问:
Seeking a geometric characterisation of a certain property of a circle, a tangent ellipse within it, and a point strictly within the ellipse.
Consider a circle $C$, an ellipse $E$ strictly within $C$ except for a tangent point $p$, and a point $b$ strictly within $E$. Consider a point $q$ on $E$, and a point $q'$ given by the intersection of the extension of $qb$ with $E$, such that both $q$ and $q'$ are distinct from $p$. Define the points $r$ and $r'$ as the intersection of $C$ with the extensions of $pq$ and $pq'$ respectively.

We are interested in whether the point $j$, the intersection of the extension of $pb$ with $rr'$, lies inside the ellipse.

We can show (algebraically using Mathematica page 1,page 2) that the point $j$ is a property only of $C$, $E$, and $b$, independent of $q$, as illustrated in the figure:

<img src="https://i.stack.imgur.com/VTl7m.png" alt="illustration of the independence of $j$ on $q$" />

So we would like some way to characterize this property (whether $j$ is interior to $E$) of $C$, $E$, and $b$, geometrically without introducing $q$.

答:
If you are interested in the locus of point $b$ when $j$ is inside ellipse $E$, you could reverse your construction: take any point $j$ on $E$, trace from $j$ any line intersecting circle $C$ at $r$, $r'$ (that line is taken parallel to $x$ axis in figure below), join $pr$, $pr'$ to find their intersections $q$, $q'$ with $E$. Point $b$ lies on segment $qq'$. Repeating the construction for another line $tt'$ through $j$ you can find a second segment, whose intersection with $qq'$ is then point $b$ (otherwise, you could also find $b$ as the intersection of $pj$ and $qq'$).

I didn't embark into a computation, but with GeoGebra I could find the locus of $b$ as $j$ varies on the ellipse: it should be another ellipse (pink in figure below), tangent at $P$ with $C$ and having its axes parallel to those of $E$.


问:
Find if $\sum_{n=0}^{\infty}\frac{(an)!}{{(n!)}^{b}}$ converges depending on $a, b$

Find if $\sum_{n=0}^{\infty}\frac{(an)!}{{(n!)}^{b}}$   converges depending on the values of parameters $a$ and $b$.

I tried to use the ratio test:

\begin{align}
R &=  \lim_{n\to\infty}\frac{x_n}{x_{n+1}}\\
&=\lim_{n\to\infty}\frac{(an)!}{{(n!)}^{b}} \frac{{((n+1)!)}^{b}}{(a(n+1))!}\\
& = \lim_{n\to\infty}\frac{{((n+1)!)}^{b}}{{(n!)}^{b}} \frac{(an)!}{(a(n+1))!}\\
& = \lim_{n\to\infty}{(n+1)}^{b}\frac{(an)!}{(an+a)!}\\
& = \lim_{n\to\infty}{(n+1)}^{b}\frac{1}{(an+1)(an+2)\dots(an+a)}\\
& = \lim_{n\to\infty}{(n)}^b(1 + \frac{1}{n})^{b}\frac{1}{n^{a}(a+\frac1{n})(a+\frac2{n})\dots(a+\frac{a}{n})}
\end{align}

From here I'm not sure how to proceed. Can I say that depending on the value of $n^{(b-a)}$  if $b\ge a$, $R\le  1$ if $b\le a$ and $R=1$ if $a=b$?

Then for $R\ge 1$ it converges, for $R\le 1$ it diverges.

Is this correct? And what do I do when $R = 1$?

Edit: I tried to use Stirling's formula, but I still couldn't reach a concrete result.

答:
So I'm assuming a is an integer greater than or equal to 0.

You can go ahead and finish the limit:

$R=(\lim_{n \to \infty}n^{b-a})(\lim_{n \to \infty}(1+\frac{1}{n})^b)(\lim_{n \to \infty}(\frac{1}{(a+\frac1{n})(a+\frac2{n})\dots(a+\frac{a}{n})}))$

For the middle limit, you can bring the middle limit inside... giving you $1^b=1$

Similarly the last limit is $\frac{1}{a^a}$

$R = \frac{\lim_{n \to \infty}n^{b-a}}{a^a}$

If $b<a$, $R<1$ so it diverges

If $b>a$, $R > 1$, so it converges

If $b=a$ and $a>1$ then $R<1$ so it diverges

For $b=a=1$, plug them into the original formula for the series.

You get
$\sum_{n=0}^{\infty}\frac{(n)!}{{(n!)}}=\sum_{n=0}^{\infty}(1)$ which diverges.

For $b=a=0$, substitute directly again to get $\sum_{n=0}^{\infty}(1)$ which diverges.

ContentLicense: CC BY-SA 4.0


问:
Making $f$ differentiable
Given the following function
$$f(x)=\left\{\begin{array}{ll} a+e^{x-1}, & x<1 \\ bx^{2}+1, x\geq1 \end{array}\right.$$
I have to find the values of $a$ and $b$ such that $f$ is differentiable for all $x\in\mathbb{R}$.

Since
$$\lim_{h\to0^{-}}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{a+e^{h}-b-1}{h}$$
and
$$\lim_{h\to0^{+}}\frac{f(1+h)-f(1)}{h}=2b$$
therefore
$$\lim_{h\to0}\frac{a+e^{h}-b-1}{h}=2b$$
How do I conclude it?

答:
$e^h = 1 + h + o(h)$, so your last equation implies $\lim\limits_{h \to 0} \left(\frac{a - b}{h} + 1 + o(1)\right) = 2b$.

There are limits of the second and the third terms, so for limit of sum to exist the first term should also have limit. But it has limit only if $a - b = 0$ - so we geet $a = b$. Substituting, we get that limit is equal to $1$, thus $2b = 1$.

Finishing, we have $a = b = \frac{1}{2}$.

Title: Making $f$ differentiable


问:
Using Stokes' theorem for evaluation of a line integral
Let C be the curve of intersection of the plane $x+y+z=1$ and the cylinder $x^2 +y^2=4$, oriented counterclockwise when viewed from above. What is the value of the line integral $\oint\limits_{C} xy^2dx +x^2ydy+z^3dz$?

I have parametrised the equations for the curve of intersection to give: $x=2cos(u)$, $y=2sin(y)$ and $z=1-2cos(u)-2sin(u)$. What do I do next?

答: The rotor is $(0,0,0)$ so you get, by Stoke's theorem, that the line integral is 0.


问:
Show that two continuous non-negative functions satisfy this relation
Assume that $f$ and $g$ are continuous non-negative functions on a compact metric space $M,$ where $\{x \in M: \ g(x) = 0\} \subset \{x \in M: \ f(x) = 0\}$

Prove that $\forall \varepsilon > 0\, \exists K(\varepsilon) \text{ s.t. } \forall x \in M$
$$f(x) \leq \varepsilon + K(\varepsilon)g(x)$$

I'm not quite sure how to generalize the subset to this formula. I see that $g(x) = 0$ in some subset, but I'm not sure how to get from that to the formula here. I initially thought that this was trivial because $g(x) = 0$, but that isn't true because there is a subset where $g(x) = 0$. So I'm not really sure where to start.

答:
Let $\varepsilon > 0$, and let $Z_f =\lbrace x \in M, f(x)=0 \rbrace$ and $Z_g =\lbrace x \in M, g(x)=0 \rbrace$.

By continuity, there exists an open set $O$ (eventually empty) containing $Z_f$ such that $f(x) \leq \varepsilon$ for all $x \in O$. Because $Z_g \subset Z_f \subset O$, then $g(x) \neq 0$ for all $x \in M \setminus O$, so by continuity of $g$ over the compact $M \setminus O$, there exists $m > 0$ such that $g(x) \geq m$ for all $x \in M \setminus O$. By continuity of $f$ over the compact $M$, there exists also $m'>0$ such that $f(x) \leq m'$ for all $x \in M$. Define then$$K=\frac{m'}{m} > 0$$
Then

if $x \in O$, one has $f(x) \leq \varepsilon \leq \varepsilon + Kg(x)$ ;

and if $x \in M\setminus O$, then $f(x) \leq m'=Km \leq Kg(x) \leq \varepsilon + Kg(x)$.

In both cases, you are done.


问:
$\mathbb{C} \setminus f(\mathbb{C})$ contains a disk of radius 1, then $f$ is constant
$f$ is a non constant entire function such that $\mathbb{C}\setminus f(\mathbb{C})$ contains a disk of radius 1, then $f$ is constant.

My thoughts: if we can show $f$ is bounded then by Liouville's theorem we are done. I'm not sure how the given condition implies that.

答: Call $c$ the center of one such disk, then $z\mapsto\frac1{c-f(z)}$ is an entire bounded function.


答:
Interesting! This problem appeared to me trivial at first, but then I quickly realized it is a bit trickier than that.

For any $\epsilon>0$, we define the following function $h$:
$$ h(x) := \frac{ f(x)-\epsilon}{g(x)} \mathbb{1}_{ f(x) \le \epsilon}$$
We can test the continuity of this function and it is continuous indeed. We just have show the continuity of $f$ at three different kind point $x_0,x_1,x_2$ such that:

$f(x_0)>\epsilon$ ( clearly true )

$f(x_1)= \epsilon$ (true because $f(x)-\epsilon$ converges to 0 at the neighborhood of $x_1$ while $g(x_1)>0$ hence $1/g$ is continous at such point)

$f(x_2)<\epsilon$ ( also true 'cause there is a neighborhood if $x_2$ such that $f(x)<\epsilon$ )

So yeah, $h$ is a continuous function on a compact metric space, thus the conclusion.

Disclaimer : Surely, a more elaborate solution is needed to show but I guess having showed the idea of proof.

Title: Show that two continuous non-negative functions satisfy this relation

Body:
Assume that $f$ and $g$ are continuous non-negative functions on a compact metric space $M,$ where $\{x \in M: \ g(x) = 0\} \subset \{x \in M: \ f(x) = 0\}$

Prove that $\forall \varepsilon > 0\, \exists K(\varepsilon) \text{ s.t. } \forall x \in M$
$$f(x) \leq \varepsilon + K(\varepsilon)g(x)$$

I'm not quite sure how to generalize the subset to this formula. I see that $g(x) = 0$ in some subset, but I'm not sure how to get from that to the formula here. I initially thought that this was trivial because $g(x) = 0$, but that isn't true because there is a subset where $g(x) = 0$. So I'm not really sure where to start.


答:
Please note in this specific case, your vector field is gradient of scalar field $F$

where $F = \frac{x^2y^2}{2} + \frac{z^4}{4}$.

$(xy^2, x^2y, z^3) = \nabla(\frac{x^2y^2}{2} + \frac{z^4}{4})$

As it is in conservative vector field, the line integral over a closed loop will be zero.

But if it was not, please note that you are interested in intersection curve of $x^2 + y^2 = 4$ and $x + y + z = 1$. The unit normal vector to the plane will be $(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}).$

You then need to find curl of the vector field $\nabla \times F$ normal to the surface (do a dot product with unit normal) and then do the double integral.

Title: Using Stokes' theorem for evaluation of a line integral

Body:
Let C be the curve of intersection of the plane $x+y+z=1$ and the cylinder $x^2 +y^2=4$, oriented counterclockwise when viewed from above. What is the value of the line integral $\oint\limits_{C} xy^2dx +x^2ydy+z^3dz$?

I have parametrised the equations for the curve of intersection to give: $x=2cos(u)$, $y=2sin(y)$ and $z=1-2cos(u)-2sin(u)$. What do I do next?


问:
diagonalizable condition
About a matrix A = 2 a
1 0

I want to find a condition for A to be diagonalizable.
I think that A needs to have 2 linearly independent eigenvectors, and also Eigenvectors associated to distinct eigenvalues are linearly independent.
But I don't know how can I access to this one.

答:
I assume this matrix $A$ would be $\begin{bmatrix} 2 & a \\ 1 & 0 \end{bmatrix}$. You first need to find the eigenvalues ($\lambda$) for this matrix such that $AX = \lambda X$, where $X$ is a vector $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ and $\lambda$ is a scalar. You can solve for $\lambda$ by manipulating $AX = \lambda X$, where $AX = \lambda X \Rightarrow AX - \lambda X = 0 \Rightarrow (A - \lambda I)X = 0$. $\lambda$ will have nonzero solutions as long as the determinant of $(A-\lambda I)$ is equal to 0, so you can solve for $\lambda$ with the equation det$(A- \lambda I) = 0$. Once you have values of $\lambda$, the eigenvector(s) for each value of $\lambda$ can be found by solving for $X$ in $(A - \lambda I)X = 0$ (remember $X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$). If you find two distinct, linearly independent eigenvectors, then you can diagonalize $A$ by constructing a matrix $P$ from its eigenvectors, where $D$ (the diagonalized matrix) = $P^{-1}AP$.

$A$ would be non-diagonalizable if (assuming you're working with real numbers) $\lambda$ has a complex result (say, $\lambda = \sqrt{-1}$). It would also be non-diagonalizable if you only found one unique eigenvalue. If the one eigenvalue yielded a singular, constant value of $X$, there is only one possible eigenvector, which is not enough. But even if it yielded infinite distinct eigenvectors, they would need to be multiples of each other-- none of these would be linearly independent from one another, meaning $A$ couldn't have a complete basis of eigenvectors and is therefore, not diagonalizable.

Title: diagonalizable condition


问:
If a power of 2 divides a number, under what conditions does it divide a binomial coefficient involving the number that it divides?
We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally gave several more examples but took TheSimpliFire's advice to shorten the list, and many other examples can still be found by looking at this question's edit history):

How can we prove that a binomial coefficient (n choose k) is divisible by the ratio of n and the gcd(n,m)?

Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$

The topic has also lead to some interesting research papers recently:

In 2014: Some divisibility properties of binomial and $q$-binomial coefficients.

In 2017: Divisibility of binomial coefficients and generation of alternating groups.

In 2019: On the divisibility of binomial coefficients.

There's also many associated theorems:

Kummer's theorem

Lucas's theorem

Granville's theorem

Désarménien's theorem

However I've come across a related problem which is expressed completely in the title, and is remarkably not covered by the above extensive body of literature. In MathJax, if $s>0$ is an integer (let's also make it the largest one for which $2^s$ divides $n$), under what conditions of $k$ do we have the following:

$$
2^s \mid n \implies 2^s \left\vert \binom{n}{k} \right. . \tag{1}
$$

Since $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ and $2^s \mid n$ we are left with determining when $\frac{q}{k} \binom{n-1}{k-1}$ is an integer ($q$ being the result when $n$ is divided by $2^s$).

The implication works for a remarkable number of cases, but not always (for example if $\binom{n}{k}$ is odd).

答:
This is a comment, but a little long comment.

Recall the definition of the $2-$adic valuation of $n,$ $\nu _2(n),$ meaning the biggest exponent of $2$ that divides $n.$

Your problem, then, can be stated as $s\leq \nu _2(n)$ implies $s\leq \nu _2 \left( \binom{n}{k}\right ).$ From legendre's theorem (i.e., $\nu _2(n!)=n-S_2(n)$ where $S_2(n)$ is the sum of the $1$'s in the $2-$adic expansion of $n$) is not hard to show that
$$\nu _2 \left( \binom{n}{k}\right )=n-S_2(n)-(k-S_2(k))-(n-k-S_2(n-k))=S_2(n-k)+S_2(k)-S_2(n).$$ and $$\nu _2(n)=\nu_2(n!)-\nu _2((n-1)!)=1+S_2(n-1)-S_2(n).$$
So your proposition becomes

$s-1\leq S_2(n-1)-S_2(n)$ implies $s\leq S_2(n-k)+S_2(k)-S_2(n).$

 this can be, at least, be satisfied if $S_2(n-1)+1\leq S_2(n-k)+S_2(k).$ This is trivial if $k=1.$ For example, if $n$ is odd, this is equivalent to $S_2(n)\leq S_2(n-k)+S_2(k)$ which is always the case, for example if the expansion of $k$ has $1'$s where there is a $1$ in the expansion of $n.$


问:
Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?
Suppose for all real $x$ that $g^{(n)}(x) = g^{(n-1)}(x)$ and $g^{(n-1)}(0) = -1$ where $g^{n}(x)$ is the nth derivative. Does the $\lim_{x \to \infty} \frac{g(x)}{e^x}$ exist?

Isn't this answer just the form $g(x) = -e^x$? and Hence $\lim_{x \to \infty} \frac{g(x)}{e^x} = -1$? How could I more rigorously show this?

答:
Weird.  Isn't that the function $h(x):= e^{-x}g(x)$ has null derivative?

Which implies $h$ is a constant function.

Is there any other constraint?


问:
Polynomial division in $\mathbb{Z}[x]$.
How is polynomial division defined in $\mathbb{Z}[x]$? For example, how to divide $x$ by $2$ or any other two polynomials in $\mathbb{Z}[x]$?</p>

答:
If you are talking about Euclidean division, it is pretty much the same as in $\mathbb Z$: If $P$ can be written $P(X)=Q(X)S(X)$ with $Q,S\in\mathbb{Z}[X]$, then $P/Q=S$. In general, the euclidean division of $P$ by $Q$ comes from writing $P(X)=Q(X)S(X)+R(X)$ with $\mathrm{deg}(R) < \mathrm{deg}(Q)$ and $Q,S,R\in\mathbb{Z}[X]$ when it is possible (see comment below).

For instance, $(X^4-1)/(X-1) = (X+1)(X^2+1)$.

ContentLicense: CC BY-SA 3.0

Title: Polynomial division in $\mathbb{Z}[x]$.


问:
Prove : $\sin^2(x)\leq\ {x^2}$
I need to prove :
$\sin^2(x)\leq\ {x^2}$

My Attempt:
(I quickly drew the graph of the functions and got the result, but after a lot of trial I am unable to prove it algebraically).

How can we prove this algebraically?

The Graph:

答:
I would split the inequality up into different cases.
$$
\sin^2 x \leq x^2
$$
For $x\leq-1$ and $x\geq1$, the inequality is certainly true, as the range of $\sin^2 x$ is $[-1,1]$, while $x^2\geq1$ when $x \leq -1$ and $x \geq 1$. When $0\leq x < 1$, we know that $x \geq \sin x$ because $\sin x$ has a gradient strictly less than $1$. Since $x \geq \sin x$, it follows that $x^2 \geq \sin^2 x$.* Also, $\sin^2(-x)=\sin^2(x)$ and $(-x)^2 = x^2$, meaning that the case $-1 < x<0$ is identical.

*It is not true in general that if $a>b$ then $a^2>b^2$. However, if $a$ and $b$ are both positive, then this statement is always true.

Title: Prove : $\sin^2(x)\leq\ {x^2}$


问:
Finding the number of sequences $(a_1, a_2, a_3, ..., a_n)$
Given that $a_i =0, 1$ for $1 \leq i \leq n,$ find the number of sequences $(a_1, a_2, a_3,..., a_n)$ such that $$a_1 \leq a_2 \geq a_3 \leq a_4 \cdots$$

I was thinking that I should somehow just arrange the $1$'s and $0$'s and be done with the problem, but I'm unsure if there's a problem with my logic here. Can someone confirm or deny my logic?

答:
The pattern repeats at $a_3$.

Let $a_2 = 0$, then $a_1$ has to be $0$.

Let $a_2 = 1$ then $a_1$ is either $0$ or $1$.

Let $a_3 = 0$ then $a_2$ is $0$ or $1$.

Let $a_3 = 1$ then $a_2$ has to be $1$.

Until $a_1$, there are $2$ solutions.

Until $a_2$, there are $3$ solutions.

Until $a_3$, there are $3 + 2 = 5$ solutions.

Can you take it from here?


问:
Pigeon hole principle to prove $a-b=9$ in subset.
I have a set of numbers
$$
[n] = \{1,2,...,n\}
$$
in my special case $n = 100$, and I have a subset of $[100]$ with the following specification
$$
A\subseteq[100]
$$
and
$$
|A| >= 55
$$
now I should prove, that this statement is true for some
$$
a,b\in A: a-b=9
$$

I thought about the problem and I realised, that if I just take the numbers $1-55$ that there are a lot of pairs $a,b$ that match the condition.

So I tried to build a set in which no pair matches the condition. Therefore I just used the even numbers from $2-100$. Because even-even=even. But there are only $50$ even numbers in $[100]$, so I have to add at least $5$ odd numbers. So as soon as I add one odd number my set matches the condition.

Using the pigeonhole principle:
$$
n,m \in \mathbb{N}, f: [n] \to [m], |f^{-1}(j)|, j \in [m], \exists j^{*} \in [m], |f^{-1}|>=\lceil\frac{n}{m}\rceil
$$
I get, that there is at least
$$
\lceil\frac{100}{55}\rceil = 1
$$
solution to my problem.

But I think that I have to specify the function for the projection to prove the problem.
And I think that I can use the modulo operator to achieve my goal, but currently I am stuck.
Could someone please help me?

答:
First consider partitioning [$100$] in following manner :

$$ \{1,2,\ldots,18 \} \, \{19,20,\ldots,36 \} \cdots \{73,74,\ldots,90 \} \, \{91,92,\ldots,100 \} $$

Now in each of first five sets we have $9$ pairs differing by $9$ like  $$ (1,10) , (2,11) , \ldots (9,18), \ldots, (89,90) $$

and in last, one pair $(91,100)$. Remaining are unpaired.

Can you complete?

We have $9\cdot5+1=46$ pairs. And $8$ : ${92,93,\ldots,99}$ are unpaired.

To make a set $A$ not having the property, choose one number from each pair and all the unpaired ones. But we can choose only $46+8=54$ such numbers. $55^{th}$ number belongs to one of previous pairs, so $a-b=9$ must satisfy.


问:
What is the Ramanujan summation for the series $\sqrt[n]{2}$
A Ramanujan summation is a

technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series

In my case, I'm interested in assigning a value to the divergent series

$$\sum_{n=1}^\infty f(n) \ \ \ \ \ \ \ \text{where}\ \ \ \ f(n)=\sqrt[n]{2}$$

According to the Wikipedia page (and my understanding), the Ramanujan summation is

$$\sum_{n=1}^\mathfrak{R} f(n)=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N f(n)-\int_{1}^N f(t)dt\Bigg]$$

Thus

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg]$$

Taking the antiderivative

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(\ln2\Big(\text{li}\ 2-\text{Ei}\frac{\ln2}{N}\Big)+N\sqrt[N]{2}-2\Bigg)\Bigg]$$

Moving some constants outside the limit

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=2-\ln2\cdot\text{li}\ 2+\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(N\sqrt[N]{2}-\ln2\cdot\text{Ei}\frac{\ln2}{N}\Bigg)\Bigg]$$

It's at this point I'm unsure of how to proceed. I'm not terribly confident what the limit converges to. From my computational estimates up to $N=10^8$, I find that

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}\approx1.6$$

But due to floating point errors or slow convergence, it deviates substantially enough for me to not be confident about any more digits.

I'd like to know if this converges at all, and if it does, is there a (reasonably) closed form / relation to other constants?

答:
I'm not sure I'm calculating the same thing, and as regularizations of divergent sums need not be unique, I will ask the OP to do some more numerical work on his formula.  The following is how I would find a closed-form expression, 'in the manner of Ramanujan summation.'

$$F_n(x) := \sum_{k=1}^n x^{1/k} = \frac{n}{n}\sum_{k=1}^n (x^{n/k})^{1/n} \to n \int_{0}^1 x^{1/(nt)} dt$$
as a Riemann sum as $n \to \infty.$  This is the first term on the right-hand side of your defining equation
$$ \sum_{k=1}^{\mathcal{R}} f(k) = \lim_{n \to \infty} \Big(\sum_{k=1}^n f(k) - \int_{1}^n f(t) dt \Big)$$
and has an explicit closed-form
$$ F_n(x) = n \ x^{1/n} + \Gamma(0,\frac{\log(x)}{n} )\log(x) . $$  I've used the incomplete gamma function, but the first argument tells us it will reduce to the Ei integral.  By the way, this is a perfectly good asymptotic approximation, giving 4 significant figures for $x=2$ and $n=1000.$  The second term also has an explicit expression,

$$ \int_1^n x^{1/t} dt = -x + n\ x^{1/n} + \log{x} \Big(-\text{Ei}(\log{x}/n)+ \text{li}(x) \Big).$$
(Both integrals have been performed in Mathematica.)  Now subtract and do an asymptotic expansion.  The singular parts cancel and you are left with

$$
\sum_{k=1}^{\mathcal{R}} x^{1/k} = -\big(x+ \log{(x)}\, \text{li}(x) \big) $$
where li$(x)$ = LogIntegral[x] in Mathematica. When I plug in $x=2,$ I get the value of -2.72445.  This is very different from the OP's answer, with not even sign agreement.  However, we should not forget the Ramanujan summation is closely aligned with zeta regularization, and with it you can 'prove' an infinite sequence of positive numbers is a negative number, e.g.,

$$ 1+2+3+... = \lim_{s \to \  -1} \sum_{k=1}^\infty k^{-s}  = \lim_{s \to \  -1}\zeta(s) = -\frac{1}{12} $$

I'm wary of regularization, though it does lead to some cute results.  Physicists are sometimes forced into such shenanigans with perturbation theory.


问:
$ \cos x\geq 1-\frac{x^2}{2} $
Prove that for $x\in\mathbb{R}$
$$
\cos x\geq 1-\frac{x^2}{2}.
$$

My try:

Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$
If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.

But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$

答:
$$\cos x-1+\frac{x^2}2\ge0$$ and equality holds at $x=0$.

Then differentiating,

$$-\sin x+x\ge 0$$ and equality holds at $x=0$.

Finally,

$$-\cos x+1\ge 0.$$

So $-\sin x+x$ grows from $0$ and is non negative, and $\cos x-1+\dfrac{x^2}2$ grows from $0$ and is non-negative.

This technique works for Taylor developments to arbitrary orders.

Title: $ \cos x\geq 1-\frac{x^2}{2} $


问:
Prove $|-a|\le a\le |a|$
Prove that $|-a|\le a\le |a|$.

Can anyone explain this to me in a step-by-step way?, I am really having trouble in understanding it.

答:
Assuming you meant
$$
-|a|\leq a\leq |a| \, ,
$$
we will prove this in the following way. The inequality $a\leq|a|$ holds because if $a\geq0$, then we know that $a \leq a$, while if $a<0$, the inequality becomes $a \leq -a$. Since $a$ is negative, this is also true. Hence
$$
a\leq|a| \, .
$$
A similar approach can be used to prove $-|a| \leq a$.

Title: Prove $|-a|\le a\le |a|$


问:
Problem with finding perfect matchings in a bipartite graph
I've been given the following problem to solve;

Let $G(X,Y)$ be a bipartite graph such that in $X$ the degree of each vertex is at least $r$ and in $Y$ the degree of each vertex is at most $r$. Prove that this graph has a perfect matching.

Now after thinking on it I was able to come up with the following:

Let $X=\{A,B,C\} ,Y=\{D,E,F,G\}$ for $r=2$ we have $\delta(X)\geq r$ and $\Delta(Y)\leq r$ but clearly this graph does not have a perfect matching. What am I missing here? Am I understanding the question wrong? Or am I misunderstanding perfect matchings? Any help would be appreciated!

