Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding a generating function of the sequence $(2,2,2,3,3,2,2,2,3,3...)$ I need to find generating function of sequence: $(2,2,2,3,3,2,2,2,3,3,2,2,2,3,3...)$
My attempt so far:
I tried splitting the sequence into 2 new sequences: $(2,2,2,0,0,2,2,2,0,0,...)$ and $(0,0,0,3,3,0,0,0,3,3...).$
Then I found that the generati... | It's periodic with period $5$, so start with
$$\frac{1}{1-x^5}$$
and the numerator is $2,2,2,3,3$, so
$$\frac{2+2x+2x^2+3x^3+3x^4}{1-x^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Do 1.$\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$ 2. $\sum_{n=1}^\infty(e^{\cos n/n}−\cos\frac{1}n)$ converge? Does
$$ 1. \hspace{8mm} \sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n} $$
and
$$ 2. \hspace{8mm} \sum_{n=1}^{\infty} \left(e^\frac{\cos{n}}{n} - \cos\frac{1}{n}\right) $$
converge? ... | For the first one you have
$$\frac{\cos(n)}{n^{2/3}+(-1)^n} = \frac{\cos(n)}{n^{4/3}-1}\,\left(n^{2/3}-(-1)^n\right) = \frac{\cos(n)}{n^{2/3}-n^{-2/3}} - \frac{(-1)^n}{n^{4/3}-1}$$
and so the first term converges by the Dirichlet-test, since $\left(n^{2/3}-n^{-2/3}\right)^{-1}$ is decreasing. The second term converges ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the reccurence relation for $T(n) = 2T(n/2) + n − 1, n > 1, T(1) = 0, n = 2^k$ Even though it's pretty similar to other questions, I'm confusing myself with the answer because I always end up with:
$$
T(2^k)=\log_2n-\log_2n
$$
which doesn't seem right at all. Plus, I'm not sure whether the final big theta notat... | You need to be more careful about your substitutions and expanding. Observe that:
\begin{align*}
A(2^k)
&= 2A(2^{k-1}) + 2^k - 1 = 2 [ 2A(2^{k-2}) + 2^{k-1} - 1] + 2^k - 1 \\
&= 2^2A(2^{k-2}) + 2(2^k) - (1 + 2) = 2^2[ 2A(2^{k-3}) + 2^{k-2} - 1] + 2(2^k) - (1 + 2) \\
&= 2^3A(2^{k-3}) + 3(2^k) - (1 + 2 + 2^2) = 2^3[ 2A(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $d \mid a^2b+1$ and $d \mid b^2a+1 \implies d \mid a^3+1$ and $d \mid b^3+1$ Prove for all $a, b, d$ integers greater than zero, that when $d$ is a divisor of both $a^2b+1$ and $b^2a+1$ then $d$ is also a divisor of both $a^3+1$ and $b^3+1$
There is a trivial case when $a = b$, but I found that there are more so... | Since $d \mid a^2\ b+1$ and $d \mid a\ b^2+1$ then $d \mid (1-b\ a^2)(a^2\ b+1)+a^3(a\ b^2+1) = a^3+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are there any ways to convert inverse trigonometric values to radicals? When we solve a cubic equation $ax^3+bx^2+cx+d=0$, the roots are supposed to be in the form of radicals in real numbers or complex realm. However, if the discriminant is less than 0, the solution is ended up with roots represented by inverse trigo... | For a cubic equation when the discriminant is less than zero, the roots may be expressed in the form of trigonometric function of an angle in inverse trigonometric form if solved by Cardano method. For example, $x^3−2x+1=0$
\begin{cases} x_1=2\sqrt{\dfrac{2}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{4}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Min-Max points with lagrange multipliers "Find the min and max points of the function
$$f(x) = x^ 2 +yz-5$$
on the sphere
$$x^2+y^2+z^2=1$$
using Lagrange multipliers".
I'm having problem with equations while finding the points:
$$\nabla f(x) = \nabla g(x) \lambda$$
$$ 2x = 2x \lambda $$
$$z = 2y \lambda $$
$$y = 2z \l... | Then, if $\lambda=0$, then $(x,y,z)=(0,0,0)$ but since $x^2+y^2+z^2-1=0$ we have a contradiction. Thus, consider $\lambda\not=0$ and $x\not=0$, then since $2x=2x\lambda$ then $\lambda=1$ and then $(x,y,z,\lambda)=(\pm 1,0,0,0)$ because $z=2y=2(2z)$ give $z=0$ but also $y=2z=2(2y)$ give $y=0$. However, for $\lambda \no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1}$?
How do I show that $\displaystyle\ln{x}=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)}\left(\frac{x-1}{x+1}\right)^{2n+1},\ x>0$?
I have tried using Mclaurin series of $\ln(1+x)$ and replacing $x$ with $\dfrac{x-1}{x+1}$ in it bu... | Simplify the expression inside the sum using an integral of a much more elementary function in the following way:
\begin{align*}
2\sum_{n=0}^{\infty}\frac 1{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}&=
2\sum_{n=0}^{\infty}\int_0^{\frac{x-1}{x+1}}y^{2n}\,\mathrm dy=
2\int_0^{\frac{x-1}{x+1}}\sum_{n=0}^{\infty}y^{2n}\,\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Turn a real number (with a complex closed form) into its trigonometric form This post outlines 'fake' complex numbers (real numbers with complex closed form that usually come from the roots of unfactorable cubics (the example I need right now), or they can come from things like $i^i = e^{-\frac{\pi}{2}}$), in his own a... | Notice that there is a clearly written complex number inside each of the cube roots, so we can express those in polar form via Euler's formula:
$$\begin{eqnarray} 1 + i\sqrt{7} & = & re^{i \theta} \\
& = & r \cos \theta + ir \sin \theta \\
r \cos \theta & = & 1 \\
r \sin \theta & = & \sqrt{7} \\
r^2(\cos^2 \theta + \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Convergence of sum with binomial coefficient I found this exercise in a book and I don't know how to start:
With the help of probabilistic methods, prove that
$$
\lim_{n\rightarrow\infty} \frac{1}{4^n} \sum_{k=0}^n \binom{2n}{k} = 1/2
$$
One cannot use the binomial theorem because the sum just goes to $n$ and not to... | We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{2n}{k}}&=\frac{1}{2}\sum_{k=0}^{n-1}\binom{2n}{k}+\binom{2n}{n}+\frac{1}{2}\sum_{k=n+1}^{2n}\binom{2n}{k}\tag{1}\\
&=\frac{1}{2}\sum_{k=0}^{2n}\binom{2n}{k}+\frac{1}{2}\binom{2n}{n}\\
&\,\,\color{blue}{=\frac{1}{2}2^{2n}+\frac{1}{2}\binom{2n}{n}}\tag{2}
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve $\dfrac{4}{x+3}-\dfrac{2}{x+1}=\dfrac{5}{2x+6}-\dfrac{2\frac{1}{2}}{2x+2}$ Solve $\dfrac{4}{x+3}-\dfrac{2}{x+1}=\dfrac{5}{2x+6}-\dfrac{2\frac{1}{2}}{2x+2}$
$\Rightarrow \dfrac{4}{x+3}-\dfrac{2}{x+1}=\dfrac{5}{2x+6}-\dfrac{5(2x+2)}{2} \ \ \ ...(1)$
$\Rightarrow \dfrac{4}{x+3}-\dfrac{2}{x+1}=\dfrac{5}{2(x+3)}-5(x+1... | Solution:
Notice that there are $2x+6=2(x+3)\ne0$ and $2x+2=2(x+1)\ne 0$, then we have $x\ne-3$, $x\ne -1$, and
\begin{align}
&\frac{4}{x+3}-\frac{2}{x+1}=\frac{5}{2x+6}-\frac{\frac{5}{2}}{2x+2}\\
\Leftrightarrow~~~~&\frac{2\times4}{2(x+3)}-\frac{2\times2}{2(x+1)}=\frac{5}{2x+6}-\frac{\frac{5}{2}}{2x+2}\\
\Leftrightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the roots of the equation $(1+\tan^2x)\sin x-\tan^2x+1=0$ which satisfy the inequality $\tan x<0$ Find the roots of the equation $$(1+\tan^2x)\sin x-\tan^2x+1=0$$ which satisfy the inequality $$\tan x<0$$
Shold I solve the equation first and then try to find which of the roots satisfy the inequality? Should I use ... | Let $\sin x = s$; the given equation reduces to:
$$2s^2-s-1=0,~(s-1)(2s+1)=0$$
we have in the first full rotation of the radius vector
$$s=1\to x= \frac{\pi}{2},~\frac{3 \pi}{2}$$
$$s=-\frac{1}{2}\to x= \frac{7 \pi}{6},~\frac{11 \pi}{6}$$
between which only $x= \dfrac{11 \pi}{6}$ has its tangent negative.
This and its ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$
However, I am not sure how to prove the above or if expansion and rear... | Use that for positive $a$ and $b$
$$a^2\ge b^3\iff \left(a^2\right)^\frac16\ge \left(b^3\right)^\frac16\iff a^{\frac13}\ge b^{\frac12}$$
then refer to the generalized mean inequality that is
$$\left(\frac{x^3+y^3}{2}\right)^\frac13≥\left(\frac{x^2+y^2}{2}\right)^\frac12\iff \left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$ Background
I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.
