问:
What is the minimum number of sets such that the product of elements in each of them is less than k?
Given a positive number n and a positive number k. How to find the minimum number of sets such that for each set s , the product of all elements in s is less than or equal to k.

And also all the sets should contain only elements from 1 to n (both inclusive).And each number from 1 to n should be in exactly one set.

My approach was to first find the square root of k (let it be t). And then divide the numbers from 1 to n into 2 parts. First part has numbers less or equal to t , second has numbers greater than t and less than equal to n. Then we can say that no two number in the second part can be in the same set , because their product will be more the k. But this does not completely answer the question. How to do it?

答: You can solve this as a bin packing problem where the $n$ items have weights $\log 1, \dots, \log n$ and each bin has capacity $\log k$.


问:
Finding the length of $DE$ given $AB = 4$ and $BE = 5$
As shown in the diagram, $ABCD$ is a parallelogram where $DC$ is tangent to the circumcircle of $\triangle ABC$ which intersects $AD$ at $E.$ If $AB = 4$ and $BE = 5,$ find the length of $DE.$

Firstly, I noted that Power of Point could be used in this problem. I let $DE = x$ and I setup the equation $$x \cdot (x+AE) = 16,$$ since $AB = CD.$ However, from here, I got stuck as I do not know how to use the fact that $BE = 5.$ Can somebody help me?

答:
Since $AE$ and $BC$ are parallel, $ABCE$ is a symmetric trapezium, implying $CE=4$ and $AC=5$. By Ptolemy's theorem
$$(x+AE)AE+4^2=5^2\implies (x+AE)AE=9$$
With the equation you already have, this produces the result $x=16/5$ and $AE=9/5$.

ContentLicense: CC BY-SA 4.0


问:
How to prove this inequality $xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$
Let $x,y,z$ is real numbers,and such that $x+y+z=1$,and in $\Delta ABC$,prove that

$$xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$$

I think this inequality maybe use $x^2+y^2+z^2\ge 2yz\cos{A}+2xz\cos{B}+2xy\cos{C},x,y,z\in R$

Thank you everyone.

答:
I take $x$ and $y$ as independent variables and put $z:=1-x-y$.

For the moment, let $a$, $b$, $c$ be arbitrary positive constants, and consider the function
$$g(x,y)= c x y + (a y + b x)(1-x-y)\ .$$
It has a single critical point $$(x_0,y_0)=\left({a(b+c-a)\over \Delta},\ {b(a+c-b)\over\Delta}\right)\ ,\quad\Delta:=2(a b+b c+c a)-(a^2+b^2+c^2)\ .$$
We therefore look at the function
$$q(u,v):=g(x_0,y_0)-g(x_0+u,y_0+v)=b u^2+(a+b-c)uv+a v^2\ .$$
We have $a>0$ and $\det(q)= ab-{1\over4}(a+b-c)^2={1\over4}\Delta$. Therefore, when $\Delta>0$, the quadratic form $q$ is positive definite:  $q(u,v)>0$ for all $(u,v)\ne(0,0)$, which implies that $g$ assumes a global maximum at $(x_0,y_0)$. It follows that when $\Delta>0$ we are sure that
$$g(x,y)\leq g(x_0,y_0)={a b c\over\Delta}\ .$$
Now in our case
$$a=\sin^2\alpha\ ,\quad b=\sin^2\beta,\quad  c=\sin^2\gamma=\sin^2(\alpha+\beta)\ .$$
A simple calculation shows that we indeed have
$$\Delta=4\sin^2\alpha\sin^2\beta\sin^2(\alpha+\beta)=4 a b c>0\ ,$$
so that we arrive at the stated inequality $g(x,y)\leq{1\over4}$.

ContentLicense: CC BY-SA 3.0


问:
Intuitive reason why $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ when $a$ and $b$ are negative
As a rule of thumb I have learnt that $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ if $a$ and $b$ are negative.

I have not understood why this was a problem and took the word of my teacher.

Is this something so fundamental that it should be taken for granted or is there a reason why we can't

I'm sorry for this basic question but I would like to understand this concept a bit more .

答:
The reason for the subtlety is that the square root is a multivalued function, i.e. for any $y$ there is more than one (two, in fact) $x$ such that $x^2 = y$. For example, if $y = 4$, then $x=2$ or $x=-2$, if $y = -9$ then $x = 3i$ or $x = -3i$ where $i$ is the imaginary unit.

In order to see how the multivalued nature of the square root undermines the product rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$, recall that every complex number can be written in the polar form as $z = re^{i\theta}$ for a non-negative $r$ and an angle $\theta$ called the argument. This notation makes it clear that raising $z$ to a power $z^a$ corresponds to multiplying the angle $\theta$ by the exponent $a$. Thus, taking a square root corresponds to halving $\theta$. Note that $\theta$ is periodic

$$
e^{i\theta} = e^{i(\theta + 2\pi)}.
$$

Halving the angles on the left hand side and the right hand side allows us to arrive at two different square roots: both $e^{i\theta/2}$ and $e^{i(\theta/2 + \pi)}$ are valid square roots of $e^{i\theta}$. Because $e^{i\pi} = -1$ (see Euler's identity), the two square roots differ in sign (c.f. $2$ and $-2$, $3i$ and $-3i$ in the examples above).

The reason that the product rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails is that due to periodicity in argument the two sides can "wind around" a different number of times. Using the example from one of the other answers

$$
\sqrt{-1}\sqrt{-1} = \sqrt{e^{i\pi}}\sqrt{e^{i\pi}} = e^{i\pi/2}e^{i\pi/2} = e^{i\pi} = -1 \\
\sqrt{(-1)(-1)} = \sqrt{e^{i\pi}e^{i\pi}} = \color{green}{\sqrt{e^{i2\pi}} = \sqrt{e^0}} = \sqrt{1} = 1.
$$

Note how in the step marked in green we have reduced the angle $2\pi$ to zero. This corresponds to taking the principal value of a multivalued function. If we chose not to do this and instead halved the angle in that step then we would have obtained the same result as in the first equation

$$
\sqrt{(-1)(-1)} = \sqrt{e^{i\pi}e^{i\pi}} = \color{red}{\sqrt{e^{i2\pi}} = e^{i2\pi/2}} = e^{i\pi} = -1.
$$

Note however that the transformation in red is in fact incorrect, because it fails to choose the principal value of a multivalued function.

Thus, the product rule fails because to use square root in algebraic expressions (and to treat it as a function more generally) we force it to take the principal value and that arbitrary choice cannot be made consistently across expressions that wind the argument different number of times, such as $\sqrt{a}\sqrt{b}$ and $\sqrt{ab}$. More concretely, in $\sqrt{ab}$ the product of $a$ and $b$ is allowed to wind a full $2\pi$ argument before being forced to the principal value by square root, but in $\sqrt{a}\sqrt{b}$ each argument is halved before they can add up to a full $2\pi$.

In summary, in an expression like $\sqrt{a}\sqrt{b} = \sqrt{ab}$ the principal value is chosen from among the set of valid square roots at different stages of the calculation and it turns out that exactly when you take the principal value affects the result.

Remark: Note that the principal value of the square root of a non-negative real number does in fact satisfy the product rule, so if $a > 0$ and $b > 0$ then $\sqrt{a}\sqrt{b} = \sqrt{ab}$.


问:
Intermediate value theorem in topological spaces
I am thinking of a proposition analogous to the intermediate value theorem for a scalar function on reals:

In a topological space $(X, \tau)$ where X is a connected set, consider a set $S \subseteq X$ and a continuous function $f:\mathbb{R} \rightarrow X$. For some $a, b \in \mathbb{R}$ with $a<b$, the function satisfies
$$
f(a) \in S^{\circ} \land f(b)\in X\setminus \bar{S}
$$
Then, there exists $c \in (a,b)$ such that
$$
f(c) \in \partial S
$$

Is there any theorem useful for proving this proposition? Or any advice will be appreciated. I am currently trying to get a hint from Wikipedia proof

答: The assumption that $X$ is connected is irrelevant; what is relevant is that the interval $[a,b]$ is connected.  Suppose no such $c$ existed.  That means that for all $x\in[a,b]$, $f(x)$ is either in $S^\circ$ or $X\setminus \overline{S}$.  That is, the sets $U=f^{-1}(S^\circ)$ and $V=f^{-1}(X\setminus\overline{S})$ cover $[a,b]$.  But $U$ and $V$ are disjoint open sets, which both have nonempty intersection with $[a,b]$, so this contradicts the fact that $[a,b]$ is connected.


问:
$P(\text{duck})$ given $P(\text{look})$ and $P(\text{swim})$ and $P(\text{quack})$

"If it looks like a duck, swims like a duck, and quacks like a duck,
then it probably is a duck"

We have the 3 probabilities:

  $P(\text{look}) = 80\%$

  $P(\text{swim}) = 70\%$

  $P(\text{quack}) = 90\%$

I would assume then that
$P(\text{duck}) = P(\text{look})·P(\text{swim})·P(\text{quack})$ $= 0.8 · 0.7 · 0.9$ $= 0.504$

This is obviously dead wrong. Common sense expects the resulted probability close to $1$. What would be right operation among the individual probabilities to represent the original saying correctly?

Update:

Giving some context: We have a problem in an industrial environment, where we have to validate if our calculated values are true or not. We can conduct a few independent measurements, each addresses a different aspect of the same setup and each measurment returns a yes or no with a probability. Then we have to summarize them and compare to the theoretical calculations. What we do is duck typing.

Update 2:

While reading up the answers did hit me: the more independent measurement events we do on properties of the duck, should prove or disprove the "duckness" of the object.
Should we have 6 independent measurements (look, swim, quack, fly, eat, walk), each around probability of 0.1, our calculation should conclude that this is certainly not a duck. Should we get confirming measurements (higher than 0.5), I expect higher "duckness" probability.

答:
The confusion, both in your question and in several of the responses, arises from improper notation for, and definitions of, the relevant events.

We first have to define appropriate conditional probabilities.  Let $L$, $S$, and $Q$ represent the events that a randomly selected object looks, swims, and quacks like a duck, respectively.  Let $D$ represent the event that a randomly selected object is in fact a duck.

Then the given probabilities are properly expressed as

$$\Pr[D \mid L] = 0.8 \\ \Pr[D \mid S] = 0.7 \\ \Pr[D \mid Q] = 0.9.$$

We will also impose another set of assumptions, namely that $$\Pr[L \mid D] = \Pr[S \mid D] = \Pr[Q \mid D] = 1,$$ which means a duck, with certainty, looks, swims, and quacks like a duck.  Hence $$0.8 = \Pr[D \mid L] = \frac{\Pr[L \mid D]\Pr[D]}{\Pr[L]} = \frac{\Pr[D]}{\Pr[L]},$$ or $\Pr[D] = 0.8 \Pr[L]$.  Similarly, $$\Pr[D] = 0.7 \Pr[S], \\ \Pr[D] = 0.9 \Pr[Q].$$  Consequently, assuming $L$, $S$, and $Q$ are independent events, $$\Pr[L \cap S \cap Q] = \frac{\Pr[D]^3}{(0.8)(0.7)(0.9)}.$$  This implies $$0 \le \Pr[D] \le \frac{\sqrt[3]{63}}{5}.$$

Now, you want to compute $$\Pr[D \mid (L \cap S \cap Q)] = \frac{\Pr[(L \cap S \cap Q) \mid D]\Pr[D]}{\Pr[L \cap S \cap Q]}.$$  We note that $\Pr[(L \cap S \cap Q) \mid D] = 1$, since given $D$, each of $L$, $S$, and $Q$ are certain events.  So we obtain $$\Pr[D \mid (L \cap S \cap Q)] = \frac{(0.8)(0.7)(0.9)}{\Pr[D]^2}.$$  We also note this implies stricter bounds on $\Pr[D]$, namely $$\frac{3}{5} \sqrt{\frac{7}{5}} \le \Pr[D] \le \frac{\sqrt[3]{63}}{5}.$$ This is the best you can do with the given information.  It implies that the unconditional probability that a randomly selected object is a duck is at least $0.70993$ and no more than $0.795811$, and the probability that an object that looks, swims, and quacks like a duck is actually a duck ranges from $0.795811$ to $1$, depending on the underlying prevalence of ducks.

However, the fact that $\Pr[D] = 0.7 \Pr[S]$ means that $\Pr[D] \le 0.7$, so we obtain a logical contradiction:  it is not possible for $L, S, Q$ to be independent events.  Therefore, one must have some kind of positive correlation between these, in which case the permissible range of $\Pr[D]$ expands.  However, without a quantitative characterization of this correlation, the question without the independence assumption becomes even more under-specified than it already is.


问:
Reverse Hölder Continuity and Hausdorff dimension
Let $f$ be a function on $[0,1]$. Say that $f$ is reverse Hölder continuous of exponent $\beta > 0$ if there is a $C >0$ such that for any $s<t\in [0,1]$, there exists $s',t'\in [s,t]$ such that $|f(s') - f(t')| \geq (t-s)^\beta$. I am trying to show that if this is the case, then the Hausdorff dimension of the graph of $f$ is at least $2-\beta$. I then want to show that standard Brownian motion satisfies this condition with probability 1 for $\beta  > 1/2$, so that the Hausdorf dimension of its graph is at least $3/2$.

This is based on an exercise from "Brownian Motion" by Morters and Peres.

