问:
Minimum of this function attained on a particular set
Have to minimize $f(x, y, z) = (x-1)^2 + (\frac{y}{x} -1)^2+(\frac{z}{y} -1)^2+(\frac{4}{z} -1)^2$  on the set $\{(x, y, z) : 1 \le x \le y \le z \le 4\}$.

Lagrange multiplier needs to be used but I'm quite lost on the particular set mentioned. I would appreciate some help.

答:
To apply Lagrange multipliers, you

first forget the bounds about $x,y,z$.

change variables to $(u_1,u_2,u_3,u_4) = \left(x,\frac{y}{x},\frac{z}{y},\frac{4}{z}\right)$

"try" to minimize $\sum\limits_{k=1}^4 (u_k-1)^2$ subject to constraint $\prod\limits_{k=1}^4 u_k = 4$.

The Lagrange multipliers will give you something like

$$u_k - 1 = \frac{\lambda}{u_k} \prod_{j=1}^4 u_j\quad\text{ for some constant }\lambda$$
which is equivalent to
$$u_1(u_1 - 1) = u_2(u_2 - 1) = u_3(u_3 - 1) = u_4(u_4 - 1)$$
This in turn implie $u_2, u_3, u_4 \in (u_1,1-u_1)$.
Since $u_1 = x$ is supposed to be $\in [1,4]$, $1 - u_1$ will be non-positive and this rules out the possibility that $u_2, u_3, u_4 = 1 - u_1$. As a result, the methods of Lagrange multipliers suggests the minimum may be located at a configuration where $u_1 = u_2 = u_3 = u_4$. Since $\prod\limits_{k=1}^4 u_k = 4$, the common value of $u_k$ should be $\sqrt{2}$.

In general, when you apply Lagrange multipliers to minimize something, there
are usually many loose ends in the logic. To verify above ansatz "all $u_k = \sqrt{2}$" indeed give us the true minimum, we apply AM $\ge$ GM to the 4 numbers $u_k$, we obtain

$$\sum_{k=1}^4 u_k \ge 4 \left(\prod_{k=1}^4 u_k\right)^{\frac14} = 4\sqrt[4]{4} = 4\sqrt{2}$$

In terms of $u_k$, we have

$$\begin{align}
f(x,y,z) = & \sum_{k=1}^4 (u_k - 1)^2\\
= & \sum_{k=1}^4 \left((\sqrt{2}-1)^2 + (u_k - \sqrt{2})^2 + 2(\sqrt{2}-1)(u_k-\sqrt{2})\right)\\
= & 4(\sqrt{2}-1)^2 +  \sum_{k=1}^4 (u_k-\sqrt{2})^2
  + 2(\sqrt{2}-1)\left(\sum_{k=1}^4 u_k - 4\sqrt{2}\right)\\
\ge & 4(\sqrt{2}-1)^2
\end{align}
$$
This means $f(x,y,z)$ is bounded from below by $4(\sqrt{2}-1)^2$.
Since this lower bound is attained at $(x,y,z) = (\sqrt{2},2,2\sqrt{2})$ which belongs to the given set. $f(x,y,z)$ does attain its minimum at a configuration where $u_1 = u_2 = u_3 = u_4 = \sqrt{2}$ with value equals to $4(\sqrt{2}-1)^2$.


问:
Finding two numbers when having their sum and product
I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad  ab=238,\quad  238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance

答:
Hint: Write $$238=2\cdot 7\cdot 17$$

What options for $a$ and $b$ does that leave you? Which of them satisfy $a+b=41$?

ContentLicense: CC BY-SA 3.0


问:
Is $|X|$ always a random variable given that $X$ is one?
Is it true that $|X|$ is always a random variable given that $X$ is a random variable?

答:
Roughly saying, if you consider a variable $X$ such that $$\Pr\{X=x\}=\begin{cases}1&,\quad x=y\\0&,\quad x\ne y\end{cases}$$
then yes, the statement is true, unless counterexamples such as
$$
\Pr\{X=1\}=\Pr\{X=-1\}=0.5
$$
exist.

ContentLicense: CC BY-SA 4.0


问:
ask a triangle inequality in complex numbers
If $|z|=2$, and $|w|=3$, then $|z \times w|=6$ and $|z+w|\leq5$ and $|z/w|=\frac{2}{3}$ and $|z-w|\leq5$

Can anyone explain why the last holds? I don't understand why $|z-w|\leq5$

答:
The triangle inequality states that
$$ |a + b| \leq |a| + |b| $$

So, plug $a=z,b=-w$ and we get

$$ |z + (-w) | \leq |z| + |-w| $$

We know that $|xy| = |x| \cdot |y|$, and $|z|=2,|w|=3$. So

$$ |z-w| \leq 3 + 2 \\ |z-w| \leq 5$$


问:
What's the bijection between scalar/inner products and (certain) almost complex structures (on $\mathbb R^2$)?
Asked on maths overflow here.

What's the bijection between (equivalence classes of) scalar products (I guess 'scalar product' is the same as 'inner product') and a.c.s. (almost complex structure/s) on $\mathbb R^2$?

From Example 1.2.12 of Daniel Huybrechts - Complex Geometry An Introduction.

Assumptions and notation:

I just pretend $V = \mathbb R^2$ literally instead of just an isomorphism.

Let $\Phi(V)$ be the set of real symmetric positive definite $2 \times 2$ matrices. This set is in bijection with inner products on $V$, I believe. We have according to this,

$$\Phi(V) = \{\begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \ | \ h+g, hg-f^2 > 0 \}_{h,f,g \in \mathbb R}$$

Let $\Gamma(V)$ be the (matrix representations of) a.c.s. on $V$. We have, according to this,

$$\{\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix}\}_{a,b \in \mathbb R, b \ne 0}=: \Gamma(V) \subseteq Auto_{\mathbb R}(V) \subseteq End_{\mathbb R}(V)$$

I understand that the 'rotation' matrices in $V$ are $SO(2) := \{R(\theta) := \begin{bmatrix}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{bmatrix}\}_{\theta \in \mathbb R}$, though I'm not sure that Huybrechts has the same usage of the term 'rotation'. (I ask about this later.)

Questions:

A. For injectivity (except for the equivalence class):

Given (equivalence class of) scalar product ($[M]$ of) $M$, choose unique $I$ that assigns $v$ to the one described. I'll call this map $\gamma: \Phi(V) \to \Gamma(V)$, $\gamma(M)=I$. (Later, $\tilde \gamma: \frac{\Phi(V)}{\tilde{}} \to \Gamma(V)$, $\tilde \gamma([M])=I$.)

It's 'rotation by $\pi/2$' or something. In what way? For $M=I_2$ (2x2 identity), then $I$ is indeed 'rotation by $\pi/2$', in the sense that it's $\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix} \in SO(2) \cap \gamma(V)$, which is the '$R(\theta)$', for $\theta = \pi/2$.

What exactly is the formula for $I=\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix} \in \Gamma(V)$ given $M = \begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \in \Phi(V)$?

I'm asking because

2a - I would exceed wolfram computation time

2b - I notice for a different $M$ I tried, $I$ isn't a 'rotation matrix' in the sense of $SO(2)$. In fact, I believe the only 'rotation' matrices that are also a.c.s. are $\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$, i.e. $SO(2) \cap \gamma(V) = \{\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}\}$. However, I think $I$ kind of 'rotates by $\pi/2$' in some other sense.

2c - I think $SO(2) \cap \gamma(V)$ isn't meant to be the image of $\gamma$

B. For surjectivity:

I'll call whatever map we would have as $\phi: \Gamma(V) \to \Phi(V)$, $\phi(I)=M$

Given an a.c.s. $I$, what are some possible scalar products $M$?

There's a comment that goes choosing the unique $M_v$ such that for some $v \in V \setminus 0$, we have $\{v,I(v)\}$ as an orthonormal basis. I tried this out (long to type!), and the only thing missing was the positively oriented. I guess either $\{v,I(v)\}$ or $\{v,-I(v)\}$ is positively oriented though. So I'll let $M_v$/$N_v \in \Phi(V)$ correspond to $\{v,I(v)\}$/$\{v,-I(v)\}$. Then by fixing $v$ (I ask about non-fixing of $v$ later), we have $\phi(I)=M_v$ or $N_v$, whichever corresponds to positively oriented basis. I'll just call this $\phi(I)=L_v$ Is this right?

Is $\phi$ supposedly an inverse (or right inverse or left inverse or whatever) to $\gamma$ (or $\tilde \gamma$ or whatever), in the sense that $\gamma(\phi(I)) = I$ for all (a.c.s.) $I \in \Gamma(V)$?

This whole thing about the $v$ makes me think there's another equivalence relation going on here. Is there?

This seems like we can have maps parametrised by the nonzero $v$, namely $\phi_v: \Gamma(V) \to \Phi(V)$. In this case, we might investigate if $\phi_v(I)=L_v=L_w=\phi_w(I)$ or at least if $[L_v]=[L_w]$ under the old equivalence relation of positive scalar $\lambda$, i.e. $L_v = \lambda L_w$. If this investigation turns out negative, then I think there's some problem like if 2 inner products are equivalent if they are from the same a.c.s. $I$ under $\phi_{\cdot}$, but for possibly different $v$ and $w$, then I think the equivalence class of $L_v$ under this new relation, which is $\{L_w\}_{w \ne 0}$, might not be the same as the equivalence class of $L_v$ under the old relation, which is $\{\lambda L_v\}_{\lambda > 0}$.

Ideas:

Perhaps there's some matrix thing here about how scalar products are in bijection with positive definite symmetric matrices and then almost complex structures are rotation matrices or something that are square roots of $-I_2$. Like given pos def symmetric $B$, there exists unique a.c.s. $J$ such that (something something).

Perhaps this is related, but I'd rather not further analyse the question or read through the answer given that I've spent over a month on almost complex structures BEFORE we even put inner products on vector spaces. Please consider spoon-feeding me here.

答:
This answer addresses the "injectivity" questions you ask.

I interpret "rotation" as "orientation preserving, and also preserving of the inner proudct".  If your inner product is the standard one on $\mathbb{R}^2$, then these corresponds to rotation matrices as you've defined them.  However, in a different inner product, the rotation matrices look different.  In this iterpretation, saying $I$ is rotation by $\pi/2$ just means that $I$ preserves lengths (as computed in the weird inner product), and the angle between $v$ and $I(v)$ (as computed in the weird inner product) is $\pi/2$.

The matrix of $I$ is $\begin{bmatrix} -\frac{f}{\sqrt{gh-f^2}} & -\frac{g}{\sqrt{gh-f^2}} \\ \frac{h}{\sqrt{gh-f^2}}  & \frac{f}{\sqrt{gh-f^2}}\end{bmatrix}.$  I found this by setting $I\begin{bmatrix} 1\\0\end{bmatrix}= \alpha \begin{bmatrix} 1\\0\end{bmatrix} + \beta \begin{bmatrix} 0\\1\end{bmatrix}$ and then using the two equations $\left\| \begin{bmatrix} 1\\0\end{bmatrix}\right\| = \left\| I\begin{bmatrix} 1\\0\end{bmatrix}\right\|$ and $\left\langle \begin{bmatrix} 1\\0\end{bmatrix}, I\begin{bmatrix} 1\\0\end{bmatrix} \right\rangle = 0$ to solve for $\alpha$ and $\beta$.  It turns out there is a sign ambiguity which is resolved using the orientation.  I believe this answers 2a; I think 1. answers 2b and 2c.


问:
Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square .
After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$
. But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?

Did I do some mistake or is the question wrong?

答: An alternative way to approach the same result: For nonnegative values of $x$, the function $f(x)=2x^2+x$ is strictly increasing, so the only way to have $f(m)=f(n)$ is if $m=n$.  And then of course $m-n=0$ is a perfect square.


问:
proof verification on $ f_{n}$ uniformly continuity
Let $f_{n}:[0,1]\rightarrow \mathbb{R}$

$f_{n}=\begin{cases}
nx & \text{ if } 0\leq x\leq \frac{1}{n}\\
 1& \text{ if } \frac{1}{n}<x\leq 1
\end{cases}$

is $f_{n}\xrightarrow[]{uni}f$?

my solution:

for $x=0 ,f_{n}=0$

for $0<x\leq 1$ it exist a $n_{0}$ such that $\frac{1}{n_{0}}<x $ so $x>\frac{1}{n}$ for every $n \geq n_{0} $
$\Rightarrow f_{n}\rightarrow f(x)=1$

now i will check if it goes uniformly to $1$

$\left | f_{n}-1 \right |\geq nx-1$ for every $n\epsilon \mathbb{N},x\epsilon X$

$\left \| f_{n}-1\right \|>nx-2$ let $n\rightarrow +\infty $

$\left \| f_{n}-1\right \|\rightarrow +\infty $

I am not sure about the inequalities, is this ok? Can someone formally prove it?

答:
No, $(f_n)_{n \in \mathbb{N}}$ is actually a bounded sequence. Indeed, $0 \leq f_n(x) \leq 1$ for all $n \in \mathbb{N}$ and $x \in [0,1]$. So your conclusion that $\|f_n-1\| \to \infty$ is incorrect.

The easiest way to see that $(f_n)_{n \in \mathbb{N}}$ does not converge uniformly is that the limit function $f$ is discontinuous (note that the $f_n$ are continuous). But the uniform limit of a sequence of continuous functions has to be continuous. Hence the convergence can't be uniform.


问:
Is it true $\operatorname{st}(^\ast A) = \overline{A}$, when $A$ is a bounded subset of $\mathbb{R}$?
Given a bounded subset of $\mathbb{R}$, let $\operatorname{st}(^\ast A)$ be the set of the standard parts of all elements in $^\ast$-transform of $A$, $\overline{A}$ is the closure of $A$.Is it true $\operatorname{st}(^\ast A) = \overline{A}$?

