问:
Observability matrix: What is its rank?
I'm writing a controller for an inverted pendulum in which the state vector isn't completely measurable.  The state vector is $\mathbf{x}\in\mathbb{R}^4$ and the output vector is $\mathbf{y}\in\mathbb{R}^2$.

The worked examples of evaluating observability that I've found all use  $\mathbf{y}\in\mathbb{R}^1$, so I'm reaching out here to ask whether I'm correctly calculating rank of the observability matrix $O$.

The output matrix $C$ is
$$
C = \begin{bmatrix}
1&0&0&0 \\
0&0&1&0
\end{bmatrix}
$$

The observability matrix $O$ is
$$
O = \begin{bmatrix}
 C \\
 CA \\
 CA^2 \\
 CA^3
 \end{bmatrix}
$$

A rounded version of the system matrix $A$ is
$$
A = \begin{bmatrix}
0&10&0&0\\
1&0&0&0\\
0&-20&0&0\\
0&0&1&0\end{bmatrix}
$$

The resulting observability matrix $O$ is then

$$
O = \begin{bmatrix}
\begin{bmatrix}1&0&0&0\\0&0&1&0\end{bmatrix}\\
\begin{bmatrix}0&10&0&0\\0&-20&0&0\end{bmatrix}\\
\begin{bmatrix}10&0&0&0\\-20&0&0&0\end{bmatrix}\\
\begin{bmatrix}0&100&0&0\\0&-200&0&0\end{bmatrix}\\
\end{bmatrix}
$$

Clarifications

1. On taking the rank of nested matrices

Wikipedia defines the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns.  Could I not say that $O$ is full rank 1 of 1 because it contains a single column of nonzero elements?

No, in this context we view this not as a 4x1 of 2x4 but as a melded 8x4 matrix.  Rank is then established by checking linear independence of the four columns.

2. On how the output matrix $C$ is represented

The way I represented the output matrix $C$ here as $q$x$n$ where $\mathbf{x}\in\mathbb{R}^n$ and $\mathbf{y}\in\mathbb{R}^q$
is in accordance with the Wikipedia article on State-Space.

However, the two controls textbooks I'm looking at:

Feedback Control of Dynamic Systems (6th Edition) (FPE)

Control Systems Engineering (7th Edition) (Nise)

strictly consider cases in which the shape of $\mathbf{x}$ and $\mathbf{y}$ is identical.  The shape of the $C$ matrix is always $1$x$n$.  Zero terms in $C$ are used to represent un-measurable terms in the output vector $\mathbf{y}$.  For example, let $\mathbf{x} = \begin{bmatrix}\dot\theta\\\theta\\\dot\phi\\\phi\end{bmatrix}$.  If only the terms $\dot\theta$ and $\dot\phi$ appear in the output vector $\mathbf{y}$, then the C matrix is given as $C = \begin{bmatrix}1&0&1&0\end{bmatrix}$

This approach feels misleading to me, as a term in the output vector $\mathbf{y}$ that's always zero is conceptually different from a term that doesn't appear at all.  Nonetheless, it's valid for the sake of calculating rank.

$$
O = \begin{bmatrix}
1&0&1&0\\
0&-10&0&0\\
-10&0&0&0\\
0&-100&0&0
\end{bmatrix}
$$

Conclusion

Both approaches show one three columns with linear independence and one without.  Therefore the rank is 3.

答: You've evaluated it correctly although it is easier to see the rank through the columns in this case. The last column of zeros places an upper bound on the rank at 3; from that alone you can conclude the system is not observable. Specifically, the first three columns are linearly independent so then you can conclude the rank is 3 and therefore you can only observe the evolution of the system in a 3 dimensional subspace of the entire state-space.  Note that last column of zeros indicate that you cannot observe the evolution of the 4th state.</p>

Title: Observability matrix: What is its rank?


问:
Orthonormal basis of $L^2( (0,a)\cup(b,c) )$
I am studying Lebesgue spaces and appear the concept of orthonormal basis.  My professor told me that a classic result is the following: the set given by $$(\sqrt{2} \sin(n\pi x))_{n\geq 1}$$ is an orthonormal basis of $L^2(0,1)$. So he ask me how is possible obtain basis for $L^2$ on union of intervals in the form:  $\mathcal{E}:=(0,a)\cup(b,c)$.

My question are: which is an orthonormal basis of $L^2(\mathcal{E})$?, the same basis works for both $L^2(0,1)$ and $L^2(\mathcal{E})$ ?.

答:
Step 1: Orthonormal basis for $L^2(a,b)$

Let $(a,b)$ be an interval . Then the inner product for $L^2(a,b)$ is given by, $$<f,g>=\frac{1}{b-a} \int_{a}^b f(t)\overline{g(t)}dt$$ (Note that we have included the factor $\frac{1}{b-a}$ just to normalize our space to be a probability space.) Now we have the natural homeomorphism $\varphi:(a,b) \to (0,1)$ given by, $\varphi(x)=\frac{x-a}{b-a}$ .Now corresponding to your favourite Orthonormal basis of $L^2(0,1)$ say $\{e_n\}_{n \in \Bbb N}$ consider $\{\tilde{e_n}\}_{n \in \Bbb N}$ given by, $\tilde{e_n}(x):=e_n\circ\varphi(x)$ .

Exercise 1: Check that $\{\tilde{e_n}\}_{n \in \Bbb N}$ is an Orthonormal basis for $L^2(a,b)$

Step 2: Look at the union of intervals

If you are considering the union of $(a_1,b_1),(a_2,b_2)$ then either their union is a single interval namely $(\min(a_1,a_2),\max(b_1,b_2))$ or a union of two disjoint intervals. If the union is a single interval, then we already have an Orthonormal basis by Step 1. Otherwise, consider $\{\tilde{e_n}\}_{n \in \Bbb N},\{\tilde{f_m}\}_{m \in \Bbb N}$ for the two intervals obtained from Step 1 .

Step 3: Looking at the direct sum

As the set is disjoint union of two open intervals, it is easy to prove that,

Exercise 2: $L^2(\mathcal{E})=L^2(a_1,b_1) \oplus L^2(a_2,b_2)$ (in fact, as an orthogonal sum)

Exercise 3: $\{(\tilde{e_n},0),(0,\tilde{f_m}):m,n \in \Bbb N\}$ is an Orthonormal basis of $L^2(\mathcal{E})$ (where you extend $\tilde{e_n}$ to be $0$ on $(a_2,b_2)$ and $\tilde{f_n}$ to be $0$ on $(a_1,b_1)$ )

Q.E.D

Let me know if more details is required !

ContentLicense: CC BY-SA 4.0


问:
Time, Speed and Distance : Returning back to pick up problem
$A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$

I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.

Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.

Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$

Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!

Please help me out.
Answer Provided : 4.8 hours

答:
This will be equivalent to the other methods, but is another way of dividing up the trip.  A and B travel together at 50 kph for a distance $ \ x \ $ ; at that point, $ \ B \ $ continues for the remaining distance at 10 kph.  The total travel time for B is then
$$ \frac{x}{50} \ + \ \frac{120-x}{10} \ \ = \ \ T \ \ \Rightarrow \ \ x \ + \ (600 - 5x)  \ \ = \ \ 600 - 4x \ \ = \ \ 50T \ \ . $$

A is the only person who makes the entire trip at 50 kph, but has to "backtrack" a distance $ \ y \ $ in order to intercept C .  So A will have traveled $ \ 120 + 2y \ $ kilometers, for a total travel time
$$ \frac{120 + 2y}{50} \ \ = \ \ T \ \ \Rightarrow \ \ 120 + 2y \ \ = \ \ 50T \ \ . $$

When C is met by A, C will have covered a distance of $ \ x - y \ $ at 10 kph, after which A and C complete the trip at 50 kph.  Thus,
$$ \frac{x-y}{10} \ + \ \frac{120-(x-y)}{50} \ \ = \ \ T \ \ \Rightarrow \ \ 5·(x-y) \ + \ 120 \ - \ (x-y) \ \ = \ \ 4·(x-y) \ + \ 120 \ \ = \ \ 50T \ \ . $$

We can put the second and third of these equations together to write $ \ 2y \ = \ 4·(x-y) \ \Rightarrow \ 6y \ = \ 4x \  $ or
$ \ y \ = \ \frac{2}{3} x \ \ . $  [Now the relation to saulspatz's solution becomes clear.]

The third and first equations together become

$$ 4·(\frac13 x) \ + \ 120 \ \ = \ \ 600 \ - \ 4x \ \ \Rightarrow \ \ \frac{16}{3}x \ \ = \ \ 480 \ \ \Rightarrow \ \ x = 90 \ , \ y = 60 \ \ , $$

and now any of the equations can be used to find $ \ T \ = 4.8 \ $ hours.

[We also have a chronology of events:  B was "dropped off" at 1.8 hours into the journey, and A met C at 3 hours.]

$$ $$

ADDENDUM: This suggests a "time-symmetric" approach somewhat related to what Jaap Scherphuis has done.  B is taken some distance along the route at 50 kph and has to complete the remainder at 10 kph, while C does the reverse, traveling at 10 kph for a portion of the trip and being driven the rest of the way at 50 kph.

As arnold points out, when C has traveled a distance $ \ d \ $ , A and B have covered $ \ 5d \ $ ; B is then left to travel $ \ 120 - 5d \ $ on their own.  C goes on for a distance $ \ s \ $ until being met by A .  So A and C come together at "mile-marker" $ \ d + s \ $ at a certain time, by which time A has gone a total distance of $ \ 5d \ + \ [ \ 5d - (d+s) \ ] \ = \ 9d - s \ . $

This amount of time is
$$ \frac{d+s}{10} \ = \ \frac{9d-s}{50} \ \ \Rightarrow \ \ 5d  +  5s \ = \ 9d - s \ \ \Rightarrow \ \ 4d \ = \ 6s \ \ , $$

as we've seen elsewhere.  Since B takes as long to travel $ \ 120 - 5d \ $ as C takes to cover $ \ d + s \ $ at the same speed, we have
$$ 120 - 5d \ = \ d + s \ \ \Rightarrow \ \ s \ = \ \frac{2}{3} d \ = \ 120 - 6d \ \ \Rightarrow \ \  \frac{20}{3}·d \ = \ 120  $$
$$ \Rightarrow \ \ d \ = \ 18 \ \ , \ \ s \ = \ 12 \ \ . $$

So B and C have each traveled alone for $ \ \frac{18+12}{10} \ = \ \frac{120-5·18}{10} \ = \ 3 \ $ hours, and each has traveled with A in the car for $ \ \frac{120 - (18+12)}{50} \ = \ \frac{5·18}{50} \ = \ 1.8 \ $ hours, making a total trip time of $ \ 4.8 \ $ hours.


问:
How to prove there are no more positive integers that are products of 2 and 3 consecutive numbers?
$6$ and $210$ share the property that both are the products of both two and three consecutive numbers. $6$ is $2\times3$ and $1\times2\times3$ and $210$ is $14\times15$ and $5\times6\times7$. It was easy enough to write a program to search for more numbers with this property, I found that there were no more up to at least $1{,}000{,}000{,}000{,}000$ ($1$ Trillion). But it is beyond me to prove that there are either no more numbers like this or to find the next one. Any ideas?</p>

答: In his book Diophantine equation,(page $257-258$) L.J.Mordell proved that the equation $$y(y+1)=x(x+1)(x+2)$$ has only the integer solutions $x=-1,-2,0,1,5.$

ContentLicense: CC BY-SA 3.0


问:
Shooting sequence in shuffled Russian Roulette to remove advantage of going first?
Alice and Bob are playing shuffled Russian Roulette. Starting with Alice, each player takes a turn to throw a dice. The first person to land a 1 (with probability $p=\frac{1}{6}$) is the winner.

But since Alice goes first, she has an advantage. It's not hard to prove her probability of winning is $\frac{6}{11}\approx 0.55$.

To combat this (and try to make it fair 50-50 odds), we adjust the sequence of turns.

Originally, it was ABABABA$\cdots$ (alternating turns).

We might think of changing it to ABBABABABA$\cdots$ (Bob gets 2 turns initially, then it goes back to alternating), but this turns out to give odds in favor of Bob.

My question is: is there a sequence of turns (with Alice's turn first) such that Bob and Alice will have equal odds of winning? Preferably, your answer should be for a generic $p$, not just $1/6$.

