问:
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$?
It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?

答:
To get things going you can calculate the inverse of $[3]_{100}$ and then try relating it to powers of $[3]_{100}$, keeping in mind that $10^2 \equiv 0 \pmod{100}$.

In $\text{modulo-}100$ you know that the inverse has the form $[10n +7]_{100}$ and solving you can write out

$\tag 1 3 \cdot 67 \equiv 1  \pmod{100}$

We can form a relation right away, suspending further ${[3^n]_{100}}$ calculations,

$\quad  3^3 \equiv 67 - 40$

and so

$\tag 2  3^4 \equiv 1 - 20 \pmod{100}$

We can now use the binomial theorem.

$\quad  \displaystyle \large 3^{1234} \equiv (3^4)^{308} \cdot 3^2 \equiv$

$\quad  \displaystyle \large  9(1-20)^{308} \equiv$

$\quad  \displaystyle \large 9(1-308 \cdot 20^1 + \Large 0  \large ) \equiv \quad \quad \normalsize \; (20^k \text{ usually has residue } 0)$

$\quad  \displaystyle \large 9 \cdot 41 \equiv 69 \pmod{100}$


问:
Ray-Lens Intersection
So imagine that I have a ray parameterized as $\vec{R} = \vec{O} + t\vec{D}$, where $\vec{O}$ = origin, $t$ = parameter and $\vec{D}$ = direction vector.

I also have a spherical lens with aperture $A$, radius of curvature $r$, and center of curvature $C$.

Assume that the lens is centered on the $z$-axis and that the ray always comes in from "one side" of the lens. So if it's a convex lens, the rays will always come from its convex direction and not from it's backside.

So this is what I'm doing to solve this:

1) Treat the lens as a full sphere with center $C$ and radius $r$.

2) Do a normal ray-sphere intersection algorithm and return the closest of the two points.

3) Once I have an intersection point $(x,y,z)$, I check to see if $ |x| \le A/2$ and $|y| \le A/2$.

The reason for step 3 is that the aperture $A$ refers to the width of the lens, so if it's centered on the $z$-axis, the range of $x$ values on the lens' surface range from $-A/2 \le x \le A/2$ (and the same for $y$) even though the range of $x,y$ values for the sphere will be from $-r \le x,y \le r$.

So my question is: Is this a reasonable way to go about solving this problem? I've sort of convinced myself in my mind that this is correct, but perhaps if there is a flaw in my reasoning I would really appreciate the feedback!

答: For step 3, you want to check that $x^2+y^2 \le (\frac A2)^2$.  If you are off $z$ axis in both $x$ and $y$ you want the total distance off axis to be less than half the aperture.  For example, if $A=2$, the point $(\frac 34, \frac 34)$ would be outside your aperture, but pass your test.  Your test would be correct for a square aperture.</p>

ContentLicense: CC BY-SA 3.0

Title: Ray-Lens Intersection


问:
The problem of the distribution function of arrival epochs $S_i$ and time $t$ difference is in the Poisson process.
Let $N(t)$ be a Poisson process, and the rate of arrival is 1. We know that at time $t=1$, $N(t)=3$.

(a) Find the distribution function of $S_4-1$.

(b) Find the $E[(S_4-1)|N(1)=3]$

(c) Find the $E[(S_4-S_3)|N(1)=3]$

I know that the distribution function of interval time $X_i$ can be expressed as:

$Pr\{X>x\}=F_X^{c}(x)=exp(-\lambda x)$.

So the answer to question A is this? $Pr\{S_4>X>1\}$

This is a hard problem for me and I hope to get your help.
I was also confused about (b) and (c) and didn't know how to do it.
It looks like in (c), the difference between $S_4$ and $S_3$ is $X_4$.

答:
The important concept to grasp is that if you know $N(1)=3$, then you can treat the process from time $t=1$ onward as a brand new Poisson process with the same rate. (That is, at time $t=1$, just start a brand new Poisson process, and this will have the same distribution as the original process conditioned on $N(1)=3$.) This is a consequence of the independent increments property of Poisson processes.

Consequently, for (a) note that $S_4-1$ is the time between $t=1$ and the first arrival after $t=1$, so in the "new" Poisson process, this is simply "the time of the first arrival," whose distribution you know to be exponential with rate $\lambda=1$. This should allow you to immediately solve (b) as well.

For (c), it suffices to compute $E[S_3 \mid N(1)=3]$. Show that
$$P(S_3 < t \mid N(1) = 3) = \frac{P(S_3 < t, N(1)=3)}{P(N(1)=3)} = \begin{cases} \frac{(e^{-\lambda t}(\lambda t)^3/3!)(e^{-\lambda(1-t)})}{e^{-\lambda} \lambda^3/ 3!} = t^3& 0 \le t \le 1 \\ 0 & \text{otherwise}\end{cases}$$
and compute $E[S_3 \mid N(1)=3] = \int_0^1 P(S_3 \ge t \mid N(1)=3) \, dt$.

ContentLicense: CC BY-SA 4.0


问:
Evaluating $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$ using L'Hospital's Rule
I'm studying normal distribution at a high school level, and I need to evaluate the limit $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$

I asked this question before here, and the answer I got was that I need to use L'Hospital's Rule. From a quick Google Search, I learnt that L'Hospital's Rule is $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$.
I tried to use it to evaluate the limit, but obviously I don't know enough about L'Hospital's Rule to do it.

I'm now just curious on how you would use L'Hospital's Rule to evaluate the limit $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$. Thank you!

答:
Take $f(x)=x$ and $g(x)=e^{\frac{1}{2}x^2}$. Then $f'(x)=1$ and $g'(x)=xe^{\frac{1}{2}x^2}$, so by L'Hopital
$$\lim_{x\to\infty}\frac{x}{e^{\frac12x^2}}=\lim_{x\to\infty}\frac{1}{xe^{\frac{1}{2}x^2}}=0.$$

I'd personally not use L'Hopital for a limit like this one -- it's a bit overkill. A simpler way would be to use the series $e^x=1+x+\frac{x^2}{2!}+\dots$, so $e^{\frac{1}{2}x^2}\geq1+\frac{x^2}{2}$ for $x>0$ and go from there.


答:
$$\lim_{z\to\infty}ze^{-\frac12z^2}=\lim_{z\to\infty}\frac z{\exp\left(\frac12z^2\right)}$$
Now that we have the function in the form $\frac{f(z)}{g(z)}$, we can take the derivative of the numerator and denominator.
$$\lim_{z\to\infty}\frac z{\exp\left(\frac12z^2\right)}=\lim_{z\to\infty}\frac {\frac d{dz}z}{\frac d{dz}\exp\left(\frac12z^2\right)}=\lim_{z\to\infty}\frac 1{\frac d{dz}\left[\frac12z^2\right]\exp'\left(\frac12z^2\right)}=\lim_{z\to\infty}\frac 1{z\exp\left(\frac12z^2\right)}=\frac1\infty=0$$

Title: Evaluating $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$ using L'Hospital's Rule

Body:
I'm studying normal distribution at a high school level, and I need to evaluate the limit $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$

I asked this question before here, and the answer I got was that I need to use L'Hospital's Rule. From a quick Google Search, I learnt that L'Hospital's Rule is $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$.
I tried to use it to evaluate the limit, but obviously I don't know enough about L'Hospital's Rule to do it.

I'm now just curious on how you would use L'Hospital's Rule to evaluate the limit $\lim _{z \rightarrow \infty} z e^{-\frac{1}{2} z^{2}}$. Thank you!


问:
Expression of a holomorphic function.
I have the following exercise:

Suppose $f:\mathbb{C^{*}}\rightarrow \mathbb{C}$ is holomorphic. Show that if $f(z)\rightarrow \infty$ when $z\rightarrow 0$ and $f(z)\rightarrow \infty$ when $z\rightarrow \infty$, then $f(z)$ can be written in the form: $f(z)=\dfrac{a_k}{z^k}+...+\dfrac{a_1}{z}+a_0+b_1z+...+b_lz^l$

Before I have solved an exercise that in my opinion is similar and said the following:

Show that if $f$ is entire and $\text{lim}_{z\rightarrow\infty}f(z)=\infty$ then $f$ is a polynomial.

I have solved this exercise using the Casorati-Weierstrass theorem, I have tried to see if the same strategy works to solve the exercise I have, but I cannot conclude anything. I hope you can help me. Thank you.

答:
The first condition $\lim_{z \to 0} f(z) = \infty$ implies that $f$ has a pole at the origin. Let $k$ be the order of that pole. Then $g(z) = z^kf(z)$ has a removable singularity at the origin and therefore can be extended holomorphically to an entire function.

Now use the other condition $\lim_{z \to\infty} f(z) = \infty$ to show that $g$ is a polynomial, so that $f(z) = g(z)/z^k$ has the desired form.

Title: Expression of a holomorphic function.


问:
How do I show that the degree $n$ Taylor polynomials of $f$ about two points are equal?
Question

Suppose that $f(x)$ is a polynomial of degree $d$, and that $n \ge d$. Let $x_0 \neq x_1$. Show that the degree $n$ Taylor polynomials of $f$ about $x_0$ and $x_1$ are equal.

Attempt

Let the polynomial be $f(x) = \sum_{k=0}^{d}a_kx^k$. Then the Taylor polynomial of $f$ about $x_0$ is $$P_n(x) = \sum_{k=0}^{n} \dfrac{f^{(k)}(x_0)(x-x_0)^k}{k!}$$

and the Taylor polynomial of $f$ about $x_1$ is $$Q_n(x) = \sum_{k=0}^{n} \dfrac{f^{(k)}(x_1)(x-x_1)^k}{k!}.$$

For them to be equal, I think it would have to be the case that $P_n(x) - Q_n(x)= 0$

I tried with a simple example. Let $d=1$ and $n=3$. Then

$$f(x) = \sum_{k=0}^{1}a_kx^k = a_0 + a_1x ,$$

$$P_3(x) = \sum_{k=0}^{3} \dfrac{f^{(k)}(x_0)(x-x_0)^k}{k!} = a_0 + a_1(x-x_0) ,$$

$$Q_3(x) = \sum_{k=0}^{3} \dfrac{f^{(k)}(x_1)(x-x_1)^k}{k!} = a_0 + a_1(x-x_1) .$$

However, $P_n(x) - Q_n(x)= 0 \Rightarrow x_0 = x_1$ or $a_1 = 0$. Where did I go wrong?

Thank you for your time.

答: The degree $n$ Taylor polynomial of a polynomial of degree $\le n$ is simply that polynomial itself - no matter what point you develop at. If you don't believe it, observe that the Taylor error term is a multiple of some $f^{({n+1})}(\xi)$, hence zero.</p>


问:
For any $f:A\to A$ and any relation $R$ on $A$, define $S$ on $A$ by $aSb$ iff $(f(a),(f(b))\in R$. Does $S$ reflexive or symmetric imply it for $R$?
Let $A = \{1,2,3,4\}$. For any function $f : A \rightarrow A$ and any relation $R$ on $A$, we define the relation $S$ on $A$ by: for any $a,b \in A,   aSb$ iff $(f(a),(f(b)) \in R$.

Prove / Disprove:

(a) $\forall$ functions $f : A \rightarrow A$ and all relations $R$ on $A$, if $S$ is reflexive then $R$ is reflexive.

(b) $\forall$ functions $f : A \rightarrow A$ and all relations $R$ on $A$, if $S$ is symmetric then $R$ is symmetric.

For both (a) and (b) I think they are false, but I am not sure on this as I do not know exactly how to start my proofs.

For (a) could I just give an example where the function $f(a)$ equals a constant for every element in $A$? i.e. $f(a) = 1,  \forall a \in A$. Then go from there to show $S$ is reflexive, but $R$ is not reflexive.

For (b) I was thinking of doing the same thing, and assigning $f(a) = 1, \forall a \in A$ and showing an example where $S$ is symmetric but $R$ isn't.

答:
A counterexample is a perfectly valid means of approach, and both of those functions work. Take $R$ to be the relation $\{(1,1),(1,2)\}$ for both of them.

In the first case, $S$ is reflexive since $(f(a),f(a))=(1,1) \in R$ for every $a \in A$. However, $(2,2) \not \in R$, so $R$ isn't reflexive.

In the second case, $S$ is symmetric trivially since $S = A \times A$. However, $(1,2) \in R$ doesn't imply $(2,1) \in R$, so $R$ is not symmetric.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Euler's totient function - inequality on even numbers?
Is there a theorem that says:

If $n$ is even then: $$\varphi(n)\le\frac{n-1}{2},$$ where $\varphi(n)$ is Euler's totient function.

If there isn't such a theorem, how would I show/prove that this is true?

答:
If you combine the two comments you'd have:

Case 1: $n=2^k m, k\ge 1$, $m>1$ odd, then $$\phi(n)=\phi(2^k)\phi(m) = 2^{k-1} \phi(m)  \le 2^{k-1} (m-1) < 2^{k-1} m - \frac 12 = \frac{n-1}{ 2}$$

Case 2: If $m=1$ then $\phi(n)=2^{k-1} =\frac n2 > \frac{n-1}{2}$.


问:
Standard Deviation Question multiplying data set by a constant
Suppose that the $25$ numbers in data set $S$ have mean $7$ and standard deviation
$3$. If each of the numbers in $S$ is multiplied by $4$ to form a new data set $T$, what is
the standard deviation of $T$?

