问:
How to find a Rotation Matrix given 2 rotated coordinates
I have been looking at this question for a while now and have not found any success.

What is the rotation matrix that relates the coordinate systems in the figure below - given that ro2a^x1y1x1 = [6 8 10]T and ro2b^(x1y1x1) = [-6 -8 10]t (the o2a and o2b is subscript and x1y1x1 is superscript). Briefly explain how you derived such a matrix

Any help would be amazing, Thanks

答:
Erik... Dude... This question comes up every year! So how far are you through the CBEPS process anyway? LOL! Struggling with this typo-filled question from the C-7 Remote Sensing exam?? (How did I know?? Hahaha!)

That is certainly an interesting question, but "interesting" in a slap-myself-in-the-face kind of way.

I'll try to word my answer in a fashion that is less focused on generalized geometry and better focused on this specific CBEPS question. You are welcome to share this message with anyone who might benefit as they move forward through the CBEPS process.

First, a bit of background:

This photogrammetry question appears EVERY YEAR on the C-7 "remote sensing" exam. It is exactly the same every year, even including the two typos.

The question always includes two vectors in the form [ a b c ] and [ -a -b c ], where the a and b values flip sign from one vector to the other, while c holds the same sign. Sometimes a, b, and c are 3, 4, 5 and sometimes they're 6, 8, 10. These small changes are trivial -- The answer to the question is exactly the same, year after year.

The superscripts in both reference vectors are typed incorrectly. Year after year, they never fix this typo. The superscript specifies the coordinate frame, which should be x1 y1 z1, not x1 y1 x1. The designers of this exam are smart, picky people, yet they let this dumb typo persist year after year.

Each year, CBEPS candidates show up on various math/compsci discussion forums asking for help... and the help never comes.

With that background in mind, I'll share my thoughts on the common WRONG answer that comes up every year.

Incorrectly, people think the following 3x3 matrix does the trick:

```<code>-1   0   0
 0  -1   0
 0   0   1
</code>```

#1) I'm happy to say that the answer is based on a solid intuition. Yes, indeed, the flipping of signs of the x and y values (while preserving the z value) would be achieved by either: (a) 180° rotation about the z axis, or (b) scaling both x and y values by -1. (Note that a negative scale factor has the effect of reflection.)

#2) Regardless of whether you choose the rotation or the reflections, the numbers work out the same. The reflection scaling is -1... The cosine of 180° is -1... whatever... It's all good. Either way, the resulting (correct) transformation matrix is:

```<code>-1   0   0   0
 0  -1   0   0
 0   0   1   0
 0   0   0   1
</code>```

This matrix can be described as "a 4x4 matrix with zeros in upper and lower triangles and a trace of [ -1 -1 +1 +1 ]."

The first and second "-1" values represent a reflection in x and y values while the next "1" represents preserving z values unchanged.
Alternatively, the "-1" values represent multiplying cos(180°) by cos(0°) and the "1" represents  multiplying cos(0°) by cos(0°). Whatever... It's all the same deal, right?

#3) ---> [The point of this post!] <---

Students often provide this (incorrect) transformation matrix:

```<code>-1   0   0
 0  -1   0
 0   0   1
</code>```

Yes, I see the intuitive appeal in the simplicity of this answer, but it is wrong for the following reason:

There are 15 parameters to a transformation in 3-space.
translate x,
translate y,
translate z,
scale x,
scale y,
scale z,
rotate around x axis,
rotate around y axis,
rotate around z axis,
shear xy,
shear xz,
shear yz,
projection xy,
projection xz, and
projection yz.

A math nerd would point out that these 15 parameters capture all possible "affine transforms" and "projection transforms," but would neglect any twisting transforms. That's okay -- There are no applications of twisting transforms in photogrammetry. (There are indeed some applications in computer graphics, but those 5x5 transformation matrices are not relevant to your average CBEPS candidate.)

Having said that, there certainly ARE applications of shearings and projections in photogrammetry.

In fact, not to put too fine of a point on it, but... I gotta say that I would be a little concerned about a student who is expected to understand photogrammetry, yet presents a 3x3 matrix as an adequate description of a photogrammetric transform.

Sure, it is true that a lot of land surveying transforms can be captured in a 3x3 matrix. This is the case only because land surveyors tend to "flatten" their data into N and E values, leaving all points on a 2-d mapping plane. All 2-d transforms can be expressed with a 3x3 matrix.

Photogrammetry, however, is necessarily a 3-d enterprise. A remote sensing student ought to know that a 3-space transformation matrix CANNOT be completely expressed in a 3x3 matrix. Any sensible answer to the C-7 question must be expressed as a 4x4 matrix.

I hope that you find good targets for this message -- I hope someone benefits from my grief over this question, typos and all, along with students' failure to realize that the 3-d affine transforms require a 4x4 matrix.

But hey... For those students who don't care about photogrammetry, but just want to push through the C-7 and never use photogrammetry in their land surveying careers, then it all boils down to this single bit of advice:

The correct answer is this matrix:

```<code>-1   0   0   0
 0  -1   0   0
 0   0   1   0
 0   0   0   1
</code>```

... along with some explanation about how "obviously the transformation is a 180° rotation about the z axis."

That was the correct answer this year... and last year... and the year before that... and the year before that...

... and that's what the correct answer will be when they give the same typo-filled question on the C-7 exam in 2022.

Buy hey... I'm just a first-year student at NAIT. I'm happy on those days when I remember how many legs are on a tripod!

Good luck on them 'zams!

Body:
Erik... Dude... This question comes up every year! So how far are you through the CBEPS process anyway? LOL! Struggling with this typo-filled question from the C-7 Remote Sensing exam?? (How did I know?? Hahaha!)

That is certainly an interesting question, but "interesting" in a slap-myself-in-the-face kind of way.

I'll try to word my answer in a fashion that is less focused on generalized geometry and better focused on this specific CBEPS question. You are welcome to share this message with anyone who might benefit as they move forward through the CBEPS process.

First, a bit of background:

This photogrammetry question appears EVERY YEAR on the C-7 "remote sensing" exam. It is exactly the same every year, even including the two typos.

The question always includes two vectors in the form [ a b c ] and [ -a -b c ], where the a and b values flip sign from one vector to the other, while c holds the same sign. Sometimes a, b, and c are 3, 4, 5 and sometimes they're 6, 8, 10. These small changes are trivial -- The answer to the question is exactly the same, year after year.

The superscripts in both reference vectors are typed incorrectly. Year after year, they never fix this typo. The superscript specifies the coordinate frame, which should be x1 y1 z1, not x1 y1 x1. The designers of this exam are smart, picky people, yet they let this dumb typo persist year after year.

Each year, CBEPS candidates show up on various math/compsci discussion forums asking for help... and the help never comes.

With that background in mind, I'll share my thoughts on the common WRONG answer that comes up every year.

Incorrectly, people think the following 3x3 matrix does the trick:

-1   0   0
 0  -1   0
 0   0   1

#1) I'm happy to say that the answer is based on a solid intuition. Yes, indeed, the flipping of signs of the x and y values (while preserving the z value) would be achieved by either: (a) 180° rotation about the z axis, or (b) scaling both x and y values by -1. (Note that a negative scale factor has the effect of reflection.)

#2) Regardless of whether you choose the rotation or the reflections, the numbers work out the same. The reflection scaling is -1... The cosine of 180° is -1... whatever... It's all good. Either way, the resulting (correct) transformation matrix is:

-1   0   0   0
 0  -1   0   0
 0   0   1   0
 0   0   0   1

This matrix can be described as "a 4x4 matrix with zeros in upper and lower triangles and a trace of [ -1 -1 +1 +1 ]."

The first and second "-1" values represent a reflection in x and y values while the next "1" represents preserving z values unchanged.
Alternatively, the "-1" values represent multiplying cos(180°) by cos(0°) and the "1" represents  multiplying cos(0°) by cos(0°). Whatever... It's all the same deal, right?

#3) ---> [The point of this post!] <---

Students often provide this (incorrect) transformation matrix:

-1   0   0
 0  -1   0
 0   0   1

Yes, I see the intuitive appeal in the simplicity of this answer, but it is wrong for the following reason:

There are 15 parameters to a transformation in 3-space.
translate x,
translate y,
translate z,
scale x,
scale y,
scale z,
rotate around x axis,
rotate around y axis,
rotate around z axis,
shear xy,
shear xz,
shear yz,
projection xy,
projection xz, and
projection yz.

A math nerd would point out that these 15 parameters capture all possible "affine transforms" and "projection transforms," but would neglect any twisting transforms. That's okay -- There are no applications of twisting transforms in photogrammetry. (There are indeed some applications in computer graphics, but those 5x5 transformation matrices are not relevant to your average CBEPS candidate.)

Having said that, there certainly ARE applications of shearings and projections in photogrammetry.

In fact, not to put too fine of a point on it, but... I gotta say that I would be a little concerned about a student who is expected to understand photogrammetry, yet presents a 3x3 matrix as an adequate description of a photogrammetric transform.

Sure, it is true that a lot of land surveying transforms can be captured in a 3x3 matrix. This is the case only because land surveyors tend to "flatten" their data into N and E values, leaving all points on a 2-d mapping plane. All 2-d transforms can be expressed with a 3x3 matrix.

Photogrammetry, however, is necessarily a 3-d enterprise. A remote sensing student ought to know that a 3-space transformation matrix CANNOT be completely expressed in a 3x3 matrix. Any sensible answer to the C-7 question must be expressed as a 4x4 matrix.

I hope that you find good targets for this message -- I hope someone benefits from my grief over this question, typos and all, along with students' failure to realize that the 3-d affine transforms require a 4x4 matrix.

But hey... For those students who don't care about photogrammetry, but just want to push through the C-7 and never use photogrammetry in their land surveying careers, then it all boils down to this single bit of advice:

The correct answer is this matrix:

-1   0   0   0
 0  -1   0   0
 0   0   1   0
 0   0   0   1

... along with some explanation about how "obviously the transformation is a 180° rotation about the z axis."

That was the correct answer this year... and last year... and the year before that... and the year before that...

... and that's what the correct answer will be when they give the same typo-filled question on the C-7 exam in 2022.

Buy hey... I'm just a first-year student at NAIT. I'm happy on those days when I remember how many legs are on a tripod!

Good luck on them 'zams!


问:
The question of linear space
Let $V$ be a vector space and $v \in V$ . Prove that

\begin{align*}
-v = (-1) \cdot v
\end{align*}

by only using the vector spaces axioms.

答:
Let $V$ be a vector space over a field $\Bbb{K}$ and $v \in V$. We first prove that $0_{\Bbb{K}} \cdot v = 0_V$. We have that

\begin{align*}
0_{\mathbb{K}} \cdot v & = (0_{\Bbb{K}} + 0_{\Bbb{K}}) \cdot v & (0_{\Bbb{K}} \in \Bbb{K} \text{ identity for $+_{\Bbb{K}}$}) \\
& = 0_{\Bbb{K}} \cdot v + 0_{\Bbb{K}} \cdot v & (\text{distributive law})
\end{align*}

Next, note that you can add $- 0_{\Bbb{K}} \cdot v$ to each side of the above equality (Why?). From this you will get that $0_{\Bbb{K}} \cdot v = 0_V$. Then we conclude that

\begin{align*}
v + (-1)\cdot v & = 1_{\Bbb{K}} \cdot v + (-1) \cdot v & (1_{\Bbb{K}} \in \Bbb{K} \text{ identity for } \cdot) \\
& = (1_{\Bbb{K}} + (-1)) \cdot v & (\text{distributive law}) \\
& = 0_{\Bbb{K}} \cdot v & (\text{existence of multiplicative inverse}) \\
& = 0_V
\end{align*}

Hence, we have that $(-1) \cdot v \in V$ is a symmetric of $v$. I will assume that you already have proved that the symmetric element of $v$ is unique (if not, you can ask), so we conclude that $-v=(-1)\cdot v$. $\square$

Any doubt on this, please let me know.

ContentLicense: CC BY-SA 4.0

Title: The question of linear space


问:
Extrema of $(1+\sin x)(1+\cos x)$

Find the extrema of $(1+\sin x)(1+\cos x)$ without using calculus.

I was able to figure out the minima by observing that each of the brackets range from $0$ to $2$. Therefore the minima has to be $0$ when either one of the brackets is zero.

However I couldnt figure out the maxima. I tried expanding it to complete the square but it didnt quite work out well.

Any hint is appreciated!

答:
Note

\begin{align}
A&=(1+\sin x)(1+\cos x)\\
&= 1+\sin x+ \cos x+ \frac12\sin2x\\
 &= 1+\sqrt2 \cos(\frac\pi4-x)+ \frac12\cos(\frac\pi2-2x)\\
&= \frac12+\sqrt2 \cos(\frac\pi4-x)+ \cos^2(\frac\pi4-x)\\
&= \left( \cos(\frac\pi4 -x) + \frac1{\sqrt2}\right)^2
\end{align}
Thus
$$0\le A \le (1+\frac1{\sqrt2})^2$$

Title: Extrema of $(1+\sin x)(1+\cos x)$


问:
The kernel of an evaluation homomorphism

We let $R$ be a commutative ring with unity, and $R[X_1,\dots, X_n]$, $a=(a_1,\dots, a_n)\in R^n$ and $\phi_a:R[X_1,\dots, X_n]\rightarrow R$, $\phi_a(f)=f(a)$. I want to show that $\ker(\phi_a)=(X_1-a_1,\dots, X_n -a_n)$.

I know that to be true for $n=1$. I want to assume that WLOG, $a=(0,\dots,0)$, but I can't see why I'm allowed to do so (in this case it'll be easy to solve) - I didn't learn about maximal ideals yet.

Is my approach correct? How can I justify assuming $a=(0,\dots,0)$?

答:
The universal property of $R[X_1,\ldots,X_n]$ says that for every commutative ring $S$, every ring homomorphism $\varphi\colon R\to S$, and every choice of elements $s_1,\ldots,s_n$, there exists a unique ring homomorphism $\Phi\colon R[X_1,\ldots,X_n]\to S$ such that $\Phi(X_i) = s_i$ for $i=1,\ldots, n$ and $\Phi(r)=\varphi(r)$ for each $r\in R$.

So take $S=R[X_1,\ldots,X_n]$, $\varphi\colon R\to S$ the standard embedding of $R$ into the polynomial ring as the constant polynomials, and take $s_i=X_i-a_i$. Let $\Phi\colon R[X_1,\ldots,X_n\to R[X_1,\ldots,X_n]$ be the corresponding morphism.

