问:
almost closed mapping
Definition: The mapping$f:X\to Y$is almost closed if for every regular-closed set of $X$ like $F$, $f(F)$ is closed in $Y$.

Definition: A subset A of space (X,τ) has $\alpha$-property if every $\tau$-open cover of A has a finite subfamily whose $\tau$-closures cover A.

Let $f:X\to Y$  be a  continuous mapping and $Y$ is Hausdorff space.then  $X$ has $\alpha$-property if only if the mapping $f$ is almost closed.

My proof for one part:

suppose  $X$ has $\alpha$-property and $F\subset X$ is regular-closed. I know that every  regular-closed subset of a space $X$ that has $\alpha$-property,has also $\alpha$-property. The $\alpha$-property  is reserved  under continuous mapping, so  $F(f)$ has  $\alpha$-property and in Hausdorff space is closed. Then f is almost closed.

I have a problem with the other side and I can not prove that $X$ has  $\alpha$-property. Can you help me?

答:
The reason why you couldn't get the other direction is because it is not true.

From your proof of the direct implication, I assume you already know that when a space $X$ satisfies the $\alpha$-property, then $X$ is called an $H$-closed space. An excellent book that deals with this kind of spaces is Porter & Woods - Extensions and Absolutes of Hausdorff Spaces; particularly, check section 4.8.

For a counterexample to the direction that you couldn't get, consider the identity map $\text{Id}:\mathbb{R}\to \mathbb{R}$, where $\mathbb{R}$ is endowed with the usual Euclidean metric. It is clear that $\text{Id}$ is an almost closed continuous map with a Hausdorff codomain. However, if we consider the open cover $\mathcal{U}=\{(-n,n):n\in\mathbb{N}\}$, and let $\mathcal{V}$ be a finite and non-empty subfamily of $\mathcal{U}$, then there must exist $m\in \mathbb{N}$ in such a way that $(-m,m)=\bigcup\mathcal{V}$. So, if we take closures in both sides of the equality, we get $[-m,m]=\overline{ \bigcup\mathcal{V}}$. It follows that $\overline{ \bigcup\mathcal{V}}$ cannot cover $\mathbb{R}$.

Title: almost closed mapping


问:
One-to-one functions on a closed interval
From 1973 AP Calculus BC: Section I page 184 for answer (and 38 for the question)

Let $g$ be a continuous function on the closed interval $\left[0, 1\right]$. Let
$g\left(0\right)=1$ and $g\left(1\right)=0$. Which of the following is NOT necessarily true?

A. There exists a number $h$ in $\left[0, 1\right]$ such that
$g\left(h\right)\geq g\left(x\right)$ for all $x$ in $\left[0, 1\right]$.

B. For all $a$ and $b$ in $\left[0, 1\right]$, if $a=b$, then
$g\left(a\right)=g\left(b\right)$.

C.There exists a number $h$ in $\left[0, 1\right]$ such that
$g\left(h\right)=\displaystyle\frac{1}{2}$.

D.There exists a number $h$ in $\left[0, 1\right]$ such that
$g\left(h\right)=\displaystyle\frac{3}{2}$

E. For all $h$ in the open interval $\left(0, 1\right)$,
$\displaystyle\lim_{x\rightarrow h}g\left(x\right)=g\left(h\right)$.

has the answer as this:

D could be false, consider $g\left(x\right)=1-x$ on $\left[0,1\right]$. A is true by the Extreme Value Theorem, B is true because g is a function, C is true by the Intermediate Value Theorem, and E is true because g is continuous.

My question is, isn't B only necessarily true if $f\left(x\right)$ is a one-to-one function?

Edit: maybe I'm confusing myself, but B is necessarily true if it's 1-to-1 but nothing in the question says that. So I think it's not necessarily true.

2nd edit: yes, I get it now. Selection B is saying that if we have the same inputs we should get the same outputs and that's because it's a function.

答:
You're almost correct! An injective function is defined as as function for which if $f(a) = f(b)$ then $a = b$ or equivalently, if $f(a)\neq f(b)$ then $a\neq b$, that is the implication is the other way around to $B$.

$B$ is in fact always true for all well-defined functions: if we input the same value, we expect the same output.

ContentLicense: CC BY-SA 4.0


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
Hint

$$f(x)=(3-\sin (2 \pi  x)) \sin \left(\pi x -\frac{\pi }{4}\right)-\sin \left(3 \pi x+\frac{\pi }{4}\right)$$ Expand and manipulate the trigonometic functions to obtain
$$f(x)=\frac{7 \big[\sin (\pi  x)- \cos (\pi  x)\big]-\big[\sin (3 \pi  x)+\cos (3 \pi  x)\big]}{2\sqrt{2}}$$


问:
Rational representation of the additive group $\mathbb{C}$ as an algebraic group
The additive group $\mathbb{C}$ is a linear algebraic group via the embedding into $SL_2(\mathbb{C})$, $$z \mapsto \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}.$$
In this way, the Lie algebra of $\mathbb{C}$ is the $\mathbb{C}$-span of the matrix $e_{12}$.

I know that the exponential map gives a bijection between nilpotent elements of $M_2(\mathbb{C})$ and unipotent elements of $GL_2(\mathbb{C})$, via $$ \begin{bmatrix} 0 & z \\ 0 & 0 \end{bmatrix} \leftrightarrow \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}.$$
I want to show the following:

if $(\pi, V)$ is any regular representation of $\mathbb{C}$, there exists a unique nilpotent matrix $A \in \text{End} V$ for which $\pi(z) = \exp(zA)$ for all $z \in \mathbb{C}$.

I know that for $A \in \mathfrak{g}$, $\exp A = I + z e_{12}$ for some $z \in \mathbb{C}$, and so for any $z \in \mathbb{C}$, the bijection given by $\exp$ says that $\pi(z)$ is effectively equivalent to $\pi(\exp A)$, where $A = I + z e_{12}$.

I also know that in this setup, $\pi(\exp A) = \exp (d\pi(A))$, so I believe I need to show that $d\pi(A) = z A'$ for some $A' \in \text{End} V$.
By what I said earlier, we know the form that the matrix $A$ takes, and in particular, $$d\pi(A) = \pi \left( \begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix} \right) d\pi \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) = z A',$$ where $A' = d\pi \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right)$.

I believe this concludes the proof - is this correct?

答:
Given any rational representation $\pi:G\to GL(V)$ (that is, $\pi$ may be written down explicitly using rational functions and matrices), the derivative $\mathrm{d}\pi:\mathfrak{g}\to\mathfrak{gl}(V)$ is also a rational map and

$$ \pi(\exp X)=\exp\mathrm{d}\pi(X). $$

In the case of $G=(\mathbb{C},+)$, we have $\mathfrak{g}=\mathbb{C}$ too with $\exp z=z$. (I assume $G$ is being treated as a complex group, so $\mathfrak{g}$ is a complex vector space.) Then $\mathrm{d}\pi:\mathbb{C}\to\mathfrak{gl}(V)$ is a linear map, it must be $\mathrm{d}\pi(z)=z\mathrm{d}\pi(1)$, so if we define $X=\mathrm{d}\pi(1)\in\mathfrak{gl}(V)$ we have $\mathrm{d}\pi(z)=zX$ and hence

$$ \pi(z)=\pi(\exp z)=\exp\mathrm{d}\pi(z)=\exp(zX). $$

There is a choice of coordinates in which $X$ is block-diagonal with Jordan blocks (i.e. canonical form). So it suffices to examine the exponential of a Jordan block and see when it's rational.

Let $J=\lambda I+N$ be a Jordan block with $N$ strictly upper triangular. By looking at the power series it is straightforward to verify $\exp(zN)$ is rational, however $\exp(zJ)=e^{\lambda z}\exp(zN)$ is then not rational unless $\lambda=0$, which forces all of $X$'s Jordan blocks and hence $X$ itself to be nilpotent.


问:
How do I factor this?
How do I factor $p^2+8pq+16q^2-9r^2$? I know how to group the first two terms, but I dont know what to do with the other half. Can someone help me with this problem?</p>

ContentLicense: CC BY-SA 3.0

Title: How do I factor this?

Body: $$(p+4q)^2=p^2+8pq+16q^2\qquad a^2-b^2=(a-b)(a+b)\qquad 9r^2=(3r)^2$$


问:
Conditional probability of joint events
Let $X, Y, Z$ be three random variables. Is the equality
$$
P(X=x, Y=y |Z=z) = P(X=x|Y=y,Z=z) P(Y=y|Z=z)
$$
true?
Note that $P(X,Y)$ denotes the joint probability of random variables $X$ and $Y$.

答:
I am guessing you meant joint densities, in which case it is true.

In general you represent it by f and we have

$$f_{X,Y}(x,y)=f_{X|Y}(x|y)f_Y(y)$$

You can condition on any number of other random variables

$$f_{X,Y|W,Z}(x,y|w,z)=f_{X|Y,W,Z}(x|y,w,z)f_{Y|w,z}(y|w,z)$$

and still be "true."

Title: Conditional probability of joint events


问:
Plotting on Matlab
I'm trying to plot the curve shown below on Matlab. $y= x \, \tan x$ and $x$ is in the range $(0, 4 \, \pi)$. The thing is

I can't seem to multiply x with tan(x) without getting an error.

I just need intersecting points between tan curve and circle (forget the cot curves in picture for now).
Any suggestions?
Figure

答: Does it have to be Matlab because Desmos does a pretty good job. See this graph.

Title: Plotting on Matlab


问:
$n$ distinct complex roots

Given a non-zero complex number $z$, prove that there $n$ distinct $x_i \in \mathbb{C}$ such that $x_i^n = z$.

Having fixed $z$, I can write $z = re^{i\theta}$ where $r = |z|$ and $\theta$ is the polar angle. By the fundamental theorem of algebra, $f(x) = x^n - z$ has exactly $n$ roots $x_1, \ldots, x_n$. But given such a root $x_i$, we have $x_i^n - z = 0$, so $x_i^n = z$.

My question is: did I trivialize the problem? I can use the fundamental theorem of algebra, but nowhere did I use the polar form or any other property of $z$, but I know that I need to in some way use it.

答:
You obtain the roots explicitly by means of the polar form, noting that the argument is undetermined to a multiple of $2\pi$:

$$x^n=|z|e^{\angle z+i2k\pi}$$ and

$$x=\sqrt[n]{|z|}e^{\angle z+i2k\pi/n}=\sqrt[n]{|z|}e^{\angle z/n}e^{i2k\pi/n}.$$

Now we can show that there are $n$ distinct values of

$$\cos\frac{2k\pi}n+i\sin\frac{2k\pi}n.$$ By periodicity all these values repeat for every other $n$ value of $k$ and there cannot be more than $n$ roots. Now

$$e^{i\theta}=e^{i\phi}\iff e^{i(\phi-\theta)}=1$$ and $\phi-\theta=2m\pi$, and this cannot happen with $k\in\{0,\cdots n-1\}$ unless $\phi=\theta$.

Title: $n$ distinct complex roots


问:
Graph Theory (Edge connectivity of a complete bipartite graph $K_{m,n}$
Find  $\lambda (K_{m,n})$, where both m and n are at least 1.

I think the answer is $\min(m,n)$ (intuitively), but I'm not sure about my answer. (Specifically, I want to prove that this graph is connected after removing less than $\min(m,n)$ edges.
Help me,please.

答:
You could also use this:

The vertex-connectivity of a graph is less than or equal to its
edge-connectivity. That is, κ(G) ≤ λ(G).. (Source: https://en.wikipedia.org/wiki/Connectivity_(graph_theory)#Bounds_on_connectivity)

As noted, $\lambda(K_{m,n}) \leq \min(m,n)$ since each vertex has degree $m$ or $n$. To show that $\lambda(K_{m,n}) \geq \min(m,n)$, the above result implies that it suffices to show $\kappa(K_{m,n}) \geq \min(m,n)$. Let $X$ be the part with $m$ vertices and $Y$ the part with $n$ vertices. The subgraph of $K_{m,n}$ induced by any $m'$ vertices of $X$ and $n'$ vertices of $Y$ is clearly $K_{m',n'}$, which is connected if $m', n'\geq 1$. Hence, in order to disconnect $K_{m,n}$ by removing vertices, we must remove either all of $X$ or all of $Y$, so $\kappa(K_{m,n}) \geq \min(|X|,|Y|) = \min(m,n)$.


问:
Manipulating Vector Identity
I would like to expand then simplify (if possible) the following quantity.

$\nabla (a \cdot (C\, a))$

Where $a = a(x)$ is a vector valued function of $x$ and $C$ is a constant matrix.

