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# Math Overflow Recent Questions

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### Published reference on the automorphism group of modular curves $X_1(N)$?

Tue, 09/25/2018 - 15:45

I wish to cite that the automorphism groups of $X_1(N)$ have already been completely calculated, and what they are, but I am having difficulty finding this calculation in the literature.

I have found in multiple [22] papers [29] on related topics a reference to "F. Momose, Automorphism groups of the modular curves X1(N). Preprint."

I cannot find it. I am wondering if this paper ever made it to publication, or where a copy of it might be? If not, is there a published reference on the topic?

### Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds?

Tue, 09/25/2018 - 15:38

Galaz-Garcia and Guijarro proved the geometrization of closed (compact, boundaryless) Alexandrov 3-spaces. Part of the strategy was to use the so-called ramified double cover $\tilde{X}$ of the space $X$. This ramified cover is a smooth $3$-manifold. Being this the case, the space $X$ would be isometric to a Riemannian $3$-orbifold.

I don't quite follow why then, it's not immediate that the geometrization of $X$ follows from the geometrization of $3$-orbifolds?

### Traveling at a rate of miles per hour [on hold]

Tue, 09/25/2018 - 15:16

Jenny is traveling at a rate that allows him to go 34 miles in 39 minutes. What is her average speed, in miles per hour?

### When does an optimization problem with symmetric constraints has symmertic optimal solution?

Tue, 09/25/2018 - 15:12

Consider the optimization problem \begin{array} \ \min\text{ / } \max &f(x_1,\cdots,x_n) \\ \text{subj. }& g(x_1,\cdots,x_n)=\text{constant} ,\end{array} where $g$ is a symmetric function, i.e. $g(x_1,\cdots,x_n)=g(\sigma(x_1,\cdots,x_n))$ where $\sigma\in\text{Symm}\{x_1,\cdots,x_n\}$. There is a principle (sometimes called Purkiss principle see here) which states, under some mild assumptions, that if $f$ is also symmetric, then the optimal solution is of form $x_1=\cdots=x_n$(see, also other post).

Q. Are there other versions of Purkiss principle that address aforementioned problem, meaning not necessarily symm. objective function but symmetric constraints?

### Hodge decomposition of the symmetric product of a curve

Tue, 09/25/2018 - 15:08

Let X be a smooth projective connected curve over $\mathbb{C}$ and let $n>1$ be an integer. Let $Y= Sym^n_X$ be the $n$-th symmetric product of $X$.

Is there, for every $i$, a nice formula for the Hodge decomposition of $H^i(Y,\mathbb{C})$?

If not, what part of the Hodge diamond can be described easily?

### Derived equivalences and the Coxeter polynomial

Tue, 09/25/2018 - 15:07

Let $A$ be a quiver algebra with an acyclic quiver and primitive idempotents $e_i$. The Cartan matrix $C_A$ of $A$ is defined as the matrix with entries $dim(e_i A e_j)$ and the Coxeter matrix $\phi_A$ of $A$ is defined as $\phi_A=-C_A^{-1} C_A^T$. The Coxeter polynomial of $A$ is defined as characteristic polynomial of $\phi_A$. The Coxeter polynomial is a derived invariant and thus derived equivalent algebras share the same Coxeter polynomial.

Is the following true:

A is derived equivalent to a path algebra $KQ$ of Dynkin type if and only if it has the same Coxeter polynomial as $KQ$?

In case the answer is no, is this true in case $A$ is additionally a Nakayama algebra with a linear quiver?

I think this is at least true when $A$ is a Nakayama algebra with a linear quiver (corresponding to a Dyck path) and there is computational way using trivial extensions and representation-finiteness of those trivial extensions to test it, but it gets very ugly when $Q$ is of type $D_n$ ($E_n$ can be done with the computer and indeed, for $E_6$ it is true. It is also true for Dynkin type $A_n$ and $n \leq 8$ and Dynkin type $D_n$ for $n \leq 6$).

### Is continuity preserved under norm operations

Tue, 09/25/2018 - 14:56

Let $F$ be a continuously differenable function over $\mathbb{R}$. Let $\Omega$ be a bounded subset of $\mathbb{R}^2$. Assume that for every $w\in L^2(\Omega)$ then $v(x)=F(w(x))$, $x\in \Omega$, is also in $L^2(\Omega)$. Can we say that $$\|F(w_1(\cdot)+\eta w_2(\cdot))\|_{L^2(\Omega)}$$ is continuous in $\eta$?

### Generating Function for x+x^2+x^4+x^8+

Tue, 09/25/2018 - 14:35

I am trying to find a closed form solution for the generating function $x+x^2+x^4+x^8+...=\sum_{i=1}^\infty x^{2^i}$. I began with the fact that the positive integers can be neatly partitioned into sets $2^i*(2\mathbb{Z}^+-1)$ for i from 0 to /infty along with some other summation manipulation to get from $$x+x+x^2+x^3+...=\sum_{i=0}^\infty\sum_{j=0}^\infty(x^{2^i*(2j+1)})$$ To $$x+x^2+x^4+x^8+...=\frac{x}{1-x}+\sum_{i=0}^\infty\frac{x^{3*2^i}}{x^{2*2^i}-1}$$ But I am unsure of how to proceed from there. To clarify, a closed form solution would be a finite summation or product.

### Classes of curves with "determinant-like operation"

Tue, 09/25/2018 - 14:11

Consider a motivating example:

Let $E\in \mathbb{Q}[y][x]$ be of degree $n=2$ (in $x$) and separable when viewed as a member of $\mathbb{Q}[x,y]$. Therefore we can calculate it's roots in $\mathbb{Q}[y][\sqrt[n]{D_E(y)}]$[1] where $D_E(y)\in\mathbb{Q}[y]$ is the discriminant of $E$

If we look for rational solutions $(x,y)$ for $E$ we must have $D_E(y)=z^n$ for some rational $z$ and therefore we can see that every such solution gives rise to a rational point $(y,z)$ on the curve $E': z^n=D_E(y)$.

Now this is not a bijection on the rational points of the curves, so we might as well make it one:

We can express $E'$ as a function of $y$ and $x(y,z)$ only (due to the fact that $D_E$ is a symmetric function of the roots) and writing $w=x(y,z)$ we can get a new curve $E''(y,w)$ with the same rational points as $E$.

[1] We can do all of the above in degrees $3, 4$ as well, only we would have to repeat the construction a number of times. We only need the solvaibilty of $S_n$ for $n\leq 4$ for the construction.

Example:

Let $E$ denote $x^2+3xy^2+2y^4+4$

$\Rightarrow D(y)=y^4-16$

$E(F)=\{(\frac{-3y^2+\sqrt{D(y)}}{2}, y) \cup (\frac{-3y^2-\sqrt{D(y)}}{2}, y): y\in F\}$ for every field $F/\mathbb{Q}(y)(\sqrt{D(y)})$

Now $(x,y)\in \mathbb{Q}^2 \iff (y, \sqrt{D(y)})\in\mathbb{Q}^2$

And we can now write $z^2 = y^4 - 16$ and we want different $z$s that give rise to the same $x$ to yield the same solution. We can see that if $x(y,z) = x(y, z')$ then we must have either $z=z'$ or $z'=-z$ and therefore we care about solutions to $z^2-y^4+16$ up to the sign of $z$ which we know how to solve (this is a pet example as we could solve both equations without doing this either way).

Now I want this this to lift this operation to an endomorphism on some subspace of curves with $E \mapsto E''$ (that preserves all rational points), the problem is that we seem to not have much control on the degree of $E''$ (in fact it seems to increase so this surely can't work in the present form).

All the above construction is just repeated application of basic Galois theory on the ground field $\mathbb{Q}(x_1,...,x_n)$

Question: Can this be replicated in some other ground field to make the following a well defined operation on some subspace of curves (and still preserve points in some "field of definition", or at least say these points lie over each other, if this field of definition is not the same between $E$ and $E''$)?

This comes from playing around with some examples, and I might be missing something basic that would not make this possible, so the answer to the question is probably "only trivial classes", but maybe there is interesting theory behind this.

I think this can only apply in some generalization of "theory of solvability by radicals" but am not sure what nature should this have.

### Tail probability of random projection

Tue, 09/25/2018 - 13:42

Suppose $v\in R^n$ is a constant unit vector. $P_l$ is a random projection matrix to an $l$ dimensional subspace of $R^n$ which is uniformly sampled from $G(l,R^n)$ which is the collection of all $l$-dimensional subspace in $R^n$. What is the upper bound of the following: $$\mathbb{P}(P_lv\leq\delta)$$ What is the order of the above probability as $\delta\to0$?

Also, when $l=n$, the above probability is the indicator function $1_{\{\delta\geq1\}}$. Can we derive an upper bound of the above probability for $l<n$ such that when $l\to n$, the upper bound tends to $1_{\{\delta\geq1\}}$?

### I am wondering what the formula (x - y) / (x + y) is commonly used for [on hold]

Tue, 09/25/2018 - 12:55

The simple formula (x - y) / (x + y) has me scratching my head, where is it used?

I don't remember seeing it during my engineering studies, it looks a bit familiar but not sure if I should know it or if it is a bit of a rare thing.

I found mention of it in a specification of a Russian analogue computational device. Many such devices might have been used for military or space programs to calculate trajectories or something that needed fast solutions to targeting systems.

It is from the family of vintage CRT storage devices and has the model designation "ЛФ2" but not very much information is to be found on the internet.

These devices were variously used to do scan conversions (between TV standards), static pattern generation, polar to Cartesian conversions (for radar), static echo background cancellation (for radar), image integration (for vector displays, film printers and scanners). The other more popular use was for digital storage memory for the earlier computers.

So I am wondering if this tube would have some interesting modern application or is it just something that a microcontroller will work out in a microsecond these days.

My first question on MO and I cannot think of any more tags so please add some if you can figure something appropriate.

### Probability Theory [on hold]

Tue, 09/25/2018 - 12:37

A private facebook group can only be joined if someone from the group invites you. Originally, the group has two members : Adam and Eve. Occasionally a random person from the group invites a new member. Someone is part of Adam's clique if they are Adam themselves or if someone from Adam's clique invited them.

a) What is the probability of Adam's clique having k members when the Facebook grouo has 4 members? ( k= 1,2,3)

b) Let X be the number of people in Eve's cliquewhen Facebook has n memberd. Give the distribution of x

I need help, for a) is it okay if i say for $K = 1$ probability is 1/4 For k=2 then P=1/2 and When k=3 then P= 1/4

Then for part b) finding the distribution, I dont know how to solve that, How can i use mathematical induction to solve it

### Almost simultaneous Dirichlet's theorem on arithmetic progressions

Tue, 09/25/2018 - 11:24

Denote the set of prime numbers by $P$. Let $u,v,m,n \in \mathbb{N}-\{0\}$ satisfy: $m \leq n$, $\gcd(u,m)=1$ and $\gcd(v,n)=1$.

Is it possible to find $L \in \mathbb{N}$, such that $u+Lm \in P$ and $v+Ln \in P$? (different primes, probably).

Of course, by Dirichlet's theorem on arithmetic progressions, we can find $L$ such that one of $\{u+Lm,v+Ln\}$ is prime. But are those $L$'s 'dense enough' to guarantee that both $\{u+Lm,v+Ln\}$ are primes?

A similar, hopefully a less difficult question, is as follows:

Is it possible to find $L \in \mathbb{N}$, such that:

(i) $u+Lm < v+Ln$;

(ii) $u+Lm = p \in P$;

(iii) $p$ does not divide $v+Ln$?

