问:
Probability exponential distribution.
May I please borrow your expertise or could anyone check if I'm on the right track please?

Consider customers arriving at a bank. The bank has $2$ types of customers - business and personal. On average, $10$ business customers arrive per hour, and $20$ personal customers. The times between arrivals of the business customers are independent of each other, and exponentially distributed, and is the same for personal customers. The two streams are independent of each other. Suppose no new customers have entered the bank in the last $5$ minutes.

1) What is the probability that no business customers arrive in the next $3$ minutes?

I'm assuming the arrival time for business customers is 60/10 = 6 customers
Thus, by using the exponential equation and also that this is a memoryless property,

$1 - e^{-3/6}$ (am I on the right track here anyone)?

2) What is the probability that no personal customers arrive in the next $3$ minutes?

Assuming the arrival time for personal customers to be 60/20 = 3 customers
Using the equation from question 1)

$1 - e^{3/3}$ (is this right)?

3) What is the probability that no customers at all arrive in the next $3$ minutes?

Assuming that this is something like $1 - ((1-e^{-3/6}) + (1-e^{-3/3}))$
by adding the no personal customer and business customer together and minus it by 1

4) What is the distribution of the time until the arrival of the next customer?

(Any hints on how I might be able to use to get the answer for this question?)

Your helps are much appreciated. Thanks,

答:
We need to decide whether to measure time in hours or minutes. Say minutes. Then the mean number of business arrivals per minute is $\frac{10}{60}$.  Thus we want to know the probability that an exponentially distributed random variable with parameter $\lambda=\frac{10}{60}$ is $\gt 3$. The probability that it is less than or equal to $3$ is $1-e^{-30/60}$. But the probability it is bigger than $3$ is $e^{-30/60}$.

A similar calculation for personal customers yields $e^{-60/60}$.

For no customers at all, by independence, we want the probability of no business customers times the probability of no personal customers.

As to the last question, if we want to calculate, let $W$ be the waiting time until the next customer. The probability that $W\gt w$ is the probability that $X\gt w$ and $Y\gt w$, where $X$ and $Y$ represent the waiting times for the two types of customers. Calculate this. You will recognize it as an exponential random variable. Indeed, it turns out that the distribution of $W$ can be written down immediately from the fact that the combined arrival rate is $30$ per hour.

Remark: Things will get clearer with practice, and when the relationship between the Poisson distribution and the exponential has been developed.

Title: Probability exponential distribution.


问:
Prove that G admits a composition series of length $a_1 + · · · + a_k$.
Let be G an abelian group of order $n = p^{a1}_1· · · p^{ak}_k$
where $p_i$ is prime such that $p_i\neq p_j$ if $i \neq j$. Prove that G admits a composition series of length $a_1 + · · · + a_k$.

My path

Let $n = p^{a1}_1· · · p^{ak}_k$ be the order of the Abelian group $G$, with $p_i$’s distinct primes.
By Sylow’s theorem it follows that $G$ has exacly one Sylow $p$−subgroup for each of the $k$
distinct primes $p_i$. Consequently $G$ is the direct product of its Sylow $p$−subgroups. [We
call the Sylow $p$−subgroups the $p$− primary parts of G.] It remains to show that an
Abelian $p$−group (corresponding to a $p$−primary part of $G$) is the direct product of cyclic
groups.

But I'm not sure if this path is correct. Can anyone help?

答:
You don't need the Sylow theorems here; you can show that a finite abelian group is a finite direct product of $p$-primary parts using the Chinese remainder theorem.

But actually you don't even need this. The argument is just by strong induction on $n$. Let $H$ be any maximal proper subgroup and show that $G/H$ has order a prime. Then apply the same argument to $H$, inductively. We don't even need the full strength of the assumption that $G$ is abelian; this argument will show that the result is still true if we just assume that $G$ is solvable, although here we need $H$ to be a maximal proper normal subgroup. Moreover this result is sharp: a finite group of order $\prod p_i^{a_i}$ admits a composition series of length $\sum a_i$ iff it is solvable.

This argument does not require knowing in advance that composition series exist, and can be used to prove that finite groups have composition series. If you already know that fact, then show that every composition series of a finite solvable group of order $\prod p_i^{a_i}$ must have length $\sum a_i$.

ContentLicense: CC BY-SA 4.0


问:
Are tensors vectors?
Take the definition of a tensor as a multilinear map $T: (V)^n \times (V^*)^m \rightarrow \mathbb R$. One may then define the addition of two tensors as the sum of their outputs on their input spaces (assuming of course the tensors have the same rank), and scalar multiplication as the output scaled by the factor used (if $T$ maps its input to $b$, then $aT$ maps its input to $ab$).

As one can easily check, tensors would then satisfy all of the vector space axioms, so then a tensor could be considered a specific type of vector. This bothered me, though, because vectors are often defined as objects with magnitude and direction in less formal contexts, and tensors can often be informally considered to have many different directions.

This leads me to the writing of this question. What distinguishes a tensor from a classical, "ordinary" vector? How can one derive the definition of a vector as something with magnitude and direction from the definition of a vector space, and where does this definition break down for tensors (if anywhere)?

答:
What distinguishes a tensor from a vector relies on  understanding what you're talking about.  Strictly speaking, tensors of a fixed rank form a vector space (over $\mathbf R$, say), and thus "tensors are vectors" for pure mathematicians who don't work in anything related to physics or differential geometry.

But nobody means anything like that when they bring up the issues of tensors vs. vectors. This is why "tensors are not vectors" for physicists, and it explains why you are getting the answers "yes" and "no" in the comments.

Look at how tensors are built. You start with a vector space $V$ and take a tensor product of tensor powers of $V$ and the dual space $V^*$, say $V^{\otimes n} \otimes (V^*)^{\otimes m}$. Elements of $V^{\otimes n} \otimes (V^*)^{\otimes m}$ are called tensors, elements of $V$ are called vectors, and elements of $V^*$ are called covectors. In this setting, where the word vector means "element of $V$", the objects in $V^{\otimes n} \otimes (V^*)^{\otimes m}$ are not vectors (unless $n = 1$ and $m = 0$). Vector is just a synonym for "element of $V$". That's all there is to it.

You ask in a comment why tensors can't be given an inner product. That's incorrect: they can, provided $V$ is given an inner product first.  If $V$ and $W$ are both real vector spaces and they are each given an inner product $\langle \cdot,\cdot\rangle_V$ and $\langle \cdot,\cdot\rangle_W$, then there is a unique inner product on $V \otimes W$ for which
$$
\langle v \otimes w,v'\otimes w'\rangle = \langle v,v'\rangle_V\langle w,w'\rangle_W
$$
on elementary tensors $v \otimes w$ and $v' \otimes w'$ in $V \otimes W$. Check that. That is, figure out why there exists exactly one inner product on $V \otimes W$ having the above behavior on pairs of elementary tensors in $V \otimes W$.

By iterating this construction, when $V$ has an inner product $\langle \cdot,\cdot\rangle_V$, each tensor power $V^{\otimes n}$ has a unique inner product that looks like
$$
\langle v_1 \otimes \cdots \otimes v_n,v_1'\otimes \cdots \otimes v_n'\rangle = \langle v_1,v_1'\rangle_V\cdots \langle v_n,v_n'\rangle_V
$$
on pairs of elementary tensors $v_1 \otimes \cdots \otimes v_n$ and $v_1'\otimes \cdots \otimes v_n'$.

When $V$ is finite-dimensional,
an inner product $\langle \cdot,\cdot\rangle_V$ on $V$ gives us an isomorphism $V \to V^*$ by $v \mapsto \langle\cdot,v\rangle_V$. Using this, we can transport the inner product on $V$ over to an inner product on $V^*$: for $\varphi$ and $\varphi'$ in $V^*$, simply declare
$$
\langle \varphi,\varphi'\rangle_{V^*} := \langle v,v'\rangle_V
$$
where $\varphi = \langle \cdot,v\rangle_V$ and $\varphi' = \langle \cdot,v'\rangle_V$. Now you can use this inner product on $V^*$ to define an inner product on $(V^*)^{\otimes m}$ by the method described above for creating an inner product on tensor powers of $V$ from an inner product on $V$ (just replace $V$ by $V^*$ everywhere). And then you can use the inner products on $V^{\otimes n}$ and $(V^*)^{\otimes m}$ to define an inner product on $V^{\otimes n} \otimes (V^*)^{\otimes m}$ by the method described up above for putting an inner product on $V \otimes W$ when $V$ and $W$ each have an inner product.  Thus, when a finite-dimensional real vector space $V$ has an inner product, every space of tensors $V^{\otimes n} \otimes (V^*)^{\otimes m}$ for fixed $m$ and $n$ gets an inner product from the inner product on $V$.  Concretely,
for elementary tensors
$$
t = v_1 \otimes \cdots \otimes v_n \otimes \varphi_1 \otimes \cdots \otimes \varphi_m
$$
and
$$
t' = v_1' \otimes \cdots \otimes v_n' \otimes \varphi_1' \otimes \cdots \otimes \varphi_m'
$$
in $V^{\otimes n} \otimes (V^*)^{\otimes m}$,
$$
\langle t,t'\rangle = \prod_{i=1}^n \langle v_i,v_i'\rangle_V \prod_{j=1}^m \langle w_j,w_j'\rangle_V,
$$
where $\varphi_j = \langle \cdot,w_j\rangle_V$ and $\varphi_j' = \langle \cdot,w_j'\rangle_V$.

Title: Are tensors vectors?


问:
Need help with problem 6(a) from Topology Without Tears
Following is Topology Without Tears by Morris

Let A,B be a space of R with Euclidean topology
If A is the set of all rational numbers,B set of irrational number

Consider these 4 sets :

(1)A $\cap$ cl(B)
(2)cl(A)$ \cap$ B. (3)cl(A)$\cap$ cl(B)(4 ) cl(A$\cap$B)

Prove  that no two sets are equal

(This is last best attempt. Please upvote me.
If it is wrong, show the proof. The proof in full
has not been given to date .When I tag it I put
#proofexplanation or something similar. This implies
I want explanation. I am having a hell of time learning
it. I might not be crying, but am close to bashing my
head against the wall in agony

In my previous attempt I thought I was dealing with dense
sets since Q and R/Q are dense in R so that is how I did
it..)

Try

0.Q and R$\setminus$ Q are the same under closure in R

1.Since A $\cap B =\emptyset$ ,cl(A $\cap B)=\emptyset$ Furthermore A and B are dense in R.

2.(From :Interior, closure and boundary of the sets of all rational and irrational numbers.)

Whenever A & B are dense disjoint subsets of a topological space X we have  cl(A)=X=cl(B) by Def.of being dense. Since  B$\subset A^{c}$ and A$\subset B^{c}$
it follows cl($A^{c})=X=cl(B^{c}$)

Since we have A being  the set of rationals (Q) ,B being
the set of irrationals (R\Q) and set intersection and taking compliments we get:

(i) A$\cap$cl(B) = Q $\cap$ R=Q

(ii)cl(A)$\cap $B= R $\cap$ R\Q= R\Q

(iii)= R $\cap$ R=R

(iv) ={}

Therefore none of them are equal

答:
You copied the problem in the book badly. (Particularly, you flipped $\cap$ as $\cup$, which makes the conclusion completely wrong.)

Here is the original one (one page 78, Chapter 3):

and the bar means "closure" (Definition 3.1.9 on page 75).

You are asking 6(a):

The set in (i) is equal to $A$

The set in (ii) is equal to $B$

The set in (iii) is equal to $\mathbb{R}$

The set in (iv) is empty.

So no two of the above four sets are equal.

Notes.

You would be able to understand the answer if you read the relevant section of the book carefully.

Example 3.1.12 tells you that $\overline{\mathbb{Q}}=\mathbb{R}$. This is also mentioned in Section 3.4.

It is a good exercise for you to work out that the set of irrational numbers is also dense in $\mathbb{R}$.

It does not make sense to write $\bar{A}\bar{\cap}\bar{B}$. A correct way to type (iv) in the problem is $\overline{A\cap B}$.


问:
Self-adjoint lift of idempotent in quotient
Let $A$ be a $C^*$-algebra and $I\subseteq A$ a closed ideal. Let $a\in A$ be such that $a^2-a\in I$.

Question: Can one construct an element $b\in A$ in terms of $a$ such that $b$ is a self-adjoint lift of $[a]$?

Remark: By a lift I mean $[b]=[a]\in A/I$.

答: If $a\in A$ is such that $[a]$ is self-adjoint, then $b=\frac{a+a^*}{2}$ is self-adjoint and satisfies $[b]=\frac{[a]+[a]^*}{2}=[a]$.  Conversely, if such a $b$ exists, then clearly $[a]=[b]$ must be self-adjoint.  So assuming that $[a]$ is idempotent is not relevant; instead, such a $b$ exists iff $[a]$ is self-adjoint.