答:
The general thing we can prove is that under these hypotheses, the graph has an $X$-perfect matching: a matching that saturates every vertex of $X$. (A perfect matching would require $|X|=|Y|$, and this is not guaranteed here.)

To prove this, you should check Hall's condition: prove that for every subset $S \subseteq X$, $N(S)$ - the set of vertices in $Y$ adjacent to a vertex in $S$ - satisfies $|N(S)| \ge |S|$.

If you've seen the proof that a regular bipartite graph has a perfect matching, this will be similar.


问:
Spivak's Inverse Function Thoerem
I've gone through numerous posts about Spivak's proof of the inverse function theorem in Calculus on Manifolds. But I can't find any that address this part.

Inverse Function Theorem
Suppose that $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable in an open set containing $a$ and det $f'(a) \neq 0$. Then there is an open set $V$ containing $a$ and open set $W$ containing $f(a)$ such that $f: V \rightarrow W$ has a continuous inverse $f^{-1}:W \rightarrow V$ which is differentiable and for all $y \in W$ satisfies $(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$

He states the following:

There is a closed rectangle $U$  containing $a$ in its interior such that $f(x) \neq f(a)$ if $x \in U$ and $x \neq a$.

Since $f$ is continuously differentiable in an open set containing $a$, we can also assume that

$det f'(x) \neq 0 $ for $x \in U$

and

$|D_jf^i(x) - D_jf^i(a)| < 1/2n^2$ for all $i,j$ and $x \in U$

Where did these two facts come from?

答:
Here are the sequence of steps Spivak makes:

The first paragraph explains why there is no loss of generality in assuming that $Df_a=\text{id}_{\Bbb{R}^n}$.

Next, by definition of $Df_a$, and our previous assumption that $Df_a=\text{id}_{\Bbb{R}^n}$, we must have $\lim\limits_{h\to 0}\dfrac{\lVert f(a+h)-f(a)- h\rVert}{\lVert h\rVert}=0$. Because of this, there exists an open set $U_1$ containing $a$ such that for all $x\in U_1$, $f(x)\neq f(a)$.

Since $f$ is by assumption continuously differentiable, and because determinant is a continuous function (actually, it is even a polynomial in the matrix entries, so this is very nice), the composition $\det \circ f':\Bbb{R}^n\to \Bbb{R}$ is continuous. Since $(\det \circ f')(a) = \det[f'(a)]\neq 0$ by assumption, continuity implies that there is an open set $U_2$ such that for all $x\in U_2$, we have $(\det\circ f')(x)\neq 0$ (in fact we can choose the $U_2$ such that the sign is the same as $\det f'(a)$).

Since $f$ is continuously differentiable, all the partial derivatives $D_jf^i$ are continuous. Thus, by the $\epsilon,\delta$ definition (with $\epsilon = \frac{1}{2n^2}$, which is a positive number), we can find open sets $V_{ij}$ (for example you can take them to be open balls centered at $a$ with sufficiently small radius) such that for all $x\in V_{ij}$, we have $\left|D_jf^i(x)- D_jf^i(a)\right|< \frac{1}{2n^2}$.

Now, let $V= U_1 \cap U_2 \cap \bigcap_{i,j}V_{ij}$. This is the intersection of finitely many open sets which contain $a$, hence $V$ is open and contains $a$. Because it is open, we can further find a small enough closed rectangle $U$ such that $a\in \text{int}(U)\subset U \subset V$.

Basically, the idea is that you find an open set small enough for all the purposes. Now, with this $U$, it is clear that

for every $x\in U$, $f(x)\neq f(a)$.

for every $x\in U$, $\det f'(x)\neq 0$

for every $x\in U$, and every $i,j$, $\left|D_jf^i(x)- D_jf^i(a)\right|< \frac{1}{2n^2}$.

Title: Spivak's Inverse Function Thoerem


答: Polynomial division of $f$ by $g$ consists of finding quotient $q$ and remainder $r$ with $f=qg+r$ and $\deg r<\deg g$. However, if $g$ is not monic, this may not be possible (while in $\mathbb Q[X]$ you have $X=2\cdot \frac X2+0$).</p>

Title: Polynomial division in $\mathbb{Z}[x]$.

Body: How is polynomial division defined in $\mathbb{Z}[x]$? For example, how to divide $x$ by $2$ or any other two polynomials in $\mathbb{Z}[x]$?</p>


问:
Diagonal Sub-Matrix of A Positive Definite Matrix
Let $A$ be a positive definite matrix partitioned in the following fashion.

$$A=\begin{bmatrix} A_{11} & A_{12} \\ A_{12}^T & A_{22}\end{bmatrix}.$$

In particular, $A_{11}$ is $m_1\times m_1$ and $A_{22}$ is $m_2\times m_2$. We need to show that both $A_{11}$ and $A_{22}$ are positive definite.

My Attempt. Let $x=[x_1^T\quad x_2^T]^T$ be an $(m_1+m_2)\times 1$ vector, where $x_1$ is $m_1\times 1$ and $x_2$ is $m_2\times 1$. Then

$$x^TAx=x_1^TA_{11}x_1+x_2^TA_{22}x_2+2x_1^TA_{12}x_2.$$

We know that $x^TAx>0$ whenever $x\neq 0$. However, we cannot make sure both $x_1^TA_{11}x_1>0$ and $x_2^TA_{22}x_2>0$, because I think it is possible that

$$x_1^TA_{11}x_1=0,\quad x_2^TA_{22}x_2=0,\quad x_1^TA_{12}x_2>0.$$

Perhaps I missed something here. Hope anyone have good ideas.

答: Hint: We know that $x^TAx > 0$ for every $x$ with $x_1 \neq 0$ and $x_2 = 0$. What does this tell you about $A_{11}$?


问:
What's the result of $\lim_{x\to 0} \dfrac{e^{-\frac{1}{x^2}}}{\sin x}$?
During a correction of an exercise in a differentiability class this limit showed up and the teacher said that it could be done with L'Hopital easily, but me and a couple of friends tried doing it and neither of us got it. How can it be solved? I know the limit is 0 thanks to Geogebra.

$$ \lim_{x\to 0} \dfrac{e^{-\frac{1}{x^2}}}{\sin x}$$

答:
Rewrite the expression as

$$\lim_{x\to0}\frac{e^{-\frac{1}{x^2}}}{x}\cdot\frac{x}{\sin x}$$

If the limits exist, the limit of the product will be the product of the limits. The limit on the right is $1$. Focusing on the left expression only, use the substitution $t=\frac{1}{x^2}$

$$\lim_{t\to+\infty}\sqrt{t}e^{-t} = \lim_{t\to+\infty} \frac{\sqrt{t}}{e^t} = 0$$


答:
we see that as $x\rightarrow -x$ we get the same inequality. Hence WLOG $x\ge 0$.
AS $$\sin (x/2)\le \frac{x}{2} \tag1$$ $$1-2\sin^2 (x/2)\ge 1-\frac{x^2}{2}$$
$$\cos x \ge 1-\frac{x^2}{2}$$

Proof of Claim $(1)$ is very famous

Title: $ \cos x\geq 1-\frac{x^2}{2} $

Body:
Prove that for $x\in\mathbb{R}$
$$
\cos x\geq 1-\frac{x^2}{2}.
$$

My try:

Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$
If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.

But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$


问:
How to prove a function is computable?
I'm reading Computation Complexity: A Modern Approach and one of the exercises is:

Prove that [the addition function is] computable by writing down a full description (including the states, alphabet, and transition function) of the corresponding Turing [machine].

I successfully wrote this description but I don't think it is a proof the addition function is computable.

According to this question, the definition of a computable function is:

If $f:\Sigma^* \to \Sigma^*$ is function, and $\exists$ a Turing machine which on the input $w\in\Sigma^*$ writes $f(w)$, $\forall w\in\Sigma^*$, then we call $f$ as computable function.

With the Turing machine I wrote, I can prove the machine is able to compute $f(w)$ for some inputs but how can I prove it can do the same for every inputs $w\in\Sigma^*$? In other words, how can I prove $f$ is computable?

答: You could potentially construct a Turing Machine that takes an two inputs, one of the form "m + n", and one of the form "x", where "m + n" is the two integers to be added and "x" is the sum. Make this Turing Machine reject if "m + n $\neq$ x" and accept if "m + n = x". If you can prove that this machine halts on all inputs (read: accepts or rejects) and accurately calculates all sums, then you could use this deciding TM to generate a language consisting of all pairs of numbers and their sums. Not sure if this is the approach the question asks for but this would be my approach.

Title: How to prove a function is computable?


问:
How do I verify if $2^{35} \equiv 1 \pmod{71}$ is true or not?
I need to know if $2^{35} \equiv 1\pmod{71}$ is true. I tried using Euler and Fermat little theorem and I got stuck. There is probably something trivial I'm not seeing so I appreciate any help, thanks.

答:
Here's another way:

$$
2^{36}\equiv 64^6 \equiv (-7)^6 \equiv (-343)^2 \equiv 12^2 \equiv 2 \pmod{71} \Rightarrow 2^{35} \equiv 1 \pmod{71}
$$


问:
How can I generate lotteries where I can simultaneously determine their expected values, variances and skewness?
First of all, thanks for taking the time to look at this. My goal is to be able to generate lotteries, where I can determine the first, second and third moments of their distributions, for a behavioural economics experiment. In other words, I'd like to be able to determine their expected values, variances and skewness.
\begin{align*}
 &\mathrm{Expected\ Value\ :} &&=\frac{\sum_{i}x_i}{n}\\
 &\mathrm{Variance:} &&=\frac{\sum_{i}(x_i-\mathrm{Mean})^2}{n}\\
 &\mathrm{Skewness:} &&=\sum_{i}\left( \frac{x_i-\mathrm{Mean}}{\sqrt{\mathrm{Variance}}}\right) ^3
\end{align*}
The lotteries will have three outcomes that are equally probable, so I believe this amounts to solving for $\{a,b,c\}$ in the following nonlinear system of equations, where $m, v$ and $s$ are just constants.
\begin{align}
 \frac{a+b+c}{3}= m\\
 \frac{
 (a-m)^2+(b-m)^2+(c-m)^2
 }{n}= v\\
 \frac{
 (a-m)^3+(b-m)^3+(c-m)^3
 }{\left( \sqrt{v}\right) ^3} = s
\end{align}
My questions is, how would you find for what values of $m,v$ and $s$ there are non-negative real solutions? I know that I can just input non-negative values for $a, b$ and $c$, but the issue with this is that, for my experiment, I would like to find triples of lotteries that satisfy one of the two following conditions. For three lotteries, say $\{1,2,3\}$, I would like either:

A. $\mathrm{Exp_1}=\mathrm{Exp_2}>\mathrm{Exp_3}$ and $\mathrm{Var_2}=\mathrm{Var_3}<\mathrm{Var_1}$ and $\mathrm{Skew_3}=\mathrm{Skew_1}<\mathrm{Skew_2}$

or

B. $\mathrm{Exp_1}=\mathrm{Exp_2}<\mathrm{Exp_3}$ and $\mathrm{Var_2}=\mathrm{Var_3}>\mathrm{Var_1}$ and $\mathrm{Skew_3}=\mathrm{Skew_1}>\mathrm{Skew_2}$

So, my question to you is, is this possible? And if so, how could I achieve it?

Many thanks and
best wishes,

Jeremy

答:
To me, it's more like a question on zeros of polynomials. It would be nice if someone provides a numerical perspective for it.

Your problem is equivalent to know if there are zeros for the following system of equation.
$$\begin{equation} a+b+c= u \\a^2+b^2+c^2= v \\ a^3+b^3+c^3= w\end{equation}$$
and that system is equivalent to:
$$\begin{equation} a+b+c= u=:x_1 \\ab+bc+ca= \dfrac{u^2-v}{2}=:x_2 \\ abc=\dfrac{2w+u^3-3uv}{6}=:x_3 \end{equation}$$

and this system of equations has three real roots( singular or not) , if the follow equation has three real roots.

$$X^3-x_1X^2+x_2X-x_3=0$$
Any further analysis that you want on $a,b,c$  can be translated to a study on the above equation.


问:
What about a module of rank $\frac{1}{2}$?
Let $R$ be a commutative ring. The possible ranks of free $R$-modules are $0,1,2,\dotsc$. But what about a generalized notion of an $R$-module where ranks may be rational numbers such as $\frac{1}{2}$?

More specifically, is there any (natural) cocomplete symmetric monoidal category which contains $\mathsf{Mod}(R)$ and a distinguished object $T$ satisfying $T \oplus T \cong R$?

Remark that for $R \neq 0$ there is no $R$-module $T$ which satisfies $T \oplus T \cong R$, because otherwise $T$ would be finitely generated projective, hence locally free of finite rank $d$ with $2d=1$, a contradiction.

Background: I try to develope some kind of "functional calculus" for monoidal categories. This example here deals with $n$th roots, for which we have to adjoin $\frac{1}{n}$ first. More precisely, if some object $T$ has the property $n \cdot T \cong 1$, where $1$ is the unit for $\otimes$, we may think of $T$ as a categorification of $\frac{1}{n}$ and define an object $(1 \oplus M)^{\frac{1}{n}} := \bigoplus_{p=0}^{\infty} \Lambda^p(T) \otimes M^{\otimes p}$, which in fact is a $n$th root of $1 \oplus M$ with respect to $\otimes$. This is a categorification of the binomial theorem $(1+z)^{\frac{1}{n}}=\sum_{p=0}^{\infty} \binom{1/n}{p} z^p$ for $z \in \mathbb{C},\,|z|<1$.

