答:
Assume $X\neq\emptyset$.

Case 1. By definition $X^\emptyset = \{ f:\emptyset \to X\}$. At least one such map exists, we'll show it's unique. Suppose $f,g\in X^\emptyset$. Then $f,g$ coincide on the $\emptyset$ (vacuously), hence they must be equal.

Case 2. $\emptyset ^X = \{f:X\to\emptyset\}$. Suppose $\emptyset^X \neq\emptyset$. Then there exists $f:X\to \emptyset$. In particular $f(x) \in \emptyset$ for any $x\in X$. Therefore, $\emptyset ^X = \emptyset$.

Title: $X^\emptyset$ versus $\emptyset^X$

Body:
I recently read that there is a difference between $X^\emptyset$ and $\emptyset^X$. I was hoping that someone could explain why this is the case.

I have interpreted $B^A$ as, "The set of all functions that have a domain of $A$ and a range of some subset of $B$". More formally, I thought that:

$f \in B^A \leftrightarrow \exists Z ( Z \in P(B)\  \land f \subseteq A \times Z) $ <-- Edit: $\color{red}{ \text{Here is the mistake} }$

Additionally, we know that $\emptyset \times X = X \times \emptyset=\emptyset \times \emptyset = \emptyset$

It is vacuously true that $\forall x [\emptyset \subseteq x]$

Also, $P(\emptyset)=\{\emptyset\}$, which means that $\emptyset \in P(\emptyset)$ and, more generally, $\forall x [\emptyset \in P(x)]$

Using this information, consider the following two cases:

Case 1: $X^\emptyset$

Case 2: $\emptyset^X$

Case 1: $X^\emptyset$

Consider if $f$ is the empty set function.

Does a $Z$ satisfy the following formula? $Z \in P(X)\  \land \emptyset \subseteq \emptyset \times Z$

Sure! Let $Z=\emptyset$.

$\emptyset$ is an element of $P(X)$ and $\emptyset$ is a subset of $\emptyset \times \emptyset = \emptyset$.

Therefore, $X^\emptyset = \{ \emptyset \}$

Case 2: $\emptyset^X$

Consider if $f$ is the empty set function.

Does a $Z$ satisfy the following formula? $Z \in P(\emptyset)\  \land \emptyset \subseteq X\times Z$

Yup. If $Z=\emptyset$, then $\emptyset \in P(\emptyset)$ and $\emptyset \subseteq X \times \emptyset=\emptyset$

Therefore, $\emptyset^X=\{\emptyset\}$

So where exactly did I go wrong?


问:
Conjecture: For almost all $n$, the sum of $n \bmod k$ for $k<n$ is unique.
It looks to me like $$f(n):= \sum_{k=1}^{n}{(n\bmod k)}$$ is almost unique to $n$, with the only systematic exceptions being $(2^m-1, 2^m)$ pairs, which share a value for any $m \in \mathbb N$. In other words, the set $\{f(n): n\in\mathbb N\land n\neq 2^m
\ \forall m\in\mathbb N\}$ shouldn't have many duplicate values.

As discussed in the answers, a handful of counterexamples have been found, but they seem more sparse than one would expect, and nobody has yet suggested why that might be, or what's special about the counterexamples found, apart from a moderate preponderance of small factors in the mean of the pair.

Although there's not much data to go on, it also looks like counterexamples may come in the form $(n,n+2i)$, or possibly even $(n,n+2^i)$, and I would be interested if anyone can find a case outside of those. The latter form would also include the $(2^m-1,2^m)$ exceptions, which seems elegant.

Also, if someone could set me straight on this: My understanding is that the term 'almost all' in this context means approaching $0$ density in the limit; so if one could show that the exceptions aren't necessarily finite, but they are less common than powers of two, would that be sufficient?

答:
Nope, not unique.

I hadn't checked far enough. The first counterexample is at $$f(38183)=f(38185)=258840858.$$

It could plausibly still have only finitely many counterexamples, though, making the 'almost all' claim true. There are three more found so far through $n< 6 \times 10^8$ with the form $(n,n+2)$: $458009$, $776111$, and $65675407$. As stated below, there was also one example of form $(n,n+4)$: $113393278$.

Note $f(x)$ grows as the square of $x$, so this may be what one would probabilistically expect. It's tough to say, but my guess is counterexamples may be very sparse, yet probably still infinite.

ContentLicense: CC BY-SA 4.0


问:
Integration by Importance sampling
I want to estimate the integral $\int_0^1\frac{1}{\sqrt x}dx$ by importance sampling method.

So, here $f(x)=1\frac{1}{\sqrt x}$

We have to introduce another probability distribution $g(x)$, then $\int_0^1\frac{1}{\sqrt x}dx = \int_0^1\frac{\frac{1}{\sqrt x}}{g(x)}g(x)dx= E_g[\frac{\frac{1}{\sqrt X}}{g(X)}] \approx \frac{1}{N}\sum_{i=1}^{N}\frac{\frac{1}{\sqrt X_i}}{g(X_i)}$

But I cannot decide what will be the function $g(x)$ so that the integration gives accurate result. Will it be a $N(0,1)$ pdf or some pdf taking values within the interval $(0,1)$ or some pdf taking positive values only?

Thanks in advance.

答:
The problem:

Estimate

$$\int_a^b p(x)dx=\int_0^1 \frac{1}{\sqrt{x}}dx=2$$

This is a very easy integral that we can exactly calculate finding 2 as a result.

In order to use Importance sampling Method we have to choose a good function that approximate our $p(x)$

A nice function is  $f(x)\propto e^{-x}$

so normalize it as to transform $f$ in a nice density....

easy...

$$f_X(x)=\frac{e}{e-1}e^{-x}\cdot \mathbb{1}_{(0;1)}(x)$$

Let's calculate its CDF....easy again (calculations omitted)

$$F_X(x)=y=\frac{e}{e-1}[1-e^{-x}]$$

Let's now invert F so that we can use the Integral Transform Theorem

We are all set!

the values of Y come from a standard uniform distribution, say $Y\sim U(0;1)$ so with any calculator you can sample from this distribution (I used Excel for the sampling) and you have only to calculate the sample mean of the ratio $\frac{p(x)}{f(x)}$

This is the beginning of my table

...the resulting mean of these few data is $\approx 1.64$ but with $n=2500$ I got a result of $1.990636$

...pretty good

Title: Integration by Importance sampling


问:
Repeating decimals in converted fractions, why these increments?
In regards to repeating decimals and cyclic numbers: I understand that many of them, while multiplying by certain integers will produce a number with the same variant of digits, however my question is in regards to the patterns already occurring within some of them.

Many fractions that convert into a number with repeating decimals, will exhibit a pattern within the repetend. This pattern is made of a consecutive list of numbers growing by the same multiple.

example: $1/31=0.032258064516129...$., the repetend is
$032258064516129$ which can be broken into the numbers $032258$, $064516$, $129032$ that are related by multiples of 2:
$$032258∗2=064516, \quad 064516∗2=129032.$$

example: $1/13=0.076923....$, the repetend is $076923$ which can be broken into $0769$, $2307$ that are related by multiplying by 3:
$$0769∗3=2307$$

example: $1/17=0.0588235294117647....$, the repetend is $0588235294117647$ which can be broken into the numbers $5882352941$, $11764705882$ that are related by multiplying by 2:
$$5882352941∗2=11764705882$$

example: $1/19=0.052631578947368421...$, the repetend is $052631578947368421$ which can be broken into the numbers $05263$ , $15789$ , $47367$ , $142101$ (think of the initial 1 in 142101 being removed from the 8 in the preceding 47368) that are related by multiples of 3:
$$05263∗3=15789, \quad 15789∗3=47367, \quad 47367∗3=142101$$

example: $1/21=0.047619...$, the repetend is $047619$ which can be broken into the numbers $0476$, $1904$ that are related by multiplying by 4:
$$0476∗4=1904$$

What are these multiples?  That is, given 1/$n$ with a repeating decimal, what multiple connects the different pieces of the repetend? (And how long are those pieces?)

答:
The first thing to think about is finding the repeating decimal by long division.  Taking $1/17$ for example, your first division starts with a remainder $1$.  You append a $0$ to it and try to divide $17$ into $10$.  It goes $0$ times, which gives the leading $0$ in the repeat.  You then pass a remainder of $10$ to the next digit.  You pull down another $0$, getting $100$, divide $17$ into it getting $5$ with a remainder of $15$.  The $5$ is the second digit of the repeat and the $15$ goes on for the next digit.

If you have a full length repeat like $17$, you must use each of the remainders from $1$ to $17-1$ once.  When you repeat a remainder you have finished the repeat.  For $13$, which has only a six digit repeat you only use six remainders.  This explains all the relationships you show.

After the $n^{th}$ digit the remainder is $10^n \mod 17$.  We can find that $10^{10} \equiv 2 \pmod {17}$  That means after $10$ digits you pass a remainder of $2$ to the next digit.  The repeat from there will clearly be twice the repeat starting from the front, where the remainder is $1$.


问:
why $\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx$ is the $min(c,d)$?
Let $\{ \tau_n (x) \}_{n=1}^\infty$ be closed orthonormal system in $L^2_{pc}[0,1]$ show that for $c,d\in (0,1)$ $$\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx = min\{c,d\}$$

答:
Consider the characteristic functions of the intervals $\chi_d : = \chi_{[0,d]}, \chi_C := \chi_{[0,c]}$. Then, we have:
$$\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx = \sum_{n=1}^\infty \int_0^1 \tau_n(x) \chi_d(x) dx \int_0^1 \overline{\tau_n(x)}\chi_c(x)dx$$
Notice now that for any $f,g$ in $L^2$ we have:

$$
\langle f,g \rangle = \sum_{n = 1}^\infty \langle f, e_n \rangle \langle g , e_n \rangle$$

In particular it follows:
$$
\sum_{n=1}^\infty \left[ \int_0^1 \tau_n(x) \chi_d(x) dx \int_0^1 \overline{\tau_n(x)}\chi_c(x)dx \right] = \int_0^1 \chi_c(x) \chi_d(x)dx = \min(c,d)
$$
as the left hand side above is the right hand side two lines above, and vice versa


问:
Solve $\sqrt{3}\tan\theta=2\sin\theta$
I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.

$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$

$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$

I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.

Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.

Where am I going wrong please?

答:
Dear PUK, you missed $\theta=0$ because you divided the equation by $\tan\theta$ in the first step which is only possible if $\tan\theta$ is nonzero and finite. That's why $\theta=0$ and $\theta=\pi$ and $\theta=-\pi$ for which $\tan\theta$ vanish must be separately checked and indeed, you will find out that the equation is satisfied because $0=0$.

On the other hand, you added the wrong solutions $\pm 5\pi/6$ because their squared cosine is $3/4$, but the cosine itself has a wrong sign, so your squaring created problems. The squaring was unnecessary, as pointed out by lhf as I was writing this sentence, but if you still want to square it, you have to check all the solutions that they work and you will find out that the $\pm 5\pi/6$ solutions don't.

Title: Solve $\sqrt{3}\tan\theta=2\sin\theta$

ContentLicense: CC BY-SA 3.0


问:
Example of a continuous function which is bounded and not contained in any $L_p$-space ($p\gt 0$)
I'm struggling to find an example of a continuous function $f:(0,\infty)\to \mathbb R$ which is bounded, not contained in any $L_p$-space ($p\gt 0$) and goes to zero when x goes to infinity.

I need help.

Thanks a lot!

Body:
Consider
$$
f(x)=\frac{1}{\log(x+3)}
$$


问:
Module of differentials of an extension of Dedekind domains
Let $A$ be a Dedekind domain with fraction field $K$, $L|K$ a finite separable field extension and $B$ the integral closure of $A$ in $L$. Assume that al the residue field extensions are separable. In the proof of Serre's "Local Fields", chapter III, Proposition 14 (page 59), it is stated that in order to show that the module of differentials $\Omega_{B|A}$ is cyclic we can assume that $A$ is local and complete. Why?

I understand the reduction to the local case of the other part of the theorem (the annihilator of the module of differentials equals the Dedekind different), since clearly two ideals are equal if they are locally equal, but being cyclic is not a local property in general, and I am not able to understand that reduction to the local case.

For convenience the above Proposition is the following:

答: $\Omega_{B|A}$ is a finitely generated $B$-module, and we know it is locally cyclic and annihilated by the different.  In particular, this means it is a finitely generated torsion $B$-module, so since $B$ is a Dedekind domain, it is a direct sum $\bigoplus B/p_i^{n_i}$ for primes $p_i$.  Now since it is locally cyclic, the $p_i$ must all be distinct (if a prime $p$ appeared twice among the $p_i$, the localization at $p$ would not be cyclic).  It now follows from the Chinese remainder theorem that $\Omega_{B|A}\cong \bigoplus B/p_i^{n_i} \cong B/\prod p_i^{n_i}$ so it is cyclic.


问:
13 cards randomly dealt to 4 players. What's the probability of 1 player being dealt quads/four-of-a-kind?
Inspired by the Vietnamese game Tiến lên where you are dealt $13$ cards. One specific rule states if you are dealt quads 2 (four 2's), you automatically win that round. I want to find the probability of that happening.