答:
No need to use index notation here; use the chain rule.

$$\begin{align*}b \cdot \nabla [a \cdot \underline C(a)] = (b \cdot \nabla a) \cdot \nabla_a [a \cdot \underline C(a)]\end{align*}$$

Let's call $\underline a(b) = b \cdot \nabla a$ and abstract that out for a moment.  $\underline a(b) \cdot \nabla_a [a \cdot \underline C(a)]$ can be attacked using the product rule.

$$\underline a(b) \cdot \nabla_a [a \cdot \underline C(a)] = ([\underline a(b) \cdot \nabla_a]a) \cdot \underline C(a) + a \cdot (\underline a(b) \cdot \nabla_a \underline C(a))$$

You get for the first term $\underline a(b) \cdot \underline C(a) = \overline a \underline C(a) \cdot b$, where $\overline a$ is the transpose.  For the second term, you get $\overline{aC}(a) \cdot b$, so the result is

$$b \cdot \overline a(\underline C + \overline C)(a)$$

$b$ is arbitrary, and we can pull it out to get

$$\nabla(a \cdot \underline C(a)) = \overline a(\underline C + \overline C)(a) = \dot \nabla(\dot a \cdot [\underline C + \overline C](a))$$

The dots denote what is to be differentiated; the second $a$ should not be differentiated and should instead be held constant.  In practice, I would compute the Jacobian transpose $\overline a$ in terms of a basis and just use that.

Title: Manipulating Vector Identity


问:
Binary translation zero issue
i'm  a student and for this semester i have to deal with binary translation. I use multiple online tools for such conversion. My question is,

The binary to text conversion for "A" is: 01000001

but as per the book "1000001" is correct answer for "A".

However, the online binary translator returns nothing when I insert "1000001".

So my question is, which is the correct answer for binary to text?

答:
I think both answers are correct. Adding zeros on the left side of a number does not change the number. For example, the numbers "0100" and "100" are same.

As long as I know binary numbers are eight-bit numbers. So, actual conversion of "A" is "1000001" but with a view to make this number an eight-bit number, they add extra required zero. On which tools u used?

Title: Binary translation zero issue


问:
Are those two theories about universes equivalent?
If we define a universe as a well founded extensional transitive set that is closed under power, union, and non-bigger than, formally this is:

$\mathbf U (X) \iff  \forall a \in X: \\ \forall m \in a (m \in X)  \\\forall b \in X :\forall z (z \in b \leftrightarrow z \in a) \to a=b   \\\forall S \subseteq X \exists m \in S: \forall n \in S (n \not \in m)  \\  \exists c,d \in X : c=\mathcal P(a),  d=\bigcup(a)  \\\forall k \subseteq X (|k| \not > |a| \implies k \in X)$

Where: $|k| > |a| \iff \exists f: a \hookrightarrow k \land \not \exists g: k \hookrightarrow a$

Where $``\hookrightarrow"$ signify injection; and $``\mathcal P, \bigcup"$ standing for powerset and set unions defined in the usual manner.

And if we add the following axioms on top of mono-sorted first order logic with equality and membership.

1. Universes: $\forall x \exists w: \mathbf U(w) \land x \in w$

Every set belongs to some universe.

2. Separation: $\forall a \exists b: b=\{x \in a: \phi\}$, where $\phi$ doesn't use $``b"$.

Then this would prove $ZFC$ over universes higher than $V_\omega$. If we want to prove $ZF+GC$ over those universes, then we'll need to modify the size condition in the definition of universes to:

$\forall k \subseteq X (|k| \neq |X| \implies k \in X)$

Where: $|k|=|X| \iff \exists f: k \hookrightarrow X \land \exists g: X \hookrightarrow k $

Call this theory as Universes Theory, to be denoted by $\sf UT$.

To get ZFC we need to add the axiom of Largeness:

3. Largeness: $\forall x \exists w: |w| > |x| \land \forall u \in w (\mathbf U(u))$

Every set is strictly smaller than a set of universes.

I think that all sentences of $\sf TG$ set theory would be provable in this system, so this is just a reformulation of it.

However, the axiom of largeness seems to be more of a fix. I think a more principled axiom would be the following:

3. Generalization: $(\forall w: \mathbf U(w) \to \phi^w) \implies \phi$,

Where $\phi^w$ is a sentence with all of its quantifiers bounded $\in w$, and $w$ not used otherwise.

In English: what is true inside all universes is true in the whole world!

Is the generalization axiom consistent with the first two?

Are both theories ($\sf UT+Larg.; UT+Gen.$) equivalent?

答:
This is a partial answer to this question.

For the second question it appears that $\sf UT + Gen$ is stronger than $\sf TG$ set theory because I think it proves the existence (among its elements) of a universe that doesn't have a set of all universes in it, more specifically a universe that is the union of inaccessibly many universes below it, this universe would satisfy all sentence of $\sf TG$ and thus would prove $\sf Con(TG)$. This very same universe would also satisfy all sentences of $\sf UT+ Larg$. However, I don't know where the consistency strength of that theory would stop, or even if its consistent either?


问:
Least squares solution of underdetermined system with constraints
With $Ax=b$, where $A \in \mathbb{R}^{3 \times 8}, x \in \mathbb{R}^{8 \times 1}$ I can find the least squares solution $x$ for a given $b$ such that I minimize $\left\|x\right\|^2_2$. The general least squares solution (According to my understanding) is given by $A^+b$ where $A^+$ is the Moore-Penrose pseudo inverse of $A$.

However I would like to find the least squares solution $x$ such that $x \geq 0$. And if it is possible, to find a general matrix $A'$ such that $A'b = x \geq 0$ ? Maybe the latter is impossible as the sign of $A'b$ depends on the sign of $b$ which in the problem I am considering is unconstrained.

答:
If I understand correctly, the problem you're trying to solve is
\begin{aligned}
\min_{\mathbf{x}}\quad & \|A\mathbf{x}-\mathbf{b}\|^2 \\
s.t. \quad & x_i\geq 0
\end{aligned}
This is known as Non negative least squares and can be solved using any convex optimization algorithm. In your case the problem is of small scale ($n=8$) so in general it's easy to solve. You can use a second order solver like quadprog or even a simple first order algorithm like Projected Gradient (with or without acceleration).

As for your last question, finding $A\,:\, A^T\mathbf{b}=\mathbf{x}\geq\mathbf{0}$, what is the reason for this requirement? I'm asking since it's different than the original question - where $A, \mathbf{b}$ are the given data (not a variables) and we look for $\mathbf{x}$ that best fits this data.


问:
If $2^x=0$, find $x$.

If $2^x=0$, find $x$.

Solution: I know range of $2^x$ function is $(0,\infty)$.

So $2^x=0$ is not possible for any real value of $x$

Hence, equation is wrong. We can't find value of $x$. Am I right?

Please help me.

Can $x$ be in $[-\infty,\infty]$?

i.e is $2^x=0$ possible for $x=-\infty$?

答:
Let $x$ be a real number $x>0$, and consider $w=2^x$ and $z=2^{-x}$. Then $wz=1$. But then neither of them can be $0$, since in either case we would reach the absurdity that $0=1$. Thus, by symmetry, and since $2^0=1$, we see that $2^x\neq 0$ for each $x\in \Bbb R$, that is, $2^x=0$ is unsolvable in $\Bbb R$. The very same proof applies when $z\in\Bbb C$.

The fact that "the range of the function $2^x$ is $(0,+\infty)$" is true because of the above, and not conversely. It is true we may define $2^x=0$ when $x=-\infty$ to extend $2^x$ continuously to $\Bbb R^*=\Bbb R\cup\{-\infty,+\infty\}$, but there is not much more to it than that.

Title: If $2^x=0$, find $x$.


问:
How to Calculate a Weighted GPA
How would I be able to calculate my overall average grade (or grade point average) for these courses?

Course
Course Credit (weight)
Course Grade

Biology 101
45
86

Psychology 210
30
74

Chemistry 250
45
88

English 200
60
77

My understanding is by simply adding the course grades column together and dividing by the total cumulative possible marks.

(86+74+88+77)/400 = 0.8125

But this number seems very strange in the 0 to 4 GPA.

答:
You calculated an unweighted GPA. To weight it, you must factor in the course credit variable.

```<code>> sum=45+30+45+60
> (86*45+74*30+88*45+77*60)/sum
[1] 81.5
</code>```

As the original units are in a 0-100 scale, your weighted gpa is about a B-.

To convert to a 4.0 scale, simply use the appropriate unit conversion

