问:
Help understand a pathological example for the Ford Fulkerson Algorithm
Consider the graph below. All edges are undirected and have capacity $\frac{1}{1-\sigma}$ where $\sigma = \frac{\sqrt{5}-1} {2}$ except for the ones indicated otherwise (those in red). I further understand that $\sigma^{n+2} = \sigma^{n}-\sigma^{n+1}$.

I was told that the Fulkerson Algorithm does not terminate on it. However, I do not understand why.  Could you explain that to me?

答:
None of the pathological examples for the Ford-Fulkerson algorithm have the property "the algorithm does not terminate for it". All networks have a maximum flow (well, all finite networks with finite capacity on each edge, and a source and sink for the problem to make sense). If we know a maximum flow, it is easy to find a sequence of augmenting paths that will let the Ford-Fulkerson algorithm find it in finitely many steps.

(In this example, the maximum flow is $\frac{4}{1-\sigma}$, and is easy to find in four augmenting steps: just take four edge-disjoint paths of length $3$ from $s$ to $t$ that don't use any of the red edges.)

So pathological examples for Ford-Fulkerson must come with a pathological choice of augmenting paths which is valid at every step, but does not terminate. Essentially, what we're saying is this: "The Ford-Fulkerson algorithm does not fully specify which augmenting path to choose at each step. So it's possible that it will choose these augmenting paths. Then it will never terminate. As a result, we conclude that it's possible for Ford-Fulkerson to never terminate, unless we come up with a rule to pick augmenting paths that avoids outcomes like this one".

Without knowing the pathological choice of augmenting paths, it's hard to reconstruct it. Presumably, however, the bad augmenting paths were specified wherever you found the example.

If not, here is a different example for you to try to understand. It is a $6$-vertex, $8$-edge example that is:

Originally described in "The smallest networks on which the Ford-Fulkerson maximum flow procedure may fail to terminate" by Uri Zwick.

Elaborated on in these lecture notes on pages 335-336.

Worked out in excruciating detail in the last section of my lecture notes.


问:
proof of the injectivity of an application
I have to solve the following exercise ?

Let $G$ and $H$ be two groups, $G = \langle g_1,\ldots,g_r\rangle$. I have the following map :

$$\begin{align*}
f \colon \mathrm{Hom}(G,H) &\longrightarrow H^r,\\
\phi &\longmapsto (\phi(g_1),\ldots,\phi(g_r)).
\end{align*}$$

I have to show that it is an injection; so I tried to find the kernel of the map but, my equation is :
$(\phi(g_1),\ldots,\phi(g_r)) = (e_H,\ldots,e_H)$.

But I don't see how i can conclude if $\phi$ is not an injection. I suppose it has a link with the definition of $G$ but I have a hard time understanding it's structure. An element $g \in  G$ can of course be $ g = (g_i)^n, i \in \{1,\ldots,r\}$ but can it be $g = g_i*g_j, j \neq i$ ?

Can you help me please ?

答:
Though the question was originally tagged "abelian groups", you seem to be saying in the comments that $G$ and $H$ are not assumed to be abelian groups. If that is indeed the case, then note that $\mathrm{Hom}(G,H)$ is not usually a group. It's not a group under composition, clearly, but it is also not a group under pointwise multiplication: if $\phi,\psi\colon G\to H$ are group homomorphisms, then in general $g(ab)h(ab) = g(a)g(b)h(a)h(b)$ is not equal to $g(a)h(a)g(b)h(b)$. (It is a group under pointwise multiplication when $H$ is abelian). Given that you cannot assume your set is a group, checking what elements map to the identity may tell you the map is not injective (if you can find two things that do), but it will not decide the issue: you can have a function that is not one-to-one but where only one thing goes to the identity.

The key here is just that if $G$ is generated by $X\subseteq G$, then two homomorphisms $\phi,\psi\colon G\to H$ are equal if and only if they agree on $X$.

One direction is clear: if $\phi=\psi$, then $\phi(x)=\psi(x)$ for all $x\in X$.

Conversely, if $\phi(x)=\psi(x)$ for all $x\in X$, then let $g\in G$. Since $G=\langle X\rangle$, there exist $x_1,\ldots,x_m\in X$ and $\epsilon_1,\ldots,\epsilon_m\in \{1,-1\}$ such that
$$ g = x_1^{\epsilon_1}\cdots x_m^{\epsilon_m}.$$
Therefore,
$$\begin{align*}
\phi(g) &= \phi(x_1^{\epsilon_1}\cdots x_m^{\epsilon_m})\\
&= \phi(x_1)^{\epsilon_1}\cdots \phi(x_m)^{\epsilon_m}\\
&= \psi(x_1)^{\epsilon_1}\cdots \psi(x_m)^{\epsilon_m}\\
&= \psi(x_1^{\epsilon_1}\cdots x_m^{\epsilon_m})\\
&= \psi(g).
\end{align*}$$

From this, you should be able to prove that if $f(\psi)=f(\phi)$, then $\psi=\phi$. Again, you can't just see whether $f(\psi)=(e,e,\ldots,e)$, unless you are dealing with the pointwise group structure of $\mathrm{Hom}(G,H)$, which you are only guaranteed is a group structure when $H$ is abelian. (Though for some pairs of groups it may work; e.g., if $G=S_5$ and $H=C_2\times K$, where $K$ is the Heisenberg group of order $3^3$; but in general, it won't.)

ContentLicense: CC BY-SA 4.0


问:
Can an outer automorphism fix the center of a group?
I have just learned about inner and outer automorphisms, and I'm having a hard time visualizing the outer automorphism as quotient classes. Since the inner automorphisms fix the center, does it mean that any automorphism that fixes the center is necessarily an inner automorphism?</p>

答:
The centre is a characteristic subgroup, and is fixed by all automorphisms. That in itself is pretty much a tautology. But note the fact that the centre has a unique definition/property, even though there may be other subgroups isomorphic to the centre as abstract groups. And that is a property preserved by homomorphism.

Note therefore that the image of the centre of a group under any homomorphism must be commutative and contained within the centre of the image, and it is easy to show from this that an automorphism must fix the centre. Note further that this does not mean that an automorphism necessarily fixes the individual elements of the centre.


问:
The commutator subgroup of a quotient in terms of the commutator subgroup and the kernel
Let $H$ be the normal subgroup of $G$.

Is it true that $(H[G,G])/H$ is isomorphic to $[G/H,G/H]$?

If so, I want to make a surjective homomorphism $\phi\colon H[G,G]\to [G/H,G/H]$ with Kernel $H$ to prove it.

But, if I define $\phi(h[g_1,g_2])=[g_1/H,g_2/H]$ then I have some trouble.

答:
The commutator subgroup is a verbal subgroup. That means that for any group homomorphism $\varphi\colon G\to K$, you have that $\varphi([G,G])\subseteq [K,K]$. If $\varphi$ is onto, then the induced map on commutator subgroups is also onto.

Applying this to the quotient map $G/H$, you have that $\pi([G,G])$ maps $[G,G]$ onto $[G/H, G/H]$. The kernel of this map is $[G,G]\cap H$, so by the isomorphism theorems we have
$$\left[\frac{G}{H},\frac{G}{H}\right] \cong \frac{[G,G]}{[G,G]\cap H} \cong \frac{H[G,G]}{H}.$$
Follow through the definitions given by the isomorphism theorems, and you'll see that the isomorphism from $\frac{H[G,G]}{H}$ to $[G/H,G/H]$ is indeed given by mapping $huH$ to $\pi(u)$ (or by mapping from $H[G,G]$ onto $[G/H,G/H]$ via $hu\mapsto \pi(u)$). This is the map you give (though you only describe it for a generating set of $H[G,G]$).

The same result holds for any verbal subgroup $\mathfrak{V}(G)$: if $N$ is normal, then
$$\mathfrak{V}\left(\frac{G}{N}\right) \cong \frac{\mathfrak{V}(G)}{\mathfrak{V}(G)\cap N} \cong \frac{N\mathfrak{V}(G)}{N}.$$
E.g., for the subgroup generated by the $n$th powers, $G^n$, we have
$$\left(\frac{G}{N}\right)^n \cong \frac{NG^n}{N}.$$

ContentLicense: CC BY-SA 3.0


问:
Monty Hall problem vs. roulette systems - how are they different?
So I got interested in the Monty Hall problem - I understand what it's about, but somehow I can't wrap my head around the idea of the final choice not being 50/50. More precisely: we all know (or maybe most of us do) the roulette progression systems which were popular a while ago - they told you that you should wait for a streak of some three-or-more consecutive reds or blacks and then bet on the contrary, always staying with this colour and eventually it has to be drawn as with every other spin, the chances become higher for the streak to be broken.

Of course the system is dumb and was shortly afterwards popularly neglected as a fallacy (though surprisingly many people still believe it works) as every spin is independent and even though you had a million reds in a row, in the next spin the odds for red, black and green are still the same as they had been these million spins ago.

And here I see an analogy to Monty Hall - sure, if we were proposed "Do you choose doors k and l (both at the same time!) or m?", it would be a straightforward 2/3 to 1/3. But why don't we cast aside the fact of some door having a goat and treat the two remaining doors as a simple 50/50 just like we do in roulette? Why aren't we interested in the past in roulette while it does matter in Monty Hall? Isn't the two remaining doors - one with a car - just "another spin"?

答:
The money hall problem is much more straightforward with 1000 doors.

Now, you choose 1, the presenter opens 998 doors which don't contain a prize, leaving 1 remaining unopened door which you didn't choose.

You can now choose to either open your door (which you picked at the start, so a 1/1000 chance its the correct one) or you can open the other door the presenter specially chose to remain un-open (a 999/1000 chance of being the prize).

