问:
Natural permutation representation definition?
I was wondering about a definition I encountered towards the beginning of Gruson-Serganova:

for G = $S_n$, and $k$ a field, and $s \in S_n$ and $(x_1, ..., x_n) \in k^n$, the natural permutation representation is defined as $\rho_s$ $(x_1, ..., x_n)$ = $(x_{s^{-1} (1)},...,x_{s^{-1}(n)})$.

I understand the permutation representation, but why do the inverses show up here? What makes this 'natural' if the permutation representation isn't? I asked my professor about why the inverses show up, and he said that otherwise it wouldn't be a homomorphism, but I see no reason why it wouldn't end up being a homomorphism.

I feel like I'm missing something pretty basic here.

答:
For a concrete example I refer to this answer I gave to a similar question. Here is an argument explaining why the definition with inverses is in fact the more natural one.

First of all, I would like to address of the relation  between permutations as values versus permutations as operations. Mathematically permutations of a set $X$ are maps $X\to X$, and hence logically compose as maps usually do, the one written on the right being applied first (unless one belongs to the school of people writing function applications with the functions to the right of their argument, which I will exclude here). While "permuting objects" has a clear informal sense as activity, this does not yet associate a particular permutation to each permutation action. It seems however natural that the action corresponding to a permutation of $X=\{1,\ldots,n\}$ that sends an element $i$ to $k$, takes the object at position $i$ and moves it to position $k$. I will therefore assume that "permuting by a permutation $\sigma$" means that for each $i$ the value initially at position $i$ ends up at position $\sigma(i)$. This rule has the natural order of composition: permuting by $\sigma\circ\tau$ can be achieved by first moving any value from position $i$ to position $\tau(i)$ (permuting by $\tau$) from where it then is moved to position $\sigma(\tau(i))$ (further permuting by $\sigma$).

If one writes that as a formula, the permutation is
$$
  (x_1,x_2,\ldots,x_n)\mapsto
  (x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},\ldots,x_{\sigma^{-1}(n)}).
$$
That may seem surprising at first. But the right hand expression has to what value arrives at each position; for the first position, the answer is "whatever value was moved to the first position", which is the value initially at position $\sigma^{-1}(1)$, in other words $x_{\sigma^{-1}(1)}$; a similar description goes for other output positions.

The $K$-linear action of $\mathbf S_n$ is a special case of this permutation action, on $n$-tuples of scalars in $K^n$. It is a morphism of groups $\mathbf S_n\to\mathbf{GL}(n,k)$ by virtue of respecting the natural order of composition, as explained. One can also characterise it by the fact that it sends the canonical basis vector $e_i$ to the basis vector $e_{\sigma(i)}$, as one would expect. One would hope that this obviously natural way of making the correspondence would inspire those who define the "permutation matrix of $\sigma$" to be the matrix for this action of $\sigma$; that matrix is $(\delta_{i,\sigma(j)})_{i,j=1}^n$, with $e_{\sigma(j)}$ as its column$~j$ for each$~j$. But this is unfortunately not the case. The Wikipedia article on Permutation Matrix is an example of a wrong choice made, while presenting this as one of two equivalent options, which it is not.

OwnerDisplayName: anon


问:
Why does $A^*A x = A^*b$ always have a solution even for rank-deficient A?
Suppose $A \in \mathbb{C}^{m \times n}$. I know that if $m \ge n$ and $A$ has full rank, then $A^*A$ is invertible so  $A^*A x = A^*b$ has a solution. However, looking at this question, I believe that this system has a solution even if $A$ is not of full rank, and even if $m \le n$.

I believe I have a solution using the SVD, but it is rather long and I'm sure that there is a better way. My question is: can you please give me a hint on how to prove this in an easier way?

Solution using SVD

Let $A^* = U \Sigma V^*$ be an SVD of $A^*$. That is $U, V$ are unitary and
$$\Sigma = \begin{bmatrix}
\sigma & 0\\
0^T & 0
\end{bmatrix}$$
where $\sigma = diag(\sigma_1, ..., \sigma_r)$ where $r$ is the rank of $A^*$.

The equation $A^*A x = A^*b$ becomes
$$U \Sigma \Sigma^TV^* x= U \Sigma V^* b$$
$$\Sigma(\Sigma^TV^* x- V^*b) = 0$$
$$\Sigma(\Sigma^Ty- V^*b) = 0$$
where $y = V^*x$. Since $V$ is invertible, we are free to select any $y$ that we want. Now,
$$\Sigma^Ty = [\sigma_1 y_1 \cdots \sigma_r y_r \ \ 0 \cdots 0]^T$$
Select $y_1, ..., y_r$ so that the first $r$ components of $\Sigma^Ty$ match the first $r$ components of $V^*b$.  Then the first $r$ components of  $\Sigma^Ty- V^*b$ are $0$, and as a result,
$$\Sigma(\Sigma^Ty- V^*b) = 0.$$

答:
Essentially, what you are trying to prove is that the column space of $A^*A$ (the set of vectors $y$ for which $A^*Ax = y$ has a solution) is equal to the column space of $A^*$ (the set of possible vectors $y = A^*b$). Because the column space of a matrix $M$ is equal to the orthogonal complement of the kernel of $M^*$ (the "left kernel" of $M$), it is equivalent to show that the kernel of $A^{**} = A$ is equal to the kernel of $(A^*A)^* = A^*A$.

It is easy to see that $\ker(A) \subseteq \ker(A^*A)$. Indeed, if $Ax = 0$, then
$$
A^*Ax = A^*(Ax) = A^*(0) = 0.
$$
On the other hand, we can see that $\ker(A) \supseteq \ker(A^*A)$ as follows. If $x$ is such that $A^*Ax = 0$, then it holds that
$$
0 = x^*0 = x^*A^*Ax = (Ax)^*Ax = \|Ax\|^2.
$$
That is, we have $\|Ax\| = 0$ and hence $Ax = 0$.

ContentLicense: CC BY-SA 4.0


问:
Is there simpler method to find $\sum_{k=1}^n(3k-2)^3$?
I have $\large\sum_{k=1}^n(3k-2)^3$. to evaluate it the normal approach is:

$$\sum_{k=1}^n(3k-2)^3=27\sum_{k=1}^nk^3-54\sum_{k=1}^nk^2+36\sum_{k=1}^nk-\sum_{k=1}^n8$$
And using the known formulas to evaluate $\sum_{k=1}^nk^3 , \sum_{k=1}^nk^2, \sum_{k=1}^nk$.

But I wonder is there other approach (preferably easier) to find this sums? I am thinking about writing $(3k-2)^3$ as $a_{k+1}-a_{k}$ then using telescoping series, but not sure how to do that.

答:
Hint Since $(n+1)^k-n^k$ is a polynomial in $n$ of degree $k-1$, you should expect that $a_k$ can be chosen to be polynomial of degree $3+1$.

So look for $a_k=ak^4+bk^3+ck^2+dk+e$.

Then,
$$
a_{k+1}-a_k=(3k-2)^3
$$
becomes a system of $4$ equations with 4 unknowns (the $e$'s cancel), which has unique solution. Solve it and you are done.

Note here that you can assume WLOG that $e=0$ from the beginning, since if $a_k$ is a solution then so is $a_k-e$.


问:
imaginary numbers - how can I understand them - especially as they occur in 'roots' of polynomials?
In another question here, about roots of equations being imaginary, the accepted answer
said something interesting about "imaginary" (as a technical word in math) not meaning "not real".  I understand that imaginary numbers are "manipulable and usable to do all kinds of things".   And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots.  In maxima, $\sqrt{-1}$ is represented as %i.

Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results:
[$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)?  (At first I thought it had reported the same root twice as I overlooked the negative sign.)

One can easily plot a real root and see how it intercepts the x axis.  How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

What uses do imaginary roots of polynomials have in math?

答: For Q2, You can imagine the absolute value of a polynomial in one complex variable, for a given point $z=x+iy$, as the height of a surface.</p>

ContentLicense: CC BY-SA 3.0


问:
Optimizing the number of robots to be used for a certain task
We have $N$ buttons numbered $1$,$2$ .. $N$. Our aim is to press each of the buttons at least once in the minimum time possible. For that purpose we choose $m$ robots. These robots can press one of the buttons in each second. But when two or more robots try to press the same button at the same time, that button won't be pressed (the robots will know that the button couldn't be pressed). Under the restriction that all the robots should be coded identically (i.e. they should execute the same instructions), what value of $m$ will you choose and how will you code them?

For example, if we choose $m=1$ and instruct the robot to press the $i$th button in the $i$th second, it will take a total of $N$ seconds before all the buttons are pressed. If we choose $m=2$ and instruct the robots to press button $1$ or $2$ uniformly at random until they have successfully pressed one of those(if one of them is successful in pressing one, then the other is automatically successful in pressing the other), and then sequentially press every other button (for example, if the robot succeeded in pressing button $1$, then it will try(successfully) to press the buttons numbered $3$, $5$,.. in the next seconds), then it will take an average time of $2+\frac{N-2}{2}$ seconds before each button is pressed.

答:
For small $N$, the details of the optimal algorithm seem likely to be complicated; they're probably best determined by dynamic programming. However, for large $N$, the following algorithm covers all buttons in $3$ seconds asymptotically almost surely.

Take a very large number of robots, i.e. $m\gg N$, and let them press any of the buttons with probability $p=1/m$ each, and no button with probability $1-Np=1-N/m$. (This last probability is non-negative since $m\gg N$.) The probability that a particular button is pressed exactly once, and thus the expected fraction of buttons that get pressed exactly once, is

$$
mp(1-p)^{m-1}=m\cdot\frac1m\left(1-\frac1m\right)^{m-1}=\left(1-\frac1m\right)^{m-1}\to_{m\to\infty}\frac1{\mathrm e}\;.
$$

For large $N$, the distribution is sharply peaked around this expected value, so the fraction of buttons successfully pressed is almost certain to be greater than $1/3$. Now let the robots that successfully pressed a button press further buttons according to some predefined scheme. (For instance, let the robots plan their actions by virtually acting in succession in the order of the numbers of the buttons they pressed in the first second, and let each robot press the button with the lowest number that hasn't been pressed either in a previous second or in a previous planning step.)


问:
Cardinality of the ranks of the constructible universe
First time I encounter the Constructible universe $L$ and the definition given in Jech is the following:

$L_0=\emptyset$,

$L_{\alpha+1}=\operatorname{def}(L_\alpha)$,

$L_\lambda=\bigcup_{\alpha<\lambda}L_\alpha$.

Where we have $\operatorname{def}(M)=\{X\subseteq M:\text{ X is definable in }(M,\in)\}$

My doubt is what's the cardinality of $L_{\omega+1}$. From the definitions it would seem it's countable since $L_{\omega}$ is countable, so is the number of $n$-tuples and so will all the finite sequences of elements of $L_{\omega}$. We also have the number of $\{\in\}$-formulas is countable so it would seem that $L_{\omega+1}$ must also be countable. Is this correct? If so since $L$ is a model of $ZFC$ which can prove $\mathcal{P}(\omega)\subseteq V_{\omega+1}$ it would seem the constructible universe thinks $L_{\omega+1}$ is uncountable. Is this also correct?

答: You are correct that $L_{\omega+1}$ is countable, but it's not correct that $L$ thinks it is uncountable.  Although $L$ satisfies $V=L$, it does not satisfy $V_\alpha=L_\alpha$ for each $\alpha$.  Every set in the $V$ hierarchy eventually shows up in the $L$ hierarchy, but typically at a later stage.  So $L_{\omega+1}$ does not contain all of the subsets of $\omega$ that are in $L$; in fact, you have to go all the way up to $L_{\omega_1^L}$ to capture all of the subsets of $\omega$ in $L$.


问:
Show that this spiral is a submanifold
I am asked to prove that the set
$$ N = \lbrace{(t \cos(1/t), t \sin(1/t)) \in \mathbb{R}^2 | t>0 \rbrace}$$
is a submanifold of $\mathbb{R}^2$.
My approach is to show that there exists a chart $(\mathbb{R}^2,h)$ of $\mathbb{R}^2$ for all $p \in N$ s.t.
$$ h(N \cap \mathbb{R}^2) = (\mathbb{R}^1 \times \lbrace{0 \rbrace}) \cap h(\mathbb{R}^2).$$
My idea was to use that $g \colon \mathbb{R}^2 \to \mathbb{R}^2,~~(x,y) \mapsto \left( x \cos\left(\frac{1}{x^2+y^2}\right),-y\sin\left(\frac{1}{x^2+y^2}\right)\right)$ is a homeomorphism that fulfills $g(N) = \mathbb{R}_{>0} \times \lbrace{0 \rbrace}$ and then use that as a chart map. With this however I get
$$ g(N \cap \mathbb{R}^2) = \mathbb{R}_{>0} \times \lbrace{0 \rbrace} \neq (\mathbb{R}^1 \times \lbrace{0 \rbrace}) \cap g(\mathbb{R}^2)$$
since $g(\mathbb{R}^2) = \mathbb{R}^2$. Can anyone give me a hint as to how I can fix this or tell me if my approach is wrong?