Would not the answer be $12$, because the book's answer is $6$? The explanation for the book answers uses the wrong formula for standard deviation,taking the difference of the new data points minus the new mean instead of the difference squared.

答: If $\sigma$ is the standard deviation of a data set, then the standard deviation becomes $k\sigma$ if each number in the data set is multiplied by $k$. In this case the standard deviation becomes $$3\times 4=12$$


问:
Is the relation $R = \{ ( 1,2 ) \}$ on $A = \{ 1,2,3,4 \}$ not reflexive, not symmetric, antisymmetric and transitive?
Given $R = \{ ( 1,2 ) \}$ over $A = \{ 1,2,3,4 \}$:

$R$ is not reflexive because there's no $(1,1)$ and $(2,2)$.

$R$ is not symmetric because there isn't $( 2,1 )$.

$R$ is anti-symmetric because there isn't $( 2,1 )$.

And $R$ is transitive because there isn't a $(2,x)$ element in $R$.

If  $(1,2) \in R \land (2,x) \in R$ is false, then

$(1,2) \in R \land (2,x) \in R \to (1,x) \in R$ is true,

and therefore $R$ is transitive.

Is this correct?

答:
The relation is not reflexive, since $(a,a) \not \in R$ for every $a \in \{1,2,3,4\}$.

The relation is not symmetric, since while $(1,2) \in R$, the reversal $(2,1) \not \in R$.

The relation is anti-symmetric in a vacuous sense: since $(1,2) \in R$ and is the only ordered pair in $R$, there is no "violation" in the sense that since there is no pair of related pairs $(a,b),(b,a) \in R$ with $a \ne b$ to begin with. Thus, anti-symmetry holds trivially.

The relation is transitive, for similar reasons. If $R$ were not transitive, then there would be pairs $(2,x),(x,y) \in R$ for which $(1,y) \not \in R$. However, no $(2,x) \in R$ for any $x$, so there is no issue there. Vacuously, in a sense, $R$ thus is transitive.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
What are the relations in this presentation of $S_n$?
The permutation group $S_n$ can be generated by an element $$(i \; \; i+1)$$ together with an $n$-cycle.

First, I would like to check that the above statement is indeed true?

My main question is, is it known what the relations are when we generate $S_n$ like this?

答:
Well, let's say that first one is on you. For the other one, one can (somewhat easily) find out that
$$ \langle t, c \,|\,t^2, c^n, (tc)^{n-1}, [t, c]^3,  [t, c^k]^2 = 1 \text{ for } 2 \leq k \leq n/2\rangle $$

is indeed the presentation you need. The best (to my knowledge) way to check its validity is first writing down Coxeter presentation as a reflection group, and then eliminating extra generators replacing them by products of conjugates of a transposition.


问:
Max/min help please
I am studying analysis, and we're just getting to the topic of maxima and minima. I'm trying to find the maximum and minimum of the set $$S := (0,4] \cap \Bbb{Q}.$$ I know the minimum DNE, and I know why it doesn’t exist. But I don’t know why the maximum is $4$ and not $> 4,$ if the universal set extends past $4$? Any help would be appreciated. Please and thank you.

答:
$\max S = 4$ since $4 \in (0, 4]$ and $4 \in \mathbb{Q}$ imply $4 \in S$, and there is no $x \in S$ for which $x > 4$.

$\min S$ does not exist. Suppose $\min S = x$. Then $y = x/2 < x$ and $y \in S$, a contradiction.

Title: Max/min help please


问:
Algorithm to determine multiplier to reach specific number
If I earn 10,000 dollars in the month of January and want to earn 1 million in the month of December by increasingly upping each month's income by a steady amount, how do I determine the multiplier? For example:

10,000 (January)

10,000 x 1.5 = 15,000 (February)

15,000 x 1.5 = 22,500 (March)

and on until 1 million is reached in December.

How do I determine the multiplier? It's around 1.5, but I'm very curious to know if there's a specific calculation that's faster than guessing and manually adding it up!

答:
Hint

$$1 + k + k^2 + \cdots + k^n = \frac{1 - k^{(n+1)}}{1-k}.$$

Edit

...until 1 million is reached in December.

The above hint assumes the goal of all monthly payments, including December's totalling 1 million.  The alternative interpretation is that you want December's payment, by itself to be 1 million.

In that case, the alternative hint is

$$k^r = s \iff e^{[r \times \ln(k)]} = s \iff [r \times \ln(k)] = \ln(s).$$


答: Let $x$ be the multiplier. This is simply equal to solving the equation $$10000x^{12}=1000000\longrightarrow x=100^{\frac{1}{12}}\approx 1.4678$$

Title: Algorithm to determine multiplier to reach specific number

Body:
If I earn 10,000 dollars in the month of January and want to earn 1 million in the month of December by increasingly upping each month's income by a steady amount, how do I determine the multiplier? For example:

10,000 (January)

10,000 x 1.5 = 15,000 (February)

15,000 x 1.5 = 22,500 (March)

and on until 1 million is reached in December.

How do I determine the multiplier? It's around 1.5, but I'm very curious to know if there's a specific calculation that's faster than guessing and manually adding it up!


问:
Pure states and minimal projections
Probably it is trivial or already discussed, but I just wanted to pose a simple question about pure states.

${\bf Lemma \; 1 \colon}$ Let $\mathcal{A}$ be a $C^*$-algebra acting irreducibly on some Hilbert space $\mathcal{H}$ and let $\varphi(x)=(\xi | x \xi)$ be some vector state. If $\xi$ is cyclic, then $\varphi$ is pure.

${\bf Proof \colon}$ It suffices to notice that by $\overline{\mathcal{A} \xi} = \mathcal{H}$ we have that the identity representation is equivalent to the GNS representation $(\pi_\varphi, \mathcal{H}_\varphi, \xi_\varphi)$. But such a representation is irreducible if and only if $\varphi$ is pure and the assertion follows.

A related lemma is the following.

${\bf Lemma \; 2 \colon}$ Let $\mathcal{M}$ be a von Neumann algebra and let $\varphi$ be some faithful normal state in $\mathcal{M}_*^+$. If $\varphi$ is pure then its support $s(\varphi)$ is minimal.

${\bf Proof \colon}$ We can identify $\mathcal{M}$ with its GNS representation associated to $\varphi$, so that $\varphi(x) = (\xi | x \xi)$. By the identity $s(\varphi) = [\mathcal{M}' \xi]$ we have that $s(\varphi)$ is a rank one projection, hence it is minimal.

${\bf Question \colon}$ Is it true that a normal state is pure if and only if its support is minimal?

Clearly it could be useful that a state $\omega$ on a $C^*$-algebra $\mathcal{A}$ is pure if and only if it is the only state vanishing on its left kernel

$$ \mathcal{L}_\omega = \{ x \in \mathcal{A} \colon \omega(x^*x)=0 \} \,.$$

I think that the decomposition of a linear functional $\varphi \in \mathcal{M}^*$ in normal and singular part could be useful too.

Any hint? Thank you in advance.

答:
The conjecture about $C^*$-algebras is true,  that is,  a state $\omega $ on a $C^*$-algebra $A$ is pure iff $\omega$ is the  only state
vanishing on $\mathcal{L}_\omega $.

Before proving this, let us observe that a state $\varphi $ vanishes on   $\mathcal{L}_\omega $ iff
$$
  \omega (x^*x)=0  \ \Rightarrow \   \varphi (x^*x)=0.
  \tag 1
  $$
The "if" part follows easily from the Cauchy-Schwartz inequality, while the "only if" part is a consequence of the
fact that
$$
  x\in \mathcal{L}_\omega \ \Rightarrow \    x^*x\in \mathcal{L}_\omega .
  $$

Summarizing, we will prove:

Theorem.  Let $\omega $ be a state on the $C^*$-algebra $A$.  Then $\omega $ is pure iff, for every state $\varphi $ satisfying
(1),
one has that $\varphi =\omega $.

Proof.
Assume that $\omega $ is not pure.  Then there are states $\varphi $ and $\psi $, both distinct from $\omega $,  and a real number
$a\in (0,1)$, such that
$
  \omega =a\varphi +(1-a)\psi .
  $
Given any $x$ in $A$ such that   $\omega (x^*x)=0$, we then have that
$$
  0\leq a\varphi (x^*x) \leq  a\varphi (x^*x)+(1-a)\psi (x^*x) = \omega (x^*x)=0,
  $$
so also $\varphi (x^*x)=0$, and we see that $\varphi $ satisfies (1).

To prove the converse, assume that $\omega $ is pure and that the state  $\varphi $ satisfies (1).

Denoting the  GNS representations associated to $\omega $ and $\varphi $ by $(\pi ,H,\xi )$ and $(\rho ,K,\eta )$, respectively,
consider the mapping
$$
  V:\pi (A)\xi ⊆H \to K
  $$
given by
$$
  V\big (\pi (a)\xi \big )=\rho (a)\eta ,\quad\forall a\in  A.
  $$
This is well defined because if   $\pi (a)\xi =0$, then
$$
  0=\langle \pi (a)\xi ,\pi (a)\xi \rangle  = \langle \pi (a^*a)\xi ,\xi \rangle  = \omega (a^*a),
  $$
so $\varphi (a^*a)=0$ by hypothesis and hence $\rho (a)\eta =0$ by a similar computation.

Observing that $\pi $ is irreducible, we have by Kadison's transitivity Theorem that in fact $\pi (A)\xi =H$, so $V$ is already
globally defined on $H$.

We next claim that $V$ is bounded.  To prove this let us first consider the mapping
$
  Q:A\to H,
  $
given by
$$
  Q(a) = \pi (a)\xi ,\quad\forall a\in  A.
  $$
As already observed, $Q$ is onto $H$, hence an open mapping thanks to  the Open Mapping Theorem.  This says that there
exists a constant $c>0$ such that,  for every $\zeta $ in $H$, there exists $a$ in $A$ with $\|a\|\leq c\|\zeta \|$, and $Q(a)=\zeta $.  We
then deduce that
$$
  \|V(\zeta )\| = \|V\big (\pi (a)\xi \big )\| = \|\rho (a)\eta \| \leq  \|a\| \leq  c\|\zeta \|,
  $$
so $V$ is indeed bounded, as claimed.

It is a simple matter to prove that
$$
  V\pi (a) =   \rho (a)V, \quad \forall a\in  A,
  $$
from where it follows that $V^*V$ lies in the commutant of $\pi (A)$.  Again because  $\pi $ is  irreducible, that commutant
must  coincide with $\mathbb C$, so $V^*V$ is necessarily a scalar,  whence $V$ is a scalar multipe of an isometric operator.
Observing that $V$ maps the unit vector $\xi $ to the unit vector $\eta $, we see that $V$ must itself be an isometry.

For every $a$ in $A$ we then have that
$$
  \omega (a) =
  \langle \pi (a)\xi ,\xi \rangle  =
  \langle V\pi (a)\xi ,V\xi \rangle  =
  \langle \rho (a)\eta ,\eta \rangle  =
  \varphi (a),
  $$
proving that indeed $\varphi = \omega $.   QED

Title: Pure states and minimal projections


问:
Divisibility relations on $\Bbb N, \Bbb Z^*$, and properties thereof
I have two problems for which I need to find whether it is reflexive, irreflexive, symmetric, antisymmetric, or transitive.

The relation $T$ on $\mathbb{N}$ as defined by $$aTb \iff a \mid b$$

The relation $U$ on the set $\mathbb{Z}^*$ is defined as $$aUb \iff a \mid b$$

For #1, the way I worked it out was:

Reflexive: Yes, because $a  \mid  a$ for any positive integer $a$

Irreflexive: No, see above

Symmetry: No, because $a \mid b$ does not necessarily mean $b \mid a$ example, $4 \mid 12$ but $12$ does not divide $4$

Antisymmetric: Yes, because if $a  \mid  b$ and $b \mid a$ then $a$ must equal $b$

Transitive: Yes, because if $a \mid b$ and $b \mid c$ then $a \mid c$

For question #2, however I am lost because the answer is reflexive and transitive.

I get that it is reflexive and transitive and not symmetric for the same reasons as #1 but I don't understand why #2 is not antisymmetric but #1 is. I am guessing that it has something to do with #2 being a relation of $\mathbb{Z}^*$, but that just means a non-zero integer. Can someone help me understand why #1 is antisymmetric and #2 isn't antisymmetric?

答:
It is always best to reference the definitions. Namely, a typical definition is

$$a \mid b \iff \exists k \in \Bbb Z \text{ such that } b = ak$$

Suppose we consider this relation on $\Bbb N$. Let's investigate the properties.

Reflexivity: Trivially true; take $k=1$.

Irreflexive: Untrue, as $1 \mid 1$ ($k=1$).

Symmetry: Untrue. $2 \mid 4$ ($k=2$) but $4 \not \mid 2$ ($k = 1/2 \not \in \Bbb Z$).

Antisymmetry: True. If $a \mid b$ and $b \mid a$, then there are $k,\ell \in \Bbb Z$ such that
$$b = ak \qquad a = b\ell$$
Substitute the latter into the former, and  vice versa, and we have
$$b = b\ell k \qquad a = ak\ell$$
Thus, $\ell k = 1$. The only integers that ensure this are $\ell = k = +1$ or $\ell = k = -1$. Since we're working in $\Bbb N$, though, $a,b > 0$, and thus $\ell = k = +1$. Thus, $a = b$.

