问:
What is the number of ring isomorphisms from $\mathbb{Z}^n$ to $\mathbb{Z}^n$.
Yesterday, I faced the problem : what is the number of ring isomorphisms from $\mathbb{Z}^2$ to $\mathbb{Z}^2$. And I got $6$. After then, I found that if $n=3$, the number is $3! \times 29$. Suddenly, I wondered what is the number of isomorphisms from $\mathbb{Z}^n$ to $\mathbb{Z}^n$. I have tried to solve it but failed. What's more, I couldn't find any answers about that on google. Is it easy to solve? I tried to solve the problem by using recursive relation. But it looks difficult.

One of my friends gave me a suggestion to find the number of $n \times n$ invertible matrices with components $1$ or $0$. At first, we thought that it is exactly $(2^n-1)(2^n-2) \dots (2^n-2^{n-1})$. But this equation doesn't work because the construction of the equation discards some cases. For example, $n=3$, the equation discards $ (\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{smallmatrix} )$. The equation only works on a finite field. I found that it is a lower bound and upper bound $((n+1)!)^2 2^\frac{n(n+1)}{2}$ by a recursive relation. But I don't know until now what is the exact number of that.

+
Due to Perter Franek, I solved it. The answer is $n!$.
Idempotent elements should go idempotent elements and if $\phi$ is an isomorhpism, then $\phi(0)$ implies $0$. The candidates of $e_i=(\delta_{ij})_j$ are each or sum of $\{e_i\}$'s. But if some $e_i$ goes to sum of $e_i's$, $0=\phi(e_i e_j)= \phi(e_i) \phi(e_j) \not =0$ for some $i \not = j$.

答:
One of my friends gave me a suggestion to find the number of $n×n$ invertible matrices with components $1$ or $0$.

This is OEIS A055165; there appears to be no simple formula for it.


问:
Test the hypothesis ....
A random sample of size 20 from a normal population give a sample mean of 42 and sample
standard deviation of 6. Test the hypothesis that the population mean is 44 at 5 % level of
significance.</p>

答:
First calculate the statistic

$$t=\frac{44-42}{6}\sqrt{20}=1.49$$

then decide, using the fact that $t\sim \mathcal{T}_{(19)}$

ContentLicense: CC BY-SA 4.0

Title: Test the hypothesis ....


问:
Indefinite symmetric quadratic form with inequality constraint
I have a scalar quantity $\Phi =u^TAu$ that I am tracking as part of a greater optimization problem.
$A=I-(1+\mu^2)nn^T$, where $u$ is a 3 x 1 vector, $I$ is the 3 x 3 identity matrix, $\mu$ is a non-negative scalar, and $n$ is a normalized 3 x 1 vector ($n^Tn=1$). The eigenvalues of $A$ are $(1,1,-\mu^2)$. Therefore, $A$ is an indefinite, symmetric matrix. Everything is real-valued.

I wish to constrain my choice of $u$ such that $\Phi\leq0$. Because $A$ is indefinite, this condition is dependent on $u$. How can I enforce this constraint? I was hoping there was a way to define a matrix $X$ such that $u=Xv$, to transform this into an unconstrained optimization problem (similar to a subspace method) but I will appreciate any advice. Thank you!

答:
Let $\frac{1}{\sqrt{1+\mu^2}}=\cos\phi$ where $\phi\in[0,\frac{\pi}{2}]$. Then $\mathbf u^TA\mathbf u\le0$ if and only if $\mathbf u=0$ or $\left|\left\langle\frac{\mathbf u}{\|\mathbf u\|},\mathbf n\right\rangle\right|\ge\cos\phi$. Therefore
$$
\mathbf u=x\left(\cos(\theta)\mathbf n+\sin(\theta)\mathbf t\right)
$$
for some $\theta\in[-\phi,\phi]$ and some unit vector $\mathbf t\perp\mathbf n$. Let $P\in\mathbb R^{3\times2}$ be any matrix whose two columns together with $\mathbf n$ form an orthonormal basis of $\mathbb R^3$. Then we can write
$$
\mathbf u=x\left(
\cos\left(\phi\cos\alpha\right)\mathbf n
+\sin\left(\phi\cos\alpha\right)P\pmatrix{\cos\beta\\ \sin\beta}
\right)
$$
where $x,\alpha$ and $\beta$ are unconstrained.


问:
What is the rank of two independent vector after linear transformation?
Suppose that an $M \times 2$ complex number matrix $\boldsymbol{Q}$ can be partitioned into $\boldsymbol{Q} = [\boldsymbol{q}_1, \boldsymbol{q}_2]$ (notion: in this post, I use square bracket to indicate the matrix), where the column vectors $\boldsymbol{q}_i \in \mathbb{C}^{M},(i=1,2,M >2)$ are independent (equivalently, $rank(\boldsymbol{Q})=2$). Given $\boldsymbol{Q}' = \boldsymbol{q}_1  \boldsymbol{q}_1^H + \boldsymbol{q}_2 \boldsymbol{q}_2^H$, prove that $rank(\boldsymbol{Q}') = 2$

My understand is that, $ \boldsymbol{Q}' = [a, b]\times[a, b]^H = \boldsymbol{Q}\boldsymbol{Q}^H $, which is a linear transformation of the rank-2 matrix $\boldsymbol{Q}$, but there is no theorem telling that after linear transformation what is the rank of $\boldsymbol{Q}$ would be. Instead, there is a lemma showing that any non-singular matrix, e.g. $\boldsymbol{P}_1$, right product to $\boldsymbol{Q}$ would preserve its rank, or $rank(\boldsymbol{Q}\boldsymbol{P}_1) = rank(\boldsymbol{Q})$. Following this lemma, I partition $\boldsymbol{Q}^H $ into $M/2$ (suppose that this is an integer) sub matrix with size $2 \times 2$, $\boldsymbol{Q}^H = [\boldsymbol{P}_1, \boldsymbol{P}_2, \dots,\boldsymbol{P}_{M/2}]$, thus  $\boldsymbol{Q}' = [\boldsymbol{Q}\boldsymbol{P}_1, \boldsymbol{Q}\boldsymbol{P}_2, \dots, \boldsymbol{Q}\boldsymbol{P}_{M/2}]$ , for each block $rank(\boldsymbol{Q}\boldsymbol{P}_i) = rank(\boldsymbol{Q})=2 , (i=1,\dots, M/2)$, but to this point, it seems that there is no lemma showing that what is the relationship for the rank between $[\boldsymbol{Q}\boldsymbol{P}_1, \boldsymbol{Q}\boldsymbol{P}_2, \dots, \boldsymbol{Q}\boldsymbol{P}_{M/2}]$ and its block matrix $\boldsymbol{Q}\boldsymbol{P}_i$, so I get stuck here, anybody has a good idea ?

答:
$Q'$ has rank $2$: Its image is spanned by $q_1,q_2$

$$Q'v=q_1(q_1^*v)+q_2(q_2^*v)\in\mathrm{Span}(q_1,q_2)$$

...and $Q'q_1$, $Q'q_2$ are independent: $$0=\alpha Q'q_1+\beta Q'q_2=\alpha (q_1^*q_1)q_1+\alpha (q_2^*q_1)q_2+\beta (q_1^*q_2)q_1+\beta (q_2^*q_2)q_2$$ $$\implies \alpha\|q_1\|^2+\beta(q_1^*q_2)=0=\alpha(q_2^*q_1)+\beta\|q_2\|^2$$ since $q_1,q_2$ are independent. But this in turn implies $\alpha=0=\beta$ since $$Q^*Qx=\begin{pmatrix}\|q_1\|^2&q_1^*q_2\\q_2^*q_1&\|q_2\|^2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$\det(Q^*Q)=\|q_1\|^2\|q_2\|^2-|q_1^*q_2|^2>0$ since equality for Cauchy-Schwarz only holds if $q_1,q_2$ are dependent.


问:
Examples of conjectures that were widely believed to be true but turned out to be false
Note: This is the same question, but it does not have enough answers and it is almost a year old. This question is similar, with many answers, but a conjecture is similar to, but not the same as a pattern. For example, one of the answers is:

"Choose n points around the circumference of a circle, and join every point to every other with a line segment. Assuming that no three of the line segments concur, how many regions does this divide the circle into?"

It is clear that this is not a conjecture.

Mathematical conjectures are proposed (usually) after being computationally tested for many cases. Because of this computational evidence, many are thought to be probably true, and many theorems are derived considering that they are true. But this not always the case. Some conjectures that were thought to be true (because of computational evidence) were false, for example that $\pi(x)-\mathrm{li}(x)$ is always positive (disproved by Littlewood). So my question is

What are some conjectures that were widely believed to be true but turned out to be false?

Like many big list question, please give only one example per answer.

Some include (these are from the answers, I wrote these because if this question gets many answers, than it would be inconvenient to scroll all the way down to see them):

Squaring the circle

Euler's sum of powers conjecture

答:
Euler's sum of powers conjecture (https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture) was an attempt to generalise the statement of Fermat's Last Theorem. No counterexamples were found until 1966.

From the article in Wikipedia:

$$27^5 + 84^5 + 110^5 + 133^5 = 144^5$$

was the first counterexample found by computer, by Lander and Parkin: the sum of $4$ fifth powers can be a fifth, whereas the conjecture says that you require at least $5$.


问:
Drawing from a deck with replacement, expected number of cards drawn to see all 4 suites
Similar questions have been asked before but I am looking for a specific way to solve this question (namely using a recurrence relation).

Assume we have a fair deck of $52$ cards. In each iteration, we will draw one card, record the suite, and then replace it in the deck. The cards are drawn uniformly and randomly from the deck. How many times would we expect to draw cards before we see at least $1$ of each suite?

I want to solve this using a recurrence relation so something of the form:

$X = 1 + \frac{3}{4}(x) + \frac{1}{4}(x+1)...$

#This is not correct just the format I want to solve the question in.

The solution is 8$\frac{1}{3}$.

As a reference, I am trying to use the same logic as presented here:
https://www.geeksforgeeks.org/expected-number-of-coin-flips-to-get-two-heads-in-a-row/

答:
As @GregMartin referenced in his comment, this is an example of the "coupon collector problem", whose solution can be found by using linearity of expectation.

Suppose we perform a binomial experiment with probability of success $p$. Let $E(S)$ be the number of expected trials until we succeed. At any point in the experiment, if we have not yet succeeded, then if the next trial fails (with probability $1-p$), the expected number of further trials until success will still be $E(S)$. This leads to the recurrence

$$E(S) = 1 + (1-p)E(S)$$

which yields $E(S) = \frac{1}{p}$.

Now applying this to the question at hand. Let $p_i$ be the probability of drawing the $i$th different suit, and $X_i$ be the expected number of draws to get the $i$th different suit. Then clearly $p_i = \frac{4-(i-1)}{4}$, so by the above result, we have that $E(X_i) = \frac{4}{4-(i-1)}$. Then by linearity of expectation, we get that

$$E \left[ \sum_{i=1}^4 X_i \right] = \sum_{i=1}^4 E(X_i) = \frac{4}{4} + \frac{4}{3} +\frac{4}{2} +\frac{4}{1} = 8\frac{1}{3}$$


答:
Ignore the ranks, then we have $13$ cards of each suit. Since this is with replacement, we might as well have a deck of only $4$ cards, one of eah suit. Then, supposing $n$ suits have not been seen, the number of cards that must be drawn to see a new suit is geometrically distributed with success probability $\frac n4$, with expectation $\frac4n$.

The expected number of draws to see all four suits is then
$$\frac41+\frac42+\frac43+\frac44=\frac{25}{3}$$

Title: Drawing from a deck with replacement, expected number of cards drawn to see all 4 suites

Body:
Similar questions have been asked before but I am looking for a specific way to solve this question (namely using a recurrence relation).

Assume we have a fair deck of $52$ cards. In each iteration, we will draw one card, record the suite, and then replace it in the deck. The cards are drawn uniformly and randomly from the deck. How many times would we expect to draw cards before we see at least $1$ of each suite?

I want to solve this using a recurrence relation so something of the form:

$X = 1 + \frac{3}{4}(x) + \frac{1}{4}(x+1)...$

#This is not correct just the format I want to solve the question in.

The solution is 8$\frac{1}{3}$.

As a reference, I am trying to use the same logic as presented here:
https://www.geeksforgeeks.org/expected-number-of-coin-flips-to-get-two-heads-in-a-row/


问:
Algebra exercise
If $a^4+a^3+a^2+a+1=0$ find the value of $a^{2000}+a^{2010}+1$

I got this problem in a book and tried to solve it.I multiplied with suitable powers of a and added and subtracted alternatively to get $a^{2010}+a^{2008}+a^{2006}+a^{2005}+a^{2004}+a^{2002}+a^{2000}=0$ but i can't figure what to do.I tried to replace $a^{2005}$ by multiplying the parent equation by $a^{2003}$ but nothing useful came.

Any help would be appreciated.Thanks in advance.

Title: Algebra exercise

Body:
I used @lab bhattacharjee's hint.
$\frac{a^5-1}{a-1}=0$
$a\not= 1$ and $a_k=e^{\frac{2\pi ik}{5}}$ for $k=1,2,3,4$

$a^{2000}+a^{2010}+1=\left(e^{2\pi i}\right)^{400k}+\left(e^{2\pi i}\right)^{402k}+1=3$


问:
Why is the SVM margin equal to $\frac{2}{\|\mathbf{w}\|}$?
I am reading the Wikipedia article about Support Vector Machine and I don't understand how they compute the distance between two hyperplanes.

In the article,

By using geometry, we find the distance between these two hyperplanes
  is $\frac{2}{\|\mathbf{w}\|}$

I don't understand how the find that result.

What I tried

I tried setting up an example in two dimensions with an hyperplane having the equation $y = -2x+5$ and separating some points $A(2,0)$, $B(3,0)$ and $C(0,4)$, $D(0,6)$ .

