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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 have:
$$
ab^2 + cd^2 = bc^2 + da^2.
$$
And more generally, if $1\le m,n\le q-1$ are two integers, how many solutions does the equation
$$
a^mb^n + c^md^n = b^mc^n + d^ma^n
$$
have?
| 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 setting $a=\frac{x+b^2}{2d}$, $c=\frac{y+d^2}{2b}$, and replacing $d$ with $-d$, we have to count the $x,y\in\mathbb F_p$ with
\begin{equation}
bx^2-dy^2=b^5-d^5.
\end{equation}
If $b^5-d^5\ne0$, we have an ellipsis with $p-1$ points if $b/d$ is a square, and $p+1$ points otherwise. The former case happens $(p-1)(p-3)/2$ times, and the latter case $(p-1)^2/2$ times. In the remaining case $b^5-d^5=0$ we have solutions $x, y$ if and only if $b/d$ is a square, and the number of solutions is $2p-1$ for each such pair $b,d$.
Indeed, carying out the count we, we find that there are all together $(p^2 + 6p - 6)p$ solutions if $p\equiv1\pmod{10}$, and $(p^2 + 2p - 2)p$ solutions if $p\not\equiv1\pmod{10}$.
| {
"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
$\sum_{j=1}^n s^{\alpha_j}=0$ and $\sum_{j=1}^n s^{\beta_j} \neq 0$ or the other hand $\sum_{j=1}^n s^{\beta_j} =0$ and $\sum_{j=1}^n s^{\alpha_j}\neq0$?
Thank you for @Peter Mueller's counter example. Since we know that the two 'polynomials'
$\sum_{j=1}^n s^{\alpha_j}$ and $\sum_{j=1}^n s^{\beta_j}$
in my question maybe have the same set of roots. But I want to ask that is it possible that all the same root has the same multiplicity? Since for the ordinary polynomials this statement is not true. But for my case, what will happen?
| 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 setting $x=s^{1/10}$ we see that
\begin{align}
\alpha_6,\dots,\alpha_1 &= 1/10,\; 2/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10\\\
\beta_6,\dots,\beta_1 &= 1/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10,\; 8/10
\end{align}
is a counterexample for $n=6$.
Added: More examples, with $n=4$, can be constructed using
\begin{align}
f(x) &= x(1+x+x^v+x^{v+1})\\\
g(x) &= x(1+x^u+x^v+x^{u+v})
\end{align}
if $v$ is odd and $1\lt u\lt v$ is a divisor of $u$. In this case, $f(x)$ and $g(x)$ again have the same complex roots.
| {
"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. But the integrands in (1) are positive: Use $\cot x > 0$ for $0 < x < \pi/2$ and $\cot(k\pi+x) = \cot x$ and $\cot(k\pi-x) = -\cot x$. Also $(1/x) - \cot x$ increases from $0$ on $(0,\pi/2)$, so $f(x) >0$ on $(0,\pi/2)$. Next
$$
f(k\pi+x)+f(k\pi-x) = \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) + \left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x
$$
and each of the two halves is positive on $(0,\pi/2)$. Recall
$$
\sum_{k=1}^\infty \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) = \csc^2 x - \frac{1}{x^2}
$$
$$
\sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)=\frac{1}{x}-\cot x
$$
$$
\sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x=\frac{\cot x}{x}-\cot^2 x
$$
Our answer is the sum of three integrals:
$$
\int_0^{\pi/2} \left[\left(\frac{1}{x^2}-\frac{\cot x}{x}\right)+\left(\csc^2 x-\frac{1}{x^2}\right)+\left(\frac{\cot x}{x}-\cot^2 x\right)\right]dx = \int_0^{\pi/2} 1\;dx = \frac{\pi}{2}
$$
| {
"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. For example, $f(k/11)$ seems infinite for
$k=2,\ldots,10$, but, e.g., $f(4/13)=f(13/4)$ is finite.
Several more examples are shown below.
$$ \frac{1}{k} \to \frac{k}{k} {=} 1 $$
$$ \frac{3}{7} \to \frac{21}{9} {=} \frac{7}{3} \to \frac{21}{9} {=} \frac{7}{3} $$
$$ \frac{5}{12} \to \frac{60}{16} {=} \frac{15}{4} \to \frac{60}{18} {=} \frac{10}{3}
\to \frac{30}{12} {=} \frac{5}{2} \to
\frac{10}{6} {=} \frac{5}{3} \to
\frac{15}{7} \to \frac{105}{21} {=} \frac{5}{1} \to \frac{5}{5} = 1 $$
$$ \frac{5}{23} \to \frac{115}{27} \to \frac{3105}{141} {=} \frac{1035}{47}
\to \frac{48645}{1081} = \frac{45}{1} \to 1 $$
$$ \frac{4}{11} \to \frac{44}{14} {=} \frac{22}{7} \to \frac{154}{28} {=} \frac{11}{2} \to
\frac{22}{12} {=} \frac{11}{6}
\to \frac{66}{16} {=} \frac{33}{8} \to \frac{264}{40} {=} \frac{33}{5}
\to \frac{165}{37} \to \cdots \to \infty ?$$
I am hoping this does not run into Collatz-like difficulties,
but it does seem straightforward to analyze...
If anyone recognizes it, or sees a way to tame it even partially, I would be interested to learn. Thanks!
| 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:
$$f\left(f\left(\frac{1}{n}\right)\right)=f\left(1\right)=1$$
$$f\left(f\left(\frac{(n-1)^2}{n^2}\right)\right)=f\left(\frac{n(n-1)}{2}\right)=1$$
$$f\left(f\left(\frac{2n-1}{2n}\right)\right)=f\left(n\right)=1$$
$$f\left(f\left(\frac{n+1}{n^2}\right)\right)=f\left(n\right)=1$$
$$f\left(f\left(\frac{3(2n+1)}{4(3n+1)}\right)\right)=f\left(4n+2\right)=1$$
If initially not much cancellation happens during the iteration and $a\approx b$, then$f^{(n)}(a/b)$ grows superexponential, probably close to $a^{\phi^{n-1}}$ which makes substantial cancellation more and more unlikely, especially that the numerator eventually hits by chance a perfect multiple of the denominator.
| {
"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 know more precisely about the behavior of $S_n$ as a function in $n$?
In this case, apparently, the Riemann sum method does not work. Indeed, if we rewrite $S_n$ as follows
\begin{align}
S_n& = \frac{1}{n^2} \sum_{k=0}^{n-1} \dfrac{(\frac{k}{n})^2}{\left(1-\left(\frac{k}{n}\right)^2\right)^2} \\
& \leq \frac{1}{n}\int_0^1 \frac{x^2}{(1-x^2)^2} dx
\end{align}
Then, we will get stuck with a divergent integral which blows up near $1$.
| 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
$$\eqalign{S_n &= 1/24\,{\pi }^{2}-1/4\,{\frac {\gamma}{n}}-1/4\,\Psi \left( 1,2\,n
\right) -1/4\,{\frac {\Psi \left( 2\,n \right) }{n}}-1/4\,\Psi
\left( 1,n+1 \right) -1/4\,{\frac {\Psi \left( n+1 \right) }{n}}+1/4
\,\Psi \left( 1,n \right) +1/4\,{\frac {\Psi \left( n \right) }{n}}\cr
&= 1/24\,{\pi }^{2}+{\frac {-\gamma/4-1/8-1/4\,\ln \left( 2 \right) -1/4
\,\ln \left( n \right) }{n}}+1/32\,{n}^{-2}+{\frac {1}{7680\,{n}^{5}}
}-{\frac {1}{32256\,{n}^{7}}}+O \left( {n}^{-8} \right)
\cr}$$
EDIT: Letting $j = n-k$, the sum becomes
$$ S_n = \sum_{j=1}^n \dfrac{(n-j)^2}{j^2 (2n-j)^2} = \sum_{j=1}^n \left(\dfrac{1}{4j^2} - \dfrac{4n-3j}{4j(2n-j)^2}\right) < \sum_{j=1}^n \dfrac{1}{4j^2}$$
so we immediately have $S_n < \frac{1}{4} \sum_{j=1}^\infty \frac{1}{j^2} = \frac{\pi^2}{24}$. On the other hand, since
$$ \dfrac{4n-3j}{4j(2n-j)^2} < \dfrac{1}{4jn}$$
we get $$S_n > \dfrac{1}{4} \sum_{j=1}^n \dfrac{1}{j^2} - \dfrac{1}{4n} \sum_{j=1}^n \dfrac{1}{j} = \dfrac{\pi^2}{24} - \dfrac{\ln (n)}{4 n} + O(1/n)$$
| {
"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:
$$
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
10 & 11 & 12 \\
\end{array}
\right) \;.
$$
Now perfectly shuffle them by selecting the top card from stack $1$,
the top card from stack $2$, and so on,
walking down the columns of the matrix above,
resulting in this shuffled deck of cards:
$$
(1,4,7,10,2,5,8,11,3,6,9,12) \;.
$$
Continue in this manner until the deck of cards returns to its initial sorting:
$$
\left(
\begin{array}{cccccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
1 & 4 & 7 & 10 & 2 & 5 & 8 & 11 & 3 & 6 & 9 & 12 \\
1 & 10 & 8 & 6 & 4 & 2 & 11 & 9 & 7 & 5 & 3 & 12 \\
1 & 6 & 11 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 12 \\
1 & 5 & 9 & 2 & 6 & 10 & 3 & 7 & 11 & 4 & 8 & 12 \\
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\end{array}
\right) \;.
$$
Here, for $n{=}12$ cards partitioned into $k{=}4$ parts, it requires $s{=}5$
perfect shuffles to
cycle.
Let us say that $f(n,k)=s$, i.e., $f(12,4)=5$.
Similarly I can calculate that
$$
f(8,2)=3,\; f(18,3)=16,\; f(33,3)=8, \; f(52,2)=8,
$$
etc. The last represents a perfect "outer-shuffle" of a standard
$52$-card deck, which is known to take $8$ shuffles to cycle.
It seems likely this function is known to combinatorialists:
Q. What is $f(n,k)$?
| 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 original configuration after the order of $m \pmod{n-1}$ times. Since $mk\equiv 1\pmod{n-1}$, this is also the order of $k \pmod{n-1}$. For more information on perfect shuffles see Diaconis, Graham, Kantor.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/187771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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:
$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2, (b^{-1}ab^{-1})^2=a^{-2}b, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
Are these groups torsion-free?
Motivation: In both of these groups $1+a+b$ as an element of the group algebra $\mathbb{F}_2[G_i]$ over the field with two elements is a zero divisor. Thus one has a counterexample for the Kaplansky zero divisor conjecture if one of $G_i$s is torsion-free!
$$(1+a+b)(b^{-1}a^{-2}ba^{-1}+a^{-1}ba^{-2}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+a^{-1}b^2a^{-1}ba^{-1}+aba^{-1}ba^{-1}+1+a^{-2}ba^{-1}+a^{-1}b^{-1}a^{-1}+b+baba^{-1}ba^{-1}+ba^{-1}ba^{-1}+a^{-1}+b^{-1}a)=0$$
$$(1+a+b)(aba^{-1}ba^{-1}+a^{-1}b^2a^{-1}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+b^{-1}a^{-2}ba^{-1}+ab+1+ba^{-1}ba^{-1}+baba^{-1}ba^{-1}+b+bab+a^{-2
}ba^{-1}+a^{-1}+b^{-1}a)=0$$
| 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",
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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 $n$ in that range. More than half were primes, while the divisibility by prime $p$ of the unsolvable composites are given by the table below. Oleg noticed a curious behavior:
\begin{array}{|l|c|c|c|c|}
\hline
p & \mbox{# divisible by } p & ...\; p^2 & ...\;p^3 \\
\hline
2 & \color{blue}{177} & 41 & 22\\
3 & 31 &-&-\\
5 & 53 &-&-\\
7 & \color{blue}{218} & 52 & 11 \\
11 & 43 &-&-\\
13 & \color{blue}{300} & 46 & 4 \\
17 & 47 &-&-\\
23 & 31 &-&-\\
31 & 34 &-&-\\
37 & \color{blue}{82} &-&-\\
43 & 39 &-&-\\
47 & 29 &-&-\\
59 & 24 &1&-\\
61 & 13 &-&-\\
73 & 16 &-&-\\
79 & 48 &4&-\\
\hline
\end{array}
Thus, in that range, there are just $31$ unsolvables of form,
$$a^3+b^3+c^3 = (3m)^3$$
but $218$ unsolvables,
$$a^3+b^3+c^3 = (7m)^3$$
and $300$,
$$a^3+b^3+c^3 = (13m)^3$$
Q: What could explain this "anomalous" behavior of $p=2^k,7^k,13^k$ for $k=1,2,3$ with respect to the cubic Diophantine equation $(1)$?
| 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).
Although the same property applies to $2, 3$ and $7$, its effect on (1) for $13$, with $n^3 = (13m)^3$ is especially restrictive in the following sense. If integers are randomly assigned to $a,b,c$ then the probability $P_0$ that none of them are divisible by $13$ is:
$$P_0 = (12/13)^3 = 1728/2197$$
The probability $P_2$ that exactly two are divisible by $13$ is:
$$P_2 = 3(1/13)^2(12/13) = 36/2197$$
So $1764/2197 ≈ 80\%$ of random assignments fail to meet the above condition. The corresponding percentages for $2, 3, 7$ are respectively $50\%$, $52\%$ and $68\%$.
The much smaller number of cases for $3$ than for $2,7,13$ is partly explained by the fact that there are solutions of (1) with $n = 6,9$, which eliminates all cases other than those of the form $18k \pm 3$, and in particular all of the form $3^2k$.
Reference
A) Bailey A (2009) Some Divisibility Properties of Cubic Quadruples, Mathematical Gazette 488 Nov 2009, Note 93.47; doi: 10.1017/S0025557200185262, jstor.
| {
"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 an ellipsoid with semiaxes $s,s,1/s^2$ and this almost completes the proof of the uniqueness hypothesis of an ellipsoid with given intrinsic volumes.
| 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 $R=\frac{1}{2}(\cos^2(\phi)\frac{1}{s}+\sin^2(\phi)s^2)$ then
\begin{equation*}
\left. \int_{0}^{\infty}
r^3 \cdot e^{-r^2R}dr= -\frac{1}{2R}\int_{0}^{\infty}
r^2\cdot de^{-r^2R}=-\frac{1}{2R}r^2e^{-r^2R}\right|_{0}^{\infty}+\frac{1}{R}\int_{0}^{\infty}
re^{-r^2R}dr=
\end{equation*}
\begin{equation*}
=\frac{1}{R}\int_{0}^{\infty}re^{-r^2R}dr=\left.-\frac{e^{-r^2R}}{2R^2}\right|_0^{\infty}=\frac{1}{2R^2}
\end{equation*}
Back to the main integral
\begin{equation*}
\int_0^{2\pi} \frac{\cos(\phi)}{R^2}d\phi=\int_0^{2\pi} \frac{\cos(\phi)}{(\frac{1}{2}\cos^2(\phi)\frac{1}{s}+\frac{1}{2}\sin^2(\phi)s^2)^2}d\phi \propto
\end{equation*}
\begin{equation*}
\propto s^2\int_0^{2\pi} \frac{\cos(\phi)}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\phi=
\end{equation*}
\begin{equation*}
=s^2\int_0^{2\pi} \frac{1}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\sin(\phi)=s^2\int_{-1}^1 \frac{1}{(1-t^2+s^3t^2)^2}dt
\end{equation*}
It is easy to chek that
\begin{equation*}
\int\frac{1}{(1+x^2a)^2}dx=\frac{x}{2(ax^2+1)}+\frac{\arctan(\sqrt{a})}{2\sqrt{a}},
\end{equation*}
So we finaly obtain for $a=s^3-1$ when $F(s)$ - is the main width of spheroid with semiaxes $\sqrt{s},\sqrt{s},\frac{1}{s}$.
\begin{equation}
F(s)=Const\cdot s^2\left(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right)=Const\cdot \left(\frac{1}{s}+\frac{s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right)
\end{equation}
And fact that $F(s)$ decrease when $s\in(0,1)$ and increase when $s\in(1,\infty)$ means that there are only two spheroids with volume$=1$ and mean width$=W>F(1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"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 {$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}, 4, 4+\frac{1}{4}, 4+\frac{2}{4}, 4+\frac{3}{4}, . . .$} and so forth. It seems like it should be easy but I can't seem to find a solution. Any suggestions?
| The sequence $a_n$ for $n\geq 1$ has the following formula:
$$a_n=\left\lfloor \sqrt{2n}+\tfrac{1}{2}\right\rfloor +\frac{\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor-\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor ^2 +2 n}{2 \left\lfloor \sqrt{2n}+\frac{1}{2}\right\rfloor }.$$
Here is the Wolfram Alpha link to check it.
It is related to OEIS A002024 and OEIS A002262.
| {
"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} $ is zero for any $n \in \mathbb{N}^{+}$.
In fact, this is my conjecture and I have tested many examples in my computer. For diagonal case, it is easy to prove it, but for the general case I do not know how to do it. One idea come to my mind is to write $A$ as the sum of a diagonal matrix and an anti-diagonal matrix, then expand $A^r$ by binomial expansion, but unfortunately they do not commute.
Could someone give me some hints?
Thanks!
| 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
$$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and for}\;\;D=\mathbf c\mathbf b^{\rm T}=\begin{pmatrix}1&-1\\1&-1\end{pmatrix},$$
I calculate
$$J(r,n)={\rm tr}\,\sigma A^r DA^{n-1-r}=$$
$$=\frac{2^{-n-1} (a+c-z)^{-r} (a+c+z)^{-r}}{(a-2 b-c) \left(a c+b^2\right)} \left[\left(z (a+c)-(a-c)^2+4 b^2\right) (a+c+z)^n (a+c-z)^{2 r}-\left(z (a+c)+(a-c)^2-4 b^2\right) (a+c-z)^n (a+c+z)^{2 r}\right],$$
with the definition $z=\sqrt{(a-c)^2-4 b^2}$.
Then I evaluate for $n\geq 1$ the sum
$$\sum_{r=0}^{n-1}J(r,n)=\frac{2^{-n-1} (a+c) \left((a-c)^2-4 b^2-z^2\right) \left((a+c-z)^n-(a+c+z)^n\right)}{z (a-2 b-c) \left(a c+b^2\right)}.$$
Substitution of the definition of $z$ finally gives the desired result
$$\sum_{r=0}^{n-1}J(r,n)=0.$$
Details of the calculation: I may assume $b\neq 0$ (otherwise $A$ is diagonal and the identity follows trivially). Then the matrix $A$ is diagonalizable when $b\neq \tfrac{1}{2}|a-c|$, in the form
$A=U\Lambda U^{-1}$ with $$U=\left(
\begin{array}{cc}
z-a+c & -z-a+c \\
2 b & 2 b \\
\end{array}
\right),\;\;\Lambda={\rm diag}\,\left(\tfrac{1}{2} \left(-z+a+c\right),\tfrac{1}{2} \left(z+a+c\right)\right)$$
With this decomposition we can readily evaluate $A^r=U\Lambda^r U^{-1}$.
If $b=\tfrac{1}{2}(a-c)\neq 0$ we instead use the Jordan decomposition $A=VJV^{-1}$ with
$$V=\left(
\begin{array}{cc}
-1 & -\frac{2}{a-c} \\
1 & 0 \\
\end{array}
\right),\;\;J=\left(
\begin{array}{cc}
\frac{a+c}{2} & 1 \\
0 & \frac{a+c}{2} \\
\end{array}
\right).$$
Then $A^r=VJ^r V^{-1}$, with $J^r=2^{-r} (a+c)^r\left(
\begin{array}{cc}
1 & 2 r (a+c)^{-1} \\
0 & 1 \\
\end{array}
\right)$.
| {
"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
$$\sum_{\vert \alpha \vert=j}(R_{\theta}z)^{\alpha} \partial^{\alpha}f(x) = \sum_{\vert \alpha \vert=j}(z)^{\alpha} (R_{-\theta}\partial)^{\alpha}f(x)$$
It is true for $j=1$, which just follows since $R_{-\theta}$ is the adjoint map of $R_{\theta}.$
| 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 $j = 2$, the LHS is:
\begin{align*}
\text{LHS} &=
(u\cos\theta - v\sin\theta)^2 \frac{\partial^2 f}{\partial x^2} \\
&\hspace{1cm} + (u\cos\theta - v\sin\theta) (u\sin\theta + v\cos\theta) \frac{\partial^2 f}{\partial x \partial y} \\
&\hspace{1cm} + (u\sin\theta + v\cos\theta)^2 \frac{\partial^2 f}{\partial y^2}
\end{align*}
and the RHS is:
\begin{align*}
\text{RHS} &=
u^2 \left(
\cos^2\theta \frac{\partial^2 f}{\partial x^2}
+ 2\sin\theta\cos\theta \frac{\partial^2 f}{\partial x \partial y}
+ \sin^2\theta \frac{\partial^2 f}{\partial y^2}
\right) \\
&\hspace{1cm} + uv \left(
-\sin\theta \cos\theta \frac{\partial^2 f}{\partial x^2}
+ (\cos^2\theta - \sin^2\theta) \frac{\partial^2 f}{\partial x \partial y}
+\sin\theta \cos\theta \frac{\partial^2 f}{\partial y^2}
\right) \\
&\hspace{1cm} + v^2 \left(
\sin^2\theta \frac{\partial^2 f}{\partial x^2}
- 2\sin\theta \cos\theta \frac{\partial^2 f}{\partial x \partial y}
+ \cos^2\theta \frac{\partial^2 f}{\partial y^2}
\right)
\end{align*}
If you fully expand out the products, you will find that these two quantities differ by:
$$ \text{LHS} - \text{RHS} = \sin\theta\cos\theta \left(
(v^2-u^2) \frac{\partial^2 f}{\partial x \partial y}
- u v \left(\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2}\right)
\right) $$
This is certainly not identically zero, so $\text{LHS} \ne \text{RHS}$ in general.
| {
"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 Legendre's (Kummer) formula $\nu_2(C_n)=s(n+1)-1$, where $s(n)$ is the sum of the binary digits of $n$.
*See E. Deutsch and B. Sagan, Congruences for Catalan and Motzkin numbers and related sequences.
*See A. Postnikov and B. Sagan, What power of two divides a weighted Catalan number?
*See O. Egecioglu, The parity of the Catalan number via lattice paths.
QUESTION. Do you know of or can provide a new proof that $C_n$ is odd if and only if $n=2^r-1$?
| 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
$$
F = \sum_{n=0}^\infty C_n x^{n+1} = x + x^2 + 2 x^3 + 5 x^4 + 14 x^5 + 42 x^6 + 132 x^7 + 429 x^8 + \cdots.
$$
Then $F = x + F^2$. Instead of solving this quadratic equation,
apply it recursively ${} \bmod 2$. Recall that for each $r=1,2,3,\ldots$
we have the congruence $(a+b)^{2^r} \equiv a^{2^r} + b^{2^r} \bmod 2$
(proof: induction, the case $r=1$ being
$(a+b)^2 = a^2 + 2ab + b^2 \equiv a^2+b^2$). We obtain:
$$
\begin{array}{rl}
F \; = \!\! & x + F^2 \cr
= \!\! & x + (x+F^2)^2 \equiv x + x^2 + F^4 \cr
= \!\! & x + x^2 + (x+F^2)^4 \equiv x + x^2 + x^4 + F^8 \cr
= \!\! & x + x^2 + x^4 + (x+F^2)^8 \equiv x + x^2 + x^4 + x^8 + F^{16} \cr
= \!\! & \cdots \cr
\equiv \!\! & x + x^2 + x^4 + x^8 + x^{16} + x^{32} + x^{64} + \cdots \cr
\end{array}
$$
so indeed $C_n$ is odd if and only if $n+1$ is a power of $2$, QED.
| {
"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:
$$
\begin{aligned}
\text{mode of x} = \arg \min_s \sum_i \lvert x_i - s \rvert^0 \\
\text{median of x} = \arg \min_s \sum_i \lvert x_i - s \rvert^1 \\
\text{mean of x} = \arg \min_s \sum_i \lvert x_i - s \rvert^2 \\
\end{aligned}
$$
Where:
$$
\begin{aligned}
x = (x_1, x_2, \ldots, x_n) \ &:\ \text{a list of rational numbers} \\
\lvert x \rvert \ &:\ \text{number of elements in the list if $x$ is a list} \\
\end{aligned}
$$
I wanted to extend this to the L4 center:
$$
\begin{aligned}
\arg \min_s \sum_i \lvert x_i - s \rvert^4 = \text{L4 center} \\
\frac{d}{ds} \sum_i (x_i - s)^4 = 0 \\
\sum_i \left[ \frac{d}{ds}(x_i-s)^4 \right] = 0 \\
\sum_i \left[ -4(x_i-s)^3 \right] = 0 \\
\sum_i (x_i-s)^3 = 0 \\
\sum_i \left[ x_i^3 -3x_i^2s + 3x_is^2 - s^3 \right] = 0 \\
\sum_i x_i^3 + \sum_i -3x_i^2s + \sum_i 3x_is^2 + \sum_i -s^3 = 0 \\
\sum_i x_i^3 -3s \sum_i x_i^2 + 3s^2 \sum_i x_i - s^3 \lvert x \rvert = 0 \\
\end{aligned}
$$
Which is a cubic with the following discriminant:
$$
\begin{aligned}
&+ 162 \lvert x \rvert \sum_i x_i \sum_i x_i^2 \sum_i x_i^3 \\
&- 108 \left(\sum_i x_i\right)^3 \sum_i x_i^3 \\
&+ 18 \left(\sum_i x_i\right)^2 \left(\sum_i x_i^2\right)^2 \\
&- 108 \lvert x \rvert \left(\sum_i x_i^2\right)^3 \\
&- 27 \left(\lvert x \rvert\right)^2 \left(\sum_i x_i^3\right)^2
\end{aligned}
$$
I was wondering if anything could be said about the roots of the L4 center given any list $x$ of rationals? For example, how many real roots, because intuitively I expect only one?
| 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 question was deleted just as I was about to post the complete answer. Since it seems a shame to waste the effort, I record the question along with the answer here.
| 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^{\dagger } -a)/\sqrt{2} $. Acting on the $\psi_{n} $, these act as $a\psi_{n} = \sqrt{n} \psi_{n-1} $, $a^{\dagger } \psi_{n} = \sqrt{n+1} \psi_{n+1} $. Now, $H$ takes the form
$$
H=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & 0 & i(a^{\dagger } -a) &\sqrt{2} \bar{z} \\ 0 & a^{\dagger } +a & \sqrt{2} z & -i(a^{\dagger } -a) \\ i(a^{\dagger } -a) &\sqrt{2} \bar{z} & -(a^{\dagger } +a ) & 0 \\ \sqrt{2} z & -i(a^{\dagger } -a) & 0 & -(a^{\dagger } +a) \end{pmatrix}
$$
One can readily verify that, in the following basis, which couples the $n$-th and $n+1$-th harmonic oscillator eigenfunctions,
$$
e_1^n = \begin{pmatrix} \psi_{n} + \psi_{n+1} \\ 0 \\ -i(\psi_{n} - \psi_{n+1} )\\ 0 \end{pmatrix} \ \ \ \ \
e_2^n = \begin{pmatrix} \psi_{n} - \psi_{n+1} \\ 0 \\ -i(\psi_{n} + \psi_{n+1} )\\ 0 \end{pmatrix}
$$
$$
e_3^n = \begin{pmatrix} 0 \\ \psi_{n} + \psi_{n+1} \\ 0 \\ i(\psi_{n} - \psi_{n+1} ) \end{pmatrix} \ \ \ \ \
e_4^n = \begin{pmatrix} 0 \\ \psi_{n} - \psi_{n+1} \\ 0 \\ i(\psi_{n} + \psi_{n+1} ) \end{pmatrix}
$$
$H$ becomes block-diagonal, with a separate $4\times 4$ block for each integer $n\ge 0$ (in addition, at the lower end of the spectrum, this is supplemented by one more two-dimensional subspace involving only $\psi_{0} $, given further below); these $4\times 4$ blocks have the form
$$
H_n = \begin{pmatrix} \sqrt{2n+2} & 0 & 0 & i\bar{z} \\
0 & -\sqrt{2n+2} & i\bar{z} & 0 \\0 & -iz & \sqrt{2n+2} & 0 \\ -iz & 0 & 0 & -\sqrt{2n+2}
\end{pmatrix}
$$
They are readily diagonalized, yielding two degenerate doublets,
$$
i (-\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_1^n + \bar{z} e_4^n \ \ \ \
\mbox{with eigenvalue} \ \ \ \ -\sqrt{2n+2+\bar{z}z}
$$
$$
i (\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_2^n + \bar{z} e_3^n \ \ \ \
\mbox{with eigenvalue} \ \ \ \ -\sqrt{2n+2+\bar{z}z}
$$
$$
-i (\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_1^n + \bar{z} e_4^n \ \ \ \
\mbox{with eigenvalue} \ \ \ \ \sqrt{2n+2+\bar{z}z}
$$
$$
-i (-\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_2^n + \bar{z} e_3^n \ \ \ \
\mbox{with eigenvalue} \ \ \ \ \sqrt{2n+2+\bar{z}z}
$$
Finally, as mentioned above, one has the additional two eigenvectors
$$
\begin{pmatrix} i\sqrt{\bar{z}z} \ \psi_{0} \\ z \ \psi_{0} \\ -\sqrt{\bar{z}z} \ \psi_{0} \\ -iz \ \psi_{0} \end{pmatrix} \ \ \ \ \mbox{with eigenvalue} \ \ \ \ -\sqrt{\bar{z} z}
$$
$$
\begin{pmatrix} -i\sqrt{\bar{z}z} \ \psi_{0} \\ z \ \psi_{0} \\ \sqrt{\bar{z}z} \ \psi_{0} \\ -iz \ \psi_{0} \end{pmatrix} \ \ \ \ \mbox{with eigenvalue} \ \ \ \ \sqrt{\bar{z} z}
$$
that become zero modes for $z=0$ (rescaling of the vectors needed in that limit).
| {
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"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_{b=1}^{r-1} c^{b} \left(A(r,b) - {r \choose b}^2 \, b\right)
\end{align*}
for the expression (1.23) from Bai and Silverstein's famous paper 'CLT for Linear Spectral Statistics of Large-Dimensional Sample Covariance Matrices', where
\begin{align*}
A(r,b) :=& \sum\limits_{m=1}^{\min(b,r-b+1)} (2m-1) {r \choose b-m}{r \choose b+m-1}\\
& + \sum\limits_{m=1}^{\min(b,r-b)} (2m+1) {r \choose b-m}{r \choose b+m} \ .