Transitivity: True. Suppose $a \mid b$ and $b \mid c$. Then $\exists p,q \in \Bbb Z$ such that
$$b = ap \qquad c = bq$$
Substitute the former into the latter, and we have
$$c = apq$$
Since $p,q \in \Bbb Z \implies pq \in \Bbb Z$, then $a \mid c$, giving transitivity.

What if we consider things on $\Bbb Z^*$ instead?

Reflexivity: Still holds for the same reason.

Irreflexive: Same deal - not irreflexive.

Symmetry: Again, same deal - not symmetric. It doesn't hold for $\Bbb N \subseteq \Bbb Z$ after all.

Antisymmetry: Now something has changed. Before, we had to completely neglect the case of $k = \ell = -1$ since that would result in $a$ or $b$ not being in $\Bbb N$. However, now that case is permissible, and gives us cases like $2 \mid -2$ and $-2 \mid 2$, but $2 \ne -2$. (Just take $k = -1$ for each divisibility relation.)

Transitivity: Still holds, nothing fundamentally changed here.

In short, the fundamental difference is that negative integers can divide positive integers, and vice versa!

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Finding double integral using change of variables
Let $R$ be a region in $xy$-plane bounded by $y=x^2$ , $y=4x^2$, $xy=1$ and $xy=5$. Then find the value of the integral $\iint _R \frac{y^2}{x} dxdy$

I am not able to get any function of $x$ and $y$ in terms of $u$ and $v$ such that I can find limit of the integrals. Is there any other way to solve this?

答:
Let $u = \frac{y}{x^2}$ and $v = xy$. This implies that $\frac{y^2}{x} = uv.$ Your bounds of integration now become $u=1$, $u = 4$, $v=1$ and $v=5$.

Now, all you need is the Jacobian and you can solve!


问:
Is an argument with all false premises and a false conclusion valid?
I have this truth table:

And I'm trying to determine the validity of the argument, which is:

(B v A), (A ^ ¬A) therefore C

I've got one row at the bottom, where (B v A) and (A ^ ¬A) are both false, but C is true.
And the one below that has all false premises and a false conclusion.

Maybe this is a dumb question, but can an argument still be valid even if either (1) all the premises are false but still has a true conclusion or (2) all premises and conclusion are false?

Furthermore, does an argument lacking any row in which all premises are true + conclusion is true automatically mean that the argument is invalid?

I would like to make it clear that I do understand that if a truth table contains a row in which all premises are true but the conclusion is false the argument is invalid.

Thanks.

(Note: the table has (A v ¬A), but it is meant to read as (A ^ ¬A). My bad!

答:
Here is the truth table of the statement you present in your question:

Source: https://www.erpelstolz.at/gateway/TruthTable.html

Notice that antecedent is always false (column 5). Therefore, by the so-called principle of explosion, the implication (last column) will always be true.

This does not mean that the consequent C (in your example) is always true. As you can see, it is false in the half the cases presented here (column 3).


问:
Convergence of infinite products flaw in this derivation of Euler product formula?
I am trying to write a derivation of the Euler product formula in a way that is accessible to younger students and those not mathematically trained at university.

Because of this I am starting from the simplest starting point that I can, and use the simplest possible steps in the derivation.

I arrive at the following which I know is wrong as it is missing the exponent $s>1$ to ensure convergence.

$$ \prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n} $$

My question is to ask where I am introducing the error in the following simple derivation.

Step 1

Start with a familiar series, valid for $|x| < 1$.

$$ \frac{1}{(1-x)}=1+x+x^{2}+x^{3}+\ldots $$

We're interested in primes, and might be tempted to set $x$ to a prime $p$, but we can't because $|x|$ needs to be $<1$. Let's try $\frac{1}{p}$ which is always $<1$.

$$ \frac{1}{(1-\frac{1}{p})}=1+\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}+\ldots $$

This series converges to a finite value.

Step 2

Let's now pick two different primes $p_1$ and $p_2$, and multiply the analogous series for each.

$$ \frac{1}{(1-\frac{1}{p_{1}})}\cdot\frac{1}{(1-\frac{1}{p_{2}})}=\left(1+\frac{1}{p_{1}}+\frac{1}{p_{1}^{2}}+\ldots\right)\cdot\left(1+\frac{1}{p_{2}}+\frac{1}{p_{2}^{2}}+\ldots\right) $$

I don't think there is a problem here.

We are multiplying two infinite series, each one known to converge to a finite value. Am I wrong?

Step 3

Now we extend from two primes, $p_1$ and $p_2$, to all primes $p_i$.

$$ \prod_{p_{i}}\frac{1}{(1-\frac{1}{p_{i}})}=\prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) $$

I think this is the flaw in my derivation. We have an infinite product of finite values.

Each factor $\frac{1}{(1-\frac{1}{p_{i}})}$ is $>1$ so the infinite product looks like it might not converge.

However each factor tends to $\rightarrow 1$ as $i \rightarrow \infty$. This suggests the infinite product is ok. Am I wrong?

Step 4

If we multiply out those brackets, each of the form $\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.

$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots  $$

Importantly, each combination occurs only once.

Step 5

The fundamental theorem of arithmetic tells us that each integer can be written as a unique product of primes. This means that each integer is represented in the terms we arrived at in the previous step.

Because each combination of primes occurs only once, each integer is represented only once.

That leads us to the Euler product formula.

$$ \boxed{\prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n}} $$

Discussion

If my suspicion is correct that the error is in the infinite product because it doesn't converge, how does changing $p$ for $p^s$ for $s>1$ help?

答:
The other answers have mentioned that your product does not converge and shown why that is. However, I don't think that ruins the value of this approach, and my answer here is to complement those by suggesting how to make it work. What I'd suggest is that, ultimately, you're going to need to get $s$ into the product somehow, because it is material to the value of the sum, so what we should instead do is actually use the fact that the simple product

$$\prod_{n=1}^{\infty} \frac{1}{1 - \frac{1}{p_n}}$$

fails, after suitable demonstration it does fail, as a pivoting point in the derivation - after all, it seems what you're asking is to "explore" for the formula, and in exploration of any type you may have to reverse course. Hence, we should then step back and figure how to introduce $s$ in somehow. And I'd say that, since we're talking $s$-th powers in our target sum, we should put that on $p_n$, i.e. consider

$$\prod_{n=1}^{\infty} \frac{1}{1 - \frac{1}{p_n^s}}$$

. And then when you do that, I would suggest you can avoid the whole business about rearranging infinite sums of terms by remembering that there's a limit definition of infinite products and sums, and stepping back to that:

$$\prod_{n=1}^{\infty} \frac{1}{1 - \frac{1}{p_n^s}} = \lim_{n_\mathrm{max} \rightarrow \infty} \prod_{n=1}^{n_\mathrm{max}} \frac{1}{1 - \frac{1}{p_n^s}}$$

(and we might even convert this to a double limit) and then convert the right-hand product to

$$\prod_{n=1}^{\infty} \frac{1}{1 - \frac{1}{p_n^s}} = \lim_{n_\mathrm{max} \rightarrow \infty} \prod_{n=1}^{n_\mathrm{max}} \left(\sum_{k=0}^{\infty} \frac{1}{p_n^{sk}}\right)$$

where now we don't have to worry about producting an infinite number of factors now. So we can now expand the sums out, and they will have terms of the form

$$\frac{1}{\underbrace{p_{a_1}^s p_{a_2}^s \cdots p_{a_n}^s}}$$

where every combination of positive integers $a_j$ of the given length $n$, $n = 1, 2, \cdots, n_\mathrm{max}$ will contribute to the sum. Because of unique factorization, we have that the denominators will thus occur exactly once and be a subset of

$$\{1^s, 2^s, 3^s, \cdots\}$$

and we can then argue that, since increasing $n_\mathrm{max}$ means the factorizations get longer, this subset must approach the full set, and so once we pass to the limit, the desired summation identity obtains.


问:
Understanding implications
I have a set of questions where to task is to add the correct implication arrow.

I was able to make sense of most of them, but for two problems, I can't figure out why I was wrong.

Here is what I put:

Statement $S1$
Statement $S2$
I answered
The answer in the book is

$\frac{x}{x+1} = 0$
$x=0$
$S1\iff S2$
$S1 \Rightarrow S2$

$(a,b)$ is a point on the line $y=2x-1$
$b=2a-1$
$S1 \iff S2$
$S1 \Rightarrow S2$

For the first one, I can't see how when $x=0$, $\frac{x}{x+1}$ can equal anything other than $0$, so surely the implication works both ways? I presume I'm missing something obvious.

For the second, the point $(a, b)$ is on line $y = 2x -1$, so $b$ must equal $2a - 1$. Again I presume I've missing something here.

Really would appreciate any help here, I've run out of ideas.

答:
First one:

Sufficiency ($\Rightarrow)$: Suppose $x/(x+1) = 0$. This expression already assumes $x\neq -1$. So we can multiply it with $x+1$ to get $x = 0$. (True)

Necessity ($\Leftarrow)$: Suppose $x = 0$. It is a trivial result that $x/(x+1) = 0$. (True)

Therefore, the first one is an N-S condition.

Second one:

In set-builder notation, 'the point is on the line' if and only if
$$
(a,b) \in \{(x,y) \in \mathbb{R}^2~|~y = 2x -1\}
$$
and it is equivalent to say that $b = 2a - 1$ given that $(a,b) \in \mathbb{R}^2$ due to Axiom schema of specification.
That is, the second one is also an N-S condition.

Title: Understanding implications


问:
How to show this matrix is invertible?
Let $f:H \times H \to \mathbb{R}$ be a mapping with $H$ a Hilbert space.

Let $A$ be a matrix with entries $a_{ij}=f(b_i, b_j)$ with
$$a_{ii}=f(b_i, b_i) \geq C\lVert b_i\rVert_{H}^2.$$
Suppose $b_i \neq b_j$ and $(b_i, b_j)_H = 0$ for $i \neq j$ .
How do I show that $A$ is invertible?

答:
If $f(u,v)$ is given by scalar product $(Bu,v)_H$, $B\in\mathcal L(H,H)$ - symetric continuous linear operator which is positive definite (because $f$ is coercive). If $b_j$ are linearly independent, then the matrix $A$ is a metric tensor on $\text{span} \{b_j\}$ and it should be invertible.

Edit I'll develop a little on this case. Suppose my hypothesis is true and $a_{i,j}=(Bb_i,b_j)_H$, $i,j=1..n$. Suppose that $A$ is singular, then there exists $u\in\mathbb R^n$ such that $(Au,u)_{\mathbb R^n} =0 $, but

$\displaystyle (Au,u)_{\mathbb R^n} =\sum_i\sum_j (Bb_i,b_j)_Hu_i u_j = \left( B\left(\sum_i   b_i u_i\right),\left(\sum_j   b_j u_j\right)\right)_H \ge C\left\|\sum_i   b_i u_i\right\|_H^2>0$. Hence $A$ is invertible. As it's easy to see, this proof relies heavily on the fact that $f$ is given by a scalar product.

Title: How to show this matrix is invertible?


问:
Bounded Self-adjoint operator with finite spectrum
if $S$ be a self-adjoint bounded operator and $\sigma(S)=\left\{\lambda_{1}, \dots, \lambda_{n}\right\} .$ show that there exist projections operator $P_{1}, \dots, P_{n} \in B(H)$ that $\sum_{j=1}^{n} P_{j}=I$ and $P_i P_j=0$ if $j \neq k$ and $S=\sum_{j=1}^{n} \lambda_{j} P_{j}$.
I can see how we can define projection operator for each term in spectrum but I can't show that they are complete in the sense $\sum_{j=1}^{n} P_{j}=I$. actually its strange for me that in general self-adjoint bounded operator with finite spectrum have a complete set of projection operator.

答:
For $S$ as described, one has $p(S)=0$ where
$$
         p(\lambda)=\prod_{k=1}^{n}(\lambda-\lambda_k).
$$
This follows from

$\;\;$ (a) $p(S)^*=p(S)$, and

$\;\;$ (b) $\sigma(p(S))=p(\sigma(S))=\{0\}$,

which forces the self-adjoint operator $p(S)$ to be the $0$ operator.

The Lagrange polynomials
$$
         p_k(\lambda)=\frac{\prod_{j\ne k}(\lambda-\lambda_j)}{\prod_{j\ne k}(\lambda_k-\lambda_j)}.
$$
satisfy $p_{k}(\lambda_j)=\delta_{k,j}$, which gives
$$
     p_1(\lambda)+p_2(\lambda)+\cdots+p_n(\lambda)=1.
$$
This forces $p_1(S)+p_2(S)+\cdots+p_n(S)=I$. And
$$
         p_k(S)p_k(S)=p_k(S) \\
         p_k(S)p_l(S)=(p_kp_l)(S)=0,\;\; k\ne l.
$$
This is because $p_k(\lambda)^2=p_k(\lambda)$ for $\lambda\in\sigma(S)$, and because $p_k(\lambda)p_l(\lambda)=0$ for $\lambda\in\sigma(S)$ whenever $k\ne l$. These have the desired properties.

