问:
Contradiction In Defining Symmetric Difference
One definition of set difference is $A\Delta B=(A-B)\cup (B-A).$ If we let $x\in A\Delta B,$ by our definition, we have that $x\in A \space\wedge\space x\notin B$ or $x\in B \space\wedge\space x\notin A$ where the or is inclusive. Since the or is inclusive we have two contradictions, namely, $x\in A$ or $x\notin A$ and $x\in B$ and $x\notin B.$

Is there a way to define the symmetric difference in a way similar to the above, that still uses the inclusive or?

答:
Notice that ($x\in A$ and $x\notin B$) or ($x\notin A$ and $x\in B$) is NOT equivalent, as you seem to imply, to ($x\in A$ and $x\notin A$) and ($x\in B$ and $x\notin B$), what would imply that
$$(A\wedge\neg B)\vee(\neg A\wedge B)\quad\text{and}\quad(A\wedge\neg A)\wedge(B\wedge\neg B),$$
for any formulas $A$ and $B$. You can, indeed, distribute the connectives to "simplify" the expression to
$$\Big((A\wedge \neg B)\vee\neg A\Big)\wedge\Big((A\wedge\neg B)\vee B\Big),$$
and then
$$(A\vee \neg A)\wedge(\neg B\vee \neg A)\wedge (A\vee B)\wedge(\neg B\vee B),$$
equivalent to
$$(A\vee B)\wedge (\neg A\vee \neg B).$$

But this alternative expression of the symmetric difference, as $(A\cup B)\cap(A^{c}\cup B^{c})$, also has no contradictions...

Your problem seems to lie, as Noah Schweber points out in a comment, in the interpretation of the inclusive disjunction: yes, a inclusive disjunction of two true formulas is indeed true, what in informal language is usually referred to as it "allowing" both formulas to be simultaneously true; but this does not mean that two formulas, connected by an inclusive disjunction, CAN be simultaneously true, as in your present case.

($x\in A$ and $x\notin B$) and ($x\notin A$ and $x\in B$) are mutually exclusive: one way you can see that is by noticing the sets they represent, $A\cap B^{c}$ and $A^{c}\cap B$, are disjoint. So there is no contradiction.


问:
Minimal number of comparisons to determine larger set
There are $2n+1$ balls in a row, on each one printed either $1$ or $0$, but we can not see what is written - we can only see the position in which they are placed.
I need to take out a ball that belong to the number printed more times, and I can only ask about two certain balls if they are the same or not. My goal is to do so in minimal number of comparisons in the worst case.
Trivial solution takes $2n$ comparisons. Any ideas to do better?

答:
You can justify $2n$ as a worst case.  If you make a graph with the balls vertices and comparisons edges it can't be connected unless there are that many edges.  If it isn't connected you can flip the identities of any component without changing the results.  If things are balanced closely, that will change the answer.

For the average, you can start with $n$ comparisons of disjoint pairs.  Any pair which comes up different can be discarded.  If the numbers of $0$s and $1$s are close, this should be about half the pairs, saving you later comparisons.  Repeating this gets you close to $n+1$ comparisons.

ContentLicense: CC BY-SA 4.0


问:
Explanation of where this trig identity comes from
I'm working on a problem but it's been a while since I last saw trig identities so I'd love some help or being pointed in the right direction.

Basically, I'd like to understand where this identity comes from;

$$\tan(2t) = \dfrac{2\tan(t)}{1 - \tan^2(t)}$$

Thanks for any help you can give - if it's useful to know the context of the problem, I'm writing a bit of code that converges on $\pi$ faster than Leibniz's series. (Please don't give too much away about the rest of the problem though, I'd like to get there myself :) )

答:
This follows from the fact that
$$
\tan(2t) = \frac{\sin(2t)}{\cos(2t)},
$$
and
$$
\sin(2t) = 2\sin(t)\cos(t),
$$
$$
\cos(2t) = \cos^2(t) - \sin^2(t).
$$
These two identities can be found using the formula
$$
e^{it} = \cos(t) + i\sin(t),
$$
so that
$$
\begin{align}
e^{i 2t} &= \cos (2t) + i\sin (2t) \\
&= \left(e^{it}\right)^2 = [\cos(t) + i\sin(t)]^2 = [\cos^2(t) - \sin^2(t)] + i[2\sin(t)\cos(t)].
\end{align}
$$</p>

ContentLicense: CC BY-SA 3.0


问:
How to calculate 3D position and update it
I'm trying to calculate the new position of an object moving over the surface of  a 3D sphere after some time Δt, from X1 to X2 always onto the surface of the sphere. The measurements/data I have of such object's displacement and sphere are:

Translational velocity (cm/seg)

Rotational velocity (deg/seg)

Direction of displacement (deg)

Radius of the sphere (cm)
Here is an example of what I mean.

Then the idea is to update for each Δt the position of the object on the sphere. I suspect some  linear algebra is needed but pretty lost at the moment. Any help would be greatly appreciated!

This is how the data frame looks like

time
translational velocity
rotational velocity
direction

0.01
2.36
6.45
0.78

0.02
1.12
1.19
0.62

0.03
1.67
4.45
1.51

0.04
1.25
5.39
1.67

答:
Let's say you have a sphere of radius $r$ centered at origin, and point $\vec{p} = (p_x, p_y, p_z)$ somewhere on the sphere so $\lVert \vec{p} \rVert = \sqrt{ \vec{p} \cdot \vec{p} } = \sqrt{p_x^2 + p_y^2 + p_z^2} = r$.

Let's say that at time $t = 0$ the point starts traveling on the sphere in direction $\vec{v}$ with velocity $\lVert\vec{v}\rVert$.  Because $\vec{v}$ is tangential to the sphere, $\vec{v} \perp \vec{p}$.  If you pick a random direction $\vec{w}$ in 3D, you can take just its tangential component,
$$\vec{v} = \vec{w} - \frac{\vec{p} \cdot \vec{w}}{\vec{p} \cdot \vec{p}} \vec{p}$$
essentially performing a single step of the Gram-Schmidt process, making $\vec{v}$ the part of $\vec{w}$ that is orthogonal to $\vec{p}$ and thus planar in the sphere tangent plane at $\vec{p}$.

If the point travels in a local straight line, it will travel along a great circle (a circle with its center at the center of the sphere, starting point on the perimeter, and the initial direction in the plane of the circle).  The simplest way to express this is by rotation around axis $\hat{a}$ perpendicular to the great circle,
$$\hat{a} = \frac{\vec{v} \times \vec{p}}{\left\lVert \vec{v} \times \vec{p} \right\rVert} \tag{1}\label{G1}$$
If $v = \lVert\vec{v}\rVert$ is the (2D) velocity, the angular velocity $\omega$ is
$$\omega = \frac{v}{2 \pi r} = \frac{v}{2 \pi \lVert \vec{p} \rVert} \tag{2}\label{G2}$$
and we can describe the trajectory of $\vec{p}$ as a function of time $t$ via
$$\vec{p}(t) = \mathbf{R}(\hat{a}, \omega t) \vec{p}_0 \tag{3a}\label{G3a}$$
where $\vec{p}_0 = \vec{p}(0)$ is the location at time $t = 0$, and
$$\begin{aligned}
\mathbf{R}(\hat{a}, \omega t) & = \left[ \begin{matrix}
c + a_x^2 u & a_x a_y u - a_z s & a_x a_z u + a_y s \\
a_y a_x u + a_z s & c + a_y^2 u & a_y a_z u - a_x s \\
a_z a_x u - a_y s & a_z a_y u + a_x s & c + a_z^2 u \\
\end{matrix} \right] \\
c & = \cos(\omega t) \\
s & = \sin(\omega t) \\
u & = 1 - \cos(\omega t) = 1 - c \\
\end{aligned} \tag{3b}\label{G3b}$$
Alternatively (and equivalently), you can use Rodrigues' rotation formula,
$$\begin{aligned}
\vec{p}(t) &= \vec{p}_0 c + \left(\hat{a} \times \vec{p}_0\right) s + \hat{a} \left( \hat{a} \cdot \vec{p}_0 \right) (1 - c) \\
c &= \cos(\omega t) \\
s &= \sin(\omega t) \\
\end{aligned}
\tag{3c}\label{G3c}$$
In both cases, we just rotate $\vec{p}_0$ around unit axis vector $\hat{a}$ by angle $\omega t$.

To find out the instant direction, the same applies to $\vec{v}$: just replace $\vec{p}_0$ by $\vec{v}_0$ in the formula you wish to use, and you get the exact local direction vector $\vec{v}(t)$ as a function of time $t$.

(By notation $\vec{p}(t)$, I am referring to a vector-valued function in a single variable $t$.  While the latter looks simpler, if you need both $\vec{p}(t)$ and $\vec{v}(t)$, the matrix form may require fewer basic arithmetic operations.)

Let's say you know the starting point $\vec{p}_0$, and the destination point $\vec{p}_1$, both on the sphere ($\lVert \vec{p}_0 \rVert = \lVert \vec{p}_1 \rVert = r$).

This time, the unit axis vector $\hat{a}$ is
$$\hat{a} = \frac{\vec{p}_0 \times \vec{p}_1}{\lVert \vec{p}_0 \times \vec{p}_1\rVert} \tag{4}\label{G4}$$

The total travel distance $L$ (measured along the surface of the sphere) is
$$L = 2 \pi r \arccos\left(\theta_\Delta\right) \tag{4a}\label{G4a}$$
where $\theta_\Delta$ is the angular separation,
$$\theta_\Delta = \frac{\vec{p}_0 \cdot \vec{p}_1}{\lVert\vec{p}_0\rVert \, \lVert\vec{p}_1\rVert} \tag{4b}\label{G4b}$$
and therefore the arrival time $t$ with (sphere surface) velocity $v$ is
$$t = \frac{L}{v} \tag{4c}\label{G4c}$$
and the angular velocity $\omega$ is
$$\omega = \frac{\theta_\Delta}{t} \tag{4d}\label{G4d}$$
Alternatively, if you know the arrival time $t$, then the (sphere surface) velocity $v$ is
$$v = \frac{L}{t} \tag{4e}\label{G4e}$$

Treating the current position as an $r$-length Cartesian vector yields a simple simulation.  However, it also means that one can easily adjust the altitude of the point also.  For example, to move $\vec{p}$ to altitude $h$ just scales it to length $r + h$,
$$\vec{p}^\prime = \frac{r + h}{\lVert \vec{p} \rVert} \vec{p} \tag{5}\label{G5}$$
and to change the altitude of $\vec{p}$ by $h$ is just
$$\vec{p}^\prime = \frac{\lVert \vec{p} \rVert + h}{\lVert \vec{p} \rVert} \vec{p} \tag{6}\label{G6}$$
where the$\,^\prime$ is there just to denote that it is the point after the altitude adjustment.

Furthermore, you can use 3D squared distance between two points to determine collisions et cetera.  The velocity $v$ is similarly measured along the sphere surface, as opposed to along the elliptical/non-spherical curve if altitude changes would be taken into account; just as we do with air travel on Earth.

Finally, if $z$ axis is towards the north pole and $x$ axis towards zero meridian on the equator, then
$$\begin{aligned}
\text{latitude} &= \displaystyle \arcsin\left(\frac{p_z}{\lVert\vec{p}\rVert}\right) \\
\text{longitude} &= \displaystyle \operatorname{atan2}\left(\frac{p_y}{\lVert\vec{p}\rVert}, ~ \frac{p_x}{\lVert\vec{p}\rVert}\right) \\
\end{aligned} \tag{7}\label{G7}$$
where $\operatorname{atan2}(y, x)$ is the two-argument version of arcus tangent provided by most programming languages; similar to $\arctan(y/x)$, but takes the signs of $y$ and $x$ into account, so that the result covers all four quadrants (full 360°).

Conversion between a 3D direction vector $\vec{v}$ and a single direction angle $\varphi$ – compass reading, with 0° towards North, 180° towards South, and 90° West – is possible, if we handle directions at north pole (where every direction is south) and at south pole (where every direction is north) specially: there, the direction angle is towards that longitude on the equator.

