Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Number of solutions of an equation over finite fields Does anyone know of any result that deals with the following problem of counting the number of solutions of a certain algebraic equation over a finite field?
Let $p$ be an odd prime and $(a,b,c,d)\in\mathbb{F}_p^4$. How many solutions does the following equation hav... | The first equation can be worked out in an elementary way. Counting the solutions where $b=0$ or $d=0$ is straightforward. So let's consider the case $bd\ne0$. The equation rewrites as\begin{equation}
d(a-\frac{b^2}{2d})^2+b(c-\frac{d^2}{2b})^2=\frac{b^5+d^5}{4bd}.
\end{equation}
So, given given $b$ and $d$, and upon s... | {
"language": "en",
"url": "https://mathoverflow.net/questions/114596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A question from complex analysis Let $n\geq2$. We assume $0<\alpha_n<\cdots<\alpha_2<\alpha_1<1$ and $0<\beta_n<\cdots<\beta_2<\beta_1<1$
, $\alpha_n=\beta_n$, and there exists $1\leq j_0\leq n$ such that $\alpha_{j_0}\neq \beta_{j_0}$.
My question: is there a complex number $s$ such that
$\sum_{j=1}^n s^{\alpha_j}=0... | The answer is no. Take the polynomials
\begin{align}
f(x) &= x^6 + x^5 + x^4 + x^3 + x^2 + x\\\
g(x) &= x^8 + x^6 + x^5 + x^4 + x^3 + x.
\end{align}
From
\begin{align}
f(x) &= x(x+1)(x^2+x+1)(x^2-x+1)\\\
g(x) &= x(x+1)(x^2+x+1)(x^2-x+1)^2
\end{align}
wee see that $f(x)$ and $g(x)$ have the same complex roots. Upon sett... | {
"language": "en",
"url": "https://mathoverflow.net/questions/127979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Principal value of integral What is the principal value of the integral $$\int \limits _0^\infty \left( \frac {1}{x^2}-\frac{\cot(x)}{x} \right) dx ?$$ Maple finds $PV \int_0^\infty \tan(x)/x dx = \pi/2.$ Such integrals arise in physics. I unsuccessfully asked it in SE.
| Yes the value $\pi/2$ can be obtained like this.
Let
$$
f(x):=\frac {1}{x^2}-\frac{\cot(x)}{x}
$$
We may compute
$$
\int_0^{\pi/2} f(x)\;dx + \sum_{k=1}^\infty \int_0^{\pi/2}\big(f(k\pi+x)+f(k\pi-x)\big)\;dx=\frac{\pi}{2}
\tag{1}$$
and this converges.
We can think of (1) as a "rearrangement" of the required integral.... | {
"language": "en",
"url": "https://mathoverflow.net/questions/130549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
A curious sequence of rationals: finite or infinite? Consider the following function repeatedly applied to a rational
$r = a/b$ in lowest terms:
$f(a/b) = (a b) / (a + b - 1)$.
So, $f(2/3) = 6/4 = 3/2$. $f(3/2) = 6/4 = 3/2$.
I am wondering if it is possible to predict when the sequence is finite,
and when infinite. Fo... | All fractions $a/b$ with $ a = b(b-1)+1 $
are fixed points because then $(a,b)=1$. Moreover, all solutions to $f^{(2)}(r)=1$ I found so far come from one of these families:
$$f\left(f\left(\frac{1}{n}\right)\right)=f\left(1\right)=1$$
$$f\left(f\left(\frac{(n-1)^2}{n^2}\right)\right)=f\left(\frac{n(n-1)}{2}\right)=1$... | {
"language": "en",
"url": "https://mathoverflow.net/questions/131715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
Estimating a sum Good morning everyone,
I would like to make a question about estimating a sum.
Consider the following sum
$$S_n:=\sum_{k=0}^{n-1} \frac{k^2}{(n-k)^2 (n+k)^2} $$
It is easy to see that this sum is bounded by $\sum_{k=0}^{n-1} \frac{1}{(n-k)^2} \leq \frac{\pi^2}{6}$ for all $n$. But i would like to k... | Actually the sum can be done in "closed form", and its asymptotics follow from the known asymptotics of the Digamma function and its derivatives. According to Maple
$$\eqalign{S_n &= 1/24\,{\pi }^{2}-1/4\,{\frac {\gamma}{n}}-1/4\,\Psi \left( 1,2\,n
\right) -1/4\,{\frac {\Psi \left( 2\,n \right) }{n}}-1/4\,\Psi
\left... | {
"language": "en",
"url": "https://mathoverflow.net/questions/185871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A perfect $(n,k)$ shuffle function Suppose you have a deck of $n$ cards; e.g., $n{=}12$:
$$
(1,2,3,4,5,6,7,8,9,10,11,12) \;.
$$
Cut the deck into $k$ equal-sized pieces, where $k|n$;
e.g., for $k{=}4$, the $12$ cards are partitioned into
$4$ piles, each of $m=n/k=3$ cards:
$$
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
4 ... | This is just my comment above, which seems to answer the question. Label the cards from $0$ to $n-1$. Then, with $m=n/k$, the shuffle in the question corresponds to multiplying card $i$ by $m$ (taken mod $n-1$). Thus repeating the shuffle $r$ times amounts to multiplying by $m^r \pmod{n-1}$, which returns us to the... | {
"language": "en",
"url": "https://mathoverflow.net/questions/187771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture Let $G_1$ and $G_2$ be the groups with the following presentations:
$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2... | Similar to the answer of Mark Sapir, for the second group let $x=ba$ and $y=b^{-1}ab^{-1}$. From the second relation we have $y^2=x^{-1}y^{-1}$. So, from the first relation we have $x=y^4$. These relations implies that $y^4=y^{-3}$ or $y^7=1$.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/231922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
What is so special about $a^3+b^3+c^3 = (13m)^3$? It seems rather surprising that, given the Diophantine equation,
$$a^3+b^3+c^3 = n^3\tag1$$
then a good $\color{red}{99.8\%}$ of $n<1000000$ are solvable in positive integers $a,b,c$. (See the discussion in this MSE post.)
Oleg567 gave a list of the $1867$ unsolvable $n... | A solution of (1) must contain $0, 2$ or $4$ terms divisible by $13$. Essentially, this is because the only cubic residues mod $13$ are $0, 1, 5, 8, 12$, and there is no combination (with or without repetition) of three of the non-zero residues with a sum $s$ such that $s \equiv 0 \pmod{13}$. A proof is in (A).
Altho... | {
"language": "en",
"url": "https://mathoverflow.net/questions/286020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral equality of 1st intrinsic volume of spheroid Computations suggest that
$$\int_{0}^{\infty}\int_{0}^{\infty} \sqrt{x+y^2} \cdot e^{-\frac{1}{2}(\frac{x}{s}+s^2y^2)}dxdy=\frac{2}{s}+\frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}.$$
The question is how to prove this equality.
Background: this is the mean width of... | First of all make a replacement $x=t^2$ and go to polar coordinates
\begin{equation*}
\int_{0}^{\infty}\int_{0}^{2\pi}
\cos(\phi)r^3 \cdot e^{-\frac{\frac{r^2\cos^2(\phi)}{s}+r^2\sin^2(\phi)s^2}{2}}drd\phi.
\end{equation*}
After than notice that we can take integral by radius. Let $R=\frac{1}{2}(\cos^2(\phi)\fr... | {
"language": "en",
"url": "https://mathoverflow.net/questions/313313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Does this deceptively simple nonlinear recurrence relation have a closed form solution? Given the base case $a_0 = 1$, does $a_n = a_{n-1} + \frac{1}{\left\lfloor{a_{n-1}}\right \rfloor}$ have a closed form solution? The sequence itself is divergent and simply goes {$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}... | The sequence $a_n$ for $n\geq 1$ has the following formula:
$$a_n=\left\lfloor \sqrt{2n}+\tfrac{1}{2}\right\rfloor +\frac{\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor-\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor ^2 +2 n}{2 \left\lfloor \sqrt{2n}+\frac{1}{2}\right\rfloor }.$$
He... | {
"language": "en",
"url": "https://mathoverflow.net/questions/323665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A problem of matrix polynomial expansion The problem is
$b = (1, -1)^\top, c = (1, 1)^\top, A \in \mathbb{R}^{2 \times 2}$, suppose the sum of reverse diagonal elements of $A$ is zero (i.e., $A_{12} + A_{21} = 0$), prove that the sum of reverse diagonal elements of $\sum\limits_{r=0}^{n-1} A^r c b^\top A^{n-1-r} $ ... | This is a bit of a brute force approach, but it's effective. Note that the sum of the reverse diagonal elements of a $2\times 2$ matrix $M$ equals ${\rm tr}\,\sigma M$ with
$$\sigma=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$
For the most general form of the matrix
$$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and f... | {
"language": "en",
"url": "https://mathoverflow.net/questions/372206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Generalized adjoint operation valid? Let $R_{\theta}$ be the rotation by an angle $\theta$.
Is it then true that for multi-indices $\alpha$ of fixed order $j$ and any smooth function $f$ we have
$$\sum_{\vert \alpha \vert=j}(R_{\theta}z)^{\alpha} \partial^{\alpha}f(x) = \sum_{\vert \alpha \vert=j}(z)^{\alpha} (R_{-\the... | If I am understanding your notation correctly (which admittedly, I may not be), I believe this is already false for $j = 2$. Let me write:
$$ R_{\theta} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \qquad f = f(x,y) \qquad z = \begin{bmatrix} u \\ v \end{bmatrix} $$
Then for $j = ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/398275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Collecting alternative proofs for the oddity of Catalan Consider the ubiquitous Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$. In this post, I am looking for your help in my attempt to collect alternative proofs of the following fact: $C_n$ is odd if and only if $n=2^r-1$.
