问:
Consequence of Hodge decomposition: What is the norm here?
Hodge Decomposition Theorem says smooth  $k$-forms on on compact oriented Riemannian (smooth) $m$−manifold $(M,g)$ (I think M need not be connected, but assume connected if need be or you want) decompose into exact, co-exact and harmonic.

About this pdf powerpoint by Ryan Vaughn,

In slides 35-36,40 does $||\alpha||^2$ necessarily refer to the inner product on slide 25 in that $||\alpha||^2 := (\alpha,\alpha):=\int_M \alpha \wedge \star \alpha$, where $\star$ is Hodge star operator?

Could it possibly be instead something like $||\alpha||^2 : = g(\alpha,\alpha)$ and then $||\alpha||^2 = 0$ if and only if $(\alpha,\alpha) = 0$?

2.1. Here, I think: we would use that $\alpha \wedge \star \alpha = g(\alpha,\alpha) vol_{(M,g)}$ and then say the integral, of a non-negative integrand, is zero if and only if the integrand is zero.

答:
First of all, I don't think there is any other reasonable interpretation using the slide besides that
$$ \| \alpha \|^2 = (\alpha, \alpha).$$
This is used in no.34$\to$no.35 and no.39$\to$no.40. And

$$(\alpha, \alpha) = \int_M \alpha \wedge \star \alpha$$

are used when applying the formal adjoint $\delta$ in several places. So 1 is the interpretation used in the slide.

To show that $\|\alpha\| = 0\Rightarrow \alpha = 0$, we use 2.1, that is

$$ \alpha \wedge \star \alpha = g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$

so if $\|\alpha\|^2 = 0$, by 1 we have

$$0 = \int_M g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$

and since $g(\alpha, \alpha)$ is non-negative, $g(\alpha, \alpha) = 0$ everywhere and thus $\alpha(x) =0$ for all $x\in M$. So $\alpha = 0$.


问:
Neukirch's motivation for $p$-adic numbers
I've started reading Neukirch's Algebraic Number Theory book and at the beginning of Chapter II he starts his motivation for the $p$-adic numbers as follows:

"The idea originated from the observation made in the last chapter that the numbers $f\in\mathbb{Z}$ may be viewed in analogy with the polynomials $f(z)\in\mathbb{C}[z]$ as functions on the space $X$ of prime numbers in $\mathbb{Z}$, associating to them their "value" at the point $p\in X$, i.e., the element

$\begin{equation}
f(p):=f \mod p
\end{equation}$

in the residue class field $\kappa(p)=\mathbb{Z}/p\mathbb{Z}$.
This point of view suggests the further question: whether not only the "value" of the integer $f\in\mathbb{Z}$ at $p$, but also the higher derivatives of $f$ can be reasonably defined."

My question is: If we define the higher derivatives of $f\in\mathbb{Z}$ at $p$ as the coefficientes of its $p$-adic expansion (as he does later on), can we give these derivatives an interpretation somehow analogous to the analytical one that goes with the $f(z)\in\mathbb{C}[z]$ polynomials?

I'm asking this because if there is no interpretation other than "coefficientes of the expansion", it seems like a very artificial analogy to me. I mean, if not, why would he even mention derivatives instead of just saying something like: "we expand functions this way, we expand integers this way and it's kind of similar."

Any explanations and/or motivations are welcome.
Thanks in advance.

答:
"My question is: If we define the higher derivatives of f∈Z at p as the coefficientes of its p-adic expansion (as he does later on), can we give these derivatives an interpretation somehow analogous to the analytical one that goes with the f(z)∈C[z] polynomials?"

Answer: At the following link

https://mathoverflow.net/questions/55244/why-must-nilpotent-elements-be-allowed-in-modern-algebraic-geometry/378811#378811

you will find an elementary construction of a "Taylor morphism"

T1. $T^l: E \rightarrow J^l(E)$

for any commutative ring $A$ and any left $A$-module $E$. If $A:=k[x_1,..,x_n]$
with $k$ a field of characteristic zero, you get a map

T2. $T^l: A \rightarrow J^l(A):=A\otimes_k A/I^{l+1}$

defined by $T^l(f(x_1,..,x_n)):=f(y_1,..,y_n)$. Let $dx_i:=y_i-x_i$. You may prove that there is an isomorphism

T3. $J^l(A)\cong A\{dx_1^{l_1}\cdots dx_n^{l_n}: \sum_i l_i\leq l\}$.

Hence the left $A$-module $J^l(A)$ is a free $A$-module on the set $\{dx_1^{l_1}\cdots dx_n^{l_n}: \sum_i l_i\leq l\}$. In this situation it follows the
element $T^l(f(x_1,..,x_n))$ is the Taylor expansion of the polynomial $f$ in the variables $x_i$.

Comment: "I think his point is that if one takes X=A1C as a variety with sheaf OX, then for any point p∈X one has that OX,pˆ=C[[z−p]] and the map C[z]=OX,x→OX,pˆ is just taking f∈C[z] and expanding it as a Taylor series at p. Similarly, one may view the p-adic case as taking the scheme X=Z, some global section f∈Z=OX(X) and mapping it to OX,pˆ=Zp."

Answer: The fiber of the $A$-module $J^l(A)$ at a $k$-rational point $\mathfrak{m}$ is the local ring $A/\mathfrak{m}^{l+1}$, hence this construction may be related to the construction mentioned in the comments above.

In Neukirch, Prop 2.6 page 114 the following isomorphism is constructed:

There is for every non-zero prime $p$ a canonical isomorphism

$T_p: \mathbb{Z}_{p}\cong \mathbb{Z}[[x]]/(x-p)$
and a canonical injective map $i_p: \mathbb{Z} \rightarrow \mathbb{Z}_p$.

There is a canonical map $T: \mathbb{Z}\rightarrow \mathbb{Z}[[x]]$ and we recover the map $i_p$ by passing to the quotient $P(p): \mathbb{Z}[[x]] \rightarrow \mathbb{Z}[[x]]/(x-p)$.

$i_p:=P(p)\circ T: \mathbb{Z}\rightarrow \mathbb{Z}[[x]]/(x-p)\cong \mathbb{Z}_p$.

Hence all the injections $i_p$ can be constructed from one "global" map $T$.
We may pass to "spectra" $C:=Spec(\mathbb{Z}[[x]])$ and $S:=Spec(\mathbb{Z})$
and the canonical map

$\pi: C \rightarrow S$.

The ideal $I(p):=(x-p) \subseteq \mathbb{Z}[[x]]$ corresponds to a closed subscheme $C_p:=Z(x-p)\subseteq C$ and we get a canonical morphism $\pi_p: C_p\rightarrow S$. There is a corresponding map of structure sheaves

$\pi_p^{\#}: \mathcal{O}_S \rightarrow (\pi_p)_*\mathcal{O}_{C_p}$

whose global sections is the canonical map $i_p:\mathbb{Z}\rightarrow \mathbb{Z}_p$. Hence all maps $i_p$ may be given a "geometric" construction using the map $\pi$: You restrict the map $\pi$ to the closed subscheme $C_p$ and pass to global sections.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Alignment of multiple bodies on circular paths
Below is a problem that I came across in which I don't know the answer. It's as follows:

$\\$

Problem:

Suppose there are $n-2$ planets currently at rest (stationary) - all positioned on the positive side of the x-axis (line) of the euclidean plane. They will soon orbit on a circular motion around the sun. Each planet $k$ will travel at the same speed and have an orbit of length $k$ where $k=3,4,...,n$ and $n>4$. All orbits will lie on the euclidean plane and will be centred at the origin (where the sun is positioned). Suppose now that all planets leave their initial position of rest at the same time. From now on, two or more planets are considered to be aligned if they lie on the positive side of the x-axis (line) at the same time in their respective orbits. Now consider all the planets that won't be aligned with planet $n$ during its first orbit. How many of them will be aligned with at least one planet that will be aligned with planet $n$ (during the first orbit of planet $n$)?

$\\$

Additional Information:

In the previous version of this problem, I realised I had made the mistake of letting planet $n$ orbit the sun more than once which made it quite obvious that every other planet would be aligned with planet $n$ after a certain number of orbits - hence you have the second comment below and the answer (from 1st January) that were driven by this mistake. So I've now corrected the mistake by letting planet $n$ orbit the sun once only - which was my original intent.

答:
Planet $j$ has radius of orbit $r_j = \frac{j}{2\pi}$. Suppose, for simplicity, the linear speed of all planets is $1$. Then, the angular speed of planet $j$ is $$\omega_j = \frac{1}{r_j} = \frac{2\pi}{j} $$ and time period $$T_j = \frac{2\pi}{\omega_j} = j$$.

Now, planet $n$ will be precisely on the positive $x$-axis at time $t=0,n,2n\dots$ i.e. $t=k_1 n$.

Planet $j$ will be on the positive $x$-axis at $t=k_2 j$.

So, planets $n$ and $j$ will be aligned when $$k_1 n = k_2 j \implies n=\frac {k_1}{k_2} j$$ But this can always be achieved by setting $k_2 =j, k_1 = n$.

In other words, there is no planet that will never be aligned with planet $n$.


问:
What kinds of "connected" components is the DFS or BFS algorithm called more than once for?
Is it correct that the result of BFS or DFS on a graph is  a forest, not necessarily a tree?

https://en.wikipedia.org/wiki/Graph_traversal#Graph_traversal_algorithms says

If each vertex in a graph is to be traversed by a tree-based algorithm (such as DFS or BFS), then the algorithm must be called at least once for each connected component of the graph. This is easily accomplished by iterating through all the vertices of the graph, performing the algorithm on each vertex that is still unvisited when examined.

For a undirected graph, is the DFS or BFS algorithm called exactly once for each connected component?

For a directed graph, is the DFS or BFS algorithm called at least once for each connected component?

What type of connectivity is used here for a directed graph? (weakly, strongly, or ...)

What kinds of "connected" components is the  DFS or BFS algorithm called more than once for?

Thanks.

答:
How many times you call DFS (or BFS) is a matter between you and DFS. What we can say is that in an undirected graph, either one of the searches done from a vertex $v$ explores the entire connected component of $v$, and nothing else. So you'd want to search from a vertex in each connected component, if you wanted to explore the entire graph.

(In particular, if you take the union of all the search trees, you get a spanning forest of the graph, with a spanning tree of each connected component.)

In a directed graph, doing a search from a vertex $v$ explores all the vertices reachable from $v$. This is not necessarily the same as a weakly connected or a strongly connected component. In general, the set of vertices reachable from $v$ is a union of multiple strongly connected components.

Tarjan's algorithm is a modification of DFS that finds the actual strongly connected components of a directed graph. Essentially, you pick a vertex $v$ and do a DFS from $v$, but do some extra book-keeping that lets you notice when you move to a different strongly connected component.

If $v$ can't reach all the vertices, then you might have to start over from a different vertex to get the rest. Alternatively and equivalently, you can add an artificial source vertex that has an edge to every other vertex, and do a single Tarjan-style DFS from there. (The artificial vertex will be its own strongly connected component, which we can disregard.)


问:
Inverse functions on $A(K)$
Let $K \subset \mathbb{C}$ be compact and let $A(K)$ be the space of functions that are continuous on $K$ and holomorphic on $K^\circ$ (the case $K = \emptyset$ is allowed). If $f$ is injective and if we put $L := f(K)$, is it true that the inverse function $f^{-1}: L \rightarrow K$ belongs to $A(L)$?

From the compactness of $K$ it follows that $f^{-1}$ is continuous. To show that $f^{-1}$ is holomorphic at $z \in L^\circ$ I need to show that $f$ is holomorphic in a neighborhood of $f^{-1}(z)$ or equivalently that $f^{-1}(z)$ is an interior point of $K$ but I don't see why this is true.

Does anyone have an idea?

答:
Let $K_o, L_o$ the interiors of $K, L$ respectively; then we show that $K_o$ is non-empty if and only if $L_o$ is so, and in this case, $f$ is an analytic isomorphism between them, so $g=f^{-1}$ is analytic on $L_o$.

By hypothesis since $K$ compact and $f:K \to L$ continuous bijective, it follows that $g:L \to K$ is continuous and bijective.

If $z \in K_o$, there is a small disc $D_z \subset K_o$ centered at $z$ and by the open mapping for holomorphic functions, it follows that $f(D_z)$ is open in the plane and it is in $L$ by our choices, so it is in $L_o$.

Conversely, if $w \in L_o$ there is a small closed disc $\overline D_w \subset L_o$ centered at $w$ and if $C_w$ is its circle boundary, $g(C_w)$ is a Jordan curve in the plane $J$ with $U$, its interior in the plane, a bounded (simply connected) open set, hence by the Jordan Curve Theorem, $g(D_w)=U$ homeomorphically. But $U$ is in $K$ hence it is in $K_o$.

This shows that $f$ restricted to $K_o$ is a homeomorphism onto $L_o$ with inverse $g$ and now by the inverse function theorem, $g$ is analytic if $f$ is so, so we are done!

Title: Inverse functions on $A(K)$


问:
Reach N from $0$ in the least number of moves where the n'th move comprises of n steps and each step is a $\pm 1$ movement
Reach the number $\text{N}$ from $0$ in the least number of moves where the $n\text{th}$ move comprises of $n$ many steps and each step is a $\pm 1$ movement. The same problem also exists here without a proof.

