问:
choosing the correct password on the $n-$th try
I'm current working on the following probability problem:

A new secretary has been given $n$ computer passwords, only one of which will permit access
to a computer file. Because the secretary has no idea which password is correct, he chooses
one of the passwords at random and tries it. If the password is incorrect, he discards it and
randomly selects another password from among those remaining, proceeding in this manner
until he finds the correct password.

and

A security system has been set up so that if three incorrect passwords are tried before
the correct one, the computer file is locked and access to it denied. If $n = 7$, what is the
probability that the secretary will gain access to the file?

One can work this out rather easily via intuition.
$$
\begin{align}
    \text{P(Correct password on first try)}=&\ \frac{1}{n}\\
    \text{P(Correct password on second try)}=&\ \text{P(wrong first try)}\cdot\text{P(correct second try)}\\
    =&\ \frac{n-1}{n}\cdot\frac{1}{n-1}\\
    =&\ \frac{1}{n}\\
    \text{P(Correct password on third try)}=&\ \text{P(wrong first try)}\cdot\text{P(wrong second try)}\cdot\text{P(correct third try)}\\
    =&\ \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdot\frac{1}{n-2}\\
    =&\ \frac{1}{n}
\end{align}
$$
Therefore the probability of getting access is $3\cdot\frac{1}{7}=\frac{3}{7}$.

My question is how can I formulate this rigorously? Particularly the conditional probability of the $i$-th attempt? Here's my attempt, but I find it rather clunky.

Denote $E$ as the event where the secretary gains access, $A_n$ as the event where the previous $(n-1)$ passwords picked were incorrect, and the $n-$th password was correct, and $k$ as the maximum attempts allowed. Thus
$$E=\bigcup^k_{i=1}A_i$$
Since each $A_{i}$ and $A_j$ is pairwise mutually exclusive,
$$P(E)=\sum^k_{i=1}P(A_i)$$
by one of the Kolmogorov axioms. Furtheremore, defining $T_i$ as the event where the $i$-th password picked is wrong, then

$$A_i=T_1\cap T_2\cap \ldots \cap T_{i-1} \cap \overline{T_i}$$
and for each $T$,
$$P(T_i)= \frac{n-i}{n -i+1}$$
and
$$P(\overline{T_i})=\frac{1}{n-i+1}$$
since each incorrect password is tossed out after the attempt. Finally,

$$A_i=\left(\frac{1}{n-i+1}\right)\cdot\prod^i_{j=1}\frac{n-i}{n -i+1}=\frac{1}{n}$$
which implies that
$$P(E)=\sum^k_{i=1}\frac{1}{n}=\frac{k}{n}$$

Relevant: Probability of guessing correct password (out of n) on k-th try when trying them at random

答:
Assume $n$ passwords, and that security lock-out after $k$ failures: $k \in \{1,2,3,\cdots, (n-1)\}.$

Assume sampling done without replacement.

Line up the $n$ passwords in a row, in the order that they will be tried, and label them $P_1, P_2, \cdots, P_n$.  Access attempt succeeds if and only if the correct password in $\{P_1, \cdots, P_k\}.$

Thus, probability is $\frac{k}{n}.$


问:
Conditional expectation of a random walk process with noise
For each $t \in \mathbb{N}$, define a process $e_t$ by $e_t = z_t + v_t$, where $v_t$ follows normal distribution $N(0, \sigma_v^2)$ independently and $z_t$ is a random walk with normal error:
$$
z_t = z_{t-1} + u_t = \sum_{s=1}^t u_s, \quad u_s \sim N(0,\sigma_u^2).
$$
$z_t$ and $v_t$ are independent.

I want to derive the conditional expectation of future $e$ at time $t$, i.e.,  $E[e_{t+k} \mid e_t, e_{t-1},\dots], k>1$.

Here is what I understand thus far:
The relevant moments are
$$
E[e_t] = 0, Var(e_t) = t\sigma_u^2 + \sigma_v^2., Cov(e_{t+k},e_t) = t \sigma_u^2
$$

For example, for a simple case such as $E[e_2 \mid e_1]$, from the result concerning the conditional expectation of normal variables, I can derive that
$$
E[e_2 \mid e_1] = \frac{Cov(e_1,e_2 )}{Var(e_1)} e_1 = \frac{\sigma_u^2}{\sigma_u^2 + \sigma_v^2} e_1.
$$
However, I am not sure how to tractably compute $E[e_{t+k} \mid e_{t}, e_{t-1}, \dots]$ (I know the formula for conditional expectation of $n$ normal variables).

答:
\begin{align}
\mathsf{E}[e_{t+k}\mid e_t,e_{t-1},\ldots,e_1]&=\mathsf{E}\!\!\left[z_t+\sum_{i=t+1}^{t+k}u_t+v_{t+k}\mid e_t,e_{t-1},\ldots,e_1\right] \\
&=\mathsf{E}[z_t\mid e_t,e_{t-1},\ldots,e_1]
\end{align}
Since $z_t, e_t,\ldots,e_1$ are jointly normal, the last conditional expectation is the linear projection of $z_t$ on $\mathbf{e}_t:=[e_t,\ldots,e_1]^{\top}$, that is,
$$
\mathsf{E}[z_t\mid e_t,e_{t-1},\ldots,e_1]=\mathbf{e}_t^{\top}\beta,
$$
where
$$
\beta=\left(\mathsf{E}[\mathbf{e}_t\mathbf{e}_t^{\top}]\right)^{-1}\mathsf{E}[\mathbf{e}_tz_t].
$$
Letting $\mathbf{z}_t:=[z_t,\ldots,z_1]^{\top}$ and $\mathbf{v}_t:=[v_t,\ldots,v_1]^{\top}$
\begin{align}
\mathsf{E}[\mathbf{e}_t\mathbf{e}_t^{\top}]&=\mathsf{E}[(\mathbf{z}_t+\mathbf{v}_t)(\mathbf{z}_t+\mathbf{v}_t)^{\top}]=\mathsf{E}[\mathbf{z}_t\mathbf{z}_t^{\top}]+\mathsf{E}[\mathbf{v}_t\mathbf{v}_t^{\top}] \\
&=\sigma_u^2U_tU_t^{\top}+\sigma_v^2I_t,
\end{align}
where $U_t$ is a $t\times t$ upper triangular matrix whose nonzero entries are ones, and
$$
\mathsf{E}[\mathbf{e}_tz_t]=\mathsf{E}[\mathbf{z}_tz_t]=\sigma_u^2U_t\mathbf{i}_t,
$$
where $\mathbf{i}_t$ is a vector of ones.

Combining these results, one gets
$$
\mathsf{E}[e_{t+k}\mid \mathbf{e}_t]=\mathbf{e}_t^{\top}(U_tU_t^{\top}+I_t\sigma_v^2/\sigma_u^2)^{-1}U_t\mathbf{i}_t.
$$

ContentLicense: CC BY-SA 4.0


问:
How to prove the following evaluation map is surjective?
Let $P_{1}, P_{2},\cdots, P_{k}\in \Bbb{C}^n$. Now define a homomorphism
$\phi: \Bbb{C}[x_{1},x_{2},\cdots,x_{n}]\rightarrow \Bbb{C}^k$ which sends $f$ to $(f(P_{1}),f(P_{2}),\cdots,f(P_{k})).$ Then how to show that $\phi$ is surjective?

I saw it here, but don't know how to prove it.

答:
Fix $P_i$,  for every $j\neq i$, you can always find a polynomial $Q_j$ which vanishes on $P_j$ and not on $P_i$ (take an hyperplane passing through $P_j$ and not through $P_i$).

Then $R_i:=\prod_{\j\neq i}Q_j$ is a polynomial which vanishes on all of the $P_j$ but not on $P_i$.

The restriction of $\phi$ to the vector space generated by the $R_i$ is surjective.


问:
Construction of the SVD explanation
I am reading the book Mathematics for Machine Learning.

Section 4.5.2 is titled "Construction of the SVD" and it explains how the SVD of a rectangular matrix $A \in \mathbb{R}^{mxn}$ is constructed.

On pages 123-124, the left and right singular vectors are computed through the exploitation of the symmetric positive semidefinite matrices $A^TA$ and $AA^T$. At the same time, the singular values are constructed too within the same procedure.

In the subsequent part, the authors define the following procedure:

The last step is to link up all the parts we touched upon so far. We have an orthonormal set of right-singular vectors in $V$. To finish the construction of the SVD, we connect them with the orthonormal vectors $U$. To reach this goal, we use the fact the images of the $v_i$ under $A$ have to be orthogonal, too. We can show this by using the results from Section 3.4. We require that the inner product between $Av_i$ and $Av_j$ must be $0$ for $i \neq j$. For any two orthogonal eigenvectors $v_i$, $v_j$, $i \neq j$, it holds that
$$(Av_i)^T(Av_j) = v_i^T (A^TA) v_j = v_i(\lambda_j v_j) = \lambda_j v_i v_j= 0 \quad(4.77)$$
For the  case $m \geq r$, it holds that $\{Av_1,...,Av_r\}$ is a basis of an $r$-dimensional subspace of $\mathbb R^m$.
To complete the SVD construction, we need left-singular vectors that are orthonormal: We normalize the images of the right-singular vectors $Av_i$ and obtain
$$ u_i := \frac{Av_i}{‖Av_i‖} = \frac{1}{\sqrt{\lambda_i}} Av_i = \frac{1}{\sigma_i}Av_i, \quad (4.78)$$
where the last equality was obtained from (4.75) and (4.76b), showing us that the eigenvalues of $AA^T$ are such that $σ^2_i = \lambda_i$.

Section 3.4 indicated in the previous excerpt defines orthogonality by means of the inner product and orthogonal matrices. Equations (4.75) and (4.76b) link together the eigenvalues of the symmetric matrices $A^TA$ and $AA^T$ with the singular values of $A$.

What is not clear to me is how this part effectively connects left and right singular vectors. How the results from 4.77 and 4.78 are used to define the final left and right singular vectors?

答:
You know that $$A^TAv_j=\sigma^2_jv_j$$
Multiplying by A $$AA^T(Av_j)=\sigma^2_j(Av_j)$$
Which says that $$Av_j \: is \:an \:eigenvector \:of \:AA^T$$
$$Also\: ||Av_j||^2=v^T_jA^TAv_j=\sigma^2_jv^T_jv_j$$
$$So\: if \:Av_j's\: are \:normalized \:then \:||Av_j||^2=\sigma^2_j$$
$$So \:\frac{Av_j}{\sigma_j}=u_j$$
$$Doing\: this\: for \:all \:u_j's \:gives\: AV=U\Sigma$$
Or $$A=U(\Sigma)V^T$$

Title: Construction of the SVD explanation


问:
Cartesian products involving empty sets
Let $(\Omega_i)_{i\in I}$ be a nonempty family of sets. For each $i \in I$, let $\mathcal{E}_i$ be a set of subsets of $\Omega_i$. Let

$$ \coprod_{i\in I} \mathcal{E}_i:=\bigg\{\prod_{i\in I}A_i : A_i\in\mathcal{E}_i \cup\{\Omega_i\} ,
 A_i\neq \Omega_i \text{ for finitely many }i\in I \bigg\}$$

Let $(I)_{\lambda \in \Lambda}$ be a partition of $I$ into disjoint nonempty sets. I am trying to prove that

$$\coprod_{i\in I} \mathcal{E}_i= \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$$

when identifying sets of the form $\prod_{i\in I} A_i$ $(A_i\subset\Omega_i)$ with $\prod_{\lambda \in \Lambda} \big( \prod_{i\in I_\lambda} A_i\big)$. I already showed $\coprod_{i\in I} \mathcal{E}_i \subset \coprod_{\lambda \in \Lambda} \bigg( \coprod_{i\in I_\lambda} \mathcal{E}_i\bigg)$, but I am not sure the reverse inclusion holds if $\Omega_i =\emptyset$ for some $i$.

Is this reasoning correct? Thanks a lot for your help.

答:
Your assumption, condition, prerequisite that:

Let $\Omega_{i_{i \epsilon I}}$  be a nonempty family of sets.

implies, garanties that

Ω=∅ for some .

is not possible. There exists no such Ω=∅ for all ∈ that Ω=∅!

$$\not \exists \Omega_{i_{i \epsilon I}} : \Omega_{i}=∅$$

or

$$\forall \Omega_{i_{i \epsilon I}} : \Omega_{i}\not = ∅$$

Use this argument and You are done. The condition is more than that I in nonempty. This website provides some inside into the meaings of the condition:
MathWorld Family. The article contains a passage that explains the details:

Title: Cartesian products involving empty sets


问:
Prove the following estimate for the spherical average
Given the spherical average of f over S(x,r)
$$S_f(x,r) = \frac{1}{\sigma_{n-1}r^{n-1}} \int_{S(x,r)} f(y) d\sigma(y) $$

and the volume average

$$V_f(x,r) = \frac{1}{\omega_n r^n} \int_{B(x,r)} f(y) dy$$

How can I prove that
$$ V_{f^2}(x,r) \leq S_{f^2}(x,r)  $$
$$ \frac{1}{\omega_n r^n} \int_{B(x,r)} f^2(y) dy \leq \frac{1}{\sigma_{n-1}r^{n-1}} \int_{S(x,r)} f^2(y) d\sigma(y) $$

答:
Fix $x \in \mathbb{R}^n$ and let $$ \phi(r) = \frac 1 {n \omega_n r^{n-1}} \int_{\partial B_r(x)} (u(y))^2 d \mathcal{H}^{n-1}_y. $$ Observe that $\phi$ is monotone increasing. The proof of this is very similar to the proof of the Mean Value Formula for Laplace's equation, see Theorem 2 in Chapter 2.2 of Partial Differential Equations by Lawrence C. Evans. Indeed, by letting $y = x+rz$\begin{align*}
\phi (r) &= \frac 1 {n \omega_n } \int_{\partial B_1(0)} (u(x+rz))^2 d \mathcal{H}^{n-1}_z,
\end{align*} so \begin{align*}
\phi '(r) &= \frac 2 {n \omega_n } \int_{\partial B_1(0)} u(x+rz) Du(x+rz) \cdot z d \mathcal{H}^{n-1}_z \\
&= \frac 2 {n \omega_n} \int_{\partial B_r(x)}  u(y) Du(y) \cdot \bigg ( \frac{y-x}{r}\bigg ) d \mathcal{H}^{n-1}_y \\
&=  \frac 2 {n \omega_n} \int_{\partial B_r(x)}  u(y) \frac{\partial u}{\partial \nu }(y) d \mathcal{H}^{n-1}_y \\
&= \frac 2 {n \omega_n} \bigg ( \int_{B_r(x)} u \Delta u d y + \int_{B_r(x)} \vert Du \vert^2 d y \bigg ) \geqslant 0 .
\end{align*} Since $\phi$ is increasing for all $0<s\leqslant r$, \begin{align*}
 \int_{\partial B_s(x)} (u(y))^2 d \mathcal{H}^{n-1}_y \leqslant \frac{s^{n-1}}{r^{n-1}} \int_{\partial B_r(x)} (u(y))^2 d \mathcal{H}^{n-1}_y.
\end{align*}

To prove your inequality use polar coordinates to find \begin{align*}
\frac 1 {\omega_n r^n} \int_{B_r(x)} (u(y))^2 d y &= \frac 1 {\omega_n r^n} \int_0^r \int_{\partial B_s (x)} (u(y))^2 d \mathcal{H}^{n-1} d s \\
&\leqslant \frac 1 {\omega_n r^{2n-1}} \int_0^r s^{n-1} \int_{\partial B_r (x)} (u(y))^2 d \mathcal{H}^{n-1} d s \\
&= \frac 1 {n\omega_n r^{n-1}}  \int_{\partial B_r (x)} (u(y))^2 d \mathcal{H}^{n-1}
\end{align*} as required.