\end{align*}
In order to show that my expression is equal to that of Bai and Silverstein, I need to show the equality
$$
A(r,b) = \frac{1}{2} {2r \choose 2b} + \left( b - \frac{1}{2} \right) {r \choose b}^2
$$
for all $r \in \mathbb{N}$ and $b<r$. I am not too well versed with combinatorial identities and so far my efforts have lead nowhere. Does someone have an idea how to prove this?
Testing with the computer shows this to be true for at least all $b<r \leq 1000$.
Technically I could just use the fact that my formula and Bai and Silverstein's formula describe the same object to show this identity, which seems rather crude unless this identity is not jet known.
| Firstly, exploit the finite support to simplify the limits of the sums. Secondly, split the second sum. We get
$$\begin{align*}A(r,b) =& \sum\limits_{m=1}^{r} (2m-1) {r \choose b-m}{r \choose b+m-1} \\
& + \sum\limits_{m=1}^{r} 2m {r \choose b-m}{r \choose b+m} \\
& + \sum\limits_{m=1}^{r} {r \choose b-m}{r \choose b+m}
\end{align*}$$
The first two are mechanical using Wilf-Zeilberger, so ask your favourite CAS (e.g. Wolfram Alpha or Sage) to get
$$\begin{align*}A(r,b) =
& \frac{b(r-b+1)\binom{r}{b-1}\binom{r}{b} + (b-2r-1)(b+r)\binom{r}{b-r-1}\binom{r}{b+r}}{r} \\
& + \frac{(b+1)(r-b+1)\binom{r}{b-1}\binom{r}{b+1} + (b-2r-1)(b+r+1)\binom{r}{b-r-1}\binom{r}{b+r+1}}{r} \\
& + \sum_{m=1}^r \binom{r}{b-m} \binom{r}{b+m}
\end{align*}$$
For the third sum, we apply $$\sum_k \binom{r}{m+k} \binom{s}{n-k} = \binom{r+s}{m+n}$$ (see e.g. identity (5.22) of Concrete Mathematics by Graham, Knuth and Patashnik). So $$\sum_{m=1}^r \binom{r}{b-m} \binom{r}{b+m} = \frac{\sum_{m=-r}^r \binom{r}{b-m} \binom{r}{b+m} - \binom{r}{b}^2}{2} = \frac{\binom{2r}{2b} - \binom{r}{b}^2}{2}$$
I've left you some cancellation, which I fully expect to be routine.
| {
"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)}}{2(x+4)}\left ( \frac{x-\sqrt{x(x+4)}}{2} \right )^n.
\end{aligned}
\end{equation*}
Using Mathematic $12.3$, when $n$ is large enough, we give the distribution of the roots of $f(x)$ in the complex plane as follows
In this figure, we can see that the closure of the real roots of $f(x)$ may be $\left [ -4,0 \right ]$.
So we have the following question
Question: all roots of $f(x)$ are real? It seems yes! But we have no way of proving it.
| (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, so a recurrence is possible. That there are no positive real zeros follows from the fact that there are no negative coefficients. The rest of your question is another matter.
| {
"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 $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).
Thanks,
Jacob
| 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 und Naturwissenschaftlichen Facultät der Kaiser Wilhelms-Universität zu Strassburg, 1903), published as “Über die Verteilung der quadratischen Reste,” Journal für die Reine und Angewandte Mathematik 127 (1904): 1–19. Karpinski proved a collection of formulae (all of which assume $p > 3$) involving sums over Legendre symbols, and showed that the most concise sums possible contain only $\lfloor p/6 \rfloor$ terms:
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \sum_{k=1}^{(p-1)/2} \left( \frac{k}{p} \right) \quad (p \equiv 3 \bmod{4});
\end{equation}
\begin{equation}
\left\{ 3 - \left( \frac{3}{p} \right) \right\} h(-p) = 2 \sum_{k=1}^{\lfloor p/3 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 3 \bmod{4});
\end{equation}
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \sum_{k=\lfloor p/4 \rfloor +1}^{(p-1)/2} \left( \frac{k}{p} \right) (p \equiv 3 \bmod{8});
\end{equation}
\begin{equation}
\left\{ 2 - \left( \frac{2}{p} \right) \right\} h(-p) = \quad \sum_{k=1}^{\lfloor p/4 \rfloor} \quad \left( \frac{k}{p} \right) (p \equiv 7 \bmod{8});
\end{equation}
\begin{equation}
\left\{ 1 + \left( \frac{2}{p} \right) + \left( \frac{3}{p} \right) - \left( \frac{6}{p} \right) \right\} h(-p) = 2 \sum_{k=1}^{\lfloor p/6 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 7, 11, 23 \bmod{24});
\end{equation}
\begin{equation}
\left\{ 1 + \left( \frac{2}{p} \right) + \left( \frac{3}{p} \right) - \left( \frac{6}{p} \right) \right\} h(-p) = -2p + 2 \sum_{k=1}^{\lfloor p/6 \rfloor} \left( \frac{k}{p} \right) \quad (p \equiv 19 \bmod{24}).
\end{equation}
| {
"language": "en",
"url": "https://mathoverflow.net/questions/16141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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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 $d_0=d_1=d_2=1$. How to show $d_i>d_{i+1}$ for all $i\ge3$?
By easy calculation, $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}$. For a recursion of 2 or 3 terms, it is easy to proceed with induction, but what about 4 terms in this case?
| 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 $6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+\frac{p-1}{p}[3p^2(i+1+2p)d_{i+1}+(2p-1)(i+3p)pd_i]$< $6p^2(i+2+p)d_{i+2}+\frac{p-1}{p} \{6p^2(i+1+p)d_{i+1}+(p-1)[3p(i+2p)+(2p-1)(i-1+3p)]d_i \}\le$ $6p^2(i+2+p)d_{i+2}+ \frac{p-1}{p} 6p^3(i+2)d_{i+2}=6p^3(i+3)d_{i+2}$, this shows $d_{i+3}< d_{i+2}$.
Moreover, $6p^3(i+3)d_{i+3}=6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_i>$ $6p^2(i+2+p)d_{i+2}+3p(p-1)(i+1+2p)d_{i+1}+(p-1)(2p-1)(i+3p)d_{i+1}=$ $6p^2(i+2+p)d_{i+2}+(p-1)[3p(i+1+2p)+(2p-1)(i+3p)]d_{i+1}$. This shows $6p^3(i+3)d_{i+3}\ge 6p^2(i+2+p)d_{i+2}+(p-1)[3p(i+1+2p)+(2p-1)(i+3p)]d_{i+1} $. Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 nor reference. It is a particular case of the question in the title: if $a$ and $b$ are quadratic residues modulo $p$, when is $a+b$ also a quadratic residue modulo $p$?
I came into this question when counting the number of diophantine $2$-tuples modulo $p$, that is, the number of pairs $\{ a,b\}\subset \mathbb{Z}^*_p$ such that $ab+1$ is a quadratic residue modulo $p$.
| 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
$$N(p)=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab(a+b),p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$
$$=\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right)\left(1+\left(\frac{a+b}{p}\right)\right)$$
$$-\frac{1}{8}\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1,p|a+b}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{b}{p}\right)\right).$$
Clearly, the second term is just
\begin{align*}&-\frac{1}{8}\sum_{\substack{a\bmod p\\(a,p)=1}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{-a}{p}\right)\right)=\frac{1}{8}-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right).\end{align*}
And for the first term, we are required to investigate the quantity
\begin{align*}L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ab(a+b)}{p}\right).\end{align*}
In fact we have
$$L:=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{ba^2+b^2a}{p}\right)
=\mathop{\sum\sum}_{\substack{a,b\bmod p\\(ab,p)=1}}\left(\frac{b(a+\overline{2}b)^2-\overline{4}b^3}{p}\right)$$
$$=\mathop{\sum\sum}_{a,b\bmod p}\left(\frac{ba^2-\overline{4}b^3}{p}\right)$$
$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-\overline{4}b^2}{p}\right)$$
$$=\sum_{b\bmod p}\left(\frac{b}{p}\right)\sum_{a\bmod p}\left(\frac{a^2-1}{p}\right)=0.$$
The other terms could be computed in a similar way. Hence we can deduce that
\begin{align*}N(p)=\frac{1}{8}(p-1)^2-\frac{p}{8}\left(1+\left(\frac{-1}{p}\right)\right)+\frac{1}{8}.\end{align*}
| {
"language": "en",
"url": "https://mathoverflow.net/questions/65183",
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"source": "stackexchange",
"question_score": "13",
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} |
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 tiny positive number, and put $c:=\ln(10)$, an important constant. We have
$$10^x=1+cx+O(x^2)$$
$$10^{10^x}=10^{1+cx+O(x^2)}=10+10c^2x+O(x^2)$$
$$10^{10^{10^x}}=10^{10+10c^2x+O(x^2)}=10^{10}+10^{11}c^3x+O(x^2)$$
$$10^{10^{10^{10^x}}}=10^{10^{10}+10^{11}c^3x+O(x^2)}=10^{10^{10}}+10^{10^{10}}10^{11}c^4x+O(x^2),$$
where $O(x^2)$ means something tiny all the way.
In the last expression we have $10^{10^{10}}10^{11}c^4x=10^{11}c^4$, which justifies my claim.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
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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 $A(t)$ are primitive, in that each has only one eigenvalue at $1$, and the other eigenvalues are $< 1$. (The matrices come from graph Laplacians.)
I'm interested in number of iterations until $x(t)$ converges to a stable point where all its entries are equal to a constant. This convergence is closely related to the product of matrices $\prod_t A(t)$, for $t = 1, 2, \ldots$.
My goal is to upper bound the convergence rate of the $x$'s as a function of the minimum second eigenvalue $\min_t \lambda_2(t)$ over all the matrices $A(t)$. This dependence seems to be a known result in the literature, but I cannot find a reference to it or a simple way of proving it.
Thanks!
| 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 doubly stochastic.
The product $\mathbf{A}\mathbf{B} \ = \ \mathbf{B}\mathbf{A}$ of the matrices
\begin{equation}
\mathbf{A} \ := \
\begin{pmatrix}
\frac{1}{3} + a^2 & \frac{1}{3} - a^2 & \frac{1}{3} \\
\frac{1}{3} - a^2 & \frac{1}{3} + a^2 & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\
\end{pmatrix}
\end{equation}
and
\begin{equation}
\mathbf{B} \ := \
\begin{pmatrix}
\frac{1}{3} + \frac{1}{4} b^2 & \frac{1}{3} + \frac{1}{4} b^2 & \frac{1}{3} - \frac{1}{2} b^2 \\
\frac{1}{3} + \frac{1}{4} b^2 & \frac{1}{3} + \frac{1}{4} b^2 & \frac{1}{3} - \frac{1}{2} b^2 \\
\frac{1}{3} - \frac{1}{2} b^2 & \frac{1}{3} - \frac{1}{2} b^2 & \frac{1}{3} + b^2 \\
\
\end{pmatrix}
\end{equation}
is
\begin{equation}
\mathbf{E} \ := \
\begin{pmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\
\end{pmatrix} \, ,
\end{equation}
so whatever the values of $a$ and $b$ and independently of the initial conditions as long as all later matrices have row and column sums equal to 1, the process converges after just two iterations.
The matrices were constructed in the following way: Take an orthogonal basis of $\mathbb{R}^3$ containing the vector $(1,1,1)$, here $x = (1,1,1); y=(1,-1,0); z=(1/2,1/2,-1)$ and construct the matrices $A$ and $B$ as $k_x*x^tx + k_y*y^ty + k_z*z^tz$ by choosing the constants $k_x, k_y,k_z$ appropriately.
The eigenvalues of $A$ are $1; 2a^2; 0$, those of $B$ are $1; 1.5b^2; 0$, but while $A$ projects onto the orthogonal complement of $(1/2,1/2,-1)$, $B$ projects onto the orthogonal complement of $(1,-1,0)$, leaving only $k(1,1,1)^{(t)}$ afterwards, where $k$ is constant.
This is clearly a more general phenomenon, so without rather specific additional assumptions it will be very difficult/essentially impossible to progress, as the above result is independent of the second eigenvalues and there is no upper bound on the convergence rate in terms of them.
One might think about two things here:
1) There should be a lower bound on the convergence rate in terms of the supremum of the second eigenvalues if this is bounded away from 1 as in the time independent case because projecting will not increase the norm of the image.
2) If all the matrices $A(t)$ are invertible and their eigenvalue $\lambda_{small}$ of smallest modulus is bounded away from zero, there should be an estimate of the rate of convergence from above involving $\inf (|\lambda_{small}|)$.
Concerning the literature on Markov chains I can not really help you, but you may wish to consult
"Non-negative Matrices and Markov Chains" by E. Seneta, Springer, reprint 2006.
The example I just constructed from scratch.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $a=kp+1$, for some $k>0$.
Could you help me about this question? If it is true how we can prove it?
Also when I checked the same fact for $p^8+1$, we get many counterexamples. What is the difference between $4$ and $8$?
| 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 the contrary. As we have just seen, $mn+1$ divides $m^4+1$ as well. Write $(m^4+1)/(mn+1)=k$, and $k=(\ell m+r)$ with $0\le r\le m-1$. Then $m^4+1=(\ell m+r)((mn+1)=mN'+r$, so $r=1$. So $(\ell m+1)$ divides $m^4+1$. Then $m$ and $\ell$ are coprime: indeed, we have $(m\ell +1)(mn+1)=m^4+1$, so $\ell(mn+1)-m^3=-n$. If a prime $p$ were dividing both $m$ and $\ell$ then it would also divide $n$, contradicting that $m$ and $n$ are coprime, so $m$ and $\ell$ are indeed coprime. We have $\ell<n$ because otherwise
$$m^4+1=(mn+1)(m\ell+1)\ge (mn+1)^2> (m^2+1)^2>m^4+1.$$ So we found a new pair $(m,\ell)$ with $\max(m,\ell)<n$, with $m\ell+1$ dividing both $m^4+1$ and $\ell^4+1$. So, assuming that $n$ is minimal, we're done unless $\ell=1$. This happens if $m+1$ divides $m^4+1$, and since $m+1$ divides $m^4-1$ as well, if this occurs then $m+1=2$, contradicting $m>1$.
(Note: without the coprime assumption the conclusion fails, as $(m,n)=(m,m^3)$ for $m\ge 2$, e.g. $(m,n)=(2,8)$, satisfies $mn+1|n^4+1$.)
Proposition If $p$ is prime then $p^4+1$ has no divisor of the form $kp+1$ except $1$ and $p^4+1$.
Proof: write $p^4+1=(kp+1)(k'p+\ell)$ with $0\le\ell\le p-1$; then $\ell=1$. So exchanging $k$ and $k'$ if necessary we can suppose $k\le k'$.
If by contradiction $k$ is divisible by $p$ then $k'\ge k\ge p$, so $p^4+1\ge (p^2+1)^2>p^4+1$, contradiction. So $k$ and $p$ are coprime, and the lemma yields a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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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, for a fixed integer $n$, if the system has an infinite number of co-prime integer solutions, then $n$ is a multiple of $3$?
Method: Eqs $(2)$ and $(3)$ can easily be given a complete solution. Incorporating $(1)$, I got,
$$(-2 p + \alpha q -\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (-2 p + \alpha q +\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag5$$
where $\alpha = n+1,\;\beta = n-1$. It is also true for $k=6$ if there is $p,q,n$ such that,
$$Poly_1:= (-3+n)(5-2n+n^2)p + 4n(1+n^2)q$$
$$Poly_2:= (-3+n)(5-2n+n^2)p^3 + 2(5+11n-5n^2+n^3)p^2q - (5+7n+15n^2-3n^3)pq^2 + 4n(1+n^2)q^3$$
and,
$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag6$$
A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,
$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}$$
Example: Let $n=12$, then,
$$((-2 p + 13 q - 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\(-2 p + 13 q + 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag7$$
which is already true for $k=1,2$. But it is also for $k=6$ if,
$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag8$$
An initial point is $p,q = -104,17.$ Hence $(8)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions to $(7)$.
Question 2. For what other positive integer $n$ below a bound can we find solutions to non-zero $(6)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)
| 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
\end{equation*}
where
\begin{equation*}
h_1=(n^2+3) \hspace{1cm} h_2=4(n^2+1)(n^2+2n+5)(n^2-2n+5)
\end{equation*}
with the reverse transformation
\begin{equation*}
\frac{p}{q}=\frac{2n\big(G+(n^2+11)H+5h_2\big)}{(3-n)\big(G+(3n^2+4n+5)H+5h_2\big)}
\end{equation*}
The elliptic curves have torsion points when $H=0$, and the torsion subgroup seems to be $\mathbb{Z}3$.
Applying the Birch and swinnerton-Dyer conjecture to these curves, the first positive rank is at $n=12$ as you state, but the second is at $n=13$, where
$H=-608000/9$ gives a rational point. Substituting backwards we arrive at the solution to the initial problem for $n=13$: $a=-1181$, $b=-6033$, $c=6302$, $d=6805$, $e=-4702$ and $f=-3015$.
I computed height estimates for $n$ up to $100$. There seems no obvious pattern and some of the heights are very large, suggesting a solution but one with enormous numbers. Since the elliptic curves do not have a point of order $2$, finding these solutions will not be easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2, (b^{-1}ab^{-1})^2=a^{-2}b, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
Are these groups torsion-free?
Motivation: In both of these groups $1+a+b$ as an element of the group algebra $\mathbb{F}_2[G_i]$ over the field with two elements is a zero divisor. Thus one has a counterexample for the Kaplansky zero divisor conjecture if one of $G_i$s is torsion-free!
$$(1+a+b)(b^{-1}a^{-2}ba^{-1}+a^{-1}ba^{-2}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+a^{-1}b^2a^{-1}ba^{-1}+aba^{-1}ba^{-1}+1+a^{-2}ba^{-1}+a^{-1}b^{-1}a^{-1}+b+baba^{-1}ba^{-1}+ba^{-1}ba^{-1}+a^{-1}+b^{-1}a)=0$$
$$(1+a+b)(aba^{-1}ba^{-1}+a^{-1}b^2a^{-1}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+b^{-1}a^{-2}ba^{-1}+ab+1+ba^{-1}ba^{-1}+baba^{-1}ba^{-1}+b+bab+a^{-2
}ba^{-1}+a^{-1}+b^{-1}a)=0$$
| 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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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
$$\displaystyle \left \lvert \frac{p}{q} - \sqrt{d} \right \rvert > \frac{c_d}{q^2}$$
holds. The existence of such a number follows from Lagrange's theorem that the continued fraction expansion of a quadratic irrational is eventually periodic.
It follows from Hurwitz's theorem that $c_d < \dfrac{1}{\sqrt{5}}$ for all $d$.
Does anyone have a reference for the size of $c_d$? One can assume that the fundamental solution $(u_0, v_0)$ to the equation $x^2 - dy^2 = \pm 4$ is known.
| 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 $a$ being the second largest, and the rest of the period is a palindrome. I believe these classical results are in Hardy and Wright, An Introduction to the Theory of Numbers.
We can represent $\sqrt{d}$ as a terminating simple continued fraction if we allow the last coefficient to be irrational, containing the rest of the simple continued fraction, instead of an integer.
$\alpha = [a_0;a_1,...,a_n,...] = [a_0;a_1,...,a_n,[a_{n+1};a_{n+2},...]]$. Let $a_{n+1}' = [a_{n+1};a_{n+2},...]$.
If $[a_0;a_1,...,a_n] = p_n/q_n$ and then $\alpha = \frac{a_{n+1}'p_n + p_{n-1}}{a_{n+1}'q_n + q_{n-1}}$ so $$\bigg|\alpha - \frac{p_n}{q_n}\bigg| = \frac{1}{q_n(a_{n+1}'q_n + q_{n-1})} = \frac{1}{q_n^2} \frac{1}{a_{n+1}' + \frac{q_{n-1}}{q_n}}$$
So, we want to determine the liminf of $\frac{1}{a_{n+1}' + q_{n-1}/q_n}$
For $\alpha = \sqrt{a^2+b}$, or equivalently, the limsup of the reciprocal $a_{n+1}' + q_{n-1}/q_n$. The only values greater than $2a$ occur just before the period ends, when the next coefficient is $2a$, where $a_{n+1}' = a+\sqrt{a^2+b}$. The continued fraction of $q_{n-1}/q_n$ is $[0;a_n,a_{n-1},...,a_1]$ so the limit is $\sqrt{a^2+b}-a$, which means $$\frac{1}{a_{n+1}' + q_{n-1}/q_n} \to \frac{1}{2\sqrt{a^2+b}} = \frac{1}{2\sqrt{d}} = c_d.$$
If we want the infimum instead of the liminf, we get slightly lower values. Then $$c_d = \frac{1}{\frac{p}{q} + \sqrt{d}}$$
where (p,q) is the smallest nontrivial solution to $p^2-dq^2=1$. This is a convergent to the simple continued fraction just before the end of the first period if the period is even, and just before the end of the second period if the period is odd.
For example, $\sqrt{3} = [1; \overline{1,2}]$ has an even period length, so $[1;1]=2 \gt \sqrt{3}$. Then $c_3=1^2|2-\sqrt{3}| = \frac{1}{2+\sqrt{3}} \lt \frac{1}{2\sqrt{3}}$. $\sqrt{2} = [1;\overline{2}]$, so $p/q = [1;2] = 3/2 \gt \sqrt{2}$, and $c_2=2^2(3/2-\sqrt{2}) = \frac{1}{3/2 + \sqrt{2}} \lt \frac{1}{2\sqrt{2}}$.
The computation above still applies. The only candidates are convergents just before the period ends. If $p_n/q_n$ is the convergent just before the period ends, then $q_n^2|p_n/q_n - \sqrt{d}| = \frac{1}{a_{n+1}' + q_{n-1}/q_n}$ and $q_{n-1}/q_n$ has a continued fraction that consists of $0$ followed by repetitions of the period of the simple continued fraction for $\sqrt{d}$, with the last $2a$ missing. This is a convergent to the simple continued fraction of the fractional part $\sqrt{d}-\lfloor \sqrt{d} \rfloor$. When $n$ is odd, $q_{n-1}/q_n > \sqrt{d}-\lfloor \sqrt{d} \rfloor$, and the odd convergents decrease toward $\sqrt{d} - \lfloor \sqrt{d} \rfloor$ so the maximum occurs at the first occurence of an odd index just before a period ends.
| {
"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 an ellipsoid with semiaxes $s,s,1/s^2$ and this almost completes the proof of the uniqueness hypothesis of an ellipsoid with given intrinsic volumes.
| 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}_0 r^3 \cos (\theta) e^{\frac{-1}{2}(\frac{1}{s} \cos ^2\theta+s^2 sin^2 \theta )r^2} drd\theta $
we pose
$A= \frac{1}{2}(\frac{1}{s} \cos^2 (\theta)+s^2 \sin^2(\theta) )$
$\int^{+\infty}_0 r^3 e^{-Ar^2} dr=\frac{-r^2}{2A} e^{-Ar^2}\Bigg |^{+\infty}_0+\int^{+\infty}_0 \frac{r}{A}e^{-Ar^2}dr
= \frac{-1}{2A^2} e^{-Ar^2} \Bigg |^{+\infty}_0=\frac{1}{2A^2}$
$$\int^{\frac{\pi}{2}}_0\frac{\cos \theta}{2 A}d\theta =2 s^2 \int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta$$
$\int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta =\int^1_0 \frac{1}{(1+(s^3-1)t^2)^2}dt$
we pose
$a=s^3-1$
$\int^1_0 \frac{1}{(1+at^2)^2}dt =\Bigg [ \frac{t}{2(1+at^2)} + \frac{\arctan(\sqrt{a}t)}{2\sqrt{a}}\Bigg ]^1_0
=\frac{1}{2}(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}})$
So
$$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\frac{2}{s}+ \frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}$$
| {
"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 $$-\int_0^1 u_{xxx}u_x \eta = \int_0^1 (u_{xx})^2\eta - \int_0^1 \frac{1}{2} (u_x)^2 \eta_{xx}$$
Can we obtain the same result using only the Fourier series
$$u(x) = \sum_{n=1}^\infty \sqrt 2 \cos (\pi n x) \, a_n, \qquad \eta(x) = \sum_{n=1}^\infty \sqrt 2 \cos (\pi n x) \, b_n $$
as a tool?
| 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
$$R_1=\int_0^1 dx\, u_{xx}^2 \eta =2^{3/2}\pi^4 \sum_{n,m,k=1}^\infty n^2m^2 a_na_mb_k \int_0^1 dx\,\cos(n\pi x)\cos(m\pi x)\cos(k\pi x)=$$
$$=2^{3/2}\pi^4 \frac{1}{4}\sum_{n,m,k=1}^\infty n^2m^2 a_na_mb_k \left(\delta_{m,n+k}+\delta_{n,m+k}+\delta_{k,n+m}\right).$$
$$R_2=-\frac{1}{2}\int_0^1 dx\, u_{x}^2 \eta_{xx} =\frac{1}{2}2^{3/2}\pi^4 \sum_{n,m,k=1}^\infty nmk^2 a_na_mb_k \int_0^1 dx\,\sin(n\pi x)\sin(m\pi x)\cos(k\pi x)=$$
$$=\frac{1}{2}2^{3/2}\pi^4 \frac{1}{4}\sum_{n,m,k=1}^\infty nmk^2 a_na_mb_k \left(\delta_{m,n+k}+\delta_{n,m+k}-\delta_{k,n+m}\right).$$
left hand side minus right hand side
$$L-R_1-R_2=2^{3/2}\pi^4 \frac{1}{4}\left(\frac{1}{2}\sum_{n=1}^\infty\sum_{m=n+1}^\infty (mn^3-nm^3)a_na_m b_{m-n}+\frac{1}{2}\sum_{m=1}^\infty\sum_{n=m+1}^\infty (mn^3-nm^3)a_{n}a_m b_{n-m}\right.$$
$$\left.+\frac{1}{2}\sum_{n,m=1}^\infty m n (m^2 -n^2)a_na_m b_{n+m}\right)=0.$$
| {
"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
$$A = \begin{pmatrix} x & -\partial_x +c\\ \partial_x+c & -x \end{pmatrix},$$
where $c$ is a real constant. It seems to me that this is a non-trivial change in the operator.
| 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 standard raising and lowering operators $a^{\dagger } $, $a$, in terms of which $x=(a^{\dagger } +a)/\sqrt{2} $ and $-\partial_{x} =(a^{\dagger } -a)/\sqrt{2} $. Acting on the $\psi_{n} $, these act as $a\psi_{n} = \sqrt{n} \psi_{n-1} $, $a^{\dagger } \psi_{n} = \sqrt{n+1} \psi_{n+1} $. Now, $A$ takes the form
$$
A=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & a^{\dagger } -a \\ -a^{\dagger } +a & -a^{\dagger } -a \end{pmatrix}
+ \begin{pmatrix} 0 & c \\ c & 0 \end{pmatrix}
$$
One has the following eigenvectors/eigenvalues:
$$
\begin{pmatrix} \psi_{0} \\ -\psi_{0} \end{pmatrix} \mbox{ with eigenvalue } \ \ \ -c
$$
$$
\left[ \frac{c\pm \sqrt{c^2 +2n+2} }{\sqrt{2n+2} } \begin{pmatrix}
\psi_{n} \\ \psi_{n} \end{pmatrix} + \begin{pmatrix} \psi_{n+1} \\ -\psi_{n+1} \end{pmatrix} \right] \ \ \ \mbox{with eigenvalue} \ \ \
\pm \sqrt{c^2 + 2n+2}
$$
For $c=0$, the spectrum reduces to the one given in the answer to the previous question.
| {
"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).