答:
Unfortunately the answer is no, at least when $2 \in R^*$. If $C$ is a cocomplete $R$-linear symmetric monoidal category with an object $T$ satisfying $1 \cong T \oplus T$, then $C=0$.

Proof: We have exterior powers in $C$ with the usual properties. Hence, $0 = \Lambda^2 (T \oplus T) \cong (\Lambda^2 T) \oplus (\Lambda^1 T \otimes \Lambda^1 T) \oplus (\Lambda^2 T)$, which means $\Lambda^2 T = 0$ and $T \otimes T = 0$. But then $T \cong 1 \otimes T \cong (T \oplus T) \otimes T \cong (T \otimes T) \oplus (T \otimes T) = 0$, hence $1 = 0$. $\square$


答:
The problem with using Taylor's theorem as you suggest is that the series only converges quickly near the origin, but we can restrict the range of $x$.  If $|x|>2$ then $1-\frac{x^2}2<-1$ and the statement is true, so we may assume $|x|\leq2$.  Then by Taylor's theorem,$$\begin{align}
\cos x-1+\frac{x^2}2
&=\frac{x^4}{4!}-\frac{x^6}{6!}\cos\theta\\
&=\frac{x^4}{4!}\left(1-\frac{x^2}{30}\cos\theta\right)\\
&\geq\frac{x^4}{4!}\left(1-\frac4{30}\right)\geq0
\end{align}$$

Title: $ \cos x\geq 1-\frac{x^2}{2} $


问:
bijection from $(a,b)$ to $\Bbb R$
I need to construct a bijection from an arbitrary interval $(a,b) \to\Bbb R$. I was thinking of somehow using the tangent function because it's asymptotic, but i'm not sure how to get started.

答:
Assuming $a<b$, you can take:
$$f(x)=\tan\left(\frac{\pi}{b-a}x+\frac{\pi}{a-b}\cdot\frac{b+a}2\right)$$
the period $P=d(a,b)$ and the function is shifted so that the center is at $x=\frac{a+b}2$. The vertical asymptotes are $x=a$ and $x=b$.

Title: bijection from $(a,b)$ to $\Bbb R$


问:
Let $D$ be an integrity domain and $F$ a subfield of $D$.
If $d$ ∈ $D - \{0\}$ is such that there exists a nonzero polynomial $f(x) ∈ F[x]$ such that $f(d) = 0$, show that there exists $d^{−1} ∈ D$.

I know that $D ⊃ F$, and $F$ is a field with respect to the restrictions of the operations of $D$, but I don't know how to follow the proof..

答:
Let $f(x)=a_nx^n+\ldots+a_1x+a_0$ (where all $a_i$'s are from $F$) be such that $f(d)=0$, and $f$ is the smallest degree polynomial from $F[x]$ with that property. This means that:

$$a_nd^n+\ldots+a_1d+a_0=0$$

Now:

If $a_0=0$, this means that already $(a_nd^{n-1}+\ldots+a_1)d=0$, but as $d\ne 0$ and $D$ is an integral domain, we conclude $a_nd^{n-1}+\ldots+a_1=0$, i.e. there is a smaller degree polynomial $g(x)=a_nx^{n-1}+\ldots+a_1$ such that $g(d)=0$ - a contradiction.

If $a_0\ne 0$, then there is $a_0^{-1}\in F$, so you can now calculate that:

$$-a_0^{-1}(a_nd^{n-1}+\ldots+a_1)d=1$$

which means $d$ is invertible and $d^{-1}=-a_0^{-1}(a_nd^{n-1}+\ldots+a_1)$.


答:
Using the Lagrange remainder,
$$ \tag1g(x)=\frac {x^4}4\cos \zeta\ge 0\qquad \text{for }|x|\le\frac\pi2$$
because that also make $|\zeta|<\frac\pi2$ and hence $\cos\zeta>0$.
Also, we trivially have
$$ g(x)\ge-2+\frac12x^2=\frac{x^2-4}2\ge 0\qquad\text{for }|x|\ge 2.$$
Hence we are only left with $\frac \pi2 <|x|<2$.
But for these we already know from $(1)$ that $\cos \frac x2\ge 1-\frac18x^2>0$, so
$$\cos x=2\cos^2\frac x2-1\ge 2\left(1-\frac18x^2\right)^2-1=1-\frac12x^2+\frac1{32}x^4\ge 1-\frac12x^2.$$

Title: $ \cos x\geq 1-\frac{x^2}{2} $


问:
If $f(x)=x^2+ax+b$ has integer roots, and $f(x+\frac1x)=f(x)+f(\frac1x)$, then find $a^2+b^2$

Let $f(x) = x^2 + ax + b$. If for all non-zero real $x$
$$f\left(x + \frac1x\right) = f(x) + f\left(\frac1x\right)$$
and the roots of $f(x) = 0$ are integers, what is the value of $a^2 + b^2$?

Here is the photo of the question if you are finding difficult to look at it here

EDIT: The answer must be in two-digits

EDIT: What I have done so far is that I added temporary values to the equation
$$f(x) = 0 = x^2 + ax + b$$ and tried to solve it in the form of $a + b$ to which then I squared and solved and got,
$$a^2 + b^2 = x^2 - 2ab$$

I don't really know what I am doing is correct, so help me out please!

答:
Hints:

Try say $x=1$ with the condition given, to immediately get $b$.

Use the fact that roots are integers and factors of $b$ to find possible values of $a$.

Even though there is more than one possible value for $a$, you should find a unique value for $a^2+b^2$.


答:
We don't need L'Hospital's Rule.  Rather, using the inequality $e^x >x$ for all $x$ we see that

$$\left|\frac{e^{-1/{x^2}}}{\sin(x)}\right|\le \left|\frac{x^2}{\sin(x)}\right|$$

Can you finish?

Title: What's the result of $\lim_{x\to 0} \dfrac{e^{-\frac{1}{x^2}}}{\sin x}$?

Body:
During a correction of an exercise in a differentiability class this limit showed up and the teacher said that it could be done with L'Hopital easily, but me and a couple of friends tried doing it and neither of us got it. How can it be solved? I know the limit is 0 thanks to Geogebra.

$$ \lim_{x\to 0} \dfrac{e^{-\frac{1}{x^2}}}{\sin x}$$


问:
Are there any non-trivial automorphisms on $(\mathbb{N}, \times)$
Obviously the identity is an automorphism. But any time other bijection I can think of does not preserve multiplication. Is the identity the only automorphism on $(\mathbb{N}, \times)$?

答:
Pick any permutation $f\colon \Bbb P\to\Bbb P$ of the set $\Bbb P$ of all primes and define
$$\phi(\prod_i p_i^{a_i}):=\prod_i f(p_i)^{a_i}. $$


问:
How can $t \delta(t) = 0$ be proven using its defining propery?
How can $t\delta(t)=0$ be proven using it's defining principle, $\int f(t) \delta(t-a) dt = f(a)$?

I have a "proof" so far but I know its not even close to rigorous or proper.

$\int f(t)\delta(t-a)dt = f(a)$

Let $f(t) = t, a = 0$:

$\int t\delta(t)dt=f(0) = 0$

Differentiate both sides:

$\frac{d}{dt} \int t\delta(t)dt = \frac{d}{dt}0$

$\Longrightarrow t\delta(t)=0$

What would be a more formal proof?

答:
This proof uses definition of convolution and Dirac delta and properties of integrals.

Define convolution of functions by
$$ h=f * g \quad \text{ iff } \quad h(x) = \int f(t)\,g(x-t)\,dt.\tag{1} $$
The identity property of the Dirac delta $\,\delta(x)\,$ is
that $\, f = f * \delta\,$ for all $\,f.$

Now given any function $\,f,\,$ define
$$ F(x) := x\,f(x),\quad
g(x) := x\,\delta(x)\quad \tag{2}$$  and $\, h=f*g \,$ as in
equation $(1)$.

By definition of convolution we have
$$ h(x) = \int f(t)\,(x-t)\,\delta(x-t)\,dt, \tag{3} $$
$$ h(x) = \int f(t)\,x\,\delta(x-t)\,dt - \int f(t)\,t\,\delta(x-t)\,dt,
\tag{4} $$
$$ h(x) = x \int f(t)\,\delta(x-t)\,dt - \int F(t)\,\delta(x-t)\,
dt. \tag{5} $$
Thus $\, h(x) = x\,f(x) - F(x)\,$ because $\,\delta\,$ is the identity of
convolution. This implies $\,h(x)=0\,$ by definition of $\,F.\,$ Thus,
we have proved that $\,x\,\delta(x)=0.$


问:
$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$
Prove:

$$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$$

where $H_n=\sum\limits_{k=1}^n\frac1k$

For the $x^k$ i tried  $x^k=((x-1)+1)^k$ and decompose it into binomial expansion but I got dual sum at the one side one sum at the other side

For the second method:

$$\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}-\sum\limits_{k=1}^n \frac{x^k}{k}=H_n$$

and try to show

$$\binom{n}{k}(x-1)^k-x^k=1$$

Either the problem is wrong or I cannot do it.

答:
This identity does hold true

express both the RHS and LHS as integrals

then interchange the integrals and summations.

Hope this helps


问:
Spivak chapter 5 exercise 36 - prove $\lim_{x \to 0^-} f(1/x) = \lim_{x \to -\infty} f(x)$
There're similar questions already, but I somewhat struggled to apply their reasoning to this particular statement.

I wanted to ask if my proof is correct (and clarify several things which I've seemingly figured out when writing it here). Here's the proof:

$$\mbox{(1) }\lim_{x \to 0^-} f\left(\frac1x\right)=l \mbox{ if } \forall\varepsilon>0(\exists\delta>0(\forall x(0<0-x<\delta \longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$

$$\mbox{(2) }\lim_{x \to -\infty} f(x)=m \mbox{ if } \forall\varepsilon>0(\exists N(\forall x(x<N \longrightarrow|f(x)-m|<\varepsilon$$

We need to show that $l=m$.

from (1) we have
$$-\delta>x>0\longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$

let $g(x)=f\left(\frac1x\right)$, then we have
$$-\delta>x>0\longrightarrow|g\left(x\right)-l|<\varepsilon$$

in (2) we suppose that $x<N$. Hence $\frac1x>\frac1N$. Let $x'=\frac1x$. Furthermore, since $-\delta$ is negative and we can assume that $N$ is negative. (If it's not, we can take $N$ to be -1 or any other negative number, because if the conclusion holds $\forall x<N$, it surely holds for a subset $(-\infty,-1)$). We can take $-\delta=\frac1N$, which gives us

$$\frac1N=-\delta<x'<0$$ and this is the hypothesis from (1). Hence we conclude that

$$|g\left(x'\right)-l|<\varepsilon$$ or $$|f\left(\frac1{\frac1x}\right)-l|<\varepsilon$$ or $$|f\left(x\right)-l|<\varepsilon$$

To summarise, $x<N\longrightarrow|f\left(x\right)-l|<\varepsilon$. By the definition of the limit $$\lim_{x \to -\infty}f(x)=l$$ but also $$\lim_{x \to -\infty}f(x)=m$$ Therefore $l=m$.

答:
WA Don nailed it, but I feel compelled to add an answer.

I'm going through Spivak myself and when first encountering problems like this, I tried approaches very similar to super.t's.

It took a while to absorb what was going on. Initially I found it a bit confusing trying to keep straight which conditions are fixed, which can vary, and what implies what.

Here I'll rework the problem, annotating with a few comments. The aim is to hopefully help clarify the thinking behind the approach. Apologies for stating things that may seem obvious to you, and generally for over-explaining. Hopefully this is of some use to someone, someday.

Show that
$$\lim_{x \to 0^-} f\left(\frac1x\right)= \lim_{x \to -\infty} f(x)$$

We can see intuitively that this makes sense: as $x$ gets very small negative on the left hand side, $1/x$ becomes very large negative, and the 2 sides both look like $f$ of some very large negative number.

Let's assume that the first limit exists, and is equal to $\ell$

$$\lim_{x \to 0^-} f\left(\frac1x\right)= \ell$$

If this is true, then from the definition of the limit at $0$ from below, we have: for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$, if

$$0 < -x < \delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$

or, multiplying the $\delta$ expression by $-1$, we have for all $x$, if
$$0 > x > -\delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$

Let's pause for a moment and consider what this says. Here, $x$ is just some number. This says that, if we have some number between $0$ and $-\delta$, then $f(\frac{1}{\text{number}})$ will be "sufficiently close" to $\ell$.

Let's try sticking $1/y$ in as "number". The motivation for this choice is that we have an expression involving $\left\vert f\left(\frac{1}{x}\right) - \ell \right\vert$ and we'd like to instead have an expression for $f(x)$. Sticking $1/y$ into the above expression for $x$ we have

$$\text{If } 0 > 1/y > -\delta \text{ then } \left\vert f\left(\frac{1}{(1/y)}\right) - \ell \right\vert < \varepsilon$$
or
$$\left\vert f(y) - \ell \right\vert < \varepsilon$$

Again, the original $x$ expression was true for all numbers $x$ that satisfied the $\delta$ condition on the left hand side, so this new expression for $f(y)$ is true for all numbers $1/y$ that satisfy $0 > 1/y > -\delta$.

Now if $0 > 1/y > -\delta$, what does this tell us about $y$?

