问:
How does one solve $\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx$
I believe this converges point wise but I am not sure how to evaluate $$\lim_{n\to\infty} \int_0^\infty \frac{\sin(x/n)}{x+x^2}dx.$$ Can someone please help?

答:
Since you received good answers for the limit, let me go beyond it.

Let $x=n t$
$$I=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}dx=\int_0^\infty\frac{\sin (t)}{n t^2+t} dt=\int_0^\infty\frac{\sin (t)}{t} dt-\frac 1n\int_0^\infty\frac{\sin (t)}{t+\frac1n} dt$$

$$\sin(t)=\sin \left(t+\frac{1}{n}-\frac{1}{n}\right)=\sin \left(t+\frac{1}{n}\right)\cos\left(\frac{1}{n}\right)-\cos \left(t+\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)$$ So, for tha last integral, there is an obvious change of variable and we face classic integrals.

Back to $x$, the antiderivative
$$J=\int\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}dx=\sin \left(\frac{1}{n}\right)
   \text{Ci}\left(\frac{x+1}{n}\right)+\text{Si}\left(\frac{x}{n}\right)-\cos
   \left(\frac{1}{n}\right) \text{Si}\left(\frac{x+1}{n}\right)$$ and
$$I=\frac{\pi}{2}   \left(1-\cos \left(\frac{1}{n}\right)\right)+\text{Si}\left(\frac{1}{n}\right) \cos
   \left(\frac{1}{n}\right)-\text{Ci}\left(\frac{1}{n}\right) \sin
   \left(\frac{1}{n}\right)$$

Using the asymptotics of the sine and cosines integrals
$$I=\frac{\log \left(n\right)+1-\gamma }{n}+\frac{\pi }{4
   n^2}+O\left(\frac{1}{n^3}\right)$$


问:
What is the solution of $x^3+x=1$?
According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:

$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$

$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way.  Am I right?

答: You can rearrange x^3 + x = 1 as x^3 = 1 – x, and let x = u + v. Notice that by the binomial theorem, (u + v)^3 = u^3 + 3vu^2 + 3uv^2 + v^3 = u^3 + v^3 + 3uv(u + v), suggesting that u^3 + v^3 = 1, while -1 = 3uv. -1 = 3uv implies v = -1/(3u), so u^3 – 1/(27u^3) = 1, implying 27u^6 – 1 = 27u^3. You can rearrange this as 27(u^3)^2 – 27u^3 – 1 = 0. This is a quadratic with respect to u^3, so you can use the quadratic formula to conclude that u^3 = [27 + sqrt(93)]/54 or u^3 = [27 – sqrt(93)]/54, although it really does not matter, since v^3 will always be the conjugate of u^3. Therefore, x = u + v = cbrt([27 + sqrt(93)]/54) + cbrt([27 + sqrt(93)]/54). This is ultimately equivalent to what was posted, it just requires some algebraic manipulations to get there. And yes, this is the simplest way you can write the answer. It is what it is. Unfortunately, simple problems do not always have simple solutions. And the laws of logic do not care about our puny concept of simplicity anyway.</p>

ContentLicense: CC BY-SA 4.0

Title: What is the solution of $x^3+x=1$?


问:
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point
If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is

A:  ${x^2} + {y^2} - x - y = 0$

B:  ${x^2} + {y^2} - 9x - y + 8 = 0$

C:  ${x^2} + {y^2} - 4x + 2y = 0$

D:  ${x^2} + {y^2} + 9x + y + 8 = 0$

I got the answer but had the choice not been provided how would I find the solution.

My steps are elaborated below

$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$

The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid  $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$

Let the centre of the circle is (h,k)

Circle passes through (2,3) and (1,1), hence the radius is

${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$

${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)

${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda  = 0$

$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$

Putting it in 2h+4k=1

$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow  - 2g - 4f + \lambda  + 2\lambda  = 11 \Rightarrow 2g + 4f = 3\lambda  - 11$

By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.

Had the option not been given how do I proceed

答:
Start with the general form of a circle:

$$x^2+y^2 + ax+by+c = 0$$

It passes through $(1, 1)$ and $(2,3)$, so $a+b+c = -2$ and $2a+3b+c = -13$. Therefore $(2a + 3b + c) - 2(a+b+c) = b-c = -9 \Rightarrow b = c - 9$.

Now, since the line $x-y+1 =0 \Rightarrow y = x+1$ is tangent to the circle, substitute in $y = x+1$.

This equation gives us the possible values of $x$, so by setting $\Delta = 0$, we ensure there is only one (real) value of $x$ common to both the circle and line, which is what a tangent is:

$$x^2 + (x+1)^2 + ax + b(x+1) + c = 0$$
$$\Rightarrow 2x^2 + (a+b+2)x + (b+c+1) = 0$$
$$\Rightarrow B^2-4AC = 0 : (a+b+2)^2 - 8(b+c+1) = 0$$
$$\Rightarrow (-c)^2 - 8(b+c+1) = 0 \tag{$a+b+c = -2$}$$
$$\Rightarrow c^2 -8(c - 9 + c + 1) = 0 \tag{$b=c-9$}$$

which gives $(a,b,c) = (-9, 1, 8)$ as expected.


问:
Proving a version of Green's formula in functional analysis
$\textbf{b}:\Omega\to\mathbb{R}^2$ is a $C^1$ vector field. Need to prove the following version of the Green's formula: $$(\nabla u,\textbf{b}v)=-(u\textbf{b},\nabla v)-(u,v\nabla\cdot \textbf{b})$$
where $(\cdot,\cdot)$ is the $L_2$ inner product.

I started from the left side $(\nabla u,\textbf{b}v)=\int_\Omega\nabla u\cdot\textbf{b}v$. My plan was to get to $$-\int_\Omega u\textbf{b}\cdot\nabla v-\int_{\partial\Omega}uv\nabla\cdot\textbf{b}$$
using integration by parts. But I am having troubles coming up with logical steps. Also I am not sure about the domain of the second integral above.

答:
Consider the following vector identity
\begin{align}
\text{div} (u (\mathbf{b}v))&= u \text{div} (\mathbf{b}v)+\nabla u \cdot \mathbf{b}v \\
&= u (\text{div}(\mathbf{b})v+\mathbf{b}\cdot\nabla v)+\nabla u \cdot \mathbf{b}v \\
&=u \text{div}(\mathbf{b})v+u\mathbf{b}\cdot\nabla v+\nabla u \cdot \mathbf{b}v
\end{align}
by Divergence Theorem
\begin{align}
\int_{\partial \Omega} uv (\mathbf{b} \cdot \mathbf{n})~dS = (uv,\text{div}~\mathbf{b})+(u\mathbf{b},\nabla v)+(\nabla u,\mathbf{b}v)
\end{align}
For smooth functions with compact support the left-hand side is zero and the conclusion follows.


问:
Remainders problem
What will be the reminder if $23^{23}+ 15^{23}$ is divided by $19$?
 Someone did this way:

$15/19 = -4$ remainder and $23/19 = 4$ remainder So $(-4^{23}) + (4^{23}) =0$ but i didn't understand it

答:
We have $23\equiv 4\pmod{19}$ and $15\equiv -4\pmod{19}$. Then for any odd positive integer $n$, we have $23^n\equiv 4^n \pmod{19}$ and $15^n\equiv (-4)^n \equiv (-1)^n4^n\equiv -(4^n)\pmod{19}$.

Add. We get $4^n -4^n\equiv 0 \pmod{19}$.

Added: We don't really need to go to $4$ and $-4$. For note that $23\equiv -15 \pmod{19}$. So if $n$ is any odd integer, then $23^n +15^n\equiv (-15)^n +15^n\equiv -15^n+15^n\equiv 0\pmod{19}$.

And we do not really need congruence notation. For recall that if $n$ is odd, then $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1}$. Let $a=23$ and $b=15$. We get that $23+15$  divides $23^{23}+15^{23}$. This yields the desired result.

ContentLicense: CC BY-SA 3.0

Title: Remainders problem


问:
$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong?
I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$

First of all I saw that it won't be possible to do that in any traditional way and actually calculate the limit, because of the form of expression. I mean - it's a sum with sqares on $n$ so I can't use Stolz lemma that easy. But, I thoght, that the solution is probably $0$, because probably every element of the sum is $0$ when $n \implies \infty$ and the limitation of sum in $\infty$ = sum of limitations in $\infty$. So I just went with that and decided to prove that using induction.

My base is:
$$\lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1}} = 0$$

My assumption:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +...+ \frac{1}{\sqrt{n^2 +2n}}\right) = 0$$

My induction:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0 + \lim_{n \to \infty} \frac{1}{\sqrt{n^2 +2n + 1}}) = 0$$

So the limit is:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0$$

But then my grader at university said that at first sight it looks totaly wrong but he actualy needs to think about it a bit. So here is my question - is that wrong? How is that wrong?

Ok, thank you for your answers. I thought I can solve that exercise that way because I asked not long ago very similar question on that forum: Find the limit of of such a sequence defined by recurrence

Because of your answers I think that the problem is actually that in this case I am dealing with a SUM of elements, am I right (or it the answer that I got in other case wrong?)?

答:
Using generalized harmonic numbers
$$a_n=\sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}=H_{n^2+2 n+1}^{\left(\frac{1}{2}\right)}-H_{n^2}^{\left(\frac{1}{2}\right)}$$

Using the asymptotics
$$H_p^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{p}}$$ and continuing with Taylor series
$$a_n=2-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$


问:
Prove that cosine distance does not satisfy the four properties of a metric over Euclidean space
How to prove that the cosine distance ($1-\text{cosine similarity}$) does not satisfy the four properties (Non-Negativity, Coincidence Axiom, Symmetry, Triangle Inequality) of a metric over Euclidean space?

If some properties are satisfied then how can I show it is satisfied?

答:
We have
$$D(x,y) = 1- S(x,y)$$
where
$$S_c(x,y) = \frac{x \cdot y}{||x|| \, ||y||}$$
First of all, $S(\cdot, \cdot)$ is not defined when either of its inputs is $0$, so $D$ cannot be a metric on $\mathbb{R}^n$. A better question to ask is whether or not $D$ is a metric on $\mathbb{R}^n \backslash \{0\}$. The answer is still no:
$$D((1,1), (2,2)) = 1 - \frac{(1,1)\cdot(2,2)}{||(1,1)|| \, ||(2,2)||}$$
$$=1- \frac{4}{\sqrt{2} \cdot 2\sqrt{2}} = 0$$
actually, the coincidence axiom is violated for any pair of nonequal vectors which are positive scalar multiples of each other. Why? Think of the definition of cosine and angles between vectors in $\mathbb{R}^n$!


问:
Definition of continuous map
Let $X,Y$ be topological spaces, let $f:X\longrightarrow Y$ a function. There are several (equivalent) ways to define continuity of $f$, one of these says: $f$ is continuous if
$$(1)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(x\in \overline{A}\Rightarrow f(x)\in\overline{f(A)}\Big)$$
where $\overline{A}$ denotes the closure of $A$ in $X$, and similarly for $f(A)$ in $Y$.

I was wondering if reversing this condition, i.e.
$$(2)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(f(x)\in \overline{f(A)}\Rightarrow x\in\overline{A}\Big)$$
also defines a continuous map. So, i'm asking if $(1)$ and $(2)$ are equivalent, if not, i'm looking for a counterexample, i.e. a continuous map $f$ not satisfying $(2)$.
Any help appreciated!

答:
I really prefer to use an intuitive interpretation when considering these problems, to help come up with guesses and examples. (You may still have to fill in details carefully to be sure.)

That is, I look at continuity informally as a statement that close points map to close points.

To interpret your two statements this way, the first one says that if $x$ is near $A$, then the image of $x$ is near the image of $A$, and that makes sense to be similar to continuity.

However, we notice right away there is something not quite right about the second one. We can informally rephrase this as, if the image of $x$ is near the image of $A$, should that imply that $x$ is near $A$? But that is not what continuity says.

In other words, continuity allows for far points to map to close points.

A counterexample, the constant map, $f(x)\equiv c$, maps lots of points to one point. Plenty of those points (in the domain) are far away from each other, yet they all land quite close to each other.

Title: Definition of continuous map


问:
Math platform for finding gaps in knowledge
I'm currently pursuing a specialization in Data Science, and in need of brushing up on some of my Algebra I and II before I take on some heftier topics in Linear Algebra and Calculus.

It's proving difficult however, to properly identify where the gaps in my knowledge are, even by taking practice tests and seeing where I've gone wrong.

Does there exist a platform where there is a central test, or a test per topic, whereby through machine learning, is able to tell me the areas in my math knowledge which are in need of improvement?

Thanks!

答:
I don't know how much machine learning it uses, but “a platform where there is […] a test per topic” that could determine “the areas in [one's] math knowledge which are in need of improvement” suggests Khan Academy (website, Wikipedia). From what I recall of how it works, Khan Academy, which offers a range of subjects,[1]

will present a range of practice exercises and use one's answers thereto to decide what one already knows and what one should study and

offers many short videos and readings with which to do that studying, although one is largely free to substitute, or to supplement with, other learning materials, if one prefers (for example, once you do reach linear algebra and calculus, you might like to try 3Blue1Brown (website, Wikipedia), whose host, Grant Sanderson, happens to have started out producing videos for Khan Academy).

