答:
$a$ is a real number, defined by the definition you have given above.

It is computable, in the sense that we can calculate it to arbitrary numerical precision via an algorithm.

However, it is not currently known whether $a=0$.

None of these statements are in conflict. In fact, there are many such cases in mathematics where there is some perfectly well-defined answer to a certain question, and where we could get successively better computational bounds that decide more and more of the digits of the number, but where we don't yet know whether those bounds will converge to a certain conjectural value.

The word "know" is doing a lot of work in this supposed paradox, and if you specify any particular notion of what it means to "know" what a number is you'll find that it is no longer an apparent contradiction.

Title: Is $a$ bigger than $0$ or not?

Body:
Consider Goldbach's original conjecture (no need worry, because we don't talk about "Goldbach" itself):

every integer $n> 2$ could be detached as a sum of three primes. (In Goldbach's age, $1$ is regarded as a prime.)

For any $n>2$, we denote the statement above as $p(n)$. Then define

$$a_n:=\begin{cases}
0, &\text{if $p(n)$ is true},\\
1, &\text{otherwise}.
\end{cases}$$

Note that $a(n)$ is definite, since every $p(n)$ could be verified, at least by machine, as long $n$ is given. However, consider the decimal number $$a:=0.00a_3a_4a_5\cdots.$$

What's $a$? Is it a real number?

If $a$ is a real number, then we can compare the magnitude of $a$ and
$0$, since the real number field is ordered.

But we can't in fact. If $a=0$, then "Goldbach" is proven, and if not, then "Goldbach" is overturned. However, it's in suspense hitherto whether Goldbach is true or not.

Probably, we may say $a$ is indefinite, that's to say, we do not know what $a$ is yet. But a new question is coming about:

Why we know any digit but do not know the number itself ? In order
to know the number, what else we need ?

Further, let's compare $a$ with another classic number
$$\pi=3.1415926\cdots.$$

Indeed, we do not know all the digits of $\pi$ untill we compute every one. But we can safely say it's $\pi$. What's the exact difference between $a$ and $\pi$ on earth? Why we think we know $\pi$ but do not know $a$? In another word, what does "knowing a number" mean exactly? And in what sense can we say we know a number?


问:
What is the probability of selecting two cards of the same suit given that you select at least one heart and one king from a modified deck?
The modified deck:The deck consists of the red J,Q,K and Aces from a standard deck.

I tried drawing TWO tree diagrams separately, one for (H is hearts) $H$ and $\overline{H}$ and one for $K$ and $\overline{K}$. Then I summed up the desirable results from $K$ to get $\frac{7}{14}$ and from $H$, I got $\frac{11}{14}$.

Now $P = \frac{\text{H,H and all K cases}}{everything} = \frac{\frac{3}{14}\left(\frac{7}{14}\right)}{\frac{11}{14}\left(\frac{7}{14}\right)}=\frac{3}{11}$ which is so close to the answer $\frac{3}{10}$. What went wrong? And I welcome and alternate solutions

Edit 1: Question: A deck consists of the red J,Q,K and Aces from a standard deck. What is the probability that Sally selects two cards of the same suit given that she selects at least one heart and one king?

答:
I can't follow your calculations, perhaps because I'm very sleepy.  Here's how I would do it.  If we draw the King of Hearts, any of the other $7$ cards works, since we already have a Heart and a King.  Otherwise, we must draw the King of Diamonds and on of the three Hearts other than the King, which make ten draws in all with a Heart and a King.

The only ones where we get two cards of the same suit are those where we draw the King of Hearts and oone of the other three hearts, so the probability of success is $\frac3{10}$

I'm guessing you counted a draw twice, probably the one with two Kings.

ContentLicense: CC BY-SA 4.0


问:
Character formula of Verma module over a Kac-Moody algebra
The character formula of a Verma module over a Kac-Moody algebra is given by

$$\textrm{ch}\ M(\Lambda)=\frac{e(\Lambda)}{\prod_{\alpha\in\Phi+}(1-e(-\alpha))^{\textrm{mult}(\alpha)}}$$

Here $\Phi_+$ are the positive root system and $e(\lambda)$ is the formal exponential.

When proving this we first consider a basis $\{e_{-\alpha,i}\}_{1\leq i\leq \textrm{mult}(\alpha)}$ of $(\mathfrak{n}^-)_{-\alpha}$. Using PBW theorem gives a basis for $U(\mathfrak{n}^-)$ consisting of products $$\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

For $h\in U(\mathfrak{g})$ then

$$h\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}=\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)(h)\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

Thus the weight space $U(\mathfrak{n}^-)_{-\lambda}$ has a basis consisting of elements satisfying
$$\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha=\lambda$$

Then

$$\textrm{ch}\ U(\mathfrak{n}_-)=\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$
Kac states that this holds since the number of times $e^{-\lambda}$ appears on the right side is the number of sets of positive integers $k_{\alpha,i}$ st. $\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}{k_{\alpha,i}}\right)\alpha=\lambda$.

This leads to my questions:

Why is $\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)$ the corresponding weight of the weight vector $\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$?

I'm not sure I understand Kac's argument that $$\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$
Why is this the case?

答:
$$h\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}=\left(\sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha\right)(h)\prod_{\alpha\in\Phi_+}\prod_{i=1}^{\textrm{mult}(\alpha)}e_{-\alpha,i}^{k_{\alpha,i}}$$

The actual weight should have $-$ in front because $[h,e_{-\alpha,i}]=-\alpha(h)e_{-\alpha,i}$.
With this, it will then be true that $U(\mathfrak{n}^-)_{\color{red}{-}\lambda}$ has a basis consisting of elements satisfying
$$\tag{1} \sum_{\alpha\in\Phi_+}\left(\sum_{i=1}^{\textrm{mult}(\alpha)}k_{\alpha,i}\right)\alpha=\lambda.$$

I'm not sure I understand Kac's argument that $$\sum_{\alpha\in\Phi_+}\dim( (U(\mathfrak{n}^-))_{-\alpha}) e(-\alpha)=\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$$

From previous argument, we know $\dim( (U(\mathfrak{n}^-))_{-\lambda})$ is number of $(k_{\alpha,i})_{\lambda\in \Phi_+, 0\le i\le \text{mult}(\lambda)}$ such that
(1) holds. Each such $(k_{\alpha,i})$ corresponds to a way to write
$e(-\lambda)=\prod_{\alpha\in \Phi_+}e(-\alpha)^{k_{\alpha,i}}$, which is a term in the product $\prod_{\alpha\in\Phi_+}(1+e(-\alpha)+(e(-\alpha))^2+\cdots)^{\textrm{mult}(\alpha)}$, hence the equality.


问:
Find two linear functions whose domain is $0 \leqslant t \leqslant 2$ and whose range is $1 \leqslant f(t) \leqslant 9$
I'm supposed to find two linear functions for the same domain and range, I have found one. How do I find the other?

The one I found:

Since Domain is $0 \leqslant t \leqslant 2$
And Range is $1 \leqslant f(t) \leqslant 9$

I took $f(0) = 1$ and considered constant $C = 1$

Then I took $f(2) = 9$ , Since linear equations are of the form $f(t) = vt + C$, v being slope and C being constant.

$$f(2) = v(2) + 1\implies 9 = v(2) + 1\implies v = 4$$

Thus the linear equation becomes $f(t) = 4t + 1$

How do I find the second linear function with the same Domain and Range?

This is the $24$th question in Strang's Calculus at section $1.1$

答:
Flip it: consider the case where $f(0) = 9$ and $f(2) = 1$ instead. You shall obtain:

$$f(t) = 9-4t$$

Graphs to illustrate the flip:


问:
Is a torus a subset of $\mathbb{R}^3$ or $\mathbb{R}^4$?
Defining a torus $T$ as $S^1 \times S^1$, it should follow that $T \subseteq \mathbb{R}^4$.   But you can also think of a torus as a bagel, which means it's a subset of $\mathbb{R}^3$.  Can anyone clarify this point?</p>

答:
Neither. The torus is generally defined as an abstract topological space, independently of any "embedding" into another space. So in the strict sense, it is incorrect to say $T\subseteq \mathbb R^4$ or $T\subseteq \mathbb R^3$.

However, we often talk about abstract spaces like the torus as subsets of other spaces, commonly $\mathbb R^n$ for some $n$. How do we resolve this? The answer is something called an embedding. We say that a space $A$ can be embedded into a space $B$ if there is a continuous function $f:A\to B$ which maps $f$ injectively onto a subset of $B$, such that $f$ restricted to $B$ has a continuous inverse in the subspace topology$^1$. What you're noticing is that there are embeddings of the torus into $\mathbb R^3$ and $\mathbb R^4$.

Note that this is for topological spaces. If we add additional structure to the space, then the embedding is required to preserve this structure in a suitable way. It is common for example to give the torus a manifold structure, and the embedding is then required to be an embedding of manifolds, which is somewhat involved to define. Depending on the additional structure imposed, the torus may fail to embedd into $\mathbb R^3$, as Matt points out.

ContentLicense: CC BY-SA 3.0


答: For the function you found, $f(0)=1$ and $f(2)=9$. It is quite clear from a geometric perspective that the other linear function has $g(0)=9$ and $g(2)=1$.

Title: Find two linear functions whose domain is $0 \leqslant t \leqslant 2$ and whose range is $1 \leqslant f(t) \leqslant 9$

Body:
I'm supposed to find two linear functions for the same domain and range, I have found one. How do I find the other?

The one I found:

Since Domain is $0 \leqslant t \leqslant 2$
And Range is $1 \leqslant f(t) \leqslant 9$

I took $f(0) = 1$ and considered constant $C = 1$

Then I took $f(2) = 9$ , Since linear equations are of the form $f(t) = vt + C$, v being slope and C being constant.

$$f(2) = v(2) + 1\implies 9 = v(2) + 1\implies v = 4$$

Thus the linear equation becomes $f(t) = 4t + 1$

How do I find the second linear function with the same Domain and Range?

This is the $24$th question in Strang's Calculus at section $1.1$


问:
When are two elements of a group equivalent in terms of their action on an element of an orbit of X?
I have strong intution that if $g_1$ and $g_2$ are two members of a group $G$ acting on $X$; then,
if $g_1(x)= g_2(x)$ for some $x$ in a $G$-orbit, $O$  of $X$; then $g_1(x)=g_2(x)$ for all $x$ in $O$. But I can't prove this.

I am using this as a preliminary result to derive that each orbit corresponds to a subgroup of G, so only want to use the definition of orbits and actions to prove this result.

Any help is appreciated.

答: Let $G=S_3$ act on $X=O=\{1,2,3\}$ Both $(23)$ and $e$ fix $1$ but do different things to the other elements of $O$.


答:
Here are all the cases:

i) Both are hearts and one of them is $K$

Probability $ = \frac{1}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{1}{7} = \frac{6}{56}$

ii) Both are $K$

Probability $ = \frac{2}{56}$

iii) $K$ of heart, other is diamond ($A, Q$ or $J$)

Probability $ = \frac{1}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{1}{7} = \frac{6}{56}$

iv) $K$ of diamond, other is heart ($A, Q$ or $J$)

Probability $ = \frac{1}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{1}{7} = \frac{6}{56}$

This adds to $\frac{20}{56}$.

Your numerator is case $(i)$ and hence the desired probability $ = \frac{3}{10}$

Title: What is the probability of selecting two cards of the same suit given that you select at least one heart and one king from a modified deck?

Body:
The modified deck:The deck consists of the red J,Q,K and Aces from a standard deck.

I tried drawing TWO tree diagrams separately, one for (H is hearts) $H$ and $\overline{H}$ and one for $K$ and $\overline{K}$. Then I summed up the desirable results from $K$ to get $\frac{7}{14}$ and from $H$, I got $\frac{11}{14}$.