For the first part, I considered a cover of the graph $G$ of $f$ by open balls and tried to show that the projections of various rescalings of these balls still cover $[0,1]$. But, this did not get me anywhere and I still fail to see where the $2-\beta$ comes in. For the second part, involving Brownian motion, I tried to break up $[s,t]$ into small independent increments and show that at least one of these increments is large if there are sufficiently many, but this did not seem to work either (the probability that an increment is large decays too quickly). Does anyone have any suggestions?

答:
With the given conditions, the graph of the function need not have Hausdorff dimension at least $2-\beta$. Consider the characteristic function on the rationals, $1_\mathbb{Q}$.

For any $s<t<1$, pick $s',t' \in [s,t]$ so that $s' \in \mathbb{Q}, t' \notin \mathbb{Q}$. Then $|f(s')-f(t')|=1\geq (t-s)^{1/2}$. Hence this is reverse Holder continuous with exponent 1/2.

However, the graph of $f$ is essentially $[0,1]\times\{0\}$ (with countably many points removed). You can cover this by $(-1,2)\times\{0\}$. Hence its 1-dimensional Hausdorff content is finite, and non-zero. Its Hausdorff dimension is definitely not > 1.5.

You might need to impose a continuity condition on $f$.


问:
Doubt about derivatives.
$$f(x+y^2) = f(x) + f(y)^2$$
Hello everybody.
I am doing this exercise, I need to find all functions real valued $f: \mathbb{R} \to \mathbb{R}$ that satisfy the condition above.
I have already found some type of functions with a little of guess and check, but I would like to go beyond.

I thought that would be a good idea to derivate both sides, but I am a little confused how to do it.

As it was said, it is a function from $\mathbb{R}$ to $\mathbb{R}$, not from $\mathbb{R}^2$ to $\mathbb{R}$, so it makes me thought that would be a simple derivative, but how to do it?

Can I say that $\frac{df(x+y^2)}{d(x+y^2)} = \frac{df(u)}{du} = \frac{df(x)}{dx}$
?

So that it would lead us to:
$f' = f' + 2f(x)f'$

答:
Yon cannot make any assumptions of continuity, differentiability etc.
Since $(f(y))^{2} \geq 0$  we see that $f(x+y^{2}) \geq f(x)$. So $f(x+t) \geq f(x)$ for all $t \geq 0$. Thus $f$ is an increasing function.

Now $x=y=0$ gives $f(0)=0$ and $x=0$ gives $f(y^{2})=(f(y))^{2}$. Now the equation beconme s $f(x+y^{2})=f(x)+f(y^{2})$ so $f(x+t)=f(x)+f(t)$ for $t \geq 0$. Putting $x=-y^{2}$ in the given equation prove that $f$ is an odd function. We now have $f(x+y)=f(x)+f(y)$ for all $x,y$ It is well known that the only increasing functions with this property are of the form $f(x)=cx$ for some constant $c \geq 0$.

EDIT: As pointed out by user bof we must have $c=0$ or $c=1$ so $f\equiv 0$ or $f(x)=x$ for all $x$.

Title: Doubt about derivatives.


问:
How to get the parametric equation of a rotated cylinder (with certain slope)
I have a basic question but I have failed in solving it. I have the equation of a cylinder which is $y^2 + z^2 = r^2$ (centered in the x-axis). The parametric equation (dependent on $L$ and $s$) is $(x,y,z) = (L, r\cos(s), r\sin(s))$.

I would like to rotate it certain angle $\theta$ (anticlockwise). Thus I have the new axis from the rotation as: $x=x'*\cos\theta + z'*\sin\theta$, $y=y'$ and $z=r*\sin\theta$. However, when rewriting the equation of the cylinder as $(y')^2 + (-x'*\sin\theta + z'*\cos\theta)^2 = r^2$ and parametrizing, I get:  $(x,y,z) = (L, r*\cos(s), z+x'*\tan\theta)$, with $z=r*\sin\theta$. When I plot this, I get a elliptic cylinder.
Does anyone know what am I doing wrong? I need such equation because I will generate multiple cylinders later computationally.

I have followed previous posts such as If I have an oblique cylinder can I trim it in to a rectilinear cylinder? but they actually obtain the elliptic cylinder.

Many thanks!

答:
You have
$$\begin{aligned}
x &= L \\
y &= L \cos \varphi \\
z &= L \sin \varphi \\
\end{aligned}$$
Rotation by $\theta$ around the $y$ axis is
$$\begin{aligned}
x^\prime &= x \cos \theta + z \sin \theta \\
y^\prime &= y \\
z^\prime &= z \cos \theta - x \sin \theta \\
\end{aligned}$$
It looks like your error is in forgetting one summand in the $z^\prime$ component.

Combining the two you get
$$\begin{aligned}
x^\prime &= L \cos \theta + L \sin\varphi \sin\theta \\
y^\prime &= L \cos \varphi \\
z^\prime &= L \sin \varphi \cos\theta - L \sin\theta \\
\end{aligned}$$

For example, at $\theta = 90°$, $\sin\theta = 1$ and $\cos\theta = 0$, and
$$\begin{aligned}
x_{90°} &= L \sin\varphi \\
y_{90°} &= L \cos \varphi \\
z_{90°} &= -L \\
\end{aligned}$$
which is the expected rotated cylinder.

This ninety-degree check is useful when examining ones own rotations.  Whenever the result gets flat (like yours with $\theta = 90°$), you know your rotation formulae was incorrect.


问:
Proving integral with finite number of values greater than any small number is $0$
I was given this question and think it is mostly wrong,there were also more mistakes than what I am going to write, but I would like to know how wrong it is.So would there be any validity to this approach, or is it completely incorrect?

Question Suppose that $f:[0,1]\rightarrow [0,1]$ satisfies the following properties:given any $\epsilon>0,f(x)<\epsilon$ except at finitely many points, prove $\int_{0}^{1}f=0$

My attempt:

Let $T=\{y_1,...,y_m\}$ be the set of points such that $f(x) \geq \epsilon$ let $P=\{x_0,...,x_n\}$ be a partition of $[0,1]$.

Let $A=\{k|[x_{k-1},x_k] \cap T\neq \emptyset\}$

Let $B=\{k|[x_{k-1},x_k]\cap T=\emptyset\}$

Then there are at most $2m$ intervals having endpoints with indices in $A$ containing points in $T$. Let $P*$ be a partition containing these intervals such that

$\Delta x_k\leq\frac{\epsilon}{4m}$

Then $U(f,P*)=\sum_{k \in A}\text{sup}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$

similarly $L(f,P*)=\sum_{k \in A}\text{inf}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$

Now consider a partition $Q$ of intervals with endpoints with indices in $B$.

Let $\Delta x_k<\frac{1}{2(n-2m)}$ Then since $f(x)<\epsilon$ for $f$ on intervals having endpoints with indices in $B$

$U(f,Q)<\frac{\epsilon}{2}$ and $L(f,Q)<\frac{\epsilon}{2}$

Then $L(f,Q)+L(f,P*)<\epsilon$ and $U(f,Q)+U(f,P*)<\epsilon$

My question is was this on the right track of a correct approach?Is there a different way to approach this? When I saw the question I felt a similarity between this question and the proof that thomae's function is integrable, but I do not think I studied/understood that proof well enough, to successfully come up for a proof for this. And I realized in that proof, that didn't come up with separate partitions for intervals having endpoints with indices in $A,B$ like I did.

答:
Your idea is correct and simple. Just deal with $U(P, f) $.

Let $k$ be the number of points where $f(x) \geq\epsilon$. And choose norm of partition less than $\epsilon/2k$. The upper sum on intervals which don't contain these points does not exceed $\epsilon$. And on the intervals which contain these points the upper sum does not exceed $2k(\epsilon /2k)=\epsilon$. Hence the upper sum $U(P, f) $ does not exceed $2\epsilon $.

I am not sure why you chose different $\Delta x_k$ for these different types of intervals. This probably makes your proof a bit difficult to comprehend (at least for me).

You can see that we don't need to deal with lower sum separately because it is already non negative and does not exceed any upper sum.


问:
Find some integral with Lebesgue measure and counting measure
If we define D as:
$$D=\{(x,y)\in \mathbb{R}^2|0<x=y<1\}$$
and let m be the Lebesgue-measure on $(\mathbb{R},\mathcal{B})$ and the counting measure $\tau$ on $(\mathbb{R},\mathcal{B})$.
Then I have to find:
$$\int\int1_{D}(x,y)dm(x)d \tau(y)$$
and
$$\int\int1_{D}(x,y)d \tau(y)dm(x)$$
I think that
$$\int\int1_{D}(x,y)dm(x)d \tau(y)=0$$
bacause the inner integral must give 0 while we have the lebesgue measue on the integral?

While the other integral must give:
$$\int\int1_{D}(x,y)d \tau(y)dm(x)=1$$
while the inner integral must give 1

But I'm not sure if it correct and how I proof these results? Can someone help me to show it correctly and not only by these intuitive thoughts?

答:
For fixed $x$ we have $\int 1_D(x,y)d\tau (y) = 1$ because $1_D(x,y)=1$ if $y=x$ and $0$ otherwise. [You are integrating the characteristic function of the set $\{x\}$ and the integral of the characteristic function of any set $E$ is $\tau (E)$].

On the other hand, for fixed $y$,  $\int 1_D(x,y)dm(x) = 0$ because you are integrating the characteristic function of $\{y\}$ w.r.t. Lebesgue measure and this function is $0$ almost everywhere w.r.t. Lebesgue measure.


答:
Note $\angle DAB = \angle DCB$, which means the arcs $EB$ and $CB$ are equal due to the shared arc $EC$. Then, $EB = CB= DA$ and, per the power of point
$$DE=\frac{DC^2}{DA} = \frac{AB^2}{EB} = \frac{16}5$$

Title: Finding the length of $DE$ given $AB = 4$ and $BE = 5$

Body:
As shown in the diagram, $ABCD$ is a parallelogram where $DC$ is tangent to the circumcircle of $\triangle ABC$ which intersects $AD$ at $E.$ If $AB = 4$ and $BE = 5,$ find the length of $DE.$

Firstly, I noted that Power of Point could be used in this problem. I let $DE = x$ and I setup the equation $$x \cdot (x+AE) = 16,$$ since $AB = CD.$ However, from here, I got stuck as I do not know how to use the fact that $BE = 5.$ Can somebody help me?


问:
Question about number of elements in $HK$ if $H$ and $K$ are finite subgroups of some group $G$
Let $H$ and $K$ be the finite subgroups of some universal group $G$. Define
$$HK = \{hk \,| \, h \in H, k \in K \}$$
I want to prove that $|HK| = \frac{|H||K|}{|H \cap K|}$, where $|Q|$ is the number of elements in set $Q$.
This was my process:

I proved that $H\cap K$ is also a subgroup.

Think of $HK$ as a collection of left cosets of $K$ defined by $H$ i.e.
$$HK = h_1 K \cup h_2 K \cup \ldots \cup h_{|H|} K $$

The set $HK$ as defined above has $|H||K|$ products. But if an element of $hk$ is repeated, we need to make sure we don't overcount.

If $$hk = h'k' \\
hk = (ht)(t^{-1}k) \implies h' = ht, k' = t^{-1} k\\ $$
where $t \in H \cap K$.
This tells me that each element of $HK$ has at least $|H\cap K|$ repetitions, each.

The problem I have is, how do I prove that if $t \in H$ but $t \notin K$, the statement $h' = ht, k' = t^{-1}k$ is untrue?

Essentially, if $a \in G, b\notin G$, is it always true that $ab \notin G$?

The closure property of groups states that if $a,b \in G$, then $ab\in G$. But it makes no mention of $a \in G$ and $b \notin G$. I feel like I am missing something obvious, but I can't place my finger on it...

答:
(The "essentially" part in the question)

Let $G\le P, a\in G, b\in P-G.$ Suppose $ab\in G$, then $a^{-1}(ab)=b\in G$, a contradiction. So $ab\notin G$. Note that we need to assume that there is some group containing both $a$ and $b$, otherwise $ab$ may not be well-defined.


问:
Consequences of uniform converegence.
Let $g_n(x)=n^2xe^{-nx}$ and given that $\int_0^2g_n(x)dx=1-a_n \rightarrow 1$ as $n \rightarrow \infty$. Then prove that $g_n$ converges uniformly on $[\delta,2]$ for any $\delta \in (0,1)$, but not on $[0,2]$ and $\int_0^2f(x)g_n(x)dx$ converges to $f(0)$ as $n \rightarrow \infty$ for any continuous function $f$ on $[0,2]$.

My attempt:
(Uniform convergence is clear for me from the supremum norm approach.)
$|\int_0^2f(x)g_n(x)dx-f(0)|=|\int_0^2f(x)g_n(x)dx-\frac{1}{1-a_n}(\int_0^2f(0)g_n(x)dx)|$

$=|\int_0^2g_n(x)\left(f(x)-\frac{f(0)}{1-a_n} \right )dx|$

Now how to proceed?

答:
To prove that the convergence is not uniform on $[0,2]$ just use the fact that $g_n(x) \to 0$ for each $x$, so uniform convergence would imply that $\int_0^{2} g_n(x) d x\to 0$. This contradicts the fact that $\int_0^{2} g_n(x) d x\to 1$.

