问:
Orthogonal Coordinates with Constant Magnitude Basis Vectors
I have been doing some work in curvilinear coordinates and I often encounter orthogonal coordinates (which are obviously super convenient). I am of course familiar with the usual suspects like cylindrical, spherical, and polar coordinates, but all of these coordinates have a particular thing in common that makes them less than ideal; their Jacobian determinant is something other than $1$, or specifically is nonconstant. Equivalently, we can say that the tangent vectors $\frac{\partial{\bf r}}{\partial y^i}$ are orthogonal, but not constant. It got me wondering about orthogonal transformations in which the Jacobian determinant is constant, or the coordinate tangent vectors are constant.

There are some obvious transformations: for $\mathbb{R}^n$, any transformation from ${\bf x}\to{\bf y}$ of the form

$${\bf y}={\bf x}_0+R{\bf x}$$

where $R$ is an orthogonal matrix is a valid coordinate transformation and the new coordinates are orthogonal coordinates. To show this, the basis vectors are

$$\frac{\partial{\bf r}}{\partial y^i}=\sum_{j=1}^n\frac{\partial {\bf r}}{\partial x^j}\frac{\partial x^j}{\partial y^i}=\sum_{j=1}^n\frac{\partial x^j}{\partial y^i}{\bf e}_j$$

where ${\bf e}_j$ is the $j^{th}$ standard basis vector $(0,...,1,...,0)$ with a $1$ in position $j$. The inverse of the transform is

$${\bf x}=R^T{\bf y}-R^T{\bf x}_0\implies\frac{\partial x^j}{\partial y^i}=R_{ij}$$

So

$$\frac{\partial {\bf r}}{\partial y^i}=\sum_{j=1}^nR_{ij}{\bf e}_j=R_i$$

where $R_{ij}$ is row $i$ column $j$ of $R$ and $R_i$ is the entire row $i$ of $R$. Since $R$ is an orthogonal matrix the rows are all orthogonal, these vectors are all orthogonal.

So it's clear that the given transformation creates orthogonal tangent vectors, since it amounts to a displacement and rotation. The question I have is whether this constitutes all such orthogonal coordinates.

The question is: suppose I have a new set of coordinates for $\mathbb{R}^n$ $(u^1,...,u^n)$ with the property that the vectors $\partial{\bf r}/\partial u^i$ are a collection of orthonormal vectors (enforcing normality for simplicity). Does this imply that these coordinates can be computed from a translation and a rotation from the standard coordinates?

I haven't really been able to show anything useful. Any thoughts are appreciated!

Edit: I can at least supply the information that I would know. In general, to have an orthonormal set of coordinates $(u^1,...,u^n)$, we require two things: orthogonality and normality, so

$$\frac{\partial{\bf r}}{\partial u^i}\cdot\frac{\partial{\bf r}}{\partial u^j}=\delta_{ij}$$

so

$$\sum_{k=1}^n\sum_{m=1}^n\frac{\partial x^k}{\partial u^i}\frac{\partial x^m}{\partial u^j}\frac{\partial{\bf r}}{\partial x^k}\cdot\frac{\partial{\bf r}}{\partial x^m}=\delta_{ij}$$

As we have $\partial{\bf r}/\partial x^k={\bf e}_k$, this is

$$\sum_{k=1}^m\frac{\partial x^k}{\partial u^i}\frac{\partial x^k}{\partial u^j}=\delta_{ij}$$

If $J$ is the Jacobian matrix for the transformation in the form

$$J=\left(\begin{array}{ccc}
\frac{\partial x^1}{\partial u^1} & \cdots & \frac{\partial x^n}{\partial u^1} \\
\vdots & \ddots & \vdots \\
\frac{\partial x^1}{\partial u^n} & \cdots & \frac{\partial x^n}{\partial u^n}
\end{array}\right)$$

This then means $JJ^T=I$, which makes sense: if the Jacobian matrix is orthogonal, then the coordinates remain orthogonal. The question then is: does there exist a nonconstant orthogonal Jacobian matrix that describes a valid coordinate system for $\mathbb{R}^n$?

Edit 2: Another thing I notice is that we can explicitly find the inverse of $J$ by noticing

$$\delta^i_j=\frac{\partial u^i}{\partial u^j}=\sum_{k=1}^n\frac{\partial u^i}{\partial x^k}\frac{\partial x^k}{\partial u^j}$$

which tells us that

$$J^{-1}=\left(\begin{array}{ccc}
\frac{\partial u^1}{\partial x^1} & \cdots & \frac{\partial u^n}{\partial x^1} \\
\vdots & \ddots & \vdots \\
\frac{\partial u^1}{\partial x^n} & \cdots & \frac{\partial u^n}{\partial x^n}
\end{array}\right)$$

Since we also know $J^T=J^{-1}$, this means

$$\frac{\partial x^i}{\partial u^j}=\frac{\partial u^j}{\partial x^i}$$

I don't know if this is helpful, but it's definitely an interesting relationship.

答: If the Jacobian is a rotation (an element of $SO(n)$ or $O(n)$), the pulled back metric is $R^T R=I$. So you have an isometry to the Euclidean space, which if you are mapping from nD to nD, means that it can only be a global rotation and translation.


问:
Rigorous book on probability theory and statistics
I would like some examples of very rigorous books on probability and statistics. I know there are lots of rigorous books on probability by itself, but I am also looking for a book that makes rigorous the theory behind statistics. From my understanding, there is a difference between probability theory and statistics.</p>

答: E. T. Jaynes is very thoughtful and addresses such matters and properly highlights the role of Bayesian probability and statistics.</p>

ContentLicense: CC BY-SA 4.0


问:
Integration but with both the transformation and differential forms of a function
Let $\int f(x) dx$ be $F(x)+C$, evaluate:
$$\int^t_0 f(x+T)F(t-x)dx$$

Where $T$ is a constant. Integration by parts come to mind but how do I deal with the additional terms inside the two functions?

Edit:
What if we let $F(x)=1-e^{-(\frac{x}{n})^b}$, where $n$ and $b$ are additional constants.

答:
Since $x+T\neq \pm (t-x)$, it is impossible to find a result without an integration sign in it. Let's do the calculations:
$$
\begin{aligned}
&\int_0^t{f}\left( x+T \right) F\left( t-x \right) dx
\\
&=\int_T^{t+T}{f}\left( x+T \right) F\left( t-x \right) d\left( x+T \right)
\\
&=\int_T^{t+T}{F\left( t-x \right) dF\left( x+T \right)}
\\
&=F\left( t-x \right) F\left( x+T \right) \mid_{T}^{t+T}-\int_T^{t+T}{F\left( x+T \right) dF\left( t-x \right)}
\\
&=F\left( t+2T \right) F\left( -T \right) -F\left( t-T \right) F\left( 2T \right) +\int_T^{t+T}{f\left( t-x \right) F\left( x+T \right) dx}
\end{aligned}
$$
And finally we cannot eliminate anything to simply further.


问:
Identity map with change of basis
Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$

and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the standard basis $(e_1,e_2,e_3)$ for the codomain

Question

Find the matrix that represents $Id:K^3 \rightarrow K^3$ with respect to the given basis for the domain and the standard basis $(e_1,e_2,e_3)$ codomain.

I have found the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain but this identity map is really difficult for me to get.

答: Hint: Given any linear transformation, the columns of the matrix that represents that map are exactly the images of the basis vectors of the domain, expressed in the basis of the codomain.</p>

Title: Identity map with change of basis


问:
how to tell whether x and y are independent or not
Suppose that $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$ Find $\operatorname{Var(X+Y)}$.

I'm having trouble with this problem the way to find $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+2\operatorname{Cov(X,Y)}$, however if $X$ and $Y$ are independent, then $\operatorname{Cov(X, Y)}=0$, the answers indicated that $X$ and $Y$ are independent since they just used $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+0$, my question is how do I tell whether $X$ and $Y$ are independent or not, based on looking only at $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$

答:
If you can factor the joint density into a product that is a function of x times a function of y, then $X$ and $Y$ are independent and their marginal densities are a constant multiple of the two functions in the product.  That is, if
$$f_{X, Y}(x, y) = g(x) h(y)$$
for some functions $g(x)$ and $h(y)$, then $X$ and $Y$ are independent and
$$f_X(x) = c g(x)$$
and
$$f_Y(y) = \frac{1}{c} h(y),$$
for some constant $c$.  In fact, this statement is an if and only if.  And, in your specific problem, you can write
$$f_{X, Y}(x, y) = (\lambda e^{-\lambda x}) \cdot (\lambda e^{-\lambda y}) = g(x) h(y)$$
so $X$ and $Y$ are independent.  That doesn't tell us what $f_X(x)$ or $f_Y(y)$ are except they are constant multiples of $g(x)$ and $h(y)$, respectively.  I broke it up like I did to be symmetrical and because each function would be the pdf of an exponential.

In this case, you can actually figure out the marginal densities with a bit of thought.  Since we know an exponential has density $\lambda e^{-\lambda x}$ we know that integrates to 1.  So, if we took away the $\lambda$, i.e., $e^{-\lambda x}$, it would integrate to $\frac{1}{\lambda}$.  Or, if we had any other constant $c$ in front, $c e^{-\lambda x}$, it would integrate to $\frac{c}{\lambda}$.  So, we must have $c = \lambda$ in front to ensure the PDF integrates to 1.  So, the $g(x)$ and $h(y)$ I picked above must in fact be the marginal PDFs for $X$ and $Y$.  Thus, if you are very familiar with some of the major densities, you can figure these out without ever integrating to get the marginal densities.  It's just factoring.

ContentLicense: CC BY-SA 3.0


问:
Is this simple proof of Archimedes' Property valid?
A simple proof of Archimedes' Property (of analysis)?

I'm working an example given on page 16 of Shilov's "Real and Complex Analysis," which is his proof of Archimedes' Property:

Given arbitrary real numbers $x > 0$ and $y, \exists n\in \mathbb{Z}$ such that $(n-1)x \le y \lt nx.$

He provides a proof which appears to be correct, but is this simpler proof valid?

Proof:
For each $n \in \mathbb{Z}, $ define the half-open interval $I_n:= [(n-1)x,nx).$ This partitions the real line, so since $y\in \mathbb{R}, y $ belongs to one of the $I_n.$ Thus $(n-1)x \le y \lt nx.$

答:
The claim

This partitions the real line

needs justification, and in fact is what you're trying to prove in disguise. So that argument doesn't work - something more complicated is needed. You need to use properties of $\mathbb{R}$ in a nontrivial way.

To elaborate on the previous sentence, there are such things as non-Archimedean ordered fields - structures where addition, subtraction, multiplication, and division make sense and behave basically as we expect them to, but which do not satisfy the Archimedean property. The crucial property of $\mathbb{R}$ is its completeness (and in fact $\mathbb{R}$ is the unique-up-to-isomorphism complete ordered field).


问:
What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/

答:
Other answers — particularly Clarinetist’s — give excellent rundowns of the most important side of the answer. Given a random variable, we can sample it, and use the sample mean (defined in the statistical sense) to estimate the actual mean of the random variable (defined in the sense of probability theory), and similarly for variance, etc.

But the connection in the other direction doesn’t seem to have been mentioned yet.  This is not as important, but it’s much more straightforward, and worth pointing out.  Given a sample, i.e. a finite multiset of values $\{x_i\}_{i \in I}$, we can “consider this as a distribution”, i.e. take a random variable $X$, with value $x_i$ for $i$ distributed uniformly over $I$.  Then the mean, variance, etc. of $X$ (in the sense of probability theory) will be precisely the mean, variance, etc. of the original multiset (defined in the statistical sense).


问:
Is it correct to define $\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega$?
In my differential geometry class the professor wrote $$\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega, \tag{1}$$
where $\operatorname{cl}$ is closure.
From Topology I recall it was $$\operatorname{cl}(\Omega) := \operatorname{Int}(\Omega) \cup \partial \Omega \tag{2},$$
where $\operatorname{Int}$ is Interior.

Is $(1)$ also correct? If it is not , what is missing in that union?
I feel the isolated points may not be taken into consideration in $(1)$.

Furthermore he said that historically the $\partial \Omega$ notation comes from the topological concept of derived set.  But I don't think  that the boundary is the same as the derived set, is it?