It seems to me it's true $\overline{A} \subseteq \operatorname{st}(^\ast A)$.For every $x \in \overline{A}$, there is a sequence $r = \{r_i\}_{i \in \mathbb{N}} \in ^\ast A$ such that $r_i \in A$ for all $i \in \mathbb{N}$ and $\lim_{k \to \infty} r_k = x$. $\operatorname{st}(r) = x$, because for an arbitary $\epsilon$, $\{ i \in \Bbb{N}:|r_i - x|>\epsilon\}$ is finite.

But what about the other direction?

答:
It appears that you’re using the ultrapower construction. Let $\mathscr{U}$ be the ultrafilter. Suppose that $x\in\operatorname{st}{}^*A$. Then there is $a\in{}^*A$ that is represented by a sequence $\langle a_n:n\in\omega\rangle$ in $A$ with the property that for each $n\in\omega$, $$U_n\triangleq\{k\in\omega:|a_k-x|<2^{-n}\}\in\mathscr{U}\;.$$

For each $n\in\omega$ fix $k(n)\in U_n$; then $\langle a_{k(n)}:n\in\omega\rangle$ is a sequence in $A$ converging to $x$, so $x\in\operatorname{cl}A$.


问:
The matrix function $\frac{e^x - 1}{x}$ for non-invertible matrices
I am trying to evaluate the matrix function
$$
f(X) = \frac{e^X - I}{X}, \tag{1}
$$
where $e^X$ is the usual matrix exponential, defined by
$$
e^X = \sum_{n=0}^{\infty} \frac{1}{n!} X^n.
$$
The problem is that all matrices I am working with are singular (if it is of relevance, $X \in \mathfrak{so}(3)$). According to this Wikipedia article, one may consider the function $f$ to be a real function, find its Maclaurin series, and use this to evaluate the function for a matrix. This seems reasonable enough, and it also seems to be how the matrix exponential is defined. The Maclaurin series for $(1)$ is given by
$$
f(x) = \sum_{n=0}^{\infty} \frac{X^n}{(n+1)!} = I + \frac{1}{2!}X + \frac{1}{3!}X^2 + \ldots,
$$
indicating that the invertibility of $X$ shouldn't matter when it comes to finding a meaningful value for $(1)$.

Now for my question:
Is there a closed form of $(1)$ that do not involve the inverse? To be clear, my goal is to implement the function in Python, and I would rather do it without computing a large partial sum and use that as the approximation. I know a closed form of the matrix exponential for matrices in $\mathfrak{so}(3)$.

答:
I suggest using the partial sum of power series expansion or
else a partial product of infinite product expansion. Here is
PARI/GP code for both methods:

```<code>/* pf(X) = pf(X/2)*(e^(X/2)+I)/2 */
pf(X, m=9) = {my(n=matsize(X)[1],
Id=matid(n), /* nxn identity matrix */
p=Id,        /* partial product matrix */
eX=Id+X/2^m  /* 1st order approx to e^(X/2^m) */);
for(k=1,m, p*= (eX+1)/2; eX=sqr(eX)); p};

/* sf(X) = partial sum of power series */
sf(X, m=9) = sum(k=1,m, X^(k-1)/k!);
</code>```

The partial sum of power series can be made more efficient by
using Horner's method. Try writing the equivalent code in Python
to see how the parameter m effects the accuracy of the result.

Body:
I suggest using the partial sum of power series expansion or
else a partial product of infinite product expansion. Here is
PARI/GP code for both methods:

/* pf(X) = pf(X/2)*(e^(X/2)+I)/2 */
pf(X, m=9) = {my(n=matsize(X)[1],
Id=matid(n), /* nxn identity matrix */
p=Id,        /* partial product matrix */
eX=Id+X/2^m  /* 1st order approx to e^(X/2^m) */);
for(k=1,m, p*= (eX+1)/2; eX=sqr(eX)); p};

/* sf(X) = partial sum of power series */
sf(X, m=9) = sum(k=1,m, X^(k-1)/k!);

The partial sum of power series can be made more efficient by
using Horner's method. Try writing the equivalent code in Python
to see how the parameter m effects the accuracy of the result.


问:
a probability/ random walk brain teaser
A random walk $S_n = X_1+\cdots+X_n$ with independent and identically distributed steps $X_i$'s.

Assume $X_i$ is uniform distributed on integer values $\{1,2,3,\cdots, L\}$, i.e $P(X_i = l) = \frac{1}{L}$ for $l\in\{1,2,3,\cdots, L\}$.

Now consider the path $S_n, n\in\mathbb{N}$, i.e each path will take infinite steps and approach to $+\infty$. Let $p_m = P(\text{the path of the random walk path reaches integer m})$

Does $\lim_{m\rightarrow\infty} p_m$ exist? If it does, then what's the value of it?

答: It does exist and the limit is $\frac 2{L+1}$.  The informal proof is that on each step you average $\frac {L+1}2$ increase in coordinate.  A more formal proof would be to formulate a Markov chain where $p_m$ is the average of $p_{m-1}$ through $p{m-L}$ because you get to $m$ with probability $\frac 1L$ from any of those.  The starting condition is $p_0=1, p_{\lt 0}=0$

Title: a probability/ random walk brain teaser


问:
Are following set functions outer measures?
The function is $\nu:\mathcal{P}(X) \rightarrow \mathbb{R}$

I want to show if a),b),c) are outer measures.

a) X is anything, $\nu(A)=\begin{cases}
0 & \text{für } A = \emptyset \\
1 & \text{für } A \neq \emptyset \\
\end{cases}$

I'd say this is an outer measure. 1.) $\nu(\emptyset)=0$ is trivial,  2) $A\subseteq B \Rightarrow \nu(A) \leq \nu(B)$ is also pretty obvious and 3) $\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i)$ also seems correct since the left side can either be 1 or 0. If it's zero, the right side is also zero. If it's 1, the right side is bigger or equal 1.

b) X is infinite, $\nu(A)=\begin{cases}
0 & \text{für } |A| < \infty \\
1 & \text{für } |A| = \infty \\
\end{cases}$

I'd say this is not an outer measure since 3) $\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i)$ isn't always true.

Assume you split the infinite set in single-element subsets. Meaning that $X = \bigcup_{i\in\mathbb{N}}^{\infty} \{i\}$ with $A_i = \{i\}$

Then...

$1=\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i) = 0$ is obviously wrong. Therefore I'd say that b) isn't an outer measure

c) X is uncountable, $\nu(A)=\begin{cases}
0 & \text{für } |A| \leq \aleph_0\\
1 & \text{für } |A| > \aleph_0\\
\end{cases}$

I'd say 1) and 2) are pretty trivial again. 3.) is clear if every set $A_i \subseteq \mathcal{P}(X)$ is finite or countable. If there exists an $A_i$ that is uncountable and without loss of generality $A_i$ is disjoint to a set $A_j$ (of course, $|A_j|=\lambda \leq \kappa = |A_i|$), then: $|A_i \cup A_j| = |A_i|+|A_j| = \max(\lambda, \kappa) = \kappa$

Therefore $1 = \nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i) = \#\{A_i: A_i \text{ is uncountable}\} \geq 1$

From that I conclude that c) is an outer measure

Is that correct?

答:
This looks fine. A remark about (c). I believe you have the right idea but the way you wrote it seems slightly suspect. A better way to do it would be to note the following fact: a countable union of countable sets is countable. This just uses the fact that if there $A_{i}$ is countable for every $i \geq 1$, then there is an obvious surjection from $\mathbb{N} \times \mathbb{N}$ onto $\cup{i}A_i$, and then $|\mathbb{N} \times \mathbb{N} | = |\mathbb{N}|$.

Given this one can do the last line of your solution (c), i.e, $\mu(\cup_{i} A_i) = 1$ iff it is uncountable iff it has at least one uncountable $A_i$, call it $A_k$ in which case, $\Sigma_i \mu(A_i) \geq \mu(A_k) = 1 = \mu(\cup_{i} A_i)$


问:
Smoothness of map of rings is preserved by base change
In stacks project, the ring map $R\rightarrow S$ is called smooth if it is of finite presentation and the naive cotangent complex $NL_{S/R}: I/I^2\xrightarrow{f}{\Omega}_{R[S]/R}\otimes_{R[S]}S$ is quasi-isomorphic to a finite projective $S$-module placed in degree $0$. So by looking at homology at dimension $0$ and $1$, this means $f$ is injective and $\text{coker}f$ is a finite projective $S$-module.

I want to learn the proof of lemma 10.136.4 which said that if $R\rightarrow S$ is smooth and $R'$ is a $R$ algebra, then $R'\rightarrow S'=R'\otimes S$ is smooth. At the bottom of the proof, I can't see why $NL(\alpha')$ is quasi-isomorphic to $S'\otimes_R\Omega_{S/R}$ from the commutative diagrm, I think this means an injectivity and $\Omega_{S'/R'}\cong S'\otimes_R\Omega_{S/R}$.

The injectivity is clear, right?

For the isomorphism, I think the coker of the upper horizontal map is $R'\otimes_R\Omega_{S/R}$, and the kernel of the lower horizontal map is $\Omega_{S'/R'}$, so this is a isomorphism of $R'$ module, tensor $S'$, we get a isomrophism of $S'$ module: $S'\otimes_R\Omega_{S/R}\rightarrow S'\otimes_{R'}\Omega_{S'/R'}$. I can't deduce $\Omega_{S'/R'}\cong S'\otimes_R\Omega_{S/R}$.

Where I am wrong?

Also, is this definition of smoothness of ring maps is standard? I haven't seen this definition anywhere else.

Thanks.

答:
Regarding the injectivity: the injectivity of the horizontal map upstairs comes from the injectivity of $I/I^2 \rightarrow \Omega_{R[\underline{x}]/R}\otimes_{R[\underline{x}]} S$ upon noting that the injection is split, hence the injectvity remains even after the base change $R'\otimes_R -$. Using this, one deduces that both the vertical maps are isomorphisms, as in the proof. Then the injectivity downstairs follows from the injectivity upstairs.

Regarding the base-change computing part: You don't need to do an additional base change: The point is that already one has $$\Omega_{S'/R'} \simeq R'\otimes_R\Omega_{S/R} \simeq S'\otimes_S\Omega_{S/R}.$$

For the first isomorphism, see Tag 00RV (actually, it follows from the proof itself also). For the second, more generally one has
for any $S$-module $M$
$$S'\otimes_S M= (R'\otimes_R S)\otimes_S M \simeq R'\otimes_R(S\otimes_S M) \simeq R'\otimes_RM.$$
The reason why $\Omega_{S'/R'}=S'\otimes_{S}\Omega_{S/R}$ is explicitely mentioned is that this isomorphism shows that the cokernel of the map downstairs, which is $\Omega_{S'/R'},$ is projective as an $S'$-module.

Regarding the standardness: Maybe it's not standard to talk about the naive cotangent complex, but note that the condition for $R \rightarrow S$ to be smooth can be also restated as

$S$ is of finite presentation over $R$ and the conormal sequence
$$0 \rightarrow I/I^2 \rightarrow \Omega_{R[\underline{x}]/R}\otimes_{R[\underline{x}]}S \rightarrow \Omega_{S/R}\rightarrow 0$$
is (exact on the left and) split-exact.

This is pretty close to saying "$R\rightarrow S$ is of finite presentation and formally smooth," which, in case this is still not the standard definition, it is at least standard characterization of smoothness.


问:
Is set $\{A \in \mathbb{R}^{2 \times 2} : \text{tr}(A) = 0\}$ is bounded and closed?
$$\{A \in \mathbb{R}^{2 \times 2} : \text{tr}(A) = 0\}$$
Inner product is defined to be  <$A, B$>$ = \text{tr}(A^TB) $
I was asked if the set
$$D = s \subset   \text{Mat}_{2×2}$$
is closed and bounded.
I showed that $D^c$ is open hence $D$ closed (is it correct?).
how do I show It's (not) bounded?
Thank you

答: Hint What is the norm of $$A = \begin{bmatrix} 0 & n \\ n& 0\end{bmatrix}$$?


答:
From where you reached:

$$b=41-a\implies 238=ab=a(41-a)=41a-a^2\implies a^2-41a+238=0\ldots\ldots$$

Title: Finding two numbers when having their sum and product

Body:
I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad  ab=238,\quad  238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance


问:
How can I generate lotteries where I can simultaneously determine their expected values, variances and skewness?
First of all, thanks for taking the time to look at this. My goal is to be able to generate lotteries, where I can determine the first, second and third moments of their distributions, for a behavioural economics experiment. In other words, I'd like to be able to determine their expected values, variances and skewness.
\begin{align*}
 &\mathrm{Expected\ Value\ :} &&=\frac{\sum_{i}x_i}{n}\\
 &\mathrm{Variance:} &&=\frac{\sum_{i}(x_i-\mathrm{Mean})^2}{n}\\
 &\mathrm{Skewness:} &&=\sum_{i}\left( \frac{x_i-\mathrm{Mean}}{\sqrt{\mathrm{Variance}}}\right) ^3
\end{align*}
The lotteries will have three outcomes that are equally probable, so I believe this amounts to solving for $\{a,b,c\}$ in the following nonlinear system of equations, where $m, v$ and $s$ are just constants.
\begin{align}
 \frac{a+b+c}{3}= m\\
 \frac{
 (a-m)^2+(b-m)^2+(c-m)^2
 }{n}= v\\
 \frac{
 (a-m)^3+(b-m)^3+(c-m)^3
 }{\left( \sqrt{v}\right) ^3} = s
\end{align}
My questions is, how would you find for what values of $m,v$ and $s$ there are non-negative real solutions? I know that I can just input non-negative values for $a, b$ and $c$, but the issue with this is that, for my experiment, I would like to find triples of lotteries that satisfy one of the two following conditions. For three lotteries, say $\{1,2,3\}$, I would like either:

A. $\mathrm{Exp_1}=\mathrm{Exp_2}>\mathrm{Exp_3}$ and $\mathrm{Var_2}=\mathrm{Var_3}<\mathrm{Var_1}$ and $\mathrm{Skew_3}=\mathrm{Skew_1}<\mathrm{Skew_2}$

or

B. $\mathrm{Exp_1}=\mathrm{Exp_2}<\mathrm{Exp_3}$ and $\mathrm{Var_2}=\mathrm{Var_3}>\mathrm{Var_1}$ and $\mathrm{Skew_3}=\mathrm{Skew_1}>\mathrm{Skew_2}$

So, my question to you is, is this possible? And if so, how could I achieve it?

Many thanks and
best wishes,

Jeremy

答:
The following Python code helps to find $a,b,c$ as a function of $m,v,s$ by transforming the set of equations eqs1 that define mean, variance and skewness into an equivalent set of equations eqs2 that is easy to solve.
In this case, it is enough to look at the last equation (which is printed by the code).

In the example I chose $m=v=s=1$ and the code produces the output
$$
a^3 - 3a^2 + \frac32a + \frac16
$$
which, by wolfram alpha, has the three solutions $-0.093199, 0.76963, 2.3236$.
By symmetry, the unique solution (up to swapping the values of $a,b,c$) is given by:
\begin{align*}
a &= -0.093199\\
b &=0.76963 \\
c &= 2.3236
\end{align*}
Which means that you cannot realize $m=v=s=1$ if you want $a,b,c >0$.