Verify that $\Phi$ is an isomorphism of $R[X_1,\ldots,X_n]$ to itself (for example, verify that the map $\Psi$ sending $X_i$ to $X_i+a_i$ is its inverse).

Now verify that $\phi_{\mathbf{0}}\circ \Phi = \phi_{a}$ and use $\Phi$ and $\Psi$ to transfer ideal information from one to the other.


问:
Why does the hypotenuse of half the triangle (formed from a circle) ≈ pi * r?
Just a while ago, while feeling bored, I took a pen and pencil, and started doing some math. After a while of doing something I found out that the hypotenuse of half the triangle (formed from a circle) ≈ pi * r
Here's how I got to that:

Convert the circle into a triangle

Now, the radius is the height of the triangle, and the circumference is the base of the triangle

Next we replace the radius with r (for simplicity), and replace the circumference with its formula (2 pi * r)

Next we divide the triangle in half (making identical 2 triangles)

Now the base of the triangle is half (pi * r), but the height is the same (r)

Now we try to calculate the hypotenuse --

h = sqrt( (r)^2 + (pi * r)^2 )

Note: sqrt means square root

Now we add in the values and calculate

For some reason, the hypotenuse ≈ pi * height
For example:
radius(h) = 3 cm
sqrt( (r)^2 + (pi * r)^2 )

= sqrt( (3)^2 + (3.14 * 3)^2 )

= sqrt( 9 + (9.42)^2 )

= sqrt( 9 + 88.7364 )

= sqrt( 97.7364 )

= 9.88617216116

Which approximately equals:
pi * r
= 3.14 * 3
= 9.42

Please tell me if I'm missing something super obvious.

答:
You appear to be forming an isosceles triangle by "unrolling" a circle a distance equal to its circumference to form the base of a triangle with height equaling the radius.

Then the length of the two congruent sides equals $\sqrt{r^2+\pi^2r^2}=r\sqrt{1+\pi^2}$.

Can you see now why this "approximately" equals $\pi r$?


问:
Show that $ \prod_{k=1}^{n}a_{k, \sigma(k)} = \prod_{k=1}^{n}a_{\sigma^{-1}(k), k}$ on any matrix
At some point, in proving that $\det{A} = \det{A^T}$, one states that

\begin{align*}
\prod_{k=1}^{n}a_{k, \sigma(k)} = \prod_{k=1}^{n}a_{\sigma^{-1}(k), k}
\end{align*}

Why is this true?

答:
Write $N=\{1,\dots,n\}$.
Index substitution $j=\sigma(k)$ (same as $k=\sigma^{-1}(j)$ inclusive $\sigma(N)=N$ because $\sigma$ bijection) yields
$$
\prod_{k=1}^n a_{k\sigma(k)}
=
\prod_{k\in N} a_{k\sigma(k)}
=
\prod_{j\in\sigma(N)} a_{\sigma^{-1}(j)j}
=
\prod_{j=1}^n a_{\sigma^{-1}(j)j}
.
$$


问:
Questions about monotone sequences and subsequences
I have the following questions about sequence:

Does every monotone sequence have a bounded subsequence?

Does every bounded sequence have a monotone subsequence?

Does every sequence have a subsequence for which the limit makes sense?

If lim sup $a_n$ is finite, then the sequence $a_n$ is bounded above?

I think the answers should be False, True, True, and True, but I cannot come up with a valid reasoning or counter examples. Can anybody help?

答:
False.  Consider the sequence $1,2,3,\dots$

True.  In fact, every sequence of real numbers has a monotone subsequence.  This is the monotone subsequence theorem.

True.  This follows from question $2$.

True.  If a sequence is not bounded above, then it has a subsequence that diverges to $\infty$, so the lim sup is infinite.


Title: How find the maximum of the complex number inequality

Body:
Let $z_{i}(i=1,2,\cdots,n)$ be complex numbers such that $|z_{i}|\le 1,z_{1}+z_{2}+\cdots+z_{n}=0$. Define
$$f_{n}(z_{1},z_{2},\cdots,z_{n})=|z^3_{1}+z^3_{2}+\cdots+z^3_{n}|$$

If $n=4$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.

For any positive $n$, find the maximum of $f_{n}(z_{1},z_{2},\cdots,z_{n})$.

I have attempted the $n=2,3$ cases, here is my try:

If $n=2$, then $f_{2}(z_1, z_2)=|z_1^3+(-z_1)^{3}|=0$.

If $n=3$, use the identity $a^3+b^3+c^3=3abc$ if $a+b+c=0$, so
$$f_{3}(z_{1},z_{2},z_{3})=3|z_{1}z_{2}z_{3}|\le 3$$
If $z_{1}=1,z_{2}=w,z_{3}=w^2$ is maximum, where $w^3=1,w\neq 1$

For the $n=4$ case, use $$(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$$
then
$$f_{4}(z_{1},z_{2},z_{3},z_{4})=3|(z_{1}+z_{2})(z_{2}+z_{3})(z_{3}+z_{1})|$$but I can't get any farther.

Body:
We have:

If $(z^*_1,z^*_2,\dots,z^*_n)\,$ is the solution of the given task, then
$\max(|z^*_1|,|z^*_2|,\dots,|z^*_n|)=1.\,$

If $(z^*_1,z^*_2,\dots,z^*_n)\,$ is the solution of the given task, then $(z^*_1e^{i\varphi},z^*_2e^{i\varphi},\dots,z^*_ne^{i\varphi})\,$ is the solution too.

This allows to assume
$$z_n=-1,\quad z_1+z_2+\dots z_{n-1}=1.\tag1$$

Сondition $(1)$ means that geometrically complex numbers are displayed as segments of a closed polyline.

On the other hand, $z^3_n=-1,$ and the triple $$z_k = \bar z_k - e^{\large-2ik\LARGE^\pi\!/\!_3},
\quad z^3_1=z^3_2=z^3_3 = -1,\quad z_1+z_2 = 1\tag2$$

gives the stable optimal solution, if $\;\mathbf{n=3}.\;$ In this way, we have
$$|s^\,_{3k+p}|\ge 3k+|s^\,_p|,\quad |s^\,_{k+1}|\ge |s_k|, \quad |s^\,_{3k}|=3k.\tag3$$

In the case $\;\color{brown}{\mathbf {n=4}}\;$ one can get
$$s_4=z_1^3+z_2^3+z_3^3 -1 = (z_1+z_2+z_3)^3-3(z_1+z_2)(z_2+z_3)(z_3+z_1)-1,$$
$$s_4 = -3(z_1+z_2)(1-z_1)(1-z_2).$$
Then, in the notation
$$z_1=p+iq,\quad z_2=u+iv,$$
$$|s_4|^2=f(p,q,u,v)=9((1-p)^2+q^2)((1-u)^2+v^2)\big((p+u)^2 +(q+v)^2\big),\tag4$$
where
$$0\le p^2+q^2\le1,\quad0\le u^2+v^2\le1,\quad 0\le (p+u-1)^2+(q+v)^2\le1\tag5$$
The inner stationary points of the $\;f(p,q,u,v)\;$ can be defined from the system $\;f'_p=f'_q=f'_u=f'_v=0.\;$ Then
\begin{cases}
(p-1)\big((p+u)^2 +(q+v)^2\big)+(p+u)((p-1)^2+q^2)=0\\
q\big((p+u)^2 +(q+v)^2\big)+(q+v)((p-1)^2+q^2)=0\\
(u-1)\big((p+u)^2 +(q+v)^2\big)+(p+u)((u-1)^2+v^2)=0\\
v\big((p+q-1)^2 +(q+v)^2\big)+(q+v)((u-1)^2+v^2)=0,
\end{cases}
$\qquad\qquad\left|(p-1)\times[1]+q\times[2],\quad q\times[1]-(p-1)\times[2]\dots\Large\mathstrut\right|$

\begin{cases}
\big((p+u)^2 +(q+v)^2\big)+(p-1)(p+u)+q(q+v)=0\\
q(p+u-1)-(p-1)(q+v)=0\\
\big((p+u)^2 +(q+v)^2\big)+(u-1)(p+u)+v(q+v)=0\\
v(p+u)-(u-1)(q+v)=0,
\end{cases}

\begin{cases}
(2p+u-1)(p+u)+(2q+v)(q+v)=0\\
(p+2u-2)(p+u)+(q+2v)(q+v)=0\\
q(p+u)-(p-1)(q+v)=0\\
v(p+u)-(u-1)(q+v)=0,
\end{cases}

$$(p+u-1)^2+(q+v)^2=0.$$
Thus, the greatest value of $\;f\;$ takes place on the edges of the area $(5).$

Case $\;\mathbf{(p+u-1)^2+(q+v)^2 = 1,\; (p-1)^2+q^2\in (0,1),\; (u-1)^2+v^2\in(0,1)}.\;$

Applying Lagrange multipliers method, we can find the greatest value of the function
$$\begin{align}
&g(p,q,u,v,\lambda) = 18(p+u)\big((p-1)^2+q^2\big)\big((u-1)^2+v^2\big)\\[4pt]
&+\lambda\big((p+u)^2+(q+v)^2-2(p+u))\big),
\end{align}\tag6$$
wherein the stationary points are defined via the system $\;g'_p=g'_q=g'_u=g'_v=g'_\lambda=0,\;$ or
\begin{cases}
18\big((u-1)^2+v^2\big)\big((p-1)^2+q^2+2(p+u)(p-1)\big) + 2\lambda(p+u-1)=0\\
18q(p+u)\big((u-1)^2+v^2\big) + \lambda(q+v)=0\\
18\big((p-1)^2+q^2\big)\big((u-1)^2+v^2+2(p+u)(u-1)\big) + 2\lambda(p+u-1)=0\\
18v(p+u)\big((p-1)^2+q^2\big) + \lambda(q+v)=0\\
(p+u)^2+(q+v)^2=2(p+u),
\end{cases}

\begin{cases}
(p-1)((u-1)^2+v^2)=(u-1)((p-1)^2+q^2)\\
q((u-1)^2+v^2)=v((p-1)^2+q^2)\\
(q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\
(p+u)^2+(q+v)^2=2(p+u),
\end{cases}

$\qquad\qquad\left|(p-1)\times[1]+q\times[2],\quad (u-1)\times[1]+v\times[2]\dots\Large\mathstrut\right|$

\begin{cases}
(u-1)^2+v^2 =(p-1)(u-1)+qv\\
(p-1)^2+q^2 =(p-1)(u-1)+qv\\
(q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\
(p+u)^2+(q+v)^2=2(p+u),
\end{cases}

\begin{cases}
(u-p)^2+(v-q)^2=0\\
(p+u)^2+(q+v)^2=2(p+u),\\
(q+v)\big((p-1)^2+q^2+2(p+u)(p-1)\big) = 2q(p+u)(p+u-1)\\
\end{cases}
$$u=p,\quad v=q,\quad p^2+q^2=p,\quad (p-1)^2+4p(p-1)+q^2 = 2p(2p-1),$$
$$u=p=\dfrac13,\quad v=q=\dfrac{\sqrt2}3,\quad g=\dfrac{16}3,\quad |s_4| = \dfrac4{\sqrt3}.$$

Case $\;\mathbf{(p+u-1)^2+(q+v)^2 = 0,\; (p-1)^2+q^2\in (0,1),\; (u-1)^2+v^2\in(0,1)}.\;$

We can find the greatest value of the function
$$\begin{align}
&h(p,q)=f(p,q,1-p,v) = 9\big((p-1)^2+q^2\big)\big(p^2+q^2\big)\\[4pt]
\end{align}\tag7$$
under the conditions
$$(p-1)^2+q^2\le1,\quad p^2+q^2\le 1.\tag8$$
Since the goal function increases by $\;q,\;$ then its greatest value achieves under the condition
$$q^2=\begin{cases}
1-p^2,\quad\text{if}\quad p\in[0,\,^1\!/_2]\\[4pt]
1-(1-p)^2,\quad\text{if}\quad p\in[\,^1\!/_2,1].
\end{cases}$$

Then
$$\max\limits_{\large p\in[0,\,^1\!/_2]} h(p,\sqrt{1-p^2})
= \max\limits_{\large p\in[0,\,^1\!/_2]} 9\big(1+(1-p)^2-p^2\big) = 9\quad\text{at}\quad p=\,^1\!/_2,$$
$$\max\limits_{\large p\in[\,^1\!/_2,1]} h(p,\sqrt{1-(1-p)^2})
= \max\limits_{\large p\in[\,^1\!/_2,1]} 9\big(1+p^2-(1-p)^2\big) = 9\quad\text{at}\quad p=\,^1\!/_2,$$
with the known solution for $\;n=3.\;$

Therefore, $\color{brown}{\mathbf{|s_4|=|s_3|= 3.}}$

$\color{brown}{\textbf{Splitting strategy.}}$

Let us find $\;|s_2(t)|,\;$ where
$$s_2=\min\big(z_1^3+(t-z_1)^3\big)\in\mathbb R,\quad i\in\mathbb R,\quad s_2<0.$$
This leads to the minimizing task for
$$f_2(t,p,q)= (t-p-iq)^3+(p+iq)^3 = t\big(3(p+iq)^2-3t(p+iq)+t^2\big),$$
which is real if $\;p=\frac12t.\;$

Therefore,.
$$g_2(t,q)=f_2\left(t,\dfrac t2,q\right) = (p-iq)^3+(p+iq)^3 = 2p^3-6pq^2
 = \frac14t^3-3q^2t,$$
$$\min g_2(t,q)=f_2\left(t,\sqrt{1-\dfrac14t^2}\right) = -3t+t^3.$$

Then
$$|s^\,_3|\ge|s^\,_2(1)-1|=3,$$
$$|s^\,_5|\ge\left|2s^\,_2\left(\dfrac12\right)-1\right|=\dfrac{15}4 > 3,$$
$$|s^\,_7|\ge\left|3s^\,_2\left(\dfrac13\right)-1\right|=\dfrac{35}9,$$
$$|s^\,_{7}|\ge\left|2s^\,_2\left(\dfrac32\right)-3\right|=\dfrac{21}4,$$
$$|s^\,_8|\ge\left|3s^\,_2\left(\dfrac23\right)-2\right|=\dfrac{64}9,$$
$$|s^\,_{10}|\ge\left|4s^\,_2\left(\dfrac12\right)-2\right|=\dfrac{15}2,$$
$$|s^\,_{10}|\ge\left|3s^\,_2\left(\dfrac43\right)-4\right|=\dfrac{80}9,$$
$$|s^\,_{13}|\ge\left|5s^\,_2\left(\dfrac35\right)-3\right|=\dfrac{273}{25},$$
$$|s^\,_{13}|\ge\left|4s^\,_2\left(\dfrac54\right)-5\right|=\dfrac{195}{16}>12\dots$$
$$|s^\,_{35}|\ge\left|12s^\,_2\left(\dfrac{11}{12}\right)-11\right|=\dfrac{5005}{144}=34\dfrac{109}{144}>34\dfrac34\dots$$

Obtained solutions allow to improve the primary estimations $(3).$


答:
Let start by fixing some notations. Let $n \in \Bbb{N}$, $X = \{i \in \Bbb{N} \colon i \leq n\}$, $\sigma \in S_n$ and $A \in \cal{M}_{n}(\Bbb{K})$. Now, consider the two products that you mention

\begin{align*}
A = \prod_{k=1}^n a_{k \, \sigma{(k)}} \qquad \text{and} \qquad B = \prod_{k=1}^{n} a_{\sigma^{-1}(k) \, k}.
\end{align*}

Let us prove that $A = B$.