Now $P = \frac{\text{H,H and all K cases}}{everything} = \frac{\frac{3}{14}\left(\frac{7}{14}\right)}{\frac{11}{14}\left(\frac{7}{14}\right)}=\frac{3}{11}$ which is so close to the answer $\frac{3}{10}$. What went wrong? And I welcome and alternate solutions

Edit 1: Question: A deck consists of the red J,Q,K and Aces from a standard deck. What is the probability that Sally selects two cards of the same suit given that she selects at least one heart and one king?


问:
How to find a derivative with respect to a matrix?
If I have $f: \mathbb{R}^{n \times m} \times \mathbb{R}^{n} \to \mathbb{R}^{m}$

And $f(K,t) = Kt + h$ where $h \in \mathbb{R}$

How would I find $\frac{\partial f}{\partial K}$ and $\frac{\partial f}{\partial t}$?

答:
$F(K+H,t) - F(K,t) = tH $, so the derivative with respect to $K$ is the map $H \mapsto t H$. This is often written as
${\partial F(K,t) \over \partial K} H = tH$. One needs to be careful in the sense that a linear map over matrices cannot always be written as matrix multiplication (as would the be case in $\mathbb{R}^n$, for example).

$f(K,t+h) - f(K,t) = hK$, so the derivative with respect to $t$ is the map
$h \mapsto hK$.
We have ${\partial F(K,t) \over \partial t} h = hK$. Since the relevant parameter in in the reals, this is usually written as ${\partial F(K,t) \over \partial t} = K$.


问:
Find the Lyapunov function for the nonlinear second order system.
Consider the system
$$
\begin{aligned}
\dot{x}_1 &= -x_1+x_2,\\
\dot{x}_2 &= -x_2^3.
\end{aligned}
$$
The origin is obviously globally asymptotically stable. Indeed, $x_2\to 0$ and the dynamics of $x_1$ is linear and thus converges to zero for the converging input $x_2$.

However, I got stuck trying to find a Lyapunov function for this system. Could you please help?

答:
Define $v=x_1-x_2$. Then $\dot{v} = -x_1+x_2+x_2^3 = -v+x_2^3$. Consider the system
$$
\begin{aligned}
\dot{v} &= -v + x_2^3, \\ \dot{x}_2 &= -x_2^3.
\end{aligned}
$$
Choose $V = v^2+\frac{1}{2}x_2^4$. Then $$\dot{V} = -2v^2 + 2vx_2^3-2x_2^6.$$ Since $2vx_2^3 \le v^2+x_2^6$ we have $$\dot{V}\le -v^2-x_2^6.$$ Thus the origin $(v,x_2)=(0,0)$ is globally asymptotically stable, and the same holds for $(x_1,x_2)$.


问:
Can an algebraic structure have indistinguishable elements?
Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.

Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.

答:
A category usually has distinct but isomorphic objects. This generalizes both of your bulleted examples: a preorder is a category in which there is at most one morphism between any two objects, and a pseudometric space is an enriched category (see Lawvere metric space). Getting rid of this extra ambiguity amounts to taking a skeleton.

Since categories are ubiquitous, this gives a wealth of examples. Here are some which are more algebraic in flavor:

A set $X$ together with an action of a group $G$ may be regarded as a category (in fact a groupoid) with a morphism $x \to y$ for every $g \in G$ such that $gx = y$. Two objects are isomorphic iff they are in the same orbit with respect to the group action.

Given a field $k$ we can consider the category of algebraic extensions of $k$. This category contains various algebraic closures of $k$, all of which are isomorphic.


问:
problem on understanding the theorem
Theorem Let $R,R'$ be a two rings, let $\phi:R\to R'$ be a ring homomorphism,
Given an element $\alpha\in R'$, there is a unique ring homomorphism $\Phi:R[x]\to R'$ which agrees with the map $\phi$ on constant polynomial and sends $x\mapsto \alpha$.

I didn't understand the way map $\Phi$ defined.
If I'll get the map then I can try to prove this theorem.
Help me to understand the map $\Phi$.

Thanks!

答:
Quick note: all rings in your question and this answer are commutative (or else the theorem statement is incorrect).

We want to produce a homomorphism $\Phi : R[x] \to R'$ such that $\Phi|_R = \phi$ and $\Phi(x) = \alpha$. An arbitrary element of $R[x]$ is a polynomial of the form $$\sum_{i=0}^n c_i x^i$$ for some $c_0, \dots, c_n \in R$. Where should this element be sent by $\Phi$? Well, $\Phi$ is supposed to be a ring homomorphism, so we should have

$$\Phi\left(\sum_{i=0}^n c_i x^i\right) = \sum_{i=0}^n \Phi(c_i x^i) = \sum_{i=0}^n \Phi(c_i) \Phi(x^i) = \sum_{i=0}^n \Phi(c_i) \Phi(x)^i$$

We're also supposed to have $\Phi(c_i) = \phi(c_i)$ and $\Phi(x) = \alpha$, so we must define

$$\Phi\left(\sum_{i=0}^n c_i x^i\right) = \sum_{i=0}^n \phi(c_i) \alpha^i$$

This shows the uniqueness: if there is any homomorphism $R[x] \to R'$ with the desited properties, it must be given by the above formula. Now you just have to show that this function actually is a homomorphism with the desired properties, and you'll get existence!

Title: problem on understanding the theorem


问:
Inspection method doubt
how do I solve (y^2 + y + x) dy - ydx = 0 using inspection method?
I reached this point but I'm not sure how to proceed because when I integrate, my answer isn't correct</p>

答:
$$(y^2+y+x)dy-ydx=0$$
$$\frac{dx}{dy}+\frac{1}{y}x=-y-1$$
Let $\quad\begin{cases}
x=Y\\
y=X
\end{cases}$
$$\frac{dY}{dX}+\frac{1}{X}Y=-X-1$$
This is a first order linear ODE. I suppose that you can solve it.

Title: Inspection method doubt


问:
Cubes inscribed into a sphere
Suppose n ≥ 3 unit cubes are inscribed into a sphere, such that every
three of them have a common vertex. Prove that all n cubes have a common vertex.

Could anyone please help me the approach?

Thanks in advance.

P.S. I edited out what I have tried since it was pointed out my approach was wrong.

答:
The following are all my thoughts on the problem:

It looks like as you say, the result possibly follows directly from Helly's theorem (although perhaps it is necessary that $n > 3$?).

As a sphere is convex, and there $n <\infty$ cubes inscribed inside, then we have a finite collection of convex subsets of $\mathbb{R}^{3}$. The
intersection number or Helley number must be $4$, but since we only have that the Helley number is $3$ I am not sure about this problem.

As I said in a comment, one might translate this to a graph theory problem using a tree where for $n = 3$, all three nodes (cubes) connect to the same point (vertex) like so: $0---Vertex(0)---0$. (It will help to draw pictures here).

Then when you increase this to $n=4$, you may notice that if this new node does not meet at the same vertex as the other $3$, you get a contradiction of the intersection number $k=3$.

Perhaps it is possible (for you or in general) to finish this off by induction. I have hopefully correctly translated this problem to something I think is more  manageable, and attempting to keep the spirit of the question.

To explain better what I am thinking of, we will consider the $n=4$ case of the graph theory problem. As above, for $n=3$, we have $0---Vertex(0)---0$. Add another node to increase the case to $n=4$ and attach this new node to either of the  two nodes that are not the $3$-vertex. Now think of what this means for vertices of cubes instead of nodes. In our graph theory example, the fourth node will be too far away to share a vertex of its cube with one of them, which implies that we must then have a $4$-vertex. If this process is repeatable, we will achieve an $n$-vertex for any $n$.

It would help others as well as myself if you gave more context to the problem, and what you have learned about geometry up to this point.

Title: Cubes inscribed into a sphere


问:
Proof of an inequality by induction: $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$
Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$

I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.

Incomplete Proof

Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$.

When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true.

Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$.

Case 1

Suppose $x_{k + 1}  = -1$.

Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$.

RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive.

Hence RHS $\le 0$ and so LHS $\ge$ RHS.

Case 2

Suppose $x_{k+1} = 0$.

Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything.

Also, adding $x_{k+1}$ to RHS does not change anything.

Therefore, LHS $\ge$ RHS still holds.

Case 3

Suppose $-1 < x_{k+1} < 0$.

Then $0 < 1 + x_{k+1} < 1$.

Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative.

If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds.

If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 < 1 + x_{k+1} < 1$.

In this case LHS $\ge$ RHS does not necessarily hold?

...

答:
Proof:

When $n=1$, it's obvious that $1+x_1 \geqslant 1+x_1$ holds.

Suppose $(1+x_1)(1+x_2)\ldots{}(1+x_k) \geqslant 1+x_1+x_2+\ldots{}+x_k$ holds when $n=k$.

\begin{align}
(1+x_1)(1+x_2)\ldots{}(1+x_k)(1+x_{k+1})
&= (1+x_{k+1})\Big[(1+x_1)(1+x_2)\ldots{}(1+x_k)\Big] \\
&\geqslant (1+x_{k+1})(1+x_1+x_2+\ldots{}+x_k) \\
&=1 + x_{k+1} + (x_1+x_2+\ldots{}+x_k) + x_{k+1}(x_1+x_2+\ldots{}+x_k)
\end{align}

Because $x_1,x_2,\ldots{},x_n \in [-1,0]$,
$$\forall i, j \in \{1,2,\ldots{},n\} :\: x_i x_j \geqslant 0$$
$$x_{k+1}(x_1+x_2+\ldots{}+x_k) \geqslant 0$$

Furthermore,
\begin{align}
(1+x_1)(1+x_2)\ldots{}(1+x_k)(1+x_{k+1})
&\geqslant 1 + x_{k+1} + (x_1+x_2+\ldots{}+x_k) + x_{k+1}(x_1+x_2+\ldots{}+x_k) \\
&\geqslant 1 + x_1 + x_2 + \ldots{} + x_k + x_{k+1}
\end{align}

Therefore, for all $n \in \mathbb{N}^{+}$, the proposition is always true.

Q.E.D.

Reference:

Bernoulli's inequality


答: A commutative ring $A$ with $1$ can contain nilpotent elements, which form an ideal of $A$, called the nilradical $nil(A)$ of $A$.  In some contexts, it makes sense to kill off these nilpotents and pass to $A_{red} := A/nil(A)$, the underlying reduced ring.  (Reduced means "all nilpotents are zero".)

Title: Can an algebraic structure have indistinguishable elements?

Body:
Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.

Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.


问:
Ambiguous solution on arranging $6$ people on $3$ indistinguishable tables.

If there must be at least one person in each table, in how many ways can $6$ people be seated around three circular tables?

(We assume tables are indistinguishable)

This problem was an introductory problem to Stirling numbers.

I did not find a solution myself, so looked into the book for it, but I feel it is ambiguous.

It says there are three possible ways to group the number of people (viz. $4 + 1 + 1$, $1 + 2 + 3$ and $2 + 2 + 2$).

Then choosing the people to form groups, and arranging them gives:
$$\tag{QED}
\underbrace{\frac{1}{2}\binom{6}{4}\binom{2}{1}\times3!\times0!\times0!}_{\text{for }4 + 1 + 1} +
\underbrace{\binom{6}{3}\binom{3}{2}\times2!\times1!}_{\text{for }1 + 2 + 3} +
\underbrace{\frac{1}{3!}\binom{6}{2}\binom{4}{2}\times1!\times1!\times1!}_{\text{for }2 + 2 + 2} = 225
$$

Looking at the first term I understood why the $\displaystyle\frac{1}{2}$ is required,  and that is because of the following:

\begin{array}
 & & &  & \text{Observer A} & & \\
^{\text{right}} & & & & & & ^{\text{left}}\\
 & & \color{red}{\text{An arrangement of 4}} & \color{green}{\text{1 person}} & \color{purple}{\text{1 person}} & & \\
^{\text{left}} & & &  & & & ^{\text{right}}\\
& & & \text{Observer B} & & & \\
\end{array}

Now, from left to right, Observer A sees the arrangement $\color{purple}{\text{1 person}}$ $\color{green}{\text{1 person}}$ $\color{red}{\text{An arrangement of 4}}$ whereas for Observer B, the same arrangement is $\color{red}{\text{An arrangement of 4}}$ $\color{green}{\text{1 person}}$ $\color{purple}{\text{1 person}}$. As we've counted both of these arrangements, while they are the same, we divide by $2$.