问:
Existence of correlation length in percolation
I saw that there exists a constant $\xi = \xi(p)$, the correlation length, such that for the percolation of parameter $p$ on $\mathbb{Z}^d$, if we note $e_1 = (1,0,...,0)$, we have for any $n\geq 1$

$$\mathbb{P}_p [0\leftrightarrow ne_1] \leq e^{-n/\xi(p)}.$$

Unfortunately, I could not find a proof of this result. I know that we have exponentiel decay of the radius of the cluster below $p_c$, but I would like to prove it without using this stronger result.

答:
Define for any $n\geq 1$ the sequence $u_n := \mathbb{P}_p (0\leftrightarrow ne_1)$. Recall that the FKG inequality says that if $A,B$ are two increasing (or decreasing) events, then $$\mathbb{P}(A\cap B)\geq \mathbb{P} (A) \cdot \mathbb{P}(B).$$

Remark now that the sequence $\left(\dfrac{1}{u_n}\right)$ is super-multiplicative in the sense that $u_{n+m} \geq u_n u_m$.
Indeed, assuming that $n < m$, we have

$$u_n u_m = \mathbb{P}(0\leftrightarrow ne_1) \mathbb{P}(0\leftrightarrow me_1)= \mathbb{P}(0\leftrightarrow ne_1)\mathbb{P}(n\leftrightarrow (n+m)e_1).$$

Now, we can use the FKG inequality to obtain

$$u_n u_m \leq \mathbb{P}\left(\lbrace 0\leftrightarrow ne_1\rbrace \cap \lbrace n\leftrightarrow (n+m)e_1\rbrace\right) \leq \mathbb{P}(0\leftrightarrow (n+m)e_1) = u_{n+m}.$$

Using the Fekete's (also known as subadditive) lemma, we know that $u_n ^{1/n}$ converges to $u:= \sup u_n^{1/n}$. Therefore, writing $u = \exp[-\frac{1}{\xi (p)}]$ we get that for all $n\geq 1$,

$$u_n \leq \exp\left[-\frac{n}{\xi (p)}\right],$$

since $u_n \leq u^n$ for any $n\geq 1$.


问:
Find the coordinate of A and B
Question:

OA: y= 4x

AB: 3y+8x=400

OB: y=0

Find the Coordinates of A and B

I am really not sure how to do this problem.

I've tried using system of equations to see which numbers will have in common between O, A and B. But I was honestly guessing which method could work.
Could someone help me on this problem?

Thanks!

答:
The point $A$ is the intersection of $OA$ and $AB$. To find $A$, we solve the linear equation system
$y = 4x$ and $3y + 8x = 400$. Inserting the first equation into the second one, you get $12x + 8x = 400$, which is $x = 20$. Hence $A$ has the coordinates $(20, 80)$.

Similarly, to find the point $B$, we solve the linear equation system $3y + 8x = 400$ and $y = 0$ and get $x = 50$. Hence $B$ has the coordinates $(50,0)$.

Title: Find the coordinate of A and B


问:
Can you tell which are correct terms of the sum of solution of the integral $\int (x^2+1)^n dx $?
According to WolframAlpha,
$$\int (x^2+1)^n dx = x \cdot _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)$$
and
$$_2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{n=0}^{\infty} \frac{1}{3}(-n)\frac{(-x^2)^n}{n!}.$$

Can you tell which are correct terms of the sum of solution of the integral  $\int (x^2+1)^n dx $?
these
$$(n=0): x, (n=1): 1/3x(x^2+3), (n=2): 1/3 x(x^4+x^2), \dots$$

or (by using $_2F_1(a,b;c;z)= \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$(1)), these which I got $$(n=0): 0, (n=1): 1/3x(x^2), (n=2): 1/3 x(-x^4+x^2), \dots$$
If mine are incorrect, can you compute some terms of the sum (1)?

答:
You are getting the wrong answer. The answer should be

$$ x\,{\mbox{$_2$F$_1$}\left(\frac{1}{2},-n;\,\frac{3}{2};\,-{x}^{2}\right)}. $$

Now, you won't have problems. For $n=0,1,2,3$, we get the corresponding answers

$$ x,\,
\frac{x}{3} \left( {x}^{2}+3 \right),
\, \frac{x}{15} \left( 3\,{x}^{4}+10\,{x}^{2}+15 \right),\, \,
\frac{x}{35} \left( 5\,{x}^{6}+21\,{x}^{4}+35\,{x}^{2}+35 \right). $$

Added: You should use different index for the sum as

$$  _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{k=0}^{\infty} \frac{(1/2)_k(-n)_k}{(3/2)_k}\frac{(-x^2)^k}{k!}.  $$

Now, you need this identity

$$ (a)_m =  \frac{\Gamma(m+a)}{\Gamma(a)}= \frac{(m+a-1)!}{(a-1)!}.  $$

ContentLicense: CC BY-SA 3.0


问:
An orthogonal matrix the given power of which is the identity matrix
What is the general form of a real orthogonal matrix of $n$th order that satisfies the condition $A^p=E$ where $p$ is a given positive integer, and $E$ is the identity matrix?

As I see, in the case of $p=2$ we have $A^2=E$ whence $A^{-1}=A$, then by the definition of orthogonality $A^T=A^{-1}$, whence $A^T=A$, that is $A$ is a symmetric matrix.

And what can one say about the general view of such matrices, for example for $p=5$ or $p=8$? Is it possible to display this general view?

答:
If $p=0$, any orthogonal $A$ will do. If $p\ne0$, the general form of $A$ is given by
$$
A=Q\left(R\left(\frac{2k_1\pi}{p}\right)\oplus R\left(\frac{2k_2\pi}{p}\right)\oplus\cdots\oplus R\left(\frac{2k_m\pi}{p}\right)\oplus I_r\oplus-I_s\right)Q^T
$$
where $Q$ is any real orthogonal matrix, the $k_i$s are any integers, $R(\theta)$ denotes the $2\times2$ rotation matrix for an angle $\theta$, and $m,r,s$ are nonnegative integers such that $2m+r+s=n$. In addition, when $p$ is odd, $s$ must be zero, or else $(-I_s)^p$ will cause a reflection.


问:
Proving Kepler's second law in a different way

Is it possible to find the linear equations for all of the 12 lines stretched from the ellipse's focus to the ellipse's perimeter in the Kepler's second law? If yes, how? (assume we have divided the year to 12 months).

Can Kepler's second law be generalized to all of ellipses (horizontal or vertical) with any given semi-major and semi-minor axes or it is only applicable to planet orbits?

I've tried to get a parametrical answer to the first question by letting one of the lines cross the ellipse focus (A point): $A=(\sqrt{a^2-b^2},0)$ and one point on the ellipse perimeter (M point): $M=(m,\frac{b}{a}\sqrt{a^2-m^2})$. my goal was to calculate $m$. so I found the area between $m$ and $a$ (semi-major axis) plus a triangle area with the base of: $m-\sqrt{a^2-b^2}$ and height of: $\frac{b}{a}\sqrt{a^2-m^2}$ with the help of integral; then I equaled the result to $\frac{\pi{ab}}{12}$. since I had $\arcsin(\frac{m}{a})$ in the result, I substituted $m$ with $a\sin\theta$ and the final result was $\frac{\pi}{3}=\theta+\cos\theta\sqrt{1-(\frac{b}{a})^2}$. then the $\theta$ was $\frac{\pi}{3}-\sqrt{1-(\frac{b}{a})^2}<\theta<\frac{\pi}{3}+\sqrt{1-(\frac{b}{a})^2}$ . now remember that $m=a\sin\theta$. so we have $a\sin(\frac{\pi}{3}-\sqrt{1-(\frac{b}{a})^2})<m<a\sin(\frac{\pi}{3}-\sqrt{1-(\frac{b}{a})^2})$.

Is this a valuable way?

答:
Let $c=\sqrt{a^2-b^2}$, $e=\frac{c}{a}$ to simplify notation. As the center of the ellipse is at the origin and the center of gravity in the focal point $(c,0)$, the angle $E$ named "eccentric anomaly" parametrizes the ellipse as $x=a\cos E$, $y=b\sin E$.

