TITLE: Software to compute generators of a module over polynomial ring
QUESTION [0 upvotes]: Let $A=\mathbb{R}[x_1,\dots,x_n]$ be the algebra of real polynomials in $n$ variables. Fix polynomials $p_1,\dots,p_k\in A$.
Consider the subset
$$M:=\{(q_1,\dots,q_k)\in A^k|\, p_1q_1+\dots+p_kq_k=0\}.$$
Clearly $M$ is an $A$-submodule of $A^k$. Necessarily $M$ is finitely generated.
I am wondering if there exists a software which allows to compute explicit generators of $M$ as an $A$-module.
In my case $n=16,k=10$, and all $p_i$'s are explicit homogeneous polynomials of second degree.

REPLY [2 votes]: What you compute is the "syzygy module" of $p_1,\ldots,p_k$.
You can try the following M2 script to check the computation times in Macaulay 2 for different $n$'s and $k$'s.
For me $n=16, k=6$ was a matter of minutes, but $k > 6$ seems to take much longer.
If you decrease the sparsity of your $p_i$ by setting Density to 0.1, 0.5 or higher, you will probably be out of luck with your computation.
n = 16
k = 10
R=QQ
--- or R=ZZ/23 or any other prime
A=R[x_1..x_n]
mat = random(A^1, A^{k:-2}, Density=>0.05)
erz = gens ker mat


TITLE: For each $n$, does there exist a prime $p$ and integer $k$ such that $p^k - 1$ has exactly $n$ prime divisors?
QUESTION [12 upvotes]: Let $n \geq 1$ be some integer. Can we always find a prime power $p^k$ such that $p^k - 1$ has exactly $n$ distinct prime divisors?
For example:

$n = 1$ example: $2^2 - 1 = 3$
$n = 2$ example: $5^2 - 1 = 2^2 \cdot 3$
$n = 3$ example: $7^3 - 1 = 2 \cdot 3^2 \cdot 19$
$n = 4$ example: $5^6 - 1 = 2^3 \cdot 3^2 \cdot 7 \cdot 31$

We can always find a prime power $p^k$ with $\geq n$ prime divisors. Let $p_1, p_2, \ldots, p_n$ be distinct primes, $p_i \neq p$ for each $i$. Then $p^k \equiv 1 \mod{p_1p_2 \ldots p_n}$ for some positive $k$, for example we can choose $$k = (p_1 - 1)(p_2 - 1)\ldots(p_n - 1)$$

REPLY [6 votes]: You might be interested in the Cunningham Project, about the factorization of numbers of the form $b^n\pm1$ and sequence A046800 in OEIS. According to the existing data, the following conjecture seems plausible: for every $n\in\mathbb{N}$ there exists $k\in\mathbb{N}$ such that $2^k-1$ has exactly $n$ distinct prime factors.
Define $f(n)$ as the smallest $k$ such that $2^k-1$ has exactly $n$ distinct prime factors. The values of $f(n)$ for $1\le n\le30$ are:
2,4,8,12,20,24,40,36,48,88,60,72,150,132,120,156,144,200,204,210,180,
324,476,288,300,432,396,480,360,468

and $f(31)>500$.


TITLE: Are there any examples of connected, dense total order spaces which are not separable?
QUESTION [0 upvotes]: Like the title: Are there any examples of connected, dense total order spaces which are not separable?
I've still haven't made much progress with this. I know that they are order complete from being order dense and connected, but anything else escapes me.
Either an example or a refutation that such spaces exists would be most enlightening.

REPLY [1 votes]: $[0,1] \times [0,1]$ in the lexicographic order (and its induced topology), (so $(x,y) < (u,v)$ iff ($x < u$ or $x=u$ and $y<v$) ) is an example of a connected (and thus order dense) compact ordered space that is not separable.
Removing $(0,0)$ and $(1,1)$ creates a non-compact example.
Another non-compact (but countably compact) space is the long line e.g.
A well-known theorem says that if $X$ is an ordered space that is connected (i.e. order dense and complete) and has no minimum nor maximum and is separable, then $X$ is homeomorphic (even order isomorphic) to the real line. This inspired the Suslin problem (whether we could weaken "separable" to ccc here, see wikipedia. And this gave rise to nice results in set theory and independence results. Both examples above are non-ccc such spaces. Under certain axioms (like $\diamondsuit$) we could use a Suslin line as a ccc example, which are even closer to the reals. In ZFC we cannot find such a space.


TITLE: Finding cumulative probability distribution
QUESTION [1 upvotes]: Question
Probability Density function:
$$f(x,y) = 3x, \quad 0 \le y \le x \le 1$$
$f(x,y) = 0$ elsewhere.
Find $P(X\le 1/2, Y\le 1/3)$
Solution
$$\int_0^\frac{1}{2}\int_0^\frac{1}{3}3x\,dx\,dy=0.1065$$
I cannot figure out why the limits of the integral are like this. I would have guess they should have been reversed, such that, limit for $x$ should have been from $\frac{1}{2}$ to $0$ and not the other way around.
Any help is much appreciated.

REPLY [2 votes]: The proposed solution is wrong; what you thought it should be is right.
$$
\require{cancel} \xcancel{
\Pr\left(X\le\frac 1 2\ \&\ Y \le \frac 1 3\right) = \int_0^{1/3} \left( \int_0^{1/2} 3x\,dx \right) \,dy=0.1065}
$$
The question has been edited since I posted this. We are now told $y\le x,$ so we have $0\le y\le x\le1.$
\begin{align}
\Pr\left(X\le\frac 1 2\ \&\ Y \le \frac 1 3\right) & = \iint\limits_{\{\,(x,y) \,:\, 0\,\le\,y\,\le 1/3\ \&\ y\,\le\,x\,\le\,1\,\}} 3x \, d(x,y) \\[10pt]
& = \begin{cases} \text{either } & \displaystyle \int_0^{1/3} \left( \int_y^{1/2} 3x\,dx \right) \, dy \\[10pt]
\text{or } & \displaystyle \int_0^{1/3} \left( \int_0^x 3x\,dy \right) \,dx + \int_{1/3}^{1/2} \left( \int_0^{1/3} 3x \,dy \right) \,dx
\end{cases} \\[20pt]
& = 0. 106\,481\,481\,481\,\ldots
\end{align}
The parentheses enclosing the inner integrals are not obligatory, but are included in order to call attention to that which needs attention in this context.