问:
f analytic in a domain contains unit disk, prove the equation of the sum of the square of coefficient modulus
$f(z)=a_{n}z^{n}$ analytic in a domain containg unit disk,prove that
$\frac{1}{2 \pi} \int_{0}^{2 \pi}|f(e^{i \theta})|^{2} d \theta= \Sigma |a_{n}|^{2} $

Thanks for the hint $|f(z)|^{2}=f \bar{f} = \Sigma a_{n}\bar{a_{m}}z^{n}\bar{z}^{m}$, but I'm still a little confused about the next steps.... I'm not sure how to get the right hand side

答:
Another hint:
$$
\int_\gamma z^n \overline{z}^m\,dz=0,\quad n\neq m
$$
where $\gamma$ is the unit circle. This can be shown easily if you use a parameterization of the circle and directly do the calculation.

Now try integration term by term.


问:
Second Fundamental Form of Torus
I want to prove that the mean curvature vector of the flat Torus $T^n=S^1\times S^1\times ...\times S^1\subset\mathbb{R}^{2n}$ is zero. But my calculations show me to the contrary that the mean curvature doesn't vanishes.

Regards

答: The usual embedding $x(s,t)=(\cos s, \sin s,\cos t,\sin t)$ does induce a flat metric. However, this does not mean that the second fundamental form vanishes. For example, the cylinder $x(s,t)=(\cos s,\sin s,t)$ in $\mathbb R^3$ is flat (i.e., has Gaussian curvature $0$) but does not have vanishing second fundamental form. Vanishing second fundamental form would mean a totally geodesic submanifold. You're asking for less ... that the mean curvature (vector) vanish. But, no, the flat torus in $\mathbb R^{2n}$ does not have zero mean curvature vector, as it is not a minimal submanifold. Now there are minimal embeddings of, for example, the torus $T^2$ in $S^3$, as opposed to $\mathbb R^4$.

ContentLicense: CC BY-SA 3.0

Title: Second Fundamental Form of Torus


问:
Angle of an ellipse point shifted up in respect to a similar at origin
I have an ellipse $(E): x = a\cdot \cos(\theta), y = b\cdot \sin(\theta)$ then I move this ellipse upwards along the y-axis (semi-minor axis) so the ellipse $E$ become an ellipse $E'$ (in green) not centered at origin $O(0,0)$ but at $O'(0,h)$ with $h < b $.

How to calculate the angle $\beta$, the new angle between the semi-axis major and point $A$ on $E'$ in terms of $\alpha$, the old angle between the semi-axis major and $A$ when $E'$ was centered at $O$.?

I tried something with eccentric angle but the shifting here is not along the foci axis.

I tried and found a relation of $\alpha$ in terms of $\beta$ but not my case thus the formula was irreversible.
Any tips? thanks in advanve.

Please see the illustration .

答:
If you just need to compute the answer for any given input,
consider the equation of the line $OA,$ which is $$y = x\tan\alpha.\tag1$$
You also have an equation of the ellipse in the form
$$ \frac{x^2}{a^2} + \frac{(y - h)^2}{b^2} = 1. \tag2 $$
Taking these as simultaneous equations, solve for $x$ and $y$.
There are two solutions, one of which is the coordinates of $A$.
The coordinates of $A$ will be the ones where the $x$-coordinate has the same sign
as $\cos\alpha.$

Another approach is to transform the ellipse to a circle.
Multiply the $y$ coordinate of every point by $\frac ba.$
This maps the ellipse to a circle of radius $a$ with its center at $O'' = (0,h')$
where $h' = \frac ab h$
and maps the point $A = (x_A,y_A)$ to $A' = (x_A,y'_A)$ where $y'_A = \frac ab y_A.$
Instead of angles $\alpha$ and $\beta$ you have angles $\alpha'$ and $\beta'$
where $\tan\alpha' = \frac ab \tan\alpha$ and $\tan\beta' = \frac ab \tan\beta.$
The problem is simpler for the circle than the ellipse, and you can work out
$\beta'$ in terms of $\alpha'$ with some trigonometry,
though if you then express $\alpha'$ in terms of $\alpha$ and $\beta$ in terms of $\beta'$ in order to have a formula for $\beta$ in terms of $\alpha,$
the formula gets quite long and complicated.

If I were programming this I think I would prefer solving Equations $(1)$ and $(2)$.


问:
Proving two versions of Pasch's Axiom are equivalent
I am trying to show that these two versions of Pasch's axiom are the same.

A1. If a line enters a triangle at a vertex, then the line intersects the opposite side.

A2. If a line enters a triangle at a side without intersecting the opposite vertex, then the line intersects one of the other two sides.

But, I can figure out how to use either axiom for proving the other axiom.

答: I'll just do one direction. Assume A1, we work to show A2. Let $ABC$ be a triangle, and assume we have a line $l$ entering triangle $ABC$ at side $AB$ (without loss of generality) at a point $D$ on $AB$ and not intersecting $C$. Then $l$ enters either triangle $DAC$ or $DBC$ at vertex $D$, and then apply A1 to whichever the case is (e.g. if $l$ enters $DAC$ then A1 says it intersects $AC$.


问:
Connection Between Étale Fundamental Group and Topological Fundamental Group
I have some problems with the understanding of the theory of fundamental groups. I hope everything I write down here is understandable and correctly written.

Let $X$ be a (smooth) complex algebraic variety. Realize the variety as a complex manifold $X(\mathbb{C})$. This question is concerned about the connection between the étale fundamental group $$\pi_1^{\text{ét}}(X,x)$$ and the topological fundamental group $$\pi_1(X(\mathbb{C}),x).$$
More precisely, I would like to understand how I can describe the étale fundamental group by using the classical fundamental group - which should be, if I'm not mistaken;
$$
\pi_1^{\text{ét}}(X,x)=\widehat{\pi_1(X(\mathbb{C}),x)},
$$
where the right hand side denotes the profinite completion of the topological fundamental group.

There are a few facts I have proven. Let me begin by mentioning them.

Proven Results

Statement 1. If we let $X$ denote a topological space with base point $x$, $\mathbf{Cov}(X)$ denote the category of covering spaces over $X$, and $\pi_1(X,x)-\mathbf{Sets}$ be the category of sets equipped by a group action by the fundamental group. Then we have the following equivalence of categories
$$
\mathbf{Cov}(X)\simeq \pi_1(X,x)-\mathbf{Sets}
$$
Also, I have proven the following two statements:

Statement 2. If $\mathbf{G}$ is a group considered as a category and $P_{\mathbf{G}}:\mathbf{G}-\mathbf{Sets}\to\mathbf{Sets}$ is the forgetful functor, then
$$
\mathbf{G}\cong \operatorname{Aut}(P_{\mathbf{G}}),
$$
where $\operatorname{Aut}(P_{\mathbf{G}})$ is the automorphism group of $P_{\mathbf{G}}$.

Statement 3. If $\mathbf{\hat{G}}$ denotes the profinite completion of $\mathbf{G}$ and $\hat{P}_{\mathbf{G}}:\mathbf{G}-\mathbf{FinSets}\to\mathbf{FinSets}$ is the forgetful functor from the category of profinite completion actions on finite sets to the category of finite sets, then
$$
\mathbf{\hat{G}}\cong\operatorname{Aut}(\hat{P}_{\mathbf{G}}).
$$
Lastly, we also have the following theorem called The Riemann Existence Theorem, which can be found in Lectures on Étale Cohomology by James Milne (see page 28):

Riemann Existence Theorem. Let $X$ be a nonsingular variety over $\mathbb{C}$. The functor sending a finite étale covering $(Y,\pi)$ of $X$ to the finite covering space $(Y(\mathbb{C}),\pi)$ of $X(\mathbb{C})$ is an equivalence of categories.

A Quest to Prove the Isomorphism

I had a discussion with a professor (not too long ago) over Zoom, he told me that the above is enough to find the connection between the fundamental groups. But I am not exactly sure how.

In the notes, this is how professor Milne gave the proof

From the Riemann Existence Theorem it follows that the étale universal
covering space $\widetilde{X}=(X_i)_{i\in I}$ of $X$ has the property
that every finite topological covering space of $X(\mathbb{C})$ is a
quotient of some $X_i(\mathbb{C})$. Let $x$ any element of
$X(\mathbb{C})$. Then

$$\pi_1^{\text{ét}}(X,x)=\varprojlim_i\operatorname{Aut}_X(X_i)=\varprojlim_i\operatorname{Aut}_{X(\mathbb{C})}(X_i(\mathbb{C})) =\widehat{\pi_1(X(\mathbb{C}),x)}.$$

I cannot see how anything I have proven comes into the picture in the above quotation.

Questions

Question 1. To use the results I have proven I guess I would like to put $\mathbf{G}=\pi_1(X(\mathbb{C}),x)$ in Statement 2 and Statement 3? This gives us
$$
\pi_1(X(\mathbb{C}),x)\cong\operatorname{Aut}(P_{\mathbf{G}})
\\
\widehat{\pi_1(X(\mathbb{C}),x)}\cong \operatorname{Aut}(\hat{P}_{\mathbf{G}})
$$
where do I go from where?

Question 2. To make The Riemann Existence Theorem more compact, let $\mathbf{FinÉtCov}(X)$ denote the category of finite étale coverings of $X$ and $\mathbf{FinCov}(X(\mathbb{C}))$ denote the finite covers of the topological space. Then the theorem says
$$
(1)\text{ }\text{ }\text{ }\mathbf{FinÉtCov}(X)\simeq\mathbf{FinCov}(X(\mathbb{C}))
$$
As given in Statement 1, I have the following equivalence
$$(2)\text{ }\text{ }\text{ }\mathbf{Cov}(X(\mathbb{C}))\simeq \pi_1(X(\mathbb{C}),x)-\mathbf{Sets}.$$
The two equivalence just given are not exactly the same, but "almost". Can I use the above equivalence somehow?

I am quite sure I cannot do it like this, but something similar to
$$\mathbf{FinCov}(X(\mathbb{C}))\simeq\pi_1(X(\mathbb{C}),x)-\mathbf{Sets}?$$
I don't think I can use the above because $(1)$ is about finite covers, when $(2)$ is not.

After having done the above, then I could replace the fundamental group, with the automorphism group of the forgetful functor. But this seems just like it is making things much more complicated for me. I'm just trying to use all of the results I have proved.

´Something which bothers me is that I want to prove that two groups are isomorphic. But right now, everything I am doing is at a category-theoretical level - which makes me confused. I want to do things at a group-theoretic level.

Question 3. At the end of the day, I would like someone to help me put everything together. Maybe someone can give me a guiding hand and tell me how this relates to what Milne did? Maybe there are more work to do before I can prove the isomorphism?

I would be really happy if someone could help me get a better understanding of this problem.

Best wishes,

Joel

答:
At some point in the argument you need to introduce a definition of the etale fundamental group. Here is one that will work fine: $\pi_1^{et}(X, x)$ is the automorphism group of the fiber functor

$$P^{et} : \text{FinEtCov}(X) \to \text{Set}$$

given by taking the fiber at the point $x$. The Riemann existence theorem produces an equivalence of categories $\text{FinEtCov}(X) \cong \text{FinCov}(X(\mathbb{C}))$ (this is the "actual hard part" of the argument and everything else is "just abstract nonsense"); importantly, this equivalence is compatible with taking fibers in the sense that there is also an analytic fiber functor

$$P^{an} : \text{FinCov}(X(\mathbb{C})) \to \text{Set}$$

given by taking the fiber at the point $x$ and the equivalence sends the etale fiber functor to the analytic fiber functor. So we get

$$\pi_1^{et}(X, x) \cong \text{Aut}(P^{et}) \cong \text{Aut}(P^{an}).$$

Now how do we compute $\text{Aut}(P^{an})$? We start from the fact that the category of covering spaces of $X(\mathbb{C})$ is equivalent to the category of actions of the topological fundamental group $\pi_1(X(\mathbb{C}), x)$ on sets. It follows that $\text{FinCov}(X(\mathbb{C}))$ is equivalent to the category of actions of $\pi_1(X(\mathbb{C}), x)$ on finite sets. Now we need the following:

Theorem: Let $G$ be a group and let $P : G\text{-FinSet} \to \text{Set}$ be the forgetful functor from the category of finite $G$-sets to sets. Then $\text{Aut}(P)$ is naturally isomorphic to the profinite completion $\widehat{G}$.

Proof. Here is a sketch. If we replaced "finite sets" with "sets" above, the automorphism group of the forgetful functor would just be $G$, and this would be because the forgetful functor in this case is representable by $G$ itself (being acted on by left multiplication).