Now $\int_0^{2} f(x)g_n(x) dx-f(0)=\int_0^{2} g_n(x)[f(x)-f(0)] dx+o(1)$. Given $\epsilon >0$ choose $\delta >0$ such that $|f(x)-f(0)| <\epsilon $ for $0 \leq x <\delta$. Split the integral into integrals from $0$ to $\delta$ and $\delta $ to $2$. For the second one use the fact that $f$ is bounded and $\int_{\delta}^{2} g_n(x) dx \to 0$ by uniform convergence.

Title: Consequences of uniform converegence.


问:
Show that a metric space $M$ is compact iff for every continuous function $f:M\rightarrow \mathbb{R}$, such that $f(x)>0$ for all $x\in M$
everybody! Could you give me feedback if my statements below $ \Leftrightarrow $ are correct or I need to change them?

$\quad$Show that a metric space $M$ is compact iff for every continuous function $f:M\rightarrow \mathbb{R}$, such that $f(x)>0$ for all $x\in M$, we have $\inf\limits_{x\in M} f(x) >0$.

($\Rightarrow$)

$\quad$If $A$ is a bounded subset of $\Bbb R$, and if $s=\sup A$ then for each $\epsilon >0$, $s−\epsilon$ is not an upper bound of $A$ (since s is the least upper bound of $A$) and therefore, there is some $a\in A$ such that $a>s−\epsilon$. But, since $s$ is an upper bound of $A$, $a\leq s$. So, $|s−a|<\epsilon$. Since this occurs for every $\epsilon>0$, $s\in \overline{A}$. For a similar reason, $\textbf{inf}$ $A \in \overline{A}$. And, since $\inf f(M)\in f(M)\subset(0,\infty)$, there is some $m\in M$ such that $\inf f(M)>0$ and, since $f(M)>0$, this proves that $\inf f(M)>0$.

($\Leftarrow$)

$\quad$ If there is a sequence $\{x_n\}$ with no convergent subsequence the $E=\{x_1,x_2,⋯\}$ is a closed set. Define $f:E\rightarrow (0,1)$ by $f(x_n)=\frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X\rightarrow (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. $\textbf{(why isn't there a infimum postive?)}$

$\square$

答:
The second part is correct but the first one looks complicated and vague. It is not even clear as to where compactness comes in.

If $M$ is compact and $f$ is continuous then $f(M)$ is compact. The infimum of any compact subset of $\mathbb R$ is attained. So there exists $t \in f(M)$ such that $t \leq f(x)$ for all $x \in M$. Since $t \in f(M)$ we can write $t=f(x_0)$ for some $x_0$. Now $\inf \{f(x): x \in N\}=f(x_0)>0$.


问:
Transposing when finding hypotenuse
Background: I have just about high school knowledge of math, I am sorry if this is a stupid question.

In school, we learned that when transposing, this:

y = x * 3

would become

x = y / 3

I was reading up on trigonometry, and as I was reading the following:

If we look at the general definition -

we see that there are three variables: the measure of the angle x, and
  the lengths of the two sides (Opposite and Hypotenuse). So if we have
  any two of them, we can find the third.

In the figure above, click 'reset'. Imagine we didn't know the length
  of the hypotenuse H. We know that the sine of A (60°) is the opposite
  side (26) divided by H.

From our calculator we find that sin60 is 0.866, so we can write

Transposing:

which comes out to 30.02

(From http://www.mathopenref.com/sine.html)

As per the rules of transposition (that I probably got wrong), isn't it supposed to read

H = 26 * 0.866

What am I missing?

答:
Transpose means to isolate the variable. In the problem given you are trying to isolate the $H$ variable. Using algebra:

our problem:
$$0.866 = \frac{26}H  $$

$$H\cdot 0.866 = \frac{26}H \cdot H$$

$$0.866H = 26$$

$$\frac{0.866H}{0.866} = \frac{26}{0.866}$$

$H = 30.02$ is the correct answer not $30.06$.

Multiplying both sides by $H$ cancels the $H$ in the quotient.
Dividing both sides by the coefficient of $H$ cancels the $0.866$ in the product.

Title: Transposing when finding hypotenuse


问:
How find real solutions to second-order linear ODE's with 3 regular singular points
I want to solve the inhomogeneous ODE

$-r(x-m)y'(x) + S(x)y''(x) = \beta,$

where $r,m,\beta > 0$ and $S$ is a quadratic polynomial with negative discriminant. For the sake of the question it may be okay consider the homogeneous case $\beta = 0$.

It seems the 'standard' method of solving this is viewing the ODE as complex (or rather on the Riemann sphere), determining the singular points and deciding whether they are regular.

A variable transformation $z := \frac{x - \lambda}{\bar \lambda - \lambda}$, where $\lambda$ is a zero of $S$ yields the hypergeometric equation

$$u''(z) + \left(\frac{\gamma}{z} + \frac{1-\gamma +\alpha}{z-1}\right) u'(z) = \frac{\beta}{sz(z-1)},$$
for some real $\alpha$ and some, not necessarily real $\gamma$.

One solution for $\beta = 0$ is of course the constant $1$. Another one is given by

$$u_h(z) := z^{1-\gamma}\,_2F_1(1-\gamma, 1-\gamma+\alpha, 2-\gamma; z).$$

A particular solution to inhomogeneous equation is given by
$$u_p(z) := -\frac{\beta}{s\gamma}z\cdot _3F_2(1,1,\alpha+1,2,\gamma+1;z),$$
and so the general solution is
$$u(z) = c_1 + c_2z^{1-\gamma}\,_2F_1(1-\gamma, 1-\gamma+\alpha, 2-\gamma; z) -\frac{\beta}{s\gamma}z\cdot\,_3F_2(1,1,\alpha+1,2,\gamma+1;z).$$

In the end we re-substitute $y(x) = u((\bar \lambda  - \lambda) z + \lambda)$ to get the general solution of our original ODE.

Now, my question:

In the end I need a real solution. We could start computing the real part of $y$ and get a new basis for the homogeneous solution space or try to pick $c_1,c_2$ to make $y$ real, but this seems not only cumbersome - I straight up don't know how to make this work. With all this information in mind:

How do I find a real solution of $-r(x-m)y'(x) + S(x)y''(x) = \beta$?

If possible I also like to see the most 'convenient' way. I'm not sure it even involves the complex domain at all.

PS: I know we can reduce the ODE to a first-order ODE with $w := y'$, but then we have to integrate a product of a "polynomial" and a hypergeometric function to get $y$. If this is just the way to go, I can be convinced of that too. Although, it would be a bit disappointing to find out that this method fails completely if we have a $q(x)y(x)$ term as well.

答:
The paper by Norbert Euler (1997) ``Transformation Properties of $\ddot{x}+f_1(t)\dot{x} + f_2(t)x + f_3(t) x^n = 0$" Journal of Nonlinear Mathematical Physics, 4:3-4, 310-337

describes in detail the "the most general conditions on the functions $f_1$, $f_2$ and $f_3$, by which the equation may be integrable, as well as conditions for the existence of Lie point symmetries."


问:
Prove the upper bound of the minimum involving trigonometric functions
Does anyone know if
$$\min\left(\begin{array}{c}
|\sin x|, |\sin y|, |\sin z|, |\cos x|, |\cos y|, |\cos z|, \\
|\sin(x-y)|, |\sin(x-z)|, |\sin(y-z)|,\\
|\cos(x-y)|, |\cos(x-z)|, |\cos(y-z)|
\end{array}\right) \leq \frac25$$
for all real $x$, $y$, $z$?

Thanks a lot!

答:
Your claim is that for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}\newcommand{ab}[1]{\lvert{#1}\rvert}
\ab{\sin x} &\leq \tfrac25, &
\ab{\sin y} &\leq \tfrac25, &
\ab{\sin z} &\leq \tfrac25, \\
\ab{\cos x} &\leq \tfrac25, &
\ab{\cos y} &\leq \tfrac25, &
\ab{\cos z} &\leq \tfrac25, \\
\ab{\sin(x - y)} &\leq \tfrac25, &
\ab{\sin(x - z)} &\leq \tfrac25, &
\ab{\sin(y - z)} &\leq \tfrac25, \\
\ab{\cos(x - y)} &\leq \tfrac25, &
\ab{\cos(x - z)} &\leq \tfrac25, &
\ab{\cos(y - z)} &\leq \tfrac25.
\end{align}

Let's prove this by proving the following stronger proposition: for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}\newcommand{sinp}{\sin\left(\tfrac\pi8\right)}
\ab{\sin x} &\leq \sinp, &
\ab{\sin y} &\leq \sinp, &
\ab{\sin z} &\leq \sinp, \\
\ab{\sin(x - y)} &\leq \sinp, &
\ab{\sin(x - z)} &\leq \sinp, &
\ab{\sin(y - z)} &\leq \sinp,
\end{align}

The second proposition is strictly stronger than the first because
$\sinp < \tfrac25$.

The second proposition can be proved by contradiction. Suppose that for some $x,y,z \in \mathbb R$
the claim is false. Then for these particular numbers $x,y,z,$ all of the following must be true:
\begin{align}
\ab{\sin x} &> \sinp, &
\ab{\sin y} &> \sinp, &
\ab{\sin z} &> \sinp, \\
\ab{\cos x} &> \sinp, &
\ab{\cos y} &> \sinp, &
\ab{\cos z} &> \sinp, \\
\ab{\sin(x - y)} &> \sinp, &
\ab{\sin(x - z)} &> \sinp, &
\ab{\sin(y - z)} &> \sinp, \\
\ab{\cos(x - y)} &> \sinp, &
\ab{\cos(x - z)} &> \sinp, &
\ab{\cos(y - z)} &> \sinp.
\end{align}

Let $k$ be an integer and $x_0$ a real number such that $x = x_0 + \frac{k\pi}{2}$
and $0 \leq x_0 < \frac\pi2.$
Note that if $\sin x_0 \leq \sinp$ then $\ab{\sin x} \leq \sinp$.
So the assumption that the claim is false implies
$\ab{\sin x} > \sinp$,
which implies that $\sin x_0 > \sinp$,
which in turn implies that
$$ \frac\pi8 < x_0 < \frac{3\pi}{8}. $$

The same logic applies to $y$ and $z$ as well. Let $m, n$ be integers and
$y_0, z_0$ be real numbers such that
$y = y_0 + \frac{m\pi}{2}$, $0 \leq y_0 < \frac\pi2$,
$z = z_0 + \frac{n\pi}{2}$, and $0 \leq z_0 < \frac\pi2.$ Then
\begin{align}
\frac\pi8 &< y_0 < \frac{3\pi}{8}, \\
\frac\pi8 &< z_0 < \frac{3\pi}{8}.
\end{align}

Without loss of generality, we can assume $x_0 \geq y_0 \geq z_0.$
Then $x_0 - y_0$ is a real number
such that $x - y = x_0 - y_0 + \frac{(k - m)\pi}{2}$ and
$0 \leq x_0 - y_0 < \frac\pi2.$
The assumption that the claim is false then implies
$\ab{\sin(x - y)} > \sinp$,
which implies that
$\sin (x_0 - y_0) > \sinp$.
In particular,
$$ x_0 - y_0 > \frac\pi8. $$

For similar reasons,
$$ y_0 - z_0 > \frac\pi8. $$

We therefore have
$$ y_0 > z_0 + \frac\pi8 $$
and
$$
\frac{3\pi}{8} > x_0 > y_0 + \frac\pi8 > z_0 + \frac{2\pi}{8}
 > \frac{3\pi}{8},
$$

in short, $\frac{3\pi}{8} > \frac{3\pi}{8},$
which contradicts the assumption that the claim was false.
Therefore the claim is true.

A similar argument can show that for every $x,y,z \in \mathbb R$, at least one of the following inequalities is true:
\begin{align}
\ab{\cos x} &\leq \sinp, &
\ab{\cos y} &\leq \sinp, &
\ab{\cos z} &\leq \sinp, \\
\ab{\cos(x - y)} &\leq \sinp, &
\ab{\cos(x - z)} &\leq \sinp, &
\ab{\cos(y - z)} &\leq \sinp.
\end{align}
Hint: let $-\tfrac\pi2 < x_0 \leq \tfrac\pi2.$


问:
What is the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$?
While working on a project, I came across the problem of finding the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$. Dirichlet's Unit Theorem stated that there should be $2$ multiplicatively independent elements among the unit group. However, aside from the generator $1+\sqrt{2}$ and $\frac{1+\sqrt{-3}}{2}$, which is not considered a generator since $\left(\frac{1+\sqrt{-3}}{2}\right)^6=1$. I cannot find any other generator. Can someone offer help or an explanation? Thanks.

答:
$K=\Bbb{Q}(\sqrt2,\zeta_3),F=\Bbb{Q}(\sqrt2)$

$\qquad $ $f(w)=|w|^2$ is an homomorphism $O_K^\times \to O_F^\times$.

Let $\sigma\in Gal(K/\Bbb{Q}), (\sqrt2,\zeta_3)\to (-\sqrt2,\zeta_3)$. Since our extension is abelian $f\circ \sigma=\sigma \circ f$, if $f(w)=1$ then $f(\sigma(w))=1$ so all the $\Bbb{Q}$-conjugates of $w$ are on the unit circle. This implies that the coefficients of the characteristic polynomial $\prod_{g\in Gal(K/\Bbb{Q})} (x-g(w^n))\in \Bbb{Z}[x]$ of $w^n$ are bounded by those of $(x+1)^4$ independently of $n$, thus for some $m>n>0$, $w^n$ and $w^m$  have the same minimal polynomial and hence $w$ is a root of unity.

