问:
Homology of Compact Surfaces
In this post, my main question is: How to calculate the $n$-th singular homology group $H_n(X)$ of the compact surfaces $X$?

My attempt 1: We know the classical theorem of the classification of compact sufaces:

Theorem: Let $X$ be a compact 2-dimensional manifold. If $X$ is orientable, then either $X \cong \mathbb{S}^2$, or $X \cong M_1 \# \cdots \# M_g$, where each $M_i$ is $\mathbb{T} := \mathbb{S}^1 \times \mathbb{S}^1$; if $X$ is nonorientable, then $X \cong M_1 \# \cdots \# M_g$, where each $M_i$ is $\mathbb{RP}^2$. Here, $\cong$ means homeomorphic and $\#$ means taking connected sums.

By the classification theorem, I'm trying to reduce the case to the basic "components" of the surfaces, namely $\mathbb{S}^2$, $\mathbb{T}$ and $\mathbb{RP}^2$. So there are mainly two steps:

Step 1: Calculate the homology groups of $\mathbb{S}^2$, $\mathbb{T}$ and $\mathbb{RP}^2$. I have already obtained these. So here is no problem.

Step 2: Calculate the homology group of the connected sums. I get stuck here. :(

So the general question may be: Let $X$ and $Y$ be two topological spaces, whose connected sum $X \# Y$ can be defined. If we know the homology groups of $X$ and $Y$, then how to calculate $H_n(X \# Y)$? I'm hoping someone to help me on this. My idea is to use the Mayer-Vietoris sequence, but the definition of connect sums are quite messy and hence I do not know how to implement it in details.

My attempt 2: In Munkres's book Topology, he calculated some 1st-homology groups in Sect. 75. By de Rham theorem, the homology groups (with real coefficients) are isomorphic to the de Rham cohomology groups of $X$, and since $X$ has dimension 2, it seems that we can worry less on the homology groups of higher $n$. However, using de Rham theorem is likely to be a little "cheaty" and I'm not sure that this is a correct way.

My attempts 3: We can use the other classification theorem, which used cutting and pasting:

Theorem: Every compact 2-dimensional manifold $X$ is homeomorphic to either $M$ or $M^{\prime}$ defined as follows:

Oriented case: Let $W_h$ be a $4h$-gon in the plane with edges labeled
$$
\alpha_1, \beta_1, \alpha_1^{-1}, \beta_1^{-1}, \ldots, \alpha_h, \beta_h , \alpha_h^{-1}, \beta_h^{-1},
$$
and let $M_h$ be the quotient space of $W_h$, in which the edges are identified according to the labels.

Non-oriented case: Let $W^{\prime}_n$ be a $2n$-gon on the plane with edges labeled
$$
\alpha_1, \alpha_1, \ldots, \alpha_n, \alpha_n,
$$
and let $M^{\prime}_n$ be the quotient space of $W^{\prime}_n$, in which the edges are identified according to the labels.

However, since I haven't learnt CW-complexes or simplicial complexes systematically, and the original question appears before introducing such tools, I don't want to use the constructions above to calculate.

Sorry for such a long post, and thank you all for commenting or answering!! :)

答:
When you take a connect sum of surfaces $X$ and $Y$, you delete a disk (call it $D$ on both surfaces) from each surface and identify the resulting boundary circles. To use Mayer-Vietoris, consider an open set $U$ that consists of $X$ together with a small portion of $Y$; if you are willing to accept the existence of a metric $d$ on $X \# Y$, you can fix $\epsilon > 0$ sufficiently small and say that this open set is the union of 1) $X$ with the disk $D$ deleted; and 2) the set of points $y$ which lie in $Y \setminus D$ such that $d(y,D) < \epsilon$. If you want to avoid metrics you can say something like "union of sufficiently small open sets centered along the boundary circle." Let $V$ be the open set constructed by the same procedure with $X$ and $Y$ swapped.

I will do the example $X = Y = T$ the torus, so $X \# Y$ is a genus $2$ surface $\Sigma_2$. The case where one of the surfaces is a projective plane is similar, and then you just induct on the number of summands. Mayer-Vietoris reads
$$
0 \to H_2(S^1) \to H_2(U)\oplus H_2(V) \to H_2(\Sigma_2) \to H_1(S^1) \to H_1(U)\oplus H_1(V) \to H_1(\Sigma_2) \to H_0(S^1) \to H_0(U)\oplus H_0(V) \to H_0(\Sigma_2) \to 0
$$

which, since $U$ and $V$ deformation retract onto the wedge of two circles (the $1$-skeleton of the standard CW structure on the torus), evaluates to

$$
0 \to 0 \to 0\oplus 0 \to H_2(\Sigma_2) \to \mathbb Z \to \mathbb Z^4 \to H_1(\Sigma_2) \to \mathbb Z \to \mathbb Z\oplus \mathbb Z \to \mathbb Z \to 0
$$

using the known $H_*(S^1)$ and the fact that $H_0(\text{connected space}) \cong \mathbb Z$. Since the $H_0$ portion of the sequence is exact on its own, we can reduce this to

$$ 0 \to H_2(\Sigma_2) \to \mathbb Z \to \mathbb Z^4 \to H_1(\Sigma_2) \to 0.
$$

Finally, you can check that the generator of $H_1(U\cap V) \cong H_1(S^1) \cong \mathbb Z$ goes to $0$ in $H_1(U)$ and $H_1(V)$, so the middle map is the $0$ map and this final sequence splits up into isomorphisms $H_2(\Sigma_2) \cong \mathbb Z$ and $H_1(\Sigma_2) \cong \mathbb Z^4$.

Title: Homology of Compact Surfaces


问:
statistics for data analysis
I am a last year math student, but I also want to learn a bit of data analysis in its theoretical aspect, many have told me that certain bases in statistics are needed, but I have only seen the course of probability and measurement theory Then I would like you to recommend a statistics book for data analysis

I want to clarify that I do not have solid knowledge in statistics, I only have knowledge in probability

答:
It appears you have some background in probability in the context of measure theory; given that, I would suggest you look at one of:

Shao, Mathematical Statistics

Keener, Theoretical Statistics

Schervish, Theory of Statistics

For a non-measure theoretic treatment:

Bickel and Doksum, Mathematical Statistics (in two volumes)

Casella and Berger, Statistical Inference

You will not only need a course in mathematical statistics, but one in linear models. Given your mathematical background, I would suggest either one of:

Agresti, Foundations of Linear and Generalized Linear Models

Monahan, A Primer on Linear Models

McCullogh, Searle, and Neuhaus, Generalized, Linear, and Mixed Models

Christiansen, Plane Answers to Complex Questions

Mathematical statistics and linear models are the core of almost everything in statistics. Once you've learned the contents of these two subjects, you could consider looking at Advanced Data Analysis from an Elementary Point of View by Shalizi, available at https://www.stat.cmu.edu/~cshalizi/ADAfaEPoV/.

ContentLicense: CC BY-SA 4.0

Title: statistics for data analysis


问:
Finding Diagonals in a Quadrilateral
This is just a problem I came up with while reading one of my geometry textbooks:

If we have a quadrilateral with side lengths $a, b, c$ and $ d$ and we know the length of one of the diagonals to be $x$, what is the length of other diagonal in terms of $a, b, c, d$ and $x$?

I understand that there is only one possible length for the other diagonal and could prove this using SSS congruence, but I could not figure out how to derive the length of the other diagonal based on the provided info. Please help with this.

答:
Since you know the length of one of the diagonals, as well as the lengths of all of the sides, you can use the law of cosines to compute all of the angles in the quadrilateral.

Having done this, you can use the law of cosines again to compute the length of the other diagonal.

Title: Finding Diagonals in a Quadrilateral


答:
Given the side lengths and the one diagonal, the law os sines gets you all the angles of the two triangles that you know. Then combine the split (by the known diagonal) angles and use the law of sines to get the length of the other diagonal.  As a check, do it for both spit angles.  The answers should be the same.

Law of cosines can also be used.

Title: Finding Diagonals in a Quadrilateral

Body:
This is just a problem I came up with while reading one of my geometry textbooks:

If we have a quadrilateral with side lengths $a, b, c$ and $ d$ and we know the length of one of the diagonals to be $x$, what is the length of other diagonal in terms of $a, b, c, d$ and $x$?

I understand that there is only one possible length for the other diagonal and could prove this using SSS congruence, but I could not figure out how to derive the length of the other diagonal based on the provided info. Please help with this.


问:
Search square factors in Lucas–Lehmer sequence
I’m interested in some sequences which have no square factor.
$$
s_i =\begin{cases}
4 & i=0; \\\\
s_{i-1}^2 - 2 & \text{otherwise}
\end{cases}$$
This is Lucas–Lehmer primality test sequence.  A003010 in OEIS.

when ${s_i}$ has a square factor? I had checked that the sequence has no square factor for i=1,2,...,7

答:
It seems plausible that there may be no square factors other than $4$, or at least that there are only finitely many primes for which $p^2$ divides $s_n$. A sketchy argument follows:

Set $\tau=2 +\sqrt{3}$. Then $s_n = \tau^{2^n} + \tau^{-2^n}$. We work in the ring $R=\mathbb{Z}[\sqrt{3}]$, where $\tau$ is a unit. So $p^2$ divides $S_n$ if and only if $\tau^{2^{n+1}} \equiv 1 \mod p^2$, where our congruence is in $R$.

Let $p$ be a prime $\geq 5$. Let $U_{p^2}$ be the unit group of $R/p^2 R$. You want to know whether the order of $\tau$ in $U_{p^2}$ is a power of $2$.

There are two cases. First, suppose there is a square root of $3$ mod $p$ (this happens when $p \equiv \pm 1 \mod 12$). This square root will lift to a solution $s$ of $s^2 \equiv 3 \mod p^2$ (we can lift by Hensel's lemma). In this case $R/p^2 \cong \mathbb{Z}/p^2 \oplus \mathbb{Z}/p^2$ and we can identify $\tau$ with the ordered pair $(2 + s, 2 - s)$. Notice that $2-s = (2+s)^{-1}$, so the two halves of the ordered pair have the same order. The unit group of $\mathbb{Z}/p^2$ is cyclic of order $p(p-1)$, and the number of elements whose orders are powers of $p$ is $2^{v_2(p-1)}$, where $v_2(k)$ is the exponent to which $2$ divides $k$. So, heuristically, I would expect the number of $p$ for which $p^2$ divides $s_n$ and there is a square root of $3$ mod $p$ to be something like
$$\sum_{p \equiv \pm 1 \mod 12} \frac{2^{v_2(p-1)}}{p(p-1)}.$$

There case where where $p \equiv \pm 5 \mod 12$ is a little trickier but, if I haven't made any mistakes, the similar heuristic estimate is
$$\sum_{p \equiv \pm 5 \mod 12} \frac{2^{v_2(p+1)}}{p(p+1)}.$$

Now, here is my point: those sums converge. In fact,
$$\sum_{n=1}^{\infty} \frac{2^{v_2(n)}}{n^2}$$
converges to $\pi^2/4$ (exercise!) and the above sums are less than this because they are restricted to just summing over primes. Moreover, the sum numerically looks pretty small.

So the "expected" number of such primes is a small finite quantity, and it wouldn't surprise me if $2$ were the only one.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 2.5


问:
Serious applications of Colouring proofs
Are there any research-level applications of proofs by colouring?

This is the kind of proof you use to show that you can't cover a mutilated chessboard with 31 dominoes. Afaik, this technique chiefly finds a market in IMO preparation classes; see  Arthur Engel's book or  this handout .

答: There is a vertex coloring argument that is used in Steve Fisk's elegant proof of the Chvatal-Klee-Art Gallery Theorem:  http://en.wikipedia.org/wiki/Art_gallery_problem  The coloring argument can be extended to handle the case where the art gallery is a rectilinear (orthogonal) polygon.</p>


问:
Simple geometry/trigonometry question
How to find the X coordinate of the red point if i know it's Y coordinate and the angle? Let's say the Y is 40 and the angle is 30 degrees:

答: Hint: think about the definition of the tangent of the angle.

Title: Simple geometry/trigonometry question


问:
Inverse Fourier Transform for $\frac{e^{-A\sqrt{(\omega^2+Bi\omega+C)}}}{\sqrt{(\omega^2+Bi\omega+C)}}$
I am trying to find the inverse Fourier transform of below:
$$\frac{e^{-A\sqrt{(\omega^2+Bi\omega+C)}}}{\sqrt{(\omega^2+Bi\omega+C)}}$$
Where A, B, and C are all positive and real.
Please let me know if you need further information ($\omega$ is the Fourier transform variable). Thanks and Merry Christmas

答:
From the lookup I have
Let:

$\mathcal{L_{\mathtt{t\rightarrow s}}}\left(f\left(t\right)\right)=g\left(s\right),\mathcal{L_{\mathtt{t\rightarrow s}}^{-\mathtt{1}}}\left(g\left(s\right)\right)=f\left(t\right)$

$\frac{e^{-\left(b\left(s^{2}-a^{2}\right)^{\frac{1}{2}}\right)}}{\left(s^{2}-a^{2}\right)^{\frac{1}{2}}}\rightarrow\begin{cases}
0 & 0<t<b\\
I_{0}\left(a\left(t-b^{2}\right)^{\frac{1}{2}}\right) & b<t
\end{cases}\,\,\,\,\,\,\,\,\,b>0,Re\left(s\right)>\left|Re\left(a\right)\right|$

I am having problems with your $i,\omega$; can you check the signs and consistency? In particular, your formula seems to give a diverging (t->s) transform. During completing the square.  Also, I have a very strong suspicion that t should be t^2 in the Bessel function argument.