Let $i \in \Bbb{N}$ such that $i \leq n$. We clearly have that $(i,\sigma(i)) \in X^2$. Since $\sigma$ is a permutation of $X$, we know that $\sigma$ is a bijective application of $X$ to $X$. Hence, it exists $\sigma^{-1} \in S_n$. Since $\sigma{(i)} \in X$, then it exists a unique (because $\sigma$ is bijective) $j \in X$ such that $j= \sigma(i)$ and $\sigma^{-1}(j) = i$. So for all $i \in X$, it exists a unique $j \in X$ such that $i = \sigma^{-1}(j)$ and $\sigma(i) = j$. Therefore, by making this substitution, and reordering the elements in the product (because multiplication is commutative), we conclude that $A = B$.

In short, it all comes from the fact that $\sigma$ is bijective and therefore we can associate those pairs $(i,\sigma(i))$ to $(\sigma^{-j},j)$ is a unique way such that, by reordering the elements in the product, we obtain the same result. (Nothing changes, it is just a substitution)

Title: Show that $ \prod_{k=1}^{n}a_{k, \sigma(k)} = \prod_{k=1}^{n}a_{\sigma^{-1}(k), k}$ on any matrix

Body:
At some point, in proving that $\det{A} = \det{A^T}$, one states that

\begin{align*}
\prod_{k=1}^{n}a_{k, \sigma(k)} = \prod_{k=1}^{n}a_{\sigma^{-1}(k), k}
\end{align*}

Why is this true?


问:
How to solve the differential equation: $y-xy'=\exp(y')$

$$y-xy'=\exp(y')$$

I want to solve this differential equation, which looks simple but hard to solve. Any method?

答:
Use Clairaut equation to find the solution.

The general solution is
$$y(x)=Cx+e^{C}.$$

Proof from Wikipedia article.

Differentiate both sides with respect to $x$
$$y'=y'+xy''+\exp\left(y'\right)y'',$$
so
$$\left[x+\exp\left(y'\right)\right]y'' = 0.$$

General solution:
$$y'' = 0 \Longrightarrow y(x)=Cx+e^C.$$

Singular solution:
$$x+\exp\left(y'\right) = 0.$$

Thank you @zwim: Desmos calculator.


问:
proving a set is a subspace of $\Bbb R^{3\times 3}$
I am working through some homework problems and I ran into one that I am not sure where to start,

Is the following set a subspace of $\Bbb R^{3\times 3}?$

$$\left\{A\in\Bbb R^{3\times 3}\,\middle|\,\dim(\operatorname{null}A)=2\right\}$$

I know that since the dimension of the null space of $A$ is $2$, $A$ must only contain $2$ pivot elements, meaning one row of $A$ is a zero vector.  Where do I go from here?

答: Hint: Does the null matrix belong to your set?


问:
What's the difference between $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$?
I noticed that $(\mathbb Z_n,+)$ and $(\mathbb Z_n,*)$ are not the same thing. For example 2 is not invertible in $(\mathbb Z_6,*)$. However, since $\bar2+\bar4=\bar0$, thus it is invertible in  $(\mathbb Z_6,+)$.

I found the following problems:

Am I right? If so, why does any book not mention this difference which is very important?

When we study groups, what's the standard operation of $\mathbb Z_n$ when the operation is omitted?

When we study rings, do we generally use $(\mathbb Z_n,+)$ to be the abelian group of the ring $(\mathbb Z_n,+,*)$?

If I work with $(\mathbb Z_n,+)$ I would lose these theorems which I proved for $(\mathbb Z_n,*)$?

An element $a$ in $(\mathbb Z_n,*)$ is an unit iff $a$ and $n$ are  coprimes

An element $a$ in $(\mathbb Z_n,*)$ is an zero divisor iff $a$ and
  $n$ aren't coprimes

I'm sure there are a lot of student with this same doubt.

I really need help.

Thanks a lot.

答:
1) Usually, $\,(\Bbb Z_n\,,\,+)\;$ would represent an additive cyclic group, which elements are usually represented as $\,\{\overline0\,,\,\overline 1\,\ldots\,\overline{n-1}\}\;$ or as $\,\{0,1,2,...,n-1\}\pmod n\,$ .

The set $\,(\Bbb Z_n\,,\,+\,,\,\cdot)\,$ represents the ring of residues modulo $\,n\,$ , with the same addition as above and multiplication modulo $\,n\,$ .

The set of invertible elements in the above ring is usually denoted by $\,\Bbb Z_n^*\;$, and sometimes $\,(\Bbb Z_n^*,\cdot)\;$ , and it is the set of elements in $\,\{0,1,2,...,n-1\}\pmod n\,$ which are coprime with $\,n\,$ (and non-zero, to avoid problems). This is a group under the usual multiplication in the above ring.

Of course, what you prove for a set under some operation may, and may not, be true for the same set, or a subset, under a different operation. Perhaps it'll be good you tell us what things were you thinking about.

ContentLicense: CC BY-SA 3.0


问:
elliptic curve $E:y^2＝x^3-x$ can be transformed to move the identity element to the origin
For example, elliptic curve $E:y^2＝x^3-x$ can be transformed
to move the identity element to the origin, $(z,w)=(0,0)$, we do the change of coordinates $z=-x/y$, $w=-1/y$. The equation of this curve then becomes
$w=z^3+zw^2\quad$

We often make this transformation to figure out formal group associated to elliptic curve.

But what do we do when $y＝0$?

I'm confused.

Thank you in advance.

答:
This is easier if you write in projective coordinates. Let $E$ be given as
$$Y^2Z = X^3 -XZ^2$$
then the rational map $(x,y) \mapsto (-x/y, -1/y)$ on the affine chart where $Z = 1$ can be extended to a morphism on the projective closure ($E$ above) by
$$(X : Y:Z) \mapsto (-X: -Z : Y)$$

In particular $(0:0:1)$ maps to $(0:1:0)$ - the point at infinity on your curve. This may also be seen in an analagous way to complex analysis by looking at the order of the poles of the functions $-x/y$ and $-1/y$ at $(0,0)$.


问:
Functions vanishing to infinite order "in $1$-mean"
When proving certain unique continuation theorems for classes of functions which satisfy some type of PDE inequality, people often talk about functions which vanish to infinite order at a point $x_0$ in the $q$-mean in the sense that, if $u: \Omega \to \mathbb{C^n}$ is an $L^q$-function, where $\Omega \subset \mathbb{R}^n$ is an open set: $$\int_{|x - x_0| \leq r} |u(x)|^q dx = O(r^k), \hspace{5pt} \forall k \in \mathbb{N}.$$By Hölder's inequality, we know that if $u$ vanishes to infinite order in the $q$-mean, and $p<q$, then $u$ also vanishes to infinite order in the $p$-mean.

Vanishing to infinite order in the sense that we're used to for smooth functions (i.e, $|u(x)| = O(|x-x_0|^k)$ when $ x \to x_0$ for all $k$; this makes sense even for non-smooth functions) is equivalent to vanishing to infinite order in the $p$-mean for $p = \infty.$

My question can be phrased in two ways:

Does the converse hold? I.e, if $p<q$, and $u$ vanishes to infinite order in the $p$-mean, does it also vanish to infinite order in the $q$-mean?

A more specific version of the question which I'm interested in, for the purpose of $J$-holomorphic curves: if $u$ is a smooth function and $n=2$ (for example, $\Omega$ can be a small open disk around $0$), does vanishing to infinite order in the $1$-mean imply vanishing to infinite order in the usual sense, i.e. $D^{\alpha}u(0) = 0$ for all multi-indices $\alpha$? This is stated in Section 2.3. of the McDuff-Salamon book "J-holomorphic curves and symplectic topology." I was able to show this for $|\alpha| = 0,1$, but I wasn't able to solve the general case, where I get stuck at an expression of the form $$\int_{|z| \leq r} |(D^k u)_0(z,\dots,z)|dz \leq M_Nr^N, \hspace{7pt} \forall N, \forall r<r_0.$$I was also able to show that it's true when $u$ is holomorphic, due to the specific nature of the $k$th derivative in this case.

答:
For simplicity, I work in two dimensions.

Smooth case : If $u$ is assumed to be smooth in a neighborhood of $0$, then indeed, its vanishing to infinite order in 1-mean (and a fortiori, in $q$-mean for $q > 1$) implies that all derivatives of $u$ at $0$ vanish. This follows from the Taylor expansion with remainder, which states that for all $N \ge 0$,
$$ u(x,y) = \sum_{0 \le k+l \le N} \frac{1}{k! \, l!} \frac{\partial^{k+l} u}{\partial^k x \partial^l y}(0,0) \; x^k y^l + O((x^2 + y^2)^{\frac{N+1}{2}}) \, . $$

Let's assume by induction that for $N \ge 0$, it is known that all derivatives of $u$ at $(0,0)$ up to order $N-1$ vanish, so that
$$ u(x,y) = \sum_{k+l = N} \frac{1}{k! \, l!} \frac{\partial^{k+l} u}{\partial^k x \partial^l y}(0,0) \; x^k y^l + O((x^2 + y^2)^{\frac{N+1}{2}}) \, =: u_N(x,y) + O((x^2 + y^2)^{\frac{N+1}{2}}). $$

We aim to prove that $u_N = 0$. For $r > 0$, set $\epsilon = r/\sqrt{2}$ and let $\lambda \in (0,1]$. Note that the rectangle $R(\epsilon, \lambda) := [0, \epsilon] \times [0, \lambda \epsilon]$ lies inside the ball $B(r)$ of radius $r$ centred at $0$. Integrating $u_N$ over this rectangle yields
$$ \iint_{R(\epsilon, \lambda)} u_N(x,y) dx dy = \epsilon^{N+2} \, \sum_{l=0}^{N} \frac{1}{(N-l+1)! \, (l+1)!} \frac{\partial^{N} u}{\partial^{N-l} x \partial^l y}(0,0) \; \lambda^{l+1} = P_N(\lambda) \, r^{N+2}  $$
where $P_N(\lambda)$ is a polynomial which vanishes identically in $\lambda$ if and only if $u_N = 0$. Next, taking into account that $u$ vanishes to order $N+3$ in 1-mean, we estimate that
$$ \begin{align} \left| P_N(\lambda)  \right| r^{N+2} &= \left| \iint_{R(\epsilon, \lambda)} u_N(x,y) dx dy \right| \\
& \le \left| \iint_{R(\epsilon, \lambda)} u(x,y) dx dy \right| + \left| \iint_{R(\epsilon, \lambda)} O((x^2 + y^2)^{\frac{N+1}{2}}) dx dy \right| \\
& \le \iint_{B(r)} \left| u(x,y) \right| dx dy  +  \iint_{B(r)} \left| O(s^{N+1}) \right| s ds d\theta \\
& \le O(r^{N+3}) + O(r^{N+3}) = O(r^{N+3}) \, .
\end{align} $$
Hence $|P_N(\lambda)|$ is $O(r)$ and also independent from $r$, so it vanishes identically in $\lambda$. This implies $u_N = 0$.

Continuous case: Assuming instead $u$ to be merely continuous, $u$ could vanish to infinite order at the origin in 1-mean but not in "$\infty$-mean". Let $\chi : \mathbb{R}^2 \to [0,1]$ be a continuous function supported in $B(1)$ and satisfying $\chi(0,0) = 1$. Set $$u(x,y) = \sum_{n=2}^{\infty} \frac{1}{n} \, \chi (2^n(x-1/n), 2^ny)$$
(This is a sum of functions supported in (disjoint) balls $B_n$ where $B_n$ is centred at $(1/n, 0)$ and has radius $1/2^n$.) It follows that $\mathrm{max}_{B(r)} |u| \sim r$, hence $u$ does not vanish to infinite order at the origin in the "$\infty$-mean". However, it $L^1$-norm over a ball of radius $1/N$ is of order $$\sum_{n \ge N} \frac{n}{4^n} \le  \sum_{n \ge N} \frac{1}{2^n} = \frac{1}{2^{N-1}} = O((1/N)^K) \qquad \forall K > 0.$$

However, to vanish to infinite order in the 1-mean implies vanishing to infinite order in all other $q$-mean with $q \in (1, \infty)$. First, continuity implies $u(0,0) = 0$. Indeed, assuming instead $|u(0,0)| = L > 0$, continuity of $u$ would imply that there is $r_0 > 0$ such that $|u| > L/2$ on $B(r_0)$ and hence $\int_{B(r)}|u| dxdy > (\pi L/2)r^2$ for $r < r_0$, in contradiction with $u$ vanishing at infinite order. Consequently, there exists $r_1 > 0$ such that $|u| < 1$ on $B(r_1)$, whence $\int_{B(r)}|u|^q dx dy \le \int_{B(r)} |u| dxdy = O(r^K)$ for all $K > 0$ and all $r < r_1$. Thus $(\int_{B(r)} |u|^q dxdy)^{1/q} = O(r^{K/q})$ for all $K > 0$.