My method:
$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times {}^{13}C_4 = 0.264\%$

The first 4 terms are the probability of being dealt quads 2 in 1 particular order. We have 13c4 of these equivalent orders, so I multiply it by 13c4.

2 questions:

Is my answer correct?

I tried to use another approach: P(I have quads 2) = 1 - P(3 other players have no 2's). I tried to calculate the subtracted probability by doing 48/52 * 47/51 * ... * 11/15 * 10/14 but it seems that it's wrong. How would I go about doing this?

Thanks!

答:
If the question is about the probability that the $13$ cards dealt to you contain the $4$ twos and $9$ other cards, then the probability would be $$\frac{\,^{4}C_4 \,^{48}C_{9}}{{\,^{52}C_{13}}} \approx 0.002641$$

This is the same as your answer of $\dfrac{\,^{13}C_4 }{{\,^{52}C_{4}}}$, as can be shown if you expand into factorials

Your alternative approach of considering the other $39$ cards also gives the same answer, though you should not subtract from $1$: $$\frac{48}{52}\times \frac{47}{51}\times \cdots\times \frac{10}{14} = \frac{ \,^{48}C_{39}}{{\,^{52}C_{39}}}= \frac{ \,^{48}C_{9}}{{\,^{52}C_{13}}}$$


问:
Irregularity of $a^ib^jc^k∣\text{ if }i=1\text{ then }j=k$ using pumping lemma
This question has been asked here several times. However none of the examples I saw actually tried to prove it's irregularity using the standard pumping lemma.

I know how to prove irregularity using the specific case of $a^1b^pc^p$ however I'm struggling to build the pumping lemma for the whole language $a^ib^jc^k$
I know this won't prove the language to be regular, but I'm more interested in how to deal with lemmas with multiple variables on the exponents.
Let's suppose I've divided the string into the following:

$ x = \epsilon$

$ y = b^\beta$

$ z = b^{p-\beta}c^m$

Let $n = 2$. I get

$xy^2z= b^{p-\beta}c^n$

But this doesn't seem like I can find a pumping length for this language as it has too many variables. Is there a way to make this pumping lemma work? (I know it can't be used to prove the language regular, but I'm trying to find at least one pumping length which works for this language)

答:
If I understand correctly, you want to explicitly show that language $\{a^i b^j c^k | i, j, k \in \mathbb{N}\}$ satisfies pumping lemma (it should be true, as this language is clearly regular).

Notation of pumping lemma: there exist pumping length $p$ s.t. any word $w$ with $|w| \geq p$ can be written as $w = xyz$ where $|y| > 0$, $|xy| \leq p$ and for all $n$, $x y^n z$ is in our language.

For our language, we can take $p = 1$, and represent word $w$ as $\epsilon . y . z$ ($.$ denotes concatenation), where $|y| = 1$.

As $w$ is in our language, $w = a^i b^j c^k$, as $|w| > 0$, at least one of $i, j, k$ is non-zero.

If $i > 0$, they $y = a$ and we have $a^{i - 1 + n} b^j c^k$ is in our language - it's true by definition.

Cases $i = 0, j > 0$ and $i = j = 0, k > 0$ are similar.


问:
Inequalities involving polynomials
Let $A_k=\{x \mid  2^k < x^k +x^{k+1} < 2^{k+1}\},\, k=1, 2, 3, \ldots$.

Find the largest positive integer $n$ such that
$$
A_1\cap A_2\cap\cdots \cap A_n \ne \emptyset.
$$

答:
Computers these days can do exact arithmetic, so there's no reason you can't have a computer do an "analytic" proof - or at least have it do the relevant computations.

The positive portions of your sets

\begin{array}{l}
 2<x^2+x<4 \\
 4<x^3+x^2<8 \\
 8<x^4+x^3<16 \\
 16<x^5+x^4<32 \\
\end{array}

look something like so

To show that the intersection of the first and last is empty, it suffices to find $x_0$ such that $x_0^2+x_0>4$ (i.e., $x_0$ lies to the right of the first interval) and $x_0^5+x_0^4<16$ (i.e., $x_0$ lies to the left of the last interval).  I propose that you choose
$$x_0 = \frac{196377467}{125086164}.$$
Then
$$x_0^2+x_0 - 4 = \frac{542019891463093}{15646548424234896} > 0$$
and
$$x_0^5+x_0^4 -16= -\frac{11888789705707430179748265315192519808033}{30622903893638169223375610368116647629824} < 0.$$

I obtained the number $x_0$ by simply rationalizing the average of numerical approximations to the two relevant endpoints, but the exact computation with rational numbers proves that $x_0$ has the desired properties.

Title: Inequalities involving polynomials


问:
Number of elements of order $n$: explain why $\Bbb{Z}_4\times\Bbb{Z}_4$ and $\Bbb{Z}_4\times\Bbb{Z}_2\times\Bbb{Z}_2$ are not isomorphic.

For the following pair of groups, explain why they are not isomorphic:
$  \mathbb{Z}_4 \times \mathbb{Z}_4$ and $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

I have started my exercises on isomorphisms and have run into this problem. In the answer section it says that these two groups are not isomorphic because they have different amounts of elements of order $2$.

I have listed the elements of each group and the effect does not have the same number of elements of order $2$. But how could this procedure be done without having to resort to listing the elements?

On the other hand, given one of these situations, what number $n$ to take? That is, can I by trial and error with the possible orders that divide the order of the group?

答:
It follows from the Fundamental Theorem of Finite(ly-generated) Abelian Groups. But this is a little glib on my part.

Yes, counting the number of elements of order two is sufficient. Trial & error will work since the groups are small enough.


问:
Let $S$ be a finite set. Show that $\text{Sym}(S)$ is a group.
Wondering if someone could check I have answered this correctly, I’ve just started a course on Group theory and don’t feel confident in ability so far. The Question asks: “Let $S$ be a finite set. Show that $\operatorname{Sym}(S)$ is a group.”

Much appreciated,
L

Answer:

Let $X,Y,W,Z\subseteq S$ be groups such that
$$\mu\colon X\to Y,\quad\Phi\colon Y\to W,\quad p\colon W\to Z$$
Then we have

$G0)$ Closure: Let $x\in X$ be arbitrary. Then
$$\Phi\circ\mu(x)=\Phi(\mu(x))=\Phi(y)\in\operatorname{Sym}(S)$$
for some $y\in Y$.

$G1)$ Associativity:
$$\rho(\Phi\circ\mu(x))=\rho(\mu\circ\Phi(x))=\rho(\Phi(\mu(x)))\in\operatorname{Sym}(S)$$
and
$$(\rho\circ\Phi)\circ\mu(x))=(\Phi\circ\mu)\circ\mu(x))=\rho(\Phi(\mu(x)))\in\operatorname{Sym}(S)$$
Thus, $\operatorname{Sym}(S)$ is closed under associativity.

$G2)$ Identity: Let $\sigma\colon S\to S,\,s\mapsto es$, then $\sigma$ is identity map $\operatorname{Id}\colon S\to S$
$$\sigma(s)=es=s$$
where $e$ is the identity element of $S$.

$G3)$ Inverse: Let $f\colon X\to Y$ and $g\colon Y\to X$ be maps where $X,Y\subseteq S$ and $f,g\in\operatorname{Sym}(S)$. Then,
$$f\circ g (y)=f(g(y))=e\quad\text{and}\quad g\circ f(x)=g(f(x))=e$$
$\Rightarrow g(y)=f^{-1}(y)$

Thus, $\operatorname{Sym}(S)$ is a group under composition.

答:
In the following, the finiteness of $S$ doesn't enter at all, so the claim holds for infinite $S$, either.

If $f\in\operatorname{Sym}(S)$, then by definition of bijection: $\forall t\in S, \exists! s\in S\mid t=f(s)$. This latter, in turn, means that there is a map $g\colon S\to S$, $t\mapsto s=g(t)$, such that $t=f(s)=f(g(t))$. Note that $g$ is bijective too, namely $g\in\operatorname{Sym}(S)$: in fact, $g(t)=g(t')\Rightarrow f(g(t))=f(g(t'))\Rightarrow t=t'$ (injectivity); moreover, for every $s\in S$, $s=g(f(s))$ (surjectivity). Suppose now that $g'\colon S\to S$ is another map such that $t=f(g'(t))$; then, $\forall t\in S$, $f(g'(t))=f(g(t))$ and hence (by the injectivity of $f$) $g'(t)=g(t)$, and finally $g'=g$. To sum up, given $f\in\operatorname{Sym}(G)$, there is a unique $g\in\operatorname{Sym}(G)$ such that:

$\forall s\in S$, $g(f(s))=s$

$\forall t\in S$, $f(g(t))=t$

Since such a map $g\in\operatorname{Sym}(S)$ is unique and depends on $f$, the symbol "$f^{-1}$" is indeed suitable to denote it.

Now, you have to prove the closure of $\operatorname{Sym}(S)$ under map composition, namely that given $f,f'\in\operatorname{Sym}(S)$ and defined $\forall s\in S$:

$$(ff')(s):=f(f'(s))$$

it turns out that $ff'\in\operatorname{Sym}(S)$. In fact:

injectivity of $ff'$: $(ff')(s)=(ff')(s')\Longrightarrow f(f'(s))=f(f'(s'))\stackrel{(f\space is\space inj.)}{\Longrightarrow}f'(s)=f'(s')\stackrel{(f'\space is\space inj.)}{\Longrightarrow}s=s'$;

surjectivity of $ff'$: $\forall t\in S, \space\space t\stackrel{(f, f'\space have\space inverse)}{=}(ff')((f'^{-1}f^{-1})(t))$.

And now, the associative law: $\forall f,f',f''\in\operatorname{Sym}(S)$, $\forall s\in S$:

$$(f(f'f''))(s)=f((f'f'')(s))=f(f'(f''(s)))=(ff')(f''(s))=((ff')f'')(s)$$

whence $f(f'f'')=(ff')f''$.

Last but one point, the identity. Define $\iota(s):=s, \forall s\in S$; it is bijective, so $\iota\in\operatorname{Sym}(S)$, and indeed $f\iota=\iota f=f$ for every $f\in\operatorname{Sym}(S)$.

Finally, the inverse of $f\in\operatorname{Sym}(S)$: as defined in the opening, it precisely gives $ff^{-1}=f^{-1}f=\iota$.


答:
This is correct for the probability that a given player has all four twos (in a random hand of $13$ cards). There is perhaps a simpler way to get the same answer, as follows. There are ${}^{48}C_9$ hands which contain all four twos (since each hand contains $9$ other cards chosen from the remaining $48$ cards), and there are ${}^{52}C_{13}$ hands in total, so the probability is $\frac{{}^{48}C_9}{{}^{52}C_{13}}$, and this simplifies to $\frac{13\times12\times11\times 10}{52\times51\times50\times 49}$.

If you want the probability that one of the four players has all the twos, then since these are disjoint events you can just add the probabilities, i.e. multiply the above by $4$.

Title: 13 cards randomly dealt to 4 players. What's the probability of 1 player being dealt quads/four-of-a-kind?

Body:
Inspired by the Vietnamese game Tiến lên where you are dealt $13$ cards. One specific rule states if you are dealt quads 2 (four 2's), you automatically win that round. I want to find the probability of that happening.

My method:
$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times {}^{13}C_4 = 0.264\%$

The first 4 terms are the probability of being dealt quads 2 in 1 particular order. We have 13c4 of these equivalent orders, so I multiply it by 13c4.

2 questions:

Is my answer correct?

I tried to use another approach: P(I have quads 2) = 1 - P(3 other players have no 2's). I tried to calculate the subtracted probability by doing 48/52 * 47/51 * ... * 11/15 * 10/14 but it seems that it's wrong. How would I go about doing this?

Thanks!


问:
Number of smooth numbers less than x
A $p$-smooth number is defined as an integer whose prime factors are all less than or equal to $p$

In the wiki article about smooth numbers it states:

Let $\displaystyle \Psi (x,y)$ denote the number of $y$-smooth integers less than or equal to $x$ (the de Bruijn function). If the smoothness bound $B$ is fixed and small, there is a good estimate for $\displaystyle \Psi (x,B)$:
$$\displaystyle \Psi (x,B)\sim \frac {1}{\pi (B)!}\prod _{p\leq B}{\frac {\log x}{\log p}}$$
where $\displaystyle \pi (B)$ denotes the number of primes less than or equal to $B$.

How was this derived? Are there any other good bounds on $\Psi(x,y)$?