Problem
Compute the integral $\int_{-\inft... | Utilize the series
$$\frac{p\sin x}{1+2p\cos x+p^2}=-\sum_{k=1}^\infty (-p)^{k}\sin kx
$$
with $p=2-\sqrt3$ to express
\begin{align}
\frac{\sin x}{2+\cos x}= -2\sum_{k=1}^\infty (\sqrt3-2)^k\sin kx
\end{align}
Then, integrate
\begin{align}
&\int_{-\infty}^\infty \frac{\sin x}x\frac{1+\cos x}{2+\cos x}dx\\
=& \ 2\int_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Relation between full elliptic integrals of the first and third kind I am working on a calculation involving the Ronkin function of a hyperplane in 3-space.
I get a horrible matrix with full elliptic integrals as entries. A priori I know that the matrix is symmetrical and that give me a relation between full elliptic i... | Have you tried MGfun from Frédéric Chyzak? It is a (Maple) package to deal with computations in multivariate Ore Algebras. The main application of this is exactly the kind of problem you pose: finding relations between (multivariate) holonomic functions. It is a generalization of WZ theory to the multi-variate case.... | {
"language": "en",
"url": "https://mathoverflow.net/questions/14898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice... | I believe the following punchline to the generating function argument doesn’t depend on whether the number of sides on each die is even or odd. Plug $z = \zeta = \exp 2 \pi i /11$ into the purported equality $1+z+z^2+\dots+z^{10} = p(z) q(z)$ with $p,q$ quintic polynomials with nonnegative coefficients. The LHS vanishe... | {
"language": "en",
"url": "https://mathoverflow.net/questions/41310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 1
} |
sums of rational squares It is a well-known fact that if an integer is a sum of two rational squares then it is a sum of two integer squares. For example, Cohen vol. 1 page 314 prop. 5.4.9. Cohen gives a short proof that relies on Hasse-Minkowski, but he attributes the theorem (without reference) to Fermat, who didn... | In his early days, Fermat realized that a natural number that can
be written as a sum of two rational squares actually is a sum of
two integral squares, but he did not come back to this claim
when eventually he discovered the proof of the Two-Squares
Theorem. The result in question can be proved with the methods
avail... | {
"language": "en",
"url": "https://mathoverflow.net/questions/88539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 3
} |
Smallest length of {0,1} vectors to satisfy some orthogonality conditions Let $n$ be a positive integer.
The output of the problem is another positive integer $r$ which must be as small as possible.
I want to construct $2n$ binary vectors $x_i\in\{0,1\}^r$ and $y_i\in\{0,1\}^r$, with $i\in\{1,...,n\}$, which must respe... | The $\log_2 n$ lower bound is actually within a constant factor of optimal for large $n$.
Now let $x_1, \dots, x_n$ be formed by randomly setting each coordinate of each $x_i$ to $1$ with probability $\frac{1}{3}$ (with the events $x_i(k)=1$ independent for all $1 \leq i \leq n$ and $1 \leq k \leq r$). Let the vecto... | {
"language": "en",
"url": "https://mathoverflow.net/questions/228514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Something interesting about the quintic $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0$ and its cousins (Update):
Courtesy of Myerson's and Elkies' answers, we find a second simple cyclic quintic for $\cos\frac{\pi}{p}$ with $p=10m+1$ as,
$$F(z)=z^5 - 10 p z^3 + 20 n^2 p z^2 - 5 p (3 n^4 - 25 n^2 - 625) z + 4 n^2 p(n^4 - 25 n^... | I think the depressed quintic in the question is what Emma Lehmer called the reduced quintic in her paper, The quintic character of 2 and 3, Duke Math. J. Volume 18, Number 1 (1951), 11-18, MR0040338. In the proof of Theorem 4, she writes that the reduced quintic is $$F(z)=z^5-10pz^3-5pxz^2-5p\left({x^2-125w^2\over4}-p... | {
"language": "en",
"url": "https://mathoverflow.net/questions/255416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Special values of the modular J invariant A special value:
$$
J\big(i\sqrt{6}\;\big) = \frac{(14+9\sqrt{2}\;)^3\;(2-\sqrt{2}\;)}{4}
\tag{1}$$
I wrote $J(\tau) = j(\tau)/1728$.
How up-to-date is the Wikipedia listing of known special values for the modular
j invariant ?
Value (1) is not on it.
Alternatively, is there ... | I do not know enough about CM (complex multiplication?) or
class field theory to tell whether Joe's answer is
sensible or feasible. And (so far) we have received no
references as I had hoped. So eventually I came up with
a proof, perhaps more elementary.
Write $\tau_1 = i\sqrt{6}/6$. Then
$$
\frac{-1}{\tau_1} = i\s... | {
"language": "en",
"url": "https://mathoverflow.net/questions/260786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Need to require a assertion for two subsets of natural numbers Let $P\ne Q$ be an arbitrary pair of primes, $M=PQ$.
$S$ = sum of all $m<M$ co-prime to $M$ such that equation $Px+Qy=m\ (1)$ has a solutions in natural numbers.
$s$ = sum of all $m<M$ co-prime to $M$ such that this equation is not solvable.
Is the assertio... | Consider the set
$$T = \{ (x,y)\ :\ 1\leq x\leq Q,\ 1\leq y\leq P,\ Px+Qy\leq PQ\}.$$
It is easy to see that
$$S = \sum_{(x,y)\in T} (Px+Qy).$$
Switching to double summation in $S$, we get
$$S = \sum_{y=1}^{P} \sum_{x=1}^{X_y} (Px+Qy) = \sum_{y=1}^{P} QyX_y + P\frac{X_y(X_y+1)}2,$$
where
$$X_y = \left\lfloor \frac{PQ-... | {
"language": "en",
"url": "https://mathoverflow.net/questions/280486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of a Combinatorial Function I need help with the following problem, proposed by Iurie Boreico:
Two players, $A$ and $B$, play the following game: $A$ divides an $n \times n $ square into strips of unit width (and various integer lengths). After that, player $B$ picks an integer $k$, $1 \leq k \leq n$, and remove... | Here is another way to get the bound obtained by Fedor Petrov.
One observation is that from the covering of the board using strips of size $1 \times i$ and $1 \times n-i$ to cover row $r$ for rows $1 \leq r \leq n$, then $l(n) \leq (n-1) + (n-1) = 2n-2$. From the Pigeonhole principle, we also have that $n \leq l(n)$. ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/298670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Strengthened version of Isoperimetric inequality with n-polygon Let $ABCD$ be a convex quadrilateral with the lengths $a, b, c, d$ and the area $S$. The main result in our paper equivalent to:
\begin{equation} a^2+b^2+c^2+d^2 \ge 4S + \frac{\sqrt{3}-1}{\sqrt{3}}\sum{(a-b)^2}\end{equation}
where $\sum{(a-b)^2}=(a-b)^2+(... | The keyword you are looking for is "quantitative isoperimetric inequalities". The case of polygons was solved in the following paper: https://arxiv.org/abs/1402.4460
The "quantitative" term in the paper above is not necessarily of the form you seek. It involves the standard deviation of the radii and central angles.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/299056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Primality test for generalized Fermat numbers This question is successor of Primality test for specific class of generalized Fermat numbers .
Can you provide a proof or a counterexample for the claim given below?
Inspired by Lucas–Lehmer–Riesel primality test I have formulated the following claim:
Let $P_m(x)=2^{-m}\c... | This is a partial answer.
This answer proves that if $F_n(b)$ is prime, then $S_{2^n-2} \equiv 0 \pmod{F_n(b)}$.
Proof :
Let $N:=F_n(b)=b^{2^n}+1$. It can be proven by induction that
$$S_i\equiv 2^{-b^{i+2}/4}(p^{b^{i+2}/4}+q^{b^{i+2}/4})\pmod N\tag1$$
where $p=a-\sqrt{a^2-4},q=a+\sqrt{a^2-4}$.
From $(1)$, we get, usi... | {
"language": "en",
"url": "https://mathoverflow.net/questions/311457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix logarithm for d-dimensional cyclic permutation matrix I want to find the matrix $\hat{H}_d$ which, when exponentiated, leads to a d-dimensional cyclic permutation transformation matrix.
I have solutions for d=2:
$$
\hat{U}_2 =\left( \begin{matrix}
0 & 1 \\
1 & 0 \\
\end{matrix} \right)=-i \exp\le... | The article below presents the general form:
https://arxiv.org/abs/2001.11909
| {
"language": "en",
"url": "https://mathoverflow.net/questions/338533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Natural number solutions for equations of the form $\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$ Consider the equation $$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}.$$
Of course, there are solutions to this like $(a,b,c) = (9,8,6)$.
Is there any known approximation for the number of ... | Here's another infinite family. Let $x,y$ be positive integers such that $x^2-2y^2=\pm1$ – there are infinitely many such pairs. Let $a=x^2$, $b=2y^2$, $c=xy$, then a little algebra will show that $(a,b,c)$ satisfy the equation in the title.
E.g., $x=3$, $y=2$ leads to $(9,8,6)$, and $x=7$, $y=5$ yields $(49,50,35)$, ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/359481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Two remarkable weighted sums over binary words This question builds off of the previous MO question Number of collinear ways to fill a grid.