TITLE: Solve integral $\int \frac{x+1}{x^2-2x+5} dx$
QUESTION [0 upvotes]: I need to solve:
$$\int \frac{x+1}{x^2-2x+5} dx$$
I cann see that $D>N$ so I tried to scompose the $D$ but I get:
$$x_{1,2} = \frac{2 \pm \sqrt{4-20}}{2}$$
So $\Delta < 0$ and I tried to use A and B.
$$\int \frac{x+1}{x^2-2x+5} dx = \frac{Ax + B}{x^2-2x+5}$$
But here I stops, on my book there is a hint, $A(2x-2)+B$.
Where should I found that $2x-2$, is it the derivate of $x^2-2x+5$?

REPLY [3 votes]: Yes, the hint in your book refers to the derivative of the denominator (see below $(*)$). For a rational function with a quadratic denominator (and numerator of degree less than 2; if not: perform long division first) with a negative discriminant, this method always work. If the discriminant is positive, go with partial fractions.
In general, if $b^2-4ac <0$, you want to do the following: because $(\color{red}{ax^2+bx+c})' = \color{blue}{2ax+b}$, you look for numbers $A$ and $B$ so you can write and split the integral as follows:
\begin{array}{rl}
\displaystyle \int \frac{px+q}{\color{red}{ax^2+bx+c}} \,\mbox{d}x
& \displaystyle = \int \frac{A(\color{blue}{2ax+b})+B}{\color{red}{ax^2+bx+c}} \,\mbox{d}x \\[8pt]
& \displaystyle = A \int\frac{\color{blue}{2ax+b}}{\color{red}{ax^2+bx+c}} \,\mbox{d}x + \int\frac{B}{ax^2+bx+c} \,\mbox{d}x \\[8pt]
& \displaystyle = A \ln\left( ax^2+bx+c \right) + \color{green}{ \int\frac{B}{ax^2+bx+c} \,\mbox{d}x }
\end{array}
And then continue completing the square in the denominator of the green integral, working towards $\arctan$.

So in this case you're looking for $A$ and $B$ such that:
$$x+1 = A(2x-2)+B$$
Clearly, $A = \tfrac{1}{2}$ and then it follows that $B=2$, so rewriting:
$$\frac{x+1}{x^2-2x+5} = \frac{\tfrac{1}{2}\left( 2x-2 \right) + 2}{x^2-2x+5} = \frac{1}{2}\frac{2x-2}{x^2-2x+5}+\frac{2}{x^2-2x+5}$$gives the integrals:
$$\int \frac{x+1}{x^2-2x+5} \,\mbox{d}x = \frac{1}{2} \int \frac{2x-2}{x^2-2x+5} \,\mbox{d}x + \int \frac{2}{x^2-2x+5} \,\mbox{d}x$$
The first gives you $\ln$ of the denominator, the second can be reworked into an $\arctan$:
$$\int \frac{2}{x^2-2x+5} \,\mbox{d}x = \int \frac{2}{(x-1)^2+4} \,\mbox{d}x =\cdots$$

$(*)$
This is a handy step because a fraction where the numerator is the derivative of the denominator, has a very simple anti-derivative (integral): the natural logarithm of the denominator! Or, symbolically:
$$\int \frac{g'(x)}{g(x)} \,\mbox{d}x = \ln g(x) + C$$You can check this by taking the derivative of $\ln g(x)$, or perform a substitution $u=g(x)$ on the initial integral.


TITLE: Can you efficiently solve a system of quadratic multivariate polynomials?
QUESTION [12 upvotes]: Given a system of 2nd-degree polynomials, $P=\{p_1,\dots,p_m\}$ where $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$, can you efficiently find a common zero of all of these polynomials?  In other words, given $P$, can you find $x_1,\dots,x_n\in\mathbb{R}$ such that $p_i(x_1,\dots,x_n)=0$ for all $i$?  I only need to find one solution.
I know that in the general case (arbitrary degree polynomials) this can't be done efficiently. However, I'm wondering if there are any good techniques in the quadratic case, specifically if there is a polynomial time solution.

REPLY [18 votes]: There is an exact algorithm that needs $n^{O(m)}$ operations (cf. http://arxiv.org/abs/cs/0403008). One cannot expect anything better than that, unless P=NP. Indeed, it is easy to formulate several NP-complete problems as testing solvability of linear equations in 0-1 variables $x_i$, and the latter can be enforced by quadratic equations $x^2_i-x_i=0$.

REPLY [11 votes]: The quadratic case is as complex as the general case (up to a polynomial-time reduction). Given a set of polynomials $\{p_1,\dots,p_m\}$, you can express each $p_i$ by a straight-line program with instructions of the form $x_i:=c$ ($c\in\mathbb R$ a constant), $x_i:=x_j+x_k$, and $x_i:=x_j\cdot x_k$, which you can in turn translate back to quadratic polynomials (e.g., the last mentioned instruction becomes $x_i-x_jx_k$). The number of variables in the new system will increase (it is linear in the size of the original system), which suggests that you might get an efficient solution if you additionally restrict the number of variables by a constant.

REPLY [8 votes]: There have been written many books and articles on this subject (see for example the book
"Sturmfels, Bernd, Solving systems of polynomial equations. American Mathematical Society, Providence, RI, 2002").
It is not the fact that the polynomials are quadratic which helps, but rather other restrictions which will sometimes result in a more effective solution.
If the Groebner basis  of the corresponding ideal in $K[x_1,\ldots ,x_n]$
can be computed, and has triangular form, then we can usually solve the system.
However, Groebner bases have large exponential complexity and cannot solve in practice systems with, say, more than $15$ variables.
For some overdetermined systems there are other techniques than Groebner bases. One of these is called "relinearization". The exact complexity of this algorithm is not
known, but for sufficiently overdetermined systems it is expected to run in
polynomial time (see papers by Nicolas Courtois; Alexander Klimov, Jacques Patarin, and Adi Shamir).


