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Does there exist a sequence $(a_i)_{i \geq 0}$ of distinct positive integers such that $\sum_{i\geq 0}\frac{1}{a_i} \in \mathbb{Q}$ and $$\{ p \in \mathbb{P} \text{ }|\text{ } \exists\text{ } i\geq 0 \text{ s.t.}\text{ } p | a_i\}$$ is infinite?

Motivation:

All geometric series (corresponding to sets $\{ 1,n,n^2,n^3,... \}$) are rational and the terms obviously contain finitely many primes. The same is true for say, sums of reciprocals of all numbers whose prime factiorisation contains only the primes $p_1, p_2, ...,p_k$ : the sum is then $\prod_{i=1}^k\left(\frac{p_i}{p_i-1}\right)$

On the other side, series corresponding to sets $\{1^2, 2^2, 3^2, ...\}, \{1^3,2^3,3^3,...\},\{1!,2!,3!,...\}$ converge to $\frac{\pi^2}{6}$, Apery's constant and $e$ respectively, which are all known to be irrationals.

I am aware of the fact that if this statement is true then it has not been proven yet (since it implies that the values of the zeta function at positive integers are irrational, which to my knowledge has not been shown yet).

Any counterexamples or other possible observations (such as, instead of requiring the set of primes to be infinite, requiring that it contains all primes except a finite set)?

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For how many positive integers $m$ the number $m^3+5m^2+3$ is a perfect cube?

What if $m$ is a negative integer?

It intuitively seems to me that no positive integer can satisfy the given condition, but I can't prove it mathematically.

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Practice Questions on Circles and Cyclic Quadrilateral

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In the remarkable book Rings of Continuous Functions, by Gillman and Jerison, I came across with the following theorem:

Urysohn's Extension Theorem (UT). A subspace $S$ of $X$ is $C^*$-embedded in $X$ if and only if any two completely separated sets in $S$ are completely separated in $X$.

The proof goes as usual, with the construction of a sequence of continuous functions that converges uniformly to the desired function. Then, after proving Urysohn's Lemma (for normal spaces), UT yields Tietze's Theorem (TT).

On the other hand, there are some ways to prove TT without mentioning uniform convergence at all. For instance, as done by Scott here.

Since UT is a generalization of TT, I wonder if there is a way to prove UT without using uniform convergence. I tried to adapt Scott's argument to the general setting of UT, but I could not avoid the need for normality.

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Let be $f : \Bbb R^n \to \Bbb R$ monotone over all lines (not affine ones, but if there is an answer over affine lines, I'm interested.)

Is it possible to find $h : \mathbb{R} \to \Bbb R$ monotone and $l : \Bbb R^n \to \Bbb R$ linear so that $f = h \circ l$ ?

I tried to look by supposing I have such a factorization, and as $\ker l$ is a hyperplane, I have $n - 1$ lines where $f$ is constant. I tried to use the monotonicity condition by trying to compare $f(0)$ and over lines, but it didn't work.

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I want to show: for any $1\leq i \leq k-1$, and $1\leq j\leq k-1$, $$c_{ij} = 0,$$ where: $$ c_{ij} = k \sum_{\ell=1}^{k-1} \frac{1}{\ell} \left(-1\right)^{\ell-i}\sum_{m=1}^{\min\{\ell,k-\ell\}}\binom{\ell}{m}\binom{k-\ell-1}{m-1}\sum_{c=0}^m\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}. $$ I think this is equivalent to: $$ c_{ij} = k\sum_{\ell=1}^{k-1}\sum_{m=0}^{\ell}\sum_{c=0}^m\frac{1}{\ell} \left(-1\right)^{\ell-i} \binom{\ell}{m}\binom{k-\ell-1}{m-1}\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}. $$ This should be numerically correct (verified by numerical method.) But I have been stuck for a month to prove this. I have tried induction on $i,j,k$, and hypergeometric functions. I think Egorychev method should help, but this equation involves 5 binomial coefficients, which I don't know how to solve. I really appreciate your help!

More details: $c_{i,j}$ is the coefficient of a polynomial.

Method 1-Egorychev method(I tried): Using Egorychev Method we have: $$ \binom{k-\ell-1}{m-1} = \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\frac{\left(1+w_2\right)^{k-\ell-1}}{w_2^{m}}d w_2 $$ $$ \binom{k-2m}{i + j -2c -\ell}= \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k-2m}}{z_2^{i + j -2c -\ell+1}}dz_2 $$ $$ \binom{m}{\ell+c-i} = \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{\left(1+z_3\right)^m}{z_3^{\ell+c-i+1}}dz_3 $$ Then, I can re-write $c_{ij}$ as: \begin{align*} c_{ij} &= k \sum_{\ell=1}^{k-1} \frac{1}{\ell} \left(-1\right)^{\ell-i} \sum_{m=1}^{\ell} \binom{\ell}{m}\binom{k-\ell-1}{m-1}\sum_{c=0}^m\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}\\ & = \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}}\sum_{\ell=1}^{k-1} \left( \frac{z_2}{z_3}\right)^{\ell}\left(-1\right)^{\ell} \\ & ~~~~ \sum_{m=0}^{\ell} \frac{k}{\ell}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \left(\frac{1+z_3}{\left(1+z_2\right)^{2}}\left(1+\frac{z_2^2}{z_3}\right)\right)^m dz_2 dz_3\\ & = \sum_{\ell=1}^{k-1} \frac{k}{\ell} \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}} \\ & ~~~~ \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\left(1+w_2\right)^{k-\ell-1}\left(1+\frac{1}{w_2} \frac{z_2^2+z_3+z_2^2z_3+z_3^2}{z_3\left(1+z_2\right)^{2}} \right)^{\ell}\left( \frac{z_2}{z_3}\right)^{\ell}\left(-1\right)^{\ell} d w_2 d z_2 d z_3\\ & = \sum_{\ell=1}^{k-1} \frac{k}{\ell} \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}} \\ & ~~~~ \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\left(1+w_2\right)^{k-\ell-1}\left(-\frac{z_2}{z_3}-\frac{z_2}{w_2} \frac{z_2^2+z_3+z_2^2z_3+z_3^2}{z_3^2\left(1+z_2\right)^{2}} \right)^{\ell} d w_2 d z_2 d z_3\\ \end{align*} Not sure how to do it next.