If $G$ is infinite the forgetful functor is no longer representable. However, it can be written as a filtered colimit of representable functors; it is the filtered colimit over the functors represented by the finite quotients $G/G_i$ of $G$ (this is the key step). Together with the Yoneda lemma, it follows that

$$\text{Hom}(P, P) \cong \text{Hom}(\text{colim}_i G/G_i, P) \cong \lim_i \text{Hom}(G/G_i, P) \cong \lim_i P(G/G_i) \cong \lim_i G/G_i$$

and although this is only an identification as sets, as written, you can show that it is also an identification as groups. $\Box$

So we get that $\pi_1^{et}(X, x)$ is the profinite completion $\widehat{\pi_1(X(\mathbb{C}), x)}$ as desired. This isomorphism is, properly understood, exactly equivalent to the Riemann existence theorem, once you internalize the idea that a group $G$ is completely determined by the pair consisting of the category of $G$-sets and its forgetful functor to sets, and similarly a profinite group $G$ is completely determined by the pair consisting of the category of continuous finite $G$-sets and its forgetful functor to (finite) sets.

You can see more examples of these kinds of profinite gadgets appearing as endomorphisms of forgetful functors in my blog post Operations, pro-objects, and Grothendieck's Galois theory which may or may not be helpful reading.


问:
To find $a,b,c$ as the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , is atmost $64$ in a direction parallel to $z$-axis?
How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $\nabla f(1,2,-1) . (1,2,\alpha)=64\sqrt{1^2+2^2+\alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance . </p>

答:
Find grad of $f$ and put $(1,2,-1)$ in it.
You will get $(4a+3c) i +(4a-b)j +(2b-2c)k$.
Since parallel to $y$ axis,
$4a+3c=0$ and $2b-2c=0$, From here we get $b=c=-4a/3$.
Put the value of $b$ in $4a-b=64$ as given in question, and
you will get $a=12, b=c=-16$. The answer will be $-20$.


问:
Infinite binary having infinite decimal
Can a finite decimal has an infinite binary representation? I have come to a conclusion that it may not be possible based on what I have read from the following:

What cannot happen is that the decimal is infinite and the binary is finite.

https://www.quora.com/Is-it-true-that-some-binary-numbers-have-infinite-digits-after-the-point-or-will-they-ever-end-for-example-decimal-0-1-to-binary

Comparing infinite binary fractions to infinite decimal fractions

number with finite binary representation and infinite decimal representation

Am I right or wrong. Any reasons or proofs to facilitate the comprehension will be helpful

Title: Infinite binary having infinite decimal

Body: $$\frac13=0.010101\ldots_{\text{two}}=0.333\ldots_{\text{ten}}$$


问:
Help evaluating an interesting series.
I'm trying to evaluate the sum
$$
s = 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\cdots.
$$

The hint I was given was $\frac{1}{2n+1} = \int_0^1 x^{2n} \, dx$. I'm familiar with the trick of writing the $n$th term of series with a definite integral and then interchanging the order of integral and sum, but the signs on the terms are making this difficult for me. If this were a normal alternating series, we could do something like
\begin{align*}
\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1} &= \sum_{n=0}^\infty\int_0^1 (-1)^n x^{2n} \, dx\\
&= \int_0^1 \left(\sum_{n=0}^{\infty} (-x^2)^{n}\right)dx\\
&= \int_0^1 \frac{1}{1+x^2} \, dx = \arctan(1)-\arctan(0) = \frac{\pi}{4}.
\end{align*}

I can't however seem to proceed with the strange alternating series. I know that $(-1)^{(n^2+n+2)/2}$ gives the right pattern of $++--++--\dots$, but that doesn't combine well with the $x^{2n}$ term in the integrand after exchanging sum and integral.

Any thoughts?

Edit: If I use the imaginary unit $i$, then I can write a related sum
$$
\tilde{s} = 1+\frac{1}{3}i-\frac{1}{5}-\frac{1}{7}i+\frac{1}{9}+\frac{1}{11}\cdots.
$$
Using the above trick gives
$$
\sum_{n=0}^{\infty}(i)^{n}\frac{1}{2n+1} = \sum_{n=0}^{\infty}\int_{0}^{1}(i)^nx^{2n}dx = \int_{0}^{1} \sum_{n=0}^{\infty}(ix^2)^n =\int_{0}^{1}\frac{1}{1-ix^2}dx
$$
Which I imagine I can do with some type of partial fraction decomposition.

My hope is that from this new sum I can recover my old sum by adding real and imaginary parts. My only concern is that this would essentially be a rearrangement of the series (adding the even indexed terms together and then the odd indexed terms separately) and this series doesn't converge absolutely so I'm worried that this isn't rigorously justified.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
You can write your series as
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 0}^{\infty}\pars{-1}^{n}
\pars{{1 \over 4n + 1} + {1 \over 4n + 3}}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}\pars{-1}^{n}
\int_{0}^{1}\pars{x^{4n} + x^{4n + 2}}\dd x
\\[5mm] = &\
\int_{0}^{1}\pars{1 + x^{2}}
\sum_{n = 0}^{\infty}\pars{-x^{4}}^{n}\,\,\dd x
\\[5mm] = &\
\int_{0}^{1}\pars{1 + x^{2}}{1 \over 1 + x^{4}}\,\dd x
\\[5mm] = &\
\int_{0}^{1}{1 - x^{4} + x^{2} - x^{6} \over
1 - x^{8}}\,\dd x
\\[5mm] = &\
{1 \over 8}\int_{0}^{1}{x^{-7/8} - x^{-3/8} + x^{-5/8} - x^{-1/8}\,\, \over 1 - x}\,\dd x
\\[5mm] = &\
{1 \over 8}\bracks{-\Psi\pars{1 \over 8} +
\Psi\pars{5 \over 8} - \Psi\pars{3 \over 8} + \Psi\pars{7 \over 8}}
\\[5mm] = &\
{1 \over 8}\braces{%
\bracks{\Psi\pars{7 \over 8} - \Psi\pars{1 \over 8}} +
\bracks{\Psi\pars{5 \over 8} - \Psi\pars{3 \over 8}}}
\\[5mm] = &\
{1 \over 8}\ \underbrace{\bracks{\pi\cot\pars{\pi \over 8} + \pi\cot\pars{3\pi \over 8}}}_{\ds{2\pi\root{2}}}
\\[5mm] = &\
\bbx{{\root{2} \over 4}\,\pi} \approx 1.1107 \\ &
\end{align}

Title: Help evaluating an interesting series.


问:
Why is the height of persons in a country a random variable of the continuous type?
If we define a measurable function on a probability space as a function that maps from the sample space to the real numbers, and treat all the people in a Country as the sample space, shouldn't the image of the random variable be a finite set? I don't understand why we would say that the random variable is countable.</p>

答:
Assuming data are normal is a practical convenience. A discrete random variable that has very many possible values requires a long list of possible values and associated probabilities.

In the US, heights of people are usually rounded to the nearest inch. (A  person's height can fluctuate by about half an inch during the day--usually taller in the AM shorter in the PM. So rounding to the nearest inch doesn't lose crucial information.)

Almost all US adults are between 50 and 90 inches tall, so in using a discrete random variable there would be about 40 different values to list along with probabilities of each.
Saying that heights are approximately normally distributed with mean 68 and standard deviation 3.5 allows us to get by with 2 parameters.

Theoretically, a normal distribution takes all values on the real line, but an interval $(56,89)$ with $99.9\%$ of
the probability is manageably short--even if it theoretically contains uncountably many values. Also one ignores the fact that this normal distribution allows a
very tiny probability of (practically impossible)
negative heights.

According to this normal distribution, what is the probability that a randomly chosen person is 68in
tall? The theoretical answer is $0.$ (if you have uncountably many points you can't assign positive probabilities to any of them.) If you mean $68.00000$ inches tall, then the practical answer is also essentially $0.$ (Who could measure that?) It is more useful to ask for
the probability that person is between
$67.5$ and $68.5$ (so that the height would be rounded to "68". Then the answer is $0.1136.$

If pilots for
a small military airplane need to be shorter than
65 inches, then almost 20% of the population (including lots of women) are still potential
candidates as pilots. All such computations are
understood to be approximate. Very few, if any,
statisticians believe any natural phenomenon
is exactly normally distributed.

Sometimes, the gulf between measure theoretic probability and applied statistics can be very wide.

