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

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1answer
58 views

Calculating Laurent Series of Complex Function

How does one alternate the Bernoulli number series expansion $$\frac x{e^x - 1}=\sum_{n=0}^{\infty}\frac{B_nx^n}{n!}$$ To calculate the Laurent Series centered at 0 in the annulus of convergence of ...
2
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0answers
38 views

Bernoulli Conjecture on $B_{2^n}$

So in a recent question I was trying to prove that $2^n-1$ will never be a Carmichael number (Can a Mersenne number ever be a Carmichael number?), I was going to prove it true as long as a certain ...
4
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1answer
76 views

Analytic Continuation of Zeta Function using Bernoulli Numbers

In my complex analysis textbook by Stein and Shakarchi, as an exercise, I am supposed to extend $\zeta(s)$ to the entire complex plane using Bernoulli numbers, but I am stuck. I can prove that $$ ...
2
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1answer
47 views

Stirling-like sum equal to zero when $k>n$

I need to prove that $$\sum_{r=0}^k\binom{k}{r}(-1)^r r^n=0$$ when $n<k$. I know that the formula above can be easily transformed into the Stirling number of the Second kind formula, which is ...
6
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2answers
99 views

Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$

How can we prove that $$\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$$ where $B_{2r}$ are the Bernoulli numbers? $$\begin{array}{c|c|c|} n & \frac{2n-1}{2n+1} & ...
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0answers
10 views

conditional mean of geometric RV

Say, there are three nodes: $S$, $R$, $D$. $S$ transmits to $R$, $R$ stores the packets, and later transmits to $D$. At any time, either $S$ or $R$ is selected to transmit according to some random ...
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1answer
72 views

Bernoulli numbers closed form.

I found this nice explicit formula for the Bernoulli numbers: $$B_n = \sum_{k \mathop = 0}^n \sum_{i \mathop = 0}^k (-1)^i \binom k i \frac {i^n} {k + 1}$$ I can't find a proof though. I want to ...
1
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1answer
74 views

Why do the even Bernoulli numbers grow so fast?

Question is in the title. We have: $$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$
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0answers
46 views

Ergodic Theory, Bernoulli Measure, Cylinder Set

Let $N\geq2$ be an integer and consider the probability space ($\Sigma^+$,B, $\mu_p$) where $\mu_p$ is the Bernoulli measure with respect to probability vector $\ p = (p_1,...,p_N)$ . Show that for ...
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1answer
20 views

Functional Choice for p in a Bernoulli Distribution

Why is the functional choice $p = \exp(x)/(1+\exp(x))$ to model $p$ a good one in a Bernoulli distribution? Is it because it is limited at $0$ as $x$ approaches $0$ and $1$ as $x$ approaches ...
1
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1answer
46 views

{Probability}: choosing keys from a pool without replacement

The OP is trying to understand the following question. The OP understand that if you can always write out the term $$P(X=k) \implies (1-\frac{1}{N})(1-\frac{1}{N-1})\cdots(1-\frac{1}{N-k+1}),$$ ...
0
votes
1answer
56 views

Probability: deviation from the mean

I am having trouble to understand the following. If $S_n=X_1+X_2+......+X_n$, where X_1,X_2 are Bernouli (p). I don't understand this. So you get an intermediate point Constant* sqrt(n). To the ...
2
votes
1answer
45 views

Probability exercise Bernoulli. [closed]

Probability random signals. Im late I have no idea to start and this is for tomorrow. I was on training and have no break to do this work. I do this.You are an Internet savvy and enjoy watching video ...
5
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1answer
72 views

Question about $ \int_{-1}^{0}\sum_{n=1}^{x}n^sdx=\zeta (-s) \forall s\in \Bbb N$

what I found from messing around was $$ \int_{-1}^{0}\sum_{n=1}^{x}n^sdx=\zeta (-s) $$ $$ s\in \mathbb{N} $$ when the partial sum is changed to an equivalent polynomial using Faulhaber's formula. ...
3
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4answers
181 views

How is the Bernoulli numbers? For example, as against $B_2$?

How is the Bernoulli numbers? For example, as against $B_2$? For example, found that in internet $$\sum_{n=0}^{\infty} \frac{B_n\;x^n}{n!}=\frac{x}{e^x-1}$$ but if I want to find $B_2$ then ...
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1answer
32 views

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 ?? ...
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0answers
21 views

Showing that Faulhaber's formula is a polynomial of degree $m+1$ [duplicate]

The Faulhaber polynomials $S_{m}(n)$ are defined as $$S_{m}(n)=\sum_{k=1}^{n}k^{m}$$ It is the case that $S_{m}(n)$ is a polynomial in $n$ of degree $m+1$; I've seen a proof of this before, but I ...
2
votes
1answer
48 views