TITLE: If $f$ is an injection, $f(S_1 \cap S_2) = f(S_1) \cap f(S_2)$
QUESTION [0 upvotes]: I need to prove that if $S_1$ and $S_2$ are subsets of a set $X$, and if $f: X \to Y$ is an injection, prove that $f(S_1 \cap S_2) = f(S_1) \cap f(S_2)$.
I know I need to show inclusion in both directions but not sure how to begin.

REPLY [1 votes]: A typical element in $f(S_1\cap S_2)$ is of the form $f(x)$ where $x\in S_1\cap S_2$. In particular, $x\in S_1$ and $x\in S_2$. So $f(x)\in f(S_1)$ since $x\in S_1$ and $f(x)\in f(S_2)$ since $x\in S_2$. Thus $f(x)\in f(S_1)\cap f(S_2)$. Since $f(x)$ was arbitrary in $f(S_1\cap S_2)$, this shows that $f(S_1\cap S_2)\subseteq f(S_1)\cap f(S_2)$.
To show that $f(S_1)\cap f(S_2)\subseteq f(S_1\cap S_2)$, take an element $y\in f(S_1)\cap f(S_2)$. Then there exists an element $x_1\in S_1$ such that $y=f(x_1)$ and there exists an element $x_2\in S_2$ such that $y=f(x_2)$. Since $f$ is injective, $y=f(x_1)=f(x_2)\implies x_1=x_2$. Put $x:=x_1=x_2$. Then $x\in S_1$ and $x\in S_2$ since $x=x_1\in S_1$ and $x=x_2\in S_2$. We have found an $x\in S_1\cap S_2$ such that $y=f(x)$. This means that $y\in f(S_1\cap S_2)$. The second inclusion is thus proved.

REPLY [0 votes]: $S_{1}\cap S_{2}\subseteq S_{i}$ so that $f\left(S_{1}\cap S_{2}\right)\subseteq f\left(S_{i}\right)$
for $i=1,2$ or equivalently $f\left(S_{1}\cap S_{2}\right)\subseteq f\left(S_{1}\right)\cap f\left(S_{2}\right)$.
Note that injectivity of $f$ has not been used for this part.
Let $y\in f\left(S_{1}\right)\cap f\left(S_{2}\right)$. Then $y=f\left(s_{1}\right)=f\left(s_{2}\right)$
for elements $s_{1}\in S_{1}$ and $s_{2}\in S_{2}$. Now use the
injectivity of $f$ to find out that...


TITLE: Confused as to how to compute this derivative
QUESTION [1 upvotes]: The question asks us to compute the derivative of the following:
$\frac{d}{dx}\frac{\left(x+10\right)^{10}}{\left(2x-4\right)^{10}}$
Now, by using the natural log and various rules related to that, I've arrived to:
$f\left(x\right)=10ln\left(x+1\right)-10ln\left(2x-4\right)$
What confuses me is the next step in my professor's notes. He taught this concept to us very quickly so I can't follow what he's doing. He ends up doing the below:
$\frac{1}{f\left(x\right)}\cdot f\:'\left(x\right)=\frac{10}{x+1}\left(1\right)-\frac{10}{2x-4}\left(2\right)$
Which cancels various things out. I don't know why he's multipled the LHS by $\frac{1}{f\left(x\right)}$ or the other terms by (1) and (2). Is he using other log rules for derivatives?

REPLY [1 votes]: $f(x) = ln(g(x))$ where $g(x) = {(x+10)^{10}\over(2x-4)^{10}}$
By the chain rule, $f'(x) = {g'(x)\over g(x)}$
Since $f'(x) = {10\over x+1} + 2{10\over2x-4}$ we get
$$
g'(x) = g(x) \left({10\over x+1} + 2{10\over2x-4} \right)
$$


TITLE: Why does the Gram-Schmidt procedure divide by 0 on a linearly dependent lists of vectors?
QUESTION [5 upvotes]: Let $v_1, \dots, v_m$ be a linearly dependent list of vectors.
If $v_1 \ne 0$, then there is some $v_j$ in the span of $v_1, \dots, v_{j-1}$
If we let j be the smallest integer with this property, and apply the gram-schmidt procedure to produce an orthonormal list $(e_1, \dots, e_{j-1})$ then $v_j$ is in the span of $(e_1, \dots, e_{j-1})$ and
$$v_j = \langle v_j, e_1\rangle e_1+ \dots + \langle v_j, e_{j-1}\rangle e_{j-1}$$
Why does this guarantee that length of $v_j$=0? I'm missing something about linear dependence that should probably be obvious sorry :\

REPLY [5 votes]: The Gram-Schmidt process goes as follows: given $v_1,\ldots,v_n$, you define
$$\begin{align*}
u_1 &= v_1\\
e_1 &= \frac{1}{||u_1||}u_1;\\
u_2 &= v_2 - \langle v_2,e_1\rangle e_1;\\
e_2 &= \frac{1}{||u_2||}u_2;\\
&\vdots\\
u_{k+1} &= v_{k+1} - \left(\langle v_{k+1},e_1\rangle e_1 + \langle v_{k+1},e_2\rangle e_2 + \cdots + \langle v_{k+1},e_k\rangle e_k\right)\\
e_{k+1}&= \frac{1}{||u_{k+1}||}u_{k+1}\\
&\vdots
\end{align*}$$
So when you construct $u_{j}$, you get $\mathbf{0}$, and when you try to construct $e_{j}$ you attempt to divide by $0$. It's not $v_j$ which has length $0$, it's $u_j$.

REPLY [2 votes]: It does not guarantee that.  What it guarantees is that $v_j - \langle v_j, e_1\rangle e_1- \dots - \langle v_j, e_{j-1}\rangle e_{j-1} = 0$, which means that you cannot define $e_j$ by dividing this vector by its norm.  Instead, you throw it away and move on to $v_{j+1}$.
E.g., consider the case $v_1\neq 0$ and $v_2=v_1$.