```<code>> 81.5*4/100
[1] 3.26
</code>```

Your original calculation was missing a multiplication by 4. The procedure for your unweighted gpa should be

```<code>> (86+74+88+77)/4*4/100
[1] 3.25
</code>```

Title: How to Calculate a Weighted GPA

Body:
You calculated an unweighted GPA. To weight it, you must factor in the course credit variable.

> sum=45+30+45+60
> (86*45+74*30+88*45+77*60)/sum
[1] 81.5

As the original units are in a 0-100 scale, your weighted gpa is about a B-.

To convert to a 4.0 scale, simply use the appropriate unit conversion

> 81.5*4/100
[1] 3.26

Your original calculation was missing a multiplication by 4. The procedure for your unweighted gpa should be

> (86+74+88+77)/4*4/100
[1] 3.25


问:
7 digit palindrome problem
This is my problem:
From {1,2,3, ..., 9}
How many palindromes of length 7 are there, where each digit can appear at most twice</p>

答:
The number should be such that $\overline{abcdcba}$, $\overline{aabcbaa}$, $\overline{abbcbba}$. For these cases, the number of palindromes are $(9*8*7*6*1*1*1)+(9*1*8*7*1*1*1)+(9*8*1*7*1*1*1)=4032$.

I'm only able to think of these cases till now. I'll update if I get more.

If you consider all the bases, then there are infinitely many 7 digit palindromes IMO. (I considered base 10 in this answer)

Title: 7 digit palindrome problem


问:
Constructing an isosceles trapezoid with a specific decomposition into triangles
A recent question asked about finding the ratio of the bases for the following isosceles trapezoid:

That problem has been solved, obtaining a result of $|CD|/|AB|=1-1/\sqrt{2}$. What I'm curious how one would construct this figure in the first place. The solution to the original question offers some hints. Let $E$ be the central point, and let $h_1,h_2$ be the lengths of the altitudes drawn from $E$ to $AB,CD$ respectively. Then the areas of the triangles $\triangle ABE,\triangle CDE$ require

\begin{align}
\frac12 h_1|AB|=4&\implies h_1=\frac{8}{|AB|}\;,\\
\frac12 h_2 |CD|=2&\implies h_2 =\frac{4}{|CD|}=\frac{4}{1-1/\sqrt{2}}\frac{1}{|AB|}=\frac{4}{|AB|}(2+\sqrt{2})
\end{align}
This determines the vertical position, but still leaves the horizontal position of $E$. Is there a straightforward way to complete this construction? (Graphical demonstrations encouraged.)

答:
Since $ABCD$ is isosceles, $\angle GEI=\angle GFH$. The two triangles are similar with ratio $IG:GH=5:3$.

So after constructing $EF$ parallel to the pair of parallel sides, divide it into ratio $5:3$.


问:
Help with inequality of rank of sum, using the rank of the product
I am getting a hard time while trying to prove the following inequality: If $A$ and $B$ are $n \times n$ matrices, then

$rank{(A + B)} \leq rank{(A)} + rank{(B)} - rank{(AB)}.$

I've tried a lot, but I can't insert the rank of $AB$ in the inequality (the others factors appears easily). I will appreciate any help.
Thank you all!

答:
When $AB=BA$, you can find a proof here: Proving $\text{rank}(A+B) \leq \text{rank}(A) + \text{rank}(B) - \text{rank}(AB)$ when $A,B$ commute

A counterexample for the non-commutative case is the following. Consider the matrices

$$ A=\left( \begin{array}{cc}
1 & 0 \\
0 & 0
\end{array} \right)\quad \text{and} \quad
%
B=\left( \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right).
$$
Then, $$ A+B=\left( \begin{array}{cc}
1 & 1 \\
0 & 1
\end{array} \right)\quad \text{and} \quad
%
AB=\left( \begin{array}{cc}
0 & 1 \\
0 & 0
\end{array} \right).
$$
Lastly, $$rank(A+B)=2>1=1+1-1 = rank(A)+rank(B)-rank(AB). $$


问:
Optimal squares to choose in the football square game
The game goes like this:

Before the (American) football game begins, $5$ players each choose $20$ squares on a $10 \times 10$ grid. After choosing, the numbers $0-9$ are chosen at random twice, once to label each row and once to label each column. At the end of the game, the last digit of each team's score is taken, and the player on the corresponding lattice wins.

For example, if rows represent the home team's final score and the columns represent the away team's score, then the winner is whoever chose the square indexed $(\text{home mod } 10, \text{away mod } 10)$. There are, indeed, most common NFL scores, so the scores themselves are not randomly distributed. My question is

Are there squares that increase the odds of winning?

I assume no, since the fact that the indices are randomized, the grid is also randomized, but I wanted to know if my intuition was wrong, and if so, why?

答:
I would instinctively assume yes because of limited modulo residues, and in this case, that is correct.

From my non-American understanding of American Football scoring, you score 6 points for a touchdown, one for a post-touchdown extra point, 3 for a field goal, and 2 for a safety.

Let's make some assumptions to help.

Safeties very rarely happen. $133$ touchdowns per safety is the figure I could find, so I will discard them from the scoring list.

$\text{# touchdowns}:\text{# field goals} \approx 3:2$

A touchdown always gets the extra point.

With these assumptions in mind, we are most of the time scoring in $7$s and $3$s with $60:40$ probability.
Conveniently, $7+3=10$, so we can identify the end digits quite easily.
Let $\tau$ be the number of touchdowns, and $\phi$ the number of field goals a given team scores. Then, the end digit $d$ is found as: $$d=\begin{cases} ... \\ 1 & \tau=\phi+3 \\  4 & \tau=\phi+2 \\  7 & \tau=\phi+1 \\0&\tau=\phi \\ 3&\tau=\phi-1 \\ 6 &\tau=\phi-2 \\9 &\tau=\phi-3 \\... \end{cases}$$

Most likely, the digits are going to end in $0,3,4,7$ (bearing in mind the $60:40$ tilt towards getting more touchdowns), sometimes in $1,6$ and almost never in $2,5,8,9$.

The data sheet you provided seems to line up very nicely with this.


答:
To choose a palindrome:

First, you have to decide which number goes in the middle ($4$-th position), so and you have 9 possibilities.

Notice that this number cannot appear anywhere else in the palindrome, otherwise it would appear at least three times. So, in the other positions, you can only pick from the 8 remaining numbers.

In order to completely specify your palindrome, you only need to choose its first three digits. Because you do not allow for any repetitions, all those digits have to be different, so you have $8 \times 7 \times 6$ choices.

This shows that there are exactly $9 \times 8 \times 7 \times 6 = \frac{9!}{5!} = 3024$ palindromes of length 7 with no more than 2 appearances of the same digit.

(However, I am not sure that this was the initial question, so ...)

Title: 7 digit palindrome problem

Body:
This is my problem:
From {1,2,3, ..., 9}
How many palindromes of length 7 are there, where each digit can appear at most twice</p>


答:
julien’s already given you the key algebraic facts, so I’ll show how you might think about the problem.

The fact that you have $p^2$, $pq$, and $q^2$ terms suggests that you should try to combine them as a product of the form $(ap+bq)(cp+dq)$. The $9r^2$ term clearly can’t be combined with the other three directly, but it’s an obvious square, $(3r)^2$; if the first three turn out actually to be the square of something of the form $ap+bq$, you’ll have a difference of two squares, which you can then factor.

Title: How do I factor this?

Body: How do I factor $p^2+8pq+16q^2-9r^2$? I know how to group the first two terms, but I dont know what to do with the other half. Can someone help me with this problem?</p>


问:
Proving from a sequence of real numbers
I am solving the following problem.

Let $a_n$ be a sequence of nonzero real numbers.

Assume that lim sup $|a_{n+1}/a_n|$ = L

a.  Let L’ be any number bigger than L. Prove that there exists N>0 such that $|a_{n+1}/a_n| < L’$ for any n > N.

b.  Prove that for any n>N we have |$a_n|<(L’)^{n-N}|a_N|$

For a. I know I want to use the definition of convergence which says for every epsilon greater than zero there exists a N>0 such that  n>N $\implies$ $|a_n-a|<\epsilon$ but my sequence is a little different and for that reason I am having a hard time with this problem. Thanks

答:
First of all, the meaning of $\lim\sup \frac{a_{n+1}}{a_n}$:

Let us define a sequence $b_n=\sup_{k\geq n}\frac{a_{k+1}}{a_k}$. Then we'll find that:
$$\begin{align}b_1&=\sup\left\{\frac{a_2}{a_1},\frac{a_3}{a_2},...\right\}\\
b_2&=\sup\left\{\frac{a_3}{a_2},\frac{a_4}{a_3},...\right\}\\
.\\
.\\
b_k&=\sup\left\{\frac{a_{k+1}}{a_k},\frac{a_{k+2}}{a_{k+1}},...\right\}\\
.\\
.\\
\end{align}$$

Then $\lim\sup\frac{a_{n+1}}{a_n}$ is defined as $\inf\left\{b_1,b_2,...\right\}$. More clearly,
$$\lim\sup\frac{a_{n+1}}{a_n}=\inf_{n\geq1}\left\{\sup_{k\geq n}\left\{\frac{a_{k+1}}{a_k},\frac{a_{k+2}}{a_{k+1}},...\right\}\right\}$$

Now, it's given that $\lim\sup\frac{a_{n+1}}{a_n}=L$. Thus, going by the definition of $\lim\sup$, we can say that, $L=\inf\left\{b_1,b_2,...\right\}$. Also, it has been given that $L'>L$. So can you now apply the definition of the infimum of a set and take it up from here?

 Hint:
 $L'$ is not a lower bound of the set $\left\{b_1,b_2,...\right\}$
 $\implies\exists r\in\mathbb{N}$ such that $b_r<L'$.
 Now just substitute $b_r=\sup_{k\geq r}\frac{a_{k+1}}{a_k}<L'$. Now as
 the supremum of this whole set is less than $L'$, thus, each and every
 element of this set is less than $L'$. Hence, we found an
 $r\in\mathbb{N}$ such that $\frac{a_{n+1}}{a_n}<L'$ $\forall n\geq r$.

Title: Proving from a sequence of real numbers


问:
How could someone conceive of using this inequality for this proof?
In James Stewart's Calculus: Early Transcendentals ($8$e), problem $90$ of Section $11.1$ asks us to prove that the sequence

$$a_n=\left(1+\frac{1}{n}\right)^n$$

has a limit. It is divided into five sub-problems, the first of which asks us to show that if $a$ and $b$ are real numbers with $0\leq a<b$, then

$$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$

After completing all five sub-problems, I can see why proving this inequality was one of them: it's useful for proving that $a_n$ satisfies the hypotheses of the Monotonic Sequence theorem. What I don't see is how one could conceive of using this particular inequality in this proof.

Let's say I wanted to prove that $a_n$ has a limit, but didn't have the guidance of the book. Are there any at-a-glance properties possessed by the sequence that would lead me to consider using the above inequality in my proof? If not, how could I guide my thinking in that direction?

答:
It's speculation on my part, but surely whoever wrote the proof knew that geometric series inequality first, found an argument that used the inequality as an intermediate step, then broke the argument into a sequence of steps for students. The mysterious step, where they "found an argument", is often hard to articulate, and in any case it's frequently not written down. In this particular case, I don't think it's a priori clear that particular inequality "should" be used, so I don't think your literal question has a great answer.

To illustrate this rather common situation more clearly, here's an alternate proof I just came up with, along with discussion afterwards about what's "really going on".

Alternate proof, exercise outline version:

Show that $\int_1^\infty \int_t^\infty \frac{1}{x(1+x)^2}dx\,dt \leq \frac{1}{2}$. (Hint: $\frac{1}{1+x} < \frac{1}{x}$.)

Let $L(t) = \int_1^t 1/x\,dx$ and let $f(t) = t L(1+1/t)$ for $t>0$. Show that $\lim_{t \to \infty} f'(t) = 0$.

Show that $f''(t) = -\frac{1}{t(1+t)^2}$. In particular, conclude that $f''(t) < 0$ for $t > 0$.

Use (2) and (3) to show that $f'(t) = \int_t^\infty \frac{1}{x(1+x)^2}dx > 0$ for all $t > 0$. In particular, conclude that $f(t)$ is increasing for all $t>0$.

Use (1) and (4) to show that $f(t)$ is bounded as $t \to \infty$. In particular, conclude that $\lim_{t \to \infty} f(t)$ exists.

Show that $a L(b) = L(b^a)$ for all $a, b > 0$. (Hint: use $u=x^a$.) In particular, conclude that $\lim_{t \to \infty} L((1+1/t)^t)$ exists.

Show that $L(t)$ is strictly increasing for $t>0$ and conclude that it has a continuous inverse.

Use (6) and (7) to conclude that $\lim_{t \to \infty} (1+1/t)^t$ exists.

What's really going on?

My first thought was that $(1+1/n)^n$ mixes bases and powers, making things hard to analyze. So, we should fix the base and use exponentials instead: $(1+1/n)^n = \exp(f(n))$ where $f(n) = n\log(1+1/n)$. For this problem, we need to show $f(n)$ monotonically increases and is bounded. Discretizing things by using integer $n$ is needlessly restrictive, so let's replace $n$ with $t$ and analyze $f(t) = t \log(1+1/t)$.

