问:
x orthogonal to y equivalent conditions
Show that for vectors $ x,y \in H $ condition $ x \bot y $ is equivalent to:

a) $ \|x+ \lambda y\|=\|x- \lambda y\|   $

b) $ \|x+ \lambda y\| \geq\|x\|   $

For every $ \lambda $

To show that $ x \bot y \Rightarrow$ a I have:

$ \|x+ \lambda y\|^2=\langle x+ \lambda y, x+ \lambda y \rangle= \langle x, x+ \lambda y \rangle+ \langle \lambda y, x+ \lambda y \rangle= \overline{\langle x+ \lambda y, x\rangle} + \overline{\langle x+ \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle + \langle \lambda y, x\rangle }+ \overline{\langle x, \lambda y\rangle + \langle \lambda y, \lambda y\rangle }= \|x\|^2+ \overline{\lambda}\langle x,y \rangle+\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $

Now because $\langle x,y \rangle=0$ we have

$ \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$

And analogously
$ \|x- \lambda y\|^2=\langle x- \lambda y, x- \lambda y \rangle= \langle x, x- \lambda y \rangle- \langle \lambda y, x- \lambda y \rangle= \overline{\langle x- \lambda y, x\rangle} - \overline{\langle x- \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle - \langle \lambda y, x\rangle }- \overline{\langle x, \lambda y\rangle - \langle \lambda y, \lambda y\rangle }= \|x\|^2- \overline{\lambda}\langle x,y \rangle-\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $

Taking $\langle x,y \rangle=0$ we have

$ \|x- \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$

So $ \|x+ \lambda y\|^2=  \|x- \lambda y\|^2$ and  $ \|x+ \lambda y\|=  \|x- \lambda y\|$

That shows that  $ x \bot y \Rightarrow a $ but how can I proof that  $ a \Rightarrow x \bot y $

I have the same problem with b. It is easy to show that $ x \bot y \Rightarrow b $:

Taking $\langle x,y \rangle=0$ we have

$  \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2 \geq  \|x\|^2$ so $  \|x+ \lambda y\| \geq \|x\|^2 $

How to prove that  $ b \Rightarrow x \bot y $?

答:
I am assuming that $H$ is a real Hilbert space but the complex case needs only obvious modifications.

Squaring both sides of a) we get $\|x\|^{2}+\lambda^{2}\|y\|^{2}+2\lambda \langle x, y \rangle=\|x\|^{2}+\lambda^{2}\|y\|^{2}-2\lambda \langle x, y \rangle$ which gives $\langle x, y \rangle=0$.

Similarly, b) gives $\lambda^{2}\|y\|^{2}+2\lambda \langle x, y \rangle \geq 0$.If this holds for all real $\lambda$ then we can divide by $\lambda$ and let $\lambda \to 0$ to see that $\langle x, y \rangle=0$

Title: x orthogonal to y equivalent conditions


问:
A circle $\Gamma_2$ centered at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$, and the segments $PQ$ and $CD$ intersect at $M$

Let $CD$ be a chord of circle $\Gamma_1$ and $AB$ be a diameter perpendicular to $CD$ at $N$ with $AN > NB$. A circle $\Gamma_2$ centered at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$, and the segments $PQ$ and $CD$ intersect at $M$. Given that the radii of $\Gamma_1$ and $\Gamma_2$ are $61$ and $60$ respectively, find $AM$.

What I Tried: Here is a picture:-

I couldn't understand the question properly, as it is written that $AB \perp CD$ but I cannot understand in what point the perpendicular line is drawn. The problem overall also seems complicated so I cannot understand anyway on how to start solving it.

Can anyone help me? Thank You.

Edit:- As @cosmos5 mentioned in the comments, my drawing as well my thinking was totally wrong. Here is the new picture :-

As usual, the problem seems complicated and I can't think of anyway to start this, I can however see that we might use Pythagorean Theorem in some form or the other but I cannot understand how since I don't know any lengths other than the radii of the circles.

答:
$CN = 60$

$ON^2 = 61^2 - 60^2 \implies ON = 11$

$AN = AO + ON = 72$.

Now we need to find $MN$ and we then know $AM$.

$MN = CN - CM = 60 - CM$

If $\angle CON = \theta$, $\sin \theta = \frac{60}{61}$

$CM = \frac{CF}{\sin \theta}$ ($\angle FCM = 90^0 - \theta)$

$CF^2 = 60^2 - PF^2$

$OF^2 = (61-CF)^2 = 61^2 - PF^2$

Solving you get $CM = MN = 30$.

$AM = \sqrt{72^2 + 30^2} = 78$

ContentLicense: CC BY-SA 4.0


问:
Alternative proof request: If $C=\{x^2,x\in S\}$, show that $\sup(C)=\max\{\sup(S)^2,\inf(S)^2\}$
This question has only an answer using theorems relying on continuity of a non-decreasing function. While I (think I) can understand the answer, I have this same exercise, but we still not studied continuity yet, we are studying real numbers and preparing to study sequences. Perhaps because of seeing this answer, the only way I see to prove this is using continuity also, but there must be a way without using those theorems about continuity. Could anyone show me the way to prove this only with real numbers/supremum/infimum/etc properties?

Any help would be appreciated.

答:
Hint

Suppose $S\subset [0,\infty )$. Let $c=\sup C$ and $s=\sup(S)$. Let $\varepsilon >0$. There is $x\in C$ s.t. $c-\varepsilon \leq x^2\leq s^2$. Since it holds for all $\varepsilon >0$, we $c\leq s^2$.

Suppose that $c<s^2$, i.e. there is $x\in S$ s.t. $c<x^2\leq s^2$. This contradict the fact that $c=\sup\{x^2\mid x\in S\}$.

Therefore $c=s^2$ as wished.

I let you adapt the proof in the case where $S\subset \mathbb R$ instead of $S\subset [0,\infty )$ only.


问:
Application of ideal class group?
I know what the class group is but I could not come up with any setting where we need to know what the class group actually is.

Does anyone know of an example where we have two number fields with the same class number but different class group, and the difference in class group lets us only use one of them for a number theory problem?

I can think of one example where different class number matter, but the class group is much more subtle.

答:
Your question is why the class group structure matters and not just the class number.

Computations of the class number, if that were all you cared about at first, can be made more efficient by simultaneously computing the class group structure. See Henri Cohen's first book on computational algebraic number theory.

Cohen and Lenstra developed a probabilistic model, based on weighting each finite abelian group by the reciprocal of the size of its automorphism group, to explain trends in numerical tables of class numbers. For example, up to isom. there are two groups of order 9: the cyclic one and a product of two groups of order 3. The first group of order 9 has 6 automorphisms and the second group of order 9 has 48 automorphisms. Reciprocating, the cyclic group gets weight 1/6 and the non-cyclic group gets weight 1/48. Note 1/48 is smaller than 1/6 and in fact it takes a lot longer in tables to find an imag. quad. field with class number 9 whose class group is not cyclic. Of course this might not seem relevant to your question, since a problem about class groups is not a satisfactory example to illustrate why we need to know about class groups rather than class numbers. So consider the following purely "class number" question: how often does a real quadratic field have class number divisible by 5? Cohen and Lenstra provide heuristics that should answer this question, and the point is that the idea behind their heuristic is to study all possible finite abelian group structures of each size and weight them in a suitable way.

As Xue says in a comment, class field theory provides a good example where class group structure matters, e.g., the fields Q(sqrt(-14)) and Q(sqrt(-30)) both have class number 4, but the first has cyclic class group and the second has class group that's a product of two groups of order 2. This means the number of quadratic unramified extensions of the two fields is not the same, since nonisom. groups of order 4 have different numbers of subgroups (equivalently, quotient groups) of order 2. More generally, class field theory tells us that class groups are Galois groups, and surely you appreciate that Galois theory is a lot more than theorems about the sizes of Galois groups! Moreover, class field theory says that any finite abelian extension of a number field $K$ has Galois group over $K$ which is (naturally) isomorphic to a generalized ideal class group of $K$. Only the abelian unramified extensions of $K$ are related to the ideal class group itself; you need generalized ideal class groups of $K$ to access the ramified abelian extensions. The moral is that you should think about the ideal class group merely as the most basic example of generalized ideal class groups, which taken all together tell you what the group structure of the maximal abelian extension of $K$ looks like. In more fancy language, the generalized ideal class groups are replaced by the group of ideles, and in the geometric setting of global fields with positive characteristic, generalized ideal class groups are replaced by generalized Jacobians to describe finite abelian extensions.

At the end of http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/matrixconj.pdf I discuss how to find an example of a 2 x 2 integral matrix which is not conjugate to its transpose over the integers. (Note that any square matrix over a field is conjugate over that field to its transpose, and that is why it's interesting to see an example where the conjugation breaks down over the integers.) The problem reduces to finding an ideal class group with an element of order greater than 2. The first imaginary quadratic field with class number greater than 2 is Q(sqrt(-14)): the class number is 4. But that's not enough to assure you that this can lead to a desired non-conjugate-transposed-matrix-over-Z example, since whether or not a group of size 4 has an element of order greater than 2 depends on the structure of the group. Of course you can simply bypass this and use the field Q(sqrt(-23)), which has class number 3, but the point is that Q(sqrt(-14)) has cyclic class group, so it really contains an ideal class of order greater than 2 and in that .pdf file I use such an ideal class to construct the desired 2 x 2 integral matrix not conjugate to its transpose.

ContentLicense: CC BY-SA 3.0

Title: Application of ideal class group?


问:
Combinations problem help
Four couples have reserved seats in a row for a concert. In how many different ways can they be seated if the two members of each couple wish to sit together?

At first I thought that this was 8!/4!*4! but then I was told by a teacher that this was wrong. I really don't understand the logic behind this problem, can someone lead me through? Why can't we use distinguish-ability?

答:
First think of each couple as a single entity: you can arrange the $4$ couples in $4!$ different orders along the row. Then the members of each couple can sit in either of $2$ orders, so there are altogether $4!\cdot 2^4$ ways to seat the couples.

Your figure of $\frac{8!}{4!}\cdot 4!$ reduces to $8!$, which is the number of ways to seat the $8$ people in the row without any restrictions on who sits next to whom. Since that includes, for instance, the arrangements in which the husbands sit in the first $4$ seats and the wives in the last $4$ seats, it cannot be right: you don’t want to count any of those arrangements.

Title: Combinations problem help


问:
Counting the total number of possible vertical seams for an image
"A vertical seam is a path of pixels connected from top to bottom in an image with one pixel in each row" - Wikipedia. For each pixel, if it is not on either edge, then it can either be connected to the pixel right below, the pixel diagonally left, or the pixel diagonally right. If it is on the edge, then the pixel can be connected to the pixel right below or diagonally to the side opposite of the edge.

I was wondering what the total number of possible vertical seams was for an n by m image (n rows and m columns, so a width of m and height of n).

So far I have an upper and lower bound, but I'm not sure how to move forward. For the lower bound, I got $2^{n-1} \cdot m$, since each pixel has at least 2 choices. For the upper bound, I initially got $3^{n-1} \cdot m$, but I reduced it to $3^{n-1} \cdot m - 2^n$.

Does anyone know a way to find an exact explicit formula?

答:
The number of vertical seams in an $n$-row-by-$m$-column image, $S(n,m)$, is the same as the number of "smooth words" of length $n$ over a size-$m$ alphabet – a smooth word has adjacent letters differing by at most one. Knopfmacher, Mansour, Munagi and Prodinger gave the following formula:
$$S(n,m)=\frac1{m+1}\sum_{j=1}^m(1+(-1)^{j+1})\cot^2\frac{j\pi}{2(m+1)}\left(1+2\cos\frac{j\pi}{m+1}\right)^{n-1}$$
All the sequences of seam counts are OEIS A188866.


问:
Looking for a function that approximates a parabola
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?

答:
I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape

An explicit function for what you want using the Heaviside Step Function :

$$y(x) = y_0\left(1-\frac{x^2}{a^2}\right)\theta(a^2-x^2)$$

giving :

$$\begin{align}
y(0) & =y_0 \\
y(\pm a) & =0
\end{align}$$

This is differentiable and is exact, not just approximate, as the Heaviside function differentiates to the Dirac Delta function.  This may be all you need without recourse to approximations depending on your precise application.

My smoothness constraint should be taken seriously, as it is a requirement for the function downstream in the thing I need it for.

Any smooth approximation for the Heaviside function should be usable as a replacement.  Wikipedia quotes the logistic function as an example :

$$\theta(t)\approx \frac 1 {1+e^{-2kt}}$$

Precisely what function you use as a smooth approximation is going to depend on your precise needs.  Mathworld gives several more limiting expressions you may be able to use.


答:
M lies on common chord of the two circles. The common chord is locus of points having same power wrt the two circles.

Formula for power of a point $X$ is $OX^2-r^2$. So we have

$$MC^2-60^2=MO^2-61^2 $$

Substituting $MN=x$, we get
$$(60-x)^2-60^2=x^2+11^2-61^2$$
$$\Rightarrow MN=30$$

$$\therefore AM=\sqrt{AN^2+MN^2}=78$$

Body:
Let $CD$ be a chord of circle $\Gamma_1$ and $AB$ be a diameter perpendicular to $CD$ at $N$ with $AN > NB$. A circle $\Gamma_2$ centered at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$, and the segments $PQ$ and $CD$ intersect at $M$. Given that the radii of $\Gamma_1$ and $\Gamma_2$ are $61$ and $60$ respectively, find $AM$.