Edit: I am attempting a proof, taking $(2)$ as definition:

$$\operatorname{Int}(\Omega) \subseteq \Omega \implies \operatorname{Int}(\Omega) \cup\partial\Omega \subseteq \Omega\cup \partial\Omega  $$

$$\implies \operatorname{cl}(\Omega) \subseteq\Omega\cup \partial\Omega $$

But don't know how to prove the oposite inclusion, any idea?

答:
Proposition

Let $(X,d)$ be a metric space, let $E$ be a subset of $X$, and let $x_{0}$ be a point of $X$. Then the following statements are logically equivalent

(a) $x_{0}$ is an adherent point of $E$.

(b) $x_{0}$ is either an interior point or a boundary point of $E$.

(c) There exists a sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to $d$.

Proof

Let us start by the implication $(a)\Rightarrow(c)$.

If $x_{0}$ is an adherent point of $E$, then for every $n > 0$ there corresponds a $x_{n}\in B(x_{0},1/n)\cap E$.

Consequently, since $d(x_{n},x_{0}) \leq 1/n$, the sandwich theorem ensures us that $E\ni x_{n}\to x_{0}$, and we are done.

Now, we are going to prove the implication $(c)\Rightarrow(b)$.

In order to do so, we are going to assume that $(c)$ is true and $(b)$ is false.

Since $x_{0}$ is an exterior point of $E$, there exists a $\varepsilon_{0} > 0$ such that $B(x_{0},\varepsilon_{0})\cap E = \varnothing$.

On the other hand, since $x_{n}\in E$ converges to $x_{0}$, for the same value $\varepsilon_{0} > 0$ there is a $n_{\varepsilon_{0}}\in\mathbb{N}$ s.t.
\begin{align*}
n\geq n_{\varepsilon_{0}} \Rightarrow d(x_{n},x_{0}) \leq \varepsilon_{0} \Rightarrow x_{n}\in B(x_{0},\varepsilon_{0})\cap E
\end{align*}
which contradicts our assumption. Hence $(c)$ implies $(b)$.

Finally, we are going to prove the implication $(b)\Rightarrow(a)$.

If $x_{0}$ is an interior point of $E$, then there exists an open ball $x_{0}\in B(x_{0},r)\subseteq E$.

Consequently, $x_{0}$ is an adherent point of $E$.

Indeed, for every positive number $r > 0$, one has that $x_{0}\in B(x_{0},r)\cap E$.

On the other hand, if $x_{0}$ is a boundary point of $E$, then it is neither an interior point nor an exterior point.

This means that every open ball $B(x_{0},r)$ is not contained in $E$ nor $E^{c}$.

More precisely, every open ball $B(x_{0},r)$ intersects $E$ and $E^{c}$.

Hence we conclude that $x_{0}$ is an adherent point of $E$, and we are done.

Solution

Since $\overline{\Omega}\supseteq\Omega$ and $\Omega\supseteq\text{int}(\Omega)$, we can conclude based on the previous result that
\begin{align*}
\overline{\Omega} = \overline{\Omega}\cup\Omega &= \bigl(\text{int}(\Omega)\cup\partial\Omega\bigr)\cup\Omega =\\
&=\bigl(\text{int}(\Omega)\cup\Omega\bigr)\cup\partial\Omega =\\ &=\Omega\cup\partial\Omega
\end{align*}
and we are done.

Hopefully this helps!


问:
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

$x,y,z >0$, prove
  $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$

Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.

Update 1: In this forum, somebody said  that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.

Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.

Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.

Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".

答:
We want to prove the inequality
$$
\sum_{cyc}\frac{x^4}{8x^3+5y^3}\geq \frac{\sum_{cyc}x}{13}\tag 1
$$
where $x,y,z>0$.

Set
$$
\Pi(x,y,z):=\sum_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}
$$
Then easily
$$
\Pi(x,y,z)= \sum_{cyc}\left(x\frac{x^3}{8x^3+5y^3}-\frac{x}{13}\right)=\sum_{cyc}x\frac{13x^3-8x^3-5y^3}{13(8x^3+5y^3)}=
$$
$$
=\frac{5}{13}\sum_{cyc}x\frac{x^3-y^3}{8x^3+5y^3}.
$$
Also (ecxept from the symetry) $\Pi(x,y,z)$ is homogeneous of degree 1. Hence $\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$.

The case $x\geq y\geq z$

If $x\geq y\geq z$, we can set $\lambda=1/z$, then
$$
\Pi(x,y,z)=\lambda^{-1}\Pi(\lambda x,\lambda y,\lambda z)=z^{-1}\Pi\left(\frac{x}{z},\frac{y}{z},1\right).
$$
Set $a=\frac{x}{z}$, $b=\frac{y}{z}$, then also $a\geq b\geq 1$ and
$$
\Pi(x,y,z)=z\Pi(a,b,1)=
$$
$$
=z\left(\frac{1-a^3}{8+5a^3}+\frac{b(-1+b^3)}{5+8b^3}+\frac{a^4-ab^3}{8a^3+5b^3}\right)\textrm{, }a\geq b\geq 1\tag 2
$$
Set now
$$
f(x,y)=\frac{x}{8x^3+5 y^3}.
$$
Then (2) becomes
$$
a^3\left(\frac{a}{8a^3+5b^3}-\frac{1}{8+5a^3}\right)+b^3\left(\frac{b}{5+8b^3}-\frac{a}{8a^3+5b^3}\right)-
$$
$$
-\left(\frac{b}{8b^3+5}-\frac{1}{8+5a^3}\right)\geq 0
$$
Hence equivalent we must show that
$$
a^3(f(a,b)-f(1,a))+b^3(f(b,1)-f(a,b))-(f(b,1)-f(1,a))\geq 0\Leftrightarrow
$$
$$
a^3f(a,b)+f(1,a)(1-a^3)\geq b^3f(a,b)+f(b,1)(1-b^3)\Leftrightarrow
$$
$$
f(1,a)(a^3-1)-f(b,1)(b^3-1)\leq (a^3-b^3)f(a,b)\tag 3
$$
But when $a\geq b\geq 1$, setting  $k=a^3-1$ and $l=b^3-1$, we have $k\geq l\geq 0$. Then (3) becomes
$$
k\frac{1}{5a^3+8}-l\frac{b}{8b^3+5}\leq (k-l)\frac{a}{8a^3+5b^3}\Leftrightarrow
$$
$$
\frac{k}{5k+13}-\frac{lb}{8l+13}\leq (k-l)\frac{a}{8k+5l+13}\Leftrightarrow
$$
$$
\frac{k}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{b}{y_1}\leq \frac{a}{8(x_1-13)/5+5(y_1-13)/8+13}
$$
Or equivalent (we set $x_1=5k+13$, $y_1=8l+13$) if:
$$
\small
f_0(x_1,y_1):=5ky_1^2/8-8blx_1^2/5+x_1y_1(al-ak+8k/5-5bl/8)-637ky_1/40+637bl x_1/40\leq 0.\normalsize
$$
This last inequality is true, as one can see using $1\leq b\leq a$ and $k\geq l\geq 0$, $k=a^3-1$, $l=b^3-1$, $x_1\geq 13$, $y_1\geq 13$. Actualy it holds
$$
f_0(x_1,y_1)<f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty)
$$
where
$$\small
x_0=-\frac{637 k (8 (5 a-8) k-5 l (8 a+5 b))}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\normalsize
$$
$$\small
y_0=\frac{637 b l (5 l (5 b-8 a)+8 (5 a+8) k)}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\normalsize
$$
are the points such that $\partial_xf_0(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$ and
$$\small
f_0(x_0,y_0)=-\frac{405769 a b k l (k-l)}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\leq 0\normalsize
$$
and $64 (8 - 5 a)^2 k^2 - 80 (8 a (-8 + 5 a) - 5 (8 + 5 a) b) k l +
 25 (8 a - 5 b)^2 l^2>0$, $a>1$, $a\geq b\geq 1$. But when $x_1\geq 13$ and $y_1\geq 13$, we have $f_0(x_1,y_1)\leq 0$.

About the case $z\geq y\geq x$

If $z\geq y\geq x$, then we have $\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$. Set $\lambda=1/x$ and $a=z/x$, $b=y/x$, $a\geq b\geq 1$. Then
$$
\Pi(x,y,z)=x^{-1}\Pi(1,b,a)=
$$

$$
b^3\left(\frac{b}{8b^3+5a^3}-\frac{1}{8+5b^3}\right)+a^3\left(\frac{a}{5+8a^3}-\frac{b}{8b^3+5a^3}\right)-
$$
$$
-\left(\frac{a}{8a^3+5}-\frac{1}{8+5b^3}\right)\geq 0\tag 5
$$
Set
$$
f(x,y)=\frac{x}{8x^3+5y^3},
$$
then (5) becomes
$$
b^3(f(b,a)-f(1,b))+a^3(f(a,1)-f(b,a))-f(a,1)+f(1,b)\geq 0\Leftrightarrow
$$
$$
(a^3-1)f(a,1)-(b^3-1)f(1,b)\geq (a^3-b^3)f(b,a).
$$
Set now $k=a^3-1$ and $l=b^3-1$. Then $k\geq l\geq 0$. Then
$$
k\frac{a}{8 a^3+5}-l\frac{1}{8+5 b^3}\geq (k-l)\frac{b}{8b^3+5a^3}\Leftrightarrow
$$
$$
\frac{ka}{8k+13}-\frac{l}{5 l+13}\geq (k-l)\frac{b}{8k+5l+13}\Leftrightarrow
$$
$$
\frac{ka}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{1}{y_1}\geq \frac{b}{x_1+y_1-13}
$$
Or equivalent ($x_1=8k+13$, $y_1=5l+13$):
$$
f_0(x_1,y_1):=aky_1^2-lx_1^2+x_1y_1(bl-l-bk+ak)-13aky_1+13l x_1\geq 0.
$$
This last inequality is true and as one can see using $1\leq b\leq a$ and $k\geq l\geq 0$, $k=a^3-1$, $l=b^3-1$, $x_1\geq 13$, $y_1\geq 13$. Actualy holds
$$
f_0(x_1,y_1)>f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty)
$$
where
$$
x_0=\frac{13ak(ak+l+b(l-k))}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}
$$
$$
y_0=\frac{13l((a+b)k+l-lb)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}
$$
are the points such that $\partial_xf(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$ and
$$
f_0(x_0,y_0)=\frac{169abkl(l-k)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}\leq 0
$$
and $a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))>0$, $a>1$, $a\geq b\geq 1$. But when $x_1\geq 13$ and $y_1\geq 13$, we have $f_0(x_1,y_1)\geq 0$. QED


问:
Suppose $|A|=n, f:A\to A$ is injective $\implies \exists k\in [n]:f^k(x)=x$

My Attempt

Case 1

Suppose $f:A \to A$ is an identity function. then $k=1\in [n]:f(x)=x$.

Case 2

Suppose $f:A \to A$ is an injective function and non identity function.

so, there exists $x\in A$ such that $f(x) \neq x.$
So, there are $n-1$ possibilities for $f(x).$
if $f(f(x))=x, $ then $k=2.$
so on. I am not able to complete the proof. Is my method correct?

答:
For any function $f$ that maps a finite set $A$ to itself, and any initial value $x_0$ in S, the sequence of iterated function values

$$x_{i+1}=f(x_i)$$

must eventually use the same value twice: there must be some pair of distinct indices $i$ and $j$ such that $x_i = x_j$.  One way to visualize this is to construct a graph and a directed arc exists from $i$ to $j$ to $x_j = f(x_i)$, eventually some node must be repeated since there are only finitely many nodes. When we construct such graph, in general, we might get  cycle part plus a chain part that is not in the cycle in general as the cycle need not start at $x_0$. Let $i^*$ be the index where the cycle begins.

Suppose $i^*$ is not equal to $0$ and the period of the cycle is $p$, then we have two preimage of $x_{i^*}$, that is $x_{i^*-1}$ and $x_{i^*-p-1}$, this would violate the injective assumption and hence we conclude that we do not have a chain that is not in the cycle. Hence $S_a = \{f^{k}(x_0): k \in \mathbb{N}\}$ is a cycle.

Hence, we can conclude that there is such a $k$ such that $f^{(k)}(a)=a$ where we just have to pick $k$ to be the period of the cycle.

