheses areOEIS: A111701 (seeConnection with Stirling numbers of the first kind).

Connection with Stirling numbers of the second kind

[edit]

Stirling numbers of the second kind,S(k,m), have the property that[25]:

xk=∑m=0kxm_S(k,m){\displaystyle x^{k}=\sum _{m=0}^{k}{x^{\underline {m}}}S(k,m)}

wherexm denotes thefalling factorial function.

TheBernoulli polynomialsBk(x) can be written as:[26]

Bk(x)=k∑m=0k−1(xm+1)S(k−1,m)m!+Bk{\displaystyle B_{k}(x)=k\sum _{m=0}^{k-1}{\binom {x}{m+1}}S(k-1,m)m!+B_{k}}

whereBk fork = 0, 1, 2,... are the Bernoulli numbers.

The following property of thebinomial coefficient:

(xm)=(x+1m+1)−(xm+1){\displaystyle {\binom {x}{m}}={\binom {x+1}{m+1}}-{\binom {x}{m+1}}}

thus implies that

xk=Bk+1(x+1)−Bk+1(x)k+1.{\displaystyle x^{k}={\frac {B_{k+1}(x+1)-B_{k+1}(x)}{k+1}}.}

One also has the following for Bernoulli polynomials,[26]

Bk(x)=∑n=0k(kn)Bnxk−n.{\displaystyle B_{k}(x)=\sum _{n=0}^{k}{\binom {k}{n}}B_{n}x^{k-n}.}

The coefficient ofx in(x
m + 1) is⁠(−1)m/m + 1⁠.

The coefficient ofx in the first expression forBk(x){\displaystyle B_{k}(x)} isk∑m=0k−1(−1)mm+1S(k−1,m)m!{\displaystyle k\sum _{m=0}^{k-1}{\frac {(-1)^{m}}{m+1}}S(k-1,m)m!} whereas in the second expression it iskBk−1.{\displaystyle kB_{k-1}.} Replacingk−1{\displaystyle k-1} byk{\displaystyle k} this yields:

Bk=∑m=0k(−1)mm!m+1S(k,m){\displaystyle B_{k}=\sum _{m=0}^{k}(-1)^{m}{\frac {m!}{m+1}}S(k,m)}

(resulting inB1 = +⁠1/2⁠) which is an explicit formula for Bernoulli numbers and can be used to proveVon-Staudt Clausen theorem.[27][28][29]

Connection with Stirling numbers of the first kind

[edit]

The two main formulas relating the unsignedStirling numbers of the first kind[n
m] to the Bernoulli numbers (withB1 = +⁠1/2⁠) are

1m!∑k=0m(−1)k[m+1k+1]Bk=1m+1,{\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{m+1 \atop k+1}\right]B_{k}={\frac {1}{m+1}},}

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

1m!∑k=0m(−1)k[m+1k+1]Bn+k=An,m.{\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{m+1 \atop k+1}\right]B_{n+k}=A_{n,m}.}

Here the numberAn,m are the rational Akiyama–Tanigawa numbers, the first few of which are displayed in the following table.

Akiyama–Tanigawa number
m
n
01234
01⁠1/2⁠⁠1/3⁠⁠1/4⁠⁠1/5⁠
1⁠1/2⁠⁠1/3⁠⁠1/4⁠⁠1/5⁠...
2⁠1/6⁠⁠1/6⁠⁠3/20⁠......
30⁠1/30⁠.........
4−⁠1/30⁠............

The Akiyama–Tanigawa numbers satisfy a simple recurrence relation which can be exploited to iteratively compute the Bernoulli numbers. This leads to the algorithm shown in the section 'algorithmic description' above. SeeOEIS: A051714/OEIS: A051715.

Anautosequence is a sequence which has its inverse binomial transform equal to the signed sequence. If the main diagonal is zeroes =OEIS: A000004, the autosequence is of the first kind. Example:OEIS: A000045, the Fibonacci numbers. If the main diagonal is the first upper diagonal multiplied by 2, it is of the second kind. Example:OEIS: A164555/OEIS: A027642, the second Bernoulli numbers (seeOEIS: A190339). The Akiyama–Tanigawa transform applied to2−n = 1/OEIS: A000079 leads toOEIS: A198631 (n) /OEIS: A06519 (n + 1). Hence:

Akiyama–Tanigawa transform for the second Euler numbers
m
n
01234
01⁠1/2⁠⁠1/4⁠⁠1/8⁠⁠1/16⁠
1⁠1/2⁠⁠1/2⁠⁠3/8⁠⁠1/4⁠...
20⁠1/4⁠⁠3/8⁠......
3−⁠1/4⁠−⁠1/4⁠.........
40............

SeeOEIS: A209308 andOEIS: A227577.OEIS: A198631 (n) /OEIS: A006519 (n + 1) are the second (fractional) Euler numbers and an autosequence of the second kind.