Write $M=\begin{bmatrix}
A_1&A_2\\
A_3&A_4
\end{bmatrix},
$ where the $A_i$ are $2$ by $2$ sumbatrix of $M$.
Let $d_i$ be the dot product of two rows of $A_i$, i.e. if $A_i = \begin{bmatrix}
a&b\\
c&d
\end{bmatrix}$, then $d_i = ac +bd$.
Let $a_i = \mathrm{det}(A_i) - d_i$.
For example if $A_1 =\begin{bmatrix}
1&2\\
3&4
\end{bmatrix}$, then $d_1 = 11$ and $a_1 = - 2 - 11 = -13$.
Consider the matrix $A = \begin{bmatrix}
a_1&a_2\\
a_3&a_4
\end{bmatrix}$.
Question: Suppose $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, can one of eigenvalues of $A$ have absolute value bigger than $1$?
Thoughts so far: It is clear that the matrix $A$ should have all entries being non-zero. I find this question hard because two similar matrices might have differnt "behaviour", e.g. if $
M_1=\Tiny\begin{pmatrix} 0 & 0 & -1 & 0 \cr
1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 0 & 1 & -3 & 4 \end{pmatrix}
$, then $
A=\begin{pmatrix} 0 & -1 \cr 1 & 2 \end{pmatrix}\in SL_2(\Bbb Z);
$
while if $
M=\Tiny\begin{pmatrix} 1 & 1 & 0 & 0 \cr
0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \end{pmatrix}
$, we have $A=\begin{pmatrix} 0 & 0 \cr 0 & 0 \end{pmatrix}$, even though $M_1$ is simliar to $M_2$.
I also look up the Block matrix determinant, but I don't think we can use the identities since our block matrices don't satisfy their assumptions.
Thank you for reading, any reference for this would be really appreciated.
| 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 $S \in \mathcal{M}_2(\mathbb{Z})$
Let
$$M = PTP^{-1} = \left(\begin{array}{cc}
aI_2 & bI_2 \\
cI_2 & dI_2
\end{array}\right)\left(\begin{array}{cc}
I_2 & S \\
0 & I_2
\end{array}\right)
\left(\begin{array}{cc}
dI_2 & -bI_2 \\
-cI_2 & aI_2
\end{array}\right),$$
Then,
$$M = \left(\begin{array}{cc}
aI_2 & aS+bI_2 \\
cI_2 & cS+dI_2
\end{array}\right)
\left(\begin{array}{cc}
dI_2 & -bI_2 \\
-cI_2 & aI_2
\end{array}\right),$$
Since $ad-bc=1$,
$$M = \left(\begin{array}{cc}
I_2-acS & a^2S \\
-c^2S & I_2+acS
\end{array}\right).$$
Now, choose $S=I_2$. Then
$$M = \left(\begin{array}{cc}
(1-ac)I_2 & a^2I_2 \\
-c^2I_2 & (1+ac)I_2
\end{array}\right).$$
Hence
$$A = \left(\begin{array}{cc}
(1-ac)^2 & a^4 \\
c^4 & (1+ac)^2
\end{array}\right).$$
and $\det(A) = (1-ac)^2(1+ac)^2 - a^4c^4 = (1-a^2c^2)^2 - a^4c^4 = 1-2a^2c^2$
and $\mathrm{Tr}(A) = (1-ac)^2+(1+ac)^2 = 2+2a^2c^2$.
If we choose $a=b=c=1$ and $d=2$, then we actually have $ad-bc=1$, $\det(A)=-1$ so $A \in GL_2(\mathbb{Z})$ and since $\mathrm{Tr}(A)=4$,
the characteristic polynomial of $A$ is $X^2-4X-1$, so the eigenvalues of $A$ are $2 \pm \sqrt{5}$.
| {
"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 and bulk-edge correspondence of magnetized cold plasmas that this can be reduced to the determinant of a $3\times3$ matrix, meaning
$$ \det(M-\omega ) = \det(N_1-N_2+N_3)=0.$$
Here
$$N_1 = \frac{aa^t}{\omega^2}, N_2 =\frac{a^ta}{\omega^2}\operatorname{id},\text{ and }N_3 = \begin{pmatrix} 1-\frac{1}{\omega^2-1} & i\frac{1}{\omega(\omega^2-1)} & 0 \\ -i\frac{1}{\omega(\omega^2-1)} &1-\frac{1}{\omega^2-1} & 0 \\ 0& 0 & 1-\frac{1}{\omega^2} \end{pmatrix}.$$
This determinant is stated as equation (1) versus the original $9\times9$ matrix is equation (11).
How can we derive the determinant of the $3\times 3$ matrix from the determinant of the $9\times 9 $ matrix without first expanding the determinant?
| 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:
$${\rm det}\,(M-\omega)=-\omega^9+2 \omega^7 \left(a_1^2+a_2^2+a_3^2+2\right)-\omega^5 \left(a_1^4+2 a_1^2 \left(a_2^2+a_3^2+3\right)+\left(a_2^2+a_3^2\right)^2+6 a_2^2+6 a_3^2+4\right)+\omega^3 \left(2 a_1^4+a_1^2 \left(4 a_2^2+4 a_3^2+3\right)+2 \left(a_2^2+a_3^2\right)^2+3 a_2^2+4 a_3^2+1\right)-\omega a_3^2 \left(a_1^2+a_2^2+a_3^2\right)$$
$$=\omega^7 \left(1-\omega^2\right)\,{\rm det}\,(N_1-N_2+N_3).$$
Link to the Mathematica notebook.
| {
"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)}}{2(x+4)}\left ( \frac{x-\sqrt{x(x+4)}}{2} \right )^n.
\end{aligned}
\end{equation*}
Using Mathematic $12.3$, when $n$ is large enough, we give the distribution of the roots of $f(x)$ in the complex plane as follows
In this figure, we can see that the closure of the real roots of $f(x)$ may be $\left [ -4,0 \right ]$.
So we have the following question
Question: all roots of $f(x)$ are real? It seems yes! But we have no way of proving it.
| 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 rational power series is rational, and I did the following computations with a Maple program I wrote using the method described here .
For any power series $A(y) = \sum_{n=0}^\infty a_n y^n$ and integers $m$ and $i$, let
\begin{equation*}
A_{m,i}(y) = \sum_{n=0}^\infty a_{mn+i}y^i,
\end{equation*}
where we take $a_n=0$ if $n<0$. Following Brendan McKay, we define the generating function.
$$F=F(y) = \sum_{n=0}^\infty f_n(x) y^n =
1+\frac{x^{2} y^{2} \left(1+y \right)^{2}}{\left(1+x y^{2}\right) \left(1-xy-x y^{2}\right)}
$$
We also define generating functions for Timothy Chow's polynomials $g_n(x)$ and $h_n(x)$:
\begin{gather*}
G=\sum_{n=0}^\infty g_n(x) y^n = \frac{1}{1-xy-xy^2}\\
H=\sum_{n=0}^\infty h_n(x) y^n = \frac{2-xy}{1-xy-xy^2}.
\end{gather*}
Then we want to prove
\begin{gather}
F_{2,0}=G*G\tag{1}\\
F_{4,1}=G_{2,2}*G_{2,-1}\tag{2}\\
F_{4,3}=G_{2,0}*G_{1,1}*H_{1,2}\tag{3}
\end{gather}
Multisections, can be computed by Hadamard products (or in other ways). For example,
$F(y)*y/(1-y^4) = yF_{4,1}(y^4)$. We find that
\begin{gather*}
F_{2,0} =\frac{1-x y }{\left(1+x y \right) \left(1-2xy -x^{2} y +x^{2} y^{2}\right)}\\
F_{4,1}=\frac{x^3y(1+3x+x^2 -x^2y)}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\
F_{4,3}=\frac{x^2(2+x -\left(3x^2+4x^3+x^4\right) y +x^{4} y^{2})}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\
G_{1,1}=\frac{x(1+y)}{1-xy-xy^2}\\
G_{2,-1}=\frac{xy}{1-2xy-x^2y+x^2y^2}\\
G_{2,0}=\frac{1-xy}{1-2xy-x^2y+x^2y^2}\\
G_{2,2}=\frac{x(1+x-xy)}{1-2xy-x^2y+x^2y^2}\\
H_{1,2}=\frac{x(2+x+xy)}{1-xy-xy^2}
\end{gather*}
We can then verify $(1)$–$(3)$ directly.
| {
"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 $d_0=d_1=d_2=1$. How to show $d_i>d_{i+1}$ for all $i\ge3$?
By easy calculation, $d_3=1, d_4=\frac{18p^3+11p^2-6p+1}{24p^3}$. For a recursion of 2 or 3 terms, it is easy to proceed with induction, but what about 4 terms in this case?
| 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 still
wonder about where it comes from). Denote
$$
F(x)=\sum_{i=0}^\infty d_ix^i=1+x+x^2+x^3+\dots
$$
the generating function of the sequence. The given recursion can
be then written as the differential equation of order 1,
$$
\frac{F'}{F}=\frac{3p(2p^2+2p(p-1)x+(p-1)(2p-1)x^2)}{6p^3-6p^2x-3p(p-1)x^2-(p-1)(2p-1)x^3}
$$
which can be explicitly solved:
$$
F(x)=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
(one checks that the expansion indeed starts $1+x+x^2+x^3+\dots$).
The desired property $d_i>d_{i+1}$ for $i\ge3$ is equivalent to showing that
the expansion
$$
1-(1-x)F(x)=c_0x^4+c_1x^5+c_2x^6+\dots
$$
has all coefficients $c_0,c_1,c_2,\dots$ positive. In turn, the latter is
equivalent to the positivity of all coefficients in the expansion
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)=4c_0x^3+5c_1x^4+6c_2x^5+\dots.
$$
We have
$$
\frac{d}{dx}\bigl(1-(1-x)F(x)\bigr)
=\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p}
$$
$$
-p(1-x)\biggl(\frac1p+\frac{(p-1)x}p+\frac{(p-1)(2p-1)x^2}{2p^3}\biggr)
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\biggl(1-\frac xp-\frac{(p-1)x^2}{2p^2}-\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^{-p-1}
$$
$$
=\frac{(p-1)(2p-1)(3p-1)}{6p^3}x^3
\cdot\Biggl(1+\sum_{n=1}^\infty\frac{(p+1)(p+2)\dots(p+n)}{n!}
\biggl(\frac xp+\frac{(p-1)x^2}{2p^2}+\frac{(p-1)(2p-1)x^3}{6p^3}\biggr)^n\Biggr),
$$
and the desired positivity follows.
| {
"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 nor reference. It is a particular case of the question in the title: if $a$ and $b$ are quadratic residues modulo $p$, when is $a+b$ also a quadratic residue modulo $p$?
I came into this question when counting the number of diophantine $2$-tuples modulo $p$, that is, the number of pairs $\{ a,b\}\subset \mathbb{Z}^*_p$ such that $ab+1$ is a quadratic residue modulo $p$.
| 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 Q & \frac{q+3}{4} & \frac{q-1}{4} \\
\hline
-1 \notin Q & \frac{q+1}{4} & \frac{q+1}{4} \\
\hline
\end{array}
Proof: If, for $x,y,c\in \mathbb{F}_q,\ c \neq 0$ we have $x^2+c=y^2$, then $c=y^2-x^2=(y-x)(y+x)$. Now for all the $q-1$ elements $d\in \mathbb{F}_q^{\ast}$, we can let $y-x=d$ and $y+x=\frac{c}{d}$. But the pairs $(d,\frac{c}{d}),(-d,\frac{c}{-d}),(\frac{c}{d},d),(\frac{c}{-d},-d)$ all give the same value of $y^2=\frac{1}{4}(d+c/d)^2$. Also, as $q$ is odd, $d\neq -d\ \forall d$. But if $c\in Q$, for $2$ values of $d$ we have $d=\frac{c}{d}$ and if $-c\in Q$, for 2 values of $d$ we have $d=\frac{c}{-d}$. So we have
$$ s_q(c) = \left\{ \begin{array}{rcll}
\frac{\frac{q-1}{2}-2}{2}+2 & = & \frac{q+3}{4} & if\ c\in Q,\ -c\in Q \\
\frac{\frac{q-1}{2}-1}{2}+1 & = & \frac{q+1}{4} & if\ c\in Q,\ -c\notin Q \\
\frac{\frac{q-1}{2}-1}{2}+1 & = & \frac{q+1}{4} & if\ c\notin Q,\ -c\in Q \\
& & \frac{q-1}{4} & if\ c\notin Q,\ -c\notin Q
\end{array} \right. $$
and hence the result.
| {
"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}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}
\end{equation}
I can prove that $f(n)=0$ for all integers $n\geq 1$ and $g(n)=0$ for all integers $n>1$ using properties of orthogonal polynomials. However, I would like to find an elementary proof, which might also be more illuminating.
While doing some experiments with calculations using Mathematica, I defined the functions $f(n)$ and $g(n)$ with the above formulas but forgot to specify that $n$ is an integer value.
When I typed $f(n)$ and $g(n)$ for a generic variable $n$, I guess that Mathematica "assumed" that the variables involved were real and provided the following simple formulas:
\begin{equation}
f(x)=\frac{\sin \pi x}{x(x+1)\pi}
\end{equation}
and
\begin{equation}
g(x)=-\frac{2\sin \pi x}{(x+1)(x-1)\pi}
\end{equation}
which makes it evident that these functions are zero for all positive (and negative) integers with the possible exceptions of $0,1,-1$.
Why are these equalities true?
I realize it might have to do, perhaps, with properties of the Euler Beta and Gamma functions, but I know too little about these functions to figure out a proof along these lines. Can anybody help? Thank you!
| 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}{k+2}\binom{2k+2}{k+1}\, .
\end{equation}
These sums can be evaluated by induction as:
\begin{equation}
(-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1}
\end{equation}
and
\begin{equation}
(-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2}\, .
\end{equation}
We can write these expression respectively as
\begin{equation}
\frac{1}{x+1}\left(1+\frac{x}{n+1} \right)\prod_{k=1}^n\left(1-\frac{x^2}{k^2}\right)
\end{equation}
and
\begin{equation}
-\frac{2x}{x^2-1}\frac{n+1}{n+2}\prod_{k=1}^{n+1}\left(1-\frac{x^2}{k^2}\right)\, .
\end{equation}
The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series :
\begin{equation}
\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)}
\end{equation}
and
\begin{equation}
\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\, .
\end{equation}
As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$.
| {
"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=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n $
But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).
| 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 $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$
(the identities are straightforward by induction).
Denote $m:=2n+1$. It is not hard to show that $\sin mx=p_m(\sin x)$ for a polynomial of degree $m$ in $\sin x$. The roots of $p_m$ are $\sin \frac{k\pi}{2n+1}$ for $k=-n,\ldots,n$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$
(the multiple $2n+1$ comes from dividing by $x$ and putting $x=0$).
Put $x=\frac{\pi}{4n+2}$. Using the inequality $\sin tx\leqslant t\sin x$ (which may be proved, for example, by induction in $t=1,2,\ldots$ from the identity $\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$) for $t=2k$ ($k=1,2,\ldots,n$)
we get
$$
1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi.
$$
Analogously we get
$$
\cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right).
$$
Dividing by $1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$ and substituting $x=\frac{\pi}{4n}$ (for computing the LHS at this point use l'Hôpital rule) we get
$$
n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi.
$$
UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $\sum 1/n^2=\pi^2/6$ and $\sum (-1)^{k-1}/(2k-1)=\pi/4$ using these trigonometric things.)
| {
"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 $10^2+11^2+12^2=13^2+14^2$. Now this Rachinsky equality can be considered as a generalization of the well-known Pythagorean triple (3,4,5), $3^2+4^2=5^2$, and in analogy with the Pythagorean triples one can define Rachinsky quintets as a set of five positive integers $(a,b,c,d,e)$ such that $a^2+b^2+c^2=d^2+e^2$. It is known that all primitive Pythagorean triples $(a,b,c)$ such that $a^2+b^2=c^2$ are generated by Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$, where $m$ and $n$ are positive integers such that $m>n$, $m$ and $n$ are coprime, and $m \not\equiv n \bmod 2$. Can one establish an analogous result for Rachinsky quintets?
|
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 conditions hold:
*
*$\ x\equiv y\equiv 1\ \mbox{mod}\ 2$
*$\ u\equiv v\equiv 0\ \mbox{mod}\ 2$
*$\ u\cdot v = c^2-x\cdot y$
i.e. we may take arbitrary $x$ and $y$ as in condition 1, and then one decomposes $c^2-x\cdot y$ (see condition 3), where $\ u\ v\ $ are as in condition 2; of course $\ 4\,|\,c^2-x\cdot y\ $ (and the expressions
under the squares are integers).
| {
"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, for a fixed integer $n$, if the system has an infinite number of co-prime integer solutions, then $n$ is a multiple of $3$?
Method: Eqs $(2)$ and $(3)$ can easily be given a complete solution. Incorporating $(1)$, I got,
$$(-2 p + \alpha q -\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (-2 p + \alpha q +\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag5$$
where $\alpha = n+1,\;\beta = n-1$. It is also true for $k=6$ if there is $p,q,n$ such that,
$$Poly_1:= (-3+n)(5-2n+n^2)p + 4n(1+n^2)q$$
$$Poly_2:= (-3+n)(5-2n+n^2)p^3 + 2(5+11n-5n^2+n^3)p^2q - (5+7n+15n^2-3n^3)pq^2 + 4n(1+n^2)q^3$$
and,
$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag6$$
A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,
$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}$$
Example: Let $n=12$, then,
$$((-2 p + 13 q - 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\(-2 p + 13 q + 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag7$$
which is already true for $k=1,2$. But it is also for $k=6$ if,
$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag8$$
An initial point is $p,q = -104,17.$ Hence $(8)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions to $(7)$.
Question 2. For what other positive integer $n$ below a bound can we find solutions to non-zero $(6)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)
| 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 $(x,y) = (0, \frac{1}{2})$ for ever prime $p \geq 5$ ?
| 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 $a+m\ne 0$, $y+1/2=(a+m)/m$ and $x^p=a(a+m)/m^2$. Since $a,a+m,m$ are mutually coprime and $p$ is odd, we get that they should be all perfect $p$-th powers, $a=A^p$, $m=B^p$, $a+m=C^p$, $A^p+B^p=C^p$.
| {
"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 polynomial factor of $\frac{x^k-1}{x-1}$. When $k$ is even there are a few more trivial pairs, namely $(x,-1),(-1,y),(x,-1-x),(x,x^n-1),(y^n-1,y)$; where $n>0$ is a multiple of $k$.
| 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,
$\frac{x^{30}-1}{x-1}=\Phi_3\Phi_5\Phi_{15}\Phi_{30}=$
$ (x^2+x+1)(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$$(x^8+x^7-x^5-x^4-x^3+x^2+1)$
So for $k=30$ one has $x,y$ where $y$ is any of the $16$ products of the factors.
Maybe you want a parametric family that is not $y=p(x).$
Interesting Variant To have $xy$ divide $(x+y)^2+1$ is quite interesting. I saw it in a movie (in another form). It may have been a math olympiad problem.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/252784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
A specific Diophantine equation restricted to prime values of variables. Consider the following Diophantine equation $$x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1).$$ Assume also that $x,a,b,c,a^2+a+1,b^2+b+1, c^2+c+1$ are all primes. We'll call such a quadruplet $(x,a,b,c)$ a triple threat.
I'd like to be able to show that there are no triple threats. I've been able to prove the following weaker result that if $(x,a,b,c)$ is a triple-threat and $x \equiv 1$ (mod 5), then none of $a,b$ or $c$ can be $1$ (mod 5). For my purposes, this is useful, but I can get a stronger result with a simpler proof if I can get that there are no triple threats.
Let me briefly sketch the idea of the proof: Most of what I have been able to do has been from looking at the more general equation
$$x^2+x+1=(a^2+a+1)(b^2+b+1)p$$ where $x,a,b,p,a^2+a+1$ and $b^2+b+1$ are all prime. Note that this equation does solutions (and probably has infinitely many). Call this an almost-triple threat.
The key approach is showing the following sort of result: If $a \leq b$, and we have some specific modulo restriction on $x,a,b,p$ then one must have $a^2+a+1<p$. If one then applies to the original triple threat equation one can in some circumstances get a contradiction by forcing that $a^2+1<c^2+c+1<b^2+b+1<a^2+a+1$.
Let me briefly sketch how in most contexts I've obtained that $a^2+a+1<p$.
My primary method of attack on this equation has been to rewrite it as
$$(x-a)(x+a+1)=((b^2+b+1)p-1)(a^2+a+1) $$ and
$$(x-b)(x+b+1)=((a^2+a+1)p-1)(b^2+b+1). $$
Since $a^2+a+1$ and $b^2+b+1$ are assumed to be prime, one has that either the pair of relations $a^2+a+1|x-a$ and $x+a+1|(b^2+b+1)p-1$, or the pair of relations $a^2+a+1|x+a+1$ and $x-a|(b^2+b+1)p-1$. One gets similar pairings using the second version above with $b^2+b+1$. One has then four cases:
I) $a^2+a+1|x-a$ and $b^2+b+1|x-b$
II) $a^2+a+1|x-a$ and $b^2+b+1|x+b+1$
III) $a^2+a+1|x+a+1$ and $b^2+b+1|x-b $.
IV) $a^2+a+1|x+a+1$ and $b^2+b+1|x+b+1$.
It is not too hard to show in Cases II and III that one always has $a^2+a+1<p$. The difficulty is primarily in Cases I and IV. In Case I one gets with a little algebra that $x+a+1|p(a+b+1) + \frac{x-b}{b^2+b+1}$. One then has $k(x+a+1)= p(a+b+1) + \frac{x-b}{b^2+b+1}$. For many choices of $x,a,b$ mod 5 one can force $k$ to be large. Say $k \geq 3$. Then one has $$a+x+1 \leq \frac{ p(a+b+1) + \frac{x-b}{b^2+b+1}}{3},$$ and one can get that $a^2+a+1<p$ from this inequality and a little work. (I can expand on this part if necessary.)
The main difficulty right now is dealing with Case IV. I can handle Case IV under some small modulus assumptions, but I cannot prove that $a^2+a+1<p$ in Case IV unconditionally.
In Case IV one has with a little work that
$x-a|p(a+b+1)-\frac{x+b+1}{b^2+b+1}$ and $x-b|p(a+b+1)-\frac{x+a+1}{a^2+a+1}$ but it seems tough to use these to get a tight enough bound to conclude what is wanted without having some extra modulus restriction on the variables. Any thoughts on either clearing out Case IV, or any other method of showing that triple threats don't exist via some other method would be appreciated.
| 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+1)(c^2+c+1)$.
These remainders are expected to have order of magnitude $(abc)^2$,
whereas x is in the vicinity of $(a+1/2)(b+1/2)(c+1/2)$.
Maybe that is a reason why triple threats are rare.
| {
"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 it simply that $|A(n)| = n$?
And similarly in $\mathbb{Z}^d$ for $d>2$?
$5 \times 5$ lattice square, $5$ collinear points.
Q2. What is the largest subset $B(n)$ of lattice points in $S_n$,
not all collinear, that have the
property that every pair of points in $B(n)$ are separated by
an integer Euclidean distance?
$5 \times 5$ lattice square, $4$ noncollinear points.
A $9 \times 9$ example with $5$ noncollinear points,
also based on $3{-}4{-}5$ right triangles,
is illustrated in the Wikipedia article
Erdős–Diophantine graph.
| 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 & 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 & 1 & 1 & 1 & 1
& 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1
& 1 & 1 & 1 & 1 & 1 & 1 & 1
& 1 & 1 & 1 & 1 & 1 & 1 & 1
& 1 & 1 & 1 \\
\end{array}
$$
You can see that $p_1$ has nine $1$'s in its coordinate representation,
$p_2$ has sixteen $1$'s,
and $p_3$ has twenty-five $1$'s:
$3{-}4{-}5$ triangles again—very clever!
| {
"language": "en",
"url": "https://mathoverflow.net/questions/312707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Cauchy's Integral with quadratic exponential term As I was studying the Cauchy's integral formula, I tried to do the integral:
\begin{equation}
I = \int\limits_{-\infty}^{\infty} \frac{1}{x - a} e^{(i A x^2 + i B x)} dx
\end{equation}
with $A>0, B>0$ and $a > 0$.
Consider an integral on a complex plan:
\begin{equation}
J = \int\limits_{C + C_R} \frac{1}{z - a} e^{(i A z^2 + i B z)} dz
\end{equation}
where $C$ is along the real axis $-\infty \rightarrow +\infty$ and $C_R$ is the upper half circle $z = Re^{i\theta}$ with $R \rightarrow \infty$ and $\theta \in [0, \pi]$.
Naively, I would expect $C_R$ part of the integral gives zero and $C$ part of the integral gives $I$, then the $I$ can be derived by Cauchy's integral formula.
