# Bernoulli numbers and polynomials

Symbol: BernoulliB $B_{n}$ Bernoulli number
Symbol: BernoulliPolynomial $B_{n}\!\left(z\right)$ Bernoulli polynomial
$B_{n} = \frac{\text{A027641}\!\left(n\right)}{\text{A027642}\!\left(n\right)}$

## Tables

Table of $B_{n}$ for $0 \le n \le 50$
Table of $B_{n}\!\left(x\right)$ for $0 \le n \le 10$

## Generating functions

$\frac{z}{{e}^{z} - 1} = \sum_{n=0}^{\infty} B_{n} \frac{{z}^{n}}{n !}$
$\frac{z {e}^{x z}}{{e}^{z} - 1} = \sum_{n=0}^{\infty} B_{n}\!\left(x\right) \frac{{z}^{n}}{n !}$

## Sum representations

$B_{n}\!\left(x\right) = \sum_{k=0}^{n} {n \choose k} B_{n - k} {x}^{k}$

## Representations by special functions

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

## Specific values

$B_{2 n + 3} = 0$
${\left(-1\right)}^{n} B_{2 n + 2} > 0$
$B_{n}\!\left(0\right) = B_{n}$
$B_{n}\!\left(1\right) = {\left(-1\right)}^{n} B_{n}$
$B_{n}\!\left(\frac{1}{2}\right) = \left({2}^{1 - n} - 1\right) B_{n}$

## Functional equations

$B_{n}\!\left(x + 1\right) = B_{n}\!\left(x\right) + n {x}^{n - 1}$
$B_{n}\!\left(1 - x\right) = {\left(-1\right)}^{n} B_{n}\!\left(x\right)$
$B_{n}\!\left(-x\right) = {\left(-1\right)}^{n} \left(B_{n}\!\left(x\right) + n {x}^{n - 1}\right)$

## Derivatives and integrals

$\int_{a}^{b} B_{n}\!\left(t\right) \, dt = \frac{B_{n + 1}\!\left(b\right) - B_{n + 1}\!\left(a\right)}{n + 1}$
$\int_{z}^{z + 1} B_{n}\!\left(t\right) \, dt = {z}^{n}$
$B'_{n}(x) = n B_{n - 1}\!\left(x\right)$

## Summation

$\sum_{k=1}^{m} {k}^{n} = \frac{B_{n + 1}\!\left(m + 1\right) - B_{m + 1}}{m + 1}$

## Denominators

$\left(B_{2 n} + \sum_{\left(p - 1\right) \mid 2 n} \frac{1}{p}\right) \in \mathbb{Z}$
$\left(B_{2 n} \prod_{\left(p - 1\right) \mid 2 n} \frac{1}{p}\right) \in \mathbb{Z}$

## Bounds and inequalities

$\left|B_{2 n}\right| < \left(1 + \frac{1}{n}\right) \frac{2 \left(2 n\right)!}{{\left(2 \pi\right)}^{2 n}}$
$\left|B_{2 n}\right| < \left(1 + \frac{1}{n}\right) 4 \sqrt{\pi n} {\left(\frac{n}{\pi e}\right)}^{2 n}$
$\left|B_{2 n}\right| > \frac{2 \left(2 n\right)!}{{\left(2 \pi\right)}^{2 n}}$
$\left|B_{2 n}\right| > 4 \sqrt{\pi n} {\left(\frac{n}{\pi e}\right)}^{2 n}$
$\left|B_{2 n}\!\left(x\right)\right| \le \left|B_{2 n}\right|$
$\left|B_{2 n}\!\left(x\right)\right| < \left|B_{2 n}\right|$

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2020-04-08 16:14:44.404316 UTC