From the answer to this question, it is clear that there exists $L \in \mathbb{N}$ such that: (i) $u+Lm < v+Ln$ and (ii) $u+Lm = p \in P$. The problem is how to guarantee that $p$ will not divide $v+Ln$?

Thank you very much!

A remark about the answer: Let us concentrate on the special case mentioned in the answer: $m=n=1$, $u=1$, $v=3$, so $A:=1+L$ and $B:=3+L$. There is a major difference between asking: 'Are there infinitely many $L$'s such that $A,B \in P$' (very very difficult question) and 'Does there exist $L$ such that $A,B \in P$', which is a very easy question that has a positive answer, for example $L=2$ yields $(A,B)=(3,5) \in P^2$. Therefore, I expect that my (first) question has a positive answer, or maybe I am missing something, and even finding only one such $L$ is a difficult task? (I do not require that $A$ and $B$ will be greater than a given number).

### A "positive diagonal plus skew-symmetric" matrix decomposition

Tue, 09/25/2018 - 10:32

Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and strictly positive eigenvalues (note that $A$ is not required to be symmetric).

My question. Do there exist an orthogonal matrix $T\in\mathbb{R}^{n\times n}$ and a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ such that $$TAPT^\top = D+S,$$ where $D\in\mathbb{R}^{n\times n}$ is a diagonal matrix with positive diagonal entries and $S\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix?

Of course, if the diagonal entries of $D$ are not required to be positive then the answer is in the affirmative (see, e.g., this related question).

### Chebyshev interpolation

Tue, 09/25/2018 - 08:00

Let's define the n-th degree Chebyshev polynomials by

$$T_{n} (x)=\cos(n\arccos(x)).$$

Find a polynomial $P$ such that

$$\mid y- P (x) \mid$$

is minimal, using the first three Chebyshev polynomials for the following data:

$$\begin{bmatrix} x & -1 & -0.5 & 0 & 0.5 & 1 \\ y & 0.6346 & 0.6565 & 1 & 1.5230 & 1.5756 \end{bmatrix}.$$

How could we manage to approach such a problem?

### Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\sin x$

Tue, 09/25/2018 - 02:29

This question is actually from MSE. I had to post it here due to the lack of response there even after placing a bounty. Here goes the question

Let tangents be drawn to the curve $y=\sin x$ from the origin. Let the points of contact of these tangents with the curve be $(x_k,y_k)$ where $x_k\gt 0 ;(k\ge 1)$ such that $x_k\in (\pi k, (k+1)\pi)$ and $$a_k=\sqrt {x_k^2+y_k^2}$$ (Which is basically the distance between the corresponding point of contact and the origin i.e. the length of tangent from origin) .

I wanted to know the value of

$$\sum_{k=1}^{\infty} \frac {1}{a_k ^2}$$

Now this question has just popped out in my brain and is not copied from any assignment or any book so I don't know whether it will finally reach a conclusion or not.

I tried writing the equation of tangent to this curve from origin and then finding the points of contact but did not get a proper result which just that the $x$ coordinates of the points of contact will be the positive solutions of the equation $\tan x=x$

On searching internet for sometime about the solutions of $\tan x=x$ I got two important properties of this equation. If $(\lambda _n)_{n\in N}$ denote the roots of this equation then

$$1)\sum_n^{\infty} \lambda _n \to \infty$$ $$2)\sum_n^{\infty} \frac {1}{\lambda _n^2} =\frac {1}{10}$$

But were not of much help.

I also tried writing the points in polar coordinates to see if that could be of some help but I still failed miserably.

On trying a bit more using some coordinate geometry I found that the locus of the points of contact is $$x^2-y^2=x^2y^2$$

Hence for third sum we just need to find $$\sum_{k=1}^{\infty} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{2\lambda _k ^2} +\sum_{k=1}^{\infty} \frac {1}{2(\lambda _k ^2 +2)} =\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2}$$

Now for the second summation I did think about it to form a series but for the roots to be $\lambda _k^2 +2$ we just need to substitute $x\to \sqrt {x−2}$ in power series of $\frac {\sin x-x\cos x}{x^3}$ and then get the result but it was still a lot confusing for me.

Using $x\to\sqrt {x-2}$ in the above power series and using Wolfram Alpha I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second summation equal to $$\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}?$$

Is this value correct or did I do it wrong?

I would also like to know if there is some other method to solve this problem

### Derived equivalences of Dyck paths

Tue, 09/25/2018 - 01:12

Call two Dyck paths $D_1$ and $D_2$ derived equivalent in case their corresponding Nakayama algebras are derived equivalent (The Dyck path of a Nakayama algebra with a linear quiver is just the top boundary of its Auslander-Reiten quiver, so we can identify Nakayama algebras with a linear quiver with Dyck paths). Derived categories of Nakayama algebras appeared in the literature (see for example https://www.sciencedirect.com/science/article/pii/S0001870813000182 ) but it seems to be wide open to classify Dyck paths with respect to derived equivalences (or?).

Call a Nakayama algebra (or the corresponding Dyck path) bouncing in case the Kupisch series is of the form $[a_1+1,a_1,...,3,2,a_2+1,a_2,...,2,...,a_r+1,a_r,...,3,2,1]$.

The truth of the conjecture in What are the periodic Dyck paths? would imply the following: In case a Dyck path $D_1$ is derived equivalent to a bouncing Dyck path $D_2$, then also $D_1$ is bouncing. Thus, it would give that being bouncing is a derived invariant.

Question 1:

Do we know whether it is true that being bouncing is a derived invariant? That is: In case a Nakayama algebra $D_1$ is derived equivalent to a bouncing Nakayama algebra $D_2$, then also $D_1$ is bouncing?

Question 2:

Is it known which bouncing Nakayama algebras with a fixed number of simple modules are derived equivalent to eachother? How many equivalence classes (up to derived equivalence) exist of bouncing Nakayama algebras?

Question 3:

Can a quiver algebra with acyclic quiver be derived equivalent to a quiver algebra with non-acyclic quiver? Can a Nakayama algebra with linear quiver be derived equivalent to a Nakayama algebra with a non-linear quiver?

edit: In case the conjecture of What are the periodic Dyck paths? is true and question 1 of this thread together with the guess that there might be only one equivalence class (answering question 2), one would have the following nice equivalences for a Dyck path $D$ of length $n$:

1.$D$ is bouncing (combinatorial characterisation).

2.$D$ is iterated tilted of Dynkin type $\mathcal{A}$ (homological characterisation).

3.The Coxeter matrix of $D$ has period $n+1$ (linear algebraic characterisation).

4.The trivial extension of $D$ is a Brauer tree algebra (representation-theoretic characterisation).

5.The coxeter polynomial is equal to $\sum\limits_{k=0}^{n}{x^k}$ (polynomial characterisation).

The equivalence of 2. and 4. is well known by a result of Happel .

Question 1 and 2 had an easy positive answer using classification results from the literature. But maybe there is a more direct proof? Question 3 remains open.

### The growth of class number in $\mathbb{Z}_p$-extensions of function fields

Mon, 09/24/2018 - 22:13

Let $X$ be a curve (proper, smooth, ...) over a finite field $\mathbb F_q$ where $q$. Suppose also that $\mathbb F_q$ contains the $p$-th roots of unity, in this case we have the following (unique) chain of field extensions: $$\mathbb F_q \subset \mathbb F_{q^p} \subset \dots \subset \mathbb F_q^{(p)}$$ where each step in the extension is degree $p$ and the union of all these extensions is defined to be $\mathbb F_q^{(p)}$.

I am interested in the $\ell$ (possibly equal to $p$) torsion of the degree $0$ Picard group of $X_{\mathbb F_q^{(p)}}$. If we consider the algebraic closure $X_{\overline{\mathbb F_q}}$, then the $\ell$ torsion here looks like $P = (\mathbb Q_\ell/\mathbb Z_\ell)^{2\lambda}$ for $\lambda \leq g(X)$, the genus.

One strategy to get information about the Picard group of $X_{\mathbb F_q^{(p)}}$ would be to consider the Galois group $\Gamma = \mathrm{Gal}(\overline{\mathbb F_q}/\mathbb F_q^{(p)}) \cong \prod_{p' \neq \ell}\mathbb Z_{p'}$ and it's action on $P$.

Can this strategy prove the following:

1) If $\ell = p$, then the $\ell$-torsion in the Picard group over $\mathbb F_q^{(p)}$ is of the form $(\mathbb Q_\ell/\mathbb Z_\ell)^r$ for $r \leq 2\lambda$.

2) If $\ell \neq p$, then the $\ell$-torsion in the Picard group over $\mathbb F_q^{(p)}$ is finite.

Ideally, I would like to avoid any knowledge of the zeta function of curves.

### Dualizable objects in homotopy category of chain complexes

Mon, 09/24/2018 - 19:53

The proposition 1.9 from "Duality, Trace and Transfer" by Dold and Puppe states that:

Given a commutative ring $R$, a chain complex of $R$-modules is strongly dualizable in $Ho(Ch(R))$, the homotopy category of chain complexes, iff it has the same homotopy type of a strongly dualizable object in $Ch(R)$.

However, they omit the proof (an I can't find the references cited there). I was wondering how can one prove such a thing.

### Alternative (easier) Proof of Ambrose Singer Holonomy theorem

Mon, 09/24/2018 - 11:48

Let $P(M,G)$ be a principal bundle. Giving a connection on $P(M,G)$ means two equivalent things. One as an assignment of subspace of $T_pP$ for each $p\in P$ and another as a $\mathfrak{g}$ valued $1$ form.

• Given connection as a $\mathfrak{g}$ valued $1$ form $\omega$ on $P(M,G)$ we have associated a $2$ form (differential of $\omega$ and modifying it little bit) which we called the curvature form for the connection, denoted by $\Omega$.

• Given connection as a distribution $p\mapsto H_pP\subseteq T_pP$, we know what it means to say a curve is horizontal curve. We write $v\sim u$ to mean that $v$ and $u$ are related by horizontal curve. Given $u\in P$ we have defined what is called a Holonomy bundle $P(u)=\{v\in P:v\sim u\}$ based at $u$ and what is called a Holonomy group $\Phi(u)=\{a\in G:u\sim ua\}$ based at $u$.

Given $p\in P$ we have $\Omega(p)(X(p),Y(p))\in \mathfrak{g}$ for all $X(p),Y(p)\in T_pP$.

As $\Phi(u)$ is a Lie subgroup of $G$, we have, Lie algebra of $\Phi(u)$ to be subalgebra of $\mathfrak{g}$.

So, from curvature form, we get a collection that is a subset of $\mathfrak{g}$ and from holonomy group/holonomy bundle we get a collection that is a subset of $\mathfrak{g}$ and considering suitable restrictions, it is natural to expect these two sets to be same. Ambrose-Singer Holonomy theorem what is the precise relation between these. The set $\Omega(p)(X(p),Y(p))$ where $p\in P(u)$ and $X(p),Y(p)\in H_pP$ generated the Lie algebra of $\Phi(u)$.

More precisely, we have the following.

