问:
Regular Conditional Bayesian Experiment
In "Elements of Bayesian Statistics" (1990), Florens, Mouchart and Rolin describe two basic forms of reduction of a Bayesian experiment: Marginalization and Conditioning (Ch. 1). I don't understand the conditioning reduction and would appreciate an explanation, if possible in measure-theoretic terms. Thanks.

EDIT: I failed to mention that what i'm struggling with is the definition of a regular conditional experiment.

答: This question is a duplicate of this one from stats.stackexchange, and has been answered there.</p>

Title: Regular Conditional Bayesian Experiment


问:
Find all points that maximizes the directional derivative
Hello got stuck on this question, would love for some insight.

Here is the image

The answer is that all points lie on the line y = x + 1 but I am unaware on how to get that outcome.

答:
Gradient of the function is
$$\nabla f = (2x-2) \hat{i} + (2y-4)\hat{j} $$

Now, we require that

$$\nabla f || (\hat{i} + \hat{j})$$

$$\implies 2y-4 = 2x-2 \implies y = x+1$$

ContentLicense: CC BY-SA 4.0


问:
Checking vanishing at cusps from the $q$-expansion
I am trying to understand cusp forms, for a congruence subgroup. As an example, take $\Gamma_0(p)$, for $p$ a prime. As I understand there are two cusps, $0$ and $i\infty$, which are the $\Gamma_0(p)$ orbits of $i(\mathbb Q \cup \infty)$.

Checking a form $f$ for $\Gamma_0(p)$ vanishes at $\tau = i\infty$ is simple as you just check it vanishes for $q=0$. So let's assume that's true so that $f = \sum_{n=1}^\infty a_n q^n$.

To check $\tau = 0$, i.e. $q = 1$ means proving that $\sum_{n=1}^\infty a_n = 0$, which is often not trivial. So how does one usually approach showing this? I had the idea you could choose a different point in the same class, e.g. $\begin{pmatrix}
1 & p\\
0 & 1
\end{pmatrix}$ takes you to $\tau = p$ but you still have to show the sum of the coefficients vanishes. Is there a trick to checking vanishing at cusps (and more generally checking holomorphicity at cusps when you have the $q$-expansion)? Also I noticed that $0$ is not in the upper half plane, so it seems there is an issue there, when it is touted as a cusp but I am assuming there is a point I am missing.

答:
It depends on the amount of information you have on $f$.

For $f(z)=\sum_{n\ge 1} a_n e^{2i\pi nz} \in M_k(\Gamma_0(p))$ with integer coefficients bounded like $|a_n|\le C n^m$,

Then $$z^{-k} f(-1/(pz))=\sum_{n\ge 0} c_n e^{2i\pi nz} \in S_k(\Gamma_0(p)),\qquad  c_0=\int_i^{i+1}z^{-k} f(-1/(pz)) dz\in \Bbb{Z}$$
Whence $c_0=0$ iff for some $N$ $$|\int_i^{i+1}z^{-k} \sum_{n=1}^N a_n e^{2i\pi(-1/(pz)) }dz|<1-\sum_{n=N+1}^\infty |\int_i^{i+1}z^{-k} e^{2i\pi(-1/(pz)) }dz|$$
In other words knowing the bound $Cn^m$ allows to check if it is a cusp form solely from the first few coefficients.

Certainly the bound $C n^m$ itself can be found from the first few coefficients, at this point I would look into https://arxiv.org/abs/1809.10908 and parigp and sage implementations of modular forms.


问:
Find out the area of a triangle with midpoints and quadrisection points

In the above $\triangle$ABC, G is the midpoint of AC, D, E and F are quadrisection points of BC, AD and BG intersect at M, AF and BG intersect at N.  The area of $\triangle$ABM is $7.2cm^2$ bigger than the area of quadrilateral FCGN.  What is the area of $\triangle$ABC?

Please provide simple and elegant answers.

I know that $\triangle$ABD and $\triangle$AFC have the same size, and $\triangle$ADF is twice as big as them. I also know that $\triangle$BAG and $\triangle$BCG have the same size. I tried building up a system of linear equations using the area of $\triangle$ABM, $\triangle$MBD, $\triangle$ANG and $\triangle$AFC as variables but failed.

答:
We just need to find the ratios $DM : MA$ and $FN:NA$. They can both be obtained using Menelaus' Theorem. We have:

$$\frac {AG}{GC} \cdot \frac {CB}{BD} \cdot \frac {DM}{MA} = 1 \leadsto \frac {DM}{MA} = \frac14$$

$$\frac {AG}{GC} \cdot \frac {CB}{BF}\cdot \frac {FN}{NA} = 1\leadsto \frac {FN}{NA} = \frac34$$

Now we see that the area of $\triangle ABM$ is $\dfrac3{16}$ that of $\triangle ABC$, and the area of quadrilateral $FCGN$ is the sum of the areas of $\triangle CFN$ and $\triangle CNG$, i.e., $\dfrac 3{28} + \dfrac 2{28}$ that of $\triangle ABC$.

Finally we solve for the area of $\triangle ABC$ using the $7.5cm^2$ condition.


问:
About continuous functions and aritmethic progression
I've try solve this question, but I haven't sucess...

The problem is the following:

A continuous functions $f:[a,b]\rightarrow \mathbb{R}$ assume positive and negative values in its domain, show that there exists $a_1,a_2,\ldots,a_k$, k numbers that are arithmetic progression and it is in the domain such that
$$f(a_1)+f(a_2)+\cdots+f(a_k)=0$$

Someone can help me with this question?

答:
Fix $k$. Let the domain of the function include the interval $[a,b]$, where $f(a) \times f(b) < 0 $.

Consider the value of $ f(a) + f( a + \frac{b-a}{k-1}) + \ldots + f(b) $. If this is 0, we are done. Otherwise, WLOG we may assume that it has different sign as $f(a)$.

Now, consider the function

$$F(a, \epsilon) = f(a) + f(a + \epsilon) + f(a + 2 \epsilon) + \ldots + f(a + (k-1)\epsilon ), $$

where $\epsilon \in [0, \frac{b-a} {k-1}] $.

Apply the Intermediate-Value Theorem, and we are done.

ContentLicense: CC BY-SA 3.0


问:
Book Recommendation on Time Series Statistics
Professionally I am analysing high-frequency data coming from motion sensors and alike.
I would like to "up my theoretical background game" in this area and am therefore looking for recommendations to books about (I guess) statistics and probability theory concerning time series pehomena.

Concerning my background: I hold a master's degree in theoretical mathematics (Did some Probability Theory, statistics und stochastical analysis, though it was not my focus)

Any hints would be greatly appreciated.
Thanks in advance!

答:
These texts are popular in Economics/Finance.

Hamilton "Time Series" -Link
It is a LONG book (816 pages) but actually does not cover THAT much material. However, instead it is very user-friendly which provides a lot of details for the material it covers. It is a standard reference in Economics PhD classes for Time Series (which are important topics for macroeconomists).

Tsay "Analysis of Financial Time Series" Link- More finance oriented text. I think this is a standard reference for Financial Time Series for Finance people. A bit expensive.

Brockwell "Time Series: Theory and Methods" Link - This used to be (still is?) the main reference for Time Series back in the day for those theoretically inclined. Heavy on mathematics, so only for people with strong mathematics/statistics background.


问:
Showing that the function $f(x) = \frac{x}{1+x^2}$ is continuous using the $\epsilon \delta$ definition
I am attempting to prove that $f(x) = \frac{x}{1+x^2}$ is continuous. Here's my attempt so far.

In order to prove that $f(x)$ is continuous, it is necessary to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|\frac{x}{1+x^2}-\frac{c}{1+c^2}|<\epsilon$ when $|x-c|<\delta$.

I first start by working with the left side of the inequality. We can say that $$|\frac{x}{1+x^2}-\frac{c}{1+c^2}| = \frac{|x-c+c^2x-cx^2|}{|(1+x^2)(1+c^2)|}$$ After this we can use the triangle inequality to say, $$\frac{|x-c+c^2x-cx^2|}{|1+x^2||1+c^2|}\leq \frac{|x-c|+|c^2x|+|-cx^2|}{|1+x^2||1+c^2|}$$
Now we can break the function into 3 terms $$\frac{|x-c|}{|1+x^2||1+c^2|}+\frac{|c^2x|}{|1+x^2||1+c^2|}+\frac{|x^2c|}{|1+x^2||1+c^2|}$$
Then we use the fact that $$\frac{c^2x}{|1+x^2||1+c^2|}<1$$ which can be applied to the last 2 terms using the same logic. This gives $$\frac{|x-c|}{|1+x^2||1+c^2|}+2$$
Now we want to isloate $|x-c|$ in order to solve for $\delta$.
I have
$$\frac{|x-c|}{|1+x^2||1+c^2|}<\epsilon-2$$. This is where I get stuck. Any help would be appreciated, thanks!

Body:
When $c \neq 0$, use the bound

$$\tag{*}\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| = \frac{|x-c+c^2x-cx^2|}{(1+x^2)(1+c^2)} =\frac{|x-c+c^2x-cx^2+c^3 - c^3|}{(1+x^2)(1+c^2)} \\ \leqslant\frac{|x-c|+c^2|x-c|+|c||x^2-c^2|}{(1+x^2)(1+c^2)}= \frac{1+c^2+|c||x+c|}{(1+x^2)(1+c^2)}|x-c|$$

If $|x-c| < \frac{|c|}{2}$, then, since $|x| - |c| \leqslant |x-c|$, we have $\frac{|c|}{2} < |x| < \frac{3|c|}{2}$ and

$$\tag{**}|x+c| \leqslant |x|+|c| <\frac{5|c|}{2}, \quad 1+x^2 > 1+ \frac{c^2}{2}$$

Bounding on the RHS of (*) with (**), we get

$$\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| < \frac{1+c^2 + \frac{5c^2}{2}}{\left(1 + \frac{c^2}{2}\right)(1+c^2)}|x-c|$$

You should be able to finish from here by finding $\delta > 0$ such that if $|x-c| < \min(\delta, \frac{|c|^2}{2})$, then $|f(x) - f(c)| < \epsilon$.

Handle the case where $c=0$ separately.


问:
Does $\sum_{k=0}^\infty \frac{16(2k+1)^2}{(4m^2 - (2k+1)^2)^2}=\pi^2$?
I think the following series has the value

$$\sum_{k=0}^\infty \frac{16(2k+1)^2}{(4m^2 - (2k+1)^2)^2}=\pi^2$$

for any positive integer $m$, but I am not sure how to go about computing it. Plugging the above sum into a computer for a few different values of $m$ seems to verify the result. I don't have much experience computing any infinite sums of the above form, so I am not sure where to begin. The context for computing this series is to show that $\cos(2mx)$ can be written as an infinite sum of the functions $\sin(kx)$. On a related note, I would also like to show that

$$\sum_{k=0}^\infty \frac{64m^2}{(4m^2-(2k+1)^2)^2} = \pi^2$$

where again, $m$ is a positive integer; this sum appears in showing the opposite relation, that $\sin(2mx)$ can be written as a sum of $\cos(kx)$. Again, I have verified on a computer that this is true for at least the first few values of $m$. This second sum would be reasonable to compute if it were an integral, but I'm not sure of a way I can immediately apply that idea (if we consider $f(x) = c^4/(c^2-x^2)^2$ then we could compute the integral of this function from $0$ to $\infty$ by substitution $x=c\tanh\theta$ fairly easily). I'm aware that there are easier ways to show these relations between sines and cosines, but I'm invested enough in this approach that I would like to find a way to compute these sums. Any tips would be appreciated.

Edit: I am actually not 100% sure that the first sum converges to $\pi^2$ for all $m$; in Wolfram Alpha, the sum comes out to be $\pi^2$ for $m=1,2$ but for any $m\geq 3$ it simply approximates the sum, and the approximated value is clearly less than $\pi^2$, so I may have made an error. The value of the second sum definitely seems to be $\pi^2$ though.

答:
In a first step, let $x=(2k+1)$ and consider
$$f(x)=\frac{16x^2}{( x^2-4m^2)^2}=\frac{16x^2}{(x-2m)^2(x+2m)^2}$$ and use partial fraction decomposition to obtain
$$f(x)=\frac{2}{m (x-2 m)}-\frac{2}{m (x+2 m)}+\frac{4}{(x+2 m)^2}+\frac{4}{(x-2 m)^2}$$ Replace $x$ by $(2k+1)$ and compute the partial sums from $k=0$ to $p$.

You will get
$$S_1=\frac{H_{p-m+\frac{1}{2}}-H_{-m-\frac{1}{2}}}{m}$$
$$S_2=\frac{H_{m+p+\frac{1}{2}}-H_{m-\frac{1}{2}}}{m}$$
$$S_3=\psi ^{(1)}\left(m+\frac{1}{2}\right)-\psi ^{(1)}\left(p+m+\frac{3}{2}\right)$$
$$S_4=\psi ^{(1)}\left(\frac{1}{2}-m\right)-\psi ^{(1)}\left(p-m+\frac{3}{2}\right)$$

Now, using the asymptotics of $(S_1-S_2+S_3+S_4)$
$$\Sigma_p=\sum_{k=0}^p \frac{16(2k+1)^2}{(4m^2 - (2k+1)^2)^2}$$
$$\Sigma_p=\frac{\pi }{m}\left(\tan (\pi  m)+\pi  m \sec ^2(\pi  m)\right)-\frac{4}{p}+\frac{4}{p^2}-\frac{8 m^2+11}{3
   p^3}+O\left(\frac{1}{p^4}\right)$$
$$\color{blue}{\Sigma_\infty=\frac{\pi }{m}\left(\tan (\pi  m)+\pi  m \sec ^2(\pi  m)\right)}\tag 1$$

So, if $m$ is an integer $\neq 0$, $\Sigma_\infty=\pi^2$ as desidered (if $m=0$, the result is $2\pi^2$).