The 12 sections of the ellipse are a representation of the 12 months of the year and serve to visualize the different speeds along the orbit, fast at the pericenter and slow at the apocenter. The connection between time and angle is given by the "Kepler equation"
$$
2\pi\frac{t}{T}=M=E-e\sin(E)
$$
What you want is to set $t_k=k\frac{T}{12}$, $k=0,1,..,11$ and solve this equation to find $E_k$
$$
\frac{k\pi}{6}=M_k=E_k-e\sin(E_k)
$$
There are multiple ways to approximate this equation. For a numerical solver a decent starting point is $E_k\approx M_k+e\sin(M_k)$. Doing just this to produce an example plot gives

```<code>from scipy.optimize import fsolve
a=5; e=0.6; c=a*e; b=(a**2-c**2)**0.5

plt.figure(figsize=(8,4))
plt.gca().set_aspect("equal")

phi = np.linspace(0,2*m.pi,300)
plt.plot(a*np.cos(phi),b*np.sin(phi),'b', lw=3)
for k in range(-6,6):
    Mk = m.pi*k/6
    Ek = fsolve(lambda E: Mk-E+e*m.sin(E),Mk+e*m.sin(Mk))
    plt.plot([c,a*m.cos(Ek)],[0,b*m.sin(Ek)],'r',lw=2)
plt.plot([c],[0],'oy', ms=8)
plt.show()
</code>```

Body:
Let $c=\sqrt{a^2-b^2}$, $e=\frac{c}{a}$ to simplify notation. As the center of the ellipse is at the origin and the center of gravity in the focal point $(c,0)$, the angle $E$ named "eccentric anomaly" parametrizes the ellipse as $x=a\cos E$, $y=b\sin E$.

The 12 sections of the ellipse are a representation of the 12 months of the year and serve to visualize the different speeds along the orbit, fast at the pericenter and slow at the apocenter. The connection between time and angle is given by the "Kepler equation"
$$
2\pi\frac{t}{T}=M=E-e\sin(E)
$$
What you want is to set $t_k=k\frac{T}{12}$, $k=0,1,..,11$ and solve this equation to find $E_k$
$$
\frac{k\pi}{6}=M_k=E_k-e\sin(E_k)
$$
There are multiple ways to approximate this equation. For a numerical solver a decent starting point is $E_k\approx M_k+e\sin(M_k)$. Doing just this to produce an example plot gives

from scipy.optimize import fsolve
a=5; e=0.6; c=a*e; b=(a**2-c**2)**0.5

plt.figure(figsize=(8,4))
plt.gca().set_aspect("equal")

phi = np.linspace(0,2*m.pi,300)
plt.plot(a*np.cos(phi),b*np.sin(phi),'b', lw=3)
for k in range(-6,6):
    Mk = m.pi*k/6
    Ek = fsolve(lambda E: Mk-E+e*m.sin(E),Mk+e*m.sin(Mk))
    plt.plot([c,a*m.cos(Ek)],[0,b*m.sin(Ek)],'r',lw=2)
plt.plot([c],[0],'oy', ms=8)
plt.show()


问:
Why can I not take the positive or negative square root while solving a quadratic equation in this instance?
I have the problem $4x^2 -36x = -81$, and I am asked to solve it by factoring. I move the $81$ to the left side of the equation, then divide by $4$, giving me $x^2 +9x + \frac {81}4 = 0$. Since I know that $\frac {81}4$ is the perfect square of $\frac 92$, I can factor this equation into $(x - \frac 92)^2 = 0$. I then take the square root, and the equation becomes $\pm$$(x-\frac 92) = 0$. I turn the equation into 2 different ones:
$(x - \frac 92) = 0$
$-1(x - \frac 92) = 0$

So, that means that $x = \frac 92$.

But, $-1(x - \frac 92) = (\frac 92 - x) = 0$,
so this means that x also equals $-\frac 92$. I put this through a python program I wrote, and it gave me $4.5$ or $4.5$. Where did the negative come from?

答:
$4x^2-36x=-81$

You added 81 to both sides:

$4x^2-36x+81 = 0$

You divided both sides by 4:

$x^2-9x+\frac{81}{4} = 0$

You factored the LHS:

$(x-\frac{9}{2})^2 = 0$

The next thing you did was incorrect. The result after taking squares roots is $x-\frac{9}{2} = ±0$, not $±(x-\frac{9}{2}) = 0$.

$x-\frac{9}{2} = ±0$

$x = \frac{9}{2} ± 0$

$x_1 = \frac{9}{2} + 0$ and $x_2 = \frac{9}{2} - 0$

$x_1 = x_2 = \frac{9}{2}$

The root is repeated.

By the way, $\frac{9}{2}$ is the solution of $\frac{9}{2} - x = 0$. Where did you get $-\frac{9}{2}$?


问:
Prove or disprove that if V is a finite set over F with a dimension 1, then F is a finite field.
I think it is a True statement, and my proof is as follows:

let $\theta $ be the additive identity in F

Given any element v in V, since v + v = $\theta$ v + $\theta$ v = 2$\theta$ v, then 2$\theta$ $\in$ F, and $\frac{\theta}{2} \in$ F, hence $\frac{\theta}{2}v \in$ V, by this argument we can construct infinite $\{2v, \frac{\theta}{2}v , 3v,  \frac{\theta}{3}v...\}$

Hence V cannot be finite, and we are done.

答:
A basic fact is that for a finite dimensional vector space $V $ over a field
$\mathbb F $ ,  we have $|V|=|\mathbb F|^{\rm {dim}V} $.

This forces $|\mathbb F|\le|V|$.

For if $\lambda\cdot v=0$ and $\lambda\ne0$, then $v=0$.  Proof: $0=\lambda^{-1}\cdot0=\lambda^{-1}\cdot(\lambda\cdot v)=(\lambda^{-1}\cdot\lambda)v=1\cdot v=v$.


问:
Drawing circles
Choose some set of $7$ points $x_1, \dots, x_7$. Find the minimum number of lines/circles that can pass through at least 3 of the 7 points

答:
Let's collect up what we know

$L\leq 7$: each line is determined by two points.  Each triple of collinear points uses up 3 pairs of points, so $L\leq\binom{7}{2}/3=7$.

$L=7$ is impossible --- each point lies on three of the 7 lines and each of the 7 lines has three points needs the extra circle in the Fano plane configuration.

$C+L\leq 11$

$\max(C+L)\geq 9$ by the equilateral triangle + medians +three circles example.

$4C+L\leq 21$.

There are no feasible nonnegative integer solutions to $C+L\geq 10$, $4C+L\leq 21$, $L\leq 6$, since this means $3C\leq 11$, so $C\leq 3$ and hence $L\geq 7$.

So the optimal bound is $C+L\leq 9$.

Title: Drawing circles


答:
Your prove is right, but remember the definition of $\mathbb{Q} (\alpha)$, the smallest field that contains $\mathbb{Q}$ and $\alpha$.
So you only need to prove that $\alpha\in \mathbb{Q}(\alpha+1)$ ($\mathbb{Q}(\alpha+1)\subset\mathbb{Q}(\alpha)$) and $\alpha+1\in \mathbb{Q}(\alpha)$ ($\mathbb{Q}(\alpha) \subset\mathbb{Q}(\alpha+1)$).

This is more easy, because $1\in \mathbb{Q}$ and  ${-1}\in\mathbb{Q}$.

Title: Is $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$?

Body:
Let $\alpha = i\sqrt[4]{2}$. Then, $1 + \alpha = 1 + i\sqrt[4]{2}$.

Then, $\mathbb{Q}(\alpha) = \{ c_0 + c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 \mid c_0, c_1, c_2, c_3 \in \mathbb{Q} \}$.

and $\mathbb{Q}(1 + a) = \{ b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 \mid b_0, b_1, b_2, b_3 \in \mathbb{Q} \}$.

Since, $b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 = b_0 + b_1 + b_1 \alpha + b_2 + 2 b_2 \alpha + b_2 \alpha^2 + b_3 + 3 b_3 \alpha + 3 b_3 \alpha^2 + b_3 \alpha^3$

$= (b_0 + b_1 + b_2 + b_3) + (b_1 + 2b_2 + 3b_3) \alpha + (b_2 + 3b_3) \alpha^2 + b_3 \alpha^3$.

Notice that $b_0,b_1,b_2,b_3 \in \mathbb{Q}$.
Then, $(b_0 + b_1 + b_2 + b_3)$, $(b_1 + 2b_2 + 3b_3)$ and $(b_2 + 3b_3)$ are all in $\mathbb{Q}$ as well.

So, let $c_0 = (b_0 + b_1 + b_2 + b_3)$, $c_1 = (b_1 + 2b_2 + 3b_3)$, $c_2 = (b_2 + 3b_3)$ and $c_3 = b_3$.

Then, $\mathbb{Q}(\alpha) = \mathbb{Q}(1 + \alpha)$.

So, $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$


问:
Approximate: $\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}$ where $\rho$ is non trivial zeros of the zeta function
I am reading Kevin Broughan's Equivalents of Riemann Hypothesis Vol. 1 (p. 38). If $\rho$ denote the non trivial zeros of Riemann zeta function in the strip $0<\Re(\rho)<1$, consider,$$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Question: Find the approximate value of $S$.

Attempt:
Riemann $\xi$ function can be written as
$$\xi(s)=\xi(0)\prod_{\Im(\rho)>0}(1-\frac{s}{\rho})(1-\frac{s}{\bar{\rho}}),$$
where $\bar{\rho}$ denotes complex conjugate.

Taking logarithm on both sides,

$$\log(\xi(s))=\log(\xi(0))+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\rho})+\sum_{\Im(\rho)>0}\log(1-\frac{s}{\bar{\rho}})$$

and differentiating both sides w.r.t. $s$,

$$\frac{\xi'(s)}{\xi(s)}=\sum_{\Im(\rho)>0}[\frac{1}{s-\rho}+\frac{1}{s-\bar{\rho}}].$$

After this I cannot think on it.
Please find an approximate value of $$S=\sum_{\Im(\rho)>0}\frac{1}{|\rho-\frac{1}{2}|^2}.$$
Edit
$$\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho})$$
Taking log we get
$$\log(\xi(s))=\log(\xi(0))+\sum_{\rho}\log(1-\frac{s}{\rho})$$
Differentiating w.r.t. s,
$$\frac{\xi'(s)}{\xi(s)} = \sum_{\rho}\frac{1}{s-\rho}$$
Differentiating w.r.t. s again,
$$\frac{d}{ds}\frac{\xi'(s)}{\xi(s)} =- \sum_{\rho}\frac{1}{(s-\rho)^2}$$
Putting $s=1/2$ and using $\xi'(1/2)=0$,
$$\frac{\xi''(\frac{1}{2})}{\xi(\frac{1}{2})}=-\sum_{\rho} \frac{1}{(\frac{1}{2}-\rho)^2}$$


问:
$Ax=0$ in a matrix
I am self-studying Linear Algebra, mainly through Khan Academy and some YouTube videos. I have encountered the following problem.

Given

$$A = \begin{bmatrix}
2 & -1 & -2\\
-4 & 2 & 4\\
-8 & 4 & 8
\end{bmatrix}$$

how can I find the solutions of $Ax = 0$? And how could these solutions be a linear combination of two vectors in $\mathbb{R}^{3}$?

As I tried reducing the matrix by the Gauss method and got $0$'s. I am quite confused as to what to do. Are there any videos that could explain this? Thank you for any help. When I reduced it with the Gauss method, I got the following:

答:
Once you reduced the matrix, you can do an extra step and divide the first row by $2$.
Then every row of zeros represents a free variable. So let $x_2=s$ and $x_3=t$, with $s,t \in \mathbb{R}$.

Now go back to a system of equations and you get $x_1-\frac{1}{2}x_2-x_3=0$, i.e. $x_1=\frac{1}{2}x_2+x_3=\frac{1}{2}s+t$. Therefore the solutions of the system are of the form
$$
\pmatrix{\frac{1}{2}s+t \\
s \\
t
}
$$
or
$$
\pmatrix{\frac{1}{2}s \\
s \\
0
}+\pmatrix{t \\
0 \\
t
}=s\pmatrix{\frac{1}{2}\\
1 \\
0
}+t\pmatrix{1 \\
0 \\
1
}
$$

Title: $Ax=0$ in a matrix


问:
Boltzmann distribution as generalization of Gaussian
I'm trying to gain some intuition on the Boltzmann distribution (also called Gibbs canonical distribution).  The definition is very broad: it's an exponential probability distribution that is a function of an energy and temperature, $$ P(x) \propto e^{-\epsilon(x)/kT(x)} $$ where $\epsilon$ and $T$ are the energy and temperature of the state $x$, and $k$ is the Boltzmann constant.

An example is the 1D Gaussian distribution, which has pdf $$P(x) \propto e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$ Apparently this is the Boltzmann distribution corresponding to the potential energy $$\epsilon(x) = \frac{(x-\mu)^2}{2\sigma^2}.$$

How do I interpret this? I can see that if a point $x$ is far from the mean, then it has high potential energy.

In statistics, the idea of "entropy" is related to the spread of a distribution. Is this saying that a high entropy distribution has high potential energy?

答:
Well, one could write pages here. This is a short glimpse from my perspective.

Let $V$ be a nice (convex, smooth, etc.) potential. We know, that the gradient flow
\begin{equation}
\dot{x}=-\nabla V(x)
\end{equation}
will relax to the equilibrium, attained at the minimum of $V$. Moreover, $V$ decreases along solutions.

The overdamped Langevin equation is a stochastic pertubation of the gradient flow dynamic above:
\begin{equation}
dX_{t}=-\nabla V(X_{t})dt+\sqrt{2T}dB_{t}
\end{equation}
which naturally arises e.g. in statistical mechanics. Classical diffusion theory now tells us that this SDE has an invariant measure of Gibbs-type
\begin{equation}
\nu_{T}(dx)=\frac{1}{Z_{T}}e^{-\frac{V(x)}{T}}dx
\end{equation}
where $Z_{T}$ is a normalising constant, often referred to as partition function in statistical mechanics. The Gaussian setting corresponds to a quadratic potential and to an Ornstein-Uhlenbeck process for the SDE. The effect of the random pertubation decreases with a decrease in temperature. On the level of the invariant measure this amounts to a large deviation result for the family $\lbrace\nu_{T}\rbrace_{T>0}$. In particular $\nu_{T}$ weakly converges to the point measure concentrated at the minimiser of $V$. This observation is used in various optimisation algorithms, e.g. simulated annealing.

And since you mentioned entropy: the laws $\mu_{t}$ corresponding to $X_{t}$ evolve according to the Fokker-Planck equation
\begin{equation}
\partial_{t}\mu=T\Delta\mu-\nabla\cdot(\mu\nabla V).
\end{equation}
It's an easy exercise to check that the relative entropy $H(\cdot||\mu_{T})$ decreases along solution, i.e. is a Lyapunov function. In fact, since the pioneering work of Felix Otto, we know that this Fokker-Planck equation can itself be interpreted as a gradient flow in an infinite dimensional setting, namely
\begin{equation}
\dot{\mu}=-\operatorname{grad}_{W}H(\mu||\nu_{T})
\end{equation}
where $\operatorname{grad}_{W}$ denotes the formal gradient within Ottos Wasserstein geometry. Since the relative entropy is convex and minimal if $\nu=\mu_{T}$, we recover again the relaxation of the laws towards the invariant Gibbs measure.


问:
When do the local homeomorphisms given by a covering map extend to deck transformations?
Let $p: Y \rightarrow X$ be a (surjective) covering map and let $G$ be the deck group of $p$. Let $U$ open in $X$ be such that the disjoint open sets $V, W \subseteq Y$ are both homeomorphic to $U$ through $p$. Then $(p \rvert_V)^{-1} \circ (p \rvert_W) : W \rightarrow V$ is a homeomorphism which also preserves fibers.

Are there any conditions on the spaces which guarantee that this homeomorphism is the restriction of some deck transformation? One obvious condition would be that $G$ has to act transitively on fibers if we want this to happen for any $V$ and $W$.

But is this enough? If not, does this at least happen if the spaces have "good" properties, for example local path connectedness, or are even topological manifolds?

答:
It is not even true for "nice" spaces. As an example consider $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$.

Take $V = (0,1) \cup (1,2)$ and $W = (1,2) \cup (2,3)$. Both are mapped by $p$ homeomorphically onto $U = S^1 \setminus \{-1, 1\}$. Then $p_{W,V} = (p \rvert_V)^{-1} \circ (p \rvert_W) : W \rightarrow V$ is the identity on $(1,2)$, but on $(2,3)$ it has the form $p_{W,V}(t) = t - 2$. Therefore it is not the restriction of some deck transformation. Note that in this example $U$ is evenly covered.

The same phenomenon will occur for any covering $p : Y \to X$ with more than one sheet if $U$ is evenly covered, but not connected. See Covering projections: What are the sheets over an evenly covered set? for a discussion.

Thus the minimal requirement will be that $U$ is a connected evenly covered open set. In that case the set $p^{-1}(U)$ decomposes uniquely into sheets which are mapped by $p$ homeomorphically onto $U$. These sheets are the components of $p^{-1}(U)$. Thus for any two sheets $V, W$ there exists exactly one homeomorphism $\phi : W \to V$ such that $p(\phi(x)) = p(x)$. We have $\phi = p_{W,V}$.

Then if $G$ acts transitively on fibers, it is true. To see that, pick $y \in V$ and let $y' =  p_{W,V} (y)$. There exists a deck transformation $h$ such that $h(y) = y'$. Clearly $h$ maps each component of $p^{-1}(U)$ homeomorphically onto some component of $p^{-1}(U)$. Since $V, W$ are such components and $h(V) = W$, we are done.

That $G$ acts transitively on fibers can be assured by suitable assumptions on $X,Y, p$.


问:
How to test the convergence of the series $\sum_{n=1}^\infty n^{-1-1/n}$?

How to test the convergence of the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{1+1/n}}?$

Help me. I'm clueless.

答:
$$
  n<\mathrm e^n\implies n^{1/n}<\mathrm e\implies n^{1+1/n}<\mathrm e\cdot n\implies \frac{1}{\mathrm e\cdot n}<x_n.
$$
Since $\sum\frac1{\mathrm e \cdot n}$ is divergent so is the series.</p>


问:
first order equivalence question
If we know that
$C\models A$ and   $C\not\models B$

Can we say that $C\models A$ and  $C\not\models B$ iff $C\models B\rightarrow A$ and   $C\not\models B$ ?

Edit:
Can we say that $C\models A$ and  $C\not\models \lnot B$ iff $C\models B\rightarrow A$ and   $C\not\models \lnot B$ ?

答:
No. One direction of the biconditional fails. From $C \vDash B \to A$ and $C \nvDash B$ it doesn't follow that $C \vDash A$.

Suppose $A$ and $B$ are the same. Then we'd trivially have $C \vDash A \to A$; but that with $C \nvDash A$ wouldn't imply $C \vDash A$!

Added: on the edited addition, no. The same direction of the biconditional fails. Again suppose $A$ and $B$ are the same. Then we'd trivially have $C \vDash A \to A$; but that with $C \nvDash \neg A$ wouldn't imply $C \vDash A$! $A$ can be independent of $C$, so $C$ entails neither it nor its negation.

Title: first order equivalence question


问:
Markov Chains: Prove that if $i$ is transient state, then $\pi_i = 0$

Setup

Let $S$ be a countable set and $(X_n)_{n \in \mathbb{N_0}}$ a Markov chain with state space $S$ and transistion matrix $P=\{p_{ij}\}_{i,j \in S}$.

Let furthermore $T_i = \inf\{n \in \mathbb{N} \colon X_n = i\}$. A state $i$ is defined to be recurrent if $P(T_i < \infty \mid X_0 = i) = 1 $ and transient if $P(T_i < \infty \mid X_0 = i) < 1 $.

A stationary distribution (for $P$) is a vector $\pi = (\pi_i)_{i \in S}$ with $\sum_{i \in S} \pi_i = 1$ satisfying that $\pi = \pi P$.

Claim

Assume $\pi = (\pi_i)_{i \in S}$ is a stationary distribution. Prove that if $i$ is transient state, then $\pi_i = 0$.

Thoughts

I know that a transient state is only visited finitely many times P-almost surely. Maybe one can prove the claim using this? If we write $P^n = \{p_{ij}^n\}_{i,j \in S}$ then another property of a transient state $\sum_{n=1}^\infty p_{ii}^n < \infty$. But I can't quite make it work using this.

答:
Here is a possible proof approach.

You say that you already know a transient state $i$ is almost surely visited only finitely many times. Can you show the stronger statement that the expectation of the number of visits to $i$ is finite?

Then, denoting the number of visits to $i$ as $N(i) = \sum_{k=0}^\infty 1(X_k=i)$,
$$\mathbb{E}[N(i)] = \sum_{k=0}^\infty \mathbb{E}[1(X_k=i)] = \sum_{k=0}^\infty P(X_k=i) = \sum_{k=0}^\infty \pi_i$$
which can only be finite if $\pi_i=0$.


问:
Connection between Trace of $A^TA$ and eigenvalues of $A$
I am visiting a course about linear algebra. Our professor proved the equality between the Frobeniusnorm and the Schatten-2-norm. For this he used the following equality for a real matrix A.

$\sqrt{(\lambda_1^2+...+\lambda_n^2)}=\sqrt{tr(A^TA)}$

where $\lambda_1, ..., \lambda_n$ are the eigenvalues of $A^TA$ and $tr(A^TA)$ denotes the trace of $A^TA$. May someone please explain, why this equality holds?

Thank you very much!

答:
It is well known that the trace of any matrix is the sum of its eigenvalues (counted with algebraic multiplicity). This is equal valid for $A^TA$ as for any other matrix.

So it really ought to be
$$
\lambda_1+\cdots+\lambda_n=\operatorname{tr}(A^TA)
$$


问:
Find Infimum and supremum of a set $A=\{ \sqrt[n^2]{(n^2+2n+1)^2}:n\in \mathbb{N}\} $
I have to find Infimum and supremum of a set $A=\{ \sqrt[n^2]{(n^2+2n+1)^2}:n\in \mathbb{N}\} $. I tied to do this by finding extrema of the function
$$
f(x)=(x+1)^{\tfrac{2}{x^2}}=\exp(\tfrac{2}{x^2}\log(x+1)) \qquad x>0
$$
and i tried to find extrema of the function inside exponent but I can't find x for which it's derivative is equal to $0$. My computations are
$$
\frac{2}{x^2(x+1)}+\frac{-4\log(x+1)}{x^3}=0
$$
$$
\frac{2}{(x+1)}-\frac{4\log(x+1)}{x}=0
$$
$$
\frac{2x-4(x+1)\log(x+1)}{(x+1)x}=0
$$
$$
2x-4(x+1)\log(x+1)=0
$$
and I don't know what to do next. Thanks in advance for any help.

答:
When $x$ is sufficiently big, for example, when $x>e-1$, we have $(\ln(f(x)))'=\frac{2x-4(x+1)\log(x+1)}{x^3(x+1)}<\frac{2x-4(x+1)}{x^3(x+1)}<0$. So $f(x)$ is monotonically decreasing when $x>e-1$. So $\sup A$ equals to one of the $a_n$ where $n\leq e-1$.
On the other hand, since $\ln(f(x))$ is monotonically decreasing when $x$ is big enough, and that $f$ is continuous, $\inf A=e^{\lim\limits_{x\rightarrow+\infty}\ln(f(x))}=e^0=1$.


问:
how to prove this trig inequality: $\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)$
The following inequality has come up in my work. I'm trying to prove that it is true:
$$
\begin{equation}
\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)
\end{equation}
$$
where
$$
\begin{align}
-\pi \leq{} &x \leq \pi \\
-\pi \leq{} &y \leq \pi \\
-1 \leq{} &z \leq 1
\end{align}
$$

The LHS of the inequality is derived from the spherical law of cosines, the RHS is derived from the law of cosines, and $z$ represents the cosine of another variable. Note that the inequality holds with equality when $z=1$.

I have also put the inequality in the following form using the product-to-sum trig identity, and rearranging the RHS:

$$
\begin{equation}
(1+z) \cos(x-y) + (1-z) \cos(x+y) \geq
2 \cos \left( \sqrt{(x-y)^2 + 2 x y (1-z)} \right)
\end{equation}
$$

I have also tried other things like applying Cauchy-Schwarz to the LHS, and applying arccosine to both sides, but I can't seem to prove that the inequality is true.

I am quite certain that the inequality holds because I have performed a Monte Carlo simulation (code shown below) which shows that the inequality holds for randomly sampled values.

python Monte Carlo random sampling code

