问:
When existence of transfer map of algebraic $K$-theory implies rational injection.
Given a map of smooth projective varieties $f:X\rightarrow Y$ over fields, there is a projection formula in algebraic $K$-theory given by $f_*(\alpha.f^*(\beta))=\beta.f_*(\alpha)$. I was wondering whether this implies that algebraic $K$-group of $Y$ rationally injects into $X$?

The natural approach is to plug in $\alpha$ to be the class of trivial line bundle $[\mathcal{O}_X]$ in $K_0(X)$. This implies that $f_*f^*\beta=\beta.f_*([\mathcal{O}_X])$. So the problem is identifying $f_*([\mathcal{O}_X])$ and seeing whether multiplication by it is rationally an isomorphism or not. This leads to GRR and some complicated formulas which lead me to think this is not true in general. Is there a situation that it becomes true (other than the famous etale case)? Does the transfer induce rational injection if the map is finite?

This is I think true for finite maps. If the generic degree is $d$ then it seems for any line bundle $L$, according to some intersect theoretic facts, $f_*L$ has chern classes $d$ times image of chern classes under $f$ (as cycles). So for trivial line bundle it is not going to have any chern classes and its rank is going to be $d$. So multiplication by $f_*([\mathcal{O}_X])$ will be like multiplying with $d$ rationally.

I'm not sure whether this is true for ramified covers.

答:
I believe the following holds: If $X/\mathbb{Z}$ is of finite type and $X_t:= X\times_{\mathbb{Z}} \mathbb{F}_p$
is the fiber of $X$ at $t:=(p)\neq (0)$ it follows $K_m(X_t)_{\mathbb{Q}}=0$ is a conjecture, the "Parshin conjecture":

https://en.wikipedia.org/wiki/Parshin%27s_conjecture

There is the canonical pull back map

$$f^*: K_m(X)_{\mathbb{Q}} \rightarrow K_m(X_t)_{\mathbb{Q}}.$$

The group $K_m(X)_{\mathbb{Q}}$ is non-zero in general.

Question: "Given a map of projective varieties $f:X→Y$, there is a projection formula in algebraic K-theory given by $f∗(α.f∗(β))=β.f∗(α)$. I was wondering whether this implies that algebraic K-group of $Y$ rationally injects into $X$?"

Answer: Take $X:=\mathbb{P}^n_{\mathbb{Z}}$. You get the map

$$f^*: K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \rightarrow K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}$$

and the left side is non-zero, the right side is zero by the projective bundle formula:

$$K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}\cong K_*(\mathbb{F}_p)_{\mathbb{Q}}\otimes \mathbb{Q}[t]/(t^{n+1})=0$$

since $K_*(\mathbb{F}_p)_{\mathbb{Q}}=0$. It follows $K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}=0$.

We get

$$K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \cong K_*(\mathbb{Z})_{\mathbb{Q}}\otimes \mathbb{Q}[t]/(t^{n+1})$$

and $K_*(\mathbb{Z})_{\mathbb{Q}} \neq 0$ hence $K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \neq 0$.

Hence $f^*$ is not injective in general as asked in the question.

Example: (The "projective bundle formula") The projection map $\pi: X:=\mathbb{P}(E^*) \rightarrow S$

where $S \subseteq \mathbb{P}^n_k$ is a smooth projective variety over $k$,
is smooth and there is - again by the projective bundle formula - an injective map

$$ \pi^*: K_*(S)_{\mathbb{Q}} \rightarrow K_*(X)_{\mathbb{Q}}$$

but the pull back map is not an injection in general.

Example: (The disjoint union). Let $Y:= \cup_i X_i$ where $X_i \cong X$ for all $i$, with $X$ smooth and where the union is the disjoint union. You get a "canonical" injection map $j: X \rightarrow Y$, and the pull back map $j^*$ is the map

$$j^*: K_*(Y)_K \cong \oplus_i K_*(X)_K \rightarrow K_*(X)_K .$$

The map $j^*$ is the projection onto one of the direct summands, and hence $j^*$ is not an injection (here $K$ is any field).


问:
What are the dimensions of a 4D cube represented in 3D?
I'm hoping to construct a physical model of a 4D cube. However, I'm struggling to work out the proper size ratio between the inner and outer cubes. From the graphics I've seen, the inner cube seems about 1/3 the size of the outer, but I haven't been able to find any documentation for that. How would you go about working that out?

Thanks for your time.

答:
Go for what looks best. There is no mathematical reason for any particular size.

A 3D model of a 4D cube is a projection. It is like perspective in a photo or painting. Things that are further away are smaller, but depending on the distance of the observer, the relative sizes of two things in the painting at different distances can vary. Similarly, the "inner" cube of a classic hypercube projection is further away than the outer cube along the unavailable fourth dimension, and hence is depicted smaller. The size difference depends on how far away you as an observer are along that same dimension, so could be anything.

ContentLicense: CC BY-SA 4.0


问:
Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$
Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$

I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that

$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$

Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and  consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.

Update:

I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try  find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.

Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.

答: I hope following answer would help. The given series is an infinite arithmetic-geometric series.


问:
Functions similar to $f(x) = x^2 \sin{\frac{1}{x}}$
I am analysing this function: $f(x) = x^2 \sin{\frac{1}{x}}$

The specific feature of this function that I am interested in is the increasing smoothness as you move away from zero. Are there similar functions to this one exhibiting continuous increase/decrease in smoothness with the change in $x$?

答:
If by 'smoothness' you mean 'number of derivatives', the function $f(x)=x^2\sin(\frac{1}{x})$ has the same degree of smoothness in every point different from $0$.

If you were talking about those small sharp edges that the graph seems to plot when $x\in (5\cdot 10^{-4}, 0.0015)$, forget about them. They are not accurate. Those sharp edges appear as a consequence of the floating-point representation of the numbers in the computer. In brief, when a computer has to perform a calculation that deals with both 'large' and 'small' numbers, these inaccurate results may appear.

This exact behavior is happening with your function. If you consider $x\in (5\cdot 10^{-4}, 0.0015)$, we could say that $x$ is small(from a 32-bits-floating-point point of view). Hence, $\dfrac{1}{x}$ is large. This is the reason why the calculation involving the numbers in this range is not 100% accurate.

This behavior can be seen with other functions where large and small numbers are involved,  such as $$f(x)=\sin\left(\dfrac{1}{x^4}\right)$$ for large values of $x$, or $$f(x)=\left(9^{-9}\right)^{1/x}$$ for small values of $x$.


问:
Probability $k$ bins are non-empty
The following problem arises in the analysis of Bloom filters.

Consider $m$ bins and $N=nk$ balls placed uniformly and independently at random into the bins. A query chooses $k$ bins uniformly and independently at random and asks if they are all non-empty.  The main question is as follows.

What is the probability that all $k$ bins in the query are non-empty?

It is assumed that this probability will be a function of $k$, $m$ and $n$.

The second question is:

For what value of $k$ (which is a function of $m$ and $n$) is this probability minimized?

The standard version of the analysis taught the world over and reproduced in the wikipedia page linked above contains a "now the magic occurs" step which ignores the non-independence of the bins. It gives $k \approx \frac{m}{n} \ln{2}$ as the answer to the second question.

Is there a clean and rigorous way of doing this analysis correctly?

This is a repost of this one which hasn't had any good answers yet.

Clarification.

The difficultly arises because the probability that all $k$ bins are non-empty is not $(1-(1-\frac{1}{m})^{kn})^k$. The reason is that the probability that one of the $k$ (not necessarily distinct) bins that is queried is non-empty is dependent on whether the other queries return non-empty bins.

An approximate answer or good bounds will suffice.

答:
(Note: The analysis in this answer is for all $k$ bins we query being distinct. This is not what we care about in the case of Bloom filters, so I've left another answer with the right analysis. The stuff in this post is still useful, so I'm not deleting this answer yet.)

To avoid the incorrect independence assumption (though it turns out to be a good approximation), we can do the analysis in the following way.

Let $E_i$, for $1 \le i \le k$, be the event that the $i$th bin (out of the $k$ bins we're interested in) is empty after the $N = nk$ balls have been inserted. For $E_i$ to happen, each of the $N$ balls must have gone to any of the other $m - 1$ bins other than bin $i$. So
$$\Pr(E_i) = \left(1 - \frac1m\right)^N$$
What about $\Pr(E_i \cap E_j)$? For both those bins to be empty, all the $N$ balls must have gone to the $m - 2$ bins other than bin $i$ and bin $j$, so
$$\Pr(E_i \cap E_j) = \left(1 - \frac2m\right)^N.$$
Similarly, the intersection of some number (say $r$) of the $E_i$s has probability $\left(1 - \frac{r}m\right)^N$.

Finally, by the inclusion-exclusion principle, we have

$$\begin{align}
&\phantom{=}\Pr(\text{all $k$ bins are non-empty}) \\
&= 1 - \Pr(\text{at least one bin is empty}) \\
&= 1 - \Pr(\bigcup_{i=1}^k E_i) \\
&= 1 - \sum_{i}\Pr(E_i) + \sum_{i<j}\Pr(E_i \cap E_j) - \sum_{i<j<k}\Pr(E_i \cap E_j \cup E_k) + \dots \\
&= 1 - k\left(1 - \frac1m\right)^N + \binom{k}{2}\left(1 - \frac2m\right)^N - \binom{k}{3}\left(1 - \frac3m\right)^N + \dots
\end{align}\\
= \sum_{r = 0}^k \binom{k}{r}(-1)^r\left(1 - \frac{r}m\right)^N
$$
as the exact probability.

From this we can see that $\displaystyle (1 - e^{-N/m})^k$ is indeed a good approximation (optimizing this gives $k = \frac{m}{n}\ln 2$ as the best value of $k$), as it's equal to (by the binomial theorem; compare with the exact expression above):
$$\sum_{r = 0}^k \binom{k}{r}(-1)^r e^{-Nr/m}.$$

The two are close, term-for-term, because for large $N$, $\left(1 + \frac{x}{N}\right)^N \approx e^x,$ and so $\left(1 - \frac{r}m\right)^N = \left(1 + \frac{-Nr/m}{N}\right)^N \approx e^{(-Nr/m)}.$ Concretely, $\left(1 - \frac{r}m\right)^N$ and $e^{(-Nr/m)}$ are, respectively
$$1 - N\frac{r}{m} + \binom{N}{2}\left(\frac{r}{m}\right)^2 +  \binom{N}{3}\left(\frac{r}{m}\right)^3 + \dots$$
and
$$1 - N\frac{r}{m} + \frac{N^2}{2!}\left(\frac{r}{m}\right)^2 + \frac{N^3}{3!}\left(\frac{r}{m}\right)^3 + \dots.$$

ContentLicense: CC BY-SA 3.0

Title: Probability $k$ bins are non-empty


问:
Let $S$ be a spanning set of a vector $V$. If another vector of $V$ is added to $S$, can the new set still be spanning set of $V$?

Let $S$ be a spanning set of a vector $V$. If another vector of $V$ is added to $S$, can the new set still be spanning set of $V$? Justify your answers.

$\underline{Attempt}$

I think that is correct,

Let $S=\{ v_1,v_2,..,v_n \}$ is spanning set of vector space $V$

If another vector$(v_k)$ in $V$ added to the $S$,then $\bar{S}=\{ v_1,v_2,..,v_n,v_k \}$

since $S$ spanning set of vector space $V$

$v_k=\alpha_1v_1+\alpha_2v_2+...+\alpha_nv_n$ where $\{\alpha_1,\alpha_2 ...\alpha_n \}\in R$

so $\bar{S}=\{ v_1,v_2,..,v_n,\alpha_1v_1+\alpha_2v_2+...+\alpha_nv_n \} \subset \{ v_1,v_2,..,v_n \}$

thus new set still be spanning set of $V$.

Is it correct if not please give some hints?

答:
I will write $v_0$ for the new vector instead of $v_k$.

You did not prove what you are supposed to prove. You have to prove that any vector $v \in V$ is  a l.c. of vectors from $\overline S$. Since we can write $v=\sum\limits_{i=1}^{n} c_iv_i$ we can write $v=\sum\limits_{i=1}^{n} c_iv_i +0(v_0)$ and that finishes the proof.


问:
Homomorphism of $R$-modules gives a commutative diagram
I am working on Problem 2.10 of Atiyah & Macdonald's Introduction to Commutative Algebra.

In this problem, we have a commutative ring $A$ and a homomorphism $u : M \longrightarrow N$ of $A$-modules.

Let $a$ be an ideal of $R$. In the solution here, they claim that the following diagram is commutative :

It's clear that $u$ induces the quotient map $w$ and the restriction map $v$, but it's unclear to me how the author makes the claim that the diagram is commutative fairly quickly. How can I see this ?

Note : It is given that $N$ is finitely generated. It is also given that $a$ is an ideal contained in the Jacobson radical of $A$. However, I don't think that either of these assumptions affect the claim being made here.