If $\hat{N}$ is a 3D unit vector towards local North, and $\hat{W}$ is a 3D unit vector towards local West, then heading $\varphi$ corresponds to 3D unit direction $\hat{d}$,
$$\hat{d} = \hat{N}\cos\varphi + \hat{W}\sin\varphi \tag{8}\label{G8}$$
If you are at 3D point $\vec{p} = (p_x, p_y, p_z)$ and not at a pole ($p_x^2 + p_y^2 \ne 0$), then
$$\hat{N} = \frac{1}{\sqrt{(p_x^2 + p_y^2)(p_x^2 + p_y^2 + p_z^2)}} \left[ \begin{matrix}
- p_x p_z \\
- p_y p_z \\
p_x^2 + p_y^2 \\
\end{matrix} \right], \quad \hat{W} = \frac{1}{\sqrt{p_x^2 + p_y^2 + p_z^2}} \left[ \begin{matrix}
p_y \\
-p_x \\
p_z \\
\end{matrix} \right] \tag{9a}\label{G9a}$$
but at the poles ($p_x = p_y = 0$),
$$\hat{N} = \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right], \quad \hat{W} = \left[ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right] \tag{9b}\label{G9b}$$

If you have direction vector $\vec{d}$ at point $\vec{p}$, calculate the north and west unit vectors $\hat{N}$ and $\hat{W}$, then ensure the direction vector is tangential to the sphere,
$$\vec{d}^\prime = \vec{d} - \displaystyle\frac{\vec{p} \cdot \vec{d}}{\vec{p} \cdot \vec{p}} \vec{p}$$
and the heading angle $\varphi$ is then
$$\varphi = \operatorname{atan2}\left( \hat{W} \cdot \vec{d}^\prime , ~ \hat{N} \cdot \vec{d}^\prime \right) \tag{G9c}\label{G9c}$$


问:
How to prove that 1/1!+1/2!+1/3!+....<3?
How can I prove that 1/1!+1/2!+1/3!...<3?
Should I use mathematical induction?
I can see that the sum is the same as the power series of e^1 and that is also less than 3. But where is the proof for that?</p>

Body: $$\sum_{k=1}^\infty\frac1{k!}= 1+\frac12+\frac12\sum_{k=3}^\infty\frac2{k!}\le1+\frac12+\frac12\sum_{k=3}^\infty\frac1{3^{k-2}}=1+\frac12+\frac12\cdot\frac12=\frac74$$


问:
Count number of total orders on a set
The question is like:
There is two sets A and B such that A and B are total different sets and |A| = m and |B| = n.
There is two total orders R on A and S on B.
C = A unite with B.
How much total orders on c contain both R and S.

Solution is like:

Choose a total order on C that contains both R and S is like order C items in a row such that A items are ordered respectively to R and B items are ordered respectively to S.
To choose an order like that, we will choose the places of the items of A in the row. there is (m+n, m) options for that, than we will order them respectively to R and than we will order B items respectively to S. Therefore, there is (m+n, m) total orders on C that contains R and S.

Problem is:

Can't see why to choose a total order on C that contains both R and S is like order C items in a row such that A items ordered respectively to R and B items are ordered respectively to S.

答:
Here is a different way to rephrase their argument. Consider the following question:

How many where are there to choose a total order on $C$?

Obviously, the answer is $(n+m)!$. However, we can answer the question in a different way. To choose a total order on $C$, you have to...

Choose a total ordering on $A$ (since every ordering on $C$ creates an ordering on $A$). This can be done in $n!$ ways. Call this ordering $R$.

Choose a total ordering on $B$, in $m!$ ways. Call this ordering $S$.

Finally, you must choose the ordering on $C$ which is consistent with these last two choices. That is, you must choose an ordering on $C$ containing $R$ and $S$. This is exactly the quantity you are trying to count.

By equating the two different ways to answer the same question, you get
$$
(n+m)!=n!\times m!\times \left(\begin{matrix}\text{# ways to choose a total ordering}\\\text{on C containing $R$ and $S$}\end{matrix}\right)
$$
Divide by $n!m!$ and you have your answer.

Title: Count number of total orders on a set


答:
I think the one of the easiest ways is starting from:

$$\tan2t = \dfrac{\sin2t}{\cos2t} = \dfrac{2\sin t\cos t}{\cos^2t-\sin^2t}$$

From there, you can divide both the numerator and denominators by something and you get the result.

$$\dfrac{2\sin t\cos t}{\cos^2t-\sin^2t}\times \dfrac{\ \frac{1}{A}\ }{\frac{1}{A}}$$

Title: Explanation of where this trig identity comes from

Body:
I'm working on a problem but it's been a while since I last saw trig identities so I'd love some help or being pointed in the right direction.

Basically, I'd like to understand where this identity comes from;

$$\tan(2t) = \dfrac{2\tan(t)}{1 - \tan^2(t)}$$

Thanks for any help you can give - if it's useful to know the context of the problem, I'm writing a bit of code that converges on $\pi$ faster than Leibniz's series. (Please don't give too much away about the rest of the problem though, I'd like to get there myself :) )


问:
Why can we assume surjectivity in short exact sequences?
This is brand new to me, so forgive me if this is simple.

I understand the structure of a short exact sequence (the image of the current homomorphism is the kernel of the next), and I can understand why $\phi$ is injective in the below example. What's confusing me is that everywhere says that $\psi$ is surjective, but I cannot see why. I've consulted some textbooks but can't find any material on it, and any online explanations don't help either.

$${0}\rightarrow N\xrightarrow{\phi} G\xrightarrow{\psi} Q\rightarrow 0$$

Thanks in advance.

答: By the definition the image of $\psi$ is the kernel of the next homomorphism, which is the zero morphism $Q\to 0$. And the kernel of the zero morphism is entire $Q$. And so $im(\psi)=ker(Q\to 0)=Q$ or in english: $\psi$ is surjective.


问:
Find all fucntions such that $f:\Bbb R \to \Bbb R$ and $f(x^{2}+yf(x))=f(x+y)$
My work :
first take $x=y=0$ we get $f(0)=f(0)=c$ , then take $x=x$ and $y=-x$ implies $f(x^{2}-xf(x))=f(0)$ now as there is another function inside the parenthesis I'm asuming there are some entries for which $x^{2}-xf(x)=0$ then we get $f(x)=x$. but I'm little confused about that assumption. If anyone can just fix my mistakes and solve that problem it'll be a great help for me.

答:
As observed in the comments, $f(x^2)=f(x)$ for all $x$ so that $f(x)=f(-x)$ for all $x$.
Thus, $f(x^2+yf(x))=f(x^2+yf(-x))$ for all $x$, and $f(x+y)=f(-x+y)$, for all $x,y$. This implies that $f$ is constant.


答:
Hint:

For some $m<n$
$$\frac1{1\cdot2\cdot3\cdot4\cdots n}<\frac1{1\cdot2\cdot3\cdot4\cdots m\cdot m\cdot m\cdot m}=\frac1{m!}\frac1{m^{n-m}}.$$

So starting from the $m^{th}$ term, you can bound the series by a geometric one of which you can get the value of the tail. Up to you to find a suitable $m$.

Title: How to prove that 1/1!+1/2!+1/3!+....<3?

Body:
How can I prove that 1/1!+1/2!+1/3!...<3?
Should I use mathematical induction?
I can see that the sum is the same as the power series of e^1 and that is also less than 3. But where is the proof for that?</p>


问:
Are small categories necessarily free?
In the same spirit that all vector spaces are free, is any small category a free category?. If not, there is a counterexample? I'm interested to know if diagrams indexed by quivers covers all cases of diagrams indexed by small categories.</p>

答: A category is small if it's collection of objects and collection of arrows are honest sincere sets. This says nothing about relations among arrows, which decides if a category is free. For a tiny counterexample consider a category with a single object $X$ and with two arrows $\mathrm{id}_X$ and $a$ where $a \circ a = \mathrm{id}_X$.

Title: Are small categories necessarily free?


问:
Introductory question to uniform continuity
This was an introductory question in my Analysis course.

Suppose $f:]0,1] \rightarrow \mathbb{R}$ is a continuous function.

Prove that if $f$ maps Cauchy sequences in $]0,1]$ to Cauchy sequences in $\mathbb{R}$, then $f$ has a continuous expansion to $[0,1]$.

We got some hints, but they were not that helpful...

One of them was that we had to prove/explain why the limit of $\phi(x_n)$ is independent of the chosen sequence $(x_n)_n$ if $(x_n)_n$ is any sequence in $]0,1]$ that converges to $0$. I intuitively get why this should be so, but I had a hard time proving this.

Could someone give this a shot?
Thanks!

答:
$f$ is bounded on $[0,1]$ as otherwise, one could find a Cauchy sequence $\{x_n\}$ in $(0,1]$ with $\{f(x_n)\}$ unbounded in contradiction with the fact that a Cauchy sequence is bounded.

Now by contradiction, if $f$ wasn't continuous, it would exist two sequences $\{x_n\}, \{y_n\}$ converging to zero with

$$\lim\limits_{n \to \infty} f(x_n) = l_x \neq l_y = \lim\limits_{n \to \infty} f(x_n).$$

The sequence

$$z_n = \begin{cases}
x_n \text{ for } n \text{ even}\\
y_n \text{ for } n \text{ odd}
\end{cases}$$

would be Cauchy while $\{f(z_n)\}$ would not be. A contradiction.

Therefore for every sequence $\{x_n\}$ converging to zero, $\{f(z_n)\}$ converges to a unique value $l$. Proving that $f$ can be extended to a continuous map on $[0,1]$ by setting $f(0) = l$.


问:
Matrix multiplication efficiency
I have a multiplication $A * B * A^T$, where $B$ is a covariance matrix (i.e. symmetric, square) and $A$ is an arbitrary matrix with compatible dimensions. Given the symmetry of $B$ and the fact that I multiply by $A$ and $A^T$, it really feels like there must be a more efficient way to compute the result than two multiplications, but I haven't been able to find anything. Does anyone have any insight on the problem, or why there isn't any better way of doing it?

Thanks!

答: If you're going to be using the same $A$ for many different $B$, it might pay to precompute the matrices $C_{ii} = A E_{ii} A^T$ and $C_{ij} = A (E_{ij} + E_{ji}) A^T$ for all $i \le j$, where $E_{ij}$ is the matrix with $(i,j)$ entry $1$ and all others $0$.  Then $A B A^T = \sum_i \sum_{j \ge i} b_{ij} C_{ij}$. </p>

Title: Matrix multiplication efficiency


问:
Show that if all points in space lie in a collection of subbasic open sets, all points lie in two of the subbasic open sets
Let $\{0,1\}^A$ be the collection of functions from an index set $A$ to $\{0,1\}$. Having subbasis $\{U(a,\delta)|a \in A,\delta \in \{0,1\}\},U(a,0)=\pi_{a}^{-1}(0),U(a,1)=\pi_{a}^{-1}(1)$.Show if a collection of subbasic open sets of $\{0,1\}^A$ has the property that every point of $\{0,1\}^A$ lies in at least one of those subbasic open sets, then there are two subbasic open sets in that collection such that every point of $\{0,1\}^A$ lies in one of those subbasic sets.

Attempt:Let $S$ be such a collection of subbasic sets. If $\{0,1\}^A \in S$. There is nothing to show. Suppose $\{0,1\}^A \notin S$.Then there are two subbasic sets $\pi_{a_1}^{-1}(b_1),\pi_{a_2}^{-1}(b_2),b_i \in \{0,1\},a_1\neq a_2$ in $S$. Then $\pi_{a_1}^{-1}(b_1)\cup\pi_{a_2}^{-1}(b_2)=\{0,1\}^A$, since the only factor of $\pi_{a_1}^{-1}(b_1)$ with points restricted to either $0$ or $1$ is the $a_1$'st factor. And the only factor of $ \pi_{a_2}^{-1}(b_2)$ with points restricted to $0$ or $1$ is the $a_2$'nd factor. So the union must contain all points.

答:
Let $\mathcal{S}$ be any cover of $\{0,1\}^A$ by subbasic sets $U(a,\delta)$.
Define for each $S \in \mathcal{S}$, $a(S) \in A$ and $\delta(S) \in \{0,1\}$ so that $S=U(a(S), \delta(S))$. Define $$A' = \{a(S)\mid S \in \mathcal{S}\}$$ the set of all $a$ that are used in $\mathcal{S}$, and for $a \in A'$, define $$r(a):=\{\delta \in \{0,1\} \mid \exists S' \in \mathcal{S}: a(S')=a, \delta(S')=\delta\}$$

the range in $\{0,1\}$ that is used, per used index $a$, always non-empty.

If $r(a)=\{0,1\}$ for some $a$ we are done, because then $U(a,0), U(a,1) \in \mathcal{S}$ and their union is $\{0,1\}^A$.