```<code>from sympy import xring, QQ, lex
from sympy.polys import groebnertools as gt

P, variables= xring('c,b,a', QQ, lex)
c,b,a = variables

# choose your values for m,v,s below
m=1 # mean
v=1 # variance
s=1 # skewness

f1 = a+b+c - 3*m
f2 = (a-m)**2 + (b-m)**2 + (c-m)**2 - 3*v
f3 = (a-m)**3 + (b-m)**3 + (c-m)**3 - s*v**(3/2)

eqs1 = [f1,f2,f3]
eqs2 = gt.groebner(eqs1, P)

#for f in eqs2:
#    print(, "= 0")
print(eqs2[-1])
</code>```

Read chapter 3.1 of Cox Little O'Shea if you want to know why this works. Note that they refer to  eqs2  as a Gröbner basis for the Ideal generated by  eqs1.

Body:
The following Python code helps to find $a,b,c$ as a function of $m,v,s$ by transforming the set of equations eqs1 that define mean, variance and skewness into an equivalent set of equations eqs2 that is easy to solve.
In this case, it is enough to look at the last equation (which is printed by the code).

In the example I chose $m=v=s=1$ and the code produces the output
$$
a^3 - 3a^2 + \frac32a + \frac16
$$
which, by wolfram alpha, has the three solutions $-0.093199, 0.76963, 2.3236$.
By symmetry, the unique solution (up to swapping the values of $a,b,c$) is given by:
\begin{align*}
a &= -0.093199\\
b &=0.76963 \\
c &= 2.3236
\end{align*}
Which means that you cannot realize $m=v=s=1$ if you want $a,b,c >0$.

from sympy import xring, QQ, lex
from sympy.polys import groebnertools as gt

P, variables= xring('c,b,a', QQ, lex)
c,b,a = variables

# choose your values for m,v,s below
m=1 # mean
v=1 # variance
s=1 # skewness

f1 = a+b+c - 3*m
f2 = (a-m)**2 + (b-m)**2 + (c-m)**2 - 3*v
f3 = (a-m)**3 + (b-m)**3 + (c-m)**3 - s*v**(3/2)

eqs1 = [f1,f2,f3]
eqs2 = gt.groebner(eqs1, P)

#for f in eqs2:
#    print(, "= 0")
print(eqs2[-1])

Read chapter 3.1 of Cox Little O'Shea if you want to know why this works. Note that they refer to  eqs2  as a Gröbner basis for the Ideal generated by  eqs1.


答:
Within the limited context of what you have provided, and from what I am inferring that you are asking, yes you are correct.

There is nothing random about the absolute value function, it is completely deterministic. Taking a variable $u$ and applying the absolute value function $|\cdot|$ to get the output $w$, that is $w = |u|$, the output $w$ is non-random.

However, the output of a function of a random variable $X$ is random, but that comes from the "randomness" of the random variable, not the function itself (which is completely deterministic). So if we take a random variable $X$, apply the absolute value function $|\cdot|$ to get the output $Y$, that is $Y = |X|$, the output $Y$ is random. Why? Because it is a function of a random variable.

Title: Is $|X|$ always a random variable given that $X$ is one?

Body: Is it true that $|X|$ is always a random variable given that $X$ is a random variable?


问:
An iterative/recursive optimization problem about minimizing the Frobenius norm
Firstly, solve the initial problem:

$$
\varphi_0 = \underset{\varphi}{\arg\min}\lVert \left( \varphi ^{\top}\varphi -I \right) X \rVert _{F}^{2},
$$

s.t.

$$
\lVert \varphi_0 \rVert _2=1,
$$

where $I$ is the $n \times n$ identity matrix, $\varphi_0\in \mathbb{R}^{1\times n}$ is a vector, $X\in \mathbb{R}^{n\times k}$ is a given matrix (dataset).

Secondly, for each $i \in \left\{ 1,2,\cdots, q  \right\} $, get the best $\varphi_i\in \mathbb{R}^{1\times n}$ by solving the following problem iteratively:

$$
\varphi_i = \underset{\varphi}{\arg\min}\lVert \left( A_i ^{\top}A_i+\varphi ^{\top}\varphi -I \right) X \rVert _{F}^{2},
$$

s.t.

$$
\lVert \varphi_i \rVert _2=1,
$$
where

$$
A_i=\left( \begin{array}{c}
 \varphi _0\\
 \cdots\\
 \varphi _{i-1}\\
\end{array} \right) _{i\times n}.
$$

The final $A_{q}$ composed of $q$ unit vectors is what I need.

答:
This amounts to an application of the EYM theorem. If $X$ has singular value decomposition $X = U\Sigma V^T$ and $U$ has columns $u_1,\dots,u_n$, then we will find that
$$
A_q = \pmatrix{u_1^T\\ \vdots \\ u_q^T}.
$$


问:
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved
How are can the above recurrence relation be solved?

I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$

And then here:

$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$

And we are given that these: $T(0) = 2, T(1) = 3$

So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.

Body:
Modify $T$ as follows:

$$
\begin{aligned}
T(n) &= T(n-1) + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + T(n-2) + 6n + 1\\
T(n) + 3n &= T(n-1) + 3n + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + 3(n-1) + T(n-2) + 3(n-2) + 10\\
T(n) + 3n + 10 &= T(n-1) + [3(n-1) + 10] + T(n-2) + [3(n-2) + 10 ]
\end{aligned}
$$

So, $S(n) = T(n) + 3n + 10$ satisfies the Fibonacci recurrence.


问:
Matrix multiplication of more than two matrices.
https://en.wikipedia.org/wiki/Matrix_multiplication

Can we perform Matrix multiplication of more than two matrices?

Example : A.B.C where A,B,C are three matrices.

Can we generalize for multiplication of n number of matrices?

答: Matrix multiplication is, by definition, a binary operation, meaning it is only defined on two matrices at a time.  That said, so long as the dimensions are compatible, you can multiply more than two matrices together by iterating the multiplication two at a time.  For example, if you have $A \in \mathbb{R}^{n\times n}$, $B\in \mathbb{R}^{n\times m}$, and $C \in \mathbb{R}^{m\times m}$ then you could form the product $ABC$ because regardless of which adjacent pair you multiplied first the dimension of the resulting product would allow for multiplication with the third.


问:
$f$-related fields using partition of unity

Let $f: M \longrightarrow \mathbb {R} $ be smooth and not singular. Prove that there is a $ X \in X (M) $ field so that it is $f$-related to $ \partial_t \in X(M)$

To show this fact, we must show equivalently that there is a field $X$ such that $X(f) = 1$.

My suggestion is to use the partition of unity.

The Existence Theorem of partition of unity guarantees me a subordinate unit partition, a family
$$(\psi_{\alpha}) _{\alpha \in A}$$

such that $\sum _{\alpha \in A}(\psi_{\alpha}) = 1$

any suggestion?

答: Let  $(U_i)_{i\in I}$ an open covering of $M$ by charts (identified to open subset of $\mathbb{R}^n$. Let $x\in M, i\in I$ such that $x\in U_i$. Since $f$ is regular, there exists $v_i\in T_{x}U_i$ such that $df_{x}(v_i)\neq 0$. Since the tangent space of $U_i$ is trivial, we can find a vector field $V_i$ defined on $U_i$. such that $V_i(x)=v_i$. The function defined on $U_i$ by $g_i(x)=df_x.V_i(x)$ is continuous, and does not vanish at $x_i$, this implies that there exists $W_{i,x}\subset U_i$ such that the restriction of $g_i$ to $W_{i,x}$ is $>0$. Write $X_i={1\over{g_i(x)}}V_i$ and take a partition of unity $\phi_i$ related to a subcover $(W'_i)$ of $W_{i,x})$ and write $X=\sum_i\phi_iX_i$.


问:
Derivatives of trigonometric functions
A line segment has one endpoint, A, on the x-axis and the other endpoint, B, on the y-axis. It passes through the point (1,2). If O is the point (0,0), for what value of $<OAB$ is the area of triangle AOB a minimum? What is the minimum area of triangle AOB?  </p>

答:
The equation of the line passing through $A=(a,0)$ and $B=(0,b)$ is $$\dfrac{x}a + \dfrac{y}b = 1$$
We are given that the line passes through the point $(1,2)$. We hence have
$$\dfrac1a + \dfrac2b = 1 \tag{$\star$}$$
The area of the triangle is given by $\Delta=\dfrac{ab}2$.
From AM-GM, we have
$$\dfrac1a + \dfrac2b \geq 2\sqrt{\dfrac1a \cdot \dfrac2b} \tag{$\dagger$}$$
$(\dagger)$ along with $(\star)$ gives us
$$1 \geq \dfrac{2\sqrt2}{\sqrt{ab}} \implies ab \geq 8 \implies \Delta\geq 4$$
Hence, the minimum area is $4$. Further, the equality occurs if the equality occurs in $\dagger$, which occurs, when $a=2$ and $b=4$. This means $$\tan\left(\angle{OAB}\right) = \dfrac{b}a = 2 \implies \angle{OAB} = \arctan(2)$$

EDIT

In the figure below the pink triangle represents the triangle with the minimum area, while the blue triangle represents a triangle where the hypotenuse passes through $(1,2)$.

Title: Derivatives of trigonometric functions


问:
Prove that lim ${a_n}$ = 0 implies lim $\sqrt {a_n}$ = 0
How do you prove that given a non-negative sequence ${a_n}$, lim ${a_n}$ = 0 implies lim $\sqrt {a_n}$ = 0?

By definition of limit in analysis I can assume that given $ϵ>0$, I have $a_n-0<ϵ$, which is just $a_n<ϵ$, for n>>1.

If I can claim that $\sqrt {a_n} < {a_n} < ϵ$ for all $a_n$ I may be able to finish the proof, but that is not true for $0\lt a_n\lt1$.

How should I approach this?

答:
That's it? This is the same thing as with functions. Square root is continuous on its domain and so we can pass limit inside square root:

$\lim \sqrt {a_n} = \sqrt {\lim a_n} = \sqrt {0} = 0$


问:
ode; zero input response: what does q(D)y = 0 mean?
my textbook on ordinary differential equations has a chapter on system modeling, and I 'm not sure if I understand what is meant by

The zero-input response is the solution to  $q(\mathbf{D})y=0$,
$y(0)=a_0$, $y'(0)=a_1$, ...

One of the examples given of how to determine the zero-input respnse from a system modeled by q(D) is

$q(\mathbf{D}) = (\mathbf{D} + 1)^2 + 4$,  $y(0)=1$,  $y'(0)=3$

I can see how the answer

$y=e^{-t}(\cos(2t) + 2\sin(2t))$

is derived by searching for the characteristic modes from the standard basis, but then I recalculated the linear equation with the derivative back from this answer, and I expected the result to be a constant 0.