Let $P(M,G)$ be a Principal bundle where $M$ is connected and paracompact. Let $\Gamma$ be a connection in $P$, $\Omega$ the curvature form, $\Phi(u)$ the holonomy group with reference point $u\in P$ and $P(u)$ the holonomy bundle through $u$ of $\Gamma$. Then the Lie algebra of holonomy group $\Phi(u)$ is equal to the subspace of $\mathfrak{g}$, Lie algebra of $G$, generated by all elements of the form $\Omega_v(X,Y)$ where $v\in P(u)$ and $X$ and $Y$ are arbitrary horizontal vectors at $v$.

This is very natural thing one can expect but the proof is not that simple (I am not saying I could have guessed the statement, I am saying that if some one show me this statement and ask me to guess if that is true or not I would have said that it is mostly true). The proof given in Kobayashi involves some thing called involutive distribution. I want to know if there is any alternative proof for this result.

We have to prove $\mathfrak{g}'$ which is Lie algebra of $\Phi(u)$ is spanned by elements of the form $\Omega(p)(X(p),Y(p))$ where $p\in P(u)$.

Let $v\in \mathfrak{g}'$ i.e., there exists a smooth curve $\alpha:[-1,1]\rightarrow \Phi(u)$ such that $\alpha(0)=e$ and $\alpha'(0)=v$. Then, I want to prove $v=\sum a_i \Omega(p_i)(X(p_i),Y(p_i))$. So, I have to look for some elements $p_i\in P(u)$ and then some horizontal vectors $X(p_i),Y(p_i)$. It is not clear what choice of $p_i$ should one make. I am not very sure if this approach works or not. Any suggestion on making this approach work are welcome. Any other alternative proof is also welcome. An exposition of proof given in Kobayashi and Nomizu is also welcome.


text:
# Is such a crypto-system available?

$E(k_1, pt) = c_1, E(k_2, c_1) = c_2, D(k_{new}, c_2) = pt$, where $k_{new} = f(k_1, k_2).$ Sharing $k_{new}$ and $k_2$ should reveal no information about $k_1$.

Clarifications:

1. Being able to decrypt the ciphertext $c_2$ knowing only $k_{new}$ is the first desired property of the system
2. One cannot derive $k_1$ knowing both $k_{new}$ and $k_2$.
3. One cannot easily derive $k_1$ knowing other pairs $k_2'$ and $k_{new}'$. See @poncho's comments.

$D(k_{new}, c_2) = pt$ or $E(k_{new}, c_2) = pt$ doesn't matter as long as you recover the plaintext.

Real world example to clarify things even more:

1. Alice has a very large database which it chooses to store at Eve's site. Because Alice doesn't want Eve to read the data, nor doesn't she know in advance which data will be shared with whom, she encrypts all the records with a single key ($k_1$).
2. Now Bob requests access to some specific information. Bob and Alice know each other so Alice gives him access to that subset. However Alice wants to do that efficiently. Also Alice doesn't want Bob to be able to read anything else except the shared subset. Therefore she cannot: a) Decrypt the shared subset at Eve's site and then re-encrypt with a different key (this would expose all the database contents to Eve), b) Retrieve the data from Eve, decrypt, re-encrypt and store back data at Eve's site - this implies a round-trip delay which is prohibitively expensive.

Thus Alice needs a cryptosystem which would allow re-encrypting the data directly at Eve's site, however Eve shouldn't be able to read any plain text (nor the one shared with Bob, nor the rest of the database records). Sharing the decryption key ($k_{new}$) is done through a direct channel between Alice and Bob (e.g. using Diffie-Hellman).

The "cryptosystem" stated in the question is just a way I saw that happening. Note that $k_1$ shouldn't become known to Bob (e.g. if Bob colludes with Eve or other people with whom Alice shares (possibly different sets of) data).

-
Knowledge of $k_{new}$ and $k_2$ will allow the decryption of any plaintext encrypted by $k_1$; what further information about $k_1$ do you want to be certain isn't revealed? – poncho Feb 5 '13 at 15:35
The value of k1 – Eugen Feb 5 '13 at 20:28
Typically, keys don't have any intrinsic value; the only reason we would prefer that people don't learn them is because of the ability that those keys would confer (for example, being able to decrypt traffic). Revealing knew and k2 allows people to decrypt; what further ability would you prefer people not to have? – poncho Feb 5 '13 at 20:33
@jug Searching for "DES is not a group" I gradually found proxy re-encryption which seems like something I am looking for. – Eugen Feb 6 '13 at 9:26
It doesn't matter for me as long as it satisfies the properties specified in the question. The originator of the data has key k1, Alice has key k_new. Even if Alice finds k2 or any other pairs (k2', k_new') she shouldn't be able to compute k1. – Eugen Feb 6 '13 at 12:13

This can be done with RSA where k1 and k2 are encryption keys and f is multiplication modulo the totient of the public modulus. The last property, that knowledge of other pairs $k_2'$ and $k_{new}'$ does not reveal $k_1$ is satisfied by using a different modulus for each pair generated.

-
You mean that f is the inverse of the multiplication of k1 and k2. In any case, this has a subtle problem; if someone learns k2, f(k1, k2), k3, f(k1, k3), they can deduce k1. Does this matter? Well, we can't tell from the question. – poncho Feb 5 '13 at 16:12
Doesn't work. 1) k_new would be an encryption key, not a decryption key 2) f is invertable. So you can calculate k1 = k_new/k2. – CodesInChaos Feb 5 '13 at 16:13
@CodesInChaos 1)No. My definition of f is correct because the post specifies D(k_new,c2)=pt not E(k_new,c2)=pt. Presumably the Decryption function magically obtains the decryption exponent from k_new! 2)Give me an example of inverting f which is not equivalent to factoring the modulus and hence breaking RSA. – Barack Obama Feb 5 '13 at 16:20
@poncho I guess I could have specified that D=E and then f is the inverse as you said. – Barack Obama Feb 5 '13 at 16:25
Checking my understanding of the suggested solution: $pt^{k_1} = c_1, c_1^{k_2} = c_2, k_{new} = (k_1 * k_2)^{-1} mod(\phi(N))$. $pt = c_2^{k_{new}} = pt^{k_1 * k_2 *(k_1 * k_2)^{-1}}$ Is that right? – Eugen Feb 5 '13 at 21:02

Actually, I believe that there is a solution, but it follows a different strategy than your proposed approach. The problem with your approach is that if Eve (with $k_{new}$) and Bob (with $k_2$) collude, there's nothing stopping them from reading the entire text (and not just the part that Alice wants Bob to read).

Instead, consider that Alice divides up the plaintext into a series of blocks; she also selects a master key (H1), and creates a tree of keys as:

               H1
/-----/ \------\
/                \
H2                  H3
/    \              /   \
H4     H5           H6    H7
/ \    /  \         / \   /  \
H8 H9  H10 H11    H12 H13 H14 H15

For each internal node $H_i$, you would use some hash function to generate the two child nodes $H_{2i}$ and $H_{2i+1}$:

$H_{2i} = F(H_i)$

$H_{2i+1} = G(H_i)$

Where both $F$ and $G$ are noninvertible functions.

You would use the values in the leaf nodes to encrypt the successive blocks (so in the example, the first block would be encrypted using H8, the second block would be encrypted using H9, etc.

If the entire database consists of $N$ blocks, this would take a depth $\log_2(N)$ tree.

Now, when Alice decided to reveal a portion of the plaintext to Bob, she would just reveal the internal nodes that would allow Bob to recompute the keys used to encrypt those sections. For example, if Alice wanted to reveal the sections protected by H10 through H14, all she would need to reveal would be H5 (which allows Bob to recompute H10 and H11), H6 (which allows Bob to recompute H12 and H13) and H14. In general, to reveal a section of $M$ consecutive blocks, Alice needs to reveal at most $O(\log(M))$ internal nodes.

Now, Bob only gets the data Alice has decided to reveal; and even if a group of Bobs (and Eve) collude, they can only read what Alice has revealed to one of the Bobs.

-
The question to which this is a good answer is quite far away from the initial question to which my answer was perfectly correct. Could you separate this into two questions for which yours would be the answer to one of them and mine the other? – Barack Obama Feb 6 '13 at 18:49
Yes, it's a ways from how the initial question was phrased. However, whether it's a good answer really depends on what Eugen is actually interested in. If he has an academic interest in this type of cipher (and gave his Real World example as a way it might be used), it's off topic. However, if his Real World Example is the problem he is really interested in (and he thought that a cipher as he described might solve it), well, he needs to be told that such a cipher doesn't have the security properties he wants, and here is a better method that does. – poncho Feb 6 '13 at 19:29
@poncho The solution satisfies the requirements although I would like to share my thoughts on its deficiencies (btw. Bob knows k_new and Eve knows k2). First is the storage overhead needed to store the hash tree. With the Barack Obama's solution I have to store k_new (and modulus) only for the portions I share (although public key crypto is much slower than symmetric encryption). If new data is added there is a problem again (if I shared H8 now Bob would know also H16, H17, right?). Regarding my interests: they are mixed indeed, but solving the real example has a bigger priority for me. – Eugen Feb 6 '13 at 20:04
@Eugen: as for the amount of storage needed, well, just storing H1 will allow you to recompute the tree as needed with $O(\log N)$ hash operations. If it becomes necessary to append data, well, you don't want to reuse external nodes as internal; you can extend the solution to make multiple trees (with each tree being bigger; this preserves the $O(\log N)$ scaling properties). One problem happens if Alice needs to update database; if she's revealed the key for a block, Bob can decrypt that block even after it has been updated. I'm not sure how to work around that. – poncho Feb 6 '13 at 20:30
@poncho Regarding the storage you are right: its either storing or recomputing it every time you need to retrieve/share some data. Making new trees requires recomputing not only the hashes but re-encrypting the database on each insertion :). I hope it is understood that this is just an open discussion. I am not nitpicking or something like that. – Eugen Feb 6 '13 at 20:59

format: plain


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Is it valid to perform Bayesian optimization and MCMC sampling with nested model parameters?

I am trying to think of a way to perform a somewhat complicated Bayesian optimization problem involving model parameters which are nested within other functions in the forward problem, and I am wondering if this is alid from a theoretical perspective or if I am violating one of the criterion for MCMC sampling methods.

To simplify the problem to it's most basic situation, I have some data vector, $$\mathbf{d}$$ (with error $$\mathbf{\sigma}$$ and length $$N$$) from which I want to ascertain some model, $$\mathbf{m}$$. I have a forward operator, $$F(\mathbf{m})$$ which takes a model and computes predicted data, $$\mathbf{d_p}$$. From this, I construct a log likelihood function using a chi-square misfit assuming Gaussian errors:

$$\chi^2 = (\mathbf{d}-F(\mathbf{m}))^T\mathbf{C_d}^{-1}(\mathbf{d}-F(\mathbf{m}))$$

where $$\mathbf{C_d}$$ is an $$N$$ x $$N$$ matrix with $$\mathbf{\sigma}$$ on the diagonal.

My question concerns the forward operator, $$F(\mathbf{m})$$.

In my scenario, the forward operator is complicated and contains many nested functions. For example, suppose I have 5 model parameters, $$m_1, m_2, ... m_5$$. In my situation I might have:

$$a = P(m_1)$$

$$b = Q(a, m_2, m_3)$$

$$c = R(a, b, m_4, m_5)$$

$$\mathbf{d_p} = S(a,b,c)$$

where $$P, Q, R,$$ and $$S$$ are some very complicated functions. In order to actually compute predicted data, $$\mathbf{d_p}$$, I need $$a,b$$, and $$c$$. But my model parameters of interest are in $$\mathbf{m}$$.

Is this nested-variables approach valid for sampling the posterior?