Proofs that I know:
*
*an application of Lege... | Apparently not mentioned yet, though surely not new:
use the quadratic equation satisfied by the generating function.
Since we look for $n+1$ to be a power of $2$, we shift the index by $1$
and consider
$$
F = \sum_{n=0}^\infty C_n x^{n+1} = x + x^2 + 2 x^3 + 5 x^4 + 14 x^5 + 42 x^6 + 132 x^7 + 429 x^8 + \cdots.
$$
The... | {
"language": "en",
"url": "https://mathoverflow.net/questions/409002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
} |
Can anything be said about the roots of the L4 center? Modes, Medians and Means: A Unifying Perspective defines the following centers based on the $L_p$ norms:
$$
\begin{aligned}
\text{mode of x} = \arg \min_s \sum_i \lvert x_i - s \rvert^0 \\
\text{median of x} = \arg \min_s \sum_i \lvert x_i - s \rvert^1 \\
\text{mea... | The function $\sum_i(x_i-s)^3$ is strictly increasing in $s$, from $-\infty$ to $\infty$. It is also continuous, so there is a unique real root.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/413246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Diagonalization of a specific Dirac operator A few hours ago, a question was posed asking for the eigenvalues and eigenvectors of the Dirac operator
$$
H=\begin{pmatrix} x & 0 & -i\partial_{x} & \bar{z} \\ 0 & x & z & i\partial_{x} \\ -i\partial_{x} & \bar{z} & -x & 0 \\ z & i\partial_{x} & 0 & -x \end{pmatrix}
$$
The... | Denote the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\lambda_{n} =2n+1$) as $\psi_{n} (x)$. Introduce also the standard raising and lowering operators $a^{\dagger } $, $a$, in terms of which $x=(a^{\dagger } +a)/\sqrt{2} $ and $-\partial_{x} =(a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/417595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is this combinatorial identity known? (of interest for random matrix theory) While playing around with random matrices and I arrived at a different formula for the mean of the limiting normal distribution for a spectral CLT for sample covariance matrices. More precisely I have the formula
\begin{align*}
& \sum\limits_{... | Firstly, exploit the finite support to simplify the limits of the sums. Secondly, split the second sum. We get
$$\begin{align*}A(r,b) =& \sum\limits_{m=1}^{r} (2m-1) {r \choose b-m}{r \choose b+m-1} \\
& + \sum\limits_{m=1}^{r} 2m {r \choose b-m}{r \choose b+m} \\
& + \sum\limits_{m=1}^{r} {r \choose b-m}{r \choose b+m... | {
"language": "en",
"url": "https://mathoverflow.net/questions/426906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
\begin{aligned}
f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
&+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4... | (This is a comment, not an answer.)
If $f_n(x)$ is your polynomial, starting with $f_0(x)=1$, then
$$ \sum_{n=0}^\infty f_n(x) y^n =
\frac{1-xy+x^2y^2+x^2y^3}{(1+xy^2)(1-xy-xy^2)}
= 1 + \frac{x^2y^2(1+y)^2}{(1+xy^2)(1-xy-xy^2)}. $$
Also, I noticed that $f_n(x)-x f_{n-1}(x)-x f_{n-2}(x)$ only has one or two terms... | {
"language": "en",
"url": "https://mathoverflow.net/questions/440962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
} |
Primes P such that ((P-1)/2)!=1 mod P I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this
implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$.
Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
Af... | Apologies for repeating some information in my reply to question 121678, which I came across before seeing this one.
Several previous answers already explain the connection to the class number. It can be added that the value of $h(-p)$ was investigated by Louis C. Karpinski in his doctoral dissertation (Mathematischen ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/16141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 1
} |
monotonicity from 4 term-recursion. In determining the monotonicity of coefficients in a series expansion (which appeared in one of my study), I come across the following problem.
Let $p\ge 2$ be an integer, and $$6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i,~~i\ge0$$ with $d_0=d_1=d... | Simple calculation shows $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}, 30p^3d_5=6p^2(4+p)d_4+(p-1)[3p(3+2p)+(2p-1)(2+3p)]d_3$, and so $d_3>d_4>d_5$.
Assume $d_i>d_{i+1}>d_{i+2}$ and $6p^3(i+2)d_{i+2}\ge 6p^2(i+1+p)d_{i+1}+(p-1)[3p(i+2p)+(2p-1)(i-1+3p)]d_i$ for all $i\ge3$ .
Then $6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+\fra... | {
"language": "en",
"url": "https://mathoverflow.net/questions/21643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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When is the sum of two quadratic residues modulo a prime again a quadratic residue? Let $p$ be an odd prime. I am interested in how many quadratic residues $a$ sre there such that $a+1$ is also a quadratic residue modulo $p$. I am sure that this number is
$$
\frac{p-6+\text{mod}(p,4)}{4},
$$
but I have neither proof n... | There is an elementary argument regarding the last problem. Denote by $N(p)$ the number of pairs of $(a,b)$ such that $a,b,a+b$ are all quadratic residues mod $p$.
Hence we have
$$N(p)=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\rig... | {
"language": "en",
"url": "https://mathoverflow.net/questions/65183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
How to calculate [10^10^10^10^10^-10^10]? How to find an integer part of $10^{10^{10^{10^{10^{-10^{10}}}}}}$? It looks like it is slightly above $10^{10^{10}}$.
| I think the number in question is $10^{10^{10}}+10^{11}\ln^4(10)$ plus a tiny positive number. That is, it starts with a digit $1$, followed by $10^{10}-13$ zeros, then by the string $2811012357389$, then a decimal point, and then some garbage (which starts like $4407116278\dots$).
To see this let $x:=10^{-10^{10}}$, a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/79217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 1,
"answer_id": 0
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Convergence rate for product of stochastic matrices Hi,
I have a system of the form $$x(t+1) = A(t + 1) x(t),$$ for $t \geq 1$, and some fixed initial condition $x(1)$. Here $A (t)$ is a time-varying $m \times m$ matrix that is stochastic at all times $t$ (so row sums are $1$).
In fact, I know that the matrices $A(t... | shIf your matrices are "time-dependent" you need additional assumptions to bound the rate of convergence and it will usually be impossible to obtain something which depends only on the minimum of the second eigenvalues.
You may choose $a$ and $b$ at will in the example below, for example such that both matrices are do... | {
"language": "en",
"url": "https://mathoverflow.net/questions/119025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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divisors of $p^4+1$ of the form $kp+1$ In group theory the number of Sylow $p$-subgroups of a finite group $G$, is of the form $kp+1$.
So it is interesting to discuss about the divisors of this form. As I checked it seems that for an odd prime $p$, there is not any divisor $a$ of $p^4+1$, where $1<a<p^4+1$ and $a=kp+1... | There's none indeed.
Lemma: if $1<m<n$ are coprime integers then $mn+1$ does not divide $n^4+1$.
First observe that for any $m,n$, of $mn+1$ divides $n^4+1$, then it divides $n^4m^4+m^4=((nm)^4-1)+1+m^4$, and since $mn+1$ clearly divides $((nm)^4-1)$, we deduce that it also divides $1+m^4$.
To prove the lemma, assume t... | {
"language": "en",
"url": "https://mathoverflow.net/questions/208645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
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For what integer $n$ are there infinitely many $-a+nb+c = -d+ne+f$ where $a^6+b^6+c^6 = d^6+e^6+f^6$? (Much revised for clarity.) I was considering the system of equations,
$$-a+nb+c = -d+ne+f\tag1$$
$$a+b+c = d+e+f\tag2$$
$$a^2+b^2+c^2 = d^2+e^2+f^2\tag3$$
$$a^6+b^6+c^6 = d^6+e^6+f^6\tag4$$
Question 1. Is it true that... | I am sorry to say that the answer to Question 1 is NO.
As you state, $q=(3-n)p/(2n)$ gives a solution to the problem. Thus the quartic must be birationally equivalent to an elliptic curve.
Using standard methods, it is possible to show that this elliptic curve is of the form
\begin{equation*}
G^2=H^3+25( h_1 H + h_2)^2... | {
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"url": "https://mathoverflow.net/questions/222719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture Let $G_1$ and $G_2$ be the groups with the following presentations:
$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2... | Denote $x=ab$, $y=a^{-1}ba^{-1}$. Then the first two relations of the first group are $x^2=y$, $y^2=(yx)^{-1}$. This implies $x^4=x^{-3}$ or $x^7=1$. So the group has torsion. I leave the second group as an exercise for the others.
| {
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"url": "https://mathoverflow.net/questions/231922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Constant related to continued fraction of quadratic irrationals Let $d$ be a positive, non-square integer, and define $c_d$ to be the smallest positive number with the following property: for all pairs of co-prime integers $(p,q)$ with $q > 0$, the inequality
$$\displaystyle \left \lvert \frac{p}{q} - \sqrt{d} \right \... | Edit: I answered a different question than the one asked. The question asks for the least $c_d = q^2|\sqrt{d}-p/q|$. I evaluated the liminf.
If $d = a^2+b$ with $1\le b \le 2a$ then the simple continued fraction for $\sqrt{d}$ is preperiodic, and the period ends with $2a$, which is the largest coefficient, with $a$ bei... | {
"language": "en",
"url": "https://mathoverflow.net/questions/247034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integral equality of 1st intrinsic volume of spheroid Computations suggest that
$$\int_{0}^{\infty}\int_{0}^{\infty} \sqrt{x+y^2} \cdot e^{-\frac{1}{2}(\frac{x}{s}+s^2y^2)}dxdy=\frac{2}{s}+\frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}.$$
The question is how to prove this equality.