Let $A(m,n)$ denote the set of binary words $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_{m+n-2})$ consisting of $m-1$ $0's$ and $n-1$ $1's$. Evidently $\#A(m,n) = \binom{m+n-2}{m-1}$... | One can perhaps look at the bivariate generating functions,
$$
\sum_{m,n \geq 0} \frac{x^m y^n mn}{(m+n-1)!}
$$
and
$$
\sum_{m,n \geq 0} \frac{x^m y^n 2^{m+n-2}}{(2m-2)!(2n-2)!}.
$$
Mathematica expresses these as
$$
\frac{x y \left(e^x x^3-e^x x^2 y+x e^y y^2-2 e^x x y+2 x e^y y-e^y y^3\right)}{(x-y)^3}
$$
and
$$
\frac... | {
"language": "en",
"url": "https://mathoverflow.net/questions/372678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of di... | Here is a way to come up with an improved lower bound for $I(n^2)$, albeit in terms of $q$ and $n$:
We write
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{I(n^2)}{q^{k+1}} = \frac{\sigma(n^2)}{q^k}\cdot\frac{1}{qn^2} > \frac{1}{qn^2},$$
from which it follows that
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}.$$
This improved ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/382050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove this high-degree inequality Let $x$,$y$,$z$ be positive real numbers which satisfy $xyz=1$. Prove that:
$(x^{10}+y^{10}+z^{10})^2 \geq 3(x^{13}+y^{13}+z^{13})$.
And there is a similar question: Let $x$,$y$,$z$ be positive real numbers which satisfy the inequality
$(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \leq(3x+... | Another way. By my previous post it's enough to prove that:
$$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7)$$ for positive $x$, $y$ and $z$.
Indeed, let $x^5+y^5+z^5=\text{constant}$ and $x^7+y^7+z^7=\text{constant}$.
Thus, by the Vasc's EV Method (see here: Cîrtoaje - The equal variable method Corollary 1.8(b)) it's enough to ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/385942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$1^2+17^2+19^2=3\cdot 7\cdot 31$ $1^2+17^2+19^2=3\cdot 7\cdot 31$.
$3$, $7$ and $31$ are the first three Mersenne primes
Let be $a$, $b$ and $c$ positive integers.
Let $M_n$ denote the n-th Mersenne prime.
Can be it proven that there are infinitely many positive a,b,c such that:
$a^2+b^2+c^2=P_n$, where $P_n$ is the pr... | For $n \geq 3$, the product of the first $n$ Mersenne primes is congruent to $3 \times (-1)^{n-1},$ (mod $8$), since every Mersenne prime other than $3$ is congruent to $7$ (mod $8$).On the other hand, Legendre's three-square theorem asserts that the only positive integers which are not expressible as the sum of three ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/394299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of 1's in binary expansion of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ My question is about the Hamming Weight (or number of 1's in binary expansion) of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ A152007
For example, $a_3 = 9709 = (10110111101001)_2 $ has nine 1's in binary expansion
I guess the answer is $3^{(n-1)}$ ... | It is, indeed, correct. Notice first that $2-(-1)=3$ is divisible by $3$, so by lifting-the-exponent lemma the number
$$
A=\frac{2^{3^{n-1}}-(-1)^{3^{n-1}}}{3^n}=\frac{2^{3^{n-1}}+1}{3^n}
$$
is an integer. Notice also that for $n>0$ it has less than $3^{n-1}$ binary digits. Assume that it has $m$ binary digits. We have... | {
"language": "en",
"url": "https://mathoverflow.net/questions/396297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Irrationality of this trigonometric function I'd like to prove the following conjecture.
Let $x = \frac{p}{q}\pi$ be a rational angle ($p,q$ integers, $q \geq 1$).
Then
$f(x) = \frac{2}{\pi} \arccos{\left(2\cos^4(2x)-1 \right)}$
is irrational if $x$ is not an integer multiple of $\frac{\pi}{8}$. Is this true? Is the ot... | Lemma. Let $n$ be a positive integer such that each number in the open interval $(n/4,3n/4)$ is not coprime with $n$. Then $n\in \{1,4,6\}$.
Proof.
*
*If $n=2m+1$ is odd, and $m\geqslant 1$, then $m\in (n/4,3n/4)$.
*If $n=4m+2$ and $m>1$, then $2m-1\in (n/4,3n/4)$
*If $n=2$, then $1\in (n/4,3n/4)$
*If $n=4m$ and $... | {
"language": "en",
"url": "https://mathoverflow.net/questions/397285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
EM-wave equation in matter from Lagrangian Note
I am not sure if this post is of relevance for this platform, but I already asked the question in Physics Stack Exchange and in Mathematics Stack Exchange without success.
Setup
Let's suppose a homogeneous dielectric with a (spatially) local dielectric response function $... | In the Lagrangian he got, there is a term that reflects the interaction of the electromagnetic field with matter which has the form of equation (2) in this article ( to within a multiplicative factor : derivation of the energy per unit of volume)) but with a different factor which contains the tensor and its Hermitian ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/404380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Combinatorics: on the number of Celtic knots in an n × m frame Everybody has seen a Celtic knot like the ones below.
Mathematically speaking a rectangular Celtic knot of size $(n, m)$ may be built as:
We draw the boundaries of a $2n \times 2m$ rectangle .
Then we draw some barriers, which are horizontal or vertical se... | Thanks to Brian, I've been able to solve the general problem.
I wrote a 9-page document to explain it all, unfortunately I cannot upload it here, so I give you the final results (the formulas could be written with just one formula but it would be a lot more confusing)
Non Square Celtic Knot
$\begin{array} {|c|r|l|}\hli... | {
"language": "en",
"url": "https://mathoverflow.net/questions/412380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Resultant of linear combinations of Chebyshev polynomials of the second kind The Chebyshev polynomial $U_n(x)$ of the second kind is characterized by
$$
U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}.
$$
It seems that
$$\operatorname*{Res}_x \left( U_n(x)+tU_{n-1}(x),\sum_{k=0}^{n-1}U_k(x) \right) =(-1)^{\frac{n(... | Since $U_n + t U_{n-1}$ is of degree $n$ and $\sum_{k=0}^{n-1} U_k$ is of degree $n-1$ with leading coefficient $2^{n-1}$, the resultant factors as
$$ 2^{n(n-1)} (-1)^{n(n-1)} \prod_{j=1}^{n-1} (U_n(x_j) + t U_{n-1}(x_j))$$
where $x_1,\dots,x_{n-1}$ are the zeroes of $\sum_{k=0}^{n-1} U_k$.
Fortunately, these zeroes ca... | {
"language": "en",
"url": "https://mathoverflow.net/questions/427585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Trivial (?) product/series expansions for sine and cosine In an old paper of Glaisher, I find the following formulas:
$$\dfrac{\sin(\pi x)}{\pi x}=1-\dfrac{x^2}{1^2}-\dfrac{x^2(1^2-x^2)}{(1.2)^2}-\dfrac{x^2(1^2-x^2)(2^2-x^2)}{(1.2.3)^2}-\cdots$$
$$\cos(\pi x/2)=1-x^2-\dfrac{x^2(1^2-x^2)}{(1.3)^2}-\dfrac{x^2(1^2-x^2)(3^... | You can derive at least some of these formulas from expansions of $\sin x\theta$ and $\cos x\theta$ as Taylor series in $\sin\theta$,
$$\sin x\theta=x\sin \theta-\frac{x(x^2-1)}{3!}\sin^3\theta+\frac{x(x^2-1)(x^2-3^2)}{5!}\sin^5\theta+-\cdots$$
$$\cos x\theta=1-\frac{x^2}{2!}\sin^2 \theta+\frac{x^2(x^2-2^2)}{4!}\sin^4\... | {
"language": "en",
"url": "https://mathoverflow.net/questions/432026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Lucas–Lehmer test and triangle of coefficients of Chebyshev's In the Lucas–Lehmer test with $ \quad p \quad $ an odd prime.
we know that $ \quad S_0=4 \quad $ and $ \quad S_i=S_{i-1}^2-2 \quad $ for $\quad i>0 \quad$
$M_p=2^p-1 \quad$ is prime if $ \quad S_{p-2} \equiv 0 \bmod {(2^p-1)}$
after some observations i found... | By the composition property of Chebyshev polynomials $T_m(T_n(x))=T_{mn}(x)$. Since $x^2-2 = 2T_2(\tfrac{x}2)$, we have $S_i = 2T_{2^i}(2)$ for all $i\geq 0$.
Furthermore, since $T_{2^k}(x) = \frac{U_{2^{k+1}-1}(x)}{2U_{2^{k}-1}(x)}$, we have
\begin{split} S_{i+1}\cdot S_{i+1}\cdots S_{i+m} &= 2^m T_{2^{i+1}}(2)\cdot T... | {
"language": "en",
"url": "https://mathoverflow.net/questions/439313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relation between full elliptic integrals of the first and third kind I am working on a calculation involving the Ronkin function of a hyperplane in 3-space.
I get a horrible matrix with full elliptic integrals as entries. A priori I know that the matrix is symmetrical and that give me a relation between full elliptic i... | Are you sure that $a$, $b$, and $c$ have no relations among them?
I ask this because I attempted to (painstakingly!) copy your proposed equation over to Mathematica (taking care of the fact that the functions in it take the modulus $m=k^2$ as argument), and I keep getting complex results instead of 0.
