问:
Basis Element of the Product topology vs. Element of the Basis
Hi i am reading Topology by Munkres and have some doubts under theorem 20.3 regarding basis element of product topology. It is written that:

Let $B=(a_1,b_1)\times...\times(a_n,b_n)$ be a basis element for the product topology and let $\bar x =(x_1,...,x_n)$ be an element of $B$.

My doubts are as follows:

I am unable to understand the above quoted statement. What does the notation $(a_1,b_1)\times(a_2,b_2)$ stand for?

What is the difference between basis element $B$ and an element of $B$? I mean a basis element for the product topology on the cartesian product was defined by $U\times V$ previously in the book where U and V are open sets in X and Y respectively. Then why is it now $(a_1,b_1)\times(a_2,b_2)$ ? And how is it different from the tuple x =(x_1,...,x_n)?

For reference i am attaching the screenshot where i have highlighted the quoted text.

答:
The $\times$ refers to the Cartesian product. Each $(a_i, b_i)$ is an (open) interval in $\mathbb{R}$. To visualise this, this is a box in the plane without its borders. So, for example,
$$
(a_1, b_1) \times (a_2, b_2) = \{ (x_1, x_2) \in \mathbb{R}^2 : a_1 < x_1 < b_1, a_2 < x_2 < b_2 \} .
$$
This pattern is then generalised to hold for $\mathbb{R}^n$.

A basis for a topology means that every set in the topology can be generated as a union of elements of the basis (these elements are themselves sets). The product topology for $\mathbb{R}^n$ is defined as the one generated by these so-called cylindrical sets - an arbitrary such set is $B = (a_1, b_1) \times \dots \times (a_n, b_n)$. Since $B$ is itself a set (it is the Cartesian product of the $n$ open intervals), that means it contains points (if it is not empty), such as $x = (x_1, \dots, x_n)$.


问:
Can two distinct sets of N numbers between -1 and 1 have the same sum and sum of squares?
Is is possible to find two different sets of numbers $\{ a_1, a_2, \dots, a_N\}$ and  $\{ b_1, b_2, \dots, b_N\}$ with $a_i,b_i\in[-1,1]$ such that $\sum a_i = \sum b_i$ and $\sum a_i^2 = \sum b_i^2$ are both true at the same time?

EDIT: Also, no number in each set may be $0$, i.e.   $a_i,b_i \neq 0$. John Omielan has posted an example for $N=3$, before I edited the question.

For $N=1$, it is obvious that this is impossible, because $a_1 = b_1$ (sum over elements in the set) and $a_1 \neq b_1$ (the sets must be different) cannot be true at the same time. Is there a minimum number of $N$ for which it is possible to find such sets, or is it never possible?

答:
As you stated, it's not possible for $N = 1$. It's also not possible for $N = 2$, where I assume the order doesn't count. To see this, assume there is a solution to get

$$a_1 + a_2 = b_1 + b_2 = c \tag{1}\label{eq1A}$$

$$a_1^2 + a_2^2 = b_1^2 + b_2^2 = d \tag{2}\label{eq2A}$$

Squaring the first $2$ sides of \eqref{eq1A} and subtracting \eqref{eq2A} gives

$$2a_1 a_2 = 2b_1 b_2 \implies a_1 a_2 = b_1 b_2 = e \tag{3}\label{eq3A}$$

Using Vieta's formulas, or just simply expanding a quadratic, i.e., $(x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$, we have that $a_1$ and $a_2$, as well as $b_1$ and $b_2$, are the roots of

$$x^2 - cx + e = 0 \tag{4}\label{eq4A}$$

Since \eqref{eq4A} has only $2$ roots (e.g., by seeing it's a parabola or, more formally, using the Fundamental theorem of algebra), this means $a_1$ and $a_2$ must be the same, up to order, as $b_1$ and $b_2$.

However, for $N = 3$, we have

$$\frac{1}{2} - \frac{1}{2} + 0 = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{3}} - \frac{1}{2\sqrt{3}} = 0 \tag{5}\label{eq5A}$$

$$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0 = \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{2\sqrt{3}}\right)^2 + \left(\frac{1}{2\sqrt{3}}\right)^2 = \frac{1}{2} \tag{6}\label{eq6A}$$

For any $N \gt 3$, as stated in Kavi Rama Murthy's question comment, examples can be constructed by just adding zeros.

Update: Since the question now states $0$ are not allowed as values, then as dm63's comment suggests, we can just simply use values which sum to $0$ and with $b_i = -a_i$. For example, with $N = 3$, we could use

$$\frac{1}{2} - \frac{1}{3} - \frac{1}{6} = -\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 0 \tag{7}\label{eq7A}$$

$$\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{6}\right)^2 = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{7}{18} \tag{8}\label{eq8A}$$

To not have the values adding to $0$, have the $a_i$ add up to a value relatively close to $0$, with $b_1$ being equal to this (e.g., have $|b_1| \lt |a_i|$ for $1 \le i \le 3$). Also, have $b_2 \gt 0$ and $b_3 = -b_2$. Then, using that the sums of squares are equal, solving for $b_2$ gives

$$b_2 = \sqrt{\frac{a_1^2 + a_2^2 + a_3^2 - b_1^2}{2}} \tag{9}\label{eq9A}$$

For example, we have

$$-\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{1}{12} + \frac{\sqrt{69}}{12\sqrt{2}} - \frac{\sqrt{69}}{12\sqrt{2}} = \frac{1}{12} \tag{10}\label{eq10A}$$

$$\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2 + \left(\frac{\sqrt{69}}{12\sqrt{2}}\right)^2 + \left(-\frac{\sqrt{69}}{12\sqrt{2}}\right)^2 = \frac{35}{72} \tag{11}\label{eq11A}$$

You can easily extend this to $N \gt 3$. Also, you can also ensure that all $|a_i|$ and $|b_i|$ are unique, but the algebra becomes more complicated and messy.

ContentLicense: CC BY-SA 4.0


问:
How to find the length of a segment which connects two triangles in a rhomboid?
The problem is as follows:

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{1 cm}\\
2.&\textrm{3 cm}\\
3.&\textrm{5 cm}\\
4.&\textrm{6 cm}\\
\end{array}$

I was only able to spot on:

$\triangle APT$ then $AT= 5$

Then:

$CT=4\cdot5=20$

But that's it. What else to do from here?.

What I attempted to find is some way to get:

$\frac{TO}{TC}=\frac{OR}{QC}$

However none of those segments seem to be known. How can this be solved relying only in euclidean geometry postulates?.

If possible please include a drawing in the answer so I can spot if a construction is needed.

答:
$APT$ right triangle

$AP=4; PT=3$

$AT = 5$ Pythagoras' theorem

$CT=4AT=20\to AC=25$

$OA=25/2=12.5$

$OT=OA-AT=12.5-5=7.5$

Triangles $ATP$ and $OTR$ are similar

$OR/OT=AP/AT$

$OR=AP/AT OT=4/5 7.5$

$OR=6$


问:
Is this an integrating factor at all and why does it work?
So I required some help for the differential equation: $y' = y\,\sin(x)+\sin(2\,x)$

I figured that it's not that easy to solve because of the integral $\int\frac{\sin(2\,x)}{e^{cos(x)}}\mathrm{d}x$ that is appearing in the premade formula.

However a kind user suggested me to multiply the particular solution $y_p = \sin(2\,x)$ with $e^{\cos(x)}$ and integrate afterward. As a result the integral becomes pretty handy. To be honest I recognized the beauty of this approach but didn't understand it yet - why this is functioning.
Can someone explain to me afterwards?

答:
Given a general, linear, first order ODE:
\begin{equation}x'(t) + p(t)x(t) = f(t).\tag{1}\end{equation}
The idea is to multiply the equation by a function $y(t)$ such that $(1)$ becomes of the form:
$$\frac{d}{dt}\bigl(y(t)x(t)\bigr) = y(t)f(t) \tag{2}$$
Then assuming that you can express $\int y(t)f(t)dt$, you get your solution by integrating $(2)$.

The factor can be found by integrating $y'(t) = p(t)y(t)$. Indeed:
$$y(t)x'(t) + y(t)p(t)x(t) = \frac{d}{dt}\bigl(y(t)x(t)\bigr) - y'(t)x(t) + y(t)p(t)x(t)$$
implies the condition:
$$y'(t)x(t) = y(t)p(t)x(t) \iff y'(t) = p(t)y(t) $$


问:
How to visualize the Standard Bounded Metric
Let $X$ be a metric space with metric $d$. Define $d'(x,y)=\min\{d(x,y),1\}$ If we assume we are working on $\mathbb{R}^2$ and $d(x,y)$ is the euclidean metric where we use open balls, how am I suppose to visualize $d'$? Basically, both metrics are equivalent for $d(x,y)$ less than or equal to $1$, but if $d(x,y)$ is greater than $1$, then $d'$ will just be one no matter how big the open balls are.

How do we draw an open ball on $\mathbb{R}^2$? Also I believe these two metrics induces the same topology, but again, I am having trouble visualizing why that would be the case.

答:
Consider a conical kitchen sink of depth $1$ unit, embedded in a flat (infinite) draining board.

Pretend that $x$ is the point of the plug hole at the bottom of the sink (the analogy falls down here because a plug hole is not a point, but you get the idea).

The depth of the water is your $\epsilon$. The open $\epsilon$-ball is the surface of the water.

When $\epsilon$ is less than $1$, the open $\epsilon$-ball is in the sink, and is round like you expect.

When $\epsilon$ is greater than $1$, the sink has overflowed, and the surface of the water has now gone everywhere.

A proof that $d'$ is equivalent to $d$ can be found here:

https://proofwiki.org/wiki/Standard_Bounded_Metric_is_Metric/Topological_Equivalence


问:
Integration of $\sin(x)/x$ from $0$ to $1$: exact value
I am trying to find the area under the curve $\sin(x)/x$ between $x=0$ and $x=1$, i.e. $\int_0^1\sin(x)/x$.

By the Taylor's series approximation (and proper justification), the answer gets to be
$$
1-\frac{1}{3\cdot 3!} + \frac{1}{5\cdot 5!} -\frac{1}{7\cdot 7!}+\cdots
$$
The values of first three partial sums of the above series go near to $0.94$.

If I want to know the exact limit of the above series, can one suggest some way?

答:
$$\int_0^1 \frac{\sin (x)}{x} \, dx=\text{Si}(1)\approx 0.946$$ where $\text{Si}$ is Sine integral, a special function  defined by $\text{Si}(x)=\int_0^x\frac{\sin t}{t}\,dt$

edit

Since
$$\text{Si}(1)=\sum _{n=0}^{\infty } \frac{(-1)^n}{(2 n+1) (2 n+1)!}$$
is an alternating series, taking the first four terms gives about $0.946083$ with a maximum error  given by the first term we neglet in absolute value, that is $3\times 10^{-7}$ which means that $6$ decimal digits are exact.

TL;DR

The integral has no solution in terms of known function, so mathematicians defined a special function to name this result. There are many others, like  Cosine integral
$$\text{Ci}(x)=-\int_x^{\infty}\frac{\cos t}{t}\,dt$$


问:
Where does the Normal Distribution of Error terms come from, and why?
In a Simple or Multiple Linear Regression Model, we are given the assumption that the error terms, $\epsilon$, are distributed like so:

$ε ∼ N(0, I_nσ^2).$

Can anybody explain where the mathematical logic comes from to derive this assumption, or what the intuition is behind it?

答:
Some of the reasons for this choice, I would say, include

Intuitively reasonable: the Normal distribution is very nice, in that it is mean zero, symmetric (i.e. there is no bias direction in the error), has a memorable bell-shape, etc. All these properties make it a very "plausible" assumption for how errors would be distributed.

Mathematically convenient: this is essentially the backbone for a lot of statistical inference, including hypothesis tests, models, etc. For example, this allows you to deduce the distributions of different test statistics, sums, etc.

Of course, it is not the only possibility; after all, it is only an assumption. For example, generalisations such as GLMs allow for other error distributions of the response variable.


问:
Proof of Hölder's inequality by weighted AM-GM inequality
I am going throw my notes, and for the proof of Hölder's inequality, I wrote the following:

Theorem. (Hölder) Let $1 \leq q,r \leq \infty $ such that $ \dfrac{1}{q} + \dfrac{1}{r} = 1 $ (here we assume $\frac{1}{\infty} = 0$), and let $a, b \in \mathbb{R}^p$ ($p\in \mathbb{N}^+$). Then we have $$ \langle a,b \rangle \leq \|a\|_q \|b\|_r,$$
where $\|x\|_t = \left(|x_1|^t + \ldots + |x_p|^t\right)^\frac{1}{t}$ and $\|x\|_\infty = \max\limits_{1\leq i \leq p} (|x_i|)$.

Proof. If $a = \mathbf{0}$ or $b = \mathbf{0}$, the proof is trivial. If $q = 1$ or $r = 1$ the proof is simple.
If $1<q,r<\infty$ and $a \neq \mathbf{0} \neq b$, then we can assume $\|a\|_q = \|b\|_r = 1 $. Then we apply the weighted AM-GM inequality to the numbers $|a_i|^q, |b_i|^r$ with weighs $\dfrac{1}{q}, \dfrac{1}{r}$, which finishes the proof.

My questions: Why is the proof simple, if $q = 1$ or $r = 1$? What I don't see either is how to finish the proof with AM-GM? I tried to apply that inequality but I don't see how it's related to Hölder's inequality...

Any help would be appreciated!

答:
The first problem is clear by recalling what is the definition of 1-norm and infinity-norm.