Thanks!

答:
The practical answer is a brief chase of elements through the diagram. The abstract nonsense answer is also quick. Consider:

On the left, $d$ is defined to be the composite of the left and bottom maps. On the right, the universal property of quotients says there exists a unique $w$ which makes the triangle commute. Obviously that requires a check: you must show $d(\mathfrak{a}M) = 0$, or equivalently $u(\mathfrak{a}M) \subset \mathfrak{a}N$, which is clear. Saying both triangles commute is precisely saying the square commutes.


答:
I doubt there is a correct answer.

It is equivalent to drawing a cube in perspective on a flat piece of paper. Just do what feels good.

Title: What are the dimensions of a 4D cube represented in 3D?

Body:
I'm hoping to construct a physical model of a 4D cube. However, I'm struggling to work out the proper size ratio between the inner and outer cubes. From the graphics I've seen, the inner cube seems about 1/3 the size of the outer, but I haven't been able to find any documentation for that. How would you go about working that out?

Thanks for your time.


问:
Show basic open sets in $\{0,1\}^{A}$ are both open and closed
Consider the topological space $\{0,1\}^A$. The set of functions from an arbitrary index set $A$ to $\{0,1\}$. Then a subbasis for the product topology is $S=\{U(a,\delta)\mid a \in A,\delta \in\{0,1\}\}$ where $U(a,0)=\pi_a^{-1}(0),U(a,1)=\pi_a^{-1}(1)$. Show that every basic open set is both open and closed.

Attempt: A basic open set is of the form $$\pi_{a_1}^{-1}(b_1) \cap \dots \cap \pi_{a_n}^{-1}(b_n)$$

where $b_i \in \{0,1\},a_i \in A$. Then $$\{0,1\}^A-(\pi_{a_1}^{-1}(b_1) \cap \dots \cap \pi_{a_n}^{-1}(b_n))=\pi_{a_1}^{-1}(c_1) \cap \dots \cap \pi_{a_n}^{-1}(c_n)$$ where $c_i \neq b_i$. Another basic open set. So basic open sets are both open and closed.

答:
The argument is simple: the subbasis open sets $\pi_a^{-1}[\{0\}], \pi_a^{-1}[\{1\}]$ are clopen because $\{0\}, \{1\}$ are clopen in $\{0,1\}$.

So their generated base also consists of clopens sets as finite intersections of open sets are open and intersections of closed sets are closed. QED.


问:
Find the expectation and variance of X
A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}

Find the value of C.

Find the expectation and variance of X.

Find the expectation of random variable $Z=\frac{X}{2X+3}$

What I tried:

Finding C is straightforward as we just need to make sure that $\int^\infty_{-\infty}f(x)dx=1$.
\begin{align*}
                1&=\int^2_0C(x+\frac{3}{2})dx\\
                &=\left[\frac{C}{2}x^2+\frac{3}{2}Cx\right]^2_0\\
                &=\left[\frac{C}{2}(2)^2+\frac{3}{2}C(2)\right]-0\\
                &=5C\\
                &C=\frac{1}{5}
            \end{align*}

Using $\mathbb{E}[X]=\int^\infty_{-\infty}xf(x)dx$,
\begin{align*}
                \mathbb{E}[X]&=\int^2_0x\left(\frac{1}{5}\left(x+\frac{3}{2}\right)\right)dx\\
                &=\left[\frac{1}{15}x^3+\frac{3}{20}x^2\right]^2_0\\
                &=\left[\frac{1}{15}(2)^3+\frac{3}{20}(2)^2\right]^2_0-0\\
                &=\frac{17}{15}
            \end{align*}
Variance is $$\text{Var}[x]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\mathbb{E}[X(X-1)]-\mathbb{E}[X]-\mathbb{E}[x]^2$$

$$\mathbb{E}[X(X-1)]=\int^2_0x(x-1)(\frac{x}{5}+\frac{3}{10})dx=\frac{7}{15}$$

Then, Var$[X]=\frac{7}{15}-\frac{17}{15}-(\frac{17}{15})^2=-\frac{439}{225}$

My concern is that the variance is negative, which should not be the case. Can I get some pointers on where I went wrong? Thank you.

This one is simple. I just need to find $\int^2_0\frac{x}{2x+3}f(x)dx$

答:
By linearity,
$$ \mathbb{E}[X(X-1)] = \mathbb{E}[X^2 - X] = \mathbb{E}[X^2] - \mathbb{E}[X]. $$
Your additional term to compensate this should be $\mathbb{E}[X]$ rather than $-\mathbb{E}[X]$ (basically you turned $x^2$ in $x (x-1) -x$ instead of $x(x-1) + x$.

Title: Find the expectation and variance of X


问:
A geometry problem with a cube (solid geometry)
I ask you to solve this problem, because in the book and I have different answers

Point $M$ is the midpoint of the edge $AB$ of the cube $ABCDA_1B_1C_1D_1$. Find the distance between the straight lines $A_1M$ and $B_1C$, given that the edge of the cube is equal to $a$.

Please help me. It's Translated from Russian and it's the full context of the problem.

答:
HINTS

Take cube edges as parallel to x-,y-, z- axes.

Use vectors to find position vectors of red lines.

Normalize for length. Minimum distance lies along their cross product to find distance between skew vectors.


问:
Need help with integral and change the variable
I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.

答:
Write x as a function of u: you have $u=\sin(x)$ so $x=\arcsin(u)$. It gets a little sticky when you have to simplify $\cos(\arcsin(u))$; remember the identity $\sin^2\theta+\cos^2\theta=1$, letting $\theta=\arcsin u$ (arcsin gives you an angle, recall) you get $\cos(\arcsin(u))=\sqrt{1-u^2}$.

Actually, this integral doesn't have a closed form; are you sure the problem wasn't $\int \sqrt{x} \cos^2 x dx$


答:
points 1 is ok

the expectation is ok

As variance is concerned

$$E[X^2]=\int_0^2 x^2f(x)dx=\frac{8}{5}$$

thus

$$V[X]=\frac{8}{5}-\left(\frac{17}{15}\right)^2=\frac{71}{225}$$

point 3, easy but incorrect

$$E[Z]=\int_0^2 \frac{x}{2x+3}f(x)dx$$

Title: Find the expectation and variance of X

Body:
A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}

Find the value of C.

Find the expectation and variance of X.

Find the expectation of random variable $Z=\frac{X}{2X+3}$

What I tried:

Finding C is straightforward as we just need to make sure that $\int^\infty_{-\infty}f(x)dx=1$.
\begin{align*}
                1&=\int^2_0C(x+\frac{3}{2})dx\\
                &=\left[\frac{C}{2}x^2+\frac{3}{2}Cx\right]^2_0\\
                &=\left[\frac{C}{2}(2)^2+\frac{3}{2}C(2)\right]-0\\
                &=5C\\
                &C=\frac{1}{5}
            \end{align*}

Using $\mathbb{E}[X]=\int^\infty_{-\infty}xf(x)dx$,
\begin{align*}
                \mathbb{E}[X]&=\int^2_0x\left(\frac{1}{5}\left(x+\frac{3}{2}\right)\right)dx\\
                &=\left[\frac{1}{15}x^3+\frac{3}{20}x^2\right]^2_0\\
                &=\left[\frac{1}{15}(2)^3+\frac{3}{20}(2)^2\right]^2_0-0\\
                &=\frac{17}{15}
            \end{align*}
Variance is $$\text{Var}[x]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\mathbb{E}[X(X-1)]-\mathbb{E}[X]-\mathbb{E}[x]^2$$

$$\mathbb{E}[X(X-1)]=\int^2_0x(x-1)(\frac{x}{5}+\frac{3}{10})dx=\frac{7}{15}$$

Then, Var$[X]=\frac{7}{15}-\frac{17}{15}-(\frac{17}{15})^2=-\frac{439}{225}$

My concern is that the variance is negative, which should not be the case. Can I get some pointers on where I went wrong? Thank you.

This one is simple. I just need to find $\int^2_0\frac{x}{2x+3}f(x)dx$


问:
Which topology is transferred by this specific metric on the rectangle $[0,\infty] \times [0 , \infty ]$?
Let $\varphi : [0,\infty]^2 \to [0,1]^2$, $(t,x) \mapsto (\frac{t}{1+t} , \frac{x}{1+x})$. The inverse is given by $\varphi^{-1}: [0,1]^2 \to [0,\infty]^2$, $(t,x) \mapsto (\frac{s}{1-s} , \frac{x}{1-x})$. By this function $\varphi$ we can induce a metric $r$ from the euclidean metric on $[0,1]^2$.

Now consider the the set $\mathcal B$ of subsets $B$ of $[0,\infty]^2$ with

$B$ is closed

$\{\infty\}\times [0,\infty] \cup [0,\infty]\times \{\infty\} \subseteq B$

$(t,x) \in B$ implies $(t,y) \in B$ for all $y>x$.

We can define a metric on $\mathcal B$ by
$$r(C,D) = \max \{  \sup_{x\in C} \inf_{y\in D} r(x,y) , \sup_{x\in D} \inf_{y\in C} r(x,y) \}$$
EDIT: I found that this metric is called Hausdorff metric.

Now my question:

I found out that there is a bijection from $\mathcal B$ to the set $\underline {\mathbb b}$ of lower semi-continuous functions $b : [0,\infty] \to [0,\infty]$ with $b(\infty) = 0$ by
$$b(t) = \inf\{x : (t,x ) \in B \} \\
B = \{(t,x) : x \geq b(t)\}$$

We can transfer the topology on $\mathcal B$ to $\underline {\mathbb b}$.

I would like to know which topology this is. For example, when does a sequence converge in $\underline {\mathbb b}$?

I tried to go step by step. For example I thought that for a sequence $B_n$ converging to $B$ at least one could show that the corresponding functions $b_n$ converge pointwise to $b$. But I have difficulties to work with this metric and came not really far.

EDIT: In the meantime I got the intuition, that $B_n \to B$ does not imply that the corresponding functions converge pointwise. For example, take a countable dense subset $D =\{d_1, \ldots \}$ of $[0,\infty]$. Then build a sequence of functions $b_n (t) = \infty$ for all $t\in [0,\infty]\setminus\{d_1 , \ldots , d_n\}$ and $b_n \equiv 1$ on $\{d_1 , \ldots , d_n\}$. Then the corresponding barrier should converge to $[1,\infty]\times [0, \infty ]$ but the functions clearly do not converge pointwise to $1$.

答:
The convergence I searched for is the so called
$\Gamma$-Convergence


问:
Leibniz question on notation, specifically using with vertical point bars?
I'm trying to understand Leibniz notation.  I have read the Wiki page on the notation and numerous others, but think I'm overlooking something crucial to understanding.

I have been told that:

$$
y=f(x)
$$
$$
f'(x) = \frac {dy}{dx}
$$

Question 1: In the above form, I'm assuming the strings of symbols $x$ (LHS) and $dx$ (RHS) should be treated as different objects, and that $dx$ represents both the denominator of the difference quotient $\lim_{h\rightarrow0}{\frac {f(a+h) - f(a)}{(a+h) - a}}$ at some point $a$ and that $x$ is the independent variable which to infinitesimally change.  Is that correct?  To eliminate symbol confusion, is the following form equivalent?

a.

$$
f'(a) = \frac {dy}{dx} \Big|_{a}= \frac {dy}{dx} = \lim_{h\rightarrow0} {\frac {f(a+h) - f(a)} {(a+h) -a}}
$$
An example of the chain rule:

b.
$$
f(x) = (f_1 \circ f_2 \circ f_3)(x)
$$
$$
y = f_1(u), u = f_2(v), v = f_3(x)
$$
$$
f'(a)
= \frac {dy} {du} \Big|_{f_2(f_3(a))}\frac {du} {dv} \Big|_{f_3(a)} \frac {dv} {dx} \Big|_{a}
$$

$$
\begin{align*}
f'(a) &= (f_1' \circ (f_2 \circ f_3))(a) \cdot (f_2' \circ f_3)(a) \cdot f_3'(a)
\end{align*}
$$

Question 2: I understand the vertical bars $\Big|_{z}$ provide a point from which to take the derivative.  I observe there is notation with and without bars and have not seen any history of Leibniz using them.  I'm assuming they help reduce back references to the function definitions.  Does anyone know why used?

答:
I think to understand Leibniz notation it is best to get into a “Leibniz mindset” because at his time the idea of a function was different from today. As far as I know, he would talk about $y$ being a a function of $x$, meaning that the value $x$ determines the value of $y$. In modern parlance: There exists a (modern) function $f : \mathbb{R} \to \mathbb{R}$ such that $y = f(x)$.