What I Tried: Here is a picture:-

I couldn't understand the question properly, as it is written that $AB \perp CD$ but I cannot understand in what point the perpendicular line is drawn. The problem overall also seems complicated so I cannot understand anyway on how to start solving it.

Can anyone help me? Thank You.

Edit:- As @cosmos5 mentioned in the comments, my drawing as well my thinking was totally wrong. Here is the new picture :-

As usual, the problem seems complicated and I can't think of anyway to start this, I can however see that we might use Pythagorean Theorem in some form or the other but I cannot understand how since I don't know any lengths other than the radii of the circles.


答:
Note that $\overline{\lambda} \langle x,y\rangle + \lambda \langle y,x\rangle = 2\Re (\lambda\langle x,y\rangle)$ so from the alternative expressions for $\|x+\lambda y\|^2$ and $\|x-\lambda y\|^2$ which you have obtained you get that $\Re(\lambda \langle x,y\rangle) = 0$  for every $\lambda \in \mathbb{C}$.

Now, if you take $\lambda = \overline{\langle x,y\rangle}$, then you get that $\Re(|\langle x,y\rangle|^2) = 0$, hence $\langle x,y\rangle = 0$.

For the second point, after squaring we end up with $|\lambda|^2\|y\|^2 + 2\Re(\lambda\langle x,y\rangle) \geq 0$, we take $\lambda = \gamma \overline{\langle x,y\rangle}$ for $\gamma<0$.

Then we divide by $\gamma$ and take the limit $\gamma\to 0^-$, similarly to what Kavi Rama Murthy suggests in his answer. We obtain that $|\langle x,y\rangle|^2 \leq 0$.

Title: x orthogonal to y equivalent conditions

Body:
Show that for vectors $ x,y \in H $ condition $ x \bot y $ is equivalent to:

a) $ \|x+ \lambda y\|=\|x- \lambda y\|   $

b) $ \|x+ \lambda y\| \geq\|x\|   $

For every $ \lambda $

To show that $ x \bot y \Rightarrow$ a I have:

$ \|x+ \lambda y\|^2=\langle x+ \lambda y, x+ \lambda y \rangle= \langle x, x+ \lambda y \rangle+ \langle \lambda y, x+ \lambda y \rangle= \overline{\langle x+ \lambda y, x\rangle} + \overline{\langle x+ \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle + \langle \lambda y, x\rangle }+ \overline{\langle x, \lambda y\rangle + \langle \lambda y, \lambda y\rangle }= \|x\|^2+ \overline{\lambda}\langle x,y \rangle+\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $

Now because $\langle x,y \rangle=0$ we have

$ \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$

And analogously
$ \|x- \lambda y\|^2=\langle x- \lambda y, x- \lambda y \rangle= \langle x, x- \lambda y \rangle- \langle \lambda y, x- \lambda y \rangle= \overline{\langle x- \lambda y, x\rangle} - \overline{\langle x- \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle - \langle \lambda y, x\rangle }- \overline{\langle x, \lambda y\rangle - \langle \lambda y, \lambda y\rangle }= \|x\|^2- \overline{\lambda}\langle x,y \rangle-\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $

Taking $\langle x,y \rangle=0$ we have

$ \|x- \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$

So $ \|x+ \lambda y\|^2=  \|x- \lambda y\|^2$ and  $ \|x+ \lambda y\|=  \|x- \lambda y\|$

That shows that  $ x \bot y \Rightarrow a $ but how can I proof that  $ a \Rightarrow x \bot y $

I have the same problem with b. It is easy to show that $ x \bot y \Rightarrow b $:

Taking $\langle x,y \rangle=0$ we have

$  \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2 \geq  \|x\|^2$ so $  \|x+ \lambda y\| \geq \|x\|^2 $

How to prove that  $ b \Rightarrow x \bot y $?


问:
What is wrong with this construction of the real numbers?
Define a real number $r$ to be

$$r = x + \sum_{k = 0}^\infty \dfrac{a_k}{10^k}$$

where $x$ is a nonnegative integer and $a_k \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.   Denote the additive inverse of $r$ by $-r$.

答: Note that even if you restrict the $a_{i}$ to come from the set $\{0,1,..,9\}$ you will still encounter issues. Consider $r_{1}=\sum^{\infty}_{k=1}\frac{a_{k}}{10^{k}}$ versus $r_{2}=1+\sum^{\infty}_{k=1}\frac{b_{k}}{10^{k}}$ where $a_{i}=9$ for all $i$ and $b_{i}=0$ for all $i$. These two expressions represent the same real number yet are defined with different $x$s and $\{a_{i}\}$s. To 'well define' the real numbers you will have to take steps (like the traditional 'equivalence classes of Chaucy sequences') to quotient out such coincidences.</p>


问:
Solving Differential Equation $y'=\cos(xy)$
My question is about solving differential equation $y'=\cos(xy)$.
I tried to changing variable $u=xy$
\begin{align*}
u &= xy \\
\frac{du}{dx} &= y+ x\frac{dy}{dx}\\
\frac{dy}{dx} &= \frac{1}{x}\frac{du}{dx}-\frac{y}{x}\\
\end{align*}
then since $u=xy\to \frac{y}{x}=\frac{u}{x^2}$
\begin{align*}
\frac{1}{x}\frac{du}{dx}-\frac{u}{x^2}-\cos(u)=\;0\\
\end{align*}
finally
\begin{align*}
x\,du-(u+x^2\cos(u))\,dx=0\\
\end{align*}
since this is not an Exact Differential Equation$($Because of the existence of $\cos(u))$, what is multiplicative "integrating" factor? Is this equation solvable or not?

答:
HINT:
$$ \log ( y')= \log (\cos (xy)) $$
differentiate
$$ y''= -\sqrt{1-y^{'2}}(xy'+y)$$
Once again differentiate
$$y^{'''}\sqrt{1-y^{'2}}= xy''(1+y{'2})+y' (xy'+y+2+2 y^{'2})$$

May be no closed form solution. Numerical integration ignoring spurious solutions.


问:
Find general equation for the following sequence
Given the following sequence: $$a_1 = 1, a_{n+1}=2*a_{n} + 1$$

what is the general equation for $a_n$?
I couldn't find it because the sequence is not in the form of $a_{n+1} = a_n * c$ neither is it
in the form of $a_{n+1} = a_n + c$, but more like a combination of the two.

答:
Hint:$a_1 = 1, a _ 2 = 3, a_3 = 7, a_4 = 15, a_5 = 31$.

What does this sequence look like? Can you guess the form?

Can you prove this is the equation for $a_n$ via induction?

If you are still stuck, write down what you've tried and where you're stuck at.


问:
Vanishing of a non-constant function on projective variety
In Gathmann's notes, there is a lemma that if $X$ is a projective variety, then $O_X(X) = k$.

There is one step in the proof saying that if $\varphi\in O_X(X)$, then we can suppose $\varphi = \frac{f}{g}$ where $f,g \in S(X)^d$ with $g\neq 0$. If $\deg g >0$, then $Z_p(g)\cap X\neq \emptyset$.

I don't understand why $Z_p(g)\cap X \neq \emptyset$. Here is my attempt:

Suppose $X = Z_p(I)\subset \mathbb{P}^n$. If $Z_p(g)\cap X = Z_p(g, I) = \emptyset$, then we have $Z_a(g, I) = \emptyset$ or $\{0\}$. The first case is impossible since $Z_a(g,I) = \emptyset$ implies $\sqrt{(g, I)} =(1)$. Then $1\in (g,I)$ and hence there is $c$ such that $cg+I = 1+I$. Therefore $g+I$ is invertible in $k[x_0,\ldots, x_n]/I$ which implies $\deg g = 0$. The second case implies $\sqrt{(g+I)} = (x_0,\ldots, x_n)$. I don't know how to get a contradiction from here.

答: Suppose that $Z_p(g)\cap X \neq \emptyset$ is false. Let $p, q$ be any two distinct points in $X$ and pick a polynomial $g'$ such that $g'(p) = 0$ and $g'(q) = 1$ (why can we find such a function?). Define the morphism $\varphi \colon X \rightarrow \mathbf{P}^1$ by $x \mapsto [g(x), g'(x)]$ (why is this a morphism?). The image of $\varphi$ is closed because projective spaces are complete and therefore a point (why?). But $\varphi(p) \neq \varphi(q)$ and we have a contradiction.


问:
$L = \begin{cases} M & \text{on }\; A\\ N & \text{on} \; A^c \end{cases}$ is a stopping time
In Durrett's probability theory and examples, it asks suppose $M \le N$ are stopping times. If $A \in\mathcal{F}_M$, then
$$L = \begin{cases} M & \text{on }\; A\\ N & \text{on} \; A^c \end{cases}$$
is a stopping time.

To show that $L$ is a stopping time, I believe what I need to show is that $ \{L= n \}$ is $\mathcal{F}_n$ measurable. I'm not sure how we can write $\{ L = n\}$ in terms of $M$ and $N$. What I have is
$$\{ L = n \} = ( \{ M = n\} \cap A) \cup ( \{ N = n \} \cap A^c)$$
Since RHS is $\mathcal{F}_n$ measurable, we see that $\{ L = n\}$ is $\mathcal{F}_n$ measurable.

However, the textbook gives us a completely different proof where it states
$$\{ N = n\} = (\{L = n\} \cap A) \cup \bigcup_{m=1}^n (\{ L = m\} \cap \{ M = n\} \cap A^c)$$
I can see how we can write $\{ N = n\} = (\{ L = n \} \cap A) \cup (\{ L = n\} \cap A^c)$ but how do we get $\bigcup_{m=1}^n (\{ L = m\} \cap \{ M = n\} \cap A^c)$?

答: Your proof assumes that $\mathcal F_M \subseteq \mathcal F_N$ which gives $A \in \mathcal F_N$ and hence $(N=n)\cap A^{c} \in \mathcal F_n$. The proof in the  book does not assume this result and it uses  only the definition of stopping times.


答: The way I think about the problem is that you have four couples $A,B,C,D$, thus $4!$ ways to arrange them in a row. We then have two different orderings for each couple, so $2^4$ arrangements hence $2^4 \cdot 4!$ different seating arrangements.</p>

Title: Combinations problem help

Body:
Four couples have reserved seats in a row for a concert. In how many different ways can they be seated if the two members of each couple wish to sit together?

At first I thought that this was 8!/4!*4! but then I was told by a teacher that this was wrong. I really don't understand the logic behind this problem, can someone lead me through? Why can't we use distinguish-ability?


Title: Find general equation for the following sequence

Body:
Given the following sequence: $$a_1 = 1, a_{n+1}=2*a_{n} + 1$$

what is the general equation for $a_n$?
I couldn't find it because the sequence is not in the form of $a_{n+1} = a_n * c$ neither is it
in the form of $a_{n+1} = a_n + c$, but more like a combination of the two.

Body:
$$
\begin{align}
a_{n+1} &= 2a_{n} + 1 = 2(2a_{n-1}+1) + 1 \\
&= 4a_{n-1} + 2 + 1 = 4(2a_{n-2} + 1) + 2 + 1 \\
&= 8a_{n-2} + 4 + 2 + 1 = ... \\
&= 2^na_1 + \sum\limits_{k=0}^{n-1} 2^k
\end{align}
$$
Using the formula for the geometric series and using the fact that $a_1 = 1$, we are able to obtain:
$$
\begin{align}
a_{n+1} = 2^na_1 + \sum\limits_{k=0}^{n-1} 2^k = 2^na_1 + 2^n-1 = 2^{n+1}-1
\end{align}
$$


问:
Can someone explain me how did they solved this limit problem?

This was the problem. And they solved it by doing this:

I don't understand what they they did. Can someone help me understand it?

答:
This is some version of the "l'Hospital" rule. Or some truncated version of Taylor expansion. I don't like it.

What you can do is this : using Taylor expansion for function $f$ at point $x$, you have :
$$f(y)=f(x)+(y-x)f'(x) + o(y-x)$$
So, putting $x=2$ and $y=2+2h+h^2$ :
$$f(2h+2+h^2) = f(2) + (2h+h^2)f'(2)+o(h) = f(2)+h((2+h)f'(2)+o(1))$$
and with $x=1$ and $y=1+h-h^2$ :
$$f(1+h-h^2)=f(1)+(h-h^2)f'(1)+o(h) = f(1)+h((1-h)f'(1)+o(1))$$
so
$$\frac{f(2h+2+h^2)-f(2)}{f(1+h-h^2)-f(1)} = \frac{h((2+h)f'(2)+o(1))}{h((1-h)f'(1)+o(1))} = \frac{(2+h)f'(2)+o(1)}{(1-h)f'(1)+o(1)}$$
If, and ONLY IF, $f'(1)\ne0$, then this fraction has limit $\frac{2f'(2)}{f'(1)}$.

The solution provided is not false, but the first equality needs a lot of justifications, so I don't like it. But maybe there's a theorem used in american teaching (l'Hospital ?) that we don't use in France, that could justify the solution.


