问:
What is this function called?
What is this function called?
Here is a graph:

```<code>          _______
         /
        /
_______/
</code>```

Sort of like:

$f(x, a, b) = 0$, if $x < a; \quad \quad \displaystyle \frac {(x - a)} {(b - a)}$ if $a < x  b$.

Closest name is Sigmoid function, but they all are smooth. Also I remember I saw similar formula for saturation, but not sure about that, googling didn't help.

Edit: Ramp function is closer:

But it is general one, while my is more specific one, used for normalization.

答: Such a function falls in the category of Elementary special functions. You may call it a (normalized) integrated Boxcar function.</p>

Title: What is this function called?


问:
Estimating volume of the solid using MidPoint Rule

If the region shown in the figure is rotated about the $x$-axis to form a solid, use Midpoint rule with $n=4$ to estimate the volume of the solid.

I am not sure what this question is asking me to do. I believe at $x=3,\,5,\,7$, and $9$ the figure will be rotated around the $x$-axis. Is it the average of the volume of these points that will be the answer?

答:
HINT:

Compute square of $y$ values by approximating midpoints and arrive at all the trapeziums. Next use Simpson's Rule as if finding areas.

ContentLicense: CC BY-SA 4.0


问:
Splitting field and intersection of subfields
Let $K/F$ be a field extension and $f \in F[x]$ with $\deg{(f)} \geq 1$. Furthermore, for a subset $L$ with $F\subseteq L \subseteq K$, let $p(L)$ denote the property that $f$ can be factored completely into linear factors in $L[x]$. Now consider the set
$$\mathcal{L}:=\bigcap\{F\subseteq L \subseteq K \mid p(L) \ \text{holds}\} $$
Question: If property $p(K)$ holds, does it follow that $p(\mathcal{L})$ holds?

In words: If $f$ can be factored completely into linear factors in $K[x]$, does it follow that $f$ can be factored completely into linear factors in $\mathcal{L}[x]$?

答: Yes, that is true. Let $\alpha$ be any root of $f$ in $K$. As $\alpha$ is contained in every $L$ such that $p(L)$ holds we have $\alpha\in\mathcal{L}$. Therefore $\mathcal{L}$ contains all roots of $f$ in $K$ and $f$ splits over $K$, so $f$ will also split over $\mathcal L$ since all the linear factors over $K$ are actually defined over $\mathcal L$.


问:
Prove that every point, satisfying given condition, lies on some line.
Segment $BC$ is the shortest side of noniscosles triangle $ABC$. Points $K$ and $L$ lie respectively on sides $AB$, $AC$ and satisfy condition: for given point $X$ on side $BC$, $BK=BX$ and $CL=CX$. Prove that midpoint of the segment $KL$ lies on some line indepedently of choosing point $X$.

I have been wondering for some time on solution for this problem but I did not come up with anything useful. Could you give some hints?

答:
Note: The condition that $BC$ is the shortest side, or that $ABC$ is non-isosceles, are not necessarily. They might have been introduced to avoid signed lengths and degenerate cases (parallel lines, gradient = 0, etc).

Set this up on the coordinate plane.

Let $ B = (0,0)$, $C = (1,0)$.

Let $D = (1+x_d, y_d)$ be a point on  $AC$ such that $CD = 1$, and $F = ((1+x_d) / 2  , y_d/2) $ be the midpoint of $ BD$.

Let $E = (x_e,y_e)$ be a point on $AB$ such that $EB = 1$, and $G = ( (1+x_e)/2, y_e/2 )$ be the midpoint of $EC$.

Assuming the problem is correct that the locus is a line, we know that the points $F,G$ lies on this line, so it must be the line $ FG$. Let's show this.

Let $X = (t,0)$, then show that

$K = (tx_e, ty_e)$,

$L  = ((1-t)(x_d) + 1, (1-t)y_d)$.

The midpoint of $ KL$ is thus $ ( \frac{ tx_e + (1-t)(x_d) + 1 } { 2} , \frac { ty_e  + (1-t)y_d } {2}) $,

Line $FG$ can be represented by $ ( \frac{1+x_d}{2} , \frac{y_d}{2 }) + t (  \frac{ x_e - x_d}{2}, \frac{y_e - y_d } {2} )$ for some dummy variable $t$.

The midpoint of $KL$ lies on line $FG$.

Hence, we are done.

Yes, using complex numbers or vectors makes the notation much cleaner, though it's the same.


问:
Let $M \in S_n(\mathbb R)$ so that $\exists d \in \mathbb N^*, M^d=I_n$. Show that $M^2=I_n$.
Let $M \in S_n(\mathbb R)$ so that $\exists d \in \mathbb N^*, M^d=I_n$. Show that $M^2=I_n$. What can we say if $d$ is odd ?

What I have done so far :
$M \in S_n(\mathbb R)$ so we can write $M=P^{-1} D P$ where D is diagonal, then $D^p=I_n$. If $d$ is odd then $D=I_n$ and so $D^2=I_n$ which means that $M^2=I_n$.

I still think I must have forgotten something or done something wrong

答: You have $M=P^{-1}DP$ with $D$ a diagonal matrix with real entries. Because $M^d=I_n$, you deduce that $D^d=I_n$. So each coefficient $\lambda$ of the diagonal of $D$ satisfies $\lambda^d=1$, and because $\lambda \in \mathbb{R}$, this means that $\lambda=\pm 1$. In particular, you have $D^2=I_n$, so $M^2=I_n$.


问:
Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$
Evaluate this definite integral:
$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$</p>

答:
$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$
$$\implies\pi\int^{441}_0\frac{\sin \pi \sqrt x}{\sqrt x} dx$$
$$\implies\pi\int^{21}_0\frac{2u \sin \pi u}{u} du$$
$$\implies2\pi\int^{21}_0 \sin{\pi u}\;du$$
$$\implies2\pi\int^{21\pi}_0 \frac{\sin{v}}{\pi}\;dv$$
$$\implies2\int^{21\pi}_0 \sin{v}\;dv$$
$$\implies -2 \left[\cos \pi \sqrt x\right]^{21}_{0}$$
$$\implies-2(\cos 21\pi - \cos 0)$$
$$\implies-2(-1 - 1)$$
$$\implies-2\times (-2)$$
$$\implies 4$$

An interesting thing is that for upper bounds that are squares of even numbers this evaluates to zero, while for squares of odd numbers (such as $21^2 = 441$) this evaluates to 4.

ContentLicense: CC BY-SA 3.0


问:
Show that $x+y$ and $x-y$ are both irrational

Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a nonzero rational, then $x+y$ and $x-y$ must be irrational.

I've made no progress on this problem, so I don't have any work to show. I suppose I can run into a contradiction, but I couldn't find any.

答: Let $a = x+y$ and $b=x-y$. Because $ab = x^2-y^2 \in \mathbb{Q}^*$, then either $a,b$ are both rational, either they are both irrational. But if they were both rational, then $x=(a+b)/2$ would also be rational, which is not true. So they are both irrational.


问:
What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu,\Sigma)$

What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$

My attempt is:

We know that distribution of $\sum^n_{i=1}\frac{1}{n}X_i\sim N_p(\mu,\frac{1}{n}\Sigma)$

So, $\mathbb{E}[X_i-\frac{1}{n}\sum^n_{i=1}X_i]=0$

And variance is $\text{Var}(X_i-\frac{1}{n}\sum^n_{i=1}X_i)=\Sigma-\frac{1}{n}\Sigma=\frac{n-1}{n}\Sigma$

Would that be correct?

edit:

suppose the first $X_i$ in $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ is $X_j$, $j\neq i$,so $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ becomes $X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j$

So,

$$\text{Var}(X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j)=\text{Var}(-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i+\frac{n-1}{n}X_j)=\frac{n-1}{n^2}\Sigma+\frac{(n-1)^2}{n^2}\Sigma=\frac{n-1}{n}\Sigma$$

答:
The question has basically been answered, but this is an attempt to prove the assumptions made in the techniques, namely that a constant times a multivariate normal random variable is itself multivariate normal and that a linear combination of multivariate normal random variables is multivariate normal. I'm not an expert on this, so it'd be interesting to know if this information is pretty interesting or completely useless.

A constant times a multivariate normal rv

Assume that $X$ is multivariate normal with mean vector $\mu$ and covariance matrix $\Sigma$. I might write this as $X\sim N_p(\mu, \Sigma)$. I would like to show that $Y=cX, c\in \mathbb R$ is mulvariate normally distributed, which isn't perhaps obvious. I use the method of taking the derivative of the cdf, and then comparing with the known density of the multivariate normal distribution. Consider the following, where $G(\textbf y)$ is the cdf of $Y$, $\Phi_X$ is the cdf of $X$, and $\phi_X$ is the pdf of $X$.

$\displaystyle G(\textbf y)=G(y_i,...,y_p)=P(Y_1\le y_1,...,Y_p\le y_p)=P(cX_1\le y_1,...,cX_p\le y_p)=P(X_1\le\frac {y_1} c,...,X_p\le\frac {y_p} c)=\Phi_X(\frac {y_1} c,...,\frac {y_p} c)=\int_{-\infty}^{y_p/c}...\int_{-\infty}^{y_1/c}\phi_X(a_1,...,a_p)da_1...da_p$.

Since the density function is the multivariable derivative of the cumulative distribution function, $\displaystyle g(\textbf y)=g(y_1,...,y_p)=\frac \partial {\partial y_1}...\frac \partial {\partial y_p}\int_{-\infty}^{y_p/c}...\int_{-\infty}^{y_1/c}\phi_X(a_1,...,a_p)da_1...da_p=\frac 1 {c^p} \phi_X(\frac {y_1} c,...,\frac {y_p} c)=\frac 1 {c^p} (2\pi)^{-k/2}\det(\Sigma)^{-1/2}e^{-\frac 1 2(\textbf y/c-\mu)^T\Sigma^{-1}(\textbf y/c-\mu)}=\frac 1 {\sqrt{c^{2p}\det \Sigma}}(2\pi)^{-k/2}e^{-\frac 1 2(\textbf y-c\mu)^T\frac 1 {c^2}\Sigma^{-1}(\textbf y-c\mu)}=\frac 1 {\sqrt{\det(c^2\Sigma)}}(2\pi)^{-k/2}e^{-\frac 1 2(\textbf y-c\mu)^T(c^2\Sigma)^{-1}(\textbf y-c\mu)}$, which is the density, according to Wikipedia article on multivariate normal distributions, of a multivariate normal random variable with mean vector $c\mu$ and covariance matrix $c^2\Sigma$. Thus $Y=cX\sim N_p(c\mu, c^2\Sigma)$.

Sum of two multivariate normal rv’s

Next I'd like to show that any linear combination of independent multivariate random variables is multivariate random. To do this, use the moment generating function again provided by Wikipedia and the fact that the sum of two independent random variables has a moment generating function that's the product of their respective mgf's.

$\psi_X(t)=\exp(\mu^Tt+\frac 1 2 t^T\Sigma t)$

Considering $X_1\sim N_p(\mu_1, \Sigma_1), X_2\sim N_p(\mu_2, \Sigma_2)$, it's relatively clear to see that $c_1X_1+c_2X_2$ has a mgf of $\displaystyle \exp(c_1\mu_1^Tt+\frac 1 2 t^Tc_1^2\Sigma t)\exp(c_2\mu_2^Tt+\frac 1 2 t^Tc_2^2\Sigma_2 t)=\exp((c_1\mu_1+c_2\mu_2)^Tt+\frac 1 2 t^T(c_1^2\Sigma_1+c_2^2\Sigma_2) t)$, the moment generating function of a multivariate random normal variable with mean vector $c_1\mu_1+c_2\mu_2$ and covariance matrix $c_1^2\Sigma_1+c_2^2\Sigma_2$. Thus when you add two multivariate normal rv's, you get a multivariate random variable with a mean of the sum of their means and a covariance of the sum of their covariances.

By applying these two principles, you correctly arrived at the answer (twice) above.


问:
Upper bound series
If I want to prove that the series $\sum _{n=0}^{\infty }\:\frac{n^3}{2^n}$ converges.

I know we can prove it by the root test, but there is a way to find an upper bound series?

答:
You can show by induction on $n$ that $n^3\le 2^{n/2}$ for $n\ge 30$. Then for $n\ge 30$ you have

$$\frac{n^3}{2^n}=\frac{n^3}{2^{n/2}}\cdot\frac1{2^{n/2}}\le\frac1{2^{n/2}}\,.$$

And $\sqrt2>1$, so

$$\sum_{n\ge 0}\frac1{2^{n/2}}=\sum_{n\ge 0}\left(\frac1{\sqrt2}\right)^n$$

converges.

Title: Upper bound series


问:
$\sigma$-algebra intuition
So, i am trying to learn measure theory, for applications in probability theory. However, i am having some issues fully understanding the definition of a $\sigma$-algebra.

I am working with the following definition of a $\sigma$-algebra:

Definition: Let $\Omega \ne \emptyset$. Then $\mathcal{B} \subseteq \mathcal{P}(\Omega)$ is a $\sigma$-algebra if

(i) $ \Omega \in \mathcal{B} $

(ii) if $B \in \mathcal{B}$, then $B^c \in \mathcal{B}$

(iii) If $B_n \in \mathcal{B}$ for all $n \in \mathbb{N}$ then $\cup_{n=1}^{\infty}B_n \in \mathcal{B}$

I do not fully understand what is going on in (iii). My intuition would say, that it should mean "If a set is in the sigma-algebra, then every possible union you could make with other sets in the sigma-algebra, is also in the sigma-algebra", is this correct? Just looking at the notation i would think that it means that the union of all sets in the sigma-algebra, should also be in the sigma-algebra, but that would just return $\Omega$ as in (i), so that cannot be correct?

答: As pointed out by @J. W. Tanner in the comments, the third criterion states that any countable union of sets in the $\sigma$-algebra is included. However, it is possible that no countable union of sets includes $\Omega$ itself. That is why we need the first criterion.

Title: $\sigma$-algebra intuition


答: The key-word which is not written in the definition is countable. A $\sigma-$algebra is intuitively a subset of $\mathcal{P}(\Omega)$ which is stable by countable "operations" (such as union, intersection and complementary) on its elements.

Title: $\sigma$-algebra intuition

Body:
So, i am trying to learn measure theory, for applications in probability theory. However, i am having some issues fully understanding the definition of a $\sigma$-algebra.

I am working with the following definition of a $\sigma$-algebra:

Definition: Let $\Omega \ne \emptyset$. Then $\mathcal{B} \subseteq \mathcal{P}(\Omega)$ is a $\sigma$-algebra if

(i) $ \Omega \in \mathcal{B} $

(ii) if $B \in \mathcal{B}$, then $B^c \in \mathcal{B}$

(iii) If $B_n \in \mathcal{B}$ for all $n \in \mathbb{N}$ then $\cup_{n=1}^{\infty}B_n \in \mathcal{B}$

I do not fully understand what is going on in (iii). My intuition would say, that it should mean "If a set is in the sigma-algebra, then every possible union you could make with other sets in the sigma-algebra, is also in the sigma-algebra", is this correct? Just looking at the notation i would think that it means that the union of all sets in the sigma-algebra, should also be in the sigma-algebra, but that would just return $\Omega$ as in (i), so that cannot be correct?


Title: Upper bound series

Body:
If I want to prove that the series $\sum _{n=0}^{\infty }\:\frac{n^3}{2^n}$ converges.

I know we can prove it by the root test, but there is a way to find an upper bound series?

Body:
hint

If $ |x|<1$ then

$$1+x+x^2+...=\sum_{n=0}^\infty x^n=\frac{1}{1-x}=F(x)$$

$$F'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$

$$G(x)=xF'(x)=\sum_{n=0}^\infty nx^n$$

$$G'(x)=\sum_{n=1}^\infty n^2x^{n-1}$$

So,

$$H(x)=xG'(x)=\sum_{n=0}^\infty n^2x^n$$
and

$$H'(x)=\sum_{n=1}^\infty n^3x^{n-1}$$

Take now $ x=\frac 12$ and compute $ xH'(x)$.


问:
Given a complex number, convert it to the form $a + bi$
So I'm trying to convert a complex number to the form of $a + bi$. The complex number in question is $$\bbox[5px,border:2px solid #C0A000]{\large 2e^{\frac{-3\pi}{4i}+\ln(3)}}$$ I'm not quite sure how to tackle this to be honest, I would appreciate some help in trying to understand how to convert these expressions into that form.

Thanks in advance!

答: First, convert it into $re^{i \theta}$. Then we have $a = r \cos \theta$ and $b = r \sin \theta$. In your case, we have $r = 2e^{\ln 3} = 6$ and $\theta = -3 \pi / 4$.


答:
The first thing you can do is factor the expression into $2e^{\ln(3)} \times e^{-3\pi/4i}$.

Then, we can use Euler's formula: $e^{ix} = \cos x + i \sin x$.

If you apply that formula to the expression, you will be able to get your answer with a little bit of arithmetic and simplification.