General case: Without any specific constraint on $u$ besides being $L^p$ for every $p \in [1, \infty]$, I don't expect its vanishing to infinite order at $(0,0)$ in 1-mean to imply its vanishing to infinite order in $q$-mean for all other $q \in (1, \infty)$. Indeed, as a clue for this, let's consider the same function $\chi$ as above and set
$$u(x,y) = \sum_{n=2}^{\infty} 2^n \, \chi (2^n(x-1/n), 2^ny).$$
Then $u$ again vanishes to infinite order in 1-mean, but it does not even have a well-defined 2-mean. So there might be a more clever construction of a function $u$ whose $q$-means exist for all $q$, and which vanishes to infinite order in $1$-mean but not in every other $q$-means.


问:
Boundness of a continuous function $u: \Omega \subset \mathbb{R^n} \rightarrow \mathbb{R}$.
I can't understand why if I have a function $u: \Omega \subset \mathbb{R^n} \rightarrow \mathbb{R}$, continuous.

(Maybe $u$ should be defined on $\overline{\Omega}$, tell me then please).

And if $\exists$ M $\gt 0$ s.t $u(x) \lt$ M, $\forall$ x $\in \Omega$,

then we can say that, and by continuity of $u$:

$u(x) \lt$ M,
$\forall$ x $\in \overline{\Omega}$ (i.e on $\partial{\Omega}$, also).

Can I have some help please! Thanks in advance.

答:
Your conclusion is slightly incomplete. If $u(x) < M$ for all $x \in \Omega$ then $u(x) \leq M$ for all $x \in \overline{\Omega}$. Because it could be that $\overline{\Omega}$ is compact and $M = \max_{x \in \overline{\Omega}} u(x)$.

To demonstrate this, note that for $ x \in \overline{O}$ there is a sequence $(x_n) \subset \Omega$ such that $ x_n \longrightarrow x$ then by continuity $u (x_n) \longrightarrow u(x)$. And remember that if given a sequence $ y_n \subset \mathbb{R}$ and $ C $ a real number such that $y_n \leq C$ then $\displaystyle \lim_{n \to \infty} y_n \leq C$. So take $y_n = u(x_n)$ and $c = M$ and you get what you want.


问:
Supremum of $B$, where $\sup B = \lambda + \sup A $
Question Setup:

Let $A$ be a set that is bounded above, where $\sup A = \lambda $
$A$ is a subset of $\mathbb{R}$ and $\lambda \in \mathbb{R} \\ $.
A set $B$ is defined as $B = \{\, x + k \ | \ x \in A, k \in \mathbb{R}\,\}$

Question:
Prove that $\sup B = \lambda + k $

My attempt:
Let $\mu \in \mathbb{R}$
$$\forall x + k \in B, \mu \geq x + k $$
i.e. $\mu$ is an upper bound of $B$
Select $\mu$ such that ,
$$\forall m \in B, m \leq \mu \implies \exists n \in B, n \geq m $$
i.e $\mu = \sup B $

We know, (everything from here on is shaky)
$$\forall x + k \in B, \mu - k \geq x $$
$$\forall x  \in A, \mu - k \geq x $$
i.e $$\mu - k \geq \sup A $$
As $ \mu = \sup B$
$$\mu - k = \sup A $$
$$\sup B = \lambda + k$$

答:
If $b\in B$, then $b=a+k$, for some $a\in A$. So, since $a\leqslant\sup A$, $b\leqslant k+\sup A$. Therefore, $k+\sup A$ is an upper bound of $B$.

And if $\mu<k+\sup A$, then $\mu-k<\sup A$. So, there is some $a\in A$ such that $a>\mu-k$, and therefore $a+k>\mu$. But $a+k\in B$. So, $\mu$ is not an upper bound of $B$. It follows that $k+\sup A$ is the least upper bound of $B$. In other words, $k+\sup B=\sup A$.


答: Note that $\sup B$ does not exist because $B = \mathbb{R}$, i. e., $B$ is unbunded. I think he lacks the hypothesis that B is bounded or something similar.

Title: Supremum of $B$, where $\sup B = \lambda + \sup A $

Body:
Question Setup:

Let $A$ be a set that is bounded above, where $\sup A = \lambda $
$A$ is a subset of $\mathbb{R}$ and $\lambda \in \mathbb{R} \\ $.
A set $B$ is defined as $B = \{\, x + k \ | \ x \in A, k \in \mathbb{R}\,\}$

Question:
Prove that $\sup B = \lambda + k $

My attempt:
Let $\mu \in \mathbb{R}$
$$\forall x + k \in B, \mu \geq x + k $$
i.e. $\mu$ is an upper bound of $B$
Select $\mu$ such that ,
$$\forall m \in B, m \leq \mu \implies \exists n \in B, n \geq m $$
i.e $\mu = \sup B $

We know, (everything from here on is shaky)
$$\forall x + k \in B, \mu - k \geq x $$
$$\forall x  \in A, \mu - k \geq x $$
i.e $$\mu - k \geq \sup A $$
As $ \mu = \sup B$
$$\mu - k = \sup A $$
$$\sup B = \lambda + k$$


问:
Determining information in minimum trials (combinatorics problem)
A student has to pass a exam, with $k2^{k-1}$ questions to be answered by yes or no, on a subject he knows nothing about. The student is allowed to pass mock exams who have the same questions as the real exam. After each mock exam the teacher tells the student how many right answers he got, and when the student feels ready, he can pass the real exam. Show that if the student is good at combinatorics he can guess all the answers after only $2^{k}$ mock exams.

For those who prefer a more formal presentation of the problem: if $E=\{1,-1\}^{N}$ we seek $n$ so that for some vectors $v_{1},v_{2}, \ldots, v_{n}\in E$
\begin{align*}
\phi\colon E &\rightarrow \mathbb{N}^{n},\\
v &\mapsto (\langle v,v_{1}\rangle, \langle v,v_{2}\rangle, \ldots, \langle v,v_{n}\rangle)
\end{align*}
is injective. Show that it is possible to find such vectors when $N=k2^{k-1}$ and $n=2^{k}$.

It is possible to use duality to transform again the problem. We seek a $n\times N$ matrix $M$ such that $v\mapsto Mv$ is injective over $E$. If $X$ is the formal polynomial vector $X=(X_{1},X_{2}, \ldots,X_{n})$ then $v\mapsto Mv$ is injective iff $v\mapsto \langle X,Mv \rangle$ also is. But $\langle X,Mv \rangle = \langle M^{T}X, v \rangle$ and it is easily seen that $v\mapsto \langle M^{T}X,v \rangle$ is injective iff the $N$ column vectors $x_{i}$ of $M$ are such that $\sum_{i\in I} x_{i}\neq \sum_{j\in J} x_{j}$ for any two different subsets $I$, $J$ of $\{1,\dots,N\}$.

This problem comes from an olympiad-like contest. The original problem was formulated with 30 questions and the aim was to prove that the student could guess with 24 trials. One of my teachers came up with the result above (which would give 16 trials for 30 questions), but I can't remember his proof or find another one by myself.

I tagged this information theory because some probabilistic arguments show that it is impossible to do better than this asymptotically. More precisely, if we choose the coordinates of the $v$ defined above randomly, then
\begin{equation*}
N =  H(\phi(v))
\leqslant  \sum_{i} H(\langle v,v_{i} \rangle) = nH(B(N,1/2)) \sim (n/2)\log_{2}(N)
\end{equation*}
where $H$ designates entropy. With $N=k2^{k-1}$ we get $n\geqslant c\frac{k2^{k}}{k-1+\log_{2}(k)}\geqslant c2^{k}$ for all $c<1$ and $k$ large enough.

答:
Thanks to DeepL translate, I was able to read the article in the Russian journal posted in the comments (https://www.mccme.ru/free-books/matprosf.html), and I can now give the solution. In fact, you can determine the answers to a test with $k2^{k-1}+1$ questions in only $2^k$ non-adaptive attempts.

First, let us reformulate the problem in terms of weighings. Suppose without loss of generality your first submission is a test with all answers being "yes." For any subset $A$ of questions, let the weight of $A$ be the number of questions in $A$ whose answer is "yes." Given the result of this all-"yes" submission, for any subset $A$ of questions, each of these two pieces of information can be deduced from the other:

The score of a test when you answer "yes" for all questions in $A$, and "no" for the other questions

The weight of $A$.

Therefore, I can and will state the strategy in terms of weights.

The fact that you can solve $k2^{k-1}+1$ questions in $2^k$ attempts follows from the following lemma.

Lemma: Suppose you can deduce the answers to a test with $M$ questions in $N$ attempts. Then it is possible to deduce the answers to a test with $2M+N-1$ questions in $2N$ attempts.

Proof: Given a test with $2M+N-1$ questions, divide the questions into three sets $A,B$ and $C$, where $|A|=|B|=M$, and $|C|=N-1$. By assumption, there exist $N-1$ sets $A_1,\dots,A_{N-1}$ such that you could solve a test whose question set is $A$ by submitting the all-"yes" test, and then weighing $A_1,\dots,A_{N-1}$. Similarly, there exist sets $B_1,\dots,B_{N-1}$ for solving a test whose question set is $B$. Finally, let $C=\{c_1,c_2,\dots,c_{N-1}\}$. Then the solution for the $(2M+N-1)$-question test is this:

Answer "yes" to all questions.

Find the weight of $B$.

For each $k\in \{1,\dots,N-1\}$: find the weights of these two sets:

$A_k\cup B_k\cup \{c_k\}$

$A_k\cup (B\setminus B_k)$.

Here is how you deduce the answers to the test. For each $k\in \{1,\dots,N-1\}$, the parity of the sum of the weights of $A_k\cup B_k\cup \{c_k\}$ and $A_k\cup (B\setminus B_k)$ and $B$ determines the answer to $c_k$. We now know the weights of the three sets $A_k\cup B_k$, $A_k\cup (B\setminus B_k)$, and $B$, which gives a simple system of equations to solve for $A_k$ and $B_k$. This gives us the information we need to deduce the answers to the $A$ questions and $B$ questions, using the assumed strategy.$\tag*{$\square$}$ That is, starting with the two-attempt solution to a two-question test, you can derive a four-attempt solution to a test with $2\cdot 2+(2-1)=5$ questions, then an eight attempt solution to a test with $2\cdot 5+(4-1)=13$ questions, and so on, in general leading to $k2^{k-1}+1$ questions in $2^k$ attempts.

The strategy can be modified extend an $N$-attempt strategy for an $M$-question test to a $2N$-attempt strategy for a $2M+r$ test for any $r\in \{0,1,\dots,N-1\}$, by ignoring some of the $C$ questions. This lets you hand the $k2^{k-1}$ question test in $2^k$ questions, as exactly asked in the OP.

Furthermore, it is easy to extend an $N$-attempt strategy for $M$ to an $(N+1)$-attempt strategy for $M+1$. Together with these two rules, you can inductively use the lemma to determine a strategy for any number of tests.


问:
Galois group $\operatorname{Gal}(f/\mathbb{Q})$ of the polynomial $f(x)=(x^2+3)(x^2-2)$

Find the Galois group $\operatorname{Gal}(f/\mathbb{Q})$ of the polynomial $f(x)=(x^2+3)(x^2-2)$.

Any explanations during the demonstration, will be appreciated. Thanks!

答:
By tower law we know that $[\mathbb Q(i\sqrt[]{3}, \sqrt[]{2}):\mathbb Q]= 4$, since $[\mathbb Q(i\sqrt[]{3}, \sqrt[]{2}):\mathbb Q(\sqrt[]{2})]=2$ and $[\mathbb Q(\sqrt[]{2}):\mathbb Q]=2$.

Let $i$ be the inclusion map $i:\mathbb Q \hookrightarrow \mathbb Q(i\sqrt[]{3}, \sqrt[]{2})$. Then $i(x^2 - 2) =x^2 - 2$ and $x^2-2$ has roots $\sqrt[]{2}$, $-\sqrt[]{2}$ in $\mathbb Q(\sqrt[]{2})$, so there is a monomorphism $j$ extending $i$ such that $j: \mathbb Q(\sqrt[]{2}) \hookrightarrow \mathbb  Q(i\sqrt[]{3}, \sqrt[]{2})$ given by $j: \sqrt[]{2} \mapsto \pm\sqrt[]{2}$.

Then $j(x^2 + 3)=x^2+3$ and $x^2 + 3$ has roots $\pm i \sqrt[]{3}$ in $\mathbb Q(i\sqrt[]{3}, \sqrt[]{2})$. Hence there is a monomorphism $\sigma$ extending $j$ such that $\sigma: \mathbb Q(i\sqrt[]{3}, \sqrt[]{2}) \hookrightarrow \mathbb Q(i\sqrt[]{3}, \sqrt[]{2})$ given by $\sigma:i\sqrt[]{3} \mapsto \pm i \sqrt[]{3}$ and $\sigma: \sqrt[]{2} \mapsto \pm\sqrt[]{2}$

But $\sigma$ is actually an automorphism since it is a $\mathbb Q$-linear map on vector spaces of the same degree. Therefore $\sigma \in Gal(f)$ and $|Gal(f)| = 4$


问:
Comparing Hamiltonians - Quantum harmonic oscillator
For standard 1D quantum harmonic oscillator we have $H\psi = E_n\psi$ with $E=(n+\frac{1}{2})\hbar\omega$ and $H = \frac{P^2}{2m} + \frac{1}{2}m\omega^2 X^2$ where $X$ is position operator and $P$ is momentum operator.

Now consider finding the energy of $H = \frac{1}{2m}((P_1+\frac{1}{2}eBX_2)^2 + (P_2-\frac{1}{2}eBX_1)^2)$ for some magnetic field B

Letting $A = P_1+\frac{1}{2}eBX_2$
and    $C = P_2-\frac{1}{2}eBX_1$

we could work this out and compare to our original H for 1D oscillator.

I was confused at this point since the original hamiltonian has momentum and position operators separate but in the compared hamiltonian we will get something like  $H = \frac{A^2}{2m} + \frac{1}{2}m\omega^2 C^2$ but A and C are combinations of position and momentum operators.

My question: How are we able to compare these 2 Hamiltonians to quote the standard result from 1d oscillator energy since the compared hamiltonian is made up of combinations of momentum and position operators

edit: im not looking for computations to solve this, since computationally I am fine, its the understanding of why we can compare that I am confused about

答:
Most good texts detail the phase-space symmetry underlying Landau quantization, and reducing phase space from 4D to 2D, as you noted. It's actually a classical structure (the i of commutators is not crucial in the canonical transformation). I am not sure you seek 4D visualization, so the corresponding canonical rotations will present algebraically, here.