But note that there is no $x$ on the left-hand side. The value of $y$ really depends on the “current” value of $x$. This is different from modern functions where writing $f(x) = x^2$ and $f(z) = z^2$ really defines the same function $f$. As such, writing $y(x_0)$ does not really make sense, as $y$ might be a function of different variables in different ways, and to express the same thing one should specify which variable one fixes and write $y|_{x=x_0}$ (i.e. mention $x$ explicitly). Writing $y|_{x_0}$ or even $y(x_0)$ is a common shorthand, though, and often does not lead to confusion.

In this context, Leibniz notation makes sense: An (infinitesemal) change of $x$ gives by the implicit dependence of $y$ on $x$ an (infinitesemal) change of $y$; their quotient is the derivative $\frac{dy}{dx}$. Now as you know, the value of this also depends on the value of $x$. In other words, $\frac{dy}{dx}$ is again a “function of $x$”, namely $\frac{dy}{dx} = f'(x)$ (if $y = f(x)$).

If we want the value of this at a specific point, we can write $\frac{dy}{dx}|_{x = x_0}$ or $\frac{dy}{dx}|_{x_0}$ for $f'(x_0)$.

For Leibniz' chain rule, we are in the following situation: The variable $u$ is a function of $x$, say $u = f(x)$, and the variable $y$ is a function of $u$, say $y=g(u)$. Substituting, we see that $y = g(f(x)) = (g \circ f)(x)$, i.e. $y$ is also a function of $x$. We can therefore try to compute $\frac{dy}{dx}$. Using the chain rule you know for the $'$-notation, we see:
$$
  \frac{dy}{dx} = (g \circ f)'(x) = g'(f(x)) f'(x) = g'(u) f'(x) = \frac{dy}{du} \frac{du}{dx}.
$$
This is the usual form of this chain rule. Note that the first factor $\frac{dy}{du}$ is a function of $u$ but also a function of $x$ because $u$ is a function of $x$.

If we want to evaluate at a specific point $x_0$, we run into the problem mentioned above that $\frac{dy}{du}$ is a function of both $x$ and $u$ in different ways. Now $x_0$ looks like is should be the value for $x$, but in your question you use $a$ where this is less clear. And $\frac{dy}{du}$ is more obviously dependent on $u$; so to be explicit, we write down the transformation from $x$ to $u$, i.e.
$$
  \frac{dy}{dx}|_{x=x_0} = \frac{dy}{du}|_{u=f(x_0)} \frac{du}{dx}|_{x=x_0}.
$$
Using the shorthands above,
$$
  \frac{dy}{dx}|_{x_0} = \frac{dy}{du}|_{u(x_0)} \frac{du}{dx}|_{x_0},
$$
where we also employed the confusion between $u$ and the function $f$.

Why do we use this notation? Aside from convention, it has two advantages:

Defining a dependence is often easier, i.e. we can write down and derive the term $x^3 + \exp(x)$ immediately instead of having to define a function $f$ with $f(x) = x^3 + \exp(x)$ first.

Once you have functions of several variables it is nice to be able to refer to them by name in the derivative notation (instead of having to use numerical indices). I think this is why physicists often still use this notion of “function”.

Last, I want to mention that if you go on to study differential geometry, you will learn how to do all of this in a way that kind of gives you the best of both worlds: Normal function (which I think are conceptually much clearer than these “functions of $x$”) combined with the ability to refer to names instead of indices.


Title: Find the expectation and variance of X

Body:
Your $C=1/5$ and $E(x)=17/15$ are correct, then $$E(x^2)=\frac{1}{5}\int_{0}^{2} x^2(x+3/2)dx=\frac{8}{5}$$
$$Var=E(x^2)-(E(x))^2=\frac{8}{5}-(\frac{17}{15})^2=\frac{71}{225}.$$


问:
What is lim n→∞ (2)^(1/n)? Why is it equal to 1?
What is lim n→∞ (2)^(1/n)? Why is it equal to 1? All proofs shown are about n^(1/n) or something like this form. What about 2?</p>

答:
$\lim _{n \rightarrow \infty} \frac{1}{n}=0$
So, $\lim _{n \rightarrow \infty} 2^{\frac{1}{n}}=2^{0}=1$
As denominator gets bigger and bigger while numerator being constant then fraction gets closer and closer to zero.


答:
Suppose you are observing a point $P=(x,y,z,w)$ in $4D$-space projected to $3D$-subspace $\mathbb R\times\mathbb R\times\mathbb R\times\{k\}$ from the point of observation being $O=(0,0,0,0)$. Then the resulting point would simply be:
$$
P'=\left(\tau x,\tau y,\tau z,k\right),\text{ where }\tau=\frac kw
$$
So the distance of the observer in the fourth dimension to each point in your $4D$-object would determine the scaling of that particular point. Hence everything depends on your choice of configuring the size of the object, the point of the observer, and the distance $k$ from the observer to the projection space.

You can play around with this until you find a setup that suits you.


问:
Identifying specific pattern in a noisy dataset
I am working on a project where I have the following forward model. There are multiple point sources sparsely distributed in 3D space with coordinate $[x_0,y_0,z_0]$ for each of them. The detected data consists of 8 points in the form of $[x,y,k]=[x_0,y_0]+r_0[\sin(k\pi/4),\cos(k\pi/4)]+\mathrm{Gauss}(0,\sigma^2), k\in\{1,...,8\}$. One challenge is that sometimes one measurement could be missing, e.g., we sometimes only detect 3 points for a certain $[x_0,y_0,z_0]$. Is there a way to correctly identify and estimate all the $[x_0,y_0,z_0]$?

答: The answer depends on what you mean by the word detect. The inability to detect points may be due to imperfection of the device or imperfection of the processing algorithm. If the device does not allow you to receive additional data on the missing points, then in your formulation of the problem the answer will be no. However, if you know any additional theory about the relative position of these points, then perhaps the solution to the problem can be obtained in the presence of incomplete information.</p>


答:
As the base $b=2$ is a constant,

$\lim_{n\rightarrow\infty} 2^{(1/n)} = 2^{\lim_{n\rightarrow\infty} (1/n)} = 2^0 = 1.$

Body: What is lim n→∞ (2)^(1/n)? Why is it equal to 1? All proofs shown are about n^(1/n) or something like this form. What about 2?</p>


问:
Prove $\cos (n\theta) \neq \frac{1}{2}$ for any $n \in \mathbb{Z}^+$, where $\cos(\theta) = -\frac{1}{4}$
Given that $\cos(\theta) = -\frac{1}{4}$.

Prove $\cos (n\theta) \neq 1$, for any $n \in \mathbb{Z}^+$.

My attempt:
If $\cos(n\theta) = 1$,
then $e^{in\theta} = 1$.

$e^{in\theta} = e^{i(n-1)\theta} \cdot e^{i \theta}$

If $e^{in\theta} = 1$, then $e^{i(n-1)\theta} = e^{i(-1)\theta}$.

Then $(-\frac{1}{4} + \frac{2\sqrt{2}}{3}i)^{n-1} = (-\frac{1}{4} - \frac{2\sqrt{2}}{3}i)$.

However, I failed to prove that $(-\frac{1}{4} + \frac{2\sqrt{2}}{3}i)^{n-1} \neq (-\frac{1}{4} - \frac{2\sqrt{2}}{3}i)$, for any $n\in \mathbb{Z}^+$.

答:
Suppose by contradiction that $\cos(n \theta)=1$.

Use the identity
$$\cos( \theta) = \frac{e^{i \theta} + e^{-i \theta}}{2}$$
Call $z=e^{i \theta}$, then $z$ satisfies the following equation
$$\frac{z + z^{-1}}{2}=- \frac{1}{3}$$
which can be rewritten as
$$z^{2} + \frac{2}{3}z+1=0 $$

Clearly, the polynomial $x^2+ \frac{2}{3}x+1$ is the minimal polynomial of $z$ over the rational numbers (because it is irreducible over the rationals). This means that $z$ is not an algebraic integer.

On the other hand, $\cos(n \theta)=1$ implies that $z^n=1$.
This means that $z$ is a root of unity, and in particular it is an algebraic integer. This is a contradiction.

EDIT: if you don't know what an algebraic integer is, look at the following theorem:

Let $\alpha \in \Bbb C$ be an algebraic number over the rationals. Then TFAE:

the monic minimal polynomial of $\alpha$ has integer coefficients

$\alpha$ is a root of a monic polynomial with integer coefficients

In such a case, $\alpha$ is called an algebraic integer.


问:
4 Color Theorem - What am I not seeing??
Let me say first that I am in no way a mathematician. Just slightly interested in mathematics. I think I may have found an exception to the 4 color theorem.
I don't claim to be smarter than those who proved the theorem, and I'll assume I'm wrong.
What am I missing here?
(forgive my quickly drawn graph!)

(I created this account just to ask this question, so the website made me embed the image instead of just adding it straight to the question.)

I may be missing some fundamental rule in setting up my problem. But I think my map fits the desired intent of the "map" motif of the theorems original question.

答: I don't think that you looked at it for long enough. By the way, you will not fault the theorem, it is proven.


问:
Binary System with Sigma Summation
Let 100110101 be a number from the binary system with the digits being $$
b_i \in \left\{0,1\right\}
$$ If we calculate this number into the decimal system, the result is 309.
What formula can be used to do this kind of calculation?

The answer:
$$
\sum_{i=0}^{n-1} b_i2^i
$$

Can someone help me understand how this formula can be used for this calculation?

Title: Binary System with Sigma Summation

Body:
You gave $B=100110101$ as an example, which amounts to saying:
$$
\begin{align}
b_0&=1\text{ (least significant digit)}\\
b_1&=0\\
b_2&=1\\
b_3&=0\\
b_4&=1\\
b_5&=1\\
b_6&=0\\
b_7&=0\\
b_5&=1
\end{align}
$$
and thus:
$$
\begin{align}
B&=\sum_{i=0}^{9-1}b_i 2^i\\
&=b_0\cdot 2^0+b_1\cdot 2^1+...+b_8\cdot2^8\\
&=1\cdot 1+0\cdot 2+1\cdot 4+0\cdot 8+1\cdot 16+1\cdot 32+0\cdot 64+0\cdot 128+1\cdot 256\\
&=309
\end{align}
$$


问:
Minimum Distance from an Arbitrary Point in Two Dimensions and a Line Using Lagrange Multipliers
I need to find the minimal distance between a point $(x_0,y_0)$ and a line $ax+by = c$ for some constants $a,b,c ∈ ℝ$ using the Lagrange Multiplier Method.

Knowing the distance between two points, $(x_0,y_0)$ and $(x,y)$ is given by the following equation, I know that I need to minimise this equation.

$D = \sqrt{(x-x_0)^2 + (y-y_0)^2)}$

I also know that $D_{min}$ = $D^2_{min}$ , which will make it easier to use the Lagrange Method since I can "discard" the square root.

Then I let:

$f(x,y) = D^2 = (x-x_0)^2 + (y-y_0)^2$

Since this distance must be on the line $ax + by = c$ , I set my constraint:

$G(x,y) = ax + by - c = 0$

Then:

$∇f = <2x-2x_0, 2y-2y_0> ; ∇G =  $

Using the Lagrange Method for some constant $λ$, we have:

$∇f = λ∇G \Rightarrow  < 2x-2x_0, 2y-2y_0 >  = λ $

Eliminating $λ$, we have:

$y = y_0 + \frac ba (x-x_0)$

Plugging this into my original constraint, $G(x,y)$, we have:

$G(x,y) = ax + b(y_0 + \frac ba (x-x_0)) = 0$

Solving for $x$, we have:

$x = \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a}$

Plugging back in once more, we have:

$y = \frac ba \left( \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a} \right) - \frac ba{x_0} + y_0$

The Hessian seems to confirm that any critical point of $f$ will be a local maximum of $f$, but the overall complexity of the point I found is making me question whether I took the right approach.

If I plug back these coordinates into $D$, I should get the minimum distance.

$D = \sqrt{\left(\frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a}-x_0\right)^2 + \left(\frac ba \left( \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a} \right) - \frac ba{x_0})^2\right)}$

This seems unnecessarily complicated and wrong. Where did I go wrong (if I did)?

答:
There isn't anything wrong with what you did; you just made the algebra more difficult than it needed to become...

From your Lagrange equations, you arrive at  $ \ a(y - y_0) \ = \ b  (x - x_0) \ \ . $   The line you found has slope $ \ \frac{b}{a} \ ,  $ while the given line $ \ ax \ + \ by \ - \ c \ = \ 0 \ $ has slope $ \ -\frac{a}{b} \ . $  So it is worth noting here that you have incidentally demonstrated the proposition from classical geometry that the shortest distance from a line to a point external to the line is along a perpendicular to the line (the so-called "perpendicular distance").