Background: this is the mean width of... | we pose $x=t^2$
$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{t^2+y^2} e^{\frac{t^2}{s}+s^2 y^2} 2tdxdt $
by change of variable ,we get:
$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy =2\int^{\frac{\pi}{2}}_0\int^{\infty}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/313313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using Fourier series to prove $-\int_0^1 u_{xxx}u_x \eta = \int_0^1 (u_{xx})^2\eta - \int_0^1 \frac{1}{2} (u_x)^2 \eta_{xx}$ Let $u, \eta$ be smooth functions and $\eta$ compactly supported in $(0,1)$. Integrating by parts, we can easily prove $$-\int_0^1 u_{xxx}u_x \eta = \int_0^1 (u_{xx})^2\eta - \int_0^1 \frac{1}{2}... | term on left hand side:
$$L=-\int_0^1 dx\, u_{xxx}u_x \eta =2^{3/2}\pi^4 \sum_{n,m,k=1}^\infty n^3m a_na_mb_k \int_0^1 dx\,\sin(n\pi x)\sin(m\pi x)\cos(k\pi x)=$$
$$=2^{3/2}\pi^4 \frac{1}{4}\sum_{n,m,k=1}^\infty n^3m a_na_mb_k \left(\delta_{m,n+k}+\delta_{n,m+k}-\delta_{k,n+m}\right).$$
two terms on right hand side
$$R... | {
"language": "en",
"url": "https://mathoverflow.net/questions/382207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues of operator In the question here
the author asks for the eigenvalues of an operator
$$A = \begin{pmatrix} x & -\partial_x \\ \partial_x & -x \end{pmatrix}.$$
Here I would like to ask if one can extend this idea to the operator
$$A = \begin{pmatrix} x & -\partial_x +c\\ \partial_x+c & -x \end{pmatrix},$$
whe... | The extended operator can be treated along similar lines as the $c=0$ case. One merely has to modify the algebra a little. Again, denote the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\lambda_{n} =2n+1$) as $\psi_{n} (x)$. Introduce also the stan... | {
"language": "en",
"url": "https://mathoverflow.net/questions/403572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Submatrices of matrices in $\mathrm{SL}(4, \mathbb{Z})$ with all eigenvalues equal to $1$ This is a follow-up question to my question from Math Stackexchange (Thank you Dietrich Burde and Michael Burr for the help).
Let $M\in \mathrm{SL}(4, \mathbb{Z})$ with all eigenvalues equal to $1$ (i.e. $M$ is a unipotent matrix)... | It is not a solution since $A$ is not in $GL_2(\mathbb{Z})$, I will try to correct it later.
Let $a,b,c,d$ be in $\mathbb{Z}$ such that $ad-bc=1$.
Set
$$P = \left(\begin{array}{cc}
aI_2 & bI_2 \\
cI_2 & dI_2
\end{array}\right) \text{ and } T = \left(\begin{array}{cc}
I_2 & S \\
0 & I_2
\end{array}\right),$$
where $S ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/425193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Reducing $9\times9$ determinant to $3\times3$ determinant Consider the $9\times 9$ matrix
$$M = \begin{pmatrix} i e_3 \times{} & i & 0 \\
-i & 0 & -a \times{} \\
0 & a \times{} & 0 \end{pmatrix}$$
for some vector $a \in \mathbb R^3$, where $\times$ is the cross product.
It is claimed in Fu and Qin - Topological phases... | The formula's in the OP contain an error: the $\omega$ in the denominator of $N_1$ and $N_2$ should be $\omega^2$, so
$$N_1 = \frac{aa^t}{\omega^2}, N_2 =\frac{a^ta}{\omega^2}\operatorname{id}.$$
Then it works out:
$${\rm det}\,(M-\omega)=-\omega^9+2 \omega^7 \left(a_1^2+a_2^2+a_3^2+2\right)-\omega^5 \left(a_1^4+2 a_1^... | {
"language": "en",
"url": "https://mathoverflow.net/questions/434565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
\begin{aligned}
f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
&+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4... | Here is a proof, using Maple calculations, of Tim Chow's empirical observations. We use Hadamard products of power series. The Hadamard product (with respect to the variable $y$) is defined by
$$
\sum_{n=0}^\infty a_n y^n *\sum_{n=0}^\infty b_n y^n= \sum_{n=0}^\infty a_n b_n y^n.
$$
The Hadamard product of two rationa... | {
"language": "en",
"url": "https://mathoverflow.net/questions/440962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
monotonicity from 4 term-recursion. In determining the monotonicity of coefficients in a series expansion (which appeared in one of my study), I come across the following problem.
Let $p\ge 2$ be an integer, and $$6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i,~~i\ge0$$ with $d_0=d_1=d... | I am a little bit upset that my previous answer to the question
was downvoted by somebody. As far as I understand the idea of MO is
not necessarily to produce final answers/solutions/responses but
mostly to provide the ideas of approaching a given problem.
It's a question of time to solve this particular problem (I sti... | {
"language": "en",
"url": "https://mathoverflow.net/questions/21643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
When is the sum of two quadratic residues modulo a prime again a quadratic residue? Let $p$ be an odd prime. I am interested in how many quadratic residues $a$ sre there such that $a+1$ is also a quadratic residue modulo $p$. I am sure that this number is
$$
\frac{p-6+\text{mod}(p,4)}{4},
$$
but I have neither proof n... | Here's a copy-paste of something I wrote up a while ago:
Lemma: Let $q$ be odd, and let $Q$ be the set of quadratic residues (including $0$) in $\mathbb F_q$. Then the number of elements $s_q(c)$ in $\{x^2+c|x \in \mathbb{F}_q\} \cap Q$ is given by
\begin{array}{|c|c|c|}
\hline
& c \in Q & c \notin Q \\
\hline
-1 \in ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/65183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
Interpolating a sum of binomial coefficients using a sin function While studying a problem about orthogonal polynomials I encountered the following
expressions
\begin{equation}
f(n)=\sum_{k=0}^{n}(-1)^k\binom{n+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\begin{equation}
g(n)=\sum_{k=0}^{n-1}(-1)^k\binom{n+k}... | Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:
\begin{equation}
\sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/109372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Upper limit on the central binomial coefficient What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ?
I just tried to proceed a bit, like this:
$ n! > n^{\frac{n}{2}} $
for all $ n>2 $. Thus,
$ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=... | We may prove without using integrals that
$$
\frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1)
$$
(1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\c... | {
"language": "en",
"url": "https://mathoverflow.net/questions/133732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
Rachinsky quintets
This 1895 painting of Nikolai Bogdanov-Belsky shows mental calculations in the public school of Sergei Rachinsky. Boys in a Russian village school try to calculate $(10^2+11^2+12^2+13^2+14^2)/365$ in their heads. One of the methods of solution is based on the equality $10^2+11^2+12^2=13^2+14^2$. No... |
The following recipe (algorithm) generates all solutions. It may be viewed as a parametrization in a general(ized) sense.
W.l.o.g. we may assume that $c$ is odd. Then
$$ \left(\frac{x-y}2\right)^2 + \left(\frac{u-v}2\right)^2 + c^2
\ =\ \left(\frac{x+y}2\right)^2 + \left(\frac{u+v}2\right)^2 $$
where three co... | {
"language": "en",
"url": "https://mathoverflow.net/questions/153558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 2,
"answer_id": 1
} |
For what integer $n$ are there infinitely many $-a+nb+c = -d+ne+f$ where $a^6+b^6+c^6 = d^6+e^6+f^6$? (Much revised for clarity.) I was considering the system of equations,
$$-a+nb+c = -d+ne+f\tag1$$
$$a+b+c = d+e+f\tag2$$
$$a^2+b^2+c^2 = d^2+e^2+f^2\tag3$$
$$a^6+b^6+c^6 = d^6+e^6+f^6\tag4$$
Question 1. Is it true that... | As asked by OP here are values of $n$ where the BSD Conjecture predicts a strictly positive rank for the elliptic curve
12,
13,
15,
16,
17,
18,
21,
23,
24,
25,
26,
27,
30,
32,
33,
35,
36,
38,
40,
41,
43,
44,
47,
49,
52,
53,
54,
56,
57,
64,
65,
66,
67,
69,
73,
75,
76,
78,
79,
80,
81,
85,
86,
88,
90,
91,
93,
94,
95,
97.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/222719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can one show that the hyperelliptic curve $y^2 = x^{p} + \frac{1}{4}$ has only one positive rational solution for every prime $p>3$? Without applying Fermat's Last Theorem, how can one show that the hyperelliptic curve $y^2 = x^{p} + \frac{1}{4}$ has only one positive rational solution $(x,y) = (0, \frac{1}{2})$ fo... | It is equivalent to FLT. Indeed, if $a^p+1=b^p$ for positive rational $a,b$, we have $x^p:=(ab)^p=a^p(a^p+1)=(a^p+1/2)^2-1/4:=y^2-1/4$.
Opposite implication (moving from the comments): if $x^p=y^2-1/4=(y-1/2)(y+1/2)$, $x\ne 0$, denote $y-1/2=a/m$ for coprime non-zero integers $a,m$. Then also $a+m\ne 0$, $y+1/2=(a+m)/m... | {
"language": "en",
"url": "https://mathoverflow.net/questions/225324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
On the divisibility of $(x+y)^k - 1$ by $xy$ For a fixed $k \geq 2$, are there infinitely many non-trivial coprime integer pairs $(x,y)$ for which $xy$ divides $(x+y)^k-1$? By trivial I mean parametrized pairs $(x,y)$ of the form $(-1,-1),(x,1),(1,y),(x,1-x),(x,1 - x^n),(1-y^n,y),(x,p(x)),(p(y),y)$, where $p(x)$ is a p... | For fixed $k$ and $y=\frac{x^k-1}{x-1}$ we have $$x^k \equiv 1 \pmod y$$and $$y \equiv 1 \pmod x$$ so also $$y^k\equiv 1 \pmod x.$$
That isn't one of your trivial examples, but maybe it should be.