Note that we are giving importance to the arrangement of people on the tables.

Giving the same importance and trying to calculate the last term, we see that we again need to divide by $2$. But it is not what is done. The author decided to divide by $3!$. Now looking at  this, we note that

$\color{blue}{\text{2 people }}\color{green}{\text{2 people } } \color{red}{\text{2 people}}$

$\color{green}{\text{2 people }} \color{blue}{\text{2 people }} \color{red}{\text{2 people }}$

$\color{blue}{\text{2 people }} \color{red}{\text{2 people }} \color{green}{\text{2 people }}$

$\color{red}{\text{2 people }} \color{green}{\text{2 people }} \color{blue}{\text{2 people }}$

$\color{red}{\text{2 people }} \color{blue}{\text{2 people }} \color{green}{\text{2 people }}$

$\color{green}{\text{2 people }} \color{red}{\text{2 people }} \color{blue}{\text{2 people }}$

are all counted as the same. This reflects that the arrangements of people on the tables does not matter.

But if this was the case, the $\displaystyle\frac{1}{2}$ in the first term should also be $\displaystyle\frac{1}{3!}$.

What am I missing? Thanks!

答: The table with four people in the first term is distinct from the tables with one; the two classes of tables cannot be swapped, but the two tables with one person can be. Thus we divide by two in the first term, representing the two ways to permute the two tables with one person.


问:
Is $a$ bigger than $0$ or not?
Consider Goldbach's original conjecture (no need worry, because we don't talk about "Goldbach" itself):

every integer $n> 2$ could be detached as a sum of three primes. (In Goldbach's age, $1$ is regarded as a prime.)

For any $n>2$, we denote the statement above as $p(n)$. Then define

$$a_n:=\begin{cases}
0, &\text{if $p(n)$ is true},\\
1, &\text{otherwise}.
\end{cases}$$

Note that $a(n)$ is definite, since every $p(n)$ could be verified, at least by machine, as long $n$ is given. However, consider the decimal number $$a:=0.00a_3a_4a_5\cdots.$$

What's $a$? Is it a real number?

If $a$ is a real number, then we can compare the magnitude of $a$ and
$0$, since the real number field is ordered.

But we can't in fact. If $a=0$, then "Goldbach" is proven, and if not, then "Goldbach" is overturned. However, it's in suspense hitherto whether Goldbach is true or not.

Probably, we may say $a$ is indefinite, that's to say, we do not know what $a$ is yet. But a new question is coming about:

Why we know any digit but do not know the number itself ? In order
to know the number, what else we need ?

Further, let's compare $a$ with another classic number
$$\pi=3.1415926\cdots.$$

Indeed, we do not know all the digits of $\pi$ untill we compute every one. But we can safely say it's $\pi$. What's the exact difference between $a$ and $\pi$ on earth? Why we think we know $\pi$ but do not know $a$? In another word, what does "knowing a number" mean exactly? And in what sense can we say we know a number?

答:
You are confusing "we can verify any number of them" with "we can verify every one of them".  yes, we can verify any of the ones we choose to but we can't verify any of the ones we don't choose to, and as there are infinitely many of them there will always be an infinitely number of them we don't choose.

So although we can know any $a_i$, we can't know every $a_i$.

Imagine I asked you.  Can you read all of Moby Dick in 5 minutes.  You say no.  I ask how much of Moby Dick can you  read in 5 minutes.  You say one page, I guess.  And I ask which page can you read.  And you say well, any page.  And I say which pages can't you read.  And you answer well I could read any page.  so I say if you can read every page and there aren't any page you can't read then you CAN read all of Moby Dick in 5 minutes.

Is that a paradox?

Title: Is $a$ bigger than $0$ or not?


问:
Does $G$ being a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is a subgroup of $GL(n, \mathbb{Z})$?
It is known that any finite subgroup of $GL(n, \mathbb{Z})$ is isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for any odd prime $p$ (see here). I am wondering if there is a converse to to this: Does $G$ being isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is isomorphic to a subgroup of $GL(n, \mathbb{Z})$?

If this does hold, I'd be curious if it could be strengthened (perhaps being a subgroup for any infinite set of primes, or all but finitely many primes, would work), and if it doesn't, I'd be curious if any weaker form could be salvaged.

答: No, the quaternion group $Q_8$ is a subgroup of ${\rm GL}(2,p)$ for all odd primes $p$, but not of ${\rm GL}(2,{\mathbb Z})$.


问:
How can I represent the sampling distribution of a random variable which has probability $\rho$ of being present?
Say we have a continuous random variable $S\in [0,1]$ which has an unknown probability distribution $p_s$. Now suppose we want to find the expected value (and distribution) of $N$ trials where in each trial we draw from $S$ with probability $\rho$ and we get $0$ with probability $1-\rho$ (we want the expected value just in terms of $E[S]$).

In other words, in each trial we have probability $\rho$ of getting something, and that something is a random variable $S$ distributed according to $p_s$. The rest of the time we get nothing at all.

We can model the number of trials which return something as a standard binomial distribution. That is, out of $N$ trials the number of times we draw from $S$ is modeled by $B(N, \rho)$ where $B$ is the standard binomial distribution.

If $S$ were always 1, we would be done. The sampling distribution would just be $B(N, \rho)$ and the expected value would be $N\rho$ (correct me if I am wrong). But what about this case where $S$ has some unknown distribution? How do we model this sampling distribution?

答:
Let $Y$ be the outcome of one trial.  The distribution of $Y$ is $(1-\rho) \cdot \delta_0 + \rho \cdot p_s$, where $\delta_0$ is a point mass at $0$.  Then after $N$ (independent, I'm assuming) trials the total outcome is $T_N := Y_1 + \dots + Y_N$, where $Y_1, \dots, Y_N$ are iid copies of $Y$.

I'm not sure exactly what you mean by "model this distribution", but in particular the expectation is easy to calculate: $\mathbb{E}(Y) = \rho \mathbb{E}(S)$, so $\mathbb{E}(T_N) = N \rho \mathbb{E}(S)$.

It's also possible to calculate the variance using the law of total variance.  Let $I$ be the random variable that indicates whether $0$ is chosen or sampling from $p_s$ is chosen.  Then
\begin{align*}
\operatorname{var}(Y) &= \mathbb{E}(\operatorname{var}(Y|I)) + \operatorname{var}(\mathbb{E}(Y|I)) = \mathbb{E}((1-\rho) \delta_0 + \rho \delta_{\operatorname{var}(S)}) + \operatorname{var}((1-\rho) \delta_0 + \rho \delta_{\mathbb{E}(S)}) \\
&= \rho \operatorname{var}(S) + \rho \mathbb{E}(S)^2 - \rho^2 \mathbb{E}(S)^2,
\end{align*}
and $\operatorname{var}(T_N) = N\cdot \operatorname{var}(Y)$.

EDIT: Here is an alternate way to calculate the same quantities.  As you have already pointed out, we can also express $T_N$ as $\sum_{i=1}^{M} S_i$ where $M \sim \operatorname{Bin}(N,\rho)$ and the $S_i$ are iid copies of $S$.  Then $\mathbb{E}(T_N)$ and $\operatorname{var}(T_N)$ can be calculated using standard results for sums of a random number of iid random variables (see for example these lecture notes)


问:
Why is WolframAlpha and WolframAlpha Widgets Giving me Different Answers?
I was just wondering in anyone's experience, if they came across an official WolframAlpha Calculation (using https://www.wolframalpha.com/) that was different from one of the widgets (which I think is unofficial and community made, for example, this).

The issue is, I don't understand if they are actually giving me different answers (thus causing me to cast doubt on ever using WolframAlpha Widgets again unless I am completely wrong on this).

WolframAlpha Official

WolframAlpha Widget

If they aren't could someone tell me if these expressions are mathematically equivalent?

答:
The two answers are identical.

Wolfram Alpha uses Wolfram Language as the underlying syntax. The correct expression in Wolfram Language would be

```<code>D[x^(rho-1) * (x^rho+y^rho)^(1/(rho-1)), x]
</code>```

leading to the same answer as in the widget.

My guess of why you are getting the same answer expressed differently is because there is ambiguity in your Wolfram Alpha Official input. In the input

```<code>derivative (x^(rho-1))*((x^rho+y^rho))^((1/(rho - 1)))
</code>```

WAO doesn't immediately know if what you want is

```<code>(derivative (x^(rho-1)) w.r.t. x) * ((x^rho+y^rho))^((1/(rho - 1)))
</code>```

or

```<code>derivative ((x^(rho-1)) * ((x^rho+y^rho))^((1/(rho - 1)))) w.r.t. x
</code>```

or one of the two above inputs, but with respect to y or to rho. Therefore it has to guess what is the most likely thing that you want - presumably this involves some internal rearrangements of the input, and once WAO figures out what it is that you most likely want, it performs the derivative with respect to the input that WAO thinks is correct.