答:
Let $S(x, B) $ be the set of $B$ smooth numbers less than or equal than $x$. You notice that $n \in S(x, B) $ iff $\log(n) \le \log(x) $. Writing

$$n = \prod_{p_i \le B} p_i^{\alpha_i} $$

For $p_i$ primes and $\alpha_i$ the exponent with which appears (possibly zero), $n\in S(x, B) $ iff

$$\sum_i \alpha_i \log p_i \le \log(x) $$

This is like asking how many integer coordinates points $(\alpha_1, \ldots, \alpha_m) $ are contained in the region

$$ A(x, B) = \{ (t_1, \ldots, t_m) \in \mathbb{R}^m : \sum (\log p_i) t_i \le \log(x), \ \ \ t_i \ge 0\} $$

Here $m =\pi(B) $ is the number of free parameters we have.
It turns out this is closely related to the volume of this region: up to small problems on the boundary, a integer coordinate point contribute with a cube of volume 1, so that

$$\# \{\text{integer coordinate points in } A(x, B) \} \sim \text{volume}(A(x, B)) $$

Since the boundary has one dimension less, the approximation gets better and the better as $x$ gets larger with respect to $B$.

Let's wrap our head around how to calculate this volume. We could do the integral but it's boring. Geometrically, it is a little pyramid of dimension $m+1$, with a nice angle at zero and then some edges departing from it. The edges are long $\log(x) /\log(p_i) $: you get this by taking the maximum possible $t_i$ while all the other parameters are zero, because we are going along an axis.

The volume is linear in the length of the edges, so that we can factor out a

$$ \prod_{p\le B} \frac{\log(x) }{\log(p) }$$

And we are left with computing the volume of

$$D =\{(t_1, \ldots, t_m) : \sum t_i \le 1, \ \ t_i \ge 0\}$$

For $m=2$, this is $1/2$. Let's show by induction that the volume of such a pyramid in dimension $k$ is $1/k! $. If we know this for $k$, we also know that a tiny piramid with edges $=s$ has volume $s^k/k! $, by the same scaling argument as above. For $k+1$, we can section along a coordinate and we get pyramids of size $s$ for all $0\le s \le 1$. Integrating we get

$$ \int_{s=0}^1 \frac{s^k}{k! } = \frac{1}{(k+1)! }$$

As desired. Getting back to our original problem, we had $m=\pi(B) $ dimensions, so that

$$ \Psi(x, B) \sim \frac{1}{\pi(B)! } \prod_{p \le B} \frac{\log x}{\log p} $$

As desired. Note also that you have estimates of both parts in term of $x, B$, so that you get a neat estimate in the end!

Title: Number of smooth numbers less than x


问:
Intuitive reason why $ \frac1j > \int_{j}^{j+1} \frac1x dx $ for $j$ < $x$
In an Econometrics - Time Series textbook (Hamilton), I saw the claim, without proof:

Consider the sequence $ \psi_j $, then, for $ x > j $, $ \frac1j > \frac1x $, and therefore:

$$  \frac1j > \int_{j}^{j+1} \frac1x dx $$ for $x$ > $j$. When I plug in numbers for $j$, I see that this is indeed satisfied. But I'd like to know the intuition behind this or a proof.

答:
If $f:[a,b]\to\mathbb R$ is decreasing (where $a<b$ are two real numbers), then for all $x\in(a,b]$, $f(x)<f(a)$, hence
$$
\int_a^bf(x)\,dx<\int_a^bf(a)\,dx=(b-a)f(a).
$$

Apply the latter with $a=j$, $b=j+1$ and $f:x\mapsto\frac1x$.


问:
Show that $Lf:=f^{'''}$ is not a generator of a Feller-Dynkin Semi-Group
Show that $Lf:=f^{'''}$ is not a generator of a Feller-Dynkin semigroup with domain $$D:=\{f \in C_{0}(\mathbb R):\;f^{'},\;f^{''},\;\;f^{'''}\in C_{0}(\mathbb R)\}.$$

I hence would like to either show that $L$ does not obey the positive maximum principle or there exists no $\lambda>0$ such that $\operatorname{range}(\lambda-L)=C_{0}(\mathbb R)$

I attempted to show the positive maximum principle does not hold by choosing $f(x)=e^{-x^2}$ but I have seen that $f^{'''}(0)\leq 0$ and I cannot seem to find another example where the positive maximum principle may not hold.

Now, let $\lambda>0$ then then for any $f\in C_{0}(\mathbb R)$ we obtain: $(\lambda-L)f$ so I need to find an $f\in D$ such that $\lambda f-f^{'''}\notin C_{0}(\mathbb R)$, which seems like I am attempting to prove an incorrect statement.

I am stuck, has anyone got an idea how I can prove the initial statement?

We thus can prove this by showing that the above is not dissipative, i.e.

$$\exists \lambda > 0,\;\exists f \in D:\; \lvert\lvert\lambda f-Lf\rvert \rvert<\lambda\lvert \lvert f\rvert\rvert$$

答: Consider $g(x) = \cos(x) +0.1x$. Then $g$ has a local maximum at $x_0=\arcsin(0.1)=0.100167..$. Also,  $g'''(x) = \sin(x)$ is strictly positive at $x_0$. You won't have trouble producing a function $f\in D$, equal to $g$ on the interval $[-1,1]$, with a global maximum at $x_0$.  This $f$ shows that $L$ doesn't satisfy the maximum principle.


问:
Defining loops: why is divisibility and identitiy implying invertibility?
Wikipedia contains the following figure (to be found, e.g. here) in order to visualize the relations between several algebraic structures. I highlighted a part that I find especially interesting.

It seems to suggest that a loop can be defined either

as a magma with identity and invertibility, or

as a magma with identity and divisibility.

It is easy to see that from 1. follows 2. But I have trouble proving the converse. Here is a proof for the case that the loop is finite:

By divisibility $xa=xb\implies a=b$ for all $x,a,b\in L$. Thus, the map $a\mapsto xa$ is injective. If the loop is finite, the map is also surjective. Thus, there is a $x'\in L$ with $xx'=e$. This $x'$ is therefore the inverse of $x$.

But what if $L$ is not finite?

答:
Having division is a stronger property than having cancellation. A magma $M$ (that is, a set with a binary operation, which we write multiplicatively) has division (and is called a quasigroup) if for all $a, b \in M$, there exist unique $x, y \in M$ such that $ax = b$ and $ya = b$. If $M$ has an identity element $e \in M$, then this trivially implies the existence of inverses: the left (resp. right) inverse of $a \in M$ is the unique $x \in M$ (resp. $y \in M$) such that $ax = e$ (resp. $ya = e$), which is exactly the definition of an inverse in a unital magma.

On the other hand, as Christoph mentions in a comment, the nonnegative integers with addition give an example of a commutative monoid that has cancellation but not inverses, and hence also doesn't have division (which we would think of as "having subtraction" in this case since it's written additively).

By the way, you've used the word "divisible" for what I've called "has division", but I've avoided the word "divisible" in this context because it has a different meaning: a commutative group $G$ is called a divisible group if for all $x \in G$ and all positive integers $n$, there exists $y \in G$ such that $y^n = x$. (If the group operation is written additively, this means that every element is "divisible by integers", hence the name.) So make sure not to mix up the notion of a magma having division (the defining property of a quasigroup) with the notion of a commutative group being divisible.


问:
Exists Lebesgue integral
Lebesgue integral $\displaystyle\int^0_1 \log x \log(1+x) dx$ exists?

I know that
$2(x-1)\leq \log x\leq x-1$ for $1/2<x<1$

$x\leq |\log(x-1)|\leq 2x$, for $0<x<1/2$

Then $|\log\log(x+1)|\leq 2x|\log x|+4(1-x)|\log(1-x)|$

答:
Yes, the lebesgue integral exists. To check it you just have to prove that the function is bounded in the interval, because a measurable, bounded function in a space of finite measure is always integrable.

That can be checked by calculating the limits of the function in $0$ and $1$ using L´Hôpital´s rule (both limits are 0).

Title: Exists Lebesgue integral


答:
The fact that $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z\not\cong\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ is implied by $\Bbb Z/4\Bbb Z\not\cong\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ (which is much easier to check: one of these groups has no element of order $>2$ while the other does) and related to the Fundamental Theorem of finite(ly generated) Abelian groups. Long story short: one can decompose (non-cyclic) abelian groups into direct products of cyclic groups of prime power order. However, splitting prime powers in this process produces non-isomorphic abelian groups.

As I am aware, this might be a bit over the top for this simple question. A simpler thing to observe is the following:
$$\operatorname{ord}((a_1,\dots,a_n))=\operatorname{lcm}(\operatorname{ord}(a_1),\dots,\operatorname{ord}(a_n))$$
It is not hard to see that $\Bbb Z/4\Bbb Z$ has $2$ elements of order $4$, $1$ of order $2$ and $1$ of order $1$ while $\Bbb Z/2\Bbb Z$ has $1$ element of order $2$ and $1$ of order $1$. Combining these gives a much more efficient way of computing the number of elements of a given order.

The process always works in the sense that if you find different numbers of elements of a given order (dividing the group order) they cannot be isomorphic as isomorphisms preserve the order of elements. However, the converse is not necessarily true, that is there are groups of the same order with the same number of elements for every order which are not isomorphic (see here for a counterexample).

Let's consider the group $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. Write $\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ and $\Bbb Z/2\Bbb Z=\{0,1\}$. As $\operatorname{lcm}(n_1,n_2,n_3)\ge\max\{n_1,n_2,n_3\}$ we conclude that for $(a_1,a_2,a_3)$ having order $2$ we have $a_1\ne1,3$. This leaves $2\cdot2\cdot2=8$ possibilites. However, if $a_1=a_2=a_3=0$ then $\operatorname{ord}(a_1,a_2,a_3)=1<2$. But this is the only exception to be found. Hence, there are exactly $7$ elements of order $2$.

For $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z$ we are immediately down to at most $4$ possibilites ($3$ remain in the end). This shows that they cannot be isomorphic.

Title: Number of elements of order $n$: explain why $\Bbb{Z}_4\times\Bbb{Z}_4$ and $\Bbb{Z}_4\times\Bbb{Z}_2\times\Bbb{Z}_2$ are not isomorphic.

Body:
For the following pair of groups, explain why they are not isomorphic:
$  \mathbb{Z}_4 \times \mathbb{Z}_4$ and $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

I have started my exercises on isomorphisms and have run into this problem. In the answer section it says that these two groups are not isomorphic because they have different amounts of elements of order $2$.

I have listed the elements of each group and the effect does not have the same number of elements of order $2$. But how could this procedure be done without having to resort to listing the elements?

On the other hand, given one of these situations, what number $n$ to take? That is, can I by trial and error with the possible orders that divide the order of the group?


答: On the interval $(j, j+1)$, $\dfrac1x < \dfrac1j$. Hence $\displaystyle\int_j^{j+1}\frac1x dx < \frac1j \cdot (j+1-j) = \frac1j$.

Title: Intuitive reason why $ \frac1j > \int_{j}^{j+1} \frac1x dx $ for $j$ < $x$

Body:
In an Econometrics - Time Series textbook (Hamilton), I saw the claim, without proof:

Consider the sequence $ \psi_j $, then, for $ x > j $, $ \frac1j > \frac1x $, and therefore:

$$  \frac1j > \int_{j}^{j+1} \frac1x dx $$ for $x$ > $j$. When I plug in numbers for $j$, I see that this is indeed satisfied. But I'd like to know the intuition behind this or a proof.


问:
Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodule.
I saw the following result somewhere with its proof:

Result: Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodules.

Proof: Let $x_1, x_2$ be two distinct element in $S$. Since $S$ is a minimal generating of $M$ we infer that $x_1R\nsubseteq x_2R$ and $x_2R\nsubseteq x_1R$. Hence by Zorn's Lemma there are two distinct maximal submodules, $N, P$ say, with $x_1\in N, x_2\notin N$ and $x_2\in P, x_1\notin P$.

I have doubt in the last line of the proof. My doubt: why $N$ and $P$ are maximal submodules of $M$? They are just maximal member of a class of submodules. Please explain it.

答:
I’m writing this under the impression that the minimal generating set is finite, because otherwise it’s not clear to me why any maximal submodules should exist.

You're right, the application of Zorn's lemma is inadequately justified.  It is not even clear what it is being applied to.  It's also unclear that just arguing with $x_1$ and $x_2$ is enough. It seems like you have to involve the (potentially very large collection of) $x_i$'s other than those two.

What's releveant is that by Zorn's lemma, any finitely generated module has a maximal submodule.

Now, to make that useful, let $x_1,\ldots x_n$ be a minimal generating set for $M$.  Now, consider the submodule generated by $\{x_2,\ldots x_n\}$. Since this is less than a minimal generating set, it must be contained in a maximal submodule $N_1$ of $M$, and $x_1$ is certainly not in $N_1$ (or else all the $x_i$'s together could not generate $M$.)

For the same reason, there's a maximal submodule $N_2$ of $M$ containing $\{x_i\mid i\neq 2\}$, and $x_2\notin N_2$.

Now it's clear that you have two distinct maximal submodules.

By this reasoning, you have at least $n$ distinct maximal submodules.

———

As the other solution now points out, using the trick of finding a maximal submodule leaving out one generator does always yield a maximal submodule, so it does work when $S$ is infinite.  That means (counterintuitively, to me) that all examples of modules with no maximal submodules do not have minimal generating sets.  I let that disbelief cloud what should have been a pretty straightforward check.

I’m leaving the answer up because it contains the feedback on the users original attempt.


