Questions tagged [bernoulli-numbers]

Questions on Bernoulli numbers, a special sequence of rational numbers that arise as the coefficients in the power series expansions of certain elementary functions.

The $n$th Bernoulli number $B_n$ is frequently defined in terms of a generating function:

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

The first few Bernoulli numbers are

\begin{align*} B_0 &=1 \\ B_1 &=\frac12 \\ B_2 &=\frac16 \\ B_3 &=0 \\ B_4 &=-\frac1{30} \\ B_5 &=0 \end{align*}

All Bernoulli numbers with $n$ odd, except for $B_1$, are zero.

Alternatively, the $n$th Bernoulli number is the constant coefficient in the $n$th Bernoulli polynomial $B_n(x)$, which can be defined in terms of a generating function as well:

$$\frac{te^{-xt}}{1-e^{-t}} = \sum_{k=0}^\infty B_n(x)\frac{t^n}{t!}$$

The Bernoulli numbers have deep connections to number theory, and frequently rise in combinatorics and asymptotic estimates of functions, as well.

407 questions
11
votes
2 answers

Proving of the multiplication theorem for Bernoulli polynomial

How the expression below can be proven: $$B_n(mx) = m^{n−1} \sum\limits_{k=0}^{m-1}B_n\left(x+\frac{k}{m}\right)$$ Where $B_n(x)$ is Bernoulli polynomial I know it is already proved by Joseph Ludwig Raabe, but I don`t know how exactly.
dehasi
  • 509
  • 4
  • 14
9
votes
4 answers

What is the simplest way to get Bernoulli numbers?

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…
Tristian
  • 311
7
votes
4 answers

Bernoulli numbers alternate signs

Michael Spivak in Calculus, Fourth ed., Chapter 27, Problem 16 (page 572) defines the Bernoulli numbers based on $$ \frac{z}{e^z-1} = \sum_{n=0}^\infty\frac{B_nz^n}{n!}. $$ He asks the reader to prove that $B_n = 0$ if $n$ is odd and $> 1$. This is…
7
votes
1 answer

Where do the Bernoulli numbers come from?

I'm trying to get an idea of how the definition of the Bernoulli numbers was settled on. I found this explanation very clear and intuitive, relating the Bernoulli numbers to the sums of a fixed power. The one part I can't quite see is how Bernoulli…
Tac-Tics
  • 2,203
6
votes
2 answers

Bernoulli Numbers

I've read that Bernoulli Numbers are defined by the series $$ \frac{z}{e^z-1}\equiv \sum\limits_{n=0}^{\infty}B_n\frac{z^n}{n!},$$ So if I start with $0$ I get $$ B_0\frac{1}{1}=B_0{1}. $$ My question is, why is there a $B_0$ in the term...is it…
bjd2385
  • 3,017
5
votes
0 answers

Are there negative Bernoulli numbers?

If not, why not? Also I don't mean Bernoulli number that are negative such as $B_4 = \frac{-1}{30}$ but numbers like $B_{-4} = ?$
3
votes
1 answer

Find all integer solutions for $ \frac{B_{2m}}m =\frac{B_{2n}}n$.

For Bernoulli number $ B_n$, prove (or disprove) that the only integer solution for $\dfrac{B_{2m}}m= \dfrac{B_{2n}}n $ is $ (m,n) = (1,7) $ for $ 1\leq m
GohP.iHan
  • 1,376
2
votes
2 answers

dual formula to Bernoulli polynomials

$$ \tilde{B}_n(x) = \frac{(-1)^{n + 1}}{n!} \left( \delta^{(n - 1)}(x - 1) - \delta^{(n - 1)}(x) \right) $$ Wikipedia says this formulae is DUAL to the Bernoulli POlynomials but dual in what sense ?? thanks EDIT: link …
Jose Garcia
  • 8,506
2
votes
1 answer

A continued fraction involving Bernoulli numbers

Let $B_n$ be the Bernoulli numbers. Then, we can write a function $F(x)$, expressed as a continued fraction involving those $B_n$, such that it gives the form, $$ \displaystyle \displaystyle F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+…
user72430
2
votes
3 answers

how the following formula or any other explicit formula for computing Bernoulli numbers can be derived?

how the following formula $$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$ can be derived? I know how to use the formula but still have not seen any proof for the given formula.
Absurd
  • 369
  • 1
  • 14
2
votes
1 answer

Can someone explain this theorem from von Staudt on denominators of Bernoulli numbers

This is an extract from Paulo Ribenboim: 13 lectures on Fermats last theorem on page 105. "In 1845, von Staudt determined some factors of the numerator $N_{2k}$. Let $2k = k_1k_2$ with $gcd(k_1,k_2)= 1$ such that $p|k_2$ if and only if $p|D_{2k}$…
2
votes
0 answers

Bernoulli number $B_{\frac{p-1}{2}}$

There is a prime number $p\equiv1\left(\bmod\,4\right)$ such that the numerator of the Bernoulli number $B_{\frac{p-1}{2}}$ is divisible by $p$?
thebalans
  • 103
2
votes
1 answer

On asymptotic properties of the sum of consecutive powers

For positive integers $p, n$, the sum of consecutive $p$-th powers is $$ S_p(n) := \sum_{k=1}^n k^p. $$ From Asymptotic behaviour of sums of consecutive powers we have that $S_p(n)/n^{p+1}$ is decreasing when viewed as a function of $p$ (with $n$…
user152169
  • 1,983
2
votes
3 answers

Explicit function for Bernoulli numbers

Is there any general explicit formula for Bernoulli numbers ? Something like: $$f(x)=B_x$$ Where $B_x$ is the $x$-th Bernoulli number ? Searching the internet I only found the so-called "generating formula" or recursive relations but can there be an…
AlienRem
  • 4,011
1
vote
0 answers

Prove the congruence $pB_{p-1} \equiv -1 \pmod p$ for Bernoulli numbers.

I need to prove that: If $p$ is prime greater than or equal to five, then $pB_{p-1}$ belongs to the p-integers and more over: $$pB_{p-1} \equiv -1 \pmod p$$ Hint:Put $N=p$ in the Faulhaber´s formula: $S_{j}(N-1)= \displaystyle\sum_{k=o}^j…
user162343
  • 3,245
1
2 3