Actually, $y$ could be any polynomial factor of $\frac{x^k-1}{x-1}.$ For example,
$\frac{x^{30}-1}{x-1}=\Phi_3\Phi_5\Phi_{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/252784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
A specific Diophantine equation restricted to prime values of variables. Consider the following Diophantine equation $$x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1).$$ Assume also that $x,a,b,c,a^2+a+1,b^2+b+1, c^2+c+1$ are all primes. We'll call such a quadruplet $(x,a,b,c)$ a triple threat.
I'd like to be able to show that th... | Let $j$ be $a$, $b$, or $c$ in $x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1)$. We have
$x^2+x+1 = 0\mod j^2+j+1$,
$x^2+x+1+(j-x-1)(j^2+j+1) = 0\mod j^2+j+1$,
$(x-j)(x-j^2) = 0\mod j^2+j+1$.
Since $j^2+j+1$ is prime, $x = j\mod j^2+j+1$ or $x = j^2\mod j^2+j+1$.
This gives 8 possible chinese remainders for $x\mod (a^2+a+1)(b^2+b... | {
"language": "en",
"url": "https://mathoverflow.net/questions/308196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Lattice points in a square pairwise-separated by integer distances Let $S_n$ be an $n \times n$ square of lattice points in $\mathbb{Z}^2$.
Q1. What is the largest subset $A(n)$ of lattice points in $S_n$ that have the
property that every pair of points in $A(n)$ are separated by
an integer Euclidean distance?
Is... | Here are @AnthonyQuas's four points $\{o,p_1,p_2,p_3\}$ in
$S(2) \subset \mathbb{Z}^{25}$:
$$
\begin{array}{cccccccccccccccc
ccccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0
& 0 & 0 & 0 & 0 & 0 & 0 & 0
& 0 & 0 & 0 & 0 & 0 & 0 & 0
& 0 & 0 & 0 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1
& 1 & 0 & 0 & 0 & 0 & 0 & 0
& 0 ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/312707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Cauchy's Integral with quadratic exponential term As I was studying the Cauchy's integral formula, I tried to do the integral:
\begin{equation}
I = \int\limits_{-\infty}^{\infty} \frac{1}{x - a} e^{(i A x^2 + i B x)} dx
\end{equation}
with $A>0, B>0$ and $a > 0$.
Consider an integral on a complex plan:
\begin{equation}... | Let me first remove the $Bx$ term by completing the square,
$$I=\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2+iBx}}{x - a}\,dx=e^{-iB^2/4A}\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a-B/2A}\,dx.$$
Mathematica evaluates the Cauchy principal value of the integral in terms of Meijer G-functions,
$$I=-\tfrac{1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/361867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this equation $a^2+3b^2c^2=7^c$ Let $a,b,c$ be poistive integers,and such $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$,fine the all $a,b,c$ such
$$a^2+3b^2c^2=7^c$$
I'm not sure that this question has been studied, but I've been trying for a long time$(a,b,c\le 100)$, and there's only one set of solutions：$(a,b,c)=(2... | We work in $\mathbb{Z}[\omega]$ where $\omega=\frac{1+i\sqrt{3}}2$. It is a factorial ring, and we factorize both sides as $(a+i\sqrt{3}bc)(a-i\sqrt{3}bc)=(2+i\sqrt{3})^n(2-i\sqrt{3})^n$. Since $2+i\sqrt{3},2-i\sqrt{3}$ are prime (and coprime), the guy $a+i\sqrt{3}bc$ can not be divisible by both (otherwise it is divis... | {
"language": "en",
"url": "https://mathoverflow.net/questions/376653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
roots of a fourth degree polynomial function (Vieta) Question
I am interested in the root of the polynomial function :
$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2
+(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)x+
abcd-abc-bcd-ad+bc=0$
Under the restriction that: $0<a,b,c,d<1$.
I tried Vietas formula but don’t kn... | There are two real solutions, the general expressions are very, very long. Even if all four coefficients are equal, $a=b=c=d$, you have an equation
$$a^4-2 a^3+(4 a-7) a^2 x+(4 a-2) x^3+6 (a-1) a x^2+x^4=0$$
with complicated solutions.
For example, if all coefficients are equal to 1/2 the two real solutions are
$$x=\... | {
"language": "en",
"url": "https://mathoverflow.net/questions/420632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the sum of general type $\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k- n + a \choose r }$?
QUESTION. How to calculate the sum of such general type?
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose r }. $$
Some particular examples
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} = {2n\choos... | Alternatively, one can get an explicit expression for a fixed $r$ via generating functions by noticing that the given sum equals
$$[y^{n+a}z^r]\ (y+1+z)^n (1+y(1+z))^n (1+z)^{-n}$$
and rewriting it as
$$[y^{n+a}z^r]\ \big((1+y)^2 + \frac{z^2}{1+z}y\big)^n = [y^{n+a}z^r]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}{i} (1+... | {
"language": "en",
"url": "https://mathoverflow.net/questions/427962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
\begin{aligned}
f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
&+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4... | (another comment, not an answer.)
Experimentally, with the following code
from sympy import *
var('x')
var('n', integer = True)
f = ( (2+(x+3) * sqrt(-x))/(2*(x+4)) *
sqrt(-x)**n + ((2-(x+3) * sqrt(-x)) / (2*(x+4))) * (-sqrt(-x))**n +
+ (x+2+sqrt(x*(x+4)))/(2*(x+4)) *
( (x+sqrt(x*(x+4)))/2 )**n +
... | {
"language": "en",
"url": "https://mathoverflow.net/questions/440962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Surprising behaviour of polynomial that generates the series 1,2,4,8,...2^(k-1) Consider the generating function $f(n)$ that produces the following values:
$$f(1) = 1$$
$$f(2) = 2$$
$$f(3) = 4$$
Obviously these values can be generated by $f(n)= 2^{n-1}$.
These values can equally well be generated by $f(n) = (n^2-n+2)/2... | Many (all?) integer series f(k), where k = 1,2,3,..K-1,K can be generated by a polynomial of order K-1.
Well, yes, all of them. Given distinct numbers $a_1, \ldots, a_n$, the polynomial
$$p_i(X)= \prod_{j\neq i}\frac{X-a_j}{a_i - a_j}$$
takes value $1$ at $a_i$ and $0$ at $a_j$ for all $j \neq i$. A linear combinat... | {
"language": "en",
"url": "https://mathoverflow.net/questions/42395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A curious sum for integers $\equiv 7\pmod 8$. For $n$ a natural integer congruent to $7$ modulo $8$, one has seemingly always
$$\sum_{k=1}^{(n-1)/2}\left(\frac{k}{n}\right)k=0$$
where $\left(\frac{k}{n}\right)$ denotes the Jacobi symbol.
First cases:
$n=7$: $1+2-3$
$n=15$: $1+2+4-7$
$n=23$: $1+2+3+4-5+6-7+8+9-10-11$
I ... | There are probably lots of ways to see this, but here's one: let $S_1$ be the sum above, and let $$S_2 = \sum_{k=(n+1)/2}^{n-1} k \left( \frac{k}{n} \right).$$ Then $$S_1 + S_2 = S = \sum_{k=1}^{n-1} k \left( \frac{k}{n} \right).$$ Now you can rewrite $S$ (since $x \to 2x$ is a bijection mod $n$) as $$S = \sum_{k=1}^{(... | {
"language": "en",
"url": "https://mathoverflow.net/questions/120521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Upper limit on the central binomial coefficient What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ?
I just tried to proceed a bit, like this:
$ n! > n^{\frac{n}{2}} $
for all $ n>2 $. Thus,
$ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=... | Since it gives tighter bounds, I will reproduce my answer from MathSE.
For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\le\frac{n+\frac13}{n+\frac43}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\bino... | {
"language": "en",
"url": "https://mathoverflow.net/questions/133732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 1
} |
Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$? Numerical evidence suggests the following.
For $c \in \mathbb{N}, c > 2$ define the sequence $a_n$ by
$a_0=0,a_1=1, \; a_n=c a_{n-1} - a_{n-2}$
For $ 5 < n < 500, \; 2 < c < 100$ there are no primes in $a_n$ though
semiprimes exist.
Is it true tha... | Put $u = (c + \sqrt{c^2-4})/2$. We have
$$a_{2n} = \frac{u^{2n}-u^{-2n}}{u-u^{-1}} = \left( \frac{ u^n-u^{-n}}{u-u^{-1}} \right) \left( \vphantom{\frac{ u^n-u^{-n}}{u-u^{-1}}} u^n + u^{-n} \right)$$
$$a_{2n+1} = \frac{u^{2n+1}-u^{-2n-1}}{u-u^{-1}} = \left( \frac{u^{n+1/2}-u^{-2n-1}}{u^{1/2}-u^{-1/2}} \right) \left( \fr... | {
"language": "en",
"url": "https://mathoverflow.net/questions/146156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
From a (not positive definite) Gram matrix to a (Kac-Moody) Cartan matrix Suppose I am given a symmetric matrix $G_{ij}$ with $G_{ii} = 2$. Can I always find an invertible integer matrix $S$ such that $(S^T G S)_{ii}=2$ and $(S^T G S)_{ij} \leq 0$ for $i \neq j$? Is there a practical algorithm to do so?
If you'd like a... | There is an algorithm, based on a 1907 article of Hurwitz I mention sometimes, based in turn on the tree of Markov numbers.
We begin with a ternary quadratic form $\langle 1,1,1,r,s,t \rangle.$ The (Lehman) discriminant of this is
$$ 4 + rst - r^2 - s^2 - t^2. $$ We would like to know whether we can find replacement... | {
"language": "en",
"url": "https://mathoverflow.net/questions/208562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Parametrizing the solutions to a diophantine equation of degree four Good evening,
Consider $x^4+y^4+z^4=2t^4$ where x,y,z,t integer.