(⁠OEIS: A164555 (n + 2)/OEIS: A027642 (n + 2)⁠ =⁠1/6⁠, 0, −⁠1/30⁠, 0,⁠1/42⁠, ...) × (⁠2n + 3 − 2/n + 2⁠ =3,⁠14/3⁠,⁠15/2⁠,⁠62/5⁠, 21, ...) =⁠OEIS: A198631 (n + 1)/OEIS: A006519 (n + 2)⁠ =⁠1/2⁠, 0, −⁠1/4⁠, 0,⁠1/2⁠, ....

Also valuable forOEIS: A027641 /OEIS: A027642 (seeConnection with Worpitzky numbers).

Connection with Pascal's triangle

[edit]

There are formulas connecting Pascal's triangle to Bernoulli numbers[c]

Bn+=|An|(n+1)!   {\displaystyle B_{n}^{+}={\frac {|A_{n}|}{(n+1)!}}~~~}

where|An|{\displaystyle |A_{n}|} is the determinant of a n-by-nHessenberg matrix part ofPascal's triangle whose elements are:ai,k={0if k>1+i(i+1k−1)otherwise{\displaystyle a_{i,k}={\begin{cases}0&{\text{if }}k>1+i\\{i+1 \choose k-1}&{\text{otherwise}}\end{cases}}}

Example:

B6+=det(120000133000146400151010501615201561721353521)7!=1205040=142{\displaystyle B_{6}^{+}={\frac {\det {\begin{pmatrix}1&2&0&0&0&0\\1&3&3&0&0&0\\1&4&6&4&0&0\\1&5&10&10&5&0\\1&6&15&20&15&6\\1&7&21&35&35&21\end{pmatrix}}}{7!}}={\frac {120}{5040}}={\frac {1}{42}}}

Connection with Eulerian numbers

[edit]

There are formulas connectingEulerian numbers⟨n
m⟩ to Bernoulli numbers:

∑m=0n(−1)m⟨nm⟩=2n+1(2n+1−1)Bn+1n+1,∑m=0n(−1)m⟨nm⟩(nm)−1=(n+1)Bn.{\displaystyle {\begin{aligned}\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}.\end{aligned}}}

Both formulae are valid forn ≥ 0 ifB1 is set to⁠1/2⁠. IfB1 is set to −⁠1/2⁠ they are valid only forn ≥ 1 andn ≥ 2 respectively.

A binary tree representation

[edit]

The Stirling polynomialsσn(x) are related to the Bernoulli numbers byBn =n!σn(1). S. C. Woon described an algorithm to computeσn(1) as a binary tree:[30]

Woon's recursive algorithm (forn ≥ 1) starts by assigning to the root nodeN = [1,2]. Given a nodeN = [a1,a2, ...,ak] of the tree, the left child of the node isL(N) = [−a1,a2 + 1,a3, ...,ak] and the right childR(N) = [a1, 2,a2, ...,ak]. A nodeN = [a1,a2, ...,ak] is written as±[a2, ...,ak] in the initial part of the tree represented above with ± denoting the sign ofa1.

Given a nodeN the factorial ofN is defined as

N!=a1∏k=2length⁡(N)ak!.{\displaystyle N!=a_{1}\prod _{k=2}^{\operatorname {length} (N)}a_{k}!.}

Restricted to the nodesN of a fixed tree-leveln the sum of⁠1/N!⁠ isσn(1), thus

Bn=∑ tree-level nN node ofn!N!.{\displaystyle B_{n}=\sum _{\stackrel {N{\text{ node of}}}{{\text{ tree-level }}n}}{\frac {n!}{N!}}.}

For example:

B1 = 1!(⁠1/2!⁠)
B2 = 2!(−⁠1/3!⁠ +⁠1/2!2!⁠)
B3 = 3!(⁠1/4!⁠ −⁠1/2!3!⁠ −⁠1/3!2!⁠ +⁠1/2!2!2!⁠)

Integral representation and continuation

[edit]

Theintegral

b(s)=2esiπ/2∫0∞sts1−e2πtdtt=s!2s−1ζ(s)πs(−i)s=2s!ζ(s)(2πi)s{\displaystyle b(s)=2e^{si\pi /2}\int _{0}^{\infty }{\frac {st^{s}}{1-e^{2\pi t}}}{\frac {dt}{t}}={\frac {s!}{2^{s-1}}}{\frac {\zeta (s)}{{}\pi ^{s}{}}}(-i)^{s}={\frac {2s!\zeta (s)}{(2\pi i)^{s}}}}

has as special valuesb(2n) =B2n forn > 0.