问:
If $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$.
My book states the following: if $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$. But I don't understand this. It is not as if $f'$ is given to be continuous, rather $f$ is given continuous and differentiable at $x=c$. So how can we possibly comment about the existence of $f'$ in the neighborhood of $c$ too? Note that f is defined on an open interval I and c belongs to I.

答:
Just start with the function $Q$ that’s the characteristic function of $\Bbb Q$, i.e., $Q(x)=1$ if $x$ is rational, zero otherwise.

Then take $f(x)=x^2Q(x)$, with $c=0$.


问:
Function that forms an Arithmetic Sequence
I've been working on this problem for a while:

Let $n > 2$ be an integer. Let $f$ be a function from the set of all integers to itself with the following property: If the integers $a_1, a_2, a_3, \dots, a_n$ form an arithmetic progression, then the numbers $f(a_1), f(a_2), \dots f(a_n)$ form an arithmetic progression in some order (that's possibly constant). Determine all values $n$ such that the only functions $f$ with the above property are the functions in this form: $f(x)=bx+c$, where $b$ and $c$ are integers.

My work is most likely trivial, but I wrote $a_1, a_2, \dots a_n$ as $a_1, a_1+x, a_1+2x, \dots, a_1+(n-1)x$. Then, applying $f(x)=bx+c$ to each term in the sequence, I got that the values of the sequence were
$$ba_1+c, b(a_1+x)+d, \dots, b(a_1+(n-1)x)+d.$$ However, I'm unable to make more progress.

Thanks for any form of help!

答:
(Fill in the gaps as needed, especially when it says "Show that". If you're stuck, explain what you've tried.)

Claim: For $n$ prime, the function $f(a) = ( a \mod n )$  satisfies the conditions.

Proof: Show that any AP of length $n$, taken modulo $n$ (prime), is either

A constant sequence, or

$\{ 1 , 2, \ldots , n-1, n \}$.

In either case, $ f(a_i)$ forms an AP. $_\square$

Suppose that for some $n$, we had such a non-linear function $f(x)$.

Suppose that the function is not injective.

Let $k$ be the minimum positive integer such that there exists $ |a-b| = k$ and $ f(a) = f(b)$. WLOG, let $ a < b$.

Case 1:  $k < n$.

Show that $ f(x) $ is a constant, which is a contradiction.

Case 2: $ k > n$.

Consider the AP's $ \{ f(a+i + j)\}_{j=0}^{n-1}$ going from $ i = 0$ to $ i = k+1 - n > 1$.

Since $ f(a) \neq f( a +n)$, and $ \{ f(a+j)_{j=0}^{n-1} \}, \{ f(a+1+j)_{j=0}^{n-1} \}$ both form an non-constant AP, show that $ \{ f(a), f(a + n) \} $ must be the minimum and maximum of $ \{ f(a+j) \} _{j=0}^n$.

WLOG, let $ f(a) < f(a+n)$, so $ f(a)$ is the minimum and $f(a+n)$ is the maximum.

Applying the same logic to $ f(a+1) \neq f(a+1+n)$, $\{ f(a+1), f(a +1+ n) \} $ must be the minimum and maximum of $ \{ f(a+j) \} _{j=1}^{n+1}$. Since $f(a+1) < f(a+n)$, hence $f(a+1)$ is the minimum and $f(a+1+n)$ is the maximum, so $ f(a+n) < f(a+n+1)$.

We thus get the chain of inequalities $ f(a) < f(a+n) < f(a+n+1) < \ldots < f(a+k)$, contradicting the choice of $k$. Thus, no such function is possible.

Case 3: $ k = n$.

Show that for any non-trivial divisor $ d \mid n$, $ 1 < d < n$ , we have $ f(a) = f( a + \frac{k}{d} )  $. Thus, if $n$ is composite, no such function is possible.

Hence, a non-injective non-linear function could exist only if $k=n$ and $n$ is prime.

Now, suppose that the function is injective.

Using the same argument as in Case 2, show that

Assuming $f(0 ) < f(n)$,

We thus get the chain of inequalities $ f(n) < f(n+1) < f(n+2) < \ldots $.

Similarly, we get the chain of inequalities $ \ldots < f(n-2) < f(n-1 ) < f(n) $.

(Show that) $f(x) = f(0) + x ( f(1) - f(0) ) $ is a linear function.

Hence, a non-linear function could exist only if $k=n$ and $n$ is prime. We have shown above that such a solution exists.

For prime $n$, we have the following solution:

Let $ g: \mathbb{Z_n} \rightarrow \mathbb{Z_n}$ be a bijection. Then, $ f(a) = b \times g ( a \mod{n} ) + c $ satisfies the conditions.
The solution at the top is the identity $ g(a \mod{n} ) = a \mod {n}$ and $b = 1, c = 0 $.

I believe that in Case 3, using the technique similar to Case 2, we can show that $f(x+n) = f(x)$ for all $x$, from which the solution follows.


问:
What is the best way to weigh multiple five-star ratings based on a signed integer from -1 to 1?
I have a certain amount of tuples, which contain a signed integer (between -1 and 1) and a five-star rating (where the higher the signed integer is, the more important the rating is).

I want to get an "average" rating, and beforehand I would just compute this, however it seems that it would be better if I weigh the rating using the accompanying integer. I tried min-max normalization and subsequently dividing by the sum, however the rating with smallest integer would get a weight of 0 (thus not count towards the final rating) and that is not what I want.

I am looking to use all ratings for the computation, but a rating in a tuple with a positive integer should be more important than a rating in a tuple with a negative integer, or a rating with a smaller integer. I can imagine there would probably be a lot of different ways to tackle this, but I'm just not sure what a nice systematic way is.

Thanks a lot, if there is anything unclear please let me know (:

答:
I suggest a simple $$\dfrac{\sum_{(s,r)} (s+2)r}{\sum_{(s,r)}(s + 2)}$$ where the sum is over all pairs with $s$ being the signed integer, and $r$ being the rating. The reason for adding $2$ to the signs is that it insures the denominator is never $0$. We have that ratings with $-1$ signs count for less than ratings with $0$ signs, which count for less than ratings with $1$ signs.

A more general method is for you to decide on positive values $w_{-1}, w_0, w_1$ representing how much weight a $-1$ rating, $0$ rating, or $1$ rating should be given. Then the average rating is
$$\dfrac{\sum_{(s,r)} w_sr}{\sum_{(s,r)}w_s}$$

For example, if you decide that a $0$ rating should be worth half as much as a $1$ rating, but a $-1$ rating should only be worth one tenth as much as a $1$ rating, you could pick
$$\begin{align}w_{-1} = 1\\w_0 = 5\\w_1 = 10\end{align}$$

Again, the reason you want all three $w_s$ to be positive is to avoid $0$ denominators. If you were to pick $w_{-1} = 0$ and all your ratings for something happened to have a sign of $-1$, the calculation would involve a division by $0$.

This gives you the freedom to pick just how much worth should be applied to each sign.


问:
Find subgroup(s) of order 2 of $D_{2p}$
Suppose that $p$ is a prime greater than 2. Dihedral group $$
D_{2p}=\left\{ 1,a,a^2,\cdots ,a^{p-1},b,ab,a^2b,\cdots ,a^{p-1}b|ord\left( a \right) =p,ord\left( b \right) =2,ab=ba^{p-1} \right\} .
$$
Now I can figure out that $\{1,b\}$ is a subgroup of order 2, and it is not normal, as $aba^{-1}=ba^{p-1}a^{-1}=ba^{p-2}\ne 1\ or\  b$. But if I consider Sylow Theorem, there exist more than one Sylow-2 subgroups since we have got a non-normal subgroup of order 2. Then what is the second Sylow-2 subgroup?
Equivalently, I am looking for another element of order 2 in $D_{2p}$ other than $b$. Any help would be appreciated! Thanks!

答: All the elements with $b$ in them are of order $2$, corresponding to a reflection of a regular polygon in some line of symmetry; this is true of any dihedral group.

Title: Find subgroup(s) of order 2 of $D_{2p}$


问:
Sequence of iteration in Newton Raphson method.
To find  positive square root of $a>0$ by solving equation $x^2-a$ by Newton-Raphson mathod, if $x_n$ denotes $n$th iterate with $x_0>0, x_0\neq\sqrt{a}$, then the sequence $(x_n) $ is

$1.$ strictly increasing.

$2.$ strictly decreasing.

$3.$ constant.

$4.$ not convergent.

According to me if I start from right of  square root of $a$ we get decreasing sequence and if started from left then increasing sequence. How can be some option correct ? For me it seems all options as incorrect. Please comment. Thank you.

答:
Concerning questions $(1)$ and $(2)$, it depends if you start higher or lower than to root of the equation.

Darboux theorem states that if $f(x_0)\times f''(x_0) >0$ the path to solution will be monotonic without any overshoot of the solution. If not, there will be an overshoot.

For illustration, let us take $a=100$.

Starting with $x_0=5$, the iterates will be
$$\{5.00000,\color{red}{12.5000},10.2500,10.0030,10.0000\}$$
Starting with $x_0=15$, the iterates will be
$$\{15.0000,10.8333,10.0321,10.0001,10.0000\}$$


答:
If the angle from the diagonal line to the $x$-axis is $\theta$, and the $y$-coordinate of the red dot is $a$, then the $x$-coordinate of the red dot is
$$\frac{a}{\tan(\theta)}.$$
In your example, $a=40$ and $\theta=30^\circ$, so the $x$-coordinate would be
$$\frac{40}{\tan(30^\circ)}=\frac{40}{\sqrt{3}}=\frac{40\sqrt{3}}{3}\approx 23.09.$$</p>

Title: Simple geometry/trigonometry question

Body: How to find the X coordinate of the red point if i know it's Y coordinate and the angle? Let's say the Y is 40 and the angle is 30 degrees:


问:
Matrices With Changing Determinants
I would love some direction and any information for some further reading about a cool thought I just had. If a square matrix has functions as the elements, would it not make the matrix depend on $x$ (or whatever variables) and hence have a changing determinant? What would be even cooler is for a matrix to sometimes have a determinant that is equal to $0$. Now that I think about it, this could be a rotation matrix, right? $A_{11} = \cos x$ and $A_{12} = -\sin x$ and $A_{21} = \sin x$ and $A_{22}= \cos x$ so if $x=45^\circ$, the determinant will be $0$ I think. This would mean it has no inverse for that value? I think I made a mistake because it is easy to visualize how to reverse a $45^\circ$ rotation.

In any case, please do comment anything that you know about this and where to find out more about it.

Edit: Yes the rotation matrix was a bad example because the determinant is always $1$. I could have used a better example with a determinant of $\cos x - \pi$ or $2x-6$

答:
The rotation matrix has a constant determinant because $\hat{i} = (\cos x, \sin x)$ and $\hat{j} = (-\sin x, \cos x)$ $ = \cos(x+\pi/2), \sin(x + \pi/2))$. The new basis vectors are just rotated by $x$ radians counterclockwise compared to the usual basis vectors $e_1 = (0, 1)$ and $e_2 = (1,0)$, hence they have the same signed area and the same determinant (for more on this, check out 3Blue1Brown's video).

The determinant can be a function, but I don't how that could be useful. Off the top of my head:

$$A = \begin{pmatrix} \cos x &  \cos(-x) \\ \sin x & \sin(-x) \end{pmatrix} = \begin{pmatrix} \cos x &  \cos x \\ \sin x & -\sin x \end{pmatrix}$$ has a determinant of $0$ when $x = k \pi, k \pi + \pi/2, k \in \mathbb Z$. Think about why this is in terms of the basis vectors and their span (when are they collinear?)

Title: Matrices With Changing Determinants


问:
How to find the following limit?

Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.

答:
Using L'Hospital rule $$\underset{x\to 0}{\text{lim}}\Bigg[\frac{\int_0^x \cos \left(t^2\right) \, dt}{x}\Bigg]=\underset{x\to 0}{\text{lim}}\Bigg[\frac{\frac d {dx}\int_0^x \cos \left(t^2\right) \, dt}{\frac d {dx}x}\Bigg]=\underset{x\to 0}{\text{lim}}\Bigg[\frac d {dx}\int_0^x \cos \left(t^2\right) \, dt\Bigg]$$

Now, using the fundamental theorem of calculus
$$\underset{x\to 0}{\text{lim}}\Bigg[\frac d {dx}\int_0^x \cos \left(t^2\right) \, dt\Bigg]=\cos(x^2)$$

Title: How to find the following limit?