答:
For $x \gt 0$ and in a small enough open subset $W$ of $(x,0)$ the map defined by

$$g(u,v) = ((u+v)\cos 1/ u, (u+v) \sin 1/u)$$ is a $\mathcal C^1$ diffeomorphism of $W \subseteq \mathbb R^2$ onto $U=g(W)$. Moreover we have $g (W \cap ( \mathbb R \times \{0\})) = N \cap U$.

This proves that $N$ is a submanifold of dimension $1$.

Title: Show that this spiral is a submanifold


问:
Simple divisibility problem

Suppose $a$ and $b$ are distinct integers and $n\mid a^n-b^n$. Prove that $n\mid\frac{a^n-b^n}{a-b}$.

Here is how I do it: Write $a=b+d$, then $a^n-b^n=(b+d)^n-b^n=\binom{n}{1}b^{n-1}d+\binom{n}{2}b^{n-2}d^2+ \cdots + \binom{n}{n-1}bd^{n-1}+d^n \equiv 0 \pmod{n}$. Since $n| \binom{n}{k}$ for $k=1,2,\cdots,n-1$, $n|d^n$, which implies $n|d^{n-1}$. Therefore $(a^n-b^n)/(a-b)=\binom{n}{1}b^{n-1}+\binom{n}{2}b^{n-2}d+ \cdots + \binom{n}{n-1}bd^{n-2}+d^{n-1} \equiv 0 \pmod{n}$ and we are done. I wanna ask why $n|d^n$ will implies $n|d^{n-1}$ and that if there are any other possible solutions, thanks in advance.

答: Write down the unique prime factorization of  $n = p_1^{k_1} \cdot p_2^{k_2} \cdot \ldots \cdot p_l^{k_l}$. Then $n \mid d^n$ implies $p_1 \cdot p_2 \cdot \ldots \cdot p_l \mid d$. Since $k_i < n$ for all $i \in \{1, \ldots , l\}$, we have $n \mid p_1^{n-1} \cdot p_2^{n-1} \cdot \ldots \cdot p_l^{n-1}$ and with $p_1^{n-1} \cdot p_2^{n-1} \cdot \ldots \cdot p_l^{n-1} \mid d^{n-1}$ our desired result $n \mid d^{n-1}$.</p>

Title: Simple divisibility problem


问:
How to prove that $n^n=O(\log_2(2^{2^{2n}}))$?
How to prove that $n^n=O(\log_2(2^{2^{2n}}))$?

I cant find the the constant that will make it true.

Thanks for the help

答: @NN2 is right. Note that $\log_2\left(2^{2^{2n}}\right)=2^{2n}=4^n\in o(n^n)$. On the other hand,$$n^n=2^{n\log_2n}\ll2^{n^2}\ll2^{2^{kn}}=\log_2\left(2^{2^{2^{kn}}}\right),\,k>0,$$so we can fix the problem statement by replacing the innermost $2n$ with $2^n$ or $2^{2n}$. The latter effect is also achieved with an extra $2^{\cdot}$ operation around the argument of $\log_2$, or equivalently deleting the $\log_2$.


问:
Basic inequality involving modulus
If $0\leq a,b\leq 1$ and $\lvert a-b\rvert\leq r$ for some $r>0$, then $a\leq b+r$ and $a\geq b-r$?

By definition, they are equivalent.

答:
Let's think it through.

If $|M| \le r$ then $-r \le -|M| \le 0$.  Now $|M|= \begin{cases}M& M \ge 0\\-M& M < 0\end{cases}$ or in other words $M =\begin{cases}|M| & M \ge 0\\-|M|& M < 0\end{cases}$.

So if $M < 0$ then we have $-r\le M < 0$.  And if $M \ge 0$ then we have $0\le M < r$.  And, in either case, it will be true that $-r < M < r$.  So

Lemma: $|M|\le r\iff -r \le M \le r$

So we have $|a-b| \le r$.  SO that means $-r \le a-b \le r$.

And so $a-b \le r$ so $a \le r+b$.

And we have $-r \le a-b$ so $b-r \le a$ so $a \ge b-r$.

So, yep, they are both true.

The condition $0 \le a,b \le 1$ is not relevant.  The result would still be true if $a,b$ where any real values.

... oh, but I guess to show the statements are equivalent I have to show they go both ways.

If we given $a\le b+r$ and $a \ge b-r$ then we conclude $a-b \le r$ and $a-b \ge -r$ and $-r \ge a-b \le r$.  (Note: we don't need to be told $r \ge 0$.  these conditions are not possible if $r < 0$.  Oh but this could hold if $r =0$.)

If $a-b \ge 0$ then $|a-b| = a-b \le r$ and if $a-b < 0$ then $-r \ge a-b = -|a-b|$ and so $|a-b| \le r$.

So yep,

For any $a,b \in \mathbb R$ and for some $r \ge 0$ then:   $|a-b| \le r \iff a\le b+r$ and $a\ge b-r$

Title: Basic inequality involving modulus


问:
Non-deterministic finite automaton-transition function
When we define:

$δ:(Q \timesΣ\to2^Q)$

$2^Q$ represents the numbers of transitions or the resulting position after the automaton get letter?

答: $2^Q$ encodes the subset of $Q$ that are active after evaluating the input string. If there are states $Q_1, Q_2, Q_3, ...$ then examples of elements of $2^Q$ are $<yes, no, yes, yes, no,...>$ which can also be written $<1, 0, 1, 1, 0,...>$ where $yes$ or $1$ represents that state being active after the string is evaluated.


问:
Slope of a curved line
I just started with calculus and I came across the slope of a curve. According to definition the slope of a curve at a point is equal to the slope of tangent at that point. Since tangent is a straight line why not just find the tan of the angle of inclination of the straight line which would give us the slope of the line and eventually the slope of the curve at that point.</p>

答:
The purpose of calculus is to give rigorous definitions to geometric concepts such as tangent lines and areas. While we all have an intuitive understanding of what a tangent line is, if we can't formalise this intuition, then finding tangents boils down to making rough approximations. With calculus, we don't need to use any measurements to say that the graph of $y=x^3$ has a slope of exactly $6$ at the point $(\sqrt{2},\sqrt{2}^3)$.

Moreover, the 'intuitive' concept of a tangent line isn't all that convincing when you take a step back and think about it. In elementary geometry, we are taught that a tangent line only intersects the graph at a single point. But the graph of $y=\cos x$ has a tangent at $x=0$ that intersects the graph at infinitely many points!

There are other curves where drawing a tangent line by hand would be near impossible. Consider, for instance, the graph of $y=x\sin(1/x)$, where $x$ is measured in radians:

Tell me, what is the slope of the tangent line to the above curve at exactly $x=1/(40\pi)$? Without calculus, not only would you be unable to find the answer, but you might consider this question to be completely meaningless.

But with calculus, this question is completely meaningful, and we can say that the answer is precisely $-40\pi$. This is something that we can't figure out simply by extending the tangent line to touch the $x$-axis, and measuring the angle of inclination. And that's the power of calculus.

Title: Slope of a curved line


问:
What does point-wise convergence in the sup norm mean?
I am starting out with functional analysis and learning about point-wise and uniform convergence. In particular, I need help with my proof that the space $$ \mathcal{B} = \left\{ f:X\rightarrow \mathbb{R} \, \middle| \, f \, \text{bounded} \right\}$$
of bounded real functions on an arbitrary space $X$ is complete in the sup norm $$ \lVert f \rVert_{\infty} := \sup_{x\in X} \lvert f\left(x\right) \rvert. $$

My book (Intro to Topology and Modern Analysis by Simmons) poses this proof as a problem before any discussion on convergence of function sequences, so I will not use terms like "point-wise" or "uniform" convergence in the proof itself.

My Proof

Let $\left(f_n\right)_{n\in \mathbb{N}} \subset \mathcal{B}$ be a Cauchy sequence.

Letting $\epsilon >0$, there is an $N\in \mathbb{N}$ such that $$ m,n\geq N \implies \lVert f_m - f_n\rVert_{\infty} < \epsilon. $$ Thus, by definition of the sup norm we have for every $x \in X$, $$ m,n \geq N \implies \lvert f_m\left(x\right) -f_n\left(x\right) \rvert \leq \lVert f_m-f_n\rVert_{\infty} < \epsilon $$ This shows that, for fixed $x$, the real sequence $\left(f_m\left(x\right)\right)_{m\in \mathbb{N}}$ is Cauchy and converges since $\mathbb{R}$ is complete. Hence, we define $f:X\rightarrow \mathbb{R}$ by $$ f\left(x\right) = \lim_{m\rightarrow \infty} f_m\left(x\right). $$

So far we have merely constructed a function $f$ to which our sequence $\left(f_m\right)_{m\in \mathbb{N}}$ may converge (see discussion after proof); it remains to show that it does converge as a bona fide sequence in the sup norm.

To this end, let $\epsilon >0$ and take $N$ in the Cauchy hypothesis so that for any $x\in X$and $m,n> N$ we have $\lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}$. Then for this same $n>N$ we have $$ \lvert f\left(x\right) - f_n\left(x\right)\rvert = \lvert \lim_{m\rightarrow \infty} f_m\left(x\right) - f_n\left(x\right) \rvert = \lim_{m\rightarrow \infty} \lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}  $$
(pulling the limit out of the absolute value required a small side lemma). Thus, we have $$  n>N \implies \lVert f-f_n\rVert_{\infty} = \sup_{x\in X} \lvert f\left(x\right) -f_n\left(x\right)\rvert \leq \frac{\epsilon}{2} < \epsilon $$ so that $\left(f_n\right)_{n\in \mathbb{N}}$ converges to our $f$ in the sup norm.

Proving the boundedness of the limit function $f$ is a whole other can of worms which is irrelevant to my confusion, so I will avoid typing it out here. Once the limit function is bounded, we know that the Cauchy sequence converges in $\mathcal{B}$ and so $\mathcal{B}$ is complete.

My Confusion

I was happy with this proof until I learned more about point-wise versus uniform convergence. My proof shows that the sequence $\left(f_n\right)_{n\in\mathbb{N}}$ converges in the sup norm to the $f$ we constructed. In doing so, I see in retrospect that I also proved that this is uniform convergence. This uniform convergence seemed to come naturally out of the proof since we used the sup norm: we found an $N\in \mathbb{N}$ for which the $\epsilon$ property holds for all $x\in X$. However, I didn't set out to prove uniform convergence, only convergence in the norm we had defined on the function space, so it looks like this convergence in this norm might imply uniform convergence.

My question is: where does the notion of point-wise convergence enter in? The construction of my limit function $f$ is done in the usual "point-wise" manner, and so I assume that by construction I can say that "$\left(f_n\right)_{n \in \mathbb{N}}$ converges point-wise to $f$," but I am bothered by when I specialize to a point $x\in X$, my "point-wise convergence" ends up using only the standard norm on $\mathbb{R}$ and not the sup norm.

Hence, I can imagine we might have function sequences which converge "point-wise" but not uniformly; how are we justified in saying that they "converge," then, since they won't converge as sequences in $\mathcal{B}$, where the norm is the sup norm?

答:
It seems to me that you are overthinking this problem. You proved that, for every $\varepsilon>0$, there is some $N\in\Bbb >N$ such that if $n\geqslant N$ and if $x\in X$, then$$\sup_{x\in X}\bigl|f(x)-f_n(x)\bigr|<\varepsilon;$$in other words, you proved that, for every $\varepsilon>0$, there is some $N\in\Bbb >N$ such that if $n\geqslant N$ and if $x\in X$, then $\|f-f_n\|_\infty<\varepsilon$, which is what you were supposed to proved.

Concerning your remark that “using only the standard norm on R and not the sup norm”, well… what's the sup norm on $\Bbb R$. On $\Bbb R^n$, you can define$$\|(x_1,\ldots,x_n\|_{\text{sup}}=\sup\{|x_1|,|x_2|,\ldots,|x_n|\}=\max\{|x_1|,|x_2|,\ldots,|x_n|\},$$but, when $n=1$, this is just the usual distance in $\Bbb R$. Besides, on $\Bbb R^n$ all norms are equivalent.


问:
How can I get the slope of a tangent to a circle from a point always oriented in the same way?
Given a circle (defined through its center $(x_c,y_c)$ and radius $r$) and an external point $(x_0, y_0)$ , I am looking for the heading of one of the two tangents (I know how to select the one I need case-by-case) such that if I write the tangent line in the parametric form

$$ x(t) = x_0 + t \cdot \sin(heading) $$
$$ y(t) = y_0 +t \cdot \cos(heading) \, , $$
I want the tangent point to be in the range $t<0$.

NOTE: the parametric equation is given to explain the desired orientation of the heading. I need the heading defined in that way.