I think Khan Academy is a generally well-made platform presenting (and slightly gamifying) a good range of mathematical exercises in such a way that it's easy to flow from one to the next.

Incidentally, since you mention linear algebra, I note that Khan Academy's linear algebra content, as of 2020, is not as developed as its calculus and lower-level content. If you decide you like Khan Academy's model of short problem sets mixed with short videos, then, for linear algebra, you could consider Lemma, which I discuss in answer #3893378. Lemma has a much more developed course in linear algebra and, unlike Khan Academy, is (as far as I'm aware) intended to be a complete resource by which to learn the subject (whereas Khan Academy is intended as a supplement to classroom instruction). If, alternatively, you prefer reading over watching, you could try the award-winning free linear algebra textbook of Jim Hefferon, which I recommend in answer #3907967.

[1]: These currently include
“Algebra 1”,
“Geometry”,
“Algebra 2”,
“Trigonometry”,
“Precalculus”,
“Statistics & probability”, …,
“Multivariable calculus”,
“Differential equations”, and
“Linear algebra”, as well as subjects outside mathematics.


问:
If $A\succ0$ and $J$ is the matrix of ones, then $A \succeq J$ if and only if $\mathrm{trace}(A^{-1}J)\leq 1$
I'm trying to prove that if $A \in \mathbb{R}^{n \times n}$ is positive definite, and $J$ is the $n \times n$ matrix of ones, then $A \succeq J$ if and only if $\mathrm{trace}(A^{-1}J) = \sum_{i,j} A^{-1}_{i,j} \leq 1$.

For the $\Rightarrow$ direction, I'm able to show that $A \succeq J \implies I - A^{-1}J \succeq 0 \implies \mathrm{trace}(I - A^{-1}J) \geq 0 \implies n \geq \mathrm{trace}(A^{-1}J)$, but this isn't the upper bound of 1 I was hoping for. I've also tried the using the eigendecomposition of $A$ but that didn't seem to get anywhere. For the $\Leftarrow$ direction I'm unsure where to start. Any pointers?

答:
proof 1

$B:= A-J = A-\mathbf {11}^T$

then for any $\mathbf x\neq \mathbf 0$, such that $\mathbf x \perp\mathbf 1$

$\mathbf x^TB\mathbf x =\mathbf x^TA\mathbf x-\mathbf x^T\mathbf {11}^T\mathbf x=\mathbf x^TA\mathbf x \gt 0$

so there is an $n-1\times n-1$ subspace where $B$ is positive definite, i.e. $B$ has signature $\big(\geq n-1,\leq 1\big)$.

Put differently $B$ has at least $n-1$ positive eigenvalues.  To check on the nth eigenvalue, compute the determinant by applying the matrix-determinant lemma

$\det\big(B\big)= \det\big( A +(-\mathbf {11}^T)\big)=\det\big(A\big)\cdot \big(1-\mathbf 1^TA^{-1}\mathbf 1\big) =\det\big(A\big)\cdot \big(1-\text{trace}(A^{-1}J)\big) $

which is $\geq 0$ iff $\text{trace}\big(A^{-1}J\leq 1\big)$ i.e. $B\succeq \mathbf 0$ iff $\text{trace}\big(A^{-1}J\big)\leq 1$

proof 2

Adapting what OP has done

This issue is that one gives up too much by directly taking the trace. A better way if going down such a path is to write

$A \succeq J \implies I - A^{-\frac{1}{2}}JA^{-\frac{1}{2}} \succeq 0$.   Now we know $\big(A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)$ is real symmetric PSD with rank one, so its trace is given by selecting a quadratic form associated with its dominant eigenpair $\lambda_1$,  $\mathbf x_1$ (with unit length)

now for any $\big \Vert \mathbf x\big\Vert_2=1$ we have

$\mathbf x^T\big( I - A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)\mathbf x$

$=1 -\mathbf x^T\big( A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)\mathbf x$

$\geq 1 -\mathbf x_1^T\big( A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)\mathbf x_1$

$=1 - \text{trace}\big(A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)$

$=1 - \text{trace}\big(A^{-1}J\big)$

which shows positive semi-definiteness if $1 - \text{trace}\big(A^{-1}J\big)\geq 0$

and for the other leg the result is immediate, i.e.
$\mathbf x_1^T\big( I - A^{-\frac{1}{2}}JA^{-\frac{1}{2}}\big)\mathbf x_1\lt 0$ implies indefiniteness


问:
How does one calculate gcd of two numbers if they are not written in base 10, without converting it to base 10 and converting back?
Let us say that I have two numbers $m$ and $n$ and $a$ is a positive number bigger than 2. Also assume that base-$a$ representations of $m$ and $n$ are:

$m = r_{M}a^{M} + r_{M-1}a^{M-1} + ... + r_{1}a + r_{0}$ and $n = s_{N}a^{N} + s_{N-1}a^{N-1} + ... + s_{1}a + s_{0} $

where all the $r_{j}$ and $s_{j}$ are in $\{0, 1, ... , a-1\}$.

I was wondering if I could calculate the quantity $gcd (m, n)$, without going back to base-$10$ representations? I have never even heard of $gcd$ in other bases before.

Is there a way to easily know when these two numbers $m$ and $n$ are relatively prime, again without converting back to base-$10$ representations?

答:
To implement the  Euclidean algorithm we require only an effective (Euclidean) "division with smaller remainder" algorithm, and this can be implemented the same way in any radix - in analogy with the grade school long-hand decimal division algorithm (for integers - unlike polynomials - it suffices to have a subtraction algorithm and, of course, a comparison test, to determine the least).

Your prior answers imply you know some algebraic number theory so you must be familiar with basic ring theory, so let's view it from that perspective. The "representation" in radix representation means we have have a ring ${\rm\color{#c00}{isomorphism}}\ h\,$ that maps integers into their radix reps, with the radix arithmetic operations.

This ring isomorphism necessarily preserves the division algorithm: $ $ in the division $\,a\div b,\,$ if $\, a = q\,b+r,\ 0\le r < b\,$ $\rm\color{#c00}{then}$  $\,h(a) = h(q)h(b) + h(r),\ 0\le h(r) < h(b).\,$ By uniqueness of the quotient and remainder, $\,h(q)\,$ and $\,h(r)\,$ are the quotient and remainder of $\,h(a)\div h(b).\,$ This implies that the Euclidean remainder sequence of $\,\gcd(h(a),h(b))\,$ is the image under $h$ of the sequence for that of $\,\gcd(a,b).\,$ Thus the Euclidean algorithm works exactly the same in any isomorphic ring representation (though the "internal" calculations in the (radix) long division algorithm will differ because they do depend on the radix digits).

Of course it's easy to directly deduce that that ring isomorphisms preserve gcds and ideals (and their generators), but your question concerns specifically how the Euclidean algorithm maps under an isomorphism, so I focused on that.


问:
What does "smallest" mean in this definition?

In first-order logic, a set $\tau$ of terms is the smallest set of strings (finite sequences of symbols) over the sets of variables, constants, and function symbols such that:

Any variable is a term.

Any constant is a term.

if $f^n$ is a function symbol of arity $n$, and $t_1,t_2,...,t_n$ are terms then $f^nt_1t_2...t_n$ is also a term.

What does the smallest set mean here? Is it really possible for more than one set satisfying the above properties to exist? If yes, an example would be great.

Thank you!

答:
As Kevin P. Barry comments, this is exactly to rule out unwanted objects. For example, consider the following definition of the set $E$ of even integers:

$E$ is the smallest set containing $0$ and closed under the maps $x\mapsto x+2$ and $x\mapsto x-2$.

The "smallest set" condition there is crucial: if we omit it, there are lots of sets satisfying the definition (e.g. $\mathbb{R}$ itself, $E\cup\{\pi+2k: k\in \mathbb{Z}\}$, and so forth). Another famous example of a definition of this form is one of the standard definitions of the natural numbers in set theory:

The set of finite ordinals is the smallest set containing $\emptyset$ and closed under the map $x\mapsto x\cup\{x\}$.

These show up all over the place in mathematics. (A useful term here is "inductive definition" or "recursive definition.") Generally we have the following useful heuristic:

Most of the time, "the smallest set $X$ such that [stuff]" is exactly the intersection of all the sets such that [stuff].

In fact, this is usually how we prove that a smallest such set exists - that's something we can't ignore! (E.g. "The smallest infinite set of natural numbers" isn't a thing.)

Incidentally, I've said a bit more about this in the specific context of formal languages (as in your OP) here.


问:
Question on positive square root of an operator
Let $T$ is a bounded linear operator on a Hilbert Space $H$. Let there exists a unique positive square root of $T^{*}T$. Let us call it as $P$

Then how can I show that the dimension of the range of $T$ is the same as that of the range of $P$?

For this, I am trying to show that their null spaces are equal.

I started with $x \in KerT$ $\rightarrow$ $Tx=0$ $\rightarrow$ $\langle Tx,Tx\rangle=0$ $\rightarrow$ $\langle T^{*}Tx,x\rangle=0$

From here, I can't imply that $Px=0$.

Help, please

答: Hint: If $\ P\ $ is a square root of $\ T^*T\ $, then $\ T^*T=P^2\ $.


问:
How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$)
The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.

This was question asked:

I don't know how to find eigenvalues of matrices ( larger than 3 $\times$ 3).

So, it's my humble request if you can tell some details or source of results and observations which are useful for dealing with such matrices.

I have studied linear algebra from David C lay.

Thanks a lot!!

答:
Let $A$ be that matrix. Then, we have $$A(x_1,x_2,x_3,x_4,x_5,x_6)^T = (x_4,x_5,x_6,x_1,x_2,x_3)^T,$$ i.e. $A$ acts on a vector by swapping the 1st/4th entries, the 2nd/5th entries, and the 3rd/6th entries.

So for a vector $v$ to be an eigenvector of $A$, swapping the 1st/4th entries, the 2nd/5th entries, and the 3rd/6th entries of $v$ should yield a scalar multiple of $v$. As an example, $$A(1,0,0,1,0,0)^T = (1,0,0,1,0,0)^T,$$ so $(1,0,0,1,0,0)^T$ is an eigenvector with eigenvalue $1$. Can you find six linearly independent eigenvectors and their corresponding eigenvalues?

Just a note: In general, it is hard/tedious to find the eigenvalues of an arbitrary large matrix. However, if a large matrix has a "pattern" to the entries, it may be possible to find the eigenvectors/eigenvalues "by inspection". That's probably what this question is expecting you to do.


问:
Find $a$ such that $\sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} = \frac{1}{2}$
Let $a > 0$ et $f$ defined by
$$ f(a) = \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2}  $$
We have

$$ \exists ! a > 0, \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} = \frac{1}{2} $$

I want to find $a$. So, I tried several thing :

1 - We know $\frac{1}{1 + x^2} = \arctan'(x)$, so

$$ \frac{1}{2} = f(a) = \sum_{k = 1}^{+ \infty} \arctan'(ak) = \sum_{k = 1}^{+ \infty} \frac{1}{\tan'(\arctan(ak))} $$

because $\arctan(y)' = \frac{1}{\tan'( \arctan(y) )}$. Now $\tan'(x) = \frac{1}{\cos^2(x)}$ so we have to find $a$ s.t

$$ \frac{1}{2} = \sum_{k = 1}^{+ \infty} \cos^2(\arctan(ak)) $$

But I don't how to deal with this sum.

2 - With a computer, I found that $a$ is close to $1.644$ and I conjecture that $a$ could be equal to a number like $ \frac{\pi^2}{6} $. But

$$ f(\frac{\pi^2}{6} ) \approx 0.4977 $$

I guess that $a$ may be a limit of the zeta function. So, I have to prove the following implication

$$ \sum_{k = 1}^{+ \infty} \frac{1}{t^2} = a \Rightarrow \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} = \frac{1}{2} $$

But I am not sure if it works.

3 - I guess that we may found $a$ like we prove $ \sum \frac{1}{n^2} = \frac{\pi^2}{6} $. But I am not sure if it's work.

Now I am in lack of idea.. so if anyone can give me a hint, I appreciate your help.

Body:
$$f(a) = \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2}=\frac{\pi  \coth \left(\frac{\pi }{a}\right)}{2 a}-\frac{1}{2}$$ Let $x=\frac \pi a$ and the problem is to find the zero of
function
$$g(x)=x \coth(x)-2$$

Using Taylor series
$$x \coth(x)=1+\frac{x^2}{3}-\frac{x^4}{45}+\frac{2 x^6}{945}-\frac{x^8}{4725}+\frac{2
   x^{10}}{93555}+O\left(x^{12}\right)$$ and series reversion gives
$$x=t+\frac{t^3}{30}+\frac{t^5}{1400}-\frac{t^7}{42000}-\frac{163
   t^9}{77616000}+O\left(t^{11}\right)$$ with $t=\sqrt{3(g(x)+1)}$.

