问:
Intuition about non-existence of non-integrable functions.
Recently in the class of integration our teacher told us that some functions like $e^{x^2}$ are non integrable , so it is not possible to obtain a function which is it's anti-derivative. But we know that integration is area under the curve so for a smooth function like $e^{x^2}$ I have an intuitive feeling that the graph of area under the curve be another smooth curve. Also every curve on the graph represents a function so isn't  there a contradiction ?

答:
The issue is in terminology.  The function $e^{-x^2}$ is definitely integrable.  It's just that the expression for the integral, $I(x) =  \int_{0}^{x} e^{-t^2} dt$ cannot be written in terms of elementary functions.  The fix is that Mathematicians simply give the function, $I$ its own name, in this case, the error function.

An example of a function that is non-integrable near zero is $I(x) = \int_{-1}^{x} \frac{dx}{x}$.  This function can only be integrated for values of $x$ less than zero.


问:
Why we assume henselian ring is local?
Henselian ring is defined as local ring in which hensel lemma holds.
Why do we assume local ring?

What is wrong with the definition that 'henselian ring is defined as hensel lemma holds' ?

答: The version of Hensel's lemma I'm familiar with says that in $\mathbb{Z}_p$, if you have a simple root of a polynomial mod the maximal ideal then it lifts to a simple root. How would you generalize this if there is more than one maximal ideal?

ContentLicense: CC BY-SA 4.0

Title: Why we assume henselian ring is local?


问:
Morse-Kelley Class Comprehension and Russell's Paradox
As I understand the ZFC solution to Russell's paradox, since $\{x\mid x\notin x\}$ must be $\{x\mid x\notin x\}\cap S$ for some set $S$, the paradox goes away, but in Morse-Kelley,  although again there must be some $M$ such that $\{x\mid x\notin x\}\cap M$, this $M$ may be a proper class, which no longer is as limiting as the ZFC version, and hence no longer gives the same solution. So  MK must handle Russell's Paradox in a different way. How? I would be grateful for enlightenment. Thanks.</p>

答:
By the axiom of regularity, no class or set is a member of itself.  So the class in question is the class of all sets, commonly denoted V, which is also called the "universe."  It is the largest class and a proper class, i.e., not a set, so it is not a candidate for membership is the paradoxical class defined by that use of
set-builder notation.</p>

ContentLicense: CC BY-SA 3.0


问:
Question about limit of integrals
I want to prove the following:

$$ \lim_{n \rightarrow \infty} \int_{0}^{b_{n}} f_{n}(x) dx= \int_{0}^{1} f(x) dx    $$ provided that $f_{n}$ is Riemann-integrable on $[0,1]$ and $f_{n} \rightarrow f$ uniformly on $[0,1]$ and $b_{1} \leq b_{2}...\leq b_{n}\le \dots$ and $b_{n} \rightarrow 1$.

I started the proof by saying that since $f_{n} \rightarrow f$ uniformly so we can exchange the limit and the integral. But, that seems to be making it almost trivial. Am I missing something very important?

Title: Question about limit of integrals

Body: Use $$\left|\int_0^{b_n}f_n-\int_{0}^{1}f\right|\leq \left|\int_0^{b_n}f_n-\int_{0}^{1}f_n\right|+\left|\int_0^{1}f_n-\int_{0}^{1}f\right|$$


问:
Relations between differential geometry and algebraic geometry
I am currently an undergrad student looking to study some algebraic geometry, I have heard that differential geometry is useful for intuition in algebraic geometry, but I have no background in that (DG).

What are some specific examples of concepts that are common in both and would help understand the other subject? I am aware "manifold -> variety", but what else?

答:
I can give one important example: what a map is. You may recall that a map between smooth manifolds $f: X \longrightarrow Y$ is smooth if it is smooth under all charts. More formally, if we have $\phi: U \longrightarrow \mathbb R^n$ and $\psi: V \longrightarrow \mathbb R^m$ in the atlases for $X$ and $Y$ respectively and such that $f[U] \subseteq V$, then we say $f$ is smooth relative to these coordinates patches if $\psi \circ f \circ \phi^{-1}: \mathbb R^n \longrightarrow \mathbb R^m$ is smooth. If this holds for all patches in the maximal atlases of $X$ and $Y$ then we say $f$ is smooth.

As you say, we want to translate some of our notions from manifolds to varieties. The definition I gave above for smoothness doesn't translate super well. I'll give you another equivalent definition that may start to ring some bells if you're familiar with algebraic geometry.

Let $C_X$ denote the "sheaf" of continuous functions from $X$ to $\mathbb R$. In other words, for all $U \subseteq X$ open, $C_X(U)$ is the set of all continuous functions $U \longrightarrow \mathbb R$. Now, given a continuous map $f: X \longrightarrow Y$, we get a "map of sheaves" $C_Y \longrightarrow f_* C_X$. Don't let this notation scare you - all I'm saying is that $f$ induces a map $C_Y(U) \longrightarrow C_X(f^{-1}[U])$ for all open subsets $U \subseteq Y$. This takes $g \mapsto g \circ f$. Now, we let $C_X^{\infty}$ be the sheaf of smooth functions to $\mathbb R$. Again, I just mean that $C_X^\infty(U)$ is the set of all smooth functions $U \longrightarrow \mathbb R$ for $U \subseteq X$ open. Smooth functions are continuous, so $C_X^\infty(U) \subseteq C_X(U)$. Of course the same holds for $X$.

So in that same sense, if we have $f: X \longrightarrow Y$ smooth and $g: U \longrightarrow \mathbb R$ smooth for $U \subseteq Y$ open. Then $g \circ f: f^{-1}[U] \longrightarrow \mathbb R$ is smooth. Using that same fancy notation, we get a map $C_Y^\infty \longrightarrow f_* C_X^\infty$. In summary, if a continuous map $f$ is in fact smooth then the map $C_Y \longrightarrow f_* C_X$ descends to a map $C_Y^\infty \longrightarrow f_* C_X^\infty$.

In fact, this last condition completely characterizes smoothness! Putting all this fancy notation away, all I'm saying is that $f: X \longrightarrow Y$ is smooth iff it is continuous and sends smooth maps on open subsets of $Y$ to smooth maps on open subsets of $X$. More formally, if for all $g: U \longrightarrow \mathbb R$ smooth, $U \subseteq Y$ open we have $g \circ f: f^{-1}[U] \longrightarrow \mathbb R$ smooth, then $f$ is smooth.

Now this is something we can work with. I won't do much in the way of rigor here; you can read something like Hartshorne for that. Take a variety $X$ over an algebraically closed field $k$. We have the intuitive notion that a regular map $X \longrightarrow k$ should look like a polynomial, and a regular map on an open subset $U \longrightarrow k$ should look like a rational function whose denominator is nowhere zero on $U$. What then is a map between varieties, say $f: X \longrightarrow Y$? Appealing to the above, it's something that takes regular maps on $Y$ to regular maps on $X$. More formally, it takes $\mathcal O_Y \longrightarrow f_* \mathcal O_X$, where $\mathcal O_X$ is the sheaf of regular functions. Put aside what exactly a regular function is. The point is that they're continuous functions defined in an algebraic fashion. To map from one variety to another and preserve this structure, we follow the analogy of smooth maps I gave above. This is how Hartshorne defines a map of varieties, modulo details.

We can take this a bit further by realizing that the set of continuous, smooth, or regular functions on an open subset actually forms a ring, and the map $g \mapsto g\circ f$ is a ring homomorphism. This will eventually lead to the definition of a map of locally ringed spaces, and hence of schemes, but that's too much for just this one answer.


问:
Basis given the rank of A is equal to n

Let $A$ be an $m \times n$ matrix with columns $C_1,C_2,\dots,C_n$. If $\operatorname{rank} (A) = n$, show that $\{ A^T C_1,\dots,A^T C_n\}$ is a basis for $\mathbb{R}^n$.

I'm really confused about how I should do this. Could anyone please help me?

答:
Hint: what are the properties of a basis in general?

Edit: some more hints

You need to show two things namely:

Linear independence

That they span $\mathbb{R^n}$

What do you know of the rank of $A^T$ given what you know of that of $A$? How could you represent $\{ A^T C_1,\dots,A^T C_n\}$ in a different way?

Alright, to be more specific:
$\{ A^T C_1,\dots,A^T C_n\}$ are the columns of $A^TA$. The rank of $rank(A^TA)=rank(A)=n$ By definition of the rank, these columns are all linearly independent and together span $\mathbb{R^n}$. Hence, they form a basis for $\mathbb{R^n}$

Title: Basis given the rank of A is equal to n


问:
The probability of $7 7$’s in $17$ digit random number?
$$205-7895777-7786742$$

What is the probability that the number $7$ appears $7$ times in a $17$ digit random number?

答:
Let us decide MSD (left most) first,then assign the remaining 7's, finally fill the rest digits.

MSD $\neq$ 7,  8 choices for this digit; $C_7^{16}$ choices for all seven 7's to take 7 digits;  $9^9$ choices for the rest 9 digits. Total  8$C_7^{16}9^9$ choices for this case.

MSD =7, only one choice for this digit; $C_6^{16}$ choices for the other six 7's to take 6 digits; $10^9$ choices for the rest 10 digits. Total  $C_6^{16}$ $10^9$  choices for this case.

Adding results from 1 and 2 , there are  8$C_7^{16}9^9$+ $C_6^{16}$ $10^9$  choices for  seven 7's in a 17 digits number. With 9.10$^{16}$  possible 17 digits numbers, the asked probability is $\frac{8C_7^{16}9^9+ C_6^{16}10^9 }{9.10^{16}}$.


问:
Trouble finding my error showing the limit of $e^{x^{e^{-x}}}$
I have found a way to proof this, but I would really like to know where the error in my initial approach is, since I am only comfortable when I can use different methods. So I don't need a solution to this problem but a correction of my (false) way. The limit is: $$\lim_{x \to \infty}  (e^{x^{e^{-x}}}) = e $$

I tried: $$ a = e^{x^{e^{-x}}}\\ ln(a) = ln (e^{x^{e^{-x}}}) \\ ln(a) = e^{-x}  ln(e^x) \\ e^{ln(a)} = e^{e^{-x}ln(e^x)} \\ a = e^{e^{-x}x}  \\ a = e^{\lim_{x \to \infty}\frac{x}{e^x}} \ \ \ (l'hopital) \\ a = e^{\lim_{x \to \infty}\frac{1}{e^x}} \\ a = e^0 = 1$$
I know it's not written down flawlessly, but I hope you get the idea of what I was trying to do..Why is my result 1 and not e?

答:
Assuming the limit exists, take two logs or log log of your limit

$$\ln \ln L = \lim_{x\to\infty} e^{-x}\ln x = 0$$

Thus $L=e^{e^0} = e$


问:
Does this ODE question have closed form solution?
These days, I am struggling with following ODE problem when I build up my research model:

$1/2f''(x)+a(b - x) f'(x) -(c+ e^{A+Bx})f(x)=0$
where f(x) is a smooth function, and $a,b,c, A,B$ are all constants.
How to get the closed form of f(x)?

I tried the Laplace transform to work on it, say $F(s) = L(f(x)) $, but because of $e^{A+Bx}$, there will be a term $F(s-B)$ in the transformed equation. How to deal with this term?

I also tried the power series method, but got some very complicate coefficients, which stops me going further.

I think the term $e^{A+Bx}$ is the difficult part.

Could anyone here tell me how to deal with this kind of problem? Does the solution exit? I tried several ODE books but cannot find similar examples. Or could any one can suggest some relevant books?

Thank you very much.

答: One thing you can do is use a change of variables $x -> p x + q$ to reduce the number of parameters by two: you could, for example, take $A=0$ and $B=1$.</p>


问:
pointwise limit of finite measures
If there is a sequence of measures $\mu_n$ such that $\mu_n(A) \overset{n}{\rightarrow} \mu(A)$ for all $A$ in the $\sigma$-field and if $\mu_n(\Omega)\leq c$ $(c<\infty)$ for all $n$, then $\mu$ is a measure.
(problem 7, section 1.5 in Robert Ash)

The above assertion is proved by the Vitali-Hahn-Saks theorem.

What I would like to know is a counter example for the case when the condition,

$\mu_n(\Omega) \leq c$ for all $n$

is not satisfied.

答:
Take $\mu_n: \mathbb{N} \to \mathbb{R}$ to be
$$
  \mu_n(A) = \#(A \cap [n,\infty)).
$$
For any bounded set $B$, you have that $\mu_n(B) \rightarrow 0$. But, for any unbounded set $U$, $\mu_n(U) = \infty \rightarrow \infty$.

The limit $\mu$ is not a measure, because
$$
  \mu(\mathbb{N})
  = \infty
  \neq 0
  = \sum_{n \in \mathbb{N}} \mu(\{n\}).
$$

This example is quite easy to extend to $\mu_n(A) = \lambda(A \setminus K_n)$, where $K_n$ is a sequence of sets such that $K_n \uparrow K$, with $\infty \neq \lambda(K_n) \rightarrow \lambda(K) = \infty$.

In this case, $\mu(K) = \infty$, while $\mu(K_n) = 0$.

Title: pointwise limit of finite measures


问:
Where is the grad trace operator ($\nabla \operatorname{trace} A$ for matrix field $A$) used?
I recently stumbled upon the linear partial differential operator $$\mathcal{L}A\colon=\nabla \mathrm{tr} A\quad\text{for matrix-valued fields } A\colon\mathbb{R}^d\rightarrow\mathbb{R}^{d\times d}.$$
In coordinates, this is $(\mathcal{L}A)_i=\partial_i(\sum_j A_{jj})$. In fact, the operator that arose was $$\nabla \mathrm{tr}-\mathrm{div},$$where the divergence is taken row wise, i.e., $(\mathrm{div}A)_i=\sum_j \partial_j A_{ij}$; also, my matrix fields are symmetric.