But you could use $(1)$ for any value of $m$ of your choice and $\Sigma_\infty $ will always be larger than $\pi^2$.


问:
measure of the set consisting of those numbers in (0,1) whose decimal expansion contains hundred consecutive 4s
It's a problem from Sheldon Axler;s Measure Theory book.

What I did so far is-

E= set of all numbers in (0,1) that have decimal expansion containing hundred consecutive 4s

$A_{n}$ = set of numbers in (0,1) such that the hundred consecutive 4s start from nth position after decimal point and before the nth position there is no hundred consecutive 4s in that expansion.

Countable union of $A_{n}$ is the set E.

So E is borel set. But I am struggling to find the measure of this set.

I tried calculating the measure of the set of points which don't contain any hundred consecutive 4s in their decimal expansion. But there also I got stuck.

I have seen the similar question has been asked here before. But I could not totally understand the probability approach to the problem.
What is the Lebesgue Measure of the set of numbers in (0,1) that contain 100 consecutve 4's?

答:
It might be easier to think about the complement of this set, i.e. the set of all numbers $x \in (0, 1)$ that don't have $100$ consecutive $4$'s lurking somewhere in their decimal expansion.

Lemma. For any base $b \geq 3$ and digit $d \in \{0, 1, ..., b-1 \}$, the Lebesgue measure of the set $$S_d := \{ x \in (0, 1): x \text{ contains no digit } d \text{ in base } b \}$$ is zero.

Proof of lemma. The set of all points that don't have $d$ as their first digit has measure $1 - 1/b$. The set of all points that don't have $d$ as their first or second digit has measure $(1 - 1/b)^2$. The set of all points that don't have $d$ as their first, second, third, ..., or $n$th digit has measure $(1 - 1/b)^n$. So $$\mu(S_d) \leq (1 - 1/b)^n \text{ for all } n,$$ which implies $\mu(S_d) = 0$.

Now think about some particular bases and digits that might apply to this problem, like, oh, I don't know, $b = 10^{100}$, $d = 4(10^{100}-1)/9$....


问:
logs Challenge between two students >>be smart
two student were given the equation  $2^{4x+6} = 3^{6x-3}$

1.steve rearranged to get  $2^{4x+6} - 3^{6x-3} =0$

then wrote $\log (2^{4x+6} - 3^{6x-3}) = \log0$

are these legal steps ? if not explain what is wrong with them

2 Ali wrote $\log( 2^{4x+6}) =\log(3^{6x-3})$  then $\log( 2^{4x+6}) - \log(3^{6x-3})=0  $

finally $2^{4x+6} / 3^{6x-3}$

are these legal steps ? if not explain what is wrong with them

答:
The first task is to get the unknown out of the exponents.

Do not rearrange the equation. Rather, make the substitutions:$$3=e^{\ln (3)}$$ $$2=e^{\ln (2)}$$which yields $$e^{(\ln2)\times (4x+6)} = e^{(\ln 3) \times (6x-3)}$$ Yoe now have two exponential terms, with the same base, equal to each other.  The exponents must then be equal. You are left with a simple linear equation with one unknown. (Two of the coefficients are sort of ugly!)


问:
existence of certain continuous function $\overline{f}$

Let $(X,d_X)$ and $(Y, d_Y)$ be complete metric spaces (so $X$ and $Y$ are nonempty). Let $A\subseteq X$ be nonempty. A function $f : A\to Y$ is uniformly continuous if given $\epsilon > 0,\exists \delta > 0$ so that $d_Y(f(x), f(y)) < \epsilon$ whenever $d_X(x,y) < \delta$ for $x,y \in A.$ Show that if $f : A \to Y$ is uniformly continuous then there is a unique continuous function $\overline{f} : \overline{A} \to Y$ so that $\overline{f}\vert_A = f$ (here $\overline{A} $ is the closure of $A$, or intersection of all closed sets containing $A$). Moreover, $\overline{f}$ is uniformly continuous.

I think I can show that $\overline{f}$, if it exists, must be unique. Suppose $g_1, g_2 : \overline{A}\to Y$ are such that $g_i\vert_A = f$ for $i=1,2.$ Then for all $x \in \overline{A}, \exists (x_n)\subset A, x_n \to x$ so $g_1(x_n)\to g_1(x)$ and $g_2(x_n)\to g_2(x).$ But $g_1(x_n) = g_2(x_n) = f(x_n)$ for all $n$ and so by the uniqueness of limits $g(x) = f(x).$

However, I'm not sure how to show the existence of $\overline{f}.$

I think I can show that if $\overline{f}$ exists, it is uniformly continuous by taking advantage of the continuity of $f,$ the definition of every element in $\overline{A}$ as the limit of a sequence in $A,$ and the uniform continuity of $\overline{f}.$

答:
You need $Y$ to be a complete metric space.

Take $x\in \overline A$ and a sequence $(x_n)\subset A:x_n\to x$, now lets prove that $f(x_n)$ converges in $Y$, take $\varepsilon>0$ and $\delta>0$ from the uniform continuity, since $x_n$ is convergent there is an $n_0$ such that for every $m,n>n_0$ $d_X(x_n,x_m)<\delta$ (every convergent sequence is a Cauchy sequence), so $d_Y(f(x_n),f(x_m))<\varepsilon$, which implies that $f(x_n)$ is convergent because of completeness.

Now take two sequences $(x_n),(y_n)\subset A$ such that $x_n\to x$ and $y_n\to x$, now lets prove that $\lim_n f(x_n)=\lim_n f(y_n)$. Since both $(x_n)$ and $(y_n)$ converges to $x$ for every $\varepsilon>0$ there is an $n_0$ such that $n>n_0$ implies $d_X(x_n,x)<\delta/2$, $d_X(y_n,x)<\delta/2$ so $d_X(x_n,y_n)<\delta$ and $d_Y(f(x_n),f(y_n))<\varepsilon$ so clearly $\lim f(x_n)=\lim f(y_n)$.

So you can define $\overline f:\overline A\to Y/\overline f(x)=\lim f(x_n)$ for any $(x_n)\subset A$ such that $x_n\to x$. Clearly $\overline f|_A=f$.

To prove that $\overline f$ is uniformly continuous (hence continuous) take $\varepsilon>0$, $\delta_0>0$ such that $d_Y(f(x),f(y))<\varepsilon/2$ when $d_X(x,y)<3\delta_0$ and $x,y\in \overline A$ such that $d_X(x,y)<\delta_0$ then by definition of $\overline f$ there exists $\delta_1,\delta_2>0$ such that if $x'\in A: d_X(x,x')<\delta_1$ then $d_Y(\overline f(x),\overline f(x'))<\varepsilon/4$ and $y'\in A:d_X(y,y')<\delta_2$ then $d(\overline f(y),\overline f(y'))<\varepsilon/4$ now take $x'.y'\in A$ such that $d_X(x,x'),d_X(y,y')<\min\{\delta_0,\delta_1,\delta_2\}=\delta$, so $d_X(x',y')\leq d_X(x,x')+d_X(x,y)+d_X(y,y')<3\delta\leq3\delta_0$ hence $d_Y(\overline f(x'),\overline f(y'))<\varepsilon/2$ and
$$d_Y(\overline f(x),\overline f(y))\leq d_Y(\overline f(x),\overline f(x'))+d_Y(\overline f(x'),\overline f(y'))+d_Y(\overline f(y),\overline f(y'))<\varepsilon$$
We conclude that $d_Y(\overline f(x),\overline f(y))<\varepsilon$ when $d_X(x,y)<\delta$.


答:
Applying Menelaus Theorem on $\triangle BGC$ considering $AMD$ will give $\frac {BM}{BG}=\frac {2}{5}$.
$\Rightarrow   Ar(\triangle ABM)=\frac {1}{5} \triangle ABC$

Now, applying Menelaus Theorem on $\triangle BGC$ considering $ANF$ as transversal will give $\frac {NG}{BG}=\frac {1}{7}$    $\Rightarrow  Ar(\triangle ANG)=\frac {1}{14}\triangle ABC$

Now, $Ar(FCGN)$

$=Ar(\triangle AFC)-Ar(\triangle ANG)$

$=\frac {1}{4} \triangle ABC-\frac {1}{14} \triangle ABC$

$=\frac {5}{28} \triangle ABC$

Now put these values in the equation and solve for the area of $\triangle ABC$.

Title: Find out the area of a triangle with midpoints and quadrisection points

Body:
In the above $\triangle$ABC, G is the midpoint of AC, D, E and F are quadrisection points of BC, AD and BG intersect at M, AF and BG intersect at N.  The area of $\triangle$ABM is $7.2cm^2$ bigger than the area of quadrilateral FCGN.  What is the area of $\triangle$ABC?

Please provide simple and elegant answers.

I know that $\triangle$ABD and $\triangle$AFC have the same size, and $\triangle$ADF is twice as big as them. I also know that $\triangle$BAG and $\triangle$BCG have the same size. I tried building up a system of linear equations using the area of $\triangle$ABM, $\triangle$MBD, $\triangle$ANG and $\triangle$AFC as variables but failed.


问:
Is there a rigorous definition for matrix derivatives?
I know that,

A function $f: \mathbb{R}^n \to \mathbb{R}$ is said to be differentiable at $x$ if there exists a vector $v$ such that,
$$
    \lim_{h \to 0} \frac{f(x+h) - f(x) - v^Th}
    {\|h\|} = 0.
$$
When $v$ exists, it is given by the "gradient" $\nabla f(x) = \left(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}\right)(x)$

Does there exist a similar definition for "matrix derivative"

https://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_matrices

答:
Yes, you can use a very similar definition. First of all, the map $h \mapsto v^\top h$ encodes an arbitrary linear map on $\mathbb R^n$. For matrices, you can substitute it by the Frobenius inner product, e.g.,
$$A \mapsto (A,B)_F := \sum_{i,j = 1}^n A_{ij} B_{ij},$$
where $B \in \mathbb R^{n \times n}$ is fixed.

Thus, for $f \colon \mathbb R^{n \times n} \to \mathbb R$ you can define $\nabla f(A)$ to be the (unique) matrix $B \in \mathbb R^{n \times n}$ (if it exists), that satisfies
$$ \lim_{H \to 0} \frac{f(A + H) - f(A) - (B,H)_F}{\|H\|} = 0.$$

This definition directly extends to Hilbert spaces (in which you use the inner product determined by the Hilbert space structure). In Banach spaces, one uses the duality product and this yields a derivative which belongs to the dual space.


问:
Troubles calculating $\int_z^\infty\mu\mathrm e^{-\mu y}(\mathrm e^{-\lambda z}-\mathrm e^{-\lambda y})\ \mathrm{d}y$
I've difficulties calculating the following integral

$$\int_z^\infty\mu\mathrm e^{-\mu y}(\mathrm e^{-\lambda z}-\mathrm e^{-\lambda y})\ \mathrm{d}y$$

I'm gonna use her to find a joint distribution of two random variables. I've try to apply the following substitution $u=e^{\lambda y}$ but I couldn't do much after that.
I appreciate any help. Thank you in advance. Sorry for my bad English.

答: Split it into two terms. $e^{-\lambda z}$ is constant in the first integral. So the answer is  $e^{-\lambda z}e^{-\mu z}-\frac {\mu} {\lambda+\mu}e^{-(\lambda+\mu)z}$


问:
If $f$ is holomorphic on some open set $G\subseteq \mathbb{C}$, then $f$ is infinitely often complex differentiable
Let $G$ be an open subset of $\mathbb{C}$, and suppose that $f$ is holomorphic on $G$. Let $a\in G$ be arbitrary. The reason why $f$ is infinitely often complex differentiable is, in my opinion (if I'm wrong, please tell me), based on Cauchy's Integral Formula for Derivatives,

If $\overline{B(a,r)}\subseteq G$, then for every $z_0\in B(a,r)$ we have
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{\partial B(a,r)}\frac{f(z)}{(z-z_0)^{n+1}}\,\mathrm{d}z,\quad n=0,1,2,\dots
$$

Clearly, one can find $\rho>0$ such that $B(a,r)\subseteq G$, because $G$ is open. But, is the existence of $\rho'>0$ satisfying that $\overline{B(a,\rho')}\subseteq G$ guaranteed?

答:
For $a\in G$ there is an $R > 0$ such that $B(a, R) \subset G$. Setting $r =R/2$ one has $\overline{B(a,r)}\subset B(a, R) \subset G$. Now one can show with induction that for all integers $n \ge 0$:

$f$ is $n$-times differentiable in $B(a, r)$, and

for all $z_0 \in B(a, r)$, $$\tag{*}f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{\partial B(a,r)}\frac{f(z)}{(z-z_0)^{n+1}}\,dz \, .
$$

For $n=0$ this is Cauchy's integral formula. For the inductive step you have to show that you can differentiate under the integral sign on the right-hand side of $(*)$.

This shows that $f$ is infinitely often differentiable in a neighbourhood of every $a \in G$, which proves the claim.

So Cauchy's formula for the derivatives shows (for each $n$) both the existence of the $n$-th derivative, and the representation of $f^{(n)}$ as an integral.


问:
Is the set of surjective functions from $\mathbb{N}$ to $\mathbb{N}$ uncountable?
I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form $\Bbb{N} \to \Bbb{N}$ is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function $p$.The problem that I'm having is that I don't know in which way to change the diagonal elements so that $p$ should, but cannot be in the matrix.

Can someone help me, or at least give me some tips?