```<code>import numpy as np

n = 10**6
x = np.random.uniform(-np.pi, np.pi, n)
y = np.random.uniform(-np.pi, np.pi, n)
z = np.random.uniform(-1, 1, n)
lhs = np.cos(x)*np.cos(y) + z*np.sin(x)*np.sin(y)
rhs = np.cos(np.sqrt(x**2 + y**2 - 2*x*y*z))
check = lhs>=rhs
print('inequality is true', np.sum(check), '/', n, 'times')
</code>```

答:
Replacing $(x, z)$ by $(-x, -z)$ or $(y, z)$ by $(-y, -z)$ does not change the inequality, therefore we can assume that
$$
 0 \le x, y \le \pi  \, , \, -1 \le z \le 1 \, .
$$

Case 1: $x+y \le \pi$. The solution for this case is inspired by Proving that $\cosh^{-1} (\cosh(x)\cosh(y) ) \geq \sqrt{x^2 + y^2}$ and the CAT(0) inequality.

The function $f(u) = \cos(\sqrt u)$ is convex on the interval $0 \le u \le \pi^2$, this is straightforward to verify, e.g. by calculating the second derivative.

We can assume that $y \le x$. The convexity condition with $\lambda = (1+z)/2$ and $1-\lambda = (1-z)/2$ then gives
$$
\begin{align}
\cos x \cos y + z \sin x \sin y &=
\frac{1+z}2 \cos(x-y) + \frac{1-z}2 \cos(x+y) \\
 &= \frac{1+z}2 f((x-y)^2) + \frac{1-z}2 f((x+y)^2) \\
&\ge f \left( \frac{1+z}2 (x-y)^2 + \frac{1-z}2 (x+y)^2\right) \\
&= f( x^2+y^2-2xyz) \\
&= \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$

$f$ is in fact strictly convex, so that equality holds exactly in the following cases:

$\lambda = 0$ or $1$, that is if $z = \pm 1$.

$(x-y)^2 = (x+y)^2$, that is if $x=0$ or $y=0$.

Case 2: $\pi < x+y \le 2\pi$. This case can be reduced to the first case, credit goes to ipst. I have only simplified the solution a bit.

Setting $x' = \pi - x$ and $y' = \pi - y$ we can apply the first case to $(x', y', z)$:
$$
\begin{align}
 \cos x \cos y + z \sin x \sin y &= \cos x' \cos y' + z \sin x' \sin y' \\
 &\ge \cos \left( \sqrt{x'^2 + y'^2 - 2 x' y' z} \right) \\
 &\overset{(*)}{\ge} \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$
The last inequality $(*)$ needs to be proved. For $\alpha, \beta \in [0, 2 \pi]$ we have
$$
 \cos \alpha  \ge \cos \beta \iff \alpha \le \beta \le 2\pi -\alpha \, .
$$
Therefore we have to show that
$$
\sqrt{x'^2 + y'^2 - 2 x' y' z} \overset{(1)}{\le} \sqrt{x^2 + y^2 - 2 x y z}\overset{(2)}{\le} 2 \pi - \sqrt{x'^2 + y'^2 - 2 x' y' z} \, .
$$
The left part $(1)$ follows from
$$
(x^2 + y^2 - 2 x y z) - (x'^2 + y'^2 - 2 x' y' z) = 2 \pi(x+y-\pi)(1-z) \ge 0
$$
and the right part $(2)$ follows from
$$
\sqrt{x^2+y^2-2xyz} + \sqrt{x'^2+y'^2-2x'y'z}
\le (x+y) + (x'+y') = 2 \pi \, .
$$

Equality holds in the following cases:

$z = \pm 1$.

$x'y' = 0$ and $x+y= \pi$, that is if $(x, y) = (\pi, 0)$ or $(x, y) = (0, \pi)$.


问:
Direct sum of Morphism ? Decomposition in irreductible representations.
Context : Finding irreductible representations such as a G module is a direct sum of these latter.

In a course i find the notation :

$p_4=p_1^{\oplus i_1} \oplus p_2^{\oplus i_2} \oplus p_3^{\oplus i_3}$ where $p_1,p_2,p_3$ are representations , seen as morphism , for example :
$p_i : G \rightarrow Gl(V_i)$.

Usually , we write the decomposition in irreductible G modules in terms of vectorial space associated ,for example $V_4=V_1^{\oplus i_1} \oplus V_2^{\oplus i_2} \oplus V_3^{\oplus i_3}$ where $V_i$ are complex vectorial spaces of irreductible representations.

Can someone explain what means the direct sum of morphism , or the notation of "direct sum of morphism" that i written ?

答:
If you have two representations $p : G \to \mathrm{GL}(V)$ and $q : G \to \mathrm{GL}(W)$,  $p \oplus q$ is the representation of $G$ on $V \oplus W = V \times W$ (i.e. $p \oplus q : G \to \mathrm{GL}(V \oplus W)$) given by
$$
  (p \oplus q)(g)(v, w) = (p(g)(v), q(g)(w)).
$$

This is really the same as your $V \oplus W$ because if you consider $V$ and $W$ as $G$-modules that basically just means that you hide the morphisms $p$ and $q$ from the notation and write $gv$ for $p(g)(v)$ if $v \in V$ and $g w$ for $q(g)(w)$ if $w \in W$. (That is, you decide whether to use $p$ or $q$ depending on which space the vector is from. Decomposing the morphism makes this implicit in the notation instead.)


问:
Properties of an alternating bilinear form its coordinate matrix
I found that I lack many basic knowledge about linear algebra, so read the wiki article about Bilinear Forms. Especially this Paragraph. I tried to proof of "Every alternating form is skew-symmetric." was quite easy. And I found a counter-example for the inverse if $char(F) = 2$.

However, I am currently trying to find a proof for this:

A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero

I looked at the following equations, to understand that.

$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix} = 0$

$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}w_1 \\ w_2\end{pmatrix} = - \begin{pmatrix}w_1 & w_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix}$

And they are both solvable for $a=0, d=0, c=-b$, thus the matrix $\begin{pmatrix}0 & b \\ -b & 0\end{pmatrix}$.

However, I am not any nearer to a proof. Can someone please point me into the right direction?

答:
Hint: Let $\beta$ be a bilinear form and $B$ its coordinate matrix. Applying $\beta$ to two basis vectors $e_i$, $e_j$ gives
$$ \beta(e_i, e_j) = e_i^t \cdot B \cdot e_j = B_{ij} $$
that is the $(i,j)$-entry of $B$. Now if $\beta$ is alternating, then it is skew-symmetric, hence $\beta(e_i, e_j) = -\beta(e_j, e_i)$ and $\beta(e_i, e_i) = 0$. Can you relate this to $B$, using the above?

Edit: As you noted in your comment, this gives $B = -B^t$ with zeros on the diagonal, that is, one direction. For the other direction, note that for arbitrary $x,y$, we have
$$ \beta(x,y) = x^t B y = -x^t B^t y = -y^t B x $$
and that
$$ \beta(x,x) = \sum_{i=1}^n B_{ii}x_i^2$$


问:
What is the geometric meaning of differentials of higher order than the first?
As the title suggests, I was wondering what could be the geometric meaning of differentials of higher order than the first, being aware that the first order differential is a linear application that approximates linearly the increment of the function itself at a point.

Thanks in advance.

答:
The second derivative of a function corresponds to the curvature or concavity of the graph. The graph of a function with a positive second derivative is upwardly concave, while the graph of a function with a negative second derivative curves in the opposite way.

Similarly, the third derivative of a function corresponds to the change of the curvature of the graph.


问:
Constraint Qualification for Lagrangian Method
I am trying to solve the problem $\max_{x,y} f(x,y)=2x^3-3x^2 \textrm{ s.t. } g(x,y)=(3-x)^3-y^2=0$ where both $f(x,y)$ and $g(x,y)$ are defined on $\mathbb{R}^2$. I would like to know if the following solution is logically sound.

From the first- and second-order conditions for $f(x,y)$, a local maximum and local minimum exist for $f(x,y)$ at $x=0$ and $x=1$ respectively. Based on the constraint, since $y^2\geq0$, $(3-x)^3 \geq 0$ and therefore $x\leq 3$. We shall then check the boundary point, stationary points and limit to infinity for $x\leq 3$. Because $\lim_{x \to -\infty} f(x,y) = -\infty$, $f(0,y)=0$, $f(2,y)=-1$ and $f(3,y)=27$, $(x,y)=(3,0)$ must maximize $f(x,y)$.