Computations using R software, where pbinom is
a continuous normal cumulative distribution function, and qnorm is its inverse (called
a quantile function),

```<code>qnorm(c(.0005,.9995), 68, 3.5)
[1] 56.48316 79.51684

pnorm(0, 68, 3.5)
[1] 2.212535e-84

diff( pnorm(c(67.5,68.5), 68, 3.5) )
[1] 0.113597

pnorm(65, 68, 3.5)
[1] 0.195683
</code>```

Body:
Assuming data are normal is a practical convenience. A discrete random variable that has very many possible values requires a long list of possible values and associated probabilities.

In the US, heights of people are usually rounded to the nearest inch. (A  person's height can fluctuate by about half an inch during the day--usually taller in the AM shorter in the PM. So rounding to the nearest inch doesn't lose crucial information.)

Almost all US adults are between 50 and 90 inches tall, so in using a discrete random variable there would be about 40 different values to list along with probabilities of each.
Saying that heights are approximately normally distributed with mean 68 and standard deviation 3.5 allows us to get by with 2 parameters.

Theoretically, a normal distribution takes all values on the real line, but an interval $(56,89)$ with $99.9\%$ of
the probability is manageably short--even if it theoretically contains uncountably many values. Also one ignores the fact that this normal distribution allows a
very tiny probability of (practically impossible)
negative heights.

According to this normal distribution, what is the probability that a randomly chosen person is 68in
tall? The theoretical answer is $0.$ (if you have uncountably many points you can't assign positive probabilities to any of them.) If you mean $68.00000$ inches tall, then the practical answer is also essentially $0.$ (Who could measure that?) It is more useful to ask for
the probability that person is between
$67.5$ and $68.5$ (so that the height would be rounded to "68". Then the answer is $0.1136.$

If pilots for
a small military airplane need to be shorter than
65 inches, then almost 20% of the population (including lots of women) are still potential
candidates as pilots. All such computations are
understood to be approximate. Very few, if any,
statisticians believe any natural phenomenon
is exactly normally distributed.

Sometimes, the gulf between measure theoretic probability and applied statistics can be very wide.

Computations using R software, where pbinom is
a continuous normal cumulative distribution function, and qnorm is its inverse (called
a quantile function),

qnorm(c(.0005,.9995), 68, 3.5)
[1] 56.48316 79.51684

pnorm(0, 68, 3.5)
[1] 2.212535e-84

diff( pnorm(c(67.5,68.5), 68, 3.5) )
[1] 0.113597

pnorm(65, 68, 3.5)
[1] 0.195683


问:
Finite Incidence Geometry Questions
Definition of finite incidence geometry given: Consists of a finite set $P$ of points and a set of nonempty subsets of $P$ called lines, that satisfy the axioms:

(F1) Two points determine exactly one line joining them;

(F2) There are at least two lines and every line contains at least two points;

(F3) There is a constant $n$ such that every line contains $n$ points.

Question 1: Prove that every finite incidence geometry is a balanced incomplete block design.

Question 2: Prove that in a finite incidence geometry, there is a number $t$ such that for any point $p$ and any line $l$ not through $p$ there are exactly $t$ lines containing $p$ that have no common point with $l$.

Update on my attempt:

We want to establish parameters $(v,b,r,k,\lambda)$. (F3) gives us $k=n$ and I think (F1) gives us $\lambda = 1$. Now, if we select any line $l$ and suppose that it contains points $a_1,\dots,a_n$ and then select a point $p$ that is not on $l$, it follows that $p$ lies on exactly $n$ lines that meet $l$, namely $pa_1,\dots,pa_n$. No lines are parallel to $l$ be the axioms (I'm not sure if this is true!) Therefore, these are all of the lines that $p$ lies on $\implies$ $r=n$. So then using the BIBD formulas, we can solve for $v$ and $b$ to arrive at a $(n^2-n+1, n^2-n+1, n, n, 1)$-design.

For problem 2, I had two ideas:  I see that t=0 for a triangle, but it also equals zero for the Fano plane (which is an example of an incidence geometry). So I suspect that t=0 for any incidence geometry (I'm not sure about that!). I'm struggling to show this. My approach was to start with n^2-n+1 lines (blocks) and subtract off r (the lines that contain p). Also, we subtract off L. Leaving us with n^2-2n lines that do not contain p and L is not one of them. Somehow, I have subtract off everything, to give me t=0.

My second idea was to say that if t>0, then this implies that there exists a line parallel to L, which implies that r = n+1 (for one parallel line), which contradicts our findings from 4ii). I can't say this for certain, because I'm not sure if incidence geometries can have parallel lines!

As always, I appreciate any thoughts, ideas, and assistance with regards to either or both problems!

答: From the axioms for an incidence geometry, show that two lines l and l ′, l ≠ l ′, either share a point in common or no points in common.</p>

Title: Finite Incidence Geometry Questions


问:
Why is the Poisson Approximation to the Binomial Distribution Useful?
My textbook says that if $n$ (number of trials / independent Bernoulli random variables) is very large and $p$ (probability of success per trial / Bernoulli random variable) is very close to $0$ or $1$, we can approximate the mass associated with the Binomial distribution, very well with the Poisson distribution.

My question is why would you use this? Why would you ever want to approximate the mass of a Binomial random variable $X$ using Poisson, when you can get the exact mass using the Binomial distribution. No approximation needed.

Can someone explain this to me?

Thank You

答:
The short answer is that the Poisson approximation is faster and easier to compute and reason about, and among other things tells you approximately how big the exact answer is.

Here's a simple example: suppose you're trying to get something to happen in a video game that is rare; maybe it happens 1% of the time you do something, independently. You'd like to know how likely it is to happen at least once if you try, say, 100 times. Here we have $p = \frac{1}{100}, n = 100$ and so the binomial distribution gives us an exact answer, namely

$$1 - \left( 1 - \frac{1}{100} \right)^{100}.$$

But how big is this? Do you know off the top of your head? Is it, say, bigger or less than 50%?

The Poisson approximation answers this question quickly and easily: in this special case, it amounts to the approximation

$$\left( 1 - \frac{1}{100} \right)^{100} \approx e^{-1} \approx 0.368 \dots $$

which gives

$$1 - \left( 1 - \frac{1}{100} \right)^{100} \approx 1 - e^{-1} \approx 0.632 \dots $$

so we get that the odds are about 63% that we'll succeed at least once, which is bigger than 50% but maybe smaller than you might hope.

We learn something else too: the Poisson approximation tells us more generally that the odds of success are approximately a function of the product $np = \lambda$ only (which is the expected number of successes), so that e.g. if we had $p = \frac{1}{1000}$ and $n = 1000$ the answer would still be about 63%. This is valuable information and not entirely obvious from the exact answer, and knowing it saves you from having to recompute a bunch of binomials.

Sometimes $n$ can get large enough that it would actually be infeasible to calculate the exact binomial answer. For example, suppose $n = 10^{25}, p = 10^{-25}$; numbers this big regularly appear in physics or chemistry since  Avogadro's number is so large. I can confidently say that the answer is still about 63% even though it is no longer feasible to exactly calculate $(1 - p)^n$ (just try it!). The funny thing here is that the larger $n$ gets the harder it becomes to exactly compute the binomials, but the more accurate the Poisson approximation gets; for numbers this large it is for all intents and purposes basically exact.


答:
One of the common reasons to use a Poisson distribution is when you don't really have a good guess at what $n$ is exactly (though you're sure it's large) - to be precise, a Poisson distribution can be described as follows:

The Poisson distribution of rate $\lambda$ is the limit of the binomial distributions with $n$ trials and an expectation of $\lambda$ successes.

This is most relevant when you want to measure an idea of events occurring independently from some sort of continuous source of independent possibilities. Maybe you never quite get to the true limit in reality, but things like:

How many neutrinos do I detect in an hour in my giant pool of water?

is exactly the sort of scenario where there's not much better than a Poisson distribution - we could be talking about a billion billion possibilities to detect something, but with many fewer successes - and we don't know exactly how many to expect anyways. Similarly, other questions such as "how many rain drops fall on a patch of land" or even "how many customer respond to a billboard" all fall into this case where, well, the intuition is pretty much that there's so many chances for success that we may as well model it as a continuous process rather than as a discrete one for some large $n$.

Otherwise said: the Poisson distribution does have some abstract interpretation due to the limit and there are scenarios where that abstraction would seem to be an appropriate model. It's not a question of approximating something we know the exact answer for, so much as modeling something where there was no exact answer to start with.

If you want more pure math reasons, the Poisson distribution has nice properties - for instance, if you add two independent Poisson distributed random variables together, you get another Poisson variable - which is not true of binomial distributions - which is related to the fact that the generating function $\mathbb E[X^k]$ where $k$ is distributed by the Poisson distribution is just $e^{\lambda (X - 1)}$, which is about as nice as you can expected a generating function to be.

You can also check that the convergence of binomial distributions to the Poisson distribution is quite fast; the probability that a binomial distribution equals $k$ is
$$\frac{n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}{k!}\cdot \left(\frac{\lambda}{n}\right)^k\cdot \left(1-\frac{\lambda}n\right)^{n-k}$$
where the Poisson distribution is
$$\frac{\lambda^k}{k!}\cdot \left(1-\frac{\lambda}n\right)^n.$$
The difference of these things decreases in proportion to $1/n$ for any fixed $k$ - which means that this is actually a good approximation for very large $n$.

Title: Why is the Poisson Approximation to the Binomial Distribution Useful?

Body:
My textbook says that if $n$ (number of trials / independent Bernoulli random variables) is very large and $p$ (probability of success per trial / Bernoulli random variable) is very close to $0$ or $1$, we can approximate the mass associated with the Binomial distribution, very well with the Poisson distribution.

My question is why would you use this? Why would you ever want to approximate the mass of a Binomial random variable $X$ using Poisson, when you can get the exact mass using the Binomial distribution. No approximation needed.

Can someone explain this to me?

Thank You


问:
How many ways to colour a tetrahedron with monochromatic triangles.
I'm trying to find how many different ways there are to colour the edges of a regular tetrahedron with n colours such that there are no monochromatic triangles.

Certainly for one triangle there must be n choose 3 ways but I'm not quite sure how to generalise this to a tetrahedron.

Any help would be much appreciated!

答:
Consider the three edges $e_1$, $e_2$, $e_3$ emanating from the top vertex $v_0$.

(i) You can color $e_1$, $e_2$, $e_3$ using three different colors in $n(n-1)(n-2)$ ways. For each of these colorings you can color the  three bottom edges $e_4$, $e_5$, $e_6$ arbitrarily, but not all of them equal. This gives $n(n-1)(n-2)(n^3-n)$ admissible colorings of type (i).

(ii) You can color $e_1$, $e_2$, $e_3$ using two different colors in $3n(n-1)$ ways. The bottom edges then can be colored in $n^2(n-1)-(n-1)$ ways, where the $-(n-1)$ term again discounts the monochromatic bottom triangles. This gives $3 n(n-1)(n^2-1)(n-1)$ admissible colorings of  type (ii).

(iii) You can color $e_1$, $e_2$, $e_3$ using a single color in $n$ ways. The bottom edges then can be colored in $(n-1)^3-(n-1)$ admissible ways. This gives $n(n-1)\bigl((n-1)^2-1\bigr)$ admissible colorings of type (iii).

Adding it all up we obtain $N:=n^6-4n^4+6n^2-3n$ admissible colorings.