Using $g(x)=0$ as desired leads to
$$x=\frac{9534829 }{8624000}\sqrt{3} \implies a=\frac{8624000 \pi }{9534829 \sqrt{3}}\approx 1.640533$$ while the solution given by Newton method is $1.640511$.

Edit

To show the impact of the level of expansion of the result, consider the expansion
$$x \coth(x)=\sum_{n=0}^{2p}\frac{2^{2 n} B_{2 n}}{(2 n)!}x^{2p}+O(x^{2p+2})$$ and use series reversion to obtain
$$\left(
\begin{array}{cc}
n & a_{(n)} \\
 1 & 1.813799364 \\
 2 & 1.648908513 \\
 3 & 1.639328024 \\
 4 & 1.640281061 \\
 5 & 1.640533429 \\
 6 & 1.640521573 \\
 7 & 1.640511406 \\
 8 & 1.640511008 \\
 9 & 1.640511422 \\
 10 & 1.640511481 \\
 11 & 1.640511466 \\
 12 & 1.640511462
\end{array}
\right)$$


问:
conformal maps between annulus

The modulus of an annulus $\{ a < \vert z - z_0\vert < b\}$ with inner radius $a$ and outer radius $b$ is defined to be

$$\frac1{2\pi} \log\left(\frac ba\right)$$

(a) Show that any conformal map from one annulus centred at the origin to another such
  annulus extends to a conformal self-map of the punctured plane.

(b) Show that there is a conformal map of one annulus onto another if and only if the annuli have the same moduli.

(c) Show that any automorphism of the annulus $ \{a < \vert z \vert < b\}$ is either a rotation $z \to e^{i\theta }z$ or a rotation followed by the inversion $z \to ab/z$

I'm totally stuck on this question

答:
Some feasible hints that you may need:

$\displaystyle{\left|\Re\left[\iint_A\dfrac{\partial f/\partial\rho}{f\cdot \rho}\right]\right|=\left|\Re\left[\int_0^{2\pi}\int_a^b\dfrac{\partial\log f}{\partial\rho}\mathrm{d}\rho\mathrm{d}\theta\right]\right|=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.

$\displaystyle{\left|\Im\left[\iint_A\dfrac{\partial f/\rho\partial\theta}{f\cdot \rho}\right]\right|}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}$.

$\displaystyle{\iint_A\dfrac{\mathrm{d}z}{|z|^2}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.

Denote the Jacobian determinant of $f$ as $\displaystyle{J_f}$. We can easily deduce that $\displaystyle{J_f=\left|\dfrac{\partial f}{\partial\rho}\right|^2=\left|\dfrac{\partial f}{\rho\cdot\partial\theta}\right|^2}$

Let $\displaystyle{f}$ be any conformal mapping from one annulus centred at the origin to another such annulus, e.g. $A(0;a;b)$, then $\displaystyle{\iint_A\dfrac{J_f\cdot\mathrm{d}z}{|f(z)|^2}=\dfrac{1}{2\pi}\cdot\log\dfrac{b}{a}}$.

Via the inequality $\displaystyle{\iint_A\dfrac{\mathrm{d}z}{|z|^2}\cdot\iint_A\dfrac{J_f\cdot\mathrm{d}z}{|f(z)|^2}\geqslant\left(\iint_A\dfrac{\sqrt{J_f}\cdot\mathrm{d}z}{|z|\cdot|f(z)|}\right)^2}$ and hints above, RHS $=\displaystyle{\left(\dfrac{1}{2\pi}\log\dfrac{b}{a}\right)^2}$, where equality holds.

It reveals that

$$
\left\{\begin{align}\dfrac{\partial f}{\partial \rho}&=C_0\cdot\dfrac{f}{\rho}\\\dfrac{\partial f}{\partial\theta}&=C_0\cdot i\cdot f\end{align}\right.
$$

Therefore, all self-maps of $A(0;a;b)$ take the form $\tilde{C}\cdot z$ or $\tilde{C}'\cdot z^{-1}$

There is also a theorem asserting that:

THEOREM (F.H. Schottky, 1877).

An annulus $A(0;r;R)$ can be mapped conformally onto the annulus $A_0(0;r_0;R_0)$ if and only if $r/R = r_0/R_0$.

Moreover, every conformal mapping $f:A\to A_0$
takes the form $f(z) = \lambda z^{\pm 1}$, where $|\lambda| = r_
0/r$ or $|\lambda| = r_0 R$ as the case may be.

Title: conformal maps between annulus


问:
If $f: X \rightarrow \mathbb{P}^n$ is a regular map and $U \subset X$ is constructible, then $f(U)$ is constructible.
I'm reading "Algebraic Geometry: A First Course" by Joe Harris, and I find myself stuck on one part of his proof for the following theorem. I copied his proof below, and bolded the part that I'm stuck on.

Let $X \subset \mathbb{P}^m$ be a quasi-projective variety, $f: X \rightarrow \mathbb{P}^n$ a regular map, and $U \subset X$ any
constructible set, then $f(U)$ is a constructible subset of
$\mathbb{P}^n.$

Proof: The key step in the proof is to establish an a priori weaker
claim: that the image $f(U)$ contains a nonempty open subset $V \subset \overline{f(U)}$ of the closure of $f(U).$ Given this, we set
$U_1 = U \cap (X - f^{-1}(V))$ and observe that the theorem for $U$
follows from the theorem for $U_1.$ We then apply the claim to $U_1$
and define a closed subset $U_2 \subset U_1$ and so on; since the
Zariski topology is Noetherian, a chain of strictly decreasing closed
subsets of a constructible set is finite, and the result eventually
follows.

I'm just not seeing what's going on right there. Why would proving the theorem for $U_1$ imply that the theorem is true for $U?$ There are a few equivalent definitions for constructible, so it's probably just that I don't understand all of them well enough.

EDIT:

A constructible set $Z \subset \mathbb{P}^n$ may be defined to be a
finite disjoint untion of locally closed subsets $U_i \subset
 \mathbb{P}^n,$ that is, a set expressible as $$Z = X_1 - (X_2 - (X_3 -
 \dots - X_n))\dots)$$ for $X_n \subset \dots \subset X_3 \subset
 X_2 \subset X_1$ a nested sequence of closed subsets of
$\mathbb{P}^n.$ Equivalently, we may define the class of constructible
subsets of $\mathbb{P}^n$ to be the smallest class including open
subsets and closed under the operations of finite intersection and
complementation.

答:
Hint: the biggest thing to do here is to figure out what the relationship between $f(U)$ and $f(U_1)$ is. The rest of this will be spoiler-tagged so you can go through it at your own pace but still have access to a full solution.

Warm up: Is $U_1$ constructible? Why?

 Yes, it is - it's the intersection of the closed subset $X\setminus f^{-1}(V)$ with the constructible set $U$.

To begin: Write down a statement relating $f(U)$ and $f(U_1)$.

 We have that $U_1=U\cap (X\setminus f^{-1}(V))$ for $V$ open. So $f(U_1)$ is $f(U)$ but without $V$. In symbols, this means $f(U_1)=f(U)\cap (Y\setminus V)$. What does that say about the constructibility of $f(U_1)$ in relation to $f(U)$?

Next:

 If $f(U)$ is constructible, then $f(U)\cap (Y\setminus V)$ is too, but this is exactly $f(U_1)$. Conversely, if $f(U_1)$ is constructible, then $f(U)$ is the (disjoint) union of $V$ and $f(U_1)$, so $f(U)$ is constructible too. So $f(U_1)$ is constructible iff $f(U)$ is, assuming we can find $V$ as claimed.


问:
Decide if it is a vector space
Given D = {p$\in$$P_n$ : p'(1)=1} decide if this is a vector space.

My thoughts is that if you have p'(1)=1 it should not be certain that we can find a p(x)=0 which is one of the criteria of vector spaces. However I'm not sure if this is correct and if so, how I can prove this.

答: This is not a vector space because even the zero element is not in this set. Also, your notation is too vague for people to understand.</p>

Title: Decide if it is a vector space


答: If $p(X)$ is the null polynomial, then $p'(0)=0\ne1$. So, $p(X)\notin D$ and therefore $D$ is not a vector subspace of $P_n$.

Title: Decide if it is a vector space

Body:
Given D = {p$\in$$P_n$ : p'(1)=1} decide if this is a vector space.

My thoughts is that if you have p'(1)=1 it should not be certain that we can find a p(x)=0 which is one of the criteria of vector spaces. However I'm not sure if this is correct and if so, how I can prove this.


问:
Conditional distribution from order statistics
I have the following data generating process:

I have two Uniform RVs $X$ and $Y$

I sample $n$ times from $X$ and $m$ times from $Y$, and then pick the max. Call this $Z$.

I'm trying to find the conditional distribution of $Z$ for each class $X$ and $Y$ separately. That is, the distribution of $Z$, given $Z$ came from $X$, and from $Y$.

$$X\sim \textit{Unif}[0,1]$$ $$Y\sim \textit{Unif}[0,b], \space  b>1$$
$$Z \sim Max\{X_1,X_2...X_n,Y_1,Y_2...Y_m\}$$

This is what I have so far.

I have the overall distribution of $Z$.

$Z$ can be written as $Z\sim Max(Max(X), Max(Y))$. Then,
\begin{align}
F_Z(z) &= Pr(Z \leq z) &\\
&=Pr(Max(X) \leq z)P(Max(Y) \leq z) \\
&=F_{Max(X)}(z)F_{Max(Y)}(z)
\end{align}

$F_{Max(X)}(x) = x^{n}$ and $F_{Max(Y)}(y) = (y/b)^{m}$

This gives

$$F_Z(z) = \begin{cases}
      (z/b)^m & z \geq 1 \\
      (z/b)^m z^n  & otherwise
   \end{cases}
$$
Note that $z$ is bounded by $[0,b]$. I checked this with simulation, and I got the same plots.

I also have $Pr(Max(X)>Max(Y))$, i.e probability that X is picked.
\begin{align}
Pr(Max(X)>Max(Y)) &= Pr(Max(Y)=x)Pr(Max(X)>x) \\
&= \int{f_{Max(X)}(1-F_{Max(Y)})dx} \\
&= \frac{b^{-m}n}{n+m}
\end{align}

I also checked this with simulation, and it looks right.

With this information, I'm trying to find the conditional distribution of Z given that it's from class $X$ and class$Y$ separately.

Here are the empirical pdfs of Z, where the different colors are $X$ and $Y$. This is what I'm trying to find analytically.

Thanks in advance for the help!

答:
Turns out, I was overcomplicating this. All I had to figure out was
$$f(Max(X)|Max(X)>Max(Y)$$ and
$$f(Max(Y)|Max(Y)>Max(X)$$

$$f_{X\mid X>Y}(x)=\frac{f_X(x)F_Y(x)}{P(X>Y)}$$

see Conditional expectation of X given X is greater than Y


问:
what are the poles of (tanz)/z
What are the poles of tanz/z and what is the best way to find them? I know what the answer is and the way I found it was to rewrite tanz/z as sinz/cosz and got 0 and (2n+1)pi/2. The one thing I am curious about though is another method where we rewrite tanz as a power series centered at 0, we get x + (x^3)/3+(x^5)/15...  by Taylor expansion and if we divide by z, the first term will always be 1. So that means we don't have any poles since nothing we plug in gives us any 0 or discontinuity, but that is impossible since these are just different ways of expressing the same thing, so what went wrong here?</p>

答: The Taylor series for $\frac{\tan z}z$ is only correct within some neighbourhood of the point it is expanded about. There will be, by definition, no poles within that neighbourhood. But there may be poles outside it.

Title: what are the poles of (tanz)/z


问:
Prove that the number of vectors in LI subset can not exceed the number of vectors in a generating set of vector space?
Any subset 'T' of vector space is the generating set of vector space if every elements of vector space can be expressed as the linear combination of elements of 'T'.

Let S be the LI subset of vector space containing 'm' elements and T be the generating set of vector space containing 'n' elements then how to prove that 'm' can not exceed 'n' or 'm' is always less than or equal to 'n'.

Is it true that;If 'T' generates vector space then adding those elements of vector space to T,which is not in T also generates the vector space??
Is it true that adding the elements to generating set of vector space which is not in generating set makes the new set linearly dependent??

答:
If $T$ spans $V$, then it has a set of maximally LI elements (= basis for $V$). Therefore, for any other set $S$ of LI elements, $\vert S \vert \leq \vert V \vert$.

If $T$ spans $V$ then any other $S$ containing $T$ also spans $V$.


问:
Sample Mean and Sample Variance are Independent, if $X_i$ is not an iid.
We know that if $X_i$ is from iid normal distribution $N(\mu,\sigma^2)$, then the sample variance ($S^2$) is independent with the sample mean ($\bar{X}$). But now, if $X_i$ is not iid, rather, $X_i$ is from a normal distribution $N(\mu, \Sigma)$, where $$\Sigma = \sigma^2
\begin{bmatrix}
    1       & \rho & \dots & \rho \\
    \rho      & 1 &  \dots & \rho \\
    \vdots \\
    \rho       & \rho  & \dots & 1
\end{bmatrix}.
$$
whether the sample mean and the sample variance are still independent? How can we show that?