TITLE: History of "natural transformations"
QUESTION [23 upvotes]: (Edit #1 after Carlo's response)
It is often claimed that the notion of natural transformations existed in mathematical vocabulary long before it had a definition. In fact, I quoted the statement in italic from [1, p. 2]. As another example, in [2, p. 70] Ralf Kromer says: The claim is that there was, at the time when [3] was written, a current informal parlance consisting in calling certain transformations natural and that Mac Lane and Eilenberg tried (and succeeded) to grasp this informal parlance mathematically.
Three years after [3], Eilenberg and Maclane discovered the fact that this notion could be mathematically defined in [4].
However, Ralf Kromer casts doubt on the above mentioned claim, due to lack of evidence (see: [2, p. 70]).
(Edit #2 after Eric's comment) My question is, can you supply an evidence of use of the phrase "natural transformations" or its variants in mathematical literature prior to [3]? I must add that in [2] Kromer gives a number of examples of use of phrases such as "natural homomorphism" or "natural projection" in the literature prior to or around the same time as [3], but in each case they turn out to have different meanings. So I am asking for example(s) of use of the phrase "natural transformations", which are really natural transformations.
References:

Peter Freyd: Abelian Categories (1964).
Ralf Kromer: Tool and Object: a history and philosophy of category theory (2007).
Samuel Eilenberg and Saunders Maclane: Group extensions and homology, Annals Math. (2) 43, p.p. 757–831 (1942).
Samuel Eilenberg and Saunders Maclane: General theory of natural transformations, Trans. AMS, 58, p.p.: 231-294 (1945).

REPLY [2 votes]: The words "natural homomorphism" and "natural isomorphism" are also used (mainly in the context related to the First Isomorphism Theorem) in Pontryagin's "Topological groups" (Russian edition 1938, English translation 1939). Google books confirms my memory of this here.


TITLE: Probability theory inequality
QUESTION [1 upvotes]: In a set of notes I have seen the following inequality. Let $A$ and $B$ be non-negative random variables. Then
$$\mathbb{P}(A+B > \epsilon) \leq \mathbb{P}\left(A > \frac{\epsilon}{2} \right)+\mathbb{P}\left(B > \frac{\epsilon}{2} \right)$$
I do not have a probability theory background so this inequality is not obvious to me. How would one derive it?

REPLY [3 votes]: The key fact is that if $A + B > \epsilon$ then either $A > {\epsilon \over 2}$ or $B > {\epsilon \over 2}$. Then use that the probability of the union of two events is less than the sum of the probability of the two events.

REPLY [1 votes]: Just think in terms of sets. If you take $\omega \in \Omega$ such that :
$$ A(\omega) + B(\omega) > \varepsilon $$
Then you have either :
$$ A(\omega)  > \frac{\varepsilon}{2} \quad \quad \text{or} \quad \quad  B(\omega)  > \frac{\varepsilon}{2} $$
since $A$ and $B$ are two positive random variables. It means that :
$$ (A+B>\varepsilon) \subset \left(A>\frac{\varepsilon}{2} \right)\cup \left(B>\frac{\varepsilon}{2} \right) $$
And finally :
$$ \mathbb P(A+B>\varepsilon) \leqslant \mathbb P\left(A>\frac{\varepsilon}{2} \right)+\mathbb P\left(B>\frac{\varepsilon}{2} \right)$$


TITLE: Give an example of a relation R on $A^2$ which is reflexive, symmetric, and not transitive
QUESTION [2 upvotes]: I am just looking for some clarification on this exercise:
Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.
I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:
$R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$
Here $R$ is reflexive, symmetric, and not transitive.
What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.
Thanks,
Tony

REPLY [2 votes]: Based on this feed back I believe the answer is:
Let $A = \{a,b,c,d\}$,
$A^2 = A \times A$ = $\{(a,a),(a,b),(a,c),(a,d),(b,a),(b,b),(b,c),(b,d),(c,a),(c,b),(c,c),(c,d),(d,a),(d,b),(d,c),(d,d)\}$
A relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive is:
$R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$
R is reflexive because each element is related to itself.
R is symmetric because when one element is related to a second element that second element is related to the first.
R is not transitive because we have $bRa$ and $aRd$ but we do not have $bRd$.


TITLE: Lipschitz maps on $[0,1]$
QUESTION [3 upvotes]: Here is just a little curiosity. Assume the $f : [0,1] \to [0,1]$ is a Lipschitz function that maps 0 to 0 and 1 to 1. If we impose that the Lipschitz constant of $f$ is $\le 1$, can $f$ be anything other than the identity map?

REPLY [4 votes]: No. For $x\in[0,1]$ you get $f(x)=\lvert f(x)-f(0)\rvert\le\lvert x-0\rvert=x$ and $1-f(x)=\lvert f(1)-f(x)\rvert\le\lvert 1-x\rvert=1-x$, so $f(x)\ge x$. Thus $f(x)=x$.

REPLY [1 votes]: If $(x,y)\in[0,1]^2$ but is not on the diagonal line, then either the line from that point to $(0,0)$ or the line from that point to $(1,1)$ has slope more than $1$.


TITLE: Can I use the spacetime interval even if origins are not equal at $t=t'=0$?
QUESTION [2 upvotes]: Can I use the relativity formula $(\Delta s)^2 = (c \Delta t)^2 - (\Delta r)^2$ even when observers did not start at the same origin at $t=t'=0$? I know this is constant, but is it constant even if the other origins of observers were not the same at $t=t'=0$?

REPLY [2 votes]: Space-Time Interval: Suppose an observer measures two events as being separated in time by $\Delta t$ and a spatial distance $\Delta x$. Then the spacetime interval $(\Delta s)^2$ between the two events that are separated by a distance $\Delta x$ in space and by  $\Delta ct =c\Delta t$ in the $ct$-coordinate is:
$$(\Delta s)^2=(\Delta ct)^2-(\Delta x)^2$$
You can see not the absolute time but the time interval between two intervals is involved in the formula and So it doesn't matter whether or not $t=t'=0$ But both observers must measure the time interval between the same events.