Firstly, nondimensionalize, as all texts should do routinely,
setting $\hbar,\omega, m, eB$ equal to 1, and reinstatable through dimensional analysis uniquely in the end, if desired (a bad idea).

You then have
$$
2H= (P_1+X_2/2)^2 + (P_2-X_1/2  )^2
$$

The following obvious rotation is a canonical transformation, that is, it preserves the original commutators,
$$
X_3=A=P_1+X_2/2, ~~~ P_3=C= P_2-X_1/2,  ~~~~ [X_3,P_3]=i, \\
P_4=D= P_1-X_2/2, ~~~ X_4= F=P_2+ X_1/2, ~~~~ [X_4,P_4]=i, \\
[A,D]=[A,F]=[C,D]=[C,F]=0.
$$

But with a flip from your expression, the hamiltonian reduces to 2D,
$$
2H= P_3^2+X_3^2,
$$
that is, the coordinates $X_4,P_4$ drop out of it, and, of course, commute with it, so they are constants of the motion: conserved charges. The extra phase-space dimensions indexed by 4 drop out of the problem, and constitute a decoupled hamiltonian piece $P_4^2+X_4^2$, a constant in an independent ector/world; however, upon changing into the 1,2 original variables, they are involved in the evident rotational symmetry of these  variables into each other.

So, your hamiltonian has the  familiar 1D oscillator spectrum, $E=n+1/2$.

The group of these two generators is, naturally, the Heisenberg group. You may observe the trivial or else translation action of these symmetry transformations on your original variables, e.g.,
$$
e^{iaX_4} X_1 e^{-iaX_4} = e^{iaX_4} (X_4-P_3) e^{-iaX_4}=X_1 ,\\
e^{iaX_4} P_1 e^{-iaX_4} = e^{iaX_4} {P_4+X_3\over 2} e^{-iaX_4}= P_1 -a/2,\\
e^{iaX_4} X_2 e^{-iaX_4} =  e^{iaX_4}  (X_3-P_4)  e^{-iaX_4}= X_2+a,\\
e^{iaX_4} P_2 e^{-iaX_4} = e^{iaX_4} {X_4+P_3\over 2} e^{-iaX_4}=P_2,
$$
etc, mutatis mutandis for $P_4$...

Culture This is a superintegrable system, that is it has the maximal symmetry allowed without restricting motion. Classically, you see that each conserved quantity, so $H,X_4,P_4$ here, represents a hypersurface in phase space. The three here intersect on a line in 4D, the unique trajectory of the oscillator.


问:
If $\xi$ is the Riemann xi function then value of $\xi'(1/2)$?
Riemann Xi function is defined as $$\xi(s)=\frac{s(s-1)}{2}\pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s)$$
Question What is the derivative of $\xi(s)$ at $\frac{1}{2}$?

My try
Logarithmic differentiation gives $$\frac{\xi'(s)}{\xi(s)}=\frac{1}{s}+\frac{1}{s-1}-\frac{log(\pi)}{2}+\frac{\Gamma'(\frac{s}{2})}{\Gamma(\frac{s}{2})}+\frac{\zeta'(s)}{\zeta(s)}$$
At, $s=\frac{1}{2}$
$$\frac{\xi'(\frac{1}{2})}{\xi(\frac{1}{2})}=-\frac{log(\pi)}{2}+\frac{\Gamma'(\frac{1}{4})}{\Gamma(\frac{1}{4})}+\frac{\zeta'(\frac{1}{2})}{\zeta(\frac{1}{2})}$$

Body: $\xi(1/2+s)$ is even, $\xi'(1/2)=0$.


答:
Unfortunately, your statement is incorrect. Consider $\Omega = (0, 1)$ and $f : \Omega \to \mathbb{R}$ defined by $f(x) = x$. Then take $M = 1$. We see that $\overline{\Omega} = [0, 1]$, and $f(1) = 1 = M$.

The correct conclusion is that if we have $u : \overline{\Omega} \to \mathbb{R}$ and $u(x) \leq M$ for all $x \in \Omega$, then $u(x) \leq M$ for all $x \in \overline{\Omega}$.

To do this, we recall that $\overline{\Omega}$ is the smallest closed set containing $\Omega$. Then consider $V = \{x \in \overline{\Omega} | u(x) \leq M\} = u^{-1}([M, \infty))$.

Because $V$ is the inverse image of a closed set under a continuous map, $V$ is closed. And it's easy to see that for all $x \in \Omega$, we have $x \in V$. Then $\Omega \subseteq V$. Then $\overline{\Omega} \subseteq V$, since $\overline{\Omega}$ is the smallest closed set containing $\Omega$. Then for all $x \in \overline{\Omega}$, $u(x) \leq M$.

Title: Boundness of a continuous function $u: \Omega \subset \mathbb{R^n} \rightarrow \mathbb{R}$.

Body:
I can't understand why if I have a function $u: \Omega \subset \mathbb{R^n} \rightarrow \mathbb{R}$, continuous.

(Maybe $u$ should be defined on $\overline{\Omega}$, tell me then please).

And if $\exists$ M $\gt 0$ s.t $u(x) \lt$ M, $\forall$ x $\in \Omega$,

then we can say that, and by continuity of $u$:

$u(x) \lt$ M,
$\forall$ x $\in \overline{\Omega}$ (i.e on $\partial{\Omega}$, also).

Can I have some help please! Thanks in advance.


问:
How to calculate inverse Laplace transform of $F(S)=\frac{se^{-3s}}{s^2+22s+125}$

$$F(S)=\frac{se^{-3s}}{s^2+22s+125}$$

My first reflex was to attempt a decomposition into partial fractions, but I am simply left with a fraction that is just as "complicated", instead of the usual decomposition into multiple simpler fractions separated by addition/subtraction/etc.

P.S. To save you some time, it is sufficient to show me how to get a decomposed fraction from this fraction, I can do the actual inverse laplace transforms from there.

Thank you!

答:
Hint:

Solve $s^2+22s+125=0$. This yields $$s_1= -11 - 2i,\quad s_2=-11+2i$$

Now
$$f(t)= \operatorname{Res}_{s=s_1}(F(s)\cdot e^{st})+\operatorname{Res}_{s=s_2}(F(s)\cdot e^{st}).$$

Calculate the residues by factoring the denominator with $s_1$ and $s_2$ and then calculate the limit, without any indeterminate form.


问:
Maclaurin series of $f(x)=\sinh(1/x)$?
As we know the formula of Maclaurin series for $f(x) = \sinh(x)$ is $f(x)=x+x^3/3! + x^5/5!+\ldots$

Could anyone tell me what is the Maclaurin series of $f(x)=\sinh(1/x)$?

答:
There is no Macluarin series for $\sinh\left(\frac{1}{x}\right)$. If you use the standard formula for Macluarin series which is:

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3...$$

And you try to evaluate $f(0)$, you will get undefined as $\frac{1}{0}$ is undefined. Thus, you cannot construct a Maclaurin series for $\sinh \left(\frac{1}{x}\right)$.

Title: Maclaurin series of $f(x)=\sinh(1/x)$?


问:
Solve $(D^2 + 3D + 2)y = e^{e^x}$ using method of variation of parameters?
This question was asked in a test and I'm stuck while solving this using method of variation of parameters. Here's an screenshot of my solution,

While finding particular integral I was not able to solve integration of e^e^x?
Can anyone please help me out a bit here. I was not able to understand how to solve it further.

答:
the final answer should be
$PI = e^{-2x} * e^{e^x}$


问:
How can I write $(\mathbb{Z}_5)^n$ as a union of finite hyperplanes
Consider $V=(\mathbb{Z}_5)^n$ a vector space over the field $F=\mathbb{Z}_5$, How can I write $(\mathbb{Z}_5)^n$ as a union of finite hyperplanes? I am just looking for an example of how I can do it.
I guess it is possible for the fact that $F$ is finite and it implies that $V$ is also finite.

答:
Well, for example, to write the $2$-dimensional plane as a union of lines, we consider the collection of lines given by their "slope":
$$\begin{align*}
L_0 &= \langle (1,0)\rangle\\
L_1 &= \langle (1,1)\rangle\\
L_2 &=\langle (1,2)\rangle\\
L_3 &=\langle (1,3)\rangle\\
L_4 &=\langle (1,4)\rangle\\
L_{\infty} &= \langle (0,1)\rangle.
\end{align*}$$

Given $(a,b)\in (\mathbb{Z}_5)^2$, if $a=0$ then $(a,b)\in L_{\infty}$; and if $a\neq 0$, then $(a,b)\in L_{a^{-1}b}$.

In $3$-dimensional space, you could do the equivalent of looking at the planes defined by their "normal lines", after enumerating the lines through the origin; the latter correspond to points in projective $2$-space, so they are of the form $[1:a:b]$ with $a,b\in F$, or $[0:1:0]$, or $[0:0:1]$.


问:
Eigenvectors of real symmetric matrices are orthogonal
Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.</p>

答:
How about Let $A$ be symmetric, then there exists a matrix $D$ such that $A=QDQ^T$, taking the transpose of $A$, namely

$$\left(A\right)^T = \left(QDQ^T\right)^T $$
  $$A^T = \left(Q^T\right)^TD^TQ^T$$
  $$A^T = QDQ^T$$

thus $A^T = A$ if and only if $A$ is symmetric.

It is noteworthy that $D^T = D$ since $D$ is diagonal and $Q$ is the matrix of normed eigenvectors of $A$, Thus $Q^T = Q^{-1}$


问:
The probability to arrange 2n socks with different colors such that every pair difference color at most one?
You have n pairs of socks, ie 2n socks, with shades of gray numbered from 1-white to n-black. You blindly remove the socks from the drawer and pair them at random.
What the probability such that every paired different at most at one color?

Note:
The colors between 1 and n, so the pair (1,2) is legal and the pair (1,3) isn't legal because the difference in the colors is 2 and not 1 or 0.

The answer: $\frac{(2^{n+1}+(-1)^n)2^nn!}{3(2n)!}$

I could not figure out how to solve the question and get to the closed expression in the answers. I would be happy if you keep the answer simple as much as possible. Thanks.

答:
Let $F(n)$ be the number of pairings where each pair differs by $1$ or $0$. One of the socks with color $n$ must be paired with either $(i)$ the other sock of color $n$ or $(ii)$ one of the two socks of color $n-1$. In case $(i)$ the remaining socks are all of colors $n-1$ and lower and may thus be paired in $F(n-1)$ ways. In case $(ii)$ since 1 sock of color $n$ is paired with $1$ sock of color $n-1$ the other sock of color $n$ must be paired with the other sock of color $n-1$. There are two ways that this can happen, but after that the remaining $2(n-2)$ may be paired in $F(n-2)$ ways. This shows that
$$F(n)=1\cdot F(n-1)+2\cdot F(n-2)$$
You can see easily that $F(1)=1$ and $F(2)=3$. Induction may then by used to show that $F(n)=\frac{2^{n+1}+(-1)^{n}}{3}$ from which the result follows because the total number of pairings is $\frac{(2n)!}{2^nn!}$.


答:
I think you'd want a union affine hyperplanes, right ? The image here will probably help: https://en.wikipedia.org/wiki/Linear_subspace

Notice that any $n$-dimensional subspace of $(\mathbb{Z}/p\mathbb{Z})^d$ has $p^d$ elements. Then you can always express elements of $(\mathbb{Z}/p\mathbb{Z})^d$ as a linear combination of an element of a subspace $F$ isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{d-1}$, and an element of $F^\perp$ the orthogonal complement to that subspace, which is isomorphic to $(\mathbb{Z}/p\mathbb{Z})$.

For this, your union should then be

$$ \bigcup_{v \in F^\perp} F \; + \; v$$

where each $F+v$ is an affine hyperplane (or just a hyperplane when $v$ is the null vector), and all $F+v$ are parallel to each other.

Body:
Consider $V=(\mathbb{Z}_5)^n$ a vector space over the field $F=\mathbb{Z}_5$, How can I write $(\mathbb{Z}_5)^n$ as a union of finite hyperplanes? I am just looking for an example of how I can do it.
I guess it is possible for the fact that $F$ is finite and it implies that $V$ is also finite.


问:
Two circles inscribed in a semicircle
Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)

答:
I must say that the first contact with this question left me so perplexed that I asked a question about one of the interrogations I had.

The answers you will find there should be read before embarking into the following solution.

Let us begin by a certain number of notations that can be  followed on figure 1.

Fig. 1. $C_0$ belongs to the circle with center $S(0,-r)$ and radius $r \sqrt{2}$. $C_1$ and $C_2$ belong to parabola $(P)$ (light blue) with equation $y=\dfrac{1}{2r}(r^2-x^2)$.

Consider the two little circles. Let $C_1(x_1,y_1)$, $C_2(x_2,y_2)$ be their resp. centers. Let $C_0(x_0,y_0)$ be their tangency point. As a consequence, the tangency points on the diameter are $T_1(x_1,0)$, $T_2(x_2,0)$. The other tangency points being, for $k=1,2$:

$$U_k \left(x'_k=\dfrac{2r^2x_k}{r^2+x_k^2}, \ y'_k=r \dfrac{r^2-x_k^2}{r^2+x_k^2}\right)\tag{1}$$

due to the fact that $S,T_k,U_k$ are aligned.

We are going to show that it is possible to write the constraints in terms of the three parameters, ultimately with the single parameter $y_0$. What are these constraints ?

a) $C_0$ belongs to the circle with center $S$ and radius $r \sqrt{2}$. Therefore:

$$x_0^2+y_0^2+2ry_0-r^2=0\tag{2}$$

giving:

$$x_0=-\sqrt{2r^2-(y_0+r)^2}\tag{2'}$$

(please note the minus sign in front of the square root, chosen in order to enforce condition $x_0<0$ which is the same as saying that the smaller circle is on the left of the bigger one).

b) The equation of the tangent line to conic curve $(P)$ in $(x_0,y_0)$ is obtained from (2) as:

$$xx_0+yy_0+r(y+y_0)-r^2=0\tag{2''}$$

As $C_1$ and $C_2$ are the intersection points of this straight line with parabola (P), their abscissas $x_1$ and $x_2$ will be the roots of the quadratic equation obtained by plugging $y=\dfrac{1}{2r}(r^2-x^2)$ in ($2'*$). Rather easy calculations give:

$$\begin{cases}x_1&=&r\dfrac{x_0-\sqrt{2}y_0}{r + y_0},\\ x_2&=&r\dfrac{x_0+\sqrt{2}y_0}{r + y_0}\end{cases}\tag{3}$$

A direct consequence of (3) is that:

$$x_0-x_1=y_0\dfrac{r\sqrt{2} + x_0}{r+y_0}\tag{4}$$

Our first goal has been achieved.