问:
Calculating probability of next event given remaining count and range for each
(Apologies if my use of terms is off - I am not a mathematician by trade and what I did learn way back was in a different language)

What I know:
the total length of the stream, the body of possible events, the pop-count for each event, and the range of each event (essentially the index of the last instance of this event).

What I want:
Given an incoming stream of events I am trying to predict what the next event is likely to be, i.e. calculate the probability of the next event being a particular one of the known population.

Example:

Let's say the incoming stream is

```<code>ABABCCCC (not known by the receiver)
</code>```

Known by the receiver:

```<code>Length=8

   Count Range
A: 2     3
B: 2     4
C: 4     8
</code>```

Ignoring the range component we can trivially calculate the probability of each event by dividing its pop-count into the total message length:

```<code>   Count Range Prob
A: 2     3     0.25
B: 2     4     0.25
C: 4     8     0.50
</code>```

When looking at the range we can calculate the local probability of each event being next (by local I mean considering only the current event).

```<code>   Count Range Local prob
A: 2     3     2/3 = 0.6666
B: 2     4     2/4 = 0.5
C: 4     8     4/8 = 0.5
</code>```

But how do I combine the two types of probability into one value without enumerating all combinations and testing each one to see if it matches the constraints?

It's obvious from inspection that in this particular case the probability of the next event being a C is 0.0 here as having anything but A or B as the first event would prevent all of the As and Bs from occurring within their range, but it's not clear to me what the formal calculation should look like.

-

Some more context:

In the real scenario, number of unique events can run into the low thousands while the stream length can be up to a few billions while the million range will be more typical.

I can pre-calculate a few incoming events essentially for free so I want to pick the most probably one(s) for that in an attempt to save overall cost.

Events can come and go - one of the static events is Define New Event - I did not include this fact initially because it does not affect the issue at hand. Sometimes all live ranges will be so long that the individual ranges don't matter - I can just go with the straight probability based on pop-count alone, but every once in a while there'll be short bursts of events that are short-lived but whose probability will locally exceed those of the longer-lived, more populous events. Calculation does not have to be exact - if I can calculate a reasonable trend then that will definitely help.

答:
Roughly using the probability of seeing a letter in that slot times the probability of that slot:

In this case the $P(A)=\frac 2 3 *\frac 3 8+ 0*\frac 5 8=1/4$.

$P(B)=1/3*3/8+1*1/8*0*4/8=1/4$

$P(C)=0*4/8+1*4/8=1/2$

You still get the same answers as the naive case!

By the way in this case there are only 3 possible sequences:

AABBCCCC

ABABCCCC

BAABCCCC

It is possible to calculate various conditional probabilities.

$P(A|A)=2/6=1/3$

$P（C|C)=1$, (excluding the last C), $P(A|C)=P(B|C)=0$. Notice the rule $P(A|C)+P(B|C)+P(C|C)=1$ holds.

$P(C|B)=1/2$

$P(A|“AB”)=1/4$

...

Not sure if you can do it without enumerating possibilities. Note: if you want an answer from a “mathematician” this is not the one.


问:
Simple formula to find number with all consecutive digits from 1 to $x$?

Given number of digits required, $x$, find an $x$-digit number such that $f(1) = 1$, $f(2) = 12$, $f(3) = 123$, $f(4) = 1,234$, and so forth.

I'm banging my head against the wall trying to logically figure out what kind of formula would give such results. I'm using this for a web app I'm developing, and I'd rather find a formula than use a for or while loop.

Can anybody help? Thanks!

P.S.: Not sure if this will help, but so far I've got this:

$$f(n) = \displaystyle\sum_{i=0}^{n-1} 10^i(n-i)$$

But is there a formula that I could easily use in code (basic algebra, so summation notation)?

答:
The formula to get what you want is:

sum{i=1 to n} i*10^{n-i} , using wolfram alpha , give the result below:

f(n) = (10^(n+1) - 9*n - 10) / 81


问:
Solve $\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}$
This question is related to the previous question I posted today. So I am trying to solve $$\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}.$$
The system is equivalent to $$\begin{cases}u^2+3u-19=0\\u^2+u-19=0\end{cases}.$$ We can actually solve the two equations for $u$ and see if they have common solution(s). But I tried to substract the the two equations to get $$2u=0\\ \Rightarrow u=0.$$ Actually $u=0$ isn't root to any of the equations. Is this a contradiction? What do we actually get when we substract the equations? (or when we substract 2 equations in general) What does this mean?

答:
You've showed that the only possible solution to your system of equations is $u=0$. But $u=0$ is not in fact a solution. So there are no solutions.

Since this is a system of two equations in only one variable, you should not find this fact surprising! You'd have to get very lucky in order for the system to have any solutions.


问:
Compact Metric Problem
while defining Compact Metric Space in one of book of HL ROYDEN's REAL ANALYSIS statement says:

If T is collection of open subsets of a metric space X, then the collection F of complements of sets in T is collection of closed sets. Moreover, T is cover if and only if F has empty intersection. Thus, "By de Morgan identities a metric space X is compact if and only if every collection of closed sets with a non empty intersection has a finite sub-collection whose intersection also is nonempty"

firstly, this statement says T is cover if and only if intersection of F is empty, then it goes on saying that X is compact if and only if every collection of closed sets with nonempty intersection has finite sub-collection whose intersection also is nonempty.

Please explain it?.

答:
Let $\{F_i \mid i \in I\}$ be a collection of closed sets with empty intersection.

Applying de Morgan:

$$\emptyset = \bigcap_{i \in I} F_i \implies X= \left(\bigcap_{i \in I} F_i\right)^\complement = \bigcup_{i \in I} F_i^\complement$$

so indeed $\{F_i^\complement \mid i \in I\}$ forms an open (the complement of a closed set is open) cover of $X$ and so has a finite subcover (if $X$ is compact), say $F_{j}^\complement, j \in F$ where $F \subseteq I$ is finite, and doing the previous backwards again:

$$X=\bigcup_{j \in F} F_j^\complement \implies \emptyset = \left(\bigcup_{j \in F} F_j^\complement\right)^\complement = \bigcap_{j \in F} F_j$$

so the intersection of the finitely many $F_j, j \in F$ is empty.

This is found in many books in a logically equivalent contrapositive formulation: they call a family of sets such that any finite subfamily has non-empty intersection a family with the FIP (finite intersection property) and then the previous can also be stated as:

If $X$ is compact and if $\{F_i\mid i \in I\}$ is a family of closed sets with the FIP, then $\bigcap_i F_i \neq \emptyset$,

which looks more "positive", a guarantee that an infinite intersecion is non-empty provided all finite intersections are. But it really comes down to complements, de Morgan and the complementary open cover argument. For, suppose that the intersection of such a FIP closed family were empty, the argument from the beginning then tells us that a finite subfamily of them had an empty intersection too, but this directly contradicts the FIP assumption on the family, so cannot happen. So the intersection is non-empty.

Title: Compact Metric Problem


答: By subtracting the two equations you  never used the fact they originally equaled $20$. You merely used the fact the expressions on LHS are equal, which is indeed true for $u=0$. The constant number on RHS got cancelled while subtracting...

Title: Solve $\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}$

Body:
This question is related to the previous question I posted today. So I am trying to solve $$\begin{cases}u+(u+1)^2=20\\u^2+(u+1)=20\end{cases}.$$
The system is equivalent to $$\begin{cases}u^2+3u-19=0\\u^2+u-19=0\end{cases}.$$ We can actually solve the two equations for $u$ and see if they have common solution(s). But I tried to substract the the two equations to get $$2u=0\\ \Rightarrow u=0.$$ Actually $u=0$ isn't root to any of the equations. Is this a contradiction? What do we actually get when we substract the equations? (or when we substract 2 equations in general) What does this mean?


答:
Not a contradiction: not every system of equations has a solution.

Your calculation shows that if the system has a solution, then $u=0$. And since $u=0$ is not a solution, it follows that your system has no solution.

But I tried to subtract the two equations to get...

When you do subtraction, you are assuming that there is a solution $u$ that satisfies both equations, which is not necessarily true.

Consider a much simpler example.

You try to solve
$$
2u+1=3,\quad u+1=3
$$

You can solve the two equations for $u$ to see if they have a common solution.

You can also subtract the second equation from the first one to get
$$
u=0
$$
which is not a solution to both equations.


问:
Combinatorics Problem on proving that a particular sum is 0
I'm having some issues with proving that the following sum is $0$ for any value of $n \geq 2$:

$$
\sum_{j=1}^{n} \frac{1}{\prod_{i=1,i\neq j}^{n}(a_{j}-a_{i})}
$$

where the $a_i$ are non-zero and positive. Just wondering if anyone had any pointers, the first few cases are easy to verify, i.e $n=2,3$.

So for $n=2$ the sum reads:

$$
\frac{1}{a_2 -a_1}+\frac{1}{a_1-a_2} = 0
$$

and for $n=3$ it would be
$$
\frac{1}{(a_{3}-a_{1})(a_{2}-a_{1})}+\frac{1}{(a_1-a_2)(a_3-a_2)}+\frac{1}{(a_1-a_3)(a_2-a_3)} = 0,
$$
it's just the general case I'm struggling to find an argument for. Thanks!

答:
The degree of the polynomial

$$q(x)=\sum_{j=1}^n \prod_{\substack{i=1 \\ i \neq j}}^{n}\frac{x-a_i}{a_{j}-a_{i}}-1$$ is $n-1$ and the $a_j$ for $1 \le j \le n$ are $n$ roots. Therefore $q$ is the always vanishing polynomial and $q$ leading coefficient which is
$$\sum_{j=1}^n \frac{1}{\prod_{\substack{i=1 \\ i \neq j}}^{n}(a_{j}-a_{i})}$$ is equal to zero.

Note: all this is closely related to Lagrange polynomials.


问:
Sum of subspaces $W$ and $W'$ of $V$ equalling $V$

Let $V$ be a vector space and let $W$ and $W'$ be subspaces such that $\dim W \geqslant\frac{3}{4}\dim V, W' \geqslant \frac34\dim V$ and $\dim(W\cap W')\leqslant\frac12\dim V$. Does it hold that $W+W' = V$?

I know that if $U$ and $W$ are subspaces of $V$, then $\dim(U+W)+\dim(U\cap W) = \dim U +\dim W$.

I have a feeling that this could be applied here, but cannot seem to figure out how?

答: HINT: We always have $\dim(U_1+U_2)\le\dim V$ and $U_1+U_2$ is naturally a subspace of $V$. Compute a lower bound for $\dim(W+W')$ using the formula mentioned. What happens if $\dim U=\dim V$ for a subspace $U\le V$?


答:
Visually, here's what's happening:

$\hskip{3cm}$

The red curve is $y=x+(x+1)^2$, and the green is $y=x^2+(x+1)$.

The curves do intersect, namely at $x=0$. But the value of $y$ at that point of intersection is not equal to $20$.


答: $$\dim(W+W')=\dim W+\dim W'-\dim(W\cap W')\ge (\frac34+\frac34-\frac12)\dim V=\dim V\,.$$

Title: Sum of subspaces $W$ and $W'$ of $V$ equalling $V$

Body:
Let $V$ be a vector space and let $W$ and $W'$ be subspaces such that $\dim W \geqslant\frac{3}{4}\dim V, W' \geqslant \frac34\dim V$ and $\dim(W\cap W')\leqslant\frac12\dim V$. Does it hold that $W+W' = V$?

I know that if $U$ and $W$ are subspaces of $V$, then $\dim(U+W)+\dim(U\cap W) = \dim U +\dim W$.

I have a feeling that this could be applied here, but cannot seem to figure out how?


答:
From the equations you immediately draw $u=u^2$, but neither $0$ nor $1$ are solutions of the system.

Substracting the equations does not introduce new solutions, but if you drop the other equations, you do.


问:
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$
Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

答:
Here is a simple strategy. Find the limits of those radicals and subtract that limit from them.

Thus for numerator the term $\sqrt{19-x}\to 4$ and hence subtract and add $4$ to get $$\sqrt{16-x}-4+4$$ The other term in numerator is $2\sqrt[4]{13+x}$ which tends to $4$ and replace it with $$2\sqrt[4]{13+x}-4+4$$ and see that that those extra $4$'s added in both terms cancel out (well such limit problems are designed in such manner that numerator and denominator tend to $0$ and thus we have the desired cancelation here). Thus numerator equals $$\sqrt{19-x}-4-(2\sqrt[4]{13+x}-4)$$ and similarly denominator equals $$\sqrt[3]{11-x}-2-(x-3)$$ Now put $x-3=h$ so that $h\to 0$ and divide each term in numerator and denominator by $h$ to get $$\lim_{h\to 0}\dfrac{\dfrac{\sqrt{16-h}-4}{h}-2\cdot\dfrac{\sqrt[4]{16+h}-2}{h}}{\dfrac{\sqrt[3]{8-h}-2}{h}-1}$$ You can now evaluate the limit of three fractions easily and get the final answer as $$\frac{(-1/8)-2(1/32)} {(-1/12)-1} =\frac{9}{52}$$


问:
construct a lebesgue integral function
Construct a Lebesgue integrable function $f$ with the property that for any interval $I$ in $\mathbb{R}$ (real line) and any $M >0$,

The measure $|{x \in I: |f(x)| > M}| > 0$

How would I go about constructing such a thing?