If you imagine the 1000 doors stretching in front of you, and you pick the first one, the presenter then runs around opening doors, leaving say no.345 closed. It should be quite clear that is much more likely that this 345 door is the winning one  rather than the no.1 you happened to choose first.

The same applies with 3 doors, but its less obvious. Lets say you can only choose door 1 to start with. Divide the doors into two groups: (1) and (2,3). If you stick you are betting door 1 has the prize, while if you switch you are betting that (2,3) has the prize, since the presenter will open one non-winning door from (2,3) leaving you with the winning one (unless 1 was the winning door).


问:
Differentiable structure of an $\mathbb{R}$-vector space
Let $ V $ an $ \mathbb{R} $-vector space, we can prove that $ V $ is a manifold. Using the topology compatible with $ \mathbb {R} ^ n $ and defining the chart $ (V, \phi_B) $ where $ \phi_B: V \to \mathbb {R} ^ n $, $\phi_B(v)=(\lambda_1,...,\lambda_n)$, $B=\{v_1,...,v_n\}$ is a base of $V$ and $v=\lambda_1
 v_1+...+ \lambda_nv_n$. I want to show that this structure does not depend on the base, that is, for another base $B'=\{v_1', ..., v_n'\}$ when we define $\phi_{B'}$ we have the compatibility of the charts $\phi$ and $\phi'$, i.e, $\phi \circ \phi'^{-1}$ and $\phi' \circ \phi^{-1}$ are smooth.

To find the composition explicitly I thought of introducing the base change, but I can't conclude that the compositions are smooth.

答:
Note that the charts you define are all globally defined, that is, they have domain $V$. Hence to prove that change of coordinates are smooth is suffices to compute

$$
\varphi_B \circ \varphi_{B'}^{-1} : \mathbb{R}^n \to \mathbb{R}^n
$$

and prove it is smooth (the other composition is automatic by interchanging $B$ with $B'$).

As you say, this is a base change. But a simpler observation suffices: this map is a composition of linear ones, hence linear itself, and so it is smooth.


问:
Factorise $3\cos(x) - 2\sin(x) + 1$
How can I find the "product form" of

$3\cos(x) - 2\sin(x) + 1$

I cannot find a way to get rid of the coefficients (3 and 2)

答:
To express $a\cos x + b\sin x$ in the form $R\cos(x-\alpha)$, just notice that by the compound angle identity,

\begin{align*}
a\cos x+ b\sin x &\equiv R\cos(x-\alpha)\\
&= \underbrace{R\cos\alpha}_{a}\cos x+\underbrace{R\sin\alpha}_b \sin x,
\end{align*}
so we can just solve the system $a=R\cos\alpha$ and $b=R\sin\alpha$ simultaneously to get that $$R = \sqrt{a^2+b^2} $$
and $$\alpha=\tan^{-1}\tfrac ba.$$

In your case, we get $$3\cos x - 2\sin x \equiv \sqrt{13}\cos(x+\tan^{-1}\tfrac23),$$
which means that $$-\sqrt{13}\leqslant 3\cos x-2\sin x\leqslant \sqrt{13}.$$
Adding $1$ throughout, we get the bounds
$$1-\sqrt{13}\leqslant 3\cos x-2\sin x+1\leqslant 1+\sqrt{13},$$
so the minimum is $1-\sqrt{13}$.

Title: Factorise $3\cos(x) - 2\sin(x) + 1$


问:
How to find the Laurent expansion for $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$?
I want to find the Laurent expansion for  $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$,

I've tried to apply this formula $\frac{1}{1-\omega}=\sum_{n=0}^{\infty }\omega^{n}$ and the usual Taylor series of the exponential function, but I don't know how to continue:

$$\begin{align}f(z)&=\frac{1}{z-1}\exp\left(\frac{1}{z^{2}}\right)\\
&=-\frac{1}{1-z}\exp\left(\frac{1}{z^{2}}\right)\\&=-\left (\sum_{n=0}^{\infty }z^{n}  \right )\left ( \sum_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )\end{align}$$
Thanks in advance.

Ps: I tried applying a Cauchy product, but I think this is not appropriate.

Edit 1: If it is useful at the end of the text, the authors say that the Laurent expansion is:

$\sum_{k=-\infty }^{\infty }a_{k}z^{k}$ with $a_{k}=-e$ if $k\geq 0$ and
$a_{k}=-e+1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(j-1)!}$if $k=-2$ or $k=-2j+1$ where $j=1,2,...$

答:
Starting with your $=-\left (\sum\limits_{m=0}^{\infty }z^{m}  \right )\left ( \sum\limits_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )$ changing one of the $n$ to $m$, you can say the coefficient of $z^k$ is

$-\sum\limits_{n=0}^{\infty} \frac1{n!} =-e$ when $k\le 0$

$-\sum\limits_{n=k/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{n=(k-2)/2} \frac1{n!}-e$ when $k\gt 0$ and even

$-\sum\limits_{n=(k+1)/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{(k-1)/2} \frac1{n!}-e$ when $k\gt 0$ and even

But that looks wrong to me:  I do not think $$\cdots -e z^{-5} -e z^{-4} -e z^{-3} -e z^{-2} -e z^{-1} -e z^{0}+ \\(1-e)z^1 +(1-e)z^2 +(2-e)z^3 +(2-e)z^4+\left(\frac52-e\right)z^5+\cdots$$ converges when $|z| \le 1$.

Meanwhile for the same question asked elsewhere, a suggested answer was in effect $$z^{-1}+z^{-2}+2 z^{-3}+2 z^{-4}+\frac{5 }{2}z^{-5}+\frac{5}{2}z^{-6}+\cdots$$ but I do not think that converges either when $|z|\le 1$


问:
Best resource for concentration of measure?
What are the best resources to learn concentration of measure phenomenon and concentration inequalities? I have heard that Talagrand's papers are good, but they are not particularly good read.

Any help is appreciated. Thanks a lot!

答: I like Roman Vershynin's book High Dimensional Probability</p>


问:
If n is multiple of 4, what is the rest of the division by 10?
I need to resolve this issue about congruence / module, but I have no idea where to start. Can anybody help me?

If n is a multiple of 4, what is the rest of the division of $1^n + 2^n +....+ 8^n + 9^n$ by 10?

Thanks.

答:
$1^{4m}$ ends with $1$, $2^{4m}$ ends with $6$, $3^{4m}$ ends with $1$, $4^{4m}$ ends with $6$, and $5^{5m}$ ends with $5$.

$(10-k)^{4n}$ ends with the same digit as $k^{4n}$.

Therefore, your sum ends with the same digit as $1+6+1+6+5+6+1+6+1$ does,

which is $3$.


问:
Finding Lagrangian of a system (Oscillations in a parabolic trough)
A particle of mass m slides down on the curve $z=1+\frac{x^2}{2}$ in the $xz$-plane without friction under the action of constant gravity.Suppose $z$-axis points vertically upward.Find the Lagrangian of given system.

The position vector of the particle is given by $\overline r=(x,0,1+\frac{x^2}{2})$.
Then the velocity vector is given by
$ \dot {\overline r}=(\dot x,0,x\dot x)$.
So the kinetic energy is given by
$T=\frac {m}{2}((\dot x)^2+x^2(\dot x)^2)$. Now I am confused in potential energy. The potential energy is given by $V=mgz$ but what will be the sign of it? What does it mean that $z$-axis points vertically upward?

答:
The Lagrangian is set up with  Mechanics energy formulation and DE set up using Euler Lagrange Equation.

Angles $ \phi$ are tangent or normal rotations positive counterclockwise.

Parabola trough coordinates  are taken  wlog $ R=1, z_{min}=0 $in parametric form:

$$ (x,z)=( R \tan \phi, R/2 \tan^2 \phi)\tag1$$
$$ ( \dot x, \dot z)= (R \sec^2 \phi\, \dot \phi,R \sec^2 \phi \tan \phi\;\dot\phi )\tag2$$

And the absolute value of velocity is

$$R \sec^3 \phi \;\dot\phi \tag3$$

$$ KE= \frac{m}{2}(R \sec^3 \phi \;\dot \phi)^2 \tag4 $$

PE is taken positive as shown  and as usual $z>0$. When PE increases KE decreases and vice-versa.

$$ PE=mgz=mg \frac{R}{2}\dot\tan^2 \phi \tag5 $$

$$\text{Lagrangian } =  L =KE -PE $$

$$ L=  \dfrac{m}{2} \cdot (R^2 \sec^6 \phi\; \dot \phi ^2)- mg \dfrac{R}{2} \tan^2 \phi \tag6$$

Euler -Lagrange Equation ( after removing constants)

$$ 6 \sec^5 \sec\phi \tan \phi\;{\dot{\phi}}^2-\frac{g}{R } 2 \tan {\phi} \sec^2\phi -\frac{d(2 \dot \phi)}{dt} \sec^6 \phi =0\tag7$$

Simplifying,

$$ \dfrac{\ddot{\phi}}{\sin \phi \cos^2 \phi}+ \frac{g}{R}\cos \phi -3 \dot\phi^2=0 \tag8 $$

I have no closed form solution of the DE or any reference. So a numerical solution with the following boundary conditions is undertaken.

$$( \phi_{max}= \frac{\pi}{3}, \dot\phi_{max}= 0, g=9.8, R=6) \tag9 $$

The parabolic trough profile and time oscillation  of slope is plotted below:

The formula for shallow $\phi_{max}$ trough as a simple pendulum of constant $L$ for verification :

$$ T= 2 \pi\sqrt{\dfrac{L}{g}} \tag {10} $$ is tallying alright.