问:
Need help understanding why formula for prime number is correct
I was reading a Mathematical paper from the '40s, and came across the following formula:

$$q = \theta\sqrt{\frac{n}{\nu}}$$

Where $q$ is a prime number, $\nu$ is a given integer, and $\theta \in (1, 1+\vartheta)$, where $\vartheta$ is a given real. $n$ is an integer.

For the argument in the paper to work, I believe this formula may only fail for finitely many $n$. The value of $\theta$ may of course depend on $n$. The only reference why this should be the case, is that: "According to the law of prime numbers, for sufficiently large $n$ such a prime $q$ exists." I was not able to find any such law, and this is were I need help.

Why should this be the case?

Link to the original paper, which is in Russian: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=5985&option_lang=eng

答:
It seems the full claim is:

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n\in\Bbb N$, there exists $\theta\in(1,1+\vartheta)$ such that $\theta\sqrt{\frac n\nu}$ is prime.

or equivalently

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n$, there exists  a prime $q$ such that $$\sqrt{\frac n\nu}<q<(1+\vartheta)\sqrt{\frac n\nu}.$$

For this it is sufficient to show

Let $\vartheta>0$. Then for all $x\gg0$,
$$ \pi((1+\vartheta) x)>\pi(x).$$

What the Prime Number Theorem gives us is: For $\alpha <1<\beta$ of our choice, we have
$$ \frac{\alpha x}{\ln x}<\pi(x)<\frac{\beta x}{\ln x}$$
for all $x\gg 0$. So we can certainly achieve $\frac \beta\alpha < 1+\vartheta$.
Then
$$\begin{align} \pi(x+\vartheta x)-\pi(x)
&>\frac{\alpha x(1+\vartheta)}{\ln x+\ln(1+\vartheta)}-\frac{\beta x}{\ln x}\\
&>\frac{\alpha x(1+\vartheta)}{\ln x+\vartheta}-\frac{\beta x}{\ln x}\\
&=\frac{\alpha x(1+\vartheta)\ln x - \beta x\ln x - \beta\vartheta x}{(\ln x+\vartheta)\ln x}\\
&=\frac{x}{(\ln x+\vartheta)\ln x}\cdot\bigl(\overbrace{(\alpha(1+\vartheta)-\beta)}^{>0} \ln x-\beta\vartheta\bigr ).
\end{align}$$
This is positive for $x\gg0$.


问:
How to neatly organize all real increasing functions?
Let $I$ be the set of increasing functions $\mathbb{R}\longrightarrow\mathbb{R}$ modulo asymptotic behavior, i.e.

\begin{equation}
f\sim g \qquad\Longleftrightarrow\qquad \lim_{x\rightarrow+\infty}\frac{f(x)}{g(x)}=1
\end{equation}

It would be super cool if one could "index" each function $f\in I$ with a real number $\lambda_f$ such that

\begin{equation}
\lambda_f < \lambda_g \qquad\Longleftrightarrow\qquad \lim_{x\rightarrow+\infty}\frac{f(x)}{g(x)}=0
\end{equation}

If this could be done, one would have for each increasing function a unique index indicating how rapidly the function is increasing; vice versa for each real number one could determine a unique function that has a particular way of increasing to $+\infty$. It would be a "universal increasing function classifier". For instance I imagine that all polynomial functions would be indexed by a bounded interval on the reals.

My question: does such a bijection exist? Can it be constructed?

答:
Can't be done. The order type of the reals is not the same as the partial order on non-decreasing functions implicit above.

Since $\Bbb R$ is order-isomorphic to $(0,\infty)$ the space $X$ is order-isomorphic to $Y$, the space of non-decreasing functions from $(0,\infty)$ to itself. The construction I have in mind comes out simpler with $Y$ in place of $X$.

Exercise If $f_0,f_1,\dots\in Y$ there exists  $g\in Y$ such that $\lim_{x\to\infty}f_n(x)/g(x)=0$ for every $n$.

(Now if your $\lambda_f$ thing exists, a sequence $f_n$ with $\lambda_{f_n}>n$ leads to a contradiction.)

Hint:$$g(x)=x\max_{n\le x}f_n(x).$$

Note I'm assuming here that the map $f\mapsto\lambda_f$ is a bijection. This seems to follow from the "vice versa" in the question. In any case, if you don't want to assume that, do this:

Say $A=\sup_f\lambda_f$. First note that there does not exist $f$ with $\lambda_f=A$, because for any $f$ there exists $g$ with $\lambda_g>\lambda_f$. Now choose $f_n$ so $\lambda_{f_n}\to A$; define $g$ as above, and $\lambda_g>\lambda_{f_n}$ leads to a contradiction.


问:
Irrationality of $\sum_{n=2}^\infty\frac{1}{\sqrt{n}(n-1)}$
I was looking at the following series $$S = \sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)}$$ which WA says that it converges to some number (WA1: 2.18401, WA2: 2.0839...). I noticed that it is (almost) equal to $$\sum_{p=3,5,...}^\infty \zeta\left(\frac{p}{2}\right) = \zeta\left(\frac{3}{2}\right) + \zeta\left(\frac{5}{2}\right) + \cdots,$$ i.e., sum of "all halves" (from 3/2) of Rieman zeta functions.

Here is my thinking:
$$\frac{\sqrt{n}}{n(n-1)} = \left(\frac{1}{n}\right)^\frac{3}{2}+\left(\frac{1}{n}\right)^\frac{5}{2}+\left(\frac{1}{n}\right)^\frac{7}{2}+\dots$$
Sum it over $n$ from 2 to $\infty$, you get e.g. for the first part $\sum_{n=2}^{\infty}\frac{1}{n^{3/2}},$ which is almost $\zeta\left(\frac{3}{2}\right)$, except that the summation does not start at $n=1$. To fix this, I do
$\sum_{n=2}^\infty\frac{1}{n^{3/2}}=\zeta\left(\frac{3}{2}\right) - 1.$

Thus, I conclude that $$\sum_{p=3,5,...}^\infty \left(\zeta\left(\frac{p}{2}\right) - 1\right) = \sum_{n=1}^\infty\left(\zeta\left(n+\frac{1}{2}\right) - 1\right)= \sum_{n=2}^\infty\frac{\sqrt{n}}{n(n - 1)}.$$
It is also known (see https://en.wikipedia.org/wiki/Riemann_zeta_function) that
$\sum_{n=2}^\infty\left(\zeta(n) - 1\right) = 1$.
We can thus combine those sums, leading to
$$\sum_{p=3}^\infty\left(\zeta\left(\frac{p}{2}\right) - 1\right)=\sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)} + 1$$

My question is. If it is correct, is this identity known? If so, can we conclude that $\sum_{n=2}^\infty\frac{\sqrt{n}}{n(n-1)}$ is an irrational number (sum of square roots "feels" like an irrational number) and thus also the sum of "Riemann halves" is irrational number?

Also, is there something known about the $\sum_{p=1,3,5,...}^\infty \zeta\left(\frac{p}{2}\right)$? Its convergence, etc.

Thanks.

答: Answer to the question if it is known: It seems that the same equality has been mentioned lately in The Spiral of Theodorus and Sums ofZeta-values at the Half-integers</p>


问:
Let $f(x)= x^3+ax^2+bx+c \in \mathbb{Q}[x]$. Show that the splitting field of $f$ over $\mathbb{Q}$ has degree 1, 2, 3 or 6 over $\mathbb{Q}$.
QUESTION: Let $f(x)= x^3+ax^2+bx+c \in \mathbb{Q}[x]$. Show that the splitting field of $f$ over $\mathbb{Q}$ has degree 1, 2, 3 or 6 over $\mathbb{Q}$.

The professor gave us this hint, but I still not understanding. I need to solve this in a step by step way. Using his tips.

HINT: The greatest difficulty would be to show that it cannot be greater than 6. Then, is enough to choose some values for $a, b$ and $c$. Try to find on the part of Galois that the extension has degree $\leq n!$. You need to find polynomials in that way that have splitting fields of degrees $1, 2, 3$ and $6$. And then show that it can't be bigger than that. It cannot be greater than 6 because this happens in the worst case ... It has a real root that has a degree $\leq3$ (it always exists since the polynomial has an odd degree, using the intermediate value theorem) and a complex one (which can also be real) of degree $\leq 2$. Then the degree of extension $\leq 6$. We use intermediate value theorem because polynomials of odd degree have a real root.

I really appreciate your help if you would take a time to help me.

答:
Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Since $\mathbb{Q}$ has characteristic zero, the extension is separable, and it is a splitting field so it is normal. Therefore $L/\mathbb{Q}$ is a Galois extension.

We know that the Galois group $G = \operatorname{Gal}(L/\mathbb{Q})$ acts faithfully on the roots of $f$ in $L$. There are three such roots $\alpha_1,\alpha_2, \alpha_3$ say, so $G$ may be viewed as a group of permutations of $\{\alpha_1,\alpha_2,\alpha_3\}$, which makes it a subgroup of the symmetric group $S_3$. Since $S_3$ has order $6$, it follows that the order of $G$ divides $6$, so it is $1,2,3$ or $6$.

It is a standard result of Galois theory that the degree of a Galois extension equals the order of its Galois group, so $[L : \mathbb{Q}] = \lvert G \rvert$ is $1, 2, 3$ or $6$.

Finally, Piquito's comment shows that each of these possibilities does actually occur.


问:
Where is the mistake? $-1=(-1)^{2/2}=\left(\left (-1\right)\displaystyle ^2\right)^{1/2}=1^{1/2}=\sqrt 1=1$
Where is the mistake?
$$-1=(-1)^{2/2}=\left(\left (-1\right)\displaystyle ^2\right)^{1/2}=1^{1/2}=\sqrt 1=1$$

答: Each $=$ but the second is clearly right. That the second $=$ is wrong proves $\color{blue}{x^{a/b}=(x^a)^{1/b}}$ admits counterexamples with $x<0$. We use $x>0$ in the manipulation$$(x^{a/b})^b=x^a\implies\color{blue}{x^{a/b}=(x^a)^{1/b}},$$which is our only reason to expect the blue part to be true.


问:
Cases for only one mixed partial derivative is continuous
During studying Clairaut's theorem(In the case of 2 variable functions), I started wondering why the condition of the theorem is written as 'both mixed partial derivative (i.e. $f_{xy}$ and $f_{yx}$) are continuous'. There can be two possibilities.

If $f_{xy}$ is continuous, then $f_{yx}$ has to be continuous.

There exist a function $f$ such that $f_{xy}$ is continuous but $f_{yx}$ doesn't.

However, I couldn't prove the first statement, nor found the counterexample, so if you know what is the right statement, please help us.

Thank you in advance

答:
If $f_{xy}$ and $f_{yx}$ are not required to exist, then obviously we can have a function where $f_{xy}$ is continuous but $f_{yx}$ does not exist, such as $f=|y|$.

Theorem. If $f_{xy}$ and $f_{yx}$ exist everywhere, and $f_{xy}$ is continuous at $(0,0)$, then $f_{yx}$ is continuous at $(0,0)$.

Proof. First, the two mixed partial derivatives are equal at $(0,0)$:
$$\begin{aligned}f_{yx}(0,0)&=\lim_{x\to0}\frac{1}{x}(f_y(x,0)-f_y(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac{1}{xy}(f(x,y)-f(x,0)-f(0,y)+f(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac1y(f_x(\theta x,y)-f_x(\theta x,0))\\&=\lim_{x\to0}f_{xy}(\theta x,0)\\&=f_{xy}(0,0).\end{aligned}$$

By continuity of $f_{xy}$ at $(0,0)$, given $\varepsilon>0$ there exists $\delta>0$ such that if $|x|+|y|<\delta$ then $$|f_{xy}(x,y)-f_{yx}(0,0)|=|f_{xy}(x,y)-f_{xy}(0,0)|<\frac{\varepsilon}{3}.\qquad(1)$$ Now fix this $(x,y)$. By definition of derivative there exists $x'$ with $|x'|<(\delta-|x|-|y|)/2$ such that $$\left|f_{yx}(x,y)-\frac{1}{x'}(f_y(x+x',y)-f_y(x,y))\right|<\frac{\varepsilon}{3}.\qquad(2)$$ Again, by definition of derivative there exists $y'$ with $|y'|<(\delta-|x|-|y|)/2$ such that $$\left|f_y(x+x',y)-f_y(x,y)-\frac{1}{y'}(f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y))\right|<\frac{\varepsilon|x'|}{3}.\quad(3)$$ Applying mean value theorem twice, there exist $\eta,\theta\in(0,1)$ such that $$f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y)=x'y'f(x+\eta x',y+\theta y').\qquad(4)$$ Combining formulas (2–4), we get $$|f_{yx}(x,y)-f_{xy}(x+\eta x',y+\theta y')|<\frac{2\varepsilon}{3}.$$ By formula (1), noticing that $|x+\eta x'|+|y+\theta y'|\le|x|+|x'|+|y|+|y'|<\delta$, we have $$|f_{yx}(x,y)-f_{yx}(0,0)|<\varepsilon.$$ Therefore $f_{yx}$ is continuous at $(0,0)$. QED


问:
If $G$ is group with $|G|=p^3$, $p$ prime, then $G'=Z(G)$
I'm trying to solve this problem from my abstract algebra text book:

Being $G$ a non-abelian group of order $p^3$, with $p$ prime. Prove
that $G'=Z(G)$

In my notation $G'$ is the derived subgroup of $G$. I guess I have to prove separatedly that $G'\subseteq Z(G)$ and  $G'\supseteq Z(G)$. My problem is that I don't know how to use the condition $|G|=p^3$. Maybe the Sylow theorems? The first thing I noticed from $|G|$ being $p^3$ is that then $Z(G)$ isn't trivial, but I don't know how to continue. Any help will be appreciated, thanks in advance.