Title: Given a complex number, convert it to the form $a + bi$

Body:
So I'm trying to convert a complex number to the form of $a + bi$. The complex number in question is $$\bbox[5px,border:2px solid #C0A000]{\large 2e^{\frac{-3\pi}{4i}+\ln(3)}}$$ I'm not quite sure how to tackle this to be honest, I would appreciate some help in trying to understand how to convert these expressions into that form.

Thanks in advance!


问:
Finding a value a so that 3 planes don't intersect in a single point
I'm currently stuck at the following question:

Three planes have equation:
$$ax+2y+z=3$$
$$-x+(a+1)y+3z=1$$
$$-2x+y+(a+2)z=k$$
where $a∈R$.

Question: Find a value $a$ for which these planes do not intersect in a point.

The easiest way would be to make one equation a multiple of another equation but this isn't possible at first glance. Then I tried to eliminate $x$ from equation $1$ and $2$:

\begin{bmatrix}a&2&1&3\\-1&a+1&3&1\\-2&1&a+2&k\end{bmatrix}
\begin{bmatrix}a&2&1&3\\2&-2a-1&-6&-2\\-2&1&a+2&k\end{bmatrix}
\begin{bmatrix}a&2&1&3\\0&-2a&a-4&k-2\\-2&1&a+2&k\end{bmatrix}
However, this did not get me very far.
The correct solution is said to be $a = 1$ but I don't understand why. Is there a generell procedure to make a system of equations have one, infinitely many or no solutions?

Edit: After having found $a = 1$, by using gaussian elimination I found a value for $k$ such that the lines intersect in a line:
\begin{bmatrix}1&2&1&3\\-1&2&3&1\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}1&2&1&3\\0&4&4&4\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}2&4&2&6\\0&4&4&4\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}2&4&2&6\\0&4&4&4\\0&5&5&k+6\end{bmatrix}

$$∴ k+6 = 5$$
$$ k = -1 $$

If $a=1$,$k=-1$ the planes therefore intersect at:
$$4y+4z=4$$
$$y+z=1$$
$$let\  z = \lambda $$
$$y = 1- \lambda$$
$$2x + 4y + 2z = 6$$
$$x = 1 + \lambda$$

$$l:\vec r = \begin{bmatrix}1\\1\\0\end{bmatrix} + \lambda \begin{bmatrix}1\\-1\\1\end{bmatrix}$$
For those struggling with a similar question :D

答:
Just calculate the determinant that those vectors define:

$$\begin{vmatrix}
a & 2 & 1 \\
-1 & a+1 & 3 \\
-2 & 1 & a+2
\end{vmatrix} = a^3+3a^2+3a-7$$

Then, the values of $a$ that will lead to the planes not intersecting at a single point will be the ones that verify
$$a^3+3a^2+3a-7=0,$$
and that just gives the value $a=1$ (considering $a\in\mathbb{R}$, since the other two roots of the polynomial are complex). The explanation to this is that if the determinant is $0$, then those vectors are not linearly independent, so the space they generate isn't of dimension $3$ and we conclude that the planes will intersect either nowhere, either at one plane, or either at one, two or three lines, so that makes it impossible for them to intersect at just one single point.

A cool interpretation of this problem is that your system of equations is indeed the representation of three different planes, and the solution to it will be the intersection between those planes. But you must notice that your system has one single solution if and only if the determinant of the system is different from $0$, so it will have either none or infinite solutions if it's equal to $0$, the case when $a=1$.


问:
Piecewise Function and Mean Value Theorem.
Here's the function:
\begin{align}
f(x) = \left\{ \begin{array}{cc}
                3 & \hspace{5mm} x=0 \\
                ax+b & \hspace{5mm} 0<x\leq 1 \\
                x^2+3x+c & \hspace{5mm} 1<x\leq 4 \\
                \end{array} \right.
\end{align}
Find the constraints a,b and c so that $f(x)$ satisfies the conditions of MVT.
Here's how I solved. First of all I write the contuinity condition of MVT for $x=1$ and $x=0$
\begin{align}
\lim_{x\to\ 1^+} f(x)=4+c
\end{align}
\begin{align}
\lim_{x\to\ 1^-} f(x)=a+b
\end{align}
\begin{align}
a+b=4+c \hspace{5mm}(1)
\end{align}
And also we can get second limit which only goes one way.
\begin{align}
\lim_{x\to\ 0^+} f(x)=b=f(0) \hspace{5mm} (2)
\end{align}
\begin{align}
b=3
\end{align}
And then I tried to use derivatives for finding other variables. By then, I write teh expressions and I get those:
\begin{align}
\lim_{\Delta x\to\ 0^{+}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=a
\end{align}
\begin{align}
\lim_{\Delta x\to\ 0^{-}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=5
\end{align}
Clearly, $a=5$ and from $(1)$ if we plug all the values that we found we get $c=4$.
Here is my questions:
1-)Is it true?
2-)Are we allowed to do the evaluation that I've done in 2. More explicitly, can I equate just one side to the actual value? Because in this question we don't have any "smaller than 0" part.
Thanks for any help.

答:
A function $f$ is continuous at a point $t$ if $$\lim_{\underset{x \in \text{dom}(f)}{x \to t}} f(x) = f(t)$$

For your function, when $t = 0$, the restriction to the domain of $f$ means we are not looking at any $x < 0$. That is, the condition of continuity becomes
$$\lim_{x \to 0+} f(x) = f(0)$$
Which is exactly the condition you examined in (2).

When $t = 1$, both sides are in the domain, so the condition of continuity is
$$\lim_{x \to 1} f(x) = f(1)$$
But for this piecewise defined function, to examine if this is true, we need to note that $\lim_{x \to 1} f(x)$ exists if and only if the two one-sided limits exist and are equal. That is, if $$\lim_{x \to 1-} f(x) = \lim_{x \to 1+} f(x)$$ which is what you examined in (1). This condition (with (2) and the similar derivative limit) was enough to determine what $a, b, c$ had to be.

But that alone doesn't show that the $f$ defined by those values of $a,b,c$ is actually continuous at $1$. For that, you must still also show that $\lim_{x \to 1} f(x)$, the common value of $\lim_{x \to 1-} f(x)$ and $\lim_{x \to 1+} f(x)$, is in fact $f(1)$ (which follows by the continuity of $x \mapsto ax+b$). I believe the intent of the problem was only to find $a,b,c$, so you don't need to actually make this argument. But all I know is what you wrote here.


问:
Five digit re-write game
In the habit of factoring numbers, a notebook I bought had a five digit item number $77076$, which factors as $2^2 3^2 2141$, which may also be $9 \cdot 8564$, and in this form the count of digits is again five. (Repeated digits count separately). [Note I thank Calvin Lin for pointing out I initially had the wrong product for $77076$.]

So I wondered how often this can work for five digit numbers, and started at $10001=73 \cdot 167$, $10002=2 \cdot 3 \cdot 1667=6\cdot 1667,$ and so on. The rules I decided to stick to were only that the five digit number has to be re-written as a product of two or more factors, where the total number (with repetitions) of digits occurring in the factors is again five.

Of course one runs into problems if the five digit number is itself a prime, for example at $10007$ and $10009$. There may be other general restrictions, statable in terms of the form of the prime factorization of the given five digit number; if so I'd be interested in that.

Sometimes the initial factorization into primes has to be juggled with. For example $10010=2 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, which as it stands is two digits over the goal of five. We can lower the digit count by $1$ if we can for example multiply a one digit by a two digit prime in the factorization, and get only a two digit result. For this case we can use $2 \cdot 11=22,\ 5 \cdot 7=35,$ to get
$$10010=7 \cdot 22 \cdot 65,$$
so the five digit requirement is met. In this same example we could instead use $2 \cdot 13=26,\ 5 \cdot 11=55$ and get another re-write:
$$10010=7 \cdot 26 \cdot 55.$$

I know this is not serious math, hence the recreational tag; maybe someone might find it amusing to look at these rewrites.

Just to make a few specific questions: Is there a nice characterization, say in terms of the numbers of digits in the primes occurring in the factorization of $n$, which would say which composite $n$ with five digits did not have five digit rewrites as above? What happens in case we increase the number of digits to say 6 or 7?

答:
Since $(10^k-1)^2=10^{2k}-2\cdot10^k+1\lt10^{2k}$ (e.g. $9\cdot9\lt100$), the number of digits in a factorization cannot decrease by further subdivision of the factors. Thus a number $n$ with $k$ digits has a non-trivial factorization with $k$ digits if and only if it has such a factorization with two factors. For a factorization of $n$ with two factors to have $k$ digits, it is necessary and sufficient that one of the factors has a higher mantissa than $n$.

Thus $n$ has a factorization with $k$ digits if and only if it has a divisor with a higher mantissa than its own. That explains why you found no counterexamples other than primes when you started with the lowest mantissas. At the other end of the spectrum, there are no factorizations with $5$ digits of any $5$-digit number greater than $99\cdot999=98901$.

Title: Five digit re-write game


问:
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$
We have a parabola and a circle with the following equations and their graph placed at the end of my question.

Parabola: $y^2 = 4x -4$

Circle: $(x-2)^2 + y^2 = 9$

My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got

$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$

$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?

What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update

When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:

$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$

$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$

$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.

2nd update

I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:

$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} +  1)^2=9 \implies
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$

答:
$$(x-2)^2 + y^2 = 9 \tag{A}$$
requires that $x \in [-1,5]$ and $y \in [-3,3]$.

$$y^2 = 4x-4 \tag{B}$$ requires that $x \in [1, \infty)$.

So, for $x$ to satisfy both (A) and (B), we must have
$x \in  [1, \infty) \cap [-1,5] =  [1,5]$.

Hence $x=-3$ will be an extraneous root.


答: If $(x-y)(x+y)$ is rational, then $x-y$ and $x+y$ are either both rational, or both irrational. If they're both rational, then their sum is rational. But their sum is $2x,$ which is irrational.

Title: Show that $x+y$ and $x-y$ are both irrational

Body:
Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a nonzero rational, then $x+y$ and $x-y$ must be irrational.

I've made no progress on this problem, so I don't have any work to show. I suppose I can run into a contradiction, but I couldn't find any.


问:
Checking whether a code is cyclic by using the generator matrix
I was wondering whether it is possible to conclude whether a code is cyclic or not from the generator matrix of a code. If so, how could one do this?
Thank you in advance!</p>

答:
In addition to leonbloy's suggestion we have the following alternative.

Undoubtedly you remember from linear algebra how to algorithmically test whether a given vector is in the row space of a matrix? Just add the vector to be tested to the matrix as an extra row and check (by reducing it to row echelon form) whether the rank of the matrix increases (negative) or stays the same (affirmative). This is particularly efficient if the matrix itself was already in the row echelon form. In particular, this is the case if the generator matrix of your code is in a systematic form.

All you need to is the following:

If the cyclic shift (by one position) of each row of the generator matrix $G$ is a codeword (i.e. in the row space of $G$) then the code is cyclic. Otherwise it is not.

It is easy to see why this works. If that condition is satisfied then, by linearity, the cyclic shift by one position of any codeword will again be a codeword. By induction it then follows that the same applies to all the cyclic shifts.

Earlier I mentioned systematic generator matrices as a particular case. Another easy case is that when the rows of $G$ correspond to the monomial multiples $x^i g(x)$ of the generator polynomial $g(x)$, $0\le i<n-k$. For example, to show that $g(x)=x^3+x+1$ generates a cyclic binary code of length $7$ we could use a generator matrix consisting of the four shifts of $g(x)=1101000$. That is
$$
G=\left(\begin{array}{ccccccc}
1&1&0&1&0&0&0\\
0&1&1&0&1&0&0\\
0&0&1&1&0&1&0\\
0&0&0&1&1&0&1
\end{array}\right).
$$
With generator matrices like this the above procedure is particularly simple.
The cyclic shifts of the top three rows are obviously codewords because the next row is the cyclic shift. All we need to do is to verify that the cyclic shift of the bottom row,
that is $1000110$ is a codeword. But we quickly realize that it is the sum of the top three rows. Just as well because this is one cyclic incarnation of the $[7,4,3]$ Hamming code.

The underlying principle in this test is the following. A cyclic code of length $n$ can be viewed as a submodule of the space $\Bbb{F}_q^n$ that we view as a module of the cyclic group $C_n=\langle c\rangle$ of order $n$, $c$ acting by cyclically shifting the components. If a code (= a subspace) $C$ is stable under the action of the generating element $c$, that is $c(C)\subseteq C$, then it is automatically also stable under the entire group (consisting of all the powers of $c$). Looking at codes of length as subspaces of the quotient ring $\Bbb{F}_q[X]/\langle x^n-1\rangle$ we recover the same test in the form: a subspace $C$ is a cyclic code if and only if $xC\subseteq C$.


问:
Improper multivariable integral calculation
Given a domain
$$D=\{(x,y)\mid 0≤x≤5,y≤x≤3y\}$$
how can I calculate the integral $$\bbox[5px,border:2px solid #C0A000]{\iint_D \frac y x \, dx\, dy}$$
using a $D_n$  domain series?

I have tried to figure out what was the domain series but it was wrong.

答:
Hint: Let
$$ u=x,v=\frac{x}{y}$$
and then
$$ x=u,y=\frac{u}{v}.$$


问:
Prove that a property holds for the roots of an equation, given some constraints.
For example, in this question:

The cubic expression $$ ax^3 + bx^2 + cx + d $$ has a pair of roots which are reciprocals of each other. Prove that $$a^2-d^2=ac-bd$$.

I understand that the correct approach is probably to use the formulae relating coefficients and roots, but I'm not exactly sure where to go for there.

Is there are a more generic framework for solving these types of problems?

答:
As a complement to @CalvinLin's answer, let us denote the polynomials
\begin{equation}
\begin{array}[l]\cr
P(x) = a x^3 + b x^2 + c x + d\cr
Q(x) = d x^3 + c x^2 + b x + a
\end{array}
\end{equation}
Explicit computation shows that the resultant of these two polynomials is
\begin{equation}
R(P, Q) = \left|\begin{matrix}
a&0&0&d&0&0\cr
b&a&0&c&d&0\cr
c&b&a&b&c&d\cr
d&c&b&a&b&c\cr
0&d&c&0&a&b\cr
0&0&d&0&0&a\cr
\end{matrix}\right| = -(a^2-d^2- a c + b d)^2 P(1)P(-1)
\end{equation}
It follows that $P$ and $Q$ have a common root if and only if $1$ or $-1$ is a root of $P$ or $a^2-d^2 = a c - b d$. Note that $1$ and $-1$ are their own reciprocals.

Here is the (python) code to prove the claim