TITLE: Question on notation: Using Conditional Sums to express the total distance traveled by a moving body with position $x=x(t)$
QUESTION [1 upvotes]: This is a question on notation
Can I express the total distance $s=s(t_1,t_2)$ that a moving body has traveled using the following (conditional) sum?
$$s(t_1,t_2) = \sum_{\forall [t_a,t_b] \subseteq [t_1, t_2]: x'(t) \neq 0} \big |x(t_a) - x(t_b) \big |$$
Note that the total space travelled is given by the summation of the absolute values of the displacements for every subspace of $[t_1,t_2]$ that the velocity has a constant sign.

REPLY [2 votes]: The presented notation (what is written under the $\sum$) is not standard, not self-explaining, and does not reflect your intention. What you mean is the following (it cannot be done without text): One has
$$s(t_1,t_2)=\sum_{k=1}^N|x(\tau_k)-x(\tau_{k-1})|$$
for any partition $${\cal P}:\quad t_1=\tau_0<\tau_1<\ldots<\tau_N=t_2$$
with $x'$ semidefinite on each subinterval $[\tau_{k-1},\tau_k]$.
Here it was tacitly assumed (as in your question) that $t\mapsto x(t)$ is real- (not vector-) valued.


TITLE: Quantum Decoherence
QUESTION [1 upvotes]: If I understand correctly, a quantum coherence is when two wave functions with same phase, frequency and period superimpose to become peaks or troughs.
In macroscopic scale, say a chair is there because by the time you saw it (measured) the wave functions collapsed and they don't longer cohere therefore leaving the object there. This happens in $10^{-31}$ seconds to be exact.
Basically, when we observe, wave functions decohere super fast, leaving us only one object to be seen. Is this a correct assumption?

REPLY [2 votes]: Let me clarify a couple of misconceptions in the question - perhaps this will be enough to answer it:

Coherence does not mean the same phase: it means that the two waves can interfere. Depending on the phase they may interfere constructively or destructively. We could take as an example the conventional light sources, such as light bulbs: they emit lots of waves with completely random phases, some of which interfere constructively, others destructively, and we observe some average. The quantum light sources, such as lasers and masers, emit coherent radiation, which explains their wonderful properties in terms of observing interference patterns, low beam divergence, transmitting energy on large distances, etc.
Interfering wave functions correspond to the states of the same object. Decoherence in this context means that we will observe only one of these states. The double-slit experiment is a good example: by trying to detect through which slit the light has passed the observer causes decoherence and destroys the interference pattern.


TITLE: Showing that $C^1[0,1]$ is a Banach space with the $||f||=||f||_\infty + ||f^\prime||_\infty$ norm.
QUESTION [8 upvotes]: So I am a bit stuck on where to begin with this one...
Show that $C^1[0,1]$ with the norm defined as
$\|f\|=\|f\|_\infty + \|f^\prime\|_\infty$ is a Banach space.
I started with an arbitrary Cauchy sequence, $f_n \to f$, so we know that for every $\epsilon > 0, \exists M \in \mathbb{N}$ such that $\|f_n - f\| \leq \epsilon , \forall n \geq M.$
Where I am getting stuck, is how do we then show that $f \in C^1[0,1]$, and that therefore $C^1[0,1]$ with the norm defined, is complete (and therefore a Banach space).

So here is my attempted solution, based on the comments below:
Assume $f_n$ is a Cauchy sequence in $C^1[0,1]$, then for every $\epsilon > 0, \exists M \in \mathbb{N}$ such that
$$\|f_n - f_m\|=\|f_n - f_m\|_\infty + \|f_n^\prime - f_m^\prime \|_\infty \leq \epsilon, \forall m,n\geq M.$$
So for any fixed $x \in C^1[0,1], \|f_n - f_m\|_\infty \leq \epsilon$ and $\|f_n^\prime - f_m^\prime \|_\infty \leq \epsilon$ and thus for this fixed $x$ $f_n$ and $f_n^\prime$ are Cauchy sequences, and therefore converge: $f_n \to f$ and $f_n^\prime \to f^\prime.$
So let $m \to \infty$ then $\|f_n - f\|\leq \epsilon, \forall n\geq M, \forall x\in [0,1]$. Then:
\begin{align}\|f_n - f\|&= \|f_n - f\|_\infty + \|f_n^\prime - f^\prime \|_\infty\\
&=\sup_{x\in [0,1]}|f_n(x)-f(x)| + \sup_{x\in [0,1]}|f_n^\prime (x)-f^\prime (x)|\\&\leq \epsilon\quad \forall n\geq M.
\end{align}
Therefore $\lim_{n\to \infty} \|f_n - f\| = 0 \implies$ $C^1[0,1]$ is complete.
How is this looking?