I have read this answer and it mostly works fine, but there are some cases in my application where the function given provides $heading +\pi$:

$$vec = [r:\mp\sqrt{(y_0 - y_c)^2+(x_0 - x_c)^2-r^2}:-\sqrt{(y_0 - y_c)^2+(x_0 - x_c)^2}] \begin{bmatrix}(y_0 - y_c)&(x_0 - x_c)&-x_c(y_0 - y_c)-y_c(x_0 - x_c)\\ -(x_0 - x_c)&(y_0 - y_c)&x_c(x_0 - x_c)-y_c(y_0 - y_c) \\ 0&0&r\sqrt{(y_0 - y_c)^2+(x_0 - x_c)^2}\end{bmatrix}$$

$$heading = \arctan_2(vec(1),vec(2))$$

Question: how can I fix this function so that I always obtain the desired value for the heading?

答:
Expanding on my comment, there is a circle with centre $C$ and radius $r$, an external point $P$, and two tangent points $T_+$ (to the right from $P$, anti-clockwise around $C$) and $T_-$ (to the left from $P$, clockwise around $C$). (Note: we are using the standard coordinate system with $\theta = 0$ starting at the $x$-axis and increasing anti-clockwise.) You want to quantify the direction of travel $\alpha_\pm$ heading away from $T_\pm$.

First we find the tangent points
$$
T_\pm = C + \underbrace{\frac{r}{\Vert P - C \Vert}}_{\text{scale}} \cdot \underbrace{R(\pm\theta)}_{\text{rotate}} \cdot (P - C),
$$
where
$$
R(\pm\theta) = \begin{bmatrix} \cos\theta & \mp\sin\theta \\ \pm\sin\theta & \cos\theta \end{bmatrix}
$$
is the general rotation matrix and
$$
\cos\theta = \frac{r}{\Vert P - C \Vert}, \quad \sin\theta = \sqrt{1 - \cos^2 \theta}, \quad \left(0 \le \theta \le \frac{\pi}{2}\right)
$$
by observing the right triangle $C T_+ P$. Then the required heading is
$$
\alpha_\pm = \text{atan2}(P - T_{\pm}).
$$

We can also make the formula more compact. Let
$$
\lambda = \tan\theta = \frac{\sin\theta}{\cos\theta} = \sqrt{\frac{\Vert P - C \Vert^2}{r^2} - 1},
$$
then some algebraic manipulation gives
$$
\alpha_\pm = \text{atan2}\left( \begin{bmatrix} \lambda & \mp 1 \\ \pm 1 & \lambda \end{bmatrix} (P - C) \right).
$$


问:
How can I integrate $\int 5\sqrt{5+4x-x^2} \, dx$ using substitution?
I know it is easier if we go through the $\sqrt{x^2+9}$ but I do not know how to integrate if we use $u=5+4x-x^2$ ? Or can we not use substitution like that?

答:
The first thing I would do is "complete the square".  $x^2+ 4x+ 5= x^2+ 4x+ 4+ 1= (x+ 2)^2+ 1$ so $5\int \sqrt{x^2+ 4x+ 5}dx= 5\int\sqrt{(x+ 2)^2+ 1}dx$.

Let u= x+ 2 so du= dx and the integral becomes $5\int \sqrt{u^2+ 1}du$.

Now use a trig or hyperbolic substitution.


答:
At a given point $P (x,y)$ you have an associated slope

$$  y'=\frac{dy}{dx} =\tan \phi$$

You refer to inclination of the straight line that you have not as yet  established.

Title: Slope of a curved line

Body: I just started with calculus and I came across the slope of a curve. According to definition the slope of a curve at a point is equal to the slope of tangent at that point. Since tangent is a straight line why not just find the tan of the angle of inclination of the straight line which would give us the slope of the line and eventually the slope of the curve at that point.</p>


问:
Time, Speed and Distance : Returning back to pick up problem
$A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$

I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.

Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.

Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$

Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!

Please help me out.
Answer Provided : 4.8 hours

答:
I have another solution that I must add because it shows the power of reframing the problem geometrically.

Let's construct a cartesian coordinate system containing order pairs of time and position, which we'll refer to as "points". There are four points of interest: O, the starting point; U, the u-turn point; R, the rendezvous point; and F, the finishing point. Note that all of these points refer to both a time and position. As the travelers are moving from point to point with constant velocity, their paths through the cartesian coordinate system can be represented by line segments. In the diagram, the line segments are labeled to refer to who takes the path; elsewhere, I will refer to the line stretching from some point P to some point Q as $\overline{PQ}$.

The big revelation of this diagram is that $\overline{OR}$ parallels $\overline{UF}$ and $\overline{OU}$ parallels $\overline{RF}$, as  the traveler(s) on each pair of paths are moving with the same velocity. That means quadrilateral $OUFR$ is a parallelogram. It is not hard to show by a similarity argument that the immediate consequence is $\overline{OR}$ is congruent to $\overline{UF}$ and $\overline{OU}$ is congruent to $\overline{RF}$.

Translating this geometric result back into the original language, we find some results that may have not been immediately obvious:

Traveler B takes the same amount of time and covers the same amount of distance to arrive at the finish line after boarding the tonga as does C to get from the starting point to the rendezvous with A.

Travelers A and B take the same amount of time and cover the same amount of distance to arrive at the U-turn point after leaving the starting point as do A and C to drive to the finish line after their rendezvous.

This is the same observation explicitly stated by Jaap Scherphuis, now made obvious if it were not initially. With this insight in mind, let's solve the problem of finding the finish time. We can immediately construct a system of equations in terms of $(t_P,x_P)$, the time and position at a point P with reference to O.

\begin{array}{lcl} x_U+x_R & = & 120 \\ x_U & = & 50t_U
\\ x_R & = & 10t_R \\ x_R-x_U & = & -50(t_R-t_U)\end{array}

The first equation says the total distance is 120km and the remaining equations are the speeds (slopes) along each path. The solution is $(t_U,x_U)=(9/5,90)$ and $(t_R,x_R)=(3,30)$. The total time is the sum of $t_U$ and $t_R$, which is 4.8 hours.


问:
How to convert an einsum expression to an equation?
I'm using Python's einsum for an operation over two arrays, and I was wondering how to correctly write this operation out in a paper.

Let's say I have two '3d' arrays $A_{i,j,k}$  and $B_{i,k,l}$, and I want a sum-product over index $k$, an outer product over indices $j$ and $l$, and I want to do that for every $i$, i.e. index $i$ is "matched" between $A$ and $B$. I would write that as einsum('ijk,ikl->ijl',A,B).

Is there an index-based mathematical notation that would let me write all these operations at once? I have looked up actual Einstein summation, but it seems to be defined more narrowly, e.g. for a sum-product but not outer product, and not in a way you could represent "for every $i$...". Specifically for the index $i$, it seems that once an index is repeated, it must be summed over, which I don't want.

Is there a set of rules for writing out derivatives with respect to arrays using the same index notation?   E.g. of outer and inner products, repeated operations, etc?

答:
Inner (sum-product) and outer product in index notation works exactly as in einsum (it worth noting that the notation and the name einsum come from Einstein notation common in tensor algebra). For example, for tensors $A_{ij}$ and $B_{ij}$:

$$
A_{ij}B_{jk} = C_{ik}
$$

Inner product on index $j$ and outer product for everything else.

However, element-wise operation (for matrices, it's called Hadamard product) is not a tensor operation (the result isn't generally a tensor), so there is no widely used notation for that. However, you are not prohibited to invent it yourself:

$$
A_{ijk}\circ_i B_{ikl} = C_{ijl}
$$

(I am once again note that multi-dimensional matrices $A,B,C$ are not tensors from the maths perspective)

Another approach is to separate indices that follow Einstein rule (e.g. lating $i,j,k$) from indices that are just indices (e.g. $\alpha,\beta,\gamma$):
$$
A_{\alpha jk} B_{\alpha kl} = C_{\alpha jl}
$$

In that case $A_{\alpha ij}$ can be tensor for each particular value of $\alpha$


问:
Why can't we write $\frac{d}{dy}(\frac{dx}{dy})=\frac{dy(\frac{dx}{dy})-dx\frac{dy}{dy}}{(dy)^2}$
I know solving $\frac{d}{dy}(\frac{dx}{dy})=\frac{d}{dy}(\frac{1}{\frac{dy}{dx}})
  =    \frac{-\frac{d}{dy}(\frac{dy}{dx})}{(\frac{dy}{dx})^2} = - \frac{d^2y}{dx^2}(\frac{dy}{dx})^{-3}$

But I like to know why its wrong to write   $\frac{d}{dy}(\frac{dx}{dy})=\frac{dy(\frac{dx}{dy})-dx\frac{dy}{dy}}{(dy)^2}$

Thanks.

答:
I do not know why you would write like that. But my guess would be you are treating $\frac{dx}{dy}$ as a ratio and are trying to use the quotient rule.

$$(\frac{u}{v})'=\frac{vu'-uv'}{v^2}$$

Which is fine, Except... you will not get what you wrote. You will instead get

$$ \frac{dy(\frac{d}{dy}(dx))-dx(\frac{d}{dy}(dy))}{(dy)^2} $$


问:
Condition for positive semi-definite or positive definite matrix, $A Q^{-1}A^T$
Suppose we have a positive definite matrix $Q \in \mathbb{R}^{n\times n}$, are there general conditions on $A \in \mathbb{R}^{m\times n}$ such that $A Q^{-1}A^T$ is positive semi-definite ($A Q^{-1}A^T \succeq 0$)? Or positive definite ($A Q^{-1}A^T \succ 0$)?

Extension: Suppose there are no general conditions on $A$ such that $A Q^{-1}A^T$ can be made positive semi-definite or positive definite given that $Q$ is positive definite. By further assuming that $Q$ is symmetric, i.e., $Q = Q^T$, are there any conditions on $A$ such that $A Q^{-1}A^T$ can be made either positive semi-definite or positive definite?

答:
If $Q$ is positive semidefinite, then $AQ^{-1}A^T$ will always be positive semidefinite.

If $Q$ is positive definite, then it can be written in the form
$$
Q = H + K,
$$
where $H = \frac 12(Q + Q^T)$ is symmetric and positive definite and $K = \frac 12(Q - Q^T)$ is skew-symmetric. Because $H$ is positive definite, there exists an invertible matrix $P$ such that $S = PP^T$, and we have
$$
P^{-1}QP^{-T} = P^{-1}HP^{-T} + P^{-1}KP^{-T} = I + P^{-1}KP^{-T}
$$
where $I$ denotes the identity matrix. Note in particular that $J:= P^{-1}KP^{-T}$ is a skew-symmetric matrix. Note that $Q^{-1}$ is positive definite if and only if the matrix
$$
P^TQ^{-1}P = [P^{-1}QP^{-T}]^{-1} = [I + J]^{-1}
$$
is positive definite. Thus, we have reduced the question to that of whether/when $[I + J]^{-1}$ is positive definite for a skew-symmetric $J$.

Note that a matrix $M$ is positive definite if and only if the symmetric matrix $M + M^T$ is positive semidefinite. Taking $M = [I + J]^{-1}$, we have
$$
M + M^T = [I + J]^{-1} + [I + J]^{-T} = [I + J]^{-1} + [I - J]^{-1}.
$$
I claim that this can be rewritten as
$$
[I + J]^{-1} + [I - J]^{-1} = 2[(I + J)(I - J)]^{-1} = 2 (I - J^2)^{-1}.
$$
Now, we note that $I - J^2$ is necessarily positive definite because it is the sum of a positive definite matrix $I$ and a positive semidefinite matrix $-J^2 = JJ^T$. Thus, $(I - J^2)^{-1}$ and hence $(I + J)^{-1}$ is positive definite.


问:
Principal Value of a divergent integral over $\mathfrak{R}$
I am trying to solve/prove this Principal Value integral
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x=\  -\ i\pi a^2\cdot\cos\left(a\right)\ ,$$
where $a$ takes a finite value.

The way I try to prove this is by noticing there pole is a first order pole at $x=a$, then by calculating the Residue at this value I find that
$$ \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x =-i\pi \cdot Res\left(\dfrac{x^2\cdot \sin\left(x\right)}{x-a},a\right)= $$

$$ =-i\pi\cdot\lim_{x\rightarrow a}\left(x^2 \cdot \sin\left(x\right)\right) = -i\pi a^2 \cdot {\sin(a)}\ ,$$
which of course is not correct.

Hence, does anyone knows what I am doing wrong, and if yes could please explain it to me?

NOTE: Just to mention that i end up to this question because a made use of Green's function for quantum many-body system. So if anyone have a clue about this they might be able to help.

Thanks in advance.

*******   EDIT:

I tried to keep it short, but the truth is that I need to prove this
$$-\dfrac{1}{a^3} \mathscr{P}\int_{-\infty}^{+\infty}\dfrac{x^3\cdot F(x)}{x-a}\dfrac{{\rm{d}}x}{2\pi} = G(a)\ ,$$
where

$$ F(x) = \frac{3}{2}\left[(1-y)\cdot \frac{\sin(x)}{x} + (1-3y)\cdot\left( \frac{cos(x)}{{x^2}} - \frac{sin(x)}{x^3}\right) \right]$$
and
$$ G(a) = \frac{3}{4}\left[-(1-y)\cdot \frac{\cos(a)}{a} + (1-3y)\cdot\left( \frac{sin(a)}{{a^2}} + \frac{cos(a)}{a^3}\right) \right],$$
where $y=const$.