答:
I've made a simple python program to calculate a sequence of turns. The idea is that $a$ and $b$ are the probabilities of Alice or Bob having won so far. $a$ is initialized to $p$ and $b$ is initialized to 0. At each step, whoever has the lower probability is given the turn. If $c$ is given the turn, their probability is updated as $c = c + p\cdot(1 - a - b)$.

```<code>    from fractions import Fraction as F # use exact Fraction class because 32-bit floats are too imprecise
    def move(a, b, p):
      if a < b:
        a += p*(1-(a+b))
        return "A", (a, b)
      else:
        b += p*(1-(a+b))
        return "B", (a, b)

    n = 500 # number of turns to compute
    p = F(1, 6) # 1/6
    turns = ["A"]
    a = p; b = 0
    for _ in range(n):
      t, (a, b) = move(a, b, p)
      turns.append(t)

    print("".join(turns))
</code>```

At the end is the turn order for the first 500 turns. Any insight about this turn order? I've found a way to compute it, but I don't really have a sense of if there's some deeper pattern emerging here.

Turn order for $p=1/6$ (first 500 turns):
ABBABAABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABBAABABBABAABABBAABBAABBAABBABAABABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABABBAABBAABBABAABABBAABBABAABABBABAABABBAABBAABBAABBABAABBAABABBAABBABAABBAABABBAABBABAABABBABAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBAABBABAABABBABAABBAABABBABAABBAABBAABABBABAABBAABBAABBAABABBAABBAABBAABBAABABBAABBAABBAABBABAABBAABABBAABBAABBAABBAABABBABAABBAABBAABABBABAABABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABABBABAA

Body:
I've made a simple python program to calculate a sequence of turns. The idea is that $a$ and $b$ are the probabilities of Alice or Bob having won so far. $a$ is initialized to $p$ and $b$ is initialized to 0. At each step, whoever has the lower probability is given the turn. If $c$ is given the turn, their probability is updated as $c = c + p\cdot(1 - a - b)$.

    from fractions import Fraction as F # use exact Fraction class because 32-bit floats are too imprecise
    def move(a, b, p):
      if a < b:
        a += p*(1-(a+b))
        return "A", (a, b)
      else:
        b += p*(1-(a+b))
        return "B", (a, b)

    n = 500 # number of turns to compute
    p = F(1, 6) # 1/6
    turns = ["A"]
    a = p; b = 0
    for _ in range(n):
      t, (a, b) = move(a, b, p)
      turns.append(t)

    print("".join(turns))

At the end is the turn order for the first 500 turns. Any insight about this turn order? I've found a way to compute it, but I don't really have a sense of if there's some deeper pattern emerging here.

Turn order for $p=1/6$ (first 500 turns):
ABBABAABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABBAABABBABAABABBAABBAABBAABBABAABABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABABBAABBAABBABAABABBAABBABAABABBABAABABBAABBAABBAABBABAABBAABABBAABBABAABBAABABBAABBABAABABBABAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBABAABABBABAABBAABABBAABBABAABABBABAABBAABABBABAABBAABBAABABBABAABBAABBAABBAABABBAABBAABBAABBAABABBAABBAABBAABBABAABBAABABBAABBAABBAABBAABABBABAABBAABBAABABBABAABABBAABBAABBAABBABAABBAABBAABBAABBAABABBAABBABAABBAABBAABABBABAA


问:
Best way to play 20 questions

Background

You and I are going to play a game. To start off with I play a measurable function $f_1$ and you respond with a real number $y_1$ (possibly infinite). We repeat this some fixed number $N$ of times, to obtain a collection $\{(f_i,y_i)\}_{i=1}^N$. Define

$\Gamma=\{\mu: \mu $ is a probability measure, and $\int f_i(x)d\mu(x)=y_i \forall i\}$

(We'll assume that $\Gamma$ is not empty, meaning you cannot give inconsistent answers. For example, if I play $f_1(x)=x^2$, then you could not play $y_1=-1$).

Next, I play a measure $P$ which must satisfy $P\in \Gamma$. You then play a measure $Q\in \Gamma$. The game is scored such that your reward is equal to $H(Q|P)$ (the cross entropy) while my reward is equal to $-H(Q|P)$. In other words, I am rewarded to the extent that I can guess your distribution, while you are rewarded to the extent that you can foil my guess.

What should my strategy be for playing this game?

Motivation

This game can be considered as a general model of scientist designing experiments (the functions $f_i$) and using the results of those experiments to develop a theory (the measure $P$). The other player is a stand-in for "Nature", which we assume acts against us at every turn.

Further, the game is a generalization of the setting considered by Grunwald and Dawid in this paper. They consider what amounts to a special case of this game in which the set $\Gamma$ is assumed to be specified ahead of time (so each participant only specifies their distribution). Very interestingly, they show that in this case the optimal strategy is to play the distribution $\text{arg max}_{P\in\Gamma} H(P)$, where $H$ denotes Shannon entropy. This amounts to a passive model of scientific inference, in which the scientist has access only to observational data in the forms of constraints $\int f_i(x)d\mu(x)=y_i$, but has no control over what the actual functions $f_i$ are. Thus I am interested in what would happen if their setting is extended to an "active" one, in which the scientist can control which statistics $f_i$ to measure in the first place.

答:
I will consider the case where all distributions are supported on a fixed finite set $\{1,2,\dots, M\}$. Hopefully someone will see how to extend to the continuous case.

All strategies for Player 1 are equivalent. Moreover, the eventual payoff does not depend on the number of questions asked.

In retrospect, this is maybe not surprising in light of the No Free Lunch theorem. To see it directly, first note that the optimal move for Player 1, conditional on any history $\{(f_i,y_i)\}$ is to play $P_{f,p}:=argmax_{P\in \Gamma_{\{(f_i,y_i)\}}} H(P)$. (This is also true when $N=0$). This is a consequence of the result of Grunwald/Dawid mentioned in the question. By a standard argument involving Lagrange multipliers, the distribution takes the form

$$P_{f,p}(x)=e^{\sum_i \lambda_i f_i(x)-A(\lambda)}$$

where $\lambda_i$ are chosen to satisfy the constraints $\sum_x f_i(x) P_{f,p}(x)=y_i$ and $A(\lambda)$ is chosen so that the total probability mass sums to 1.

Now let $Q\in \Gamma_{(f,p)}$. By the definition of cross entropy we have

\begin{eqnarray*}
H(Q|P_{f,y})&=&-\sum_x Q(x)\log P_{f,y}(x)\\
&=&-\sum_x Q(x) \left(\sum_i \lambda_i f_i(x)-A(\lambda)\right)\\
&=&-\sum_i \lambda_i \sum_x f_i(x)Q(x)+A(\lambda)\\
& = & -\sum_i \lambda_i y_i+A(\lambda)
\end{eqnarray*}

(In the last equality, we used the assumption that $Q\in \Gamma_{(f,y)}$). So in fact the cross entropy does not depend on $Q$, and is always equal to $H(P_{f,y})$. Said another way, the payoff to Player 2, conditional on the moves $\{(f_i,y_i)\}$ is equal to $H(P_{f,y})$.

The best case scenario for Player 2 would be if $P_{f,y}$ is the uniform distribution $Unif_M$ (since this has the highest entropy among all distributions on $\{1,2,\dots, M\}$). Player 2 can achieve this by following the strategy of playing $y_i=\sum_x f_i(x)/M$ in response to any move $f_i$ made by Player 1. It is easy to see that if $y_i$ are defined in this way, then $Unif_M\in \Gamma_{(f,y)}$, no matter what the functions $f_i$ are, which implies that $Unif_M=P_{f,y}$.

So no matter which or how many functions Player 1 plays, Player 2 can always force Player 1 to play $Unif_M$, for a resulting score of $log M$.

Unbounded Case

Here we assume that the distributions are supported on a non-compact set, either the integers or the real line. In this case, there exist distributions that have infinite entropy (Can the entropy of a random variable with countably many outcomes be infinite?). Now in general $H(Q|P)\ge H(Q)$, since KL divergence is always non-negative. This means that, as long as there is some distribution in $\Gamma$ that has infinite entropy, Player 2 can play that distribution, and thus obtain infinite reward. But making sure that $\Gamma$ contains such a distribution is easy, since Player 2 can just choose some distribution $Q$ with infinite entropy ahead of time, and then respond to any function $f_i$ with $y_i=E_{x\sim Q}f_i(x)$. So the game as stated is degenerate in this case.

Title: Best way to play 20 questions


问:
Distribution of number of trials before $k$ distinct successes
I am familiar with the Negative Binomial distribution $NB(p, k)$, which gives the number of failures before $k$ successes occur in a Bernoulli process with parameter $p$. I am wondering, however, if there the distribution of the number of failures before getting $k$ distinct successes is a well-known distribution.

More precisely, suppose we have some set of $N$ elements; without loss of generality, we may assume this set is $[N] = \{1, \cdots, N\}$. We have a subset $S \subseteq [N]$ which contains the successes, while every element in $[N] \setminus S$ is a failure. Now in our process, we select a random element uniformly at random from $N$, and keep on doing this until we have attained some $k \leq |S|$ successes. Let $X$ be the total number of trials performed before we stop the process. What is the distribution of $X$, in terms of $k$, N, and p = $|S| / N$? If we can't concisely describe this distribution, can we at least come up with some (relatively tight) tail bounds? And how valid is the approximation $X \approx NB(p, k)$?

I know this problem can simplify to the coupon collector's problem in the case that $S = [N]$ and $k = N$, but I'm more interested in the case where $0 < p < 1$ and $k \ll |S| = pN$.

答:
$X$ can be written as a sum of $k$ independent geometric variables
$X=\sum_{i=0}^{k-1} G_i$ where $G_i$ is geometric with parameter $\frac{|S|-i}{N-i}$. This representation allows to easily calculate the mean, variance and moment generating function of $X$. In particular, for $k$ large, $X$ is quite concentrated near its mean. If you subtract the mean and normalize by the standard deviation, it is asymptotically Gaussian as $k \to \infty$. If $k=o(N)$ as $N \to \infty$ then the law of $X$ is well approximated by a negative binomial law.


问:
If $f(x)=1-f(x-a_1)f(x-a_2)...f(x-a_n)$, prove that $f$ is eventually periodic. (Czech and Slovak 2001)
Let $a_1,a_2,...,a_n$ be positive integers.  Suppose that $f:\Bbb Z\to \Bbb R$ is given by
$$f(x)=\begin{cases}1&x<0,\\
1-f(x-a_1)f(x-a_2)...f(x-a_n)&x\ge 0.\end{cases}$$
Show that there exist positive integers $s,t$ st  for all integers $x>s$, $f(x+t)=f(x)$.

Source : https://imomath.com/othercomp/Czs/CzsMO01.pdf

My attempt :

By induction, we can see that $f(x)=0$ or $1$ for all $x$.  Therefore, $f(x)^2=f(x)$.  Hence we can assume that $a_1<a_2<...<a_n$.

If $n=1$, then $f(x+2a_1)=1-f(x+a_1)=1-(1-f(x))=f(x)$ for every $x\ge -a_1$.  So we can take $s=1$ and $t=2a_1$ in this case.

If $n=2$, I think we can take $t=a_1+a_2$.  Not sure yet...

Edit to add : Is it possible to find an exact formula for the minimum period $t$ in terms of $a_1<a_2<...<a_n$?

答:
Hint: Any consecutive sequence of $\max(a_i)$ values on positive integers will uniquely determine the subsequent sequence.

Hint: Show that there is some sequence of $\max(a_i)$ integers that repeats by Pigeonhole Principle on $2^{\max(a_i)} + \max(a_i)$ terms

Hence, the sequence is eventually periodic.


问:
Flux through a surface and divergence theorem
Find the flux through the surface limited by  the plane $y+z=2$, the cylinder $x^2+y^2=1$ and the plane $z=0$, for the vector field $\vec{F}(x,y,z)=(4x,-3y,2z)$.

I'm trying to find the flux  through this surface, it's a truncated cylinder, so I tried to find the flux trought three diferent parts of the surface, the top $S_1$, the cylinder $S_2$, and the bottom $S_3$, and for each one there is a diferent normal vector ¿true? then applied the formula $flux=\displaystyle{\int\int_S \vec{F}\cdot \vec{n}d\sigma}$.

The normal vector for $S_1: \vec{n_1}=(0,1,1), \quad$ for $S_2: \vec{n_2}=(2x,2y,0),\quad $ and for $S_3: \vec{n_3}=(0,0,-1) $

Flux of $S_1$: $\displaystyle{\sqrt{2}\int_0^{2\pi}\int_0^1(-5r^2sin(\theta)+4r)drd\theta=4\sqrt{2}\pi }$

Flux of $S_2$: $\displaystyle{\int_0^{2\pi}\int_0^{2-sin(\theta)}(8cos^2(\theta)-6sin^2(\theta))dzd\theta =4\pi }$

And flux of $S_3$ is zero because $z=0$. So the total flux is the sum of each flux, $(4+4\sqrt{2})\pi$.

Is all this correct ? I'm asking about this exercise because I want verify the divergence theorem

$\displaystyle{\int\int_S=\vec{F}\cdot\vec{n}d\sigma=\int\int\int_D \nabla\cdot\vec{F}dV}$.

I calculated the right side and I got:
$\displaystyle{\int\int\int_D \nabla\cdot\vec{F}dV=3\int_0^{2\pi}\int_0^1\int_0^{2-rsin(\theta)}rdzdrd\theta=6\pi}$

It's not the same for the left side, Please could you explain me what is wrong?

答:
a) Flux through $S_1$ (top disk in plane $y+z=2$):

Mistake: you multiplied by $\sqrt2$. Dot product with normal vector $(0, 1, 1)$ takes care of the factor for surface area element. If you are multiplying by $\sqrt2$, then you also need to make sure you have dot product with unit normal vector which is $\hat{n}=\frac{1}{\sqrt2}(0, 1, 1)$.

Parametrization of the surface: $(r \cos t, r \sin t, 2 - r \sin t), 0 \leq r \leq 1, 0 \leq t \leq 2\pi$
and $\vec{n} = (0, 1, 1)$.

$I_{S_1} = \displaystyle \int_0^{2\pi} \int_0^1 (4 r \cos t, -3r \sin t, 4 - 2 r \sin t) \cdot (0, 1, 1) \ r \ dr \ dt$

b) Flux through $S_2$ (cylindrical surface $x^2+y^2 = 1$):

Mistake: you hade a mistake in your normal vector and that is doubling the flux.

Parametrization of the surface: $s(z, t)=(\cos t, \sin t, z), 0 \leq t \leq 2\pi, 0 \leq z \leq 2 - \sin t$
and $\vec{n} = (x, y, 0) = (\cos t, \sin t, 0)$.

$I_{S_2} = \displaystyle \int_0^{2\pi} \int_0^{2-\sin t} (4 \cos t, -3 \sin t, 2z) \cdot (\cos t, \sin t, 0) \ dz \ dt$

[For outward normal vector to the cylinder it is easy to see that from axis of the cylinder $(0, 0, z)$ to a point on the cylinder orthogonal to the surface $(x, y, z)$, the vector is $(x, y, 0)$. Otherwise you can take partial derivatives of $s(z, t)$ and do the cross product $s'_t \times s'_z$ and that would give you the same.]