Suppose $k$ is the largest number such that $S :=1 + 2 + \cdots + k \leq \text{N}$. If it equals $\text{N}$ we are done and if the distance left after going from $S$ to $N$ is even, we are still done (and our answer will be $k+1$) since we can then repetitively go back from $N$ to $\text{N}-1$ and back to $\text{N}$ in an even number of steps. Now if the distance left is odd, we do the exact same thing, and depending on the parity of $k$, the answer will be $k+2$ or $k+3$. But the point is, when the distance left $(k+1 - (N - S))$ is odd, I do not have a good enough reasoning to conclude that "it" is the answer. I only have that as the upper-bound and $k+1$ the lower bound, nothing more.

Rewrite every move $j$ as $a_j + b_j = j$ (both non-negative) such that the problem can be re-written as :
Find the least $j \in \mathbb{N}$ such that $\sum(a_j) + \sum(b_j) = N$  subject to $0 \leq \max(|a_j|, |b_j|) \leq j$.

May I get a solid proof for this?

答:
The remaining case is that all the following are true:

$S:= 1 + 2 + \cdots + k < N$,

$S+(k+1) > N$ and

$S+(k+1) - N$ is odd

Condition $3$ means $S+(k+1)$ and $N$ are in different parity,

$$S+(k+1) \not\equiv N \pmod 2$$

and so the minimum number of moves is strictly greater than $k+1$.

Then there are two cases depending on the parity of $k+2$:

If $k+2\equiv 0 \pmod 2$, then

$$S+(k+1)+(k+2)\not\equiv N\pmod 2$$

so the minimum number of moves is still strictly greater than $k+2$. But considering the $(k+3)$th move,

$$\begin{align*}
(k+2)+(k+3) &\equiv 1 \pmod 2\\
S+(k+1)+(k+2)+(k+3) &\equiv N
\end{align*}$$

Otherwise if $k+2 \equiv 1\pmod 2$, then

$$S+(k+1)+(k+2) \equiv N \pmod 2$$

This shows that the minimum number of moves that matches the parity of $N$ is $k+2$ or $k+3$ (depending on the parity of $k$), and this rejects $k+1$ as the answer.


问:
Continuous (Smooth) Fractal Zoom
This is my first ever post, I made an account to ask this question. Could have put it on a code forum but thought this challenge would be better suited to a mathematician with a programming foundation.

I'm looking for an implementation of a fractal rendering program that efficiently allows for smooth zooming. I imagine this would require some components of the frame (those already rendered that are still at pixel scale) to simply be scaled linearly in the frame, while new components (those either out of frame or less than pixel size) to be rendered dynamically.

This seems like an interesting problem in dynamic programming, if nothing else. I kind of hope someone can find an existing implementation and share it - if not I'd love some help in constructing one! Ideally python or Julia.

Thanks in advance

答:
https://xaos-project.github.io xaos is a fractal zoomer which caches previously calculated pixels and refines the image when zooming in.  It does this by pushing pixels apart and filling in the gaps.  The exact details can be viewed in its open source code. It does not support deep zooms, and the 53-64 bits of (long) double precision are quickly exhausted when zooming in (you need at least enough bits to distinguish neighboring pixels' C values).

Mandelbrot set deep zooming with arbitrary high precision (emulated in software using libraries like MPFR) for every pixel is prohibitively expensive. In 2013 K.I.Martin popularized perturbation techniques, which allow using low (machine) precision numbers for most of the calculations, and series approximation reduces the calculations required still further: https://web.archive.org/web/20130701075021/http://www.superfractalthing.co.nf/sft_maths.pdf I currently maintain some software that uses these techniques: https://mathr.co.uk/kf/kf.html It does have a progressive display (low resolution preview is gradually refined) but that isn't smooth zooming like xaos.

http://retinamandelbrot.com retinamandelbrot is an iOS app (which has just left beta) that claims to combine smooth zooming with deep zooms, but I haven't tried it myself.  Though, reading the documentation it may pause occasionally to recache some reference orbits or so.  Some nonspecific information on how it works was posted by its author at https://fractalforums.org/meet-and-greet/4/retinamandelbrot-beta-testing/3486/msg20961#msg20961.  There is a beta version for Windows x86_64: http://retinamandelbrot.com/winapp.html (I tried it in Wine on Linux but it did not work properly).

The perturbation techniques can be applied to other escape time fractals like the Burning Ship, but series approximation is tricky with non-analytic folding from $|.|$.

Title: Continuous (Smooth) Fractal Zoom


问:
Relationship of Hom and tensor for vector spaces
Let $U,V$ be two vector spaces over $K$ and $L$ is a field extension of $K$.

Is it true that for any finite dimension $U,V,L$ over $K$ that we have natural isomorphisms

$$\mathrm{Hom}_K(U,V)\otimes L\cong \mathrm{Hom}_K(U\otimes L,V)\cong\mathrm{Hom}_K(U,V\otimes L)?$$

It looks that it is related to the Hom-tensor adjunction, but slightly different.

Is this still true for infinite dimensional?

Bacially, I want to show this on page 26 lemma 1.2.6.

$$\mathrm{Hom}_\mathbb{R}(V,\mathbb{R})\otimes \mathbb{C}\cong \mathrm{Hom}_\mathbb{R}(V\otimes \mathbb{C},\mathbb{R})\cong\mathrm{Hom}_\mathbb{R}(V,\mathbb{R}\otimes \mathbb{C}).$$

The natural isomorphism between the first and third is clear via $V^*\otimes \mathbb{C}\cong\mathrm{Hom}_{\mathbb{R}}(V,\mathbb{C})$. But what about the second one?

答:
Not a full answer, but in a more general context (modules over a commutative ring $R$), we have a natural homomorphism:
$\DeclareMathOperator{\Hom}{Hom}$
$$\Hom_R(U,V)\otimes_R\Hom(M,L)\longrightarrow\Hom_R(U\otimes_R M,V\otimes_R L)$$
which is an isomorphism is any of the pairs $(U,M)$ or $(U,V)$ or$(M,L)$ is made up of projective modules of finite type.

Considering the particular case when $M=R$, which implies $\Hom_R(R,L)\simeq L$, we have that
$$\Hom_R(U,V)\otimes_R L\longrightarrow\Hom_R(U,V\otimes_R L)$$
is an isomorphism is $U$ or $L$ is a projective module of finite type – which, if the base ring is a field $K$, means $U$ or $L$ is a finite dimensional vector space.


问:
Why demonstrations are important in mathematics?
Good evening, I'm studying math and would like to know how important are mathematical proofs in the world and particularly in a school of mathematics

Thanks for your help

答:
If you are not

creating your own proofs

working out examples and doing computations

(with the goal of gaining intuition and understanding, to better create your own proofs)

reading other people's proofs

(with the goal of gaining intuition and understanding, to better create your own proofs)

I would say you are not doing mathematics. Now, there are other things that people often confuse with mathematics, such as

blindly following someone else's directions for doing a computation

blindly giving directions to a calculator or computer to make it do a computation

both of which have their purpose in the world, but they are not mathematics. Since proofs are an intrinsic part of mathematics, I would say they are of paramount importance (or, more accurately, the inextricable combination of intuition, understanding, and proofs is paramount).

To put it simply: math is really interesting, beautiful, and fun to think about! There's just a fundamental intellectual curiosity, a desire to understand. When we can condense that understanding into a statement which we are confident is true, and explain our justification for it, we have stated a theorem and given a proof. To me at least, and I expect to most other mathematicians, that is why proofs are important!

"Mathematicians create by acts of insight and intuition. Logic then sanctions the conquests of intuition." - Morris Kline

What is the importance of mathematical proofs in the world? Well, lots of the mathematics that humans do is, or originally was, motivated by our desire to understand the world. So far, mathematics has been very useful for this. To quote Paul Dirac:

The physicist, in his study of natural phenomena, has two methods of making progress:

the method of experiment and observation, and

the method of mathematical reasoning.

The former is just the collection of selected data; the latter enables one to infer results about experiments that have not been performed. There is no logical reason why the second method should be possible at all, but one has found in practice that it does work and meets with reasonable success.

So, we have observed that, when we use mathematical objects as a model for the world around us, theorems about the mathematical objects seem to correspond to true statements about the corresponding real-world objects. Very useful indeed!

It is also well-known that mathematics has a very close relationship with computers (e.g., via cryptography, information theory, computational complexity theory). Proving things in mathematics can give us information and ideas for how to get our computers to do things most efficiently, or send signals with minimal chance of garbling the message, etc. Also very useful!

It's true that not all of mathematics has direct applications like these right now. But many surprising connections have been made where computer scientists or physical scientists realized that some bit of mathematics, which had previously been considered just an academic curiosity, was just what they needed to model their science. So there's always a chance that any mathematical theorem will end up being useful in the world, and a theorem is nothing without a proof telling you it's true.


问:
How to find the order of the group of rigid motions of platonic solids in $\mathbb{R}^3$?
The following appear as exercises in Dummit and Foote's Algebra (Section $1.2$ - Dihedral Groups):

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a tetrahedron. Show that $|G| = 12$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a cube. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an octahedron. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a dodecahedron. Show that $|G| = 60$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an icosahedron. Show that $|G| = 60$

From this answer I figured that rigid motions are orientation-preserving isometries, i.e. reflections are not allowed.

So, for a tetrahedron, I thought of symmetry axes passing through a vertex and the centroid of the opposite face. There are four such axes (let's call them $A,B,C,D$). Along every axis, we can define $1_i, r_i, r_i^2$ as three rotations with $r_i^3= 1$, the identity element ($i=A,B,C,D$). Since there are four such axes, $|G| = 3\times 4 = 12$. Is this fine, or am I missing something? I'm slightly concerned about the fact that $1_A,1_B,1_C,1_D$ may all possibly be the same (since they are identity transformations), and that I am over-counting?

Minor question (detour): Are identity transformations corresponding to different axes different, or the same?

For the cube, I did the following:

For every pair of opposite faces, we have an axis of symmetry. There are $3$ such pairs, hence $3$ such axes (say $A,B,C,D$). About each axis we define $1,r_i,r_i^2,r_i^3$ with $r_i^4 = 1$ where $i=A,B,C,D$.

There are four body diagonals (say $E,F,G,H$), and about each diagonal (symmetry axis) we define $1,r_j,r_j^2$ with $r_j^3= 1$ where $j=E,F,G,H$.

In view of the above calculations, we have $|G| = 3\times 4 + 4\times 3 = 24$.

Using this method becomes difficult hereafter, for larger solids. It is not easy to identify all the axes of symmetry by hand. Moreover, the only group I have learned about in some detail at this point is $D_{2n}$, so please do not give solutions such as "the required group $G$ is isomorphic to a known and well-studied group $X$, and we know $|X| = ?$ so $|G| = ?$"

I think it boils down to having a good way to count all the distinct rigid motions. Could someone help me with this?

I came across James Ha's solutions here, but I don't understand how the solutions presented in the PDF are equivalent to mine even for the tetrahedron and cube cases. It'd be nice if someone can help me see the equivalence, and also tell me how to proceed with the other platonic solids! Thanks a lot!

答:
There is no substitute for cutting four equal equilateral triangles out of cardboard and taping them together to make a tetrahedron. Once you've done this, put a fingertip in the center of an edge and another fingertip at the center of the opposite edge. Then spin the tetrahedron about the axis joining your fingertips. You should find that a $180^\circ$ rotation brings the tetrahedron back to itself. In my experience, this is hard to visualize until you've done it physically.

There are three such pairs of opposite edges and hence three such $180^\circ$ rotations. These, together with the identity and eight rotations of $\pm120^\circ$ about various axes joining the centroid of a face to the opposite vertex account for all the rotational symmetries of the tetrahedron.

The other Platonic solids have similar $180^\circ$ rotations. But if you just want a count, you can do something simpler. Start with one face of the solid facing you with fixed orientation (say one edge horizontal). If it's an $m$-sided face, there are $m$ edges that can be the horizontal one, and these $m$ orientations can all be obtained from one another by rotating about the center of the face. Now if the solid has $f$ faces, any of the $f$ can be brought to the "facing you" position by a rotation. So there should be $mf$ rotational symmetries. This accounts for everything.

The answer of orangeskid is similar, but even simpler, than this one. Start with an edge facing you, oriented horizontally. Let the horizontal plane containing this edge be such that it bisects the dihedral angle between the two faces that meet along that edge. (In other words, from your perspective, these two faces, which slant away from you, will appear equal.) Now you can do the $180^\circ$ rotation discussed above, but you can also bring any other edge of the solid to the "facing you" position by a rotation. So there are $2e$ symmetries.


问:
Simplify, $\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$ where $x > 3$.

Simplify, $$\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}$$ where $x > 3$.

What I Tried: I tried to rationalize the denominator by multiplying with $x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9})$ to get :-
$$\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}} * \frac{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$$
$$\rightarrow \frac{2(x - 1)(x^2 + 4x\sqrt{x^2 - 9}x - 9)}{(x^2 + 4x + 3)^2 - (x - 1)^2(x + 3)(x - 3)}$$

I was only able to get to this, and I am stuck on how to move next. Can anyone help me?