答:
Suppose $X=(X_1,X_2,\ldots,X_n)^T$ has a multivariate normal distribution $N_n(\mu\mathbf1_n,\Sigma)$ where $\Sigma=\sigma^2\left[(1-\rho)I_n+\rho\mathbf1_n\mathbf1_n^T\right]$. Here $\rho\in \left(-\frac1{n-1},1\right)$ and $\mathbf1_n$ is a column vector of all ones.

One way to prove independence of $\overline X=\frac1n\sum\limits_{i=1}^n X_i$ and $S^2=\frac1{n-1}\sum\limits_{i=1}^n (X_i-\overline X)^2$ is pretty much the same as was done here  for $\rho=0$ (the particular case when $X_i$'s are i.i.d $N(\mu,\sigma^2)$).

Here is a brief proof sketch:

Transform $X\mapsto Y=AX$ where $A$ is an $n\times n$ orthogonal matrix given by

$$A=
\begin{bmatrix}\frac1{\sqrt n} &\frac1{\sqrt n} & \frac1{\sqrt n}& \frac1{\sqrt n}& \cdots &\frac1{\sqrt n}
\\ \frac1{\sqrt 2} & \frac{-1}{\sqrt 2} &0 & 0 & \cdots & 0
\\ \frac1{\sqrt 6} & \frac1{\sqrt 6} &\frac{-2}{\sqrt 6} &0 &\cdots & 0
\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots
\\ \frac1{\sqrt {n(n-1)}} & \frac1{\sqrt {n(n-1)}} & \frac1{\sqrt {n(n-1)}} & \frac1{\sqrt {n(n-1)}} & \cdots & \frac{-(n-1)}{\sqrt {n(n-1)}}
\end{bmatrix}$$

Then $X\sim N_n(\mu\mathbf1_n,\Sigma)$ would imply $Y\sim N_n(\mu A\mathbf1_n,A\Sigma A^T)$.

Verify that $A\Sigma A^T$ turns out to be a diagonal matrix:

$$A\Sigma A^T=\sigma^2 \operatorname{diag}\left(1+(n-1)\rho,1-\rho,\ldots,1-\rho\right)$$

So if $Y=(Y_1,Y_2,\ldots,Y_n)^T$, then $Y_i$'s are independent normal with $Y_1=\sqrt n\overline X$ and $\sum\limits_{i=2}^n Y_i^2=(n-1)S^2$. Hence conclude.

This would also give you the exact distribution of $\overline X$ and $S^2$ in this setup.


问:
Do we come to a contradiction if we accept completeness axiom for rational numbers?
I found the following formulation of the completeness axiom:

If two non-empty sets ($A$ and $B$) are such that for any two elements
from the first and second sets ($a$ and $b$, respectively) it is true that $a \le b$,
then there is a number $c$ such that $a \le c$ and $c \le b$.

It is also said that this axiom does not hold for rational numbers. However, if we take two non-empty sets ($A$ and $B$) such that $a \le b$ and find among the elements of the two sets such $a$ and such $b$ that $a = b$, then we can say that we found such $c$ ($c = a = b$) that $a \le c$ and $c \le b$. So, for that case it seems to me that the axiom holds also for rational numbers.

Shouldn't it be the case that in the formulation of the axiom we should requests that for any $a$ and $b$ it should be the case that $a < b$ (instead of $a \le b$)? But even in this case we can also find another rational number $c$ (c = (max(a) + min(b)) / 2) such that $a \le c$ and $c \le b$.

So, it looks to me that the completeness axiom holds for rational numbers if we stay in the domain of rational numbers. Do I misunderstand something?

答:
While between any two numbers there is a rational number, this doesn't save completeness. Let $A=\{q\in\mathbb{Q}: q^2<2\}$ and $B=\{q\in\mathbb{Q}: q^2>2,q>0\}$. Then every element of $A$ is less than every element of $B$, but there is no rational which is simultaneously $\ge$ every element of $A$ and $\le$ every element of $B$.

The point is that completeness looks at the interaction between sets of numbers, not just individual numbers themselves. So while $\mathbb{Q}$ and $\mathbb{R}$ look similar in some ways (namely, each is a dense linear order without endpoints), completeness totally breaks for $\mathbb{Q}$ but holds for $\mathbb{R}$.

This "points-to-sets" shift can be subtle at first, and easy to miss. In your OP for example you start talking about sets $A$ and $B$ but wind up talking about individual numbers $a$ and $b$, and this is what makes it look like some semblance of completeness holds in $\mathbb{Q}$: while for any specific $a\in A$ and $b\in B$ there will be a rational number between them, there is no rational number which is between the sets $A$ and $B$ in the appropriate sense - no single rational $c$ which works for all possible choices of $a$ and $b$.


问:
Sheaf of rings on a discrete set.
I was reading through some notes for an exam and one exericse asks me to prove the following

There is a unique sheaf of rings making a topological set $X$ with discrete topology a ringed space.

I tried doing it but I feel I'm missing something, using the definition of presheaf and than of sheaf doesn't seem to bring me any result.
How can I solve such a problem? I leave you my definitions of presheaf and sheaf.

A presheaf $\mathfrak{F}$ (of rings) on a topological space X consists of the data:

for every open set $U\subset X$ a ring $\mathfrak{F}(U)$ (think of this as the ring of functions on U),

for every inclusion $U\subset V$ of open sets in X a ring homomorphism $\rho_{V,U}:\mathfrak{F}(V)\to \mathfrak{F}(U)$ called the restriction map (think of this as the usual restriction of functions to a subset),
such that

$\mathfrak{F}(\varnothing)=0$

$\rho_{U,U}$ is the identity map of $\mathfrak{F}(U)$ for all U,

for any inclusion $U\subset V\subset W$ of open sets in X we have $\rho_{V,U}\circ \rho_{W,V}=\rho_{W,U}$

The elements of $\mathfrak{F}(U)$ are usually called the sections of $\mathfrak{F}$ over U, and the restriction maps $\rho_{V,U}$
are written as $\varphi\to \varphi|U$.

A presheaf $\mathfrak{F}$ is called a sheaf of rings if it satisfies the following gluing property:

if U$\subset X$ is
an open set, $\{U_i : i \in I\}$ an arbitrary open cover of U and $\varphi_i\in \mathfrak{F}(U_i)$ sections for all i such that $\varphi_i|_{U_i\cap U_j}=\varphi_j|_{U_i\cap U_j}$
for all $i, j \in I$, then there is a unique $\varphi \in \mathfrak{F}(U)$ such that $\varphi|_{U_i} = \varphi_i$ for all i.

EDIT: added picture of the exercise text for reference in case I explained myself wrong

EDIT2: This is what is given as the definition of a K-ringed space:

A ringed spaces equipped with a sheaf of rings such that the elements of $\mathfrak{O}_X(U)$ are actual functions from U to a fixed ring K;

EDIT3: It turns out that the actual definition is

A ringed spaces equipped with a sheaf of rings such that the elements of $\mathfrak{O}_X(U)$ are actual functions from U to a fixed ring K and $\mathfrak{O}_X(U)$ is not only a subring of the ring of functions from $U\to K$ but a $sub-K-algebra$ of it;

What does this change?

答:
Being a $K$-ringed space has a very specific meaning, namely that the associated sheaf $\mathcal{O}_X(U)$ is a $K$-subalgebra of the $K$-algebra $\operatorname{Mor}(U, K)$. In particular this means that $\mathcal{O}_X(U)$ contains the constant functions.

A set $X$ with the discrete topology has the trivial open covering $$X = \bigcup_{x \in X} x, $$ where we identify $x$ with the set $\{x\}$. The sheaf axioms then give that we can decompose any sheaf $\mathcal{F}$ over an open subset $U$ as $$\mathcal{F}(U) \cong \prod_{x \in U} \mathcal{F}(x).$$ After all the intersection of $x \cap y$ for any distinct elements is empty. For $\mathcal{F}(x)$ to sheaf for a $K$-ringed space we need it to be a $K$-subalgebra of $K_x := \operatorname{Mor}(x, K)$ and must contain the constant functions. But all elements of $\operatorname{Mor}(x, K)$ are constant functions and the set can be identified with $K$. We must thus have $\mathcal{F}(x) = K_x$ for all $x \in X$. It follows that for a $K$-ringed space with the specified conditions we must have $$\mathcal{O}_X(U) \cong \prod_{u \in U} K.$$ It is easy to see existence of a presheaf defined that way and the sheaf axioms  hold because the restriction maps come from dropping terms in the product.

Title: Sheaf of rings on a discrete set.


问:
Game of sets. Limit points
Let $A, B \subset \mathbb{R}$. We know that $A' = B' = \varnothing$ and $(A + B)' = [0, 1]$, where $$A + B = \{a + b \; | \; a \in A, b \in B\}$$
Find $A, B $.

I am trying to solve that problem for the second day in a row and the only thing that seems reasonable to me is taking $A = B = \frac{1}{n}$. If so, we can get a fraction any close to each point from $[0, 1]$. But in this case we have $A' = B' = \{0\}$ and that ruins the solution.

Would be thankful for any help!

答:
I managed to get the answer only a few days ago. The idea was to take the set $\mathbb{Q}$ and some function, that grows fast. For example $2^n$. Let set $A$ consist of $a_i$, where $a_i \in 2^n$ and $i$ is an index of the sequence.
Let $B$ equal to $-a_i + j$, where $j \in \mathbb{Q}$ is a rational number with index j.

Title: Game of sets. Limit points


问:
Proof of a compound proposition being a tautology
The question states the following compound proposition is a tautology, which I have to prove.

$$(p \rightarrow q) \wedge (q \rightarrow r) \rightarrow (p \rightarrow r)$$

My attempt is as follows:

$$\equiv (\neg p \vee q) \wedge (\neg q \vee r) \rightarrow (\neg p \vee r)$$
$$\equiv \neg((\neg p \vee q) \wedge (\neg q \vee r)) \vee (\neg p \vee r)$$

$$\equiv \neg[((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee (q \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee F )\vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv \neg[(\neg p \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$
$$\equiv [((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [(F \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Negation law]}$$

This goes on for quite a while, but I don't get the desired $T$ result. I'm guessing I've done something horribly wrong in the above calculation. I'll be so grateful for your help here. Anything regarding this text, from calculation error to using some incorrect terms. Thank yoou.

答:
Apply implication equivalence as much as possible, then deMorgan's rules (with DNE if you need to) as much as possible, before association and commutation to something useful.  The path should become clear from there.

$$\begin{align}&((p\to q)\land(q\to r))\to(p\to r)\\&\vdots&&\text{conditional equivalence}\\&\lnot((\lnot p\lor q)\land(\lnot q\lor r))\lor(\lnot p\lor r)\\&\vdots&&\text{de Morgan's}\\&((p\land\lnot q)\lor(q\land\lnot r))\lor(\lnot p\lor r)\\&\vdots&&\text{association and commutation}\\&((p\land \lnot q)\lor \lnot p)\lor((q\land\lnot r)\lor r)\\&\vdots\\&\top\end{align}$$


答:
Denote the given $6 \times 6$ matrix by $A$.

Hint One option is to observe that the square of the given matrix is the identity matrix, or, rearranging, that $A^2 - I = 0$. So, the minimal polynomial of $A$ divides $t^2 - 1 = (t + 1) (t - 1)$. (Alternatively, any eigenvector $v$ of $A$, say of eigenvalue $\lambda$, satisfies $v = A^2 v = \lambda^2 v$.)

Hint for another method Consider as usual the polynomial $\lambda I - A$; the first and fourth columns of this matrix are, respectively
$$\pmatrix{\lambda\\\cdot\\\cdot\\-1\\\cdot\\\cdot}, \pmatrix{-1\\\cdot\\\cdot\\\lambda\\\cdot\\\cdot}.$$
What values of $\lambda$ make these two columns linearly dependent?

Remark Both of these methods are specific to this matrix and ones that have a similar structure. Like JimmyK45542 wrote in his good answer, the question is really testing your ability to find and execute a method for computing those eigenvalues that takes advantage of the features that make the given matrix special.

Title: How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$)

Body:
The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.

This was question asked:

I don't know how to find eigenvalues of matrices ( larger than 3 $\times$ 3).

So, it's my humble request if you can tell some details or source of results and observations which are useful for dealing with such matrices.

I have studied linear algebra from David C lay.

Thanks a lot!!


问:
What's the bijection between scalar/inner products and (certain) almost complex structures (on $\mathbb R^2$)?
Asked on maths overflow here.

What's the bijection between (equivalence classes of) scalar products (I guess 'scalar product' is the same as 'inner product') and a.c.s. (almost complex structure/s) on $\mathbb R^2$?

From Example 1.2.12 of Daniel Huybrechts - Complex Geometry An Introduction.

Assumptions and notation:

I just pretend $V = \mathbb R^2$ literally instead of just an isomorphism.

