问:
Prove that the map $A \mapsto (B \mapsto \operatorname{trace}(AB))$ is an isomorphism $\operatorname{Mat}_{n,m}(K) \to \operatorname{Mat}_{m,n}(K)^*$
For $A \in \operatorname{Mat}_{n,m}(K)$ and $B \in \operatorname{Mat}_{m,n}(K)$, let $\tau_A(B) := \operatorname{trace}(AB)$. I want to show that the map $\operatorname{Mat}_{n,m}(K) \to \operatorname{Mat}_{m,n}(K)^*, A \mapsto \tau_A$ is an isomorphism.

I already showed that $\tau_A: \operatorname{Mat}_{m,n}(K) \rightarrow K$ is linear, maybe I can do something with this information.

答:
For the linearity of the map $A\mapsto \tau_A$ just work through the definitions. It is similar to the linearity of $\tau_A$.

Let us show that the given map is injective. So let $A\in K^{n\times m}$ with $\tau_A=0$, i.e. $\operatorname{Tr}(AB)=0$ for all $B\in K^{m\times n}$. We want to show that $A=0$. For $i=1,\dots,m,j=1,\dots,n$ define the matrix $E_{ij}\in K^{m\times n}$ so that $E_{ij}$ has only zero entries except in the $i$-$j$-index where it has a $1$. Verify that $$\operatorname{Tr}(AE_{ij})$$ is just the entry of $A$ at the position $j,i$ (notice the indices are swapped). Conclude that $A=0$.

So we proved the injectivity of the map. Now argue via a dimension argument why this map is then already an isomorphism.


问:
Proving an equality involving a product $\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$
How can I prove that:

$$\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$$

When $n\in\mathbb{N}$ such that $n\ge2$.

I think that we need to prove that all terms are monotonically increasing functions. But how do I prove that?

ContentLicense: CC BY-SA 4.0

Body:
Note that

$$n\geq 2 \Rightarrow n^2-1 \geq 3$$

Hence,

\begin{eqnarray*} \frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}
& \geq & \frac{(3n^2-3)(3n^2-3)(3n^3-3n)}{2} \\
& = & \frac{3^3(n^2-1)^3 n}{2} \\
& \stackrel{n^2-1\geq 3}{\geq} & 3^6 \\
& \geq  & 1
\end{eqnarray*}


问:
Confirmation on Proofs
Q1: Suppose $(x_n)$ satisfies $-1$ $\leq$ $x_n$ $\leq$ $1$ for all $n$ $\in$ $\Bbb{N}$. Suppose $(x_{n_k})$ is a subsequence of $(x_n)$ converging to $(x_0$) $\in$ $\Bbb{R}$. Prove that it is always true that $-1$ $\leq$ $x_0$ $\leq$ $1$

Just for confirmation, is this true because we know (x_n) is bounded, thus by Bolzano-Weierstrass Theorem (x_n) has a monotone subsequence $(x_{n_k})$ and this monotone sequence is either non-increasing or non-decreasing. If it is non-increasing then the sequence $x_0$ is equal to the infimum of the set, and if it is non-decreasing then $x_0$ is equal to the supremum of the set?

Q2: Suppose $(x_n)$ satisfies $-1$ $\leq$ $x_n$ $\lt$ $1$ for all $n$ $\in$ $\Bbb{N}$. Suppose $(x_{n_k})$ is a subsequence of $(x_n)$ converging to $(x_0$) $\in$ $\Bbb{R}$. Prove that it is always true that $-1$ $\leq$ $x_0$ $\leq$ $1$.

How can this be the case?

Q3 Is the statement $\alpha$ $\in$ $\Bbb{R}$ and x $\in$ $\Bbb{R^d}$, then $\Vert{\alpha x}$$\Vert$ $=$ $\alpha$ $\Vert{x}$$\Vert$ true or false?:

Why must the RHS of the equation have $\vert{\alpha}$$\vert$ and not just $\alpha$

Q4: Let $x$, $y$ $\in$ $\Bbb{R^d}$ Prove using the triangle inequality $\Vert{x-y}$$\Vert$ + $\Vert{x - 2y}$$\Vert$ $\geq$ $\Vert{y}$$\Vert$.

Not sure how to prove this?

答:
Q1 and Q2 appear to be the same. They are a consequence of the following fact:

Lemma. If $x_n \leq M, \forall n \in \mathbb{N}$ for some real $M$ and $(x_n)$ converges to a limit $l$, then $l \leq M$.

Proof. Suppose for a contradiction that $l > M$. Then $\epsilon = \frac{l-M}{2}>0$. Since $(x_n)$ converges to $l$, there is an $N > 0$ such that
$$\mid x_n-l\mid < \epsilon, \forall n > N.$$

This implies that $x_n > l - \epsilon = \frac{l+M}{2} > M$, which is false. Hence, we need $l \leq M$.

We get a similar result for a lower bound.

Q3. The LHS is always positive so we need an absolute value on the RHS to have it always be positive.

Q4. $\Vert y \Vert = \Vert (x-y) - (x-2y) \Vert$. Now apply Triangle Inequality.

Title: Confirmation on Proofs


问:
Solving Second Order Non-Homogeneous ODE
I haven't solved an ODE in quite some time and I'm finding myself a bit rusty! If someone could help me solve the following ODE that would be great:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$

I attempted using the method of undetermined coefficients and found the complementary solution as follows:
$$r^2 + \frac{1}{c_1}r = 0 \implies r = 0 \text{ or } r = -\frac{1}{c_1} \implies x_C(t) = A + Be^{-\frac{1}{c_1}t} \text{ , } A,B \in \Bbb{R}$$
From there I attempted to find the particular solution with a guess of $x(t)= C \in \Bbb{R}$ which would have resulted in $c_2 = 0$ which I don't believe to be right.

Any help would be great!

答:
$$
x_h(t)=A+B\mathrm{e}^{-t/c_1}
$$
is the general solution of the homogeneous equation
$$
x''+c_1x'=0.
$$
In order to find the general solution of the inhomogeneous one
$$
x''+c_1x'=c_2, \tag{1}
$$
you need first to find a particular solution of it. One such is
$$
x_p(t)=(c_2/c_1)t.
$$
Then the general solution of (1) is
$$
x(t)=x_h(t)+x_p(t)=A+B\mathrm{e}^{-t/c_1}+(c_2/c_1)t.
$$

Note. In general, if $x_h=c_1\varphi_1+\cdots+c_n\varphi_n$ is the general solution of
$$
x^{(n)}+a_{n-1}(t)x^{(n-1)}+\cdots+a_0(t)x^{(0)}=0,
$$
and $x_p$ is a particular solution of
$$
x^{(n)}+a_{n-1}(t)x^{(n-1)}+\cdots+a_0(t)x^{(0)}=h(t), \tag{2},
$$
then $x=x_h+x_p$ is the general solution of (2).


问:
Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$
The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$ as $x \rightarrow + \infty$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.

So, what are available ways to solve this problem?

答: Hint: let $f(x):=\arctan x-\tfrac{x}{1+2x/\pi}$, whose unique maximum on $x>0$ you're welcome to locate. For $0<x\ll1$, $f(x)\sim(\tfrac13+\tfrac{2}{\pi})x^3$; for $x\gg1$,$$f(x)=\tfrac{\pi}{2}-\arctan\tfrac1x-\tfrac{\pi/2}{1+\pi/(2x)}\sim\tfrac{\pi^2-4}{4x}.$$Now sketch $f$.


问:
How to evenly scale down measurements in a recipe to meet certain weight by gram?
Very new to this.

I have a recipe that has 9 ingredients, with varying weights that produces a total of 191grams.