For example,b(3) =⁠3/2⁠ζ(3)π−3i andb(5) = −⁠15/2⁠ζ(5)π−5i. Here,ζ is theRiemann zeta function, andi is theimaginary unit. Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

p=32π3(1+123+133+⋯)=0.0581522…q=152π5(1+125+135+⋯)=0.0254132…{\displaystyle {\begin{aligned}p&={\frac {3}{2\pi ^{3}}}\left(1+{\frac {1}{2^{3}}}+{\frac {1}{3^{3}}}+\cdots \right)=0.0581522\ldots \\q&={\frac {15}{2\pi ^{5}}}\left(1+{\frac {1}{2^{5}}}+{\frac {1}{3^{5}}}+\cdots \right)=0.0254132\ldots \end{aligned}}}

Another similarintegral representation is

b(s)=−esiπ/22s−1∫0∞stssinh⁡πtdtt=2esiπ/22s−1∫0∞eπtsts1−e2πtdtt.{\displaystyle b(s)=-{\frac {e^{si\pi /2}}{2^{s}-1}}\int _{0}^{\infty }{\frac {st^{s}}{\sinh \pi t}}{\frac {dt}{t}}={\frac {2e^{si\pi /2}}{2^{s}-1}}\int _{0}^{\infty }{\frac {e^{\pi t}st^{s}}{1-e^{2\pi t}}}{\frac {dt}{t}}.}

The relation to the Euler numbers andπ

[edit]

TheEuler numbers are a sequence of integers intimately connected with the Bernoulli numbers. Comparing theasymptotic expansions of the Bernoulli and the Euler numbers shows that the Euler numbersE2n are in magnitude approximately⁠2/π⁠(42n − 22n) times larger than the Bernoulli numbersB2n. In consequence:

π∼2(22n−42n)B2nE2n.{\displaystyle \pi \sim 2(2^{2n}-4^{2n}){\frac {B_{2n}}{E_{2n}}}.}

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

Bernoulli numbers can be expressed through the Euler numbers and vice versa. Since, for oddn,Bn =En = 0 (with the exceptionB1), it suffices to consider the case whenn is even.

Bn=∑k=0n−1(n−1k)n4n−2nEkn=2,4,6,…En=∑k=1n(nk−1)2k−4kkBkn=2,4,6,…{\displaystyle {\begin{aligned}B_{n}&=\sum _{k=0}^{n-1}{\binom {n-1}{k}}{\frac {n}{4^{n}-2^{n}}}E_{k}&n&=2,4,6,\ldots \\[6pt]E_{n}&=\sum _{k=1}^{n}{\binom {n}{k-1}}{\frac {2^{k}-4^{k}}{k}}B_{k}&n&=2,4,6,\ldots \end{aligned}}}

These conversion formulas express a connection 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 toπ. These numbers are defined forn ≥ 1 as[31][32]

Sn=2(2π)n∑k=0∞(−1)kn(2k+1)n=2(2π)nlimK→∞∑k=−KK(4k+1)−n.{\displaystyle S_{n}=2\left({\frac {2}{\pi }}\right)^{n}\sum _{k=0}^{\infty }{\frac {(-1)^{kn}}{(2k+1)^{n}}}=2\left({\frac {2}{\pi }}\right)^{n}\lim _{K\to \infty }\sum _{k=-K}^{K}(4k+1)^{-n}.}

The magic of these numbers lies in the fact that they turn out to be rational numbers. This was first proved byLeonhard Euler in a landmark paperDe summis serierum reciprocarum (On the sums of series of reciprocals) and has fascinated mathematicians ever since.[33] The first few of these numbers are

Sn=1,1,12,13,524,215,61720,17315,2778064,622835,…{\displaystyle S_{n}=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 } (OEIS: A099612 /OEIS: A099617)

These are the coefficients in the expansion ofsecx + tanx.

The Bernoulli numbers and Euler numbers can be understood asspecial views of these numbers, selected from the sequenceSn and scaled for use in special applications.

Bn=(−1)⌊n2⌋[n even]n!2n−4nSn ,n=2,3,…En=(−1)⌊n2⌋[n even]n!Sn+1n=0,1,…{\displaystyle {\begin{aligned}B_{n}&=(-1)^{\left\lfloor {\frac {n}{2}}\right\rfloor }[n{\text{ even}}]{\frac {n!}{2^{n}-4^{n}}}\,S_{n}\ ,&n&=2,3,\ldots \\E_{n}&=(-1)^{\left\lfloor {\frac {n}{2}}\right\rfloor }[n{\text{ even}}]n!\,S_{n+1}&n&=0,1,\ldots \end{aligned}}}

The expression [n even] has the value 1 ifn 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 ofRn =⁠2Sn/Sn + 1⁠ whenn is even. TheRn are rational approximations toπ and two successive terms always enclose the true value ofπ. Beginning withn = 1 the sequence starts (OEIS: A132049 /OEIS: A132050):

2,4,3,165,258,19261,427136,43521385,124653968,15872050521,…⟶π.{\displaystyle 2,4,3,{\frac {16}{5}},{\frac {25}{8}},{\frac {192}{61}},{\frac {427}{136}},{\frac {4352}{1385}},{\frac {12465}{3968}},{\frac {158720}{50521}},\ldots \quad \longrightarrow \pi .}

These rational numbers also appear in the last paragraph of Euler's paper cited above.