Also from that $K\subset \Bbb{Q}(\zeta_{12})$ and $[\Bbb{Q}(\zeta_l):\Bbb{Q}]=\varphi(l)$ we get that the roots of unity of $K$ are generated by $\zeta_6$, ie. $\ker(f)=\langle \zeta_6\rangle$.

Thus $$[O_K^\times / \langle \zeta_6 \rangle:O_F^\times/ \langle -1\rangle] = [f(O_K^\times):f(O_F^\times)]=1$$ which implies that $$O_K^\times = \langle \zeta_6\rangle O_F^\times=\langle \zeta_6\rangle (1+\sqrt2)^\Bbb{Z}$$
($O_F^\times=\pm (1+\sqrt2)^\Bbb{Z}$  because $1+\sqrt2$ is the least unit $>1$, we know it from testing the roots of $x^2+ax\pm 1$ for $|a|\le 4$)


问:
Properties of this family of function
Consider the family $F$ of all $f$ differentiable functions $f: \mathbb R\to\mathbb R$, that satisfy, for any pair of real numbers $x$ and $y$, the condition:

$$\frac{f(x)-f(y)}{x-y} = f'\left(\frac{x+y}{2}\right)$$

Now there is a lot of alternatives t and $f$, and the only true is:
"All functions of $F$ are class $C^{\infty}$"

I can not understand why. The function $ax + b$ satisfy the properties (1). But this type of function is not smooth.

(1):
$$\frac{f(x)-f(y)}{x-y} = \frac{ax + b - ay - b}{x-y} = a\\f' = a$$

答:
The $C^\infty$ function space is defined as the set of functions $f$ being $C^n$ for all $n$. If $f$ has a $n$-th derivative (over its whole definition domain) for any integer $n$, then it is $C^\infty$. If a function has a derivative it is necessarily continuous so talking about continuity of the derivative in this definition is not mandatory. In practice we rather say that $f$ is indefinitely derivable or smooth. Note that the former can have different meanings depending on the context.

For $f(x)=ax+b$ is $C^\infty$, we can compute the $n$-th derivative explicitely for arbitrary large ranks $n$. The first derivative, $n=1$, is $f'(x)=a$ and from there $f^{(n)}(x)=0$ for all $n\ge2$. So it is $C^\infty$.

Title: Properties of this family of function


问:
Axiom of infinity exceptions?
It is my understanding that the axiom of infinity, or axioms that depend on the axiom of infinity, are the only axioms that are not representable in finite First-Order Predicate Logic.  Is this correct?  If so, how can the axiom of infinity be justified?  Furthermore, it is my understanding that the axiom of infinity requires infinite induction, which requires the axiom of infinity.  How is this circular definition resolved?  Is there a way for the axiom of infinity to be defined so that it does not require infinite induction?</p>

答:
While it is correct that generally it is impossible to write a sentence in first-order logic whose content is "there are infinitely many objects in the universe", we can still - under some circumstances - write a sentence which can prove that there are infinitely many objects.

For example, in a language which has a binary relation symbol $<$ we can write the following axioms (and take their conjunction, which is one sentence):

$\forall x(x\nless x)$ ($<$ is irreflexive).

$\forall x\forall y\forall z(x<y\land y<z\rightarrow x<z)$ ($<$ is transitive).

$\forall x\forall y(x<y\rightarrow\exists z(x<z\land z<y)$ ($<$ is dense).

From the conjunction of the three sentences we have one axiom whose implications are, amongst other things, that there are infinitely many elements in the universe. Formally, of course, we don't prove just one sentence, we prove a whole schema. We prove that for every $n\in\Bbb N$, these three prove that $\forall x_1,\ldots,\forall x_n\exists y(\bigwedge_{i=1}^n x_i\neq y)$. So no matter what $n$ elements we take, there is one more. Therefore there can't be finitely many elements in the universe.

Similarly, the axiom of infinity is a first-order axiom which states that there exists a set with certain properties. From the existence of that set we can prove that the universe contains a set which is not finite.

Also see Set Notation (Axiom of Infinity).

Title: Axiom of infinity exceptions?


问:
Continuous if and only if graph of every path connected subspace is path connected?
It is clear that if $f : X \to Y$ is continuous then the graph of $f$ over every path connected subspace is path connected.

Is the converse true? That is, if the graph of $f : X \to Y$ over every path connected subspace is path connected, does it follow that $f$ is continuous?

The topologist's sine curve is a well known counterexample if we say "connected" instead of "path connected." Is there a standard counterexample in the path connected case? Is there a counterexample that works when $X = Y = \mathbb{R}$?

Part of why I ask is that continuity is often informally defined in introductory calculus classes as a function whose graph you could "walk along" as though it were a path. This, however, is quite different from the formal definition and, I suspect (but have not been able to prove), is not even true.

答:
For arbitrary spaces it certainly isn't true.  For instance, if $X$ is totally path-disconnected (e.g., $X=\mathbb{Q}$), then your condition is trivial, but not every map out of $X$ is continuous unless $X$ is discrete.

To characterize the spaces for which it is true, consider the following properties of a topological space $X$:

$X$ is well-graphed if it has your property for arbitrary $Y$: that is, a map $X\to Y$ is continuous iff its graph over every path-connected subspace of $X$ is path-connected.

$X$ is path-generated if for any $Y$, a map $X\to Y$ is continuous iff its composition with every continous path $[0,1]\to X$ is continuous. (For more about this property see this answer of mine; in particular, a path-generated space must be locally path-connected and sequential.)

$X$ is arc-generated if for any $Y$, a map $X\to Y$ is continuous iff its composition with every embedding $[0,1]\to X$ is continuous.

We then have the following theorem:

Theorem: Every arc-generated space is well-graphed, and the converse holds for Hausdorff spaces.  Every well-graphed space is path-generated.

Proof:  Suppose $X$ is well-graphed and $f:X\to Y$ has the property that the composition of $f$ with every path in $X$ is continuous.  Then the graph of $f$ over every path-connected subspace is path-connected, since every path in $X$ lifts to a path in the graph.  Thus $f$ is continuous, so $X$ is path-generated.  Moreover, if $X$ is Hausdorff, then it would suffice to assume the composition of $f$ with every embedding $[0,1]\to X$ is continuous, since in a path-connected Hausdorff space any two points are connected by a path which is an embedding, so $X$ would be arc-generated.

Conversely, suppose $X$ is arc-generated.  Let $f:X\to Y$ be a map whose graph over every path-connected subset is path-connected.  To show $f$ is continuous, it suffices to show $fg$ is continuous for every embedding $g:[0,1]\to X$.  Since $g$ is an embedding, the graph of $fg:[0,1]\to Y$ is homeomorphic to the graph of $f$ restricted to the image of $g$.  By hypothesis, the latter graph is path-connected, so the graph of $fg$ is path-connected.  Now take a path $h$ from $(0,f(g(0)))$ to $(1,f(g(1)))$ in this graph.  The first coordinate of $h$ is a continuous surjection $p:[0,1]\to[0,1]$, which is automatically a quotient map since the domain is compact and the codomain is Hausdorff.  The second coordinate of $h$ is then the map $fgp:[0,1]\to Y$ and is continuous since $h$ is.  But since $p$ is a quotient map, this implies $fg$ is continuous, as desired. $\blacksquare$

In particular, $\mathbb{R}$ is arc-generated and thus well-graphed, so your statement is true for the motivating example of maps $\mathbb{R}\to\mathbb{R}$.

As a counterexample to show that a Hausdorff path-generated space need not be well-graphed (or equivalently, arc-generated), let $X$ be the 1-point compactification of the set $$A=[0,\infty)\times\{0\}\cup\mathbb{N}\times[0,1]\subset\mathbb{R}^2.$$  Then $X$ is path-generated (indeed, it is a quotient of $[0,1]$, by a path that goes from $(0,0)$ off to $\infty$ taking a detour on each $\{n\}\times[0,1]$ along the way).  However, I claim $X$ is not well-graphed.  Consider the map $f:X\to[0,1]$ that is the first projection on $A$ and maps $\infty$ to $0$.  Then $f$ is not continuous at $\infty$, since the points $(n,1)\in A$ approach $\infty$ as $n$ gets large but $f(n,1)=1$ and $f(\infty)=0$.  However, the graph of $f$ over any path-connected set is path-connected, since any path-connected subset that contains both $\infty$ and a point $(n,t)$ for $t>0$ must also contain the path from $(n,t)$ down to $(n,0)$ and then the horizontal path from $(n,0)$ out to $\infty$, and $f$ is continuous on both of these paths.

I don't know whether there is a (necessarily non-Hausdorff) well-graphed space that is not arc-generated.


问:
Proving the solution of an ODE exists for $t \geq 0$
Consider the system
$$x'(t) = y(t)$$
$$y'(t) = -y(t)^3-p(x(t)),$$
where $p(x(t))$ is a polynomial of odd degree whose leading coefficient is positive. I'm interested in proving that the solution exists for all $t \geq 0$. I was thinking of using the Lyapunov function $L(x,y)= \frac{1}{2}y^2+P(x)$, where $P(x)$ is the antiderivative of $p(x)$. However, I'm not sure how to prove it. Any help would be appreciated.

答:
Consider the function
$$
V(x,y)=P(x)+\frac12 y^2,
$$
where $P(x)$ is the antiderivative of $p(x)$ such that (to be specific) $P(0)=0$. $P(x)$  is a polynomial of even degree whose leading coefficient is positive, thus,
$$
\lim_{x\to\infty} P(x)=+\infty,
$$
thus,
$$
\lim_{\|(x,y)\|\to\infty} V(x,y)=+\infty.
$$

The only posibility for the solution of the system to not exist is to go to infinity. Suppose that there exists some solution $(x(t),y(t))$ such that
$$
\lim_{t\to A} \|(x(t),y(t))\|=+\infty,
$$ where $A$ is some finite number or $+\infty$. This implies that for this solution
$$
\lim_{t\to A} V(x(t),y(t))=+\infty.
$$
But this is impossible because
$$
\frac{d}{dt}V(x(t),y(t))= p(x(t))y-y^4(t)-p(x(t))y=-y^4(t)\le 0,
$$
thus, $V(x(t),y(t))$ is nonincreasing.
This also proves that all solutions of the system are bounded.


问:
The ring $\Bbb Q[x,y,z]/(x^a,y^b,z^c)$ is noetherian
How to show the ring $\Bbb Q[x,y,z]/(x^a,y^b,z^c)$ is noetherian?
I know this ring is artin,and we can conclude the ring is artin because every ration ring is Noether.But I want to show this by definition of Noether ring.Thank you for your kind help.

答: Hint: Every ideal of $R=\mathbb Q[x,y,z]/(x^a,y^b,z^c)$ is a linear subspace of $R$, viewed as a vector space over $\mathbb Q$. Now, observe that $$\dim_{\mathbb Q}R=abc < \infty$$


问:
Vectors that form a triangle!
I have a problem here.

How can I prove that sum of vectors that form a triangle is equal to 0 $(\vec {AB}+\vec {BC}+\vec {CA}=\vec 0)$ ?

答:
from the triangle law : $\vec {AB}+\vec{BC}=\vec{AC}$

$\vec {AC}$ will be resultant vector of addition of other two vectors.

$\vec {AB}+\vec{BC}=\vec{AC}$

$\vec {AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}$ (add $\vec{CA}$ on both side)

$\vec {AB}+\vec{BC}+\vec{CA}=\vec{AC}-\vec{AC}$

(because when AC and CA are same in magnitude but they are opposite in direction)

$\vec {AB}+\vec{BC}+\vec{CA}=\vec0$

hence proved

Title: Vectors that form a triangle!


问:
Find the MVUE for a uniform parameter

Let $Y_{(n)}=\max(Y_{1},Y_{2},\ldots,Y_{n})$, where $Y_{1},Y_{2},\ldots,Y_{n}$ is a sample uniform distribution in $(0,\theta)$. Find the MVUE for $\theta$.

My approach:

I know that $U=Y_{(n)}$ is a sufficient statistic for $\theta$. I could prove this using the factorization theorem by Neymann.

I know that $\displaystyle h(U)=\left(\frac{n+1}{n}\right)U$ satisfy that $\mathbb{E}[h(U)]=\theta$.

So, $$\left(\frac{n+1}{n}\right)Y_{(n)}$$
is a (or the?) MVUE for $\theta$.

My approach is correct? Can I find other MVUE for $\theta$?

答:
As suggested, to confirm that $\frac{n+1}{n}Y_{(n)}$ is UMVUE you have only to prove that $Y_{(n)}$ is CSS (Complete and Sufficient Statistic)

Sufficiency is easy proved observing that

$$L(\theta)=\underbrace{1}_{=h(\mathbf{y})}\times\underbrace{\frac{1}{\theta^n}\mathbb{1}_{(y_{(n)};\infty)}(\theta)}_{g[t(\mathbf{y}),\theta]}$$

To prove completeness, as the model does not below to the Exponential family, you must use the definition. Say you have to prove that, for all $g$ measurable and for all $\theta$

$$\mathbb{E}_{\theta}g(T)=0\rightarrow \mathbb{P}_{\theta}[g(T)=0]=1$$

In your case you have

$$0=\mathbb{E}_{\theta}g(T)=\int_0^{\theta}g(t)\frac{nt^{n-1}}{\theta^n}dt=\frac{1}{\theta^n}\underbrace{\int_0^{\theta}g(t)nt^{n-1}dt}_{=0}$$

Now take the derivative w.r.t. $\theta$ obtaining

$$0=\frac{1}{\theta^n}g(\theta)n\theta^{n-1}=\frac{ng(\theta)}{\theta}$$

Thus actually expectation =0 implies also $g=0$ and $Y_{(n)}$ is complete

Now you have only to apply the following theorem

Title: Find the MVUE for a uniform parameter


问:
Evaluate $\int \frac{1}{(1+x^2)\sqrt{1-x^2}}\text dx$
The standard operating procedure for such questions is substituting $x$ with $\frac 1t$

So
$$\int \frac{-t}{(t^2+1)\sqrt {t^2-1}} \text dt$$

Which gives more or less the same expression. How do I solve this question?