问:
Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer
Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer

I said that since $a+b+c=0$ then $c=-a-b$ and hence factoring $K=a^3b+b^3c+c^3a$ we have that $K=-(a^2+ab+b^2)^2$

This is where I got stuck. I can't work out how to finish off the solution. Could you please explain to me the full solution and how you intuitively thought of each step?

答:
you have $$ 2017 + \left( a^2 + ab + b^2 \right)^2 = w^2 $$  or
$$ w^2 - \left( a^2 + ab + b^2 \right)^2 = 2017 $$
Difference of squares on the left, factors, meanwhile 2017 is prime. Thus
$$  w + a^2 + ab + b^2 = 2017 , $$
$$  w - a^2 - ab - b^2 = 1 , $$
or
$$ w = 1009$$
$$ a^2 + ab + b^2 = 1008 = 2^4 3^2 \cdot 7 = 7 \cdot 144 $$
You can check, this means $a,b$ are both divisible  by $12.$ Let $\alpha = \frac{a}{12}$ and $\beta = \frac{b}{12},$ then
$$  \alpha^2 + \alpha \beta + \beta^2 = 7$$
which has just a few integer solutions.


问:
A closed-form expression for the integral $\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx$
Is there a closed-form expression for this integral:
$$\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx,$$
where $\text{Ci}(x)=-\int_x^\infty\frac{\cos z}{z}\mathrm dz$ is the cosine integral?

$\text{Ci}(x)$ and $\text{Ci}^{2}(x)$ have primitives/antiderivatives that can be expressed in terms of the trigonometric integral functions.

So it's not too difficult to show that  $$\int_0^\infty\text{Ci}(x) \, \mathrm dx =0$$ and $$\int_0^\infty\text{Ci}^{2}(x) \, \mathrm dx = \frac{\pi}{2}.$$

But $\text{Ci}^{3}(x)$ doesn't appear to have a primitive that can be expressed in terms of known functions.

答:
The answer is
$$\int_0^{\infty}\text{Ci}^3x\,dx=-\frac{3\pi\ln 2}{2}.$$
I would like to trade the method of evaluation for convincing story about what made this integral interesting for you. The story should be longer than "a friend of mine told it could be calculated in a closed form".

Update: Not that I was really convinced by the comment below... but for those who would eventually like to figure it out:

Using that $\int\mathrm{Ci}\,x\,dx=x\,\mathrm{Ci}\,x-\sin x$, integrate once by parts. This yields two integrals:

$\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx=-\frac{\pi}{2}\ln 2$ (computable by Mathematica),

$\displaystyle \int_0^{\infty}\cos x \,\mathrm{Ci}^2x\,dx$

Integrating the 2nd expression once again by parts (with $u=\mathrm{Ci}^2x$, $v=\sin x$), one again reduces the problem to computing $\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx$.


问:
Choosing the best deck - Strategy Game
$\def\argmax{\operatorname{argmax}}$
One mate told this problem about a game and its optimal strategy:

You have infinite cards with 1, 2, 3, 4, or 5 as their value. You can make a deck with any of this cards you like. Now, the other player says a number from 1 to 5. The deck is randomly shuffled and the first card is taken, if it is the said number you pay as much as this number to the other player. Otherwise you lose nothing. The question is, what is the optimal way of forming the deck?

I've reached the conclusion that having 120 ones, 60 twos, 40 threes, 30 fours, 24 fives seems to be optimal. The reason is that the second player is going to always say the number
$$\argmax_j\{jPr(X=j)\} = \argmax_j\left\{j\frac{x_j}{\sum_{i=1}x_i}\right\} = \argmax_j\{ jx_j\}$$
where $X$ is the first card, and $x_j$ is the number of j's in the deck. Given that, having all the quantities $jx_j$ equal seems to be optimal but I can't see why.

答:
With the composition of the deck as you proposed, you are losing on average 60/137 in each round no matter what player B says (if he says 1, he will win 1 with a probability of 60/137, if he says 2, he will win 2 with a probability of 30/137 and therefore the expectation value is again 60/137. The same for 3, 4 and 5).

Let's assume you change the composition. For instance you replace a 2 by a 3. Now you have 41 threes out of 274 cards. Player B realizes this and he will always say "3" to increase his chances. The expectation value for his win will now be 3*41/274 which is a bit more than 60/137.

So whatever number you increase in the deck, it will increase your losses.

Title: Choosing the best deck - Strategy Game


问:
How to turn bound on $ne^{-n}$ into bound for n?
I have a bound of the form $ne^{-n}\leq C$, where $C$ is a constant greater than 0. I was wondering if anyone had ideas on how to turn this into a bound of the form $n\leq C'$?

答:
$ne^{-n}$ has a maximum at $n=1$, so $ne^{-n}\leq C$ is always true if $C \ge e^{-1}$.
Now for $0 < C < e^{-1}$, it is clear that all $n \le 0$ satisfy this condition.

For positive $n$, you can indeed find a corresponding condition but this is $n \color{red}{\ge} C'$, since $ne^{-n}$ is falling with  $n$. You find $C'$ by equating $C'e^{-C'}= C$.

Now solving this equation is not possible with standard algebraic functions. There is however the Lambert W-function which adresses exactly this type of equation (see here). It is given as $y = W(x)$ where $y e^y = x$.

So you can bring it into that form by writing it as ${-C'}e^{-C'}= -C$, hence $-C' = W(-C)$ or  $C' = -W(-C)$.


问:
How to determine the rotation angles of a telescope
I have two control buttons on my telescope, one to increase/decrease the right ascension (rotation around z-axis) and one to increase/decrease the declination (rotation around y-axis). My first target has a right ascension of 10 deg and declination of 20 deg. First I command a 20-deg rotation around the y-axis and then a second rotation of 10 deg about the z-axis. Now my telescope (original pointing direction to (1, 0, 0)) points to
$$ (0.925416 \hspace{3mm} 0.163176 \hspace{3mm} 0.342020)$$
The second target has a right ascension of 20 deg and declination of 40 deg. If I repeat the same matrix multiplications as before (20-deg rotation around the y-axis followed by a rotation of 10 deg about the z-axis) I am not ending up exactly at right ascension of 20 deg and declination of 40 deg. This is the result:
$$ (0.712860 \hspace{3mm} 0.291390 \hspace{3mm} 0.637905)$$
But I should point to:
$$ (0.719846 \hspace{3mm} 0.262003 \hspace{3mm} 0.642788)$$

My question is, how do I calculate the correct rotation angles for the third and the fourth rotation? I could numerically solve for the unknown rotation angles, but I wonder if there exists an analytical solution where I do not need a root finder. I am grateful for any help.

答:
The reason why it does not work for you is that matrix multiplication is not commutative. Let's call $\vec v_0=(1,0,0)$. Then you apply a rotation first around the $y$ axis $R_y(20)$, and then a rotation around $z$ axis $R_z(10)$. In matrix notation this is $$\vec v_1=R_z(10)R_y(20)\vec v_0$$
When you repeat the same procedure you get $$\vec v_2=R_z(10)R_y(20)R_z(10)R_y(20)\vec v_0$$
But this is different than $R_z(20)R_y(40)\vec v_0$ because
$$R_z(10)R_y(20)\ne R_y(20)R_z(10)$$
So how to fix this? Start from $\vec v_1$ and you want to get to some desired $\vec v_d$. Write the general rotation matrix consisting of a rotation around $z$ by an angle $\alpha$ and a rotation around $y$ by an angle $\beta$.
You get $$\vec v_d=R_z(20)R_y(40)\vec v_0=R_z(\alpha)R_y(\beta)\vec v_1=R_z(\alpha)R_y(\beta)R_z(10)R_y(20)\vec v_0$$
So $$R_z(20)R_y(40)=R_z(\alpha)R_y(\beta)R_z(10)R_y(20)$$
Noting that $R^{-1}(p)=R(-p)$, multiply both sides on the right first by $R_y(-20)$, then by $R_z(-10)$, and you get:
$$R_z(\alpha)R_y(\beta)=R_z(20)R_y(40)R_y(-20)R_z(-10)=R_z(20)R_y(20)R_z(-10)$$
If you wrote explicitly the general rotation matrix you can now determine the angles.


问:
Multiplicative inverse of a gaussian integer
So, I had to prove that Gaussian integers had an identity element and ended up with it being $(1 + 0i)$. Now I have to see if any $(a+bi)$ (except $(0 + 0i)$)has a multiplicative inverse. Then I ended up with this:
\begin{align*}
 ac-bd &= 1\\
 ad + bc &= 0\\
\end{align*}
But I don't really know if I can prove that for any given $a,b\in\mathbb{Z}$ I can find $c,d\in \mathbb{Z}$ such that that happens.
Am I on the wrong path?
Can I really prove it?

答:
Hint :

Consider the application $N : a+ib \mapsto a^2+b^2$

Show that for every $a+ib$ and $c+id$, one has $N((a+ib)(c+id))=N(a+ib)N(c+id)$.

Deduce that if $a+ib$ has a multiplicative inverse, then $N(a+ib)=\pm 1$.

Deduce all the Gaussian integers which have a multiplicative inverse.


问:
How many triangles can be formed from the vertices of a polygon of $n$ sides if the triangle and the polygon may not share sides?
Note: not a duplicate of this question as I'd like to know what's wrong with my approach.

My attempt: Choose any 3 of the $n$ vertices of the polygon, and let the number of points between the three be $x$,$y$ and $z$. Then, $x+y+z=n-3$ and $x, y, z\ge 1$. The number of possible solutions is a sticks-and-stars problem, and can be shown to be $n-4 \choose 2$. From the top answer to the question linked above though, the correct answer is $\binom{n}{3}-(n-4)n-n$.