If I take a vector $\mathbf{w}(-2,-1)$ normal to that hyperplane and compute the margin with $\frac{2}{\|\mathbf{w}\|}$ I get $\frac{2}{\sqrt{5}}$ when in my example the margin is equal to 2 (distance between $C$ and $D$).

How did they come up with $\frac{2}{\|\mathbf{w}\|}$ ?

答:
Let $\textbf{x}_+$ be a positive example on one gutter, such that
$$\textbf{w} \cdot \textbf{x}_+ - b = 1$$

Let $\textbf{x}_-$ be a negative example on another gutter, such that $$\textbf{w} \cdot \textbf{x}_- - b = -1$$

The width of margin is the scalar projection of $\textbf{x}_+ - \textbf{x}_-$ on unit normal vector , that is the dot production of $\textbf{x}_+ - \textbf{x}_-$ and $\frac{\textbf{w}}{\|\textbf{w}\|}$

\begin{align}
width & = (\textbf{x}_+ - \textbf{x}_-) \cdot \frac{\textbf{w}}{\|\textbf{w}\|} \\
      & = \frac {(\textbf{x}_+ - \textbf{x}_-) \cdot {\textbf{w}}}{\|\textbf{w}\|} \\
      & = \frac{\textbf{x}_+ \cdot \textbf{w} \,{\bf -}\, \textbf{x}_-\cdot \textbf{w}}{\|\textbf{w}\|} \\
      & = \frac{1-b-(-1-b)}{\lVert \textbf{w} \rVert} \\
      & = \frac{2}{\|\textbf{w}\|}
\end{align}

The above refers to MIT 6.034 Artificial Intelligence


问:
Problem about Probability theory : marginal density.
I'm studying 'Probability theory' and I cannot figure out the way to find marginal density.

The problem is :

Take A to be the square with corners at (0,1), (1,0), (2,1), (1,2). Find the marginal densities of f = indicator function of A.

The solution is :

Fix $x \in [0,2], \int_0^2 \mathbf{1}_A (x,y)\, dy = m(A_x)$, hence $$
f_X(x) =
\begin{cases}
x, ~\text{for}~x \in [0,1] \\
2-x, ~\text{for}~x \in [1,2] \\
0, \text{otherwise}
\end{cases}
$$(triangle distribution). By symmetry, the same holds for $f_Y$.

I cannot figure out how to find f_X(x) with integral part!

I think f_X(x) should be '2x' for x in [0,1], 4-2x for x in (1,2], and 0 for otherwise.

Please give me some advice!!

答:
The joint distribution is uniform on the square $A$,

Thus

$$f_{XY}(x,y)=\frac{1}{2}\mathbb{1}_{[(x,y)\in A]}$$

Then do a drawing and calculate the marginal $f_X(x)=\int_{-\infty}^{+\infty}f(x,y)dy$

When $0<X<1$ the integral is the following

$$f_X(x)=\frac{1}{2}\int_{1-x}^{x+1}dy=x$$

When $1<X<2$ the integral is the following

$$f_X(x)=\frac{1}{2}\int_{x-1}^{3-x}dy=2-x$$

The result is a triangular distribution that you can write in a more compact way

$$f_X(x)=[1-|1-x|]\cdot\mathbb{1}_{[0;2]}(x)$$

Same result for $f_Y(y)$ using symmetry properties


问:
Relationship between nullspaces of A when $rank(A) = rank(A^2)$
I have a square matrix A such that $rank(A) = rank(A^2)$. I am first tasked to prove that nullspace of $A$ is equal to the nullspace of $A^2$. For this, my reasoning is:

Let $v$ be a solution to the nullspace of $A$. As such, $v$ will also be a solution to the nullspace of $A^2 = A*A$. Hence nullspace $A \subseteq \ $subset of $A^2$.Then, because $rank(A) = rank(A^2)$, $dim(A) = dim(A^2)$. Hence nullspace of $A$ = nullspace of $A^2$. How can I make this more airtight as a proof?

In addition, how can I possibly extend this concept to show that (nullspace of $A$) $\cap$ (column space of $A$) = $\{0\}$? I don't quite understand the relationship between the nullspace and column space.

Thank you!

答:
I don't understand what $\dim A$ means. Perhaps you want to say nullity of $A$. Then your proof is "air-tight" because $\text{rank}(A)=\text{rank}(A^2)\implies \text{nullity}(A)=\text{nullity}(A^2)=n-\text{rank}(A)$ by the rank-nullity theorem, where $n$ is the size of $A$.

The column space of $A$ is the range space of $A$. Let $v$ belong to both the range space and null space of $A$. Then $\exists u|Au=v$ and $Av=0$. This gives $A(Au)=A^2u=0$, thus $u$ belongs to the null space of $A^2$. But since the null space of $A$ is the same as the null space of $A^2$, $u$ belongs to the null space of $A$. Thus $v=Au=0$.


问:
Argument Valid or Not
Premise $1$ : $\neg p \wedge q$

Premise $2$ : $\neg p \implies \neg r$

Premise $3$ : $\neg r \implies s$

Premise $4$ : $s \implies t$

Conclusion : $t$

How to check this argument is valid or not?

Thanks

答:
Consider the premisses written in argument form as follows.

$$\begin{align}
(1) \quad & \neg p \wedge q \\
(2) \quad & \neg p \implies \neg r \\
(3) \quad & \neg r \implies s \\
(4) \quad & s \implies t\\
\end{align}$$

Now we will use some Rules of Inference to deduce the conclusion from the premisses. And we do it as follows.

$$\begin{align}
& (5) \quad \neg p & \text{($(1),$ Simplification)}\\
& (6) \quad \neg p \implies s & \text{($(2),(3),$ Hypothetical & Syllogism)}\\
& (7) \quad \neg p \implies t & \text{($(6),(4),$ Hypothetical Syllogism)}\\
\therefore \quad & (8) \quad t & \text{($(7),(5),$ Modus Ponens)}\\
\end{align}$$

We have just prove that

$$\big( (\neg p \wedge q) \wedge (\neg p \implies \neg r) \wedge (\neg r \implies s) \wedge (s \implies t) \big) \implies t \quad \square$$

Title: Argument Valid or Not


问:
Are all multivariate functions on finite fields equivalent to a unique polynomial of smallest degree?
Consider a field field $\mathbb{F}$ and a function $f:\mathbb{F}^n\rightarrow\mathbb{F}$. Let $P$ be the set of all polynomials that agree with $f$ on all inputs, that is, $P=\{p:\forall x\in\mathbb{F}^n,p(x)=f(x)\}$. Because there always exists some n-variate polynomial $p$ such that $p(x) = f(x)$, we know that $P\neq\emptyset$. Therefore we can define a set $L$ consisting of all elements of $P$ with lowest degree, that is, $L=\{p\in P:\forall q\in P,deg(p)≤deg(q)\}$.

Must it be the case that $|L|=1$?

Here is my attempt at proving so:

Assume $p,q$ are different polynomials, both of lowest degree $d$. Their difference is a polynomial of degree $d$ or lower, and as a function, takes all elements of $\mathbb{F}^n$ to $0$. I'm not sure what to do next.

NOTE

If it is possible for there to be multiple polynomials of lowest degree, (equivalently, $|L|>1$), I would be interested in knowing for which finite fields and values of n this is the case.

答:
If $A_1$, $A_2$, $\ldots$, $A_n$ are finite subsets of a field $\mathbb{F}$, then any function $f\colon A_1\times \cdots \times A_n\to \mathbb{F}$ is given by a unique polynomial $p\in \mathbb{F}[x_1, \ldots, x_n]$, with $\deg_{x_i}p \le |A_i|-1$. This is basically the Lagrange interpolation polynomial.

In the case of  a finite field $\mathbb{F}$ of cardinality $q$, your unique minimal polynomial will have the degree in each variable $\le q-1$.

How to get the minimal polynomial from a polynomial? Note that we can substitute any $x_i^q$ with $x$. Thefore, we can substitute any $x_i^n$ $n\ge q$ with $x_i^m$, $m\in \{1,2,\ldots, q-1\}$, and $n\equiv m \mod (q-1)$. This should give the minimal polynomial, and also show the uniqueness.


答:
Any ring homomorphism $\phi$ sends idempotents (elements $e$ satisfying $e^2=1$) to idempotents. The idempotents of $\mathbb{Z}^n$ are the vectors with coordinates all $0$ or $1$. These correspond to subsets of $\{1,\cdots,n\}$ (to describe where the $1$s occur in the coordinates). Thus, if $e_1,\cdots,e_n$ refer to the standard basis vectors of $\mathbb{Z}^n$, there are subsets $A_1,\cdots,A_n\subset\{1,\cdots,n\}$ for which

$$ \phi(e_i)=\sum_{a\in A_{\Large i}}e_a. $$

Moreover, ring homomorphisms preserve zero division. That is, $e_ie_j=0$ (when $i\ne j$) implies

$$ 0=\phi(e_i)\phi(e_j)=\Big(\sum_{a\in A_{\Large i}}e_a\Big)\Big(\sum_{b\in A_{\Large j}} e_{b}\Big)=\sum_{c\in A_{\Large i}\cap A_{\Large j}}e_c.  $$

This implies the $A_1,\cdots,A_n$ are disjoint. They must be nonempty, else $\phi$ would have a kernel, so

$$ n=1+\cdots+1\le |A_1|+\cdots+|A_n|=|A_1\cup\cdots\cup A_n|\le n. $$

The only way this inequality is squeezed so tightly is if $|A_i|=1$ for each $i$. Thus, $\phi$ simply permutes the standard basis vectors $e_1,\cdots,e_n$, and there are $n!$ such permutations!

Title: What is the number of ring isomorphisms from $\mathbb{Z}^n$ to $\mathbb{Z}^n$.

Body:
Yesterday, I faced the problem : what is the number of ring isomorphisms from $\mathbb{Z}^2$ to $\mathbb{Z}^2$. And I got $6$. After then, I found that if $n=3$, the number is $3! \times 29$. Suddenly, I wondered what is the number of isomorphisms from $\mathbb{Z}^n$ to $\mathbb{Z}^n$. I have tried to solve it but failed. What's more, I couldn't find any answers about that on google. Is it easy to solve? I tried to solve the problem by using recursive relation. But it looks difficult.

One of my friends gave me a suggestion to find the number of $n \times n$ invertible matrices with components $1$ or $0$. At first, we thought that it is exactly $(2^n-1)(2^n-2) \dots (2^n-2^{n-1})$. But this equation doesn't work because the construction of the equation discards some cases. For example, $n=3$, the equation discards $ (\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{smallmatrix} )$. The equation only works on a finite field. I found that it is a lower bound and upper bound $((n+1)!)^2 2^\frac{n(n+1)}{2}$ by a recursive relation. But I don't know until now what is the exact number of that.

+
Due to Perter Franek, I solved it. The answer is $n!$.
Idempotent elements should go idempotent elements and if $\phi$ is an isomorhpism, then $\phi(0)$ implies $0$. The candidates of $e_i=(\delta_{ij})_j$ are each or sum of $\{e_i\}$'s. But if some $e_i$ goes to sum of $e_i's$, $0=\phi(e_i e_j)= \phi(e_i) \phi(e_j) \not =0$ for some $i \not = j$.


问:
Counting probability of disease
There are 2 types of people in a big factory: prone to COVID (30% of workers), and nonprone to COVID (70% of workers). For a COVID-prone person, p(COVID) = $0.4$. For a nonprone to COVID person, p(COVID) = $0.2$.

Compute:

a) the probability a random factory worker will have a COVID

b) given a person who has a COVID, what is p(person is COVID-prone).

I'm not sure, should I use binomial probability or usual, or how I should count in general. I suppose in (a) I need to calculate $0.3*0.4 + 0.7*0.2 = 0.26$

答:
For a) you are correct. The probability that someone has Covid is the sum of the probability for a covid-prone person to get it, plus the probability for a non-prone person to get it, or
$$0.4\times 0.3 +0.2\times 0.7 = 0.26$$

For b) you want to know the probability for a person who has covid to be covid-prone. The probability to have covid if you are prone is
$$0.4\times 0.3$$and since you know that the person has already covid you divide by the probability of the known event calculated above (see conditional probability)
$$\dfrac{0.4\times 0.3}{0.26} \approx 0.4615$$

Title: Counting probability of disease


答:
If we write everything we know, we got :

$P(Covid~\vert~prone~Covid) = 0.4 $

$P(prone~Covid) = 0.3$

$P(non-prone~Covid) = 0.7$

$P(Covid~\vert~non-prone~Covid) = 0.2$

We can deduce : $P(Covid) = P(Covid\bigcap prone~Covid) + P(Covid \bigcap non-prone~ Covid)$

This expression corresponds to the answer you found for a)

For b) : You have to calculate $P(prone~Covid \vert ~Covid)$ using what you found previously

Title: Counting probability of disease

Body:
There are 2 types of people in a big factory: prone to COVID (30% of workers), and nonprone to COVID (70% of workers). For a COVID-prone person, p(COVID) = $0.4$. For a nonprone to COVID person, p(COVID) = $0.2$.

Compute:

a) the probability a random factory worker will have a COVID

b) given a person who has a COVID, what is p(person is COVID-prone).

I'm not sure, should I use binomial probability or usual, or how I should count in general. I suppose in (a) I need to calculate $0.3*0.4 + 0.7*0.2 = 0.26$


问:
Find number of words of length $n$ which can be written using letters: $\{A,B,C,D,E\}$, but letter $A$ has to appear even number of times.
Find number of words of length $n$ which can be written using letters: $\{A,B,C,D,E\}$, but letter $A$ has to appear even number of times.