Two-term Binomial-Bernoulli Transform

The binomial transform states that if one has two real sequences $\{a_k \}$ and $\{b_k \}$ satisfying $b_n = \sum_{k = 0}^{n} \binom{n}{k} a_k$, then $a_n = \sum_{k = 0}^{n} (-1)^{n-k} \binom{n}{k} ...
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0answers
38 views

Find the closed form of $S(m, n)=\sum_{j=1}^{m}\sum_{i=1}^{n}i^j$

I know that the closed form of the sum $\sum\limits_{i=1}^{n}i^j$ can be written using Bernoulli numbers $B_k$. It is the famous Faulhaber's formula: $$\sum\limits_{i=1}^{n}i^j =\frac{1}{j+1} ...
0
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2answers
44 views

Finding a Correlation between Bernoulli Variables?

Let X and Y be Bernoulli random variables. We don't assume independence or identical distribution, but we do assume that all 4 of the following probabilities are nonzero. Let a := P[X = 1, Y = 1], b ...
2
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2answers
70 views

How does one derive this formula $\zeta(-n) =\frac{B_{n+1}}{n+1}$?

$$\zeta(-n) =-\frac{B_{n+1}}{n+1}$$ What is the motivation or the derivation of this formula?
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0answers
84 views

Is there a closed form formula for the Bernoulli numbers?

A while ago I found this algorithm. Today I read in wikipedia that Euler zig zag numbers can be used for computing the Bernoulli numbers. This Mathematica program computes the Euler zig zag numbers ...
7
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1answer
289 views

Convergence of $\sum_\lambda \frac{1}{1-\lambda x}$ where $p(\lambda)=0$ for a certain polynomial $p$

The powers of the roots $\lambda$ of these polynomials $$p_n(x):=\sum_{k=1}^{n-1}\frac{n!}{(n-k)!}x^{k-1}$$ (compare with the $p_n$ here) sum to these values $$\sum_\lambda ...
3
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0answers
54 views

References mentioning the relationship between cumulants of uniform distribution and the Bernoulli numbers?

Is there anyone knows where is some official reference mentioning the relationship between cumulants of uniform distribution and the Bernoulli numbers ...
4
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1answer
159 views

Roots of some modified Bernoulli polynomials

Update The polynomials are generated as follows: Where $B_n(x) = \sum_{k=0}^n {n \choose k} b_{n-k} x^k$ is used to generate standard Bernoulli polynomials, top plot is generated as follows: ...
38
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5answers
773 views

Generalizing the sum of consecutive cubes $\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$ to other odd powers

We have, $$\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$$ $$2\sum_{k=1}^n k^5 = -\Big(\sum_{k=1}^n k\Big)^2+3\Big(\sum_{k=1}^n k^2\Big)^2$$ $$2\sum_{k=1}^n k^7 = \Big(\sum_{k=1}^n ...
4
votes
1answer
146 views

Is $\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$ a rational number for $m,n\ge 2\in\mathbb N$?

Question : Is $$\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$$ a rational number for $m,n\ge 2\in\mathbb N$ where $\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$? Motivation : We know that $$\zeta ...
5
votes
2answers
97 views

Verifying a relation involving Bernoulli polynomials

I would appreciate help, please, as to how to verify this relation from Kato's "Fermat's Dream" p.96. He say: By the definition of $B_n(x)$, the Bernoulli polynomial, we have ...
0
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0answers
55 views

Probability of Runs of Heads of Length N [duplicate]

For example: $“THHTHTTHHHTHTHTTHHTHT”$ contains 1 run of heads of length 3, 2 runs of length 2, and 4 runs of length 1. Assuming $P(H) = p$ and $P(T) = (1-p)$, calculate (using properties such as ...
2
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3answers
129 views

Bernoulli numbers: comparison to factorials

I am trying to understand the behaviour of the Bernoulli numbers with respect to factorials, specifically I'd like to know whether it is true that, for all $n \in N$ with $n \ge 2$ we have $$ ...
3
votes
1answer
91 views

show that $\lim_{x\to \frac14} \frac{d^{2k-1}}{dx^{2k-1}}\cot(\pi x)=-(2\pi)^{2k-1}2^{2k}(2^{2k}-1) \frac{\left | B_{2k} \right |}{2k} $

I try to prove the relation between Polygamma function and Bernoulli numbers but I faced this problem,is how to show that $$\lim_{x\to \frac14} \frac{d^{2k-1}}{dx^{2k-1}}\cot(\pi ...
2
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0answers
78 views

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
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0answers
83 views

Is there an explicit formula for the Bernoulli numbers that doesn't implicitly recapitulate Faulhaber's formula?