To show the second, apply the generalized AM-GM so we get
$$
\frac{1/p|a_i|^p + 1/q|b_i|^q}{1} \geq |a_i||b_i|
$$
Now sum the equation through all $i$. Then we get
$$
\frac{1}{p}\sum_i |a_i|^p + \frac{1}{q} \sum_j|b_j|^q \geq \sum_i |a_i||b_i| \geq \langle a,b \rangle
$$
Now as you assume $||a||_p = ||b||_q = 1$, $\sum_i|a_i|^p = \sum_j|b_j|^q = 1$, so
$$
\frac{1}{p}\sum_i |a_i|^p + \frac{1}{q} \sum_j|b_j|^q = \frac{1}{p} + \frac{1}{q} = 1 = ||a||_p||b||_q
$$
so the inequality is proved as required.


问:
Probability of total number of event A is less than the total number of event B
Imagine an experiment which outputs either  $A$ or  $B$ or $C$. $A$ happens with the probability $p_a$, $B$ happens with probability $p_b$ and $C$ happens with probability $p_c$.   Here, $p_a > \frac{1}{2}$ and $p_b > p_c$ and $p_a + p_b + p_c = 1$. After running the experiment $n$ times, what is the probability of having $n_a$ times $A$ and $n_b$ times event $B$ such that $n_a<n_b$.

What I tried is the following:

$Pr[n_a < n_b] \leq \sum_{i=0}^{n_b-1}\binom{n}{i}p_{a}^i(1-p_{a})^{n-i}\leq \sum_{i = 0}^{n_b-1}\binom{n}{i}p_{a}^n$ since $p_a > p_c + p_b$. Then, I could not upper bound this with a nice formula which is what I need and also I am not sure if it is the best approach or correct.

答:
In your solution you completely disregard the event $C $. The correct computation shall be:
$$
\sum_{n_c=0}^{n-1}\sum_{n_a=0}^{\left\lceil\frac {n-n_c}2\right\rceil-1}\frac {n!}{n_a!n_c! (n-n_a-n_c)!}p_a^{n_a}p_b^{n-n_a-n_c}p_c^{n_c}.
$$
I do not think the expression can be simplified in general case.


问:
Mathematics behind parallel resistors
Could someone explain the mathematical logic behind the summation of total resistance for parallel resistance?
$$\frac{1}{R_{tot}} = \frac{1}{R_1}+\frac{1}{R_2}+\dots+\frac{1}{R_n}$$
I saw the similar way of computing many times with other physics phenomena, for example when calculating spring coeficient of more springs connected in series, but I don't really understand it.

答: The electrical conductance is given by $\frac{1}{R}$, meaning the inverse value of resistance. The bigger conductance is, the better it conducts electricity. If you plug two resistors in parallel, their conductance is summed together, which is quite intuitive.

Title: Mathematics behind parallel resistors


问:
Does the set intersection operation has unit
I'm trying to get better understanding of binary operations, and I came across this problem: namely on one online discussions I saw that set intersection as binary operation doesn't have a unit, however I think it has:

The set intersection operation for any set universe $U$ is defined on the $\mathcal P(U) \times \mathcal P(U)$ set. From the definition of unit for binary operation, wouldn't it be true that the $U$ set is the unit of this operation. Any set in $PU$ intersected with $U$ will give back the set, and it works in both directions.

Am I doing something wrong, or the discussion on that place was wrong?

答:
For a given set $U$, the intersection indeed has $U$ as the identity element, among the subsets of $U$.

On the other hand, seeing the intersection as an operation on all sets, its identity element would be the biggest set (the union of all sets), and such a thing doesn't exist (at least it's not a set).


问:
A function that is measurable on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$
This seems to be a result that is used often, which I am having difficulty proving. There may be a more general statement, but I'll write out the statement that I was trying to prove:

Let $f:(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n)) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be a function that is continuous on a set $A \in  \mathcal{B}(\mathbb{R}^n)$, which may not necessarily be continuous on $A^c$. Then, $f \chi_{A}$ is measurable in $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, where $\chi$ is the indicator function.

My attempt was to start on the set $A$ and artificially 'extend' $f \chi_{A}$ continuously to all of $\mathbb{R}^n$. But I met many difficulties, as the set $A$ may not be convex, and in general borel sets are hard to visualize. Does this have a neat solution?

答:
A direct approach:

EDITED ANSWER

Since $f\chi_A(x)=\begin{cases}f(x)&x\in A\\0&x\notin A\end{cases}$, take $a\in\Bbb  R$.

If $a> 0$, then $$\{f\chi_A(x)< a\}=\{x\in A:\, f(x)< a\}\cup A^c$$
Since $f$ is continuous on $A$, the first set is a (relative) open set on $A$, so measurable ($A$ is measurable). So both sets are measurables and we are done.

Now, if $a\le 0$, then $$\{f\chi_A(x)< a\}=A\cap\{f(x)<a\}$$ and again is measurable using the same argument above.

PREVIOUS (not correct) ANSWER

If $a\ge 0$, then $$\{f\chi_A(x)\le a\}=\{x\in A:\, f(x)\le a\}\cup A^c=(A\cap\{f(x)\le a\})\cup A^c$$ and this set is measurable since both $A$ and $f$ so are.

Now, if $a<0$, then $$\{f\chi_A(x)\le a\}=A\cap\{f(x)\le a\}$$ and again is measurable since $A$ and $f$ are.


Title: Can two distinct sets of N numbers between -1 and 1 have the same sum and sum of squares?

Body:
Is is possible to find two different sets of numbers $\{ a_1, a_2, \dots, a_N\}$ and  $\{ b_1, b_2, \dots, b_N\}$ with $a_i,b_i\in[-1,1]$ such that $\sum a_i = \sum b_i$ and $\sum a_i^2 = \sum b_i^2$ are both true at the same time?

EDIT: Also, no number in each set may be $0$, i.e.   $a_i,b_i \neq 0$. John Omielan has posted an example for $N=3$, before I edited the question.

For $N=1$, it is obvious that this is impossible, because $a_1 = b_1$ (sum over elements in the set) and $a_1 \neq b_1$ (the sets must be different) cannot be true at the same time. Is there a minimum number of $N$ for which it is possible to find such sets, or is it never possible?

Body:
Using the fact that $9^2+2^2=6^2+7^2$ it is easy to see that for $N=4$ the two sets

$$\{a_i\} = \{-0.9, -0.2, 0.2, 0.9\} \\
\{b_i\} = \{-0.7, -0.6, 0.6, 0.7\} \\ $$

satisfy all the requirements.


答:
Solve
$$y'=y \sin x\tag{1}$$
$\frac{y'}{y}=\sin x\to \log y=-\cos x +C\to y=ke^{-cos x}\tag{2}$

$$y'-y \sin x=\sin 2x\tag{3}$$

Then guess $y_p=a\cos x+b\to y_p'=-a\sin x$ and plug into $(3)$
$$-a\sin x -(a\cos x+b)\sin x=2\sin x \cos x$$
$$-(a+b)\sin x -a\sin x \cos x  =2\sin x \cos x$$
therefore must be $a=-2;\;b=2$ and $y_p=-2\cos x +2$

General solution to $(3)$ is
$$y=ke^{-cos x}+2-2\cos x$$

Title: Is this an integrating factor at all and why does it work?

Body:
So I required some help for the differential equation: $y' = y\,\sin(x)+\sin(2\,x)$

I figured that it's not that easy to solve because of the integral $\int\frac{\sin(2\,x)}{e^{cos(x)}}\mathrm{d}x$ that is appearing in the premade formula.

However a kind user suggested me to multiply the particular solution $y_p = \sin(2\,x)$ with $e^{\cos(x)}$ and integrate afterward. As a result the integral becomes pretty handy. To be honest I recognized the beauty of this approach but didn't understand it yet - why this is functioning.
Can someone explain to me afterwards?


答:
When resistances are connected in parallel, their two electrodes are perforce respectively at the same potential. The current that flows in them is (by definition of the resistance) inversely proportional to their resistance. Hence the total current, which is inversely proportional to the equivalent resistance, is proportional to the sum of the inverses of the individual resistances.

$$I=\sum_k I_k,\\\frac UR=\sum_k\frac{U}{R_k}.$$

With series resistances,

$$U=\sum_k U_k,\\RI=\sum_kR_kI.$$

Title: Mathematics behind parallel resistors

Body:
Could someone explain the mathematical logic behind the summation of total resistance for parallel resistance?
$$\frac{1}{R_{tot}} = \frac{1}{R_1}+\frac{1}{R_2}+\dots+\frac{1}{R_n}$$
I saw the similar way of computing many times with other physics phenomena, for example when calculating spring coeficient of more springs connected in series, but I don't really understand it.


问:
A nice enumeration of $R(\omega)$
Define $R(0)=0=\emptyset, R(n+1)=P(R(n))$ and $R(\omega) = \cup_{n < \omega} R(n)$. Thus $R(\omega)$ is the set of all sets, which are build out of finitely many braces and $0$.

Consider the following relation $E$ on $\omega$: If and only if the $n$th number in the binary representation of $m$ is $1$, then $n E m$. Now I want to construct an isomorphism $(R(\omega),\in) \cong (\omega,E)$ [This is an exercise in Kunen's set theory]. After some playing around I've come up with the following definition:

$g : R(\omega) \to \omega, g(x) = \sum_{y \in x} 2^{g(y)}$.

This is a well-defined recursion, since $rank(y) < rank(x)$. If $g$ is injective, then it is easy to see that $x \in y \Leftrightarrow g(x) E g(y)$. However I don't see this; neither why $g$ is surjective.

答:
It is easier to build the isomorphism in the other direction, and indeed we can see that the map is forced upon us. There is a unique isomorphism.

Specifically, you want a map $h:\mathbb{N}\to R(\omega)$ such that $n\mathrel{E} m\iff h(n)\in h(m)$. This very equivalence tells you that you must define $h(m)=\{ h(n) \mid n\mathrel{E} m\}$. This function is defined by recursion, and obviously preserves $E$ to $\in$. Thus it is injective. It is surjective because your function $g$ is the inverse. QED

If one knows about the Mostowski collapse, then you can see immediately that $h$ is precisely the Mostowski collapse of $(\mathbb{N},E)$, which is a well-founded extensional relation. This provides another way to see that $h$ is an isomorphism.

ContentLicense: CC BY-SA 2.5

Title: A nice enumeration of $R(\omega)$

ContentLicense: CC BY-SA 3.0


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答:
Depending on every data your given, you might want to choose some polynomial interpolation.

You will notice that there are different approaches to find such a polynomial, but since it is unique, you'll always find the same.

I'm not an expert either in numerical methods nor in programming, so there might be other ways I'm not aware of.

Also, like Arturo says, this will be best if your scale of levels doesn't go too far to infinity (see his answer for more insight).

Title: How to "invent" a function?


问:
Improper Integral Question $\int_0^1 \ln\sqrt{x}dx $
I'm trying to compute this integral and check if it's an improper integral.

What I did so far is to write the limit.

$$\begin{align*}
\int_0^1\ln\sqrt x\, dx &= x\ln\sqrt x - \int \frac12\, dx = x\ln\sqrt x - \frac12x + C \\
&=\lim_{\epsilon\to0^+}\Bigl(x\ln\sqrt x - \frac12x \Bigr)\Bigr\rvert_\epsilon^1
\end{align*}$$
 after I set the epsilon and the value 1, I got the expression : $\epsilon \ln \sqrt{x}$ and I want to know if this expression say that this integral is not improper.

Thanks.

答:
It would have been slightly more pleasant to rewrite the integrand as $\frac{1}{2}x\ln x$.

After the integration, the issue is whether $\lim_{\epsilon\to 0^+}\epsilon\ln \epsilon$ exists. It does, and is equal to $0$.

One way of showing it is by writing $\epsilon=e^{-t}$. Then we are interested in the limit as $t\to\infty$ of $e^{-t}(-t)$.

The fact that
$$\lim_{t\to\infty} -\frac{t}{e^t}=0$$
can be proved in various ways. It follows that $\lim_{\epsilon\to 0^+}\epsilon\ln \epsilon=0$.


问:
Degree of a pmap
Today I have started reading stable homotopy. I have came across the notion of a pmap which is basically equivalence class of maps from cofinal sub-spectra. My query is what do we mean by degree of such map?</p>

答: I'm sorry you got no response. By 'degree' they mean a shift in index in the underlying maps of spaces. So a 'map of spectra of degree $r$' is the data consisting of maps of CW complexes $E_n \to F_{n-r}$ satisfying certain properties. (I hope I got the signs right.) Equivalently, it's just a map of spectra $E \to \Sigma^{-r} F$.

Title: Degree of a pmap


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

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

Any hints would be greatly appreciated.
Thanks in advance!

答:
I think Ljung's System Identification: Theory for the user is the authoritative reference, although it was a bit difficult for me.

I have found there are several "schools of thought" in the field that pervade theory, notation, and approach to essentially the same methods: control (ODEs), statistics (ARIMA methods), signal theory (Kalman filters)...

You can also differentiate between continuous/discrete time approaches. As far as I recall, Ljung's book was mainly discrete.


问:
Asymtotic convergence between a Sum and an Integral.
Let $ X_1,\ldots,X_n $ be a series on independent random variables that follow the Poisson distribution with mean $1$. Then, define Y as follows:
$$ Y_0 =1 \text{ and } Y_i = Y_{i-1} + X_i -1$$

And condition $Y$ on the event that $Y_n=0$ and $Y_i > 0$ for all $0<i<n$.

Let $g(i/n)= \frac{1}{\sqrt{n}}Y_i $.
Interpolating linearly between the obtained values creates a curve $g(s)$ on $[0,1]$.
Now let $$A = \int^1_0 g(s)\,ds $$

And define the sum
$S_n = \sum^{n-1}_{i=1}(Y_i -1) $.