Method 2: Induction on $i,j,k$ $$ h_{m,\ell,i,j,k} = \sum_{c=0}^m \binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i} $$ And $$ h_{m,\ell,i,j,k+1} =h_{m,\ell,i,j,k}+h_{m,\ell,i,j-1,k} $$ When $\ell > m$, we have the: $$ h_{m,\ell,i+1,j} = h_{m,\ell-1,i,j} $$ Thus, we have: \begin{align*} c_{i,j,k} & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell,i,j,k}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \end{align*} \begin{align*} c_{i+1,j,k} & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell-1,i,j}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \\ & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell-1,i,j}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \\ \end{align*} More complicated.... if we look into the difference between the two equations.

I really appreciate the help!!!!

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Practice Problems on Cyclic Quadrilateral

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The corners of a fixed convex (but not necessarily regular) n-gon are labeled with distinct letters. If an observer stands at a point in the plane of the polygon, but outside the polygon, they see the letters in some order from left to right, and they spell a “word” (that is, a string of letters; it doesn’t need to be a word in any language).

Determine, as a formula in terms of $n$, the maximum number of distinct $n$-letter words which may be read in this manner from a single $n$-gon. Do not count words in which some letter is missing because it is directly behind another letter from the viewer's position.

//My attempt//

• First I notice that there is a bijection between number of words and {sides and diagonals}:
• On extending each edge to both its sides, outside the polygon, the plane is divided into $2n$ parts. Standing in each of these sides, gives a different word.
• Further, extending each diagonal to both its sides adds more division of the plane, equal to twice the number of diagonals of $n$-gon, which is $$2(\binom{n}{2}-n)$$
• Adding these two gives $$2\binom{n}{2}$$
• I don't know, but this seems to work for triangles, quadrilaterals and maybe even pentagons.

On the official site's solution page, two solutions are given: One is ugly and well, seems complicated. The other one is given to be: $$2\binom{n}{2}+2\binom{n}{4}$$ which my solution is closer to. Any help to get it would be appreciated. And, please point out if I've done something wrong.

Thanks a lot, good signiors, for thy help I shall be eternally obliged ;-)

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I wonder if someone can help me get a number for possible positions for this childs game of 15 sliding pieces in a 4x4 board.

The normal game has numbers 1-15 that need to be sorted. The board I want to calculate has only three kind of pieces: 6 red, 6 green, 3 black, and of course the empty slot you can move through.

I referred to this post and came up with this solution $\binom{16}{6}\binom{10}{6}\binom{4}{3}=6.726.720$. Does that look correct, or did I miss something?

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Area and Perimeter Worksheet

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Practice Questions on Circles for Grade 9

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Practice Questions of Transversal Worksheet

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Problem Solving with Quadrilaterals Worksheet

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Practice Problems in Geometry for 9th Grade

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Flag of the United States from 1777 to 1795.

In honor of Independence Day we thought it would be fun to look at one story about the first American Flag.

The rough design of the flag was drawn by a committee of George Washington, Robert Morris, and George Ross in 1776.  When approached by the committee, George Ross's niece, a respected seamstress, Betsy Ross, suggested some important changes.  One change was to use a 5-pointed star to represent each of the 13 colonies. The committee objected that a pentagram would be too hard to make. Betsy demonstrated that she could create the desired star by simply folding fabric and making just one scissors cut.  So, the design was changed.

How did she do that fold and cut?

The activity: BetsyRossStar.pdf

CCSS: 7.G.B, 8.G, HSG.CO

For members we have an editable Word docx and one solution to the folded line task.

BetsyRossStar.docx       FoldLines-solution.pdf

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We also have two previous activities for the Fourth of July.

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I do an increasing amount of work in primary schools and with primary school trainees developing maths subject knowledge and pedagogy. Over the last few years I’ve managed to blow my own mind many times just imagining the beginnings of concepts being developed in children. At secondary level (11+), most concepts at their basic root are pretty much embedded and understood or at least taken as ‘the way it is’. For example, a square is called a square, and the number after ten is called eleven. When I wrote the book ‘Yes, But Why?’, I really enjoyed putting myself in the position of a student being told all these strange and mystical maths facts and I tried my best to explain them to make them make sense. At a younger age, concepts must be even more mind boggling at first. An example I’ve used before is counting.

We count in base-10 (ie we use 10 symbols (0-9) and then start combining them. Imagine learning this for the first time, and getting to the number 9, then being told we start to use them all again. It’s weird right? And why do we use ten symbols? Why not five, or sixty? Does everyone count using the same system? (spoiler: no). These are thoughts we as adults probably don’t have, but they fascinate me. Who decided to call a square a square and not a quadrangle or a a tetragon? What were they thinking? Why are we using 360 degrees, and why are they called degrees? Where do these weird mathematical words come from? We’re used to using terms like polygon and acute angle, but these words are weird! “I’m going to find the mean of this data set”. The what?? Who decided to call it that? I “can’t subtract 3 from 1 so I need to borrow…” – you’re doing what now?? Can I borrow from elsewhere? Can I just move everything around? Also who decided that we should stack numbers like that to add or subtract them? What were they thinking? How did that idea become so popular?

Trying to think ‘like they do’ is a good first step into empathising with how abstract even the little details can be in mathematics, and how easy it is to misunderstand or misinterpret things. Furthermore, it’s a great exercise in testing your own subject knowledge, and figuring out what might be beneficial to incorporate into your own teaching and explanations to prioritise what must be prioritised – making maths make sense.

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Is there a way you can get the number 6 from the numbers 6, 7, 8, and 9 using only addition, subtraction, multiplication, and division, without combining two numbers e.g. using the 6 and 7 to create 67. You may use parentheses and you can have to use each of the 4 numbers once.

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I need a mathematical algorithm for finding the angle, formed by three points, which is open toward a fourth point.

For example, in Fig1 below I desire angle $\theta$ because it is "facing" point $P$. However, the formula for the angle between two vectors, $$\theta=\cos^{-1}\Big(\frac{\vec{u}\cdot\vec{v}}{uv}\Big)$$ gives angle $\alpha$ because the above relation always returns the angle which is less than 180 degrees.

(Fig1)

In contrast, in Fig2 I desire angle $\alpha$ because this time it is "facing" point $P$.