So striving for a contradiction we assume that for all $a \in A'$ we have $|r(a)|=1$. But then define

$f \in \{0,1\}^A$ by

$$f(a)= \begin{cases} 0 & a \notin A'\\
                      1-\delta & r(a)=\{\delta\}\\
        \end{cases}$$

But then $f \notin \bigcup \mathcal{S}$ and we have the required contradiction already. So some $r(a)=\{0,1\}$ etc.

Sidenote: note that this shows that $\{0,1\}^A$ is compact by Alexander's subbase lemma, and even supercompact:

A space $X$ is called supercompact if it has a subbase $\mathcal{S}$ for its topology such that every cover from $\mathcal{S}$ has a subcover of $2$ elements.

This notion was studied in the 70's and 80's (a Dutch hobby perhaps, van Douwen and van Mill wrote about it); it is well-known that products of supercompact spaces are supercompact, and all metrisable compact spaces and ordered compact spaces are supercompact. $\beta \Bbb N$ is not however.


问:
Pigeonhole principle application sums and differences
Let $A \subset \{1,2,...,99\}$, prove or disprove the following:

a. For $|A| = 27$

b. For $|A| = 26$

There are $2$ different numbers in $A$ that their sum or their difference can be divided with $50$.

I am sure the question points me to use Pigeonhole Principle, but I'm not quite sure which are the pigeons and which are the holes because the question deals with or so proving that a sum (or a difference) can be divided with $50$ closes the proof without handling the other case.

I've started with saying that there are $(99+98)-(1+2)+1=195$ different sums,
and $(99-1)-(1-99)+1=197$ differences (including negatives) in $\{1,2,...,99\}$.

Also, when dividing with $50$ you can get maximum 50 remainders (let $r \in \{0,1,...,49\}$ mark the reminders)

Can anyone hint me with how to proceed from here?

答:
Every element of your set {1... 99} is either in A or A' (complement of A) You can pigeonhole more efficiently by considering which elements cannot be in the same set, then demonstrating that the requirements for how many elements are in A are violated.

So for example: If 1 is in A then 99 and 49 and 51 must be in A'. Try to continue this line of argument.


问:
Find the joint pdf of $X_{(n/2)}$ and $M$.
Let $X_1, X_2, \ldots , X_n$ be a random sample from a Uniform$(0, θ)$ population with
$θ > 0$, where $n$ is an even integer. Let $M=\frac{X_{\left(\frac{n}{2}\right)}+X_{\left(\frac{n}{2} + 1\right)}}{2}$ be the median of the random sample. Find the
joint pdf of $X_{(n/2)}$ and $M$.

Here I find that $$f_{X_{(n/2)}} = \frac{n!}{\left(\frac{n-2}{n}\right)!\left(\frac{n}{2}\right)!}\left(\frac{x}{\theta}\right)^{\frac{n-2}{2}}\frac{1}{\theta}\left(1-\frac{x}{\theta}\right)^{n/2}.$$

But I not sure how to find the rest. Can I have some guidance?

答:
First note that you can set $n=2k$, $k=1,2,3,...$

Using the Order Statistics you have that the joint density $(X_{(k)};X_{(k+1)})$ is

$$f_{X_{(k)},X_{(k+1)}}(x,y)=\frac{(2k)!}{[\theta(k-1)!]^2}\left(\frac{x}{\theta}\right)^{k-1}\cdot \left(1-\frac{y}{\theta}\right)^{k-1}\cdot\mathbb{1}_{(x;\theta)}(y)$$

Given this joint density, you are requested to calculate the joint density of

$$\left( X_{(k)}; \frac{X_{(k)}+X_{(k+1)}}{2}\right)$$

I think you can proceed by yourself using standard method of transformation to solve the problem (i.e. the jacobian method)

To use the jacobian, simply set (remember that $0<x<y<\theta$

$$\begin{cases}
z=\frac{x+y}{2} \\
u=x
\end{cases}$$

easy calculate the jacobian that is $|J|=2$ and then just substituting in the joint density you get

$$f_{UZ}(u,z)=\frac{2(2k)!}{[\theta(k-1)!]^2}\left(\frac{u}{\theta}\right)^{k-1}\cdot \left(1-\frac{2z-u}{\theta}\right)^{k-1}\cdot\mathbb{1}_{(?;?)}(u)\cdot\mathbb{1}_{(?;?)}(z)$$


问:
Transformation in area element using Jacobian
Recently I gave an exam for Physics masters entrance where they asked the following question

Let $(x, y)$ denote the coordinates in a rectangular Cartesian coordinate system $\mathcal{C}$. Let $(x', y')$ denote the coordinates in another coordinate system $\mathcal{C'}$ defined by:
$$\begin{align}x' &= 2x + 3y\\ y' &= -3x + 4y\end{align}$$
The area element in $\mathcal{C'}$ is:
$$(\text{A})\ \frac{1}{17}\ dx'dy'\quad\quad(\text{B})\ 12\ dx'dy'\quad\quad(\text{C})\ dx'dy'\quad\quad(\text{D})\ x'\ dx'dy'$$

To solve it, I first calculated the Jacobian and found it to be 17. But, then I saw that the question asked for the area element in the C' coordinate system, so I thought that the answer would simply be $dx'dy'$ which I gave as my answer.

But, today the answer keys came out and it says that the answer is $\frac 1{17}dx'dy'$

Where did I go wrong/Is the answer key wrong?

答:
The answer key is not wrong (the answer is not necessarily $dx'~dy'$ because the transformed coordinate system is not rectangular). The change of variables formula for a transformation $x=g(x',y')$ and $y=h(x',y')$ is given by
$$\iint_R f(x,y)~dx~dy=\iint_S f(g(x',y'),h(x',y'))\left|\frac{\partial(x,y)}{\partial(x',y')}\right|~dx'~dy'.$$
Therefore you have to invert your coordinate transformation to get the correct answer. With this, one has that
$$x=\frac{4x'-3y'}{17},\quad y=\frac{3x'+2y'}{17},$$
which results in
$$\left|\frac{\partial(x,y)}{\partial(x',y')}\right|=\frac{1}{17}.$$
But in your simple case, you could of course use the fact that
$$\left|\frac{\partial(x,y)}{\partial(x',y')}\right|=1/\left|\frac{\partial(x',y')}{\partial(x,y)}\right|,$$
which follows from the fact that the determinant of the inverse of a matrix is the reciprocal of the determinant and the inverse function theorem.


问:
Why is this operator called a lift in a PDE context?
Let $L$ be some differential operator and consider the PDE $Lu = 0$ with some boundary conditions. Let $W$ denote the space that solutions live in and $B$ the space that boundary conditions live in.

Suppose I've constructed a bounded operator $J:B\to W$ such that for $g\in B$, $Jg$ solves the PDE. Suppose also that I have a bounded trace operator $T:W\to B$.

My question is as follows: I've seen $J$ being referred to as a lift operator, particularly on page 8 here. But I'm struggling to reconcile this with the definition given in Wikipedia (I know nearly nothing of category theory). The only commutative diagram I can construct that makes $J$ a lift is somewhat trivial:

Here $I$ is just the identity. Is this the right way to think about $J$ as a lifting?

答:
Yes, this is the right way to think about this from the completely abstract perspective of lifts in the sense of category theory.  From the perspective of PDE and function spaces, "lift" is often the name we give to an operator that serves as a right inverse to the trace operator, i.e. a $J$ such that $T \circ J = I$.  This is exactly the commutative diagram you've written above.  In this context it's also reasonable to use the word "extension" in place of "lift" as we can think about the operator as extending the boundary data into the bulk.

When we construct such lifting operators, it is frequently the case that we begin in the canonical model problem where we want to take a function $f$ defined on $\mathbb{R}^{n-1}$ and construct a function $F$ on the upper half-space $\mathbb{R}^n_+ = \{x \in \mathbb{R}^n : x_n >0\}$ in such a way that $F$ is nice enough for us to use our trace operator and $TF = f$.  Here there is a strong natural language motivation for describing the operator as a lifting because in some sense we lift the data "up" from the set $\{x_n =0\}$ into the set above, $\{x_n >0\}$.  Personally, this is what I have in mind when I call such operators a lifting.

There is obviously no uniqueness in building such a lifting, and this is where we get nice connections to PDE and related areas.  We can try to single out a particular way of lifting such that the resulting lifted / extended function satisfies some auxiliary conditions.  In particular, we can ask that it satisfies some PDE.  I can't help but mentioning one of the most striking examples in which we want to construct a right inverse to the trace operator $T : H^1(\mathbb{R}^n_+) \to H^{1/2}(\mathbb{R}^{n-1})$.  In this case it's natural to look for a lifting $u \in H^1(\mathbb{R}^n_+)$ of $f \in  H^{1/2}(\mathbb{R}^{n-1})$ that minimizes the $H^1$ norm of $u$.  In other words, we want to find $u$ such that
$$
\int_{\mathbb{R}^n_+} |\nabla u|^2 + |u|^2 = \min\left\{ \int_{\mathbb{R}^n_+} |\nabla v|^2 + |v|^2 : v \in H^1(\mathbb{R}^{n}_+) \text{ and } Tv =f \right\}.
$$
It turns out that we can do this with the calculus of variations (and some more Sobolev theory), and the resulting $u$ then satisfies the elliptic Dirichlet-type problem
$$
\begin{cases}
-\Delta u + u =0 &\text{in } \mathbb{R}^n_+ \\
u = f & \text{on } \mathbb{R}^{n-1}.
\end{cases}
$$
The map $J$ defined by $f \mapsto Jf = u$ is then exactly as in your post when the linear operator is $L= -\Delta + I$.


问:
Is this type of a transformation possible in propositional logic?
Let $\Phi_1(A,B)$ be a propositional logic formula consisting only of propositions $A$ and $B$,
similarly we have $\Phi_2(C,D)$. Now lets say we have a formula as follows:
$$(M \leftrightarrow \Phi_1(A,B)) \land (N \leftrightarrow \Phi_2(C,D)) $$

Where $M$ and $N$ are propositional variables as well.
Is there any transformation possible which converts this form into a formula of the following form :
$$\Phi_1'(M,N) \land \Phi_2'(A,B,C,D)$$

The transformed formula must be equivalent to the original formula.
Where $\Phi_1'(M,N)$ consists only of $M$ and $N$ and similarly for $ \Phi_2'(A,B,C,D)$.
No new propositional variables should be introduced.

Corrections made in the problem, thanks to @Alex Kruckman's comment

答:
This is possible if and only if for each $i \in \{1,2\}$, $\Phi_i$ is a tautology or a contradiction.

Let $\Theta$ be the formula $(M \leftrightarrow \Phi_1(A,B)) \land (N \leftrightarrow \Phi_2(C,D))$.

Suppose that for each $i \in \{1,2\}$, $\Phi_i$ is a tautology or a contradiction. Then $\Theta$ is equivalent to $((\lnot)M\land (\lnot) N)\land \top$, where we put the $\lnot$ in front of $M$ if $\Phi_1(A,B)$ is a contradiction and we put the $\lnot$ in front of $N$ if $\Phi_2(C,D)$ is a contradiction. Here $\Phi_2'$ is the tautology $\top$ (but if $\top$ is not in your language, you could use any tautology involving just $A,B,C,D$).

On the other hand, suppose that there is some $i\in \{1,2\}$ such that $\Phi_i$ is neither a tautology nor a contradiction. Let's assume $i = 1$, since the proof in the other case is similar.  Assume for contradiction that there is some formula $\Psi = \Phi_1'(M,N) \land \Phi_2'(A,B,C,D)$ equivalent to $\Theta$. Fix some valuation $v_0$  on $\{N,C,D\}$ so that $v_0$ makes $(N\leftrightarrow  \Phi_2(C,D))$ true (pick the truth values of $C$ and $D$ arbitarily, and then pick the truth value of $N$ to agree with $\Phi_2(C,D)$).

Case  1: There is a valuation $v_1$ extending $v_0$ and making $\Phi'_2(A,B,C,D)$ false. Let $v_1'$ be the valuation obtained from $v_1$ by changing the truth value of $M$ (if necessary) to agree with $\Phi_1(A,B)$. Under $v_1'$, $\Phi'_2(A,B,C,D)$ is still false, so $\Psi$ is false, but $\Theta$ is true, contradiction.