I wish to show that $n^{-3/2}S_n \thicksim A$ as $n$ approaches infinity.

I have used numerical methods to check if this is true, and to my knowledge it is. However, I cannot seem to figure out a nice way to show this. Any help would be much appreciated.

答:
You are interpolating linearly, so for $s \in \left[\frac{i}{n},\frac{i+1}{n}\right]$ we get $$g(s) = \frac{g\left(\frac{i+1}{n}\right) - g\left(\frac{i}{n}\right)}{\frac{i+1}{n} - \frac{i}{n}}\left(s - \frac{i}{n}\right) + g\left(\frac{i}{n}\right)$$
With the given definitions this simplifies to
$$g(s) = \sqrt{n}(Y_{i+1} - Y_i)\left(s - \frac{i}{n}\right) + \frac{1}{\sqrt{n}}Y_i$$

So we get: \begin{align*}
A &= \int^1_0 g(s)\,ds \\ &= \sum_{i=0}^{n-1} \int_\frac{i}{n}^\frac{i+1}{n} g(s)\, ds \\ &= \sum_{i=0}^{n-1} \left(\sqrt{n}(Y_{i+1} - Y_i)\int_\frac{i}{n}^\frac{i+1}{n}\left(s - \frac{i}{n}\right)\,ds + \frac{1}{n\sqrt{n}}Y_i\right)\\ &= \sum_{i=0}^{n-1} \left(\sqrt{n}(Y_{i+1} - Y_i)\frac{1}{2n^2}\,ds + \frac{1}{n\sqrt{n}}Y_i\right) \\&= \frac{1}{2}n^{-3/2} \sum_{i=0}^{n-1}(Y_{i+1} - Y_i + 2Y_i) = \frac{1}{2}n^{-3/2}\sum_{i=0}^{n-1}(Y_{i+1} + Y_i) \end{align*}

But we can rewrite: $$
\sum_{i=0}^{n-1}(Y_{i+1} + Y_i)  = 2S_n + Y_n + Y_0 + 2(n-1)$$

So in total we get by considering $$Y_n - 1 = \sum_{j=1}^n \left(X_n - 1\right) = \sum_{j=1}^n \left(X_n - E[X_n]\right)$$ that
\begin{align*}
A &= n^{-3/2}S_n +\frac{1}{2}n^{-3/2}(Y_n + 2n - 1) \\ &= n^{-3/2}S_n +\frac{1}{2}n^{-1/2}\left(\frac{1}{n}\sum_{j=1}^n \left(X_n - E[X_n]\right) + 2\right)\end{align*}

But by the law of large numbers we have $\frac{1}{n}\sum\limits_{j=1}^n \left(X_n - E[X_n]\right) \to 0$ hence the desired result.


问:
A problem on construction of a function
Given that $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, it is to be proved that there exists an  $\xi \in (a,b)$ such that
$$f'(\xi)=\dfrac{2}{a-\xi}\cdot (f(\xi)-f(b)).$$

答:
Let
$$F(x)=(x-a)^2(f(x)-f(b)),x\in[a,b].$$
Then $F(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$ and
$$F(a)=F(b)=0.$$
By Rolle's MVT, we know there exists $\xi\in(a,b)$ such that $F'(\xi)=0.$
It is easy to see that
$$F'(x)=2(x-a)(f(x)-f(b))+(x-a)^2f'(x).$$
So
$$F'(\xi)=0\iff f'(\xi)=\dfrac{2}{a-\xi}\cdot (f(\xi-f(b)).$$

Title: A problem on construction of a function


问:
Only one month left, need help with establishing a preparation plan for these topics
I am going back to school and preparing my first-semester course which is so-called "engineering maths". I used to be a physics major 5 years ago and had ok-ish knowledge in calculus, but now the part of my calculus brain is shiny as a piece of metal; I have forgotten 90% of it.

I have exactly a month left to prepare for it, but I am a bit confused about where to start and where are these topics lie in;

*Vector calculus, including Gauss’ and Stokes’ Theorems;

*sequences and series;

*Fourier series, Laplace transforms;

*systems of homogeneous ordinary differential equations, including phase plane and linearization for nonlinear systems;

*second-order partial differential equations and separation of variables.

Given only one month time, where is the best to start and what is the best materials for learning and practising? I am doing the single variable course from MIT, but it's taking too long and makes me want to hurry to vector calculus; ironically I worry about the sort of gap among these topics and a single variable. I know it is not realistic given a short period but at least I would want to study in a correct order that covers all the topic above.

I am both excited and worried about studying maths again because I want to do well instead of just passing with Bs as I did in undergraduate.

Thanks for reading and you're much appreciated!

答:
If you can find a copy of Engineering Mathematics by A.J.M. Spencer (and others) (Van Nostrand, 1977), don't hesitate to get it. It was my undergraduate text in my 3rd year of my elec. eng. degree course back in the early 1980s. It has everything you need, not just for your 1 month re-acquaintance with the material, but should be a constant companion over the rest of your course.

If it had not been for this book, I would have failed my exam on 2nd-order partial differential equations.

I recently got a second-hand copy of it, replacing the old one I had, and it is just as good as I remember it.

There may be newer (and hence more modern) books available nowadays, but I won't be able to comment on them.


问:
Improvement of $W^{1,p}$ regularity of a elliptic equation solution.
I'm looking for some reference for results like
$$ \mbox{div}(A(x) \nabla u) = 0, \ \ u \in H^1=W^{1,2} \Rightarrow u \in W^{1,p}, p>2 $$
where $A(x)$ is elliptic, this is, $Id\lambda \le A(x) \le \Lambda Id $ with $0< \lambda<\Lambda$  are constant matrices and $Id$ is the identity matrix.
I do not know if this is exactly true. Then if you know something like this, as adding hypothesis, please tell me or send me. I will be very grateful.</p>

答: Have a look for this paper by Gröger. There, $W^{1,p}$-regularity for your problem is shown for a suitable domain $\Omega$ and mixed boundary conditions. This also applies to the case where the matrix $A$ is only $L^\infty(\Omega)$.</p>


答:
You are interested in finding the least period $q$ such that, for all $n$,

$mv(n)\equiv mv(n+q)\pmod {18}$

$\Leftrightarrow  2^{mv(n)}\equiv 2^{mv(n+q)}\pmod {27}$

$\Leftrightarrow  n^m\equiv (n+q)^m\pmod {27},$ providing $n$ and $n+q$ are coprime to $3$.

The period  has to be a factor of $27$ and so $q\in \{ 1,3,9,27 \}$.

Using the Lemma given below,  the period of $f_{m}(n)$ is $\frac {27}{(m,27)}$ for $1\le m\le 17$.

The general case for $p$ odd

The crucial equation is $ n^m\equiv (n+q)^m\pmod {p^k}.$

For $1\le m<\varphi(p^{k})$ the period is $\frac {p^k}{(m,p^k)}$.

Lemma (Hensel lifting)

For an odd prime $p$, if $a$ and $b$ are coprime to $p$ and $a\equiv b\pmod p$, then the power of $p$ dividing $a^m-b^m$ is the sum of the powers of $p$ dividing $a-b$ and $m$.

Title: Least period of functions

Body:
For $n$ coprime with $3$ define $v(n)$ as the number such that $n\equiv 2^{v(n)}\mod{27}$, and if $(n,3)>1$ define $v(n)=0$. With this we have $v(n)$ is periodic with least period $27$.
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & \cdots \\
\hline
v(n) & 18 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 &\cdots\\
\hline
\end{array}

Now for each $m\geq 1$ define $f_{m}(n):=m\cdot v(n)\mod{18}$. I've made some calculations and I found that $f_{m}$ has least period $27$ if $(m,3)=1$, also $f_{3}$, $f_{6}$, $f_{12}$ and $f_{15}$ have least period $9$ and finally $f_{9}$ has least period $3$.

My question is: How can you find the least period of $f_{m}$ as a formula in terms of $m$? (without explicitly calculating all the values). And in general, if $g$ is a primitive root mod $p^{k}$, $v(n)$ is such that $n\equiv g^{v(n)}\mod{p^{k}}$, and define $f_{m}(n):=m\cdot v(n)\mod{\varphi(p^{k})}$. How can we find the least period of each $f_{m}$?

Edit: Additional info:

\begin{array}{|c|c|c|c|}
\hline
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & \cdots \\
\hline
v(n) & 18 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 &\cdots\\
\hline
f_{1}(n) & 0 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 & \cdots \\
\hline
f_{2}(n) & 0 & 2 & 0 & 4 & 10 & 0 & 14 & 6 & 0 & 12 & 8 & 0 & 16 & 16 & 0 & 8 & 12 & 0 & 6 & 14 & 0 & 10 & 4 & 0 & 2 & 0 & 0 & \cdots \\
\hline
f_{3}(n) & \color{red}{0} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{3} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{15} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{9} & \color{red}{0} & \cdots \\
\hline
f_{4}(n) & 0 & 4 & 0 & 8 & 2 & 0 & 10 & 12 & 0 & 6 & 16 & 0 & 14 & 14 & 0 & 16 & 6 & 0 & 12 & 10 & 0 & 2 & 8 & 0 & 4 & 0 & 0 & \cdots \\
\hline
f_{5}(n) & 0 & 5 & 0 & 10 & 7 & 0 & 8 & 15 & 0 & 12 & 11 & 0 & 4 & 13 & 0 & 2 & 3 & 0 & 6 & 17 & 0 & 16 & 1 & 0 & 14 & 9 & 0 & \cdots \\
\hline
f_{6}(n) & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{0} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{0} & \cdots \\
\hline
f_{7}(n) & 0 & 7 & 0 & 14 & 17 & 0 & 4 & 3 & 0 & 6 & 1 & 0 & 2 & 11 & 0 & 10 & 15 & 0 & 12 & 13 & 0 & 8 & 5 & 0 & 16 & 9 & 0 & \cdots \\
\hline
f_{8}(n) & 0 & 8 & 0 & 16 & 4 & 0 & 2 & 6 & 0 & 12 & 14 & 0 & 10 & 10 & 0 & 14 & 12 & 0 & 6 & 2 & 0 & 4 & 16 & 0 & 8 & 0 & 0 & \cdots \\
\hline
f_{9}(n) & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \cdots \\
\hline
f_{10}(n) & 0 & 10 & 0 & 2 & 14 & 0 & 16 & 12 & 0 & 6 & 4 & 0 & 8 & 8 & 0 & 4 & 6 & 0 & 12 & 16 & 0 & 14 & 2 & 0 & 10 & 0 & 0 & \cdots \\
\hline
f_{11}(n) & 0 & 11 & 0 & 4 & 1 & 0 & 14 & 15 & 0 & 12 & 17 & 0 & 16 & 7 & 0 & 8 & 3 & 0 & 6 & 5 & 0 & 10 & 13 & 0 & 2 & 9 & 0 & \cdots \\
\hline
f_{12}(n) & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{0} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{0} & \cdots \\
\hline
f_{13}(n) & 0 & 13 & 0 & 8 & 11 & 0 & 10 & 3 & 0 & 6 & 7 & 0 & 14 & 5 & 0 & 16 & 15 & 0 & 12 & 1 & 0 & 2 & 17 & 0 & 4 & 9 & 0 & \cdots \\
\hline
f_{14}(n) & 0 & 14 & 0 & 10 & 16 & 0 & 8 & 6 & 0 & 12 & 2 & 0 & 4 & 4 & 0 & 2 & 12 & 0 & 6 & 8 & 0 & 16 & 10 & 0 & 14 & 0 & 0 & \cdots \\
\hline
f_{15}(n) & \color{red}{0} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{15} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{3} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{9} & \color{red}{0} & \cdots \\
\hline
f_{16}(n) & 0 & 16 & 0 & 14 & 8 & 0 & 4 & 12 & 0 & 6 & 10 & 0 & 2 & 2 & 0 & 10 & 6 & 0 & 12 & 4 & 0 & 8 & 14 & 0 & 16 & 0 & 0 & \cdots \\
\hline
f_{17}(n) & 0 & 17 & 0 & 16 & 13 & 0 & 2 & 15 & 0 & 12 & 5 & 0 & 10 & 1 & 0 & 14 & 3 & 0 & 6 & 11 & 0 & 4 & 7 & 0 & 8 & 9 & 0 & \cdots \\
\hline
\end{array}


问:
Why does the intersection of two parametric curves have different t results for each curve?
I'm slightly confused after reading the problem posed here: Points of intersection of two parametric curves

Why is it that the "t" result at the intersection point of the two curves is not the same? I.e. the solution involves the answer "t1" for the first curve and "t2" for the second curve, but if the two curves are being made equal, why is t1 != t2?

Also, is there any way we can "manufacture" two different curves (say two parametric straight lines) that, at a specific t value for BOTH curves, yield the SAME point in space? How would you go about doing that? (given, for example, 3 points in space)

答:
Maybe a simple example helps to see what's going on. Let's say you have two curves $$c_1:I \rightarrow \mathbb{R}^2, t \mapsto (\sin(t), \cos(t))$$
$$c_2:I \rightarrow \mathbb{R}^2, t \mapsto (\cos(t), \sin(t)),$$
where $I = [0, 2\pi]$.

Both these curves are the unit circle around the origin. So if you ask, where do these two curves intersect, the answer should basically be 'everywhere'. So that means that for any point $p$ on the unit circle, you can find $t_1$ and $t_2$ such that $c_1(t_1) = c_2(t_2) = p$. But only rarely, if ever, will you find a value $t$ such that $c_1(t) = c_2(t) = p$, that's just too much to ask. So in the exercise you link, the first question asks for $t_1$ and $t_2$, where the second one asks for a single $t$.

Hope this helps.