Let $\Phi(V)$ be the set of real symmetric positive definite $2 \times 2$ matrices. This set is in bijection with inner products on $V$, I believe. We have according to this,

$$\Phi(V) = \{\begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \ | \ h+g, hg-f^2 > 0 \}_{h,f,g \in \mathbb R}$$

Let $\Gamma(V)$ be the (matrix representations of) a.c.s. on $V$. We have, according to this,

$$\{\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix}\}_{a,b \in \mathbb R, b \ne 0}=: \Gamma(V) \subseteq Auto_{\mathbb R}(V) \subseteq End_{\mathbb R}(V)$$

I understand that the 'rotation' matrices in $V$ are $SO(2) := \{R(\theta) := \begin{bmatrix}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{bmatrix}\}_{\theta \in \mathbb R}$, though I'm not sure that Huybrechts has the same usage of the term 'rotation'. (I ask about this later.)

Questions:

A. For injectivity (except for the equivalence class):

Given (equivalence class of) scalar product ($[M]$ of) $M$, choose unique $I$ that assigns $v$ to the one described. I'll call this map $\gamma: \Phi(V) \to \Gamma(V)$, $\gamma(M)=I$. (Later, $\tilde \gamma: \frac{\Phi(V)}{\tilde{}} \to \Gamma(V)$, $\tilde \gamma([M])=I$.)

It's 'rotation by $\pi/2$' or something. In what way? For $M=I_2$ (2x2 identity), then $I$ is indeed 'rotation by $\pi/2$', in the sense that it's $\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix} \in SO(2) \cap \gamma(V)$, which is the '$R(\theta)$', for $\theta = \pi/2$.

What exactly is the formula for $I=\begin{bmatrix}
a & b\\
\frac{-1-a^2}{b} & -a
\end{bmatrix} \in \Gamma(V)$ given $M = \begin{bmatrix}
h & f\\
f & g
\end{bmatrix} \in \Phi(V)$?

I'm asking because

2a - I would exceed wolfram computation time

2b - I notice for a different $M$ I tried, $I$ isn't a 'rotation matrix' in the sense of $SO(2)$. In fact, I believe the only 'rotation' matrices that are also a.c.s. are $\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$, i.e. $SO(2) \cap \gamma(V) = \{\pm \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}\}$. However, I think $I$ kind of 'rotates by $\pi/2$' in some other sense.

2c - I think $SO(2) \cap \gamma(V)$ isn't meant to be the image of $\gamma$

B. For surjectivity:

I'll call whatever map we would have as $\phi: \Gamma(V) \to \Phi(V)$, $\phi(I)=M$

Given an a.c.s. $I$, what are some possible scalar products $M$?

There's a comment that goes choosing the unique $M_v$ such that for some $v \in V \setminus 0$, we have $\{v,I(v)\}$ as an orthonormal basis. I tried this out (long to type!), and the only thing missing was the positively oriented. I guess either $\{v,I(v)\}$ or $\{v,-I(v)\}$ is positively oriented though. So I'll let $M_v$/$N_v \in \Phi(V)$ correspond to $\{v,I(v)\}$/$\{v,-I(v)\}$. Then by fixing $v$ (I ask about non-fixing of $v$ later), we have $\phi(I)=M_v$ or $N_v$, whichever corresponds to positively oriented basis. I'll just call this $\phi(I)=L_v$ Is this right?

Is $\phi$ supposedly an inverse (or right inverse or left inverse or whatever) to $\gamma$ (or $\tilde \gamma$ or whatever), in the sense that $\gamma(\phi(I)) = I$ for all (a.c.s.) $I \in \Gamma(V)$?

This whole thing about the $v$ makes me think there's another equivalence relation going on here. Is there?

This seems like we can have maps parametrised by the nonzero $v$, namely $\phi_v: \Gamma(V) \to \Phi(V)$. In this case, we might investigate if $\phi_v(I)=L_v=L_w=\phi_w(I)$ or at least if $[L_v]=[L_w]$ under the old equivalence relation of positive scalar $\lambda$, i.e. $L_v = \lambda L_w$. If this investigation turns out negative, then I think there's some problem like if 2 inner products are equivalent if they are from the same a.c.s. $I$ under $\phi_{\cdot}$, but for possibly different $v$ and $w$, then I think the equivalence class of $L_v$ under this new relation, which is $\{L_w\}_{w \ne 0}$, might not be the same as the equivalence class of $L_v$ under the old relation, which is $\{\lambda L_v\}_{\lambda > 0}$.

Ideas:

Perhaps there's some matrix thing here about how scalar products are in bijection with positive definite symmetric matrices and then almost complex structures are rotation matrices or something that are square roots of $-I_2$. Like given pos def symmetric $B$, there exists unique a.c.s. $J$ such that (something something).

Perhaps this is related, but I'd rather not further analyse the question or read through the answer given that I've spent over a month on almost complex structures BEFORE we even put inner products on vector spaces. Please consider spoon-feeding me here.

答:
DISCLAIMER: posting as answer instead of comment since too long for comment. You could think of this as a Cunningham's law thing or whatever, but I'm really doing this just because it's too long for a comment. I'm also going to do community wiki if this makes any difference.

TL;DR I think the bijection can be summed up in this formula (based on Jason DeVito's explicit computation of $I$)

$$\frac{MI}{\sqrt{\det(M)}} = J$$

Edit: Adding my intuition:

Given $M$, we want unique $I$ such that $I$ is something like '(anti-clockwise) rotation by $\frac{\pi}{2}$' but 'with respect to $M$'. In precise terms, this is the unique $I$ s.t. $\frac{MI}{\sqrt{\det(M)}} = J$, where $J$ is literally (anti-clockwise) rotation by $\frac{\pi}{2}$ ('with respect to $I_2$' or something)

Similarly, given $I$, we want an $M$ such that $I$ is '(anti-clockwise) rotation by $\frac{\pi}{2}$' but 'with respect to $M$'. Turns out there are several $M$'s that satisfy this condition, where this condition is stated in precise terms as $\frac{MI}{\sqrt{\det(M)}} = J$.

The bijection is:

From $M$ to $I_M$: Given $M$, choose unique $I_M = M^{-1}J\sqrt{\det(M)}$

From $I$ to $[M_I]$: Given $I$, choose unique equivalence class $[M_I]$ given by all $M_I$ such that $\frac{M_I}{\sqrt{\det(M_I)}} = JI^{-1} = -JI$

Injectivity (of map $M$ to $I_M$): Given $M$ and $I_M=I_N$, I believe Joppy's $\lambda$ and $\mu$ divide to give ratio or square root of ratio of determinants

Surjectivity (of map $M$ to $I_M$): Given $I$, there exists a lot of $M$, namely $[M_I]$.

Joppy and Jason DeVito, I've done some analysis of your answers. Follow-up questions:

FUP1 - What do you think of the above formula, with $M$ as inner product, $I$ as almost complex structure and $J$ as rotation matrix by $\frac{\pi}{2}$? I think this encapsulates all 3 properties of I. In particular...

FUP2 - I think the positively oriented is given in choosing $J$ instead of $-J$ (rotation matrix by $\frac{-\pi}{2}$)?

FUP2a - In this case, I think the 1st 2  properties of unique $I$ given $M$ (respectively, $v^TMIv=0$ and $v^T(I^TMI-M)v=0$, for all nonzero $v$)  is equivalent to $\frac{MI}{\sqrt{\det(M)}} = \pm J$?

FUP3 - We can see the that the $\lambda$ between $M$'s is given by square root of ratio of determinants:

FUP3a - For $\frac{M}{\sqrt{\det(M)}} = \frac{N}{\sqrt{\det(N)}} \implies M=\lambda N, \lambda=\frac{\sqrt{\det(M)}}{\sqrt{\det(N)}}$?

FUP3ai - I think in this part we do not assume that each matrix $\frac{M}{\sqrt{\det(M)}}$, $\frac{N}{\sqrt{\det(N)}}$ is equal to $JI^{-1} = -JI$?

FUP3b - Conversely for $M=\lambda N$ and $\lambda > 0$, we can take $\det$ of both sides to get $\frac{M}{\sqrt{\det(M)}} = \frac{N}{\sqrt{\det(N)}}$?

FUP3bi - In particular, we can the relevance of $2$ dimensions here: taking determinant of both sides of $M=\lambda N$ gives us a $\lambda^2$?


问:
Conformally Map Region between tangent circles to Disk
Suppose we are given two circles, one inside the other, that are tangent at a point $z_0$. I'm trying to map the region between these circles to the unit disc, and my thought process is the following:

I feel like we can map $z_0$ to $\infty$, but I'm not really sure about this. If it works, then I get a strip in the complex plane, and I know how to handle strips via rotation, translation, logarithm, power, and then I'm in the upper half-plane (if I've thought this through somewhat properly). My problem really lies in what point $z_0$ can go to, because I thought the point symmetric to $z_0$ (which is $z_0$ in this case) had to be mapped to $0$. Is this the right idea, and if what other details should I make sure I have? Thanks!

答: As it turns out, writing down an explicit formula is very long, but describing the general process is easy (and is similar to Chris C's, although I still feel his solution has a hole). Take $z_0$ to be the point of tangency, $z_1$ to be the antipodal point on the inner circle, and $z_2$ to be the antipodal point on the outer circle (in particular, these points are on the boundaries of the circles). Assuming we're given equations for the circles, we can explicitly calculate these antipodal points. From here, rotate (if necessary) so that these points are all on the real axis and translate so that $z_2$ is at $0$. Now, given the equations of the circles, we can ensure $z_0$ and $z_1$ lie on the positive real axis. From here, we can use a Moebius transformation to send $z_2\mapsto 0$ (so $0$ is fixed), $z_1\mapsto 1$, and $z_0\mapsto\infty$. Since angles are preserved and this maps the circles to lines, this maps the region to a vertical strip in the complex plane of width $1$. Now rotate this by $\pi/2$, dilate it by $\pi$, and apply the exponential map. This brings us to the upper half-plane, and now we can use the standard map to go the unit disc.</p>


问:
If two functions are one-to-one, will the sum of it also be one-to-one?
If $f:\mathbb{R}→\mathbb{R}$ and $g:\mathbb{R}→\mathbb{R}$ are both one-to-one functions, is $f + g$ also one-to-one? How do I show it whether $f+g$ is one-to-one or not?

答:
A counterexample suffices. Take $f,g:\mathbb{R}\to\mathbb{R}$ such that $f(x)= x$ and $g(x)=-x$ (as mentioned in the comments).

Clearly, $f(x) + g(x) = 0\forall x\in\mathbb{R}$, and $f+g$ is not one-one.

However, if $f, g$ are both strictly increasing or strictly decreasing, then the statement holds. WLOG consider the case where $f,g$ are differentiable and such that $f'(x)>0$ and $g'(x)>0$ for all $x\in\mathbb{R}$. Obviously, $(f+g)'(x) = f'(x) + g'(x) > 0$.


问:
We know that $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. Prove ${x_n}$ is convergent
Suppose $a>0, x_0>0$ and $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. How to prove ${x_n}$ is convergent? Can someone helps me?

I tried to prove that $x_n$ is monotonically decreasing and have a lower bound: $a^\frac{1}{3}$.

答:
Easy to see that all terms of $(x_n)$ are positive.
Now by AM-GM, we have the lower bound:
$$x_{n+1} = \frac 13\left(x_n + x_n + \frac{a}{x_n^2}\right)\ge \sqrt[3] a, \quad (\forall n \in \mathbb N) \tag{1}$$
Also, from the given formula we have
$$x_{n+1}-x_n = -\frac{1}{3}x_n + \frac {a}{3x_n^2} = \frac{-x^3_n + a}{3x_n} \tag{2}$$
But from $(1)$ we have $$-x_n^3 \le -a$$
Plugging this into $(2)$, we have
$$x_{n+1}-x_n \le \frac{- a + a}{3x_n^2} = 0 \\ x_n \ge x_{n+1}$$
So, the sequence is decreasing and lower-bounded and, therefore, convergent.

Finally, let the limit be $x$. Then, we have
$$3x = 2x + \frac{a}{x^2} \\ x = \frac{a}{x^2} \implies x = \sqrt[3]a \\ \bbox[5px,border:2px solid #C0AA00]{\lim_{n\to\infty}x_n = \sqrt[3]a}$$


问:
What do we call this small area in integration in math?

In the above image we can see to find area of circle they consider the very small annulus with area of $\delta A=2\pi r \delta r$. and in order to to find $\int u^3dA$ and it plugged in equation of $u_r$ into integral.

My question is :  We consider the $\delta A$  as area of annulus and find the integral with bound $r\in [0,R]$. which give the area of circle. I wonder what we call this small area $\delta A$ in math? Is this riemenn?

答:
That kind of quantity is known as a differential.

Basically, a differential of a quantity-A dependent on another quantity-B is the amount of increase of quantity-A as we nudge a little of quantity B i.e: a relation between increments.

This video discusses it see here


问:
Based on beta function, show that $\int_{-\pi/2}^{\pi/2} (\cos{\theta})^{n-1} d{\theta} = β(n/2, 1/2)$
I am trying to solve this problem but seems that I am stuck. In order to show that $\int_{-\pi/2}^{\pi/2} (\cos{\theta})^{n-1} d{\theta} = β(n/2, 1/2)$.