I tried this in matlab:

```<code>syms x

y = exp(-x) * (cos(2*x) + 2*sin(2*x));

qD = (diff(y) + y)^2 + 4*y;

diffy = diff(y);

hold on

fplot(y, [0 5])
fplot(qD, [0 5])
fplot(diffy, [0 5])

%fplot(hn)

legend ('y', 'qD', 'diffy')
</code>```

but I got anything but a constant 0 for $q(\mathbf D)$.

so is my matlab formula for $q(\mathbf D)$ wrong? or what else is meant by $q(\mathbf{D})=0$?

regards, Danny.

Body:
$$q(\mathbf{D}) = (\mathbf{D} + 1)^2 + 4$$
$q(D)y=0$ (and not $q(D)=0)$  means:
$$( (\mathbf{D} + 1)^2 + 4)y=0$$
$$(D^2+2D+5)y=0$$
$$y''+2y'+5y=0$$
Characteristic polynomial is:
$$(r+1)^2-4i^2=0 \implies r=-1 \pm 2i$$
$$\implies y(t)=e^{-t}(c_1 \cos(2t)+c_2 \sin (2t)$$


答:
One has
\begin{equation}
(f_{2n} - f_{n})\left(\frac{1}{2n}\right) =  1-\frac{1}{2} = \frac{1}{2}\quad \Longrightarrow
\quad \|f_{2n}-f_{n}\|_{\infty}\ge \frac{1}{2}\end{equation}
Hence the convergence cannot be uniform.

Title: proof verification on $ f_{n}$ uniformly continuity

Body:
Let $f_{n}:[0,1]\rightarrow \mathbb{R}$

$f_{n}=\begin{cases}
nx & \text{ if } 0\leq x\leq \frac{1}{n}\\
 1& \text{ if } \frac{1}{n}<x\leq 1
\end{cases}$

is $f_{n}\xrightarrow[]{uni}f$?

my solution:

for $x=0 ,f_{n}=0$

for $0<x\leq 1$ it exist a $n_{0}$ such that $\frac{1}{n_{0}}<x $ so $x>\frac{1}{n}$ for every $n \geq n_{0} $
$\Rightarrow f_{n}\rightarrow f(x)=1$

now i will check if it goes uniformly to $1$

$\left | f_{n}-1 \right |\geq nx-1$ for every $n\epsilon \mathbb{N},x\epsilon X$

$\left \| f_{n}-1\right \|>nx-2$ let $n\rightarrow +\infty $

$\left \| f_{n}-1\right \|\rightarrow +\infty $

I am not sure about the inequalities, is this ok? Can someone formally prove it?


问:
About idempotent and invertible matrix
So I get a square matrix $A$ is idempotent when $A^2 = A$.

How can you prove that if $A$ is an $n \times n$ matrix that is idempotent and invertible, then
$A = I$??

I'm quite confused on this.

Any help much appreciated, Thanks.

答: Note that $A^2 = A$ gives us $(A^2-A) = 0$ , i.e. $A(A- I) = 0$. Now pre-multiplying $A^{-1}$ on both side we get $$A^{-1}A(A- I) = 0 \Rightarrow (A-I) = 0 \Rightarrow A =I$$

Title: About idempotent and invertible matrix


问:
Understanding a proof of the local version of nakayama's lemma
Let $A$ be a local ring with maximal ideal $\mathfrak{m}$. Let $M$ be a finitely generated $A$ module. If $x_1,...,x_n$ are in $M$ such that their image forms a basis of $M/\mathfrak{m}M$, then the $x_i$'s generate $M$.

I am trying to understand this proof: First we let $N$ be the submodule of $M$ generated by the $x_i$'s. By assumption, the image of $N$ in the quotient map $\pi$ from $M$ to $M/\mathfrak{m}M$ is all of $M/\mathfrak{m}M$. Hence $\pi^{-1}(\pi(N))=M$.

My notes say that we also have $\pi^{-1}(\pi(N))=N+\mathfrak{m}M$ since it is the smallest submodule of $M$ generated by $N$ and $\mathfrak{m}M$. I don't understand this equality. Is it a definition on or some consequence of inverse image of module homomorphism?

I also do not understand this part. Since $N+\mathfrak{m}M=M$, and since $A$ is local, we have $\mathfrak{m}\subset \operatorname{Jac}(A)$ and hence $N=M$. I know that if $A$ is local then its maximal ideal is in the Jacobson radical but why does this give that $\mathfrak{m}M=0$?

答:
About your first question: If $f:X \to Y$ is a module homomorphism and $X' \subset X$ is a submodule, then $f^{-1}(f(X'))=X'+\mathrm{ker}(f)$

Let's prove this. Clearly $X' \subset f^{-1}(f(X'))$ and $\mathrm{ker}(f) \subset f^{-1}(f(X'))$, so $X'+\mathrm{ker}(f) \subset f^{-1}(f(X'))$.
For the other inclusion, let $v \in f^{-1}(f(X'))$. This means that $f(v) \in f(X')$, so we can find somet $v' \in X'$ with $f(v)=f(v')$. This means that $v-v' \in \ker(f)$, as $f(v-v')=f(v)-f(v')=0$. Thus we see that $v=v'+(v-v') \in X' + \mathrm{ker}(f)$.
To apply this to the problem, note that the kernel of the projection $\pi:M \to M/\mathfrak{m}M$ is $\mathfrak{m}M$, so $\pi^{-1}(\pi(N))=N+\mathfrak{m}M$

About the second question: If $M$ is a finitely generated $A$-module over a ring $A$ with Jacobson radical $J(A)$ with a submodule $N \subset M$ such that $M=N+J(A)M$, then $M=N$

The proof goes like this: consider $M/N$, then by assumption $J(A)(M/N)= (N+J(A)M)/N=M/N$, thus by Nakayama's lemma we have $M/N=0$. This implies $N=M$


问:
Solving anagrams using math.
On Hacker Rank, there is Sherlock and Anagrams coding test.

The opening statement of the coding test is:

Two strings are anagrams of each other if the letters of one string can be rearranged to form the other string. Given a string, find the number of pairs of substrings of the string that are anagrams of each other.

There is this example:

kkkk

There are 6 anagrams of the form [k,k] at positions [[0],[1]], [[0],[2]], [[0],[3]], [[1],[2]],[[1],[3]] and [[2],[3]]

There are 3 anagrams of the form [kk,kk] at positions [[0,1],[1,2]],[[0,1],[2,3]] and [[1,2],[2,3]]

There are 1 anagrams of the form of [kkk,kkk] at positions [[0,1,2],[1,2,3]]

Equally 10 anagrams.

Can this be solved with one or more mathematical equations?

From Googling around, I have found this reference that this is a simple combinatoric question with a known answer: binominal coefficient of $(n,2)$ which is $n(n-1)/2$.

I am not a mathematician so can someone tell me if it is possible please and show the maths equations involved?

答:
The number of substrings of a string of the form $x^n=\underbrace{xxx\cdots xx}_{n\text{ times}}$ is $\binom{n+1}{2}$ cause essentially you are choosing inclusively(they can be the same, a substring of length $1$) the initial position and the end position of your substring. Now, every substring is an anagram of any other substring iff the length is the same, as you have noticed so it is not enough to take pairs in the space counted by $\binom{n+1}{2,}$ notice that the number of substrings of length $k$ is $n-k+1,$ just go left to right and you will see that. Here, we can take any other string, either itself but we can not repeat the pair, so we will have,notice that the plus one comes from choosing the same substring, $$\binom{n-k+1+1}{2}.$$
For example, if $k=1$ we will have $\binom{n+1}{2}=\binom{n}{2}+\binom{n}{1}$ which means take $2$ positions or the same position.(So, in your example, you are missing taking $([0],[0]),([1],[1]),\cdots$ )

The problem when the string is not of this form I suspect, requires some kind of careful recursion.DP?

Title: Solving anagrams using math.


问:
$\dim W + \dim W^\perp = n$
Let $W$ be a subspace of $\mathbb{R}^n$. We know that $W^\perp$ is also a vector subspace of $\mathbb{R}^n$. How can I show that $\dim W + \dim W^\perp = n$?

I know that for a matrix $A$, $(Lgn A)^\perp = Ker A$ and $(ImaA)^\perp = Ker A^T$. I think it might help, but I am really not sure how I can use it.

答:
You should prove that given $V$ vector space over $F$ then
$V=W \oplus W^{\perp}$  this is  you should prove that
$$\forall v\in V v=w+w^{\prime}$$ where $w\in W$ and $w^{\prime}\in W^{\perp}$
$$W \cap W^{\perp}= \lbrace 0 \rbrace$$
Of this fact we can deduce
$\dim V= \dim W + \dim W^{\perp}$

Title: $\dim W + \dim W^\perp = n$


问:
Finding two numbers when having their sum and product
I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad  ab=238,\quad  238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance

Body:
$$(t-a)(t-b)=t^2-(a+b)t+ab=t^2-41t+238$$
whose roots are: $7,34$


问:
Boundedness and closedness properties in euclidian space
Every undergrad knows that in (finite dimensional) euclidian space "compactness=boundedness+closedness". I've been thinking: if we're given a connected, bounded, but neither closed nor open subset A of euclidian space is it always possible to find a closed but not bounded subset B (different from whole space) in the same space, s.t. there is a homeomorphism between them? (I.e. can we interchange these properties?) That seems very wrong, but I wasn't able to come up with a concrete example of subset, s.t. it is impossible. I know that A can't be open, because of the Brouwer's theorem, also if A has only one bounady point NOT in it, then we can always perform an inversion centered at that point, to get the desired set B. Can somebody please ruin my dreams with an example where it is (more or less) obviously impossible?</p>

答: Any connected open set $U$ "almost" works: Any homeomorphic image $f(U)$ on the same Euclidean space has to be open (this result is called invariance of domain). The only issue is that $f(U)$ could be the entire space, which is the only open and closed set (not counting the empty set). So we only have to construct an connected open set that's not homeomorphic to the entire space; for instance in dimension 2 you can take the annulus $U=\{ x:  1<|x|<2\}$.


答:
Let $A$ be a matrix whose columns are a basis for $W$. If $W = \operatorname{Col}(A)$ then $W^\perp = \operatorname{Null}(A^\top)$.

By Rank-Nullity, we know that
$$\dim \operatorname{Null}(A^\top) + \dim \operatorname{Col}(A^\top) = n$$

We also know that

$$\dim \operatorname{Null}(A^\top) = \dim W^\perp, \text{ and } \dim \operatorname{Col}(A^\top) = \dim \operatorname{Col}(A) = \dim W$$

since $A$ and $A^\top$ have the same rank. The result follows.

Title: $\dim W + \dim W^\perp = n$

Body:
Let $W$ be a subspace of $\mathbb{R}^n$. We know that $W^\perp$ is also a vector subspace of $\mathbb{R}^n$. How can I show that $\dim W + \dim W^\perp = n$?

I know that for a matrix $A$, $(Lgn A)^\perp = Ker A$ and $(ImaA)^\perp = Ker A^T$. I think it might help, but I am really not sure how I can use it.


问:
$\lim_{x \to 0}x^2 (a^{1/x} - a^{1/(x-1)}); a > 0$
First post in this forum. This is from my first semester calculus course.... I'm completely lost.

$$\lim_{x \to 0}x^2 (a^{1/x} - a^{1/(x-1)}); a > 0$$

Any hints or steps would be cool

Thanks!

答:
As to $x \to \infty$ mentioned by @Z Ahmed, by the mean value theorem, there exists a $\theta_x$ between $\frac{1}{x}$ and $\frac{1}{x-1}$ such that

$$x^2 (a^{1/x} - a^{1/(x-1)}) = x^2 a^{\theta_x} \ln a \left(\frac{1}{x}-\frac{1}{x-1}\right) = -a^{\theta_x} \ln a\frac{x^2}{x(x-1)}. $$

Since $\theta_x \to 0$ as $x \to \infty$, then

$$\lim_{x \to \infty} x^2 (a^{1/x} - a^{1/(x-1)}) = -\ln a.$$


问:
Number of non-identity elements of order $2$ in this set. (Related to abelian group of order $34.$)
This question was asked in a bachelor exam for which I am preparing and I was unable to solve it.

Let $G$ be an abelian group of order $34$ and $S=\{ g \in G \mid g=g^{-1}\}$ . Then what is the  number of non- identity elements in $S$?

I used sylow theorem: There is $1$ sylow subgroup of order $17$ and $17$ sylow subgroups of order $2$ but $17$ is not the answer ( I'm not even close!) .

What is wrong in my approach? Can you please tell?

答: In a finite abelian group $G$, there's only one $p$ Sylow subgroup for all $p$. This follows from the Sylow theorems: all $p$ Sylow subgroups are conjugate, but in an abelian group, conjugation doesn't do anything. Thus an abelian group of order $34=2 \cdot 17$ will have one $2$-Sylow subgroup and hence one non-identity element $g \in G$ satisfying $g=g^{-1}$.


答:
Start with an orthonormal basis of $W$, say $\{v_1, v_2, ... v_k \}$ and extend it to an orthonormal basis of $\mathbb{R}^n$. Since it has dimension $n$, this would then be $\{v_1, ... v_k, v_{k+1}, ..v_n \}$. Then by definition we have that $\{v_{k+1}, .. v_n\}$ is a linearly independent set in $W^{\perp}$. And since $\{v_1 .. v_n \}$ is a basis for $\mathbb{R}^n$, we simply have that for any $v \in \mathbb{R}$, $v = a_1 v_1 + ... a_kv_k + a_{k+1}v_{k+1} + ... a_{n}v_n  = (a_1v_1 + .... a_kv_k) + (a_{k+1}v_{k+1} + ... a_nv_n)$. The former is a vector in $W$, and the latter in $W^{\perp}$. Thus every vector in $\mathbb{R}^n$ is the sum of vectors in $W$ and $W^{\perp}$. Thus $\mathbb{R}^n = W + W^{\perp}$, and clearly $W \cap W^{\perp} = \emptyset$ since if $v \in W \cap W^{\perp}$, then $<v, v> = 0$ which means $v = 0$.

So we have $\mathbb{R}^n = W + W^{\perp}$ and $W \cap W^{\perp} = 0$, from which the result follows.

Title: $\dim W + \dim W^\perp = n$


问:
Proof of bases of fundamental subspaces
Could anyone help me to prove the following theorems? Here SVD means singular value decomposition

i.e. $U$ is a $m\times m$ matrix with $\mathbf u_i=\frac{Av_i}{||Av_i||}$, V is $n\times n$ matrix with $v_i$ is the eigenvector of $A^TA$, and $\Sigma$ is kind of block matrix with left upper block a diagonal matrix with singular value of A and $0$ otherwise.

答:
Sketch of the proof of the first claim:

$V^\top$ is invertible so $\text{Col}(U\Sigma V^\top) = \text{Col}(U\Sigma)$

$\Sigma$ is diagonal and the nonzero diagonal entries are the first $r$ diagonal entries. Thus $U\Sigma$ has columns $\sigma_1 u_1, \ldots, \sigma_r u_r$, and then the remaining $m-r$ columns are zero.

Title: Proof of bases of fundamental subspaces


问:
Addition in polar form
$$u_{1}(t) = 120\sqrt{2}e^{j5000t}$$
$$u_{2}(t) = -j160\sqrt{2}e^{j5000t}$$

I need to add these two values, so:

$u(t) = u_{1}(t) + u_{2}(t) = (120 - j160)\sqrt{2}e^{j5000t} = ...$

What next? How can I do this without changing the form of the numbers?

答:
Write $120-j160$ in polar form; that is, find the $r>0$ and $\theta\in[0,2\pi)$ such that
$$120-j160=re^{j\theta}.$$
Then you'll have
$$u(t)= (120 - j160)\sqrt{2}e^{j5000t} = re^{j\theta}\cdot\sqrt{2}e^{j5000t}=(r\sqrt{2})e^{j(\theta+5000t)}.$$</p>

Title: Addition in polar form


问:
Convergence in $\mathcal{S}'$ and $L^p$ spaces
I expect that the following statement is true:

Let $p \in (0, \infty], f \in L^p(\mathbb{R}^n), \{f_j\} \subset L^1(\mathbb{R}^n)$ and $f_j \to f \in \mathcal{S}'(\mathbb{R}^n)$. In addition, if $\{f_j\}$ is $L^1$-bounded, then, $f \in L^1(\mathbb{R}^n)$.

I can neither prove nor disprove this claim.
I would be grateful if anyone could give me some advice.

My English is not very good. Please forgive me if there are any grammatical errors or confusing parts.

答:
This is not true if you only assume $L^1$ boundedness of the sequence $\{ f_j\}$: Take any function $g\in L^1(\mathbb{R}^n)$ and set $f_j(x)=j^ng(jx)$. You can check that $f_j\to c\delta$ in $\mathcal{S}'$, where $c=\int_{\mathbb{R}^n} g$ and $\delta$ is the Dirac mass at the origin.

On the other hand, if you ask for $L^p$ boundedness for some $p>1$ the result is true by reflexivity of these spaces.


答:
Hint: Since $\lvert G\rvert=34=17\times 2$, by the classification theorem for groups of order $2p$ for prime $p$ greater than two,${}^\dagger$ we have that either $G\cong \Bbb Z_{2p}$ or $G\cong D_p$, the dihedral group of order $2p$. (Let $p=17$.)

But $G$ is abelian, so . . .

$\dagger$: Theorem 7.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".

Title: Number of non-identity elements of order $2$ in this set. (Related to abelian group of order $34.$)

Body:
This question was asked in a bachelor exam for which I am preparing and I was unable to solve it.

Let $G$ be an abelian group of order $34$ and $S=\{ g \in G \mid g=g^{-1}\}$ . Then what is the  number of non- identity elements in $S$?

I used sylow theorem: There is $1$ sylow subgroup of order $17$ and $17$ sylow subgroups of order $2$ but $17$ is not the answer ( I'm not even close!) .

What is wrong in my approach? Can you please tell?


问:
Can every SU(4) matrix be written as a product of of these 4 matrices
Consider a general $SU(4)$ matrix
$$U=e^{i a^j T^j},$$
where the generators $T^i$ are chosen in the basis
$$\{T\} = 1_2 \times \sigma^i, \quad \sigma^i \times 1_2, \quad \sigma^i \times \sigma^j, $$
for $i,j = 1,2,3$. How would I go about proving or disproving that every $U$ can be written as the following product.
$$U = e^{i(1_2 \times \sigma^j)b^j } \; e^{i(\sigma^k \times 1_2)c^k }\; e^{i((\sigma^l \times \sigma^1)d^{l} + (\sigma^m \times \sigma^2)e^{m} )}\;e^{i(\sigma^n \times \sigma^3)f^n },$$
for some $b,c,d,e,f$.

Edit: I am aware of the general Lie theory property (see for example page 3 of this paper) that if one can divide the generators $\{T\}$ into two subsets $L(K)=\{k^m\}$ and $L(P)=\{ p^n \}$ with the following commutation relations:
$$[k^{m_1},k^{m_2}]\in L(K), \quad [k^m,p^n]\in L(P), \quad [p^{n_1},p^{n_2}]\in L(K),$$
then one can decompose any $U$ as
$$ U = e^{i \alpha^m k^m} e^{i \beta^n p^n} .$$
This allows me to pull out the the first two exponentials in my desired decomposition above, by setting
$$L(K) = \{1_2 \times \sigma^i, \quad \sigma^i \times 1_2\}, \quad L(P) = \{ \sigma^i \times \sigma^j\}.$$
However, the same `trick' does not allow me to decompose any further. I don't know if this means it cannot be done at all though...

答:
The generators of su(4) in the fundamental are 4×4 traceless hermitian matrices, so 15 independent ones.

The matrices in your basis {T} are manifestly traceless and hermitian! Convince yourself of the obviousness of the fact. They are 3+3+9=15 in number, and, again, their independence should be obvious.

They are then a good basis of su(4), so the generic group element of SU(4) may be written as the exponential whose phase is their generic linear combination.

So you must convince yourself that your quadruple exponential product does not represent a mere subgroup. The exponents of the first two exponentials commute, so you may combine them into one, by the trivial limit of the CBH identity. I'm not sure how much rigor you require, but, to lower nontrivial orders of CBH-combining the 3 remaining exponentials you have a full span.

NB. Near the identity,  to linear order of your 15 ("angle") parameters, you completely cover all 15 directions of your manifold and tangent space.  Your original question asked whether higher orders could conspire to leave  "uncovered  regions" of that manifold. Presentation theory addresses these issues.


问:
A geometry problem with the reflection of the incenter
Consider $\triangle ABC$ and its incenter $I$. Let $M$ be the midpoint of arc $BC$ (not containing $A$) and $P$ be the second intersection point of the circumcircle of $\triangle ABC$ and the circle with center $I$ and radius $IA$. Prove that if $I'$ is the reflection of $I$ with respect to $BC$, then $I', P, M$ are collinear (lie on the same line).

Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur.

For help, here is a figure:

答:
Denote circumcircle of $\triangle ABC$ as $\Omega$ and circle with center $M$ and radius $MI$ as $\Gamma$. It's well-known that $\Gamma$ is the circumcircle of $\triangle BIC$.

Let $\Phi$ be the inversion with respect to the circle $\Gamma$. Then, for any point $X\neq M$ of the plane denote its image under the $\Phi$ as $X^*=\Phi(X)$.

Since $I$ and $J$ (here $J=I'$) are symmetric with respect to the line $BC$ their images $I^*$ and $J^*$ are symmetric with respect to $\Phi(BC)=\Omega$ (because inversion preserves symmetry eith respect to generalized circles; $\Phi(\{I,B,C\})=\{A,B,C\}$). However, $\Phi(I)=I$, so $J^*$ is the image of $I$ under the inversion with respect to the circle $\Omega$. In particular, $J^*$ lies on the line $OI$.

It's clear that $A$ and $P$ are symmetric with respect to $OI$. Hence, if $D$ is the point symmetric to $M$ with respect to $OI$, then $D\in\Omega$, $D, I, P$ are collinear and lines $MP$, $AD$ and $OI$ are concurrent. Denote the intersection of $MP$, $AD$ and $OI$ as $K$.

Now note that the collinearity of $P$, $M$ and $J$ is equivalent to the collinearity of $P$, $M$ and $J^*$ (since $M$ is the center of $\Gamma$). Thus, we need to prove that $J^*=K$.

Indeed, we have cyclic quadrilateral $ADMP$ with $AP\parallel MD$, where $K=AD\cap PM$, $I=AM\cap PD$. It means that $I$ and $K$ are inverse images of each other with respect to $\Omega$. Therefore, $J^*=K$, as desired.


问:
Every undirected graph with $n$ vertices and $2n$ edges is connected?
I should prove this claim:

Every undirected graph with n vertices and $2n$ edges is connected.

If it is false I should find a counterexample.
I was thinking to consider the complete graph with $n$ vertices. Such a graph is connected and contains $\frac{n(n-1)}{2}$ nodes. Considering that $2n > \frac{n(n-1)}{2}\implies$ my graph is connected too. But I'm not sure this could be a solution because even my graph has $2n$ edges it doesn't have to be complete.
Can anybody help me?

答:
The claim is true for up to $6$ vertices.  The smallest counterexample (unique up to isomorphism) is this graph on $7$ vertices:

The vertices are ascribed their degree, so we can easily verify there are $14$ edges via the Handshaking Lemma.

For $n \geq 8$ there always exists counterexamples; we add $n-7$ vertices and connect them to the two blue vertices above, thereby adding $2(n-7)$ edges.


问:
How to calculate inverse function of a "weird" function like $y=x+\cos x$
As the title said, how to calculate the inverse function of $y = f(x) = x + \cos x$ ?

I found a super function $x = g(y)$ like this:
$x = y -\cos(y-\cos(y-\cos(y-cos(...(y-\cos y)...))))$, with infinite nested loop.
And I use matlab to test it, find that the more nests, the lower approximation error.
So I'm conjecturing if $g(y)$ is the true inverse function of $f(x)$.

If yes, why? And can we reduce the form into some "simpler" finite style?

If not, what should it be? And also why?

###############

Note of changes: Sorry for my poor math knowledge, I cannot find a proper symbol describe what "..." exatly means. how about I use programming language?