答:
Among the surjections $\mathbb N\to\mathbb N$ you have the maps
$$f(n)=\begin{cases}k&\text{if }n=2k\\g(k)&\text{if }n=2k-1\end{cases}$$
where $g\colon\mathbb N\to\mathbb N$ is one of the uncountable arbitrary maps $\mathbb N\to\mathbb N$.


问:
Explain the difficulty of find primes with the same last digit
Refer to https://oeis.org/A340800 to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase.  Is there an explanation for this?  How can the size of primes have any influence on the last digit of the following primes?</p>

答:
Of course the general trend of A340800 must be increasing (although it is not monotonic).  For any $M$, the maximum of $A072971(k)$ for $k \le M$ is some finite
value $N$, and so if $n > N$ and $A340800(n)$ exists at all it must be $prime(k)$ for some $k > M$.  The existence of $A340800(n)$ follows from the prime $k$-tuples conjecture.


问:
Prove or disprove that equations $x^2+y^2+z^2=3$, $xy+tz=2$, $xz+ty+e^t=0$ can be solved of $t$ near $(x,y,z,t)=(-1,-2,1,0)$

Prove or disprove that equations $x^2+y^2+z^2=3$, $xy+tz=2$, $xz+ty+e^t=0$ can be solved of $t$ near $(x,y,z,t)=(-1,-2,1,0)$

I thought that I can apply the implicit function theorem, but I found that $f_1(x,y,z,t)=x^2+y^2+z^2-3$, $f_1(-1,-2,1,0)\ne 0$. What can I do? Appreciate any suggestion.

答:
HINT:

To apply the implicit function theorem you need to check that the jacobian
$$\frac{\partial (f,g,h)}{\partial(x,y,z)}_{|(−1,−2,1,0)} \ne 0$$
where $f(x,y,z,t) = x^2 + y^2 + z^2 -3$, $\ \ g(x,y,z,t)= x y + z t - 2$, $\ \ h(x,y,z,t) = x z + y t + e^t$.


问:
Combinatorics of the Zeta function of a variety
I want to know if there is a good combinatorial interpretation of what the Zeta function of a variety $X$ over a finite field $\mathbb{F}_p$ counts. It is defined as $$\exp\sum N_j/j\,t^j,$$ where $N_j$ is the number of points in $X$ over $\mathbb{F}_{p^j}$.

I hope an answer will be something along the following lines: the coefficients of $\sum N_j/j\,t^j$ count the orbits of the solution sets under iterations of the Fröbenius automorphism, and the exponential of the sequence somehow counts representatives.

I reckon this is very elementary combinatorics. A concise source I could use to produce such a combinatorial statement would be very much appreciated!

答:
The zeta function counts effective divisors, e.g. objects of the form $\sum n_i p_i$ where

$p_i$ is a point of $X$ over a finite extension of $\mathbb{F}_p$,

$n_i \ge 0$ are non-negative integers, and

the expression $\sum n_i p_i$ is Galois-invariant.

If $X$ is affine, this is equivalent to saying that it counts ideals of the coordinate ring of $X$; in fact in this setting the zeta function is more or less the Dedekind zeta function of the coordinate ring.

This is a corollary of a version of the exponential formula which asserts that

$$\exp \sum \frac{z_j}{j} t^j = \sum_{n \ge 0} Z_{S_n}(z_1, z_2, ...) t^n$$

where $Z_{S_n}$ are the cycle index polynomials of the symmetric groups. See this blog post and this blog post for details.


问:
Using quadratic residue
show that $2b^2+ 3$ cannot be a divisor of $a^2-2$

I have tried working with residue of $2$ modulo $p$ where $p$ is a divisor of $2b^2+3$ but couldn't stablish the result.

答:
You have the right idea. Consider there are integers $a$ and $b$ with $2b^2 + 3 \mid a^2 - 2$. With $2b^2 + 3$, it's always odd and $\gt 1$, so it has at least one prime factor and all of these prime factors are odd. Let $p$ be any of these prime factors. Since $p \mid 2b^2 + 3$, then $p \mid a^2 - 2$. Thus, as you stated, $2$ must be a quadratic residue modulo $p$. The table in the Law of quadratic reciprocity section shows $2$ is a quadratic residue mod $p$ if and only if

$$p \equiv 1, 7 \pmod{8} \tag{1}\label{eq1A}$$

Note since $7 \equiv -1 \pmod{8}$, the product of any number of these factors will only be congruent to either $1$ or $7$ mod $8$.

However, if $b$ is even, then $2b^2$ has a factor of $2 \times 2^2 = 8$, so $2b^2 + 3 \equiv 3 \pmod{8}$. Alternatively, if $b$ is odd, then $b^2 \equiv 1 \pmod{8}$, so $2b^2 + 3 \equiv 2 + 3 \equiv 5 \pmod{8}$. In both cases, it's not congruent to $1$ or $7$ mod $8$, meaning $2b^2 + 3$ must have at least one odd prime factor for which $2$ is not a quadratic residue. Thus, this means $p \not\mid a^2 - 2$, so

$$2b^2 + 3 \not\mid a^2 - 2 \tag{2}\label{eq2A}$$

Title: Using quadratic residue


问:
What is the meaning of Matlab's ODE solver naming conventions?
MatLab has a number of built-in ODE solvers and are named using the convention odepq, for exampleL ode45, ode23, etc.

I understand that $p$ here stands for the order of the numerical method. For example, when using Euler's method, this is an order 1 because you only need to substitute the value of the function once before going on the update equation. As opposed to 2 times for Heun's and midpoint method (hence order 2). Classical RK method is order 4.

What I don't understand is the $q$ part. What is it?

答:
Actually, ode45 is Dormand-Prince method with updates of 4th and 5th order, using the difference to steer the step size controller, and using the 5th order updates to advance the integration, which is the "extrapolation mode" of embedded methods.

All ode1q are implicit linear multi-step methods with not only step size control but also adaptation of the order. For stiff equations it can be useful to lower the order where the dynamic becomes too contrarian, and then increase order and step size where it is "smooth sailing". q is the maximum order, the documentation says what class of methods is actually used


问:
Integrating the derivative of a function
I have this question of calculus, my answer was d but my teacher told me it was wrong. What am I not seeing? What would be the answer and why:

If $f$ is a continuous function and if $F'(x) = f(x)$ for all real numbers $x$, then $\int_{1}^3 f(2x)dx$ is

a) $2F(3) − 2F(1)$

b) $0.5F(3) − 0.5F(1)$

c) $2F(6) − 2F(2)$

d) $F(6) − F(2)$

e) $0.5F(6) − 0.5F(2)$

答:
$$I = \int_1^3f(2x)dx = \frac{1}{2}\int_{2}^{6}f(u)du$$

If $u=2x$, the limits will double $x =1 \to u=2, x=3 \to u = 6$, and $du = 2dx$

Using fundamental theorem of calculus

$$I = 0.5F(6)-0.5F(2)$$


问:
Finding the Inverse Cumulative Distribution function for the Rice distribution
I'm interested in implementing an algorithm for producing random samples according to some Rician distribution (sometimes known as the Ricean distribution, see).

I intend to produce these psuedo-random numbers using the Inverse Transform Sampling technique. That is, given the Cumulative Distribution Function (CDF) $C(x)$, on some PDF $P(x)$, it is possible to pick psuedo-random numbers from $P(x)$ by first generating some $U \in  \text{Unif}(0,1) $, then our random number, $n$, is given by $n = C^{-1}(U)$.

My trouble is in determining $C^{-1}(x)$ for the Rice distribution. Surprisingly, I'm unable to find it online.

What I know:

The Rice distribution is characterized by some $\sigma$ and $\nu$. It's PDF is given by:

$$\frac{x}{\sigma^2}\exp\bigg(\frac{-(x^2+\nu^2)}{2\sigma^2}\bigg)I_0\bigg(\frac{x\nu}{\sigma^2}\bigg)$$

where $I_0$ is the modified Bessel function, with order $0$.

Further, the CDF of the Rician distribution is given by:

$$1 - Q_1 \bigg(\frac{\nu}{\sigma},\frac{x}{\sigma}\bigg)$$

Where $Q_1$ is the Marcum Q-function. Expanding this, I believe we have:

$$C(x) = 1 - \exp \bigg(-\frac{\frac{\nu}{\sigma}^2+\frac{x}{\sigma}^2}{2}\bigg)\sum_{k=1}^\infty \bigg(\frac{\frac{\nu}{\sigma}}{\frac{x}{\sigma}}\bigg)^k I_k(\frac{x}{\sigma} \cdot \frac{\nu}{\sigma})$$

The issue, now, is determining $C^{-1}(x)$. Unfortunately (and not surprisingly, perhaps), WolframAlpha, etc. choke up on this quickly, and I'm unsure how to start with this (dealing with $I_0$ seems particularly tricky).

What is the inverse, and how can it be found?

答:
If all you want to do is generate random variates with the desired distribution, then the most efficient way to do so is to generate pairs of independent normal random variates as follows:  let $$X_i \sim \operatorname{Normal}(\nu, \sigma^2), \quad Y_i \sim \operatorname{Normal}(0, \sigma^2)$$ and then $R_i = \sqrt{X_i^2 + Y_i^2}$ will obey $$R_i \sim \operatorname{Rice}(|\nu|, \sigma).$$  This is because this process is how the Rice distribution arises.  See https://en.wikipedia.org/wiki/Rice_distribution#Related_distributions.

In the same section, there is another method that is described:  let $$P_i \sim \operatorname{Poisson}\left(\lambda = \frac{\nu^2}{2\sigma^2}\right), \\ X_i \mid P_i \sim \chi^2 (2P_i + 2).$$  Then $R_i = \sigma \sqrt{X_i}$ is also Rice-distributed.

Either of these methods will be much faster than the inverse transform sampling, but between the two, I prefer the first.  You can modify the Box-Muller method in a very simple way:  just generate the pair $$(Z_0, Z_1) = \left(\sqrt{-2 \log U_1} \cos (2\pi U_2), \sqrt{-2 \log U_2} \sin (2\pi U_2)\right)$$ where $U_1, U_2$ are independent uniform on $(0,1)$, and then $$(X_i, Y_i) = (\nu + \sigma Z_0, \sigma Z_1)$$ or more directly $$\begin{align}
R_i &= \sqrt{(\nu + \sigma Z_0)^2 + (\sigma Z_1)^2} \\
&= \sqrt{\nu^2 + 2 \sigma \nu \sqrt{-2 \log U_1} \cos (2 \pi U_2) - 2\sigma^2  \log U_1}
\end{align}$$
will generate the desired variate.

I should report that I was able to use the first method (two normal variates via Box-Muller) successfully, but I have not been able to get the second method (chi-squared with Poisson hierarchical model) to work.


答:
Hint:

$$\int_u^\infty e^{-at}dt=-\left.\frac{e^{-at}}a\right|_u^\infty=\frac{e^{-au}}a.$$

Your integral is

$$\mu e^{-\lambda z} \int_z^\infty e^{-\mu y}dy-\mu\int_z^\infty e^{-(\mu+\lambda) y}dy.$$

Title: Troubles calculating $\int_z^\infty\mu\mathrm e^{-\mu y}(\mathrm e^{-\lambda z}-\mathrm e^{-\lambda y})\ \mathrm{d}y$

Body:
I've difficulties calculating the following integral

$$\int_z^\infty\mu\mathrm e^{-\mu y}(\mathrm e^{-\lambda z}-\mathrm e^{-\lambda y})\ \mathrm{d}y$$

I'm gonna use her to find a joint distribution of two random variables. I've try to apply the following substitution $u=e^{\lambda y}$ but I couldn't do much after that.
I appreciate any help. Thank you in advance. Sorry for my bad English.


答:
This is not an answer, just an elaboration.

Let $A$ be the set of sequences that do not contain a run of one hundred $4$s.

Let $A_k$ be the set of sequences such that at least one of $x_k,...,x_{k+99}$ is not $4$. Note that $P A_k = 1-{1 \over 10^{100}} <1$.

It is clear that $A \subset A_1 \cap A_{101} \cap A_{201} \cap ...$ and so  we have $PA \le P(A_1 \cap A_{101} \cap \cdots \cap A_{100n+1} )$ for any $n$.

By independence we have $P(A_1 \cap A_{101} \cap \cdots \cap A_{100n+1} ) =(P A_1)^n$ and so $PA = 0$.

Title: measure of the set consisting of those numbers in (0,1) whose decimal expansion contains hundred consecutive 4s

Body:
It's a problem from Sheldon Axler;s Measure Theory book.

What I did so far is-

E= set of all numbers in (0,1) that have decimal expansion containing hundred consecutive 4s

$A_{n}$ = set of numbers in (0,1) such that the hundred consecutive 4s start from nth position after decimal point and before the nth position there is no hundred consecutive 4s in that expansion.

Countable union of $A_{n}$ is the set E.

So E is borel set. But I am struggling to find the measure of this set.

I tried calculating the measure of the set of points which don't contain any hundred consecutive 4s in their decimal expansion. But there also I got stuck.

I have seen the similar question has been asked here before. But I could not totally understand the probability approach to the problem.
What is the Lebesgue Measure of the set of numbers in (0,1) that contain 100 consecutve 4's?


问:
Does this polynomial factorize further?
I just did a national exam and this question was in it; I am convinced this does not work:

Given that $(x - 1)$ is a factor of $x^3 + 3x^2 + x - 5$, factorize this cubic fully.

My attempt

```
1 | 1  3  1 -5
  |    1  4  5
  |____________
    1  4  5  0
```

$$(x - 1)(x^2 + 4x + 5) = 0$$

That's all I got. I gave up and couldn't factorize further. Am I missing something?

