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6 votes

Compute the integral $\int_0^1\int_0^1\ldots\int_0^1 f(x_1 + x_2 + \ldots + x_n)\,dx_1\,dx_2\ldots dx_n $

Here is a detailed solution motivated by @Arthur's. Let $\mathbf w \in \mathbb R^n$, non of whose coordinates is $0$. For any real $z$, define $$G_{\mathbf w,z} := \{\mathbf x \in \mathbb R^n | \...
dohmatob's user avatar
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6 votes

Proof that $e^{i\theta}/e^ {i\phi} = e^{i(\theta - \phi)}$

Hint: The identity you provide is useful, the other key idea is : $$\frac{a+b}{c+d} = \frac{a+b}{c+d} \frac{c-d}{c-d} = \frac{(a+b)(c-d)}{c^2-d^2} $$
SagarM's user avatar
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5 votes
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References/Proof of the conjectured identity for the Stirling permutation number $\left\{{n\atop n-k}\right\}$

"Concrete Mathematics (what else?) - Eulerian Numbers" - says: "Second-order Eulerian numbers are important chiefly because of their connection with Stirling numbers" Eq. (6.43) therein gives $$ \...
G Cab's user avatar
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5 votes
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The relation of the Bernoulli numbers to the Catalan numbers

We seek to show with Bernoulli numbers, Catalan numbers, Eulerian numbers and the Factorial counterpart of Leibniz's harmonic triangle: $$\frac{B_n}{C_n} = \sum_{k=0}^n (-1)^k \frac{k! (n-k)!}{(n+1)^\...
Marko Riedel's user avatar
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4 votes
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Second-order Eulerian numbers, Lambert's W function, and Schröder's fourth problem

We seek to show that the following identity holds: $$2^{n+1} \sum_{k=0}^n \left\langle\!\! \left\langle n \atop k \right\rangle\!\! \right\rangle \frac{1}{2^k} = n! [x^n] \frac{1}{1 + W(-\exp((x-1)/2)...
Marko Riedel's user avatar
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4 votes

Compute the integral $\int_0^1\int_0^1\ldots\int_0^1 f(x_1 + x_2 + \ldots + x_n)\,dx_1\,dx_2\ldots dx_n $

I propose the following: If $A(c)$ is the hyperarea of the region of the hyperplane $x_1+\cdots + x_n = c$ bounded by the hypercube $0\leq x_1, \ldots,x_n\leq 1$, the integral is equal to $\frac 1{\...
Arthur's user avatar
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4 votes

A closed form of Eulerian numbers

(We count descents instead of ascents. Let $A(n, k)$ denote the number of permutations on $[n]$ with $k-1$ descents) Here's a combinatorial proof via inclusion-exclusion the essence of which can be ...
Aryaman Jal's user avatar
  • 1,543
4 votes

Bounds on the difference between the polylogarithm with negative base and the gamma function

To compare the series and the integral, we will use the Abel-Plana formula which reads \begin{equation} \sum_{j=0}^\infty f(j)=\int_0^\infty f(x)\,dx+\frac{1}{2}f(0)+i\int_0^\infty\frac{f(it)-f(-it)}...
Paul Enta's user avatar
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3 votes

How to prove Eulerian Identity?

By way of enrichment and not necessarily following the text we start from the bivariate generating function of the Eulerian numbers, which seems like a reasonable starting point and which is $$\...
Marko Riedel's user avatar
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3 votes
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Half of the binomial theorem

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
Felix Marin's user avatar
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3 votes
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Simplify binomial sum

Suppose we seek an alternate representation of $$\sum_{p=q}^k (-1)^p {k\choose p} (q-p)^k.$$ This is $$\sum_{p=0}^k (-1)^p {k\choose p} (q-p)^k - \sum_{p=0}^{q-1} (-1)^p {k\choose p} (q-p)^k.$$ We get ...
Marko Riedel's user avatar
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3 votes

Simplify binomial sum

This answer is a(nother) supplement to @MarkoRiedel's great derivation. This time we look at some details of his proof of \begin{align*} \color{blue}{\sum_{p=0}^{q-1} \left\langle k \atop p \right\...
Markus Scheuer's user avatar
3 votes
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A Stirling number identity representing the second-order Eulerian numbers.

We seek to show that with $0\le k\le n$ the following identity holds: $$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} = \sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$ We ...
Marko Riedel's user avatar
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3 votes

Difference of the Stirling cycle numbers and the Stirling set numbers

We can start from Concrete Mathematics, page 271, Equations (6.43) and (6.44). They are said to be obtained by induction on $n$, the second order Eulerian numbers are supposed to be defined by their ...
René Gy's user avatar
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3 votes
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The value of the second-order Eulerian polynomials at $x = \frac{-1}{2}$.

Here are some comments. Seeking to invert $$-\frac{1}{9} + \frac{2}{3} x + \frac{1}{9} \exp(3x) = z$$ we consult Wikipedia on LambertW to find that the closed form solution to $$x = a + b \exp(cx)$$ ...
Marko Riedel's user avatar
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3 votes
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A recurrence of the second-order Eulerian polynomials

This is the same as in the Tree function Eulerian identity. Observe that $$x (1-x)^{2n} \left(\frac{1}{(1-x)^{2n-1}} \sum_{k=0}^{n-1} \left\langle\!\! \left\langle n-1 \atop k \right\rangle\!\! \right\...
Marko Riedel's user avatar
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2 votes
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An identity related to the second-order Eulerian numbers.