We may alternatively express the result in this manner.  The given line may be parameterized using

$$ ax \ + \ by \ = \ c \ \ \rightarrow \ \ a·(X \ + \ bt) \ + \ b·(Y \ - \ at) \ = \ c $$
$$ \Rightarrow \ \ x \ = \ X \ + \ bt \ \ , \ \ y \ = \ Y \ - \ at \ \ , $$

with $ \ (X,Y) \ $ being a "reference point" on the line.  So a direction vector for the given line is $ \ \overrightarrow{v} \ = \ \langle \ b \ , \ -a \ \rangle \ \ . $  If we make $ \ (X \ , \ Y) \ $ the point on the given line that is closest to $ \ (x_0 \ , \ y_0) \ $ , so that the vector from the external point to that point is $ \ \langle \ (X-x_0) \ , \ (Y-y_0) \ \rangle \ $ , then your result may be understood as

$$ b(X-x_0) \ - \ a(Y-y_0) \ \ = \ \ 0 \ \ = \ \ \overrightarrow{v} \ · \ \langle \ (X-x_0) \ , \ (Y-y_0) \ \rangle \ \  , $$

this dot product telling us that these vectors are perpendicular.

As for the distance formula, if we don't need to have the actual coordinates of the closest point on the line, we can start with the distance-squared equation $ \ D^2 \ = \  (x-x_0)^2 \  + \ (y-y_0)^2 \ $ and use your result $ \ (y - y_0) \ = \ \frac{b}{a} ·  (x - x_0) \ $ to write

$$ \ D^2 \ = \  (x-x_0)^2 \  + \ \left( \frac{b^2}{a^2} \right) · (x-x_0)^2 \ \ = \ \ \left( \frac{a^2 \ + \ b^2}{a^2} \right) · (x-x_0)^2 \ \ . $$

The given line equation may be written as $$ \ ax \ + \ by \ = \ c \ \ \rightarrow \ \ a·(x \ - \ x_0) \ + \ b·( y \ - \ y_0) \ = \ c \ - \ ax_0 \ - \ by_0 \ \ .   $$

We may again apply your result to obtain

$$ a·(x \ - \ x_0) \ + \ b·( y \ - \ y_0) \ \ = \  \ a·(x \ - \ x_0) \ + \ b·\frac{b}{a} ·  (x \ - \ x_0)  $$
$$  = \ \  \left( \frac{a^2 \ + \ b^2}{a} \right) ·  (x \ - \ x_0) \ \ =  \ \  c \ - \ ax_0 \ - \ by_0 \ \ . $$

Upon squaring both sides of this equation, we find

$$    \left( \frac{(a^2 \ + \ b^2)^2}{a^2} \right) ·  (x - x_0)^2 \ \ = \ \ (a^2 \ + \ b^2)·D^2  \ \ = \ \  (c \ - \ ax_0 \ - \ by_0)^2  $$

$$ \Rightarrow \ \ D^2  \ \ = \ \  \frac{(c \ - \ ax_0 \ - \ by_0)^2}{a^2 \ + \ b^2} \ \ \Rightarrow \ \ D  \ \ = \ \  \frac{| \ c \ - \ ax_0 \ - \ by_0 \ |}{\sqrt{a^2 \ + \ b^2}}  \ \ . $$

[Should you require the coordinates of the closest point on the given line, the relations  $ \ \left( \frac{a^2 \ + \ b^2}{a} \right) ·  (X - x_0) \ \ =  \ \  c \ - \ ax_0 \ - \ by_0 \ $ and $ \ (Y - y_0) \ = \ \frac{b}{a} ·  (X - x_0) \ $ will deliver them.]


问:
Analysis - Prove a function that maps the unit Euclidean ball to R with bounded partial derivatives is uniformly continuous
I am stuck with this problem from my textbook and I cannot see the solution. I'm certain the solution is fairly simple and I am just missing the mark somehow. Any help would be appreciated.

Suppose the function $f: B_1(0) \rightarrow \mathbb{R}$ has continuous bounded partial derivatives on the unit Euclidean ball $B_1(0)$ in $\mathbb{R}^n$. Prove that $f$ is uniformly continuous on $B_1(0)$

Edit: Here is my take using Lipschitz and multivariable MVT, can someone please let me know if I am on the right track? Is it as simple as this?

let $x,y \in B_1(0)$ then by MVT we have $|f(x) - f(y)| = \nabla f(c)|x-y| \leq M|x-y|$ for some c on the line segment connecting x and y, thus Lipschitz thus uniformly cts?

答:
Let us prove that $f$ is Lipschitz continuous, with Lipschitz constant
$M := \sup_{x\in B_1} |\nabla f(x)|$.

Given two points $x,y\in B_1$, let
$$
g(t) := f((1-t)x + ty), \qquad t\in [0,1].
$$
Since $f$ is of class $C^1$, then also $g$ is, and by the chain rule formula
$$
g'(t) = \nabla f((1-t)x + t y) \cdot (y-x).
$$
By the mean value theorem there exists $\tau \in (0,1)$ such that
$g(1) - g(0) = g'(\tau)$, hence
$$
|f(y) - f(x)| = |g(1) - g(0)| = |g'(\tau)| \leq |\nabla f((1-t)x + ty)| \, |x-y|
\leq M |x-y|.
$$
(The first inequality follows from the Cauchy-Schwarz inequality.)


答:
Hint: Try this if you want:
$$x\sqrt { x } =u\\ x={ u }^{ \frac { 2 }{ 3 }  }\\ \sqrt { x } ={ u }^{ \frac { 1 }{ 3 }  }\\ dx=\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 }  }du\\ \int _{ 0 }^{ \pi  } \sqrt { x } \cos  x~ dx=\int _{ 0 }^{ \pi  }{ u } ^{ \frac { 1 }{ 3 }  }\cos { u } ^{ \frac { 2 }{ 3 }  }\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 }  }du=\frac { 2 }{ 3 } \int _{ 0 }^{ \pi\sqrt\pi  }{ \cos { u } ^{ \frac { 2 }{ 3 }  }du } $$

$\int { \cos { u } ^{ \frac { 2 }{ 3 }  }du } $ does not have an analytical expression. You can try on wolframalpha.com .

Title: Need help with integral and change the variable

Body:
I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.


问:
The notion of event, experiment and sample space
I'm learning basics of the probability theory and trying to map the probability concepts I learnt to the following example from my book (Introduction to Algorithms (or CLRS):

Suppose that we have a fair coin and a biased coin that always comes up heads. We run an experiment consisting of three independent events: we choose one of the two coins at random, we flip that coin once, and then we flip it again. Suppose that the coin we have chosen comes up heads both times. What is the probability that it is biased?

I'm confused with what the author here means by an experiment and by an event? What's the sample space in this problem?

My thoughts: the experiment here is actually "choose a coin, flip it twice" meaning that a possible outcome of this experiment is, for example, "biased coin, two heads" or "fair coin, head, tail" etc. Therefore this means that this "biased coin, two heads" and this "fair coin, head, tail" are elementary events.

Consequently our sample space is something like

$$ S = \{ \text{(F, TT), (F, HT), (F, TH), (F, HH), (B, HH)} \} $$

Therefore events should be subsets of the $S$, but author's event is like "choose a coin" or "flip coin first time" which are definitely can't be subsets of $S$.

Where am I wrong and what's exactly the author means by his experiment and events and how they are related to the sample space concept.

答:
The experiment is what you actually do: "you choose one of the two coins at random, you flip that coin once, and then you flip it again"

The sampe space is exactly what you stated

An Event is each element of the sample space

Of course any event of you sample space is composed by 3 elementary events; the first related to the choice of the coin and the other two related to the coin's outcomes

Thus the solution is

$$P(B|HH)=\frac{P(B,HH)}{P(F,HH)+P(B,HH)}=\frac{\frac{1}{2}\times1}{\frac{1}{2}\times \frac{1}{4}+\frac{1}{2}\times1}=\frac{4}{5}$$


问:
Open ball with respect to the Euclidean metric is open ball with respect to advance French railroad metric?
Let $x\neq0$ in $\mathbb{R}^2$, and consider the open ball (i.e. open disk) of radius $r$ around $x$, with respect to the Euclidean metric. Then is it true that this is an open set with respect to the advanced French railroad metric? I think it is, because for every point $y$ in the disk, we can define an advanced French open ball of appropriate radius $l$ (i.e. a line segment of length $l$ with $y$ as its centre, on the straight line passing from $y$ and the origin) such that the line segment is completely contained in the Euclidean disk. Is this correct? Sorry if the description is a bit confusing, but I do not have a picture available atm. Thanks in advance!

Edit: the advanced French railway metric is, per my notes, defined by: $d(x,y)=|x-y|$, if $x=ay$, for some $a\in \mathbb{R}$, and $|x|+|y|$ otherwise.

答:
As for the question in the title, the French-railroad open ball $B_F(x,r)$ is:

the open euclidean ball $B(x,r)$ if $x=0$;

the segment $x_{(r)}:=\left\{\frac t{\lVert x\rVert}\,:\,t\in(\lVert x\rVert-r,\lVert x\rVert+r)\right\}$ if $x\ne 0$ and $r\le\lVert x\rVert$;

$B(0,r-\lVert x\rVert)\cup x_{(r)}$ if $r>\lVert x\rVert$ and $x\ne0$.

Therefore I would contend that an Euclidean open ball $B$ is a French-railroad open ball if and only if $0$ is the center of $B$. It is clear that an Euclidean ball does not fall into the case (1) unless its center is $0$, and it can never fall into case (2) or (3) because they aren't Euclidean open (case (3) always has an "antenna" protruding out of the ball in the direction of $x$).

As for $B$ being French-railroad open, this is always the case because you can see it as union of open segments and possibly a small Euclidean ball centered in $0$, and all of those can be chosen to be French-railroad balls.


问:
Compute the solution of the IVP and determine the largest interval
Compute the solution of the IVP

$$
\begin{cases}
x'(t) = - \frac{t}{x(t)},\\
\\
x(t_0)=x_0,  x_0>0,
\end{cases}
$$

and determine the largest interval where the solution
exists and satisfies the IVP,. depending on $t_0$ and $x_0$.

I have worked out I have to use the method for separable equations but am unsure of how it works if anyone can help me out it would be deeply appreciated sorry this isn't as progressed as it really should be but I have hit a bit of a wall with this subject.

答:
The method for separable equations gives the general solution for the ODE:

$$x(t)=\sqrt{-2t+C},$$

where $C \in \mathbb R.$

If $x(t_0)=x_0$, then

$$x_0=\sqrt{-2_0t+C},$$

Hence $C=x_0^2+2t_0$. Therefore the solution of the IVP is given by

$$x(t)=\sqrt{-2t+x_0^2+2t_0},$$

therefore we must have $-2t+x_0^2+2t_0 >0.$ This means $ t < \frac{x_0^2+2t_0}{2}.$

Consequence: largest interval where the solution exists and satisfies the IVP is

$$(- \infty, \frac{x_0^2+2t_0}{2}).$$


问:
The existence of a contradiction between the two cases related to division
I came across the following problem:

We know that by comparing coefficients.

Case A: It is impossible for a non-zero polynomial $p(y)+q(z)+ax+b=F(x,\,y,\,z)$ to be divided by the polynomial $x^2+1$ because the degree of $F$ (in $x$) is $1$. See: https://mathoverflow.net/questions/384455/what-are-conditions-such-that-the-polynomial-x21-divides-pyqzaxb-fx

Here $p$ and $q$ are also polynomials and $a$, $b$ are real numbers.

Now passing to the values of these polynomials and consider the logical case:

Case B: There exist values of $x,y,z,a,b,c$ such that $x^2+1$ divides $p(y)+q(z)+ax+b=F(x,\,y,\,z)$

I am asking about the existence of a contradiction between the two cases.

答:
There is no contradiction here. If Case A holds then Case B also holds. This is better illustrated by an example: Let $F(x,y,z)=x+1$,
which cannot be devided by $x^2+1$. However, you can see that $(1)^2+1=2$ divides $F(1,y,z)=2$.


问:
Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>

答:
Recall the definition of the derivative:
$$
f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.
$$
Now let's first apply this to $f(x)=x^a$, with $a$ real:
$$
(x^a)' = \lim_{h\rightarrow 0}\frac{(x+h)^a - x^a}{h} = \lim_{h\rightarrow 0}x^a\frac{(1+h/x)^a - 1}{h}.
$$
Using the Taylor expansion
$$
(1+y)^a = 1 + ay + a(a-1)\frac{y^2}{2} + \ldots\qquad \text{for $y<1$},
$$
we get
$$
(x^a)' = \lim_{h\rightarrow 0}x^a\frac{1 + ah/x + \mathcal{O}(h^2/x^2) - 1}{h} = x^{a-1}\left[a + \lim_{h\rightarrow 0}\mathcal{O}(h/x)\right] = ax^{a-1}.
$$

Now, let's do the same for $f(x)=e^x$:
$$
(e^x)' = \lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h} = \lim_{h\rightarrow 0}e^x\frac{e^{h}-1}{h}.
$$
Since
$$
e^h = 1 + h + \frac{h^2}{2} + \ldots,
$$
we get
$$
(e^x)' = \lim_{h\rightarrow 0}e^x\frac{1 + h + \mathcal{O}(h^2) -1}{h} = e^x\left[1 + \lim_{h\rightarrow 0}\mathcal{O}(h)\right] = e^x.
$$


问:
What equation to use when finding 6 variables?
I'm programming an application which will need to calculate and invert 6 variables.

e.g.