(Fig2)

As a final example, in Fig3 below I need angle $\theta$, because it is still technically "facing" point $P$.

(Fig3)

This problem can occur in any orientation, making it difficult to say, for example, "If point $P$ is to the left of points $A, B, C$, use the leftmost angle, otherwise use the right" or something like that.

Any help would be appreciated!

---

My Attempted Solution

If anyone's interested, my current solution is to split the angle facing the point $P$ in half and add those two together.

In Fig4 below, I use the grey line to split the angle in half, find $s$ and $t$ using dot product, then add $s$ and $t$ to get the angle facing $P$.

(Fig4)

This works until I reach a situation like that in Fig5. What I need is $s+t$, but what I get is $q+t$ because the dot product gives $q$ (since $s$ is greater than $180$ degrees).

(Fig5)

It's a conundrum.

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In the fridge there is a box containing 10 expensive high quality Belgian chocolates, which my mum keeps for visitors. Every day, when mum leaves home for work, I secretly pick 3 chocolates at random, I eat them and replace them with ordinary cheap ones, that have exactly the same wrapping. On the next day I do the same, obviously risking to eat also some of the cheap ones. How many days on average will it take for the full replacement of the expensive chocolates with cheap ones?

I would say $10/3$ but this is very simplistic. Also, the total number of ways to pick 3 chocolates out of 10 is $\binom {10} 3=\frac {10!}{3!7!} = 120$ which means that after 120 days I will have replaced all chocolates but I don't think it is correct.

Any help?

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By the help of Mathematica numeral calculations, I find the following formula holds

$$\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)\quad ?$$

$m>1$ is a positive integer. But I can't prove it.

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$$\text{Find : }\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}$$

I know that $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-4x^{2}}}$ so $\displaystyle\sum_{n=1}^{\infty}n\binom{2n}{n}x^{2n}=\frac{2x^2}{\sqrt{1-4x^{2}}}$.

But I don't know he to complete this work because I find hypergeometric function.

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In the algebraic formulation of quantum physics/information, states $\omega: \mathcal{A}\rightarrow \mathbb{C}$ are defined as linear functionals on a $C^*$-algebra $\mathcal{A}$ (algebra of observables, representable as $\mathcal{B}(H)$ for some Hilbert space $H$ via the GNS construction) that are positive ($\omega(A^*A)\geq 0\,\forall A\in \mathcal{A}$) and normalized ($\omega(I)=1$ for $\mathcal{A}$ with unit element $I$ or an equivalent condition for non-unital $\mathcal{A}$). These quantum states are then usually represented as density operators defined via $\omega(A)=:\text{Tr}(\omega A)$, but it is well-known that in infinite dimensions there are so-called non-normal states that are not representable this way. Is this due to the fact that for infinite dimensions $\mathcal{B}(\mathcal{H})\simeq H\otimes H^*$ does not hold?

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I’ve been reading Mathematics Galore! by James Tanton. He has a brief elegant and accessible proof of the existence of a power of 3 that ends in 001. He then asks if a power of 3 ever ends in 007.

This is equivalent to asking if $3^n = 7 \bmod1000$.

I can brute force the answer using WolframAlpha but cannot figure out an elegant proof.

This question has been driving me nuts for about a month now. Any help is greatly appreciated.

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In this picture: a runner wants reach B by starting at A. Velocity in White space is $10 m/s$ and in brown space is $5 m/s$. what is the minimum time that he need?

$$a)\sqrt{26}$$ $$b)\sqrt{20}$$ $$c)5$$ $$d)\sqrt{30}$$ $$e)\sqrt{34}$$

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Let $f$ be a polynomial of total degree at most three in $(x,y,z)\in\mathbb{R}^3$. Prove that $$\int\limits_{x^2+y^2+z^2\leq1}f(x,y,z)\,dx\,dy\,dz = \frac{4\pi f((0,0,0))}{3} + \frac{2\pi(\Delta f)((0,0,0))}{15}$$ Here $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ is the Laplacian operator on $\mathbb{R}^3$.

I even don't know how to approach this problem. Since the LHS of the equality is given by a volume integral, I thought about applying the divergence theorem, but wasn't able to do it. Any help?

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So I was reading this, about the practical application of topology.

I wanted to ask what exactly is in it---as a subject what is studied in this field.

I have seen videos wherein they oversimplify tell us that in topology, you squeeze and stretch but don't cut, or how a donut and a mug is equivalent. (I think it must be like simplifying calculus and saying that it's just fancy addition.)

How does it help in mathematics, because it is studied as a full fledged course, it must have its perks and used too.... [ Also considering that there are so many tags under this topic on SE ] ??

(I am still in high school so I don't have a lot of information, but I consider myself to be a curious math enthusiast)

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We could have written = as "equals", + as "plus", $\exists$ as "thereExists" and so on. Supplemented with some brackets everything would be just as precise.

$$\exists x,y,z,n \in \mathbb{N}: n>2 \land x^n+y^n=z^n$$

could equally be written as:

ThereExists x,y,z,n from theNaturalNumbers suchThat 
     n isGreaterThan 2 and x toThePower n plus y toThePower n equals z toThePower n

What is the reason that we write these words as symbols (almost like a Chinese word system?)

Is it for brevity? Clarity? Can our visual system process it better?

Because not only do we have to learn the symbols, in order to understand it we have to say the real meaning in our heads.

If algebra and logic had been invented in Japan or China, might the symbols actually have just been the words themselves?

It almost seems like for each symbol there should be an equivalent word-phrase that it corresponds to that is accepted.

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Angles at the Center and Circumference Examples

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Let $R$ be an Euclidean domain with the degree function $d$. Let $A\in R^{n\times n}$ be an $n\times n$-matrix with entries in $R$ such that det$(A)=0$. As a module map $A:R^n\rightarrow R^n$, there always exists a kernel element $v\in R^n$ since det$(A)=0$.

Assuming $d(A_{ij})\leq m$ for all $i,j$, is there an explicit bound $k(m,n)$ such that there exists a kernel element $v\in R^n$ satisfying $d(v_i)\leq k(m,n)$?

Edit : $v$ is assumed to be nonzero.

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Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4.$ Prove that: $$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}+\sqrt{c+11ab+6}\geq 9\sqrt2.$$

The equality occurs for $(a,b,c)=(1,1,1)$ and again for $(a,b,c)=(2,2,0)$ and for the cyclic permutations of the last.