On the other hand... | {
"language": "en",
"url": "https://mathoverflow.net/questions/14898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
solution of the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ i am wondering if there is a complete solution for the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ in which $a,b,c,d,k$ are integer(not all zero) and $p$ is odd prime.
| For the equation:
$$a^2+pb^2+(p+k^2)z^2=2c^2+2kcz$$
If the number $k$ is the problem any, and $p$ is such as this: $p=\frac{t^2}{2}-1$
Then the solution can be written:
$$a=\pm{t}n^2+2(tpr\mp(p+1)kj)ns-(2p(p+1)kjr\pm{t}((p+1)(p+k^2)j^2+pr^2))s^2$$
$$b=\pm{t}n^2-2(tr\pm(p+1)kj)ns+(2(p+1)kjr\mp{t}((p+1)(p+k^2)j... | {
"language": "en",
"url": "https://mathoverflow.net/questions/38354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice... | You can write a polynomial that encodes the probabilities for each die:
$$ P(x) = p_1 x^1 + p_2 x^2 + p_3 x^3 + p_4 x^4 + p_5 x^5 + p_6 x^6 $$
and similarly
$$ Q(x) = q_1 x^1 + q_2 x^2 + q_3 x^3 + q_4 x^4 + q_5 x^5 + q_6 x^6. $$
Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of you... | {
"language": "en",
"url": "https://mathoverflow.net/questions/41310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 2
} |
Is the set of undecidable problems decidable? I would like to know if the set of undecidable problems (within ZFC or other standard system of axioms) is decidable (in the same sense of decidable). Thanks in advance, and I apologize if the question is too basic.
| For every consistent recursively axiomatizable theory $T$ (one of which $ZFC$ is widely believed to be) there exists an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this fact cannot be proved in $T$:
\begin... | {
"language": "en",
"url": "https://mathoverflow.net/questions/81429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Question on a Basel-like sum Hello all,
I have happened upon the following sum:
$ 1^2 + \Big(1 \times \frac{1}{3} + \frac{1}{3} \times 1 \Big)^2 + \Big(1 \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{5} \times 1 \Big)^2 + \Big(1 \times \frac{1}{7} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{5} \t... | Interesting problem. Here is my version.
$$\begin{align}
f(u) &= u+ \frac{u^3}{3}+\frac{u^5}{5}+\dots = \frac{1}{2}\log\frac{1+u}{1-u}
\cr
f(u)^2 &= u^2 + \left(1\cdot\frac{1}{3}+\frac{1}{3}\cdot 1\right)u^4 +
\left(1\cdot \frac{1}{5}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{5}\cdot 1\right)u^6+\dots
\end{align}$$
out of ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/102974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Building on this question scaling the imaginary part of $\rho$s in infinite products, I like to conjecture that:
$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1... | Below is a proposed proof, that assuming the RH, the following equation is true:
$$\displaystyle \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{\overline{\mu_n}} \right)$$
where $\mu_n = a + i x \gamma_n$ and $\gamma_n ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/117874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Writing a matrix as a sum of two invertible matrices Let $n\geq 2$. Is it true that any $n\times n$ matrix with entries from a given ring (with identity) can be written as a sum of two invertible matrices with entries from the same ring ?
| This may help. Any $2\times 2$ matrix is a sum of four units:
$$ \begin{pmatrix} x & y \\ z & t \end{pmatrix} = \begin{pmatrix} x & 1 \\ -1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ 1 & t \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ z & -1 \end{pmatrix} + \begin{pmatrix} -1 & y \\ 0 & 1 \end{pmatrix}$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/141382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
On integers as sums of three integer cubes revisited It is easy to find binary quadratic form parameterizations $F(x,y)$ to,
$$a^3+b^3+c^3+d^3 = 0\tag{1}$$
(See the identity (5) described in this MSE post.) To solve,
$$x_1^3+x_2^3+x_3^3 = 1\tag{2}$$
in the integers, all one has to do is to check if one term of $(1)$ c... | Perhaps if you start with my three-rational-cubes identity
$$
ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3}
- \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\biggr)^{\!3}
- \biggl(\frac{(a^2-3b^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3}
$$
you might be able to find something?
| {
"language": "en",
"url": "https://mathoverflow.net/questions/142654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$ I'm working on solving the quartic Diophantine equation in the title. Calculations in maxima imply that the only integer solutions are
\begin{equation}
(r,s) \in \{(-3, -2), (-2, 3), (-1, 0), (0, -1), (0, 1), (1, 0), (2, -3), (3, 2)\}.
\end{equatio... | Inspired by @PeterMueller, I believe I found a proof that $r = 3$.
Because of how this equation was obtained in the first place, I can assume $s \ge 2$ is even, and $r \ge s+1$ is odd. Writing $s=2v$ and $r=2v+2t+1$, substituting, and factoring yields
\begin{align}
2v(v-2t-1)(2v+2t+1)^2 &= t(t+1)(2t^2+2t+1),
\end{alig... | {
"language": "en",
"url": "https://mathoverflow.net/questions/143599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Taylor series coefficients This question arose in connection with A hard integral identity on MATH.SE.
Let
$$f(x)=\arctan{\left (\frac{S(x)}{\pi+S(x)}\right)}$$
with $S(x)=\operatorname{arctanh} x -\arctan x$, and let
$$f(x)=\sum_{n=0}^\infty a_nx^n=\frac{2}{3\pi}x^3-\frac{4}{9\pi^2}x^6+\frac{2}{7\pi}x^7+\frac{16}{81\p... | $\newcommand{\Catalan}{\operatorname{Catalan}}$
I made an attempt for $\int_0^1 S^2\;dx/x$, but with limited success.
Let
$$
q_1 := \frac{1}{16}\left(
\ln \left( 1-i \right) {\pi }^{2}+16\,\zeta \left( 3 \right) -4\,i
\ln \left( 1+i \right) \pi \,\ln \left( 2 \right) +i{\pi }^{3}+4\,i
\ln \left( 1-i \right) \pi ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/155263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Inequality for an integral involving $ \exp $, $ \sin $ and $ \cos $ Let $ t > 0 $ and $ k \in \{ 0,1,2,\ldots \} $. Does the following inequality hold?
$$
\int_{k + 1/2}^{k + 3/2}
\frac{x \sin(2 \pi x)}{1 + 2 e^{2 \pi t} \cos(2 \pi x) + e^{4 \pi t}}
\mathrm{d}{x}
\leq \frac{1}{2 \pi} \cdot \frac{1}{(1 ... | The inequality is true, and follows upon integrating by parts. The integral is
$$
\int_{k+1/2}^{k+3/2} x d\Big( -\frac{\log (1+2 e^{2\pi t} \cos(2\pi x) +e^{4\pi t}}{4\pi e^{2\pi t}} \Big)
$$
and integration by parts gives
$$
= \frac{1}{4\pi e^{2\pi t}} \int_{k+1/2}^{k+3/2} \log \frac{1+2e^{2\pi t} \cos (2\pi x) + ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/184745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Sum of Stirling numbers with exponents I have a trouble with the following sum
$\sum_{i=0}^n\binom{n}{i}S(i,m)3^i$, where $S(i,m)$ is the Stirling number of the second kind (the number of all partitions of $i$ elements into $m$ nonempty sets).
Below it was obtained the sum
$f(n,m)=\frac{1}{m!}\sum_{k=0}^m(-1)^{m-k}\bin... | here is a table of $f(m,n)$ for small values of $m$ and $n\geq m$ (note that $f(m,n)=0$ for $n<m$):
$f(n,1)=-1+4^n$
$f(n,2)=\tfrac{1}{2}\left(1-2\cdot 4^n+7^n\right)$
$f(n,3)=\tfrac{1}{6}\left(-1+3\cdot 4^n-3\cdot 7^n+10^n\right)$
$f(n,4)=\tfrac{1}{24}\left(1-4\cdot 4^n+6\cdot 7^n-4\cdot 10^n+13^n\right)$
$f(n,5)=\tfra... | {
"language": "en",
"url": "https://mathoverflow.net/questions/220479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rate of Convergence of Borwein Algorithm for computing Pi In a book "Pi and the AGM" in 1987, authors, Jonathan Borwein and Peter Borwein, introduced a magical algorithm to compute $\pi$. However there is a problem that I couldn't understand and couldn't find any proof or explaination. It is related to rate of converge... | Even for different starting values around $1$, both $x_n$ and $y_n$ rapidly approach $1$.
Define $\alpha_n = x_n-1$ and $\beta_n = y_n-1$. These will be very small quantities. $\beta_5 \lt 10^{-40}$ and $\alpha_5$ is smaller. We want to estimate $\alpha_n/\beta_n$.
$$\sqrt{x_n} = \sqrt{1+\alpha_n} = 1+\frac{1}{2}\alph... | {
"language": "en",
"url": "https://mathoverflow.net/questions/228068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A surprising conjecture about twin primes Just for fun, I began to play with numbers of two distinct ciphers. I noticed that most of the cases if you consider the numbers $AB$ and $BA$ (written in base $10$), these have few common divisors: for example $13$ and $31$ are coprime, $47$ and $74$ are coprime. Obviously thi... | Suppose $n-1$ and $n+1$ are both primes.
$\gcd(an+b,bn+a)$ divides $an+b - (bn+a) = (a-b)(n-1)$.
There are two cases. If $n-1$ divides $\gcd(an+b,bn+a)$ then $b=n-1-a$ so $an+b= (n-1) (a+1)$ and $bn+a=(n-1)(b+1)$, so $\gcd(an+b,bn+a) = (n-1)\gcd(a+1,b+1)$.