问:
How does the probability of reaching a threshold of D6 scale with the size of the dice pool?
Assume the following: you have a pool of $N$ d6s. You roll the dice, tally up the number of 6s rolled, and tally up the number of 1s rolled. Then, subtract the number of 1s from the number of 6s. Call this result $X.$ How does the probability that $X$ is greater than or equal to a given threshold ($T$) scale with the size of the pool, $N$? I'm also interested in how $T$ changes the distribution. Ignore the cases where $N < T.$

I'm looking for a simple summary of how it scales, for example, "For $T=5,$ the probability that $X\geq T$ scales linearly with $N.$"

答:
$X$ is the sum of $N$ independent variables $D_i$, each having pmf
$$f_D(x)=\begin{cases}\frac16&x=+1\\\frac16&x=-1\\\frac23&x=0\end{cases}$$
(A $+1$ result corresponds to rolling a six and a $-1$ result to rolling a one.) Each $D_i$ has mean $0$ and variance $\frac13$. Therefore, by the central limit theorem, $X$ tends to a normal distribution with mean $0$ and variance $\frac N3$ as $N$ increases, and
$$\Pr(X\ge T)\to\frac12(1-\operatorname{erf}(T\sqrt{3/2N}))\to\frac12$$
This is true for all $T\in\mathbb R$. $T$ is something you choose and is therefore independent of $X$.


问:
In a integral domain every prime element is irreducible
I'm trying to understand a proof of Hungerford's book which says that in a integral domain every prime element is irreducible:

I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?

I'm a little confused

Thanks in advance

答:
Prime $\Rightarrow$ irreducible is obvious if we employ a definition of irreducible in associate (vs. $\rm\color{#0a0}{unit}$) form. Then $\,\color{#c00}{ p=ab\,\Rightarrow\,  p\mid ab}\,$ immediately yields the sought inference, as follows.

Theorem $\ \ $ In the following, $\,\ (1)\,\Rightarrow\,(2)\!\iff\! (3)$

$(1)\ \ \ \color{#c00}{p\ \mid\ ab}\  \Rightarrow\  p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is prime]

$(2)\ \ \ \color{#c00}{p=ab}\ \Rightarrow\ p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is irreducible, in associate form]

$(3)\ \ \ p=ab\ \Rightarrow\ a\:|\:1\ \ {\rm or}\ \ b\:|\:1\quad$ [Definition of $\:p\:$ is  irreducible, in $\rm\color{#0a0}{unit}$ form]

Proof $\ \ \ (1\Rightarrow 2)\,\ \ \ \color{#c00}{p = ab\, \Rightarrow\, p\mid ab}\,\stackrel{(1)}\Rightarrow\,p\mid a\:$ or $\:p\mid b.\ $ Hence prime $\Rightarrow$ irreducible.

$(2\!\!\iff\!\! 3)\ \ \ $ If $\:p = ab\:$ then $\:\dfrac{1}b = \dfrac{a}p\:$ so $\:p\:|\:a\iff b\:|\:1.\:$ Similarly $\:p\:|\:b\iff a\:|\:1.$

Beware that factorization theory is more complicated in non-domains. Basic notions such as associate and irreducible bifurcate into a few inequivalent notions. See here for more on that.


问:
How to find where a set of parametric equations crosses itself
The set of parametric equations:

$x(t) = t^2 − 2$

$y(t) = t^3 − 9t + 5$

Compute $(x,y)$ where the crossing occurs.

The answer key says that $(x,y) = (7,5)$, which means that $t=3$. That's really all I have, I have no idea how to solve this problem. Can someone point me in the right direction?

答:
This is a special case of the self-intersection problem for cubic Bézier curves; I wrote an answer to this general problem here. Using the specific values in this problem, the relevant linear equation is
$$\begin{bmatrix}0&1\\1&-9\end{bmatrix}\begin{bmatrix}\lambda^2+\lambda\mu+\mu^2\\\lambda+\mu\end{bmatrix}=\begin{bmatrix}0\\9\end{bmatrix}$$
from which we get $\lambda^2+\lambda\mu+\mu^2=9$ and $\lambda+\mu=0$. Here $\lambda,\mu$ are the parameters of self-intersection; solving the last two equations we get $\lambda,\mu=\{-3,3\}$ and the point of self-intersection as $(7,5)$.


答:
The $t$ of interest must be of the form of $\pm t_0, t_0 \ne 0$ to make $x$ takes the same value.

$$t_0^3-9t_0+5 = -t_0^3+9t_0+5$$

$$t_0^3-9t_0=0$$

$$t_0(t_0^2-9)=0$$

Hence we know $|t_0|=3$.

Title: How to find where a set of parametric equations crosses itself

Body:
The set of parametric equations:

$x(t) = t^2 − 2$

$y(t) = t^3 − 9t + 5$

Compute $(x,y)$ where the crossing occurs.

The answer key says that $(x,y) = (7,5)$, which means that $t=3$. That's really all I have, I have no idea how to solve this problem. Can someone point me in the right direction?


问:
Value of an integral
A function  $f$ is defined as
$$f(t)=t^{r-1}(2t-t^2)^{n-1}(2-2t)$$
where $r,n$ are natural numbers. Can we somehow find an a closed expression for the inegral $\int_0^1 f(t)\,dt$ or even an asymptotic expression as $n$ goes to infinity? One could substitution $u=2t-t^2$ but for the extra term $t^{r-1}$.

Mathematica yields the following value :
$$2^{-1 + 2 n + r} (B[1/2, -1 + n + r, n] - 2 B[1/2, n + r, n])$$
where $B(x,a,b)=\int_0^x u^{a-1}(1-u)^{b-1} du$ is the incomplete beta integral

OwnerDisplayName: sajjad veeri

Title: Value of an integral

Body:
Since
$$f(t)=t^{r-1}(2t-t^2)^{n-1}(2-2t)\\
\qquad\qquad\qquad=2t^{r-1+n-1}(1-t)\sum_{i=0}^{n-1}(-1)^i2^{n-i}{n-1 \choose i}t^i$$
we get
$$\int_0^1f(t)\,\text dt=2\sum_{i=0}^{n-1} (-1)^i 2^{n-i}{n-1 \choose i} B(r+n+i-1,1)\\
\qquad =\sum_{i=0}^{n-1} (-1)^i 2^{n-i+1}{n-1 \choose i}\Big/(r+n+i-1)$$


问:
The complex version of the chain rule
I want to prove the following equality:

\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = (\frac{\partial g}{\partial z} \frac{\partial f}{\partial z}) + (\frac{\partial g}{\partial \bar{z}} \frac{\partial \bar{f}}{\partial z})
\end{eqnarray}

So I decide to do the following:

\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = \frac{1}{2}[(\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x}) + \frac{1}{i}(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y})]
\end{eqnarray}

but the thing is that I am doing something wrong here since I don't get any conjugate function and any derivative with respect to $\bar{z}$ so Can someone help me to see where I am wrong and fix it please?

In fact I don't see what to do next, so I appreciate your help.

Thanks a lot in advance.

Edition:

What I've got so far is the following:

$$\frac{1}{2}[(\frac{\partial g}{\partial x} \circ f + \frac{\partial g}{\partial y} \circ f)\frac{\partial f}{\partial z} ]$$

but I'm still stuck.

答:
Define $z=x+iy$, then$$\begin{cases}
   x=\frac{z+\overline{z}}{2}\\
   y=\frac{z-\overline{z}}{2i}
  \end{cases}.$$ We consider $x$ and $y$ are functions of $z$ and $\overline{z}$, thus we have
$$\frac{\partial }{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{1}{i}\frac{\partial}{\partial y} \right) ;\quad \frac{\partial }{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial}{\partial y} \right)$$which shows $z$ is exactly what we have already known. We consider $h$, $f$, $g$ all are the functions of $z$ and $\overline{z}$, then we have$$dh=\frac{\partial h}{\partial z}dz+\frac{\partial h}{\partial \overline{z}}d\overline{z}.$$Also we can express$g\circ f$ as
$$g\circ f=g(f(z,\overline{z}),\overline{f}(z,\overline{z})).$$Then,$$d(g\circ f)=\left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial z}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial z}\right)dz+
 \left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial \overline{z}}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial \overline{z}}\right)d\overline{z}. $$Compared with $dh$ and $d(g\circ f)$, the complex version of chain rule is proved.

Title: The complex version of the chain rule


问:
How to prove this summation?
How to prove this summation without mathematical induction? Thanks a lot!
I can't figure out a way to reduce n to $\sqrt{n}$.

$$
\sum_{i=1}^{n}\left\lfloor \frac{n}{i} \right\rfloor = 2\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor - \lfloor \sqrt{n} \rfloor ^2
$$

UPD: I think I got one proof, thanks to all, especially peter's idea.

let f(n) denote the numbers of n, then $\sum_{i=1}^{n}\left\lfloor \frac{n}{i} \right\rfloor=\sum_{i=1}^{n}f(i)$, because $\left\lfloor \frac{n}{i} \right\rfloor$ can be viewed as $i,2i...\left\lfloor \frac{n}{i} \right\rfloor i \leq n$, i's multiple, we can also consider i as some number's divisors, it appears the same time which is $\left\lfloor \frac{n}{i} \right\rfloor$.

then we calcuate f(n), if $a\vert n$ then $\frac{n}{a}\vert n$, it's one to one, then $f(n)=2\sum_{i\leq \sqrt{n}, i\vert n}1$. But there is one exception, that is when n is a square number, it counted twice, let's denote g(n)=1 if n is square, else g(n)=0. then:

$$
f(n) = 2\sum_{i\leq \sqrt{n}, i\vert n}1 - g(n)
$$

thus we get

$$
\sum_{i=1}^{n}\left\lfloor \frac{n}{i} \right\rfloor
=\sum_{i=1}^{n}f(i)
=\sum_{i=1}^{n}(2\sum_{j\leq \sqrt{i}, j\vert i}1 - g(i))
=2\sum_{i=1}^{n}(\sum_{j\leq \sqrt{i}, j\vert i}1) - \sum_{i=1}^{n}g(i)
$$

we do subscript changing and got:

$$
\sum_{i=1}^{n}(\sum_{j\leq \sqrt{i}, j\vert i}1)
=\sum_{j=1}^{\left\lfloor \sqrt{n} \right\rfloor}\sum_{i\geq j^2,j\vert i}1
=\sum_{j=1}^{\left\lfloor \sqrt{n} \right\rfloor}\sum_{k=j}^{\left\lfloor \frac{n}{j} \right\rfloor}1
=\sum_{j=1}^{\left\lfloor \sqrt{n} \right\rfloor}(\left\lfloor \frac{n}{j} \right\rfloor-j+1)\\
=\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor-\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}(i-1)
=\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor-\left\lfloor \sqrt{n} \right\rfloor(\left\lfloor \sqrt{n} \right\rfloor-1)/2
$$

so we got:

$$
2\sum_{i=1}^{n}(\sum_{j\leq \sqrt{i}, j\vert i}1) - \sum_{i=1}^{n}g(i)
=2(\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor-\left\lfloor \sqrt{n} \right\rfloor(\left\lfloor \sqrt{n} \right\rfloor-1)/2)-\left\lfloor \sqrt{n} \right\rfloor\\
=2\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor - \lfloor \sqrt{n} \rfloor ^2
$$

答:
You can see why this holds from the graph of the hyperbola:

The sum $$\sum_{i=1}^{n}\left\lfloor \frac{n}{i} \right\rfloor$$ counts the area under this curve while the sum $$2\sum_{i=1}^{\left\lfloor \sqrt{n} \right\rfloor}\left\lfloor \frac{n}{i} \right\rfloor - \lfloor \sqrt{n} \rfloor ^2$$ counts the area below the square and one leg of the hyperbola twice, then takes away the square which was double counted.

The prove this notice we can write the sum as a sum over divisors $$\sum_{i \le n} d(i) = \sum_{i \le n} \sum_{d|i} 1 = \sum_{dk \le n} 1 = \sum_{i \le n}\left\lfloor \frac{n}{i} \right\rfloor$$

it is the second last form which shows the same symmetry we noticed geometrically, so we can use that one to prove the identity: Just split the sum up like this $$\sum_{dk \le n}  = \sum_{\begin{array}{c}d \le \lfloor \sqrt{n} \rfloor \\ dk \le n\end{array}}+\sum_{\begin{array}{c}k \le \lfloor \sqrt{n} \rfloor \\ dk \le n\end{array}}-\sum_{\begin{array}{c}d \le \lfloor \sqrt{n} \rfloor \\ k \le \lfloor \sqrt{n} \rfloor\end{array}}$$ and we are done

Title: How to prove this summation?


问:
What are the meanings of $\operatorname{trig}(x)^n$ and $\operatorname{trig}^n(x)$?
When a trigonometric function has an exponent does that mean multiply itself or apply itself to the result recursively?

For example, does $\sin(x)^2$ denote $\sin(x)\sin(x)$ or does it denote $\sin(\sin(x))$?  What about $\sin^2x$?

答:
The currently accepted answer claims that $$ \color{blue}{\sin(x)^2}
   := \sin(x^2), \tag{1}$$ but in fact, possibly the most common interpretation
is that $$\color{blue}{\sin(x)^2} := \left[\sin(x)\right]^2 \tag{2}$$
(the standard argument being that it is silly to move the exponent
out of the parenthesis if one actually means $\sin(x^2)$).

However, I avoid writing $\text“\color{blue}{\sin(x)^2}\text”$ altogether, as I instinctively parse it as
$(1)$ rather than $(2).$

For $\left[\sin(x)\right]^2,$ I prefer to write $\text“\color{violet}{\sin^2(x)}\text”,$ knowing that this is the most conventional choice, is unlikely to be interpreted as the rarely-occurring function composition $\sin\left(\sin(x)\right),$ and is at least less potentially controversial than $\text“\sin^{-1}(x)\text”,$ which has clear alternatives $\arcsin(x)$ and $\operatorname{cosec}(x).$

Addendum to restore the OP's Accepted (green-ticked) Answer by PObdr  (referenced above; to be clear: I disagree with its tone of finality), which, together with the comments under it, have been deleted by the community:

$$
\color{blue}{\sin(x)^2} = \sin((x)^2) = \sin(x^2)\\
\color{violet}{\sin^2(x)} = (\sin(x))^2 = \sin(x)\sin(x)\\
\sin(\sin(x)) \text{ is forever alone and never simplified}
$$

Xander Henderson: $\quad$ This is simply not the way that most of the mathematical community interprets $\color{blue}{\sin(x)^2}$—indeed, I only see this interpretation when grading student work, and I mark it down.  This answer is simply incorrect.

me: $\quad$ @XanderHenderson Unless your required interpretation has been stressed in class, it's unfair to mark students down for being unaware that a significant subset of the mathematical community opts to understand $\text“\color{blue}{\sin(x)^2}\text”$ as the square of a function output.

After all, reading $\text“\color{blue}{\sin(x)^2}\text”$ as $\sin(x^2)$ is neither nonsensical (reading it as $\sin\cdot\sin\cdot xx$) nor outre (reading it as $\sin(\sin(x))$ ) nor more unnatural than reading $\text“t(p)^2\text”$ as $t(p^2)$ instead of $[t(p)]^2.$

Since $\text“\color{violet}{\sin^2(x)}\text”$ is by far the most prevalent expression for $[\sin(x)]^2,$ and since reading $\text“\color{blue}{\sin(x)^2}\text”$ as $[\sin(x)]^2$  is not clearly more logically/mathematically valid than as $\sin(x^2)$ (for example, no precedence convention says whether function application or exponentiation binds stronger), $\text“\color{blue}{\sin(x)^2}\text”$ is arguably more ambiguous than $\text“\color{violet}{\sin^2(x)}\text”.$ Certainly, there is no firm convention for interpreting $\text“\color{blue}{\sin(x)^2}\text”;$ for example, see the discussions here and here and here.


问:
Ivy/APL primes algorithm
The Google’s Ivy interpreter is a APL-like math language.
This is an example of first N primes numbers algorithm in Ivy/APL