Edit: The answer that is given in my side is $\sqrt{\frac{x^2 - 9}{x + 3}}$ , so for now none of the answers match with it.

Body:
$$\frac{(x-3)(x-1) +(x-1)\sqrt{(x-3)(x+3)}}{(x+3)(x+1) +(x-1)\sqrt{(x-3)(x+3)}}\\
= \frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}} \frac{\sqrt{x-3} +\sqrt{x+3}}{(x+1)\sqrt{x+3}+(x-1)\sqrt{x-3}}$$

Multiply top and bottom by $\sqrt{x+3}-\sqrt{x-3}$.

$$=  \frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}}\frac{6}{8x-2\sqrt{x^2-9}} \\ =  3\frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}} \frac{4x+\sqrt{x^2-9}}{15x^2+9} \\ =  \frac{(x-1)\sqrt{x^2-9}}{x+3} \frac{4x+\sqrt{x^2-9}}{5x^2+3} \\ =\frac{(x-1)(x^2-9 +4x\sqrt{x^2-9})}{(5x^2+3)(x+3) } \\ =\frac{(x-1)(x-3)}{5x^2+3} +\frac{4x(x-1)}{(5x^2+3)(x+3)} \sqrt{x^2-9}$$

I doubt this can be simplified any further.


问:
Question about notation in a theorem about Galois theory from Lang's Algebra (chapter 6 §1, corollary 1.16)
I have a question about the notation in an assertion in Lang's Algebra, chapter 6 §1, corollary 1.16:

Let $K/k$ be finite Galois with group $G$, and assume that $G$ can be written as a direct product $$G=G_1 \times \cdots \times G_n.$$ Let $K_i$ be the fixed field of $G_1 \times \cdots  \times \{1 \} \times \cdots  \times G_n$ where the group with $1$ element occurs in the $i$th place. Then $K_i/k$ is Galois, and $K_{i+1} \cap (K_1 \cdots K_i) = k$...

Very well then; suppose $n=3$ in the above situation. Then $K_1/k$ is Galois and $K_2 \cap K_1 = k$. Also, $K_3/k$ is Galois and... Well, what would $K_{3+1}$ be? In the case $K_1$ I suppose is meant that $K_2 \cap K_1 K_3 = k$, but in the case $K_3$ I have no idea. What does he mean? Thank you.

答:
The index $i$ should be restricted to $1 \le i < n$ (so there is no $i=3$ in the case $n=3$).

In the case $n=3$, the two cases are $i=1$ and $i=2$.  If $i=1$, the theorem is saying that $K_2 \cap K_1 = k$.  If $i=2$, the theorem is saying that $K_3 \cap K_1 K_2 = k$.

The result is a consequence of the correspondence theorem between subgroups of the Galois group of $K/k$, and subfields of $K$ containing $k$.  If subgroups $H, H'$ have fixed fields $K, K'$ then the intersection $H \cap H'$ has fixed field $KK'$ and the subgroup $\langle H, H' \rangle$ has fixed field $K \cap K'$.


Title: Simplify, $\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$ where $x > 3$.

Body:
Simplify, $$\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}$$ where $x > 3$.

What I Tried: I tried to rationalize the denominator by multiplying with $x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9})$ to get :-
$$\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}} * \frac{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$$
$$\rightarrow \frac{2(x - 1)(x^2 + 4x\sqrt{x^2 - 9}x - 9)}{(x^2 + 4x + 3)^2 - (x - 1)^2(x + 3)(x - 3)}$$

I was only able to get to this, and I am stuck on how to move next. Can anyone help me?

Edit: The answer that is given in my side is $\sqrt{\frac{x^2 - 9}{x + 3}}$ , so for now none of the answers match with it.

Body:
$(x^2+4x+3)^2 - (x-1)^2(x+3)(x-3) = (x+1)^2(x+3)^2 - (x-1)^2(x+3)(x-3) = (x+3)(10x^2 + 6)$

And $x^2 + 4x^2\sqrt{x^2-9} -9 = \sqrt{x^2-9}(\sqrt{x^2-9}+4x^2)$.

So you get $$\frac{\sqrt{x^2-9}(\sqrt{x^2-9}+4x^2)}{(x+3)(10x^2 + 6)} = \frac{\sqrt{x-3}(\sqrt{x^2-9}+4x^2)}{\sqrt{x+3}(10x^2 + 6)}$$

Or $$\frac{(x-3)(x+3)+4x^2\sqrt{x^2-9}}{(x+3)(10x^2 + 6)}  = \frac{x-3}{10x^2+6} + \frac{4x^2\sqrt{x^2-9}}{(x+3)(10x^2 + 6)}$$


问:
Confusion about spherical coordinates
This answer to my previous post confuses me: I have always thought about spherical coordinates as a triad of mutually orthogonal vectors $\hat{r},\hat{\theta},\hat{\phi}$ disposed as in the following figure copied from here

but the answer says that the correct representation of a vector $\vec{r}$ is $r \hat{r}$, not something like this:

$$\left(\begin{array}{c}
r\\
\theta\\
\phi
\end{array}\right)$$

Can anyone explain in a simple way where my interpretation is wrong?

答:
The vector $\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$ tells you how to go from the origin to the point $(x,y,z)$. And that's the same thing as going $r$ steps from the origin in the $\hat{r}$ direction, right?

Here $r=r(x,y,z)=\sqrt{x^2+y^2+z^2}$ of course, and “the $\hat{r}$ direction” means $\hat{r}(x,y,z)$, the $\hat{r}$ direction associated with that specific point $(x,y,z)$. You could write the formula as
$$
\vec{r}(x,y,z) = r(x,y,z) \, \hat{r}(x,y,z)
$$
to make this clearer.

Title: Confusion about spherical coordinates


问:
Find $E[X]$, when $X$ is a random number from Uniform$(0,1/k^2)$
Question:

We have a Random Number Generator defined as:

$\displaystyle U_{K} \sim \operatorname{Uni}\left[ 0,\frac{1}{k^{2}}\right] ,\ k\geqslant 1$.

We will take a random number from $I\sim \operatorname{Poi}( \lambda )$, by throwing a Poisson cube, add $1$ to the output, this will be the $k$ of our generator.

We will generate a random number with the $U_{k}$, let's call it $X$.

Find $E[ X]$.

My solution:

First, let's find what is the expected value from $I\sim \operatorname{Poi}(\lambda )$.

By definition, it is: $E[ I] =\frac{1}{\lambda}$.

Let's now add $1$ to it: $ \dfrac{1}{\lambda } +1=\dfrac{1+\lambda }{\lambda }$.

Hence, our generator is expected to look like:

$U_{\frac{1+\lambda }{\lambda }} \sim \operatorname{Uni}\left[ 0,\dfrac{1}{\left(\dfrac{1+\lambda }{\lambda }\right)^{2}}\right]$.

Since we add $1$ to the expectation $E[ I] \geqslant 0$, we are guaranteed to get a number $k\geqslant 1$.

Now, let's calculate what is the expected value from $\operatorname{Uni}\left[ 0,\dfrac{1}{\left(\dfrac{1+\lambda }{\lambda}\right)^{2}}\right]$.

This is a uniform  distribution, hence, the expectation will be:

\begin{align}
E[ X] &=\frac{\cfrac{1}{\left(\frac{1+\lambda }{\lambda }\right)^{2}}}{2}=\frac{1}{2\cdotp \left(\frac{1+\lambda }{\lambda }\right)^{2}}\\
&=\frac{1}{\cfrac{2\left( 1+\lambda ^{2} +2\lambda \right)}{\lambda ^{2}}}
=\frac{\lambda ^{2}}{2+2\lambda ^{2} +4\lambda }
\end{align}

Someone sent me this solution:

$$E[ X] =\sum ^{\infty }_{0}\frac{e^{-\lambda } \lambda ^{k-1}}{( k-1) !} \cdotp \frac{1}{2k^{2}}$$

Which is completely different from mine. What did I do wrong?

答:
The mistake in your reasoning is as follows: if $Y$ is a random variable which can be expressed in terms of another random variable $X$ in the form of an equation
$Y = f(X)$, it is not true in general that $\mathbb{E}(Y) = f(\mathbb{E}(X))$.


问:
Irreducible finite Markov chain and its stochastic matrix
Let $P$ be a stochastic matrix on a finite set $I$. Show that a distribution $\pi$ is invariant for $P$ if and only if $\pi(I-P+A)=a$, where $A=(a_{ij}:i,j\in I)$ with $a_{ij}=1$ for all $i$ and $j$, and $a=(a_i:i\in I)$ with $a_i=1$ for all $i$. Deduce that if $P$ is irreducible then $I-P+A$ is invertible.

My efforts:

$\pi(I-P+A)=a\iff\pi-\pi P+\pi A=a\iff\pi=\pi P$.

I have no idea about the invertible part.

答:
Theorem Every irreducible Markov chain with a finite state space is positive recurrent and thus has a stationary distribution (unique probability solution to $\pi=\pi P$).

Let $Q=I-P+A$. Then $Q^Tx=a^T$ has a unique solution $\pi^T$. So $Q^T$ is invertible.


问:
An Olympiad geometry question; Angle chase, Orthocenter, Bash, Circles.
Points $O$ and $H$ are the circumcenter and orthocenter of acute triangle $ABC$, respectively. The perpendicular bisector of segment $AH$ meets sides $AB$ and $AC$ at $D$ and $E$, respectively. Prove that $\angle DOA= \angle EOA$.

After drawing the diagram; I was stuck
so,  I figured I knew the coordinates of all points so I should coord bash this
but it was a very tedious computation and I must have made a mistake somewhere as I couldn't get the right solution. (P.S. if you want me to share my failed coordinate bash, do let me know though my handwriting is not very good).

My work

A elementary solution/trigonometrical/coordinate bash would be ideal but any solution would mean a lot.

Thanks!

答:
First see that $\triangle ADH\sim\triangle AOC$, so $\triangle ADO\sim\triangle AHC$ (either by spiral similarity, or look at length ratios).

Similarly, $\triangle AEO\sim\triangle AHB$. So $\angle DOA=\angle HCA=\angle HBA=\angle EOA$.


问:
corollary to the completeness axiom
The corollary states "Every nonempty subset $S$ of $\mathbb{R}$ that is bounded below has a greatest lower bound inf S.

The part I don't get in the proof is from where they came up with the set $-S$ where $-S=(-s : s\in S)$? Did they create it or used it from somewhere?

Proof:
Let $-S$ be the set $(-s : S\in S)$; $-S$ consists of the negatives of the numbers in $S$. Since $S$ is bounded below there is an $m$ in $\mathbb{R}$ such that $m < s$ for all $s\in S$. This implies that $-m \geq -s$   for all $s\in S $, so $m > u$ for all u in the set $-S$ Thus $-S$ is bounded above by $m$.. The Completeness Axiom 4.4 applies to $-S$, so $sup(S)$ exists... There is some more but those parts are left as an exercise.

答: Basically they created it.  You are given a set $S$ that is bounded below.  The completeness axiom refers to sets that are bounded above, so you need to find a way to use that.  Since negation changes the sense of an inequality, it is a way to bring it into the picture.</p>

Title: corollary to the completeness axiom


问:
how to solve $\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$

how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$

i used the differentiation under the integral and got

\begin{align}
\mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x
\\[3mm]
\mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x
\\[3mm]
&=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x
\\[3mm]
&=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2}
\end{align}
But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$
\begin{align}
\int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b
\\
&-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b
\\[3mm]
\mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}}
\end{align}
but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work

答:
Suppose $\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$, and set

$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\theta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Then, the following logarithmic integral can be evaluated in terms of Clausen functions using the general result derived here:

$$\begin{align}
\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}
&=\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\theta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}\\
&~~~~~+\left(\theta-\alpha\right)\ln{\left(b^{2}\sec^{2}{\left(\gamma\right)}\right)},\\
\end{align}$$

where the Clausen function (of order 2) is defined for real arguments by the integral representation

$$\operatorname{Cl}_{2}{\left(\vartheta\right)}:=-\int_{0}^{\vartheta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\vartheta\in\mathbb{R}}.$$

Applying the formula to the particular case where $a=0\land c=1\land z=1$, and with a little help from the duplication formula for the Clausen function, we obtain the following delightfully compact result:

$$\begin{align}
\forall b\in\mathbb{R}_{>0}:\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x+1\right)}}{x^{2}+b^{2}}
&=\frac{4\theta\ln{\left(\csc{\left(\theta\right)}\right)}-\operatorname{Cl}_{2}{\left(4\theta\right)}}{4b};~~~\small{\theta:=\arctan{\left(\frac{1}{b}\right)}\in\left(0,\frac{\pi}{2}\right)}.\\
\end{align}$$

It should be clear that Clausen functions are the natural tool to use if you're trying to stick to real methods.