If we want to prove that a common $k$ can be chosen for all elements, we just have to pick the least common multiple of all the periods of the cycles.

Here is the wikipedia page of Cycle Detection Reference.


答:
[This is, at bottom, the answer Daniel gave in the comments]

First note that, since $f$ is injective, there is a function $g:A\to A$ such that for all $x$ in $A$, $(g\circ f)(x)=x$. [In fact, since $A$ is finite, $g$ is actually the inverse function to $f$, but all we need is that $g$ is a left inverse to $f$.]

Now given $a$ in $A$, consider the $n+1$ terms, $a,f(a),f(f(a)),f^{(3)}(a),\dots,f^{(n)}(a)$. These are all elements of the $n$-element set $A$, so two of them must be equal. Let $f^{(i)}(a)=f^{(j)}(a)$ for some $i<j$. Applying $g$ $i$ times, we get $g^{(i)}\circ f^{(i)}(a)=g^{(i)}\circ f^{(j)}(a)$, so $a=f^{(k)}(a)$, where $k=j-i$, and we're done.

Title: Suppose $|A|=n, f:A\to A$ is injective $\implies \exists k\in [n]:f^k(x)=x$

Body:
My Attempt

Case 1

Suppose $f:A \to A$ is an identity function. then $k=1\in [n]:f(x)=x$.

Case 2

Suppose $f:A \to A$ is an injective function and non identity function.

so, there exists $x\in A$ such that $f(x) \neq x.$
So, there are $n-1$ possibilities for $f(x).$
if $f(f(x))=x, $ then $k=2.$
so on. I am not able to complete the proof. Is my method correct?


问:
Is $R = \{(1,1),(2,2)\}$ a reflexive relation on $A = \{1,2,3,4,5\}$?
Suppose I have a set $\mathrm{A}=\{1,2,3,4,5\}$ and a reflexive relation $\mathrm R$ defined on $\mathrm A$, i.e.,

$$R\colon A\mapsto A\quad\textrm{and}\quad \mathrm R\textrm{ is reflexive.}$$

Is the relation quoted below a possible candidate for being $\mathrm{R}$ ?

$$\mathrm R=\{(1,1),(2,2)\}$$

My reasoning:

Since the definition of a reflexive relation $\mathrm R$ on $\mathrm A$ is that $a\mathrm Ra~\forall~a\in\mathrm A$, we have that the given relation is not reflexive on $\mathrm{A}$ and hence not a possible candidate for being $\mathrm R$. The main fact used here is that $3,4,5\in\mathrm A$ but $\require{cancel} a\cancel {\mathrm R} a$ for $a=3,4,5$

答:
Your answer is correct, and there's nothing really further to comment on. $R$ would be reflexive if and only if $(a,a) \in R$ for every $a \in A$; however, since that doesn't hold for $a=3,4,5$, $R$ is not a reflexive relation on $A$.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Mistake in Hatcher
A $1$-dimensional cell complex, $X$, is called a $\textit{graph}$.  A subset of $X$ is closed iff its intersection with the closure $\overline{e_\alpha}$ of each edge $e_{\alpha}$ ($1$-dimensional cell) is closed in $\overline{e_\alpha}$.  So every sequence in $X$ of points belonging to the interiors of disjoint edges is closed and doesn't converge.  This is correct but then Hatcher makes the following incorrect claim.  "This cell complex topology cannot be a metric topology if $X$ contains a vertex that is a common vertex of infinitely many edges; otherwise take a sequence of points in the disjoint interiors of these edges that gets closer and closer to the common vertex."  The common vertex is not in the closure of this sequence and hence there is no contradiction with the fact that, in a metric space, every point in the closure of a set is the limit of a sequence of points in the interior of the set.

答:
Hatcher's point is correct. Suppose $\alpha$ is a vertex common to infinitely many edges $E_i$ ($i\in\mathbb{N}$), and suppose $d$ were a metric on $X$ which induced the given topology. Let $x_i$ be a point on $E_i$ with $d(x_i, \alpha)<2^{-i}$.

On the one hand, we know by the definition of the topology on $X$ that $\{x_i:i\in\mathbb{N}\}$ is closed, and so $\alpha$ is not a limit point of $\{x_i:i\in\mathbb{N}\}$. On the other hand, for every $\epsilon>0$ there is some $i\in\mathbb{N}$ with $d(x_i, \alpha)<\epsilon$, so $\alpha$ is a limit point of $\{x_i:i\in\mathbb{N}\}$.

This gives us a contradiction, and so our original assumption - that $X$ is metrizable - was wrong.

Title: Mistake in Hatcher


答: Hatcher is correct. The contradiction is that in the CW topology no such sequence of points in the interior of distinct edges converges, but if we had a metric topology we could pick a point on the interior of the nth edge that was distance $1/n$ away from the common vertex. Since the distance between the nth point and the common vertex goes to 0, this converges which contradicts our earlier statement.

Title: Mistake in Hatcher

Body: A $1$-dimensional cell complex, $X$, is called a $\textit{graph}$.  A subset of $X$ is closed iff its intersection with the closure $\overline{e_\alpha}$ of each edge $e_{\alpha}$ ($1$-dimensional cell) is closed in $\overline{e_\alpha}$.  So every sequence in $X$ of points belonging to the interiors of disjoint edges is closed and doesn't converge.  This is correct but then Hatcher makes the following incorrect claim.  "This cell complex topology cannot be a metric topology if $X$ contains a vertex that is a common vertex of infinitely many edges; otherwise take a sequence of points in the disjoint interiors of these edges that gets closer and closer to the common vertex."  The common vertex is not in the closure of this sequence and hence there is no contradiction with the fact that, in a metric space, every point in the closure of a set is the limit of a sequence of points in the interior of the set.


问:
How can i define the width of a curve (RS)
I would like to determine the width of the curve - as suggested in this paper on page 133

paper

I am very confused, since this doesnt look like a Gaussian Curve... what is this RS? Can someone explain it to me in simple terms?

Best Regards and many thanks

答:
The "relative spread" appears to be obtained by fitting a Normal($\mu,\sigma^2$) distribution to the sigmoid curve, then computing the ratio of $\sigma$ to $\mu$.

Here is another paper that discusses relative spread. Equation (3) in that paper is
$$p(n)=\frac12\left(1+\operatorname{erf}\left(\frac{V_{\rm stim}(n)-V_{\rm thr}(n)}{\sqrt 2\sigma}\right)\right).\tag3$$
Using the identity $\operatorname{erf}(x)=2\Phi(\sqrt 2 x)-1$ where $\Phi$ is the standard normal cumulative, equation (3) is the same as
$$p(n)=\Phi\left(\frac{V_{\rm stim}(n)-V_{\rm thr}(n)}{\sigma}\right).$$ So viewed as a function of $V_{\rm stim}$, this is the cumulative distribution function of a normal distribution with mean $\mu=V_{\rm thr}$ and standard deviation $\sigma$.
The authors of that paper define the dimensionless quantity
$$
\operatorname{RS}:=\frac{\sigma}{V_{\rm thr}(n)}\tag4$$ which is the same as $\sigma/\mu$. In other words, to compute this version of RS, you fit a Normal($\mu,\sigma^2$) distribution to your sigmoid, then define ${\rm RS}:=\sigma/\mu$.

The paper that you are looking at omits the factor of $\sqrt 2$.
If that's not a typo, it seems there is disagreement about the exact definition of RS. At any rate we can say RS is proportional to $\sigma/\mu$.

FYI, here is one of the earliest papers (1960) that discusses relative spread. This paper (1963), by the same author, explicitly states that relative spread is the coefficient of variation.

EDIT: Regarding your comment about estimating the RS from the "linear part" of the sigmoid: Yes, this approach gets you within a factor of RS. Specifically, if you construct the tangent line to the sigmoid
$$
F(x):=\Phi\left(\frac{x-\mu}\sigma\right)
$$
at $x=\mu$, and extend the tangent line to intersect the $x$-axis at the point $(x_0,0)$, then the distance $\mu-x_0$ from the $x$-intercept $x_0$ to the threshold $\mu$, after dividing by $\mu$, is proportional to RS.

Here is the calculation: The equation of the tangent line to $F$ through the point $(\mu,0.5)$ is
$$
\frac{y-0.5}{x-\mu} = F'(\mu) = \frac1{\sigma{\sqrt{2\pi}}}.\tag a
$$
Set $y=0$ in (a) to find that the $x$-intercept $x_0$ satisfies
$$
\mu-x_0 = 0.5\sigma\sqrt{2\pi}
$$
and therefore the ratio
$$\frac{\mu-x_0}\mu=\sqrt{\frac\pi2}\frac\sigma\mu\approx 1.2533\frac\sigma\mu\tag b$$
is about 25% higher than the coefficient of variation. To recover exactly the coefficient of variation, multiply (b) by $\sqrt{\frac2\pi}$.


问:
How good can a "near-miss" polyomino packing be?
Given a polyomino $P$ with $n$ cells, we can ask about its maximal packing density $\delta_P$ in the plane (perhaps the limsup if we are concerned about convergence issues, though I don't think this actually comes up).

If $\delta_P=1$, then $P$ tiles the plane: having arbitrarily good packings implies that $P$ can cover arbitrarily large $N\times N$ squares, and from there the result follows via a compactness argument.

We can then ask: among polyominoes on $n\ge 7$ cells that do not tile the plane, which one achieves the highest density? Call this maximal value $\Delta_n$. For each $n$, $\Delta_n<1$, though of course in the limit for large $n$ it approaches $1$.

As a simple lower bound, we always have $\Delta_n\ge n/(n+1)$, as exhibited by the polyomino given by taking all but the first square in a spiral of length $n+1$ around the origin; once we plug the hole, it tiles the plane without gaps, but the hole cannot be filled. For instance, with $n=7$:

In contrast, putting remotely nontrivial upper bounds on $\Delta_n$ in general should be exceedingly difficult or impossible, since any computable upper bound less than 1 would yield an algorithm to decide the tiling problem for polyominoes, which is suspected not to exist. (Non-computable upper bounds on the order of $1-1/\text{BB}(k\cdot n)$ can be done, of course, but these are rather silly.)

However, improved lower bounds and exact values for small $n$ seem pretty tractable, and I'm curious what is known in this direction. Some questions:

Is the sequence $\Delta_7,\Delta_8,\ldots$ monotonically increasing? I suspect not.

When does $\Delta_n$ first exceed $n/(n+1)$? I don't actually know that it does, but I strongly suspect so for reasons described above. When $n=7$ I have not found any packings of density greater than $7/8$, although only for one of the four non-tiling heptominoes (the one pictured above) have I proven this is optimal. (All four obtain $7/8$ by adding one square to yield a tiling octomino.)

Can $\Delta_7$ be proven to equal $7/8$, if indeed it does so? Exhausting all tilings of an $N\times N$ square by each of the four non-tiling polyominoes and counting the max density therein would at least yield an upper bound; if $\Delta_7$ exceeds $7/8$, I would guess it does so via finding a tiling $15$-omino which is the union of two copies of a non-tiling heptomino and an additional cell.

[Subjective] What are some examples of interesting "near-miss" tilings?

答: Here's a tiling of the plane using a $29$-omino that can be partitioned into $4$ heptominoes plus a single cell, giving that $\Delta_7\geq 28/29$.


问:
Can each root equation be rearranged to a corresponding polynomial equation?
Let $n\in\mathbb{N}_+$.

Can each root equation of $n$ unknowns be rearranged to a polynomial equation of $n$ unknowns whose solution set contains the solution set of the root equation?

If not, is this true at least for root equations of one or two unknowns?

I already know the solution method with raising both sides of the root equation to the same power. But I guess this method is limited to simple root equations because raising to a power of a sum of at least three summands on one side of the equation does not reduce the number of roots on this side of the equation. Take e.g. the equation $\sqrt{A(x)}+\sqrt{B(x)}+\sqrt{C(x)}=\sqrt{D(x)}+\sqrt{E(x)}$.

I already know the method of introducing new unknowns. Does this method answer my question?