$(a+1)+(b+1)=n+1$. Because $n+1$ is prime, two numbers that sum... | {
"language": "en",
"url": "https://mathoverflow.net/questions/257822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Periodic tilings of the plane by regular polygons Let $A$ be a tiling of $\mathbb{R}^{2}$ using regular polygons. Assume that the tiling is edge-to-edge. Assume also that there are two directions of periodicity, so that $\mathbf{u},\mathbf{v}\in \mathbb{R}^{2}$ are linearly independent vectors, and $A+\mathbf{u}=A+\mat... | I claim that the tiling https://upload.wikimedia.org/wikipedia/commons/6/66/5-uniform_310.svg does not admit a orthogonal period.
The basis for the period lattice is given by the two vectors $$ v_1 = \begin{pmatrix} 3 + \sqrt{3} \\ -1 \end{pmatrix}, v_2 = \begin{pmatrix} 1/2 \\ (2+\sqrt{3})/2 \end{pmatrix}. $$ So if t... | {
"language": "en",
"url": "https://mathoverflow.net/questions/263692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Dividing squares by sums This question is out of curiosity and came to me thinking about another MO question which is linked below.
Question: Do there exist positive integers $a,b,c$ such that $\gcd(a,b,c) =1 $ and each of $\frac{a^2}{b+c},\frac{b^2}{a+c},$ and $\frac{c^2}{a+b}$
are also integers?
My question was i... | First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$).
As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime.
Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/264405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$ In my research, I found this identity and as I experienced, it's surely right. But I can't give a proof for it.
Could someone help me?
This is the identity:
let $a$ and $b$ be two positive inte... | Let us denote
$$S=\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2.$$
First, let $s=i+j$ so that
$$S = \sum_{s\geq 0}\sum_{i=0}^s \binom{s}{i}^2 \binom{a+b-s}{a-i}^2.$$
Consider the generating function
$$F(x,y) = \sum_{s,i} \binom{s}{i}^2 x^i y^s = (1-2y+y^2-2xy-2xy^2+x^2y^2)^{-1/2}.$$
Then $S$ is nothing e... | {
"language": "en",
"url": "https://mathoverflow.net/questions/283540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 3,
"answer_id": 2
} |
Algebraic inequalities on different means If $a^2+b^2+c^2+d^2=1$ in which $a,b,c,d>0$, prove or disprove
\begin{equation*}
\begin{aligned}
(a+b+c+d)^8&\geq 2^{12}abcd;\\
a+b+c+d+\frac{1}{2(abcd)^{1/4}}&\geq 3.
\end{aligned}
\end{equation*}
Can you tell any general algorithm for this type of problems? Thanks.
| Put
\begin{align*}
f(a,b,c,d) &= \frac{(a+b+c+d)^8}{2^{12}abcd} \\
g(a,b,c,d) &= \frac{a+b+c+d}{3} + \frac{1}{6(abcd^{1/4})}
\end{align*}
so the conjecture is that $f,g\geq 1$. I used Maple to search randomly for places where $f$ is as small as possible, then used Maple's fsolve() to find a local minimimum of $f$ n... | {
"language": "en",
"url": "https://mathoverflow.net/questions/302192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Bounding a sum of products of binomial coefficients I am trying to understand the following sums for $k\le n$ :
$$
\sum_{s=0}^{k} \begin{pmatrix} 2n-s/2\\ s\end{pmatrix}\begin{pmatrix} 2n-3s/2\\ k-s\end{pmatrix}
$$
$$
\sum_{s=0}^{k} \begin{pmatrix} 2n-s\\ s\end{pmatrix}\begin{pmatrix} 2n-s\\ k-s\end{pmatrix}
$$
More pr... | Suppose that $k$ is fixed. The sums represent polynomials of degree $k$ in $n$. To answer your question, it's enough to compute two first leading terms of these polynomials. Let me do that for the second sum (the first sum is treated similarly), replacing $2n$ with $n$.
We have
\begin{split}
S(n,k)&:=\sum_{s=0}^{k} \bi... | {
"language": "en",
"url": "https://mathoverflow.net/questions/304151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that this expression is greater than 1/2 Let $0<x < y < 1$ be given. Prove
$$4x^{2}+4y^{2}-4xy-4y+1 + \frac{4}{\pi^2}\Big[
\sin^{2}(\pi x)+ \sin^{2}(\pi y) + \sin^{2}[\pi(y-x)] \Big] \geq \frac{1}{2}$$
I have been working on this problem for a while now and I have hit a wall. I have plotted this and it seems to b... | By the Max Alekseyev's hint we need to prove that $\sum\limits_{cyc}f(a)\geq\frac{3}{4},$
where $f(x)=x^2+\frac{2}{\pi^2}\sin^2\pi x,$ $a$, $b$ and $c$ are positives such that $a+b+c=1.$
We have $$f''(x)=4\left(\frac{1}{2}+\cos2\pi x\right),$$
which gives that $f$ is a convex function on $\left[0,\frac{1}{3}\right]$ a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/336472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Natural number solutions for equations of the form $\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$ Consider the equation $$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}.$$
Of course, there are solutions to this like $(a,b,c) = (9,8,6)$.
Is there any known approximation for the number of ... | Above equation shown below, has solution:
$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$
$a=9w(2p-1)(18p-7)$
$b=4w(72p^2-63p+14)$
$c=3w(72p^2-63p+14)$
Where, w=[1/(36p^2-7)]
For, $p=0$ we get:
$(a,b,c)=(9,8,6)$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/359481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Faster than Euler's substitution. How to derive this formula? I wish someone could help me derive this expression. ($K$ is a constant coefficient. $P_n(x)$ is a polynomial function of degree n.)
$$
\int\frac{P_n(x)\mathrm{d}x}{\sqrt{ax^2+bx+c}} \equiv P_{n-1}(x) \cdot\sqrt{ax^2+bx+c} + K\cdot\int\frac{\mathrm{d}x}{\sq... | This is a special case of a technique known as Hermite reduction (here is the original article from 1872). It can be derived by means of the following identity,
$$\frac{d}{dx}\left(x^{n-1}\sqrt{ax^2+bx+c}\right)=\frac{c (n-1) x^{n-2}+b(n-1/2) x^{n-1}}{ \sqrt{a x^2+b x+c}}+\frac{a n x^n}{\sqrt{a x^2+b x+c}}.$$
So for $a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Three squares in a rectangle One of my colleagues gave me the following problem about 15 years ago:
Given three squares inside a 1 by 2 rectangle, with no two squares overlapping, prove that the sum of side lengths is at most 2. (The sides of the squares and the rectangle need not be parallel to each other.)
I couldn't... | Since the squares are convex, we can draw lines which separate them. In particular, if two separating lines go from $(b-a,0)$ to $(b,1)$ and from $(c,1)$ to $(c+d,0)$, then we can prove the result in terms of those lines and those variables.
So: let the rectangle go from $(0,0)$ to $(2,1)$. Let $A$ be the leftmost squa... | {
"language": "en",
"url": "https://mathoverflow.net/questions/396776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Question on OEIS A000085 The OEIS sequence A000085 is defined by
$$ a_n \!=\! (n-1)a_{n-2} + a_{n-1} \;\text{with }\; a_0\!=\!1, a_1\!=\!1.$$
If $n$ of the form $b^2-b+1, b \in \mathbb{N}, b > 2, \;\text{then: }\;$ $$ \left\lfloor \frac{a_n}{a_{n-1}} \right\rfloor > \left\lfloor \frac{a_{n-1}}{a_{n-2}} \right\rfloor$$... | $\newcommand{\fl}[1]{\lfloor #1 \rfloor}\newcommand\N{\mathbb N}$User LeechLattice gave a complete answer to the original post.
This post is to complement that answer by confirming the empirical observation, made in my previous comment, that the inequality
$$ \left\lfloor \frac{a_n}{a_{n-1}} \right\rfloor > \left\lfloo... | {
"language": "en",
"url": "https://mathoverflow.net/questions/412605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
An inequality about e I happen to encounter the following inequality which I need to prove:
$$\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right)>1,$$ for $k\in\mathbb{Z}^{+}$.
My current idea is to expand $$k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\righ... | We can actually prove this directly for $k \ge 7$ and it should be easy to check it for $k<7$
We note that for $x < 1$ we have $x/2-x^2/3+x^3/4-x^4/5... \ge x/2-x^2/3$ (series converges absolutely and grouping in pairs the remainder is positive).
For $y<1$ we also have $1-e^{-y} \ge y-y^2/2$ again by using the Taylor s... | {
"language": "en",
"url": "https://mathoverflow.net/questions/441848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is $ \frac{\pi^2}{12}=\ln(2)$ not true? This question may sound ridiculous at first sight, but let me please show you all how I arrived at the aforementioned 'identity'.
Let us begin with (one of the many) equalities established by Euler:
$$ f(x) = \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2... | Maybe I'm too late to be of much use to the original question-asker,
but I was surprised to see that all of the previous answers seem to
not quite address the real point in this question.
*
*While it is important to be aware of the dangers of rearranging conditionally convergent series, it not true that any rearr... | {
"language": "en",
"url": "https://mathoverflow.net/questions/27592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 2
} |
Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice... | I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are... | {
"language": "en",
"url": "https://mathoverflow.net/questions/41310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
} |
Length of Hirzebruch continued fractions Suppose $a,b$ are two natural numbers relatively prime to $n$ and to each other. Assume $n\geq ab+1$. Suppose further that $\frac{a}{b}\equiv k \pmod{n}$ for some $k\in \lbrace 1,2,\dots, n-1\rbrace$ and $\frac{a}{b}\equiv k'\pmod{n+ab}$ for some $k'\in \lbrace 1,2,\dots, n+ab-1... | Forgive me, this should be in the comments, but I am still building my reputation up to comment. If a=5, b=4, n=7, then k=3, k'=8, and n+ab=27.