I could swear I've seen a fluid mechanics reference where the second operator appeared, but I cannot remember what it was. I could, of course, be wrong.

Could you please let me know if you've encountered any of these operators in applications or in any place in the literature?

Thank you!

答:
$\def\tr{\mathrm{tr}}$
$\def\div{\mathrm{div}}$
$\def\grad{\mathrm{grad}}$
$\def\d{\mathrm{d}}$
$\def\R{\mathbb{R}}$

I will suppose your divergence operator is
$$(\div A)_i = \sum_{j=1}^d \partial_j A^j{}_{i}.$$

Admittedly, this answer is rather convoluted, but you can interpret your matrix field $A$ as a vector-valued differential $1$-form, since at any point $x\in\R^d$ you have an operator $A(x)$ which acts linearly on a vector $v\in\R^d$ and returns a vector $A(x)(v)\in\R^d$.

Then the components of the exterior derivative of $A$ are
$$(\d A)^k{}_{ij}=2\partial_{[i} A^{k}{}_{j]}
= \partial_iA^k{}_j-\partial_jA^k{}_i$$
where, the brackets are the antysimmetrization operator. Then your operator $\grad\circ\tr-\div$ is the exterior derivative followed by the trace (taken on the second index):
$$\begin{aligned}
(\tr_2(\d A))_i
&= \sum_{j=1}^d(\d A)^j{}_{ij} \\
&= \sum_{j=1}^d\partial_iA^j{}_j - \sum_{j=1}^d\partial_jA^j{}_i \\
&= \partial_i(\tr A) - (\div A)_i.
\end{aligned}$$
Hence,
$$\tr_2(\d A) = \grad(\tr A) - \div A.$$

I seriously doubt this is what you want, though?


问:
Understanding Proof that generalized eigenvectors are linearly independent (axler)
There is one part of the proof that I don't understand which I have marked in blue

I thought it was the case that $Tw=\lambda_{1}w$ for this very specific $w$ that we have constructed, so only the term that is being applied to $w$ is it true that $(T-\lambda_{1}I)^{n}w=(\lambda_{1}-\lambda)^{n}w$

So for the step marked in blue, it should be $=(T-\lambda_{2}I)^{n}...(\lambda_{1}-\lambda_{m})^{n}w$ How are we allowed to turn all the $T$ into $\lambda_{1}$ if they are not operating on w? The reason I think this is the case is because for the step right before this one, we had to commute (according to the powerpoint by Axler himself) $(T-\lambda_{1}I)^{k}$ to where it is just in front of $v_{1}$ to get $w$

答:
Let's focus on two terms and see if it is easier to understand

\begin{align}(T-\lambda_2I)^n (T-\lambda_3I)^n w &=(T-\lambda_2I)^n (\lambda_1-\lambda_3)^n w \\
&=(\lambda_1- \lambda_3)^n[(T-\lambda_2I)^n w]\\
&=(\lambda_1- \lambda_3)^n[(\lambda_1-\lambda_2)^n w]\\
&=(\lambda_1- \lambda_2)^n(\lambda_1-\lambda_3)^n w\end{align}

If necessary, we can have an induction proof.


问:
Difference between $P(Y \ge y)$ and $P(Y > y)$.
In the book "Probability: Theory and Examples", there is this theorem:

Lemma 2.2.8 If $Y \ge 0$ and $p > 0$ then $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y > y) \,  \mathrm dy$.

The proof is by using Fubini's Theorem as follows

\begin{align*}
\int_0^\infty py^{p-1} \mathbb P (Y > y)  \mathrm dy
& = \int_0^\infty \int_\Omega py^{p-1} 1_{(Y > y)} \, \mathrm dP \, \mathrm dy \\
& = \int_\Omega \int_0^\infty py^{p-1} 1_{(Y > y)} \, \mathrm dy \, \mathrm dP \\
& = \int_\Omega \int_0^Y py^{p-1} \mathrm dy \mathrm dP  \\
& = \int_\Omega Y^p \mathrm dP = \mathbb E Y^p \\
\end{align*}

For me it seems totally fine to replace $Y > y$ with $Y \ge y$ in this proof, but the conclusion would be $\mathbb E(Y^p) = \int_0^\infty py^{p-1} P (Y \ge y) \,  \mathrm dy$. So if $Y$ can only take nonnegative integer values, we have $\mathbb E Y = \sum_{y \ge 0} P(Y \ge y)$, which is wrong.

So why we can not replace $Y > y$ with $Y \ge y$?

答:
Lemma 2.2.8 is equally valid when one replaces $P[Y\gt y]$ in the integral by $P[Y\geqslant y]$ because $P[Y\gt y]=P[Y\geqslant y]$, for almost every $y$ with respect to the Lebesgue measure, since the set of values of $y$ such that $P[Y\gt y]\ne P[Y\geqslant y]$ is at most countable.

This holds for every nonnegative random variable $Y$, either continuous or discrete or neither continuous nor discrete.


问:
Are the set of direction ratios for a given line unique?
I perfectly understand the concept of direction cosines for a Vector.

Corresponding to an axis, the direction cosine is just the cosine of the angle that the Vector makes with the positive direction of the axis.

But I have problems in uniquely defining the direction cosines of a Line. According to me there is no way to uniquely specify direction cosines of a Line since it has no unique direction. There should always be 2 sets of direction cosines (of opposite sign).

There was a question in my textbook that read: " If a line is parallel to $-18i+12j-4k$   then what are its direction cosines?"

In the solution they have simply calculated the direction cosines of the Vector given, Why isn't the negative counterpart of the direction cosines also given as the solution?

Another example, the direction cosines of the $x$ axis is always given as $(1,0,0)$.  But $(-1,0,0)$ is also valid according to me.

Is there some convention we have to follow?

答: The convention followed is usually this: Direction cosines are cosines of angles with positive axes measured counterclockwise. Let $\alpha_x, \alpha_y, \alpha_z$ be angles made by a line with positive x, y, z axes respectively in counterclockwise sense, then DCs are $\cos \alpha_x, \cos \alpha_y, \cos \alpha_z$.


问:
How do we integrate $\int \frac{1}{x\sqrt{x^2+a^2}}\mathrm{d}x$?
I would appreciate it if someone explained to me how to integrate

$$\int \frac{1}{x\sqrt{x^2+a^2}}\mathrm{d}x$$

For context, I am a physics major who is integrating an electric field expression for a straight line of charge, but I do not know the correct method to integrate such an expression.

答:
HINT

You can either apply the substitution $x = a\sinh(u)$ or $x = a\tan(u)$. Let us stick with the first option:
\begin{align*}
\int\frac{\mathrm{d}x}{x\sqrt{x^{2}+a^{2}}} = \frac{1}{a}\int\frac{\mathrm{d}u}{\sinh(u)} = \frac{1}{a}\int\frac{\sinh(u)}{\sinh^{2}(u)}\mathrm{d}u = \frac{1}{a}\int\frac{\sinh(u)}{\cosh^{2}(u) - 1}\mathrm{d}u
\end{align*}

Then you can apply the partial fraction method to the last integral.

Can you take it from here?


答:
Pull out an $x$ from the square root

$$\int \frac{dx}{x^2\sqrt{1+\frac{a^2}{x^2}}} = \int \frac{-1}{\sqrt{1+a^2\left(\frac{1}{x}\right)^2}}\:d\left(\frac{1}{x}\right)$$

From here the substitution becomes obvious

$$\frac{1}{x} = \frac{\sinh t}{a} \implies d\left(\frac{1}{x}\right) = \frac{\cosh t}{a}\:dt$$

which means the integral is simply

$$\int \frac{-\cosh t}{a\cosh t}\:dt = -\frac{t}{a} + C = -\sinh^{-1}\left(\frac{a}{x}\right)+C$$

Title: How do we integrate $\int \frac{1}{x\sqrt{x^2+a^2}}\mathrm{d}x$?

Body:
I would appreciate it if someone explained to me how to integrate

$$\int \frac{1}{x\sqrt{x^2+a^2}}\mathrm{d}x$$

For context, I am a physics major who is integrating an electric field expression for a straight line of charge, but I do not know the correct method to integrate such an expression.


答:
I suspect this is the same thing as the other answer but I lack the fluency in differential geometry to verify. Note that if you apply trace to a jacobian, you end up with div. So in the context of fluid mechanics, I bet it arose from the double curl formula
$$ \nabla\times(\nabla\times u) = [\nabla,\text{div}]u = \nabla \operatorname{div} u - \Delta u=  \nabla\operatorname{tr} (\nabla u) - \operatorname{div} (\nabla u)$$
For divergence free fields one can invert the Laplacian to recover the "velocity" $u$ from the "vorticity" $\omega=\nabla \times u$ (this is called the "Biot-Savart law")

Title: Where is the grad trace operator ($\nabla \operatorname{trace} A$ for matrix field $A$) used?

Body:
I recently stumbled upon the linear partial differential operator $$\mathcal{L}A\colon=\nabla \mathrm{tr} A\quad\text{for matrix-valued fields } A\colon\mathbb{R}^d\rightarrow\mathbb{R}^{d\times d}.$$
In coordinates, this is $(\mathcal{L}A)_i=\partial_i(\sum_j A_{jj})$. In fact, the operator that arose was $$\nabla \mathrm{tr}-\mathrm{div},$$where the divergence is taken row wise, i.e., $(\mathrm{div}A)_i=\sum_j \partial_j A_{ij}$; also, my matrix fields are symmetric.

I could swear I've seen a fluid mechanics reference where the second operator appeared, but I cannot remember what it was. I could, of course, be wrong.

Could you please let me know if you've encountered any of these operators in applications or in any place in the literature?

Thank you!


问:
square cake with raisins
Alice bakes a square cake, with $n$ raisins (= points).

Bob cuts $p$ square pieces. They are axis-aligned, interior-disjoint, and each piece must contain at least $2$ raisins.

Note that a single raisin can be shared by two pieces (if it is on their boundary), or even four pieces (if it is on their corner).

Bob tries to maximize the number of pieces $p$, and Alice tries to minimize $p$ by a sophisticated placement of the raisins.

For a given $n$ (number of raisins), how should Alice place the raisins such as to minimize $p$, and what is the minimum?

Here are some simple cases:

For n=0, 1, obviously p=0.

For n=2, 3, 4, Alice can make p=1, by placing the raisins on the corners of the cake, since the only square piece that can contain 2 raisins is the entire cake.

For n=5, the minimum p is 2: there is at least one quarter of the cake that contains 2 raisins. Bob can cut a piece that contains these 2 raisins on its boundary, and is entirely contained within that quarter of the cake. Now there is enough room for another square, that contains one of these raisins and another raisin.

What is the minimum $p$ for a given $n$?

-

Some insights:

Given two raisins, the minimum length of a square that contains both of them is the chessboard distance between them.

Therefore, it seems that Alice's best strategy is to maximize the  chessboard distances, and so force Bob to use large squares. This makes sense, but I cannot prove this.

If n is a square number ($n = k^2$), then Alice can place the raisins in a k-by-k grid, such that the smallest chessboard distance between raisins is $1/(k-1)$. In this case, Bob can cut at most $(k-1)^2$ squares. But is this really Alice's best strategy?

Bob can always do best with pieces that have raisins only on their boundary (He never has to cut a piece which has a raisin in the interior). PROOF: For any two raisins, there is a square that contains both of them in the boundary. If a piece contains two raisins in the interior, or one in the interior and one in the boundary, Bob can shrink that piece to get a square that has these two raisins in the boundary. This square will be entirely contained in the original square, and so will not affect the other pieces.

答:
Edit (31 May 2013): The proof of the upper bound below is wrong (see comments), the lower bound should still hold.

Let $p(n)$ be the optimal value of $p$ if both players play optimally. I will prove that $p(n)\leq (n+2)/3$ and will sketch a proof of $p(n)\geq C n$ for an absolute constant $C>0$.

Alice can attain $p(n)\leq (n+2)/3$. Proof is by induction on $n$. Identify the cake with $[0,1]^2$ in the Cartesian plane. Place three raisins at $(0,1)$, $(1,0)$, and $(1,1)$. Then take an optimal placement of $n-3$ raisins, scale by a factor $1/3$ to make a placement in $[0,1/3]^2$. The result is a set of $n$ raisins. Now, if $p\geq 2$, then at most one of the three corner raisins belong to a square. We thus obtain that either $p(n)<2$ or $p(n)\leq 1+p(n-3)$. In either case, the induction step is complete.

Next I explain Bob's strategy. Assume that instead of $[0,1]^2$ square the whole game happens on the infinite plane $\mathbb{R}^2$. So, the squares are allowed to extend beyond cake's boundary. Before cutting the squares out, which is an irreversible operation, Bob should mentally choose which squares to cut, as follows. He starts with no squares at all. For each raisin $r$ there is a unique smallest square $S(r)$ that is centred at $r$ and contains at least one other raisin. (I assume that $n\geq 2$ for $S(r)$ to be well-defined.) At each step Bob does the following:

1) If square $S(r)$ does not intersect any of the squares in his mental list, then Bob add $S(r)$ to the list.

2) If square $S(r)$ meets some existing squares, then $S(r)$ is added only if it is smaller than the squares that it intersects. In that case, the square it intersects are removed from the list.