Note: I have tried implementing the problem as written using a known model and I run into various problems. First and foremost the MCMC sampler gets "stuck" in an area of low probability (even though I know that there is a better solution with higher probability since I am using a known model). This does not happen if I try to sample $$a, b$$ and $$c$$ as model parameters. So I am wondering if there is some theoretical violation of the method.

• (a) the $\chi^2$ expression is unlikely to be a likelihood function. Do you mean log- or negative log-likelihood? (b) I do not understand the distinction between $F(\mathcal m)$ and $\mathcal{d_p}$. (c) The model parameter is $\mathcal m$ and thus should be the one simulated, while $a,b,c$ are deterministic functions of $\mathcal m$. This seems unrelated with the use of MCMC or another simulation technique. Apr 16, 2021 at 16:57
• (a) Yes it should be log likelihood as you say. (b) $d_p = F(m)$ (i.e. in order to compute predicted data ($d_p$), I need to pass my model parameters through some function $F(m)$). (c) $m_1, m_2, m_3, ...$ are the unknown model parameters. As such $a, b,$ and $c$ are also unknown (and depend on $m$). I am trying to sample the posterior distribution (of e.g. $m$) given the likelihood (and some prior which at this point are just upper and lower limits on the modelled parameters). Apr 16, 2021 at 17:15

Considering a likelihood function $$\exp\{-(\mathbf d-F(\mathbf m))^2/\sigma^2\}$$ is equivalent to assume the observation is Normal. The posterior distribution is hence $$\pi(\mathbf m|\mathbf d) \propto \exp\{-(\mathbf d-F(\mathbf m))^2/\sigma^2\} \pi(\mathbf m)$$ which can be simulated by an MCMC algorithm, no matter how complex the transform $$F(\cdot)$$ is. Simulating directly and jointly the triplet $$(a,b,c)$$ is feasible, but this requires the projected prior $$\pi^\bot(a,b,c)$$ to be available in close form.


text:
is the presheaf category of a locally small category locally small? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:58:14Z http://mathoverflow.net/feeds/question/24540 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24540/is-the-presheaf-category-of-a-locally-small-category-locally-small is the presheaf category of a locally small category locally small? apk 2010-05-13T20:52:34Z 2010-05-13T22:52:15Z
e.g. is $\widehat{\mathbf{SET}}$ locally small?
 http://mathoverflow.net/questions/24540/is-the-presheaf-category-of-a-locally-small-category-locally-small/24549#24549 Answer by Todd Trimble for is the presheaf category of a locally small category locally small? Todd Trimble 2010-05-13T22:37:45Z 2010-05-13T22:37:45Z
There's a theorem due to Freyd and Street that says that if $C$ is locally small and the presheaf category $Set^{C^{op}}$ is locally small, then $C$ is equivalent to a small category. See this paper.
 http://mathoverflow.net/questions/24540/is-the-presheaf-category-of-a-locally-small-category-locally-small/24551#24551 Answer by Andrej Bauer for is the presheaf category of a locally small category locally small? Andrej Bauer 2010-05-13T22:52:15Z 2010-05-13T22:52:15Z
No, not in general. Todd gave a useful general answer and here is a concrete counter-example.

Let $\mathbf{C}$ be a category whose objects are sets and where the only morphisms are the identities. Consider the functors $F, G : \mathbf{C} \to \mathbf{Set}$ given by $F(X) = 1 = \lbrace 0 \rbrace$ and $G(X) = 2 = \lbrace 0,1\rbrace$. Because $\mathbf{C}$ is discrete, every family of maps $(\eta_X : F(X) \to G(X))_{X \in \mathbf{C}}$ is natural. Therefore, natural transformations $F \to G$ are in bijective correspondence with classes:

a natural transformation $\eta : F \to G$ corresponds to the class $\lbrace X \in \mathbf{Set} \mid \eta_X(0) = 1\rbrace$.

a class $C$ corresponds to the natural transformation $\eta : F \to G$ given by

$\eta_X(0) = 1$ if $X \in C$ and $\eta_X(0) = 0$ if $X \not\in C$.

Thus the presheaf category $\widehat{\mathbf{C}}$ is not locally small (note that $\mathbf{C}^{op} = \mathbf{C}$).

I cannot think at the moment of concrete presheaves $F$ and $G$ on $\mathbf{Set}$ for which the natural transformations $F \to G$ form a proper class. Someone please help.


text:
# Exposure with binomial responses

Last week, we’ve seen how to take into account the exposure to compute nonparametric estimators of several quantities (empirical means, and empirical variances) incorporating exposure. Let us see what can be done if we want to model a binomial response. The model here is the following: ,

• the number of claims $N_i$ on the period $[0,1]$ is unobserved
• the number of claims $Y_i$ on $[0,E_i]$ is observed (as well as $E_i$)

that can be visualize below

Consider the case where the variable of interest is not the number of claims, but simply the indicator of the occurrence of a claim. Then we wish to model the event $\{N=0\}$ versus $\{N>0\}$, interpreted as non-occurrence and occurrence. Given the fact that we can only observe $\{Y=0\}$ versus $\{Y>0\}$. Having an inclusion is not enough to derive a model. Actually, with a Poisson process model, we can get easily that

$\mathbb{P}(Y=0) = \mathbb{P}(N=0)^E$

With words, it means that the probability of not having a claim in the first six months of the year is the square root of not have a claim over a year. Which makes sense. Assume that the probability of not having a claim can be explained by some covariates, denoted $\boldsymbol{X}$, through some link function (using the GLM terminology),

$\mathbb{P}(N=0|\boldsymbol{X})=h(\boldsymbol{X}^{\text{\sffamily T}}\boldsymbol{\beta})$

Now, since we do observe $Y$ – and not $N$ – we have

$\mathbb{P}(Y=0|\boldsymbol{X},E)=h(\boldsymbol{X}^{\text{\sffamily T}}\boldsymbol{\beta})^E$

The dataset we will use is always the same

> sinistre=read.table("http://freakonometrics.free.fr/sinistreACT2040.txt",
> sinistres=sinistre[sinistre$garantie=="1RC",] > sinistres=sinistres[sinistres$cout>0,]
> T=table(sinistres$nocontrat) > T1=as.numeric(names(T)) > T2=as.numeric(T) > nombre1 = data.frame(nocontrat=T1,nbre=T2) > I = contrat$nocontrat%in%T1
> T1= contrat$nocontrat[I==FALSE] > nombre2 = data.frame(nocontrat=T1,nbre=0) > nombre=rbind(nombre1,nombre2) > sinistres = merge(contrat,nombre) > sinistres$nonsin = (sinistres$nbre==0) The first model we can consider is based on the standard logistic approach, i.e. $\mathbb{P}(Y=0|\boldsymbol{X},E)=\left(\frac{\exp(\boldsymbol{X}^{\text{\sffamily T}}\boldsymbol{\beta})}{1+\exp(\boldsymbol{X}^{\text{\sffamily T}}\boldsymbol{\beta})}\right)^E$ That’s nice, but difficult to handle with standard functions. Nevertheless, it is always possible to compute numerically the maximum likelihood estimator of $\boldymbol{\beta}$ given $(Y_i,\boldsymbol{X}_i,E_i)$. > Y=sinistres$nonsin
> X=cbind(1,sinistres$ageconducteur) > E=sinistres$exposition
> logL = function(beta){
+ 	pi=(exp(X%*%beta)/(1+exp(X%*%beta)))^E
+ 	-sum(log(dbinom(Y,size=1,prob=pi)))
+ }
> optim(fn=logL,par=c(-0.0001,-.001),
+ method="BFGS")
$par [1] 2.14420560 0.01040707$value
[1] 7604.073
$counts function gradient 42 10$convergence
[1] 0
$message NULL > parametres=optim(fn=logL,par=c(-0.0001,-.001), + method="BFGS")$par

Now, let us look at alternatives, based on standard regression models. For instance a binomial-log model. Because the exposure appears as a power of the annual probability, everything would be fine if $h$ was the exponential function (or $h^{-1}$ was the log link function), since

$\mathbb{P}(Y=0|\boldsymbol{X},E)=\exp(E+\boldsymbol{X}^{\text{\sffamily T}}\boldsymbol{\beta})$

Now, if we try to code it, it starts quickly to be problematic,

> reg=glm(nonsin~ageconducteur+offset(exposition),
Error: no valid set of coefficients has been found: please supply starting values

I tried (almost) everything I could, but I could not get rid of that error message,

> startglm=c(0,-.001)
> names(startglm)=c("(Intercept)","ageconducteur")
> etaglm=rep(-.01,nrow(sinistresI))
> etaglm[sinistresI$nonsin==0]=-10 > muglm=exp(etaglm) > reg=glm(nonsin~ageconducteur+offset(exposition), + data=sinistresI,family=binomial(link="log"), + control = glm.control(epsilon=1e-5,trace=TRUE,maxit=50), + start=startglm, + etastart=etaglm,mustart=muglm) Deviance = NaN Iterations - 1 Error: no valid set of coefficients has been found: please supply starting values So I decided to give up. Almost. Actually, the problem comes from the fact that $\mathbb{P}(Y=0)$ is closed to 1. I guess everything would be nicer if we could work with probability close to 0. Which is possible, since $\mathbb{P}(Y>0)=1-\mathbb{P}(Y=0) = 1-[1-\mathbb{P}(N>0)]^E$ where $\mathbb{P}(N>0)$ is close to 0. So we can use Taylor’s expansion, $\mathbb{P}(Y>0)\sim1-1+E\cdot \mathbb{P}(N>0)]=E\cdot \mathbb{P}(N>0)]$ Here, the exposure does no longer appears as a power of the probability, but appears multiplicatively. Of course, there are higher order terms. But let us forget them (so far). If – one more time – we consider a log link function, then we can incorporate the exposure, or to be more specific, the logarithm of the exposure. > regopp=glm((1-nonsin)~ageconducteur+offset(log(exposition)), + data=sinistresI,family=binomial(link="log")) which now works perfectly. Now, to see a final model, perhaps we should get back to our Poisson regression model since we do have a model for the probability that $\mathbb{P}(Y=\cdot)$. > regpois=glm(nbre~ageconducteur+offset(log(exposition)), + data=sinistres,family=poisson(link="log")) We can now compare those three models. Perhaps, we should also include the prediction without any explanatory variable. For the second model (actually, it does run without any explanatory variable), we run > regreff=glm((1-nonsin)~1+offset(log(exposition)), + data=sinistres,family=binomial(link="log")) so that the prediction is here > exp(coefficients(regreff)) (Intercept) 0.06776376 This value is comparable with the logistic regression, > logL2 = function(beta){ + pi=(exp(beta)/(1+exp(beta)))^E + -sum(log(dbinom(Y,size=1,prob=pi)))} > param=optim(fn=logL2,par=.01,method="BFGS")$par
> 1-exp(param)/(1+exp(param))
[1] 0.06747777

But is quite different from the Poisson model,

> exp(coefficients(glm(nbre~1+offset(log(exposition)),
(Intercept)
0.07279295

Let us produce a graph, to compare those models,

> age=18:100
> yml1=exp(parametres[1]+parametres[2]*age)/(1+exp(parametres[1]+parametres[2]*age))
> plot(age,1-yml1,type="l",col="purple")
> yp=predict(regpois,newdata=data.frame(ageconducteur=age,
+ exposition=1),type="response")
> yp1=1-exp(-yp)
> ydl=predict(regopp,newdata=data.frame(ageconducteur=age,
+ exposition=1),type="response")
> plot(age,ydl,type="l",col="red")
> lines(age,yp1,type="l",col="blue")
> lines(age,1-yml1,type="l",col="purple")
> abline(h=exp(coefficients(regreff)),lty=2)

Observe here that the three models are quite different. Actually, with two models, it is possible to run more complex regression, e.g. with splines, to visualize the impact of the age on the probability of having – or not – a car accident. If we compare the Poisson regression (still in red) and the log-binomial model, with Taylor’s expansion, we get

The next step is to see how to incorporate the exposure in a tree. But that’s another story…

## One thought on “Exposure with binomial responses”

1. Arthur, nice post. Why not use cloglog link instead of exponential?

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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Go back to the page of Warning 3.4.0.1.