答:
First of all, there are $2$ standard results which I will use here: (both very easy to prove)

$1$. If $G/Z(G)$ is cyclic then $G$ is abelian.

$2$. If $p$ is a prime then groups of order $p$ and $p^2$ are abelian.

So let $G$ be a non abelian group of order $p^3$. Then $|Z(G)|=p$. Indeed, if we had $|Z(G)|=p^2$ then $G/Z(G)$ would be cyclic and then $G$ would be abelian, a contradiction. So $|Z(G)|=p$, and so $|G/Z(G)|=p^2$. Thus $G/Z(G)$ is abelian, and this implies $G'\leq Z(G)$. By Lagrange's theorem it follows that either $G'=\{e\}$ or $G=Z(G)$. Since $G$ is not abelian it can't be the first option. So $G'=Z(G)$.


答:
We use a fundamental theorem of Galois theory, that the degree of a Galois extension equals the order of the Galois group of that extension. Note that extensions obtained by adding roots of a polynomial with coefficients in the field are automatically Galois extensions.

The logic is that since $f(x) \in \mathbb{Q}[x]$ is a cubic, its Galois group (i.e. the Galois group of a splitting field) will be a subgroup of $S_3$ which has order $6$.

More explicitly, let $x_1, x_2, x_3$ be the (complex) roots of $f$. Then certainly $K=\mathbb{Q}(x_1, x_2, x_3)$ is a splitting field. The Galois group $G$ is the set of those automorphisms of $K$ that fix $\mathbb{Q}$, and so are determined by how they act on the roots. However, since any automorphism fixes $f$, the image of a root under any automorphism is still a root, so $G$ permutes the roots and hence $G$ is a subgroup of $S_3$.

Now the second part is actually finding polynomials that have Galois groups $1$, $C_2$, $C_3 = A_3$ and $S_3$.

$1$ is easy enough: just take the product of three linear polynomials such as $(x-1)(x-2)(x-3)$.

For $C_2$, you need a quadratic polynomial with non-rational roots, for instance $(x-1)(x^2+1)$.

For $S_3$, you can repeat the idea in $C_2$ but this time giving a non-rational root to the linear part, e.g. $x^3 -2$.

Getting a polynomial with $C_3$ is perhaps the hardest one, but with a bit of trial-and-error or some additional insight on an object called "the discriminant" $x^3 -3x+1$ is an example.

Body:
QUESTION: Let $f(x)= x^3+ax^2+bx+c \in \mathbb{Q}[x]$. Show that the splitting field of $f$ over $\mathbb{Q}$ has degree 1, 2, 3 or 6 over $\mathbb{Q}$.

The professor gave us this hint, but I still not understanding. I need to solve this in a step by step way. Using his tips.

HINT: The greatest difficulty would be to show that it cannot be greater than 6. Then, is enough to choose some values for $a, b$ and $c$. Try to find on the part of Galois that the extension has degree $\leq n!$. You need to find polynomials in that way that have splitting fields of degrees $1, 2, 3$ and $6$. And then show that it can't be bigger than that. It cannot be greater than 6 because this happens in the worst case ... It has a real root that has a degree $\leq3$ (it always exists since the polynomial has an odd degree, using the intermediate value theorem) and a complex one (which can also be real) of degree $\leq 2$. Then the degree of extension $\leq 6$. We use intermediate value theorem because polynomials of odd degree have a real root.

I really appreciate your help if you would take a time to help me.


问:
Differential equation
$$x^2(x^2+1)y''-2x^3y'+2(x^2-1)y=0$$

Anyway my tutor assumes that one of the solutions must be $y=x^2$, yet I can't seem to prove this, I tried using $y=x^n$ but I can't seem to get an answer that cancels down. I know soon as I get this solution I can use Abel's forumla to get the other solution.

Any help appreciated.

答:
Let $r=x^2$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{dr}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+2\dfrac{dy}{dr}=2x\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+2\dfrac{dy}{dr}=2x\dfrac{d^2y}{dr^2}2x+2\dfrac{dy}{dr}=4x^2\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}=4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}$

$\therefore r(r+1)\left(4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}\right)-4r^2\dfrac{dy}{dr}+2(r-1)y=0$

$4r^2(r+1)\dfrac{d^2y}{dr^2}+2r(r+1)\dfrac{dy}{dr}-4r^2\dfrac{dy}{dr}+2(r-1)y=0$

$4r^2(r+1)\dfrac{d^2y}{dr^2}-2r(r-1)\dfrac{dy}{dr}+2(r-1)y=0$

$\because$ The latter ODE has trivial solution $y=r$

$\therefore$ The former ODE has trivial solution $y=x^2$

Title: Differential equation


问:
Probability of guessing 4-digit code
Super-simple question, I just want to be sure that I'm right. We want to find a 4-digit code, so we can choose numbers from 0 to 9, but repetition are not allowed and the order does not matter. We want to find the probability of guessing right the number after $t$ trials.

The probability in the first trial is

$\frac{1}{\frac{n!} {k!(n − k)!} } = 1/210$

can we say then that the probability in $t$ trials is $1/(210-(t-1))$? Does someone want to propose a more general formula? Thanks.

答:
Failing the $k$ first trials has probability
$$\frac{209}{210}\frac{208}{209}\dots\frac{209-k+1}{210-k+1}=\frac{210-k}{210}$$
So the probability to find the code in $k$ trials or less is $\frac{k}{210}$.

Title: Probability of guessing 4-digit code


问:
How to find Sylow 3-subgroups of $SL_n(F_3)$?
I've enumerate all 24 elements in $SL_2(F_3)$. As $24=2^3\times 3$, I need to find groups of order 3. Sylow Theorem tells me that the number of Sylow 3-subgroups $n_3 \equiv 1\ \text{mod}\ 3$ and divides $8$, so $n_3= 1 \text{ or } 4$.

I've found 2 of them: upper/lower triangular matrices with 1 on the diagonal, i.e.,
$$
S_1=\left\{ \left( \begin{matrix}
 1&  0\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  1\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  2\\
 0&  1\\
\end{matrix} \right) \right\} ,S_2=\left\{ \left( \begin{matrix}
 1&  0\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  0\\
 1&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  0\\
 2&  1\\
\end{matrix} \right) \right\} .
$$
The remaining ones are given by my friend: Groups generated by $
\left( \begin{matrix}
 0&  2\\
 1&  2\\
\end{matrix} \right) ,\left( \begin{matrix}
 2&  2\\
 1&  0\\
\end{matrix} \right)
$ respectively. However, this process is rather experimental than theoretical. I wonder if there is a logical explanation to this, or a method to figure them out more efficiently.

Any idea or help would be appreciated! Thank you!

答: Sylow groups are conjugate. So once you have found one, you may try to conjugate it with suitable elements, in particular generators, of $G$, such as $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ (or both combined)


问:
Bilinear forms on normed vector spaces
For a positive definite bilinear form a: $\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ defined by:
$$a(x,y)=\sum^n_{i,j=1}a_{ij}x_{i}y_{j}$$
$a_{ij}\in \mathbb{R}\ x,y,$ two vectors of length $n$, prove that:
$$a(x,x)\ge \alpha\lVert x \rVert_2^2 \quad \forall x\in \mathbb{R}^n\quad\alpha>0$$
After rewriting a bit we can see, that it is equivalent to saying that
$$a\left(\frac{x}{\lVert x\rVert_2},\frac{x}{\lVert x\rVert_2}\right) \ge \alpha$$
Doesn't this already show that due to the bilinearity, such an $\alpha$ exists, since we can always pick it such a way, that:$$\alpha = \frac 1{\lVert x\rVert_2^2}a(x,x)$$or if need be:
$$\alpha = \frac 1 {\lVert x\rVert_2^2+1}a(x,x)$$
to ensure that it is less than the bilinear form? Since all factors are $>0$ it seems to me like it would work but it is so simple that I'm uncertain about it. Is there anything I overlooked/missed?

答:
Reducing the problem to vectors of norm $1$ is the first step. The second step is to notice that your bilinear form has a positive minimum on the unit sphere
$\mathbb{S}^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}$, because it is continuous and the unit sphere is compact.

Title: Bilinear forms on normed vector spaces


问:
Let $(X,T_{p})$ be the particular point topology with $p \in X$. Let $H \subseteq X$. Prove that if $p \notin H$, then $(H,T_{p|H})=(H,T_D)$
I have tried to prove that every subset of $H$ is open in $T_{p|H}$ in this case but I've been thinking for hours and I really don't know how to start.

答:
Hint: A subset $S\subseteq H$ is open in $T_{p|H}$ if and only if there is an open set $U$ in $T_p$ such that $S=U\cap H$.

So, given any subset $S\subseteq H$, find $U\in T_p$ such that $S=U\cap H$.


问:
Does the categorical notion of a coequalizer coincide with the usual quotient objects considered in algebra?
I heard that the category-theoretic formulation of "quotient object" is that of a coequalizer. Instead of taking the quotient of, say, a group $G$ by some normal subgroup, one takes the quotient (really, the coequalizer) by some diagram of the form $H\rightrightarrows G$ consisting of two arrows from some group $H$ to $G$.

I wonder: Can each quotient $G/N$ of $G$ by some normal subgroup $N$ be written as the coequalizer of some diagram $H\rightrightarrows G$? What about the converse: Can each coequalizer be written as the quotient by some normal subgroup?

Of course, I'm not only interested in groups, but I also want to know whether the same things are true for rings/ideals, modules/submodules, and so on.

答:
Given two algebraic objects like you describe (group and normal subgroup, ring and ideal, and so on), their quotient is the coequalizer of the inclusion map and the trivial (zero) map.

For instance, given a group $G$ and a normal subgroup $N$, there is an incision map and a trivial map $N\to G$. Their coequalizer is $G/N$.

This special case of a coequalizer (between the zero map and another, not necessarily injective / mono map) is called a cokernel.


问:
What is wrong with this construction of the real numbers?
Define a real number $r$ to be

$$r = x + \sum_{k = 0}^\infty \dfrac{a_k}{10^k}$$

where $x$ is a nonnegative integer and $a_k \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.   Denote the additive inverse of $r$ by $-r$.

答:
If you confine $0\le a_k \le 9$ - nothing.

However, the definition is dependent on base 10, and various properties about limits/completeness need to be proved. I think Prof T Gowers has a note somewhere about how this works (couldn't find it immediately, to my shame).

The normal characterisations of the reals have been fine-tuned to the properties which seem to matter in further mathematical development. They can be shown to be equivalent (under appropriate conditions).

[Note Kaya's answer, which I know the aforementioned Prof Gowers has dealt with in his characteristically efficient fashion]


答:
The given solution is not correct - although its final result is right - because we do not know whether $f'$ exists around $2$ and $1$, resp.

So, we should only use the existence of the two given derivatives and this is possible:

We know

$$\lim_{t\to 0}\frac{f(2+t)-f(2)}{t}=f'(2)=6 \text{ and } \lim_{t\to 0}\frac{f(1+t)-f(1)}{t}=f'(1)=4$$

Hence, only using this, rewrite
\begin{eqnarray*}\frac{f(2+2h+h^2)-f(2)}{f(1+h-h^2)-f(1)}
& = & \frac{f(2+2h+h^2)-f(2)}{2h+h^2}\cdot\frac{h-h^2}{f(1+h-h^2)-f(1)}\cdot\frac{2h+h^2}{h-h^2}\\
& \stackrel{h\to 0}{\longrightarrow} & f'(2)\cdot \frac 1{f'(1)}\cdot 2 = 3
\end{eqnarray*}

Title: Can someone explain me how did they solved this limit problem?

Body:
This was the problem. And they solved it by doing this:

I don't understand what they they did. Can someone help me understand it?


问:
Are two lines in the exact same position one line?
I was just now reading the answers to the question "Is a line parallel with itself?" and that lead me to the question are two lines in exactly the same position in fact just one line?

One can think of two lines that happen to be in the same place. Taking two parallel lines, keeping one fixed, moving the other to it, then have it change sides.  Are there always two lines?  Or do we have an instant when there is only one?

答:
I think so, yes. If two lines are in the exact same position then certainly they have the same set of points. The same line can be parameterised in two different ways, but as As Ryan G pointed out in the comments, the lines represented by the two different parameterisations is in fact the same line.

So in a non-parameterised (usual) context, $y = 2x+1$ and $4x-2y=-2$ are considered to be the "same line" because they represent the same set of points.

With parameterisation: $(x_1(t),y_1(t)): x_1(t) = t$ and $y_1(t) = t $ corresponds to the same set of points as $(x_2(t),y_2(t)): x_2(t) = 2t$ and $y_2(t) = 2t $, despite the value of $(x_1, y_1)$ being different to the value of $(x_2, y_2)$ for a given value of $t$. This means that, despite the two lines having different paths, the two lines are still considered to be the same.