The process terminates because if we sort the squares sizes, then the resulting sequence becomes lexicographically smaller at each step.

Now if the point $r'$ is not in the resulting list of squares because $S(r')$ intersects some smaller (or equal) $S(r)$, then we say that the point $r'$ blames the point $r$. I claim a point $r$ can be blamed by at most $8$ other points. Sadly, I do not see a clean proof, though it is fairly obvious from drawing the pictures. Assuming the claim, it follows that $n$ is at most $9$ times the number of squares (each square contains the centre, and at most eight other points can blame it).

In the above I assumed that the game is played on $\mathbb{R}^2$. I do not think it should be a major difficulty, but again I do not see a clean way to deal with it either.

Title: square cake with raisins


问:
Product and intersection of comaximal ideals in noncommutative ring
So, exercise 4.5. of Chapter III of Aluffi's "Algebra: Chapter 0" states the following:

Let $I, J$ be ideals in a ring, such that $I + J =
(1)$. Prove that $IJ = I\cap J$.

(We say that ideals $I,J$ such that $I + J = (1)=R$ are comaximal).

Now, I have been able to prove the statement for the commutative case, but is it possible to prove it for noncommutative rings? If not, are there counterexamples? The curious thing is that the questions I found here about this problem all assume that $R$ is commutative, so I am wondering if there is actually something missing in Aluffi's exercise.

答: The result does not hold in general in the non-commutative case. Let $R$ be the ring of upper triangular $2\times 2$ matrices over $\mathbb{Q}$, and for each $i\leqslant j\in\{1,2\}$ let $E_{ij}$ denote the matrix whose entries are $1$ in the ${ij}$-th position and $0$ everywhere else. Now, let $I$ be the two-sided ideal $\{\lambda E_{12}+\mu E_{22}:\lambda,\mu\in\mathbb{Q}\}$ and let $J$ be the two-sided ideal $\{\lambda E_{11}+\mu E_{12}:\lambda,\mu\in\mathbb{Q}\}$. Certainly $I+J=R$, since $1_R=E_{22}+E_{11}$, and we have $I\cap J=\{\lambda E_{12}:\lambda\in\mathbb{Q}\}$. However, we have $IJ=\{0\}$, so this gives the desired counterexample.


问:
Suppose $X$ be the random variable that denotes the number of heads minus the number of tails when $10$ unbiased coins are tossed. Find the variance.
Suppose an unbiased coin is tossed $10$ times. Let $X$ be the random variable that denotes the number of heads minus the number of tails. What is the variance of $X$?

This seems to have the solution but I do not understand what it says. I am looking for an explanation of the link I shared.

I am still sharing my solution so as to show I am not posting it without trying it at all.

$\text{E}[X] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))\right) = \frac{1}{2^{9}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5) = 0$

$\text{E}[X^2] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))^2\right) = \frac{1}{2^{8}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5)^2 = 10$

This gives $\text{Var}[X] = \text{E}[X^2] - \text{E}[X]^2 = 10 - 0^2 = 10$. This is how I did, considering the PMFs.

答:
$E[X] = \sum x \cdot P(x)$
The probability is as you seem to indicate, $${10 \choose h}(\frac{1}{2})^n \frac{1}{2}^{10-h} = \frac{{10 \choose h}}{2^{10}}$$
If you wanted the number of heads that should be multiplied by $h$ and summed.  As you want the number of heads minus tails, your first calculation is correct, $E[X] = 0$, which makes sense by symmetry anyway.
I repeated your calculation of $E[X^2]$ and got the same answer you did, which appears to be correct according to the other answers.


答:
We’ll assume we know $$\lim_{y\to\infty}y^{1/y}=1.\tag 1$$

Then note that since $e^x>x$ for any real $x,$ then $$0<e^{-x}<1/x$$ for $x>0.$

But then $$1<x^{e^{-x}}<x^{1/x}$$ for $x>1.$ So by the squeeze theorem and (1), $$\lim_{x\to\infty} x^{e^{-x}} =1.$$

From this we see:

$$\lim_{x\to\infty} e^{x^{e^{-x}}} =e^1=e.$$

Your error is that you interpreted the expression as $$\left(e^x\right)^{e^{-x}}$$ which if you let $y=e^x$ is $y^{1/y},$ and gives a limit of $1,$ as you calculated, again from (1).

Title: Trouble finding my error showing the limit of $e^{x^{e^{-x}}}$

Body:
I have found a way to proof this, but I would really like to know where the error in my initial approach is, since I am only comfortable when I can use different methods. So I don't need a solution to this problem but a correction of my (false) way. The limit is: $$\lim_{x \to \infty}  (e^{x^{e^{-x}}}) = e $$

I tried: $$ a = e^{x^{e^{-x}}}\\ ln(a) = ln (e^{x^{e^{-x}}}) \\ ln(a) = e^{-x}  ln(e^x) \\ e^{ln(a)} = e^{e^{-x}ln(e^x)} \\ a = e^{e^{-x}x}  \\ a = e^{\lim_{x \to \infty}\frac{x}{e^x}} \ \ \ (l'hopital) \\ a = e^{\lim_{x \to \infty}\frac{1}{e^x}} \\ a = e^0 = 1$$
I know it's not written down flawlessly, but I hope you get the idea of what I was trying to do..Why is my result 1 and not e?


答:
Hint: Let $I=\int \frac{1}{x\sqrt{x^2+a^2}}\mathrm{d}x$; Substitute $x=a\tan \theta\implies dx=a\sec^2\theta d\theta$ so

$I=\frac 1a \int \operatorname {cosec} \theta  d\theta=\frac 1a \ln|\tan \frac \theta 2|+c$

Can you take it from here?


问:
Find the covariance of $Y=X_{1}+3X_{2}$ and $W=X_{1}+X_{3}+2X_{4}$
My problem:

Suppose that a random sample of size $n=4$ is taken without replacement from a finite population with $N=10$. Now suppose that $c_{i}=3i+4, i=1,2,...,N$, where $c_i$ is the label on the $i^{th}$ ball in the urn.

Let $Y=X_{1}+3X_{2}$ and $W=X_{1}+X_{3}+2X_{4}$.

Find the covariance of $Y$ and $W$.

My attempt:

So far I have found the variance of $c_{i}$,
$$\sigma^{2}= \frac{\sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}}{N}=\frac{742.5}{10}=74.25$$

After this is where I start to get lost.

I know that for integers $r, s \in \{1,2,...,n\}$ with $r\ne s$, the covariance can be calculated as
$$cov\left(X_{r}, X_{s}\right)= -\frac{\sigma^{2}}{N-1}$$

However, I am not sure how to proceed with the multiple variables. Any hints or tips would be greatly appreciated. Thanks.

Body:
Remember that $Cov$ is linear in each argument.  Letting $\sigma^2=74.25$, you have

$Cov(Y,W)=Cov(X_1+3X_2,X_1+X_3+2X_4)\\=Cov(X_1,X_1)+Cov(X_1,X_3)+2Cov(X_1,X_4)+3Cov(X_2,X_1)+3Cov(X_2,X_3)+6Cov(X_2,X_4)\\=\sigma^2-\frac{1}{9}(1+2+3+3+6)\sigma^2\\=-\frac{2}{3}\sigma^2\\=-49.5$


问:
What is the sum from 0 to infinity of the reciprocal of the rising factorial with both parameters having the index as the argument?
What is:

$$\sum_{n=0}^{\infty} \frac{1}{(n)_n}$$

where $(x)_n$ is the Pochhammer symbol (rising factorial), defined as:

$$\frac{\Gamma(x+n)}{\Gamma(x)} = \frac{(x+n-1)!}{(x-1)!}$$
?

I'm really more interested in the process of finding the closed form because just plugging it in to Wolfram Alpha is easy enough.

答:
One possible solution

$$f(x)=\sum_{n=0}^\infty \frac{x^n}{(n)_n}$$
$$\int f(x)\,dx=x \, _2F_2\left(1,1;\frac{1}{2},2;\frac{x}{4}\right)$$
$$\frac d{dx}\int f(x)\,dx=-\frac{1}{4} e^{x/4} \sqrt{x} \left(2 \sqrt{\pi }
   \left(1-\text{erf}\left(\frac{\sqrt{x}}{2}\right)\right)-\frac{4
   e^{-x/4}}{\sqrt{x}}-2 \sqrt{\pi }\right)$$ Expand and simplify
$$f(x)=1+\frac{1}{2} \sqrt{\pi } e^{x/4} \sqrt{x}
   \text{erf}\left(\frac{\sqrt{x}}{2}\right)$$ Now, compute $f(1)$.


问:
Every open covering of a subset of $\Bbb R^n$ has a countable subcover
I want to show the title. More formally,
Let $X\subset\Bbb R^n$ be a subset and $\{U_\alpha\}_{\alpha\in I}$ be a open covering of $X$. Then there is a subcover $\{U_\alpha\}_{\alpha\in I'\subset I}$ that covers $X$.

My basic plan is that define $S:=\Bbb Q^n$. Then for each $r\in S$, if there is $U_r\in\{U_\alpha\}_{\alpha\in I}$ that contains $r$ then choose only one such $U_r$. Collect them all for each $r$ (if there's such). I want to show this $\{U_r\}$ covers $X$. Any help? or any other idea?

答:
Your idea will, unfortunately, not quite work. Take for instance $X = \mathbb R$. Enumerate the rationals $\mathbb Q$ as $q_n$ and let $U_n = (q_n - 2^{-n}, q_n + 2^{-n})$. Take also $U_* = \mathbb R$ itself. Now, your construction is to take exactly one open neighborhood of every rational point. This won't work for any arbitrary choice. For instance, each $q_n \in U_n$ but the union $\bigcup U_n$ is not all of $\mathbb R$. The sum of the lengths of all the $U_n$ is $\sum 2^{-n} = 2$, so they cannot cover $\mathbb R$ even though we have a neighborhood of every rational point.

But as I said, your intuition to use $\mathbb Q$ is good. Recall that the definition of an open subset $U$ of $\mathbb R$ is that for all $x \in U$ there is a ball containing $x$ that is in $U$. Symbolically, $x \in B \subseteq U$. Such a ball is determined by two parameters - its center $p \in \mathbb R^n$ and its radius $r > 0$, denoted $B = B_p(r)$. We can perturb this a bit to show that $U$ is open iff every point $x \in U$ is contained in a rational ball $B \subseteq U$. We say that $B = B_p(r)$ is rational if $p \in \mathbb Q^n$ and $r \in \mathbb Q$. Importantly, there are only countably many rational balls!

Here's the proof. If $U$ is covered by rational balls then it's in particular covered by balls and is therefore open. Suppose conversely that $U$ is open and take $x \in U$. Then by assumption there is a ball $x \in B_p(r) \subseteq U$. By density, we can find rational points arbitrarily close to $x$. Then indeed we can find a rational point $q \in \mathbb Q^n$ such that $x \in B_q(\varepsilon) \subseteq B_p(r)$ Imagine this as nudging $p$ ever so slightly to a very nearby rational point $q$. We can additionally find some rational $s > \varepsilon$ that is just ever so slightly larger, so that $B_q(s) \subseteq B_p(r)$. Then $x \in B_q(s) \subseteq U$ is a rational ball. I didn't do this in great detail, but if you draw a picture you should be able to prove this more rigorously.

Anyways, take now your open cover $\{U_\alpha\}_{\alpha \in I}$. Then for each $x \in X$ there is some rational ball $x \in U_{\alpha}$. By the above, we can find a rational ball $x \in B^\alpha \subseteq U_{\alpha}$. There are only countably many rational balls $B^\alpha$, and for each ball pick exactly one $U_\alpha$ containing it. Then we have chosen countably many elements of the cover. Furthermore, we have that each $x$ is in such a rational ball. Hence, these chosen $U_\alpha$ are a countable subcover.


问:
Relation with same extension with different properties
Suppose there are two relations where  $R' \subseteq A' \times A'$ and $R \subseteq A \times A$ where $A \subset A'$ and $R =  R'$ . For $R$ the condition of reflexivity holds, since $\forall x \in A; xRx$, but $\exists x \in A'; \neg(xR'x)$, so reflexivity does not hold for R'.

Given a principle for equality, $R = R' \Rightarrow \forall \mathfrak{F} (\mathfrak{F}R' \Leftrightarrow \mathfrak{F}R)$. This should hold for the specific instance of reflexivity of a relation.

Are there different properties for extensionally identical objects?

答:
There are two different ways we can think about a relation: as simply a set of ordered pairs, or as a set of ordered pairs together with an explicit domain and codomain. Some properties - like symmetry and transitivity - make sense for the first approach, while others explicitly refer to domain/codomain and so only make sense for the second approach. Reflexivity on the other hand is in this latter category, for exactly the reason you observe. It doesn't make sense to say that a given set of ordered pairs is or is not a reflexive relation.

(This is similar to the situation regarding functions, where we have a choice of whether or not to include an explicit codomain. Some properties of functions, such as surjectivity, only make sense if we include this extra information.)

If we need to keep things straight, we can introduce distinguishing terminology (although this is enough of a non-issue that I don't think there's a standard here). For instance:

A bare binary relation is just a set of ordered pairs.

A detailed binary relation is triple $(A,B,R)$ where $R\subseteq A\times B$.

Then the point is this:

It makes sense to ask whether a detailed binary relation is reflexive, but it does not make sense to ask whether a bare binary relation is reflexive.