Is it known how to find all parametrisation of this equation ?
If you have any parametrisation or reference of this equation, please post it
Thank you.
| Ramanujan gave two parametrizations: if $a+b+c=0$ then
$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$
and
$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$
(listed in Mathworld, equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth... | {
"language": "en",
"url": "https://mathoverflow.net/questions/221658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An identity involving a product of two binomial coefficients I'm trying to find a closed formula (in the parameters $q,N$) for the following sum:
$$ \sum_{k=q}^{N} {{k-1}\choose{q-1}} {{k}\choose {q}} $$
Can anybody give me a lead?
Lior
| This is a hypergeometric sum (the ratio of two consecutive terms is a rational function) and is therefore susceptible to Gosper-type algorithms, of which you can read more in the book A = B.
I believe Mathematica has got Gosper's algorithm implemented. It says:
In[38]:= s[n_, q_] := Sum[Binomial[k - 1, q - 1] * Binomia... | {
"language": "en",
"url": "https://mathoverflow.net/questions/221853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Generalizing a pattern for the Diophantine $m$-tuples problem? A set of $m$ non-zero rationals {$a_1, a_2, ... , a_m$} is called a rational Diophantine $m$-tuple if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties. The problem is to generalize the relations below ... | There is also an equation for extending quintuples to sextuples.
$(abcde+abcdf+abcef-abdef-acdef-bcdef+2abc-2def+a+b+c-d-e-f)^2 =
4(ab+1)(ac+1)(bc+1)(de+1)(df+1)(ef+1)$
This can be solved for $f$ with two rational roots when $\{a,b,c,d,e\}$ is a rational Diophantine quintuple (except in a few exceptional circumstances... | {
"language": "en",
"url": "https://mathoverflow.net/questions/233367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Correspondence between $SBT (n)$ and $W(B_n)$ Let $W(B_n)$ be a Weyl group of type $B_n$ and $SBT (n)$ the set of standard bitableaux of size $n$. Similar to Robinson-Schensted correspondences, I know that there exists a map $W(B_n) \to SBT (n) \times SBT (n)$ such that the image of $W(B_n)$ is the set of the same shap... | Devra Garfinkle developed such a correspondence to pairs of domino tableaux in a series of Compositio Mathematica papers in the early 1990s. (The first & third, but curiously not the second, are available through EUDML: https://eudml.org/doc/90031 and https://eudml.org/doc/90244.) More succinct summaries are given in... | {
"language": "en",
"url": "https://mathoverflow.net/questions/251751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Nested trigonometric integral I am trying to solve the following:
$$\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\sin(a\cos\phi)}{1+b\cos\phi}$$
with $-1 < b < 0$.
I started with $\cos\phi = \operatorname{Re}[z]$, but it led to nowhere as I had to find the residue at 0, which doesn't have a closed form.
| Mathematica finds
Integrate[Sin[a*Cos[t]]*b^n*Cos[t]^n,{t, 0, 2*Pi},Assumptions->n \[Element] Integers&&n>0]
$$-\frac{1}{2} \pi a \left((-1)^n-1\right) b^n \Gamma \left(\frac{n}{2}+1\right) \, _1\tilde{F}_2\left(\frac{n+2}{2};\frac{3}{2},\frac{n+3}{2};-\frac{a^2}{4}\right) $$
and
Integrate[Cos[t]^n/(1 + b*Cos[t]), ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/290324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A specific Diophantine equation restricted to prime values of variables. Consider the following Diophantine equation $$x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1).$$ Assume also that $x,a,b,c,a^2+a+1,b^2+b+1, c^2+c+1$ are all primes. We'll call such a quadruplet $(x,a,b,c)$ a triple threat.
I'd like to be able to show that th... | Pace Nielsen and Cody Hansen just put this preprint on the Arxiv which shows that no triple threats exist.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/308196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Maximum of a quantity for two normal orthogonal vectors in $\mathbb{R}^n$ Let's define for every pair of vectors $u,v\in\mathbb{R}^n$, a quantity as follows:
$$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$
I want to find:
$$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$
An easy estimate using th... | Numerical approximations suggest $M(n)=\sqrt{2n^2-1}$ when $n$ is odd, with the max reached for some $u=(a,b,a,b,\dots,a)$, $v=(c,d,c,d,\dots,c)$. Solving exactly for such $u,v$ is easy and only implies a quadratic equation in one variable, $x=b^2$ for instance. Assuming $b>d>0$ without loss of generality, one finds $b... | {
"language": "en",
"url": "https://mathoverflow.net/questions/313936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Polynomial inequality $n^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^na_i\right)^3$ Let $n\ge 3$ be an integer. I would like to know if the following property $(P_n)$ holds: for all real numbers $a_i$ such that $\sum\limits_{i=1}^na_i\geq0 $ and $\sum\limits_{1\leq i<j<k\leq n}a_ia_ja_k\geq0$, we have
$$n^2\sum_{i=1}^na_i^... | A sort of (partial) explanation for what happens:
Let $N \ge 3$ the degree and let $A=\sum{a_k}, B=\sum_{j<k}a_ja_k, C=\sum_{j\ne k\ne m \ne j}a_ja_ka_m $. We are given that $A \ge 0, C \ge 0$ and we need to prove that $N^2(A^3-3AB+3C) \ge A^3$. Now we can assume wlog $A =1$ since if $A=0$ the inequality is obvious and... | {
"language": "en",
"url": "https://mathoverflow.net/questions/357847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Lagrange’s interpolation formula: Theoreme and Example I would like to know where they come up with the formula of Lagrange interpolation (Lagrange’s interpolation formula),Lagrange_polynomial because I did some research, but I find a different definition of Lagrange interpolation. Beside would someone explain and elab... | You would do well to study the idea of Lagrange interpolation. There is a unique polynomial of the minimal reasonable degree fitting data. In your setting, $t^4=t$ so any of $x+y,x^4+y,x+y^4,x^4+y^4$ or $(x+y)^4=x^4+x^3y+xy^4+x^4$ agree at all points.
Here are a few observations about your example: Consider the $9$ fun... | {
"language": "en",
"url": "https://mathoverflow.net/questions/368237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Two equivalent matrices? By coincidence I noticed that the following two matrices yield the same eigenvalues
\begin{pmatrix} A & B \\ B^* & A \end{pmatrix} and \begin{pmatrix} 0& A+b1_{\mathbb C^{2 \times 2}} \\ A+b^* 1_{\mathbb C^{2 \times 2}} & 0 \end{pmatrix}
where $A = \begin{pmatrix} 0 & a \\ a^* & 0 \end{pmatrix}... | Define the unitary matrix
$$U=\left(
\begin{array}{cccc}
i e^{i \pi /4} & 0 & 0 & 0 \\
0 & 0 & 0 & -e^{-i \pi /4} \\
0 & 0 & i e^{i \pi /4} & 0 \\
0 & -e^{-i \pi /4} & 0 & 0 \\
\end{array}
\right),$$
then
$$U\begin{pmatrix} A & B \\ B^* & A \end{pmatrix}U^{-1}=\begin{pmatrix} 0 & A+b\mathbb{1} \\ A+b^\ast\mathbb{1}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/373425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify $f(y) = \displaystyle \sum^{\lfloor\frac{1}{y} \rfloor}_{x= 1} xy\left(\frac{1-y}{1-(x-1)y}\right)^{x-1}$, where $0 < y < 0.5$ Kindly simplify $f(y) = \displaystyle \sum^{\lfloor\frac{1}{y} \rfloor}_{x= 1} xy\left(\frac{1-y}{1-(x-1)y}\right)^{x-1}$, where $0 < y < 0.5$.
| $$f(y) = \displaystyle \sum^{\lfloor\frac{1}{y} \rfloor}_{x= 1} xy\left(\frac{1-y}{1-(x-1)y}\right)^{x-1},\;\;0<y<1/2.$$
The nontrivial regime is when $y\rightarrow 0$. Then the sum is dominated by the last term, so we can approximate
$$f(y)\approx y\lfloor\frac{1}{y} \rfloor \left(\frac{1-y}{1+y-y\lfloor\frac{1}{y} \... | {
"language": "en",
"url": "https://mathoverflow.net/questions/383290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Identity involving double sum with binomials (asked previously in MSE here)
In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that
$$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A... | Rewriting the l.h.s. of the conjectured identity and using the properties of beta function, we have:
\begin{split}
\text{l.h.s.} &= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{1}{(B+b)\binom{B+b-1}b (x+B-1) \binom{x+B-2}{B+b-1} }\\
&= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1} \int_0^1 \int_0^1 (1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/414885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Another plausible inequality. I come across the following problem in my study.
Consider in the real field. Let $ 0\le x\le1 $, $a_1^2+a_2^2=b_1^2+b_2^2=1$.Is it true
$ (a_1b_1+xa_2b_2)^2\le\left(\frac{(1-x)+(1+x)(a_1b_1+a_2b_2)}{(1+x)+(1-x)(a_1b_1+a_2b_2)}\right)^{2}(a_1^2+xa_{2}^{2})(b_1^2+xb_{2}^{2})$?
| I think your inequality is false, dear miwalin. Please check the case when $a_1=b_2=\frac{\sqrt{3}}{2}$ and $a_2=b_1=-\frac{1}{2}.$ But I think it is true when $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative numbers.