Case 2: Every valuation extending $v_0$ makes $\Phi'_2(A,B,C,D)$ true, but there is a valuation $v_2$ extending $v_0$ and making $\Psi$ false. Then $v_2$ must make $\Phi_1'(M,N)$ false. Since $\Phi_1(A,B)$ is neither a tautology or a contradiction, we can make a new valuation $v_2'$ by changing the truth values of $A$ and $B$ (if necessary) to make the truth value of $\Phi_1(A,B)$ equal $v_2(M)$. Under $v_2'$, $\Phi_1'(M,N)$ is still false, so $\Psi$ is false, but $\Theta$ is true, contradiction.

Case 3: Every valuation extending $v_0$ makes $\Psi$ true. We can pick a valuation extending $v_0$ and making $\Theta$ false, just by picking the truth values of $A$ and $B$ arbitrarily and then picking the truth value of $M$ to disagree with the truth value of $\Phi_1(A,B)$. Contradiction.


问:
Nonzero vector such that $Xv = av$ for some scalar $a$
I have a matrix $X$ such that $X = \begin{bmatrix}
2 & 0\\
10 & 1
\end{bmatrix}$

How do I go about finding the nonzero vector such that $Xv = av$ for some scalar $a$?

答:
You need to find the eigenvalues and eigenvectors of the matrix (the eigenvalues are 1 and 2 - so we can find vectors for $a=1$ and $a=2$).
Another option would be to see from the form of the matrix that every vector of the form $\begin{pmatrix} 0 \\ y \end{pmatrix}$ is mapped to itself by the multiplication with the matrix.
Finding the vectors for $a=2$ is easy to find as well: Just solve $Xv=2v$ and you will find all vectors of the form $\begin{pmatrix} y \\ 10y \end{pmatrix}$


问:
Does a monotone function on an arbitrary subset of $\mathbb R$ always have at most countable number of discontinuity?
I know a monotone function of a closed and bounded interval can have at most countably many point of discontinuity. And hence a monotone function on $\mathbb R$ can have at most countably many point of discontinuity since $\mathbb R=[0,1]\cup[1,2]\cup[2,3]\cup...\cup[-2,-1]\cup[-3,-2]\cup...$

Now my question is

Does a monotone function on an arbitrary subset of $\mathbb R$ always have at most countable number of discontinuity?

Does the funtion $f:(\mathbb R-\mathbb Q)\to\mathbb R:x\mapsto x$ serve our purpose?

答:
Suppose $f: D \rightarrow \mathbb{R}$ is a monotone function with cardinality $C$ discontinuities, where $D \subset \mathbb{R}$. Define the extended function $\tilde{f}$ by setting $\tilde{f}(x) = f(x)$ when $x \in D$, and for each $x \notin D$
$$\tilde{f}(x) = \inf_{y \in D, y > x} f(y).$$
Thus $\tilde{f}$ extends $f$ by interpolating horizontally across subsets of $\mathbb{R} \setminus D$. Then $\tilde{f}: \mathbb{R} \rightarrow \mathbb{R}$ is monotone and has cardinality $C'$ discontinuities, where $C' \geq C$. Since $C' \le \aleph_0$, then also $C \le \aleph_0$, so $f$ has at most countably many discontinuities.</p>


问:
Prove or disprove that $f\left(x\right)\leq 1,$ which is a continuous function so that $\int_0^x tf\left(2x-t\right){\rm d}t=\frac{\tan^{-1}x^2}{2}$

Prove or disprove that $f\left ( x \right )\leq 1,$ where $f$ is a continuous function so that
$$\int_{0}^{x}tf\left ( 2x- t \right ){\rm d}t= \frac{\tan^{-1}x^{2}}{2}$$
Source: AoPS/@Ji123_ https://artofproblemsolving.com/community/q1h144652p819322

I edited the original problem a little bit, my friend@twelve_sakuya also has a completed disproof in case $f\left ( x \right )< 1,$ we substitute
$$t\mapsto 2x- t\Rightarrow\int_{0}^{x}tf\left ( 2x- t \right ){\rm d}t\equiv\int_{x}^{2x}\left ( 2x- t \right )f\left ( t \right ){\rm d}t$$
$$\frac{{\rm d}}{{\rm d}x}\int_{x}^{2x}\left ( 2x- t \right )f\left ( t \right ){\rm d}t= \frac{{\rm d}}{{\rm d}x}\left ( 2x\int_{x}^{2x}f\left ( t \right ){\rm d}t- \int_{x}^{2x}tf\left ( t \right ){\rm d}t \right )= 2\int_{x}^{2x}f\left ( t \right ){\rm d}t- xf\left ( x \right )$$
On the other hand
$$\frac{{\rm d}}{{\rm d}x}\frac{\tan^{-1}x^{2}}{2}= \frac{x}{x^{4}+ 1}\quad\therefore 2\int_{x}^{2x}f\left ( t \right ){\rm d}t- xf\left ( x \right )= \frac{x}{x^{4}+ 1}$$
When $x> 0,$ we have
$$f\left ( x \right )= \frac{2}{x}\int_{x}^{2x}f\left ( t \right ){\rm d}t- \frac{1}{x^{4}+ 1}= 2f\left ( c \right )- \frac{1}{x^{4}+ 1}\quad\left ( x< \exists c< 2x \right )$$
(According to Mean-value theorem), if $x\rightarrow +0\Rightarrow c\rightarrow +0\Rightarrow f\left ( 0 \right )= 2f\left ( 0 \right )- 1\quad\left ( x\rightarrow +0 \right ),$ so $f\left ( 0 \right )= 1,$ we have contractdiction.

答:
Using Leibnitz integral rule:

$xf(x) = \dfrac{x}{x^4+1}$

Let $x \in \mathbb{R} - \{ 0\}.$ Now we can divide the above expression by $x$ as $x\ne0.$

$f(x) = \dfrac{1}{x^4+1}$ which is clearly less than one as denominator is always greater than numerator. Now lets include zero into the domain...

Substituting zero into the original equation we have:

$0f(0) = \dfrac{0}{0^4+1} =0 \implies f(0) \in \mathbb{R}$. But its given $f(x)$ is continuous. Hence LHL must equal RHL.  $\lim_{x \to 0} \dfrac{1}{x^4+1} = 1$. Hence f(0) must equal 1.

Now we have: $f(x) \leq 1 \forall x \in \mathbb{R}$


问:
Projected normal distribution: projecting 3D vectors perturbed by Gaussian noise.
I am interested in the following problem.

Consider a unit vector $\hat{u}$ in 3D and the perturbed vector $\vec{v} = \hat{u} + \vec{x}$, where $x_i \sim \mathcal{N}(\mu=0,\sigma)$, that is, the vector is perturbed by a random vector sampled from a normal distribution with zero mean and variance $\sigma^2$. I am now interested in the distribution of
\begin{equation}
\frac{\vec{v}\cdot\hat{u}}{\|\vec{v}\|} = \cos\theta,
\end{equation}
where $\theta$ is the angle between the original and perturbed vector. In particular I am interested in the case where $\sigma \ll 1$.

By running numerical simulations I find that $1-\cos\theta$ is distributed according to an exponential distribution when $\sigma\ll 1$. I am interested in a derivation of, or argument for, this result. If there is a way to relate the variance of the normal distribution to the mean of the exponential distribution that would be even better!

I believe this problem is related to the projected normal distribution. See e.g. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3773532/
However, they consider the general case an in particular when effectively $\sigma>1$. From this general result I am having a hard time finding the limit $\sigma \ll 1$.

答:
Let's change the name of the variables because I want to use $x_i$ for the coordinate axes.

Very briefly: Assuming independence of your $x_i$, which I'm going to call $\varepsilon_i$.  Since $\varepsilon\sim N(0,\sigma^2I_3)$ is radially symmetric, we may rotate and pick $u$ to be on the positive $x_1$-axis.  Then $v=u+\varepsilon$ gives
$$
\cos\theta=\frac{u\cdot v}{\lvert v\rvert}=\frac{1+\varepsilon_1}{\sqrt{(1+\varepsilon_1)^2+(\varepsilon_2^2+\varepsilon_3^2)}}
$$
or equivalently (since $\sigma\ll 1$)
\begin{align*}
\tan\theta&=\frac{\sqrt{\varepsilon_2^2+\varepsilon_3^2}}{1+\varepsilon_1}=\sqrt{\varepsilon_2^2+\varepsilon_3^2}(1+O_p(\sigma))\\
\implies\theta&=\sqrt{\varepsilon_2^2+\varepsilon_3^2}(1+O_p(\sigma))
\end{align*}
So
$$
1-\cos\theta=2\sin^2(\theta/2)=\frac12(\varepsilon_2^2+\varepsilon_3^2)(1+O_p(\sigma)).
$$
Standard result says the sum of squares of two independent $N(0,1)$s is the exponential distribution with mean $2$.  So $\frac12(\varepsilon_2^2+\varepsilon_3^2)$ is $\frac12\sigma^2$ times an exponential of mean $2$.  Since exponential distribution is closed under scaling by positive factor, this gives exponential with mean $\sigma^2$ as the approximate distribution of $1-\cos\theta$.


问:
The minimum size of an edge set that contains an edge of every cycle
The question is like this:

There is a connected graph $G=(V,E)$ with $n$ vertices and $m$ edges such that $m \ge n$.
Let's say that the graph has a spanning tree, meaning there is a tree $T$ such that $T = (V, E_t)$ and $E_t$ contained in $E$.
Find the minimum size $k$ such that a set $F$ exists such that $F \subseteq E$ and these two conditions hold:

$|F| = k$

Every cycle in $G$ contains an edge in $F$.

My solution is like:

I know that $G$ has a spanning tree $T$.
A tree has $n-1$ edges so for going from $G$ to $T$ i need to erase a set of edges (call it $F$) that their size is $m - (n - 1)$ so after i erase them i have a subgraph of $G$ that has $n-1$ edges as needed for it to be a tree.

Let's say that there is a cycle in $G$ that doesn't contain an edge in $F$ so after I erase $F$ from $G$ this cycle will remain as it was and that is a contradiction to the assumption that $T$ is a tree. So, as a result of the contradiction it means that $F$ has to be a set of edges that $|F| = k$ and also every cycle in $G$ contain a set in $F$, now we need to prove that $k$ is the minimum size of a set like this. (Comment: $k = m - (n-1)$)

Let's say $m - (n-1)$ is not the minimum size, so we have $k < m - (n - 1)$ such that $k$ is the size of a set $F$ that every cycle in  $G$ contains a set in $F$. If we erase $F$ from $G$ we will get a subgraph $G_t$ which has $m - k$ edges so we can say that the number of edges in $G_t$ is bigger than $n-1$ and we know that the number of vertices is $n$.

Because $G$ is a connected graph and every edge that we erase is contained in a cycle in $G$, we can infer that $G_t$ is a connected graph. Because $G_t$ has more than $n-1$ edges and only $n$ vertices we can infer that $G_t$ is a connected graph that is not a tree and therefore $G_t$ has at least one cycle.
If $G_t$ has a cycle, its obvious that there is a cycle in $G$ that does not contain an edge in $F$. Therefore every group of edges that has more than $m - (n-1)$ edges does not satisfy the conditions.

The question is about the last step, if I am erasing an edge of a cycle, couldn't it stay a cycle without the edge? (Maybe if its stays a cycle it means that there is a subcycle in $G$ that doesn't contain an edge in $F$ and a subcycle is a cycle and there is the answer).

Just want to make sure, and I already wrote all of this so I will post it maybe it will help other people who faced this question.

答:
Your argument is solid. To summarize: a spanning tree $T$ of $G$ has $n-1$ edges, and this is the most edges that a graph without cycles can have.

We can delete the $m-n+1$ edges of $G$ which are not in $T$, and that is a valid solution (there are no surviving cycles, so we must have deleted at least one edge from every cycle).

If $F$ is a set of fewer than $m-n+1$ edges, then $G-F$ has at least $n$ edges left, so it has at least one cycle: a cycle disjoint from $F$. So no such set can be a valid solution.

Now, on to your concern: if we are erasing an edge of a cycle, could it stay a cycle without that edge?

Well, not the cycle of those exact vertices in that exact order. If you have a cycle with vertices $a,b,c,d,e$ and edges $ab,bc,cd,de,ea$, and you delete one of those edges, that's not a cycle anymore. You might, in general, have some related cycles:

A cycle like $a,c,e,b,d$ with edges $ac, ce, eb, bd, da$ which has those vertices in a different order, so it uses different edges.

A cycle like $a,b,c,e$ with edges $ab, bc, ce, ea$ which has most of the same vertices in the same order, but skips some, so some edges are different. (If you delete edge $cd$ or $de$ from the first cycle, this cycle survives.)