I understand that beta function follows β(a, b)= $\int_0^1 x^{a-1}(1-x)^{b-1} dx $, and it can also be expressed as β(a, b)= Γ(a)Γ(b)/Γ(a+b).

I have tried to solve the problem from both ways, I got Γ(1/2)=$\sqrt{\pi}$, and also tried the option of letting $\sqrt{x}=cos{\theta}$. But I can't process further to find the solution.

So please help, thank you.

答:
Put $x=\cos \theta$ then $dx = -\sin \theta d \theta = -(1-x^2)^{\frac 12}d \theta$. However, before doing this we must ensure that the substitution is $1-1$ , so we use the fact that $\cos$ is an even function, to write $\int_{- \frac \pi 2}^{\frac \pi 2} (\cos \theta)^n d \theta = 2\int_{0}^{\frac \pi 2} (\cos \theta)^n d \theta $. Therefore, we get :
$$
\int_{0}^{\frac \pi 2} (\cos \theta)^n d \theta = \int_0^1 x^n(1-x^2)^{- \frac 12} dx
$$

now put $y = x^2$, so that $dy = 2xdx $ so $dx = \frac{dy}{2 \sqrt y}$. Now we get :
$$
\int_0^1 x^n(1-x^2)^{- \frac 12} dx = \frac 12\int_0^1 y^{\frac n2} (1-y)^{- \frac 12} y^{- \frac 12}dy\\ =
\frac 12\int_0^1 y^{\frac {n+1}2 - 1} (1-y)^{- \frac 12}dy = \frac{\beta(\frac{n+1}{2},\frac 12)}{2}
$$

this $2$ cancels out with the $2$ we got from the even break up at the start, leading to the final answer being $$
\int_{- \frac \pi 2}^{\frac \pi 2} (\cos \theta)^n d \theta = \beta\left(\frac{n+1}{2},\frac 12\right)
$$

Your problem is solved by changing $n$ to $n-1$ in this argument. Also, since you have apparently attempted this approach but failed, it is possible that some of the steps need to be elaborated further, so kindly let me know.


问:
Minimal generating sets of free modules, and endomorphisms of free modules
I know that it seems very loose as a title but I hope this post will be beneficial to all the forum members.

One thing I like about free modules is that they help one define maps directly as we do in a vector space by just defining the images of the elements of a base (if it exists).

My questions are:

I have read somewhere that two minimal generating sets for a free module do not necessarily have the same cardinality, except if the corresponding ring is local. Is that true? What is the intuition behind a ring being "local" then?

"A map (module map of course) from our free module to itself is bijective iff it is injective." In which general setting is that statement true?

I hope that post end up containing many examples and counterexamples that are certainly beneficial to beginners like myself.

Regards

答:
1 In $\mathbb Z$ seen as a free module of rank one over itself, the sets consisting of $\{1\}$ on one hand and  $\{2,3\}$ on the other are minimal generating sets of different sizes.

2 is false for every domain $A$ , unless it is a field. Indeed if injective implied surjective for the free module $A^1$, then  for every non zero $a\in A$, the map $A^1 \to A^1:x\mapsto ax$ , being injective,  would be surjective and so $a$ would have as multiplicative inverse the $x\in A$ mapping to $1$.

However a theorem of Vasconcelos states that an endomorphism of a finitely generated module (not assumed free) over any commutative ring is bijective if and only if it is surjective.

Full Disclosure My main motivation for answering this question is that it gives me an opportunity to advertise Vasconcelos's theorem , proved on page 9, Theorem 2.4,  of Matsumura's Commutative Ring Theory . Even in  Atiyah-Macdonald's excellent Introduction to commutative Algebra  this result  is given, in Exercise 6.1, only with the superfluous hypothesis that the finitely generated module be noetherian.


问:
Definition of continuous map
Let $X,Y$ be topological spaces, let $f:X\longrightarrow Y$ a function. There are several (equivalent) ways to define continuity of $f$, one of these says: $f$ is continuous if
$$(1)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(x\in \overline{A}\Rightarrow f(x)\in\overline{f(A)}\Big)$$
where $\overline{A}$ denotes the closure of $A$ in $X$, and similarly for $f(A)$ in $Y$.

I was wondering if reversing this condition, i.e.
$$(2)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(f(x)\in \overline{f(A)}\Rightarrow x\in\overline{A}\Big)$$
also defines a continuous map. So, i'm asking if $(1)$ and $(2)$ are equivalent, if not, i'm looking for a counterexample, i.e. a continuous map $f$ not satisfying $(2)$.
Any help appreciated!

答:
Looks like everyone else has covered a continuous counterexample, and Hagen has given a nice counterexample for the other direction.

If you want a non-trivial counterexample, let $X=\{0\}\cup[1,\infty)$ and $Y=[0,1]$ be considered as subspaces of $\Bbb R$, and define $f:X\to Y$ by $$f(x)=\begin{cases}0 & x=0\\1/x & x\ge1\end{cases}$$ This is a continuous bijection without the mentioned property. Its inverse is a counterexample the other direction.

Title: Definition of continuous map


问:
Vector differentiation of Mean Squared Error
I have a set of notes where it states the following,

S = $(\mathbf{y}- \mathbf{X} \boldsymbol{\theta})^{\text{T}}(\mathbf{y}- \mathbf{X} \boldsymbol{\theta})$

$\frac{\partial S}{\partial \boldsymbol{\theta}} = -2\mathbf{X}^{\text{T}}(\mathbf{y}-\mathbf{X}\boldsymbol{\theta}) $

(Assume $\mathbf{X}$ is a matrix, $\mathbf{y}$ and $\boldsymbol{\theta}$ are vectors all of suitable dimensions). So essentially, S is the squared error, $\mathbf{X}$ is the design matrix and $\mathbf{y}$ is a vector of observations and I need to find the $\theta$ that minimises S, but I'm having trouble with differentiation. My working is as follows,

$(\mathbf{y}- \mathbf{X} \boldsymbol{\theta})^{\text{T}}(\mathbf{y}- \mathbf{X} \boldsymbol{\theta})= (\mathbf{y}^{\text{T}} - \boldsymbol{\theta}^{\text{T}}\mathbf{X}^{\text{T}})(\mathbf{y}-\mathbf{X}\boldsymbol{\theta}) = \mathbf{y}^{\text{T}}\mathbf{y} -\mathbf{y}^{\text{T}}\ \mathbf{X} \boldsymbol{\theta}- \boldsymbol{\theta}^{\text{T}}\mathbf{X}^{\text{T}} \mathbf{y} +\boldsymbol{\theta}^{\text{T}}\mathbf{X}^{\text{T}} \mathbf{X}\boldsymbol{\theta} \\ \text{Diff. w.r.t.} \ \boldsymbol{\theta} \text{ yields}; \\
\frac{\partial {S}}{\partial \boldsymbol{\theta}} = - \mathbf{y}^{\text{T}}\ \mathbf{X} - \mathbf{X}^{\text{T}} \mathbf{y} +2\mathbf{X}^{\text{T}} \mathbf{X}\boldsymbol{\theta} $

答:
When dealing with vectors and matrices, you have to be careful about the dimensions of quantities. Assuming $y$ and $\theta$ are column vectors, then $X^Ty$ and $X^TX\theta$ are column vectors, but $y^TX$ is a row vector. So the expression $-y^TX-X^Ty+2X^TX\theta$ doesn't make sense.

You should get $\dfrac{\partial S}{\partial \theta} = -2X^Ty + 2X^TX\theta$. Now, simply set this equal to $\vec{0}$ and solve for $\theta$.


问:
Verifying that $x^{x}$ is continuous
Using the following definition $x=\ln(y)\iff y=e^{x}$ I will show $x^{x}$ is continuous at $(0,\infty)$.

Attempt :
We note that $x^{x}=e^{x\ln(x)}$. We have that $x$ is continuous on $(0,\infty)$ and $\ln(x)$ is continuous on $(0,\infty)$. Thus, $x\ln(x)$ is continuous at $(0,\infty)$. Therefore, the composition $e^{x\ln(x)}$ is continuous at $(0,\infty)$.

Is this proof correct? because I feel like some necessary details are missing.

答: It looks good; the only thing I would add is that you also use the fact that $e^x$ is continuous on $(-\infty,\infty)$ (you need this for the full range since $x\ln x$ can be negative).

Title: Verifying that $x^{x}$ is continuous


答:
Rewrite$$x_{n+1}=x_n-x_n+\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$$ that is to say
$$x_{n+1}=x_n-\frac{x_n^3-a}{3x_n^2}$$ and recognize the iterates of Newton method for solving $x^3-a=0$

Title: We know that $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. Prove ${x_n}$ is convergent

Body:
Suppose $a>0, x_0>0$ and $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. How to prove ${x_n}$ is convergent? Can someone helps me?

I tried to prove that $x_n$ is monotonically decreasing and have a lower bound: $a^\frac{1}{3}$.


问:
Distribution of waiting time of "final" customer in finite capacity $M/M/2$ queue with $\mu_1 = 1, \mu_2 = 2, \lambda = 3$
A queue with a maximum capacity of $5$ customers has $2$ servers operating at rates
$\mu_1 = 1$ customer/hour and $\mu_2 = 2$ customers/hour respectively. Service times are exponentially distributed. Moreover, assume that all customers are served on a first-come, first-served (FCFS) basis. Customers arrive according to a Poisson process at rate $λ = 3$ customers/hour. Suppose that there are $4$ customers in the salon ($2$ being served, $2$ waiting) and new customer arrives. What is the distribution of the waiting time of this new customer (the time spent until the customer is served)?

I do not know how to account for the fact that the two servers have different rates. If it were only $1$ server with $3$ other waiting customers, then the distribution would just be the sum of four independent $Exp(\mu)$ variables, which is just an $Erlang(4, \mu)$ variable. Also, the question kind of doesn't make sense to me because we are not told how long the $2$ customers who are currently being served have been there.

答:
I was not able to find an analytical solution to this question, but I hope the following analysis is useful in your attempts.

Since the question did not specify, I will assume that customers arriving to the system when there are multiple available servers have equal probability of being served by each available server. We can model this system with a continuous-time Markov chain. Let $\lambda$ be the arrival rate of customers to the system, $\mu_A$ the service rate of the slower server, and $\mu_B$ the service rate of the faster server. Let $S = \{0, 1_A, 1_B, 2, 3, 4, 5\}$ and define transition rates $q_{ij}$ for each pair of states $i$ and $j$ by
$$
\gamma_{ij} = \begin{cases}
0,& i=j\\
\frac\lambda2,& (i,j) \in \{0\}\times\{1_A,1_B\}\\
\lambda,& (i,j)\in \{1_A,1_B\}\times \{2\}\\
\lambda,& i\in\{2,3,4\} \text{ and } j=i+1\\
\mu_A,& (i,j) \in\{(1_A,0), (2,1_B)\}\\
\mu_B,& (i,j) \in\{ (1_B,0), (2,1_A)\}\\
\mu_A+\mu_B,& i\in\{3,4,5\}\text{ and } j=i-1.
\end{cases}
$$
Define a $|S|\times|S|$ matrix $Q$ by
$$
Q_{ij} = \begin{cases}
\gamma_{ij},& i\ne j\\
-\sum_{j\in S\setminus\{i\}} \gamma_{ij},& i=j.
\end{cases}
$$
We call $Q$ the transition rate matrix of the process, as we shall see shortly.