So, I'm familiar with the "standard" explicit formula for the Bernoulli numbers: $$B_m (n) = \sum^m_{k=0}\sum^k_{v=0}(-1)^v {k \choose v} {(n+v)^m \over k+1}$$ where choosing $n=0$ gives the ...
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0answers
31 views

Hypergeometric distribution, $Hg(1,a,b)$ follows Bernoulli with $Be(\frac{a}{a+b})$

The probability function of Hypergeometric distribution , $Hg(n,a,b)$ is $$P(X=m)=\frac{\binom{a}{m}\binom{b}{n-m}}{\binom{a+b}{n}}$$ I have to show $Hg(1,a,b)$ follows Bernoulli with ...
1
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1answer
37 views

Bernoulli Trial variables?

You are given n=10,000 light bulbs. Each has a reliability of p=99.99% Suppose you select a batch of r=190 light bulbs to light your warehouse. Can you use the equation to predict the ...
1
vote
1answer
98 views

On the numerators of Bernoulli numbers

Von Staudt-Clausen theorem implies that $pB_{2n} \in \mathbb{Z}_{p}$ for all primes $p$ and for all $n \in \mathbb{N}$. It means that the highest power of any prime that can occur in the denominator ...
4
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1answer
185 views

Expressing an integral in terms of the Bernoulli numbers

In Ahlfors' Complex Analysis text, the Bernoulli numbers, $B_k$, are defined as the coefficients in a Laurent development: $$(e^z-1)^{-1}=\frac{1}{z}-\frac{1}{2}+ \sum_1^\infty (-1)^{k-1} ...
3
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2answers
110 views

how to prove that $\psi_1(x)=\sum_{n=0}^{\infty}\frac{B_n}{x^{n+1}}$ ?

how to prove that $$\psi_1(x)=\sum_{n=0}^{\infty}\frac{B_n}{x^{n+1}}$$ where $\psi_1(z)$ is Trigamma function and $B_n$ is Bernoulli number
0
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1answer
127 views

Infinite sum and equality between coefficients of the same index

I have two infinite sums that forms an equality: $$\sum_{n=1}^{\infty} \left(\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right) = \sum_{n=1}^\infty ...
7
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1answer
694 views

Bernoulli numbers generating function

Consider the following generating formula: $$\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$$ There is some intuitive explanation about it? I want to know because I need to proof to myself ...
6
votes
3answers
247 views

The vanishing of the $\bf B$s

Define $B_0=1$ and recursively $$\tag 1 \sum_{k=0}^{n}\binom {n+1}k B_k=[n=0]$$ How can I prove $B_{2n+1}=0$ for $n\geqslant 1$ using this definition? Note the above means that ...
0
votes
1answer
212 views

Probability - rolling a fair die 10 times, what is the probability you would match a separate set of 10 numbers?

Having some trouble with this problem... Say someone is rolling a fair die 10 times, and using that roll as an attempt to guess what number (1-6) someone else has written down on a piece of paper for ...
14
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2answers
963 views

Explicit formula for Bernoulli numbers by using only the recurrence relation

It is not hard to show, by induction on $m\in\mathbb N$, that there exist a sequence $(B_n)_{n\geq0}$ of rational numbers such that ...
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0answers
70 views

Sum of power and Bernoulli intuitive discover

I really looked up to the 10th page of google, and the PDF's I find aren't complete, and definetively have NOT intuitive explanation about Bernoulli's discover. I know that he observed the formulas of ...
3
votes
4answers
128 views

Telescoping sum of powers

$$ \begin{array}{rclll} n^3-(n-1)^3 &= &3n^2 &-3n &+1\\ (n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\ (n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\ ...
1
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1answer
183 views

Proving the Bernoulli number relation $(1+B)^n=B^n$

We know that we can generate the Bernoulli numbers using the relation $(1+B)^n=B^{[n]}$ where $B_n$ is $n$th Bernoulli number. But how we can prove this works? Thanks to all. Edit 2: is there a ...
4
votes
2answers
170 views

Radius of convergence of the Bernoulli polynomial generating function power series.

The generating function of the Bernoulli Polynomials is: $$\frac{te^{xt}}{e^t-1}=\sum_{k=0}^\infty B_k(x)\frac{t^k}{k!}.$$ Would it be right to say that the radius of convergence of this power series ...
2
votes
1answer
367 views

Radius of convergence of a power series with Bernoulli numbers

Say, we use the definition: Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$ and then derive power series representations of the ...
3
votes
0answers
122 views

Bernoulli formula

The sum: $$S_m(n) = 1^m + 2^m + 3^m + 4^m + 5^m...+ n^m$$ Can be calculated by this formula, called the "Bernoulli formula" in wikipedia $$S_m(n) = \frac{1}{m+1}\sum_{k=0}^m {m+1\choose k}B_k ...
8
votes
5answers
263 views

When are we (not) allowed to replace $x$ by $ix$?

It seems to be quite a common manipulation to replace $x$ by $ix$. Every time I see it's being done in a textbook, I blindly trust the author without really understanding when are we allowed to do so ...