Consider the Akiyama–Tanigawa transform for the sequenceOEIS: A046978 (n + 2) /OEIS: A016116 (n + 1):

01⁠1/2⁠0−⁠1/4⁠−⁠1/4⁠−⁠1/8⁠0
1⁠1/2⁠1⁠3/4⁠0−⁠5/8⁠−⁠3/4⁠
2−⁠1/2⁠⁠1/2⁠⁠9/4⁠⁠5/2⁠⁠5/8⁠
3−1−⁠7/2⁠−⁠3/4⁠⁠15/2⁠
4⁠5/2⁠−⁠11/2⁠−⁠99/4⁠
58⁠77/2⁠
6−⁠61/2⁠

From the second, the numerators of the first column are the denominators of Euler's formula. The first column is −⁠1/2⁠ ×OEIS: A163982.

An algorithmic view: the Seidel triangle

[edit]

The sequenceSn has another unexpected yet important property: The denominators ofSn+1 divide the factorialn!. In other words: the numbersTn = Sn + 1n!, sometimes calledEuler zigzag numbers, are integers.

Tn=1,1,1,2,5,16,61,272,1385,7936,50521,353792,…n=0,1,2,3,…{\displaystyle T_{n}=1,\,1,\,1,\,2,\,5,\,16,\,61,\,272,\,1385,\,7936,\,50521,\,353792,\ldots \quad n=0,1,2,3,\ldots } (OEIS: A000111). See (OEIS: A253671).

Theirexponential generating function is the sum of thesecant andtangent functions.

∑n=0∞Tnxnn!=tan⁡(π4+x2)=sec⁡x+tan⁡x{\displaystyle \sum _{n=0}^{\infty }T_{n}{\frac {x^{n}}{n!}}=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)=\sec x+\tan x}.

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

Bn=(−1)⌊n2⌋[n even]n2n−4nTn−1 n≥2En=(−1)⌊n2⌋[n even]Tnn≥0{\displaystyle {\begin{aligned}B_{n}&=(-1)^{\left\lfloor {\frac {n}{2}}\right\rfloor }[n{\text{ even}}]{\frac {n}{2^{n}-4^{n}}}\,T_{n-1}\ &n&\geq 2\\E_{n}&=(-1)^{\left\lfloor {\frac {n}{2}}\right\rfloor }[n{\text{ even}}]T_{n}&n&\geq 0\end{aligned}}}

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbersE2n are given immediately byT2n and the Bernoulli numbersB2n are fractions obtained fromT2n - 1 by some easy shifting, avoiding rational arithmetic.

What remains is to find a convenient way to compute the numbersTn. However, already in 1877Philipp Ludwig von Seidel published an ingenious algorithm, which makes it simple to calculateTn.[34]

Seidel's algorithm forTn
  1. Start by putting 1 in row 0 and letk denote the number of the row currently being filled
  2. Ifk is odd, then put the number on the left end of the rowk − 1 in the first position of the rowk, and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper
  3. At the end of the row duplicate the last number.
  4. Ifk is even, proceed similar in the other direction.

Seidel's algorithm is in fact much more general (see the exposition of Dominique Dumont[35]) and was rediscovered several times thereafter.

Similar to Seidel's approach D. E. Knuth and T. J. Buckholtz gave a recurrence equation for the numbersT2n and recommended this method for computingB2n andE2n 'on electronic computers using only simple operations on integers'.[36]

V. I. Arnold[37] rediscovered Seidel's algorithm and later Millar, Sloane and Young popularized Seidel's algorithm under the nameboustrophedon transform.

Triangular form:

1
11
221
2455
161614105
163246566161
27227225622417812261

OnlyOEIS: A000657, with one 1, andOEIS: A214267, with two 1s, are in the OEIS.

Distribution with a supplementary 1 and one 0 in the following rows:

1
01
−1−10
0−1−2−2
55420
0510141616
−61−61−56−46−32−160

This isOEIS: A239005, a signed version ofOEIS: A008280. The main andiagonal isOEIS: A122045. The main diagonal isOEIS: A155585. The central column isOEIS: A099023. Row sums: 1, 1, −2, −5, 16, 61.... SeeOEIS: A163747. See the array beginning with 1, 1, 0, −2, 0, 16, 0 below.

The Akiyama–Tanigawa algorithm applied toOEIS: A046978 (n + 1) /OEIS: A016116(n) yields:

11⁠1/2⁠0−⁠1/4⁠−⁠1/4⁠−⁠1/8⁠
01⁠3/2⁠10−⁠3/4⁠
−1−1⁠3/2⁠4⁠15/4⁠
0−5−⁠15/2⁠1
55−⁠51/2⁠
061
−61

1. The first column isOEIS: A122045. Its binomial transform leads to:

110−20160
0−1−2216−16
−1−1414−32
0510−46
55−56
0−61
−61

The first row of this array isOEIS: A155585. The absolute values of the increasing antidiagonals areOEIS: A008280. The sum of the antidiagonals is−OEIS: A163747 (n + 1).

2. The second column is1 1 −1 −5 5 61 −61 −1385 1385.... Its binomial transform yields:

122−4−1632272
10−6−1248240
−1−6−660192
−506632
56666
610
−61

The first row of this array is1 2 2 −4 −16 32 272 544 −7936 15872 353792 −707584.... The absolute values of the second bisection are the double of the absolute values of the first bisection.