For example:

```<code>ingredient   weight(grams)
a              15
b              10
c              5
d              5
e              145
f              8
g              1
h              1
i              1
</code>```

I need to scale down the weights of all these ingredients, evenly, to 135 grams.

How do I calculate a ratio for all the ingredients that evenly scales down?

答: Multiply each weight by $135/191.$

Body: Multiply each weight by $135/191.$


问:
Primitive roots of a finite field
I understand the definition of a primitive root of integers; however, I am quite confused trying to find the primitive roots of $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$. I know that $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$ has elements of the form $ax^2+bx+c$, with $a,b,c \in \mathbb{Z}/3\mathbb{Z}[x]$, so it has $27$ elements, and $\mathbb{F}_{27} \simeq \frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$.

Also, it is not clear to me if I must find an element of order 26 or an element of order $\varphi(27) = 18$, where $\varphi$ is the Euler Totient Function.

Could someone give me a tip?

答:
As Leoli1 commented, you have to find an element of order $26$,

since the multiplicative group of $\mathbb F_{27}$ has order $26$.

You may consider this brute force, but using $x^3\equiv x-1$ we have

$x^4\equiv x^2-x, $ $x^5\equiv -x^2+x-1$, $x^6\equiv x^2+x+1$, $x^7\equiv x^2+2x+2$, $x^8\equiv2x^2-1$,

$x^9\equiv x+1, x^{10}\equiv x^2+x$, $x^{11}\equiv x^2+x-1$, $x^{12}\equiv x^2-1$, and $x^{13}\equiv-1$.

It follows that $x$ is a primitive root.

(I could have taken a short cut:  $x^{12}\equiv (x^4)^3\equiv (x^2-x)^3\equiv(x^2)^3-x^3= x^6-x^3$

$=x^3(x^3-1)\equiv (x-1)(x-2)\equiv (x-1)(x+1)=x^2-1$.)

Title: Primitive roots of a finite field


问:
Value of $i^\sqrt3$

Find all values of $i^\sqrt3$.

I am trying to apply de Moivre's formula here but cannot find a way to do so. I am not sure if i am approaching this wrong.

答:
Consider: from deMoivre's

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

with

$\theta = 2n \pi + \dfrac{\pi}{2}, \; n \in \Bbb Z, \tag 2$

we have

$e^{i(2n\pi + \pi/2)} = i; \tag 3$

thus

$i^{\sqrt 3} = e^{i(2n\pi + \pi/2)\sqrt 3}$
$= \cos ((2n\pi + \pi/2)\sqrt 3) + i \sin ((2n\pi + \pi/2)\sqrt 3), \ n \in \Bbb Z. \tag 4$

We may check for consistency: (4) yields

$-i = i^3 = (i^{\sqrt 3})^{\sqrt 3} $
$= (e^{i(2n\pi + \pi/2)\sqrt 3})^{\sqrt 3} = e^{i(2n\pi + \pi/2)(3)}$
$= e^{6\pi i}e^{3\pi i /2} = -i. ✓ \tag 5$

Title: Value of $i^\sqrt3$


问:
Hyperbolic half plane and the fith postulate
I have just done some fairly common exercises around geodesics in the hyperbolic half-plane. In particular, I showed that vertical lines and semicircles form geodesics.

If I take a vertical line geodesic, and a point not on it and sufficiently far from the $x$-Axis (to prevent semicircle-geodesics from reaching it), then there is exactly one other geodesic, the vertical line through that point, that doesn't intersect the original geodesic.

Doesn't this violate the fifth postulate that generates hyperbolic geometry from absolute geometry? Don't I need to be able to find two or more geodesics?

Semicircle-geodesics don't have this problem, there are infinitely many of those.

答:
There are other geodesics. Suppose the vertical line geodesic is given by $x = a$ and the point by $(b, c)$, with $b \ne a$. Choose $d \ne b$. Then there is a geodesic semicircle with endpoint $(d, 0)$ and passing through $(b, c).$ To find it, note that if the center of the semicircle is $(e, 0)$, then we must have
$$(e - d)^2 = (e - b)^2 + c^2.$$
Solving this equation, we obtain
$$e = \frac{b^2 + c^2 - d^2}{2(b - d)}.$$
Now, if we further suppose that $d$ is between $a$ and $b$, then the semicircle will not intersect the vertical line.


答:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$
For the particular solution try:
$$x_p'=A \implies A=c_2c_1$$
$$ \implies x_p=c_1c_2t+K$$
You can choose $K=0$. Note that you can integrate directly the DE:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$
$$\frac{dx}{dt} + \frac{1}{c_1} x(t) = c_2t+K$$
Then it's just a first order differential equation that is easy to solve.

Title: Solving Second Order Non-Homogeneous ODE

Body:
I haven't solved an ODE in quite some time and I'm finding myself a bit rusty! If someone could help me solve the following ODE that would be great:
$$\frac{d^2x}{dt^2} + \frac{1}{c_1} \frac{dx}{dt} = c_2$$

I attempted using the method of undetermined coefficients and found the complementary solution as follows:
$$r^2 + \frac{1}{c_1}r = 0 \implies r = 0 \text{ or } r = -\frac{1}{c_1} \implies x_C(t) = A + Be^{-\frac{1}{c_1}t} \text{ , } A,B \in \Bbb{R}$$
From there I attempted to find the particular solution with a guess of $x(t)= C \in \Bbb{R}$ which would have resulted in $c_2 = 0$ which I don't believe to be right.

Any help would be great!


问:
Basic probability questions
How is P(A , B) different from P(A $\cap$ B)? I'm genuinely curious as to why one might prefer one over the other.

Also as far as some probability P (A | B , C) goes, what is the order of operations i.e., is it P (A | (B , C)) or P ((A | B) , C)? Or are they both similar?

答:
Note that  $P(A,B)$ is equal to $P(A\cap B)$. </p>

<p>P((A|B), C) is meaningless. </p>

ContentLicense: CC BY-SA 3.0

Title: Basic probability questions

Body:
Note that  $P(A,B)$ is equal to $P(A\cap B)$.

P((A|B), C) is meaningless.


问:
Intersection of Ker of two linear maps

Let $T,S: \mathbb{R}^n\to \mathbb{R}$ be two linear maps with:

\begin{equation} T(e_1)=1,T(e_2)=0,  S(e_1)=0, S(e_2)=1.\end{equation}
I have to compute $\dim(\ker T \cap \ker S)$.

Let $v \in \ker T \cap \ker S$ with $v=a_1e_1+...+a_ne_n$. So if $v \in \ker T$ we have $a_1T(e_1)+a_2T(e_2)+...+a_nT(e_n)=0$ and if $v \in \ker S$ we have $a_1S(e_1)+a_2S(e_2)+...+a_nS(e_n)=0$. So $a_1+a_3T(e_3)+...+a_nT(e_n)=0$ and $a_2+a_3S(e_3)+...+a_nS(e_n)=0.$ How could I conclude?

答:
Perhaps consider the map $A:\mathbb{R}^n \to \mathbb{R}^2$ given by $Ax = (Tx, Sx)$. Then $x \in \ker A$ iff $s \in \ker T \cap \ker S$.

Then we have $n = \dim {\cal R A} + \dim \ker A = 2 +  \dim \ker A$ and
so
$ \dim \ker A = \dim (\ker T \cap \ker S) = n-2$.

Title: Intersection of Ker of two linear maps


问:
Is multiplication $\sigma$-weakly continuous on bounded subsets?
Let $M \subseteq B(H)$ be a von Neumann algebra. Let $S\subseteq M$ be a bounded subset. Is the multiplication map
$$S \times S \to M: (x,y) \mapsto xy$$
$\sigma$-weakly (= weak$^*$) continuous?

Attempt: Assume $x_i \to x$ and $y_i \to y$ in the $\sigma$-weak topology where $(x_i)_i$ and $(y_i)_i$ are bounded nets. We must show that
$x_iy_i \to xy$ in the $\sigma$-weak topology. Let $z \in L^1(H)$ be a trace class operator. We need to show that
$$|Tr(z(xy-x_iy_i)| \to 0$$

I tried
$$|Tr(zxy-x_iy_i)| = |Tr(zxy-zxy_i + zxy_i - zx_i y_i)|$$
$$\le |Tr(z(xy-xy_i)| + |Tr(z(xy_i-x_iy_i))|$$

Now I would like to use something like
$$|Tr(zx)| \leq |Tr(z)|\|x\|$$
but I'm not sure that holds.

答:
As written the inequality cannot hold (consider the case $x=z^*$, $\operatorname{Tr}(z)=0$). But, with $|z|$, it is standard (it's Hölder for $1,\infty$, if you think about it).

Let $z=v|z|$ be the Polar Decomposition. Using Cauchy-Schwarz,
\begin{align}
|\operatorname{Tr}(zx)|
&=|\operatorname{Tr}(|z|^{1/2}\,(x|z|^{1/2}v))|
\leq\operatorname{Tr}(|z|)^{1/2}\,\operatorname{Tr}(v^*|z|^{1/2}x^*x|z|^{1/2}v)^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(v^*|z|v)^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(|z|^{1/2}v^*v|z|^{1/2})^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)\,\|x\|
\end{align}


问:
On the Possibility of Godel Numbering
I've gotten to the point of the book Godel, Escher, Bach which really gets into demonstrating the mechanisms of Godel's incompleteness proof. I've had to go back and forth through it a few times, and have been bouncing it off of Cantor's diagonal proof, as the author intends. But I think I've reached a point that is unintended and not sure how to resolve it.

If any natural number is itself a well-formed expression (right? [1]), then there are at least as many well-formed expressions as there are natural numbers. And clearly there are even more well-formed expressions than that, since for example, each $i + 1$ is also a well formed expression.

So piggy backing off the diagonal idea, it seems similarly that the number of well-formed expressions in a formal system is greater than the number of natural numbers. And wouldn't that mean that Godel-numbering itself is impossible in any given formal system?

It seems to be a similar point reached in each proof leads to different outcomes. Why doesn't Godel's proof similarly lead to the result that "Godel-numbering a formal system is impossible"?

Please excuse that I am not a mathematician and only have kind of a tenuous grip on both proofs, or on the idea of "distinct infinities".

[1] I'm not sure if an natural number by itself is a well-formed expression, but it doesn't matter, since each $i = i$ is clearly a well-formed expression.

答:
You must be very careful when arguing with infinities.  Consider this part of your argument:

If any natural number is itself a well-formed expression (right?), then there are at least as many well-formed expressions as there are natural numbers. And clearly there are even more well-formed expressions than that, since for example, each $i+1$ is also a well formed expression.

Everything you said is right, but we cannot conclude from this argument that the number of well-formed expressions is greater than the number of natural numbers.  For finite sets, your argument would be correct, but not in the case of infinite sets.  Remember that if $A$ and $B$ are sets, then the definition of $|A| < |B|$ (i.e., there are more elements in $B$ than $A$) is that there is an injective map from $A$ to $B$, but there is no map from $A$ to $B$ which is both injective and surjective.  (Please look up the definitions of those terms if you are not familiar with them.)

Now let's state your argument formally.

Let $A$ be the set of natural numbers.

Let $B$ be the set of well-formed expressions.

Your argument states that there is an injective map $f : A \to B$, sending each natural number $n$ to a well-formed expression $n$.  And then you have stated that there are even more well-formed expressions which are not in the image of $f$, i.e., $f$ is not surjective. That is all correct.

If $A$ and $B$ were finite sets, this would be enough to conclude that $|A| < |B|$, but not in the case of infinite sets.  For in the infinite case, the existence of the injective map $f$ is not enough to preclude the existence of a different map $g: A \to B$ which is both injective and surjective.

To see an example of this, let $\mathbb{N} = \{0,1,2,\ldots\}$ be the natural numbers. Consider a map $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = n+1$.  Then $f$ is clearly injective, and $f$ is not surjective because 0 is not the image of $f$.

However, it is obviously not true that $|\mathbb{N}| < |\mathbb{N}|$, because there is another map $g : \mathbb{N} \to \mathbb{N}$ which is both injective and surjective, namely $g(n) = n$.

Title: On the Possibility of Godel Numbering


问:
How quickly can you mess up a domino tiling in 3D?
Suppose that we are trying to tile $\mathbb{Z}^3$ with dominoes, i.e., two face-adjacent cubes. We start going about this haphazardly, laying dominoes in random spots. How long can we keep this up before it becomes impossible to extend existing domino placements to a complete tiling?

Obviously, with $6$ tiles we can make it impossible to extend: surround every face of a single-cube hole that cannot be filled. I think I've got a proof that any $3$ dominoes can be extended to a tiling of space, but I don't know how to resolve the cases of $4$ or $5$ without extremely involved casework.

In two dimensions, it is relatively easy to show that any $3$ dominoes extend to a tiling of the plane, but $4$ dominoes obviously do not in general. This makes me weakly suspect that the answer is $6$ in three dimensions, and $2n$ in dimension $n$ more generally, but I don't trust my intuition to be particularly reliable here.

Edit: I can prove that any $4$ dominoes can be extended to a tiling of $\mathbb{Z}^3$, as follows:

Define a blob to be any domino or single cell. First, we prove that the complement of any three disjoint blobs in the plane can be tiled by dominoes (this is a bit of work; I found it helpful to first prove any one blob in a quadrant extends to a tiling of that quadrant, then that any two blobs extend to a tiling of the half-plane, then the planar case).

Then, take an arrangement of four dominoes. Choose any pair of them; there is at least one axis-aligned plane separating that pair. Then, look at the unit-width "slabs" parallel to this plane; each such slab contains at most three blobs (because a domino's intersection with a planar slab is either a domino, a single cell, or empty), and so can be tiled by dominoes. (In fact, this proof generalizes to four blobs in $\mathbb{Z}^3$, not just four dominoes.)

So only the case of $5$ dominoes remains.

答:
Consider a large 3d torus with even side lengths that contains the configuration of 5 starter dominoes. Colour the cells making up the torus white and black in an alternating 3d checkerboard pattern. (Each white cell has 6 black neighbours, and vice-versa. This is possible because the torus has even side lengths.)

Let $X$ be the set of black squares not covered by any of the 5 starter dominoes and $Y$ be the set of white squares (including the 5 covered by starter dominoes). Note that $|X|+5 = |Y|$. For every set $B\subset X$, define $N(B)$ to be the set of all neighbours of the elements of $B$. (So $y\in Y$ will be an element of $N(B)$ if and only if it is neighbours with some $x\in B$.)

According to Hall's marriage theorem, it will be possible to tile the torus with dominoes (and thus the whole plane, since the torus tiling can just be repeated) if the following condition is met: "For all sets $B\subset X$, $|N(B)| \geq |B| + 5$." i.e. each subset of $n$ black cells has at least $n+5$ white neighbours. (This is so that even if all 5 starter dominoes happen to be covering our neighbours, we still have $n$ neighbours left over to satisfy the Hall marriage criterion. This is a sufficient condition, but not a necessary one.)

This seems obviously true after a bit of thought, but I found it very difficult to come up with a proof. Here is a messy one with 3 separate cases. Maybe it's possible to come up with a better proof.

Proof:

Make the torus very large: at least $4\cdot 21600$ cells long in each direction. (It may need to be bigger in order to contain the full configuration of starter dominoes which could be spaced very far apart, but that's fine for our proof.)

We will show that $|N(B)| \geq |B|+5$ for each of 3 cases.

Case 1: $|B| < 216000$

Write down the $x$ coordinates of all elements of $B$ and all neighbours of $B$ (i.e. elements of $N(B)$). If the $x$ coordinates of the elements of $B$ are $x_1, x_2, \dots$, then the coordinates of the elements and their neighbours are $x_1, x_1-1, x_1+1, x_2, x_2-2, x_2+2, \dots$. We've just written down less than $3|B| < 3\cdot 216000$ unique $x$ coordinates. Since the side length of the torus is at least $4\cdot 216000$, we can find an $x$ coordinate that belongs to no elements of $B$ and no neighbour of $B$. This defines of plane of cells that all have that as their $x$ coordinate and are neither contained in $B$ nor neighbours of $B$.

Using the same method, find unused $y$ and $z$ coordinates, which also define planes that do not contain any elements of $B$ or neighbours of $B$. Remove these 3 planes from the torus, leaving behind a large block, without the wraparound structure of the torus.