答:
We have
$$\frac{1}{2} = .5 = (.1)_2$$
$$\frac{1}{3} = .333\ldots = (.010101\ldots)_2$$
$$\frac{1}{5} = .2 = (.001100110011\ldots)_2$$

So it can happen that

both expansions are finite

both are infinite

the decimal is finite and the binary is infinite.

What cannot happen is that the decimal is infinite and the binary is finite.
May be that's what you are trying to prove.

In other words we have the implication:

Decimal infinite $\Rightarrow$ binary infinite.

Or equivalently

Binary finite $\Rightarrow$ decimal finite.

To prove this observe that if $x = (.b_1b_2\ldots b_n)_2$ then
$$x = \frac{m}{2^n} = \frac{m\, 5^n}{10^n}$$ which has a finite decimal expansion.

Title: Infinite binary having infinite decimal

Body:
Can a finite decimal has an infinite binary representation? I have come to a conclusion that it may not be possible based on what I have read from the following:

What cannot happen is that the decimal is infinite and the binary is finite.

https://www.quora.com/Is-it-true-that-some-binary-numbers-have-infinite-digits-after-the-point-or-will-they-ever-end-for-example-decimal-0-1-to-binary

Comparing infinite binary fractions to infinite decimal fractions

number with finite binary representation and infinite decimal representation

Am I right or wrong. Any reasons or proofs to facilitate the comprehension will be helpful


答:
This is basically contained in KCd's answer but it's worth saying explicitly: what almost no one will say explicitly about tensors is that they are a fundamentally relative notion, and what they are defined relative to is a "base" vector space $V$. There is in an important sense no such thing as a "tensor" in isolation; the grammar of "tensor" is "a tensor of type $(m, n)$ with respect to a vector space $V$," namely an element of the tensor product $V^{\otimes m} \otimes (V^{\ast})^{\otimes n}$. Often the vector space $V$ is understood from context and left implicit.

This can be formalized category-theoretically using the language of functors, which here amounts to making precise what people mean when they talk about tensors transforming in such-and-such a way under change of coordinates - they mean coordinates of the base vector space $V$.

It is true that every tensor product of vector spaces is another vector space, and hence every tensor is technically a tensor of type $(1, 0)$ (in other words, a vector) with respect to this tensor product of vector spaces. But this is irrelevant; when we talk about tensors we have in mind a particular choice of base vector space $V$, and in that context "vector" means an element of $V$, as KCd says.

(So, in other words, the word "vector" is overloaded; it refers to several related but actually somewhat different concepts. This sort of thing is unfortunately common.)

Title: Are tensors vectors?

Body:
Take the definition of a tensor as a multilinear map $T: (V)^n \times (V^*)^m \rightarrow \mathbb R$. One may then define the addition of two tensors as the sum of their outputs on their input spaces (assuming of course the tensors have the same rank), and scalar multiplication as the output scaled by the factor used (if $T$ maps its input to $b$, then $aT$ maps its input to $ab$).

As one can easily check, tensors would then satisfy all of the vector space axioms, so then a tensor could be considered a specific type of vector. This bothered me, though, because vectors are often defined as objects with magnitude and direction in less formal contexts, and tensors can often be informally considered to have many different directions.

This leads me to the writing of this question. What distinguishes a tensor from a classical, "ordinary" vector? How can one derive the definition of a vector as something with magnitude and direction from the definition of a vector space, and where does this definition break down for tensors (if anywhere)?


问:
Proof of Basis Representation Theorem; How do we know b_k(1) = 1?
I am currently making my way through George E. Andrews' Number Theory and am having some trouble with his proof for the basis representation theorem.

I follow the proof up until he applies Corollary 1.1, which states that "If $m$ and $n$ are positive integers and if $m>1$, then $n<m^n$." To be clear, I understand the corollary and I am pretty sure I know how/why it applies in the proof. However it is followed by something that is never explained clearly.

To start, let me introduce the proof. I will more or less copy it directly from the text to reduce potential for miscommunication.

Theorem 1.3 (Basis Representation Theorem): Let k be any integer larger than 1. Then, for each positive integer n, there exists a representation

$$n=a_0k^s+a_1k^{s-1}+...+a_s, \ \ \ \ \ \ (1.2.1)$$

where $a_0 \neq0$, and where each $a_i$ is nonnegative and less than k. Furthermore, this representation of n is unique; it is called the representation of n to the base k.

Remark: For each base $k$, we can also represent $0$ by letting all the $a_i$ be equal to $0$.

Proof: Let $b_k(n)$ denote the number of representations of $n$ to the base $k$. We must show that $b_k(n)$ always equals $1$.

If some of the coefficients $a_i$ in a particular representation of $n$ are equal to zero, they may be excluded without affecting representation. Therefore,

$$n=a_0k^s+a_1k^{s-1}+...+a_{s-t}k^t,$$

where $a_0\neq0$ and $a_{s-t}\neq0$. Then,

$$n-1=a_0k^s+a_1
k^{s-1}+...+a_{s-t}k^t-1$$
$$=a_0k^s+a_1k^{s-1}+...+(a_{s-t}-1)k^t+k^t-1$$
$$=a_0k^s+a_1k^{s-1}+...+(a_{s-t}-1)k^t+\sum_{j=0}^{t-1}(k-1)k^j,$$

by virtue of Theorem 1.2, which states $\frac{x^n-1}{x-1}=\sum_{j=0}^{n-1}x^j$. Here $x=k$ and $n=t$. From this we see that for each representation of $n$ to the base $k$, we can find a representation of $n-1$. If $n$ has another representation to the base $k$, the same procedure will yield a new representation of $n-1$. Consequently,

$$b_k(n)≤b_k(n-1),\ \ \ \ \ \ (1.2.2)$$

This inequality implies the following inequalities:

$$b_k(n+2)≤b_k(n+1)≤b_k(n),$$
$$b_k(n+3)≤b_k(n+2)≤b_k(n+1)≤b_k(n),$$

and, in general, if $m≥n+4$,

$$b_k(m)≤b_k(m-1)≤b_k(m-2)≤...≤b_k(n+1)≤b_k(n).$$

My problem emerges here:

Since $k^n>n$ by Corollary 1.1, and since $k^n$ clearly has at least one representation (namely, itself), we see that

$$1≤b_k(k^n)≤b_k(n)≤b_k(1)=1.$$

The extreme entries in this set of inequalities are ones, so that all of the interediate entries must be equal to 1. Thus $b_k(n)=1$, and Theorem 1.3 is established.

I am going to describe how I understand this.

Clearly, as $n$ increases, $b_k(n)$ must either decrease or remain the same, as shown by the inequality $b_k(m)≤b_k(m-1)≤b_k(m-2)≤...≤b_k(n+1)≤b_k(n).$ Since, $k^n > n$, we know that $b_k(k^n)≤b_k(n).$ And all of this must be greater than or equal to one, since we know $k^n$ has at least one representation. But where have we established that $b_k(n)≤b_k(1)$? Clearly $1$ also has at least one representation, but how have we shown that it does not have more? Similarly, $b_k(n)≤b_k(1)$ implies that $n ≥ 1$; where did we establish this?

答:
That $b_k(n)\le b_k(1)$ for all $n\in\Bbb Z^+$ follows from $(1.2.2)$ by a straightforward induction argument: if $b_k(n)\le b_k(n-\ell)$ for some $\ell\ge 0$, and $n-\ell>1$, then

$$b_k(n)\le b_k(n-\ell)\le b_k\big((n-\ell)-1\big)=b_k(\big(n-(\ell+1)\big)$$

by $(1.2.2)$, so by induction $b_k(n)\le b_k(1)$.

It is clear that $1$ has only one representation: the righthand side of $(1.2.1)$ is greater than $1$ if $s>0$, or if $s=0$ and $a_0>1$.


问:
Finding convex conjugate of a bounded function
The convex conjugate of a function $f:\mathcal{X}\mapsto \mathbb{R}$ is formally defined as $$f^\star\left(y\right)=\sup_{x\in\mathcal{X}}\ \left\langle x,y\right\rangle-f\left(x\right).$$

In cases where $f\left(\cdot\right)$ is bounded, the supremum can become unbounded (e.g., $\mathcal{X}=\mathbb{R}^n$). To avoid this problem, $x$ is further constrained to a bounded set as I've read in textbooks and papers.

My question is that if there's a principle on how to bound $x$, because different constraints result in different convex conjugate functions. For instance, to find convex conjugate of $f\left(x\right)=\left\Vert x\right\Vert_0=\sum_{i=1}^n 1(x_i\neq 0)$ with $\mathcal{X}=\mathbb{R}^n$, it does make a difference to choose $\left\Vert x\right\Vert_{+\infty}\leq 1$ or $\left\Vert x\right\Vert_2 \leq 1$ as the bounding constraint.

答:
The trick is to find a set/constraint that is necessary and sufficient for the supremum to be finite. You mentioned Fazel's thesis; look on p58 in her proof of Theorem 1 that you are pointing out. She shows that for $\|Z\|>1$, we can choose $Y$ such that $\phi^{**}(Z)\to\infty$. On the other hand, for $\|Z\|\leq 1$, $\phi^{**}(Z) = \|Z\|_*$, which is always finite.

There is no general method here, other than to manipulate $\phi^{**}(Z)$ to see what such condition is necessary and sufficient for the supremum to be finite.


问:
Tough Number Theory mod problem
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.

I tried using mod rules to simplify but I got a very strange answer which was divisible by 7 and did not fit the answer criteria. I noted that 15 is 1 mod 7 and even 8 is 1 mod 7, but I do not know how to go ahead from there.

EDIT: I also noted that 6 is -1 mod 7 and 30^12 is 1 mod 7. But that didn't help either.

答:
You might try a smaller exponent to start with, say $30^{2}$ or $30^{3}$.

Making out a multiplication table would be a good first step.  This also makes a lot more sense after plotting out some of the paths toward 1, for example:

$$
\begin{align}
2 * 2 * 2 = 1\\
2 * 3 * 2 * 3 = 1 \\
2 * 5 * 5 = 1 \\
\end{align}
$$

I prefer using $-2$ instead of $5$.  Thus we have instead that $2 * -2 * -2 = 1$.  If you get a $-1$, you'll have to make another $-1$ to make $1$.

I've got a method I like.

Start by laying out a 13x13 two dimensional grid.  Label your x axis 2 and y axis 3.  Ticks on the axes correspond to the exponent for that base.  The point (3,0) then would correspond to $2^3*3^0$ which is a solution.  Everything on this grid has no factors of 5.  Starting with a 2d grid and adding layers works better for counting up points than just starting with a 3d grid.

For each solution point (a,b) there is a corresponding solution point at (a+3, b).  If I remove a factor of two, I can add two factors of three to replace it.  That is, if (a,b) is a solution point, then (a-1, b+2) is a solution point.

This all makes a very regular repeating pattern.  The amount to add to the sum for any point (a, b, c) is simply a+b+c.  The sum for the $5^0$ level is 382.

When we go up to the $5^1$ level, that multiplies in a 5 mod 7, and we need an additional 3 mod 7 factor to bring the solution point back into true.  So on the $5^1$ level, each solution point is shifted y-wise by one.  For every point, we gain two in the exponent (one for the 3 and one for the 5), for 26 of the 31 points, adding $26*2$ to the total, and removing the sum for the points entirely that shift off: $12+15+18+21+24$

Continue this way.  If you're careful, you can reuse many of your sums, and it isn't too tedious.  Make sure you get all 13 levels (0-12)

Title: Tough Number Theory mod problem


问:
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that

$$\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$$

rather annoying that the powers of denominators and numerators are not the same and can't think of easy way to flatten them..

答:
Hint: Use holders inequality

$$\left(\sum_{cyc} \frac{a^3}{b+c+d}\right)(1+1+1+1)\left(\sum_{cyc} b+c+d\right)\ge {\left(\sum_{cyc} a\right)}^3$$

 Apply AM-GM inequality :$$a+b+c+d\ge 2\sqrt{(a+c)(b+d)}$$ to finish


问:
Integral cohomology. follows from GAGA?
Let $X$ be a projective variety over $\Bbb C$. One can compute the integral cohomology groups $H^p(X,\Bbb Z)$ by looking at the constant sheaf $\Bbb Z$ in the Zariski topology on $X$, but one can do the same with respect the Euclidean topology on $X$ as well.

Question: what guarantees that the two definitions will give the same cohomology groups? My initial reaction was that this should follow from GAGA, but $\Bbb Z$ is not a coherent $\mathcal O_X$-module, so not sure what to do.

I am sure this is addressed in the literature, so a reference would be fine as well.

答:
These are not at all the same. If $X$ is irreducible, then $H^q_{Zar}(X, \mathbb{Z})$ will be $0$ for $q>0$.

Proof: If $U_1$, $U_2$, ..., $U_n$ are nonempty Zariski open subsets of $X$, then $\bigcap_{i=1}^n U_i$ will be nonempty and connected. So the Cech complex of $\mathbb{Z}$ with respect to any cover with $N$ open sets will be $\mathbb{Z}^{\binom{N}{k+1}}$ in degree $k$. This is the complex which computes the cohomology of an $N$-1 dimensional simplex, and is thus acyclic.