In trying to verify the identity $$\sum_{j=0}^{k} {n-j \choose n-k} \left\langle\!\! \left\langle n\atop j \right\rangle\!\! \right\rangle = \sum_{j=0}^k (-1)^{j+k} {n+k \choose n+j} \left\{ n+j \...
Marko Riedel's user avatar
  • 62.1k
2 votes

Does $x\left(\frac{d}{dx}\left(\cdots x \left(\frac{d}{dx} \left( \frac{x}{1-x}\right)\right)\cdots\right)\right)$ have a closed form expression?

Note : $$\operatorname{Li}_{-m}(z)= \displaystyle \sum_{k = 1}^{\infty} k^m z^k=\frac 1{(1-z)^{m+1}}\sum_{k=0}^{m-1} E(m,k)\, z^{m-k}$$ with $E(m,k)$ the Eulerian numbers. This relationship is ...
JJacquelin's user avatar
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2 votes
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Difference of the Stirling cycle numbers and the Stirling set numbers

We seek to show that with $0\le k\le n$ the following identity holds: $${n\brack n-k} - {n\brace n-k} = \sum_{j=0}^k \left({n+j-1\choose 2k} - {n+k-j\choose 2k}\right) \left\langle\!\! \left\langle k\...
Marko Riedel's user avatar
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2 votes
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An associated Stirling number identity related to the second-order Eulerian numbers.

We seek to show that with $0\le k\le n$ the following identity holds: $$ \sum_{j=0}^{k} (-1)^{k-j} {n-j \choose k-j} \left\{ \!\! \left\{ n+j\atop j\right\} \!\! \right\} = \sum_{j=0}^{n-k+1} (-1)^{...
Marko Riedel's user avatar
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2 votes
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A combinatoric identity involving Eulerian and Stirling numbers

Here is another proof for the curious. We seek to verify that (with $n\ge 1$, $n=0$ holds by inspection) $$\sum_{k=0}^n \left\langle n \atop k \right\rangle x^{n-k} = (1-x)^n \sum_{k=0}^n {n\brace k} ...
Marko Riedel's user avatar
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2 votes

Simplify binomial sum

This answer is a supplement to @MarkoRiedel's interesting derivation. We provide a somewhat different approach to derive the identity \begin{align*} \color{blue}{\sum_{p=0}^{q-1} \left\langle k \atop ...
Markus Scheuer's user avatar
2 votes

Positivity of a certain sum of Stirling numbers

For the moment at least, I can just individuate the first step of an approach which might be possibly interesting. The sum can be rewritten as $$ \eqalign{ & S(q,n,m) = \sum\limits_{\left( {...
G Cab's user avatar
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2 votes

Recurrence formula for the Eulerian and derangement polynomial

I don't know whether this question is still of interest, but I just spotted a recurrence formula for the derangement polynomial in "On derangement polynomials of type B. II" by Chak-On Chow, ...
Balazs's user avatar
  • 129
2 votes
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The ratio between Central Eulerian Numbers and the sum of Eulerian Numbers at a fixed level converges to zero

Using the asymptotic provided on the OEIS page $$C(n)\sim\frac{2^{2n}n^{2n-1}\sqrt3}{e^{2n}}$$ and comparing it with the asymptotic of $(2n-1)!$ $$(2n-1)!\sim\sqrt{2\pi(2n-1)}\left(\frac{2n-1}e\right)^...
Parcly Taxel's user avatar
2 votes

Proof that the Eulerian Numbers $A(n,m)$ are asymptotic to $(m+1)^n$ when $m$ is fixed and $n\rightarrow\infty$

Let us prove that $$\lim_{n\rightarrow\infty}\frac{A(n,m)}{(m+1)^n}=1.$$ We use the formula for the Eulerian Numbers provided in the book Concrete Mathematics: $$A(n,m)=\sum_{k=0}^m\binom{n+1}{k}(m+1-...
user559748's user avatar
2 votes

An identity related to the second-order Eulerian numbers.

A sketch in three steps: First, we need a definition for $ \operatorname{W}_{n}(x)$. With this, we do not want to assume the second-order Eulerian numbers. So we take the RHS of the identity. The next ...
Peter Luschny's user avatar
2 votes

Power series of $x/(1-ae^{-x})$.

This is "almost" the generating function of the Eulerian polynomials: for $|a|<1$ we have \begin{align*}\frac1{1-ae^{-x}}&=\sum_{k=0}^\infty a^k\sum_{n=0}^\infty\frac{(-kx)^n}{n!}=\...
metamorphy's user avatar
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2 votes

Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

A formal consideration. In terms of the Eulerian polynomials $$ g_k (x) = \frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}, $$ whence $$ G(x) = \sum\limits_{k = 0}^\infty {\frac{{xA_k (x)}}{{(1 - x)^{k + 1} }}} ...
Gary's user avatar
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2 votes
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Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

For $0 < x < 1$ and all $k \ge 0$ is $$ g_k(x) =\sum_{n=1}^{\infty} x^nn^k \ge \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$ so that $$ G(x)=\sum_{k=0}^{\infty} g_k(x) $$ is not convergent (the ...
Martin R's user avatar
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