问:
Show that $|\mathcal{M}(\mathbb{R})|=|\mathcal{P}(\mathbb{R})|$
I have to prove that $\mathcal{M}(\mathbb{R})=\{E \subset \mathbb{R}:\text{ E is meaurable}\} $ has the same number of elements of $\mathcal{P}(\mathbb{R})$
($|\mathcal{M}(\mathbb{R})|=|\mathcal{P}(\mathbb{R})|$). How could I start to show this property?
Thank you in advance.

答:
Let $C$ be the cantor set. It's measure is $0$. Since Lebesgue measure is complete every subset of $C$ is again a measure zero set. Also $C$ is uncountable. Hence there is exist a surjective map from $P(C)$ to $P(\mathbb{R})$. Can you complete from here?

$P(C) \subset M(\mathbb{R}) \subset P(\mathbb{R})$

$P(C)$ and $P(\mathbb{R})$ have the same cardinality by the above argument. Hence $|M(\mathbb{R})|=|P(\mathbb{R})|$

$\mathbf{Edit}$: By $C$ is uncountable I meant $C$ and $\mathbb{R}$ have the same cardinality. That is there is a bijection between $C$ and $\mathbb{R}$.


问:
Solving for the value of the limit as n approaches infinity
So, I got a test that contains the question...

I need to find the value of the limit, the answer I inputted was

The teacher said I got it wrong, can someone help me find where I messed up?

答: $$\underset{n\to \infty }{\text{lim}}\prod _{k=2}^n \left(1-\frac{1}{k^2}\right)!=\underset{n\to \infty }{\text{lim}}\Gamma \left(\frac{n+1}{2 n}+1\right)=\Gamma \left(\underset{n\to \infty }{\text{lim}}\left(\frac{n+1}{2 n}+1\right)\right)=\Gamma\left(\frac32\right)=\frac{\sqrt{\pi }}{2}$$


问:
Why does $\delta$ depend on $f$?
In my textbook on metric spaces by N.L. Carothers, the defintion for a continuous map is as follows (slightly modified by my professor from my lecture notes):

Let $(M,d)$ and $(N,\rho)$ be metric spaces and $f : M \to N$. $f$ is continuous at $x \in M$ $\iff$ $\forall$ $\epsilon >0$ $\exists$ $\delta >0$ (which depends on $f$, $x$, and $\epsilon$) such that $d(x,y) < \delta$ $\implies$ $\rho(f(x),f(y)) < \epsilon$.

This definition is pretty standard, however, there is something that has recently been of question to me as I have gotten further in my studies: what does the author (and my professor) mean when they say $\delta$ depends on $f$? I understand that it depends on your choice of $x$ and $\epsilon$, but this is the only literature I have read where the author says that $\delta$ depends on $f$ as well. I have taken this for granted up until now, but I have been thinking about it a lot.

While uniform continuity is much stronger than continuity, this dependence of $\delta$ on $f$ makes an appearance again in their definitions (while on wikipedia, they state $\delta$ only depends on $\epsilon$):

$f : M \to N$ is uniformly continuous $\iff$ $\forall \epsilon > 0$ $\exists$ $\delta >0$ (which depends on $f$ and $\epsilon$ alone) such that $d(x,y) < \delta$ $\implies$ $\rho(f(x),f(y)) < \epsilon$.

If someone could clear this up for me, it would be much appreciated.

答:
The "usual" definition is

Let $(M,d)$ and $(N,\rho)$ be metric spaces and $f : M \to N$. $f$ is continuous at $x \in M$ $\iff$ $\forall$ $\epsilon >0$ $\exists$ $\delta >0$  such that $d(x,y) < \delta$ $\implies$ $\rho(f(x),f(y)) < \epsilon$.

You say it is clear that $\delta$ will (in general) depend on $x$ and $\epsilon$. However, it is also clear that $\delta$ will depend on all parameters occurring in
$$d(x,y) < \delta \implies \rho(f(x),f(y)) < \epsilon .$$
These are $x, \epsilon, f, d, \rho$. That is, if you take other metrics $\bar d$ on $M$, $\bar \rho$ on $N$ and another function $\bar f : M \to N$, you cannot expect that the same $\delta$ will do.

Usually the dependency on $f, d, \rho$ is not explicitly mentioned because the first sentence of the definition is "Let $(M,d)$ and $(N,\rho)$ be metric spaces and $f : M \to N$." This fixes these three parameters.

However, I am sure that you aware that this dependency exists. Think about the proof that the sum of two continuous real-valued functions $f_1, f_2 : M \to \mathbb R$ is again continuous. You certainly chose $\delta(x,\epsilon/2,f_1)$ and $\delta(x,\epsilon/2,f_2)$ and define $\delta(x,\epsilon,f_1+f_2) = \min (\delta(x,\epsilon/2,f_1),\delta(x,\epsilon/2,f_1))$.

Title: Why does $\delta$ depend on $f$?


问:
Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$ , find $[\triangle BDE]$ .

Let $B$ be a point on the circle centered at $O$ with diameter $AC$. Let $D$ and $E$ be the circum-centres of the triangles $OAB$ and $OBC$ respectively. Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$, find $[\triangle BDE]$ .

What I Tried: Here is a picture in Geogebra :-

From the given information, as I am not that good at Trigonometry I could only conclude that $\sin \angle BOC = \frac{4}{5}$ probably means $\angle BOC$ is around $53^circ$ or so, in considering $\angle BOC$ to be acute. The value however, is not exactly $53^\circ$, but varies.

The other thing which is known is the radius of the circle. I don't seem to find a way to use it, especially I can see that$\angle ABC$ is $90^\circ$ but they come to no use, and maybe to find $[\triangle BDE]$, I have to use the formula like $\frac{1}{2}ac \sin \beta$ or so? I also could not understand if the circum-centres $D$ and $E$ can be used in some way or not.

Can anyone help me here? Thank You.

答:
Let $(ABC)=r$, $(BAO)=r_1$ and $(BOC)=r_2$

By sine rule, $$2r_1=\dfrac{r}{\sin \theta} , \quad 2r_2=\dfrac{r}{\sin (\pi/2-\theta)} =\dfrac{r}{\cos \theta}$$

As $\triangle COB$ is isosceles, it can be shown $OE$ is its internal angle bisector and $OD$ exterior angle bisector. So $\angle DOE=90$.

Also $\triangle DOE \cong \triangle DBE \, $ by SSS. $\angle DBE=90$.

$$\therefore [BDE] = \dfrac{1}{2}\cdot BD \cdot BE$$
$$= \dfrac{1}{2}\cdot r_1 \cdot r_2$$
$$= \dfrac{r^2}{4 \sin 2\theta}$$

Given is $r=12$ and $\sin 2\theta = 4/5$


问:
$z_1,z_2,z_3\in \mathbb{S} (1)$ form an equilateral triangle iff the sum is zero
Let $z_1,z_2,z_3$ be complex numbers of module 1. Prove that they form an equilateral triangle if, and only if, $z_1+z_2+z_3=0$.

I know that if they form am equilateral triangle, then
$$ \frac{z_3-z_2}{z_2-z_1}=
   \frac{z_1-z_3}{z_3-z_2}.
 $$
I would appreciate any hint or idea to show that the sum is 0. Thank you.

Body: Hint: $z_1+z_2=-z_3$, so $|z_1+z_2|=1$.


问:
Show that the sequence $x_{n+1}=1+\sqrt {x_n}$ is convergent

Let, $x_1=1$ and for all $n\in \Bbb N$, $x_{n+1}=1+\sqrt {x_n}$. Prove that the sequence is convergent.

I'm unable to how the sequence is monotonic increasing !
We have, $x_{n+1}-x_n=1+\sqrt{x_n}-x_n$. How can I say this is $\ge 0$ ?n Or any other way to prove this ?

答:
This is easier than you think. Note that all terms are positive, so
$$x_n \le x_{n+1} \implies \sqrt{x_n} \le \sqrt{x_{n+1}} \implies 1+ \sqrt{x_n} \le 1 + \sqrt{x_{n+1}}.$$


问:
What would be the answer to this question? We were instructed to use Polya’s Four Steps in Problem Solving
Vacation Money

Last January 2018, my family went on vacation to Cagwait White Beach Park at Cagwait, Surigao del Sur. We also celebrated the 70th birthday of our grandmother. We saved money for the said vacation. We spent 10,000 on the catering services, 5,000 on gasoline, 2,000 on room accommodation, and half of the remaining money for transportation meal. After grandmother remitted 20,000 for the income of
the farm, we were left with 25,000. How much money did we save for the vacation?

Step 1. Understand the Problem

Step 2. Make a Plan

Step 3. Carry-out the Plan

Step 4. Look back on your Work/ Review your Solution.

答: We can solve this problem chronologically in the sense that we account for the information given to us in the order it arrives. First off, calculate how much money was spent in total: $$10,000+5,000+2,000=17,000.$$ We are then told that half of the remaining money was spent on transportation. Clearly, the said amount is unknown, so we declare a variable for it, say $x$. Therefore $\frac{1}{2}x$ has been spent on transportation, and $\frac{1}{2}x$ still remained. $20,000$ was then remitted to the grandmother, leaving us with $25,000$. That is, $$\frac{1}{2}x - 20,000 = 25,000.$$ This is a fairly simple equation that I trust you'll be able to solve for $x$. The only thing left to do would then be to sum up all the quantities to find out how much money we had started with.


答:
$BO$ is common chord of both circumcircles. So $DE$ is perpendicular to $BO$.

$\angle BED = \angle BCO \,$ (as $\angle BEO = 2\angle BCO)$

$\angle BDE = \frac{1}{2} \angle BDO = \angle BAO$

So, $\triangle BDE$ is right angled triangle.

$If \angle BOC = 2 \theta$,

As per extended Sine law,

$\frac{12}{\cos \angle \theta} = 2r$ ($r$ is circumradius of $\triangle BOC$, $\angle BCO = 90^0 - \theta$).

Similarly $\frac{12}{\sin \theta} = 2R$ ($R$ is circumradius of $\triangle AOB$, $\angle BAC = \theta$).

Desired area = $\frac{1}{2} R \times r = \frac{36}{\sin 2\theta} = 45$

Title: Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$ , find $[\triangle BDE]$ .

Body:
Let $B$ be a point on the circle centered at $O$ with diameter $AC$. Let $D$ and $E$ be the circum-centres of the triangles $OAB$ and $OBC$ respectively. Given that $\sin \angle BOC = \frac{4}{5}$, $AC=24$, find $[\triangle BDE]$ .

What I Tried: Here is a picture in Geogebra :-

From the given information, as I am not that good at Trigonometry I could only conclude that $\sin \angle BOC = \frac{4}{5}$ probably means $\angle BOC$ is around $53^circ$ or so, in considering $\angle BOC$ to be acute. The value however, is not exactly $53^\circ$, but varies.

The other thing which is known is the radius of the circle. I don't seem to find a way to use it, especially I can see that$\angle ABC$ is $90^\circ$ but they come to no use, and maybe to find $[\triangle BDE]$, I have to use the formula like $\frac{1}{2}ac \sin \beta$ or so? I also could not understand if the circum-centres $D$ and $E$ can be used in some way or not.

Can anyone help me here? Thank You.


问:
Finding number of elements $(x,y)$ such that $y^2=x^3+1$ in $\mathbb{Z}_p\times \mathbb{Z}_p,$ where $p$ is an odd prime and $p\equiv 2(\mod 3)$
I have been trying to solve this problem for different values of $p.$ For instance, for $p=5,$
$$(x,y)=(0,1),(0,4),(2,2),(2,3),(4,0)$$ satisfies the problem. So we can say tha the given problem has $5$ solutions for $p=5.$

My problem is what can be said about all other primes. Is there any general method to calculate such problem. Intuitively, it seems that for all prime $p$ satisfying the given condition, $(x,y)$ must have at least $2$ solutions and at most $2p$ solutions. But I am unable to prove it.

答:
Since $3\ \nmid \ p-1$ then $f(z)=z^3$ is bijective $\Bbb{F}_p\to \Bbb{F}_p$ (use $z^p = z$ to check that its inverse is $z\to z^{(2p-1)/3}$)  thus
$$E(\Bbb{F}_p) = O \cup \{ (f^{-1}(y^2-1),y),y\in \Bbb{F}_p\}$$
$O$ is the point $(\infty,\infty)$ that we add to make it a group (the discriminant is $-27$)


问:
If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that $\frac{1}{1-z}= \frac{1}{2}\left(1 + i\cot(\theta/2)\right)$

Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$

I have been trying to rearrange the right-hand-side in order to make it look like the required form, but have been struggling to do so and want to see if anyone is able to show me what steps need to be taken to solve this question.

答:
Note that $e^{i\theta}=\cos\theta+i\sin\theta$.

$$\frac{1}{1-e^{i\theta}}$$
$$=\frac{1}{2}\times \frac{-2}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \frac{e^{i\theta}-1-(e^{i\theta}+1)}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \left(1-\frac{e^{i\theta}+1}{e^{i\theta}-1}\right)$$
Note that $e^{i\theta}+1=e^{\frac{i\theta}{2}}\cdot 2\cos(\frac{\theta}{2})$ and $e^{i\theta}-1=e^{\frac{i\theta}{2}}\cdot 2i\sin (\frac{\theta}{2})$.