Let $Z_t$ be the state of the system at time $t$, then $\{Z_t: t\geqslant 0\}$ is a continuous-time Markov chain; write $P(t)$ for the matrix with entries $p_{ij} = \mathbb P(Z_t = j\mid Z_0=i)$. Then $P(t)$ satisfies the backward Kolmogorov equation
$$
P'(t) = QP(t)\tag 1
$$
with initial condition $P'(0) = Q$. From general theory of differential equations we know that there exists a unique solution to $(1)$, namely $P(t) = e^{Qt}$, where $$e^{Qt} = \sum_{n=0}^\infty \frac{t^nQ^n}{n!} $$ denotes the matrix exponential function. It is a bit unwieldy to write $P(t)$ explicitly, but we need not do so to answer the question at hand. Set $T_0=0$ and for $n\geqslant$ let
$$
T_n = \inf\{t>T_{n-1} : Z_t\ne Z_{T_{n-1}}\}
$$
be the jump times of the process. We can define an embedded Markov chain $\{X_n:n=0,1,\ldots\}$ by setting $X_n = Z_{T_n}$. Let $R$ be the transition matrix of the embedded chain. Recall that if $U\sim\mathsf{Expo}(\alpha)$ and $V\sim\mathsf{Expo}(\beta)$ are independent, then $\mathbb P(U<V) = \frac{\alpha}{\alpha+\beta}$ (this can be verified by integrating over the joint density of $(U,V)$). It follows then that
$$
R_{ij} = \begin{cases}
0,& i=j\\
\frac{\gamma_{ij}}{\sum_{k\in S}\gamma_{ik}},& i\ne j.
\end{cases}
$$
In our example,
$$
R = \left(
\begin{array}{ccccccc}
 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\
 \frac{\mu _A}{\mu _A+\lambda } & 0 & \frac{\lambda }{\mu _A+\lambda } & 0 & 0 & 0 & 0 \\
 \frac{\mu _B}{\mu _B+\lambda } & 0 & 0 & \frac{\lambda }{\mu _B+\lambda } & 0 & 0 & 0 \\
 0 & \frac{\mu _B}{\mu _A+\mu _B+\lambda } & \frac{\mu _A}{\mu _A+\mu _B+\lambda } & 0 & \frac{\lambda }{\mu _A+\mu _B+\lambda } & 0 & 0 \\
 0 & 0 & 0 & \frac{\mu _A+\mu _B}{\mu _A+\mu _B+\lambda } & 0 & \frac{\lambda }{\lambda+\mu _A+\mu _B} & 0 \\
 0 & 0 & 0 & 0 & \frac{\mu _A+\mu _B}{\mu _A+\mu _B+\lambda } & 0 & \frac{\lambda }{\mu _A+\mu _B+\lambda } \\
 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)
$$
We wish to find the distribution of $$\tau = \inf\{t>0: Z_t \in\{1_A,1_B\}\mid Z_0 = 5\}. $$

This is not an easy task. Consider the simpler question of finding the distribution of $$N= \inf\{n>0: X_n\in\{1_A,1_B\}\mid X_0=5\}. $$ This is the number of jumps (services and arrivals) until one of the servers becomes available to serve the new customer. $N$ has a discrete phase-type distribution, i.e. it is the distribution of the time hitting time of the absorbing state in a terminating Markov chain (one in which all but the absorbing state are transient). For this terminating Markov chain, lump together the states $1_A$ and $1_B$ into the state $1$, and reorder the states as $(5,4,3,2,1)$. Now consider the substochastic matrix of $R$, call it $R_a$, obtained by this lumping and reordering process (as well as removing the state $0$, as it is not relevant). Then
$$
R_a = \left(
\begin{array}{ccccc}
 0 & 1 & 0 & 0 & 0 \\
 \frac{\lambda }{\mu _A+\mu _B+\lambda } & 0 & \frac{\mu _A+\mu _B}{\mu _A+\mu _B+\lambda } & 0 & 0 \\
 0 & \frac{\lambda }{\mu _A+\mu _B+\lambda } & 0 & \frac{\mu _A+\mu _B}{\mu _A+\mu _B+\lambda } & 0 \\
 0 & 0 & \frac{\lambda }{\mu _A+\mu _B+\lambda } & 0 & \frac{\mu _A+\mu _B}{\mu _A+\mu _B+\lambda } \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right).
$$
We can write
$$
R_a = \begin{pmatrix}
T & \mathbf T^0\\
\mathbf 0 & \mathbf 1,
\end{pmatrix}
$$
where $T$ is the substochastic matrix corresponding to transitions between transient states and $\mathbf T^0 + T\mathbf 1=\mathbf 1$. It can be shown that $p_n:=\mathbb P(N=n)$ is given by
$$
p_n = \tau T^{n-1}\mathbf T^0,
$$
where $\tau$ is the initial distribution (in our case, $\tau = \begin{pmatrix}1&0&0&0&0\end{pmatrix}^T$). There does not appear to be a concise closed form for the $n^{\mathrm{th}}$ power of $T$, although we can compute $\mathbb E[N]$ without much trouble. Define
$$
N_a := \sum_{k=0}^\infty T^k = (\mathbf{I_4}-T)^{-1},
$$
where $\mathbf{I_4}$ is the $4\times 4$ identity matrix. (Note that this series converges because there is a row which sums to strictly less than one.) Then the expected number of steps until absorption when starting in state $5$ is the first entry of the vector $N\mathbf 1$. Doing the computations, we find that
$$
\mathbb E[N] = \frac{2 \left(\mu _A+\mu _B+\lambda \right) \left(\left(\mu _A+\mu _B\right) \left(2 \mu _A+2 \mu _B+\lambda \right)+\lambda ^2\right)}{\left(\mu _A+\mu _B\right){}^3},
$$
which is equal to $16$ when $\lambda=3$, $\mu_A=1$, and $\mu_B=2$.

Unfortunately, even the exact distribution of $N$ would not lend much insight as to the distribution of $\tau$. This is because the holding time in state $5$ is exponential with rate $\mu_A+\mu_B$, but the holding time in states $4$ and $3$ have a hyperexponential distribution with density
$$
f(t) = \frac{\lambda(\mu_A+\mu_B)}{\lambda+\mu_A+\mu_B}\left(e^{-(\mu_A+\mu_B)t} + e^{-\lambda t} \right)
$$
and mean
$$
\frac{2 \mu _A \mu _B+\mu _A^2+\mu _B^2+\lambda ^2}{\lambda  \left(\mu _A+\mu _B\right) \left(\mu _A+\mu _B+\lambda \right)}.
$$
Although your question concerns the transient behavior of the process $\{Z_t\}$, it may be worthwhile to compute the stationary distribution (the rows of $\lim_{t\to\infty} e^{Qt}$), and proceed from there. To my knowledge there is no general way to find the distribution of the hitting time of a state in a continuous-time Markov chain (but if there is, then feel free to correct me!)


问:
When is the sum of consecutive squares a prime?
For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?

I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$

答:
The sum of $x$ consecutive squares starting from $n$ is $$n^2+(n+1)^2+...+(n+x-1)^2=xn^2+2n\sum_{k=1}^{x-1} k +\sum_{k=1}^{x-1} k^2 $$
$$\Rightarrow n^2+(n+1)^2+...+(n+x-1)^2=xn^2+nx(x-1)+\frac{(x-1)x(2x-1)}{6}$$
If $x\equiv 1,5\ (mod\ 6)\ $ then we have $(x-1)(2x-1)\equiv0\ (mod\ 6)$
which means that $\frac{(x-1)(2x-1)}{6}\ $ is an integer so the sum becomes:
$$n^2+(n+1)^2+...+(n+x-1)^2=x(n^2+nx(x-1)+\frac{(x-1)(2x-1)}{6})$$ meaning that if $x\equiv 1,5\ (mod\ 6)\ $ the sum of the square of x consectuive integers will never be a prime.

If  $x\equiv 0\pmod6\ \text{and}\ x>6$ then $\frac {x}{6}$ will be an integer greater than 1
so we will have: $$n^2+(n+1)^2+...+(n+x-1)^2=\frac {x}{6}(6n^2+6nx(x-1)+(x-1)(2x-1))$$ and again no pirme. In the same fashion if $x\equiv2\ (mod\ 6)$ and $x>2\ $ we take $\frac{x}{2}$ as a factor, and if $x\equiv4\pmod6$ we take $\frac{x}{2}$ again as a factor and finally if $x\equiv3\pmod6\ $ and $x>3\ $ we take $\frac{x}{3}$ as a factor.
Now we need to consider tha cases in which x = 2, 3, 6:
if $x = 2$ then we have $1^2+2^2=5$ which is a prime
if $x = 3 $ then we have $2^2+3^2+4^2=29$ which is prime
and if $x = 6$ then we have $2^2+3^2+4^2+5^2+6^2+7^2 = 139$ which is again a prime.
So the answer for this problem will be 2, 3 and 6


问:
Lower bound for sum of reciprocals of positive real numbers
I am reading an article where the author seems to use a known relationship between the sum of a finite sequence of real positive numbers $a_1 +a_2 +... +a_n = m$ and the sum of their reciprocals. In particular, I suspect that
\begin{equation}
\sum_{i=1}^n \frac{1}{a_i} \geq \frac{n^2}{m}
\end{equation}

with equality when $a_i = \frac{m}{n} \forall i$. Are there any references or known theorems where this inequality is proven?

This interesting answer provides a different lower bound. However, I am doing some experimental evaluations where the bound is working perfectly (varying $n$ and using $10^7$ uniformly distributed random numbers).

答:
The author is using the Arithmetic Mean - Harmonic Mean ("AM-HM") inequality:
https://en.wikipedia.org/wiki/Harmonic_mean#Relationship_with_other_means

This is a popular inequality in the math olympiad world; you can find a proof here:
https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality


问:
Is the image of a continuous functions always a topological manifold
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be continuous.  Then, for some sufficiently large $M\geq m$, can we always guarantee that $f(\mathbb{R}^n)$ is locally-homeomorphic to some $\mathbb{R}^N$, where $N\leq n$?

My intuition is no, since I can imagine a curve crossing itself; but I can't write down an explicit example.

答: No. Take, for instance,$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R^2\\&t&\mapsto&\left(\frac{\cos(t)}{1+\sin^2(t)},\frac{\sin(t)\cos(t)}{1+\sin^2(t)}\right).\end{array}$$Its range is a Lemniscate of Bernoulli, which is not locally homeomorphic to $\Bbb R$ (precisely because of the existence of a self-intersection).


答:
by Cauchy schwarz inequality$$\left(\sum_{i=1}^{n}{\sqrt{a_i}}^2\right)\left(\sum_{i=1}^{n}\frac{1}{{\sqrt{a_i}}^2}\right)\ge (\sum_{i=1}^{n} 1)^2=n^2$$

WLOG $a_1\ge a_2...\ge a_n$ then $\frac{1}{a_1}\le \frac{1}{a_2}..\le \frac{1}{a_n}$

So  by chebyshev's inequality $$n^2=n\left(a_1\frac{1}{a_1}+a_2\frac{1}{a_2}..+a_n\frac{1}{a_n}\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}\frac{1}{a_i}\right)$$

Title: Lower bound for sum of reciprocals of positive real numbers

Body:
I am reading an article where the author seems to use a known relationship between the sum of a finite sequence of real positive numbers $a_1 +a_2 +... +a_n = m$ and the sum of their reciprocals. In particular, I suspect that
\begin{equation}
\sum_{i=1}^n \frac{1}{a_i} \geq \frac{n^2}{m}
\end{equation}

with equality when $a_i = \frac{m}{n} \forall i$. Are there any references or known theorems where this inequality is proven?

This interesting answer provides a different lower bound. However, I am doing some experimental evaluations where the bound is working perfectly (varying $n$ and using $10^7$ uniformly distributed random numbers).


问:
Arc length meaning
I am currently studying about Parametric integration and very confuse about the arc length meaning.

In my understanding, the arc length is the total length that an object traveled along a curve in a given interval, in other words the length of the curve in that interval.
Let say an object travels on a circle motion with  the parametric equations.

$x=3\cos(3t)$.   $y=\sin(3t)$

in the interval $[0,2\pi]$

using the formula for arc length which is $ L= \int_{0}^{2\pi} \sqrt(y’^2+x’^2)dx$
the answer is $18\pi$ which meets my understanding that arc length is the whole distance travelled so 3 times around  the circle.

However if we given the parametric curve
$x=\sin^2(t)$  $y=\cos^2(t)$  for the interval $[0,3\pi]$

although the object should be going 6 times back and fourth between 2 points however when using L formula with limits 0 and $3\pi$ the answer is different from when we integrate in the interval $[0,\pi/2]$ and then multiply by 6, so why in this case the arc length formula is not working if we integrate over the full interval $[0,3\pi]$ as I thought that it would give us the total distance without changing the limits.

Also is my understanding correct that the formula would give the total distance traveled and not the displacement

答:
$$\int_0^{\pi/2}\sqrt{4\sin^2t\cos^2x+4\cos^2t\sin^2t}\,dt=\sqrt2\int_0^{\pi/2}|\sin 2t|\,dt.$$
and using periodicity and symmetry,
$$\int_0^{3\pi}|\sin 2t|\,dx=3\int_0^{\pi}|\sin 2t|\,dx=6\int_0^{\pi/2}|\sin 2t|\,dt.$$

Title: Arc length meaning


问:
Some general method to find limit, accumulation and isolated points of a set.
I have a hard time finding limit, accumulation and isolated points of sets so I was wondering if there is some general way to approach such problems.

So for example consider a simple subset of the reals $$S= \{1+\frac{1}{n+1} | n\in \mathbb{N} \}$$
Can you explain how I would find the limit, accumulation and isolated points in this set?

The way I would do it is this, for limit points I would just find the limit of $1+\frac{1}{n+1}$ which is 1, for isolated points I would consider open sets around each point and realized that not every open set contains points of $S$ therefore every point is an isolated point, now for accumulation points I have no idea how how to find them.

Can you explain?

答:
Let $A \subseteq X$. Then, $x \in X$ is an accumulation point of $A$ if $\forall r > 0, B(x, r) \cap (A\backslash \{x\}) \ne \emptyset$. Also note that accumulation point is also called limit point.

So here, if you can deduce that every point in $S$ is an isolated point, then they cannot be accumulation point (It is important to note that we exclude the point itself from the intersection). Other than that, it is always good to check whether the limit is accumulation point or not since limit is one of the best candidates for being an accumulation point.

Here, I assume our bigger set is $\mathbb{R}$. Then, $\forall r > 0$, we can check the intersection $B(1, r) \cap (S\backslash \{1\})$. But note that since we are in $\mathbb{R}$, we have $B(1,r) = (1-r, 1+r)$. Here, by using $\mathbb{Q}$ being dense in $\mathbb{R}$, one can show that we can find an element of the form $1+\frac{1}{n}, n \in \mathbb{N}$ in the interval $(1,1+r)$, for any $r > 0$ (because it is equivalent to find a rational number of the form $\frac{1}{n}$ in the interval $(0,r)$, but this is easy to show once you know what dense means). Then, the intersection will not be empty for any such $r > 0$, and thus by definition, $1 \in \text{Acc}(S)$.