I was thinking of stars and bars method so I started this way:

$x_1+x_2+x_3+x_4+x_5=n$, where $x_i\geq 0$ and $x_1$ is an even number. I dont know if in this task $A$ can appear $0$ times since this is a question from an old test. Let's say that it can appear $0$ times. Now when I try to substitute $y_1=\frac{x_1}{2}$ and $y_i=x_i, i=2,3,4,5$, I do not know what to do with $n$ on the right side of the equation, if this is the right approach in the first place. Since these are combinations I would have to permute everything at the end...

答:
For the first $n-1$ slots, you have $5$ choices each. For the last slot, you must fill $A$ if the first $n-1$ slots have odd number of $A$s, and anything except $A$ if there are even number of $A$s in the first $n$ slots. Let $f(n)$ be the number of ways of filling $n$ slots such that there are even number of $A$s. Then:
$$f(n)=(5^{n-1}-f(n-1))(1)+f(n-1)(4)=5^{n-1}+3f(n-1)$$

Then, expanding this recursion:
$$f(n)=5^{n-1}+3f(n-1)=5^{n-1}+3\cdot5^{n-2}+3^2f(n-2)=\cdots$$
$$f(n)=5^{n-1}+3\cdot5^{n-2}+\cdots+3^{n-1}+3^nf(0)$$
Since $f(0)=0$, we have:
$$f(n)=(5^{n-1}+3\cdot5^{n-2}+\cdots+3^{n-1})+3^n=\frac{5^n-3^n}{2}+3^n$$
$$f(n)=\frac{5^n+3^n}{2}$$


答:
Take, for example $\mathbb{F} = \mathbb{Z}/2\mathbb{Z}$, $n=2$, and $f(x, y) = xy$. It is easy to check that no polynomial of degree $\leq 1$ agrees with this function on all inputs, i.e., the minimal degree is $2$. But $xy + x(x-1)$ is another polynomial of degree $2$ which agrees with the function on all inputs.

A similar example can be constructed whenever $n\geq |\mathbb{F}|$.

Edit: actually, $n=2$ seems to be enough for any finite field: take $p(x,y) = \left(\prod\limits_{a\in\mathbb{F}\backslash \{0\}} (x-a)\right) y$ and $q(x,y) = p(x,y) + \prod\limits_{a\in\mathbb{F}} (x-a)$.

Title: Are all multivariate functions on finite fields equivalent to a unique polynomial of smallest degree?

Body:
Consider a field field $\mathbb{F}$ and a function $f:\mathbb{F}^n\rightarrow\mathbb{F}$. Let $P$ be the set of all polynomials that agree with $f$ on all inputs, that is, $P=\{p:\forall x\in\mathbb{F}^n,p(x)=f(x)\}$. Because there always exists some n-variate polynomial $p$ such that $p(x) = f(x)$, we know that $P\neq\emptyset$. Therefore we can define a set $L$ consisting of all elements of $P$ with lowest degree, that is, $L=\{p\in P:\forall q\in P,deg(p)≤deg(q)\}$.

Must it be the case that $|L|=1$?

Here is my attempt at proving so:

Assume $p,q$ are different polynomials, both of lowest degree $d$. Their difference is a polynomial of degree $d$ or lower, and as a function, takes all elements of $\mathbb{F}^n$ to $0$. I'm not sure what to do next.

NOTE

If it is possible for there to be multiple polynomials of lowest degree, (equivalently, $|L|>1$), I would be interested in knowing for which finite fields and values of n this is the case.


问:
Calculate the joint PDF of the following random variables.
Suppose $\theta$ and $R$ are independent random variables, with $\theta$ being Uniform$(−\pi/2,\pi/2)$ and $R$ having the probability density function given by:

$$f_R(r)=\frac{2r}{(1+r^2)^2} \:\text{for } r>0, \quad f_R(r)=0 \:\text{otherwise}.$$

Determine the joint PDF for $X=R\cos\theta$ and $Y=R\sin\theta$.

It is simple enough to calculate the cumulative distribution functions of $\theta$ and $R$, having

$$F_\theta(x)=\frac{1}{\pi}x+\frac{1}{2} \:\text{ on }\: (-\pi/2,\pi/2),\quad \text{and } \:F_r(x)=\frac{x^2}{x^2+1}\:\text{ on }\:(0,\infty),$$

with my aim being eventually to calculate $F_{XY}(x,y)$ from which, by taking partial derivatives, I would be able to obtain $f_{XY}(x,y).$

However, $F_{XY}(x,y)=\mathbb{P}[X=R\cos\theta\le x,Y=R\sin\theta\le y],$ and no matter how much I try to interpret this expression, I can't seem to find a way to bring it into terms of $F_r$ and $F_\theta$ as would lead (I imagine) to the solution.

Is this the correct way to tackle this sort of problem? If so, any hints?

答:
Let me give a different exposition of essentially the same answer as Graham Kemp's.

You want to find
$$F_{XY}(x,y) = P[X≤x,Y≤y] = \int_{-\infty}^x\int_{-\infty}^y f_{XY}(\xi, \eta) d\xi d\eta$$
and, because you know the joint density function $f_{R \theta}$ of $R$ and $\theta$, you can compute the probability of an event of the form $\{(R, \theta) \in A\}$ as $$P[(R, \theta) \in A] = \iint_A f_{R\theta}(r, t) dr dt.$$

The key now is to use the change of variables formula, stated for an integrable function $h$ and a domain $U$ as
$$\iint_{\phi(U)} h(\mathbf v) d\mathbf v = \iint_{U} h(\phi(\mathbf u)) \det(D\phi)(\mathbf u) d\mathbf u,$$
to calculate the probability of the event $\{X≤x,Y≤y\}$ using the known density $f_{R\theta}$. Thus,

Let $\mathbf v = (r, t)$  and $\mathbf u = (\xi, \eta)$

Let $U = (-\infty, x] \times (-\infty, y]$, so that $P[X≤x,Y≤y] = P[(X, Y) \in U]$

Let $\phi : \mathbb R^2 \to \mathbb R^2$ be such that $\phi(X,Y) = (R, \theta)$. That is, $\phi(\xi, \eta) = (\sqrt{\xi^2 + \eta^2}, \arctan(\eta / \xi))$

Let $h=f_{R\theta}$.

Then, $P[X≤x,Y≤y] = P[(X, Y) \in U] = P[\phi(X,Y) \in \phi(U)] = P[(R, \theta) \in \phi(U)]$ and, by definition of density and the change of variables formula, this last expression is equal to
$$\iint_{\phi(U)} f_{R\theta}(r, t) dr dt = \iint_{U} f_{R\theta}(\phi(\xi, \eta)) \det(D\phi)(\xi, \eta) d\xi d\eta.$$

Finally, since $U$ is arbitrary, you can conclude that $$f_{XY}(\xi, \eta) = f_{R\theta}(\phi(\xi, \eta)) \det(D\phi)(\xi, \eta) \\ =
f_{R}(\sqrt{\xi^2 + \eta^2}) f_\theta(\arctan(\eta / \xi)) \det(D\phi)(\xi, \eta)
,$$ and now all that is left is to compute the Jacobian $\det(D\phi)$.

Note that the description of the set $\phi(U)$ in terms of the variables $r$ and $t$ is not easy to write, but it is not necessary to do so, as it is possible to express probabilities of arbitrary events $P[(R, \theta) \in A]$ as integrals of the density over abstract sets $A$.


问:
Open maps failing separation axioms
Can one give an example of a surjective open mapping between topological spaces with domain satisfying the $T_0$ separation axiom, whose range is not $T_0$? Similarly, for $T_2$...</p>

答:
Hint: Consider the mapping $f : \mathbb{R} \to \{ 0 , 1\}$ defined by $$f(x) =
\begin{cases}
0, &\text{if }x\text{ is rational} \\
1, &\text{if }x\text{ is irrational.}
\end{cases}$$
Find a non-T$_0$-topology on $\{0,1\}$ for which $f$ is a continuous open surjection.

ContentLicense: CC BY-SA 3.0

Title: Open maps failing separation axioms


问:
Showing that, if $\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$, then $\sqrt2\sin\beta=\sin\alpha-\cos\alpha$

If
$$\tan\beta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$$
then prove that
$$\sqrt2\sin\beta=\sin\alpha-\cos\alpha$$

I have been trying to solve this exercise but I don't get it. I need help. Thanks.

答:
You can use the fact that
$$
\sin^2\beta=\frac{\sin^2\beta}{\cos^2\beta+\sin^2\beta}=\frac{\tan^2\beta}{1+\tan^2\beta}
$$
and this will show that
$$
\sin^2\beta=\frac{(\sin\alpha-\cos\alpha)^2}{2}\tag{*}
$$
On the other hand, the statement is generally false. Take $\alpha=\pi/3$ and $\beta=\pi+\arctan(2-\sqrt{3})$, after having observed that
$$
\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}=2-\sqrt{3}
$$
Then $\sin\beta<0$, but $\sin\alpha-\cos\alpha>0$.

Therefore (*) is the best you can prove.


问:
Relation between the function image and function divergence
Let $D = \{x\in \mathbb{R}~|~x\geq 0\}$ and $f:D \rightarrow \mathbb{R}$ be a continuous function with $f[D] = D$ ($f[D]$ is the image of $D$ via $f$). Then, I am going to show that $f(x) \rightarrow \infty $ as $x \rightarrow \infty$.

Let us start by assuming $f(x)$ converges to $c \geq 0$.
$$
\forall \epsilon > 0 \ \exists a \in D : \forall x \in D: x > a \implies |f(x) - c | < \epsilon \implies f(x) \leq c + \epsilon
$$
Then, $a = a(\epsilon)$. Meanwhile, by the extreme value theorem,
$$
\forall a \in D \ \exists b \in D : \forall x \in [0,a] : f(x) \leq b
$$
Then, $b = b(a)$.
Therefore, we have
$$
\forall \epsilon > 0 \ \forall x \in D : f(x) \leq \max \{ c + \epsilon, b(a(\epsilon)) \} < \infty
$$
hence $f[D] \neq D$, which is a contradiction.

Is this correct?

答:
No, what you want to prove is false, if you do not assume also that $f$ is monotonic increasing. Indeed, there exists a (non-monotonic) continuous function $f \colon D \to \mathbb{R}$ such that $f[D] = D$ but $f(x) \not\to \infty$ as $x \to \infty$. For instance, consider
$$\tag{1} f(x) = e^x |\sin(x)|$$
The $\sin(x)$ component gives an oscillating behavior to $f$, and the amplitude of these oscillations grows as $x \to \infty$ because of the $e^x$ component. Since $|\sin(x)| \geq 0$ for every $x \in \mathbb{R}$ and in particular for every $x \in D$, we have that $f[D] = D$.

What is wrong in your proof is the first step. You want to prove, by contradiction, that for every continuous function $f \colon D \to \mathbb{R}$, if $f[D] = D$ then $f(x) \to \infty$ as $x \to \infty$. So, you assume that there exists a continuous function $f \colon D \to \mathbb{R}$ such that $f[D] = D$ but $f(x) \not\to \infty$ as $x \to \infty$ and you want to show that this leads to a contradicton. But (and this is your error) the fact that $f(x) \not\to \infty$ as $x \to \infty$ does not imply that $f(x)$ converges to some $c$ in $f[D]$. Indeed, there is the possibility that $f(x) \not\to \infty$ as $x \to \infty$  because the limit of $f(x)$ as $x \to \infty$ does not exist! And this is what happens in the function $(1)$.

Actually, your proof is correct but for another statement: for every continuous function $f \colon D \to \mathbb{R}$, if $f[D] = D$ then it is impossible that $f(x)$ converges to $c$ (for any $c \geq 0$) as $x \to \infty$.


答:
We can use reccurrence relations. Let's say $a_n$, number of even numbers of times of letter $A$ with $n$ letters. Also $b_n$ number of odd numbers of times of letter $A$ with $n$ letters. Therefore for $n\geq 1$, $$a_n +b_n=5^n \tag{1}$$.

Otherhands, for $a_{n+1}$; if last letter is $A$ then number of this sub-case: $b_n$, if last letter is $B,C,D$ or $E$ then number of this sub-case: $4a_n$. Hence we yieldes for $n\geq 1$,
$$ a_{n+1}=4a_n + b_n \tag{2}$$

By $(1)$ and $(2)$, we find $a_{n+1}-3a_n=5^n$. Easly we can see that $a_1=4$, $a_2=17$. By $a_{n+1}-3a_n=0$ homogen form and with term $5^n$; roots of characteristic polynom of this reccurrence relation are $3$ and $5$. That is $a_n$ will be form:

$$ a_n =C_15^n +C_23^n \tag{3}$$

By using $a_1=4$, $a_2=17$ values at $(3)$: we yields $C_1=C_2=\dfrac{1}{2}$. Therefore,
$$ a_{n} = \dfrac{1}{2}(5^n + 3^n).$$

Body:
Find number of words of length $n$ which can be written using letters: $\{A,B,C,D,E\}$, but letter $A$ has to appear even number of times.

I was thinking of stars and bars method so I started this way:

$x_1+x_2+x_3+x_4+x_5=n$, where $x_i\geq 0$ and $x_1$ is an even number. I dont know if in this task $A$ can appear $0$ times since this is a question from an old test. Let's say that it can appear $0$ times. Now when I try to substitute $y_1=\frac{x_1}{2}$ and $y_i=x_i, i=2,3,4,5$, I do not know what to do with $n$ on the right side of the equation, if this is the right approach in the first place. Since these are combinations I would have to permute everything at the end...


答:
Well, this is embarassing.  I am going to leave a comment to the OP suggesting that she un-accept this answer and accept one of the other answers.  The answer below is wrong because I misinterpreted the question.