Multiplying by $-1$ to make everything nonegative, to avoid any sign confusion:

$$0< -1/y < \delta$$
Inverting
$$0<1/\delta<-y$$

Finally, multiplying again by $-1$ we have
$$0 > -1/\delta > y$$

Let's summarize what this all implies:

We know that if the first limit exists, then for any $\varepsilon >0$ there exists some $\delta>0$ such that for all $y$ if
$$y < -1/\delta \text{ then } \left\vert f(y) - \ell \right\vert < \varepsilon$$

Or, substituting $N = -1/\delta$ we have, for any $\varepsilon >0$ there exists some $N<0$ such that for all $y$, if
$$y < N \text{ then } \left\vert f(y) - \ell \right\vert < \varepsilon$$

Another way of writing this is:
$$\lim_{y \to -\infty} f(y) = \ell$$

Thus, if the first limit exists, then so too does the second, and they are equal.

A key point here is the equivalence of the conditions $y < -1/\delta < 0$ and $0 > 1/y > -\delta$. The first implies the second and vice versa.

Finally to complete the proof, we can begin with the second limit and similarly show that if it exists, then so too does the first, and they are equal.

The steps will be very similar to what we've already done.

Alternate justification for substituting in $y = 1/x$

Going back to the beginning, we see that the first limit involves $f(1/x)$. What do the $\delta$-restrictions tell us about $1/x$?

If the first limit is $\ell$ then, for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$ if
$$0 < -x < \delta, \text{then} \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$

If
$$0 < -x < \delta$$
then
$$0 < 1/\delta <\frac{1}{-x}$$ or
$$0 > -1/\delta >\frac{1}{x}$$

Thus, for any $\varepsilon > 0$ there exists some $N = -1/\delta < 0$ such that for all $x$, if
$$\frac{1}{x}< N \text{ then } \left\vert f\left(\frac{1}{x}\right) - \ell \right\vert < \varepsilon$$

We can see here we're almost to the second limit. We just need to replace the number $\frac{1}{x}$ with $x$ (or $y$, or any other name).


问:
On an upper bound of the diagonal entries of an orthogonal projection
Let $X$ be an $n\times p$ matrix of full column rank and $H=X(X^TX)^{-1}X^T$ be the orthogonal projection onto the column space of $X$. If $X$ has $c$ identical rows for some $c\leq n$, I want to prove that $h_{ii} \leq c^{-1}$ whenever $i$ is among the row indices of the $c$ identical rows.

I guess I need to start with the fact that correspoding rows in $H$ are also identical, but I don't really know how to prove that. Then it would be suitable to make use of the fact that $H$ is symmetrical and idempotent, so:
$$h_{ii}=\sum_{j=1}^nh_{ij}^2\ .$$
But I can't really make any progress from there. Could you give me a hint?

答:
Assume that the first $c$ rows of $X$ are identical to each other. Then we can express $X$ in the form of
$$
X=\pmatrix{ev^T\\ Y},
$$
where $e$ is the all-one vector of length $c$. Hence
$$
H=X(X^TX)^{-1}X^T=\pmatrix{ev^T\\ Y}(cvv^T+Y^TY)^{-1}\pmatrix{ve^T&Y^T}.
$$
Therefore, for each $i\le c$, we have
\begin{align}
h_{ii}
&=v^T(cvv^T+Y^TY)^{-1}v\\
&=\lim_{\varepsilon\to0^+}v^T(\varepsilon I_p+cvv^T+Y^TY)^{-1}v\\
&\le\lim_{\varepsilon\to0^+}v^T(\varepsilon I_p+cvv^T)^{-1}v\\
&=\lim_{\varepsilon\to0^+}\frac{\|v\|^2}{\varepsilon+c\|v\|^2}\\
&\le c^{-1}.
\end{align}
(I use an inequality sign instead of an equality sign on last line because because $v$ can be zero.)


问:
What is the constant term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ ? and the middle term in it?
I want to determine the constant term and middle term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$, I know that the constant term in the expansion of $(x+a)^n$ is always $a^n $, We can see that the general term becomes constant when the exponent of variable $x$ is $0$ ,Therefore the condition for the constant term is $n-2k=0\implies k=\frac n2$, Really I can't use that fact to determine the constant term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ since both terms depend of $x$ however if we assume that $y=\dfrac 1x  $ this means that y can be expressed as a ratio using the variable $x $ , In the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ the constant term according to its definition in the expansion of $(x+a)^n$ would be $-\dfrac{1287}{x^4}$ or $x^{26}$ the latter mixed me then What is the constant term in the expansion ${(x^2-\dfrac{1}{x^4})}^{13}$ ? and the middle term in it ?

Body:
$$\binom{13}{r}(x^2)^{13-r}\times (-x^{-4})^{r}=\binom{13}{r}x^0$$

$$26-2r-4r=0$$

$$3r=13$$

$r \notin Z$, there is no constant term.


问:
Let $X$ and $Y$.A function $f: X \longrightarrow Y$ is continous if and only if $f^{-1}(C)$ is closed in $C$ for every closed set $C \subset Y$

4.2 Let $X$ and $Y$ topological spaces.A function $f: X \longrightarrow Y$ is continous if and only if $f^{-1}(C)$ is closed in $C$ for every closed set $C \subset Y$

答:
Remark:

Theorem1.2:  Let $f: A \longrightarrow B$, $Y \subset B$ then $f^{-1}(B^{c})=(f^{-1}(B))^{c}$

proof

Let $x\in f^{-1}(B^{c})$ then $f(x)\in B^{c}$, then $f(x)\not \in B$ and therefore $f(x)\not \in f^{-1}(B) \therefore x \in (f^{-1}(B))^{c}$.\
Conversely let $x\in (f^{-1}(B))^{c}$ by definition of pre image $f(x)\in (f^{-1}(B))$ then $f(x) \not \in B$ and $f(x)\in B^{c} \therefore x\in f^{-1}(B^{c})$.

Proof of theorem1.2

Suppose $f: X \longrightarrow Y$ is continuous, by definition consider $W$ open in $Y$ then $f^{-1}(W)=V$ is open in $X$, consider $W/Y$ is close in $Y$ and notice that $f^{-1}(W/Y)=f^{-1}(W)/X=V/X$ (by Remark) which is closed in $X$.

Conversely suppose $C$ is arbitrary closed set in $Y$ and $f^{-1}(C)$ closed set in $X$.\

Since $C$ is closed in $Y$ then $W=C/Y$ is open in $Y$ and $f^{-1}(W)=f^{-1}(C/Y)=f^{-1}(C)/X=V$ which is open in $X$ since $f^{-1}(C)$ is closed in $X$  then their complement is open. Since $C$ was arbitrary closed in $Y$ $f$ is a continuous function.


问:
Proving that $7n+3$ is prime infinitely often?
The Question is:
"Prove that there are infinite values of natural $n$ for which $7n+3$ is prime.

Edit: I have realised that this is a direct consequence of Dirichlet's Theorem, but the proof of this theorem is beyond me at the moment.

Is there any simple/elementary way to prove this statement?

答:
An elementary proof along the lines of Euclid's proof exists for the arithmetic progression $a \bmod n$ iff $a^2 \equiv 1 \bmod n$. Since $3^2 \not\equiv 1 \bmod 7$, there is no such proof.

Here are some references:

Primes in Certain Arithmetic Progressions by Murty and Thain.

How I discovered Euclidean proofs by Murty.

Euclidean proofs of Dirichlet's theorem by Keith Conrad.


问:
Why does Rudin say $V_\alpha \subset G_\alpha$?
I have some confusion about something in Baby Rudin.

My doubt is marked below in the red box.

My  doubt: Why does Rudin say that $V_\alpha \subset G_\alpha$?

My attempt:  Here  both $V_\alpha$  and $G_\alpha$  are open covers of $K$. So it must be that $V_\alpha = G_\alpha$.

Both $V_\alpha$ and $G_\alpha$ are open covers that contain finite subcovers.  Why do we not have $V_\alpha = G_\alpha$?

答: Note that in that portion of the proof Rudin defines $\lbrace G_{\alpha}\rbrace \subset X$ as an open cover of $K$, and defines $V_{\alpha} = G_{\alpha}\cap Y$.  Then, since $V_{\alpha}$ is only the part of $G_{\alpha}$ that is also in $Y$, we can't conclude that $V_{\alpha} = G_{\alpha}$, but we can conclude that $V_{\alpha}\subset G_{\alpha}$.


答: Maybe an example will help illustrate. Imagine the $x$-axis $Y = \mathbb R\times\{0\}$ inside $X = \mathbb R^2$. The interval $K = [0,1]\times\{0\}$ inside $Y$ is compact relative to $Y$. The interval $K$ can be covered by sets of the form $G_\alpha$, where for every $\alpha$, $G_\alpha$ is an open ball in $\mathbb R^2$. Then put $V_\alpha = Y\cap G_\alpha$. The $V_\alpha$ are all open intervals contained in the $x$-axis $Y$, so they are not themselves equal to the balls $G_\alpha$.

Title: Why does Rudin say $V_\alpha \subset G_\alpha$?

Body:
I have some confusion about something in Baby Rudin.

My doubt is marked below in the red box.

My  doubt: Why does Rudin say that $V_\alpha \subset G_\alpha$?

My attempt:  Here  both $V_\alpha$  and $G_\alpha$  are open covers of $K$. So it must be that $V_\alpha = G_\alpha$.

Both $V_\alpha$ and $G_\alpha$ are open covers that contain finite subcovers.  Why do we not have $V_\alpha = G_\alpha$?


问:
Show that a matrix is positive definite
Let $U$ be a symmetric,positive definite matrix and let $Q \ne I$ be an orthogonal matrix s.t. $\det(Q)>0$. I want to show $$(Q-I)^tU(Q-I)$$ is positive definite

I want to show that $$\langle (Q-I)^tU(Q-I) v,v\rangle \geq 0$$ for every $v \in \mathbb{R}^n$ and equality holds iff $v=0$.

So I wrote $$\langle (Q-I)^tU(Q-I) v,v\rangle = \langle U(Q-I) v,(Q-I)^ v\rangle $$

Let's set $w = (Q-I) v$, then I have

$$\langle Uw,w \rangle \geq 0$$

for every $w \in \mathbb{R^n}$ since $U$ is positive definite and equality holds iff $$w=0$$

Now, $w=0$ iff $Qv = v$

but this does not imply that $v=0$... I don't know how to move now

答:
Take
$$
Q = \begin{pmatrix}
0 & 1 & 0\\
-1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}
$$
and $U=I$ as a counterexample.

Title: Show that a matrix is positive definite


问:
Weak convergence of probability measures.
Theorem. Let $E$ be a metric space. Let $\{\mathbb{P}_n\}_{n\ge 1}$ a probability measure sequence. If $\{\mathbb{P}_n\}_{n\ge 1}$ weakly converges to the probability measure $\mathbb{P}$ in $(E,\mathcal{B}(E))$, then $$\limsup_n \mathbb{P}_n(A)\le \mathbb{P}(A),$$ where $A\in\mathcal{B}(E)$ is a closed set.

Proof. Let $\rho$ be a metric on $E$. Define $$f_A^{(\varepsilon)}(x):=\bigg(1-\frac{\rho(x,A)}{\varepsilon}\bigg)\vee 0\quad\forall\varepsilon >0$$ and $$A_\varepsilon:=\text{supp}f_A^{(\varepsilon)}=\{x\in E\;|\;f_A^{\varepsilon}(x)>0\}=\{x\in E\;|\;\rho(x,A)<\varepsilon\}.$$ Therefore, $$\mathbb{P}_n(A)=\int_E\mathbb{P}_n(dx)\chi_A(x)\le\int_E\mathbb{P}_n(dx)f_A^{(\varepsilon)}(x).$$ Then we have $$\limsup_n\mathbb{P}_n(A)\le\limsup_n\int_E\mathbb{P}_n(dx)f_A^{(\varepsilon)}(x)=\int_E\mathbb{P}(dx)f_A^{(\varepsilon)}(x)\le\mathbb{P}(A_\varepsilon)\xrightarrow[\varepsilon\downarrow0]{}\mathbb{P}(A).$$

Question.

Where did we use the fact that $A$ is a closed set? What if it isn't?

Why $\mathbb{P}(A_\varepsilon)\xrightarrow[\varepsilon\downarrow0]{}\mathbb{P}(A)$

答:
Take $\mathbb{P}_n=\delta_{1/n},$ note that $\mathbb{P}_n\to \delta_0$ weakly. Let $A=(0, \infty).$ It is clear that $\mathbb{P}_n(A)=1$ for all $n,$ and $\delta_0(A)=0.$ This shows that the conclusion is false if $A$ is not closed.

If you go through your proof, nothing stops you from defining $f^{\epsilon}_A.$ And $\chi_{A}\le f^{\epsilon}_A$ holds no matter what set $A$ you choose. But what property of $f^{\epsilon}_A$ is required to get $\limsup\int d\mathbb{P}_nf^{\epsilon}_A = \int d\mathbb{P}f^{\epsilon}_A$? Does that property hold if $A$ is open?

For your second question, note that $A\subseteq A_{\epsilon}$ for all $\epsilon.$ And, $A_{\epsilon_1}\subseteq A_{\epsilon_2}$ if $\epsilon_1\ge \epsilon_2.$ That is, $A_{\epsilon}$ decreases to $A.$ Now, you can use a version of MCT to obtain $\mathbb{P}(A_{\epsilon})\to \mathbb{P}(A).$ (And, in fact this also fails if $A$ is open.)


问:
Figuring out notation of random variables

Hi, I'm currently working on probability and statistics question and have become a little stuck on understanding this question.

Could someone explain to me what an N(3,4) variable is as I haven't come across it in my notes at all.

Thanks!

答:
Note that if $Y = 5X - 4,$ then $Y > 18.5$ if and only if
$5X - 4 > 18.5$, which in turn is true if and only if $X > \frac{18.5 + 4}{5} = 4.5.$

So really you are just being asked to find the value of $P(X > 4.5).$
You don't actually need to figure out the distribution of $Y,$
although for other purposes it is good to know how to do that.