If you clarify your intent by enclosing the expression in brackets and specifying the variable with respect to which the derivative should be computed, e.g.

```<code>derivative ((x^(rho-1))*((x^rho+y^rho))^((1/(rho - 1)))) wrt x
</code>```

the output will be identical to that of the widget.

Note that the widget doesn't have these ambiguities because from f(x,y)=expression it knows what the variables are (because they're specified in f(x,y)), and it knows what the intended expression is.

Body:
The two answers are identical.

Wolfram Alpha uses Wolfram Language as the underlying syntax. The correct expression in Wolfram Language would be

D[x^(rho-1) * (x^rho+y^rho)^(1/(rho-1)), x]

leading to the same answer as in the widget.

My guess of why you are getting the same answer expressed differently is because there is ambiguity in your Wolfram Alpha Official input. In the input

derivative (x^(rho-1))*((x^rho+y^rho))^((1/(rho - 1)))

WAO doesn't immediately know if what you want is

(derivative (x^(rho-1)) w.r.t. x) * ((x^rho+y^rho))^((1/(rho - 1)))

or

derivative ((x^(rho-1)) * ((x^rho+y^rho))^((1/(rho - 1)))) w.r.t. x

or one of the two above inputs, but with respect to y or to rho. Therefore it has to guess what is the most likely thing that you want - presumably this involves some internal rearrangements of the input, and once WAO figures out what it is that you most likely want, it performs the derivative with respect to the input that WAO thinks is correct.

If you clarify your intent by enclosing the expression in brackets and specifying the variable with respect to which the derivative should be computed, e.g.

derivative ((x^(rho-1))*((x^rho+y^rho))^((1/(rho - 1)))) wrt x

the output will be identical to that of the widget.

Note that the widget doesn't have these ambiguities because from f(x,y)=expression it knows what the variables are (because they're specified in f(x,y)), and it knows what the intended expression is.


问:
$\int^{\pi/2}_{0}\log|\sin x| \,dx = \int^{\pi/2}_{0}\log|\cos x| \,dx $
Prove that :

$$\int^{\pi/2}_0 \log|\sin x| \,dx = \int^{\pi/2}_0 \log|\cos x| \,dx $$

I tried to cut the integral into a sum of parts and changing variable but it didn't work out right, i dont know how to solve this kind of problems in any other way, any hint will be much appreciated!

答:
$$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx\cdots \tag 1 $$
use this property:$$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$
$$I=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx\cdots \tag 2$$

Add both eqn:
$$2I=\int_0^{\dfrac{\pi}{2}}\log\sin x+\log\cos x\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}\log({\sin x\cdot\cos x})\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}\log(\dfrac{2\sin x\cdot\cos x}{2})\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}(\log\sin 2x-\log2)\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}\log\sin 2x\,dx-\int_0^{\dfrac{\pi}{2}}\log2\,dx$$
In first integration put $2x=t$
$$2I=\dfrac12\int_0^\pi{\log\sin t}\,dt-\dfrac {\pi}{2}\log 2$$
use this property:
If $f(2a-x)=f(x)$ then
$$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$$
$$2I=\dfrac22\int_0^\dfrac{\pi}{2}{\log\sin t}\,dt-\dfrac {\pi}{2}\log 2$$
$$2I=\int_0^\dfrac{\pi}{2}{\log\sin x}\,dx-\dfrac {\pi}{2}\log 2$$
$$2I=I-\dfrac {\pi}{2}\log 2$$
$$I=-\dfrac {\pi}{2}\log 2$$
$$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx=-\dfrac {\pi}{2}\log 2$$


问:
Indepence of Two Events
Suppose X and Y are r.vs and $(X-Y)$ and $Y$ are independent. Does this imply the independence of $X$ and $Y$? Since $P((X-y)|Y=y) = P(X-y)$, it seems $X$ and $Y$ are independent. How do you formally show this?

答:
Let $Y,Z$ be independent, identically distributed random variables, each uniformly distributed on $\{0,1\}$, and let $X=Y+Z$.

Then $X-Y=Z$, so $X-Y,Y$ are independent.

But $X,Y$ are not independent, since for example, $P(X=0)={\large{\frac{1}{4}}}$,$\;$whereas $P(X=0{\,\mid\,}Y=1)=0$.

Title: Indepence of Two Events


问:
Is the relative order of elements in column matrices $Mf(x)$ and $Mx$ the same if $f$ is strictly increasing?
Suppose we have an $n$ by $1$ matrix A. We define $\arg\max A$ as the $k$ such that the largest element of A is $A_{k, 1}$.

Consider a $m$ by $n$ matrix $M$ with nonnegative entries and an $n$ by $p$ matrix $p$. Let $x = \arg\max Mp$.

Now take a strictly increasing function $f$, and let $x' = \arg\max M f(p)$, where $f$ is applied elementwise.

Is $x = x'$? Every example I try works, but I'm not sure how to prove it. I feel like a proof needs to leverage the relationship between increasing functions and rank-preservation, but I can't seem to put my finger on it.

答:
The answer is no, because the matrix mixes up the elements.

For example, let $$M:=\begin{pmatrix}1&0\\0&3\end{pmatrix},\qquad p=\begin{pmatrix}1\\0\end{pmatrix},\qquad f(x):=2^x,$$ then $$Mp=\begin{pmatrix}1\\0\end{pmatrix},\qquad Mf(p)=\begin{pmatrix}2\\3\end{pmatrix}$$


问:
Find least value of $9\tan\phi+4\cot\phi+3$ without derivatives
The problem is as follows:

In a pharmaceutical lab in Daegu, a group of scientists need to find
the power of a visible blue light lamp to get through a sucrose
crystal in order to measure its purity for a certain medication.

The power for such light is given by:

$$J=\frac{1}{H}$$

Assuming that $H$ has to be the minimum value of the equation from
below:

$$H=9\tan\phi+4\cot\phi+3$$

$\theta$ is an acute angle which lies in the first quadrant.

What would be the value of $J$ so that the scientists could set up
their lamp?

The alternatives given in my book are:

$\begin{array}{ll}
1.&\frac{1}{5}\\
2.&\textrm{1}\\
3.&\textrm{15}\\
4.&\frac{1}{15}\\
\end{array}$

This problem is making me feel confused. Should derivatives be used here?

I've attempted to do so and derivate the equation, by doing so this yields:

$H'=9\sec^2\phi-4\csc^2\phi$

But equating this to zero doesn't seem to be the intended approach.

Then I came up with the idea of adding the intervals from where $\phi$ in the first quadrant lies.

$0\leq \cot \phi < +\infty$

Tangent is also in the same interval:

$0\leq \tan \phi < +\infty$

Then multiplying by $9$ and by $4$ and adding $3$ to the second equation.

This results into:

$3\leq 9\tan\phi+4\cot\phi+3 < +\infty$

But without any further constrain there isn't any way to tell, which would be the least value?

If it would had been the integer value, wouldn't had been $4$?

Running out of idead, what it came to my mind was to equate both functions:

$\tan\phi=\cot\phi$

This corresponds to $\phi=\frac{\pi}{4}$

Assuming this: I'm guessing that the result would be

$9\tan\phi=1$

$4\cot\phi=1$

then $1+1+3=5$

But this is more like guessing and not a serious way to justify the answer. I cannot be sure if that's the answer. Can someone help me here and explain me why derivatives doesn't seem to work?

How to solve this without derivative?. Help me please, I'm confused.

答:
A bit late this answer but I think worth mentioning it.

It is just AM-GM - the inequality between arithmetic and geometric mean:

For $\theta \in \left(0,\frac{\pi}2\right)$ you have $\tan \theta >0$ and AM-GM gives
$$9\tan \theta + 4\frac 1{\tan \theta}+3 \stackrel{AM-GM}{\geq} 2\sqrt{9\tan \theta \cdot 4\frac 1{\tan \theta}}+3 = 12+3 = 15$$

The minimum is reached for
$$9\tan \theta = 4\frac 1{\tan \theta} \stackrel{\theta \in \left(0,\frac{\pi}2\right)}{\Leftrightarrow} \theta = \arctan \frac 23$$


问:
resolvent clause theory
This is about correctness of resolution lemma

Let R be a resolvent of two clauses $C_1$ and $C_2$. Then $C_1, C_2 \models R$.

Proof By definition $R = (C_1 − \{L\}) \cup (C_2 −\{\bar{L}\}) $ (for some L).

Let $A \models C_1$ and $A \models C_2$. There are two cases.

If $A \models L$ then $A \models C_2 − \{\bar{L}\}$  (because $A \models C_2)$, thus $A \models R$.

If $A \not\models L $ then $A \models C_1 − \{L\}$ (because $A \models C_1)$, thus $A \models R$ \

Can someone explain why If $A \models L$ then $A \models C_2 − \{\bar{L}\}$  ?

答:
There is something wrong...

The resolution rule in propositional logic is a single valid inference rule that produces a new clause implied by two clauses containing complementary literals.

$C_1$ and $C_2$ are clauses i.e. disjunctions of literals, and a literal is an atom or the negation of an atom.

In order to resolve $C_1$ and $C_2$ with literal $L$ we must have that in one of the clause the literal $L$ occurs negated, i.e. $C_1 := D_1 \lor \ldots \lor L$ and $C_2 := D_2 \lor \ldots \lor \lnot L$.

Thus, the cases are: either $A \vDash L$ or not.

If $A \vDash L$ , due to the fact that $A \vDash C_2$ and $L$ occurs negated in $C_2$, we must have $A \vDash C_2 \setminus \{ \lnot L \}$.

Title: resolvent clause theory


问:
Spherical Trigonometry: Spherical triangle
ABC is an equilateral spherical triangle in which small displacements are made, in the sides and angles, of such a nature that the triangle remains equilateral. Prove that
$$
\frac{da}{dA} = \cos\left({\frac{A}{2}}\right) \cot\left({\frac{a}{2}}\right)
$$

I would have mentioned my progress. But all I have been able to done is differentiate different trigonomeric formulae to get an expression which is of no use to simplify, since it never matches with answer on a numerical verification.

Please help. A hint on how to start might be enough.

NOTE: This question is Q9 of Execrcise 1 from W.M. Smart's Spherical Astronomy.

答:
On page 7 there is a "fundamental formula" that for an equilateral triangle is

$$ \cos a = \cos^2 a + (\sin^2 a) \cos A$$

The rest is a differential calculus problem of implicit differentiation on the equation to write a relation between $da$ and $dA$ and thence to calculate $\frac{da}{dA}$; followed by a (plane) trigonometry problem of using half- or double-angle formulas to compare the formula for $\frac{da}{dA}$ from differentiation to the formula given in the question.

One reason to use the half-angle expressions is that if the $a$ and $A$ terms are moved to opposite sides of the equation, the $A$ side is well behaved, just a cosine, but on the $a$ side there is a removable singularity at $\sin a = 0$, that can be thought of as coming from factors of $\sin^2 (a/2)$ in the numerator and denominator.  It could be easier or more illuminating to first remove those factors and then do the implicit differentiation.


问:
Dense and locally compact subset of a Hausdorff space is open
Let $X$ be a Hausdorff space and let $D \subseteq X$ be locally compact and dense in $X$. Why is $D$ open?

I can see that $D$ is regular but don't see why $D$ is in fact open.

答:
Here is a straightforward proof inspired by Theorem 2.70 in Aliprantis and Border's Infinite Dimensional Analysis (3rd ed.), p.56). Let $p \in D$. Since $D$ is locally compact, there is a neighborhood of $x$ in $D$ which is compact in $D$ and a neighborhood $V$ of $x$ in $D$ which is open in $D$ and $V \subset U$.

First, it is easy to see that $U$ is also compact in $X$. Since $X$ is Hausdorff, it implies that $U$ is closed (see, for example, Proposition 4.24 in Folland's Real Analysis (2nd ed.), p.128), and consequently $\overline{U}=U$.

By definition, there is an open set $W$ in the topology of $X$ such that $V = W\cap D$. Since $D$ is dense in $X$, it follows that $$W \subset \overline{W} = \overline{W \cap E} = \overline{V} \subset \overline{U} = U \subset D.$$ Hence for every  $p \in D$ there is a neighborhood of $p$ open in $X$ which is included in $D$, i.e., $D$ is open in $X$.


问:
Generalizing Heron's Formula for Cyclic $n$-gons
Consider the following extension of Heron's Formula.
For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, \dots, x_n$ and semi-perimeter $P = \frac{1}{2} \left( x_1 + x_2 + \dots + x_n\right)$ define:
$$ M = \sqrt{P^{4-n} (P-x_1)(P-x_2) \dots (P-x_n)} $$

After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?

To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $\displaystyle \lim_{n \rightarrow \infty} M(R_n)/A(R_n) = \pi/e$.

Observe that $M$ is a homogeneous function $M(\lambda C) = \lambda^2 M(C)$. Area is also homogenous in the same degree $A(\lambda C) = \lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?

答:
Not a full answer, but some assorted observations:

When $n=4$, the formula is exact; this is Brahmagupta's formula.

For a given cyclic ordering of side lengths in a polygon, the area is maximized by taking the polygon to be cyclic (which uniquely specifies it up to congruence), so lower bounds for the ratio $M/A$ in the cyclic case are actually lower bounds for all polygons.

There are, for each $n$, cyclic polygons with $M$ arbitrarily close to $A$; simply take any triangle with side lengths $a,b,c$ and copy one of its vertices $n-3$ times. (To avoid degenerate cases, just take the neighboring points to be extremely close to one of the vertices - the limits work out fine.) Then the area is that of the original triangle, and the Heron-like formula is
$$\sqrt{s^{4-n}(s-a)(s-b)(s-c)(s-0)(s-0)\ldots(s-0)}=\sqrt{s(s-a)(s-b)(s-c)}$$
In fact, this same process shows that the interval of possible ratios for a given $n$ is a subset of the possible ratios for higher $n$; in particular, a bound for $n$-gons implies the same bound must hold for all $k$-gons with $k<n$.

In all experimental data I have collected thus far (tens of thousands of randomly-generated cyclic polygons), $M/A$ appears to be bounded below by $1$ and above by $\pi/e$. For large random cyclic polygons (convex hull of many independently chosen random points on the circle), "typical" such polygons have a ratio very close to $\pi/e$. This follows from the fact that a generic large polygon has $M\approx s^2/e, A\approx s^2/\pi$, with each approximate equality being a $\le$ inequality in general.

For all $n$, the regular $n$-gon appears to maximize the ratio $M/A$, though this is based on experimental data only. (I suspect that proving this is the best route to showing the upper bound of $\pi/e$.)

Some plots of the ratio $M/A$ for different numbers of independently chosen points on the circle are shown below.

$n=5$, $100000$ samples:

(The regular pentagon has $M/A \approx 1.013$.)

$n=6$, $100000$ samples:

$n=10$, $10000$ samples:

$n=100$, $10000$ samples:

($M/A=1$ is of course possible here, it's just vanishingly unlikely.)