问:
Counting The Number Of K-Bounded Good Lists Of Length N
Problem Statement:

A k-bounded list of length n is a sequence [x1, x2, . . . , xn] where each xj
is an
integer between 0 and k, inclusive. A good list is one that does not contain three
consecutive values 0, 1, 0. For example, for n = 4 and k = 2, [2, 0, 0, 1] and [1, 0, 0, 1]
are good lists while [0, 1, 0, 2] and [0, 0, 1, 0] are not.
Let good(n, k) be the number of good k-bounded lists of length n. For instance,
the good 1-bounded lists of length 2 are {[0, 0], [0, 1], [1, 0], [1, 1]}, so good(2, 1) = 4.
Similarly, you can check that good(4, 1) = 12 and good(3, 2) = 26.
Your task is to compute good(n, k) for the following values of n and k.

(a) good(7, 1)

(b) good(7, 3)

(c) good(20, 1)

Note: This is a question from a past contest and the official answers for the parts a, b and c are as follows:

(a) good(7, 1) = 65

(b) good(7, 3) = 15163

(c) good(20, 1) = 97229

My Approach: I came up with the following recurrence relations,

$$p(i,0) = p(i-1,0) + p(i-1,1) - p(i-2,0)$$ $$p(i,1) = [p(i-1,0) + p(i-1,1)]\times k$$
where $p(i,0)$ is the number of good lists of length $i$ ending with 0 and $p(i,1)$ is the number of good lists of length i not ending with 0. The idea behind this is that every good list of length $i$ is an extension of a good list of length $i-1$ but there can also be some bad extensions in the process, hence it is better to handle them separately. Therefore, in case a good list ends with 0,1 then we need to take care of the fact that while extending this list we don't count the lists where we place a 0 at the end, otherwise we will have 0,1,0 in the list. So what we do is that we just count all the lists of length $i$ ending with 0 and then subtract the lists of length $i-2$ that end with a 0. Whereas the case where we do not use a 0 at the end is a bit simpler.

I just filled the table for the above recurrence manually with the initial values as $p(0,0) = 0$, $p(0,1) = 0$, and $p(1,0) = 1$,  $p(1,1) = k$ (since it was a pen and paper exam) and could arrive at the answers which is the sum of $p(n,0)$ and $p(n,1)$ [C++ code for reference: https://ideone.com/wSVhVL]. But then I have been thinking whether this problem has a closed form solution, some function in n and k. I couldn't arrive at one until now. Can anyone provide some combinatorial insight? Is a closed form solution possible?

答:
I believe that this is from a hand-written Computing contest, where you're expected to use algorithms to evaluate these finite cases. If so, the goal is to come up with a simple equation to describe the scenario, and then iterate to the corresponding $n,k$ values.

Fill in the gaps as needed, especially when it says "Show that". If you're stuck, explain what you've tried and why it doesn't seem to work.

Let

$A_k (n)$ be the number of $k-$bounded lists of length $n$ that end with $0$.

$B_k (n)$ be the number of $k-$bounded lists of length $n$ that end with $0, 1$.

$C_k (n)$ be the number of $k-$bounded lists of length $n$ that end with "not $0$, and not $0,1$".

We are after $A_k (n) + B_k(n) + C_k(n)$.

Show that we have the recurrence relations

$A_k(n+1) = A_k(n) + C_k(n)$.

$B_k(n+1) = A_k(n)$

$C_k (n+1) = (k-1) A_k (n)  +  (k) B_k (n) + (k) C_k(n)  $

Show also that we have the initial values

$A_k(2) = k+1 $

$B_k(2) = 1 $

$C_k(2) = (k+1)^2 - (k+1) - 1 = k^2 + k -1$.

For example, here is good$(7,3)$.


答: Take a look at Trigonometric_functions#Right-angled_triangle_definitions at wikipedia...</p>

Title: Simple geometry/trigonometry question


问:
Extreme points of the unit ball
I want to show that for $X$ compact and Hausdorff, any $f\in C(X)$ satisfying $|f(x)|=1$ for all $x\in X$ is an extreme point of the unit ball of $C(X)$. Here, $C(X)$ is the space of continuous complex-valued functions on $X$.

I tried to prove it by contradiction, writing $f = \frac12(f_1+f_2)$ for distinct $f_1,f_2$ in the unit ball of $C(X)$. Then by the triangle inequality, $|f_1(x)|=|f_2(x)|=1$ for all $x\in X$. But I cannot see how this leads anywhere. Am I missing something obvious?

Anyway, I am quite stuck on this and would appreciate a hint.

答:
Hints:

$1).\ $ This seems to be more or less your idea: first prove the more general fact that a point $x$ is an extreme point of the unit ball in a Banach space if and only if for all $t\in X$ we have $\|x\pm t\|\le 1\Rightarrow t=0.$ For one direction, set $y=x+t$ and $z=x-t$ and apply the definition. For the other, write $x=\alpha z+(1-\alpha)w$ and suppose $\alpha\neq 0,1.$

$2).\ $ Suppose $|f(x)|=1.$ Take a $g\neq 0$ so there is an $x_0$ such that $g(x_0)\neq 0$. Now consider what happens if $\|f\pm g\|\le 1.$

Title: Extreme points of the unit ball


问:
Products of Manifolds
Let $M$ and $K$ be manifolds, and let $(f_{\alpha})_{\alpha \in I}$ and $(h_{\gamma})_{\gamma \in J}$ be partitions unity for $M$ and $K$ respectively. Show that the collection of mappings on $M \times K$ given by all products $f_{\alpha}(a)h_{\gamma}(b)$ is a partition of unity for $M \times N$.

I've fooled around with this and still I can't find the idea on how to show this. Any hints and help would do. Thank you very much!!

答:
Fix $(m,k)\in M\times K$. Then there are neighbourhoods $M_m\subseteq M$ and $K_k\subseteq K$ of $m$ and $k$ such that finitely many $f_\alpha$ and $h_\gamma$ are nonzero, respectively. Take a look at $M_m\times K_k$.

Lastly, note that $$\sum_{\alpha\in I}\sum_{\gamma\in J} f_\alpha(m)h_\gamma(k)=\left(\sum_{\alpha\in I} f_\alpha(m)\right)\left(\sum_{\gamma\in J} h_\gamma(k)\right).$$

Title: Products of Manifolds


问:
Base of topologic space
Let $X$ be an infinite set and $x_{0}$ a point of $X$. Let $B = \{\{x\} : x \neq x_0\} \cup \{A \subseteq X : x_0 \in A$ and $X - A$ is finite$\}$. Show that $B$ generates a topology over $X$.

For $B$ to be a base I must take a set of $X$ and there must be a set of $B$ such that given $U$ in $\tau$ and $S \in B$, we can write $U= \cup S$. My difficulty is to know which subset I take from $X$.
Why use base properties?

答:
First off, every family $S$ of subsets of some non-empty set $X$ generates a topology on $X$ (if $S$ is used as a so-called subbase). So, there is nothing to show, the way the question is stated.

Note that a topology on $X$ is not given in your question. You cannot say "given $U$ in $\tau$", because there is no $\tau$ here, yet. There is only $B$, for the moment, and $\tau$ will come later, it will be the topology generated by the base $B$.

You would want to define $\tau$ as follows. A set $U$ is going to be open, if for every $x\in U$ there is $B_x\in B$ such that $x\in B_x\subseteq U$. (Equivalently, there is $B_U\subseteq B$ such that $U=\cup B_U=\cup\{C:C\in B_U\}$.)

But, if you define $\tau$ as above, it is not necessarily the case (in general) that $\tau$ will be a topology, because in order for $\tau$ to be a topology it must satisfy certain conditions:

(t1) $X$ and $\varnothing$ are in $\tau$ (i.e. $X$ and $\varnothing$ are open),

(t2) the union of any subfamily of $\tau$ is a member of $\tau$ (i.e. the union of any family of open sets is open), and

(t3) the intersection of any finite subfamily of $\tau$ is a member of $\tau$ (i.e. the intersection of any finite family of open sets is open).

So, if you start with $B$ and define $\tau$ as above, in order for $\tau$ to satisfy conditions (t1), (t2), (t3), the family $B$ itself must satisfy certain conditions. These are:

(b1) $\cup B=X$ (also written as $\cup\{C:C\in B\}=X$), and

(b2) If $B_0,B_1\in B$ and if $x\in B_0\cap B_1$ then there must be $B_2\in B$ such that $x\in B_2\subseteq B_0\cap B_1$.

Before you could talk about $\tau$ you need to verify that the family $B$ satisfies conditions (b1) and (b2) above. Consider this part of my answer so far to be a hint (that is, I am just trying to clarify to you your question, since I believe I know what you are asking, and you didn't state your question the way one would expect), and please try to verify (b1) and (b2) on your own. Report your progress, and ask more questions (you could use comments for your question or write comments to my answer for that), and I may provide the rest of the details, if you don't get them, but it would be better for you if you try to get the details on your own.

Here are the details. Condition (b1) is easy to verify (e.g. use that $X\in B$ since $X\setminus X=\varnothing$ which is finite). In order to verify (b2) take any  nonempty $A_0,A_1\in B$ and
consider cases. If $x_0\in A_0\cap A_1$ then $X\setminus A_0=F_0$ and $X\setminus A_1=F_1$, where each of $F_0$ and $F_1$ is finite, and hence the set $F=F_0\cup F_1$ is finite (and $x_0\not\in F$). Hence $A_0\cap A_1=(X\setminus F_0)\cap(X\setminus F_1)=X\setminus(F_0\cup F_1)=X\setminus F\in B$ (so we are done with this case, and it was not necessary to consider specific elements of $A_0\cap A_1$).

If $x_0\not\in A_0$ then it must be that $A_0=\{x\}$ for some $x\not=x_0$.
Then either $A_0\cap A_1=\{x\}\in B$ or
$A_0\cap A_1=\varnothing$ (in which case there is noting to prove). The case when
$x_0\not\in A_1$ is similar.

Title: Base of topologic space


问:
Cannot find limit using epsilon delta definition
Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.

I try as below.

Let $\varepsilon>0$. We choose $\delta>0$.

Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}

Now I don't know how to continue this answer. I confused with $6x$.

Anyone can help me?

EDIT: I have tried as below as JC12's answer.

Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$

Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$

Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
 \vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}

Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
 \frac{\varepsilon}{19} = \varepsilon. \end{align}

So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.

答:
You'll find the same approach with many other questions of a similar type, though you first notice that $|x-2|$ is the "part you want" since it is the same form as $0<|x-a|<\delta$. Say we restrict $|x-2|<1$ (this could be any other number but we just choose $1$ for sake of simplicity). Then $|x|-|2|<|x-2|<1$ and thus $|x|<3$. Then $|(x-2)^2+6x|<|(3-2)^2+6\times3|=19$. Thus you can choose your delta as $\min(1,\frac{\varepsilon}{19})$.

I leave you to finish and formalise the proof. As an added bonus, you could also try proving that $\lim_{x\rightarrow a}x^3=a^3, \forall a\in\mathbb{R}$ using a similar method.


答:
Let $\epsilon > 0$ be arbitrary and start off with $\delta_{1} := 1$ so that we have $|x-2| < \delta_{1} = 1$.

Then $x>1$ and $x < 3$ so that $(x-2)^{2} < 1$ and $6x < 18$.

Then $|(x-2)^{2} + 6x|<19$ and set $\delta _{2}$ so that $|x-2| < \delta_{2}$ where $\delta_{2} := \frac{\epsilon}{19}.$

Hence, if $\delta := \min\{\delta_{1},\delta_{2}\}$, we have the result.

The reason why it is necessary to end with a single $\delta$ is that, for this argument to work, we need $|x-2| < 1$ and $|x-2| < \frac{\epsilon}{19}$.

Then to find a fixed value that will work all the way through, choose the smaller one.

Title: Cannot find limit using epsilon delta definition

Body:
Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.

I try as below.

Let $\varepsilon>0$. We choose $\delta>0$.

Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}

Now I don't know how to continue this answer. I confused with $6x$.

Anyone can help me?

EDIT: I have tried as below as JC12's answer.

Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$

Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$

Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
 \vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}

Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
 \frac{\varepsilon}{19} = \varepsilon. \end{align}

So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.


问:
Decision variable must lie in union of multiple disjoint intervals
In my Linear Program, I'm trying to express that a decision variable $x \in R$ can only lie in certain intervals, e.g. $x \in [0,2] \cup [5,8] \cup [9,15]$.

I'm aware of that you can model either constraint 1 OR constraint 2 with a Big M trick (e.g. explained here in section 7.3 and asked here, but do not see directly how this could solve my question. Any ideas?

答:
With the same binary variables $z_i$ as in @MichalAdamaszek's answer, a tighter formulation is
\begin{align}
\sum_i a_i z_i \le x &\le \sum_i b_i z_i \\
\sum_i z_i &= 1
\end{align}


问:
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$
Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$

It has two poles at $\pm i$ and  branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.

Body:
\begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\
&= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx
\end{align*}

For the first integral, we plug

$$ u = \frac{1-x}{1+x}, \quad dx = - \frac{2}{(u+1)^2} \, du. $$

Then it is easy to find that

$$ \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx = \int_{0}^{1} \frac{\log 2 - \log (u + 1)}{u^2 + 1} \, du = \frac{\pi}{4}\log 2 - \int_{0}^{1} \frac{\log (u + 1)}{u^2 + 1} \, du $$

and hence

$$ \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx = \frac{\pi}{8}\log 2. $$

For the second integral, we plug $x = e^{-t}$ and we have

\begin{align*}
\int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx
&= - \int_{0}^{\infty} \frac{t e^{-t}}{1 + e^{-2t}} \, dt
 = - \sum_{n=0}^{\infty} (-1)^{n} \int_{0}^{\infty} t \, e^{-(2n+1)t} \, dt \\
&= - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = - G,
\end{align*}

where $G$ is the Catalan constant.