```<code>import sympy as sp

a, b, c, d = sp.symbols(['a','b','c','d'])

M = sp.Matrix([
    [a, b, c, d, 0, 0],
    [0, a, b, c, d, 0],
    [0, 0, a, b, c, d],
    [d, c, b, a, 0, 0],
    [0, d, c, b, a, 0],
    [0, 0, d, c, b, a],
])
p = a**2 - d**2 - a * c + b * d
q = (a + b + c + d) * (a - b + c - d)
R = M.det()
s = R - p**2 * q
assert s.simplify() == 0
</code>```

Body:
As a complement to @CalvinLin's answer, let us denote the polynomials
\begin{equation}
\begin{array}[l]\cr
P(x) = a x^3 + b x^2 + c x + d\cr
Q(x) = d x^3 + c x^2 + b x + a
\end{array}
\end{equation}
Explicit computation shows that the resultant of these two polynomials is
\begin{equation}
R(P, Q) = \left|\begin{matrix}
a&0&0&d&0&0\cr
b&a&0&c&d&0\cr
c&b&a&b&c&d\cr
d&c&b&a&b&c\cr
0&d&c&0&a&b\cr
0&0&d&0&0&a\cr
\end{matrix}\right| = -(a^2-d^2- a c + b d)^2 P(1)P(-1)
\end{equation}
It follows that $P$ and $Q$ have a common root if and only if $1$ or $-1$ is a root of $P$ or $a^2-d^2 = a c - b d$. Note that $1$ and $-1$ are their own reciprocals.

Here is the (python) code to prove the claim

import sympy as sp

a, b, c, d = sp.symbols(['a','b','c','d'])

M = sp.Matrix([
    [a, b, c, d, 0, 0],
    [0, a, b, c, d, 0],
    [0, 0, a, b, c, d],
    [d, c, b, a, 0, 0],
    [0, d, c, b, a, 0],
    [0, 0, d, c, b, a],
])
p = a**2 - d**2 - a * c + b * d
q = (a + b + c + d) * (a - b + c - d)
R = M.det()
s = R - p**2 * q
assert s.simplify() == 0


答:
there are two standard forms for complex numbers.

Rectangular form $z =a+bi$ where $a = Re(z)$ and $b = Im(z)$.

And, if $z \ne 0$ polar form  $z = r e^{\theta i}$ where $r = |z|$ and $\theta =\arg(z)$.

$e^{\theta i}$ is defined to be $e^{\theta i} = \cos \theta + i \sin \theta$.

So there is a conversion between these two forms.

If $z = r e^{\theta i} = r\cos \theta + (r \sin \theta) i$.

And if $z = a + bi= \sqrt{a^2 + b^2} (\cos (\arctan \frac ba) + \sin (\arctan \frac ba)i) = \sqrt{a^2 + b^2} e^{\arctan \frac ba i}$.

....

Now the only trick this problem has done is it has combined then $r=|z|$ and $e^{\theta i}$ by replacing $r$ with $e^{\ln r}$ and combining $r e^{\theta i} = e^{\ln r} e^{\theta i} = e^{\theta i + \ln r}$ to get it as a single exponent.

.....

So $2e^{-3\pi/4i+ln(3)}=$

$2e^{-\frac {3\pi}4i}e^{\ln 3} =$

$6e^{-\frac {3pi}4i} = 6(\cos(-\frac {3\pi}4)+ i\sin (-\frac {3\pi} 4))=$

$6(-\frac {\sqrt 2}2 + \frac {\sqrt 2}2i) =$

$-3\sqrt 2 +3\sqrt 2 i$.

That's it.


问:
Find $ \lim_{n \to \infty}n \log\left(1+ \left(\frac{f(x)}{n}\right)^p\right)$
Find $$\lim_{n \to \infty}n \log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)$$  where $0<p<1$

my attempt :By using L-Hospital Rule i got

$$\lim_{n\to \infty} \dfrac  {\dfrac{1}{1+\left( \dfrac{f(x)}{n}\right)^p}
\cdot\dfrac {f(x)^p}{-n^{2p}} }{\dfrac{-1}{n^2}}$$

After that I’m   not able to   proceed further

答:
There are problems of defintion in $f(x)^p$ if $f(x)<0,$ so we should assume $f(x)\ge 0.$

If $f(x)=0,$ then the expression equals $0$ for all $n,$ so the limit is $0.$

If $f(x)>0,$ we can use

$$\lim_{u\to 0}\ln(1+u)/u= 1,$$

which is nothing more than the definition of $\ln'(1),$ which equals $1.$

Our expression equals

$$n(f(x)/n)^{-p}\cdot\frac{\ln(1+(f(x)/n))^p}{(f(x)/n)^p}.$$

The first factor equals $n^{1-p}f(x)^{-p},$ which has limit $\infty.$ The second factor $\to 1$ by $(1).$ Thus the limit equals $\infty$ in the case $f(x)>0.$


问:
Bipartite graph with even degrees
Given a bipartite graph with an equal number of vertices in its parts. Each vertex in the first part is connected with an even number of vertices in the second part. Prove that we can remove some (maybe none, but NOT all) vertices from the first part so that every vertex in the second part will have an even degree?

I tried to build an eulerian subgraph of such graph using some traverses like in Kuhns algorithm but got nothing good.

答:
This is not true.
Consider the complete bipartite graph $K_{2,2}$ and remove one vertex so the resulting graph is $K_{1, 2}$ and hence the vertices in the second part have degree 1.

Title: Bipartite graph with even degrees


问:
Prove that every point, satisfying given condition, lies on some line.
Segment $BC$ is the shortest side of noniscosles triangle $ABC$. Points $K$ and $L$ lie respectively on sides $AB$, $AC$ and satisfy condition: for given point $X$ on side $BC$, $BK=BX$ and $CL=CX$. Prove that midpoint of the segment $KL$ lies on some line indepedently of choosing point $X$.

I have been wondering for some time on solution for this problem but I did not come up with anything useful. Could you give some hints?

答:
Here's a synthetic solution, which has roots in the coordinate solution.

Let $ D$ be a point on $AC$ such that $ DC  = BC$.

Let $E$ be a point on $AB$ such that $ BE = EC$.

For a given point $X$ on $BC$ such that $ BX : XC = t : 1 - t$, define $Y$ on $ED$ such that $DY : YC = t : 1 - t$.

Show that

$KY \parallel BD \parallel XL$

$YL \parallel CE \parallel KX$

$KXLY$ is a parallelogram

Mid point of $KL$ is the midpoint of $XY$.

Locus of midpoint of $XY$ is thus a straight line.

In fact, Midpoint of $XY$ lies on the line connecting the midpoint of $CE$ and $BD$.

Note: To be fair, I'm not certain of a purely synthetic solution to the last statement. It's clearly true by coordinates. I feel that it should be true by some similarity / homothety argument, but that's eluding me for the time being.


问:
Odds on winning a 100 club draw two consecutive months
I have been lucky enough to win a 100 club draw for two consecutive months running. What are the odds of this occurring?
There are 100 persons in the draw each month, each with one ticket, therefore my chances of winning are I assume 1 in a 100 each month. There are four monetary prizes each month.
However, I have won first prize in the draw in December and January. What are the odds of this event occurring?. Thank you.</p>

答:
It's very important, when asking "What's the probability of this event?", to be specific about what event you're talking about and what space of possible events you're considering. This is because people are very bad at understanding probability intuitively and while we tend to notice rare things occurring we ignore the fact that with enough rare events the chance that at least one of then happening can be reasonably high.

The probability that you win the prize in any one given month is, as you correctly say, $\frac{1}{100}$.

Assuming independence of the draws (which seems fair), the probability that you win the prize in any two consecutive months (e.g. December and January) is $\frac{1}{100} \times \frac{1}{100} = \frac{1}{10000}$, that is one in ten thousand.

However, if it is already a given that you won the prize in December, then the probability that you win in January as well doesn't depend on the probability of winning in December, so it's $\frac{1}{100}$ again. This is a natural consequence of the draws being independent.

Similarly, in any given month, the probability that the person who won it last month wins it again is still $\frac{1}{100}$, same as anyone else.