Comment #1037 by Elizabeth Goldfinch on

There is a repitition "of of" in the first sentence.

Comment #1041 by Kerodon on

Yep. Thanks!

There are also:

• 5 comment(s) on Chapter 3: Kan Complexes
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text:
Global existence without solving a differential equation

I have the following cauchy problem:

$$y(t)' = (y(t) - t^3 - 1)^3$$ $$y(0) = \alpha$$ Discuss the global existence when

$1)\,\, \alpha < 1$

$2) \,\,\alpha = 1$

I tried the following:

If you let be $F(y,t) = (y - t^3 - 1)^3$, the function $F(y,t)$ exist for every $(y,t) \in \Re \times \Re$, being continuous and then a lipschitz function. That means the equation has a unique solution given the initial conditions for every $[a,b] \subseteq \Re$ with $a < b$. I know the function isn't sublinear, but this is a sufficent condition. If I would be able of solve the problem I also would find the domain of the solution, but I can't solve it.

I had attemp making $z = y -t^3 - 1$, so I have

$$z(t)' = z^3 - 3t^2$$ $$y(0) = \alpha - 1$$

but I can't solve neither this. That's why I believe that I need to do this without solving it.

• $F$ is not globally Lipschitz with respect to $y$. Although being merely locally Lipschitz with respect to $y$ and continuous with respect to $t$ does guarantee a unique local solution, you can still have blowup in finite time if you do not have a global Lipschitz property. For example, $y'=y^2,y(0)=1$ has a solution on $(-1,1)$ but blows up at $1$. Looking at your equation involving $z$, it should be clear that $z$ is a decreasing function if $z(0) \leq 0$, and so by comparison to my example you should expect blowup in finite time whenever $\alpha \leq 1$. – Ian Aug 2 '14 at 19:24
• That's what I mean, I can see that $y′=y2,y(0)=1$ blow up at t = 1, but I need to solve it. Neither I can't see how ( or why ) the function blow up, only for being a decreasing function. Maybe I'm forgotten a theorem? – Leonhard Aug 2 '14 at 20:18
• Suppose $z'=z^3-3t^2$, $y'=y^3$, and $z(0)=y(0)<0$. Then it should not be too hard to see that $z(t) \leq y(t)$ for all $t \geq 0$ (where both are defined). So because $y$ blows up in finite time, $z$ also blows up in finite time. – Ian Aug 2 '14 at 20:58
• Unfortunately I do not know of a nice theorem for guaranteeing blowup in finite time. – Ian Aug 2 '14 at 21:04
• Oh, that's make sense. Thanks you very much. – Leonhard Aug 2 '14 at 21:05


text:
# Prove that this finite set is a group

## Homework Statement

Let G be a nonempty finite set with an associative binary operation such that:
for all a,b,c in G
ab = ac => b = c
ba = ca => b = c
(left and right cancellation)
Prove that G is a group.

2. The attempt at a solution
Let a $\in$ G, the set <a> = {$a^k$ : k $\in$ N} is a finite closed subset of G. So, $\exists$($k_1$, $k_2$)$\in$N, such that:
$a^{k_1}=a^{k_2}$ using the cancellation property I found that $a=a.a^{k_2-k_1}$.
So,
$a^{k_2-k_1}$ is the identity but the problem is in this reasoning every cyclic subgroup will generate a different identity. And the identity is supposed to be unique. I don't know how to proceed now, any help would be appreciate it.
NAta

Related Calculus and Beyond Homework Help News on Phys.org
Deveno
suppose our set is X = {x1,x2,...,xn}

these have to be all distinct, since if:

x1xi = x1xj, by (left) cancellation we have: xi = xj

so for SOME xi, we must have x1xi = x1.

in a similar fashion (from right cancellation), we know that all products xjx1 are distinct.

so for any element in X, say xk, we know that xk = xjx1, for some xj in X.

so, if xi is a particular element of X with x1xi = x1, then for any xk in X:

xkxi = xjx1xi = xjx1 = xk

this means that xi is a right-identity for X.

now, for any xk in X, consider all products xkxj in X. again, these products are all distinct, so one of these products is xi, our right-identity.

that means every xk in X has some xj with xkxj = xi.

thus X has a right-identity and right-inverses.

now, note that xixk = (xixi)xk (since xi is a right-identity)

= xi(xixk) by associativity. so by left-cancellation:

xk = xixk, that is, xi is a left-identity as well. this means that xi is the (unique) identity for X.

now if xj is a (who knows, we might have more than one) right-inverse for xk, then:

xj = xjxi= xj(xkxj) = (xjxk)xj. but since we now know xi is also a left-identity,

xixj = (xjxk)xj, and by right-cancellation xi = xjxk, so xj is a left-inverse for xk as well.

since 2-sided inverses are necessarily unique, we have a group.

Thank you very much! Your explanation is splendid! :)


text:
# Math Help - A 42.0 kg block of ice

1. ## A 42.0 kg block of ice

A 42.0 kg block of ice slide down a plane with an incline of 34 degree assuming friction is negligible, what is the acceleration of the block down the incline?

2. Originally Posted by Refujoi
A 42.0 kg block of ice slide down a plane with an incline of 34 degree assuming friction is negligible, what is the acceleration of the block down the incline?
Set up your Free-Body Diagram. I am going to assume the incline is down and to the right. For a coordinate system I am going to say that +x is down the incline and +y is perpendicular to and coming out of the plane of the incline.

There are two forces in the FBD: the weight, w, acting straight downward, and the normal force, N, acting in the +y direction.

The only force having a component in the x direction is the weight, so:
$\sum F_x = w sin \theta = ma$
where $\theta$ is the angle of incline. (Note that it is not cosine!)

Since w = mg we may solve this equation for a and we get
$a = g sin( \theta )$.

(Note that the acceleration is less than g as it should be.)

-Dan


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# Fisher-Yates shuffle in JavaScript

This is my first time making a function just from reading about it on Wikipedia, so I'm sure there's room for a lot of improvement.

function fisherYates(str) {
let result = '';
for (let i = 0, len = str.length; i < len; i++) {
let rand = (Math.random()*str.length)|0;
result += str[rand];
str = str.slice(0, rand) + str.slice(rand+1);
}
return result;
}

• I know it is 10 years old and pertains to arrays instead of strings, but you mind find this post interesting – Sᴀᴍ Onᴇᴌᴀ Oct 17 '17 at 17:49

## Mostly just style

Why add the variable len inside the for loop. It is a constant so put it outside the loop as one.

The variable result is in function scope. Use var rather than let.

The variable rand does not change inside the loop so it should be a const.

The expression to get the random number does not need the (...) as operator precedence ensures that the | is applied correctly.

Add some spaces between operators for readability. eg (Math.random()*str.length)|0; is better as (Math.random() * str.length) | 0

### One issue

The only problem is that you have one to many iterations. The last random value will always be to select from a 1 character string, so it is not needed. You need only iterate the string length minus one.

### The rewrite

How I would rewrite your code. I personally don't like putting the let inside the for loop, so I put the declaration in the scope above. This is because for loops can have a lot of noise (clutter) and that reduces it a bit. Also the function is so short and simple variable names can be shortened without problem, I would make rand just the r.

function fisherYates(str) {
var i, result = '';
const len = str.length - 1;
for (i = 0; i < len; i ++) {
const r = Math.random() * str.length | 0;
result += str[r];
str = str.slice(0, r) + str.slice(r + 1);
}
return result + str;
}

• I'd go the other way and change rand to randomIndex, but each to his own, I guess. – Tamoghna Chowdhury Oct 18 '17 at 4:20
• I'd also declare each variable in the smallest lexical scope, using let. – Tamoghna Chowdhury Oct 18 '17 at 4:21
• @TamoghnaChowdhury Block scope has its place but without a clear need its should not be used. You use scope to protect a variable's name, but good function design (no rambling 100+ line functions) eliminates that as an issue (no risk in a 8 line function of using the same variable name by mistake). So i go on the side of efficiency, each scope declaration requires a heap assignment and additional context stack work. In nested loops this can contribute considerable overhead for no good reason but that of "protecting against incompetent programmers" – Blindman67 Oct 18 '17 at 5:04
• I didn't know that let was that inefficient in JavaScript... Anyway, your comment should IMO be a part of your answer. Especially given that almost every article about ES6+ online is preaching about how var is outdated and let should be used everywhere. – Tamoghna Chowdhury Oct 18 '17 at 8:22
• @TamoghnaChowdhury That is standard in the language since day one (1995). Implementation method is not part of the language spec, V8 uses a "context" object which carries the scope chain. The scope is mutable so I guess it is similar to Python (though I am not familiar with python's inner workings) – Blindman67 Oct 18 '17 at 15:27


text:
# Calculate the integral of $1/(2x^2+2x+1)$

I need to calculate the integral of $1 / (2x^2+2x+1)$.

I used WolframAlpha and get this answer: $$\tan^{-1}(2x+1)$$ but I don't understand how to get there.

Can you help?

-
complete the square in the denominator. –  GEdgar Mar 15 '12 at 2:22

First note that $\displaystyle \int \frac{dy}{a^2+y^2} = \frac1a \tan^{-1}\left(\frac{y}{a} \right)$. This is obtained by substituting $y = a \tan( \theta)$ in the integrand and integrating.

For your problem, $I = \displaystyle \int \frac{dx}{2x^2 + 2x+1}$. First complete the square in the denominator i.e. rewrite $2x^2 + 2x + 1$ as $2 \left(x^2 + x + \frac12 \right) = 2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)$. Hence, we get $$I = \displaystyle \int \frac{dx}{2x^2 + 2x+1} = \int \frac{dx}{2 \left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)} = \frac12 \int \frac{dx}{\left( \left( x + \frac12 \right)^2 + \left(\frac12 \right)^2 \right)}.$$ Setting $y = x + \frac12$, we get $$I = \frac12 \int \frac{dy}{y^2 + \left(\frac12 \right)^2} = \tan^{-1}(2y) = \tan^{-1}(2x+1).$$

-
Thanks. That was what I had done but I made a mistake after setting $y$. –  Justin D. Mar 15 '12 at 2:29

$2/((4x^2+4x+1)+1)$ just multiply the numerator and denominator by 2

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Somebody had flagged this answer for low quality. I think that it is ok. –  Jyrki Lahtonen Mar 15 '12 at 8:06
Well, I have little knowledge of Latex. Thank you for your comment. –  eccstartup Mar 15 '12 at 9:39
@Jyrki: That was not "somebody", but rather the automatic process which flags certain posts which it thinks to have a low quality score. Often those are short messages which contain LaTeX code (which the machine probably recognizes as junk of some sort). –  Asaf Karagila Mar 15 '12 at 11:26
@AsafKaragila, thanks for the explanation. This was one of the first times SW managed to call my attention to a list of flagged posts (+10k users have the privilege to reinforce/water down such messages to moderators), and I'm inexperienced at that. It does sound like I did the right think in directing the moderators' attention elsewhere? –  Jyrki Lahtonen Mar 15 '12 at 11:29

Here's the method to completion. If it still doesn't make sense, post your progress and we can proceed from there.