I also notice that the Lagrange method for this problem does not work because it fails the constraint qualification. In particular, the first-order conditions for $L=2x^3-3x^2-\lambda[(3-x)^3-y^2]$ are $\frac{\partial L}{\partial x}=6x^2+3\lambda(3-x)^2=0$ and $\frac{\partial L}{\partial y}=-6y+2\lambda y=0$, and yield no solution. In that case, is checking the points that fail the constraint qualification ($\nabla g(x^*,y^*) \neq (0,0))$ the only way to solve the problem?

答:
Assuming $f(x,y) = 2x^3-3y^2$, the Lagrange method of multipliers is based on regularity to the point of non null gradient existence in the involved functions. This method relies on the possibility of tangency between the objective function level surfaces and the restriction. Those points when they exist are called stationary points. In the present case we have

$$
\cases{
\nabla f = (6x^2,-6y,0)\\
\nabla g = (-3(3-x)^2,-2 y,0)
}
$$

making now $\nabla f\times\nabla g = (0,0,108x y-162 y-30 x^2 y)$ we can verify that the only point in which $\nabla f = \lambda \nabla g$ is at $y=0$ but at this point $\nabla g=(0,0,0)$ which invalidates the method. Resuming: the method doesn't fail, because it can't be applied here.

NOTE

If $f(x,y) = 2x^3-3x^2$ then

$$
\nabla f\times\nabla g = (0,0,-12x(x-1)y)
$$

and we have linear dependency at $x = 0, x = -1, y = 0$. If $y = 0$ then $\nabla g=(0,0,0)$ and if $x=0$ or $x = 1$ we have $\nabla f=(0,0,0)$


答:
Instead of substituting the roots later, you could have substituted them right at the start:

$(-1)^2+p·(-1)+q = 0$.

$(4)^2+p·(4)+q = 0$.

And it is easy from there.

Title: Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.

Body:
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.

Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 - 4p
\\
-1 - p = - 16 - 4p
\\
3p = -15
\\
p = -5
\\
q = -1 - (-5)
\\
q = 4
$$

According to the book the answer should be $-3$ & $-4$


问:
Distances between measures
Suppose that $\mu_n$ and $\nu_n$ are two probability measures with
$$
\lim_{n \to \infty} \sup_{a \in \mathbb{R}} | \mu_n(-\infty,a)-\nu_n(-\infty,a)|=0.
$$

Then is it true that $\mu_n(B)- \nu_n(B) \to 0$ for all Borel sets $B$?

The usual approach to this sort of thing is to define
$$
\mathcal{B}:=\{B \subset \mathbb{R}: \mu_n(B)-\nu_n(B) \to 0 \},
$$
and then prove that $\mathcal{B}$ is a $\sigma$-algebra that contains all $(-\infty,a)$, or something generating the Borel sets.

$\mathcal{B}$ does contain all $(-\infty,a)$ and is closed under taking complements because $\mu_n$ and $\nu_n$ are probability measures, but I am not sure how to show closure under countable and unions and intersections without assuming some uniformity accross all $B$.  (Maybe I just need to add that hypothesis to $\mathcal{B}$?)  One could also rephrase my question in terms of the cdf's for $\mu_n$ and $\nu_n$.

答: This is not true. By CLT there exist discrete distributions $\mu_n$ converging to standard normal distribution. The fact that standard normal distribution $\nu$ is continuous implies that the hypotheis is satisfied with $\nu_n=\nu$ for all $n$. But there is a countable set $C$ such that $\mu_n(C)=1$ for all $n$ whereas $\nu(C)=0$.

Title: Distances between measures


问:
Basis of field extension of rational number that is closed square root without axiom of choice
Consider a field extension $F$ of the rational numbers. $F$ contains all numbers that can be constructed using square root. In this field, there are the regular rational numbers such as 1, 2, 3, $\frac{42}{13}$, etc. There are square roots such as $\sqrt{2}$, $\sqrt{3}$, $\sqrt{\frac{42}{13}}$, etc. It does not* contain cubic roots, i.e., $\sqrt[3]{2}$, $\sqrt[3]{3}$, $\sqrt[3]{\frac{42}{13}}$. The square root can be applied iteratively. This means that $\sqrt[4]{3} = \sqrt{\sqrt{3}}$ is in the set. The square root operation can only be applied a finite number of times, i.e., constructions that involve the limit of $a_i=\sqrt{42+a_{i-1}}$ or similar sequences are not considered.

(*I assume that F does not contain cubic roots. If one can show that a cubic root can somehow be written as combination of square roots then F would contain cubic roots. This would be valuable information to me, if my assumption is wrong here.)

Can somebody provide a constructive (i.e. no axiom of choice) construction of a basis of this field as rational vector space? If $F$ has no basis without axiom of choice would also be an interesting result.

My objective is to perform exact geometric computations that involve length computations in a computer. These computations may be very slow. The idea is to represent the numbers of $F$ by storing the coefficients of this basis. This is the reason why I do not want to use the axiom of choice. I need to know how the basis actually looks like to be able to perform computations on it. A statement of the form "if you believe the axiom of choice, then the basis exists but nobody knows how it looks" is not helpful. That is not something one can implement in a computer.

答:
There are two fine points here.

There is always a canonical, countable, field which is the smallest subfield of $\Bbb C$ closed under $\sqrt\cdot$. Since this field is countable, simply enumerate it, and recursively construct a basis. Done and done.

It is consistent with $\sf ZF$ that there is a field which is a closure of $\Bbb Q$ under $\sqrt\cdot$, but it is not well-orderable. I'm not sure if this field doesn't have a basis, but that stands to reason to expect that it doesn't. If it does, though, this basis is certainly not well-orderable. So working with it from a computational point of view is ... odd, at best.

Since you're saying "this field", it seems that you have in mind the first option. To see that it is countable more explicitly, note that every polynomial of the form $x^2-p=0$ has at most two roots and there are countably many such polynomials; using the fact that $\Bbb C$ can be linearly ordered (as a set!), this is a countable union of pairs; this defines a canonical enumeration of the smallest field in which $\sqrt q$ exists for each rational $q$. Now repeat this, using said canonical enumeration to kickstart the next step in the recursion.

After going through all the finite steps of the recursion, you are left with a countable sequence of countable sets that are equipped with canonical enumerations, and therefore their union is indeed countable.


问:
What are the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$?
I would like to calculate the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$.
I know that the elements of $\mathbb{Q}(\sqrt[3]{2})$ have the form of ${a+b\sqrt[3]{2}+c\sqrt[3]{4}}$, where a,b,c $\in \mathbb{Q}$ and the elements of $\mathbb{Q}(\sqrt{3})$ have the form of $a+b\sqrt{3}$, where a,b $\in \mathbb{Q}$.
I also calculated the minimal polynomial of $(\sqrt[3]{2}+\sqrt{3})$ over $\mathbb{Q}$ , which is: $x^6−9x^4−4x^3+27x^2−36x−23$.

Can you help me to calculate the form of the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$? I have to find the elements of the linear combinations which form the higher powers of $(\sqrt[3]{2}+\sqrt{3})$ which are also powers of $(\sqrt[3]{2}+\sqrt{3})$?
Could you give me a proper method to find the solution? Also, could you write down your calculation in your answer? Thank you for helping me!

答:
Let $\beta=\sqrt[3]{2}+\sqrt{3}$.

Your problem becomes easy if $\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.
$\mathbb{Q}(\beta)\subset\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$ ($\beta\in\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$). Let's see that $\sqrt{3}\in\mathbb{Q}(\beta)$:

$$\beta-\sqrt{3}=\sqrt[3]{2}$$
$$(\beta-\sqrt{3})^3=2$$
$$\beta^3-3\sqrt{3}\beta^2-9\beta+3\sqrt{3}=2$$
$$\sqrt{3}=(\beta^3-9\beta^2+1)/3(\beta^2-1)\in\mathbb{Q}(\beta)$$

Thus, $\sqrt[3]{2}\in\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$ too, so $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3}) =\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.


问:
can we say $x=aV\sqrt{a}$ in Kepler's third law? What else can we say?
we know that Kepler's third law says that the Period of the planet (time elapsed for a planet to perform a complete rotation around its sun) to the power of 3 is proportional to the orbit's semi-major axis to the power of 2.
so we can say $P^2=a^3$ or $T^2=a^3$.

we know $T=\frac{x}{V}$ where $x$ is the orbit's perimeter and $V$ the speed of the planet.

so the Kepler's third law can be rewritten as $x=aV\sqrt{a}$

$a$ and $V$ are constant for each planet and we can find the orbit's perimeter for each planet with the above equation. right?

another question I have is that can we use the last equation to get an approximately good perimeter for each given ellipse? (for example we draw an ellipse with the semi-major and semi-minor axes of 5 and 3 meters respectively; then we put an object like a ping pong ball on the ellipse to move with the speed of 0.1 $\frac{m}{s}$ on it. can we use that equation to find the perimeter of that ellipse?)

答:
Keplers second law tells you that
$$
r^2\dot\varphi=ab\omega
$$
where $r,φ$ are the polar coordinates around the gravity center, $a,b$ are the half-axes of the ellipse (relative to the center of the ellipse) and $ω=\frac{2\pi}{T}$ is the average angular speed. If one knew a nice formula for the perimeter, one could make a proportion law as proposed, but it would be artificial.

The velocity along the orbit can now be compactly computed from
$$
\dot x+i\dot y=\frac{d}{dt}(re^{iφ})=(\dot r+ir\dotφ)e^{iφ},
$$
Using the formula of the first law, with $b^2+c^2=a^2$, $c=ea$,
$$
r=\frac{b^2}{a+c\cosφ}
\implies
\dot r=\frac{b^2c\sinφ\dotφ}{(a+c\cosφ)^2}
=\frac{acω}{b}\sinφ
$$
gives the square of the speed as
$$
\dot r^2+r^2\dotφ^2=\frac{(acω)^2}{b^2}\sin^2φ+\frac{(aω)^2}{b^2}(a+c\cosφ)^2
=\frac{a^2ω^2}{b^2}\left(c^2+a^2+2ac\cosφ\right)
$$
so that
$$
v=\frac{aω}{b}\sqrt{c^2+a^2+2ac\cosφ}
$$

The third Kepler law follows from the gravity equation $\ddot z=-\frac{GMz}{|z|^3}$, which has the radial component
$$
\ddot r+\frac{GM}{r^2}-\frac{(abω)^2}{r^3}=0
$$
which integrates after multiplication with $2\dot r$ to the first integral
$$
\dot r^2-\frac{2GM}{r}+\frac{(abω)^2}{r^2}
=-\frac{2GM}{r_{\rm peri}}+\frac{(abω)^2}{r_{\rm peri}^2}
=-\frac{2GM}{r_{\rm apo}}+\frac{(abω)^2}{r_{\rm apo}^2}
$$
so that
$$
2GM\left(\frac1{a(1-e)}-\frac1{a(1+e)}\right)
=b^2ω^2\left(\frac1{(1-e)^2}-\frac1{(1+e)^2}\right)
\\~\\
GM=a^3ω^2
$$


问:
Amalgamation of matching family for empty cover
Let $\mathcal{J}$ be a Grothendieck topology on a small category $\mathbb{C}$. If $C$ is an object of $\mathbb{C}$ for which $\mathcal{J}(C)$ contains the empty cover/sieve, then if $F : \mathbb{C}^{\mathsf{op}} \to \mathsf{Set}$ is any $\mathcal{J}$-sheaf, it is my understanding that $F(C)$ must be a singleton, since any element of $F(C)$ will be an amalgamation for the (empty) matching family for the empty cover of $C$, and such amalgamations are unique. Is it a 'common' situation for an object to have an empty cover in a Grothendieck topology? Is it possible to assume without loss of generality that no object is covered by the empty sieve, in the sense that there is another site $(\mathbb{D}, \mathcal{K})$ with this property for which the topos of $\mathcal{J}$-sheaves is equivalent to the topos of $\mathcal{K}$-sheaves?

答:
There is already an answer about how common empty covers are. I will focus on your last question. In short the answer is yes. You can simply leave out all the objects from $\mathbb{C}$ that are covered by the empty sieve (in the Grothendieck topology $\mathcal{J}$). This will give you a full subcategory $\mathbb{D}$ of $\mathbb{C}$ with an induced Grothendieck topology $\mathcal{K}$, which consists of the same sieves as in $\mathcal{J}$ but only for the objects in $\mathbb{D}$ of course. Then the categories of sheaves on $(\mathbb{C}, \mathcal{J})$ and $(\mathbb{D}, \mathcal{K})$ are equivalent.

This is in a simple application of the comparison lemma. That lemma is much stronger, so it is probably overkill and you could probably also write a simple direct proof using the insight you had (about $F(C)$ being a singleton).


问:
Biggest (centered) polydisc of holomorphy
I have trouble defining the biggest centered polydisc of holomorphy (where I can apply cauchy's inequality) of a multivariate complex holomorphic function. As an example, suppose a function of 2 variables with 3 poles:

$$f(z_1,z_2) = \frac{1}{\left(1-\frac{z_1}{2}\right)\left(1-\frac{z_2}{2}\right)\left(1-z_1z_2\right)}$$

The function has singularities:

On $\left\{(z_1,z_2) \in \mathbb C^2:\, z_1 = 2\right\}$

On $\left\{(z_1,z_2) \in \mathbb C^2:\, z_2 = 2\right\}$

On $\left\{(z_1,z_2) \in \mathbb C^2:\, z_1 = \frac{1}{z_2}\right\}$

I have two questions:

For each of the three singularities, if only this one occured, what would be the polyradius of the biggest polydisc of holomorphy ?

What is the polyradius of the biggest polydisc of holomorphy for the full function ?

答:
The OP asks for the largest polydisk contained in a certain set, as though the existence of a largest polydisk was clear. It's been commented that the existence of a largest polydisk is not at all clear to at least one reader, while of course a maximal polydisk is much more plausible. When the OP replies "if you prefer to call [largest] [maximal] be my guest" it seems like an example clarifying the difference may be a good idea.
$\newcommand{\R}{\mathcal R}$
Say $D=\{(x,y):x,y\ge0,xy\le1\}$. Say a rectangle is a set of the form $[a,b]\times[c,d]$, let $\R$ be the set of all rectangles contained in $D$, and let $R_0=[0,1]\times[0,1]$.