Additionally, the probability that the prize is won by someone other than last month's winner is always $\frac{99}{100}$. So if we keep track of the winners for, say, a year, then there are 11 chances to have a doubling-up of winners, so the probability that this doesn't happen is $\frac{99}{100} \times \frac{99}{100} \times \ldots \times \frac{99}{100} = \left( \frac{99}{100} \right)^{11} \approx 0.895$, so there's a $10.5\%$ chance that there is someone who wins two months in a row, which is not particularly implausible.


问:
What is the name for the maximum power of a variable in the terms of a multivariable polynomial?
Consider a multivariable polynomial, for example, $f(x, y, z) = 1 + 2x + 3xy + 5zy + 3xy^2z + 2x^2y$. The degree of this polynomial is $4$ because of the $x y^2z$ term.

I am looking for the name for the maximum power of any one variable. Here it would be $2$ because of both the $x^2y$ term, and the $x y^2 z$ term.

Another way to put it is if we consider the set of polynomials expressible as
$$ f(\vec x) = \sum_{i_1 = 0}^{K} \sum_{i_2 = 0}^{K} ... \sum_{i_N = 0}^{K} a_{\vec{i}}\prod_{j=1}^{N} x_{j}^{i_j} $$
where $\vec{x} \in \mathbb{R}^{N} $, $\vec{i} = [i_1, i_2, ..., i_N] \in \mathbb{N}^{N}$, and the $a_{\vec{i}}$ are the various coefficients in $\mathbb{R}$ for each term, then I would like to know the name for $K$.

Tangentially the name for the $i_j$ in a particular term would be good too, when I need to talk about the power of a particular variable in a particular term.

Most searches for this name return only discussions of single variable polynomials, for which it is obviously equivalent to the degree of the polynomial and thus not given a separate name. I found this reference which appears to call $K$ the "max degree", but I am not certain whether that is a widely used term or specific to the context in which it appears there.

答: Since a polynomial is defined under a ring $R$, you can look at a polynomial of the form $f(x,y)$ as a polynomial $f(x)$ under the ring $\mathbb R[y]$, i.e the coefficients of the polynomial in $x$ are polynomials from $\mathbb R[y]$. So the degree with respect to $x$ is defined in the usual way.


问:
How to find the intersection point of two moving spheres?
I'm writing a gas dynamics simulation. I created a two degree of freedom simulation, where all the gas molecules are represented as hard spheres and their collisions are perfectly elastic. However, I want to take that into three dimensions. I have the XYZ coordinates of all of the moving spheres, along with their vectors. By subtracting the difference between two vectors, I can essentially simplify the problem to a moving sphere colliding with a stationary sphere. We can assume that the spheres will collide because the moving sphere is calculated to approach closer than the sum of the spheres' radii.

How can I find when and where the spheres will collide?

Any thoughts would be much appreciated, and let me know if there is anything I can do to clarify the problem.

答: The spheres will collide when their centers are a distance apart equal to the sum of their radii. The collision point is a point on the line between the two centers. If you have the vector coordinates of these centers that should be relatively easy to figure out.</p>


问:
How to express $2 \cos X = \sin X$ in terms of $\sin X$?
The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

答:
As already mentioned by the users , you must use the identity :

$$\large \sin^2 x + \cos^2 x = 1 ........... \boxed{1} $$

Here is , how to start solving the problem :

$$\large \text{Given : } \quad 2\cos x = \sin x ......... \boxed{2}$$
Now, simplifying equation 1 further :

$$\large \cos^2 x = 1 - \sin^2 x $$
Solve for $\cos x $ further by taking square root both sides and put that value in equation 2 , you will get your answer.


问:
Rule of Three with different origin
I know that this may by a very newbie question but I am really struggling with it.

If I have 2 dimensions that are directly proportial I can use a rule of three to transform from one magnitud to another. But what if each dimension has a different origin?

Example:

```<code>1 is a 3
as 5 is a 10
as 3 is a X
</code>```

How can I calculate X?

答:
I found this:

```<code>(b - a) is a (d - c)
as (e - a) is a (x - c)
</code>```

Then I apply the rule of three:

```<code>(e - a)(d - c)
-------------- = x - c
   (b - a)
</code>```

Or:

```<code>    (e - a)(d - c)
c + -------------- = x
        (b - a)
</code>```

Applying this to our example we have

```<code>    (3 - 1)(10 - 3)
3 + -------------- = x
        (5 - 1)

x = 6.5
</code>```

Which looks valid

Can someone confirm my reasoning is valid? and if I can simplify the above formula?

Title: Rule of Three with different origin

Body:
I found this:

(b - a) is a (d - c)
as (e - a) is a (x - c)

Then I apply the rule of three:

(e - a)(d - c)
-------------- = x - c
   (b - a)

Or:

    (e - a)(d - c)
c + -------------- = x
        (b - a)

Applying this to our example we have

    (3 - 1)(10 - 3)
3 + -------------- = x
        (5 - 1)

x = 6.5

Which looks valid

Can someone confirm my reasoning is valid? and if I can simplify the above formula?


Title: How to find the intersection point of two moving spheres?

Body:
I'm writing a gas dynamics simulation. I created a two degree of freedom simulation, where all the gas molecules are represented as hard spheres and their collisions are perfectly elastic. However, I want to take that into three dimensions. I have the XYZ coordinates of all of the moving spheres, along with their vectors. By subtracting the difference between two vectors, I can essentially simplify the problem to a moving sphere colliding with a stationary sphere. We can assume that the spheres will collide because the moving sphere is calculated to approach closer than the sum of the spheres' radii.

How can I find when and where the spheres will collide?

Any thoughts would be much appreciated, and let me know if there is anything I can do to clarify the problem.

Body:
$$\|(\vec c_0+t\vec v_0)-(\vec c_1+t\vec v_1)\|=r_0+r_1$$

or

$$(\vec{\Delta c}+t\vec{\Delta v})^2=\vec{\Delta c}^2+2\vec{\Delta c}\vec{\Delta v}\,t+\vec{\Delta v}^2\, t^2=(r_0+r_1)^2.$$

Solve for $t$.


问:
Laurent series of $\dfrac{\sin(z)}{z\log(z+1)}$
I want to calculate the Laurent series of $f(z)=\dfrac{\sin(z)}{z \cdot \log(z+1)}$ about $z=0$.

I know $\lim_{z \to 0} \dfrac{\sin(z)}{\log(1+z)}=1$ so f has a pole of order $1$ at $z=0$. Then $a_{-1}=1$ and $a_{-k}=0$ for $k\geq2$

¿What should I do next? I was thinking about multipling $x^{n}f(z)$ and taking  derivatives so I can get $a_{0}, a_{1}, a_{2}...$ but I'm not sure that's a good aproach. Any suggestion? Thanks.

答: Write $\frac{\sin z}{z\ln(1+z)}=\sum_{k\ge0}c_kz^{k-1}$ so $\sum_{k,\,l\ge0}c_k\frac{(-1)^{l+1}}{l+1}z^{k+l}=\frac{\sin z}{z}$ has $z^{2m}$ coefficient $\frac{(-1)^m}{(2m+1)!}$. This gives simultaneous equations in the first few coefficients. The Laurent series begins $\tfrac1z+\tfrac12-\tfrac14z+\tfrac{z^3}{240}+\tfrac{23z^5}{1440}$. I'm not sure if a closed form for the $z^{2n-1}$ coefficient with $n\ge0$ is obtainable.


问:
Minimal steps to reach a natural number by a signed arithmetic progression
$F:=\big\{f\,\big|\,\text{function }f: \mathbf N\rightarrow \{1,2\}\big\}$.
\begin{align}
&\min_{f\in F,\,k\in\mathbf N} k, \\
&\sum_{i=1}^k (-1)^{f(i)}i = n \in\mathbf N.
\end{align}

Is there an analytical or an asymptotic solution to this problem for general $n$?

For a specifically given $n$, what would a good algorithms to solve this problem? The dynamic programming?

答:
Essentially, for any given $n$, the task is to select $+$'s and $-$'s to reach $n$ as fast as possible (minimize $k$):
$$\pm 1 \pm 2 \pm 3 \pm \dots \pm k = n$$

If we don't worry about overshooting $n$, the obvious strategy is to only add and never subtract. If $n$ is a triangular number, this will give us the answer. This would necessarily be the optimal solution since we did not subtract at all.

If $n$ is not a triangular number, at some point $k'$, we will overshoot $n$:
$$1 + 2 + 3 + \dots + k' > n$$

Since we need $k$ at least $k'$ to even reach $n$, we know that $k=k'$ is optimal if it works.

Consider the value $1 + 2 + 3 + \dots + k' - n$. We will denote this amount as $a$. We know that $a < k'$. If $a$ is even, then we can switch the operator of $a/2$ to $-$, and we have our optimal solution.

Figuring out the solution for when $a$ is odd may be trickier.

EDIT:

Figured out the "odd" case!

If $a$ is odd, then we cannot have $k = k'$ because there is no expression that will give us exactly $n$ using $k'$ terms. For any signs we switch in the $k'$ terms, the total sum will retain the same parity, so we cannot reach exactly $n$. Thus, if we can find a solution with $k'+1$ terms, it is necessarily optimal.

Consider $1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k'+1$.

If $k'$ is even, then consider the value $\frac{a+k'+1}{2}$. We have
$$\frac{a+k'+1}{2} < \frac{2k'+1}{2} = k' + 1/2 \,.$$

So, $\frac{a+k'+1}{2} \leq k'$. Thus, it is in our series, so we can assign it the operator $-$, and we have achieved exactly a sum of exactly $n$.

Now, to consider what happens when both $a$ and $k'$ are odd...

EDIT 2:

First off, is it possible to find a solution with $k=k'+1$ when $k'$ is odd?

Well if we use all $+$'s, we have $1 + 2 + 3 + \dots + k' + (k'+1) = n + a + k' + 1$. As before, switching the signs of any of our terms will preserve the parity of the total sum. Since $a$ and $k'$ are odd, $n+a+k'+1$ has opposite parity from $n$. So, no rearrangement of signs will enable a series of $k'+1$ terms to work.

Our next candidate is $k=k'+2$. Let us consider the following sum:
$$1 + 2 + 3 + \dots + k' + (k'+1) + (k'+2) = n + a + (k'+1) + (k'+2)$$

We are too high by the quantity $a + 2k' + 3$. We can switch the sign of $k'$ and switch the sign of $\frac{a+3}{2}$, and we arrive at exactly $n$ with $k=k'+2$.

And that concludes the proof! No dynamic programming needed - a purely analytic solution.

TL;DR

Let $k'$ be the least integer such that $T_{k'} \geq n$. If $T_{k'}$ and $n$ have the same parity, then $k=k'$. Else, $k$ is equal to the least odd integer greater than $k'$.

See this OEIS sequence.


问:
Conditional expectation $\mathbb E(Y \mid X \in B)$ instead of $\mathbb E(Y \,|\, X = x)$ (generalization of Shorack's PFS Notation 7.4.1)
In Shorack's Probability for Statisticians Notation 7.4.1, he notes that the conditional expectation (defined in the measure theoretic way) $\mathbb E(Y\mid X)$ is $g(X)$ for some measurable $g :(\mathbb R, \mathcal B_\mathbb R) \to (\mathbb R, \mathcal B_\mathbb R)$. He then defines $\mathbb E(Y \mid X=x)$ as simply $g(x)$. For conditional probabilities, I'm pretty sure this means that $P(A \mid X=x)$ will be defined as $\mathbb E(1_A \mid X=x)$. I'm not entirely confident that this measure-theoretic definition of conditional probability conditioned on $X=x$ matches with the classical notion of conditional probability, so if someone could shed some light on that too that'd be great.

Here's a picture of the relevant section from the book.

My question is: is there a generalization of this definition to $\mathbb E(Y\mid X\in B)$ for some Borel set $B \in \mathcal B_\mathbb R$?

Looking at this question/answer When do the measure-theoretic and elementary definitions of conditional probability/expectation coincide?, it seems like the generalization would require dividing $P(X\in B)$, which may not be possible since it could be $0$. In that case, why is it that we can have a general definition of $\mathbb E(Y\mid X=x)$ but not of $\mathbb E(Y\mid X\in B)$?

答:
Why not continue the analogy and set $\mathbb{E}(Y \, \mid \, A) = \mathbb{E}(Y , \mid \, 1_{A} = 1)$?  In this case, you can check that $\mathbb{E}(Y \, \mid \, 1_{A} = 1) = \frac{\mathbb{E}(Y : A)}{\mathbb{P}(A)}$ so that it is consistent with the naive approach.  Further, it's worth noting that $\mathbb{E}(Y \, \mid \, 1_{A})$ has the form
\begin{equation*}
\mathbb{E}(Y \, \mid \, 1_{A}) = \left\{ \begin{array}{r l} \frac{\mathbb{E}(Y : A)}{\mathbb{P}(A)}, & \text{in} \, \, A, \\
\frac{\mathbb{E}(Y : A^{c})}{\mathbb{P}(A^{c})}, & \text{in} \, \, A^{c}.
\end{array} \right.
\end{equation*}
I leave the confirmation of these identities as exercises, which are not too hard since $\sigma(1_{A}) = \{\phi, A, A^{c}, \Omega\}$.

An interesting fact that's worth mentioning here is that if $Y,X$ are two real random variables (you could also use random vectors, but I won't), if $\mu_{X}$ is the law of $X$, and if $Y$ is integrable, then
\begin{equation*}
\mathbb{E}(Y \, \mid \, X = x) = \lim_{\epsilon \to 0^{+}} \frac{\mathbb{E}(Y : X \in [x - \epsilon, x + \epsilon])}{\mathbb{P}\{X \in [x- \epsilon, x + \epsilon]\}} \quad \text{for} \, \, \mu_{X}\text{-a.e.} \, \, x \in \mathbb{R}
\end{equation*}
This follows from the Besicovitch Differentiation Theorem (cf. Chapter 5 of Sets of Finite Perimeter and Geometric Variational Problems by Maggi); I don't know an easier way to prove it in general.  Replacing $Y$ by $1_{A}$ for suitable events $A$, we can also use this to compute probabilities conditioned on $\{X = x\}$.

On the topic of "conditioning on sets of measure zero," there is yet another fact worth mentioning.  Let $U$ be a bounded open subset of $\mathbb{R}^{d}$ with smooth boundary, fix $x \in U$, and let $B^{x}$ be be a standard Brownian motion with $B^{x}_{0} = x$.  Let $\tau_{U}$ be the first time $B^{x}$ reaches $\partial U$.  It turns out that there is another process $\tilde{B}^{x}$ such that, for each $t \geq 0$, $B^{x}_{t}$ conditioned on $\{\tau_{U} \geq T\}$ converges in distribution to $\tilde{B}^{x}_{t}$ as $T \to \infty$.  That is,
\begin{equation*}
\mathbb{E}(f(\tilde{B}^{x}_{t})) = \lim_{T \to \infty} \frac{\mathbb{E}(f(B_{t}^{x}) : \tau_{U} \geq T)}{\mathbb{P}(\tau_{U} \geq T)}.
\end{equation*}
Note that $\tau_{U} < \infty$ almost surely so here, as in the last paragraph, we have "(asymptotically) conditioned on a set of measure zero."  I don't know if it's possible to interpret the statement "$\tilde{B}^{x}$ equals $B^{x}$ conditioned on $\tau_{U} = \infty$" in a more rigorous way.  (More generally, in the theory of stochastic process, the previous construction is called a Yaglom limit.)


答:
Consider the incidence matrix $M$, where entry $M_{ij} = 1$ if the ith vertex in the first part is connected to the ith vertex in the second part, 0 otherwise.

Work in $ F_2$.

We are given that $ M \times 1 = 0$. (Translation: Each vertex in the first part is connected to an even number of vertices, which is 0 in $F_2$).

So $ \det(M) = 0$.

So there is a non-trivial left vector $a$ such that $ a M = 0 $.

Translating back, restricting to the vertices in the first part that correspond to a non-zero entry in the vector $a$, then every vertex in the second part is connected to an even number of these.

Title: Bipartite graph with even degrees

Body:
Given a bipartite graph with an equal number of vertices in its parts. Each vertex in the first part is connected with an even number of vertices in the second part. Prove that we can remove some (maybe none, but NOT all) vertices from the first part so that every vertex in the second part will have an even degree?

I tried to build an eulerian subgraph of such graph using some traverses like in Kuhns algorithm but got nothing good.


问:
¿Can you help me with axiom of regularity?
I understand that if we define a set such that $A=\{A\}$ we have then a contradiction because of regularity axiom, thus, a set can not be a member of itself. My question is; how can it be so, if I define a set $A=\{A,b\}$ ($b$ disjoint from $A$)?

Thanks.

答: In general, if we suppose that $A$ is any set such that $A\in A,$ then let $C=\{A\}$. That will give you your contradiction from the Axiom of Regularity.</p>


问:
Draw 3 card simultaneously from a standard deck without replacement.
I have this the following question with 4 parts, but I don't know how the numbers are calculated for parts 1,3,4. Can anyone help me.

We simulteneously draw three cards:
A-) Probability of observing exactly 2 diffrent shapes: ANSWER: 0.551-->HOW

B-) Probability of Observing at least one king and one queen. --> I know how to do this part.

C-) Probability of observing 2 jacks and diamond as the third card. -->ANSWER: 0.0034 -->HOW

D-)Probability of observing 3 different face value. --> ANSWER: 0.83  --> NO IDEA HOW

For part A, I remember there was a method with combinations but I don't remember the method. Does anyone know how?

I tried 39/51 but it is not correct. I thought 1st could be anything 2nd should be one of the remaining 39 cards from shapes that are not the same as the shape of first.

For C I did:
(4x3x12)/(52x51x50) + (3x2x13)/(52x51x50) --> 2 jacks including  diamond + 2 jack no diamond
But the answer is 4 times the number I found. Where does *4 coming from.

答:
A. Probability of only picking one suit is $4\cdot\frac{13\cdot12\cdot11}{52\cdot51\cdot50}$. Probability of picking $3$ suits is $1\cdot\frac34\cdot\frac12$. You can finish from here

C. Case 1: The last card is a jack: Then, there is one (non-diamond) jack left to pick, and two possible places it could be picked. So, we have $2\cdot\frac{48}{52}\cdot\frac3{51}\cdot\frac1{50}$. Case 2: No jack of diamonds is picked: the first two cards must be jacks. So, we have $\frac3{52}\cdot\frac2{51}\cdot\frac{13}{50}$. Case 3: A jack of diamonds is picked, but not as the last card: $2\cdot\frac3{52}\cdot\frac1{51}\cdot\frac{12}{52}$. The sum of these cases is the answer.

D. $1\cdot\frac{48}{51}\cdot\frac{44}{50}$


问:
Simple composition of random variables.
If I have a (discrete) random variable $X$ defined as

\begin{equation}
X = \begin{cases}0 \qquad p_1=0.5\\
1 \qquad p_2=0.2 \\
2 \qquad p_3=0.3
\end{cases}
\end{equation}

If now I have $Y=2^X$ is it correct to write that

\begin{equation}
Y = \begin{cases}
1 \qquad p_1=0.5 \\
2 \qquad p_2=0.2 \\
4 \qquad p_3=0.3
\end{cases}
\end{equation}

and then calculating mean/variance etc.. thank you.

答: yes, to see why note that $\mathbb P_X(\{0\}) = \mathbb P(\{X^{-1}(0)\}) = \mathbb P(\{\omega\in\{0,1,2\}\mid \omega = 0\}) = 0.5$ by definition. Now what is $\mathbb P_Y(\{1\}) = \mathbb P(\{Y^{-1}(1)\}) = \mathbb P(\{\omega\in\{0,1,2\} \mid 2^\omega = 1\})$? Clearly, $\omega = 0$ is the only $\omega\in\{0,1,2\}$ such that $2^\omega = 1$. Thus, $\mathbb P_Y(\{1\}) = \mathbb P_X(\{0\})$.