Consider the Akiyama-Tanigawa algorithm applied toOEIS: A046978 (n) / (OEIS: A158780 (n + 1) = abs(OEIS: A117575 (n)) + 1 =1, 2, 2,⁠3/2⁠, 1,⁠3/4⁠,⁠3/4⁠,⁠7/8⁠, 1,⁠17/16⁠,⁠17/16⁠,⁠33/32⁠....

122⁠3/2⁠1⁠3/4⁠⁠3/4⁠
−10⁠3/2⁠2⁠5/4⁠0
−1−3−⁠3/2⁠3⁠25/4⁠
2−3−⁠27/2⁠−13
521−⁠3/2⁠
−1645
−61

The first column whose the absolute values areOEIS: A000111 could be the numerator of a trigonometric function.

OEIS: A163747 is an autosequence of the first kind (the main diagonal isOEIS: A000004). The corresponding array is:

0−1−125−16−61
−1033−21−45
130−24−24
2−3−240
−5−2124
−1645
−61

The first two upper diagonals are−1 3 −24 402... =(−1)n + 1 × OEIS: A002832. The sum of the antidiagonals is0 −2 0 10... = 2 × OEIS: A122045(n + 1).

−OEIS: A163982 is an autosequence of the second kind, like for instanceOEIS: A164555 /OEIS: A027642. Hence the array:

21−1−2516−61
−1−2−1711−77
−1184−88
27−4−92
5−11−88
−16−77
−61

The main diagonal, here2 −2 8 −92..., is the double of the first upper one, hereOEIS: A099023. The sum of the antidiagonals is2 0 −4 0... = 2 × OEIS: A155585(n +1).OEIS: A163747 − OEIS: A163982 = 2 × OEIS: A122045.

A combinatorial view: alternating permutations

[edit]
Main article:Alternating permutations

Around 1880, three years after the publication of Seidel's algorithm,Désiré André proved a now classic result of combinatorial analysis.[38][39] Looking at the first terms of the Taylor expansion of thetrigonometric functionstanx andsecx André made a startling discovery.

tan⁡x=x+2x33!+16x55!+272x77!+7936x99!+⋯sec⁡x=1+x22!+5x44!+61x66!+1385x88!+50521x1010!+⋯{\displaystyle {\begin{aligned}\tan x&=x+{\frac {2x^{3}}{3!}}+{\frac {16x^{5}}{5!}}+{\frac {272x^{7}}{7!}}+{\frac {7936x^{9}}{9!}}+\cdots \\[6pt]\sec x&=1+{\frac {x^{2}}{2!}}+{\frac {5x^{4}}{4!}}+{\frac {61x^{6}}{6!}}+{\frac {1385x^{8}}{8!}}+{\frac {50521x^{10}}{10!}}+\cdots \end{aligned}}}

The coefficients are theEuler numbers of odd and even index, respectively. In consequence the ordinary expansion oftanx + secx has as coefficients the rational numbersSn.

tan⁡x+sec⁡x=1+x+12x2+13x3+524x4+215x5+61720x6+⋯{\displaystyle \tan x+\sec x=1+x+{\tfrac {1}{2}}x^{2}+{\tfrac {1}{3}}x^{3}+{\tfrac {5}{24}}x^{4}+{\tfrac {2}{15}}x^{5}+{\tfrac {61}{720}}x^{6}+\cdots }

André then succeeded by means of a recurrence argument to show that thealternating permutations of odd size are enumerated by the Euler numbers of odd index (also called tangent numbers) and the alternating permutations of even size by the Euler numbers of even index (also called secant numbers).

Related sequences

[edit]

The arithmetic mean of the first and the second Bernoulli numbers are the associate Bernoulli numbers:B0 = 1,B1 = 0,B2 =⁠1/6⁠,B3 = 0,B4 = −⁠1/30⁠,OEIS: A176327 /OEIS: A027642. Via the second row of its inverse Akiyama–Tanigawa transformOEIS: A177427, they lead to Balmer seriesOEIS: A061037 /OEIS: A061038.

The Akiyama–Tanigawa algorithm applied toOEIS: A060819 (n + 4) /OEIS: A145979 (n) leads to the Bernoulli numbersOEIS: A027641 /OEIS: A027642,OEIS: A164555 /OEIS: A027642, orOEIS: A176327OEIS: A176289 withoutB1, named intrinsic Bernoulli numbersBi(n).

1⁠5/6⁠⁠3/4⁠⁠7/10⁠⁠2/3⁠
⁠1/6⁠⁠1/6⁠⁠3/20⁠⁠2/15⁠⁠5/42⁠
0⁠1/30⁠⁠1/20⁠⁠2/35⁠⁠5/84⁠
−⁠1/30⁠−⁠1/30⁠−⁠3/140⁠−⁠1/105⁠0
0−⁠1/42⁠−⁠1/28⁠−⁠4/105⁠−⁠1/28⁠

Hence another link between the intrinsic Bernoulli numbers and the Balmer series viaOEIS: A145979 (n).

OEIS: A145979 (n − 2) = 0, 2, 1, 6,... is a permutation of the non-negative numbers.