However, as I tried to check the $C_R$ part of the integral, I found that ($z = Re^{i\theta}$):
$$
\begin{split}
I_R &= \int\limits_0^{\pi} d\theta \frac{iRe^{i\theta}}{Re^{i\theta} - a} \exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big) \\
|I_R| &\leq \int\limits_0^{\pi} d\theta\left |\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \Big|\exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big)\Big|
\end{split}
$$
where the first term
\begin{equation}
\left|\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \leq \frac{R}{R-a} \rightarrow 1 \ as\ R \rightarrow \infty
\end{equation}
and the second term
\begin{equation}
\left|\exp(iAR^2e^{2i\theta}+iBRe^{i\theta})\right| \leq e^{-AR^2\sin(2\theta) - BR\sin(\theta)}
\end{equation}
will not approach to zero because of $e^{-AR^2\sin(2\theta)}$.
Is there anything wrong in my approach? And is there any other way I can perform this integral $I$?
Thanks a million for advises!
| Let me first remove the $Bx$ term by completing the square,
$$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}{8} \pi ^{-5/2} e^{-iB^2/4A}\biggl\{G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
0,\frac{1}{4},\frac{3}{4} \\
0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
\frac{1}{4},\frac{1}{2},\frac{3}{4} \\
0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\
\end{array}
\right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl|
\begin{array}{c}
\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\
\end{array}
\right)\biggr\},$$
with
$$\alpha=\left(a+\frac{B}{2A}\right)^4\frac{A^2}{4}.$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/361867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this equation $a^2+3b^2c^2=7^c$ Let $a,b,c$ be poistive integers,and such $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$,fine the all $a,b,c$ such
$$a^2+3b^2c^2=7^c$$
I'm not sure that this question has been studied, but I've been trying for a long time$(a,b,c\le 100)$, and there's only one set of solutions：$(a,b,c)=(2,1,1)$,But I can't prove it. I may need your help. Thank you
| We work in $\mathbb{Z}[\omega]$ where $\omega=\frac{1+i\sqrt{3}}2$. It is a factorial ring, and we factorize both sides as $(a+i\sqrt{3}bc)(a-i\sqrt{3}bc)=(2+i\sqrt{3})^n(2-i\sqrt{3})^n$. Since $2+i\sqrt{3},2-i\sqrt{3}$ are prime (and coprime), the guy $a+i\sqrt{3}bc$ can not be divisible by both (otherwise it is divisible by $(2+i\sqrt{3})(2-i\sqrt{3})=7$ that contradicts to our assumption that $\gcd(a,b)=\gcd(a,c)=1$), we see that $a+i\sqrt{3}bc=\varepsilon(2\pm i\sqrt{3})^c$ where $\varepsilon$ is a unit in $\mathbb{Z}[\omega]$. If $\varepsilon\ne \pm 1$, we get $\varepsilon=\frac{\pm 1\pm i\sqrt{3}}2$ and therefore $$\varepsilon(2\pm i\sqrt{3})^c= \left(\varepsilon (\pm i\sqrt{3})^n+(\text{something in } \mathbb{Z}[i\sqrt{3}])\right)\notin \mathbb{Z}[i\sqrt{3}],$$
a contradiction.
So we get $\varepsilon=\pm 1$ and $a+i\sqrt{3}bc=\pm (2\pm i\sqrt{3})^c$. Consider two cases.
*
*$c$ is even, $c=2^\alpha \beta$ for odd $\beta$. Then
$$2i\sqrt{3}bc=\pm\left((2+i\sqrt{3})^{c}-(2-i\sqrt{3})^{c}\right)$$
is divisible by
$$
(2+i\sqrt{3})^{2^\alpha}-(2-i\sqrt{3})^{2^\alpha}=2i\sqrt{3}\cdot 4\\ \cdot((2+i\sqrt{3})^{2}+(2-i\sqrt{3})^{2})
((2+i\sqrt{3})^{4}+(2-i\sqrt{3})^{4})\ldots
((2+i\sqrt{3})^{2^{\alpha-1}}+(2-i\sqrt{3})^{2^{\alpha-1}})
$$
which is divisible by $2i\sqrt{3}\cdot 2^{\alpha+1}$, thsu $b$ is also even, a contradiction with $\gcd(b,c)=1$.
*3 divides $c$. Since $(2+i\sqrt{3})^3-(2-i\sqrt{3})^3=18i\sqrt{3}$, we analogously get that 3 divides $b$ (the guys $A^2+AB+B^2$ for $A,B=(2\pm i\sqrt{3})^{3^t}$ are divisible by 3.)
*$c$ is odd and not divisible by 3. Let $3<p_1<p_2<\ldots<p_k$ be distinct prime divisors of $c$ (over $\mathbb{Z}$). Then $p_1$ divides (in $\mathbb{Z}[\omega]$) the number $(2+i\sqrt{3})^c-(2-i\sqrt{3})^c=\pm 2i\sqrt{3}bc$. In particular $p_1\ne 7=(2+i\sqrt{3})(2-i\sqrt{3})$.
3.1) If $p_1$ is prime in $\mathbb{Z}[\omega]$, then by Lagrange theorem it divides $(2 \pm i\sqrt{3})^{p_1^2-1}-1$, thus it divides $(2 + i\sqrt{3})^{p_1^2-1}-(2 - i\sqrt{3})^{p_1^2-1}$. Therefore $p_1$ divides $(2 + i\sqrt{3})^{\gcd(p_1^2-1,c)}-(2 - i\sqrt{3})^{\gcd(p_1^2-1,c)}=2i\sqrt{3}$, a contradiction. (We used that $p_1^2-1=4\cdot \frac{p_1-1}2\cdot \frac{p_1+1}2$ has prime divisors less than $p_1$, thus is coprime with $c$.)
3.2) If $p_1=q\bar{q}$ for a prime $q\in \mathbb{Z}[\omega]$, then by Lagrange theorem $q$ divides $(2 + i\sqrt{3})^{p_1-1}-(2 - i\sqrt{3})^{p_1-1}$ and we analogously get a contradiction.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/376653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
roots of a fourth degree polynomial function (Vieta) Question
I am interested in the root of the polynomial function :
$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2
+(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)x+
abcd-abc-bcd-ad+bc=0$
Under the restriction that: $0<a,b,c,d<1$.
I tried Vietas formula but don’t know how to come to a decomposition
$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$
| There are two real solutions, the general expressions are very, very long. Even if all four coefficients are equal, $a=b=c=d$, you have an equation
$$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=\frac{4+6^{2/3}\pm\sqrt{20 \sqrt{4+6^{2/3}}-6 \sqrt[3]{6}+4\ 6^{2/3}+32}}{4 \sqrt{4+6^{2/3}}}.$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/420632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the sum of general type $\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k- n + a \choose r }$?
QUESTION. How to calculate the sum of such general type?
$$\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\choose n+a},$$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} \binom{2 k - n + a}1 = 0, $$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 2} = -\frac{(a+ n)(a-n)}{2 (2n-1)} {2n\choose n+a}, $$
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 3} = \frac{(a+ n)(a-n)}{2 (2n-1)} {2n\choose n+a}. $$
I am struggling to calculate at least
$$\sum_{k=0}^n {n\choose k} {n\choose k+a} {2 k - n + a \choose 4} $$
Can someone help me with this one?
| Alternatively, one can get an explicit expression for a fixed $r$ via generating functions by noticing that the given sum equals
$$[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+y)^{2(n-i)} y^i\left(\frac{z^2}{1+z}\right)^i,$$
which we evaluate as
$$[y^{n+a}]\ \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}i\binom{-i}{r-2i} (1+y)^{2(n-i)}y^i = (-1)^r \sum_{i=0}^{\lfloor r/2\rfloor} \binom{n}i \binom{r-i-1}{r-2i} \binom{2(n-i)}{n+a-i}.$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/427962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
\begin{aligned}
f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
&+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x-\sqrt{x(x+4)}}{2} \right )^n.
\end{aligned}
\end{equation*}
Using Mathematic $12.3$, when $n$ is large enough, we give the distribution of the roots of $f(x)$ in the complex plane as follows
In this figure, we can see that the closure of the real roots of $f(x)$ may be $\left [ -4,0 \right ]$.
So we have the following question
Question: all roots of $f(x)$ are real? It seems yes! But we have no way of proving it.
| (another comment, not an answer.)
Experimentally, with the following code
from sympy import *
var('x')
var('n', integer = True)
f = ( (2+(x+3) * sqrt(-x))/(2*(x+4)) *
sqrt(-x)**n + ((2-(x+3) * sqrt(-x)) / (2*(x+4))) * (-sqrt(-x))**n +
+ (x+2+sqrt(x*(x+4)))/(2*(x+4)) *
( (x+sqrt(x*(x+4)))/2 )**n +
(x+2-sqrt(x*(x+4)))/(2*(x+4)) *
( (x-sqrt(x*(x+4)))/2 )**n
)
pprint(f)
for i in range(10):
eq = simplify(f.subs(n,i))
print ('========== n = ', i)
pprint(eq)
print ('have solutions')
sols = solve(eq)
pprint (sols)
pprint ('approx. = ')
pprint ([s.evalf() for s in sols])
we see the following:
$$
\begin{array}{l}
f_{0} = 1\\
f_{1} = 0\\
f_{2} = x^{2}\\
f_{3} = x^{2} \left(x + 2\right)\\
f_{4} = x^{2} \left(x^{2} + 2 x + 1\right)\\
f_{5} = x^{3} \left(x^{2} + 3 x + 1\right)\\
f_{6} = x^{4} \left(x^{2} + 4 x + 4\right)\\
f_{7} = x^{4} \left(x^{3} + 5 x^{2} + 7 x + 3\right)\\
f_{8} = x^{4} \left(x^{4} + 6 x^{3} + 11 x^{2} + 6 x + 1\right)\\
f_{9} = x^{5} \left(x^{4} + 7 x^{3} + 16 x^{2} + 13 x + 2\right)\\
\end{array}
$$
And it seems that already $f_9$ have some imaginary roots, albeit very small.
Do these imaginary roots really converge to 0 when $n\to \infty$ or they stay on the same magnitude ?
| {
"language": "en",
"url": "https://mathoverflow.net/questions/440962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Surprising behaviour of polynomial that generates the series 1,2,4,8,...2^(k-1) Consider the generating function $f(n)$ that produces the following values:
$$f(1) = 1$$
$$f(2) = 2$$
$$f(3) = 4$$
Obviously these values can be generated by $f(n)= 2^{n-1}$.
These values can equally well be generated by $f(n) = (n^2-n+2)/2$, a second order polynomial.
Many (all?) integer series $f(k)$, where $k = 1,2,3,...,K-1,K$ can be generated by a polynomial of order $K-1$.
The integer series $2^{n-1}$, where $n = 1,2,3,...,K$ can also be generated by a polynomial of order $K-1$.
The following interesting thing happens.
If we describe the series $1,2,4$ by $f(n) = \frac{n^2-n+2}{2}$ then $f(4) = 7$
For $1,2,4,8$, $f(n) = \frac{n^3-3n^2+8n}{6}$ and $f(5) = 15$
For $1,2,4,8,16$, $f(n) = \frac{n^4-6n^3+23n^2-18n}{24}$ and $f(6)=31$
For $1,2,4,8,16,32$, $f(n) = \frac{n^5-10n^4+55n^3-110n^2+184n}{120}$ and $f(7)=63$
For $1,2,4,8,16,32,64$, $f(n) = \frac{n^6-15n^5+115n^4-405n^3+964n^2-660n+720}{720}$ and $f(8)=127$
I have verified this till order 14.
Lets add the series "1" and "1,2" for completeness:
For $1$, $f(n) = 1$ and $f(2) =1$. $f(2) = 2 \cdot f(1)-1$
For $1,2$, $f(n) = n$ and $f(3) = 3$. $f(3) = 2 \cdot f(2)-1$
This suggests that $f(k+1) = 2 \cdot f(k) -1$ when $f(n)$ is the $k-1$ th order polynomial function that generates the values $1,2,4,...2^{k-1}$.
This raises the question if this is true for all integer series of this type and if so, how to prove it. I hope that I am not overlooking the obvious.
Another observation is that if you write the polynomials that describe the series $1,2,4,8,...$ in a fractional form where all coefficients of $n^k$ in the numerator are integers, then the denominator always seems to be $(K-1)!$ ($1,1,2,6,24,120,$ etc.)
Can anybody shine some light on these observations please?
| Many (all?) integer series f(k), where k = 1,2,3,..K-1,K can be generated by a polynomial of order K-1.
Well, yes, all of them. Given distinct numbers $a_1, \ldots, a_n$, the polynomial
$$p_i(X)= \prod_{j\neq i}\frac{X-a_j}{a_i - a_j}$$
takes value $1$ at $a_i$ and $0$ at $a_j$ for all $j \neq i$. A linear combination of those gives you your desired outcome.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/42395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A curious sum for integers $\equiv 7\pmod 8$. For $n$ a natural integer congruent to $7$ modulo $8$, one has seemingly always
$$\sum_{k=1}^{(n-1)/2}\left(\frac{k}{n}\right)k=0$$
where $\left(\frac{k}{n}\right)$ denotes the Jacobi symbol.
First cases:
$n=7$: $1+2-3$
$n=15$: $1+2+4-7$
$n=23$: $1+2+3+4-5+6-7+8+9-10-11$
I do not see any reason for this. Did I miss something obvious?
| There are probably lots of ways to see this, but here's one: let $S_1$ be the sum above, and let $$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}^{(n-1)/2} 2k \left( \frac{2k}{n} \right) + \sum_{k=(n+1)/2}^{n-1} (2k-n) \left( \frac{2k}{n} \right) = 2S - n \sum_{k=(n+1)/2}^{n-1} \left( \frac{k}{n} \right), $$ where we used $(2/n) = 1$. Finally, we have $$S = 2S + n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right),$$ by changing $k$ to $n-k$ in the summation and using $(-1/n) = -1$. So $S = -n\sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right)$.
On the other hand, we can switch the index in $S_2$ from $k$ to $n-k$ as well, to get
$$S_2 = \sum_{k=1}^{(n-1)/2} (n-k) \left( \frac{n-k}{n}\right) = -n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right) + \sum_{k=1}^{(n-1)/2} k \left( \frac{k}{n} \right) = S + S_1 = 2S_1 + S_2.$$
(We used $(-1/n) = -1$ again).
Therefore $S_1 = 0.$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/120521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Upper limit on the central binomial coefficient What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ?
I just tried to proceed a bit, like this:
$ n! > n^{\frac{n}{2}} $
for all $ n>2 $. Thus,
$ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n $
But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).
| Since it gives tighter bounds, I will reproduce my answer from MathSE.
For $n\ge0$, we have (by cross-multiplication)
$$
\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}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2}
\end{align}
$$
Inequality $(2)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}
$$
For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\ge\frac{n+\frac14}{n+\frac54}\tag{4}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5}
\end{align}
$$
Inequality $(5)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}
$$
Note that the formula in $(3)$, which is decreasing, is bigger than the formula in $(6)$, which is increasing. Their ratio tends to $1$; therefore, they tend to a common limit, $L$.
Theorem $1$ from this answer says
$$
\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}
$$
which means that
$$
\begin{align}
L
&=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\
&=\frac1{\sqrt\pi}\tag8
\end{align}
$$
Combining $(3)$, $(6)$, and $(8)$, we get
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9
$$
| {
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} |
Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$? Numerical evidence suggests the following.
For $c \in \mathbb{N}, c > 2$ define the sequence $a_n$ by
$a_0=0,a_1=1, \; a_n=c a_{n-1} - a_{n-2}$
For $ 5 < n < 500, \; 2 < c < 100$ there are no primes in $a_n$ though
semiprimes exist.
Is it true that $a_n$ is always composite for $n > 5$
If yes is there explicit partial factorization?
Searching OEIS solved the case $c=6$
with a Pell equation.
Counterexamples are welcome.
| Put $u = (c + \sqrt{c^2-4})/2$. We have
$$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( \frac{u^{n+1/2}+u^{-n-1/2}}{u^{1/2}+u^{-1/2}} \right) =$$
$$\left( \frac{u^{n+1/2}-u^{-n-1/2}}{u^{1/2}-u^{-1/2}} \right) \left( \frac{u^{n+1/2}+u^{-n-1/2}}{u^{1/2}+u^{-1/2}} \right)=\left( \frac{u^{n+1}-u^{-n}}{u-1} \right) \left( \frac{u^{n+1}+u^{-n}}{u+1} \right).$$
Put $v_{n} = u^n + u^{-n}$, $x_{n} = \frac{u^{n+1}-u^{-n}}{u-1} $, $y_{n}=\frac{u^{n+1}+u^{-n}}{u+1} $ so $a_{2n} = a_n v_n$ and $a_{2n+1} = x_n y_n$. I claim that each of $v$, $x$ and $y$ are integer valued sequences which are greater than $1$ for $n \geq 2$, thus proving the claim.
The fastest way to see this is to note the recursions:
$$v_n = c v_{n-1} - v_{n-2} \quad v_0=2 \quad v_1 =c $$
$$x_n = c x_{n-1} - x_{n-2} \quad x_0=1 \quad x_1 =c+1 $$
$$y_n = c y_{n-1} - y_{n-2} \quad y_0=1 \quad y_1 =c-1 $$
A more conceptual way to see the rationality is to note that the Galois symmetry $\sqrt{c^2-4} \mapsto - \sqrt{c^2-4}$ takes $u$ to $u^{-1}$ and takes each of $v_n$, $x_n$ and $y_n$ to themselves.
| {
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} |
From a (not positive definite) Gram matrix to a (Kac-Moody) Cartan matrix Suppose I am given a symmetric matrix $G_{ij}$ with $G_{ii} = 2$. Can I always find an invertible integer matrix $S$ such that $(S^T G S)_{ii}=2$ and $(S^T G S)_{ij} \leq 0$ for $i \neq j$? Is there a practical algorithm to do so?
If you'd like a particular challenge, I'd like to know the answer for
$$G = \begin{pmatrix}
2 & -4 & 3 \\
-4 & 2 & -2 \\
3 & -2 & 2 \\
\end{pmatrix}.$$
| There is an algorithm, based on a 1907 article of Hurwitz I mention sometimes, based in turn on the tree of Markov numbers.
We begin with a ternary quadratic form $\langle 1,1,1,r,s,t \rangle.$ The (Lehman) discriminant of this is
$$ 4 + rst - r^2 - s^2 - t^2. $$ We would like to know whether we can find replacement values of $r,s,t$ so that all are nonpositive. Note first that this means the maximum discriminant we can allow is $4;$ anything bigger and we are out of luck as far as getting nonpositive off-diagonal coefficients.
Next note that we can always negate two $(rst)$ coefficients at a time. We can also permute $rst$ as we like.
What we actually do is an operation on the $(r,s,t)$ triples. Suppose that $r$ has opposite sign to $st,$ so that $|st-r| < |r|.$ If $r$ is the largest entry (in absolute value) for which this is true, we replace $r$ by $st-r$ and keep the same discriminant. Give me a few minutes to fiddle with matrices and find out what $3$ by $3$ matrix, of the type that David calls $S,$ that corresponds to this Hurwitz flip. Hurwitz gave no name to the operation; the high schoolers on MSE call it Vieta jumping. Well; in order to have $rst$ negative, or nonpositive, we must have their absolute values fairly small.
Later: the jump specified above goes with the matrix product
$$
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
s & 0 & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & t & s \\
t & 2 & r \\
s & r & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & s \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}
\right) =
\left(
\begin{array}{rrr}
2 & t & s \\
t & 2 & st-r \\
s & st-r & 2
\end{array}
\right)
$$
Here is another, which could be found from the first with some permutations on both sides.
$$
\left(
\begin{array}{rrr}
-1 & 0 & s \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & t & s \\
t & 2 & r \\
s & r & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
-1 & 0 & 0 \\
0 & 1 & 0 \\
s & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{rrr}
2 & rs- t & s \\
rs-t & 2 & r \\
s & r & 2
\end{array}
\right)
$$
Third:
$$
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & r & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
2 & t & s \\
t & 2 & r \\
s & r & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & r \\
0 & 0 & -1
\end{array}
\right) =
\left(
\begin{array}{rrr}
2 & t & rt-s \\
t & 2 & r \\
rt-s & r & 2
\end{array}
\right)
$$
| {
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Parametrizing the solutions to a diophantine equation of degree four Good evening,
Consider $x^4+y^4+z^4=2t^4$ where x,y,z,t integer.
Is it known how to find all parametrisation of this equation ?
If you have any parametrisation or reference of this equation, please post it
Thank you.
| Ramanujan gave two parametrizations: if $a+b+c=0$ then
$$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$$
(listed in Mathworld, equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract]
[summary]
The F. Ferrari Identity (1909) gives
$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$
There is also K. Ford's Theorem: if
$S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then
$$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$
For $k=2$ and $k=4$, the Ford parametrizations are the same as the two Ramanujan ones. For $k=2$, this is simple, and for $k=4$, $S_0 = a^4-4 a b^3$, $S_1 = b^4-4 a^3 b$, $S_2 = 6 a^2 b^2$, and
$$\begin{align}
S_0 - S_1 = -(c^3+2abc)(a-b)\\
S_1 - S_2 = -(b^3+2abc)(c-a)\\
S_2 - S_0 = -(a^3+2abc)(b-c)\\
\end{align}$$
where $c=-a-b$.
Edit: A complete discussion of these identities, by S. Ramanujan, F. Ferrari and Kevin Ford, plus further identities by S. Bhargava, may be found on pages 96 to 101 of Bruce Berndt's Ramanujan's Notebooks, Part IV (1994), at e.g. http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20IV.pdf.
| {
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} |
An identity involving a product of two binomial coefficients I'm trying to find a closed formula (in the parameters $q,N$) for the following sum:
$$ \sum_{k=q}^{N} {{k-1}\choose{q-1}} {{k}\choose {q}} $$
Can anybody give me a lead?
Lior
| This is a hypergeometric sum (the ratio of two consecutive terms is a rational function) and is therefore susceptible to Gosper-type algorithms, of which you can read more in the book A = B.
I believe Mathematica has got Gosper's algorithm implemented. It says:
In[38]:= s[n_, q_] := Sum[Binomial[k - 1, q - 1] * Binomial[k, q], {k, q, n}]
In[40]:= FullSimplify[s[n, q]]
Out[40]= Gamma[-2 q]/(Gamma[1 - q] Gamma[-q]) - Binomial[n, -1 + q] Binomial[1 + n,
q] HypergeometricPFQ[{1, 1 + n, 2 + n}, {2 + n - q, 2 + n - q}, 1]
The answer is that "this is some hypergeomteric series". This makes me think there is no closed form solution. We can plug in specific values of $q$ in which case we do get specific closed forms:
\begin{array}{l|l}
q & \text{closed form} \\ \hline
1 & \frac{n^2}{2}+\frac{n}{2} \\
2 & \frac{n^4}{8}-\frac{n^3}{12}-\frac{n^2}{8}+\frac{n}{12} \\
3 & \frac{n^6}{72}-\frac{7 n^5}{120}+\frac{n^4}{18}+\frac{n^3}{24}-\frac{5
n^2}{72}+\frac{n}{60} \\
4 & \frac{n^8}{1152}-\frac{17 n^7}{2016}+\frac{17 n^6}{576}-\frac{29
n^5}{720}+\frac{n^4}{1152}+\frac{13 n^3}{288}-\frac{n^2}{32}+\frac{n}{280} \\
5 & \frac{n^{10}}{28800}-\frac{31 n^9}{51840}+\frac{n^8}{240}-\frac{179
n^7}{12096}+\frac{773 n^6}{28800}-\frac{301 n^5}{17280}-\frac{47 n^4}{2880}+\frac{83
n^3}{2592}-\frac{53 n^2}{3600}+\frac{n}{1260} \\
\end{array}
I am not a combinatorialist, so these numbers seem completely arbitrary to me.
| {
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} |
Generalizing a pattern for the Diophantine $m$-tuples problem? A set of $m$ non-zero rationals {$a_1, a_2, ... , a_m$} is called a rational Diophantine $m$-tuple if $a_i a_j+1$ is a square. It turns out an $m$-tuple can be extended to $m+2$ if it has certain properties. The problem is to generalize the relations below to $m=5$.
I. $m=2$. Given $a,b$:
$$ax_i+1 = \big(a\pm\sqrt{ab+1}\big)^2\tag1$$
then {$a,b,x_1$} is a triple for any choice of $x_i$ . However, {$a,b,x_1,x_2$} is a quadruple if,
$$2(a^2+b^2)-(a+b)^2-3 = y^2$$
Ex. From $a,b = \frac{1}{16},\frac{17}{4}$, we get $x_1,x_2 = \frac{33}{16},\frac{105}{16}$, a quadruple first found by Diophantus.
II. $m=3$. Given $a,b,c$:
$$ax_i+1 = \big(a\sqrt{bc+1}\pm\sqrt{(ab+1)(ac+1)}\big)^2\tag2$$
then {$a,b,c,x_1$} is a quadruple. However, {$a,b,c,x_1,x_2$} is a quintuple if,
$$2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$
Ex. From $a,b,c = \frac{28}{5},\frac{55}{16},\frac{1683}{80}$, we get $x_1,x_2 = \frac{3}{80},1680$.
III. $m=4$. Given $a,b,c,d$:
$$\small(ax_i+1)(abcd-1)^2 = \big(a\sqrt{(bc+1)(bd+1)(cd+1)}\pm\sqrt{(ab+1)(ac+1)(ad+1)}\big)^2\tag3$$
then {$a,b,c,d,x_1$} is a quintuple. However, {$a,b,c,d,x_1,x_2$} is a sextuple if,
$$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$
Ex. From $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4}$, we get $x_1,x_2 = \frac{3213}{676},\;\frac{665}{1521}$, one of first sextuples found by Gibbs in 1999.
$^{\color{red}\dagger}$ These two can be satisfied by the parametric example in the variable $t$ in Dujella's website.
IV. Notes:
In general, an $n$-tuple can be extended to a $n+1$ (unconditional) and $n+2$ (conditional) for $n=2,3,4$. Also, one root $x_i$ is equal to zero if,
$$(a-b)^2 = 4\\
(a+b-c)^2 = 4(ab+1)\\
(a+b-c-d)^2 = 4(ab+1)(cd+1)$$
for relations $(1), (2), (3)$, respectively.
V. Question:
For $m=5$, given $a,b,c,d,e$:
$$\text{LHS}? = \text{RHS}?\tag4$$
*
*Can we find $(4)$, analogous to the first three? If yes, then maybe we can use known $5$-tuples or $6$-tuples to generate $7$-tuples, of which there is yet no known example.
*The pattern is suggestive. But, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$?
| 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)
It does not always work because it only guarantees that the new number $a_6$ times numbers from the quintuple plus one will be squares times some number, i.e.
$a_6 a_i + 1 = qx_i^2, i<6 $
If we get lucky and $q$ itself is a square we get a rational Diophantine sextuple
Here is an example of a regular sextuple that can be generated this way from any of its quintuples.
249/2048 3720/6241 715/384 369/128 38/3 920/3
| {
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"source": "stackexchange",
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"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 shape pairs of standard bitableaux.