The most obvious thing to do is show that $f'(t) > 0$. By the product rule, $f'(t)$ will still have a log left over (by itself!), and in any case will be "mixed" with a rational function of $t$, and it probably won't be obvious that the result is positive. But, if we differentiate again, the log will die, so without even doing the actual calculations, I knew $f''(t)$ would be a rational function. Such rational functions are frequently "obviously" positive when they need to be, and anyway I know a lot about rational functions, so I figured I'd be able to get it to work. Indeed such is the case here in step (3).

Once we've got $f''(t) < 0$, it's just tying up loose ends with the Fundamental Theorem of Calculus to get in this case that $f'(t) > 0$. To get that $f(t)$ is bounded as $t \to \infty$ requires glancing at the polynomial growth rates of the integrals involved for computing "$f(\infty)$" in terms of the rational function $f''(t)$; the necessary bound is step (1).

As I was typing it out, it occurred to me that assuming the existence of logarithms is probably not fair for a problem that is basically defining $e$, so I added the stuff about $L(t)$ like steps (6) and (7) to derive its basic properties without making the whole argument circular.

A lot of learning more advanced mathematics boils down to being able to understand what the author didn't write as much as what they did write, or in any case being able to come up with your own ways to fill gaps. Gauss was famous for "covering his tracks". Modern standards on the amount of "derivation" (as opposed to "proof") vary depending on the discipline and author.

If you're really interested in this stuff, I'd suggest a book more focused on rigor than Stewart--Rudin's Principles of Mathematical Analysis comes to mind.


问:
Example of complete formula on constants.
I have question considering example of complete formula, that was presented in Keisler/Chang - Model Theory:

2.3.1. EXAMPLES

(1). Let $T$ be a complete theory and let $c_0,c_1,c_2,\ldots$ be constant symbols of $\mathscr{L}$. Then any formula of $\mathscr{L}$ of the form
$$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$
is complete in $T$. If $\mathfrak{U}$ is a model of $T$ such that every element of $A$ is a constant, then $\mathfrak{U}$ is an atomic model.

Why so? Am I right, when I think, that if theory is complete than -> every of it's formulas is complete too? So that formula with conjunction too?

Also, I pin basic definitions about it:

Consider a complete theory $T$ in $\mathcal{L}$. A formula $\varphi(x_1\ldots x_n)$ is said to be complete (in $T$) iff for every formula $\psi(x_1,\ldots x_n)$ exactly one of
$$T\vDash\varphi\rightarrow\psi,\quad T\vDash\varphi\rightarrow\neg\psi$$
holds. A formula $\theta(x_1\ldots x_n)$ is said to be completable (in T) iff there is a complete formula $\varphi(x_1\ldots x_n)$ with $T\vDash\varphi\rightarrow\theta$. If $\theta(x_1\ldots x_n)$ is not completable it is said to be incompletable.

A theory $T$ is said to be atomic iff every formula of $\mathcal{L}$ which is consistent with $T$ is completable in $T$. A model $\mathfrak{U}$ if is said to be an atomic model iff every $n$-tuple $a_1,\ldots,a_n\in A$ satisfies a complete formula in $\mathrm{Th}(\mathfrak{U})$.

答:
I changed my answer to myself:

Thus $T$ is complete, therefore it's consistent therefore it has model. So, it's sufficient to show, that
$T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ also will be consistent (and from that complete).

Suppose it isn't true.

Then, $T \models \forall x_0...\forall x_n \neg(x_0 \equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n)$
or $T \models \forall x_0...\forall x_n (\neg x_0 \equiv c_0\vee \neg x_1\equiv c_1 \vee \cdots\vee \neg x_n\equiv c_n)$

Interpretation $I^{\mathfrak{A}}_{}$ of $c_{i} := t_{2i}$, as a term from signature in the model $\mathfrak{A} = $ is $I^{\mathfrak{A}}_{}[c_i] = a_{i}$.

Some labeling $\gamma_{j}$ of $x_{i} := t_{1i}$, as a term, in model $\mathfrak{A} =$ is $t^{\mathfrak{A}}_{1i}[\gamma_{j}] = \gamma_{j} x_{i} = a_{1i}$.

So, thus in the last formula with all variables bounded, it's must be true on all labelings $\gamma_{j}$ and on all of our variables.

But for all $i \in {0,...,n}$, $x_{i}$ takes all values from $A$, and eventually it takes $a_i$ in some labeling $\gamma_{j}$.

The resulting contradiction completes the proof.

Body:
I changed my answer to myself:

Thus $T$ is complete, therefore it's consistent therefore it has model. So, it's sufficient to show, that
$T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ also will be consistent (and from that complete).

Suppose it isn't true.

Then, $T \models \forall x_0...\forall x_n \neg(x_0 \equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n)$
or $T \models \forall x_0...\forall x_n (\neg x_0 \equiv c_0\vee \neg x_1\equiv c_1 \vee \cdots\vee \neg x_n\equiv c_n)$

Interpretation $I^{\mathfrak{A}}_{}$ of $c_{i} := t_{2i}$, as a term from signature in the model $\mathfrak{A} = <A,I^{\mathfrak{A}}_{}>$ is $I^{\mathfrak{A}}_{}[c_i] = a_{i}$.

Some labeling $\gamma_{j}$ of $x_{i} := t_{1i}$, as a term, in model $\mathfrak{A} =<A,I^{\mathfrak{A}}_{}>$ is $t^{\mathfrak{A}}_{1i}[\gamma_{j}] = \gamma_{j} x_{i} = a_{1i}$.

So, thus in the last formula with all variables bounded, it's must be true on all labelings $\gamma_{j}$ and on all of our variables.

But for all $i \in {0,...,n}$, $x_{i}$ takes all values from $A$, and eventually it takes $a_i$ in some labeling $\gamma_{j}$.

The resulting contradiction completes the proof.


问:
How many ways can the students be selected if they can serve on both committees?
Question: There are 10 students. Four are needed for a committee A and three are needed for committee B. How many ways can the students be selected if they can serve on both committees?

Textbook's answer: 25,200.

My approach:

Selecting 4 students out of 10

$\binom{10}{4}$ = 210

Arranging the four students into the committee A:

$4\cdot3\cdot2\cdot1$ = 24

Selecting three students out of those 4 and arranging them:

$\binom{4}{3}$ * 3! = 24

Answer: 210 * 24 * 24 = 120,960

My logic might be wrong, so if you could provide me a correct approach, then it would be appreciated.

答:
The answer is $25\,200$, because that's $\binom{10}4\binom{10}3$.

You seem to have aimed at the situation in which all students from committee $B$ also belong to committee $A$.


答:
As the students can be on either or both committees, we choose members independently for both committees from $10$ students.

Choice for first committee = $10 \choose 4$

Choice for second committee = $10 \choose 3$

So total number of combinations for two committees = ${10 \choose 4}{10 \choose 3} = 25200$

Please note there is no way to arrange students in a committee unless there is more information, such as there are specific roles within the committee that selected members play.

Body:
Question: There are 10 students. Four are needed for a committee A and three are needed for committee B. How many ways can the students be selected if they can serve on both committees?

Textbook's answer: 25,200.

My approach:

Selecting 4 students out of 10

$\binom{10}{4}$ = 210

Arranging the four students into the committee A:

$4\cdot3\cdot2\cdot1$ = 24

Selecting three students out of those 4 and arranging them:

$\binom{4}{3}$ * 3! = 24

Answer: 210 * 24 * 24 = 120,960

My logic might be wrong, so if you could provide me a correct approach, then it would be appreciated.


问:
Prove that $x^2+xy$ is odd if and only if x is odd and y is even.
I have a question on proof by cases. I need to prove the statement on title.

First, I can think of proof by contrapositive :  x is even and y is odd if and only if $x^2+xy$ is even. Therefore, I let x be even integer and y be odd integer. $x^2+xy$ becomes $2(2k^2+2kl+k)$ where $k,l \in \mathbb{Z}$.
Then, I'm stuck on thinking about other one, but my guess is direct proof, which I can let x be odd integer and y be even integer. I was wondering if I have a right idea or not. Thanks in advance!

答:
Hint:

$$x^2+xy=x(x+y)$$ will be odd $\iff x, x+y$ are both odd

Now as $x$ is odd, $x+y$ will be odd $\iff y$ is even


问:
"Zero set of finitely many polynomials." but what polynomials are we talking about?
This is a follow-up question of Zero set of finitely many polynomials.

When it says "As the zero set of finitely many polynomials,  is a closed subset of ×"
I understand we are saying that determinant of submatrix is all zero. But I'm not sure how it is associated with polynomials. It just looks like a bunch of computations. What polynomials are we talking about? Could you give me any concrete example or formula for this? I would really appreciate the answers.

答:
Let $(a, b, c)$ and $(d, e, f)$ be two vectors in $\mathbb{R}^3$: there is a linear dependence between these vectors if and only if all of the 2x2 subdeterminants of the matrix $M$ vanish, where

$$ M = \begin{pmatrix} a & d \\ b & e \\ c &  f \end{pmatrix}.$$

There are three subdeterminants:

$$ M_{1, 2} = ae - bd, \quad M_{1, 3} = af-cd, \quad M_{2, 3} = bf-ce.$$

So to check that $(a, b, c)$ and $(d, e, f)$ are colinear, we just need to check that $M_{1, 2}$, $M_{1, 3}$, and $M_{2, 3}$ are all zero. You can find many proofs of this fact (or facts like this) around the place.

The connection with polynomials is that each $M_{i, j}$ can be treated as a polynomial in the six variables $a, b, c, d, e, f$, for example
$$ M_{1, 2}(a, b, c, d, e, f) = ae - bd.$$

We can then imagine $M_{1, 2}$ as a function on the space $\mathbb{R}^6$, and we can look at the set
$$ V(M_{1, 2}) = \{(a, b, c, d, e, f) \in \mathbb{R}^6 \mid M_{1, 2}(a, b, c, d, e, f) = 0\} \subseteq \mathbb{R}^6.$$
This set $V(M_{1, 2})$ is called the vanishing of the polynomial $M_{1, 2}$. In order to find colinear vectors, we are looking at the simultaneous vanishing $V(M_{1, 2}) \cap V(M_{1, 3}) \cap V(M_{2, 3})$ of three polynomials (finitely many polynomials).


答:
To calculate weighted GPA, multiple each grade on a 4.0 scale by the weight of the course and then find the average.

The conversion of 0-100% grades to a 0.0-4.0 scale may be calculated differently at different institutions. You need to check with your institution to see exactly how they do it. However, almost no institution directly scales 0-100% to 0.0-4.0 proportionally. Instead, 0-100% is divided into discrete sections, each of which is then mapped to a specific point value.

The most common system in the United States uses sections of size 10% for grades of 60% or higher, and gives no points for grades lower than 60%, which are considered failing:

90-100% -> 4.0

80-89% -> 3.0

70-79% -> 2.0

60-69% -> 1.0

0-59% -> 0.0

Under this system, your grades would convert as follows:

86% -> 3.0

74% -> 2.0

88% -> 3.0

77% -> 2.0

And your weighted GPA would then be:

(45 * 3.0 + 30 * 2.0 + 45 * 3.0 + 60 * 2.0) / (45 + 30 + 45 + 60) = 2.5

Another common system uses sections of size 5% instead of 10%:

95-100% -> 4.0

90-94% -> 3.5

85-89% -> 3.0

80-84% -> 2.5

75-79% -> 2.0

70-74% -> 1.5

65-69% -> 1.0

60-64% -> 0.5

0-59% -> 0.0

Under this system, your grades would convert as follows:

86% -> 3.0

74% -> 1.5

88% -> 3.0

77% -> 2.0

And your weighted GPA would then (rounded to three digits after the decimal point) be:

(45 * 3.0 + 30 * 1.5 + 45 * 3.0 + 60 * 2.0) / (45 + 30 + 45 + 60) ≈ 2.417

Title: How to Calculate a Weighted GPA

Body:
How would I be able to calculate my overall average grade (or grade point average) for these courses?

Course
Course Credit (weight)
Course Grade

Biology 101
45
86

Psychology 210
30
74

Chemistry 250
45
88

English 200
60
77

My understanding is by simply adding the course grades column together and dividing by the total cumulative possible marks.

(86+74+88+77)/400 = 0.8125

But this number seems very strange in the 0 to 4 GPA.


问:
Stationary processes versus processes with stationary increments?
what is the difference between stationary processes and processes with stationary increments?
I know that independent increments is a solid property to increments of a process, and not the process itself.

But Stationary has a different story I guess! Since we have stationary processes and process with stationary increments. It is two property that can be defined to same process.

My question is on the relation between these two properties. Does a stationary process have stationary increments or vice versa! Or these two can't be related this way or another?

I think having stationary increments is broader property and cant be narrowed down to stationary processes. I guess there are processes that are not stationary processes but have stationary increments.

Is there a way to figure out a relationship between these two?

答:
Stationary processes have stationary increments. To see roughy why this is true suppose $(X_t)_{t \geq 0}$ is stationary and consider $X_{t+s}-X_t$. Since $(X_t,X_{t+s})$ has the same distribution as $(X_0,X_s)$ it follows that the increments $X_{t+s}-X_t$ and $X_s-X_0$ have the same distribution. This can be generalized to more than two increments.