We don't have to care about that in this proof, because we make sure to delete an edge from every cycle. When we consider cycle $a,b,c,d,e$, we conclude that we've deleted an edge from that cycle. When we consider cycle $a,b,c,e$, we conclude that we've deleted an edge from that cycle, too. It might be the same edge, or a different one; we don't particularly care.


问:
Square Root of $9a + 36\sqrt{3ax} + 108x$
An old mathematics book found on Google asserts that:

Square root of $9a + 36\sqrt{3ax} + 108x$

Equals to $3\sqrt{a} + 6\sqrt{3x}$

I find this incorrect, but am not 100% sure. My derivation is:

Let $m = \sqrt{a}$ and $n = 2\sqrt{3x}$, then

$
\begin{align}
  9a + 36\sqrt{3ax} + 108x & = 3\sqrt{a + 4\sqrt{3ax} + 12x} \\
& = 3(m^2 + 2mn + n^2) \\
& = 3(m+n)^2 \\
& = 3(\sqrt{a} + 2\sqrt{3x})^2
\end{align}
$

Which is definitely a different expression.

Is the book being correct or me? Thanks in advance for answering this secondary school level question. I am an adult trying to make up my broken mathematics since teenage.

Thanks,

JF

(Update)

I found my mistake rather stupid. Derivation should be:

$
\begin{align}
  \sqrt{9a + 36\sqrt{3ax} + 108x} & = 3\sqrt{a + 4\sqrt{3ax} + 12x} \\
& = 3\sqrt{m^2 + 2mn + n^2} \\
& = 3\sqrt{(m+n)^2} \\
& = 3(m+n) \\
& = 3(\sqrt{a} + 2\sqrt{3x}) \\
& = 3\sqrt{a} + 6\sqrt{3x}
\end{align}
$

Which is same as the book's answer. Maybe Friday night isn't good time to re-picking up math. But I will keep on.

答:
do $(3\sqrt{a}+6\sqrt{3x})^2$ and you get $9a+36\sqrt{3ax}+108x$

$\therefore \sqrt{9a+36\sqrt{3ax}+108x} = \sqrt{(3\sqrt{a}+6\sqrt{3x})^2} = |3\sqrt{a}+6\sqrt{3x}|$

Since $3\sqrt{a}+6\sqrt{3x}$ is always greater than zero, $|3\sqrt{a}+6\sqrt{3x}|=3\sqrt{a}+6\sqrt{3x}$

Hence it is true.


问:
paradoxical derivation involving tall orthonormal matrix
I am misunderstanding something very fundamental about linear algebra.

Say A is a tall orthonormal matrix, C is some arbitrary matrix and I want to find X.

$$AX=C$$

Since A is tall orthonormal matrix, $A^TA=I$. Therefore, multiplying $A^T$ from the left, we get

$$X=A^TC$$

But if we plug this back into the first equation,

$$AA^TC \ne C $$

since $AA^T \ne I$.

答:
Answering my own question... For the first equality to be satisfied, C should have direction in A (The column space of C should be the same as the column space of A). In other words it should have the form $C=AB$ where B is some arbitrary matrix. Plugging this into the last equation,

$AA^TAB=AIB=AB=C$


问:
Permissable to define a fork function with two x values mapping to different functions but whose y-values are identical?
Say we got this simple example

$$f(x) = \begin{cases}
p_1(x),\text{ for }5 \leq x \leq 8,\\
p_2(x),\text{ for }8 \leq x \leq 10
\end{cases}$$

Assuming $p_1(8)=p_2(8)$, is this permissable?

Arguments for and against:

Yes, because $f(x)$ maps each x-value to exactly one y-value.

No, because it becomes unclear whether to choose $p_1$ or $p_2$ when $x=8$.

Counter: well, it doesn't matter what function you choose, they yield identical y-values.

答:
I would say it is an acceptable definition, in the sense its formalization describes a function. However, some people may be displeased, but again some people will always be displeased...

A function $f:X\rightarrow Y$ is a subset $F$ of $X\times Y$ such that:

for every $x\in X$, there exists a $y\in Y$ for which $(x,y)\in F$;

and, if $(x, y_{1}), (x, y_{2})\in F$, then $y_{1}=y_{2}$.

When we define a function by cases, that is not absolutely precise in the sense of this definition, but the correspondent idea can always be formalized: in your case, the set $F$ corresponding to the function may be defined as the subset of $[5, 10]\times\mathbb{R}$ (I am assuming here that your function is real-valued and not defined outside of $[5,10]$) such that
$$F=\{(x,y)\in[5,10]\times\mathbb{R} : \text{if $x\in[5,8]$, $y=p_{1}(x)$, and if $x\in [8,10]$, $y=p_{2}(x)$}\}.$$
Does that define a function?

For every $x\in [5,10]$, there indeed exists an $y\in\mathbb{R}$ such that $(x,y)\in F$, equal to $p_{1}(x)$ if $x\in[5,8)$, equal to $p_{2}(x)$ if $x\in(8,10]$, and equal to $p_{1}(x)=p_{2}(x)$ if $x=8$.

If $(x, y_{1}), (x, y_{2})\in F$, there are three cases: if $x\in[5,8)$, since $p_{1}$ is a function $y_{1}=y_{2}$; if $x\in (8,10]$, since $p_{2}$ is a function $y_{1}=y_{2}$; and if $x=8$, given both $p_{1}$ and $p_{2}$ are functions and $p_{1}(8)=p_{2}(8)$, $y_{1}=y_{2}$.

So $F$ indeed describes a function, and your definition is indeed acceptable.


问:
Explanation of where this trig identity comes from
I'm working on a problem but it's been a while since I last saw trig identities so I'd love some help or being pointed in the right direction.

Basically, I'd like to understand where this identity comes from;

$$\tan(2t) = \dfrac{2\tan(t)}{1 - \tan^2(t)}$$

Thanks for any help you can give - if it's useful to know the context of the problem, I'm writing a bit of code that converges on $\pi$ faster than Leibniz's series. (Please don't give too much away about the rest of the problem though, I'd like to get there myself :) )

答:
If we agree on the two identities:

$$\sin \left( 2 \theta \right) = 2 \sin \left(\theta \right) \cos \left(\theta \right)\\
\cos \left(2 \theta \right)=\cos^2\left(\theta \right)-\sin^2\left(\theta \right)
$$
then the rest of it is straight-forward.

$$\tan \left(2\theta \right)=\frac{\sin\left(2\theta \right)}{\cos\left(2\theta \right)}=\frac{ 2 \sin \left(\theta \right) \cos \left(\theta \right)}{\cos^2\left(\theta \right)-\sin^2\left(\theta \right)}=\frac{ 2 \sin \left(\theta \right) \cos \left(\theta \right)}{\cos^2\left(\theta \right)}\frac{1}{1-\tan^2\left(\theta \right)}=\frac{2 \tan{\left(\theta \right)}}{1-\tan^2\left(\theta \right)}$$

Those two identities can be proved in one step using Complex-Numbers. It is a well known identity that $e^{i \theta}=\cos{\left(\theta \right)}+i\sin{\left(\theta \right)}$.

Now consider this:

$$e^{2i\theta}=\cos{2\theta}+i\sin{2\theta}$$

On the other hand:

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

Bingo!


问:
Proof that cube has 24 rotational symmetries
I was doing a combinatorics problem which states this definition of symmetry: for a subset $S$ of $\mathbb{R}^3$ a symmetry is "rigid motion" $f:\mathbb{R^3}\rightarrow \mathbb{R^3}$ such that any $x\in S$ has its image $f(x)$ also in S. (To provide some fodder for the problem, the author dumbs it down for me that a symmetry is some operation which leaves $S$ in the same place although the individual points may be rearranged)

From this working definition it is asked to find the number of rotational symmetries of the cube. So far I have seen two different arguments and I still dont know because drawing on paper is getting confusing. Also my solution was incorrect and I dont know where I undercounted.

Ignore my method.<s>Drawing 4 axis perpendicular with the faces, each with 4 rotational symmetries $(0^\circ,90^\circ,180^\circ,270^\circ)$ and I get $4\times 4 =16$ symmetries. Plus the four body diagonals$=20$. This is wrong and the answer is $24$.</s>

The book gives the short solution: "move any vertex to any other and rotate the edges leading to it in three ways". I don't get it. There are 8 vertices, and 3 ways of what?

I google and get another argument from wikipedia: "The group of orientation-preserving symmetries is S4, or the group of permutations of four objects, since there is exactly one such symmetry for each permutation of the four pairs of opposite sides of the octahedron." Indeed $4!=24$ but I cannot convince myself that aall permutations of the pair of opposite edges is all the symmetries there can be (I clearly undercounted when I got 16 and still have no insight why, so this does not help me as I dont know for sure there cant be more)

My imagination and visualization skills are very weak. Is there a good online software that allow me to visualize and play around with a labelled cube?

答: The point-group symmetry of a cube is m-3m, which has 48 symmetry operations. These include the mirror reflections. If we insist on rigid-body operations only, with no change of "handedness" (inversions), then the mirror operations must be removed  and the point-group symmetry is reduced to 432 – four-fold rotation axes about each face (parallel to the edges), 3-fold rotation axes parallel to each body-diagonal and 2-fold rotation axes parallel each face-diagonal; all axes pass through the cube center. This gives 4x3x2 = 24 orientations. The 432 group is a subgroup of m-3m. The latter includes the additional 24 orientations that arise after each orientation of the rigid cube is reflected.</p>


问:
How to interpret this psychological construct mathematically?
I'm looking for some guidance as to how a psychological construct I have learned about could be rigorously formulated.

The idea is that when a decision maker begins to favor a particular choice early on, this tentative preference often shifts the evaluation of subsequent information in a manner that benefits on the early leader (called predecisional information distortion). There is what is called proleader distortion, a positive distortion of the tentatively leading option, and antitrailer distortion, the tendency to negatatively distort information about the trailing option.

To be clear:

Feedback loop:

Interpretation of new information biased to favor emerging preference

Emerging preference is biased in the direction of interpretation

Information arrives and individuals evaluate the information and subconsciously update their preferences.

A "dynamical probability system" is a tuple $S$ $(X,P,F)$ where $S_X$ is a set of outcomes, $X_P$ is a probability distribution over all events in $X$, and $S_F$ is an update rule for $S_P(x_i\in X)$.

As mentioned, we have two probability trajectory systems. The first is the interpretation system($I$) which describes how interpretation probabilities evolve over time, and the second is the ultimate choice system ($C$)which describes how the probability distribution over ultimate choices evolves.

We can begin with some initial probability distributions $I_P$ and $C_P$ which should be close to each-other.

Stipulate that $\sum{P(x_i)}=1$

The probability that any choice will be made is responsive to how information at time $n$ is interpreted.

We have some update rule $I_F$ for $I_P$ as a function of $C_P$ and new information corresponding to each choice $x \in C_X$ that arrives the environment.

We then use a second update rule $C_F \iff C_P(x \in C_X)_{n+1}=k_iI_P(x_i)-\sum_{j\ne i}{q_jI(x_j)}$ where $1<k_i,0<q_j<1 \in \mathbb{N}$ are, respectively, proleader and antitrailer learning rates relating each $C_P(x \in C_X)$ with each $I_P(x \in I_X)$. This allows for nuanced relationships between all the updates. I hope, but have no idea how to set things up well enough for a proof, that eventually $C_P$ will converge into a long tail.

This seems a little clunky and not well defined. There is no guarantee that $\sum {C_{Pi}}=1$, etc.

I'm not an expert and have not studied probability theory. That said, I will appreciate a simple mathematical formulation of this idea to a great degree, but not too complex a formulation. And I know I may have committed some great fallacy. Making definitions of things is very easy. Really understanding them is hard and rigorous.

答:
Coincidentally, I just read this question about a robot that flips coins and thought that it might provide an insight.

Someone walks into your room and dumps a huge bag of quarters all over the floor. They spread them out so no quarters are on top of any other quarters. a robot then comes into the room and is programmed such that if it sees a head, it flips it to tails. If it sees a tail, it throws it in the air.

My thinking is that the robot resembles a decision maker who has developed a bias towards tails.  Think of the decisionmaker receiving a sequence of coins;

if it is tails (aligning with their bias) they accept it and move onto the next packet of information.

If it is heads (against their bias) they reject it.

The modification to your application could be if the decisionmaker has a slightly different rule:

If it is tails I will accept with choice probability $p_{\text{Accept T}}$ otherwise reject (and new coin flip)

If it is head I will accept with choice probability $p_{\text{Accept H}}$ otherwise reject (and new coin flip).