The terms of the first row are f(n) =⁠1/2⁠ +⁠1/n + 2⁠. 2, f(n) is an autosequence of the second kind. 3/2, f(n) leads by its inverse binomial transform to 3/2 −1/2 1/3 −1/4 1/5 ... = 1/2 + log 2.

Consider g(n) = 1/2 – 1 / (n+2) = 0, 1/6, 1/4, 3/10, 1/3. The Akiyama-Tanagiwa transforms gives:

0⁠1/6⁠⁠1/4⁠⁠3/10⁠⁠1/3⁠⁠5/14⁠...
−⁠1/6⁠−⁠1/6⁠−⁠3/20⁠−⁠2/15⁠−⁠5/42⁠−⁠3/28⁠...
0−⁠1/30⁠−⁠1/20⁠−⁠2/35⁠−⁠5/84⁠−⁠5/84⁠...
⁠1/30⁠⁠1/30⁠⁠3/140⁠⁠1/105⁠0−⁠1/140⁠...

0, g(n), is an autosequence of the second kind.

EulerOEIS: A198631 (n) /OEIS: A006519 (n + 1) without the second term (⁠1/2⁠) are the fractional intrinsic Euler numbersEi(n) = 1, 0, −⁠1/4⁠, 0,⁠1/2⁠, 0, −⁠17/8⁠, 0, ... The corresponding Akiyama transform is:

11⁠7/8⁠⁠3/4⁠⁠21/32⁠
0⁠1/4⁠⁠3/8⁠⁠3/8⁠⁠5/16⁠
−⁠1/4⁠−⁠1/4⁠0⁠1/4⁠⁠25/64⁠
0−⁠1/2⁠−⁠3/4⁠−⁠9/16⁠−⁠5/32⁠
⁠1/2⁠⁠1/2⁠−⁠9/16⁠−⁠13/8⁠−⁠125/64⁠

The first line isEu(n).Eu(n) preceded by a zero is an autosequence of the first kind. It is linked to the Oresme numbers. The numerators of the second line areOEIS: A069834 preceded by 0. The difference table is:

011⁠7/8⁠⁠3/4⁠⁠21/32⁠⁠19/32⁠
10−⁠1/8⁠−⁠1/8⁠−⁠3/32⁠−⁠1/16⁠−⁠5/128⁠
−1−⁠1/8⁠0⁠1/32⁠⁠1/32⁠⁠3/128⁠⁠1/64⁠

Arithmetical properties of the Bernoulli numbers

[edit]

The Bernoulli numbers can be expressed in terms of the Riemann zeta function asBn = −nζ(1 −n) for integersn ≥ 0 provided forn = 0 the expression−nζ(1 −n) is understood as the limiting value and the conventionB1 =⁠1/2⁠ is used. This intimately relates them to the values of the zeta function at negative integers. As such, they could be expected to have and do have deep arithmetical properties. For example, theAgoh–Giuga conjecture postulates thatp is a prime number if and only ifpBp − 1 is congruent to −1 modulop. Divisibility properties of the Bernoulli numbers are related to theideal class groups ofcyclotomic fields by a theorem of Kummer and its strengthening in theHerbrand-Ribet theorem, and to class numbers of real quadratic fields byAnkeny–Artin–Chowla.

The Kummer theorems

[edit]

The Bernoulli numbers are related toFermat's Last Theorem (FLT) byKummer's theorem,[40] which says:

If the odd primep does not divide any of the numerators of the Bernoulli numbersB2,B4, ...,Bp − 3 thenxp +yp +zp = 0 has no solutions in nonzero integers.

Prime numbers with this property are calledregular primes. Another classical result of Kummer are the followingcongruences.[41]

Main article:Kummer's congruence
Letp be an odd prime andb an even number such thatp − 1 does not divideb. Then for any non-negative integerk
Bk(p−1)+bk(p−1)+b≡Bbb(modp).{\displaystyle {\frac {B_{k(p-1)+b}}{k(p-1)+b}}\equiv {\frac {B_{b}}{b}}{\pmod {p}}.}

A generalization of these congruences goes by the name ofp-adic continuity.

p-adic continuity

[edit]

Ifb,m andn are positive integers such thatm andn are not divisible byp − 1 andm ≡n (modpb − 1 (p − 1)), then

(1−pm−1)Bmm≡(1−pn−1)Bnn(modpb).{\displaystyle (1-p^{m-1}){\frac {B_{m}}{m}}\equiv (1-p^{n-1}){\frac {B_{n}}{n}}{\pmod {p^{b}}}.}

SinceBn = −nζ(1 −n), this can also be written

(1−p−u)ζ(u)≡(1−p−v)ζ(v)(modpb),{\displaystyle \left(1-p^{-u}\right)\zeta (u)\equiv \left(1-p^{-v}\right)\zeta (v){\pmod {p^{b}}},}

whereu = 1 −m andv = 1 −n, so thatu andv are nonpositive and not congruent to 1 modulop − 1. This tells us that the Riemann zeta function, with1 −p−s taken out of the Euler product formula, is continuous in thep-adic numbers on odd negative integers congruent modulop − 1 to a particulara ≢ 1 mod (p − 1), and so can be extended to a continuous functionζp(s) for allp-adic integersZp,{\displaystyle \mathbb {Z} _{p},} thep-adic zeta function.