Putting in some values for $n$, I can see that I'm undercounting. How?

答:
How are you setting up the bijection?

IE Say $n = 10$ and we chose vertices 2, 4, 6. What does $x, y, z$ correspond to?

What you're going for is something along the lines of

Fix a vertex of the triangle, let it be vertex 1. There are $n$ possibilities.

Let $x \geq 2 $ correspond to the difference between the index of the second vertex and vertex 1.

Let $y \geq 2$ correspond to the difference in the index of the third and second vertex.

Let $z \geq 1$ correspond to the difference between $n$ and the index of the third vertex

We have $x+y+z = n-1$ and $(x-1) + (y-1) + z = n-3$ as positive integer solutions. Thus, there are ${n-4 \choose 2 }$ possibilities.

Hence, there are $ n \times { n - 4 \choose 2 } $ possibilities, but we double-counted 3 times, so it is $ \frac{ n (n-4)(n-5) } { 6}$.


问:
Factorial of infinity
So, I've read in this article that:

$$\zeta'(0) = \log\sqrt\frac 1 {2\pi}$$

And that:
$$e^{-\zeta'(0)} = 1\cdot2\cdot3\cdots \infty = \infty! = \sqrt{2\pi}$$

I found this result very strange and tried to simplify if but i couldn't. Can somebody help me?

Thanks!

答:
I am riemannium, author of the above quoted entry and the blog The Spectrum Of Riemannium (TSOR).
I have just corrected the typos in my article quoted above. I mean, I wrote $\eta (1)=\log 2$. It was a typo while I was writing the article. I think there are no additional mistakes but if you find more "errors", let me know. I am taking my blog as a serious specialized "job", so I don't like to input errors. I am also a perfectionist.
I also thank you for the links so some pretty references about regularized infinite products I did not know till now! :).</p>

Title: Factorial of infinity


问:
Is it true that any continuous injective map from a circle to circle is surjective?
Here is my idea:

Suppose $f : \mathbb S^1 \to \mathbb S^1$ is continuous and injective.

If $f$ is not surjective, its image is homeomorphic to a subset of $\mathbb S^1 \setminus \{ c\} \cong \mathbb R$ (by considering the projection).

Then, the restriction of $f$'s codomain $g : \mathbb S^1 \to f(\mathbb S^1)$ is injective and surjective, so it is bijective.

Also, since $\mathbb S^1$ is compact and connected, its image under the continuous map is also compact and connected, so $f(\mathbb S^1)$ is homeomorphic to some closed interval $[a, b]$, which is simply connected.

Noting $f(\mathbb S^1)$ is Hausdorff and $\mathbb S^1$ is compact, by the closed map lemma, it follows $g$ is a homeomorphism. This contradicts that $\mathbb S^1$ is not simply connected, since $[a, b]$ is simply connected. Thus, $f$ must be surjective.

Does it work?

答: This is basically the same argument, but you can be a little bit shorter by noticing that any non-surjective continuous map $f:X \rightarrow \mathbb{S}^1$, where $X$ is a metric space, can be lifted to a map $\tilde{f} : X \rightarrow \mathbb{R}$. Of course, if $X=\mathbb{S}^1$, then $\tilde{f}$ cannot be injective, so $f$ is not injective.


答:
Your approach is nearly correct, you just need to tweak it because you are only determining the spacing between the vertices, but not the vertices concretely.

For instance, if $n=10$, then the solution $3 + 2 + 3$ means you picked 3 vertices spaced out by $3,2$ and $3$, but concretely we don't know which they are. (You basically need to pick one of the vertices, because then the others will be fixed.)

In fact, if you multiply your result by $n/3$, you'll see that you get the correct result:

$$\frac n3\binom{n-4}2 = \frac{n(n-4)(n-5)}{6} = \binom n3-(n-4)n-n.$$

Title: How many triangles can be formed from the vertices of a polygon of $n$ sides if the triangle and the polygon may not share sides?

Body:
Note: not a duplicate of this question as I'd like to know what's wrong with my approach.

My attempt: Choose any 3 of the $n$ vertices of the polygon, and let the number of points between the three be $x$,$y$ and $z$. Then, $x+y+z=n-3$ and $x, y, z\ge 1$. The number of possible solutions is a sticks-and-stars problem, and can be shown to be $n-4 \choose 2$. From the top answer to the question linked above though, the correct answer is $\binom{n}{3}-(n-4)n-n$.

Putting in some values for $n$, I can see that I'm undercounting. How?


问:
For what finite cardinalities of $Y$ is $\mathbb{R}^2 - Y$ a topological group?
I am working on the following problem: for what cardinalities (of $Y$) can $\mathbb{R}^2 - Y$ be a topological group if $Y$ must be finite? My intuition is telling me this somehow involves considering a wedge of circles, but I am not 100% sure on the details. Would anyone be able to confirm this, and if I am wrong, to put me on the right track? Thanks.

答:
Hint: The fundamental group of a topological group is abelian

The question in the body of the question is different from the title though do you want $\mathbf R^3$ or $\mathbf R^2$.


问:
Integer solutions to $z^2=10x^2-2y^2$
I need to show that a $(67, 12, 2)$-design does not exist. I tried to use the Bruck, Ryser and Chowla theorem which in this case states that if a $(67, 12, 2)$-design exists, then the equation
$$z^2=10x^2-2y^2,$$
has a non-trivial integer solution $(x,y,z)\neq \overline{0}$. Checking from Wolfram I found that it does not have any, which would prove my original problem, but I would like to see proof why it doesn't have any.

It is easy to see that $z$ should be even so I tried to plug in $z=:2k$ and split it in some cases but it did not go anywhere.

答:
I believe the following is correct after seeing from Daniyar that one should resubstitute:

By Jyrki's hint, $z^{2} = 10x^{2} - 2y^{2} \equiv -2y^{2} \pmod 5$.
It is easily verified that $y^{2} \equiv 0,1,4 \pmod 5$ so that $-2y^{2} \equiv 0,-2,-8 \pmod 5\equiv 0,2,3 \pmod 5$.

Clearly then as we have $z^{2}$ on the $LHS$, we have both $z,y \equiv 0 \pmod 5 \implies 25 | z^{2},y^{2} \implies 25| z^{2},2y^{2} \implies $

$25|z^{2}+2y^{2}\implies 25|10x^{2} \implies 5|2x^{2} \implies 5|x \implies x \equiv 0 \pmod 5$

Suppose $a=z,b=y,c =x$ is the solution to $z^{2} = 10x^{2} - 2y^{2}$ where one of $a,b,c$ is smallest.

Hence we may find a "smaller" solution by checking with $a=5k, b=5j$, and $ c = 5l$:

As Daniyar has shown me, it is important to substitute this back into the original equation to get
$25k^{2} = 250l^{2}  -50j^{2} \Longleftrightarrow k^{2} = 10l^{2} -2j^{2}$ where $k,j,l < a,b,c$ respectively, contradicting the assumption that $(a,b,c)$ is the smallest solution.

To clarify, this leaves the case where $a=k, b=j, c=l$, which from above is true if and only if $a,b,c=0$. Since we have found that $(0,0,0)$ satisfies $z^{2} = 10x^{2} - 2y^{2}$, this is the only solution.

Title: Integer solutions to $z^2=10x^2-2y^2$


问:
$A=2^3×5^2×7^3$ find number of divisors of A
My attempt :

$A=2^3×5^2×7^3$

Let's determine number of numbers primes with A, and
Smaller than A

$\rho (A) = 2^2 ×4×5×6×7^2 =23520$

23520 is a number of numbers primes with A and smaller than A

$\forall k$ $ \in {23520} $ $gcd(A, k)=1$

So : number of numbers doesn't prime white A is $A-23520=45080$

$\forall k $ $\in {45080} $ $gcd(A, k) ≠1$

let's search   about  The number of numbers that divide A

$A=0[k]$
$\Leftrightarrow$ $68600=0[k]$

In the first we can see  K can take the following values ${1,2,4, 5,7,8,10,20,40,50,70,100}$

After using this theory $(a=b[n] $and $a=b[m] $ and $gcd(m, n) =1$$\Rightarrow$ $a=b[MN] $)

We can see that

$K=(1, 2,4,5,7,8,10,20,40,50,70,100,35,56,28,14,140,280,350,700)$

So igot 20 value of k

But I know, there are 48 values of k

I want to know how can I continue for find 48 value of k

Note : I don't want any
other  solution, I want to develop of my attempt

答:
There is a simpler way. We can use the theorem for the number of divisors, where the number of divisors of $$p_1^{e_1}\cdot{p_2^{e_2}}\cdot{p_3^{e_3}}\ldots$$ is $$(e_1+1)(e_2+2)(e_3+3)\ldots$$

Putting that into your problem, $(3+1)(2+1)(3+1)=48$.

$A$ has $48$ divisors.


问:
About Lebesgue integral and Lebesgue measure-how to show if $f(x)$ is measurable+simple then $f(x+n)$ is.
Let $(R,B(R),m)$ a Lebesgue measurable space on $R$. Let $f: R-> [0, \infty]$ a measurable function.
Denote $F(x)=\sum_{n=1}^{\infty} f(x+n)$.

Show that if $\int_{R} F(x) dm$ is finite, then $f=0$ almost everywhere.

Hint: first show that if $\phi$ is a measurable, simple function so $\phi_n(x)=\phi(x+n)$ is measurable and simple too, and $\int_R \phi dm=\int_R \phi_n dm$.