答:
You are almost there !

we have $$\int \frac{-t}{(t^2+1)(\sqrt {t^2-1})} dt$$
let $t^2=u$ :then we have $$\int \frac{-du}{2(u+1)\sqrt{(u-1)}}$$  Now take $\sqrt{u-1}=y$ we have $$\int \frac{-dy}{y^2+2}$$ which should be easy


问:
Let $X$ be a metric space with bolzano weistrass property then $X$ is separable.
My attempt:

Let $X$ be a metric space in which every infinite set has a limit point .We fix $\delta >0$ and we pick $x_1 \in X$ .Having chosen $x_1,...,x_j \in X$., we choose $x_j+1 \in X$ so that $d(x_i,x_j) \ge \delta$ where $i=1,2,3,...,j$.This process must stop after a finite number .If not then we can find a sequence $(x_n)$ such that it has no limit point ,which is a contradiction to our assumption.

We pick $\delta =1/n$($n=1,2,...$).Then we can find $(x_1(1/n),...,x_k(1/n))$ for each $1/n$ such that the above condition holds true .Now, $A=\cup B_d(x_i(1/n),1/n)$ will cover $X$.
We try to show that this is a countable dense subset , let $x \in X$ be a point then  we take the neighborhoods $B_d(x,r)$ where $r>0$ .Now , we choose $1/N<r$ and we see whether $B_d(x,r)$ contains any point of $x_1(1/n),...,x_k(1/n)$ for $n \ge N$.If not then we can conclude that $d(x_i,x) \ge 1/N$ then $x \in A$.

So this has been my attempt.I am sorry if my notations are confusing but I have tried my best to make it comprehensive. Can someone point out my mistake in the proof or how should I continue from here.

答: Your idea is good. What you have to do get a complete proof is to choose  a maximal set $(x_1(\frac 1 n).,..., x_k(\frac 1 n))$ for each $n$. In other words choose $k$ to be the largest integer such that a finite set with $k$ elements has points separated by distance $\frac 1 n$. Then  your last step will lead to a  contradiction, finishing the proof.


Title: Evaluate $\int \frac{1}{(1+x^2)\sqrt{1-x^2}}\text dx$

Body:
The standard operating procedure for such questions is substituting $x$ with $\frac 1t$

So
$$\int \frac{-t}{(t^2+1)\sqrt {t^2-1}} \text dt$$

Which gives more or less the same expression. How do I solve this question?

Body:
Let $x=\sin(t)$, so $dx=\cos(t)dt$. Then $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}(t)}=\cos(t)$ and $t=\arcsin(x)$, so we can see that $$\int \frac{1}{\sqrt{1-x^{2}}(x^{2}+1)}dx=\int \frac{1}{\sin^{2}(t)+1}dt=\int \frac{1}{\sin^{2}(t)+1}\left(\frac{\csc^{2}(t)}{\csc^{2}(t)}\right)dt$$
now, let $s=\cot(t)$ then $ds=-\csc^{2}(t)$. Therefore, $$\int \frac{1}{\sin^{2}(t)+1}\left(\frac{\csc^{2}(t)}{\csc^{2}(t)}\right)dt=\int-\frac{1}{s^{2}+2}ds=-\frac{1}{\sqrt{2}}\arctan \left(\frac{s}{\sqrt{2}}\right)+c$$
So, we have $$\boxed{\int \frac{1}{\sqrt{1-x^{2}}(x^{2}+1)}dx =\frac{1}{\sqrt{2}}\arctan \left(\frac{\sqrt{2}x}{\sqrt{1-x^{2}}} \right)+c}$$


问:
Is there an algebraic number which has all possible combinations of numbers?
Today i saw this question. A similar question just came into my mind. Is there any irrational algebraic number so that it contains all possible number combinations in its digits? I'm really curious about it so if you had any idea how to find a number like this it will be good to share xD</p>

答:
I'm not sure if this directly answers your question, but from the Wikipedia entry,

a normal number is a real number whose infinite sequence of digits in every base $b$ is distributed uniformly in the sense that each of the $b$ digit values has the same natural density $1/b$, also all possible $b^2$ pairs of digits are equally likely with density $b^{−2}$, all $b^3$ triplets of digits equally likely with density $b^{−3}$, etc.

Also,

It has been conjectured that every irrational algebraic number is normal; while no counterexamples are known, there also exists no algebraic number that has been proven to be normal in any base.


问:
Given $X=\left \{ a,b,c,d,e,f \right \}$ prove if the subset collection of $X$ defined as $Y$ is a topology
Given $X=\left \{ a,b,c,d,e,f \right \}$ prove if the subset collection of $X$ defined as $Y=\left \{ X,\varnothing,\left \{ c \right \},\left \{ b,d,e \right \}, \left \{ b,c,d,e \right \} ,\left \{ b \right \} \right \}$ then $Y$ is a topology otherwise prove why not.

well as the title says , I need to profe if $Y$ define a topology, otherwise prove why not,

I have little knowledge of topology and I don't know how to proceed.

so any insight or solution would be of  great help for me,

regards

答:
Recall the definition of a topology:

Let $X$ be a non-empty set. A set $\mathcal{T}$ of subsets of $X$ is
said to be a topology on $X$ if

(i) $X$ and the empty set, $\emptyset,$ belong to $\mathcal{T}$

(ii) the union of any (finite or infinite) number of sets in $T$ belongs to $\mathcal{T}$, and

(iii) the intersection of any two sets in $\mathcal{T}$ belongs to $\mathcal{T}$.

Since $\{c\} \cup \{b\}=\{c,b\}$ is not in $Y$, $Y$ is not a topology.


问:
Speed Time and distance question
A person walks from his house to school at 2.5 km/h and arrives 12 minutes late. When he comes back from school to home at a speed of 4km/h, he reaches 15 minutes early. What is the distance?
The answer is 3km

I have no idea how to solve this question. Can somebody please explain to me how to get the answer?

答:
Hint :

distance = rate $\times$ time.

Assume distance = $D$, morning trip takes time $(T + 0.2)$ and afternoon trip takes
time $(T - 0.25).$

Morning rate was $R_1 = 2.5$ km/h and afternoon rate was $R_2 = 4$ km/h.

Then $D = R_1 \times (T + 0.2)$ and $D = R_2 \times (T - 0.25)$.

You have two (non-linear) equations in two unknowns, $D$ and $T$.

You can solve for these two variables.

Title: Speed Time and distance question


问:
Questions on (subring/ submodule) of a graded (ring/ module)
I have a question which seems a bit silly...

If we have $R$ a graded ring, does it follow that every subring of $R$ is also graded?

Because I have a problem here as such: I have a graded ring $R$ and a subring $A$. Does it follow that $R$ is a graded $A$-module?

Thanks for your help!

答: I think you need to have that $A$ is a graded subring to have a canonical $A$-module structure on $R$. (I'm supposing you want to have a grading on A which is inherited by that of $R$, otherwise you just trivially grade $A$ and you get it).</p>


问:
How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$
The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$

My attempt

I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.

答:
Too long for a comment.

Making the problem more general where you have
$$\int \frac{A(x)}{[B(x)]^2}\,dx$$ because of the square in denominator is to guess, as a first attempt, that it could be
$$\int \frac{A(x)}{[B(x)]^2}\,dx=\frac{P(x)}{B(x)}$$ Differentiating both sides gives
$$A(x)=B(x)P'(x)-B'(x)P(x)$$ and try to indentify.

Quite often during your studies, you will face this kind of situations in particular when $A(x)$ and $B(x)$ are polynomials or linear combinations of sines and cosines.

This is exactly what @lab bhattacharjee did in his answer.


问:
Prove that $C(x+y,y)$ is compositive number if $x,y\in \mathbb{Z}_{>1}^{+}$

Prove that $$x,y \in \mathbb{Z}^{+}, x,y>1 \implies  \binom{x+y}{y} \quad \text{is compositive.}$$

My approach: Let $x,y\in \mathbb{Z}^{+}$ such that $x,y>1$ so we need to prove that $\binom{x+y}{y}$ is compositive. Now, since that $$\binom{x+y}{y}=\frac{(x+y)!}{x!(x+y-x)!}=\frac{(x+y)!}{x!y!}$$
Now, we can see that $\displaystyle \frac{(x+y)!}{x!y!}$ is compositive if $x!y!| (x+y)!$, but how can I continue from here?

答:
It is clear that we have:
$${{x+y} \choose y}>1 $$
since $x,y>1$. Assuming the contrary, let ${{x+y} \choose y}=p$ where $p$ is a prime. Since $p \mid (x+y)!$, we must have $p \leqslant (x+y)$. Thus, we have:
$${{x+y} \choose y} \leqslant x+y \implies (x+y-1)! \leqslant x!y!$$
$$(x+y-1)! \leqslant (2 \cdot 3 \cdots x)(2\cdot3\cdots y)$$
$$(x+y-1)! < (2 \cdot 3 \cdots x)((x+1)\cdot(x+2)\cdots (x+y-1))$$
$$(x+y-1)!<(x+y-1)!$$
which is a contradiction. Hence, ${{x+y} \choose y}$ is a composite number.


问:
How many permutations of the string "000011112222" contain the substring "2020”?
So far, this is what I know. There are 9 ways to place 2020 and $\frac{8!}{4!2!2!}$ ways to arrange the remaining numbers. Im having a problem tackling the overcounting case.. If there are 2 seprate 2020s then i need to subtract $6C4$? Thanks in advance!

Is there a solution using PIE or recursion?

答:
Unfortunately, the solution is not so straight-forward. There's no problem with the first step:

Total number of permutations including all over-counts = $\frac{9!}{4! 2! 2!} = 3780$

Next, we consider where we can get over-counts, so as to get rid of them from the total number of permutations. Over-counts arise from the following scenarios.

202020201111: All four pairs of 20 are together; Every sequence like this has been counted 3 times. For e.g. 202020201111, 202020201111 and 202020201111 are all considered different permutations in the initial step. So each is over-counted by 2 times.

202020111021: Three pairs of 20 are together; Every sequence like this has been counted 2 times.

202011112020: Two pairs of separate 2020; Every sequence like this has been counted 2 times.

How many permutations of each scenario are there?

5 permutations

We are permutating 7 elements, 4 of which are 1s, so $\frac{7!}{4!}$. But note that we need to exclude from these cases of 4 pairs of 20 being together (already accounted for in scenario 1), of which there are 10. So in total $\frac{7!}{4!} - 10 = 200$

$\binom{5}{2} = 10$

The final answer is then $3780 - 5 \times 2 - 200 - 10 = 3560$.


问:
Solving Dudeney's "A Question of Cubes": sum of consecutive cubes is a square
The following is a puzzle of Dudeney:

Professor Rackbrane pointed out one morning that the cubes of successive numbers, starting from $1,$ would sum to a square number... He stated that if you are forbidden to use the $1,$ the lowest answer is the cubes of $23,24,25,$ which together equal $204^2.$  He proposed to seek the next lowest number using more than three consecutibe cubes and as many more as you like excluding $1.$

There is an answer provided but a proof is not included:

 The cubes of $14,15,$ up to $25$ inclusive (twelve in all) add up to... the square of $312.$ The next lowest answer is the five cubes of $25,26,27,28,$ and $29,$ which together equal $315^2.$

My query is more general: Classify all finite sets of consecutive positive integers, the sum of whose cubes is a square. Any idea how we can do this? If no one manages to answer this question, I will accept an answer that shows how Dudeney came to the minimal solutions.

Here are my thoughts on the matter so far:
Clearly the sum of first $n$ positive cubes works. The sum of the cubes of $m+1,m+2,\ldots,n$ for positive $n$ and non-negative $m$ is $$(1^3+2^3+\cdots+n^3)-(1^3+2^3+\cdots+m^3) = \left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2.$$
If this is equal to a square, it is equivalent to seeking all Pythagorean triples such that at least two of the elements of the triple are triangular numbers. At that point, I tried to use the classification of all (primitive) Pythagorean triples, but that went nowhere.

答:
We show the solutions of $y^2 = x^3 + (x+1)^3 +...+ (x+n-1)^3$ using Stroeker's method.

We consider below diophantine equation.

$$y^2 = nx^3 + \frac{3}{2}n(n-1)x^2 + \frac{1}{2}n(n-1)(2n-1)x + \frac{1}{4}n^2(n-1)^2\tag{1}$$

$$\sum_{k=0}^{n-1} k = \frac{1}{2}n(n-1)$$
$$\sum_{k=0}^{n-1} k^{2} = \frac{1}{6}n(n-1)(2n-1)$$
$$\sum_{k=0}^{n-1} k^{3} = (\frac{1}{2}n(n-1))^2$$

Hence we get equation $(1)$.