Those choice probabilities would initially both start at 1/2, but would adjust based on the coins that they initially see.  In particular, if they were calculated from the proportion of tails observed in a sample. My conjecture is that any local variations in the sample proportion would cause the choice probabilities to move away from 1/2.


问:
Interior points and closure

Let $A$ be a subset of the metric space $(X,d)$. Is it true that $$\left(\left(A'\right)'\right)'\ =\left(A'\right)' \tag{1}$$where $A'$ denotes the set of the limit points?
Assuming that $(1)$ is true, can we prove the following equations?$$X\setminus A^\circ = \overline{X\setminus A} \tag{2}$$ $$\\ X\setminus\overline{A} = (X \setminus A)^\circ \tag{3}$$Note that $A^\circ$ denotes the interior points of $A$.

My try: For proving $(1)$, I tried to use $(A')'\subseteq\operatorname{cl}A$ which implies $\left(\left(A'\right)'\right)' \subseteq \operatorname{cl}A'$ but I couldn't continue.

If $A$ be a finite subset then both sides of the equation are $\emptyset$, so the assumption is true. Also $A^\circ = \emptyset$ which implies $X\setminus A^\circ = X$. Using the fact $\overline{X\setminus A}\subseteq \overline{X}\setminus \overline{A}$ and $\overline{X} = X,\overline{A} = A$ we get $X\subseteq X\setminus A$ which is false, so $(2)$ is incorrect.

I tried to use the same example for $(3)$. Since $A = \overline{A}$ we have $X\setminus A = (X\setminus A)^\circ$ which shows $X\setminus A$ is open. We know that $X$ and $A$ are closed set, so if we can show that a closed set minus another closed set can't be open, then we can disprove $(3)$.

Edit: Thanks to Berci, $(2)$ and $(3)$ are proved in general. The only remaining problem is $(1)$.

答:
There are metric spaces with finite open sets (e.g. the discrete metric on any set).

Also, the complement of a closed set is always open.

(2) and (3) follow directly from the definitions:

$x\in X\setminus A^\circ$ means that every open neighborhood $U$ of $x$ is not entirely in $A$, that is, intersects $X\setminus A$.

And the closure $\overline B$ of any set $B$ consists exactly of those points $x$ such that every neighborhood of $x$ intersects $B$.

Similarly, $x\in X\setminus\overline A$ means that there's a neighborhood of $x$ which doesn't intersect $A$, that is, it's fully in $X\setminus A$.

Title: Interior points and closure


答:
The issue here is a common misconception about techniques for solving equations, in more than just linear algebra. Your logic shows that
$$AX = C \implies X = A^\top C.$$
This is a logically valid implication. What you are not getting is the converse:
$$X = A^\top C \not\Rightarrow AX = C.$$
Your logic showed that if there is a solution to $AX = C$, then it must be $X = A^\top C$. So, there is at most one solution to the equation. But, it doesn't guarantee that there is a solution. What you are observing is that, for many $C$ (in particular, $C$ such that $\operatorname{colspace}(C) \not\subseteq \operatorname{colspace}(A)$), there is no solution for $X$.

What can we learn from this? Whenever solving an equation of any kind, always check the result in the original equation.

Title: paradoxical derivation involving tall orthonormal matrix

Body:
I am misunderstanding something very fundamental about linear algebra.

Say A is a tall orthonormal matrix, C is some arbitrary matrix and I want to find X.

$$AX=C$$

Since A is tall orthonormal matrix, $A^TA=I$. Therefore, multiplying $A^T$ from the left, we get

$$X=A^TC$$

But if we plug this back into the first equation,

$$AA^TC \ne C $$

since $AA^T \ne I$.


问:
Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $
I have the conjecture for the integral

$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$

where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.

I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral

$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$

is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.

Edit:

Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$

which may not reduce the difficulty, albeit appearing compact.

答:
Here is an attempt at an elementary solution. The identity is readily seen to be the same as $$\int_{\pi/4}^{3\pi/4}\log\cot\frac v9\,dv+\int_{5\pi/4}^{7\pi/4}\log\cot\frac v9\,dv=4G.$$ We can combine both integrals into one through the linear substitution $2u=v-3\pi/4\pm\pi/2$ to get $$\int_0^{\pi/4}\log\cot\left(\frac{2u}9+\frac\pi{36}\right)+\log\cot\left(\frac{2u}9+\frac{5\pi}{36}\right)\,du=2G.$$ Taking advantage of symmetry over $t=2u/9+\pi/12$, the integral can be written as \begin{align}\frac49G&=\int_{\pi/12}^{5\pi/36}\log\left(\cot\left(t-\frac\pi{18}\right)\cot\left(t+\frac\pi{18}\right)\right)\,dt\\&=\int_{\pi/12}^{5\pi/36}\log\frac{-\cos2t-\cos\pi/9}{\cos2t-\cos\pi/9}\,dt\end{align} or through a Taylor series expansion,$$\frac89G=\int_{\pi/6}^{5\pi/18}\log\frac{\cos\pi/9+\cos x}{\cos\pi/9-\cos x}\,dx=\int_{\pi/6}^{5\pi/18}\sum_{k=1}^\infty\frac2{2k-1}\left(\frac{\cos x}{\cos\pi/9}\right)^{2k-1}\,dx.$$ We can exchange operators to get \begin{align}\frac49G&=\sum_{k=1}^\infty\frac{(\sec\pi/9)^{2k-1}}{2k-1}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx\end{align} and since \begin{align}\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx&=\int_{\pi/6}^{5\pi/18}\frac1{2^{2k-2}}\sum_{m=0}^{k-1}\binom{2k-1}m\cos(2k-2m-1)x\,dx\\&=\frac4{4^k}\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{5(2k-2m-1)\pi}{18}-\sin\frac{(2k-2m-1)\pi}6}{2k-2m-1}\end{align} the identity is equivalent to $$\frac1{18}G=\sum_{k=1}^\infty\sum_{m=0}^{k-1}\binom{2k-1}m\frac{\sin\frac{(2k-2m-1)\pi}9\cos\frac{4(2k-2m-1)\pi}9}{4^k(2k-1)(2k-2m-1)(\cos\pi/9)^{2k-1}}.$$ The trig expressions should reduce to something simpler (and not involving trig functions) — need help!

Remark: if $\int_{\pi/6}^{5\pi/18}(\cos x)^{2k-1}\,dx=a+b\sin\pi/18+c\cos\pi/9+d\cos2\pi/9$ for some rational $a,b,c,d$ then $$\frac49G=\sum_{k=1}^\infty\frac{a+(b-d)\sin\pi/18+(c+d)\cos\pi/9}{(2k-1)(\cos\pi/9)^{2k-1}}$$


问:
Countability of removable discontinuities
I saw a problem of proving that number of removable discontinuities in a function is countable. I was not able to prove it and tried many things. Can anyone do it?
Thanks in advance.</p>

答:
For $n\in \{1,2,\dots,\}$, let $E_n$ be the set of removable discontinuities $x_0$ where the difference between $f(x_0)$ and $\lim_{x\to x_0}f(x)$ is at least $1/n$. If we can prove $E_n$ is countable for all $n$, we will be done, since the set of removable discontinuities is the union of the sets $E_n$.

Given $x_0\in E_n$, let $\lim_{x\to x_0}=L$. There is a neighborhood around $x_0$ where points in the neighborhood are all with $\frac1{4n}$ of $L$. Therefore, they are all within $\frac1{2n}$ of each other. This means there cannot be any removable discontinuities with a gap of $\frac1n$ in that neighborhood, so this neighborhood is disjoint from the rest of $E_n$. Therefore, the points in $E_n$ are all isolated from each other, and therefore countable.


问:
Arranging Drivers and Passengers from IB Math AA HL

A group of $12$ people want to go a concert.  They can travel in a small car that takes on driver and one passenger and two cars each taking one driver and four passengers.  If there are five drivers in the group, in how many ways can they travel?

The solution in the textbook is $6300$ but I have no idea how they got to that answer.

So far, here is what I've tried: $nPr(5,3) \cdot nCr(9,1) \cdot nCr(8,4) \cdot nCr(4,4) = 37800$

I'm not sure whether my work is correct or not, but here's my chain of logic:

There are $5$ drivers going into $3$ cars, and the order that we pick them does matter because the drivers are in different cars. That's where $nPr(5,3)$ comes from.

There are $9$ people left total. In the first car, we are choosing $1$ person out of the $9$. Then, in the second car, we are choosing $4$ people out of the $8$ people left. Then, in the third car, we are choosing $4$ people out of the $4$ people left.

答:
The drivers can be placed in $5\cdot4\cdot3 = 60$ ways

There are spaces for $1, 4,4$ people in the three cars. Since no seating arrangement has been specified, they can be placed in $\binom9 1 \binom 84 \binom 4 4 = 630$ ways

Multiply to get $60\cdot630 = 37800$

Your answer seems correct to me


问:
Determine if vectors are generator
How do you determine if vectors are generator?

I have $⃗ = (−3,3,2)$, $⃗ =(-3,0,3)$, $⃗ = (−2,1,1)$

Thanks

答:
How do you determine if vectors are generator? I have $⃗ = (−3,3,2)$, $⃗ =(-3,0,3)$, $⃗ = (−2,1,1)$.

Just use the definition of generator on $S=\{(-3,3,2),(-3,0,3),(-2,1,1)\}\subseteq \mathbb{R}^{3}$. It's to say, you need to solve:

$$ \operatorname{span}(S)=\{(x,y,z)\in \mathbb{R}^{3}: (x,y,z)=\alpha_{1}(-3,3,2)+\alpha_{2}(-3,0,3)+\alpha_{3}(-2,1,1); \alpha_{i}\in \mathbb{R}\}$$

Title: Determine if vectors are generator


问:
Median of $f(x)$ = $f($median of $x)$ if $f$ is monotonic?
Suppose $X \in \mathbb{R}$ is a continuous random variable, $f: \mathbb{R} \Rightarrow \mathbb{R}$ is monotonic. Let $Y = f(X)$, can we claim that $median(Y)=f(median(X))$, or median of a monotonic function of a R.V. is equal to the monotonic function at median of the R.V.?

If so, how do we formally show this?

If we cannot, are there any further restrictions that we can use to make the claim true?

Intuitively, I think this is true. When $X$ is discrete (and number of possible $X$ is odd?), I think we can show that $x_i>x_j$ iff $f(x_i)>f(x_j)$ so there are as many $f(x)<f(x_{median})$ as $f(x)>f(x_{median})$ since there are as many $x<x_{median}$ as $x>x_{median}$. This should be able to generalize to when there are infinitely many possible values of $X$ i.e. when $X$ is continuous?

(I suppose we want to somehow play with CDF of $X$ and $Y$?)

答:
Let's call the function $g: \mathbb{R} \Rightarrow \mathbb{R}$

$F(x_{median}) = 0.5$ by definition, also since g is bijective $y=g(x)$ and $x = g^{-1}(y)$.