答:
Say you had two flies, fly A and fly B, which trace out respective parametric curves, but fly B is not as fast as fly A. We are looking for a shared point, i.e. the curves pass through the same coordinate in space, so fly A might get to the shared point quicker than fly B. They arrive at the shared point at different times. It’s not necessary that they arrive to the shared point at the same time, although it may happen in special cases as well.

When the intersection happens at the same time, that is when the two particles would collide. Two particles equidistant from a point travelling at the same speed in the direction of the that point would be an example when $t_1 = t_2$. Given lines $\mathcal{L}_1,\mathcal{L}_2, \mathcal{L’}_1$

$ \mathcal{L}_1 =
\begin{cases}
x=1-t \\
y=0
\end{cases}  ,  \mathcal{L’}_1 =
\begin{cases}
x=1-\frac{t}{2} \\
y=0
\end{cases} $

$ \mathcal{L}_2 =
\begin{cases}
x=0\\
y=1-t
\end{cases}$

$\mathcal{L_1,L_2}$ cross the origin $(0,0)$ at time $t_1=t_2=1$. $\mathcal{L’_1,L_2}$ cross the origin $(0,0)$ at times $t_1’=2$ and $t_2 =1$.

Title: Why does the intersection of two parametric curves have different t results for each curve?

Body:
I'm slightly confused after reading the problem posed here: Points of intersection of two parametric curves

Why is it that the "t" result at the intersection point of the two curves is not the same? I.e. the solution involves the answer "t1" for the first curve and "t2" for the second curve, but if the two curves are being made equal, why is t1 != t2?

Also, is there any way we can "manufacture" two different curves (say two parametric straight lines) that, at a specific t value for BOTH curves, yield the SAME point in space? How would you go about doing that? (given, for example, 3 points in space)


答:
As for your second question, given two parametric curves $$C_1 :\ (a(t),b(t)) \\ C_2 :\ (c(t), d(t)) $$

they can be made to intersect at $t=t_0$ by choosing functions $a,b,c,d$ such that $$a(t_0) = c(t_0) \\ b(t_0)= d(t_0) $$


问:
well separated points on sphere
Is there a way to generate k points on a n-sphere, say, $x_1,\dots,x_k$ such that $\min_{ i \neq j } \| x_i - x_j \| $ is as large as possible? Approximate solutions are also OK, I just need well separated points on a sphere.</p>

答:
There are lots of papers for nearly uniform distribution of points on 2-sphere, however, the question is for general n-sphere, for which the only systematic way I have seen is Fisher's 1986 PVQ: pyramid vector quantizer, used in data compressors e.g. OPUS audio codec.

It starts with uniform point distribution on l1 sphere ("pyramid"): set of integer points of fixed sum of absolute values. Then projects these points to l2 sphere. To get more uniform point distribution we can use some deformation like power before the projection.

Title: well separated points on sphere


问:
"Old-Fashioned" O-level Maths Book
I'm looking for recommendations for an "old-fashioned" style of maths textbook for O-level/GCSE, i.e., one which is concise (not full of pictures) with plenty of exercises. Ideally something which is available in PDF format online. I don't necessarily mean an old book, more a book written in a style which is typical of the early 20th century or earlier. (For instance, Gwynne's Latin is a modern book for learning Latin which I would consider to fall under this category.)

I am quite fond of this one which I own a physical copy of, but am struggling to find anywhere online in PDF format.

Also I would appreciate if it is just one book rather than many volumes. A lot of maths textbooks made for school come in many volumes with titles like "School Maths 1A, 1B, 2A, 2B", etc.

To clarify, by "O-level/GCSE", I mean a book with a table of contents which should at least tackle all of the following:

答:
For part 1 and 2--Higher algebra by hall and knight
For part 3-Coordinate geometry by sl looney

Well these were books which take from beginner level to advanced level of maths,(nothing off topic but deeper in the topics you mentioned). They being pretty concise (plenty of questions included). are ideal for someone who has the knack of solving problems. (I am sure you do :~)

I really do not know what level of difficulty you want, but the theory is really ideal for your description, even if some of the questions (especially from higher algebra) might be a little too difficult to crack for school students, in case you want it for teaching.  The questions are really fun to solve, but not intended for someone who is scared of difficult problems being thrown at him.

For part 4, here is the school textbook statistics
and probability, which is old fashioned in all but the language......

Since the curriculum varies slightly from place to place, it is indeed difficult to get every subtopic in one place........but youtube is always there.

Being widely popular books, I am sure that you will be able to find the solutions of all the books mentioned online too

You might want to check these out
disclaimer- I don't own the  copyrights :~)

New Syllabus Mathematics for O-Level 1

New Syllabus Mathematics for O-Level 2

Title: "Old-Fashioned" O-level Maths Book


问:
Generating functions to solve recurrence relation
Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you

答:
You want to solve
$a_{n} = 3a_{n−1} + 2$
and you said that you could do it
if you could bring it to the form
$a_{n} = 3a_{n−1}$.

Here is one way to do that:

Let $a_n = b_n+c$,
where $c$ is a constant to be determined.
We want to choose $c$ so that
$b_n = 3 b_{n-1}$.

Substituting the expression for $a$,
$b_n+c = 3(b_{n-1}+c)+2
= 3 b_{n-1} + 3c+2
$,
or $b_n = 3 b_{n-1}+2c+2$.

To make $b_n = 3 b_{n-1}$,
we need $2c+2 = 0$, so $c = -1$.
Then $b_n = 3 b_{n-1}$,
so $b_n = 3^n b_0$.

Writing $b_n$ in terms of $a_n$,
$a_n-c = 3^n(a_0-c)$
or
$a_n = 3^n a_0 - 3^n c + c
= 3^n a_0-c(3^n-1)
= 3^n a_0 + 3^n - 1
$
(since $c = -1$).

This idea of modifying a recurrence
so that it becomes easier to solve
is a very general problem solving technique.
It looks like this:

You have a hard problem.

Convert it to an easier problem.

Solve the easier problem.

Convert that solution to a solution of the hard problem.

Think of multiplying by using logs, for example.

In your case,
your problem was converted to
$b_n = 3 b_{n-1}$
and, when the solution to this was found,
the $b$'s were converted back to
the original $a$'s.


答:
Let $A$ be a set which is not Borel. Let $f=\chi_A$. Then $f$ is continuous on $A$ (w.r.t the realtive topology from $\mathbb R^{n}$ or any other topology!). But $f\chi_A=\chi_A$ is not Borel measurable since the inverse image of $\{1\}$ is  not  Borel set. Hence, neither $f$ nor  $f\chi_A$ has to be  measruable.

However, if $A$ is  a Borel set then $f$ and $f\chi_A$ are Borel meaurable: For any open set $U$ in the real line $f^{-1}(U)$ is an open set in $A$ which means there is an open set $V$ in $\mathbb R^{n}$ such that $f^{-1}(U)=A\cap V$. It folows that $f^{-1}(U)$ is Borel.

Title: A function that is measurable on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$

Body:
This seems to be a result that is used often, which I am having difficulty proving. There may be a more general statement, but I'll write out the statement that I was trying to prove:

Let $f:(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n)) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be a function that is continuous on a set $A \in  \mathcal{B}(\mathbb{R}^n)$, which may not necessarily be continuous on $A^c$. Then, $f \chi_{A}$ is measurable in $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, where $\chi$ is the indicator function.

My attempt was to start on the set $A$ and artificially 'extend' $f \chi_{A}$ continuously to all of $\mathbb{R}^n$. But I met many difficulties, as the set $A$ may not be convex, and in general borel sets are hard to visualize. Does this have a neat solution?


答:
Check the second example in the "examples section in this wikipedia article.

I think that this is an improper integral by definition, because $\ln(\sqrt x) $ is not defined when $x=0$, so you have to use the limit to approach this point during the integration. Hope, this helps...

Title: Improper Integral Question $\int_0^1 \ln\sqrt{x}dx $

Body:
I'm trying to compute this integral and check if it's an improper integral.

What I did so far is to write the limit.

$$\begin{align*}
\int_0^1\ln\sqrt x\, dx &= x\ln\sqrt x - \int \frac12\, dx = x\ln\sqrt x - \frac12x + C \\
&=\lim_{\epsilon\to0^+}\Bigl(x\ln\sqrt x - \frac12x \Bigr)\Bigr\rvert_\epsilon^1
\end{align*}$$
 after I set the epsilon and the value 1, I got the expression : $\epsilon \ln \sqrt{x}$ and I want to know if this expression say that this integral is not improper.

Thanks.


问:
Bound on Av/v by eigenvalues
I have a $2 \times 2$ matrix $A$. I want to bound $\frac{|Av|}{|v|}$ from above. I thought that the upper bound will be equal to the largest eigenvalue, but somehow it turned out not to be true: example I found is $A = \begin{pmatrix} 1&5\\1&6 \end{pmatrix}$ and $v = (13,20)^T$. Then $Av = (113,133)^T$, $|v| \approx 23.85$, $|Av| \approx 174.52$ and $\frac{|Av|}{|v|} \approx 7.32$, while eigenvalues of $A$ are $\phi^4,\ \phi^{-4}$ (bigger one is equal to $\approx 6.85$).

So my questions are:

Why geometric intuition, that eigenvector is the vector whose length increases the most, fails here?

How to obtain some other bound on $\frac{|Av|}{|v|}$? My hypothesis is to take square root of the largest eigenvalue of $AA^T$ and it intuitively works but, well, my intuition had already failed once.

答:
To find $v$ which maximizes $\frac{|Av|}{|v|}$ is the same as maximizing $\frac{|Av|^2}{|v|^2}$. Now $|Av|^2$ is a quadratic form when expanded out. So the problem is the equivalent to maximizing some quadratic form $q(v)$ subject to the condition $|v|=1$.

Quadratic forms, like $ax^2+bxy+cy^2$ in two variables, are nicely behaved functions whose contours are ellipses or hyperbolas. On the unit circle/sphere, they take a maximum and minimum on perpendicular axes. (These are the eigenvectors of $A^TA$.)

The eigenvectors of $A$ will be some 'random' directions, not necessarily perpendicular, that have nothing to do with these max/min directions, unless the matrix $A$ is symmetric so that its eigenvectors agree with those of $A^TA$.

Here are two specific examples that illustrate this. Take $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $v=(x,y)$. Then $|Av|^2=y^2$ which has max/min directions $(0,1)$ and $(1,0)$; but it has only one eigenvector $(1,0)$ with eigenvalue $0$.

Take $A=\begin{pmatrix}1&-20\\0&-1\end{pmatrix}$. Its quadratic form is elliptical, $(x-20y)^2+y^2=x^2-40xy+401y^2\approx 401y^2$, whose perpendicular max/min directions are $(-0.05, 0.999), (0.999, 0.05)$; yet the eigenvectors are $(1,0)$ and $(1,0.1)$, totally unrelated. What's going on here is that the eigenvectors are pointing in almost the same direction, yet their eigenvalues are opposite in sign; the direction which is being multiplied out the most is being almost cancelled by another close-by eigenvector. The maximum is in an almost perpendicular direction.

Note that for any given quadratic form $q(v)=v^TSv$, there are many matrices such that $A^TA=S$.

Title: Bound on Av/v by eigenvalues


问:
$S^{n + m}$ is not homeomorphic to $M \times N$
Let $M$ and $N$ be two oriented manifolds such that $m = \dim M$ and $n = \dim N$. Using the de Rham cohomology I have to show that $S^{n + m}$ is not homeomorphic to $M \times N$. I have no idea how to proceed, but maybe the next statement could be helpful: if $M$ is an oriented and compact manifold such that $m = \dim M$, then $\dim H^m(M) \geq 1$. I think I have to use just geometry and topology statements, not analysis theorems like Fubini.

答: The Künneth formula is the general mechanism at play here, but this particular case can be proved ad hoc without much effort. Take a volume form $\omega\in\Omega^m(M)$ and pull it back along the projection $\pi\colon M\times N\rightarrow M$ to a form $\pi^{\ast}\omega\in\Omega^m(M\times N)$. Pick a point $p\in N$ and consider the integral $\int_{M\times\{p\}}\pi^{\ast}\omega$. Can you relate this integral to $\int_M\omega$? What could you say about this integral if $H^m(M\times N)=0$? This will lead to a contradiction with $H^m(S^{m+n})=0$.


问:
Why one cannot equate a sphere and a plane but a sphere with a sphere?
This is a general question about the intersection of a sphere with a plane or sphere which is confusing me.

To find the intersection between two spheres K1 and K2, you can equate them, solve the equation K1=K2, and find the intersection plane.
But the same procedure is not possible between a sphere (K) and a plane (P). One cannot just solve the equation K=P and gets the circle. But I don't understand why? Can anyone explain this?

Edit:
As an example:
Let K : (x-1)² +(y-9)² + (z-4)² -85=0 be the sphere in 3d and P: 6x-2y+3z-49=0 be a plane. Why can't I just solve the equation K=P and get the intersection circle?

Thank you!

答:
The problem has nothing to do with geometry. The real problem is that saying $K=0$ and $P=0$ is NOT equivalent to saying $K=P$.

Example: In the real plane, the system $x=0$ and $y=0$ yields a single point $(0,0)$, while $x=y$ yields a line.

The fact that your false recipe $K=P$ works in a particular case is a pure coincidence (the proof is that tomi gave you a counterexample where your method  does not work).