I tried Holder: $$\sum_{cyc}\sqrt{a+11bc+6}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+11bc+6}\right)^2\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}\geq$$ $$\geq \sqrt{\frac{\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}$$ and it's enough to prove that $$\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3\geq162\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3,$$ which is true for $(a,b,c)=(2,2,0)$, $(a,b,c)=(1,1,1)$, but it's wrong for $(a,b,c)=\left(8,\frac{1}{2},0\right).$

Thanks to River Li for this counterexample.

Also, I tried a substitution $a=\frac{2x}{y+z},$ $b=\frac{2y}{x+z}$, $c=\frac{2z}{x+y}$ and SOS, but it seems very complicated.

Also, I tried the following estimation. By Minkowcki: $$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}\geq\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}.$$

Now, for $c=\min\{a,b,c\}$ it's enough to prove that $$\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}+\sqrt{c+11ab+6}\geq9\sqrt2,$$ which not so helps.

Also, LM does not help.

Thank you!

Update

Also, there is the following.

We need to prove that: $$\sum_{cyc}\sqrt{\frac{2x}{y+z}+\frac{44yz}{(x+y)(x+z)}+6}\geq9\sqrt2$$ or $$\sum_{cyc}\sqrt{\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3}\geq9,$$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0.$

Now, by Holder $$\left(\sum_{cyc}\sqrt{\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3}\right)^2\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3\geq$$ $$\geq\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3,$$ where $k$, $m$ and $n$ are reals such that the expression $kx^2+y^2+z^2+myz+nxy+nxz$

is non-negative for all non-negatives $x$, $y$ and $z$.

Thus, it's enough to choose values of $k$, $m$ and $n$ for which the following inequality is true. $$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3\geq$$ $$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3$$ From the equality case we can get that should be $$2k-5m+2n=8.$$ For $k=1$, $m=0$ and $n=3$ we need to prove that:

$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\geq$$ $$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3,$$ which is true for $y=z$ and it's true for $z=0$, but I have no a proof for all non-negative variables.

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I already told few questions ago that I'm currently reading an abstract about the Lotka Volterra differential equations. But now I have a proof, where I need explanations. Consider: $$ \dot{x} = -xy\frac{\delta H}{ \delta y} , x(0) = \hat{x} $$ $$ \dot{y} = xy\frac{\delta H}{ \delta x} , y(0) = \hat{y} $$where $H(x,y) = x + y - ln(x) -ln(y)$.

I have to show that this System preserve the weighted area $(dx \wedge dy)/xy$. I marked my Questions in the proof below.

Proof:

Let $\Omega_0$ be a subset of $\mathbb{R}^2$ at time $t_0$ and $ \Omega_1$ the set into which $\Omega_0$ is mapped by the system above at time $t_1$. Preservation of $(dx \wedge dy)xy$ is equivalent to $$ \int_{\Omega_0} \frac{1}{xy}dxdy = \int_{\Omega_1} \frac{1}{xy} dxdy $$ first Question: why is this equivalent? We now look at the Domain $D$ in x,y,t space with bondary $\delta D$ given by $\Omega_0$ at $t_0$, $\Omega_1$ at $t_1$ and the set of trajectories emerging from the boundary of $\Omega_0$ and ending on the boudnary of $\Omega_1$. Consider the vector field $$ v := \frac{1}{xy}(\dot{x},\dot{y},1)^T $$ in $x,y,t$ space. Integrating this vector field over the boundary $\delta D$ of $D$, we obtain $$ \int_{\delta D} v \cdot n = \int_{\Omega_0} v \cdot n_0 + \int_{\Omega_1} v \cdot n_1 = \int_{\Omega_0} \frac{1}{xy} dxdy - \int_{\Omega_1} \frac{1}{xy}dxdy $$ where $n_0 =(0,0,-1)^T$ denote the unit outward normal of $\Omega_0$ and $\Omega_1$.Second question & Third question: Can you explain why we integrate $v \cdot n$ ? I thought we integrate $v$ and can you explain the first equation above? There is no other contribution to the surface integral, because the vector field $v$ is by contruction parallel to the trajectories, which form the rest of the bondary $\delta D$. Forth question: Can you explain why vector field is parallel to the trajectories? Applying the divergence theorem to the left hand side of the same equation, we get $$ \int_{\delta D} v \cdot n = \int_D \nabla v = \int_D - \frac{\delta H^2}{\delta x \delta y} + \frac{ \delta H^2}{\delta x \delta y} + 0 = 0 $$ which concludes the proof.

I hope that my questions are not to easy, but I'm a beginner.

from Hot Weekly Questions - Mathematics Stack Exchange

A particle is allowed to move in the $\mathbb{Z}\times \mathbb{Z}$ grid by choosing any of the two jumps:

1) Move two units to right and one unit up

2) Move two units up and one unit to right.

Let $P=(30,63)$ and $Q=(100,100)$, if the particle starts at origin then?

a) $P$ is reachable but not $Q$.

b) $Q$ is reachable but not $P$.

c) Both $P$ and $Q$ are reachable.

d) Neither $P$ nor $Q$ is reachable.

I could make out that the moves given are a subset of that of a knight's in chess. I think that it'd never be able to reach $(100,100)$ but I'm not sure of the reason. It has got to do something with the move of the knight but I cannot figure out what.

I don't have a very good idea about chess, so I'd be glad if someone could answer elaborately.

from Hot Weekly Questions - Mathematics Stack Exchange

A new field of research has just been created by Misha Klin, Misha Muzychuk and Sven Reichard: proper Jordan schemes.

They answered a question which I posed some time ago (I don’t remember when), about whether such objects exist. I would not be interested in “empty set theory”, but now we know that they do exist, so we can go ahead and study them.

A little background. You can find more here or here (the second of these references gives some further reading).

Jordan algebras

A Jordan algebra is a vector space (here over the real numbers) with a multiplication ∗ satisfying

• A∗B = B∗A;
• (A∗B)∗(A∗A) = A∗(B∗(A∗A)).

These algebras were introduced by Pascual Jordan as a mathematical foundation for quantum mechanics. While they have not caught on for this purpose, they are used in some parts of statistics, especially for estimation of variance components.