Now construct the set $B$ in layers, starting from a single cell that has no other elements of $B$ below it. This single cell has $6$ white neighbours. $6 \geq 5+1$, so this will serve as our base case. Add the elements of $B$ one at a time, always making sure to complete 1 layer before moving on to the next. Each black cell we add to $B$ always adds at least 1 new white cell to $N(B)$, namely the cell just above it. So we always have $|N(B)| \geq 5+|B|$ at each step of the process.

Case 2: $216000 \leq |B| \leq |Y|-216000$

Divide the space up into $2\times 1\times 1$ blocks. (Do this so that the edges of the blocks are all lined up: We map $(x,y,z)$ to $(x, y, \lfloor z/2 \rfloor)$, and any two cells mapped to the same final coordinates belong to the same block.) We say a block is full if its black cell is an element of $B$, and empty otherwise. An empty cell may or may not contain a white neighbour of $B$, but all full cells contain one. We just need to show that there are at least 5 empty cells containing white neighbours of $B$. Since there are between 216000 and $|Y|-216000$ full blocks, the boundary between full blocks and empty blocks must contain at least 60 faces. (Faces are just the sides of the blocks. A face can have a surface area of 1 or 2. If we shrink the blocks by a factor of 2 in the $z$ direction, so that the blocks become cubes, we're claiming that any set of between 216000 and $|Y|-216000$ unit cubes in a torus of size at least $216000\times216000\times108000$ must have a surface area of at least 60. This really needs a proof of its own, but it seems likely enough to be true, so I'll leave it as an "exercise for the reader".)

Since each empty block that neighbours a full block can have at most 6 of these faces as neighbours, there must be at least 10 empty blocks with full blocks as neighbours. Any empty block with a full block to the north, south east, or west must have its white cell be a neighbour of $B$ (its white cell touches the black cells of all horizontally adjacent blocks). As for vertical neighbours: In some columns of blocks, the white cell in a block is below the black cell. In that case all empty blocks with a full block underneath will contain a white neighbour of $B$. In other columns, the white cell is above the black cell in each block, so all empty blocks with a full block above will contain a white neighbour of $B$. Note that in each column, there will be exactly as many empty blocks with full blocks above them as there are empty blocks with full blocks below them. (Because runs of empty and full blocks must alternate and the torus loops around.) Therefore, if there are $n$ empty blocks with full blocks as neighbours, at least $n/2$ empty blocks must contain white cells neighbouring $B$.

We now know that there are at least $10/2=5$ empty blocks containing white neighbours of $B$, which completes the proof for this case.

Case 3: $|Y|-216000 < |B|$

By the same argument as case 1, we can split the torus along 3 planes so that none of the planes contain any black cells not in $B$ or their neighbours. Then we start with the set of all black cells, and remove elements from it 1 at a time until we are left with $B$. As before, we proceed in layers. After removing 5 black cells, we have $|Y|$ neighbours, and $|Y|-5$ black cells. (We still have |Y| neighbours at this point because we started with all white cells being neighbours to the set of all black cells, and to get rid of a white cell as a neighbour, we'd need to remove all 6 black cells surrounding it, not just the 5 we have actually removed. Note that $|B| \leq |X| = |Y|-5$, so we always have to remove at least 5 black cells to get to $B$. So this will serve as our base case.

We continue removing cells 1 layer at a time. Each black cell we remove can only remove 1 neighbour when we take it away, namely the white cell directly below it, since to eliminate a white cell as a neighbour, we certainly must eliminate the black cell just above it first. And because we are proceeding in layers, the cell above will be the last of a white cell's neighbours to be removed. So the desired property is preserved with each cell we remove.

Q.E.D.

Commentary: This proof looks like it can be generalized to prove your conjecture in higher dimensions too.


问:
Properties of Continuous Functions
Prove that there is no continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for $c \in \mathbb{R}$ the equation $f(x)=c$ has exactly two solutions.
This is what I have so far.

Proof by contradiction, suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous and the equation $f(x)=c$ has exactly two solutions, $a$ and $b$ with $a<b$.  On the interval $[a,b], f(t)\leq c$ for all $t$, or $f(t) \geq c$ for all $t$.  Assume the latter.  Then $f(t)<c$ for $t<a$ and $f(t)<c$ for $t>b$.  $f$ has a maximum value on $[a,b]$ at exactly two points, $d$ and $g$, assume $d<g$.  Now choose any $t$ such that $d<t<g$.  Then $f(a)=c<f(t)<f(d)$ and $f(b)=c<f(t)<f(g)$.  So the equation $f(x)=f(t)$ has three solutions, $t, c_{1}, c_{2}$ where $a<c_{1}<d$ and $g<c_{2}<b$ by the intermediate value theorem.

答:
Assuming you mean, as Alex Becker suggested in the comments, that for any $c$ the equation $f(x)=c$ has exactly two solutions, then your proof is complete as you found three distinct solutions to one of those equations.

Moreover, you could also significantly shorten your proof by noting that $f$ is unbounded both from above and from below. After showing $f(t)<c$ whenever $tb$, we see that $f$ is unbounded in the closed interval $[a,b]$, which is a contradiction to the continuity of $f$.

Title: Properties of Continuous Functions

Body:
Assuming you mean, as Alex Becker suggested in the comments, that for any $c$ the equation $f(x)=c$ has exactly two solutions, then your proof is complete as you found three distinct solutions to one of those equations.

Moreover, you could also significantly shorten your proof by noting that $f$ is unbounded both from above and from below. After showing $f(t)<c$ whenever $t<a$ or $t>b$, we see that $f$ is unbounded in the closed interval $[a,b]$, which is a contradiction to the continuity of $f$.


问:
Functional of BM and time is martingale
We are confronted with an exercise that asks us to find a function $h(t)$ such that:
$$M_t = [B_t + h(t)] \cdot e^{-B_t - \frac{t}{2}}$$
is a martingale wrt to its natural filtration. After applying Ito's lemma using the function
$$f(t, x) = [x + h(t)] \cdot e^{-x - \frac{t}{2}}$$
we noted that if $h(t) = t$ we obtain a nice result:
$$[B_t + t] \cdot e^{-B_t - \frac{t}{2}} = \int_0^t(1 - B_s - s) \cdot e^{-B_s - \frac{s}{2}} dB_s$$
So our aim is to show that the stochastic integral is a Martingale. To do that, we observed ($\epsilon_t = e^{-B_t - \frac{t}{2}}$):

$E\left[ \int_0^t \epsilon_s^2 ds \right] < \infty$

$E\left[ \int_0^t s^2 \cdot \epsilon_s^2 ds \right] < \infty$

We are then stuck in showing $E\left[ \int_0^t B_s^2 \cdot \epsilon_s^2 ds \right] < \infty$

If we show 3. we get that the stochastic integral is sum of martingales, which will prove $M_t$ to be a martingale under $h(t) = t$. Are we on the right track? Any ideas to show 3.??

答:
To show (3) Exchange expectation and integration using Tonelli's Theorem. All that is left is to denote $s=\sigma^2$ and bound

$$ E[B_s^2 \cdot \epsilon_s^2] \le   \int_{-\infty}^{\infty} (\sigma x)^2 e^{-2\sigma x-\sigma^2} e^{-x^2/2} \, dx  \, \, ,
$$
then integrate the result over $s$.

Title: Functional of BM and time is martingale


问:
wrong answer in calculating the principal value of integral using complex analysis
I have read the answer to this question and it has helped a lot but for$\int_{-\infty}^{\infty}\frac{1}{x-2}dx$, I get the wrong answer.
I used this contour
\begin{align}
P.V\int_{-\infty}^{\infty}\frac{1}{x-2}dx&=\int_{C\ \cup P\ \cup L_1\cup L_2}^{}\frac{1}{z-2}dz-\int_{C}^{}\frac{1}{z-2}dz-\int_{P}^{}\frac{1}{z-2}dz\\
&=0-0-(-i\pi)=i\pi
\end{align}
As you can see the answer is wrong because it's complex even though it's the result of a real integral.
And I suppose that we can't be using $P.V=\text{Re}(i\pi)$ because there's no $\cos(ax)$ in the numerator.

答:
Your problem comes from the fact that the integral over $C$ is wrong.  It not zero.

Suppose that $C$ were a semicircle given by $z-2=100\exp(i\theta), 0\le\theta\le\pi$.  You evaluate the integral over this path, what do you get?  Can you see that the result will be independent of the radius of the large semicircle?


问:
Radius of curvature for a plane curve
Given the plane curve $(x(t),y(t)) = (2 \cos t, \sin t)$, the task is to find the radius of curvature at $(0,1)$. (The given point corresponds to time $t=\pi/2$.)

The radius $R$ is given by $1/\kappa$ where $\kappa$ is the curvature,
$$ \kappa = \frac{|v \times a |}{ |v|^3}, $$
with $v$ and $a$ being the velocity and acceleration vectors, respectively. Taking the first and second derivative, one can obtain $v=(-2\sin t, \cos t)$ and $a=(-2\cos t, -\sin t)$. At the given point, $v = (-2,0)$ and $a=( 0,-1)$, so that $v\times a = (0,0,2)$ and
$$ \kappa = \frac{2}{8} = \frac{1}{4} \Rightarrow R = 4.$$

However, my solution manual says that $R = 2$ with no technical explanation. Can you help me figure out what's wrong?

答: Let's do it in general: $\alpha(t) = (2\cos t, \sin t)$ implies $\alpha'(t) = (-2\sin t, \cos t)$ as well as $\alpha''(t) = -\alpha(t)$, and thus $$R(t) = \frac{1}{|\kappa(t)|} = \frac{\|\alpha'(t)\|^3}{\det(\alpha'(t),\alpha''(t))} = \frac{(4\sin^2t+\cos^2t)^{3/2}}{2\cos^2t +2\sin^2 t} = \frac{1}{2}(4\sin^2t+\cos^2t)^{3/2}.$$For $t = \pi/2$, we have $\cos \pi/2 = 0$ and $\sin \pi/2 = 1$, thus $$R(\pi/2) = \frac{4^{3/2}}{2} = 4.$$I think you have the right solution and the textbook has a mistake.

Title: Radius of curvature for a plane curve


答:
A possible way is to set $x = \tan t$ with $t \in \left(0,\frac{\pi}{2}\right)$.

So, the inequality is equivalent to

$$\frac{\tan t}{1+\frac 2{\pi} \tan t}< t \text{  for } t \in \left(0,\frac{\pi}{2}\right)$$

or after rearranging

$$ \tan t < \frac t{1-\frac 2{\pi}t} \text{  for } t \in \left(0,\frac{\pi}{2}\right)$$

But this is true because for $t \in \left(0,\frac{\pi}{2}\right)$ we have $\sin t < t$ and because of the concavity of $\cos$ on this interval we have $\cos t > 1-\frac 2{\pi} t$. Hence,

$$\tan t = \frac{\sin t}{\cos t} < \frac t{1-\frac 2{\pi} t} \text{  for } t \in \left(0,\frac{\pi}{2}\right)$$

Title: Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$

Body:
The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$ as $x \rightarrow + \infty$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.

So, what are available ways to solve this problem?


问:
Let $G$ be an abelian group s.t $H$ is a subgroup of index $n$. Prove that $\forall g \in G, g^n \in H$.
Let $G$ be an abelian group s.t $H$ is a subgroup of index $n$. Prove that $\forall g \in G, g^n \in H$.

I'm trying to split into some cases, where both cases that $ g \in H$ and case where ${\rm ord}(g)\mid n$ are clear.

答: Recall that in any finite group $A$, $a^{|A|}$ is the identity of the group for every $a \in A$. So, since $|G/H| = n$, $(gH)^n = H$ for every $g \in G$.


问:
How can i calculate this contour integral?
Here is my function $f(z)=\frac{1}{(z+1)(z+2)(z+3)...(z+100)}$
I need to find this $\int f(z)dz$ where $C:|z|=150$ counterclockwise

I tried to use second residue theorem here. So i get
$\int f(z)dz=2\pi i Res_{z=0}[1/z^2 f(1/z)$
so my function here:
$f(1/z)=\frac{1}{(1/z+1)(1/z+2)(1/z+3)...(z+100)}$
$1/z^2 f(1/z)=1/z^2((1/z+1)(1/z+2)(1/z+3)...(1/z+100))$

But after that i tried to use $Res_{z=0}[1/z^2 f(1/z)=lim_{z\to 0}\frac{1}{1/z^2(z+1)(z+2)(z+3)...(z+100)}$
So i get 100 times $1/0$ and that not make sense

So any what should i do to solve this integral?

答: Since $f(1/z)=\prod_{j=1}^{100}\tfrac{1}{1/z+j}=\prod_j\frac{z}{z+j}$, the correct calculation is$$z^{-2}f(1/z)=\tfrac{z^{98}}{\prod_j(z+j)}\stackrel{z\to0}{\to}0,$$making the integral $0$. As a sanity check, compare with $R>2$ to$$\oint_{|z|=R}\tfrac{dz}{(z+1)(z+2)}=\oint(\tfrac{1}{z+1}-\tfrac{1}{z+2})dz=2\pi i-2\pi i=0.$$


问:
Does zero vector in the span
This is seems to be an trivial question, I just want to know does zero vector always in the span since we could produce zero vector by multiply zero in each component of the basis set.</p>

答:
You observe correctly that the zero vector is always in the span (of a set of vectors) since it is the "zero combination" of the vectors in that set.

By definition, given a set of vectors, $S=\{v_1,\cdots, v_n\}$ where $v_k$ are vectors in a vector space $V$ (over a field $F$),
$$
\textrm{span}(S):=\{\sum_{k=1}^n a_kv_k:a_k\in F\}
$$
So
$$
0=0v_1+\cdots+0v_n\in \textrm{span}(S)
$$

Title: Does zero vector in the span


问:
Free action on integral homology spheres
Motivation: The cyclic group with two elements can freely act on $\mathbb{S}^1$. Note that $\mathbb{S}^1$ is also the Eilenberg–MacLane space $K(\mathbb{Z}, 1)$.

Question Let $n\geq 3$. Is there any integral-homology $n$-sphere such that it admits a free $\mathbb{Z}_2$-action and it is also an Eilenberg–MacLane space $K(G,1)$ for some group $G$?

答:
I will assume that by a (integral) homology $n$-sphere you mean an $n$-dimensional manifold $M$ which has $H_*(M)\cong H_*(S^n)$.  (One can consider instead of manifolds more general CW complexes.) According to Thurston's version of the Hopf Conjecture, if $M$ is a closed aspherical manifold of dimension $2k$ then
$$
(-1)^k \chi(M)  \ge 0.
$$
In particular, if $k$ is odd, then, conjecturally, $\chi(M)\le 0$. Such a manifold cannot be a homology sphere.

On the other hand, there is an abundant supply of closed aspherical 3-manifolds which are integral 3-spheres. Early examples of these are given by Seifert manifolds whose base-orbifold has nonpositive Euler characteristic. For every Seifert manifold $M$ whose Seifert fibration is orientable, we have a canonical $S^1$-action on $M$, corresponding to the rotation of the fibers. This action is not free if there are singular fibers. However, suppose that Seifert invariants $(\alpha_i,\beta_i)$ of singular fibers are all such that $\alpha_i$ is coprime with  $n$, then the order $n$ rotation of the fiber acts freely on $M$. It remains to find examples $M$ which are integral homology spheres. These are given, for instance, by Brieskorn manifolds $M(p,q,r)$ for which $p, q, r$ are all coprime. A very nice exposition of Brieskorn 3-manifolds is given in Milnor's paper

Milnor, John W., On the 3-dimensional Brieskorn manifolds M(p,q,r), Knots, Groups, 3-Manif.; Pap. dedic. Mem. R. H. Fox, 175-225 (1975). ZBL0305.57003.

Taking $p, q, r$ which are all odd and coprime (say, $3, 5, 7$) yields an example of a 3-dimensional integral homology sphere which is aspherical and admits a free involution.

I am sure there are other examples in all odd dimensions, but sorting this out would take some work.

I am not sure what happens if you allow arbitrary CW complexes as your $K(G,1)$'s, I suspect that you get examples in even dimensions as well.

There is a separate interesting question regarding which finite groups admit free actions on integral homology spheres. For some results in this direction, see

Milnor, John W., Groups which act on (S^ n) without fixed points, Am. J. Math. 79, 623-630 (1957). ZBL0078.16304.

and

Wall, C. T. C., Free actions of finite groups on spheres, Algebr. geom. Topol., Stanford/Calif. 1976, Proc. Symp. Pure Math., Vol. 32, Part 1, 115-124 (1978). ZBL0407.57031.


答:
Your solution is correct.

The given curve is an ellipse $(A\cos t, B\sin t)$.

At the extremities of axes, $\vec{v}$ and $\vec{a}$ are perpendicular. Then using $$a=\frac{v^2}{R}$$

known to high school students, it can be easily calculated
$$R(t=\pm \frac{\pi}{2})=\frac{|(-A,0)|^2}{|(0,B)|}=\frac{A^2}{B}$$
$$R(t=0, \pi)=\frac{|(0,B)|^2}{|(-A,0)|}=\frac{B^2}{A}$$

A pleasing and symmetrical result, I first came across in a Physics book.

Title: Radius of curvature for a plane curve

Body:
Given the plane curve $(x(t),y(t)) = (2 \cos t, \sin t)$, the task is to find the radius of curvature at $(0,1)$. (The given point corresponds to time $t=\pi/2$.)