Finally,
\begin{align}
           S &= S(p_1(S)+p_2(S)+\cdots+p_n(S)) \\
             &= \lambda_1 p_1(S)+\lambda_2 p_2(S)+\cdots+\lambda_n p_n(S)
\end{align}
The projections you want are $P_k = p_k(S)$.


问:
Relation on the set of polynomials
Consider a relation defined on the set of polynomials, such that two polynomials are related if and only if their indefinite integrals are equal. Is this an equivalence relation?

I'm slightly confused with one thing in this question - since every function has infinitely many indefinite integrals, is something going wrong here? In case the answer is no, then I feel its both reflexive and symmetric, but I'm not sure how to prove its transitivity (or lack of it). So please help me with this question, so that I can strengthen my concepts further on relations.

答:
As noted, the relation is painfully vague and ambiguous. After all,

$$\frac 1 2 x^2 + 2 \qquad \frac 1 2 x^2 + 3$$

are both indefinite integrals for $f(x) = x$. Thus within a certain frame of mind, a function doesn't even share an antiderivative with itself. So it would be better to define the notion of this relation as so:

$$f \sim g \iff \int f(x)dx - \int g(x)dx = \text{constant}$$

That is possibly more in line with the intention of the exercise.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue. Posting as Community Wiki since I don't have much to add.

Title: Relation on the set of polynomials


问:
Operator that is the limit of a sequence of operators
Let $X$ be Banach and $Y$ be a reflexive space with $T_n \in \mathscr{L}(X,Y)$ so that $\Lambda (T_nx) $ converges for all $x \in X$ and $\Lambda \in Y^{*}$.

Then there exists a  $T \in \mathscr{L}(X,Y)$ so that $\Lambda (Tx) = \lim_{n \rightarrow \infty} \Lambda(T_nx)$ for all $x \in X$ and $\Lambda \in Y^{*} $.

I have already proven that for a Banach space $X$ and normed $Y$ the pointwise limit of a linear continous operator is again linear and continous but I am just not getting forward here. Thanks in advance.

答:
For each $x$ in $X$, consider the linear map
$$
  \varphi _x:\Lambda \in  Y^*\mapsto \lim_{n\to \infty }\Lambda \big (T_n(x)\big )
  $$
By the  Banach-Steinhaus theorem we have that $\varphi _x$ is continuous, so $\varphi _x\in Y^{**}$.  Since $Y$ is reflexive there must
be some vector $T(x)$ in $Y$ such that
$$
  \varphi _x(\Lambda ) = \Lambda \big (T(x)\big ),\quad \forall \Lambda \in  Y^*.
  $$
It then follows that
$$
  \Lambda \big (T(x)\big ) = \varphi _x(\Lambda ) = \lim_{n\to \infty }\Lambda \big (T_n(x)\big ).
  $$
Can you pick up from here and prove that $T$ is bounded?


答:
$A\land\lnot A\vDash C$ is valid.   There is no interpretation of the literals where the antecedent is true and the consequent is false.

After all, $A\land\lnot A$ is false in every valuation of $A$.   It is a contradiction.

$$\begin{array}{cc|c}A&C&A\land\lnot
A\\\hline\top&\top&\bot\\\top&\bot&\bot\\\bot&\top&\bot\\\bot&\bot&\bot\end{array}$$

So there are clearly no valuations where $C$ is false yet $A\land\lnot A$ is true.

This will remain so no matter what other antecedents you add to the sequent.

Title: Is an argument with all false premises and a false conclusion valid?

Body:
I have this truth table:

And I'm trying to determine the validity of the argument, which is:

(B v A), (A ^ ¬A) therefore C

I've got one row at the bottom, where (B v A) and (A ^ ¬A) are both false, but C is true.
And the one below that has all false premises and a false conclusion.

Maybe this is a dumb question, but can an argument still be valid even if either (1) all the premises are false but still has a true conclusion or (2) all premises and conclusion are false?

Furthermore, does an argument lacking any row in which all premises are true + conclusion is true automatically mean that the argument is invalid?

I would like to make it clear that I do understand that if a truth table contains a row in which all premises are true but the conclusion is false the argument is invalid.

Thanks.

(Note: the table has (A v ¬A), but it is meant to read as (A ^ ¬A). My bad!


问:
The representation of closed set
We already know that closed set is the complement of open set.

But I want to know the representation of closed set without the fact above in the usual topology.

For instance, the integers $\mathbb{Z}$ is closed due to the fact that $\mathbb{Z}$ can be represented by the union of one point sets $\{n\}$
(I don't want to use the fact that "the union of open intervals $(n, n+1)$ is open").

答: Hint: You can look at Kuratowski closure axioms.

Title: The representation of closed set


问:
If $p, q, r$ are primes such that $p^2+q^2+r^2-222pqr=2017$ and $p\geq max(q,r)$, evaluate $(p+1)/{(q+1)(r+1)}$
I figured that since Odd-Even=Odd, $p^2+q^2+r^2$ has to be odd. I came up with $p^2\equiv1\pmod4, q^2\equiv1\pmod4, r^2\equiv1\pmod4$ but I cannot proceed further. What is the trick in finding the actual values of $p, q, r$?

答:
(Fill in the gap as needed. If you're stuck, show your work.)

Hint: Work modulo 3.

Show that $ q = r  = 3$.

(I previously assumed that the primes were distinct.)

Hence show that $ p = -1, 1999$, of which we reject $-1$.


问:
A possibly elegant way to compose and write proofs
I have always observed proofs requiring complicated structure of proposition, either as the premise, as the conclusion or in the middle of proof construction. I have also felt that readers feel difficult understanding the logical structure of the proofs, as they are written in daily English. The constructor may have experienced the logical flow, but he/she has to present the proof in daily language. From my observations, I felt it necessary to indicate the use of logical rules at least implicitly in proof presentation. For instance, when we see an existential proposition, we would like to eliminate the existential quantification and enter a subproof with a much simpler proposition as the premise. The introduction of a new constant symbol should be implied in the proof presentation. I'd like to use an example to show such an implicit logical style.

Our task is to prove the following proposition, which is a step in proving the implicit function theorem in real analysis:
\begin{equation}
\begin{aligned}
&\textrm{Given an open set $S \subseteq \mathbb{R}^{n+m}$, suppose that $\left(x_{0},u_{0}\right) \in S$, where $x_{0} \in \mathbb{R}^{n}$ and $u_{0} \in \mathbb{R}^{m}$.}\\
&\textrm{Then there exists a neighborhood $N$ of $x_{0}$ such that for all $x \in N$, $\left(x,u_{0}\right) \in S$}.
\end{aligned}
\end{equation}
Below is my proof.
The fact that $S$ is open leads to the following premise:
\begin{equation}
\forall x \in S, \exists \delta > 0, \forall y \in \mathbb{R}^{n+m}, \lVert y-x \rVert < \delta \longrightarrow y \in S.
\end{equation}
We also have the following premise:
\begin{equation}
x_{0} \in \mathbb{R}^{n} \wedge u_{0} \in \mathbb{R}^{m} \wedge \left(x_{0},u_{0}\right) \in \mathbb{R}^{n+m}.
\end{equation}
We wish to deduce the following conclusion:
\begin{equation}
\exists \delta > 0, \forall x \in \mathbb{R}^{n}, \lVert x-x_{0}\rVert < \delta \longrightarrow \left(x,u_{0}\right) \in S.
\end{equation}
First of all, replacing $x$ with $\left(x_{0},u_{0}\right)$ in the first premise and the fact that $\left(x_{0},u_{0}\right) \in S$, we get:
\begin{equation}
\exists \delta > 0, \forall x \in \mathbb{R}^{n}, \forall u \in \mathbb{R}^{m}, \lVert \left(x-x_{0},u-u_{0}\right)\rVert < \delta \longrightarrow \left(x,u\right) \in S.
\end{equation}
Introducing a constant symbol $\delta_{0}$, we introduce the following premise:
\begin{equation}
\delta_{0} > 0 \wedge \forall x \in \mathbb{R}^{n}, \forall u \in \mathbb{R}^{m}, \lVert \left(x-x_{0},u-u_{0}\right)\rVert < \delta_{0} \longrightarrow \left(x,u\right) \in S.
\end{equation}
This proposition can be decomposed into two separate deductions:
\begin{equation}
\delta_{0} > 0,
\end{equation}
and
\begin{equation}
\forall x \in \mathbb{R}^{n}, \forall u \in \mathbb{R}^{m}, \lVert \left(x-x_{0},u-u_{0}\right)\rVert < \delta_{0} \longrightarrow \left(x,u\right) \in S.
\end{equation}
Replacing $u$ with $u_{0}$, we have the following proposition:
\begin{equation}
\forall x \in \mathbb{R}^{n}, \lVert x - x_{0} \rVert < \delta_{0} \longrightarrow \left(x,u_{0}\right) \in S.
\end{equation}
Thus, we have proved the following proposition:
\begin{equation}
\exists \delta > 0, \forall x \in \mathbb{R}^{n}, \lVert x - x_{0} \rVert < \delta \longrightarrow \left(x,u_{0}\right) \in S
\end{equation}

This proof introduces implicitly where we used the rule of elimination of universal quantification and the elimination and introduction of existential quantification. I think with proper training in logic and rigorous use of English, both the conciseness of English and preciseness of logical language can be combined. Has anyone thought about this type of training for self improvement or for teaching students?

答:
In my case, I feel much better at formal proofs in logical symbols than the informal proofs in English because the former is much concise and does not cause any confusion. Rules of inferences apply very neatly to them. Note that 'informal' does not imply 'not logically strict.'

But the proof can be too verbose when we keep writing statements in symbols, while each line has only slight changes as you demonstrated. In that sense, the Fitch-style formalization will be a good tool for teaching how to prove propositions in first-order logic. The Fitch-style proof will prevent the proof from being unnecessarily long. In case we have to show the result to the others including publication, we can just translate it into English.

Conclusion: It is a good strategy to prove a proposition in formal proving methods (e.g., Fitch-style) to exclude any ambiguity and translate the result into English.

I recommend you to read 2011 - Barker-Plummer - Language, Proof, and Logic


问:
Serre's Trees Exercise 1.2
I'm having some confusion over Serre's Trees section on Amalgam.

Let $A = \mathbb{Z}$, $G_1 = PSL(2, \mathbb{Q})$ and $G_2 = \mathbb{Z}/2\mathbb{Z}$. We take $f_1: A \to G_1$ and $f_2: A \to G_2$ to be a surjection. Show that $G_1 \ast_{A} G_2 = \{1\}$.

In the text, we haven't formally constructively define what it means to have an amalgam over $A$.

So are we simply take the  disjoint union of generators of $A, G_1$ and $G_2$ and mod out the relations?

The relations I got is simply identifying all the even numbers with $0$ in $G_2$, identifying all $n \in \mathbb{Z}$ with the matrix
$$ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}.$$
But since $\mathbb{Q}$ has infinitely many generators, I'm not sure why we can reduce this group to a trivial one.

Any help is appeciated!

答:
Have you done exercise 1.1? I'll quote it for convenience:

Let $f_1 : A \to G_1$ and $f_2 : A \to G_2$ be two homomorphisms  and let $G_1 *_A G_2$ be the corresponding amalgam. We define subgroups of $A^n$, $G_1^n$ and $G_2^n$ of $A$, $G_1$ and $G_2$ recursively by the following conditions:
$$
A^1  = \{1 \}, \ \ \ \ G_1^n = \{1\}, \ \ \ \ G_2^1 = \{1\}
$$
$$
A^n = \text{subgroup of $A$ generated by} \ f_1^{-1}(G_1^{n-1}) \ \text{and} \ f_2^{-1}(G_2^{n-1})
$$
$$
G_i^n = \text{subgroup of $G_i$ generated by $f_i(A^n)$}.
$$
Let $A^\infty$, $G_i^\infty$ be the union of the $A^n$, $G_i^n$ respectively. Show that $f_i$ defines an injection $A/A^\infty \to G_i/G_i^\infty$ and that $G$ may be identified with the amalgam of $G_1/G_1^\infty$ and $G_2/G_2^\infty$ along $A/A^\infty$.

I should say that I think we need to assume that $G_i^n$ is the normal subgroup generated by $f_i(A^n)$ in order to guarantee that the $G_i^\infty$ are normal subgroups (anyways, $f_i(A^n)$ is already a subgroup, so it would be strange to talk about the subgroup it generates).

In view of this exercise, it suffices to show that $A = A^\infty$ and $G_i = G_i^\infty$ in our current example. It is very easy to show that $G_1^2 = (\mathrm{PSL}(2,\mathbb{Q}))^2$ is non-trivial. Since $\mathrm{PSL}(2,\mathbb{Q})$ is a simple group, we must have that $G_1^\infty = G_1^2 = G_1$. The desired result then follows very quickly using the recursive formulas defining $A^n$ and $G_i^n$. By the result of the exercise $G_1 *_A G_2$ is isomorphic to the amalgam of trivial groups over a trivial groups, which is of course trivial.

Title: Serre's Trees Exercise 1.2


问:
Number of subsets the cardinality of whose intersections with some other subsets are known
$A$ is a non-empty finite set. $A_1,A_2,\ldots,A_n$ are subsets of $A$. How many subsets $B$'s of $A$ are there that satisfy that $|B\cap A_i|=a_i,\forall 1\leq i\leq n$, where $a_i\geq 0$'s are given integers? (Assume that we know $|A_{i_1}\cap A_{i_2}\cap \ldots A_{i_k}|$, where $1\leq i_j\leq n,\forall j$.)