Let me prove it in the case $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative real numbers. Write the inequality as... | {
"language": "en",
"url": "https://mathoverflow.net/questions/26665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Resultant probability distribution when taking the cosine of gaussian distributed variable I am trying to do a measurement uncertainty calculation. I have a gaussian distributed phase angle (theta) with a mean of 0 and standard deviation of 16.6666 micro radians. The variance is the square of the standard. The formu... | For $y=\cos(x)$, the CDF of $y$
\begin{array}{l}
F_Y \left( y \right) = \left\{ {\begin{array}{*{20}l}
{0,y < - 1} \\
{P\left( {2k\pi + \arccos y \le x \le 2\left( {k + 1} \right)\pi - \arccos y} \right),k \in \Bbb Z, - 1 \le y \le 1} \\
{1,y > 1} \\
\end{array}} \right. \\
P\left( {2k\pi + \arccos y... | {
"language": "en",
"url": "https://mathoverflow.net/questions/35260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
Surprising behaviour of polynomial that generates the series 1,2,4,8,...2^(k-1) Consider the generating function $f(n)$ that produces the following values:
$$f(1) = 1$$
$$f(2) = 2$$
$$f(3) = 4$$
Obviously these values can be generated by $f(n)= 2^{n-1}$.
These values can equally well be generated by $f(n) = (n^2-n+2)/2... | The second observation is true of all polynomials which interpolate an integer sequence. This is the subject of the method of finite differences, the "main theorem" of which is this: if we define $\Delta f(n) = f(n+1) - f(n)$, then the unique polynomial of degree $n$ which interpolates the sequence $f(0), f(1), ... f(... | {
"language": "en",
"url": "https://mathoverflow.net/questions/42395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
On bounding the average cost of top-down merge sort Let $A_n$ be the average number of comparisons to sort $n$ keys by merging them in a top-down fashion (see any algorithm textbook). It can he shown that
$$
A_0 = A_1 = 0;\quad A_n = A_{\lfloor{n/2}\rfloor} + A_{\lceil{n/2}\rceil} + n - \frac{\lfloor{n/2}\rfloor}{\lcei... | Inductively, if $A_p \le p \log_2 p + \alpha p + \beta$, then
$A_{2p} \le ((2p) \log_2 (2p) + \alpha (2p) + \beta) + (\beta -2 + \frac {2}{p+1})$
so we'll want $\beta-2 + \frac{2}{p+1} \le 0.$
The odd case is a little harder. Assume the inequality is true for $A_p,A_{p+1}.$
$A_{2p+1} = A_p + A_{p+1} + 2p - 1 + \frac... | {
"language": "en",
"url": "https://mathoverflow.net/questions/105328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is every positive integer a sum of at most 4 distinct quarter-squares? There appears to be no mention in OEIS: Quarter-squares, A002620. Can someone give a proof or reference?
Examples:
quarter-squares: ${0,1,2,4,6,9,12,16,20,25,30,36,...}$
2-term sums: ${2+1, 4+1, 6+2, 6+2,...,90+9,...}$
3-term sums: ${12+2+1, 16+2+... | The quarter-squares are defined by $Q(a) = \lfloor a/2\rfloor\lceil a/2\rceil$. So $Q(2a) = a^2$.
First, write every number as a sum of four squares. We can assume they're not all the same (by induction: if this is a problem, the sum is a multiple of 4; we can produce a better representation by representing n/4 as a su... | {
"language": "en",
"url": "https://mathoverflow.net/questions/202903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Identities involving sums of Catalan numbers The $n$-th Catalan number is defined as $C_n:=\frac{1}{n+1}\binom{2n}{n}=\frac{1}{n}\binom{2n}{n+1}$.
I have found the following two identities involving Catalan numbers, and my question is if anybody knows them, or if they are special cases of more general results (referenc... | The case $k=n+1$ can be verified with generating functions:
$$ C(z) = \sum_{i=0}^\infty c_i z^i . $$
For the Catalan numbers such function is given by
$$ C(z) = \frac{1-\sqrt{1-4z}}{2z} . $$
Given any two generating functions $A$ and $B$ their product is the generating function for the convolution of the
two sequences ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/210709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Generalizing a pattern for the Diophantine $m$-tuples problem? A set of $m$ non-zero rationals {$a_1, a_2, ... , a_m$} is called a rational Diophantine $m$-tuple if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties. The problem is to generalize the relations below ... | (Too long for a comment, but may help in a generalization.)
After some sleuthing around, it turns out $(1),(2),(3)$ can be encapsulated in the single equation,
$$(a b c d e + 2a b c + a + b + c - d - e)^2 = 4(a b + 1)(a c + 1)(b c + 1)(d e + 1)\tag1$$
which I think is by Dujella. For example,
*
*Let $a,b,c,d =1,3,0,... | {
"language": "en",
"url": "https://mathoverflow.net/questions/233367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
On random divisor sums modulo $2^k$ Let $k,n,\ell$ be positive integers with $k,n\ge 2$ and $0\le \ell \le k-1$. For each integer $2\le j \le n$, choose a divisor $d_j$ of $j$, uniformly at random from the divisors of $j$. We denote by $P(n,k,\ell)$ the probability that $$d_2 + d_3 + \cdots + d_n \equiv \ell \pmod{k}.$... | This is a consequence of the Catalan conjecture.
$'x^8+1'$ is a 'cube plus 1', hence 'misses a beat' when it factors $x+x^3+x^9+x^{27}$.
Because $27-3$ has $8$ as a factor, the factorization is:
$$x+x^3+x^9+x^{27}=(1+x^8)(x+x^3-x^{11}+x^{19})$$ as $11$ and $19$ are both $3\mod8$.
And this only works because $9=3^2=1+8=... | {
"language": "en",
"url": "https://mathoverflow.net/questions/234468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Sum of multinomals = sum of binomials: why? I stumbled on the following identity, which has been checked numerically.
Question. Is this true? If so, any proof?
$$\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j}
=\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-k-2j-1}{k-2j}.$$
Here, $\binom{m}{a,b,c}$... | For convenience set $m=n-2k$. Then
\begin{equation}
\begin{split}
\binom{n-2k+j}{j,k-2j,n-3k+2j} &= \binom{m+j}{j,k-2j,m-k+2j} \\
&= \binom{m+j}{m} \binom{m}{k-2j} \\
&= [t^j](1-t)^{-(m+1)} \cdot [t^{k-2j}](1+t)^m \\
&= [t^{2j}](1-t^2)^{-(m+1)} \cdot [t^{k-2j}](1+t)^m
\end{split}
\end{equation}
where $[t^... | {
"language": "en",
"url": "https://mathoverflow.net/questions/271286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Prove this conjecture inequality $x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k}\le \frac{1}{(k+2)^{k-1}}$ let $x\in (0,1)$, and $k$ be postive intgers,such $k\ge 2$,
I conjecture following inequality maybe hold?
$$x\cdot \dfrac{(1-x)^{k-1}}{(k+1)^{k-2}}+\dfrac{(1-2x)^k}{k^k}\le \dfrac{1}{(k+2)^{k-1}}$$ ... | Consider only $k>2$. Denote $$f(x)=x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k},$$
we need to prove that $f(x)\leqslant f(\frac1{k+2})$ for $x\in [0,1]$. We have $f'(\frac1{k+2})=0$ and $$f'(0)=(k+1)^{2-k}-2k^{1-k}=k^{2-k}\left(\left(1+\frac1k\right)^{2-k}-\frac2k\right)>0$$
by Bernoulli inequality $(1+x... | {
"language": "en",
"url": "https://mathoverflow.net/questions/284345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Expectation inequality for sampling without replacement Is the following proposition correct?
$X_1, X_2, X_3$ are uniformly at random sampled from a finite set $\mathcal X$ without replacement.
$f : \mathcal X^2 \rightarrow \mathbb R_{\ge0}$ is symmetric:
$
f(x, y) = f(y, x)
$, then:
$$
\mathbb E_{X_1, X_2, X_3} f(X_1... | Let $\lambda_1,\dots,\lambda_n$ be eigenvalues of the symmetric matrix $(f_{ij})$, where $f_{ii}=0$ by definition. They are real, $\sum \lambda_i=0$ and the inequality rewrites as
$$
\left(\frac{\sum \lambda_i^3}{n(n-1)(n-2)}\right)^2\leqslant
\left(\frac{\sum \lambda_i^2}{n(n-1)}\right)^3,
$$
or $(\sum \lambda_i^3)... | {
"language": "en",
"url": "https://mathoverflow.net/questions/306273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Can a minimal generating set for an ideal always be made into a Groebner basis? Let $I\subseteq k[x_0,\ldots,x_n]$ be an ideal, generated by some polynomials $F_1,\ldots,F_r$, all homogeneous and of the same degree. Suppose $r$ is the smallest number of generators that will suffice to generate the ideal.
Can one choos... | No. Let $k$ not have characteristic $2$, let $I = \langle x^2, xy, y^2 \rangle$ and consider the generating set $x^2$, $(x+y)^2$, $y^2$. After a linear change of coordinates, these are $(a_1 x + b_1 y)^2$, $(a_2 x + b_2 y)^2$ and $(a_3 x + b_3 y)^2$ for some $a_j$, $b_j$. But the leading term of $(ax+by)^2$ will always... | {
"language": "en",
"url": "https://mathoverflow.net/questions/328134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Inequality in a triangle associated with Golden ratio
Let $ABC$ be arbitrary triangle, $D$, $E$, $F$ are the midpoints of $BC$, $CA$, $AB$ respectively. Define points, segments in the figure below. I am looking for a proof that:
$$DE+EF+FD \le (DG+DH+EI+EJ+FJ+FQ).\frac{\varphi}{2}$$
Wher $\varphi=\frac{\sqrt{5}+1}{2... | Let $DH=x$, $DG=y$, $FK=z$, $FQ=t$, $EG=u$ and $EI=v$.
Thus, in the standard notation we obtain:
$$u\left(y+\frac{c}{2}\right)=\frac{b^2}{4},$$$$y\left(u+\frac{c}{2}\right)=\frac{a^2}{4},$$
Which gives $$u-y=\frac{b^2-a^2}{2c},$$
$$y\left(y+\frac{b^2-a^2}{2c}\right)+\frac{c}{2}y=\frac{a^2}{4},$$
$$y=-\frac{b^2+c^2-a^2}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/392857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Latin squares with one cycle type? Cross posting from MSE, where this question received no answers.