Plug in these results above,
$$\frac{1}{1-e^{i\theta}}=\frac{1}{2}\times \left(1-\frac{\cot(\frac{\theta}{2})}{i}\right)$$

$$=\frac{1}{2}\times \left(1-\frac{i\cot(\frac{\theta}{2})}{i^2}\right)$$

$$\fbox{$\huge{\dfrac{1}{1-e^{i\theta}}=\frac{1}{2}\times \left(1+i\cot\left(\frac{\theta}{2}\right)\right)}$}$$

I used some results above. See how I got it below.

$$e^{i\theta}+1=\cos \theta+i\sin \theta+1$$

$$=2\cos^2 \frac{\theta}{2}-1+2i\sin\frac{\theta}{2} \times \cos\frac{\theta}{2}+1$$

$$=2\cos\frac{\theta}{2}\cdot e^{\frac{i\theta}{2}}$$

The other result:

$$e^{i\theta}-1=\cos \theta+i\sin \theta-1$$

$$=1-2\sin^2 \frac{\theta}{2}+2i\cos\frac{\theta}{2}\cdot \sin\frac{\theta}{2}-1$$

$$=2i\sin\frac{\theta}{2}\left(\frac{-\sin\frac{\theta}{2}}{i}+\cos\frac{\theta}{2}\right)$$

$$=2i\sin\frac{\theta}{2}\cdot e^{\frac{i\theta}{2}}$$


问:
Convergence Iteration matrix
Let $u_0:=v \in R^d$ and $u_{k+1}:=Au_k, A \in M_d(\mathbb{R})$   being a symmetrical matrix with non-negative coefficients such that the sum of its row elements is $1$ for each row. Moreover, one eigenvalue is $1$, the other eigenvalues are $\lambda_i<1$ . I need to prove the following: $$\lim_{k\to \infty} u_k=\left(\frac{1}{d}\sum_{i=1}^{d} v_i~, \cdots ,~\frac{1}{d}\sum_{i=1}^{d} v_i\right)$$

Can somebody provide some hint or prove the statement ?

答:
By Perron-Frobenius, $1$ is also the spectral radius. If the condition could be sharpened to $|λ_i|<1$, then from the properties of the power iteration we know that $u_k$ converges to the eigenvector of the largest eigenvalue. By the matrix properties, $\newcommand{\one}{1\!\!1}\one$, the one-vector, is that eigenvector. But then also
$$
\one^Tu_{n+1}=\one^TAu_n=\one^Tu_n=...=\one^Tv.
$$
All these properties combined give the claim.

Title: Convergence Iteration matrix


问:
How do I calculate Big Omega Notation for a function?
I was looking at the definition of Big Omega:
\begin{align}
\Omega(g(n)) &= \{ f(n): \text{ there exist positive constants }c \text{ and }n_0 \\
           &\ \ \ \ \ \ \ \  \text{ such that }0 \le cg(n) \le f(n) \text{ for all }n \ge  n_0 \}
\end{align}

I have a function $\frac{n^2+n}2$ to prove that it belongs to $\Omega(n^3)$.

I started like this

$cn^3 \le \frac{n^2+n}2$ but I am not quite sure how do I get $c$, if I pick $n=1$ I get $c\le 1$ but what is the better way of proving this?

答:
In order to have $\frac{n^2 +n}{2} \in \Omega(n^3)$, you need a positive constant $c$ s.t. for $n\geq n_0$ you have

$$0<cn^3 \leq \frac{n^2 +n}{2}$$

But

$$\frac{n^2 +n}{2n^3} =\frac 12\left(\frac 1n + \frac 1{n^2}\right)\stackrel{n \to \infty}{\longrightarrow}0$$

So, there is no such $c$. Hence, $\frac{n^2 +n}{2}\not \in \Omega(n^3)$


问:
Circle theorems question I can't solve involving dodecagon...
This is my last resort, as I can't figure it out at all!

https://www.drfrostmaths.com/getfile.php?fid=655

This worksheet was given to us (GCSE Maths, top set, yr 11, UK) about a month or two ago in a lesson on circle theorems. The question we all had trouble with was the second-from-bottom one on the second page, marked skull-and-cross-bones (b). Nobody, not even the teacher could solve it. The closest I got was by assuming that the non-circle shape was a regular dodecagon and therefore had interior angles of 150 degrees. Thus, I was able to conclude that the shape invloving the portion of the dodecagon inside the circle, as well as angle X, was a hexagon. I then worked out that a+b+x = 270 degrees, where a and b are the angles at those points between the lines of the dodecagon inside the circle and the line linking to angle x. However, I was not able to work out any more! Any help would be greatly appreciated!

答: Draw lines from $A$ and $B$ to the circle centre $O$. This encloses a hexagon to the left of the newly drawn lines; its internal angles must sum to $720^\circ$. Three of those angles are internal angles of the dodecagon, hence $150^\circ$ each; two more are at $A$ and $B$ and are $60^\circ$ each – $150^\circ$ minus the right angle that appears because the new lines are circle radii. This leaves the last angle as $150^\circ$. As this is the angle subtended by a chord from the centre, the angle subtended at any point at the edge – the desired $x$ – is half that, $75^\circ$.


问:
Examples of products of G-complexes
Suppose that $K$ is a $G$-complex and $K'$ is a $G'$-complex. Then $K\times K'$ with the product cell-structure and the weak topology is a ${\bf \mathbf{GxG'}-complex}$

Can someone show me some examples?

I have just started learning about equivariant cohomology theory.

答:
If you ask about examples of $G$-complexes, a nice class arises as follows:
If $V$ is an orthogonal $G$-representation, then its unit sphere has a structure of a finite $G$-complex (A $G$-CW-structure is described in: Lück, Transformation Groups and Algebraic K-theory on page 9, available on the webpage of the author).

For example, the finite cyclic groups $\mathbb{Z}/n$ act on $S^1$ and you can imagine a $G$-CW structure by taking $\mathbb{Z}/n$ act on the $n$-gon by rotations.

Now if you take the product of two such $S^1$'s, one with a $\mathbb{Z}/n$-action, the other with a $\mathbb{Z}/m$-action, you get by the process you described a $\mathbb{Z}/n \times \mathbb{Z}/m$-complex structure on the torus which you can even draw.

Another very important example is of course the following: The universal cover of a connected CW-complex $X$ has a structure of a $\pi_1(X)$-CW complex. If $X$ and $Y$ are connected CW-complexes, their product has fundamental group isomorphic to $\pi_1(X)\times \pi_1(Y)$ and its universal cover is homeomorphic to the universal covers $\widetilde{X}, \widetilde{Y}$ of $X$ resp. $Y$. On this, there is a $\pi_1(X\times Y) = \pi_1(X)\times \pi_1(Y)$-action. Now it can be equipped by a $\pi_1(X)\times \pi_1(Y)$-CW structure by the process you mentioned.

One has to be a little careful with products of $CW$-complexes, when not working in compactly generated weak hausdorff spaces.

Title: Examples of products of G-complexes


问:
Integrating left to right versus right to left.
OK, I understand that when integration is done left to right with respect to x increasing left to right (dx is positive), that the answer is positive, and vice versa when integrating right to left. But, as a reality check, there is something that I have never understood (or, understanding, I find so counterintuitive as to want to question it): Interpreting the result as area, one and the same area is either positive or negative not depending on what it is, but on how it is counted. I recognize that it is a logical extension of, say, -1 versus +1 on the number line, but at least -1 and +1 occupy different positions! Whenever I've brought this up, I tend to get blank looks, like, "how could you be so stupid", or "huh", but, clearly it does not keep me from being bothered by it, so I'll ask here ... once again! To me, apart from what the mechanics of integration require, a more logical "positioning" of the negative area would be its reflection across the x-axis.</p>

答:
In short, the answer to your question is simply a matter of convention.

Suppose we have a rectangle of height $4$, whose base lies on the $x$-axis, beginning at $x=2$ and ending at $x=2.5$.  If I asked you to find the area of this rectangle, you'd probably do the following computation:

$$4\cdot(2.5-2)=2\tag{$*$}$$

and you'd be correct.  By convention, our number line increases as we move to the right, so to find the positive value of the base of the rectangle, we subtract its leftmost point from its rightmost point.  If we changed the convention of our number line, we would also have to change our subtraction to obtain a positive result.

The Riemann integral is first defined only for intervals $[a,b]$ with $a<b$ (that is, with $a$ to the left of $b$).  This definition is:

$$\int_a^bf(x)dx=\lim_{|P|\to 0}\sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$$

where $|P|$ is the mesh of the partition $P=\{x_0,\ldots,x_n\}$, with $a=x_0<x_1<\ldots<x_n=b$, and $x_i<x_i^*<x_{i+1}$.  The motivation for this definition, and not one in which we take $x_{i}-x_{i+1}$ in the sum is found precisely in the convention that caused us to calculate as we did in $(*)$.  Then, having defined the Riemann integral for intervals $[a,b]$, we define the symbol $\int_{b}^af(x)dx$ to be:

$$\int_b^af(x)dx=-\int_a^bf(x)dx$$

Distributing the negative sign through the sum, you can think of this as making a definition

$$\int_b^af(x)dx=\lim_{|P|\to 0}\sum_{i=0}^{n-1} f(x_i^*)(x_{i}-x_{i+1})$$

It would be completely consistent to do calculus on a 'flipped' number line, where positive numbers are to the left of $0$ so that $a<b$ if and only if $a$ is to the right of $b$.  Then we could integrate from right to left and obtain positive areas.


问:
Find general equation for the following sequence
Given the following sequence: $$a_1 = 1, a_{n+1}=2*a_{n} + 1$$

what is the general equation for $a_n$?
I couldn't find it because the sequence is not in the form of $a_{n+1} = a_n * c$ neither is it
in the form of $a_{n+1} = a_n + c$, but more like a combination of the two.

Body:
$$a_1 = 1, a_{n+1}=2a_{n} + 1\tag{1}$$
Substitute $n$ with $n+1$
$$a_{n+2}=2a_{n+1}+1;\;a_2=2a_1+1=2+1=3\tag{2}$$
Subtract $(2)-(1)$
$$a_{n+2}-a_{n+1}=2a_{n+1}+1-(2a_{n} + 1)$$
$$a_{n+2}-3a_{n+1}+2a_n=0;\;a_1=1;\;a_2=3$$
solve
$$\lambda^2-3\lambda+2=0\to \lambda=1;\;\lambda=2$$
solutions have the form
$$a_n=a\cdot 1^n+b\cdot 2^n$$
plug the index $n=1,n=2$. Get the system
$$
\begin{cases}
a+2b=1\\
a+4b=3\\
\end{cases}
$$
$a=-1;\;b=1$ is the solution, so the solution of the recurrence is
$$a_n=2^n-1$$


问:
Probability of guessing 4-digit code
Super-simple question, I just want to be sure that I'm right. We want to find a 4-digit code, so we can choose numbers from 0 to 9, but repetition are not allowed and the order does not matter. We want to find the probability of guessing right the number after $t$ trials.

The probability in the first trial is

$\frac{1}{\frac{n!} {k!(n − k)!} } = 1/210$

can we say then that the probability in $t$ trials is $1/(210-(t-1))$? Does someone want to propose a more general formula? Thanks.

答: Since order does not matter, the problem is exactly equivalent to asking the Pr of winning a $10/4$ lottery if you have purchased $k$ tickets, $$thus\;\; \frac{k}{\binom{10}{4}} = \frac{k}{210}$$

Title: Probability of guessing 4-digit code


问:
Intuition for $\{X>0\}=\cup_{n\in\mathbb{N}}\{X\geq\frac1n\}$?
What is the intuition for the event $\{X>0\}$ being equivalent to $\cup_{n\in\mathbb{N}}\{X\geq\frac1n\}$? I have seen rigorous expositions on it, but have yet to develop a good inuition of this. Why just the natural numbers, i.e. why not $\cup_{n\in\mathbb{R}_{\geq0}}\{X\geq\frac1n\}$, and why don’t we use strict inequalities? When is it useful to write it this way? I have seen it in multiple proofs (e.g. for proving certain quantities are stopping times, proving non-negative r.v. with zero expectation is zero a.s.). but I haven't really seen why only including the natural numbers works. Cheers! (Also, please edit the tags if need be, not sure which ones to include!)

答:
The reason one uses natural numbers and not real numbers is to make sure we have a countable union. The countable union of measurable sets is measurable, but this need not be true for uncountable unions.

Also, you can also use strict inequalities, it doesn't change anything.

Ultimately, the 'intuition' here is the archimedian property: If $x >0$, you can choose $n$ so large that $x > 1/n$.


问:
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$?
Please help!

Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?

答: Hint: If $f(0) = 0$ and $f$ is not identically zero, we can find another entire function $g$ so that $f(z) = z^k g(z)$ for some positive integer $k$ and $g(0) \ne 0$.