答:
By irrational equation, I assume you meant one with one or more variable under a radical as assigned here. The technique of raising both sides to the same power may be invoked repeatedly if needed until there are no radicals. Take, for example.

$$A=\sqrt{x}+\sqrt{y}\implies A^2=\big(\sqrt{x}+\sqrt{y}\big)^2 = 2 \sqrt(x) \sqrt(y) + x + y$$

$$A^2 = 2 \sqrt(x) \sqrt(y) + x + y\implies A^2-x-y=2\sqrt{x}\sqrt{y}$$

$$\big(A^2-x-y\big)^2=\big(2\sqrt{x}\sqrt{y}\big)^2\implies
 A^4 - 2 A^2 x - 2 A^2 y + x^2 + 2 x y + y^2 = 4 x y$$

$$A^4 - 2 A^2 x - 2 A^2 y + x^2 - 2 x y + y^2 = A^4-2A^2(x+y)+(x-y)^2=0$$
The last is an algebraic equation with no radicals an no loss of the "solutions" of the original. The example has been limited to two unknowns and of the same $[(1/2)]$ power. We can, to an extent, also mix the powers of the radicals as in:

$$A=\sqrt{x}+\sqrt[3]{y}\implies (\sqrt[3]{y})^3=\big(A-\sqrt{x}\big)^3
=A^3 - 3 A^2 (x)^{1/2} + 3 A x - x^{3/2}$$
$$\implies y = A^3-3Ax-\sqrt{x}(3A^2+x)$$

$$\big(\sqrt{x}(3A^2+x)\big)^2=\big(A^3-3Ax-y\big)^2$$

$$\implies 9 A^4 x + 6 A^2 x^2 + x^3=A^6 - 6 A^4 x - 2 A^3 y + 9 A^2 x^2 + 6 A x y + y^2$$
$$\implies A^6 - 15 A^4 x - 2 A^3 y + 3 A^2 x^2 + 6 A x y - x^3 + y^2 = 0$$


问:
Topology of naturals
For $a, r \in \mathbb{N}$, be $B_{a,r} = \{a + kr : k\in \mathbb{N}\}$.

(a) Show that $B = \{B_{a,r} : a, r \in \mathbb{N}, r\neq 0, < r\}$ is a basis for a
topology on $\mathbb{N}$.

(b) Show that in this topology, each $B_{a,r}$ is a closed one.

(c) Write $\mathbb{N}-\{1\}$ as a closed meeting.

(d) Conclude that there are infinite primes.

First who are the $\mathbb{N}$ topology open?

To prove that the $B_{a,r}$ collection is a basis for the $\mathbb{N}$ topology should I show that joining base elements generates any element sofa in the topology?

Does it make sense $B_{a,r}$ to be closed, because it is elements that joined generate the elements of the topology that are opened?

Why proving the letter (a),(b) and (c) can at the end conclude that there are infinite prime numbers? I didn't understand anything.

答:
I'm gonna leave (a) and (b) to you because I think you should be able to prove such things by yourself (also I'm too lazy to do it). Just to facilitate (d), here's the answer for (c):

$\mathbb{N} \setminus \{1\} = \cup \{B_{0,p} : p$ is a prime $\}$. This is because every natural number that is not $1$ is a product of primes.

By (b), we know also that each $B_{0,p}$ is closed.

Suppose $A \subseteq \mathbb{N}$ is open. Then by (a), as the $B_{a,r}$ form a basis, there must be some $B_{a,r}$ such that $B_{a,r} \subseteq A$ and thus $A$ is infinite (clearly the $B_{a,r}$ are infinite). This tells us that finite sets can't be open. Now we can put all the pieces together:

Suppose there were finitely many primes. Then $\cup \{B_{0,p} : p$ is a prime $\}$ is a finite union of closed sets, and thus closed- meaning $\mathbb{N} \setminus \{1\}$ is closed. Thus the complement is open- $\{1\}$ is open. But we just estabilished that finite sets can't be open. Thus we have reached a contradiction.

Title: Topology of naturals


问:
Is the relation $xPy$ iff $ y = x + n\pi$ an equivalence relation on $\Bbb R$?
I am trying to prove that the relation $P$ on $\mathbb{R}$ given by the rule

$$\forall x, y \in \mathbb{R}, xPy \text{ if and only if } \exists n \in \mathbb{Z} \text{ such that }y = x+ n\pi$$

From what I can see, $P$ fails the reflexive test, i.e. when showing xPx:
$x \neq x + n\pi$.

But I was told by somebody that $P$ is an equivalence relation.

Could someone please confirm whether it is indeed an equivalence relation or not?

Thanks heaps
C :)

答:
As you realized in the comments, your issue was in the quantifier: we only need some integer $n$ such that $x = x +n \pi$ for reflexivity (not all $n$); therefore the choice of $n=0$ ensures reflexivity for you.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Having trouble using eigenvectors to solve differential equations
The question asked to solve $$\frac{dx}{dy} = \begin{pmatrix}
        5 & 4 \\
        -1 & 1\\
        \end{pmatrix}x$$ ,where $$ x = \begin{pmatrix} x_1 \\
x_2 \\ \end{pmatrix}$$

I went ahead an found the determinant of matrix $$ |A - I\lambda| = \lambda^2 - 4\lambda + 9$$
And found $\lambda = 3$

Then the $\alpha$ matrices was found to be

$$ \begin{pmatrix}
5 & 4\\
-1 & 1\\
\end{pmatrix} \alpha = 3\alpha$$ where $$\alpha = \begin{pmatrix}
\alpha_1 \\
\alpha_2 \\
\end{pmatrix}$$

Ultimately $-\alpha_1 = 2\alpha_2$ so I write $$\alpha = \begin{pmatrix}
-1 \\
2 \\
\end{pmatrix}$$

Then because $\lambda$ is a repeated root I know the solution is supposed to look something like this:

$$x = c_1\alpha e^{\lambda t} + c_2( \alpha t + \beta) e^{\lambda t}t$$

And then this is where it gets tricky for me. I know we find the $\beta$ matrix by figuring this out:

$$(A - I\lambda)\beta = \alpha$$

Now when I multiply all that out I get

$$2\beta_1 + 4\beta_2 = -1$$
$$\beta_1 + 2\beta_2 = -2$$

This is the system of equations I can't seem to solve to get a suitable $\beta$. One option I have is to make $$\beta = \begin{pmatrix}
0\\
-1\\
\end{pmatrix}$$

But this doesn't work for the second system of equations. Help please.

答:
The characteristic polynomial is:

$$|A - \lambda I| = 0 \rightarrow \lambda^2-6 \lambda+9 = 0 \rightarrow \lambda_{1,2} = 3$$

Substituting in the first eigenvalue to find the first eigenvector:

$$[A - \lambda I]v_1 = 0 \rightarrow \begin{pmatrix}
        2 & 4 \\-1 & -2\\\end{pmatrix}v_1 = 0$$

After RREF, for the first eigenvector, I would have chosen:

$$a + 2b = 0 \rightarrow b = 1 \rightarrow a = -2$$

So, the first eigenvector is $v_1 = (-2, 1)$.

Since we have a repeated eigenvalue, we need a generalized eigenvector and you did the right approach, we have:

$$[A - \lambda I]v_2 = v_1$$

$$\begin{pmatrix}
        2 & 4 \\-1 & -2 \end{pmatrix}v_2 = \begin{pmatrix}-2\\1 \end{pmatrix}$$

The RREF is:

$$\begin{pmatrix}1 & 2 \\ 0 & 0 \end{pmatrix}v_2 = \begin{pmatrix}-1\\0 \end{pmatrix} $$

This yields:

$$a + 2b = -1 \rightarrow b = 0 \rightarrow a = -1$$

From this, the second eigenvector is $v_2 = (-1, 0)$.


答:
By definition, the open sets of your topology will be (possibly infinite) unions of elements $B_{a,r}$ of your given basis.

Yes, it makes sense for each $B_{a,r}$ to be also closed (such a subspace is sometimes called a clopen set, as it's both closed and open). Notice though that this does not imply that every open set is also closed, because infinite unions of closed sets are not necessarily closed.

The exercise is basically an excuse to get to the final point, which is known as the
Furstenberg's proof of the infinitude of primes. Basically, with such a topology on the set of natural numbers, any (non-empty) open set contains infinite elements, so no finite set can be open. If you assume that prime numbers are finite, then the union of all (closed sets) $B_{p,0}$ such that $p$ is prime would still be closed. But, as you'll find at c),
$$ \bigcup_{p\text{ prime}} B_{p,0}=\mathbb N\smallsetminus\{1\},$$
which can't be closed because the finite set $\{1\}$ cannot be open. I'll leave you to work out the details!

Title: Topology of naturals

Body:
For $a, r \in \mathbb{N}$, be $B_{a,r} = \{a + kr : k\in \mathbb{N}\}$.

(a) Show that $B = \{B_{a,r} : a, r \in \mathbb{N}, r\neq 0, < r\}$ is a basis for a
topology on $\mathbb{N}$.

(b) Show that in this topology, each $B_{a,r}$ is a closed one.

(c) Write $\mathbb{N}-\{1\}$ as a closed meeting.

(d) Conclude that there are infinite primes.

First who are the $\mathbb{N}$ topology open?

To prove that the $B_{a,r}$ collection is a basis for the $\mathbb{N}$ topology should I show that joining base elements generates any element sofa in the topology?

Does it make sense $B_{a,r}$ to be closed, because it is elements that joined generate the elements of the topology that are opened?

Why proving the letter (a),(b) and (c) can at the end conclude that there are infinite prime numbers? I didn't understand anything.


问:
Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds?
I asked this question on Mathoverflow and received 6 downvotes so far and no answer.

In the comment section of this question several people seemingly freely assume that fundamental theorem of calculus should still hold for divergent integrals with improper bounds.

That is, they assume

$$\int_a^b f(\varphi(x))\varphi'(x) dx = \int_{\varphi(a)}^{\varphi(b)} f(u)du$$

holds even if $a$ or $b$ are $\pm\infty$ and the integral in the left-hand side is divergent to infinity.

I wonder, what can justify this assumption. In my view it is totally unjustified. Why do I think so? Because it allows the following:

$$I=\int_0^\infty1dx=2\int_0^\infty1du=2I$$
or (with substitution $u=2x$)
$$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$

In the second case the indegrals even have different regularized values, the left-hand side integral has the regularized value $0$ while the right-hand side integral has the regularized value $-\ln2$. In other words, they are different divergent integrals with different properties.

In the comment section the user Johannes Hahn justified the relation with words "because it's true" but I do not see in what sense this equality even can be true for divergent integrals?

Is it equality of the values? If so, what value can have a divergent integral without extension of real numbers? Or an extension is assumed?

Is it equality of some other set of properties? In that case we see that the regularized values of these two integrals before and after substitution are different.

Something else?

答:
I think @Rivers McForge gave you a perfect answer to this, but I thought I'd try to maybe rephrase it a different way.

Ok the whole answer has been edited.

The question you asked (which I think is a completely valid and good question) is why do we say
$$
\int_{a}^{b}f(\phi(x))\phi'(x)dx = \int_{\phi(a)}^{\phi(b)}f(u)du
$$
if the integral diverges? Don't you end up with nonsense? The answer is yes - you do end up with nonsense. We can only really technically use an equals sign on numbers (or similar set of objects), so if neither integral converges, then using an equals sign is meaningless.

The point to make is, that if the left hand side is nonsense, the right hand side will also be nonsense, and vice versa. In this sense, it is consistent (by consistent what I mean is you don't end up with number = nonsense, you can only ever get number = number or nonsense = nonsense). So people continue (and are comfortable with) using the equals sign still.

What you did with your two examples is manipulate expressions that are essentially "nonsense = nonsense", which doesn't make sense. You can only algebraically manipulate or do things with expressions that are well-defined quantities, which they were not to begin with. The point is that if we know one side of the formula is defined, so is the other. And that is all. Nothing more, nothing less.

Last thing to mention - regularised values for divergent sums and integrals play by a completely different set of rules, so please don't use them for arguments like this. There's no reason to think that substitution should preserve regularised values.


问:
How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$?
Consider Numbers:

\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is congruent to $10^m \bmod 41$ for some $m$?

答:
Below I thought it through and  that is why I can conclude that:

$10^4 \equiv -4 \pmod {41}$. $245\equiv -1\pmod {41}$. $239 \equiv -7 \pmod {41}$.  And $9^{-1}\equiv -9$.  That is if $k=9k'$ is divisible by $9$ then $(-9)k =(-9)\cdot 9k'= -81k'\equiv k'$.