It's time now to involve the information about the areas. We will consider the blue and yellow areas with resp. values $33$ and $36$:

The area of the blue triangle is clearly ("half base times height"):

$$\frac12(x_2-x_1)y_0=33$$

Using (3), we can transform it into:

$$\dfrac{\sqrt{2}y_0^2r}{r+y_0}=33 \ \iff \ r=\dfrac{33y_0}{\sqrt{2} y_0^2 - 33}\tag{5}$$

Besides, the area of the yellow triangle $C_0U_1T_1$ (we have described the triangle with a positive orientation) can be expressed under the following determinant form:

$$\tfrac12 \begin{vmatrix}x_0&x'_1&x_1\\
y_0&y'_1&0\\
1&1&1\end{vmatrix}=\tfrac{1}{2 (r^2+x_1^2)} \begin{vmatrix}x_0&2r^2x_1&x_1\\
y_0&r(r^2-x_1^2)&0\\
1&(r^2+x_1^2)&1\end{vmatrix}=\tfrac{1}{2 (r^2+x_1^2)} \begin{vmatrix}x_0-x_1&2r^2x_1&x_1\\
y_0&r(r^2-x_1^2)&0\\
0&(r^2+x_1^2)&1\end{vmatrix}$$

(we have used formulas (1) and replaced column 1 by the difference between column 1 and column 3), giving rise to the following equation:
$$ \begin{vmatrix}x_0-x_1&2r^2x_1&x_1\\
y_0&r(r^2-x_1^2)&0\\
0&(r^2+x_1^2)&1\end{vmatrix}=2 (r^2+x_1^2) 36\tag{6}$$

in which $x_0-x_1$ can be replaced by its expression in (4).

We have now the possibility to deduce from all these equations, by "cascade substitutions" a polynomial equation in $y_0$ (that I haven't explicited, using a CAS for these last computations: see remark below). This equation gives the only possible solution in conformity with the figure:

$$y_0=6\sqrt{\dfrac{11 \sqrt{2}}{17}}\approx 5.7396$$

from which, using (5), we get:

$$r=\dfrac67\sqrt{187\sqrt{2}}\approx 13.9390$$

(which is the same value as that found by g.kov).

Remarks: One could object that we haven't used the two other values of the two other triangles. I must say that I haven't attempted to use them because

I was tired by all these endless calculations [not entirely done by hand; I have relied for the final computation upon the Computer Algebra System available under the Matlab environment].

One can await that the two other values attest the values already obtained. Otherwise... one could question the author... Besides, what is the origin of this problem ?

Title: Two circles inscribed in a semicircle


问:
Is this partial derivative correct
For

$$
a^{2} = \sigma(w^2 a^{1}+b^2) = \sigma(z^2)\\
\\ \text{where} \;
z^l =w^l a^{l-1} +b^l
$$
If we write the derivate

$$
{\frac{\partial a^{l} }{\partial w^{l}} = \frac{\partial \sigma (z^{l}) }{\partial w^{l}} = \sigma' (z^{l}) \quad \rightarrow  ( {a})} \\
$$
Then is the below proper
$$
\frac{\partial(a^2)}{\partial(a^1)} = \frac{\partial(\sigma(w^2 a^{1}+b^2))}{\partial(a^1)} =  w^2.\sigma'(a^1)  \; \rightarrow  ( {b})\\ \\
$$

答: Assuming that $a^2$ and $a^1$ are distinct variables, then you are mostly correct.  It should be $$\frac{\partial a^2}{\partial a^1} = w^2\sigma'(w^2 a^1 + b^2) $$ rather than $w^2\sigma'(a^1)$.

Title: Is this partial derivative correct


问:
Dual space of $H^1(\Omega)$
I'm a bit confused, why do people not define $H^1(\Omega)^*$? Instead they only say that $H^{-1}(\Omega)$ is the dual of $H^1_0(\Omega).$ </p>

<p>$H^1(\Omega)$ is a Hilbert space so it has a well-defined dual space. Can someone explain the issue with this?</p>

答: As far as I remember, one usually defines $H^{-1}(\Omega)$ to be the dual space of $H^1(\Omega)$.   The reason for that is that one usually does not identify $H^1(\Omega)^*$ with $H^1(\Omega)$ (which would be possible) but instead works with a different representation. E.g. one works with the $L^2$-inner product as dual pairing between $H^{-1}(\Omega)$ and $H^1(\Omega)$ (in case the element in $H^{-1}(\Omega)$ is an $L^2$-function).

Title: Dual space of $H^1(\Omega)$


问:
Showing no solution exists for small system of multivariate quadratics over a finite field
I would like to show for the system of polynomials
$$
f_j(\vec{a})=\sum_{i=1}^8 a_i a_{i+j \pmod{8}} = 0, \quad j=1,2,3,4
$$

with $a_i \in \{-1,+1\}$, no common solution can exist. It is easy to confirm this by straightforward brute force calculation of all possible variable combinations.

I would like to find some clearer argument however that does not rely on computational methods, something I think ought to be possible given the highly structured nature of the problem. It is possible to find a common solution for any three out of four polynomials as in $\vec{a} = \{1,1,1,-1,1,1,1,-1\}$ which is a root for all except when $j=4$ but I can only show this in the specific instance and generalizing the inconsistency has proven difficult.

Is there a way to show this inconsistency directly from the above system of equations? The best approach I can see is via Hilbert's Nullstellenstatz which would entail finding a set of any polynomials $q_j(\vec{a})$ such that
$$
\sum_{j=1}^4 q_j(\vec{a})f_j(\vec{a}) = 1.
$$
I have been quite unsuccessful in this however - such a set of $q_j(\vec{a})$ certainly exist given the iff criteria of the theorem but they may be very complicated. The nice behavior of the $f_j(\vec{a})$ lead me to believe they may be quite nice.

答:
Here I might try the following approach.

Let's denote
$$
\Theta(j):=\sum_{i=1}^8a_ia_{i+j}
$$
with index arithmetic done modulo $8$ everywhere. As $a_i\in\{\pm1\}$ for all $i$, we clearly have $\Theta(0)=8$. Also, the obvious symmetry says that $\Theta(j)=\Theta(8-j)$ for $j=1,2,3,4$.

Write
$$
S=a_1+a_2+\cdots+a_8.
$$
If $\Theta(j)=0$ whenever $j=1,2,3,4$, then
$$
\begin{aligned}
S^2&=\sum_{i=1}^8\sum_{j=1}^8a_ia_j\\
&=\sum_{i=1}^8\sum_{\ell=0}^7a_ia_{i+\ell}\\
&=\sum_{\ell=0}^7\Theta(\ell)\\
&=\Theta(0)+2\Theta(1)+2\Theta(2)+2\Theta(3)+\Theta(4)\\
&=\Theta(0)=8.
\end{aligned}
$$
But $S$ is an integer and $8$ is not a perfect square, so this is a contradiction.


问:
The dimension of a matrix as a subspace of $\cal{M}_{3 \times 3}$
I was going through Linear Algebra homework and ran into another problem that I struggled with.

Determine the dimension of

$$\left\{A \in M_{3\times3} : A\begin{pmatrix}
    1 & 1 & 1 \\
    1 & 0 & -1 \\
    0 & 1 & 2 \\
    \end{pmatrix}
=   \begin{pmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0\\
    0 & 0 & 0 \\
    \end{pmatrix}\right\}
$$

I am really unsure of where to start from here. I appreciate any help.

答: Let$$B=\begin{bmatrix}1&1&1\\1&0&-1\\0&1&2\end{bmatrix}.$$Then $A.B=0\iff B^T.A^T=0^T=0$. But\begin{align}B^T.(x,y,z)=(0,0,0)&\iff\left\{\begin{array}{l}x+y=0\\x+z=0\\x-y+2z=0\end{array}\right.\\&\iff z=y=-x\end{align}So, $B^T.A^T=0$ if and only if $A$ is a matrix of the form$$\begin{bmatrix}a&-a&-a\\b&-b&-b\\c&-c&-c\end{bmatrix}.$$Can you take it from here?


问:
$\det(\varphi + \psi)$ of 2 A-endomormphisms $\varphi,\psi: M \to M$ with A: commutative ring with unit, M: free A-module or rank 2
As is said in the title, we consider A to be a commutative ring with unit, M a free A-module of a finite rank r = 2, and $\varphi$,$\psi: M \to M$ A-endomorphisms. With this data I would like to show that

$$det(φ+ψ) + det(φ−ψ) = 2(det(φ) + det(ψ))$$

However the only thing related to this that came to my mind was the fact that $det(λφ) =λ^2det(φ)$ for $\lambda \in A$, but I'am not sure it could be usefull here.

Also, after some research I think that this problem can be shown using external product, but I have not seen it. Is there another way to show this? I just want to find where to starts, and know wether I have seen enough in class to solve this problem or not.

Thanks for your help!

答:
Let $u_1,u_2$ be the free generators of $M$, then every element of $M$ can be uniquely expressed as $\alpha_1u_1+\alpha_2u_2$ with $\alpha_1,\alpha_2\in A$.

Also, any homomorphism from $M$ is determined by the images of $u_1$ and $u_2$ which can be arbitrary.

So, an endomorphism $\varphi:M\to M$ can be encoded in a $2\times 2$ matrix with entries in $A$, namely let
$$[\varphi]=\pmatrix{\alpha&\beta\\ \gamma&\delta}$$
mean that $\varphi(u_1)=\alpha u_1+\gamma u_2$ and $\varphi(u_2)=\beta u_1+\delta u_2$.

Then $\det\varphi=\alpha\delta-\beta\gamma$, and addition of endomorphisms simply correspond to addition of matrices.

 We simply open up the brackets here:
 $$\det\pmatrix{\alpha+\alpha'&\beta+\beta'\\ \gamma+\gamma'&\delta+\delta'}\ +\ \det\pmatrix{\alpha-\alpha'&\beta-\beta'\\ \gamma-\gamma'&\delta-\delta'}\ = $$
 $$=\ (\alpha+\alpha')(\delta+\delta')-(\beta+\beta')(\gamma+\gamma')\ +\ (\alpha-\alpha')(\delta-\delta')-(\beta-\beta')(\gamma-\gamma')\ =\ \dots $$


问:
$|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$
Let $x,y,w,p,q \geq 0$ such that $p,y \leq 1.$
Is it true that $$|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$$

Tried to come with a proof but it didn't work, it seems the statement is false.

Any idea-example to disprove it?

答:
Since all variables are nonnegative (in case $y\ne 0$, $p=0$, exchange $p,y$ and $w,q$  in the following treatment; the case  $p=y=0$ is trivial) we can rewrite
$$
|p\min(x,q)-y\min(x,w)| \leq x |p-y|
$$
as
$$
|\frac{1-\frac{y/p}{\min(x,q) / \min(x,w)}}{1 -y/p}| \leq \frac{x}{\min(x,q)}
$$
Denote $z = y/p$. As $0 \le y,p \le 1$ we have that $0\le z< \infty$.

Further, denote $a = \min(x,q) / \min(x,w)$. Then the question translates to
$$
|\frac{1-\frac{z}{a}}{1 -z}| \leq \frac{x}{\min(x,q)}
$$

First, consider $ a\ne 1$. This happens when $w > x \ge q$ or $q > x \ge w$. Then the LHS takes values from $0$ (for $z=a$) to $\to \infty$ (for $z=1$) so the LHS can, for all $z$, never be smaller than anything, which fails the assertion.

Now, consider $ a =  1$. Then we have to show $1 \leq \frac{x}{\min(x,q)} $. This falls into two subcases.

2.1. $a = \min(x,q) / \min(x,w) =1$ is generated for $x >  w=q$. Then we have the condition $1 \leq \frac{x}{\min(x,q)} = \frac{x}{q} > 1$ which holds, with a generic "$<$" in the assertion.

2.2. $a = \min(x,q) / \min(x,w) =1$ is generated for $x \le q$, $x \le  w$. Then we have the condition $1 \leq \frac{x}{\min(x,q)} = \frac{x}{x}= 1$ which holds true, so in this  case  the assertion holds with equality.

Title: $|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$


问:
Lebesgue measure of a triangle
"Proof that the Lebesgue measure of a triangle is the euclidean area."

Well, I know firstly that a triangle is measurable with finite measure, because is closed and bounded, and that the Lebesgue measure is invariant by translations. But I don't know how to prove it basing on what I've just said.

Can someone give me a hand on this problem?

Thanks in advance.

答:
Hint: I presume you know the measure of a rectangle is its area.  Two congruent triangles form a parallelogram, and a parallelogram can be dissected and the pieces
rearranged to form a rectangle.

EDIT: Another way to do it (assuming you know the measure of a square is its area): bound above and below the number of nonoverlapping squares of side $\epsilon$ that can fit inside the triangle.

Title: Lebesgue measure of a triangle


问:
12 students into groups of 2. Why doesn't my solution work?
I have the following problem:

There are 12 students in the class. How many ways are there to split them into working groups of size 2 to work on the same assignment?

I know the solution but I don't understand why my solution doesn't work.
My solution is:

Let's choose the first members of each group, it's combination ${12 \choose{} 6}=924$ (we have no order and no repetitions, n = 12 = number of students, k = 6 = number of groups). Then, let's choose the second members of each group, it's permutation $\frac{6!}{(6-6)!}=720$ (now order matters because we already have first members and still no repetitions, n = 6 = number of remaining students, k = 6 = number of groups). Also, we need to divide the result by 2 to account that the order inside a group doesn't matter ({1, 3} = {3, 1}). So it's $924*720/2=332640$

答:
You want to enumerate the number of ways of forming $6$ distinct pairs.

That is
$$\frac{(12)!}{(2!)^6} = \binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}.$$

However the above computation overlooks overcounting, re in any specific partion into 6 pairs, there are $(6!)$ ways of ordering when each pair is created.