```<code>y = a; % a is a constant
x = y - cos(y);
for i = 1:10000
   x = y - cos(x);
end
disp(x)
</code>```

答:
The sequence in your code is defined by $x_0 = a$ and $x_{n+1} = a - \cos(x_n)$, and you want to set $g(a) = \lim_{n \to \infty} x_n$. Assume for the moment that the limit exists.  Then of course the limit of $x_{n+1}$ is also $g(a)$, so taking the limit of both sides of $x_{n+1} = a - \cos(x_n)$ gives $g(a) = a - \cos(g(a))$, or $g(a) + \cos(g(a)) = a$, or $f(g(a)) = a$.  That is what we would expect for the inverse of $f$, so this is promising.

To complete the proof you must check 2 more things: (1) that the limit exists, and (2) $g(f(a)) = a$ as well.


问:
Combinatorics / fractal geometry.
I have a question on combinatorial fractal geometry.

I drew the dragon curve on a piece of graph paper. For this particular figure the total distance of the sides increases by a factor of squareroot(2) each iteration, and the number of full boxes at each level is: 1, 4, 11, 28, 67, 152, 335, 724, 1539, 3232, 6727, 13900, 28555...

Here are two resources on this fractal:

https://en.m.wikipedia.org/wiki/Dragon_curve

https://mathworld.wolfram.com/DragonCurve.html

as you follow  the curve, at no point did a line create two boxes at the same time. Rather, it was always possible to tell which box would be created and numbered next, even when the figure ran up against itself and started creating boxes rapidly.

The mean of all the numbers for a given translation of the dragon figure increased in a progression that corresponded to what I think is the equation of the curve: 1, 3, 8, 20, 47, 107, 238...
a_n = a_(n-1) + a_(n-2) + a_(n-3) + 2^(n-3).

It seems to me that some of these numbers might correspond in other interesting ways. For example, 1 could fold into 4, then 7, then 20. I’m imagining numbering the boxes by drawing the fractal using the lindenMayer system and then folding it to see if there’s an interesting combination of numbers. I tried checking in the online encyclopedia of integer sequences oeis but I didn’t find anything.

My basic question is - is numbering the boxes in that way useful? I haven’t seen any resources with these boxes numbered like this

答:
"My basic question is - is numbering the boxes in that way useful?"

No.  I see no use for this numbering.  It is arbitrary and gives no insight, especially because whatever question that might be related is unclear and unmotivated.

Title: Combinatorics / fractal geometry.


问:
If $f,g:U \longrightarrow R $ be twice differentiable function, $f(a)= g(a)$, $df(a)=dg(a)$ and $d^2f(x)=d^2g(x)$, then $f=g$
Let $f,g:U \longrightarrow R $ be twice differentiable function and $U \subset \mathbb{R}^m$ and $U$ a convex open set, such that $f(a)= g(a)$, $df(a)=dg(a)$ and $d^2f(x)=d^2g(x)$ for all $x \in U$, then $f=g$.

Note:
if $v=\left(\alpha_1,\dots ,\alpha_n \right) $ then
$d^2f(a) v = \Sigma_{i,j}\frac{\partial{f} }{{\partial x_i}\partial x_j}\left( a\right) \alpha_i \alpha_j $

My approach:

For Taylor

$f(a+u)= f(a)+df(a)+r(u)$

$g(a+u)= g(a)+dg(a)+h(u)$

where

$\lim_{u \to 0}\frac{r(u)}{|u|}=\lim_{u \to 0}\frac{h(u)}{|u|}=0$

Now I use Lagrange Remainder: $U$ is a convex set, then $\left[a, a+v \right] \subset U $, $f,g \in C^1$ and are twice differentiable function in $\left(a,a+u \right) $, then there are $\theta_1, \theta_2 \in \left( 0,1\right)  $    such that

$r(u)= \frac{1}{2}d^2f(a+\theta_1 u)u^2= \frac{1}{2}d^2g(a+\theta_1 u)u^2$

$h(u)= \frac{1}{2}d^2g(a+\theta_2 u)u^2$

anyone have a tip to continue?

答:
We have to prove the following claim:

$$d^2 h(x)\equiv0\quad\Rightarrow\quad h(x)=p\cdot x+ q\tag{1}$$
with constant vectors  $p$ and $q$. When $(1)$ is established we can apply it to $f-g$ in your question. When $df(a)=dg(a)$ then $p=0$, and when also $f(a)=g(a)$ then $q=0$.