问:
Question about proving differentiability
I'm asked to prove $f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}$ is differentiable for every x > 1
and calculate it's deriviate.

I have no idea how to approach this question.. Do I need to define a function g that will be the inverse of f?

i.e. $g\left(x\right)=\left(\frac{e^{x}}{x}\right)^{e^{x}}$, prove it's continuous, differential and that it's derivate is not 0 and then use the chain rule?

My attempt:

$f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}=e^{\ln\left(x\cdot\ln x\right)^{\ln x}}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$

Note that by the algebra of differential functions $e^x,ln(x)$ are differential for every $x > 1$. Therefore $g(x)={\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$ is also differential for every $x > 1$, and $f$ as a whole is differential for every $x > 1$. Therefore by the chain rule we have

$f'\left(x\right)=\left(e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\right)^{'}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{1}{x}\cdot\ln\left(x\cdot\ln x\right)\right)+\left(\frac{1}{x\cdot\ln x}\cdot\left(\ln x+1\right)\right)\cdot\ln x\right)$

$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\cdot\ln x\right)}{x}\right)+\left(\frac{\ln^{2}x+\ln x}{x\cdot\ln x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)}{x}\right)+\left(\frac{\ln x+1}{x}\right)\right)$

$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)+\ln x+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{2\ln\left(x\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\cdot\ln x\right)+1}{x}\right)\right)$

答:
if $x \ge e$ then $\ln x \ge 1$ so $x\cdot \ln  x > 0$ and $\ln (x\cdot \ln  x)$ exists and $x\cdot \ln x = e^{\ln (x\cdot \ln  x)}$.

And then $f(x) = (x\cdot \ln x)^{\ln x}= [e^{\ln (x\cdot \ln  x)}]^{\ln x} = e^{\ln x\dot \ln(x\cdot \ln x)}$.

And that can be differentiated by the using the chain rule and the product rule many times.

If we use the notation $exp(x)= e^x$ and $exp'(x) = $ derivative of $exp(x) =$ derivative of $e^x = e^x=exp(x)$ then

First use of chain rule:  $[exp(\ln x\dot \ln(x\cdot \ln x))]' = exp'(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))' =$

$exp(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))'=$

$(x\cdot \ln x)^{\ln x} \times (\ln x\dot \ln(x\cdot \ln x))'=$

Now apply the product rule;

$(x\cdot \ln x)^{\ln x} \times [\ln' x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$

Now apply the chain rule again:

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln'(x\cdot \ln x)\times (x\cdot \ln x)')=$

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot \ln x)')=$

And use the product rule again

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot (\ln x)'+ (x)'\ln x))=$

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot\frac 1x+ 1\cdot\ln x))=$

$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (1+ \ln x))=$

$(x\cdot \ln x)^{\ln x}[\frac {\ln(x\cdot\ln x)}x + \ln x(\frac 1{x\ln x} + \frac 1x)]$.

Sof $1 < x < e$ we have $x\ln x < 0$ so $\ln (-x\cdot \ln x)$ exists and $x\cdot \ln x= e^{-\ln(-x\ln x)}$.

And do what we did above.


问:
If $X$ is separable then so is $B_X.$

Let $X$ be a separable normed linear space. Then show that $B_X = \{x \in X\ |\ \|x\| = 1\}$ is also separable.

Since $X$ is separable it has a countable dense subset. Let $D = \{x_n\ |\ n \in \Bbb N \}$ be a countable dense subset of $X.$ WLOG we may assume that $0 \not\in D$ for otherwise we could take $D \setminus \{0\}$ which continues to be dense in $X.$ Then I guess that $D' = \left \{\frac {x_n} {\|x_n\|}\ \bigg |\ n \in \Bbb N \right \}$ is a dense subset of $B_X.$ Let $y \in X$ with $\|y\| = 1.$ Let us choose $\varepsilon \gt 0$ arbitrarily. Then I need to show that $B(y,\varepsilon)$ intersects $D'.$ If $y = x_n$ then we are through. If not then there exists a sequence $\{x_{k_n}\}_{n \geq 1}$ in $D$ converging to $y.$ Since norm is a continuous function it follows that $\|x_{k_n}\| \to \|y\| = 1$ as $n \to \infty.$ But this shows that $\frac {x_{k_n}} {\|x_{k_n}\|} \to y$ as $n \to \infty,$ proving that $D'$ is dense in $B_X.$

Does my proof make sense? Can anybody plaese check my argument above?

Thanks in advance.

答: A shorter proof could be : If $X$ is metric and separable then has a countable basis. Any subspace of a countable basis space is countable basis (take the trace on the subspace), and in the end recall that if a space has a countable basis is separable.

Title: If $X$ is separable then so is $B_X.$


问:
Question about quadratic maximum/minimum word problem with linear modeling
I have the below problem from a textbook:

It goes through the solution. I don't get this part here:

This is a simple part, as this is just describing a linear equation. I thought I knew how to model these. Why is it not Q = 84,000 - 2,500p? In English, I can say we are starting at 84,000 subscribers, and deceasing 2,500 subscribers for every $1 we increase the price. It seems so easy! But in this problem they make me solve for b. I've done other linear equation problems where I didn't have to do this, so I'm a little confused.

*I guess what I'm confused about is, when reading a word problem, how do I know something is a starting value or a regular y/output value? knowing that will help me determine where I need to plug in these figures.

答:
Your approach without calculating $b$ via an equation is also possible but you need to $\color{blue}{\text{cater for the case $p=30$}}$.

So, your approach should lead directly to

$$Q(p) = 84000 - 2500(p\color{blue}{-30})$$


问:
Notation for changing a component of a vector
Is there a notation for changing a component of a vector in a specific index to a different number? Or is such an operation not used in maths? E.g. in programming languages you can do something like vector = [1, 2, 3]; vector[0] = 4.

If there is no way of showing changing a component by changing the vector itself, is there a notation for showing a new vector with the same components as the other vector apart from the component that is being changed? So e.g. in the example above the new vector would be [4, 2, 3].

答:
This is odd, but you may simply define a vector $\mathbf{x} = [x_0, x_1, \cdots, x_{n-1}]^T$ where $x_i$ denotes the $i$th element of $\mathbf{x}$. For example, you may then modify individual element as either

$$x_i = \alpha$$
or
$$x_i \leftarrow \alpha$$

where $\alpha$ is a number.


问:
Qing Liu, Exercise 2.4.10
The problem is the following: Let $f:X\to\operatorname{Spec}(A)$ be a quasi-compact morphism. Let $I\subset A$ be an ideal. Show that $f(X)\subset V(I)$ if and only if $f^\#(\operatorname{Spec}(A))(I)$ is nilpotent in $\mathcal{O}_X(X)$. For simplicity, I'll put $\varphi:=f^\#(\operatorname{Spec}(A))$.

Since $X$ is quasi-compact, I'd like to use the fact that the canonical mapping $\mathcal{O}_X(X)_h\to\mathcal{O}_X(X_h)$ is injective (where $X_h = \{x\in X\;:\; h_x\in\mathcal{O}_X(X)^\times\}$). Using this condition, we have that $h$ is nilpotent if and only if $\mathcal{O}_X(X)_h = 0$, so if I can show $\mathcal{O}_X(X_h) = 0$ for all $h\in\varphi(I)$ if and only if $f(X)\subset V(I)$, then I'm all set.

Unfortunately I'm unable to prove their equivalence. Any help would be much appreciated!

答:
$\mathcal{O}_X(X_h)=0$ iff for all affine opens $U=\operatorname{Spec}B$ of $X$ we have that $h$ is nilpotent in $B$. So we can reduce to the case $X=\operatorname{Spec}B$ affine. if $f(X)\subseteq V(I)$ we get $$X=f^{-1}(f(X))\subseteq f^{-1}(V(I))=V(\varphi(I))$$
The other direction is similar.

Title: Qing Liu, Exercise 2.4.10


问:
An integral problem to compute $\iint_{y\ge x^2+1} \frac{1}{x^4+y^2}\,dx\,dy.$
How to compute $$\iint_{y\ge x^2+1} \frac{1}{x^4+y^2}\,dx\,dy.$$ I've tried using the transformation $x = \sqrt{r\cos(\theta)},y=r\sin(\theta)$. But I found it hard to compute the integral with $\theta$.

答: @metamorphy's tip is to first evaluate$$\int_{x^2+1}^\infty\frac{dy}{x^4+y^2}=[x^{-2}\arctan(yx^{-2})]_{x^2+1}^\infty=x^{-2}(\tfrac{\pi}{2}-\arctan(1+x^{-2}))$$so the double integral is$$\begin{align}\int_{-\infty}^\infty x^{-2}(\tfrac{\pi}{2}-\arctan(1+x^{-2}))dx&=\int_0^\infty x^{-2}(\pi-2\arctan(1+x^{-2}))dx\\&\stackrel{z=1/x}{=}\int_0^\infty(\pi-2\arctan(1+z^2))dz\\&=\left[\pi z-2z\arctan(1+z^2)+\sqrt{2}\sum_\pm e^{\pm3\pi i/8}\arctan\frac{ze^{\pm\pi i/8}}{\sqrt{2}}\right]_0^\infty\\&=\sqrt{2(\sqrt{2}-1)}\pi.\end{align}$$


问:
Degree of a splitting field
I came across something related to the degree of a splitting field for a polynomial over a field $K$.  Let's suppose $p \in K[x]$ with degree $n$, and $p$ has irreducible factors $f_{1}, \ldots, f_{c}$ with respective degrees $d_{1}, \ldots, d_{c}$.

Ok, I know we can construct the splitting field as a tower of extensions.  BUT here is the question: According to Wikipedia, the degree of the splitting field is $\leq n!$, but why $n!$? Is there any way to achieve this bound? I would guess that the degree would be bound by $\prod_{i} d_{i}$, which could never be this big no matter the values of the $d_{i}$.  Where is the error in my thinking?

答:
Let $L_i$ be a splitting field of $f_i$ over $L_{i-1}$, whence $L_0$ is defined to be $K$ and $L_c=L$. Then $[L:K]=[L_c:L_{c-1}]\cdots [L_2:L_1][L_1:K]$. For each term in the product, $[L_j:L_{j-1}]\le [L_j:K]$, the degree of any splitting field of $f_j$ over $K$.

If $f_j$ has degree $1$, the result is obvious, so assume $f_j$ has degree $\gt 1$. By induction, consider $K[x]/(f_j(x))$ which is a field, then we have an extension  $K_1/K$ such that $f_j$ has a root $\alpha$ in $K_1$ and thus $f_j(x)=(x-\alpha)g_j(x)$ in $K_1[x]$, which may be taken as $K(\alpha)[x]$. The degree of $g_j$ is $d_j-1$ and so the degree of any splitting field of $g_j$ over $K(\alpha)$ is $\le(d_j-1)!$ by the induction hypothesis. $f_j$ and the minimal polynomial of $\alpha$ over $K$ differ only by the leading coefficient, and hence have the same degree which is $[K(\alpha):K]=d_j$. Hence, $[L_j:K]=[L_j:K(\alpha)][K(\alpha):K]\le d_j!$

Therefore, $[L:K]\le \prod _{i=1}^c d_i!\le (\sum _{i=1}^c d_i)!=n!$. Note that the equality in place of the second inequality may occur when $p$ is itself irreducible, in which case we only have $d_1$ which is exactly $n$.

Title: Degree of a splitting field


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
If you want to be more precise you could write:

  $f'(x) = \lim_{h→0} \frac{(x+h)^2-x^2}{h}$ if the limit exists

    $= \lim_{h→0} (2x+h)$ if the limit exists

    $= 2x$.

Meaning that each line only holds "if the limit exists". But we do not have to actually bother to do so in most cases for two reasons:

It is usually easy enough to mentally add such conditions and check that we did not at any point rely on the existence of the limit.

If we allow expressions to attain an "undefined value", and define that every expression with an "undefined" subexpression is itself undefined, then we do not even have to write the condition "if the limit exists"! If the limit is not defined, then the "$\lim \cdots$" expression would simply have value "undefined", which will not lead to any incorrect conclusions.


问:
On the uniform convergence of the series of the following functions
prove that $$\sum_{n=1}^{\infty}\int_{0}^{x}t^n\sin\pi tdt$$Uniform convergence on$[0,1]$.
I try to use Weierstrass discriminant
$$\left|\int_{0}^{x}t^n\sin\pi tdt\right|\leqslant\int_{0}^{x}t^ndt=\frac{x^{n+1}}{n+1}$$
Uniform convergence when $x < 1$.I don't know if this is wrong,I want to know how to judge the convergence and divergence at 1.Thanks.

答: I got it,let $$u_{n}(x)=\int_{0}^{x}t^n\sin\pi tdt$$ be aware$$0\leqslant u_{n}(x)\leqslant u_n(1)=\int_{0}^{1}t^n\sin\pi tdt=\left|\frac{\pi}{n+1}\int_{0}^{1}t^{n+1}\cos\pi tdt\right|<\frac{\pi}{(n+1)(n+2)}.$$according to Weierstrass discriminant,$$\sum_{n=1}^{\infty}\int_{0}^{x}t^n\sin\pi tdt$$Uniform convergence on$[0,1]$.


问:
Extreme values of a multivariable function lying on a helix
Suppose I've a function $f(x,y,z)$ and a helix. There are parameterised points on the helix (using the parameter $t$) such that-
  $\:\:\:\:\:\:\:\:\:$ $x=cos(t)$ , $y=sin(t)$ , $z=t$
 Now, the partial derivatives of $f$ at the given points are given by-
 $\:\:\:\:\:\:\:\:\:$ $f_x=cos(t)$ , $f_y=sin(t)$ , $f_z=t^2$
 So, at what points on the helix can $f$ take an extreme value?

My approach:

If the helix is represented by the curve $c(t)$, then $f(x,y,z)$ must be lying on $c(t)$. As a result, $f(cos(t),sin(t),t)$ must satisfy the helix $c(t)$, where each input in $f$ is a function of $t$. In that case, $f'(t)=\nabla f c'(t)$, thus fetching me $t^2$ as the output of $f'(t)$. For an extrema, $f'(t)=0$ is valid, so setting $t^2$ equal to zero, I get the required points as $(1,0,0)$.

My doubt:

In reality, $f(x,y,z)$ doesn't have an extremum. But, how is this justifiable, when I do get a value of $t(=0)$ corresponding to an extreme value of $f$? Is it somewhat, an inflection point at $(1,0,0)$? Can someone help me out in this with a lucid explanation? Thanks.

答:
Your situation involves a function $(x,y,z)\mapsto f(x,y,z)$ and a helix $t\mapsto {\bf c}(t):=(\cos t,\sin t,t)$ in ${\mathbb R}^3$. You are told to study the composed function
$$g(t):=f\bigl({\bf c}(t)\bigr)\ .$$
You can view $f(x,y,z)$ as temperature at the point $(x,y,z)$ and $t\mapsto{\bf c}(t)$ as orbit of a space craft. The function $t\mapsto g(t)$ is then the temperature measured on the space craft.

By the chain rule and information given about $f$ one has
$$g'(t)=\nabla f\bigl({\bf c}(t)\bigr)\cdot{\bf c}'(t)=(\cos t,\sin t, t^2)(-\sin t,\cos t,1)=t^2\ .$$
This shows that $g'(t)\geq0$ for all $t$, and $=0$ only at $t=0$. It follows that $g$ is monotonically increasing. In fact $g(t)={1\over3}t^3+C$ for some constant $C$. When you draw the graph of $t$ you have an inflection point at $t=0$.

By the way: The function $f(x,y,z):={1\over2}(x^2+y^2)+{1\over3}z^3$ is an example for an $f$ described in the story.


问:
Prove that if the altitude and median of a triangle form equal angles with sides then the triangle is right.
Problem statement:

Prove that if the altitude and median drawn from the same vertex of a nonisosceles triangle lie inside the triangle and form equal angles with its sides, then this is a right triangle.