So, you have to multiply the above enumeration by the overcounting scalar of $\frac{1}{(6)!}.$


问:
A beautiful limit involving primes and composites
I observed the following limit empirically. Let $p_n$ be the $n$-th prime and $c_n$ be the $n$-th composite number then,

$$
\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{p_n c_n}{p_n c_n + p_i c_i} = \frac{\pi}{4}.
$$

I am looking for a proof.

答:
We have that

$$\lim_{n\to\infty}\,\frac1n\,\sum_{k=1}^n\frac n{n+k}=\lim_{n\to\infty}\,\frac1n\,\sum_{k=1}^n\frac1{1+\frac kn}=\int\limits_0^1\frac{dx}{1+x}=\log 2$$

Your sum is not, of course, the above one, but it ressembles it a little, so I'd expect its sum to be closer to $\,\log 2\,$ than to $\,\pi/4\,$ , in particular since the difference between these two numbers is less than $\,1/100\,$ , yet I cannot tell for sure.

A simple program that can add automatically can make some checkings...


问:
Question about graphs
A random graph is a graph $\mathcal{V} = (V,R)$ which is non directed and has no loops and such that for every pair of finite disjoint subsets $A$ and $B$ of vertices satisfying that there exists a vertex $c \in V$ such that $(c, a) \in R$ for every $a \in A$ and $(c, b) \in R$ for every $b \in B$.

I need to prove that the point $c$ can be specifically chosen in $V \setminus (A \cup B)$ (notice that the set $V$ is infinite). I've been looking on the web and every site I have found about random graphs said that this is either the definition or trivial.

What I've tried to show that if $c \in A \cup B$ then we have a contradiction, but I cannot find the contradiction... Also, the proof by induction seems to be quite difficult. Also, I am supposed to prove this without probability theory ideas.

Any help would be appreciated. Thanks in advance.

答:
In my opinion, unless I'm missing a simpler argument, this is not trivial; although the argument is short, it does take a trick. The idea is to apply randomness twice, with the first application being "preparatory" and designed to force the second application to choose something outside $A\cup B$.

Suppose $\mathcal{G}$ is a random graph and $A,B$ are disjoint finite subsets of $\mathcal{G}$. Applying randomness to the pair $(A\cup B,\emptyset)$ we get a vertex $v$ connected to all vertices in $A$ and $B$; by the no-loops condition we must have $v\not\in A\cup B$. Now apply randomness again to $(A,B\cup\{v\})$. A vertex $w$ which is connected to everything in $A$ but nothing in $B\cup\{v\}$ cannot lie in $A\cup B$ since everything in $A\cup B$ is connected to $v$. Forgetting about $v$, such a $w$ is then a vertex in $\mathcal{G}$ connected to everything in $A$, not connected to anything in $B$, and not in $A\cup B$.

Note that the above relies on the no-loops condition. This may seem a bit odd; in fact, a different two-step argument shows that it isn't necessary. First, you can show that every random graph must be infinite (if you haven't done this yet it's a good exercise). Now let $X$ be a set of vertices of size $\vert A\cup B\vert$ which is disjoint from $A\cup B$. Since $\vert A\cup B\vert<\vert\mathcal{P}(A\cup B)\vert=2^{\vert A\cup B\vert}$, there is some $Y\subseteq X$ such that no element of $A\cup B$ is connected to everything in $Y$ and not connected to anything in $X\setminus Y$. Now apply randomness to the pair $(A\cup Y, B\cup (X\setminus Y))$.

Title: Question about graphs


问:
How fast decrease the Fourier coefficients
Given are the almost every where smooth, $2\pi $ periodic, absolutely integrable  function $f:\mathbb R \to \mathbb C$. In this case we have finite many points $\gamma_1,\dots ,\gamma_n\in [-\pi,\pi)$ where $f$ is not continuous, but every where else is $f$ smooth.  We can also assume that the limits $\lim_{h\to 0} f(x\pm h),\: \lim_{h\to0}f'(x\pm h)$ always exists for every $x\in \mathbb R$.

Now I want to show that $$c_k(f):=\frac 1 {2\pi} \int_{-\pi}^{\pi} f(x)e^{-ikx}dx$$
does not decrease faster than $\tfrac 1{|k|}$.

So this is what I tried:

We can define the "sawtooth" function $$V_\gamma(x):=\cases {\tfrac 1 {2\pi}(\gamma -\pi-x)\quad,-\pi<x<\gamma\\\tfrac 1 {2\pi} (\gamma+\pi-x)\quad,\gamma\leq x\leq\pi}$$
which has a jump at $\gamma$ of magnitude $1$. The Fourier coefficients of $V_\gamma$ are given by
$$c_0(V_\gamma)=0,\qquad c_k(V_\gamma)=\tfrac 1 {2\pi i k}e^{-ik\gamma},\quad k=\pm1,\pm2,\dots$$
Let $A_j$ the magnitude of the jump of $f$ at $\gamma_j$, then we can define the $2\pi$ periodic, almost everywhere smooth, absolutely integrable function $g:\mathbb R\to\mathbb C$
$$g(x):=f(x)-\sum_{j=1}^n A_j V_{\gamma_j}(x).$$
We have now that $g$ is also continuously, so we know that $c_k(g)$ decrease faster to zero than $\tfrac 1{|k|}$. We get now by using the Fourier operator (and some small  equation manipulation)
$$c_k(f)=c_k(g)+\tfrac 1 {2\pi i k} \sum_{j=1} A_j e^{-ik\gamma_j}.$$
My Problem is now, that I don't have a good way to show that $\sum_{j=1}^nA_j e^{-ik\gamma_j}$ does not vanish for big $k$. It should be clear since $e^{-ik\gamma _j}$ is a kind of periodic and the $\gamma_j$ are all different to each other, but I don't want to go this way.

So does anyone have an idea how I could show that  $\sum_{j=1}^nA_j e^{-ik\gamma_j}$ does not vanish or maybe a total different way to show this?

答:
Take $$H(x)=\sum_k c_k(H) e^{i kx}$$ $2\pi$-periodic, $C^1$ and non-zero at $\gamma_j$ only for $j=1$.

Let $h_m(x)=e^{i m x}H(x)$ so that $$h_m(x)=\sum_k e^{ikx} c_{k-m}(H)$$

If $\lim_{k\to \infty}\sum_{j=1}^nA_j e^{ik\gamma_j}=0$ you'd get that
$$|A_1 H(\gamma_1)|=\lim_{m\to \infty}|\sum_{j=1}^n A_j h_m(\gamma_j)|=\lim_{m\to \infty} |\sum_k c_{k-m}(H) \sum_{j=1}^nA_j e^{ik\gamma_j}|=0$$
which is a contradiction.


答:
For simply a counterexample, you could use $p=y=1$, $x=w=1$, $q=0$.

If the inequality was true, this would give $1\le0$.

Title: $|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$

Body:
Let $x,y,w,p,q \geq 0$ such that $p,y \leq 1.$
Is it true that $$|p\min(x,q)-y\min(x,w)| \leq x |p-y|?$$

Tried to come with a proof but it didn't work, it seems the statement is false.

Any idea-example to disprove it?


问:
if $\log_xa+\log_ya=4\log_{xy}a$ prove that $x=y$
Let $x,y$ be numbers in the interval $(0,1)$ with the property that there exists a positive number $a\ne1$ such that
$$\log_xa+\log_ya=4\log_{xy}a$$
I have used the property
$$\log _ba=\frac{\log a}{\log b}$$

and I have
$$\log_xa+\log_ya=\frac{\log a}{\log x}+\frac{\log a}{\log y}$$

$$\log a\left(\frac{\log x+\log y}{\log x \log y}\right)=4\frac{\log a}{\log(xy)} \implies (\log(xy))^2=4 \log x \log y$$

I don't see how this implies $x=y$

答:
HINT: From $(\log xy)^2=4 \log x \log y$, write out the right hand side as

$$(\log x + \log y)^2=  (\log x)^2 + 2 \log x \log y + (\log y)^2 \; .$$


问:
Derive alternate representation of parametric equations of first quadrant of a circle.
In a math book I am reading, the author mentions that the first quadrant of a circle can be defined by the parametric equations
\begin{align*}
x(u)&=\cos{u}\\
y(u)&=\sin{u},
\end{align*}
for $0\leq u\leq\frac{\pi}{2}$.  The author then goes on to say that by setting $t=\tan(u/2)$, one can derive the alternate representation
\begin{align*}
x(t)&=\frac{1-t^2}{1+t^2}\\
y(t)&=\frac{2t}{1+t^2},
\end{align*}
for $0\leq t\leq 1$.

Let's consider the function $x$ for a moment.  I understand, that if $t=\tan(u/2)$, then $u=2\arctan{t}$, and $\cos{u}=\cos(2\arctan{t})=(1-t^2)/(1+t^2).$  What I can't see is how this is equal to $x(t)$.  It seems like it should be equal to $x(2\arctan{t})$!  When I actually try to compute $x(t)=x(\tan(u/2))$, I inevitable get stuck and cannot arrive at $(1-t^2)/(1+t^2)$.

Can someone please help me to connect the dots here?

答:
I think your confusion is coming from the fact that $x\mapsto x(u)$ and $x\mapsto x(t)$ are not the same function, despite having the same name.  We can be a bit more precise:

Consider $x(u) = \cos u$, and let $u = 2\arctan t$.  Then $$x(u) = x(2\arctan t) = \cos(2\arctan t) = \frac{1-t^2}{1+t^2}.$$  Now, let's define $x_1(t)$ by $$x_1(t) = x(2\arctan t) = \frac{1-t^2}{1+t^2}.$$  In the same way, we can start with $y(u)$ and get to $$y_1(t) = y(2\arctan t) = \frac{2t}{1+t^2}.$$  So the functions $x,y$ and $x_1,y_1$ are in fact distinct functions, although we can think of $x_1$ and $y_1$ as compositions of $x$ and $y$ with $2\arctan t$.

The important thing here is that
$$x_{1}^2(t)+y_{1}^2(t) = \left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 = \frac{1 -2t^2 + t^4 + 4t^2}{1+2t^2 + t^4} = \frac{1+2t^2+t^4}{1+2t^2+t^4} = 1,$$ and so this parameterization does in fact give the unit circle.


问:
Prove that if the element $a\in G$, where $G$ is a group, has order $n$ and its homomorphic image $f(a)$ has order $m$, then $m\mid n$

Let $G$ be a group. If $a\in G$ has order $n$ and its homomorphic image $f(a)$ has order $m$, then $m\mid n$.

So I have a confusion with this question, what does it mean the order of $a$? wouldn't the order be of the group and the homomorphic group of $f(a)$?

besides from that i have tried using Lagrange's theorem but the theorem says that

The order of the subgroup of $G$ divides the group $G$
But the homomorphic image of $f(a)$ does not mean that it's a subgroup of $G$ or does it?

I appriciate all help!
Thanks in advance!

答:
An important distinction for you to know is the difference between the order of a group and the order of an element of a group. It seems like you already understand the former, so the latter:

Let $G$ be a group with identity element $e$, and let $a \in G$. The order of $a$ is the smallest positive integer $n$ such that $a^n = e$.

Does knowing this help?


问:
Conditional distribution of $X$ given $Y$

Let $Y$ be an exponential random variable with mean $\frac{1}{\theta}$, where $\theta\gt0$. The conditional distribution of $X$ given $Y$ has Poisson distribution with mean $Y$. Then, the variance of $X$ is

(A)$\frac{1}{\theta^2}\quad\quad$(B)$\frac{\theta+1}{\theta}\quad\quad$(C)$\frac{\theta^2+1}{\theta^2}\quad\quad$(D)$\frac{\theta+1}{\theta^2}$

The thing is since $Y$ a continuous random variable , is it possible that the conditional distribution of $X$ given $Y$ follow Poisson distribution (which is discrete in nature) and moreover we don't know whether $X$ is a discrete or a continuous random variable.

How do we find the variance of $X$?

答:
You can use the law of total expectation

$$E[X]=E[E[X|Y]]=E[Y]=\frac{1}{\theta}$$

$$V[X]= E[E[X^2|Y]]  -E^2[X]=E[Y^2+Y]-\frac{1}{\theta^2}$$

$$V[X]=\frac{2}{\theta^2}+\frac{1}{\theta}-\frac{1}{\theta^2}=\frac{1+\theta}{\theta^2}$$

Thus the correct answer is D)


问:
Nice underestimated elementary topology problem
There is a nice elementary topology problem (proposition) that is often
missing from the introductory books on the topic.

PROBLEM. Let $\varphi:\mathbb{S}^{1}\rightarrow\mathbb{S}^{1}$ be a continuous
self-map of the circle of degree $\deg(\varphi)=d$. Then $\varphi$ has at
least $|d-1|$ fixed points. (For example, if $\varphi$ is an orientation
reversing homeomorphism, then it has at least 2 fixed points - the 'two monks
walking in opposite directions' problem.)

Its place should be just after the notion of degree, fundamental group etc. In
my opinion, it is a very good exercise, as it combines basic notions, such as
degree, fixed point, $\pi_{1}(\mathbb{S}^{1})$ and has useful applications.
Paradoxically, I don't see it in the appropriate place ("degree", "fundamental
group"), but in the more heavy advanced context of Nielsen theory. Nielsen
theory, in its turn, is often missing from elementary topology books. The
available to me ones are dealing with almost one and the same list of problems
(nice, indeed), but this one seems not to be present there.

So my question is: Does anyone know a good elementary proof of this
problem/proposition (without referring to advanced things such as Nielsen index
theory or so). Any references are welcome as well. Thanks in advance.

答:
I thought I'd better make my comment into an answer:

Lift $\varphi$ to a continuous map $f: \mathbb{R} \to \mathbb{R}$ and look at its graph. The condition on the degree forces $f(x+1) = f(x) + d$ for all $x \in \mathbb{R}$. But this implies that the graph over $[0,1)$ must intersect at least $d-1$ among the graphs  $y = x + k$ with $k \in \mathbb{Z}$ by the mean value theorem (to be specific, the ones with $k$ between $[f(0), f(0)+d)$ of which there are at worst $d-1$). I let you flesh out the details, but that's what I call completely elementary.


问:
Collecting proofs $SO(3)\cong\Bbb R P^3$
There are a few questions on SE about the well-known homeomorphism $SO(3)\cong \Bbb R P^3$, but I thought it would be fun to collect as many proofs as we can find in one spot.

