Questions on Bernoulli polynomials and their series expansions.
Questions on Bernoulli polynomials and their series expansions, also in combination with the Riemann zeta function.
Questions tagged [bernoulli-polynomials]
92 questions
11
votes
1 answer
Formula for a sequence defined on $K_1(x,y) := y+0$ if $x \geq y$ and $y-1$ otherwise
Define $K_1:[0,1]^2\rightarrow\mathbb{R}$ as
$$K_1(x,y) := x - \frac{1}{2} - \begin{cases}
\ +(x - y - \frac{1}{2}) & \text{if $x \geq y$},\\
\ -(y - x - \frac{1}{2}) & \text{otherwise}
\end{cases}$$
then with
$$K_n(x,y) :=…
user11260
8
votes
2 answers
The sum of fractional powers $\sum\limits_{k=1}^x k^t$.
This post is a continuation of Generalization of the Bernoulli polynomials ( in relation to the Index ), the definition of the Bernoulli polynomial $B_t(x)$ with $|x|<1$ has an extension through $B_t(x+1)=B_t(x)+t x^{t-1}$.
Two equivalent…
user90369
- 11,696
7
votes
0 answers
Sum $\sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, \,\, n_1+n_2+n_3=N}k_1^{n_1}\times k_2^{n_2} \times k_3^{n_3}$
Let $k_1, k_2, k_3$ be natural non-negative numbers such that $k_1+k_2+k_3=K$. Let $n_1, n_2, n_3 \in \{0, \ldots, N\}$ and such that $n_1+n_2+n_3=N$.
Calculate
$$
S=\sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, \,\, n_1+n_2+n_3=N}k_1^{n_1}\times k_2^{n_2}…
user4164
- 301
5
votes
2 answers
Sum with Bernoulli polynomial
I'm trying to prove the following identity:
$$\sum_{k=0}^n \dfrac {\binom n k B_k(x)} {(n-k+1)} = x^n$$
I transformed this identity as follow:
$$\dfrac{1}{(n+1)}\sum_{k=0}^n \binom {n+1} k B_k(x) = x^n$$
Also I tried to do the…
Katy
- 149
5
votes
1 answer
How to prove that: $2e^{-2x}(e^x -1)= \displaystyle\sum_{n=1}^\infty \frac{\mu(n)}{\sinh(nx)}$
I'm quite lost at this, I've tried to express the $csch(nx)$ as a sum like this:
$$\frac{csch(nx)}{2}= \frac{1}{e^{nx}-e^{-nx}}=\frac{1}{2nx}\sum_{k=0}^{\infty} \frac{(2nx)^k B_k(1/2)}{k!}$$ Where $B_k(x)$ are the Bernoulli polynomials. Then I put…
Rafa
- 576
4
votes
2 answers
closed form for $(n+a)$th derivative of $(e^x-1)^{n}$
I worked on the summation for natural $a$
$$ \Omega_a(n)=\sum_{k=0}^n (-1)^{n-k} \binom{n}{k}k^{n+a}$$
and since
$$ k^{n+a}=\lim_{x\to 0} \frac{d^{n+a}}{dx^{n+a}}e^{k x}$$
So I got
$$ \Omega_a(n)=\lim_{x\to 0} \frac{d^{n+a}}{dx^{n+a}}…
Faoler
- 2,568
- 6
- 18
4
votes
1 answer
Proving that $\lim_{ n \to \infty} \left\{ \frac{1}{2^{m n}} \sum_{r=1}^{2^n-1}(-1)^r r^m\right\}=-\frac{1}{2}$ independently of the value of $m$.
It seems that, independently of the value of $m$, we have
$$
\lim_{ n \to \infty} \left\{ \frac{1}{2^{m n}} \sum_{r=1}^{2^n-1}(-1)^r r^m\right\}=-\frac{1}{2}
$$
I've tested it numerically but I have no idea how to prove it.