The radius $R$ is given by $1/\kappa$ where $\kappa$ is the curvature,
$$ \kappa = \frac{|v \times a |}{ |v|^3}, $$
with $v$ and $a$ being the velocity and acceleration vectors, respectively. Taking the first and second derivative, one can obtain $v=(-2\sin t, \cos t)$ and $a=(-2\cos t, -\sin t)$. At the given point, $v = (-2,0)$ and $a=( 0,-1)$, so that $v\times a = (0,0,2)$ and
$$ \kappa = \frac{2}{8} = \frac{1}{4} \Rightarrow R = 4.$$

However, my solution manual says that $R = 2$ with no technical explanation. Can you help me figure out what's wrong?


问:
Is something wrong with my proposed proof of the sequential continuity of probability?
The proof I came up with is different from what I found everywhere else so I'm trying to find my error.

Given a probability triple $(\Omega,P,\mathcal{U})$ and the events $A_1 \subseteq A_2 \subseteq A_3 \subseteq A_4 \ldots $ show that:

$$ P( \bigcup\limits_{n=1}^\infty A_n ) = \lim_{m \to \infty} P(A_m)$$

The proofs I've come across consider disjoint sets and use the axiom of additivity. These are fine but I came up with something else:

If $A_1 \subseteq A_2 $ then $ A_1 \cup A_2 = A_2 $. Hence,

$$ P( \bigcup\limits_{n=1}^\infty A_n ) = P( \lim_{m \to \infty} A_m) = \lim_{m \to \infty} P(A_m)$$

Thoughts?

Thanks.

答:
The proof you have is probably like this:

Let
$$
B_1=A_1,\quad B_m=A_m\setminus A_{m-1},\quad m>1
$$

Then you have a disjoint union $\cup B_m$ which is equal to $\cup A_m$.

Now you have
$$
P(\cup A_m)=P(\cup B_m)=\lim_{N\to\infty}\sum_{m=1}^N P(B_m)=
\lim_{N\to\infty}P(A_N)
$$

As user251257 pointed out in the comment, you need to define $\lim_{m\to\infty} A_m$. But it is usually defined that
$$
\lim_{m\to\infty} A_m:=\cup_{m=1}^\infty A_m
$$
for $A_1\subset A_2\subset \cdots$.

You are using what you are supposed to prove.

In general,
$$
\limsup_{n\to\infty} A_n:=\cap_{n=1}^\infty\cup_{m\ge n} A_m,\quad
\liminf_{n\to\infty} A_n:=\cup_{n=1}^\infty\cap_{m\ge n} A_m,\quad
$$


问:
Regular rational functions over real projective lines and projective spaces.
In Algebraic Geometry one usually says that "working with algebraically closed fields makes life easier". Today I stumbled over one instance of this: suppose $X$ is an irreducible projective variety over a field $K$. Over algebraically closed fields it is known that there are no non-constant (ie, $\not\in\,K$) rational functions regular on $X$. In symbols $$\Gamma(X,\mathcal{O}_X)=K.$$

It turns out that this seems to be false if $K$ is not algebraically closed: take $K=\mathbb{R}$ and look at the projective line $X=\mathbb{P}^1_{\mathbb{R}}$ and the rational function $f=\dfrac{x_0^2}{x_0^2+x_1^2}$ where $(x_0:x_1)$ are projective coordinates on $X$. It is non-constant and regular everywhere.

On the other hand, looking at Hartshorne's book on AG, I discovered Theorem III, 5.2 from where one deduces that under same assumptions on $X$ we have $$\Gamma(X,\mathcal{O}_X)=K^r,$$ for some $r\geqslant 1$ because, after all $K\subseteq \Gamma(X,\mathcal{O}_X)$ is still valid. What I was not able to find is how to compute this integer $r$.

Question: is $r=\dim_K\,\Gamma(X,\mathcal{O}_X)$ computable in terms of known invariants of $X$ for (at least) projective, irreducible curves over an arbitrary field $K$?

A generalization of course would be (in case of curves) how to compute $\dim_K\,\Gamma(X,\mathcal{O}_X(nP))$ where $P$ is a rational point of X. If relevant, I'm interested in $K=k(t)$ where $k$ is algebraically closed.

Thanks!

答:
The given example is not a regular function on the scheme $\operatorname{Proj} \Bbb R[x_0,x_1,x_2]$ because it has a pole at $(x_0^2+x_1^2)$. As has been discussed recently, the naive definition of regularity for a function on a "classical" variety over an field which is not algebraically closed is incorrect and must be modified. One more reason to work with schemes instead of classical varieties!

Now to your main question: how do we compute $\dim_k \Gamma(X,\mathcal{O}_X)$ for $X$ an irreducible projective variety over a field $k$? By the answer here and the basic properties of the tensor product, we have that $\dim_{\overline{k}} \Gamma(X_{\overline{k}},\mathcal{O}_{X_{\overline{k}}})=\dim_k \Gamma(X,\mathcal{O}_X)$ where $k\subset \overline{k}$ is the algebraic closure. Now we can apply the result that for a connected projective scheme over an algebraically closed field $\overline{k}$, the only global sections are $\overline{k}$ (proof: a global section gives a map to $\Bbb A^1_{\overline{k}}$, the image must be closed, connected, and proper, so it is a closed point of $\Bbb A^1_{\overline{k}}$). Thus $r$ is the number of connected components of $X_{\overline{k}}$.


问:
How do you solve $\cos(θ) = \sin(θ) = a$ non graphically?
If
$\cos(θ) = a$, and
$\sin(θ) = a$,
and if $a$ is between $-1$ and $1$, how could we find $θ$. I could do this graphically by finding the intersection of $\sin(x), \cos(x)$ and $y=a$. But how could I find out with algebra, maybe the sign rules for each function in each quadrant.

答:
$$\cos(\theta)=\sin(\theta)\iff\sin\left(\frac\pi2-\theta\right)=\sin(\theta).$$

Hence

$$\frac\pi2-\theta=\theta+2k\pi$$ or $$\frac\pi2-\theta=\pi-\theta+2k\pi.$$

The second equation is impossible and finally

$$\theta=\frac\pi4+k\pi.$$

From this,

$$\sin(\theta)=\pm\frac1{\sqrt2}=a$$

which shows the condition for a solution.


问:
Cardinality of Set and its Power Set Singleton

I know that I should prove that there is a bijection between the two sides but I do not know how to do it.

答:
Your question is asking the followings (in the standard terminology): if $X$ is Dedekind-infinite, then $2^{2^{|X|}}=\left(2^{2^{|X|}}\right)^2$. You can prove it by using simple cardinal arithmetic:
observe that $$\left(2^{2^{|X|}}\right)^2=2^{2^{|X|}\cdot 2}=2^{2^{|X|+1}}.$$ By the assumption, we have $2^{2^{|X|+1}}=2^{2^{|X|}}$.


答: You're right both of those cases are pretty clear.  But now consider the subgroup $\langle gH\rangle$ of $G/H$ and apply Lagrange.

Body:
Let $G$ be an abelian group s.t $H$ is a subgroup of index $n$. Prove that $\forall g \in G, g^n \in H$.

I'm trying to split into some cases, where both cases that $ g \in H$ and case where ${\rm ord}(g)\mid n$ are clear.


答:
Since $\sin(t) = \cos(\pi/2-t)$,
if $0 \le t \le \pi$ then
$t = \pi/2-t$
or
$t = \pi/4$.

If $\pi \le t \le 2\pi$ then,
letting $t = \pi+s$
so $0 \le s \le \pi$,
since
$\sin(t) = \sin(\pi+s)
= -\sin(s)
$
and
$\cos(t)
=\cos(\pi+s)
=-\cos(s)
$,
we have,
again,
$\sin(s)=\cos(s)
$
so $s = \pi/4$
and
$t = 5\pi/4
$.

The value of $a$ is
$\pm \sqrt{2}/2$.

Title: How do you solve $\cos(θ) = \sin(θ) = a$ non graphically?

Body:
If
$\cos(θ) = a$, and
$\sin(θ) = a$,
and if $a$ is between $-1$ and $1$, how could we find $θ$. I could do this graphically by finding the intersection of $\sin(x), \cos(x)$ and $y=a$. But how could I find out with algebra, maybe the sign rules for each function in each quadrant.


问:
Find the expected number of of questions
I was reading Introduction to Probability, 2nd Edition and the following problem appears on page 123:

A prize is randomly placed in one of ten boxes, numbered from 1 to 10.

You search for the prize by asking yes-no questions.

Find the expected number of questions until you are sure about the
location of the prize, under each of the following strategies:

An enumeration strategy: you ask questions of the form "is it in box k?".

(The second strategy omitted because it's not relevant to my question).

The answer in the book is $5.5$ which presumes that we need to ask 10 questions sometimes, but that's not true, we only need 9 questions at most since after asking 9 questions, either we find the box in one of the first 9 questions or not and then the prize is in the remaining box which makes the expected value equal to $5.4$.

Am I wrong, or either the solution of the book is wrong, or the question itself is not clear?

答:
The book answer is incorrect.

The probability that you are sure after $1$ question is $0.1$

The probability that you are sure after precisely $2$ questions is $0.1$

The probability that you are sure after precisely $3$ questions is $0.1$

and so on until the end when

The probability that you are sure after precisely $9$ questions is $0.2$

The calculation then gives $5.4$ as you say.


问:
Deduce plus and minus with Cross Product in 3th and 4th Maxwell equations
The laws:

$\nabla \times \bar{E} = \bar{I}_{m} - \frac{\partial \bar{B}}{\partial \bar{t}}$

$\nabla \times \bar{H} =  \bar{J}_{f} + \frac{\partial \bar{D}}{\partial \bar{t}}$

so how can I remember with fingers or any other deduction method which is minus and which is plus? Since there is now the nabla, I am a bit lost how the cross product hand-rule work.

The term $\bar{I}_{m} = 0$ if magnetic monopoles do not exist. It is there to show that the formulas are of the same structure, cannot just remember which is minus and which is plus.

[Update]

Is it easier to deduce the laws if I suppose that the circuit moves and not the field?

$\bar{v} \times \bar{E} = \bar{I}_{m} - \frac{\partial \bar{B}}{\partial \bar{t}}$

$\bar{v} \times \bar{H} =  \bar{J}_{f} + \frac{\partial \bar{D}}{\partial \bar{t}}$

I am unsure whether the formulae are right so please check it. Could this way result in some easy deduction?

答:
Dear hhh, first of all, magnetic "monotones" are called "monopoles".

Second, you can't replace $\nabla \times$ by $\vec v\times$ in Maxwell's equations because these two objects don't even have the same units.

Third, you must just remember the signs. The Ampere's law, $\nabla\times H= j$, has the plus sign. It's the oldest law mixing electricity and magnetism, so it has the plus sign. It's about the magnetic fields around electric wires. The corresponding currents and magnetic fields around them are described by the right-hand rule.

You may also remember that in this oldest law - magnetic fields of wires - Maxwell's correction $\partial D/\partial t$ also comes with a plus sign.

It then follows that Faraday's law must have the opposite sign - so there is $-\partial B / \partial t$ on the right hand side for the equation for $\nabla \times E$.

But more generally, it's meaningless for you to incorporate "right hand rules" into the form of Maxwell's equations. Right hand rules are not supposed to help you to remember signs in a written form of the equations - on the paper; the "plus rules" and "minus rules" will do.

Right hand rules are supposed to help you to find the directions of vectors in various situations in the real world where there is no God-given understanding which direction in a given setup is "plus" and which direction is "minus". As some people have mentioned before me, "hands" are only useful to define "rules" when you deal with a physical world, so hands are not useful to write down equations in mathematics.


问:
Two problems on analytic function and Mapping of elementary functions
Let $G$ be a region and let $f$ and $g$ be analytic functions on $G$ such that $f(z)g(z)=0$ for all $z \in G$. Show that either $f$ or $g$ is identically zero on $G$.

Here is how I do it: Assume $f$ is non zero on $B(a,R)$, then $fg=0$ implies $a$ is a root of $g$ (of order $n>0$). Therefore $g(z)=(z-a)^nF(z)$ for some non zero analytic function $F(z)$. And then I can't move any more...

The second question is to determine the image of $\{z=x+iy : -\pi < x < \pi,
y=3\}$ under the mapping $u+iv = w = \sin z$.

答:
This is a consequence of the identity theorem: Assume $f \not= 0$, then $$N := \{z \in G; f(z) = 0 \}$$ has no accumulation point (otherwise, the identity theorem would imply $f=0$ since $G$ is a region). Clearly, $g|_{G \setminus N} = 0$ since $f \cdot g=0$. Since $G \setminus N$ has an accumulation point, the identity theorem states $g=0$.

By definition, $$\sin z = \frac{1}{2\imath} (e^{\imath \, z} - e^{-\imath \, z})$$ Now set $z=x+3\imath$, $x \in (-\pi,\pi)$, to calculate the image of the given set.


问:
Idea of compact operators
I am studying functional analysis and in particular Hilbert spaces and operators, I would like some clarification on the idea of compact operators.

The definition that was presented to me is:

Given $\mathcal{H}_1$ and $\mathcal{H}_2$ Hilbert spaces on the field $\mathbb{K}$, an operator $T\in \mathcal{L}(\mathcal{H}_1,\mathcal{H}_2)$ is said to be compact if the immage of the ball $\overline{B_1}$ in $\mathcal{H}_1$ has compact closure in $\mathcal{H}_2$.

Also, how would one show that an equivalent definition is that, in the same environment, $T$ is compact if and only if: for all bounded sequences $\{x_n \}_{n\in \Bbb N}$ in $\mathcal{H}_1$, then the sequence of the immages $\{ Tx_n\}_{n\in \Bbb N}$ has a convergent subsequence in $\mathcal{H}_2$?

Thanks in advance.

答:
If the image of the ball has compact closure, then the image of any bounded set has a compact closure as well (try to understand why). But in a metric space, being precompact (having compact closure) means that any sequence has a subsequence which converges (maybe to a point in the closure).

For the other direction - if the image of any bounded sequence has a convergent subsequence, then this is in particular true for the ball. Then the image of the ball is sequentially precompact, which in a metric space, is equivalent to being precompact (so the closure is compact).

Bottom line - since any Hilbert space is in particular a metric space, being (pre)compact and sequentially (pre)compact is the same, and there is no real difference between the ball and any other bounded set.

Title: Idea of compact operators


问:
How can I generate cyclic whole number?
My goal was to create a function that would generate number from 1->N->1. 1 2 3 4 3 2 1.

Easiest approach I was going for was create a function where using cosine i'd generate integers, and where x -> 2N.

So if N is 4, then x would go upto 2x4 = 8.

1 2 3 4 5 6 7 8

then output of ƒ would be

1 2 3 4 4 3 2 1

or

1 2 3 4 3 2 1

either would have been great.

Title: How can I generate cyclic whole number?

Body: $$\frac{2(N-1)}\pi\arcsin\left(\sin\left(\pi\frac{x-1}{2(N-1)}\right)\right)+1.$$


问:
Function with $f(f(n))=f(n-1)f(n+1)-f(n)^2$
Let $\mathbb{N}$ denote the set of positive integers. Does there exist a function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that  $$f(f(n))=f(n-1)f(n+1)-f(n)^2$$ for all $n\geq 2$?

If $f$ is linear, plugging in $f(n)=an+b$, we get
$$a(an+b)+b=(a(n-1)+b)(a(n+1)+b)-(an+b)^2$$
which is
$$a^2n+ab+b=-a^2$$

This means $a=0$. But then also $b=0$, hence $f\equiv 0$, but this is not allowed. So there is no such linear function.

答:
Since the values of $f$ must be positive, $f(n-1)f(n+1) > f(n)^2$ for all $n \geq 2$, so $\frac{f(n+1)}{f(n)} > \frac{f(n)}{f(n-1)}$.

Therefore there exist $M, R > 1$ such that for all $n > M$, $f$ is increasing and $f(n) > nR^n > n + 2$. For such $n$,

$$f(f(n)) < f(n-1)f(n+1) < f(n+1)^2$$

On the other hand,
$$f(f(n)) > f(n + 2) > (f(n+2)f(n))^{1/2} > (f(f(n+1)))^{1/2}$$
$$> f(n+1)^{1/2}R^{f(n+1)/2}$$

Therefore
$$f(n+1)^{1/2}R^{f(n+1)/2} < f(n+1)^2$$
$$R^{f(n+1)/3} < f(n+1)$$

But since $R > 1$ and $f$ is increasing for $n > M$, this cannot hold for all such $n$. Therefore no such $f$ exists.


问:
$I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$
Suppose $X_1, X_2\subset \mathbb{A}^n$ are algebraic sets. Why does $I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$?

In particular, why do we have the inclusion $I(X_1\cap X_2)\subset\sqrt{I(X_1)+I(X_2)}$?

答:
We have$$I(X_1\cap X_2)=I(V(I(X_1))\cap V(I(X_2)))=I(V(I(X_1)+I(X_2)))=\sqrt{I(X_1)+I(X_2)}$$
The only non-trivial equation is the last one which follows from the Nullstellensatz ($I(V(J))=\sqrt{J}$)

Title: $I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$


问:
Smallest asymmetric relation such that each $k$ elements are dominated by a common element.
How large does a non-empty set $X$ need to be if there exists an asymmetric relation $R \subset X^2$ such that for any $k$ elements $x_1,\dots,x_k$ of $X$ there exists a $x \in X$ such that $(x,x_i) \in R$ for all $i$ from $1$ to $k$?