Ramanujan's congruences

[edit]

The following relations, due toRamanujan, provide a method for calculating Bernoulli numbers that is more efficient than the one given by their original recursive definition:

(m+3m)Bm={m+33−∑j=1m6(m+3m−6j)Bm−6j,if m≡0(mod6);m+33−∑j=1m−26(m+3m−6j)Bm−6j,if m≡2(mod6);−m+36−∑j=1m−46(m+3m−6j)Bm−6j,if m≡4(mod6).{\displaystyle {\binom {m+3}{m}}B_{m}={\begin{cases}{\frac {m+3}{3}}-\sum \limits _{j=1}^{\frac {m}{6}}{\binom {m+3}{m-6j}}B_{m-6j},&{\text{if }}m\equiv 0{\pmod {6}};\\{\frac {m+3}{3}}-\sum \limits _{j=1}^{\frac {m-2}{6}}{\binom {m+3}{m-6j}}B_{m-6j},&{\text{if }}m\equiv 2{\pmod {6}};\\-{\frac {m+3}{6}}-\sum \limits _{j=1}^{\frac {m-4}{6}}{\binom {m+3}{m-6j}}B_{m-6j},&{\text{if }}m\equiv 4{\pmod {6}}.\end{cases}}}

Von Staudt–Clausen theorem

[edit]
Main article:Von Staudt–Clausen theorem

The von Staudt–Clausen theorem was given byKarl Georg Christian von Staudt[42] andThomas Clausen[43] independently in 1840. The theorem states that for everyn > 0,

B2n+∑(p−1)∣2n1p{\displaystyle B_{2n}+\sum _{(p-1)\,\mid \,2n}{\frac {1}{p}}}

is an integer. The sum extends over allprimesp for whichp − 1 divides2n.

A consequence of this is that the denominator ofB2n is given by the product of all primesp for whichp − 1 divides2n. In particular, these denominators aresquare-free and divisible by 6.

Why do the odd Bernoulli numbers vanish?

[edit]

The sum

φk(n)=∑i=0nik−nk2{\displaystyle \varphi _{k}(n)=\sum _{i=0}^{n}i^{k}-{\frac {n^{k}}{2}}}

can be evaluated for negative values of the indexn. Doing so will show that it is anodd function for even values ofk, which implies that the sum has only terms of odd index. This and the formula for the Bernoulli sum imply thatB2k + 1 −m is 0 form even and2k + 1 −m > 1; and that the term forB1 is cancelled by the subtraction. The von Staudt–Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid forn > 1).

From the von Staudt–Clausen theorem it is known that for oddn > 1 the number2Bn is an integer. This seems trivial if one knows beforehand that the integer in question is zero. However, by applying Worpitzky's representation one gets

2Bn=∑m=0n(−1)m2m+1m!{n+1m+1}=0(n>1 is odd){\displaystyle 2B_{n}=\sum _{m=0}^{n}(-1)^{m}{\frac {2}{m+1}}m!\left\{{n+1 \atop m+1}\right\}=0\quad (n>1{\text{ is odd}})}

as asum of integers, which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. LetSn,m be the number of surjective maps from{1, 2, ...,n} to{1, 2, ...,m}, thenSn,m =m!{n
m}. The last equation can only hold if

∑odd m=1n−12m2Sn,m=∑even m=2n2m2Sn,m(n>2 is even).{\displaystyle \sum _{{\text{odd }}m=1}^{n-1}{\frac {2}{m^{2}}}S_{n,m}=\sum _{{\text{even }}m=2}^{n}{\frac {2}{m^{2}}}S_{n,m}\quad (n>2{\text{ is even}}).}

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.

A restatement of the Riemann hypothesis

[edit]

The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of theRiemann hypothesis (RH) which uses only the Bernoulli numbers. In factMarcel Riesz proved that the RH is equivalent to the following assertion:[44]

For everyε >⁠1/4⁠ there exists a constantCε > 0 (depending onε) such that|R(x)| <Cεxε asx → ∞.

HereR(x) is theRiesz function

R(x)=2∑k=1∞kk¯xk(2π)2k(B2k2k)=2∑k=1∞kk¯xk(2π)2kβ2k.{\displaystyle R(x)=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\left({\frac {B_{2k}}{2k}}\right)}}=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\beta _{2k}}}.}

nk denotes therising factorial power in the notation ofD. E. Knuth. The numbersβn =⁠Bn/n⁠ occur frequently in the study of the zeta function and are significant becauseβn is ap-integer for primesp wherep − 1 does not dividen. Theβn are calleddivided Bernoulli numbers.

Generalized Bernoulli numbers

[edit]

Thegeneralized Bernoulli numbers are certainalgebraic numbers, defined similarly to the Bernoulli numbers, that are related tospecial values ofDirichletL-functions in the same way that Bernoulli numbers are related to special values of the Riemann zeta function.