答:
The discriminant of your polynomial is $$\Delta=4^2-4\cdot 5<0$$ so this has no real roots, that is, factorization over $\Bbb R$ is finished. On the other hand, factorization over $\Bbb C$ is possible, yielding

$$x^2+4x+5=(x+2)^2+1=(x+2-i)(x+2+i)$$

Title: Does this polynomial factorize further?


问:
Let a graph $G$ have a cycle that contains a vertex covering of the graph. Prove that $L(G)$ is Hamiltonian
Suppose a graph $G$ have a cycle that contains a vertex covering of the graph. Prove that $L(G)$ is Hamiltonian

答:
Let $c_1,c_2, ... , c_n$ be successive edges of the cycle $C$ of $G$. Then vertices $c_1,c_2, ... , c_n$ form a cycle in $L(G)$

Let $c_1$ and $c_2$ meet at vertex $v$ of $C$. Let $e_1,e_2, ... , e_k$ be the other edges (if any) of $G$ which meet at $v$.

In $L(G)$, the vertices $c_1,c_2,e_1,e_2, ... , e_k$ are then all adjacent and so the cycle $c_1,c_2, ... , c_n$ can be enlarged into the cycle $c_1,e_1,e_2, ... , e_k,c_2, ... , c_n$.

This process can be repeated with each pair of successive edges of $C$ except that we ignore any edges already added into the cycle of $L(G)$. Since $C$ is a vertex covering of $G$, the cycle of $L(G)$ then contains every edge of $G$ and $L(G)$ is therefore Hamiltonian.


问:
Trying to show that an equation in two variables implies that the variables are the same

Let $a ,b$ be non-zero integers such that $\frac{a^2 + \sqrt{2}}{a} = \frac{b^2 + \sqrt{2}}{b}$. Prove that $a=b$.

My proof:

Assume that $a \neq b$. Then, $a - b \neq 0$, implying that one can divide by $a-b$.

The given expression simplifies like so

$$ a^2b +b\sqrt2=ab^2+a\sqrt2 \\ ab(a-b) = \sqrt2(a-b) \\ ab = \sqrt2 $$

In the third step here, we have divided by $a-b$.

The conclusion $ab = \sqrt2$ is absurd. The product of two non-zero integers cannot be $\sqrt2$.

$\implies$ the initial assumption is false. $a = b \text{ QED.}$

I think something is wrong with my proof.

Have I taken into account during the assumption that $a$ and $b$ are integers?

答:
Indeed, as the comments indicated, you actually did use the fact that $a, b$ are integers in your proof, since the product of two integers cannot equal $\sqrt{2}$.

To illustrate, note that if $a = \sqrt{2}$ and $b = 1$, the equality should hold, and in fact it does, since $$\frac{(\sqrt{2})^2 + \sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2} + 1)}{\sqrt{2}} = 1 + \sqrt{2} = \frac{1^2 + \sqrt{2}}{1}.$$  So if the condition that $a, b$ are integers is relaxed, it is possible for $a \ne b$.


问:
What could be the matrix of Transformation that has no effect on vectors? The vectors are $T(x)=(x)$.
We call this matrix Identity and denote it by $I$. But I am not clear with how to assume one matrix to proceed with its solution. The Form of Transformation here is $\mathbb R^2 \to \mathbb R^2$ subspaces. I am quoting the exact question here:

All of the transformations here have the form $T: \mathbb R^2 \to \mathbb R^2$.
Find the matrix of the transformation that has no effect on vectors, that is $T(x)= x$. We call this matrix the identity matrix and denote it by $I$.

答:
Any transformation is represented by a matrix. Now, for a $\mathbb{R}^2 \to \mathbb{R}^2$ transformation, this would be a $2 \times 2$ matrix

$$T = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$

Now, we are given that

$$T\vec{x} = \vec{x}$$

$$\implies \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}ax+by \\ cx + dy \end{bmatrix} = \begin{bmatrix}x \\ y\end{bmatrix}$$

Since this is true for all $x,y \in R$, we have

$$a = 1, b = 0, c = 0, d =1$$


问:
Find out the area of a triangle with midpoints and quadrisection points

In the above $\triangle$ABC, G is the midpoint of AC, D, E and F are quadrisection points of BC, AD and BG intersect at M, AF and BG intersect at N.  The area of $\triangle$ABM is $7.2cm^2$ bigger than the area of quadrilateral FCGN.  What is the area of $\triangle$ABC?

Please provide simple and elegant answers.

I know that $\triangle$ABD and $\triangle$AFC have the same size, and $\triangle$ADF is twice as big as them. I also know that $\triangle$BAG and $\triangle$BCG have the same size. I tried building up a system of linear equations using the area of $\triangle$ABM, $\triangle$MBD, $\triangle$ANG and $\triangle$AFC as variables but failed.

答:
Here is an elementary method just building the linear equations as you mentioned, though it is of course not the most efficient.

say, area of $\triangle ABC = a$, $AN = m \times AF$ and $AM = n \times AD$.

Then you have,

$FCGN = \triangle ACF - \triangle AGN = \frac{a}{4} - \frac{a m}{8}$

$\triangle BNF = \frac{3a}{4} (1-m) = \frac{a}{2} - FCGN = \frac{a}{4} + \frac{am}{8} \implies m = \frac{4}{7}$.

Similarly, $\triangle ABM = \frac{an}{4}$

$\triangle AMG = \frac{n}{2} \times \frac{3a}{4} = \frac{3an}{8}$

$\triangle ABM + \triangle AMG = \frac{5an}{8} = \frac{a}{2} \implies n = \frac{4}{5}$

So, $\frac{a}{5} - \frac{5a}{28} = 7.2 \implies a = 336$


答:
Great job. here is a more calculus approach.

Let $f(x) = x + \frac{\sqrt2}{x}$ which is an odd function, it is positive when $x$ is positive and negative when $x$ is negative.

$f'(x) =1-\frac{\sqrt2}{x^2}$, it increases on $(\sqrt{2}, \infty)$.

We just have to check that $f(1) \le f(\sqrt{2})$ which is true since $f(1)=f(\sqrt2)=1+\sqrt2$

Title: Trying to show that an equation in two variables implies that the variables are the same

Body:
Let $a ,b$ be non-zero integers such that $\frac{a^2 + \sqrt{2}}{a} = \frac{b^2 + \sqrt{2}}{b}$. Prove that $a=b$.

My proof:

Assume that $a \neq b$. Then, $a - b \neq 0$, implying that one can divide by $a-b$.

The given expression simplifies like so

$$ a^2b +b\sqrt2=ab^2+a\sqrt2 \\ ab(a-b) = \sqrt2(a-b) \\ ab = \sqrt2 $$

In the third step here, we have divided by $a-b$.

The conclusion $ab = \sqrt2$ is absurd. The product of two non-zero integers cannot be $\sqrt2$.

$\implies$ the initial assumption is false. $a = b \text{ QED.}$

I think something is wrong with my proof.

Have I taken into account during the assumption that $a$ and $b$ are integers?


问:
Problem to understand the divergence theorem in the exercise
For the vector field $\mathbf{F}=(xy^2,yz^2,x^2z)$ use the divergent theorem to evaluate
$$\iint_S\mathbf{F}\cdot d\mathbf{S}$$
where $S$ is the sphere of radius $1$ centered at the origin.

We use the divergence theorem and get that the divergence is, $div\mathbf{F}=y^2+z^2+x^2$ and hence the surface integral is equal to the triple integral
$$\iiint_B(y^2+z^2+x^2)dV$$
where $B$ is the ball of radius $1$.
Now to evaluate the tripple integral, we can change to variables of spherical coordinates. In our case
$$0 \leq \rho \leq 1, 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi$$

Now why we get that the integral is simply $x^2+y^2+z^2=\rho^2$? Here I got stucked.

EDIT: We than know that the Jacobian determinant is $dV=\rho^2 \cdot sin(\phi)\cdot d\phi \cdot  d \theta\cdot d \rho$

答:
In spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho\cos\phi$

So $x^2+y^2+z^2 = \rho^2 \ $ where $0 \leq \rho \leq 1$

The region is a unit ball and the equation is written as $x^2 + y^2 + z^2 \leq 1$. Note the inequality sign. When you do surface integral, your integral is over the surface and hence $\rho = 1$. But in this case, you are applying divergence theorem which is volume integral and hence $0 \leq \rho \leq 1$

So your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4  \sin \phi \ d\rho \ d\phi \ d\theta = \frac{4\pi}{5} \ $ ($\rho^2 \sin\phi$ is the Jacobian)


问:
Finishing of basic proof of Wilson's theorem
I've been trying to simplify my proof of Wilson's theorem so that I can easily understand it, but I'm running into some trouble. I reached the point where I proved $(p-1)! \equiv -(p-2)! \mod p$, but I'm wondering how I can push this further. I understand that each number in $(p-2)!$ has an inverse that will cancel it out to equal 1, but is there a way of proving that $(p-2)! \equiv 1 \mod p$ in a more elementary fashion? Thanks

答:
Outline to prove that $(p-2)! \equiv 1 \pmod p$ (thanks to above comment from Gio and this thread):

For prime $p$, every positive integer $a \in Z_p$ has a unique "multiplicative inverse" $x \in Z_p$, that is, $ax \equiv 1 \pmod p$.

So for each integer $a$ in 1 to $(p-2)$, which has its multiplicative-inverse $b$ different from itself, you can combine $a$ and $b$ and we will have $(ab) \bmod p = 1$.

Now we are left with integers $a$ which are their own multiplicative-inverse. That is: $a^2 \equiv 1 \pmod p$.
But that means,
$p | (a-1)(a+1)$.
Since $p$ is prime, this is only possible for $a=1$ and $a=p-1$. So, among 1 to $(p-2)$, there is only one such $a=1$.

(For Wilson's Theorem, we can directly use above arguments to say that:
$(p-1)! \bmod p = (p-1) \bmod p = (p-1)$.)


问:
Can't establish a lower bound on a supremum
I have a sequence of functions $f_{k,j}:[0,1]\to\mathbb{R}$ defined by
$$f_{k,j} = k^{\frac{1}{p}}\chi_{(\frac{j-1}{k},\frac{j}{k})},$$
for all $k\geq 1,1\leq j\leq k$.

This serves as an example of a sequence that converges to $0$ in measure, but not in $L^{p,\infty}$.

It is easy to verify convergence in measure, but I am trying to check that $\|f_{k,j}\|_{p,\infty}\geq 1$ for all $k\geq 1$.  The following proof is given which I cannot follow:

(Take $\mu$ to be the Lebesgue measure.)
\begin{eqnarray*}
\|f_{k,j}\|_{p,\infty} &=& \sup_{\alpha > 0}\alpha\cdot\mu(\{x\in [0,1] : f_{k,j}(x) > \alpha\})^{\frac{1}{p}}\text{, (by definition)}\\
&\geq& \sup_{k\geq 1}\frac{(k - \frac{1}{k})^{\frac{1}{p}}}{k^{\frac{1}{p}}}\\
&=& 1
\end{eqnarray*}

I am stuck trying to verify the $(\geq)$ step.  At first I thought it was just restricting the supremum to the range of $k\geq 1$ which is more specific than $\alpha > 0$, but $k$ is already being used as an index.

Any ideas?

答:
As David Mitra pointed out
$$
\mu(\{x\in[0,1]:f_{k,j}(x)>\alpha\})=
\begin{cases}
0\quad&\text{ if }\quad \alpha>k^{1/p}\\
k^{-1}\quad&\text{ if }\quad \alpha\leq k^{1/p}
\end{cases}
$$
So
$$
\alpha\mu(\{x\in[0,1]:f_{k,j}(x)>\alpha\})^{1/p}=
\begin{cases}
0\quad&\text{ if }\quad \alpha>k^{1/p}\\
\alpha k^{-1/p}\quad&\text{ if }\quad \alpha\leq k^{1/p}
\end{cases}
$$
Hence supremum is attained for $\alpha_0=k^{1/p}$ and now we conclude
$$
\sup\limits_{\alpha>0}\alpha\mu(\{x\in[0,1]:f_{k,j}(x)>\alpha\})^{1/p}=\alpha_0k^{-1/p}=1
$$</p>


问:
How are the hyperoperations defined?
For the first few operations such as addition and multiplication they follow rules such as
$$A(x,y+1)=A(x,y)+1$$
$$A(x,0)=x$$
For multiplication
$$M(x,y+1)=M(x,y)+x$$
$$M(x,1)=x$$
So for a general hyper operation you would think that
$$\Pi_n(x,y+1)=\Pi_{n-1}(\Pi_n(x,y),x)$$
But then this doesnt work for tetration because
$$\Pi_4(x,y+1)=\Pi_{3}(\Pi_4(x,y),x)$$
$$^{y+1}x =(^{y}x)^x$$
But this implies
$$^{y}x =(x)^{x^y}$$
Because exponents multiply
When it should be
$$^{y}x =x^{x^{x^{x^{x^{{...}}}}}}$$
What should be the correct formula for the nth hyperoperation.
Thanks.

答:
Just a comment, not an answer.

I've once given it a try, but gave it also an operator-notation (=adapted to look more like the binary $+$,$*$ - notation). Also looked at neutral and annihilating elements. But didn't fill all table-entries. Didn't took the final conclusion what pleased me most... See this

Title: How are the hyperoperations defined?


问:
measure of the set consisting of those numbers in (0,1) whose decimal expansion contains hundred consecutive 4s
It's a problem from Sheldon Axler;s Measure Theory book.

What I did so far is-

E= set of all numbers in (0,1) that have decimal expansion containing hundred consecutive 4s

$A_{n}$ = set of numbers in (0,1) such that the hundred consecutive 4s start from nth position after decimal point and before the nth position there is no hundred consecutive 4s in that expansion.

Countable union of $A_{n}$ is the set E.