Thanks again for the effort.

答:
Integral $I=P.V.\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(x\right)}{x-a}{\rm{d}}x$ makes sense as the function acting in the space of smooth and finite function (in the sense of tempered distribution).
https://ncatlab.org/nlab/show/tempered+distribution#:~:text=Formally%2C%20a%20tempered%20distribution%20is,%F0%9D%92%AE%E2%80%B2(%E2%84%9Dn).

Lets evaluate
$I(k)=\int_{-\infty}^{+\infty}\dfrac{x^2\cdot \sin\left(kx\right)}{x-a}{\rm{d}}x=\Im{J}(k) $, where the integral is considered in the sense of principal value.

$$J(k)=\int_{-\infty}^{+\infty}\dfrac{x^2e^{ikx}}{x-a}{\rm{d}}x=\int_{-\infty}^{+\infty}\dfrac{(t+a)^2e^{ik(t+a)}}{t}{\rm{d}}t=e^{ika}\int_{-\infty}^{+\infty}e^{ikt}\Bigl(\dfrac{a^2}{t}+2a+t\Bigr)dt$$

$$\int_{-\infty}^{+\infty}\frac{e^{ikt}}{t}=\pi{i}\,sgn(k)$$

$$\int_{-\infty}^{+\infty}e^{ikt}=2\pi\delta(k)$$
$$\int_{-\infty}^{+\infty}te^{ikt}=-2\pi{i}\frac{d}{dk}\delta(k)$$

All these have meaning in the sense of distribution, i.e. if we consider $\int_{-\infty}^{+\infty}J(k)f(k)dk$, where $f(k)$ is a finite function with smooth derivatives:

$$\int_{-\infty}^{+\infty}\delta(k)f(k)dk=f(0)$$
$$\int_{-\infty}^{+\infty}f(k)\Bigl(\frac{d}{dk}\delta(k)\Bigr)dk=-\frac{d}{dk}f(k)|_{k=0}=-f'(0)$$

Therefore,
$$J(k)=\pi{i}{a}^2e^{ika}\,sgn(k)+4\pi{a}e^{ika}\delta(k)-2\pi{i}e^{ika}\frac{d}{dk}\delta(k)=$$
$$I(k)=\Im{J}(k)=\pi{a^2}\cos(ka)sgn(k)+4\pi{a}\sin(ka)\delta(k)(=0)-2\pi\cos(ka)\frac{d}{dk}\delta(k)$$
The last term should be understood in the following way:

$$-2\pi\int_{-\infty}^{+\infty}f(k)\cos(ka)\Bigl(\frac{d}{dk}\delta(k)\Bigr)dk=2\pi\int_{-\infty}^{+\infty}\frac{d}{dk}\Bigl(f(k)\cos(ka)\Bigr)\delta(k)dk=$$ $$=2\pi{f'}(0)-a\lim_{k\to0}f(k)\sin(ak)=2\pi{f'}(0)$$

if the function $f(k)$ (as supposed) is smooth and does not have singularities.

Finally
$$I(k)=\Im{J}(k)=\pi{a^2}\cos(ka)sgn(k)-2\pi\frac{d}{dk}\delta(k)$$
where the second term makes sense only in the sense of tempered distribution.

We can put $k=1$ and (formally) get $I(1)=\pi{a^2}\cos(a)$, but the meaning of $I(1)$ depends on what we are going to do with it further.


问:
Doubts: All ideals of $M_n(R)$ are of the form $M_n(I)$, where $I$ is an ideal of $R$.
I know the result is true, however, if we are to consider $E_{ij} = \left\{\begin{pmatrix} 0 & 0 & ... & 0\\
. & . & ...& . \\
0 & 0 & r_{ij} & 0 \\
. & . & ...& . \\ \end{pmatrix} : r_{ij}\in R  \right\}$.

$E_{ij}$ is an ideal of $M_n(R)$ but doesn't appear to be of the type $M_n(I)$.

Can someone please explain what I am doing wrong here?

答:
Your set is not closed by multiplication on the right.

You can multiply a nonzero element like that by a matrix with a 1 in the $j,1$ position and zeros elsewhere, and that is not in your set.


问:
Let $\alpha\in[0,9].$ Is the set of real numbers such that the average of the digits of each of these numbers $=\alpha,$ countable or uncountable?
Let $\alpha \in [0,9]\subset\mathbb{R}.$ Is the following set countable or uncountable:

The set of all (nonnegative) real numbers which, when written in decimal expansion form

$$k_1\ldots k_m\cdot k_{m+1}\ldots$$

where the $k_i$ are digits between $0$ and $9$, satisfy

$$\lim_{n\to\infty}\left( \frac{\sum_{j=1}^n k_j}{n} \right) = \alpha ?$$

答:
If you are willing to allow a little heavy machinery, one can show that the set is uncountable for every $\alpha$, even just restricting to the set of numbers of the form $0.x_1 x_2 x_3 \dots$.  In fact the argument below will give more precise information than just uncountability.

Setup.  Rather than work with the averages of the digits $\frac{1}{n} \sum_{j=1}^{k} x_j$ directly, we will instead work with the frequencies of occurrences of specific digits, and make the connection at the end.  If $x = 0.x_1 x_2 \dots$ and $r \in \{0,1,2,\dots, 9\}$, let us define
$$
D_x(r) = \lim_{n \to \infty} \frac{1}{n} \#\{1 \leq j \leq n : x_j = r\}
$$
whenever the limit exists.

Black box theorem.  Let $p = (p_0, p_1, \dots, p_9)$ be a probability vector over the set $\{0,1,2,\dots, 9\}$ (that is, all $p_i \geq 0$ and $\sum p_i = 1$).  Then the Hausdorff dimension of the set
$$
\{x \in [0,1] : D_x(r) \text{ exists and is equal to } p_r \text{ for each } r \in \{0,1,2,\dots,9\} \}
$$
is given by $\frac{H(p)}{\log 10}$, where $H(p) = \sum_{i=1}^{10} -p_i \log p_i$ is the Shannon entropy of the vector $p$.  The relevant fact about $H$ that we'll need later is that $H(p) > 0$ whenever at least two of the $p_i$ are $>0$.

Essentially this theorem specifies the "size" (in the sense of Hausdorff dimension) of the set of numbers that have specified frequencies of digits in their base 10 expansions.  It was proved originally by Patrick Billingsley, I think.  I can't find an exact reference right now but I'd be happy to look a little more if you are curious.  The proof uses a little ergodic theory and a little geometric measure theory (hence the aforementioned "heavy machinery").

Connection. Fix $\alpha \in [0,9]$.  There are two cases, depending on if $\alpha \in \mathbb{Z}$ or not.  If $\alpha \in \mathbb{Z}$, let $x = 0.x_1 x_2 \dots$ be any number satisfying $D_x(\alpha - 1) = D_x(\alpha + 1) = 1/2$.  Then
$$
\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} x_j = \lim_{n \to \infty} \sum_{r=0}^9 r \cdot \frac{1}{n} \#\{1 \leq j \leq n : x_j = r \} = \sum_{r=0}^9 r \cdot D_x(r) = (\alpha - 1) \cdot \frac12 + (\alpha + 1) \cdot \frac12 = \alpha.
$$
By Billingsley's theorem, the set of $x$ satisfying this has strictly positive Hausdorff dimension, which implies that it is uncountable.

Finally, consider the case where $\alpha$ is not an integer.  Then write $\alpha = A + t$ where $A \in \mathbb{Z}$ and $t \in (0,1)$.  Then we let $x = 0.x_1x_2 \dots$ be any number satisfying $D_x(A) = 1-t$, $D_x(A+1) = t$.  A similar calculation as above will then show that any such $x$ satisfies $\frac{1}{n} \sum_{j=1}^{n} x_j = A+t = \alpha$, and again applying Billingsley's theorem shows that the set of such $x$ has positive Hausdorff dimension and is therefore uncountable.


问:
addition with a variable (mod)
Given $2+x \equiv 7 \pmod 3$.

$2 + 0 = 2$

$2 + 1 = 3$

$2 + 2 = 4$

.
.
.

$2 + 5 = 7$

so, the answer will be $x = 5, 8, 11, 14, 17,\dots$

Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$

答:
We are given that:

$(2+x) = 7 (\mod 3)$

Thus, $2+x-7 = 3k$ where $k$ is an integer.

Simplifying, yields:

$x-5 = 3k$

Therefore, $x\in (5,8,..)$

Title: addition with a variable (mod)


问:
Is there an equation for the number of iterations required to escape the Mandelbrot Set's bounding circle?
The Mandelbrot Set is the set of all complex values $c$ that cause the recurrence relation $z_n = z_{n-1} + c$ to remain bounded within a circle of radius 2 when iterated from $z_0 = 0$. In general, there is no way to prove that a given value of $c$ is a member of the set. Instead, we iterate the recurrence relation some large number of times until we either find some value of $n$ where $|z_n| > 2$ (proving that the current value of $c$ is not in the set) or give up and declare that the current value of $c$ is probably part of the set (or at least very close to it).

When the value of $c$ is far from the set, $z_n$ quickly escapes the bounding circle. As $c$ gets closer to the set, the number of iterations $n$ required to exceed the bounding circle grows without bound. If we look at values of $c$ on the positive real axis only and plot how many iterations it takes to exceed the bounding circle, we get a graph like this:

If $c$ is greater than two, it only takes one iteration to escape the circle because $z_0$ is always zero and $z_1$ is always $c$. When $1 < c ≤ 2$ it takes two iterations. The next twelve transition values are approximately 0.68233, 0.5352, 0.4537, 0.4034, 0.36999, 0.34659, 0.32952, 0.31666, 0.30672, 0.29887, 0.29256, and 0.28741 — that's as far as I can get with WolframAlpha. Positive real values of $c$ less than or equal to 0.25 are part of the Mandelbrot Set and will never escape even after infinitely many iterations.

To me, that looks very much like a graph of $1/x$ with some scaling and translating (for example, the vertical asymptote is at $x = 0.25$), but I can't find an equation that fits perfectly.

Is there a formula that fits these points exactly?

UPDATE based on responses from PM 2Ring and Claude

As the distance $\epsilon$ from the boundary of the set tends towards zero on the positive real axis, the number of iterations required for escape tends towards $\frac{π} {\sqrt{ϵ}}$. As $ϵ$ increases to 1.75 along the positive real axis, the relationship changes to $\frac{2\sqrt{1.75}} {\sqrt{ϵ}}$. It seems like I need some function $v(ϵ)$ such that $\frac{v(ϵ)} {\sqrt{ϵ}}$ fits my points exactly. Based on the points I have, it looks like this:

UPDATE 2

I calculated the iteration counts for $c = -0.75 + ϵ i$ and discovered that approaching the boundary of the set from this direction is more closely approximated by $\frac{π}{ϵ}$ than by $\frac{π}{\sqrt{ϵ}}$. As in the first approach, the actual iteration count starts out far from the function and approaches it as $ϵ$ goes to zero, though it wobbles around more than it does when approaching from the positive real axis. It seems my hope for a single function to calculate iteration counts may go unfulfilled.

答:
Mark McClure's comment references a nice historical overview by G. Edgar.

The $c = -\frac{3}{4} \pm \epsilon i$ and $c = \frac{1}{4} + \epsilon$ relationships were found by D. Boll.

The $c = \frac{1}{4} + \epsilon$ relationship is proven by A. Klebanoff:

The $\pi$-Theorem

Choose $\epsilon>0$ and let $N(\epsilon)$ be the number of iterations required for the orbit of zero, under the map $Q_{\frac{1}{4}+\epsilon}(x)=x^2+\frac{1}{4}+\epsilon$, to exceed $2$, i.e. $$N(\epsilon) = \min_n Q_{\frac{1}{4}+\epsilon}(0)>2$$ Then $$\lim_{\epsilon \to 0^+} \sqrt{\epsilon}N(\epsilon) = \pi$$

-- "$\pi$ in the Mandelbrot set" by Aaron Klebanoff, Fractals, Vol. 9, No. 4 (2001) 393-402 https://doi.org/10.1142/S0218348X01000828

A. Klebanoff's paper also mentions Jay Hill's $\pi$-result for $-\frac{5}{4}-\epsilon^2 \pm \epsilon i$.  G. Edgar's page says Jay Hill also mentioned a $\pi$-result for $-\frac{7}{4} + f(\epsilon)$ for some unspecified $f$.  Another point to investigate with an easy route is $-2 - \epsilon$, though I suspect the relationship here will not involve $\pi$.

Tabulating:

$c(\epsilon)$
$N(\epsilon)$

$\frac{1}{4} + \epsilon$
$\frac{\pi}{\sqrt{\epsilon}}$

$-\frac{3}{4} \pm \epsilon i$
$\frac{\pi}{\epsilon}$

$-\frac{5}{4} - \epsilon^2 \pm \epsilon i$
$\frac{\pi}{2 \epsilon}$

$-\frac{7}{4} + f(\epsilon)$
?

$-2 - \epsilon$
?

All of these are parabolic apart from the last, which is a Misiurewicz point with aymptotically geometric scaling.