For the purpose of this question we say that a relation is asymmetric if $(x,y) \in R$ implies $(y,x) \not \in R$.

Example: for $k=1$ the answer is $|X| \ge 3$, because the 'rock-paper-scissors' relation does the trick: $X = \{r,p,s\}$ and $R = \{(r,p),(p,s),(s,r)\}$ does the trick, and you can check that it is not possible with $|X| \le 2$. For $k=2$, I believe the answer is $|X| \ge 7$.

Context: I followed the `Topics in Combinatorics' lecture series by Tim Gowers on YouTube [1] (which I warmly recommend) and this problem was the second part of a problem on a question sheet [2]. The first part of the question is to prove that for each $k$, there exists a relation $R \subset X^2$ with this property for some $X$. You can solve this part with an averaging argument.

[1] https://www.youtube.com/watch?v=WcLrT263p2w&list=PLOft35kj95aYT2SgFbYwEOavJ7gagmCVZ

[2] https://drive.google.com/file/d/1OFgVrk1n2zxSF5tcfSe5Fxrva0YVhXEl/view

答:
Let $f(k)$ be the smallest number such that there exists an asymmetric relation $R$ on a set $X$ of size $f(k)$, so that for all $\{x_1,\dots,x_k\}\subseteq X$, there will exist $x\in X$ so that $(x,x_i)\in R$ for all $i\in \{1,\dots,k\}$ (what a mouthful!). One can show that
$$
f(k)\ge 2^{k+1}-1 \, .
$$
Furthermore, this bound is tight for $k=1$ and $k=2$. Erdös himself conjectured that $f(k)=2^{k+1}-1$ for all $k\ge 1$. However, the following paper (1) by Szekeres and Szekeres further shows that
$$
f(k)\ge (k+2)2^{k-1}-1
$$
So $f(3)\ge 19$, and $f(3)\neq 15$. In the same paper, they prove $f(3)\le 19$. Here, the asymmetric relation on $19$ vertices which works is the quadratic residue tournament modulo $19$. That is, $X=\mathbb Z/19\mathbb Z$, and $(x,y)\in R$ whenever there exists $k\in \mathbb Z/19\mathbb Z$ so that $x\equiv y+k^2\pmod{19}$.

Finally, Erdös showed that there exists a constant $C$ for which
$$
f(k)\le Ck^22^k,\qquad \text{for all }k\ge 1
$$
using a probabilistic argument. These are the best bounds I know on $f(k)$.

(1) E. Szekeres, and G. Szekeres. “On a Problem of Schütte and Erdös.” The Mathematical Gazette, vol. 49, no. 369, 1965, pp. 290–293. JSTOR, www.jstor.org/stable/3612854. Accessed 17 Jan. 2021.


问:
How to find all the maximal ideals that contain $(X+2Z+1,Y-Z,Z^2+Z+1)$ in $\mathbb{C}[X,Y,Z]$

I need to find all the maximal ideals that contain $\mathfrak{a}=(X+2Z+1,Y-Z,Z^2+Z+1)$ in $\mathbb{C}[X,Y,Z]$.

I have tried doing the following:

Let $\mathfrak{b}$ be a maximal ideal which contains $\mathfrak{a}$.
Therefore, we have that $X+2Z+1,Y-Z,Z^2+Z+1 \in \mathfrak{b}$.

Because of this, I have defined $\mathfrak{b}$ in the following way: $$\mathfrak{b}=(X+2Z+1,Y-Z,Z^2+Z+1,f_1(X,Y,Z),f_2(X,Y,Z),...)$$ where $f_i(X,Y,Z) \in  \mathbb{C}[X,Y,Z]$, $\forall i \geq 1$.

Apart from that, we have these congruences: $$ X \equiv -2Z-1 \bmod(X+2Z+1), Y \equiv Z \bmod(Y-Z) $$
So $f_i(X,Y,Z)\equiv f_i(-2Z-1,Z,Z)=g_i(Z) \in \mathbb{C}[Z]$, $ \forall i \geq 1$.
Therefore, $$\mathfrak{b}=(X+2Z+1,Y-Z,Z^2+Z+1,g_1(Z),g_2(Z),...) $$
But I don't know which polinomials I should choose so that $\mathfrak{b}$ is maximal...

I would be really thankful if someone could help me.

答:
As leoli says in the comments, this is a problem we should solve geometrically. The maximal ideals containing $\mathfrak{a}$ are precisely the ideals $(X-x_0, Y-y_0, Z-z_0)$ where $(x_0, y_0, z_0)$ is a point of the variety $$V = \{(X,Y,Z) \in \mathbb{C}^3 : X+2Z+1 = Y-Z = Z^2 + Z + 1 = 0\}.$$
So we just have to find this set! I'll leave this up to you -- to check your work, $V$ has exactly two elements.

If you haven't yet proved this general fact about maximal ideals of $\mathbb{C}[X,Y,Z]$, the key fact is that every maximal ideal of $\mathbb{C}[X,Y,Z]$ is of the form $(X-x_0, Y-y_0, Z-z_0)$ for some $(x_0, y_0, z_0) \in \mathbb{C}$. This requires a bit of Nullstellensatz: you need to know that the variety defined by a proper ideal $I$ is nonempty. Then take a point $(x_0,y_0,z_0)$ in the variety defined by a maximal ideal $\mathfrak{m}$, and show that $X-x_0 \in \mathfrak{m}$...


问:
How to tell if a base for a number system is consistent (a number has only one representation)?
Let's say I have a number system with an array defining the bases at each position. For example, [1,1,1] would be a binary number system (each digit can be 0 or 1) in which any number from 0 to 3 can be represented. Similarly, I can have [3,2,1] where the units digit can be 0 or 1, the tens digit can be 0,1 or 2 and the hundreds digit can be 0,1,2 or 3. if these digits are $u$, $t$ and $h$ respectively the number becomes: $u+2t+3h$.

Let's define consistency as the property where every integer less than the max permissible value has one and only one representation in the base system.

Now, the second number system in the examples above isn't consistent. If I take 4, there are two ways to represent it: $[1,0,1] = 3*1+2*0+1*1$ or $[0,2,0] = 3*0+2*2+1*0$. On the other hand, [6,2,1] is consistent. Is there a way by looking at the array of bases weather or not the number system is consistent.

答:
A result I proved about 50 years ago is that
a representation
$ n
=\sum_{j=0}^{\infty} a_jB_j
$
where the
$a_j$ and $B_j$ are integers,
$B_0=1$,
$1 \le B_j < B_{j+1}$,
and
$0 \le a_j \lt \dfrac{B_{j}}{B_{j-1}}
$
is unique and exists iff
$\dfrac{B_{j+1}}{B_j}$
is an integer for all $j$.

This is why
$B_j = m^j$
for integral $m\ge 2$
and $B_j = j!$
have a unique representation
and $B_j = F_j$
(Fibonacci numbers)
does not.

The representation or $n$ is gotten
by the standard method of
starting with the largest
$B_k \le n$,
setting
$n_k = n$,
$a_k
=\lfloor \dfrac{n_k}{B_k} \rfloor
$,
$n_{k-1}
=n_k-a_kB_k$,
and iterating with
$k = k-1$
until $k = 0$.


问:
Counting possible ways for a group of objects
A drawing with n different objects is being coloured. For every object a painter uses one of k different colors.

a) for which values of k is sure that no matter how is the drawing coloured, at least one color will be used for at least four objects?

b) for how many different colourings of the objects in the drawing every color is used at least once?

答:
HINT: For (a) it’s easier to turn the problem around: given $k$ colors, what values of $n$ will guarantee that at least one color is used for at least $4$ objects?

If $k=1$, the result is guaranteed whenever $n\ge 4$. If $k=2$, $n=6$ is too small, since you might have three objects of each color, but $n\ge 7$ works: no matter how you split $7$ or more objects into two parts, one part contains at least $4$ objects. If $k=3$, $n=9$ is too small, since you could have three objects of each color, but $n\ge 10$ works: no matter how you split $10$ or more objects into three parts, one part contains at least $4$ objects. Now generalize this reasoning to find the general principle. It may help to realize that you’re using a form of the pigeonhole principle. Once you have the minimum $n$ that ‘works’ for each possible $k$, it’s not hard to turn it back around and get the maximum $k$ that ‘works’ for any given $n$.

For (b), observe that there are $k^n$ ways to color $n$ objects with $k$ colors. Now use an inclusion-exclusion argument to subtract the colorings that don’t use all $k$ colors.


问:
Recursive integral equation with different variables
$$S = \frac{1}{\int_1^2 \frac{1}{\int_1^2 \frac{1}{\int_1^2 \ddots}}}$$

I know that these kind of recursive questions are one inside another - meaning that:

$$ S = \frac{1}{\int_1^2 S}$$

The question is what is the variable? $x$ ? $ S$ ? whatever I would like it to be?

If the variable is $x$ (or anything else than $S$ ) then the answer is:

$$ S (x \mid _1^2) = S = 1 \Rightarrow S = 1$$

If the variable is $S$ than the answer is:

$$ S ( \frac{S^2}{2}  \mid _1^2 ) = S[ 2 - 0.5] = 1.5S = 1 \Rightarrow S = \frac{2}{3}$$

Which is correct? I assume there is only one value that is correct and not $2$ because, this integral keeps on forever so we don't actually know if it is $dx$ or $dS$

答:
$$S = \frac{1}{\int_1^2 \frac{1}{\int_1^2 \frac{1}{\int_1^2 \ddots}}}$$
if the right side is anything, it is a value at least in $[-\infty,+\infty]$, so $S$ is not a variable.
Said this I would procede this way:

$$S = \frac{1}{\int_1^2 Sdx} \iff S^2 = \frac{1}{\int_1^2 dx}= 1 \iff S=\pm 1.$$
I actually have no idea if the negative solution is acceptable.


问:
Rule of calculation of residue
Let $U \subset \Bbb C$ be open and $f,g:U \to \Bbb C$ be holomorphic. I wanna show that if $g$ has a simple zero in $z_0\in U$, then
$$\operatorname{Res}\biggl(\frac{f}{g},z_0\biggr)=\frac{f(z_0)}{g'(z_0)}$$
The proof I use is: Because $f$ and $g$ are holomorphic we can expand them as entire serie
$$f(z)=b_0+b_1(z-z_0)+b_2(z-z_0)+ \dots$$
and
$$g(z)=c_1(z-z_0)+c_2(z-z_0)+ \dots$$
Where we already used that $g$ has a simple zero in $z_0$.
$$\frac{b_0+b_1(z-z_0)+b_2(z-z_0)^2+ \dots}{c_1(z-z_0)+c_2(z-z_0)^2+ \dots}=\frac{1}{z-z_0}(\frac{b_0+b_1(z-z_0)+b_2(z-a)^2+ \dots}{c_1+c_2(z-z_0)+\dots})$$
The right fraction is nothing else than the holomorphic function $h(z):=\frac{(z-z_0)f(z)}{g(z)}$ with $h(z_0)=\frac{b_0}{c_1}=\frac{f(z_0)}{g'(z_0)}$. But this should be the end of the proof. But why? I don't understand the end...

EDIT: Is it not wrong? Should it not be that $f$ can be represented as a Laurent serie near $z_0$ and not as an entire function?

答: The quotient$$\frac{b_0+b_1(z-z_0)+b_2(z-z_0)^2+\cdots}{c_1+c_2(z-z_0)+c_3(z-z_0)^2+\cdots}$$is the quotient of two holomorphic functions and therefore it is holomorphic; besides, it maps $z_0$ into $\frac{b_0}{c_1}$. It can be written, near $z_0$, as $d_0+d_1(z-z_0)+d_2(z-z_0)^2+\cdots$, with $d_0=\frac{b_0}{c_1}$. So, near $z_0$,$$\frac{f(z)}{g(z)}=\frac{d_0}{z-z_0}+d_1+d_2(z-z_0)+\cdots$$Therefore, by the definition of residue,$$\operatorname{Res}\left(\frac fg,z_0\right)=d_0=\frac{b_0}{c_1}.$$

Title: Rule of calculation of residue


问:
Question about generating functions about an exercise
I have a question about generating functions,
I've understood that generating functions are also representing some series of numbers, but is this series could be formed by 2 different functions?

For example, an exercise that I've encountered goes like this:
Find the generating function that represents the number of solutions to the equation:

$$ x_1 + x_2 + \dots + x_k=n$$

Where the variables are even numbers that are not divisible by 3,
So I assume we should find a function that represents the series:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...

0 1 0 1 0 0 0 1 0  1  0  0  0  1  0  1  0  0 ..

The zero's and one's stands for the coefficient of $x^n$

So to get to that series we have few options the first one i've thought about is to take the generating function: $1 + x + x^2 + x^3 + \dots$

$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x} $$

Which represents the series of 1 1 1 1 1 ...

and then to subtract those (subtract odd numbers and multiples of 6):
$$ f(x) = ((1 + x + x^2 + x^3 + ...) - x(1 + x^2 + x^4 + ...) - (1 + x^6 + x^{12} +...))^k $$

Which gives us the series we wanted.

But the solution to this exercise shows different answer and the generating function goes like this:

$$ g(x) = ((x^2 + x^4) + (x^8 + x^{10}) + (x^{14} + x^{16}) + ... )^k $$

which represents the same series as well but looks differently, does those 2 functions coefficients represent the same number of solutions to the equation?

Thanks alot !

答:
Your approach is fine. We can show the generating functions are equal: $f=g$.

We have
\begin{align*}
\color{blue}{g(x)}&=\left(\left(x^2 + x^4\right) + \left(x^8 + x^{10}\right) + \left(x^{14} + x^{16}\right) + ... \right)^k \\
&=\left(x^2\left(1+x^2\right)+x^8\left(1+x^2\right)+x^{14}\left(1+x^2\right)+\cdots\right)^k\\
&=\left(1+x^2\right)^k\left(x^2+x^8+x^{14}+\cdots\right)^k\\
&=\left(1+x^2\right)^kx^{2k}\left(1+x^6+x^{12}+\cdots\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}

On the other hand we obtain

\begin{align*}
\color{blue}{f(x)}&= \left(\left(1 + x + x^2 + x^3 +\cdots\right) - x\left(1 + x^2 + x^4 + \cdots\right)
- \left(1 + x^6 + x^{12} +\cdots\right)\right)^k\\
&=\left(\frac{1}{1-x}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1+x}{1-x^2}-\frac{x}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{1}{1-x^2}-\frac{1}{1-x^6}\right)^k\\
&=\left(\frac{\left(1-x^6\right)-\left(1-x^2\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&=\left(\frac{x^2\left(1-x^4\right)}{\left(1-x^2\right)\left(1-x^6\right)}\right)^k\\
&\,\,\color{blue}{=\left(\frac{x^2\left(1+x^2\right)}{1-x^6}\right)^k}
\end{align*}
and the claim follows.


答:
The functions $f$ and $g$ are identical.

You have taken all the powers then deleted all odd ones and then the multiples of $6$.

The book has started with the even powers and then deleted the multiples of $6$.

It's just the same!

Title: Question about generating functions about an exercise

Body:
I have a question about generating functions,
I've understood that generating functions are also representing some series of numbers, but is this series could be formed by 2 different functions?

For example, an exercise that I've encountered goes like this:
Find the generating function that represents the number of solutions to the equation:

$$ x_1 + x_2 + \dots + x_k=n$$

Where the variables are even numbers that are not divisible by 3,
So I assume we should find a function that represents the series:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...

0 1 0 1 0 0 0 1 0  1  0  0  0  1  0  1  0  0 ..

The zero's and one's stands for the coefficient of $x^n$

So to get to that series we have few options the first one i've thought about is to take the generating function: $1 + x + x^2 + x^3 + \dots$

$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x} $$

Which represents the series of 1 1 1 1 1 ...

and then to subtract those (subtract odd numbers and multiples of 6):
$$ f(x) = ((1 + x + x^2 + x^3 + ...) - x(1 + x^2 + x^4 + ...) - (1 + x^6 + x^{12} +...))^k $$

Which gives us the series we wanted.

But the solution to this exercise shows different answer and the generating function goes like this:

$$ g(x) = ((x^2 + x^4) + (x^8 + x^{10}) + (x^{14} + x^{16}) + ... )^k $$

which represents the same series as well but looks differently, does those 2 functions coefficients represent the same number of solutions to the equation?

Thanks alot !


问:
generic way of reducing the rank of a matrix by adding $vv^T$, $v$ from eigenspace of $A$
Say I have a matrix $A$ and consider $B$ of the form $A+vv^T+ww^T$ for some vectors $v,w$.

How do I have to chose $v$ and $w$ in order for $\text{rank}B=\text{rank}A-2$?

I was thinking that $v$ and $w$ should have something to do with eigenvectors of $A$ but I don't want to restrict $B$, I want any rank rank$A-2$ matrix $B$ to be constructible from this method.

答:
I infer you are working over $\mathbb R$.

your stated goal of

"I don't want to restrict $B$, I want any rank rank$A-2$ matrix $B$ to
be constructible from this method."

is doomed to fail.  Suppose I select $A$ to be rank 2 and nilpotent.  Then the only possibility is $B=\mathbf 0$.

By subadditivity of rank, this means that $\mathbf v \neq \mathbf 0$ and $\mathbf w \neq \mathbf 0$.

Thus

$\mathbf 0 = B = A+\mathbf {vv}^T+\mathbf {ww}^T$  but

$0 = \text{trace}\big(\mathbf 0\big) = \text{trace}\big( B\big) = \text{trace}\big(A+\mathbf {vv}^T+\mathbf {ww}^T\big) = 0 + \text{trace}\big(\mathbf {vv}^T\big) + \text{trace}\big(\mathbf {ww}^T\big)\gt 0 $

more generally: if you instead try for

$\mathbf 0 = B = A+ \alpha_1\mathbf {vv}^T+\alpha_2\mathbf {ww}^T$

you'll find this is still impossible, e.g. because it implies a symmetric matrix given by $-\big(\alpha_1\mathbf {vv}^T+\alpha_2\mathbf {ww}^T\big)$ is equal to $A$ which is (orthogonally similar to) a non-zero strictly upper triangular matrix.


问:
Reference request: Cartier operator paper
Feel free to delete this post if it is inappropriate.

Is there anyway one could access the paper by [3] of Cartier?