Here, $\frac{n}{k}=\frac{7}{3}=3-\frac{1}{2-\frac{1}{2}}$ but $\frac{n+ab}{k'}=\frac{27}{8}=4-\frac{1}{2-\frac{1}{3-\frac{1}{2}}}$. Am I missing something or is there a furthe... | {
"language": "en",
"url": "https://mathoverflow.net/questions/111312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
What is known about a^2 + b^2 = c^2 + d^2 Could you state or direct me to results regarding the Diophantine equation $a^2+b^2=c^2+d^2$ over integers? Specifically, I am looking for a complete parametrization. In the case that a complete parametrization does not exist, I would be interested in seeing complete parametriz... | This equation is quite symmetrical so formulas making too much can be written:
So for the equation:
$X^2+Y^2=Z^2+R^2$
solution:
$X=a(p^2+s^2)$
$Y=b(p^2+s^2)$
$Z=a(p^2-s^2)+2psb$
$R=2psa+(s^2-p^2)b$
solution:
$X=p^2-2(a-2b)ps+(2a^2-4ab+3b^2)s^2$
$Y=2p^2-4(a-b)ps+(4a^2-6ab+2b^2)s^2$
$Z=2p^2-2(a-2b)ps+2(b^2-a^2)s^2$
$R=p^... | {
"language": "en",
"url": "https://mathoverflow.net/questions/130131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
can we say that $(p^2+1)/2\ne p_0^2$ where $p$ is a Mersenne prime Let $p=2^a-1>7$ be a Mersenne prime and so $a$ is an odd prime.
Can we say that $(p^2+1)/2$ is not equal to the square of a prime number?
Many thanks for your help
BHZ
| Suppose, $p^2-2p_1^2=-1.$ Substituting $p=2^a-1,$ we arrive at
$$2^a(2^{a-1}-1)=(p_1-1)(p_1+1).$$ Observe, $(p_1-1,p_1+1)=2,$ so we must have the following options: $p_1-1=2^{a-1}k$ and $p_1+1=2l$ and $kl=2^{a-1}-1.$ This is impossible unless $k,l$ and thus $a$ are small. Indeed, if $k\ge 2,$ then $p_1\ge 2^a+1$ and $... | {
"language": "en",
"url": "https://mathoverflow.net/questions/131570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sequences with integral means Let $S(n)$ be the sequence whose first element is $n$, and from then onward,
the next element is the smallest natural number ${\ge}1$ that ensures that the
mean of all the numbers in the sequence is an integer.
For example, the second element of $S(4)$ cannot be $1$ (mean $\frac{5}{2}$),
b... | Let $k$ be sufficiently large .Denote the sequence $n,a_1,...a_{k-1}$ with $a_i\leq i+1$ where $i\leq k-1$.
The sum of the first $k$ elements including $n$ (denote it by $S_+(a_k)$) must be always a multiple of $k$ .
So let $d$ be a positive integer with $S_+(a_k)=d\cdot k=n+a_1+...+a_{k-1}\leq n+2+...+k=n-1+k\fra... | {
"language": "en",
"url": "https://mathoverflow.net/questions/146733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 4
} |
Taylor series coefficients This question arose in connection with A hard integral identity on MATH.SE.
Let
$$f(x)=\arctan{\left (\frac{S(x)}{\pi+S(x)}\right)}$$
with $S(x)=\operatorname{arctanh} x -\arctan x$, and let
$$f(x)=\sum_{n=0}^\infty a_nx^n=\frac{2}{3\pi}x^3-\frac{4}{9\pi^2}x^6+\frac{2}{7\pi}x^7+\frac{16}{81\p... | Here is a possible approach toward proving (1). Let $i=\sqrt{-1}$. Let $g(x)=f(x) +f(ix) +f(-x)+f(-ix)$. We need to
show that $g(x)=0$. Unfortunately, Maple is unable to do this directly. Thus write each
$\arctan u$ as $\frac i2\log\frac{1-iu}{1+iu}$ and $\mathrm{arctanh}\,u$ as $\frac 12\log\frac{1+u}{1-u}$
and simpli... | {
"language": "en",
"url": "https://mathoverflow.net/questions/155263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
A possibly surprising appearance of Fibonacci numbers Let $H(n) = 1/1 + 1/2 + \dotsb + 1/n,$ and for $i \leq j,$ let $a_1$ be the least $k$ such that
$$H(k) > 2H(j) - H(i),$$
let $a_2$ be the least k such that
$$H(k) > 2H(a_1) - H(j),$$
and for $n \geq 3,$ let $a_n$ be the least $k$ such that
$$H(k) > 2H(a_{n-1}) - H(... | The statement is true. Write $F_n$ for the $n$-th Fibonacci (my indexing starts at $(F_0, F_1, F_2, F_3, \dots) = (0,1,1,2,\dots)$). We are being asked to show that
$$\frac{1}{F_{j+1}} > \sum_{m=F_{j}+1}^{F_{j+1}} \frac{1}{m} - \sum_{m=F_{j-1}+1}^{F_{j}} \frac{1}{m} > 0\ \mbox{for}\ j \geq 6.$$
Computer computations... | {
"language": "en",
"url": "https://mathoverflow.net/questions/196681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
} |
Prove $4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}$ Wolstenholme's theorem is stated as follows:
if $p>3$ is a prime, then
\begin{align*}
\sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\
\sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}.
\end{align*}
It is also not hard to prove th... | This goes way back to Emma Lehmer, see her elementary paper on Fermat quotients and Bernoulli numbers from 1938.
Assume $p\ge 7$. First, I reformulate your congruence. I replace
$$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}$$
with the equivalent
$$\sum_{k=1}^{p-1} \frac{1}{k^2} \e... | {
"language": "en",
"url": "https://mathoverflow.net/questions/226804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Mutation of valued quivers Mutations of valued quivers are defined in cluster algebras II, Proposition 8.1 on page 28. I have a question about the number $c'$. For example, let $a = 2, b=1, c=1$ and consider the quiver $Q$:
$1 \overset{a}{\to} 2 \overset{b}{\to} 3$ and there is an arrow $1 \overset{c}{\to} 3$ in this ... | I don't think your diagram is coming from a skew-symmetrizable matrix. Let $B$ be a skew-symmetrizable matrix and assume $Q = \Gamma(B)$ so your quiver is the diagram of $B$ as defined in Definition 7.3 of the linked paper. Then we must have
$$B = \begin{bmatrix}0 & x & y \\ \frac{-1}{x} &0 & z \\ \frac{-1}{y} & \frac{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/248083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $m$ such $2^m+1\mid5^m-1$ The problem comes from a problem I encountered when I wrote the article
Find all positive integer $m$ such
$$2^{m}+1\mid5^m-1$$
it seem there no solution. I think it might be necessary to use quadratic reciprocity knowledge to solve this problem.
If $m$ is odd then $2^m+1$ is div... | Here is a proof.
Theorem. $2^m+1$ never divides $5^m-1$.
Assume that there is some $m$ such that $2^m+1$ divides
$5^m-1$. We already know that $m$ must be divisible by $4$.
Let $m = 2^n a$ with an odd integer $a$ and $n \ge 2$.
The $n$th Fermat number $$F_n = 2^{2^n} + 1$$ is congruent
to $2$ mod $5$ (this uses $n \g... | {
"language": "en",
"url": "https://mathoverflow.net/questions/317643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$ It is well-known that one can evaluate the sum
$$\sum_{k =1}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)=\frac{N^2-1}{3}.$$
The answer to this problem can be found here
click here.
I am now interested in the more d... | Start from the well known formula
\begin{equation}
2^{N-1} \prod _{k=1}^N \left[\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)\right]=\cos N(x-y)-\cos N(x+y),
\end{equation}
take logarithmic derivative
\begin{equation}
\sum _{k=1}^N \frac{\sin(x-y)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{N\sin ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/349776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Diophantine representation of the set of prime numbers of the form $n²+1$ A polynomial formula for the primes (with 26 variables) was presented by Jones, J., Sato, D., Wada, H. and Wiens, D. (1976). Diophantine representation of the set of prime numbers. American Mathematical Monthly, 83, 449-464.
The set of prime numb... | Call your polynomial $P$. I propose the following polynomial:
$$
P' = (\xi^2+1)(1 - (\xi^2+1-P)^2)
$$
Proof (that the positive values of $P'$ are exactly the primes of the form $N^2+1$):
Let $P_0$ be one of the values of $P$, and let $\xi_0$ be any integer.
Case (i). Suppose $P_0 = \xi_0^2+1$. Then the value of the a... | {
"language": "en",
"url": "https://mathoverflow.net/questions/350089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A probability question from sociology We know that $\frac{1}{2} \leq a \leq p \leq 1$. And, $n \geq 3$ is a positive odd number, and $t$ is an integer. $a$ satisfies the equation below.
\begin{equation} \small
\begin{aligned}
&\sum_{t=0}^{n-1} \left( {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}]... | If I expand the left-hand-side of your equation around $a=1/2$ I find
$$
\sum_{t=0}^{n-1} {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}] $$
$$\times\left[\frac{a^{t+1} (1-a)^{n-t-1}}{a^{t+1} (1-a)^{n-t-1}+(1-a)^{t+1} a^{n-t-1}}-\frac{a^{t} (1-a)^{n-t}}{a^{t} (1-a)^{n-t}+(1-a)^{t} a^{n-t}}\right]... | {
"language": "en",
"url": "https://mathoverflow.net/questions/363761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this high-degree inequality Let $x$,$y$,$z$ be positive real numbers which satisfy $xyz=1$. Prove that:
$(x^{10}+y^{10}+z^{10})^2 \geq 3(x^{13}+y^{13}+z^{13})$.