Title: Simple composition of random variables.


答:
For A)

Exactly 2 shapes mean 2 out of the 3 cards is of the same suit. The number of ways we can get 2 same cards out of any suit is ${13 \choose 2}$. After the 2 cards, we can choose any cards from any suit, and there are 39 of them. And since we have 4 suits, we multiply by $4$.

Diving by the total number of ways we can get 3 cards, we get

$\frac{4({13 \choose 2} 39)}{{52 \choose 3}}$

Title: Draw 3 card simultaneously from a standard deck without replacement.

Body:
I have this the following question with 4 parts, but I don't know how the numbers are calculated for parts 1,3,4. Can anyone help me.

We simulteneously draw three cards:
A-) Probability of observing exactly 2 diffrent shapes: ANSWER: 0.551-->HOW

B-) Probability of Observing at least one king and one queen. --> I know how to do this part.

C-) Probability of observing 2 jacks and diamond as the third card. -->ANSWER: 0.0034 -->HOW

D-)Probability of observing 3 different face value. --> ANSWER: 0.83  --> NO IDEA HOW

For part A, I remember there was a method with combinations but I don't remember the method. Does anyone know how?

I tried 39/51 but it is not correct. I thought 1st could be anything 2nd should be one of the remaining 39 cards from shapes that are not the same as the shape of first.

For C I did:
(4x3x12)/(52x51x50) + (3x2x13)/(52x51x50) --> 2 jacks including  diamond + 2 jack no diamond
But the answer is 4 times the number I found. Where does *4 coming from.


答:
I'll help you out with the first part, and then hopefully you will be able to try a similar process for the rest.  To get the probability of exactly two of the three drawn cards having different suits I'm going to think about the situation as follows.  There are two possibilities:

Draw a card, now there are $51$ cards left.

The second card is the same suit as the first: probability $\frac{12}{51}$, $50$ cards left.

The third card is a different suit from the first two: probability $\frac{39}{50}$

or

Draw a card, now there are $51$ cards left

The second card is a different suit from the first: probability $\frac{39}{51}$, $50$ cards left.

The third card is either of the first two suits, of which there are $12$ left in the deck each: probability $\frac{24}{50}$

Probability for the first option:$$\frac{12}{51} \cdot \frac{39}{50}.$$

Probability for the second option:$$\frac{39}{51} \cdot \frac{24}{50}.$$

Total probability that either event occurs:$$\frac{12}{51} \cdot \frac{39}{50} + \frac{39}{51} \cdot \frac{24}{50} = \frac{234}{425} \approx 0.551.$$


问:
Trouble understanding proof of the Cauchy-Schwarz inequality
Hope you're doing well.

Well I'm stuggling to understand the proof of Cauchy-Schwarz inequality, especially with the following equation: $$
\frac{1}{\|\mathbf{v}\|^{2}}\|\| \mathbf{v}\left\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\right\|^{2}=\|\mathbf{u}\|^{2}\|\mathbf{v}\|^{2}-|\langle\mathbf{u}, \mathbf{v}\rangle|^{2}
$$

As it says in wikipedia: if you simplify the right side of the equation you'll get the left side easily. I tried and it doesn't work for me.

Thanks in advance for your answers.

答: Note that\begin{align}\bigl\|\| \mathbf{v}\|^{2} \mathbf{u}-\langle\mathbf{u}, \mathbf{v}\rangle \mathbf{v}\bigr\|^2&=\|\mathbf v\|^4\|\mathbf u\|^2-2\|\mathbf v\|^2\langle\mathbf u,\mathbf v\rangle^2+\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\|\mathbf v\|^2\\&=\|\mathbf v\|^2\bigl(\|\mathbf v\|^2\|\mathbf u\|^2-\bigl|\langle\mathbf u,\mathbf v\rangle\bigr|^2\bigr).\end{align}Now, divide both sides by $\|\mathbf v\|^2$ and you will get your equality.


问:
Why is Hilbert's Seventeenth Problem important?
I'm self-learning about Model Theory and I just got to the proof of Hilbert's 17th Problem via Model Theory of Real Closed Fields. The 17th problem asks to show that a non-negative rational function must be the sum of squares of rational functions.

It seems to me that I lack a strong enough understanding of the context of the problem to understand its significance. Why did Hilbert think it was important enough to include it as one of his 23 problems?

答:
I am not an expert on the history of math nor on the mathematics around Hilbert's 17th problem, so I welcome corrections and other insights from others more knowledgeable than me. Also, most of what I'm going to say already appears on the Wikipedia page for the 17th problem, but hopefully I can turn it into a useful answer.

The first connection between squares and positivity is the observation that a real number $r$ is a square if and only if $r \geq 0$. This is no longer true for integers and rational numbers, but it remains true that a number is $\geq 0$ if and only if it is a sum of squares (trivially for the integers by writing $n = \underbrace{1^2 + \dots + 1^2}_{n \text{ times}}$ and slightly less trivially for the rationals by writing $\frac{a}{b} = \underbrace{\left(\frac{1}{b}\right)^2 + \dots + \left(\frac{1}{b}\right)^2}_{ab\text{ times}}$). Lagrange famously improved this in 1770 by showing that every non-negative integer is the sum of four squares, from which it follows easily that the same is true for non-negative rational numbers.

Now it's natural to ask analogous questions for other (partially) ordered rings. Let's say a function $f\colon\mathbb{R}^n\to \mathbb{R}$ is non-negative if $f(x)\geq 0$ for all $x\in \mathbb{R}$. If $f$ is a sum of squares of real-valued functions, then it is obviously non-negative. Does the converse hold? Well, yes, $f$ is the square of the function $\sqrt{f}$. But in analogy to restricting to integers or rational numbers above, the question becomes more interesting if we want to represent $f$ as a sum of squares of polynomials or rational functions.

Hilbert proved in 1888 that there are non-negative polynomials over $\mathbb{R}$ which are not sums of squares of polynomials (but the first explicit example of such a polynomial was apparently not found until 1967!). On the other hand, in 1893, Hilbert showed that any non-negative polynomial over $\mathbb{R}$ in at most $2$ variables is a sum of squares of rational functions.

It's then a very reasonable conjecture that we can remove the restriction on the number of variables. In the years between 1893 and Hilbert's address in 1900, I would be surprised if Hilbert didn't try to extend his result at least to the case of polynomials in $3$ variables - but evidently even this incremental improvement was out of reach. So as a natural question, related to Hilbert's previous work, and apparently resistant to the current methods of 1900, it's not surprising that Hilbert would have included it on his list.

And indeed, the resolution of the problem (both the original purely algebraic solution and its elegant reformulation using model theory) only came after the introduction of bold new ideas, namely the Artin-Schreier theory of real closed fields.


问:
Write the contrapositive of "if $n$ is an even integer such that n + 1 is a square then n is divisible by 8"
For the purpose of clarity, I do not want anybody to prove this statement, I am just looking to get some help translating it into the contrapositive using symbolic logic. Right now I have the following translation:

Original Statement:

$$(\exists k \in \mathbb{Z} \quad n = 2k) \quad \land \quad (\exists m \in \mathbb{Z} \quad n+1 =m^2) \quad \Rightarrow \quad \text{n is divisible by 8}$$

Contrapositive Statement:

$$ \text{n is not divisible by 8} \quad \Rightarrow \quad (\exists k \in \mathbb{Z}  \quad n = 2k+1) \quad \lor \quad (\forall m \in \mathbb{Z}  \quad n+1 \neq m^2) $$

A couple of points of clarification:

I wasn't sure how to write $n$ is divisible by 8. Perhaps it could have been stated as $\exists p \in \mathbb{Z} \quad n = 8p$?

For the negation of $\exists k \in \mathbb{Z} \quad n = 2k$, I know that typically you might want to reverse the $\exists$ to an $\forall$, but I think what I wrote makes sense since the negation of even numbers is odd numbers.

Eventually I will try to prove this statement by contrapositive, but I want to get some practice in with translating mathematical statements into a more formal structure.

My question(s) are following:

Is the original translation correct? How might it be improved?

How could I more formally translate '$n$ is divisible by 8'?

Is the contrapositive stated correctly?

I also am not able to use $mod$ notation, $a|b$ notation, or anything that is unique to Abstract Algebra.

答: If $n$ is not divisible by 8, then either $n$ is not an even integer or  $n+1$ is not a square.


问:
Applying the chain-rule to the function $u(x+sb, t+s)$ where $b \in \mathbb{R}^n$ is a constant vector and $s \in \mathbb{R}$
I want to apply the chain-rule to the function $z(s) := u(x + sb, t+s)$ where $x = (x_1, \cdots, x_n) \in \mathbb{R}^n$ and $b \in \mathbb{R}^n$, $s \in \mathbb{R}$ are constants. I am supposed to get this:

\begin{align}
z(s) &= u(x+sb,t+s)\\[5pt]
z'(s) &= u_{x_{1}}(x+sb,t+s)b_{1} + \dotsb + u_{x_{n}}(x+sb,t+s)b_{n} + u_{t}(x+sb,t+s)\cdot 1
\end{align}

but I do not know how this is done concretely.

Here is my attempt. I set $\xi(s) = x+sb$ and $\gamma(s) = t+s$. Then, $z(s) : = u(\xi(s), \gamma(s))$. The chain-rule then says: $$\frac{dz}{ds} = \frac{du}{d\xi}\frac{d\xi}{ds} + \frac{du}{d\gamma}\frac{d\gamma}{ds}.$$

What is $\frac{du}{d\xi}$ ?

Since $\xi$ is a vector then I think it is the sum $u_{\xi_1} + \cdots + u_{\xi_n}$. Then what is $u_{\xi_1}$?

This is my confusion that I hope someone can clear for me.

答:
Based on the solution, it seems that $u(x + sb, t+s)$ is shorthand for $$u(x_{1}+sb_{1},x_{2} + sb_{2}, \dotsc, x_{n} + sb_{n},t+s).$$  So, we should really be defining $\xi_{i}(s) = x_{i} + sb_{i}$ and so $z(s)$ becomes $$z(s) = u(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s)).\tag{1}$$  Now, because $u\colon \mathbb{R}^{n+1}\to \mathbb{R}$ we think of $$u = u(x_{1},x_{2},\dotsc,x_{n},t).$$

That means that when we apply the chain rule to $(1)$ we get $$\frac{dz}{ds} = \frac{du}{dx_{1}}\frac{d\xi_{1}}{ds} + \frac{du}{dx_{2}}\frac{d\xi_{2}}{ds} + \dotsb + \frac{du}{dx_{n}}\frac{d\xi_{n}}{ds} + \frac{du}{dt}\frac{d\gamma}{ds}.$$

Then:
\begin{align}
\frac{du}{dx_{i}}\frac{d\xi_{i}}{ds} &= u_{x_{i}}(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s))\frac{d}{ds}(x_{i} + sb_{i})\\
&= u_{x_{i}}(\xi_{1}(s), \xi_{2}(s),\dotsc,\xi_{n}(s),\gamma(s))b_{i}\\
&=u_{x_{i}}(x_{1}+sb_{1},x_{2} + sb_{2}, \dotsc, x_{n} + sb_{n},t+s)b_{i}\\
&=u_{x_{i}}(x+sb,t+s)b_{i}.
\end{align}


问:
Product of two compact spaces with the product topology is compact
The way I wanted to go about it was this. We have $X\times Y$ with the product topology, which in our case is equivalent to the box topology. If we take an open cover of $X\times Y$ ,$\{G_k\}_{k\in K}$ this will be made of unions of products of open sets in $X$ and $Y$. Let's take $F=\{U_i\times V_i | i\in I\}$ , where $U_i\times V_i$ are the products that make up the unions . Take $x\in X$ , then we know that $\{x\}\times Y$ is omeomorphic to $Y$ , meaning it is compact, so we can find a finite index set $I_x \subseteq I$ such that  $A=\{U_i\times V_i | i\in I_x\}$ is a finite subcover of $\{x\}\times Y$ . At this point, my book does it in a slightly different way, but what I did was this, in the same way, take $y\in Y$ , we can find a finite index set $I_y\subseteq I$ such that $B=\{U_j\times V_j | j\in I_y\}$ is a finite subcover of $X\times \{y\}$ . Now, if we take $A\cup B$ , this is a finite cover of $X\times Y$, and we also know that every element of $A\cup B$ is contained in some open set of the original open cover  $\{G_k\}_{k\in K}$ , so there are  a finite number of $k(i,j)\in K$ such that all of the $G_{k(i,j)}$ cover $X\times Y$ , and they are our finite subcover. Is this proof flawed in some way?

答:
I’m afraid that it is flawed. Your set $A$ is a finite open cover of $\{x\}\times Y$, and your set $B$ is a finite open cover of $X\times\{y\}$, so $A\cup B$ is a finite open cover of

$$\big(\{x\}\times Y\big)\cup\big(X\times\{y\}\big)\,,$$

but there is no guarantee that it covers all of $X\times Y$, and in general it won’t. At this point in your argument you really have a different $A_x$ for each $x\in X$ and $B_y$ for each $y\in Y$.


问:
Show that $\alpha$ is a homomorphism. Also, find $\text{ker}(\alpha)$, and decide if $\alpha$ is onto.

Let  $\alpha: \mathbb{Z}_9 \times \mathbb{Z}_{27} \rightarrow \mathbb{Z}_{27}$ be given by $\alpha((a, b)) = 3b$, for all $a\in \mathbb{Z}_9,b\in \mathbb{Z}_{27}$.
Show that $\alpha$ is a homomorphism. Also, find $\text{ker}(\alpha)$, and decide if $\alpha$ is onto.

My attempt: Let $(a_1,b_1), (a_2,b_2)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}$.

Then $$\alpha((a_1,b_1)+(a_2,b_2))=\alpha(a_1+a_2,b_1+b_2)=3(b_1+b_2)=3b_1+3b_2 =\alpha((a_1, b_1))+\alpha((a_2, b_2))$$

thus, $\alpha$ is a homomorphism.

$$\text{ker}(\alpha)=\{(a,b)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}:\alpha((a,b))=0\}=\mathbb{Z}_9 \times\{0\}$$ (In particular $\alpha$ is not one-to-one).

My question is in determining if $\alpha$ is onto.

If $b\in \mathbb{Z}_{27}$, then $\alpha\left ( \left (a,\frac{1}{3}b  \right ) \right )=b$

Does this argument help me to say that $\alpha$ is onto?

答:
Your answer for the kernel is wrong -- $(a,9)$ and $(a,18)$ are also in the kernel, so $$\ker\alpha=\mathbb Z_9\times\{0,9,18\}\cong\mathbb Z_9\times\mathbb Z_3.$$
So by the isomorphism theorem, $$\operatorname{im}\alpha\cong(\mathbb Z_9\times\mathbb Z_{27})/(\mathbb Z_9\times\mathbb Z_3),$$
which (looking at orders) cannot possibly be $\mathbb Z_{27}$.

Your last claim doesn't really make sense -- how to interpret $\frac{1}{3}$ in $\mathbb Z_{27}$? You can't really, because $3$ and $27$ aren't coprime.


答:
Let's use a coordinate system where one of the spheres is at origin.  The other travels along a straight line in direction $\vec{p}_\Delta = (x_\Delta, y_\Delta, z_\Delta)$, at $\vec{p}_0 = (x_0, y_0, z_0)$ at time $t = 0$, and at $\vec{p}_0 + t \vec{p}_\Delta = (x_0 + t x_\Delta, y_0 + t y_\Delta, z_0 + t z_\Delta)$ at time $t$.  Let $R$ be the sum of the two spheres radiuses, i.e. the distance of the two spheres centerpoints we are interested in.  At the moment when the two spheres touch, we have