答:
Well, for starters, I think you can make your life easier in evaluating the limit (and in integrating) by rewriting $\ln(\sqrt{x})$ as $\frac{1}{2}\ln x$. $\frac{1}{2}x$ is obviously easy to deal with, so you are interested in the term $$\lim_{x \rightarrow 0^{+}} \frac{1}{2}x\ln x$$
This is a standard indeterminate form $0 * -\infty$, so we apply L'Hospital's rule to conclude:

$$ \lim_{x \rightarrow 0^{+}} \frac{1}{2}x\ln x = \lim_{x \rightarrow 0^{+}} \frac{1}{2} * \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 0^{+}}\frac{1}{2} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} = \lim_{x \rightarrow 0^{+}} -\frac{1}{2}x = 0$$


问:
Define random variable with desired property
I'm reading Casella's Statistical Inference (2nd ed), lemma 11.2.7

In proof, he claims followings:

If we define random variables $B$ and $V$ by $P(B= \frac{b_i}{c_i}, V=v_i) = \frac{c_i}{C}, i=1,\cdots,k$, where $C=\sum c_i$. Then $EB=\sum \frac{b_i}{c_i} \frac{c_i}{C} = 0$

(last equality comes from $\sum{b_i}=0$, which we assume in proof)

I have two questions:

how can we actually define these variables? how can we argue the existence of such random variables?

we define probability when both $B$, $V$ are such values. However, how can we claim that $P(B=\frac{b_i}{c_i}) = \frac{c_i}{C}$? I think he use this equation to calculate expectation of$B$. But we define probability only when both variables are simultaneously given. How can I understand this?

In first question, I think we can define variable $B$ so that $P(B=\frac{b_i}{c_i})=\frac{c_i}{C}$, which is obvious. But, as similar reason of second question, we define only probability when both variables are simultaneously given. So I think there is some incompleteness in definition.

答:
The condition $C=\sum c_i$ implies that $\sum P(B=\frac {b_i} {c_i},V=v_i)=1$. If construct $B$ we can take $V=v_i$ when $B=\frac {b_i} {c_i}$ (or $V=\sum v_i I_{\{B=\frac {b_i} {c_i}\}}$).

It follows now that $(B=\frac {b_i} {c_i},V=v_i)$ is same as the event $B=\frac {b_i} {c_i}$ so $B$ takes the values $\frac {b_i} {c_i}$ with probability $\frac {c_i} {C}$. Hence, $EB =\sum \frac {b_i} {c_i} \frac {c_i} {C}$.


问:
Numbers that can be expressed as the sum of two cubes in exactly two different ways
It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: Number Theory Taxicab Number).

We know that

$$1729 = 10^3+9^3 = 12^3 + 1^3,$$

and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in exactly two ways?

In fact, are there even any other such numbers?

EDIT:
As provided by MJD in the comments section, here are other examples:
$$4104 = 2^3+16^3 = 9^3+15^3,$$
$$13832 = 20^3+18^3=24^3+2^3,$$
$$20683 = 10^3 +27^3 = 19^3 +24^3.$$