In order to prove $(1)$ we argue coordinatewise. Choose a $j\in[n]$, and let
$$\psi(x)=\bigl(\nabla h(x)\bigr)_j={\partial h\over\partial x_j}\ .$$
We then have
$$\nabla\psi(x)=\left({\partial^2 h\over\partial x_j\partial x_1},\ldots,{\partial^2 h\over\partial x_j\partial x_n}\right)\equiv0\ .$$
It follows that $x\mapsto\psi(x)$ is a constant $p_j$ , and this implies $\nabla h(x)\equiv p$ for a constant vector $p=(p_1,\ldots,p_n)$. A second such step then proves that $h(x)=p\cdot x+q$, since $h(x)-p\cdot x$ has zero derivative.


问:
Differentiate an Inverse Function, Two Methods?
I would like to take the derivative of this inverse function at $\pi$: $f(x) = 2x + \cos{x}$, given that ${f}^{-1}(\pi) = \frac{\pi}{2}$.

I know that there are two methods of doing it. Let me demonstrate the method that I have down pat, using the fact that $\frac{d}{dx}\left[{f}^{-1}(x)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(x)\right)}$.

Method 1:

$f(x) = 2x + \cos{x}$

${f}^{\prime}(x) = 2 - \sin{x}$

Given: ${f}^{-1}(\pi) = \frac{\pi}{2}$

$\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(\pi)\right)}$

$\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left({f}^{-1}(\pi)\right)}}$

$\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left(\frac{\pi}{2}\right)}}$

$\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - 1}$

$\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = 1$

This method make sense. It is this next method that I am a little sketchy on. For the most part it utilizes some algebra for inverse functions...

Method 2:

$f(x) = 2x + \cos{x}$

$y = 2x + \cos{x}$

$x = 2y + \cos{y}$

The next few steps involve finding the inverse function (can it be done with a function like this?), taking the derivative of that, and plugging in $\pi$ for the answer...

My problem is that I am stuck after this point:

Am I going about this process correctly?

Can I find the inverse function of this crazy looking function? It is one-to-one, as shown in the graph below.

Thank you for your time.

答:
I got a similar question:

How to calculate inverse function of a “weird” function like $y=x+cosx$


问:
Invariant $C^\ast$ subalgebras of $C(\mathbb R)$
On the bounded continuous functions on $\mathbb R$, $C(\mathbb R)$,  define the $\mathbb R$-group action of translations $\alpha: \mathbb R \times C(\mathbb R) \to C(\mathbb R)$ as $\alpha_x(f)(y) := \alpha(x, f)(y) = f(y - x)$. Further, let us consider $\beta: C(\mathbb R) \to C(\mathbb R), ~\beta(f)(y) = f(-y)$.

Of course, there are plenty of $\alpha$- and/or $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$. It is not difficult to come up with subalgebras which are $\alpha$- but not $\beta$-invariant, such as $\{ f \in C(\mathbb R): ~f(x) \to 0, ~x \to \infty\}$. Somehow, all examples of $\alpha$- but not $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$ I can imagine are of the following kind:

They are ideals in some $\alpha$- and $\beta$-invariant subalgebra $B\subseteq C(\mathbb R)$;

They are of the form $A = A_0 \oplus \mathbb{C} 1$, where $A_0$ is an ideal in some $\alpha$- and $\beta$-invariant subalgebra $B \subseteq C(\mathbb R)$.

To possibly clarify this: In the above example we have

$\{ f \in C(\mathbb R): f(x) \to 0, x \to \infty\}$ is $\alpha$- but not $\beta$-invariant and it is of the first kind; it is an ideal of $C(\mathbb R)$,

The algebra $\{ f \in C(\mathbb R): f(x) \to c, x \to \infty$ for some $c \in \mathbb C\}$ is of the form $A_0 \oplus \mathbb C1$ with $A_0$ the ideal of $C(\mathbb R)$ from 1.

So my question is:

Are there examples of $\alpha$- but not $\beta$-invariant $C^\ast$ subalgebras of $C(\mathbb R)$ which do not arise in one of the two ways? Or conversely:

If $A \subseteq C(\mathbb R)$ is an $\alpha$-invariant $C^\ast$ subalgebra (say, without unit), is there always some $\alpha$- and $\beta$-invariant subalgebra $B \subseteq C(\mathbb R)$ such that $A$ is an ideal of $B$?

答:
Here is an example.

Consider the closure of graph $\Gamma$ of
$$\sin\left(\frac{1}{|x|-1}\right), \qquad x\in(-1,1)$$

This is a compact space which looks sort of like this:

Basically there are two intervals and that are connected via some line. Chose some homoemorphism of the line in the middle with $\Bbb R$ to embed the algebra $C(\overline\Gamma)$ into $C_b(\Bbb R)$. Now the points on the intervals are explicit examples of "points at infinity" (basically $\overline\Gamma$ is a special compactification of $\Bbb R$ that we have chosen). The $\alpha$-action leaves the points at infinity unchanged and shifts the points on the connecting line forwards and backwards in a way depending on our homeomorphism (but we don't care about the details).

We can choose the identification of the line with $\Bbb R$ in such a way that $\beta$ becomes the map
$$\beta(f)= [(x,y)\mapsto f(-x,y)].$$
Now look at a sub-algebra $A$ of $C(\overline\Gamma)$ defined by:

$$A=\{f\in C(\overline\Gamma) \mid f(1,1) = f(1,-1),  \quad f(-1,-1) = 0\}$$

In other words we want the point at the top right to evaluate to the same thing as the point of the bottom right and the point at the bottom left should evaluate to $0$. This condition is $\alpha$-invariant but not $\beta$-invariant.

Let $g$ be some function on $\overline\Gamma$ with $g(1,1)=1=g(1,-1)$, and $g(-1,1)=1$, $g(-1,-1)=0$. Then
$$(\beta(g)\cdot g) (1,1) = 1\cdot 1, \quad (\beta(g)\cdot g)(1,-1) = 0\cdot 1$$
and clearly $\beta(g)\cdot g\notin A$. But $g\in A$, hence $A$ is not an ideal of any $\beta$-invariant sub-algebra of $C_b(\Bbb R)$.


问:
Does every infinite group have a subgroup?
This question was asked in abstract algebra quiz and  I am confused about it. So, I am asking for help here.

Does every infinite group have a subgroup?

I thought about $(\mathbb{Z},+)$ and $(\mathbb{Q}, +)$ but they have subgroup and the former implies that every infinite cyclic group have subgroup.

So, No progress could be made on this particular question.

Any help please!!

答:
There always is a proper non-trivial subgroup in any infinite group.

To see why, first note that $\mathbb Z$ has a proper non-trivial subgroup, $2\mathbb Z$. So any group isomorphic to $\mathbb Z$ has a non-trivial proper subgroup.

Let $G$ be an infinite group not isomorphic to $\mathbb Z$.

Note that for any $g\neq0\in G$, there exists a group homomorphism $\mathbb Z\to G$ given by $z\mapsto g^z$. Hence, $(g)$ is isomorphic to a quotient group of $\mathbb Z$, and $(g)\neq(0)$, which means that $(g)$ is proper and non-trivial.


问:
How to find the nth binary number?
I was wondering if there is mathematical formula or a way to get the nth binary number where n is given. Obviously, this is a very easy problem for a programmer and I can construct a program (using loops) to find it but am interested in a mathematical way to obtain it.

To give an example if we assume 0 is the 0th binary number, and 1 is the first, I would like to get 1001 when I ask about the 9th number. Another way to put this is, does a fast way exist to convert decimal numbers into binary.

What about the general case? Does a formula exist that can quickly convert a number from a numerical system in base x to one in base y?

Thank you in advance!

答:
The $n$-th binary number is just the number $n$ written in the binary number system.

And then... there is a well-known algorithm how to obtain the binary representation of $n$.

But OK... if you mean if there's a formula for the binary digits of $n$, given $n$ itself - no, not a good/easy one at least. So you just need to keep dividing by $2$ (in a loop), and keep collecting the remainders (i.e. the binary digits).

The number of times to repeat the operation (dividing by $2$ and getting the remainder) is basically $\log_2 n$. Because each natural number $n$ has about $log_b\ n$ digits in a number system with base $b$.

Title: How to find the nth binary number?


问:
Connection Between Finite Separable Extensions and Galois Group Actions
I am trying to understand the proof of Lemma 1.5.1 given in "Galois Groups and Fundamental Groups" by Tamás Szamuely.

Let $k$ be a base field and $k_s\subset \bar{k}$ a separable and algebraic closure, and let $\operatorname{Gal}(k):=\operatorname{Gal}(k_s/k),$ the absolute galois group. Let us furthermore define $\mathbf{FinSep}_k$ be the category of finite separable extensions of $k$ and $\operatorname{Gal}(k)-\mathbf{FinSetCon}$ be the category of finite sets with continuous and transitive left $\operatorname{Gal}(G)$-action. I want to understand the construction of the functor $$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCon}.$$

So, I have to figure out where to send objects.

Let $L\in \mathbf{FinSep}_k$, and make the following assignment $L\mapsto \operatorname{Hom}_k(L,k_s)$, which is the set of $k$-algebra homomorphisms. We have a natural action by the Galois group, on the set, by
$$\varphi:\operatorname{Gal}(k)\times \operatorname{Hom}(L,k_s)\to \operatorname{Hom}(L,k_s)\\(\sigma,\phi)\mapsto \sigma\circ\phi.$$
Group action axioms.

The axioms of a group action is checked as follows: $e(x)=x$ is an element of $\operatorname{Gal}(k)$, and serves as the identity transformation in the definition of group actions: $e\circ\phi=\phi$. That $\varphi(\sigma\circ \tau,\phi)=\varphi(\sigma,\varphi(\tau,\phi))$ follows since composition of functions are associative.

Transformation is continuous.

$\operatorname{Hom}_k(L,k_s)$ is equipped with the discrete topology and so, the action is continuous if and only if the stabilizer $\operatorname{Hom}_k(L,k_s)_{\phi}$ is open for all $\phi\in \operatorname{Hom}_k(L,k_s)$. The stabilizer consists of all $\sigma\in\operatorname{Gal}(k)$, such that $\sigma\circ\phi=\phi.$ That is, all elements fixing $\phi(L)$.

By using Theorem 1.3.11, one can conclude that $\operatorname{Hom}(L,k_s)_{\phi}$ is $\text{open}^{\textbf{[1]}}$. If that is the case, we have that $\operatorname{Gal}(k)$ acts continuously.

Transitivity.

By the primitive element theorem $L$ is generated by a primitive element $\alpha\in k$, with minimal polynomial, $f$, say. An element $\phi\in\operatorname{Hom}_k(L,k_s)$ is given by mapping $\alpha$ to a root of $f$ in $k_s^{\textbf{[2]}}$. The Galois group permutes roots transitively, which shows that the Galois group gives a transitive action.

Next, they claim that the above argument shows that the map $\xi:\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$, induces an isomorphism of $\operatorname{Hom}(L,k_s)$ and the left coset space $\operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)$, $$\xi_*:\operatorname{Hom}(L,k_s)\xrightarrow{\cong} \operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)^{\textbf{[3]}}.$$
If $U$ is normal, we obtain the quotient $\operatorname{Gal}(k)/\operatorname{Hom}(L,k_s)_{\phi}$. By Theorem 1.3.11, this arises if and only if $L$ is Galois over $k$.

Questions.

[1] I'm not really sure how I apply Theorem 1.3.11 to conclude that the stabilizer is open. First things first, in this section, we do not assume $L$ is a subextension of $k_s$. But in Theorem 1.3.11, we do need to assume it, at least that is how I understand it (which is probably wrong). Even if I understood that it is possible to apply it, I do not see how.

Theorem 1.3.11 tells us what the closed subgroups looks like. There is also Corollary 1.3.9 in the book which tells me that the open subgroups of a profinite group are precisely the closed subgroups of finite index - which might be something I could use to conclude what we want.

I'm just confused about The Krull Theorem, and how I might apply it, do you have any idea?

[2] I haven't been doing Galois Theory for a while, so I am a little rusty. But I know that the Galois group sends root of the minimal polynomial to roots of the same minimal polynomial. But they are claiming that an arbitrary $k$-morphism, $\phi:L\to k_s$ does so too.

Why does it do that, or which result might be applied to conclude that? (This is probably obvious, and I am sorry for that).

[3] Between which objects does the map $\sigma\circ\phi\mapsto \sigma U$ operate? I'm stuck on the word "induce", which map does it induce from? Is it the following map we are considering
$$
\xi:\operatorname{Hom}(L,k_s)\to \{(\sigma\operatorname{Hom}(L,k_s)_{\phi})\}_{\sigma\in\operatorname{Gal}(k)},
$$
and from that one define the map $\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$?

When we are talking about isomorphisms now, I guess we are thinking about everything as $\operatorname{Gal}(k)$-sets? Maybe I shouldn't say $\sigma$ is in the Galois group in the above map? Maybe $\sigma$ is in the set $\operatorname{Hom}(L,k_s)$? Do I fix any of $\sigma$ or $\phi$ when I define the above map?

Sorry, that was a lot of questions. But I am a little bit confused about this construction. There are quite a few things to keep track of, I think.

I would be just as happy if you took your time to answer just one question, as answering all of the questions! If anything is unclear, I am happy to edit.