问:
Related rates of change involving trig functions

Question: A train is moving along a straight track at
$75 \text{km h}^{-1}$ heading east. A camera is positioned $2$km north of the train track and is focused on the train.

a) Find the rate of change of the distance between the camera and the train when the train is $4$km from the camera.

b) Find the rate at which the camera is rotating when the train is 4km from the camera. Give your answer in degrees per second correct to the nearest tenth of a degree.

This is a question that has caused my extension maths group a bit of confusion! We have solved part a of the question and we believe the answer is $65$km/h (2sf). This is correct.
Part b is the one we are unsure about. We have set up the problem using a couple of variables and applied various differentiation techniques. The answers we are getting, however, are varied and even our teacher has said that he needs a bit more time to work out which one is correct! I am quite eager to find out whether my technique is correct. The different answers we have got are:

$2.6 \times 10^{-3}$ radians/second (then we can convert this to degrees per second as the question asks)

$2.6$ degrees/second

I appreciate that by putting this question out there, I may generate different responses, so I will wait to see if a several people start getting the same answer before accepting if it is correct.

Best wishes!

答:
Firstly, $75 \text{km h}^{-1} = \frac1{48} \text{km s}^{-1}$

Draw a right angled triangle with the triangle 'pointing' right.
Label the top vertex $A$, and then $B$ and $C$ will follow in clockwise fashion.

$AC = 2$km. The rate of change of $BC$ is $\frac1{48}  \text{km s}^{-1}$ and $BC$ is $\frac1{48} t $ km (I assume this comes from speed = distance*time?).

Let $\angle BAC = \theta$.

Then $\tan(\theta) = \frac1{96}t$ and, differentiating both sides with respect to $t$, we get $\frac1{cos^2\theta}\frac {d\theta}{dx} = 1/96$

We can know divide both sides and work out an expression.
We can also work out the value of $\cos \theta$ by plugging in that the hypotenuse is $4$ and so work out $\cos^2 \theta$. Thus, we can plug this value into our expression for $\frac {d\theta}{dt}$, which gives them $1/4 \times 1/96 = 1/384 = 2.6$ degrees per second (?)

Sorry I don't have the complete working.


问:
What does it mean for a stationary distribution to have imaginary values?
Let's say I have a transition matrix P
$$
\begin{bmatrix}
0.4 & 0.6 & 0. & 0.&  0.\\
 0.5 &0. & 0.5 &0. & 0. \\
 0.6 &0. & 0. & 0.4 &0. \\
 0.7 &0. & 0. & 0. & 0.3\\
 1. &  0.&  0.&  0.&  0.
\end{bmatrix}
$$

The stationary distribution of this matrix, which describes a real-world process, has imaginary values. The Perron-Frobenius eigenvalue is 1. I know for advection processes or processes where stuff stays in the system forever i.e. $\lambda=1$, we deal with imaginary values because those don't disappear. However, I don't know enough about transition matrices to know if the same thing is going on here.

Also, if I wanted to use the stationary distribution as a probability distribution in order to calculate something in expectation, but I end up having imaginary values in the stationary distribution, were I to calculate my expected value as normal, can I still treat it as a proper expectation or is it now meaningless?

Thanks.

答:
Here's some code to compute the eigenvector using the standard scipy linear algebra library:

```<code>import numpy as np
import scipy.linalg as la

mat = [[0.4, 0.6, 0., 0.,  0],
 [0.5, 0.,  0.5, 0., 0.],
 [0.6, 0.,  0.,  0.4, 0.],
 [0.7, 0.,  0.,  0.,  0.3],
 [1.,  0.,  0.,  0.,  0. ]]
print(la.eig(np.transpose(mat))[1][:,0])
</code>```

The output:

```<code>[0.8259962 +0.j 0.49559772+0.j 0.24779886+0.j 0.09911954+0.j
 0.02973586+0.j]
</code>```

The eigenvectors of the matrix are the columns of the second array returned, not the rows.

Body:
Here's some code to compute the eigenvector using the standard scipy linear algebra library:

import numpy as np
import scipy.linalg as la

mat = [[0.4, 0.6, 0., 0.,  0],
 [0.5, 0.,  0.5, 0., 0.],
 [0.6, 0.,  0.,  0.4, 0.],
 [0.7, 0.,  0.,  0.,  0.3],
 [1.,  0.,  0.,  0.,  0. ]]
print(la.eig(np.transpose(mat))[1][:,0])

The output:

[0.8259962 +0.j 0.49559772+0.j 0.24779886+0.j 0.09911954+0.j
 0.02973586+0.j]

The eigenvectors of the matrix are the columns of the second array returned, not the rows.


问:
When can we end up on the same seat?
You and your friend  sit at the  opposite side of  a round table with 10 seats. You  play a seat moving game by  throwing in turn three dices of blue, green and red simultaneously. $(x, y, z)$  represent the outcome of each throw.

Rules:

Blue: Move CW if $x$ is even, otherwise CCW.

Green: Move to the $y^{th}$ seat(s) next to you.

Red:  Reset back to original seat if $z$ is 1.

Question:

On average after how many throws both of you will sit on the same chair?

答:
This can be modelled as finite-state, absorbing Markov chain.
Let us number the chairs from $0$ to $9$, with you originally in chair $0$ and your friend in chair $5$.  A state is of the form $(y, f, t)$ where $y$ is the number of the chair you are sitting in, $f$ is the number of the chair your friend is sitting in, and $t$ indicates whose turn it is, say $0$ if it your turn, and $1$ if it is your friend's turn.  The initial state is $(0,5,0)$.

All states with $y\neq f$ are transient, and we collapse all the states with $y=f$ into a single absorbing state.  That gives $180$ transient states and one absorbing state, so we have to deal with a $180\times180$ matrix, and hand-calculation is impossible.  It shouldn't be difficult to do this with a computer program, though.  In each state, there are $216$ rolls of the dice, but they have essentially the same effect in every each state, depending on whose turn it is.  You need to number the states so that you can relate them to the matrix.

EDIT

Sorry. I misunderstood what you said about there being no memory.  Of course you are right.  It's just a matter of figuring out what the states are.  Then you can compute the average time to absorption in the usual way.

Title: When can we end up on the same seat?


问:
area to the right of $r\cos\theta=6$ and $r=10$
I have been trying to solve this question for a while and I seem to get stuck although it does not seem hard Given $r\cos\theta=6$ and $r=10$ we need to find the area inside the circle and on the right of $r\cos\theta=6$.
I am self-learning calculus with analytic geometry so  I don't know if my ways are correct.

first I took $r=10$ and made it $x^2+y^2=100$ the center is $(0,0)$ and radius is $10$ and also $r\cos\theta=6$ to $x=6$, I found where they intersect and got $6^2+y^2=100$ and we get $y=8$ so the intersection point is $(6,8)$.
since we know that we have $x=6$ and the radius of the circle is $10$ we can use Pythagorean theory to find the area of a triangle inside the circle, we get that the area of the triangle is $0.5\times8\times4=16$.

which leaves a small area between the triangle and the circle which I cannot seem to find. I am still not sure how to find the needed $\theta$ for the integral and I am not sure how to use it.

The final answer according to the book should be $100\left(\cos^{-1}\left(\frac3{5}\right)\right)-48$ thanks for any tips and help!

答:
Let $A_\Delta=\frac{1}{2} \cdot 6 \cdot 8 = 24=$ area of triangle
and $A_\text{circle}=100\pi$ be the area of the circle.

The area of the circle to the left of the line is thus

$$A_\text{L} = 100\left(\pi - \cos^{-1} \frac{3}{5} \right) + 2 A_\Delta$$

The area to the right is

$$A_\text{circle} - A_\text{L}  = 100 \cos^{-1}\frac{3}{5} - 48.$$


问:
Is it possible to isolate p in $"a = 2bpq + 2apq + a(p^2) + 2aqq + bqq"?$
Is it possible to simplify this equation so that $p$ is isolated on one side instead of $a$? I tried factoring out $p$ on the right side, but I get stuck with the $ap^2$.

$$a = 2bpq + 2apq + ap^2 + 2aq^2 + bq^2$$

Body:
By solving p in this equation you get:

$p = \frac{\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$

or

$p = \frac{-\sqrt{a^2(-q)^2+a^2+abq^2+b^2q^2} -aq - bq}{a}$

EDIT:

Steps as requested:

$ap^2+2apq+2bpq+2aq^2+bq^2=a$

$p^2 + \frac{p(2aq+2bq)}{a }+ \frac{2aq^2+bq^2}{a} = 1$

$p^2 + \frac{p(2aq+2bq)}{a} = 1 - \frac{2aq^2+bq^2}{a}$

$p^2 + \frac{p(2aq+2bq)}{a} + \frac{(2aq+2bq)^2}{4a^2} = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$

$(p + \frac{(2aq+2bq)}{2a})^2 = 1 + \frac{(2aq+2bq)^2}{4a^2} - \frac{2aq^2+bq^2}{a}$

Can you take it from here?


问:
Comprehensive text on Wreath Products
I am trying to learn about wreath products in the context of group theory. I was following Rotman's book on the subject but I find it very difficult to follow and he makes some mistakes (for instance here).

Is there a concise and relatively simple text on this topic? I'm looking for a text which actually provides some intuition if possible.

答:
Hopefully this is still interesting to you after 3 years, but I think I can give a graphical approach to wreathproducts that may help to give an intuition. I'm not too sure if this is completely correct, so read with caution:

Let $M$, $N$ be Monoids and $X$ an set. Further let $M^X \rtimes N$ be the semidirect product.
Then there is $M^X = \{f \mid f: X \to M\}$ with $f, g \in M^X$ and $m, m' \in M$. Last there is $n,n' \in N$.

We now give an graphical interpretation of the "first" part of the tupel:

$$(f, n)(g,n') = (f+ng,nn')$$
If we apply this to an $x$ we get
$$(f(x)+n\cdot g(x), nn') = (m + g(x'), nn') = (m + m', nn') = (\overline{m}, nn')$$

Title: Comprehensive text on Wreath Products


答: If the stationary distribution has imaginary values, then there is no stationary distribution. Any Markov chain's state transition matrix will have a complex eigenvector with corresponding eigenvalue 1, but not every Markov chain has a stationary distribution.</p>

Title: What does it mean for a stationary distribution to have imaginary values?

Body:
Let's say I have a transition matrix P
$$
\begin{bmatrix}
0.4 & 0.6 & 0. & 0.&  0.\\
 0.5 &0. & 0.5 &0. & 0. \\
 0.6 &0. & 0. & 0.4 &0. \\
 0.7 &0. & 0. & 0. & 0.3\\
 1. &  0.&  0.&  0.&  0.
\end{bmatrix}
$$

The stationary distribution of this matrix, which describes a real-world process, has imaginary values. The Perron-Frobenius eigenvalue is 1. I know for advection processes or processes where stuff stays in the system forever i.e. $\lambda=1$, we deal with imaginary values because those don't disappear. However, I don't know enough about transition matrices to know if the same thing is going on here.

Also, if I wanted to use the stationary distribution as a probability distribution in order to calculate something in expectation, but I end up having imaginary values in the stationary distribution, were I to calculate my expected value as normal, can I still treat it as a proper expectation or is it now meaningless?

Thanks.


问:
Is there simpler method to find $\sum_{k=1}^n(3k-2)^3$?
I have $\large\sum_{k=1}^n(3k-2)^3$. to evaluate it the normal approach is:

$$\sum_{k=1}^n(3k-2)^3=27\sum_{k=1}^nk^3-54\sum_{k=1}^nk^2+36\sum_{k=1}^nk-\sum_{k=1}^n8$$
And using the known formulas to evaluate $\sum_{k=1}^nk^3 , \sum_{k=1}^nk^2, \sum_{k=1}^nk$.

But I wonder is there other approach (preferably easier) to find this sums? I am thinking about writing $(3k-2)^3$ as $a_{k+1}-a_{k}$ then using telescoping series, but not sure how to do that.

答:
Maybe not the most simple approach but elementary and generalising the original problem.

Defining the generating function

$$g(t)=\sum_{k=1}^{n} e^{t (k-2)}\tag{1}$$

we can obtain sums of the type

$$s(p) = \sum_{k=1}^{n} (k-2)^p\tag{2}$$

by the p-fold derivative with respect to $t$ as

$$s(p) = \frac{1}{p!}\left(\frac{d}{dt}\right)^p g(t) |_{t\to 0}\tag{3}$$

Now we can calculate the finite geometric sum over $k$ in $g$ to get

$$g(t)= \frac{e^{2 t}-e^{(2-3 n) t}}{e^{3 t}-1}\tag{4}$$

The Taylor expansion of $g$ at $t=0$ starts linke this

$$\begin{align}
g(t)\simeq
& n + \frac{1}{2} \left(n-3 n^2\right) t+\frac{1}{4} \left(6 n^3-3 n^2-n\right) t^2\\
& +\frac{1}{24} \left(-27 n^4+18 n^3+9 n^2-4 n\right) t^3\\
& +\frac{1}{240} \left(162 n^5-135 n^4-90 n^3+60 n^2+13 n\right) t^4\\
& +\frac{1}{480} \left(-162 n^6+162 n^5+135 n^4-120 n^3-39 n^2+20 n\right) t^5+O(t^6)\end{align}\tag{5}$$

Now we can easily find the first few sums $s(p)$ from $(5)$.