Then $R_0$ is not the largest element of $\R$, because for example if $R_1=[0,2]\times[0,1/2]$ then $R_1\in\R$ but $R_1\not\subset R_0$.

But $R_0$ is a maximal element of $\R$; this means that, as you can easily verify, if $R_1\in\R$ and $R_0\subset R_1$ then $R_0=R_1$.


问:
Find the coordinate of A and B
Question:

OA: y= 4x

AB: 3y+8x=400

OB: y=0

Find the Coordinates of A and B

I am really not sure how to do this problem.

I've tried using system of equations to see which numbers will have in common between O, A and B. But I was honestly guessing which method could work.
Could someone help me on this problem?

Thanks!

Title: Find the coordinate of A and B

Body:
Your problem is

A is at $OA=AB$.

That is $4 x= (400-8x)/3=400/3-8 x/3$.
$20 x/3=400/3$ or $20 x=400$ or $x=20$.
$y=(400-8x)/3=(4000-8*20)/3=80$.

So A is the point, coordinate tuple A=$(20,80)$.

B is at $OB=AB$.

That is  $0=(400-8x)/3$.
$0=400-8x$ or $x=50$.

So B is the point, coordinate tuple B=$(50,0)$.

The solution in a plot is:


问:
Claim and Proof of Claim within Proof of a Theorem
While writing a proof of a theorem for a journal can we use the Claim environment within the Proof environment?

Suppose I want to break a step in the proof as a Claim and proof the claim, but the claim cannot be posed separately since it is very specific to a step of the proof a theorem. In that case, can we write the Claim and the proof within the proof itself?

If someone shows me an example where there are such instances it will be great.

Please let me know if I am allowed to post this question here? Any suggestions are welcome.

答:
It is normal to state and prove lemmas which are then used to prove a main theorem. It is common to state the main theorem first, then state some further definitions and state/prove several lemmas about the things that were just defined, essentially setting up all the machinery needed for a proof, and then finally give a proof of the main theorem (where you use the lemmas).

If the claim/lemma really is too specific to build up to before starting the main proof, you could instead merely state it at the point you use it in the main proof, and then restate and prove the lemma afterwards. That way you keep the main proof more concise and readable.


问:
Is there a way to write logarithm as a function of the inverse of $\cos$?
How can I prove that

$\begin{equation*}
    \log x=2i\cdot \cos^{-1}\left(\frac{x+1}{2\sqrt{x}}\right)
\end{equation*}$

Any reference?

Body: $$\cos\left(\frac{\log x}{2i}\right)=\frac{e^{\log x/2}+e^{-\log x/2}}2=\frac{x+1}{2\sqrt x}.$$


问:
What is wrong with this fake proof that $\lim\limits_{n\rightarrow \infty}\sqrt[n]{n!} = 1$?
$$\lim_{n\rightarrow \infty}\sqrt[n]{n!}=\lim_{n\rightarrow \infty}\sqrt[n]{1}*\sqrt[n]{2}\cdots\cdot\sqrt[n]{n}=1\cdot1\cdot\ldots\cdot1=1$$
I already know that this is incorrect but I am wondering why. It probably has something to do with the fact that multiplication in $n!$ is done infinite number of times.</p>

答: Another way of explaining this is that for infinite $n$, each of the factors $\sqrt[n]{1}$, $\sqrt[n]{2}$, $\sqrt[n]{3}$, ... $\sqrt[n]{n}$ will be infinitely close to $1$, but this is not enough to conclude anything about the product because there are infinitely many factors in the product.</p>


答:
The Vieta's formulas tells you that

$$p=-(r+s),\\q=rs$$ (where $r,s$ are the roots).


问:
contradiction to commutative tensor product.
I am studying now abstract algebra and my professor is always assuming that we have a commutative ring with unity and she told us that the tensor product is commutative. But I am now was trying to solve the following problem:

Let $k$ be a field and $B = k[x,y].$ Define $R-$modules $J = (x,y), N = J/J^2$ and $M = J \otimes_B J.$

Define $m \in M$ by $m = x \otimes y - y \otimes x.$ Show that $m \neq 0$

And this was the answer given to me by @Eric Wofsey:

It is not correct that $x \otimes y - y \otimes x = (x-y) \otimes (y-x)$ in this case.  Indeed, note that $N$ is just a 2-dimensional vector space over $B/J\cong k$, with $\{x,y\}$ as a basis.  So, $N\otimes_B N=N\otimes_{B/J} N$ is just a tensor product of vector spaces and so it is 4-dimensional with $\{x\otimes x,x\otimes y,y\otimes x,y\otimes y\}$ as a basis.  This shows that $x \otimes y - y \otimes x \neq (x-y) \otimes (y-x)$, but it also just directly shows that $x\otimes y-y\otimes x$ is nonzero.

My question is:

Why the tensor product does not commute in this case? i.e., why  this element $x\otimes y$ from the basis is not considered the same as this element $y\otimes x$?

答: The phrase "tensor product is commutative" most probably refers to the fact that for any modules $M$ and $N$ there exists an isomorphism between $M\otimes N$ and $N\otimes M$ defined by $m\otimes n\mapsto n\otimes m$. One usually says that tensor product of modules (not their elements) is commutative up to an isomorphism. Note that it says nothing about $m\otimes n$ and $n\otimes m$ being actually equal (in fact, they even belong to different modules, namely $M\otimes N$ and $N\otimes M$, respectively).


问:
If the sequence $a_k$ satisfies that $a_1=1,a_2 = 2;a_k-4a_{k-1}+3a_{k-2} =0$, ($k \ge 3$)
I have applied $k=3,k=4,k=5$ and found out that $a_3=5,a_4=14,a_5=41$. I cant find $a_k$, $(k>=1)$
please help.

答:
For $$a_k-4a_{k-1}+3a_{k-2} =0$$ the characteristic equation is
$$r^2-4r+3=0=(r-1)(r-3)$$ So,as usual, the solution is
$$a_k=c_1 (1)^k+c_2(3)^k$$ Use the conditions to get $c_1$ and $c_2$


问:
$(0,1) \times \{0\}$ is an open subset of $[0,1] \times \{0\}$ but not of $[0,1]^2$
I'm new to Real analysis and reading the chapter on metric spaces in the book 'Real Analysis
with Economic Applications'. It is stated in the book that $(0,1) \times \{0\}$ is an open subset of $[0,1] \times \{0\}$ but not of $[0,1]^2$. Is the singleton $\{0\}$ open in itself?

答:
You can prove the following:

If $A$ is open in $X$ and $B$ is open in $Y$, then $A \times B$ is open in $X \times Y$.

It follows that $(0,1)\times \{0\}$ is open in $[0,1]\times \{0\}$.

To see that this set is not open in $[0,1]^2$, note that any open ball inside $[0,1]^2$
centered at $(1/2,0)$ has points that are not in $[0,1]\times \{0\}.$


问:
What is the use of Spectral Theorem?
Obviously the version for compact and self-adjoint linear operators on Hilbert Spaces is very useful since it decomposes the operators into orthogonal projections.

However, the following more general version for commutative $C^*$ subalgebras of $\mathcal{L}(\mathcal{H})$ do not seem so handy to me.

Let $\mathcal{A}\in\mathcal{L}(\mathcal{H})$ be a $C^*$ subalgebra containing $I$ and $\Sigma$ be its spectrum. Then for $u,v\in \mathcal{H}$, $f\mapsto \langle T_f u,v\rangle$ is a bounded linear functional on $C(\Sigma)$, where $T_f$ is the operator mapped to $f$ by Gelfand transformation, hence defines a measure $\mu_{u,v}$ on $\Sigma$.

Then there is a regular projection-valued measure $P$ on $\Sigma$ such that $T=\int \hat{T}dP$  for all $T\in\mathcal{A}$ and $T_{f}=\int fdP$ for all $f\in B(\Sigma)$, where $\hat{T}$ is the Gelfand transformation of $T$ and $T_{f}$ is defined by $\langle T_{f}u,v\rangle=\int fd\mu_{u,v}$. Moreover, if $S\in\mathcal{L}(\mathcal{H})$, the followings are equivalent:

i.$S$ commutes with every $T\in\mathcal{A}$.

ii. $S$ commmutes with $P(E)$ for every Borel $E$

iii. $S$ commutes with $\int f dP$ for every $f\in B(\Sigma)$.

Although it is also somewhat decomposition, but they involve integrals and regular measures that exist, but are not constructed in a step-by-step fashion like in the simpler version and about which we do not actually know much.

Thus I wonder how useful this theorem is and it would be great if some examples can be provided.

Thanks!

答:
I like the application where one uses the spectral theorem to show that any irreducible unitary representation of an abelian group $G$, $\pi:G\to U(H)$, must be $1$ dimensional. I guess you would look at the spectrum of any fixed $\pi(x_0)$ and argue that if it had more than one point, you could use $P$ to find non-trivial projections which commute with every $\pi(x)$ (thus giving invariant subspaces). This last part uses the equivalence stated in the theorem.
I don't know how to prove this without using the spectral theorem.

Also, the $L^2$-ergodic theorem has a neat proof using spectral theory. You can show $\frac{1}{n}(x+Ux+\dots U^{n-1}x)$ converges by looking at it as an integral on the spectrum of $U$ and using the dominated convergence theorem.

Title: What is the use of Spectral Theorem?


问:
Alternative way to write combinatory probabilities
Suppose we have a bag of $20$ balls, each ball is uniquely numbered from $1$ to $20$.

Say that we choose any $5$ unique numbers at random between $1$ and $20$ inclusive, before any ball has been drawn from the bag.

$5$ Balls are then drawn without replacement.

I know the probability of all $5$ being correctly guessed can be expressed as:

$$5/20 \times 4/19 \times 3/18 \times 2/17 \times 1/16$$

I also know the probability of correctly guessing $4$ balls out of the $5$ is:

$$\frac{C(5, 4) \times C(20-5, 1)}{C(20, 5)},$$

which is the number of combinations of $4$ numbers in every $5$ ball draw, over the total.

What I'm looking for in particular: is there a way to express $P(4$ balls guessed out of $5)$ without using combinations as I have done just above this?

Every topic I've found on the subject only shows answers using combinations.

答:
A mild variant of your "all right" calculation will do it. Imagine drawing the balls one at a time. Write Y if the ball you get matches one of your numbers, and N if it doesn't. Then we can get exactly $4$ right in $5$ ways: (i) NYYYY, (ii) YNYYY, (iii) YYNYY, (iv) YYYNY, or (v) YYYYN.

Let us calculate  the probability of NYYYY. The probability the first is a N is $\frac{15}{20}$. Given that the first was a N, the probability the second is a Y is $\frac{5}{19}$. Given the first was a N and the second was a Y, the probability the third is a Y is $\frac{4}{18}$. and so on. So the probability of NYYYY is
$$\frac{15}{20}\cdot\frac{5}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}.$$

Now we find the probability of YNYYY. There is an easy way to do it (all sequences of $5$ balls are equally likely, so the answer will be the same as for NYYYY). Or there is the hard way, computing just as for NYYYY. If we do that, we get
$$\frac{5}{20}\cdot\frac{15}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}.$$
This is the same number as for NYYYY.

The other $3$ cases each give the same probability. So the probability of $4$ right and $1$ wrong is
$$(5)\left(\frac{15}{20}\cdot\frac{5}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}\right).$$


问:
Connection between Trace of $A^TA$ and eigenvalues of $A$
I am visiting a course about linear algebra. Our professor proved the equality between the Frobeniusnorm and the Schatten-2-norm. For this he used the following equality for a real matrix A.

$\sqrt{(\lambda_1^2+...+\lambda_n^2)}=\sqrt{tr(A^TA)}$

where $\lambda_1, ..., \lambda_n$ are the eigenvalues of $A^TA$ and $tr(A^TA)$ denotes the trace of $A^TA$. May someone please explain, why this equality holds?

Thank you very much!

答:
What You do have though is $$\sqrt{tr(A^{\perp}A)}=\sqrt{\sum_{j=1}^n\sigma_j²}$$
where $\sigma_j,j=1,...,n$ are the singular values of $A$, i.e. the eigenvalues of $\sqrt{A^{\perp}A}$, $\sigma_j^2$ are the eigenvalues of $A^{\perp}A$ and the right hand side is the $2$-Schatten norm of $A$ by definition. Now choose the standard basis $e_j,j=1,...,n$ and calculate:
$$tr(A^{\perp}A)=\sum_{j=1}^n\langle A^{\perp}Ae_j,e_j\rangle=\sum_{j=1}^n\langle Ae_jAe_j\rangle=\sum_{j=1}^n|Ae_j|^2=$$
$$\sum_{j=1}^n\sum_{k=1}^n|\langle Ae_j,e_k\rangle|^2=\sum_{j=1}^n\sum_{k=1}^n|a_{jk}|^2$$
using Parseval's equality and the right hand side is the square of the Frobenius-norm $||.||_F$ of the matrix $A$. If $||.||_2$ denotes the $2$-Schatten-norm this shows:
$$||A||_2=||A||_F.$$
An analogous result holds for compact linear operators on an infinite dimensional Hilbert-space that belong to the Hilbert-Schmidt-class.


问:
Does the following set spans the real vector space $\mathbb{C} $
Does the set $\beta = \{2i-1,2i+3,4 \} $ spans the real vector space $\mathbb{C} $ ?

What I have done:

$span(2i-1,2i+3,4) = \mathbb{C} $

Let $a,b,c,d,e \in \mathbb{R}$ then

$a(2i-1)+b(2i+3)+4c = d+ie$

$-a+3b+ 4c + (2a+2b)i = d+ie $

$\implies$ $-a+3b+ 4c =d $ and  $2a+2b = e $

Which gives

$b = \dfrac{e}{2} - a$ and  $c = \dfrac{d}{4} + a - \dfrac{(3 e)}{8}$

As $b,c$ are dependent on $a$ as well. So, $span(2i-1,2i+3,4) \neq \mathbb{C} $

Is this solution correct ?

答:
Your working is good, up until the final line, where you draw a false conclusion. Dependence on $a$ signifies that the solution is not unique, not that there are no solutions. It just means that $a$ can be chosen arbitrarily, while $b$ and $c$ depend on the choice of $a$.