问:
Sum of a non arithmetic nor geometric sequence
Express the following sum as a common fraction:
$\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}$

This is from a timed competition, so fastest answers are best.
I tried rearranging it to
$$\frac{7\cdot10\cdot13\cdot...\cdot100+1\cdot10\cdot13\cdot...\cdot100+...+1\cdot4\cdot7\cdot...\cdot 94}{1\cdot4\cdot7\cdot10\cdot13\cdot...\cdot100}$$but I don't know what to do after this. Any help?

Body:
Write the sum as $$\begin{align}
\sum_{k=0}^{32}\frac1{3k+1}\frac1{3k+4}&
=\frac13\sum_{k=0}^{32}\left(\frac1{3k+1}-\frac1{3k+4}\right)\\
&=\frac13\left(1-\frac1{100}\right)\\
&=\frac{33}{100}
\end{align}$$


Title: Sum of a non arithmetic nor geometric sequence

Body:
Express the following sum as a common fraction:
$\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}$

This is from a timed competition, so fastest answers are best.
I tried rearranging it to
$$\frac{7\cdot10\cdot13\cdot...\cdot100+1\cdot10\cdot13\cdot...\cdot100+...+1\cdot4\cdot7\cdot...\cdot 94}{1\cdot4\cdot7\cdot10\cdot13\cdot...\cdot100}$$but I don't know what to do after this. Any help?


问:
How to show that $u \in W^{1,p}(\Omega)$
I have the following question, please help me.

Let $1 < p \leq \infty$ and $u \in L^p(\Omega)$, $\Omega \subset \mathbb{R}^n$ an open set. Show that if there exists a constant $C >0$ such that
\begin{equation}
\left|\,\int\limits_{\Omega} u \frac{\partial \phi}{\partial x_j}\mathrm{d}x\right| \leq C|\phi|_{0,q,\Omega}, \label{1} \tag{1}
\end{equation}
for every $\phi \in \mathcal{D}(\Omega)$ and $1 \leq j \leq n$, $q$  the conjugate exponent of $p$, then $u \in W^{1,p}(\Omega)$.

My work: to prove that  $u \in W^{1,p}(\Omega)$ we need to show that $u \in L^p(\Omega)$ which is already given in the question, and we also need to show that $\frac{\partial u}{\partial x_j} \in L^p(\Omega)$. I tried to apply Holder's inequality with the aid of equation \eqref{1} but I am not sure how to work with $\phi$ and eliminate it from the calculations.

Any help is appreciated!

答: Note that $\int u \phi_j = - \langle u_j, \phi\rangle_{D' \times D}$ where $u_j \in D'$ is the distributional derivative of $u$. You know $|\langle u_j, \phi \rangle_{D' \times D}| \leq C|\phi|_{L^q}$ for all $\phi \in D \subset L^q$, which means $\sup_{\phi \in L^q} \frac{|\langle u_j, \phi \rangle_{D' \times D}|}{|\phi|_{L^q}} \leq C$. Now, note that this is just the norm of $(L^q)^*=L^p$. Hence $u_j \in L^p$.


问:
Embeddings of $\mathrm{GL}(n-1,q)$ into $\mathrm{GL}(n,q)$
Let $n>2$ and $q$ be a prime power. I want to find all the embeddings of $\mathrm{GL}(n-1,q)$ into $\mathrm{GL}(n,q)$. An embedding I first thought of is as following. Let $V$ be an $n$-dimensional vector space over $\mathrm{GF}(q)$ and $H$ is a subgroup of $\mathrm{GL}(V)$ fixing a nonzero vector in $V$. Then $H$ is isomorphic to $\mathrm{GL}(n-1,q)$.

答:
So the question should be: if $H$ is a subgroup of ${\rm GL}(n,q)$ isomorphic to ${\rm GL}(n-1,q)$, then does $H$ necessarily fix two complementary subspaces of $V$ of dimensions $1$ and $n-1$?

This is a nontrivial question. Let $K$ be the field of order $q$. Then the only simple $KH$-modules of dimension less than $n$ have dimensions 1 and $n-1$, so the constituents of $V$ as $KH$-module must have dimensions $1$ and $n-1$, but then the question is whether this module is completely reducible, and that depends on the 1-cohomology of $H$ on the $(n-1)$-dimensional constituent.

There are some instances in which there are embeddings that are not completely reducible, so the answer to the question is no. These include $n=3$, $q=2^k$ for $k \ge 2$, and also $n=4$, $q=2$. I think the answer is yes in all other cases, but I would not bet on it!

Added later: here is a rough description of the non-natural examples when $n=3$ and $q=2^k$. Consider first the subgroup of matrices of the form

$$\left( \begin{array}{ccc}a&b&0\\c&d&0\\x&y&1\end{array} \right).$$

in which $ab-cd=1$. This is a semidirect product of $K^2$ with ${\rm SL}(2,q)$ with the natural action. It turns out that $H^1({\rm SL}(2,2^k),K^2)$ is nonzero, and has dimension 1 over $K$, when $k \ge 2$.

Let $C \cong {\rm SL}(2,q)$ be the a complement of $K^2$ corresponding to a nonzero element of $H^1({\rm SL}(2,q),K^2)$. Then $C$ does not act completely reducibly on $V$, and the subgroup $H$ gerenrated by $C$ and scalar matrices is isomorphic to ${\rm GL}(2,q)$. Then (depending on whether your matrices act on the left or right) $H$ fixes a 1-dimensional or a 2-dimensional subspace of $V$, but not both, and the transpose of $H$ does the opposite. So there are just two conjugacy classes of non-natural ${\rm GL}(2,q)$s in ${\rm GL}(3,q)$.


问:
Is $\nleq$ equivalent to $>$?
This seems to be the case, but if so, how can we write that an inequality is not true in general.

For example, what if I want to say, $f(x,a)$ is not always less than $f(x,b)$ for all $x$ where $a$ and $b$ are constants. In other words, $f(x,a) \leq f(x,b)$ is a false statement.

Basically, I have a hypothesis that ended up being false, and I am deciding to how write the result.

答:
When it comes to two real numbers $a$ and $b$: $a \not <b$ iff $a \geq b$

But, your

$f(x,a)$ is not always less than $f(x,b)$

is not equivalent to (in case that that is what you're thinking):

$f(x,a)$ is always greater or equal to $f(x,b)$

Rather, it is equivalent to:

$f(x,a)$ is sometimes greater or equal to $f(x,b)$

To be a little more formal:

I interpret your:

$f(x,a)$ is not always less than $f(x,b)$

as:

it is not the case that $f(x,a)$ is always less than $f(x,b)$

and that as:

it is not the case that for all $x$: $f(x,a)$ is less than $f(x,b)$

which is equivalent to:

for some $x$, it is not the case that $f(x,a)$ is less than $f(x,b)$

which by the basic equivalence between $\not <$ and $\geq$ is equivalent to:

for some $x$, $f(x,a)$ is greater or equal to $f(x,b)$

which you can read as:

$f(x,a)$ is sometimes greater or equal to $f(x,b)$

But this is not the same as:

for all $x$, $f(x,a)$ is greater or equal to $f(x,b)$

which you can read as:

$f(x,a)$ is always greater or equal to $f(x,b)$

Title: Is $\nleq$ equivalent to $>$?


问:
Proving convegence of sequence
How do I prove

$w_{k+1} = \frac{-(2 (3 - 4 e + 2 e^2) g )}{(-5 + 4 e)} \times w_k$

converges to $0$ over $k\rightarrow \infty$ and say $w_0=1$

Possible bounds are: $0<e<1$ and $0<g<1$.

Also $e$ is just another variable, not Euler number

答:
For $g=0.9$ and $e=0.1$, then $$\left| \frac{-(2 (3 - 4 e + 2 e^2) g )}{(-5 + 4 e)}\right| >1$$

So the sequence diverges.

Title: Proving convegence of sequence


问:
Proving a space is not Banach in $\ell^1$
I am aiming to prove that with the norm on $\ell^{1}$ given by $\left\|\left(a_{1}, a_{2}, \ldots\right)\right\|=\sup \left\{\left|a_{n}\right|: n \geq 1\right\}$, $\left(\ell^{1},\|\cdot\|\right)$ is not a Banach space.

So I need to prove that every Cauchy sequence in $\left(\ell^{1},\|\cdot\|\right)$ converges to some element of $\left(\ell^{1},\|\cdot\|\right)$.

I tried by starting with a sequence $(x^{(n)})_{n=1}^{\infty}$ with $x^{n}=(0,0,\dots,1/n,0,0,\dots)$ where the value $1/n$ is the $n$-th element. But, i didn't manage to get anywhere with this. Any help appreciated. Thanks

答:
Take $x^n=(1,1/2,\ldots,1/n,0,\ldots)$.

FIRST: This sequence is $\|\cdot\|$-Cauchy.

Indeed, assuming $n>k$, we have that
$$\|x^n-x^k\|=\|(0,\ldots,0,1/(k+1),\ldots,1/n,0,\ldots)\|=$$ $$=\sup\{1/j,\ k+1\le j\le n\}=\frac{1}{k+1}\to0\ (k\to\infty).
$$

SECOND: We can prove that $x^n\overset{\|\cdot\|}{\longrightarrow} x:=(1,1/2,1/3,\ldots)$.

Indeed $\|x^n-x\|=\|(0,\ldots,0,1/(n+1),1/(n+2),\ldots)\| =\sup_{j>n}\frac{1}{j}=\frac{1}{n+1}\to0\ (n\to\infty)$.

THIRD It is clear that $x^n\in\ell^1$ but $x\notin\ell^1$.

So you have a $\|\cdot\|$-Cauchy sequence $(x^n)_n\subset\ell^1$ that is $\|\cdot\|$-convergent to something not belonging to $\ell^1$.


问:
On the definition of the structure sheaf attached to $Spec A$
Let $A$ be a ring (commutative with $1$),if $X=Spec A$ we want to attach a sheaf of rings to $X$. If  $f\in A$, $D(f)=X\setminus V(f)$ is an element of the base and we define
$$\mathcal O_X(D(f)):=A_f$$

I've problems to show that $\mathcal O_X(\cdot)$ is well defined on the set $\mathcal B=\{D(f)\,:\, f\in A \}$. If $D(f)=D(g)$, then it is easy to show that $A_f\cong A_g$, but these two rings are not the same, so we have that $\mathcal O_X(D(f))\neq \mathcal O_X(D(g))$ even if they are isomorphic. Textbooks say "we can identify  $A_f$ with $ A_g$" but what they exactly mean with the word "identify"?

答:
Since $A_f,A_g$ are localizations of $A$, there is almost one homomorphism of $A$-algebras between $A_f$ and $A_g$. Therefore all identifications between them are compatible with each other. So there is no harm to identify $A_f$ with $A_g$ when $D(f)=D(g)$.

However, I always wonder why this approach to the structure sheaf is so popular. As you observe, it is not clear at all why this is well-defined in the usual sense of this word. And it is also quite clumsy! Instead, you can do the following:

1) Define $\mathcal{O}_X(U)$ to be the set of all functions $s$ on $U$ with $s(\mathfrak{p}) \in A_{\mathfrak{p}}$, such that locally $s$ is equal to some fraction in $A_f$. Details can be found in Hartshorne's book, for example.

2) Let $S_U = A \setminus \cup_{\mathfrak{p} \in U} \mathfrak{p}$ and $\mathcal{O}'(U) = A[S_U^{-1}]$. Then $\mathcal{O}'$ is a presheaf of rings, and define $\mathcal{O}$ to be the associated sheaf. Details can be found in MO/80548.


问:
Expected value of 3 uniform random variables
You got $3$ random variables (iid), uniformly distributed between $\left[0, 1\right]$.
$$
\mbox{What is}\,\,\, \mathbb{E}\left[\max\left\{X1, X2, X3\right\}
\min\left\{X1,X2, X3\right\}\right]\ {\large ?}.
$$

Attempt:

I found the link: Expected value of Max times Min of 2 uniform random variables. Can we extend this to $3$ variables $?$.

I know the pdf of $\max\left\{X1, X2, X3\right\}$ and $\min\left\{X1, X2, X3\right\}$. But the expected value can't by factor out because these two quantities are not independent.

Do I have to go through the expected value calculation through the pdfs $?$.

答:
You're absolutely right about being cautious regarding multiplying non-independent PDFs. However, what you can instead do is write down the joint distribution of the order statistics (in this case the first order statistic is the minimum of your three variables and the third order statistic is the max).

The joint PDF is surprisingly simple in this case $f_{\min(X_1,X_2,X_3),\max(X_1,X_2,X_3)}(u,v)=3!(v-u)$, it's essentially given by the number of possible permutations (that's the $3!$ term) times the probability that the "middle" variable is somewhere between $u$ and $v$. Check out https://en.wikipedia.org/wiki/Order_statistic#The_joint_distribution_of_the_order_statistics_of_the_uniform_distribution for more details.

From this, one approach could be to compute the expected value using multivariable calculus $$E[\min(X_1,X_2,X_3) \max(X_1,X_2,X_3)] = 3! \int_{0<u<v<1} u v (v-u) dvdu = 3! \int_0^1 \left(\int_u^1 u v (v-u) dv\right)du.$$


问:
Size of the least common multiple
Suppose $k \in \mathbb{N}$ is fixed and that $n, n_1, \dots, n_k \in \mathbb{N}$ are given such that $n = n_1 + \dots + n_k$.

Are there any known upper bounds for $\text{lcm}(n_1, \dots, n_k)$ in terms of n, that take $k$ into account as well?