Letχ be aDirichlet character modulof. The generalized Bernoulli numbers attached toχ are defined by

∑a=1fχ(a)teateft−1=∑k=0∞Bk,χtkk!.{\displaystyle \sum _{a=1}^{f}\chi (a){\frac {te^{at}}{e^{ft}-1}}=\sum _{k=0}^{\infty }B_{k,\chi }{\frac {t^{k}}{k!}}.}

Apart from the exceptionalB1,1 =⁠1/2⁠, we have, for any Dirichlet characterχ, thatBk,χ = 0 ifχ(−1) ≠ (−1)k.

Generalizing the relation between Bernoulli numbers and values of the Riemann zeta function at non-positive integers, one has the for all integersk ≥ 1:

L(1−k,χ)=−Bk,χk,{\displaystyle L(1-k,\chi )=-{\frac {B_{k,\chi }}{k}},}

whereL(s,χ) is the DirichletL-function ofχ.[45]

Eisenstein–Kronecker number

[edit]
Main article:Eisenstein–Kronecker number

Eisenstein–Kronecker numbers are an analogue of the generalized Bernoulli numbers forimaginary quadratic fields.[46][47] They are related to criticalL-values ofHecke characters.[47]

Appendix

[edit]

Assorted identities

[edit]

See also

[edit]

Notes

[edit]
  1. ^Translation of the text: " ... And if [one were] to proceed onward step by step to higher powers, one may furnish, with little difficulty, the following list:
    Sums of powers
    ∫n=∑k=1nk=12n2+12n{\displaystyle \textstyle \int n=\sum _{k=1}^{n}k={\frac {1}{2}}n^{2}+{\frac {1}{2}}n}
    ⋮
    ∫n10=∑k=1nk10=111n11+12n10+56n9−1n7+1n5−12n3+566n{\displaystyle \textstyle \int n^{10}=\sum _{k=1}^{n}k^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n}
    Indeed [if] one will have examined diligently the law of arithmetic progression there, one will also be able to continue the same without these circuitous computations: For [if]c{\displaystyle \textstyle c} is taken as the exponent of any power, the sum of allnc{\displaystyle \textstyle n^{c}} is produced or
    ∫nc=∑k=1nkc=1c+1nc+1+12nc+c2Anc−1+c(c−1)(c−2)2⋅3⋅4Bnc−3+c(c−1)(c−2)(c−3)(c−4)2⋅3⋅4⋅5⋅6Cnc−5+c(c−1)(c−2)(c−3)(c−4)(c−5)(c−6)2⋅3⋅4⋅5⋅6⋅7⋅8Dnc−7+⋯{\displaystyle \textstyle \int n^{c}=\sum _{k=1}^{n}k^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c(c-1)(c-2)}{2\cdot 3\cdot 4}}Bn^{c-3}+{\frac {c(c-1)(c-2)(c-3)(c-4)}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}+{\frac {c(c-1)(c-2)(c-3)(c-4)(c-5)(c-6)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}+\cdots }
    and so forth, the exponent of its powern{\displaystyle n} continually diminishing by 2 until it arrives atn{\displaystyle n} orn2{\displaystyle n^{2}}. The capital lettersA,B,C,D,{\displaystyle \textstyle A,B,C,D,} etc. denote in order the coefficients of the last terms for∫n2,∫n4,∫n6,∫n8{\displaystyle \textstyle \int n^{2},\int n^{4},\int n^{6},\int n^{8}}, etc. namely
    A=16,B=−130,C=142,D=−130{\displaystyle \textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}}."
    [Note: The text of the illustration contains some typos:ensperexit should readinspexerit,ambabimus should readambagibus,quosque should readquousque, and in Bernoulli's original textSumtâ should readSumptâ orSumptam.]
  2. ^TheMathematics Genealogy Project (n.d.) shows Leibniz as the academic advisor of Jacob Bernoulli. See alsoMiller (2017).
  3. ^this formula was discovered (or perhaps rediscovered) by Giorgio Pietrocola. His demonstration is available in Italian language (Pietrocola 2008).

References

[edit]
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  6. ^Arfken (1970), p. 278.
  7. ^Donald Knuth (2022),Recent News (2022): Concrete Mathematics and Bernoulli.

    But last year I took a close look at Peter Luschny's Bernoulli manifesto, where he gives more than a dozen good reasons why the value of $B_1$ should really be plus one-half. He explains that some mathematicians of the early 20th century had unilaterally changed the conventions, because some of their formulas came out a bit nicer when the negative value was used. It was their well-intentioned but ultimately poor choice that had led to what I'd been taught in the 1950s. […] By now, hundreds of books that use the “minus-one-half” convention have unfortunately been written. Even worse, all the major software systems for symbolic mathematics have that 20th-century aberration deeply embedded. Yet Luschny convinced me that we have all been wrong, and that it's high time to change back to the correct definition before the situation gets even worse.

  8. ^Peter Luschny (2013),The Bernoulli Manifesto
  9. ^abKnuth (1993).
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  13. ^Jordan (1950) p 233
  14. ^Ireland and Rosen (1990) p 229
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