Equation $(1)$ is birationally equivalent to an elliptic curve below.

$$Y^2 = X^3 + \frac{1}{4}n^2(n^2-1)X\tag{2}$$

$X=nx+\frac{1}{2}n(n-1), Y=ny$

Multiply both sides of the equation $(1)$ by $n^2$ and substitute $\frac{1}{2n}(2X-n^2+n)$ for $x$, then we get equation $(2)$.

$1$. Equation $(1)$ always has the integer solutions for odd values of $n$

According to Stroeker, we know equation $(2)$ has a solution $(X,Y) = ( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) ).$

Hence we obtain $(x,y)=( \frac{1}{2}(n-1)(2n^2+2n-1), \frac{1}{2}n(n-1)(n+1)(2n^2-1) ).$

Thus $(x,y)$ is always the integer for odd values of $n$.

We show how this solution is derived from group law.

First, equation $(2)$ has a solution $(X,Y)=( 0,0 )$.

Substitute $X=\frac{1}{2}n(n+1)$ to equation $(2)$, then we get $Y=\frac{1}{2}n^2(n+1)$.

Hence we know equation $(2)$ has a solution $(X,Y)=( \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$.

Let $P1(X,Y)=( 0,0 )$ and $P2(X,Y)=(  \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$ and computing $P1-2P2$ using group law.

We get a solution $P1-2P2=( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) )$.

$2$. Equation $(1)$ has infinitely many integer solutions for even values of $n=2m^2$

Vladimir Pletser had already got this result.

We rediscovered it by brute force search as follows.

As a candidate of solution, $(X,Y^2)=(\frac{1}{2}n^3-\frac{1}{2}n, \frac{1}{8}n^5(n-1)^2(n+1)^2)$ was found.

Hence $\frac{1}{8}n^5(n-1)^2(n+1)^2)$  needs to be a square number.

Let $n=2m^2$, then equation $(2)$ has a solution $(X,Y) = (4m^6-m^2, 2m^5(2m^2-1)(2m^2+1))$.

Hence we obtain $(x,y)=( m^2(2m^2-1), m^3(2m^2-1)(2m^2+1) )$.

$m$ is arbitrary.

```<code>           m<20
           [  n       x         y ]

           [  8      28        504]
           [ 18     153       8721]
           [ 32     496      65472]
           [ 50    1225     312375]
           [ 72    2556    1119528]
           [ 98    4753    3293829]
           [128    8128    8388096]
           [162   13041   19131147]
           [200   19900   39999000]
           [242   29161   77947353]
           [288   41328  143325504]
           [338   56953  250991871]
           [392   76636  421651272]
           [450  101025  683434125]
           [512  130816 1073737728]
           [578  166753 1641349779]
           [648  209628 2448874296]
           [722  260281 3575480097]
</code>```

$3$. Numerical solutions

We extened the search range n to 99.

Search results of the integer points for equation $(1)$ using Online Magma Calculator.

```<code>           [  n         x          y ]

           [  3         23        204]
           [  5         25        315]
           [  5         96       2170]
           [  5        118       2940]
           [  7        333      16296]
           [  8         28        504]
           [  9        716      57960]
           [ 11       1315     159060]
           [ 12         14        312]
           [ 13        144       6630]
           [ 13       2178     368004]
           [ 15         25        720]
           [ 15       3353     754320]
           [ 15      57960   54052635]
           [ 17          9        323]
           [ 17        120       5984]
           [ 17       4888    1412496]
           [ 18        153       8721]
           [ 18        680      76653]
           [ 19       6831    2465820]
           [ 21         14        588]
           [ 21        144       8778]
           [ 21       9230    4070220]
           [ 23      12133    6418104]
           [ 25      15588    9742200]
           [ 27      19643   14319396]
           [ 28         81       4914]
           [ 29      24346   20474580]
           [ 31      29745   28584480]
           [ 32         69       4472]
           [ 32        133      10296]
           [ 32        496      65472]
           [ 33         33       2079]
           [ 33      35888   39081504]
           [ 35        225      22330]
           [ 35      42823   52457580]
           [ 37      50598   69267996]
           [ 39        111       9360]
           [ 39      59261   90135240]
           [ 40       3276    1196520]
           [ 41      68860  115752840]
           [ 42         64       5187]
           [ 43      79443  146889204]
           [ 45        176      18810]
           [ 45      91058  184391460]
           [ 47     103753  229189296]
           [ 48         64       5880]
           [ 48        410      62628]
           [ 48      19881   19455744]
           [ 48      60040  101985072]
           [ 49     117576  282298800]
           [ 50       1225     312375]
           [ 51     132575  344826300]
           [ 53     148798  417972204]
           [ 54        265      36729]
           [ 54       1272     343917]
           [ 55     166293  503034840]
           [ 57       1625     507471]
           [ 57     185108  601414296]
           [ 59     205291  714616260]
           [ 60        118      14160]
           [ 61     226890  844255860]
           [ 63        217      31248]
           [ 63        837     203112]
           [ 63       1121     310464]
           [ 63     249953  992061504]
           [ 64        105      13104]
           [ 64      34272   50827392]
           [ 65     274528 1159878720]
           [ 67     300663 1349673996]
           [ 69         81      10695]
           [ 69     328406 1563538620]
           [ 71     357805 1803692520]
           [ 72       2556    1119528]
           [ 73      16864   18771220]
           [ 73      26937   37849113]
           [ 73     388908 2072488104]
           [ 75     421763 2372414100]
           [ 76        295      53200]
           [ 77        144      22022]
           [ 77     456418 2706099396]
           [ 79     492921 3076316880]
           [ 81     531320 3485987280]
           [ 82        144      23247]
           [ 83     571663 3938183004]
           [ 85     613998 4436131980]
           [ 87        232      43065]
           [ 87     658373 4983221496]
           [ 89     704836 5583002040]
           [ 91       4785    3202290]
           [ 91     753435 6239191140]
           [ 92       4992    3429530]
           [ 93     804218 6955677204]
           [ 94        400      91979]
           [ 95     857233 7736523360]
           [ 97      23668   35970704]
           [ 97      33660   60951696]
           [ 97     912528 8585971296]
           [ 98         25       7497]
           [ 98         97      18333]
           [ 98        216      43309]
           [ 98        745     221697]
           [ 98        760     227997]
           [ 98       3961    2513511]
           [ 98       4753    3293829]
           [ 99     970151 9508445100]
</code>```

Body:
We show the solutions of $y^2 = x^3 + (x+1)^3 +...+ (x+n-1)^3$ using Stroeker's method.

We consider below diophantine equation.

$$y^2 = nx^3 + \frac{3}{2}n(n-1)x^2 + \frac{1}{2}n(n-1)(2n-1)x + \frac{1}{4}n^2(n-1)^2\tag{1}$$

$$\sum_{k=0}^{n-1} k = \frac{1}{2}n(n-1)$$
$$\sum_{k=0}^{n-1} k^{2} = \frac{1}{6}n(n-1)(2n-1)$$
$$\sum_{k=0}^{n-1} k^{3} = (\frac{1}{2}n(n-1))^2$$

Hence we get equation $(1)$.

Equation $(1)$ is birationally equivalent to an elliptic curve below.

$$Y^2 = X^3 + \frac{1}{4}n^2(n^2-1)X\tag{2}$$

$X=nx+\frac{1}{2}n(n-1), Y=ny$

Multiply both sides of the equation $(1)$ by $n^2$ and substitute $\frac{1}{2n}(2X-n^2+n)$ for $x$, then we get equation $(2)$.

$1$. Equation $(1)$ always has the integer solutions for odd values of $n$

According to Stroeker, we know equation $(2)$ has a solution $(X,Y) = ( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) ).$

Hence we obtain $(x,y)=( \frac{1}{2}(n-1)(2n^2+2n-1), \frac{1}{2}n(n-1)(n+1)(2n^2-1) ).$

Thus $(x,y)$ is always the integer for odd values of $n$.

We show how this solution is derived from group law.

First, equation $(2)$ has a solution $(X,Y)=( 0,0 )$.

Substitute $X=\frac{1}{2}n(n+1)$ to equation $(2)$, then we get $Y=\frac{1}{2}n^2(n+1)$.

Hence we know equation $(2)$ has a solution $(X,Y)=( \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$.

Let $P1(X,Y)=( 0,0 )$ and $P2(X,Y)=(  \frac{1}{2}n(n+1), \frac{1}{2}n^2(n+1) )$ and computing $P1-2P2$ using group law.

We get a solution $P1-2P2=( n^2(n-1)(n+1), \frac{1}{2}n^2(n-1)(n+1)(2n^2-1) )$.

$2$. Equation $(1)$ has infinitely many integer solutions for even values of $n=2m^2$

Vladimir Pletser had already got this result.

We rediscovered it by brute force search as follows.

As a candidate of solution, $(X,Y^2)=(\frac{1}{2}n^3-\frac{1}{2}n, \frac{1}{8}n^5(n-1)^2(n+1)^2)$ was found.

Hence $\frac{1}{8}n^5(n-1)^2(n+1)^2)$  needs to be a square number.

Let $n=2m^2$, then equation $(2)$ has a solution $(X,Y) = (4m^6-m^2, 2m^5(2m^2-1)(2m^2+1))$.

Hence we obtain $(x,y)=( m^2(2m^2-1), m^3(2m^2-1)(2m^2+1) )$.

$m$ is arbitrary.

           m<20
           [  n       x         y ]

           [  8      28        504]
           [ 18     153       8721]
           [ 32     496      65472]
           [ 50    1225     312375]
           [ 72    2556    1119528]
           [ 98    4753    3293829]
           [128    8128    8388096]
           [162   13041   19131147]
           [200   19900   39999000]
           [242   29161   77947353]
           [288   41328  143325504]
           [338   56953  250991871]
           [392   76636  421651272]
           [450  101025  683434125]
           [512  130816 1073737728]
           [578  166753 1641349779]
           [648  209628 2448874296]
           [722  260281 3575480097]

$3$. Numerical solutions

We extened the search range n to 99.

Search results of the integer points for equation $(1)$ using Online Magma Calculator.

           [  n         x          y ]

           [  3         23        204]
           [  5         25        315]
           [  5         96       2170]
           [  5        118       2940]
           [  7        333      16296]
           [  8         28        504]
           [  9        716      57960]
           [ 11       1315     159060]
           [ 12         14        312]
           [ 13        144       6630]
           [ 13       2178     368004]
           [ 15         25        720]
           [ 15       3353     754320]
           [ 15      57960   54052635]
           [ 17          9        323]
           [ 17        120       5984]
           [ 17       4888    1412496]
           [ 18        153       8721]
           [ 18        680      76653]
           [ 19       6831    2465820]
           [ 21         14        588]
           [ 21        144       8778]
           [ 21       9230    4070220]
           [ 23      12133    6418104]
           [ 25      15588    9742200]
           [ 27      19643   14319396]
           [ 28         81       4914]
           [ 29      24346   20474580]
           [ 31      29745   28584480]
           [ 32         69       4472]
           [ 32        133      10296]
           [ 32        496      65472]
           [ 33         33       2079]
           [ 33      35888   39081504]
           [ 35        225      22330]
           [ 35      42823   52457580]
           [ 37      50598   69267996]
           [ 39        111       9360]
           [ 39      59261   90135240]
           [ 40       3276    1196520]
           [ 41      68860  115752840]
           [ 42         64       5187]
           [ 43      79443  146889204]
           [ 45        176      18810]
           [ 45      91058  184391460]
           [ 47     103753  229189296]
           [ 48         64       5880]
           [ 48        410      62628]
           [ 48      19881   19455744]
           [ 48      60040  101985072]
           [ 49     117576  282298800]
           [ 50       1225     312375]
           [ 51     132575  344826300]
           [ 53     148798  417972204]
           [ 54        265      36729]
           [ 54       1272     343917]
           [ 55     166293  503034840]
           [ 57       1625     507471]
           [ 57     185108  601414296]
           [ 59     205291  714616260]
           [ 60        118      14160]
           [ 61     226890  844255860]
           [ 63        217      31248]
           [ 63        837     203112]
           [ 63       1121     310464]
           [ 63     249953  992061504]
           [ 64        105      13104]
           [ 64      34272   50827392]
           [ 65     274528 1159878720]
           [ 67     300663 1349673996]
           [ 69         81      10695]
           [ 69     328406 1563538620]
           [ 71     357805 1803692520]
           [ 72       2556    1119528]
           [ 73      16864   18771220]
           [ 73      26937   37849113]
           [ 73     388908 2072488104]
           [ 75     421763 2372414100]
           [ 76        295      53200]
           [ 77        144      22022]
           [ 77     456418 2706099396]
           [ 79     492921 3076316880]
           [ 81     531320 3485987280]
           [ 82        144      23247]
           [ 83     571663 3938183004]
           [ 85     613998 4436131980]
           [ 87        232      43065]
           [ 87     658373 4983221496]
           [ 89     704836 5583002040]
           [ 91       4785    3202290]
           [ 91     753435 6239191140]
           [ 92       4992    3429530]
           [ 93     804218 6955677204]
           [ 94        400      91979]
           [ 95     857233 7736523360]
           [ 97      23668   35970704]
           [ 97      33660   60951696]
           [ 97     912528 8585971296]
           [ 98         25       7497]
           [ 98         97      18333]
           [ 98        216      43309]
           [ 98        745     221697]
           [ 98        760     227997]
           [ 98       3961    2513511]
           [ 98       4753    3293829]
           [ 99     970151 9508445100]


问:
Coefficient of $\frac{1}{1-x-x^2-..-x^{d-1}}$ and its asymptotic

Let $C_n$ be the set of all compositions of n such that each letter is one of ${1,2,\ldots,d-1}$.