So:

$245\cdot\underbrace{22222....2}_m11 -239\equiv$

$245\cdot 200\cdot\frac {10^m-1}9 + 245\cdot 11 - 239\equiv$

$-1\cdot (-9)\cdot 200\cdot(10^m - 1)-11 + 7\equiv$

$9\cdot (-5)(10^m-1)-4\equiv$

$-4(10^m-1) - 4\equiv -4(10^m) + 4 - 4 \equiv -4(10^m) \equiv 10^4\cdot 10^m \equiv 10^{m+4}\pmod {41}$

==== Now here is how a figured that out: ====

Let's work backwards. $10^{40} \equiv 1 \pmod {41}$.

$10^2 \equiv 18\pmod {41}$

$10^3 \equiv 180\equiv 16 \pmod {41}$

$10^4 \equiv 160 \equiv 37\equiv -4 \pmod {41}$

$10^5 \equiv -40 \equiv 1 \pmod {41}$.  So there are only $5$ powers of $10$ in $\mod 41$. (because $10^{5k + j} \equiv (10^5)^k\cdot 10^j \equiv 1\cdot 10^j\pmod {41}$.

Now bear with me:

As there are only $5$ possible values of $10^m$ there are only $5$ values of $10^m -1 = \underbrace{999...9}_m$ and of $\frac {10^m-1}9 =\underbrace{1111...1}_m$ and of $245\cdot 200\cdot{10^m-1}9 + 245\cdot 11 - 239=245\cdot\underbrace{22222....2}_m11 -239$.

(Note: $9$ is invertible in $\mod 41$ so dividing by $9$ doesn't add any extraneous values.)

So lets just calculate them all.

It helps to realize $245\equiv -1 \pmod{41}$ and that $9^{-1}\equiv -9$ as $9\cdot (-9)\equiv -81 \equiv 1 \pmod {41}$ and $239 \equiv -7$.

$245\cdot\underbrace{22222....2}_m11 -239\equiv$

$245\cdot 200\cdot\frac {10^m-1}9 + 245\cdot 11 - 239\equiv$

$-1\cdot (-9)\cdot 200\cdot(10^m - 1)-11 + 7\equiv$

$9\cdot (-5)(10^m-1)-4\equiv$

$-4(10^m-1) - 4\equiv -4(10^m) + 4 - 4 \equiv -4(10^m) \equiv 10^4\cdot 10^m \equiv 10^{m+4}\pmod {41}$

oh.... cool.  We didn't have to calculate any of them.


问:
Set Theory Ordinal Problem
I have this problem from Enderton's Elements of Set Theory which says, "Assume that $\lt$A, < $\gt$ is a well-ordered structure with ordinal number $\alpha$ and that B$\subseteq$A. Let $\beta$ be the ordinal number of the well-ordered structure $\lt$B, <$^\circ$$\gt$ and show that $\beta \in \alpha$ or $\beta=\alpha$".

My first approach was to let $a$ be the least element of A-B and say that E(seg($a$))=$\beta$ (where E is the epsilon-image function) and continue from there, but now I realize this won't work, as B isn't necessarily an initial segment of A; we could have A={a,b,c,d,e} (where a< b< c< d< e) and B={b,c,d,e}. Then B$\subseteq$A, but E(seg($a$))$\neq\beta$. So now I'm not sure what to try. It seems obvious that it's true, but I'm having trouble turning that into an actual proof.

I'm thinking now that, using trichotomy of ordinals, I really just have to show that it's NOT true that $\alpha\in\beta$, so I might be able to do a proof by contradiction. If anyone might be able to give me a hint as to where I should look next, I'd appreciate it.

Edit: Could I argue something like this?

Suppose $\alpha\in\beta$. Then $\lt\alpha, \in_\alpha\gt$ $\simeq$ $\lt A$, <$\gt$ $\simeq$ $\lt$ seg(a), <$^\circ\gt$ for some a$\in$B, but B$\subseteq$A so a$\in$A, and we have A is isomorphic to a segment of itself, which is a contradiction?

答:
I found problem in Rene's answer. I don't think it can be proved that every order preserving map has the property $\mu \le f(\mu)$. Maxime's hint is the right direction and helped me finish my proof:

Prove by contradiction. Assume $\alpha \in \beta$. Then there exists $t \in B$ that $\langle A, \lt\rangle \cong \langle seg_B(t), \lt^t\rangle$, which means we have an order preserving map $f : A \to seg_B(t)$.

Since $B \subseteq A$ and then $t \in A$, we have $f(t) \in seg_B(t)$, which means $f(t) \lt t$. Let $\{x \in B|f(x) \lt x\}$ be C. Then C is nonempty as we have shown that t fits in. So by well order principle there is minimum m in C. But that means $f(m) \lt m$ which will violate the minimality of m if we can show that $f(m) \in C$.

To show $f(m) \in C$, first notice that $f(m) \in seg_B(t)$ so $f(m) \in B$. And the last thing we have to show is $f(f(m)) \lt f(m)$. It can be obtained from assumption $f(m) \lt m$ and the order preserving property of f.

Title: Set Theory Ordinal Problem


问:
Definition of twisted sheaf on projective scheme
Let $X$ be a closed subscheme of some $\mathbb{P}_A^n$ where $A$ is a ring. Then I know we can define twist sheaf on $\mathbb{P}_A^n$ by $\mathcal{O}(k):=\widetilde{R(k)}$ where $R=A[x_0,...,x_n]$.

First question: I'd like to ask if there exists a definition of twisted sheaf on $X$, maybe it can be defined as the pullback of $\mathcal{O}(k)$ under the closed immersion $i: X\rightarrow\mathbb{P}_A^n$. But $X$ maybe also embedding into other $\mathbb{P}_A^m$, we denote such a morphism $j: X\rightarrow\mathbb{P}_A^m$, then does $i^*\mathcal{O}(k)\cong j^*\mathcal{O}(k)$ hold?

Second question: How to define the twisted sheaf $\mathcal{O}(k)$ over $\mathbb{P}_S^n$ where $S$ is a general scheme? Is it defined by pulling back the sheaf $\mathcal{O}(k)$ over $\mathbb{P}_{\mathbb{Z}}^n$ under the projection $\mathbb{P}_S^n\rightarrow\mathbb{P}_{\mathbb{Z}}^n$?

答:
Suppose $X = \operatorname{Proj} R_\bullet$, where $R_\bullet$ is a graded ring with $R_0 = A$, finitely generated in degree 1. Then we can define $\mathscr O(n)$ as the quasicoherent sheaf associated to the graded module $R(n)_\bullet$ (whose degree $k$ component is $R_{k+n}$). The sections of this sheaf over $D_+(f)$ are by definition the degree-n component of the localization $(R(n)_\bullet)_f$. For each $n$, this sheaf is a line bundle because of the assumptions on the graded ring. Now for any quasicoherent sheaf $\mathcal F$ on $X$, we can define the twist simply by $\mathcal F(n) = \mathcal F\otimes \mathscr O(n)$.

For the second question, you can globalize the construction above, by using the global $\operatorname{Proj}$ construction, which associates to every quasicoherent sheaf of graded $\mathscr O_S$-algebras over a scheme $S$ an $S$-scheme locally glued out of $\operatorname{Proj}$ of the graded $A$-algebras of sections over affine opens $\operatorname{Spec}A \subset S$. Assuming finite generation in degree 1, the global $\operatorname{Proj}$ construction comes with an $\mathscr O(1)$, locally glued out of the $\mathscr O(1)$'s from the previous paragraph. Then you can define $\mathscr O(n) = \mathscr O(1)^{\otimes n}$. Note that for the case of $\mathbb P^n_S$, the relevant sheaf of $\mathscr O_S$-algebras is $\mathscr O_S^{\oplus n+1}$.

These constructions are discussed in Vakil's FOAG 15.2 and 17.2, or in Hartshorne II.7 (starting page 160).


问:
Where, if ever, is it proper (expected) to use a space? Where by convention is it [preferred, or preferred not] but allowed, or forbidden?
The only place I see a space (aside from within tables) inserted with some regularity is surrounding an equals sign to more clearly separate a right-hand from left-hand side of an equation. Aside from that, I have encountered similar places where it has been both inserted and left out, particularly within some containerized object of at least somewhat distinct constituents ⟨[{(a, b, c, ...)}]⟩, but more commonly without. Besides maintaining consistency within single handling, is there a standard correct usage (and non-usage) of a using a space in mathematical statements?

Logically I would think that a space should be used between more distinct parts, such as solutions within a set as contrasted to coordinates assigned to a solution. One place where absence of a space could hinder clarity is with an especially long number, particularly one with an absval greater than or equal to 10^6 that in normal entry often has comma[,]s (sometimes space[ ]s or period[.]s) for readability. Are there any places where spacing is used without a comma? Ever intentional different lengths of spacing (outside of hand-written text)?

答:
The best comment above may have been the one about whatever makes it easier to read. Using Euclid's formula as an example, which of these to you find easier?
$$\text{Euclid's formula is } A=m^2-n^2, B=2mn, C=m^2+n^2\text{ with commas}.$$
$$\text{Euclid's formula is }\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2\quad\text{ with "\quad"s}.$$
Another case is the right ceiling function after a radical
$$\lceil\sqrt{X}\rceil\quad \text{vs}\quad \lceil\sqrt{X}\space\rceil\quad \text{the latter with "\space" in front of }"\rceil"$$
There must be a hundred places where "space" decisions need to be made.

For more info, you might go to

https://academia.stackexchange.com/

or

https://tex.stackexchange.com/


问:
Matrix multiplication related to complex numbers?
Evaluate and simplify the product

$\begin{bmatrix} r\cos(\alpha) & -r\sin(\alpha) \\ r\sin(\alpha) & r\cos(\alpha)\\ \end{bmatrix}$ $\begin{bmatrix} s\cos(\beta) & -s\sin(\beta) \\ s\sin(\beta) & s\cos(\beta)\\ \end{bmatrix}$

where $\alpha$, $\beta$ are any real numbers and $r$, $s$ are nonnegative real numbers. Relate your answer to multiplication of complex numbers. Is there any connection with addition of complex numbers?

I was able to multiply through and I will post my answer below, but I can't see the connection to complex numbers. Could someone please explain? Thanks.

答: Using the fact that $e^{i\theta}=\cos\theta+i\sin\theta$, what do you get when you multiply $$re^{i\alpha}\cdot se^{i\beta}?$$Given the matrix $$\left[\begin{array}{rr} a & -b \\ b & a \end{array}\right]$$ what complex number would you associate it with?</p>


问:
"Numerical results" from category theory
I want to keep a mini class on category theory, and I would be very satisfied to have a class of examples in which some theorems - or just the reformulation in categorical terms - yields numerical results or answers.

To be explicit, I would have such an example, but it is too advanced for my target. Here it is: suppose $G$ is a finite group acting on a finite set $X$. Let $V(X)$ be the vector space freely generated by elements in $X$. What is the dimension of $V(X)^G$? Is it for example equal to $V(X^G)$?

Answer. Recall that orbits of $G$ writes as $G/H$ for some $H$, so that we can write $X= \cup_{i\in I} G/H_i$. Also, we have that
$$V(X) = \bigoplus_{i \in I} V(G/H_i) = \bigoplus_{i \in I} \text{Ind}_{H_i}^G( \mathbb{C} ) $$

And now
$$ \dim_{\mathbb{C}} V(X)^G = \dim_{\mathbb{C}} Hom_G(V(X),\mathbb{C}) = \sum_{i \in I} \dim_{\mathbb{C}} Hom_G( \text{Ind}_{H_i}^G(\mathbb{C}), \mathbb{C}) =  $$
$$ = \sum_i \dim_{\mathbb{C}}Hom_{H_i}(\mathbb{C}, \mathbb{C}) = |I| = \#\{\text{orbits}\} = |X/G| $$

So that the frobenius adjunction yields a very numerical result, and indeed all this adjunction can be reformulated "numerically".

In general, a number of examples can arise when you consider adjoint functors between categories that have finite dimensional vector spaces.

I look for something that could be ok for people that know one or two years of university math. Something combinatorial, or number theoretical, would be perfect, for example; something about K-theory, definitely not. Hope I have not been too vague!