After many attempts, I came up with this: let $ABC$ be the triangle where $CH$ is the altitude, $CM$ is the median, and name the angles $ACH = MCB = \theta$, $CAH = \alpha$, $CBM = \beta$ and $HCM = \gamma$. Using the sine theorem in the triangle $CMB$ we get $$\frac{MB}{\sin(\theta)} = \frac{MC}{\sin(\beta)},$$ while using the sine theorem in the triangle $ACM$ we get $$\frac{MC}{\sin(\alpha)} = \frac{MA}{\sin(\theta + \gamma)}.$$ Combining these equations I found $$\sin(\alpha) \sin(\theta) = \sin(\beta) \sin(\theta + \gamma).$$ Applying the identity $$\sin(a)\sin(b) = \frac{\cos(a-b) - \cos(a+b)}{2}$$ I concluded $$\cos(\alpha - \theta) = \cos(\theta + \gamma - \beta).$$ Since all angles are within $[0, \pi]$ range I concluded $$\alpha - \theta = \theta + \gamma - \beta,$$ thus $2 \theta + \gamma = \alpha + \beta$. From the original triangle we know $$2 \theta + \gamma + \alpha + \beta = \pi,$$ and combining the equations proves the triangle is right.

Is this correct? Is there a synthetic way to do it?

答:
Extend median $\overline{AM}$ of $\triangle ABC$ to meet the circumcircle at  $M'$, and let $D$ be the foot of the altitude from $A$. (Note that $D$ and $M$ are distinct —and, thus, $\overline{AM'}\not\perp\overline{BC}$— because we assume $\triangle ABC$ is non-isosceles.)

Since $\angle M' \cong\angle B$ (by the Inscribed Angle Theorem), we conclude $\angle ACM'\cong\angle ADB=90^\circ$, so that $\overline{AM'}$ is a diameter. Now, a diameter can only bisect a non-perpendicular chord at its own midpoint (aka, the circumcenter); consequently, $M$ is that circumcenter, $\overline{BC}$ is a diameter, and $\angle BAC$ is a right angle. $\square$


问:
Thin sets in free ultrafilters
Let's all a set $A\subseteq \omega$ (where $\omega$ denotes the set of non-negative integers) thin if $$\lim\sup_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}=0.$$

Is it true that there are free ultrafilters on $\omega$ containing thin sets, and there are also free ultrafilters on $\omega$ containing no thin sets?

答:
Let $X$ be thin and infinite, let $F$ be the cofinite filter. Then $F\cup\{X\}$ has the fip, so it extends to an ultrafilter.

To avoid all thin sets remember that upper density is subadditive, so the family of upper density zero sets is closed under finite unions, and the family of their complements has the fip and contains $F$ again.

Title: Thin sets in free ultrafilters


问:
There is concept of finite sets that can have only one "interpretation"?
In our mind we have a naive idea of what a set is, and in nature we can only observe something that behave like a finite set, ZFC (or set theories in general) tries to catch these properties in axioms. Like Peano axioms describe the nature of natural numbers, and all the "things" that behave in that way are "isomorphic to the naturals, that property seems is called categoricity (in my naive understanding is that all models of the naturals are isomorphic).

Well, I think that all agree on how sets behave, in the sense that in our experience of the word, we can see only a kind of finite set, or like a bag full of something (or empty) or as a property (comprension axiom), and all sets are finite.

My question is there is a concept of set (axioms), that have a "unique interpretation" and are exactly the naive concept of set?

答:
If we really want to consider finite sets, we want not only that they will be finite but also that their elements will be finite, and the elements of their elements will be finite, and so on. In short, we want that the transitive closure of the set will be finite. (Where the transitive closure of a set $x$ is the smallest set $y$ such that $x\subseteq y$ and $t\in s\in y\rightarrow t\in y$.)

In a model of $\sf ZFC$ we can define the set of all hereditarily finite sets. This is commonly denoted by $V_\omega$. It is a countable set, and in fact it is exactly the [increasing] union of $\mathcal P^n(\varnothing)$, the iterated power set operation of the empty set.

To add more on this, $V_\omega$ satisfies all the axioms of $\sf ZFC$ except the axiom of infinity, and it does satisfy its negation: there is no inductive sets, and therefore there are no infinite sets either. So does this theory, $\sf ZFC$ with the axiom of infinity replaced by its negation good enough to decide what are finite sets?

Generally, it seems that yes. But as any first-order theory with infinite models, we can produce models of the same theory which are as large as we want, and that's not very good. But much like with the real numbers, or with $\sf PA$, we can move to second-order logic (with full semantics) and have a categorical theory!

We slightly modify our theory, and instead of the schema of replacement for definable sets, we add a full-blown second-order axiom of $\in$-induction:

$$\forall A(\varnothing\in A\land\forall x(\forall y(y\in x\rightarrow y\in A)\rightarrow x\in A)\rightarrow\forall x(x\in A))$$
  Every class $A$ such that $\varnothing$ is a member of $A$, and for every $x$, whenever all the members of $x$ are in $y$ then $x$ is itself a member of $A$, then $A$ is the entire universe.

This should be, if I am not mistaken, to a second-order formulation of the replacement schema, although I don't recall the proof at this moment.

Why does this guarantees that there is only one model? Suppose that $M$ was a model of our modified theory, then it has a naturally defined power set operation and an empty set. Construction by recursion $V_\omega$ as the union of the finite iterations of $\mathcal P^n(\varnothing)$, and we have a class satisfying the assumption of this induction axiom. Therefore it must be equal to the whole universe!

Note that the above proof happens outside the model $M$. Very much like how the proof that there is only one model of $\sf PA_2$ occurs outside $\Bbb N$.

It seems to me that this theory is reasonable to capture what is truly finite in the universe of sets.


问:
for which $\alpha$ is this function integrable
Let's say $P\in \mathbb{R[X]}$ a polynomal. For witch $\alpha \in \mathbb{R}$ is the function

$f:(0,+\infty)\to \mathbb{R}:f(x)=\frac{P(\sin(x))}{x^{\alpha}}$  integrable.

My idea/ solution:

First we see for $x \to +\infty$

we know that $\frac{1}{x^{\alpha}}=\Theta(\frac{1}{x^{\alpha}})$ for $x\to 0$. This means that for $x\to 0$ that $\frac{1}{x^{\alpha}}$ is integrable if $\alpha <1$.

I found that $\sin(x)=\Theta(x)$. So the function $\frac{\sin(x)}{x^{\alpha}}=\theta(\frac{1}{x^{\alpha -1}})$ and is integrable for $\alpha <2$

Now I watch $P(\sin(x))$. If it's zero then there is no problem and is it integrable. If it's not zero I'ts still integrable because $P(\sin(x))$ is a polynomal.

Now here is where the problem starts. If $P(\sin(x))$ is always integrable then the fact that $f$ is integrable depends only at $\frac{1}{x^{\alpha}}$. But for $x\to 0$ , $\frac{1}{x^{\alpha}}$ is integrable if $\alpha <1$ but for for $x\to +\infty$ that $\frac{1}{x^{\alpha}}$ is integrable if $\alpha >1$. Can someone help me out?

答:
If you only need Lebesgue (i.e. absolute) integrability, just complete the analysis of the behavior of $f(x)$ at $x\to 0$ and $x\to\infty$. Let $P(x)=\sum_{k=0}^n c_k x^k$ and assume it is not identically zero.

As $x\to 0$ we have $f(x)=\Theta(x^{d-\alpha})$ with the smallest $d$ such that $\color{blue}{c_d\neq 0}$. Hence $\int_0^{2\pi} f(x)\,dx$ (say) converges if and only if $d-\alpha>-1$, that is, iff $\alpha<d+1$. As for $\int_{2\pi}^\infty|f(x)|\,dx$, it converges if $\alpha>1$ (because of the comparison with $x^{-\alpha}$ as you have noticed) and diverges otherwise (hints: $x^\alpha|f(x)|>c$ for some $c>0$ on some interval of values of $x$; this function is $2\pi$-periodic; comparison with $\sum_n n^{-\alpha}$). In total, the answer is "integrable iff $\color{blue}{1<\alpha<d+1}$".

More interesting would be to consider the convergence of $\int_{2\pi}^\infty f(x)\,dx$ in the improper sense (that is, as $\lim_{A\to\infty}\int_{2\pi}^A f(x)\,dx$). The answer here is: it converges for $\alpha>0$ if $\int_0^{2\pi}P(\sin x)\,dx=0$, and for $\alpha>1$ otherwise (the main hint here is to use Dirichlet's test). In terms of $c_k$, we have $$\int_0^{2\pi}P(\sin x)\,dx=0\iff\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(2k-1)!!}{(2k)!!}c_{2k}=0\qquad\color{LightGray}{[0!!=(-1)!!=1]}$$


问:
Number of way choose ordered distinct quadruples
I think in following manner: ∀x∈{1,2,...,n} if A included it then it must be in B∨C∨D or if x not in A it can be at any B∨C∨D such a way give us 7n .Any one can verify my solution?</p>

答: Hint: Numerator is $e^{\frac 1 n \ln (999)}-1$. And $\frac {e^{x}-1}x \to 1$ as $x \to 0$. Compare with $\sum \frac 1 {n^{1+\alpha}}$.


问:
Does $G$ being a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is a subgroup of $GL(n, \mathbb{Z})$?
It is known that any finite subgroup of $GL(n, \mathbb{Z})$ is isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for any odd prime $p$ (see here). I am wondering if there is a converse to to this: Does $G$ being isomorphic to a subgroup of $GL(n, \mathbb Z/p\mathbb Z)$ for all odd primes $p$ imply it is isomorphic to a subgroup of $GL(n, \mathbb{Z})$?

If this does hold, I'd be curious if it could be strengthened (perhaps being a subgroup for any infinite set of primes, or all but finitely many primes, would work), and if it doesn't, I'd be curious if any weaker form could be salvaged.

答:
Here's one thing we can say. Suppose a finite group $G$ embeds into $GL_n(\mathbb{F}_p)$ for all $p$; actually we can ask for the much weaker condition that $G$ embeds into $GL_n(\prod \mathbb{F}_p)$, and also weaken $G$ to be finitely generated. Then $G$ embeds into $GL_n(K)$ where $K$ is the quotient of $\prod \mathbb{F}_p$ by any of its maximal ideals containing the ideal of elements with finite support; these correspond to ultraproducts $\prod \mathbb{F}_p/U$, and importantly are fields of characteristic $0$.

Since $G$ is finitely generated, this embedding actually takes values in the finitely generated subring $R$ of $K$ generated by the matrix entries of a finite set of generators of $G$. So $G$ embeds into $GL_n(R)$ where $R$ is a finitely generated integral domain of characteristic $0$. This sort of manipulation is a key step in the proof of Malcev's theorem that a finitely generated linear group is residually finite.


问:
A question on the distributional derivative of $y^{-1/3}$
Let $\Omega := \{(x,y) \in \mathbb{R}^2 \ : \ 0<x<1, \ \sqrt x < y < 1\}$ and the function $u:\Omega \to \mathbb{R}$ defined as $u(x,y) := y^{-1/3}$.

Prove that $u$ belongs to $W^{1,p}(\Omega)$ for all $p \in [1,9/4)$

This is a problem that was gave to me and I had no problem in showing that $u \in L^p(\Omega)$ for all $p<9$. Then considering the derivatives of first order of $u$ the solution given by the text says that $u \in C^1(\Omega)$ so the distributional derivatives are the usual derivatives and the solution comes from the integral of $\frac{\partial u}{\partial y}(x,y)$ etc etc..

My question is: Why is $u\in C^1(\Omega)$? for $(0,y) \to (0,0)$ doesn't explode $u$'s common derivative?
My thought was to consider the distributional derivative near $(0,0)$ using dominated convergence theorem as

$$ \langle\frac{\partial u}{\partial y} , \phi\rangle = \lim_{\varepsilon \to 0^+} - \int_\varepsilon^1 dy\int_0^{y^2} dx \ \ y^{-1/3} \frac{\partial \phi}{\partial y}$$

And then going on to establish the distributional derivative..

Let me know what you think about it. Thank you.

答:
I suspect your text is using the convention that $C^1(\Omega)$ and $C^1(\overline\Omega)$ are not the same thing; the latter is a Banach space, but the first may not be. So since the domain avoids points where $y=0$, the function is positive and finite everywhere it is defined (without a uniform upper  bound).

(Since you can check which $L^p$ $u$ belongs to, you should be able to finish the question)


问:
Efficient way of explicitly writing Sylow groups of $S_5$
I am trying to write explicitly the Sylow 5-subgroups of the symmetric group $S_5$. Using the properties of the cardinal of the set of Sylow subgroups, I have come to the conclusion that there must be $6$ Sylow 5-subgroups. I know one of them is $H_1=\langle (1,2,3,4,5)\rangle$ and the other five subgruops should also be generated by a cycle of order 5. My problem is that $H_1$ contains several cycles of order 5 and I would like to write the rest of Sylow 5-subgroups making sure that I do not make the mistake of writing two subgroups that are actually equal but just "look different" because the generator is different.

The only way I can think of for approaching this problem is writing all of the elements of each subgroup and making sure that they are all different, but this is not very efficient. I was wondering if there might be some an efficient way to do this or a clever way of choosing the generator of each subgroup so as to make sure that all subgroups are different.

Any trick you can come up with would be very useful. Thanks in advance.

答:
Let's generalize to $S_p$.

The map $\sigma\mapsto\langle\sigma\rangle$ from $p$-cycles to Sylow $p$-subgroups in $S_p$ is a $(p-1)$-to-$1$ map, since every such subgroup has $p-1$ generators. There are $p!/p=(p-1)!$ elements of order $p$, so there are $(p-2)!$ Sylow subgroups.

Sylow theorems says the Sylow subgroups are conjugate, which by orbit-stabilizer means they correspond to cosets of the normalizer of one of them. If we let $\sigma=(12\cdots p)$ then any $\tau$ which fixes adjacent numbers but no others cannot be in the normalizer (consider how conjugation affects cycle types), so in paticular $S_{p-2}$ is a set of coset representatives for $N_{S_p}(\langle\sigma\rangle)$. Thus, the set of $p$-cycles $\pi$ fixing $p-1$ and $p$ form an irredundant list of generators for the $p$-Sylows.


问:
What is the equation to determine an element's index in a jagged array given it's index from the equivalent flattened array?
If one has a jagged array of N dimensions and an index of an element in the flattened representation of that array, is there an equation that would retrieve the index of that element in the original, jagged array?

Example:

Assume the jagged_array is:

```<code>[[0, 1, 2, 3].
 [4, 5],
 [6]
 [7, 8, 9, 10, 11]]
</code>```

The element 7 has an index of jagged_array_flattened[7]. If I have it's position in the flattened array, how can I get it's position in the original array (it would be jagged_array[3][0].

Would be cool if it did not matter how many dimensions you had, but I am mostly dealing with 2D arrays. If there is no equation, perhaps some pseudocode?

Thanks.

Edit: I feel like the N-dimensions makes it too difficult to generalize, so unless I am wrong, perhaps the focus should be on 2D arrays.

Edit 2: Here is a quick write up of my solution in code.

```<code>function getValue(list, flattenedIndex) {
    let firstIndexOfSubList = 0;
    let lastIndexOfSubList = 0;
    for (let i = 0; i < list.length; i++) {
        lastIndexOfSubList += list[i].length - 1;
        if (lastIndexOfSubList === flattenedIndex) return list[i][lastIndexOfSubList - firstIndexOfSubList];
        else if (Math.max(flattenedIndex, lastIndexOfSubList) === flattenedIndex) {
            lastIndexOfSubList++;
            firstIndexOfSubList = lastIndexOfSubList;
        } else return list[i][flattenedIndex - firstIndexOfSubList];
    }
}
</code>```

Edit 2: A solution with binary search if you have a precomputed list of lengths:

```<code>function binaryWhatever(jsonb, flattenedIndex) {
    let lo = 0;
    let hi = jsonb.length

    while (lo < hi) {
        let mid = Math.floor((lo + hi) / 2);
        if (flattenedIndex < jsonb[mid]) hi = mid;
        else lo = mid + 1;
    }
    let index1 = lo - 1;
    let index2 = flattenedIndex - jsonb[lo - 1];