Alternate (probably better) proof: If $X$ is irreducible and $U \subseteq V$ are nonempty open sets, then $U$ and $V$ are both connected, so the restriction map $\underline{\mathbb{Z}}(U) \longrightarrow \underline{\mathbb{Z}}(V)$ is an isomorphism, and thus the sheaf $\underline{\mathbb{Z}}$ is flasque.

If you want to do something algebraic and get the classical betti numbers as an answer, you either have to use something like etale cohomology or else something like the de Rham complex.

EDIT TO ADD: If you want to get cohomology with $\mathbb{Z}$ coefficients as an answer, you honestly have to work analytically, no algebraic method will work. Consider elliptic curves with complex multiplication by $\mathbb{Z}[\sqrt{-5}]$. Over $\mathbb{C}$, there are two isomorphism classes, with $j$-invariants $-632000 \pm 282880 \sqrt{5}$. Let $E_+$ and $E_-$ be the ones with $+$ and $-$ signs. We have $H^1(E_+, \mathbb{Z}) \cong H^1(E_-, \mathbb{Z}) \cong \mathbb{Z}^2$ as abelian groups. However, the complex multiplication structure makes them into modules for $\mathbb{Z}[\sqrt{-5}]$ and they are not isomorphic as such: $H^1(E_-, \mathbb{Z})$ is free of rank $1$, where as $H^1(E_+, \mathbb{Z})$ is isomorphic to the nonprincipal ideal $\langle 2, 1+\sqrt{-5} \rangle$.

Why is this bad news? We can choose a model $\mathcal{E}$ for $E_+$ over $\mathbb{Q}(\sqrt{5})$, with complex multiplication $\phi$ given by a morphism defined over $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$. Let $\sigma$ be a Galois symmetry of $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$ with $\sigma(\sqrt{5}) = - \sqrt{5}$. Then $\sigma(\phi)$ is an endomorphism of $\sigma(\mathcal{E})$, and $\sigma(E_+)$ is $E_-$. If we had an algebraic way of describing the functor $H^1(\ \ , \mathbb{Z})$, then we would have an isomorphism between $H^1(\mathcal{E}, \mathbb{Z})$ and $H^1(\sigma(\mathcal{E}), \mathbb{Z})$, conjugating $\phi^{\ast}$ to $\sigma(\phi^{\ast})$. (Because everything algebraic commutes with Galois symmetries.) But $\sigma(\mathcal{E})$ is $E_-$ as a complex manifold, so this contradicts that $H^1(E_-, \mathbb{Z})$ and $H^1(E_+, \mathbb{Z})$ are not isomorphic as $\mathbb{Z}[\sqrt{-5}]$ modules.

Title: Integral cohomology. follows from GAGA?


问:
The pressure of a fluid as the change of internal energy.
I never do fluid mechanics or physics so go easy on me!

I am trying to understand the description of pressure given by Villani in his book Optimal Transport Old and New. This is the discussion right at the start of Chapter 15 Otto Calculus.

We have a (probability) density $\rho$ of some fluid (say over $\mathbb{R}^3$), which has internal energy given by the functional

$$U(\rho):=\int_{\mathbb{R}^3}u(\rho(x))dx,~~~\text{of course}~u:\mathbb{R}\to \mathbb{R}. $$

Then the pressure of the fluid is defined as the rate of change of the energy with respect to the volume, for instance https://physics.stackexchange.com/questions/216342/what-is-pressure-energy#:~:text=The%20pressure%20energy%20is%20the,as%20well%20as%20the%20pressure.

Villani says the pressure is then $$p(\rho):=\rho u'(\rho)-u(\rho).~~~~~~~~~~(1)$$

Call the volume$=V$, here I guess this can loosely be defined as the Lebesgue measure of the space in which fluid is present? Then as Villani says if the density is evenly distributed $\rho=\frac{1}{V}$ then clearly

$$U(\rho)=Vu(\frac{1}{V})$$

and chain rule says $-\frac{d}{dV}U(\rho)=p(\frac{1}{V})=\rho u'(\rho)-u(\rho)$. Ok so defining pressure as $(1)$ works when the density is uniform, but what about if the density is not uniform, we still use this $(1)$ to mean pressure?

答:
Before getting to your question, note that $\rho$ is not probability density. It is just density, i.e., it is a measure of concentration of mass per volume.

Now, in general, $\rho$ can have various values in various locations in the system. Therefore, properties such as pressure ($p$) and internal energy ($u$), which depend on $\rho$, also take different values in different locations in the system. So, to answer your question: yes, equation (1) gives you local pressure.

Also note that the internal energy of the whole system ($U$) is the sum of internal energies of infinitesimal elements that make the system, hence the integral that you mentioned.


问:
Evaluate $\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$
I'm trying to evaluate the integral $$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$$
where $\chi_{[0,1]}(x)=1$ is the characteristic function, i.e. equals $1$ for $x \in [0,1]$ and $0$ otherwise. How do I evaluate this integral? So far I've done
$$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y = \int_{0}^1 \chi_{[0,1]} (x-y) \, \mathrm{d}y.$$
I've tried substitution but I don't find it yielding anything useful. The answer should be the function $$f(x)= \left\{
  \begin{array}{l l}
    x & \quad \text{if $x \in [0,1]$}\\
    2-x & \quad \text{if $x \in [1,2]$}\\
    0 & \quad \text{otherwise.}
  \end{array} \right.$$</p>

答:
You're off to a good start. Next note that
$$
\chi_{[0,1]}(x-y) = \begin{cases} 1, & \text{for $x-1 \le y \le x$} \\ 0, & \text{otherwise}\end{cases}.
$$
Draw some graphs!

Hence, if $0< x < 1$, then $\chi_{[0,1]}(x-y)$ on the whole of $[0,1]$. Thus the integral is $0$.

If $0 < x < 1$, then $\chi_{[0,1]}(x-y) = 1$ on $[0,x]$ and $=0$ on the rest of $[0,1]$. In this case, the integral will be
$$ \int_0^x 1\,dy = x. $$

Third case, if $1 < x < 2$, then $\chi_{[0,1]}(x-y) = 1$ on $[x-1,1]$ and $=0$ on the rest of $[0,1]$, so the integral will be
$$ \int_{x-1}^1 1\,dy = 2-x.$$

Finally, if $x > 0$, then $\chi_{[0,1]}(x-y) = 0$ on $[0,1]$ again.


问:
Maximizing the area of a cyclic pentagon made up of triangle centres.
Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA \le 90^\circ$, $BC=1$, and $AC \ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

This is problem 25 from the 2011 AMC12A. My friend sent me the following solution:

Angle formulas give that $BCOIH$ is cyclic. Further $I$ and $H$ are both on the arc to the "left" of $O$, with $H$ further "left" than $I$. To maximize the area, we want to distribute $I$ and $H$ evenly over arc $BO$. This gives that $\angle{IBC}=\angle{IBO}+\angle{OBC}=10+30=40$. Multiplying by two gives that the answer is $80$.

I agree with this solution up to the point where they claim they want to distribute evenly abut BO. Why not distribute evenly over the arc BC, doesn't this make the area even greater? If anyone could explain why we would only distribute on arc BO that would be great.

答:
Begin with a circle with center $O$, and the three triangle points $A$, $B$, and $C$ have to lie on the circle.  How are you going to get the incenter point $I$ on the arc $BC$?  The incenter is formed by the bisectors of the angles of the triangle, and thus you would have to have a degenerate null triangle to have the bisector ray moving along one of the triangle edges.  And if you mean the circular arc $BC$, that doesn't work either, since the incenter will always be in the triangle's interior.

Could you give an example triangle to consider that illustrates better what you have in mind?


问:
Proving $\sqrt{2}$ is a real number when the set is bounded from below.
My professor proved there exists a real number $\alpha\in\mathbb{R}$ such that $\alpha^2=2$. He proved the case where the set is bounded from above, namely $S=\{s\in\mathbb{R}:s^2<2\}$. He said a good exercise (not for hw/extra credit/assignment, i.e., for fun)
would be to try to prove it when it is bounded from below, so here is my  attempt but I get stuck very fast.

Proof:

Consider $S=\{s\in\mathbb{R}:s^2>2\}$, for any $x\in\mathbb{R}$, $x^2\geq 0$, thus $S\not=\emptyset$ and is a nonnegative set, so it has a lower bound. By the Axiom of Completeness, there exists an $\alpha=\inf S$, we claim $\alpha^2=2$. We will proceed by contradiction by exhausting the cases $\alpha^2>2$ and $\alpha^2 <2$.

(If my logic is correct showing $\alpha^2>2$ contradicts $\alpha$ being a lower bound, and $\alpha^2<2$ contradicts $\alpha$ being the infimum.)

But this is where I get lost, I am not sure if for $\alpha^2>2$ do I look for an element in $S$ smaller than $\alpha$ namely ($\alpha -\frac{1}{n}$) that is larger than 2? As that would show $\alpha$ is not a lower bound, a contradiction to our original assumption.

What I mean if my above assumption is correct am I looking for

$$
\begin{align*}
\left(\alpha-\frac{1}{n}\right)^2=& \cdots\text{stuff}\\
>&\cdots\text{less stuff}
\end{align*}
$$
And then show that $(\dots\text{less stuff})>2$?

Your help, advice, and expertise would be appreciated.

答:
Proof:

Consider $S=\{s\in\mathbb{R}:s^2>2\}$, for any $x\in\mathbb{R}$, $x^2\geq 0$, thus $S\not=\emptyset$ and is a nonnegative set, so it has a lower bound. By the Axiom of Completeness, there exists an $\alpha=\inf S$, we claim $\alpha^2=2$. We will proceed by contradiction by exhausting the cases $\alpha^2>2$ and $\alpha^2 <2$.

for the first case we assume $\alpha^2>2$ then
$$
\begin{align*}
\left(\alpha-\frac{1}{n}\right)^2=&\alpha^2 -\frac{2\alpha}{n}+\frac{1}{n^2}\\
>& \alpha^2-\frac{2\alpha}{n}
\end{align*}
$$
Since $\alpha^2>2$,  let $\alpha^2-2=\epsilon >0$, then we have $\displaystyle\alpha^2-\frac{2\alpha}{n}-2=\epsilon -\frac{2\alpha}{n}$

$\displaystyle\epsilon -\frac{2\alpha}{n}>0\Leftrightarrow\epsilon>\frac{2\alpha}{n}\Leftrightarrow\frac{\epsilon}{2\alpha}>\frac{1}{n}$

By the Archimedean property, such $n\in\mathbb{N}$ exists, thus $\left(\alpha-\frac{1}{n}\right)\in S$, but  $\left(\alpha-\frac{1}{n}\right) < \alpha$, contradicting our assumption that $\alpha$ is a lower bound.

For the second case we assume $\alpha^2<2$ then

$$
\begin{align*}
\left(\alpha+\frac{1}{n}\right)^2=&\alpha^2 +\frac{2\alpha}{n}+\frac{1}{n^2}\\
<& \alpha^2+\frac{2\alpha}{n}+\frac{1}{n}\\
=&\alpha^2+\frac{2\alpha+1}{n}
\end{align*}
$$
Since we assumed $\alpha^2<2$ we can fix the $\displaystyle\frac{2\alpha+1}{n}$ term such that we are less than 2, that is by the Archimedean property we can find an $n_0\in\mathbb{N}$ such that:
$$
\frac{1}{n_0}<\frac{2-\alpha^2}{2\alpha+1},\text{ which implies } \frac{2\alpha+1}{n_0}<2-\alpha^2
$$

Then $$ \left(\alpha+\frac{1}{n}\right)^2 < \alpha^2+(2-\alpha^2)=2$$

Thus $\displaystyle\left(\alpha+\frac{1}{n}\right)$ is a lower bound for $S$, but $\displaystyle \left(\alpha+\frac{1}{n}\right)>\alpha$, a contradiction to our assumption that $\alpha =\inf A$.


问:
Proof of greatest element of a non-empty set
I am working from Elliott Mendelson's "Number Systems and the Foundations of Analysis," which I am thoroughly enjoying thus far.

The proof I am struggling with comes at the end of a section on order relations in the chapter on natural numbers:

By a greatest element of a set $B \subseteq P$ we mean an object $z$ such that $z\in B$ and $(\forall u)(u\in B \implies u \leq z)$.

If $\emptyset \neq A \subseteq P$ and $A$ is bounded above (that is, $(\exists w)(\forall u)(u \in A \implies u \leq w)$), then $A$ has a greatest element. (Hint: Let $B = \{w: (\forall u)(u\in A \implies u \leq w)\}$. By hypothesis, $B \neq \emptyset$. Apply the least number principle.)

The hint basically prescribes how to satisfy the requirement of a greatest element possessing the property that there exists some $z$ such that $(\forall u)(u\in A \implies u \leq z)$. In particular, define $B$ as given in the hint; since $A$ is bounded above, $B$ is non-empty. Since $B$ is, in turn, non-empty, it possesses a least element by the least number principle. This least element of $B$ is clearly the "$z$" referred to above, i.e., the greatest element of $A$. I have been struggling for quite a while with demonstrating that this "$z$" is, indeed, contained in $A$. I have attempted the method of proof by contradiction considering the alternative case that $z \notin A$, but can't seem to figure a means to arrive at a contradiction. I've also considered induction proofs as a means of reaching a contradiction, but likewise can't reach any contradictions. Can anyone provide an extra tip or approach to explore, please?

答: Prove by contradiction. Suppose $z \notin A$. Then, for any $a \in A$, we have $a\leq z$ (because $z \in B$) but $a \neq z$. So $a <z$ or $a \leq z-1$. This proves that $z-1 \in B$ but this contradicts the fact that $z$ is the smallest element of $B$.


问:
definition of generalized eigenspace
If $G_\lambda$ is the generalized eigenspace corresponding to the eigenvalue $\lambda$, and $E_\lambda$, the eigenspace, then why is both of the following true?

$G_\lambda = \ker(T-\lambda I)^{\dim V}$