答:
I give several examples involving groupoid cardinality in this MO answer. Here's another example which is not in that answer. First I'll quote the brief introduction to groupoid cardinality from over there:

To say it very tersely, if $X$ is a groupoid with at most countably many isomorphism classes of objects, such that each automorphism group $\text{Aut}(x)$ is finite, we can define its groupoid cardinality

$$|X| = \sum_{x \in \pi_0(X)} \frac{1}{|\text{Aut}(x)|}$$

if this sum converges, where $\pi_0(X)$ denotes the set of isomorphism classes of objects. Groupoids such that this sum converges are called tame. Groupoid cardinality has the following properties:

Normalization: $|\bullet| = 1$, where $\bullet$ denotes the terminal groupoid.

Additivity: $|X \sqcup Y| = |X| + |Y|$.

Multiplicativity: $|X \times Y| = |X| |Y|$.

Covering: If $f : X \to Y$ is an $n$-fold covering map of groupoids then $n |X| = |Y|$.

These properties (in fact just normalization, additivity and covering) uniquely determine $|X|$ for finite groupoids (finitely many objects and morphisms). $|X|$ is even countably additive and this together with normalization and covering determines it for tame groupoids.

Now here is the example. Dobinski's formula for the Bell numbers $B_n$, which count the number of partitions of an $n$-element set, is

$$B_n = \frac{1}{e} \sum_{k \ge 0} \frac{k^n}{k!}.$$

This formula can be given a "bijective proof" using groupoid cardinality as follows; what we will prove is that

$$B_n \sum_{k \ge 0} \frac{1}{k!} = \sum_{k \ge 0} \frac{k^n}{k!}$$

by calculating the groupoid cardinality of a specific groupoid in two different ways. Fix a set $N$ of size $n$. The groupoid we'll consider is the groupoid whose objects are maps $N \to K$ from $N$ to another finite set $K$, and whose morphisms are isomorphisms between these.

On the one hand, if the cardinality $|K| = k$ of $K$ is fixed, the groupoid of maps $N \to K$ and isomorphisms has a $k!$-fold cover given by the groupoid of maps $N \to K$ where $K$ is equipped with a total order and isomorphisms are required to preserve this order. This groupoid is discrete (has no nontrivial automorphisms) and has exactly $k^n$ isomorphism classes of objects, and the groupoid of maps $N \to K$ is the quotient of it by the action of $\text{Aut}(K)$, so the groupoid cardinality is $\frac{k^n}{k!}$. Summing over all $k$ produces the RHS.

On the other hand, a map $f : N \to K$ produces a partition of $N$ given by its fibers, together with a finite set given by the complement of the image of $f$. Our groupoid is in fact equivalent to the groupoid whose objects are pairs consisting of a partition of $N$ and a finite set, and whose morphisms are isomorphisms of the finite set only (the morphisms involving the partition are "canceled out" by the fact that, if the image $K' \subseteq K$ of $f$ is fixed, $\text{Aut}(K')$ acts freely on the set of surjections $N \to K'$, with quotient the set of partitions of $N$ into $|K'|$ nonempty blocks). This groupoid has groupoid cardinality $B_n \sum_{k \ge 0} \frac{1}{k!}$ which is the LHS as desired.

Performing this second identification more carefully gives a bijective proof (in the ordinary, non-groupoid sense) of the identity

$$k^n = \sum_{j=0}^n \left\{ n \atop j \right\} (k)_j$$

where $(k)_j = k(k-1) \dots (k-(j-1))$ is the falling factorial and $\left\{ n \atop j \right\}$ is the Stirling number of the second kind counting the number of partitions of $n$ into $j$ parts; this identity corresponds exactly to counting functions $f : N \to K$ based on the size $j$ of the image. Substituting this expression into $\sum_{k \ge 0} \frac{k^n}{k!}$ gives a "traditional" proof of Dobinski's formula, but to my mind the groupoid cardinality argument is cleaner because it explains the meaning of this proof.


问:
The uniform convergence of $\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x \quad w.r.t \ a$

For fixed $n\in \mathbb{N}$, study the uniform convergence of
$$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x  \quad w.r.t \ a\in(0,\infty)$$.

Actually I want to compute the integral by differentiating
$$
\int_0^\infty \mathrm{e}^{-ax^2}\mathrm{d}x=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}},
$$
but we need to prove $$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n+2}\mathrm{d}x $$
converges uniformly w.r.t $a>0$. Then the procedure is legal. The Weierstrass, Dirichlet and Abel criterion all seem do not work.

Appreciate any help!

Body:
Hint

$$I_n=\int_0^\infty {e}^{-ax^2}x^{2n}\,dx$$
$$x=\sqrt t \implies I_n= \frac{1}{2}\int_0^\infty e^{-a t} t^{n-\frac{1}{2}}\,dt=\frac{1}{2} a^{-(n+\frac{1}{2})} \Gamma \left(n+\frac{1}{2}\right)$$


问:
Does the series $\sum_{k = 1}^{\infty}\frac{\sin(\frac{2k + 1}{2}\pi)}{\sqrt{k+1}}$ converge or diverge?
Determine whether the alternating series
\begin{align*}
\sum_{k = 1}^{\infty}\frac{\sin(\frac{2k + 1}{2}\pi)}{\sqrt{k+1}}
\end{align*}

absolutely converges or conditionally converges or diverges. Please show all steps

答: HINT. Note $\sin\left(\frac{2k+1}{2}\pi\right) = (-1)^k$, thus it is an alternating series. All you need to do is to check $\lim\limits_{n\to\infty}a_n=0$ and $|a_{n+1}|\le |a_n|$.


问:
PDF - Probability density function
Fred lives in Blissville, where buses
always arrive exactly on time, with the time between successive buses fixed at $10$ minutes. Having lost his watch, he arrives at the bus stop at a uniformly random
time on a certain day (assume that buses run $24$ hours a day, every day and that
the time that Fred arrives is independent of the bus arrival process).

Question: What is the average time and the PDF that Fred has to wait?

I find that the average time is in the range $[0,10]$ so the mean is $5$ minutes but I couldn't find the PDF.

答:
Let's start by letting $T$ represent the amount of time, in minutes, Fred has to wait at the bus stop. One thing we know right away is $T\ge0$, since Fred can't possibly wait a negative amount of time at the stop. We also know that the bus arrives every 10 minutes on a fixed schedule, so the longest he could possibly wait is 10 minutes. This tells us $T\le 10$. So far, we have:

$$ P[T < 0] = 0 , $$
$$ P[T > 10] = 0 . $$

Now, in that range of $ 0 \le T \le 10 $, we also know that Fred's arrival is uniformly random and independent of the bus's arrival. This means that all possible wait times between $T = 0$ and $T = 10$ are equally likely - this is how we know Fred's wait time is uniformly distributed:

$$ T \sim \mathcal{U}(0,10) $$

Broadly, for a random variable $ X \sim \mathcal{U}(a,b) $, the density function is given by

$$ f(x) = \begin{cases}\frac{1}{b-a} & x\in[a,b]\\ 0 & x\notin[a,b] \\ \end{cases} .$$

The reason this is the functional form we want is it's a constant value across the domain, which is what we're looking for when we say that any time between 0 and 10 is "equally likely."

Note, for a continuous distribution, $P[X=x]=0$, so we're not saying that the probability of waiting 1 minute is equal to the probability of waiting 2 minutes - both of those values are 0 - it's meant to be a general description of what the density looks like - a horizontal line.

In this case, $a=0$ and $b=10$, so we have $b-a=10$ and

$$ f(t) = \begin{cases}\frac{1}{10} & t\in[0,10]\\ 0 & t\notin[0,10] \\ \end{cases} .$$

Title: PDF - Probability density function


问:
Probability of particular order of occurrence of independent exponential distributions
So I have three independent Poisson processes $X, Y, Z$ with $\lambda_1, \lambda_2, \lambda_3$ respectively as the rate parameters. I'm interested in the probability that the first event comes from $X$, the second from $Y$ and the third from $Z$ (in that order exactly).

How would I go about doing that?

答:
Comment: Possible hints:

(1) If $\lambda_1=\lambda_2=\lambda_3,$ then by symmetry the answer is $1/6.$

This can be illustrated by a simulation of $10^6$ iterations in which we ask whether $X = \min(X,Y,Z)$ and
$Z = \max(X,Y,Z).$ (With a million iterations, one can
expect two or three place accuracy.)

```<code>lam.1 = 1; lam.2=1; lam.3=1; m = 10^6
set.seed(2020)
x = rexp(m,lam.1); y = rexp(m,lam.2); z=rexp(m,lam.3)
v = pmin(x,y,z); w = pmax(x,y,z)
mean(x==v & z==w)
[1] 0.166596  # aprx 1/6
</code>```

(2) Here is a simulation for unequal $\lambda_i.$

```<code>lam.1 = 1; lam.2=2; lam.3=3; m = 10^6
set.seed(2020)
x = rexp(m,lam.1); y = rexp(m,lam.2);  z=rexp(m,lam.3)
v = pmin(x,y,z);  w = pmax(x,y,z)
mean(x==v & z==w)
[1] 0.066714
</code>```

[Even if this doesn't help you see how to get a
general analytic solution, you can check your
analytic solution against these results.]

(3) Maybe start by finding $P(X<Y).$

Body:
Comment: Possible hints:

(1) If $\lambda_1=\lambda_2=\lambda_3,$ then by symmetry the answer is $1/6.$

This can be illustrated by a simulation of $10^6$ iterations in which we ask whether $X = \min(X,Y,Z)$ and
$Z = \max(X,Y,Z).$ (With a million iterations, one can
expect two or three place accuracy.)

lam.1 = 1; lam.2=1; lam.3=1; m = 10^6
set.seed(2020)
x = rexp(m,lam.1); y = rexp(m,lam.2); z=rexp(m,lam.3)
v = pmin(x,y,z); w = pmax(x,y,z)
mean(x==v & z==w)
[1] 0.166596  # aprx 1/6

(2) Here is a simulation for unequal $\lambda_i.$

lam.1 = 1; lam.2=2; lam.3=3; m = 10^6
set.seed(2020)
x = rexp(m,lam.1); y = rexp(m,lam.2);  z=rexp(m,lam.3)
v = pmin(x,y,z);  w = pmax(x,y,z)
mean(x==v & z==w)
[1] 0.066714

[Even if this doesn't help you see how to get a
general analytic solution, you can check your
analytic solution against these results.]

(3) Maybe start by finding $P(X<Y).$


问:
Study the uniform convergence of $ I(\alpha)=\int_0^{\frac{\pi}{2}}\ln(\alpha^2-\sin^2\varphi)\mathrm{d}\varphi, $

Study the uniform convergence w.r.t. $\alpha$ of the following integral with parameters:
$$
I(\alpha)=\int_0^{\frac{\pi}{2}}\ln(\alpha^2-\sin^2\varphi)\mathrm{d}\varphi,
$$
where $\alpha\geq 1$.

I just learned integral with parameters. When $\alpha=1$, $\varphi=\frac{\pi}{2}$ is a singularity point (we can actually compute the integral), when $\alpha>1$, the integral is proper. I do not how to solve the combined situation.

Appreciate any help!

答:
Why not to use Feynman trick ?
$$I(\alpha)=\int_0^{\frac{\pi}{2}}\ln(\alpha^2-\sin^2(\varphi))\,d\varphi$$
$$I'(\alpha)=\int_0^{\frac{\pi}{2}}\frac {2\alpha}{\alpha^2-\sin^2(\varphi)}\,d\varphi=-\frac{\pi }{\sqrt{\alpha^2-1}}$$


问:
Moving from a rational generating function to recurrence relation
I am trying to understand the following segment of my notes:

.

I will try to explain what I think I understand and point out the parts I don't understand.

First, there is a result in my notes (not shown here) proving that a rational generating function defines a homogeneous recurrence relation. Then, I consult the following section of my notes:

.

This is a procedure for turning a homogeneous recurrence relation into a generating function (in its rational form). Given the rational form $f(x) = \frac{1 + 7x}{1 + x - 6x^2}$, we see that the coefficients that define the homogeneous recurrence relation (i.e. the $b_1, \dots, b_k$) are located in the denominator, so we conclude that $a_n = 6 a_{n - 2} - a_{n - 1}$.