Thank you very much for your kindness.

答:
If the $A_i$ are disjoint, it is easy.  Let $k$ be the number of elements of $A$ that are not members of any $A_i$.  Then you need to pick $a_1$ elements out of $A_1$ and so on then any collection out of the rest, so there are $$2^k\prod_{i=1}^n {|A_i| \choose a_i}$$
If there is overlap between the $A_i$ I don't think there is an easy formula.  For example, if $A_1 \subset A_2$ and $a_2 \lt a_1$ there is no solution.</p>


问:
Probably of Fukushima like event
I was reading an article where an author was discussing the relative safety of power plants. In the article there was statement that a tsunami with a wave of 15 meters happens once every 5000 years.

One of the commenter said that if we assume that the lifetime of an reactor is 50 years the probability of similar event is 50/5000 = 0.01 or 1%. And knowing that the are 23 nuclear power plants that operate in a high risk tsunami areas, the probability that it happens in one of them is 1 - (1-0.01)^23 = 20%

It seems to be too high to be true. What is wrong and what would be more accurate estimation?

答:
The probability of having a tsunami with a wave of 15 meters in a certain year is $\frac{1}{5000}=2\times 10^{-4}$. If $X$ is the number of such waves in the lifetime of a reactor (50 years), then we can model $X$ as a Poisson-distributed random variable with rate $50\times 2\times 10^{-4}=0.01$. The probability that a reactor experiences at least one such event is $$\mathbb P(X\ge 1)=1-\mathbb P(X=0)=1-e^{-0.01}=0.009950$$
In calculating the probability that the event happens to one of the 23 reactors, you assumed that the events are independent, i.e. a tsunami affecting one of the reactors is not in any way related to another tsunami affecting another one, which is not necessarily true.

Title: Probably of Fukushima like event


问:
prove MLE don't exist
so the pdf is
$$f(x)=\frac{1}{2}\frac{1}{\sqrt{2\pi}}\exp^{-\frac{x^2}{2}}+\frac{1}{2}\frac{1}{\sqrt{2\pi}\sigma}\exp^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$
$\mu$ and $\sigma$ are unknown,prove that they don't have MLE

it is a gaussian mixture model I guess? And I know there are no analytic solve for the derivative of log likehood function. But the text seems to mean that there are no numerical solve either. And I really don't know any methodology for prove MLE don't exist

答:
One way to prove that MLE doesn't exist is to show that the density or mass function doesn't have a global maxmium.

Here is the joint density function :
$$
\boldsymbol{f}_{\boldsymbol{X}_1,..,\boldsymbol{X}_{\boldsymbol{n}}}\left( \boldsymbol{x}_1,...,\boldsymbol{x}_{\boldsymbol{n}}|\boldsymbol{\mu },\boldsymbol{\sigma }^2 \right) =\prod_{\boldsymbol{i}=1}^{\boldsymbol{n}}{\left( \frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}}\exp \left\{ -\frac{\boldsymbol{x}_{\boldsymbol{i}}^{2}}{2} \right\} +\frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}\boldsymbol{\sigma }}\exp \left\{ -\frac{\left( \boldsymbol{x}_{\boldsymbol{i}}^{2}-\boldsymbol{\mu } \right) ^2}{2\boldsymbol{\sigma }^2} \right\} \right)}
$$

After observing the value of X, we have the Log likelihood:

$$
\boldsymbol{L}\left( \boldsymbol{\mu },\boldsymbol{\sigma }^2|\boldsymbol{x}_1,...,\boldsymbol{x}_{\boldsymbol{n}} \right) =\sum_{\boldsymbol{k}=1}^{\boldsymbol{n}}{\ln \left( \frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}}\exp \left\{ -\frac{\boldsymbol{x}_{\boldsymbol{i}}^{2}}{2} \right\} +\frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}\boldsymbol{\sigma }}\exp \left\{ -\frac{\left( \boldsymbol{x}_{\boldsymbol{i}}^{2}-\boldsymbol{\mu } \right) ^2}{2\boldsymbol{\sigma }^2} \right\} \right)}
$$
if μ happens to be the value of some x, for example, μ = x1, then we have:
$$
\begin{align}
\boldsymbol{L}=&\,\,\underset{\boldsymbol{part}\,\,\boldsymbol{A}}{\underbrace{\ln \left( \frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}}\exp \left\{ -\frac{\boldsymbol{x}_{\boldsymbol{i}}^{2}}{2} \right\} +\frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}\boldsymbol{\sigma }} \right) }}
\\
&\,\,  +\underset{\boldsymbol{part}\,\,\boldsymbol{B}}{\underbrace{\sum_{\boldsymbol{k}=2}^{\boldsymbol{n}}{\ln \left( \frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}}\exp \left\{ -\frac{\boldsymbol{x}_{\boldsymbol{i}}^{2}}{2} \right\} +\frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}\boldsymbol{\sigma }}\exp \left\{ -\frac{\left( \boldsymbol{x}_{\boldsymbol{i}}^{2}-\boldsymbol{x}_1 \right) ^2}{2\boldsymbol{\sigma }^2} \right\} \right)}}}
\end{align}\\
$$

if we set σ → 0+:
$$
\begin{align}
\boldsymbol{part}\,\,\boldsymbol{A}&\rightarrow +\infty
\\
\boldsymbol{part}\,\,\boldsymbol{B}&\rightarrow \sum_{\boldsymbol{k}=2}^{\boldsymbol{n}}{\ln \left( \frac{1}{2}\frac{1}{\sqrt{2\boldsymbol{\pi }}}\exp \left\{ -\frac{\boldsymbol{x}_{\boldsymbol{i}}^{2}}{2} \right\} \right)}
\end{align}\\
$$

Thus you can say the Log likelihood doesn't have a global maxmium and the density function doesn't have a global maxmium and MLE equivalently.

Title: prove MLE don't exist


问:
Is $y = -x$ injective?
Is $y = -x$ an  injective function?

I assume it is injective because $x_1=x_2$ then $f(x_1)=f(x_2)$.

Is this an injective function? I would like to ask for more proper and specific proof.

答:
Injective means $f(a)=f(b) \implies a =b$. And not the other way around: $a=b \implies f(a)=f(b)$. The other way around obviously applies to every function so it's pointless.

It's injective since
$$f(a)=f(b)$$
$$\implies -a=-b$$
$$\implies a=b$$

Title: Is $y = -x$ injective?


问:
Why is Lie derivative smooth?
Let $G$ be a linear Lie group, say $\mathrm{SL}(2,\mathbb{R})$. Suppose $X\in\mathfrak{g}$ and $f:G\to\mathbb{R}$ is smooth. The Lie derivative of $f$ with respect to $X$ is the function $\mathcal{L}_X f:G\to\mathbb{R}$ defined as $$\mathcal{L}_X f(y):=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}f\left(ye^{tX}\right).$$ Why is $\mathcal{L}_X f$ smooth? In his book on $\mathrm{SL}(2,\mathbb{R})$, Lang just says so without any comment (p. 90), but I failed to provide a rigorous justification up to now. A reference is also welcome.</p>

答:
Because the map $$\phi:G\times\Bbb R\to G,~(g,t)\mapsto g\exp(tX)$$ is smooth, as it is obtained from the smooth map $\Bbb R\to G,~t\mapsto \exp(tX)$ and the smooth map $\mu:G\times G\to G,~(g,g')\mapsto gg'$. The Lie derivative of $f$ is the derivative with respect to $t$ of the smooth map $$f\circ\phi:G\times\Bbb R\to\Bbb R,~(g,t)\mapsto f(\phi(g,t))$$ and it is thererfore smooth.

To be $100\%$ precise (and pretty pedantic), it is obtained as the composite
$$\begin{array}{c}
G&\to &T(G\times\Bbb R)&\to& T\Bbb R\simeq \Bbb R\times\Bbb R&\to&\Bbb R\\
g&\mapsto& {\frac{\partial}{\partial t}}\bigg|_{(g,0)}&\mapsto & T_{(g,0)}(f\circ\phi)\bigg({\frac{\partial}{\partial t}}\bigg|_{(g,0)}\bigg)= \underbrace{\lambda}_{=\mathcal L_Xf(g)}\frac{\partial}{\partial t}\bigg|_{f(g)}&\mapsto&\lambda
\end{array}$$

Title: Why is Lie derivative smooth?


答: More interesting is the calculus method: $y=f(x)$ is injective if either negative definite:$f'(x)<0$ in the domain or it is positive definite: $f'(x)>0$ in the domain. So here it is $f'(x)=-1$ for all real numbers. so it is injective for all real numbers.

Title: Is $y = -x$ injective?

Body:
Is $y = -x$ an  injective function?

I assume it is injective because $x_1=x_2$ then $f(x_1)=f(x_2)$.

Is this an injective function? I would like to ask for more proper and specific proof.


答: The function $f(x)=-x$ has an inverse (which is in fact itself), so it is injective.

Title: Is $y = -x$ injective?


问:
Conditional pdf $f(y|x,w)$ where $Y = XW$
Title is more of a generalization of my question.

Consider $Y=XW$, where $X$ is a positive constant, $W$ is from the standard normal distribution $N(w;0,1)$. I wonder what the pdf $f(y|x,w)$ is.

My answer is $f(y|x,w) = \delta(y-xw)$; however, this answer implies:
$$f(y,x|w) = f(y|x,w)f(w) = \delta(y-xw) \cdot N(w;0,1)$$

When I integrate this joint distribution over $w$ to obtain the pdf $f(y|x)$, the result becomes $f(y|x) = xN(y;0,x^2)$, which is obviously wrong, since we know that $f(y|x)=N(y;0,x^2)$, clearly.

I feel very confused and hopeless right now. If you can help me, that'd be great. Thanks!

答:
The dirac-delta impulse function has the property, for constant $c$ and function $g$, that $$\int_\Bbb R \delta_{w-c}\, g(w)\,\mathrm d w= g(c)$$

Also for non-zero $x$, we have $\delta_{y-xw}=\delta_{w-y/x}$ ... when integrating over $w$ the impulse occurs when $y-xw=0$, which is exactly when $w-y/x=0$ .

Thus for a positive constant $X$...

$$\begin{align}f_{Y\mid X}(y\mid x)&=\mathbf 1_{\small x=X}\,\int_{\Bbb R} \delta_{y-xw}\,\mathcal N(w;0,1^2)\,\mathrm d w\\[1ex]&=\mathbf 1_{\small x=X}\,\int_{\Bbb R}\delta_{w-y/x}\,\mathcal N(w;0,1^2)\,\mathrm d w\\[1ex]&=\mathcal N(y/x;0,1^2)\,\mathbf 1_{\small x=X}\\[2ex]&=\mathcal N(y;0,x^2)\,\mathbf 1_{x=X}\end{align}$$


问:
how to calculate the induced metric
Consider the mapping $f:\mathbb R^3\to\mathbb R^4$ between two Euclidean spaces:
$$f(\mathbf x)=(x_1,x_2,x_3,1/\|\mathbf x\|)^T$$
How to calculate the induced metric by $f$ on $\mathbb R^3$ (i.e. the pullback metric)?

答:
Assuming $f$ is defined by

$$f(\mathbf x) = f(x_1,x_2,x_3) = (x_1,x_2,x_3, \frac{1}{\sqrt{x_1^2+x_2^2+x_3^2}})$$

and the metric on $\mathbb R^4$ is defined by the Euclidean norm, it looks like the distance between $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ according to the pullback metric  is

$\sqrt{ (x_1 - y_1)^2 + (x_2 - y_2)^2 + (x_3 - y_3)^2 + ( \frac{1}{\sqrt{x_1^2+x_2^2+x_3^2} }-\frac{1}{\sqrt{y_1^2+y_2^2+y_3^2}})^2}.$

Title: how to calculate the induced metric


问:
Is $\mathfrak{so}(2n)$ contained in $\mathfrak{sl}(2n)$?
Let $\mathfrak{so}(2n) = D_n  =\left\{ \begin{pmatrix} A&B \\ C&-A^T \end{pmatrix}:  A,B,C\in M_n(K), B=-B^T, C=-C^T \right\}$ . Of course the matrices in $D_n$ have trace zero. This could lead me to conclude that $\mathfrak{so}(2n) \subseteq \mathfrak{sl}(2n)$, but this doesn’t seem right as for the inclusion of the respective Dynkin diagrams: in fact, it is not true that $D_n$ is contained in $A_n$. What am I missing here?

答:
Although the other answer is correct and it is important to learn that one semisimple Lie algebra being included in another does not at all imply inclusion the respective root systems / Dynkin diagrams, I'd like to point out that something is going on here.

Namely, if the subalgebra in question is given as fixed points of a certain automorphism, then under certain conditions which are not entirely clear to me, the root system of the subalgebra can be found (almost) by looking at the quotient (people say "folding") of the root system of the big algebra. (In principle this is sound: the roots are elements of the dual of a CSA, so if there were any hope for functoriality from Lie algebras to root systems, it should be contravariant. But see what comes for weird mixes of quotients and subobjects.)