答:
Let

$$g(x) = \frac{1}{x^2 + 1}.$$

Then $g$ is Lebesgue integrable on $\Bbb{R}$ and $\{ g(x) > \tfrac{1}{2} \} = (-1, 1)$. Now let $(r_n)$ be an enumeration of $\Bbb{Q}$ and define

$$f(x) = \sum_{n=1}^{\infty} 2^{n} g\left( 4^{n} (x - r_n) \right). $$

Since each summand is non-negative, $f \geq 0$ and

$$ \int_{\Bbb{R}} f \, dx = \sum_{n=1}^{\infty} \int_{\Bbb{R}} 2^{n} g\left( 4^{n} (x - r_n) \right) \, dx = \sum_{n=1}^{\infty} \frac{\pi}{2^{n}} < \infty. $$

Now pick any $M > 0$ and any interval $I$. We may assume $I$ is open by taking the interior instead. Then there exists $n$ such that $2^{-n} M < \frac{1}{2}$ and $r_n \in I$. For this choice of $r_n$,

\begin{align*}
\{ f > M \} \cap I
&\supset \{ 2^{n} g\left( 4^{n} (x - r_n) \right) > M \} \cap I \\
&\supset \{ g\left( 4^{n} (x - r_n) \right) > \tfrac{1}{2} \} \cap I \\
&= \left( r_n - 4^{-n}, r_n + 4^{-n} \right) \cap I
\end{align*}

Since the set in the last line is non-empty and open, it has positive measure. Thus the same is true for $\{ f > M \} \cap I$ and the proof is complete.

Title: construct a lebesgue integral function


答:
If we introduce the variable $v$ and the constraint $v=u+1$, then the system of equations may be written as

\begin{align}u+v^2&=20\\u^2+v&=20\\ v-u&=1 \end{align}

Plotting these equations (below) we obtain two parabolas and a line. The parabolas happen to intersect in four points, none of which happen to land on the line. As such there's no solution to all three equations.


问:
volume of hyeprcube with sum of coordinates larger than x
Consider an hypercube in $N$ dimensions with one vertex in the origin and another vertex in the point with all coordinates equal to 1, with edges alignes with axes.
Is there a general analytic expression, close form, not in terms of an integral or of an intinite series, for the function $f(x)$ that measures the volume of the part of the hypercube where the sum of the $N$ coordinates is larger than $x$, as a function of $x$ and $N$? In other words

\begin{equation}
f_N(x) = \int_0^1 dx_1 \int_0^1 dx_2 ... \int_0^1 dx_N \Theta\left(x_1+x_2+...+x_n-x\right)
\end{equation}

My first guess would be no, but I am reading on a paper whose result, if correct, would imply a yes to this question.

答:
The desired volume is naturally equivalent to the probability that the Irwin–Hall distribution on $N$ $U(0,1)$ random variables is larger than $x$, or equivalently smaller than $N-x$. The CDF is given in the link:
$$P(X\le x)=\frac1{N!}\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom Nk(x-k)^N$$
Replace $x$ with $N-x$ and you have your formula.


问:
Asymptotics for an expression
I have a rather silly question, I guess.
Suppose $(1+x)^{2y}\sim 1$ as $y\to\infty$.

Does this imply $x\sim 1/y$ as $y\to\infty$?

答:
Suppose that $x=x(y) \to 0$ as $y\to +\infty$. Then
$$
(1 + x)^{2y}  = \exp (2y\log (1 + x)) = \exp (2yx + \mathcal{O}(yx^2 )).
$$
So if you want
$$
\mathop {\lim }\limits_{y \to  + \infty } (1 + x)^{2y}  = 1,
$$
you need that
$$
\mathop {\lim }\limits_{y \to  + \infty } xy = \mathop {\lim }\limits_{y \to  + \infty } x(y)y = 0,
$$
i.e., that $x=o(1/y)$ as $y\to +\infty$. So $x$ must tend to $0$ faster than $1/y$ does.

Title: Asymptotics for an expression


问:
Pythagoras theorem proof for n-dimensions using the L2-norm

The Pythagoras theorem asserts for a set of $n$ orthogonal vectors $x_i \in \mathbb{R}^m$ for $i = 1,\dots, n$ that
$$ {\Bigg \lvert\Bigg \lvert \sum_{i=1}^{n} x_{i} \Bigg \lvert \Bigg \lvert_{2}^{2}} = \sum_{i=1}^{n} \lvert \lvert x_{i} \rvert \rvert_{2}^{2}.\tag{1}$$
Establish the result for general $n$ by induction.

For  $n=1$:

This case is trivial since in both sides of $(1)$ the sum vanishes to $1$ such that the equality is held.

For $(n-1) \implies n$ :

\begin{align}
\Bigg \lvert\Bigg \lvert \sum_{i=1}^{n-1} x_{i} \Bigg \lvert \Bigg \lvert_{2}^{2} &= \lvert \lvert x_{1} + x_{2} + \dots + x_{n-1} \rvert \rvert_{2}^{2} \\
&=\sum_{i=1}^{n-1} \lvert \lvert  x_{i} \rvert \rvert_{2}^{2}  + \underbrace{\sum_{i \neq j} x_{i} x_{j}}_{=0 \ \text{since }\ x_{i} \perp x_{j}} \\
&= \sum_{i=1}^{n-1} \lvert \lvert  x_{i} \rvert \rvert_{2}^{2}
\end{align}

I'm not sure if this proof holds and/or if some steps are missing. Could anyone perhaps confirm ?

答:
Your proof is good, but it does not rely on induction. Here is the Inductive Step.

Assume the theorem holds for $n-1$ vectors, and let $x_1, \ldots, x_n \in \mathbb{R}^m$ be orthogonal. Then note that $x_1 + \ldots + x_{n-1} = y \in \mathbb{R}^m$ is orthogonal to $x_n$. Thus,
$$
\begin{split}
\|x_1+\ldots x_{n-1}+x_n\|_2^2
 &= \|y+x_n\|_2^2 \quad \text{apply theorem for 2 vectors} \\
 &= \|y\|_2^2 + \|x_n\|^2 \\
 &= \|x_1+\ldots x_{n-1}\|_2^2 + \|x_n\|^2 \quad \text{apply theorem for $n-1$ vectors}\\
 &= \|x_1\|_2^2 + \ldots + \|x_{n-1}\|_2^2 + \|x_n\|^2
\end{split}
$$


问:
Can we always "locate" a transitive model inside some larger transitive model?
We work in $\mathsf{ZFC}$ + "Every set is an element of a transitive model of $\mathsf{ZFC}$."

Suppose $M$ is a transitive set model of $\mathsf{ZFC}$. Say that $M$ is locatable iff there is some transitive set $N$ such that:

$N\models\mathsf{ZFC}$,

$M\in N$, and

$M$ is parameter-freely-definable in $N$.

The requirement that $N\models\mathsf{ZF}$ prevents coding $M$ into $N$ in a "silly" way. To see how this can be an issue, let $\alpha=\min\{\gamma: L_\gamma\models\mathsf{ZFC}\}$. If $N$ is a transitive model of $\mathsf{ZFC}$ with $L_\alpha\in N$, then $N$ sees that $L_\alpha$ is countable and so contains lots of (say) reals which are Cohen-generic over $L_\alpha$. Suppose $r,s$ are such reals; how is $N$ to distinguish between $L_\alpha[r]$ and $L_\alpha[s]$ in a parameter-free way? This seems to be an obstacle in general to the locatability of $L_\alpha[c]$ for $c$ Cohen-generic over $L_\alpha$. However, such an $L_\alpha[c]$ is nonetheless locatable: force over $L_\beta$, where $\beta=\min\{\gamma: \gamma>\alpha, L_\gamma\models\mathsf{ZFC}\}$, to code $c$ into the continuum pattern appropriately.

Indeed, such coding tricks give:

Every countable $M$ is locatable.

Note that the locating model is very simple - it's an element of $L(M)$. For uncountable $M$s this seems like a nontrivial further condition, and in particular none of the absoluteness theorems I'm familiar with seem to directly apply.

Similarly, we have:

If $L_\theta\models\mathsf{ZFC}$ then $L_\theta$ is locatable, even if $\theta$ is uncountable.

(For example, every ordinal $\theta$ is parameter-freely-definable in the $(\vert\theta\vert^++\theta+1)$th level of $L$ satisfying $\mathsf{ZFC}$ since $(i)$ that level of $L$ can "count" the $\mathsf{ZFC}$-satisfying levels below it, and $(ii)$ we can recover $\theta$ from $\vert\theta\vert^++\theta$ as there is a unique way to write the latter in the form $\kappa+\gamma$ for $\kappa$ a cardinal and $\gamma<\kappa$.)

However, the general situation is not clear to me:

Question: is every transitive set model of $\mathsf{ZFC}$ locatable?

Of course it is quite plausible that this will depend on additional set-theoretic hypotheses; for example, I think an affirmative answer is more likely assuming $\mathsf{V=L}$ than assuming the existence of large cardinals. But I don't immediately see how to show any of this.

答:
Great question! It has a positive answer in general:

Let $M$ be a transitive set-sized model of $\mathrm{ZFC}$. We may now find some regular cardinal $\kappa$ and some $X\subseteq\kappa$ so that $M\in L_\kappa[X]$. From now on we will work in $L[X]$. Let $\beta$ be the ordinal so that $M$ is the $\beta$-th set according to $<_X=:$ the canonical wellorder on $L[X]$. If both $X$ and $\beta$ were parameter-free definable then we could define $M$ without parameters. Unfortunately, neither of this is necessarily the case so we will try to fix that. Note that $\mathrm{GCH}$ holds in $L[X]$ at and above $\kappa$ and by our assumption on $V$, there are arbitrarily large $\lambda$ so that $L_\lambda[X]\models\mathrm{ZFC}$.

Claim: There is some $\kappa<\lambda<\kappa^+$ with

$X\in L_\lambda[X]\models\mathrm{ZFC}$

$L_\lambda[X]^{<\kappa}\subseteq L_\lambda[X]$

Proof: Let $\theta>\kappa^+$ so that $L_\theta[X]\models\mathrm{ZFC}$. Construct a continuous increasing sequence $\langle X_\alpha\mid \alpha\leq\kappa\rangle$ of elementary submodels of $L_\theta[X]$ of size $\kappa$. Make sure thate $\kappa\cup\{X\}\subseteq X_0$ and that $X_\alpha^{<\kappa}\subseteq X_{\alpha+1}$ for any $\alpha<\kappa$. Then the collapse $L_\lambda[X]$ of $X_\kappa$ is as desired.$\square$

We aim to find a generic extension of $L_\lambda[X]$ (inside $L[X]$) so that $M$ is definable there w/o parameters.

In $L_\lambda[X]$, let $\mathbb P$ be the/an  ${<}\kappa$-closed Easton-style forcing that codes $\{X, \beta\}$ (observe that $\beta<\kappa<\lambda$) into the $\mathrm{GCH}$-pattern on the successor cardinals in the interval $(\aleph_\kappa, \aleph_{\kappa+\kappa})^{L_\lambda[X]}$ and also makes $\mathrm{GCH}$ fail at $\aleph_{\kappa+\kappa+1}^{L_\lambda[X]}$. Now the following holds:

$\mathbb P$ is ${<}\kappa$-closed in $L_\lambda[X]$

$L_\lambda[X]$ has size $\kappa$

$L_\lambda[X]^{<\kappa}\subseteq L_\lambda[X]$

Thus we can construct a filter $G\subseteq\mathbb P$ (in $L[X]$!) that is generic over $L_\lambda[X]$.

Finally, $N=L_\lambda[X][G]\in L[X]\subseteq V$ is a transitive model of $\mathrm{ZFC}$ that locates $M$: First of all $\kappa$ is definable w/o parameters in $N$ since
$$N\models \aleph_{\kappa+\kappa+1}\text{ is the largest cardinal at which }\mathrm{GCH}\text{ fails}$$
Using $\kappa$, $N$ can decode $X$ and  $\beta$ from the $\mathrm{GCH}$-pattern on the successors in the interval $(\aleph_\kappa, \aleph_{\kappa+\kappa})^N$ and so can define $M$ as the $\beta$-th set according to $<_X$.


问:
Why are intersections of rings each integrally closed in the fraction field of their intersection integrally closed
Specifically, Lorenzini writes (in "An invitation to arithmetic geometry):

Let $\{ A_i \}_{i \in I}$ any set of subrings of $K$ the field of fractions of a commutative domain $A$, such that all the $A_i$ are integrally closed in $K$, and such that $K$ is also the field of fractions of $\bigcap\limits_{i \in I} A_i$. Then the ring $\bigcap\limits_{i \in I} A_i$ is also integrally closed in $K$.

It seems like a very basic fact, but I really want to get it down since it is so relevant to the rest of the book.