$$\begin{aligned}
\lVert \vec{p}_0 + t \vec{p}_\Delta \rVert &= R \\
\iff \quad \lVert \vec{p}_0 + t \vec{p}_\Delta \rVert^2 &= R^2 \\
\iff \quad (x_0 + t x_\Delta)^2 + (y_0 + t y_\Delta)^2 + (z_0 + t z_\Delta)^2 &= R^2 \\
\iff \quad (x_\Delta^2 + y_\Delta^2 + z_\Delta^2) t^2 + 2 (x_0 x_\Delta + y_0 y_\Delta + z_0 z_\Delta) t + (x_0^2 + y_0^2 + z_0^2 - R^2) &= 0 \\
\end{aligned}$$
This is a quadratic equation in $t$, and easily solved:
$$\begin{aligned}
t = &- \frac{x_0 x_\Delta + y_0 y_\Delta + z_0 z_\Delta}{x_\Delta^2 + y_\Delta^2 + z_\Delta^2} \\
~ &\pm \frac{\sqrt{R^2 ( x_\Delta^2 + y_\Delta^2 + z_\Delta^2) - (x_0 y_\Delta - y_0 x_\Delta)^2 - (x_0 z_\Delta - z_0 x_\Delta)^2 - (y_0 z_\Delta - z_0 y_\Delta)^2}}{x_\Delta^2 + y_\Delta^2 + z_\Delta^2} \\
\end{aligned}$$
Obviously, only positive $t$ are interesting, since only those correspond to collisions in the future.

The denominator, $\lVert \vec{p}_\Delta \rVert^2 = x_\Delta^2 + y_\Delta^2 + z_\Delta^2$, is only zero when the two spheres are stationary with respect to each other (i.e., travelling in the same direction with the same velocity).

The expression within the square root is nonnegative only if the trajectories are such that the two spheres do touch/intersect at some point.

If you have two spheres traveling along straight lines, $\vec{a}_0 + t \vec{a}_\Delta$ and $\vec{b}_0 + t \vec{b}_\Delta$, the position of the second sphere ($\vec{b}_0 + t \vec{b}_\Delta$) in a coordinate system where the first sphere is at origin is
$$\left\lbrace ~ \begin{aligned}
\vec{p}_0 &= \vec{b}_0 - \vec{a}_0 \\
\vec{p}_\Delta &= \vec{b}_\Delta - \vec{a}_\Delta \\
\end{aligned} \right.$$
Calculating the expression in the square root above,
$$D = R^2 ( x_\Delta^2 + y_\Delta^2 + z_\Delta^2) - (x_0 y_\Delta - y_0 x_\Delta)^2 - (x_0 z_\Delta - z_0 x_\Delta)^2 - (y_0 z_\Delta - z_0 y_\Delta)^2$$
is enough to tell you if the two spheres' trajectories may intersect: If $D \lt 0$, there is no intersection.  If $D = 0$, there is one intersection.  If $D \gt 0$, there are two intersections.  You do need to calculate $t$ as above, to find out if the intersection(s) occur in the future, though.


问:
Best bijection between $\mathbb{N}^\mathbb{N}$ and $[0,1] \subset \mathbb{R}$.
Cantor's diagonal argument proves as a special case that the set of functions from $\mathbb{N}$ to $\{0,1\}$ has the same cardinality as the interval $[0,1].$ There is an 'obvious' mapping from functions to real numbers that almost provides a bijection, except very few pairs of functions happen to map to the same reals. This can then be dealt with by using the Cantor-Bernstein-Schröder Theorem to prove the existence of a bijection.
For an application in number theory I would like to apply similar bijections between $\mathbb{N}^\mathbb{N}$ and $[0,1].$ The construction which is 'obvious' is to use dove-tailing, as follows. Let $M$ be a matrix with rows and columns indexed by $\mathbb{N}$ so that if $f$ is a function, then the binary representation of $f(n)$ is contained in the $n$'th row of $M.$ Now let $f$ correspond to the real number which has as its binary digits the entries (0,0),(0,1),(1,1),(0,2),(1,1),(2,0),(0,3) etc. of $M.$ Clearly each function produces a unique real number between $0$ and $1.$ Again it may be necessary to apply the Cantor-Bernstein-Schröder Theorem to eliminate double occurrences of real numbers. In my particular application I do not worry about this, since all my functions map only a finite number of natural numbers to zero.
But my question is whether other bijections are also known, other than this 'obvious' one.

答: Assuming you're talking about $\Bbb{N}$ starting at $1$ rather than $0$, the bijection I like best is $$(n_1, n_2, n_3, ...) \mapsto \frac{1}{2^{n_1}} + \frac{1}{2^{n_1 + n_2}} + \frac{1}{2^{n_1 + n_2 + n_3}} + ...$$ The first number of the sequence tells me how many digits after the binary point to count to see the first $1$ in the binary expansion of $x$, and every number after that counts the number of binary digits to the next $1$. Although $0$ is not in the image of this map as a result, it is a true bijection of $\Bbb{N}^{\Bbb{N}}$ with $(0, 1]$--nothing gets double counted because this function can never output an infinite tail of zeroes.


答:
You have fixed the homomorphism proof.

Continuing from there:

α((a,b))=e

Whenever 3b = e

This is true for b=0,9,18.

So $ker(\alpha) = Z_{9}\times (0,9,18)$

Body:
Let  $\alpha: \mathbb{Z}_9 \times \mathbb{Z}_{27} \rightarrow \mathbb{Z}_{27}$ be given by $\alpha((a, b)) = 3b$, for all $a\in \mathbb{Z}_9,b\in \mathbb{Z}_{27}$.
Show that $\alpha$ is a homomorphism. Also, find $\text{ker}(\alpha)$, and decide if $\alpha$ is onto.

My attempt: Let $(a_1,b_1), (a_2,b_2)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}$.

Then $$\alpha((a_1,b_1)+(a_2,b_2))=\alpha(a_1+a_2,b_1+b_2)=3(b_1+b_2)=3b_1+3b_2 =\alpha((a_1, b_1))+\alpha((a_2, b_2))$$

thus, $\alpha$ is a homomorphism.

$$\text{ker}(\alpha)=\{(a,b)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}:\alpha((a,b))=0\}=\mathbb{Z}_9 \times\{0\}$$ (In particular $\alpha$ is not one-to-one).

My question is in determining if $\alpha$ is onto.

If $b\in \mathbb{Z}_{27}$, then $\alpha\left ( \left (a,\frac{1}{3}b  \right ) \right )=b$

Does this argument help me to say that $\alpha$ is onto?


问:
How to choose ultraproducts to prove that the class of finite ($p$, torsion) groups is not elementary?
I want to use Łoś's theorem to show that finite groups, $p$-groups, and torsion groups do not form elementary classes. Thus, I have to construct the ultraproduct, say, of some finite groups that is not finite. Can you please tell me what groups and ultrafilters I have to choose? Can I restrict myself with ultrafilters over $\omega$? Shall I use principal, or non-principal ultrafilters?

答: For each $n\in \omega$, let $G_n = \mathbb{Z}/p^n\mathbb{Z}$. Note that each $G_n$ is a finite group, a torsion group, and a $p$-group. Let $U$ be any non-principal ultrafilter on $\omega$. Let $G = \prod_U G_n$ be the ultraproduct of the $G_n$ by $U$. The element represented by $(1,1,1,\dots)$ in $G$ has infinite order. This shows that $G$ is infinite, not torsion, and not a $p$-group.


问:
When does $V=L$ becomes inconsistent?
In a wonderful course I'm taking with Magidor we are finishing the proof of the Covering Theorem for $L$.

The theorem, in a nutshell, says that $V$ is very close to being $L$ if and only if $0^\#$ does not exist.

It is consistent that no large cardinals exist and $V\neq L$, and it is consistent that there are inaccessible, Mahlo, weakly-compact, ineffable and so on and yet $V=L$.

On the other hand, if there exists a measurable cardinal then $V\neq L$, as Scott tells us, and in fact a Ramsey cardinal is enough for that. This way we slowly fine tune the demands until we reach the definition of $0^\#$.

My question is, basically, if there is a weaker notion than $0^\#$ in terms of consistency of large cardinal axioms, which is inconsistent with $V=L$?

I am aware of the theorem that a supercompact implies $\forall A(V\neq L[A])$. Are there any other ways to measure how far are we from $L$ or $HOD$ or even $HOD[A]$ for some or all $A$?

答:
The borderline seems to be very near the $\omega_1$-Erdős cardinals.

The Wikipedia page on $0^\sharp$ explains it thus:

The existence of $\omega_1$-Erdős cardinals implies the existence of $0^\sharp$. This is close to being best possible, because the existence of $0^\sharp$ implies that in the constructible universe there is an $\alpha$-Erdős cardinal for all countable $\alpha$, so such cardinals cannot be used to prove the existence of $0^\sharp$.

In particular, an $\omega_1$-Erdős cardinal is inconsistent with $V=L$, but having $\alpha$-Erdős cardinals for countable $\alpha$ is fine with $V=L$. All the smaller large cardinals in the hierarchy (see the Wikipedia large cardinal list) also appear to relativize down to $L$.

Title: When does $V=L$ becomes inconsistent?


问:
Questions about $\mathrm{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$

Consider the extension $\mathbb{Q} \subset\mathbb{Q} (\sqrt{2}, \sqrt{3})$. How many elements are there in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})?$ Describe all elements in $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$. Find all subgroups of $\text{Aut}(\mathbb{Q} (\sqrt{2}, \sqrt{3})/\mathbb{Q})$ and their fixed fields.

I'm using Dummit & Foote Chapter 14 (Galois Theory) if it needs to be referenced. Chapter 13 was a breeze for me so I just feel like I need someone to explain how easy of a connection this is and it'll just click (or maybe it's not and I need to be told that, too). I really just need direction.

答: Hint: An automorphism fixes the integers, indeed the rationals. So it has to carry $\sqrt{2}$ to $\pm\sqrt{2}$, and $\sqrt{3}$ to $\pm\sqrt{3}$. There are not many candidates.


问:
Chain of elementary submodels of $L_{\omega_2}$
I have a chain of elementary submodels defined inductively in the following way.

Pick $N_0\prec L_{\omega_2}$ s.t $|N_0|=\aleph_0$ and $N_0$ is minimal in $\leq_L$. For every $\alpha<\omega_1$ let $N_{\alpha+1}\prec L_{\omega_2}$ be the minimal (with respect to $\leq_L$) elementary submodel s.t $N_\alpha \in N_{\alpha+1}$ and $|N_{\alpha+1}|=\aleph_0$. For $\alpha<\omega_1$ a limit let $N_\alpha = \bigcup_{\beta<\alpha}N_\beta$.

I want to show $N_\alpha \subseteq N_{\alpha+1}$.

In a previous discussion done in this question I asked I understood the argument should be something similar to

Let $\alpha<\omega_1$ be some ordinal, from construction $|N_\alpha|=\aleph_0$ hence $\exists f:\omega \rightarrow N_\alpha$ a bijection using the $L$ global well order

$$L_{\omega_2}\models f \text{ is the} \leq_L \text{minimal bijection from } \omega \text{ to } N_\alpha$$

$N_{\alpha+1} \prec L_{\omega_2}$ and $N_\alpha,\omega\in N_{\alpha+1}$ thus

$$N_{\alpha+1}\models f \text{ is the} \leq_L \text{minimal bijection from } \omega \text{ to } N_\alpha$$

Hence $\forall n<\omega,\ f(n)\in N_{\alpha+1}$, $f$ is bijective giving $N_\alpha \subseteq N_{\alpha+1}$.

Doesn't the transition to $N_{\alpha+1}$ affects the chosen function? Is it because $L_{\omega_2}$ thinks $N_\alpha$ is $\aleph_0$ so $N_{\alpha+1}$ thinks it too? Or $f$ is described uniquely using a single ordinal via $\leq_L$?

答:
This is really an instance of a more general fact.

Suppose $\kappa$ is uncountable and regular and $M\prec H_\kappa$ is an
elementary submodel. If $p\in M$ and $p$ is countable, then
$p\subseteq M$.

To start, notice that for uncountable regular $\kappa$, $H_\kappa$ gets countability facts correctly. That is, if $p$ is countable, then $H_\kappa\vDash \exists f:\omega\to p \text{ surjective}$. Since $\omega, p\in M$, this sentence is true in $M$ as well. We now show that for any $q\in p$, we have also $q\in M$.

First fix $f_p\in M$ such that $M\vDash f_p:\omega\to p \text{ is surjective}$. Since $M$ thinks $f_p$ is a function, by elementary $H_\kappa$ thinks $f_p$ is a function, so $f_p$ is really a function. Now let $q\in p$ be arbitrary. We want to show that $q$ is an element of $M$. In $V$, we know that there is some $n$ such that $f_p(n)=q$. Also, for each $n\in\omega$, $n\in M$ by definability of the natural numbers. So we can ask for what $M$ thinks is $f_p(n)$. I claim that this object is $q$.

To see this, notice that $H_\kappa\vDash (\exists! x)(x\in p\wedge x=f_p(n))$. By elementarity, we have $M \vDash (\exists! x)(x\in p\wedge x=f_p(n))$. Now if $M$ is transitive then we are done: the unique $x$ is just $q$. But $M$ might not be transitive. We work around this obstacle by appealing to elementarity.

Fix the element $z\in M$ such that $M \vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. Passing this up to $H_\kappa$ by elementarity, we have $H_\kappa\vDash z \text{ is the unique element in } p \text{ such that } z=f_p(n))$. But then $H_\kappa$ also thinks that $q$ is the unique element satisfying the description. Hence $H_\kappa\vDash q=z$, and so $q$ and $z$ are really identical. This proves $q\in M$.

As to the question: notice that we have in general $(L_\kappa=H_\kappa)^L$. This means the general fact applies in this case, if we apply it in $L$. So just take $N_{\alpha+1}$ to be $M$ and take $p$ to be $N_\alpha$ and apply the fact.


问:
Reliability of an $l$ pod $k$ of $n$ system.
We have $n$ machines that form a system. The system works if $k$ or more of the $n$ machines are running. The machines are hosted in pods. They are evenly distributed across $l$ of the pods (best effort). For example, if there are $3$ pods and $7$ machines, one pod will have $3$ machines and the other two will have $2$ each.

If a pod is down, all machines hosted on it will go down as well. The probability that any machine is running at a given time is $p$ and the probability that any of the pods will be running is $q$. The machines and pods function independently of each other.

What is the probability that the system will be running at any given time (or that $k$ or more of the machines are functional)? Let's call this $P_{l,n,k}$.

I have a solution that involves simply listing out all the possibilities but it's highly inefficient for large $l$ and not very pretty either.

I also tried creating a recurrence relation for $P_{l,n,k}$ but haven't gotten very far.

One observation: some pods will have $\left[\frac{n}{l}\right]+1$ machines (where $[x]$ is the greatest integer less than or equal to $x$) and some will have $\left[\frac{n}{l}\right]$. Let's call the former "heroes" and the latter "joes". Perhaps doing some combinatorics over heroes and joes will provide an elegant, efficient solution?

Some special cases:

If $l\geq n$, it becomes a $k$ of $n$ system with reliability: $pq$. The probability in this case becomes:

$$\sum_{j=k}^{n} {n \choose k}(pq)^j{(1-pq)^{n-j}}$$