Best wishes,

Joel

Theorem 1.3.11 (Krull) Let $L$ be a subextension of the Galois extension $K|k$. Then $\operatorname{Gal}(K|L)$ is a closed subgroup of $\operatorname{Gal}(K|k)$. Moreover, the maps
$$
L\mapsto H:=\operatorname{Gal}(K|L)\qquad \text{ and }\qquad H\mapsto L:=K^H
$$
yield an inclusion-reversing bijection between subfields $K\supset L\supset k$ and closed subgroups $H\subset G$. A subextension $L|k$ is Galois over $k$ if and only if $\operatorname{Gal}(K|L)$ is normal in $\operatorname{Gal}(K|k)$; In this case there is a natural isomorphism $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$

答:
Here are some comments:

Note that even though $L$ may not be a priori contained in $k_s$, it is nonetheless true that $\phi(L) \subset k_s$ (and $\phi(L)$ is isomorphic to $L$ since $\phi$ is injective, since $L$ is a field). Then the point is that the stabilizer is exactly isomorphic to $\mathrm{Gal}(K/\phi(L))$, which by Theorem 1.3.11 is a closed subgroup. But it's also a finite index subgroup, because $L$ is assumed to be a finite separable extension (one slightly complicated way to see this is to note that $\phi(L)$ is contained in some minimal finite Galois extension $M$ of $k$ such that $M \subset k_s$: this is called its Galois closure. Then you have a surjection of sets $$\mathrm{Gal}(M/k) \cong \mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/M) \twoheadrightarrow \mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/\phi(L))$$ and then note that $\mathrm{Gal}(M/k)$ is a finite group since $M$ is finite Galois. This implies that $\mathrm{Gal}(k_s/\phi(L))$ has finite index). Therefore, as you mentioned, Corollary 1.3.9 implies that $\mathrm{Gal}(K/\phi(L))$ is open, and you get continuity of the Galois action.

Note that a $k$-morphism $\phi: L \to k_s$ (which is always injective) preserves $k$ by definition. So if we take any polynomial $f(x) = \sum c_nx^n$ with $c_n \in k$ and pick $\alpha \in L$ such that $f(\alpha) = 0$, then this implies that $$f(\phi(\alpha)) = \sum c_n\phi(\alpha)^n = \phi(\sum c_n\alpha^n) = \phi(f(\alpha)) = \phi(0) = 0.$$

This is an application of the orbit-stabilizer theorem to a transitive group action: if a group $G$ acts on a set $X$ transitively (i.e. for every $x,y \in X$ there exists $g \in G$ such that $g \cdot x = y$) then there is a bijection between the set $X$ and $G_x \backslash G$ where $G_x$ is the stabilizer of any element in $X$ (it doesn't matter which element you pick, the stabilizers are all canonically isomorphic). The bijection is given by defining the map $\rho_x: G \to X$ as $\rho(g) = g \cdot x$, and then noting that the map acts as the identity on $G_x$. So now you just need to apply this formalism when $G = \mathrm{Gal}(k)$ and $X = \mathrm{Hom}_k(L, k_s)$.

Hope this helps!


问:
How to show Warsaw circle is non-contractible?
The Warsaw circle is defined as a subset of $\mathbb{R}^2$：
$$\left\{\left(x,\sin\frac{1}{x}\right): x\in\left(0,\frac{1}{2\pi}\right]\right\}\cup\left\{(0,y):-1\leq y\leq1\right\}\cup C\;,$$
where $C$ is the image of a curve connecting the other two pieces.

A map from Warsaw circle to a single point space seems to be a well-known example showing weak homotopy equivalence is indeed weaker than homotopy equivalence. I am trying to see why the Warsaw circle is non-contractible. It seems intuitively reasonable since two 'ends' of it are connected in some sense, but I failed to give a proof. Any hint would be appreciated. Thank you very much.

答:
Let $W$ denote the Warsaw circle and $C$ is the space as in the question.  By collapsing $C$ down to a point we find the map $f:W\rightarrow W/C \cong S^1$.

$\underline{Claim}$: f is not null homotopic.

If we assume that the $\underline{Claim}$ holds we find the thesis. Indeed, any continuous $f:X\rightarrow Y$ is null homotopic if its domain is contractible; so this will show that $W$ is not contractible.

So, we have done if we prove the $\underline{Claim}$.

$\textit{Proof}(\underline{Claim})$ Assume by absurd that $f$ is null homotopic. Let $e:\mathbb{R}\to S^1:r\mapsto (cos2\pi r,sin2\pi r)$ be the usual helic covering space of $S^1$ and let $H:W\times I\to S^1$ (with I the interval $[0,1]$) be an homotopy between $f$ and a constant map $c:W\to S^1:w\mapsto (0,1)$. Since $e$ is surjective there exists a lifting $c_0 :W\to \mathbb{R}$ of $c$ along $e$.
Now since every covering space is a Hurewicz fibration (see Proposition 1.30 Hatcher Algebraic Topology https://pi.math.cornell.edu/~hatcher/AT/AT.pdf) there is a lifting of $H_0:W\times I\to \mathbb{R}$ of $H$ along $e$ such that $e\circ H_0=H$ and $c_0=i\circ H_0$ where $i:W\cong W\times\{0\}\to W\times I$ is the obviously inclusion. In particular $f_0(-):=H_0(-,1):W\to \mathbb{R} $ is a lifting of $f$ along $e$.

Recall that $C$ is the space as in the question.Let $C$ be the space as in the question. The map $f_{|C}~$ is injective and the commutative equation $ f_{|C} =e \circ f_{0|C}~$ implies that $f_{0|C}~$ is injective too. Moreover $f_{0|W/C}$$ ~:W/C\to \mathbb{R}~$ has image in a connected component of $e^{-1}\{(0,1)\}~$, then it is a constant map. Hence by the universal property of quotient $f_0~$ defines a unique, continuous and injective map $\hat{f}_0:S^1\cong W/C\to \mathbb{R}$. Here we get the absurd since there isn't such map. Indeed if by absurd such map $\hat{f}_0~$ exists then $Im(\hat{f}_0)~$ is a compact, path connected subspace of $\mathbb{R}$ because $S^1~$ is a path connected, compact space (with the topology of subspace of $\mathbb{R}^2$). But compact, connected subspace of $\mathbb{R}$ is simply a closed, path connected interval of $\mathbb{R}~$ $[a,b]$. Finally, we find this second absurd since this implies that $S^1~$ is homeomorphism to a closed, connected interval of $\mathbb{R}$ (i.e.  $S^1\cong Im{\hat{f}_0}\cong [a,b]$): this is impossibile since $\pi_1(S^1)\cong\mathbb{Z}\neq 0\cong\pi_{1}([a,b])$.


答:
[1] While $L$ may not be a subextension of $k_s$, $\phi(L)$ is. The stabilizer is, as you mentioned $\mathrm{Gal}(k_s/\phi(L))$. This has finite index because $\phi(L)/k$ is finite. Since it's not mentioned in theorem 1.3.11 let's prove this, as it doesn't seem to be mentioned in the theorem of Krull. Let $L'$ be the Galois closure of $\phi(L)$ inside $k_s$, then $L'/k$ is finite. We have that $\mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/L') \cong \mathrm{Gal}(L'/k)$, so $\mathrm{Gal}(k_s/L')$ has finite index. Because $\mathrm{Gal}(k_s/L') \subset \mathrm{Gal}(k_s/\phi(L))$, it follows that $\mathrm{Gal}(k_s/\phi(L))$ has finite index as well. As you mentioned, a closed subgroup of finite index is open (this implication holds in any topological group).

[2] Let $\varphi:A \to B$ be a homomorphism of $k$-algebras and let $f= \sum_{i=0}^n \lambda_ix^i\in k[x]$ be a polynomial and suppose that $f(a)=0$ for some $a \in A$.
Then we obtain using that $f$ is $k$-linear and a ring homomorphism: $f(\varphi(a))=\sum_{i=0}^n \lambda_i \varphi(a)^i=\varphi(\sum_{i=0}^n \lambda_i a^i)=\varphi(f(a))=\varphi(0)=0$. The proof is exactly the same as a for automorphism of field extensions

[3] This follows from the orbit-stabilizer theorem.

Title: Connection Between Finite Separable Extensions and Galois Group Actions

Body:
I am trying to understand the proof of Lemma 1.5.1 given in "Galois Groups and Fundamental Groups" by Tamás Szamuely.

Let $k$ be a base field and $k_s\subset \bar{k}$ a separable and algebraic closure, and let $\operatorname{Gal}(k):=\operatorname{Gal}(k_s/k),$ the absolute galois group. Let us furthermore define $\mathbf{FinSep}_k$ be the category of finite separable extensions of $k$ and $\operatorname{Gal}(k)-\mathbf{FinSetCon}$ be the category of finite sets with continuous and transitive left $\operatorname{Gal}(G)$-action. I want to understand the construction of the functor $$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCon}.$$

So, I have to figure out where to send objects.

Let $L\in \mathbf{FinSep}_k$, and make the following assignment $L\mapsto \operatorname{Hom}_k(L,k_s)$, which is the set of $k$-algebra homomorphisms. We have a natural action by the Galois group, on the set, by
$$\varphi:\operatorname{Gal}(k)\times \operatorname{Hom}(L,k_s)\to \operatorname{Hom}(L,k_s)\\(\sigma,\phi)\mapsto \sigma\circ\phi.$$
Group action axioms.

The axioms of a group action is checked as follows: $e(x)=x$ is an element of $\operatorname{Gal}(k)$, and serves as the identity transformation in the definition of group actions: $e\circ\phi=\phi$. That $\varphi(\sigma\circ \tau,\phi)=\varphi(\sigma,\varphi(\tau,\phi))$ follows since composition of functions are associative.

Transformation is continuous.

$\operatorname{Hom}_k(L,k_s)$ is equipped with the discrete topology and so, the action is continuous if and only if the stabilizer $\operatorname{Hom}_k(L,k_s)_{\phi}$ is open for all $\phi\in \operatorname{Hom}_k(L,k_s)$. The stabilizer consists of all $\sigma\in\operatorname{Gal}(k)$, such that $\sigma\circ\phi=\phi.$ That is, all elements fixing $\phi(L)$.

By using Theorem 1.3.11, one can conclude that $\operatorname{Hom}(L,k_s)_{\phi}$ is $\text{open}^{\textbf{[1]}}$. If that is the case, we have that $\operatorname{Gal}(k)$ acts continuously.

Transitivity.

By the primitive element theorem $L$ is generated by a primitive element $\alpha\in k$, with minimal polynomial, $f$, say. An element $\phi\in\operatorname{Hom}_k(L,k_s)$ is given by mapping $\alpha$ to a root of $f$ in $k_s^{\textbf{[2]}}$. The Galois group permutes roots transitively, which shows that the Galois group gives a transitive action.

Next, they claim that the above argument shows that the map $\xi:\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$, induces an isomorphism of $\operatorname{Hom}(L,k_s)$ and the left coset space $\operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)$, $$\xi_*:\operatorname{Hom}(L,k_s)\xrightarrow{\cong} \operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)^{\textbf{[3]}}.$$
If $U$ is normal, we obtain the quotient $\operatorname{Gal}(k)/\operatorname{Hom}(L,k_s)_{\phi}$. By Theorem 1.3.11, this arises if and only if $L$ is Galois over $k$.

Questions.

[1] I'm not really sure how I apply Theorem 1.3.11 to conclude that the stabilizer is open. First things first, in this section, we do not assume $L$ is a subextension of $k_s$. But in Theorem 1.3.11, we do need to assume it, at least that is how I understand it (which is probably wrong). Even if I understood that it is possible to apply it, I do not see how.

Theorem 1.3.11 tells us what the closed subgroups looks like. There is also Corollary 1.3.9 in the book which tells me that the open subgroups of a profinite group are precisely the closed subgroups of finite index - which might be something I could use to conclude what we want.

I'm just confused about The Krull Theorem, and how I might apply it, do you have any idea?

[2] I haven't been doing Galois Theory for a while, so I am a little rusty. But I know that the Galois group sends root of the minimal polynomial to roots of the same minimal polynomial. But they are claiming that an arbitrary $k$-morphism, $\phi:L\to k_s$ does so too.

Why does it do that, or which result might be applied to conclude that? (This is probably obvious, and I am sorry for that).

[3] Between which objects does the map $\sigma\circ\phi\mapsto \sigma U$ operate? I'm stuck on the word "induce", which map does it induce from? Is it the following map we are considering
$$
\xi:\operatorname{Hom}(L,k_s)\to \{(\sigma\operatorname{Hom}(L,k_s)_{\phi})\}_{\sigma\in\operatorname{Gal}(k)},
$$
and from that one define the map $\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$?

When we are talking about isomorphisms now, I guess we are thinking about everything as $\operatorname{Gal}(k)$-sets? Maybe I shouldn't say $\sigma$ is in the Galois group in the above map? Maybe $\sigma$ is in the set $\operatorname{Hom}(L,k_s)$? Do I fix any of $\sigma$ or $\phi$ when I define the above map?

Sorry, that was a lot of questions. But I am a little bit confused about this construction. There are quite a few things to keep track of, I think.

I would be just as happy if you took your time to answer just one question, as answering all of the questions! If anything is unclear, I am happy to edit.

Best wishes,

Joel

Theorem 1.3.11 (Krull) Let $L$ be a subextension of the Galois extension $K|k$. Then $\operatorname{Gal}(K|L)$ is a closed subgroup of $\operatorname{Gal}(K|k)$. Moreover, the maps
$$
L\mapsto H:=\operatorname{Gal}(K|L)\qquad \text{ and }\qquad H\mapsto L:=K^H
$$
yield an inclusion-reversing bijection between subfields $K\supset L\supset k$ and closed subgroups $H\subset G$. A subextension $L|k$ is Galois over $k$ if and only if $\operatorname{Gal}(K|L)$ is normal in $\operatorname{Gal}(K|k)$; In this case there is a natural isomorphism $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$


问:
Can a single random variable have a covariance matrix?
This link about the closed-form analytical solution for the Wasserstein distance says:

Let $\mu_1\in \mathbb{R}^N$ be a normally distributed random variable with expected value $m_1$ and covariance matrix $C_1$.