The standard Brownian Motion BM $(B_t)$ has stationary increments: the distrbution of $B_s-B_t$ depends only on the difference $s-t$. But it is not stationary since the distribution of $B_t$ is not independent of $t$ (Its variance is $t$).


问:
$\ln{2n+1\over n+1}<{1\over n+1}+{1\over n+2}+\cdots +{1\over 2n}<\ln 2$
Using the fact: ${x\over 1+x}<\ln (1+x)<x$, when $x>0$; deduce that:
\begin{equation}
\ln{2n+1\over n+1}<{1\over n+1}+{1\over n+2}+\cdots +{1\over 2n}<\ln 2
\end{equation}

This problem also asked to prove the fact I have given. I could prove it using MVT, but the deduction which should be easier is out of my hand.

Body:
Make a change of variables $y = \frac{x}{1+x}$, we have: $x = \frac{y}{1-y}$. So, from $\frac{x}{1+x}<\ln(x+1)$, we have :
$$y<\ln(\frac{1}{1-y}) = -\ln(1-y) \tag{1}$$

And because $\ln(1+x)<x\tag{2}$

From (1) and (2), we deduce that

$$\ln(1+x)<x<-\ln(1-x)$$

Replacing $x$ by $\frac{1}{n+k}$ for $k=1,...,n$
$$\ln(n+k+1)-\ln(n+k) < \frac{1}{n+k}< -(\ln(n+k-1)-\ln(n+k)) $$
Hence
$$\sum_{i=1}^n(\ln(n+k+1)-\ln(n+k))<\sum_{i=1}^n \frac{1}{n+k}<\sum_{i=1}^n(\ln(n+k)-\ln(n+k-1))$$
or
$$\ln(2n+1)-\ln(n+1)<\sum_{i=1}^n \frac{1}{n+k}<\ln(2n)-\ln(n)$$
or
$$\ln \left(\frac{2n+1}{n+1} \right)<\sum_{i=1}^n \frac{1}{n+k}<\ln(2)$$


答:
$$(x^2+xy)\bmod2=((x\bmod2)^2+(x\bmod2)(y\bmod2))\bmod2$$ and it suffices to try all parities.

$$(0,0)\to 0,\\(1,0)\to 1,\\(0,1)\to 0,\\(1,1)\to 0.$$

This will work for any integer polynomial.

Title: Prove that $x^2+xy$ is odd if and only if x is odd and y is even.

Body:
I have a question on proof by cases. I need to prove the statement on title.

First, I can think of proof by contrapositive :  x is even and y is odd if and only if $x^2+xy$ is even. Therefore, I let x be even integer and y be odd integer. $x^2+xy$ becomes $2(2k^2+2kl+k)$ where $k,l \in \mathbb{Z}$.
Then, I'm stuck on thinking about other one, but my guess is direct proof, which I can let x be odd integer and y be even integer. I was wondering if I have a right idea or not. Thanks in advance!


问:
Estimate a sum of fractions
Are there any easy ways to prove that:
$$
\sum_{j=1}^{k-1} \left({j \over k - j}\right)^{k - 2j - 1} = O(\sqrt k)
$$
or give some other estimates to the sum.

Title: Estimate a sum of fractions

Body:
Using @Svyatoslav 's approach and using one more term in the expansions, we need to compute

$$\int_{-\infty}^{+\infty} \exp\Bigg[\left(\frac{4}{3 k^3}+\frac{4}{k}\right) l-\frac{8 \left(k^2+1\right) }{k^3}l^2 \Bigg]\,dl$$ and the result is

$$ \sqrt{\frac{\pi \,  k^3}{8 \left(k^2+1\right)} }\,\,\exp\Bigg[\frac{\left(3 k^2+1\right)^2}{18 k^3\left(k^2+1\right)} \Bigg]$$

Some results
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
 10 & 2.07259 & 2.05718 \\
 20 & 2.86980 & 2.85546 \\
 30 & 3.48807 & 3.47548 \\
 40 & 4.01192 & 4.00063 \\
 50 & 4.47477 & 4.46447 \\
 60 & 4.89400 & 4.88448 \\
 70 & 5.28003 & 5.27114 \\
 80 & 5.63969 & 5.63132 \\
 90 & 5.97774 & 5.96981 \\
 100 & 6.29767 & 6.29011 \\
 200 & 8.88434 & 8.87890 \\
 300 & 10.8721 & 10.8676 \\
 400 & 12.5488 & 12.5449 \\
 500 & 14.0265 & 14.0230 \\
 600 & 15.3627 & 15.3595 \\
 700 & 16.5916 & 16.5887 \\
 800 & 17.7356 & 17.7328 \\
 900 & 18.8101 & 18.8075 \\
 1000 & 19.8265 & 19.8241
\end{array}
\right)$$


问:
Group equals union of two subgroups
Suppose $G=H\cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$.

What I did: For finite $G$, if $H\neq G$ and $K\neq G$, then $|H|,|K|\le |G|/2$. But they clearly share the identity element, so $H\cup K\neq G$.

How can I extend it to $G$ infinite?

答:
We show that if $H\cup K$ is a group, then either $H\subseteq K$ or $K\subseteq H$.

If neither $H\subseteq K$ nor $K\subseteq H$ is true, then there is an $h\in H$ which is not in $K$, and a $k\in K$ which is not in $H$.

If $H\cup K$ is a group, then the product $hk$ is in $H\cup K$. Thus the product is in $H$ or in $K$. If $hk\in H$, then $k\in H$, contradicting the choice of $k$. We get a similar contradiction if $hk\in K$.

Title: Group equals union of two subgroups


问:
Prove the infinite product topology where each factor are compact subspace of [0,1] is a compact space
"If $(X_i,\tau_i)_{i\in\mathbb{N}}$, are compact subspaces of $[0,1]$, deduce from Theorem 9.3.8 and Exercises 9.2 #1, that $\Pi(X_i,\tau_i)$ is compact."

Theorem 9.3.8 is the hilbert cube is compact. Exercise 9.2 #1 is that the infinite product of closed sets is a closed set in the product topology.

So every compact subspace of $[0,1]$ are simply closed intervals in $[0,1]$ and closed intervals are all homeomorphic to each other. So we have an infinite product where each factor is homeomorphic to $[0,1]$ which is by definition the Hilbert Cube so that is a compact space.

Now I did not had to use the fact that the infinite product of closed sets is a closed set in the product topology, which I think I'm suppose to use. Is there a mistake in my logic?

答:
As Kavi pointed out in the comments, it is not true that all compact subsets of $[0,1]$ are closed intervals.

What you can use is that since $(X_i,\tau_i)$ are compact subspaces of the compact set $[0,1]$, they are closed.
So, by Exercise 9.2 #1, $\Pi_i(X_i,\tau_i)$ is closed.
Then, by compactess of the Hilbert cube, any closed subset is also compact.

By the way, it is true that any product of compact topological spaces is compact in the product topology - this is called Tychonoff's Theorem.


问:
Why the integral does not converge?
Why the integral $\int_1^{ + \infty } \frac{dx}{x - 1}$ is not a finite number? I know that when you compute that its infinite but from the looks of the graph it looks like finite, the graph comes very close to $x=1$ and $y=0$, isn't it the same with $0.99999\ldots=1$?

答:
Not close enough.

Consider

$$\int_{\epsilon}^{2\epsilon}\frac{dx}{x}.$$

By a rescaling of the variable $z=\epsilon x$, this is

$$I:=\int_1^2\dfrac{dz}{z},$$ a finite quantity, independent of $\epsilon$.

So the integral on

$$\left[\frac1{2^n},1\right]=\left[\frac1{2^n},\frac1{2^{n-1}}\right]\cup \left[\frac1{2^{n-1}},\frac1{2^{n-2}}\right]\cup\cdots\left[\frac1{2},1\right]$$ is $nI$. Even if the slices get thinner and thinner, the height compensates and the total diverges.

Similarly, the integral on

$$[1,2^n]=[1,2]\cup[2,4]\cup\cdots[2^{n-1},2^n]$$ is $nI$. The slices are lower and lower, buth their width compensates.

Convergence depends on how fast the quantity decreases to zero.

Title: Why the integral does not converge?


问:
Concrete Bezout's Theorem Calculation
For context: I'm taking an algebraic geometry class and TAing for a calculus class. My students had the classic problem of finding the area between $x^3$ and $x^2$, but primed by algebraic geometry, I had a thought while I was grading:

Shouldn't $x^2$ and $x^3$ intersect in $2 \cdot 3 = 6$ points by Bezout's Theorem?

Obviously for the purpose of the calculus class the answer is "no", the only intersections are $0$ and $1$, but there are 3 reasons why this might be the case:

We're working over $\mathbb{R}$ not $\mathbb{C}$

We're not working in a projective setting

$x^2$ and $x^3$ actually intersect twice at $0$, so our curves aren't suitably "generic" (whatever that means).

So I tried to solve these problems by considering (in $\mathbb{C}$) the intersection of

$yz = x^2 + \alpha z^2$

$yz = x^3 + \beta z^3$

Here we've projectivized, and the $\alpha$ and $\beta$ were added to try and solve the genericness issue while keeping the equation somewhat simple.

So surely these equations will have 6 solutions, right? Well...

If we look in the $x=1$ plane, then we see $1 + \alpha z^2 = yz = 1 + \beta z^2$, so either $z=0$ or $z = \frac{\alpha}{\beta}$. I think the solution $z=0$ doesn't actually exist, though, since
$y0 = 1 + \alpha 0^2$ has no solutions for $y$. So we pick up 1 solution.

If we look in the $y=1$ plane, nothing interesting happens since we can always just eliminate $y$, so any solutions will already by in the $x=1$ or $z=1$ planes. Let's move on.

If we look in the $z=1$ plane, then we get $x^2 + \alpha = y = x^3 + \beta$, so we pick up 3 complex solutions.

So then we end up with 4 solutions instead of the expected 6. Where did I go wrong? I suspect the issue is either

adding $\alpha$ and $\beta$ alone don't make these curves "generic enough", whatever that means (so the double root $z=0$ that we killed off in step 1 should really be counted, bringing our total to 6)

I've misunderstood something about solving equations where we can eliminate some variables, and there are some extra roots in the $y=1$ plane that I shouldn't have ignored.

Any clarification is appreciated! Thanks in advance!

答:
You're leaving out an important part of Bezout's theorem: you need to count the intersections with multiplicity. The intuition is then that if two curves intersect with multiplicity $m$ at a point $P$, we can slightly jiggle our curves to see $m$ actual intersections there.

To start, let's calculate some intersection multiplicities to see what should happen. Remember that the intersection multiplicity at a point $P$ of curves cut out by equations $f,g\in\mathcal{O}_{\Bbb P^2,P}$ is $\dim_k \mathcal{O}_{\Bbb P^2,P}/(f,g)$.

$P=[0:0:1]$: the ring in question is $k[x,y]_{(x,y)}/(y-x^3,y-x^2)$. Making the change of coordinates $y\mapsto y+x^2$, this is isomorphic to $k[x,y]_{(x,y)}/(y+x^2-x^3,y)\cong k[x]_{(x)}/(x^3-x^2)$, and as $x^3-x^2=(x-1)x^2$ and $x-1$ is a unit, we have that the dimension of this ring as a $k$-vector space and thus the intersection multiplicity is two.

$P=[1:1:1]$: the ring in question is $k[x,y]_{(x-1,y-1)}/(y-x^3,y-x^2)$. Making the same coordinate change as last time, we have that this ring is isomorphic to $k[x,y]_{(x-1,y)}/(y+x^2-x^3,y)\cong k[x]_{(x-1)}/(x^2(x-1))\cong k$, so the intersection multiplicity is one.

$P=[0:1:0]$: the ring in question is $k[x,z]_{(x,z)}/(z-x^2,z^2-x^3)$. This is isomorphic to $k[x]_{(x)}/(x^4-x^3)$ by the map sending $z\mapsto x^2$, so we get that the intersection multiplicity is three.

At the first point, it's not so hard to see what happens when we wiggle: considering $y=x^3$ and $y=x^2-c$ for small positive $c$, we get two roots. The second point needs no perturbation, and the third is a little interesting. Considering $z=x^2$ and $z^2=x^3+c$ for small $c$, we see that our equation reduces to $x^4=x^3+c$, which for small $c$ is approximately $(x-1)(x^3+c)$, which can be seen to have exactly three roots near $x=0$. So everything is as it should be.

Title: Concrete Bezout's Theorem Calculation


问:
Abstract Geometry?
Are there similar terms in other areas for the idea the "angle" conveys in geometry ? I find that functions for abstract things such as pressure,electrical currents ( nothing geometric there ) on AC circuits are described by the sin function ( argument is an angle, there fore an geometric entity ) .Do we really have that abstract things are necessarily described by geometric entities ?   I understand that we can relate time to angle with the concept of angular frequency, but it seems too unintuitive and unnecessary to have to relate a sin-like variation generally ( a variation that decelerates smoothly towards the peaks and accelerates smoothly towards the center ) with geometry and angles.

Maybe the concept of angle we know is a simply an example of a greater idea ( like a  a measure of difference between instances of the same thing ) applied to geometry or am i tripping ?

答:
These things are all cycles with a certain sort of simple behaviour. If you're looking for a picture then angles and circles are a very good way to go, since they also have this behaviour, but that doesn't mean there's actually a physical angle lurking there somewhere. It's less

$$\text{Angles}\rightarrow \text{Cycles}\rightarrow \text{Currents}$$

and more

$$\text{Cycles}$$
$$\downarrow\ \ \ \ \ \ \ \ \ \ \downarrow$$
$$\text{Angles}\ \ \ \  \text{Currents}$$

Sinusoidal waves describe a particularly simple sort of cycle, and so they show up in physical angles. But they also show up in other places where simple cycles occur.

So you are correct in saying that it is unintuitive to relate sine and cosine to geometry and angles all the time, because plenty of times the geometry isn't there. In many cases angles aren't the things that are actually doing the cycling. But the sinusoidal functions still describe cycling behaviour regardless, so they show up in plenty of cases where physical angles are totally irrelevant.

Title: Abstract Geometry?


问:
Taking derivatives of a vector
I stumbled across the following question in a machine learning textbook:

Let $x \in \mathbb{R}^{n}$. Let $y = $sin$(Ax) \in \mathbb{R}^{m}$. What is $\frac{\partial y}{\partial x}?$ What is its dimension?