I don't understand how the initial conditions $a_0 = 1$ and $a_1 = 6$ are determined through long division. I am only vaguely familiar with polynomial division so I may be missing something obvious, but isn't $\frac{1 + 7x}{1 + x - 6x^2}$ simply equal to $1 + 7x$, because the degree of the numerator is less than the degree of the denominator? How does this help me determine the initial conditions?

According to my notes (not shown here), the "term formula" as I call it for the $i$th coefficient $a_i$ of the generating function can be found by factoring the denominator of $\frac{1 + 7x}{1 + x - 6x^2}$ into a suitable form, applying the method of partial fractions and using a result regarding the term formula for $\frac{1}{(1 - cx)^k}$ for $c \in \mathbb C$. Eventually I get that $a_i = \frac{9}{5} \cdot 2^i - \frac{4}{5} (-3)^i$, which is the same result as the one in my notes, and from this we can determine the initial conditions $a_0$ and $a_1$.

I appreciate any help.

答:
How polynomial long division shows that $a_0=\color{blue}1$ and $a_1=
\color{green}6$:

$$
\require{enclose}
\begin{array}{r}
                &\color{blue}1+\color{green}6x\dots  \\[-3pt]
1+x-6x^2 &\enclose{longdiv}{1+7x} \\[-3pt]
     &\underline{\color{blue}{1+x-6x^2}}\phantom{2} \\[-3pt]
                &6x+6x^2  \\[-3pt]
     &\underline{\color{green}{6x+6x^2-36x^3}}
\end{array}
$$

[Editing for better alignment is welcomed.]


问:
Metrics whose sum induces a topology different from components
Does there exist a metric $d$ defined as the sum of two metrics $d_1$ and $d_2$ (i.e. $d(x,y) = d_1(x,y) + d_2(x,y)$), but $d$ generates a different topology than both $d_1$ and $d_2$? If so, what are some examples of this?

答:
Sure. A sequence converges with respect to the sum metric $d(x, y) = d_1(x, y) + d_2(x, y)$ iff it converges with respect to both $d_1$ and $d_2$, so it suffices to construct $d_1$ and $d_2$ such that some sequence $x_n$ converges with respect to $d_1$ but not $d_2$ and some sequence $y_n$ converges with respect to $d_2$ but not $d_1$; then neither sequence converges with respect to $d$, so the topology generated by $d$ must be different from both.

This can be done as follows. We'll take our underlying set to be $\mathbb{N}$, and we'll pick metrics on $\mathbb{N}$ by embedding in $\mathbb{N}$ into $\mathbb{R}$ in various ways and considering the induced metric. First, consider the embedding

$$f(n) = \begin{cases} 0 & \text{if } n = 1 \\ \frac{1}{n} & \text{if } 2 \mid n \\ n & \text{otherwise}\end{cases}.$$

With respect to the induced metric, the sequence $x_n = 2n$ converges to $1$, but the sequence $y_n = 2n+1$ is not Cauchy. On the other hand, consider the embedding

$$g(n) = \begin{cases} 0 & \text{if } n = 1 \\ n & \text{if } 2 \mid n \\ \frac{1}{n} & \text{otherwise}\end{cases}.$$

With respect to the induced metric, this time the sequence $y_n = 2n+1$ converges to $1$, but the sequence $x_n = 2n$ is not Cauchy. So with respect to the sum metric, neither sequence converges.


答:
$\frac{1+7x}{1+x-6x^2}$ is certainly not equal to $1+7x$, since $1+x-6x^2$ is not identically equal to $1$. The fact that the degree of the numerator is less than that of the denominator is not a bar to doing long division. If you divide $1+7x$ by $1+x-6x^2$ using ordinary polynomial long division, but starting at the low order end, for the first four terms you get:

$$\begin{array}{ccc}
&&&&&&1&+&6x&&&+&36x^3\\
&&&&&&-&-&-&-&-&-&-\\
1&+&x&-&6x^2&)&1&+&7x\\
&&&&&&1&+&x&-&6x^2\\
&&&&&&-&-&-&-&-\\
&&&&&&&&6x&+&6x^2\\
&&&&&&&&6x&+&6x^2&-&36x^3\\
&&&&&&&&-&-&-&-&-\\
&&&&&&&&&&&&36x^3
\end{array}$$

If you keep going, you can get as much of the power series as you want. You know that the power series is $\sum_{n\ge 0}a_nx^n$, so this already shows that $a_0=1$, $a_1=6$, $a_2=0$, and $a_3=36$.

You are correct, however, in thinking that you can also determine $a_0$ and $a_1$ from the closed form that you get from power series expansions of the partial fractions.

Title: Moving from a rational generating function to recurrence relation

Body:
I am trying to understand the following segment of my notes:

.

I will try to explain what I think I understand and point out the parts I don't understand.

First, there is a result in my notes (not shown here) proving that a rational generating function defines a homogeneous recurrence relation. Then, I consult the following section of my notes:

.

This is a procedure for turning a homogeneous recurrence relation into a generating function (in its rational form). Given the rational form $f(x) = \frac{1 + 7x}{1 + x - 6x^2}$, we see that the coefficients that define the homogeneous recurrence relation (i.e. the $b_1, \dots, b_k$) are located in the denominator, so we conclude that $a_n = 6 a_{n - 2} - a_{n - 1}$.

I don't understand how the initial conditions $a_0 = 1$ and $a_1 = 6$ are determined through long division. I am only vaguely familiar with polynomial division so I may be missing something obvious, but isn't $\frac{1 + 7x}{1 + x - 6x^2}$ simply equal to $1 + 7x$, because the degree of the numerator is less than the degree of the denominator? How does this help me determine the initial conditions?

According to my notes (not shown here), the "term formula" as I call it for the $i$th coefficient $a_i$ of the generating function can be found by factoring the denominator of $\frac{1 + 7x}{1 + x - 6x^2}$ into a suitable form, applying the method of partial fractions and using a result regarding the term formula for $\frac{1}{(1 - cx)^k}$ for $c \in \mathbb C$. Eventually I get that $a_i = \frac{9}{5} \cdot 2^i - \frac{4}{5} (-3)^i$, which is the same result as the one in my notes, and from this we can determine the initial conditions $a_0$ and $a_1$.

I appreciate any help.


问:
Is there a mathematical notation for restricting the depth of a factorial?
Say I only want to go $3$ "levels" deep for $5!$ and end up with just $5\times 4\times 3$. Is there a notation for this?

答:
This is also sometimes called "permutations of 3 out of 5 things" and denoted 5P3. You can find this on TI-8x calculators, at least, as nPr in the MATH->PRB menu (so you'd type 5 nPr 3).

I have also seen the falling factorial denoted  with brackets as $[5]_3$. I'd say any notation for this will have to be clarified when you use it, though.


问:
Are there any good examples from other math fields or intuition supporting $\int_0^1\frac1xdx=\int_1^\infty\frac1xdx$?
This question is related to the potential possibilities of classification of divergent integrals more precisely than just "divergent to infinity" and the like. Improper divergent integrals can have many distinguishing properties, like germ at infinity, rate of grow and regularized value.
A separate question is whether we can talk about equivalence (in whatever sense) of integrals divergent at infinity and those divergent at a singularity or pole.

For instance, formally applying the Laplace transform we can establish the following formal relations (for $n>0$):

$$\int_0^1 x^{-1+n} dx=\int_1^\infty n!x^{-1-n}dx$$

and

$$\int_1^\infty x^{-1+n} dx=\int_0^1 n!x^{-1-n}dx$$

This works for convergent integrals:

$$\int_0^1 dx=\int_1^\infty \frac1{x^2}dx=1$$

If generalized to divergent integrals, it also seems to work well, the equality

$$\int_1^\infty dx=\int_0^1 \frac1{x^2}dx$$

seems to be supported well by the hyperfunction theory.

But with $n=0$ we have (in the both formulas):

$$\int_0^1 \frac1x dx=\int_1^\infty \frac1x dx$$

This looks counter-intuitive at the first glance, as the figures under the integrals are identical up to a reflection relative the $y=x$ axis, except the first one has an additional $1\times 1$ square.

Of course, since the integrals are divergent, it should not be an obstacle to the identity (as the positions of the figures and the limits of integration are different). The both integrals are regularizable to the value $0$.

So, I wonder, if there is any intuition or example from other fields or mathematics or physics where these integrals demonstrate equivalence in a more strict sense than being both divergent to infinity?

For instance, from geometry or probability theory.

UPDATE

It seems, the equality should not hold, and the more correct one would be $$\int_0^1 \frac1x dx=\gamma+\int_1^\infty \frac1x dx$$

答:
Your paradox, as I understand it, concerns the correspondence between the divergent integral and the geometric area.  Let's be a little more careful and define finite regions which are symmetric about $y=x$ before considering the limit as the regions approach the $x$- and $y$-axes.

There are two such areas we can define. First, there is $A_1$, between $y=1$ and $y=1/x$ on the interval $\epsilon < x< 1$, given by
$$A_1 = \int_\epsilon^1 dx\left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x} +\epsilon-1.$$ Here, $0<\epsilon<1$ is a small number.

If we reflect this region about $y=x$, we get
$$ A_1 = \int_1^{\epsilon^{-1}} dx\left(\frac1x-\epsilon\right) = \int_1^{\epsilon^{-1}} \frac{dx}{x} + \epsilon-1.$$

A second area can be defined as that between $x=1,$ $x=\epsilon^{-1}$, $y=1/x$, and the $x$-axis:
$$
A_2 = \int_1^{\epsilon^{-1}} \frac{dx}{x}.$$

The reflection about $y=x$ is
$$
A_2 = \int_0^\epsilon dx (\epsilon^{-1}-1) + \int_\epsilon^1 dx \left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x}.$$

So, in both cases, the geometric areas agree after reflection about $y=x$ if and only if the divergent integrals agree as claimed.

I believe the confusion arises because there are two relevant geometric areas, and which one you get depends on how you take the limit. If you imagine replacing the integral with a finite sum of rectangles, the height of the rectangles may or may not diverge depending on how they are oriented, and this leads to two different answers.


问:
How to solve this integral easily: $\int \frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$
I am trying to solve this integral $$\int\frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$$
I can do it by brute force (means using a substitution then long division and then substitutions again) but it's too long (suspiciously long solution). Is there any better way to solve it? I guess it should be. If you could provide me with the process that leads to the answer that would really help.</p>

答:
Substitute $\sqrt[3]{x+2}=u$ therefore $\frac{du}{dx} =\dfrac{1}{3\sqrt[3]{(x+2)^2}}$

Your integral will become : 3$\int\dfrac{u^3(u^3-2)}{u^3+u-2}du $

= $3\int [u^3+ \dfrac{5u-2}{4(u^2+u+2)}-u-\dfrac{1}{4(u-1)}]du$

=$3\int u^3du + \dfrac{3}{4}\int(\dfrac{5(2u+1)}{2(u^2+u+2)}-\dfrac{9}{2(u^2+u+2)})du - \dfrac{3}{4}\int\dfrac{1}{u-1}du -3\int u.du$

Now you can  easily see that differentiation of $u^2+u+2 = 2u+1$ so you can substitute $u^2+u+2 = t$ then proceed you will get your result.

In case of further clarification do let me know...


问:
Function composition for proving polynomials are continuous
I am attempting to prove that polynomials $\mathbb{R} \to \mathbb{R}$ are continuous in the topological sense. Let $f$ be such a polynomial.

The answer here suggests showing that linear functions are continuous (easy) and that the composition of continuous functions is continuous (easy).

However - and perhaps it's because I've been looking at this stuff too long - I fail to see why
$$\begin{align}
p_0(x) &= a_n \\
p_1(x) &= xp_0(x) + a_{n-1} \\
p_2(x) &= xp_1(x) + a_{n-2} \\
&\vdots \\
p_n(x) &= f(x)
\end{align}$$
is, per the answer in the question, "just composing a linear function with a continuous function."

From my perspective, $p_0$ is a constant function, so it's continuous.

Let's suppose I have a function $g: \mathbb{R} \to \mathbb{R}$ given by $g(y) = cy + d$ for some $c, d \in \mathbb{R}$ fixed. I have shown that $g$ is continuous.

If I assume $c = x$ and $d = a_{n-1}$, then I get $g(p_0(x)) = xp_0(x) + a_{n-1}$.

However, while setting $d = a_{n-1}$ seems like fair game (they are both constants), $x$ is definitely not a constant, so I don't feel comfortable setting $c = x$.