I said

$\int_R F(x) dm =\int_R \sum_{n=1}^{\infty} f(x+n) dm$ =(by hint f(x+n) is measurable, and using a theory to change sum and $\int$) = $\sum_{n=1}^{\infty} \int_R f(x+n) dm= \sum_{n=1}^{\infty} f(x+n) \int_R \chi_R dm= \sum_{n=1}^{\infty} f(x+n) m(R)= \sum_{n=1}^{\infty} f(x+n) * \infty= F(x)* \infty$.

So, assuming that $\int_R F(x) dm$ is finite so $F(x)$ must be 0.

$F(x)=\sum_{n=1}^{\infty} f(x+n)=0$ therefore $f(x+n)=0$ for every n
, since $f(x+n):R\to [0, \infty]$.

So, $\int_R f(x+n) dm =\int_R 0 dm= 0* \int_R \chi_R dm=0* \infty=0$.

Can you explain how to show the hint?

答:
Your statement $$\int_R f(x+n) \, dm = f(x+n) \int_R \chi_R \, dm$$ is egregiously wrong.

Instead you should use the translation invariance of the integral (which follows from the hint you were given) to write $$\int_R F(x) \, dm = \int_R \sum_n f(x+n) \, dm = \sum_n \int_R f(x+n) \, dm = \sum_n \int_R f(x) \, dm.$$ The last expression on the right is, by definition,
$$\lim_{n \to \infty} \left[ n \int_R f(x) \, dm \right]$$ which is not finite if $0 < \int_R f(x) \, dm \le \infty$.


问:
How do you call a partition of a totally ordered set that....
A partition of a set S is a collection P={B1,B2,...,Bk}  of subset of S such that for each i, j, ∈ {1,2,...,k}:

Bi ≠ ∅

Bi ⋂ Bj ≠ ∅

B1 ⋃ B2 ⋃...⋃ Bk = S

now if S is a totally ordered set and P is also equipped with a partial order ≼ how do you call a partition that meets the following additional condition?

for x ∈ Bi, y x ∈ Bj, if x<y then i≤j (or B i ≼ Bj)

For example if S= {1, 2, 3, 4} we can have the following partitions of S into two subsets:

1/234, 12/34 and 123/4 (but not 13/24 nor 14/23)

I've come accross the following definitions: monotone partition, consecutive partition...but I don't seem to find a commonly accepted terminology nor any body of litterature that would study these ...  (I know it is subset of partitions and non-crossing partitions)

Any directions, hints for me?

Many thanks

P.S: to give a bit of context, I wanted to study this type of partitions in details (their lattice structure, total number given binomials as suggested by Donald S. etc...)

答:
By your p.s I think you would appreciate the following discussion.

Notice that your partitions, because there is an order inherit from the line, are completely characterize by the size of the sets. So, a partition of your interest say $\pi =\{B_1,B_2,\cdots ,B_k\}$ is such that $B_i=[x_1+\cdots +x_{i-1}+1,x_1+\cdots +x_{i-1}+x_i]\cap \mathbb{Z}$ where $x_0=0$ so the size of $|B_i|=x_i.$ in this way, the number of these partitions is $$|\{(x_1,\cdots ,x_k):\sum _{i=1}^kx_i=n,x_i>0\}|=\binom{n-1}{k-1},$$
which adds up to $2^{n-1}$ if you add over the number of parts. That is why is not very interesting to compute them and also the motivation to call them compositions.

In terms of the order structure that they inherit from the partitions you can check that the refinement looks like the Boolean lattice for $[n-1].$

Notice that a very similar concept that is very well studied is the concept of flag. The partitions that you describe are what happens to basis elements in flags of $\mathbb{R}^n.$


问:
Calculating the area of part of the surface area $x^2+y^2+z^2=4$ which lies inside the surface $x^2+y^2=2x$
The area of the part of the surface $x^2+y^2+z^2=4$ which lies inside the surface $x^2+y^2=2x$ is equal to $n(\pi-2)$ for an integer n. What is the value of n?

What I know so far:

I get that my answer is $4(\pi-2)$ hence n=4?

答:
One is a sphere of radius $R=2$, and the other is a cylinder of radius $1=R/2$.

Keeping the ratio of the radii, the area intercepted on the sphere shall be proportional to $R^2$ at varying $R$ : thus $n=2^2$


问:
Limit of $\frac{1}{M} \int_1^M M^{\frac{1}{x}} \textrm{d} x$ when $M \to +\infty$
The exercise

Find the limit of
$$
\frac{1}{M} \int_1^M M^{\frac{1}{x}} \textrm{d} x
$$
when $M \to +\infty$.

My try

I tried a change-of-variable but as the lower bound of the integral is $1$ and not $0$, it wasn't successful. I showed that the limit, if it exists, is $\geq 1$. We also have: $\frac{1}{M} \int_1^M M^{\frac{1}{x}} \textrm{d} x = \int_1^M M^{\frac{1}{x}-1} \textrm{d} x$ and in the integral, $\frac{1}{x}-1 < 0$.

Any help is welcome.

答:
Here is another proof you could use, which is more like the standard analysis proof.

Observe that $$\lim_{x\to \infty} M^{\frac{1}{x}} = 1$$
Hence there exists $K(\epsilon)$ such that $$\forall x > K : \left | M^{\frac{1}{x}} - 1 \right | < \epsilon$$
Also, note that: $$\sup_{1\leq x \leq K} \left | M^{\frac{1}{x}} - 1 \right | = c < \infty$$
for some constant $c(K)$.

Now, combining the above, we have that:
\begin{align}
 0 &\leq \frac{1}{M} \left | \int_1^M M^{\frac{1}{x}}dx -  \int_1^M 1 dx \right| \\ &\leq \frac{1}{M} \int_1^M \left | M^{\frac{1}{x}} - 1 \right| dx \\
&= \frac{1}{M} \int_1^K \left | M^{\frac{1}{x}} - 1 \right| dx + \frac{1}{M} \int_K^M \left | M^{\frac{1}{x}} - 1 \right| dx \\
&\leq \frac{c(K-1)}{M} + \frac{\epsilon (M- K)}{M} \\
&\leq \frac{c(K-1)}{M} + \epsilon \\
&\to 0 \text { as } M\to \infty
\end{align}

Hence, we can conclude that:
$$\frac{1}{M} \int_1^M M^{\frac{1}{x}}dx= \frac{1}{M}\int_1^M 1 dx = 1$$


问:
$x_n\to x$ in a metric space if and only if the Dirac measures $\delta_{x_n}$ converge weakly to $\delta_x$
Let $E$ be a metric space and $\delta_x$ denote the Dirac measure concentrated on $x$ for $x\in E$.

Now let $(x_n)_{n\in\mathbb N}\subseteq E$ and $x\in E$. I would like to show that

$x_n\xrightarrow{n\to\infty}x$ in $E$;

$\delta_{x_n}\xrightarrow{n\to\infty}\delta_x$ weakly$^1$

are equivalent. In a second step, I would like to show that if $\delta_{x_n}\xrightarrow{n\to\infty}\mu$ for some probability measure $\mu$ on $E$, then $\mu=\delta_z$ for some $z\in E$.

(1.) implies (2.) is a triviality, but I'm not sure how to conclude the other implication.

Assume (2.). It's not enough to conclude, but we may observe the following: Assume $$x_{n_k}\xrightarrow{k\to\infty}y\tag2$$ for some increasing $(n_k)_{k\in\mathbb N}$ and some $y\in E\setminus\{x\}$. Then, since $C_b(E)$ is separating, there is a $f\in C_b(E)$ with $f(x)\ne f(y)$. But now, since $f$ is continuous and $(1)$, $$f(y)=\lim_{k\to\infty}f(x_{n_k})=\lim_{k\to\infty}=\delta_{x_{n_k}}=\delta_xf=f(x)\tag3;$$ in condradiction to $x\ne y$.

So, how can we prove that (2.) implies (1.) and what do we need to do, to show the second step?
$^1$ i.e. $$\delta_{x_n}f=f(x_n)\xrightarrow{n\to\infty}f(x)=\delta_x(f)\tag1\;\;\;\text{for all }f\in C_b(E).$$

答:
So as I showed in our discussion and function of form
$$ f(y) = 1 \wedge d(y,x)$$
would prove the implication (2) --->(1).

Now for your second statement, we can do as follows:

Main reply

Let
$$S =\{ x_n | n \in \mathbb{N}\}$$
Firstly, if $(x_n)$ admit a convergent subsequence with a limit $x$, our question is done.

So assume there is no such property, then $\overline{S}=S $ and $(x_n)$ seperating

Now, let's consider the following function :
$$ g(y)= 1 \wedge d( y, \overline{S})$$

So as we know, this function is bounded and continuous.Morever :
$$g(y)=0 \quad \iff \quad y \in \overline{S}$$
Thus we imply that $supp(\mu) \subset \overline{S}=S$

Thus $\mu$ must have the following form:
$$ \mu = \lambda_1 \delta_{x_1}+\lambda_2 \delta_{x_1}+...$$
Where $(\lambda_n)$ is a nonnegative real sequence such that $ \sum_n \lambda_n=1$

Using the separation of $(x_n)$ to construct some functions to show that $$\lambda_n=0 \quad \forall n$$ Therefore,  our conclusion

$\square$

Comments

I hope this one solution does not have any error.

Perhaps, ( if mine is true), there is a more elegant solution.


问:
Compactly supported hamiltonian diffeomorphisms
I'm having trouble with an Exercise (12.3.6) from McDuff's and Solomon's book, "Introduction to Symplectic Topology" (3rd Ed.).

The goal is to prove the monotonicity of the symplectic displacement energy. But in simpler terms, here is the question.

