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
1
vote
1 answer

Why do the even Bernoulli numbers grow so fast?

Question is in the title. We have: $$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$
Superbus
  • 2,136
1
vote
1 answer

Bernoulli random variables choice of values

It seems that typically the standard is to define a bernoulli random variable $X$ as $$ X = \begin{cases} 1 & \text{with probability } p \\ 0 & \text{with probability } 1 - p \end{cases} $$ My question is, why do we choose $1$ and $0$ instead of $2$…
1
vote
1 answer

The denominator of a Bernoulli number is always **an even** integer. Why?

Apparently, the denominator of a Bernoulli number is always an even integer. Where does this come from?
1
vote
0 answers

Sign convention for Bernoulli numbers

I can't seem to find an answer as to why there are two conventions for Bernoulli numbers. I have done a fair amount of research and can't find a definitive answer. Maybe I am just not reading between the lines but I would greatly appreciate any…
1
vote
3 answers

How to get the explicit formula of Bernoulli number using its generating function?

How to get this Bernoulli number explicit formula: $$B_k=\sum_{n=0}^k\frac{1}{n+1}\sum_{j=0}^{n}(-1)^j\binom nj j^k$$ by using Bernoulli number's generating function: $$G(k)=\frac{t}{e^t-1}=\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}$$ Thanks for your any…
1
vote
1 answer

Faulhaber's Formula proof

I'm going through the proof from the book 'Bernoulli Numbers and Zeta Functions' and there's a step that I don't understand. My best guess at the minute is Taylor series. (It's the line highlighted in yellow in the image) Thanks Similarly, taking…
Jacamo
  • 21
1
vote
1 answer

Absolute value of numerator of $2k$-th Bernoulli number $> 1$ for $k >4$.

Let $B_n$ be $n$-th Bernoulli number. And let $N_n$ be the numerator of $B_n$. For example, $|N_0| = 1,$ $|N_2| = 1,$ $|N_4| = 1,$ $|N_6| = 1,$ $|N_8| = 1,$ $|N_{10}| = 5.$ Is that $|N_{2k}| > 1$ for $k > 4$ true? If it is true, please tell me the…
TOM
  • 1,479
1
vote
1 answer

Prove $\left(1 + 1/\sqrt{n}\right)^n > \sqrt{n}$ for all natural $n$

$$\left(1 + 1/\sqrt{n}\right)^n > \sqrt{n}$$ I'm trying to use Bernoulli's inequality So $\left(1 + 1/\sqrt{n}\right)^n \ge 1 + n/\sqrt{n}$, but I'm not sure what to do from there. Could I say that $1 + n/\sqrt{n} > \sqrt{n}$ for all natural numbers…
0
votes
2 answers

Bernoulli Trial Help!

Assume that $n = 9$, and $p = \frac{4}{5}$ . Find the probability of at least 3 successes and at least 2 failures. What I have so far: $c(9,3)\cdot (\frac{4}{5})^3\cdot (\frac{1}{5})^6=.002753$ $c(9,4)\cdot (\frac{4}{5})^4 \cdot…
Eric
  • 17
0
votes
2 answers

Bernoulli Process Help!

The professor who sometimes forgets to bring her briefcase to the office, but assume that, each day, the probability that she forgets the briefcase is 1 /8 . Assume that her forgetting is a Bernoulli process. (1) What is the probability that she…
Eric
  • 17
0
votes
0 answers

The efficient computation of the zeta function at even integers

It is commonplace (see for instance this paper by McGown) to calculate the Riemann $ζ$ function at even integer values by using an approximation to the real value and the Staudt--von Clausen theorem. My question is this: For the first item, why do…
Cloudscape
  • 5,124
0
votes
0 answers

Efficiency of two different estimators (Bernoulli)

I am using a book, where the following exercise appears: Two estimators for a random sample of size $n$ (Bernoulli population): $T_1=\frac{\sum\limits_{i=1}^n X_i + 2X_n}{n+2}$ $T_2=\frac{\sum\limits_{i=1}^{n-2} X_i + 2X_n}{n+2}$ I want to compare…
Sinval
  • 101
0
votes
1 answer

Getting $B_4$ from the recursive definition $B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$

Background I've been learning the basics of Bernoulli numbers and this is a reference to What is the simplest way to get Bernoulli numbers? and vadim123's succinct reply, where he stated: The simplest way to calculate them, using very few fancy…
Jessie
  • 1,463
0
votes
0 answers

Relationship between Bernoulli numbers of first and second type

I am interested in an expresion that allows to calculate Bernoulli numbers of the first kind $B_n$ from the values of the Bernoulli numbers of the second kind $b_n$ (also known as Cauchy numbers of the first kind, reciprocal logarithm numbers or…
0
votes
1 answer

Probability of a voting majority given people changing their votes (Bernouilli trials)

In a community, $a$ persons are pro-choice, $b$ $(b < a)$ are pro-life, and $n$ $(n > a – b)$ are undecided. Suppose that there will be a vote to determine the will of the majority with regard to legalizing abortion. If by then all of the undecided…