问:
Formal Proof of "No Largest Number"
I need to create a formal proof of there not being the largest number, with only the definition of "less than" being given with Peano arithmetic. $s(x)$ means successor of $x$ or $x + 1$.

$$
\text{Premise}\\
\forall x\forall y (x \lt y \iff \exists z(x + s(z) = y)\\
\text{Prove}\\
\forall x \exists y (x < y)
$$

I've tried proof by contradiction, but the resources online don't seem to translate well into a formal proof. In particular, the strategy of "assume there is a largest number, add one, done" doesn't really translate with formal proof steps.

答:
Using the idea of Barry Cipra in the comments:

$$\forall x \exists y~(y=s(x)) \\
\forall x \exists y \exists w ~(x+y=x+s(x)=w) \\
(\forall x \exists z (=x)~(x +s(z)=w)) \Rightarrow x \lt w \Rightarrow \exists w~(x \lt w)$$

Title: Formal Proof of "No Largest Number"


问:
Criterion for Simpleness of Modules over an Algebra
Let $k$ be a (if necessary algebraically closed) field. Let $A$ be a finite-dimensional $k$-algebra and $M$ be an $A$-left-module. Consider $M$ to be a right-module over $B = \mathrm{End}_A(M)$. Assume that $N$ is a simple $B$-left-module. Is it true that $M\otimes_B N$ is a simple $A$-left-module?

In particular, is it true, that if $\mathrm{End}_A(M) \cong k$, we have $M$ is a simple $A$-module?

答:
No. Let $A$ be the three dimensional path algebra of the quiver $1 \to 2$, then the projective cover $M$ of the first simple module is not simple but has $\operatorname{End}_A(M) \cong k$.

More explicitly, let $$
A = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} : a,b,c \in k \right\}, \qquad
M = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} : x,y \in k \right\}, \qquad
N = \left\{ \begin{bmatrix} x \\ 0 \end{bmatrix} : y \in k \right\}$$

Then $M$ has exactly three submodules, $0$, $N$, and $M$, so $M$ is indecomposable. The general element of $A$ acts as multiplication by $a$ on $N$, and as multiplication by $c$ on $M/N$, so $N$ and $M/N$ are not isomorphic. Hence every endomorphism is 0 or an isomorphism, and so $B$ is a division ring. While it is not too hard to see $B\cong k$ already (especially if $k$ is algebraically closed), one can also give a direct calculation:

Consider $\operatorname{End}_k(M) = \left\{ \begin{bmatrix} d & e \\ f & g \end{bmatrix} : d,e,f,g \in k \right\}$ and its subring $B = \operatorname{End}_A(M) = \left\{ F \in \operatorname{End}_k(M) : FG=GF \text{ for every } G \in A \right\}$.  Finding $B$ is then just a matter of solving the equations $FG=GF$ for $d,e,f,g$. One gets $e=f=0$ by taking $a=1$ and $b=c=0$, and one gets $d=g$ by taking $a=c=0$ and $b=1$. Hence $$B = \left\{ \begin{bmatrix} d & 0 \\ 0 & d \end{bmatrix} : d \in k \right\} \cong k$$


问:
Find the distance between intersection points of the bisectors of the base angle and perpendiculars from vertex
Let $ABC$ be a triangle where $AB = 10$ and $AC = 12$. The bisectors of ${\angle}B$ and ${\angle}C$ intersect $AC$ and $AB$ at points $D$ and $E$ respectively. We draw perpendiculars $AM$ and $AN$ from vertex $A$ on line segments $CE$ and $BD$ respectively. Knowing that the length of $MN$ is $4$, how would one find the length of $BC$?

答:
Let $\overrightarrow {AN} \cap \overline {BC} = \{F\}, \; \overrightarrow {AM} \cap \overline {BC} = \{G\}$.

$$\text{In} \ \Delta \text{BNA}, \Delta \text{BNF}: \\ \ m(\angle ABN) = m(\angle FBN), \\ m(\angle ANB) = m(\angle FNB), \\ \overline {BN} \text{ is a common side.} \\ \therefore \Delta \text{BNA} \equiv\Delta \text{BNF} \qquad \text{(1)}$$

$$\text{In} \ \Delta \text{CMA}, \Delta \text{CMG}: \\ m(\angle ACM) = m(\angle GCM), \\ m(\angle AMC) = m(\angle GMC), \\ \overline {CM} \text{ is a common side.} \\ \therefore \Delta \text{CMA} \equiv\Delta \text{CMG} \qquad \text{(2)}$$
From $(1), (2)$ we deduce that:
$$AB=BF=10 \qquad \text{(3)} \\ AC=CG=12 \qquad \text{(4)}$$
$$\text{In} \Delta \text{AMN}, \Delta \text{AGF}: \\ AN=NF, \\ AM=MG, \\ \angle A \text{ is common.} \\ \therefore \Delta \text{AMN} \sim \Delta \text{AGF}$$

$$AM = MG, \\ AN = NF, \\  MN =4 \\ \therefore FG = 2MN = 8 \qquad \text{(5)}$$
From $(3), (4)$ and $(5)$ we conclude that: $\fbox{BC = 14}$


问:
Find the angle in the triangle
The triangle $ABC$ is isosceles at $A$ with $\angle BAC=20^o$. The points $D$ and $E$ are on the sides $[AB]$ and $[AC]$, respectively. We know that $\angle EBC=40^o$ and $\angle BCD=50^o$. What is the measure of $\angle BED$?

What I did: since $ABC$ is isosceles at $A$ and $\angle BAC=20^o$, we get that $\angle ABC=\angle BCA=80^o$. Therefore, we have $\angle DBE=40^o$ and $\angle BCD=50^o$. Moreover, the lines $(EB)$ and $(CD)$ are perpendicular. It follows that $\angle BEC=60^o$ and $\angle BDC=50^o$.

All the angles of the "butterfly" BDCE are known, and we know the angle $\angle EBC$. So the $\angle BED$ should not be too complicated to get, but I am stuck. Any idea?

答:
This is an easier version of the classical problem since one of the lines is an angle bisector and can be solved using congruences.

Let $BE$, $CD$ intersect in $O$. $\triangle BOD  \cong \triangle BOC$ using $ASA$ since $BE$ is angle bisector. Hence $OD=OC$.

Thus $\triangle DOE \cong \triangle COE$ using $SAS$. Hence
$$\angle BED = \angle BEC=60^{\circ}$$

Title: Find the angle in the triangle


问:
Partitioned positive definite matrix property
I am interested in the following problem:

Let $X$ be a real symmetric positive definite matrix partitioned into four submatrices as follows:
$$
X = \begin{pmatrix}
     A&B\\
     B^T & C
\end{pmatrix}
$$
Define the symmetric matrix $Y$ by:
$$
Y = C - B^TA^{-1}B
$$
Is $Y$ always positive definite?

I suspect that the answer is yes. I have attempted using the definition of positive definite matrices to prove this and also tried looking for counterexamples, but have so far been unsuccessful. I would greatly appreciate any help or recommendations.

答:
Yes.

Recognizing that $Y$ is the Schur complement of $A$ in $X$, we can use the following condition for positive definiteness of a Schur complement:

Let $X$ be a symmetric matrix of real numbers given by
$$
X=\begin{pmatrix}A&B\\B^T&C\end{pmatrix}
$$
Then if A is invertible, then X is positive definite if and only if A and its complement X/A are both positive definite:
$$
{\displaystyle X\succ 0\Leftrightarrow A\succ 0,X/A=C-B^{\mathsf {T}}A^{-1}B\succ 0}
$$

Since $X$ is positive definite, $A$ is positive definite and is therefore invertible. Then by the above equivalence statement, we know that $Y=C-B^TA^{-1}B$ is positive definite.


问:
Cauchy product on exponential-looking power series
Original posting by dioxen here: Double summation including power and factorial

I am finding some trouble in computing the following sum:

$$\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$

Could you please provide a result?

Thanks in advance

After a failed attempt at getting to the binomial expansion, I tried to play around with it a little bit. I came up with two ideas of a solution, but can't see it through. Both involve applying the Cauchy product.
\begin{equation}
\Bigg(\sum_{n=0}^\infty a_n \Bigg) \Bigg(\sum_{n=0}^{\infty} b_n \Bigg) = \sum_{n=0}^{\infty} c_n \space \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k} \space \space \space \space \space \space \space \space  \space \space \space \space (1)
\end{equation}

1: In this problem, we cound treat $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!}=\frac{1}{n!}\sum_{k=0}^n\frac {(xy)^{n-k}}{(n-k)!}x^{k}=\frac{1}{n!}\sum_{k=0}^n a_k b_{ n-k}$$

So if we're thinking within the framework of the first two summations in (1), our $a_n$ and $b_n$ are as follows: $a_n=x^n$ and $ b_n = \frac {(xy)^{n}}{n!}\\$

our only problem is the we're left with the $\frac{1}{n!}$ in front of the summation. If we thought up some way to get rid of it, we'd have a nice geometric series and $e^{xy}$. But I can't think of a way...any leads here?

2: Alternatively, we could leave the $\frac{x^n}{n!}$ outside and treat it as a number (because it is, before we choose to sum it up over $n$). $$ c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!} 1^{n-k}=\frac{x^n}{n!}\sum_{k=0}^n a_k b_{ n- k}$$ which gives us $a_n=\frac{y^n}{n!}$ and $b_n=1$. Could we do something about $\frac{x^n}{n!}$?

$$
\Bigg(\sum_{n=0}^\infty \frac{x^n}{n!} \Bigg) \Bigg(\sum_{n=0}^{\infty} \frac{x^n}{n!} \frac{y^n}{n!} \Bigg)
$$
Does this even make sense? Because otherwise I can't think of a way this sum would be computable.

答:
We start by Laplace Transformation of the inner sum argument, this way we will get rid of the factorial in the denominator and end up with a simple series:

$$\mathcal{L}\left(\frac{y^m}{m!} \right)=s^{-(m+1)} $$

Where $\mathcal{L}$, stands for Laplace transform from $y$ to $s$(something tells me, this will deeply assume y is positive). Now we can do the inner summation easily:

$$\sum_{m=0}^k{s^{-(m+1)}}=\frac{s^{-k-1} \left(s^{k+1}-1\right)}{s-1}=\frac{1-s^{-k-1}}{s-1} $$

Now putting this back into the initial sum, we will have:

$$\sum_{k=0}^\infty \frac{x^k}{k!} \frac{1-s^{-k-1}}{s-1}=\frac{1}{s-1} \sum_{k=0}^\infty\left(\frac{x^k}{k!}-\frac{(\frac{x}{s})^k}{s\ k!} \right)$$

These sums are just the expansion of $e^z$, so we have:

$$\frac{e^x}{s-1}-\frac{e^\frac{x}{s}}{s(s-1)}=\frac{e^x}{s-1}-\frac{e^{x/s}}{s-1}+\frac{e^{x/s}}{s}$$

Now, we have to calculate the inverse Laplace transform from $s$ to $y$. I could produce explicit expressions for all but the middle term(which I will write as an integral)

$$\sum_{k=0}^\infty \frac{x^k}{k!}\ \sum_{m=0}^k\frac {y^m}{m!}=\mathcal{L}^{-1}\left(\frac{e^x}{s-1}-\frac{e^{x/s}}{s-1}+\frac{e^{x/s}}{s} \right) \\ =e^{x+y}+J_0 (2\sqrt{xy})-\frac{1}{2 \pi i} \lim_{T\to\infty}\int_{ \gamma - i T}^{ \gamma + i T} e^{sy} \frac{e^{x/s}}{s-1}\,ds$$
where $\gamma$  is a real number so that the contour path of integration is in the region of convergence of $\frac{e^{x/s}}{s-1}$.

This is how far I could come. Hope it's a useful result for you.


问:
Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of converges to the same limit $a\in\mathbb{R}$
Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit  $a\in\mathbb{R}$. Show that $(a_n)$ must converge to $a.$

I have tried to prove this by contradiction. That is, assume $(a_n)$ does not converge to $a.$ Then there are two cases:

$(a_n)$ converges to some other value $b.$

$(a_n)$ diverges.

For case1, if $(a_n)$ converges to $b,$ then all the subsequences of $(a_n)$ should converge to $b,$ which contradicts that all convergent subsequences converge to $a.$

For case2, if $(a_n)$ diverges, then I can find counterexamples to show $(a_n)$ does not have the property that all convergent subsequences converge to the same limit. For example $(-1)^n$ has two subsequences with different limits. $(a_n)$

Then I figured out that I did not use the hypothesis that $(a_n)$ is bounded. So there might be something wrong in my proof, but I could not find it out.

答:
You did not handle case 2) properly.

If $(a_n)$ does not converge to $a$ then there exists $\epsilon >0$ and integers $n_k$ such that $n_1<n_2<...$ and $|a_{n_k}-a| \geq \epsilon$ for all $k$. Now $(a_{n_k})$ is bounded sequence and hence it has a convergent subsequence. But the inequality $|a_{n_k}-a| \geq \epsilon$  holds for this subsequence also. Do you see a contradiction now?


答:
As $\angle BCD = 50^0, \angle CBE = 40^0$, we have $\angle BFC = 90^0$.

Please note as $\angle DBF = \angle CBF$ and $BF \perp CD$, $BF$ is perpendicular bisector of $CD$. So, $DF = FC$ and $EF \perp CD$.

That leads to $\triangle DFE \cong \triangle CFE$. So, $\angle CDE = \angle DCE = 30^0$ and $\angle BED = \angle BEC = 60^0$.