Let $\phi \in Ham_c(\mathbb{R}^{2n},\omega_0)$ be a hamiltonian diffeomorphism with compact support. We define $\phi_\lambda(z) = \lambda \phi(\frac{1}{\lambda}z)$ for $\lambda>0$, and we want to prove that $\phi_\lambda \in Ham_c(\mathbb{R}^{2n}, \omega_0)$.

The book provides a hint: consider the hamiltonian functions $H_\lambda (z) = \lambda^2 H(\lambda^{-1}z)$.

So, following the hint, we consider $H:[0,1]\times \mathbb{R}^{2n}\to\mathbb{R}$ the hamiltonian that generates $\phi$ as the time one map of its flow: $\partial_t \phi_t^H = X_{H_t}\circ \phi_t^{H}$, where $X_{H_t}$ is the hamiltonian vector field of $H_t = H(t,.)$. So, $\phi = \phi_1^H$. Then, given $H$, we consider $H_\lambda$ as defined before.

At this stage, two approaches seem reasonable: either we try to write $X_{H_\lambda}$, or we try to guess an isotopy that generates $\phi_\lambda$.

When I try to compute $X_{H_\lambda}$:

$d_pH_\lambda(v_p) = \lambda^2 d_pH(\lambda^{-1}v_p) \implies d_pH_\lambda(v_p) = \lambda^2\omega_0(X_H(p), \lambda^{-1}v_p) = \omega_0(\lambda^2X_H(p), \lambda v_p)$.

But I am not able to explicitly write $X_{H_\lambda}$ from there!

Also, if I try to guess an isotopy that generates $\phi_\lambda$, I tried to use $\psi_t(z)  = \lambda \phi_t^{H}(\lambda^{-1}z)$, but differentiating w.r.t time:

$\partial_t \psi_t(z) = \lambda\partial_t\phi_t^H(\lambda^{-1}z) = \lambda X_{H_t}(\phi_t^H(\lambda^{-1}z)) = \lambda X_{H_t}(\lambda^{-1}\psi_t(z)) \implies \partial_t\lambda^{-1}\psi_t(z) = X_{H_t}(\lambda^{-1}\psi_t(z))$.

But by uniquiness, this would imply that $\lambda^{-1}\psi_t = \phi_t^H$ which makes no sense for me.

I am surely commiting some mistake in the calculations of this derivatives, but I can't seem to find it.

答:
For clarity, I am taking $H_\lambda(z,t) := \lambda^2 H(\lambda^{-1}z,t)$, and suppressing the $t$.

There are a couple of problems in your derivation. First of all the definition of $H_\lambda$ implies
$$
d_zH_\lambda(v_z) = \lambda^2 d_{\lambda^{-1}z}H (\lambda^{-1}v_z)
$$
(note the different basepoint in the derivative). Also, the next step should be
$$
\lambda^2\omega_0(X_H(\lambda^{-1}z),\lambda^{-1}v_z) = \omega_0(\lambda^2 X_H(\lambda^{-1}z),\lambda^{-1}v_z) = \omega_0(\lambda X_H(\lambda^{-1}z),v_z).
$$
Here I'm using the fact the $\omega_0$ is bilinear, not linear in $\mathbb{R}^{2n}\oplus\mathbb{R}^{2n}$ as you take it to be. From this, we deduce
that $X_{H_\lambda}(z) = \lambda X_H(\lambda^{-1}z)$. It's then not too difficult to show that $\phi^{t,0}_{H_\lambda}$ and $z\mapsto \lambda \phi^{t,0}_H(\lambda^{-1}z)$ are the same diffeomorphism, since they both satisfy the ode
$$
\frac{\partial}{\partial t}\psi^t = X_{H_\lambda}\circ \psi^t \quad\text{and} \quad \psi^{0} = \operatorname{id}_{\mathbb{R}^{2n}}.
$$
Evaluating at $t=1$, we get
$$
\phi^{1,0}_{H_\lambda}(z) = \lambda\phi_H^{1,0}(\lambda^{-1}z) = \lambda\phi(\lambda^{-1}z) =:\phi_\lambda(z).
$$
So $\phi_\lambda \in \operatorname{Ham}_c(\mathbb{R}^{2n},\omega_0)$, as claimed.


问:
Localization as dual to quotienting?
Why does localization look so close to quotienting? Consider the ismorphism theorem:

For a ring map $\phi: R_1 \rightarrow R_2$, let $\ker\phi \equiv \{ r_1\in R_1 : \phi(r_1) = 0 \}$. Then, $\ker \phi$ is an ideal of $R_1$, and there exists an epi-mono factorization of $\phi$ into $R_1 \twoheadrightarrow R_1 /\ker\phi \hookrightarrow R_2$  where $\hookrightarrow$ is injective and $\twoheadrightarrow$ is surjective.

Now consider the similar theorem for localization, where I denote by $R  \hat \times S \equiv S^{-1}R$ for notational suggestivity. Now I write down:

For a ring map $\psi : R_1 \rightarrow R_2$, Let $\operatorname{rek}\psi \equiv \{ r_1 \in R_1 : \psi(r_1) = 1 \}$. Then $\operatorname{rek}\psi$ is a multiplicative subset of $R_1$ and there exists a mono-epi factorization of $\psi$ into $R_1 \hookrightarrow R_1 \hat \times \psi \twoheadrightarrow R_2$

I can build a table:

$\phi \leftrightarrow \psi$.

$\ker \leftrightarrow\operatorname{rek}$.

injection $\leftrightarrow$ surjection.

$/ \leftrightarrow \hat\times$

ideal $\leftrightarrow$ multiplicative subset.

$0 \leftrightarrow 1$.

to convert from the quotienting into localization. Is there some "deep" going on here for this duality? This $0 \leftrightarrow 1$ business makes me hopeful that there might be something deeper / categorical lurking in the background.

EDIT: I had only commutative rings in mind when I wrote this. Please feel free to take assumptions on $R$ as required (Commutative, Noetherian, for example), if that allows us to explain this "duality".

答:
Fabio gives a very nice answer to your question, but doesn't directly address an important point of confusion in your original post/comments, so I'm adding this answer for posterity. In general, a map $\psi:R_1\rightarrow R_2$  will absolutely not induce an epimorphism $\text{rek}(\psi)^{-1}R_1\twoheadrightarrow R_2$, even if we take the stronger definition $\text{rek}(\psi)=\psi^{-1}(R_2^\times)$ given by Fabio. For instance, if every element of $\text{rek}(\psi)$ is already a unit in $R_1$, then we will just have $\text{rek}(\psi)^{-1}R_1=R_1$, and so using this it is easy to come up with examples where the induced map is not epi.

For instance, take $R_1=\mathbb{Q}$, and $R_2$ any field extension of $\mathbb{Q}$ with a non-trivial automorphism $\alpha$ that fixes $\mathbb{Q}$ pointwise, with $\psi:R_1\hookrightarrow R_2$ the inclusion map. Then $\text{rek}(\psi)=\mathbb{Q}^\times$ , so $\text{rek}(\psi)^{-1}\mathbb{Q}=\mathbb{Q}$ and the induced map to $R_2$ is just $\psi$, which is certainly not an epimorphism. (E.g. $\alpha\circ\psi=\text{id}_{R_2}\circ\psi$ but $\alpha\neq\text{id}_{R_2}$).

Indeed, the polynomial ring example you give in the comments of your post does not hold in general either. If we let $R_1=\mathbb{R}[x]$ and $R_2=\mathbb{R}[x,y]$, with $\psi:R_1\hookrightarrow R_2$ again the inclusion map, then once again $\text{rek}(\psi)=R_1^\times$ but $\psi$ is certainly not epi.

The problem in all of these examples is that $R_2$ can be very big compared to the image of $R_1$; hopefully the above examples clarify that point. (Note however, that – provided $R_2\neq\{0\}$ – the map $R_1\hookrightarrow \text{rek}(\psi)^{-1}R_1$ will still be injective, even if we use Fabio's stronger definition of $\text{rek}(\psi)$, because no element of $\text{rek}(\psi)$ can be a zero-divisor in $R_1$.)

Title: Localization as dual to quotienting?


答:
Note that $n^2 \equiv 0,1,4 \ (mod \ 5)$, which means that $-2n^2 \equiv 0,2,3 \ (mod \ 5)$. Therefore if $z^2 = 10x^2-2y^2 \equiv -2y^2 \ (mod \ 5)$, then $z$ and $y$ are both divisible by 5, so $z=5z_1$ and $y=5y_1$ for some integers $y_1$ and $z_1$. Then equation becomes $5z_1^2=2x^2-10y_1^2$, which means that $x$ is divisible by 5, so $x=5x_1$ for some $x$.

Now we get $z_1^2=10x_1^2-2y_1^2$ which is of the same form as the original one, so we can repeat these steps infinitely many times and get $(x,y,z)=(5x_1,5y_1,5z_1)=(5^2x_2,5^2y_2,5^2z_2)=...=(5^nx_n,5^ny_n,5^nz_n)=...$ . However, the only number which is divisible by any power of 5 is zero, which means that $(x,y,z)=(0,0,0)$, so there are no non-trivial solutions.

Title: Integer solutions to $z^2=10x^2-2y^2$

Body:
I need to show that a $(67, 12, 2)$-design does not exist. I tried to use the Bruck, Ryser and Chowla theorem which in this case states that if a $(67, 12, 2)$-design exists, then the equation
$$z^2=10x^2-2y^2,$$
has a non-trivial integer solution $(x,y,z)\neq \overline{0}$. Checking from Wolfram I found that it does not have any, which would prove my original problem, but I would like to see proof why it doesn't have any.

It is easy to see that $z$ should be even so I tried to plug in $z=:2k$ and split it in some cases but it did not go anywhere.