And there is a similar question: Let $x$,$y$,$z$ be positive real numbers which satisfy the inequality
$(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \leq(3x+... | This is to give an alternative proof of the inequality in Fedor Petrov's nice answer
$$(2 x^{10} + x^{-20})^2\ge3 (2 x^{13} + x^{-26})$$
for all $x > 0$.
We have
\begin{align*}
&2x^{13+1/3} + x^{-26-2/3} - (2x^{13} + x^{-26})\\
={}& 2(x^{13+1/3} - x^{13}) + (x^{-13-1/3} - x^{-13})
(x^{-13-1/3} + x^{-13})\\
... | {
"language": "en",
"url": "https://mathoverflow.net/questions/385942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Calculation of a series It seems that we have:
$$\sum_{n\geq 1} \frac{2^n}{3^{2^{n-1}}+1}=1.$$
Please, how can one prove it?
| This is the special case $q=3$ of a formula
$$
\qquad\qquad
\sum_{n=1}^\infty \frac{2^n}{q^{2^{n-1}}+1} = \frac{2}{q-1}
\qquad\qquad(*)
$$
which holds for all $q$ such that the sum converges, i.e. such that $|q|>1$.
This follows from the identity
$$
\frac{1}{x-1} - \frac{2}{x^2-1} = \frac{1}{x+1}.
$$
Substitute $q^{2^{... | {
"language": "en",
"url": "https://mathoverflow.net/questions/389313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Computing global maximum For $\lambda\in\mathbb{R}$, I want to find the expression of $f(\lambda)$:
$$f(\lambda)=\max_{E\in\mathbb{C}}arccosh{\frac{|E^2+i\lambda E|+|E^2+i\lambda E-4|}{4}}-arccosh{\frac{|E^2-i\lambda E|+|E^2-i\lambda E-4|}{4}}$$
that is
$$A={\sqrt{a^2+b^2} \sqrt{a^2+(b+\lambda )^2}+\sqrt{\left(a^2-b^2... | As $g(t)=\log{t}$ is a strictly increasing function, what you really needs is just the maximum of $\frac{A}{B}.$
Also, after some experiments, it seems that, for pure imaginary numbers only, the maximum occurs when the imaginary part is equal to $\lambda$. That is (in your notation), when $a=0$ the maximum is obtained ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/407264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Two interpretations of a sequence: an opportunity for combinatorics The sequence that is addressed here is resourced from the most useful site OEIS, listed as A014153, with a generating function
$$\frac1{(1-x)^2}\prod_{k=1}^{\infty}\frac1{1-x^k}.$$
In particular, look at these two interpretations mentioned there.
$a(n)... | For what it's worth, here is a non-combinatorial proof.
That $\frac{1}{(1-x)^2}\prod_{k=1}^{\infty} \frac{1}{1-x^k}$ is the generating function for partitions of $n$ with three flavors of $1$'s is clear.
So I will focus on the sum of all the irredundant parts of partitions of $n$. Let me use $||\lambda||$ to denote thi... | {
"language": "en",
"url": "https://mathoverflow.net/questions/408194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
\begin{aligned}
f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
&+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4... | Assume that we have $x = - (t + 1/t)^2 = -t^2 - 1/t^2 - 2$ for some $t \in \mathbb{C}$. Then
$$
\sqrt{-x} = t + \frac{1}{t} = \frac{t^2 + 1}{t},\quad x + 4 = -\left(t - \frac{1}{t}\right)^2, \quad \sqrt{x(x + 4)} = t^2 - \frac{1}{t^2},
\\
x + \sqrt{x(x + 4)} = -2\left(1 + \frac{1}{t^2}\right) = -2\frac{1 + t^2}{t^2}, \... | {
"language": "en",
"url": "https://mathoverflow.net/questions/440962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 0
} |
Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice... | You can write this as a single polynomial equation
$$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$
where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$.
So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$
where the factors satisfy some extra conditions (coefficients positive,
$p(1)=1$ etc.... | {
"language": "en",
"url": "https://mathoverflow.net/questions/41310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 4
} |
What is the independence number of hamming graph? Hamming graph H(d,q) is Cartesian product of d complete graphs Kq. We know the independence number of direct product of d complete graphs Kq. What is the independence number of hamming graph?
| Experiments suggest it might be $q^{d-1}$.
Here is data from sage:
d= 2 q= 2 alpha= 2 factor= 2 alpha - q^(d-1) 0
d= 2 q= 3 alpha= 3 factor= 3 alpha - q^(d-1) 0
d= 2 q= 4 alpha= 4 factor= 2^2 alpha - q^(d-1) 0
d= 2 q= 5 alpha= 5 factor= 5 alpha - q^(d-1) 0
d= 2 q= 6 alpha= 6 factor= 2 * 3 alpha - q^(d-1) 0
d= 3 q=... | {
"language": "en",
"url": "https://mathoverflow.net/questions/141842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
On $a^4+nb^4 = c^4+nd^4$ and Chebyshev polynomials In a 1995 paper, Choudhry gave a table of solutions to the quartic Diophantine equation,
$a^4+nb^4 = c^4+nd^4\tag{1}$
for $n\leq101$. Seiji Tomita recently extended this to $n<1000$ and solved all $n$ except $n=967$ (which was later found by Andrew Bremner).
It can b... | Question 2:
The reason must be that Chebyshev polynomials solve Fermat-Pell equations.
The difference between the two sides of equation (2),
$$
(n+nx+y)^4 + n(1−nx−xy)^4 = (n−nx+y)^4 + n(1+nx+xy)^4,
$$
factors as $8nx(n+y) \phantom. ((x^2-1)y^2 + 2n(x^2-1)y + 1 - n^2)$.
The last factor is a quadratic in $n$ with leadin... | {
"language": "en",
"url": "https://mathoverflow.net/questions/142192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Taylor series coefficients This question arose in connection with A hard integral identity on MATH.SE.
Let
$$f(x)=\arctan{\left (\frac{S(x)}{\pi+S(x)}\right)}$$
with $S(x)=\operatorname{arctanh} x -\arctan x$, and let
$$f(x)=\sum_{n=0}^\infty a_nx^n=\frac{2}{3\pi}x^3-\frac{4}{9\pi^2}x^6+\frac{2}{7\pi}x^7+\frac{16}{81\p... | Here is a simple dérivation of the Taylor expansion of $~\arctan(\frac{x}{1+x})~:$
Since $$~\arctan(u)=\dfrac{i}{2}\ln\big(\dfrac{1-iu}{1+iu}\big),~~\arctan(\frac{x}{1+x})=\dfrac{i}{2}\ln\Big(\dfrac{1-i\frac{x}{1+x}}{1+i\frac{x}{1+x}}\Big)$$
$$=
\dfrac{i}{2}\ln\Big(\dfrac{1+(1-i)x}{1+(1+i)x}\Big)=\dfrac{i}{2}\ln\Big(\d... | {
"language": "en",
"url": "https://mathoverflow.net/questions/155263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Two questions about Whittaker functions I am watching the video: Modeling p-adic Whittaker functions, Part I. I have two questions about Whittaker functions in the video.
From 33:00 to 37:00, it is said that after changing of variables, we have
\begin{align}
W^0(t_{\lambda}) = \int_{U} f^0(w_0 u t_{\lambda}) \psi^{-1}... | In the case of $SL_2$, we have
$$
U = \left\{ \left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right) : x \in F \right\}
$$
and
\begin{align}
& \int_U f^0\left( w_0 u t_{\lambda} \right) \psi^{-1}(u) du \\
& = \int_U f^0\left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} 1 & x \\ 0 & 1... | {
"language": "en",
"url": "https://mathoverflow.net/questions/207825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Do there exist infinitely many $n$ such that $n^3+an+b$ is squarefree? Question: Assume that $a,b\in Z$, and $4a^3+27b^2\neq 0$. Prove that there exist infinitely many positive integers $n$ such $n^3+an+b$ is square-free.
I have following
There exist infinitely many postive integers $n$ such $n^2+1$ is squarefree.
... | Erdos was the first to show that cubic polynomials (meeting obvious necessary conditions) do take on square-free values. More generally he considered polynomials of degree $\ell$ taking on $\ell-1$-th power-free values infinitely often. Granville has shown that the ABC conjecture allows one to find square-free values... | {
"language": "en",
"url": "https://mathoverflow.net/questions/235330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Norm of Moore-Penrose inverse of a product I had asked this question in math.stackexchange (link: https://math.stackexchange.com/questions/1902276/bounds-on-the-moore-penrose-inverse-of-a-product ) but I did not get any response so I am trying my luck here.