Can anyone do it? Or…
Neves
- 5,701
4
votes
0 answers
How to derive this polylogarithm identity (involving Bernoulli polynomials)?
How can one derive the following identity, found here, relating the polylogarithm functions to Bernoulli polynomials?
$$\operatorname{Li}_n(z)+(-1)^n\operatorname{Li}_n(1/z)=-\frac{(2\pi i)^n}{n!}B_n\!\left(\frac12+\frac{\ln(-z)}{2\pi…
WillG
- 7,382
4
votes
1 answer
FUN with f̶̶l̶̶a̶̶g̶̶s̶ Newton Cotes Quadrature formula and Bernoulli polynomials of the second kind
I was told to phrase my question in a more exciting way when I asked it last time. The following is a preliminary consideration. If you don't need it, just scroll down to START HERE. Here we go then:
Imagine you have a really cool…
Physics
- 195
4
votes
1 answer
Solution containing Riemann Zeta function for an integral involving the EGF of the Bernoulli/Euler polynomials
In this post, the first of the following integrals is questioned. I added the second one.
$$
\begin{align}
&2\int_{0}^{\infty}\left(\sum_{k=0}^{n}\frac{\left(-1\right)^{k}B_{k}(1)}{k!}x^{k-n-1}-\frac{1}{x^{n}\left(e^{x}-1\right)}\right)dx\\\\
=&\…
tyobrien
- 3,667
4
votes
0 answers
How to describe $\overset{\sim}{B}_n(x):=\sum_{k=0}^n\binom{n}{k}B^-_{n-k}H_kx^k$ and in particular $\overset{\sim}{B}_n(1)$?
Denote by $B_n(x)=\sum_{k=0}^n\binom{n}{k}B^-_{n-k}x^k$ the $n$-th Bernoulli polynomial, where $B^-_0=1,B^-_1=-\frac{1}{2},B^-_2=\frac{1}{6},...$ are the Bernoulli numbers. Im interested in describing the polynomial $\overset{\sim}{B}_n$ given…
Redundant Aunt
- 12,473
4
votes
1 answer
Integral identity involving Bernoulli polynomials
I found the following identity on Wikipedia, and I am having a difficult time proving it.
For $m,n\in\Bbb N$,
$$I(m,n):=\int_0^1B_n(x)B_m(x)\mathrm{d}x=(-1)^{n-1}\frac{m!n!}{(m+n)!}b_{n+m}$$
Where $B_n(x)$ is the $n$-th Bernoulli polynomial, and…
clathratus
- 18,002
4
votes
0 answers
Riemann zeta meromorphic cont. using Abel summation formula
In Stein&Shakarchi, Complex Analysis, chapter 6, problem 2-3 (p. 180), they hint at a method to meromorphically continue the zeta function to the entire complex plane. I can see from Abel's summation formula that
$$ \zeta(s) = \frac{s}{s-1} -…
JKoolstra
- 51
3
votes
0 answers
An integral inequality involving the Bernoulli polynomials
The classical Bernoulli polynomials $B_j(t)$ are generated by
\begin{equation*}
\frac{z\operatorname{e}^{t z}}{\operatorname{e}^z-1}=\sum_{j=0}^{\infty}B_j(t)\frac{z^j}{j!}, \quad |z|<2\pi.
\end{equation*}
For $\alpha, \beta\in\mathbb{R}$ such that…
qifeng618
- 2,218
3
votes
3 answers
How to find the correct constant term with Euler-Maclaurin formula, $\sum_{j=1}^n j\log j$
Question: Is there a way to find the complete asymptotic expansion (by deriving the correct $O(1)$ term) via the Euler-Maclaurin formula?
Let $$s_n=\sum_{j=1}^n j\log j$$
Fix $m\ge 1$. By the Euler-Maclaurin formula
$$
\begin{align*}
s_n&=\int_1^n…
bob
- 3,079