By converting the question into counting the number of solutions to $2x_1 + x_2 + x_3 + x_4 + x_5 = n$, I overlooked that the OP was not asking for how many possible satisfying combinations of n characters there were that used the 5 letters.  Instead, the OP was asking what are the total number of ways of permuting any satisfying combination into a $n$ character string.  That is, the OP was asking how many $n$ character words could be formed, that used the 5 characters (with A used an even # of times).

There is no simple way to convert my answer into the requested answer, because various individual solutions to $2x_1 + x_2 + x_3 + x_4 + x_5 = n$ will permit a variable number of permutations (i.e. words).  Because of this, Stars and Bars analysis, which is what I used, is totally inappropriate here.

I am going to leave this answer in, as another example of going off the rails.

There are two approaches.  One, as suggested by N. F. Taussig's comment, is:

compute $c = \left\lfloor \frac{n}{2}\right\rfloor.$

let $x_1$ range from $\{0,1,\cdots, c\}.$

for each value of $x_1$, determine the number of solutions as a function of both $x_1$ and $n$.

express the total number of solutions as a summation.

The alternate approach involves generating functions.

Both approaches may be viewed by delving into links that start with this.

I suspect that when the Stars and Bars problem gets this complicated, generating functions may be preferable.  Unfortunately, I don't know generating functions.  Therefore, I will map out the Stars and Bars approach.

Let $c \equiv \left\lfloor \frac{n}{2}\right\rfloor.$

For each $x_1$ in the range from $\{0, 1, \cdots, c\}$,

let $f(x_1,n) \equiv$ the # of non-negative integer solutions to

$x_2 + x_3 + x_4 + x_5 = (n - 2x_1).$

From Stars and Bars analysis,

$$f(x_1,n) = \binom{[n - 2x_1] + [4-1]}{4-1} = \binom{n + 3 - 2x_1}{3}.$$

Then the overall answer will be

$$\sum_{x_1 = 0}^c f(x_1,n) ~=~ \sum_{x_1 = 0}^c \binom{n + 3 - 2x_1}{3}.$$

Title: Find number of words of length $n$ which can be written using letters: $\{A,B,C,D,E\}$, but letter $A$ has to appear even number of times.


问:
why $|A_{E}(f) - \alpha| \le r $ give contradiction?
I have  some confusion in RCA book

My confusion given below  and marked in red box

My attempt  : Here  we  have already taken  $r >0 $ that is $r$ is positive

$|A_{E}(f) - \alpha|$is also $+ve$  since Mod  is always $+ve$

My doubt :  why $|A_{E}(f) - \alpha| \le r $ give  contradiction ?

答:
Since $S$ is closed, we have that $S^\mathsf c$ is open. We want to prove $f(x)\in S$ a.e. by proving $\mu(f^{-1}(S^\mathsf c))=0$.

$S^\mathsf c$ is open (remember we're in the topology for the complex plane associated to the euclidean metric), and we have a countable base of closed discs centred at points of the form $p+qi$, such that $p,q\in\Bbb Q$, with radius $r\in\Bbb Q$. So if we prove that for any such closed disc $V$ in $S^\mathsf c$ we have $f^{-1}(V)$ has zero measure, then we'll have $\mu(f^{-1}(S^\mathsf c))=\mu(f^{-1}(\bigcup_i V_i))=\mu(\bigcup_i f^{-1}(V_i))\le \sum_i\mu(V_i)=0$.

In fact, it'll be proved the zero measure of any closed disc $V$ in $S^\mathsf c$ (let's say it's centred at $\alpha$ with radius $r$).  Denote $E=f^{-1}(V)$, which is measurable. Then $|f-\alpha|\le r$ on $E$, since $E$ is precisely the set of points whose images lie in the closed disc $V$. If we had $\mu(E)>0$, then by hypothesis

$\left|A_E(f)-\alpha\right|=\left|\frac{1}{\mu(E)}\int_E f\,\text{d}\mu-\alpha\frac{\mu(E)}{\mu(E}\right|=\frac{1}{\mu(E)}\left|\int_E f\,\text{d}\mu-\alpha\int_E 1\,\text{d}\mu\right|=\frac{1}{\mu(E)}\left|\int_E(f-\alpha)\,\text{d}\mu\right|\le$

$\le\frac{1}{\mu(E)}\int_E\left|f-\alpha\right|\text{d}\mu\le\frac{1}{\mu(E)}\int_E r\,\text{d}\mu=\frac{r}{\mu(E)}\int_E 1\,\text{d}\mu=r$, which means $A_E(f)$ is in the closed disc $V$, since the distance to its center $\alpha$ is less or equal to its radius $r$. But $V$ was a closed disc in $S^\mathsf c$, so we are saying that $A_E(f)\in V\subset S^\mathsf c$, which is impossible since we had $A_E(f)\in S$ for every measurable set $E$ with $\mu(E)>0$.

We conclude we actually had $\mu(f^{-1}(V))=\mu(E)=0$, which completes the proof.


问:
Given the unitary matrix $X = ABCD$, is there a way to recover $DCBA$ only using operations on $X$?
Suppose I have square unitary matrices $A, B, C, D$. I also have the unitary matrix $X = ABCD$ (or more generally, $X = X_1X_2\cdots X_n$). Is there a way to recover $Y=DCBA$ (or more generally, $Y = X_n\cdots X_2 X_1$) only using operations on $X$?

I'm also interested in knowing how much incremental information about $X$ is needed before we learn something about $Y$. Specifically, if we only know the first column of $X$, can we say anything about what $Y$ looks like?

答:
No. $DCBA$ is not a function of $ABCD$. Let $A$ and $B$ be any two non-commuting unitary matrices, such as
$$
A=\pmatrix{0&-1\\ 1&0},\ B=\pmatrix{1&0\\ 0&-1}.
$$
Now let $(A_1,B_1,C_1,D_1)=(A,B,I,I)$ and $(A_2,B_2,C_2,D_2)=(AB,I,I,I)$. Then $A_1B_1C_1D_1=AB=A_2B_2C_2D_2$ but $D_1C_1B_1A_1=BA\ne AB=D_2C_2B_2A_2$.


问:
How to show that $\sum_{i\ge i_0} \frac{i^2}{4^i}\to 0$
How to show that: for all $\varepsilon>0$, take a $i_0=i_0(\varepsilon)$ such that

$$\sum_{i\ge i_0} \frac{i^2}{4^i}\le \varepsilon.$$

I know that is convergence to zero by the ratio test but how to choose $i_0$?

答:
$$S_x=\sum_{i=x}^\infty \frac{i^2}{4^i}=\frac{ 4^{1-x}}{27} \left(9 x^2+6 x+5\right)$$ If I properly understook you want to know $x$ such that $S_x \leq \epsilon$ that is to say
$$4^x \geq \frac{4 \left(9 x^2+6 x+5\right)}{27 \epsilon }$$ It is easy to show that, as soon as $x >3$, $9 x^2+6 x+5 < 11x^2$. consider that an upper bound will be given by the solution of
$$4^x=\frac {44}{27} \frac {x^2}{\epsilon }$$

The only analytical solution is given by
$$x=-\frac 1 {\log(2)} W_{-1}\left(-\frac{3}{2} \sqrt{\frac{3}{11}}  \log (2)\,\sqrt{\epsilon }\right)$$ where $W_{-1}(t)$ is the second branch of Lambert function.

Using $\epsilon=10^{-6}$ gives, as real, $x=14.14$, then $\lceil x \rceil=15$ and computing the exact value,

$$S_{15}=\frac{265}{905969664} \sim 2.93 \times 10^{-7} <10^{-6}$$ while
$$S_{14}=\frac{1853}{1811939328} \sim 1.02 \times 10^{-6}>10^{-6}$$

In the linked page, you will find good approximations for the computation of $W_{-1}(t)$ when $t$ is small.


答:
I see a basic "obstruction" that begins with the product of two matrices $AB$:

As $X=AB$ is identical to $(AC)(C^{-1}B)$ for any invertible $C$, being able to retrieve the product $Y=BA$  in a unique way from $X$ is deemed to fail because it would mean in particular that  $Y=(C^{-1}B)(AC)=C^{-1}(BA)C$, which is similar but in general non identical to $BA$.

Title: Given the unitary matrix $X = ABCD$, is there a way to recover $DCBA$ only using operations on $X$?

Body:
Suppose I have square unitary matrices $A, B, C, D$. I also have the unitary matrix $X = ABCD$ (or more generally, $X = X_1X_2\cdots X_n$). Is there a way to recover $Y=DCBA$ (or more generally, $Y = X_n\cdots X_2 X_1$) only using operations on $X$?

I'm also interested in knowing how much incremental information about $X$ is needed before we learn something about $Y$. Specifically, if we only know the first column of $X$, can we say anything about what $Y$ looks like?


问:
Is the naturality condition a generalized commutativity law?
If objects and functors are equal ( $F=G$ and $a=b$) then $\alpha_a \circ Ff = Gf \circ \alpha_b$ translates to $\alpha_a \circ Ff = Ff \circ \alpha_a$.

If associativity is more basic than commutativity (even to the definition of category), is this the most basic or "deep" occurrence of a (the?) commutativity law?

答:
According to nlab:

"The term “natural” is used for traditional reasons, cf. EilenbergMac Lane45, and is an example of the common phenomenon that concepts branded natural, normal or regular tend to be rather non-generic or non-random. In fact, Eilenberg and MacLane comment (cf. p. 233, first paragraph in loc. cit) on the term, associating it with an intuition of simultaneity, rather than genericity. This intuition is in tune with conceiving of naturality as generalized commutatitivity."


问:
How to show that a certain value function is convex?
So I have the following:

Assume $S \subseteq \mathbb{R}^2$ , we have the following function: $v_s: \mathbb{R}^2 \rightarrow  \mathbb{R}$ that is defined as follows:

$$\max_{\epsilon_1 , \epsilon_2}(3−2\epsilon_1+\epsilon_2)x_1+ (1 +\epsilon_1−2\epsilon_2)x_ 2  \\ \text{s.t.} (\epsilon_1, \epsilon_2) \in S$$

How can I prove that this function is convex? What steps should I take, because at the moment I do not know what to do with $\epsilon_1$ and $\epsilon_2$

答:
Note that for any $\epsilon_1,\epsilon_2\in S$,
$$f_{\epsilon_1,\epsilon_2}(x_1,x_2)\to(3−2\epsilon_1+\epsilon_2)x_1+ (1 +\epsilon_1−2\epsilon_2)x_ 2$$
is convex because it is linear. Therefore
$$v_s(x_1,x_2)=\max_{\epsilon_1,\epsilon_2\in S}f_{\epsilon_1,\epsilon_2}(x_1,x_2)$$
is still convex because Proving that the maximum of two convex functions is also convex


问:
Convergence in Probability not implying Convergence in Distribution
I would immensely appreciate a little help here.

I am suffering with the following statement: "Convergence in probability implies Convergence in distribution"

Let U $\sim$ Uniform($0$,$1$). Let $X_n$ = \begin{cases}
      n  &  U\leq \frac{1}{n} \\
      0 &otherwise&
   \end{cases}

Clearly $X_n \to 0  $

However: E($X_n$) = $(\frac{1}{n} \times n  ) + (\frac{n-1}{n} \times 0 ) = 1$ $ \forall n \in \mathbb{N}
$

So E($X_n$) $  \not \to  $ E($0$) = $0$ Despite $X_n \to 0  $

Any help would really help me out lots.
Oskar :)

答:
To show that the given sequence converges in Law it is enough to calculate

$$F_{X_n}(t)=\begin{cases}
0,  & \text{if $t<0$} \\
1-\frac{1}{n},  & \text{if $0\leq t<n$} \\
1, & \text{if $t\geq n$}
\end{cases}$$

and its limit

$$\lim\limits_{n \to \infty} F_{X_n}(t)=\begin{cases}
0,  & \text{if $t<0$} \\
1, & \text{if $t\geq 0$}
\end{cases}$$

Thus $X_n \xrightarrow{d}0$


问:
Change the integral from spherical coordinates to cylindrical coordinates
This is the integral to evaluate the volume of the cone $z=\sqrt{x^2+y^2}$ inside the sphere $x^2+y^2+z^2=z$ How can I rewrite this integral in cylindrical coordinates. Thank you.

$$\int_{0}^{2 \pi} \int_{0}^{ \frac{\pi}{4} } \int_{0}^{\cos{\phi}} \rho^2sin(\phi)d\rho d \phi d \theta  $$

答:
The sphere and cone intersect at $(0,0,0)$ and the curve $x^2+y^2=1/4,z=1/2$.
The volume looks like a cone with a hemispherical top. In cylindrical coordinates, the lower boundary is $z=r$ and the upper boundary is the sphere $z^2-z+r^2=0$, which gives $z=\frac12+\sqrt{\frac14-r^2}$. The intersection curve is $r=z=\frac12$. The required integral is$$\int_0^{2\pi}\int_0^{1/2}\int_r^{\frac12+\sqrt{\frac14-r^2}}r~dz~dr~d\theta$$


问:
Wronskian and Linear Independence
I am having trouble showing this claim and honestly I was not able to get very far:

Suppose we have $n$ solutions $X_1,...X_n \in C([a,b],\mathbb{C}^n)$ to $X'=A(t)X$ where each entry in $A(t)$ is continuous and there exists a value $t_0$ on $[a,b]$ for which the Wronskian $W[X_1,..,X_n](t_0)\not=0$.  Show that $W[X_1,..,X_n](t)=0$ for all $t$ in $[a,b]$.

My attempt: Since we know that there exists a $t_0$ in $[a,b]$ such that the $$det([X_1(t_0),...,X_n(t_0)])\not = 0$$Which means that $\{X_i(t_0)\}$ are linearly independent. From here I was hoping to use uniqueness of solutions and maybe argue that this determinant is independent of $t$, but honestly I do not know how to approach it. If some can give me a small hint so I can get started in the correct direction I would appreciate it very much.