Therefore we have

$$ \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx = \frac{\pi}{4} \log 2 + G. $$


问:
Extension field of degree 2 contains an element whose square is in the field
Little question.
Let $F$ be a field with $\operatorname{char}(F) \neq 2$. Let $E$ be a field extension such that $[E:F]=2$

I need to show there exists $y \in E$ such that $y^2 \in F$ and $E=F[y]$.

Let $x \in E$ and not in $F$. If $[E:F]=2$, then $\deg(m_x(T))=2$ and so $m_x(T)=T^2+bT+c$ for some $b,c \in F$. So for all $x$ not in $F$: $x^2+bx+c=0$ (still for some $b(x),c(x) \in F$ )

The idea now is to set $y=x-\frac{b}{2}$ which
$\textbf{if} y \notin F$ leads to $y^2=\frac{b^2}{4}-c \in F$ and clearly $y$ satisfies $E=F[y]$

My questions is: why (if true) may I assume $y$ is not in $F$?

Thank you.

答:
Let's see . . .

$[E:F] = 2 \Longrightarrow F \subsetneq E \Longrightarrow \exists x \in E \setminus F; \tag 1$

$x \in E \setminus F \Longrightarrow F \subsetneq F(x) \subset E \Longrightarrow [E:F(x)][F(x):F] = [E:F] = 2 \Longrightarrow [F(x):F] \mid 2; \tag 2$

$[F(x):F] \mid 2 \Longrightarrow [[F(x):F] = 1] \vee [[F(x):F] = 2]; \tag 3$

$[F(x):F] = 1 \Longrightarrow x \in F \Rightarrow \Leftarrow x \in E \setminus F; \tag 4$

therefore,

$[F(x):F] = 2, \tag 5$

which implies that there exist

$a, b, c \in F, \text{not all zero}, \tag{6}$

with

$ax^2 + bx + c = 0; \tag 7$

we have

$a \ne 0, \tag 8$

lest

$bx + c = 0, \tag 9$

which prohibits

$b = 0, \tag{10}$

for this forces

$c = 0; \tag{11}$

but then

$a, b, c = 0, \tag{12}$

in contradiction to (6); we thus infer that (8) binds and hence we may write (7) in the form

$x^2 + \alpha x + \beta = 0, \tag{13}$

where

$\alpha = \dfrac{b}{a}, \beta = \dfrac{c}{a}; \tag{14}$

now, from (13) we find

$x^2 + \alpha x = -\beta, \tag{15}$

and since $\text{char} \; F \ne 2, \tag{16}$

$\dfrac{\alpha}{2} \in F, \tag{17}$

whence

$\left (x + \dfrac{\alpha}{2} \right)^2 = x^2 + \alpha x + \dfrac{\alpha^2}{4} = \dfrac{\alpha^2}{4} -\beta \in F; \tag{18}$

setting

$y = x + \dfrac{\alpha}{2} \in E \setminus F, \tag{19}$

we see that

$y^2 \in F, \tag{20}$

and that

$F(y) = F[y] = F[x] = F(x) = E, \tag{21}$

which binds by virtue of the fact that

$F[y] = F(y), F[x] = F(x) \tag{22}$

when $x$ and $y$ are algebraic over $F$, and that (19) implies

$F(y) = F(x). \tag{23}$

$OE\Delta$.


问:
Why do proof authors use natural language sentences to write proofs?
I haven't read very many proofs. The majority of the ones that I've read, I've read in my first-year proofs textbook. Nevertheless, its first chapter expatiates on the proper use of English in mathematical proofs, so I suspect that most proof authors do use both English (or another natural language) and formal proof systems for writing their proofs. However, I can't imagine many circumstances where mathematical notation and a formal proof system wouldn't suffice to convey the author's pertinent thoughts. </p>

<p>I suspect that proof authors have good reasons for making use of both English and formal expressions in their proofs, but I can't think of what those reasons are. Why do proof authors use English and formal expressions to write proofs? Are there stylistic rules (or guidelines) that state how, and in what parts of the proof, an author should, and shouldn't, use English? </p>

答:
I write software, and I would like to draw an analogy using programming languages.

Math proofs can be traced at the bottom level to formal systems. The systems can be compared to the bottom level of computer hardware composed of CPUs, memory, buses, and possibly GPUs for heterogeneous systems. Rules of deduction in the formal system are analogous to electrical behaviors of devices. Theoretically, an excellent engineer knowing everything about computer hardware should be able to use sequences of 0 (low voltage) and 1 (high voltages) to control computers for specific purposes. But in reality, engineers typically don't do that. It is ridiculous for software engineers at Google, Amazon, Facebook and Microsoft to write programs using hardware languages, even if they can, as there exist programming languages much more efficient, like C/C++, JAVA, etc. These high-level languages are designed so that when applied, they generate common results on different hardware. Engineers do not have to know exactly what happens in hardware, but they only need to trust that each sentence written in a high-level language corresponds to a specific behavior in hardware, which will finally generate the CORRECT result ensured by the documentation of the high-level language. Experts in programming languages design languages so that they are of behaviors described in the documentation. These experts have to understand hardware thoroughly to design correct hardware behaviors of languages.

Similarly, different formal systems are like different hardware. Daily languages such as English and Chinese are like high-level programming languages. Proofs from experienced mathematicians, whatever language they are written in, have to correspond to formal languages in some way. We are confident that the correspondence exists, but we do not specify what it is. Logicians have to design and verify such correspondence. But all mathematicians are not logicians. They are "engineers" who construct proofs.

The question "why don't we construct proofs in formal languages" is analogous to the question "why don't we write computer programs in machine language". The answer is, it is inefficient. But engineers who knows machine languages may be more efficient in debugging, just as mathematicians who have formal systems in mind are less likely to make wrong deductions for complicated propositions.


问:
Doubt showing commutativity of Pushforward and Lie Bracket
In this journey of understanding some of differential geometry I started to do some proof of results. And now I on doubt about this one which is the commutativity between Pushforward and the Lie bracket without draw upon local coordinates:

Let $M$ and $N$ two differential manifolds, and
$$ \varphi: M \longrightarrow N $$
A diffeomorphism between the two.

So if I have a map $g: N \rightarrow \mathbb{R} $ then I can define a function $\varphi^*g: M \rightarrow \mathbb{R} $ the Pullback of $g$.

And with the same philosophy, given a vector field on $M$, $X: C^\infty(M) \rightarrow C^\infty(M)$ the Pushforward vector field on $N$ is defined as: $\varphi_*X(g) = X(\varphi^*g), \ g \in C^\infty(N)$

(I made all this intro because I'm not even 100% sure about if I got this concepts right, so any comment is welcome).

So my idea of proof is first start with the pushformward of the Lie bracket, with $f \in C^\infty(N)$:
$$ \varphi_*[X,Y](f) = [X,Y](\varphi^*f) = X(Y(\varphi^*f)) - Y(X(\varphi^*f))$$
Since the two terms are the same under a swaping of $X$ and $Y$ I will continue with the first one.

So, by the definition of pushforward $Y(\varphi^*f) = \varphi_*Y(f)$, but $\varphi_*Y(f) \in C^\infty(N)$ so it cannot be an argument for $X(\cdot{})$ therefore the expression
$$ X(Y(\varphi^*f)) = X(\varphi_*Y(f))$$
Cannot the be true.

My intuition tells me that the solution of this is that, since $\varphi_*Y(f)$ is another map on $N$ I can first "pull-it back" to $M$ before feed it into $X$. Having:
$$X(Y(\varphi^*f)) = X(\varphi^*(\varphi_*Y(f))) = \varphi_*X(\varphi_*Y(f))$$
Letting me completing my "proof" after a few straightforward steps.

Is this correct? Is so, what is the justification for incorporate that last pullback? How can I know that the pullback of $\varphi_*Y(f)$ if the map on $M$ that I'm looking for?

Sorry for the bad English and thanks in advance!

答:
(I made all this intro because I'm not even 100% sure about if I got this concepts right, so any comment is welcome).

So far, the definitions and setup seem fine.

Is this correct? Is so, what is the justification for incorporate that last pullback?

Yes, it is correct that $Y(\varphi^*f) = \varphi^*(\varphi_*Y(f))$.

Here is a rough proof.

I'm mostly copying this proof from Proposition 8.16 of John Lee's book "Smooth Manifolds"

To prove it, take a point $p \in M$. Then $Y(\varphi^*f)$ is a function on $M$, which at the point $p$ is:
$$ Y(\varphi^*f)(p) = (\varphi_*Y)_{\varphi(p)}(f) $$
Here, $(\varphi_*Y)_{\varphi(p)}$ is the pushforward vector field evaluated at $\varphi(p)$. This last expression is equal to $\varphi_*Y(f)(\varphi(p))$. That is, $\varphi_*Y(f)$ is a function on $N$, and evaluating it at $\varphi(p)$ is equal to $(\varphi_*Y)_p(f)$. Now, this last expression, $\varphi_*Y(f)(\varphi(p))$, is the composition $\Big( \varphi_*Y(f) \Big) \circ \varphi$, evaluated at $p$. By definition of pullback of functions, this is equal to $\varphi^*(\varphi_*Y(f))$.


Body:
More generally,
if you want to show that
$\lim_{x \to c} x^n=c^n$,
note that
$x^n-c^n
=(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}
$.

If
$c-r < x < c+r$
where
$0 < r< \min(1, x)$
then
$\sum_{k=0}^{n-1} x^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^k(c+r)^{n-1-k}
=n(c+r)^n
$
so

$\begin{array}\\
|x^n-c^n|
&=|(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&=|x-c|\,|\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&<|x-c|n(c+r)^{n-1}\\
\end{array}
$

so if
$|x-c|
\le \dfrac{\delta}{n(c+r)^{n-1}}
$
then
$|x^n-c^n|
\lt \delta$.


问:
Primes in solutions to Pell-type equations
What is known about primes in solutions to Pell-type equations?

In particular, consider the negative Pell equation $x^2 - 5 y^2 = -1$.
As far as I've been able to check
(in the first $4000$ solutions) the only positive-integer solution with $y$ prime  is $x=38$, $y=17$,
but I don't see any obvious reason why this should be the case.

答:
The general solution of the negative Pell equation $\;x^2-5y^2=-1\;$ is the following

$\begin{cases}
x_n=\dfrac12\bigl[(2+\sqrt5)^{2n+1}+(2-\sqrt5)^{2n+1}\bigr]\\
y_n=\dfrac1{2\sqrt5}\bigl[(2+\sqrt5)^{2n+1}-(2-\sqrt5)^{2n+1}\bigr]
\end{cases}$

$\forall n\in\mathbb{N}\cup\{0\}\;.$

Since

$\begin{align}
2+\sqrt5&=\dfrac{16+8\sqrt5}8=\dfrac{1+3\sqrt5+15+5\sqrt5}8=\\
&=\left(\dfrac{1+\sqrt5}2\right)^3\;,
\end{align}$

$\begin{align}
2-\sqrt5&=\dfrac{16-8\sqrt5}8=\dfrac{1-3\sqrt5+15-5\sqrt5}8=\\
&=\left(\dfrac{1-\sqrt5}2\right)^3\;,
\end{align}$

it follows that

$y_n=\dfrac1{2\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^{6n+3}-\left(\dfrac{1-\sqrt5}2\right)^{6n+3}\right]$

$\forall n\in\mathbb{N}\cup\{0\}\;.$

Let $\;F_n =\dfrac1{\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^n-\left(\dfrac{1-\sqrt5}2\right)^n\right]\;$

be the $\;n^{\text{th}}\;$ Fibonacci number, then it results that

$\begin{align}
y_n&=\dfrac1{2\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^{6n+3}-\left(\dfrac{1-\sqrt5}2\right)^{6n+3}\right]=\\
&=\dfrac1{2\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^{2n+1}-\left(\dfrac{1-\sqrt5}2\right)^{2n+1}\right]\cdot\\
&\quad\quad\cdot\left[\left(\dfrac{1+\sqrt5}2\right)^{4n+2}-1+\left(\dfrac{1-\sqrt5}2\right)^{4n+2}\right]=\\
&=\dfrac1{2\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^{2n+1}-\left(\dfrac{1-\sqrt5}2\right)^{2n+1}\right]\cdot\\
&\quad\quad\cdot\left[\left(\dfrac{1+\sqrt5}2\right)^{4n+2}+2+\left(\dfrac{1-\sqrt5}2\right)^{4n+2}-3\right]=\\
&=\dfrac1{2\sqrt5}\left[\left(\dfrac{1+\sqrt5}2\right)^{2n+1}-\left(\dfrac{1-\sqrt5}2\right)^{2n+1}\right]\cdot\\
&\quad\quad\cdot\left\{\left[\left(\dfrac{1+\sqrt5}2\right)^{2n+1}-\left(\dfrac{1-\sqrt5}2\right)^{2n+1}\right]^2-3\right\}=\\
&=\dfrac12 F_{2n+1}\biggl(5F_{2n+1}^2-3\biggr)\quad\forall n\in\mathbb{N}\cup\{0\}\;.
\end{align}$

Hence,

$y_n\;$ is prime $\iff F_{2n+1}=2\iff n=1\;.$

Moreover,

$y_1=\dfrac12 F_3\biggl(5F_3^2-3\biggr)=\dfrac12\cdot 2\biggl(5\cdot 2^2-3\biggr)=17\;.$


问:
Solution verification: the minimal/maximal elements of a relation on $\mathcal{P}(\{1,2,3,4\})$
Consider the power set $\mathcal P ( \{ 1 , 2 , 3 , 4 \} ) $ and the partial order $ S $ on $ $ such that $ A \mathrel S B $ is defined as

$$A \; S \; B \iff A \cap \{ 1 , 2 \} \subset B \cap \{ 1 , 2 \} $$

$S$ is a partial order and I`m trying to find the minimum and maximum elements of it. Are my results correct?