答:
I found a number recently which can be expressed as a sum of two cubes in exactly two different ways.</p>
```<code>  65673928=164³+394³=103³+401³
</code>```

Body:
I found a number recently which can be expressed as a sum of two cubes in exactly two different ways.

  65673928=164³+394³=103³+401³


答:
Well, it's not true that the intersection between two spheres is a plane.
E.g., consider the two spheres
$$x^2+y^2+z^2=1$$ and $$(x-1)^2+y^2+z^2=1,$$ the intersection gives
$$
\left\{\begin{array}{l}
x^2+y^2+z^2=1\\
2x=1
\end{array}\right.
$$
which is not a plane, but the a circonference that lies in the plane $2x=1$ and is contained in the spheres $x^2+y^2+z^2=1$; sure, the system
$$
\left\{\begin{array}{l}
(x-1)^2+y^2+z^2=1\\
2x=1
\end{array}\right.
$$
defined the same circumference.

Now, if you consider the plane $2x=1$ and the sphere $x^2+y^2+z^2=1$, their intersection is again
$$
\left\{\begin{array}{l}
(x-1)^2+y^2+z^2=1\\
2x=1
\end{array}\right.
$$
i.e., the same circumference as above. You can write the same system as
$$
\left\{\begin{array}{l}
y^2+z^2=\frac{3}{4}\\
2x=1
\end{array}\right.
$$
namely, you can eliminate the $x$ from the first equation. Now, what represents the equation $y^2+z^2=\frac{3}{4}$? Not a circumference: indeed, it is a cylinder (remember that we are working in $\mathbb{R}^3$), to be precise the cylinder having as directrix the circumference
$$
\left\{\begin{array}{l}
(x-1)^2+y^2+z^2=1\\
2x=1
\end{array}\right.
$$
and axis perpendicular to the plane $2x=1$.

Summing up: an equation as $y^2+z^2=1$ is not a circumference in $\mathbb{R}^3$ but a cylinder, the equation of a circumference in $\mathbb{R}^3$ is always given by a system, the same holds for a line in $\mathbb{R}^3$, indeed $2x=1$ is a plane $\mathbb{R}^3$, and a line in $\mathbb{R}^2$.

Title: Why one cannot equate a sphere and a plane but a sphere with a sphere?

Body:
This is a general question about the intersection of a sphere with a plane or sphere which is confusing me.

To find the intersection between two spheres K1 and K2, you can equate them, solve the equation K1=K2, and find the intersection plane.
But the same procedure is not possible between a sphere (K) and a plane (P). One cannot just solve the equation K=P and gets the circle. But I don't understand why? Can anyone explain this?

Edit:
As an example:
Let K : (x-1)² +(y-9)² + (z-4)² -85=0 be the sphere in 3d and P: 6x-2y+3z-49=0 be a plane. Why can't I just solve the equation K=P and get the intersection circle?

Thank you!


问:
Showing that gcd(a,b) = gcd(c,d) in this case
We have a $2\times2$ matrix with $\mathbb{Z}$ entries, $$M =\begin{bmatrix}i&j\\k&l\end{bmatrix}$$
with $\det(M) = 1$.  If $(c\; d) = (a\; b)M$ then how do we show that $\gcd(a,b) = \gcd(c,d)$ ?

Multiplying out we get $ai + bk =c$ and $aj + bl = d$. assuming $a,b,c,d$ non-zero integers then by Bezout $\gcd(a,b) |c$ and $\gcd(a,b) |d$ so $\gcd(a,b) |\gcd(c,d)$

since $\det(M) = 1$ then $il - kj = 1$. by Bezout $\gcd(i,k) = \gcd(j,l) = 1$

That's all I have really, any help would be appreciated thanks.

答:
The key point here is that if the determinant is $1$ then $$\begin{bmatrix}i&j\\k&l\end{bmatrix}^{-1}=\begin{bmatrix}l&-j\\-k&i\end{bmatrix}.$$
You've already shown that $\gcd(a,b)\mid\gcd(c,d)$; this fact allows you to run the same argument backwards.


问:
On the HOMFLY polynomial of a split link
Basically my question has to do with the HOMFLY  polynomial.

In its wikipedia page, http://en.wikipedia.org/wiki/HOMFLY_polynomial, I see that it says

$$P(L_1 \cup L_2) = \left(\frac{-l-l^{-1}}{m}\right)P(L_1)P(L_2) $$
where $L_1 \cup L_2$ is the split link of the two links

Can anyone show a proof of this fact from the skein relation of the HOMFLY polynomial?
I get stuck when I have to prove that
$$P(L_1 \# L_2 ') = P(L_1)P(L_2) $$
where # is the knot sum operation and L' is the mirror image of L.

答:
Note, that is is actually just $P(L_1 \# L_2) = P(L_1) P(L_2)$.

Given a link $L$, we can change crossings using the skein relation to write $L$ as a linear combination (where the coefficients are polynomials in $m$ and $l$) of unlinks with various numbers of components.  So, if $U_k$ represents the unlink with $k$ components, and $L_1,L_2$ are diagrams each with $\le n$ crossings, we can write
$$L_1 = \sum_{k=1}^n a_k U_k$$
$$L_2 = \sum_{k=1}^n b_k U_k$$
where each $a_k$ is a polynomial in $l,m$.  It is not hard to show by induction that
$$P(U_k) = \left[\frac{-(l + l^{-1})}{m}\right]^{k-1}$$
so that $P(U_i)P(U_j) = P(U_{i+j-1})$.  It follows immediately that
$$P(L_1)P(L_2) = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i}P(U_{k-1})$$

Now, it remains to look at the connected sum of the diagrams $L_1 \# L_2$.  If we remove all the crossings in $L_1$ by the skein relations, then we get
$$L_1 \# L_2 = \sum_{k=1}^n a_k (U_{k-1} \cup L_2)$$
because one of the unknotted components at the end will be stuck together with the diagram for $L_2$.  Expanding $L_2$ gives
$$L_1 \# L_2 = \sum_{k=1}^n a_k \left(U_{k-1} \cup \sum_{k=1}^n b_k U_k\right) = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i} U_{i-1} \cup U_{k-i} = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i} U_{k-1}$$
so
$P(L_1 \# L_2) = P(L_1) P(L_2)$


答:
$(c,d)=(a,b)M$ implies $c,d \in \mathbb Z a + \mathbb Z b = \mathbb Z \gcd(a,b)$.

Therefore, $\gcd(c,d) \in  \mathbb Z \gcd(a,b)$ and so $ \mathbb Z \gcd(c,c) \subseteq \mathbb Z \gcd(a,b)$.

Repeat the argument with $M^{-1}$ (which has integer entries) to conclude the reverse inclusion.

Title: Showing that gcd(a,b) = gcd(c,d) in this case

Body:
We have a $2\times2$ matrix with $\mathbb{Z}$ entries, $$M =\begin{bmatrix}i&j\\k&l\end{bmatrix}$$
with $\det(M) = 1$.  If $(c\; d) = (a\; b)M$ then how do we show that $\gcd(a,b) = \gcd(c,d)$ ?

Multiplying out we get $ai + bk =c$ and $aj + bl = d$. assuming $a,b,c,d$ non-zero integers then by Bezout $\gcd(a,b) |c$ and $\gcd(a,b) |d$ so $\gcd(a,b) |\gcd(c,d)$

since $\det(M) = 1$ then $il - kj = 1$. by Bezout $\gcd(i,k) = \gcd(j,l) = 1$

That's all I have really, any help would be appreciated thanks.


问:
Show that any matrix product $AB$ can be expressed in the form $U\Lambda V$ where $\sum_{i}U_{ij}^2 = 1, \sum_{j}V_{ij}^2 = 1$

Show that any matrix product $AB$ can be expressed in the form $U\Lambda V$, where $U$ is
a matrix in which the sum of the squares of the entries in each column is $1$, $V$ is a
matrix in which the sum of the squares of the entries in each row is $1$, and $\Lambda$ is an
appropriately chosen diagonal matrix with nonnegative entries on the diagonal.

So, we have that  $\sum_{i}u_{ij}^2 = 1, \sum_{j}V_{ij}^2 = 1$. $(AB)_{ij}= \sum_{r}a_{ir}b_{rj}$.
The i-th row of a $\Lambda V = (\lambda_iv_{i1},\lambda_iv_{i2},...,\lambda_iv_{id})$ and the $j$-th column is $(\lambda_1v_{1j},\lambda_2v_{2j},...,\lambda_nv_{nj})$. So, the $ij$-th entry of $(U\Lambda V)_{ij}  = (u_{i1},u_{i,2},...,u_{in})*(\lambda_1v_{1j},\lambda_2v_{2j},...,\lambda_nv_{nj})^t = \lambda_1v_{1j}u_{i1}+\lambda_2v_{2j}u_{i2}+...+\lambda_nv_{nj}u_{in} = \sum_{k}\lambda_k u_{ik}v_{kj}$. And we should show that $\sum_{r}a_{ir}b_{rj} = \sum_{k}\lambda_k u_{ik}v_{kj}$ for some $\lambda_k, u_{ik},v_{kj}$.

答:
I suppose the matrices have real number entries. The result follows immediately from the singular value decomposition of $AB$.

Alternatively, let $AB$ be $m\times n$. Perform a rank decomposition $AB=XY$, so that $X$ is $m\times r$ and $Y$ is $r\times n$ with $r=\operatorname{rank}(X)=\operatorname{rank}(Y)=\operatorname{rank}(AB)$. Let $x_i$ be the Euclidean norm of the $i$-th column of $X$ and $y_i$ be the Euclidean norm of the $i$-th row of $Y$. Then $AB=U\Lambda V$ where
$$
U=X\operatorname{diag}(x_1,\ldots,x_r)^{-1},\ \Lambda=\operatorname{diag}(x_1y_1,\ldots,x_ry_r),\ V=\operatorname{diag}(y_1,\ldots,y_r)^{-1}Y.
$$
If you want $U$ and $V$ to be square matrices, just append some columns of unit norms to $U$, some rows of unit norms to $V$ and some zeroes to the diagonal of $\Lambda$.


问:
Similarity of extension of prime ideal and extension of valuation
$\textbf{[Extension of prime ideal]}$

Let $L/K$ be a separable extension is given by a primitive element $\theta \in B$ so that $L=K(\theta)$.
( $A$ is a dedekind domain , $K=Frac(A)$, $B$ is integral closure in $L$ of $A$ ) Let $f(x) \in A[x]$ be a minimal polynomial of $\theta$ ,$\frak{p}$ be a prime ideal of $A$ which is relatively prime to the conductor of $A[\theta]$, and let $$ \bar{f}(X)=\bar{f}_1(X)^{e_1}\cdots \bar{f}_r(X)^{e_r}$$
be the factorization of $f(x)$ in $A/\frak{p}$$[X]$. Then $$ \frak{p}=\frak{P}_1^{e_1}\cdots \frak{P}_r^{e_r}$$

$\textbf{[Extension of valuation]}$

Let $L=K(\theta)$, $f(x)$ be a minimal polynomial of $\theta \in K[x]$ and $v$ be a valuation of $K$
and $$ f(X)=f_1(X)^{m_1} \cdots f_r(X)^{m_r}$$ be the decomposition of $f(x)$ into irreducible factors over the completion $K_v$. Then there is the valuation $w_1, \ldots , w_r$ extending $v$ to $L$ (corresponding the irreducible factors $f_i(X)$)

[My question]

I have some intuition that each prime ideal and valuation correspond.

So, I know that the two Theorems above make sense to some extent.

However, one is to observe the factorization of f(x) in the residue field, and the other is to observe the factorization of f(x) when the base change is made to completion.

These two concepts are completely different. ( quotient and base change to completion)

Is there any reason for this connection?

答:
There are two steps involved in the translation.

First is the step from prime ideals to valuations. Let $\mathfrak{o}$ be a Dedekind domain with field of fractions $K$. Then to every nonzero prime ideal $\mathfrak{p}$ of $\mathfrak{o}$ we may associate a $\mathfrak{p}$-adic valuation $v_{\mathfrak{p}}$ on $K$, as follows: starting with $a \in K$, we have a factorisation of the fractional ideal $(a)$ into a product $\prod_{\mathfrak{p}} \mathfrak{p}^{v_{\mathfrak{p}}}$; we then set $v_{\mathfrak{p}}(a)$ to be that integer $v_{\mathfrak{p}}$.

If $L / K$ is now a finite extension, and $\mathfrak{p}$ decomposes as $\mathfrak{P}_1^{e_1} \cdots \mathfrak{P}_r^{e_r}$, then the valuations $w_i = e_i^{-1} v_{\mathfrak{P}_i}$ are precisely the extensions of $v_{\mathfrak{p}}$ to $L$. It is easily seen that the notions of ramification index and inertia degrees, defined for extensions of primes and extensions of valuations, match up.

The valuation-theoretic approach has some advantages. For one, we have the Archimedean valuations, which have no obvious analogue in terms of prime ideals. But this is beyond the scope of this answer. Another advantage is that we can now pass from the extension $L / K$ to the completions $L_{w} / K_v$ (called a 'local-to-global principle' even though you're going global to local here). The world of complete fields is easier in some respects, so it's advantageous to be able to do this.

Moving to the second step, which is from quotients to completions. As remarked in the comments, the key here is Hensel's lemma. Let $K_v$ (the slightly awkward notation is deliberately suggestive) be a field which is complete with respect to a non-Archimedean valuation, and let $\mathfrak{o}_{v}$ be the valuation ring with residue field $\mathfrak{o}_{v} / \mathfrak{p}$ at the maximal ideal $\mathfrak{p}$. If $f(x)$ is a primitive polynomial in $\mathfrak{o}_{v}[x]$ which factorises as $\overline{f_1}(x)^{e_1}\cdots \overline{f_r}(x)^{e_r}$ modulo $\mathfrak{p}$, then in fact $f(x)$ already factorises in $K_v$ as $f_1(x)^{e_1} \cdots f_r(x)^{e_r}$ --- and $f_i(x)$ modulo $\mathfrak{p}$ is precisely $\overline{f_i}(x)$.


问:
How to prove the formula for the Reciprocal Multifactorial constant?
https://mathworld.wolfram.com/ReciprocalMultifactorialConstant.html

The sum of reciprocal multifactorials can be given in closed form by the
beautiful formula
\begin{align}
m(k) &= \sum_{n=0}^\infty\frac1{n\underset{k}{\underbrace{!\dots!}}}    \\
&= 1+\frac{e^{1/k}}k\sum_{r=1}^k k^{r/k}\gamma\Big(\frac rk,\frac 1k\Big)  \\
&= \frac{e^{1/k}}k\left[k+\sum_{r=1}^{k-1}k^{r/k}\gamma\Big(\frac rk,\frac 1k\Big)\right]
\end{align}  where $\gamma(a,x)$ is a lower incomplete gamma function (E. W. Weisstein, Aug. 6, 2008). This gives the special cases
\begin{align}
m(2) &= \sqrt e \left[1+\sqrt{\frac\pi2}\operatorname{erf}\Big(\frac{1}{\sqrt2}\Big)\right] \\
m(3) &= \frac13 e^{1/3}\left[3+3^{1/3}\gamma\Big(\frac13,\frac13\Big)+3^{2/3}\gamma\Big(\frac23,\frac13\Big)\right].
\end{align}

I saw the formula on the above Mathworld link.

Since the reciprocal of $n$ (from $n=0$ to $\infty$) with $k$ number of factorials converge. This page gives a general formula to find the value of the constant for any $k$.

I've tried searching online for the "Reciprocal Multifactorial Constant" and how to derive its formula but I cannot find it anywhere.

Does anyone know of a proof for this formula? Or at least how one can approach proving it?

答:
Let's "decipher" the notation of "multifactorial". Assume $k>0$. Any $n>0$ can be written in the form $n=kq+r$ with $q\geqslant 0$ and $1\leqslant r\leqslant k$; then $n\underbrace{!\ldots!}_{k}:=\prod_{j=0}^q(kj+r)$, and $0\underbrace{!\ldots!}_{k}:=1$. But $$\prod_{j=0}^q(kj+r)=k^{q+1}\prod_{j=0}^q\left(j+\frac rk\right)=k^{q+1}\frac{\Gamma(q+1+r/k)}{\Gamma(r/k)}=\frac{k^{q+1}q!}{\mathrm{B}(r/k,q+1)}$$ using the gamma and beta functions. Using the integral representation of the latter, we get

\begin{align*}
m(k)&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{(kq+r)\underbrace{!\ldots!}_{k}}
\\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\mathrm{B}\left(\frac rk,q+1\right)
\\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\int_0^1 t^{r/k-1}(1-t)^q\,dt
\\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}\sum_{q=0}^\infty\frac1{q!}\left(\frac{1-t}{k}\right)^q dt
\\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}e^{(1-t)/k}\,dt\quad\color{gray}{[t=kx]}
\\&=1+\frac{e^{1/k}}k\sum_{r=1}^k k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx,
\end{align*}

which is the first of the two closed forms. The second one is then easy to obtain: at $r=k$ $$k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx=k\int_0^{1/k}e^{-x}\,dx=k(1-e^{-1/k}).$$


问:
Are the quotient groups in a composition sequence necessarily subgroups?
Does there exist a finite group G and a normal subgroup N of G so that G/N is a simple group and G/N is not isomorphic to any subgroup of G ? </p>

答:
Yes, the perfect group $\operatorname{SL}(2,5)$ of order 120 is one. By Cauchy's theorem you won't find an example with $G/N$ abelian, so this is more or less the typical example. You want a non-split extension, and you probably want $N$ to be small to avoid coincidentally containing $G/N$. A good way to do this are the non-simple quasi-simple groups.

If $G/Z(G)$ is non-abelian simple, and $H \leq G$ is such that $H \cong G/Z(G)$, then $H \cap Z(G) \leq Z(H) = 1$, so we get a semi-direct product of $H$ and $Z(G)$, but such a semi-direct product is direct since $Z(G)$ is central, hence $G \cong G/Z(G) \times Z(G)$.

All we need then is for $Z(G) \leq [G,G]$, that is, for $G$ to be perfect, so that $G$ is quasi-simple.

There are examples that are not quasi-simple. It'd be nice to see a proof that they all have a (non-simple) quasi-simple subquotient, so that this example really is “typical.”


问:
Questions about Hopf-Rinow theorem
I am reading about Hopf-Rinow theorem using the Jonh M. Lee "Riemannian manifolds: an introduction to curvature",page 108, Theorem 6.13. My doubts are the next ones:

In the book, there is an exercise asking to prove that $\mathbb{H}_{\mathbb{R}}^n$, the hiperbolic space, is complete. How can I do this? I know I could just try to compute the geodesics of the space, but I am trying to deduce this from Hopf-Rinow theorem.

In the statement of the main theorem, one of the hypothesis is for $M$ being connected. I am looking for an example where the completeness fails due to $M$ being not connected, and where is used this hypothesis of being connected in the proof of Hopf-Rinow theorem.

答:
Well, as the Hopf-Rinow theorem suggests, it would suffice to show that $\mathbb{H}^n$ is complete as a metric space; if you want to do this by hand, you have to take a Cauchy sequence in the hyperbolic space, and show that it converges to a point in $\mathbb{H}^n$ (endowed with the hyperbolic metric, of course; perhaps using the upper half-plane model is a good idea for this).

To see why you need this, the easy answer is that even defining what geodesically complete means in the disconnected case is somewhat strange; for instance, in the context of Riemannian manifolds, your distance function is given by the infimum of all curves joining two given points. If no such curve exists, what is the distance in this case?

As far as I know there are some alternative definitions for these cases, but they are far less interesting than the usual connected cases.

Title: Questions about Hopf-Rinow theorem


问:
Is this an integrating factor at all and why does it work?
So I required some help for the differential equation: $y' = y\,\sin(x)+\sin(2\,x)$

I figured that it's not that easy to solve because of the integral $\int\frac{\sin(2\,x)}{e^{cos(x)}}\mathrm{d}x$ that is appearing in the premade formula.

However a kind user suggested me to multiply the particular solution $y_p = \sin(2\,x)$ with $e^{\cos(x)}$ and integrate afterward. As a result the integral becomes pretty handy. To be honest I recognized the beauty of this approach but didn't understand it yet - why this is functioning.
Can someone explain to me afterwards?

答:
It works because this example is very carefully chosen. Its structure then allows even a solution without integrating factor, at least in the construction of the particular solution. Apply the double-angle identity to get
$$
y'(x)=\sin(x)(y(x)+2\cos(x)).
$$
Set $u(x)=y(x)+2\cos(x)$. Then
$$
u'(x)=y'(x)-2\sin(x)=\sin(x)(u(x)-2),
$$
which is separable and implies
$$
u(x)-2=(u(0)-2)e^{1-\cos(x)}\implies y(x)=2(1-\cos(x))+y(0)e^{1-\cos(x)}
$$


问:
An integral involving determinant.
I learned the following identity (without proof) from the book "Lectures on Elliptic Partial Differential Equations" by L. Ambrosio et.al.
$$\int_\Omega \det(A+\nabla\varphi)=\det(A)|\Omega|,$$
where $A$ is $n\times n$ matrix, $\varphi\in C^2_0(\Omega)$ and $\Omega\subset\mathbb{R}^n$ is bounded smooth domain.
After some consideration, I have an elegant proof of this result (using divergent theorem and Hadamard identity). My question is:

What is the name of this identity? I believe that it should be discovered by some mathematician(s), thus should be named by the mathematician(s).

How other people proved this result?

答:
The following proof is inspired by the transgression formula of Chern classes.

Note that $I(t)=\int_{\Omega}\det(A+t\,\nabla \varphi)d x=\int_{\Omega}d y^1\wedge\cdots\wedge d y^n$, where $y=Ax+t\,\varphi$, so we have
$$I'(t)=\int_{\Omega}d\left(\varphi^1\wedge d y^2\wedge\cdots\wedge d y^n+\cdots+d y^1\wedge\cdots\wedge dy^{n-1}\wedge \varphi^n\right)=0,$$
Thus $I(t)=I(0)=\det(A) |\Omega|$.

Title: An integral involving determinant.


问:
Flipping an unfair coin
An coin has probability $p<0.5$ of landing heads up. We flip it infinitely many times, after $n$ tosses, let $S_n$ be the number of heads minus the number of tails. What is the expected value of $max\{S_0,S_1,S_2,S_3,...\}$?</p>

答:
Let $M=\max\{S_k;k\geqslant0\}$ and, for every $n\geqslant0$, $T_n=\inf\{k\geqslant0\mid S_k=n\}$. Thus $T_0=0$, if $n\geqslant1$ then $T_n$ may be infinite with positive probability, and $[M\geqslant n]=[T_n\lt\infty]$ for every $n\geqslant0$. Furthermore, conditionally on $[T_n\lt\infty]$, $T_{n+1}\lt\infty$ means that the path after time $T_n$ reached level $+1$ above level $n$. By Markov strong property, this implies that $P[T_{n+1}\lt\infty\mid T_n\lt\infty]=u$ with $u=P[T_{1}\lt\infty]$. Thus, $P[T_n\lt\infty]=u^n$ for every $n\geqslant0$ and
$$
E[M]=\sum_{n\geqslant1}P[M\geqslant n]=\sum_{n\geqslant1}P[T_n\lt\infty]=\sum_{n\geqslant1}u^n=\frac{u}{1-u}.
$$
To compute $u$, let us condition on $X_1$. If $X_1=1$, $T_1\lt\infty$. If $X_1=-1$, to hit $1$ starting from $-1$ one must climb two stairs. Hence, $u=p\cdot1+(1-p)\cdot u^2$. Since $p\lt\frac12$, $u\lt1$ hence $u=p/(1-p)$. Finally,
$$
E[M]=\frac{p}{1-2p}.
$$</p>

Title: Flipping an unfair coin


问:
Proof of general divisibility rule
I am looking for proof of, or a suggestion on how to start on a proof of, or any general insights into, the following statement:

$$n\equiv0~~(\text{mod}~d) \implies f(n,d,s)\equiv0~~(\text{mod}~d)$$

$f$ is the following:

Separate both $n$ and $d$ into groups of digits of size $s$, starting from the right

$s$ must be assigned so that $d$ is split into two groups

$n$ will now be split into at least 2 groups and $d$ will be split into exactly 2 groups, $d_1$ and $d_2$ which are, respectively, the left and right parts of $d$

Multiply each $n$ group by $d_1^l$ where $l$ is the groups distance from the leftmost $n$ group

Multiply each $n$ group by $d_2^r$ where $r$ is the groups distance from the rightmost $n$ group

Take the absolute value of the alternating sum of the new $n$ groups

An example:

Start with $n = 39445434288 = 2843529 * 13872$, $d = 13872, s = 4$

Separate $n$ into $n_1=394,n_2=4543,n_3=4288$

Separate $d$ into $d_1 = 1$, $d_2 = 3872$

Now $f(n,d,s) = |(n_1*d_1^0*d_2^2) - (n_2*d_1^1*d_2^1) + (n_3*d_1^2*d_2^0)|$

$f(n,d,s) = |394*1*64269752490-4543*1*3872+4288*1*1|$

$f(n,d,s) = |5889413088| = 5889413088 = 424554 * 13872$

Repeating this on the subsequent values of $f(n,d,s)$ yields a sequence that decreases to zero:

```<code>39445434288, 5889413088, 834941808, 106412112, 12512544, 4841328, 1872720, 721344, 277440, 97104, 27744, 0
</code>```

Every number in the sequence is divisible by $d = 13872$

This function doesn't, however, have the property $n\equiv f(n,d,s)~~(\text{mod}~d)$ as the remainder will change if $n$ is not divisible by $d$:

Below is some ugly python code for the function that I made (I lost the original from months ago):

```<code>def getParts(num:int, n:int) -> list:
    s = str(num)
    l = []
    while len(s) % n != 0:
        s = "0" + s
    for i in range(0, len(s), n):
        l.append(int(s[i:(i+n)]))
    return l

def testFunc(n:int, d:int, size:int=-1) -> int:
    if n < d:
        return n
    if size == -1:
        size = int(round(len(str(d)) / 2))
    nParts = getParts(n, size)
    dParts = getParts(d, size)
    d1 = dParts[0]
    d2 = dParts[1]
    for i in range(len(nParts)):
        nParts[i] *= d1 ** i
    for i in range(len(nParts)):
        nParts[len(nParts) - i - 1] *= (d2 ** i) * (-1) ** i
    return abs(sum(nParts))
</code>```

A sequence of numerators from the function does not always reach 0. Sometimes it oscillates between a few values forever.

I am, however, fairly certain that a sequence of numerators from the function will never diverge to infinity.

If needed, I can post data on how the function performs with different $s$, as well as detailed data on how many iterations of the function it requires to reach 0 (when $n$ is a multiple of $d$)

This function is extremely interesting to me and I would greatly appreciate any insights into it.

答:
This equivalence arises by working in radix $\,t = 10^4\,$ so $\bmod d = d_1 t + d_0\,$ we have $\,t \equiv -d_2/d_1\,$ is congruent to a $ $ fraction, $ $ so $\, 0\equiv n = n_1 t^k + \cdots + n_{k+1} t^0 =: f(t)$ $\iff\! 0\equiv d_1^k f(t),\,$ where scaling by $\,d_1^k\,$ clears the denominators after evaluating $\,f(t)\,$ at $\,t\equiv -d_2/d_1\,$ (which uniquely exists $\bmod d\iff d_1^{-1}$ uniquely exists $\!\bmod d$ $\!\iff\! \color{#0a0}1 = (d_1,d) = (d_1,d_1 t+d_2) = \color{#0a0}{(d_1,d_2)},\,$ i.e. $\!\iff\!$ the radix $\,t\,$ digits $\,d_i$ are $\rm\color{#0a0}{coprime}),\,$ e.g. for your  $\,n\,$ with $\,3\,$ digits $\,(k = 2)\,$ we have

$$\begin{align}
\bmod\, d \:\!=\:\! d_1 \:\! t + d_2\!: &\ \ \ \,\color{#0a0}{d_1\:\! t}\equiv \color{#c00}{-d_2}\\[.2em]
{\rm thus}\ \ \, 0 \equiv\ &\!n\! =\! n_1\, t^2\, +\, n_2\, t \,+\, n_3\\[.2em]
\iff 0\equiv\ &d_1^2 \,(n_1\, t^2 \,+\, n_2\, t \,+\, n_3),\, \ {\rm by}\ \ (d_1,d) = 1\\[.2em]
\equiv\ & n_1 (\color{#0a0}{d_1 t})^2 \,+ n_2 d_1 (\color{#0a0}{d_1 t})\ \,+ d_1^2\, n_3\\[.2em]
\equiv\ & n_1 (\color{#c00}{-d_2})^2\! + n_2 d_1 (\color{#c00}{-d_2}) + d_1^2\, n_3\\[.2em]
\equiv\ & n_1\, d_1^0\,d_2^2\ -\ n_2\, d_1^1\, d_2^1\ +\ d_1^2\, d_2^0\, n_3
\end{align}\qquad\qquad$$

which is precisely the divisibility test that you discovered. All of the common divisibility tests of this form arise via this general method. Arithmetical inferences like this become very easy once one masters the basics of congruences (or, equivalently, residue or quotient rings), esp. if one masters the fractional viewpoint (equivalently localizations).

Remark $\ d_1^k f(d_2/d_1)$ is a  homogenization of $\,f(t).\,$ For geometric intuition behind such see e.g. the answers to this question.

Title: Proof of general divisibility rule


问:
Given $\frac{A}{4B}+\frac{2a^3A}{B^3}-\frac{a^2A}{B^3}-\frac{2a^2A^2}{B^3}=0$, what does $\frac{a}{B}$ tend to as $\frac{A}{B}\to \infty$?
In this problem, all of $A, a, B$ can vary.

My attempt was to let $\frac{A}{B}=C$ and $\frac{a}{B}=c$, which gives $ \frac{C}{4}+2Ac^3-c^2C-2Ac^2C=0$.

Now $C\to \infty $ means either $A\to \infty $ or $B\to 0 $. At this point I'm not sure at all how to proceed.

答:
You can rewrite the equation to get a function $c=f(C)$ and then calculate $\lim_{C\to\infty}f(C)$. It requires you to solve a cubic equation. However, there is an easier solution:

You rewrite the equation to this form:$$
C=\frac{-2Ac^3}{\frac14-c^2-2Ac^2}=\frac{2Ac^3}{(2A+1)c^2-\frac14}
$$
If $2A+1>0$ then:
$$
C=\frac{2Ac^3}{\left(c\sqrt{2A+1}-\frac12\right)\left(c\sqrt{2A+1}+\frac12\right)}
$$
You need to make something with $c$ to make $C$ go to infinity. Let's assume that $A$ is positive. Now what do you have to do to make $C=\infty$ if you can change only $c$? If $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right, $C$ goes to infinity. And vice-versa, if $C$ goes to infinity, $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right. And if $A$ is negative, we just flip the sign.

If $2A+1\leq0$ then $(2A+1)c^2-\frac14$ is always nonpositive. It also means that $A<0$. In this case, $c$ must go to $\infty$.

We have up to three results, so what now? It is because the equation gives up to three sollutions for each $C$. So the sollution varies on which of these solutions you pick.

To sum it up:
$$
\lim_{\frac{A}{B}\to\infty}=
\begin{cases}
\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}},  & \text{if $\liminf A>0$} \\
-\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}},  & \text{if $\liminf A>-\frac12$ and $\limsup A<0$} \\
\infty, & \text{if $\limsup A\leq-\frac12$}
\end{cases}
$$
Note that $\frac{A}{B}$ can't go to $\infty$ if $A=0$.


问:
Formula for Sum of Logarithms $\ln(n)^m$
As you know $\sum_{n=1}^k \ln(n) =\ln(k!)$ is there a formula for $\sum_{n=1}^k \ln(n)^m$?</p>

答:
The Euler-Maclaurin Sum Formula gives the asymptotic approximation:

$$
\begin{align}
\sum_{n=1}^k\log(n)^m
&\sim k\left(\log(k)^m-m\log(k)^{m-1}+m(m-1)\log(k)^{m-2}-\dots+(-1)^mm!\right)\\
&+\frac12\log(k)^m+C+\frac{m}{12k}\log(k)^{m-1}+O\left(\frac{\log(k)^{m-1}}{k^3}\right)
\end{align}
$$
The constant $C$ depends on $m$ and needs to be determined separately. For $m=1$, Stirling's approsimation says that $C=\frac12\log(2\pi)$.


答:
my attempt, starting from your attempt:
$$\frac{C}{4} + 2Ac^3 - c^2C-2Ac^2C = 0$$
divide by $c^2$:
$$\frac{C}{4c^2} + 2Ac - C-2AC = 0$$
$$2Ac = -\frac{C}{4c^2} + C + 2AC$$
$$2Ac = C(-\frac{1}{4c^2} + 1 + 2A)$$
and isolate $C$:
$$\frac{2Ac}{-\frac{1}{4c^2} + 1 + 2A} = C$$
as we know, $C$ approaches infinity. Therefore, the LHS of the equation should too. Assuming $2Ac$ does not approach infinity, the following can be deduced:
$$-\frac{1}{4c^2} + 1 + 2A \rightarrow 0$$
$$-1 + 4c^2 + 8Ac^2 \rightarrow 0$$
$$ c^2(4 + 8A) = 1$$
and therefore:
$$c = \sqrt{\frac{1}{4+8A}}$$

Title: Given $\frac{A}{4B}+\frac{2a^3A}{B^3}-\frac{a^2A}{B^3}-\frac{2a^2A^2}{B^3}=0$, what does $\frac{a}{B}$ tend to as $\frac{A}{B}\to \infty$?

Body:
In this problem, all of $A, a, B$ can vary.

My attempt was to let $\frac{A}{B}=C$ and $\frac{a}{B}=c$, which gives $ \frac{C}{4}+2Ac^3-c^2C-2Ac^2C=0$.

Now $C\to \infty $ means either $A\to \infty $ or $B\to 0 $. At this point I'm not sure at all how to proceed.


问:
Challenging problem for me in General topology
Suppose that $X=\{(0,0)\}\cup \{(x,sin\frac{1}{x}:x\in\Bbb{R}- \{0\}\}$ and $Y=\{(0,0)\}\cup \{(x,xsin\frac{1}{x}:x\in\Bbb{R}- \{0\}\}$ are metric spaces with metrics induced by the Euclidean metric of $\Bbb{R}^2$. Let $B_X$ and $B_Y$ be the open unit balls around $(0,0)$ in $X$ and $Y$ , respectively.

Prove or disprove:

The closure of $B_X$ in $X$ is compact.

The closure of $B_Y$ in $Y$ is compact.

Let $(t_n,t_nsin\frac{1}{t_n})$ be the convergent sequence in $Y$. and $(t_n,t_nsin\frac{1}{t_n})\to (p,q)$. We can choose $p=0,q=0$. therefore $(p,q)\in Y$. How to show that $Y$ is closed? It seems 'yes it is'. But I am uncomfortable at writing it formally.

I have no idea about first.

Any help will be appreciated. Thanks!

答:
The sequence $\{X_n\}$ with

$$X_n= \left(\frac{1}{2 n \pi + \frac{\pi}{4}},\frac{\sqrt 2}{2}\right)$$ is a sequence of $B_X \cap X$ for $n$ large enough. However$\{X_n\}$ has no converging subsequence in the closure of $B_X$ in $X$. Hence the closure of $B_X$ in $X$ is not compact.

Regarding $Y$, the map defined on $\mathbb R$

$$f(x) = \begin{cases}
(x, x\sin \frac{1}{x}) & x \neq 0\\
(0,0) & x = 0
\end{cases}$$
is continuous. Therefore, the image of $[-2,2]$ under $f$ is compact as $[-2,2]$ is a compact interval. As the closure of $B_Y$ in $Y$ is a closed subset of $f([-2,2])$, it is compact.


问:
Why one cannot equate a sphere and a plane but a sphere with a sphere?
This is a general question about the intersection of a sphere with a plane or sphere which is confusing me.

To find the intersection between two spheres K1 and K2, you can equate them, solve the equation K1=K2, and find the intersection plane.
But the same procedure is not possible between a sphere (K) and a plane (P). One cannot just solve the equation K=P and gets the circle. But I don't understand why? Can anyone explain this?

Edit:
As an example:
Let K : (x-1)² +(y-9)² + (z-4)² -85=0 be the sphere in 3d and P: 6x-2y+3z-49=0 be a plane. Why can't I just solve the equation K=P and get the intersection circle?

Thank you!

答:
You have to be a little careful in your formulation.

First, as mentioned in comments, it isn't true that $K_1=K_2$ gives the [circle of] intersection of the two spheres. Rather, it gives the plane of the circle of intersection. (This plane is the analogue of a radical axis, the line containing the points of intersection of two circles.)

Second, it's not even necessarily true that $K_1=K_2$ gives the equation of that plane. That magic works when (and only when) $K_1$ and $K_2$ have matching coefficients on $x^2$, $y^2$, $z^2$, because then (and only then) do those terms cancel in $K_1=K_2$; what remains is a linear equation in $x$, $y$, $z$: the target plane.

A problem, though, is that there's no a priori reason for those coefficients to match. Both $x^2+y^2+z^2-3=0$ and $4x^2+4y^2+4z^2-12=0$ represent the same sphere; either could be "$K_1$". And neither will yield a sphere when simply set equal to a "$K_2$" of the form $3(x-4)^2+3(y+5)^2+3(z-6)^2-7=0$.

Almost-certainly, you're assuming $K_1$ and $K_2$ to be in "standard form", with necessarily-matching coefficients of $1$ on their second-degree terms; that's not an unreasonable assumption (standard form is common), but it is an assumption and so must be stated explicitly. An alternative approach is to restate things thusly:

If $K_1(x,y,z)=0$ and $K_2(x,y,z)=0$ represent two spheres, then there are non-zero values $k_1$ and $k_2$ such that $k_1 K_1 = k_2 K_2$ is the equation of the plane containing the circle of intersection (if any).

(Specifically, we take $k_1$ to be the coefficient of $x^2$ (and $y^2$ and $z^2$)  in $K_2$, and $k_2$ to be the corresponding coefficient in $K_1$. This guarantees that those terms cancel.)

The advantage of the restatement is that it applies when, say, $K_1=0$ represents a plane instead of a sphere. Since we end up taking $k_2=0$ (as $K_1$ has no second-degree terms), the equation $k_1 K_1=k_2 K_2$ reverts to $K_1=0$, as expected; after all, the plane containing the circle of intersection is that original plane.


问:
Cardinality of atomless sigma-algebras