Any associative algebra gives rise to a Jordan algebra, on setting A∗B = (AB+BA)/2. Certain subsets of matrix algebras are also closed under this product and define Jordan algebrasi. Most significantly, the symmetric matrices form a Jordan algebra. An important theorem, the Jordan–von Neumann–Wigner theorem, asserts that apart from some infinite families arising in this way, the only simpleJordan algebra is an exceptional 27-dimensional algebra related to the octonions and the Lie group E₆.

Coherent configurations and Jordan schemes

A coherent configuration is a set C of zero-one matrices satisfying

• the sum of the matrices in C is the all-one matrix J;
• the identity is the sum of some of the matrices in C;
• the set C is closed under transposition;
• the linear span of C is an algebra (closed under matrix multiplication).

The axioms, of course, have a combinatorial interpretation, which you can work out with sufficient diligence.

The configuration is homogeneous if the second condition is replaced by the stronger version asserting that the identity is one of the matrices in C. If in addition all the matrices in C are symmetric, we speak of an association scheme. (I am aware that different terminology is used by different authors here; I have discussed this elsewhere, but let me use my own preferred conventions here.)

For much more on this, see several talks at last year’s Pilsen conference, which can be found here.

Now the definition of a (homogeneous) Jordan scheme is obtained from that of an association scheme by replacing matrix multiplication by the Jordan product A∗B = (AB+BA)/2.

It is easy to see that, if we take a homogeneous association scheme and “symmetrise” it (by replacing a non-symmetric matrix in C and its transpose by their sum) is a homogeneous Jordan scheme.

My question, which Klin, Muzychuk and Reichard have answered negatively, was:

Is every homogeneous Jordan scheme the symmetrisation of a homogeneous coherent configuration?

So now it makes sense to say that a homogeneous Jordan scheme is proper if it is not the symmetrisation of a homogeneous coherent configuration, and to develop the theory of such objects, as Klin, Muzychuk and Reichard have begun to do.

The example

Actually they have many examples, but I will briefly describe the first one, based on a presentation by Sven Reichard at the Slovenian graph theory conference today.

Start with the alternating group A₅ acting transitively on 15 points, the stabiliser of a point being the Klein group V₄. Because this subgroup is contained with index 3 in A₄, the group is imprimitive, with five blocks of size 3. So apart from equality, there are two invariant relations forming five ordered triangles and the reverse, and three further symmetric relations.

One of the triangles is the island, and the other four make up the continent. The edges joining the island to the continent are called bridges, and are of three colours (corresponding to the three further symmetric relations). Now swap two of the colours on the bridges, leaving the remaining edges alone. Symmetrising the resulting structure gives a homogeneous Jordan scheme, which is proper.

Questions

Here are a few questions which could be looked at.

• Does the Jordan–von Neumann–Wigner theorem have any relevance to proper Jordan schemes? In particular, is there one whose Jordan algebra involves the exceptional simple Jordan algebra?
• Given a connected simple graph, think of it as an electric circuit, where each edge is a one-ohm resistor. The effective resistance between pairs of terminals defines a metric, called resistance distance, on the vertex set. This is a refinement of the graph structure, similar to that produced by the symmetric version of Weisfeiler–Leman stabilisation. What is the precise relation between these concepts?

Pascual Jordan

Curiously enough, the name of Pascual Jordan (who introduced Jordan algebras) came up in a completely different context yesterday; I would like to say a bit about him.

Jordan is described as one of the unsung heroes of quantum mechanics. It was in a joint paper by Born, Heisenberg and Jordan that the matrix mechanics approach to quantum mechanics was first published (as opposed to Schrödinger’s wave function approach). It is said that the mathematics of matrices in the paper is Jordan’s work. The Nobel Prize was awarded to Heisenberg, Schrödinger and Dirac. Jordan also invented Fermi–Dirac statistics, but because of an unfortunate publication delay he was beaten into print.

According to the MacTutor biography, the reason for the neglect may have been in part his membership of the Nazi party. He wrote in support of the party, but strongly opposed the more extreme views of Ludwig Bieberbach, who believed that there was a real difference between say “French mathematics” and “German mathematics”, and that teaching “German mathematics” to children would increase their “Germanness” (to put it rather crudely).

Anyway, the context in which Jordan’s name came up was a very entertaining lecture given by one of this year’s St Andrews honorary graduands, Jim Al-Khalili, on the new subject of quantum biology. He pointed out that the first paper ever written on quantum biology was by Pascual Jordan, although nobody took it very seriously at the time. It was generally thought that quantum systems would decohere so rapidly in the messy, hot surroundings of a living cell that no effects would be observed. Jim put the opposite “spin” on it: rather than the environment interfering with they system, we can regard the system as exporting information to the environment. It is possible that European robins detect the Earth’s magnetic field (for navigational purposes) by a quantum effect in the bird’s eye, where the collapse of entanglement gives a signal which can be transmitted to the brain by the optic nerve.

from Peter Cameron's Blog https://ift.tt/2ZYRvsC

I have been trying to get my head round ZF set theory and Peano's axioms, but I have hit some confusion over Peano's definition of the successor function, or more accurately von Neumann's model.

Why did von Neumann use $S(x) := x \bigcup \{x\} $ and not just plain old $S(x) := \{x\} $? The latter seems a lot more simple and easy to work with, so am I missing some major advantage of the former, or is my function incompatible in some way?

Edit: is the only advantage just that the cardinality of the set is equal to the value it represents?

from Hot Weekly Questions - Mathematics Stack Exchange

I need to solve with basic methods this simple Shallow Water Model:

$$\begin{bmatrix}h\\ hv\end{bmatrix}_t+\begin{bmatrix}hv\\ hv^2+\frac{1}{2}gh^2\end{bmatrix}_x=\begin{bmatrix}0\\ 0\end{bmatrix}$$

where $h$ is the heigth of the water, $v$ is the horizontal velocity of the fluid and $g$ is the gravitacional constant.