I care up to coarse equivalence. Given two non-decreasing functions $f, g \colon \mathbb{N} \to \mathbb{N}$, we say that $g$ coarsely dominates $f$, write $f \preceq g$, if there exists $C>0$ such that $f \leq C g(Cn)$ for all $n \in \mathbb{N}$. We say that $f,g$ are coarsely equivalent, write $f \approx g$ if $f \preceq g$ and $g \preceq f$. Note that this notion is really coarse, for example $n2^n \preceq 2^n$.

EDIT:Landau's function does work, but it is rather general, as it allows any number of summands.

Ideally, for each $k \geq 1$ I would like a to have a function $B_k(n)$ given by a closed form formula that coarsely dominates $\text{lcm}(n_1, \dots, n_k)$ and is tight up to coarse equivalence. For example, $B_1(n) = n$, $B_2(n) = n^2$.

答:
With $k$ fixed, an asymptotically tight upper bound is
$$B_k(n) = \frac{n^k}{k^k}\,.$$
First, it is immediate that $\operatorname{lcm}(n_1, \dotsc, n_k) \leqslant n_1\cdot \ldots \cdot n_k$, and
$$\prod_{j = 1}^{k} x_j$$
with the constraints $x_j \geqslant 0$ and $\sum x_j = n$ is maximised for $x_j = \frac{n}{k}$, $1 \leqslant j \leqslant k$. Thus $B_k(n)$ as given above is an upper bound.

For $k = 1$ it is trivial that this is not only asymptotically tight, but in fact an equality. Thus consider $k \geqslant 2$. Let $\varepsilon > 0$ be small. For sufficiently large $n$, we can find at least $k-1$ primes between $e^{-\varepsilon}\cdot \frac{n}{k}$ and $\frac{n}{k}$. Pick the largest $k-1$ of these primes as $n_1, \dotsc, n_{k-1}$ and put $n_k = n - n_1 - \ldots - n_{k-1}$. Then
$$\frac{n}{k} \leqslant n_k \leqslant n\cdot\biggl(1 - e^{-\varepsilon}\frac{k-1}{k}\biggr) = e^{-\varepsilon}\cdot \frac{n}{k}\cdot \bigl(1 + k(e^{\varepsilon}-1)\bigr) < 2e^{-\varepsilon}\cdot \frac{n}{k}$$
(we need $e^{\varepsilon} < 1 + \frac{1}{k}$ for this), and hence the $n_j$ are coprime. Therefore
$$\operatorname{lcm}(n_1,\dotsc,n_k) = \prod_{j = 1}^{k} n_j \geqslant e^{-(k-1)\varepsilon}\cdot \frac{n^k}{k^k}\,.$$
Since we can choose $\varepsilon > 0$ arbitrarily small, it follows that $B_k(n)$ is asymptotically tight.

Title: Size of the least common multiple


问:
Register Machine on Fibonacci Numbers
This problem is easy to understand but I am struggling to come up with any solutions.

According to Wikipedia a register machine is a generic class of abstract machines used in a manner similar to a Turing machine. A (processor) register is a small amount of storage available as part of a CPU or other digital processor.

We can simplify the problem by modelling the registers as empty buckets, and we can let marbles or rocks to be put into the register ("bucket"). The rule is to add or remove marbles from the bucket to perform computations.

The rules are:

1. A register machine uses of a finite number of buckets, and an unending supply of marbles.

2. Each bucket can be individually identified. The marbles need not be distinguishable.

3. A register machine program is a finite set of instructions:

- To add marble to a bucket (and then to go on to the next instruction).

- Or to take a marble away from a bucket (and to go on to one next
instruction if you can, or another one, if you can’t).

4. The program can be written in a flowchart or in a list of instructions.

Here is an example of a register machine program that performs addition.

Let A, B, C be buckets.

1. (-B; 2,4) means take away one marble from bucket B, go to instruction 2 if can, or 4 if cannot

2. (+A; 3) means add one marble to bucket A, then go to instruction 3

3. (+C; 1) means add one marble to bucket C, then go to instruction 1

4. (-C; 5,-) means take away one marble from bucket C, go to instruction 2 if can, or exit if cannot

5. (+B; 4) means add one marble to bucket B, then go to instruction 4

It is easily shown that suppose we have 3 marbles in bucket A and 2 marbles in bucket B, and 4 marbles in bucket C. After performing this algorithm, there will be |A|+|B| = 3+2 = 5 marbles in bucket A and |B|+|C| = 2+4 = 6 marbles in bucket B.

Now, I would like to write a register machine code which when given an input of n in bucket A, returns (also in bucket A) the nth Fibonacci number. The Fibonacci numbers are 0 (the 0th), 1 (the 1st), 1 = 0 + 1 (the 2nd), etc. We can use any number of buckets, the goal is to write the code as simple as possible (i.e. with fewest instructions).

Here is my attempt and questions:

1. I know that it is necessary to use a bucket for the "counter".

2. The part that I am struggling is to consider the number of buckets needed. If the input digit is small, we can use less buckets. Since Fibonacci sequence is recursive, if the input digit is large, that means we have to recursively perform the addition and hence need more temporary storage? Or is there any simpler way?

3. I tried to analyse and mimic the algorithm from a programming language, but I found that any programming language is too "high level" and it is difficult to extract it to such simple operations.

4. Suppose we list the Fibonacci numbers with the input number:

0, 1, 1, 2, 3, 5, 8, 13

0, 1, 2, 3, 4, 5, 6, 7

I tried to observe a pattern (which is obvious): If n is 0 or 1, the Fibonacci number is the n itself. Suppose n=2, then the Fibonacci number is 0+1 = 1, where 0 and 1 is the Fibonacci number of n=0 and n=1 respectively. Now again suppose, n=3, the Fibonacci number is 1+1 = 2, where 1 and 1 is the Fibonacci number of n=1 and n=2 respectively. But when n is large to get the Fibonacci number of n-1 and n-2, we need to calculate from the base case. Is there any formula that give the Fibonacci number directly? Besides using golden ratio or similar approaches which I think are hard to program it into a register machine.

Many thanks in advance.