So my initial approach to this question was the following. If $x$ is an n-dimensional vector, then $x$ has the following form $<x_{1}, x_{2}.....x_{n}>$. Now since, sin$(Ax)$ is an m-dimensional vector, it must be the case that A is an m X n matrix. But after this how do I proceed? How are we taking the derivative of this vector? And what is the dimension for this? If someone has a concrete example they could present, that would be great! Thanks!

答:
According to the dimension of x and y, the dimension of the derivative you are looking for is $ m \times n $. This derivative (of vector with respect to vector) is called Jacobian Matrix, J, where:

$$ J_{ij} = \frac{\partial y_{i}}{\partial x_{j}} $$

Example: two input $x_{1}$ and $x_{2}$ and the outputs are $y_{1} = {x_{1}}{x_{2}}$ and $y_{2} = {x_{1}} - {x_{2}}$.

$$ J =
\begin{bmatrix}
 x_{2} &   x_{1}\\
 1 & -1 \\
\end{bmatrix}
\quad
$$

Further explanation can be found here:

short intro on vector-to-vector derivatives

The Matrix Calculus You Need For Deep Learning

Title: Taking derivatives of a vector


问:
Proving the product topology is discrete iff each component is discrete

Let $(X_i,T_i),i\in\Bbb N$, be a countably infinite (non-empty) family of topological spaces. Prove that $\prod(X_i,T_i)$ is a discrete space iff each $(X_i,T_i)$ is discrete and all but a finite number of the $X_i,i\in\Bbb N$ are singleton sets.

I think the backward direction is quite trivial so I will just do the forward direction here.

If $(X_i,T_i),i\in\Bbb N$ is discrete, then all singletons or points are open. Now all basic open sets are of the form $U=O_1\times O_2\times\cdots\times O_n×X_{n+1}\times X_{n+2}$ where all but a finite number of the open sets are of the form $X_i$.

How do I go from here to showing that each component is discrete and furthermore all but a finite number of the components are singleton sets?

答:
In general, if $f: X \to Y$ is open and onto, and $X$ is discrete then $Y$ is discrete too. This is rather trivial to prove.

We can apply this to $X=\prod_{i \in I} X_i$ and the open projections $\pi_i$ to conclude that $X$ discrete implies that all factor are.

Moreover, if $(a_i)_i$ is any point of $X$, then $\{a\}$ is open and so by the definition of the product topology there is a basic open subset $O=\prod_{i \in I} O_i$ of $X$, so all $O_i$ are open and there is a finite subset $F \subseteq I$ such that $O_i = X_i$ for all $ i \notin F$ (I say that $O$ depends on the finite set $F$, usually), and such that $O = \{a\}$ (or $\{a\}$ would not be open).

But then for all $i \in I\setminus F$ (so almost all $i$) we have  $\{a_i\}=\pi_i[\{a\}]= \pi_i[O] = O_i = X_i$ and so all but finitely many of the $X_i$ are singleton sets.

In fact no infinite product of non-singleton spaces can have isolated points.


答:
The two facts combined tell us that $\prod_{n \in \Bbb N^+} X_n$ is a closed subset of $\prod_{n \in \Bbb N^+} [0,1]$, which is compact. $X_i$ can be much more general than a closed interval, like middle third Cantor set, any finite subset or a convergent sequence with limit, e.g. So this fact is more general than the Hilbert cube fact.

And in general, a closed subspace of a compact space is compact. That’s the complete proof.

The general Tychonoff theorem is only discussed later in the text, but this would have been an immediate consequence of it.

Title: Prove the infinite product topology where each factor are compact subspace of [0,1] is a compact space

Body:
"If $(X_i,\tau_i)_{i\in\mathbb{N}}$, are compact subspaces of $[0,1]$, deduce from Theorem 9.3.8 and Exercises 9.2 #1, that $\Pi(X_i,\tau_i)$ is compact."

Theorem 9.3.8 is the hilbert cube is compact. Exercise 9.2 #1 is that the infinite product of closed sets is a closed set in the product topology.

So every compact subspace of $[0,1]$ are simply closed intervals in $[0,1]$ and closed intervals are all homeomorphic to each other. So we have an infinite product where each factor is homeomorphic to $[0,1]$ which is by definition the Hilbert Cube so that is a compact space.

Now I did not had to use the fact that the infinite product of closed sets is a closed set in the product topology, which I think I'm suppose to use. Is there a mistake in my logic?


问:
On index notation and matrix multiplication
In my linear algebra class today, the following calculation about bilinear forms was performed:

Let $b$ be a bilinear form on $V$ and $x,y \in V$. Suppose also $\{v_1,...,v_n\}$ is a basis for V. Then we can write:
$$b(x,y) = b(x^iv_i, y^jv_j)\\
= x^iy^jb(v_i,v_j)\\$$
Then if you let $M_{ij} = b(v_i,v_j)$, we can write this as:
$$b(x,y) = x^i M_{ij}y^j$$

I don't get the last step, intuitively I think it should be $$b(x,y) = x^iy^jM_{ij}$$$. Are the two formulas equivalent? Also why are we allowed to do that?

答: Since $x$ and $y$ are in $V$ they can be written as linear combinations of the basis vectors s.t $x=\sum_{i=1}^nx^iv_i$ and analogously for $y$ you can write $y=\sum_{j=1}^ny^jv_j$.($x^i , y^j$ are scalars and $v_i,v_j$ are basisvectors). Plugging this to formulas in the bilienar form you get $$b(x,y)=b(\sum_{i=1}^nx^iv_i,\sum_{j=1}^ny^jv_j)=\sum_{i=1}^n\sum_{j=1}^nx^iy^jb(v_i,v_j)$$where in the last steps bilinearity is used a couple of times. With the representation of $b(v_i,v_j)$ as a matrix $M_{i,j}$ you get the desired equality $$\sum_{i=1}^n\sum_{j=1}^nx^iy^jM_{i,j}$$ .Furthermore you can interchange the multiplication since the multplication of two scalaras and a scalar with a matrix is commutative.


问:
Solubility From Row Echelon Form
Here is the question I am attempting to solve

Determine which values of $k$, if any, will give:

a) A unique solution,
 b) No solution,
 c) Infinitely many solutions to the system of equations.

$$\begin{align} x+y+kz& =2 \\ \\ 3x+4y+2z& =k \\ \\ 2x+3y-z& =1\end{align}$$

I transformed this into row-echelon form. I got.

\begin{pmatrix}
1 & 1 & k &2 \\
0 & 1 & 2-3k &k-6 \\
0 & 0 & -3+k & 3-k
\end{pmatrix}

The answers say for a unique solution $k \neq 3$

a) So my reasoning is that if $k$ is not equal to $3$.Then we obviously have a unique solution, because if $k=3$, we would have everything in row $3$ equal to $0$. Hence we need to introduce parameters to solve the equation thus making us have infinitely many solutions.

b) No solutions occurs if we have in the last row $(0,0,0|\alpha)$, where $alpha \neq 0$. We can't have any values of k that can attain this result.

c) See a).

Is this the right reasoning?

答:
With corrections in your REF, your answers are indeed correct. I suspect that the errors in your initial REF clears up the initial concern you had.

I would include the unique solution in part $(a)$, and include your reasoning for $(c)$ in part $(c)$: in $(a)$, mention that "at $k = 3$, there are infinitely many solutions, (i.e., $0\cdot z = 0 \implies$ any $z$ in the domain is a solution, hence no single unique solution: see part(c))."

I would mention the word inconsistent in part $(b)$: the system of linear equations would be inconsistent (there would be no solution for $z$ such that $0\times z = c$ where $c\neq 0$ - hence no solution for the system of equations), and so since there is no value of $k$ for which that can happen, the system must either have a unique solution ($k \neq 3$) or infinitely many solutions, if and only if $k = 3$.

Title: Solubility From Row Echelon Form


问:
Riccati equation solution as a ratio of Bessel functions
I'm trying to find a general solution to this Riccati equation:

$f'(x) = -a f(x)^2 + g(x)$

where $a$ is a constant, $g$ is a smooth real-valued function of the real independent variable $x$, and $f$ is the unknown real-valued function of $x$.

I know that the general Riccati equation can't be solved in closed form for arbitrary coefficient functions, but the one I have involves one constant coefficient $a$ and one arbitrary function $g$. Does that change anything?

More specifically, is this equation solvable in closed form (or at all)? The author of this simulation claims that Riccati solutions are ratios of Bessel functions... but that seems a bit too easy because if that were true, some mathematician would have probably worked that into a general form by now since these things have been around for 300 years:

https://demonstrations.wolfram.com/RiccatiDifferentialEquationWithContinuedFractions/

Any help is appreciated, thanks!

答:
$$f'(x) = -a f(x)^2 + g(x)$$
Change of function :
$$f(x)=\frac{y'(x)}{a\:y(x)}\quad\implies\quad f'(x)=-\frac{(y')^2}{a\:y^2}+\frac{y''}{a\:y}$$
$$-\frac{(y')^2}{a\:y^2}+\frac{y''}{a\:y} = -a \left(\frac{y'}{a\:y} \right)^2 + g(x)$$
$$\frac{y''}{a\:y}=g(x)$$
$$y''(x)-ag(x)y(x)=0$$
This is a second order linear ODE. This ODE can be of various kind depending on the function $g(x)$.

Of course for some functions $f(x)$ the result involves Bessel functions. In this case $f(x)=\frac{y'(x)}{a\:y(x)}$ is a ratio of Bessel functions.

For other kind of functions $g(x)$ the solution of the above second order linear ODE involves functions which are not of Bessel kind. For example if $g(x)=a+bx+cx^2$ the solution involves parabolic cylinder functions.

Another example : if $g(x)=(b+\frac{c}{x})$ the solution involves confluent hypergeometric functions.

Thus one cannot conclude definitively if the function $g(x)$ is not of a specified kind.


问:
multiplication by invertible matrix preserves open-ness?
How do I show that, if $U$ is an open subset of some $\mathbb{R^{m\cdot n}}$ then $Ug$ where $g$ is an invertible matrix is also open?

答: Hint : If $g$ is an invertible matrix, then the map $x \mapsto gx$ is a homeomorphism.


问:
Why do we care about the Hilbert Cube?
The Hilbert Cube is defined to be the countable infinite Cartesian products of the interval $[0,1]$ or anything homeomorphic to $[0,1]$. Why do we care about this object?

答:
Every compact metrisable space (of any dimension ) is homeomorphic to a closed subset of the Hilbert cube. Any separable metrisable space is homeomorphic to a subspace of the Hilbert cube. Every separable metrisable topological vector space is homeomorphic to the pseudo-interior $(0,1)^{\Bbb N} \simeq \Bbb R^{\Bbb N}$ of the Hilbert cube.

There is a very nice theory of Z-sets and homeomorphisms of the Hilbert cube (see van Mill’s books on infinite-dimensional topology, or Bessaga and Pelczynski’s book for more on these theorems). It’s a fundamental object in infinite-dimensional topology. The hyperspace of any Peano continuum (in the Hausdorff metric) is homeomorphic to it too, e.g.

Title: Why do we care about the Hilbert Cube?


问:
Construction of a rhombus with given conditions

given two parallel lines $P$, $Q$ and two points $X$,$Y$ how to construct a rhombus $ABCD$ passing through $X$,$Y$  and opposite sides lie on $P$ and $Q$.

I solved a special case of this problem where one point $Y$ lies on the parallel line. any hints or ideas?

If $Y$ lies on line $q$, draw a circle with center $Y$ and radius $XY$. The circle will intersect line $p$ at point $Z$. If I draw a perpendicular bisector of $ZX$ it will intersect line $p$ at $B$. Then it is easy to construct the required rhombus.

答:
In a rhombus the distance between one pair of parallel sides is same as distance between another pair of parallel opposite sides.

So first find perpendicular distance $AB$ between $p$ and $q$.

Next with $X$ as center and radius $AB$, draw a circle. From $Y$ drop tangents to this circle. Extend a suitable tangent to intersect $p,q$ in $Q$ and $R$.

Now construct a line through $X$ parallel to $QR$ to intersect $p,q$ in $P,S$. $PQRS$ is our desired rhombus.

We used the fact that $AB=XT$. For this construction, $X,Y$ need not lie between $p,q$.


问:
How do I factor this?
How do I factor $p^2+8pq+16q^2-9r^2$? I know how to group the first two terms, but I dont know what to do with the other half. Can someone help me with this problem?</p>

Title: How do I factor this?

Body:
Using Quadratic Equation formula

$$p=\frac{-8q\pm\sqrt{(8q)^2-4(16q^2-9r^2)}}2=-4q\pm 3r$$

$$\implies p^2+8pq+16q^2-9r^2=\{p-(-4q+3r)\}\{p-(-4q-3r)\}=(p+4q-3r)(p+4q+3r)$$


问:
Differential of a rational representation of algebraic groups
Let $(\pi,V)$ be a rational (regular) representation of $G \subset GL(n,\mathbb{C})$, $\pi^*$ the dual representation, and set $\rho = \pi \otimes \pi^*$.

Let $T : W \otimes W^* \to \text{End}(W)$ denote the canonical isomorphism for any vector space $W$.
I know that for $Y \in \text{End}(W)$, $T \circ (Y \otimes I) = Y \cdot T$, and supposedly $T \circ (I \otimes Y^t) = T Y$ (this last identity is written in the textbook - Goodman & Wallach - but I'm not convinced it's correct).

If we set $$ \sigma(g) = T \rho(g) T^{-1},$$ then $(\sigma, \text{End}(V))$ is another rational representation of $G$, and $ \sigma(g) (Y) = \pi(g) Y \pi(g)^{-1}$.