问:
Is there a value of $\alpha$ such that $\cot\alpha=-2\sqrt{3}\sin\alpha?$
Is there a value of $\alpha$ such that $\cot\alpha=-2\sqrt{3}\sin\alpha?$

How can I think to solve the problem? We can find that $\dfrac{\sin\alpha}{\cot\alpha}=-\dfrac{1}{2\sqrt{3}},$ so $\sin\alpha$ and $\cot\alpha$ should be with different signs. I looked at the trig ratio table and saw that $\dfrac{\sin150^\circ}{\cot150^\circ}=-\dfrac{1}{2\sqrt{3}},$ but I really don't like that solution. Can you come up with something better?

What if we have the following problem:
Is there a value of $\alpha$ such that $2\tan\alpha=-\sqrt{2}\sin\alpha?$

Body:
Hint:
$$\cos(\alpha)=-2\sqrt{3}\sin^2{\alpha}=-2\sqrt{3}(1-\cos^2{\alpha})$$


问:
Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}?
Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?

Obviously the set $L$ depends on where $C$ is, which is why I've written $L$ as a function of $C.$

If $C$ is the midpoint of $AB$, or on the perpendicular bisector of $AB$, it is not clear to me that $L= \{$ the perpendicular bisector of $AB\},$ because maybe $L$ contains more points. [Edit: Karan Elangovan has answered this part of the question.]

答:
Let $M$ be the midpoint of $AB$.

$C$ lies on the perpendicular bisector of $AB$, hence $AC = BC$

$\angle AsC = \angle BsC$

$sC$ is common.

Hence either the angles $\angle sCA$ and $\angle sCB$ are equal or supplementary. If they're supplementary, then $ACB$ is a straight line, meaning $C$ is the midpoint of $AB$. We will consider this case at the end. Otherwise though, we have $\angle ACs = \angle BCs$, and so $\triangle ACs \equiv \triangle BCs$, so $As = Bs$, so $s$ must lie on the perpendicular bisector of $AB$, so $L$ = the perpendicular bisector of $AB$, as required.

Finally consider the case that $C$ is the midpoint of $BC$, and the angles are supplementary.

$$ \frac{\sin\angle sBC}{sC} = \frac{\sin\angle BsC}{BC} \text{(Sine law)}$$
$$ = \frac{\sin\angle AsC}{AC} \text{(AC = BC})$$
$$ = \frac{\sin\angle sAC}{sC} \text{(Sine Law)}$$

So we have $\sin\angle sBC = \sin\angle sAC$, so the two angles are either equal or supplementary. Since they are angles of a triangle, they cannot be supplementary, so they are equal, so $\triangle sAB$ is isoceles with base $AB$, so $As = sB$, so again we have that $s$ lies on the perpendicular bisector, as required.

So yes, the only points in $L$ are those on the perpendicular bisector of $AB$.


答:
$\cot\alpha = \frac{\cos \alpha}{\sin \alpha}=-2\sqrt{3}\sin\alpha$ is equivalent to $\cos \alpha = -2\sqrt{3}\sin ^2 \alpha$ and so to $$\cos \alpha = -2\sqrt{3} (1-\cos ^2 \alpha)$$

If we note $X = \cos \alpha$, then $$2\sqrt{3} X^2 -X -2\sqrt{3}=0$$

The discriminant is $\Delta = 49>0$.

So the solutions of the polynomial equations are $x_{1,2}=\frac{1\pm7}{4\sqrt{3}}$ and $x_2 = -\frac{\sqrt{3}}{2}$ is the only one that is inferior to $1$ in absolute value (this is important because $\forall \alpha, |\cos \alpha | \leqslant 1$)

So there is only one solution to your equation, which are $\alpha = \pm \cos ^{-1} -\frac{\sqrt{3}}{2} +2k\pi$ so $$\alpha = \pm \frac{5\pi}{6} + 2k\pi$$

Body:
Is there a value of $\alpha$ such that $\cot\alpha=-2\sqrt{3}\sin\alpha?$

How can I think to solve the problem? We can find that $\dfrac{\sin\alpha}{\cot\alpha}=-\dfrac{1}{2\sqrt{3}},$ so $\sin\alpha$ and $\cot\alpha$ should be with different signs. I looked at the trig ratio table and saw that $\dfrac{\sin150^\circ}{\cot150^\circ}=-\dfrac{1}{2\sqrt{3}},$ but I really don't like that solution. Can you come up with something better?

What if we have the following problem:
Is there a value of $\alpha$ such that $2\tan\alpha=-\sqrt{2}\sin\alpha?$


问:
If a series is absolutely convergent, then does it have to pass the ratio test or the root test?
That being said. I learned in the lecture, that if a series passes the ratio test and/or the root test, then it has to converge absolutely.

Is the reversal also true, that a series will pass the ratio test and/or the root test if it converges absolutely.

答: No. Every series of the form $\sum_{n=1}^\infty n^\alpha$, with $\alpha<-1$, is absolutely convergent. But both the ratio test and the root test are inconclusive when applied to it.


问:
bound on the leading coefficient of a polynomial
Given a polynomial with real coefficients, that satisfies $\forall x \in \left[ -1,1 \right]: \ |p(x)|\leqslant 1$, I have to show, that its leading coefficient, $a_m$ satisfies $a_m \leqslant 2^{m-1}$.

I have obtained some easy partial results: for example under the given conditions the sum of all even coefficients has to be in $\left[ -1,1 \right]$ (and the sum of the odd in $\left[ -2,2 \right]$), but that didn't help me much...

答: This is a property of Chebyshev polynomials. See the section "Minimal $\infty$-norm" there, where there's also a proof.</p>


答: If you know that there exists a (unique) least upper bound of any set of reals that is bounded above, and you want to prove that there exists a (unique) greatest lower bound of a particular set of reals that is bounded below, then it's only natural to reflect the particular set about $0$ and apply what you already know to the reflected set.</p>

Title: corollary to the completeness axiom

Body:
The corollary states "Every nonempty subset $S$ of $\mathbb{R}$ that is bounded below has a greatest lower bound inf S.

The part I don't get in the proof is from where they came up with the set $-S$ where $-S=(-s : s\in S)$? Did they create it or used it from somewhere?

Proof:
Let $-S$ be the set $(-s : S\in S)$; $-S$ consists of the negatives of the numbers in $S$. Since $S$ is bounded below there is an $m$ in $\mathbb{R}$ such that $m < s$ for all $s\in S$. This implies that $-m \geq -s$   for all $s\in S $, so $m > u$ for all u in the set $-S$ Thus $-S$ is bounded above by $m$.. The Completeness Axiom 4.4 applies to $-S$, so $sup(S)$ exists... There is some more but those parts are left as an exercise.


问:
Almost sure convergence of a series with a given distribution
I have a sequence $\{X_{n}\}$ of independent random variables with law
\begin{equation} P\left(X_{n}=-n^{2}\right)=\frac{1}{n^{2}} ; \quad \quad P\left(X_{n}=\frac{n^{2}}{n^{2}-1}\right)=1-\frac{1}{n^{2}}.
\end{equation}

\begin{equation}S_{n}=X_{1}+...+X_{n}
\end{equation}
I have to prove that $\frac{S_{n}}{n} \to 1 \quad a.s. \quad $
Can anyone help me? I have already tried in different ways

答: Show that $X_n\to 1$ a.s using Borel Cantelli and apply stolz-cesaro to deduce the result


问:
Washington 2.18d, elliptic curves change of variables in characteristic 3
I got stuck on the Exercise 2.18d in the Book: Elliptic Curves: Number Theory and Cryptography.
The problem is:

Show that if $y^2 = x^3 +a_4 x+a_6$ and $y^2 = x^3 +a'_4x^2 + a'_6$ are two elliptic curves (in characteristic 3), then there is a change of variables $ y \mapsto ay , x \mapsto bx +c$, with $a,b \in \overline{K}^{\times}$ and $c \in \overline{K}$, that changes one equation into the other.

My Problem is transforming the $x$ into the $x^2$, I guess you should somehow use that we operate in char 3. Since $(bx +c)^3 = b^3x^3 + c^3$ we can't get our $x^2$ from here.

Edit:
It seems my version of the Book has a spelling mistake and the exercise should be $y² = x³ +a_4x +a⁶$ and $y²=x³ + a'_4x + a'_6$. I use the version "Elliptic Curves: Number Theory and Cryptography, 2nd Edition, CRC Press, Versiondate:20131121". If someone should stumble on the same problem.

答:
This is a spelling mistake in the Book. I found an errata for the Book.

Show that if $y^2 = x^3 +a_4 x+a_6$ and $y^2 = x^3 +a'_4{\color{red}{x}} + a'_6$ are two elliptic curves (in characteristic 3), then there is a change of variables $ y \mapsto ay , x \mapsto bx +c$, with $a,b \in \overline{K}^{\times}$ and $c \in \overline{K}$, that changes one equation into the other.


问:
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.

That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?

I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.

But from there I am stuck, I am not able to see how to link all the concepts.

Could someone give me a hint?

Thank you and Happy New Year 2021 :)

答:
You are almost there.

Since $f(0) = f(+\infty) = +\infty$ and there is a unique solution $x_0 > 0$ such that $f'(x_0)=0$ (because $2x^2(1+x)$ is an increasing function of $x$) we only need to show that $f(x_0) > \frac{1+2 \ln 2}{4}$.

Since $x_0 \in \left(\frac 12, 1\right)$, then
$$f(x_0) = x_0^2 + \ln \left( 1+\frac{1}{x_0}\right)> \left(\frac 12\right)^2 + \ln \left( 1 + \frac{1}{1}   \right) = \frac{1+4\ln 2}{4}> \frac{1+2\ln2}{4}.\blacksquare$$


问:
How can I determine if the group $G=(\Bbb Z_{16}\times\Bbb Z_{18},+)$ is isomorphic to $(\Bbb Z_4\times\Bbb Z_{72},+)$?
How can I determine if the group $G=(\Bbb Z_{16}\times\Bbb Z_{18}, +)$ is isomorphic to $(\Bbb Z_4\times\Bbb Z_{72}, +)$?

I came up with that $(1,1)$ in $G$ has order $144$ because $\operatorname{lcm}(16,18)=144$.

Highest possible order for an element in $(\Bbb Z_4\times\Bbb Z_{72}, +)$ is $72$ because $\operatorname{lcm}(4,72)=72$.

Therefore they cant be isomorphic because there is no element with order $144$ in $(\Bbb Z_4\times\Bbb Z_{72}, +)$.

Is there any other way I can prove it?

答:
Of course, you provided an excellent and elementary way to see this. In general, you can use the following, but it requires a classification result, which is non-trivial to prove:

You can invoke the fundamental theorem of finite abelian groups. Decompose
$$\mathbb{Z}_{16} \times \mathbb{Z}_{18}\cong \mathbb{Z}_{2^4} \times \mathbb{Z}_2 \times \mathbb{Z}_{3^2}$$
$$\mathbb{Z}_{4} \times \mathbb{Z}_{72}\cong \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{2^3}$$
and such a decomposition is unique (up to permutation of the factors). Since the decompositions do not agree, the groups are not isomorphic.

Here, I used the fact that $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$ when $m$ and $n$ are coprime.


问:
Evaluating $x^3−6x^2+6x$, with $x=2+2^{2/3}+2^{1/3}$. Where did I go wrong here?
This question has been answered by @Ak and Z Ahmed by posting their answers.

If $x=2+2^{2/3}+2^{1/3}$, then evaluate $x^3−6x^2+6x$.

So , how I solved is

$$x-2 = 2^{2/3}+2^{1/3}$$

I got
$$x^3 - 8 - 6x^2 +12x = 6+6x-12 $$

Then I got
$$ x^3 - 6x^2 -6x = 2 $$

But not $+6x$.

Yes , answer is $2$. Where did I go wrong?

Body:
$$x=2+2^{2/3}+2^{1/3} \implies (x-2)=2^{2/3}+2^{1/3}$$
Cubing both sides we get
$$(x-2)^3=2^2+2+3.2(2^{2/3}+2^{1/3}) \implies x^3-8-6x(x-2)=6+6(x-2)$$
$$\implies x^3-6x^2+12x-8=6+6x-12\implies x^3-6x^2+6x=2$$


问:
Annihilator of an element is the entire ring implies the element is $0$
I am currently reading the chapter on finitely-generated modules over PIDs in Jacobson's Basic Algebra and at several points he uses the property that if the annihilator of an element $x$ of the module (i.e. the elements $r$ of the base ring $R$ s.t. $rx=0$) is the entire base ring $R$, then the element is zero (i.e. $\text{ann}(x)=R\Rightarrow x=0$). I seem to have missed the part where he actually proves this or states why it is true, so I would like to ask what the reason for this might be in this specific case (finitely-generated modules over PIDs) and whether it holds in more general cases (non-finitely-generated modules over more general rings).

Thank you in advance.

答:
It's enough to assume that the ring has identity element (and the module is unitary: $1x=x$).

Then in particular for $1\in R$ we get $0=1x=x$.


问:
Kähler differentials and ramification of infinite places
Let $L/K$ be an extension of number fields (i.e. finite extensions of $\mathbb{Q}$). The ramification of finite places, i.e. prime ideals, is controlled by the module of Kähler differentials $\Omega_{L/K}$. But what about infinite places? We say that an infinite place $\sigma: K\to \mathbb{R}$ ramifies if it extends to two distinct places $\sigma, \bar \sigma: L\to \mathbb{C}.$ So my question is, can we somehow enhance Kähler differentials so that they tell us something about ramificatioin of those infinite places? Or do we have some other algebraic gadget for that?

