10
votes
Accepted
Evaluation of an interesting Integral
The conjecture $f(n)=1$ is only correct for $n\leq 55$, see H. Schmid, Two curious integrals and a graphic proof. For $n=56$ an analytical calculation using the Poisson summation formula gives
$$f(56)=...
8
votes
Accepted
Numerical Calculations
In 1939 Rosser proved that $p_n>n\log n$. In 1999 Dusart proved
$p_n>n(\log n+\log\log n-1)$ and also
$$p_n\ge n\Bigl(\log n+\log\log n-1+\frac{\log\log n-2.25}{\log n}\Bigr),\qquad
n\ge 2.$$
...
6
votes
Accepted
Is there a 1/poly(n) or 1/polylogn upper-bound for this tail bound?
Let $A:=\Bigg\{\bigg\vert\dfrac{\sum_{j=1}^n(\sum_{i=1}^n a_{ij})^2}{n^2} -1\bigg\vert > \epsilon\Bigg\}\,$ denote the event in question. We will show that
$\operatorname{P}(A)\le C_{\epsilon}/n$ ...
3
votes
Solving equation for higher degree of composition
To solve $f_n(x)=x$ (where $f_n$ is the $n$-fold composition of $f$), write $f_n(x)-x$ as a rational function, take the numerator (which is a polynomial, I think of degree $2^n-2$), and find its roots....
2
votes
Accepted
lower bound volume of a set
Such a constant $c$, not depending on $k$ and $x_1,\dots,x_k$, does not exist.
Indeed, suppose $k\ge2$. Let $x_i:=(i-1)h$ for $i=0,\dots,k+1$, where $h:=\frac1{k-1}$ -- so that $x_1=0$ and $x_k=1$.
...
2
votes
Accepted
Showing equality of Eberlein polynomials
Multiply each sum by $x^i y^n$. Sum on $n$, then $i$, then $r$. In both cases we get
$$y^{k+j}(1-y)^{j-k-1}(1-x)^j(1-y+xy)^{k-j}.$$
1
vote
Time ordered integral involving beta function:
I presume with "the case $n=2$" you mean the integral
$$\beta_2(n,m,p,q)=\int _0^1\int _0^t(1-t)^m t^n s^p (1-s)^qdsdt=\frac{m!(n+p+1)!}{(p+1) (m+n+p+2)!}\,\, _3F_2(p+1,n+p+2,-q;p+2,m+n+p+3;1)$$
which ...
1
vote
Accepted
packing with special sets in high dimensional Euclidean space
[EDIT] This is an affirmative answer to the original question, with $C=2$. The negative answer to the version where $1/2$ is replaced by a smaller constant is kept below.
For every $\mathbf x\in[0,1]^...
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