问:
Why does this double integral have the value zero?
Why does the following double integral have the value zero (no calculation needed)
$$\iint_Dx^8y^7\,dxdy=0$$ where $D={x^2+y^2 \le 2, x \ge0}$.

Does this have anything to do with symmetry, odd and even functions? Can anyone help me with how to think and understand integrals more, especially how to use symmetry to facilitate integral calculations.

Any useful links or videos would be appreciated as well-

答:
Because the map $y \mapsto y^7$ is odd and the domain $D$ symmetric with regards to the $y$-axis.

In general for an odd map, $\int_{-a}^a f(x) \ dx=0$ for $a \gt 0$.


问:
How is the property of having no infinitely large or small elements derived from the original statement of the archimedean property?
There is already a question asked related to this.

Here's the link:Does this definition of archimedean property guarantee that a set has no infinitely large or infinitely small elements?

That question asks that, does his/her statement guarantees the property of having no infinitely large or small elements. But there is no proper answer and I  also don't know is his/her derived statement truly guarantees the property mentioned in the wikipedia page of the Archimedean property:

Roughly speaking, it is the property of having no infinitely large or
infinitely small elements

My question is kind of a generalized version of that question.

From the original statement

$\forall x,y \in \mathbb{R}_{+}$,
$\exists n \in \mathbb{N}:nx>y$

How to derive the property of having no infinitely large or infinitely small element?

(I am not a mathematician so please don't used too advanced mathematics in the answers)

答:
My statement is based on a simpler version of the Archimedean property, which states $\forall y \in \mathbb{R} \exists n \in \mathbb{N} (n > y)$. This is easy to prove from OP's definition: we note that $|y| + 1 \in \mathbb{R}_+$ and $1 \in \mathbb{R}_+$. Then, we take some $n \in \mathbb{N}$ such that $n \cdot 1 > |y| + 1$. Then $n \cdot 1 > |y| + 1 > |y| \geq y$. Then $n > y$.

What is an "infinitely large number?" We need to define this term precisely before we can answer your question.

A sensible definition of "$x$ is an infinitely large number" is $\forall n \in \mathbb{N} (|x| > n)$. Obviously, the Archimedean property immediately refutes the existence of an infinitely large $x$,  since it states that there exists $n \in \mathbb{N}$ such that $|x| < n$.

What about an "infinitely small number"? I would define "x is infinitely small" to be the statement $|x| > 0 \land \forall n \in \mathbb{N} (|x| < \frac{1}{n})$.

There can be no infinitely small $x$. For if there were such an $x$, take some $n$ such that $n > \frac{1}{|x|}$. In that case, we would have $\frac{1}{n} < |x|$.


问:
Interesting behavior of the $2$-tag system $\{a\rightarrow abc,b\rightarrow ab,c\rightarrow c\}$
Consider the following $2$-tag system:

$$\begin{cases}a\rightarrow abc\\b\rightarrow ab\\c\rightarrow c\end{cases}$$

In brief, a $2$-tag system is a Turing-complete string substitution system wherein you check the current first letter against the rules given, append the corresponding letters (e.g. $a$ maps to $abc$) to the end of the string, and then remove the first two letters from the string. Repeat until string disappears or cycle is detected.

I assert that for any starting string $(abc)^{k}$, this system will always cycle back to $(abc)^k$.