    return [index1, index2]
}
<span class="math-container">```
```

答:
Your attempt can be improved slightly by combining the first and third cases together:

if (lastIndexOfSubList >= flattenedIndex) return list[i][lastIndexOfSubList - firstIndexOfSubList];

Note that if you only want to perform this conversion operation once, then you cannot do better than such a linear for-loop because you at least have to read the lengths of all the sublists up to the point where the answer is.

However, if you want to perform this conversion multiple times, then there is in fact a much more efficient way. To understand it, you will have to learn and understand binary search. The gist of the efficient solution is that you precompute the total length of the first $k$ sublists, for each $k$ from $1$ to list.length. Then you binary search that precomputed array (of cumulative lengths) to find the sublist in which the answer lies. And then you can extract the answer easily from there.

Have a try first and let me know in the comments if you need further help!


问:
Equivalence between two definitions of affine algebraic set.
In my algebraic geometry course, we have seen two definitions of an affine algebraic set. The first says that $X$ is an affine algebraic set if
$$X = \{x \in \mathbb{A}^n(k)~|~f_i(x) = 0, \forall i \in I\}$$
where $f_i \in k[X_1, \ldots, X_n]$ for all $i \in I$. The second one defines $X$ as a representable functor from $\text{Alg}_k$ to $\text{Set}$, i.e. there exists $A \in \text{Alg}_k$ such that
$$X = \text{Hom}_{\text{Alg}_k}(A, -).$$
Now the two definitions should match for $k$ algebrically closed and $A \in \text{Alg}_k$ finitely generated and reduced (= nilpotent-free).

There is a step in this equivalence that I don't understand: We take $X$ as in the first
definition, we consider the associated coordinate algebra
$$\mathcal{O}(X) = k[X_1, \ldots, X_n]/I(X)$$
and then we should have
$$\text{Hom}_{\text{Alg}_k}(\mathcal{O}(X), k) \cong X.$$
I don't really get this isomophism, does anyone know how to prove this or just a way to understand it intuitively ?

答:
The intuition of this so-called functor of points viewpoint is the following: An $K$-algebra homomorphism $\varphi: \mathcal{O}(X) \to k$ is determined by its action on $X_1, \dots, X_n$! More precisely, a homomorphism is fully determined by $\varphi(X_1), \dots, \varphi(X_n)$ since $\varphi$ is a $K$-algebra homomorphism and moreover, $X_1, \dots, X_n$ satisfy the relations given by $I(X)$, so hence $\varphi(X_1), \dots, \varphi(X_n)$ also satisfy those. In other words, we give coordinates $x_1, \dots, x_n$ of a point on $X$.

This also formalizes to a proof of your assertion.

Let‘s illustrate this with an example.

Example. Consider $\mathcal{O}(X) = k[X_1,X_2]/(X_1 - X_2)$. An element $\varphi \in \operatorname{Hom}(\mathcal{O}(X), k)$ is then fully determined by $\varphi(X_1), \varphi(X_2)$. Moreover, we have $$\varphi(X_1) - \varphi(X_2) = \varphi(X_1 - X_2) = \varphi(0) = 0.$$ So $(\varphi(X_1), \varphi(X_2)) \in V(X_1 - X_2) = X$.

Of course, the exact same thing can be done with any other affine variety. For me, understanding this viewpoint was what helped me better understand and further appreciate the Yoneda Lemma.


问:
Is it possible to find the control points of a quadratic bezier curve, given the starting, ending and stationary point?
I know from this question on SO that it is possible to get the stationary point of a bezier curve given the control points, but I want to know wether the opposite is true:

If I have the start and end points of a Parabola, and I have the maximum point, is it possible to express this a quadratic bezier curve?

I have seen this answer explaining that you usually would need 6 points on a cubic to effectively describe a bezier curve, but looking at the question itself, it describes a few special cases where you can find the control points on a cubic, and I am curious on wether a quadratic bezier curve has such points, and wether a stationary point is one of them

The variable t of the stationary point is not avalible

I am coming from a computer science background, and although im not too weak in math, any answers/articles that don't use complicated maths and keep it simple would be appreciated

If this is not possible, would it help of the stationary point is always a maximum point and the x coordinate if the max point is the mean of the x coordinates of the start and end points?

By maximum point I mean the point with the maximum value for its y coordinate and by stationary I meant the point with the maximum or minimum value for y

答:
Define the following:

The quadratic Bézier has control points $(x_0,y_0)$ (start), $(x,y)$, $(x_2,y_2)$ (end)

The third known point is at $(x_1,y_1)$ with parameter $t$ (in this part of the answer, this is the point with extremal $y$-coordinate)

The curve's equation is then
$$x(t)=(1-t)^2x_0+2t(1-t)x+t^2x_2\tag1$$
$$y(t)=(1-t)^2y_0+2t(1-t)y+t^2y_2\tag2$$
and the constraints are $x(t)=x_1$, $y(t)=y_1$ and $y'(t)=0$. Eliminating $y$ in the last two equations gives
$$\Delta y_2t^2-2\Delta y_1t+\Delta y_1=0\qquad(\Delta c_i=c_0-c_i)$$
$$t=a\pm\sqrt{a(a-1)},a=\frac{\Delta y_1}{\Delta y_2}\qquad(\Delta y_2\ne0)$$
$$\Delta y_2=0\implies t=\frac12$$
There is a solution in general iff $a\not\in(0,1)$ and only one choice of sign will give $t\in[0,1]$. Once $t$ has been found, it may be substituted back into $(1)$ and $(2)$ to find $x$ and $y$ respectively.

If $(x_1,y_1)$ is the parabola's vertex, the third equation may be obtained from this diagram:

The curve's tangent at the vertex is perpendicular to the line passing between the midpoint of the curve endpoints and the third control point, i.e.
$$\begin{bmatrix}\frac{x_0+x_2}2-x\\\frac{y_0+y_2}2-y\end{bmatrix}\cdot\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}=0$$

A resultant computation followed by factoring out bogus solutions then shows that $t$ solves the cubic
$$\begin{align}
(\Delta x_2^2+\Delta y_2^2)&t^3\\
-3(\Delta x_1\Delta x_2+\Delta y_1\Delta y_2)&t^2\\
+(\Delta x_1(2\Delta x_1+\Delta x_2)+\Delta y_1(2\Delta y_1+\Delta y_2))&t\\
-(\Delta x_1^2+\Delta y_1^2)&=0
\end{align}$$


问:
For what $\alpha$ do $\alpha$-universally connected spaces exist?
Let's call a topological space $X$ connected iff it can not be represented as a disjoint union of two non-empty open sets.

Now, suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Those two variants of connectivity are connected in the following ways:

If $X$ is $Y$-connected and $Y$ is connected then $X$ is also connected.

Proof: Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Converse is however generally false - for example, $\{(t, \sin(\frac{1}{t})|t \in \mathbb{R})\} \cup \{(0, 0)\}$ is connected, but not $[0;1]$-connected.

If $Y$ is not connected, then all topological spaces are $Y$-connected.

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected, Q.E.D.

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Suppose $x_0, x_1 \in X$. Thes $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now suppose $\alpha$ is a cardinal. let's call a topological space $Y$ of cardinality $\alpha$ $\alpha$-universally connected iff a topological space of cardinality $\leq \alpha$ being $Y$-connected is equivalent to it being connected (equivalently, $Y$ is connected and all connected spaces of cardinality at most $\alpha$ are $Y$-connected)

My question is:

For what $\alpha$ do $\alpha$-universally connected spaces exist?

This question is a follow up of that one: Do universally connected spaces exist?

答:
Partial answer.

For a given cardinal $\alpha$ the collection of all connected topological spaces with cardinality at most $\alpha$ (up to homeomorphism) is a set. Let $C$ be that set. Then $Y=\prod C$ is $\alpha$-universally connected. There's even stronger property here: every connected topological space of cardinality up to $\alpha$ is a quotient of $Y$.

Except that $Y$ itself may be (and probably is) of bigger than $\alpha$ cardinality, I'm not sure.


问:
Matrix representation of a symmetric positive definite matrix $X$
Suppose $X$ is positive semidefinite with eigenvalues $\lambda_i\ge 0$. We can write $X=Q\Lambda Q^T$ for a diagonal matrix $\Lambda$. But why is this equivalent to

$$X=\sum_i \lambda_iq_iq_i^T$$

答:
It's just a matter of carrying out the matrix multiplication. If the $q_i$ are the columns of $Q$, then $Q_{ij} = (q_j)_i$, the $i$-th entry of $q_j$. Also $\Lambda_{ij} = \delta_{ij}\lambda_j$. Hence,

$$ X_{ij} = \sum_{k,m} Q_{ik}\lambda_{k}\delta_{km}Q_{jm} = \sum_m \lambda_m (q_m)_i(q_m)_j.$$

The right hand side is also exactly the $ij$-element of $$ \sum_m \lambda_m q_m q_m ^T.$$


问:
Evaluation of a complex numbers partial sum
Let $w = e^{i\frac{2\pi}5}$. I would like to evaluate

$$w^0 + w^1 + w^2 + w^3 +...+ w^{49}$$

Can anyone please give me an idea how to evaluate the expression?

Thanks in advance

答:
Recall geometric series:
$$1+w+w^2 + \cdots + w^n = \dfrac{1-w^{n+1}}{1-w}$$
Also, make use of the fact that
$$e^{i(2k\pi + \theta)} = e^{i \theta},\,\,\,\,\, \text{where } k \in \mathbb{Z}$$
I trust you can finish it off from here. </p>


问:
Find two linear functions whose domain is $0 \leqslant t \leqslant 2$ and whose range is $1 \leqslant f(t) \leqslant 9$
I'm supposed to find two linear functions for the same domain and range, I have found one. How do I find the other?

The one I found:

Since Domain is $0 \leqslant t \leqslant 2$
And Range is $1 \leqslant f(t) \leqslant 9$

I took $f(0) = 1$ and considered constant $C = 1$

Then I took $f(2) = 9$ , Since linear equations are of the form $f(t) = vt + C$, v being slope and C being constant.

$$f(2) = v(2) + 1\implies 9 = v(2) + 1\implies v = 4$$

Thus the linear equation becomes $f(t) = 4t + 1$

How do I find the second linear function with the same Domain and Range?

This is the $24$th question in Strang's Calculus at section $1.1$

答: Once you get $f(t)=4t+1$, notice that $$1 \le f(t) \le 9 \iff 1 \le 10-f(t) \le 9$$ so $g(t)=10-f(t)=9-4t$ is the other linear function.


答:
Let's work in any general $S_p$ and look for Sylow $p$-subgroups, which must all be generated by $p$-cycles.  Any such subgroup must have a unique element that permutes $1$ to $2$.  Use that element to generate the subgroup, and (written in cycle notation) each permutation of the remaining elements gives you a unique subgroup.

Thus, $\sigma = (1~2~\tau)$ generates a unique Sylow subgroup where $\tau$ is any permutation of $\{3, 4, \ldots p \}$.  This also demonstrates that there are $(p-2)!$ such subgroups.

Title: Efficient way of explicitly writing Sylow groups of $S_5$

Body:
I am trying to write explicitly the Sylow 5-subgroups of the symmetric group $S_5$. Using the properties of the cardinal of the set of Sylow subgroups, I have come to the conclusion that there must be $6$ Sylow 5-subgroups. I know one of them is $H_1=\langle (1,2,3,4,5)\rangle$ and the other five subgruops should also be generated by a cycle of order 5. My problem is that $H_1$ contains several cycles of order 5 and I would like to write the rest of Sylow 5-subgroups making sure that I do not make the mistake of writing two subgroups that are actually equal but just "look different" because the generator is different.

The only way I can think of for approaching this problem is writing all of the elements of each subgroup and making sure that they are all different, but this is not very efficient. I was wondering if there might be some an efficient way to do this or a clever way of choosing the generator of each subgroup so as to make sure that all subgroups are different.

Any trick you can come up with would be very useful. Thanks in advance.


问:
Double integral $\int_{0}^{1}\int_{\sqrt{x}}^{1} \sin(\pi \ y^3)dy\ dx$
In this integral i have tried some substitution and have changed the order of integration too but i couldn't solve this

$$ \int_{0}^{1}\int_{\sqrt{x}}^{1} \sin(\pi \ y^3)dy\ dx$$
any suggestion would be great

Thanks

答:
Let's take a look at the graphs of the functions that describe the domain of integration:
We can easily observe that the area can be also written as $A=\{(x,y)\in\mathbb R^2:0\le x\le y^2,0\le y\le1\}$, so the integral becomes:$$\int_0^1\int_{0}^{y^2}\sin(\pi y^3)dxdy=\int_0^1y^2\sin(\pi y^3)dy=-\dfrac{\cos(\pi y^3)}{3\pi}\Bigg|_0^1=\dfrac{1}{3\pi}+\dfrac{1}{3\pi}=\dfrac{2}{3\pi}.$$


问:
Ordered pairs in sets
So I am asked to explain why $(x,y,z)=(x,(y,z))$ is a reasonable definition of an ordered triple. However, I am in a limbo at the moment let alone with trying to explain an ordered pair,(x,y), intuitively and why it is so except it is excepted by Mathematicians. So an order pair states that $(a,b)=\{\{a\},\{a,b\}\}$ with an unordered pair $\{a,b\}$. I understand the definition but not intuitively. I have looked at some postings on M.SE such as How can an ordered pair be expressed as a set? but the answer provided seemed a little vague, for me at least, since I am now beginning to learn Set Theory. So if anyone could explain it in English and demonstrate your point with notations, that would be great.

Also, the way I tried viewing or explaining it to myself "intuitively" is that since $(x,y)$ is an ordered pair on a coordinate plane, $\{\{a\}\}$ can be viewed as a set of points up to the x-coordinate and $\{\{a,b\}\}$ as counting the elements of the set up to the x-coordinate and upwards up to the y-coordinate. Not sure if this makes sense to anyone else but somehow this explanation is not satisfying even if it is correct.

答:
What would be considered a reasonable definition of an ordered pair, or a triplet? We would like it to have the following property:

$$(x,y)=(u,v)\iff x=u\textbf{ and } y=v.$$
(For triplets, of course, we want equality in all three coordinates.)

Assuming that we have a reasonable definition of an ordered pair, i.e. one that satisfies the above, then defining $(x,y,z)=(x,(y,z))$ is a good definition for a triplet. To see why, you need to unwind the definition of an ordered pair twice and conclude that two triplets are equal if and only if their elements are [pointwise-]equal -- which is a task I will leave to you to verify.

Title: Ordered pairs in sets