问:
If $p>3$ and $p+2$ are twin primes then $6\mid p+1$
I have to prove that if $p$ and $p+2$ are twin primes, $p>3$, then $6\ |\ (p+1)$.
I figure that any prime number greater than 3 is odd, and therefore $p+1$ is definitely even, therefore $2\ |\ (p+1)$. And if I can somehow prove $3\ |\ (p+1)$, then I would be done. But I'm not sure how to do that. Help!</p>

答:
Consider a triple {m,n,o} of consecutive natural numbers.

Then, at least one of these numbers must be divisible by 2 - an even number, and at least one of these numbers - even or odd - must be divisible by 3 (not necessarily but possibly the same number).

The triple must either be odd-even-odd or even-odd-even.

Since all twin primes are odd primes, only an odd-even-odd triple can contain a twin prime tuple.

So, because of (2.), for primes > 3, the middle number n between twin primes must be both divisible by 2 and 3.

In other words, in the triple {p, n, p+2} (| p+2 prime),  n ≡ 0 (mod 6). ■


问:
surface integral with normal unit vector
Let  $\vec f(\vec r) = (y,-x, zxy)$ and let $S$ be the surface $x^2+y^2+3z^2=1$ , $z≤0$, with unit normal vector $\vec n$ pointing in the positive $z$-direction. The value of the surface integral $$\iint_S(\nabla \times \vec f) \cdot \vec n\,\mathrm dS$$ is $n\pi$ where $n$ is an integer. What is the value of $n$?

what i know:

I know you need to use stokes theorem when calculating this. I have calculated the curl of the vector field as $(xz, -yz, -2)$. I dont know where to go from this, any help would be appreciated.

答:
If you want to use Stokes' theorem, then there's no need to actually compute the curl of $\vec f$. The theorem lets you exchange integrating the curl of $\vec f$ over $S$ with the line integral of $\vec f$ over the boundary of $S$, which is the ellipse/circle $x^2+y^2=1$ in the plane $z=0$.

So you have

$$\begin{align}
\iint_S(\nabla\times\vec f)\cdot\vec n\,\mathrm dS&=\oint_{\partial S}\vec f\cdot\mathrm d\vec r\\[1ex]
&=\oint_{x^2+y^2\le1}(y,-x,xyz)\cdot\mathrm d\vec r\\[1ex]
&=\int_0^{2\pi}(\sin t,-\cos t,0)\cdot(-\sin t,\cos t,0)\,\mathrm dt\\[1ex]
&=-\int_0^{2\pi}\mathrm dt
\end{align}$$

where we convert to polar coordinates $(x,y,z)=(\cos t,\sin t,0)$ in the third line. Determining the value of $n$ from this point is trivial.


问:
Is the ring of the polynomials on a finite algebraic set always a product of local rings?
A proposition from Fulton's book 'Algebraic Curves':

Let $I$ be an ideal in $k[X_1, ..., X_n]$ ($k$ algebraically closed), and
suppose $V(I) = \{ P_1, \ldots, P_N \}$ is finite. Let
$\mathcal{O}_i = \mathcal{O}_{P_i}(\mathbb{A}^n)$. Then there is a natural
isomorphism of $k[X_1, \ldots, X_n] \mathbin{/} I$ with
$\prod_{i = 1}^{N} \mathcal{O}_i \mathbin{/} I\mathcal{O}_i$.

But if $k$ is not algebraically closed, is it still true that, if $V(I)$ is
finite, $k[X_1, \ldots, X_n] \mathbin{/} I$ is a direct product of local
rings? And if so, then some factors are most likely given by localizations at
the points of $V(I)$, whereas other factors are not; how to describe these
latter factors?

The reason for this question is the observation that field extensions of $k$
are fields and thus local rings. For instance,
$\mathbb{Q}[X] \mathbin{/} (X^2 + 1)$ is a field, but $V(I)$ is empty and
$\mathbb{Q}$ is not algebraically closed. Hence, by the Chinese remainder
theorem, it seems to me that, for any $f$ in $k[X]$, $k[X] \mathbin{/} (f)$ is
the direct product of local rings anyway. Is it at least that correct?

答:
If you upgrade to schemes and take $V(I)$ to be the closed subscheme $\operatorname{Spec} k[X_1,\cdots,X_n]/I \subset \operatorname{Spec} k[X_1,\cdots,X_n] = \Bbb A^n_k$, then your claim that $V(I)$ finite implies a natural isomorphism between $k[X_1,\cdots,X_n]/I$ and $\prod \mathcal{O}_i/I\mathcal{O}_i$ is still true. Without upgrading to schemes or enforcing the requirement that $k$ is algebraically closed, you cannot guarantee this result will still hold.

In the case of a scheme, the reason is that the former is the global sections of the structure sheaf on $V(I)$, while the latter is the product of the stalks, and on any discrete space these two things are naturally isomorphic. The sheaf condition gives that the global sections are the product of sections over each open set $\{p_i\}$, while the fact that the $\{p_i\}$ are open means they're the final object among open sets containing $p_i$ and thus the stalk at $p_i$ is the value of the sheaf on $\{p_i\}$.


问:
Interval of unique solution of differential equation $y’=x^2+y^2, y(0)=1 .$
Consider the following initial value problem  $$y’=x^2+y^2, y(0)=1$$

Which of the following statements is/ are

$1.$ there exist a unique solution in $[0,\frac{\pi}{4}]$.

$2.$ every solution is bounded on $[0,\frac{\pi}{4}]$.

$3.$ The solution has a singularity at some point in $[0,1]$.

$4.$ The solution because unbounded at some sub interval of $[\frac{\pi}{4}, 1]$.

I tried to find interval for unique solution by using existence and uniqueness theorem as

For $$|x|\leq a$$ $$|y-1|\leq b$$ we have

$$|f(x,y)|=|x^2+y^2|\leq a^2+(1+b^2)$$
$$h=min\{a,\frac{b}{M}\}=min\{a,\frac{b}{1+a^2 +b^2+2ab}\}$$ Now to find maximum of $$g(b)= \frac{b}{1+a^2 +b^2+2ab}$$ we have $$g’(b)= \frac{1+a^2-b^2}{(1+a^2 +b^2+2ab)^2}=0\iff b=\pm\sqrt{1+a^2}$$ So $$h=min\{ a,\frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}\}$$ $$a= \frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}$$ gives $$ 2a(1+a^2)=(1-2a^2)\sqrt{1+a^2}$$ $$a=\frac{1}{2\sqrt{2}}$$ Now it seems no conclusions. If it is unique( that I don’t know how ) in $[0,\frac{\pi}{4}]$ then it will be bounded also in this interval as continuous functions are bounded on compact set and if solution can’t be extended to $[0,1]$ then it is because of unboundedness of solution as I guess . I don’t know how to handle last two portions. Please help. Thanks.

答:
As you found out, this is a duplicate of checking uniqueness and boundedness of an initial value problem.

As to your solution attempt, the minimum in the determination of $h$ from the "butterfly" condition is just the minimum of two numbers, not of a function. Thus
$$
M=a^2+(1+b)^2
$$
gives besides $h\le a$ the condition $h\le \frac{b}{a^2+(1+b)^2}$. You can see that you made some transcription errors up to this point.

Next you want to ensure contractivity of the Picard iteration, to that end determine the Lipschitz constant $L=\max2|y|=2(1+b)$. Contractivity is ensured as long as $Lh<1$.

If you try out values for $a,b$ like $a=b=1$, you will find rather small admissible values for $h$, thus this approach is not very suited for this task.


问:
How does $(abc) = (ac)(ab)$?
I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots  \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$

However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.

Thank you in advance!

答:
We can also conveniently calculate the product of transpositions in cycle notation by switching to the two-line notation of permutations.

Considering the first identity $(abc)=(ac)(ab)$ we have the representation
\begin{align*}
(abc)\equiv\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\qquad\qquad&
(ac)\equiv\begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\\
&(ab)\equiv\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
\end{align*}

We obtain
\begin{align*}
\color{blue}{(ac)(ab)}&\equiv \begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}\circ\begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}\\
&=\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\tag{1}\\
&\,\,\color{blue}{\equiv(abc)}\tag{2}
\end{align*}

Comment:

In (1) we use the convention to multiply out from right to left:
\begin{align*}
&a\to b\to b\\
&b\to a\to c\\&
c\to c\to a
\end{align*}

In (2) we switch back to the cycle notation: $a\to b\to c\to a$.

In the same way we can calculate e.g. the next identity with four elements:
\begin{align*}
\color{blue}{(ad)(ac)(ab)}&
\equiv  \begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\left(\begin{pmatrix}a&b&c&d\\c&b&a&d\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&a&c&d\end{pmatrix}\right)\\
&=\begin{pmatrix}a&b&c&d\\d&b&c&a\end{pmatrix}
\circ\begin{pmatrix}a&b&c&d\\b&c&a&d\end{pmatrix}\\
&=\begin{pmatrix}a&b&c&d\\b&c&d&a\end{pmatrix}\\
&\,\,\color{blue}{\equiv (abcd)}
\end{align*}
and the claim follows.

Title: How does $(abc) = (ac)(ab)$?


问:
What is the probability of when drawing 10 items from a set of 100 with replacement that you draw the same 5 items twice in the 10 draws?
If I am drawing 10 items from a set of 100, each item is unique i.e $X_1$ to $X_{100}$, what is the probability that in the 10 draws I draw 5 items exactly twice. A success for example would be $\lbrace X_1,X_1,X_2,X_2,X_3,X_3,X_4,X_4,X_5,X_5 \rbrace$. Order does not matter.