Title: Find the angle in the triangle

Body:
The triangle $ABC$ is isosceles at $A$ with $\angle BAC=20^o$. The points $D$ and $E$ are on the sides $[AB]$ and $[AC]$, respectively. We know that $\angle EBC=40^o$ and $\angle BCD=50^o$. What is the measure of $\angle BED$?

What I did: since $ABC$ is isosceles at $A$ and $\angle BAC=20^o$, we get that $\angle ABC=\angle BCA=80^o$. Therefore, we have $\angle DBE=40^o$ and $\angle BCD=50^o$. Moreover, the lines $(EB)$ and $(CD)$ are perpendicular. It follows that $\angle BEC=60^o$ and $\angle BDC=50^o$.

All the angles of the "butterfly" BDCE are known, and we know the angle $\angle EBC$. So the $\angle BED$ should not be too complicated to get, but I am stuck. Any idea?


问:
Prove if $\operatorname{adj}(A) = O$ then $\operatorname{rank}A \leq n-2$
Prove if $\operatorname{adj}(A) =0$, then $\operatorname{rank}A \leq n-2$

My initial idea is to setup an induction. Base case $A \in M_{2×2}$ which obviously implies that $A = [O]_{2×2} \rightarrow \operatorname{rank}A \leq 0 = n-2$. However, I can't seem to reason how assuming it worked in the $M_{n\times n}$ case allows for the $M_{(n+1)×(n+1)}$ case to work....

答:
One approach is to use the notion of "determinantal rank", as explained in this post. In particular: if the adjoint of $A$ is zero, then all size $n-1$ submatrices of $A$ have determinant zero which means that $A$ has rank at most $n-2$.

For a direct proof: we are given that $\operatorname{adj}(A) = 0$. Suppose for the purpose of contradiction that $\operatorname{rank}(A) \geq n-1$. It follows that we may select $n-1$ linearly independent rows of $A$. This submatrix has rank $n-1$, which means that we may select $n-1$ linearly independent columns. These columns form a square, size $(n-1)$ submatrix of $A$.

Because this submatrix is invertible, its determinant is non-zero. However, this means that one of the entries of $\operatorname{adj}(A)$ is non-zero, contradicting our premise.


问:
Classification of groups $G$ of order 24.
I supposed $n_3=4$ and $n_2=3$, I want to show that $G\cong S_4$. Then I made $G$ act by conjugation on $\text{Syl}_4 (G)=\{X_1,X_2,X_3,X_4\}$. This action defines a representation $\varphi:G\rightarrow S_4$. we have $\ker(\varphi)=\cap_{i=1}^{4}N_G(X_i)$ (the intersection of the normalizers), By Sylow theorem : $N_G(X_i)=6$ for all $i$. then $|\ker(\varphi)|\in\{1,2,3,6\}$ i want to show that $|\ker(\varphi)|=1$. By Sylow theorem : for all $i$ $N_G(X_i)$ has a single group of order 3 which is $X_i$ then

if $|\ker(\varphi)|=3$ then $X_1=X_2=X_3=X_4$ contradiction $(n_3=4) $

if $|\ker(\varphi)|=6$  then $X_1=X_2=X_3=X_4.$

Problem if $X$ and $Y$ two $3-$ Sylow  such that $X=Y$ then $N_G(X)=N_G(Y)$

答:
Since normal Sylow subgroups are unique, if $X$ and $Y$ are $3$-Sylows and $X\ne Y$, then $N(X) \ne N(Y)$ and $N(X) \cap N(Y)$ is a $3'$-group, and so has order $1$ or $2$ as each normalizer has order $6$.  If their intersection has order $1$ then the action of $G$ on its four $3$-Sylow subgroups is faithful and so since $|G|=24$, $G$ must be isomorphic to $\frak{S}_4$.  Now, suppose $|N(X)\cap N(Y)|=2$.  The kernel of the conjigation action must be contained in it, so we still have $G$ acting faithfully on its four $3$-Sylow subgroups, and thus isomorphic to $\frak{S}_4$, unless  $N(X)\cap N(Y)$ is a normal subgroup (call it $K$).  But if that were the case, then $G/K$ would have order $12$ and would have four $3$-Sylows since the $3$-Sylows would have distinct images in $G/K$.  ($K$ is central of order $2$, so we can't have $x\in yK$ for $x \ne y$ elements of order $3$.)  But that leaves room only for a unique $2$-Sylow in $G/K$, and its inverse image is clearly a unique $2$-Sylow in $G$, contradicting that there are three $2$-Sylows in $G$.

Note that this also shows that if a group of order $24$ with multiple $3$-Sylows is not $\frak{S}_4$, then it must be either $2\times \frak{A}_4$ or a double cover of $\frak{A}_4$, i.e., it must be $SL(2,3)$ or $2\times \frak{A}_4$.

Thus, except for abelian groups, $\frak{S}_4$, $2\times\frak{A}_4$, and $SL(2,3)$, all groups of order 24 are realized as semidirect products $3\rtimes H$, with $H$ of order $8$ acting nontrivially on the group of order $3$.


问:
What should the bounds of integration be for calculating the marginal distribution function from a joint distribution function?

I calculated $c=\dfrac{1}{8}$. To find $f_y(y)$, I integrated $f(x,y)$ from $-y$ to $y$. What should the bounds be to find $f_x(x)$?

I'm confused because x depends on y. Should the bounds just be $x$ to $\infty$?

答:
A figure indicating the range of $(X, Y)$ often helps to determine the integral bounds and solve problems like this.

In this problem, the range of $(X, Y)$ is given by
$$-y \leq x \leq y, \ 0 < y < \infty,$$
which corresponds to the highlighted area (in light blue).

As such, the marginal density of $Y$ is given by
$$f_Y(y) = \int_{-y}^y f(x,y) dx = \int_{-y}^y c(y^2 - x^2)e^{-y} dx;$$
and the marginal density of $X$ is given by
$$f_X(x) = \begin{cases}\int_x^\infty f(x,y) dy = \int_x^\infty c(y^2 - x^2)e^{-y} dy & \text{for } x \geq 0;\\ \int_{-x}^\infty f(x,y) dy = \int_{-x}^\infty c(y^2 - x^2)e^{-y} dy & \text{for } x < 0.\end{cases}$$
Remark 1: The integral bounds to get $f_X(x)$ depend on the value of $x$ (being positive or negative), which can be illustrated straightforwardly from the figure.

Remark 2: One may notice that the resulting $f_X(x)$ could also be summarized into a "one-liner" formula, given by
$$f_X(x) = \int_{\left|x\right|}^\infty f(x,y) dy = \int_{\left|x\right|}^\infty c(y^2 - x^2)e^{-y} dy \quad \text{for } x \in \mathbb{R}.$$
These integral bounds are implied by the fact that
$$-y \leq x \leq y \Longleftrightarrow y \geq x \text{ and } y \geq -x \Longleftrightarrow y \geq \left|x\right|.$$
The resulting $f_X(x)$ would then be a function of $\left|x\right|$.


问:
Can every even integer greater than four be written as a sum of two twin primes?
Thinking of Goldbach conjecture I arrived at this

$\mathrm{Conjecture}$: Every even integer greater than four can be written as a sum of two twin primes.

What do you think?

I hope this is true.
I tried to verify this up to some extent.

答: I have found by the conjecture of goldbach that there are two unique prime numbers whose sum results in an even number . example- 64=53+11</p>


问:
If $f$ is a continuous function and $q \neq 0$ such that $f(f(x)) = p f(x) + q x$ $\forall$ $x \in \mathbb{R}$, what can we say about $f$?
If $f(f(x)) = p f(x) + q x$ for all $x \in \mathbb{R}$ such that $p \in \mathbb{R}, q \in \mathbb{R}\backslash\{0\}$ and $f$ is continuous, what can we say about $f$?

We can clearly say that $f$ is unbounded since the relation $\sup \{f\} \geq p \cdot \sup\{f\} + q\cdot\sup\{x\}$ will be inconsistent if $f$ is bounded.

Now that we know $f$ is unbounded, we need to find out if $f$ is monotonous or not. I need help with this part.

答:
we need to find out if $f$
is monotonous or not.

If $f(a)=f(b)$ then $f(f(a)) = f(f(b))$ so
$$ 0 = f(f(b)) - f(f(a)) =p\cdot (f(b)-f(a)) + q\cdot (b-a) = q(b-a)$$
since $q\neq 0$, $a=b$. So $f$ is injective. An injective continuous function is monotone.


问:
INMO 2021: A convex quadrilateral with lots of nonstandard angles
From INMO 2021 (held 7 March):

In a convex quadrilateral ABCD, $\angle ABD = 30^{\circ}$, $\angle BCA = 75^{\circ}$, $\angle ACD = 25^{\circ}$ and $CD = CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE = BD$.

Lots of angle information makes it suitable for a trigonometric attack. And I know a solution using trigonometry.

Any other approaches? Especially pure geometry proofs will be surprising to me here.

答:
Select a point $O$ on $AB$ such that $BC=OC$ (it also follows that $OC=CD$).

Via angle chase, observe that $\bigtriangleup OCD$ is equilateral.

Via angle chase, observe that $\bigtriangleup AOC$ is isosceles, hence O is the circumcenter of $\bigtriangleup ADC$.

Finally, note that $\triangle EOC  \cong \triangle BCD$, hence $EC=BD$.

As pointed out by @cosmo5 , $O$ must lie between $A$ and $B$ due to angles given in the problem.


问:
How to show $P−λI$ is invertible for every $λ≠0,1$
I'm trying to show that for the projection $P$, $P−λI$  is invertible for every $λ≠0,1$.

The eigenvalues of $P$ are $0$ and $1$ and to determine the spectrum of $P$, I am trying to show that $P−λI$  is invertible, for $λ≠0,1$. Can someone help me with this? I am not sure how to show this invertibility. Thanks!

答:
It is easy to verify that for all $\lambda\ne 0,1$, one has
$$
  (\lambda I-P)\left(\frac{1}{\lambda}(I-P)+\frac{1}{(\lambda-1)}P\right)
   \\ = (\lambda(I-P)+(\lambda-1)P)\left(\frac{1}{\lambda}(I-P)+\frac{1}{\lambda-1}P\right) \\
      = (I-P)+P=I.
$$
So $\lambda I-P$ is invertible for $\lambda\ne 0,1$, with
$$
         (\lambda I-P)^{-1}=\frac{1}{\lambda}(I-P)+\frac{1}{\lambda-1}P.
$$


问:
Plotting the graph of $x(t)$ when the graph of $x'(t)=f(x)$ is given
I am not sure how to present this problem clearly. Please feel free to suggest edits to this problem so that this will be useful to someone who will look at this in the future.

Here's my question:

Suppose $x'(t)=f(x)$ and suppose the graph of $f(x)$ is given. How can we draw the graph of $x(t)$ using the graph of $f(x).$

What I have tried so far: I have found equilibrium points and their stabilities using the graph of $f(x).$ But I don't know how this can be used when plotting $x(t).$

Thanks in advance!

答:
You cut the graph in as small slices as is reasonable for you to follow. Then instead of a graph you will have a series of bars. However, definite integrals, and this is the only one you can draw, differ by a constant from each other.

Now

$$\int_{a}^{b} f(x) dx = F(b)-F(a)$$

So if $F(a)=0$ then you have

$$\int_{a,F(a)=0}^{b} f(x) dx = F(b)$$

so you can ignore this at first as after you draw an integral you can just shift the axis appropriately to the location you want to have.

So, in order to draw a graph you just stack bars (respecting the direction of each bar) and moving as you go along to the right, bar by bar. It is very easy to do this using compass and triangle albeit tedious.

Notice that your choice of step size will affect the final scale of y-axis.


问:
A consequence of Van der Corput Lemma
Let $\phi (x) = x^2$, $k,\lambda\in \mathbb{R}$, and $[a,b]\subset\mathbb{R}$ and define the integral $$I(\lambda;k)=\int_{a}^{b}e^{i(\lambda\phi(x)-kx)}dx$$
I am trying to prove that $|I(\lambda;k)|\leq c_\phi|\lambda|^{-\frac{1}{2}}$ such that $c_{\phi}$ is a constant depending only on $\phi$.

Van der Corput Lemma states that for $\phi:[a,b]\rightarrow \mathbb{R}$, a smooth function such that $|\phi^{(n)}(x)|\geq 1$ $\forall x \in (a,b)$ and $\psi$ a complex valued differentiable function we have,
$$\Bigg|\int_{a}^{b}e^{i\lambda \phi(x)} \psi(x)dx \Bigg| \leq c_n|\lambda|^{-\frac{1}{n}}\Big[|\psi(b)|+\int_{a}^{b}|\psi'(x)|dx\Big]$$ where $c_n$ is an absolute constant.

I have tried to use this by taking $n=2$ and $\psi(x)=e^{-kx}$ but on the RHS I am left with $c_2|\lambda|^{-\frac{1}{2}}(1+|k|(b-a))$ so the bound is clearly not dependent only on $\phi$.

This is actually a special case of Lemma 2.7 (page 42) in https://www.jstor.org/stable/24896258?seq=1#metadata_info_tab_contents

Any help will be very much appreciated.

答: I would complete the square, $$I(\lambda;k)= \int_a^b e^{i[\lambda(x-\frac{k}{2\lambda})^2-\frac{k^2}{4\lambda}]}dx = e^{-i\frac{k^2}{4\lambda}}\int_{a-\frac{k}{2\lambda}}^{b-\frac{k}{2\lambda}}e^{i\lambda y^2}dy.$$ Then $$|I(\lambda;k)| \leq \sup_{\alpha<\beta}\bigg\vert\int_\alpha^\beta e^{i\lambda y^2}dy\bigg\vert,$$ and the supremum can be bounded using Van der Corput's lemma.