```<code>op primes N = (not T in T o.* T) sel T = 1 drop iota N
</code>```

Could someone describe the notation for the primes algorithm here?

Ivy/APL doc is here.

答:
It can be directly translated to APL from the vocabularies provided in the doc.

```<code>{(~T∊T∘.×T)/T←1↓⍳⍵}
</code>```

someone who have learned the notation of APL might already figured it out, but since APL isn't very popular, here's an attempt of explanation:

it means assign T with dropping first element of iota N, which is a sequence of numbers 2..N, then filtering out those that belongs to `T∘.×T that is the outer product, what been left are the primes.

Title: Ivy/APL primes algorithm

Body:
It can be directly translated to APL from the vocabularies provided in the doc.

{(~T∊T∘.×T)/T←1↓⍳⍵}

someone who have learned the notation of APL might already figured it out, but since APL isn't very popular, here's an attempt of explanation:

it means assign T with dropping first element of iota N, which is a sequence of numbers 2..N, then filtering out those that belongs to `T∘.×T that is the outer product, what been left are the primes.


问:
Show that any group of order 294 is solvable.
Show that any group of order 294 is solvable.

Proof: Suppose $G$ is a group of order $294=2\cdot 3\cdot 7^2$. By Sylow's Theorem,

$n_2\equiv 1 \mod 2$ and $n_2|147$ $\Rightarrow$ $n_2=1,3,7,21,49,147$.

$n_3\equiv 1 \mod 3$ and $n_3|98$  $\Rightarrow$ $n_3=1,7$.

$n_7\equiv 1 \mod 7$ and $n_7|6$   $\Rightarrow$ $n_7=1$.

Hence there is a normal Sylow $7$-subgroup, $H$, with order $49$.

So where am I suppose to go from here?

答: The normal Sylow $7$-subgroup $H$, being a $p$-group, is solvable. The quotient $G/H$ has order $6$ and is thus solvable (the first non-solvable group has order $60$). Hence $G$ is solvable.


问:
Can I say $|f(x)g(x)|=||fg||$
Let $f,g:[0,1]\to \Bbb{R}$ be continuous functions. Show that $$||fg||\le||f||\space||g||$$
What I have got so far :

$|f(x)| \le\max|f(x)|=$ norm of $f$, $||f||$.$\forall x\in[0,1]$. (Note: I have replaced supremum with maximum.)

$|f(x)||g(x)| =|f(x)g(x)|\le \max |f(x)|g(x)=||f|| \space\space |g|\le \max|f(x)|\space \max|g(x)|=|f||\space\space||g||$

$|f(x)g(x)|\le||f||\space\space||g||$

As I have to show that $||fg||\le||f||\space||g||$
:

Can I say $|f(x) g(x)|=||fg||$? I'm not sure about that  Because $|f(x)g(x)| \le\max|f(x)g(x)|=\max|f(x)| \space \max|g(x)|$

I feel I am missing concept to prove $|f(x) g(x)|=||fg||$, through which I think I finally can prove $||fg||\le||f||\space||g||$

please If you guys could clarify.

答:
I think OP you need to be clear on your definitions.

By definition, $$||f|| = \max_{x \in [0,1]} |f(x)|.$$
And similarly, $$||g|| = \max_{x \in [0,1]} |g(x)|.$$
And as $fg(x) \doteq f(x)g(x)$, also by definition: $$||fg|| = \max_{x \in [0,1]} |f(x)g(x)|.$$

So with definitions in hand, let us now show the desired result. At each point $x_0 \in [0,1]$:
$$fg(x_0) \doteq f(x_0)g(x_0) \le \max_{x \in [0,1]}|f(x)|\max_{x' \in [0,1]} |g(x')| =||f||\cdot ||g||,$$

where the last equation in this above string of relations follows from the def'ns of $||f||, ||g||$. So the above string of relations says in particular:
$$fg(x_0) \le ||f||\cdot ||g||  \ \forall x_0 \in [0,1].$$

And so:

$$\max_{x_0 \in [0,1]} fg(x_0) \le ||f||\cdot ||g||.$$

Thus, the above definition for $||fg||$ implies
$$||fg|| \doteq \max_{x_0 \in [0,1]} fg(x_0) \le ||f||\cdot||g||,$$
which is what you want to prove.

Title: Can I say $|f(x)g(x)|=||fg||$


问:
Random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Find the pdf of $Z$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_Z(z)= P(Z \le z)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Then $Z$'s image is also $(0,1)$ (or $[0,1)$ or whatever). (more details in part D.)

B. Answer (that I found online):

Edit: it says $Z$ is uniform, but I think this is wrong.

C. My intuition for the answer (but I would like to know precisely please):

Edit: Nvm. I think the answer is wrong.

For parts D,E,F,G: Edit: nvm this is wrong.

H. New section: What exactly is the formula of $Z$ in relation to $A,B,C$? Supposedly $f_Z(z)=2(1-z)1_{(0,1)}(z)$. But what exactly is $Z$? $Z=|A-C|$? $Z=\min\{?,?\}$

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
From symmetry we can fix the point $B$ and consider the distribution of the other two point with respect to it. Let assume that the next point is at distance $z$. Then the remaining configuration space for the other point  is $1-z$.

Therefore the pdf in question is
$$
2(1-z)
$$
where the factor $2$ accounts for the number of ways to choose the closest point.

Title: Random points on a circle


问:
Affine mapping and hyperplanes
Let $E$ be an affine space attached to a $K$-vector space $T$ and let $u:E\rightarrow K$ be a non-constant affine mapping. Take $\lambda\in K$ and write $S:=\{x\in E\ |\ u(x)=\lambda\}$. We want to show that $S$ is a hyperplane of $E$: i.e. that $S\ne\emptyset$, $S$ is an affine subset and $\text{codim}_K(D)=1$, where $D$ is the direction $S$.

First of all, why should $S$ be nonempty? Supposing $S$ is indeed nonempty, we can show that $S$ is affine.

Now, how about $\text{codim}_K(D)=1$? Take $a\in S$. Then $D=\{(x-a)+(y-a)\ |\ x,y\in S\}$. How should I proceed? Any suggestions?

答:
Let's let $T$ be a $K$-vector space, and $E$ be the associated affine space. If $u: E \to K$ is a non-constant affine mapping, then there exist ${\bf x}, {\bf y} \in E$ with $u({\bf x}) \neq u({\bf y})$, let us put
$$u({\bf x}) - u({\bf y}) = \alpha \neq 0.$$
By the definition of an affine mapping, for any $c\in K$,
$$u(c {\bf x} + (1-c) {\bf y}) = c u({\bf x}) + (1-c) u({\bf y}) = u({\bf y}) + c(u({\bf x}) - u({\bf y})) = u({\bf y}) + c\alpha$$
so putting $c = \frac{\lambda - u({\bf y})}{\alpha}$ gives us a vector $\bf v$ with $u({\bf v}) = \lambda$, and so $S:=\{x\in E\ \mid u(x)=\lambda\}$ is nonempty.

Showing we have a hyperplane is straightforward; there is a unique linear mapping $v: T \to K$ such that $u({\bf x} + {\bf y}) = u({\bf x}) + v({\bf y})$ for all ${\bf x}, {\bf y} \in T$ (this is slightly different from Bourbaki's wording, but is equivalent; notice that $u$ and $v$ can both be considered as maps $E \to K$ or $T \to K$, but $u$ is affine and $v$ is linear). Since $u$ is nonconstant, $v$ is a nonzero linear functional, and so $H = \{ {\bf x} \mid v({\bf x}) = 0 \}$ is a hyperplane of $T$.

Title: Affine mapping and hyperplanes


问:
Extrapolation of linear transformations
So, assuming that I had two linear transformations $\mathbb{R^n}$ -> $\mathbb{R^m}$, P and K, and there are n linearly independent vectors v for which P(v) = K(v), does P(v) = K(v) for all vectors v?

答: Since any vector in $\Bbb R^n$ is a linear combination of the $n$ linearly independent vectors $\{v_i\}_{i=1}^n$ i.e. $v=\sum_i c_iv_i$, the image of $v$ under $P,K$ is a linear combination of the images of $v_i$, i.e. $P(v)=\sum_i c_iP(v_i)=\sum_i c_iK(v_i)=K(v)$.

Title: Extrapolation of linear transformations


问:
How to calculate the total surface area of the remaining solid?
A solid cube has sides of length 4 cm. A hemisphere of radius 1.5 cm is removed from the cube as shown in the figure. Calculate the total surface area, in cm^2 to 3 significant figures, of the remaining solid.

I got the total surface area of the cube = 6(a)^2 = 96 cm^2
The total surface area of a hemisphere = 3pi(r)^2 = 21.2 cm^2
If I subtract them both to find the tsa of the remaining solid = 96 - 21.2 = 74.8 cm^2 but THE ANSWER IS APPARENTLY 103 cm^2. HOW??

答:
We are talking about surface area and not volume. It is a solid cube and if you cut out a hemisphere, the surface area of resulting shape in fact increases, instead of decreasing.

Net surface area ($S$) = Surface Area of cube ($S_1$) - area of circle in the plane where hemisphere is cut out ($S_2$) + surface area of hemisphere ($S_3)$.

If $a$ is the side of the cube and $r$ is the radius of the hemisphere cut out (where $r \leq \frac{a}{2}$ if the center of the hemisphere is at the center of one of the faces of the cube),

$S = 6 a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$


问:
Eliminate parameter $t$ in a set of parametric equations

Question: Given the set of parametric equations:

$x = 2 \tan t$

$y = 2 \cos^2 t$

Eliminate the parameter $t$ and find the rectangular equation for the curve.

How would I go about solving this? I know that when the equations involve $\sin$ and $\cos$ I can simply square and add them to $1$ but I don't know of any comparable identity for $\tan$ and $\cos^2$.

答:
Hint $\to$

Given (1)$\cos^2t$ +$\sin^2t$=1 , (2) $\tan t$ =$\frac{sint}{cost}$

x=2$\tan t$ =2$\frac{sint}{cost}$, $x^2$=4$\tan ^2t$=4$\frac{sin^2t}{cos^2t}$

with y=2$\cos^2t$ , you should be able to finish the rest from here.


问:
Finding the general solution of the differential equation $\left[ 1+\left( x+y\right)^2\right]dx+\left[1+x(x+y)\right]dy=0 $.
Find the general solution of $\left[ 1+\left( x+y\right)^2\right]dx+\left[1+x(x+y)\right]dy $.

I am stuck at this problem. Here is my solution:

Let $u=x+y$

$du=dx+dy$

$dy=du-dx$

$\left(1+u^2 \right)dx+\left(1+ux\right)(du-dx)=0$

$\left(u^2-ux\right)dx+(1+ux)du=0$

Let $x=uz; dx=udz+zdu$

$\left(u^2-u^2z \right)(udz+zdu)+\left(1+u^2z\right)du=0$

$u^3\left(1-z\right)dz+\left(2u^2z-u^2z^2+1\right)=0$

I do not know what to do next.

[Update]

Using the hint that was provided:

$\left(x+y\right)^2dx +x(x+y)dy=-(dx+dy)$

$u^2dx+ux(du-dx)=-du$

$xdx-udx-xdu+u^{-1}du=0$

$(x-u)dx+(u^{-1}-x)du=0$

Let $M(u,x)=x-u$ and $N(u,x)=u^{-1}-x$

$\frac{\partial M}{\partial u}=-1$ and $\frac{\partial N}{\partial x}=-1$

By partial integration:

$ \displaystyle\int (x-u)dx+ \displaystyle \int \left( u^{-1}+x\right)du=0$

$2ux-x^2+2\ln\vert u\vert +\ln \vert c\vert=0$

$\ln \left[c(x+y)^2\right]=x^2-2(x+y)x$

$-\ln \left[c(x+y)^2\right]=x^2+2xy$

答:
Hint:

$$(x+y)^2dx+x(x+y)dy=-d(x+y)$$

For $x+y\ne0,$

$$-\dfrac{d(x+y)}{x+y}=(x+y)dx+xdy=x\ dx+y dx+x\ dy$$

$$\iff-\dfrac{d(x+y)}{x+y}=x\ dx+d(xy)$$

Now integrate both sides


问:
Range of an expression related to coefficients of a quadratic equation
Let $ax^2+bx+c=0$ be a quadratic equation such that both the roots lie in [0,1]. What is the range of the expression $\frac{(a-b)(2a-b)}{a(a-b+c)}$ ?

I tried using the fact that $\frac{c}{a} \leq 1$ and tried to substitute $2a - b \geq a + b - c$, but got stuck

答:
$$\frac{(a-b)(2a-b)}{a(a-b+c)}=\frac{(1+s)(2+s)}{1+s+p}$$
where $s=-\frac ba$ is the root sum and $p=\frac ca$ is the root product.

$s\in[0,2]$ and for a fixed $s$, $p$ attains a maximum of $s^2/4$ (when the roots are equal) and a minimum of $0$ (if $s\le1$) or $s-1$ (if $s\ge1$). Accordingly, to minimise the expression we must $p=s^2/4$, whereupon it becomes
$$4\cdot\frac{(1+s)(2+s)}{(s+2)^2}=4\cdot\frac{1+s}{2+s}$$
and clearly this is minimised at $s=0$, i.e. the lower bound on range is $2$. To maximise the expression we consider two cases:

If $s\in[0,1]$ we set $p=0$ and the expression simplifies to $2+s$, whose maximum is $3$.