For example, we could make the executive decision that $a = 0$. From this arbitrary choice, we conclude that
\begin{align*}
b &= \frac{e}{2} \\
c &= \frac{d}{4} - \frac{3e}{8}.
\end{align*}
This gives us a linear combination of the three vectors to produce our arbitrary vector $d + ei$. Thus, every $d + ei$ lies in the span of our three vectors.

What the infinitely many choices implies is that the set of vectors is not linearly independent. This is to be expected: we have a two-dimensional space, so any three vector set must be linearly dependent.


问:
quadratic with complex solutions including a conjugate z within equation how to solve?
$z^2+3\bar{z}+2=0$
, this is my question I solved the real roots of -1 and -2 but am not sure how to compute these into a complex solution in wolfram alpha the solutions were1/2 (3 - i sqrt(35)) and 1/2 (3 + i sqrt(35)) which is where I am confused how to reach , thank you

答:
Hint

$z = x+iy$.

Express the equation as two equations, one representing the real portion and one representing the imaginary portion.  Then, you will have two equations in the two unknowns, $x$ and $y$.

Focus on the equation representing the imaginary portion first.  You should see two cases to explore.  Explore them separately.


问:
A Subset of Abelian group with order of elements either $1\ or\ even$ is not subgroup?
From Abstract Algebra By Gallian:

Let $G$ be an Abelian group and $H=\{x\in G | |x|=1\ or\ even\}$.Give an example to show that $H$ need not be a subgroup of $G$.

$e\in H,\ so H\neq \phi$

Suppose $a,b\neq e$ such that $a,b\in H$ since $ab=ba,o(ab)=lcm(o(a),o(b))$.But since $o(a)\ and\ o(b)$ both are even ,its lcm,$lcm(o(a),o(b))$ must also be even.So $ab \in H$

And since $a\ and\ a^{-1}$ have same order implies for $a\in H,a^{-1}\in H$.Which imply $H$ is always a subgroup.
So,why they are saying $H$ need not be a subgroup ?
Am I wrong somewhere?

Thanks in advance!

答: If $G$ is cyclic of order $6$, and $a$ generates $G$, what are the orders of $a$ and $a^2$?


问:
How does BDiff$(F)$ classify smooth $F$-bundles and why is it topologized with the Whitney $C^\infty$-topology?
We consider a smooth fiber bundle $p:E \rightarrow B$ with fiber $F$. I want to understand how such a fiber bundle is classified by a map $B\rightarrow $BDiff$(F)$. Namely, we should have a bijection
$$
[B,\text{BDiff}(F)] \rightarrow \{\text{smooth $F$-bundles }E\to B\}/\text{iso}
$$
given by the pullback of some universal bundle EDiff$(F)\to$BDiff$(F)$. However, i can't seem to figure out the details or locate them in the literature.

Furthermore, how does the topology of Diff$(F)$ come into play? I know that we topologize Diff$(F)$ with the Whitney $C^\infty$-topology, but why is it exactly this topology that makes BDiff$(F)$ classifying such $F$-bundles? Also, in the case where $F$ is not compact, we have two different non-equivalent versions of the Whitney $C^\infty$-topology on Diff$(F)$ namely the strong and weak one. Which one of these are used or does it matter?

Any help or helpful references would be appreciated.

答:
Since asking the question i have learned that the canonical reference for this is the very detailed book "The convenient setting of global analysis" by Kriegl and Michor. In this book, they introduce the theory of infinite dimensional smooth manifolds. Let me sketch the argument.

They work with convinient vector spaces which are locally convex vector space satisfying a condition called $c^\infty$-completeness and a notion of smooth mappings between them. It turns out that one has to introduce a new topology called the $c^\infty$-topology on the locally convex vector spaces, because in the locally convex topology, smooth mappings are not necessarily continuous. Then infinite dimensional manifolds are defined analogously to ordinary smooth manifolds using convenient vector spaces equipped with the $c^\infty$-topology instead of Euclidian space. The set $C^\infty(M,N)$ equipped with the Whitney $C^\infty$-topology becomes an infinite dimensional manifold, and $\text{Emb}(M,N)\subset C^\infty(M,N)$ and $\text{Diff}(F)\subset C^\infty(F,F)$ are submanifolds, and $\text{Diff}(F)$ is a Lie group.

Let $F$ be a closed smooth finite dimensional manifold, and let $\ell^2(\mathbb{N})$ denote the Hilbert space of sequences bounded in the norm $\lvert \lvert \cdot \rvert \rvert_2$. The Lie group $\text{Diff}(F)$ acts on the embedding space $\text{Emb}(F,\ell^2(\mathbb{N}))$ by composition and yields a principal $\text{Diff}(F)$-bundle
$$
\text{Emb}(F,\ell^2(\mathbb{N})) \rightarrow \text{Emb}(F,\ell^2(\mathbb{N}))/\text{Diff}(F).
$$
The total space is contractible which implies that the base space is homotopy equivalent to $\text{BDiff}(F)$. Denote the base space with $B_\infty(F)$. As they mention in the book, the base space can be thought of as the nonlinear version of the infinite Grassmannian. Consider the associated bundle
$$
\text{Emb}(F,\ell^2(\mathbb{N})) \times_{\text{Diff}(F)}F \rightarrow B_\infty(F),
$$
and denote its total space by $E_\infty(F)$. This associated bunde has fiber $F$ and is the universal $F$-bundle (Theorem 44.24 in the book). Since $B_\infty(F)\simeq B\text{Diff}(F)$, we get the wanted bijection in the question.


问:
Find all solutions $z\in \mathbb C$ for the following equation: $z^2 +3\bar{z} -2=0$
The question is

Find all solutions $z\in \mathbb C$ for the following equation: $z^2 +3\bar{z} -2=0$

I have attempted numerous methods of approaching this question, from trying to substitute $x+iy$ and $x-iy$ respectively, in addition to substituting $z^2$ for $z\bar z$, but with no luck. I would really appreciate if you were able to provide some direction so I know where to start. Thank you!

答:
Plugging in $x+iy$ for $z$ seems to be a good idea actually. You get
$$(x^2-y^2+3x-2)+i(2xy-3y)=0$$
For a complex number to be zero both the real and the imaginary part have to be zero, so we get $2xy=3y$ and $x^2-y^2+3x-2=0$. The first equation is easy for $y \neq 0$. Then we get $x=1.5$ and from there we can conclude $y= \pm \sqrt{4.75}$ with the second equation.

If $y=0$ the second equation is a quadratic in $x$ with two solutions that can easily be calculated with the quadratic formula. All in all we will have four solutions: $$(1.5, \sqrt{4.75}), (1.5, -\sqrt{4.75}), (..., 0), (..., 0)$$


问:
References for trivial extension of a ring by a module
I'm reading a note having this definition: Let $R$ be a ring amd $M$ an $R$-module. The trivial extension of $R$ by $M$, written by $R\ltimes M$, is the $R$-algebra formed by endowing the direct sum $R\oplus M$ with the multiplication:
$$(r,m)(s,n)=(rs,rn+sm).$$

Can you give me some references about this definition so that I can understand it better? This note also states some facts like if $R$ is Noetherian and $M$ is finitely generated then $R\ltimes M$ is Noetherian and has the same dimension as $R$. To prove these facts myself, I need to understand the ring structure on it, but I'm having some troubles understand its ideals and prime ideals.

答: A good reference is the survey "Idealization of a module" by Anderson and Winders, which you can find here (idealization is another name for trivial extension). The results you need are proved in Proposition 2.2 and Theorem 3.2.</p>


问:
Simplifying Factorials to solve for a variable
I was solving a problem in discrete math and i got the answer down to:

$$26^8 = (n-7)(n-7)!$$

I was wondering how you would solve this for n, if that's even possible. Or do you think I made a mistake getting to this point?

答:
Hint

Is it possible, if you are forcing $n$ to be a positive integer $\geq 7$?

$(26)^8$ is divisible by both $2$ and $13$, but is relatively prime to each of $3,5,7,11$.  Can a positive integer $m$, be found such that $m!$ satisfies these constraints?


问:
The validity of these two proofs (real analysis).
This problem is from An Introduction to Proof through Real Analysis by D. J. Madden and J. A. Aubrey.

Prove that for all natural numbers $n$, $n^2 \geq n$.

I propose two different proofs.

Proof 1. This proof is induction on $n$.

We let $n = 1$, then $n^2 = 1 \cdot 1 = 1 = n$. So the statement is true for $n = 1$.

Now we let $n = n_0$, and assume that the statement holds. So $n_0^2 \geq n_0$.

Let $n = n_0 + 1$, then we have
\begin{align}
(n_0 + 1)^2 = n_0^2 + 2n_0 + 1 &\geq n_0 + 1 \\
n_0^2 + 2n_0 &\geq n_0 \tag{cancellation of addition} \\
n_0^2 + n_0 &\geq 0 \tag{cancellation of addition}.
\end{align}
Which is true. So the statement holds for $n_0 + 1$. Hence the statement is true for all natural numbers.

Proof 2. We use the same information and verify that it works for $n = 1$. Now for $n > 1$, we can express $n$ as an eventual successor of $1$, which is
$$n = ((( \cdots (1 + 1)+1)+1)$$
By the definition of multiplication,

$n = ((( \cdots (1 + 1) + 1) + 1) + 1).$

$n \cdot m = ((( \cdots (m + m) + m) + m) + m).$

So $n^2 \equiv n \cdot n$ can be written as
\begin{align}
n \cdot n &= ((( \cdots (n + n) + n) + n) + n) \\
&=  ((( \cdots (n + ((( \cdots (1 + 1) + 1) + 1) + 1)) + n) + n) + n) \\
&= \vdots
\end{align}
which is obviously larger than $n$, because $n^2$ can be written as a successor of $n$.
Hence, we claim that for $n > 1$, $n^2 > n$.
The logical union of these two cases is $n^2 \geq n$.

Are my proofs valid? If they aren't, which parts do not make sense? Thank you.

答:
Your ideas appear to be correct, but the way you have written up the proofs make them rather non-understandable, and you need to fix your presentation and logic.

In proof 1, the idea of induction is: to prove $P(n)$ for all positive integers $n$, we show $P(1)$, and then show $P(n)\implies P(n+1)$ for all $n$. You have shown $P(1)$ correctly, very nice. But your logic in showing that $P(n)\implies P(n+1)$ is faulty. In particular, $(n_0+1)^2\geq n_0+1$ is what you need to show, but you have started with that statement without explanation, so your proof is not correct. If you instead note that each line is an equivalent statement, that is,
$$\begin{split}&(n_0+1)^2\geq n_0+1\\\iff &n_0^2+2n_0\geq n_0\\\iff& n_0^2+n_0\geq0 \end{split}$$
and then observe the last line is true, then you will in fact have a correct proof, becuse then the chain of logical implications goes back up: starting from a true statement you deduce back the true first statement.

But realise that in your entire proof, you did not ever use the inductive hypothesis that $n_0^2\geq n_0$. So it is not even an induction proof at all. In reality your proof $1$ should be written more succintly as follows:
$$n^2\geq n\iff n^2-n\geq 0\iff n(n-1)\geq0$$ which is true since $n\geq 1$, so both $n$ and $n-1$ are $\geq 0$.

In proof 2, you might have the right idea, but writing an integer as the sum of a bunch of $1$s only obfuscates whatever you are trying to say. There is literally zero point in doing this. What you are really trying to say is that for $n\geq 2$, $$n^2=n\cdot n\geq 2n>n.$$

In conclusion, both proofs might have roughly the right idea, but the way you wrote it makes the ideas difficult if not impossible to discern. So I would say your proofs are both incorrect.


问:
Conjugate elements of $SO(3)$ group
Composition of two rotations in 3d space yields another rotation $$R_1 R_2 = R_3, $$
and I can understand this by help of some figures in my book.

So, the rotations in 3d space forms group. Then in my book, the author says that for two rotations $R$ and $R'$ with same angle but different rotation axis, they are connected by
  $$ R' = QRQ^{-1}.$$
  Where $Q$ rotates the rotation axis of $R$ into $R'$. How do we prove this?

Graphically, I can understand this. $Q$ just changes rotation axis $R$ into $R'$, so after all steps, the rotation axis comes to initial position. Next, since $R$ and $R'$ has the same rotation angle, any rigid body is rotated same angle. So I think $R'$ and $QRQ^{-1}$ would give same result.

But How do I prove this by mathematically?

答:
For me, the easiest way to prove this is to use matrices and eigenvalues.
Rotations around an axis are  exactly  3x3 real matrix satisfying $R^T=R^{-1}$ and $\det R=1$.

To see that, note that a rotation transform the standard basis in a new one, keeping vector ortogonal and with length=1, so $R^T=R^{-1}$, and keeps to orientation of the basis, so $det R=1$.

Conversely, if $R$ is a 3x3 real matrix such that $R^T=R^{-1}$, then they eigenvalues (real or complex) have always modulus 1  (complex eigenvalues must be conjugate pairs since $R$  is real).

Therefore, for 3x3 real matrices with $R^T=R^{-1}$ and $\det R=1$ the eigenvalues must be $a+ib$,$a-ib$, 1 with $a^2+b^2=1$ (the case a=1 is the identity, all eigenvalues are 1,  the case $a=-1$ the eigenvalues are -1,-1,1).

Then, there is a basis of eigenvectors of $R$, $v,u,w$ associated to eigenvalues  $a+ib$,$a-ib$, 1.
If $b\not=0$ (a pair of non real eigenvalues), you can take $v^*=\frac{u+v}{2}$,
$u^*=\frac{u-v}{2}$ and check that they satisfied $Rv^*=av^*-bu^*$ and $Ru^*=av^*+bu^*$, so the matrix in the basis $v^*,u^*,w$ is
$$Q=\begin{bmatrix}a&-b&0\\
b&a&0\\
0&0&1
\end{bmatrix}$$
which is a rotation of angle $\theta$, $\cos\theta=a$, around the axis $w$ (you can pick $v^*,u^*,w$ of norm=1).

Indeed, taking $P$ the matrix with columns $v^*,u^*,w$ (or $v^*,u^*,-w$ to get a basis with an correct orientation) you prove indeed that  $R=P^{-1}QP$, with $P$ again a rotation ...