答:
At the following link you will find some info on Kahler differentials and rings of integers in number fields:

Why isn't every finite locally free morphism etale?

Note: Any finite extension $K \subseteq L$ is separable, hence $\Omega^1_{L/K}=0$ is zero.

Citation: "For any finite extension  $K \subseteq L$ of number fields, it follows the morphism $\pi: S:=Spec(\mathcal{O}_L) \rightarrow T:=Spec(\mathcal{O}_K)$ is always integral, locally free (and ramified when $K=\mathbb{Q}$). Hence it is seldom etale. There is an open subscheme $U \subseteq T$ with the property that $\pi_U:\pi^{-1}(U) \rightarrow U$ is etale. The fiber  $\pi^{-1}((p))$ in formula F1 above is unramified iff $l_i=1$ for all $i$.
The module of Kahler differentials $\Omega:=\Omega^1_{\mathcal{O}_L/\mathcal{O}_K}$ is zero on the open subscheme $\pi^{-1}(U)$. Note moreover that the field extension

$\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$

is always separable since it is a finite extension of finite fields (here the ideals $\mathfrak{p}, \mathfrak{q}$ are maximal ideals). The ramified prime ideals in $\mathcal{O}_K$ are related to the discriminant of the extension $L/K$ (see in "Algebraic number theory", page 49, by Neukirch). In Neukirch he defines the discriminant using a basis for $L$ over $K$, but it can also be defined using the module of Kahler differentials $\Omega$."


问:
$\int_{0}^{\infty} x^3 e^{-x^8 \sin^2 x} dx$ and $\int_1^{\infty} \frac{1} {1+ x^2 \sin^2 x} dx$ are not Lebesgue integrable
This question is from Tom Apostol Mathematical Analysis.

Prove that $\int_{0}^{\infty} x^3 e^{-x^8 \sin^2 x} dx$ and $\int_1^{\infty}  \frac{1} {1+ x^2 \sin^2 x} dx$ are not Lebesgue integrable.

The problem is that I proved that both of them are Lebesgue integrable by using theorem 10.33

For (1) : Consider $\int_{0}^{\infty} x^3 e^{-x^8 \sin^2 x} dx\leq\int_{0}^{b} x^3 dx$ and the integral on the right is less than M for some M >0 and that M depends upon b.( I used property of $e^{-x}$ over when $x\in [0,\infty]$).

Similarly for (2):  $\int_1^{\infty}  \frac{1} {1+ x^2 \sin^2 x}dx \leq \int_{1}^{b} 1$= b-1=M .

So, what I am doing wrong and what is the right way of proving it?

Do you mind giving your time to this question?

答:
Apostol's definition is as follows:
$f$ is Lebesgue integrable on $I$ if it can be written as $f= u-v$ where $u \in U(I)$ and $v \in V(I)$. The set $U(I)$ is the set of "upper functions" which he defines to be the functions $u$ which are limits of increasing sequences of step functions $s_n$ defined on $I$  such that $s_n \nearrow u$ almost everywhere on $I$ and  $ \lim_{n\to \infty} \int_I s_n  < \infty.$ He defines $V(I)$ similarly. In particular, $\int_I f = \int_I u  - \int_I v := \lim s_n - \lim t_n < \infty.$ The latter condition is what you're missing; the integral must be finite.

Now the functions you use as upper bounds do not have a finite itnegral and thus are not Lebesgue integrable, which is what is wrong with your argument.

What you need to do is to prove the integral is infinite. To do so, it suffices to find an appropriate lower bound whose integral is infinite, then monotonicity insures that your function has infinite integral. You need to analyze the quantitative behavior of both functions, for example: the function $\frac{1}{1+x^2 \sin^2(x)}$ is close to $1$ for $ x$ close to $ \pi + n \pi.$ You would be done if you prove that the integral around these points is big enough to make the integral infinite.


问:
Trying to figure out a fitting function for a curve where the fitting function always passes through the origin (0,0)
I have a curve that looks like this.

Acceleration v Velocity

I am trying to figure out a function I can use to fit to this curve. I have tried using a curve of the form $Ax^2 + Bx + C$ and while that does a good job of fitting the points it does not go through the origin. I am trying to construct a function of some sort that fits the points in the image and also goes through or is as close to the origin as possible.

I'd really appreciate some guidance!

答: If you want the graph of a polynomial to go through the origin, omit the constant term. For your quadratic polynomial, omit $C$.


问:
Dominant morphism, equal dimensions: always finite?
Let $f:X\to Y$ be a dominant morphism of varieties (integral separated schemes of finite type over an algebraically closed field) such that dim $X$ = dim $Y$.

Question: must f be finite?

It seems that f must have finite fibers by looking at dimensions.  So perhaps it is the same question to ask if $f$ must be proper, because finite fibers + proper = finite.  I am not familiar enough with standard non-examples of properness to have a good intuition on this.

Finally, if it makes any difference to assume X and Y are quasi-projective, please do so.

Edit: After Steve's answer to the original question, I would like to ask the same question for a self-morphism $f:X\to X$.

答:
No, $f$ might be an inclusion of an open set; for example, the morphism corresponding to the (not module finite) extension $\mathbb{C}[x] \subseteq \mathbb{C}[x,x^{-1}]$.

Edited in light of edit to question:

Here's another type of thing that can happen: look at the map $(x,y) \mapsto (x,xy)$ from $\mathbb{C}^2$ to itself. This is evidently dominant, but it does not have finite fibers (since the fiber over the origin is a line). I bet one can find an example with finite fibers which is not finite, but I can't think of one at the moment.


问:
Surface integrals in spherical coordinates
If I am given a surface in spherical coordinates $(r,\theta,\varphi)$, such that it is parametrised as:

$$
\begin{align}
r&=r(\theta,\varphi)\\
\theta&=\theta\\
\varphi&=\varphi
\end{align}
$$

What is the area $S$ of such surface? Or more specifically, can you show how to get the result:
$$
S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{r^2+\left(\frac{\partial r}{\partial \theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\frac{\partial r}{\partial \varphi}\right)^2}\;r\sin\theta\;{\rm d}\theta\,{\rm d}\varphi
$$

Some definitions that I am using:

$k$-surface:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is called a $k$-surface, if there exists a non-empty open set $E\subset \mathbb{R}^k$ and a map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, such that: (i) $\varphi(E)=M$, (ii) $\varphi\in C^1(E;\mathbb{R}^N)$, and (iii) the rank of Jacobi matrix of $\varphi$ is equal $k$ everywhere on $E$.
The surface is called simple if $\varphi$ is also injective on $E$ and $\varphi^{-1}$ is continuous of $\varphi(E)$.

Surface integral of the first kind:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is a simple $k$-surface parametrized by the map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, from the open set $E\subset \mathbb{R}^k$ and $f:\mathbb{R}^N\to\mathbb{R}$ is defined on $M$. The surface integral of the first kind is defined by:
$$
\int_M f\,\mathrm{d}S:=\int_E f(\varphi(t))\sqrt{\det{G(D_\varphi(t))}}\,\mathrm{d}t\,,
$$
if the integral on the right exists in the Lebesgue sense and is finite. Here, $G(A)$ denotes the Gramm matrix made from columns of $A$ and $D_\varphi$ is the Jacobi matrix of the map $\varphi$. The numeric value of:
$$
S_k(M):=\int_M f\,\mathrm{d}S\,,
$$
is called the $k$-dimensional surface area of the $k$-surface $M$.

Motivation for the question

Now these definitions can be used to calculate e.g. the surface of a unit sphere. When one describes the sphere by the map (omitting one longitudinal line): $\varphi: (\eta,\psi)\mapsto(\cos\psi\cos\eta,\cos\psi\sin\eta,\sin\psi)$, where $(\eta,\psi)\in E=(0,2\pi)\times(-\frac{\pi}{2},\frac{\pi}{2})$. The Gramm matrix looks like:
$$
\begin{pmatrix}
\cos^2\psi & 0 \\
0 & 1
\end{pmatrix}
$$

one ends up with the (here $\lambda_2$ denotes the Lebesgue measure):
$$
S_2(M) = \int_E 1\sqrt{\det{G(D_\varphi(\eta,\psi))}}\,\mathrm{d}\lambda_2(\eta,\psi)=\int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\psi\,\mathrm{d}\psi\,\mathrm{d}{\eta}=4\pi
$$

I did not do anything else but blindly followed the definitions.

The wrong result:

If I do the same approach to the above problem at hand. My map is $\varphi:\,(\theta,\phi)\mapsto (r(\theta,\phi),\theta,\phi)$, the Jacobian is then:
$$
\begin{pmatrix}
\frac{\partial r(\theta,\phi)}{\partial \theta} & \frac{\partial r(\theta,\phi)}{\partial \phi} \\
1 & 0 \\
0 & 1
\end{pmatrix}
$$

The rank of this matrix is 2 as needed for a 2-surface. The Gramm matrix is then:
$$G(D_\varphi)=
\begin{pmatrix}
1+ \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 & \frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} \\
\frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} & 1+ \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
\end{pmatrix}
$$
with determinant evaluated to:
$$
\det(G(D_\varphi)) = 1 + \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 + \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
$$

which leads to the surface area:

$$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{1+\left(\frac{\partial r}{\partial \theta}\right)^2 + \left(\frac{\partial r}{\partial \varphi}\right)^2}\;{\rm d}\theta\,{\rm d}\varphi$$

Which is incorrect. The question is why, and please show the correct way with explanation because as suggested by the example with unit sphere, I did not have to do any transformations and it did work with curvilinear coordinates right away. So the answer provided by Quanto does not address this at all.

Note to the edit:

I have added quite a lot of details since the answer by Quanto, but the root problem is the same - to see by calculation explicitly how one arrives to the correct result and to understand why the my calculation for a sphere works (where i am not using Cartesian coordinates) but fails here.

答:
Your surface corresponds to the following idea: $$S:\quad r(\theta,\phi):=R + h(\theta,\phi)\ ,\tag{1}$$
where $R$ is the earth radius at sea level, and $(\theta,\phi)$ is the height over sea level of the point with geographical coordinates $(\theta,\phi)$. You want to know the area of $S$. In order to find this area we have to resort to cartesian coordinates, because $S$ inherits its area from the euclidean ${\mathbb R}^3$, and not from the "practical" representation. This means that we have to replace $(1)$ by
$$S:\quad (\theta,\phi)\mapsto{\bf r}(\theta,\phi)=r(\theta,\phi)\,\bigl(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta\bigr)\ ,\tag{2}$$
where $(\theta,\phi)\in B:=\bigl[-{\pi\over2},{\pi\over2}\bigr]\times[-\pi,\pi]$. Note that $$(x,y,z)=\bigl(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta\bigr)$$ is the point on the unit sphere $S^2$ having geographical coordinates $(\theta,\phi)$.

In calculus 102 it is taught that the area of the surface $(2)$ is computed as
$${\rm area}(S)=\int_B \bigl|{\bf r}_\theta\times {\bf r}_\phi\bigr|\>{\rm d}(\theta,\phi)=\int_{-\pi/2}^{\pi/2}\int_{-\pi}^\pi \Psi(\theta,\phi)\>d\phi\>d\theta\ ,$$
where $\Psi(\theta,\phi):=\bigl|{\bf r}_\theta\times {\bf r}_\phi\bigr|_{(\theta,\phi)}$
has to be computed carefully from $(2)$.

In theory one arrives at this expression $\bigl|{\bf r}_\theta\times {\bf r}_\phi\bigr|_{(\theta,\phi)}$ by noting that a tiny $[\theta,\theta+\Delta\theta]\times[\phi,\phi+\Delta\phi]$ rectangle  in the parameter plane is mapped by $(2)$ onto a tiny parallelogram of area $\bigl|{\bf r}_\theta\times {\bf r}_\phi\bigr|_{(\theta,\phi)}\>\Delta\theta\Delta\phi$.


问:
Find a constant $c$ such that $\omega \wedge \nu \wedge \gamma = cdx\wedge dy \wedge dz$
Let $\omega = dx +5dy - dz, \nu = 2dx-dy+dz, \gamma = -dx+dy +2dz$

Find a constant $c$ such that $\omega \wedge \nu \wedge \gamma = cdx\wedge dy \wedge dz$

I multiplied it and ended up at $(-11dx\wedge dy+4d\wedge dz-3dz\wedge dx)\wedge (-dx+dy+2dz)$ but I haven't been taught the definition of multiplying 1-forms with 2-forms yet.. I do not know how to proceed or  how I should have done this.