The following Latin square
$$\begin{bmatrix}
1&2&3&4&5&6&7&8\\
2&1&4&5&6&7&8&3\\
3&4&1&6&2&8&5&7\\
4&3&2&8&7&1&6&5\\
5&6&7&1&8&4&3&2\\
6&5&8&7&3&2&4&1\\
7&8&5&2&4&3&1&6\\
8&7&6&3&1&5&2&4
\end{bmatrix}$$
has the property ... | There are also "pan-Hamiltonian" Latin squares, see Perfect Factorisations of Bipartite Graphs and Latin Squares Without Proper Subrectangles by I. M. Wanless, Electronic J. Combin. 6 (1999), R9.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/415514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Existence of solution for a system of quadratic diophantine equations / symmetric quadratic froms I am interested in solving, or even just deciding the existence of a solution, for a system of quadratic diophantine equations.
Let $p$ be a prime congruent to 1 modulo 8, so $ p =17$ is the first case. We want to solve th... | UPDATE. Using factorization $-2(x+1)^2x^{p-3}$ over the corresponding number field, I established that there are no solutions for $p=17$. Furthermore, I computationally verified that for primes $p<30$ we have solutions for all $p\equiv 3\pmod{4}$ and do not have any for $p\equiv 1\pmod{4}$.
This is just an extended co... | {
"language": "en",
"url": "https://mathoverflow.net/questions/421510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A question about generalized harmonic numbers modulo $p$ Let $p \equiv 1 \pmod{3}$ be a prime and denote $H_{n,m} = \sum_{k = 1}^n 1/k^m$ as the $n,m$-th generalized harmonic number. I'm interested in computing $H_{(p-1)/3,\, 2}$ and $H_{(p-1)/6,\,2}$ modulo $p$. From this paper I know
\begin{align*}
H_{(p-1)/6,2} \e... | Here is what I've learned:
\begin{align*}
2\sum_{n = 0}^{\infty} B_{2n+1}\left(\frac{1}{3}\right) \frac{x^{2n+1}}{(2n+1)!} & = \frac{xe^{\frac{1}{3}x}}{e^x -1} + \frac{xe^{-\frac{1}{3}x}}{e^{-x} -1} = \frac{xe^{\frac{1}{3}x}}{e^x -1} - \frac{xe^{\frac{2}{3}x}}{e^{x} -1}
\\
&= \frac{x(e^{\frac{1}{3}x}-e^{\frac{2}{3}x}... | {
"language": "en",
"url": "https://mathoverflow.net/questions/424966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An elementary inequality of operators Suppose $a,b$ are two positive-definite linear operators on (say) $\mathbb R^n$. For $p\in(0,1)$, do we then have $(a+b)^p\leq a^p+b^p$ (with respect to the Loewner order)?
| No. E.g., (identifying linear operators with matrices in a standard manner) let
$$a=\left(
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right),\quad
b=\frac12
\left(
\begin{array}{cc}
1 & 1 \\
1 & 1 \\
\end{array}
\right).
$$
Then (by straightforward but somewhat tedious calculations) for all $p\in(0,1)$
$$d(p... | {
"language": "en",
"url": "https://mathoverflow.net/questions/433656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If $\left(1^a+2^a+\cdots+n^{a}\right)^b=1^c+2^c+\cdots+n^c$ for some $n$, then $(a,b,c)=(1,2,3)$? Question : Is the following conjecture true?
Conjecture : Let $a,b(\ge 2),c,n(\ge 2)$ be natural numbers. If $$\left(\sum_{k=1}^nk^a\right)^b=\sum_{k=1}^nk^c\ \ \ \ \ \cdots(\star)$$
for some $n$, then $(a,b,c)=(1,2,3).$
R... | This is a partial solution. I've just been able to get the following theorem:
Theorem : If $(\star)$ for some $n=8k-5,8k-4\ (k\in\mathbb N)$, then $(a,b,c)=(1,2,3)$."
I wrote the proof for this theorem on MSE.
PS: This idea (using mod $8$) does not seem to work for the other $n$. Another idea would be needed.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/144208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
Combinatorial identities I have computational evidence that
$$\sum_{k=0}^n \binom{4n+1}{k} \cdot \binom{3n-k}{2n}= 2^{2n+1}\cdot \binom{2n-1}{n}$$
but I cannot prove it. I tried by induction, but it seems hard. Does anyone have an idea how to prove it?
What about
$$\sum_{k=0}^n \binom{4n+M}{k} \cdot \binom{3n-k}{2n}$$... | According to sage your first sum is:
$$ \frac{8^{n} 2^{n} \left(n - \frac{1}{2}\right)!}{\sqrt{\pi} n!} $$
According to Maple your second sum is:
$${3\,n\choose 2\,n}{2F1(-n,-4\,n-M;\,-3\,n;\,-1)}-{4\,n+M\choose n+1}{
2\,n-1\choose 2\,n}{3F2(1,1,-3\,n-M+1;\,n+2,-2\,n+1;\,-1)}$$
For $M=2$:
$$\frac{2 \, {\left(2 \, n \Ga... | {
"language": "en",
"url": "https://mathoverflow.net/questions/150093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Records in $Z$-numbers and a relaxation A Z-number
is a (non-zero) real number $x$ such that the fractional parts
$$\left\lbrace x \left(\frac 3 2\right)^ n \right\rbrace $$
are less than $\frac12$ for all natural numbers $n$.
It is not known whether $Z$-numbers exist.
First, I am interested in finite records, i.e. lar... | Suppose the fractional parts of $x, \frac{3}{2} x, (\frac{3}{2})^2 x, ... (\frac{3}{2})^kx$ are under $\frac{1}{2}$, but the fractional part of $(\frac{3}{2})^{k+1} x$ is between $\frac{1}{2}$ and $1$. Then consider $y = x + 2^k$. For $n \le k, \lbrace (\frac{3}{2})^n y\rbrace = \lbrace (\frac{3}{2})^n x \rbrace$ whil... | {
"language": "en",
"url": "https://mathoverflow.net/questions/153815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Dihedral extension of 2-adic number field Sorry if the question is too long and maybe elementary.
I am reading a paper by Hirotada Naito on "Dihedral extensions of degree 8 over the rational p-adic fields". To generate dihedral extension $K(\sqrt{\epsilon},\sqrt{\epsilon^\sigma})$ in part 2-1, he said let $K=\mathbb{Q}... | Answer to your first question: The numbers contain a set of representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$. (Naito said earlier that "we examine a representative system of $K_i^*/(K_i^*)^2$", and you copied his equation that shows representatives of this quotient group can be obtained from representa... | {
"language": "en",
"url": "https://mathoverflow.net/questions/199192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is every positive integer a sum of at most 4 distinct quarter-squares? There appears to be no mention in OEIS: Quarter-squares, A002620. Can someone give a proof or reference?
Examples:
quarter-squares: ${0,1,2,4,6,9,12,16,20,25,30,36,...}$
2-term sums: ${2+1, 4+1, 6+2, 6+2,...,90+9,...}$
3-term sums: ${12+2+1, 16+2+... | The quarter-squares are the numbers of form $k^2$ and $k(k+1)$.
Start by expressing $N$ as the sum of four squares.
If you used some square four times, i.e. $N=x^2 + x^2 + x^2 + x^2$, then $N=(2x)^2$ is a quarter-square.
If you used some square three times but not four times, i.e. $N=x^2 + x^2 + x^2 + y^2$, then $N=(x-... | {
"language": "en",
"url": "https://mathoverflow.net/questions/202903",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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A property of 47 with respect to partitions into five parts Is 47 the largest number which has a unique partition into five parts (15, 10, 10, 6, 6), no two of which are relatively prime?
| Here is a quick demonstration that it is effectively solvable for any number of parts.
A sufficiently large number that has exactly one partition with the property must be prime. Otherwise we can write it as $n = a \cdot b$ with $1 < a \leq b$ and partition $b$ into $k$ parts in two different ways, then multiply them ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/289084",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
The product of $\frac{b^i-a}{b^i-1}$ lies in a special ring (conjecture) Consider any three positive integers $a, b, n$. Is it true that $$\frac{(b-a)(b^2-a)\cdot\dotsb\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\dotsb\cdot(b^n-1)}\in\mathbb{Z}\left[\frac{a-1}{b-1},\frac{a^2-1}{b^2-1},\dotsc ,\frac{a^n-1}{b^n-1}, 1/2 \right]?$$
Mor... | This may be understood $p$-adically as is done in my answer to your question on math.SE. It would be nice to see the algebraic formula proving the same statement (that is, the explicit polynomial not depending on $a$ and $b$ with integer coefficients for which $$
P\left(b,\frac{a-1}{b-1},\dots,\frac{a^n-1}{b^n-1}\right... | {
"language": "en",
"url": "https://mathoverflow.net/questions/324115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Frequency of digits in powers of $2, 3, 5$ and $7$ For a fixed integer $N\in\mathbb{N}$ consider the multi-set $A_2(N)$ of decimal digits of $2^n$, for $n=1,2,\dots,N$. For example,
$$A_2(8)=\{2,4,8,1,6,3,2,6,4,1,2,8,2,5,6\}.$$
Similarly, define the multi-sets $A_3(N), A_5(N)$ and $A_7(N)$.