答:
If $W(t_1)=0$, then for this point there exist non-trivial coefficients so that
$$
0=c_1X_1(t_1)+c_2X_2(t_1)+...+c_nX_n(t_1).
$$
Then build
$$
Y(t)=c_1X_1(t)+c_2X_2(t)+...+c_nX_n(t).
$$
By linearity, this is also a solution of the ODE. Because of $Y(t_1)=0$ and uniqueness, $Y(t)=0$ everywhere. This is a contradiction to linear independence at $t_0$.

Title: Wronskian and Linear Independence


问:
Why do we name equations more in applied math?
In pure math, equations aren't often named. For instance, we might define that $f$ is a (real) polynomial function iff there exists a finite sequence of real numbers $a_i$ such that $$f(x)=a_0 + a_1 x + \cdots a_n x^n.$$

Notice, though, that we haven't defined an equation, but rather a kind of function.

Furthermore, we might define that a solution multiset of a polynomial function $f$ is a multiset of real numbers $x$ such that $f(x)=0$, with the additional requirement that the multiplicities are correct. Again, note that we've defined a way of getting a multiset from a function; technically, this is another function! Its certainly not an equation.

However, in applied math, I've noticed a lot more naming of equations / syntactical objects. Like, a text might define that a polynomial equation is an equation of the form $$a_0 + a_1 x + \cdots a_n x^n = 0.$$

Why is this the case?

答:
Your question is a little confused & confusing.

I would say that applied mathematics is generally more concerned with model building, so there are large numbers of things which in some way describe either a particular model or class of models, or a method of solution of models.

Since solutions can generally be written as thing you care about = function of input, and since models are generally described by a governing equation in applied mathematics, this means you might end up with a fair few equations named after people.

In pure mathematics, you tend not to have governing equations so much (due to the absence of physics, such as time evolution), and tend not to be so intently focused on getting out quantities as answers. Even when one does come across formulae, pure mathematicians tend to call them identities or refer to them as theorems. There are huge numbers of named theorems in pure mathematics.

Anyhow, I think this is not a very deep observation. It's just a convention for nomenclature, and a suggestion of where the emphasis lies in each of these (vague) subdivisions of mathematics.


答:
We assume that the field $\Bbb F$ is finite and $|\Bbb F|=q$. Litho’s example shows that it can happen that $|L|>1$.

On the other hand, we can achieve an uniqueness of polynomials of $L$, imposing a natural restriction on their degrees. Indeed, given $f$, by induction with respect to $n$ we can construct a multidimensional Lagrange interpolation polynomial for $f$, which has degree at most $q-1$ with respect to each variable (and so a total degree at most $(q-1)n$). It follows that the set $L$ is non-empty.

Since $x^q=x$ for each $x\in\mathbb F$, given any polynomial $p\in L$ represented as a sum of monomials, if we substitute, as orangeskid suggested, in each of the monomials a factor $x_i^{n_i}$ by $x_i^{m_i}$, where $m_i\in \{1,2,\ldots, q-1\}$, and $n_i\equiv m_i \mod (q-1)$, we obtain a reduced polynomial $\bar p$ which has degree at most $q-1$ with respect to each variable and $\bar p(x)=p(x)$ for each $x\in \Bbb F^n$.

For any polynomials $p,r\in L$, a polynomial $\bar p-\bar r$ has degree at most $q-1$ with respect to each variable. So it is zero by the following

Theorem (Combinatorial Nullstellensatz II). [A] Let $\Bbb F$ be a ﬁeld and $f\in \Bbb F[x_1,\dots, x_n]$. Suppose $\deg f =\sum_{i=1}^n t_i$ for some nonnegative integers $t_i$ and the coefficient of $\prod_{i=1}^n x_i^{t_i}$ is nonzero. If $S_1,\dots, S_n\subset \Bbb F$ such that $|S_i| > t_i$ then there exists $s_1\in S_1,\dots, s_n\in S_n$ such that $f(s_1,\dots,s_n)\ne 0$.

References

[A] N. Alon, Combinatorial Nullstellensatz, Combinatorics, Probability and Computing 8 (1999), 7–29.

See (3) in this answer for more references.


问:
Solve the initial value problem $y′′+36y=e^{−t}$

Solve the initial value problem $y′′+36y=e^{−t}$

$y(0)=y_0$,
$y'(0)=y_0'$

Suppose we know that $y(t)\to0$ as $t\to\infty$. Determine the solution and the initial conditions.

So far I have come up with
$m^2+36=0$

$m=6i$

$m=-6i$

Guess solution is $y(t)=c_1 \sin(6t)+c_2\cos(6t)+e^{-t}$

答:
When you do $m^2+36=0$ and you get $m=\pm6i$, you're actually looking for the solutions of the homogeneous associated differential equation; in this case, $y''+36y=0$. So what you got are two independent solutions that generate any solution of the homogeneous equation: $c_1\sin(6t)+c_2\cos(6t)$. Now you have to look for a particular solution $\phi_p(t)$ of the differential equation you're given, since all solutions are of the form $\phi_p(t)+c_1\sin(6t)+c_2\cos(6t)$.

To obtain $\phi_p(t)$ suppose it is of the form $\phi_p(t)=a(t)\sin(6t)+b(t)\cos(6t)$, where $a(t),b(t)$ are some suitable functions.

If we impose two conditions $\begin{cases}a'(t)\sin(6t)+b'(t)\cos(6t)=0\\a'(t)(\sin(6t))'+b'(t)(\cos(6t))'=e^{-t}\end{cases}$

we'll get that $\phi_p(t)$ is indeed a solution (this is a general method; you can check it works).

Then we need $a(t),b(t)$ such that $\begin{cases}a'(t)\sin(6t)+b'(t)\cos(6t)=0\\6a'(t)\cos(6t)-6b'(t)\sin(6t)=e^{-t}\end{cases}$

Note that this can be written as $\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}\pmatrix{a'(t)\\b'(t)}=\pmatrix{0\\e^{-t}}$, so you can solve for $a'(t),b'(t)$ with the inverse of the matrix.

$\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}^{-1}=A^{-1}=\frac{1}{\det A}\text{adj}(A^\mathsf t)$

We have

$\det A=-6\sin^2(6t)-6\cos^2(6t)=-6\\A^\mathsf t=\pmatrix{\sin(6t)&6\cos(6t)\\\cos(6t)&-6\sin(6t)}\\\text{adj}(A^\mathsf t)=\pmatrix{-6\sin(6t)&-\cos(6t)\\-6\cos(6t)&\sin(6t)}$

So we get

$\pmatrix{\sin(6t)&\cos(6t)\\6\cos(6t)&-6\sin(6t)}^{-1}=\frac{-1}{6}\pmatrix{-6\sin(6t)&-\cos(6t)\\-6\cos(6t)&\sin(6t)}=\frac{1}{6}\pmatrix{6\sin(6t)&\cos(6t)\\6\cos(6t)&-\sin(6t)}$

Then $\pmatrix{a'(t)\\b'(t)}=A^{-1}\pmatrix{0\\e^{-t}}$, so

$\pmatrix{a'(t)\\b'(t)}=\frac{1}{6}\pmatrix{6\sin(6t)&\cos(6t)\\6\cos(6t)&-\sin(6t)}\pmatrix{0\\e^{-t}}=\frac{1}{6}\pmatrix{e^{-t}\cos(6t)\\-e^{-t}\sin(6t)}\Rightarrow\begin{cases}a'(t)=\frac{e^{-t}\cos(6t)}{6}\\b'(t)=\frac{-e^{-t}\sin(6t)}{6}\end{cases}$

So $a(t)=\int a'(t)\,\text{d}t$ and $b(t)=\int b'(t)\,\text{d}t$ (I think this is the most tedious part). Once you have obtained them, you get the particular solution $\phi_p(t)$ (to be honest, at this point I used Wolfram Alpha, but when I first learned this I had to do quite a few integrals). We get $\phi_p(t)=\dfrac{e^{-t}}{37}$.

So a general solution $\varphi(t)$ is of the form $\varphi(t)=\dfrac{e^{-t}}{37}+c_1\cos(6t)+c_2\sin(6t)$.

If we know that $\phi(t)\xrightarrow{t\to+\infty}0$ then we impose $\varphi(t)=\dfrac{e^{-t}}{37}+c_1\cos(6t)+c_2\sin(6t)\xrightarrow{t\to+\infty}0$. But $\dfrac{e^{-t}}{37}\xrightarrow{t\to+\infty}0$, and $\cos(6t)$ and $\sin(6t)$ doesn't stop oscillating; so we need to get rid of them in order to get to $0$ when we do $t\to+\infty$. The only possibility is $c_1=c_2=0$.

Therefore the solution is $\varphi(t)=\dfrac{e^{-t}}{37}$, so the initial conditions are $\varphi(0)=\frac{1}{37},\varphi'(0)=-\frac{1}{37}$.


问:
$\Psi:G/N \to (\mathbb R^+,\times)$ with $aN \mapsto x$ is isomorphism
Assume $G=\left\{a+ib\in\mathbb{C} \mid a+ib \ne 0 \right\}$ is a multiplicative group and $N=\left\{a+ib \mid a^2+b^2=1\right\}$,then show that the quotient $G/N$ is isomorphic to the group $(\mathbb R^+,\times)$.

I guess we need to show that there is an isomorphism $\Psi:G/N \to (\mathbb R^+,\times)$ with $aN \mapsto x$

But for example how will we show that the function is homomorphism?
As I know if $N \trianglelefteq G$ then for $a_1N,a_2N \in G/N :(a_1N)(a_2N)=a_1a_2N$.

If we show that $N \trianglelefteq G$ then: $$\Psi(a_1N)=\Psi(a_2N)$$

$$\Psi(a_1N)\Psi(a_2N)^{-1}=1$$
$$\Psi(a_1N)\Psi(a_2^{-1}N)=1$$
$$\Psi(a_1a_2^{-1}N)=1$$
However we first need to show thet the function is homomorphism.

But does that help? Or maybe the mapping is not useful?

答:
(First Isomorphism Theorem)

Let $G,H$ be groups and $f:G\rightarrow H$ be a homomorphism. If $f$ is surjective, then $G/{\ker f}$ is isomorphic to $H$.

Hint:

Define $f:G\rightarrow \Bbb{R}^+$ by $$a+ib\mapsto a^2+b^2$$
where $a+ib\in G$.

(i) Show that $f$ is a homomorphism.

(ii) Show that $f$ is surjective.

(iii) Show that $\ker f=N$.

Then by the First Isomorphism Theorem, we have $G/N\cong \Bbb{R}^+$.


问:
Identifying the sequence $1,2,4,8,16,23,28,38,49,62,70, \ldots$, in which increasing a term by the sum of its digits yields the next term

The number $1$ is written on a blackboard. After every minute the number written on the blackboard is increased by the sum of its digits. The first few numbers, therefore, will be
$$1,2,4,8, 16,23,28,38,49,62,70, \ldots$$
Prove that at some point of time $2011$ will be written on board.

I am actually trying to find a particular rule in those numbers but failed to find any such. Somehow, I observed that if I get $1991$ in the sequence, then $2011$ will be the next number. But can't guarantee the existence of $1991$ in this sequence.

Any help is appreciated. Thanks in advance.

答:
As Greg pointed out in the comment, the digit sum sequence you have mod 9 is equivalent to the repeat of $1, 2, 4, 8, 7, 5, \dots$.
If you apply the 'digit-sum' operator to the added number repeatedly, you can make the sequence $1, 2, 4, 8, 7, 5, \dots$.
For instance, from $1991$ to $2021$, you add $20$ (the digit sum of $20$ itself is $2$). From $2011$, you add $4$, from $2015$, you add $8$, and so $2, 4, 8, \dots$ the next added part will have digit sum $7$, $5$, and so on...

The sequence is listed at https://oeis.org/A004207.
In fact, you can find the existence of $2011$ from the table at https://oeis.org/A004207/b004207.txt.

I don't think there exists a closed-form solution to find the $n$th number in this sequence. You may write a code to generate the sequence and check the existence of $2011$ which will not take long.


问:
Find $\arg\max_x \operatorname{corr}(Ax, Bx)$ for vector $x$, matrices $A$ and $B$
This is similar to, but not the same as, canonical correlation: For $(n \times m)$ matrices $A$ and $B$, and unit vector $(m \times 1)$ $x$, is there a closed-form solution to maximize the correlation between $Ax$ and $Bx$ w.r.t. $x$? Note that I am optimizing over just one vector (in contrast to canonical correlation).</p>

答:
Here is an answer for the case $m>n$.

Write $x=(x_1,\ldots,x_m)^T,A=(a^{1},\ldots,a^{m}),B=(b^{1},\ldots,b^{m})$, so $Ax=\sum_{i\le m} x_ia^i$, $Bx=\sum_{i\le m} x_i b^i$. Since $m>n$, columns $a^i - b^i$ of the matrix $A-B$ are linearly dependent, i.e. there is $x$ such that $Ax=Bx$. For this $x$ we have ${\rm corr}(Ax,Bx)=1$, i.e. is maximal.


问:
If $\mathrm i\overline A^\top=A$, then the eigenvalues of $A$ have same real and imaginary part
Can we show that if $\mathrm i\overline A^\top=A$, then the eigenvalues of $A$ have same real and imaginary part?

I came across this when studying random matrices $A$ of the form $A=X+\mathrm iX^\top$ for real (random) square matrices $X$. Their eigenvalues $\lambda$ all seem to satisfy $\Re(\lambda)=\Im(\lambda)$.

I was able to show that if $\lambda_1+\mathrm i\lambda_2$ is an eigenvalue of $A$, then so is $\lambda_2+\mathrm i\lambda_1$. How can one continue?