You should complete the square on the bottom, use a u-substitution on the squared part, and use the regular arctan integral.

Does that make sense?

-
I don't understand how Wolfram goes from $1 / (1+2x+2x²)$ to $1 / (sqrt(2)x + 1/(sqrt(2)))² + 1/2$. –  Justin D. Mar 15 '12 at 2:22
@JustinDomingue: See my answer. –  Samuel Reid Mar 15 '12 at 2:23

Well, there are a few general methods that you want to employ when you see a rational function like that. Remember to look for completing the square, inverse trigonometric substitutions, direct substitutions, integration by parts, or partial fractions. In this case, it seems like completing the square or maybe even just factoring it might work, I'll give it a shot.

$$\int \frac{1}{2x^2 + 2x +1} dx = \int \frac{1}{2(x^2 + x + \frac{1}{2})} dx= \int \frac{1}{2((x+\frac{1}{2})^2 + \frac{1}{4})} dx$$

From here I would recommend using a direct substituion of $u=x + \frac{1}{2}$ and continuing from here as suggested in the other answers.

-


text:
# Linear regression and assumptions about response variable

Wikipedia states:

Ordinary linear regression predicts the expected value of a given unknown quantity (the response variable, a random variable) as a linear combination of a set of observed values (predictors). This implies that a constant change in a predictor leads to a constant change in the response variable (i.e. a linear-response model). This is appropriate when the response variable has a normal distribution.

So, Wikipedia makes an assumption about the response variable, namely that it is normally distributed. However, in other sources and here in stack exchange the normality is required for the error terms. If they are not normally distributed we should go for some generalized linear model.

What is the Wikipedia article referring to or is it wrong?

The Wikipedia statement

This is appropriate when the response variable has a normal distribution.

is wrong.

OLS does NOT have assumptions on response variable. But has assumptions on residual (See Gauss–Markov theorem). Also see this post for details.

Why linear regression has assumption on residual but generalized linear model has assumptions on response?

I am stealing @Cliff AB 's example here. The following distribution on $y$ and residual does not violate OLS assumption!

Related posts:

What is a complete list of the usual assumptions for linear regression?

How does linear regression use the normal distribution?

What if residuals are normally distributed, but y is not?


text:
# PSP Bibliography

 Notice: Clicking on the DOI link will open a new window with the original bibliographic entry from the publisher. Clicking on a single author will show all publications by the selected author. Clicking on a single keyword, will show all publications by the selected keyword.

Observations of Energetic-particle Population Enhancements along Intermittent Structures near the Sun from the Parker Solar Probe

 Author Bandyopadhyay, Riddhi; Matthaeus, W.; Parashar, T.; Chhiber, R.; Ruffolo, D.; Goldstein, M.; Maruca, B.; Chasapis, A.; Qudsi, R.; McComas, D.; Christian, E.; Szalay, J.; Joyce, C.; Giacalone, J.; Schwadron, N.; Mitchell, D.; Hill, M.; Wiedenbeck, M.; McNutt, R.; Desai, M.; Bale, Stuart; Bonnell, J.; de Wit, Thierry; Goetz, Keith; Harvey, Peter; MacDowall, Robert; Malaspina, David; Pulupa, Marc; Velli, M.; Kasper, J.; Korreck, K.; Stevens, M.; Case, A.; Raouafi, N.; Keywords Astrophysics - Solar and Stellar Astrophysics; Parker Data Used; parker solar probe; Physics - Plasma Physics; Physics - Space Physics; Solar Probe Plus Abstract Observations at 1 au have confirmed that enhancements in measured energetic-particle (EP) fluxes are statistically associated with "rough" magnetic fields, i.e., fields with atypically large spatial derivatives or increments, as measured by the Partial Variance of Increments (PVI) method. One way to interpret this observation is as an association of the EPs with trapping or channeling within magnetic flux tubes, possibly near their boundaries. However, it remains unclear whether this association is a transport or local effect; i.e., the particles might have been energized at a distant location, perhaps by shocks or reconnection, or they might experience local energization or re-acceleration. The Parker Solar Probe (PSP), even in its first two orbits, offers a unique opportunity to study this statistical correlation closer to the corona. As a first step, we analyze the separate correlation properties of the EPs measured by the Integrated Science Investigation of the Sun (IS☉IS) instruments during the first solar encounter. The distribution of time intervals between a specific type of event, i.e., the waiting time, can indicate the nature of the underlying process. We find that the IS☉IS observations show a power-law distribution of waiting times, indicating a correlated (non-Poisson) distribution. Analysis of low-energy (̃15 - 200 keV/nuc) IS☉IS data suggests that the results are consistent with the 1 au studies, although we find hints of some unexpected behavior. A more complete understanding of these statistical distributions will provide valuable insights into the origin and propagation of solar EPs, a picture that should become clear with future PSP orbits. Year of Publication 2020 Journal The Astrophysical Journal Supplement Series Volume 246 Number of Pages 61 Section Date Published 02/2020 ISBN URL https://iopscience.iop.org/article/10.3847/1538-4365/ab6220 DOI 10.3847/1538-4365/ab6220


text:
# How to Reduce the Size of a VirtualBox Virtual Machine and Boost Disk Space

1267

Here we can see, “How to Reduce the Size of a VirtualBox Virtual Machine and Boost Disk Space”

VirtualBox creates dynamic drives by default, which grow over time as you add data. If you later delete data from the virtual computer, you’ll find that the disc does not shrink automatically. However, a hidden command can be used to shrink a dynamic disc manually.

Remember that this will only shrink the disc if it has grown in size and you have eliminated data since then. This will not shrink a dynamic disc that you have just made. However, if you construct a dynamic drive, download 10 GB of data into it, and then delete the data, you should be able to decrease the disc by around 10 GB.

## Step 1: Double-check that you’re using a dynamic disc.

This method is only applicable to dynamic discs, which can expand and contract in size. Dynamic discs can be up to a given maximum capacity—50 GB, for example—but only when they hold that much data do they truly grow to that amount. Disks with a fixed size will always be at their maximum size.

If you want to shrink a fixed-size disc, convert it to a dynamic disc and follow the methods below. If you have a 50 GB fixed drive with just 20 GB of data on it and you convert it to a dynamic disc, you should be able to reduce it down to only 20 GB of Space.

• In VirtualBox, right-click the virtual machine that utilizes the disc and select “Settings” to see if it is a dynamic or fixed size. Select the disc from the “Storage” menu. Next to “Details,” you’ll notice what type of disc it is.
• “Dynamically allocated storage,” for example, indicates that this is a dynamic disc.
Also See:  How to: Fix Windows Defender Error 0x800b0100

## Step 2: In the Virtual Machine, write zeros to the disc.

Before continuing, make sure to erase any files you no longer desire from the virtual machine’s drive to free up space. Delete unnecessary files, delete apps you don’t use, and empty your Recycle Bin. After that, you’ll need to fill in the blanks with zeroes.

### Guest Operating Systems for Windows

• If the virtual machine has Windows installed, you should now boot it up and defragment the discs. Search for “Defragment” in the virtual machine’s Start menu and run the “Disk Defragmenter” or “Defragment and Optimize Drives” program.
• Click “Defragment disc” after selecting the disc you want to shrink.

After the defragmentation procedure is complete, you must write zeros to the virtual machine’s vacant Space. VirtualBox cannot automatically shrink the drive because destroyed data is still saved on the disc when you delete files. However, if you write zeroes over the removed files, VirtualBox will notice a high number of zeroes—Space—and will be able to condense the disc.

• To do so, go to Microsoft’s website and download the SDelete application. The sdelete.exe File should be extracted to a location on your PC.
• To get started, open a Command Prompt window. To do so, go to the Start menu and type “Command Prompt” into the search box.
• By typing cd, pressing Space, entering the path to the directory, then pressing Enter, you may go to the directory containing the sdelete.exe File. If the path contains a space character, it must be enclosed in quotation marks. This is how it should look:
cd "C:\path\to\folder"

cd "C:\Users\bob\Downloads"

Type cd into the Command Prompt window, press Space, and drag and drop the folder icon from the file manager’s address bar to fill in the directory path.

Execute the command below:

sdelete.exe c: -z
• This will fill all of the free disc space on drive C: with zeros. If you want to shrink a secondary drive in the virtual machine with a different drive letter, then c: the type that drives letter instead of c: This is exactly what the tool was made for. The -z option is “useful for virtual disc optimization,” according to Microsoft’s SDelete page.
• Before continuing, you’ll be required to accept the tool’s license agreement. Select “Agree” from the drop-down menu.
• Wait for the procedure to finish. When it’s finished, use the “Shut Down” option in the virtual machine’s Start menu to shut it down. It’s now time to condense it.

### Guest Linux Operating Systems

You can bypass the defragmentation procedure and use built-in instructions to zero the free space on the drive if you have Linux installed in the virtual machine (a Linux guest operating system rather than a Windows guest operating system, in other words). The technique will be comparable to other Linux distributions. However, we’ll use Ubuntu as an example.

You’ll need to install the zero-free utility inside the virtual computer first, and it should be available in the software repository of your Linux system. For example, you may install it on Ubuntu by entering the following command into your virtual machine’s terminal:

sudo apt install zerofree

When you’ve booted into the conventional Linux environment, you can’t utilize zerofree on your / partition. Instead, you’ll want to boot into a unique recovery mode that disables the mounting of your standard root disc.

• To access the Grub menu on Ubuntu, restart your virtual machine and continually hit the “Esc” key while it is booting. Select “*Advanced options for Ubuntu” from the Grub menu and click Enter.
• Select the “(recovery mode)” option associated with the most recent Linux kernel—that is, the option towards the top of the list with the highest version number—and press Enter.
• To boot to a root shell prompt, select “root” from the recovery menu.
• When the message “Press Enter for maintenance” displays on your screen, press “Enter.” A terminal prompt will appear.

Run the following command from the command line to determine which virtual disc you wish to zero:

df

We can see that /dev/sda1 is our lone physical disc device in the output below. It’s the only disc with /dev/ in the leftmost column, so we know.

• If you use the default options, your virtual machine will only have /dev/sda1, the first partition on the first hard drive.
• You may need to zero another partition or many partitions if you’ve set things up differently with numerous discs or partitions.
• Now it’s time to zero the disc. Replace /dev/sda1 with the partition’s device name you want to write zeros to in the following command. The majority of people will just set the /dev/sda1 device to zero.
zerofree -v /dev/sda1

Run the following command to shut down the virtual machine once the zero process is complete:

halt

The system has ceased, and you can now shut down your virtual machine when you see the “System halted” sign on your screen. Select “Power of the virtual machine” after closing the virtual machine’s window.

Also See:  How to Convert Between Fixed and Dynamic Disks in VirtualBox

## Step 3: Locate the VBoxManage command.

The remaining steps will be completed on your host operating system outside of the virtual computer. If you had Windows 10 on your PC and Windows 7 in a virtual machine, for example, you’d complete the rest of the process on Windows 10.