Title: Conjugate elements of $SO(3)$ group


问:
a well-defined algebra isomorphism
How to prove that this map is well-defined, surjective and 1-1:

$$1\otimes a+i\otimes b+j\otimes c+ij\otimes d \mapsto a \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix} + b \begin{pmatrix}
i & 0\\
0 & -i
\end{pmatrix} + c \begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix} + d \begin{pmatrix}
0 & i\\
i & 0
\end{pmatrix}$$

Where $a,b,c,d$ are complex numbers and this map gives the isomorphism $\Bbb{H\otimes_R C }\to M_2(\mathbb C)$

Another question: Why a map that maps a basis to a basis must be an isomorphism?is there a proof or a theorem that say this? Should I use that theorem here?

答:
First, you ask: "Why a map that maps a basis to a basis must be an isomorphism?" Well, remember these are vector spaces over $\mathbb C$ in particular. So the correct statement is that a linear map that maps a basis to a basis is a (linear) isomorphism. This does not by itself guarantee that they are isomorphic as rings too. Okay, on to the problem.

Well-defined

$\mathbb H$ is an $\mathbb R$-algebra with basis $\{1, i, j, ij\}$. We then know by "extension of scalars" that $\mathbb H\otimes_{\mathbb R} \mathbb C$ is a $\mathbb C$-algebra with basis $\{1\otimes 1, i\otimes 1, j\otimes 1, ij\otimes 1\}$. But that's the same as saying that an $x\in \mathbb H\otimes_{\mathbb R} \mathbb C$ can be written uniquely as
$$
x = 1\otimes a+i\otimes b+ j\otimes c+ ij\otimes d
$$
with $a,b,c,d\in \mathbb C$, so your map $\varphi:\mathbb H\otimes_{\mathbb R} \mathbb C \to M_2(\mathbb C)$ is well-defined.

Homomorphism of algebras

It is easy to see that $\varphi$ is linear ($a,b,c,d$ are scalars). What remains is to show $\varphi(1)=1$ (which is evident) and $\varphi(xy)=\varphi(x)\varphi(y)$. Using linearity of $\varphi$, we only need to check this on base vectors. For example:
$$
\varphi(i\otimes 1)\varphi(j\otimes1)
= \begin{pmatrix} i&0\\0&-i\end{pmatrix} \begin{pmatrix} 0&1\\-1&0\end{pmatrix}
= \begin{pmatrix} 0&i\\i&0\end{pmatrix}
= \varphi(ij\otimes 1)
$$
and so on. (This amounts to showing that the matrices generate the quaternion group.) Alternatively, we could take the product of two general elements and do it all in one go without appealing to linearity.

Bijective

Now we just need to show $\varphi$ is bijective. Since $\varphi$ is a linear map, we only need to show that the four matrices are a basis of $M_2(\mathbb C)$. It's obvious that another basis for $M_2(\mathbb C)$ is
$$
\mathscr B = \left\{ \begin{pmatrix} 1&0\\0&0\end{pmatrix},
\begin{pmatrix} 0&1\\0&0\end{pmatrix},
\begin{pmatrix} 0&0\\1&0\end{pmatrix},
\begin{pmatrix} 0&0\\0&1\end{pmatrix} \right\}
$$
In particular $M_2(\mathbb C)$ is $4$-dimensional over $\mathbb C$. And now we can prove linear indepence by writing our matrices as (column) vectors over $\mathscr B$ and take the determinat:
$$
\det \begin{pmatrix} 1&i&0&0\\0&0&1&i\\0&0&-1&i\\1&-i&0&0 \end{pmatrix}
= 4 \ne 0
$$
Done! We conclude that $\varphi$ maps a basis to a basis, so it is bijective and thus an isomorphism.

Title: a well-defined algebra isomorphism


问:
probability: Relays used in the construction of electric circuits function properly with probability .9

Question: Consider circuit B. If we know that current is flowing, what is the
probability that switches 1 and 4 are functioning properly?

First I have calculated the probability of current flowing in B =0.9639 (it's correct).

The sample answer is below

P （R1 ∩ R4|B）= P （R1 ∩ R4 ∩ B）/P（B).
（R1 ∩ R4 ∩ B） = (R1 ∩ R4 )∩(R1 ∩ R3 )∪(R2 ∩ R4 ) = ( R1 ∩ R4 ∩ R3） ∪ （R2 ∩ R4 ）.
The probability of the above event is .9^3 + .9^2- .9^4 = 0.8829. So, the conditional probability in question is .8829/.9639 = 0.916.

I am not very sure how to derive the bold parts, and I think it should be ( R1 ∩ R4 ∩ R3） ∪ （R1∩R2 ∩ R4 ） (distributive law)

Can I get some help here? Appreciate it!

答:
You calculated the probability of circuit working correctly

$P(B) = 0.81 + 0.81 - 0.81^2 = 0.9639 \ ,$ where $B$ is the event of circuit working correctly.

We are now trying to find $P(A | B)$ and for that we need to
find $P(A \cap B) \ \big(= P(B | A) \cdot P(A) \ \big)$ where $A = R_1 \cap R_4$.

$P(A) = 0.9^2$

If $R_1$ and $R_4$ function properly, the circuit will function properly so long as, at least one of $R_2$ and $R_3$ function properly.

$P(B|A) = P(R_2 \cup R_3) = 1 - P(R'_2 \cap R'_3) = 1 - (1-0.9)^2 = 0.99$

$P(A \cap B) = 0.9^2 \times 0.99 = 0.8019$

So $P(A | B) = \displaystyle \frac{0.8019}{0.9639}$


问:
Showing that the matrix transformation $T(f) = x*f'(x)+f''(x)$ is linear
I want to show that the following matrix transformation is linear.

$T(f) = x*f'(x)+f''(x)$

I know I have to show that $T(f+g) = T(f) + T(g)$ but I don't understand what $T(f+g)$ will look like.

Is $T(f+g) = x*(f'(x)+g'(x))+(f''(x)+(g''(x))$ the correct interpretation?

答:
By the definition of $T$, you have for every function $h$ in your domain of $T$, that
$$ T(h) = x\cdot h'(x) + h''(x) $$
If now $h = f+g$, we get
$$ T(f+g) = x \cdot (f+g)'(x) + (f+g)''(x), $$
which equals your calculated result, which moreover used the linearity of differentiation.</p>


问:
Factorising a principal ideal where its norm is $p^k$
I was considering the following example:

Let $K=\mathbb{Q}(\sqrt{-29})$ and so $O_K=\mathbb{Z}[\sqrt{-29}].$ So by Dedekind's theorem, I wanted to decompose the principal ideal $(5)$ and I obtained $(5)=(5,\sqrt{-29}-1)(5,\sqrt{-29}+1):=P_1P_2 $ and that both prime ideals have norm $5$.

Then my book said

"Consider the ideal $(3+2\sqrt{-29})$, this has norm $125$ and so must be either $P_1^3, P_1^2P_2, P_1P_2^2$ or $P_2^3.$"

Why does this follow? I know that we can prime decompose this ideal and that $P_1$ and $P_2$ are the only ideals with norm $5$. But why did we rule out the possibility that $(3+2\sqrt{-29})$ is already prime and the possibility that $(3+2\sqrt{-29})=QP_1$ where $Q$ is a prime ideal with norm $25$, say?

I am thinking I must have missed some small details, many thanks in advance for lending a hand!

答:
Note that the norm of an ideal is defined by the size of $O_K/I$ for $I$ an ideal. Thus by Lagrange's Theorem, we must have $N(I).x\in I$ for each $x\in O_K$, where $N(I)$ is the norm of the ideal $I.$

Now, this implies that $N(I)\in I$ for each ideal $I.$ Thus we have $I|(N(I)).$

So apply to this case, we have $(3+2\sqrt{-29})|(125)=(5)(5)(5)=P_1^3P_2^3.$ Now, by considering the norm on the ideal $(3+2\sqrt{-29})$, we deduce it must have the required form.


问:
What is regularity of solution of Dirichlet problem with Dirac distribution as boundary data?
I was thinking if we have Dirac distribution as boundary condition, then what will be regularity of solution. Problem is following,
$$
\left\{\begin{matrix}
\nabla_x(\gamma(x)\nabla_x u(x,y))=0 & in~\Omega\\
u(x,y)=\delta_y(x) & on ~\partial \Omega
\end{matrix}\right.
$$
where $x,y\in \mathbb R^n , \gamma(x)>c>0 $. $\Omega$ is with smooth boundary.

Any help or  reference is greatly appreciated.

答:
For a regular enough boundary function $\psi$ the solution of the first BVP can be written as
$$
u(x)=\int_{\partial \Omega}\frac{\partial G(x,y)}{\partial \bar n_y}\gamma(y)\psi(y)\,dy,
$$
where $G(x,y)$ is the Green function of the corresponding first BVP and $\bar n_y$ is a unit normal vector to $\partial\Omega$. It follows from the Green's third identity written out for your operator.

Substituting $\psi(x)=\delta_y(x)$ gives $u(x)=\gamma(y)\frac{\partial G(x,y)}{\partial n_y}$. In some cases this function  is also called the Poisson kernel. Estimates for the Green function are known (for smooth enough coefficients and bounded domains they are the same as for a fundamental solution) and it follows that
$$
|u(x,y)|\le C|x-y|^{1-n} \quad x\in \Omega,\ y\in \partial \Omega.
$$


问:
Proving $\frac{1}{2}+...+\frac{1}{n} \leq \ln(n)$ by induction
I'm currently learning about mathematical induction right now and I'm stuck on this question:

Use induction to prove
$\frac{1}{2}+...+\frac{1}{n} \leq \ln(n)$

I know $n=2$ is true, then for $n=k$, $\frac{1}{2}+...+\frac{1}{k} \leq \ln(k)$

Now for $n=k+1$,

$\frac{1}{2}+...+\frac{1}{k}+\frac{1}{k+1} \leq \ln(k)+\frac{1}{k+1}$

This is where I am unsure how to proceed. Should subtract $\ln(k)$ from both sides as $\ln(k)>0, k>1$?

答:
Then note that,

$$\ln k+ \frac{1}{k+1} ≤\ln(k+1).$$

$\large{\underline{\text{Proof:}}}$

$$\frac {1}{k+1}≤\ln \left(\frac{k+1}{k}\right)$$

Let  $\dfrac{k+1}{k}=1+\dfrac 1k=e^t, t>0$

$$k+1-ke^t=0 \Longrightarrow k(1-e^t)=-1$$

$$k(e^t-1)=1 \Longrightarrow  k=\frac{1}{e^t-1}$$

$$\frac{1}{k+1}=\frac{1}{\frac{1}{e^t-1}+1}$$

$$\frac{e^t-1}{e^t}≤t$$

$$f(t)=te^t-e^t+1≥0$$

$f(t)$ is the strictly increasing function for $∀t>0$, because $f'(t)=te^t>0, ∀ t>0$ and if $t=0$, then $f(t)=0$. Therefore, for all $t>0$, we have

$$\begin{align}&f(t)=te^t-e^t+1≥0 \\ &\Longleftrightarrow
\ln k+ \frac{1}{k+1} ≤\ln(k+1).\end{align}$$


答: You want to prove that$$\frac12+\frac13+\cdots+\frac1k+\frac1{k+1}\leqslant\log(k+1)\tag1$$assuming that$$\frac12+\frac13+\cdots+\frac1k\leqslant\log k.\tag2$$Note that, in order to prove $(1)$, it is enough to prove that $\log(k)+\frac1{k+1}\leqslant\log(k+1)$, since you are assuming $(2)$. But\begin{align}\log(k)+\frac1{k+1}\leqslant\log(k+1)&\iff\frac1{k+1}\leqslant\log(k+1)-\log(k)\\&\iff\frac1{k+1}\leqslant\int_k^{k+1}\frac{\mathrm dx}x,\end{align}which is true, since$$(\forall x\in[k,k+1]):\frac1{k+1}\leqslant\frac1x.$$

Title: Proving $\frac{1}{2}+...+\frac{1}{n} \leq \ln(n)$ by induction

Body:
I'm currently learning about mathematical induction right now and I'm stuck on this question:

Use induction to prove
$\frac{1}{2}+...+\frac{1}{n} \leq \ln(n)$

I know $n=2$ is true, then for $n=k$, $\frac{1}{2}+...+\frac{1}{k} \leq \ln(k)$

Now for $n=k+1$,

$\frac{1}{2}+...+\frac{1}{k}+\frac{1}{k+1} \leq \ln(k)+\frac{1}{k+1}$

This is where I am unsure how to proceed. Should subtract $\ln(k)$ from both sides as $\ln(k)>0, k>1$?


问:
Two succeeding integers in $\left(\mathbb{Z}/n\mathbb{Z}\right)^{*}$ for an odd n, and the Jacobi symbol of the latter one
Given an odd integer $n$, I want to find out if there exists two succeeding integers, $1\leq m-1<m\leq n-1$ s.t both are invertible (i.e $m,m-1\in \left(\mathbb{Z}/n\mathbb{Z}\right)^{*}$) and also $\left(\frac{m}{n}\right)=1$ when $\left(\frac{m}{n}\right)$ is the Jacobi symbol.

For example: if i take $n=9$ and $m=8\in \left(\mathbb{Z}/n\mathbb{Z}\right)^{*}$ then i get $m-1=7\in \left(\mathbb{Z}/n\mathbb{Z}\right)^{*}$ and also $\left(\frac{m}{n}\right)=1$ which is exactly what i want.

Is there a criterion for a given odd $n$, to find such an $m$, or even just to know that it exists?

Thanks!

答: It might not be exactly what you want, but since 1 always has Jacobi symbol 1, we have a solution with $m=2$ whenever 2 has Jacoby symbol 1, i.e. when $n$ is 1 or 7 modulo 8.</p>


答:
Use the definition of $\ln(n)$ instead. To be clear here: Let $k'$ be an integer. Then $\ln (k')$ is as defined as follows:
$$\ln (k') =  \int_1^{k'} \frac{dx}{x}.$$

So from this definition, for all positive integers $k$:
$$\ln (k+1) = \ln (k) + \int_k^{k+1} \frac{dx}{x}  \ \ge \ \ln (k) + \frac{1}{k+1}.$$ [The last inequality following because $\frac{1}{x} \ge \frac{1}{k+1}$ for all $x \in [k,k+1]$.] Now are you able to plug this in to your inductive hypothesis to finish.