Minimum elements:
$$   \{\text{Ø}\},\{3\},\{4\},\{3,4\}  $$
Maximum elements:
$$\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4\} $$

答:
$
\newcommand{\set}[1]{ \left\{ #1 \right\} }
\newcommand{\P}{ \mathcal{P}( \set{1,2} ) }
\newcommand{\PP}{ \mathcal{P}( \set{1,2,3,4} ) }
$
Part of the problem with this, well, problem is that it's hard to intuitively take on except under brute force. However, one thing you might be able to notice off the bat is that the relation $S$ on $\PP$ is the same as the subset order $\subseteq$ on $\P$ in some sense, due to the intersections with $\{1,2\}$. Any $X \in \PP$ satisfies $X \cap \set{1,2} \in \P$, in other words.

In that light, we can reduce this problem somewhat. If we can find the maximal/minimal elements of $\P$ under $\subseteq$, then the sets that generate those extrema in $\PP$ are its maximal/minimal elements under $S$.

Of course, the extreme elements of $\P$ are easily determined. We can even examine the situation visually with a graph like below, where $X$ points to $Y$ if and only if $X \subseteq Y$:

From this, it is clear that the maximal element of $\P$ is $\set{1,2}$, and the minimal element of $\P$ is $\varnothing$.

So the question remains: what sets of $\PP$ generate these under intersection with $\set{1,2}$? Well,

$X \cap \set{1,2} = \set{1,2}$ if and only if $\set{1,2} \subseteq X$. That is, $1,2 \in X$.

$X \cap \set{1,2} = \varnothing$ if and only if, similarly, $1,2 \not \in X$.

From these and an enumeration of $\PP$, it is clear that

$$X \cap \set{1,2} = \set{1,2} \iff X \in \Big\{ \set{1,2},\set{1,2,3},\set{1,2,4},\set{1,2,3,4} \Big\}$$

and

$$X \cap \set{1,2} = \varnothing \iff X \in \Big\{ \varnothing, \set{3},\set{4},\set{3,4} \Big\}$$

Thus, we have that

\begin{align*}
\text{the set of maxima of } S &= \Big\{ \set{1,2},\set{1,2,3},\set{1,2,4},\set{1,2,3,4} \Big\} \\
\text{the set of minima of } S &= \Big\{ \varnothing, \set{3},\set{4},\set{3,4} \Big\}
\end{align*}

This is mostly in line with your answer. However, be cognizant of the difference between $\set{\varnothing}$ and $\varnothing$ -- in particular, $\varnothing \in \mathcal{P}(X)$ for any set $X$, but we can't always say that of $\set \varnothing$.

But otherwise, your answer is correct.

...granted, this question is a fair bit old (over a month), so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


答:
Yes, a matrix with functions as elements definitely has some other function as its determinant in general case.
If you're interested as to where would you find such matrices without just artificially constructing one, then you could read some fluid mechanics.

For example, determinant of the strain rate tensor shows the volume rate of change of a
piece of fluid at that point. You can imagine a compressible fluid flowing through a bottleneck and getting compressed along the path and that means that this tensor changes its determinant from point to point.

Title: Matrices With Changing Determinants

Body:
I would love some direction and any information for some further reading about a cool thought I just had. If a square matrix has functions as the elements, would it not make the matrix depend on $x$ (or whatever variables) and hence have a changing determinant? What would be even cooler is for a matrix to sometimes have a determinant that is equal to $0$. Now that I think about it, this could be a rotation matrix, right? $A_{11} = \cos x$ and $A_{12} = -\sin x$ and $A_{21} = \sin x$ and $A_{22}= \cos x$ so if $x=45^\circ$, the determinant will be $0$ I think. This would mean it has no inverse for that value? I think I made a mistake because it is easy to visualize how to reverse a $45^\circ$ rotation.

In any case, please do comment anything that you know about this and where to find out more about it.

Edit: Yes the rotation matrix was a bad example because the determinant is always $1$. I could have used a better example with a determinant of $\cos x - \pi$ or $2x-6$


问:
$p(x)=0$ with real coefficient has purely Imaginary roots.Then the equation $p(p(x)) = 0$ has
If the Quadratic equation $p(x)=0$ with real coefficient has purely Imaginary roots.Then the

equation $p(p(x)) = 0$ has

$\bf{OPTIONS::}$

$(a)\;\; $ Only purely Inaginary Roots.  $\;\;\;\;\;\;(b)$ all real roots.

$(c)$ Two real and Two purely imaginary roots.$\;\;\;\;\;\;(d)$ neither real nor purely imaginary roots.

$\bf{My\; Try}::$ Let $p(x) = x^2+1.$ Then $p(p(x)) = (x^2+1)^2+1 = x^4+2x^2+2$.

Now Roots of $\displaystyle p(p(x))=0\Rightarrow x^4+2x^2+2=0\Rightarrow x^2 = \frac{-2\pm\sqrt{4-8}}{2} = \frac{-2\pm 2i}{2} = -1\pm i$

So $x^2=-1+i\;\;,-1-i$. Means $x$ must be in the form of $a+ib\;,$ where $a,b\neq 0$ and $i=\sqrt{-1}$

So option $(d)$ mut be Correct.

But my question is how can we solve the given question in Analytical method. means not by guessing.

Help me

Thanks

答:
Another version on @Brad's answer :

$p(p(x)) = 0$

$a(ax^2+b)^2 + b = 0$

$ax^2 + b = ki$ (where $k=\sqrt\frac{b}{a}$ )

But, since $a,b\in R$,

Therefore

LHS will always be real (if $x\in R$) or Complex (if $x \in C$)
However RHS is purely imaginary.

Therefore the only solutions of the equation $p(p(x)=0)$ will be when x is a complex number.


问:
Find the length of AB in the given question.

In a rectangle $ABCD$, $E$ is the midpoint of $AB$; $F$ is a point on $AC$ such that $BF$ is perpendicular to $AC$; and $FE$ perpendicular to $BD$. Suppose $BC = 8\sqrt3$. Find $AB$.

Note: I know that this question can be easily solved by trigonometry as mentioned in this post on Sarthaks.com.  However I need a solution in which normal Euclidean Geometry is used and not Trigonometry.

答:
Probably not the best solution, but this is what I've come up with. Let $AC$ and $BD$ intersect at $O$. Let $EF$ and $BO$ intersect at $G$.

First, $\angle OEB+\angle OFB=90+90=180^{\circ}$ and thus $EOFB$ is cyclic. Then $\angle OBF=\angle OEF$ and $\angle EBG=90-\angle GEB=90-(90-\angle OEG)=\angle OEG=\angle OEF=\angle OBF=\angle GBF$. Then $\triangle EFB$ is iscocele with $EB=FB$.

Let $\angle BOF=\alpha$. Then $\angle OBF=90-\alpha$. Since $\triangle AEO\cong \triangle OEB$, we also have $\angle AOB=180-\alpha\implies \angle EOB=90-\frac{\alpha}{2}\implies \angle EBO=\frac{\alpha}{2}$. Now, $\frac{\alpha}{2}=\angle EBO=\angle EBG=\angle GBF$. Then since $\angle EBG=\angle GBF$ from above, $90-\alpha=\frac{\alpha}{2}\implies\alpha=60^{\circ}$. Now, $OB=OC$ since $ABCD$ is a rectangle so $\angle OBC=\angle OCB=\frac{180-60}{2}=60^{\circ}$. Then $\triangle OCB$ is equilateral.

Finally, $BF$ is perpendicular to $OC$ so $BF$ is the altitude of triangle $OBC$. Keep in mind $EB=FB$ from above. Then $AB=2EB=2FB=2(\frac{BC\sqrt{3}}{2})=2(\sqrt{3}\times \frac{8\sqrt{3}}{2})=24$.

Now the rule that states the altitude of an equilateral triangle is equal to $\frac{a\sqrt{3}}{2}$ where $a$ is one of its side lengths can be proved without trigonometry, such as the first proof here: https://www.quora.com/In-an-equilateral-triangle-with-side-a-prove-that-altitude-is-a-root-3-2-and-area-root-3-a-square-2. However, this is pretty well known anyway. Note this is how we got $2BF=2(\frac{BC\sqrt{3}}{2})$.


问:
If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is
Below are the steps I have done so far.

Find $a$ in terms of $c$

$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$

Find $b$ in terms of $c$

$$
ab = 432 \\
\implies 4.5c \cdot b = 432 \\
\implies b = \frac{96}c.
$$

Substituting, we get

$$
a + b + c = 4.5c + \frac{96}c + c = 5.5c + \frac{96}c.
$$

How should I proceed from here?

答:
The factors of $96$ that are  less than $9$ are $\{1,2,3,4,6,8\}$

Plug these values of $c$ into your equation above to find the set of $a+b+c$


问:
Cannot solve the non-exact ODE, I don't know how to find the integrating factor such that make the exact ODE.
Solve the ODE $xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$.

I have tried as below. I'm using exact/non exact method.

We change the ODE into form
$$\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) dx-xdy = 0.$$

Let $M(x,y)=y-x\cos ^2\left(\dfrac{y}{x}\right)$ and $N(x,y)=-x$.
\begin{align*}
 \dfrac{\partial M(x,y)}{\partial y}&=
 \dfrac{\partial }{\partial y}\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) =
 1+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right).\\
 \dfrac{\partial N(x,y)}{\partial x}&=
 \dfrac{\partial }{\partial x}(-x) = -1.
\end{align*}
Since $\dfrac{\partial M(x,y)}{\partial y}\neq \dfrac{\partial N(x,y)}{\partial x}$ then that ODE is non exact ODE.

Now I want to change it into exact ODE. We must find integrating factor to make the exact ODE. But now I don't know how to do it.
\begin{align*}
 \dfrac{\partial M(x,y)}{\partial y}
 -
 \dfrac{\partial N(x,y)}{\partial x}
 =
 2+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right)
\end{align*}

Anyone can help me?

答:
$$xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$$
The DE is homogeneous:
$$y' = \dfrac yx-\cos ^2\left(\dfrac{y}{x}\right)$$
Substitute $y=tx \implies y'=t'x+t$
$$t'x=- \cos ^2 t$$
It's now separable.
$$\int \dfrac {dt}{ \cos^2 t}=-\int \dfrac {dx}{x}$$


问:
Showing that a relation tied to prime factorization is an equivalence relation

Consider the relation $≈$ on $ℕ$:

If $x = {p_1}^{x_1} \times {p_2}^{x_2} ... \times {p_k}^{x_k}$ and $y = {p_1}^{y_1} {p_2}^{y_2} ... {p_k}^
{y_k}$ , where $p_i$ are distinct primes and $x_i$ and $y_i$ are nonnegative integers, then
$x ≈ y$ iff $x_1 + x_2 + \ldots + x_k = y_1 + y_2 +\ldots + y_k$.

Show $≈$ is an equivalence relation

I know I need to show reflexivity, symmetry, and transitivity.

For reflexivity I have $x ≈ x \iff x_1 + x_2 + \ldots + x_k = x_1 + x_2 + \ldots + x_k$, which is clearly true.

And for symmetry I have $x ≈ y \implies y ≈ x$,

so

\begin{align*}
&x_1 + x_2 + \ldots + x_k = y_1 + y_2 +\ldots + y_k \\
\implies &y_1 + y_2 +\ldots + y_k = x_1 + x_2 + \ldots + x_k
\end{align*}

which is also clearly true.

This seems too easy and I feel like I'm doing this wrong.

答:
$
\newcommand{\p}[1]{\prod_i p_i^{#1_i}}
\newcommand{\s}[1]{\sum_i #1_i}
$
This is more or less correct; the main remaining issue, as noted in the comments, is to use that the fundamental theorem of arithmetic ensures the uniqueness of the prime factorization. (Your relation is not only relative to the sums of the exponents, but a priori also relative to the primes $p_i$ you use.) Then each property follows easily:

Reflexivity: Since $x = \prod_i p_i^{x_i}$ (and this is the only representation of $x$ for the primes $p_i$, up to having $x_i = 0$ for some $i$'s) and $\sum_i  x_i = \sum_i x_i$ trivially, $x \approx x$.