答:
$a\wedge a=0$, so $dx\wedge dy\wedge dx=0$ and so on. So we have
$$\begin{aligned}&(-11dx\wedge dy+4dy\wedge dz-3dz\wedge dx)\wedge (-dx+dy+2dz)\\&=-22dx\wedge dy\wedge dz-4dy\wedge dz\wedge dx-3dz\wedge dx\wedge dy\\
&=-22dx\wedge dy\wedge dz+4dy\wedge dx\wedge dz+3dx\wedge dz\wedge dy\\
&=-22dx\wedge dy\wedge dz-4dx\wedge dy\wedge dz-3dx\wedge dy\wedge dz\\
&=-29dx\wedge dy\wedge dz \end{aligned}
$$


问:
How to estimate $\sum_{q \leq Q, n \in \mathbb{Z}} w(n/q)$ with $w$ a smooth function?
Let $w$ be a smooth function with compact support. I am interesting in estmating $\sum_{q \leq Q, n \in \mathbb{Z}} w(n/q)$ as $Q \to \infty$. By applying the Poisson summation formula we can deduce
$$
\sum_{q \leq Q, n \in \mathbb{Z}} w(n/q) = \frac12 \hat{w}(0) + O(Q).
$$
I was wondering is it possible to obtain a secondary term for a very nice of choice of $w$?
Any comments are appreciated. Thank you!

Body:
$$\sum_n w(x+n)=\sum_k \hat{w}(k) e^{2i\pi kx}$$ then $$\sum_{q=1}^Q \sum_n w(n/q)=\sum_{q=1}^Q \sum_{m=0}^{q-1}\sum_k \hat{w}(k) e^{2i\pi k m/q}=\sum_{q=1}^Q \sum_k \hat{w}(k)  \sum_{m=0}^{q-1} e^{2i\pi k m/q}$$

$$= \sum_{q=1}^Q \sum_{k, q|k} \hat{w}(k) q= \frac{Q(Q+1)}{2}\hat{w}(0)+\sum_{k\ne 0} \hat{w}(k) \sum_{q| k,q\le Q}q$$
$$ = \frac{Q(Q+1)}{2}\hat{w}(0)+\sum_{k\ne 0} \hat{w}(k) \sigma_1(k)+O(\sum_{|k|\ge Q} |\hat{w}(k)| \sigma_1(k))$$
Since $\hat{w}$ is rapidly decreasing, for any fixed $r>1$ it is
$$ = \frac{Q(Q+1)}{2}\hat{w}(0)+\sum_{k\ne 0} \hat{w}(k) \sigma_1(k)+O(\sum_{|k|\ge Q} k^{-r})$$ $$=\frac{Q(Q+1)}{2}\hat{w}(0)+\sum_{k\ne 0} \hat{w}(k) \sigma_1(k)+O(Q^{1-r})$$


问:
How to calculate $|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle$ for $\mathfrak{su}(2)$?
At $\mathfrak{su}(2)$, tensor product $3 \otimes 3$ can be decomposed to
$3\otimes3=5\oplus3\oplus1$.

In this space, $|1,1\rangle \otimes |1,-1\rangle$ has weight $0$.

So we should be able to write down
$$|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle.$$

How to calculate $\alpha,\beta,\gamma$?

答:
You are given, 4th relation,
$$  \langle  1 \, 1 \, 1 \, (- 1) | J \, 0 \rangle
  = 2  \sqrt{\frac{2 J + 1}{(J + 2   + 1)! (2   - J)!}}~~, $$
so read off, as instructed,
$$
\alpha=1/\sqrt{6}, \qquad \beta=1/\sqrt{2},\qquad \gamma=1/\sqrt{3}.
$$


问:
Is it okay to raise to the power of $n$ (where $n\in\mathbb{R}$) on both sides in Euler's Identity?
So I was playing around with the identity and did this:
$$e^{i\pi}=-1$$
$$e^{(2\pi)i}=(-1)^{2}=1$$
$$e^{(2n\pi)i}=1^{n}=1\ \ \ \ \ (n\in \mathbb{R})$$
$$\Rightarrow\ e^{(2n\pi)i}=e^{(2\pi)i}\ \ (\forall n\in\mathbb{R})$$

Then, $$2n=2$$ So,$$n=1$$

Which is quite contradicting as any value of n should satisfy the condition. Which means I have made a mistake somewhere. Can anyone tell me where am I going wrong? Help would be appreciated! Thanks in advance.

答:
$$1^m =1^n $$ does not imply $$m=n$$

The positive  real number base $b$ should be different from 1  to have $$b^m=b^n $$implies $$m=n$$


问:
Why am I getting wrong answer for this?
If the product of the roots of the equation $2x^2+6x+ y^2 + 2 = 0$ is $−y$ ,then the value of $y$ is.

How I solved:

$xy = -y$ is given

$x = -1$

How I did further:

I put value of $x$ as $-1$ and solved it.
Putting in equation I got $y  = \sqrt{3}$

It also satisfies the equation ,

-1*$\sqrt{3} = -\sqrt{3}$.

But the answer in my book is $2$. I am not getting how they solved it that way

答:
Well, the roots of quadratic polynomial can be found using:

$$\text{a}x^2+\text{b}x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}\tag1$$

So, in your case we get:

$$2x^2+6x+\text{y}^2+2=0\space\Longleftrightarrow\space x=\frac{-6\pm\sqrt{6^2-4\cdot2\cdot\left(\text{y}^2+2\right)}}{2\cdot2}=$$
$$\frac{-3\pm\sqrt{5-2\text{y}^2}}{2}\tag2$$

So, the product gives:

$$\frac{-3+\sqrt{5-2\text{y}^2}}{2}\cdot\frac{-3-\sqrt{5-2\text{y}^2}}{2}=\frac{2+\text{y}^2}{2}\tag3$$

Now, we get:

$$\frac{2+\text{y}^2}{2}=-\text{y}\space\Longleftrightarrow\space\text{y}=-1\pm i\tag4$$

Where $i^2=-1$

Title: Why am I getting wrong answer for this?


问:
Prove that a linear transformation exists with certain conditions

Let $V$ be a finite vector space and $U$ a sub-vector space to $V$

Prove that a linear transformation $T:V\rightarrow V$ exists, where:

$\ker T = U$ & $\ker T \cap\operatorname{Im}T = \{0\}$

I understand that each vector in $U$ is a vector that follows:
$v \in V$ & $T(v)=0$

And technically $U$ can just be the zero vector space, and $T(v) =v$

But this doesn't feel like a proof to me, if anybody can help me understand better how to approach such a question, that would be highly appreciated.

答:
Hint: Let $W$ be an arbitrary complement of $U$ (say, extend a basis of $U$ to a basis of $V$ and keep only the new basis elements).

Then you can apply the identity on $W$.


问:
Graph minus a point has a separation
(It's wrong there is a separation even for for a graph with a basic Hamiltonian path.)

Let G(V,E) be a connected (no separation) directed graph, such that, $S_x$=V∖{x}, ∀x∈V is a subset of vertices of G. Then the induced subgraph G[$S_x$] is the graph whose vertex set is $S_x$ and whose edges set consists of all of the edges in E that have both endpoints in $S_x$.

Then a necessary condition (not sufficient see link Graph minus a point has at most 2 connected components for a counterexample) for a graph G(V,E) to have a Hamiltonian path (not necesarily a cycle) is that the induced subgraphs ∀x∈V, G[$S_x$] have at most 2 connected components.

Now, let G[$S_x$] have 2 disjoint nontrivial connected components, let's say that $S_x$ = $A_x \cup B_x$, then we say that:

$E_{-}$={$(a,b) \in E: \forall a \in A_x \forall b \in B_x \ | \ (a,x) \in E, (x,b) \in E$}

$E_{+}$={$(b,a) \in E: \forall a \in A_x, \forall b \in B_x \ | \ (b,x) \in E, (x,a) \in E$}

$E_x$={$(y,z) \in E: \forall y,z \in V, y \neq x, z \neq x$}

Then I want to either proof or find a counterexample of the following statement, if $\forall x \in V$ none of the graphs,

G(V\ {x}, $E_x \cup E_{+}$)

G(V\ {x}, $E_x \cup E_{-}$)

have a separation then G(V,E) has a Hamiltonian cycle.

(A separation of V are 2 disjoint nontrivial sets C and D, such that C $\cup$ D = V. I define a separation between two points in a graph G(V,E) $x, y \in V$ when I can't find a direct/undirect path from x to y in E) i.e.

$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix}$

Can't find a path from cities (1 and 2) to cities (3 and 4)

$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix}$

Can't find a path from city 1 to cities (2, 3 and 4)

notice that the power of those matrices M^2, M^3 preserve the zeroes, so it should be easy to find a separation (hopefully in Polynomial time).

答:
I'm assuming that you aim to find a polynomial time algorithm for deciding whether a graph has a Hamiltonian path.

Consider the undirected graph obtained from a cycle graph $C_n$ (with $n \geq 3$) and modifying it by taking two distinct vertices of the cycle and attaching (to each one) a new vertex.
The resulting graph has no HP but is strongly connected and the removal of any vertex still leaves us with a strongly connected graph and thus has no "separation", leading to your algorithm falsely outputting that the original graph has a HP.

Title: Graph minus a point has a separation


问:
Show that $\mathcal{O}_k/(m)$ is finite.
Let $K$ be a number field and $\mathcal{O}_K$ be its ring of integers. It is easy to show that there is some nonzero rational integer $m$ in a prime ideal $\mathfrak{p}$. Now, I know that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n=[K:\mathbb{Q}]$.

I am now having trouble understanding why $\mathcal{O}_K/(m)$ is finite with size $m^n$. I can't really see why $(m)$ would be full rank here, and I think I'm missing something kind of obvious.

答:
Question: "I am now having trouble understanding why OK/(m) is finite with size mn. I can't really see why (m) would be full rank here, and I think I'm missing something kind of obvious."

Answer: Any non-zero prime ideal $\mathfrak{p}\subseteq \mathcal{O}_K$ "restricts" to a non-zero prime ideal  $(p)= \mathfrak{p}\cap \mathbb{Z} \subseteq \mathbb{Z}$, since the intersection may be defined as the inverse image under the canonical map of rings $i: \mathbb{Z}\rightarrow \mathcal{O}_K$. There is a canonical decomposition of the extended ideal

D1. $(p)\mathcal{O}_K =\mathfrak{p}_1^{l_1} \cdots \mathfrak{p}_i^{l_i}\mathcal{O}_K$

where the ideals $\mathfrak{p}_i$ are non-zero prime ideals, and an isomorphism

D2. $\mathcal{O}_K/(p)\mathcal{O}_K \cong \oplus_j \mathcal{O}_K/\mathfrak{p}_j^{l_j}.$

The rings $\mathcal{O}_K/\mathfrak{p}_j^{l_j}$ are Artinian local rings for all $j$.
Your original prime ideal satisfies $\mathfrak{p}=\mathfrak{p}_j$ for some $j$ and this gives rise to a finite extension of finite fields

D3. $\mathbb{F}_p:=\mathbb{Z}/(p)\mathbb{Z} \subseteq \mathcal{O}_K/\mathfrak{p}$

and from this it follows $\mathcal{O}_K/\mathfrak{p}\cong \mathbb{F}_{p^r}$ for some integer $r \geq 1$. Hence you can choose your integer $m$ to equal a non-zero prime $p$.


问:
British Maths Olympiad Round 2 2011, Question 3: Is there a more efficient method?
I've been practicing for the second round of the BMO and I've come across a question for which I think there must be a more efficient solution than I found.

Question 3:
https://bmos.ukmt.org.uk/home/bmo2-2011.pdf

To paraphrase the question, it asks for the number of positive integers $n < 2011$ such that $f(n) = f(2011) = 21$, where $f$ is defined on the positive integers as:

$f(1) = 1$

$f(4n) = f(2n)$

$f(4n+1) = 2f(2n) + 1$

$f(4n+2) = 2f(2n+1)$

$f(4n+3) = f(2n+1)$

My method was to note that the only "path" to $21$ is $1,2,5,10,21$. The only positive ints that are mapped to $1$ are of the form $2^a - 1$, and so the only ints mapped to $2$ are of the form $2^b(2^a-1)$, and continuing in this fashion and expanding out, we have the only integers that map to $21$ are those of the form $2^{a+b+c+d+e}-2^{a+b+c+d}+2^{a+b+c}-2^{a+b}+2^e$, where $a,b,c,d,e \geq 1$.

So the number of solutions is equal to the number of sequences $y_1 > y_2 > y_3 > y_4 > y_5 \geq 1$ such that
$$ 2^{y_1} - 2^{y_2} + 2^{y_3} - 2^{y_4} + 2^{y_5} - 1 < 2011$$.

This in turn is equal to the number of sequences $x_1 > x_2 > x_3 > x_4 > x_5 \geq 0$ such that
$$ 2^{x_1} - 2^{x_2} + 2^{x_3} - 2^{x_4} + 2^{x_5} \leq 1005 $$.

Then I defined the function $F_k(n)$ as equal to the number of solutions to the more general problem, where the sum of powers must be $\leq n$ and there are $k$ variables instead of $5$, so the answer is $f_5(1005)$.

Because of the self similarity of the constraint, we have the following recurrence relation:

$$ F_k(n) = \sum_{m\geq 0}{F_{k-1}\left\lfloor\frac{n + (-1)^k2^m}{2^{m+1}}\right\rfloor} $$

With base cases of:

$$ F_1(n) = \lfloor \log_2{n} \rfloor $$
$$ F_k(n) = 0 \text{ when } n \leq 0$$

I implemented both the original problem and my solution in Python to check, and they both gave the same answer, $455$, so my solution is correct. However the paper is intended to be done without a calculator or computer, and it computing the desired value of $F$ by hand seems to need far too many calculations (these questions rarely require drawn out calculations).