The CDF of y is $F_y(y) = F(g^{-1}(y))$ since g preserve the order. Then $F_y(y_{median}) = F(g^{-1}(y_{median})) = F(x_{median}) = 0.5 $

$g^{-1}(y_{median}) = x_{median} $

$y_{median} = g(x_{median})$


问:
How do I compare two expressions in Maple to see if they’re equivalent?
I have to compare two expressions as an assignment but the professor didn’t explain how to do it (I doubt he actually knows how to do the stuff he asks us to do). The two expression pairs in question:

```<code>tg(x) + tg(y) = (sin(x+y))/cos(x)*cos(y)

arcsin(x) + arcsin(y) = pi/2
</code>```

Now I know these are true, but whenever I use evalb or is I invariably get false. I tried many different things but none of them work and I’ve searched the internet for help but I didn’t find anything useful. The most similar thing to mine was this but I tried the solution and it didn’t work. Please help, I’m losing my sanity.

答:
For the first equation, this worked:

For the second equation, it appears that the same method didn't simplify it to 0 automatically. Converting arcsin() to arccos() solved the problem:


问:
MILP formulation of constraint
I'm very new to MILP, and I'm testing python-mip.
I'm trying to model a simple power distribution system with some constraints, but I'm stuck....

I want to find a valid power balance between a device, a power loss and a power input, that maximisizes, if possible, power_provided_to_device, like:

max(power_provided_to_device)

power_available = power_provided_to_device + unused_power

with:
0. 0 < lb <= power_provided_to_device <= ub

if power_available < lb: power_provided_to_device = 0, unused_power = power_available

if lb <= power_available <= ub: power_provided_to_device = power_available, unused_power = 0

if ub < power_available: power_provided_to_device = ub, unused_power = power_available - ub

I made it work for power_available >= lb.
However, I just can't figure out how to fulfill the condition 1. . I tried some exemples with the bigM and binary variables, piecewise functions, but with no success...

Sorry for my poor formating and problem description.
Thanks a lot for your kind help!
Max.

答:
The objective and linear constraints naturally satisfy conditions 2 and 3.  For condition 1, you want to enforce
$$\text{power_available} < \text{lb} \implies \text{power_provided_to_device} = 0$$
Introduce a binary variable $x$ and big-M constraints
\begin{align}
\text{lb} - \text{power_available} &\le M_1 x \tag1\\
\text{power_provided_to_device} &\le M_2 (1-x) \tag2\\
\end{align}
Constraint $(1)$ enforces
$$\text{power_available} < \text{lb} \implies x=1$$
Constraint $(2)$ enforces
$$x=1 \implies \text{power_provided_to_device} = 0$$
You can take $M_1=\text{lb}$ and $M_2=\text{ub}$.

Title: MILP formulation of constraint


问:
Covering of a compact set, where the sets are not necessarily open.
Let $K = [-1,1]$; for example, and lets say there exists a cover of $K$, namely:
$$K \subseteq \bigcup_{i=1}^{\infty}S_i.$$
I did not specify if each $S_i$ for $i \in \mathbb{N}$ is open, does there exist a finite sub cover of $K$ such that
$$K \subseteq \bigcup_{i=1}^{k}S_i', $$

where $S_{i}' = S_{j}, i \in 1..k,j\in \mathbb{N}?$

答:
No, not at all. Take any injective sequence $(x_n)$ in $K$. Define
$$S_{n+1} = \{x_n\}$$
for $n \ge 1$ and
$$S_1 = K \setminus \bigcup_{n=2}^\infty S_n.$$
Then the $S_n$ sets partitition $K$, hence they cover $K$, but no proper subcover will cover $K$.


问:
How many combinations of groups are there where no member of a group has been with another member before?
I found this hard to word in the title, so let me give an example.

I have 16 students, and I want to split them up into 4 groups of 4. However, I want to make sure that every time I have a new combination of groups, that a student has never been with any of their new groupmates before.

I found info on how to separate objects into groups, but not in only unique ways as I've described above.

I think I figured this out the rote way, by assigning each student a number and then finding the unique combinations, and got 3. But I'd like to know the math behind it, and how it could be applied to other configurations (20 students w/ 4 groups of 5?).

Thanks for any help you can give!

答:
I think there are $5$ ways to do it, not $3$.

Identify each student with one of the $16$ points of the affine plane over the field with $4$ elements.  To divide the students into $4$ groups, we can group them according to a pencil of $4$ parallel lines. Then the question becomes, "How many parallel directions are there in the plane?"

Of course, this is just the number of lines through the origin.  We can join the origin to any of the other $15$ points, but each line is counted $3$ times, once for every point on it other than the origin, so there are $5$ lines.

In general, this is a problem in combinatorial designs, and there are a lot of open questions, I believe.

EDIT

Here is a python script that computes the answer.  I don't know how to draw a comprehensible picture of the plane over $\mathbb{F}_4$.  I think of a $4\times4$ array of dots for the points.  The problem is that there are $20$ lines, and I don't see how to draw them without making a mess.

Here is my script:

```<code>plus = { }

for x in range(4):
    plus[x,0] = plus[0,x] = x
for x in range(4):
    plus[x,x] = 0
plus[1,2]=plus[2,1] =3
plus[1,3]=plus[3,1] = 2
plus[2,3]=plus[3,2] = 1

times = { }
for x in range(4):
    times[x,0]=times[0,x] = 0
    times[x,1]=times[1,x] = x
    times[2,2] = 3
    times[3,3] = 2
    times[2,3]=times[3,2] = 1

points = set(range(1,16))  # points that have to be with 0

def add(x,y):
    x0, x1 = divmod(x,4)
    y0, y1 = divmod(y,4)
    a = plus[x0, y0]
    b = plus[x1,y1]
    return 4*a+b

def mult(k, x):
    x0, x1 = divmod(x,4)
    a = times[k, x0]
    b = times[k, x1]
    return 4*a+b

def plane(pt):
    answer = 4*[None ]
    answer[0] = sorted([0, pt, mult(2,pt), mult(3,pt)])
    used = set(answer[0])
    for n in range(1,4):
        for x in range(16):
            if x not in used: break
        answer[n] = sorted([add(x, y) for y in answer[0]])
        used |= set(answer[n])
    return answer

planes = []
while points:
    x = points.pop()
    p = plane(x)
    planes.append(p)
    for pt in p[0]:
        points.discard(pt)

for plane in planes:
    for line in plane:
        print(line)
    print()

for x in range(16):
    for y in range(16):
        if x==y: continue
        found = False
        for plane in planes:
            if found:
                break
            for line in plane:
                if x in line and y in line:
                    found = True
                    break
else: # not found
    print(f'Did not find {x}, {y}')