Symmetry: Assume $x \approx y$. Since $x = \prod_i p_i^{x_i}$ and $y = \p y$, then $\s x = \s y$. Equality is a symmetric relation, so $\s y = \s x$, and the prime decomposition is the only one, so $y \approx x$.

Transitivity: Assume $x \approx y$ and $y \approx z$. Then we may write $x,y,z$ uniquely by $x = \p x$, $y = \p y$, and $z = \p z$. By assumption, $\s x = \s y$ and $\s y = \s z$. By the transitivity of equality, $\s x = \s z$, and, alongside the uniqueness of the prime decomposition, this ensures $x \approx z$.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


答:
From similar triangles BFG and BFO, $\angle$BFG = $\angle$BOF = $2\alpha$. Note that the midpoint E is the circumcenter of the right triangle ABF. So, the triangles AEF and BEF are isosceles and $\angle$EBF = $\angle$BFE = $\angle$BEF =  $2\alpha$. Also, from isosceles triangle BCO and similar triangles BFC and BCA, $\angle$CBO = $\angle$BCO = $\angle$FBA = $2\alpha$.

Thus, both triangles BEF and BCO are equilateral, and
$$AB = 2EB = 2BF =2\cdot \frac{\sqrt3}2BC = 24$$

Title: Find the length of AB in the given question.

Body:
In a rectangle $ABCD$, $E$ is the midpoint of $AB$; $F$ is a point on $AC$ such that $BF$ is perpendicular to $AC$; and $FE$ perpendicular to $BD$. Suppose $BC = 8\sqrt3$. Find $AB$.

Note: I know that this question can be easily solved by trigonometry as mentioned in this post on Sarthaks.com.  However I need a solution in which normal Euclidean Geometry is used and not Trigonometry.


Title: If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is

Body:
Below are the steps I have done so far.

Find $a$ in terms of $c$

$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$

Find $b$ in terms of $c$

$$
ab = 432 \\
\implies 4.5c \cdot b = 432 \\
\implies b = \frac{96}c.
$$

Substituting, we get

$$
a + b + c = 4.5c + \frac{96}c + c = 5.5c + \frac{96}c.
$$

How should I proceed from here?

Body: $bc=96$ and $c<9$ constrains $c$ to be one of $1,2,3,4,6,8$. Since $\gcd(432,96)=48$, $c$ cannot be $1$ or $3$, leaving four cases: $(a,b,c)=(9,48,2),(18,24,4),(27,16,6),(36,12,8)$. Of these, $(18,24,4)$ achieves the smallest $a+b+c$ of $46$.


答:
Hint

Considering $$xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$$ first let $y=x \,z$ to make
$$x^2 z'+x \cos ^2(z)=0$$ Assume $x\neq 0$ to make
$$x z'+\cos ^2(z)=0$$ Switch variables
$$\frac x {x'}+\cos ^2(z)=0 \implies \frac  {x'} x=-\frac 1 {\cos ^2(z)}$$ which seems to be simple.

Title: Cannot solve the non-exact ODE, I don't know how to find the integrating factor such that make the exact ODE.

Body:
Solve the ODE $xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$.

I have tried as below. I'm using exact/non exact method.

We change the ODE into form
$$\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) dx-xdy = 0.$$

Let $M(x,y)=y-x\cos ^2\left(\dfrac{y}{x}\right)$ and $N(x,y)=-x$.
\begin{align*}
 \dfrac{\partial M(x,y)}{\partial y}&=
 \dfrac{\partial }{\partial y}\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) =
 1+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right).\\
 \dfrac{\partial N(x,y)}{\partial x}&=
 \dfrac{\partial }{\partial x}(-x) = -1.
\end{align*}
Since $\dfrac{\partial M(x,y)}{\partial y}\neq \dfrac{\partial N(x,y)}{\partial x}$ then that ODE is non exact ODE.

Now I want to change it into exact ODE. We must find integrating factor to make the exact ODE. But now I don't know how to do it.
\begin{align*}
 \dfrac{\partial M(x,y)}{\partial y}
 -
 \dfrac{\partial N(x,y)}{\partial x}
 =
 2+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right)
\end{align*}

Anyone can help me?


问:
Finding the transitive closure of a given relation on $X = \{1,5,7,9\}$
What is the transitive closure of the relation $\{(1,5),(5,7),(7,9),(1,7),(1,9),(5,9)\}$ on $X=\{1,5,7,9\}$?

I think this relation is already transitive.

$(1,5),(5,7) \implies (1,7)$ already exist

$(1,5),(5,9) \implies (1,9)$ already exist

$(1,7),(7,9) \implies (1,9)$ already exist

Did I miss something or this relation is already transitive so my answer which is $\{(1,5),(5,7),(7,9),(1,7),(1,9),(5,9)\}$ correct?

答:
You are indeed correct that the relation is already transitive. This is easy to see when you visualize the relation as a directed graph like below, where $x$ points to $y$ iff $(x,y) \in R$ (where $R$ is your relation on $X$).

In diagrams like these, transitivity amounts to, whenever you can travel from node $x$ to $y$ and then to node $z$, then there is a path letting you cut across from $x$ to $z$. Basically whenever two sides of a triangle exist, the third is also filled in (though the orientation of the third side depends on the other two).

Using the graph above, follow every path of length two from every node, and you'll always end up finding a "shortcut" path of length one exists from the start to end. Thus, the relation is transitive, and is its own transitive closure.

...granted, this question a bit old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Showing a map is well defined
Problem: Let $A$ be any $\Bbb{Z}$-module, let $a$ be any element of $A$ and let $n\in \Bbb{Z}^+$. Prove that the map $\psi_a:\Bbb{Z/nZ}\to A$ given by $\psi_a(\overline{k})=ka$ is a well-defined $\Bbb{Z}$-module homomorphism if and only if $na=0$.

Attempt: Suppose $\psi_a$ is a well-defined $\Bbb{Z}$-module homomorphism.
Let $\psi_a(\overline{x}+n\overline{y})=\psi_a(\overline{x}+n\overline{z})$, where, $\overline{z}\neq \overline{y}$.
Therefore $$\psi_a(\overline{x})+n\psi_a(\overline{y})=\psi_a(\overline{x})+n\psi_a(\overline{z})$$ $(\because \psi_a$ is a $\Bbb{Z}$-module homomorphism).
$$xa+nay=xa+naz$$

Only possible if $na=0$.

Am i on right track?

答:
Module homomorphisms maps $0$ to $0$. Thus, if $\psi_a$ is well-defined, $0_A = \psi_a(\overline{0}) = \psi_a(\overline{n}) = na$.

If $na=0_A$, then for any $k_1,k_2 \in \mathbb Z$ the following holds: $k_1 \equiv k_2 \!\pmod{n}$ implies that $k_1-k_2 = kn$ for some $k \in \mathbb Z$, and then $$\psi_a(\overline{k_1})-\psi_a(\overline{k_2}) = k_1a-k_2a = kna = k0_A = 0_A$$ which means: $\overline{k_1}=\overline{k_2}$ implies that $\psi_a(\overline{k_1})=\psi_a(\overline{k_2})$.

Title: Showing a map is well defined


问:
Invertibility of Fractional Laplacian in Lebesgue spaces
I'm trying to digest the following statements, which are taken from §2 of Carlo Morpurgo's paper Sharp inequalities for functional integrals and traces of conformally invariant operators.

For $0< \alpha < n$, up to a positive constant multiple, one can define the fractional Laplacian of order $n$ on the Schwartz class $\mathcal S(\mathbb R^n)$ as
\begin{equation*}
\mathcal F((-\Delta)^{\alpha/2}\varphi)(\xi)
=
|\xi|^\alpha \mathcal F\varphi(\xi),
\qquad \text{ for }\varphi\in \mathcal S(\mathbb R^n),
\end{equation*}
where $\mathcal F$ is the Fourier transform. This operator can be extended by duality to the space of tempered distributions $\mathcal S'(\mathbb R^n)$.

If $p\geq 1$ and $f\in L^p(\mathbb R^n)$ is given, the equation $(-\Delta)^{\alpha/2} u = f$ has a unique solution in $L^q(\mathbb R^n)$, where $\frac 1 q = \frac 1 p - \frac \alpha n$. Up to constant multiple, this solution is given explicitly by convolution with the Riesz potential:
\begin{equation*}
u(x) = I_\alpha (f):=\int_{\mathbb R^n}\frac{f(y)}{|x - y|^{n - \alpha}}\; \mathrm d y.
\end{equation*}

My first question is in regards to the extension of $(-\Delta)^{\alpha/2}$ to $\mathcal S'(\mathbb R^n)$ via duality. Heuristically, for $v\in \mathcal S'(\mathbb R^n)$ one would like to define
\begin{equation*}
\langle (-\Delta)^{\alpha/2}v, \varphi\rangle
:=
\langle v, (-\Delta)^{\alpha/2}\varphi\rangle
=
\langle \mathcal F v , |\xi|^\alpha \mathcal F\varphi\rangle
\end{equation*}
whenever $\varphi \in \mathcal S(\mathbb R^n)$. The problem with this heuristic is that $\mathcal F v$, being a tempered distribution, acts on functions in $\mathcal S(\mathbb R^n)$. However, $|\xi|^\alpha \mathcal F\varphi\notin \mathcal S(\mathbb R^n)$ due to the lack of differentiability at $\xi = 0$. What is the correct interpretation of the action of $\mathcal F v$ on $|\xi|^\alpha \mathcal F\varphi$?

My second question is in regard to the statement that $I_\alpha f$ satisfies $(-\Delta )^{\alpha/2} u = f$ whenever $f\in L^p$ (of particular interest is $p = 2n/(n + \alpha)$ and $q = 2n/(n - \alpha)$, but that doesn't matter for the question at hand). The Hardy-Littlewood-Sobolev inequality guarantees that $I_\alpha f\in L^q$, so $((-\Delta)^{\alpha/2}\circ I_\alpha )(f)$ should be interpreted in the distributional sense. In particular, for $\varphi\in \mathcal S(\mathbb R^n)$ I would like to perform the following computation (up to constant multiple):
\begin{eqnarray*}
\langle (-\Delta)^{\alpha/2}\circ I_\alpha f, \varphi\rangle
& = &
\langle I_\alpha f, (-\Delta)^{\alpha/2}\varphi\rangle\\
& = &
\langle \mathcal F(I_\alpha f), \mathcal F((-\Delta)^{\alpha/2} \varphi)\rangle\\
& = &
\langle |\xi|^{-\alpha}\mathcal F f, |\xi|^\alpha \mathcal F\varphi\rangle\\
& = &
\langle\mathcal F f, \mathcal F\varphi\rangle\\
& = &
\langle f, \varphi\rangle.
\end{eqnarray*}
The main problem is the interpretation of the expression $\langle \mathcal F(I_\alpha f), \mathcal F((-\Delta)^{\alpha/2} \varphi)\rangle$. Since $I_\alpha f\in L^q$, we can interpret $\mathcal F(I_\alpha f)$ as a tempered distribution. However, with this interpretation, $\mathcal F (I_\alpha f)$ can not act on $\mathcal F((-\Delta)^{\alpha/2}f)$ since the latter of these two quantities is not in the Schwartz class. What is the correct interpretation of the action of $\mathcal F(I_\alpha f)$ on $\mathcal F((-\Delta)^{\alpha/2}\varphi)$?

答:
To the first question, I don't think there is a way to extend the fractional Laplacian to all tempered distributions. The difficulty is exactly as you stated, that multiplication by $|\xi|^{\alpha}$ doesn't preserve the Schwartz class $\mathcal{S}$. That said, if the distribution is such that its Fourier transform agrees with a measure or function in a neighborhood of the origin, we can make things work by keeping the multiplication by $|\xi|^{\alpha}$ on the distribution rather than the Schwartz function, at least in that neighborhood.

To the second question, there's a slightly different way of looking at things that avoids the difficulty in the calculation. To begin, even though the fractional Laplacian $(-\Delta)^{\alpha/2}$ doesn't preserve the Schwartz space, it does map this space into $L^r$ for $1 \leq r \leq \infty$. (By looking at the inverse Fourier transform of $|\xi|^{\alpha} \mathcal{F} \varphi(\xi)$ and using integration by parts, we can obtain generic decay at infinity on the order of $|x|^{-n - \alpha}$.) Then for $f \in L^p$, and letting $u = I_{\alpha} f \in L^q$ (with $1/p = 1/q - \alpha/n$), one (weak type) sense in which $(-\Delta)^{\alpha/2} u = f$ is that, for $\varphi \in \mathcal{S}$,
$$
\int u(x) \overline{(-\Delta)^{\alpha/2} \varphi(x)} \, dx
= \int f(x) \overline{\varphi(x)}
\, dx.
$$
Heuristically the same thing is going on as in the not-quite-formally-correct distributional calculation, of course. Perhaps the key thing to keep in mind is the basic fact that distributions given by $L^p$ functions act via integration, and so the integration point of view can help with some of the formalities.