Now I know that $d\rho(A) = d\pi(A) \otimes I - I \otimes d\pi(A)^t$, where $A \in \mathfrak{g}$ the Lie algebra of $G$.
Somehow I am supposed to see that $$ d \sigma(A) (Y) = d \pi(A) Y - Y d\pi(A),$$ but I cannot figure it out.

I have tried expanding the left hand side using the previous two identities, but this only yields
$$ d\sigma(A)(Y) = d\pi(A)(Y) - T d\pi(A) T^{-1}(Y),$$ which is close-but-not-quite-there.
I don't think going back to first principles using the definition of the differential via derivative of the exponential and/or vector fields will work here, but I cannot think of what else to try.

答:
$\newcommand{\End}{\operatorname{End}}$
I'm going to remove $T$ from consideration by instead defining actions of $\End(V)$ on the left and right of both $\End(V)$ and $V \otimes V^*$, which commute with $T$. The endomorphism $Y \in \operatorname{End}(V)$ acts on the left of the vector space $\End(V)$ by postcomposition $Y \cdot_L X = YX$, and on the left of $V \otimes V^*$ by action on the first term: $Y \cdot_L (v \otimes f) = Yv \otimes f$. As for right actions, we have $X \cdot_R Y = XY$ (precomposition), and $(v \otimes f) \cdot_R Y = v \otimes (f \circ Y) = v \otimes Y^t f$.

Now we have a group homomorphism $\pi \colon G \to \End(V)$ defining a representation, which we use to define another representation $\sigma: G \to \End(\End(V))$ by

$$\sigma(g)(X) = \sigma(g) \cdot_L X \cdot_R \sigma(g^{-1}) = \sigma(g) X \sigma(g^{-1}).$$

This action was defined using the left and right actions: left action by $\sigma(g)$, and right action by $\sigma(g^{-1})$. We can do the same to get a representation $\rho: G \to \End(V \otimes V^*)$ by

$$ \rho(g)(v \otimes f) = \sigma(g) \cdot_L (v \otimes f) \cdot_R \sigma(g^{-1}) = \sigma(g) v \otimes \sigma(g^{-1})^t f.$$

Note that $\sigma(g^{-1})^t = \sigma^*(g)$ by definition. While it is not so clear how to write down $\sigma(g)$ as an endomorphism of $\End(V)$, it is clear to write down $\rho(g)$ as an endomorphism of $\End(V \otimes V^*)$, we simply have $\rho(g) = \pi(g) \otimes \pi^*(g)$.

Knowing the rules for duals and tensor products, one can mechanically differentiate $\rho$ to get

$$ d\rho(g) = d \pi(g) \otimes 1 - 1 \otimes d \pi(g)^t,$$

which one could write down in terms of actions:

$$ d \rho(g)(v \otimes f) = d \pi(g) \cdot_L (v \otimes f) - (v \otimes f) \cdot_R d\pi(g).$$

Now just transport this action description back to $\End(V)$:

$$ d \sigma(g)(X) = d \pi(g) \cdot_L X - X \cdot_R d\pi(g) = d \pi(g) X - X d \pi(g).$$


问:
Calculation of $\int_a^b f(x)\,dx$
Let $f:\Bbb R\to \Bbb R$ a continous function and $a,b\in \Bbb R,a<b$. Calculate $\int_a^b   f(x)\,dx$ if $f(tx)\ge f(x)$ for any $t>0$ and $x\in \Bbb R$

I obtained just that $f(x)\ge f(0)$ but I do not think it is helpful.

答: Put $t=\frac y x$ to get $f(y) \geq f(x)$ for all $x,y>0$. Can you show now that $f$ is a constant on $(0,\infty)$? A similar argument would show that $f$ is  a constant on $(-\infty,0)$. By continuity, $f$ is a constant.

Title: Calculation of $\int_a^b f(x)\,dx$


问:
How to solve for $\theta$ in the equation $\sin(2\theta) = \sin(\theta)$ using complex numbers?
I know how to solve for $\theta$ using the general solution method, by transposing $\sin(\theta)$ to the other side and applying the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ to get the equation $\sin(\theta)(2\cos(\theta) - 1) = 0$ and then taking the general solution for $\sin(\theta) = 0$ and $2\cos(\theta) - 1 = 0$.

But I want to know how to use complex numbers to solve this question and get the same answers. I tried using Euler's formula and writing both $\sin(\theta)$ and $\sin(2\theta)$ as $(e^{i\theta}-e^{-i\theta})/2i$ and $(e^{2i\theta}-e^{-2i\theta})/2i$ respectively and I even got an equation of the form $y^4 - y^3 + y - 1 = 0$, where $y = e^{i\theta}$. However, solving that only yielded one of the principle solutions i.e $\theta = \pi$, I should also get $\theta = \pi/3$, but I haven't been able to get it.

Any help on this is greatly appreciated!

答: So, you tried to solve$$y+\frac1y=y^2+\frac1{y^2}.$$This is indeed equivalent to$$y^4-y^3+y-1=0.$$But\begin{align}y^4-y^3+y-1=0&\iff(y-1)(y^3+1)=0\\&\iff y=1(=e^0)\vee y=-1\left(=e^{\pi i}\right)\vee y=e^{i\pi/3}\vee y=e^{5\pi i/3}.\end{align}And so, yes, you do get the solutions $0$, $\pi$, $\frac\pi3$, and $\frac{5\pi}3$. Actually the whole set of solutions of the equation $\sin(\theta)=\sin(2\theta)$ is$$\pi\Bbb Z\cup\left\{\frac\pi3+2n\pi\,\middle|\,n\in\Bbb Z\right\}\cup\left\{\frac{5\pi}3+2n\pi\,\middle|\,n\in\Bbb Z\right\}.$$


问:
More than 5 but less than 50 rational solutions
Can there be system of irreducible polynomial equations over $\mathbb{Q}$
which has more than 5 but less than 50 $\mathbb{Q}$-solutions?

I somehow feel that I have to use the fact that polynomials in more than one variables over an infinite field have infinity many solutions. I'm thinking in terms of algebraic sets but I really can't come up with examples (or counterexamples).

答:
For any $a \in \mathbb{Q}^n$, define $P_a(x_1,\ldots,x_n)=(x_1-a_1)^2+\ldots+(x_n-a_n)^2$. Given a finite subset $S \subset \mathbb{Q}^n$, let $P_S=\prod_{a \in S}{P_a}$. Then the zeroes of $P_S$ are exactly the roots of $S$.

Edit: I was recalled that the question was about irreducible polynomials. Then, natural objects to consider are the equations of “elliptic curves” ($y^2=x^3+Ax+B$ with $x^3+Ax+B$ separable as well) and “hyperelliptic curves” ($y^2=P(x)$, $P$ separable of degree at most five).

For elliptic curves, it is a well-known result of Mazur that, when this equation has finitely many solutions, its number of solutions is $15$, or $11$, or at most $9$ (and all cases occur). See for instance https://en.m.wikipedia.org/wiki/Torsion_conjecture .

For hyperelliptic curves (or any curve really but the formulation is a bit trickier), it is a theorem of Faltings (again, an important and difficult one) that the number of rational solutions is finite – and a difficult problem to provide in general a somewhat tight upper bound of the number of solutions (see however the very striking Theorem 1.1 of https://arxiv.org/pdf/2001.10276.pdf ).

Here is an explicit equation (a hyperelliptic curve): $y^2=x(x-1)(x-2)(x-5)(x-6)$ (from http://www-math.mit.edu/~poonen/papers/chabauty.pdf ). It has exactly nine solutions, that are: the five where $y=0$, $(3,\pm 6)$ and $(10,\pm 120)$. The proof in the reference that these are the only solutions is somewhat involved though.


答: Take $x,y \in \mathbb{R^+}$. You have that $f(x)=f(\frac{x}{y}y) \geq f(y)$ and $f(y)=f(\frac{y}{x}x) \geq f(x)$. So $f$ is constant on $\mathbb{R^+}$. In the same way you obtain that $f$ is constant on $\mathbb{R^-}$. Since $f$ is continuos $f(x)=k \;\; \forall x \in \mathbb{R}$. So you have that $\int_a^bf(x) dx=k(b-a)$

Title: Calculation of $\int_a^b f(x)\,dx$

Body:
Let $f:\Bbb R\to \Bbb R$ a continous function and $a,b\in \Bbb R,a<b$. Calculate $\int_a^b   f(x)\,dx$ if $f(tx)\ge f(x)$ for any $t>0$ and $x\in \Bbb R$

I obtained just that $f(x)\ge f(0)$ but I do not think it is helpful.


问:
Continous uniform variable and a dependent discrete bernoulli variable
Let $X\sim\text{Uniform}(\left[2,4\right])$, and $Y\sim\text{Ber}\left(\frac{1}{X}\right)$, I wonder how I can work with these variables (mainly with Y), let's say I want to calculate
$$
P\left(Y=1\right)
$$
My intuition is that I can calculate this value using integraion (something like extending the law of total probability):
$$
P\left(Y=1\right)=\int^4_2\frac{1}{2}\frac{1}{x}dx=0.5\left[\ln\left(x\right)\right]_{2}^{4}=\frac{\ln\left(4\right)-\ln\left(2\right)}{2}
$$
And it looks correct (is it?), but I am not sure this is correct, and if it is correct, why is it.

Another (very similar) problem (and a solution) I though of is
$$
P\left(Y=1\wedge X\in\left[3,4\right]\right)=\int_{3}^{4}\frac{1}{2}\cdot\frac{1}{x}dx=\frac{1}{2}\int_{3}^{4}\frac{1}{x}dx=\frac{1}{2}\left[\ln\left(x\right)\right]_{3}^{4}=\frac{\ln\left(4\right)-\ln\left(3\right)}{2}
$$

If my solutions are correct, what is the reasoning for their correctness? Is there another way to get to them (in the second one, maybe Bayes?)?

答:
If my solutions are correct, what is the reasoning for their correctness?

Your first solution is correct and here is the explanation you asked for:

To be precise your $Y$ is a conditional rv, thus you have

$$\left(Y|X=x\right) \sim B\left(\frac{1}{x}\right)$$

First observe that in a Bernulli model,

$$\mathbb{P}[Y=1]=\mathbb{E}[Y]$$

now using Conditional expectation's properties you get

$$\mathbb{E}[Y]=\mathbb{E}\left[\mathbb{E}[Y|X]\right]=\mathbb{E}\left[\frac{1}{x}\right]=\int_2^4 \frac{1}{2x}dx=\frac{\log 4-\log 2}{2}$$

At present I did not check you second one


问:
how to find shortest simple cycles in directed graph that contain two given nodes
*this problem can be NP-hard

I am looking for algorithm to find shortest simple cycle (every node in cycle is traversed exactly once) in directed (unweighted for simplicity) graph that contain two given nodes, also it would be nice to have polynomial time instead bruteforce-like solutions (although some NP solutions with good optimizations may be valuable)

one idea is to use Dijkstra and in second step "inverted" Dijkstra although, after normal Dijkstra removing traversed arcs/edges may disconnect graph which should contain cycle for given nodes,

any ideas would be welcome

答:
Input: A digraph $G=(V,E)$ and two vertices $u,v \in V$.

Output: The shortest simple directed cycle in $G$ containing $u$ and $v$.

I am not sure if we could use a 2-disjoint shortest path approach for the directed case as well (a version in which we find a shortest path from $u$ to $v$ and one from $v$ to $u$ and these paths are disjoint). This problem seems to be $\mathcal{NP}$-hard in general for directed graphs, but I am not sure. There might be a neat solution.

However, we could use Johnson's algorithm. It numbers the vertices from $1$ to $|V|$ and finds all cycles containing the smallest numbered vertex, and continues with the next one and so on. So we modify it such that our vertex $u$ receives the smallest number and we only iterate over the cycles containing $u$. While doing so, we keep track of the lengths of each such cycle and if a cycle contains $v$ we temporarily store it to some variable $C$. In each iteration we compare the length of each new match (that is a cycle containing $u$ and $v$) with the length of $C$  and if it is smaller, we replace the value of $C$ with the new cycle. We do that for each cycle. Then return $C$.

Johnson's algorithm is implemented in NetworkX and you could fork their implementation from here. Johnson's algorithm is implemented in the method "simple_cycles(G)".

Input: An undirected graph $G=(V,E)$ and two vertices $u,v \in V$.

Output: The shortest simple cycle in $G$ containing $u$ and $v$.

This works for undirected graphs. We translate the undirected graph into its directed version (replace each undirected each by two directed edges being inverse to each other) and use Suurballe‘s algorithm exploiting the fact that the union of two shortest disjoint paths with the same endpoints form a shortest cycle in an undirected graph. It finds two edge disjoint shortest paths between two vertices $u$ and $v$ by running Dijsktra‘s algorithm twice.

There is a nice running example in this article.

Notice that standard Suurballe is not necessarily vertex disjoint but we can modify it as described in both provided articles:

The version of Suurballe's algorithm as described above finds paths that have disjoint edges, but that may share vertices. It is possible to use the same algorithm to find vertex-disjoint paths, by replacing each vertex by a pair of adjacent vertices, one with all of the incoming adjacencies u-in of the original vertex, and one with all of the outgoing adjacencies u-out.


问:
Integer Part Equations...again
This is a plea for readers solutions!

Solve $[20 \, x−3] = [15 \, x + 27]$, where $[x]$ is the integer part of $x$.

I have my solution which is spread over five intervals and can solve any $[a \, x+b] = [c \, x + d]$ if $a,b,c,d$ are known integers.

What I wish to do is set down a general solution for integer $a, b, c, d, ....$ this is proving 'awkward'. So, I would welcome readers methods to solve the equation in order that it may furnish me with ideas for a general solution. The number of intervals in which the solutions lie varies depending on the values of $a, b, c, d$.

答:
$$\lfloor 20x−3 \rfloor = \lfloor 15x + 27 \rfloor$$

Let's start by wondering if there is a solution where $20x-3$ and $15x + 27$ are both integers. If that were the case, then we would get

\begin{align}
   20x−3 &= 15x + 27 \\
   5x &= 30 \\
   x & = 6
\end{align}

This is the union of these sets
\begin{align}
\end{align}