Writing in quasilinear form, I've got

$$\begin{bmatrix}h\\ hv\end{bmatrix}_t+\begin{bmatrix}0&1\\ gh-v^2&2v\end{bmatrix}\begin{bmatrix}h\\ hv\end{bmatrix}_x=\begin{bmatrix}0\\ 0\end{bmatrix}$$

The Jacobian can be diagonalized as $J=RDR^{-1}$ where

$$R=\begin{bmatrix}\dfrac{1}{v-\sqrt{gh}}&\dfrac{1}{v+\sqrt{gh}}\\ 1&1\end{bmatrix}$$

$$R^{-1}=\begin{bmatrix}\dfrac{v^2-gh}{2\sqrt{gh}}&\dfrac{1}{2}-\dfrac{v}{2\sqrt{gh}}\\ \dfrac{gh-v^2}{2\sqrt{gh}}&\dfrac{1}{2}\bigg(\dfrac{v}{\sqrt{gh}}+1\bigg)\end{bmatrix}$$

$$D=\begin{bmatrix}v-\sqrt{gh}&0\\0&v+\sqrt{gh}\end{bmatrix}$$

I've done the change of variables

$$\begin{bmatrix}\overline{h}\\ \overline{k}\end{bmatrix}=R^{-1}\begin{bmatrix}h\\ hv\end{bmatrix}$$

$$\begin{bmatrix}\overline{h}\\ \overline{k}\end{bmatrix}=\begin{bmatrix}\frac{h}{2}(v+\sqrt{gh})\\ \frac{h}{2}(v-\sqrt{gh})\end{bmatrix}$$

Geting the system

$$\begin{bmatrix}\overline{h}\\ \overline{k}\end{bmatrix}_t+\begin{bmatrix}v-\sqrt{gh}&0\\0&v+\sqrt{gh}\end{bmatrix}\begin{bmatrix}\overline{h}\\ \overline{k}\end{bmatrix}_x=\begin{bmatrix}0\\ 0\end{bmatrix}$$

ie

$$\overline{h}_t+(v-\sqrt{gh})\overline{h}_x=0\\ \overline{k}_t+(v+\sqrt{gh})\overline{k}_x=0$$

Trying to solve this conservation laws by the charactheristic method, I've got to the first

$$x(t)=(v_0(x_0)-\sqrt{gh_0(x_0)})t+x_0\qquad (i)$$

and to the second

$$x(t)=(v_0(\overline{x}_0)+\sqrt{gh_0(\overline{x}_0)})t+\overline{x}_0\qquad (ii)$$

where $v_0,h_0$ are the initial conditions.

To find the solution at time $(x,t)$, I need to solve $(i)$ for $x_0$ and take $\overline{h}(x,t)=\overline{h}_0(x_0)$ and solve $(ii)$ for $\overline{x}_0$, taking $\overline{k}(x,t)=\overline{k}_0(\overline{x}_0)$.

The solution in time $(x,t)$ may be the solution of the system

$$\begin{bmatrix}\overline{h}_0(x_0)\\ \overline{k}_0(\overline{x}_0)\end{bmatrix}=\begin{bmatrix}\frac{h}{2}(v+\sqrt{gh})\\ \frac{h}{2}(v-\sqrt{gh})\end{bmatrix} $$

That is

$$hv(x,t)=\overline{h}_0(x_0)+\overline{k}_0(\overline{x}_0)$$

$$h(x,t)=\bigg(\dfrac{\overline{h}_0(x_0)-\overline{k}_0(\overline{x}_0)}{\sqrt{g}}\bigg)^{2/3}$$

Or better

$$hv(x,t)=\frac{h_0(x_0)(v_0(x_0)+\sqrt{gh_0(x_0)})+h_0(\overline{x}_0)(v_0(\overline{x}_0)-\sqrt{gh_0(\overline{x}_0)})}{2}$$

$$h(x,t)=\bigg(\dfrac{h_0(x_0)(v_0(x_0)+\sqrt{gh_0(x_0)})-h_0(\overline{x}_0)(v_0(\overline{x}_0)-\sqrt{gh_0(\overline{x}_0)})}{2\sqrt{g}}\bigg)^{2/3}$$

I would like to know:

1) If my ideia is correct;

2) If these calculus are sufficient to "solve" the system. In other words, what mean the "solution" of the system? The question that I am dealing is only the following:

Find the solution to the model.

I think this is something vague once I don't have any data. What do you understand as the "solution of the system"?

3) And if the charactheristics crosses or occur some problem with this method? Or if (i) and (ii) don't have solutions? Without the initial conditions, how can I know about possible troubles?

Many thanks.

from Hot Weekly Questions - Mathematics Stack Exchange

The paper “Inertial manifolds for the hyperbolic relaxation of semilinear parabolic equations” co-authored by  Vladimir Chepyzhov (Russian Academy of Sciences, Moscow), Anna Kostianko, and Sergey Zelik, has been published in the AIMS journal Discrete and Continuous Dynamical Systems B. The paper gives a comprehensive study of inertial manifolds for hyperbolic relaxations of an abstract semilinear parabolic equation in a Hilbert space.  A link to the published version is here. A link to the arXiv version is here.  The photo left shows a cascade of tori (picture credit here).

from Surrey Mathematics Research Blog https://ift.tt/320wUWj

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Problem 733

Suppose that three fair coins are tossed. Let $H_1$ be the event that the first coin lands heads and let $H_2$ be the event that the second coin lands heads. Also, let $E$ be the event that exactly two coins lands heads in a row.

For each pair of these events, determine whether they are independent or not.

Definition of Independence

Solution.

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Click here if solved 0

from Problems in Mathematics https://ift.tt/320pMt1

I found this question recently looking around the internet, apparently it was on the IMO shortlist many years back. I haven't been able to solve it, and I am looking for hints and/or full solutions. Apparently it can be done very elegantly.

Consider N non overlapping spheres of equal radius placed in 3D space. Let $S$ be the set of points on the surface of these spheres which are not visible from any other sphere. Show that the total area of $S$ is equal to the surface area of one sphere.

Thoughts so far: the problem is trivial in one dimension, and likely equivalent in difficulty in 2D and 3D (it looks like it still holds in 2D, hence I imagine it works for any number of dimensions). It is also obviously true for 2 spheres, and can be verified with some effort for 3. I've had many ideas but none of them have yet led me anywhere I thought promising.

from Hot Weekly Questions - Mathematics Stack Exchange

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to […]

from Mean Green Math

Peter and I were discussing in a chat room and I thought that it would be nice to test the sequence $$n!+{p_n}!+1$$ for primality.

Then I wrote Peter that I expect much of primes in this sequence, well, at least the expression is formed in such a way that there could be much of them.