Title: A consequence of Van der Corput Lemma


问:
Closed sets in $~\mathbb R^3~$, topology
A singleton set is closed in $~\mathbb R~$ and $~\mathbb R^2~$, but it is not closed in $~\mathbb R^3~$. Why? We can say a singleton set is not closed in $~\mathbb R^3~$ as its complement is not open. But how do we prove this?
Also I can say since in $~\mathbb R~$ complement of every singleton set is open thus the singleton set is closed and same goes for  $~\mathbb R^2~$. (I hope I am correct). As in space $~\mathbb R~,~\mathbb R$ \ $\{x\}~$ can give union of two open sets.

答: For any positive integer $n$, in standard topology of $\mathbb{R}^n$, i.e. one induced by the Euclidean metric, a singleton set $\{x\}$ is the intersection of all closed sets containing $x$. Therefore, the singleton sets are closed in $\mathbb{R}^n$.


问:
Exponentials are not constant multiples of each other?
For logarithms, there is the change of base formula $\log_a x = \frac{\log_b x}{\log_b a}$. So, all logarithms are constant multiples of each other, because the logarithm in base $a$ can be expressed as $\frac{1}{\log_b a}$ times the logarithm in base $b$.

Now, it seems the corresponding result isn’t true for exponential functions? Is there an intuitive explanation for why not? Intuitively, one might reason “the exponential is just the inverse of the logarithm, and so if logarithms are within multiples of each other, so are exponentials”. What is wrong with this reasoning?

We can ask the general question: If a family of invertible functions are within, say, constant multiples of each other (or some other relation), can we say anything about a corresponding relationship within the family of inverses?

My attempt:

I claim that exponentials are not within constant multiples of each other. Indeed, suppose there are two bases $a, b$ and a constant $k$ such that $a^x = kb^x$ for all $x$. Solving for $k$ gives $\displaystyle k = \frac{a^x}{b^x} = \frac{e^{x \ln a}}{e^{x \ln b}} = e^{x (\ln a - \ln b)}$. But this is not a constant function.

Now, this derivation shows we can multiply the variable $x$ by the constant $\ln a - \ln b$, so in some sense there is still “multiplication by a constant”, but I don’t really understand why we multiply the argument $x$ by a constant, whereas for logarithms it is just a simple universal constant.

答:
Suppose you have two functions $f(x)$ and $g(x)$ which are both invertible, and they differ by multiplication by a constant:

$$ g(x) = k \cdot f(x) \tag{for all $x$} $$

Then if the point $(x,y)$ is on the graph of $f$, then the point $(x,ky)$ is on the graph of $g$.

If $f$ is a function and $(x,y)$ is on its graph, then $(y,x)$ is on the graph of $f^{-1}$. So in the example above, $(y,x)$ is on the graph of $f^{-1}$ and $(ky,x)$ is on the graph of $g^{-1}$.

So comparing the graphs of $f^{-1}$ and $g^{-1}$, points with the same $y$-coordinate have their $x$-coordinates differ by a constant multiple. This is what you are seeing the the arguments of the exponential functions differing by a constant multiple (rather than the functions themselves).

Intuitively, passing from a function to its inverse switches the roles of the $x$ and $y$ coordinates on the graph, so the roles of inputs and outputs of the function are changed. For logarithms, the outputs differ by a constant multiple, so for exponentials, the inputs differ by constant multiple.


问:
$5 \mid n^2 - m^2$ is an equivalence relation
How can I show this is an equivalence relation:

$$
n \operatorname{R} m \Longleftrightarrow n^2 - m^2 \textrm{ is divisible by } 5
$$

I know equivalence relations are symmetric, reflexive and transitive.  I'm just not sure how to use this knowledge to prove it.

答:
We need to show that for all integers $x, y, z$, the relation $$R = \{(m, n): m^2 - n^2 \;\;\text{is divisible by 5}\}$$ is reflexive, symmetric, and transitive.

We will use the notation $\;5 \mid m^2 - n^2$ to denote "$m^2 - n^2$ is divisible by $5$."

Let $x, y, z \in\mathbb{Z}$

Reflexive:
We need to show that for any $x \in \mathbb Z$, $x\,R\, x$. Now $x^2-x^2=0$. Since $0$ is divisible by every integer, it is also divisible by 5, so the relation is reflexive.

Symmetric
We need to show that for any $x, y \in \mathbb Z$, if $x\,R\,y$ then $y\,R\,x$. So, suppose $x\,R\,y$. This would mean that $\,5 \mid x^2 - y^2$. Then $\,5 \mid -(x^2-y^2)=y^2-x^2$. So $\,5\mid y^2 - x^2\,$, hence $y\,R\,x$ and the relation $R$ is therefore symmetric.

Transitive

We need to show that if $x\,R\,y\,$ and $\,y\,R\,z\,$, then $x\,R\,z.\,$ So: suppose $x\,R\,y$ and $y\,R\,z$. It follows that $\,5\mid (x^2 - y^2)$ and $\,5\mid y^2 - z^2$. But then $$5 \mid \left[(x^2 - y^2) + (y^2 - z^2)\right] = x^2 - y^2 + y^2 - z^2 = x^2 - z^2.$$ So $5 \mid x^2 - z^2$, which means $x\,R\,y$, and the relation is therefore transitive.

Since our choice of $x, y, z \in \mathbb Z$ was arbitrary, it holds that $R$ is an equivalence relation on the set of integers, since it is reflexive, symmetric, and transitive.


问:
Suppose $X$ be the random variable that denotes the number of heads minus the number of tails when $10$ unbiased coins are tossed. Find the variance.
Suppose an unbiased coin is tossed $10$ times. Let $X$ be the random variable that denotes the number of heads minus the number of tails. What is the variance of $X$?

This seems to have the solution but I do not understand what it says. I am looking for an explanation of the link I shared.

I am still sharing my solution so as to show I am not posting it without trying it at all.

$\text{E}[X] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))\right) = \frac{1}{2^{9}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5) = 0$

$\text{E}[X^2] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))^2\right) = \frac{1}{2^{8}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5)^2 = 10$

This gives $\text{Var}[X] = \text{E}[X^2] - \text{E}[X]^2 = 10 - 0^2 = 10$. This is how I did, considering the PMFs.

答:
Let $Y$ be the random variable that denotes the number of heads out of the 10 coin tosses. Since the coin is unbiased and different tosses are independent of each other, we can model $Y$ as
$$Y \sim Bin(n, p),$$
where $n = 10$ and $p = 0.5$.

Since $X$ is the random variable that denotes the number of heads minus the number of tails, we have that
$$X = \underbrace{Y}_{\text{number of heads}} - \underbrace{(n - Y)}_{\text{number of tails}} = 2Y - n.$$
To get the variance of $X$, first we would like to recap two basic properties of variance: if $a$ is a constant, then

(1) $\text{Var}(X + a) = \text{Var}(X)$;

(2) $\text{Var}(aX) = a^2\text{Var}(X)$.

Back to this problem, we have
\begin{align*}
\text{Var}(X) &= \text{Var}(2Y - n)\\
&= \text{Var}(2Y) \quad & \text{by property (1)}\\
&= 2^2\text{Var}(Y) \quad & \text{by property (2)}\\
&= 4\text{Var}(Y)\\
&= 4np(1-p),
\end{align*}
where the last step follows from the variance of Binomial distribution. Plugging in $n = 10$ and $p = 0.5$, we have
$$\text{Var}(X) = 4np(1-p) = 4 \times 10 \times 0.5 \times (1-0.5) = 10.$$


Title: Suppose $X$ be the random variable that denotes the number of heads minus the number of tails when $10$ unbiased coins are tossed. Find the variance.

Body:
Suppose an unbiased coin is tossed $10$ times. Let $X$ be the random variable that denotes the number of heads minus the number of tails. What is the variance of $X$?

This seems to have the solution but I do not understand what it says. I am looking for an explanation of the link I shared.

I am still sharing my solution so as to show I am not posting it without trying it at all.

$\text{E}[X] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))\right) = \frac{1}{2^{9}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5) = 0$

$\text{E}[X^2] = \displaystyle\sum_{h=0}^{10} \left(\frac{{10 \choose h}}{2^{10}} (h - (10-h))^2\right) = \frac{1}{2^{8}} \displaystyle\sum_{h=0}^{10} {10 \choose h} (h-5)^2 = 10$

This gives $\text{Var}[X] = \text{E}[X^2] - \text{E}[X]^2 = 10 - 0^2 = 10$. This is how I did, considering the PMFs.

Body:
Argument: $n = H+T,$ implies $T = n - H,$ so $X = H- T = H - (n-H) = 2H - n.$ Also, $E(X) = E(H-T) = E(H) - E(T) = n/2 - n/2 = 0.$

Then $Var(X) = Var(2H-n) = 4Var(H) = 4(n/4) = n = 10.$

Simulation with 10 million 10-toss sessions.

set.seed(2021)
n = 10;  p = 1/2
h = rbinom(10^7, n, p)
t = n-h;  x = h-t;
mean(x);  var(x)
[1] 0.0010386    # aprx E(X) = 0
[1] 10.00229     # aprx Var(X) = 10

summary(x)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max.
 -10.000000  -2.000000   0.000000   0.001039   2.000000  10.000000

hist(x, prob=T, br = (-11:10)+.5, col="skyblue2",
     main="Simulated Dist'n of H - T")


问:
On positive self adjoint unbounded operator
This is a theorem from Rudin Functional Analysis.

$T$ is a self adjoint operator in $H$ (a Hilbert space).
$T$ is not necessarily bounded and $\mathscr D(T)$ denotes the domain of $T$.
We are to show that -

If $\langle Tx,x \rangle \ge 0$ for all $x\in \mathscr D(T)$ then
$\sigma (T)\subset [0,\infty )$.

Infact the converse is also true. It is written in book that the proof works similar to the one for bounded case.
I tried it the following way-
Let $\lambda\ge 0$ observe,
$$0\le\lambda\|x\|^2= \langle \lambda x,x \rangle \le \langle Tx+\lambda x,x\rangle\le \|Tx+\lambda x\|\cdot \|x\|$$
$\implies\lambda\|x\|\le \|(T+\lambda I) x\|$ for all $x\in \mathscr D(T)$.

Therefore $(T+\lambda I)$ is one-one on $\mathscr D(T)$.

I want to show that $T+\lambda I$ has bounded inverse, for that I must have range of $(T+\lambda I)$ is all of $H$. How do I show that ? Please help.

答:
Since $T$ is self-adjoint its spectrunm is contained in $\mathbb R$. We only have to show that if $\lambda >0$ then $T+\lambda I$ is invertible. We have $ \langle Tx+\lambda x,x\rangle \geq \lambda \|x\|^{2}$. This gives $\|Tx+\lambda x\| \geq \lambda \|x\|$. From this you conclude that $T+\lambda I$ is one-to-one and its range is closed. [See details below]. To show that the range is dense take any $y$ orthogonal  to the range. You get $ \langle Tx+\lambda x, y \rangle=0$ for all $x$. In particular, $ \langle Ty+\lambda y, y \rangle=0$. But LHS is a sum of two non-negative numbers. so  each of them must be $0$. In particular $\|y\|^2=0$, so $y=0$.

To show  that the range of $T+\lambda I$ is closed let $Tx_n+\lambda x_n \to  y$. Then the inequality proved above shows that $\|x_n-x_m\| \to 0$  as $n, m \to \infty$. Hence, $x_n \to x$ for some $x \in H$. Now $Tx_n\to y-\lambda x$. The sequence $(x_,Tx_n)$ in the graph of $T$ converges to $(x,y-\lambda x)$. But the graph of $T$ is closed [Theorem 13.9 in Rudin's book]. It follows that $x \in D(T)$ and $y-\lambda x=Tx$. Thus $y=(T+\lambda I)x$.


答:
A singleton is closed in $\Bbb{R}^n$ because, if $p \in \Bbb{R}^n$, then
$$\Bbb{R}^n \setminus \{p\} = \bigcup_{x \in \Bbb{R}^n \setminus \{p\}} B(x; \|x - p\|),$$
where $B(x; \|x - p\|)$ is the open ball, centred at $x$, with radius $\|x - p\|$.

To see why this is true, any $x \in \Bbb{R}^n \setminus \{p\}$ is contained in the right hand side, because it is the centre of one of the balls in the union. On the other hand, $p$ is not in the right hand side, because it doesn't lie in any of the open balls in the union (it lies in their boundary spheres, but since the balls are open, it doesn't lie in the balls themselves).

This means that $\Bbb{R}^n \setminus \{p\}$ is the union of a family of open sets, which implies it is open, and its complement $\{p\}$ is closed.

Title: Closed sets in $~\mathbb R^3~$, topology

Body:
A singleton set is closed in $~\mathbb R~$ and $~\mathbb R^2~$, but it is not closed in $~\mathbb R^3~$. Why? We can say a singleton set is not closed in $~\mathbb R^3~$ as its complement is not open. But how do we prove this?
Also I can say since in $~\mathbb R~$ complement of every singleton set is open thus the singleton set is closed and same goes for  $~\mathbb R^2~$. (I hope I am correct). As in space $~\mathbb R~,~\mathbb R$ \ $\{x\}~$ can give union of two open sets.