REPLY [6 votes]: Since $f_n,f'_n\in C([0,1])$ and both sequences converge in the sup-norm, there exist $f,g\in C([0,1])$ such that $$\|f_n-f\|_\infty \rightarrow 0 \text{ and } \|f'_n-g\|\rightarrow0.$$
It needs to be shown that $f$ is differentiable in $f'=g$. This means showing that $$\lim_{h\to 0}\frac{f(x+h)-f(x)-g(x)h}{h}=0 \ \forall x\in [0,1].$$
Fix $x_0\in [0,1]$. Let $\epsilon>0$ be given. Since $g$ is continuous, there exists $\delta>0$ such that $|x_0-x|<\delta$ implies $|g(x_0)-g(x)|<\frac\epsilon4$. Let $h$ be arbitrary but fixed, such that $|h|<\delta$. Choose $n$ such that $\|f_n-f\|_\infty<\frac{|h|\epsilon}{4}$ and $\|f'_n-g\|<\frac\epsilon4$. By the mean value theorem, there exists $\xi$ such that $|x_0-\xi|<|h|$ and $\frac{f_n(x_0+h)-f_n(x_0)}{h}=f'_n(\xi).$ It follows that
\begin{align}
\left|\frac{f(x_0+h)-f(x_0)-g(x_0)h}{h}\right|&\le
\left| \frac{f(x_0+h)-f_n(x_0+h)}{h}\right|+\\
&+\left|\frac{f_n(x_0+h)-f_n(x_0)}{h}-g(x_0)\right|+\left|\frac{f_n(x_0)-f(x_0)}{h}\right|\le\\
&\le\frac{\epsilon}{2}+\left|\frac{f_n(x_0+h)-
f_n(x_0)}{h}-g(x_0)\right|\le\\
&\le\frac\epsilon2+\left|\frac{f_n(x_0+h)-f_n(x_0)}{h}-g(\xi)\right|+\left|g(\xi)-g(x_0)\right|=\\
&=\frac\epsilon2+|f'_n(\xi)-g(\xi)|+|g(\xi)-g(x_0)|<\epsilon.
\end{align}
Since $h,x_0$ and $\epsilon$ were arbitrary, the conclusion follows.


TITLE: Evaluate $\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$
QUESTION [2 upvotes]: Evaluate :
$$\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$$

This is my attempt in below ... but what I want is simplify more to answer....thanks.
My own attempt:

REPLY [2 votes]: I really do not see how you could simplify the monster. I really wonder if a closed form expression could exist. I would try numerical integration.
We can get an approximate solution using the following series expansions at $x=0$.
$$\log (\Gamma (x))=-\log (x)-\gamma  x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1-x))=\gamma  x+\frac{\pi ^2 x^2}{12}-\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1+x))=-\gamma  x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$ which would make the integrand to be
$$\log ^4(x)+4 \gamma  x \log ^3(x)+4 \gamma ^2 x^2 \log ^2(x)-\frac{2}{3} x^3
   \left(\psi ^{(2)}(1) \log ^3(x)\right)-\frac{4}{3} x^4 \left(\gamma  \psi
   ^{(2)}(1) \log ^2(x)\right)+O\left(x^5\right)$$ and the antiderivative
$$x \left(\log ^4(x)-4 \log ^3(x)+12 \log ^2(x)-24 \log (x)+24\right)+\frac{1}{2}
   \gamma  x^2 \left(4 \log ^3(x)-6 \log ^2(x)+6 \log (x)-3\right)+\frac{4}{27}
   \gamma ^2 x^3 \left(9 \log ^2(x)-6 \log (x)+2\right)+\frac{1}{192} x^4 \psi
   ^{(2)}(1) \left(-32 \log ^3(x)+24 \log ^2(x)-12 \log (x)+3\right)-\frac{4}{375}
   x^5 \left(\gamma  \psi ^{(2)}(1) \left(25 \log ^2(x)-10 \log
   (x)+2\right)\right)+O\left(x^6\right)$$ Then, for the definite integral
$$\gamma  \left(\frac{16 \zeta (3)}{375}-\frac{3}{2}\right)-\frac{\zeta
   (3)}{32}+24+\frac{8 \gamma ^2}{27}\approx 23.2249$$ while the numerical integration leads to $23.2372$ that is to say in error by $0.05$%.


text:
\DOC AP_TERM_TAC

\TYPE {AP_TERM_TAC : tactic}

\SYNOPSIS
Strips a function application from both sides of an equational goal.

\KEYWORDS
tactic.

\DESCRIBE
{AP_TERM_TAC} reduces a goal of the form {A ?- f x = f y} by stripping away
the function applications, giving the new goal {A ?- x = y}.
{
    A ?- f x = f y
   ================  AP_TERM_TAC
     A ?- x = y
}

\FAILURE
Fails unless the goal is equational, with both sides being applications
of the same function.

\SEEALSO
ABS_TAC, AP_TERM, AP_THM_TAC, BINOP_TAC, MK_COMB_TAC.

\ENDDOC

subset_name: curated


TITLE: Set up integral in spherical coordinates outside cylinder but inside sphere
QUESTION [2 upvotes]: I have the equation of a cylinder and the equation of a sphere given:
Cylinder: $x^2+y^2=4$
Sphere: $x^2+y^2+z^2=25$
I'm asked to set this up in cylindrical and spherical coordinates. Cylindrical was easy enough, but I'm having some trouble figuring out the limits for spherical.
My first thought:
$\int\limits_0^{2\pi}\int\limits_0^\pi\int\limits_2^5\rho^2\sin\phi\ d\rho\ d\phi\ d\theta$
I thought that in spherical $\rho$ is basically the radius so setting it from 2 to 5 would exclude the area of the cylinder but apparently it is not that simple.  Below is the answer my teacher gave and I'm not exactly sure what she did.
Correct integral in spherical:
$2\int\limits_0^{2\pi}\int\limits_{cos^{-1}(\frac{\sqrt{21}}{5})}^{\pi/2}\int\limits_{2\csc(\phi)}^5\rho^2\sin\phi\ d\rho\ d\phi\ d\theta$
The picture when graphed is just a sphere with radius 5 and a cylinder that cuts through the middle.  I'm not sure how my teacher got limits like $cos^{-1}(\frac{\sqrt{21}}{5})$ or $2\csc(\phi)$

REPLY [2 votes]: The sphere and the cylinder intersect where $r=2$ and $\rho=5$, so $r=\rho\sin\phi\implies5\sin\phi=2\implies$
$\phi=\sin^{-1}\frac{2}{5}=\cos^{-1}\frac{\sqrt{21}}{5}$.
(Alternatively, $x^2+y^2=4$ and $x^2+y^2+z^2=25\implies$
$z^2=21\implies\rho^{2}\cos^{2}\phi=21\implies25\cos^{2}\phi=21\implies\cos\phi=\frac{\sqrt{21}}{5}\implies\phi=\cos^{-1}\frac{\sqrt{21}}{5}.)$
A line segment starting at the origin and passing through the cylinder intersects the cylinder where
$r=2$, so $\rho\sin\phi=2\implies \rho=\frac{2}{\sin\phi}=2\csc\phi$, so
this gives the lower limit for $\rho$.