But, I don't know what the map $W(B_n) \to SBT (n) \times SBT (n)$ is. In particular, $W(B_2) \to SBT (2) \times SBT (2)$? Here
\begin{align*}
W(B_2) \cong & \left\{
\begin{pmatrix}
1 & 2 & -1 & -2 \\
1 & 2 & -1 & -2
\end{pmatrix}, \begin{pmatrix}
1 & 2 & -1 & -2 \\
2 & 1 & -2 & -1
\end{pmatrix}, \begin{pmatrix}
1 & 2 & -1 & -2 \\
-1 & 2 & 1 & -2
\end{pmatrix}, \right. \\
& \left. \begin{pmatrix}
1 & 2 & -1 & -2 \\
1 & -2 & -1 & 2
\end{pmatrix},
\begin{pmatrix}
1 & 2 & -1 & -2 \\
-2 & 1 & 2 & -1
\end{pmatrix}, \begin{pmatrix}
1 & 2 & -1 & -2 \\
2 & -1 & -2 & 1
\end{pmatrix}, \right. \\
& \left. \begin{pmatrix}
1 & 2 & -1 & -2 \\
-1 & -2 & 1 & 2
\end{pmatrix}, \begin{pmatrix}
1 & 2 & -1 & -2 \\
-2 & -1 & 2 & 1
\end{pmatrix}\right\}.
\end{align*}
| 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 the work of McGovern, van Leeuwen, Shimizono, Pietraho, Taskin, etc.
The $W(B_2)$ example you request doesn't involve any of the horizontal/vertical domino rotations that make this theory tricky, just a little bumping. Here are the 8 domino tableaux pairs in the order you gave the $W(B_2)$ elements. Note that all but the 5th and 6th are involutions, so their left and right tableaux are the same in all but those. [If someone knows how to TeX these into nice domino pictures, please do so and let me know how you do it.]
\begin{gather*}
\left(\begin{matrix} 0 & 1 & 1 & 2 & 2 \end{matrix}, \quad \begin{matrix} 0 & 1 & 1 & 2 & 2 \end{matrix}\right), \quad
\left(\begin{matrix} 0 & 1 & 1 \\ 2 & 2 \end{matrix}, \quad \begin{matrix} 0 & 1 & 1 \\ 2 & 2 \end{matrix}\right), \\ \\
\left(\begin{matrix} 0 & 2 & 2 \\ 1 \\ 1 \end{matrix}, \quad \begin{matrix} 0 & 2 & 2 \\ 1 \\ 1 \end{matrix}\right), \quad
\left(\begin{matrix} 0 & 1 & 1 \\ 2 \\ 2 \end{matrix}, \quad \begin{matrix} 0 & 1 & 1 \\ 2 \\ 2 \end{matrix}\right), \\ \\
\left(\begin{matrix} 0 & 1 & 1 \\ 2 \\ 2 \end{matrix}, \quad \begin{matrix} 0 & 2 & 2 \\ 1 \\ 1 \end{matrix}\right), \quad
\left(\begin{matrix} 0 & 2 & 2 \\ 1 \\ 1 \end{matrix}, \quad \begin{matrix} 0 & 1 & 1 \\ 2 \\ 2 \end{matrix}\right), \\ \\
\left(\begin{matrix} 0 & 2 \\ 1 & 2 \\ 1 \end{matrix}, \quad \begin{matrix} 0 & 2 \\ 1 & 2 \\ 1 \end{matrix}\right), \quad
\left(\begin{matrix} 0 \\ 1 \\ 1 \\ 2 \\ 2 \end{matrix}, \quad \begin{matrix} 0 \\ 1 \\ 1 \\ 2 \\ 2 \end{matrix}\right).
\end{gather*}
(The domino tableaux for $W(C_2)$ also have dominoes labeled 1 and 2, but no 0 square.)
| {
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"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]), {t, 0, 2*Pi},
Assumptions -> n \[Element] Integers && n >= 0 && b > -1 && b < 0]
$$-\left(2 \sqrt{\pi } (-i b)^{-n} \left(\cos \left(\frac{\pi n}{2}\right)+i \sin \left(\frac{\pi n}{2}\right)\right) \left((-i b)^n \cos \left(\frac{\pi n}{2}\right) \Gamma \left(\frac{n+1}{2}\right) \Gamma \left(\frac{n-1}{2}\right) \, _2F_1\left(\frac{1}{2},1;\frac{3-n}{2};\frac{1}{1-b^2}\right)+i b (-i b)^n \sin \left(\frac{\pi n}{2}\right) \Gamma \left(\frac{n}{2}\right)^2 \, _2F_1\left(\frac{1}{2},1;1-\frac{n}{2};\frac{1}{1-b^2}\right)+2 \sqrt{\pi -\pi b^2} \Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{n+1}{2}\right)\right)\right. $$
$$\left(\left(b^2-1\right) \Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{n+1}{2}\right)\right)^{-1}$$
Therefore, the integral under consideration can be expressed as a series.
| {
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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 there are no triple threats. I've been able to prove the following weaker result that if $(x,a,b,c)$ is a triple-threat and $x \equiv 1$ (mod 5), then none of $a,b$ or $c$ can be $1$ (mod 5). For my purposes, this is useful, but I can get a stronger result with a simpler proof if I can get that there are no triple threats.
Let me briefly sketch the idea of the proof: Most of what I have been able to do has been from looking at the more general equation
$$x^2+x+1=(a^2+a+1)(b^2+b+1)p$$ where $x,a,b,p,a^2+a+1$ and $b^2+b+1$ are all prime. Note that this equation does solutions (and probably has infinitely many). Call this an almost-triple threat.
The key approach is showing the following sort of result: If $a \leq b$, and we have some specific modulo restriction on $x,a,b,p$ then one must have $a^2+a+1<p$. If one then applies to the original triple threat equation one can in some circumstances get a contradiction by forcing that $a^2+1<c^2+c+1<b^2+b+1<a^2+a+1$.
Let me briefly sketch how in most contexts I've obtained that $a^2+a+1<p$.
My primary method of attack on this equation has been to rewrite it as
$$(x-a)(x+a+1)=((b^2+b+1)p-1)(a^2+a+1) $$ and
$$(x-b)(x+b+1)=((a^2+a+1)p-1)(b^2+b+1). $$
Since $a^2+a+1$ and $b^2+b+1$ are assumed to be prime, one has that either the pair of relations $a^2+a+1|x-a$ and $x+a+1|(b^2+b+1)p-1$, or the pair of relations $a^2+a+1|x+a+1$ and $x-a|(b^2+b+1)p-1$. One gets similar pairings using the second version above with $b^2+b+1$. One has then four cases:
I) $a^2+a+1|x-a$ and $b^2+b+1|x-b$
II) $a^2+a+1|x-a$ and $b^2+b+1|x+b+1$
III) $a^2+a+1|x+a+1$ and $b^2+b+1|x-b $.
IV) $a^2+a+1|x+a+1$ and $b^2+b+1|x+b+1$.
It is not too hard to show in Cases II and III that one always has $a^2+a+1<p$. The difficulty is primarily in Cases I and IV. In Case I one gets with a little algebra that $x+a+1|p(a+b+1) + \frac{x-b}{b^2+b+1}$. One then has $k(x+a+1)= p(a+b+1) + \frac{x-b}{b^2+b+1}$. For many choices of $x,a,b$ mod 5 one can force $k$ to be large. Say $k \geq 3$. Then one has $$a+x+1 \leq \frac{ p(a+b+1) + \frac{x-b}{b^2+b+1}}{3},$$ and one can get that $a^2+a+1<p$ from this inequality and a little work. (I can expand on this part if necessary.)
The main difficulty right now is dealing with Case IV. I can handle Case IV under some small modulus assumptions, but I cannot prove that $a^2+a+1<p$ in Case IV unconditionally.
In Case IV one has with a little work that
$x-a|p(a+b+1)-\frac{x+b+1}{b^2+b+1}$ and $x-b|p(a+b+1)-\frac{x+a+1}{a^2+a+1}$ but it seems tough to use these to get a tight enough bound to conclude what is wanted without having some extra modulus restriction on the variables. Any thoughts on either clearing out Case IV, or any other method of showing that triple threats don't exist via some other method would be appreciated.
| 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 the triangle inequality gives $M(n) <2n$, but it seems there should be better upper bounds.
Remark. 1) Let $a = \cos(\pi/8), b = \sin(\pi/8)$, then if for even numbers $n$ we define the vectors $u = \sqrt{\frac{2}{n}}(a,b,a,b,\dots)$ and $ v = \sqrt{\frac{2}{n}}(b,-a,b,-a,\dots)$, we have $f(u,v)=\sqrt{2}n\leq M(n)$.
2)A similar construction shows that for every positive integers $n,k$, $\frac{M(n)}{n}\leq \frac{M(nk)}{nk}.$
| 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 $b^2 = \frac{1}{n-1}(1+\frac{n}{\sqrt{2n^2-1}})$ and deduce $a,c,d$ from the constraints $|u|=|v|=1$ and $u\cdot v=0$, yielding $M(n)\ge \sqrt{2n^2-1}$ when $n$ is odd. Now we still must show that more $a$ in $u$ or more than two distinct components in $u$ cannot improve the bound.
EDIT: Here are elementary details on an analysis for $u$ and $v$ restricted to
$$u=(a,a,\dots,a,b,b,\dots,b)\in\{a\}^{k}\times\{b\}^{\ell}$$
$$v=(c,c,\dots,c,d,d,\dots,d)\in\{c\}^{k}\times\{d\}^{\ell}$$
with $k+\ell=n$ odd or even and $0\lt d\le b$ and $0\lt a\le -c$.
$\langle u|u\rangle = \langle v|v\rangle = 1$ and $\langle u|v\rangle = 0$ translate into
$$ka^2+\ell b^2 = 1$$
$$kc^2+\ell d^2 = 1$$
$$kac+\ell bd = 0$$
hence
$$(kac)^2 = (\ell bd)^2 = (\ell b^2)(\ell d^2) = (1-ka^2)(1-kc^2)$$
simplifies into
$$ka^2+kc^2 = 1$$
and likewise
$$\ell b^2+\ell d^2 = 1$$
Furthermore
$$\begin{matrix}
k\ell(ab)^2 &=& (ka^2)(\ell b^2) &=& (1-\ell b^2)(\ell b^2) & &\\
k\ell(cd)^2 &=& (kc^2)(\ell d^2) &=& (1-\ell d^2)(\ell d^2) &=& (\ell b^2)(1-\ell b^2)\\
\end{matrix}$$
hence $0\lt 1-\ell b^2$ and
$$ab = -cd = \sqrt{\frac{b^2-\ell b^4}{k}}$$
This all enables to write
$$f(u,v) = k^2(c^2-a^2)+\ell^2(b^2-d^2)+2k\ell(ab-cd)$$
$$= k(1-2ka^2)+\ell(2\ell b^2-1)+4k\ell ab$$
$$= k(2\ell b^2-1)+\ell(2\ell b^2-1)+4k\ell ab$$
and finally
$$f(u,v) = g(x) := 2n\ell x -n + 4\ell\sqrt{k}\sqrt{x-\ell x^2}$$
where $0\lt x=b^2\lt \frac{1}{\ell}$. Actually $b^2$ cannot be arbitrarily close to $0$ because from our assumptions
$$1 = \ell b^2+\ell d^2 \le 2\ell b^2$$
thus
$$\frac{1}{2\ell}\le x=b^2\lt \frac{1}{\ell}$$
Now
$$g'(x)=2n\ell + 2\ell\sqrt{k}\frac{1-2\ell x}{\sqrt{x-\ell x^2}}$$
starts positive and goes toward $-\infty$ on the mentioned interval for $x$, therefore $g(x)$ and $f(u,v)$ will be maximal when $g'(x)=0$, which happens when
$$ 2\ell x-1 = \frac{2n\ell}{2\ell\sqrt{k}}\sqrt{x-\ell x^2}$$
or equivalently on the prescribed interval
$$ (2\ell x-1)^2 = \frac{n^2}{k}(x-\ell x^2)$$
which is the quadratic equation
$$x^2-\frac{1}{\ell}x+\frac{k}{\ell(4k\ell+n^2)} = 0$$
whose solutions are
$$x_\pm = \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{1}{\ell^2}-\frac{4k}{\ell(4k\ell+n^2)}}$$
$$= \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{4k\ell+n^2-4k\ell}{\ell^2(4k\ell+n^2)}}$$
$$= \frac{1}{2\ell} \left(1 \pm \frac{n}{\sqrt{4k\ell+n^2}}\right)$$
where only $x_+$ is in $[\frac{1}{2\ell},\frac{1}{\ell})$. The maximum is then
$$g(x_+) = \frac{n^2}{\sqrt{4k\ell+n^2}} + 4\ell\sqrt{k}\sqrt{x_+-\ell x_+^2}$$
The constant coefficient in the quadratic equation is also $x_+x_-$ thus
$$x_+-\ell x_+^2=x_+(1-\ell x_+)=x_+(\ell x_-)=\ell x_+x_- = \frac{k}{4k\ell+n^2}$$
yields
$$g(x_+) = \frac{n^2+4\ell k}{\sqrt{4k\ell+n^2}} = \sqrt{4k\ell+n^2}$$
In the end, when $u$ and $v$ are restricted to the shapes above, $f(u,v)$ is maximal for
$$a = +\sqrt{\frac{1}{2k} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad b = \sqrt{\frac{1}{2\ell} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\
c = -\sqrt{\frac{1}{2k} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad d = \sqrt{\frac{1}{2\ell} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\
$$
and the maximum is
$$f(u,v)=\sqrt{4k\ell+n^2}$$
Now allowing $k$ and $\ell=n-k$ to vary for $n$ fixed, one gets that $f(u,v)=\sqrt{4k\ell+n^2}$ is maximum over $k$ for $k=n/2$ when $n$ is even, in which case $f(u,v)=\sqrt{2n^2}$, and $k=(n\pm1)/2$ when $n$ is odd, in which case $f(u,v)=\sqrt{2n^2-1}$. We still have to show that other shapes for $u$ and $v$ cannot improve $f(u,v)$.
| {
"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
$$n^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^na_i\right)^3.$$
I have a proof that $(P_n)$ holds for $3\leq n\leq8$, but for $n\geq9$ my method does not work and I did not see any counterexample for $n\ge 9$.
Is the inequality $(P_n)$ true for all $n$? Or otherwise, what is the largest value of $n$ for which it holds?
Thank you!
| 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 otherwise we can divide by $A>0$ and consider $c_j=\frac{a_j}{A}$ and prove the inequality for them etc
So we need to prove $1-3B+3C \ge \frac{1}{N^2}$ under the hypothesis as above (the polynomial $X^N-X^{N-1}+BX^{N-2}-CX^{N-3}+...$ has real roots and $C \ge 0$
Then if we let $b_j=a_j-\frac{1}{N}, A_1,B_1,C_1$ the corresponding symmetric polynomials in $b_j$ we have $A_1=0, B_1=B-\frac{N-1}{2N}=B-\frac {1}{2}+\frac{1}{2N}, C_1=C-B+\frac{2B}{N}+\frac{1}{3}-\frac{1}{N}+\frac{2}{3N^2}$ so the inequality becomes ($C-B=C_1+...$ from the last equality)
$1+3C_1-\frac{6B}{N}-1+\frac{3}{N}-\frac{2}{N^2}\ge \frac{1}{N^2}$ and since
$\frac{6B}{N}=\frac{6B_1}{N}+\frac{3}{N}-\frac{3}{N^2}$ all reduces to
$3C_1-\frac{6B_1}{N} \ge 0$
But now the polynomial $X^N+B_1X^{N-2}-C_1X^{N-3}+...$ has real roots too as they are just $b_k$ and hence $B_1 \le 0, B_1=-B_2, B_2 \ge 0$ so the inequality reduces to $2B_2+NC_1 \ge 0$ and we know that $C_1=C+\frac{N-2}{N}B_2-\frac{(N-1)(N-2)}{6N^2}$
So we need $C_1$ negative but $C \ge 0$
By differentiating $N-3$ times and using Gauss Lucas/Rolle (so the cubic that results which is in standard form) has real roots so $4(-p)^3 \ge 27q^2$, we get some constraints on $B_2, -C_1$ which are enough to give the result for $N \le 6$ with some crude approximations
Then if we try easy counterexamples for the $b_k$ of the type $N-1$ $a$ and one $-(N-1)a$ we can solve at $N=10$, $a > \frac{3}{80}$ close enough to it to satisfy the inequality $C>0$ (which is satisifed at $a=\frac{3}{80}$ that one giving equality in the OP inequality as normalizing to integers we get $11$ taken $9$ times, $-19$ taken once and it is easy to see that the constraints are good and $S_1=80, S_3=5120$ and obviously $100\cdot 5120=80^3$
So as noted in the comments taking $9$ of $111$ and one of $-199$ gets a counterexample with a positive sum of cubes (corresponding to $a=\frac{3}{80}+\frac{1}{800}$ normalized to integers)
| {
"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 elaborate more how they apply the formula 5.1 in example 5.6. because I don't know how to apply it. for example, what is $c_{i1}$
Thanks in advanced.
Theorem $5.5 ([26])$.
Let $f : k^{r} \to k$ be any function on a finite field $k$. Then there exists a unique polynomial $g : k^{r} \to k$, such that $\forall x \in k^{r}, f(x)=g(x).$
Any such mapping over a finite field can be described by a unique polynomial.
Using Lagrange interpolation, we can easily determine the polynomial. Let $f: k^{r}\to k$ be any function on $k$. Then
$$g(x)=\sum_{\left(c_{i1},\ldots,c_{ir}\right)\in k^{r}}f\left(c_{i1},\ldots,c_{ir} \right)\prod_{j=1}^{r}\left(1-\left(x_{j}-c_{ij} \right)^{p-1} \right)\quad \mbox{(5.1)}$$
is the unique polynomial that defines the same mapping as $f$.
Example $5.6$
Suppose $k=\mathbb{F}_{3},r=2,$ and the mapping $f$ is defined on $\mathbb{F}_{3}^{2}=\{0,1,2\}\times\{0,1,2\}$ as follows:
$$f(0,0)=0,$$
$$f(0,1)=1,$$
$$f(0,2)=2,$$
$$f(1,0)=1,$$
$$f(1,1)=2,$$
$$f(1,2)=0,$$
$$f(2,0)=2,$$
$$f(2,1)=0,$$
$$f(2,2)=1.$$
Then the polynomial $g$ that defines the same mapping as $f$ is constructed as follows:
$$\begin{align*}
g(x,y)&=0\\
&+1\left((1-x^{2})(1-(y-1)^{2}) \right)\\
&+2\left((1-x^{2})(1-(y-2)^{2}) \right)\\
&+1\left((1-(x-1)^{2})(1-y^{2}) \right)\\
&+2\left((1-(x-1)^{2})(1-(y-1)^{2}) \right)\\
&+0\\
&+2\left((1-(x-2)^{2})(1-y^{2}) \right)\\
&+0\\
&+1\left((1-(x-2)^{2})(1-(y-2)^{2})\right)\\
&=x+y.
\end{align*}$$
Reference:
Mathematical Concepts and Methods in Modern Biology 1st Edition Using Modern Discrete Models
| 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 $9$ function values you specify, but as integers. The standard Lagrange interpolation would be
$$\begin{align*}
g(x,y)&=0\\
&+1 \frac {x \left( x-2 \right)
\left( y-2 \right) \left( y-1 \right) }{-2}
\\
&+2\frac {x \left( x-1
\right) \left( y-2 \right) \left( y-1 \right) }{4}
\\
&+1\frac {y
\left( x-2 \right) \left( x-1 \right) \left( y-2 \right) }{-2}
\\
&+2\frac{xy \left( x-2 \right) \left( y-2 \right)}{1} \\
&+0\\
&+2\frac {y \left( x-2 \right)
\left( x-1 \right) \left( y-1 \right) }{4}
\\
&+0\\
&+1\frac {xy \left( x-1
\right) \left( y-1 \right) }{4}
\\
&=\frac{9\,{x}^{2}{y}^{2}-15\,{x}^{2}y-15\,x{y}^{2}+21\,xy+4\,x+4\,y}{4}.
\end{align*}$$
And that is the unique polynomial of that degree over the reals, which does that. Then, $\mod 3,$ we have $4=1$ and the other coefficients are $0$, so indeed, $x+y.$
What you (and your book) wrote, treated as over the integers, comes out $9\,{x}^{2}{y}^{2}-18\,{x}^{2}y-18\,x{y}^{2}+6\,{x}^{2}+24\,xy+6\,{y}^{
2}-2\,x-2\,y-3
$
The values at the appropriate integer points are $$[0, 0, -3], [1, 0, 1], [2, 0, 17], [0, 1, 1], [1, 1, 2], [2, 1, -3], [0, 2, 17], [1, 2, -3], [2, 2, -11]$$ but again,$\mod 3,$ everything comes out fine.
| {
"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 $A = \begin{pmatrix} 0 & a \\ a^* & 0 \end{pmatrix}$ and $B=\begin{pmatrix} b & 0 \\ 0 & b^* \end{pmatrix}$ for any complex numbers, $a,b \in \mathbb{C}.$ Can one understand this somehow?
| 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
$$U\begin{pmatrix} A & B \\ B^* & A \end{pmatrix}U^{-1}=\begin{pmatrix} 0 & A+b\mathbb{1} \\ A+b^\ast\mathbb{1} & 0 \end{pmatrix}$$
has the desired form.
Note also that the eigenvalues $\lambda$ of this matrix come in inverse pairs $\pm \lambda$, because it anticommutes with $\begin{pmatrix} \mathbb{1} & 0 \\ 0& -\mathbb{1} \end{pmatrix}$ (chiral symmetry).
| {
"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
$$f(y)\approx y\lfloor\frac{1}{y} \rfloor \left(\frac{1-y}{1+y-y\lfloor\frac{1}{y} \rfloor}\right)^{\lfloor\frac{1}{y} \rfloor-1}.$$
This is actually quite accurate already for $y\lesssim 0.2$, see the plot where the exact result (red) is almost indistinguishable from the approximation (green).
The upper envelope of the zigzag curve (orange) is given by the points where $1/y$ is an integer, so by setting $x=1/y$,
$$f_{\rm upper-envelope}(y)=(1/y-1)^{1/y-1}.$$
The lower envelope (blue) is obtained by setting $x=1/y-1$,
$$f_{\rm lower-envelope}(y)=(2y)^{2-1/y} (1-y)^{1/y-1} .$$
| {
"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
$$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A-1)!a!}{(A+a)(x-A+1)^{(A+a)}},$$
where $x$ is some variable and $(x)^{(a)}=x(x+1)\cdots(x+a-1)$ is the rising factorial. This is what I would like to prove.
Notice how the left hand side in principle has poles at all integer values of $x$, while the right hand side only has poles at integers smaller than $A-1$.
| 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-y)^{x-b-1} y^{B+b-1} (1-z)^bz^{B-1}\,{\rm d}y\,{\rm d}z\\
&= \int_0^1 \int_0^1 (1-y)^{x} y^{-2} (z(1-z))^{-1} \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1} (y(1-z)/(1-y))^{b+1} (yz)^{B}\,{\rm d}y\,{\rm d}z\\
&= -\int_0^1 \int_0^1 (1-y)^{x} y^{-2}(z(1-z))^{-1} \left( \frac{y(1-z)}{1-yz} \right)^{a+1} \left(\frac{yz}{1-yz}\right)^{A}\,{\rm d}y\,{\rm d}z\\
&= -\int_0^1 \int_0^1 (1-y)^{x} (yz)^{A-1} (y-yz)^a (1-yz)^{-(A+a+1)}\,{\rm d}y\,{\rm d}z.
\end{split}
Substituting $(u,v)=(\frac{1-y}{1-yz},yz)$, or $(y,z) = (1-u(1-v),\frac{v}{1-u(1-y)})$ we further get
\begin{split}
... &= \int_0^1 \int_0^1 u^x (1-u)^a v^{A-1} (1-v)^{x-A} \frac1{1-u(1-v)}\,{\rm d}u\,{\rm d}v\\
&= \sum_{i\geq 0}\int_0^1 \int_0^1 u^{x+i} (1-u)^a v^{A-1} (1-v)^{x-A+i}\,{\rm d}u\,{\rm d}v \\
&= \sum_{i\geq 0} \frac{1}{(a+1)A\binom{x+i+a+1}{a+1}\binom{x+i}{A}} = \sum_{i\geq 0} \frac{1}{(a+1)A\binom{A+a+1}A \binom{x+i+a+1}{A+a+1}},\\
&= \frac{x+a+1}{(a+1)A\binom{A+a+1}A (A+a) \binom{x+a+1}{A+a+1}} = \frac{1}{(A+a)^2\binom{A+a-1}{a}\binom{x+a}{A+a}},
\end{split}
which matches the r.h.s.
QED
| {
"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
$$\frac{(a_1^2+xa_2^2)(b_1^2+xb_2^2)}{(a_1b_1+xa_2b_2)^2} -1 \ge \left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1.$$
Since
$$(a_1^2+xa_2^2)(b_1^2+xb_2^2)-(a_1b_1+xa_2b_2)^2=x(a_1^2b_2^2+a_2^2b_1^2-2a_1a_2b_1b_2)= x[(a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2]= x[1-(a_1b_1+a_2b_2)^2]$$
and
$$\left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1=\frac{4x[1-(a_1b_1+a_2b_2)^2]}{[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2},$$
the above inequality is equivalent to (notice that $x[1-(a_1b_1+a_2b_2)^2] \ge 0$)
$$[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2 \ge 4(a_1b_1+xa_2b_2)^2,$$
or
$$(1-x)+(1+x)(a_1b_1+a_2b_2) \ge 2(a_1b_1+xa_2b_2),$$
or
$$(1-x)(1-a_1b_1+a_2b_2) \ge 0,$$
which is obvious.
| {
"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 formula for the measurment uses cos(theta) in the calculation. I need to know the mean, the variance and the distribution function that result from taking the cosine of theta in order to do my calculations correctly.
| 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 \le x \le 2\left( {k + 1} \right)\pi - \arccos y} \right) \\
= \sum\limits_{k = - \infty }^{ + \infty } {\int_{2k\pi + \arccos y}^{2\left( {k + 1} \right)\pi - \arccos y} {f_X (x)} } dx \\
f_Y (x) = \sum\limits_{k = - \infty }^{ + \infty } {\left[ { - \left( { - \frac{1}{{\sqrt {1 - y^2 } }}} \right)f_X \left( {2\left( {k + 1} \right)\pi - \arccos y} \right) - \left( { - \frac{1}{{\sqrt {1 - y^2 } }}} \right)f_X \left( {2k\pi + \arccos y} \right)} \right]} \\
= \frac{1}{{\sqrt {1 - y^2 } }}\sum\limits_{k = - \infty }^{ + \infty } {\left[ {f_X \left( {2\left( {k + 1} \right)\pi - \arccos y} \right) + f_X \left( {2k\pi + \arccos y} \right)} \right]} \\
\end{array}
| {
"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 $f(n) = (n^2-n+2)/2$, a second order polynomial.
Many (all?) integer series $f(k)$, where $k = 1,2,3,...,K-1,K$ can be generated by a polynomial of order $K-1$.
The integer series $2^{n-1}$, where $n = 1,2,3,...,K$ can also be generated by a polynomial of order $K-1$.
The following interesting thing happens.
If we describe the series $1,2,4$ by $f(n) = \frac{n^2-n+2}{2}$ then $f(4) = 7$
For $1,2,4,8$, $f(n) = \frac{n^3-3n^2+8n}{6}$ and $f(5) = 15$
For $1,2,4,8,16$, $f(n) = \frac{n^4-6n^3+23n^2-18n}{24}$ and $f(6)=31$
For $1,2,4,8,16,32$, $f(n) = \frac{n^5-10n^4+55n^3-110n^2+184n}{120}$ and $f(7)=63$
For $1,2,4,8,16,32,64$, $f(n) = \frac{n^6-15n^5+115n^4-405n^3+964n^2-660n+720}{720}$ and $f(8)=127$
I have verified this till order 14.