I can't be sure if I have se... | It is not an answer but some numercal data for $A_2(N)$. Here digits are ordered according to their frecuences ($2$, $4$ and $6$ look like most frequent):
$$
\begin{array}{rl}
N=1000: & 2, 1, 4, 6, 8, 9, 5, 3, 0, 7\\
N=2000: & 2, 1, 6, 4, 8, 5, 3, 9, 7, 0 \\
N=3000: & 6, 2, 1, 4, 3, 8, 0, 5, 7, 9 \\
N=5000: & 2, 0, 4, ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/325136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to do a multinomial theorem sum faster For example we have this question :
Find the coefficient of $x^6$ in the following
$\frac{\left(x^{2}+x+2\right)^{9}}{20}$
So using multinomial Theorem which is this :
$\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}=\sum_{b_{1}+b_{2}+\cdots+b_{k}=n}\left(\begin{array}{c} n \\ b_{1... | Extracting a сomplete square can save some time. For example, from $x^2+x+2=\frac{(2x+1)^2+7}{4}$, we get
$$[x^6]\, (x^2+x+2)^9 = \frac{1}{4^9} \sum_{k=0}^9 \binom{9}{k} \binom{2k}{6} 2^6 7^{9-k}.$$
Alternatively, we can employ the factorization $x^2+x+2 = (x-\alpha_1)(x-\alpha_2)$, where $\alpha_{1,2}=\frac{-1\pm I\s... | {
"language": "en",
"url": "https://mathoverflow.net/questions/362394",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can each natural number be represented by $2w^2+x^2+y^2+z^2+xyz$ with $x,y,z\in\mathbb N$? It is well known that each $n\in\mathbb N=\{0,1,2,\ldots\}$ can be written as $2w^2+x^2+y^2+z^2$ with $w,x,y,z\in\mathbb N$. Furthermore,
$$\{2w^2+x^2+y^2:\ w,x,y\in\mathbb N\}=\mathbb N\setminus\{4^k(16m+14):\ k,m\in\mathbb N\}.... | The answer to question 1 is yes - the other questions seem to me to be more difficult.
If $n$ is odd, then there are non-negative integers $w$, $x$ and $y$ so that $n = 2w^{2} + x^{2} + y^{2}$. One way to see this is that the class number of this quadratic form $Q_{1} = 2w^{2} + x^{2} + y^{2}$ is $1$, and so every loca... | {
"language": "en",
"url": "https://mathoverflow.net/questions/416344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? Let $p_n$ be the $n$th degree polynomial that sends $\frac{k(k-1)}{2}$ to $\frac{k(k+1)}{2}$ for $k=1,2,...,n+1$. E.g., $p_2(x) = (6+13x -x^2)/6$ is the unique quadratic polynomial $p(x)$ satisfying $p(0) = 1$, $p(1) = 3$, and $p(3) =... | According to Mathematica your polynomials satisfy the recurrence relation
$$
(2 n+1) p(n) \left(n^2+3 n-2 x+2\right)+p(n+1) \left(-4 n^3-18 n^2+4 n x-27 n+2 x-14\right)+(2 n+3) (n+2)^2 p(n+2)=0
$$
with initial conditions $p_1=1+2x,p_2=1-x(x-13)/6$. The solution to this equation is
$$
p_n(x)=\frac{1}{4} \left(\frac{\Gam... | {
"language": "en",
"url": "https://mathoverflow.net/questions/441935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 2,
"answer_id": 0
} |
Up to $10^6$: $\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ (Number of partitions with no even part repeated ) Up to $10^6$:
$\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$
A001935 Number of partitions with no even part repeated
Is this true in general?
It would mean relation between restricted partitions of $n$ and ... | Let's call A001936(n) by $a(n)$. Here is a sketch of why $$a(n)\equiv \sigma(4n+1)\pmod{4}$$
Firs note that the generating function of $a(n)$ is
$$A(x)=\sum_{n\geq 0}a(k)x^n=\prod_{k\geq 1}\left(\frac{1-x^{4k}}{1-x^k}\right)^2$$ for $\sigma(2n+1)$ the generating function is $$B(x)=\sum_{k\geq 0}\sigma(2k+1)x^k=\prod_{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/59178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
the following inequality is true，but I can't prove it The inequality is
\begin{equation*}
\sum_{k=1}^{2d}\left(1-\frac{1}{2d+2-k}\right)\frac{d^k}{k!}>e^d\left(1-\frac{1}{d}\right)
\end{equation*}
for all integer $d\geq 1$. I use computer to verify it for $d\leq 50$, and find it is true, but I can't prove it. Thanks ... | [Edited mostly to fix a typo noted by David Speyer]
The following analysis simplifies and completes the
"routine but somewhat unpleasant" task of
recovering the actual inequality from the asymptotic analysis.
The idea is that once we've obtained the asymptotic expansion
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \rig... | {
"language": "en",
"url": "https://mathoverflow.net/questions/133028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 3,
"answer_id": 0
} |
Roots of the derivative as symmetric functions of the roots of the polynomial Let $p(t)=(t^2-a_1^2)\ldots(t^2-a_n^2)$ be an even polynomial with distinct real non-zero roots. Can the roots of its derivative $p'(t)$ be expressed nicely (e.g. as rational symmetric functions) in terms of the roots of $p(t)$? How?
| Rational symmetric functions of the roots of the derivative can be expressed as rational symmetric functions of the roots of the polynomial, because they are rational functions of the coefficients of the derivative, thus rational functions of the coefficients.
But the individual roots cannot be:
$\frac{d}{dt}(t^2-a_1^2... | {
"language": "en",
"url": "https://mathoverflow.net/questions/144949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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zeta-function regularized integrals I gather that the following two identities about $\xi(3)$ hold via some notion of zeta-function regularized integrals.
$\xi(3) = \frac{(2\pi)^3}{3}\int _0 ^\infty d\lambda \frac{\sqrt{\lambda} }{1 + e^{2 \pi \sqrt{\lambda} } } = - \frac{4 \pi^3 }{3} \int _0^\infty d\lambda \sqrt{\la... | We have
$$\tanh(x) = \dfrac{1 - e^{-2x}}{1 + e^{-2x}} = (1-e^{-2x}) \sum_{k=0}^{\infty}(-1)^k e^{-2kx} = 1 + 2 \sum_{k=1}^{\infty}(-1)^ke^{-2kx}$$
Now we have
$$\sqrt{x} \tanh(\sqrt{x}) = \sqrt{x} + 2 \sum_{k=1}^{\infty}(-1)^k \sqrt{x}e^{-2k\sqrt{x}}$$
Now throwing away the divergent part, i.e., $\sqrt{x}$, as every go... | {
"language": "en",
"url": "https://mathoverflow.net/questions/150867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Representation of rationals by quadratic form In one paper about number theory author stated 2 lemmas
Lemma 1. If $p$ is a prime $\equiv3(mod $ $4)$ then $x^2+y^2-pz^2$ represents a non-zero rational number $m$ if and only if $m$ is not of the form $kps^2$ with $\left(\frac{k}{p}\right)=1$ or $ks^2$ with $k\equiv p(mod... | Lemma B (for binary) (completing the square and a few cases to check): Given integers, $f(x,y) = a x^2 + b x y + c y^2 ,$ with discriminant $\Delta = b^2 - 4 a c $ not a square. Given a (always positive) prime $r$ with Legendre $(\Delta|r) = -1,$ so that $\Delta \neq 0 \pmod r$ in particular. IF $f(x,y) \equiv 0 \pmod ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/154368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Genus of a simple graph Let $G$ is a finite simple undirected graph. Suppose there exist subgraph $G_1,G_2,\dots,G_n$ of $G$, such that $G_i \cong K_5$ or $K_{3,3}$, $E(G_i)\cap E(G_j) = \emptyset$ and $|V(G_i)\cap V(G_j)| \leq 3$, for $i\neq j$. Then, is it true that genus of $G$ is greater than or equal to $n$?
Than... | Second try. I believe $K_{6,3}$ is a counterexample.
It is genus $1$ and contains $2$ edge disjoint $K_{3,3}$s sharing only
$3$ vertices.
Explicitly:
K_{6,3}=[(0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (3, 8), (4, 6), (4, 7), (4, 8), (5, 6), (5, 7), (5, 8)]
first K_{3,3}=[(3... | {
"language": "en",
"url": "https://mathoverflow.net/questions/157181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Congruence equation for Apery numbers Does the system of congruence equations
\begin{eqnarray}
A_{17k}&\equiv& 0 \pmod {17^2}, \nonumber \\
A_{17k+1}&\equiv& 0 \pmod {17^2}, \tag{1}
\end{eqnarray}
has solutions other than $k=3$? Here $A_n$ are Apery numbers:
$$A_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2.$$
... | Results of Gessel can be used to solve this question. Gessel showed, in Theorem 1 here, that for any prime $p$, if $n=\sum d_i p^i$ is the base-$p$ expansion of $n$, then
$$(*) A_n \equiv \prod A_{d_i} \bmod p.$$
In Theorem 4 here he essentially refined this as follows. He proved that there is an (explicit) sequence $\... | {
"language": "en",
"url": "https://mathoverflow.net/questions/168608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integer points on $y^2=x^2-x^3+x^4$ Does the Diophantine equation $y^2=x^2-x^3+x^4$ have solutions other than
$x=1,y=1$? Interestingly, the Diophantine equation $y^2=x^2-x^3+x^5$ has such solutions: $x=3,y=15$, $x=5,y=55$, $x=56,y=23464$. However I don't know if there are any other solutions. The solution $x=3,y=15$ is... | The only integer solutions $(x,y)$ are $(0,0)$ and $(1,\pm 1)$.
Your equation is $y^2=x^2(1-x+x^2)$. This can only have integer solutions if $p(x)=1-x+x^2$ is a square or $y=0$. For $x>1$ we have $(x-1)^2<p(x)<x^2$, so $p(x)$ is not a square. For $x<0$ we have $x^2<p(x)<(x-1)^2$, so $p(x)$ is not a square. The only pos... | {
"language": "en",
"url": "https://mathoverflow.net/questions/176865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.