e.g. for $k=1$:

```<code>abc
  cabc
    bcc
      cab
        bc
          ab
            abc
</code>```

I am fairly certain this is true but can't prove it.

If anyone can provide a proof, counterexample, or persuasive argument for or against its validity, I'll consider this answered.

I also suspect there's no easy way to calculate the period of the cycle and would welcome input on that too. For reference, the first 25 periods (courtesy of r.e.s.) are:

$\{6, 24, 78, 96, 63, 1030, 6558, 3936, 5922, 75756, 40474, 1594320, 79630, 7166907, 837353,\\1311364, 9620370, 6835198, 577674197, 2092070640, 784506057, 1116341675, 282429536478, 11933733565, 169161110854\}$

My gut sense is that this is not really a Collatz-level problem; the fact that it's in a fairly well-behaved bounded cycle suggests to me that by gluing a few of these rules together properly, it should be provable that $(abc)^k$ necessarily repeats. That said, my gut is often bad at number theory.

Some observations

The number of $b$'s is fixed at $k$, and they're always preceded by $a$, unless it's a $b$ at the start of the string. At each end of the string, and between every $b$ and $a$, there may be between $0$ and $2$ $c$'s, and the way they float around and interact with parity seems like what drives the cycles. Those constraints can be confirmed by establishing that, bottom line, the only unique substrings you can generate within this system are $\{ab,abc,abcc\}$. All possible strings reachable from $(abc)^k$ will be some combination of those pasted together, possibly with a wraparound truncation to start. You can confirm this via inspection or with the "table method" described in De Mol (2010).

This provides a strict lower bound of $2k/3$ for the length of the shortest word in the cycle (note that $(ab)^k$ always appears), with $4k/3$ as the upper bound on word length. We can take a stab at an upper bound on period by treating $b$ as dividing the string into $k+1$ partitions, each of which contains $0$ to $2$ $c$'s, yielding at most
$3^{k+1}$ unique configurations. Since the combination of leading and trailing $c$'s can't exceed two, we eliminate the possibilities $\{ccScc,cScc,ccSc\}$. This leaves us with an upper bound of $$f(k) := 3^{k+1}-3.$$

This seems to work well for small $k$ and provides the exact period for $k \in \{1,2,3,7,12,23\}$, which is as high as I can confirm.

答:
We write $U = ab, V = abc, W = abcc$.

Definition:

A nonempty string will be called "regular" if it is the concatenation of finitely many copies of $U, V, W$.

If $X$ is any regular string, then we write $X'$ for the string obtained from $X$ by removing the leading $a$. Such a string will be called "quasi-regular".

A "normal" string is a string that is regular or quasi-regular.

The system can then be translated to a system on normal strings.

More precisely, we have the following rules (*):
\begin{eqnarray}
UX &\rightarrow& XV\\
VX &\rightarrow& X'W\\
WX &\rightarrow& XW\\
U'X &\rightarrow& X'U\\
V'X & \rightarrow& XU\\
W'X & \rightarrow& X'V
\end{eqnarray} where $X$ denotes any regular string.

Note that this rule is injective: from the resulting string, we can deduce what the original string is.

Definition:

The "weight" of a regular string is the total number of $U, V, W$ in the string.

The weight of a quasi-regular string is the weight of the regular string it comes from.

(Equivalently, the weight of any normal string is just the number of $b$'s appearing in it.)

It is obvious from the above rules that the weight of a normal string does not change after applying the rules.

Therefore, starting from any normal string, the generated system can have only finitely many possible strings.

Since the procedure continues forever, the system is eventually periodic.

However, the rule being injective, the system must be purely periodic, hence will eventually return to the initial status.

This works for any normal string. In particular, it works for the string $V^k$.

(*)There is an exception when $X$ is the empty string, which we exclude here.


答: Because for every point (x,y) in your D, there exists another point (x,-y) for which your integral would yield the exact negative value. So everything gets canceled out.</p>

Title: Why does this double integral have the value zero?

Body:
Why does the following double integral have the value zero (no calculation needed)
$$\iint_Dx^8y^7\,dxdy=0$$ where $D={x^2+y^2 \le 2, x \ge0}$.

Does this have anything to do with symmetry, odd and even functions? Can anyone help me with how to think and understand integrals more, especially how to use symmetry to facilitate integral calculations.

Any useful links or videos would be appreciated as well-


问:
More random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.

Find the pdf of $W$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_W(w)= P(W \le w)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Let $A$ be the anti-clockwise neighbour and $C$ clockwise.

Question:

Well, I hope to find the pdf of $W$ via its cdf, which I think I'm able to compute if I know what $W$ is. Is it $W = \min\{B-A,A-C\}$? Or is it $W=\min\{Z,U\}$, where $Z$ is the distance from $B$ to $A$ (anti-clockwise of course) and $U$ is the distance from $B$ to $C$ (clockwise of course; which I think is equivalent to saying 'distance from $C$ to $B$ anti-clockwise') ? Or something else? (Note: I asked about $Z$ here, but I hope I made this post self-contained.)

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
Let $X$ be the anticlockwise distance along the circle's circumference from $B$ to $A$.
Let $Y$ be the anticlockwise distance from $B$ to $C$.

Since the three points are independently uniformly distributed along the circumference of the circle, $X$ and $Y$ are iid variables with uniform distributions on $[0,1).$

Since the clockwise arc from $B$ to $A$ is $1 - X,$
the distance from $B$ to $A$ is the smaller of $X$ and $1-X$.
That is, if we let $X_m = \min\{X, 1 - X\}$,
then $X_m$ is the distance from $B$ to $A$.

Similarly, let $Y_m = \min\{Y, 1 - Y\}$,
then $Y_m$ is the distance from $B$ to $C$.

By symmetry, $X_m \in [0,\frac12]$ and $X_m$ is uniformly distributed
(glossing over the fact that the zero-probability events $X_m=0$ and $X_m=\frac12$ can each happen in only one way).
Likewise, $Y_m$ also is uniform on $[0,\frac12]$.
Moreover, $X_m$ and $Y_m$ are iid.

The nearest neighbor of $B$ is $A$ if $X_m < Y_m$, but is $C$ if $X_m > Y_m$.
So $W = \min\{ X_m, Y_m \}.$

So we want the pdf (or cdf) of the minimum of two iid variables uniformly distributed on $[0,\frac12]$.
For $0 \leq w \leq \frac12$ we have

\begin{align}
P(W < w) &= 1 - P(W \geq w)\\
& = 1 - P(X_m \geq w) P(Y_m \geq w) \\
& = 1 - \left(2\left(\frac 12 - w\right)\right)^2
\end{align}

with $P(W < 0) = 0$ and $P(W < 1) = 1.$
The cdf is therefore
$$ F_W(w) = 1 - (1 - 2w)^2. $$

The pdf is easily found from this.

Title: More random points on a circle


问:
Zero-Divisors & Units in $\mathbb{Z}/n\mathbb{Z}$
Let $n\in\mathbb{Z}_{\geq2}$. Considering the ring $\mathbb{Z}/n\mathbb{Z}$, we denote $(\mathbb{Z}/n\mathbb{Z})^*$ to be its ring of units. Furthermore, the map $\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z};\,a\mapsto\overline{a}$ is meant to be the canonical projection modulo $n$.

claim 1: $\overline{a}\in\mathbb{Z}/n\mathbb{Z}$ zero-divisor $\iff$ $1<\gcd(a,n)<n$.

I am wondering if claim 1 is true or not. I was able to prove $''\implies''$ using Bézout as follows:

$$\gcd(a,n)=1\stackrel{\text{Bézout}}\iff\exists x,y\in\mathbb{Z}: ax+ny=1\iff ax\equiv 1\mod n \iff \overline{a}\in(\mathbb{Z}/n\mathbb{Z})^*.$$
So in particular $\overline{a}$ is not a zero-divisor.

If claim 1 is true, would it then imply the following proposition?

claim 2: $\overline{a}\in\mathbb{Z}/n\mathbb{Z}$ zero-divisor $\iff \overline{a}\notin(\mathbb{Z}/n\mathbb{Z})^*$

答:
Let $R$ be a finite ring. Then every non-zero element of $R$ is either a zero-divisor or a unit, but not both.

Proof: suppose that $a$ is a zero-divisor. Then clearly, $a$ cannot be a unit. For if $ab = 1$, and if we have $c \neq 0$ such that $ca = 0$, then we would have $cab = c1 = c = 0$. This is a contradiction.

On the other hand, suppose $a$ is not a zero-divisor. Then the map $x \mapsto a \cdot x$ is injective. For suppose we have $ax = ay$; then $a(x - y) = 0$; then $x - y = 0$; then $x = y$. Since this map is injective and its domain and codomain are both $R$, which is a finite set, the map must be surjective. Then there must exist $x$ such that $a \cdot x = 1$.

Thus, for $\overline{a} \neq 0$, we have $\overline{a}$ zero-divisor iff $\overline{a} \notin (\mathbb{Z} / n \mathbb{Z})^*$.

Now note that $\overline{a} \in (\mathbb{Z} / n \mathbb{Z})^*$ iff $gcd(a, n) = 1$. Clearly, if we have $\overline{a} \overline{x} = 1$, then $ax = 1 + bn$ for some $b \in \mathbb{Z}$, and thus $ax - bn = 1$, and thus $gcd(a, n) = 1$. And conversely, if $gcd(a, n) = 1$, then we have some $x, b$ such that $ax + bn = 1$. Then $ax = 1 - bn$. Then $\overline{a} \overline{x} = 1$.

Then $\overline{a}$ is a zero-divisor iff ($\overline{a}$ is not a unit and $\overline{a}$ is not zero) iff ($gcd(a, n) \neq 1$ and $gcd(a, n) \neq n$) iff $1 < gcd(a, n) < n$.


Title: addition with a variable (mod)

Body:
Given $2+x \equiv 7 \pmod 3$.

$2 + 0 = 2$

$2 + 1 = 3$

$2 + 2 = 4$

.
.
.

$2 + 5 = 7$

so, the answer will be $x = 5, 8, 11, 14, 17,\dots$

Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$

Body:
Hint: $\ x\equiv 7-2 \equiv \color{#c00}5 \pmod 3 = \color{#c00}5 + 3\,\Bbb Z = \{\,\ldots,-4, -1, 2, \color{#c00}5, 8, 11,14,\ldots\}$

Or: $\ x\equiv 5\pmod{3}\!\iff\! 3\mid x\!-\!5\!\iff\! 3n = x\!-\!5 \!\iff\! x = 5\! +\! 3n,\ $ for some $\, n\in \Bbb Z$


答:
I'll illustrate a general solution to the problem since you seem to be enthusiastic about this topic. Below is a diagram of the situation:

$x$ is the distance of the train to the perpendicular, $d$ is the distance to the observer, $v$ is the train's speed (constant), $s$ is the distance of the train to the observer, and $\theta$ is the angle the train makes with the perpendicular, some number in the open interval $(-\pi/2,\pi/2)$. I'll use dots to denote time derivatives. We know that the train moves at speed $v$, in other words
$$\dot{x}=v$$
We also know, from Pythagoras, that
$$x^2+d^2=s^2$$
So,
$$2s\dot{s}=2x\dot{x}\implies \dot{s}=\frac{x\dot{x}}{s}=\frac{x\dot{x}}{\sqrt{x^2+d^2}}=\frac{xv}{\sqrt{x^2+d^2}}$$
So, if $d=2\text{ km}$, $v=75\text{ km h}^{-1}$ and $x=\pm 4\text{ km}$ we get
$$\dot{s}=\frac{\pm 4\cdot 75}{\sqrt{4^2+2^2}}\text{ km h}^{-1}=\frac{\pm 4}{\sqrt{20}}\cdot 75\text{ km h}^{-1}\approx 67\text{ km h}^{-1}$$

Onto the second part. For this part I'll take positive $x$ on the right. With trigonometry we see that
$$\tan(\theta)=\frac{x}{d}\implies \theta=\arctan\left(\frac{x}{d}\right)$$
Taking time derivatives of both sides we get
$$\sec^2(\theta)\dot{\theta}=\frac{\dot{x}}{d}$$
Now using our expression for $\theta$ and the fact that $\dot{x}=v$,
$$\dot{\theta}=\frac{v}{d~\sec^2\left(\arctan\left(\frac{x}{d}\right)\right)}=\frac{v}{d\left(\frac{x^2}{d^2}+1\right)}=\frac{vd}{x^2+d^2}=\frac{vd}{s^2}$$
I'll leave it as an exercise to prove that
$$\sec(\arctan(t))=\sqrt{t^2+1}$$
Using $v=75\text{ km h}^{-1}$, $s=4\text{ km}$, and $d=2\text{ km}$ we get
$$\dot{\theta}=\frac{75\cdot 2}{4^2}\text{ radians h}^{-1}\approx 0.149\text{ degrees s}^{-1}$$

I don't think either of the provided answers are correct.

Title: Related rates of change involving trig functions

Body:
Question: A train is moving along a straight track at
$75 \text{km h}^{-1}$ heading east. A camera is positioned $2$km north of the train track and is focused on the train.

a) Find the rate of change of the distance between the camera and the train when the train is $4$km from the camera.

b) Find the rate at which the camera is rotating when the train is 4km from the camera. Give your answer in degrees per second correct to the nearest tenth of a degree.

This is a question that has caused my extension maths group a bit of confusion! We have solved part a of the question and we believe the answer is $65$km/h (2sf). This is correct.
Part b is the one we are unsure about. We have set up the problem using a couple of variables and applied various differentiation techniques. The answers we are getting, however, are varied and even our teacher has said that he needs a bit more time to work out which one is correct! I am quite eager to find out whether my technique is correct. The different answers we have got are:

$2.6 \times 10^{-3}$ radians/second (then we can convert this to degrees per second as the question asks)

$2.6$ degrees/second

I appreciate that by putting this question out there, I may generate different responses, so I will wait to see if a several people start getting the same answer before accepting if it is correct.

Best wishes!


问:
What is the coefficient of the second term in $ xyz+(x+1)(y)+4x$?
The definition I found for a coefficient is 'the number multiplied by a variable.'
For the expression
$$
xyz + \left(x+1\right)\left(y\right) + 4x
$$
are both the factors $\left(x + 1\right)$ and $y$ considered variables?

In this case, I think the coefficient of the second term is $1$. Am I right $?$.

答:
Typically, when people order the terms (and refer to "the second term" or "the first term") they are ordering the terms from highest degree to lowest degree. To do this, you should expand the expression and look at the degrees of the terms. The variables here are $x$, $y$, and $z$. Note that the degree of the term is the sum of the exponents of the variables (in polynomials) or, in simpler words, the number of variables (or copies) being multiplied together. You expression expands to this:
$xyz+xy+y+4x$
Where the terms are $xyz$ with degree 3, $xy$ with degree 2, $y$ with degree 1, and $4x$ with degree 1. If we want the second highest degree, we would be talking about the term $xy$ which, as you said, has a coefficient of $1$.


问:
Incremental Addition?
I am not sure why I have become so puzzled by this, but I am pretty sure I am completely overcomplicating everything here.

The equation (or set of equations) I am trying to make has a starting price, ending price, multiplier, and a quantity. I am trying to get the ending price only.

So if I give it a price of 40, a quantity of 4 and a multiplier of 0.0001 it would return an ending price of 40.0006

(40 + 40.0001 + 40.0002 + 40.0003) = 160.0006

It starts at 40 because that is my price, it does it 4 times because that is my quantity, and it goes up 0.0001 each time because that is my multiplier.

I have tried summation formulas, but it never seems to end up right

答:
$\underbrace{start + (start + mult) + (start + 2\cdot mult) + ..... +(start + \{quant-1\}\cdot mult)}_{quant \text{ times}}=$

$\underbrace{start + start + .... + start}_{quant\text{ times}} + mult(0+ 1+2+3+...... +\{quant-1\}) =$

$quant\cdot start + mult\cdot \sum_{k=1}^{quant-1} k$

....

Now there is a well known trick that everyone remembers where they where when they first learned it[1] (and everyone is secretly envious of the kid who aged less than ten years old who figured it on his own in a matter of a half a minute) that

$1 + 2 + 3+ ..... +n = \frac {n(n+1)}2$.

So:  You want:

$quant\cdot start + mult\cdot \frac {(quant -1)\cdot quant}2$

.....

[1] It's the mathematician's equivalence of JFK....

JFK?... He was a president of the United States a while back?....  Apparently he died and it was big deal for the people at the time... I dunno, I don't remember him either.

Title: Incremental Addition?


问:
The topology of a $k$-dimensional ball in $\mathbb R^N$ and the definition of attaching a $k$-cell
Let $B_k$ be the $k$-dimensional ball in $\mathbb R^N$:
$$
B_k = \left\{x \in \mathbb R^N \ : \ \sum_{i = 1}^k x_i^2 < 1, \ x_{k + 1} = \ldots = x_N = 0 \right\}.
$$
In the context of Morse Theory, it is called a $k$-cell.

The operation of attaching a $k$-cell to a topological space $Y$ is defined in my book (Morse Index of Solutions of Nonlinear Elliptic Equations, Damascelli and Pacella, 2019) as follows:

Take a continuous function $g: \overline{\partial B_k} \to Y$ which is homeomorphic to its image. The topological space obtained by taking the union of $Y$ and the closed $k$-cell $\overline{B_k}$ with the equivalence relation which identifies $x \in \overline{\partial B_k}$ with its image $g(x) \in Y$ will be denoted by $Y \cup_g B_k$ ($Y$ with a $k$-cell attached).