But, to my and Peter´s surprise, he computed and tested this sequence for all $1\leq n \leq 100$ and found that for only $n=3$ in that range the expression is prime.

He is pushing his computations even further, even at this moment, and, for me almost unbelievable is that he passed $n=800$ and found that only for $n=3$ the expression is prime and we are curious why there is (so far) only one prime although small factors are impossible for large values of $n$.

So, we agreed that a question about this unusuality should be asked, and I ask three of them, very much interrelated and connected and answerable.

Do we have any explanation of why this sequence has, for the range computed, so a small number of primes? Is it naive to expect that this sequence has an infinite number of primes? What "should" be expected to happen in some very large ranges?

from Hot Weekly Questions - Mathematics Stack Exchange

Problems Based on Circles Examples

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Quadrilateral Practice Problems for 9th Grade

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If $ABCD$ is a parallelogram such that $\angle BAD=\frac{\pi}{4}$, $AB=\sqrt{2}$, $AD=1$ then we shall say that $ABCD$ is a good parallelogram. Parallelograms can be rotated by any angle.

a) Prove that in a rectangular container $2 \times n$ ($n \ge 2$) cannot be packed more than $2n-2$ good parallelograms (packing must be without overlaps between parallelograms).

b) Prove that in a rectangular container $4 \times 4$ cannot be packed more than $12$ good parallelograms (packing must be without overlaps between parallelograms).

My work. It is easy to pack $2n-2$ good parallelograms into rectangular container $2 \times n$. The area of the good parallelogram is $1$. The area of the rectangular container $2 \times n$ is $2n$. Obviously, parallelograms will not be able to cover every region of the rectangular container. Therefore, in a rectangular container $2 \times n$ ($n \ge 2$) cannot be packed more than $2n-1$ good parallelograms. But I have no idea how to improve the evaluation for one parallelogram.

from Hot Weekly Questions - Mathematics Stack Exchange

Area and Perimeter

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Prove that $$ \frac{1}{1.2.3.4}+\frac{4}{3.4.5.6}+\frac{9}{5.6.7.8}+\frac{16}{7.8.9.10}+\dots=\frac{1}{6}\log2-\frac{1}{24} $$

My Attempt $$ T_n=\frac{n^2(2n-2)!}{(2n+2)!}=\frac{n^2}{(2n+2)(2n+1)(2n)(2n-1)}\\ =\frac{1}{24}\Big[\frac{-2}{n+1}+\frac{3}{2n+1}+\frac{2}{2n-1}\Big]\\ $$ $$ S=\frac{1}{24}\Big[{-2}\big[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots\big]+{3}\big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\big]+{2}\big[1+\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\\ =\frac{1}{24}\Big[{2}\big[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots\big]+{3}\big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\big]-{2}\big[\frac{1}{3}+\frac{1}{5}+\dots\big]\Big]\\ =\frac{1}{24}\Big[2\log2+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\Big]=\frac{1}{12}\log 2+\frac{1}{24}\Big[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots\Big] $$ I dont think that the last term series converges so I think I'm stuck, and what does it mean? How do I proceed further and evaluate the infinite series ?

from Hot Weekly Questions - Mathematics Stack Exchange

In Quanta Erica Klarreich recently wrote up Yaroslav Shitov’s new counter example which disproves Stephen Hedetniemi’s 50 year old conjecture, original dissertation, that the number of colors required to color the tensor product of two graphs is the lesser of the numbers used to color the original graphs. These colorings have applications in areas from scheduling to seating plans, and it is clear from Klarreich’s reporting that mathematicians are excited about this result. In fact, Hedetniemi responded very positively when asked by Klarreich about the counter example, saying it “has a certain elegance, simplicity and definitive quality to it.” The counter-example may show Hedetniemi’s conjecture is not true, but Klarreich points out that we do not yet know just how false it is. So, while Shitov has closed one door on this problem, there are still many which are open.

via Thomas Lin on Twitter.

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     Shashi Thutupalli is a scientist -- in Bangalore, India -- who enjoys poetry and often explores the connections between poetry and mathematics.   He has shared with me several samples of his work and I offer below the opening page (of three) of Thutupalli's poem, "Out of Nothing I Have Created a Strange New Universe."  (The full poem, together with artwork, is available in Visual Verse -- at this link.)

Page 1 of 3  -- the entire poem is found here at VisualVerse.org.

from Intersections -- Poetry with Mathematics

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It was a busy summer. We’d taken our first overseas vacation. We moved from California to Texas, spending three days in the car with just enough belongings to start our household before the moving truck arrived several days later. We arrived in Austin in time for our daughter to experience her first summer T-storm where the air is warm, but its pouring rain—quite a phenomenon to a six-year-old from Southern California.

Combating Summer Slide

My daughter was excited and nervous about starting first grade in a new school. She came home from that first day thrilled about her teacher, Miss Reneau, and her classmates, but frustrated that she’d had some challenges remembering how to do things she knew she’d learned in kindergarten. 

Ten years later, I don’t recall what those things were, but I do remember noticing effects of the “summer slide” as I looked over the work she brought home, and in the little homework assignments she had those first few weeks of school. I knew then that it would take effort to keep her both in the habit of school, and help her retain what she’d learned during the summer months.

That particular “summer slide” wasn’t terrible; we’d read plenty of books together, and had all kinds of conversations about the things we experienced together. There are 528 steps in St Paul’s Cathedral. Custard has different ingredients than ice cream, but it’s still sold in a cone at the bottom of those 528 steps. Scottish broadswords are longer than she was tall, and the Mons Meg cannon could fire cannon balls that weighed 400 pounds! There are 1,364 miles, and two whole other states (our route took us through Arizona and New Mexico) between Orange County, CA and Austin, TX, and it was only a half-mile walk from our new home to her new school.

Having these sorts of discussions with our kids, and answering their endless “why” questions is one way that we can support their learning, their curiosity, and intrinsic motivation to continue learning. These discussions don’t require extra planning, or preparation, they just require attention and focused listening. They also don’t require any special destination. I remember curious conversations about tomato worms (hornworms), because they looked like the plant they were eating. There is curious stuff all around us.