So I assume that there is a more efficient method, either in calculating the values of $F$ or by looking at the problem from a completely different angle, both of which I would find extremely helpful.

答:
Write everything in binary. Then we can see that $f(n)$ is the result when we collapse blocks of identical digits in $n$ to one such digit, e.g. $$f(2011)=f(11111011011_2)=10101_2=21.$$
So we want the number of $11$ bit numbers that have $5$ "switches" in the sequence of binary digits, and end in $1$. This is $\binom{11}{5}=462$. However, $7$ of these numbers are at  least $2011$, so the final answer is $455$.


答:
The product of roots of $$ax^2+bx+c=0$$ is $$c/a$$
In your case we have $$ c/a = (y^2+2)/2 =-y $$
$$y^2+2y+2=0$$

Solving for $y$ you get $$ y=-1\pm i$$

Title: Why am I getting wrong answer for this?

Body:
If the product of the roots of the equation $2x^2+6x+ y^2 + 2 = 0$ is $−y$ ,then the value of $y$ is.

How I solved:

$xy = -y$ is given

$x = -1$

How I did further:

I put value of $x$ as $-1$ and solved it.
Putting in equation I got $y  = \sqrt{3}$

It also satisfies the equation ,

-1*$\sqrt{3} = -\sqrt{3}$.

But the answer in my book is $2$. I am not getting how they solved it that way


答:
Your mistake was that you assumed $x$ and $y$ are the roots of the equation you wrote.
The given equation is quadratic in terms of $x$ and it has 2 roots, that is, there are two values of $x$ that make that quadratic expression equal to zero. The question refers to those two values, and their product. In this sense, at the beginning of your solution you should consider $x$ as a variable and $y$ as just a number, or a parameter.

In general, the product of the roots of the equation $ax^2+bx+c=0$ equals $\frac{c}{a}$. You can verify that by multiplying the roots $\frac{-b+\sqrt{b^2 - 4ac}}{2a}$ and $\frac{-b-\sqrt{b^2 - 4ac}}{2a}$ .

With this consideration, you can see that the text of the problem gives an additional piece of information about $y$. Then, you can consider $y$ as an unknown variable that should satisfy that additional piece of information. The calculations are as shown by Derik.

Title: Why am I getting wrong answer for this?


问:
Interesting question about functionals in measure theory, any suggestion please.

Let $(X,A,\mu)$ be a measurable space. $1<p,q<\infty$, $\frac{1}{p}+\frac{1}{q}=1$.
$Z\subset L^p$ a subspace of $L^p$ .Let $\phi: Z\to C$ a linear functional.
Assume that there's $m<\infty$ such that for all $f\in Z$ such that $||f||_p<1$ : $|\phi(f)|\leq m$.
Show that there's $g_0\in L^q$ such as $||g_0||\leq m$ and $\phi(f)=\int_X fg_0 d\mu$ for all $f\in Z$.

I tried to use the dual theorem.
What i did:
Let $g_0\in L^q$ (under the given conditions)
Define $\phi: Z\to C$ as:

For $f\in L^p$ , $\phi(f)=\int_X fg_0 d\mu$.
It is clear that $\phi$ is linear sincd the  integral is linear.(and $\phi \in (L^p)^*$).

$||\phi(f)||_C=|\int_X fg_0 d\mu|\leq \int_X |fg_0|d\mu=||fg_0||_1\leq ||f||_p||g_0||_q$.
Here we used Holder inequality.

Definition of dual norm gives:

$||\phi(f)||=sup |\phi(f)|_{||f||\leq 1}\leq m$

So in 1 we get by using 2 that:

$||\phi(f)||_C=|\int_X fg d\mu|\leq \int_X |fg|d\mu=||fg||_1\leq ||f||_p||g||_q \leq 1*||g_0||_q \leq m$.

Unfortunately, this does not show that there is $g_0$ that satisfy the given conditions.

答:
If $\phi$ is a linear functional $Z\to\Bbb C$ so that $|\phi(f)|≤m$ whenever $\|f\|<1$ then $\phi$ is necessarily continuous and the operator norm of $\phi$ is $≤m$. You may then apply Hahn-Banach to extend $\phi$ to a linear functional $\tilde \phi: L^p(X)\to\Bbb C$, the Hahn Banach extension will also satisfy $\|\tilde \phi\|≤m$.

Now if $p\in [1,\infty)$ then the dual of $L^p(X)$ is equal to $L^q(X)$ where $q$ is so that $\frac1p+\frac1q=1$. Hence you may identify $\tilde\phi$ with an element $g_0\in L^q(X)$. The dual action of $L^q(X)$ on $L^p(X)$ is given by integrating, for $f\in L^p(X)$ and $g\in L^q(X)$ you have:
$$g(f) := \int_X g(x)\cdot f(x)\ d\mu(x)$$
Hence you find for all $f\in L^p(X)$ that
$$\tilde \phi(f) = \int_X g_0(x)\cdot f(x)\ d\mu(x)$$
in particular since $\tilde\phi\lvert_{Z}=\phi$ you have that  for all $f\in Z$ you've got:
$$\phi(f) = \int_X g_0(x)\cdot f(x)\ d\mu(x)$$
One more statement about the duality $L^p(X)^*\cong L^q(X)$ needs to be referenced, namely that the norms
$$\|g\|_{op}:= \sup_{f\in L^p(X), \ \|f\|≤1} |g(f)|\quad \text{ and }\quad\|g\|_q := \sqrt[q]{\int_X |g(x)|^q\ d\mu(x)}$$
are equal. Hence $\|g_0\|_q =\|\tilde\phi\|_{op} ≤ m$.

The statements about the duality $L^p(X)^* = L^q(X)$ can be found eg on wikipedia or also on this site (here is something that came up in a search: Duality of $L^p$ spaces)


问:
a few questions about $L_2$ and Fourier transform
I need to proove or refute these claims.

if $f ∈ L_2[-π,π]$ is integrable then $f$ is Piecewise Continuous

Fourier transform is well defined for every $f ∈ L_2[-π,π]$

if $f:R\rightarrow F$ is Piecewise smooth on $R$ then  $f ∈ L_1[R]$

I think 1 is true but I don't know how to show that.
about the rest I am just not sure :(
Any help will be great

答:
No.  Neither Riemann integrable nor Lebesgue integrable functions need to be piecewise continuous.  See the Thomae function.

Yes. $L_2[-\pi,\pi] \subset L_1[-\pi,\pi]$, and the Fourier transform is defined for any element of $L_1$.  (Note: $L_2(\mathbb R) \not\subset L_1(\mathbb R)$, so this argument does not work there.  As mathreader said, the Plancherel transform extends the Fourier transform to $L_2(\mathbb R)$.)

Proof for $L_2[-\pi,\pi] \subset L_1[-\pi,\pi]$.  Let $f \in L_2[-\pi,\pi]$.
Note $|f(x)| \le |f(x)|^2 + 1$, which we can see by considering the two cases $|f(x)|\le 1$ and $|f(x)| > 1$.  So
$$
\int_{-\pi}^{\pi} |f(x)|\;dx \le \int_{-\pi}^{\pi} |f(x)|^2\;dx + 2\pi
< +\infty
$$
and $f \in L_1[-\pi,\pi]$

No. $f(x) = 2+\sin x$ is smooth on $\mathbb R$, but $f$ is not in $L_1(\mathbb R)$.


问:
Constrained optimization problem with multiple variables.
I am looking for the solution to a constrained maximization problem,

\begin{align}
f(w_1,w_2,\dots,w_n) = &\prod_{i=1}^{m}(C_{i1}w_1+C_{i2}w_2+\dots+C_{in}w_n)\notag\\
\text{such that}\;\; &\sum_{j=1}^{n}w_j = 1\notag
\end{align}

where $m, n, C_{ij}$ are all constants.

I want to find the $w_j (j=1,\dots,n)$ to maximize $f$.

Now I have 2 thoughts:

If $f$ is convex, use gradient descent. I'm not sure if it's possible with the constraint on $w$.

$f$ looks similar to the likelihood function for maximum likelihood estimation. Not sure if it's possible to use related algorithm like EM. I'm struggling to connect $f$ to it.

If both ways don't work, is there any other method to solve it?

答:
According to the Lagrange multipliers technique, calling

$$P(\omega) = \Pi_{i=1}^m\sum_{j=1}^n c_{ij}\omega_j$$

and forming the lagrangian

$$
L(\omega,\lambda) = P(\omega)+\lambda \left(\sum_{j=1}^n \omega_j-1\right)
$$

the stationary points are the solutions to

$$
\cases{
\displaystyle\sum_{k=1}^m\frac{c_{k \nu}}{\sum_{j=1}^n c_{k j}\omega_j}-\mu=0\\
\displaystyle\sum_{j=1}^n \omega_j-1=0
}
$$

with the new multiplier $\mu = \frac{\lambda}{P(\omega)}$


问:
Find the minimum value of 6a.

If $4\sin^2x + \csc^2x , a , \sin^2 y+ 4 \csc^2 y$ are in arithmetic progression, then find the minimum value of $6a$.

My Work:

Well just at a glance this problem seems to be related to AM-GM inequality. However whenever we apply am-gm we get the minimum value of $4\sin^2 x+\csc^2x$ and $\sin^2 y+4 \csc^2y$; but we do not get anything related to the arithmetic progression or anything related to $a$. So I need help as to any other way to attack this problem.

答:
$a$ has to be the arithmetic mean of the terms before and after it so we can apply AM-GM:

\begin{align}
6a &= 3\left(4\sin^2 x + \frac1{\sin^2 x} + \sin^2 y + \frac4{\sin^2 y}\right)
\\
&= 12\left(\sin^2 x+\frac1{\sin^2 y}\right) + 3\left(\sin^2 y+\frac1{\sin^2 x}\right)\\
&\ge 24\sqrt{\frac{\sin^2 x}{\sin^2 y}} + 6\sqrt{\frac{\sin^2 y}{\sin^2 x}}\\
&\ge 2\sqrt{24\sqrt{\frac{\sin^2 x}{\sin^2 y}} \cdot 6\sqrt{\frac{\sin^2 y}{\sin^2 x}}}\\
&= 2\sqrt{24 \cdot 6}\\
&= 24
\end{align}
with equalities for $\sin^2 x = \sin^2 y = 1$ so $a = 4$ is indeed the minimum.

Alternatively, you can bound $$a  =\frac{4\sin^2 x + \frac1{\sin^2 x} + \sin^2 y + \frac4{\sin^2 y}}2 \ge \sqrt{4\sin^2 x \cdot \frac1{\sin^2 x}} + \sqrt{\sin^2 y \cdot \frac4{\sin^2 y}} = 2+2 = 4$$
but it is not immediately visible how to attain the bound.

Title: Find the minimum value of 6a.


问:
Counting question and cardinality
A football league contains $6$ teams. During the season each team plays two matches against each other team. The result of each match is a draw or a win for one or other team. How many matches are played in the season? How many possibilities are there for full list of results for all matches played in the season?

My answer - total matches played is $\dfrac{6\times 10}{2}= 30$ or

How would I also calculate the different possibilities? Is it $6 \times 5$?

答:
Each team plays 5 other teams twice each (one home, one away). So each team plays 10 matches. So there are 30 matches played in total (not 60, because otherwise you will double count each match).

The three possibilities for each match are a home win, an away win, or a draw.

The number of total possibilities for the full list of results $3^{30}.$

Title: Counting question and cardinality


问:
Given $G = \mathbb{Z}_6 \times\mathbb{Z}_6$ with a particular operation, find $Z(G)$ and $G/Z(G)$
Given the group $G = \mathbb{Z}_6 \times\mathbb{Z}_6$ and the following operation:
$$(a,b)\cdot (c,d)=(a+(-1)^bc,b+d)$$
find (or classify) $Z(G)$ and $G/Z(G)$.

For the first part, I found that $(a,b) \in Z(G) \Leftrightarrow(a,b)\cdot(c,d)=(c,d) \cdot (a,b) \text{ for every }(c,d) \in G \Leftrightarrow a(1+(-1)^{d+1})=c(1+(-1)^{b+1}) \Rightarrow Z(G)=\langle(0,2)\rangle \simeq \mathbb{Z}_3$.

I'm having troubles on the second part though; I know that $|G/Z(G)|=12=2^23$, and I also know that there's either a normal $2$-Sylow or a normal $3$-Sylow, but at this point I'm stuck. Some hints on how to proceed?

答:
$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$The two subsets $A = \Set{(x, 0) : x \in \Z_{6}}$ and $B = \Set{(0, y) : y \in \Z_{6}}$ are subgroups, and $B$ normalises $A$, as
$$\begin{align}
(0, 1)^{-1} (1, 0) (0, 1)
&=
(0, -1) (1, 0) (0, 1)\\
&=
(-1, -1) (0, 1)\\
&=
(-1, 0).
\end{align}$$
So $G$ is a semidirect product of $A$ by $B$, with $b = (0, 1)$ inducing the inversion automorphism on $A$. It follows that $Z(G) = \Span{(3, 0), (0, 2)}$ has order $6$. A Sylow $3$-subgroup is normal, while Sylow $2$-subgroups are not, as $a = (1, 0)$ conjugates $\Set{(0, 0), (3, 0), (0, 3), (3, 3)}$ to $\Set{(0, 0), (3, 0), (1, 3), (4, 3)}$.