```<code>Une nouvelle opération sur les formes différentielles. C. R. Acad. Sci. Paris, 244 (1957), 426-428.
</code>```

I have been searching for quite a while.

答: This is available freely online from the French National Library here (direct link). Back issues of the Comptes Rendus from 1835 through 1965 are available at the same location - Wikipedia notes that all issues through 1996 are available freely online through the French national library, with links to the appropriate section of the website depending on the time period. Starting in 1997, the journal was published commercially by Elsevier - articles from 1997 to present will require different tactics.


问:
If a graph has no K4 or K2,3 subdivision then it is outerplanar
Is this a valid argument to prove the claim? I put in bold the part I am doubting the most:

If a graph has no $K_{4}$ or $K_{2,3}$ subdivision then it is outerplanar

Suppose G has no $K_{4}$ or $K_{2,3}$. Add a vertex $v$ to the exterior face of $G$ (outside of $G$) and connect it to every vertex in $G$. Call this graph $G^{\prime}$. Assume $G^{\prime}$ is not planar and by Kuratowski theorem contains a $K_{5}$ or $K_{3,3}$ subdivision. Therefore $G$ must contain a $K_{4}$ or $K_{2,3}$ subdivision, as every $K_{5}$ subdivision contains a $K_{4}$ and every $K_{3,3}$ contains a $K_{2,3}$ subdivision. However, by assumption $G$ does not have a $K_{4}$ or $K_{2,3}$ subdivision, hence we have a contradiction. Hence $G^{\prime}$ must be planar. As $G^{\prime}$ is planar and $v$ is exterior to all of $G$ and adjacent to all vertices in $G$, $G$ must be outerplanar.

答:
First, you begin by adding vertex $v$ to the "exterior face of $G$", and later using the fact that $v$ is "exterior to all of $G$".

This is not okay, since at this point, we don't know that $G$ is planar, and don't have a fixed plane embedding of $G$. Even if you invoke Kuratowski's theorem to say that $G$ is planar and pick a plane embedding, it might be the wrong embedding for us to use later! We will prove that $G'$ has a plane embedding later on, but that plane embedding might not be compatible with the plane embedding we picked for $G$.

(Note also that even if $G$ is outerplanar, not all planar embeddings of it are outerplanar embeddings. Some embeddings might have all vertices along a face that is not the exterior face; others might have no such face at all.)

So in the first step, all we can do is say "contruct $G'$ by adding a new vertex $v$ adjacent to all vertices of $G$". This will be fine.

To justify the bolded step, consider a subdivision of, say, $K_5$ in $G'$.

If it does not contain the new vertex $v$, we're fine; it's a subdivision of $K_5$ in $G$, and this contains a subdivision of $K_4$.

If it contains $v$ along a subdivided edge, then forget about that entire subdivided edge. We get a subdivision of $K_5 - e$ in $G$, which still contains a subdivision of $K_4$.

If it contains $v$ as one of the $K_5$ vertices, then forget about that vertex and all the subdivided edges out of it. We get a subdivision of $K_4$ in $G$.

The same argument works for $K_{3,3}$.

We have to be more careful with the last step: just because $G'$ is planar, can we conclude that $G$ is outerplanar?

Let's take a planar drawing of $G'$. Deleting $v$ and all its edges gives a planar drawing of $G$. Let $S$ be the subset of the plane where $v$ and all its edges were embedded. Then

$S$ is a connected subset of the plane which intersects no vertices or edges of $G$, so it's contained entirely in some face of $G$.

Since $S$ contains points arbitrarily close to where vertices of $G$ are embedded, that face of $G$ contains all the vertices.

So we have an embedding of $G$ where one of the faces contains all the vertices. If it is already exterior, fine. If not, we can invert the plane about a point contained in this face, and get a new embedding of $G$ where it is the exterior face.


问:
Fourier Transform on product of 2 functions
Using Euler's Formula, e^ix = cos(x) + i*sin(x), or otherwise, derive the Fourier transform of the function:

g(x) = cos(ax)f(x)

in terms of a, x and F(w), the Fourier transform of f(x).

Simply stuck on this question, where do you simplify down. My initial idea was substituting in (e^axi + e^-axi)/2 for cos(ax) as that formula was shown in one of our lectures but I just get stuck with:

F[g]= 1/(2 * ROOT(2 * pi)) * ∫  f(x) * e^((a-w)ix) dx + ∫ f(x) * e^(-(a+w)ix) dx

and don't know what to do from here, I'm very new to Fourier transforms and Euler's formula and the only examples given to us are like for F[Af(x) + BG(x)] not when they're multiplied together so I'm struggling with this. Am I suppose to integrate or substitute back to trig or am I completely off?

The basic Fourier Transform example they gave us is in the lecture is:

F[f] (w) = 1/(ROOT(2 * pi)) * ∫  f(x) * e^(-iwx) dx

答:
You were on the right track and very close.  You just missed that you can treat $\omega\pm a$ as a constant during the integration.