答: If $iA^\ast=A$, then $AA^\ast=A^\ast A$, i.e. $A$ is normal. Therefore $A$ can be unitarily diagonalised as $UDU^\ast$. The condition $iA^\ast=A$ hence gives $iD^\ast=D$. Therefore $\Re(d_{ii})=\Im(d_{ii})$ for each $i$.


答:
Your guess solution is incorrect because $e^{-t}$ does not satisfy the ODE.$$y''+36y=e^{-t}+36e^{-t}=37e^{-t}\ne e^{-t}$$which suggests the particular solution is $\frac{e^{-t}}{37}$, giving our guess solution as $y(t)=c_1\sin 6t+c_2\cos 6t+\frac{e^{-t}}{37}$.

As $t\to\infty$, we want that $y$ should vanish. The third term $e^{-t}/37$ vanishes but we can't say $c_1\sin 6t+c_2\cos 6t$ vanishes unless coefficients of both $\sin6t,\cos 6t$ are $0$. Thus the solution we desire is $e^{-t}/37$. Can you find $y_0,y'_0$ now?

Body:
Solve the initial value problem $y′′+36y=e^{−t}$

$y(0)=y_0$,
$y'(0)=y_0'$

Suppose we know that $y(t)\to0$ as $t\to\infty$. Determine the solution and the initial conditions.

So far I have come up with
$m^2+36=0$

$m=6i$

$m=-6i$

Guess solution is $y(t)=c_1 \sin(6t)+c_2\cos(6t)+e^{-t}$


问:
Transforming Euclidean figures
By their nature, Euclidean theorems do not specify lengths. Where a line has zero length a consistent interpretation is (almost?) always possible and is frequently instructive. Is there a theorem which enshrines (or constrains) this principle?</p>

答:
As I understand you, when you use the verb "enshrines", you are thinking more to axioms than theorems, isn't it ?

I don't think that any valuable axiom/result can be edicted regarding a zero distance between points: saying that $A$ is at distance 0 from $B$ means plainly that $A$ and $B$ are two namings for the same point.

But it can become interesting in the context of the distance from a point to a straight line. Let us consider in particular the context of pole-polar line duality. There is a result saying the fact that a pole is at distance zero from its polar line with respect to a conic curve $C$ iff the pole is on $C$ and its polar line is the tangent line at this point to $C$.

I see another direction of thought: as you entitle your question "transforming figures", have you considered infinitesimal transformations ?

Title: Transforming Euclidean figures


问:
Number theory question including concepts of primes and divisibility.
prove that $(n!)^{(n-1)!}$ divides (n!)!

I know this question already exists but i'm looking for a purely number theory proof, no combinatorics.

My attempt: I tried to go about the concept of largest prime power that divides n! , which is given by [n/p]+[n/$p^2$] +...... upto infinity (where [.] is the greatest integer function). So i tried to prove that largest power of prime p that divides (n!)! ≥ largest power of p that divides $(n!)^{(n-1)!}$, but i ended up with an ugly inequality with no idea how to proceed further.

Any help would be appreciated, cheers!

答:
You can partition $n!$ elements into $(n-1)!$ groups of $n$ elements.

Therefore,
the group
$S_n \times S_n \times \cdots \times S_n$ (with $(n-1)!$ factors) is a subgroup of $S_{n!}$.

The claim follows by Lagrange's theorem of group theory.


问:
What is the exact return of argmax or argmin operator?
In many journals, I saw $\text{argmin}$ and $\text{argmax}$ used as follows:
$$
x^* = \underset{i}{\text{argmin}}~f(x_i).\tag{1}\label{1}\\
$$

I think that Eq. \eqref{1} is incorrect, so it should be changed into
\begin{align}
i^* &= \underset{i}{\text{argmin}}~f(x_i),\tag{2}\label{2}\\
x^* &= x_{i^*}.\tag{3}\label{3}
\end{align}

Is it okay if I use Eq. \eqref{1} instead of Eqs. \eqref{2} and \eqref{3}?

答:
Perhaps an equally succint yet less confusing way to express this is:

$$x^* = \underset{x_i}{\text{argmin}}~f(x_i).$$

But ideally, it would be good to have a set that you're enumerating over rather than just some arbitrary expression:

$$X = \{x_1, \ldots, x_n\},$$

$$x^* = \underset{x \in X}{\text{argmin}}~f(x).$$

This way, you actually have a collection of objects to check rather than specifying an arbitrary expression which can't actually be looped over.


问:
Is there a formal explanation of the concept of "improper prior" in Bayesian statistics?
The Bayesian concept of "improper prior" seems to be surrounded with magic. Even formal, Bayesian-oriented books, such as Schervish's "Theory of Statistics", treat it with the heuristic hand waving ambiguity usual in less rigorous textbooks. Is there a book or article that deals with this concept/technique rigorously? Schervish mentions a couple formal attempts at tackling this concept, but also notes that they are radical in that they go beyond the standard axiomatization of probability theory, and hence open a whole "can of worms" (in his words). However, Schervish's book was published almost 20 years ago. Perhaps some advances in the field have been achieved in the meanwhile?</p>

答:
In section 1.2.6, A Remark Regarding So-called "Improper" Prior Distributions of their text Elements of Bayesian Analysis. Marcel Dekker. 1990, Florens, Mouchart and Rolin cite the following references as a sample of works where more detailed analyses of "improper" or "noninformative" prior distributions may be found.

Bernardo, J. M. (1979), Reference posterior distributions for Bayesian inferences (with discussion). Journal of the Royal Statistical Society, Series B,41, 113-147.

Hartigan, J. A. (1964), Invariant prior distributions. The Annals of Mathematical Statistics, 35, 836-845.

Jeffreys, H. (1961), Theory of Probability. Third Edition. London: Oxford University Press.

Villegas, C.

(1971), On Haar priors. In: Foundations of Statistical Inference, edited by V. P. Godambe and D. A. Sproot. Toronto: Holt, Rinehart, and Winston.

(1972), Bayesian inference in linear relations. Annals of Mathematical Statistics, 43, 1767-1791.

(1977a), Inner statistical inference. Journal of the American Statistical Association, 72, 453-458.

(1977b), On the representation of ignorance. Journal of the American Statistical Association, 72, 653-654.

Zellner. A. (1971), An Introduction to Bayesian Inference in Econometrics. New York: John Wiley.

In the first paragraph of said section, the following two works are mentioned.

Dawid, A. P., Stone, M. and Zidek, J. V. (1973), Marginalization paradoxes in Bayesian and structural inference. Hournal of The Royal Statistical Society, Series B, 35 189-233.

Mouchart, M. (1976), A note on Bayes theorem. Statistica, 36(2), 349-357.

Schervish (Theory of Statistics. Springer. 1995 (1st printing)) cites the following works (pp. 20-21) as expositing the two existing approaches to improper priors.

DeFinetti, B. (1974), Theory of Probability, Vols. I and II. New York: Wiley.

Hartigan, J. (1983), Bayes Theory. New York: Springer-Verlag.

He also mentions the following works in the same context.

Berti, P., Regazzini, E. and Rigo, P. (1991), Coherent statistical inference and Bayes theorem. Annals of Statistics, 19, 366-381.

Heath, D. and Sudderth, W. D. (1989), Coherent inference from improper priors and from finitely additive priors. Annals of Statistics, 17, 907-919.

Kadane, J. B., Schervish, M. J. and Seidenfeld, T. (1985), Statistical implications of finitely additive probability. In P. Goel and A. Zellner (Eds.), Bayesian Inference and Decision Techniques with Applications: Essays in Honor of Bruno DeFinetti (pp. 59-76). Amsterdam: Elsevier Science Publishers.

Schervish, M. J., Seidenfeld, T. and Kadane, J. B. (1984), The extent of non-conglomerability of finitely additive probabilities. Zeitschrift fuer Wahrscheinlichkeitstheorie, 66, 205-226.

Stone, M. (1976), Strong inconsistency from uniform priors. Journal of the American Statistical Association, 71, 114-125.

Stone, M. and Dawid, A. P. (1972), Un-Bayesian implications of improper Bayes inference in routine statistical problems. Biometrika, 59, 369-375.

Schervish also writes (ibid, p. 21):

An alternative to using improper priors is to do a robust Bayesian analysis

This "robust Bayesian analysis" is described in section 8.6.3 of Schervish's book.


问:
Homotopy cardinality of weak quotients
Let $G$ be a group (regarded as a one-object groupoid) acting on a groupoid $X$ (i.e.: a functor $F: G \to \mathsf{Groupoids}$ sending the unique object of $G$ to $X$). Denote with $X//G$ the “weak quotient” of such an action, i.e. the Grothendieck construction of the functor $F$.

It seems to be well-known that the homotopy cardinality of the weak quotient $X//G$ is given by the formula
$$
|X//G| = \frac{|X|}{o(G)}\text{,}
$$
where $|X|$ is the homotopy cardinality of $X$ and $o(G)$ is the order of $G$ (of course this makes sense only when $|X|$ and $o(G)$ are finite). However, I wasn't able to find any recorded proof of this fact (maybe it's very simple and I can't see it). Can someone sketch a proof or give some reference where I can find one?

答:
If $p : E \to X$ is an $n$-sheeted covering of groupoids, then $|E| = n\cdot |X|$; apply this result to the case of the fibration of elements $p : X/\!\!/G \to G$, just reminding that that this map isn't a covering map unless the action of $G$ is free.

However, you can replace $p$ by a covering map $\tilde p$ that is sort of a fibrant replacement for $p$, in the category ${\bf Gpd}/X$; now $|G|=1/o(G)$ and $\tilde p$ has $|X|$ sheets!

Title: Homotopy cardinality of weak quotients


问:
Eigenvalue problem!!
I just want to know is there any short trick to get all the eigenvalues of the below matrix ? I am expecting some easy way due to its blocks structure, which I don't know much about.

$A=\begin{bmatrix}
a & b&e&0\\
c& d&0&e\\
f & 0&0&0\\
0& g&0&0
\end{bmatrix}$

Thanks in advance. Any help/hint will be appreciated.

答:
I don't think there are any tricks, but the calculation can be made slightly shorter by using block matrices. Let $X=\pmatrix{a&b\\ c&d}$ and $Y=\pmatrix{f&0\\ 0&g}$. Then $\lambda I-A=\pmatrix{\lambda I-X&-eI\\ -Y&\lambda I}$. Since the two sub-blocks on the bottom row commute, we have
\begin{aligned}
\det(\lambda I-A)
&=\det\left((\lambda I-X)(\lambda I)-(-eI)(-Y)\right)\\
&=\det(\lambda^2 I-\lambda X-eY)\\
&=(\lambda^2-\lambda a-ef)(\lambda^2-\lambda d-eg)-(-\lambda b)(-\lambda c).
\end{aligned}
This is a quartic polynomial in $\lambda$. There is, in theory, an explicit radical formula for its zeroes.

Title: Eigenvalue problem!!


答:
With the inner product on $\mathbb{C}^n$:

If $x \in \mathbb{C}^n$ is a unit eigen vector of $A$ with eigenvalue $\lambda = a + i b$:
\begin{align*}
\lambda = \langle x, Ax \rangle = \langle \bar{A}^T x , x \rangle = \langle -i \left( i \bar{A}^t\right)x , x \rangle = \langle-i \lambda x , x \rangle = \overline{-i \lambda} .
\end{align*}
Hence we have $a + ib = \overline{-ia + b} = b + ia$. Thus $a = b$.

Title: If $\mathrm i\overline A^\top=A$, then the eigenvalues of $A$ have same real and imaginary part

Body:
Can we show that if $\mathrm i\overline A^\top=A$, then the eigenvalues of $A$ have same real and imaginary part?

I came across this when studying random matrices $A$ of the form $A=X+\mathrm iX^\top$ for real (random) square matrices $X$. Their eigenvalues $\lambda$ all seem to satisfy $\Re(\lambda)=\Im(\lambda)$.

I was able to show that if $\lambda_1+\mathrm i\lambda_2$ is an eigenvalue of $A$, then so is $\lambda_2+\mathrm i\lambda_1$. How can one continue?


问:
Compact notation to express all permutation of a set
I would like to express $$\text{The set of vectors}\,\{\boldsymbol p_1, \boldsymbol p_2, \ldots, \boldsymbol p_{24}\}\,\text{is the permutations of}\,\begin{bmatrix}A&B&C&D\end{bmatrix}$$ $$\text{e.g.:} \boldsymbol p_1=\{A,B,C,D\}, \boldsymbol p_2 = \{A, B, D, C\}, \text{etc.}$$in a more compact manner.  Is there a special notation for this?

答: As suggested in the comments, you could do $S_4$. However, if it's important to you that these are permutations of $A, B, C, D$ specifically and not just any set with 4 elements, then you could also say $\Sigma(\{A, B, C, D\})$. In general, $\Sigma(X)$ refers to all the permutations of the set $X$.


问:
Singular points in ODE's
It is generally said we need to bring an ODE to the form:

$$ y'' +P(x)y' + Q(x)y = 0$$

And then check where $P(x)$ and $Q(x)$ are not analytic to determine singular points. My question is why? Why do we check it only after dividing the equation and getting $y''$ to "stand alone". What's the reason behind this?

答:
Because the usual existence theorems are only formulated for explicit ODE.

It is a relatively easy consequence of the Picard-Lindelöf-Cauchy-Lipschitz theorem(s) that
$$
y^{(n)}(x)+a_{n-1}(x)y^{(n-1)}(x)+...+a_0(x)y(x)=r(x)
$$
has solutions with domain $I$ whenever the coefficients and right side are continuous on $I$.

That the DE is singular at some point does not necessarily mean that any solution has to be singular there, only that the examination of the point requires extended means. But usually the family of solutions shows singular behavior at singular points, either generally diverging or converging to some proper subset of the state space (remember, a regular DE is defined on all of the state space).