A. Calculate the generating  function of the number of compositions in $C_n$.

B. For each composition $\pi \in C_n$ define  $f(\pi)$ as the number of letters that equal $1$.
Find yhe generating function of $\sum_{n\geq 0} \sum_{\pi \in C_n}  q^{f(\pi)} x^n$.

C.Find the expectation of the number of letters that equal 1 (1's appearances) in these compositions, and analyze it asymptotically.

*Generally: Composition of $n$: a sequence of numbers in $N$ such that their sum equals $n$. For example:
$C_0={\epsilon}$ , $C_1={1}$, $C_2={11,2}$ , $C_3={111,12,21,3}$.

In this question, a composition of a number $n$ , consists of numbers from the set {$1,2,\ldots, d-1$} where $d$ is a given number.

This is what I did:

A. Every compisition of n = empty compisition $\cup$ (1 / every compisition) $\cup$ (2 / every compisition) $\cup$....$\cup$((d-1) / every compisition).
so if x counts the length of a number, then the  generating function $C(x)$ for the problem satisfy:

$C(x)=1+xC(x)+x^2C(x)+\cdots+x^{d-1}C(x)$

(1 for the empty compisition)

$C(x)=1+C(x)(x+x^2+\cdots+x^{d-1})$

Therefore, $C(x)=\frac{1-x}{1-2x+x^d}=\frac{1}{1-x-x^2-...-x^{d-1}}$

B.Let $q$ counts the appearances of 1 and $x$ as we mentioned counts the length of a number.
Then:

$C(x)=1+qxC(x,q)+x^2C(x,q)+\cdots+x^{d-1}C(x,q)$

$C(x,q)=1+C(x,q)(qx+x^2+\cdots+x^{d-1})$

So, $C(x,q)=\frac{1-x}{x^d+(q-1)x^2-(q+1)x+1}$

C. We have to calculate $[x^n]C(x,1)=[x^n] C(x)$ and $[x^n] ([(d/dq) C(x,q)]_{q=1})=\frac{x}{(1-x-x^2-...-x^{d-1})^2}=xC(x)^2$ so the expectation will be the proportion of them.

But I could not see how the coefficients can be calculated!
I did not succeed to connect it to Fibonacci series which I know (the case of d=2).

I would be thankful for any help to show part C.

Thanks in advance

答:
The analysis of this sort of generating function's asymptotics is an exercise in partial fraction decomposition. Suppose we're trying to find the asymptotics of $P_n$ whose generating function $\sum_i P_ix^i$ is $\dfrac1{P(x)}$. If we can factor the polynomial $P(x)$ as $(x-r_1)(x-r_2)\cdots(x-r_n)$ (and all the roots are distinct), then we can write $\dfrac1{P(x)}$ as $\sum_i\dfrac{m_i}{x-r_i}$ for some constants $m_i$ (more on this later). We can then go the other direction, using the geometric series to write $P_n$ as $\sum_i \frac{m_i}{r_i}(r_i^{-1})^n$. If $r_1$ is the smallest root in absolute value, then this implies that $P_n=\frac{m_1}{r_1}(r_1^{-1})^n(1+O(\epsilon^n))$ for some $\epsilon\lt 1$. (There are very minor complications when the smallest roots are complex conjugates but that's not relevant here.)

In this case, we know that the polynomial $P(x)=1-(x+x^2+\ldots+x^{d-1})$ is positive at $x=0$ and negative at $x=1$ (as long as $d\gt 2$; if $d=2$ the analysis is trivial), so it has a root in $(0,1)$. What's more, since $P'(x)$ is negative on this interval, it has a single, simple root $\rho_{d-1}$ in the interval. (And in fact, all roots of $P()$ are simple since $\gcd(P, P')=1$.) Using Rouché's Theorem applied to $(1-x)P(x) = 1-2x+x^d$, we can show that there are no other roots of $P(x)$ in the disc $x\lt 1$, so this root dominates the asymptotics; we have that $C_n=C(\xi_{d-1})^n$ for some constant $C$, where $\xi_{d-1}=\rho_{d-1}^{-1}$. Note that we can say this even without being able to write an explicit formula for $\xi_{d-1}$ (except in the first few cases, such as the Fibonacci numbers).

The next piece of the puzzle is to compute $C$; fortunately, there's a standard tool for this, the residue method — I'll leave the details of this to you.

Also, one small additional note: if you're curious, you can also examine the asymptotics of $\xi_d$ itself — specifically, the asymptotics as $d$ goes to infinity. It should be straightforward to find the limiting value (hint: what $x$ has $x+x^2+x^3+\ldots=1$?) but the rate of approach is a really nice problem.

A similar analysis can be done in principle for $[x^n]\left(xC(x)^2\right)$ $=[x^{n-1}]\left(C(x)^2\right)$, but it's substantially complicated by the fact that the root is no longer simple. Fortunately, there's another way of finding the asymptotics: you can use the equation for the coefficients of $C(x)^2$ in terms of the $C_n$ and then applying the asymptotic analysis of the $C_n$ to those terms. (You should find that each of the summands in the equation for the coefficients of $C(x)^2$ is comparable in size — see if you can figure out why.)


问:
Show that some group must have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$
I have two different questions, but they are related.

The first question is, Let $G$ be a finite abelian group. show that if $G$ contains (atleast) $2^n-1$ distinct elements of order 2, then there must be a subgroup of $G$ isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.

The second question is, Let $H$ be a finite abelian group.  show that if $H$ contains $\mathfrak{p}_1,...,\mathfrak{p}_n$ which are n distinct elements of order 2, such that for any $I\subsetneq\{1,2,\cdots,n\}$ we have $\prod_{i\in I}\mathfrak{p}_i\neq 1$ Show that $G$ contains a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.

Some background and context: This may be a fairly easy question to do by induction, but I haven't done group theory in a little bit, and I was never that good at it in the first place. The point of this question is that I was trying to show that a certain quadratic number field had a (ring of integers with a) class number divisible by $2^n$.

edit
It was pointed out that the second question follows fairly easily from the first one. I believe there should be a relatively easy way to show the first question by induction, but I am struggling with that.

I think the base case is trivial, if a group contains 1 element of order 2, then that element will generate a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

So now we assume any (finite abelian) group with (atleast) $2^n-1$ elements of order 2, will have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$. And let $G$ be a finite abelian group with $2^{n+1}-1$ elements of order 2. We want to show that $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ is isomorphic to a subset of $G$.

I think we can also show pretty easily that, since we have more than $2*(2^n-1)$ elements of order 2, we can find 2 distinct injective homomorphisms $\varphi_1$ and $\varphi_2$ from $(\mathbb{Z}/2\mathbb{Z})^n$ into G, such that their images are disjoint.

So can we just take $\varphi_3$ from $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ to G, and then define it somehow based on the last 2 functions?

答:
For the first question.

Note that in an abelian group $o(xy)|o(x)o(y)$, so that in $G$ the elements whose order is a power of $2$ form a subgroup, and this subgroup has at least $2^n-1$ elements of order $2$. That means we can assume $G$ is a $2$-group.

Now let $\Omega(G)=\{x\in G\mid x^2=1\}$. If $x,y\in g$ we have $(xy)^2=x^2 y^2=1$, so that $\Omega(G)$ is a subgroup and contains all the elements of order $2$. If $\Omega(G)$ has order $2^k$ then it contains $2^k-1$ elements of order $2$, and so $k\geqslant n$.

We can now pick recursively a sequence of non-trivial elements of $\Omega(G)$, $x_1,x_2,\dots, x_k$ such that $x_{j+1}\not\in \langle x_1,x_2,\dots x_j\rangle$ for each $j$: our hypotheses says there are always enough possible choices. Then   $\langle x_1,x_2,\dots x_n\rangle$ is easily seen to be $(\mathbb{Z}_2)^n$.


问:
Limit of a recursive sequence with $u_n$
It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.

The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.

答: I just found out how to do it. Assume that the limit is equal to a. Therefore, the limit of u_n+1 and u_n are both equal to a. Using the recursive definition, place a back in to get. a=1+1/a which is equivalent to a^2-a-1=0. Solving for a gives two answers. But, since u_n is bounded between 1 and 2; (1+sqrt(5))/2 is the accepted limit.</p>


问:
Vakil 5.5.F Associated primes/points behaves well with respect to localizing
In the notes of Vakil's 'The rising sea', he attempted to explain the concepts of associated points geometrically. He discussed associated points of $M$ where $M$ is a finitely generated $A$ module (and $A$ is Noetherian ring). And he stated the important property (rather than defining it) first:

(A) The associated primes/points of $M$ are precisely the generic points of irreducible
components of the support of some element of $M$ (on Spec $A$).

And the exercise 5.5.F asks:

Show that the definition in (A) of associated primes/points behaves well with respect to localizing: if $S$ is a multiplicative subset of $A$, then the associated primes/points of $S^{-1}M$ are precisely those associated primes/points of $M$ that lie in Spec $S^{-1}A$, i.e., associated primes of $M$ that do not meet $S$.

This has being asked here (which is closed as a duplicate). After reading these answers I got a slightly different one following Vakil's approach, and wanted to share here. Any comments are appreciated.

答:
First suppose $p\in Supp\ m$ and $p\cap S=\emptyset$. Being in the support means that $m_p\ne 0$, that is
\begin{align*}
  rm\ne 0, \forall r\notin p.
\end{align*}
That is, $p$ contains the annihilator of $m$, $p\supset Ann_A(m)$. Now since $p$ is disjoint from $S$, for any $t\in S$ we have $rt\notin p$, hence $rtm\ne 0$. In other words
\begin{align}
    r t m\ne 0, \forall r\notin p, \forall t\in S,
\end{align}
which is nothing but the condition for $p\in Supp\frac{m}{s}$ (for any $s\in S$). This completes the first half.

On the other side, now suppose $p\in Supp \frac{m}{s}$ is an associated point of $S^{-1}M$. Let $t=1$ in the above equation we see that $rm\ne 0, \forall r\notin p$, which shows that $m_p\ne 0, p\in Supp\ m$, and similarly let $r=1$ we conclude that $Ann_A(m)\cap S=\emptyset$.

Then there must be a prime ideal $q\supset Ann_A(m)$ and being disjoint from $S$ (e.g. refer to here), so $q\in Supp\frac{m}{s}$ too.

But $p$ as an associated point, is minimal (among primes containing $Ann_A(m)$). Hence $p\subset q$ thus $p\cap S=\emptyset$ too.


问:
A problem on almost sure convergence
Consider a sequence of random variables defined on the standard unit interval probability space :
$$X_n  =
\begin{cases}
  2^n & \text{when}  \quad \frac{1}{2^n} \leq \omega \leq \frac{1}{2^{n-1}};  \\
   0  & \mbox{otherwise}.
\end{cases}
$$
Is $\{X_n\}$ convergent in a almost sure sense ?

I think it is not, it converges in probabilistic sense. Is this correct ?

答:
The sequence $\{X_n\}$ actually converges pointwise to $0$, because for each $\omega$, there is an integer $N(\omega)$ for which $X_n(\omega)=0$ whenever $n\geqslant N(\omega)$.

In particular, it does converge to $0$ in probability.

Title: A problem on almost sure convergence


问:
Does limit exist at zero?
We know for a limit to exist the LHL and RHL should both be finite and equal or if they are infinite then both should tend towards positive infinity or both towards negetive infinity.

So my question is, if I define a function as
$$
F(x) =
\begin{cases}
|\ln(-x)|, & \text{for }x<0\\
1/x,& \text{for }x>0
\end{cases}
$$
Then does $\lim_{x\to 0}F(x)$ exist?

答:
$$\lim_{x\to 0} \lvert \ln (x) \rvert = \infty$$
$$\lim_{x\to 0^+} \frac{1}{x} = \infty$$

Whether this limit really exists is a matter of convention. Remember that $\infty$ is just a convenient shorthand for saying that the function grows indefinitely, i.e. it doesn't "tend" to any particular point which is what we define a limit to be. This is why most mathematicians would say that the limit of your function does not exist. In some fields, however, (and judging by the criteria you mention in your post this seems to be the case for you) infinite limits are said to exist. Your personal mileage may vary, so I suggest you stick with the definitions that have been provided to you and abide by that.

Title: Does limit exist at zero?


问:
Showing the infinite product space of $[0, 1]^\omega$ is not compact with respect to the box topology using the notion of open covers
I wish to show that the space $[0,1]^\omega$ in the box topology is not compact using the notion of open covers. The box topology has been a topic where I have struggled in my studies of topology, and I am having trouble trying to come up with an open cover that has no finite sub-cover for this space (I understand that there are other ways to prove that this space is not compact, but my goal is to use the notion of open covers and finite sub-covers in the proof). Any help would be greatly appreciated

答: Show that the product topology on $[0,1]^\omega$ is strictly coarser than the box topology. Since a compact topology is minimal among T2 topologies, the box topology is therefore not compact.