So E is borel set. But I am struggling to find the measure of this set.

I tried calculating the measure of the set of points which don't contain any hundred consecutive 4s in their decimal expansion. But there also I got stuck.

I have seen the similar question has been asked here before. But I could not totally understand the probability approach to the problem.
What is the Lebesgue Measure of the set of numbers in (0,1) that contain 100 consecutve 4's?

答:
Here is an argument for which you can fill in the details.

Look at the decimal expansion of $x\in (0,1)$ and break the expansion into contiguous blocks, each of length $d=100$. Let $B_k(x)$ be the function that maps $x\in(0,1)$ to the $k$th such block.

Let $S$ be the set of all sequences of $d$ decimal digits, so $S$ has $10^d$ elements. Argue that for any $s_1,\ldots,s_n \in S$ (with the $(s_k)$ not necessarily distinct):
$$
\lambda\left(\bigcap_{k=1}^n\left\{x\in(0,1):B_k(x)=s_k\right\}\right)=\frac1{10^{dn}}.\tag1
$$
In words, the set of numbers whose decimal expansion begins with $0.s_1s_2\cdots s_n$ has Lebesgue measure $1/10^{dn}$. It follows that for any subsets $A_1,\ldots,A_n$ of $S$, we have
$$
\lambda\left(\bigcap_{k=1}^n\left\{x:B_k(x)\in A_k\right\}\right)=\prod_{k=1}^n\lambda\left(\left\{x:B_k(x)\in A_k\right\}\right).\tag2
$$
(Proof: Express the set on the LHS of (2) as a disjoint union of sets of the form appearing in the LHS of (1).) In probabilistic terms, this establishes that the $B_k$ are independent random variables.

To finish the proof, argue that the complement of your set $E$ is a subset of the set
$$\bigcap_{k=1}^n\left\{x: B_k(x)\ne 44\cdots 4\right\}.\tag3$$
Compute the measure of (3) using (2), giving an upper bound for $\lambda(E^c)$. This upper bound holds for every $n$, so we are done.


答:
Alternative proof :

If you want another way of thinking about Wilson's theorem, here is an expeditive proof :

Check that in $\mathbb{Z}/p\mathbb{Z}[X]$,
$$X^p-X = \prod_{x \in \mathbb{Z}/p\mathbb{Z}} X-x$$
(these two polynomials are unitary, share the same degree and the same roots in the field $\mathbb{Z}/p\mathbb{Z}$)

Compare the coefficients of $X$.

Title: Finishing of basic proof of Wilson's theorem

Body: I've been trying to simplify my proof of Wilson's theorem so that I can easily understand it, but I'm running into some trouble. I reached the point where I proved $(p-1)! \equiv -(p-2)! \mod p$, but I'm wondering how I can push this further. I understand that each number in $(p-2)!$ has an inverse that will cancel it out to equal 1, but is there a way of proving that $(p-2)! \equiv 1 \mod p$ in a more elementary fashion? Thanks


问:
Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$?
Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$

Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?

I start with $AM-GM$

So $A \geq 8xyz$

Then I need to prove $xyz\geq 1$ right?

How to do that and if it's a wrong way, please help me to solve it !!!

答:
Let $A(x,y,z)=(1+x^2)(1+y^2)(1+z^2)$, defined on $\{(x,y,z)\in\mathbb{R}^3{\,\mid\,}x+y+z=3\}$.

Since $\max(|x|,|y|,|z|)\ge 3$ would yield $A > 8$, it follows $A$ has a global minimum value, $a$ say.

Since $(x,y,z)=1$ yields $A=8$, it follows that $a\le 8$.

Suppose $A(x,y,z)=a$.

Without loss of generality, assume $x\le y\le z$.

From $x+y+z=3$, it follows that $z\ge 1$.

We must have $\max(|x|,|y|,|z|)< 3$, hence $z < 3$.

Let $s=x+y$.

From $x+y+z=3$ and $1\le z < 3$, we get $0 < s \le 2$.

Let $u=(1+x^2)(1+y^2)$ and let $v=\left(1+\left({\Large{\frac{s}{2}}}\right)^2\right)^2$.

Replacing $y$ by $s-x$, we get
$$
u-v
=
\Bigl({\small{\frac{1}{16}}}\Bigr)
\!
\left((s-2x)^2+2(4-s^2)\right)
(s-2x)^2
$$
hence $u\ge v$ with equality if and only if $s=2x$.

But we can't have $u > v$, else $A(x,y,z) > A\Bigl({\Large{\frac{s}{2}}},{\Large{\frac{s}{2}}},z\Bigr)$, contrary to $A(x,y,z)=a$.

Thus $u=v$, hence $s=2x$, so $y=x$.

It follows that $a$ is the minimum value of
$$
f(x)
=
A(x,x,3-2x)
=
(1+x^2)^2\left(1+(3-2x)^2\right)
$$
where $f$ can be restricted to the interval $0\le x\le 1$.

At the endpoints we have $f(0)=10$ and $f(1)=8$.

Identically we have
$$
f'(x)
=
12(2x-1)(1+x^2)(x-1)^2
$$
which on the interval $(0,1)$ has the unique critical point $x={\large{\frac{1}{2}}}$.

Noting that $f\Bigl({\large{\frac{1}{2}}}\Bigr)={\large{\frac{125}{16}}}=7.8125 <  8$, it follows that $a={\large{\frac{125}{16}}}$, with $(x,y,z)=\Bigl({\large{\frac{1}{2}}},{\large{\frac{1}{2}}},2\Bigr)$.


问:
A simple 2 grade equations system
If we have:
$$x^2 + xy + y^2 = 25 $$
$$x^2 + xz + z^2 = 49 $$
$$y^2 + yz + z^2 = 64 $$

How do we calculate $$x + y + z$$

答:
There are solutions $(x,y,z)=(-5,0,8), \ (5,0,-8)$ having different sums $3,-3$, and two others I found on maple, numerically roughly $(-2.2,-3.5,-5.6)$ with sum around $-11.35$, and the same with all positive signs. So my guess is there's not a clever way to get at the sum, however the thing is at worst quadratic for the third answer.

ADDED: By subtracting equations and factoring, one can get three expressions for $x+y+z$ of the type constant over a difference of two of the coordinates. Setting these equal in pairs one finds that either some two of the varibles are equal (I didn't check that leads nowhere), or else in each case the same relation $8x-13y+5z=0$ results. Solving this for $y$ and plugging into the equations, and messing around a bit, leads to an equation for $z$ which factors as $(z-8)(z+8)(129z^2-4096).$ So besides the two integer solutions there may be others for which $z=\pm (64/\sqrt{169})$.

Going back to the other equations then gives $(x,y,z)=(25,40,64)/\sqrt{129}$ [notation meaning deivide each by the radical] as a solution, as well as its opposite obtained by making all signs negative. Looking at other combinations for values of $x,y$ from the equations (given this $z$) did not produce other solutions, in agreement with what maple found.

Title: A simple 2 grade equations system


问:
How to find $\left(\sqrt{\theta^2+1}\right)\sin \beta$ when a rotating wheel goes up in an incline?
The problem is as follows:

A wheel of radius $r$ given in inches starts moving over an incline from point $O$ which is a tangential point. The wheel moves up in the incline due an exterior force. Assuming that the wheel stopped at point $T$ as indicated in the figure and the angle swept by the wheel is $\theta$.

Find:

$$\left(\sqrt{\theta^2+1}\right)\sin \beta$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\cos\alpha-\theta\sin \alpha\\
2.&\sin \alpha+\theta\cos \alpha\\
3.&\cos\alpha+\theta\sin \alpha\\
4.&\sin\alpha - \theta \cos\alpha\\
\end{array}$

What's exactly the right way to approach this problem?
The part which confuses me the most is how to use the information of the angle $\beta$ from the drawing?

The reason for that is angle $beta$ is in the dotted line but not in the slope of the incline. Thus this is confusing.

It seems that the intended strategy is to use the radius of the wheel as an auxiliary variable to get the whole thing in terms of $\alpha$.

But how to do this methodically?. Can someone help me here?. What's exactly the right way to do this without much fuss?.

答:
As the wheel swept an angle $\theta$, $OT = r\theta$.

Now draw perpendiculars, $PT \perp OT$, $PR \perp OR$ .

We can observe that the angles $\angle POR = \pi - \beta$ and $\angle TPS = \alpha$.

The hypotenuse $OP = \sqrt{r^2+(r\theta)^2} = r\sqrt{1+\theta^2}$.

From triangle, PTS

$PT = r$, $ST = r\tan \alpha$, $PS = r\sec\alpha$

So, $OS = r\theta - r\tan\alpha$

Now

$\begin{align}\sin(\beta) &= \sin(\pi-\beta) = \frac{PR}{OP} = \frac{PS + SR}{OP} = \frac{PS + OS\sin\alpha}{OP}\\& = \frac{r(\sec\alpha+(\theta - \tan\alpha)\sin\alpha)}{r\sqrt{a+\theta^2}} = \frac{\theta\sin\alpha + \frac1{\cos\alpha} - \frac{\sin^2\alpha}{\cos\alpha}}{\sqrt{1+\theta^2}} \\&=\frac{\theta\sin\alpha + \cos\alpha}{\sqrt{1+\theta^2}} \\
\text{Thus }&\boxed{\sqrt{1+\theta^2}\sin(\beta) = \cos\alpha+\theta\sin\alpha}
\end{align}$


问:
Perpendicular Bisector of the segment joining
Find the perpendicular bisector of the segment joining $A(1,2)$ $B(1,-4)$.
In this question, first I found the midpoint of $[AB]$ $M(1,-1)$
Then I found $N= B-A = (0,-6)$. Then we may take $u=(6,0)$.
vector equation $P= M + tu$

$P= (1,-1)+t(6,0)$
$P= (6t+1, 0t-1)$
In here, $x=6t+1 y=0t-1$
Here I wanted to make the t equal to each other, but this is not possible. I will be grateful if you could help me.

答:
Think about the problem geometrically, which is always a good idea before jumping to any equation-solving.

The two points given lie on the same (vertical) line $x=1$, with midpoint $M(1,-1)$, as you gave. Hence, the perpendicular line will be parallel to the $x$-axis, which must hence be $y=-1$.

You can see this too in your set of points $P(6t+1,-1)$, which cover every possible $x$-coordinate but always have $y$-coordinate $-1$.


问:
Please help me to prove this inequality
Let $a$ and $b$ be positive real numbers. Prove that
$$\frac{(a+b)^3}{a^2b}\geq \frac{27}{4}$$

答:
Use $AM \ge GM$ for $a/2,a/2,b$
$$\implies \frac{a/2+a/2+b}{3} \ge (\frac{a}{2} \frac{a}{2} b)^{1/3}$$
$$\implies \left(\frac{a+b}{3}\right)^3\ge \frac{a^2b}{4}\implies \frac{(a+b)^3}{a^2b}\ge \frac{27}{4}. $$
the equality holds when $a=2b$.

Title: Please help me to prove this inequality


问:
Proving a certain theorem in L1 logic.
Description of Logic $L_1$:

Primary connectives: $\lnot$ , $\lor$

$A\rightarrow B$ is $\lnot A \lor B$

Axioms:

(A1) $B \lor B \rightarrow B$

(A2) $B \rightarrow B \lor C$

(A3) $B \lor C \rightarrow C \lor B$

(A4) $(C \rightarrow D ) \rightarrow (B \lor C \rightarrow B \lor D)$

The only rule of inference is $MP$

With these , I was able to prove the following theorem about $L_1$

a. $B \rightarrow C \vdash_{L_1} D \lor B \rightarrow D \lor C$

b. $\vdash_{L_1} ( B \rightarrow C ) \rightarrow ((D \rightarrow B) \rightarrow (D \rightarrow C))$

c. $D \rightarrow B , B \rightarrow C \vdash_{L_1} D \rightarrow C$

d. $\vdash_{L_1} B \rightarrow B$

e. $\vdash_{L_1} B \lor \lnot B$

f. $\vdash_{L_1} B \rightarrow \lnot \lnot B$

g. $\vdash_{L_1} \lnot B \rightarrow (B \rightarrow C)$

h. $\vdash_{L_1} B \lor (C \lor D) \rightarrow ((C \lor (B \lor D)) \lor B)$

But I got stumped at the next question:

i. $\vdash_{L_1} ((C \lor (B \lor D)) \lor B) \rightarrow (C \lor (B \lor D))$

I tried every theorem and axiom relentlessly (except the ones with negation) to prove this theorem.My best guess is that it is either so easy that I am missing out. Or , it imvolves theorems with negation.

Can someone give me a clue about this?

答:
We have to use (A4) to "import" all the needed transformations:

$(D \lor B) \to (B \lor D)$ --- (A3)

Thus, using (A4), we have:

$[C \lor (D \lor B)] \to [C \lor (B \lor D)]$.

Consider now: $[(C \lor D) \lor (B \lor B)]$. By (A1) using (A4) we get:

$[(C \lor D) \lor (B \lor B)] \to [(C \lor D) \lor B]$

We need also the Associativity Lemma:

$\vdash [(A \lor B) \lor C)] \to [A \lor (B \lor C)]$ to get, by (b):

$[(C \lor D) \lor (B \lor B)] \to [C \lor (D \lor B)]$,

Using 2) above and (b) again we get:

$[(C \lor D) \lor (B \lor B)] \to [C \lor (B \lor D)]$.

Now, the same approach must be used to derive:

$[(C ∨ (B ∨ D)) ∨ B] \to [(C \lor D) \lor (B \lor B)].$

Proof of Associativity:

Lemma 1: if $\vdash A \to C$ and $\vdash B \to C$, then $\vdash A \lor B \to C$.