Milnor, in his classical book, takes $g$ defined on $S^{k - 1}$

My questions are:

What is $\overline{\partial B_k}$, and why do we take it as the domain of $g$?

For the first part, as I understand, $\overline{\partial B_k} = \overline{B_k}$. Indeed, the boundary is obtained by taking the closure and excluding the interior. But the interior of $\overline{B_k}$ is empty. What am I missing?

What does it mean to take the equivalence relation $x \sim g(x)$ in $Y \cup_g B_k$? Intuitively, what are we doing?

I'm quite clueless about this.

答:
If that is all what the authors say, it is poor notation.

One should emphasize that $B_k$ is topologically an open ball. But note that it is not an open subset of $\mathbb R^N$ unless $N = k$. Anyway, the closure $\overline{B_k}$ in $\mathbb R^N$ is always topologically a closed $k$-dimensional ball.

They authors do not define $\partial B_k$.

For $N > k$ the interpretation of $\partial B_k$ as the topological boundary of $B_k$ in $\mathbb R^N$ produces in fact $\partial B_k = \overline{B_k}$. Only for $N = k$ we get $\partial B_k = \overline{B_k} \setminus B_k$.

An interpretation of $\partial B_k$ which depends on $N$ does not make sense. Therefore the only reasonable interpretation is $\partial B_k = \overline{B_k} \setminus B_k$. But then $\partial B_k$ is closed and $ \overline{\partial B_k} = \partial B_k$. So it remains open why the authors write $\overline{\partial B_k}$. Anyway, $\partial B_k \approx S^{k-1}$. This agrees with Milnor.

The notation $Y \cup_g B_k$ is unusual because $g$ is defined on $\partial B_k$ which is disjoint from $B_k$. The standard notation is $Y \cup_g  \overline{B_k}$


问:
Rank of $AA^T$ if $A$ is a full rank matrix
Assume we have $A \in \mathbb{R}^{m \times n}$, $rg(A) = m$. Will $\det(AA^T)$ be non-zero? Same question is for $\det(\Phi^T\Phi)$, where $\Phi \in \mathbb{R}^{n \times m}, rg(\Phi) = m$.

答:
Yes, it is true: if $A$ is $m \times n$ with rank $m$, then $\det(AA^T)$ will be non-zero. Similarly, if $A$ is $m \times n$ with rank $n$, then $\det(A^TA)$ will be non-zero.

There are a few ways you could go about proving this. One is to use the fact that
$$
\operatorname{rank}(A) = \operatorname{rank}(A^TA) = \operatorname{rank}(AA^T),
$$
and a square matrix will have non-zero determinant if and only if its rank is equal to its size. Alternatively, you could use the Cauchy Binet formula to express either determinant as a sum of squares.


问:
What is the Idea of a Relative Open Set?
My real-analysis text gave the following defintion:

Let U be a subset of E.  U is open relative to E if

for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.

Although the idea that U is open in $\mathbb R$ follows the definition,

I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.

U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.

Intuitively, each point, t, is contained in a "bubble".

So, what is the significance of specifying the intersection?

Apparently, I can't think of a relative open set in terms like interior points and "bubbles".

答:
As you said, a subset $U$ of $\mathbb{R}$ is open, if for every point $x$ of $U$, there is some $\epsilon>0$ such that $N_\epsilon(x) \subset U$, which means "every point of $\mathbb{R}$ that is closer to $x$ than $\epsilon$ lies in $U$".

Now we want to talk more generally about open subsets of $E$, where $E \subset \mathbb{R}$, but using the same distance function; i.e. keeping as much the same as possible.

So we have $U \subset E$ and we want to define what it means to be open in $E$. The easiest way is to say: for every $x \in U$ there is some $\epsilon > 0$ such that "every point of $E$ that is closer to $x$ than $\epsilon$ lies in $U$". The statement between quotes is just $E \cap N_\epsilon(x) \subset U$, as your text's definition. It makes no sense to consider points not in $E$; we consider $E$ to be the new "universe of discourse", as we only talk about its subsets, no longer about all subsets of $\mathbb{R}$.


问:
Number of solutions of $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ where $x\in\mathbb{Q} , x\notin\mathbb{Z}$
How many solutions the equation $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ has in rational non integer sets of numbers?

$1)\text{zero}\quad\quad\quad\quad\quad\quad\quad2)1\quad\quad\quad\quad\quad\quad\quad3
)2\quad\quad\quad\quad\quad\quad\quad4)87\quad\quad\quad\quad\quad\quad\quad5)\text{infinity}$

So the solution can't be an integer therefore $\lfloor x\rfloor\ne x$ and generally we can say $\lfloor x\rfloor< x$. also denominator of the fractions can't be zero so  $\lfloor x\rfloor \neq0$ hence $x\notin[0,1)$

I wrote the equation as:

$$x-\lfloor x\rfloor=\frac{87}{\lfloor x \rfloor}-\frac{87}{x}=\frac{87(x-\lfloor x\rfloor)}{x\lfloor x\rfloor}$$

But can't see a way to continue.

答:
You're on the right track.

Hint 1: Divide both sides by $(x - \lfloor x \rfloor)$

Hint 2: Note that for $\lfloor x\rfloor \geq 0$,
$$
\lfloor x\rfloor^2 \leq \lfloor x\rfloor x \leq \lfloor x\rfloor(\lfloor x\rfloor + 1).
$$


问:
Weak containment of unitary representations and group $C^*$-algebras
Let $G$ be a locally compact group. In order to make things simpler, let us say it is discrete. Let $\pi$ and $\rho$ be unitary representations of $G$ on Hilbert spaces $H_\pi$ and $H_\rho$ respectively. Let $C^*_\pi(G)$, resp. $C^*_\rho(G)$ be the corresponding group $C^*$-algebras, i.e. the $C^*$-algebras generated by $\{\pi(g): g\in G\}\subseteq B(H_\pi)$, and analogously for $\rho$.

It seems to be a folklore fact that the following two facts are equivalent

$\rho\prec \pi$, i.e. $\rho$ is weakly contained in $\pi$;

the map $\pi(g)\to\rho(g)$ extends to a ${}^*$-homomorphism from $C^*_\pi(G)$ onto $C^*_\rho(G)$; or equivalently, for every $x\in \mathbb{C}G$, $\|\pi(x)\|_{B(H_\pi)}\geq \|\rho(x)\|_{B(H_\rho)}$.

It looks quite clear on the first sight, however when trying to write down a precise proof, I get stuck.

I would be grateful for either writing down the proof or providing a reference where this is proved exactly as stated.

答: The result is Theorem 7 in de la Harpe's paper "On simplicity of Reduced $C^*$-Algebras of Groups" (here is a link).  The proof is skimmed there, and the details are not hard to fill in.


问:
Transforming an elliptic integral with arbitrary coefficients and imaginary roots
I'm trying to solve the integral
$$\int_0^\pi \dfrac{\text{d}\phi}{\sqrt{\cos^2(\phi) - a \cos(\phi) + b}}
= \int_{-1}^1 \dfrac{\text{d}x}{\sqrt{1-x^2}\sqrt{x^2-ax+b}} $$
where $a$ and $b$ are arbitrary positive real numbers, and $a^2-4b<0$.

I know that in theory the solution can be written in terms of elementary functions and standard elliptic integrals, but I'm struggling to put it in that form. I've tried working out a change of variables like
$$x = \dfrac{rt+s}{ut+v}$$
to simplify the radicand for comparing against tables (Byrd and Friedman or Gradshteyn and Ryzhik), but since my polynomial has imaginary roots I keep getting stuck. Computer algebra like Wolfram Alpha also seems to fail.

Is this just impossible, or am I missing something? Thanks.

答:
You didn't look hard enough. The relevant formula is Byrd & Friedman 259.00.

Since $x^2-ax+b$ has complex roots, rewrite it as $(x-c)^2+d$. Define
$$A^2=(1-c)^2+d\qquad B^2=(1+c)^2+d\qquad g=\frac2{\sqrt{AB}}\qquad m=\frac{4-(A-B)^2}{4AB}$$
Then the given integral evaluates to $gK(m)$ where $K$ is the complete first-kind elliptic integral and $m$ is the parameter.


问:
The distribution of the uniform random tree
I have a question about how to interpreted the following:

Proposition 2.3. The uniform random tree $\mathbb{T}_{n}$ has the same distribution as a tree generated
as follows:

Take a Galton-Watson tree with Poisson(1) offspring distribution;

Condition it to have total progeny precisely (n);

Assign the vertices random labels chosen from [n] and forget the original ordering.

Proof. Recall the standard labelling of the vertices of a Galton-Watson tree $\mathbf{T}$, and let $\mathbf{t}$ be a particular tree with $\#(\mathbf{t})=n$ and numbers of offspring $c\left(v_{1}\right)=c_{1}, \ldots, c\left(v_{n}\right)=c_{n}$. Then
$$
\mathbb{P}(\mathbf{T}=\mathbf{t})=\prod_{i=1}^{n} \frac{e^{-1}}{c_{i} !}=e^{-n} \prod_{i=1}^{n} \frac{1}{c_{i} !}
$$
Now observe that $\mathbb{P}(\#(\mathbf{T})=n)$ is a function only of $n$. Hence,
$$
\mathbb{P}(\mathbf{T}=\mathbf{t} \mid \#(\mathbf{T})=n)=f(n) \prod_{i=1}^{n} \frac{1}{c_{i} !}
$$
for some function $f$. Now consider labelling the vertices of $\mathbf{T}$ with $[n]:=\{1,2, \ldots, n\}$. There are $n!$ different ways to do this, of which $\prod_{i=1}^{n} c_{i} !$ give rise to the same unordered labelled tree, once we forget the ordering. Hence, the probability of obtaining a particular labelled unordered tree $t$ is $f(n) / n !$. Since this depends only on $n$, and not on any other feature of the tree, it must be the case that
the tree is uniformly distributed on $\mathbb{T}_{n}$.

(From An introduction to random trees by Christina Goldschmidt )

I believe I fully understand how the tree is generated, but what I have trouble with is: $\mathbb{T}_{n}$ has the same distribution as a tree generated
as above. What does that mean exactly?

Is it, $$
\mathbb{P}(\mathbf{T}=\mathbf{t} \mid \#(\mathbf{T})=n)=\mathbb{P}(\mathbf{t} \in \mathbb{T}_{n})
$$ and what would $\mathbb{P}(\mathbf{t} \in \mathbb{T}_{n})$ be?

答:
Despite the varied notation, the following must be true in order for the statements in the proof to make sense:

$\mathbb T_n$ is a random variable: it is a labeled tree uniformly sampled from the set of all labeled $n$-vertex trees.

$\mathbf T$ is a random variable: it is an ordered tree generated by the Galton-Watson process.

$\mathbf t$ is a fixed $n$-vertex ordered tree.

Now, $\Pr[\mathbf t \in \mathbb T_n]$ doesn't make sense: $\mathbb T_n$ is not a set. We are closer to wanting to say $\Pr[\mathbb T_n = \mathbf t]$, but that also doesn't make sense: $\mathbb T_n$ is a labeled tree, and $\mathbf t$ is an ordered tree.

To make the statement of the theorem precise, let's define new variables:

$\mathbf T'$ is a labeled tree constructed from $\mathbf T$ by forgetting the ordering, and giving the vertices random labels.

$t$ is a fixed $n$-vertex labeled tree.

Then our goal is to say that $\Pr[\mathbf T' = t \mid \#(\mathbf T) = n] = \Pr[\mathbb T_n = t]$. Here, $\Pr[\mathbb T_n = t] = \frac1{n^{n-2}}$, because that's how many labeled $n$-vertex trees we are, but for the proof, all we need to know is that $\Pr[\mathbb T_n = t]$ is a constant independent of $t$ (because $\mathbb T_n$ has a uniform distribution). Therefore if $\Pr[\mathbf T' = t \mid \#(\mathbf T) = n]$ also does not depend on $t$, we obtain the result we want.

The proof shows that $\Pr[\mathbf T = \mathbf t \mid \#(\mathbf T) = n]$ is proportional to $\prod_{i=1}^n \frac1{c_i!}$, which does depend on $\mathbf t$. But also, if $t$ is a particular labeling of $\mathbf t$, then the probability of obtaining $t$ by labeling $\mathbf t$ is proportional to $\prod_{i=1}^n c_i!$, so the two effects cancel.


问:
How can I prove that if $a < b$ then $a^2$ < $b^2$ for $a,b \in \Bbb N$
How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.

答: Write $b=a+d$ so, for $c\in\Bbb N$, $ac<ac+dc=bc$. In particular, $ab<b^2$. By the same logic, $a^2<ab$. Now verify your definition of $<$ (which is what it is, if you take @AtticusStonestrom's advice) implies $<$ is transitive. Alternatively, note $a^2+D=b^2$ with $D:=d(2a+d)$.


答: Suppose $a < b$. Take $c$ such that $a + c = b$. Then we have $b^2 = (a + c)^2 = a^2 + (2c + c^2)$. Then $a^2 < b^2$.

Title: How can I prove that if $a < b$ then $a^2$ < $b^2$ for $a,b \in \Bbb N$

Body:
How can I prove that:

If $a < b$ then $a^2 < b^2$ for $a,b \in \Bbb N$, using the ring properties on the integers and the following property:

If $m < n$ then exist $d \in \Bbb N$ such that $n = m + d$.


问:
Prove that a set E is not complete by exhibiting a sequence of points in E that is Cauchy but does not converge to an element of E
Prove that the set $E=\{(x_1, . . . , x_d) :x_1, . . . , x_d>0\}$ in $\mathbb{R}^d$ is not complete by exhibiting a sequence of points $(x_n)_{n \in \mathbb{N}}$ in $E$ that is Cauchy but does not converge to an element of $E$.

I know that $E$ is complete if for every sequence $(x_n)_{n \in \mathbb{N}}$ of points in $E$ that is Cauchy, there exists an $x \in E$ such that $x_n \rightarrow x$. So incompleteness would be that there does not an exist such an $x$. I am not sure how to prove this.

Body: $\{x_n\}$ with $x_n =(1/n , 1/n, \dots,1/n)$.


问:
Help understanding two theorems on primes and residue systems
My textbook lists two theorems and I'm not sure how I'm supposed to interpret them. I don't need a proof; I'm only trying to figure out what information I'm being told by each theorem.

Let $p$ be a prime and let $a$ be an integer not divisible by $p$; that is, $gcd(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$

So for this first theorem, I believe the canonical complete residue system $p$ would be $\{0,1,2,...,p-1\}$. So the set $\{a,2a,3a,...,pa\}$ would be the integers that satifsy the congruences, $a\equiv0 \space (mod \space p), 2a\equiv1 \space (mod \space p), 3a\equiv2 \space (mod \space p),...,pa\equiv (p-1) \space (mod \space p)$.

Is my interpretation correct?

Let $p$ be a prime and let $a$ be an integer not divisible by $p$. Then,

$a\cdot2a\cdot3a\cdot...\cdot(p-1)a\equiv1\cdot 2 \cdot 3 \cdot...\cdot(p-1) \space (mod \space n)$

For this second theorem, is this theorem telling me that I can decompose congruence integers into products of primes?

Thanks for any help.

答:
Okay, what does it mean.

A complete residue system $\mod p$.  What's that?

If you compare any integer $n \pmod p$ that are only $p$ options.  Either $n\equiv 0\pmod p$ or $n\equiv 1 \pmod p$ or $n\equiv 2 \pmod p$ or ...... or $n\equiv p-1\pmod p$.

So the list $\{0,1,2,.....,p-1\}$ is called a complete residue system because it contains a representation, a residue, for every case.

But that there nothing special about choosing just those (the smallest non-negative) residues.  We could have chosen any $\{r_1, r_2, ....., r_p\}$ so that every integer is equivalent to one of them (and we can list them in any order).

For example to find a complete residue system $\mod 6$ we can represent all the integers that have remainder of $0$ by $78$.  (Because $6|78$).  And we can represent all the integers that have a remainder of $2$ with $-22$ (because $-22 = -4*6 + 2$).  And we can represent all the integers that have remainder of $5$ with $647$ (because $647 = 642 + 5=6*107 + 5$) and so on.

So $\{78, -22, 647, 3001, -2,10\}$ is a complete residue system $\pmod 6$ because every interger is congruent to exact one of those $\pmod 6$. (test them: $$0\equiv 78\pmod 6; 1\equiv 3001 \pmod 6; 2\equiv -22\pmod 6; 3\equiv -2\pmod 6; 4\equiv 10\pmod 6; 5\equiv 647\pmod 6; 6\equiv 78\pmod 6; 7\equiv 3001\pmod 6$$ and so on. [they just cycle and cycle through].

(Admittedly, it's a fairly irrelevant and awkward way to represent them... but it does represent all $6$ possible congruences.)

So the theorem is saying:  If $p$ is a prime and $a$ is a number that .... isn't a multiple of $p$... then the set $\{a, 2a, 3a,.... , pa\}$ is a complete residue system $\mod p$ and that every integer is congruent to exactly one of those number.

For example:  $\mod 7$ and the number $3$ we'd have $\{3,6,9,12,15,18,21\}$ is a complete residue system.

And if we test it.  $$0\equiv 21 \pmod 7; 1\equiv 15\pmod 7; 2\equiv 9\pmod 7; 3\equiv 3 \pmod 7; 4\equiv 18\pmod 7; 5\equiv 12\pmod 7; 6\equiv 6\pmod 7; 7\equiv 21 \pmod 7; 8\equiv 15\pmod 7$$ and so on....

We could to it with any number (as long as it's not a multiple of $7$).  For example...... $93$....

$\{93, 186, 279, 372, 465, 558,651\}$ should work.  Does it?  I'm sure it does because I have complete faith in the theorem but looking at that I'd have no idea if it weren't for the theorem but:

$651 = 7*93$ that was a gimme; $372=53*7 + 1$ and $93=7*13+2$ and $465=7*67+3$ and $186=7*26+4$ and $558 = 7*79+5$ and $279=7*39 +6$ and so on.

It does work.