If $s\in[1,2]$ we set $p=s-1$ and the expression simplifies to $\frac{(1+s)(2+s)}{2s}=1/s+3/2+s/2$, which is decreasing on $[1,\sqrt2]$ and increasing otherwise. Thus we check the value at $s=2$, which is also $3$, and conclude that the maximum is still $3$.

Hence the expression range is $[2,3]$.


问:
Show that $\mu(E_a) \leq \frac{1}{a^p} \int_E |g|^pd \mu$
Let $(\Omega, \mathcal{M}, \mu)$ be a measure space, $E \subset \Omega$ and $g: \Omega \to \mathbb{R}$ are measurable. Let $E_a= \{x \in E \space | \space |g(x)| \geq a \}$ for $a>0$.

Show that for $p>0$: $$\mu(E_a) \leq \frac{1}{a^p} \int_E |g|^pd \mu \space .$$

It's probably not that difficult but I don't really know how to do it. I have a lot of similar problems, hopefully I can solve the others if I see one example.

答:
Using the hint given by Kavi, it is simple to see that:
$$a^p\mu(E_a) =\int\limits_{\Omega} \, a^p \mathbb{1}_{E_a} \, \mathrm{d} \mu  =\int\limits_{E} \, a^p \mathbb{1}_{E_a} \, \mathrm{d} \mu  \leq \int\limits_{
E} \,  |g|^p \mathbb{1}_{E_a} \, \mathrm{d} \mu \leq \int\limits_{E} \, |g|^p \, \mathrm{d} \mu $$
Then divide both sides by $a^p$ and then we are done


问:
Sum of residue systems is complete iff they are both complete
Let $p$ and $q$ be distinct prime numbers. The integers $a_1, \ldots, a_p$ and $b_1,\ldots,b_q$ are such that the sums $a_i + b_j$ form a complete residue system modulo $pq$ (that is, there is precisely one sum which is $0$ mod $pq$, precisely one which is $1$ mod $pq$, etc.). Prove that $a_1,\ldots,a_p$ form a complete residue system mod $p$ and $b_1,\ldots,b_q$ form a complete residue system mod $q$.

Any idea on how to approach this? Any help appreciated!

答:
Let's assume $p<q$. You can only conclude that the the $a_i$'s form  a complete system of residues modulo $p$. Let's note  that $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv 0 \pmod{p}$ if and only if $p-1\nmid t$ and $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q\pmod{p}$ for $p-1|t$. Same goes for switching $p$ and $q$.

Inductively one can show that $\displaystyle \sum_{i=1}^{p} a_i^{n}\equiv 0 \pmod{p}$ for $n\leq p-2$;this formalizes mathworkers post . More precisely if we denote with $\displaystyle A_n=\sum_{i=1}^p a_i^n$ and $\displaystyle B_n=\sum_{i=1}^q b_i^n$ we have
$$\sum_{i,j} (a_i+b_j)^n=\sum_{k=1}^{n-1} A_k B_{n-k} \dbinom{n}{k}+qA_n+pB_n$$

Now for $n=p-1$ we have $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q \pmod{pq}$ and thus $A_{p-1}\equiv -1\pmod{p}$. Now Fermat's little theorem gives $a_i^{p-1} \equiv \{0,1\} \pmod{p}$ and thus we must have one number divisible by $p$ and $p-1$ coprime with $p$ amongst the $a_i$. Wlog $p|a_1$.

Now considering the polynomial $\displaystyle f(X)=\prod_{i=2}^{p} (X-a_i)$ in $\mathbb{F}_p[X]$ we can show that all the coefficients are zero; these are symmetric sums in the $a_i$ and using Newton's relations between power sums and symmetric sums. Thus $f(X)=X^{p-1}-a$ and since $a_{i}^{p-1}\equiv 1 \pmod{p}$ we have $a=1$. Since $\displaystyle X^{p-1}-1=\prod_{j=1}^{p-1} (X-j) $ in $\mathbb{F}_p[X]$ we have shown that $\{a_i\}_{i=1}^p$ form a complete system of residues modulo $p$.

Note that the same strategy guarantees that $q|B_n$ for $n<q-1$ using the same recurrence relation. By the same argument as above one gets that $\{b_i\}$ form a complete system of residues mod $q$ also.


问:
Cardinalities of $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$ and $\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$

Consider $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$, and $\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$. $\text{Hom}(A,B)$ is the set of all homomorphisms from $A$ to $B$, where $A,B$ are groups. Find $|\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)|$, and $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)|$.

So, here are my thoughts and questions. First, $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$. Consider some $f\in \text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$. $f$ is completely determined by $f(1) = a$, since $f$ is a homomorphism. What do elements in $\Bbb Q/\Bbb Z$ look like? I am not too comfortable with quotient groups yet. They'd be of the form $q + \mathbb Z$ where $q\in \Bbb Q$, right? Assuming this holds, I feel that $|\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)| = \infty$, because there are infinitely many choices of $q\in \mathbb Q$ for $q + \mathbb Z$. I know that if $q_1-q_2\in \mathbb Z$, then $q_1 + \mathbb Z = q_2 + \Bbb Z$ in $\Bbb Q/\Bbb Z$, but even then, the choices are infinite indeed - the representatives are rational numbers in $[0,1)$.

About $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)|$, consider $f\in \text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$. This seems tougher! For $q\in [0,1)$, I must assign values to $f(q + \mathbb Z)$ from $\{0,1,...,n-1\} \in \Bbb Z/n\Bbb Z$. I'm unable to figure out the number of ways of doing this.

Could you please drop any hints to point me in the right direction? Solutions are good too. Thank you!

Update:

Could it be the case that $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z) \cong \text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Z/n\Bbb Z)$ in general, or at least for some $n$?

Also, on second thoughts, feels like $\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$ is a singleton, i.e. $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)| = 1$ for all $n$ (consisting of only the homomorphism which maps all elements of $\Bbb Q/\Bbb Z$ to $0\in \Bbb Z/n\Bbb Z$. Is that right, and how do I show it?

答:
It is much simpler than that. We have $|Hom(\mathbb{Z}/n\mathbb{Z},\mathbb{Q}/\mathbb{Z})|=n$ and $|Hom(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})|=1$.
Indeed, in first case $f_k(1+n\mathbb{Z})=\frac{k}{n}+\mathbb{Z}$ is homomorphism for all $0\leq k\leq n$. Conversely, if $f$ is homomorphism and $f(1+nZ)=\frac{a}{b}+\mathbb{Z}$, where $a$ relatively
prime to $b$, then $n\frac{a}{b}\in\mathbb{Z}$ and $b$ is divisor of $n$ and $\frac{a}{b}+\mathbb{Z}=\frac{k}{n}+\mathbb{Z}$, where $k\equiv a\frac{n}{b} (\rm mod\, n)$.

In second case, if $f:\mathbb{Q}/\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ is homomorphism and $f(\frac{1}{mn})=k+n\mathbb{Z}$, then $f(\frac{1}{m})=f(n\cdot\frac{1}{mn})=nk+n\mathbb{Z}=n\mathbb{Z}$ for all $m\in\mathbb{Z}$. So, $f(\mathbb{Q}/\mathbb{Z})=0$, where $0=n\mathbb{Z}$. That is, in this case there is a single homomorphism equal to zero


问:
Complete sequence metric space proof
Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Given that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric, show that $(X, d)$ is a complete metric space.

My attempt:

Consider $\{x^j\}_{j\geq 1}$ Cauchy in $d$. Fix $n \geq 1$. Then for all $0 < \epsilon < \frac{1}{2}$, there is an $j_{\epsilon} \in \mathbb{N}$ so that $d(x^{(j)}, x^{(k)}) < \frac{\epsilon}{2^n}$ for all $j, k \geq j_{\epsilon}$. So for $j, k \geq j_{\epsilon}$ we have

$\sum_{m = 1}^{\infty} 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies |x_m^{(j)} - x_m^{(k)}| < \frac{\epsilon}{1-\epsilon} \leq 2\epsilon$

So $\{x^{(j)}\}_{j\geq 1}$ is cauchy $\implies x_n^{(j)} \rightarrow x_n$ as $x \rightarrow \infty$.

Is this correct so far? I have come this far but I'm not sure how to complete the proof. Any assistance is much appreciated.

答:
Your $j_{\epsilon,n}$ should be just $j_{\epsilon}$. Take $m=n$ in your argument. Rest is fine.

You now have $x_n=\lim_{j \to \infty} x_n^{j}$ for each $n$. Now let $x=(x_n)$. Then $d(x,x^{j}) \leq \sum_{m=1}^{N} 2^{-m} \frac {|x_m^{j}-x_m|} {1+|x_m^{j}-x_m|} + \sum_{m=N+1}^{\infty} 2^{-m} \frac {|x_m^{j}-x_m|} {1+|x_m^{j}-x_m|}$. The second term here is at most  $\sum_{m=N+1}^{\infty} 2^{-m} <\epsilon$ if $N$ is large enough. The first term is a finite sum and each term tends to $0$. I will let you write out a proof in terms of $\epsilon$ and $n_0$.

Title: Complete sequence metric space proof


问:
Coercive bilinear form satisfying $Az=\Phi'(z)$
Let $X$ be a Hilbert space, $A$ compact such that there exist $\alpha$ with
$<Au,u>  \ge \alpha||u||^2$ $\forall u $ (i.e coercive)
Let $\Phi \in C^1(X, \mathbb{R})$ such that $\Phi$ satisfies
$lim_{||u|| \to \infty}\frac{|\Phi(u)|}{||u||^2}=0$
Assume that $\Phi$ is weakly continuous. I want to show there exist $z \in X$ such that $Az=\Phi'(z)$

I'm not really sure where to start. I don't see how the assumptions lead to conclusion, any nudge in the right direction would be appreciated

答:
If $A$ is self-adjoint then you should be able to show existence of solutions for
$$
\min_u \frac 12\langle Au,u\rangle - \Phi(u).
$$
Every solution $z$ of this problem then satisfies $Az=\Phi'(z)$.

(Compactness is not needed here)


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

答:
Hint: Use Taylor series of  $\ln(1+x)$ at the point $x=0$ and see what you get. Here is the Taylor series

$$ \ln(1+x) = x-{\frac {1}{2}}{x}^{2}+O \left( {x}^{3} \right) . $$

I think it is clear now.

Added:

$$\ln( 1+\frac{1}{n} )= \frac{1}{n}-{\frac {1}{2}}{\frac{1}{n^2}}+{\frac {1}{2}}{\frac{1}{n^2}}-\dots$$

$$ \implies  n^2\ln( 1+\frac{1}{n} )= {n}-{\frac {1}{2}}{}+{\frac {1}{3}}{\frac{1}{n}}-\dots$$

$$ \implies \lim_{n\to \infty} n^2\ln( 1+\frac{1}{n} ) = \infty .$$


问:
$L^p$ norm equivalence
I am trying to show $A = ||Du||_{L^p}$ is equivalent to $B = \sum_{|\alpha| = 1} ||D^{\alpha} u ||_{L^p}$ in the sense of norms where $ p \geq 1$ by using Jensen's inequality.

$B^p = \sum_{i=1}^n \left( \int_U |u_{x_i}|^p \right)^{1/p} = \left[ \left(\int_U |u_{x_1}|^p\right)^{1/p} + \cdots + \left(\int_U |u_{x_n}|^p\right)^{1/p}\right]= ( a_1 + \cdots + a_n)^p $

and it would be nice if I can somehow show $ A^p = \int_U |Du|^p = \int_U \left(\sqrt{u_{x_1}^2 + \cdots + u_{x_n}^2}\right) ^p$ is greater than $n^{p-1} (a_1^p + \cdots + a_n^p)$ in order to use Jensen's inequality
$$ (a_1^p + \cdots a_n^p) \leq (a_1+\cdots a_n)^p \leq  n^{p-1} (a_1^p + \cdots a_n^p)$$

Any tips please?

Title: $L^p$ norm equivalence

Body:
Assume $2\le p$ (a reverse argument holds for $p\le 2$). By the arithmetic mean inequalities, $$\frac{|a_1|+\cdots+|a_n|}{n}\le\sqrt{\frac{|a_1|^2+\cdots+|a_n|^2}{n}}\le\sqrt[p]{\frac{|a_1|^p+\cdots+|a_n|^p}{n}}\le\max_i|a_i|\tag{1}$$ Hence
$$|a_1+\cdots+a_n|\le n^{\frac{1}{2}}\sqrt{|a_1|^2+\cdots+|a_n|^2}\le n^{1-\frac{1}{p}}\sqrt[p]{|a_1|^p+\cdots+|a_n|^p}\tag{2} $$
$$\sqrt[p]{|a_1|^p+\cdots+|a_n|^p}\le (|a_1|+\cdots+|a_n|)\le n^{\frac{1}{2}}\sqrt{|a_1|^2+\cdots+|a_n|^2}\tag{3}$$

Therefore
\begin{align}A=\|Du\|_p&=\sqrt[p]{\int(|u_1|^2+\cdots+|u_n|^2)^{p/2}}\\&\le n^{\frac{1}{2}-\frac{1}{p}}\sqrt[p]{\int(|u_1|^p+\cdots+|u_n|^p)}\qquad\textrm{by (2)}\\&\le n^{\frac{1}{2}}\max_i\|u_i\|_p\le n^{\frac{1}{2}}\sum_i\|u_i\|_p=n^{\frac{1}{2}}B\qquad\textrm{by (1)}\end{align}

Conversely, \begin{align}B=\sum_i\|u_i\|_p&\le n^{1-\frac{1}{p}}\sqrt[p]{\|u_1\|_p^p+\cdots+\|u_n\|_p^p}\qquad\textrm{by (2)}\\
&=n^{1-\frac{1}{p}}\sqrt[p]{\int|u_1|^p+\cdots|u_n|^p}\\
&\le  n^{\frac{3}{2}-\frac{1}{p}}\sqrt[p]{\int\left(|u_1|^2+\cdots+|u_n|^2\right)^{p/2}}\qquad \textrm{by (3)}\\
&=n^{\frac{3}{2}-\frac{1}{p}}\|Du\|_p=n^{\frac{3}{2}-\frac{1}{p}}A\end{align}


问:
For square matrices $ (AB)^k=I_n$ for some $k$ implies $(BA)^k=I_n$
Let $A$ and $B$ be $n \times n$ square real matrices.

Suppose $(AB)^k=I_n$ for some $k$ then $(BA)^k=I_n$

I know the result holds for $k=1$. I believe this should be easy but I cannot see it. Any ideas?

答:
First, try to prove that $B$ is invertible. To do this, suppose that $B$ is not invertible, i.e. $Bx = 0$ for some non-zero vector $x$, and try to arrive at a contradiction. There might be an easier way to do this too.

Then, take the equation $(AB)^k = I_n$, i.e. $$\underbrace{ABAB\cdots ABAB}_{k \ AB\text{'s}} = I_n$$ and multiply both sides by $B$ on the left and $B^{-1}$ on the right. What do you get?


问:
$f_n\to f$ in $L^2$ implies $(f_n)^k\to f^k$ in the sense of distributions for all $k\in\mathbb{N}$?
I have a very naive question: Let $\{f_n\}_{n\in\mathbb{N}}\subset L^2(\mathbb{R})$ a sequence of functions satisfying $f_n\to f$ in $L^2$ for some $f\in L^2(\mathbb{R})$. Does this implies that, for all $k\in\mathbb{N}$, we have $$
(f_n)^k\to f^k \quad \hbox{in the sense of distributions }?
$$
Is this somehow direct? or maybe directly false? I have the feeling that it might be false since, for example, even on compact sets we could have functions belonging to $L^2$ such that they are not in $L^3$. However, with this naive intuition we could guess that the above convergence should hold for $k=1$, right?

答:
This is false for $k>2$ as you already mentioned. For $k\in [1,2]$ it is true.
Let $\phi$ be smooth with compact support, then by Hoelder inequality
$$
\int_{\mathbb R} (f_n-f)^k \phi = \int_{supp \ \phi} (f_n-f)^k \phi
\le \|\phi\|_{L^\infty} \|f_n-f\|_{L^2}^k |supp \ \phi|^{1-\frac k2}
= c(\phi) \|f_n-f\|_{L^2}^k
$$
with $\frac1q +\frac1k = \frac12$. Here, it is important that the support of $\phi$ is compact and so has finite measure.


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答:
Look at the university websites - Cambridge has a reading list here which has good general section.

There is also the Princeton Companion to Mathematics, if you can get hold of it from a library, which has a range of articles across a variety of mathematical topics at different levels of sophistication. The articles have a bibliography - so when you find a subject that interests you, you can do some further reading.

I would suggest pursuing what interests you rather than getting too focussed on the curriculum.

Also, though challenging, you might explore Hardy's classic Course of Pure Mathematics, which isn't exactly a modern treatment. But there are lots of things in there (particularly in some of the exercises) which are worth knowing.

Title: What should I learn next for pure math?


问:
Question about the representability of a relation
I am trying to prove lemma 9.3,But I cant understand how if we have $T \vdash \phi(\bar{n_1},..\bar{n_r})$. Therefore $T\vdash \psi(\bar{n_1},...\bar{n_r},v) \iff v=\bar{0}$. I don't know how to prove this statement, it will be great if you can help me! Any help is appreiciated. Thanks!

答:
I cant understand how if we have $T ⊢ \varphi(\overline n)$, therefore $T ⊢ \psi(\overline n,v) ⟺ v =\overline 0$ [simplified the number of arguments].

The formula $\psi$ has been defined- as: $(\varphi(\overline n) \land v=\overline 0) \lor  (\lnot \varphi(\overline n) \land v=\overline 1)$.

Suppose that $P(n)$ holds, that means that $K_P(n)=0$.

By representability we have $T⊢ \varphi(\overline n)$ and thus the first disjunct of $\psi$ will hold iff $v=\overline 0$.

We have a predicate $P$ which is represented by formula $\varphi$, and we have the corresponding characteristic function $K_P$.

The gist of the Lemma is to define a new formula $\psi(n,v)$ such that if $P(n)$ holds, and thus $K_P(n)=0$,  then $T \vdash \psi(n,0)$ and if $¬P(n)$ holds, then $T \vdash \psi(n,1)$.

If so, formula $\psi$ will represent the characteristic function $K_P$.


问:
morphism between two princial bundle over the same mainfold?
Let $ \pi : P \rightarrow B$ and $\pi' ; Q \rightarrow B$  be two principal  G- bundles.

Why this is true:

If $f : P \rightarrow Q $ is a morphism of the pricipal G-bundles P and Q ( i.e. $f(p.g)= f(p).g, \forall p \in P, \forall g \in G $), then $ \pi'\circ f = \pi $ ?

答:
This isn't true as stated. For an easy counter example, let $P=Q=B \times G$, the trivial bundle. Then, given any nonidentity map $f:B \to B$ you get a map $P \to Q$ defined by $(x,g) \mapsto (f(x),g)$. This is certainly equivariant, but it doesn't satisfy the property you ask for.

There is a condition missing from the source you cite. Indeed, the answer at the linked page says that a morphism is an equivariant fiber bundle map, this means that it maps over the identity (i.e. satisfies the condition you ask for). The notes you mention just forgot to say this, its implicit in everything happening over the base space.


问:
Isomorphism between quotient of polynomial ring and polynomial ring $\Bbb Z[x]/\Bbb Z \cong \Bbb Z[x]$
The question is as follows:

Consider the group $(\Bbb{Z}[x], +)$. Is the group $\Bbb Z[x]/\Bbb Z$ isomorphic to $\Bbb Z[x]$, where $\Bbb Z[x]$ is a polynomial ring?

I am not able to develop any particular isomorphism for this.

I would appreciate any help.

答:
Consider the principal ideal $(x) \subseteq \mathbb{Z}[x]$ as an abelian group. Then, the homomorphism $\phi: (x) \to \mathbb{Z}[x]/\mathbb{Z}$ given by $\phi(f) = [f]$ is an isomorphism. If $\phi(f) = 0$, then given some expansion $f = \sum_{k = 0}^n c_k x^k $ we have that all $c_k$ are zero except $c_0$. Hence, $f \in \mathbb{Z}$. Since $\mathbb{Z} \cap (x) =\{0\}$, it follows that $f = 0$. Thus $\phi$ is injective.

Conversely, if some $[g] \in \mathbb{Z}[x]/\mathbb{Z}$ we can expand $g = \sum_{k = 0}^n b_kx^k$. Then, let $h = \sum_{k = 1}^n b_kx^k \in (x)$. It follows from their definitions that $g - h = b_0 \in \mathbb{Z}$. Hence $\phi(h) = [g]$ proving that $\phi$ is bijective.


问:
Minimum of $|z_1| + |z_2|+ \dots + |z_n|$ when $z_1 + \dots + z_n = 1$ and all the arguments of $z_m$ are prescribed.
Let $-\frac{\pi}{2} \le \phi_1 < \dots < \phi_k < 0 < \phi_{k+1} < \dots < \phi_N < \frac{\pi}{2}$ be $N$ prescribed phases. Let $a_1, \dots, a_N$ be $N$ real numbers such that
$$
a_1 e^{i \phi_1} + \dots + a_N e^{i \phi_N} = 1.
$$
What is the minimum of $|a_1|+\dots+|a_N|$?

The minimum is achieved when $a_m=0$ for $m \ne k, k+1$. The proof can be easily understood in a geometrically point of view, as kindly suggested in the following answer. Many thanks to all the answers and comments.

答:
A proof with painting a drawing with words. Not very rigorous. :)

Let's look at the problem geometrically. Suppose you chose three values of $\phi$ and built a three-segment path with vertices $0, P, Q, 1$ on the complex plane. Then the middle segment $PQ$ has a slope somewhere between the other two segments'.

Suppose then $P$ is closer to the real axis than $Q$. Let's replace segments $0P$ and $PQ$ with a segment $0Q$. This makes the path shorter at a cost of loosing the slope.

However, we can restore the slope of $0Q$ to the original one of $PQ$ by shitfting $Q$ towards $1$ to some appropriate $R$ – and at the same time we shorten the path again.

This applies to any sequence of $\phi$ arguments: if you choose three or more, you can get rid of the one (or more) initial or terminal values and improve the result, as long as you keep at least one positive and one negative term. Finally, the shortest path is reached with the two slopes closest to zero.

Edit: In the above I meant 'closest' as a neighboorhood in the $(\phi_i)$ sequence, not an arithmetic 'distance'. That is, as you proposed, $a_k<0$ and $a_{k+1}>0$.

Hopefully it can be translated from a a simple geometry to an obfuscated exponential notation of complex analysis, but I really don't feel like doing that now. :)


问:
A definite integration problem on law of large number
The problem is given as follows:

If $g(x),h(x)$ are continuous function on $[0,1]$, satisfying $0\le g(x) <M h(x)$, where $M$ is a nonzero constant. Prove that
  $$\lim_{n\to \infty} \int_{0}^1 \int_0^1\cdots\int_0^1\frac{g(x_1)+g(x_2)+\cdots +g(x_n)}{h(x_1)+h(x_2)+\cdots +h(x_n)}dx_1dx_2\cdots dx_n=\frac{\int_{0}^1g(x)dx}{\int_0^1h(x)dx}$$

I have some idea of the solution, but I think there must be some problem. Here is my solution:

Let $T_i$ be the random variables with density function $\frac{g(x)}{\int_{0}^1 g(t)dt}$ , where $x\in[0,1]$.Similarly, we define $S_i$, with density function $\frac{h(x)}{\int_{0}^1 h(t)dt}$.

By the law of large numbers, $$\lim_{n\to\infty} A_n=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n T_i=1 \mbox{  a.s,}\lim_{n\to\infty} B_n=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n S_i=1 \mbox{  a.s}$$

But I got stuck here, because I cannot combine the two result to get the solution, it does not mean anything since the density function cannot be divided directly. Is there any method to tackle this problem?

Any hint of a solution is welcomed, thanks!

答:
(This was already posted on the site.) Consider an i.i.d. sequence $(U_n)$ uniform on $(0,1)$, then the $n$th integral on the LHS is $E[W_n]$, where
$$
W_n=\frac{g(U_1)+\cdots+g(U_n)}{h(U_1)+\cdots+h(U_n)}.
$$
You already know that $W_n\to w$ almost surely, where $w$ is the RHS. Furthermore, $0\leqslant W_n\leqslant M$ almost surely hence $W_n\to w$ in $L^1$, which implies that $E[W_n]\to w$.</p>


问:
More random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.

Find the pdf of $W$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_W(w)= P(W \le w)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Let $A$ be the anti-clockwise neighbour and $C$ clockwise.

Question:

Well, I hope to find the pdf of $W$ via its cdf, which I think I'm able to compute if I know what $W$ is. Is it $W = \min\{B-A,A-C\}$? Or is it $W=\min\{Z,U\}$, where $Z$ is the distance from $B$ to $A$ (anti-clockwise of course) and $U$ is the distance from $B$ to $C$ (clockwise of course; which I think is equivalent to saying 'distance from $C$ to $B$ anti-clockwise') ? Or something else? (Note: I asked about $Z$ here, but I hope I made this post self-contained.)

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
I will assume that the unit circle means a circle of unit length. Otherwise you should rescale the quantities by the factor $2\pi$.

Let the distance to the closest point be $w$. It leaves the other point the configuration space of $1-2w$.

Therefore the pdf of the least distance is:
$$
\begin{cases}
4(1-2w),&w<0.5\\
0,&w>0.5
\end{cases}
$$

Title: More random points on a circle


答: Given any $e>0,$ take $m$ large enough that $2^{-m}<e/2.$ You have shown that each $(x^j_n)_j\to x_n$ as $j\to\infty.$ So for $n\le m,$ take $j_n $ such that $(j\ge j_n\implies |x^j_n-x_n|<e/2)$. And let $k=\max_{n\le m}j_n.$ Now we have $$ j\ge k \implies d(x^j,x)=\sum_{n=1}^m 2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}+\sum_{n=m+1}^{\infty}2^{-n}\frac {|x^j_n-x_n|}{1+|x^j_n-x_n|}\le$$ $$\le\sum_{n=1}^m2^{-n}(e/2)+\sum_{n=m+1}^{\infty}2^{-n}<$$  $$<e/2+2^{-m}<e.$$

Title: Complete sequence metric space proof

Body:
Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Given that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric, show that $(X, d)$ is a complete metric space.

My attempt:

Consider $\{x^j\}_{j\geq 1}$ Cauchy in $d$. Fix $n \geq 1$. Then for all $0 < \epsilon < \frac{1}{2}$, there is an $j_{\epsilon} \in \mathbb{N}$ so that $d(x^{(j)}, x^{(k)}) < \frac{\epsilon}{2^n}$ for all $j, k \geq j_{\epsilon}$. So for $j, k \geq j_{\epsilon}$ we have

$\sum_{m = 1}^{\infty} 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$

$\implies |x_m^{(j)} - x_m^{(k)}| < \frac{\epsilon}{1-\epsilon} \leq 2\epsilon$

So $\{x^{(j)}\}_{j\geq 1}$ is cauchy $\implies x_n^{(j)} \rightarrow x_n$ as $x \rightarrow \infty$.

Is this correct so far? I have come this far but I'm not sure how to complete the proof. Any assistance is much appreciated.


问:
show this $\prod_{i=1}^{\frac{p-1}{2}}a_{i}\equiv\prod_{i=1}^{\frac{p-1}{2}}b_{i}\pmod {p^3}$
let $p=8k+1$ prime number, show that there exist two set $A=\{a_{1},a_{2},\cdots,a_{\frac{p-1}{2}}\},B=\{b_{1},b_{2},\cdots,b_{\frac{p-1}{2}}\}$ such that
$A\bigcup B=\{1,2,3,\cdots,p-1\},A\bigcap B=\varnothing $,and
$$\prod_{i=1}^{\frac{p-1}{2}}a_{i}\equiv\prod_{i=1}^{\frac{p-1}{2}}b_{i}\pmod {p^3}$$

This problem should be treated with the knowledge of quadratic residue.

答:
This problem appeared in Kvant (M2300). I guess it is more classical than that.
I will quote the construction given in the solution.

Let $p-1=4k$. Let us note that the number of odd quadratic residues is equal to the number of even quadratic residues; same goes for non residues. This is because $x$ and $p-x$ have the same Legendre symbol; $-1$ is a square for $p\equiv 1 \pmod{4}$. There are $2k$ nonzero quadratic residues and $2k$ quadratic residues.
We denote by $a_1, \ldots, a_k$ the odd quadratic residues, $b_1=p-a_1,\ldots, b_k=p-a_k$ the even quadratic residues, $c_1,\ldots, c_k$ the odd quadratic nonresidues, and finally $d_1=p-c_1,\ldots, d_k=p-c_k$ the even quadratic nonresidues. The claim is we can take $\mathcal{A}=\{a_1,\ldots, a_k, d_1,\ldots, d_k\}$ and $\mathcal{B}=\{b_1,\ldots, b_k, c_1,\ldots, c_k\}$. Note that this is consistent with the above numerical constructions given in the comments.

Let $A=a_1\ldots a_k$ and similarly we define $B,C,D$. Let us obtain first a relation between $A$ and $B$. Note that $2^{2k}AB$ is a product of all even numbers which are quadratic residues, but this time between $1$ and $2p-1$. These can only be $b_1, b_2,\ldots, b_k, p+a_1, \ldots, p+a_k$ in some order . Thus
$$2^{2k} AB=\prod_{i=1}^k (p-a_i)(p+a_i)=\prod_{i=1}^k (p^2-a_i^2)\equiv A^2\left(1-p^2\sum_{i=1}^k \frac{1}{a_i^2}\right) \pmod{p^3}.$$ Now note that $2\displaystyle \sum_{i=1}^k \frac{1}{a_i^2}\equiv \sum_{i=1}^{k} \frac{1}{a_i^2}+\sum_{i=1}^{k} \frac{1}{b_i^2}\pmod{p}$. If we invert a quadratic residues it is still a quadratic residue thus $\displaystyle \sum_{i=1}^{k} \frac{1}{a_i^2}+\sum_{i=1}^{k} \frac{1}{b_i^2}\equiv \sum_{i=1}^k a_i^2+\sum_{i=1}^k b_i^2\pmod{p}$. If we take a primitive root $g$ mod $p$ the quadratic residues are $g_i^{2j}$ for $j=0,\ldots, 2k-1$. Thus
$$\sum_{i=1}^k a_i^2+\sum_{i=1}^k b_i^2\equiv \sum_{j=0}^{2k-1} (g^4)^{j}=\cfrac{g^{8k}-1}{g^4-1}\equiv 0\pmod{p} $$ since $p>5$.
Putting it all together $2^{2k} AB\equiv A^2 \pmod{p^3}$ and thus $B\equiv 2^{-2k} A\pmod{p^3}$.
A similar argument shows that $C\equiv 2^{2k} D \pmod{p^3}$ and we are done.


问:
What is the fastest way to find the characteristic polynomial of a matrix?
Finding the characteristic polynomial of a matrix of order $n$ is a tedious and boring task for $n > 2$.

I know that:

the coefficient of $\lambda^n$ is $(-1)^n$,

the coefficient of $\lambda^{n-1}$ is $(-1)^{n-1}(a_{11} + a_{22} + \dots + a_{nn})$,

the constant term is $\det{A}$.

When finding the coefficient of the linear term $\lambda$ of the characteristic polynomial of a $3\times 3$ matrix, one has to calculate the determinant of the matrix $A - \lambda I_n$ anyway. (But you don't have to sum all the terms, only the linear terms.)

Does anybody know a faster way?

答: This may or may not make you happy. The coefficient of $\lambda^{n-k}$ is $(-1)^{n-k}$ times the sum of all the principal $k\times k$ minors, i.e., the determinants of all $k\times k$ submatrices built using the same rows and columns. Note that this generalizes your observations for $k=n$ and $k=1$.</p>


问:
Is every axiom in the definition of a vector space necessary?

Definition: A vector space over a field $K$ consists of a set $V$ and two binary operations $+: V \times V \to V$ and $\cdot: K \times V \to V$ satisfying the following axioms:

Commutativity of $+$.

Associativity of $+$.

Existence of an identity element $\mathbf{0}$ for $+$.

Existence of inverses for $+$.

Compatibility of $\cdot$ with multiplication in $K$.

Distributivity of $\cdot$ over $+$.

Distributivity of $\cdot$ over addition in $K$.

$1_K$ is a left identity of $\cdot$.

Question: Are all seven of the previous axioms necessary (in the sense that weakening any one of them permits a structure which is not a vector space)? If not, which can be weakened (or removed)?

EDIT: user7530 has quite cleverly shown that the commutativity of $+$ can be derived from axioms 2-8. Supposing we throw this out, can the remaining axioms all be proven necessary?

EDIT 2: It was pointed out that axiom 3 cannot simply be thrown out, as the definition of an inverse in axiom 4 depends on the existence of $\mathbf{0}$. What if we tweak the statement of axiom 4 to axiom 4': "For every $x \in V$, there exists $y \in V$ such that $(x+y)+x = x$ and $(y+x)+y = y$"? Is this weakened version equivalent to the original, and if so, does it allow the removal of axiom 3?

答:
Here is an even more trivial example showing why Axiom 8 is necessary (I wanted to post this as a reply under the remark of Jyrki Lahtonen, but the site did not allow me to do that).

Take any abelian group $V$  with any field $F$, and for any $v \in V$ and $a \in F$ set $av=0$ to be the trivial element of $V$ (written additively).


问:
questions about the pumping lemma
Below is the Pumping lemma as stated in Automata and Computability by (Dexter C. Kozen)

Let $A$ be a regular set. Then the following property holds of $A$:

There exist $k≥0$ such that for any string $x,y,z$ with $xyz \in A$ and $|y|≥k$, there exist strings $u,v,w$ such that $y=uvw$,$v\not=ε $, and for all $i≥0$, the string $xuv^iwx\in A$

Here are my questions:

Let  $∑=\{1\}$ $A=\{1\}$. Clearly $A$, is regular. Now, applying the pumping lemma, let $x=1,y=ε,z=ε$ and so $1εε=1 \in A$, and $|y|=0$ so $k=0$  but there is no way to split $y$ such that $v \not = ε$. So, how is the pumping lemma valid for this regular set?

Also, we can think of $k$ as being the number of states, but a DFA cannot have $0$ states. So why is $k$ not required to be strictly greater than 0.

Perhaps you can help my understanding by applying the pumping lemma above to the language $\{1\}$ as well the language of the empty string. What do value does k have in these situations? What about x,y,z and u,v,w?

答:
Suppose that $A$ is regular. The pumping lemma says that there is a $k$ such if $w\in A$ has length at least $k$, then $w$ can be pumped; it does not say that $A$ necessarily has any words of length $k$ or more. In fact it’s clear that if $A$ actually does have a word of length at least $k$, then pumping it will produce infinitely many words. Thus, if $A$ is finite, every word of $A$ must be shorter than $k$. In particular, for $A=\{1\}$ we can take $k$ to be any integer greater than $1$: $A$ will then satisfy the conclusion of the pumping lemma vacuously, since $A$ has no words of length $k$ or more. More generally, if $A$ is finite, and $k>\max\{|w|:w\in A\}$, then $A$ vacuously satisfies the conclusion of the pumping lemma, since it has no words long enough to be pumped.

The pumping lemma is really only interesting when $A$ is infinite.

Title: questions about the pumping lemma


问:
Question regarding surjectivity of a nonzero function satisfying $ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) $
I was practicing that how to prove surjectivity of functions in a functional equation and I came across the following question.

Let $ f : \mathbb R \to \mathbb R $ be a function such that
$$ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) $$
for all $ a , b \in \mathbb R $. Also, suppose that $ f $ is not the all-zero function. Prove that $ f $ is surjective.

I really don't have any idea about how to prove it. A pedantic and easiest possible proof will be highly appreciated and would be helpful.

答:
Consider a function $ f : \mathbb R \to \mathbb R $ such that
$$ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) \tag 0 \label 0 $$
for all $ a , b \in \mathbb R $ which is not the constant zero function. Setting $ a = b = - \frac 1 2 $ in \eqref{0} gives
$$ f \left( - \frac 1 2 + 2 f \left( - \frac 1 2 \right) ^ 2 \right) = 0 \text . \tag 1 \label 1 $$
For any $ a \in \mathbb R $ with $ f ( a ) = 0 $, \eqref{0} gives $ a f ( b ) = 0 $. In case $ a \ne 0 $, that shows that $ f $ is the constant zero function, contrary to the assumption. Thus such $ a $ must be equal to $ 0 $, which by \eqref{1} not only shows that $ f ( 0 ) = 0 $, but also $ - \frac 1 2 + 2 f \left( - \frac 1 2 \right) ^ 2 = 0 $, or equivalently $ f \left( - \frac 1 2 \right) = - \frac k 2 $ where $ k ^ 2 = 1 $, i.e. $ k \in \{ - 1 , 1 \} $. Now, setting $ b = - \frac 1 2 $ in \eqref{0} shows that
$$ f \big( a - k f ( a ) \big) = f ( a ) - k a \text , \tag 2 \label 2 $$
while plugging $ a = - \frac 1 2 $ in \eqref{0} gives
$$ f \left( - \frac 1 2 - k f ( b ) \right) = - \frac k 2 - f ( b ) \text . \tag 3 \label 3 $$
By setting $ a = \frac 1 2 $ in \eqref{2} we have $ f \Big( \frac 1 2 - k f \left( \frac 1 2 \right) \Big) = f \left( \frac 1 2 \right) - \frac k 2 $, which shows that letting $ b = \frac 1 2 - k f \left( \frac 1 2 \right) $ in \eqref{3} we get
$$ f \Bigg( - k f \left( \frac 1 2 \right) \Bigg) = - f \left( \frac 1 2 \right) \text . \tag 4 \label 4 $$
By plugging $ a = \frac 1 2 $ and $ b = - k f \left( \frac 1 2 \right) $ in \eqref{0} and using \eqref{4} we get $ f \Big( \frac 1 2 - 2 f \left( \frac 1 2 \right) \Big) = 0 $, which means that we must have $ f \left( \frac 1 2 \right) ^ 2 = \frac 1 4 $, or equivalently $ f \left( \frac 1 2 \right) = \pm \frac k 2 $. But by \eqref{4}, $ f \left( \frac 1 2 \right) = - \frac k 2 $ leads to $ f \left( \frac 1 2 \right) = - f \left( \frac 1 2 \right) $, contradicting $ f \left( \frac 1 2 \right) ^ 2 = \frac 1 4 $. Thus we have $ f \left( \frac 1 2 \right) = \frac k 2 $.

Now, assume that for some $ c \in \mathbb R $ we have $ f ( c ) = - k c $. By \eqref{3} we have $ f \left( - \frac 1 2 + c \right) = k \left( - \frac 1 2 + c \right) $. Then, setting $ x = \frac 1 2 $ and $ y = - \frac 1 2 + c $ in \eqref{0}, we get $ f ( c ) = k c $, which means that we must have $ c = 0 $. By \eqref{2}, we know that for any $ a \in \mathbb R $, letting $ c = a - k f ( a ) $ we have $ f ( c ) = - k c $, and therefore we must have $ k f ( a ) = a $, or equivalently $ f ( a ) = k a $. Hence, $ f $ has to either be the identity function or its opposite. It's straightforward to verify that those are both surjective solutions.