问:
Prove $\coth\;2v = \frac{x^2 + y^2 + 1}{2y}$
Q:
Given that $x + jy =\tan (u + jv)$, prove that

$$\coth\;2v = \frac{x^2 + y^2 + 1}{2y}$$

I would like to ask this question, how can we prove it?
I had tried to expand the equation $x + jy =\tan (u + jv)$ to the following equations

$$x = \frac{\tan\;u(1 +\tanh^2v)}{1-\tan^2u\;\tanh^2v}$$

and
$$y = \frac{\tanh\;v(1 -\tan^2u)}{1-\tan^2u\;\tanh^2v}$$

I tried to relate to the question but lastly I stuck at here

$$\coth\;2v = \frac{x(1-\tan^2u)}{2y\;\tan\;u}$$

Hope somebody can give me some tips to solve it, thank you : )

答: Since $\frac{x}{y}=\frac{\tan u(1+\tanh^2v)}{\tanh v(1-\tan^2u)}$, you want to prove $\coth2v=\frac{1+\tanh^2v}{2\tanh v}$, which is well-known.


Body:
$$a_{n+1}+1=2(a_n+1) \implies a_n+1=2^{n-1}(a_1+1)=2^n \implies a_n=2^n-1$$

It works for any $a_{n+1} = ka_n+b$.


问:
Find the radius of convergence of $\sum\limits_{n=1}^\infty n!x^{n^2}$
I've got difficulties finding the radius of convergence of $\sum_{n=1}^\infty n!x^{n^2}$. I've already tried presenting it as $\sum_{n=1}^\infty b_n x^n$, where $b_n = (\sqrt{n})!$ if $n$ is a square and $b_n=0$ otherwise. I wanted to use the Cauchy–Hadamard formula and ended up with something like this: $$\frac{1}{R}=\limsup\limits_{n\rightarrow\infty}\sqrt[n]{\lvert b_n \rvert} = \lim\limits_{n\rightarrow\infty}\sqrt[n]{(\sqrt{n})!}$$ And I'm not sure if it's correct. Even if it was, I wouldn't know how to calculate this limit.

答:
The ratio test gives

$$L=\lim_{n\to+\infty}\frac{|a_{n+1}|}{|a_n|}$$

$$=x.\lim_{n\to+\infty}(n+1)e^{2n\ln(|x|)}$$

If $ |x|<1 $, $ L=0 $ it converges.

if $ |x|>1 $ , $L=+\infty$, it diverges.

thus the radius is $ R=1$.


问:
How many functions can be defined from $\textrm{A}$ to $\textrm{B}$?
Help me with the confusion here.

Suppose that set ${A}$ has $4$ elements and set ${B}$ has $2$ elements.

Let $f:{A}\to {B}$ be a function.

If each element from ${A}$ can make $2$ possible choices, shouldn't the number of functions be $2\times 4=8$?

Why does the formula suggest $2^4 = 16$? ($n^m$)

If possible, please give an elaborate answer.

答:
If we label the elements in set $A$ as $a_1,a_2,a_3,a_4$ and the elements in set B as $b_1,b_2$, then it might be easier to see why there are $16$ possible functions. Recall that by definition, each element in $A$ should be mapped to one and only one element in set $B$ (otherwise $f$ wouldn't be a function). Since each element of $A$ can be mapped to either $b_1$ or $b_2$, there are
$$
2 \times 2 \times 2 \times 2 = 2^4 = 16
$$
possible combinations.


问:
If $\alpha$ is an n-cycle, then $\alpha^k$ is a product of $(n,k)$ disjoint cycles, each of length $n/(n,k)$
If $\alpha$ is an n-cycle, then $\alpha^k$ is a product of $(n,k)$ disjoint cycles, each of length $n/(n,k)$

What I tried is let $a_i$ be an element of the cycle and $l$ be length of each disjoint cycles.
Then $$a_{i+lk}=a_i$$
which means that $i+lk\equiv i$ and hence $lk\equiv n$ where $l$ is the smallest positive integer satisfying this equation.
So how can I show that $$l=n/(n,k)$$

答:
Lemma: Let $a$ be an element of a group $G$ such that $|a|=n>0$. Hence $|a^k|=\frac{n}{(n,k)}$.
Write $d:=(n,k)$ and consider $0<q_1,q_2\in \mathbb{Z}$ such that $dq_1=k$ and $dq_2=n$. So we have $(a^k)^{q_2}=a^{kq_2}=a^{q_1n}=1$. If $t\in \mathbb{Z}_{>0}$ is such that $t<q_2$  then it can not be $(a^k)^t=1$ because $d=xk+yn$ (Bezout theorem) whence $dt=x(kt)+ytn$ with $dt<n$. Therefore $|a^k|=\frac{n}{(n,k)}$.

In the initial problem we have $\alpha^k$ is a regular permutation, i.e., $\alpha^k=\beta_1 \beta_2\ldots\beta_m$, $\beta_i$ are $l$-cycles disjoints. From of this we have $lm=n$ and $|a^k|=$ least commom multiple of $\{ l \}$. The lemma implies $l=\frac{n}{(n,k)}$. Hence $\frac{n}{(n,k)}\cdot m=n$ and therefore $m=(n,k)$.


问:
How to solve integral of rational function with logarithm
I have no idea how to solve this integral:
$$
\int\frac{dx}{x\sqrt{1−\ln^2(x)}}
$$
What should I substitute?

答:
$$u=\ln x \implies du=\frac{dx}{x}$$

$$∫\dfrac{dx}{x\sqrt{(1−ln^2(x))}}=∫\dfrac{du}{\sqrt{1−u^2}}=\arcsin u +c= \arcsin (\ln x) +c$$


问:
How to determine whether a planar graph can be generated by a degree sequence?

For the following degree sequences of graphs,

$2,2,2,3,3,3,3,4,5,5\\1,1,1,1,2,2,2,2,3,3$

which can be the degree sequence of a simple undirected
graph that is connected and planar?

I think only the second one can be a valid sequence. Is there a general rule to judge whether a degree sequence can generate a planar graph?

答:
In general it will require some thought on whether a degree sequence is that of a planar graph. For example, see this question for some possible strategies, which include using Kuratowski's Theorem, or the well known edge bound $3n - 6$. You can also compute the average degree, which for a planar graph must be strictly less than 6. (See Planarity Criterion.) You'll notice a lot of the time we check to see if a graph violates any of the criterion.

Both sequences may represent a planar graph.

For $2,2,2,3,3,3,3,4,5,5$, you may notice several things. Since we're assuming $G$ is connected, $G$ cannot be a tree (there are no degree $1$ vertices) and thus has a cycle. However, $|E(G)| = 16 \le 3(10) - 6 = 24$, so no luck there (remember, we can only use this bound to prove a graph is non-planar.) Perhaps after playing around you might think that there is a planar graph, and you'd be right. Using the Havel--Hakimi Algorithm, stopping short when I obtained the degree sequence of the path on $8$ vertices, we find the following graph:

From this example we see that it's not necessarily the case that $G$ has a cycle but no triangle (which would have enabled us to use the improved bound of $2n-4$). Even then, $|E(G)| = 16 = 2(10) - 4$.

For $1,1,1,1,2,2,2,2,3,3$, you can quickly notice that this graph cannot contain $K_5$ or $K_{3,3}$ as a subdivision and thus must be planar. This uses Kuratowski's Theorem. It also shouldn't be too hard to find a tree with this degree sequence.

For the first sequence, you may immediately notice that $G$ cannot have a $K_5$ subdivision, however $K_{3,3}$ may require more reasoning. I.e., unlike the second sequence, we've shown the first sequence can represent a planar graph, we haven't shown that it must.


答:
The Rule of Product is sometimes written as $|A\times B| = |A|\times |B|$.

Alternatively, the Rule of Product as it applies to counting can be phrased as the following:

If you have a scenario you wish to count the number of possible outcomes and each and every possible outcome can be described uniquely via a sequence of steps where the number of possible results for each step does not depend on previously made choices for earlier steps (though the choices themselves may vary so long as the quantity of choices does not), then the total number of outcomes is equal to the product of the number of choices available for each step.

Here, we count the number of functions by breaking into the following steps:

Pick which element of $B$ is the output of the function for the first element of $A$ (You have two choices)

Pick which element of $B$ is the output of the function for the second element of $A$ (You have two choices)

Pick which element of $B$ is the output of the function for the third element of $A$ (You have two choices)

Pick which element of $B$ is the output of the function for the fourth element of $A$ (You have two choices)

Multiplying (not adding) these number of choices together gives the total count as being $2^4$

It is worth noting that some authors choose the very convenient notation of $Y^X$ to denote the set of functions from a domain $X$ to a codomain $Y$ at which point you have the convenient result that $|Y^X| = |Y|^{|X|}$

Title: How many functions can be defined from $\textrm{A}$ to $\textrm{B}$?

Body:
Help me with the confusion here.

Suppose that set ${A}$ has $4$ elements and set ${B}$ has $2$ elements.

Let $f:{A}\to {B}$ be a function.

If each element from ${A}$ can make $2$ possible choices, shouldn't the number of functions be $2\times 4=8$?

Why does the formula suggest $2^4 = 16$? ($n^m$)

If possible, please give an elaborate answer.


问:
Oddity: Determinant of Skew-Symmetric $n\times n$ Matrices
Let $A \in M_{n×n}(\mathbb{R})$ be a skew-symmetric matrix, i.e., $A^t = −A$. Prove that if $n$ is odd, then $\det{A} = 0$.</p>

答: Hint : remember that $\det A= \det A^t$ and that $\det kA= k^n \det A$.</p>


问:
Approximate $X$ by a sequence of simple random variables $X_n$ in $l_1$ norm such that $|X_n|<|X|$
Given a random variable $X$ with finite expectation, I know that $$X_n\to X, a.s.\text{and}  |X_n| \leq X\implies \mathbb{E}|X-X_n|\to 0 \text{ by DCT.}$$

I am wondering if it is possible to approximate $X$ (with finite expectation) by a sequence of simple random variables:
$$\forall \varepsilon, \exists \text{  a simple r.v.  } X_{{\varepsilon}} \text{ such that } |X_{{\varepsilon}}|\leq |X| \text{ and } \mathbb{E}|X-X_{{\varepsilon}}|< \varepsilon.$$

Any help will be greatly appreciated!

答:
Decompose $X$ as $X=X^+-X^-$, where $X^+=\max\{X,0\}$ and $X^-=\max\{-X,0\}$.
By definition of Lebesgue integral, there exist simple non-negative random variables $X_\varepsilon^+$ and $X_\varepsilon^-$ such that $X_\varepsilon^+\leqslant X^+$, $X_\varepsilon^-\leqslant X^-$ and $\max\{\mathbb E\left[X^+-X_\varepsilon^+\right],\mathbb E\left[X^--X_\varepsilon^-\right]\}\lt \varepsilon/2$.

Let $X_\varepsilon:=X_\varepsilon^+-X_\varepsilon^-$; then $X_\varepsilon$ is a simple random variable. Observe that
$$
X_\varepsilon\leqslant X_\varepsilon^+\leqslant X^+\leqslant X
$$
and
$$
-X_\varepsilon\leqslant X_\varepsilon^-\leqslant X^-\leqslant -X
$$
hence $\lvert X_\varepsilon\rvert\leqslant \lvert X\rvert$. Moreover,
$$
\mathbb E\lvert X-X_\varepsilon\rvert\leqslant \mathbb E\lvert X^+-X^+_\varepsilon\rvert+\mathbb E\lvert X^--X^-_\varepsilon\rvert\lt\varepsilon.
$$


答:
At least for positive random variables it is very close to how the Lebesgue integral (expectation) is defined. Note that if $(\Omega,\mathcal{F},P)$ is the underlying probability space, then
$$\mathbb{E}X = \int_{\Omega} X(\omega) dP(\omega).$$

If $X\geq 0$ is measurable, then the integral of $X$ is defined as the supremum over the simple functions which approximate $X$ from below. An approximating sequence can be constructed as follows.

Let $n\in \mathbb{N}$ and consider a simple function:
$$\begin{cases} X_n(\omega) = n,\quad &{\rm if }\ X(\omega)\geq n,\\ X_n(\omega) = 2^{-n}(k-1),\quad &{\rm if} \ X(\omega)\in [2^{-n}(k-1),2^{-n}k),\ k=1,2,\ldots,n2^n. \end{cases}$$
We have $X_n\nearrow X$ almost everywhere and the convergence of integrals follows from the monotone convergence theorem.

Usually in the definition of the integral for a signed function $X$, you decompose into the positive and the negative part: $X = X_+ - X_-$. If at least one of the functions $X_+$, $X_-$ is integrable, then the integral of $X$ makes sense as the difference of the two integrals. Knowing that $|X|$ has finite expectation we actually know that both $X_+$ and $X_-$ have finite first moments. Thus we may slightly modify the above construction for the non-negative functions in order to obtain an approximating sequence $|X_n|\leq |X|$ for signed r.v. $X$.

Title: Approximate $X$ by a sequence of simple random variables $X_n$ in $l_1$ norm such that $|X_n|<|X|$

Body:
Given a random variable $X$ with finite expectation, I know that $$X_n\to X, a.s.\text{and}  |X_n| \leq X\implies \mathbb{E}|X-X_n|\to 0 \text{ by DCT.}$$

I am wondering if it is possible to approximate $X$ (with finite expectation) by a sequence of simple random variables:
$$\forall \varepsilon, \exists \text{  a simple r.v.  } X_{{\varepsilon}} \text{ such that } |X_{{\varepsilon}}|\leq |X| \text{ and } \mathbb{E}|X-X_{{\varepsilon}}|< \varepsilon.$$

Any help will be greatly appreciated!