</code>```

This produced the output

```<code>[0, 1, 2, 3]
[4, 5, 6, 7]
[8, 9, 10, 11]
[12, 13, 14, 15]

[0, 4, 8, 12]
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]

[0, 5, 10, 15]
[1, 4, 11, 14]
[2, 7, 8, 13]
[3, 6, 9, 12]

[0, 6, 11, 13]
[1, 7, 10, 12]
[2, 4, 9, 15]
[3, 5, 8, 14]

[0, 7, 9, 14]
[1, 6, 8, 15]
[2, 5, 11, 12]
[3, 4, 10, 13]

Did not find 15, 15
</code>```

Where the last line is a spurious error message due to sloppy programming.

EDIT

In response to user's comments.

I thought of a way to present this that might make it a little clearer.  In each group, the $16$ numbers represent the points of the plane, and points of the same color lie on the same line.  In the first two groupings, this looks pretty dull; the lines are rows or columns.  In the last three groupings, the lines don't look like what we normally think of as lines, but they satisfy the same kind of linear equations as the lines you're used to.

This isn't the place to try to explain that.  You can search the Web for "finite geometries", but you'll have to learn a little bit about finite fields before it makes sense.

Body:
I think there are $5$ ways to do it, not $3$.

Identify each student with one of the $16$ points of the affine plane over the field with $4$ elements.  To divide the students into $4$ groups, we can group them according to a pencil of $4$ parallel lines. Then the question becomes, "How many parallel directions are there in the plane?"

Of course, this is just the number of lines through the origin.  We can join the origin to any of the other $15$ points, but each line is counted $3$ times, once for every point on it other than the origin, so there are $5$ lines.

In general, this is a problem in combinatorial designs, and there are a lot of open questions, I believe.

EDIT

Here is a python script that computes the answer.  I don't know how to draw a comprehensible picture of the plane over $\mathbb{F}_4$.  I think of a $4\times4$ array of dots for the points.  The problem is that there are $20$ lines, and I don't see how to draw them without making a mess.

Here is my script:

plus = { }

for x in range(4):
    plus[x,0] = plus[0,x] = x
for x in range(4):
    plus[x,x] = 0
plus[1,2]=plus[2,1] =3
plus[1,3]=plus[3,1] = 2
plus[2,3]=plus[3,2] = 1

times = { }
for x in range(4):
    times[x,0]=times[0,x] = 0
    times[x,1]=times[1,x] = x
    times[2,2] = 3
    times[3,3] = 2
    times[2,3]=times[3,2] = 1

points = set(range(1,16))  # points that have to be with 0

def add(x,y):
    x0, x1 = divmod(x,4)
    y0, y1 = divmod(y,4)
    a = plus[x0, y0]
    b = plus[x1,y1]
    return 4*a+b

def mult(k, x):
    x0, x1 = divmod(x,4)
    a = times[k, x0]
    b = times[k, x1]
    return 4*a+b

def plane(pt):
    answer = 4*[None ]
    answer[0] = sorted([0, pt, mult(2,pt), mult(3,pt)])
    used = set(answer[0])
    for n in range(1,4):
        for x in range(16):
            if x not in used: break
        answer[n] = sorted([add(x, y) for y in answer[0]])
        used |= set(answer[n])
    return answer

planes = []
while points:
    x = points.pop()
    p = plane(x)
    planes.append(p)
    for pt in p[0]:
        points.discard(pt)

for plane in planes:
    for line in plane:
        print(line)
    print()

for x in range(16):
    for y in range(16):
        if x==y: continue
        found = False
        for plane in planes:
            if found:
                break
            for line in plane:
                if x in line and y in line:
                    found = True
                    break
else: # not found
    print(f'Did not find {x}, {y}')

This produced the output

[0, 1, 2, 3]
[4, 5, 6, 7]
[8, 9, 10, 11]
[12, 13, 14, 15]

[0, 4, 8, 12]
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]

[0, 5, 10, 15]
[1, 4, 11, 14]
[2, 7, 8, 13]
[3, 6, 9, 12]

[0, 6, 11, 13]
[1, 7, 10, 12]
[2, 4, 9, 15]
[3, 5, 8, 14]

[0, 7, 9, 14]
[1, 6, 8, 15]
[2, 5, 11, 12]
[3, 4, 10, 13]

Did not find 15, 15

Where the last line is a spurious error message due to sloppy programming.

EDIT

In response to user's comments.

I thought of a way to present this that might make it a little clearer.  In each group, the $16$ numbers represent the points of the plane, and points of the same color lie on the same line.  In the first two groupings, this looks pretty dull; the lines are rows or columns.  In the last three groupings, the lines don't look like what we normally think of as lines, but they satisfy the same kind of linear equations as the lines you're used to.

This isn't the place to try to explain that.  You can search the Web for "finite geometries", but you'll have to learn a little bit about finite fields before it makes sense.


答:
If vectors are LI, the can be generators. For LI the determinant
$$\begin{vmatrix} -3 & 3 &2\\ -3 & 0 & 3 \\ -2 & 1 & 1\end{vmatrix}$$
does not vanish, so here it being -6. the given three vectors are generators. That means any 3 dimensional vector can be expresses as a linear combination of these
three vectors called bases.

Title: Determine if vectors are generator

Body:
How do you determine if vectors are generator?

I have $⃗ = (−3,3,2)$, $⃗ =(-3,0,3)$, $⃗ = (−2,1,1)$

Thanks


问:
Find all fucntions such that $f:\Bbb R \to \Bbb R$ and $f(x^{2}+yf(x))=f(x+y)$
My work :
first take $x=y=0$ we get $f(0)=f(0)=c$ , then take $x=x$ and $y=-x$ implies $f(x^{2}-xf(x))=f(0)$ now as there is another function inside the parenthesis I'm asuming there are some entries for which $x^{2}-xf(x)=0$ then we get $f(x)=x$. but I'm little confused about that assumption. If anyone can just fix my mistakes and solve that problem it'll be a great help for me.

答:
Let $x=x$ and $y=0$ $ \implies f(x^2)=f(x)$

Let $x=-x$ $\implies f(-x)=f(x)$ i.e $f(x) $ is an even function

$f(x^2+yf(x))=f(x^2+yf(-x)$ since $f(x)=f(-x)$

$\implies f(x+y)=f(-x+y)$

If we let $y=-x$ then $f(0)=f(-2x)$ which implies that $f(x)=f(0)$. We are not given value of $f(0)$ nor we are able to find it. Since $f(0)$ is constant which means that $f(x)$ is an constant function.


问:
If G = (V,A) a connected graph, planar, proof that... $q\leq 3p -6$
sorry, just started to see graphs, but how can you proof that,

If G = (V,A) a connected graph, plannar, with p verices and q edges.

Proof that...

$q\leq 3p -6$

Any help or orientation would be deeply appreciated. Thank you!

(I made a drawing where p = 3, but I doubt this is the correct answer...):

答: Hint: Assume that $G$ is a maximal planar graph. By maximality of $G$, we can not add more edges without destroying planarity. Observe that each face in this maximal planar graph is a triangle. Otherwise, we could always add diagonal edges (convince yourself that this is true; draw a few planar graphs). Now, denote $p$ as the number of vertices, $q$ as the number of edges and $f$ as the number of faces in $G$. And every edge bounds two faces, hence $2q=3f$. Use Euler's formula $p-q+f=2$, and the other observations and you quickly get the upper bound.


答: There are at least 4 ways to do it for 4 groups of 4. I'm not sure if it's possible to do more, or about how to extend to larger groups.

Title: How many combinations of groups are there where no member of a group has been with another member before?

Body:
I found this hard to word in the title, so let me give an example.

I have 16 students, and I want to split them up into 4 groups of 4. However, I want to make sure that every time I have a new combination of groups, that a student has never been with any of their new groupmates before.

I found info on how to separate objects into groups, but not in only unique ways as I've described above.

I think I figured this out the rote way, by assigning each student a number and then finding the unique combinations, and got 3. But I'd like to know the math behind it, and how it could be applied to other configurations (20 students w/ 4 groups of 5?).

Thanks for any help you can give!


问:
Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$

Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

答: I would do this by using L'Hopitals rule.  Because $\lim_{x \to 0} \frac{e^{x}-1}{x}= \frac{0}{0}$, we can take the the derivative of the numerator and denominator individually, then plug in 0 again to the new equation.  After taking the derivative of the numerator and denominator we will have $\lim_{x \to  0}\frac{e^x}{1}$, and from here we see why the answer is $1$.</p>


问:
Can a contingent argument be proved invalid with natural deduction? (without truth tables)
I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning:

$$P\vdash P\land Q$$

I don't know how to refute it.

EDIT: What I mean is how someone can figure out that for example the fallacy of Affirming the consequent $P\rightarrow Q,Q\vdash P$ is invalid reasoning without making truth tables. A mathematician could use a contingency in his reasoning inadvertently, so how could he be aware of his error?

Here are similar questions but I don't understand the answers.

Propositional logic: Proving contingency without truthtable

Proving Expression is Contingency with Logical Equivalences

答:
Natural deduction offers a system for proving sequents, but to show that  $\Gamma\not\vdash \phi$ you need an interpretation for the formulae such that $\Gamma$ is true (or the conjunction of the formulae in $\Gamma$) and $\phi$ is false.

In your example you can just put $P$ as true and $Q$ as false; since $P \land Q$ is therefore false we see that $P \not\vdash P \land Q$.

If you are looking for a more systematic method for finding counterexamples we have truth trees (semantic tableaux).

Edit: a proof theoretic approach using cut-free sequent calculus.

Suppose $\Rightarrow P \to \, (P \land Q)$ were derivable with $P$ and $Q$ as atomic formulae.

The last rule used must have been $R \to$, (introduction of $\to$ on the right). That is, we have:

$$\frac {P \Rightarrow P \land Q}  {\Rightarrow P \to (P \,\land \,Q)}$$

And before that we must have had an instance of $R \, \land$ (introduction of $\land$ on the right) :

\begin{align*} \overline{ P \Rightarrow P} \; \;\;\;\; P\Rightarrow Q \\ \hline P \Rightarrow P \land Q \ \ \ \ \ \\ \hline \Rightarrow P \to (P \,\land \,Q) \end{align*}

$P \Rightarrow P$ is axiomatic, which is fine, but there's no rule with which we can derive $Q$ from $P$ as atomic formulae, so the sequent above is underivable.


问:
$u(x)$ satisfies the integral equation $u(x) = \int_0^x \sin(u(t))u(t)^p dt$ on $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval
Suppose that $u(x)$ is continuous and satisfies the integral equation

\begin{equation}\label{1.4.1}
        u(x) = \int_0^x \sin(u(t))u(t)^p dt
    \end{equation}

on the interval $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval if $p \geq 0$.

This is what I have:

Since $\sin(u(t))u(t)^p$ is continuous, it follows from the integral definition that $u(x)$ is differentiable. Let us differentiate both sides of equation above with respect to $x$. This yields:

\begin{equation}
    u'(x) = \sin(u(x))u(x)^p
    \end{equation}

This ODE is separabale and becomes:

\begin{equation}
        \frac{1}{\sin(u(x))u(x)^p}du(x) = dx
    \end{equation}

However, this doesn't seem easily solvable so I'm not sure how to show that $u(x) = 0$ from this.

答:
Use @Matthew Pilling's setting. Assume there exists $t\in(0,1)$ such that $u(t)>0$. Because $u$ is continuous and $u(0)=0$, we can identify $a,b\in [0,1]$ such that $u(a)=0,a<t<b,$ and $u>0$ on $(a,b)$. Since $\sin\big(u(t)\big)$ is also continuous, we can find $c\in (a,b)$ such that $\sin \big(u(t)\big)>0$ on $(a,c)$. Now consider the IVP:
$$\frac{dy}{dx}=y^p\sin(y),x\in(a,c).$$
So
$$ y'=y^p\sin(y)\le y^{p+1}, x\in(a,c)$$
and hence
$$ y^{-p-1}y'\le 1. \tag1$$
Integrating (1) from $x_0$ to $x$ ($x_0,x\in(a,c)$, $x_0<x$), one has
$$ -\frac1p\bigg[y^{-p}(x)-y^{-p}(x_0)\bigg]\le x-x_0\le1 $$
from which
$$ y^{-p}(x)\ge y^{-p}(x_0)-p. \tag2$$
Choose $x_0$ close to $a$ such that
$$ y^{-p}(x_0)> p $$
and then from (2), one has
$$ 0< y^p(x)\le \frac{y^p(x_0)}{1-py^p(x_0)}. \tag3$$
Letting $x_0\to a^+$ in (3) gives
$$ 0<y^p(x)\le0, x\in(a,c)$$
which is absurd for $x\in(a,c)$.


问:
Can we always define the projection in a ball of raius $R$ for any Polish metric space $X$?
I'm generalizing some theorems to Polish metric spaces, which were originally proved for $\mathbb R^n$. Hence, as I'm adapting the proofs, I'm checking whether where I can reuse the same arguments. In one of my proofs, I had the following:

"Define $\pi_R : \mathbb R^n \to B(R)$ to be the projection operator on the closed ball with radius $R$ and centered at zero".

Now I'd like to use this for an $X$ metric Polish space. Can I "safely" assume that there exists such projection operator $\pi_R: X \to B(R)$ such that the closed ball is centered at some
$x_0 \in X$? How can I know that this map is well defined?

答:
The answer is "no". Consider the separable real Banach space $\ell^2$. Any closed subspace of $\ell^2$ is Polish, a fact we shall soon use.

Let $e_n$ be the $n$th basis vector, i.e. $e_n$ is the sequence such that the $n$th entry in the sequence is $1$, while the other entries are $0$. Let
$$C = \left\{\frac{1 + \frac{1}{n}}{4}e_{n+1} : n \ge 1\right\}$$
and
$$X = C \cup \{2e_1\}.$$
Note that every point in $X$ is isolated, and hence $X \subseteq \ell^2$ is closed, and hence a Polish space in its own right. If we have distinct $\frac{n + 1}{2n} e_n$ and $\frac{m + 1}{2m} e_m$ in $C$, then
$$\left\|\frac{n + 1}{4n} e_n - \frac{m + 1}{4m} e_m\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + \left(\frac{m + 1}{4m}\right)^2 \le \frac{1}{2} + \frac{1}{2} = 1,$$
but
$$\left\|\frac{n + 1}{4n} e_n - 2e_1\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + 4 \ge 4.$$
Therefore, $C$ is the closed unit ball in $X$, centred at any point in $C$. We can now show that $2e_1$ has no closest point, i.e. the projection map simply does not extend to this point in $X$. We see that

$$\left\|\frac{n + 1}{4n} e_n - 2e_1\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + 4  \to \frac{17}{4},$$

while remaining strictly greater than $\frac{17}{4}$ for every $n \ge 2$. Therefore, there is no point in $C$ that achieves the infimal distance of $\frac{\sqrt{17}}{2}$ to any unit ball centred in $C$.


问:
Showing an LDU of a LU matrix is unique.
How do I show that every LU matrix A∈K$_n$$_×$$_n$ admits a unique LDU factorisation, that is a triple L∈L$_1$(Kn),D∈D(Kn) and U∈U$_1$(Kn) such that A=LDU? I'm new to these factorisations and I'm just trying to get to grips with the composition of each factorisation.

答: If a matrix $A$ has a $L_1U_1$ factorization it also has an $L_2D_2U_2$ factorization with $D_2U_2$ = $U_1$. $U_2$ is just the normalised version of $U_1$ where the diagonal is all one. $L_1$, $L_2$, $U_2$ are all unitriangular matrices.


问:
Proving that $||A-B||=||A+B||\Leftrightarrow AB=0$
I have to prove that

\begin{equation*}
||A-B||=||A+B||\Leftrightarrow AB=0
\end{equation*}

and I was wondering if this approach is correct, or if there's a better/more elegant way to prove this.

Given n-dimensional vectors A and B, we can write $||A-B||=||A+B||$ as:

\begin{equation*}
\sqrt{\sum\limits_{j=1}^{n}(a_j-b_j)^2}=\sqrt{\sum\limits_{j=1}^{n}(a_j+b_j)^2}.
\end{equation*}

Squaring both sides and expanding the binomials:

\begin{equation*}
\sum\limits_{j=1}^{n}a^2_j-2a_jb_j+b_j^2=\sum\limits_{j=1}^{n}a^2_j+2a_jb_j+b_j^2.
\end{equation*}

Simplifying:

\begin{equation*}
-\sum\limits_{j=1}^{n}a_jb_j=\sum\limits_{j=1}^{n}a_jb_j,~\text{which holds true if and only if}~\sum\limits_{j=1}^{n}a_jb_j=0.
\end{equation*}

Since $AB$ is equivalent to $\sum\limits_{j=1}^{n}a_jb_j$, then $||A-B||=||A+B||\Leftrightarrow AB=0$

Thanks in advance.

答:
What you've done is correct, but I think it's better to work without coordinates; just with the definition of norm in terms of the dot product:

$$
\| A \| = +\sqrt{A\cdot A} \ .
$$

Then you may observe that, since $\|A \| \geq 0$,

$$
\|A+B\| =  \|A -B\| \  \Longleftrightarrow  \  \|A +B\|^2  =  \|A-B\|^2 .
$$

Now, for instance, compute the difference

\begin{align}
 \|A +B\|^2  -  \|A-B\|^2   &= (A+B)\cdot (A+B) - (A-B)\cdot(A-B)    \\
               &= A\cdot A + A\cdot B + B\cdot A + \cdots
\end{align}

EDIT. I forgot to point out an obvious geometric interpretation of this result: if you draw a parallelogram with sides $A$ and $B$, then $A+B$ and $A-B$ are the diagonals of the parallelogram, right? These diagonals are equal if and only if...?

ContentLicense: CC BY-SA 2.5


问:
Cardinality of the set of all subsets of $E$ equipotent to $E$
I'm trying to prove the following statement (an exercise in Bourbaki's Set Theory):

If $E$ is an infinite set, the set of subsets of $E$ which are equipotent to $E$ is equipotent to $\mathfrak{P}(E)$.

As a hint, there is a reference to a proposition of the book, which reads:

Every infinite set $X$ has a partition $(X_\iota)_{\iota\in I}$ formed of countably infinite sets, the index set $I$ being equipotent to $X$.

I don't have any idea how that proposition might help.

If $E$ is countable, then a subset of $E$ is equipotent to $E$ iff it is infinite. But the set of all finite subsets of $E$ is equipotent to $E$. So its complement in $\mathfrak{P}(E)$ has to be equipotent to $\mathfrak{P}(E)$ by Cantor's theorem. Hence the statement is true if $E$ is countable. Unfortunately, I don't see a way to generalize this argument to uncountable $E$.

I'd be glad for a small hint to get me going.

答:
Another approach would make use of the fact that $\kappa_1 + \kappa_2 = \max(\kappa_1,\kappa_2)$ when $\kappa_1,\kappa_2$ are cardinals at least one of which is infinite. From this it follows that, for any subset of $S \subset X$, either $S$ or its complement has cardinality $|X|$. Since more straightforward cardinal arithmetic shows there are as many complementary pairs of subsets as there are subsets of $X$, we get the result.

By the way, JDH's approach could allow you to show the (slightly) stronger statement that there are $2^X$ sets $S \subset X$ with both of $S$ and $X - S$ equipotent to $X$. Simply write $X$ as a disjoint union $X = X_0 \cup X_1 \cup X_2$ with all parts equipotent to $X$ and consider subsets of the form $X_0 \cup A$ where $A \subset X_1$.

An interesting follow-up question might be to see if you can produce $2^X$ sets $S \subset X$ so that $|S|=|X|$, but $|X-S| < |X|$.