Let $A^{\dagger}$ denote the Moore-Penrose inverse of a real m... | Here’s a small counterexample involving two rank-deficient matrices $A$ and $B$: $$A = \left( {\begin{array}{*{20}c}
0 & { - 1} & 0 \\
0 & 0 & 0 \\
0 & { - 1} & 0 \\
\end{array}} \right), B = \left( {\begin{array}{*{20}c}
4 & 0 & 2 \\
2 & 0 & 1 \\
4 & 0 & 2 \\
\end{array}} \right)$$
The correspo... | {
"language": "en",
"url": "https://mathoverflow.net/questions/248367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to reduce the integral into hypergeometric function? The equation is
$\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi
^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},%
\frac{1}{2};2;\Pi ^{2})$,
where $\Pi \le 1$.
How to reduce the left integral into the right hyp... | Letting $\beta=\frac{2\Pi}{1+\Pi^2}$, write the integrand as
$$\sqrt{1+\Pi^2}\left(\sqrt{1+\beta \sqrt{1-y^{2}}}-\sqrt{1-\beta \sqrt{1-y^{2}}}\right).\tag1$$
Use the binomial expansion to express (1), after combining two infinite series, in the form
$$2\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\beta^{2n+1}\int_0^1(1-... | {
"language": "en",
"url": "https://mathoverflow.net/questions/256947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Efficient method to write number as a sum of four squares? Wikipedia states that there randomized polynomial-time algorithms for writing $n$ as a sum of four squares
$n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$
in expected running time $\mathrm {O} (\log^{2} n).$
My question is can someone give the efficient algorithm(... | One of their methods for $n=4k+2$ is as follows:
Randomly select an even number $a$ and an odd number $b$ such that $a^2+b^2 < n$. Then, we hope $p=n-a^2-b^2$ is a prime (you can show there's about a $1/(A \log n \log \log n)$ chance of $p$ being prime ); $p$ is of the form $4r+1$, so if $p$ is prime there's a solutio... | {
"language": "en",
"url": "https://mathoverflow.net/questions/259152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
"Nearly" Fermat triples: case cubic Suppose $a^2+b^2-c^2=0$ are formed by a (integral) Pythagorean triple. Then, there are $3\times3$ integer matrices to generate infinitely many more triples. For example, take
$$\begin{bmatrix}-1&2&2 \\ -2&1&2 \\ -2&2&3\end{bmatrix}\cdot
\begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bm... | Daniel Loughran: I apologize for the confusion. You were right. I was only focusing on a single family of solutions to $x^3+y^3-z^3=\pm1$, for which of course there is tedious experimental finding that does a parametrization. Not for all solutions.
Let's see if we can prove that it works: start with $[9,10,12]^T$ with... | {
"language": "en",
"url": "https://mathoverflow.net/questions/269799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Algebraic inequalities on different means If $a^2+b^2+c^2+d^2=1$ in which $a,b,c,d>0$, prove or disprove
\begin{equation*}
\begin{aligned}
(a+b+c+d)^8&\geq 2^{12}abcd;\\
a+b+c+d+\frac{1}{2(abcd)^{1/4}}&\geq 3.
\end{aligned}
\end{equation*}
Can you tell any general algorithm for this type of problems? Thanks.
| There is a general algorithm for this type of problems.
We can use the Vasc's EV Method: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
For example, your second inequality is wrong, but the following is true.
Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^2+b^2+c^2+d^2=1$. Prove that:
... | {
"language": "en",
"url": "https://mathoverflow.net/questions/302192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Probability distribution optimization problem of distances between points in $[0,1]$ Let $\mathcal{D}$ be a probability distribution with support $[0,1]$. Let $X, Y, Z$ three i.i.d. random variables with distribution $\mathcal{D}$, and $T$ a random variable uniformly distributed in $[0,1]$ independent from $X$, $Y$ and... | Sorry, my computation in the comments was wrong. I think it leads to something with $\rho < \frac{16}{17}$.
Namely, let $\mathcal{D}$ be the distribution with $\mathrm{Pr}(\mathcal{D}=0)=\mathrm{Pr}(\mathcal{D}=3/4)=1/N$, and $\mathrm{Pr}(\mathcal{D}=1)=(N-2)/N$, where $N$ is large.
Then the possibilities for $(x,y,z)$... | {
"language": "en",
"url": "https://mathoverflow.net/questions/372341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Irrationality of $e^{x/y}$ How to prove the following continued fraction of $e^{x/y}$
$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$
Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalize... | I think this might be a solution.
The Continued Fraction Expansion of the hyperbolic tanh function discovered by Gauss is
$$\tanh z = \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\\\$$
We also know that the hyperbolic tanh function is related to the exponential function with the following formula
$$\t... | {
"language": "en",
"url": "https://mathoverflow.net/questions/383273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Given an integer $N$, find solutions to $X^3 + Y^3 + Z^3 - 3XYZ \equiv 1 \pmod{N}$ Given an integer $N > 0$ with unknown factorization, I would like to find nontrivial solutions $(X, Y, Z)$ to the congruence $X^3 + Y^3 + Z^3 - 3XYZ \equiv 1 \pmod{N}$. Is there any algorithmic way to rapidly find such tuples? One specia... | Since, $X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)((X-Y)^2+(Y-Z)^2+(Z-X)^2)$, taking $X,Y,Z$ close to each other give some non-trivial and cheap solutions.
For instance $(k+1,k,k)$ for $N=3k$, $(k+1,k+1,k)$ for $N=3k+1$, $(k+1,k,k-1)$ for $N=9k-1$, etc.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/387018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)? Let $\ n\ $ be an arbitrary natural number ($\ 1\ 2\ \ldots).\ $ Then
*
*$\ n\ $ is coarse $\ \Leftarrow:\Rightarrow\ $ there exists a prime divisor $p$ of $\ n\ $ such that $\ p^3>n.$;
*$\ n\ $ is a p-cube $\ \Leftarrow:\Rightarrow\ $ t... | For any $k\in\mathbb{N}$, the number $n=2^3\cdot3^{72k}$ works:
Obviously, $n$ itself is fine.
For $n^2-1 = 2^6\cdot3^{144k}-1 = (2^2\cdot 3^{48 k} + 2\cdot 3^{24 k} + 1)(2^2\cdot 3^{48 k}-2\cdot 3^{24 k} + 1)(2\cdot3^{24 k} + 1) (2\cdot3^{24 k} - 1)$, the first factor is divisible by $7$, the others are small, hence $... | {
"language": "en",
"url": "https://mathoverflow.net/questions/396310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Modular arithmetic problem Here's the problem: start with $2^n$, then take away $\frac{1}{2}a^2+\frac{3}{2}a$ starting with $a=1$, and running up to $a = 2^{n+1}-2$, evaluating modulo $2^n$. Does the resulting sequence contain representatives for all the congruence classes module $2^n$?
Some examples of these sequences... | This is correct. Denote $f(x)=x(x+3)/2 \pmod {2^n}$. Then $f(x)=f(y)$ if and only if $x(x+3)-y(y+3)=(x-y)(x+y+3)$ is divisible by $2^{n+1}$. Since $x-y$ and $x+y+3$ have different parity, this in turn happens if and only if one of them is divisible by $2^{n+1}$. So, if $x=1,2,\ldots,2^{n+1}$, then $f(x)$ takes each val... | {
"language": "en",
"url": "https://mathoverflow.net/questions/416699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are the number of solutions to $ax^2+bxy+cy^2\equiv u\pmod{p}$, $(x,y)\in\{0,\dotsc,p-1\}$, the same for all units $u$? Let $p$ be an odd prime and $F={\mathbb Z}/p{\mathbb Z}$. With $a,b,c\in F$, let $Q(x,y)=ax^2+bxy+cy^2$ where $p\nmid b^2-4ac$. I wish to prove that the number of solutions $(x,y)\in F^2$ of
$$ax... | Inspired by @JoeSilverman's answer, I've updated this answer to handle any odd-characteristic field $F$, and to remove a few ugly case distinctions from the original argument; but I do assume that the characteristic is not $2$, as @KConrad points out.
Put $\theta = \dfrac{b + \sqrt{b^2 - 4a c}}2$, where we have chosen ... | {
"language": "en",
"url": "https://mathoverflow.net/questions/424856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
On the equation $a^4+b^4+c^4=2d^4$ in positive integers $a\lt b\lt c$ such that $a+b\ne c$ Background: The equation
$$a^4+b^4+c^4=2d^4$$
has infinitely many positive integral solutions if we take $c=a+b$ and $a^2+ab+b^2=d^2$ further assuming that $GCD(a,b,c)=1$.
Main problem: Find some positive integral solutions to th... | $(a, b, c, d) = (32, 1065, 2321, 1973), (2156, 5605, 8381, 7383)$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/438196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Does anyone know a polynomial whose lack of roots can't be proved? In Ebbinghaus-Flum-Thomas's Introduction to Mathematical Logic, the following assertion is made:
If ZFC is consistent, then one can obtain a polynomial $P(x_1, ..., x_n)$ which has no roots in the integers. However, this cannot be proved (within ZFC).... | For every consistent recursively axiomatizable theory $T$ there exists (and can be effectively computed from the axioms of $T$) an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this fact cannot be proved in $... | {
"language": "en",
"url": "https://mathoverflow.net/questions/32892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 5,
"answer_id": 1
} |
Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice... | You can't even solve this with two-sided dice. Consider two dice with probabilities p and q of rolling 1, and probabilities (1-p) and (1-q) of rolling 2. The probability of rolling a sum of 2 is pq, and the probability of rolling a sum of 4 is (1-p)(1-q). These are equal only if p=(1-q). Hence they are equal to one... | {
"language": "en",
"url": "https://mathoverflow.net/questions/41310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
} |
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