TITLE: What is known about polyhedra nets that allow overlapping?
QUESTION [7 upvotes]: It is an open problem that the net of any convex polyhedron can be unfolded onto a flat plane with no overlapping. Is anything known if we allow x faces to overlap? For example, is it known if any convex polyhedron can be unfolded with a maximum of 2 faces overlapping? What about the more general case of a polyhedron that is topologically convex (that is, its graph is isomorphic to the graph of a convex polyhedron)?
This paper provides an example of an open polyhedron without a net that is topologically convex, and the closed case appears to be taken care of by this paper (mentioned in an answer below). One solution to the topologically convex case would then to be to find a procedure to modify either polygon so that the number of overlapping sides increases without bound. I have been unable to do so without breaking the topologically convex property, but it seems a reasonable task.

REPLY [1 votes]: I would like to point out a closely related question, which is described in our book that Joe Malkevitch mentioned (p.308).  We called it the Fewest Nets problem.  What is the fewest number of non-overlapping nets into which you can partition the surface of a convex polyhedron, cutting along polyhedron edges?  The answer might be 1.  But our ignorance leaves us with just fractions of F, the number of faces of the polyhedon.  The best fraction obtained to date (as far as I know) is (1/2)F.


TITLE: System of particles in classical mechanics and classical statistical mechanics
QUESTION [5 upvotes]: $\bullet$ Both classical mechanics and classical statistical mechanics can describe the properties of a system of classical particles.
$\bullet$ In classical statistical mechanics, we assume that we do not know the microstate of a system at any given time. Therefore, we consider all possible microstates at any time, an ensemble.
$\bullet$ In classical mechanics, we do not consider such an ensemble i.e., a large number of copies of the same system. We consider only one system, and still get some correct results.
Below are a few questions.

Do you assume that we know the microstate (i.e., position coordinates and momenta of each particle) for a system of particles in classical mechanics? But since that is not true, why does the classical mechanical description work? At least it describes some properties of the system if not all.
Why doesn't the notion of temperature arise in the discussion of system of particles in classical mechanics?
Why is it that some properties of a system of particles can be described by the classical mechanical approach i.e., no ensemble? Why are some properties cannot be described by classical mechanical approach i.e. use of ensembles and microstates?

For me the holistic picture is missing. If someone could help! I am not currently asking about quantum systems just to simplify matters.

REPLY [0 votes]: Ok to answer your first question:
in principle Classical Mechanics is capable of describing the system completely(at leas as long as we do not reach the scale of quantum world) the reason we use statistical mechanics is that when we increase the number of particles in a CM system thereby some new feature rise for instance some behavior that only depands on the interaction between the particles of our manybody system , also another difficulty is related to the chaos phenomena you can wikipedia that , and naivley speaking that occurs when we try to asymptotically use CM so these are some issues regarding to use CM that motivate people to use SM instead.
to your second question the notion of temperature arise when we talk about heat and again naively speaking heat is the energy in transfer not a specific kind of energy that most of the time it is related to the concept of friction so we use heat and therefore temperature when we do not want to specifically talking about the friction itself that one of the strength of SM that we substitute the difficult concept of friction with heat and temperature so thats why we dont talk about it in CM because there we do not seek for asymptotic answer but rather the exact one so we use friction itself .
and to your last question ensemble theory serve as a helpful tool when we cannot solve the system exactly and that is either because the equations are quite large that we do not have the computation power to do so so we use SM and ensemble theory to have some estimation of the answer the point is in principle CM is enough but because of the reasons I provide and so many others we use SM for estimation.


TITLE: Why are parabolic subgroups called "parabolic subgroups"?
QUESTION [51 upvotes]: Over the years, I have heard two different proposed answers to this question.

It has something to do with parabolic elements of $SL(2,\mathbb{R})$. This sounds plausible, but I haven't heard a really convincing explanation along these lines.
"Parabolic" is short for "para-Borelic," meaning "containing a Borel subgroup."

Which answer, if either, is correct?
A related question is who first introduced the term and when.  Chevalley perhaps?

REPLY [15 votes]: The naming (attributed by Borel as quoted by @GjergjiZaimi) happened quite publicly, in Roger Godement : Groupes linéaires algébriques sur un corps parfait, Sémin. Bourbaki 13 (1960/61), Exp. No. 206, 22 p. (1961). ZBL0119.27206, first page (my bold & link):

When $G_A/G_{\mathbf Q}$ is not compact, it is equally easy to conjecture that one must be able to define something like Poincaré’s classical “parabolic cusps”, which must correspond to nontrivial unipotent subgroups of $G_{\mathbf Q}$ (...) We shall, in this talk, define and study “parabolic subgroups” by methods of algebraic group theory.

Godement describes this at length in his Analyse mathématique IV, e.g. p. 441: in the Poincaré upper half-plane,

a parabolic cusp with vertex $\xi$ is the part of a horocycle with center $\xi$ comprised between two arcs of circle orthogonal to the real axis at $\xi$; see the figure (...)
$\hspace{9em}$


TITLE: Prove: Any prime of the form $3n+1$ is also in the form $6m+1$
QUESTION [2 upvotes]: Prove that any prime of the form $3n+1$ is also in the form $6m+1$.

So first I found an example... $3(2)+1=7$ and $6(1)+1=7 $. How do I go about a formal proof though?

REPLY [3 votes]: It's well known fact that every prime number, expect for $2$ and $3$ is of the form $6n \pm 1$. It's easy to prove using contradiction.
Assume the contrary. Let $p=6k-1=3n + 1$
Work modulo $3$ and you should get:
$$-1 \equiv 1 \pmod 3$$
A contradiction. Hence the proof.
Another way is to check both cases $p \equiv 0 \pmod 2$ and $p\equiv 1 \pmod 2$. Use Chinese remainder and you'll get for the first case that $p = 6k + 4$, clearly not a prime. Now you need to check the second case.
EDIT:
A little late, but here's the proof for the initial claim. Checking the remaining cases we have:
$$6k + 2 = 2(3k + 1)$$
$$6k + 3 = 3(2k + 1)$$
$$6k + 4 = 2(3k + 2)$$
Since $k$ is natural numbers it's clear that these numbers are composite. $6k$ is trivial. Hence the proof.