[4][20][15][16][0][1] -> (perform calculation) -> 450

450 -> (perform calculation) -> [4][20][15][16][0][1]

What would be the best direction to go to calculate and invert the calculation for these variables or nth variables if I wish to expand later on? I was looking at Matrices, linear equations, etc, but not sure the best and easiest method for programmatically making this a bidirectional equation. Edit: I’m looking to get the smallest number possible, and the range is 0 to 55 for my variables.

答:
With 6 variables $x_0, \ldots, x_5$ all in the range $0 - 55$ you could form
$$n = x_0 + 56 x_1 + 56^2 x_2 + \cdots + 56^5 x_5.$$
To revert to the variables you can do (pseudocode):

```<code>x = [0, 0, 0, 0, 0, 0]
i = 0
while n > 0 {
  x[i] = n % 56
  n := floor(n / 56)     # integer division
}
</code>```

Body:
With 6 variables $x_0, \ldots, x_5$ all in the range $0 - 55$ you could form
$$n = x_0 + 56 x_1 + 56^2 x_2 + \cdots + 56^5 x_5.$$
To revert to the variables you can do (pseudocode):

x = [0, 0, 0, 0, 0, 0]
i = 0
while n > 0 {
  x[i] = n % 56
  n := floor(n / 56)     # integer division
}


问:
Finding a general form of the inverse of a function
I have obtained the asymptotic expansion of some function at infinity and the leading orders are of the following form
$$\rho (r)=\frac{r^q}{2 q}+\frac{3 (q-1) r^{2 q-1}}{8 q (2 q-1)}+\frac{3 r^{2 q-1}}{8 q (2 q-1)}-\frac{\sqrt{\pi } \Gamma \left(\frac{q}{q-1}\right)}{\Gamma \left(\frac{1}{2}+\frac{1}{q-1}\right)}+r $$
and I want to obtain its inverse, i.e. $r(\rho)$. What technique can I use to explicitly write the inverse for a general $-1\leq q <1$? Thanks

答:
First you have to make sure that there exists an inverse for $-1\leq q<1$, which you will have to find out (hint: a strictly monotonous polynomial in a given range is invertible in this range). Supposing that an inverse exists and baring in mind that this function is rather complicated, I would suggest that you use some computer software to find the inverse, like Mathematica.

You can do it by solving the equation $\rho=\rho(r)$ and obtain the solution(s) $r=r(\rho)$.


问:
Error propagation in a simple linear model (asked by a non math-major researcher).
I have the following linear problem in a matrix form $Ax=Y$, where $A$ is a coefficient matrix, $x$ is a vector of unknown parameters and $Y$ is a vector of observed data. Matrix $A$ is usually over-determined and often rank deficient; it has 1000s of rows and columns. I find a solution by applying the pseudo-inverse (SVD): $x=A^{-1}Y$. I compute a conditioning number (equal about 5) to learn how the input errors propagate to the output errors. Can I determine by looking at matrix $A$ if each unknown parameter in vector $x$ is resolved well and if there is an unwanted correlation between some $x_{i}$ and $x_{j}$. I have done tests using some synthetic data and they are good, but here I would like to know if anything can be determined from the matrix $A$ alone.

答:
The covariance has the following transformation property
\begin{equation}
Cov[A\vec{x}] = A Cov[\vec{x}]A^{\intercal},
\end{equation}
where $Cov[.]$ is a covariance matrix. This property can be proved as follows
\begin{equation}
Cov[A\vec{x}] = E[(A\vec{x} - E[A\vec{x}])(A\vec{x} - E[A\vec{x}])] \\
= E[A(\vec{x} - E[\vec{x}])(\vec{x} - E[\vec{x}])A^{\intercal}] \\
= A E[(\vec{x} - E[\vec{x}])(\vec{x} - E[\vec{x}])]A^{\intercal} \\
= A Cov[\vec{x}]A^{\intercal},
\end{equation}
where $E[.]$ is the expectation.
I.e. in your case
\begin{equation}
A Cov[\vec{x}]A^{\intercal} = Cov[\vec{y}],
\end{equation}
so the covariance matrix for $\vec{x}$ has the following form
\begin{equation}
Cov[\vec{x}] = A^{+}Cov[\vec{y}](A^{\intercal})^{+},
\end{equation}
where $A^{+}$ is pseudoinverse of $A$
Sometimes even better to write the answer in terms of inverse covariance matrices:
\begin{equation}
Cov^{-1}[\vec{x}] = A^{\intercal}Cov^{-1}[\vec{y}]A.
\end{equation}

There is also well-known inequality  that links relative perturbations of the solution $\vec{x}$ and relative perturbations of the right hand side $\vec{y}$:
\begin{equation}
\frac{\|\Delta\vec{x}\|}{\|\vec{x}\|} \leq \frac{cond(A)}{1 - cond(A)\frac{\|\Delta A\|}{\|A\|}}\left(\frac{\|\Delta\vec{y}\|}{\|\vec{y}\|} + \frac{\|\Delta A\|}{\|A\|}\right),
\end{equation}
where $\Delta\vec{y}$ is the perturbation of the vector $\vec{y}$, $\Delta\vec{x}$ is the perturbation of $\vec{x}$ and $\Delta A$ is the perturbation of the matrix $A$. If the matrix perturbation is zero, the inequality has a simpler form
\begin{equation}
\frac{\|\Delta\vec{x}\|}{\|\vec{x}\|} \leq cond(A)\frac{\|\Delta\vec{y}\|}{\|\vec{y}\|}.
\end{equation}

The last two inequalities can be simply obtained using the definition of the condition number $cond(A)=\|A^{+}\|\|A\|$.


问:
How to show that $\int_{-\infty}^{\infty}|\frac{\sin{nx}\sin{x}}{x^2}|dx \to \infty$ as $n \to \infty$?
We now have a sequence of functions
\[
f_n = \frac{\sin{nx}\sin{x}}{x^2},
\]
and we need to show that $\lVert f_n \rVert_1 \to \infty$ as $n \to \infty$, which means
\[
\lim_{n \to \infty}\int_{-\infty}^{\infty}|f_n(x)|dx = 0.
\]

How could we do that? It's easy to see that $\int_{-\infty}^{\infty}f_ndx=\pi$, but there is an absolute value function. I managed to compute
\[
J_n = \int_{0}^{2\pi}|\sin{nx}\sin{x}|dx=\frac{2n}{n^2-1}\frac{\sin\frac{(n+1)\pi}{2n}}{\sin\frac{\pi}{2n}}
\]
and wanted to multiply it with $\sum_{j=1}^{\infty}\frac{1}{(2\pi j)^2}$, but the sequence $\{J_n\}$ converges (with respect to $n$), so if my calculation is not wrong, this estimation is too floopy.

For those who are interested about the source of this problem, you can find it on W. Rudin's Real and Complex Analysis, Chapter 9, exercise 2. It is the last step of proving the range of the Fourier transform is a proper subset of $C_0$. Define $g_n = \chi_{[-n,n]}$, we can show that $g_n \ast g_1$ is the Fourier transform of $f_n$. It remains to prove that $\lVert f_n \rVert_1 \to \infty$ as $n \to \infty$. And this is where I stuck.

Update: think I find the solution in another question: A proof of the fact that the Fourier transform is not surjective. So I'm OK now. But I think random people in the future may search this function so I think this title may help people. Still if you have good method it's a good idea to put it down.

答: $$\int_{-\infty}^{\infty} |\frac  {\sin (nx) \sin x}{ x^{2}}|dx$$ $$\geq \int_0^{1} |\frac  {\sin (nx) \sin x}{ x^{2}}|dx$$ $$ \geq \frac 2 {\pi} \int_0^{1} |\frac  {\sin (nx) }{ x}|dx $$ since $\sin x\geq \frac 2 {\pi} x$ for $0 <x<1$. Now put $y=nx $ and use the fact that $\int_0^{\infty}  |\frac {\sin y} y| dy =\infty$.


问:
Finding radius of convergence for this power summation $\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} dt\right) x^n$
I have been given this tough power summation that its' general $c_n$ has an integral.
I am asked to find the radius of convergence $R$

$$\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} \, dt\right) x^n $$

Do I first calculate the integral?

Any help would be appreciated!

答:
we have

$$\left|\frac{\sin^2t}{\sqrt[3]{t^7+1}}\right|\leq \frac{1}{\sqrt[3]{t^7+1}}\sim_\infty t^{-7/3}$$
and since  $\frac{7}{3}>1$ then the improper integral
$$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$
is convergent and the value of the integral is $\ell\neq 0$  so if we denote by
$$a_n=\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$
then by ratio test we have
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$$
hence we have
$$R=1$$


问:
Making sure $a$ and $b$ are relatively prime
I came across this interesting problem in the Olympiad Maths challenge practice problem, and it is really fascinating:

Some $n$ numbers are selected randomly from the integers $1$ to $420$. $2$ integers $a$ and $b$ are chosen from that set of $n$ numbers such that $a$ and $b$ are relatively prime. Find the minimum value of $n$ that ensures there exists $2$ such integers $a$ and $b$

If you don't know what relatively prime is it means that the two numbers don't share any factors greater than one (therefore the HCF of $a$ and $b$ is $1$)

My evaluation is that two numbers is relatively prime if at least $1$ of the number is prime, therefore not sharing any factors with the other number except $1$ (since primes have the factors only $1$ and itself).

This evaluation should work! UNLESS the second number is the prime number doubled, or tripled, or quadrupled, etc.

First of all, I have to deal with the fact that we're randomly picking from the set of $1$ to $420$, so I have to find the number of prime numbers between $1$ and $420$.

This post almost answers my prime question, but I tried $\pi(420) $ and I got about $69$ (Very nice coincidence by the way, $69420$). And the pool of answers is no near that (Multiple choice: $208, 211, 215, 220$) So I am now stuck in this hole precisely here.

Any help would be greatly appreciated!

答:
This can be easily solved by using the pigeon hole principle.

Observe that if two numbers are consecutive, then they are relatively primes. Now one can pick every second number starting from $2$ till $420$ (all the even numbers). Now as every alternate number is picked already, the pigeonhole principle guarantees that picking any more numbers would result in a pair of consecutive numbers, which will be relatively prime.

$\therefore $ The minimum value of $n$ to guarantee that there will be $2$ relatively prime numbers in the chosen $n$ numbers is $211$.

Note: It is not necessary that one of the numbers is prime for the pair to be relatively prime. For example $(9,14)$ is such a pair of relatively prime numbers with none of them being prime. Any numbers of the form $a=p_1^{a_1}\cdot p_2^{a_2}\cdots p_n^{a_n}$ and $b=q_1^{b_1}\cdot q_2^{b_2}\cdots q_m^{b_m}$ are relatively prime $(p_i\neq q_i\space \forall\ i\in \mathbb N)$ with none of them being primes.


问:
Is each power of a prime ideal a primary ideal?
I want to show that each power of a prime ideal is a primary ideal or I have to think about a counterexample?</p>

答: One may even find such an example in the polynomial ring $k[x,y,z]$: the ideal $I=\langle x^3-yz, y^2-xz, z^2-x^2y\rangle$ is prime but $I^2$ is not primary as explained in Northcott, Ideal Theory, Example 3, p. 29.


问:
How to prove this propositional logic equivalence?
Problem : $\lnot[(\lnot p∧q)∨\lnot q]∨[\lnot(p∧\lnot q)]≡\lnot p∨q$  Prove this equility without using truth table.

I have managed to solve this question until this part :

$$\begin{align}\lnot[(\lnot p∧q)∨[\lnot(p∧\lnot q)] &≡ [\lnot(\lnot p∧q)∧q]∨(\lnot p∨q)\\& ≡ [(p∧q)∨(\lnot q∧q)]∨(\lnot p∨q)]\\&  ≡ (p∧q)∨(\lnot p∨q)\end{align}$$

I stucked here. I tried to $(p∧q)∨\lnot(p∧\lnot q)$ like this for de morgan rule but it doesnt work. What do I need to do?

答:
Let's go on...

$\begin{align}& [(p ∧ q) ∨ (\lnot p ∨ q)]\\ \equiv~& [((p ∧ q) ∨ \lnot p) ∨ q] \\ \equiv~& [((p ∨ \lnot p) ∧ (\lnot p ∨ q)) ∨ q]\\ \equiv~& [(\text T ∧ (\lnot p ∨ q)) ∨ q]\\ \equiv~& [(\lnot p ∨ q) ∨ q]\\\equiv~& [\lnot p ∨ (q ∨ q)]\\ \equiv~& (\lnot p ∨ q)\end{align}$


问:
Given the function,check the continuity at $(0,0)$ and find $f_x(0,0),f_y(0,0)$ if they exist
Given a function defined by :
$$f(x,y) =
  \begin{cases}
                                   \frac{x^{3}+y^{3}}{x^{2}+y^{2}} & (x,y) \ne (0,0) \\
  0 & (x,y) = (0,0)
  \end{cases}$$

Find $f_x(0,0),f_y(0,0)$ is they exist ,is the function continuous at $(0,0)$?

$f_x(0,0)$ and $f_y(0,0)$ exist if the following limits exist (respectively):

$$\lim_{\Delta x \to 0}\frac{f\left(\Delta x,0\right)-f\left(0,0\right)}{ \Delta x}=\lim_{\Delta x \to 0}\frac{\Delta x-0}{\Delta x}=1$$
$$\lim_{\Delta y \to 0}\frac{f\left(0,\Delta y\right)-f\left(0,0\right)}{ \Delta y}=\lim_{\Delta y \to 0}\frac{\Delta y-0}{\Delta y}=1$$

So $f_x(0,0) =1=f_y(0,0)$

About the continuity of the function at the given point we have that $$\lim_{(x,y) \to (0,0)} \frac{x^{3}+y^{3}}{x^{2}+y^{2}}=0$$

Because for $\left|x\right|,\left|y\right|<\delta$ :