If $l=1$, we need the pod to be up. And beyond that, it's a $k$ of $n$ system with component reliability, $p$. So the probability in this case becomes:

$$q\sum_{j=k}^{n} {n \choose k}(p)^j{(1-p)^{n-j}}$$

答:
I came up with an efficient solution. If the machines are distributed among $l$ pods as evenly as possible, the pods will be divided into two sets. The pods in the first set will each have one more machine assigned to them than the ones in the second set. Let's call the first set of pods "hero pods" and the second set, "joe pods". It's easy to see that the number of machines assigned to each hero pod will be:

$$h = n-l \left\lfloor \frac{n}{l} \right\rfloor$$
and the number of machines assigned to each joe pod will be:
$$j = l \left\lceil \frac{n}{l} \right\rceil-n$$

Next, we can simply condition on all possibilities of the number of hero and joe pods that are working. Within each of those conditions, we're left with a simple k-of-n system. So, the probability that the overall system is UP becomes:

$$P = \sum\limits_{\eta=0}^{h}\sum\limits_{\nu=0}^{l} P(H=\eta)P(J=\nu) P(X_{\eta \left\lceil \frac{n}{l} \right\rceil + \nu \left\lfloor \frac{n}{l} \right\rfloor,p}\geq k)$$

Here, $X_{n,p}$ is a Binomial random variable with parameters $n$ and $p$, $H$ is the random variable representing the number of hero pods that work (binomial with parameters $h$ and $q$) and $J$ represents the number of joe pods (binomial with parameters $j$ and $q$).


问:
3 Toast Problem From "Thinking Mathematically": Solution Check
I am working through "Thinking Mathematically" by Mason, Burton and Stacey. One of the questions goes as follows:

Three slices of bread are to be toasted under a grill. The grill can hold two slices at once but only one side is toasted at a time. It takes 30 seconds to toast one side of a piece of bread, 5 seconds to put a piece in or take a piece out and 3 seconds to turn a piece over. What is the shortest time in which the three slices can be toasted?

I am wondering if I am overlooking something in my solution method, so I have come here. Some sources online say 130 seconds and others say 139. My solution, if non-fallacious, should be able to get it done in 118 seconds. The process goes as follows. Allow $T_n$ to represent the first side of the $n$th piece of toast and $T_n'$ to represent the opposite side of the $n$th piece of toast. Then, assuming a piece of toast is cooking as soon as it is placed, then the suggested sequence of events

Sorry for the convoluted timeline in the picture, but essentially I place the first piece of toast and then the second. $T_1$ and $T_2$'s cooking time will overlap, but there is necessarily a five-second window where they do not. You flip $T_1 \to T_1'$ where $T_1'$ officially begins toasting at 38 seconds. This gives another thirty-second window (until $t = 68$) to get a lot of the time-wasting flipping and removing done. For example, $T_2$ finished at $t=40$ and the most original part of this solution is to remove $T_2$ rather than flip it over and begin cooking the other side immediately. As soon as we are done removing $T_2$ the running total will be $t=45$ at which point we immediately begin placing $T_3$ bringing us to a running total of $t=50$. Note, that $T_3$ will be done at $t = 80$. At this point, we wait until $t=68$ to remove $T_1'$ which brings us to a running total of $t = 73$, at which point we immediately begin placing $T_2'$ which we put aside before. When we are done placing $T_2'$ we will be at $t = 78$ and at $t = 80$ it will be time to flip $T_3$, which brings us to $t = 83$ and which point $T_3'$ will begin toasting. $T_3'$ will thus be done at $t = 113$ and in that time we will have to remove $T_2'$ which began cooking at $t = 78$. Finally, we remove $T_3'$ at $t = 113$ and ending at $t = 118$.

Are there any problems with this solution that I am not detecting? I think it works well, indeed if it works at all, because it uses the long 30-second windows of cooking to do all the time-consuming activities of flipping etc.

答:
That is smart approach but I see 2 mistakes:

You do not add up the 5 seconds to add the Toast 2. You assume that since the "clock" is running once the toast 1 started to grill Toast 2 magically starts to cook at time 10 implying some parallelism there in the actions which does not seem to be justified from the problem.

You remove T2 at time 40 and place T3 at time 45 but you didn't add either the 5  secs to remove T2 or to add T3. I.e. you should be adding 10 hence T3 starts to cool at 55 and not 45. There is no mention of being able to remove slices/add slices at one movement.


问:
Lineer operators continuty problem in normed space
X normed space $T,S:X \to X$ are two operator.If $$ST-TS=I$$ then T or S is not continous prove that

I is unit operator $Iv=v $,$\forall v \in X$ ,
$T.S=ToS$

答:
Let's assume that both $T$ and $S$ are bounded. We need to prove that their commutator cannot be the identity operator.

We will use the following facts:

For a natural number $n$, $[A,B^n]=nB^{n-1}[A,B]$

The operator norm is subadditive: $\lVert A + B \rVert \leqslant \lVert A \rVert + \lVert B \lVert$

The operator norm is submultiplicative: $\lVert AB \rVert \leqslant \lVert A \rVert\lVert B \lVert$

Let's assume that $[S,T]=I$. Then we must have that
$$[S, T^n]=nT^{n-1}[S,T]=nT^{n-1}$$
But:
\begin{align}
\lVert [S, T^n] \rVert
&= \lVert ST^n-T^nS\rVert \\
&\leqslant \lVert ST^n \rVert + \lVert T^n S \rVert \\
&= \lVert S T T^{n-1}\rVert + \lVert T^{n-1} T S \rVert \\
&\leqslant 2\lVert S \rVert \lVert T \rVert \lVert T^{n-1}\rVert
\end{align}
Taking the norm of the commutation relation would imply that
$$\lVert [S, T^n] \rVert = n \lVert T^{n-1}\rVert$$
So we must have:
$$n\lVert T^{n-1} \rVert \leqslant 2\lVert S \rVert \lVert T \rVert \lVert T^{n-1}\rVert$$
Which is absurd!

Note: This works only if $\lVert T^{n-1} \rVert \neq 0$ for all $n$. But what can we do if it does not hold?

Connection to physics: This proves that the momentum and position operators cannot be bounded at the same time, as they need to satisfy the canonical commutation relation $[x, p]=i\hbar$.


问:
How to express $2 \cos X = \sin X$ in terms of $\sin X$?
The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

答:
Of course, since $\sin^2(x)+\cos^2(x)=1$, if $2\cos(x)=\sin(x)$, squaring and adding $4\sin^2(x)$ to both sides yields
$$
4=5\sin^2(x)\tag{1}
$$
Of course, $(1)$ also has solutions where $2\cos(x)=-\sin(x)$. Knowledge of
$\sin(x)$ fully determines $|\cos(x)|$, but says nothing about the sign of $\cos(x)$. For this reason, the equation $2\cos(x)=\sin(x)$ cannot be fully characterized in terms of $\sin(x)$ alone.

On the other hand, $2\cos(x)=\sin(x)$ is equivalent to $\tan(x)=2$.


问:
Relation between $G$ bundle over free $G$ space $X$ and vector bundles over $X/G$ (Atiyah)
In Atiyah K theory there is a result : If $X$ is  $G$-free  $G$-vector bundles over $X$ corresponds bijectively to vector bundles over  $X/G$ by $E \to E/G$.

I need some clarity in the proof :

Let $p: E \to X$ be a $G$-bundle. Then  $E$ is also $G$-free. We need to prove that the orbit space $E/G$ forms a vector space over $X/G$. The mapping $p': E/G \to X/G$ is $Ge \mapsto Gp(e)$. This map is continuous.

The fiber of $E/G$ is  $(E/G)_{Gx}= \{ Ge \in E/G : Gp(e)=Gx \}$. If we are able to prove that this set is "same" as (I think I can't  use the term  'isomorphic') $E_{x}/G$ then there will be a vector space structure on the fiber $(E/G)_{Gx}$ induced from $E_{x}/G$. Are we supposed to use the fact that $E$ is free $G$ space ? then how ?

Author says that $p': E/G \to X/G$ is locally isomorphic to $p :E \to X$. Not sure about the following  proof :

Let $U$ be an open subset of $X$. Then $\pi(U)$ is an open subset of $X/G$ as $\pi : X \to X/G $, the natural map is the quotient map. Suppose $p^{-1}(U) $  is isomorphic to $U \times V$, then $ p'^{-1}(\pi(U)) $ is isomorphic to $\pi(U) \times V$. That is how local triviality of $E$ gives the local triviality of $E/G$. Can any one suggest better methods ?

The converse part is clear to me.

答:
For your first question, the fiber $(E/G)_{Gx}$ is in bijection with $E_x$ (under the quotient map) and inherits the vector space structure from $E_x$. This choice is not unique, but if we choose to instead give $(E/G)_{Gx}$ the vector space structure from $gx$, then the identity on $(E/G)_{Gx}$ descends from the vector space isomorphism $g:E_x\rightarrow E_{gx}$ (which comes from the definition of a $G$-bundle over $X$), and therefore identifies the two vector space structures on $(E/G)_{Gx}$. This gets rid of any ambiguity in the definition of the vector space structure on $(E/G)_{Gx}$.

For your second question, you are right in your general approach to trivialization. But we need to be careful about the $U$ we pick (and use Atiyah's assumptions that $X$ is Hausdorff and $G$ is finite). Fix an arbitrary $x\in X$. By freeness of the group action, for each $g\in G$ not the identity, we have $x\neq gx$ and thus we can find neighborhoods $U_g\ni x$ and $V_g\ni gx$ with $U_g\cap V_g=\emptyset$ (using the Hausdorff-ness of $X$). Also let $U_e=V_e$ be a neighborhood of $x$ over which $E$ is trivializable (where $e\in G$ is the identity). Now, using the finiteness of $G$, define the open set $$W=\bigcap_{g\in G}(U_g\cap g^{-1}V_g).$$ Since $x\in U_g$ and $gx\in V_g$ for every $g\in G$, this $W$ is a neighborhood of $x$. Also $W\subset U_e$ and thus $E$ is trivializable over $W$.  Finally, notice that $$W\cap gW\subset U_g\cap V_g=\emptyset$$ for every $g\in G$ not the identity. This implies that $\pi:W\rightarrow\pi(W)$ is a homeomorphism, which enables the identification of the open set $W$ with the set it covers in the quotient. The analogous identification in the total space gives you the desired trivialization: $$p'^{-1}\big(\pi(W)\big)\cong p^{-1}(W)\cong W\times \mathbb C^n.$$ (Of course, there's nothing special about $\mathbb C$ here, I'm just writing it for concreteness. The bundle could also be real, quaternionic, or various other things.)


问:
Generating function for a combinatorics problem
I have this combinatorics problem:

How many n strings are there of letters of english alphabet in which there are no consecutive z's?

I want to solve this problem using generating functions

Generating functions of single letters other than z is obvious but i can't find the generating function for z. I know that its function will be a polynomial with degree of n/2 or (n+1)/2 depending on whether n is even or odd but i can't find the coefficients of the function. Can anyone lead me to a solution?

Thanks in advance.

答:
Or,
we can build a good word by adding, step by step, a new component

$sA + sB + \cdots + sY + sZ + s^2.c.ZZ + s^3.c^2.ZZZ + \cdots \ $   where

$c$ is a counter for the at least number of occurrences of a pair ZZ in the word.

$s$ is a counter for the length of the added component.

overall the strings are depicted by

$$1 \over 1 - (sA + sB + \cdots + sY + sZ + s^2.c.ZZ + s^3.c^2.ZZZ + \cdots )\ $$

we are interested in words that have exactly $ \ 0 \ $ occurences of ZZ, so we have to replace $ \ c \ $ by $ \ c-1 \ $ for the PIE reason, than take $ \ c=0 \ $.

the generating expression becomes $${1 \over 1 -(  sA + sB + \cdots + sY + {sZ \over 1+sZ}) }={ 1+sZ \over 1 -(  sA + sB + \cdots + sY )(1 + sZ)}$$

when we are interested only in length ( the sort of letter does not matter)  we get the generating function

$$  \frac {1+s} {1 - 25s.(1 + s)} $$


问:
Choosing with replacement with minimal allocation
If I have $r$ items and I am making $n$ distinct choices with replacement, then there are $r^{n}$ possible selections. However, how could I count with the added constraint that each of the $r$ items must be selected at least once?

This problem implies that I want to count all $r^n$ possibilities, but I want to exclude cases where any of the $r$ items were selected zero times. That is, if I have items $(r_1,r_2,r_3)$ and choices $(n_1,n_2,n_3,n_4,n_5)$, the following would be a VALID selection because each $i$ was chosen at least once: $\{r_1:\{n_1,n_2,n_3\}, r_2:\{n_4\}, r_3:\{n_5\}\}$. However, the following would be INVALID to include: $\{r_1:\{n_1,n_2,n_3\}, r_2:\{n_4,n_5\}, r_3:\{\}\}$ because $r_3$ was not chosen.

I'd like to use this as a step in a derivation for another question.

答:
Since each item must be selected at least once the possible counts can be expressed using Stirling number of second kind:
$$
r!{n \brace r}=\sum_{i=0}^r (-1)^i\binom ri (r-i)^n.
$$
The right hand side expression can be obtained using inclusion-exclusion principle.


问:
Union set theory question.
Is it true that
$$
A \subseteq B, C \subseteq D \implies A\cup C \subseteq B \cup D.
$$
If not would it change anything if $A \cap C, B\cap D = \emptyset$?

答: Yes it is true. To see why note that $A \subseteq A\cup X$ for any set. So $A\subseteq A\cup D\subseteq B\cup D$ and $C\subseteq C\cup D\subseteq B\cup D$. Morevoer, $A\subseteq X$ and $B\subseteq X$ implies that $A\cup B\subseteq X$. Taking all together we arrive at your statement (there is no need for the sets to be disjoint)

Title: Union set theory question.


答:
It is true, here's a proof that uses the rule $\varphi \to \psi \iff \lnot\psi \to \lnot\varphi$, called contraposition. $\lnot\psi \to \lnot\varphi$ is called the contrapositive.

Suppose by way of contrapositive $A \cup C \not\subset B \cup D$.

That means

$$ \exists x \in A \cup C \mathop. x \not\in B \land x \not\in D $$

And therefore

$$ \exists x \in A \mathop. x \not\in B \land x \not\in D \;\;\text{or}\;\; \exists x \in C \mathop. x \not\in B \land x \not\in D $$

And therefore

$$ A \not\subset B \;\;\text{or}\;\; C \not\subset D $$

Applying the contrapositive, we get the following as desired.

$$ \text{If $A \subset B$ and $C \subset D$, then $A \cup C \subset B \cup D$} $$

Title: Union set theory question.

Body:
Is it true that
$$
A \subseteq B, C \subseteq D \implies A\cup C \subseteq B \cup D.
$$
If not would it change anything if $A \cap C, B\cap D = \emptyset$?