Here's the follow-through of what you started, using intermediate substitutions for $\omega\pm a$ to hopefully make things more obvious:

$$\begin{align*}\mathcal{F}\left\{g(x)\right\} &= \mathcal{F}\left\{\cos(ax)f(x)\right\}\\
\\
&=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \cos(ax)f(x)e^{-i\omega x}dx\\
\\
&=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-iax}f(x)e^{-i\omega x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{iax}f(x)e^{-i\omega x}dx\right]\\
\\
&=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i(\omega+a) x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i(\omega-a)x}dx\right]\\
\\
&=\dfrac{1}{2}\left[\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-iu x}dx+\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ivx}dx\right]\\
\\
&=\dfrac{1}{2}\left[F(u)+F(v)\right]\\
\\
&=\dfrac{1}{2}\left[F(\omega+a)+F(\omega-a)\right]\\
\end{align*}$$


问:
Fundamental group of discs with boundaries identified
Let $D^2\subseteq\mathbb{R}^2$ be the closed disc with center at the origin and radius 1 and let $S^1$ be its boundary. Also, let $F_n=\{1,2,\ldots,n\}$ be a $n$-point set with the discrete topology. Define, for $n\geq 2$
$$ X_n = \left(D^2\times F_n\right)/\sim$$
where $\sim$ is the equivalence relation described by
$$ \left(x,a\right)\sim\left(y,b\right)\iff x=y\in S^1$$
Prove (by induction on $n$) that $X_n$ is simply connected for all $n\geq 2$.

I've quite easily proven the base case ($n=2$), since $X_2$ is homeomorphic to the $S^2$ sphere, which is simply connected. For the induction step, proving that $X_{n+1}$ is path-connected is also fairly straight-forward. However, I'm struggling to prove that $X_{n+1}$ has the trivial fundamental group. I've tried using the Seifert–van Kampen theorem, yet I'm unable to find two path-connected open set that would satify van Kampen's conditions.

Any hint would be appreciated.

答:
The idea here is that $D^2 \times F_n = \bigsqcup_{i=0}^n D^2$ are $n$ disjoint disks, and the quotient glues them all at their boundary. This is a bit hard to imagine when $n > 2$, but we can extrapolate the following idea.

The sphere $S^2$ can be thought as two semi-spheres seamed at their boundary; that being the sphere's equator. Each semi-sphere is homeomorphic to a disk.

One way to see that a - path connected, say - space $X$ satisfies $\pi_1(X) = 0$ is to show that it consists of the union of two simply connected open sets $U,V$ with path connected intersection. In that case, by van Kampen we have a surjection

$$
1 = \pi_1(U) \ast \pi_2(V) \to \pi_1(X)
$$

and so $X$ is again simply connected.

In the case of the sphere, we can take away the north pole and set $U := S^2 \setminus \{N\}$, and then put $V = B$ an open disk around $N$. The latter is easily seen to be contractible. Now we are left to show that $U$ is simply connected, but this follows from the fact that it deformation retracts to a disk (moreover, it is contractible, but that won't generalize as smoothly).

Can you make this work for higher $n$? Try to translate this 'movements' in terms of the "open disk presentation" for $S^2$.

Edit: spoilers seem not to be working :/

Consider $U = D^2 \times F_{n+1} \setminus \{p\}$, where $p \in D^2 \times \{n+1\}$ is a point in one of the disks, and $V$ a small disk around it. You can define the open sets in the disjoint union and check that they descend to open sets $\widetilde{U}$ and $\widetilde{V}$ in the quotient.

In the same fashion, the hotomopy that contracts $V$ to a point induces a homotopy that proves $\widetilde{V} \simeq \ast$, and the
homotopy equivalence $D^2 \times \{n+1\} \setminus \{p\} \times \{n+1\}\simeq S^1 \times \{n+1\}$ shows that $\widetilde{U} \simeq X_n$ and $\pi_1(U) = \pi_1(X_n) = 1$.


问:
Does AM-GM Follow From the Convexity of Some Function
The AM-GM for $n = 2$ is
$$\frac{x+y}{2} \ge (xy)^{1/2}$$
This is very easy to prove with algebra. However, I am wondering if there is a proof using convexity. That is, can we find a convex function $f:\mathbb{R} \to \mathbb{R}$ so that for some $t = t(x, y)$ between $0$ and $1$ we have
$$tf(x) + (1-t)f(y) = \frac {x+y}{2}$$
and
$$f(tx + (1-t)y) = (xy)^{1/2}$$

Is there such a proof?

答:
No. The first equation implies (by setting $x = y$) that $f(x) = x$ is the identity function, which then forces $t(x,y) = \frac{1}{2}$ for $x \neq y$. Then the second equation cannot be satisfied unless $x = y$.

Even if we only impose the first equation when $x \neq y$, we get the same conclusion. Every convex function $\mathbb{R} \to \mathbb{R}$ is continuous, so $f$ is continuous, so
$$f(x) = \lim_{h \to 0} (t f(x) + (1-t) f(x+h)) = \lim_{h \to 0} \frac{x + x + h}{2} = x.$$

To justify the first equality, note that $t f(x) + (1-t) f(x+h)$ lies between $f(x)$ and $f(x+h)$ for all $h$. The intersection of all these intervals $[f(x), f(x+h)]$ (or $[f(x+h),f(x)]$, depending on the value of $h$) is just $\{f(x)\}$ (this is by continuity of $f$). Thus, the limit exists and equals $f(x)$.


问:
Complete square values of quadratic formula
For Which values of $x \in [0,n]$ the polynomial $P(x)=ax^2+bx+c$  $\ $ gives a complete square value.
For instance the polynomial $P(x)=3x^2+5x+7$ on the interval $x \in [0,100]$ has only two values for $x=3  \rightarrow P(3)=49 $ and for $x=54  \rightarrow P(54)=9025 $.
I was wondering if there is a method to find such values of $x$ directly.

答:
I don't know how useful it gets with calculations, but if you let $k\in \Bbb N$, and set
$$P(x)=ax^2+bx+c = k^2 ,$$
and rewriting it as an equation equaling zero
$$ax^2+bx+c-k^2=0,$$
you can use the quadratic formula to find
$$x= \frac{-b\pm \sqrt{b^2-4a(c-k^2)}}{2a}.$$
So in theory you should get solutions only of this type with $k$ natural number, in particular you need
$$ak^2 \ge \frac{4ac-b^2}{4}.$$
I hope I answered you question, or at least helped a bit.


问:
Michael line is normal direct proof
I'm trying to do a direct proof to show that Michael line is normal. Michael line is the topological space of $\mathbb{R}$ with the topology $\tau*=\{U \cup B : U \in \tau, B \subseteq \mathbb{R} - \mathbb{Q} \}$
Where $\tau$ Is the usual topology

Now let $F, K \subseteq \mathbb{R}$ be disjoint closed sets and we contruct for each $x \in F$ the sets $A_{x}$ as follows: if $x\notin \mathbb{Q}$ then $A_{x} = \{x\}$  Otherwise, since $F, K $ are disjoint then $F \subseteq \mathbb{R} - K$ and as $K$ is closed then, there exists $W \in \tau,  B\subseteq \mathbb{R}-\mathbb{Q} $ such that $\mathbb{R} - K = W \cup B$ and as $x \in \mathbb{Q}$ It follows $x \in W$ and then $\exists \delta_{x} >0$ such that $B(x; \delta_{x}) \subseteq W$ in this case we choose $A_{x} = B(x; \delta_{x})$ obseve that every $A_{x}$ is open

Similarly we contruct for each $k \in K$ open sets $B_{k}$ as we did with every $A_{x}$

My claim is that $U = \bigcup_{x \in F} A_{x}, V = \bigcup_{k \in K} B_{k} \in \tau*$ are disjoint, I would appreciate any hint for that because I tried contradiction but didn't know how to continue the case in wich I have some $w \in B(x; \delta_{x}) \cap B(k; \delta_{k})$ with $x \in F, k \in K$

答: I suggest taking a slightly different approach. Let $F_0=F\cap\Bbb Q$ and $K_0=K\cap\Bbb Q$. Then $F_0$ and $K_0$ are closed in the topology that $\Bbb Q$ inherits from the Michael line, but that is simply its usual topology, so $F_0$ and $Q_0$ are disjoint closed sets in the metric space $\Bbb Q$. Thus, they have disjoint open nbhds $U_0$ and $V_0$ in $\Bbb Q$. For each $x\in F_0$ there is an $\epsilon_x>0$ such that $B_{\Bbb Q}(x,\epsilon_x)\subseteq U_0$ and $B_{\Bbb R}(x,\epsilon_x)\cap K=\varnothing$, where $B_{\Bbb R}$ denotes a ball in the usual metric on $\Bbb R$. Similarly, for each $x\in K_0$ there is an $\epsilon_x>0$ such that $B_{\Bbb Q}(x,\epsilon_x)\subseteq V_0$ and $B_{\Bbb R}(x,\epsilon_x)\cap F=\varnothing$. Let $U_1=\bigcup_{x\in F_0}B_{\Bbb R}(x,\epsilon_x)$ and $V_1=\bigcup_{x\in K_0}B_{\Bbb R}(x,\epsilon_x)$; clearly $U_1$ and $V_1$ are open in the Michael line, $U_1\cap V_1=\varnothing$, $F_0\subseteq U_1$, and $K_0\subseteq V_1$. Moreover, $U_1\cap K=V_1\cap F=\varnothing$, so $U=U_1\cup(F\setminus\Bbb Q)$ and $V=V_1\cup(K\setminus\Bbb Q)$ are disjoint open nbhds of $F$ and $K$ in the Michael line.

Title: Michael line is normal direct proof


问:
Find $u\in\mathbb{R}$ such that $\mathbb{Q}(u) = \mathbb{Q}(2^{1/2}, 5^{1/3})$.
I am having trouble finding such a $u$.  My instincts at first told me to do the obvious thing and let $u = 2^{1/2}5^{1/3}$ but $u^{2} = \left(2^{1/2}5^{1/3}\right)^{2} = 2\cdot5^{2/3}$ but we want $\frac{1}{2}u^{2} = 5^{1/3}$ right??  Any hints on how to approach the exercise?

Then the next part ask to describe how you would find all $w\in\mathbb{Q}(2^{1/2}, 5^{1/3})$ such that $\mathbb{Q}(w) = \mathbb{Q}(2^{1/2}, 5^{1/3})$ I have no approach on this one except to possible look at linear combinations of the basis terms but dont know how to go about it.

答: Hint: Use the proof of the primitive element theorem.


答:
An idea for you: since $\;z_0\;$ is a simple zero of $\;g(z)\;$ , we have that $\;g(z)=(z-z_0)h(z)\;$ , where $\;h\;$ is analytic and non-zero in some neighborhood of $\;z_0\;$ ,
and $\;f(z)\neq0\;$ in the same neighborhood. Observe also that $\;g'(z)=h(z)+(z-z_0)h'(z)\implies g'(z_0)=h(z_0)\;$

$$\text{Res}_{z_0}\,\left(\frac fg\right)=\lim_{z\to z_0}(z-z_0)\frac{f(z)}{g(z)}=\lim_{z\to z_0}\frac{f(z)}{h(z)}=\frac{f(z_0)}{h(z_0)}=\frac{f(z_0)}{g'(z_0)}$$

Title: Rule of calculation of residue

Body:
Let $U \subset \Bbb C$ be open and $f,g:U \to \Bbb C$ be holomorphic. I wanna show that if $g$ has a simple zero in $z_0\in U$, then
$$\operatorname{Res}\biggl(\frac{f}{g},z_0\biggr)=\frac{f(z_0)}{g'(z_0)}$$
The proof I use is: Because $f$ and $g$ are holomorphic we can expand them as entire serie
$$f(z)=b_0+b_1(z-z_0)+b_2(z-z_0)+ \dots$$
and
$$g(z)=c_1(z-z_0)+c_2(z-z_0)+ \dots$$
Where we already used that $g$ has a simple zero in $z_0$.
$$\frac{b_0+b_1(z-z_0)+b_2(z-z_0)^2+ \dots}{c_1(z-z_0)+c_2(z-z_0)^2+ \dots}=\frac{1}{z-z_0}(\frac{b_0+b_1(z-z_0)+b_2(z-a)^2+ \dots}{c_1+c_2(z-z_0)+\dots})$$
The right fraction is nothing else than the holomorphic function $h(z):=\frac{(z-z_0)f(z)}{g(z)}$ with $h(z_0)=\frac{b_0}{c_1}=\frac{f(z_0)}{g'(z_0)}$. But this should be the end of the proof. But why? I don't understand the end...

EDIT: Is it not wrong? Should it not be that $f$ can be represented as a Laurent serie near $z_0$ and not as an entire function?


问:
Which of the following sets are subspaces of V?
Let V be a vector space and let U, W, X $\subset$ V be sub-vector spaces of V.

Which of the following sets are again subspaces of V?

a) U $\triangle$ W

b) U $\cap$ W

c) U $\cup$ W

d) v + U for a v $\in$ V

*So a subspace of a Vector V is a subset U that satisfies three Axioms(Conditions):

(UV0) The zero Vector is in U

(UV1) U is closed under addition

(UV2) U is closed under multipication of scalars*

-For a) I don`t know how to prove if it is a vector subspace of V or not! Need some help

-For b) I have proved this way but I´m not sure: We have to check if the three Axioms are satisfied:

(UV0): Is 0 $\in$ U and 0 $\in$ W, and because that both of them are subspaces $\Rightarrow$ 0 $\in$ U $\cap$ W $\Rightarrow$ U $\cap$ W $\neq$ $\varnothing$

(UV1): Let u,v $\in$ U $\cap$ W, and because that U and W are subspaces, follows:
(u,v $\in$ U $\Rightarrow$ u+v $\in$ U) and (u,v $\in$ W $\Rightarrow$ u+v $\in$ W) $\Rightarrow$ u+v $\in$ U $\cap$ W

(UV2): Let u $\in$ U $\cap$ W and $\lambda \in$ K $\Rightarrow$ (u $\in$ U, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ U ) and (u $\in$ W, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ W )
$\Rightarrow$ $\lambda$u $\in$ U $\cap$ W.