\eqref{0} is a variant of the functional equation
$$ f \big( a + f ( a ) f ( b ) \big) = f ( a ) + a f ( b ) \text , $$
which has appeared in The American Mathematical Monthly, with an official answer here. This question has been asked many times on AoPS, and you can for example find solutions here, here, here, here and here. The official answer and several others first prove surjectivity and then use that to characterize the solutions of the equation. The answer I gave above is similar to the one in the last linked page. Not only it doesn't appeal to surjectivity first, but also it's shorter and simpler than all others.

Wu Wei Chao, and The University of Louisiana at Lafayette Math Club, Two Functional Equations: 11053, The American Mathematical Monthly 112 (2005), no. 7, 661--662.


问:
Showing that $X_{1:n}$ is sufficient for $\eta$, by factorization
I'm asked to show that $X_{1:n}$ (the minimum order statistic) is sufficient for $\eta$, in the case of a random sample $(X_1, ... , X_n)$ where $X_i\sim EXP(1,\eta$) (this is the two-parameter exponential distribution $EXP(\theta,\eta):$ $(1/\theta)\exp(-(x-\eta)/\theta)$, $x>\eta$; in this case $\theta=1$), by using the "factorization method", that is, writing $f(x_1,...,x_n;\eta)$ as $g(s,\eta)h(x_1,...,x_n)$, where $S$ is the statistic ($X_{1:n}$, in this case), $g(s;\eta)$ does not depend on $x_1,...,x_n$ except through $s$, and $h(x_1,...,x_n)$ does not involve $\theta$.

I have $f(x_1,...,x_n;\eta)=\exp(n\eta)\exp(-\sum_{i=1}^n x_i)$. The sum in the exponential is the same as $\sum_{i=1}^n x_{i:n}$, but I need an expression that involves $x_{1:n}$ in one factor and the $x_i$'s in the other. I don't know how to do this.

I know there are formulas for joint pdf's of any set of order statistics (quite lengthy), but I really don't know how to proceed.

Thank you.

答:
I'm going to make an educated guess that $X \sim \mathrm{Exp}(1,\eta)$ means that $X$ has density
$$\newcommand{\Ind}[1]{\mathbf{1}_{(#1)}}
f(x) \equiv f_X(x) = e^{-(x-\eta)}\Ind{x\geq\eta} \>,
$$
where $\Ind{A}$ is the indicator function of the event $A$.

If $X_1, X_2, \ldots$ is an iid sample from this distribution, then the joint density is
$$
\prod_{i=1}^n f(x_i) = \prod_{i=1}^n e^{-(x_i - \eta)} \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \prod_{i=1}^n \Ind{x_i \geq \eta} = e^{-\sum_{i=1}^n x_i} e^{n \eta} \Ind{\min_i x_i \geq \eta} \> ,
$$
where the last equality follows since the the product of the indicators $\Ind{x_i \geq \eta}$ is one if and only if the indicator $\Ind{\min_i x_i \geq \eta}$ is one and otherwise is zero.

Now, take $h(x_1,\ldots,x_n) = \exp(-\sum_i x_i)$ and $g(s,\eta) = e^{n \eta} \Ind{s \geq \eta}$ with $s = \min_i x_i$ and apply the factorization theorem.

Hence $S = \min_{1 \leq i \leq n} X_i$ is sufficient for $\eta$.


问:
Picard-Lindelöf Existence & Uniqueness Help
I'm not quite positive where to even begin with this problem.  Can somebody help me interpret what it is I need to do and/or how to find the correct solution?

By the Picard-Lindelöf existence and uniqueness theorem (also known as Picard's existence theorem or Cauchy–Lipschitz theorem), we can say that $x^\prime = x^{2}$, $x(0) = x_{0}$ has a unique solution on an interval $[-b/M , b/M]$ where $$ \sup_{x \epsilon B_{b} (x_0 )} \parallel f(x) \parallel $$.  For $x_0 > 0$, show that $M = (x_0 + b)^2$ and then find a value of $b$ to maximize this interval.

Thanks!

答:
The first thing you want to do is explicitly solve this ODE. I will write $y' = y^2$ and $y(0) = x_0$, using the variable $y$ for the sake of convention. The solution to this is

$$y(x) = \frac{1}{c - x}$$

when $x=0$, you have $1/c$, so set $c = 1/x_0$. You can see that this will not have a solution that exists for all time, since this function blows up at $x=c = 1/x_0$. Thus you will want to assert that the solution exists for $|x| < c$. Now try and rewrite this condition in the form you are asked.


问:
Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$

$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$

Find $x$.

I am not sure how to proceed.

Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?

Any help would be sincerely appreciated.

答:
This is just an empirical answer. Consider

$$f(n) = \sqrt{t+\sqrt{t+1 - \sqrt{t+2 + \sqrt{t+3 - \cdots}}}}$$

We need to find $f(1)$. But

\begin{align}
f(1) &= \sqrt{1+\sqrt{2 - f(3)}}\\
f(3) &= \sqrt{3+\sqrt{4 - f(5)}}\tag{Similarly moving on }\\
f(n) &= \sqrt{n+\sqrt{n+1 - f(n+2)}}\\
\end{align}

After squaring and rearranging, we get

$$f(n+2) = n+1 - (f(n)^2 - n)^2$$

I used the wolfram alpha recurrence solver to solve the recurrence and it gave me this:

$$\begin{array}{|c|c|c|c|}
\hline
n& 0 & 1 & 2 & 3 & 4 \\ \hline
 g(n) & 0  & 1 & 1 & 2& 2\\ \hline
\end{array}$$

I don't know if this helps, but $x = 1$.


问:
Unclear how to prove this logarithmic reduction of a sequence, in reference to the birthday problem
On page 46 of Rouse Ball and HSM Coxeter's Mathematical Recreations and Essays, 13th ed., appears this step, in reference to the birthday problem
$$
(1-\frac{1}{365})(1-\frac{2}{365})...(1-\frac{n-1}{365})=1/2
$$
"by taking logarithms, we obtain approximately"
$$
\frac{1}{365}+\frac{2}{365}+...+\frac{n-1}{365}=log_e 2
$$

I'm unfamiliar with the step taken, and after some thinking and searching, haven't found any justification for the step.

答:
They use the linear approximation
$$
\log(1-x) \approx -x
$$
which is good for $x$ close to $0$. And of course $\log(xy) = \log x + \log y$.


问:
Dirac delta derivative identity
I have found the following identity online $f(t)\delta'(t) = f(0)\delta'(t) − f'(0) δ(t)$ however I am quite stuck on to how I would start this to prove this identity, I have an idea that I should use the distribution product rule however I can't seem to find it. Any help on how I can start this will be greatly appreciated hopefully I can continue and prove this identify.

答:
By the product rule:
$$
(f\delta)' = f'\delta + f\delta'
$$

But $f\delta = f(0)\delta$ and $f'\delta = f'(0)\delta$ so
$$
(f(0)\delta)' = f'(0)\delta + f\delta'
$$
i.e.
$$
f\delta' = (f(0)\delta)' - f'(0)\delta = f(0)\delta' - f'(0)\delta.
$$

Title: Dirac delta derivative identity


问:
Finding parameter interval for closed curve in folium of Descartes
What would be an analytical approach to showing that the parameter interval of the parametrized closed curve $\left(\frac{3t}{1+t^3}, \frac{3t^2}{1 + t^3}\right)$of folium of Descartes is $t \in [0, \infty)$? I can only come up with an argument that the starting point is the origin $(0, 0)$, and as the parametrized curve is closed, we can examine for which $t$ the parametrized point reaches/approaches the origin.

答:
The notion of the closed loop portion in the first quadrant as corresponding to $t \in [0,\infty)$ certainly seems to suggest letting $\theta \in [0, \pi/2)$ and $t = \tan \theta$.  Then if we can find a function $r(\theta)$ such that $$(r(\theta) \cos \theta, r(\theta) \sin \theta) = \left(\frac{3 \tan \theta}{1 + \tan^3 \theta}, \frac{3 \tan^2 \theta}{1 + \tan^3 \theta}\right),$$ we can express the folium as a polar curve whose distance from the origin is $r(\theta)$ for the respective angle $\theta$.  Such a condition is easily met, since $$r^2 (\theta) =  \frac{9 (\tan^2 \theta + \tan^4 \theta)}{(1 + \tan^3 \theta)^2} = \left(\frac{3 \tan \theta \sec \theta}{1 + \tan^3 \theta} \right)^2.$$
All that remains is to show that $$r(\theta) = \frac{3 \tan \theta \sec \theta}{1 + \tan^3 \theta} = \frac{3 \sin \theta \cos \theta}{\sin^3 \theta + \cos^3 \theta}$$ on the interval $[0, \pi/2]$ has zeroes only at $0$ and $\pi/2$, is strictly positive otherwise, and has but one critical point.  The first claim is obvious.  The second is also straightforward, since $\sin \theta$ and $\cos \theta$ are positive for $0 < \theta < \pi/2$.  The last claim follows from the derivative:
$$r'(\theta) = \frac{(\cos \theta - \sin \theta)(\cos^4 \theta + \cos^3 \theta \sin \theta + 3 \cos^2 \theta \sin^2 \theta + \cos \theta \sin^3 \theta + \sin^4 \theta)}{(\cos^3 \theta + \sin^3 \theta)^2},$$ and the numerator is zero if and only if $\sin \theta = \cos \theta$ or $\theta = \pi/4$.  This shows the loop has a unique maximum distance from the origin.  Finally, there is symmetry, since $r(\theta) = r(\pi / 2 - \theta)$.


问:
Prove that $\{1, 11, 1001,\dots\}$ is an irregular language
Let $L:=\{1, 11, 1001,\dots\}$ be the language with alphabet $\{0,1\}$ which is formed by all powers $3^n, n=0,1,\dots$ written in binary notation. How to prove that $L$ is not regular?</p>

答:
Use the pumping lemma:

Assume in contradiction that $L$ is regular, so any word $w$ in $L$ longer than $n$, can be divided into $w=xyz$ such that:

$y$ is not empty word

$|xy| \le  n$

and for each $k$, the word $xy^kz$ is in $L$ too.

Now, let $n=3$, and $w=1001$ such that $x = \emptyset, y=100$ and $z=1$. Take $k = 2$, we'll get a new word $w' = 1001001$, which is in binary $73$, which is NOT in $L$.

It's the second time I give an answer, I'll be glad to have a feedback from the others.


问:
A cardinal greater than $2^{\aleph_0}$ of countable cofinality
I am trying to find a cardinal $k\ge \mathfrak c$ such that $\text{cof}(k)=\aleph_0$.
It is know that $\text{cof}(\aleph_\omega)=\aleph_0$. Obviously under CH $2^{\aleph_0}<\aleph_\omega$. In general (that is, without assuming CH), is is true that there is a cardinal of countable cofinality bigger than the continuum?
The only thing that comes to my mind is to use the sequence $\aleph_\omega,\aleph_{\aleph_\omega},\dots,$ but I am not sure on how to prove that this sequence eventually surpasses $\mathfrak c$ (if this is true)

答:
You have two easy options:

We know that $2^{\aleph_0}=\aleph_\alpha$ for some $\alpha$. Then take its $\omega$th successor, which is $\aleph_{\alpha+\omega}$, which can also be written as $(2^{\aleph_0})^{+\omega}$.

Use $\beth$ numbers, where $\beth_0=\aleph_0$, $\beth_{\alpha+1}=2^{\beth_\alpha}$, and for a limit ordinal $\alpha$, $\beth_\alpha=\sup\{\beth_\gamma\mid\gamma<\alpha\}$. Then $2^{\aleph_0}$ is always $\beth_1$ by definitions, and $\beth_\omega$ is where your answer lies.


问:
Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$. Show that $n$ divides $k!$
Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$ with $H\ne G$. Show that $n$ divides $k!$.</p>

答:
A pity such a question will remain unanswered. Define a group action

$$X:=\left\{\;gH\;;\;g\in G\;\right\}\;,\;\;G\times X\to X\;,\;\;x(gH):=(xg)H$$

Prove the above indeed is an action of $\,G\,$ on $\,X\,$ , and thus we get the induced homomorphism $\,\phi:G\to \text{Sym}(X)\,$ . The kernel of this homomorphism, also known as the core of $\,H\,$ in $\,G\,$ , is characterized as the largest normal subgroup of $\,G\,$ contained in $\,H\,$ .

But since $\,G\,$ is simple we get then that $\,\ker\phi=1\,$ , and this means we can embed $\,G\,$ into $\,\text{Sym}(X)\,$ , and this means, by Lagrange's Theorem, that $\,|G|\,\mid\,[G:H]!\,$ ...