Title: Singular points in ODE's


问:
Why is $\frac X2$ a standard normal variable if it's density is $\frac 12$ the density of the standard normal distribution?
I don't understand the algebra of scaling a normal variable. Say I have $X\sim N(0,2^2)$ and an $ Y\sim N(0,1)$ , then for the density function I get:
$$p_Y(t) = P(Y=t) = \frac 1{\sqrt{2\pi}} \exp{-\frac {t^2} 2} = 2P(X=2t) = 2p_{X/2}(t)$$

Now everyone knows that $X/2\sim N(0,1)$, by why is it just allowed to "correct" the density function with the factor $2$? Where does this factor come from? It seems "unmathematical" to say that we need the factor $2$ simply because otherwise, we wouldn't have a probability density. Could someone shed some light, please?

答:
So if  you want to derive the density of $Y=\frac{X}{2}$ you have just to apply the Fundamental Transformation Theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

In your case you have

$$2\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(2y)^2}{2\cdot4}}=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$$


问:
Probability in Minesweeper
Suppose I click a random tile during a Minesweeper game. It is a 1. During each time I click an adjacent square, what are the chances of hitting a mine? How would this change if it were a 2 or another number?</p>

答:
For any tile in a MineSweeper game, the number revealed when clicked (if not a mine) tells the player the number of mines adjacent to that tile.

For example, assuming your tile is not on the edge of the playing field, it will have 8 adjacent tiles- three above, one on the left, one on the right, and three below.

So, if our tile $T$ reveals some $x \in {1, 2, 3, 4, 5, 6, 7, 8}$ , then the probability a randomly-selected adjacent tile to $T$ is a mine would be $\frac{x}{8}$ ..

The probability calculations for the number of edge cases (on the playing field) is similar.

EDIT

The key strategy of the game then, is to deduce locations of mines from the already-revealed values of neighboring tiles to T.  Then you can 'flag' correct tiles as mines.

Title: Probability in Minesweeper


答: If $c >0$ then $P(cX \leq x)=P(X \leq \frac x c)$ or $F_{cX} (x)=F_X(\frac  x c)$. To find density function you have to differentiate both sides. When you differentiate using chain rule you get $(F_X'(x)) \frac d {dx} (\frac x c )=\frac 1c F_X'(x)$ on the right side. That is how you get the factor $\frac 1 c$. Nothing special about normal distribution.

Title: Why is $\frac X2$ a standard normal variable if it's density is $\frac 12$ the density of the standard normal distribution?

Body:
I don't understand the algebra of scaling a normal variable. Say I have $X\sim N(0,2^2)$ and an $ Y\sim N(0,1)$ , then for the density function I get:
$$p_Y(t) = P(Y=t) = \frac 1{\sqrt{2\pi}} \exp{-\frac {t^2} 2} = 2P(X=2t) = 2p_{X/2}(t)$$

Now everyone knows that $X/2\sim N(0,1)$, by why is it just allowed to "correct" the density function with the factor $2$? Where does this factor come from? It seems "unmathematical" to say that we need the factor $2$ simply because otherwise, we wouldn't have a probability density. Could someone shed some light, please?


问:
If a polynomial equation is satisfied by a matrix .Is the Polynomial it's characteristics equation????
If a n order square matrix Satisfy any Arbitrary n degree Polynomial..
Can we say it is the character Polynomial Of the Matrix..???

If NO then what is necessary condition for the Matrix or Polynomial to be The statement true????

答:
No. If $p(A)=0$ then the only thing you can say is that $\mu_A\mid p$, where $\mu_A$ is the minimal polynomial of $A$. In particular, the characteristic polynomial $\chi_A$ has the same property $\chi_A(A)=0$, so $\mu_A\mid\chi_A$.

There isn't much more you can say in general. However, for a lot of matrices (especially those that have all the eigenvalues different) you have $\mu_A=\chi_A$ so in that case you can replace $\mu_A\mid p$ with $\chi_A\mid p$.

In the latter case, you can still not conclude that $\chi_A=p$, though, without an additional information about $p$. For example, if, as in your example, $A$ is of order $n$ (so the degree of $\chi_A$ is $n$) and $p$ is also of degree $n$, then knowing $p\mid\chi_A$ lets you conclude that $p$ is "essentially" the same as $\chi_A$ (i.e. they are the same up to a constant multiplier).


问:
Inverse image of $e^z$ is bounded for finite circle or not?
I got struck on this option  of a question in my complex analysis quiz and couldn't correctly solve it.

So, I am asking for help here.

Consider entire function $g(z)=e^z $ $ z\in \mathbb{C}$. Then is $g^{-1}${($z \in \mathbb{C}$; $|z|\leq R$)} bounded for every R>0 or not?

Answer is its not but I am getting that it will be bounded. CAn you please tell how to prove it's not bounded!!

答: Indeed, the answer is negative. If$$a\in g^{-1}\bigl(\{z\in\Bbb C\mid|z|\leqslant R\}\bigr),$$then$$(\forall n\in\Bbb Z):a+e^{2\pi in}\in g^{-1}\bigl(\{z\in\Bbb C\mid|z|\leqslant R\}\bigr)$$too, since $n\in\Bbb Z\implies e^a=e^{a+2\pi in}$. Therefore, $g^{-1}\bigl(\{z\in\Bbb C\mid|z|\leqslant R\}\bigr)$ is either empty or unbounded. And it is clearly non-empty.


问:
Find a particular solution to the non-homogeneous differential equation $y''+4y'+5y=−15x+e^{−x}$
Find a particular solution to the non-homogeneous differential equation

$$y''+4y'+5y=−15x+e^{−x}$$

I got the homogeneous solution as :
$$
c_1e^{-2x}\cos(x)+c_2e^{-2x}\sin(x)
$$
I am having trouble finding the particular solution

答:
Thinks of a function $p$ such that when you apply the operator $(D^2+4D+5I)$ to it, you end up with $-15x+e^{-x}$. Surely something like this would have to have the form
$$p(x) = ae^{-x} + bx + c.$$
Then just plug it in to $y''+4y'+5y$, you'll get
$$(D^2+4D+5I)p = 2ae^{-x}+5bx+4b+5c.$$
If you put $a=\frac12$, $b=-3$ and $c=\frac{12}5$, you'll end up with $-15x+e^{-x}$, thus the particular integral you should take is
$$\boxed{p(x) = \tfrac12e^{-x}-3x+\tfrac{12}5.}$$

This method (and why it works) is explained on pages 2-6 of these notes.


问:
What is the meaning of an "irreducible representation"?
What does it mean to talk about the "irreducible representatives of SO(3)"?  I'm struggling to understand the concept of irreducible representations.  Could someone give a concrete example for someone unfamiliar with group theory or representation theory?</p>

答:
A representation of the group $G$ means a homomorphism from $G$ into the group of automorphisms of a vector space $\mathbf{V}$. Essentially, you are trying to interpret each element of $G$ as an invertible linear transformation $\mathbf{V}\to\mathbf{V}$, in order to try to understand the group $G$ by how it "acts on $\mathbf{V}$."

If you have an action $\rho_1$ of $G$ on a vector space $\mathbf{W}$ (that is, one representation), and you have some other action $\rho_2$ of $G$ on another vector space $\mathbf{Z}$ (another representation), then you can use these two actions to construct an action of $G$ on the vector space $\mathbf{W}\oplus\mathbf{Z}$: just let $G$ act on the first coordinate using the old action on $\mathbf{W}$, and let it act on the second coordinate using the old action on $\mathbf{Z}$.

The point to observe, however, is that the action of $G$ on $\mathbf{W}\oplus \mathbf{Z}$ defined this way does not give you any new insights into the structure of $G$: anything you can glean about $G$ from this action, you can learn about $G$ by considering the original actions $\rho_1$ and $\rho_2$. So this new action does not give us anything new.

Conversely, suppose you have one representation $\rho$, with $G$ acting on $\mathbf{V}$, and that there are proper subspaces $\mathbf{W}$ and $\mathbf{Z}$ of $\mathbf{V}$ that satisfy the following properties:

$\mathbf{V}=\mathbf{W}\oplus\mathbf{Z}$; and

The action of every $g\in G$ on $\mathbf{V}$ maps $\mathbf{W}$ to itself; and

The action of every $g\in G$ on $\mathbf{V}$ maps $\mathbf{Z}$ to itself.

Then you can look at the restriction of the action of $G$ on $\mathbf{W}$ to get a representation, and the restriction on $\mathbf{Z}$ to get another representation; and these two representations will give you all the information from the original representation, the same way we had before. The advantage being that since $\mathbf{W}$ and $\mathbf{Z}$ are proper subspaces of $\mathbf{V}$, they have smaller dimension and, presumably, it's easier to understand a subgroup of linear automorphisms for them than for $\mathbf{V}$.

So the moral is that we want to find representations that cannot be "broken up" into smaller ones, because there's no point in trying to understand ones that do break up, we can focus our attention on those that don't, because all the other representations can be built up in terms of the ones that cannot be broken up.

The irreducible representations are precisely the ones that cannot be broken up into smaller pieces (at least for finite groups). Maschke's Theorem says that if you have a representation $\rho$ of a finite group $G$ acting on $\mathbf{V}$, and $\mathbf{W}$ is a subspace of $\mathbf{V}$ such that for all $g\in G$, the image of $\mathbf{W}$ under the action of $g$ is $\mathbf{W}$ itself, then you can find a subspace $\mathbf{Z}$ of $\mathbf{V}$ such that $\mathbf{V}=\mathbf{W}\oplus\mathbf{Z}$ and every $g\in G$ maps $\mathbf{Z}$ to itself (that is, in order to break up $\rho$ into two smaller pieces, it is enough to find a single proper piece on which $\rho$ acts; then you can find a complement for it).

With this in mind, we say:

Let $\rho\colon G\to \mathrm{Aut}(\mathbf{V})$ be a representation of $G$. We say that $\rho$ is irreducible if and only if $\mathbf{V}$ is not the zero vector space, and the only subspaces of $\mathbf{V}$ that are mapped to themselves under the action of every $g\in G$ are $\{\mathbf{0}\}$ and $\mathbf{V}$ itself.

An irreducible representation of $SO(3)$ will be a representation of $SO(3)$ that is irreducible. $SO(3)$ acts naturally on the vector space $\mathbb{R}^3$: it consists of all automorphisms of $\mathbb{R}^3$ that respect the inner product, so this is itself a representation of $SO(3)$ (which is irreducible, because no proper subspace of $\mathbb{R}^3$ is sent to itself by all elements of $SO(3)$).

(Maschke's Theorem holds if the vector space is over a field of characteristic $0$, or if the characteristic does not divide the order of the group; there are similar theorems for certain kinds of infinite groups, but it does not hold for arbitrary infinite groups in general.)


问:
The weighted distribution function for combination of two variables
For example, we have two random variables $a$ and $b$. And they have cumulative distribution function $F(x)$ and $H(x)$. We have number $0 < p < 1$.

Suppose, some machine get this random variables.
So, what algorithm should it use to these variables to get new variable $c$ with cumulative distribution function $G(x) = p*F(x) + (1 - p)*H(x)$?

Sadly, I have no idea. :(

答:
Hint: Try
$$
Z=UX+(1-U)Y
$$
where $X\sim F$, $Y\sim H$ and $U\sim \mathrm{bin}(1,p)$ is a binomial distributed random variable such that $X,Y$ and $U$ are independent


问:
Is Gower's Distance a metric?
A novice here

My previous question was closed due to inadequate details

So here I've added more details

A metric should basically satisfy 3 properties

Distance is equal to zero if and only if $x$ is equal to $y$ ($d(x,y)=0 ⇔ x=y$))

Distance from $x$ to $y$ is the same as distance from $y$ to $x$ ($d(x,y)=d(y,x)$)

Distance should satisfy the triangular inequality ($d(x,y)\leq d(x,z) +d(z,y)$)

I already know that Gower's distance satisfy the first 2 properties to be a metric, but I want to know whether it satisfies the triangular inequality property.

The reason I want to know this, is because all metric spaces are Hausdorff spaces, and I want use the Gowers distance in order to find the Hausdorff distance for 2 sets of points.

In my case, a point contains data of mixed types (logical, categorical & numeral), and therefore I have to use the Gowers distance.

Any help would be appreciated.

Thank You!

Edit:

According to a suggestion on the comments, here is the formal definition of a metric

A metric on a set X is a function (called the distance function or simply distance)

$d : X × X → R$

(where R is the set of real numbers). For all $ x, y, z $ in $X$, this function is required to satisfy the following conditions:

$d(x, y) ≥ 0$ (non-negativity)

$d(x, y) = 0$   if and only if   $x = y $

$d(x, y) = d(y, x) $    (symmetry)

$d(x, z) ≤ d(x, y) + d(y, z) $

Note that the first condition is implied by the others.

答:
Referring to the answer in https://stats.stackexchange.com/questions/15287/hierarchical-clustering-with-mixed-type-data-what-distance-similarity-to-use

"Gower distance". Without ordinal variables present (i.e. w/o using the Podani's option) √1−GS behaves as Euclidean distance, it fully supports euclidean space. But 1−GS is only metric (supports triangular inequality), not Euclidean. With ordinal variables present (using the Podani's option) √1−GS is only metric, not Euclidean; and 1−GS isn't metric at all."

Therefore it can be used a metric

Papers

1 Gower J. C. A general coefficient of similarity and some of its properties // Biometrics, 1971, 27, 857-872

2 Podani, J. Extending Gower’s general coefficient of similarity to ordinal characters // Taxon, 1999, 48, 331-340

Title: Is Gower's Distance a metric?


问:
Showing that a function is a (family of) copula(s)
I am currently working with the book An Introduction to Copulas by Roger Nelsen. I am having trouble solving one of the exercises in the first chapter. The exercise reads as follows:

Let $C$ be a copula, and let $(a,b)$ be a point in $[0,1]\times [0,1]$. For $(u,v)$ in $[0,1]\times [0,1]$, define $$K_{a,b}(u,v) = V_{C}([a(1-u), u+a(1-u)]\times [b(1-v), v+b(1-v]).$$ Show that $K_{a,b}$ is a copula.