$G_\lambda = \ker(T-\lambda I)^m$

Where $m$ is the algebraic multiplicity? $\dim V$ is not necessarily equal to the algebraic multiplicity an eigenvalue?

This is really confusing since Axler's Linear Algebra Done Right book on pg. 255 defines the algebraic multiplicity $m = \dim(G_\lambda) = \dim(\ker(T-\lambda I)^{\dim V})$. On the other hand, Friedberg's Linear Algebra on pg. 486 theorem 7.2 defines $G_\lambda$ as in $2.$

Am I missing something? I know that dimension of kernel of any power of $T$ is at most $\dim V$. I may have gone as far as thinking it has something to do with the restriction of $T$ to $G_\lambda$.

答:
Both 1 and 2 in the original question are true, and the two definitions of the generalized eigenspaces given by 1 and 2 lead to the same subspace. The two definitions arise from the two different approaches taken by the two books mentioned in the question.

Suppose $T$ is a linear operator on a finite-dimensional vector space $V$. In Linear Algebra Done Right, the generalized eigenspace $G_\lambda$ corresponding to an eigenvalue $\lambda$ is defined to be the set of vectors $v \in V$ such that $(T - \lambda I)^j v= 0$ for some positive integer $j$. The (algebraic) multiplicity of $\lambda$ as an eigenvalue of $T$ is then defined to be $\dim G_\lambda$. This definition is equivalent to the usual definition but it is much more geometric, defining the (algebraic) multiplicity as the dimension of a natural subspace instead of defining the (algebraic) multiplicity to be the multiplicity of $\lambda$ as the root of a mysterious polynomial.

Title: definition of generalized eigenspace


问:
Is this function decreasing at its points of discontinuity?
I have a real valued function $V (t)$. It is continuous at all points except a countable of number of points $t_k $ at which it has jump discontunities. Now this sequence $\{t_k\} $ is strictly increasing. And $V (t) $ is right continuous at all these points such that $V (t_k)\leq V (t_k^-) $ for all $k\in\mathbb {N}.$ Then can i say that $V (t_k)\leq V (t_{k-1}) $ for all $k\in\mathbb{N}?$ The reasons for this intuition are  $V (t) $ is continuous at all points except $t_k,$ the left limit of $V (t)$ at $t_k $ is greater than $V (t_k)$ and $t_{k-1} $ lies to left of $t_k $?

答: Your intuition fails. A simple counter-example is $V(x)=x^{2}-[x]$ with $t_k=k$ for all $k$. In this case $V(t_2) >V(t_1)$.


问:
For two sets of points find the plane about which they are reflected
For two sets of points in vector-3 space that are mirrored transformations of each other about a given plane that doesn't go through the origin. And where the reflection matrix is not known. How could one calculate the normal and the origin of the said plane?</p>

答: Since the plane is a reflection plane that means it has to be perpendicular to the line connecting the two points. This means that a normal vector would be $v - v'$ where the $v$ and $v'$ are the two points. I'm not sure what you mean by the origin of a plane but the midpoint, $(v + v')/2$, between $v$ and $v'$ would have to be in the plane. A plane is uniquely described by a normal vector and a point in the plane so this would describe the reflection plane. Let $n = v - v'$ and $r_0 = (v + v')/2$, a point $w$ is in the reflection plane if and only if $n \cdot (w - r_0) = 0$.


问:
Show that $8^3>9^{8/3}$.
I am trying to show algebraically that $8^3>9^{8/3}$. This came from trying to complete the base case of an induction proof.

I have struggled because $8$ and $9$ cannot be manipulated to be the same base. Otherwise I could just argue that $3>\dfrac{8}{3}$.

I tried raising both sides to the third power and got $8^9>9^8$. I can rewrite this as $8^9>9^{9-1}$ but I am not sure if this is the right direction.

答:
We want to compare $x^y $ vs. $y^x$, for $x, y > e$. Take log base $y$ of both sides, we get $y \log_y(x) = y \ln(x) / \ln(y)$ vs. $x$.

Note that if $x = y$, then $y ln(x) / ln(y) = x$. Consider $y$ as a constant and take derivatives. $\frac{y}{x ln(y)}$ vs $1$.

Therefore, if $x > y$, then $\frac{y}{x \ln(y)} < 1$ (Here is where we use $y > e$, so that $ln(y) > 1$. Without this assumption, it could be possible that $\frac{y}{x \ln(y)} > 1$). Therefore, 1 is bigger. Thus, for $x > y$, $x > y \ln(x) / \ln(y)$, so $y^x$ is bigger than $x^y$.

In general, $small^{big} > big^{small}$ for numbers $> e$

Title: Show that $8^3>9^{8/3}$.


问:
Integral of $\int \frac{x^2+5x-3}{x+2}dx$ using integration by parts?
Does anyone know how I might solve this integral using integration by parts? Thanks!
$$
\int \frac{x^2+5x-3}{x+2}dx
$$

Body:
We have
$$\int \frac{x^2+5x-3}{x+2}dx=\int\frac{(x+2)(x+3)-9}{x+2}dx=\int(x+3)dx-9\int\frac{1}{x+2}dx$$
$$=\frac{x^2}{2}+3x-9\ln(x+2)+C.$$


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
Here is a simple but cute one:

The expected number of instances of consecutive pairs of cards of the same suit is $12$ in a shuffled standard deck ($4$ suits of $13$ cards each).

The proof is trivial once you use linearity of expectation, since each of the $51$ pair-positions contributes $1$ to the total count with probability $12/51$ and $0$ otherwise. It is a little cute that the $51$ cancels.


问:
How is $y'=x^2+1$ a differential equation?
From James Stewart Essential Calculus Early Transcendentals Textbook,

A differential equation is an equation that contains an unknown function and one or more of its derivatives.

But the equation
$y'=x^2+1$
doesn't contain the unknown function $y$, it only contains $y'$ and $x$. And yet my textbook says its a DE of order 1.

Even Wikipedia says a DE "relates one or more functions and their derivatives", but $y'=x^2+1$ doesn't contain the function $y$ so how is it a DE ?

答:
That definition is not a very good one, for the reason you're pointing out here. To be precise, an ordinary differential equation of positive integer order $n$ is one which can be written in the form

$$ F\left(x, y, y',\dots,y^{(n)}\right) = 0 $$

where $y$ is an unknown function of the single variable $x$, and $\frac{\partial F}{\partial y^{(n)}} \not\equiv 0$. The upshot of that condition is that the degree will be equal to the highest order of derivative present in the equation, barring cancellations.

For your example given,

$$ F\left(x,y, y'\right) = y' - x^2 - 1 $$

with $\frac{\partial F}{\partial y'} = 1 \neq 0$, so it is a first order ordinary differential equation. This should align with your instinct upon inspection, even if it technically disagrees with Stewart's definition.


问:
A possible upper bound on the fractional part of exponential functions

Does there exist a real number $a$ bigger than $1$ that's not a rational power of an integer such that $\displaystyle\lim_{n\to\infty}${$a^n$}$=0$ where values for $n$ are positive integers?

(P.S. $\{ x \} = x - \lfloor x \rfloor$ or namely the fractional part function)

I just saw this problem today. I've just started studying about the fractional part approximations so unfortunately I've made no progress solving this.

*I made a slight change in the problem to avoid any confusions about small values

答:
It is hard to find such $a$. Indeed, according to Wikipedia, the set of such $a$ is countable. (Remark that Corollary 3 of Zeldich’s attraction theorem states that a set $\{a\in (1,\infty): \{a^n\}\mbox{ is not dense in }[0,1]\}$ is meager in $(1,\infty)$.)

Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then there is no such $a$. Indeed, for a contradiction pick such $a$. The affirmative answer to Pisot-Vijayaraghavan problem implies that $a$ is a Pisot–Vijayaraghavan number. Let $a_1=a$ and $a_2, \dots, a_m$ be the conjugates of $a$. Then for each natural $n$, $\sum_{k=1}^m a_k^n$ is integer.

For each $k=1,\dots, m$ let $a_k=r_ke^{\varphi_ki}$ for some positive $r_k$ and $\varphi_k$. Let $G=\Bbb T^m=\{z\in\Bbb C:|z|=1\}$ be a (multiplicative) topological group and $g=(e^{\varphi_1i}, e^{\varphi_2i},\dots, e^{\varphi_m i})$ be an element of $G$. Since the group $G$ is compact, it is well-known and easy to show that $G$ topologically periodic, so for any neighborhood $U$ of the identity of $G$ there exists arbitrarily big natural number $n$ such that $g^n\in U$. Pick $U_0=\{(x_1,\dots,x_m)\in\Bbb T^m: \forall i (\operatorname{Re} x_i\ge 0)\}$.

Let $N>0$ be any number. Since $a$ is a  Pisot-Vijayaraghavan number, we have $r_k<1$ for each $k=2,\dots, m$. So  there exists $M>0$  such that $\sum_{k=2}^m r_k^M<1/2$. Pick $n>N,M$ such that $g^n\in U_0$. Then for some integer $K$ we have

$$K=\sum_{k=1}^m a_k^n=\sum_{k=1}^m r_k^ne^{n\varphi_ki} =\sum_{k=1}^m \operatorname{Re} r_k^n (e^{n\varphi_ki})=
a^n+\sum_{k=2}^m r_k^n \operatorname{Re} e^{n\varphi_ki}.$$

So $a^n\le K\le a^n+\sum_{k=2}^m r_k^n\le a^n+\sum_{k=2}^m r_k^M<a^n+1/2$. Thus  $\{a^n\}>1/2$, a contradiction.


问:
Bilinear forms on R^n: What is the exact meaning of the indices i and j for <ei,ej>

Let $f:\mathbf{R}^2\to\mathbf{R}^2$, $f\left(\begin{bmatrix}x_1 \\ y_1\end{bmatrix},\begin{bmatrix}x_2 \\ y_2\end{bmatrix}\right) = 2x_1x_2 + 3x_1y_2 + x_2y_1$.
As an example we want to represent $f$ with the standard basis. It follows that: $$f(e_1,e_1)=2 \ ;\ f(e_1,e_2)=3 \ ; \ f(e_2,e_1) = 1 \ ; \ f(e_2,e_2)=0 $$

Hey,

Can someone explain to me how e.g. $f(e_1, e_1) = 2$, according to the definition of bilinear forms and matrices? I think i am confused about the indices. I do not really understand what $e_1$ is referring to: is it the first row of the identity-matrix?

I added the example of my class and my question as a picture (i dont know how to write mathematical expression on my computer)

Q:

What does $f(e_1,e_1)$ mean?

How is it related to the theorem of bilinear forms:
$$\boxed{f(v,w) = v\ ^t\!Aw = \sum_{i=1}^n\sum_{j=1}^n a_{ij}v_iw_j}$$

答:
As you will see, $\beta=\{e_1,e_2\}$ is a basis for $R^2$.
So, what you define bilinear form $f$ as $[f]_\beta$ and correspondingly you find the respective coordinates of the matrix that is, for example,
You say $f(e_1,e_1)$ is first element of first row and you calculated that $f(e_1,e_1)=2$. Likewise, you calculate other entries!!
Hope it helps!!


问:
Existence of a path between two sides of a square, which is contained in the intersection of two open subsets of it
Denote by $I$ the interval $[0,1]$. Let $A,B$ be two open subsets of the square $I\times I=I^2$ such that $A\cup B=I^2$, $\{0\}\times I\subseteq A$ and $\{1\}\times I\subseteq B$. In more geometric terms, the left edge of the square is contained in $A$ and the right edge is contained in $B$.

My question is: does there exist a continuous path $\gamma:I\to A\cap B$ such that $\gamma(0)\in I\times\{0\}$ and $\gamma(1)\in I\times\{1\}$? In other words, is there a curve connecting the two horizontal sides of the square, whose trace is contained in $A\cap B$ ?

答:
Since $(\{0\} \times I) \cup B^c$ and $(\{1\} \times I) \cup A^c$ are closed and disjoint, they can be precisely separated by a continuous function $f: I^2 \to I$.  Let $C$ be the connected component of $(I \times \{0\}) \cup f^{-1}([\frac13, \frac23])$ that contains $I \times \{0\}$.  Being a connected component of a closed set, $C$ is closed.  If we suppose that $C$ and $I \times \{1\}$ are disjoint, they too can be precisely separated by a continuous function $g: I^2 \to I$.

The map $(f, g): I^2 \to I^2$ sends each edge of the square to the same edge, so it sends a path $p$ that winds once around the square’s edges to another such path.  Continuously contracting $p$ to a point also contracts $(f, g)(p)$ to a point, and such a contraction must sweep through every point in the square.  So $(f, g)$ is surjective.

Choose a sequence of points $x_n \in I^2$ such that $(f, g)(x_n) = (\frac12, \frac{1}{n})$.  By Bolzano–Weierstrass, some subsequence of $x_n$ converges to a point $x \in I^2$.  Then $(f, g)(x) = (\frac12, 0)$, so $x \in C$, and in fact a connected neighborhood of $x$ in the open set $f^{-1}((\frac13, \frac23))$ lies in $C$.  But since $g(x_n) \ne 0$, the sequence $x_n$ can never reach that neighborhood, which contradicts its convergence to $x$.

Therefore, $C$ and $I \times \{1\}$ are not disjoint, which means there’s some connected component of $f^{-1}([\frac13, \frac23])$ that intersects both $I \times \{0\}$ and $I \times \{1\}$.  Enlarging it to a connected component of the open set $f^{-1}(A \cap B)$ gives an open component, which is therefore path-connected.


问:
Prove or Disprove $A \subseteq C \land B \subseteq D $ is equivalent to $A \times B \subseteq C \times D$
Can this statement be proved? because if i insert values, LHS is not equal to RHS.
For example,
let  $A=\{1\}, B=\{1,2\}, C=\{1\}, D=\{1\}$ then
$A\times B = \{1,2\}, C\times D=\{1\}$

答:
We only have $\Rightarrow$.