And we see that $20x-3 = 15x + 27 = 117$, so both sides are both equal to the integer $117$.

So lets move our origin to the point $x=6$.

Let $x = t+6$. Then we get

\begin{align}
   \lfloor 20x−3 \rfloor &= \lfloor 15x + 27 \rfloor \\
   \lfloor 20t+120−3 \rfloor &= \lfloor 15t + 90 + 27 \rfloor \\
   \lfloor 20t+117 \rfloor &= \lfloor 15t + 117 \rfloor \\
   \lfloor 20t \rfloor + 117 &= \lfloor 15t \rfloor + 117 \\
   \lfloor 20t \rfloor &= \lfloor 15t \rfloor
\end{align}

Let $n$ be an integer. Then

$\lfloor 15t \rfloor = n
 \implies  n \le 15t < n + 1
 \implies \frac{n}{15} \le t \lt \frac{n+1}{15} $
$\lfloor 20t \rfloor = n
 \implies  n \le 20t < n + 1
 \implies \frac{n}{20} \le t \lt \frac{n+1}{20}$

For non negative $n$,  the intersection
$$\bigg[\frac{n}{20}, \frac{n+1}{20}\bigg) \cap
 \bigg[\frac{n}{15}, \frac{n+1}{15}\bigg)
= \bigg[\frac{n}{15}, \frac{n+1}{20}\bigg)$$
must be non empty.

This will work until

\begin{align}
   \frac{n}{15} &\ge \frac{n+1}{20} \\
   20n &\ge 15n + 15 \\
   n &\ge 3
\end{align}

For negative $n$, the interserction
$$\bigg[\frac{n}{15}, \frac{n+1}{15}\bigg) \cap
 \bigg[\frac{n}{20}, \frac{n+1}{15}\bigg)
 = \bigg[\frac{n}{20}, \frac{n+1}{15}\bigg)$$
must be non empty.

This will work until

\begin{align}
   \frac{n}{20} &\ge \frac{n+1}{15} \\
   15n &\ge 20n + 20 \\
   n &\le -4
\end{align}

So the solution set is
$$\bigcup_{n=-3}^{-1}\left[6+\frac{n}{20}, 6+\frac{n+1}{15} \right)
 \cup \bigcup_{n=0}^{2}\left[6+\frac{n}{15}, 6+\frac{n+1}{20} \right)
$$

This is the union of these six intervals

\begin{array}{l}
   \left[6-\frac{3}{20}, 6-\frac{2}{15} \right) \\
   \left[6-\frac{2}{20}, 6-\frac{1}{15} \right)\\
   \left[6-\frac{1}{20}, 6               \right)\\
   \left[6+\frac{0}{15}, 6+\frac{1}{20} \right) \\
   \left[6+\frac{1}{15}, 6+\frac{2}{20} \right) \\
   \left[6+\frac{2}{15}, 6+\frac{3}{20} \right)
\end{array}

Title: Integer Part Equations...again


问:
Hatcher, topology of cell complexes
In the appendix to Algebraic Topology, Hatcher briefly goes over why

A subset $A$ of a CW complex $X$ is open in $X$ $\iff$ $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\Phi_\alpha$, where $\Phi_\alpha:D_\alpha^n \to X$ is a characteristic map (the map which attaches a particular $n$-disk to $X$ continuously.)

The proof in the $(\implies)$ direction is quite easy, since the $\Phi_\alpha$ are continuous.

However, for the proof in the $(\impliedby)$ direction, I do not follow Hatcher completely. Quoting:

...and in the other direction, suppose $\Phi_\alpha^{-1}(A)$ is open in $D^n_\alpha$ for each $\Phi_\alpha$, and suppose by induction on $n$ that $A ~\cap~X^{n-1}$ is open in $X^{n-1}$. Then since $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\alpha$, $A \cap X^n$ is open in $X^n$ by the definition of the quotient topology on $X^n$.

For reference, $X = \bigcup_n X^n$, and $X^n = (X^{n-1} \sqcup_{\alpha}D_\alpha^n)/(x \in \partial D_\alpha^n \sim \Phi_\alpha(x))$.

It is the bolded portion where I do not precisely follow. In more detail, what is Hatcher's reasoning? I believe what is going on is that because $\Phi_\alpha^{-1}(A\ \cap X^n)$ is open in every $D^n_\alpha$, and $\Phi_\alpha$ is just the quotient map $q:X^{n-1} \sqcup D_\alpha^n \to X^n$ composed appropriately with inclusions, we get$q^{-1}(A \cap X^n)$ is open. Is this approximately correct? Where can my understanding be made more precise?

答:
A CW complex $X$ is endowed with the weak topology, meaning that, in order to show that $A$ is open in $X$, it suffices to show that $A \cap X^n$ is open in $X^n$. Now, the idea is to work your way up inductively. Suppose you know already that $A \cap X^{n-1}$ is open in $X^{n-1}$. How can we show $A \cap X^n$ is open in $X^n$?

We invoke the fact that $X^n$ is formed from $X^{n-1}$ as a quotient space $(X^{n-1} \sqcup_\alpha D^n_\alpha) / (x \in \partial D^n \sim \Phi_\alpha^n(X))$: it suffices to verify, by the quotient property, that $A$ is open in $X^{n-1}$ (but this we know by induction) and that its pre-images under $\Phi^n_\alpha$ are open in the $D_\alpha^n$. But this is our starting assumption.

Title: Hatcher, topology of cell complexes


问:
To determine the existence of a map from a set to another set.
Let  $A = \{ x^2| 0<x<1\}$ and $B = \{x^3| 1<x<2\}$. Does there exist a one to one and onto map from A to B?

My attempt:- For a one to one and onto map (basically a bijection) to exist between two sets, their cardinalities must be same. But I do not know how to proceed. Any help will be much appreciated.

答:
As it can be seen $ 0< x <1 \Rightarrow 0 < x^2 < 1 $

$0 < x^2 < 1 \Rightarrow A=(0,1)$

$1<x<2 \Rightarrow 1 < x^3< 8$

$1 < x^3< 8 \Rightarrow B=(1,8)$

let $f(x)=7x+1$

$f(a)=f(b) \Rightarrow 7a=7b \Rightarrow a=b$

let $n \in (1,8)$

$n=f(x)=7x+1 \Rightarrow x=\frac{n-1}{7}$

so $f(x)$ is a bijection between $A$ and $B$


问:
Find the supremum and infimum (if exist) of the set $\{\frac{m}{n} + \frac{4n}{m} \mid m,n \in \Bbb N\}$ and $\{x:x^2+x+1>0\}$.
Find the supremum and infimum (if exist) of the set
$S = \{\frac{m}{n} + \frac{4n}{m} \mid m,n \in \Bbb N\}$ and $T = \{x:x^2+x+1>0\}$.

Attempt:
First, notice that $S = \{x \in \Bbb R \mid x \gt 4\}$.

For the set $S$, by $AM-GM$ inequality, we have
\begin{align*}
\frac{m}{n} + \frac{4n}{m} &\ge 2\sqrt{\frac{m}{n} \cdot \frac{4n}{m}} = 2\cdot \sqrt{4} = 2\cdot 2 = 4.
\end{align*}
Thus, $4$ is a lower bound of $S$. We claim that $\inf(S)=4$. Let $m$ be an any lower bound of $S$.
We'll show that $m \le 4$. For the sake of contradiction, suppose $m \gt 4$. Let $r$ be an arbitrary rational number such that $4 < r < m$. Then, we have $r \in S$ and $m \gt r$. This is impossible since $m$ is a lower bound of $S$. Hence, $m \le 4$ and therefore, $\inf(S) = 4$.

Next, set $n=1$ and $m \to \infty$. Then, $\frac{m}{n} + \frac{4n}{m} \to \infty$.
Hence, $\sup(s)$ is doesn't exist.

Am I true?

${}$

For the set $T$, notice that $x^2+x+1$ has two distinct complex roots. How to process next?
What if $x$ there is in real set? What if $x$ there is in complex set?

Thanks in advanced.

答:
Once you showed that $4$ is a lower bound, all that needs to be demonstrated is that the lower bound is attained for some $m, n \in \mathbb N$ in order for $\inf S = 4$.  This is easily accomplished since AM-GM implies that equality is attained when $$\frac{m}{n} = \frac{4n}{m},$$ or $m = 2n$.  So $(m,n) = (2,1)$ gives the desired lower bound, hence this must be the greatest lower bound.

For $x^2 + x + 1$, simply complete the square.  $$x^2 + x + 1 = (x+1/2)^2 + 3/4 \ge 3/4,$$ and so all real numbers satisfy the condition $x^2 + x + 1 > 0$.  Thus I suspect you intended to write instead $$T = \{x^2 + x + 1 : x \in \mathbb R\},$$ in which case $\inf T = 3/4$ and $\sup T = \infty$.


问:
True or false statements odd integers
One is false one is true write the negation for the false one and prove both of them

a) for all integers a, there exists an integer b so that a+b=odd

b) there exists an integer b such that for all integers a, a+b=odd

I said (a) is false and (b) is true. I wrote the negation for (a) then I split it in two cases. Case 1: b is even. Let b=2k and let a=0 then a+b=2k which is even.
Case 2: b is odd. Let b=2k+1 and let a=-1 then a+b=2k

Then for (b) I set up two cases again. Case 1: a is even. A=2k and b=1 then a+b=2k+1 which is odd
Case 2: a is odd, let a=2k+1 and b=0 then a+b=2k+1 which is odd

Input in whether I'm right or not would be great thanks

答:
I’m afraid that you got them exactly backwards. Let’s take a look.

(a) For each integer $a$ there is an integer $b$ such that $a+b$ is odd.

Think of this in terms of a game. I give you an integer $a$, and you win if you can find an integer $b$ such that $a+b$ is odd; if you cannot find such a $b$, I win. The assertion (a) says that you can always win, no matter how I choose my $a$.

Suppose I give you the integer $101$; can you find an integer to add to it to make an odd integer? Sure: it’s already odd, so just add $0$. Suppose instead that I give you the integer $100$; can you find an integer to add to it to make an odd integer? $0$ won’t work, but $1$ will: $100+1=101$, which is certainly odd. Was there anything very special here about $101$ and $100$? No: all I used was the fact that $101$ is odd, so that adding $0$ was bound to give me an odd total, and the fact that $100$ is even, so that adding $1$ would give me an odd total. If I give you any odd integer $a$, you can use $b=0$ and be sure that $a+b=a+0=a$ will be odd. And if I give you any even integer $a$, you can use $b=1$ and be sure that $a+b=a+1$ is odd. Every integer is either even or odd, so no matter what integer $a$ I give you, you’re covered: you know how to pick a $b$ such that $a+b$ is odd. In other words, (a) is true: you do have a winning strategy.

(Of course there are other choices that work besides the ones that I’ve mentioned; mine are just the simplest.)

(b) There is an integer $b$ such that for all integers $a$, $a+b$ is odd.

Again you can think of this in terms of a game, with you picking $b$ and me picking $a$. The difference is that in this game you have to play first: you pick some integer $b$, then I pick my $a$, knowing what your $b$ is. You win if $a+b$ is odd, you lose if it isn’t, and (b) says that you have a winning strategy: there is some $b$ that you can pick that will make $a+b$ odd no matter what integer I pick for $a$.

But that’s clearly not true: if your $b$ is even, I’ll just let $a=0$, so that $a+b=b$ is even, and you lose. And if your $b$ is odd, I’ll pick $a=1$, so that $a+b=1+b$ is even, and again you lose. No matter how you play — no matter what integer you choose for your $b$ — I can beat you by choosing $a$ to make $a+b$ even.

Title: True or false statements odd integers


问:
Singular Value Decomposition proof - Why does U need to be an orthonormal basis?
So in the textbook they have the following theorem:

With the following proof:

What I don't understand is why they extend ${\textbf{u}_1, ..., \textbf{u}_r}$ to an orthonormal basis. In fact, if you just define
$$U = [\textbf{u}_1, ... , \textbf{u}_r, \textbf{0}, ..., \textbf{0}]$$

The answer should be the same...? Ie. $U\Sigma = AV$

答:
I have chosen to turn my brief comment into a more elaborate answer. If I understand the question correctly, we should really answer why it is interesting that there exists an ortogonal $U$ as opposed to just an arbitrary $U$ that satisfies the equality $U \Sigma = AV$.

First of all, the $U$ suggested in the question is singular, as pointed out in the comments. That choice of $U$ would therefore make the theorem alot more "one-sided" in the sense that you would not have $U^T A = \Sigma V^T$. When both $U$ and $V$ are invertible, the theorem is more "symmetric" and easier to utilize.

Secondly, when both $U$ and $V$ are ortogonal, the singular value decomposition has a nice interpretation. By finding such a decomposition, we can better understand the map $x \mapsto Ax$. Since $A$ is not necessarily neither invertible, nor at all square, it can be difficult to get a grip of what multiplying by $A$ does. The singular value decomposition provides an answer. Since $A = U \Sigma V^T$, we can look at the maps $V^T$, $\Sigma$ and $U$ successively. Here it is really helpful that both $U$ and $V^T$ are ortogonal because multiplication by ortogonal matrices can be thought of as rotations. The multiplication by $\Sigma$ can be thought of as a scaling.

For an even more elaborate answer, please consult the Wikipedia article about SVDs (https://en.wikipedia.org/wiki/Singular_value_decomposition), which explains this interpretation well and provides some illustrative figures and animations.


问:
What is the automorphism group of a complete bipartite graph isomorphic to?
I couldn't quite figure out what the automorphism group of a complete bipartite graph $\Gamma (K_{r,s})$ is isomorphic to. I did some back-of-the-envelope calculations and found that $\Gamma (K_{r,s})$ has an order of $(s-1)!$ when $r = 1$ and $s \geq 2$, and has an order of $2(r!s!)$ when $ r, s \geq 2$ (I don't have a proof of that. I might be wrong.) but I'm not aware of any groups that might be isomorphic to $\Gamma (K_{r,s})$ as my knowledge of group theory is rather elementary.

答:
Your computations of the number of automorphisms aren't quite right in general, though they are close.

Hints: Clearly, the $r$ nodes on one side are all equivalent to each other, and can be freely permuted; what is the symmetry group of $r$ identical objects? Given that similarly the $s$ nodes on the other are also all equivalent, we could consider an arbitrary permutation of those as well; how do we combine these two symmetry groups together to get a larger one? And finally, are there any special values of $r,s$ for which there is an extra symmetry?