For reference, the C-volume $V_{C}$ is defined as $V_{C}([u_{1},u_{2}]\times [v_{1},v_{2}]) = C(u_{2},v_{2}) - C(u_{2},v_{1}) - C(u_{1},v_{2}) + C(u_{1},v_{1})$.

The only thing I have not been able to show in this exercise is that $V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}]) \geq 0$ for any rectangle $[u_{1},u_{2}]\times [v_{1},v_{2}] \subset [0,1]\times [0,1]$ , which is the last defining property of copulas given in the book. Any help/hints for this problem would be greatly appreciated.

For reference, it has already been demonstrated in the book that copulas are continuous. The following inequalities has also been established for any copula $C(u,v)$: For every $(u,v)$ in the domain of $C$, we have $\text{max}(u+v-1,0)\leq C(u,v)\leq \text{min}(u,v)$. Further, it has been established that bivariate copulas are nondecreasing in both arguments.

I have tried writing out the whole expression for $V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}])$ and use the property that copulas are nondecreasing in both arguments to establish that $V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}]\geq 0$. I have also found out (if it is not wrong), that one can write

$V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}]) = V_{C}([a(1-u_{2}), u_{2} + a(1-u_{2})]\times [b(1-v_{2}),v_{2} + b(1-v_{2})]) - V_{C}([a(1-u_{2}), u_{2} + a(1-u_{2})]\times [b(1-v_{1}),v_{1} + b(1-v_{1})]) - V_{C}([a(1-u_{1}), u_{1} + a(1-u_{1})]\times [b(1-v_{2}),v_{2} + b(1-v_{2})]) + V_{C}([a(1-u_{1}), u_{1} + a(1-u_{1})]\times [b(1-v_{1}),v_{1} + b(1-v_{1})])$,

but I have not been able to use this to solve the exercise. However, the first and fourth term must be non-negative, due to $C$ being a copula. Therefore, I have tried to show that the sum of the first and fourth term must be greater than or equal to the sum of the second and third term, without success. I have also tried using the bounds above, but quickly got lost when splitting into cases based on the assumed relative sizes of the variables involved.

Thanks in advance to anyone that can provide any hints/help, and thanks to all that took the time to read this post.

Update: Using the fact that every $V_{C}$ in the above expression maps to $[0,1]$, one can see that $V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}]) \geq 0$ if $V_{C}([a(1-u_{2}), u_{2} + a(1-u_{2})]\times [b(1-v_{2}),v_{2} + b(1-v_{2})]) +  V_{C}([a(1-u_{1}), u_{1} + a(1-u_{1})]\times [b(1-v_{1}),v_{1} + b(1-v_{1})]) \geq 2$

but this restriction is too strict to be useful. I am therefore currently looking for looser restrictions that might be useful. I have also tried to check whether or not the C-volume is monotonic in the size of the rectangle in $[0,1]\times [0,1]$ (if C is a copula), but have not been able to conclusively demonstrate this. In this pursuit, I also found this Is $H$-measure actually monotonic (at least on hyperrectangles)? , but I am not entirely convinced of the answer provided in the link.

Second update:
I probably should have mentioned that not only has continuity been established, but something a bit stronger, namely that if $C$ is a copula, then for every $(u_{1},u_{2}), (v_{1},v_{2})$ in the domain of $C$, we have $|C(u_{2},v_{2}) - C(u_{1},v_{1})| \leq |u_{2}-u_{1}| + |v_{2}-v_{1}|$. I have been trying to use this to estimate relevant bounds on different terms of $V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}])$, but still without any luck.

Third update: I may have found a looser restriction than in the first update. If this is correct, it should be enough to show that $V_{C}([a(1-u_{2}), u_{2} + a(1-u_{2})]\times [b(1-v_{2}),v_{2} + b(1-v_{2})]) + V_{C}([a(1-u_{1}), u_{1} + a(1-u_{1})]\times [b(1-v_{1}),v_{1} + b(1-v_{1})]) + C(a(1-u_{2},v_{1}+b(1-v_{1})) - C(a(1-u_{2}), b(1-v_{1})) + C(a(1-u_{1}), v_{2}+b(1-v_{2})) - C(a(1-u_{1}), b(1-v_{2})) \geq v_{1} + v_{2}$.

This is because the first and second term has to be non-negative due to $C$ being a copula, and the difference between the third and fourth term and the difference between the fifht and the sixth term has to be non-negative due to $C$ being nondecreasing in both arguments. The remaining four terms of $V_{K_{a,b}}$ gives a net negative contribution to $V_{K_{a,b}}$, the upper bound of which I have estimated using the information in the second update.

答:
It believe I have solved it now. If this is correct, it turns out that the answer was quite simple all along. Also, everything in the three updates above turned out to lead to a dead end as I was able to show using a special case that it was impossible in general to get a "lowest possible" bound on a C-volume that was also strictly positive. Hence, it should not be possible to verify the inequality I presented in the third update using only the information we have available. However, playing around with proving special cases led me to notice that if we write out $V_{K_{a,b}}([u_{1}, u_{2}]\times [v_{1}, v_{2}])$ in a particular manner, the solution presents itself quite nicely. Write all the four terms associated with $u_{i}, v_{j}$, where $i=1,2, j=1,2$, in one row for each pair $i,j$. Visually, this gives a $4\times 4$ matrix. One may now notice that each column of the matrix represents a C-volume. More specifically, one can write

\begin{align}
V_{K_{a,b}}([u_{1},u_{2}]\times [v_{1},v_{2}]) &= V_{C}([u_{1}+a(1-u_{1}), u_{2}+a(1-u_{2})]\times [v_{1}+b(1-v_{1}), v_{2}+b(1-v_{2})]) \\
&+ V_{C}([u_{1}+a(1-u_{1}), u_{2}+a(1-u_{2})]\times [b(1-v_{2}), b(1-v_{1})]) \\
&+ V_{C}([a(1-u_{2}), a(1-u_{1})]\times [v_{1}+b(1-v_{1}), v_{2}+b(1-v_{2})]) \\
&+ V_{C}([a(1-u_{2}), a(1-u_{1})]\times [b(1-v_{2}), b(1-v_{1})]) \\
&\geq 0
\end{align}

because each C-volume is non-negative due to $C$ being a copula.


问:
Explanation for these transformations of integrals
I've recently found the following transformations:

$$\int _{a} ^{\infty} \frac{\ln x}{x^2 + a^2}\,dx = \int _{a} ^{0} \frac{\ln x}{x^2 + a^2}\,dx$$

$$\int _{0} ^{\pi/2} \log(\sin(\tfrac{\pi}{2}-u))\,du = \int _{0} ^{\pi/2}\log(\sin u)\, du = \int _{\pi/2} ^{\pi}\log(\sin u)\, du$$

Could you explain to me why that is?

答:
For the first set $t:=\frac 1x$ then
\begin{align}
\int _{a} ^{\infty} \frac{\ln x}{x^2 + a^2}\,dx &= -\int _{\frac1a} ^{0} \frac{\ln\left(\frac 1t\right)}{\frac 1{t^2} + a^2}\,\frac{dt}{t^2}\\
&=\int _{\frac 1a}^{0} \frac{\ln\left(t\right)}{1 + (at)^2}\,dt\\
&=\int _a^{0} \frac{\ln\left(z\right)-\ln(a^2)}{1 + \frac {z^2}{a^2}}\frac{dz}{a^2}\\
&=\int _a^{0} \frac{\ln\left(z\right)}{a^2 + z^2}\;dz-2\;\ln(a)\int _a^{0} \frac{dz}{a^2 + z^2}\\
&=\int _a^{0} \frac{\ln\left(z\right)}{z^2 + a^2}\;dz+2\;\ln(a)\frac{\pi}{4a}\\
\end{align}
(setting $z:=a^2t$)

So that we get an additional term $\ \displaystyle\ln(a)\frac{\pi}{2a}$ here.

To be complete...

For the second (setting $v:=\frac{\pi}2-u$):
\begin{align}
I=\int _{0} ^{\pi/2} \log\left(\sin\left(\tfrac{\pi}{2}-u\right)\right)\,du&=-\int_{\pi/2}^{0} \log(\sin(v))\;dv\\
&=\int _{0} ^{\pi/2} \log(\sin(v))\,dv\\
&= \int _{\pi/2} ^{\pi}\log(\sin w)\, dw\\
\end{align}
(wtih the substitution $w:=\pi-v$)


问:
Is this considered a regular expression?
I need to design an NFA that recognizes the language $L$, with alphabet $\{a,b\}$, that accepts strings that:

have length at least 4

two letters before the final letter of the string there is always a $b$

For example the language accepts $abba$ but neither $aba$ nor $abbaba$.   I designed this:

If I wanted to provide a regular expression also I was wondering if I could write :

$(a|b)^*b(a|b)^2$

I have never seen an exponent as $2$ that's why I am concerned.

答:
Denoting by $A$ the alphabet, your language is $A^*AbA^2$.
If $L$ is any language, $L^n$ is defined by induction as follows: $L^0 = \{1\}$
(where $1$ denotes the empty word, which is the identity of the free monoid $A^*$) and $L^{n+1} = L^nL$. Thus $L^2 = LL$.

You can remove state $1$ in your non-deterministic automaton.


问:
Cipolla's algorithm for finding square roots - why is first step fast?
I'm reading about the Cipolla algorithm to find the square root of some quadratic residue n in $\mathbb{F}_p$ - https://en.wikipedia.org/wiki/Cipolla%27s_algorithm.

The first step of the algorithm is to keep picking a random $a \in \mathbb{F}_p$ until you find one with $a^2 - n$ not a square.

The couple of references I've found online suggest this should be quick, and that about half of all possible $a$ will satisfy this. I can't see why this is necessarily the case, as the distribution of $a^2 - n$ isn't going to be uniform across $\mathbb{F}_p$, so might not be distributed as an even split across quadratic residues / non-residues.

Can anyone give me a hint as to why this holds? Thanks.

答: Given a nonzero square $n$ in $\mathbb{F}_p^\times$ (with $p$ odd), the number of pairs $(a,b) \in \mathbb{F}_p \times \mathbb{F}_p$ satisfying $a^2-n=b^2$ is equal to $p-1$, as can be seen by putting $x=a+b$, $y=a-b$. From this one can deduce that the number of $a \in \mathbb{F}_p$ such that $a^2-n$ is a square is equal to $(p+1)/2$. So trying random values of $a$, we succeed with about $2$ tries.


问:
Show $\log 15$ is irrational and that $\log 3$+ $\log 5$ is also irrational.
I am trying to show that $\log 15$ and $\log 3$ + $\log 5$ is irrational.

For $\log 15$ I feel like I have no issues showing this is irrational.

By contradiction assume $\log 15$ is rational then

$\log 15 = \frac{m}{n}$ where $m,n\in\mathbb{N}$ and $n\not= 0$

$15 = 10^{\frac{m}{n}}\Rightarrow 15^n = 10^m \Rightarrow 3^n5^n = 5^m2^m$

The left side of the equation has an odd multiple of $5$ but the right side has an even multiple of $5$ and we have arrived at a contradiction.

But for  $\log 3$ + $\log 5$  I do not even know where to begin, I am assuming I cannot use the fact that   $\log 15 = \log 3$ + $\log 5$ as that would be trivial. Should I show they are individually irrational? But that leads me to my next thought as the sum of two irrational numbers is not always irrational (even though in this case I know the sum is irrational). And then would that lead me to breaking it into a case by case where I assume one is rational and the other is not, etc.  I am still an undergrad so I do not know much advanced number theory so I would appreciate a more intuitive approach if possible.

答:
Since $\log_{10}3+\log_{10}5=\log_{10}(3\times5)=\log15$ and since $\log_{10}15$ is irractional…

On the other hand, the problem with the equality $3^n5^n=5^m2^m$ lies in the fact that $3^n5^n$ is odd, whereas $5^m2^m$ is even.


答:
The sum of two irrational numbers can be rational. Take $a$ irrational, and $1-a$, which is also irrational (in fact $-a$ is even simpler, but I guess that this would not satisfy the readers). So there is no way to prove that the sum of two general irrationals is irrational, as this is false.

If you are not allowed to use that $\log 3+\log 5=\log 15$ and only that $\log 3$ and $\log 5$ are irrational, I don't see a way to obtain a proof. You could claim that these irrationals are "special" in that they are logarithms of integers, but this brings you back to the sum/product property. If they have no special property, then no proof.

Title: Show $\log 15$ is irrational and that $\log 3$+ $\log 5$ is also irrational.

Body:
I am trying to show that $\log 15$ and $\log 3$ + $\log 5$ is irrational.

For $\log 15$ I feel like I have no issues showing this is irrational.

By contradiction assume $\log 15$ is rational then

$\log 15 = \frac{m}{n}$ where $m,n\in\mathbb{N}$ and $n\not= 0$

$15 = 10^{\frac{m}{n}}\Rightarrow 15^n = 10^m \Rightarrow 3^n5^n = 5^m2^m$

The left side of the equation has an odd multiple of $5$ but the right side has an even multiple of $5$ and we have arrived at a contradiction.

But for  $\log 3$ + $\log 5$  I do not even know where to begin, I am assuming I cannot use the fact that   $\log 15 = \log 3$ + $\log 5$ as that would be trivial. Should I show they are individually irrational? But that leads me to my next thought as the sum of two irrational numbers is not always irrational (even though in this case I know the sum is irrational). And then would that lead me to breaking it into a case by case where I assume one is rational and the other is not, etc.  I am still an undergrad so I do not know much advanced number theory so I would appreciate a more intuitive approach if possible.