To prove $\Rightarrow:$ If $A\times B=\emptyset$, we are done. Suppose
that $A\times B\neq\emptyset$. Let $(x,y)\in A\times B$, then $x\in A$
and $y\in B$. $x\in A$ and $A\subseteq C$ implies that $x\in C$.
$y\in B$ and $B\subseteq D$ implies that $y\in D$. Therefore $(x,y)\in C\times D$.
This shows that $A\times B\subseteq C\times D$.

We do not have $\Leftarrow$.

Counter-example. Let $A=\{1,2\}$, $B=\emptyset$, $C=\{1\}$, $D=\{3\}$.
Since $B=\emptyset$, we have that $A\times B=\emptyset$. It follows
that $A\times B\subseteq C\times D$. On the other hand, clearly $A\not\subseteq C$.


问:
Finding the sum of a geometric binomial progression
By considering $x^{n}+x^{\left(n-1\right)}\left(1+x\right)+x^{\left(n-2\right)}\left(1+x\right)^{2}+...+\left(1+x\right)^{n}$, find the sum of $\begin{pmatrix}n\\ r\end{pmatrix}+\begin{pmatrix}n-1\\ \:r-1\end{pmatrix}+\begin{pmatrix}n-2\\ \:\:r-2\end{pmatrix}+...+\begin{pmatrix}n-r+1\\ \:\:1\end{pmatrix}$.

I have found the sum of the Sum GP to be $$\frac{x^{n+1}-\left(1+x\right)^{n+1}}{2x-1}$$

Now, if I consider $x^{n}+x^{\left(n-1\right)}\left(1+x\right)+x^{\left(n-2\right)}\left(1+x\right)^{2}+...+\left(1+x\right)^{n}$, I tried to find the term $x^r$ which works well EXCEPT for $\begin{pmatrix}n-r+1\\ \:\:1\end{pmatrix}$. The term of this is not $x^r$, but $x^{n+2}$ which is out of place. By my way, the final term should be $\begin{pmatrix}n-r\\ r-r\end{pmatrix}$. What am I doing wrong?

Body:
The required can be re-written as
$$S=\sum_{j=1}^{r} {n-(j-1) \choose r-(j-1)}=\sum_{j=1}^{r} {n-(j-1) \choose n-r}=\sum_{k=0}^{r-1} {n-k \choose n-r}$$
$$\implies S=[x^{n-r}]~~ \sum_{k=0}^{r-1} (1+x)^{n-k}=[x^{n-r}]\left((1+x)^n \frac{(1+x)^{-r}-1}{(1+x)^{-1}-1}\right)$$
$$S=[x^{n-r}]~~ \left( (1+x)^n \frac{(1+x)^{-r+1}-(1+x)}{-x}\right)$$
$$S=[x^{n-r+1}] ~~ ((1+x)^{n+1}-(1+x)^{n-r+1})$$
$$S={n+1 \choose n-r+1}-1= {n+1 \choose r}-1$$

Here $[x^j]~ F(x)$ means the coefficient of $x^j$ in $F(x)$.


问:
How to find the shaded angle of a right angled isosceles triangle?

Hey everyone,

Just another question. I’m trying to figure out how to find the shaded angle of an isoceles triangle. All I know is the ratios between the sides and of course the value of the angles (as you can see in the image).

Can anyone help with this question? Thank you very much!

Body:
We may arbitrarily place $B$ at the origin, $A=(0,1)$ and $C=(1,0)$. Then $X=(0,1/2)$, $Y=(1/3,0)$, $\vec{AY}=(1/3,-1)$ and $\vec{CX}=(-1,1/2)$. The angle between $\vec{AY}$ and $\vec{CX}$ may be found through the dot product:
$$\cos\theta=\frac{(1/3,-1)\cdot(-1,1/2)}{\|(1/3,-1)\|\|(-1,1/2)\|}=\frac{-5/6}{\sqrt{10/9}\sqrt{5/4}}=\frac{-5}{\sqrt{50}}=-\frac1{\sqrt2}$$
Hence the required angle is $\cos^{-1}-\frac1{\sqrt2}=\frac{3\pi}4$.


问:
Find the general form of the $n$th derivative of $\ln^2(x+1)$
$f(x)=\ln^2(x+1)$.

I tried to write $f(x)=\ln (x+1)\cdot\ln (x+1)$ and then apply Leibniz but i think maybe i did something wrong.

Could you explain me the right method?

答:
Ok we talked a bit about it in the comments: we can start with the chain rule to find $f'$. Then we calculate a few more derivatives, and eventually we find a nice form that looks like a good guess. That guess can be proven by induction. I will leave the guessing and the proof by induction for you to do. Alright, here we go:
$$f(x) = \ln^2(x+1) $$
$$f'(x) = 2 \times\frac{\ln(x+1)}{x+1} $$
Now we can use $(\frac{u}{v})'=\frac{u'v-v'u}{v^2}$ and the fact that $(\ln(x+1))'=(x+1)^{-1}$ :
$$f''(x) = 2 \times \frac{(x+1)^{-1+1}-\ln(x+1)}{(x+1)^2}=2 \times \frac{1-\ln(x+1)}{(x+1)^2}$$
$$f'''(x) = 2 \times \frac{-(x+1)^{-1+2}-2(x+1)(1-\ln(x+1))}{(x+1)^4}=2 \times \frac{-1-2+2\ln(x+1)}{(x+1)^3}$$
$$f^{(4)}(x)=2\times\frac{2(x+1)^{-1+3}-3(x+1)^2(-1-2+2\ln(x+1))}{(x+1)^6}$$
$$ = 2 \times \frac{2+3\times1+3\times2-3\times2\ln(x+1)}{(x+1)^4}$$
Almost feeling like there's a pattern, right? Let's keep going:
$$f^{(5)}(x) = 2 \times \frac{-3\times2(x+1)^{-1+4}-4(x+1)^3(2+3\times1+3\times2-3\times2\ln(x+1))}{(x+1)^8} $$
$$= 2 \times \frac{-3\times2-4\times2-4\times3\times1-4 \times 3 \times 2 + 4\times3\times2 \ln(x+1)}{(x+1)^5} $$
Curious...let's organize:
$$f^{(5)}(x) = 2 \times \frac{-(4\times3\times2 + 4\times3\times1 + 4\times2\times1+3\times2\times1)+4\times3\times2\times1\times\ln(x+1)}{(x+1)^5} $$
$$= 2 \times \frac{-4!(\frac11+\frac12+\frac13+\frac14)+4!\times\ln(x+1)}{(x+1)^5} $$
Now the general form is evident. I still recommend that you do $f^{(6)}(x)$ yourself before moving on to writing the general form and proving by induction. If you find difficulties with induction, let us know and we will help you. Good luck!


问:
How to prove this identity related to dedekind sums
I am studying analytic number theory and this question was asked in my Mid-semester question paper.

Edit 1: Definition of dedekind sum
https://mathworld.wolfram.com/DedekindSum.html

Question:

10. If $\,p\,$ is prime prove that
$$ (p+1)\,s(h,k) = s(p\,h,k)+\sum_{m=0}^{p-1} s(h+m\,k,p\,k). $$

I used the definition of s(h, k) in both s(ph, k) and sum involving s(h+mk,pk) in hope of getting LHS.but couldn't.

Another thing to note is that there are p terms in sum and 1 in RHS and s(h, k) is p+1 times on LHS.

But I was not able to use that also.

It is my request to you to help in proving this identity.

Original image: https://i.stack.imgur.com/bVqDv.jpg

答:
Definition:
$$((x)) = \left\{\begin{array}{ll}
                   0 & x\ \mathrm{is\ an\ integer}, \\
                   x - \lfloor x\rfloor - \frac{1}{2} & \mathrm{otherwise}
                 \end{array}
\right.$$
and
$$s(h,k) = \sum_{r=1}^{k-1} \left(\left(\frac{r}{k}\right)\right)\left(\left(\frac{hr}{k}\right)\right).$$

Fact 1: $s(ph, pk) = s(h, k)$.

(See: An exercise related to properties of dedekind sums)

Fact 2: If $p$ is a prime, $q$ is an integer, $p \nmid q$, and $x$ is rational, then
$$\sum_{t=0}^{p-1} \left(\left( \frac{x + qt}{p}\right)\right) = ((x)). $$
(The proof is given at the end.)

Now, we have
\begin{align}
&\sum_{m=0}^{p-1} s(h + mk, pk)\\
=\ &
\sum_{m=0}^{p-1} \sum_{r=1}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)\Big(\Big(\frac{(h + mk) r}{pk}\Big)\Big)\\
=\ & \sum_{r=1}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)\\
=\ & \sum_{r = 1, \, p\, \nmid\, r}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)
+ \sum_{r = 1, \, p\, \mid \, r}^{pk - 1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)\\
=\ & \sum_{r = 1, \, p\, \nmid\, r}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\Big(\Big(\frac{hr}{k}\Big)\Big)
+ \sum_{v=1}^{k-1} \Big(\Big(\frac{v}{k}\Big)\Big)
p\Big(\Big(\frac{hv}{k}\Big)\Big) \tag{1}\\
=\ & \sum_{r=1}^{pk - 1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\Big(\Big(\frac{hr}{k}\Big)\Big) - \sum_{v=1}^{k-1} \Big(\Big(\frac{vp}{pk}\Big)\Big)
\Big(\Big(\frac{hvp}{k}\Big)\Big)
+ \sum_{v=1}^{k-1} \Big(\Big(\frac{v}{k}\Big)\Big)
p\Big(\Big(\frac{hv}{k}\Big)\Big)\\
=\ & s(ph, pk) - s(ph, k) + ps(h,k)\\
=\ & s(h,k) - s(ph,k) + ps(h,k) \tag{2}\\
=\ & (p+1)s(h,k) - s(ph,k)
\end{align}
where we have used Fact 2 in (1), Fact 1 in (2).

We are done.

$\phantom{2}$

Proof of Fact 2: Since $\{1, 2, \cdots, p-1\} = \{q, 2q, \cdots, (p-1)q\}\ (\mathrm{mod}\ p)$,
we have $\sum_{t=0}^{p-1} \left(\left( \frac{x + qt}{p}\right)\right) = \sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right)$.

Let $f(x) = \sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right)$.
Clearly, $f(x+1) = f(x)$. Since $((0))=0$ and $((y)) = y - \frac{1}{2}$ for $y\in (0, 1)$, we have
$f(0) = \sum_{t=0}^{p-1} \left(\left( \frac{t}{p}\right)\right)
= \sum_{t=1}^{p-1} \left( \frac{t}{p} - \frac{1}{2}\right) = 0$.
For $0< x < 1$, we have
$\sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right) =
\sum_{t=0}^{p-1} \left(\frac{x + t}{p} - \frac{1}{2}\right) = x - \frac{1}{2} = ((x))$. We are done.


问:
A question about Sylow's theorems and index of groups.
I'm trying to prove this exercise:

Let be $G$ a finite group, p a prime number, and $H$ a normal subgroup. If $p \nmid [G:H]$, then $\{ x \in G : o(x)=p^{n_x}, \text{ with } n_x \in \mathbb{N} \} \subseteq H$.

I tried by contradiction. If $B=\{x \in G : o(x)=p^{n_x}, \text{ with } n_x \in \mathbb{N} \} \nsubseteq H$. Then, $\exists \; y \in B$ such that $y \notin H$ and $o(y)=p^{n_y}$. I take $\langle y \rangle \in Syl_p(n_y, G)$ and I tried to use Sylow's theorems but didn't get to much...

I think that $B$ is a Sylow $p$-subgroup, but I can't prove this.

I appreciate any help you can give me.

I use the next:

$o(x):=$ order of $x$

Definition: Let be $G$ a finite group, $p \in \mathbb{N}$ a prime and $n \geq 1$ such that $p^n \mid |G|$ and $p^{n+1} \nmid |G|$. For every $r \in [1,n]$ define:

$a)$ $Syl_p(r, G)= \{H \leq G : |H|=p^r \}$.

$b)$ the set of Sylow p-subgroups  $Syl_p(G):=Syl_p(n, G)$.

答:
Hints:

If $p$ does not divide $[G:H]$, then every Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of $G$. (Why?)

If $y$ has order a power of $p$, then $\langle y\rangle$ is a $p$-subgroup, hence contained in a Sylow $p$-subgroup $P$ of $G$. (Why?)

In particular, there is a conjugate of $y$ that is contained in a Sylow $p$-subgroup of $H$. (Why?)

But $H$ is normal, which implies that $y$ itself is contained in a Sylow $p$-subgroup of $H$ (Why?)

Thus, $y\in H$.


问:
golden section search algorithm: why are there two 'update' equations?
I understand that the golden section search algorithm (for finding minimum points) is loosely based on the bisection method (for finding roots). In both methods, we first assign an upper and a lower bound ($x_u$, $x_l$, respectively) as an interval where we bracket our root (or minimum for the golden section search).

My question: why is it enough for the bisection method to only have one 'update equation', that is, finding the midpoint of the upper and lower bounds. In golden section search, we have two equations: $x_1 = x_l -d$ and $x_2 = x_u - d$, where $d=(\phi-1)(x_u-x_l)$.

The two equations above give a very 'asymmetric feel' to the update equations. What is going on here?

答:
Suppose that we have points $x_{l} < x_{m} < x_{r}$ with $x_{m}=(x_{l}+x_{r})/2$ and $f(x_{m}) < f(x_{l})$ and $f(x_{m})<f(x_{r})$. We assume that $f$ is unimodal so that there must be a minimum somewhere between $x_{l}$ and $x_{r}$.

This information isn't enough to tell us whether the minimum point lies in the interval from $x_{l}$ to $x_{m}$ or in the interval from $x_{m}$ to $x_{r}$.  This is fundamentally different from using bisection to find a root of $f$.

However, if we look at two points in between $x_{l}$ and $x_{r}$, with $x_{l} < x_{1} < x_{2} < x_{r}$, then depending on whether $f(x_{1}) < f(x_{2})$ or $f(x_{2}) < f(x_{1})$, we can determine whether the minimum is between $x_{l}$ and $x_{2}$ or between $x_{1}$ and $x_{r}$.

You could, for example, implement the search using $x_{1}=x_{l}+(1/3)(x_{r}-x_{l})$ and $x_{2}=x_{l}+(2/3)(x_{r}-x_{l})$.

The golden section search uses the golden ratio $\phi$ to determine the points $x_{1}$ and $x_{2}$ because when the interval is reduced in length, either $x_{1}$ or $x_{2}$ will end up as one of the two interior points in the next iteration.


问:
Is there any way of calculating this long formula?
$$\left\{\frac{-x^4+y^4-y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}\left\{1+\left\{\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}+\left\{\frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}^{\:}\left\{1+\left\{\frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}-2\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\cdot \frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{y^2+x^2}\sqrt{y^2+x^2-1}}$$

I'm sorry if this question is not proper here, but is there any way(website) to calculate this formula? In principle, this should be zero, but when I do it by hand, it doesn't seem like it. And I want to verify it. Thanks!

答:
Your question is equivalent to asking whether the last term (without the minus sign) and the sum of the first two terms are equal. Expressing these two in the format of one big fraction, the denominators on both sides of the equation are the same: $(x^2+y^2)^{7/2} (x^2+y^2-1)^{7/2}$, therefore comparing the numerators suffices. Make the substitution $p=x^2$ and $q=y^2$ for simplification.

The numerator on the L.H.S. is

$
(-p^2+q^2-q)\big[ (p+q)^2 (p+q-1)^2 + q^2 \big] +(-q^2+p^2-p)\big[ (p+q)^2 (p+q-1)^2 + p^2 \big]
$

which can be simplified into

$-p^3-q^3-(p+q)^3(p+q-1)^2$.

The numerator on the R.H.S. is $2pq(p^2-q^2-p)$.

Notice how all terms when expanded on the R.H.S. must be in the form of $p^m q^n$ where $m,n \in \mathbb{Z} \neq 0$. But on the L.H.S., there are at least a $-p^5$ and $-q^5$ term.