• The graphical interface of VirtualBox hides this option. You must instead use the VBoxManage.exe command.
• To proceed, look for this command. It’s in the VirtualBox program directory on Windows, which is C:\Program Files\Oracle\VirtualBox by default. Instead, look in the place where you installed VirtualBox.
• To get started, open a Command Prompt window. To do so, go to the Start menu and put cmd into the search box.
• In the Command Prompt, type cd, then the path to the folder where the VBoxManage command is located. You’ll have to put it in quotations.
• Type cd into the Command Prompt window, then drag and drop the folder icon from the file manager’s address bar into the Command Prompt.

If you’re following the default path, it should look something like this:

cd "C:\Program Files\Oracle\VirtualBox"

Note: These instructions are based on the assumption that you’re running VirtualBox on Windows. If you’re running VirtualBox on macOS or Linux, open a Terminal window and type the vboxmanage command like any other command.

## Step 4: Find the path to the disc you’d like to compact.

To get a list of all the virtual hard discs on your computer, type the following command in the command prompt window:

VBoxManage.exe list hdds

Look through the list to find the virtual disc file path you wish to condense. Let’s imagine we wish to change the virtual disc connected with the “Windows 7” virtual machine. The path to that virtual disc on our machine is C:\Users\chris\VirtualBox VMs\Windows 7\Windows 7.vdi.

## Step 5: Make the Disk Compact

To compress the disc, run VBoxManage.exe with the appropriate command. Run the command below, replacing the path to the disc VDI file with the disc on your system that you discovered with the command above. If the File’s path contains a space anywhere in it, make sure to wrap it in quotes.

This is how it should look:

VBoxManage.exe modifymedium disk "C:\path\to\disk.vdi" --compact

For example, since the path to the File is C:\Users\chris\VirtualBox VMs\Windows 7\Windows 7.vdi in our example, we’d run the following command:

VBoxManage.exe modifymedium disk "C:\Users\chris\VirtualBox VMs\Windows 7\Windows 7.vdi" --compact

VirtualBox will compact the disc, and, if desired, you can start the virtual computer right away. The amount of space you save is determined by how much vacant Space there was to restore.

## Consider deleting snapshots as well.

You may make snapshots for each virtual machine in VirtualBox. These backups provided a complete image of the virtual machine when the snapshot was taken, allowing you to restore it to a previous state. These can eat up a lot of room.

Delete photos you don’t use to free up space. Select a virtual machine in the main VirtualBox window and click the “Snapshots” button to the right of Details on the toolbar to see the snapshots you’ve saved.

If you don’t need a snapshot anymore, right-click it in the list and choose “Delete Snapshot” to clear the Space.

## Conclusion

I hope you found this information helpful. Please fill out the form below if you have any queries or comments.

## User Questions:

1. What is the procedure for resizing a VirtualBox disc?

In the main VirtualBox window, go to File> Virtual Media Manager to get there. Select a virtual hard disc from the list and adjust its size using the “Size” slider at the bottom of the window. When you’re finished, click “Apply.”

1. Is it possible to minimize the size of the VMDK?

Virtual discs can be flattened such that they only take up as much space as the data they contain. VMware Converter and hosted products, such as VMware Workstation, are currently capable of thinning. See the vCenter Converter Standalone Documentation for your VMware product version for further information.

1. How do I expand the size of my VirtualBox VM?

Go to View in the VM Window menu and ensure the Auto-resize Guest Display option is turned on. Change the size of the VM window by moving the mouse pointer over the corner and pressing the left mouse button.

Also See:  A Fan Created A Justice League Game Because Warner Bros. "Won't"
1. The dynamically-sized discs in VirtualBox only grow in size.

1. How can I keep the virtual HDD consumption of a Windows 10 VM in VirtualBox to a minimum?


text:
Plane (geometry)/Citable Version

Main Article
Discussion
Related Articles  [?]
Bibliography  [?]
Citable Version  [?]

This version approved either by the Approvals Committee, or an Editor from the listed workgroup. The Mathematics Workgroup is responsible for this citable version. While we have done conscientious work, we cannot guarantee that this version is wholly free of mistakes. See here (not History) for authorship.
Help improve this work further on the editable Main Article!

In Euclidean geometry, a plane is an abstract concept that models the common notion of a flat surface — without depressions, protrusions, holes and a boundary — that for any two of its points entirely contains the straight line joining them.

Assuming a common (intuitive, physical) idea of the geometry of space, "plane" can be defined in terms of distances, orthogonality, lines, coordinates etc. In a more abstract approach (vector spaces) planes are defined as two-dimensional affine subspaces. In an axiomatic approach, basic concepts of elementary geometry, such as "point", "line" and "plane", are undefined primitives.

(CC) Image: Dina Tsirelson
The surface of this metallic body consists of rectangles situated in various planes.

Non-axiomatic approach

Definitions

A remark

To define a plane is more complicated than it may seem.

It is tempting to define a plane as a surface with zero curvature, where a surface is defined as a geometric object having length and breadth but no depth. However, this is not a good idea; such definitions are useless in mathematics, since they cannot be used when proving theorems. Planes are treated by elementary geometry, but the notions of surface and curvature are not elementary, they need more advanced mathematics and more sophisticated definitions. Fortunately, it is possible to define a line via more elementary notions, and this way is preferred in mathematics. Still, the definitions given below are tentative. They are criticized afterwards, see axiomatic approach.

The definitions of "plane" given below may be compared with the definition of a circle as consisting of those points in a plane that are a given distance (the radius) away from a given point (the center). A circle is a set of points chosen according to their relation to some given parameters (center and radius). Similarly, a plane is a set of points chosen according to their relation to some given objects (points, lines etc). However, a circle determines its center and radius uniquely; for a plane, the situation is different.

Four equivalent definitions of "plane" are given below. Any other definition is equally acceptable provided that it is equivalent to these. Note that a part of a plane is not a plane. Likewise, a line segment is not a line.

Below, all points, lines and planes are situated in the space (assumed to be a three-dimensional Euclidean space), and by lines we mean straight lines.

Definition via distances

Let two different points A and B be given. The set of all points C that are equally far from A and B — that is,

${\displaystyle |AC|=|BC|}$

— is a plane.

This is the plane orthogonal to the line AB through the middle point of the line segment AB.

Definition via right angles (orthogonality)

Let two different points A and B be given. The set of all points C such that the lines AB and AC are orthogonal (that is, the angle BAC is right) is a plane.

This is the plane orthogonal to the line AB through the point A.

Every point F on the line DE belongs to the plane through given points A, B and C, provided that D belongs to AB, E belongs to BC, and D, E do not coincide with B.

Definition via lines

Let three points A, B and C be given, not lying on a line. Consider the lines DE for all points D (different from B) on the line AB and all points E (also different from B) on the line BC. The union of all these lines, together with the point B, is a plane.

This is the plane through A, B and C.

In other words, this plane is the set of all points F such that either F coincides with B or there exists a line through F that intersects the lines AB and BC (in distinct points).

Definition via Cartesian coordinates

In terms of Cartesian coordinates x, y, z ascribed to every point of the space, a plane is the set of points whose coordinates satisfy the linear equation

${\displaystyle ax+by+cz=d}$.

Here real numbers a, b, c and d are parameters such that at least one of a, b, c does not vanish.

Some properties of planes

Most basic properties

For any three points not situated in the same straight line there exists one and only one plane that contains these three points.

If two points of a straight line lie in a plane, then every point of the line lies in that plane.

If two planes have a common point then they have at least a second point in common.

Every plane contains at least three points not lying in the same straight line, and the space contains at least four points not lying in a plane.

Further properties

Two planes either do not intersect (are parallel), or intersect in a line, or coincide.

A line either does not intersect a plane (is parallel to it), or intersects it in a single point, or is contained in the plane.

Two lines perpendicular to the same plane are parallel to each other (or coincide).

Two planes perpendicular to the same line are parallel to each other (or coincide).

Axiomatic approach

What is wrong with the definitions given above?

The definitions given above assume implicitly that the 3-dimensional Euclidean space is already defined, together with (at least one of) such notions as distances, angles, straight lines, Cartesian coordinates, while planes are not defined yet. However, this situation never appears in mathematical theory.

In the axiomatic approach points, lines and planes are undefined primitives.

The modern approach (below) defines planes in a completely different way.

How does it work

Axiomatic approach is similar to chess in the following aspect.

A chess piece, say a rook, cannot be defined before the whole chess game is defined, since such a phrase as "the rook moves horizontally or vertically, forward or back, through any number of unoccupied squares" makes no sense unless it is already known that "chess is played on a square board of eight rows and eight columns" etc. And conversely, the whole chess game cannot be defined before each piece is defined; the properties of the rook are an indispensable part of the rules of the game. No chess without rooks, no rooks outside chess! One must introduce the game, its pieces and their properties in a single combined definition.

Likewise, Euclidean space, its points, lines, planes and their properties are introduced simultaneously in a set of 20 assumptions known as Hilbert's axioms of Euclidean geometry.[1] The "most basic properties of planes" listed above are roughly the plane-related assumptions (Hilbert's axioms), while "further properties" are the first plane-related consequences (theorems).

Modern approach

The modern approach defines the three-dimensional Euclidean space more algebraically, via linear spaces and quadratic forms, namely, as a real affine space whose difference space is a three-dimensional inner product space. For further details see Affine space#Euclidean space and space (mathematics).

In this approach a plane in an n-dimensional affine space (n ≥ 2) is defined as a (proper or improper) two-dimensional affine subspace.

A less formal version of this approach uses points, vectors and scalar product (called also dot product or inner product) of vectors without mentioning linear and affine spaces. Optionally, Cartesian coordinates of points and vectors are used. See algebraic equations below. There, in particular, equivalence between the definition via right angles (orthogonality) and the definition via Cartesian coordinates is explained.

Plane geometry

Plane geometry (also called "planar geometry") is a part of solid geometry that restricts itself to a single plane ("the plane") treated as a geometric universe. In other words, plane geometry is the theory of the two-dimensional Euclidean space, while solid geometry is the theory of the three-dimensional Euclidean space.

Plane geometry studies the properties of plane figures (and configurations). Plane figures in elementary geometry are sets of points, lines, line segments and sometimes curves that fall on the same plane. For example, triangles, polygons and circles. In plane geometry every figure is plane, in contrast to solid geometry.

Algebraic equations

In analytic geometry several closely related algebraic equations are known for a plane in three-dimensional Euclidean space. A few algebraic representations will be discussed.

Point-normal representation

PD Image
The equation for a plane. P is an arbitary point in plane; ${\displaystyle \scriptstyle {\vec {d}}}$ and ${\displaystyle \scriptstyle {\vec {n}}_{0}}$ are collinear.