Now, I'm leaving to you to check whether there are other accumulation points in $\mathbb{R}\backslash S$ or not.


问:
How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$)
The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.

This was question asked:

I don't know how to find eigenvalues of matrices ( larger than 3 $\times$ 3).

So, it's my humble request if you can tell some details or source of results and observations which are useful for dealing with such matrices.

I have studied linear algebra from David C lay.

Thanks a lot!!

答:
In addition to the existing answers I'd like to say that it's not hard to calculate the characteristic polynomial using Laplace expansion:

\begin{align}p(x)&=\det\begin{bmatrix} x & 0 & 0 & -1 & 0 & 0\\
0 & x & 0 & 0 & -1 & 0\\
0 & 0& x&  0 & 0& -1\\
-1 & 0 &0&x& 0 & 0\\
0 & -1 & 0 & 0&x &0\\
0 & 0& -1 & 0 & 0 &x
 \end{bmatrix}\\
&=x\det\begin{bmatrix}
 x & 0 & 0 & -1 & 0\\
 0& x&  0 & 0& -1\\
 0 &0&x& 0 & 0\\
 -1 & 0 & 0&x &0\\
 0& -1 & 0 & 0 &x
 \end{bmatrix}+1\det\begin{bmatrix} 0 & 0 & -1 & 0 & 0\\
 x & 0 & 0 & -1 & 0\\
 0& x&  0 & 0& -1\\
 -1 & 0 & 0&x &0\\
 0& -1 & 0 & 0 &x
 \end{bmatrix}\\
&=x\cdot x\det\begin{bmatrix}
 x & 0  & -1 & 0\\
 0& x&  0 & -1\\
 -1 & 0 &x &0\\
 0& -1 &  0 &x
 \end{bmatrix}-1\det\begin{bmatrix}
 x & 0  & -1 & 0\\
 0& x & 0& -1\\
 -1 & 0 &x &0\\
 0& -1  & 0 &x
 \end{bmatrix}\\
&=(x^2-1)\det\underbrace{\begin{bmatrix}
 x & 0  & -1 & 0\\
 0& x & 0& -1\\
 -1 & 0 &x &0\\
 0& -1  & 0 &x
 \end{bmatrix}}_{=M}
\end{align}
Now it's your turn to prove $\det(M)=(x^2-1)^2$ using the same method (Laplace expansion) as above.

The final result is

$p(x)=(x^2-1)^3$


问:
How to show: if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$?
A little stumped on this problem, any help would be greatly appreciated.

Show that for all $a,b,c \in \mathbb{Z}$, if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$.

答:
We are given $a=nb$, $a=mc$, $1=ub+vc$.
Then $n=nub+nvc=ua+nvc=(um+nv)c$, hence $a=(um+nv)bc$.</p>


问:
Convergence of a composite infinite series + comparison to brute force solution.
I have the following series:

$C_N = C_{N-1} + \dfrac{1}{2N\sum\limits_{i=1}^{N}\frac{1}{\sqrt{i}}}$

Assuming $C_0 = 0$, the first few terms are as follows:

$C_N = 0+\dfrac{1}{2}+\dfrac{1}{4(1+\frac{1}{\sqrt{2}})}+\dfrac{1}{6(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})} +\cdots + \dfrac{1}{2N\sum\limits_{i=1}^{N}\frac{1}{\sqrt{i}}}$

This series seems to be slightly smaller than $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +\cdots$ which we know diverges.

Is there a valid way to verfiy whether or not this series converges? If so, is there a reasonable estimate for its value?

FYI - After running this sum through 4000 iterations ($C_{4000}$), its value seems to be around 1.037058.

答:
From $$\frac1{2\sqrt{i+1}}\le\frac1{\sqrt{i+1}+\sqrt{i}}=\sqrt{i+1}-\sqrt{i}\le\frac1{2\sqrt{i}},$$ we see that
$$\sum^{n+1}_{i=2}\frac1{\sqrt{i}}\le2(\sqrt{n+1}-1)\le\sum^{n}_{i=1}\frac1{\sqrt{i}},$$ i.e.
$$\sum^{n}_{i=1}\frac1{\sqrt{i}}=2\sqrt{n}+O(1).$$ This means that your sum converges, because $$\sum^\infty_{n=1}\frac1{n^{3/2}}$$ converges.
One can use those inequalities (and the numerical value of $\zeta(3/2)$) to give rough estimates from above and below of the sum. Calculating it with higher precision is tricky, though.


问:
How to find a value of a point in a new coordinate system?

The known data: length of $AO$, $OB$, angles $\alpha$, $\beta$, point value of $B= \left(b_1,b_2\right)$ in the Cartesian coordinate system where $O$ is the origin. It is also known that point value $O= \left(O_1,O_2\right)$ in the Cartesian coordinate system where $A$ is the origin

How can I find the value of a point $B= \left(b_{1new}, b_{2new}\right)$ in Cartesian coordinate system where $A$ is the origin? First I tried to use matrix and map points, but it did not work.

The Cartesian coordinate system with origin $O$ was moved from point $A$ to point $O$ and rotated clockwise through an angle $\alpha$.

答:
We provide you with a diagram depicting the coordinate system transformation described in your problem statement. The coordinate system, which has $O_0 = A=(0,0)$ as its origin, is shown in $\mathrm{\mathbf{\color{red}{red}}}$. The same coordinate system after translation, which is also drawn in $\mathrm{\mathbf{\color{red}{red}}}$, has its origin at $O=(o_\mathrm{x}, o_\mathrm{y})$. This coordinate system is then subjected to a clockwise rotation to obtain the coordinate system drawn in $\mathrm{\mathbf{\color{green}{green}}}$.

We leave it to OP to obtain the sought expressions for the coordinates of point $B=\left(b_\mathrm{xnew}, b_\mathrm{ynew}\right) $ with respect to a coordinate system having its origin at $A=\left(0, 0\right)$. First, you have to find the angle $\measuredangle BON$ in terms of $\alpha$ and $\beta$. Then, express $QB$ and $BN$ in terms of $OA$, $OB$, $\alpha$, and $\beta$.


问:
circles in a square
N circles of maximum possible equal radius are contained in a square of side 1 with no overlap among the circles. What is the % of the area of the square outside the circle for N=3? (would appreciate a way of thinking about solving this for the general case; also, is there an easy way to extrapolate this into higher dimensions? i.e. spheres of maximum possible equal radius inside a cube of side 1).

This is similar to the question here, but I am looking for a generalized way of thinking about this, for circles and cubes:
Maximum area of $2$ circles in a square
Thank you.

答:
Hint: This figure is drawn with condition that all three circles are mutually tangent.
As can be seen maximum r is when three circle are symmetric about DB. In this case DB bisects $\angle SPT$ which is $60^o$. BD makes $45^o$ angle with PQ which is parallel with AD. Therefore:

$\angle QPS=45-30=15^o$

You can use this figure to find the ratio.

$AD=2 r +2r(1+ \cos (\angle QPS)=2r(1+ \cos (\angle QPS)$;

With $\angle (QPS)=15^o$  we get  $r=0.25435$

Can do the rest?

Title: circles in a square


问:
Combinatorics problem - choosing $6$ cards out of $32$-card deck so that there are exactly three different suits (Inclusion-Exclusion)
Suppose we have a deck of $32$ cards with $8$ cards of each of the four suits. In how ways we can choose six cards such that there are cards of exactly three different suits among the chosen cards?

I believe inclusion-exclusion principle is the way to solve it, where we first count the number of total ways to choose $6$ cards out of $32$ (which is $\binom{32}{6}$), then exclude the number of combinations where exactly two of the suits are missing (which is $\binom{4}{2}\binom{16}{6}$) and then by inclusion-exclusion formula add the combinations where all three suits are missing (which is $\binom{4}{3}\binom{8}{6}$). The number of combinations of all $4$ suits missing is, of course, zero.

My question is - where is my logic incorrect? I know it is, but can't seem to spot the error.

答:
It is better to count the hands that are void in exactly one suit.

If we try $\binom{24}6 - \binom4 2 \binom{16}6 + \binom4 3\binom8 6$, we will get the number of hands void in at least one void suit, since we are only  subtracting the overcount of the multiply void ones, whereas to get the number with exactly one void suit, we need to subtract the entire count of the multiply void ones

This number is 4 times the number of hands void in,say,$\;$$ \spadesuit$ which can
be combined in $3$ ways to form hands void in $2$ suits, and eliminate the over count by adding the $3$ ways in which the hand can be void in $3$ suits in  combination with the $\spadesuit$ to give us $4[\binom{24}6 - 3\binom{16}6 +3\binom8 6]$


问:
A complex function which is differential only in the closed unit disc.
Let $f:\mathbb C\to\mathbb C$ be a complex function. As usual, we say that $f$ is differentiable at $z_0$ if the following limit exists
$$
\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}
$$
Can $f$ be differentiable only in the closed unit disc? I cannot find any simple example.

答:
Let
$$f(z)=(|z|-1)^2\chi_{\mathbb{C}\backslash\overline{\mathbb{D}}}(z)$$
where $\chi_E(z)$ is one if $z\in E$ and otherwise is $0$

It is clear that $f$ is complex differentiable on the open unit disc and it is not outside the closed unit disc (since $(|z|-1)^2$ is not complex differentiable). It remains to prove that it is differentiable on the boundary.  It suffices to prove that

$$\lim_{z\to z_0\in\partial\mathbb{D}\\z\in \mathbb{C}\backslash\overline{\mathbb{D}}}\frac1{z-z_0}(|z|-1)^2=0\\
\lim \left|\frac1{z-z_0}(|z|-1)^2\right|\le \lim \frac{1}{|z|-1}(|z|-1)^2=0$$


问:
Continuity on the diagonal of a certain function
I have just started my course in multivariable calculus and I am doing several exercises to practice. In particular, I have not very clear how to do this one, so any help will be welcome.

Given $\Psi\in C_c^\infty(\mathbb{R}^d,\mathbb{R}^d)$, then the map defined by

$$F_{\beta}(x,y)= \left\{ \begin{array}{cc}
             \dfrac{\Psi(x)-\Psi(y)}{|x-y|^\beta} & x\neq y \\
             \\ 0  & x=y
             \end{array}
   \right.$$
is continuous for $\beta\in (0,1)) $ but not for $\beta=1$.

My attempt:

Out of the diagonal $x=y$ the  vector field is continuous. Furthermore,
by mean value theorem, we know that $|F_{\beta}(x,y)|\leq{M}|x-y|^{1-\beta}$, so we can concluded that $F_\beta$ is continuouus when $\beta \in (0,1)$.

However I don't know how to prove that $F_1$ is not continuous.

答: When $\Psi \equiv 0$ the function $F_{\beta} $ is continuous for any $\beta$, so we cannot prove that $F_{\beta} $ can never be continuous for $\beta=1$.  But we can given an example to show that $F_{\beta} $ need not be continuous when $\beta=1$. For this take $d=1$ and take $\Psi$ such that the function is equal to the identity function  in some interval around $0$. In this case $F_{\beta} $ is not continuous at $(0,0)$.


问:
Is it possible to rewrite $exp(x)$ with $x$ as the base
Simple question, rewriting $e^x$ with $x$ as the base- or more generally, trying to remove $x$ from the exponent.
I don't know if it's possible, as I played around with it for a little while and got nothing. If it is, I suspect it has something to do with the Lambert W function. If it's not possible, is there an intuitive reason as to why not or is it just something that can't be done algebraically?

I'm using it to simplify a string of terms into a shorter string of terms with $x$ as the base for each one. It's just convenient if the answer could be given in this polynomial-like form.

Thanks, and sorry if I was a bit vague or missed details.

Edit: It really is what's on the label; $e^x = x^?$- it's meant to be general.

If you're solving for $?$ in the situation above,

$e^x=x^y$

$y=\log_x(e^x)$

$y=x*log_x(e)$

$y=\frac{1}{\log_{e^x}(x)}$

And from there, I don't know how to go about reducing the equation to a single $x$, let alone removing the log and putting it in the form of $x^y$. Any ideas?

答:
There's no $y$ such that $e^x = x^y$ for all $x$, because in the long term, the exponential beats any polynomial.

One way to see it, is when defining the exponential as a power series: $e^x := \displaystyle\sum_{n \geq 0} \dfrac{x^n}{n!}$. In particular, if the expression you're trying to simplify has $x \approx 0$, you can use the expression above and neglect terms above a certain order (that's called a MacLaurin/Taylor series). For example, an approximation of order $2$ would be $e^x = 1 + x + \dfrac{x^2}{2} + O(x^3)$, where $O(x^3)$ denotes the remainder, which is of order $x^3$.

Another expression, which is not in polynomial form but might come in useful, is Euler's relation $e^{ix} = \cos x + i\sin x$.