I know that an infinite $\sigma$-algebra $\mathcal{A}$ has at least cardinality $\mathfrak{c}=2^{\aleph_0}$.

Suppose now that $\mathcal{A}$ is $\alpha$-generated ($\alpha>\omega$). Clearly, $\mathcal{A}$ is atomless. But what is its cardinality precisely?

Edit: By $\alpha$-generated I mean that the set of generators of $\mathcal{A}$ has size $\alpha$.

答:
An $\alpha$-generated sigma-algebra $\mathcal A$ has
$$
\alpha \le |\mathcal A| \le \alpha^{\aleph_0}
$$
In case $\alpha=\alpha^{\aleph_0}$, this answers the question.

The fact you refer to in the question is: there is no sigma-algebra with
$$
\aleph_0 \le |\mathcal A| < \mathfrak c = \aleph_0^{\aleph_0}
$$
so we conclude that an infinite countably-generated sigma algebra has power exactly $\mathfrak c$.

Let $\mathcal A$ be a sigma-algebra on a set $X$ generated by $\mathcal G$ with $|\mathcal G| = \alpha$.  Here is a proof for $|\mathcal A| \le \alpha^{\aleph_0}$

Let $\Omega$ be the set of all countable ordinals.  We recursively define
sets $\mathcal G_\lambda \subseteq \mathcal P(X)$
for $\lambda \in \Omega$ as follows.

$\bullet\;\mathcal G_0 = \mathcal G$

$\bullet\;\mathcal G_{\lambda+1}$ is the set of all countable unions $\bigcup_{n=1}^\infty E_n$ where, for each $n$, either $E_n \in \mathcal G_\lambda$ or $X\setminus E_n \in \mathcal G_\lambda$.

$\bullet\;$If $\lambda$ is a limit ordinal, then
$$
\mathcal G_\lambda = \bigcup_{\gamma < \lambda}\mathcal G_\gamma .
$$

Then:
$$
\sigma(\mathcal G) = \bigcup_{\lambda \in \Omega}\mathcal G_\lambda
$$
and we may prove recursively that
$|\mathcal G_\lambda| \le \alpha^{\aleph_0}$ for all $\lambda \in \Omega$.

Next, $|\Omega| = \aleph_1 \le \mathfrak c = 2^{\aleph_0} \le \alpha^{\aleph_0}$, so
$$
|\sigma(\mathcal G)| = \left|\bigcup_{\lambda \in \Omega}\mathcal G_\lambda\right| \le \aleph_1\cdot\alpha^{\aleph_0}
\le \alpha^{\aleph_0}\cdot \alpha^{\aleph_0} = \alpha^{\aleph_0} .
$$

Title: Cardinality of atomless sigma-algebras


问:
Prove that there are no positive real numbers $x$, $y$, $z$ such that $ (12x^2 + yz)(12y^2 + xz)(12z^2 + xy) = 2021x^2y^2z^2.$
Prove that there are no positive real numbers $x, y, z$ such that : $(12x^2 + yz)(12y^2 + xz)(12z^2 + xy) = 2021x^2y^2z^2$.
I managed to simplify it to $12(x^3z^3 + y^3z^3+x^3y^3) + xyz (x^3+y^3+z^3) = (73/3)(x^2y^2z^2$) but i still can't figure it out.
Can someone help?

答: Following what you have left off, apply the AM-GM inequality to both factors : $x^3z^3+z^3y^3+x^3y^3 \ge 3x^2y^2z^2, x^3+y^3+z^3 \ge 3xyz$ and simplify the LHS  you see that LHS > RHS . Done


问:
Check for sufficiency
We have the function

$$f(x) = \frac{1}{2x\sqrt{2\pi}}e^{-\frac{(\ln x - \theta)^2}{8}}$$ for the I.I.D (identically independent distributed) sample $X_1,\dots, X_n$.

I have to show that $\sum_{i=1}^n{ln X_i}$ is a sufficient statistic.

My approach was to take the product of the pdf and then to split the product up in two parts, one with the sufficient statistic and $\theta$ and the second part containing $X_1,\dots, X_n$.

But for me it's an impossible task to split the product up and find the sufficient statistic, if someone could help, that would be great!

答:
Well, for better understanding how to find the sufficient statistic one cat see this very nice lecture.
Anyway, you can start with the joint probability density function of $\vec{X}=(X_1,\dots,X_n)$. So it will be $$f_{\vec{X}}(x_1\ldots x_n)=\prod_{i=1}^n\frac{1}{2x_i\sqrt{2\pi}}e^{-\frac{(\ln x_i - \theta)^2}{8}}=\prod_{i=1}^n\frac{1}{2\sqrt{2\pi}}e^{-\ln (x_i)}e^{-\frac{(\ln( x_i) - \theta)^2}{8}}$$
$$f_{\vec{X}}(x_1\ldots x_n)=\left(\frac{1}{2\sqrt{2\pi}}\right)^n e^{-\sum_{i=1}^n\left(\ln (x_i)+ \frac{(\ln( x_i) - \theta)^2}{8}\right)}$$
Expanding the power one can obtain:
$$f_{\vec{X}}(x_1\ldots x_n)=\left(\frac{1}{2\sqrt{2\pi}}\right)^n e^{-\sum_{i=1}^n\left(\ln (x_i)+ \frac{\ln^2( x_i)}{8}-\frac{\ln(x_i)\theta}{4}+\frac{\theta^2}{8}\right)}$$
Or
$$f_{\vec{X}}(x_1\ldots x_n)=\left(\frac{1}{2\sqrt{2\pi}}\right)^n e^{\sum_{i=1}^n\left(-\frac{\ln^2( x_i)}{8} +\ln (x_i)\left(\frac{\theta}{4}-1\right)-\frac{\theta^2}{8}\right)}$$
By the way one can omit two last steps. The sufficient statistic can be obtained from the second step.
Then one can rewrite it in the following manner:
$$f_{\vec{X}}(x_1\ldots x_n)=\left(\frac{1}{2\sqrt{2\pi}}\right)^n e^{-\sum_{i=1}^n \left(\frac{\ln^2( x_i)}{8}+\ln( x_i)\right)}e^{-\frac{n\theta^2}{8}}e^{\frac{1}{4}\theta\sum_{i=1}^n \ln(x_i)}$$
According to the Fisher–Neyman factorization theorem:

"If the probability density function is $f_\theta(x)$, then $T$ is sufficient for $\theta$ if and only if nonnegative functions $g$ and $h$ can be found such that $f_\theta(x)=h(x) \, g_\theta(T(x))$".

So in the power of the last multiplier the parameter $\theta$ depends on the sample $\vec{X}$ only in terms of statistics $T=\sum_{i=1}^n \ln(x_i)$.

Title: Check for sufficiency


问:
Formula for finding $\cos(\theta)$?
Suppose we have that $\cos(\theta) = \sqrt{3} / 2$. Now I know that I can find the value of $\theta$ easily by referring to a trigonometric table but I want to ask is there any other way to find it's value without the table ? A sort of relationship or formula etc.

答:
I think for your purposes,series expansion of $\cos^{-1}(x)$ would do :
\begin{align}
\cos^{-1}(x) &= \frac{\pi}{2} - \sin^{-1}(x)\\
&=\frac{\pi}{2} - \sum_{n=0}^{\infty} \frac{2n!}{(2^nn!)^2}\frac{x^{2n+1}}{2n+1}; |x|\leq 1
\end{align}
Obviously the formula is much more complicated than the values. But I reckon, given that $|x|\leq1$, and the exponent grows "exponentially" quick,the first few terms should do.

Title: Formula for finding $\cos(\theta)$?


问:
General solution of differential equation with tan
I'm trying to learn this type of question but this one is hardest one that I tried.

Find the general solution of the following differential equation:

$$y'' + y = \tan(t)$$

答:
Multiply with the cosine and add a cancelling zero
$$
[\cos(x)y''(x)-\sin(x)y'(x)]+[\sin(x)y'(x)+\cos(x)y(x)]=\sin(x)
$$
which can be integrated in all terms
$$
\cos(x)y'(x)+\sin(x)y(x)=-\cos(x)+C
$$
This linear DE now has integrating factor $\cos(x)^{-2}$, resulting in
$$
\frac{y(x)}{\cos(x)}=C\tan(x)+D-\int\frac{dx}{\cos x}.
$$
Now you can apply the "trick" $\frac1{\cos x}=\frac{\cos x}{1-\sin^2(x)}=\frac{u'(x)}{1-u(x)^2}$ or something different.


答:
Without using the table you have to derive the values.

In general, you can build a construction of geometric shapes that demonstrate that the angle is a certain proportion of the angle swept out by a full revolution. Keeping in mind that the angle of a full revolution is known to be 360 degrees or 2π radians you can obtain the angle.

The exact value of $\text{arccos}( \sqrt{3 }/ 2 )$ is $\frac{\pi}{6}$ radians.

So take a look at the construction of the Dodecagon.

If that still isn't very intuitive there is an even nicer construction which demonstrates this particular angle which comes from the bisection of an equilateral triangle.

In an equilateral triangle all of the sides have the same length and so all of the angles are the same. Call their value A radians. We know the interior angles of a triangle add to be $\pi$ radians, and so $A+A+A = \pi$. Therefore the angle on an equilateral triangle is $\frac{\pi}{3}$ radians. When you bisect an equilateral triangle the angle is $\frac{\pi}{6}$ radians. If you bisect an equilateral triangle and flip the pieces to make a rectangle, the diagonal across opposite corners has length equal to the original side length. Let's give that original side a value of 1 for simplicity. The diagonal is length 1, The short side of the rectangle is length $\frac{1}{2}$ since we bisected the triangle. So, by the Pythagorean theorem, the length of the long side of the rectangle $L$ satisfies the equation:

$(1/2)^2 + L^2 = 1^2 $

So $L$ is $\sqrt{3 }/ 2$. We can then use the relationship between the sin and cos of angles and the length of the sides of triangles to find theta so that

$\cos( \theta ) = \sqrt{3 }/ 2$

Title: Formula for finding $\cos(\theta)$?

Body: Suppose we have that $\cos(\theta) = \sqrt{3} / 2$. Now I know that I can find the value of $\theta$ easily by referring to a trigonometric table but I want to ask is there any other way to find it's value without the table ? A sort of relationship or formula etc.


问:
When is it okay to take limits in inequalities?

$a_n$ is a convergent sequence in $\mathbb{R}$, and $$a_n \stackrel{n\to\infty}{\longrightarrow}a$$
Show that $$\inf_n a_n \le a\le \sup_n a_n$$

I think it's okay to conclude that $$\inf_n a_n \le a_i\le \sup_n a_n$$ for all $i\in\mathbb{N}$. Now, can we just take $i\to\infty$ and conclude our result? When is it okay/not okay to take limits in inequalities like this? I haven't come across rules for the same before.

答:
I think that $ \inf a_n $ means $ \inf \{a_n: n \in \mathbb N$ and $ \sup a_n $ means $ \sup \{a_n: n \in \mathbb N\}$ . If yes, then

$$\inf a_n \le a_i\le \sup a_n$$

for all $i$ is trivial.

(Definition of $ \inf $ and $ \sup$)

This holds for each bounded sequence.

If $(a_n)$ is convergent to $a$, then we get

$$\inf a_n \le a\le \sup a_n$$

squeeze theorem !


问:
Reduction of order method - Differential Equations
I need help with this question please.

Find the general solution by using reduction of order method to

$$ty'' -(t+1)y'+y = 0$$

given that $y_1(t) = e^t$ is a solution

答: Hint: Let $y=e^tu$ and then you will get an equation of $u$.


问:
Numerically Solve PDE
I am looking for some help finding a numerical solution to a pde of the form:

$$C_t=f(x)C_x+\alpha C_{xx}$$

with initial condition for $C(x,t)$:

$$C(x,0)=\delta(x)$$

and boundary condition

$$C(\pm\infty,0)=0$$

I am using Mathematica's NDSolve function, but it doesn't seem to be able to find a solution for some chosen $f(x)$ and $\alpha$. I think this may be due to the condition's at infinity? Another idea I have is to turn it into a ODE via separation of variables. Any advice on resources or methods would be helpful.

答: You can use a standard method, like the Crank-Nicolson method, on a truncated spatial domain. The conditions on this artificial boundary can be approximated by  setting the solution to be zero, or some function of $(x,t)$, if some adicional information on the asymptotics is available.

Title: Numerically Solve PDE


问:
How can two r.v. be independent if one's density is a function of the other?
Suppose you have the following joint density function and marginal densities:
image

Apparently, X and Y are independent, since their joint density is the product of their densities. However, I do not understand how this is possible, when the marginal density of X is a function of y.

Thanks in advance

答:
The product you are describing is, in fact, the chain rule:
$$
P(X,Y) = P(X \mid Y)P(Y)
$$

In this case, we have that the density associated with $Y$ is $f_Y(y) = e^{-y}$ while the density of the conditional distribution $X \mid Y$ is $f_{X \mid Y} = y^{-1}  e^{\frac{-x}{y}}$. This tells us that both $X$ and $Y$ are not independent for this joint model, because $P(X,Y) \neq P(X)P(Y)$.