Lets add the series "1" and "1,2" for completeness:
For $1$, $f(n) = 1$ and $f(2) =1$. $f(2) = 2 \cdot f(1)-1$
For $1,2$, $f(n) = n$ and $f(3) = 3$. $f(3) = 2 \cdot f(2)-1$
This suggests that $f(k+1) = 2 \cdot f(k) -1$ when $f(n)$ is the $k-1$ th order polynomial function that generates the values $1,2,4,...2^{k-1}$.
This raises the question if this is true for all integer series of this type and if so, how to prove it. I hope that I am not overlooking the obvious.
Another observation is that if you write the polynomials that describe the series $1,2,4,8,...$ in a fractional form where all coefficients of $n^k$ in the numerator are integers, then the denominator always seems to be $(K-1)!$ ($1,1,2,6,24,120,$ etc.)
Can anybody shine some light on these observations please?
| 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 $f(0), f(1), ... f(n)$ is
$$f(x) = \sum_{i=0}^{n} \Delta^i f(0) {x \choose i}.$$
(You should think of this as analogous to Taylor expansion. The proof uses the identity $\Delta {x \choose i} = {x \choose i-1}$.) In particular, the $\Delta^i f(0)$ are all integer if and only if $f(0), f(1), ... f(n)$ are all integers, which is the second pattern you observe.
For the powers of $2$ we have $\Delta^i f(0) = 1$ for $1 \le i \le n$, which gives
$$f(x) = \sum_{i=0}^{n} {x \choose i}.$$
It follows that $f(n+1) = 2^{n+1} - 1$, which I think is the first pattern you observe.
| {
"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
$$
A_0 = A_1 = 0;\quad A_n = A_{\lfloor{n/2}\rfloor} + A_{\lceil{n/2}\rceil} + n - \frac{\lfloor{n/2}\rfloor}{\lceil{n/2}\rceil+1} - \frac{\lceil{n/2}\rceil}{\lfloor{n/2}\rfloor+1}.
$$
(See Knuth's AOCP, for instance.)
Flajolet and Golin in 1993 used complex analysis (Mellin transforms) and Fourier analysis to find a precise asymptotic approximation of $A_n$. I am interested in finding lower and upper bounds on $A_n$ of the form $n\lg n + \alpha n + \beta$, where $\lg n$ is the binary logarithm, not using these powerful but complicated analytical approaches.
By distinguishing on the parity of $n$, we simply get
$$
A_{2p} = 2 A_{p} + 2p - 2 + \frac{2}{p+1};\quad A_{2p+1} = A_{p} + A_{p+1} + 2p - 1 + \frac{2}{p+2}.
$$
I tried difference equations, by letting $\Delta_n := A_{n+1} - A_{n}$, yielding
$$
\Delta_{2p} = \Delta_{p} + 1 + \frac{2}{p+2} - \frac{2}{p+1};\quad \Delta_{2p+1} = \Delta_p + 1.
$$
Then, I am stuck.
The same study for the maximum number of comparisons leads to simpler difference equations: $\Delta_{2p} = \Delta_{2p+1} = \Delta_{p} + 1$, which implies $\Delta_n = \lfloor{\lg n}\rfloor + 1$, to wit, the number of bits in the binary expansion of $n$. From there, a closed form for the maximum cost $\sum_{k=1}^{n-1}\Delta_k$ follows relatively easily (see Flajolet and Sedgewick, for instance).
Any idea how to bound $\Delta_k$ and $\sum_{k=1}^{n-1}\Delta_k$ in the present case?
| 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 $A_p,A_{p+1}.$
$A_{2p+1} = A_p + A_{p+1} + 2p - 1 + \frac{2}{p+2}$
$ \le (p+1)\log_2 (p+1) + p ~\log_2 p + \alpha(2p+1) + \beta + \beta + 2p - 1 + \frac{2}{p+2}.$
Using $\log_2 x \le \frac{(x-(p+1/2))}{(p+1/2)\log 2} + \log_2 (p+1/2)$ at $x=p, p+1$:
$$A_{2p+1} \le (p+1)\log_2 (p+1/2) + p~\log_2p + \frac{p+1}{(2p+1)\log 2} - \frac {p}{(2p+1)\log 2} \\\ + \alpha(2p+1) + \beta + \beta + 2p - 1 + \frac{2}{p+2}$$
$$A_{2p+1} \le \bigg((2p+1)\log_2(2p+1) + \alpha(2p+1) + \beta \bigg) + \\\ \beta -2+ \frac{2}{p+2}+\frac{1}{(2p+1)\log 2}.$$
So, we'll also want $\beta - 2 + \frac{2}{p+2} + \frac{1}{(2p+1)\log 2} \le 0.$ Then if we choose $\alpha, \beta$ so that the base of the induction is satisfied, we get $A_n \le n \log_2 n + \alpha n + \beta.$
I think that you can get slightly better bounds by starting the induction higher, but let's start it at $p=1, A_2 = 1, A_3 = \frac 8 3.$ Then we want $4$ inequalities to be satisfied:
$1 \le 2 \log_2 2 + 2\alpha + \beta$
$\frac 8 3 \le 3 \log_2 3 + 3\alpha + \beta$
$\beta -2 + 1 \le 0$
$\beta -2 + \frac 23 + \frac{1}{3 \log 2} \le 0$
and we want to minimize $\alpha$ and then minimize $\beta$ for that $\alpha$ subject to these constraints. If I calculate correctly, then the solution is $\alpha = -\frac76 + \frac{1}{6 \log 2} = -0.9263, \beta = \frac43 - \frac 1{3\log 2} = 0.8524,$ and $A_n \le n \log_2 n + \alpha n + \beta$ should be satisfied for all $n \ge 2$. You can get a slightly lower value of $\alpha$ if you don't need the formula to work for $n=2$.
| {
"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
} |
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: ${12+2+1, 16+2+1,...,72+6+2,...}$
4-term sums: ${240+12+2+1,...,6480+72+6+2,...}$
| 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 sum of four squares not all the same and multiplying everything by $2^2$).
Now, if two or three are the same, then replace two using the identity $a^2 + a^2 = Q(2a+1) + Q(2a-1)$. The only situation where this fails is where our expression is of the form $a^2 + a^2 + (a+1)^2 + (a+1)^2$ (since we'd use $Q(2a+1)$ twice). But this is $4a^2 + 4a + 2$, which is $(2a+1)^2 + 1^2$, so we have an alternative here too.
PS. I notice that Emil has suggested an alternative way of dealing with repeats, while this was being edited into existence.
| {
"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 (references?):
(1) For any $n\geq 1$ we have
\begin{equation}
\binom{2n}{n} + \binom{2n}{n-1} = \sum_{i=0}^{n-1} (4i+3)C_i C_{n-i-1} \enspace. \quad (1)
\end{equation}
(2) For any $n\geq 1$ and $k=n,n+1,\ldots,2n-1$ we have
\begin{equation}
\frac{k-n+1}{n}\binom{2n}{k+1}=\sum_{i=0}^{2n-1-k} \frac{2k-2n+1}{k-i}\binom{2(n-i-1)}{k-i-1} \cdot C_i \enspace. \quad (2)
\end{equation}
For the special case $k=n$ equation (2) is the well-known relation
\begin{equation} C_n=\sum_{i=0}^{n-1} C_{n-1-i} C_i \enspace. \quad (2')
\end{equation}
For the special case $k=n+1$ equation (2) yields
\begin{equation*}
C_n=\frac{n+2}{2(n-1)} \sum_{i=0}^{n-2} \frac{3(n-1-i)}{n+1-i} \cdot C_{n-1-i} C_i \enspace, \quad (2'')
\end{equation*}
a weighted sum with one term less than (2').
I appreciate any hints, pointers etc.!
| 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 $(a_i)$ and $(b_i)$
$$
A(z)\,B(z)
= \sum_{i=0}^\infty a_i z^i \sum_{j=0}^\infty b_j z^j
= \sum_{n=0}^\infty z^n \sum_{i=0}^n a_i b_{n-i} .
$$
The terms in the summation in (2'') look like a convolution of $c_i$ and $\frac{j}{j+2}c_j$.
To get the necessary form one can differentiate and integrate. For instance
\begin{multline*}
z^2 \frac{d}{dz} \left( \frac{1}{z^2} \int_0^z t C(t)\, dt \right)
= z^2 \frac{d}{dz} \left( \frac{1}{z^2} \int_0^z t \sum_{i=0}^\infty c_i t^i \, dt \right)
= z^2 \frac{d}{dz} \left( \sum_{i=0}^\infty \frac{1}{z^2} \int_0^z c_i t^{i+1} \, dt \right) \\
= z^2 \frac{d}{dz} \left( \sum_{i=0}^\infty \frac{1}{z^2} \frac{c_i}{i+2} z^{i+2} \right)
= z^2 \frac{d}{dz} \left( \sum_{i=0}^\infty \frac{c_i}{i+2} z^{i} \right)
= \sum_{i=1}^\infty \frac{i}{i+2} c_i z^{i+1} \enspace .
\end{multline*}
First few terms are
$$
\frac{1}{3}\cdot z^2 + 1\cdot z^3 + 3\cdot z^4 + \frac{28}{3}\cdot z^5
+ 30\cdot z^6 + 99\cdot z^7 + \frac{1001}{3}\cdot z^8 + \cdots .
$$
Thus
\begin{multline*}
\sum_{i=0}^\infty c_i z^i \sum_{j=1}^\infty \frac{j}{j+2} c_j z^{j+1}
= c_0 \frac{1}{3} c_1 z^2 + \left(c_0 \frac{2}{4} c_2+ c_1 \frac{1}{3} c_1 \right) z^3 + \cdots \\
= \sum_{n=2}^\infty z^n \sum_{k=0}^{n-2} c_k c_{n-k-1} \frac{n-k-1}{n-k+1} .
\end{multline*}
Continuing in this fashion we arrive at the following equation, equivalent to identity (2'')
$$
\frac{3}{2z}\cdot \frac{d}{dz} \left( z^3
\int_0^z \left( C(z) \cdot\frac{d}{dz} \left( \frac{1}{z^2} \int_0^z z\cdot C(z) \, dz \right) \right) \, dz \right)
= C(z) - z - 1 ,
$$
which can be easily verified.
| {
"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 to $m=5$.
I. $m=2$. Given $a,b$:
$$ax_i+1 = \big(a\pm\sqrt{ab+1}\big)^2\tag1$$
then {$a,b,x_1$} is a triple for any choice of $x_i$ . However, {$a,b,x_1,x_2$} is a quadruple if,
$$2(a^2+b^2)-(a+b)^2-3 = y^2$$
Ex. From $a,b = \frac{1}{16},\frac{17}{4}$, we get $x_1,x_2 = \frac{33}{16},\frac{105}{16}$, a quadruple first found by Diophantus.
II. $m=3$. Given $a,b,c$:
$$ax_i+1 = \big(a\sqrt{bc+1}\pm\sqrt{(ab+1)(ac+1)}\big)^2\tag2$$
then {$a,b,c,x_1$} is a quadruple. However, {$a,b,c,x_1,x_2$} is a quintuple if,
$$2(a^2+b^2+c^2)-(a+b+c)^2-3 = y^2\,^{\color{red}\dagger}$$
Ex. From $a,b,c = \frac{28}{5},\frac{55}{16},\frac{1683}{80}$, we get $x_1,x_2 = \frac{3}{80},1680$.
III. $m=4$. Given $a,b,c,d$:
$$\small(ax_i+1)(abcd-1)^2 = \big(a\sqrt{(bc+1)(bd+1)(cd+1)}\pm\sqrt{(ab+1)(ac+1)(ad+1)}\big)^2\tag3$$
then {$a,b,c,d,x_1$} is a quintuple. However, {$a,b,c,d,x_1,x_2$} is a sextuple if,
$$2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcd+(abcd)^2 = y^2\,^{\color{red}\dagger}$$
Ex. From $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4}$, we get $x_1,x_2 = \frac{3213}{676},\;\frac{665}{1521}$, one of first sextuples found by Gibbs in 1999.
$^{\color{red}\dagger}$ These two can be satisfied by the parametric example in the variable $t$ in Dujella's website.
IV. Notes:
In general, an $n$-tuple can be extended to a $n+1$ (unconditional) and $n+2$ (conditional) for $n=2,3,4$. Also, one root $x_i$ is equal to zero if,
$$(a-b)^2 = 4\\
(a+b-c)^2 = 4(ab+1)\\
(a+b-c-d)^2 = 4(ab+1)(cd+1)$$
for relations $(1), (2), (3)$, respectively.
V. Question:
For $m=5$, given $a,b,c,d,e$:
$$\text{LHS}? = \text{RHS}?\tag4$$
*
*Can we find $(4)$, analogous to the first three? If yes, then maybe we can use known $5$-tuples or $6$-tuples to generate $7$-tuples, of which there is yet no known example.
*The pattern is suggestive. But, like quintics, is there a Galois-theoretic restriction on five variables $a,b,c,d,e$ that prevent generalization for $m>4$?
| (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 $a,b,c,d =1,3,0,0,\,$ yields $e_1, e_2 = 0,8$.
*Let $a,b,c,d =1,3,8,0,\,$ yields $e_1, e_2 = 0,120$.
*Let $a,b,c,d =1,3,8,120,\,$ yields $e_1, e_2 = 0,\frac{777480}{8288641}$.
The last was also found by Euler, so he must have a version of $(1)$. Thus, in general, an $m$-tuple can be extended to a quintuple. However, if $e_1 e_2+1 =\square$, then it yields a sextuple as in the example above,
*
*Let $a,b,c,d = \frac{5}{4},\;\frac{5}{36},\;\frac{32}{9},\;\frac{189}{4},\,$ yields $e_1,e_2 = \frac{3213}{676},\;\frac{665}{1521}$.
a solution unnoticed by Euler and only found in 1999.
Tinkering with $(1)$, it can be expressed by the elementary symmetric polynomials $\alpha_i$ in a much simpler form,
$$(\alpha_1-\alpha_5)^2=4(\alpha_2+\alpha_4+1)\tag2$$
where,
$$\begin{aligned}
\alpha_1 &=a + b + c + d + e\\
\alpha_2 &=a b + a c + b c + a d + b d + c d + a e + b e + c e + d e\\
\alpha_4 &=a b c d + a b c e + a b d e + a c d e + b c d e\\
\alpha_5 &=abcde\\
\end{aligned}$$
So if $(2)$ can be generalized, then the question can be rephrased as: Is there a version of $(2)$ using the elementary symmetric polynomials $\alpha_i$ for $a,b,c,d,e,f$?
| {
"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 answer on MSE shows that $P(n, k, \ell) \to 1/k$ as $n\to\infty$, regardless of $\ell$.
The proof given in the MSE answer is "Fourier analytic." Alternatively, we could argue "algebraically", as follows:
Sketch when $k=p$ is an odd prime: using Dirichlet's theorem, take a sequence of primes $\{a_i\}$ such that each $a_i$ is a primitive root modulo $p$. Consider the sequence $q_i = a_{i}^{p-2}$. The divisors of each $q_i$ are equidistributed modulo $p$, so roughly, whenever $n$ is one of these $q_i$, the probability of the divisor sum hitting what we want will be about $1/p$. Doing a sort of interpolation between these points (the details aren't too tricky) gives the result.
But when $k = 2^j$, this argument won't work, since powers of $2$ don't have primitive roots. However, the following generating function/"Fourier analytic" argument will work:
Consider the polynomial $$p_n (x) = \prod_{k=2}^{n} \left(\sum_{d\vert n} x^d \right) = (x+x^2)(x+x^3)(x+x^2+x^4) \cdots$$ Taking $\omega$ a $2^j$-th root of unity, a roots of unity filter gives $$P(n, 2^j, 0) = \frac{1}{2^j p_n (1)}\sum_{i=0}^{2^j - 1} p_n (\omega^i)$$ Now, note that for each $i$, by Dirichlet's theorem, we can find a prime $p$ of the form $$2^{i+1} m + 2^i + 1 = 2^i (2m+1)+1$$ so that $x^{2^i} + 1$ divides $x^p + x$. It follows that there exists $n_0$ such that for all $n\ge n_0$, $$(x^{2^{k-1}} + 1)(x^{2^{k-2}}+1) \cdots (x^4 + 1)(x^2 + 1)(x+1)\, \big \vert\, p_n (x)$$ and hence $p_n (\omega) = p_n (\omega^2) = \cdots p_n (\omega^{2^j - 1}) = 0$. Accordingly, $P(n, 2^j,0)$ is exactly $1/2^j$ for $n\ge n_0$.
My first question is what can be said about $n_0$? For instance, when $k=32$, then using Dirichlet's theorem to compute the smallest primes that satisfy the desired divisibility condition will show that $p_{41} (x)$ is divisible by $\prod_{i=0}^{4} (x^{2^i} + 1)$. But actually, in this case, $n_0 = 27 < 41$, because $x^8 + 1$ "anomalously" divides $x+x^3 + x^9 + x^{27}$.
So we are asking the question: for each $i$, what is the smallest $k$ for which $x^{2^i} + 1$ divides $q_k (x) = \sum_{d\vert k} x^d$? This also suggests the more general problem: how does the polynomial $q_k (x)$ factor?
| 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 $9=3^2=1+8=1+2^3$, so it is the only 'anomaly'.
| {
"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, $\binom{m}{a,b,c}$ is understood as $\frac{m!}{a!\,b!\,c!}$.
| 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 $[t^a]p$ is the coefficient of $t^a$ in the formal power series $p=p(t)$.
Note that $[t^{2j+1}](1-t^2)^{-(m+1)} = 0$. So the left hand side is
\begin{equation}
\begin{split}
[t^k](1-t^2)^{-(m+1)}(1+t)^m &= [t^k](1-t)^{-(m+1)}(1+t)^{-1} \\
&= \sum_{j=0}^k [t^{k-j}](1-t)^{-(m+1)} \cdot [t^j](1+t)^{-1} \\
&= \sum_{j=0}^k \binom{m+k-j}{m} (-1)^j \\
&= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j}{m} (-1)^{2j} + \binom{m+k-2j-1}{m} (-1)^{2j+1} \\
&= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j}{m}-\binom{m+k-2j-1}{m} \\
&= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j-1}{m-1}
\end{split}
\end{equation}
as desired.
| {
"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?
$$x\cdot \dfrac{(1-x)^{k-1}}{(k+1)^{k-2}}+\dfrac{(1-2x)^k}{k^k}\le \dfrac{1}{(k+2)^{k-1}}$$ creat by wang yong xi
This is my attempt
when $k=2$,then inequality can be written as
$$x\cdot (1-x)+\dfrac{(1-2x)^2}{4}\le\dfrac{1}{4}$$
it is obviously true.
when $k=3$then inequality can be written as
$$\dfrac{x(1-x)^2}{4}+\dfrac{(1-2x)^3}{27}\le\dfrac{1}{25}$$
or $$-\dfrac{(5x-1)^2(5x+8)}{2700}\le 0$$ it is clearly true.
when $k=4$ it's equivalent
$$\dfrac{x(1-x)^3}{25}+\dfrac{(1-2x)^4}{256}-\dfrac{1}{216}\le 0$$
or
$$\dfrac{(6x-1)^2(108x^2+12x-125)}{172800}\le 0$$
it is clearly
when $k=5$, it's equivalent
$$x\cdot\dfrac{(1-x)^4}{216}+\dfrac{(1-2x)^5}{5^5}-\dfrac{1}{7^4}=-\dfrac{(7x-1)^2(185563x^3-181202x^2-127589x+156384)}{1620675000}<0$$
But I can't prove for any postive intgers $k$.and I have found
$$LHS-RHS=[(k+2)x-1]^2\cdot h(x,k)$$.so we must prove $h(x,k)\le 0$ Thanks
| 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 $(1+x)^a>1+ax$ for $a=2-k$, $x=1/k$. Also $f(1)$ equals $\pm f(0)$. It means that the maximal value of $f$ on $[0,1]$ is attained at an interior point $a\in (0,1)$, and thus $f'(a)=0$.
I claim that $f'$ has unique root (multiplicity counted) on $(0,1/2]$, and this root is $\frac1{k+2}$, and at most one root on $[1/2,1)$. In both cases the only possible maximum point is $\frac{1}{k+2}$ (the second extremal point would be a local minimum)
We have $$f'(x)=(1-x)^{k-2}(1-kx)(k+1)^{2-k}-2(1-2x)^{k-1}k^{1-k}=0.$$
Denote $y=\frac{1-x}{1-2x}$, then $x=\frac{1-y}{1-2y}$, $1-kx=\frac{(k-2)y-(k-1)}{1-2y}$. Our equation $f'(x)=0$ in terms of $y$ rewrites as $$y^{k-2}((k-2)y-(k-1))(k+1)^{2-k}+2k^{1-k}=0.$$
LHS is monotone in $y$ for $y\in (1,\infty)$ (corresponds to $x\in (0,1/2)$) and for $y<0$ (corr. $x\in (1/2,1)$), that implies the above claim.
| {
"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:
$$
\mathbb E_{X_1, X_2, X_3} f(X_1, X_2) f(X_1, X_3) f(X_2, X_3)
\le ( \mathbb E_{X_1, X_2} f^2(X_1, X_2) )^{3/2}
$$
I tried to use Hoeffding's result
$$
\mathbb E f\left( \sum_{i = 1}^n X_i \right) \le \mathbb E f\left( \sum_{i = 1}^n Y_i \right)
$$
($X_i$ are uniformly at random sampled without replacement, $Y_i$ are uniformly at random sampled with replacement, $f$ is convex and continuous) by combining two elements from set $\mathcal X$ to form a new set: $\{ ( X_i, X_j ) : i \ne j, X_i, X_j \in \mathcal X \}$.
However, the sampling process for new set is no longer uniformly at random so I cannot use Hoeffding's result.
Since items are sampled uniformly, this is equivalent to:
$$
\left(
\dfrac{
\sum_{1 \le i < j < k \le n} f_{ij} f_{ik} f_{jk}
}{\binom{n}{3}}
\right)^2
\le
\left(
\dfrac{
\sum_{1 \le i < j \le n} f^2_{ij} }
{\binom{n}{2}}
\right)^3
$$
For $n = 3$, this is:
$$
\left(f_{12} f_{13} f_{23}\right)^2
\le
\left(
\dfrac{ f_{12}^2 + f_{13}^2 + f_{23}^2}{3} \right)^3
$$
which follows from the inequality between the geometric mean and the root-mean-square:
$$
\left(abc\right)^{1/3}
\le
\sqrt{\dfrac{a^2 + b^2 + c^2}{3}}
$$
For $n=4$, this is:
$$
\left(\frac{f_{12}f_{13}f_{23}+f_{12}f_{14}f_{24}+f_{13}f_{14}f_{34}+f_{23}f_{24}f_{34}}{4}\right)^2
\leq
\left(\frac{f_{12}^2+f_{13}^2+f_{14}^2+f_{23}^2+f_{24}^2+f_{34}^2}{6}\right)^3
$$
which follows from https://artofproblemsolving.com/community/user/12908:
\begin{align}
\left(abd+ace+bcf+def\right)^2
&= \Big(a(bd+ce)+(bc+de)f\Big)^2 \\
&\le \left(a\sqrt{(b^2+c^2)(d^2+e^2)}+f\sqrt{(b^2+d^2)(c^2+e^2)}\right)^2 \\
&\le \left(\sqrt{(a^2+f^2)\big((b^2+c^2)(d^2+e^2)+(b^2+d^2)(c^2+e^2)\big)}\right)^2 \\
&= (a^2+f^2)(b^2+c^2)(d^2+e^2)+(a^2+f^2)(b^2+d^2)(c^2+e^2)\\
&\le \left(\frac{a^2+f^2+b^2+c^2+d^2+e^2}{3}\right)^3+\left(\frac{a^2+f^2+b^2+d^2+c^2+e^2}{3}\right)^3 \\
&= 16\left(\frac{a^2+b^2+c^2+d^2+e^2+f^2}{6}\right)^3
\end{align}
| 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 $(\sum \lambda_i^3)^2\leqslant \frac{(n-2)^2}{n(n-1)}(\sum \lambda_i^2)^3$.
Now fix $\sum \lambda_i^2=S$ and $\sum \lambda_i=0$ and maximize $\sum \lambda_i^3$. When $\sum \lambda_i^3$ is maximal, the gradient vectors
$(1,1,\dots,1),2(\lambda_1,\lambda_2,\dots,\lambda_n),3(\lambda_1^2,\lambda_2^2,\dots,\lambda_n^2)$ must be linearly dependent by Lagrange multipliers theorem. In other words, there should exist number $A,B,C$ not all equal to 0 such that $A+B\lambda_i+C\lambda_i^2=0$ for all $i$. Therefore $\lambda$'stake at most 2 different values. Without loss of generality $a+b=n$, $a$ $\lambda$'s are equal to $b$, $b$ $\lambda$'s are equal to $-a$ (for some $a\in \{1,2,\dots,n-1\}$), and the inequality rewrites as $(ab^3-ba^3)^2\leqslant \frac{(n-2)^2}{n(n-1)}(ab^2+ba^2)^3$, or $(b-a)^2\leqslant \frac{(n-2)^2}{n-1}ab$. This is clear from $|b-a|\leqslant n-2$, $ab\geqslant n-1$.
| {
"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 choose a monomial order or change coordinates to ensure that this generating set is also a Groebner basis for the ideal?
For example, consider the ideal $$I = (xyz, (x + y + 3z)(x + 7z - y)(y - z + 2x), xy(x - y)).$$
With respect to the degree reverse lexicographic ordering, it has as a reduced Groebner basis
$$\{y^4z - 5y^3z^2 - 17y^2z^3 + 21yz^4, xy^3 - y^4 + 5y^3z +
17y^2z^2 - 21yz^3, x^3 - \frac{1}{2}xy^2 - \frac{1}{2}y^3 + \frac{19}{2}x^2z +
\frac{5}{2}y^2z + 16xz^2 + \frac{17}{2}yz^2 - \frac{21}{2}z^3, x^2y - xy^2, xyz\},$$ which includes nontrivially more elements than just the $F_j$.
However, after changing coordinates with the matrix (here I just generated several invertible matrices at random until one had the desired property)
$\begin{pmatrix}0 && -31 && 1\\ 0 && -\frac{5}{6} && 1\\ 1 && -2 && 0\end{pmatrix}$,
under the same monomial order, the transformed original generators (up to a constant multiple) now make up a reduced Groebner basis for the new ideal:
$$(x^3 + \frac{5281}{126}x^2y - \frac{7}{3}x^2z - \frac{4108577}{3348}xyz +
\frac{769192975}{140616}y^2z + \frac{22027}{558}xz^2 - \frac{4095575}{23436}yz^2, xy^2 -
\frac{191}{155}xyz + \frac{6}{155}xz^2, y^3 - \frac{191}{155}y^2z + \frac{6}{155}yz^2).$$
Does anyone know whether this is always possible, or of references where this sort of question is addressed? Thanks!
| 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 be either $x^2$ or $y^2$, so these three quadratics cannot have distinct leading terms.
| {
"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 $\varphi=\frac{\sqrt{5}+1}{2}$ the golden ratio.
Equality if only if $ABC$ is the equilateral triangle.