I never knew there could be a covariance matrix for univariate data, only multivariate data. How is it possible for a single random variable to have a covariance matrix?

答:
A covariance matrix is something which exists for any random vector, where a random vector is nothing more than $N$ random variables which are possibly correlated. Since $\mu_1\in \mathbb R^N$, $\mu_1$ is a random vector, and has a covariance matrix. Specifically, each realization of $\mu_1$ is some vector:
$$
(X_1,X_2,\dots,X_N),
$$
and the $(i,j)$ entry of the covariance matrix are $\text{Cov}(X_i,X_j)=E[X_iX_j]-E[X_i]E[X_j]$, for $i,j\in \{1,\dots,N\}$.

In the other answer, you seem to be saying that there cannot be a covariance matrix, since "there is only one random variable." However, I would say that there are $N$ random variables. Just like a vector in $\mathbb R^N$ has $N$-coordinates, so $N$ real numbers, a random vector in $\mathbb R^n$ will have $N$ random variables as coordinates. You seem to be referring to these as "sample observations," but that seems to gloss over the fact that the coordinates may be correlated.

Let us specialize to $N=2$ for a second. Just like a continuous random variable in one dimension should be thought of in terms of its pdf, which is a function defined on $\mathbb R^1$, a random vector in $\mathbb R^2$ should be thought of in terms of its pdf, which is a function defined on the plane $\mathbb R^2$. Here is an example of such a pdf for a normal distribution. Each realization of this random vector is some ordered pair of points, like $(52, 45)$, which gives rise to two random variables, $X_1=52$ and $X_2=45$. The numbers $52$ and $45$ are not different sample observations; they are coordinates of the same single sample observation. These two coordinate may have different means and variances; in the picture, the $X$ coordinate of that distribution has a mean of around $50$, while the $Y$ coordinate has a means near $40$.


问:
Derivatives of trigonometric functions
A line segment has one endpoint, A, on the x-axis and the other endpoint, B, on the y-axis. It passes through the point (1,2). If O is the point (0,0), for what value of $<OAB$ is the area of triangle AOB a minimum? What is the minimum area of triangle AOB?  </p>

答:
The equation of any line passing through $(1,2)$ is $$\frac{y-2}{x-1}=m$$ where $m$ is the gradient

On re-arrangement , $$\frac x{\frac{m-2}m}+\frac y{-(m-2)}=1$$

So, the area of  $\triangle OAB=\frac12\left| \frac{m-2}m \cdot (2-m)\right|=\frac{(m-2)^2}{2|m|}$

Clearly, $m\ne0$

If $m<0,m=-n$(say) (where n>0) and  we need to minimize  $\frac{(m-2)^2}{-2m}=\frac{(n+2)^2}{2n}$

$$\text{Now,}\frac{(n+2)^2}{2n}=\frac{\left(\sqrt n-\frac2{\sqrt n}\right)^2+4+4}2\ge \frac82=4$$

The minimum value $=4$ occurs if $\sqrt n-\frac2{\sqrt n}=0\implies n=2$

$\implies \tan\angle OAB=-m=+2$ as $m=\tan(\pi-\angle OAB)=-\tan \angle OAB$

$\implies \angle OAB=\arctan(+2)$

If $m>0,\triangle OAB\ge0;$ the equality occurs if $m=2$

Title: Derivatives of trigonometric functions


问:
Expressing "does not imply''
In ordinary discourse, when we say that "$A$ implies $B$", we shall formalize it by writing the following:
$$A\rightarrow B$$
But when we say that "$A$ does not imply $B$", we cannot formalize it as the following:
$$\neg(A\rightarrow B)$$
Because by material implication, we have the following equivalences for the second formula:
$$\neg(A\rightarrow B)\Longleftrightarrow\neg(\neg A\vee B)\Longleftrightarrow A\wedge\neg B$$
But the ordinary meaning of the sentence "$A$ does not imply $B$" is that $B$ does not follow from $A$: if $A$ is true, $B$ is either false or undecidable. I wonder how "$A$ does not imply $B$" is formally expressed in the object language (not at the meta-level). Thanks!

答:
The formalization of a sentence in ordinary discourse claiming that "$A$ implies $B$" or "$A$ does not imply $B$" is outside the languages of propositional or first-order or higher-order logic. It is in the language of logical consequences, which roughly consists of pairs of (sets of)$-$propositional of first-order or higher-order$-$formulas.

When we use the expression "$A$ does not imply $B$" in ordinary discourse, we can mean two distinct things:

$\models \lnot (A \to B)$, which means that $\lnot (A \to B)$ is a valid (or provable) formula. This amounts to say $\models A \land \lnot B$, i.e. every structure satisfies $A$ but not $B$.

$A \not \models B$, which means that $B$ is not a logical consequence of (or is not provable from) $A$, so there exists a structure where $A$ is true and $B$ is false. This amounts to say $\not \models A \to B$.

The difference is essentially whether we mean "not" as part of the language of propositional logic, or as a negation of the relation of logical consequence.
Usually, the intended meaning is 2.

Note that meaning 1 is stronger than meaning 2, in the sense that, given the formulas $A$ and $B$, if Point 1 holds then in particular Point 2 holds.

Title: Expressing "does not imply''


问:
What is the difference between $:=$ and $:\iff$?
What is the difference between
$:=$ and $:\iff$? Which one is used when (and how ist the latter called)?

I've seen the following: $$ X \subset Y :\iff \forall x \in X: x \in Y$$

Why cannot it be written with like this/What difference does this formulation make?: $$ X \subset Y := \forall x \in X: x \in Y$$

Thank you!

EDIT I'm asking specifically about the $:\iff$ Operator and not about the $\iff$ without the colon.

答:
First there is a difference between equality and equivalence. Equality holds between objects while equivalence holds between formulas, sentences, propositions. Objects ( numbers, functions, curves, etc.) are not true or false, only propostions  are, and they are logically equivalent just in case they have the same truth value in all possible situation.

Example . The proposition  (A&~B) is not the same proposition as ~ ( A$\rightarrow B$); however they are equivalent.

So, properly speaking, one should not use an equal symbol between two expressions ( though one can see this in old logic books).

Second, some propositions are logically equivalent just because we decide them to be so, that is, we use one proposition as a definition of the other. In that case they are equivalent by definition , and we can write " $  X:\iff Y$ or " $X\equiv_{Df} Y$. "

To put it briefly, equivalence-by-definition is a special case of logical equivalence ( and is not the same relation as equality).


问:
A power series expression for $\sum_{j=1}^\infty \left(\frac{1+z}{1-z}\right)^j $
I know that:

$$\sum_{j=1}^\infty \left(\frac{z}{1-z}\right)^j = \sum_{j=1}^\infty \sum_{k=0}^\infty {j+k-1\choose k}z^{j+k} = \sum_{m=1}^\infty \left(\sum_{j=1}^m {m-1\choose j-1}\right)z^m $$

and I am now looking for a similar expression for:

$$\sum_{j=1}^\infty \left(\frac{1+z}{1-z}\right)^j $$

and would be grateful for any help (hope it is trivial).

答: $\sum_{j=1}^\infty \left(\frac{1+z}{1-z}\right)^j$ converges for $\Re(z) <0$, there we can use the geometric series to simplify it. You'll find that it is a rational function with only one pole at $z=0$, thus no chance to find a power series expansion at $z=0$ for it. It has a unique Laurent series valid on the whole of $\Re(z) <0$.


问:
Prove that limit of $\sqrt{\frac{n^2+3}{2n+1}}$ is $\infty$ as $n$ tends to infinity.
I need to prove that $\lim_\limits{n\to \infty}$$\sqrt{\frac{n^2+3}{2n+1}} = \infty$ (series) by using the definition:

"A sequence $a_n$ converges to $\infty$ if, for every number $M$, there exists $N∈N$ such that whenever $n≥N$ it follows that $a_n>M$."

I came with a proof which I'm not sure is valid, because I just learned the definition and maybe I didn't grasp it yet.

Please let me know what do you think of the following proof:

I'll find an $N$ such that whenever $n≥N$ it follows that $a_n>M$ for every $M$:

$M < \sqrt{\frac{n^2+3}{2n+1}} \Rightarrow M^2 < \frac{n^2+3}{2n+1} < \frac{n^2+n^2}{2n} = \frac{2n^2}{2n} = n$

Hence, we got $n > M^2$, therefore we can say that when $N = M^2$ then $a_n>M$ for every $n > M$.
End of proof.

Do you think it is valid or am I missing something? Any suggestion for a better proof using the definition?

Thanks!

答:
Your proof is mostly okay but you do three things wrong.  One is linguistically wrong.  You say, and quote,

"I'll find an $N$ such that whenever $n≥N$ it follows that $a_n>M$ for every $M$"

As stated, no such $N$ exists or can exist.  $a_n >M$ for some $M$ but not for every $M$.  For any $a_n$ we can always have an $M = a_n + 1 > a_n$ so there can't be any $a_n> M$ for all $M$.

The actually thing you wish to prove is that for any $M$ that we can choose as an arbitrary test limit we can find a point $N$ where suddenly, once we look at the values $a_n$ after $n=N$ we find they are all bigger than that $M$ we choose.

For example if we say $\lim_{n\to \infty} 2n = \infty$ what does that even mean?  It means if we want to find a point where are the $a_n = 2n$ are bigger than $1000$ we can by looking at all $a_n$ where $n > 500$.  In those cases $a_n > 1000$.  But $a_{735} > 1000$ but we don't have $a_{735} >  1,000,000$.  But if we want to find a point where all $a_n > 1,000,000$ we can do that by looking at all $a_n$ where $n > 500,000$.  In those cases $a_n > 1,000,000$.  But we don't have $a_{678,923} > 10^{100}$.  Bue if we want to find a point where all $a_n > 10^{100}$ we just need $n > M = 5\cdot 10^{99}$.  Then if $n > N$ then $a_n > 10^{100}$.

But we will never have a $N$ where $n>N$ means $a_n > M$ for all $M$.  The best we can do, and it is a heck of a lot and it is all we actually want.  Is that for any target we want, we can always find an $N$ for that target.

If $M$ is an number as big as we want, it could be $10^{(10^{100000}}$... or it could be $39$, but we can't pick an $M$ that is actually "infinite".  But if $M$ is "as big as we like", then if $N = \frac M2$ thenfor every $n>\frac M2$ we will find that all $a_{n; n>N} = 2n > M$.  That means whatever limit we want, we can always find a point where all $a_n$ past that point will all be bigger than that limit.

I'm picky that you understand what that means.

... okay... of my high horse

Secondly, you have you implication backwards.

You prove that if $a_n > M$ then $n > M^2$ but you didn't prove they other way around, if $n > M^2$ then $a_n > M$.

Thirdly you found an upper limit rather than a lower limit.

We want

$\sqrt{\frac {n^2 + 3}{2n + 1}} > M$.

And if we find anything $K > M$ we have to have $\sqrt{\frac {n^2 + 3}{2n + 1}}\ge K > M$.  If we have $K > \sqrt{\frac {n^2 + 3}{2n + 1}}$ and $K > M$ that won't help us at all!  (Consider we want to prove that $5 > 7$ and we get that $2*4 > 5$ and $2*4 > 7$ and therefore $5 > 7$ .... well, that's just wrong.)

$\sqrt {\frac {n^2 +3}{2n+1}} > M$ means

$\frac {n^2 + 3}{2n+1} > M^2$ (assuming $M > 0$).  Now we must keep in mind this is what we WANT to show.  THis is not what we actually know.  We CAN'T conclude $\frac {n^2 + 3}{2n+1} > M^2\implies n >\frac {n^2 + 3}{2n+1} > M^2$ because we don't  KNOW that $\frac {n^2 + 3}{2n+1} > M^2$.

we need to work the other way around $\frac {n^2 + 3}{2n+1} > M^2\Leftarrow ... something ...$.

And we have $\frac {n^2 + 3}{2n+1} > \frac {n^2}{2n+1} > \frac {n^2}{2n+n} = \frac n3$ (assuming $n\ge 1$)

And $\frac n3 > M^2 \Leftarrow$ (is implied BY) $n > 3M^2$.

So if  $n > \max(1,3M^2)$ then $\frac {n^2 + 3}{2n+1} > \frac {n^2}{3n} = \frac n3 > M^2$ and so $a_n =\sqrt{\frac {n^2+3}{2n+1}} > |M| \ge M$.

And we are done.


Title: A power series expression for $\sum_{j=1}^\infty \left(\frac{1+z}{1-z}\right)^j $

Body:
I know that:

$$\sum_{j=1}^\infty \left(\frac{z}{1-z}\right)^j = \sum_{j=1}^\infty \sum_{k=0}^\infty {j+k-1\choose k}z^{j+k} = \sum_{m=1}^\infty \left(\sum_{j=1}^m {m-1\choose j-1}\right)z^m $$

and I am now looking for a similar expression for:

$$\sum_{j=1}^\infty \left(\frac{1+z}{1-z}\right)^j $$

and would be grateful for any help (hope it is trivial).

Body:
Let $\frac{z+1}{1-z}=x$. We have
$$\sum _{j=1}^{\infty } \left(\frac{z+1}{1-z}\right)^j=\sum _{j=1}^{\infty } x^j=\frac{x}{1-x};\;|x|<1$$
Therefore if $\left|\frac{z+1}{1-z}\right|<1\to \text{Re}(z)<0$ the series is
$$\sum _{j=1}^{\infty } \left(\frac{z+1}{1-z}\right)^j=\frac{\frac{z+1}{1-z}}{1-\frac{z+1}{1-z}}=-\frac{z+1}{2 z};\;\text{Re}(z)<0$$