Curiosity doesn’t have to slip away as our kids grow older. My daughter recently shared alarming statistics on corporate farming that she’d uncovered for a presentation in her English class. Once she finishes her final exams, I’ll share an article I found on the difference one person can make with some good research, and a strong message focused on the right recipients. I also know she intends to spend many summer hours at the beach where I hope she’ll learn to read and ride the waves, and check out all the creatures living in the rocks along the jetty.

Some school work habits are harder to maintain over the summer. Summer is for fun and for downtime, especially unplanned, unorganized, unscheduled fun.

But having been a teacher, been raised by a teacher, having siblings who are teachers, and conducting research with teachers and students, I do know that kids who have to spend time ramping back up, can fall quickly behind come September.

What I found though, after that summer between kindergarten and first grade, is that it doesn’t take much for kids to keep their work habits in gear for an easy back to school transition. We had a fair-trade rule in our house. Extra screen time was traded for extra book time. Play Wii or a game on mom’s phone for 20 minutes, read for 20 more minutes. Watch a 30-minute TV show, read for 30 minutes.

Curiosity Driven Learning

We also had summer “homework” folders. At the completion of each homework folder was a little treat—a visit to a favorite park, a yummy sweet, or a special activity with mama. These folders had tasks in them appropriate to her age. She might play Yahtzee to practice her multiplication skills, or bake to work on her fraction skills. When she struggled with reading comprehension, she had to find things she was interested in to read about, and then tell her dad or I about them.

A visit to a discovery center sparked an interest in electricity and how it was made, how it made its way into our house, and how it was used by the things in our house. After limitless “how” and “why” questions—the answers to which were waaaaaay beyond our knowledge, and honestly, even our interest—my husband and I decided that her summer homework would be to answer her own questions and share her discoveries with us.

One summer she became super interested in photography, so she learned about light and shadow, and how to put together a concise, interesting PowerPoint presentation—a skill necessary to those next few years of school and beyond.

Some years, she continued working in a workbook or writing journal that came home on the last day of school with pages yet unfilled. Some years she read every book in a series she loved. Always there were conversations about what she was learning—even what she learned from reading fictional stories about fantastic characters. In recent years, she’s shared some of her favorite books with me, and we’ve had engaging discussions about choices the characters made, or the societal wrongs in the story.

Strengthening Relationships & Mathematical Understanding

Through the years, my small efforts to thwart the summer slide have become wonderful relationship builders with my kid. I have a teenager and we get along. We don’t always agree, but we can discuss, negotiate, strategize, and solve together.

By attending to, and encouraging her problem-solving abilities early on, we’ve given her tools she needs to navigate the pitfalls and pathways of being an American teenager.

My daughter spent time in three elementary schools between kindergarten and 5th grade. Unfortunately, none of these, nor her middle school, used ST Math. In fact, she was in high school before I learned about MIND Research Institute and ST Math. As MIND Research Institute supports the 1.2M+ students who have ST Math accounts, we remind teachers to turn on access to all content so kids can continue to play over the summer months.

I can’t help but think how much fun it would have been for my daughter to play ST Math games as one of her “homework folder” tasks. I know it would have deepened her problem-solving skills and her persistence (though that could’ve been a double-edged sword as persistence is not something she lacks!). Maybe the fun activity with mama would have been for mama to try to figure out the challenge games my daughter beat!

Teachers, please turn on all content so your students can continue to play ST Math in the summer months.

Parents, consider checking with your kids’ teachers to make sure they have summer access.

Wishing you all a fun, MINDful, and curious summer!

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If water is poured into a cone at a constant rate and if $\frac {dh}{dt}$ is the rate of change of the depth of the water, I understand that $\frac {dh}{dt}$ is decreasing. However, I don't understand why $\frac {dh}{dt}$ is non-linear. Why can't it be linear?

EDIT: I am NOT asking whether or not the height function is linear. Many are telling me that the derivative of height is not a constant so thus the height function is not linear but this is not what I am asking. This is my mistake because I had used $h(x)$ originally to denote the derivative of height which is what my book used. Rather I am asking if $\frac {dh}{dt}$ is linear or not and why. What would be nice if someone could better explain what my book is telling me: "At every instant the portion of the cone containing water is similar to the entire cone; the volume is proportional to the cube of the depth of the water. The rate of change of depth (the derivative) is therefore not linear."

from Hot Weekly Questions - Mathematics Stack Exchange

Given an equation: $$ x^5 - 3x = 1 $$ Show that:

• It has at least $1$ root on $(1, 2)$;
• It has at least $3$ roots on $\Bbb R$

I've started with considering a function $f(x)$ for $x\in [1, 2]$: $$ f(x) = x^5 - 3x - 1 $$

Then calculating its value on the left and right sides of the closed interval yields: $$ f(1) = -3\\ f(2) = 25 $$

Applying the Intermediate Value Theorem yields that there exists a point for $x_0 \in [1, 2]$ such that $f(x_0) = 0$. Which means that indeed at least one root exists.

However, for the second part of the question if we consider $f(x)$ for $x \in \Bbb R$, the only way I see is to try and guess the intervals where the function changes its sign and then apply IVT again. Consider for example $f(x)$ for $x \in \{-2, -1, 1, 2\}$.

I see how derivatives could be to the rescue here, the problem is that I'm not allowed to use derivatives.

Is there a rigorous way to prove what's stated without guessing and without using derivatives? Thank you!

from Hot Weekly Questions - Mathematics Stack Exchange

Yuri Latushkin, a Professor of Mathematics at the University of Missouri (link here) is visiting the department this week (24-28 June).  His hosts for the visit are Gianne Derks and Claudia Wulff, and the visit is funded by the Research in Pairs programme of the London Mathematical Society. Yuri has wide ranging research interests in dynamical systems, nonlinear PDEs, and fluids.  He gave a talk to the department on Wednesday on the Maslov index (announcement here). The photo left shows Gianne and Yuri in discussion after the seminar, and the photo below shows Yuri during the seminar.

from Surrey Mathematics Research Blog https://ift.tt/2XdPdZB

     
This series gives examples of some of the unusual things that can happen in math. Some examples are an infinite field that cannot be ordered, a bounded function having no relative extrema, a function only continuous at irrational points, and a sequence of functions with derivatives that diverge everywhere.


from The Center of Math Blog https://ift.tt/2FCaroZ

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from The Center of Math Blog https://ift.tt/2YeITxv