$B \to C$

$(A \lor B) \to (C \lor A)$ --- from 1) and (A4) with (A3), by MP

$A \to C$

$(C \lor A) \to (C \lor C)$ --- as above

$(A \lor B) \to C$ --- from 2) and 4) and (A1) using (b).

Lemma 2: if $\vdash A \to B$, then $A \to (C \lor B)$ and $A \to (B \lor C)$

$A \to B$

$A \to (B \lor C)$ --- from 1) and (A2) using (b)

$A \lor (C \lor B)$ --- from 2) and (A3).

Lemma:

$A \to [A \lor (B \lor C)]$ --- (A2)

$B \to (B \lor C)$ --- (A2)

$B \to [A \lor (B \lor C)]$ --- from 2) by Lemma 2

$(A \lor B) \to [A \lor (B \lor C)]$ --- from 1) and 3) by Lemma 1

$C \to  [A \lor (B \lor C)]$ --- from $\vdash C \to C$ [your (d)] and Lemma 2 twice.

$\vdash [(A \lor B) \lor C] \to [A \lor (B \lor C)]$ --- from 4) and 5) by Lemma 1

Title: Proving a certain theorem in L1 logic.


问:
Failing a basic integration exercise; where did I go wrong?
(This is a basic calculus exercise gone wrong where I need some feedback to get forward.)

I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?

Here are my steps:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$

Applying integration by parts formula

$$\int uv' \, dx = uv - \int u'v \, dx$$

where $u=\sqrt{x}$, $v'=\sin\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$ and $v=-\cos\sqrt{x}$, resulting in:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$

Next, I tried substituting $x$ with $g(t)=t^2$ in the remaining integral, i. e. replacing each $x$ in the integral with $t^2$ and the ending $dx$ with $g'(t)=2t\,dt$. I would later bring back the $x$ by substituting $t=\sqrt{x}$ after integration. Continuing from where we left by substituting:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
$$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$

Then I pulled out the constant multipliers from the integral:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

which turned out to eliminate each other (resulting in just $\cdot1$), so we end up with:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

Reducing the integrand by reducing $\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$, which again is eliminated by multiplying with the integrand's $t$, and reducing $\cos\sqrt{t^2}\Rightarrow\cos t$. Therefore resulting in:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$

where the integral can be solved as $\int \cos t \, dt = \sin t + C$. So now we are at:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$

The correct answer however is
$$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$

So something somewhere in my process went horribly wrong. What?

答:
After the substitution $u = \sqrt{x}$, the integral becomes $2\int u^2 \sin(u) \ du$ as Adam Rubinson has said. We can proceed using tabular integration:

$$\begin{array}{c|c} u^2  & \sin(u) \\ \hline 2u & -\cos(u) \\ \hline \ 2 & -\sin(u) \\ \hline \ 0 & \cos(u)\end{array}$$

where we differentiate on the left and integrate on the right, and multiply the terms diagonally (the signs alternate).

Hence $2\int u^2 \sin(u) \ du = 2 \left(-u^2 \cos(u) - -2u \sin(u) + 2\cos(u) \right) + C$, which is:

$$2 \left( -x \cos( \sqrt x )+2\sqrt x \sin(\sqrt x) + 2\cos(\sqrt x)\right)+C = 4 \sqrt{x} \sin(\sqrt x) - 2(x-2)\cos (\sqrt x) +C$$


答:
Please note the straight line going through points $A$ and $B$ is parallel to $y-$ axis which is clear from the fact that $x$ coordinate is $1$ for both points $A$ and $B$. So the line perpendicular to it will be parallel to $x$ axis which means $y$ remains constant and is not a function of $t$. In this case, $y = -1$.

So points on the perpendicular bisector is given by  $(6t+1, -1)$. So your working is correct and $y$ coordinate cannot be written as a function of $t$.

Title: Perpendicular Bisector of the segment joining

Body:
Find the perpendicular bisector of the segment joining $A(1,2)$ $B(1,-4)$.
In this question, first I found the midpoint of $[AB]$ $M(1,-1)$
Then I found $N= B-A = (0,-6)$. Then we may take $u=(6,0)$.
vector equation $P= M + tu$

$P= (1,-1)+t(6,0)$
$P= (6t+1, 0t-1)$
In here, $x=6t+1 y=0t-1$
Here I wanted to make the t equal to each other, but this is not possible. I will be grateful if you could help me.


问:
Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:
Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:

$|Z(G/Z(G))| = 1$

$G/Z(G)$ has a subgroup of order 17

$G$ has a subgroup of order 34

I am preparing for my Algebra exam and I have been stuck on this question for a bit.

1.
So the only thing I know is that $|Z(G/Z(G))| = 1$, which means that $Z(G) \in G/Z(G)$ is the only element for which $A \cdot Z(G) = Z(G) \cdot A$ for every $A \in G/Z(G)$. But I don't know how to prove that there are no other subgroups for which that holds. At first I thought it might be a general property, but apparently it isn't (according to this example: $Z(G/ Z(G)) = 1$ Prove or disprove.).

So I know that it has to do something with the order of the groups, but I don't know how to use the information I have to get the conclusion. So far what I know is that with Lagrange I get $|G/Z(G)| = 102/2 = 51 = 3 \cdot 17$. I see that both are prime but I don't know how this is supposed to help me any further.

2.
I thought of maybe using the fact that $|Cl(A)|$, for every $A \in G/Z(G)$, has to be a divisor of $|G/Z(G)|$, but then I have that $|Cl(A)|$ for every $A$ has to be either 3 or 17, but how to prove that there exists one of 17 I don't know. Or maybe I can use the Class equation but I'm not sure.
I unfortunately don't know how to get any further with this question.

3.
I suppose that this will follow from (1) and (2), since if $G/Z(G)$ has a group of order 17, since $|Z(G)| = 2$, it follows with Lagrange that $G$ has a subgroup of order 34.

Can someone please help me by giving hints that will help me solve this problem? This is a rather introductory course, I've seen other people solve this with 'Sylow theory' but we haven't seen that yet in this course.

答:
A proof for 1) without Sylow: Note that for any Group $H$ if $H/Z(H)$ is cyclic then $H$ is abelian.

So if $|Z(G/Z(G))|>1$ we get $[G/Z(G):Z(G/Z(G))]=1,3$ or $17$. By the above remark $G/Z(G)$ will be abelian. Since $|G/Z(G)|=3\cdot 17$ it follows that $G/Z(G)$ is cyclic and hence $G$ is abelian, a contradiction.

Since $|Z(G)|=2$ claim 3) will follow from 2).

For 2) notice that by Cauchy's theorem there is an element and therefore also a subgroup of order $17$ in $G/Z(G)$.

Edit: A self contained proof that $G/Z(G)=:H$ contains a subgroup of order $17$ (similar to the proof of Cauchy's theorem): Let $x_1,\dots,x_r$ be a system of representatives for the non-trivial conjugacy classes of $H$ (i.e. those with more than one element). Then we have $$H=\underbrace{Z(H)}_{\{e\}}\cup\bigcup_{i=1}^r[x_i]$$ where $[x_i]$ denotes the conjugacy class of $x_i$ and the union is disjoint. This gives us the class equation: $$3\cdot 17=1+\sum_{i=1}^r |[x_i]|$$ Since $|[x_i]|$ divides $|H|$ and $|[x_i]|>1$ there has to be some $i$ with $|[x_i]|=3$. It follows that the centralizer $C_H(x_i)$ has order $17$ which is our subgroup of order $17$.


问:
Really easy algebra problem
Can somebody show me how to go about solving the following (really easy) equation for $\alpha$ please?
$$\displaystyle 0 = \sum_{n=1}^{N}\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$</p>

答:
$$\displaystyle 0 = \sum_n\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$
$$0 = \sum_n\left( \frac{y_n}{\alpha} \right) + \sum_n\left(\frac{1-y_n}{1-\alpha}\right)$$
$$0 = \frac{1}{\alpha} \sum_n y_n + \frac{1}{1-\alpha} \sum_n {(1-y_n)}$$

Then you can solve it.

Title: Really easy algebra problem


问:
Does this polynomial factorize further?
I just did a national exam and this question was in it; I am convinced this does not work:

Given that $(x - 1)$ is a factor of $x^3 + 3x^2 + x - 5$, factorize this cubic fully.

My attempt

```
1 | 1  3  1 -5
  |    1  4  5
  |____________
    1  4  5  0
```

$$(x - 1)(x^2 + 4x + 5) = 0$$

That's all I got. I gave up and couldn't factorize further. Am I missing something?

答:
You've gotten as far as you can in factoring:

$$x^3 + 3x^2 + x - 5 = (x - 1)(x^2 + 4x + 5)$$

When the discriminant is less than zero, as in the case of the quadratic $x^2 + 4x + 5$, there are no real roots. Recall, the $\color{blue}{\bf \text{discriminant}}$ is the portion of the quadratic formula inside the square root $$\sqrt{\Delta} =  \sqrt{\color{blue}{b^2 - 4ac}} $$

In your case, we have $a = 1, \; b = 4, \; c = 5,$ so $\Delta = b^2 - 4ac = 16 - 20 = -4 \lt 0$.

Whenever you need to factor a quadratic over the reals, take the time to check the value of the discriminant $\Delta$. If it evaluates to less than 0, don't waste  time trying to factor it, especially when you are taking an exam and time is limited!

Title: Does this polynomial factorize further?


问:
Find sequence of functions such that $||f_n||_1\to0 , ||f_n||_{\infty}\to \infty$
Is it possible to  Find sequence $\{f_n\}$ of functions such that $\underbrace{||f_n||_1}_{n \to \infty} $ tends to zero,and$ \underbrace{||f_n||_{\infty}}_{n \to \infty}$ tends to ${\infty}$?  Really I got stuck on this, an d I don not have a clue "how to start". I think about $$f_n=n^2\chi_{[0,\frac1n]}$$but I am not sure, that I am in a right way.Can someone please help me?

答:
Take $f_{n}=n χ[0,\frac{1}{n^{2}}]$.

Supremum of $f_{n}$ is n which goes to infinity.

The 1 norm is  $\frac{n}{n^{2}}$ which goes to 0.


问:
Different logical outcomes based on two similar statements with different bracketing
In trying to understand some predicate calculus notes, let the domain for $x,y$ be buildings, and the domain for $z$ be cities.

Let $H(x,y)$ signify that $x$ is higher than $y$

Let $O(x,y)$ stand for $x$ and $y$ are not the same building

Let $L(x,z)$ stand for $x$ is in $z$,

Then my prof states:

"A building is higher than every other building only if this building
is in Toronto"

He writes: this can be formulated as:

$1) \forall x (\forall y(O(x,y) \implies H(x,y)) \implies L(x, "Toronto"))$,

But it cannot be formulated as:

$2) \forall x \forall y ((O(x,y) \implies H(x,y)) \implies L(x, "Toronto"))$

In trying to understand this, I am trying to come up with an example where $1$, $2$ have different truth values for the same $x,y$ values since I believe this is the strategy to show that they are not logically equivalent, but I am stumped and need a hint to continue.

答:
In the second formulation, we are considering all of the pairs of buildings. What it says, is that for any two buildings in which one ($x$) is higher than the other ($y$), then the former building must be in Toronto.

In ($1$) however, we are restricting to the case where $x$ is a particular building that is higher than every other building.


答:
See Prenex Normal Form: $(\forall x\phi )\rightarrow \psi$ is equivalent to $\exists x(\phi \rightarrow \psi )$ under the assumption that $x$ is not free in $\psi$.

In your example, $y$ is not free in $L(x , \text {Toronto})$.

Why the above equivalence (in classical logic)?

Transform $(\forall x\phi )\rightarrow \psi$ into $\lnot (\forall x\phi )\lor \psi$ and then in $(\exists x \lnot \phi )\lor \psi$.

If $x$ is not free in $\psi$ we have $\exists x (\lnot \phi \lor \psi)$ and this in turn is $\exists x (\phi \to \psi)$.

Title: Different logical outcomes based on two similar statements with different bracketing

Body:
In trying to understand some predicate calculus notes, let the domain for $x,y$ be buildings, and the domain for $z$ be cities.

Let $H(x,y)$ signify that $x$ is higher than $y$

Let $O(x,y)$ stand for $x$ and $y$ are not the same building

Let $L(x,z)$ stand for $x$ is in $z$,

Then my prof states:

"A building is higher than every other building only if this building
is in Toronto"

He writes: this can be formulated as:

$1) \forall x (\forall y(O(x,y) \implies H(x,y)) \implies L(x, "Toronto"))$,

But it cannot be formulated as:

$2) \forall x \forall y ((O(x,y) \implies H(x,y)) \implies L(x, "Toronto"))$

In trying to understand this, I am trying to come up with an example where $1$, $2$ have different truth values for the same $x,y$ values since I believe this is the strategy to show that they are not logically equivalent, but I am stumped and need a hint to continue.


问:
Show that every totally bounded metric space is a bounded metric space
I might be overthinking this, but if a metric space is totally bounded, then the distance between any two point is less than epsilon for all epsilon. That implies that the distance between any two point is 0. Now does this mean that the space has only one point? Because according to the axioms of metric space, distance between two point is 0 iff they are the same point. So if the case has only one point, then it trivially bounded.</p>

答:
Total boundedness only tells you that there is an $\epsilon -$ net for each $\epsilon >0$. But the number of points in the net depends on $\epsilon$. So it doesn't follow that the space has only one point.

If $\{B(x_1,\epsilon), B(x_2,\epsilon),...,B(x_n,\epsilon)\}$ cover $X$ then, given any $x,y \in X$, we can pick $i,j$ such that $x \in B(x_i,\epsilon),$ and $y \in B(x_j,\epsilon)$. It follows that $d(x,y) \leq d(x,x_i)+d(x_i,x_j)+d(x_j,y)\leq 2\epsilon+\max \{d(x_p,x_q): 1\leq p,q \leq n\}$. So $X$ is bounded. But we have no control over the behavior of the bound as $\epsilon \to 0$.