Background:

$SO(3)$ is the group of orientation-preserving orthogonal transformations of $\Bbb R^3$, or equivalently the group of $3\times 3$ real matrices $A$ with $AA^\top=I$ and $\det A=1$

$\Bbb R P^3$ is the space of lines going through the origin in $\Bbb R^4$ topologized equivalently as

$\left(\Bbb R^4\setminus\{0\}\right)/\sim$, where $x\sim cx$ for all $x\in\Bbb R^4\setminus\{0\}$ and $c\in\Bbb R\setminus\{0\}$ or

$D^3/\sim$, where $p\sim -p$ for all $p\in S^2$

Both spaces are oriented $3$-manifolds with smooth structures

$SO(3)$ is diffeomorphic (and thus homeomorphic) to $\Bbb R P^3$

As varieties, $SO(3)$ is affine while $\Bbb R P^3$ is projective. Thus, they are not isomorphic as varieties.

You can find some proofs of the homeomorphism $SO(3)\cong \Bbb R P^3$ here, here, and in Hatcher $\S3.D$ (currently at the bottom of page $293$).

I will start with Hatcher's proof along with a version I've written using quaternions.

One direction I'm curious about is to think of $SO(3)$ as the Stiefel manifold $V_2(\Bbb R^3)$, and to somehow turn a pair of orthonormal vectors in $\Bbb R^3$ into a line in $\Bbb R^4$.

答:
Hatcher's proof:

We define a map $\varphi\colon D^3\to SO(3)$ which takes a vector $p\in D^3$ to the rotation about the line spanned by $p$ by $|p|\pi$ radians (see $\ast$ below). If $p\in S^2$, then $\varphi(p)=\varphi(-p)$, so we have an induced $\tilde\varphi\colon \Bbb RP^3\to SO(3)$. It's clear $\tilde\varphi$ is injective since we can recover the axis of rotation and the rotation angle. Further, since every matrix $A\in SO(3)$ has either one or three real eigenvectors, every transformation must be rotation about some axis. It follows that $\tilde\varphi$ is surjective. Since $SO(3)$ is compact, $\tilde\varphi$ is a homeomorphism.

$\ast$: We use the right-hand-rule orientation convention here, so if your thumb points along $p$, your fingers wrap around in the rotation direction. This is akin to the direction of a magnetic field around a cable with an electric current.

Quaternionic proof:

Identify $\Bbb R^4$ with $\Bbb H$ via $(a,b,c,d)\mapsto a+bi+cj+dk$.

Given a line $\ell\in\Bbb RP^3$, pick a unit quaternion $h$ which spans $\ell$ over $\Bbb R$. We define $\psi_\ell\colon \Bbb R^4\to\Bbb R^4$ by $x\mapsto hxh^*$ (where $h^*$ is the quaternionic conjugate of $h$). Since the only other choice of unit quaternion is $-h$, $\ell\mapsto \psi_\ell$ is well-defined.

Now $\psi_\ell(1)=h1h^*=|h|^2=1$, so in the basis $1,i,j,k$, the matrix of $\psi_\ell$ is
$$\begin{pmatrix}1 & 0\\0 & A \end{pmatrix}.$$

On the other hand, the standard inclusion $Sp(1)\to GL(4;\Bbb R)$ lands in $SO(4)$, and this shows that the $3\times 3$ matrix $A$ is in $SO(3)$. We hence obtain $\tilde\psi\colon \Bbb RP^3\to SO(3)$. Some more work is required to show $\tilde\psi$ is a bijection, but this is at least a familiar fact to anyone familiar with computer graphics, for example.


问:
Is the line measure operator bounded in $W^{1,p}(\mathbb{R}^2)$
Consider the linear funtional $S_l: W^{1,p}(\mathbb{R}^2) \to \mathbb{R}$ for some segment $l \in \mathbb{R}^2$ such that $S_l (\mu)$ measures the length of $l$ with respect to $\mu dx$. In other words, it performs the line integral of the function $\mu$ on $l$. I'm curious if $S_l$ is a bounded functional for all finite $l$ for some $p > 1$? I know that for point evaluation operator it is true, since for $p > n$, there is continuous injection of $W^{1,p}(\mathbb{R}^n)$ to $L^\infty(\mathbb{R}^n)$. I wonder if it's also the case for evaluating the function on a line segment.

答: I just realized it suffices to use the trace operator $T:W^{1,p}(\Omega) \to L^p(\partial\Omega )$ for this, and as $\mu(l)$ is finite apply Holder to get $|Tu|_{L^1(\partial\Omega)} \leq C|u|W^{1,p}(\Omega)$ for all $p \geq 1$. So as long as the set where the integral is taken is some bounded boundary then the result holds.


问:
What was the $2021$st number in this new system?
This question came up recently:

John III, the third king of Johnland was very angry with John II, who killed the founder of Johnland, the beloved John I. By his new law the digit $2$ was forbidden to use. The numbers were listed as $1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, $$ 15, 16, 17, 18, 19, 30, 31, 33, \dots $ What was the $2021$st number in this new system?

I then came up with this:

```<code>Sol.:

First, I studied how many numbers would be ‘skipped’ in the system within a power of 10;
100: 0
101: 1
102: 20
103: 300
104: 4000
105: 50000
The pattern is quite clear.
Skipped numbers before 2021st number: 300 x 2 + 1 x 2 + 0 x 1 = 602.
However, in 602 there are some skipped numbers:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) = 602 + 120
Also in 120 there are some more:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) + (20 x 1 + 1 x 2) = 602 + 120 + 22
And in 22:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) + (20 x 1 + 1 x 2) + (1 x 2 + 0 x 2) = 602 + 120 + 22 + 2
There are none in two.
</code>```

From there I didn't know what to do with myself and just got the answer in python, but that's cheating, so I would like to know how I can hope to solve this question with pure maths.

Cheers!

答:
First convert $2021$ base ten to base nine, thus $2685$.

Now map the digits from the base nine representation so that $2$ is excluded instead of $9$, to wit $(0,1,2,3,4,5,6,7,8)\to(0,1,3,4,5,6,7,8,9)$.  Thus $2685\to\color{blue}{3796}$.


问:
Showing $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$
I want to show $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$ rigoursly.

How I can prove rigorously? (Of course suppose $\sum_{k=0}^{\infty}a_k < \infty$, $\sum_{k=0}^{\infty}b_k < \infty$ condition is required)

答:
The weakest condition under which this holds is with at least one of the series $\sum a_k$ and $\sum b_k$ converging absolutely.

Suppose that $\sum_{k=0}^\infty a_k$ is absolutely convergent and $\sum_{k=0}^\infty b_k$ is convergent, so that there exist bounds $A,B > 0$ such that for all $n \in \mathbb{N}$,

$$\tag{1}\sum_{k=0}^n |a_k| < A, \quad \left|\sum_{k=0}^n b_k\right| < B$$

Define the partial sums $A_n = \sum_{k=0}^n a_k$, $B_n = \sum_{k=0}^n b_k$, and note that

$$C_n = \sum_{k=0}^n\sum_{i+j=k}a_ib_j = \sum_{k=0}^n\sum_{j=0}^ka_jb_{k-j} $$

The objective is to prove that $\lim_{n \to \infty}C_n = \lim_{n \to \infty}A_nB_n = \sum_{k=0}^\infty a_k \sum_{k=0}^\infty b_k. $

We have (looking at a table with entries $a_kb_j$ in row $k$ and column $j$ will make this clear)

$$\tag{2}|C_{2n}- A_nB_n| = \left|\sum_{k=0}^{n-1}a_k\sum_{j=n+1}^{2n-k}b_j + \sum_{k=n+1}^{2n}a_k\sum_{j=0}^{2n-k}b_j\right| \\ \leqslant \sum_{k=0}^{n-1}|a_k|\left|\sum_{j=n+1}^{2n-k}b_j\right| + \sum_{k=n+1}^{2n}|a_k|\left|\sum_{j=0}^{2n-k}b_j\right|$$

By the Cauchy criterion, for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ and all $m \geqslant 0$, we have

$$\tag{3}\sum_{k=n+1}^{n+m} |a_k| < \frac{\epsilon}{A+B}, \quad \left|\sum_{j=n+1}^{n+m} b_j\right| < \frac{\epsilon}{A+B}$$

Applying (3) with $m = n-k$ and the bounds in (1) to (2), we get for all $n \geqslant N$

$$|C_{2n}- A_nB_n|\leqslant A \cdot\frac{\epsilon}{A+B} + \frac{\epsilon}{A+B} \cdot B= \epsilon$$

Since $|C_{2n} - \lim_{n \to \infty}A_nB_n| \leqslant |C_{2n} - A_nB_n|+ |A_nB_n - \lim_{n \to \infty}A_nB_n|$ this implies that the subsequence $C_{2n}$ converges to $\lim_{n \to \infty} A_nB_n$. By a similar argument we can show that the subseqence $C_{2n+1}$ converges to $\lim_{n \to \infty} A_{n+1}B_n$.
Q.E.D

The result does not hold if both series $\sum a_k$ and $\sum b_k$ are conditionally convergent.  The standard counterexample is

$$a_k = b_k = \frac{(-1)^{k+1}}{\sqrt{k+1}}$$


问:
Changing the interval for alternating series test
Say I am doing an alternating series test, on a series from one to infinity, but after finding the first derivative of my sequence and drawing a number line, I find that it is only decreasing on the interval 2 to infinity. Would it be acceptable to say that the series has the same end behavior as a series from 2 to infinity, and continue with checking if the limit as n approaches infinity of that is 0? If I find that the second one is convergent, does that mean the first one must be?</p>

答: Yes, you are correct. If a series meets the conditions for the alternating series test after the 2nd term, then it converges. Indeed, even if a series meets these conditions only after the 10th, 100th, or 1,000th term, it still converges. The first $n$ terms of a series, no matter how large $n$ is (so long as $n$ is finite), do not affect the convergence of an infinite series.


答:
I remember a solution to an old olympiad problem which I had seen in Titu's Number theory book.

It is a really cool solution and I have remembered it ever since, I think we can apply it here.

As the digit 2 is skipped, we can think of it as counting in base 9 , where every digit after 2, corresponds to the digit before it, ie. 0-> 0, 1->1, 3->2 , 4 -> 3 ....9 - > 8,

So when we look at it like this, then 11 corresponds to 10, 13 corresponds to 11 and so on...

So the $2021^{th}$ number will correspond to 2021 in base 9, ie. 2685, but as per our rule, that is $3796$

Body:
This question came up recently:

John III, the third king of Johnland was very angry with John II, who killed the founder of Johnland, the beloved John I. By his new law the digit $2$ was forbidden to use. The numbers were listed as $1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, $$ 15, 16, 17, 18, 19, 30, 31, 33, \dots $ What was the $2021$st number in this new system?

I then came up with this:

```<code>Sol.:

First, I studied how many numbers would be ‘skipped’ in the system within a power of 10;
100: 0
101: 1
102: 20
103: 300
104: 4000
105: 50000
The pattern is quite clear.
Skipped numbers before 2021st number: 300 x 2 + 1 x 2 + 0 x 1 = 602.
However, in 602 there are some skipped numbers:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) = 602 + 120
Also in 120 there are some more:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) + (20 x 1 + 1 x 2) = 602 + 120 + 22
And in 22:
(300 x 2 + 1 x 2 + 0 x 1) + (20 x 6 + 0 x 2) + (20 x 1 + 1 x 2) + (1 x 2 + 0 x 2) = 602 + 120 + 22 + 2
There are none in two.
</code>```

From there I didn't know what to do with myself and just got the answer in python, but that's cheating, so I would like to know how I can hope to solve this question with pure maths.

Cheers!


答: Your solution does work, you just made a mistake at the end. ({1,2},{3,4},...), ({1,2},{4,3},...), ({2,1},{3,4},...) and so on are all equivalent, but counted separately by 924*720. Since each group of two can either be flipped or not flipped, there are $2^6$ ways to do this type of rearrangement, and the answer is $\frac{924*720}{2^6}=10395$. This agrees with user2661923’s answer, which uses a more common method.

Title: 12 students into groups of 2. Why doesn't my solution work?

Body:
I have the following problem:

There are 12 students in the class. How many ways are there to split them into working groups of size 2 to work on the same assignment?

I know the solution but I don't understand why my solution doesn't work.
My solution is:

Let's choose the first members of each group, it's combination ${12 \choose{} 6}=924$ (we have no order and no repetitions, n = 12 = number of students, k = 6 = number of groups). Then, let's choose the second members of each group, it's permutation $\frac{6!}{(6-6)!}=720$ (now order matters because we already have first members and still no repetitions, n = 6 = number of remaining students, k = 6 = number of groups). Also, we need to divide the result by 2 to account that the order inside a group doesn't matter ({1, 3} = {3, 1}). So it's $924*720/2=332640$


问:
Barycentric Coordinates of the circumcenter of an arbitrary triangle
Given points $A(1, 0, 0), B(0, 1, 0), C(0, 0, 1)$ in barycentric coordinates, and points $P(x_P, y_P, z_P), Q(x_Q, y_Q, z_Q), R(x_R, y_R, z_R)$, what would the barycentric coordinates of the circumcenter of $\triangle PQR$ be? The circumcircle can be found in a straightforward manner by using the equation for a circle in barycentric coordinates.</p>

答:
This answer explains how to get the barycentric coordinates of the circumcenter of an arbitrary $n$-simplex. For triangles, it boils down to the trilinear coordinates of the circumcenter being
$$
\begin{split}
\lambda_0 &= \frac{\|v_0 - \frac{1}{2}(v_1 + v_2)\|^2 - \frac{1}{4}\|v_1 - v_2\|^2}{2h_0},\\
\lambda_1 &= \frac{\|v_1 - \frac{1}{2}(v_2 + v_0)\|^2 - \frac{1}{4}\|v_2 - v_0\|^2}{2h_1},\\
\lambda_2 &= \frac{\|v_2 - \frac{1}{2}(v_0 + v_1)\|^2 - \frac{1}{4}\|v_0 - v_1\|^2}{2h_2},
\end{split}
$$
where the $h$ are the respective altitudes. Multiply those by the edge length to get the barycentric coordinates.

(For the general simplex, the $ \frac{1}{2}(v_1 + v_2)$ are the circumcenters of the faces and the $\frac{1}{4}\|v_1 - v_2\|^2$ are the squared circumradii.)