答:
Start with $A=n$, $B=0$, $C=1$, $D=0$. (Or, if by definition all but $A$ are empty, preceed the code with a +C).
We want to keep $B=F_{n-A}$, $C=F_{n+1-A}$, $D=0$.

```<code>1  while -A {
2     while -B {
3        +D
      }  // that is: D:=B, B:=0
4     while -C {
5        +B
6        +D
      } // that is:  D:=D+C, B:=C, C:=0
7     while -D {
8        +C
      } // that is: C:=D, D:=0
   }
9  while (-B) {
10    +A
   } // that is: A:=B, B:=0
11
</code>```

Each numbered line corresponds to a state. Lines with while -X mean that we perform -X and go to the next line on success, but to the line following the closing } on failure.
Lines with +A mean that we perform +X and go to the next line of the same block, except for the last line before a } go to the enclosing while.

Title: Register Machine on Fibonacci Numbers

Body:
Start with $A=n$, $B=0$, $C=1$, $D=0$. (Or, if by definition all but $A$ are empty, preceed the code with a +C).
We want to keep $B=F_{n-A}$, $C=F_{n+1-A}$, $D=0$.

1  while -A {
2     while -B {
3        +D
      }  // that is: D:=B, B:=0
4     while -C {
5        +B
6        +D
      } // that is:  D:=D+C, B:=C, C:=0
7     while -D {
8        +C
      } // that is: C:=D, D:=0
   }
9  while (-B) {
10    +A
   } // that is: A:=B, B:=0
11

Each numbered line corresponds to a state. Lines with while -X mean that we perform -X and go to the next line on success, but to the line following the closing } on failure.
Lines with +A mean that we perform +X and go to the next line of the same block, except for the last line before a } go to the enclosing while.


问:
Calculating the fundamental group of $X \setminus \mathbb S^1$, for $n\ge 3$

$X = \mathbb S^n$

I think I got this one very roughly. Since we had that $\mathbb S^n\setminus\{N\}\cong\mathbb R^n$ by stereographic projection, we can suppose that $N$ is sent to the point of infinity in $\mathbb R^n$ and say that $\mathbb S^n\setminus\mathbb S^1\cong \mathbb R^n\setminus\{$straight line$\}(\cong\mathbb R^n\setminus\{$axis$\})$. Since $\Pi_1(\mathbb R^n\setminus\{$axis$\})\cong \Pi_1(\mathbb S^{n-2})$ (all of this has been proved beforehand, I just don't want to make my proof any longer), we have that
\begin{equation*}
\Pi_1(\mathbb S^n\setminus\mathbb S^1) =
\begin{cases}
\mathbb Z & \quad \text{if $n=3$} \\
\{0\} & \quad \text{if $n>3$}
\end{cases}
\end{equation*}
The different cases rub me the wrong way, so I don't know whether it is correct or not.

X = $\mathbb R^n$

In page 46 of Hatcher's Algebraic Topology, we see an example of an application of Seifert-Van Kampen Theorem where they get the fundamental group of $\mathbb R^3\setminus\mathbb S^1$, which is $\mathbb Z$. In this case I haven't finished the proof, since I have no idea how to move on from here. Maybe $\mathbb R^n\setminus\mathbb S^1$ is homotopic to $\mathbb R^n\setminus\{$straight line$\}$, and we get the same result as before? Or maybe we get that it deformation retracts to $\mathbb S^{n-1}\vee \mathbb S^1$ (or is homotopic, at the very least).

Could anyone, please, help me out?

答:
Ok, I think I found the answer to the 2nd case.

If we reason similarly to the 1st case, we have that if we have that $\mathbb R^n\cong \mathbb S^n\setminus\{N\}$, and locate $N\in\mathbb S^1$, then we have that $\mathbb R^n\setminus \mathbb S^1 \cong \mathbb S^n\setminus\mathbb S^1 (*)$, hence $\Pi_1(\mathbb R^n\setminus \mathbb S^1) \cong \Pi_1(\mathbb S^n\setminus\mathbb S^1)$. We computed in 1. the latter fundamental group, so by the isomorphism,
\begin{equation*}
\Pi_1(\mathbb R^n\setminus\mathbb S^1) =
\begin{cases}
\mathbb Z & \quad \text{if $n=3$} \\
\{0\} & \quad \text{if $n>3$}
\end{cases}
\end{equation*}

I don't know if $(*)$ is very precise. Is this answer correct?

P.S.: Thanks to Tabes Bridges for confirming that 1. was correct.


问:
Forced Duffing equation $\ddot x +x+\varepsilon(bx^3+k \dot x+ax−F\cos(t))=0$ bifurcation analysis
For the forced Duffing oscillator in the limit where the forcing, detuning, damping, and nonlinearity are all weak:
$$\ddot x +x+\varepsilon(bx^3+k \dot x+ax−F\cos(t))=0$$
where $0<\varepsilon<<1$, $b>0$ is the nonlinearity, $k>0$ is the damping, $a$ is the detuning,
and $F>0$ is the forcing strength. This system deals with saddle-node bifurcations of cycles arise in its analysis. The averaged equations of this system are
$$r'=-\frac{1}{2}(kr+\sin(\phi)) \\\phi'=-\frac{1}{8}(4a-br^2+\frac{4F}{r}\cos(\phi))$$
Show that fixed
points for the averaged system correspond to phase-locked periodic solutions for the original forced oscillator. Show further that saddle-node bifurcations of fixed
points for the averaged system correspond to saddle-node bifurcations of cycles
for the oscillator.

So far, all I have done is find the fixed points of the averaged system, which occur when
$$r=\frac{-F}{k}\sin(\phi)\\\phi=2\pi c-\cos^{-1}(\frac{r(3br^2-4a)}{4F})$$
I don't exactly know how to show that these correspond to phase-locked periodic solutions for the original Duffing oscillator, though. Even more, I struggle to understand how I can classify the fixed points of the averaged equations as saddle-node bifurcations. I found that the Jacobian $J$ of the averaged system is$$J=\begin{bmatrix} -\frac{1}{2}k & -\frac{1}{2}cos(\phi)\\\frac{1}{4}br-\frac{4F}{8}\ln
|r|\cos(\phi) & \frac{F}{2r}\sin(\phi)\end{bmatrix}$$
but this seems way too complicated to be a good way to classify these fixed points. Any help would be extremely appreciated!

答: There's really no reason to try to analytically find the fixed points of the phase-locked periodic solutions. Instead, recognize that at the fixed points of the average system, $r'=\phi '=0$. This means the radius is constant and the angular velocity is constant, so this corresponds to a closed orbit being swept out at some constant period. Thus, what you're left with is phase-locked periodic solutions in the original system. Similarly, for saddle-node bifurcations of the averaged system, this must mean two limit cycles are colliding and annihilating each other in the original system, and you then have a saddle-node bifurcation of cycles by defintion.


问:
Why do we *restrict* to universes instead of *surrounding* us with them?
In set theory and category theory one easily runs into the problem of size. For example Russell's paradox tells us that it is impossible to have consistent set theory allowing a set of all sets. Similarly a category with ALL limits/colimits is a poset, so a consistent theory allowing interesting categories should rather talk about having small limits/colimits.

The "formal" way around this problem is to add the axiom of universes to our favorite set theory, which allows us to restrict to certain small sets, which basically form a model of set theory within set theory. I do not doubt that this is useful or that it solves these problems! However it does not address the following issue:

When introducing material set theories (say ZFC) we usually define it as a first order theory satisfying some axioms. Likewise when introducing structural set theories (say ETCS) we define it speaking of a category of objects (called sets), which satisfies some axioms. In either way the surrounding metatheory (in which we define the notion of first order theory / category) makes use of notions such as collection, equal, distinct, exists, for all, finite, infinite etc. So in some sense our definitions make use of themselves, which makes me feel very uncomfortable, as it somewhat invalidates all formalism.

The only way around this problem I see is to state another axiom, that every model of our set theory is an internal model of another (larger) model ie. something like

Every model of ZFC is a universe in a model of ZFC

Every category satisfying ETCS is an internal category (ie. small category) in a category satisfying ETCS.

I could not find a treatment of this problem (maybe because I am unfamiliar with the literature), or someone mentioning my axiom. Why is this?

PS: Again, on the practical side it doesn't really matter, whether universes "go inside" or "go outside". Yet I think philosophically it does make a difference...

答:
Let me address the set theoretic part of your question.

First of all, the term "universe" is a bit loaded. It can mean either (1) just a mathematical universe, understood as a universe of sets, i.e. a model of set theory; or (2) a Grothendieck universe, specifically, especially in the context of category theory, which is a model of second-order set theory, which means that it is isomorphic with $V_\kappa$ for an inaccessible cardinal $\kappa$, if you're unfamiliar with these notions, the key point to take forward is that this is a very special model of set theory which agrees with its metatheory on things like power sets, etc.

So right off the bat, we get that in the specific interpretation as a Grothendieck universe, it cannot be that every model of set theory is a universe in a larger model of set theory. But let's focus on the first meaning, simply "a model" (which raises the obvious question: why not just refer to it as a model?)

Nevertheless, these things are indeed studied, mostly in philosophy of set theory. You can read about the multiverse axioms, specifically in Hamkins' approach to the multiverse. Hamkins also has work on set theoretic potentialism, which you might want to read about.

Overall, if there is a proper class of inaccessible cardinals, you immediately get an ever growing hierarchy of universes that are even Grothendieck universes as well. This can be made stronger (e.g. require that "$\mathrm{Ord}$ is Mahlo") or weaker (e.g. require that simply every set is a transitive model of $\sf ZFC$", or even that there is a proper class of cardinals $\alpha$, such that $V_\alpha$ is a model of $\sf ZFC$; both of which are weaker than even a single inaccessible cardinal as far as consistency strength goes).

If my understanding of Grothendieck universes is correct, we actually do use them as you suggest. To make something "small", you must first envelope it by a larger universe. And since both universes agree on things like power sets, we get exactly what you're asking for.

One last thing worth pointing out, though, is that we can easily formalise all of the theories you mentioned, and more, in something like Peano arithmetic, or even Primitive Recursive Arithmetic, where there is no innate notion of "collection", rather this is meaning we assign—as people, as the readers and executors of mathematics—to the objects of set theory.


问:
How to find sample mean
The standard deviation of the starter salary for economics students graduation is 4200.
Calculate the probability that sample mean of 40 student(x.bar) will be bigger than mean of the total population (mu)

In general I need to calculate what is P(x.bar>mu)

Aside of calculating the SED I don't know how to continue.

答:
It is just as likely the mean of $40$ students is above the population mean as below the population mean, and hence the probability is $50\%$.

(By the way, the title is extremely misleading, given you never have to "find sample mean.")

Title: How to find sample mean


问:
Show that $h:X \longrightarrow \times_{i=1}^{n}{Y}$ defined by $h(x)=(f_1(x),f_2(x), \cdots, f_n(x))$ is continuous function

Let $f_{i}:X \longrightarrow Y_{i}$ for $i\in \lbrace 1,2,3, \cdots,n\rbrace$ continuous functions.Show that $h:X \longrightarrow \times_{i=1}^{n}{Y}$ defined by $h(x)=(f_1(x),f_2(x), \cdots, f_n(x))$ is continuous function

答:
We use induction over $i$, when $i=1$ is obviously that $f$ is continuous function, when $i=2$ we prove that the assertion is true.

Let $x\in X$ and consider $h(x)\in  U_1 \times U_2$ an open set in $Y_1 \times Y_2$ since $U_1$ is open
neighbourhood of $f_1(x)$ in $Y_{1}$ and $f_1$ is a continuous function there exist $V_1$ open neighbourhood of $x$ such that $f_1(V) \subset U_1$, analogue for $f_2(x)$ there exist $U_2$ open neighbourhood of $f_2(x)$ and $V_2$ open neighbourhood of $x$ such that $f(V_2) \subset U_2$.

Consider $U=U_1 \times U_2$ a neighbourhood of $h(x)$ such that exist $V=V_1 \cap  V_2$ neighbourhood of $x$ such that $f(V)=F(V_1 \cap V_2)=f(V_1) \cap f(V_2) \subset V_1 \subset U_1 \subset U_1 \times U_2=U$ since $x$ was arbitrary $h$ is a continuous function as well.

Assume that the result is true for $i=n$ and we should prove for $i=n+1$.

Let $f_{i}:X \longrightarrow Y_{i}$continuous functions for $i\in \lbrace 1,2,3 \cdot, n \rbrace$  and $f_{n+1}:X\longrightarrow Y_{n+1}$ continuous function, by our assumption $h_1:X \longrightarrow \times_{i=1}^{n}Y_{i}$ is a continuous function and by hypothesis $h_{2}:X \longrightarrow Y_{n+1}$ is a continuous function, applying  the case for $i=2$ we deduce that $h:X \longrightarrow (\times_{i=1}^{n}Y_i) \times Y_{n+1}$ is a continuous function, but this is equal to put $h:X \longrightarrow (\times_{i=1}^{n+1}Y_i)$ and therefore the result is true for $i\in \lbrace 1,2,3 \cdots, n \rbrace$


问:
How to formulate ordinary least squares regression in component formalism?
Matrix formulation is straightforward:

$\mathbf{y} = \mathbf{X} \boldsymbol{\hat{\beta}} +\boldsymbol{\hat{\varepsilon}}$

cost function:
$E = {\boldsymbol{\hat{\varepsilon}}}^T{\boldsymbol{\hat{\varepsilon}}} = {(\mathbf{y} - \mathbf{X}\boldsymbol{\hat{\beta}})}^T(\mathbf{y} - \mathbf{X}\boldsymbol{\hat{\beta}})$

... differentiating wrt $\boldsymbol{\hat{\beta}}$ and searching for extremum:

$\frac{\partial E}{\partial \boldsymbol{\hat{\beta}}} = 2 \mathbf{X}^T\mathbf{X} \boldsymbol{\hat{\beta}} - 2 \mathbf{X}^T \mathbf{y} = 0$

thus the OLS estimate of $\boldsymbol{\hat{\beta}}$ is:
$\boldsymbol{\hat{\beta}} = (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T \mathbf{y}$

So, there is probably some limitation to the previous relation (e.g. $(\mathbf{X}^T\mathbf{X})^{-1}$ have to exist) ... am I right?

If I try to make the same with component notation, there is a problem in the same formula (I'll come back later to this).

In component formalism (using Einstein's summation convention):

$E = (X_{ij} \beta_j - y_i)^2 = (X_{ij}\beta_j)^2 - 2 X_{ij}\beta_j y_i + y_i^2$

$\frac{\partial E}{\partial \beta_j} = 2X_{ij} \beta_j X_{ij} - 2X_{ij} y_i = 0$

$X_{ij} \beta_j X_{ij} = X_{ij} y_i$

Now, every term is just scalar, so it's tempting to cancel $X_{ij}$ on both sides.
However, this just leads to trivial relation: $y_i = X_{ij} \beta_j$

Can someone help me to enlighten this, please? Isn't it somehow connected to the use of only lower indices? When I have to consider both lower and upper indices (tensors and duals)?

Thank you!

答:
What you have is essentially $\boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}^T\boldsymbol{y}$, regardless of which notation you use. As you pointed out, if you cancel $\boldsymbol{X}^T$ from both sides, you are left with $\boldsymbol{X}\boldsymbol{\beta}=\boldsymbol{y}$. However, the point of doing least squares in the first place is that $\boldsymbol{X}\boldsymbol{\beta}=\boldsymbol{y}$ cannot be solved for $\boldsymbol{\beta}$, i.e., $\boldsymbol{X}$ is not invertible.
By multiplying each side by $\boldsymbol{X^T}$, you get $\boldsymbol{X}^T\boldsymbol{X}$ on the left, which is invertible.


问:
Computing product of two cosets
Let in $S_3$, $H=\{(1),(12)\}$ show that $(13)H(23)H \neq (13)(23)H$

The solution claims $(13)H(23)H=\{(132),(23),(12)\}$ however I calculated $(13)H(23)H=\{(132),(23),(12),(1)\}$

The solution seems to omit the identity from the set. Am I correct here or is the solution correct?

Title: Computing product of two cosets

Body:
If we just take it one element at a time, we have
$$
(23)H = \{(23)(1), (23)(12)\} = \{(23), (132)\}\\
H(23)H = \{(1)(23), (1)(132), (12)(23), (12)(132)\} = \{(23), (132), (123), (13)\}\\
(13)H(23)H = \{(13)(23), (13)(132), (13)(123), (13)(13)\} = \boxed{\{(132), (23), (12), (1)\}}
$$
There is no reason to omit the identity element from this, and doing so would give a different subset of $S_3$. So I agree with you in this case.


问:
Polynomial division in $\mathbb{Z}[x]$.
How is polynomial division defined in $\mathbb{Z}[x]$? For example, how to divide $x$ by $2$ or any other two polynomials in $\mathbb{Z}[x]$?</p>

答:
Given two polynomials $f=\sum_{k=0}^na_kx^k$ and $g=\sum_{k_0}^mb_kx^k$ in $\mathbb{Z}[x]$, the principal problem when we try to divide $f$ by $g$ is that if $g$ is not monic then maybe there are problems when we divide coefficients by the leading coefficient and maybe the algorithm of division maybe go outside $\mathbb{Z}[x]$. In order to prevent that, a modification is made by multiplying $f$ by $b_m^n$ and in this way there is not problem of going outside $\mathbb{Z}$ when dividing by $g$ by the usual algorithm.
In this way we obtain uniquely determined $q,r\in\mathbb{Z}[x]$ with $\deg r<m$ and
$$b_m^nf=qg+r$$
This method can be extended to arbitrary rings different from $\mathbb{Z}$, although you can see that has the problem of multiplying by the leading coefficient of $g$ in order to don't go outside the ring you are working in. </p>

Title: Polynomial division in $\mathbb{Z}[x]$.


问:
Confusion when applying the Empty Set to the Axiom Schema of Replacement
In Kenneth Kunen's "The Foundations of Mathematics", one of the ZFC Axioms (The Axiom Schema of Replacement) reads as follows:

$$\forall A \Big ( \forall x \in A \  \exists !y \varphi(x,y) \rightarrow \exists B \ \forall x \in A \ \exists y \in B \ \varphi (x,y) \Big)$$

which can be reformulated more strictly as:

$$\forall A \Big ( \forall x \big(x\in A \rightarrow \exists!y \varphi(x,y)\big) \rightarrow \exists B \ \forall x \big(x \in A \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)\Big)$$.

Consider when $A = \emptyset$, which has the property that $\forall x (x \notin \emptyset)$.

Looking at the more strict version of the axiom, we have:

$$\forall x \big(\color{red}{x\in \emptyset \rightarrow \exists!y \varphi(x,y)}\big) \rightarrow \exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$$

The "embedded" implication $\color{red}{x\in \emptyset \rightarrow \exists!y \varphi(x,y)}$ is vacuously true for any $x$. Therefore, the antecedent of the axiom when $A = \emptyset$ is true. Because we now have "access" to the consequent, this means that there exists a set of the following form:

$$\exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$$

Now, my question is, what exactly does $\exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$ mean?

Suppose $B'$ is a set that satisfies the formula: $\forall x \Big (x \in \emptyset \rightarrow \exists y \big (y \in B' \land \varphi(x,y) \big ) \Big )
$

Choose an arbitrary $x'$.

We know that $x' \in \emptyset \rightarrow \exists y \big ( y \in B' \land \varphi(x',y) \big )$ is a true statement.

However, because the antecedent of the implication is false, the truth of the implication can occur in two situations involving the consequent:

$\exists y \big ( y \in B' \land \varphi(x',y) \big )$ is false

$\exists y \big ( y \in B' \land \varphi(x',y) \big )$ is true

Situation 1. can be false in two (meaningfully) different ways:

1A. $\neg \exists y (y \in B')$. This would mean, by definition, that $B'$ is the empty set.

1B. $\neg \exists y \big (\varphi(x',y) \big )$ but $\exists y (y \in B')$. This would mean that $B'$ is not empty...but we cannot claim that any $y$ in the universe of all sets has a relationship with $x'$ as described by $\varphi$.

Situation 2. can be true in only one way:

2A. where $B'$ is non-empty and at least one $y$ has the property that $\varphi(x',y)$

Now, realize that the above comments were only made with respect to a particular $x$... (which we called $x'$)... in our domain of discourse (which is the universe of all sets). Obviously, because of the universal quantifier, claims will also be made for an x'', an x''', etc.

So, it seems as though the sets we can prove exist from this axiom when $A = \emptyset$ are the empty set and non-empty sets that may or may be related to an $x$ in the universe of all sets through $\varphi$.

Is this the correct interpretation?

If so, could some please help me understand what it means to prove the existence of a non-empty set that "may or may not" have particular properties? It's almost as if I am saying, "Here. I proved that an ambiguous object that I cannot describe with any clarity exists." I am having difficulties grasping what this means.

答:
what exactly does $\exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$ mean?

This formula means that there exists a set $B$ satisfying the condition $\forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$. Since for each set $x$, $ x \in \emptyset $  is false, the implication $x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))$ is true. So the condition is true (for each set $x$) and therefore the formula means that there exists (some) set $B$.