问:
Using Quadratic Residue to show that $2^m-1$ doesn't divide $3^n-1$
Show that if $m$ is congruent to $n$ modulo $2$ and $m>1$ then
$2^m-1$ cannot be a divisor of $3^n-1$

What I have tried:

I showed that if $p$ is a divisor of $2^m-1$ then
if $m$ and $n$ are odd then $2$ and $3$ are quadratic residue of $p$ if $p$ is congruent to $1$ or $-1$ modulo $12$ but I cannot see how to proceed.

答:
With both $m$ and $n$ odd, since $m \gt 1$ so $2^m - 1 \gt 1$, for any prime factor $p \mid 2^m - 1$, since $p \mid 3^n - 1$ as well, then both $2$ and $3$ must be quadratic residues modulo $p$ (as you already stated). The table in the Law of quadratic reciprocity section shows $2$ is a quadratic residue mod $p$ if and only if

$$p \equiv 1, 7 \pmod{8} \tag{1}\label{eq1A}$$

and $3$ is a quadratic residue mod $p$ if and only if

$$p \equiv 1, 11 \pmod{12} \tag{2}\label{eq2A}$$

Since $11 \equiv 3 \pmod{8}$ and $13 \equiv 5 \pmod{8}$, combining the requirements in \eqref{eq2A} and \eqref{eq3A} gives

$$p \equiv 1, 23 \pmod{24} \tag{3}\label{eq3A}$$

Since $23 \equiv -1 \pmod{24}$, the product of any number of these factors will be $1$ or $-1$ mod $24$. However, for odd $m \gt 1$, you can easily show, such as by induction using $(2^2)(8) \equiv 8 \pmod{24}$, that

$$2^m - 1 \equiv 7 \pmod{24} \tag{4}\label{eq4A}$$

Thus, there must be at least one prime factor of $2^m - 1$ which is not congruent to either $1$ or $-1$ modulo $24$. This therefore means

$$2^m - 1 \not\mid 3^n - 1 \tag{5}\label{eq5A}$$


问:
Algebraic closure of field extension
Easy question on field theory. I imagine this has already been asked before but I couldn't find it on the search. Feel free to redirect me in case.

Let's consider a field $K$ and one algebraic closure $\overline{K}$. Then for any algebraic extension $L$ of $K$, $\overline{K}$ is the algebraic closure of $L$.

I feel this is is true. By algebraically extending $K$ we "add" roots of polynomials with coefficients in $K$ and these roots must lie in $\overline{K}$ by definition.
However I'm not sure if by allowing coefficients in the extension $L$ one could get roots outside of $\overline{K}$. As I said, this sounds unlikely to me but I can't prove it.

Thank you for any help.

答:
You have to be a bit careful.  If $\overline{K}$ is an algebraic closure of $K$, and $L$ is an algebraic extension of $K$, it is not necessarily true that $L$ is contained in $\overline{K}$.  However, $L$ is isomorphic to a (not necessarily unique) subfield of $\overline{K}$.  But if I understand your question correctly, you are assuming that $L$ is an algebraic extension of $K$ which is already contained in $\overline{K}$, and you then want to show that $\overline{K}$ is an algebraic closure of $L$.

To say that $\overline{K}$ is an algebraic closure of $L$ is to say that $\overline{K}$ is algebraic over $L$, and if $f \in L[X]$ is any polynomial, then all of its roots lie in $\overline{K}$.  Since $\overline{K}$ is algebraic over $K$, it is definitely algebraic over $L$.  Let's show the second claim, that all the roots of $f$ lie in $\overline{K}$.

Since $L[X] \subset \overline{K}[X]$, we can think of $f$ as a polynomial in $\overline{K}[X]$, and let $E$ be its splitting field over $\overline{K}$.  That is, $E$ is an algebraic extension of $\overline{K}$, $f$ has all its roots in $E$, and $E = \overline{K}(\alpha_1, ... , \alpha_n)$, where $\alpha_1, ... , \alpha_n$ are all the roots of $f$ in $E$.

Now each field $\overline{K}(\alpha_i)$ is algebraic over $\overline{K}$.  Since $\overline{K}$ is also algebraic over $K$, the transitivity property of algebraic extensions implies that $\overline{K}(\alpha_i)$ is an algebraic extension of $K$.  Therefore $\alpha_i$ is algebraic over the field $K$.  Since $\overline{K}$ is an algebraic closure of $K$, this forces us to have $\alpha_i \in \overline{K}$.  This is what we wanted to show.

Title: Algebraic closure of field extension


问:
Second theorem of Ascoli: is it true?
I've read a paper, and in the proof of Theorem 2, the authors cited a theorem of Ascoli (as the second theorem of Ascoli).
I'd like to see the full statement of the cited theorem, but the reference in the paper is written in (maybe) French so I cannot demonstrate it.

Can anybody help me?

Is it true, and can I find a reference (written in English)?:
pointwise convergence of real-valued equicontinuous function sequence $(f_j)_{j \geq 1}$ defined on $X$ to a target function $f$ implies that uniform convergence to the same function on the closure of $X$.

答:
Here is a translation of the theorem cited in Laurent Schwartz's book :

Theorem : Let $E$ be a topological space and $F$ a semi-metric space. On an equicontinuous set $\mathcal{F}$ of functions from $E$ to $F$, the semi-metric structures of the pointwise convergence on a dense subset $E_0 \subset E$, of the pointwise convergence on $E$, and of the uniform convergence on every compact of $E$, are uniformly equivalent (which means that the corresponding uniform structures are the same, and in particular the corresponding topologies are the same).

(and because it is written by Laurent Schwartz, I guess you can assume it is true...)

Title: Second theorem of Ascoli: is it true?


问:
A question based on intermediate fields , separable elements , inseparable elements
Please consider the following question:

Let F be an algebraic extension field of K , S the set of all elements of F
which are separable over K, and P the set of all elements of F which are purely inseparable
over K .Let E be an intermediate field. Then , if  $E  \cap S = K  $ then show that $E\subset P$.

using the given information I can see that $ K \subset S$ and $ K \subset E$ but I am unable to see how P will co relate to E and unable to do so which is necessary.

SO, Can you please help me with this?

答:
The statement is trivial in characteristic zero so we may assume that $\operatorname{char} K=p$. Take any $a\in E$. then we can write the minimal polynomial of $a$ over $K$ as $g(x^{p^r})$ where $g\in K[x]$ is irreducible and separable. Hence $a^{p^r}$ is a zero of $g$ and therefore separable. But by assumption $E\cap S=K$, so we get $a^{p^r}\in K$. Therefore $a$ is purely inseparable.

Note that this proof actually shows a bit more, namely that if $E\setminus K$ contains only inseparable elements, then it is already a purely inseparable extension.


问:
How do I know when a form represents an integral cohomology class?
Suppose $M$ is an $n$-dimensional manifold, and $\omega \in \Omega^p(M)$ is a closed $p$-form. Moreover, assume that $d\omega = 0$, so that it represents a de Rham cohomology class.

I would like to understand the meaning of the following sentence that I read in a book about geometric quantisation: $\omega$ represents an integral cohomology class.

Here's what I thought: we have the Poincaré isomorphism $\Phi : H_{dR}^p(M) \rightarrow {H_p}(M, \mathbb{R})^*$, where:

$$H_{dR}^p(M) = \text{$p$th de Rham cohomology group;}$$
$${H_p}(M, \mathbb{R}) = \text{$p$th differentiable singular homology group;}$$

and the star denotes the dual. This isomorphism is given by integration of the given form over the corresponding $p$-cycle.

Now, since we have a natural inclusion ${H_p}(M, \mathbb{Z}) \subseteq {H_p}(M, \mathbb{R})$, I could say that $\omega \in \Omega^p(M)$ represents an integral cohomology class when $\Phi(\omega) \in {H_p}(M, \mathbb{Z})$. But this is basically saying that the integral of $\omega$ over any differentiable $p$-cycle is an integer. Surely this cannot hold, except for trivial cases. So what does it mean to represent an integral cohomology class?

I would also like to know: when does the volume form of a closed manifold represents an integral cohomology class, and why?

Thank you.

答:
On $M = \mathbb{R}^2 \setminus \{ (0, 0) \}$ consider the differential form

$$\omega = \frac{1}{2\pi} \frac{x \, dy - y \, dx}{x^2 + y^2}.$$

This is a closed $1$-form. Its integral over a path in $M$ is its winding number; in particular, it is always an integer.

For nontrivial examples of integral $2$-forms you just need to find examples of nontrivial complex line bundles. More precisely, any such form is the curvature form of a connection on a complex line bundle $L$, and the corresponding integral cohomology class is the first Chern class $c_1(L)$ of $L$. This is a special case of Chern-Weil theory. This example comes up in physics when discussing Dirac's quantization argument; see, for example, this paper by Bott.


问:
Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$?
Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$

Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?

I start with $AM-GM$

So $A \geq 8xyz$

Then I need to prove $xyz\geq 1$ right?

How to do that and if it's a wrong way, please help me to solve it !!!

答:
With Lagrange multipliers: consider $f(x,y,z)=(x^2+1)(y^2+1)(z^2+1)$ and
$$
L(x,y,z,t)=f(x,y,z)-t(x+y+z-3)
$$
Then you have
\begin{align}
\frac{\partial L}{\partial x}&=2x(y^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2y(x^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2z(x^2+1)(y^2+1)-t
\end{align}
These should vanish together, so we obtain, for instance,
$$
2x(y^2+1)(z^2+1)=2y(x^2+1)(z^2+1)
$$
that yields $xy^2+x=x^2y+y$, hence $xy(x-y)=x-y$. Thus either $xy=1$ or $x=y$. Similarly, $xz=1$ or $x=z$. Thus we have four cases to examine.

The case $x=y$ and $y=z$ yields $x=y=z=1$.

The case $xy=1$ and $xz=1$ yields $y=z$ and so $x=2,y=1/2,z=1/2$.

Since the function $f$ as well as the constraint are symmetric in $x,y,z$, the remaining two cases will provide $x=1/2,y=2,z=1/2$ and $x=1/2,y=1/2,z=2$.

We have $f(1,1,1)=8$ and $f(2,1/2,1/2)=125/16$.


答:
The formula is homogeneous, so with $t:=\dfrac ba$,

$$\frac{(t+1)^3}{t}\ge\frac{27}4.$$

You can compute the minimum by differentiating.

The numerator of the derivative is $2t^3+3t^2-1=(t+1)^2(2t-1)$. The positive root  yields the value of the minimum, $\dfrac{27}4$.

Title: Please help me to prove this inequality

Body:
Let $a$ and $b$ be positive real numbers. Prove that
$$\frac{(a+b)^3}{a^2b}\geq \frac{27}{4}$$


问:
Finding a generalized inverse
I am trying to find a generalized inverse for all real $n \times n$ matrices such that it is 1 on the diagonal as well as 1 on the two entries directly above the diagonal, but 0 otherwise. Is there a "slick" way of doing it or does it amount to calculating until a pattern if found?

答:
Yes, denote $J$ the matrix with 1 above the diagonal and $0$ otherwise.
$$X=I+J+J^2$$ $$X^{-1}=(I+J+J^2)^{-1}=(I-J)(I-J^3)^{-1}=(I-J)(I+J^3+J^6+\cdots)=I-J+J^3-J^4+J^6-J^7+\cdots$$
Note that $J^n=0$. The sum above has only finite terms which is not $0$.
Wish it helps.

Title: Finding a generalized inverse


问:
Is the set of surjective functions from $\mathbb{N}$ to $\mathbb{N}$ uncountable?
I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form $\Bbb{N} \to \Bbb{N}$ is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function $p$.The problem that I'm having is that I don't know in which way to change the diagonal elements so that $p$ should, but cannot be in the matrix.

Can someone help me, or at least give me some tips?

答:
Here is an idea for how to use the diagonalization method after all.

Suppose $f_n$ is a list of surjections. We define $g$ in the following way:

$$g(k)=\begin{cases}\frac k2 & k\text{ is even}\\1+\sum_{m\leq k}f_m(k) &\text{otherwise}\end{cases}$$

So $g$ is clearly surjective (its restriction to the even numbers is surjective). To see that $g$ is not one of the $f_n$'s, we need to show that whenever $n\in\Bbb N$ there is some $k$ such that $f_n(k)\neq g(k)$.

Let $n\in\Bbb N$, and let $k$ be any odd number such that $n\leq k$, then we have that $$f_n(k)\leq\sum_{m\leq k}f_m(k)<1+\sum_{m\leq k}f_m(k)=g(k).$$

Therefore $f_n(k)\neq g(k)$ so $f_n\neq g$, and so we have that $g$ is not on the list.

If you are willing to use something other than diagonalization then we can give the following argument:

For every infinite subset of $\Bbb N$, we can write the following surjection:

$$f_A(n) =|\{a\in A\mid a<n\}|$$

Then $f_A$ is surjective because $A$ is infinite, so for every $n$ we can find $k$ such that $f_A(n)=k$.

Note that if $A\neq B$ then $f_A\neq f_B$: if $n=\min A\triangle B$, and suppose that $n\in A$, then $f_A(n+1)>f_B(n+1)$.

Therefore there are at least $2^{\aleph_0}$ surjections. Moreover all of them are non-decreasing!


答: You can use the classical method consisting in writing $A=I+N$ where $I$ stands for the identity and $N$ is nilpotent or order $n$ (check it, $N^{n}=0$). Then, Use Newton's binomial expansion on $A^{n}$ to obtain a formula for the inverse by gathering all powers of $A$ on one side and identity on the other side.