问:
Relationship between two elements from two dual bases
I am trying to figure out the relationship between two elements from two dual bases of a vector space, i.e. say ${e_1,...,e_n}$ and ${f_1,...,f_n}$ are 2 bases of $V$ with ${w^1,...,w^n}$ and ${y^1,...,y^n}$ are their dual bases. Write $f_i = \sum_j A_{ij}e_j$. I want to express $y^i$ in terms of the $w^j$.

My attempt. Write $y^m = \sum_k B^{mk} w^k$.
Then
$$
\delta_i^m = y^m(f_i) = \sum_k B^{mk} w^k (\sum_j A_{ij} e_j) = \sum_k B^{mk} A_{ik}
$$
which implies that $B = (A^{-1})^T$. So I guess I've reduced the problem down to expressing the entries of the transpose inverse of a matrix in terms of its entries, but I found this to be quite difficult to do. Can someone give a hint or offer a better way of approaching this? Thanks in advance!

答:
It seems like you're looking for a closed-form expression for the entries of $B = (A^{-1})^T$ in terms of the entries of $A$. The most common way to accomplish this is to use the adjugate formula for the inverse of a matrix,
$$
A^{-1} = \frac 1{\det(A)}\cdot \operatorname{adj(A)}.
$$
Taking the transpose of both sides (noting that $(A^T)^{-1} = (A^{-1})^T$) yields
$$
B = A^{-T} = \frac 1{\det(A)} \operatorname{cof}(A)
$$
where $\operatorname{cof}(A)$ denotes the cofactor matrix of $A$ (see the linked article). To be a bit more explicit, we could write
$$
B_{ij} = \frac{\det(A^{(ij)})}{\det(A)},
$$
where $A^{(ij)}$ denotes the matrix $A$ with its $i$th row and $j$th columns removed.

Note: the adjugate formula is rarely used as a basis for computation due to the prohibitive computational cost of determinants of large matrices. In practice, one would compute $A^{-1}$ using a standard method such as the Gauss-Jordan algorithm.


问:
Inequalities Problem : Cancelling out the variables
$\frac{(x+4)}{(x^2-9)} - \frac{2}{(x+3)} < \frac{4x}{(3x-x^2)}$

Below is my solution apprach for this solution :-

$\frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4x}{3x-x^2}
\\ \Rightarrow \frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4}{3-x} \\ \Rightarrow \frac{x+4}{(x+3)(x-3)} - \frac{2}{x+3} < \frac{-4}{x-3} \\ \Rightarrow \frac{x+4}{x+3} - \frac{2(x-3)}{x+3} < -4 \\ \Rightarrow \frac{10-x}{x+3} <-4 \\ \Rightarrow x<-\frac{22}{3}
$

But this is not the correct answer. Can someone please tell me what am I missing here?

Answer : $x \in (-\infty,-7.33) \cup (-3,3)-\{0\}$

答:
When you multiply by $x+3$, or $x-3$, it matters whether that is positive or negative.  If it is negative, you have to change greater-than to less-than.

Break it into three separate cases: $x\lt-3,-3\lt x\lt3,3\lt x$.


答:
When you are to find all possible values of $x$ meeting the inequality, you should rather expand as is.

$\displaystyle \small \frac{x+4}{x^2-9} - \frac{2}{x+3} \lt \frac{4x}{3x-x^2}$ simplifies to,

$\displaystyle \small \frac{x+4-2x+6}{x^2-9} \lt \frac{4x}{3x-x^2}$

$\displaystyle  \small \mathbf {\frac{10-x}{x^2-9} \lt \frac{4}{3-x}} \ $ when $\mathbf {x \ne 0}$

i) When $x \gt 10$, both LHS and RHS are negative. So the inequality becomes

$\displaystyle \small \frac{x-10}{x^2-9} \gt \frac{4}{x-3} \implies 3x \lt -22 $ which is contradictory.

ii) When $3 \lt x \lt 10$, RHS is negative and LHS is positive and hence there are no solutions.

iii) When $-3 \lt x \lt 3$, RHS is positive and LHS is negative so inequality holds.

iv) when $x \lt -3$, both LHS and RHS are positive.

$\displaystyle \small \frac{10-x}{(-x-3)} \lt 4 \implies x \lt -\frac{22}{3} \ \big[ $ note $(-x-3)$ is positive as $\displaystyle \small x \lt -3 \big] $.

Title: Inequalities Problem : Cancelling out the variables

Body:
$\frac{(x+4)}{(x^2-9)} - \frac{2}{(x+3)} < \frac{4x}{(3x-x^2)}$

Below is my solution apprach for this solution :-

$\frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4x}{3x-x^2}
\\ \Rightarrow \frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4}{3-x} \\ \Rightarrow \frac{x+4}{(x+3)(x-3)} - \frac{2}{x+3} < \frac{-4}{x-3} \\ \Rightarrow \frac{x+4}{x+3} - \frac{2(x-3)}{x+3} < -4 \\ \Rightarrow \frac{10-x}{x+3} <-4 \\ \Rightarrow x<-\frac{22}{3}
$

But this is not the correct answer. Can someone please tell me what am I missing here?

Answer : $x \in (-\infty,-7.33) \cup (-3,3)-\{0\}$


问:
Proofs Involving Disjunctions: Sets
I want to focus on Exercise 4 of Section 3.5 of Velleman's book. I am aware of the other questions that have been asked, so this one is not a duplicate. I would like to discuss a little more considering the answer that was given in:
For all sets $A$, $B$, and $C$, if $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$, then $A \subseteq B$.

Suppose $A\cap C \subseteq B\cap C$ and $A \cup C \subseteq B\cup C$. Prove that $A \subseteq B$.

So the assumptions are clear: $x \in A$, $A\cap B \subseteq B\cap C$, and $A \cup C \subseteq B\cup C$. My question is on the following reasoning: Since $x\in A$ and $A \cup C \subseteq B\cup C$, we conclude that $x\in B\cup C$, so our cases to analyze would be $x\in B$ and $x\in C$, instead of $x\in C$ and $x\notin C$ as suggested in the question I quoted earlier. Is my reasoning correct? Ohterwise, I would like to know if there is some special reason why the $x\in C$ and $x\notin C$ cases could be the preferred path to follow.

Thank you for your feedback.

答:
You can start by observing that if $x\in A$, then $x\in A\cup C\subseteq B\cup C$, but that does not force you to split the rest of the argument into cases $x\in B$ and $x\in C$: there is a perfectly good continuation that starts by splitting it into cases $x\in C$ and $x\notin C$. That argument goes as follows:

If $x\notin C$, then clearly $x\in B$, and we’re done. If $x\in C$, then $x\in A\cap C$, so $x\in B\cap C$, so $x\in B$.

Splitting it your way leads to essentially the same argument:

If $x\in B$, we’re done. If $x\in C$, then $x\in A\cap C$, so $x\in B\cap C$, so $x\in B$.

It’s pretty much six of one and half a dozen of the other.

Title: Proofs Involving Disjunctions: Sets


问:
Facts or Theorems Similar to Pigeonhole Principle
Are there any facts or theorems in mathematics having similar energy with pigeonhole principle? What I mean by similar energy is the statement is simple, "trivial" as you don't need to prove it to understand why it's true, but has many useful and powerful applications to prove many results. Thank you.</p>

答:
There is no integer (strictly) between zero and one. Looks pretty stupid, but many irrationality and transcendence proofs rely on it. [and in some sense it is equivalent to the pigeonhole principle]

Example: the proof of the transcendence of $e$ given in http://www.math.leidenuniv.nl/~evertse/dio15-4.pdf – note particularly the statement after Corollary 4.3, "Our aim is to show that for a suitable choice of $f$, the quantity $M := q_0F(0) +\cdots+ q_nF(n)$ is a non-zero integer with $|M| < 1$."

The proof of Lindemann-Weierstrass later in that same document is much more elaborate but also comes down to the non-existence of an integer $M$ with $0<M<1$.

This non-existence crops up again in the middle of the proof of the Gel'fond-Schneider Theorem, in the middle of page 64 of the same document.


问:
Can anyone tell me if my proof is right?
Let $0<a<b,$ compute $\lim_{n\to \infty}{\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n}}$

My proof are followings :

Assuming that $r=b-a>0,$ so $b=a+r.$ So the original formula equals to $\frac{a\cdot a^{n}{+}{(a+r)\cdot}b^{n}}{a^n+b^n}=a+\frac{r\cdot b^n}{a^n+b^n}=a+\frac{r}{1+(a/b)^n},$ so it can limit to $a+r=b.$

From my point of view, I think that the results should contain $a,$ however my result contains no $a.$ I am a self learner, I don't know wheather my result is right. Hope that someone can help me out.

答:
What you made is $\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n}=a+\frac{b-a}{1+(a/b)^n}$ and using $\left( \frac{a}{b}\right)^n\to 0$, when $n\to\infty$, you obtain answer $b,$ so everything is correct.

Another possible way is
$$\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n} = b\frac{1+\left( \frac{a}{b}\right)^{n+1}}{1+\left( \frac{a}{b}\right)^{n}}\to b,\text{ when }n\to\infty.$$


答:
Your proof is correct. Here's another way to see why the limit is $b$:

Notice that

\begin{align*}
\frac{a^{n+1}+b^{n+1}}{a^n+b^n} &= a^n\cdot\frac{a}{a^n+b^n}+b^n\cdot\frac{b}{a^n+b^n}\\
&= \frac{1}{\frac{1}{a^n}}\cdot\frac{a}{a^n+b^n}+\frac{1}{\frac{1}{b^n}}\cdot\frac{b}{a^n+b^n}\\
&= \frac{a}{1+\left(\frac{b}{a}\right)^n}+ \frac{b}{1+\left(\frac{a}{b}\right)^n}
\end{align*}

Since $b>a$, it's easy to see that $\left(\frac{b}{a}\right)^n\to\infty$ and $\left(\frac{a}{b}\right)^n\to 0$ as $n\to\infty$, so

$$\lim\limits_{n\to\infty}\left(\frac{a}{1+\left(\frac{b}{a}\right)^n}+ \frac{b}{1+\left(\frac{a}{b}\right)^n}\right)=0+\frac{b}{1}=b$$

Title: Can anyone tell me if my proof is right?

Body:
Let $0<a<b,$ compute $\lim_{n\to \infty}{\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n}}$

My proof are followings :

Assuming that $r=b-a>0,$ so $b=a+r.$ So the original formula equals to $\frac{a\cdot a^{n}{+}{(a+r)\cdot}b^{n}}{a^n+b^n}=a+\frac{r\cdot b^n}{a^n+b^n}=a+\frac{r}{1+(a/b)^n},$ so it can limit to $a+r=b.$

From my point of view, I think that the results should contain $a,$ however my result contains no $a.$ I am a self learner, I don't know wheather my result is right. Hope that someone can help me out.


问:
Reflexivity, James' theorem and a strengthening of a theorem of Riesz
Let us consider only Banach spaces.

A classical theorem of Riesz says that for every proper closed subspace $Y$ of a Banach space $X$ there is a sequence of vectors $x_n\in X$ with $\Vert x_n\Vert=1$ for all $n$ and such that $\Vert x_n+Y\Vert\to 1$, where this latter norm is the usual norm in the quotient space $X/\!Y$: $$\Vert x+Y\Vert=\inf_{y\in Y}\Vert x+y\Vert.$$

I will call the validity of the natural strengthening of this theorem the strong Riesz property (SRP):

A Banach space has the SRP iff for every proper closed subspace $Y$ of $X$, there is $x\in X$ with $\Vert x\Vert=\Vert x+Y\Vert=1$.

Two applications of Hahn-Banach (on $X$ and on the dual $X^*$) imply that every reflexive space has the SRP.

A closely related property is given by James' theorem, which I will also expres as a property:

A Banach space $X$ has James' property (JP) iff for every continuous linear functional $f\in X^*$ attains its norm at some unit vector; i.e., there exists $x\in X$ with $\Vert x\Vert=1$ such that $|f(x)|=1$.

James' theorem states that JP and reflexivity are equivalent.

I would like to know whether reflexivity and the SRP are equivalent. It seems that passing through JP would be a natural way to prove it.

The implications (reflexivity)$\Rightarrow$(JP)$\Rightarrow$(SRP) are easy. For example, the last one: Let $Y$ be a proper closed subspace of a Banach space with JP. By Hahn-Banach, take $f\in X^*$ with $f(Y)=0$ and $\Vert f\Vert=1$. By JP, take $x\in X$ with $\Vert x\Vert=|f(x)|=1$. Then
$$\Vert x+Y\Vert\leq\Vert x\Vert=1=|f(x)|=\inf_y|f(x+y)|\leq\inf_y\Vert x+y\Vert=\Vert x+Y\Vert,$$
which yields SRP.

The "obvious way" to prove the converse would be: Assume $f\in X^*$ has norm $1$. By SRP, take $x\in X$ with $\Vert x\Vert=\Vert x+\ker(f)\Vert=1$. Then proceed to prove that $|f(x)|=1$.

However, this very last step does not seem to be valid, or at least not in any easy manner. Everything that I could obtain is that $|f(x)|\leq 1$.

tl;dr Does SRP as above imply reflexivity?

答:
The "obvious way" to prove the converse would be: Assume $f\in X^*$ is
nonzero. By SRP, take $x\in X$ with $\Vert x\Vert=\Vert
x+\ker(f)\Vert=1$. Then proceed to prove that $|f(x)|=1$.