问:
Right triangles with integer sides
Most of you know these triples:

$3: 4 :5$

$5: 12 :13$

$8: 15 :17$

$7: 24 :25$

$9: 40 :41$

More generally we can construct such triangles such as
$$2x:x^2-1:x^2+1$$

My question is why one of the sides seems to be always prime? (When there is no common divisor)

答: The pattern you see (one side always having prime length) fails for larger $x$ and is a good example of the strong law of small numbers. Consider, for example, $x=12$.</p>

Title: Right triangles with integer sides


问:
Characterisation of surjectivity of a linear application in finite dimension with open sets

Let's $f\in \mathcal{L}(\mathbb{R^n},\mathbb{R^p})$.
Prove that $f$ is surjective if and only if the image of each open set is an open set.

My attempt :

The second implication is easy. If the image of each open set is an open set, it's especially true for $\text{Im} f$, and so it's $\mathbb{R}^p$ because the only open subvectorial space that is not empty is the vectorial space (because each open ball centred on $0$ contains a base of $\mathbb{R}^n$).

I've got problems proving the direct implication. Here is what I tried :

If $U$ is an open set, $y\in U$, $x$ such that $f(x)=y$, then the only thing I noticed is for all $r$, $f(B(x,r)) = f(x+rB)=y+rf(B)$ where $B$ is $B(0,1)$, so that I just need to show $f(B)$ is a neighbourhood of $0$ to conclude.

Can someone help me?

Thanks in advance.

答:
Thanks to @user3482749 's hint, I can answer my own question :

$f(B)$ contains a basis of $\mathbb{R}^n$. Let's not $B=(e_1,...,e_n)$. We can suppose without loosing generality that $(f(e_1),...f(e_p))$ is a basis of $\mathbb{R}^p$. Then $B$ contains all $\lambda_1 f(e_1) + \cdot\cdot\cdot \lambda_p f(e_p)$, with the $\lambda_i\in ]-1,1[$ which is clearly a neighbourhood of $0$.


问:
Expected values for a sum of Bernoulli random variables
For a Bernoulli indicator, why is $$(\sum \sum)_ {i \in \{1,..n\} j\in \{1,..,n \}/i}E(X_iX_j)=n(n-1)\frac{1}{n(n-1)} ?$$ I've seen it here: matching problem - find variance using indicator variables

As an example: $$E(X_1X_2)+E(X_1X_3)+E(X_2X_1)+E(X_2X_3)+E(X_3X_1)+E(X_3X_2)=6\frac{1}{3}=2$$, but  this is not correct

答:
In the link, it is claimed that $$\sum_{i\neq j}\mathbb E[X_iX_j]=1,$$ where the $X_i$ are Bernoulli with parameter $1/n$. This result though is specific to the way that the $X_i$ are defined.

In the link (which essentially concerns fixed points of a permutation), when $i\neq j$ we have $\mathbb E[X_iX_j]=\frac{1}{n(n-1)}$. This is because $X_iX_j=1$ iff both $i$ and $j$ are fixed by the permutation; otherwise it is $0$. There are $(n-2)!$ permutations that fix both $i$ and $j$, and $n!$ permutations in total. So
$$\mathbb E[X_iX_j]=0+1\cdot\mathbb P(X_iX_j=1)=\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}.$$
The number of pairs $(i,j)$ with $i\neq j$ is precisely $n(n-1)$. So the sum evaluates to $1$.

This result though is false when the $X_i$ are pairwise independent (in the link the $X_i$ depend on each other). In that case,
$$\mathbb E[X_iX_j]=\mathbb E[X_i]\cdot\mathbb E[X_j]=\frac{1}{n^2}.$$
So the sum evaluates to
$$\sum_{i\neq j}\mathbb E[X_iX_j]=\frac{n(n-1)}{n^2}=\frac{n-1}{n}.$$


答:
$x, y$ are not that roots of the polynomials.  $x$ is just a variable that expesses the polynomial and $y$ is a fixed constant.

Okay, the person who gave the question should have pointed that out but that is the only interpretation that makes sense.

If $x$ and $y$ are both unfixed variables but then a "root" is an ordered pair $(c_x, c_y)$ that makes the statement true.  But talking of a product of ordered pairs doesn't make sense and even if it didn't the result would be a constant.  It couldn't be the negative of a variable $y$.  So to say the product of the roots is $-y$ means that $y$ is not a variable but some constant value.

Alternative if $x$ and $y$ are both fixed constants the question doesn't make sense.  You just have and equation involving number like $2(-2)^2 + 6(-2)+(\sqrt 2)^2 + 2= 0$.  It doesn't have any "roots".

So the only way to think of this is that we have a variable $x$ and $y$ is constant number, call it $c$.  and we have the equation $2x^2 + 6x + c^2 +2 = 0$.  That equation will have two roots.  Call them $x_1$ and $x_2$.

We are give $x_1x_2 = -c$ so what is $c$.

Well be the quadratic equation if $2x^2 +6x + c^2 + 2 = 0$ then the two roots are $x_1 = \frac {-6+ \sqrt{6^2 - 4*2*(c^2+2)}}{2*2}$ and $x_1= \frac {-6 - \sqrt{6^2-4*2*(c^2+2)}}{2*2}$.

So we need to solve $\frac {-6+ \sqrt{6^2 - 4*2*(c^2+2)}}{2*2} \frac {-6 - \sqrt{6^2-4*2*(c^2+2)}}{2*2}= -c$.

Can you do that?  Hint: notice that because $(m-n)(m+n) = m^2 - n^2$ we have

$\frac {-6+ \sqrt{6^2 - 4*2*(c^2+2)}}{2*2} \frac {-6 - \sqrt{6^2-4*2*(c^2+2)}}{2*2}=$

$\frac {(-6)^2 - \sqrt{6^2-4*2*(c^2+2)}^2}{2*2*2*2}=$

$\frac {6^2 - (6^2 -4*2*(c^2+2))}{16} = $

$\frac {4*2*(c^2+2)}{16} = \frac {c^2+2}2$

So you must solve $\frac {c^2 + 2}2 = -c$. So

$c^2 + 2 = -2c$ and $c^2 +2c + 2 = 0$ and so $c = \frac {-2 \pm \sqrt{2^2 - 4*2}}2=\frac {-2\pm \sqrt{-4}}2 = -1 \pm \sqrt {-1} = -1 \pm i$.

I do not get any real value.

Someone can double check my work.

Title: Why am I getting wrong answer for this?


问:
Is there any name or symbol for rational numbers with a bounded amount of decimals?
As the title says, is there any name or symbol for rational numbers with a bounded amount of decimals?

For example, for the set of rational numbers that are multiples of $0.01$ (with two digits after the point; or equivalently, the subset of rational numbers that can be written using 100 as denominator), is there any specific mathematical symbol or concise way to name that?

答:
Correct me if I'm wrong, but I think in your question you meant to use the phrase, finite number of digits rather than bounded amount of decimals.

As amWhy pointed out in the comments, a terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point.

As for your final paragraph, I would refer to that set as "The set of all numbers with exactly two digits after the decimal point".

Or, if you don't mind $5.670$ being in your set, seeing as it has the same value as $5.67$, then as JMoravitz pointed out, you could instead write this set more concisely as $\frac{1}{100} \mathbb{Z} = 0.01\mathbb{Z}.$ However, if you consider $5.670$ and $5.67$ to be different numbers because you care about accuracy for example in chemistry, or you really only want to allow numbers with exactly two numbers after the decimal point in your set for whatever reason you may have, then this definition will not do, because $5.670 \in \frac{1}{100} \mathbb{Z},$ but $5.670 \notin$ { all numbers with exactly two digits after the decimal point }.


问:
How can you prove $\mathbb Q$ is Archimedean directly from its axiomatization as the minimal ordered field?
The Archimedean property for $\mathbb R$ follows from the axioms of complete ordered field.

$\mathbb Q$ is often characterized by stating it to be the minimal ordered field (with $0\ne 1$). But this (or any other equivalent axiomatic characterization, which pins it down to an isomorphism like the Peano axioms pin down $\mathbb N$) is rarely used for proving the Archimedean property for $\mathbb Q$. What is used (like here) is that they are formal quotients of integers, which is just a model of $\mathbb Q$.

I insist on proving Archimedean property for $\mathbb Q$ directly from its axiomatic characterization since particular models might in addition contains junk theorems (like the common set theoretic model of $\mathbb N$ requires $2\subset 3$).

答:
Well, this depends on exactly what you mean by "the minimal ordered field".  Let's say that we axiomatically define $\mathbb{Q}$ as an ordered field which contains no proper subfields.

Now, given such an ordered field $\mathbb{Q}$, let $K\subseteq\mathbb{Q}$ be the set of elements that can be written as a quotient of two integers (where we identify integers with elements of $\mathbb{Q}$ by taking sums of copies of $1$ or $-1$).  Then it is easy to verify that $K$ is a subfield of $\mathbb{Q}$, and thus $K=\mathbb{Q}$.  So for any $x\in\mathbb{Q}$, there are integers $a$ and $b$ such that $x=\frac{a}{b}$.  Replacing $a$ and $b$ with their negatives if necessary, we may assume $b>0$.  Since $b$ is an integer, this implies $b\geq 1$ (since $b$ is a sum of copies of $1$ and $1>0$).  Thus $x=\frac{a}{b}\leq \frac{a}{1}=a$.  Since $x\in\mathbb{Q}$ was arbitrary and $a$ is an integer, this proves $\mathbb{Q}$ is Archimedean.


问:
Consider the quadratic integer ring $A=\Bbb Z[\sqrt-2]$. Show that $U(A)=\{-1,1\}$.
Consider the quadratic integer ring $A=\Bbb Z[\sqrt-2]$. Show that $U(A)=\{-1,1\}.$

$\bf{Remark:}$ $U(A)$ indicates the set of inversible elements of $A$.

I know that $\alpha = a + b\sqrt{-2}$ is inversible iff $N(\alpha)=a^2+2b^2=1$.

but I don’t know how to go on. Can someone help me?

答:
$a^2+2b^2=1 \Rightarrow 2b^2=1-a^2=(1-a)(1+a)$

$2b^2 \geq 0 \Rightarrow -1 \leq a \leq 1$

but we know the fact that $a$ is an integer so

$ -1 \leq a \leq 1 $ and $a \in \Bbb Z \Rightarrow $ $a=-1$ or  $a=1$ or $a=0$

let $a=0$ then

$2b^2=1$  which does not have an integer solution so $a \neq 0$

$a=1$ , $b=0$

$a=-1$ , $b=0$ are only solutions


问:
Is there a value of $\alpha$ such that $2\tan\alpha=-\sqrt{2}\sin\alpha?$
Is there a value of $\alpha$ such that $2\tan\alpha=-\sqrt{2}\sin\alpha?$

We have $2\tan\alpha=-\sqrt{2}\sin\alpha \iff 2\cdot\dfrac{\sin\alpha}{\cos\alpha}=-\sqrt{2}\sin\alpha.$ If we divide by $\sin\alpha$, we'll get $2=-\sqrt{2}\cos\alpha,$ or $\cos\alpha=-\dfrac{2}{\sqrt2}=-\sqrt{2}$ which isn't in the inverval $\left[-1;+1\right].$ So there isn't a value of $\alpha$ such that  $2\tan\alpha=-\sqrt{2}\sin\alpha$. Are we allowed to divide by $\sin\alpha$ and should I see what's going on when $\sin\alpha=0?$

答: You cannot divide by $\sin\alpha$ if $\sin\alpha=0$. Any solution of $\sin\alpha=0$ is a solution of $2\tan\alpha=-\sqrt{2}\sin\alpha$, since both sides are then zero.


问:
Method to discover linear independence of binary code words
How to check in the easiest way possible the linear independence of these code words?

They are supposed to be linear independent, but I cannot find out why...

```<code>1st) 10101 00101 100
2nd) 00111 01100 101
3rd) 01001 10101 111
4th) 10011 01011 101
5th) 01100 11100 100
</code>```

I've been told that, just by looking and the ones on each of them, I can discover quickly that they are linear independent, but how?

NB: There are similar questions but none of them address the down-to-earth method doable by a human to check linear independence, so, as far as I know, this is not duplicated

答:
So, suppose you had a dependence, $$\sum_{i=1}^5a_i\vec {v_i}$$

We want to prove that all the $a_i$ are $0$.

Looking at the second to last column, we see that $a_3=0$.

Looking at the sixth column shows that $$a_3+a_5=0\implies a_5=0$$

Looking at the first column we see that $$a_1+a_4=0$$

Looking at the last column we see that $$a_2+a_4=0\implies a_2=a_1=-a_4$$

Looking at the third column tells us that $$a_1+a_2+a_5=0\implies a_1+a_2=0\implies a_1=a_2=-a_4=0$$ and we are done.