If we have a root system of type $A_{2n-1}$ with basis $(\alpha_1, ..., \alpha_{2n-1})$ and "mod out" the relation $\alpha_i = -\alpha_{2n-i}$, i.e. look at the coinvariants of the non-trivial Dynkin diagram automorphism, we get a root system of type $C_n$. If we lift this to the "right" automorphism of the Lie algebra, this corresponds to the inclusion of split forms $\mathfrak{sp}(2n) \subseteq \mathfrak{sl}(2n)$. (Case $n=2$ was asked here; note that $B_2=C_2$.)

In the same setup, a different lifting of that root same system automorphism to an automorphism of the Lie algebra actually kills the root spaces that belong to the long roots in $C_n$ and leaves the short roots therein, which form a system of type $D_n$. This corresponds to the inclusion of split forms $\mathfrak{so}(2n) \subseteq \mathfrak{sl}(2n)$ in your question.

If we have a root system of type $A_{2n}$ with basis $(\alpha_1, ..., \alpha_{2n})$ and "mod out" the relation $\alpha_i = -\alpha_{2n+1-i}$, i.e. look at the coinvariants of the non-trivial Dynkin diagram automorphism, we get a non-reduced root system of type $BC_n$. The corresponding "correct" automorphism in $\mathfrak{sl}(2n+1)$ kills the root spaces that would belong to the roots of the form $2\alpha$ in that root system, and what remains is a root system of type $B_n$. This corresponds to the inclusion $\mathfrak{so}(2n+1) \subseteq \mathfrak{sl}(2n+1)$.

Big Caveat: All this is not entirely clear to me and as far as I know, there is no good written literature on these things. I would love to learn more about this, or see somebody smarter than me work it out. One case in which things worked, this time "folding" $D_4$ to $G_2$, is How can the generators of subalgebra $\mathfrak g^{\sigma}$ of $\sigma$-stable elements be expressed through generators of Lie algebra $\mathfrak g$?. Compare the comments to the answer to MathOverflow post https://mathoverflow.net/q/244893/27465 for more links and the apparently widely used folding $E_6 \twoheadrightarrow F_4$. I also tried to express what I say here in my answer to Map for roots of a Lie group to roots of a special subalgebra?, but apparently nobody took that seriously, maybe rightly so because I have not really figured out these things clearly enough for myself.


问:
An recursive formula for the n-th prime?
A very short recursive formula for the $n$-th prime is (by a combination of Euclids proof and the sieve of Erathostenes):

$$p_{n+1} = \min_{x>1,\gcd(x,p_1\cdots p_n)=1} x$$

Using an approximation for the $\min$ function, we find:

$$p_{n+1} = \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log(\sum_{x>1,\gcd(x,p_1\cdots p_n)=1} \exp(-\rho x))$$

My idea is to sum over all numbers $x>1$ but to weight them:

$$p_{n+1} =^? \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log \left( \sum_{x\ge 2}^\infty \exp(-\rho x)\cdot \exp(-\gcd(p_1 \cdots p_n,x)^{\rho})\right )$$

Can this last equality be proven?
(I have checked it numerically with SAGEMATH).

Is there any benefit in using this formula to deduce any property of the primes?

Thanks for your help.

答:
Since $exp$ is positive for all real values we have that $ \sum \limits_{x \geq 2}^{\infty} exp(-\rho x) exp(-gcd(p_1 \cdots p_n ,x)^{\rho}) \geq exp(-p_{n+1} \rho) exp(-1)$ and $ \lim \limits_{\rho \to \infty} \frac{-1}{\rho} \log(exp(-p_{n+1} \rho) exp(-1)) = \lim \limits_{\rho \to \infty} \frac{-1}{\rho} ( -\rho p_{n+1}-1) = p_{n+1}$

Also $ \sum \limits_{x \geq 2}^{\infty} exp(-\rho x) exp(-gcd(p_1 \cdots p_n ,x)^{\rho}) \leq \sum \limits_{x=2}^{p_{n+1}-1} exp(-\rho x)exp(-2^{\rho}) + \sum \limits_{x = p_{n+1}}^{\infty} exp(-\rho x)exp(-1) \leq \sum \limits_{x=2}^{p_{n+1}-1} exp(-2^{\rho}) + \sum \limits_{x = p_{n+1}}^{\infty} exp(-\rho x)exp(-1) \leq 2p_{n} exp(-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{1}{e^{\rho}}+ \frac{1}{e^{2\rho}}+...) \leq \frac{2p_n}{e^{2^{\rho}}} + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) $ for $e^\rho >> 2p_n$ we have that $\frac{2p_n}{e^{2^{\rho}}} + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) \leq
 exp(\rho-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho}) $ and so $ \lim \limits_{\rho \to \infty} \frac{-1}{\rho} \log( exp(\rho-2^{\rho}) + exp(-\rho p_{n+1}-1) (1+\frac{2}{e^\rho})) = p_{n+1}$.

And by the squeeze theorem we have that $ \lim_{\rho \rightarrow \infty} -\frac{1}{\rho} \log \left( \sum_{x\ge 2}^\infty \exp(-\rho x)\cdot \exp(-\gcd(p_1 \cdots p_n,x)^{\rho})\right ) = p_{n+1}$

In the end your equality is correct


问:
computing Ryser's formula for the permanent
I am a bit confused on how to actually use Ryser's formula for the calculation of permanents of a square matrix. Wikipedia states the formula as:

$$ \mathrm{perm}(A)= (-1)^n \sum_{S \subset \{1,2,...,n\}}(-1)^{|S|} \prod _{i=1}^{n} \sum_{j \in S} a_{ij}$$

for a matrix, such as
$A=\begin{pmatrix}a & d & g\\
b & e & h\\
c & f & i
 \end{pmatrix}$, I am a bit confused on how to compute the permanent according to the above equation. Here A has nine different elements, but I am a bit unsure specifically on the how to calculate it explicitly. The wiki article mentions something about calculating row sums, but it is not obvious to me if $|S|$ would be referring to specific subsets. Just need to see an example of how Ryser's equation works and I think I will understand.

答:
The permanent is defined to be a sum of all products, where you pick one element from each row, and every choice has to come from distinct rows.

For this 3 x 3 matrix, that would be

$ aei + ahf + dbi + dhc + gbf + gec $

Computing this quantity for a large matrix requires a lot of computation. Ryser is trying to speed that up.

Sometimes, you can compute a sum of product by doing a product of sum, much faster, for example:

$ (a + b + c)(d + e + f) $ requires 4 adds and 1 multiplication

is the same as

$ ad + ae + af + bd + be + bf + cd + ce + cf $ which requires 8 additions and 9 multiplications.

Sadly, there are no good factorization for just the permanant sum, not one that can give significant speed up anyway.

That being said, it is easy to over-compute, for example, $ (a + d + g)(b + e + h)(c + f + i) $ will contain all terms in the permanent, we just created a lot more terms, which we need to figure out a way to cancel them.

That explains the product-of-sum part of the formula above, the goal is to compute a sum-of-product faster through product-of-sum. The alternative signs is meant for cancellation.

That hopefully gives the intuition of the formula. Now let's get to the details.

To avoid clutter, let's just consider the only the terms starting with $ a $, we have
$$
(a + d + g)(b + e + h)(c + f + i) = \\
abc + abf + abi + \\
aec + aef + aei + \\
ahc + ahf + ahi + \cdots
$$

Out of all the terms starting with $ a $, there are only two terms we wanted, $ aei $ and $ ahf $. The remaining term have chosen a value coming from the same column. Now let's consider this term

$$
(a + d)(b + e)(c + f) = \\
abc + abf + \\
aec + aef  + \cdots
$$

By considering terms only coming from the first two columns, now we can cancel all of them from the equation. We will also want to do the same for column 1 and 3 (note that we should consider column 2 and 3 as well, but they won't contain any term starting with $ a $, so we can ignore it for it now)

$$
(a + g)(b + h)(c + i) = \\
abc + abi + \\
ahc + ahi + \cdots
$$

That's great, we get 4 more terms to cancel, but wait, out of the 9 terms in the full expansion, 2 terms are useful, and now we can cancelling 8 of them? (4 of them from each pair of columns)

The problem is that $ abc $ is cancelled twice, this is because $ abc $ coming from column 1 only, that is why we need to add back $ abc $.

That is why you would want to have alternating sign, for product-of-sum consisting two columns, you would want to subtract. For product-of-sum consisting of just one column, you would like to add it back.

This is the intuition behind that formula, the rigorous proof for this is done by modifying the proof of the inclusion-exclusion principle, which this illustration looks very much alike anyway.

Here is a link where I learn the idea from, I hope it helps.


问:
Axis of symmetry parallel to the y-axis and passes through the points (1, 1), (2, 2) and (-1, 5)
For this problem, we are given a hint that we should apply the general form of the equation which is:

$$x^2+Dx+Ey+F=0$$ or $$y^2+Dx+Ey+F=0$$

how should I solve this one? is this simply substituting or not?

Thank you very much.

答: Take the parabola as $y=ax^2+bx+c$, stisfy it with points $(1,1), (2,2), (-1,5)$ to get $a,b,c$ from three equations.


问:
Is the following a Wiener process?
This is a worked example on Wiener processes.

Question: Pick a normally distributed random variable $Z \sim N(0,1)$, then define $W(t) = Z\sqrt{t}$. Is $W(t)$ a Wiener process?

Answer:

It is continuous.

$W(0) = 0$.

Therefore two required properties are satisfied.

However, $W(t+s) - W(s) = Z(\sqrt{t+s}-\sqrt{s})$; which has variance $(\sqrt{t+s}-\sqrt{s})^2$ so it is not a Wiener process as the incremental change in such a process should be $Z \sim N(0,t)$ also.

I don't understand how the example arrived at $(\sqrt{t+s}-\sqrt{s})$ as the number evaluated within the normal distribution (why the square root?) nor how the variance of the increment is $(\sqrt{t+s}-\sqrt{s})^2$. I think I am missing some knowledge about manipulating $t + s$ and $t$. Can someone kindly break down the manipulation into simple steps, I would be so grateful! Or otherwise explain how to arrive at the conclusion based on the third property? Yeah, I know I am not the best at maths!

答:
I know this is a very old question but I recently came upon this very same example and I don't agree with the discussion above, although I agree with the conclusion that W(t) is not a Weiner process.

Assuming two random variables X,Y are independent we have these facts:
Var (X + Y) = Var(X) + Var(Y), and also:
Var (X - Y) = Var (X) + Var(Y)
Also if a is a constant, then Var(aX) = a^2*Var(X)

We also know that a normal random variable Z multiplied by a constant "a" is another normal random variable with variance a^2*Var(Z).

According to these facts, we can determine:
Each of W(t + s) and W(s) are normal random variables with variances of (sqrt(t + s))^2 = t + s and (sqrt(s))^2 = s, respectively.

In other words, Var( W(t + s) ) = t + s, and
Var ( W(t) ) = t

Next we know that Var ( W(t + s) - W(s) ) = Var (W(t + s)) + Var (W(s)) = t + s + s = 2s + t

So W(t) is not a Weiner process because its variance is "2s + t", not because its variance is (sqrt(t + s) - sqrt(s))^2.

Title: Is the following a Wiener process?


答:
Here is an example where the program you gave fails. This gives details from my comment above.  You can plug in the following data into the Python program you linked here:
https://www.geeksforgeeks.org/minimum-cost-maximum-flow-from-a-graph-using-bellman-ford-algorithm/

```<code>s = 6
t = 2

cap = [ [ 0, 1, 1, 1, 0, 0, 0],
        [1, 0, 1, 1, 1, 0, 0],
        [1, 1, 0, 0, 0, 0, 0],
        [1, 1, 0, 0, 1, 1, 0],
        [0, 1, 0, 1, 0, 1, 0],
        [0, 0, 0, 1, 1, 0, 1],
        [0, 0, 0, 0, 0, 1, 0]]

cost = [ [ 0, -1, -1, -1, 0, 0, 0],
         [-1, 0, -1, -1, -1, 0, 0],
         [-1, -1, 0, 0, 0, 0, 0],
         [-1, -1, 0, 0, -1, -1, 0],
         [0, -1, 0, -1, 0, -1, 0],
         [0, 0, 0, -1, -1, 0, -1],
         [0, 0, 0, 0, 0, -1, 0]]
</code>```

This is a graph with 7 nodes labeled {0, 1, ..., 6}. The graph is symmetric, so for every link (i,j) there is a link (j,i). All link capacities are 1 and all costs are -1. Node 6 is only connected to node 5 and so the max flow to or from node 6 is 1 unit.

I used source s=6 and destination t=2. The program gives an output of 1,  -5. This means a flow of 1 unit and a cost of -5 (5 hops). This is incorrect because there is a 6-hop Hamiltonian path that the program does not find:

$$ 6\rightarrow 5\rightarrow 4 \rightarrow 3 \rightarrow 1 \rightarrow 0 \rightarrow 2 $$

On the other hand, if we reverse the path and use s=2, t=6, the program gives a correct output of 1,  -6.

Unfortunately the program only outputs the cost of the path, not the actual path, so I don't know what paths it is proposing in either case.

Title: Using Bellman-Ford to find a Hamiltonian cycle? (NP-complete)

Body:
Let $G(V,E)$ be a directed graph, where $V=\{a_1,\ldots,a_n\}$ is a set of vertices and $E$ is a set of ordered pairs of $V$, with $|V|=n$.

Now, let be $G(W,F)$ be a graph where $W$ is a set of vertices, such that $W=\{a_1,\ldots,a_{2n}\}$ with $|W|=2n$, and $F$ is a set of ordered pairs of $W$ defined as follows:
$$\forall i, j \in \{1, \ldots, n\} : (a_i,a_{j+n}) \in F \text{ if } (a_i,a_{j}) \in E \\
\forall i \in \{1, \ldots, n\} : (a_{i+n},a_i) \in F$$

Now we define the capacities:
$$c(a,b)=1 \text{ if } (a,b) \in F$$

And we define a cost:
$$\forall i \in \{1, \ldots, n\} : p(a_{i+n},a_i)=-1 \text{ if } (a_{i+n},a_i) \in F \\
p=0 \text{ else}$$

The source as $a_{n+1}$ and the sink as $a_{n+1}$.

Does the graph $G(W,F,c,p)$ have a minimum cost flow of $-n$ if and only if the graph $G(V,E)$ has a Hamiltonian cycle?

(Comment to an answer in the comments section)

Michael, for your matrix with capacities G(V,E) the matrix G(W,F) is:

```<code>cap = [[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0]]

cost = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0]]
</code>```

Body:
Here is an example where the program you gave fails. This gives details from my comment above.  You can plug in the following data into the Python program you linked here:
https://www.geeksforgeeks.org/minimum-cost-maximum-flow-from-a-graph-using-bellman-ford-algorithm/

s = 6
t = 2

cap = [ [ 0, 1, 1, 1, 0, 0, 0],
        [1, 0, 1, 1, 1, 0, 0],
        [1, 1, 0, 0, 0, 0, 0],
        [1, 1, 0, 0, 1, 1, 0],
        [0, 1, 0, 1, 0, 1, 0],
        [0, 0, 0, 1, 1, 0, 1],
        [0, 0, 0, 0, 0, 1, 0]]

cost = [ [ 0, -1, -1, -1, 0, 0, 0],
         [-1, 0, -1, -1, -1, 0, 0],
         [-1, -1, 0, 0, 0, 0, 0],
         [-1, -1, 0, 0, -1, -1, 0],
         [0, -1, 0, -1, 0, -1, 0],
         [0, 0, 0, -1, -1, 0, -1],
         [0, 0, 0, 0, 0, -1, 0]]

This is a graph with 7 nodes labeled {0, 1, ..., 6}. The graph is symmetric, so for every link (i,j) there is a link (j,i). All link capacities are 1 and all costs are -1. Node 6 is only connected to node 5 and so the max flow to or from node 6 is 1 unit.

I used source s=6 and destination t=2. The program gives an output of 1,  -5. This means a flow of 1 unit and a cost of -5 (5 hops). This is incorrect because there is a 6-hop Hamiltonian path that the program does not find:

$$ 6\rightarrow 5\rightarrow 4 \rightarrow 3 \rightarrow 1 \rightarrow 0 \rightarrow 2 $$

On the other hand, if we reverse the path and use s=2, t=6, the program gives a correct output of 1,  -6.

Unfortunately the program only outputs the cost of the path, not the actual path, so I don't know what paths it is proposing in either case.


问:
Special arrows for notation of morphisms
I've stumbled upon the definition of exact sequence, particularly on Wikipedia, and noted the use of $\hookrightarrow$ to denote a monomorphism and $\twoheadrightarrow$ for epimorphisms.

I was wondering whether this notation was widely used, or if it is common to define a morphism in the general form and indicate its characteristics explicitly (e.g. "an epimorphism $f \colon X \to Y$").

Also, if epimorphisms and monomorphisms have their own special arrows, are isomorphisms notated by a special symbol as well, maybe a juxtaposition of $\hookrightarrow$ and $\twoheadrightarrow$?

Finally, are there other kinds of morphisms (or more generally, relations) that are usually notated by different arrows depending on the type of morphism, particularly in the context of category theory?

Thanks.

答: There is a Unicode character 2916 (⤖) for "bijective mapping". My algebraic topology instructor, educated in Canada and New York, wasn't familiar with it when I employed it.</p>


问:
Choose distinct elements that satisfy certain conditions in $\Bbb{Z}_n$
I am trying to understand the proof of a proposition in a paper as below:

I cannot understand how and why the condition $km<n$ lead to choosing $a_1,\dots,a_m\in \Bbb{Z}_n$ such that the listed elements are all distinct. I guess that Axiom of Choice applies here, but the method of choosing those elements are not stated in the proof.

Next, I am thinking whether such result can be generalized. For example, if $p+q+r<n$ where $p,q,r$ are positive integers, can we choose $a_1,a_2,a_3\in \Bbb{Z}_n$ such that
\begin{align*}
&a_1,a_1+1,a_1+2,\dots,a_1+(p-1);\\
&a_2,a_2+p,a_2+2p,\dots,a_2+(q-1)p;\\
&a_3,a_3+pq,a_3+2pq,\dots,a_3+(r-1)pq
\end{align*}
are all distinct in $\Bbb{Z}_n$.

Remark:

$A_1,\dots,A_m$ is a complete decomposition of $\Bbb{Z}_n$ if $A_1+\dots+A_m=\Bbb{Z}_n$ and $A_1,\dots,A_m$ are pairwise disjoint.

答:
I assume by $\mathbb Z_n$ you meant $\mathbb Z/n\mathbb Z$, the integers mod $n$. Correct me if I'm wrong.

This proof seems dubious to me. Take $k=3, n=54$, then $m=4$ and $n=(k-1)k^{m-1}$. As such $a_m\equiv a_m+(k-1)k^{m-1}\pmod n$ and you would never be able to find such distinct numbers.

However, I can say one thing: this has absolutely nothing to do with the Axiom of Choice. It is used when you are choosing infinitely many things from infinite sets. Here you are choosing finitely many integers. Even if you were to choose infinitely many representatives from $\mathbb Z$, it can be easily well-ordered (e.g. by absolute value) and requires no choice whatsoever.


问:
show a function is Borel measurable
Prove the continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ is Borel measurable in the following two steps:

First let $G = \{U \subset \mathbb{R} \vert g^{-1}(U) \text{is Borel} \}$

(a) show all open sets are in $G$.

(b) show $G$ is a $\sigma$-algebra.

So for (a) at least, take an arbitrary open set in $\mathbb{R}$ call it $V$, then $V$ can be written as a countable union of standard basis elements, that is,

$V = \bigcup_{j=1}^\infty (a_j,b_j)$ where $a_j \leq b_j$ are real numbers. Then by continuity of $g$, we have that the pullback $g^{-1}(V)$ is open in $\mathbb{R}{^n}$. Then this pullback is the union of open balls in $n$ dimension. How do I show something is a Borel set? I know my set can be generated by open sets and I know Borel sets are elements of the Borel $\sigma$-algebra which is sort of like a topological space (in that it has an axiomatic definition of certain closures of its elements) except that now the entire space need be in the collection of subsets, together with countable unions and intersections, and complements. Where do I go from here? So for (b) and I supposed to show these axioms hold for members of $G$?

答:
The question asks to prove the continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ is Borel measurable in the following two steps:

First let $G = \{U \subset \mathbb{R} \:\vert\: g^{-1}(U) \text{ is Borel } \}$

(a) show all open sets are in $G$.

(b) show $G$ is a $\sigma$-algebra.

Proof: Let us prove (a).
Since $g: \mathbb{R}^n \rightarrow \mathbb{R}$ is continuous, we have that, for all $A$ open subset of $\mathbb{R}$, $g^{-1}(A)$ is an open subset  of $\mathbb{R}$. So, in particular  $g^{-1}(A)$ is Borel. In other words, all open sets are in $G$.

Let us prove (b).

(b.1) Since $g^{-1}(\emptyset)=\emptyset$  and $\emptyset$ is Borel, we have that $g^{-1}(\emptyset)$ is Borel. So $\emptyset \in G$.

(b.2) Suppose that $E \in G$. Then $g^{-1}(E)$ is Borel. So $g^{-1}(E)^c$ is Borel. Since
$$ g^{-1}(E^c)= g^{-1}(\mathbb{R} \setminus E) = g^{-1}(\mathbb{R}) \setminus g^{-1}(E)= \mathbb{R}^n \setminus g^{-1}(E) = g^{-1}(E)^c$$
we have that $g^{-1}(E^c)$ is Borel. So, $E^c \in G$.

(b.3) Let $\{E_j\}_j$ be a sequence of sets in $G$. Then, for all $j$, $g^{-1}(E_j)$ is Borel. So $\bigcup_j g^{-1} (  E_j  )  $ is Borel. Since
$$ g^{-1} \left ( \bigcup_j E_j  \right)= \bigcup_j g^{-1} (  E_j  )  $$
we have that $g^{-1} \left ( \bigcup_j E_j  \right)$ is Borel. So $ \bigcup_j E_j  \in G$.

So $G$ is a $\sigma$-algebra.

Title: show a function is Borel measurable


问:
What are cohomology rings good for?
I am studying some concepts of algebraic topology myself, and I read lately a bit about cohomology rings (created by the direct sum of cohomology groups) but besides all definitions I could not find any information.

What are they good for?

How do we use them?

Could someone give me a very short descriptive example?

Thank you

答: Just to throw out another example of the use of cup-products, there is an invariant called the cup length, which is the largest $k$ such that $\alpha_1\cup\ldots\cup \alpha_k\neq 0$ for some cohomology elements $\alpha_i$ of degree $\geq 1$. The cup length gives a lower bound for the so-called Lusternik-Schnirelmann category.

Title: What are cohomology rings good for?


问:
Eigenvectors and differential equations

I was able to find part (a), and I got 4 and -1 for the eigenvalues and from these values I got eigenvectors of [1,1] and [-3,2], but I don't know what to do for part (b) and (c)

答: Let ${\bf y}=(x,y)$. Then the coupled system can be written as $${\bf y}'=A{\bf y}$$ The general solution is $${\bf y}=c_1e^{4t}(1,1)+c_2e^{-t}(-3,2)$$ assuming that you did the eigenvector/eigenvalue calculations correctly. </p>

Title: Eigenvectors and differential equations


问:
Reformulation of a theorem about the lower bound on no. of facets of a symmetric polytope in $\mathbb{R}^n$

Preliminaries

The distance $d(K, L)$ between symmetric convex bodies $K$ and $L$ is the least positive $d$ for which there is a linear image $\tilde L$ of $L$ such that $\tilde L ⊂ K ⊂ d\tilde L$. Note that this distance is multiplicative, not additive: in order to get a metric (on the set of linear equivalence classes of symmetric convex bodies), we would need to take $\log d$ instead of $d$. In particular, if $K$ and $L$ are identical then $d(K, L) = 1$.

More details on this MathSE post.

Question

I have trouble understanding the reformulation of the following theorem which provides a lower bound on the number of facets of a symmetric polytope in $\mathbb{R}^n$. The proof of the theorem and the discussion hereafter (in the article I am reading) depends heavily on this reformulation - so I need help in understanding it. I quote:

Theorem 2.1. Let $K$ be a (symmetric) polytope in $\mathbb{R}^n$ with $d(K,B_2^n) = d$. Then
$K$ has at least $\exp\left(\frac{n}{2d^2}\right)$ facets. On the other hand, for each $n$, there is a polytope with $4^n$ facets whose distance from the ball is at most $2$.

The distance being referred to in the above discussion is defined in the Preliminaries section of this post.

The arguments in this lecture, including the result just stated, go back to the
early days of packing and covering problems. A classical reference for the subject
is [Rogers 1964]. Before we embark upon a proof of Theorem 2.1, let’s look at a reformulation
that will motivate several more sophisticated results later on. A symmetric
convex body in $\mathbb{R}^n$ with $m$ pairs of facets can be realized as an $n$-dimensional
slice (through the center) of the cube in $\mathbb{R}^m$. This is because such a body is
the intersection of $m$ slabs in $\mathbb{R}^n$, each of the form $\{x: |\langle x, v\rangle| \le 1\}$, for some nonzero vector $v$ in $\mathbb{R}^n$. This is shown in Figure $8$ (below).

I'm able to decipher that $\{x: |\langle x, v\rangle| \le 1\}$ is a collection of hyperplanes of the form $\{x: \langle x, v\rangle = k\}$ where $-1\le k\le 1$, hence forming a slab. However,

Why is the slice $n$-dimensional? What's the cube they are talking about?

How did we change our perspective from $\mathbb{R}^n$ to $\mathbb{R}^m$? I didn't get that.

In the figure above, where is the cube?, where is the polytope?, are we working with $m=3$ (since there seem to be $3$ pairs of opposite facets)?

Thus $K$ is the set $\{x: |\langle x, v_i\rangle| \le 1, 1 ≤ i ≤ m\}$, for some sequence $(v_i)_1^m$ of vectors in $\mathbb{R}^n$. The linear map $$T:x\mapsto (\langle x,v_1\rangle,\ldots \langle x,v_m\rangle)$$ embeds $\mathbb{R}^n$ as a subspace $H$ of $\mathbb{R}^m$, and the intersection of $H$ with the cube $[−1, 1]^m$ is the set of points $y$ in $H$ for which $|y_i| \le 1$ for each coordinate $i$. So this intersection is the image of $K$ under $T$. Conversely, any $n$-dimensional slice of $[−1,1]^m$ is a body with at most $m$ pairs of faces. Thus, the result we are aiming at can be formulated as follows:
 "The cube in $\mathbb{R}^m$ has almost spherical sections whose dimension $n$ is roughly $\log m$ and not more."