答:
This's a homogeneous equation, so we can solve this ODE using the sustitution $y=xv$. Indeed,
\begin{eqnarray*}
xy'=y-x\cos^{2}\left(\frac{y}{x}\right) &\overset{y=xv}{\underset{y'(x)=xv'(x)+v(x)}{\iff}}  & x\left(xv'(x)+v(x)\right)=x\cos^{2}(v(x))+xv(x)\\
&\iff& v'(x)=-\frac{\cos^{2}(x)}{x}\\
&\iff& \int \sec^{2}(v(x))dv(x)=\int -\frac{1}{x}dx\\
&\iff& \tan(v(x))=-\ln|x|+c\\
&\implies& v(x)=\arctan(-\ln|x|+c)\\
&\overset{y(x)=xv(x)}{\implies}& \boxed{y(x)=x\arctan(-\ln|x|+c)} &
\end{eqnarray*}


问:
Atiyah-Macdonald: Exercise 3.6
Q: Let $A$ be a nonzero ring and let $\Sigma$ be the set of multiplicatively closed subsets of $A$ such that $0 \notin S.$ Show that $\Sigma$ has a maximal element, and that $S \in \Sigma$ is maximal iff $A \setminus S$ is a minimal prime ideal of $A$.

It can be shown using Zorn's Lemma $\Sigma$ has a maximal element, be showing every chain has an upper-bound. (for this $0 \notin S$ condition is not required.)

$\underline{Claim}$: $S$ maximal in $\Sigma$ iff $A \setminus S$ is a minimal prime ideal. I proceeded as follows:

$"\impliedby"$

As $A\setminus S$ is a prime ideal, $S \in \Sigma$.

Now how do I show that $S$ is maximal?

$"\implies"$

$S$: a maximal element in $\Sigma$.

$0 \in A\setminus S$, and if $xy \in A\setminus S$ then either $(x \in A
   \setminus S)$ or $(y \in A \setminus S)$ as $S$ is multiplicatively closed.

Also, $A \setminus S$ does not contain any prime ideal properly.

If it does, let $\mathfrak{p}$ be one such prime ideal. Then $A \setminus
   \mathfrak{p} \in \Sigma$
containing S properly, a contradiction.

It remains to show is $A \setminus S$ is a subgroup of the additive group $A.$ How can I show this?

Edit

Using @SteveD's comment I have completed the above proof, which I have posted as an answer. But the proof is a bit set-theoretic. I would like to see a proof using tools from Commutative Algebra.

Is there any nice application of this result?

答:
Using Steve D's suggestion I am trying to complete the above proof.

$"\implies"$

Let $\mathfrak{P}:=\{\text{set of all ideals contained in} \ A\setminus S\}$. $\mathfrak{P} \neq \phi$ as $\{0\} \in \mathfrak{P}$.

For any chain $\{\mathfrak{p}_i:i \in I\} \subset \mathfrak{P}, \cup \mathfrak{p_i}$ is an upperbound of that chain in $\mathfrak{P}$. i.e it contains a maximal element say $\mathfrak{p}$.

If $\left(\mathfrak{p} \subsetneq A\setminus S\right)$ then $\left(S \subsetneq A\setminus \mathfrak{p}(\in \Sigma) \right)$, contradicting the maximality of $S \implies \mathfrak{p=}A\setminus S$ $\left(as \ \mathfrak{p} \subset A\setminus S.\right)$ $\implies A\setminus S$ is an ideal.

From 1. and 2. in the question we can conclude:

$A\setminus S$ is a minimal prime ideal.

$"\impliedby"$

Let $\mathfrak{M:=}\{\text{Set of all multiplicative subset of A containing S}\}$. $\mathfrak{m} \subset \Sigma$.

From Zorn's Lemma, $\mathfrak{M}$ contains a maximal element say $\mathfrak{m}.$ Now $\mathfrak{m}$ also is maximal in $\Sigma$,(if not then it would contradict the maximality of $\mathfrak{m}$ in $\mathfrak{M}$.)

If $S \subsetneq \mathfrak{m}$ then $A\setminus \mathfrak{M} \subsetneq A\setminus S$. As $\mathfrak{m}$ is maximal in $\Sigma$ we have $A\setminus \mathfrak{m}$ is a (minimal) prime ideal, contradicting the minimality of $A\setminus S$. Hence $S=\mathfrak{m}$

i.e $S$ is maximal in $\Sigma.$

Title: Atiyah-Macdonald: Exercise 3.6


问:
Focal points in abstract sense
Guillemin and Pollack define a focal point of a hypersurface $X \subset \mathbb{R}^n$ to be a critical value of the normal bundle map $h: N(X) \to \mathbb{R}^n$ defined by $h(x,v) =x+v$.

What does this really mean intuitively?

I'm asked to find the focal points of the parabola $y=x^2$ in $\mathbb{R}^2$. At any point $(x,x^2) \in X$ (the parabola), the normal to the tangent space should be the line $y= -x/2$. But I'm not sure where to go from here.

答:
The idea given by @Ted Shifrin is correct. I just give some illustration here.

The key point is to comprehend the derivative of the normal bundle map. Although it has a linear form, $dh_{x,v}$ is not equal to $h$ since the parametrization of $N(X)$ is not simply the identity( even though $dimN(X)=n$. See p.71 of Pollack).
In order to clarify the structure of $dh_{x,v}$, let $x=(t,t^2)$. Then for $t\neq 1$, $v$ can be expressed as $s\cdot(1,-\frac{1}{2t})$, which implies
$$
h(t,s)=(t,t^2)+(s,-\frac{s}{2t})=(t+s,t^2-\frac{s}{2t})
$$
In this form, the domain $h$ is set to be open in $R^2$, which leads to the derivative
$$
dh_{t,s}=(1,1;2t+\frac{s}{2t^2},-\frac{1}{2t})
$$
Hence the critical point occurs iff $2t+\frac{s}{2t^2}+\frac{1}{2t}=0$. Incorporate it into $h(t,s)$, we know the critical value can be expressed as
$$
h(t,s(t))=(t-(t+4t^3),t^2+\frac{t+4t^3}{2t})=(-4t^3,\frac{1}{2}+3t^2)
$$
that is the curve we requires, with some appropriate modification on $t=0$.

Title: Focal points in abstract sense


问:
Probability about cards
Eight hands of five cards each are dealt from a shuffled deck of cards without replacement so that there are twelve undealt cards. Find the probability that at least one of these hands has no diamonds.

The solution uses the inclusion-exclusion theorem. Is there any other way to solve this problem without using that theorem?

Thanks.

答:
Comment. A simulation of a million iterations
of this game approximates the answer as
$0.9248\pm 0.0005.$