TITLE: Random Walk Probability in a Circle
QUESTION [5 upvotes]: Consider a circular arrangement of 8 people as shown below:

Let's say person 1 has a candy which is passed around with equal probability either to the left or right. All the other people do the same.
a) Find the probability that the person 3 gets the candy before person 6.
b) Find the expected number of times the candy will be passed before the person 5 gets it for the first time.
For a) I could just break the circle between 3 and 4 and consider them siting in a straight line as,
$3\hspace{10pt}2\hspace{10pt}1\hspace{10pt}8\hspace{10pt}7\hspace{10pt}6$
Now how do I go about calculating the probability? Do I take different cases? I think using conditional probability I could get some recurrence relation. Not sure how to go about b).
Any help is appreciated. Thanks!

REPLY [3 votes]: We can solve this using the theory of absorbing markov chains. It may be over-complicating the question a little but I think it's a useful and general tool worth using. For question a, the absorbing states will be 3 & 6 while state 4 & 5 have been removed. Our transition matrix will then be
$$
    P =
    \begin{pmatrix}
        0 & 1/2 & 1/2 & 0 & 0 & 0 \\
        1/2 & 0 & 0 & 0 & 1/2 & 0 \\
        1/2 & 0 & 0 & 1/2 & 0 & 0 \\
        0 & 0 & 1/2 & 0 & 0 & 1/2 \\
        0 & 0 & 0 & 0 & 1 & 0 \\
        0 & 0 & 0 & 0 & 0 & 1 \\
    \end{pmatrix}
    =
    \begin{pmatrix}
        Q & R \\
        0 & I_2
    \end{pmatrix}
$$
where our indices are in the order $(1,2,8,7,3,6)$.
We then find the fundamental matrix (using some CAS) which is
$$
    N = (I_4 - Q)^{-1} =
    \frac25
    \begin{pmatrix}
        6 & 3 & 4 & 2\\
        3 & 4 & 2 & 1\\
        4 & 2 & 6 & 3\\
        2 & 1 & 3 & 4\\
    \end{pmatrix}
$$
Finding the probability of being absorbed can be done just by examing the (1,1) entry in
$$
    B = NR =
    \frac15
    \begin{pmatrix}
        3 & 2 \\
        4 & 1 \\
        2 & 3 \\
        1 & 4 \\
    \end{pmatrix}
$$
So the probability of 3 getting the candy before 6 is $3/5$.
For question b, the method is similar but with a larger transition matrix and 5 as the only absorbing state. We are looking for the number of states visited before being absorbed when starting from position 1, which is just the sum of the first row of N.
If we order the states $(1,2,3,4,6,7,8,5)$ in $P$ then we will get
$$Q=
    \frac12
    \begin{pmatrix}
        0 & 1 & 0 & 0 & 0 & 0 & 1 \\
        1 & 0 & 1 & 0 & 0 & 0 & 0 \\
        0 & 1 & 0 & 1 & 0 & 0 & 0 \\
        0 & 0 & 1 & 0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 & 0 & 1 & 0 \\
        0 & 0 & 0 & 0 & 1 & 0 & 1 \\
        1 & 0 & 0 & 0 & 0 & 1 & 0
    \end{pmatrix}
$$
And then
$$
    N = (I_7 - Q)^{-1} =
    \frac14
    \begin{pmatrix}
        16 & 12 &  8 & 4 & 4 &  8 & 12 \\
        12 & 15 & 10 & 5 & 3 &  6 &  9 \\
        8  & 10 & 12 & 6 & 2 &  4 &  6 \\
        4  &  5 &  6 & 7 & 1 &  2 &  3 \\
        4  &  3 &  2 & 1 & 7 &  6 &  5 \\
        8  &  6 &  4 & 2 & 6 & 12 & 10 \\
        12 &  9 &  6 & 3 & 5 & 10 & 15
    \end{pmatrix}
$$
If we sum all values on the first row, because we are starting from position 1, we find that the expected number of times the candle is passed is 16.


TITLE: Is my proof correct of: If $A$ is a complete proper subspace of an inner product space $X$, then there exists a non-zero vector in $A^{\perp}$?
QUESTION [0 upvotes]: $A$ is complete iff $A$ is closed, then we have $X = A \oplus A^{\perp}$. Suppose $A^{\perp}=\{0\}$, then $X=A$ which is a contradiction?

REPLY [0 votes]: If $A$ is complete and $x \notin A$, then you can choose $\{ a_{n} \} \subset A$ such that $d_n=\|a_n-x\|$ converges to $d=\mbox{dist}(x,A)$. Then it follows that $\{ a_n \}$ is a Cauchy sequence. Because $A$ is complete, then $\lim_n a_n = a$ for some $a \in A$. Furthermore, $\|a-x\|=d$. Because $a$ is the closest element of $A$ to $x$, then $x-a \perp A$.


TITLE: How to factor 3 out of $10^n-1$
QUESTION [1 upvotes]: There is a statement $$3|10^n-1 \quad \quad  \textrm{for} \quad n \in \mathbb{N}:n \geq 0$$
that can be easily proven with mathematical induction.
However, if that number is divisible by $3$, we should be able to write $10^n-1$ as $3\cdot x$. And the question is how to do that. I have been toiling over that since weeks and I haven't found any solution. Even WolframAlpha doesn't show any other form of that equation, but I believe we must be able to factor $3$ out of that.

REPLY [5 votes]: $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots +x+1)$$ so $$(10^n-1)=3 \times 3 \times (10^{n-1}+10^{n-2}+\cdots +10+1)$$
This is really saying for example $$10000000000 - 1 = 9999999999 = 3\times  3333333333 = 3\times 3 \times 1111111111$$ and it does not matter how many repeated digits are involved