$$\left|\frac{x^{3}+y^{3}}{x^{2}+y^{2}}-0\right|\le\frac{\left|x\right|^{3}+\left|y\right|^{3}}{x^{2}+y^{2}}=\frac{\left(\left|x\right|+\left|y\right|\right)\left(x^{2}-\left|xy\right|+y^{2}\right)}{x^{2}+y^{2}}\le\frac{\left(\left|x\right|+\left|y\right|\right)\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$$$$=\left|x\right|+\left|y\right|<2\delta$$

So taking $\delta \le \frac{\epsilon}{2}$ proves the claim.

I would like to know if my solution is true or not,and see the errors (if exists),so if you have an alternative solution then please first check mine and then provides yours.

答:
Your solution is fine. For the continuity of $f$ in $(0,0)$ you can use polar coordinates:

With $x= r \cos t$ and $y=r \sin t$ you get

$$|f(x,y)-0| \le r (| \cos t|+ |\sin t|) \le 2r=2 \sqrt{x^2+y^2}.$$


答:
We show that the radius of convergence is $1$.

Let $f(x)=\frac{\sin^2 x}{\sqrt[3]{x^7+1}}$. The integral $\int_0^\infty f(x)\,dx$ converges. Suppose it has value $B$.

Let $a_n =\int_0^n f(x)\,dx$. Then $a_n \lt B$ for all $n$.

By comparison with the series $\sum B|x|^n$, the series $\sum a_n |x|^n$ converges whenever $|x|\lt 1$. So the radius of convergence of our series is $\ge 1$.

We complete the solution by showing that the radius of convergence is $\le 1$.
Note that $a_n \gt \int_0^1 f(x)\,dx$. Let $b$ be the value of this integral. Note that $b$ is positive.

If $|x|\gt 1$, then $b|x|^n\to\infty$ as $n\to\infty$. So $a_n|x|^n\to\infty$ as $n\to\infty$. It follows that if $|x|\gt 1$, then $\sum a_n x^n$ diverges.

Title: Finding radius of convergence for this power summation $\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} dt\right) x^n$

Body:
I have been given this tough power summation that its' general $c_n$ has an integral.
I am asked to find the radius of convergence $R$

$$\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} \, dt\right) x^n $$

Do I first calculate the integral?

Any help would be appreciated!


问:
Finding the closed form of a summation when i is 0
I am trying to find the closed form summation of this:

$$\sum_{i=0}^{n-1} (5 + 2n + i) $$

Here I tried to break up the terms but I know that I'm doing something wrong by not accounting for $i=0$

I did this to begin and I know it is wrong...

$$\ 5(n-1) + 2n(n-1) + \frac{(n(n-1)}{2}$$

If I could get help getting started with this first step by converting in terms of n I think I could figure the rest out.

答:
You've done pretty well. The only problem is that there are $n$ terms (counting the $i = 0$ term as the first term), so the sum should be
$$5n + 2n^2 + \frac{n(n - 1)}{2}.$$


问:
Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $
I have the conjecture for the integral

$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$

where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.

I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral

$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$

is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.

Edit:

Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$

which may not reduce the difficulty, albeit appearing compact.

答:
Using the third of these equations we can express your integrals in terms of the Clausen function:
\begin{align}
I &\equiv \int \limits_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \log (\cot(t))\, \mathrm{d} t + \int \limits_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \log (\cot(t))\, \mathrm{d} t \\
&= \frac{1}{2} \left[- \operatorname{Cl}_2 \left(\vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)}\frac{\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{3\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{5\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{7\pi}{18}\right) \right. \\
&\phantom{= \frac{1}{2} \left[ \vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)} \right.} \left.+ \operatorname{Cl}_2 \left(\frac{11\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{13\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{15\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{17\pi}{18}\right)\right] \\
&= \frac{1}{2} \left[\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) - \sum \limits_{k=0}^{8} (-1)^k \operatorname{Cl}_2 \left(\frac{(2k+1)\pi}{18}\right) \right] \, .
\end{align}
The Fourier series of $\operatorname{Cl}_2$ and its special value $\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) = \mathrm{G}$ allow us to write this result as
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{n=1}^\infty \frac{\alpha(n)}{n^2}\right]$$
with $\alpha \colon \mathbb{N} \to \mathbb{R}$ given by
$$\alpha (n) = \sum \limits_{k=0}^8 (-1)^k \sin \left(\frac{(2k+1)n\pi}{18}\right) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \sum \limits_{k=0}^{8} \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right) k} \right] \, .$$
Clearly, we have
$$\alpha (9 + 18 m) = \operatorname{Im} \left[\mathrm{i} (-1)^m \sum \limits_{k=0}^{8} 1 \right] = 9 (-1)^m $$
for $m \in \mathbb{N}_0$, while for all other $n \in \mathbb{N}$ we find
$$ \alpha (n) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \frac{1 - \mathrm{e}^{\mathrm{i} \pi (9+n)}}{1 - \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right)}} \right] = \operatorname{Im} \left[2 \frac{1 + (-1)^n}{\left \lvert 1 + \mathrm{e}^{\mathrm{i} \pi n/9} \right \rvert^2} \cos \left(\frac{n \pi}{18}\right)\right] = 0 \, .$$
Therefore,
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{m=0}^\infty \frac{9 (-1)^m}{(9 + 18 m)^2}\right] = \frac{1}{2} \left[\mathrm{G} - \frac{1}{9}\sum \limits_{m=0}^\infty \frac{(-1)^m}{(2m+1)^2}\right] = \frac{4}{9} \mathrm{G} \, .$$

The same method leads to the more general result
\begin{align}
\sum \limits_{\mu = 0}^{\nu - 1} (-1)^{\nu - 1 - \mu} \int \limits_0^{\frac{2\mu +1}{2\nu+1} \frac{\pi}{4}} \log(\cot(t)) \, \mathrm{d} t &= \frac{1}{2} \left[\mathrm{G} + (-1)^{\nu - 1} \sum \limits_{k=0}^{2\nu} (-1)^k \operatorname{Cl}_2 \left(\frac{2k+1}{2\nu+1} \frac{\pi}{2}\right)\right] \\
&= \frac{\nu + 1_{2\mathbb{N} - 1} (\nu)}{2\nu+1} \mathrm{G}
\end{align}
for $\nu \in \mathbb{N}$ ($1_{2\mathbb{N}-1}$ is the indicator function of the odd positive integers). The first three cases are
\begin{align}
\int \limits_0^{\frac{\pi}{12}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{3} \mathrm{G} \, , \\
\int \limits_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{5} \mathrm{G} \, , \\
\int \limits_0^{\frac{\pi}{28}} \log(\cot(t)) \, \mathrm{d} t + \int \limits_{\frac{3\pi}{28}}^{\frac{5\pi}{28}} \log(\cot(t)) \, \mathrm{d} t &= \frac{4}{7} \mathrm{G}
\end{align}
and $\nu = 4$ corresponds to the original question.


问:
If $\beta$ is an $\mathcal{L}$ formula and $\beta_P$ is a tautology then $\beta $ is valid
I am reading A Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen. Let $\beta$ be a $\mathcal{L}$-formula in first order language $\mathcal{L}$. In the first line of Page 52, the authors claim that if $\beta_P$, a formula in propositional logic, is a tautology then $\beta$ is valid.

$\beta_P$ is obtained from $\beta$ in the following fashion (which is written is Page 51):

Find all subformulas of $\beta$ of the form $\forall x\alpha$ that are not in the scope of another quantifier. Replace them with propositional variables in a systematic fashion. This means that if $\forall yQ(y,c)$ appears twice in $\beta$, it is replaced by same letter both times, and distinct subformulas are replaced with distinct letters.

Find all atomic formulas that remain, and replace them systematically with new propositional variables.

At this point, $\beta$ will have been replaced with a propositional formula $\beta_{P}$.

I was trying to prove this by induction on the structure of formula $\beta$ but I got stuck in the base case. Here's my attempt: Let $\mathfrak{U}$ be any $\mathcal{L}$-structure and $s$ be any variable assignment function to $\mathfrak{U}$. Suppose that $\beta\equiv=t_1t_2$ (written in Polish notation) where $t_1 , t_2$ are terms of the language $\mathcal{L}$. Then $\beta_P$ is $A$ where $A$ is a propositional variable. Let's assume that $\beta _P$ is a tautology. We need to prove that $\beta$ is valid, i.e., $\mathfrak{U}\models \beta [s]$.

Here's my question: Since $\beta_P$ is a tautology, are not allowed to make the substitution $A=F$ because if we do, then $\beta_P$ would not longer be a tautology. How do we remedy this situation?

答:
As I see it, the problem is that the final $\beta_P$ might be a tautology while its parts you obtain in your induction might not be.
You are right that the partial $\beta_P$ you get from e.g. $t_1 = t_2$ would be $A$ and thus not a tautology, but the complete $\beta_P$ might be $A \lor \neg A$, which is a tautology.


问:
Find the probability of placing $5$ dots on an $8 \times 8$ grid.
$5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.

What is the probability that no row or column has more than $1$ dot?

I thought about this in the following way, the number of ways to place $5$ dots on $64$ cells is $$\begin{pmatrix} 64 \\ 5 \end{pmatrix} $$
Now if I want the $8 \times 8$ to have no row or column with more than $1$ dot then I reasoned that the dots must be placed in the diagonal which has $8$ cells, so the number of ways I can place $5$ dots into $8$ cells is $$\begin{pmatrix} 8 \\ 5 \end{pmatrix} $$

So the desired probability would be $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$

However I know that this is not the answer, the correct answer is $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4  }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
but I can't see what the term $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ counts, can you explain how this term came?

答:
${64 \choose 5}$ looks OK for the denominator

For distinct rows, there are ${8 \choose 5}$ ways of choosing the five rows.

The dot in the first occupied row can be in any of the $8$ columns

The dot in the second occupied row can be in any of the $7$ remaining columns

The dot in the third occupied row can be in any of the $6$ remaining columns

The dot in the fourth occupied row can be in any of the $5$ remaining columns

The dot in the fifth occupied row can be in any of the $4$ remaining columns

So I would get a probability of $$\dfrac{{8 \choose 5}\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}{{64 \choose 5}}$$


问:
simplify pseudo inverse of matrix with paired rows
In remote sensing we often have a linear problem that can be written in a matrix form as $aAx+bAy=B$, where $a$ and $b$ are scalar constants, $A$ is a matrix; and  $x$, $y$, and $B$ are vectors (i.e. 1-column matrix). This can be further re-written as $\hat{A} \hat{x} = B$, where $\hat{A}$ is a matrix with 2 times more columns than $A$ (i.e. $\{aAbA\}$) and $\hat{x}$ is a vector with two times more rows than $x$ and $y$. Then we solve this problem by inverting the matrix $\hat{A}$ using the pseudoinverse (i.e. SVD): $\hat{x}=\hat{A}^{-1}B$.

The problem is that the computation of $\hat{A}^{-1}$ for each pixel is expensive. Knowing that the matrix $\hat{A}$ has a particular column structure, can the problem be reduced to inverting the matrix $A$ instead, which has half the amount of columns. In general, the matrix $A$ is rank deficient and overdetermined (i.e. more rows than columns, but rank is less or equal than the number of columns).

答:
I think that yes due to the following equation (this equation should be true for block-diagonal matrices)
\begin{equation}
\big(\begin{smallmatrix}
aA & 0\\
0 & bA
\end{smallmatrix}\big)\big(
\begin{smallmatrix}
a^{-1}A^{+} & 0\\
0 & b^{-1}A^{+}
\end{smallmatrix}\big) =
\big(\begin{smallmatrix}
aA a^{-1}A^{+} & 0\\
0 & bA b^{-1}A^{+}
\end{smallmatrix}\big) =
\big(\begin{smallmatrix}
I & 0\\
0 & I
\end{smallmatrix}\big),
\end{equation}
where $A^{+}$ is the pseudoinverse of $A$ and $I$ is the identity matrix. So, If I'm right, the pseudoinverse of $\hat{A}$ can be calculated using the following formula
\begin{equation}
\hat{A}^{+} =
\big(\begin{smallmatrix}
a^{-1}A^{+} & 0\\
0 & b^{-1}A^{+}
\end{smallmatrix}\big).
\end{equation}


问:
Prove that the sequence $(x_n)$ defined by $x_n = \frac{1}{3^{n+5}}$ is bounded, find its supremum and its infimum.
Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.

Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows.
To show: $x_{n+1} < x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that's, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$

Now, back to the problem. To show: A is bounded. Since $A$ is a decreasing sequence and tending to $0$, then $A$ is bounded below by $0$. Next, it's clear that for all $s \in A$, we have
$s \le \frac{1}{3^6}$, which is happenned if and only if $n=1$. Hence, by definition, $A$ is bounded above by $\frac{1}{3^6}$. Therefore, $A$ is bounded.

Next, we find the supremum and the infimum of $A$. Indeed, $\inf A = 0$ and $\sup A = \frac{1}{3^6}$.
To show: $\inf A = 0$. Let $m$ be the another lower bound of $A$ and suppose that $0 \lt m$.
Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $r$,
such that $r \lt m$. A contradiction with the definition of a lower bound. Hence, we must have
$m \le 0$ and thus, $\inf A = 0$.

To show: $\sup A = \frac{1}{3^6}$. Let $M$ be the another upper bound of $A$ and suppose that $M \lt \frac{1}{3^6}$. Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $s$, such that $M \lt s$. A contradiction with the definition of a upper bound. Hence, we must have
$\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.

Am I true? Anyone please correct me. Thanks in advanced.

答:
The proof of the infimum is wrong, as the existence of such an $r$ does not follow from decreasing and bounded. A better proof would be that if $m > 0$ is a lower bound of the sequence, then choose $n > - \log_{3}(m) - 5$ which leads to $x_n < m$, a contradiction to $m$ being a lower bound. This $n$ can easily be found by working backwards. Since $0$ is obviously a lower bound, $0$ is the infimum.

Likewise, the existence of such an $s$ for the supremum is not guaranteed as written. This one is easy; since the sequence is decreasing, $x_n \leq x_1 = \frac{1}{3^6}$, meaning $\frac{1}{3^6}$ is an upper bound and is the best such one as it is the value of $x_1$. The rest of the proof is fine.


答:
For induction part, base case is OK. But then inductive hypothesis case should also mention that $n=k\gt 1$. Then Induction part is OK.

To prove $\inf A=0$, you have mentioned that "Let $m$ be the another lower bound of A and suppose that $0<m$. Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $r$, such that $r<m$..." Question is why does such $r$ exist? Example: The sequence $(y_n)=(2+1/n)$ is also decreasing and $0$ and $2$ are lower bounds but there is no $k\in \mathbb N$ such that $y_k\lt 2$. Similarly for supremum case.

In order to find $\sup A$ and $\inf A$, you may proceed as below also:

Let $m\gt 0$ be a lower bound for $A$, then by Archimedean property of real numbers $\exists N $ such that $1/N\lt m$. Now choose $n$ so large that $3^{n+5}\gt N$ and therefore, $1/3^{n+5}\lt m$, which is a contradiction. Therefore, $\inf A=0$.

Let $l\lt 1/3^6$ be an upper bound for $A$. Hence, $1/3^{n+5}\le l\lt 1/3^6$ for all $n$. Clearly, $n=1$ gives $l=1/3^6$, which is a contradiction.Therefore, $\sup A=1/3^6$ as $1/3^6$ is an upper bound for $A$.

Title: Prove that the sequence $(x_n)$ defined by $x_n = \frac{1}{3^{n+5}}$ is bounded, find its supremum and its infimum.

Body:
Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.

Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows.
To show: $x_{n+1} < x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that's, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$

Now, back to the problem. To show: A is bounded. Since $A$ is a decreasing sequence and tending to $0$, then $A$ is bounded below by $0$. Next, it's clear that for all $s \in A$, we have
$s \le \frac{1}{3^6}$, which is happenned if and only if $n=1$. Hence, by definition, $A$ is bounded above by $\frac{1}{3^6}$. Therefore, $A$ is bounded.

Next, we find the supremum and the infimum of $A$. Indeed, $\inf A = 0$ and $\sup A = \frac{1}{3^6}$.
To show: $\inf A = 0$. Let $m$ be the another lower bound of $A$ and suppose that $0 \lt m$.
Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $r$,
such that $r \lt m$. A contradiction with the definition of a lower bound. Hence, we must have
$m \le 0$ and thus, $\inf A = 0$.

To show: $\sup A = \frac{1}{3^6}$. Let $M$ be the another upper bound of $A$ and suppose that $M \lt \frac{1}{3^6}$. Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $s$, such that $M \lt s$. A contradiction with the definition of a upper bound. Hence, we must have
$\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.

Am I true? Anyone please correct me. Thanks in advanced.


问:
a funktion that behaves in the way of acceleration at start, and a constant force after a set time..
How id like the funktion to behave, acceleration at start, and a constant force after a set amount of time

All i want is a funktion that behave this way, il edit it myself if need be to fit for its purpose..

答: You can choose your function to be $y(x)=\tanh{x}$, for $x\geq 0$. This function is bounded by the line $y=1$.


问:
Field extension of $\Bbb Q$ of degree $4$ has reals
Suppose $L / \mathbb{Q}$ is a field extension with $[L : \mathbb{Q}]$ = 4, with $L \not\subset \mathbb{R}$. Is it true that $L \cap \mathbb{R} \neq \mathbb{Q}$? If not, is it true if $L / \mathbb{Q}$ is normal?

I have tried to suppose $a \in L \cap \mathbb{R} \implies a \in \mathbb{Q}$, then from $z = a + b i \in L$ we get, $z + \bar{z} = 2a$ ($\bar{z} \in L$ if $L$ is normal at least). So $a \in \mathbb{Q}$. But I don't know how to proceed from there. I guess one should be able to show that $L = \mathbb{Q}(i)$ which contradicts degree $4$.

答:
Hint: As an example, consider a primitive 8th root of unity $\xi$. It obeys the equation $\xi^4=-1$.

Then $L = \{a+b\xi+c\xi^2+d\xi^3\mid a,b,c,d\in\Bbb Q\}$.


答:
Pick $5$ of $8$ rows to occupy. Sort those in ascending order. Hence the order they were picked in will be forgotten. This gives:
$$
8C5=\binom 85
$$

Pick $5$ of $8$ columns to occupy. Pair those with the ordered set of chosen rows in the order you pick them. Hence order counts this time, and we have:
$$
8P5=8!/(8-5)!=8\cdot 7\cdot 6\cdot 5\cdot 4
$$

which is where the extra product comes from.

Title: Find the probability of placing $5$ dots on an $8 \times 8$ grid.

Body:
$5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.

What is the probability that no row or column has more than $1$ dot?

I thought about this in the following way, the number of ways to place $5$ dots on $64$ cells is $$\begin{pmatrix} 64 \\ 5 \end{pmatrix} $$
Now if I want the $8 \times 8$ to have no row or column with more than $1$ dot then I reasoned that the dots must be placed in the diagonal which has $8$ cells, so the number of ways I can place $5$ dots into $8$ cells is $$\begin{pmatrix} 8 \\ 5 \end{pmatrix} $$

So the desired probability would be $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$

However I know that this is not the answer, the correct answer is $$\frac{ \begin{pmatrix} 8 \\ 5 \end{pmatrix} 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4  }{\begin{pmatrix} 64 \\ 5 \end{pmatrix} }$$
but I can't see what the term $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ counts, can you explain how this term came?


答:
Too long for comments.

Let
$$f(a)=\int_{5a}^{7a}\log (\cot (t))\,dt+\int_{a}^{3a}\log (\cot (t))\,dt$$ leads to
$$f(a)=a \Big[-\log (\cot (a))+3 \log (\cot (3 a))-5 \log (\cot (5 a))+7 \log (\cot (7 a))\Big]+$$
$$\frac{\pi}{2}  \Big[ \log (\cot (a))-\log (\cot (3 a))+\log (\cot (5 a))-\log (\cot
   (7 a))\Big]+$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (a))-\text{Li}_2(-i \cot (a)))-$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (3a))-\text{Li}_2(-i \cot (3a)))+$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (5a))-\text{Li}_2(-i \cot (5a)))-$$
$$\frac{i}{2}  (\text{Li}_2(i \cot (7a))-\text{Li}_2(-i \cot (7a)))$$
At this point, I am stuck.

Expanding $f(a)$ as a (long) Taylor series around $a=0$ and making $a=\frac \pi {36}$ leads to
$$f\left(\frac{\pi}{36}\right)=
0.407095819634319562246490451081059604788510834$$ which is recognized as $\frac{4 }{9}C$ by the inverse symbolic calculator (notice that all digits are exact).

One point on the side : $\frac{4 }{9}C$ is the maximum value of $f(a)$.

Title: Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $

Body:
I have the conjecture for the integral

$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$

where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.

I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral

$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$

is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.

Edit:

Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$

which may not reduce the difficulty, albeit appearing compact.


问:
Interpolation with Regularized Linear Squares
I am fundamentally missing something and would appreciate some clarification. I am reading the book Statistical Image Processing and Multidimensional Modeling on Regularized Least Squares. This text has a problem where I have to find z, a set of unknown elements. z in this case is a column vector of five 2D points.

To compute $\hat{z}$, I use the following equation (provided by the text):

$$\hat{z} = (C^TR^{-1}C+\lambda LL^T)^{-1}C^TR^{-1}\vec{m}$$

where C is the transformation matrix, L is the constraint matrix, $\vec{m}$ is the measurement column vector and R is an Identity matrix.

$$\vec{m} = C\vec{z} + \vec{v} = \begin{bmatrix}z_{1}\\z_{5}\end{bmatrix} +\vec{v}$$

where $\vec{v}$ ~ I and $\vec{m} = \begin{bmatrix}10\\50\end{bmatrix}$

Here's the thing this problem is set on a 2D plane, so z is a vector of 2D (x,y) points. However, the text does not give m, a set of measurements on the 2D plane, in the same type. Instead m is a column vector of [10;50]. This reduces z to a column vector as well.

The question then asks us to plot z. How can I plot z if I have only a column vector?

What am I missing here?

答:
I think you can set the vector $\vec{z}$ in the following form
\begin{equation}
\vec{z} =
\begin{pmatrix}
x_1\\
y_1\\
x_2\\
y_2\\
\vdots\\
x_n\\
y_n
\end{pmatrix},
\end{equation}
or, for example, in the following form
\begin{equation}
\vec{z}=
\begin{pmatrix}
x_1\\
x_2\\
\vdots\\
x_n\\
y_1\\
y_2\\
\vdots\\
y_n
\end{pmatrix}.
\end{equation}


答:
Assume that $L/\Bbb Q$ is normal. Let $\sigma$ be the field automorphism given by complex conjugation (which is a field automorphism because the extension is normal). Then the subgroup $H$ of $\operatorname{Aut}(L)$ generated by $\sigma$ has order $2$, so $L$ has degree $2$ over the fixed field $L^H$. We get $[L^H:\Bbb Q]=4/2=2>1$ and $L^H\subset \Bbb R$, i.e. $L\cap \Bbb R\ne \Bbb Q$.

Without the normal assumption it is not necessarily true, take a polynomial of degree $4$ with Galois group $S_4$ and a nonreal root $\alpha$. Then $L=\Bbb Q(\alpha)$ doesn't have any non-trivial intermediate fields at all, so since $L\nsubseteq\Bbb R$ we get $L\cap \Bbb R=\Bbb Q$.

Title: Field extension of $\Bbb Q$ of degree $4$ has reals

Body:
Suppose $L / \mathbb{Q}$ is a field extension with $[L : \mathbb{Q}]$ = 4, with $L \not\subset \mathbb{R}$. Is it true that $L \cap \mathbb{R} \neq \mathbb{Q}$? If not, is it true if $L / \mathbb{Q}$ is normal?

I have tried to suppose $a \in L \cap \mathbb{R} \implies a \in \mathbb{Q}$, then from $z = a + b i \in L$ we get, $z + \bar{z} = 2a$ ($\bar{z} \in L$ if $L$ is normal at least). So $a \in \mathbb{Q}$. But I don't know how to proceed from there. I guess one should be able to show that $L = \mathbb{Q}(i)$ which contradicts degree $4$.


问:
exterior angle of a reflex angle
How are exterior angles for reflex angles defined? Are they negative? This doubt came to my mind while thinking about the sum of the exterior angles of a concave polygon. I think it should be negative because this would allow for the sum of the exterior angles to add up to 360 degrees, just like in the convex polygon counterpart.</p>

答:
I worry about Joseph's language (turn angle and external angle) and maybe it's different to UK. We use interior and exterior only. Interior is self explanatory- sum of interior = (n-2)x pi rad/180 degrees. Interior + Exterior = pi rad/180 degrees at each vertex, ie is the angle made by extending a side at one vertex. Sum of exterior angles =2pi rad/360 degrees.

As you say, Apekshik, this must mean that the exterior angle of reflex angles is negative. If you remember the derivation of the rule (form triangles at the centre of the shape, find angle at base, then find exterior by angles on a straight line- works best for regular shapes but can be done for any polygon- angles at centre sum to 2pi/360 so the exterior angles do as well), what do the triangles look like and where are the exterior angles? Pretty strange because the exterior angles are inside the shape (and negative in this sense). So that the red, exterior, angle is negative in the sense that is inside not outside the shape. I add the beginning of the proof for sum of exteriors but each exterior is half of two central angles and when summed they make the sum of the central angles.

Title: exterior angle of a reflex angle


问:
How to calculate 3D position and update it
I'm trying to calculate the new position of an object moving over the surface of  a 3D sphere after some time Δt, from X1 to X2 always onto the surface of the sphere. The measurements/data I have of such object's displacement and sphere are:

Translational velocity (cm/seg)

Rotational velocity (deg/seg)

Direction of displacement (deg)

Radius of the sphere (cm)
Here is an example of what I mean.

Then the idea is to update for each Δt the position of the object on the sphere. I suspect some  linear algebra is needed but pretty lost at the moment. Any help would be greatly appreciated!

This is how the data frame looks like

time
translational velocity
rotational velocity
direction

0.01
2.36
6.45
0.78

0.02
1.12
1.19
0.62

0.03
1.67
4.45
1.51

0.04
1.25
5.39
1.67

答:
You need to solve Euler-Lagrange equation of motion along the geodesic curve in your metric space:
\begin{equation}
\frac{d^2x^{\mu}}{dt^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt} = 0,
\end{equation}
where t is the time, $x^{\mu}$ are coordinate components, $\Gamma^{\mu}_{\alpha\beta}$ is the Christoffel symbols of the metric.
The Cristoffel symbols $\Gamma^{\mu}_{\alpha\beta}$ are given by the following formula
\begin{equation}
\Gamma^{\mu}_{\alpha\beta} = \frac{1}{2}g^{\mu\nu}(\frac{\partial g_{\nu\alpha}}{\partial x^{\beta}} + \frac{\partial g_{\nu\beta}}{\partial x^{\alpha}} - \frac{\partial g_{\alpha\beta}}{\partial x^{\nu}}),
\end{equation}
where $g_{\alpha\beta}$ is the metric tensor. $g_{\alpha\beta}$ and $g^{\beta\gamma}$ are linked by the following equation
\begin{equation}
g_{\alpha\beta}g^{\beta\gamma} = \delta_{\alpha}^{\,\gamma} = \delta^{\gamma}_{\,\alpha},
\end{equation}
where $\delta^{\gamma}_{\,\alpha}$ is the Kronecker symbol:
\begin{equation}
\delta^{\gamma}_{\,\alpha} =
\begin{cases}
1,\text{ if }\gamma=\alpha\\
0,\text{ otherwise}
\end{cases}.
\end{equation}
So the matrix $g^{\alpha\beta}$ is inverse of the matrix $g_{\alpha\beta}$. The metric tensor is the tensor, which links the length $ds$ of differential minor arc with the coordinate differentials $dx^{\alpha}$:
\begin{equation}
ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}.
\end{equation}
In the case of the sphere
\begin{equation}
g =
\begin{pmatrix}
1 & 0\\
0 & \sin^2\theta,
\end{pmatrix},
\end{equation}
so $ds^2$ is given by the following formula
\begin{equation}
ds^2 = d\theta^2 + \sin^2\theta d\phi,
\end{equation}
where $\theta$ is a polar angle, $\phi$ is axial angle. $\theta$ and $\phi$ are the components of the vector $x^{\mu}$ in this case. The differential $dx^{mu}$ can be written in the following form $dx^{\mu} = \frac{dx^{\mu}}{dt}dt$, where $dt$ is the time differential and $\frac{dx^{\mu}}{dt}$ is the generalized speed. Let us consider the Lagrange functional of the following form
\begin{equation}
\mathcal{L}[x^{\mu}] = \int^{s_1}_0 ds = \int^{s_1}_{0}\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}} = \int^{t}_{0}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{dt}dt\frac{dx^{\nu}}{dt}dt} = \int^{t}_{0}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}}dt.
\end{equation}
We can use the calculus of variations and find a function $dx^{\mu}(t)$ that minimizes our functional. To do this, you need to write down the Euler-Lagrange equation. In the case of this functional, it has the form
\begin{equation}
\frac{d^2x^{\mu}}{dt^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt} = 0.
\end{equation}
Since the functional $\mathcal{L}[x^{\mu}] $ has the meaning of length, the solution to the previous equation is a curve of shortest length connecting two points in a given metric space. Such curves are called geodesic curves.
The non-zero Christoffel symbols in the case of sphere metric are (see the following link http://einsteinrelativelyeasy.com/index.php/general-relativity/34-christoffel-symbol-exercise-calculation-in-polar-coordinates-part-ii)
\begin{equation}
\Gamma^{\phi}_{\phi\theta}=\Gamma^{\phi}_{\theta\phi} = \frac{\cos\theta}{\sin\theta}, \\
\Gamma^{\theta}_{\phi\phi} = -\sin\theta\cos\theta,
\end{equation}
so the system of motion equation has the following form
\begin{equation}
\begin{cases}
\frac{d^2\theta}{dt^2} - \sin\theta\cos\theta\left(\frac{d\phi}{dt}\right)^2 = 0,\\
\frac{d^2\phi}{dt^2} + 2\frac{\cos\theta}{\sin\theta}\frac{d\theta}{dt}\frac{d\phi}{dt} = 0.
\end{cases}
\end{equation}
Let us introduce $\theta$-speed $u_{\theta}$ and $\phi$-speed $u_{\phi}$:
\begin{equation}
u_{\theta} = \frac{d\theta}{dt},\\
u_{\phi} = \frac{d\phi}{dt}.
\end{equation}
In this terms the system of motion equations has the following form:
\begin{equation}
\begin{cases}
\frac{du_{\theta}}{dt} - \sin\theta\cos\theta u^2_{\phi} = 0,\\
\frac{du_{\phi}}{dt} + 2\frac{\cos\theta}{\sin\theta}u_{\theta}u_{\phi} = 0.
\end{cases}
\end{equation}

The useful information about motion along the sphere geodesics can be found here Geodesics of the Unit Sphere using Christoffel symbols.

The simplest way to simulate motion along geodesic curve is to rotate sphere in a such way that the motion will be at a constant $\phi^{\prime}$, for example, $\phi^{\prime} = 0$. In this case only $\theta^{\prime}$ changes with the time. Then calculate $\theta^{\prime}$ as $\theta^{\prime} = \omega t$. Where $\omega$ is a cyclic frequency. Then rotate sphere back into initial basis. There $\theta^{\prime}$ and $\phi^{\prime}$ are coordinates of the point object in the rotated basis.


问:
Countable vs uncountable infinitesimals in calculus
There is a general understanding that some infinities are bigger than others, in particular the infinity of the integers, the countable set of infinite discrete elements, and that larger infinity of the uncountable continuum.

Do mathematical limits take this distinction into consideration? Having an epsilon approach zero could be via an iterated mechanism of subdivision or via some continuous shrinking process, and because there are different types of infinity it feels like the way an epsilon becomes infinitesimal should matter.

For a concrete example, I was curious about the distinction between the limit of adding countably infinite edges/vertices to a regular polygon and the traditional notion of a continuous circle. There's a meme about starting with a circle's bounding square and recursively inverting the right angles, approaching something that bounds the area of the circle in the limit but still has the perimeter of the original bounding square. That breaks down because the boundary, that first derivative, is no longer smooth, it is a discontinuous fractal of right angles.

I was curious if there was something more subtle but similarly broken in taking the limit of a regular polygon with countably infinite sides. Such a polygon would only have vertices at rational divisions of the angle around the perimeter, and in theory would be overwhelmed by all of the irrational points lying on straight edges with zero curvature. Could there perhaps be a difference in the second derivative, where a real circle is continuously curved everywhere and the second derivative of an infinite regular polygon is a dust with curvature zero almost everywhere and curvature 1/(countably infinite) just at the rational vertices?

答:
The underlying problem here is that the natural numbers are all of the following (and more):

cardinals,

ordinals,

real numbers.

Because people are exposed to limits a long time before they are exposed to any other concept of infinity, there's a tendency to think that all of these things are the same, and that "a limit is a limit is a limit", so $\aleph_0$ and $\omega$ should somehow connect to the real numbers. Maybe not as real numbers directly, but maybe in its "infinite relative" in the form of infinitesimals.

But this is not how it works. These systems have a different notion of limits. The notion of infinitesimals is not related to cardinals, and cardinals are not real, hyperreal, or surreal numbers.

Now, it's true that you can ask "how many sides does a circle have", and you'd expect the answer to be in the form of a cardinal number (because those are the answers to "how many objects"). Now, we need to define what constitutes as a "side" here, and then we can have a proper answer.

The real answer, though, is that there is a lot of subtlety here, and a flagrant lack of continuity when it comes to limits. So even if a circle is somehow a limit of polygons, the number of sides of a circle need not be the limit of the number of sides of the polygon. That's just not how it works.


问:
The angle between the pair of tangents drawn from a point P to the circle and its locus
The angle between the pair of tangents drawn from a point P to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha  + 13{\cos ^2}\alpha  = 0$ is $2\alpha$ . Then the equation of the locus of the point P is

A- $x^2+y^2+4x−6y+4=0$

B- $x^2+y^2+4x−6y-9=0$

C- $x^2+y^2+4x−6y-4=0$

D- $x^2+y^2+4x−6y+9=0$

My approach is as follow

${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha  + 13{\cos ^2}\alpha  = 0 \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} = 13{\sin ^2}\alpha  - 9{\sin ^2}\alpha  = {\left( {2\left| {\sin \alpha } \right|} \right)^2} = {R^2}$

${X^2} + {Y^2} = {R^2}$

$Y = mX \pm \sqrt {{{\mathop{\rm R}\nolimits} ^2} + {R^2}{m^2}} $ is the general equation of the tangent

$\Rightarrow Y - mX =  \pm R\sqrt {1 + {m^2}}  \Rightarrow {Y^2} + {m^2}{X^2} - 2mYX = {R^2} + {R^2}{m^2}$

$ \Rightarrow {m^2}\left( {{X^2} - {R^2}} \right) - 2mYX + {Y^2} - {R^2} = 0$

$\tan 2\alpha  = \frac{{{m_1} + {m_2}}}{{1 - {m_1}{m_2}}} = \frac{{\frac{{2YX}}{{\left( {{X^2} - {R^2}} \right)}}}}{{1 - \frac{{\left( {{Y^2} - {R^2}} \right)}}{{\left( {{X^2} - {R^2}} \right)}}}} = \frac{{2YX}}{{{X^2} - {Y^2}}}$

$2\sin \alpha  = R \Rightarrow \sin \alpha  = \frac{R}{2} \Rightarrow \tan \alpha  = \frac{R}{{\sqrt {4 - {R^2}} }}$

$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2YX}}{{{X^2} - {Y^2}}} = \frac{{2\left( {y - 3} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2} - {{\left( {y - 3} \right)}^2}}}$

not able to proceed from here

答:
The choices should alert you that a simpler method is possible.

On drawing a diagram, we see that $P$ is at a constant distance from center $O=(-2,3)$. You can now finish.


答:
There are several problems in what you wrote, and discussing the growth and the limit is not really necessary.

Obviously $x_n\ge\color{blue}0$, and this is a lower bound. Also $x_n=\dfrac1{3^{n+5}}=\dfrac1{3^{n-1}}\dfrac1{3^{1+5}}\le\color{red}{\dfrac1{3^{1+5}}}$ and this is an upper bound.

It turns out that these bounds are at the same time the infimum and supremum.

Infimum: we have $\color{blue}0\le x_n$. Now let $t>0$. We certainly have $x_n=\dfrac1{3^{n+5}}<t$ for some $n$ large enough, because this condition is equivalent to $-(n+5)<\log_3 t$.

Supremum: we have $\color{red}{\dfrac1{3^{1+5}}}\ge x_n$ and $\dfrac1{3^{1+5}}= x_1$. The supremum is also a maximum.