问:
$|G| =231$, all Sylow p subgroups are normal then, is G cyclic?
$|G|=231=3 * 7 * 11$.
Now it is given that all the Sylow p subgroups are normal so unique.
My attempt:
I have uniquely one Sylow 11 subgroup of order 11 which is isomorphic to $Z11$ say $P_1$, similarly I have uniquely one Sylow 7 subgroup which is isomorphic to $Z_7$ say $P_2$ , now say $P_1=<x>$
and $P_2=<y>$ now the order of $(xy)$ is $7,11 or 77$ , since there are uniquely one Sylow 7 and 11 subgroups so $|xy|=77$ otherwise $<xy>$ will be a group of order 7 or 11 otherthan $P_1$ and $P_2$.
So, suppose $<xy>= P_3$  where $P_3$ is isomophic to $Z77$ . Now say $h$ is a generator of $P_3$ and $k$ is a generator of the unique Sylow 3 subgroup say $P_4$, so, $P_4=<k>$ and $P_3=<h>$ where $|h|=77 , |k|=3$, so now the $|hk|= 3,77 or 231$ , it can't be 3 because there is uniquely one subgroup of order 3 otherwise $<hk>$ will be a group of order 3 otherthan $P_4$. Now $|hk|= 77 or 231$ , now if $|hk|=231$ then we are done because we have found a generator of this group so G is cyclic.
Now my question is if $|hk|=77$ can we get a contradiction or it is possible so in that case the G is not cyclic, then what is the examples of these type of groups.
Now does the generalisation hold?
$|G|=P_1*P_2*....*P_n$ where all $P_i$'s are primes and all Sylow $P_i$ subgroups are normal then is the group G cyclic? If not then give example.

答:
(Note if $N$ is a normal subgroup and $H$ is any subgroup, $NH$ is also a subgroup. Moreover, if $H$ is also normal, then the subgroup $NH$ is normal too. Good exercise.)

Suppose $A=\langle a\rangle$ and $B=\langle b\rangle$ are normal cyclic subgroups. Then the commutator $[a,b]$ may be written as a product in two ways, $(aba^{-1})b^{-1}$ or $a(ba^{-1}b^{-1})$, showing it is in both $A$ and $B$. Thus, if $A$ and $B$ intersect trivially (e.g. when $|A|,|B|$ are coprime) we must have $[a,b]=e$, or in other words $a$ and $b$ commute. You can use this to show $AB=\langle a,b\rangle$ is an internal direct product of $A$ and $B$. (Actually, the hypothesis that $A$ and $B$ are cyclic isn't necessary - can you see why?)

(More generally, if $N_1,N_2\trianglelefteq G$ are normal subgroups, then $[N_1,N_2]\subseteq N_1\cap N_2$. Exercise.)

You can generalize this to multiple subgroups, not just two. If $N_1,\cdots,N_k$ are normal subgroups of $G$ with coprime order, then $N_1\cdots N_k$ is an internal direct product. Moreover, if $|G|=|N_1|\cdots|N_k|$ then all of $G$ is an internal direct product $N_1\times\cdots\times N_k$. In particular this applies for the collection of Sylow subgroups, and moreover if all the Sylow subgroups are cyclic we can use the Chinese Remainder theorem to show that $G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ (where $n=|G|$ is the product of its own prime factors).


问:
Axiom in Steven Lay's Book on Real Analysis
In Steven Lay's book "Analysis with an introduction to proof", in the chapter on Real Numbers, Section 2 Ordered Fields, (on p. 120 in the 5th edition of the book) the axioms for the "operation" $+$ ${\mathbb R}$ are listed, and the first one is

$A_1$:  For all $x,y\in {\mathbb R}$, $x+y\in {\mathbb R}$, and if $x=w$ and $y=z$, then $x+y=w+z$

It seems to me that, if we start by assuming that by "operation" on a set $X$ we mean a function $X\times X\rightarrow X$, there is no need for this axiom because it just says that if $p=q$ than $f(p)=f(q)$. On the other hand, if something else is meant by "operation" then it should be addressed. So my question is, is there something I am missing here that requires the statement of that axiom?

答:
My understanding is since you are defining R, you have to ignore your current knowledge and hence this is required.

You are essentially saying that I will have the + operation on my structure (R) and it will behave in this manner.


问:
Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$
I don’t know how to begin solving. Can I get a hint?

My failed attempt

Let $t =\tan (x/2)$

Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve

答:
Since $1=\sin^2x+\cos^2x$, $1+2\cos x=\sin^2x+\cos x(2+\cos x)$, so the integrand is $\frac{\sin^2x}{(2+\cos x)^2}+\frac{\cos x}{2+\cos x}$. By the product rule, one antiderivative of this integrand is $\frac{\sin x}{2+\cos x}$, so $I=\frac{\sin\pi/2}{2+\cos\pi/2}=\frac12$.

If you'd prefer a solution using $t=\tan\tfrac{x}{2}$, follow it up with $t=\sqrt{3}\tan y$, reducing the problem to $\int_0^{\pi/6}\tfrac{2}{\sqrt{3}}\cos2ydy=\tfrac12$.


问:
Determine the remainder when
$1025^{1005}\equiv x(1023)$

I did:

$1025\equiv 2(1023)$

$1025^{1005}\equiv 2^{1005}(1023)$

I don't know the next step.

Title: Determine the remainder when

Body:
\begin{align}
2^{1005} &\equiv (2^{10})^{100}\cdot 2^5\pmod{1023} \\
&\equiv (1)^{100}\cdot 32 \pmod{1023} \\
&\equiv 32 \pmod{1023}
\end{align}


问:
Pushforward of Coherent Sheaf Along Projective Morphism
This is Vakil 18.1 A, self-study, and intended to be a consequence of the basic properties of the cohomology of quasi-coherent sheaves.

We are to show that if $\pi: X \to Y$ is projective, and $\mathcal F$ is coherent on $X$, then $\pi_*\mathcal F$ is coherent on $Y$ (if $X$ and $Y$ are locally Noetherian). We do not yet know this holds for proper morphisms, so we cannot use that.

Facts we know that seem relevant (in fact, this was proven almost immediately before this exercise): If $\mathcal F$ is coherent on a projective $A$-scheme $X$, with $A$ Noetherian, then $H^i(X, \mathcal F)$ is a coherent $A$-module for all $i$. In particular, $\mathcal F(X)$ is finitely generated as an $A$-module.

My best attempt so far is to factor $\pi$ through some $\mathbb P \mathcal E \to Y$ via closed immersion $X \hookrightarrow \mathbb P \mathcal E$, and use the fact that the composition of pushforwards is the pushforward of the composition, but I do not see how this gives us the result.

答: Let me recap the solution developed in the comments. Start by reducing to the case where $Y=\operatorname{Spec} A$ is the spectrum of a noetherian ring. By Vakil's theorem 16.2.1, the pushforward of the quasicoherent sheaf $\mathcal{F}$ on $X$ along a projective morphism $\pi:X\to Y$ is quasicoherent on $Y$. This means it's the sheaf associated to $(\pi_*\mathcal{F})(Y)=\mathcal{F}(X)$, which by corollary 18.1.5 is coherent. As the sheaf associated to a coherent module is coherent, we are done.


问:
How to find an irreducible polynomial of degree $p$ over finite field $F_p$ where $p$ is prime?
I have learnt a theorem that there always exists an irreducible polynomial of an arbitrary degree $m$ over a finite field. This theorem, of course, ensures the existence of an irreducible polynomial
of degree $p$, but how to actually find such a polynomial? Any help would be appreciated!

答:
An irreducible polynomial $f(x)$ of degree $n$ over $\mathbb{F}_p$ is necessarily a divisor of $x^{p^n}-x$.

Which gives a known criterion: $f(x)\in\mathbb{F}_p[x]$ is irreducible if and only if $f(x)\mid x^{p^n}-x$, where $n=\deg f$, and for each prime $d\mid n$ we have $\gcd\big(f(x),x^{p^{n/d}}-x\big)=1$. In terms of roots of $f$ (in an algebraic closure of $\mathbb{F}_p$), these must be distinct (this is checked by $\gcd(f,f')=1$), and any root $\alpha$ must satisfy $\alpha^{p^n}=\alpha$ but $\alpha^{p^{n/d}}\neq\alpha$ for each prime $d\mid n$.

This gives an easy way to check that $g(x):=x^p-x-1\in\mathbb{F}_p[x]$ is irreducible. Since $g'=-1$, the condition $\gcd(g,g')=1$ holds trivially. Let $\alpha$ be a root of $g(x)$. Then, by induction, $\alpha^{p^k}=\alpha+k$ for any natural $k$, hence $\alpha^{p^p}=\alpha$. Since $d=p$ is the only possibility here, and $\alpha^{p^1}=\alpha+1\neq\alpha$, we're done.


问:
Help me prove $K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$ where $K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$
Hi this problem is posed in Sean M. Stewart's How to Integrate It book pg.143

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$$ where $ 0 \leq k < 1$

Based on the substitution:

$$ \tan{u} = \frac{\sin{2\theta}}{k+k\cos{2\theta}} =\frac{\tan{\theta}}{k}  $$

Show that:

$$K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$

$K(k) $ is some function defined in term of definite-integral called complete elliptic integral which is not easily computable, so attacking $K(k)$ gave no fruitful result.

Well then proceeding with substitution ( It is easy to lose track as it  involved a lot of trigonometric manipulations )

I reduced the integral into this form :
$$K(k) = k\int^{\pi/2}_0 \frac{\sec^2{\theta}d\theta}{\sqrt{k^2 + \tan^2{\theta}}\sqrt{k^2 + \tan^2{\theta} - k^2\tan^2{\theta}}}$$

Using substitution : $\tan{\theta} = p$

$$K(k) = k\int^{\infty}_0 \frac{dp}{\sqrt{k^2 +p^2}\sqrt{k^2 +p^2 - (kp)^2}}$$

To my naive eyes this looked manageable but WolframAlpha says otherwise.

Any hint or ideas on how to attack this problem?

I thought  the trig manipulations to find a more suitable form but in the I end up transform $\sin {u}$ term into $\tan{u}$ to use the given substitution

EDIT:  It has been suggested in commens that there is typo in the substituion(given in book) , it should be :

$$ \tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}} $$

答:
Here is my solution based on Paramanand Singh's hint:

First convert integral from [0,π/2] to [0,π] and then the interval
[0,π] of u maps to [0,π/2] of θ

Using the definite integral property :

$\int^{2a}_0 f(x) dx = 2\int^a_0 f(x)dx$  if $f(2a-x) = f(x)$

We change the limits to : Upper Limit $ = \pi$ and lower limit $= 0$ which will allow us to use the substitution given in the book [ $$ ]

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}} = \frac{1}{2}\int^{\pi}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}} $$

As the substitution is in the form of $\tan u $ we will express the integrand in terms of $\tan u$

We arrive at :

$$\frac{1}{2}\int^{\pi}_0 \frac{\sec{u}du}{\sqrt{1 + \tan^2{u}-(k\tan{u})^2}} $$

Substituting : $\tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}}$ [ $$ ]

$$\frac{1}{2}\int^{\pi}_0 \frac{(k+ \cos{2\theta})}{k\cos{2\theta} +1}\sec{u}du $$

Now,

$$\sec^2{u}du = \frac{2(k\cos{2\theta}+1)d\theta}{(k + cos{2\theta})} $$

Leading to a cleaner integral:

$$\int^{\pi/2}_0 \frac{d\theta}{(k+\cos{2\theta})\sec{u}}$$

As

$\sec u = \sqrt{1+ \tan^2{u}} = \frac{\sqrt{k^2 + 1 + 2k\cos{2\theta}}}{(k + \cos{2\theta})} = \frac{\sqrt{(k+1)^2 - 4k\sin^2{\theta}}}{(k + \cos{2\theta})} $

Integrand reduces to desired form:

$$ \int^{\pi/2}_0 \frac{d\theta}{\sqrt{(k+1)^2-(2\sqrt{k} \sin{\theta})^2}} $$

I think it is trivial to take from here to the desired form :)

[ $$ ] If you are coming here from the book, this problem passage had lot of minor typos, rely on this post as a reference.

For.eg

For this substitution it can be shown that the limits of integration
change in the following manner. When u = $0$, $\theta$ = $0$, and when u =
$\pi/2$, $\theta= \pi/2$

This is incorrect


问:
Vectors that form a triangle!
I have a problem here.

How can I prove that sum of vectors that form a triangle is equal to 0 $(\vec {AB}+\vec {BC}+\vec {CA}=\vec 0)$ ?

答:
well, define $A := (a_1,a_2, \dots ,a_n)$, then $B := (b_1,b_2, \dots ,b_n)$ and $C := (c_1,c_2, \dots ,c_n)$.

Then $\overrightarrow{AB}=(b_1-a_1,b_2-a_2, \dots ,b_n-a_n)$ similarly $\overrightarrow{AC}=(c_1-a_1,c_2-a_2, \dots ,c_n-a_n)$ and $\overrightarrow{BC}=(b_1-c_1,b_2-c_2, \dots ,b_n-c_n)$.

let's make $\overrightarrow{AB}+\overrightarrow{BC}= (b_1-a_1 + c_1-b_1,b_2-a_2+c_2-b_2, \dots ,b_n-a_n+c_n-a_n) = \overrightarrow{AC}=-\overrightarrow{CA}$
and hence the result.

NB if you don't know how to work with vectors, then you may want to take a look here Wiki - vectors. I've worked in $\mathbb{R}^n$ with and affine structure on it.

Title: Vectors that form a triangle!