I do see that $$H = \{(\langle x,v_1\rangle,\ldots \langle x,v_m\rangle)\in\mathbb{R}^m:x\in\mathbb{R}^n\}$$ is a subspace of $\mathbb{R}^m$. So the intersection of $H$ with $[-1,1]^m$ is the set of points $y\in H$ such that for every $i$, we have $|y_i|\le 1$. So we have $|\langle x,v_i\rangle|\leq 1$ for every $i$ (since the points are in $H$) - which gives $x\in K$. As a result, the intersection of $H$ with $[-1,1]^m$ is the image of $K$ under $T$.

Why is any $n$-dimensional slice of $[−1,1]^m$ a body with at most $m$ pairs of faces?

What is meant by almost spherical sections in the cube?

I need help in understanding the reformulation of Theorem 2.1 to the last statement above, via the process outlined in the quoted extract. I would appreciate a detailed explanation, and/or anything that might help me! Thanks a lot for your patience in reading this long post!

References:

Lecture 2 (Pg. 8-9) of these notes.

答:
Why is the slice $n$-dimensional? What's the cube they are talking about?

The cube, which is just $[-1,1]^m\subseteq\mathbb R^m$, is not visible in the diagram since they have only shown the polytope itself (in $\mathbb{R}^n$). The "$n$-dimensional slice" is explained in the subsequent paragraph. The corresponding embedding is being described by a linear transformation, namely the map $T$.

To understand this better, perhaps you could think of the polytope in $2$ dimensions given by $v_1=\left(\frac{1}{2},0\right)$, $v_2=\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)$, and $v_3 = \left(\frac{1}{4},-\frac{\sqrt{3}}{4}\right)$ - this is just a hexagon that is inscribed in the unit ball $B_2^2$. So what does the map $T$ do? It embeds this hexagon in $\mathbb{R}^3$ as the intersection of $[-1,1]^3$ with some subspace:

Similarly, you can view the intersection of any subspace of $\mathbb{R}^m$ with $[-1,1]^m$ as a polytope in a lower dimension.

Why is any $n$-dimensional slice of $[−1,1]^m$ a body with at most $m$ pairs of faces?

Note that each face of the polytope formed by a slice corresponds to an $(m-1)$-dimensional face of $[-1,1]^m$. Since there are $2m$ such faces, (the polytope corresponding to) any slice can have at most $2m$ faces.

What is meant by almost spherical sections in the cube?

Roughly, it just means a slice of the cube whose distance from the Euclidean ball (in that dimension) is not too large. The theorem you are reading says that a polytope that is at a small distance from the Euclidean ball has exponentially many facets. Alternatively, if you consider a slice of the cube, it can only be close to the Euclidean ball if it has exponentially many facets, which can only be true if the dimension you take is very small (logarithmic).

The exact statement regarding $\log m$ is actually made more precise in Dvoretzky's Theorem which is discussed in the final lecture of the notes you are reading - you could look ahead at the statement of this theorem right now. It states

There is a positive number $c$ such that for every $\varepsilon>0$ and every natural number $n$, every symmetric convex body of dimension $n$ has a slice of dimension
$$k\geq \frac{c\varepsilon^2}{\log(1+\varepsilon^{-1})}\log n$$
that is within distance $\varepsilon$ of the $k$-dimensional Euclidean ball.

which, sans constant, is more or less what the author has said here.

The proof you are interested in does not really use this $\log m$ formulation, it just uses the fact that you can think of a polytope in $\mathbb{R}^n$ with $m$ pairs of faces as an $n$-dimensional slice of $[-1,1]^m$.


问:
Interconverting different equations for spherical caps in $\mathbb{R}^n$
I need help in interconverting between these two definitions of a spherical cap in $\mathbb{R}^n$:

Figure 9 (Left): For a fixed unit vector $v$ and $\epsilon \in [0,1)$, the set $\theta\in S^{n-1}$ for which $$\langle\theta,v\rangle \ge \epsilon$$ is called the $\epsilon$-spherical cap about $v$. It is denoted by $C(\epsilon,v)$. Here $S^{n-1}$ is the sphere of radius $1$, i.e. $$S^{n-1} = \{x\in\mathbb{R}^n: \|x\| = 1\}$$
Figure 9 (Right): The cap of radius $r$ about $v$ is given by $$\{\theta \in S^{n-1}: \| \theta - v\|\le r\}$$

So how do I express $\epsilon$ in terms of $r$? This will help me switch between the two definitions of the spherical cap. Note that we are in $\mathbb{R}^n$ and not $\mathbb{R}^3$.

My try:

I started with the second definition and saw that $$\| \theta - v\|\le r \implies \langle \theta - v,\theta-v\rangle \leq r^2 \implies \|\theta\|^2 - \|v\|^2 - 2\langle\theta,v\rangle \leq r^2$$
Since $\theta,v\in S^{n-1}$ we have $\|\theta\| = \|v\| = 1$ so that $$\| \theta - v\|\le r\implies \langle\theta,v\rangle \geq 1-\frac{r^2}{2} $$

Now, looking at the first definition, we see that $\epsilon =-\frac{r^2}{2}$ but we had assumed $\epsilon\in[0,1)$ so this does not make sense!

Where am I going wrong, what am I missing?

Addendum:

It was a silly blunder. The right inequality is:
$$\| \theta - v\|\le r \implies \|\theta\|^2 + \|v\|^2 - 2\langle\theta,v\rangle \leq r^2 \implies \langle\theta,v\rangle \geq 1 - \frac{r^2}{2}$$
So we get $$\epsilon = 1 - \frac{r^2}{2}$$

答:
$$\langle \theta - v,\theta-v\rangle \leq r^2 \implies \|\theta\|^2 - \|v\|^2 - 2\langle\theta,v\rangle \leq r^2$$

It should be

$$\langle \theta - v,\theta-v\rangle \leq r^2 \implies \|\theta\|^2 + \|v\|^2 - 2\langle\theta,v\rangle \leq r^2.$$


问:
Prove an inequality in the space of $L^{2}(\Omega)$.
EDIT: First of all, I thank Calvin Khor's exhaustive comments on the inequality that I proposed regarding the semi-norm in $L^{2}(\Omega)$, which after his teachings and his final comments allowed me to realize that this inequality did not make sense for any semi-norm of $L^{2}(\Omega)$. For that reason I have modified my problem, changing from the semi-norm $|\cdot|_{L^{2}(\Omega)}$ in $L^{2}(\Omega)$ to the norm $||\cdot||_{L^{2}(\Omega)}$ in $L^{2}(\Omega)$. Again, thank you very much Calvin Khor.

Fixed issue: My goals is to prove that there exists a positive-strictly constant $C_{0}>0$ such that for all $v\in H^{1}(\Omega)$, we have that $$||v||^{2}_{L^{2}(\Omega)} \leq C_{0} \left[ ||v||^{2}_{L^{2}(\Phi)}+||\nabla v||^{2}_{L^{2}(\Omega)}\right]$$
where $||\cdot||_{L^{2}}$ be norms in $L^{2}(\Omega)$ and $\Omega=(0,1)\times (0,1)$ and $\Phi=\{0\}\times (0,1)$.

My approach:  Maybe I should use the Poincaré's inequality:

Poncaré's inequality: Let $\Omega$ a bounded open set of $\mathbb{R}^{n}$. Then, there exists a positive $C_{p}(\Omega)$ such that: $$\color{blue}{\boxed{||v||_{L^{2}(\Omega)}\leq C_{p}(\Omega)|v|_{H^{1}(\Omega)}=C_{p}(\Omega)||\mathbf{\nabla v}||_{L^{2}(\Omega)}}}$$

We have that $$ ||v||_{L^{2}(\Omega)}^{2}\leq C_{p}(\Omega)^{2}||\mathbf{\nabla v}||_{L^{2}(\Omega)}^{2} $$
how can I continue from here? maybe the Rellich's compactness theorem help here?

答:
You are right that your inequality is reminiscent of Poincaré. In fact, you can just follow the proof by contradiction of the normal Poincaré inequality. By usual density arguments we can assume $v\in C^\infty(\overline \Omega)$. Suppose such a $C_0>0$  did not exist. Thus for each $k\in\mathbb N$ we can find $v_k\in C^\infty(\overline\Omega)$ such that $\|v_k\|_{L^2(\Omega)} = 1$ and

$$\|v_k\|_{L^2(\Phi)}^2 + \|\nabla v_k \|_{L^2(\Omega)} \le \frac1k.$$

Then $v_k$ are bounded in $H^1$ and therefore after passing to a subsequence using Rellich compactness theorem, there is some $v\in L^2$ such that
$$ v_k|_{\Phi} \to 0 \text{ in } L^2(\Phi), \ \nabla v_k \to 0 \text{ in } L^2(\Omega),\ v_k\to v \text{ in } L^2(\Omega). $$  In
particular $v_k\to v$ in $H^1$ with $\nabla v \equiv 0$, and hence $v\equiv c$ for some $c\in\mathbb R$ ($\Omega$ is connected). Therefore, $v|_{\Phi} \equiv c$ in the sense of trace. By continuity of the trace map, $v_k|_\Phi\to v|_{\Phi}$ in $L^2(\Phi)$. But also $\|v_k\|_{L^2(\Phi)} \to 0$, so $c=0$. This is in contradiction with the assumption that $\|v\|_{L^2}=1$.


问:
How to analyze the following complex number?
$f(T) = \int_{0}^{1}e^{-iT\lambda(s)}ds$, where $\lambda(s)$ increases monotonously. I want to show that  $|f(T)|$ has an upper bound that decreases with $T$ for $T \gg 1$.

For example, if $\lambda(s) = s$ then $|f(T)| \le 2/T$.

This question comes from numerical analysis of my research. It's the expression of error analysis in adiabatic evolution. More specifically, the digital error has expression : $\sum_{k=1}^{L}e^{-iT\sum_{j<k}\lambda_{j}}/L$, where $\lambda_{j}$ equals $\lambda_{j} = \lambda(j/L)$. From numerical result I am sure it has an upper which bound decays with $T$. However, I don't know how to prove that. I wonder if there is related formulas in complex analysis and fourier series.

答:
I make here two assumptions on $\lambda$

it has derivative bounded from below by m > 0

derivative is monotonous (so $\frac{1}{\lambda^{'}}$ is also monotonous)

First off, if f(t) is a smooth function then $$f(s_{k}) - f(s_{k-1}) = f^{'}(s_{k-1})*(s_{k}-s_{k-1}) + \epsilon(s_{k}-s_{k-1})$$
This is just a mean value theorem.
Here $s_{1} < s_{2} < s_{3} <... <s_{n}$ is a partition of [0,1] and $\epsilon$ stands for something which tends to zero (uniformly) when $s_{k} - s_{k-1}$ tends to zero.
Using this for $f(s) = e^{iT\lambda(s)}$ one gets
$$e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}= iT\lambda^{'}(s_{k-1})e^{iT\lambda(s_{k-1})}(s_{k}-s_{k-1}) + \epsilon*(s_{k}-s_{k-1})$$
so
$$\frac{1}{iT\lambda^{'}(s_{k-1})}e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})} = e^{iT\lambda(s_{k-1})}(s_{k}-s_{k-1}) + \frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}$$
Now, writing the integral $\int_{0}^{1}e^{iT\lambda(s)}ds$ as the limit of Riemann' sums we have
$$\int_{0}^{1}e^{iT\lambda(s)}ds = \lim_{\Delta s \to 0}\sum_{k}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})})(s_{k}-s_{k-1})$$
But from the formula above we see that the sum can be estimated by
$$\sum_{k}\frac{1}{iT\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) - \sum_{k}\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}$$
The second sum is easy to bound from above (i.e. assuming that $\lambda^{'}$ is bound from below by, say m) we get
$$|\sum_{k}\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}| <= $$
$$\sum_{k}|\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}| <= $$
$$\frac{1}{Tm}\sum_{k}|\epsilon*(s_{k}-s_{k-1})| <=$$
$$\frac{max(\epsilon)}{Tm}\sum_{k}|(s_{k}-s_{k-1})| <=$$
$$\frac{max(\epsilon)}{Tm}$$
Now we have to estimate the first sum
$$\frac{1}{iT}\sum_{k}\frac{1}{\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) $$
When we change order of summation we get that the sum is
$$|\frac{1}{iT}\sum_{k}\frac{1}{\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) | <= $$
$$|\frac{1}{iT}\sum_{k}(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) e^{iT\lambda(s_{k-1}}) | + 2\frac{1}{Tm} <= $$
$$\frac{1}{T}\sum_{k}|(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) e^{iT\lambda(s_{k-1}}) | + 2\frac{1}{Tm} <= $$
$$\frac{1}{T}\sum_{k}|(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) | + 2\frac{1}{Tm} <= $$
And since $\frac{1}{\lambda^{'}}$ is by assumption monotone we can skip absolute values under the last sum to get
$$\frac{1}{T}|\frac{1}{\lambda^{'}(0)} - \frac{1}{\lambda^{'}(1)} | + 2\frac{1}{Tm} $$