On paper, what is the simplest way to generate the Bernoulli fractions like $\frac{-1}{30} $and $\frac{7}{6}$?
Basically I'm trying to find and understand $B_n =$ (the stuff on this side) and I've seen something using $i$ and a contour integral with
$$\frac{z}{e^z-1}\frac{ dz}{z^{n+1}}$$
and I don't pretend to understand that at all.
$$B_n=\frac{n!}{2\pi \color{red}{i}}\color{red}{\oint}\color{red}{\frac{z}{\color{black}{e\color{red}{^z}}}}\color{red}{\frac{dz}{\color{black}{e^{n+1}}}}$$
$$B_n=\sum^n_{\color{red}{k}=0}\frac{1}{\color{red}{k}+1}\sum^\color{red}{k}_{\color{red}{v}=0}(-1)^\color{red}{v}\color{red}{\binom{k}{v}}\color{red}{v}^n$$
and then there's this variant I see used a lot called a Generating function where
$$B_n = \sum \frac{1}{k+1} \sum (-1)v (k\dots v)v^n$$
but I don't understand double sums either. I need to be able to reliably get Bernoulli numbers for Taylor series stuff like tangent and the hyperbolic variants. I've reached a limit of my understanding and made clear in red the parts I'm confused about. Like I get that the $i$ is imaginary and somehow related to rotation of $\pi$, probably positive and negative, but I don't know what the $d$ or $z$ mean in that equation.
In the second formula, I don't understand why it switched from $n$ to $k$, then from $k$ to $v$, though I suspect I'm supposed to increment $1:1$ increases in $n$ for $k$, which also increases my increments for $v$ by $1:1$ (creating a $\frac{+1}{-1}$ sequence) but I don't understand the vertical parentheses at all. I'm not even going to pretend I took a class above calculus 1.
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$\begingroup$The simplest way to calculate them, using very few fancy tools, is the following recursive definition:
$$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$ in other words $$B_n=1-{n\choose 0}\frac{B_0}{n-0+1}-{n\choose 1}\frac{B_1}{n-1+1}-\cdots -{n\choose n-1}\frac{B_{n-1}}{n-(n-1)+1}$$
Here, ${a\choose b}$ denotes a binomial coefficient. So, we begin with $B_0=1, B_1=\frac{1}{2}$, and we can calculate $B_2$ using the above recursive definition. That is, $B_2=1-{2\choose 0}\frac{B_0}{3}-{2\choose 1}\frac{B_1}{2}=1-\frac{1}{3}-2\frac{\frac{1}{2}}{2}=\frac{1}{6}$.
Now, with $B_2$ in hand, we can calculate $B_3$. And so on.
$\endgroup$ 9 $\begingroup$I sort of favor the $\cot$ Laurent series as a starting point for calculating the Bernoulli numbers. Although the recursive formula still goes over all previous Bernoulli numbers, it includes $2$ per term, so only has half as many terms. If we let $$y(x)=\cot x=\frac1x+\sum_{n=1}^{\infty}\frac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}=\frac1x+\sum_{n=1}^{\infty}a_nx^{2n-1}$$ Then $$\begin{align}y^{\prime}&=-\frac1{x^2}+\sum_{n=1}^{\infty}(2n-1)a_nx^{2n-2}=-\csc^2x=-1-\cot^2x\\ &=-1-y^2=-1-\frac1{x^2}-2\sum_{n=1}^{\infty}a_nx^{2n-2}-\sum_{n=1}^{\infty}\sum_{n=1}^{\infty}a_na_mx^{2n+2m-2}\end{align}$$ So $$\sum_{n=1}^{\infty}(2n+1)a_nx^{2n-2}=\sum_{n=0}^{\infty}(2n+3)a_{n+1}x^{2n}=-1-\sum_{n=1}^{\infty}\left(\sum_{m=1}^{n}a_ma_{n-m+1}\right)x^{2n}$$ Comparing coefficients of like powers of $n$ on both sides we get $$a_1=-\frac13$$ $$a_{2n}=-\frac1{4n+1}\left(a_n^2+2\sum_{m=1}^{n-1}a_ma_{2n-m}\right)$$ $$a_{2n+1}=-\frac2{4n+3}\sum_{m=1}^{n}a_ma_{2n-m+1}$$ Sample program:
program bernoulli implicit none integer, parameter :: wp = kind(1.0d0) integer, parameter :: N = 10 real(wp) a(N) ! a_n real(wp) b(N) ! b_2n real(wp) prefactor ! (-1)**n*(2n)!/2**(2n) integer i prefactor = 1 do i = 1, N if(i == 1) then a(1) = real(-1,wp)/3 else if(mod(i,2) == 0) then a(i) = -(a(i/2)**2+2*dot_product(a(1:i/2-1),a(i-1:i/2+1:-1)))/(2*i+1) else a(i) = -2*dot_product(a(1:i/2),a(i-1:i/2+1:-1))/(2*i+1) end if prefactor = -2*i*(2*i-1)*prefactor/4 b(i) = prefactor*a(i) write(*,'(*(g0))') 'a(',i,') = ',a(i) write(*,'(*(g0))') 'B(',2*i,') = ',b(i) end do
end program bernoulliOutput:
a(1) = -.3333333333333333
B(2) = .1666666666666667
a(2) = -.2222222222222222E-01
B(4) = -.3333333333333333E-01
a(3) = -.2116402116402116E-02
B(6) = .2380952380952380E-01
a(4) = -.2116402116402116E-03
B(8) = -.3333333333333333E-01
a(5) = -.2137779915557693E-04
B(10) = .7575757575757573E-01
a(6) = -.2164404280806397E-05
B(12) = -.2531135531135530
a(7) = -.2192594785187377E-06
B(14) = 1.166666666666666
a(8) = -.2221460878997967E-07
B(16) = -7.092156862745096
a(9) = -.2250784651680898E-08
B(18) = 54.97117794486213
a(10) = -.2280515120459217E-09
B(20) = -529.1242424242421Chowla and Hartung, An "exact" formula for the $n$th Bernoulli number, Acta Arithmetica 22 (1972) 113-115, give the following formula, quoted in Comtet, Advanced Combinatorics: Let $$\phi_n={2(2^{2n}-1)(2n)!\over2^{2n-1}\pi^{2n}}\sum_{k=1}^{3n}k^{-2n}$$ and write $[x]$ for the greatest integer not exceeding $x$. Then $$B_{2n}=(-1)^{n-1}{1+[\phi_n]\over2(2^{2n}-1)}$$
You might also want to look at the paper, Kevin J. McGown, Computing Bernoulli numbers quickly.
My friend, David Harvey, at UNSW, may be the current record holder for fast computation of Bernoulli numbers for large values of $n$. See Richard P. Brent and David Harvey, Fast computation of Bernoulli, Tangent and Secant numbers; see also A multimodular algorithm for computing Bernoulli numbers, Math. Comp. 79 (2010), no. 272, 2361–2370. MR 2684369 (2011h:11019)
Note: I've corrected an error I made in copying the formula out of Comtet.
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