Am I missing something here?

答:
As MPW says in the comments, the quoted answer is wrong; it is treating $x$ as a constant which it, of course, is not. An approach that will work and is, to my mind, the conceptual and straightforward thing to do here, is the following: prove that addition and multiplication

$$+, \times : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$

are continuous, that the diagonal map

$$\Delta : X \ni x \mapsto (x, x) \in X \times X$$

is continuous, and that if $f : X \to Y, g : Z \to W$ are two continuous functions, then their product $f \times g : X \times Z \to Y \times W$ is continuous.

Using these facts (and that composition of continuous functions is continuous), prove that if $f(x), g(x) : X \to \mathbb{R}$ are two continuous functions, then their pointwise sum $f(x) + g(x)$ and pointwise product $f(x) g(x)$ are continuous. Now starting from the constant and identity functions $\mathbb{R} \to \mathbb{R}$ and repeatedly taking pointwise products and sums you get exactly the polynomials.


问:
How to find the profit earned from getting right a set of logic statements?
The problem is as follows:

Diana offers her sister Maria $3$ bitcoins for each true statement and $4$ bitcoins for each false that she finds correctly in the statements shown, where $\mathbb{Z}$ is the universal set for the variables $x$ and $y$.

I. $\forall y; \forall x: x^2+y^2>0$

II. $\exists x; \exists y: x+y<2xy$

III. $\forall x; \exists y: x+y=0$

How many bitcoins did Maria receive, after answering all of them correctly?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{11 bitcoins}\\
2.&\textrm{12 bitcoins}\\
3.&\textrm{10 bitcoins}\\
4.&\textrm{9 bitcoins}\\
\end{array}$

How exactly this problem should be approached?. I'm stuck at the part where it comes the inequalities, from then on I don't know what to do. Can someone help me?.

The thing which I'm getting is that the set from where $x$ and $y$ belong to is $\mathbb{Z}$ hence we're talking about integers but I don't know how to use this information to further solve this problem. Can someone help me from this part in the most detailed way as possible and step by step?.

答:
Try some numbers:

The first is False since $0^2+0^2=0$.

For the second one, let $x$ and $y$ takes some values. For example, you can set $x$ to be $0$ and determine if you can find your $y$.


问:
How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?
One  way to solve these types of questions is to go one by one and figure out all the possibilities starting from 111. And I have found till now 19 possibilities for this question. But the answer is somewhat different. It seems that I might have missed some of the possibilities.
I was wondering if there is any other way to approach these types of questions that makes sure that I do not miss any of the possibilities. Any simpler or logical or diagrammatical way to solve these types of questions, that would be really helpful.

Thanks!

答:
A more organized approach is to note that $2^3=8>7$, so the only combinations that will work (in any order of digits) are $11x$ for $x \in \{3,4,5,6\}$, and also $122$ and $123$.

Now you must determine how many numbers are contributed by each. For example, $123$ contributes $3!=6$ to the overall sum.


问:
What's the probability that you find a working pen?
You find a container of $27$ old pens in your
school supplies and continue to test them (without replacement), until you find
one that works. If each individual pen works $25$% of the time (regardless of the
other pens), what is the probability that you find one that works within the
first four tries?

My best guess was

$0.25 = \cfrac{1}{4}$ is the Probability of us having the 1st pen work ($0.25$ = probability of success)

$0.75 \cdot 0.25 = \cfrac{3}{16}$ is the probability of us having the 2nd pen work where $0.75$ is the probability of the 1st pen failing

$0.75 \cdot 0.75 \cdot 0.25 = \cfrac{9}{64}$ is the probability of us having the 3rd pen work where $0.75$ is the probability of the 1st and 2nd pen failing

$0.75 \cdot 0.75 \cdot 0.75 \cdot 0.25 = \cfrac{27}{256}$ is the probability of us having the 4th pen work $0.75$ is the probability of the 1st and 2nd and 3rd pen failing

So probability of us having a success within four tries $= \cfrac{1}{4} + \cfrac{3}{16} + \cfrac{9}{64} + \cfrac{27}{256} = \cfrac{175}{256}$

What I wanted to ask this website is if I had this right? I ask because the fact that they emphasize the fact that there's exactly 27 pens and that you test them without replacement makes me think that I'm missing a key step/have a logical flaw somewhere. Can someone tell me if I am a missing a step, because it feels like I am.

答:
You are correct. The number $27$ is a bit of a red herring; it doesn't really matter how many there are (so long as there are at least four).

Another way to do it: suppose that all four pens fail. The chance of this is $$\left(\frac34\right)^4 = \frac{81}{256}.$$

Now, if that didn't happen, then at least one of them worked: $$1-\frac{81}{256}=\frac{175}{256}.$$

This agrees with your answer; the advantage is that it's a much simpler calculation, no matter how many pens you have to test (try both methods with all $27$ instead of $4$ pens...).


问:
Graphs from Seifert surfaces
Given a Seifert surface if we make the disks and bands infinitely small and thin it becomes a graph where the disks are vertices and the bands are edges. Can we say that following theorem,

For every Seifert surface there is an unique graph and vice versa.

is trivial?

答: I believe that twisting a region to get more bigons doesn't change the graph but gives you a seifert surface that is not ambient-isotopic to the original surface.</p>

Title: Graphs from Seifert surfaces


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

Body:
Given $n$, let $L_n=\|l_{ij}\|$, $L’_n=\|l’_{ij}\|$, and $U_n=\|u_{ij}\|$, be $n\times n$ matrices such that $l_{ij}$ equals $1$, if $i\ge j$, and equals $0$, otherwise; $l’_{ij}$ equals $-1$, if $n-1\ge i\ge j$, equals $n$, if $i=n$, and equals $0$, otherwise;
$u_{ij}$ equals $1$, if $j\ge i$, and equals $0$, otherwise.  It is easy to check that $A_n=L_nU_n$ and $B_n=U_nL’_n$. Since $L_n$, $L’_n$, and $U_n$ are triangular matrices with $1$’s on the main diagonal, $\det L_n=\det U_n=1$ and $\det L’_n=(-1)^{n-1}n$, so $\det A_n=1$ and $\det B_n=(-1)^{n-1}n$. Alternatively, we can observe that $A_n J_n=-L_n$ and $J_nB_n=-L’_n$, where $J_n=-U_n^{-1}$ is an $n\times n$ Jordan cell with the eigenvalue $-1$. Since $\det J_n=(-1)^n$, we obtain the same conclusions.

Let us show the decompositions $A_n=L_nU_n$, $B_n=U_nL_n’$, $A_nJ_n=-L_n$, and $J_nB_n=-L’_n$
for $n=4$:
$$
\begin{pmatrix}
1 & 1 & 1 & 1\\
1 & 2 & 2 & 2\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{pmatrix},
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 2 & 3 & 4\\
3 & 3 & 3 & 4\\
4 & 4 & 4 & 4
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 & 0\\
-1 & -1 & 0 & 0\\
-1 & -1 & -1 & 0\\
4 & 4 & 4 & 4
\end{pmatrix},
$$

$$\begin{pmatrix}
1 & 1 & 1 & 1\\
1 & 2 & 2 & 2\\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4
\end{pmatrix}
\begin{pmatrix}
-1 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & -1 & 1\\
0 & 0 & 0 & -1
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 & 0 & 0\\
-1 & -1 & 0 & 0\\
-1 & -1 & -1 & 0\\
-1 & -1 & -1 & -1
\end{pmatrix},$$
$$\begin{pmatrix}
-1 & 1 & 0 & 0\\
0 & -1 & 1 & 0\\
0 & 0 & -1 & 1\\
0 & 0 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 2 & 3 & 4\\
3 & 3 & 3 & 4\\
4 & 4 & 4 & 4
\end{pmatrix}
=\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
-4 & -4 & -4 & -4
\end{pmatrix}.$$


答:
Hint:let the number be $\overline{xyz}$

Now if $x\cdot y\cdot z=p$ where $p=3,5$ then we $x,y,z$ must be a permutation of $1,1,p$ Hence number of such numbers is $2\cdot \frac{3!}{2!}$

Now if $x\cdot y\cdot z=4,6$ then $x,y,z$ must be a permutation of either $1,1,4$ or $2,2,1$ or $1,1,6$ or $2,3,1$ .Each of which the number of cases can be easily found......

Title: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Body:
One  way to solve these types of questions is to go one by one and figure out all the possibilities starting from 111. And I have found till now 19 possibilities for this question. But the answer is somewhat different. It seems that I might have missed some of the possibilities.
I was wondering if there is any other way to approach these types of questions that makes sure that I do not miss any of the possibilities. Any simpler or logical or diagrammatical way to solve these types of questions, that would be really helpful.

Thanks!


问:
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$
I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.

One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)

Body:
$$I_n=\int_{0}^{1}\frac{dx}{1+x+x^2+\ldots+x^n}=\underbrace{\int_{0}^{1}\frac{1-x}{1-x^{n+1}}dx}_{x\rightarrow y^{\frac{1}{n+1}}}=\frac{1}{n+1}\int_{0}^{1}\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y(1-y)}dy$$

$$=\frac{1}{n+1}\int_{0}^{1}\left(\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y}+\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{1-y}\right)dy$$

$$1-\frac{n+1}{2(n+2)}+\frac{1}{n+1}\int_{0}^{1}\left(\frac{y^{\frac{1}{n+1}-1}\log{\left(1-y\right)}}{n+1}-\frac{y^{\frac{2}{n+1}-1}\log{\left(1-y\right)}}{n+2}\right)dy$$

$$=\frac{n+3}{2(n+2)}+\frac{1}{n+1}\left[\frac{\mathfrak{B}\left(\frac{1}{n+1},1\right)\left(\psi^{\left(0\right)}\left(1\right)-\psi^{\left(0\right)}\left(1+\frac{1}{n+1}\right)\right)}{n+1}-\frac{\mathfrak{B}\left(\frac{2}{n+1},1\right)\left(\psi^{\left(0\right)}\left(1\right)-\psi^{\left(0\right)}\left(1+\frac{2}{n+1}\right)\right)}{n+2}\right]$$

$$=\frac{n+3}{2(n+2)}+\frac{\psi^{\left(0\right)}\left(1+\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(1+\frac{1}{n+1}\right)}{n+1}=\frac{1}{2\left(n+2\right)}+\frac{\psi^{\left(0\right)}\left(\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(\frac{1}{n+1}\right)}{n+1}$$

Therefore:
$$I_n=\int_{0}^{1}\frac{dx}{1+x+x^2+\ldots+x^n}=\frac{1}{2\left(n+2\right)}+\frac{\psi^{\left(0\right)}\left(\frac{2}{n+1}\right)-\psi^{\left(0\right)}\left(\frac{1}{n+1}\right)}{n+1}$$
Notes:
$\mathfrak{B}(x,y)$ stands for the Beta Function and $\psi^{(0)}(z)$ stands for the Digamma Function.


问:
What is the simplest way to do this combinatorics problem?
Question: Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to each other?

Here are the solutions:

In this case, we take the first case of all possible combinations: $\frac{8!}{2^4}$
But we see that we counted the possibilities with at least one pair together
so we have to subtract $\frac{7!}{2^3}\cdot \binom{4}{1}$ if we tie the two offending shirts together.

^This I don't get - why and what is being double-counted?

However, we see now that we double subtracted ones that have two pairs thus we must add the chance of two pairs so we must add $\frac{6!}{2^2}\cdot \binom{4}{2}$ and so on

Thus the final answer is equal to $864$

Is there a better way that doesn't involve BOTH adding AND subtracting?

答:
The method in your solutions is actually the Inclusion–exclusion principle. Since $${\displaystyle \left|\bigcup _{i=1}^{n}A_{i}\right|=\sum _{i=1}^{n}|A_{i}|-\sum _{1\leqslant i<j\leqslant n}|A_{i}\cap A_{j}|+\sum _{1\leqslant i<j<k\leqslant n}|A_{i}\cap A_{j}\cap A_{k}|-\cdots +(-1)^{n-1}\left|A_{1}\cap \cdots \cap A_{n}\right|.}$$ the additions and substractions you see are exactly the coefficients $\pm1$ in the formula.

To avoid substractions, you may need to divide the orginal problem into distinct ones, and the number of conditions could be really large. I don't think it's a better way tho.