See also:
*
*Golden ratio as a property of conic section (is it known?)
| 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 $$u-y=\frac{b^2-a^2}{2c},$$
$$y\left(y+\frac{b^2-a^2}{2c}\right)+\frac{c}{2}y=\frac{a^2}{4},$$
$$y=-\frac{b^2+c^2-a^2}{4c}+\sqrt{\left(\frac{b^2+c^2-a^2}{4c}\right)^2+\frac{a^2}{4}},$$
$$u=-\frac{a^2+c^2-b^2}{4c}+\sqrt{\left(\frac{b^2+c^2-a^2}{4c}\right)^2+\frac{a^2}{4}}$$ and by C-S we have:
$$u+y=-\frac{c}{2}+\sqrt{\left(\frac{b^2+c^2-a^2}{2c}\right)^2+a^2}=$$
$$=-\frac{c}{2}+\sqrt{\frac{(a^2-b^2)^2+c^2(c^2+2a^2+2b^2)}{4c^2}}\geq-\frac{c}{2}+\frac{1}{2}\sqrt{c^2+2a^2+2b^2}=$$$$=-\frac{c}{2}+\frac{1}{2\sqrt5}\sqrt{(1+2+2)(c^2+2a^2+2b^2)}\geq-\frac{c}{2}+\frac{c+2a+2b}{2\sqrt5}.$$
By the same way we obtain:
$$v+t\geq-\frac{a}{2}+\frac{a+2b+2c}{2\sqrt5}$$ and
$$x+z\geq-\frac{b}{2}+\frac{b+2a+2c}{2\sqrt5}.$$
Can you end it now?
| {
"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 that for all pairs of two different rows $a$ and $b$, the permutations $ab^{-1}$ have the same cycle type (one 2-cycle and one 6-cycle).
What is known about Latin squares with the property that all $ab^{-1}$ have the same cycle type (where $a$ and $b$ are different rows)? For example, do they have a particular structure, for which cycle types do they exist, are there any infinite families known, do they have a name, etc?
The only example of an infinite family I'm aware of are powers of a single cyclic permutation when $n$ is prime, for example:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&4&5&1\\
3&4&5&1&2\\
4&5&1&2&3\\
5&1&2&3&4
\end{bmatrix}$$
| 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 the following equations inside the integers.
Let $\alpha_1,\alpha_2,...,\alpha_p$ and $\beta_1,\beta_2,...,\beta_p$ be unknowns. We understand the indices of the $\alpha$'s and $\beta$'s modulo $p$, so that e.g. $\alpha_{-1} = \alpha_{p-1}$. We have the linear dependence between the unknowns:
\begin{align}
& \sum_{j = 1}^p \alpha_j = 0, \\
& \sum_{j = 1}^p \beta_j = 0
\end{align}
Moreover for all $i$ between $1$ and $p-1$ we have two quadratic equations given as:
\begin{align}
& \sum_{j=1}^p \alpha_j\beta_{j-i} + \alpha_{j-i}\beta_j = 0, \\
& \sum_{j=1}^p - \alpha_j\alpha_{j-i} + \beta_j\beta_{j-i} = \left\{ \begin{array}{ll} 2, & i = \pm 2 \\ 0, & i \neq \pm 2 \end{array}\right.
\end{align}
It is easy to see that the equations for $\pm i$ are actually the same.
Additionally we have conditions on the parities:
\begin{align}
& \alpha_1 \equiv \alpha_{p-1} \equiv \beta_1 \equiv \beta_{p-1} \equiv 1 \mod 2, \\
& \alpha_i \equiv \beta_i \equiv 0 \mod 2, \ \ \text{if} \ \ i \neq \pm 1
\end{align}
Alternative formulation: The problem can also be formulated using symmetric quadratic forms. For this let $A$ be the $p \times p$-permutation matrix
$$\begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix} $$
Note that $A$ has order $p$. Let $B_i = A^i + (A^i)^T$ for each $1 \leq i \leq \frac{p-1}{2}$. Writing in block form, in particular $0$ for a $0$-matrix, define for $1 \leq i \leq \frac{p-1}{2}$ the quadratic forms
$$Q_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} 0 & B_i \\ B_i & 0 \end{pmatrix}x $$
and
$$R_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} -B_i & 0 \\ 0 & B_i \end{pmatrix}x $$
Then we can formulate the equations as
$$Q_i(x) = 0 \ \ \text{and} \ \ R_i(x) = 4\delta_{i,2} $$
for all $1 \leq i \leq \frac{p-1}{2}$ where $\delta_{i,j}$ is the Kronecker delta.
If this is of any help, we can completely solve the equations modulo $4$. This gives that
\begin{align}
& \alpha_1\beta_{p-1} + \alpha_{p-1}\beta_1 \equiv 0 \mod 4, \\
& \alpha_i + \beta_i + \alpha_{-i} + \beta_{-i} \equiv 0 \mod 4 \ \ \text{if} \ \ i \neq 0, \pm 1 \\
& \alpha_0 + \beta_0 \equiv 2 \mod 4
\end{align}
By the origin of the problem from a question on group rings, we also know that solutions exist when p is congruent to 3 modulo 4, but we have no clue when p is congruent to 1 modulo 8. We have tried some computer experiments, but found no solution and the system seems too big for a complete solution by the programs we tried.
| 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 comment, giving reformulation of the problem and reducing it to just $p-1$ unknowns and $p-1$ quadratic equations over the Gaussian integers.
Consider the generating polynomials:
\begin{split}
A(x) &:= \sum_{i=0}^{p-1} \alpha_i x^i, \\
B(x) &:= \sum_{i=0}^{p-1} \beta_i x^i.
\end{split}
The linear equations $\sum_j \alpha_j = \sum_j \beta_j = 0$ are equivalent to $A(1)=B(1)=0$, i.e., both $A(x)=(x-1)\bar A(x)$ and $B(x)=(x-1)\bar B(x)$ are multiples of $x-1$.
Viewing indices modulo $p$ is equivalent to viewing the polynomials modulo $x^p - 1 = (x-1)\Phi_p(x)$, where $\Phi_p(x) := 1 + x + \dots + x^{p-1}$ is $p$-th cyclotomic polynomial.
For reciprocal polynomials (of fixed degree $p-1$) we have $A^\star(x):=x^{p-1}A(x^{-1})\equiv x^{p-1}A(x^{p-1})\pmod{x^p-1}$ and $B^\star(x):=x^{p-1}B(x^{-1})\equiv x^{p-1}B(x^{p-1})\pmod{x^p-1}$. Then the quadratic equations (under the condition $A(1)=B(1)=0$) translate into
$$\begin{cases}
A(x)B^\star(x) + A^\star(x)B(x) \equiv 0 \pmod{x^p-1},\\
-A(x)A^\star(x) + B(x)B^\star(x) \equiv -4x^{p-1} + 2x + 2x^{p-3} \equiv 2(x^2-1)^2x^{p-3} \pmod{x^p-1}
\end{cases}
$$
Dividing both congruences by $(x-1)x(\frac1x-1)=-(x-1)^2$, we get
$$\begin{cases}
\bar A(x)\bar B^\star(x) + \bar A^\star(x)\bar B(x) \equiv 0 \pmod{\Phi_p(x)},\\
-\bar A(x)\bar A^\star(x) + \bar B(x)\bar B^\star(x) \equiv -2(x+1)^2x^{p-3} \pmod{\Phi_p(x)}.
\end{cases}
$$
In terms of polynomials over Gaussian integers, we have
$$F(x)F^\star(x) \equiv -2(x+1)^2x^{p-3}\pmod{\Phi_p(x)},$$
where
$$F(x) := \bar B(x) + I\cdot \bar A(x)$$
is a polynomial of degree $p-2$ over the Gaussian integers.
The last congruence can be viewed as a system of $p-1$ quadratic equations on the coefficients of $F(x)$ as unknowns.
Alternatively, it can also be viewed as the identity of palindromic polynomials:
$$F(x)F^\star(x) + 2(x+1)^2x^{p-3} = G(x)\cdot \Phi_p(x),$$
where the left-hand side, $G(x)$, and $\Phi_p(x)$ are palindromic polynomials of degree $2p-4$, $p-3$, and $p-1$, respectively.
| {
"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} \equiv -\frac{B_{2p-3}(1/6)}{2p-3}\pmod p
\end{align*}
and
\begin{align*}
H_{(p-1)/3,2} \equiv -\frac{B_{2p-3}(1/3)}{2p-3}\pmod p
\end{align*}
where $B_n(x) = \sum_{k = 0}^n {n \choose k}x^{n-k}B_k$ is the $n$-th Bernoulli polynomial.
It is well known, I think, that
\begin{align*}
B_n(1/6) = (1 - 3^{1-n})(1 - 2^{1-n})\frac{B_n}{2\cdot 3^{n-1}}
\end{align*}
and
\begin{align*}
B_n(1/3)=(1 - 3^{1-n})\frac{B_n}{2\cdot 3^{n-1}}
\end{align*}
but only for even $n$. From this we can eventually express $H_{(p-1)/3} = H_{(p-1)/3, 1}$ and $H_{(p-1)/6} = H_{(p-1)/6,1}$ modulo $p$ in terms of the Fermat quotients $q_p(2)$ and $q_p(3)$ where $q_p(a) = (a^{p-1} - 1)/p$ for $a$ co-prime to $p$.
It is mentioned in the paper above that there is some kind of expression for for $B_n(1/3)$ and $B_n(1/6)$ when $n$ is odd in terms of "$I$ numbers", but I can't find the source, or anything else that is relevant.
Is it possible to write $H_{(p-1)/3,2}$ and $H_{(p-1)/6,2}$ in terms of Fermat quotients modulo $p$ analogous to $H_{(p-1)/3}$ and $H_{(p-1)/6}$? What are these "$I$ numbers?"?
I've asked a similar question here but I thought I would ask here.
| 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})}{e^x -1} = \frac{xe^{\frac{1}{3}x}(1-e^{\frac{1}{3}x})}{(e^{\frac{1}{3}x} -1)(e^{\frac{2}{3}x}+e^{\frac{1}{3}x}+1)}
\\
& = \frac{-xe^{\frac{1}{3}x}}{e^{\frac{2}{3}x}+e^{\frac{1}{3}x}+1} = \frac{-x}{e^{\frac{1}{3}x}+e^{-\frac{1}{3}x}+1}
\end{align*}
The $n$-th $I$ number is defined by
\begin{align*}
\frac{3/2}{e^x + e^{-x} + 1} = \sum_{n = 0}^{\infty} I_n \frac{x^n}{n!}
\end{align*}
Notice that
\begin{align*}
2 \sum_{n = 0}^{\infty} I_{2n+1} \frac{x^{2n+1}}{(2n+1)!} &= \sum_{n = 0}^{\infty} I_n \frac{x^n}{n!} - \sum_{n = 0}^{\infty} I_n \frac{(-x)^n}{n!}
\\
& = 0.
\end{align*}
Therefore, $I_n = 0$ whenever $n$ is odd, and $ \sum_{n = 0}^{\infty} I_n \frac{x^n}{n!} = \sum_{n = 0}^{\infty} I_{2n} \frac{x^{2n}}{(2n)!}$. Now,
\begin{align*}
\frac{-x}{e^{\frac{1}{3}x}+e^{-\frac{1}{3}x}+1} &= -\frac{2}{3}x \frac{3/2}{e^{\frac{1}{3}x}+e^{-\frac{1}{3}x}+1}
\\
& = -\frac{2}{3}x \sum_{n = 0}^{\infty} I_{2n} \frac{\left( \frac{1}{3}x\right)^{2n}}{(2n)!}
\\
& = 2\sum_{n = 0}^{\infty} -\frac{(2n+1)I_{2n}}{3^{2n+1}} \frac{x^{2n+1}}{(2n+1)!}.
\end{align*}
Therefore,
\begin{align*}
\sum_{n = 0}^{\infty} -\frac{(2n+1)I_{2n}}{3^{2n+1}} \frac{x^{2n+1}}{(2n+1)!} = \sum_{n = 0}^{\infty} B_{2n+1}\left(\frac{1}{3}\right) \frac{x^{2n+1}}{(2n+1)!}
\end{align*}
which implies,
\begin{align*}
B_{2n+1}\left(\frac{1}{3}\right) = -\frac{(2n+1)I_{2n}}{3^{2n+1}}.
\end{align*}
| {
"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 $p\in(0,1)$
$$d(p):=\det(a^p+b^p-(a+b)^p) \\
=2^{-p-3/2} \\
\times\left(\sqrt{2} \left(2^p+2\right)-2 \left(\sqrt{2}-1\right)
\left(2+\sqrt{2}\right)^p-2 \left(1+\sqrt{2}\right) \left(2-\sqrt{2}\right)^p\right)<0 \tag{1}\label{1},$$
and hence $(a+b)^p\not\le a^p+b^p$.
To prove the inequality in \eqref{1}, let
$$d_0(p):=d(p)2^{p+3/2},\quad
d_1(p):=\frac{d_0'(p)}{(2+\sqrt2)^p},\\
d_2(p):=d_1'(p)(1+1/\sqrt2)^p.
$$
Then
$$d_2'(p)=\left(1+\sqrt{2}\right) 2^{1-p} \left(2-\sqrt{2}\right)^p \ln\left(3-2
\sqrt{2}\right) \ln\left(2-\sqrt{2}\right) \ln\left(2+\sqrt{2}\right)>0,$$
$d_2$ is increasing (on $(0,1)$), $d_2(1)=\sqrt{2} \ln\left(2-\sqrt{2}\right) \ln\left(6+4 \sqrt{2}\right)<0$, $d_2<0$, $d_1$ is decreasing, $d_1(1)=0$, $d_1>0$, $d_0$ is increasing, $d_0(1)=0$, $d_0<0$, $d<0$. $\quad\Box$
Here is the graph $\{(p,d(p)(1-p)^{-2})\colon0<p<1\}$:
| {
"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 $(a,b,c)=(1,2,3).$
Remark : This question has been asked previously on math.SE without receiving any answers.
Motivation : This question comes from
$$\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3.$$
This got me interested in $(\star)$. I've got the followings :
1. If $(\star)$ for any $n\in\mathbb N$, then $(a,b,c)=(1,2,3).$
We can easily prove this by considering the limitation $n\to\infty$ of the both sides of
$$\frac{n^{(a+1)b}}{n^{c+1}}\left\{\sum_{k=1}^n\frac 1n\left(\frac kn\right)^a\right\}^b=\sum_{k=1}^n\frac 1n \left(\frac kn\right)^c.$$
2. If $(\star)$ for $n=2$, then $(a,b,c)=(1,2,3).$
3. If $(\star)$ for $n=3$, then $(a,b,c)=(1,2,3).$
Since both 1 and 2 are easy to prove, I'm going to prove 3.
Proof : Supposing $c\le ab$, since $b\ge 2$, we get
$$(1+2^a+3^a)^b=1+2^{ab}+3^{ab}+\cdots\gt 1+2^c+3^c.$$ This is a contradiction. Hence, $c\gt ab$. Supposing $b\ge 3$, we get $c\gt ab\ge 3$.
Here, since $3^c+1\equiv 4,2$ (mod $8$) for any $c\in\mathbb N$, $3^c+1$ is not a multiple of $8$. By the way, since $1+2^a+3^a$ is even, $(1+2^a+3^a)^b$ is a multiple of $8$. Since $2^c$ is a multiple of $8$, this leads that $3^c+1$ is a multiple of $8$, which is a contradiction. Hence, $b=2, c\gt 2a$.
If $a\ge 3$, since
$$\left(\frac 23\right)^a+\left(\frac 13\right)^a\le\left(\frac 23\right)^3+\left(\frac 13\right)^3=\frac13,$$
$2^a+1\le \frac{3^a}{3}.$ Hence,
$$3^c\lt 1+2^c+3^c=(1+2^a+3^a)^2\le \left(\frac{3^a}{3}+3^a\right)^2=3^{2a}\left(\frac 43\right)^2=3^{2a}\cdot\frac {16}{9}\lt 3^{2a+1}.$$
$3^c\lt 3^{2a+1}$ leads $0\lt c-2a\lt 1$, which means that $c-2a$ is not an integer. This is a contradiction. Hence, we know $a=1$ or $a=2$.
The $(a,b)=(1,2)$ case leads $c=3$.
The $(a,b)=(2,2)$ case leads $c\ge5\Rightarrow 1+2^c+3^c\gt 196$, which is a contradiction. Now the proof is completed.
After getting these results, I reached the above conjecture. Can anyone help?
| 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
$$\sum_{k=0}^n \binom{4n+M}{k} \cdot \binom{3n-k}{2n}$$
where $M\in\mathbb{Z}$? Is there a similiar identity?
| 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 $M=2$:
$$\frac{2 \, {\left(2 \, n \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right) + \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right)\right)} 8^{n} 2^{n} \left(n - \frac{1}{2}\right)!^{2} - \pi 4^{2 \, n + 1} \left(n + \frac{1}{4}\right)! \left(n - \frac{1}{4}\right)!}{\sqrt{\pi} \left(n + 1\right)! \left(n - \frac{1}{2}\right)! \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right)}
$$
For $M=3$:
$$
\frac{2 \, {\left(2 \, {\left(6 \, n^{2} \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right) + 7 \, n \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right) + 2 \, \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right)\right)} 8^{n} 2^{n} \left(n - \frac{1}{2}\right)!^{2} - \pi 4^{2 \, n + 2} \left(n + \frac{3}{4}\right)! \left(n + \frac{1}{4}\right)!\right)}}{\sqrt{\pi} \left(n + 2\right)! \left(n - \frac{1}{2}\right)! \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{4}\right)}
$$
| {
"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. large $k$
such that for explicit $x > 1$ the fractional part is $ < \frac12$
for $n = 1 \ldots k$.
So far got $k=12$ for $x \approx 2.81365$ and don't think $ x < 2$ is possible
in this case.
Second, consider the following relaxation. Let $f$ be a strictly increasing
function $\mathbb{N} \to \mathbb{N} $
Is it possible:
$$\left\lbrace x \left(\frac 3 2\right)^ {f(n)} \right\rbrace < \frac12 $$
for all natural $n$?
| 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 $n \le k, \lbrace (\frac{3}{2})^n y\rbrace = \lbrace (\frac{3}{2})^n x \rbrace$ while $\lbrace (\frac{3}{2})^{k+1} y \rbrace = \lbrace (\frac{3}{2})^{k+1} x \rbrace - \frac{1}{2} \lt \frac{1}{2}.$ So, if we start with an arbitrary $x$, then one of the $2^{k}$ numbers $x, x+1, ... x+2^{k}-1$ will have fractional parts under $\frac{1}{2}$ when you multiply by $\frac{3}{2}, ..., (\frac{3}{2})^k$.
Here is some Mathematica code which implements this:
Clear[j];
j[1] = 1;
j[n_] := j[n] =
If[FractionalPart[j[n - 1] (3/2)^n] < 1/2,
j[n - 1],
j[n - 1] + 2^(n - 1)]
For the choice $j[1]=1$, this computes the integer from $1$ to $2^n$ which has the right fractional parts when multiplied by $\frac{3}{2}, ..., (\frac{3}{2})^n$. $j[4]=1, j[5]=17, j[20] = 386737, j[100] = 719590229933913224019274229425$.
Yes, we can take $f(n)=3n$.
Lemma: For any integer $a$, $[(\frac{3}{2})^3 a, (\frac{3}{2})^3(a+\frac{1}{2})]$ contains at least one interval $[b,b+\frac{1}{2}]$ for some integer $b$.
This follows because $\frac{1}{2}(\frac{3}{2})^3 \gt 1+\frac{1}{2}$.
Start with an interval $[a_1,a_1+\frac{1}{2}]$. Inductively pick $a_{n+1}$ so that $[a_{n+1},a_{n+1}+\frac{1}{2}] \subset (\frac{3}{2})^3 [a_n,a_n+\frac{1}{2}].$ Then the intervals $(\frac{3}{2})^{-3n}[a_n,a_n+\frac{1}{2}]$ are nested hence their intersection contains an element $x$ so that $\lbrace x (\frac{3}{2})^{3n} \rbrace \lt \frac{1}{2}$ for all $n$. You can choose $x\ne 0$.
| {
"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 $K=\mathbb{Q}_2(\sqrt{m})$ for $m=2,-2, 10$ or $-10$
$$K^*/K^{*2}\cong (\langle\sqrt{m}\rangle/\langle m\rangle)\times (\mathcal{O}^\times/\langle 1+m+2\sqrt{m}, 1+ \mathfrak{p}^5 \rangle)$$
for $\mathfrak{p}= \langle\sqrt{m}\rangle$.
And after that, he said "it is sufficient to examine $\epsilon$ and $\epsilon\sqrt{m}$ where $\epsilon=a+b\sqrt{m}$ for $a=1,3,5,7$ and $b=0,1,2,3$."
My first question is "why it is enough to check these numbers?"
In the following, he said "we take $\epsilon$ (resp. $\epsilon\sqrt{m}$) such that $\epsilon$, $\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $\epsilon^\sigma (1+m+2\sqrt{m})$ (resp. $\epsilon$, $-\epsilon^\sigma$, $\epsilon (1+m+2\sqrt{m})$ and $-\epsilon^\sigma (1+m+2\sqrt{m})$) are different modulo $\mathfrak{p}^5$ each other." Where $\sigma$ is the generator of Galois group of $K/\mathbb{Q}_2$.
And in the following, for $m=2$ he took $1+\sqrt{2}$, $3+\sqrt{2}$, $\sqrt{2}$ and $3\sqrt{2}$. I couldn't get to these numbers following his method explained above, can anyone show me the calculations? And can you show me any other possible numbers other than the given ones?
Here is the paper: Dihedral extensions of degree 8 over the rational p-adic fields.
Thank you.
| 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 representatives of $\mathfrak{o}^\times/(\mathfrak{o}^\times)^2$ and their multiples by $\sqrt{m}$.)
It is easy to see that these numbers contain a set of representatives if you note that $\mathfrak{p}^5 = 4\mathfrak{p}$. (This is obvious for $m = \pm 2$. For $m = \pm 10$, note that $5$ is a unit in $\mathfrak{o}$.)
For any $x \in \mathfrak{o}^\times$, write $x^{-1} = a' + b' \sqrt{m}$, and choose $a \equiv a' \pmod{8}$ and $b \equiv b' \pmod{4}$. Then it is easy to see that $x(a + b \sqrt{m}) \in 1 + 8 \mathfrak{o} + 4\mathfrak{p} \subseteq \mathfrak{o}^{\times 2}$ (since $1 + 8 \mathfrak{o} \subseteq (\mathfrak{o}^\times)^2$).
Partial answer to your second question: We are supposed to check all 16 possible combinations of $a$ and $b$ (and also consider multiplying each possibility by $\sqrt{m}$). I will only do two of them (and I won't multiply by $\sqrt{m}$).
For $a = b = 1$ (with $m = 2$), we have:
*
*$\epsilon = 1 + \sqrt{2}$,
*$\epsilon^\sigma = 1 - \sqrt{2}$,
*$\epsilon(1+m+2 \sqrt{2}) = (1 + \sqrt{2})(3+2 \sqrt{2}) = 7 + 5 \sqrt{2}$,
*$\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - \sqrt{2})(3+2 \sqrt{2}) = -1 - \sqrt{2}$.
It is easy to check that none of these are congruent modulo $4 \sqrt{2}$:
$$x + y \sqrt{2} \equiv x' + y' \sqrt{2} \pmod{4 \sqrt{2}} \iff \text{$x \equiv x' \pmod{8}$ and $y \equiv y' \pmod{4}$} .$$
So this gives an example, which means that $1 + 2 \sqrt{2}$ should be on Naito's list. It is indeed there.
For $a = 1$ and $b = 2$ (with $m = 2$), we have:
*
*$\epsilon = 1 + 2\sqrt{2}$,
*$\epsilon^\sigma = 1 - 2\sqrt{2}$,
*$\epsilon(1+m+2 \sqrt{2}) = (1 + 2\sqrt{2})(3+2 \sqrt{2}) = 11 + 8 \sqrt{2}$,
*$\epsilon^\sigma(1+m+2 \sqrt{2}) = (1 - 2\sqrt{2})(3+2 \sqrt{2}) = -1 - 2\sqrt{2}$.
Since
$$\epsilon - \epsilon^\sigma = (1 + 2\sqrt{2}) - (1 - 2 \sqrt{2}) = 4 \sqrt{2} \equiv 0 \pmod{4 \sqrt{2}} ,$$
this does not give an example. So $1 + 2 \sqrt{2}$ should not be on Naito's list. It is indeed not there.
By checking each of the 32 possibilities for each of the four possible values of $m$, it should be easy to determine whether his list is correct.
| {
"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: ${12+2+1, 16+2+1,...,72+6+2,...}$
4-term sums: ${240+12+2+1,...,6480+72+6+2,...}$
| 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 $N=(x-1)x + x^2 + x(x+1) + y^2$ is a sum of four distinct quarter-squares.
If you used some square twice combined with two other distinct squares, i.e. $N=x^2 + x^2 + y^2 + z^2$, then $N=(x-1)x + x(x+1) + y^2 + z^2$ is a sum of four distinct quarter-squares.
If you used two distinct squares twice each, i.e. $N=x^2 + x^2 + y^2 + y^2$, then $N=(x+y)^2 + (x-y)^2$ is a sum of two distinct quarter-squares.
Finally, if you used four distinct squares, then $N$ is clearly the sum of four distinct quarter-squares.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/202903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
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 by $a$ to get two different partitions of $n$ with the property.
Now suppose $n = 6 \cdot m + 35$. A sufficiently large $m$ has two different partitions into $k - 1$ parts such that each part of each partition is divisible by either $5$ or $7$. Multiplying by $6$ and appending $35$ then gives two different partitions of $n$ with the property.
On the other hand, suppose $n = 6 \cdot m + 55 = 6 \cdot m + 5 \cdot 11$. The same logic above applies, and that covers all the cases. Filling in some more details results in a simple formula for an upper bound in terms of $k$: the first part says if $n$ is composite then $n \lt (k+2)^2$, and the second part chases down the primes relatively more quickly placing the largest one in $6 \cdot k + O(1)$.
More generally still, we can require that the pairwise $\text{gcd}$ of the parts be greater than or equal to some $d \ge 2$ (so the question of $47$ corresponds to the case $d = 2$ and $k = 5$). An effective upper bound on unique solutions can still be obtained: we just have to carry out the above argument for all congruence classes mod $d$, in each case representing the remainder as a semiprime with prime factors larger than $d$.
EDIT: That led me to the following even simpler version, working mod $2$ instead of mod $6$: if $m$ has two different partitions into $5$ parts, then $n = 2 \cdot m$ has two different partitions into $5$ parts, no two parts of which are relatively prime. So $n < 2 \cdot (5+2) = 14$ if $n$ is even. Otherwise, suppose $n = 2 \cdot m + 15$, and observe that
$18 = 3+3+6+6 = 3+3+3+9$,
$19 = 3+5+5+6 = 3 + 3 + 3 + 10$, and
$20 = 3+3+5+9 = 3+5+6+6$
to conclude $m \le 17$ and $n \le 49$.
| {
"language": "en",
"url": "https://mathoverflow.net/questions/289084",
"timestamp": "2023-03-29T00:00:00",
"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 $$\frac{(b-a)(b^2-a)\cdot\dotsb\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\dotsb\cdot(b^n-1)}\in\mathbb{Z}\left[\frac{a-1}{b-1},\frac{a^2-1}{b^2-1},\dotsc ,\frac{a^n-1}{b^n-1}, 1/2 \right]?$$
Moreover, if the first question is true, then what is the minimal value of the degree $\deg_{t_{n+1}}$ of the polynomial $F(t_1, t_2,\ldots , t_{n+1})$ with $n+1$ variables and the integer coefficients, for which $$\frac{(b-a)(b^2-a)\cdot\dotsb\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\dotsb\cdot(b^n-1)}=F\left(\frac{a-1}{b-1},\frac{a^2-1}{b^2-1},\dotsc,\frac{a^n-1}{b^n-1}, 1/2 \right)?$$
This conjecture is a generalization of my post here. And I will be interested in any comments on it.
| 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 $$
P\left(b,\frac{a-1}{b-1},\dots,\frac{a^n-1}{b^n-1}\right)=2^{[n/2]+[n/4]+\dots}\cdot \frac{(b-a)(b^2-a)\dots (b^n-a)}{(b-1)(b^2-1)\dots(b^n-a)}.
$$
| {
"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 seen any discussion of the below question. If you do, please do let me know of a reference.