Title: Finding a generalized inverse

Body: I am trying to find a generalized inverse for all real $n \times n$ matrices such that it is 1 on the diagonal as well as 1 on the two entries directly above the diagonal, but 0 otherwise. Is there a "slick" way of doing it or does it amount to calculating until a pattern if found?


问:
Smooth 1-forms and smooth sections
The following statement is from the book introduction to manifolds by Loring Tu.

" We define a 1-form $\omega$ on a manifold M to be smooth if $\omega : M \longrightarrow T^*M $ is smooth as a section of the cotangent bundle $\pi : T^*M \longrightarrow M $ "

Well I know what is the definition of a smooth map between manifolds, but never knew about  a map which is smooth "as a section" of a fiber bundle!  Could you please explain to me this definition.

I'm just a begginer in differential geometry and while trying to teach my self more about it, I've notice the large use of the expression "smooth sections" but I'm not aware of its importance and why do people emphasis on  the property of smootheness of sections ?

Thanks a lot!

答: You might not be familiar with the concept of vector bundles but this is related to it. So when you have the cotagent bundle or the tangent bundle we have a canonical smooth map $\pi : T^*M\rightarrow M$ and a smooth section will be a smooth map $s:M\rightarrow T^*M$ such that $\pi\circ s=id$, which is exactly saying that at each point we get an element in $T_p^*M$.

Title: Smooth 1-forms and smooth sections


答:
Smooth sections are defined to be maps $\sigma : M \to E$ between the base space of a smooth fiber bundle (which is a smooth manifold) and the total space of the bundle (which is another smooth manifold) which are smooth as maps in the usual sense, and satisfy some particular properties. In other contexts you might want to restrict to bundles and manifolds of lower regularity, maybe $C^k$ or even $C^0$, and in that case the sections will be $C^k$ or $C^0$ as well. So their regularity is really just a sanity check (in the world of smooth manifolds, everything had better be smooth!) – what really matters is the fundamental property of "preserving fibers",
$$\pi \circ \sigma = \operatorname{id}_M $$
which distinguishes sections from more general maps $M \to E$.

Title: Smooth 1-forms and smooth sections

Body:
The following statement is from the book introduction to manifolds by Loring Tu.

" We define a 1-form $\omega$ on a manifold M to be smooth if $\omega : M \longrightarrow T^*M $ is smooth as a section of the cotangent bundle $\pi : T^*M \longrightarrow M $ "

Well I know what is the definition of a smooth map between manifolds, but never knew about  a map which is smooth "as a section" of a fiber bundle!  Could you please explain to me this definition.

I'm just a begginer in differential geometry and while trying to teach my self more about it, I've notice the large use of the expression "smooth sections" but I'm not aware of its importance and why do people emphasis on  the property of smootheness of sections ?

Thanks a lot!


问:
Trace of a matrix using property
If

${P_1} = I = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right];{P_2} = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&0&1\\
0&1&0
\end{array}} \right];{P_3} = \left[ {\begin{array}{*{20}{c}}
0&1&0\\
1&0&0\\
0&0&1
\end{array}} \right];{P_4} = \left[ {\begin{array}{*{20}{c}}
0&1&0\\
0&0&1\\
1&0&0
\end{array}} \right];{P_5} = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
1&0&0\\
0&1&0
\end{array}} \right];{P_6} = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]$

and

$X = \sum\limits_{i = 1}^6 {{P_i}\left[ {\begin{array}{*{20}{c}}
2&1&3\\
1&0&2\\
3&2&1
\end{array}} \right]{P_i}^T} $

where $P_k^T$ denotes the transpose of the matrix $P_k$. Then which of the following options is are correct ?

(A) $X\left[ {\begin{array}{*{20}{c}}
1\\
1\\
1
\end{array}} \right] = \alpha \left[ {\begin{array}{*{20}{c}}
1\\
1\\
1
\end{array}} \right]$, then $\alpha =30$

(B) X is symmetric Matrix

(C) The sum of diagonal entries of X is 18

(D) $X-30I$ is an invertible matrix

The official answer is A,B,C . I have solved the option A,B and D and have proved that A&B are correct and D is incorrect.

For Choice C I need some help

$X = {P_1}Q{P_1}^T + {P_2}Q{P_2}^T + {P_3}Q{P_3}^T + {P_4}Q{P_4}^T + {P_5}Q{P_5}^T + {P_6}Q{P_6}^T$

${P_1} = I = {P_1}^T$

$X = \underbrace Q_{Traace = 3} + {P_2}Q{P_2}^T + {P_3}Q{P_3}^T + {P_4}Q{P_4}^T + {P_5}Q{P_5}^T + {P_6}Q{P_6}^T$

In most of the solution they only solved $Trace(P_1QP_1^T)=3$ and they multipled by 6 to get the answer 18. How do we solve it using property.

答:
For each given $P_i$, pairwise dot product of column vectors (or row vectors) is zero and magnitude of each column (or row) vector is $1$. Thus each $P_i$ is orthogonal matrix.

$$P_iP_i^{T}=I$$

Using the hint given by @macton in comments,

$$\text{Tr}(P_iQP_i^{T})=\text{Tr}(QP_iP_i^{T})=\text{Tr}(Q) $$
$$\therefore \text{Tr}(X)=6 \, \text{Tr}(Q)=18$$

Title: Trace of a matrix using property


问:
Which of the following is/are correct? Related to Limits, Differentiability, Continuity
Which of the following is/are correct?

$(A)$ If a function is differentiable then its derivative is a continuous function.

I couldn't find any example which makes derivative, not continuous. So I need help here.

$(B)$ If LHD and RHD does not exist finitely at $x=a$ function is discontinuous at $x=a$

Taking the example of $y=|x|$ it is not differentiable at $x=0$ but is continuous. So this option is incorrect.

So Differentiability $\rightarrow $Continuity but (Not Differentiability) $\nrightarrow$ (Not Continuity )?

$(C)$ Every function which is defined for all $x$  can be thought of as a derivative of some function.

$(D)$Every continuous function(need not to be differentiable) has an antiderivative.

I couldn't find any counter-example for the last two as well, makes them correct or I lack some knowledge please clear my doubts regarding these options.

答:
The standard example for (A) is
$$
f(x)=\begin{cases} x^2\sin(1/x) & x\ne0 \\[6px] 0 & x=0 \end{cases}
$$
The function is obviously differentiable for $x\ne0$, with
$$
f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}
$$
It is also differentiable at $0$, because
$$
\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}h\sin\frac{1}{h}=0
$$
However the derivative is not continuous at $0$.

Your example for (B) is incorrect, because your function has one-sided derivatives at $0$. A correct example would be $f(x)=\sqrt[3]{x}$.

For (C) one can consider Darboux's theorem, according to which, if a function is differentiable over an interval, its derivative satisfies the intermediate value property. A function that doesn't satisfy this property cannot be a derivative. Example: the sign function.

For (D), every function that is continuous over an interval has an antiderivative. This is precisely the so-called “fundamental theorem of calculus”.


问:
$P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$

Prove that $P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$, where $I$ is a countable index set.

$P$ is a probability measure, and we have generalized Boole's inequality, to begin with:
$$0\le P\left(\bigcup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty P(E_i)$$

Is the following proof okay?

$P(E_i) = 0$ for all $i\in I$ then $P(\cup_{i\in I}E_i)=0$ by Boole's inequality above. If $P(\cup_{i\in I}E_i)=0$ holds, then Boole's inequality is probably of no use. I tried a proof by contradiction. Let's suppose there exists $j\in I$ such that $P(E_j) \neq 0$. Then $P(\cup_{i\in I}E_i)=0$ is definitely absurd - but how do I put this in mathematical terms? $(\star \star \star)$

In the other case, we have $P(E_i^c) = 0$ for all $i\in I$, which tells us (with the help of the first part) that $$P\left(\bigcup_{i\in I}E_i^c\right)=0 \implies P\left(\bigcap_{i\in I}E_i\right)^c=0 \implies P\left(\bigcap_{i\in I}E_i\right) = 1$$ and the proof is complete.

If everything above sounds fine, I only need help with the line marked $(\star \star \star)$. Thanks!

答:
$P(A) \leq P(B)$ if $A \subseteq B$. [Because $P(B)=P(A)+P(B\setminus A)$]. Hence, $0\leq P(E_j) \leq P(\bigcup E_i)=0$ for each $j$.

Rest of your arguments are fine if $I$ is countable.

Counter-example when $I$ is uncountable: Let $P$ be Lebesgue measure on $(0,1)$. Let $E_x=\{x\}$ for all $x \in (0,1)$. Then $P(E_x)=0$ for ll $x$ but $P(\bigcup E_x)=1$.


问:
$C^r$ submanifold of $C^r$ manifold
Let $M$ be a $C^r$ manifold $r, \geq 1$, and $A \subset M$ a connected subset .Suppose that there is a $C^r$ retraction $f: M \rightarrow A$, i.e. $f|A=$ identity. Then $A$ is a $C^r$ manifold.

I'm having trouble understanding where the retraction plays avpart. Why can't I just take charts $(\varphi_i, U_i), (\varphi_j, U_j)$ of $M$, restrict to $A$, then we have $\varphi_j\varphi_i^{-1}$ is of class $C^r$ which is a chart of $A$ thus making $A$ a $C^r$ manifold?

答: Recall that in the definition of a submanifold you need that $A\cap U=\phi^{-1}(\mathbb{R}^k)$ for a fixed $k\in \mathbb{N}$, and so here is where we need the retraction and connected conditions, in general you will not have that this $k$ is fixed. And also recall that you need the charts to be homeomorphisms between open sets, if $A$ is not open you might not have that $\phi(U\cap A)$ is open in $\mathbb{R}^n$ , however if $A$ is open your reasoning works and you have that all open subsets are $n-$dimensional submanifolds.

Title: $C^r$ submanifold of $C^r$ manifold


问:
Isomorphism over $\mathbb{Q}$ but not isomorphism in $\mathbb{C}$
Sorry for my bad English.

Is there example as follows?

$Let$ $K,L$ be subfield of $\mathbb{C}$ such that $L\neq K$ as subfield of $\mathbb{C}$,
but $L\cong K$ as field (ignoring emmbding to $\mathbb{C}$) .

答: Take any $x,y\in \mathbb{C}$ which are transcendental over $\mathbb{Q}$. Then $\mathbb{Q}(x)\simeq \mathbb{Q}(y)$ are isomorphic to the field of rational fractions $\mathbb{Q}(X)$, but they are in general not equal.


问:
Area of an Ellipse using U-Substitution
I am supposed to determine the area of an ellipse using $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1.

I start by setting up an integral for the first quadrant, so I solved the above equation for y and got the integral $\int b\sqrt{1-\frac{x^2}{a^2}} $

The next step is to do a change of variables using u = $\frac xa$, and this is where I get a little confused.  I then have to finish up by using geometry to evaluate the integral.  Could someone explain how to do this? Thanks for any help in advance!

答:
I think I understand what you want, but if I am mistaken please tell me.

We currently have the integral
$$\int_0^a b\sqrt{1-\frac{x^2}{a^2}}dx$$
Make the substitution $u=\frac{x}{a}$. Hence $dx=a~du$. When $x=0$, $u=0$, and when $x=a$, $u=1$. Applying this to our integral we are left with
$$\int_0^1 ab\sqrt{1-u^2}~du=ab\int_0^1 \sqrt{1-u^2}~du$$
Let's focus on the expression
$$\int_0^1 \sqrt{1-u^2}~du$$
Using 'geometry', we can see that this is the area of a quarter-circle with radius $1$, ie it is equal to $\frac{\pi}{4}$. Hence one quarter of the area of an ellipse has area
$$\frac{\pi ab}{4}$$
so the total area of the ellipse is, by symmetry, equal to
$$\pi ab$$
Does that help? Please tell me if you have any questions.

Title: Area of an Ellipse using U-Substitution


问:
If $\operatorname{char} K =p$ and $[F:K]$ is finite and not divisible by $p$, then $F$ is separable over $K$
Consider the following question,

If $\operatorname{char} K =p$ and $[F:K]$ is finite and not divisible by $p$, then prove that $F$ is separable
over $K$.

I am really sorry but I am not able to use the given information to solve the problem. I am not able to deduce anything using $p$ doesn't divides $[F:K]$ .

It is my humble request to you to kindly give some hints.

I am following Abstarct Algebra by Joseph Gallian.

答:
Assume that $F/K$ is not separable.

There is $a\in F$ with $K$-minimal polynomial $f\in K[x]$ such that $K(a)/K$ is not separable, ie. $f'(a)=0$.

But then $f(a)=f'(a)=0$ means that $\gcd(f,f')\ne 1$, if also $\gcd(f,f')\ne f$ we get that $\gcd(f,f')$ divides $f$ which isn't irreducible. Therefore $f'(a)=f(a)=0$ and $f$ irreducible implies $\gcd(f,f')=f$ which implies $f'=0$ ie. $f \in K[x^p]$.

Whence $p$ divides $\deg(f)=[K(a):K]$ which divides $[F:K]$.


问:
Are singletons limit points?
So I just started learning topology and I came across this problem.

Consider the topological space $(X,\tau)$where the set $X=\{a,b,c,d,e\}$ and $\tau=\{X,\emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e\}\}$

Determine the limit points of

i) $\{a\}$

Is $a$ a limit point for itself since the singleton a is an open set in the topology.
From the definition of limit point of a set  every neighbourhood of the point must contain a point of the set that id  different from a however when considering the singleton there are no different points of a so what is the correct answer im very confused.

Thanks in advance

答: The set of limit points of $\{a\}$ is empty.

Title: Are singletons limit points?