I wrote down:

$\forall i \in I \forall a \in K \exists b_1, ..., b_{n-1} \in A_i [a^n + a^{n-1} b_{n-1} + ... + ab_1 + a = 0] $ Since $A_i$ integrally closed in $K$ $\forall i \in I$, $a \in A_i$. Since $K = Frac(\bigcap\limits_i A_i)$, $a \in Frac(\bigcap\limits_i A_i)$, but since it's in $A_i$ it's not really a fraction, so it's properly in $\bigcap\limits_i A_i$. So $\bigcap\limits_i A_i$ is integrally closed in $K$ and we are done.

Is this proof ok? It doesn't seem like it. What is wrong?

答: If $a\in K$ is integral over $\cap_{i\in I}A_i$, then surely $a$ is integral over each $A_i$, so that $a\in A_i$ for all $i$. It follows that $a\in\cap_{i\in I}A_i$, so that $\cap_{i\in I}A_i$ is integrally closed. (I think this is roughly what you said, more simply stated.)


问:
Symmetric distributions in $\mathbb{R}$.
Let $X$ be a real random variable central symmetric with respect to $c\in\mathbb{R}$, that is, $X - c$ is equal in distribution to $c - X$, that is,

$$
\mathbb{P}(X - c\leq t) = \mathbb{P}(c - X\leq t),
$$
for all $t\in\mathbb{R}$.

It is clear that $c$ is unique and $c$ is a median of $X$.

In https://en.wikipedia.org/wiki/Symmetric_probability_distribution they say that $c$ also coincides with $\text{E}[X]$ (if it exists). Why?

I've tried to use the definition of the expectation using cummulative distribution function, but i don't see clear the proof.

Thanks in advance.

答:
If $X-c$ and $c-X$ have the same distribution, then
$$\mathbb E[X]-c=\mathbb E[X-c]=\mathbb E[c-X]=c-\mathbb E[X]\implies\mathbb E[X]=c.$$


问:
Show that the coefficients $c_1$, …, $c_5$ are uniquely determined. What is $\left(M^{-1}\right)_{ij}$?
Consider $5$ functions $f_1, …, f_5$ defined on the interval $[0,5]$, and $\textbf{assume they have the following property:}$ given any numbers $b_1, …, b_5$ there always exist coefficients $c_1, ..., c_5$ such that

$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle \sum_{j=1}^5 c_jf_j(i) = b_i$

Then, $\textbf{show}$ that under these circumstances that once the number $b_1, …, b_5$ are fixed the coefficients $c_1, ..., c_5$ that solve the above problem are uniquely determined, this will in particular mean there is a linear function that determines the $c_j's$ from the $b_i's$, which means there is a matrix $M_{ij}$ such that

$\,\,\,\,\,\,\,\,\,\,$ $\displaystyle c_j = \sum_{i=1}^5 M_{ij}b_i$

Without knowing anything else about $M$, say what is $\left(M^{-1}\right)_{ij}$.

I am not sure if I am setting this up correctly or taking the proper approach. First, I wrote out the first equation as a system of equations and got:

$c_1f_1(1)+c_2f_2(1)+c_3f_3(1)+c_4f_4(1)+c_5f_5(1)=b_1$

$c_1f_1(2)+c_2f_2(2)+c_3f_3(2)+c_4f_4(2)+c_5f_5(2)=b_2$

$\vdots$

$c_1f_1(5)+c_2f_2(5)+c_3f_3(5)+c_4f_4(5)+c_5f_5(5)=b_5$.

So, I set up a matrix out of these systems:

$\begin{pmatrix}
c_1f_1(1) & c_2f_2(1) & ... & c_5f_5(1)\\
\vdots \\
c_1f_1(5) & ... & ... & c_5f_5(5)
\end{pmatrix} = \begin{pmatrix}
b_1 \\
\vdots \\
b_5
\end{pmatrix}$ $\implies$
$\begin{pmatrix}
f_1(1) & f_2(1) & ... & f_5(1)\\
\vdots \\
f_1(5) & ... & ... & f_5(5)
\end{pmatrix}\begin{pmatrix}
c_1 \\
\vdots \\
c_5
\end{pmatrix} = \begin{pmatrix}
b_1 \\
\vdots \\
b_5
\end{pmatrix}$

Let $M = \begin{pmatrix}
f_1(1) & f_2(1) & ... & f_5(1)\\
\vdots \\
f_1(5) & ... & ... & f_5(5)
\end{pmatrix}$, $\overline{\textbf{c}} = \begin{pmatrix}
c_1 \\
\vdots \\
c_5
\end{pmatrix}$, and $\overline{\textbf{b}} = \begin{pmatrix}
b_1 \\
\vdots \\
b_5
\end{pmatrix}$.

So, we have $M \,\overline{\textbf{c}} = \overline{\textbf{b}}$.

**Here I wasn't too sure what to do next. I wanted to show that $M$ is invertible, but the only way I know is that the determinant is not zero. I didn't know how to do that with this $5 \times 5$ matrix. I wanted to show this becuase I figured it would help me show that $\overline{\textbf{c}} = M^{-1} \overline{\textbf{b}}$ which would mean that $\left(M^{-1}\right)_{ij} = M^T$. Or am I mistaken about the inverse being the transpose? As far as uniqueness goes for $\overline{\textbf{c}}$, I thought if we assume there is a $\overline{\textbf{c}}^{'}$, then we could obtain $\overline{\textbf{b}}$ which int turn means $\overline{\textbf{c}}^{'} = \overline{\textbf{c}}$.

答:
Note that the matrix that you have called $M$ does the reverse of what is being requested: given the vector $\mathbf c$, you have found a matrix $A$ for which $A\mathbf c = \mathbf b$. What is required is a matrix "that determines the $c_j$'s from the $b_i$'s". In other words, $M$ should satisfy $\mathbf c = M \mathbf b$.

However, your work so far makes this easy. Note that
$$
A \mathbf c = \mathbf b \implies \mathbf c = A^{-1} \mathbf b.
$$
In other words, the desired matrix $M$ is the inverse of the matrix you found (which I have called $A$). Note that this requires the inverse of $M$ to exist, but we can conclude that this must hold because the $c_j$'s are uniquely determined from the $b_i$'s.

With that, the last part of the question is easy: the inverse of $M$ is simply $(A^{-1})^{-1} = A$. In other words, we have
$$
(M^{-1})_{ij} = f_j(i).
$$


问:
Letter sequences in a series
Let's suppose we have a set of letters $\{a,b,c,x,y,z\}$. How many series (6-letter words, for example $abbxaz$) can we create provided that we have at least one sequence $xy$?

At first I tried to split it into three cases, with only one $\textbf{xy}$ (e.g. $ab\textbf{xy}az$), two (e.g. $\textbf{xy}ya\textbf{xy}$) and three (easiest part, as there is only one combination $\textbf{xy}\textbf{xy}\textbf{xy}$), but it doesn't seem to be the right idea, as I have to consider also all of the subsets and dependencies between them (e.g. when I have to have $\textbf{xy}$ once then $\_ \ \_\ \textbf{xy} \_ \ \_$ would differ from $\_ \ \textbf{xy} \_ \ \_ \ \_$ and I also need to exclude all the intersections with all the cases when $\textbf{xy}$ occurs twice or thrice). I could also try permuting and using symmetry wherever it's possible, but still, it probably doesn't vary much from what I wrote above. Is there an elegent way to do that? I somehow can't believe that I have to write all the cases possible.

答:
BIG HINT

If you let $P_i$ be the  words  with 'xy' in the $i$th and $i+1$th places, then $|P_i|=6^4$ for $1\le i \le5$.

Then the inclusion-exclusion principle takes care of all your concerns about double counting. You only have to calculate

$$\sum |P_i|-\sum|P_i\bigcap P_j|+|P_1\bigcap P_3\bigcap P_5|.$$

Title: Letter sequences in a series


问:
Rings with noncommutative addition
I was wondering if "rings" with noncommutative addition are studied at all? Of course, if a ring $R$ has a $1$, then for all $a, b\in R$, $a+a+b+b=(1+1)a+(1+1)b=(1+1)(a+b)=(a+b)+(a+b)=a+b+a+b$, from which it follows from cancellation that $a+b=b+a$. Thus, rings with $1$ automatically must have commutative addition.

So, are there interesting, necessarily non-unital, "rings" with noncommutative addition? Of course, you can take any non-abelian group and make such a "ring" by having all multiplications give the additive identity. Are there more interesting examples?

答: Vectors on a sphere? Their addition is essentially multiplication of rotations, which is non-commutative.</p>

Title: Rings with noncommutative addition


问:
Why are all knots trivial in 4D?
A classical knot is defined to be an embedding $S^1 \to \mathbb{R}^3$ where $S^1$ is a 1-sphere or circle.

Embeddings $S^1 \to \mathbb{R}^4$ are usually not considered knots because they are trivial knots, i.e., they can be continuously deformed to $S^1$.

How can I show that such embeddings are indeed trivial knots?

答:
The following theorem is implicit in the paper by Blankinship, where it is proven that for each $n\ge 3$ and $q\in [1, n]$, there exists a subset $A\subset {\mathbb R}^n$ homeomorphic to the (closed) $q$-dimensional cube and whose complement is not simply-connected. Note that this result also shows that not all 2-dimensional topological disks in ${\mathbb R}^4$ are isotopic to each other.

William A. Blankinship, Generalization of a construction of Antoine, Ann. Math. (2) 53, 276-297 (1951). ZBL0042.17601.

Theorem. For every $n\ge 4$ there exists a nontrivial (wild) 1-dimensional knot $K\subset {\mathbb R}^n$.

Proof. Blankinship proves that for each $n\ge 4$, there exists a subset $C\subset R^n$ homeomorphic to the classical cantor set such that $R^n\setminus C$ is not simply-connected (the complement is necessarily connected by the Alexander duality). Now, take such $C$. I proved in my answer here that for every totally disconnected compact $C$ in ${\mathbb R}^n$ (e.g. one as above), there exists a subset $K\subset {\mathbb R}^n$ homeomorphic to $S^1$ and containing $C$. Moreover, in the construction $K$ can be chosen such that each component of $K\setminus C$ is a smooth (open) arc in ${\mathbb R}^n$. (There are at most countably many of these components.)

Lemma. $\pi_1({\mathbb R}^n \setminus K)\ne 1$.

Proof. Since $\pi_1({\mathbb R}^n \setminus C)\ne 1$, there is a smooth loop $c: S^1\to {\mathbb R}^n \setminus C$ which is not null-homotopic in ${\mathbb R}^n \setminus C$. A priori, the image of $c$ intersects $A=K \setminus C$. However, by perturbing $c$ to be transversal to $A$, since $n\ge 2$, we can find a loop $c_1$ homotopic to $c$ in $R^n \setminus K$ and whose image is disjoint from $A$. Hence, $c_1$ is a homotopically nontrivial loop in ${\mathbb R}^n
\setminus K$ (since it is already homotopically nontrivial in the complement to $C$). qed

On the other hand, if $T\subset {\mathbb R}^n$, $n\ge 4$, is a trivial knot, $\pi_1({\mathbb R}^n\setminus T)=1$. Thus, the knot $K$ constructed above nontrivial. qed

What Andrew's answer proves that every tame 1-dimensional knot in ${\mathbb R}^4$ (and, more generally, ${\mathbb R}^n, n\ge 4$) is trivial, i.e. is ambient isotopic to a round circle. The knot constructed in my proof is wild.

In particular, step 1 in Andrew's answer breaks down for general knots.

On the positive side:

Theorem. If $K\subset {\mathbb R}^3$ is a topological knot, then $K$ is unknotted in ${\mathbb R}^4$.

Proof. The key is to prove that $K$ is tame (or locally flat) as a knot in ${\mathbb R}^4$, i.e. for every $x\in K$ there is a neighborhood $U$ of $x$ in ${\mathbb R}^4$ and a homeomorphism $U\to V\subset {\mathbb R}^4$ sending $U\cap K$ to a subset of a straight line $L\subset {\mathbb R}^4$.

Once it is done, it follows from the main theorem of

H. Gluck, Unknotting $S^ 1$ in $S^ 4$, Bull. Am. Math. Soc. 69, 91-94 (1963). ZBL0108.36503.

that $K$ is unknotted in ${\mathbb R}^4$.

Let $f: K\to [0,1]$ with exactly two point of minimum and maximum, $f(a)=0, f(b)=1$. Let $\alpha, \beta$ denote the closures of the components of $K\setminus \{a, b\}$. The function $f$ defines an embedding $K\to {\mathbb R}^4$,
$$
\phi: x\mapsto (x, f(x)), x\in K,
$$
whose image is a knot $K'$ in ${\mathbb R}^4$. Let $a':= \phi(a)=a, b':= \phi(b)$.

Step 1.  $K'$ is ambient-isotopic to $K$ in ${\mathbb R}^4$.