QUESTION. For $N$ large, is it true that the most frequent digit in $A_x(N)$ is $x$, where $x\in\{2,3,5,7\}$?
| 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):
$$
\begin{array}{rl}
N=1000: & 2, 1, 4, 6, 8, 9, 5, 3, 0, 7\\
N=2000: & 2, 1, 6, 4, 8, 5, 3, 9, 7, 0 \\
N=3000: & 6, 2, 1, 4, 3, 8, 0, 5, 7, 9 \\
N=5000: & 2, 0, 4, 1, 8, 6, 5, 3, 7, 9 \\
N=6000: & 4, 8, 2, 1, 9, 7, 0, 6, 5, 3 \\
N=7000: & 4, 9, 8, 7, 1, 2, 3, 6, 5, 0 \\
N=8000: & 4, 8, 7, 1, 9, 2, 3, 6, 5, 0 \\
N=10000: & 4, 8, 7, 1, 2, 9, 6, 3, 5, 0 \\
N=15000: & 4, 2, 8, 1, 6, 7, 3, 0, 5, 9 \\
N=20000: & 6, 4, 3, 8, 2, 1, 7, 9, 0, 5 \\
N=25000: & 6, 4, 2, 3, 8, 1, 9, 7, 5, 0 \\
N=30000: & 6, 9, 4, 3, 2, 8, 1, 0, 7, 5 \\
\end{array}$$
| {
"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 :
$\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}=\sum_{b_{1}+b_{2}+\cdots+b_{k}=n}\left(\begin{array}{c} n \\ b_{1}, b_{2}, b_{3}, \dots, b_{k} \end{array}\right) \prod_{j=1}^{k} x_{j}^{b_{j}}$
I have to first find out the combinations of power for which I am going to get $x^6$
then i have to calculate the multinomial coefficient multiplied by the $2$ raised to some power for all of those combinations.
Is there an easier way to somehow change this question into some other question then use Binomial or some other theorem to solve it quickly ?
| 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 $\alpha_{1,2}=\frac{-1\pm I\sqrt{7}}2$, to get
$$[x^6]\, (x^2+x+2)^9 = \sum_{i=0}^6 \binom9i\binom9{6-i}(-\alpha_1)^{9-i} (-\alpha_2)^{3+i}.$$
Yet another approach is to represent $x^2+x+2=(x^2+x)+2$ and notice that $[x^6]\,(x^2+x)^k = \binom{k}{6-k}$. Then
$$[x^6]\, (x^2+x+2)^9 = \sum_{k=0}^9 \binom9k \binom{k}{6-k}2^{9-k}.$$
| {
"language": "en",
"url": "https://mathoverflow.net/questions/362394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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,
$$\{2w^2+x^2+y^2:\ w,x,y\in\mathbb N\}=\mathbb N\setminus\{4^k(16m+14):\ k,m\in\mathbb N\}.$$
Motivated by this, here I pose the following novel question.
Question 1. Can each $n\in\mathbb N$ be written as $2w^2+x^2+y^2+z^2+xyz$ with $x,y,z\in\mathbb N$?
I guess that the answer is positive, which has been verified for $n\le 10^6$.
Similarly, I have the following four questions.
Question 2. Are $7$ and $487$ the only natural numbers which cannot be written as $w^2+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
Question 3. Is it true that each $n\in\mathbb N$ with $n\not\equiv 3\pmod4$ can be written as
$4w^2+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
I also believe that the answers to Questions 2 and 3 should be positive, which have been verified for $n\le 10^6$. It is easy to see that $$x^2+y^2+z^2+xyz\not\equiv3\pmod4$$ for any $x,y,z\in\mathbb Z$.
Question 4. Is $23$ the only natural number which cannot be written as $w^2+x^2+y^2+z^2+3xyz$ with $w,x,y,z\in\mathbb N$?
I guess that the answer is positive. I have checked this for natural numbers up to $2\times10^6$.
Question 5. Is it true that each $n\in\mathbb N$ with $n\not\equiv3\pmod4$ can be written as $4w^3+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$? Are $7,\,87$ and $267$ the only natural numbers which cannot be written as $w^3+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
It seems that Question 5 should also have a positive answer; I have checked this for natural numbers up to $10^5$.
Any ideas to the above new questions? Your comments are welcome!
| 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 locally represented integer is represented. The only local obstructions are at $2$, and this implies that every integer not of the form $14 \cdot 4^{k} \pmod{4^{k+2}}$ for $k \geq 0$ is represented by $Q_{1}$.
If $n$ is even, then there are non-negative integers $w$, $x$ and $y$ so that $n = 2w^{2} + x^{2} + xy + y^{2} + 1$. The quadratic form $Q_{2} = 2w^{2} + x^{2} + xy + y^{2}$ also has class number $1$, and it represents all integers not of the form $10 \cdot 4^{k} \pmod{4^{k+2}}$ for some $k \geq 0$. In particular, $Q_{2}$ represents every odd number and so every even $n$ can be written in the form $2w^{2} + x^{2} + xy + y^{2} + 1$.
Note that if $m \in \mathbb{N}$ and $m = x^{2} + xy + y^{2}$ for some $x, y \in \mathbb{Z}$, then there are $x'$ and $y'$ in $\mathbb{N}$ so that $m = (x')^{2} + (x')(y') + (y')^{2}$. This can be seen by applying one of the six automorphisms of the quadratic form $x^{2}+xy+y^{2}$.
| {
"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 $p(3) = 6$. Then it appears that $p_n(x)-x$ always has precisely one negative real root, and moreover this root (as a function of $n$) appears to approach $-0.577$ as $n$ gets large. Do these roots indeed approach a limit, and is this limit indeed the negative of the Euler-Mascheroni constant?
For $n=60$, the root is about $-0.580$; for $n=120$, the root is about $-0.577$. That's as far as I've gone.
(If the question seems unmotivated, I'll give some background, though it might not help anyone find an answer. One way to think about the "values" of certain divergent series is to view them as fixed points of an associated function; e.g., for the series $1+2+4+8+\dots$, the linear map $x \mapsto 2x+1$ sends the sum of the first $n$ terms of the series to the sum of the first $n+1$ terms, and the fixed point of this map is $-1$, which is indeed the natural value to assign to the divergent series. The problem I'm posting arose from trying to analyze $1+2+3+4+\dots$ in a similar way.)
| 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
$$
p_n(x)=\frac{1}{4} \left(\frac{\Gamma \left(n+\frac{1}{2}\right) \cos \left(\frac{1}{2} \pi \sqrt{8 x+1}\right) \Gamma \left(n+\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2}\right) \Gamma \left(n-\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2}\right) \, _4\tilde{F}_3\left(1,n+\frac{1}{2},n-\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2},n+\frac{1}{2} \sqrt{8 x+1}+\frac{3}{2};n+\frac{3}{2},n+2,n+2;1\right)}{\pi }+2 \, _3F_2\left(-\frac{1}{2},\frac{1}{2}-\frac{1}{2} \sqrt{8 x+1},\frac{1}{2} \sqrt{8 x+1}+\frac{1}{2};\frac{1}{2},1;1\right)+4 x+2\right)
$$
For $n\to\infty$, Mathematica claims that this becomes
$$
p_\infty(x)=\frac{1}{2} \left(\, _3F_2\left(-\frac{1}{2},\frac{1}{2}-\frac{1}{2} \sqrt{8 x+1},\frac{1}{2} \sqrt{8 x+1}+\frac{1}{2};\frac{1}{2},1;1\right)+2 x+1\right)
$$
which solves $p(x)=x$ at around
$$
x=-0.573825523080029241015952733\dots
$$
| {
"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 divisors of $8n+1$.
Another one up to $10^6$ is:
$\sigma(4n+1) \mod 4 = A001936(n) \mod 4$
A001936 Expansion of q^(-1/4) (eta(q^4) / eta(q))^2 in powers of q
$\sigma(n)$ is sum of divisors of $n$.
sigma(8n+1) mod 4 starts: 1, 1, 2, 3, 0, 2, 1, 0, 0, 2, 1, 2, 2, 0, 2, 1, 0, 2, 0, 2, 0, 3, 0, 0, 2, 0, 0, 0, 3, 2
sigma(4n+1) mod 4 starts: 1, 2, 1, 2, 2, 0, 3, 2, 0, 2, 2, 2, 1, 2, 0, 2, 0, 0, 2, 0, 1, 0, 2, 0, 2, 2
Update
Up to 10^7
A001935 mod 4 is zero for n = 9m+4 or 9m+7
A001936 mod 4 is zero for n = 9m+5 or 9m+8
Question about computability
| 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 $$B(x)=\sum_{k\geq 0}\sigma(2k+1)x^k=\prod_{k\geq 0}(1-x^k)^4(1+x^k)^8$$
So $$B(x)\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2\equiv \prod_{k\geq 1}(\frac{1-x^{8k}}{1-x^{2k}})^2\equiv A(x^2)\pmod{4}$$
Now the proof is complete once we know that $$B(x)\equiv \sum_{k\geq 0} \sigma(4n+1)x^{2n}\pmod{4}$$ this is an other way of saying $\sigma(4n-1)$ is divisible by $4$, which can be shown by pairing up the divisors $d+\frac{4n-1}{d}\equiv 0\pmod{4}$.
The proof for the other congruence is similar, but slightly longer, I might update this post later to include it.
Let's prove that $\sigma(8n+1)\equiv q(n)\pmod{4}$, where $q(n)$ is the number of partitions with no even part repeated. The generating function is $$Q(x)=\sum_{n\geq 0}q(n)x^n=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k}$$
Since we know from above that $$\sum_{n\geq 0}\sigma(4n+1)x^{2n}\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2 \pmod{4}$$ we conclude that $$L(x)=\sum_{n\geq 0}\sigma(4n+1)x^n\equiv Q(x)^2 \pmod{4}$$
so that $$\sum_{n\geq 0} \sigma(8n+1)x^{2n}\equiv \frac{L(x)+L(-x)}{2}\pmod{4}$$
So to finish off the proof we need the following
$$\frac{Q(x)^2+Q(-x)^2}{2}\equiv Q(x^2)\pmod{4}$$ which I will leave as an exercise Actually let me write the proof, just to make sure I didn't mess up calculations. This reduces to proving
$$\frac{\prod_{k\geq 1}(1+x^{2k})^4(1+x^{2k-1})^2+\prod_{k\geq 1}(1+x^{2k})^4(1-x^{2k-1})^2}{2}$$ $$\equiv \prod_{k\geq 1}(1+x^{4k-2})(1+x^{4k})^2 \pmod{4}$$ and since $$(1+x^{2k})^4\equiv (1+x^{4k})^2 \pmod{4}$$ this reduces to
$$\frac{\prod_{k\geq 1}(1+x^{2k-1})^2+\prod_{k\geq 1}(1-x^{2k-1})^2}{2}\equiv \prod_{k\geq 1} (1+x^{4k-2})\pmod{4}$$ but we can write $$\prod_{k\geq 1}(1-x^{2k-1})^2\equiv \left(\prod_{k\ geq 1}(1+x^{2k-1})^2\right) \left(1-4\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{8}$$ therefore now we have to show
$$\prod_{k\geq 1}(1+x^{2k-1})^2\left(1-2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\equiv \prod_{k\geq 1}(1+x^{4k-2})\pmod{4}$$ Now everything is clear since $$\prod_{k\geq 1}(1+x^{2k-1})^2\equiv \prod_{k\geq 1}(1+x^{4k-2})\left(1+2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{4}$$
| {
"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 for your answer.
| [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
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!}
\sim e^d \left( 1 - \frac1{d} + \frac1{d^2} - \frac2{d^3} \cdots \right)
$$
by expanding $1 - 1/(2d+2-k)$ in a power series about $k=d$,
we should be able to replace $1 - 1/(2d+2-k)$ by something smaller
that can be summed exactly and is close enough that the result
is within a small enough multiple of $e^d$ to maintain the
desired inequality.
Because it takes about $2m$ terms of the power series in $k$
to get within $O(1/d^m)$, I had to match the power series to within $O(k-d)^6$.
Let
$$
A_6(k) =
\frac{d+1}{d+2}
- \frac{k-d}{(d+2)^2}
- \frac{(k-d)^2}{(d+2)^3}
- \frac{(k-d)^3}{(d+2)^4}
- \frac{(k-d)^4}{(d+2)^5}
- \frac{(k-d)^5}{(d+2)^6}
- \frac{(k-d)^6}{2(d+2)^6},
$$
so the final term has denominator $2(d+2)^6$ instead of $(d+2)^7$.
Then
$$
1 - \frac1{2d+2-k}
= A_6(k) + \frac{(k-d)^6(2d-k)}{2(d+2)^6(2d+2-k)} \geq A_6(k)
$$
for all $k \leq 2d$. Hence
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!} >
\sum_{k=1}^{2d} A_6(k) \frac{d^k}{k!}.
$$
On the other hand, since $A_6(k)$ is a polynomial in $k$,
the power series $\sum_{k=0}^\infty A_6(k) d^k/k!$ is elementary
(see my earlier answer for the explanation; David Speyer implicitly
used this too in the calculation "with the aid of Mathematica").
I find
$$
\sum_{k=0}^\infty A_6(k) \frac{d^k}{k!}
= \frac{2d^6 + 22d^5 + 98d^4 + 102d^3 + 229d^2 + 193d + 64}{2(d+2)^6} e^d
$$ $$
= \left(
1 - \frac1d + \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6}
\right) \cdot e^d > \left(1 - \frac1d\right) e^d.
$$
We're not quite finished, because we need a lower bound on
$\sum_{k=1}^{2d} A_6(k) d^k/k!$, not $\sum_{k=0}^\infty$.
However, once $d$ is at all large the terms with $k=0$ and $k>2d$
are negligible compared with our lower bound
$$
\sum_{k=0}^\infty A_6(k) d^k/k! - \left(1 - \frac1d\right) e^d
\geq \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6} e^d >
\frac{d^4}{(d+2)^6} e^d.
$$
Indeed the $k=0$ term is less than $1$, and for $k>2d$ we have
$A_6(k) < A_6(2d) = 1/2$ while $d^k/k!$ is exponentially smaller than $e^d$:
$$
\sum_{k=2d+1}^\infty \frac{d^k}{k!} <
2^{-2d} \sum_{k=2d+1}^\infty \frac{(2d)^k}{k!} <
2^{-2d} \sum_{k=0}^\infty \frac{(2d)^k}{k!} = (e/2)^{2d}.
$$
So we're done once
$$
1 + \frac12 \left(\frac{e}{2}\right)^{2d} < \frac{d^4}{(d+2)^6} e^d,
$$
which happens once $d \geq 14$. Since the desired inequality
has already been verified numerically up to $d=50$, we're done. QED
| {
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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:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2) = 4t^3 - 2 (a_1^2 + a_2^2)t$, so the nonzero roots are $\pm \sqrt{\frac{a_1^2 + a_2^2}{2} }$.
For higher degrees, we will get more complicated algebraic functions. For instance:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2)(t^2-a_3^2) = 6 t^5 - 4 (a_1^2+a_2^2+a_3^2)t^3 + 2 (a_1^2a_2^2 + a_2^2a_3^2 + a_1^2a_3^2)t $ so the nonzero roots are:
$$\pm \sqrt{ \frac{a_1^2+a_2^2+a_3^2\pm \sqrt{ \left(a_1^2+a_2^2+a_3^2\right)^2 - 3\left(a_1^2a_2^2+a_1^2a_3^2+a_2^2a_3^2\right)}}{3} }$$
and it just gets worse as the degree increases.
| {
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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{\lambda}\text{ } tanh (\pi \sqrt{\lambda}) $
This comes up in the context of Quantum Field Theory but I haven't been able to locate any QFT resource either which proves these.
*
*I would like to know the proof of the above.
*I would like to know if there is any generalization of these to $\xi(n)$
*I am hoping that there is generalization of the second identity to cases like $\int _0 ^\infty \sqrt{\lambda + \frac{1}{4}} tanh (\pi \sqrt{\lambda})$
| 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 good QFT person does, we get
$$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = 2 \sum_{k=1}^{\infty}(-1)^k \int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}} \tag{$\star$}$$
Now note that$$\int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}}dx = \dfrac1{2k^3}$$which is obtained by setting $\sqrt{x}=t$ and integrating by parts. Plugging it back into $\star$ gives us
$$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = \sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac34 \zeta(3)$$
I will let you fix the constant that scale during the integration process.
Note that we landed up with $\zeta(3)$, since the integral was of the form $\sqrt{x} \tanh(\sqrt{x})$. If we were to start with the integral of the form
$x^{1/n} \tanh(x^{1/n})$ and mimic the process above, we will get $\zeta(n+1)$.
Added on OP's request
We have
$$\zeta(3) = 1 + \dfrac1{2^3} + \dfrac1{3^3} + \dfrac1{4^3} + \cdots$$
Note that
$$\dfrac1{2^3} + \dfrac1{4^3} + \dfrac1{6^3} + \cdots = \dfrac1{2^3}\left( 1 + \dfrac1{2^3} + \dfrac1{3^3} + \dfrac1{4^3} + \cdots\right) = \dfrac{\zeta(3)}8$$
Hence,
$$1 + \dfrac1{3^3} + \dfrac1{5^3} + \cdots = \zeta(3) - \dfrac{\zeta(3)}8 = \dfrac78 \zeta(3)$$
Therefor, the sum
$$\sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac1{1^3} + \dfrac1{2^3} - \dfrac1{3^3} + \dfrac1{4^3} \mp = -\dfrac78 \zeta(3) + \dfrac18 \zeta(3) = -\dfrac34 \zeta(3)$$
| {
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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 $k\equiv p(mod$ $8)$.
Lemma 2. If $p$ and $q$ are odd primes with $p = 1(mod$ $4)$ and $\left(\frac{q}{p}\right)=-1$ then $x^2+qy^2-pz^2$ reperesents a non-zero rational number if and only if $m$ is not of the form $pks^2$ with $\left(\frac{k}{p}\right)=-1$ or $qks^2$ with $\left(\frac{k}{q}\right)=-1$.
To prove these lemmas author refers to general theorem of Hasse-Minkowski. Can they be proved without using such strong results?
| 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 $f(x,y) \equiv 0 \pmod r,$ THEN $x,y \equiv 0 \pmod r$ and $f(x,y) \equiv 0 \pmod {r^2}.$
It suffices to find out what numbers are represented integrally by these forms. Robinson was very careful about picking these.
Given $g(x,y,z) = x^2 + y^2 - p z^2,$ with prime $p \equiv 3 \pmod 4.$ If $g(x,y,z) \equiv 0 \pmod 4,$ it follows that $x,y,z$ are all even. We say that $g$ is anisotropic in $\mathbb Q_2;$ this is the most immediate version of anisotropy available for ternary forms. At the same time, if $p \equiv 3 \pmod 8,$ it is easy to check that $g \neq 3 \pmod 8.$ Similarly, if $p \equiv 7 \pmod 8,$ it is easy to check that $g \neq 7 \pmod 8.$
So, as far as integers go, $x^2 + y^2 - 7 z^2 \neq 4^k (8n+7).$ This is exactly the same behavior as $x^2 + y^2 + z^2$ as regards the prime 2.
Next, $p \equiv 3 \pmod 4,$ so $p$ does not divide $-4$ and $(-4|p)= -1.$ As a result, if $x^2 + y^2 \equiv 0 \pmod p,$ then $x,y \equiv 0 \pmod p$ and $x^2 + y^2 \equiv 0 \pmod {p^2}.$ Thus, if $x^2 + y^2 - p z^2 \equiv 0 \pmod {p^2},$ we then get $x,y,z \equiv 0 \pmod p.$
Then what happens at the first power but short of the second? If $p \parallel g, $ which is a popular symbol that means $p|g$ but not $p^2 | g,$ we know that $x,y$ are divisible by $p$ but not $z$. The result is $p$ times $-z^2.$ Now, $p \equiv 3 \pmod 4,$ so $-z^2$ is a quadratic nonresidue; it is not possible to integrally represent $p$ times a quadratic residue $\pmod p.$ So,
$$ x^2 + y^2 - p z^2 \neq 4^k (8n + p); \; \; \; x^2 + y^2 - p z^2 \neq p^{2k+1} (\mbox{residues} \pmod p); $$
In the expression on the left the $n$ may be postive or negative, and so we can replace $p$ by 3 if $p \equiv 3 \pmod 8,$ or by 7 if $p \equiv 7 \pmod 8.$ Both expression take on a different look if the numbers considered are negative...
Now, Minkowski showed that the form $x^2 + y^2 - p z^2$ integrally represents every other integer. Hasse extended such results to number fields. Leonard Eugene Dickson, in his books, wrote several complete proofs of this more annoying direction for indefinite ternary forms, that they do represent all numbers not ruled out by "conguence considerations." The numbers ruled out are usually referred to as "the progressions" by Dickson, B. W. Jones, and Gordon Pall. And i still encourage you to get the 1950 book of B.W. Jones. As I wrote by email, I do not know the books on this material in Russian. Dickson takes a very elementary approach in Modern Elementary Theory of Numbers (1939). For example, on page 159, exercise 3 is: $x^2 + y^2 - C z^2=0$ has integral solutions, not all $0,$ if and only if $-1$ is a quadratic residue of $C.$
FOUND IT, no Minkowski or Hasse. Dickson (1939), Theorem 118 on page 164. Let $p$ be an odd prime $p \neq 1 \pmod {24}.$ Then every "T-form" of Hessian $-p$ represents every integer $a$ which is none of the types $4^k(8l+p), p^{2k+1}(pl+R),$ where $k \geq 0$ and $R$ ranges over the $(1/2)(p-1)$ quadratic residues of $p.$ But it never represents an integer of one of those types.
Dickson proves it the way I or Minkowski would, given "eligible" number $a,$ construct a quadratic form $$ F = a x^2 + b y^2 + c z^2 + 2 r y z + 2 s z x $$ of the correct determinant (Hessian) and showing that $F$ is integrally equivalent to $x^2 + y^2 - p z^2.$
Alright, Robinson's second form is $x^2 + q y^2 - p z^2,$ with $p \equiv 1 \pmod 4$ and $(q|p) = -1. $ Since $p \equiv 1 \pmod 4,$ it follows that $(-4q|p) = -1.$ By Lemma B above, if $x^2 + q y^2 \equiv 0 \pmod p,$ then $x,y \equiv 0 \pmod p$ and $x^2 + q y^2 \equiv 0 \pmod {p^2}.$ If $p \parallel x^2 + q y^2 - p z^2,$ the result is $(-p)$ times a residue. But again $p \equiv 1 \pmod 4,$ so this is just $p$ times a quadratic residue. So, the ternary cannot represent $p$ times a NON-residue. The part about $p^2$ applies as before. Next, $(q|q) = -1. $ If $x^2 -p y^2 \equiv 0 \pmod q,$ then $x,y \equiv 0 \pmod q$ and $x^2 -p y^2 \equiv 0 \pmod {q^2}.$ If $q \parallel x^2 + q y^2 - p z^2,$ the result is $q$ times a residue. So $q$ times a NON-residue is impossible.
All together,
$$ x^2 + q y^2 - p z^2 \neq p^{2k+1} (\mbox{NonRes} \pmod p); \; \; \; x^2 + y^2 - p z^2 \neq q^{2k+1} (\mbox{NonRes} \pmod q). $$
Every other number is so represented. In one case, this is Theorem 119 on page 168 of Dickson 1939. The bad news is that he restricts the prime we are calling $p,$ because he wants to give a complete list of all such "T-forms." Otherwise everything is there.
VERY finally, Robinson uses lower case letters to mean integers, while upper case mean rational numbers. Since she chose forms with such strong anisotropy properties, the numbers represented rationally are simply those divided by the squares of integers, as we can replace any $(x,y,z)$ by $(x/m,y/m,z/m).$
That is probably enough. It seems you want to see Dickson (1939) or a translation.
| {
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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 $n$?
Thanks in advance.
| 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, 6), (3, 7), (3, 8), (4, 6), (4, 7), (4, 8), (5, 6), (5, 7), (5, 8)]
second K_{3,3}=[(0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8)]
shared vertices 8,6,7
Added $K_{3,3n}$ are counterexamples too.
$g(K_{3,3n})=\lceil (3n-2)/4 \rceil < (3n-2) / 4 + 1$
$K_{3,3n}$ has $n$ disjoint $K_{3,3}$ sharing only $3$ vertices:
connect the $3$ partition to $3$ distinct vertices from the $3n$ partition.
| {
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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:
$$A_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2.$$
Thanks to the recurrence relation
$$n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}- (n-1)^3 A_{n-2},$$
if $k$ is a solution of (1), then all seventeen consecutive numbers beginning from $A_{17k}$ are congruent to zero modulo $17^2$. In particular,
$$A_{51},A_{52},A_{53},\ldots,A_{66},A_{67}\equiv 0 \pmod {17^2}.$$
| 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 $\{b_n\}_{n \ge 0}$ such that $$(**) A_{pn+k} \equiv A_{n} (A_{k} + pn b_{k}) \bmod {p^2},$$
where $p$ is a prime, $n \ge 0$ and $0 \le k < p$. The first values of $b_n$ are given by $b_0=0,b_1=12$.
Specializing $(*)$ to $p=17$, we see that $A_n$ is divisible by $17$ iff one of the digits $d_i$ in the base-17 expansion of $n$ satisfies $A_{d_i} \equiv 0 \bmod 17$. A finite check shows that the solutions to $A_d \equiv 0 \bmod 17$ in the range $0 \le d < 17$ are $d=3,13$.
Specializing $(**)$ to $p=17$ and $k \in \{0,1\}$, we find that
$$A_{17n} \equiv A_{n} \bmod {17^2},$$
$$A_{17n+1} \equiv A_{n} (A_1 + 17\cdot 12\cdot n) \bmod {17^2}.$$
This immediately shows that the first equation in your system implies the second one, and that your system is equivalent to
$$(***)A_n \equiv 0 \bmod {17^2}.$$
We now proceed to solve $(***)$. Let $S$ be the set of solutions to $(***)$. Writing $n$ as $17m+m'$ ($0 \le m' < 17$), we find that $(***)$ is equivalent to
$$A_{m} (A_{m'}+17mb_{m'}) \equiv 0 \bmod {17^2}.$$
There are three possible cases.
*
*$A_m \equiv 0 \bmod {17^2}$.
*$A_m \equiv 0 \bmod {17}$, $A_{m'}+17mb_{m'} \equiv 0 \bmod 17$.
*$A_{m'} + 17mb_{m'} \equiv 0 \bmod {17^2}.$
The first case says: if $n \in S$ then $17n+n' \in S$ ($0 \le n' < 17$).
The second case says: $m$ has $3$ or $13$ as one of its base-17 digits, and $m' \in \{3,13\}$.
The third case says: Either $m'=3$ and $m \equiv 0 \bmod 17$, or $m'=13$ and $m \equiv -1 \bmod 17$.
(Here I've used the values $A_3=1445=17^2\cdot 5$, $A_{13} = 364713572395983725$, $b_{3}=4438$, $b_{13}=7475161875743183448469/6006$.)
This characterizes the set $S$.
| {
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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 related to the Mordell's equations $y^3=x^2+2$ and $y^3=x^2+18$ which have well known sole solutions $x=5,y=3$ and $x=3,y=3$. Indeed it follows from $3^3=5^2+2$ and $3^3=3^2+18$ that $3^5-(15)^2-(3^3-3^2)=0$.
| 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 possibilities are $x=0$ and $x=1$, and they indeed give solutions.
| {
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Subsets and Splits