I suppose  the OP forgot to require  that $\|f\|=1$, above.  Assuming this, one has for every $y$ in $X$ that
$$
  k:=f(x)y-f(y)x\in \ker(f).
  $$
Therefore
$$
  |f(y)| =
  |f(y)| \|x+\ker(f)\| =
  \|f(y)x+\ker(f) \| = $$$$ =
  \|f(x)y+\ker(f)\| \leq
  |f(x)|\|y\|.
  $$
So,
$$
  1 = \|f\| = \sup_{\|y\|\leq 1}|f(y)| \leq  |f(x)| \leq \|f\|\|x\| = 1.
  $$


问:
Putnam 1967 problem, integration
I have some problems understanding this solution of this problem:

Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that

$$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$

It says there that if we split this integral into $n$ parts to get

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$

we will have

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$

because for large $n$, $f$ is roughly constant over the range.

Could you tell me why $f$ is roughly constant over the range for $n$ large enough?

I would really appreciate all your insight.

答:
I hesitate to post an answer after an accepted answer is there, but somebody should say this.

$f$ is roughly constant on small intervals because $f$ is uniformly continuous.

The fact that $f$ is continuous means you can make $f$ "roughly contstant" on a small interval about a specified point.  But intervals small enough to make $f$ roughly constant near one point may fail to be small enough to make $f$ roughly constant near other points.  That's where "uniformly" helps: $f$ is continuous on a closed bounded interval $[0,1]$ and that makes it uniformly continuous, so that once you've specified that you want $f$ not to change by more than a certain amount, the intervals that are small enough will be small enough at all points.

Title: Putnam 1967 problem, integration


问:
I know it's true, but why?
$$A ∪ (B' ∩ C' ∩ A) = A$$

Does anyone know of a law that stipulates this?  I can't find anything in my textbook.

答: I don't fully understand the question, but $B'\cap C' \cap A \subset A$ and for any subset $X$ of $A$ ($X \subset A$) it holds that $X\cup A=A$.

Title: I know it's true, but why?


问:
The complement of the double countable union nowhere dense closed sets.
I am working on an exercise of functional analysis, which has been partially discussed here: Condensation of Singularities, Principle of Uniform Boundedness and here: The Principle of Condensation of Singularities.

I have finished most of it, but to conclude, I got stuck in topology. The following describes my question.

Say, I have proved that $Int(F_{k,n})=\varnothing$ for all $k,n\in\mathbb{N}$ and $F_{k,n}$ is closed for all $k,n\in\mathbb{N}$. Consider the set defined by $$H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}\ \ \text{and}\ \ O:=\bigcup_{k=1}^{\infty}H_{k}.$$

The space we are working with is a Banach space $V$, and the set I am considering is $E$ that satisfies $E=V\setminus O$.

I want to use  Baire category theorem to say that $E$ is dense in $V$.

So I start to argue that $O=\bigcup_{k,n=1}^{\infty}F_{k,n}$ is a countable union of closed dense sets, so that $E$ is a countable intersection of open dense sets. Therefore, the Baire category theorem implies that $E$ is dense.

However, I got confused about why $O$ is a countable union of closed dense sets. If we separate the double countable union, we will have $$O=\bigcup_{k=1}^{\infty}\Bigg(\bigcup_{n=1}^{\infty}F_{k,n}\Bigg),$$ but $\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily closed right? And in fact $H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily dense anymore. So is this still safe to say $O$ is a countable union of closed dense sets?

Perhaps I am overthinking this and I believe there must be a way to prove the density of $E$ in $V$. Please let me know if you think I should provide more functional setup of this question. But for now I want to keep it as concrete as possible.

Thank you so much!

答: The collection $\{F_{k,n}:k, n \geq 1\}$ is a countable collection of closed nowhere dense sets and $O$ is its union. (You can always arrange this family in a single sequence $(C_n)$).


答: Your sets $F_{k,n}$ are far from being dense: if they have empty interiors, they are nowhere dense. It’s true that the sets $H_k$ may not be nowhere dense, but they are meagre, and a countable union of meagre sets is meagre. Your set $E$ is therefore co-meagre and hence by the Baire category theorem dense in $V$.

Title: The complement of the double countable union nowhere dense closed sets.

Body:
I am working on an exercise of functional analysis, which has been partially discussed here: Condensation of Singularities, Principle of Uniform Boundedness and here: The Principle of Condensation of Singularities.

I have finished most of it, but to conclude, I got stuck in topology. The following describes my question.

Say, I have proved that $Int(F_{k,n})=\varnothing$ for all $k,n\in\mathbb{N}$ and $F_{k,n}$ is closed for all $k,n\in\mathbb{N}$. Consider the set defined by $$H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}\ \ \text{and}\ \ O:=\bigcup_{k=1}^{\infty}H_{k}.$$

The space we are working with is a Banach space $V$, and the set I am considering is $E$ that satisfies $E=V\setminus O$.

I want to use  Baire category theorem to say that $E$ is dense in $V$.

So I start to argue that $O=\bigcup_{k,n=1}^{\infty}F_{k,n}$ is a countable union of closed dense sets, so that $E$ is a countable intersection of open dense sets. Therefore, the Baire category theorem implies that $E$ is dense.

However, I got confused about why $O$ is a countable union of closed dense sets. If we separate the double countable union, we will have $$O=\bigcup_{k=1}^{\infty}\Bigg(\bigcup_{n=1}^{\infty}F_{k,n}\Bigg),$$ but $\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily closed right? And in fact $H_{k}:=\bigcup_{n=1}^{\infty}F_{k,n}$ is not necessarily dense anymore. So is this still safe to say $O$ is a countable union of closed dense sets?

Perhaps I am overthinking this and I believe there must be a way to prove the density of $E$ in $V$. Please let me know if you think I should provide more functional setup of this question. But for now I want to keep it as concrete as possible.

Thank you so much!


问:
Every open covering of a subset of $\Bbb R^n$ has a countable subcover
I want to show the title. More formally,
Let $X\subset\Bbb R^n$ be a subset and $\{U_\alpha\}_{\alpha\in I}$ be a open covering of $X$. Then there is a subcover $\{U_\alpha\}_{\alpha\in I'\subset I}$ that covers $X$.

My basic plan is that define $S:=\Bbb Q^n$. Then for each $r\in S$, if there is $U_r\in\{U_\alpha\}_{\alpha\in I}$ that contains $r$ then choose only one such $U_r$. Collect them all for each $r$ (if there's such). I want to show this $\{U_r\}$ covers $X$. Any help? or any other idea?

答:
$\mathbb R^n $ is second countable:  the open balls with centers with rational coordinates and rational radii form a basis for the topology.

The property you are referring to is called the Lindelöf property.   Any second countable space is Lindelöf.


问:
What method should I use to solve this integral?
I have no clue how to solve it. Any help is appreciated. Thanks in advance.

$$ \int\dfrac{1}{\sqrt{\cos^3\left(x\right)\sin^5\left(x\right)}}dx$$

答:
Hint:

Express both $\sin x$ and $\cos x$ through $\tan x$ and $\sec x$.
$$\sin x = \frac{\tan x}{\sec x}, \qquad \cos x = \frac{1}{\sec x}$$
After some simplifications, you can set $t = \tan x$.


问:
Motivation for the Mapping Class Group
Question: What is the motivation for studying the mapping class group?  In particular, what types of questions does it attempt to answer and what kind of invariant is it?

Motivation for this Question: Recently I've seen a number of references to things which assume knowledge of the mapping class group.  I've attempted to page through Dan Margalit and Benson Farbs' book on the subject as well as check out the wikipedia page for it, but both sources give the motivation as just "what it is" as opposed to, bluntly, "why should I care about this."

答:
A couple of things spring to mind. One is that if you are interested in $3$-manifolds, then the study of the mapping class group is the most natural thing in the world. This is because any compact connected $3$-manifold has a Heegard decomposition as the union of two handlebodies glued along their boundary. Thus the homeomorphism type of the $3$-manifold is controlled by the way the boundaries of the two handlebodies are identified. I.e. the $3$-manifold depends on the automorphism $\phi\colon \Sigma_g\to\Sigma_g$ of the boundary $\Sigma_g=\partial H_g$ of a genus $g$ handlebody. In fact, the homeomorphism type only depends on the isotopy class of $\phi$. So every mapping class gives rise to a $3$-manifold!

The other motivation that comes to mind is that a mapping class group of an $n$-punctured disk is isomorphic to the pure braid group on $n$ strands, which maybe is more obviously of interest. (I say "a" mapping class group, because you have to specify what happens to the boundary of the punctured disk, giving rise to different versions of the mapping class group, one of which is the pure braid group.)

Title: Motivation for the Mapping Class Group


问:
The total number of prime factors of this expression
The total number of prime factors of this number:
$2^{22}$ x $7^5$ x $11^2$

Should be only 3, which are 2, 7 and 11 itself. However the answer states that there are total $(22+5+2) = 29$ prime factors. What? How?

答:
When you are asked

How many prime factors does the number $n$ have?

this doesn't mean

How many factors / divisors does $n$ have that are prime?

which is what you answered, but rather

If you write $n$ as a product of primes, how many factors do you get?

In this case we get twenty-two 2's, five 7s and two 11s for a total of 29.


Body:
You solved as if you were solving an equation for $x$, but from the looks of the solution, the question would have said " find the range of $x$ "
$$\frac{(x+4)}{(x^2-9)} - \frac{2}{(x+3)} \lt \frac{4x}{(3x-x^2)}$$
$ x \in \{ -\infty \} $

$$\frac{ (x+4)(x+3)-2(x^2-9)}{(x+3)(x^2-9) } \lt \frac{4x}{(3x-x^2)}$$
$$(3x-x^2) \big( (x+4)(x+3)-2(x^2-9) \big) \lt 4x(x+3)(x^2-9) $$
$$x(3-x)(x+4)(x+3) -2x(3-x)(x+3)(x-3)\lt 4x(x+3)^2(x-3)$$
$$-x(x-3)(x+4)(x+3) +2x(x-3)^2(x+3) \lt 4x(x+3)^2(x-3)$$
$$ (x-3)(x+3) \big[ -x(x+4) +2x(x-3) \lt 4x(x+3) \big] $$
$$ (x-3)(x+3)(\dots \dots )$$
$ x \in \{ 3, -3 \}$
$$ -x^2-4x+2x^2-6x \lt 4x^2+12x $$
$$ 0 \lt 3x^2+22x$$
$$ 0 \lt x(3x+22)$$
$ x \in \{ 0, -\frac{22}{3} \}$


问:
Are finitely generated subrings of $\mathbb{H}$ Noetherian?
Let $R$ be a finitely generated subring of the ring of real quaternions $\mathbb{H}$, that is, $R$ is the subring generated by a finite subset of $\mathbb{H}$. I want to show (or find a counterexample) that $R$ is Noetherian.

My attempt is the same as the attempt in this question. However, this attempt is not correct for general rings as pointed out in the answers. I wonder if there is some special property of $\mathbb{H}$ that would make this attempt work.

Any help is much appreciated.

答:
Let $a,b,c,d\in\mathbb{R}$ be algebraically independent over $\mathbb{Q}$, let $x=a+bi$ and let $y=c+dj$.  Then the subring $R\subset\mathbb{H}$ generated by $x$ and $y$ is not Noetherian.

To prove this, consider the left ideal $I\subseteq R$ generated by the elements $xy^n$ for each $n\in\mathbb{N}$.  I claim that for each $n$, $xy^n$ is not in the left ideal generated by the $xy^m$ for $m<n$, and so $I$ cannot be finitely generated.

Note first that $R$ is a subring of the quaternion algebra $S$ over the polynomial ring $\mathbb{Z}[a,b,c,d]$.  We may consider $\mathbb{Z}[a,b,c,d]$ as a bigraded ring where the first degree is respect to $a$ and $b$ and the second degree is with respect to $c$ and $d$.  This then makes $S$ a bigraded ring, with $x$ being homogeneous of bidegree $(1,0)$ and $y$ being homogeneous of bidegree $(0,1)$.  Since $x$ and $y$ are homogeneous, the ring $R$ is bigraded as well.

Now suppose we could write $xy^n$ as an $R$-linear combination of the $xy^m$ for $m<n$.  The only way to take a multiple of $xy^m$ that has the same bidegree as $xy^n$ is to take an integer multiple of $y^{n-m}xy^m$, so this in fact means that we can write $$xy^n=\sum_{m<n}k_my^{n-m}xy^m$$ where each $k_m$ is an integer.  Now consider the homomorphism $S\to\mathbb{H}$ given by mapping $a$ to $0$, $b$ to $1$, $c$ to $1$, and $d$ to $2$ (and $i,j,k$ to themselves).  This turns the equation above into $$i(1+2j)^n=\sum_{m<n}k_m(1+2j)^{n-m}i(1+2j)^m.$$  Multiplying both sides by $-i$ and noting that $-i(u+vj)i=u-vj$, we get $$(1+2j)^n=\sum_{m<n}k_m(1-2j)^{n-m}(1+2j)^m.$$  But this now violates unique factorization in the Gaussian integer ring $\mathbb{Z}[j]$, since the right-hand side is divisible by $1-2j$ and the left-hand side is not.


答:
HINT:

Generalization:

$$\int\dfrac{dx}{\sin^mx\cos^{2n-m}x} =\int\dfrac{\sec^{2n-2}x}{\tan^mx}\sec^2x\ dx$$

Now choose $\tan x=u$

Can you recognize $m,n$  here?

Title: What method should I use to solve this integral?

Body:
I have no clue how to solve it. Any help is appreciated. Thanks in advance.

$$ \int\dfrac{1}{\sqrt{\cos^3\left(x\right)\sin^5\left(x\right)}}dx$$