-For c) I need an Idea aswell and

-For d) I think it should be proved with to cases 1.Fall v $\notin$ U and 2.Fall v$\in$U, I´m not sure if this is the right Idea and if it is how to prove it

Thank you very much:)

答:
All your ideas are right. The proof is right. Adapt it to prove that $v+U$ is a subspace when $v \in U$. Can $0 \in v+U$ if $v \notin U$?

For union think of lines in the plane for a counterexample.

For $U \Delta V$, can $0$ be in it?

Proof for $v+U$. If $v \notin U$, then suppose for a contradiction that $v+U$ is a subspace. Then $0 \in v+U$ i.e. there is an $u \in U$ such that $0 = v + u$. But then $v=-u \in U$, which is a contradiction.

If $v \in U$, then $\forall u \in U$, we have that $v+u \in U$. So $v+U \subseteq U$. Moreover, $\forall u \in U$ we have $(-v)+u \in U$, so $u = v + (-v) + u \in v + U$. Hence, also $U \subseteq v + U$.

By double inclusion, $v + U = U$, and thus is a subspace.


答:
Proposition: $\;U\cup V\;$ is a vectorial subspace iff $\;U\subset V\;$ or $\;V\subset U\;$

Hint for the proof: Direction $\;\Longleftarrow\;$ is trivial. For the other diection, suppose that both $\;U\not\subset V\;$ and also $\;V\not\subset U\;$ . Then there exists $\;u\in U\setminus V\;$ nad $\;v\in V\setminus U\;$ ...well, your turn

Title: Which of the following sets are subspaces of V?

Body:
Let V be a vector space and let U, W, X $\subset$ V be sub-vector spaces of V.

Which of the following sets are again subspaces of V?

a) U $\triangle$ W

b) U $\cap$ W

c) U $\cup$ W

d) v + U for a v $\in$ V

*So a subspace of a Vector V is a subset U that satisfies three Axioms(Conditions):

(UV0) The zero Vector is in U

(UV1) U is closed under addition

(UV2) U is closed under multipication of scalars*

-For a) I don`t know how to prove if it is a vector subspace of V or not! Need some help

-For b) I have proved this way but I´m not sure: We have to check if the three Axioms are satisfied:

(UV0): Is 0 $\in$ U and 0 $\in$ W, and because that both of them are subspaces $\Rightarrow$ 0 $\in$ U $\cap$ W $\Rightarrow$ U $\cap$ W $\neq$ $\varnothing$

(UV1): Let u,v $\in$ U $\cap$ W, and because that U and W are subspaces, follows:
(u,v $\in$ U $\Rightarrow$ u+v $\in$ U) and (u,v $\in$ W $\Rightarrow$ u+v $\in$ W) $\Rightarrow$ u+v $\in$ U $\cap$ W

(UV2): Let u $\in$ U $\cap$ W and $\lambda \in$ K $\Rightarrow$ (u $\in$ U, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ U ) and (u $\in$ W, $\lambda$ $\in$ K $\Rightarrow$ $\lambda$u $\in$ W )
$\Rightarrow$ $\lambda$u $\in$ U $\cap$ W.

-For c) I need an Idea aswell and

-For d) I think it should be proved with to cases 1.Fall v $\notin$ U and 2.Fall v$\in$U, I´m not sure if this is the right Idea and if it is how to prove it

Thank you very much:)


问:
How to relate Molien's Theorem to this result.

Let $G$ be a finite group and $(\pi, V )$ a finite dimensional $\mathbb{C}G$-module. For $n \ge 0$, let
$\mathbb{C}[V]_n$ be the space of homogeneous polynomial functions on $V$ of degree $n$. For a simple
$G$-representation $\rho$, denote by $a_n(\rho)$ the multiplicity of $\rho$ in $\mathbb{C}[V]_n$. Show that

$$
\sum_{n\ge 0} a_n(\rho) t^n = \sum_{g \in G} \frac{1}{|G|}\frac{\chi_\rho(g)}{\det(id_V-\pi(g)t)}.
$$

This seems to be very similar the the proof of Molien's Theorem, I just do not see how the multiplicity of the simple function gives a factor of $\chi_\rho(g)$ on each term of the sum on the right.

答:
The proof of Molien's theorem carries through for this generalization with no issues.

The degree $n$ homogeneous polynomial functions on a rep $V$ can be identified with the $n$th symmetric power of the dual rep, $\mathbb{C}[V]_n=\mathrm{Sym}^nV^{\ast}$. The only difference between the characters of $\mathrm{Sym}^nV$ and $\mathrm{Sym}^nV^{\ast}$ is that they are complex conjugates.

The multiplicity of an irrep $U$ within a rep $W$ is $\dim\hom_G(U,W)$, which is calculable as

$$ \dim\hom_G(U,W)=\dim(U^{\ast}\otimes W)^G=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_U(g)}\chi_W(g). $$

Therefore, the multiplicities may be expressed as

$$ a_n(U)=\dim\hom_G(U,\mathrm{Sym}^nV^{\ast})=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_U(g)}\, \overline{\chi_{\mathrm{Sym}^nV}(g)}. $$

Since this expression is real, we may discard the complex conjugations.

If $d=\dim V$ and the eigenvalues of $\rho_V(g)$ are $\lambda_1,\cdots,\lambda_d$ (imagine these are functions of $g$) then the eigenvalues of $\rho_{\mathrm{Sym}^nV}(g)$ are $\lambda_1^{k_1}\cdots\lambda_d^{k_d}$ with $k_1+\cdots+k_d=n$ (all counted with multiplicity).

Therefore we may write the function as

$$ \begin{array}{ll} M(t) & \displaystyle = \sum_{n=0}^\infty a_n(U)t^n \\[5pt]
& \displaystyle =\sum_{n=0}^\infty t^n\cdot\frac{1}{|G|}\sum_{g\in G} \chi_U(g)\chi_{\mathrm{Sym}^nV}(g) \\[5pt]
& \displaystyle =\frac{1}{|G|}\sum_{n=0}^\infty t^n\sum_{g\in G}\chi_U(g)\sum_{k_1+\cdots+k_d=n}\lambda_1^{k_1}\cdots\lambda_d^{k_d} \\[5pt]
& \displaystyle =\frac{1}{|G|}\sum_{g\in G}\chi_U(g)\sum_{k_1=0}^\infty\cdots\sum_{k_d=0}^\infty (\lambda_1t)^{k_1}\cdots(\lambda_dt)^{k_d} \\[5pt]
& \displaystyle = \frac{1}{|G|}\sum_{g\in G}\frac{\chi_U(g)}{(1-\lambda_1t)\cdots(1-\lambda_dt)} \\[5pt]
& \displaystyle = \frac{1}{|G|}\sum_{g\in G}\frac{\chi_U(g)}{\det(I_V-\rho_V(g)t)}. \end{array} $$


答:
$y^2=3x^2+5x+7$.

$12y^2=36x^2+60x+84=(6x+5)^2+59$.

$u^2-3v^2=-59$ where $u=6x+5$, $v=2y$.

Given one solution of $u^2-3v^2=-59$, and a fundamental solution of $u^2-3v^2=1$, you can get infinitely many solutions of $u^2-3v^2=-59$, and from those infinitely many solutions of $y^2=3x^2+5x+7$.

Now a fundamental solution of $u^2-3v^2=1$ is $u=2$, $v=1$. A solution of $u^2-3v^2=-59$ is given by $u=4$, $v=5$. If you let $a_n+b_n\sqrt3=(2+\sqrt3)^n(4+5\sqrt3)$, $n=1,2,3,\dots$, then you'll get $a_n^2-3b_n^2=-59$. Then you can see which of those solutions yield integers for $x,y$.

Title: Complete square values of quadratic formula

Body:
For Which values of $x \in [0,n]$ the polynomial $P(x)=ax^2+bx+c$  $\ $ gives a complete square value.
For instance the polynomial $P(x)=3x^2+5x+7$ on the interval $x \in [0,100]$ has only two values for $x=3  \rightarrow P(3)=49 $ and for $x=54  \rightarrow P(54)=9025 $.
I was wondering if there is a method to find such values of $x$ directly.


问:
If $x_1, x_2$ are roots of $a\cos x+b\sin x+c=0$ for $x_1+x_2\ne 2k\pi$ show that $\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}.$
If $x_1, x_2$ are roots of $a\cos x+b\sin x+c=0$ for $x_1+x_2≠2kπ$ show that $\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$

What I've done till now:

$$a\cos x_1+b\sin x+c-(a\cos x_2+b\sin x_2+c)=0$$

$a\cos x_1+b\sin x_1-a\cos x_2-b\sin x_2=0$

$a(\cos x_1-\cos x_2)+b(\sin x_1-\sin x_2)=0$

$a(-2\sin(\frac{x_1+x_2}{2})\sin(\frac{x_1-x_2}{2}))+b(2\sin(\frac{x_1-x_2}{2})\cos(\frac{x_1+x_2}{2}))=0$

$-a\sin(\frac{x_1+x_2}{2})+b(\frac{\cos x_1+x_2}{2})=0$

That's it. Does anyone know how to solve this?

答:
Just rewrite this equation in terms of complex exponentials:

$$\frac{a-bi}{2}e^{ix}+\frac{a+bi}{2}e^{-ix}+c=0$$

Setting $z=a+bi$ one can rewrite the equation as a quadratic in $w=e^{ix}$
$$\bar{z}w^2+cw+z=0$$
From Vieta's formulae we know that the product of the two roots of the equation is

$$w_1w_2=e^{i(x_1+x_2)}=\frac{z}{\bar{z}}$$

which, upon separating real and imaginary parts yields

$$\cos(x_1+x_2)=\frac{a^2-b^2}{a^2+b^2}~,~\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$$


问:
A question about direct sums of subspaces...

Let $W_1$ be a subspace of a finite dimensional vector space $V$. Prove that there exists a subspace $W_2\subset V$ such that $V=W_1\oplus W_2$.

EDIT$^1$:

This may be of use here:

What does
$$ W_1 \cap W_2 = \{0\}$$
even mean? The zero vector is the only thing shared in common by the two subset?

答: Consider the set $\mathcal W$ of subspaces $W\subseteq V$ with $W_1\cap W=\{0\}$. Let $W_2$ be a maximal element of $\mathcal W$ (with respect to inclusion). Why does such maximal element exist and why does that solve the problem?</p>


问:
Is this proof of Carathéodory's theorem valid?
I'm working through some discrete geometry and am trying to get a more rigorous grasp of convex geometry. In particular, I would like to try proving this theorem without the use of Radon's Lemma, which is the classic route. Is this proof valid?

Carathéodory's Theorem: Given a point set $X \subset \mathbb{R}^d$, any point in conv$(X)$ may be expressed as a convex combination of at most $d+1$ points in $X$.

Proof: Let $X$ be as in the statement; if $|X| \leq d+1$, we are done, so assume that $|X| > d+1$, and assume for contradiction that $c \in$ conv$(X)$ may not be expressed as a convex combination of $d+1$ points in $X$. Since the convex hull are those points which may be expressed as convex combination, we know that $c$ is a convex combination of $q>d+1$ points, say $\{x_i\}_{i=1}^{q} = Z \subseteq X$.

As the largest affine independent point set in $\mathbb{R}^d$ is of size $d+1$, we know that $c$ may be expressed as an affine combination of $d+1$ points $D$ in $Z$, since the convex hull is a subset of the affine hull. That is:

$$c = \sum_{n=1}^{d+1} b_n x_{i_n}$$

where $\sum_{n=1}^{d+1} b_n = 1$ and the $b_i \in \mathbb{R}$. By the supposition, at least one of the $b_i$ must be negative; suppose that it is $b_1$. I claim we can find a convex set which includes $D$ and excludes $c$. Let $\lambda_1 = b_1/2$ and let:

$$\lambda_n = b_n + \frac{b_1}{2d}$$

It follows that $\sum_{n=1}^{d+1} \lambda_n = 1$. Let $l = \sum_{n=1}^{d+1} \lambda_n x_{i_n}$; I claim that the set described above is conv$(D \cup\{l\}) = K$. It is clear that $K \supseteq$ conv$(D)$, we must show that $c \notin K$. Indeed, assume for contradiction that $c \in K$, which implies $(c+l)/2$ is a convex combination of $K$. This means that $c+l = 2\sum_{n=1}^{d+1} \Lambda_n x_{i_n} + 2\Lambda_{d+2}l$, where $\sum \Lambda_i =1$ and $\Lambda_i \geqslant 0$. Then:

$$c + (1-2\Lambda_{d+2})l = 2\sum_{n=1}^{d+1} \Lambda_n x_{i_n}$$

which implies:

$$\frac{b_1 (3-2\Lambda_{d+2})}{2} = 2\Lambda_1$$

but by the convexity condition, $0 \leqslant \Lambda_{d+2} \leqslant 1$, and the supposition gives $b_1 < 0$, hence we have:

$$0 > \frac{b_1}{2} \geqslant \frac{b_1 (3-2\Lambda_{d+2})}{2} = 2\Lambda_1 \geqslant 0$$

which is a contradiction. Then $c \notin K$, where $K$ contains $D$. In other words, there is a small-enough convex set that includes any $D$ but not $c$; by definition $c$ is not in the convex hull of $D$. Any decomposition of $\text{conv}(X)$ into simplices shows that this is impossible ($c$ must be in some simplex), so we are done. $\square$

EDIT: For clarity on my main concern, I do not know if the last sentence is justified, or is in fact identical to Caratheodory's Theorem in a different wording. Am I begging the question?

答: Carathéodory's theorem can be rephrased as "given a set $X \subseteq \mathbb{R}^d$, the set of simplices $\{\text{conv}(Y) : Y \subseteq X, |Y| \leq d+1\}$ covers $\text{conv}(X)$", i.e. the union if these simplices is exactly $\text{conv}(X)$. So it seems, indeed, that you use the result that you are trying to prove.