In the R code below, in deck Diamonds are 1s, cards of
other suites are 0; dlt is a 40-vector of cards dealt; h1 is $0$ if there are no Diamonds in Hand 1, and so on for the other seven hands. Thus. nd
has a million elements: $0$ if one or more hands
has no Diamond. Finally, mean(nd) is the proportion of iterations in which one or more hands had no Diamond. With a million iterations we can
expect simulation error of $\pm 0.001.$

```<code>deck = c(rep(1,13),rep(0,39))
set.seed(2020)
m = 10^6;  nd = numeric(m)
for(i in 1:m) {
 dlt = sample(deck, 40)
 h1 = sum(dlt[1:5]);    h2 = sum(dlt[6:10])
 h3 = sum(dlt[11:15]);  h4 = sum(dlt[16:20])
 h5 = sum(dlt[21:25]);  h6 = sum(dlt[26:30])
 h7 = sum(dlt[31:35]);  h8 = sum(dlt[36:40])
 nd[i] = h1*h2*h3*h4*h5*h6*h7*h8 }
mean(nd==0)
[1] 0.924804
2*sd(nd==0)/1000
0.0005274149    # aprx 95% margin of sim error
</code>```

Note: Consider the following very rough combinatorial approximation. The probability that
one 5-card hand avoids Diamonds is
${{13\choose 0}{39\choose 5}}/{{52\choose 5}} \approx 0.22.$ Then, if the eight hands were independent, the probability of at least one hand
without Diamonds would be about $1 - .78^8 \approx 0.86.$

Title: Probability about cards

Body:
Comment. A simulation of a million iterations
of this game approximates the answer as
$0.9248\pm 0.0005.$

In the R code below, in deck Diamonds are 1s, cards of
other suites are 0; dlt is a 40-vector of cards dealt; h1 is $0$ if there are no Diamonds in Hand 1, and so on for the other seven hands. Thus. nd
has a million elements: $0$ if one or more hands
has no Diamond. Finally, mean(nd) is the proportion of iterations in which one or more hands had no Diamond. With a million iterations we can
expect simulation error of $\pm 0.001.$

deck = c(rep(1,13),rep(0,39))
set.seed(2020)
m = 10^6;  nd = numeric(m)
for(i in 1:m) {
 dlt = sample(deck, 40)
 h1 = sum(dlt[1:5]);    h2 = sum(dlt[6:10])
 h3 = sum(dlt[11:15]);  h4 = sum(dlt[16:20])
 h5 = sum(dlt[21:25]);  h6 = sum(dlt[26:30])
 h7 = sum(dlt[31:35]);  h8 = sum(dlt[36:40])
 nd[i] = h1*h2*h3*h4*h5*h6*h7*h8 }
mean(nd==0)
[1] 0.924804
2*sd(nd==0)/1000
0.0005274149    # aprx 95% margin of sim error

Note: Consider the following very rough combinatorial approximation. The probability that
one 5-card hand avoids Diamonds is
${{13\choose 0}{39\choose 5}}/{{52\choose 5}} \approx 0.22.$ Then, if the eight hands were independent, the probability of at least one hand
without Diamonds would be about $1 - .78^8 \approx 0.86.$


问:
A sequence that converges normally but not uniformly
Consider the sequence of complex-valued functions $f_n(z)=z^n$, for $n=1,2,3,\dots$. How to show that
this sequence does not converge uniformly to $f(z)=0$ on all of the unit disc. However it converges
uniformly in every compact subset of the unit disc to $f(z)=0$?

答:
To see that $f_n$ doesn't converge to $0$ uniformly on the entire disc, use the fact from real analysis that the corresponding real functions $f_n(x) = x^n$ don't converge uniformly to $0$ on $x\in(0,1)$. (This is a famous example from real analysis. And it's not too painful to prove: pick $\epsilon>0$ small enough so that the $x^n$ "poke out" of the $\epsilon$-tube surrounding $0$, for $x$ close to $1$.) This implies that uniform convergence doesn't hold in the complex case, because uniform convergence of the real functions is necessary (but not sufficient) to ensure uniform convergence of the complex functions.

On compact subsets of the unit disc, I'd try comparing $z^n$ with the partial sums of the geometric series: $$|z^n| < |1|+|z|+|z^2|+ \dots + |z^n|. $$ Since the geometric series converges absolutely and uniformly on compact subsets of the unit disc, the RHS converges absolutely as $n \to \infty$; so the left side converges as well. To see that it is uniform, use the Weierstrauss M-test. (The terms $M_n$ can come from the above bound, using the formula for a finite geometric sum.)


问:
Show that $\left(\sum_{n=1}^{\infty} a_n^q \right)^{1/q}$ is a decreasing function of $q$ for $q>0$

Let $\{a_n\} $ be a sequence of non-negative real number such that $\sum_{n=1}^{\infty} a_n^q$ is convergent . Then show that $\left(\sum_{n=1}^{\infty} a_n^q \right)^{1/q}$ is a decreasing function of $q$ for $q>0$

Since given that $\sum_{n = 1}^{\infty} a_n^q$ is convergent  therefore $\sum_{n=1}^{\infty} a_n^q=c, c>0$ then   $\left(\sum_{n = 1}^{\infty} a_n^q \right)^{1/q}=c^{1/q}$ .
Now if $c>1 $ then $ c^{1/q} $ is decreasing function of $q$ but if $0<c<1 $ then $c^{1/q}$ is increasing function of $ q $.

答:
As D F said in the comment,
$$\sum_{j = 1}^\infty a_j^q = c(q),$$
so your argument doesn't work. For $p < q$, we have
$$\frac{|a_j|}{\|a\|_{q}} \le 1, \quad \text{for }~\|a\|_{q} = \left(\sum_{j = 1}^\infty a_j^q\right)^{1/q}.$$
Therefore,
$$\left|\frac{a_j}{\|a\|_q}\right|^q \le \left|\frac{a_j}{\|a\|_q}\right|^p,$$
and you conclude the statement by summation.


问:
If $A$ is an idempotent matrix i.e, $A^2=A$, and given that $|A| \ne 0$, prove that $A=I$, where $I$ is the identity matrix.

If $A$ is an idempotent matrix i.e, $A^2=A$, and given that $|A| \ne 0$, prove that $A=I$, where $I$ is the identity matrix.

Now, clearly if $|A| \ne 0$, $A^{-1}$ exists and multiplying with $A^{-1}$ in the equation $A^2=A$ yields the result. But I wanted to proceed differently.

If we take the determinant on both sides of $A^2=A$, we get either $|A|=0$ or $|A|=1$ and if we rewrite the equation as $A^2-A=A(A-I)=O$ and then taking the determinant on both sides yields $|A|=0$ and $|A-I|=0$ but since $|A| \ne 0$, we get from above two conditions i.e $|A|=1$ and $|A-I|=0$.

So my question is, do these two conditions together imply that $A=I$? Or are they still insufficient?

答: They are insufficient. Just take the diagonal matrix with diagonal elements $1,2,1/2$. Its determinant is one and $1$ is an eigenvalue, so $|A-I|=0$. But $A\neq I$.


答:
The constraint on $c$ mentioned in the other answers can be sharpened by noting that $a =4.5c$ is only an integer when $c$ is even, so you only have to consider $c \in \{2,4,6,8\}$. Unfortunately $b= \frac{96}c$ does not permit further constraint, so you have to test those four values.

If you're permitted calculus, you can observe that $f(c) = 5.5c + \frac{96}c$ has a single turning point and reaches a global minimum for real $c$ at $c = \sqrt{\frac{192}{11}} \approx 4$, so this is very likely to be the correct solution based on the "nice behaviour" of such functions. But you really do have to test at least $c=4$ and $c=6$ (either side of the turning point) to be certain.


问:
Inertia degree of primes in p-adic extensions
I'm reading through some number theory and ran across a theorem where the proofs referenced were incomprehensible to me, and I was hoping there might be a simpler proof than slogging through another $\approx$100 pages of background.

The statement is that if you have a finite extension $K$ of the $p$-adic rationals $\mathbb{Q}_p$, and $\pi$ is a uniformizer of $\mathcal{O}_K$, then

$$[\mathcal{O}_K/\pi\mathcal{O}_K : \mathbb{Z}_p/p\mathbb{Z}_p] = v_p(\vert N(\pi)\vert)$$

In the non-$p$-adic case, where $K'$ is some extension of $\mathbb{Q}$ and where $\pi \vert p$, showing that $[\mathcal{O}_{K'}/\pi\mathcal{O}_{K'}:\mathbb{Z}/p\mathbb{Z}] = v_p(\vert N(\pi)\vert)$ is more straightforward to me: Since $\mathcal{O}_{K'}$ is a $\mathbb{Z}$-module, I can imagine it as $\mathbb{Z}^n$, and then the quotient by $\pi$ is like a lattice, with $\pi$ as matrix giving a basis of that lattice, and the volume of the fundamental parallelepiped is the determinant of that matrix, which is $N(\pi)$. The number of integer points in the fundamental parallelepiped will be equal to its volume, showing that $\vert \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}\vert = \vert N(\pi)\vert$. Since $\vert\mathbb{Z}/p\mathbb{Z}\vert =p$, that gives the result.

But this doesn't seem to extend to the $p$-adic case. $\mathbb{Z}_p^n$ doesn't look like a lattice to me, and even if it were, I'm not sure how to relate the volume to the number of points in it.

Maybe I could argue that $\mathbb{Z}$ is a subring of $\mathbb{Z}_p$, and that $\mathbb{Z}_p/p\mathbb{Z}_p$ can take representatives of each coset from $\mathbb{Z}$, so I can treat $\mathcal{O}_K/\pi\mathcal{O}_K$ as a $\mathbb{Z}$-module. But then I can't seem to treat $\pi$ as a linear operator without causing problems with the determinant.

Or maybe, since $\mathbb{Z}_p/p\mathbb{Z}_p \cong \mathbb{Z}/p\mathbb{Z}$, I could show that $\mathcal{O}_K/\pi\mathcal{O}_K\cong \mathcal{O}_{K'}/\pi\mathcal{O}_{K'}$. Certainly I can say that every linear operator over $\mathbb{Z}_p/p\mathbb{Z}_p$ is a linear operator over $\mathbb{Z}/p\mathbb{Z}$, but I'm not sure that every element of $\mathcal{O}_K$ will form a linear operator that is equal to one from $\mathcal{O}_{K'}$. In fact this is almost certainly false, so the quotient by $\pi$ must be crucial, but I don't know how to incorporate this fact.

Is there a relatively simple proof of this, or any way to complete my attempts?

答:
The problem is that you haven't really told us what you think is too much theory. I'll give a proof below with some annotations, hopefully this is enough.

Consider the function $\phi:K^\times \to \mathbb{R}$ given by

$$\phi(x):=\left|N(x)\right|_p^{\frac{1}{n}}$$

where $N:L^\times\to \mathbb{Q}_p^\times$ is the usual field norm, $n=[K:\mathbb{Q}_p]$, and $|\cdot|_p$ is the usual $p$-adic absolute value.

Since $\pi$ is a uniformizer of $K$ we know that $p=\pi^e u$ for some unit $u$ of $K$ $\color{red}{(1)}$. So then, we have that

$$\phi(\pi)^e\overset{{\color{red}{(2)}}}{=}\phi(\pi^e)\overset{{\color{red}{(3)}}}{=}\phi(\pi^e u)=\phi(p)=|p^n|_p^{\frac{1}{n}}=p^{-1}$$

From this, we see that

$$|N(\pi)|_p^{\frac{1}{n}}=p^{\frac{-1}{e}}$$

which implies that

$$|N(\pi)|=p^{\frac{-n}{e}}\overset{\color{red}{(4)}}{=}p^{-f}$$

But,

$$|N(\pi)|=p^{-v_p(N(\pi))}$$

so we deduce that

$$v_p(N(\pi))=f$$

where $f=[\mathcal{O}_K/\pi\mathcal{O}_K:\mathbb{F}_p]$.

$\color{red}{(1)}$ This is, essentially, a definition. Namely, we know that $p=\pi^e u$ for some $e$, and so I'm just naming this power--it's e.

$\color{red}{(2)}$ I am using the fact here that $\phi$ is multiplicative. This is clear since $N$ is multiplicative (since it's just a determinant!) and $|\cdot|_p$ is multiplicative.

$\color{red}{(3)}$ I've snuck in a sneaky $u$ here. The reason this is ok is that $\phi(u)=1$. Indeed, note that $N$ sends elements of $\mathcal{O}_K$ into elements of $\mathbb{Z}_p$. The reason is that $N(x)$ can be thought of as the constant term of the minimal polynomial for $x$, but if $x$ is $\mathcal{O}_K$ these minimal polynomial coefficients are in $\mathbb{Z}_p$. So, since $u$ is a unit it has an inverse $v$ in $\mathcal{O}_K$ and since $1=N(uv)=N(u)N(v)$ you know that $N(u)$ is not just in $\mathbb{Z}_p$ but $\mathbb{Z}_p^\times$ and thus $|N(u)|=1$ so $\phi(u)=1$.

$\color{red}{(4)}$ This is the only truly hard part I'm using. Namely, I am using the fac that $n=ef$. The proof of this is pretty simple though. Namely, note that $\mathcal{O}_K$ is a torsion free $\mathbb{Z}_p$ module and since $\mathbb{Z}_p$ is a PID this implies (by the structure theory of modules over PIDs) that $\mathcal{O}_K=\mathbb{Z}_p^r$ for some $r$. But, note that

$$\mathbb{Q}_p^r\cong \mathbb{Z}_p^r\otimes_{\mathbb{Z}_p}\mathbb{Q}_p\cong \mathcal{O}_K\otimes_{\mathbb{Z}_p}\mathbb{Q}_p=K$$

as $\mathbb{Q}_p$-modules. But, since $K=\mathbb{Q}_p^n$ where $n=[K:\mathbb{Q}_p]$ we know that $n=r$. So, we see that $\mathcal{O}_K$ is a free $\mathbb{Z}_p$-module of rank $n=[K:\mathbb{Q}_p]$. Thus, we see that

$$\mathcal{O}_K/p\mathcal{O}_K\cong \mathcal{O}_K\otimes_{\mathbb{Z}_p}\mathbb{Z}_p/p\mathbb{Z}_p$$

is a free module over $\mathbb{Z}_p/p\mathbb{Z}_p\cong \mathbb{F}_p$ of rank $n$. But, we can count the dimension of the $\mathbb{F}_p$-module $\mathcal{O}_K/p\mathcal{O}_K$ in another way. Namely, writing $p=\pi^e u$ for $u$ a unit it's easy to see using the PIDness that $\mathcal{O}_K/p\mathcal{O}_K$ has a filtration

$$\mathcal{O}_K/p\mathcal{O}_K\supseteq \pi\mathcal{O}_K/p\mathcal{O}_K\supseteq\cdots p^{e-1}\mathcal{O}_K/p\mathcal{O}_K\supseteq p^e\mathcal{O}_K/p\mathcal{O}_K=0$$

and that each $i$ we have an isomorphism of $\mathbb{F}_p$-modules

$$(\pi^i\mathcal{O}_K/p\mathcal{O}_K)/(\pi^{i+1}\mathcal{O}_K/p\mathcal{O}_K)\cong \mathcal{O}_K/\pi\mathcal{O}_K$$

So, since this filtration has length $e$ we see that

$$n=\dim_{\mathbb{F}_p} \mathcal{O}_K/p\mathcal{O}_K=e\dim_{\mathbb{F}_p}\mathcal{O}_K/\pi\mathcal{O}_K=ef$$

as desired.


问:
Relation between incenter, circumcenter and orthocenter of a triangle
Let $ABC$ be a triangle with $\widehat{ABC}=60°$ such that $O$, $I$, $H$ are its circumcenter, incenter and orthocenter respectively. Show that $OI=IH$.

By using the laws of sines and cosines, it's rather simple to obtain that $BH=BO$, but from there I'm not sure how to proceed. It's sufficient to prove that $\widehat{OBI}=\widehat{HBI}$ in which case $BI$ would be the bisector of angle $\widehat{OBH}$ and since $OBH$ is isoceles it's also a mediator which gives us $OI=IH$. The problem is proving this angle equality...

答:
Triangle ABC

Draw perpendiculars from vertex $B$ & $C$ on side $CA$ & $AB$ meeting them at points $E$ & $F$ respectively. They intersect at $H$. Draw $BO$,$OH$,$OC$ and $BI$. Extend $BI$ to meet $OH$ at $L$. Also, drop a perpendicular from the circumcentre on side $BC$ meeting it at point $M$.

Observe that, in $\triangle BOM$ & $\triangle BHF$,

$\left(i\right)$ $\angle BMO=\angle BFH=90^\text{o}$

$\left(ii\right)$ $\angle OBM=90-\frac{\angle BOC}{2}=90-\angle BAC=\angle ABE=\angle HBF$

$\left(iii\right)$ Since $\triangle BFC$ is a  $30-60-90$ triangle, $BF=\frac{1}{2}BC=BM$

Hence, $\triangle BHF\cong \triangle BOM$ by $A-S-A$ criterion of congruence.

Thus, $BO=BH$.

Since $\angle HBF=\angle OBM$ and $BI$ bisects $\angle ABC$, $BI$ must also bisect $\angle OBH$.

In $\triangle BOH$, $BI$ bisects $\angle  OBH$ and $BO=BH$; Hence, $BI$ is the perpendicular bisector of $OH$. Since $I$ lies on this perpendicular bisector, $\boxed {OI=IH}$.


问:
Expectation of two cards
I have a question and I will be happy for help.
there is a normal deck of cards (ace = 1...king=13, 52 cards, 4 cards for each value)
someone takes out without return two cards.
what is the expectation of the sum of the cards?
I tried to seperate it to odd sums and even sums but it is very difficult and I think there is an easier way.
does someone have an Idea how to solve it?

答:
Hint

See Linearity of Expectation.

Title: Expectation of two cards

Body:
Hint

See Linearity of Expectation.


问:
Understanding $(r,\theta)$ in $\mathbb{R}^n$ - how does $\theta$ represent $n-1$ coordinates?
I'm reading an article on convex geometry, and there is a part that attempts to calculate the volume of the Euclidean ball in $\mathbb{R}^n$. The first part of the argument talks about spherical polar coordinates in $\mathbb{R}^n$. I'm unable to understand a few things, which I have written down after quoting an excerpt:

We shall need to know the volume of the ball: call it $v_n$ . We can calculate the
surface “area” of $B^n_2$ very easily in terms of $v_n$: the argument goes back to the
ancients. We think of the ball as being built of thin cones of height $1$: see Figure
4, left. Since the volume of each of these cones is $1/n$ times its base area, the
surface of the ball has area $nv_n$. The sphere of radius $1$, which is the surface of
the ball, we shall denote $S_{n−1}$. To calculate $v_n$, we use integration in spherical polar coordinates. To specify a point $x$ we use two coordinates: $r$, its distance from $0$, and $θ$, a point on the sphere, which specifies the direction of $x$. The point $θ$ plays the role of $n − 1$ real coordinates. Clearly, in this representation, $x = rθ$: see Figure 4, right.

Questions:

How does $\theta$ represent $n-1$ real coordinates?

Why do we have $x = r\theta$ here? It's not as obvious as it is in $\mathbb{R}^3$.

答:
Well.. a point on the $n$-dimensional unit sphere is a vector with $n$ real coordinates subjected to the constraint $\|\Theta\|=1$, which leaves only $n-1$ free parameters. For example, in 2D you have $x=r(\cos\theta,\sin\theta)$ (polar coordinates) so $\Theta=(\cos\theta,\sin\theta)$. In 3D you have $x=r(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi)$ (here $\theta$ and $\phi$ are spherical coordinates), etc. There are general "polar" coordinates in any dimension. You can check

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwi3oO3TievtAhUbCTQIHYtMBU0QFjACegQIARAC&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FPolar_coordinate_system&usg=AOvVaw31Elzt4lc3jtRDOkQuZj7_