First, extend $f$ to a continuous function $f: {\mathbb R}^3\to [0,1]$.
For each $t\in [0,1]$  define the "shear"
$$
h_t(x, s)= (x, s+tf(x)), x\in {\mathbb R}^3, s\in {\mathbb R}
$$
which is clearly a homeomorphism. This defines an ambient isotopy of the knot $K$ to the knot $K'=f_1(K)$ in ${\mathbb R}^4$.

Step 2. I will now prove a slightly easier result than unknottedness of $K'$ in ${\mathbb R}^4$, I will show that the arc $\alpha'=\phi(\alpha)$ (and, similarly, $\beta'=\phi(\beta))$ is tame, equivalently, unknotted in $R^4$. Note that $\alpha'$ intersects each horizontal slice ${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$ in exactly one point, namely,
$(f^{-1}(s), s)$. The straight-line segment $I':=a'b'$ intersects each slice ${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$ also at exactly one point, $p_s$.
Thus, the shear
$$
h^\alpha: (x, s) \mapsto (x + (p_s- f^{-1}(s)), s), s\in [0,1]
$$
is a homeomorphism ${\mathbb R}^3\times [0,1]\to {\mathbb R}^3\times [0,1]$ sending $\alpha'$ to $I'$; this homeomorphism is the identity on the boundary of the slab ${\mathbb R}^3\times [0,1]$. Hence, $h$ extends to a homeomorphism ${\mathbb R}^4\to{\mathbb R}^4$ (by the identity outside of the slab). Linear interpolation again yields an isotopy $h^\alpha_t, t\in [0,1]$, of $h$ to the identity map. Hence, the arc $\alpha'$ is unknotted in ${\mathbb R}^4$. Note that $\{h^\alpha_t(\alpha'): t\in [0,1]\}$ lies inside the convex hull of the arc $\alpha'$ in ${\mathbb R}^4$.

The same construction applies to the arc $\beta'$ resulting in an isotopy $h^\beta$. The trouble is that we cannot combine these "straightening" maps of $\alpha', \beta'$. We obtain, however, that the knot $K'$ is tame in ${\mathbb R}^4$, except possibly at the points $a', b'$. More precisely, $K'$ is the concatenation of two tame arcs, $\alpha', \beta'$. Since $K$ is ambient-isotopic to $K'$ in ${\mathbb R}^4$, it follows that $K$ was tame in ${\mathbb R}^4$ as well, except possibly at the points $a, b$. But the points $a, b$ in $K$ were chosen arbitrarily.  Hence, $K\subset {\mathbb R}^4$ locally flat at every point, i.e. is tame.

Title: Why are all knots trivial in 4D?


问:
Given an ode $x'=e^{-x}\sin(x)$, initial condition is $x(0)=-2\pi$. Why the x(t) is not a straight line in the graph x vs t?
Question:

Given an ode $x'=e^{-x}\sin(x)$, initial condition is $x(0)=-2\pi$. Why the x(t) is not a straight line in the graph x vs t?

Thoughts:

I found out the fixed points, $x=-2\pi$ is one of the unstable fixed points. Which means the velocity is 0 at that point.

Which means if its initially there, it should never move. Then it should be a straight line in the graph.
Am I correct?

Situation:

But when I draw it in python its not the case::

```<code>import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

# function that returns dx/dt
def dxdt(x, t):
    return np.exp(-x)*np.sin(x)

# time points
t = np.linspace(0,20, 100)

plt.rcParams['figure.figsize'] = [15, 6]
fig, ax = plt.subplots(1)

# initial condition
x0 = -2*np.pi
# solve ODE
x = odeint(dxdt,x0,t)

# plot results
ax.plot(t, x)

ax.set(xlabel='time', ylabel='x(t)')
import matplotlib.ticker as tck
ax.yaxis.set_major_formatter(tck.FuncFormatter(
   lambda val,pos: '{:.0g}<span class="math-container">$\pi$</span>'.format(val/np.pi) if val !=0 else '0'
))
ax.yaxis.set_major_locator(tck.MultipleLocator(base=np.pi))

</code>```

For more initial conditions, x(t) is actually not in -2pi, -4pi, ... when time is positive :

Why is that?

答:
The issue is with numerical errors:

```<code>np.exp(2*np.pi)*np.sin(-2*np.pi)=1.31157628381704e-13
</code>```

This is not zero. You can do something like:

```<code># function that returns dx/dt
def dxdt(x, t):
    r = np.exp(-x)*np.sin(x)
    if np.abs(r)<1e-10:
       r=0
    return r
</code>```

Body:
The issue is with numerical errors:

np.exp(2*np.pi)*np.sin(-2*np.pi)=1.31157628381704e-13

This is not zero. You can do something like:

# function that returns dx/dt
def dxdt(x, t):
    r = np.exp(-x)*np.sin(x)
    if np.abs(r)<1e-10:
       r=0
    return r


问:
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.

My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.

There are $2$ cases to consider:

Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:

\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) +  abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}

Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?

答:
For $a,b,c$ sides of a triangle, let $$\begin{matrix}a=y+z\\b=z+x\\c=x+y\end{matrix}$$

Then the inequality becomes $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ This is true by Cauchy's inequality: $$x+y+z\le\sqrt{\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}}\sqrt{y+z+x}$$


问:
Scale transformation of gamma variable
we know that $(n −1)S^2/\sigma^2 ∼ \chi^2(n−1) = gamma((n − 1)/2, 2)$.

Now I want to know the distribution of $S^2$. How to do that?

答:
If $\frac{n −1}{\sigma^2}S^2 \sim \mathrm{gamma}\left(\frac{n − 1}2, 2\right)$ using a shape-scale parametrisation

then $S^2 \sim \mathrm{gamma}\left(\frac{n − 1}2, \frac{2\sigma^2}{n-1}\right)$ by rescaling

Title: Scale transformation of gamma variable


问:
Is $f^{-1}$ a manifold?
Given  $M = f^{-1}(0)$ for $f(x,y) = x^2-y^2$ in $\mathbb{R}^2$.

It is to be shown whether this is a manifold or not.

Pardon my bad understanding of manifolds, but what's the set here?

Is it $M = \{(x,y)\in\mathbb{R}^2|f^{-1}(0) = 0\}$ where $f = x^2-y^2$

Well this runs into various problems: first of $f^{-1}(0)$ is a single point and not a space, secondly I don't happen to find $f^{-1}$ because $f$ is a vector valued function with just one row.

So what am I messing up this time?

答:
If $M=f^{-1}(0)$, then $f(M)=0$. Therefore

\begin{equation}
M=\{(x,y)\in\mathbb{R}^2| x^2-y^2=0\}.
\end{equation}

You need to show whether this set is a manifold or not.

Title: Is $f^{-1}$ a manifold?


问:
Is there a connection between the Whitney embedding theorem and Cayley's theorem?
Background: I've been working through Guillemin and Pollack's "Differential Topology." They take the approach of defining smooth manifolds as "concrete" submanifolds of some ambient $\mathbb R^N$, as opposed to "abstract" topological spaces with charts and atlases. In my opinion, this approach leads to better intuition about the subject. Moreover, even in the standard "abstract" approach we can establish the Whitney embedding theorem, meaning that we aren't losing generality by working "concretely."

With this viewpoint of "abstract" vs "concrete" definitions, I realized that the Whitney embedding theorem with its ambient $\mathbb R^N$, and Cayley's theorem with its ambient symmetric group $S_N$, are of the same flavor. I know that there are many other representation theorems in math like these two, and that many can be viewed as generalization of the Yoneda embedding.

However, I can't find any sources that include the geometric representation theorems like Whitney and Nash in this Yoneda or categorical framework. So, I'm asking:

Question: Can the Yoneda lemma or other category theory be used to unify the Whitney embedding theorem with Cayley's theorem and/or other representation theorems in math?

My background consists of the category theory that showed up in the first year grad school, e.g. in an algebraic topology sequence and commutative algebra. However, any comments on this circle of ideas is greatly appreciated. Thank you in advanced!

答:
It's really a different kind of embedding. The Whitney embedding theorem embeds the manifold within $\mathbb{R}^N$, whereas the Yoneda lemma 'embeds' the functor of points into the functor category. The reason why Cayley's lemma fits within the Yoneda lemma is because groups happen to be reinterpretable as one-object categories, and there's no such thing for smooth manifolds (well, as far as I know).

That is not to say that you can't apply the Yoneda lemma to the category of (smooth) manifolds. You certainly can, and in this way a smooth manifold $M$ can be captured completely by the functor $h_M \colon \operatorname{Man}^{\mathrm{op}} \to \operatorname{Set}$ sending a manifold $N$ to $\operatorname{Hom}(N,M)$. I'm sure people have tried to consider manifolds from this functorial point of view ('diffeological spaces', 'smooth sets', are some buzzwords that come up for me). But it's a different thing.

Side remark: A vaguely related question, to which I don't know the answer, is whether you can detect smooth embeddings in the category of smooth manifolds entirely in categorical terms.


问:
Exists Lebesgue integral
Lebesgue integral $\displaystyle\int^0_1 \log x \log(1+x) dx$ exists?

I know that
$2(x-1)\leq \log x\leq x-1$ for $1/2<x<1$

$x\leq |\log(x-1)|\leq 2x$, for $0<x<1/2$

Then $|\log\log(x+1)|\leq 2x|\log x|+4(1-x)|\log(1-x)|$

答:
It can actually be integrated explicitly as follows
\begin{align} \int^0_1 \log x \log(1+x) dx &\overset{IBP}=
\int_0^1 \left( \ln(1+x) + \ln x-\frac{\ln x}{1+x} \right)dx\\ &=2 \ln2 -2+ \frac{\pi^2}{12}
\end{align}
where the result $\int_0^1 \frac{\ln x}{1+x} = -\frac{\pi^2}{12}$ is used. Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$

Title: Exists Lebesgue integral


答:
The only problem with the integrand happens at near $0$. Since $\lim_{x\rightarrow0}\sqrt{x}\log(x)=0$ We get that
$$|\log(x)\log(1+x)|\leq \frac{M}{x^{1/2}}$$
for some $M>0$ and all $0<x\leq 1$.

From there you see that your integral indeed exists.

As it has been pointed out by @Quanto, you can then use standard tricks to evaluate the inetegral.

Title: Exists Lebesgue integral

Body:
Lebesgue integral $\displaystyle\int^0_1 \log x \log(1+x) dx$ exists?

I know that
$2(x-1)\leq \log x\leq x-1$ for $1/2<x<1$

$x\leq |\log(x-1)|\leq 2x$, for $0<x<1/2$

Then $|\log\log(x+1)|\leq 2x|\log x|+4(1-x)|\log(1-x)|$


问:
Are bits defined in a Boolean ring?
Premise: I'm not a mathematician, please be patient.

Anyway, if I consider the Galois field $GF(2)$ and two operations $+$ (inclusive or, also denoted with $\lor$) and $\neg$ (negation), where, given $x,y \in GF(2)$:
$$
x + y := (x \cdot y) \oplus (x \oplus y),
$$
$$
\neg x:= x \oplus 1,
$$
I've defined a boolean ring $(GF(2),+,\neg)$.

Note: I know that in information theory, code theory and cryptography a bit is usually defined in $GF(2)$, and in these cases there's no need to use Boolean rings, but for example in some context in computer science or digital electronics you use boolean algebra, since you also need inclusive or and "negated" variables or signals.

Does this definition work in both cases? Is it safe to say that mathematically bits are defined in a Boolean ring?

Another question is: if I need a string of $n$-bit
$
\boldsymbol{b} = [b_1, b_2, \dots, b_n],
$
is it safe to say that $\boldsymbol{b} \in (GF(2^n),+,\neg)$?

答:
$GF(2)$ forms a boolean ring.
It follows that $GF(2)^n$ forms a boolean ring, because the operations are defined row by row and the rows are independently considered (as in linear algebra).

Title: Are bits defined in a Boolean ring?


答:
Method 1:As Chrystomath used Ravi's substituition we are left to prove $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ Now

$x\ge y\ge z\implies x^2\ge y^2\ge z^2,\frac{1}{x}\le \frac{1}{y} \le \frac{1}{z}$ $$x^2\left(\frac{1}{x}\right)+y^2\left(\frac{1}{y}\right)+z^2\left(\frac{1}{z}\right)\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$ $$\iff x+y+z\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$

$x\ge z\ge y$ is analogous can you finish it off...

Method 2: As I have already given an answer by rearrangement here is a far more reliable way

WLOG $c=max(a,b,c)$

so let  $a=x,b=x+u,c=x+u+w$ where $x,u,w \ge 0$ also by triangle inequality $w< x$ we have to prove $$-ux^2(x+u)-w{(x+u)}^2(x+u+w)+{(x+u+w)}^2x(u+w)\ge 0$$ $$\iff u^3(x-w)+u^2(x^2-w^2)+w^3x+w^2x^2+u^2x^2+uwx^2\ge 0$$ which is obvious!!

If $a=x+u,b=x,c=x+u+w$ then again the inequality is after expanding $$u^3x+u^2wx+u^2x^2+2uw^2x+w^3z+w^2x^2\ge 0$$ which is true

Title: Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$

Body:
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.

My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.

There are $2$ cases to consider:

Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:

\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) +  abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}

Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?


问:
discrete system of equations on a torus
Im interested in solving the system of equations $C_{i,j} = C_{i,j-1} + C_{i,j+1} + C_{i-1,j} + C_{i+1,j}$ for $0 \leq i < N$ and $0 \leq j < N$, where $C_{i,j} \in \mathbb{C}$. One could think of this a regular grid where each entry of a cell corresponds to the sum of the neighboring cells. Additionally i want to have periodic boundary conditions in both indices, i.e. $C_{i+N, j} = C_{i, j}$ and $C_{i, j+N} = C_{i, j}$, hence we can think of the grid mentioned before as a torus. Problem: find all possibile non-trivial solutions of this system of equations. My first attempt was to substract $- 4C_{i,j}$ on both sides such that we can think of the problem as $- 3 C_{i,j} = C_{i,j-1} - 2 C_{i,j} + C_{i,j+1} + C_{i-1,j} - 2C_{i,j} + C_{i+1,j}$ which is essentially a discrete version of theHelmholtz equation on a unit grid, i.e. $\frac{\partial^{2} C_{i,j}}{\partial i^2} + \frac{\partial^{2} C_{i,j}}{\partial j^2} + 3 C_{i,j}$. Due to lack of knowledge i did not find any other good ansatz. Has this problem already been solved?

Thanks in advance!

答:
One could use the fact that the associated coefficient matrix is block-circulant.

Let $\otimes$ denote the Kronecker product. Let $\mathbf e_0, \mathbf e_1,\dots,\mathbf e_{N-1}$ denote the standard basis of $\Bbb R^{N}$. Let $P$ denote the size-$N$ permutation matrix
$$
P = \pmatrix{0&\cdots & 0 & 1\\1&0&\cdots&0\\\vdots & \ddots & \ddots & \vdots\\
0 & \cdots & 1 & 0}.
$$
Let $\mathbf c$ denote the column-vector
$$
\mathbf c = (C_{0,0},C_{0,1},\dots,C_{0,N},C_{1,0},C_{1,1},\dots) = \sum_{i,j = 0}^{N -1} C_{ij}\, \mathbf e_i \otimes \mathbf e_j.
$$
Your system of equations can be written in the form $M \mathbf c = 0$, where
$$
M = I_{N} \otimes (P + P^T) + (P + P^T) \otimes I_{N} - I_{N^2}.
$$
We see that this matrix can be diagonalized. In particular, we have
$$
(F \otimes F) M (F \otimes F)^*  = I_{N} \otimes D + D \otimes I_{N} - I_{N^2},
$$
where $D$ is the diagonal matrix
$$
D = \operatorname{diag}(2\cos(0 \cdot \theta),2\cos(1 \cdot \theta), \dots, 2 \cos((N-1) \cdot \theta)), \quad \theta = \frac{2 \pi }{N}.
$$

An alternative, simpler approach. Use the Ansatz
$$
C_{p,q} = \omega_1^p \omega_2^q
$$
where $\omega_1,\omega_2$ are complex numbers. We have
$$
C_{p,q} = C_{p,q-1} + C_{p,q+1} + C_{p-1,q} + C_{p+1,q} \implies\\
\omega_1^p\omega_2^q =
\omega_1^p\omega_2^{q+1} +
\omega_1^p\omega_2^{q-1} +
\omega_1^{p+1}\omega_2^q +
\omega_1^{p-1}\omega_2^q  \implies\\
\omega_1^p \omega_2^q =
(\omega_1 + \omega_1^{-1} + \omega_2 + \omega_2^{-1})\omega_1^p \omega_2^q \implies\\
\omega_1 + \omega_1^{-1} + \omega_2 + \omega_2^{-1} = 1.
$$
Now, due to the periodic boundary condition, we must have $\omega_1^{N} = \omega_2^{N} = 1$, so that we have
$$
\omega_j = \exp(i k_j\theta), \quad \theta = \frac{2 \pi}{N}, \quad 0 \leq k_j \leq N-1, \quad j = 1,2.
$$
For this value of $\theta$, there is a non-trivial solution to the system for every solution to the equation
$$
\omega_1 + \omega_1^{-1} + \omega_2 + \omega_2^{-1} = 1 \implies\\
2[\cos(k_1 \theta) + \cos(k_2 \theta)] = 1.
$$
As can be shown using the matrix representation of the system from the first part, the system will have a solution if and only if this system has a solution for some integer values $k_1,k_2$.

An alternative formulation: let $\alpha = \theta/2$. With the sum to product formula, we can rewrite this as
$$
\cos([k_1 + k_2]\alpha)\cos([k_1 - k_2]\alpha) = \frac 14.
$$

Title: discrete system of equations on a torus


问:
Solve $\sqrt{3}\tan\theta=2\sin\theta$
I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.

$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$

$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$

I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.

Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.

Where am I going wrong please?

答:
When you square both sides of a equation you get a new equation that has all the solutions of the first but may admit other solutions. That was in the step $\sqrt{3}=2\cos\theta \Rightarrow 3=4\cos^2 \theta$.

Added: Also you can only divide your initial equation by $\sin \theta\neq 0$. But for $\theta=0$ or $\theta=\pm\pi$, $\sin\theta=\tan\theta =0$, which are solutions of your initial equation.

Added 2: In general the equation $A^n=B^n$ has all the solutions of the equation $A=B$ but may have additional ones. This is a consequence of the algebraic identity $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$

Title: Solve $\sqrt{3}\tan\theta=2\sin\theta$


问:
Degree measure of a missing angle in a polygon
Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$. She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?

so I just kind of guessed and checked like trying to find the total measure of angles of a 16-gon,15,14 and then I got the measure of the 14-gon which was 2160 and subtracted 2017 from it and got the answer 143 which is correct. How do I solve this using modular arithmetic ???? How do I solve this without guess and checking? Any help would be greatly appreciated.

答:
The sum of all angles should be a multiple of $180^\circ$, so if the missing angle is $\theta^\circ$ then

$$\begin{align*}
\theta + 2017 &\equiv 0 &&\pmod{180}\\
\theta &\equiv -2017\\
&\equiv 143
\end{align*}$$

(and also the polygon is convex, so $0^\circ < \theta^\circ < 180^\circ$)


答: You can solve this using modular arithmetic by considering the fact that the sum of the angles in a polygon is always a multiple of 180. Therefore, the sum should be 0 mod 180. Since 2017 is 37 mod 180, we know that the answer would be 180-37 = 143</p>

Title: Degree measure of a missing angle in a polygon

Body:
Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$. She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?

so I just kind of guessed and checked like trying to find the total measure of angles of a 16-gon,15,14 and then I got the measure of the 14-gon which was 2160 and subtracted 2017 from it and got the answer 143 which is correct. How do I solve this using modular arithmetic ???? How do I solve this without guess and checking? Any help would be greatly appreciated.


答:
$2017$ plus the missing angle must add up to a multiple of $180$ (an $n$-gon has angles summing to $180(n-2)$ degrees). Thus
$$2017+x\equiv0\equiv37+x\bmod180$$
so $x=143$, since it lies in $(0,180)$ by the polygon's convexity.


问:
How can we compute this quantity, constructed using motivation from Vandermonde's identity?
Vadermonde's identity states that

$$ \sum_{k=0}^{r} {m \choose k}{n \choose r-k} = {m+n \choose r} $$

I wonder if we can use this formula or otherwise to calculate:

$$ \sum_{k=0}^{r}k^l {m \choose k}{n \choose r-k}, l \in \mathbb{N} ?$$

or if not,  at least when $l=1,$ namely:

$$ \sum_{k=0}^{r}k {m \choose k}{n \choose r-k}?$$

I'd appreciate the relevant calculations or some links where I can find the above?

答:
Let $k_{(r)}$ denote the falling factorial $k(k-1)\cdots(k-r+1)$. Recall that $k^n=\sum_{r=1}^n {n\brace r}k_{(r)}$, where ${n \brace r}$ is a Stirling number of the second kind.

Expand $k^l$ into falling factorials, use the identity $k_{(j)}\binom{m}k=m_{(j)}\binom{m-l}{k-l}$, then apply Vandermonde's identity:
\begin{align}
\sum_kk^l\binom{m}k\binom{n}{r-k}
  &=\sum_k\sum_{j=1}^l{l \brace j} k_{(j)}\binom{m}k\binom{n}{r-k}
\\&=\sum_{j=1}^l{l \brace j}\sum_k m_{(j)}\binom{m-j}{k-j}\binom{n}{r-k}
\\&=\sum_{j=1}^l{l \brace j} m_{(j)}\binom{m+n-j}{r-j}
\\&=\boxed{\binom{m+n}{r}\sum_{j=1}^l{l \brace j} \frac{\binom{m}j}{\binom{m+n}j}\cdot r_{(j)}}
\end{align}
I have replaced your summation with a another summation. However, this new summation only has $l$ terms. For example:

$l=1$: $$\binom{m+n}r\cdot \left(1\cdot \frac{m}{m+n}\cdot r\right)$$

$l=2$: $$\binom{m+n}r\cdot \left(1\cdot \frac{m}{m+n}\cdot r+1\cdot \frac{\binom{m}2}{\binom{m+n}2}\cdot r(r-1)\right)$$

$l=3$: $$\binom{m+n}r\cdot \left(1\cdot \frac{m}{m+n}\cdot r+3\cdot \frac{\binom{m}2}{\binom{m+n}2}\cdot r(r-1)+1\cdot \frac{\binom{m}3}{\binom{m+n}3}\cdot r(r-1)(r-2)\right)$$

The $l=1$ equations represents the familiar fact that the the mean of the hypergeometric distribution is $\frac{m}{m+n}\cdot r$, and the other equations represent higher moments of the hypergeometric distribution.


问:
Example of $\pi$-metrizable space

A tychonoff space $X$ is $\pi$-metrizable if and only if it has a $\sigma$-locally finite
  $\pi$-base.

Please help me to find some example of $\pi$-metrizable space.

Is it true that every $\pi$-metrizable space is a moscow space ?

For a space $X$,

A collection of nonempty open sets $\mathcal{A}$ , is called a $\pi$-base if for every nonempty
open set $O$, there exists $U \in\mathcal{A}$  such that $U\subset O$.

答:
Some other examples:

Example 1: Every discrete space $X$ has a $\sigma$-locally finite $\pi$-base.

Example 2: $\mathbb R$ with the uaual topology has a $\sigma$-locally finite $\pi$-base.

Title: Example of $\pi$-metrizable space


问:
How to find a tangent when you have an implicit equation and the e function
Consider this:

Find the tangent to the function $f$ at point $(2,f(2))$ where $f(2)=0$ and the function $f:(1,\infty)\to\mathbb{R}$ using the implicit equation:

$$xe^{xf(x)} = (x-2)^3 + e^{f(x)}+1.$$

Now I've found tried to find the $g'(x)$ and $h'(x)$.

$$g(x) = xe^{xf(x)} \implies g'(x) = e^{xf(x)} + xe^{xf(x)'} * f(x)';$$
$$f(x) = (x-2)^3 + e^{f(x)} +1 \implies h'(x) = 3(x-2)^2 + e^{f(x)'}. $$

Now I am not sure if I derived correctly, and if not please let me know. But the other important thing is I'm not sure what to do next? Am I supposed to input $2$ into $f(x)$ to "simplify". A tip would be great.

Thanks!

答:
You're wrong in a couple of points:

$\frac{d}{dx}xe^{xf(x)}=e^{xf(x)}+ x\bigg(e^{xf(x)}\cdot \big(xf'(x)+f(x)\big)\bigg)=$

$\frac{d}{dx}\bigg((x-2)^3+e^{f(x)}+1\bigg)=3(x-2)^2+e^{f(x)}\cdot f'(x)$

So finally we can compute the derivative: (Just factorize $f'(x)$ out of both sides of the equality) $$f'(x)=\frac{e^{xf(x)}\big(xf(x)+x+1\big)-3(x-2)^2}{(1-x^2)e^{f(x)}}$$

Now because $f(2)=0$, clearly $f'(2)=\frac{3}{-3}=-1.$ So the slope of the tangent line is $-1$ and it also passes the point $(2,0)$ so the equation of the tangent line is: $y=2-x$ and the problem is solved.


问:
Quick Integration with Vectors Question
So I was given the following prompt:

The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.

I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!

Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.

答: Take the indefinite integral of $r(t)$ to get $g(t)+c$, where $c$ is a constant vector. If you know $g(7)$, then you know $c$, and then you can find $g(0)$.

Title: Quick Integration with Vectors Question


答:
Hint

$$\begin{align*}\vec g(t)=\int{{\vec r\left( t \right)\,dt}}  & = \left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}}} \right\rangle \color{red}{+\vec C}\end{align*}$$

To integrate $\sqrt{t^2+1}$, use trig-substitution, then Integration by Parts.

$\int{\sin{t^2}}dt$ is not an elementary integral; it requires a $u$-substitution that leads to a Fresnel Integral.

Title: Quick Integration with Vectors Question

Body:
So I was given the following prompt:

The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.

I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!

Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.