问:
Are two spaces obtained by a quotient of the other homeomorphic?
I was thinking about an exercise on quotient topology that asks the following:

If $Y$ and $X$ are two topological spaces such that $Y$ is a quotient of $X$ and $X$ is a quotient of $Y$, need $X$ and $Y$ to be homeomorphic?

I think the answer is, in general, no, because quotient maps between spaces do not give all the requesites of an homeomorphism between spaces. But I don't see any counterexample to this. Could you give me any hint?

Thanks in advance.

答: The answer is no. $S^1$ is a quotient of $[0, 1]$ by identifying $0$ and $1$. Likewise, $[0, 1]$ is a quotient of $S^1$ by identifying the upper and lower semi-circles together. Yet, $[0, 1]$ and $S^1$ are not homeomorphic.


答:
You should substitute $u=\ln(x)$ because $du=1/x \, dx$ appears elsewhere in the integrand. If we write the integral as
$$
\int \frac{1}{\sqrt{1-\ln(x)^2}} \cdot \frac{1}{x} \, dx \,
$$
then it is clear that when we make the substitution $u=\ln(x)$ it simplifies nicely
$$
\int \frac{1}{\sqrt{1-u^2}} \, du \, .
$$

Integration by substitution comes from reversing the chain rule. Recall that if $y=f(g(x))$, then
$$
\frac{dy}{dx}=f'(g(x))g'(x) \, .
$$
Thus,
$$
\int f'(g(x))g'(x) \, dx = f(g(x))+C \, .
$$
On the other hand, if we make the substitutions $u=g(x)$ and $du = g'(x) \, dx$, then  the integral becomes
$$
\int f'(u) \, du=f(u)+C=f(g(x))+C \, .
$$
This shows that when we make a substitution of the form $u=g(x)$, ideally $g'(x)$ should appear in the integrand. It doesn't have to, though. Consider the integral
$$
I = \int \frac{1+e^x}{1-e^x} \, dx \, .
$$
At first sight, it doesn't seem like the substitution $u=e^x$ will work, or is even mathematically correct. However, if we rewrite $I$ as
$$
I = \int \frac{1+e^x}{1-e^x} \cdot \frac{1}{e^x} \cdot e^x \, dx
$$
then $I$ can be expressed in the form $\int f'(g(x))g'(x) \, dx$! So every substitution ultimately boils down to the reverse chain rule.

There are even more algebraic tricks you can use to evaluate integrals; if you are interested in hearing them, just ask.

Title: How to solve integral of rational function with logarithm

Body:
I have no idea how to solve this integral:
$$
\int\frac{dx}{x\sqrt{1−\ln^2(x)}}
$$
What should I substitute?


问:
$\lim\limits_{n \to\infty}\dfrac{n^{2}-n+2}{3n^{2}+2n-4}=\dfrac{1}{3}.$ Change $n$'s by $x$ and find $M$.
$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$.

With epsilon definition I get my answer as $N=\left[ \dfrac{5}{9\varepsilon }\right] +1$. But then I thought that how can I evaluate this sequence, in functions $\delta$-$\varepsilon$ definition by substituting $n$ by $x$. But I can't make it.

So here's my question:

What can I do is I have function like this one?
$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$.

I'm thinking about that function. But how to find.

$\forall \varepsilon>0$ be given, then $\exists M>0$ such that $x>M$ implies $| f\left( x\right) -l|<\varepsilon$

For function:
$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$

And this is where I stuck. Because $x\in \mathbb{R}$, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!

Body:
If you use the long division, you have
$$\frac{n^{2}-n+2}{3n^{2}+2n-4}=\frac{1}{3}-\frac{5}{9 n}+\frac{40}{27
   n^2}+O\left(\frac{1}{n^3}\right)$$ Therefore
$$ \frac{1}{3}-\frac{5}{9 n}<\frac{n^{2}-n+2}{3n^{2}+2n-4}<\frac{1}{3}-\frac{5}{9 n}+\frac{40}{27
   n^2}$$


问:
How should I start answering this chemical substance question?
Two  substances,  A  and  B,  are  being  converted  into  a  single  compound C. In  the  laboratory  it  has  been  shown  that,  for  these  substances,  the following  law  of  conversion  holds:    the  time  rate  of  change  of  the amount x of compound C is proportional to the product of the amounts of unconverted substances A & B.  If at time t = 0 there are CAO mols/li of substance A and CBO mols of substance B, and none of compound C present,  find  x(t).    Assume  that  1  mole  of  C  is  formed  from  the combination of 1 mole of A and one mole of B.

答:
Start by setting up equations.

I prefer to write  $A_0$ and $B_0$ for $C_{AO}$ and $C_{BO}$, respectively, and $C$ for $x$.

$\dfrac{dC}{dt}=-\dfrac{dA}{dt}=-\dfrac{dB}{dt}=kAB$

$A+C=A_0$

$B+C=B_0$

Therefore $\dfrac{dC}{dt}=k(A_0-C)(B_0-C)=k(A_0B_0-(A_0+B_0)C+C^2)$.

This is a separable differential equation, which can be solved with standard methods.


问:
$6!\cdot 7!=10!$. Is there a natural bijection between $S_6\times S_7$ and $S_{10}$?
Aside from $1!\cdot n!=n!$ and $(n!-1)!\cdot n! = (n!)!$, the only product of factorials known is $6!\cdot 7!=10!$.

One might naturally associate these numbers with the permutations on $6, 7,$ and $10$ objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how $1^2+2^2+\ldots+24^2=70^2$ can be viewed as an ingredient that makes the Leech lattice work.

The most natural thing to hope for would be a product structure on the groups $S_6$ and $S_7$ mapping to $S_{10}$, but as this MathOverflow thread shows, one cannot find disjoint copies of $S_6$ and $S_7$ living in $S_{10}$, so a product structure seems unlikely.

However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of $1,2,\ldots 7$ in a permutation of size $10$, and then biject $_{10}P_{3}=720$ with $S_6$ in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.

I tried doing something with orderings on polar axes of the dodecahedron ($10!$) and orderings on polar axes of the icosahedron ($6!$), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length $7$.

I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.

答:
This family of bijections (of sets) $S_6\times S_7 \to S_{10}$ has already been suggested in comments and linked threads, but it is so pretty I wanted to spell it out:

There are $10$ ways of partitioning the numbers $1,2,3,4,5,6$ into two (unordered) pieces of equal size: $P_1,P_2,\cdots,P_{10}$.  Thus we have a canonical embedding $S_6\hookrightarrow S_{10}$, coming from the induced action on the $P_i$.

Any distinct pair $P_i,P_j$ will be related by a unique transposition.  For example $\{\{1,2,3\},\{4,5,6\}\}$ (denoted hereafter $\left(\frac{123}{456}\right)$)  is related to $\left(\frac{126}{453}\right)$ via the transposition $(36)$.

There are two types of ordered (distinct) triples $P_i, P_j,P_k$:

They may be related pairwise via transpositions $(ab),(cd),(ef)$ with $a,b,c,d,e,f$ distinct and each of $\{a,b\}, \{c,d\},\{e,f\}$ not on the same side of any of $P_i, P_j,P_k$:$$
\left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{ade}{bcf}\right).$$

Here, there are $10$ choices for $P_i$, $9$ choices for $P_j$ and $4$ choices for $P_k$, giving $360$ triples in total.

They may be related pairwise via transpositions $(ab),(ca),(bc)$ with $a,b,c$ distinct:
$$
\left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{abe}{cdf}\right).$$

Again, there are $10$ choices for $P_i$, $9$ choices for $P_j$ and $4$ choices for $P_k$, giving $360$ triples in total.

An element of the stabiliser (in $S_6$) of a type 1 ordered triple (written as above) must preserve the pairs $\{a,b\}, \{c,d\},\{e,f\}$. Further if it swaps any of these pairs it must swap all of them, so the only non-trivial element of the stabiliser is an odd permutation: $(ab)(cd)(ef)$.

An element of the stabiliser (in $S_6$) of a type 2 ordered triple (written as above) must preserve the sets $\{d,f\}, \{e\},\{a,c,b\}$. Further it must fix each of $a,b,c$. Thus the only non-trivial element of the stabiliser is an odd permutation: $(df)$.

As $|A_6|=360$, in particular this means there is a unique element of $A_6$ taking the ordered triple $P_1,P_2,P_3$ to a specified ordered triple $P_i,P_j,P_k$ of the same type as $P_1,P_2,P_3$.

Fix $t\in S_{10}$ a permutation taking $P_1,P_2,P_3$ to an ordered triple of the other type.  Then there is a unique element in $A_6$ which composed with $t$ takes the ordered triple $P_1,P_2,P_3$ to a specified ordered triple $P_i,P_j,P_k$ of the other type to $P_1,P_2,P_3$.

Let $S_7$ denote the group of permutations of $P_4,P_5,\cdots,P_{10}$.  Then any permutation in $S_{10}$ may be written uniquely as an element of $S_7$ followed by an element of $(A_6\sqcup tA_6)$, where the latter is determined by where $P_1,P_2,P_3$ are mapped to.

Thus we have established a bijection of sets $$S_{10}\to (A_6\sqcup tA_6)\times S_7.$$
Once we fix an odd permutation $t'\in S_6$, we may identify the sets $$(A_6\sqcup t'A_6)\to S_6.$$
Composing we get: $$S_{10}\to (A_6\sqcup tA_6)\times S_7\to (A_6\sqcup t'A_6)\times S_7\to S_6\times S_7.$$

That is for any choice of the permutations $t,t'$ we have the required bijection of sets.


问:
What an implication means when stating the solutions of an equation?
I am having trouble understanding what the following statement means: $$x^2=4 \implies x=2 \hspace{1em} \text{or} \hspace{1em} x=-2.$$

How can we relate this statement to a material conditional?

I am trying to see it as a material conditional. So on the left hand side we have the equation $A(x)$ which takes different truth values for different $x$. The same goes for the right hand side which we can denote it as $B(x)$. Now does the original statement means that $$\forall x \in \mathbb{R}: A(x) \implies B(x).$$

If I substitute the values of $x$ in these two expressions then the conditionals are all true. So how can such a statement make sense when we are trying to solve an equation or when stating the solutions of the equation? Does it mean that in the case that the left hand is true then the value "we substituted" was $2$ or $-2$ in order to make the right hand side also true?

答:
When solving an equation, the underlying chain of logic actually involves $\color{#00F}\iff$ ("if and only if") instead of merely $\color{#C00}\implies$ (or $\color{#C00}\impliedby$) between steps.

This reversibility ensures that extraneous solutions are rejected $$\begin{pmatrix} \;\sqrt{x+3}=x+1 \color{#C00}\implies x+3=(x+1)^2 \color{#00F}\iff x=-2\;\text{or}\;1\;\\ \text{versus}\\ \sqrt{x+3}=x+1 \color{#00F}\iff x=1 \end{pmatrix}$$ and that no solution is inadvertently discarded $$\begin{pmatrix} x^2=x \color{#C00}\impliedby x=1\\ \text{versus}\\ \;x^2=x \color{#00F}\iff x=0\;\text{or}\;1\; \end{pmatrix}.$$

In practice, because trying to ensure invertibility at every step is tedious and prone to carelessness, when solving equations, it is best to just work in a forward direction like this $$\sqrt{x+3}=x+1 \color{#C00}\implies x+3=(x+1)^2 \color{#C00}\implies x=-2\;\text{or}\;1,$$ then plugging the candidate solutions into the given equation to sift out any extraneous ones.

Here is a more visual explantion.

P.S. It is not instructive to interpret $\implies$ (implication) as the material conditional $(\rightarrow)$ or to interpret $\iff$ (equivalence) as the material biconditional $(\leftrightarrow).$ In any case, it isn't necessary to delve into formal logic when solving equations.


问:
Show that E is measurable?
Suppose $E_1= [1, 1 \frac12] , E_2 = (2, 2\frac14), E_3 = [3, 3\frac18], E_4 = (4 ,  4 \frac{1}{16}) , \dots , E=  \bigcup_{n=1}^{\infty}E_n
$

i) Show $E$ is measurable

ii) Compute $m(E)$

Here is what I know from reading:
If $E$ is an element in $M$, we say $E$ is Lebesgue measurable and $M$ will be called Lebesgue measure.

I was looking at this problem in a book which explained some things about Lebesgue measures in my self study class. I asked my TA but he said he would not do this since we are just learning some proofs. This topic was also not covered.

I would like to see how we would solve a numerical proof like this. I did not do anything like this yet. It seems like a nice proof of a problem to know for other problems.

答:
$E$ is the countable union of intervals, which are measurable  . Therefore, since the Lebesgue-measurable sets are a $\sigma$-algebra, we conclude that countable unions of measurable sets are measurable, i.e. $E$ is measurable.

Now, regarding the second question, all sets $E_i$ are disjoint, therefore from countable additivity of the Lebesgue measure:
$$m(E)=m\left(\bigcup\limits_{n=1}^\infty E_n\right)=\sum\limits_{n=1}^\infty m(E_n)=$$
$$=m(E_1)+m(E_2)+\dots=$$
$$=\frac12+\frac14+\frac18+\dots=$$
$$=1$$

where $m(E_1)=1\frac12-1=\frac12$, $m(E_2)=2\frac14-2=\frac14,\dots$

Title: Show that E is measurable?