One such equation is illustrated in the figure. Point P is an arbitrary point in the plane and O (the origin) is drawn outside the plane. The point A in the plane is chosen such that vector

${\displaystyle {\vec {d}}={\overrightarrow {OA}}}$

is orthogonal to the plane. The collinear vector

${\displaystyle {\vec {n}}_{0}={\frac {1}{d}}{\vec {d}}\quad {\hbox{with}}\quad d=\left|{\vec {d}}\,\right|}$

is a unit (length 1) vector normal (perpendicular) to the plane which is known as the normal of the plane in point A. Note that d is the distance of O to the plane. The following relation holds for an arbitrary point P in the plane (according to the definition via right angles):

${\displaystyle \left({\vec {r}}-{\vec {d}}\;\right)\cdot {\vec {n}}_{0}=0\quad {\hbox{with}}\quad {\vec {r}}={\overrightarrow {OP}}\quad {\hbox{and}}\quad {\vec {r}}-{\vec {d}}={\overrightarrow {AP}}.}$

This equation for the plane can be rewritten in terms of coordinates with respect to a Cartesian frame with origin in O. Dropping arrows for component vectors (real triplets) that are written bold, we find

${\displaystyle \left(\mathbf {r} -\mathbf {d} \right)\cdot \mathbf {n} _{0}=0\Longleftrightarrow \mathbf {r} \cdot \mathbf {n} _{0}=\mathbf {d} \cdot \mathbf {n} _{0}\Longleftrightarrow xa_{0}+yb_{0}+zc_{0}=d}$

with

${\displaystyle \mathbf {d} =(a,\;b,\;c),\quad \mathbf {n} _{0}=(a_{0},\;b_{0},\;c_{0}),\quad \mathbf {r} =(x,\;y,\;z),}$

and

${\displaystyle \mathbf {d} \cdot \mathbf {n} _{0}=\mathbf {d} \cdot {\frac {1}{d}}\mathbf {d} ={\frac {1}{d}}\mathbf {d} ^{2}={\frac {d^{2}}{d}}=d={\sqrt {a^{2}+b^{2}+c^{2}}}.}$

The definition via Cartesian coordinates is thus derived from the definition via right angles.

Moreover, the Hesse normal form for the plane (called after the 19th century mathematician Ludwig Otto Hesse) is obtained,

${\displaystyle \mathbf {r} \cdot \mathbf {n} _{0}=d;}$

it is characterized by the use of a unit-length vector ${\displaystyle \mathbf {n} _{0}}$ rather than an arbitrary vector orthogonal to the plane.

Conversely, the definition via right angles can be derived from the definition via Cartesian coordinates as follows. Given a linear equation for a plane

${\displaystyle ax+by+cz=e,\,}$

we write

${\displaystyle \mathbf {r} =(x,\;y,\;z),\quad \mathbf {f} =(a,\;b,\;c),\quad {\hbox{and}}\quad \mathbf {d} =\left({\frac {e}{a^{2}+b^{2}+c^{2}}}\right)\mathbf {f} .}$

It follows that

${\displaystyle \mathbf {f} \cdot \mathbf {r} =e=\mathbf {f} \cdot \mathbf {d} .}$

Hence we find the same orthogonality relation,

${\displaystyle \mathbf {f} \cdot (\mathbf {r} -\mathbf {d} )=0\;\Longrightarrow \;(\mathbf {r} -\mathbf {d} )\cdot \mathbf {n} _{0}=0\quad {\hbox{with}}\quad \mathbf {n} _{0}={\frac {1}{\sqrt {a^{2}+b^{2}+c^{2}}}}\mathbf {f} }$

where f , d, and n0 are collinear. The equation may also be written in the following mnemonically convenient form

${\displaystyle \mathbf {d} \cdot (\mathbf {r} -\mathbf {d} )=0,}$

which is the equation for a plane through a point A perpendicular to ${\displaystyle {\overrightarrow {OA}}={\vec {d}}}$.

Three-point representation

PD Image
Plane through points A, B, and C.

The figure shows a plane that by definition passes through three different points A, B, and C that are not on one line. The point P is an arbitrary point in the plane and the reference point O is again drawn outside the plane, but the case that the plane passes through O is not excluded. Referring to figure 2 we introduce the following definitions

${\displaystyle {\vec {a}}={\overrightarrow {OA}},\quad {\vec {b}}={\overrightarrow {OB}},\quad {\vec {c}}={\overrightarrow {OC}},\quad {\vec {r}}={\overrightarrow {OP}}.}$

Clearly the following two non-collinear vectors belong to the plane

${\displaystyle {\vec {u}}={\overrightarrow {AB}}={\vec {b}}-{\vec {a}},\quad {\vec {v}}={\overrightarrow {AC}}={\vec {c}}-{\vec {a}}.}$

Because a plane (an affine space), with a given fixed point as origin is a 2-dimensional linear space and two non-collinear vectors with "tails" in the origin are linearly independent, it follows that any vector in the plane can be written as a linear combination of these two non-collinear vectors. (This is also expressed as: Any vector in the plane can be decomposed into components along the two non-collinear vectors.) In particular, taking A as origin in the plane,

${\displaystyle {\overrightarrow {AP}}={\vec {r}}-{\vec {a}}=\lambda {\vec {u}}+\mu {\vec {v}},\qquad \lambda ,\mu \in \mathbb {R} .}$

The real numbers λ and μ specify the direction of ${\displaystyle {\overrightarrow {AP}}}$. Hence the following equation for the position vector ${\displaystyle {\vec {r}}}$ of the arbitrary point P in the plane:

${\displaystyle {\vec {r}}={\vec {a}}+\lambda {\vec {u}}+\mu {\vec {v}}}$

is known as the point-direction representation of the plane. This representation is equal to the three-point representation

${\displaystyle {\vec {r}}={\vec {a}}+\lambda ({\vec {b}}-{\vec {a}})+\mu ({\vec {c}}-{\vec {a}}),}$

where ${\displaystyle {\vec {a}}}$, ${\displaystyle {\vec {b}}}$, and ${\displaystyle {\vec {c}}}$ are the position vectors of the three points that define the plane.

Writing for the position vector of the arbitrary point P in the plane

${\displaystyle {\vec {r}}=(1-\lambda -\mu )\,{\vec {a}}+\lambda \,{\vec {b}}+\mu \,{\vec {c}}\;\equiv \;\xi _{1}\,{\vec {a}}+\xi _{2}\,{\vec {b}}+\xi _{3}\,{\vec {c}}\;,}$

we find that the real triplet (ξ1, ξ2, ξ3) with ξ1 + ξ1 + ξ1 = 1 forms a set of coordinates for P. The numbers {ξ1, ξ2, ξ3 | ξ1+ ξ2+ ξ3 = 1 } are known as the barycentric coordinates of P. It is trivial to go from barycentric coordinates to the "three-point representation",

${\displaystyle {\vec {r}}=\xi _{1}{\vec {a}}+\xi _{2}{\vec {b}}+\xi _{3}{\vec {c}}\quad {\hbox{with}}\quad \xi _{1}=1-\xi _{2}-\xi _{3}\;\Longleftrightarrow \;{\vec {r}}={\vec {a}}+\xi _{2}({\vec {b}}-{\vec {a}})+\xi _{3}({\vec {c}}-{\vec {a}}).}$

Beyond mathematics

In industry, a surface plate is a piece of cast iron or other appropriate material whose surface (or rather a part of it) is made as close as possible to a geometric plane (or rather a part of it, usually a square). An old method of their manufacturing is the three-plate method: three roughly flat surfaces become more and more flat when rubbing against each other: first and second; second and third; third and first; first and second again, and so on. It is possible to achieve a surface close to a plane up to 10–5 of its size.

Reference

1. D. Hilbert, Grundlagen der Geometrie, B. G. Teubner, Leipzig (1899) 2nd German edition


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# The fifth period of modern periodic table includes

$\begin{array}{1 1}\text{8 elements}\\\text{18 elements}\\\text{32 elements}\\\text{unknown elements}\end{array}$

The fifth period includes 18 elements due to filling of 5s,4d and 5p orbitals.


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# Active

 ID 834536 Comments 2 Status Active Workarounds 2 Type Bug Repros 4 Opened 3/15/2014 10:06:55 AM Access Restriction Public Primary Feedback Item 808375

### Description

As a University, we have used MathML to deliver typeset mathematics over the Web for many years.  Until the advent of Design Science's dramatically expanded high-quality support for enabling technologies in its MathPlayer IE plugin, we were unable to support the many Americans (and others around the world) who, through no fault of their own, require special technologies to use the web at all.

Microsoft has provided excellent support for these enabling technologies within its more recent Windows operating systems; however, because of the quite recent lack of support for a corresponding delivery of web-based mathematical tools, we are forced to revert to19th century methods (e.g., e-mailed paper copies) to accommodate people who clearly deserve appropriate consideration and better treatment.  (Among other uses, we have been providing on-line tests to people all over the world using MathML technologies.)

Current alternative technologies (e.g., MathJax) for displaying MathML are too slow and incomplete to be taken seriously, AND THEY PROVIDE NO SUPPORT WHATSOEVER FOR THE MANY PEOPLE WITH SPECIAL NEEDS.

We truly hope that the current situation (i.e., an IE which simply no longer supports MathML) can be addressed in the near future.
Sign in to post a comment.
Posted by Microsoft on 3/27/2014 at 10:02 AM
Thank you for your feedback.

We are tracking Connect Feedback #808375 with the same issue. Please follow the progress of this bug at http://connect.microsoft.com/IE/feedback/details/808375/mathml-support .

Best regards,
The Internet Explorer Team
Posted by Lionel Fourquaux on 3/26/2014 at 5:01 PM
I strongly support this!

Here are a few points to keep in mind:
* MathML is not only useful for education! many people (scientists, engineers...) use math in their line of work and sometimes wish to publish technical content on the web (scientists at CERN gave you the web, please give some support back to them!);
* while workarounds exist, please note that MathJax developers are strongly in favor of native MathML support (source: mozilla.dev.platform), because MathJax without MathML support is not accessible, much slower, and less precise;
* Microsoft already has an excellent math rendering engine in the PTL Services library used by IE (cf. Murray Sargent's blog), the main problem is parsing MathML (which is admittedly not quite as easy as it seems at first glance);
* MathML is included in HTML5 and EPUB3 (with good reasons);
* Wikipedia is ready to switch the MathML once it is supported well enough (and has experimented with it and alternative rendering methods).

This are active development efforts for better MathML support, mostly for Gecko but also in WebKit, see e.g.
http://www.ulule.com/mathematics-ebooks/
If you need sample documents for testsuites, or contacts with people involved in developing MathML rendering code, I'm sure they would be glad to help and add Trident to the list of rendering engines with MathML support. (Although this sounds like a plug for this project, I'm not involved in it at all).

So, can we hope for some MathML in IE 12?


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# Question on elementary statics multiple choice

Tags:
1. May 7, 2016

### RoboNerd

1. The problem statement, all variables and given/known data

2. Relevant equations
sum of torques = 0

3. The attempt at a solution
So I tried to put the torques into the equation:

Sum torques = mgL - T *sin(theta) * (L-x) = 0.

Solving for the T gave me [mgL]/[ (L-x) * sin(theta) ]. This is not an option in the answers.

2. May 7, 2016

### SteamKing

Staff Emeritus
What if answer 4) has a typo, and it should read: $\frac{mgL}{(L-x)sin\, \theta}$ ?

3. May 8, 2016

### RoboNerd

If yes, then that is the answer. The test questions are written by some crummy authors [thanks a lot whoever you are!]

4. May 8, 2016

### RoboNerd

Thanks for the help!


text:
image credit
Graph theory
In mathematics graph theory is the study of graphs, which are mathematical structures used to model pairwise relations between objects. A graph in this context is made up of vertices, nodes, or points which are connected by edges, arcs, or lines. A graph may be undirected, meaning that there is no distinction between the two vertices associated with each edge, or its edges may be directed from one vertex to another; see Graph (discrete mathematics) for more detailed definitions and for other variations in the types of graph that are commonly considered. Graphs are one of the prime objects of study in discrete mathematics. MORE
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