So far I am thinking its 1*$\frac{99}{100}*\frac{98}{100}*\frac{97}{100}*\frac{96}{100}*\frac{5}{100}*\frac{4}{100}*\frac{3}{100}*\frac{2}{100}*\frac{1}{100}$. As we need to draw 5 unique items and then we need to draw the five unique items again.

答:
The five duplicates can be chosen ${100 \choose 5} = 75287520$ ways.  For each of these sets of $5$ there are $\frac{10!}{(2!)^5}$ equivalent rearrangements.

Thus there are $8537604768000$ ways to make ten selections of which there are exactly two repetitions from each of five categories (objects).

To get the probability divide this by the total number of ways you can choose 10 items without restriction:  $100^{10}$.

I get:  $p = 8.5376 \cdot 10^{-8}$.


问:
Meaning of a set of measure zero
I have problem with the meaning of a set of measure zero in the following sentence:

The set of initial data with the property $X$ that gives rise to some solutions is of measure zero in the space of all initial data with the property $X$.

Is this related to measure theory? If yes, how and what does it mean?

答:
Yes, this is a term from measure theory.  I encountered it while studying Hausdorff and Lebesgue measure, but it can be used with other measures as well.

A set with measure zero means just that -- it has no effective value when measured.  A canonical example with Lebesgue measure is the Cantor set.  While it is easy to define the measure of the interval $[0,1]$ on the number line to have measure 1, the Cantor set removes one-third of that interval in the first iteration, and one third of each segment in following iterations, such that with each iteration, the measure gets smaller, and the limit of such activities is that the measure of the Cantor set is zero.  This is not to say that the Cantor set is empty; indeed, the Cantor set has infinitely many endpoints that remain even after infinite iterations.

What I believe the quote is trying to suggest, then, is that the initial data that gives rises to your solutions exists, but the probability of selecting it at random is zero.  To use another wording from measure theory, the set of all initial data gives rise to your solutions almost nowhere.

Title: Meaning of a set of measure zero


问:
Using the comparison test to determine the convergence of these series
I'm given the following problem:

$\sum_{k=1}^\infty \frac{\sqrt{k}}{1+k^2}$

$\sum_{k=1}^\infty \frac{1}{(k-\pi)^2}$

I'm asked to solve these using the comparison test, however I'm unable to determine a suitable comparison. How does one choose such a series? Is there a technique that can be used, in general?

答:
For the first one, we have that

$$\frac{\sqrt{k}}{k^2+1}<\frac{\sqrt{k}}{k^2}=\frac{1}{k^{3/2}}$$

For the second one, we have that

$$\frac{1}{(k-\pi)^2}=\frac{1}{k^2-2k\pi +\pi^2}<\frac{1}{k^2-2k\pi}$$

note that for $k\geq 13$ we have
$$\frac{k}{2}>2\pi $$

This implies

$$\frac{1}{k^2-2k\pi}<\frac{1}{k^2-k\frac{k}{2}}=\frac{2}{k^2}$$


问:
Parametric form of square
What is the appropriate parametric equation of the boundary of a square? For example, the unit circle has a parametric equation $x(t)=\cos(t)$ and $y(t)=\sin(t)$.</p>

答:
You can just break it down into 4 line segments:
$c(t)=(x(t),y(t))=$
$(t,0); 0\leq t\leq 1$

$(1, t-1); 1\leq t\leq 2$

$(3-t,1); 2\leq t\leq 3$

$( 0, 4-t); 3 \leq t\leq 4. $

Title: Parametric form of square


问:
Divisibility and remainders
I know that if $d|a$ exists then for there is some $q$ such that $dq+r=a$ where $q$ is our quotient and $r$ is our remainder. Now I know that IF $d$ divides $a$ exactly then we have a remainder of $0$, and so $r\ge 0$.

My textbook however, expands to say $r$ must ALSO be $<d$, so $0\le r<d$. Where does the restriction $r<d$ come from?

答: In our usual division algorithm, we choose largest $q$ at each step. We ensure that $a - dq < d$ at every step. This is why $ r =a - dq < d$.

Title: Divisibility and remainders


问:
Rearranging elements in a matrix
It's been a while since I studied linear algebra. I am trying to get the elements of a matrix to be in a certain order to follow a convention used in optics. In that context, the elements of the matrix are partial derivatives. Here, imagine the simpler scenario where I have the equations:

$2x+y=1$

$x-y=0$

I can represent these as:

\begin{bmatrix}
2 & 1 & x & 1\\
1 & -1 & y &0
\end{bmatrix}

In the convention, I want the $x$ to be next to the $1$ and the $y$ next to the $0$. Therefor, the system can be equally represented as:
\begin{bmatrix}
-1 & 1& y & 0\\
1 & 2 & x & 1
\end{bmatrix}

If I were to generalize this to a $3\times 3$ (or higher order) matrix (i.e. add another equation to the list), how could I change change the order of the elements while still keeping the $x$ next to $1$ (and $y$ next to $0$)?

答:
The best way to understand how to use elementary operations in linear algebra is to discard conventions such as that in your question. Yes, discard!

You can use it for communication. And that should be the end of it. Never use it for understanding.

What

$$
\begin{equation}\tag{1}
\begin{bmatrix}
-1 & 1& y & 0\\
1 & 2 & x & 1
\end{bmatrix}
\end{equation}
$$

truly means is as follows.

$$
\begin{bmatrix}
-1 & 1\\
1 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
0\\
1
\end{bmatrix}
$$

In the above matrix-vector multiplication, you can clearly see your original equations.

Applying elementary row operations means to left-multiple both sides of the equation by the corresponding elementary matrix. And that corresponding matrix can be obtained by performing the same elementary row operation on the identity matrix.

For example, suppose you want to make row 2 add row 1. Perform that to the identity matrix, you get:

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

The row operation on the original equation is thus

$$
\begin{bmatrix}
1 & 0\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
1 & 0\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
0\\
1
\end{bmatrix}
$$

which simplifies to

$$
\begin{equation}\tag{2}
\begin{bmatrix}
-1 & 1\\
0 & 3
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
0\\
1
\end{bmatrix}
\end{equation}
$$

(1) is true if and only if (2) is true. Thus, you can solve (2) and get the answer for (1). You do that because the answer for (2) is (more) obvious.

To the specific question of "how could I change change the order of the elements while still keeping the x next to 1?"

It is the same idea. Switching row 1 and row 2 on identity matrix first. (1 and 2 are ofc arbitrary choices. You can generalize it to any pair of rows.)

In 2D, you get

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

from here, perhaps you can try simplifying the multiplication and see what you get.

Now that you understand what elementary row operations are. You should be able to state any linear system of equations in matrix-vector multiplication.

With the understanding, you should be able to re-derive the rule for switching rows any time, which is: you also need to switch the corresponding rows on the RHS of the equation.

Title: Rearranging elements in a matrix


问:
Showing a property for a set of rewriting rules
Let $\to$ be a relation over the set of binary strings of 0 and 1. $\to$ is defined by the following rules:

R1. $x10y \to x0001y $

R2. $x01y \to x1y $

R3. $x11y \to x0000y $

R4. $x00y \to x0y$

Where $x,y$ are arbitrary strings. The relation is interpreted as a rewriting system, mapping strings to stirings.

The exercise I am trying to solve ask for a function $f$ mapping strings of 0 and 1 into naturals such that $ x \to y $ implies $f (x) > f(y)$.

If the rewriting system would not have rules R1 and R3 one could simply take the lenght of the input string as the function f.

However it seems hard to find a function that yields the required inequality considering also the rules R1 and R4 that amplify the initial string.

Anyone happens to see such a function?

Edit: After the comment of Peter I would be interested in knowing if there is any easy way to prove that the rewriting system is locally confluent. That is, if a -> b and a -> c then b ⇝ d and c ⇝ d

答:
Here maybe an answer that doesn't really give any more insight to the problem, but as I see it, I is only suppose to show, that there are only finitely many possibilities to rewrite a string.

Taking a string wie will assign a natural number to each digit and the value of the string will be the sum of the assigned natural numbers. We will first count the number of 1's to the left of that digit, say the number is $n$, and then assign $4^n$ to the digit.

Let us now prove that $x\to y$ indeed implies $f(x)>f(y)$. For this we will consider $x10y$ and so on, hence let $x$ include $n$ 1's. Then
$$ f(x10y) - f(x0001y) = 4^{n+1} - 3\cdot 4^n = 4^n,$$
$$ f(x11y) - f(x0000y) =  4^n + 4^{n+1} - 4\cdot 4^n = 4^n,$$
$$ f(x01y) - f(x1y) = 4^n,$$
$$ f(x00y) - f(x0y) = 4^n.$$

My answer seems to have a strong similarity to the one given by quanta, but is carried out more explicitely.


问:
Can we assume that this separation is formed of connected sets?
Suppose $Y$ is a subset of $\mathbb{R}^n$ and we want to prove that $Y$ is connected. By contradiction we assume this is not the case, and so there are sets $A$ and $B$ such that $Y=A\cup B$ and the closure of each set does not interset the other set. My question is: can we ASSUME that $A$ and $B$ are connected?

答: No, consider the space $Y := (0,1) \cup (2,3) \cup (4,5) \subseteq \mathbb{R}$. This space is disconnected since there exists closed sets $(0,1) \cup (2,3)$ and $(4,5)$ whose union is the whole space and whose intersection is empty. But one of the two is not connected.


问:
What's the process between this two steps of this limit?
I'm reading a proof and when calculating a complex residue, this limit has to be computed:
$$\lim _{z \rightarrow(n+x)} \frac{(z-(n+x)) R(z)}{e^{2 \pi i z}-e^{2 \pi i x}}=\frac{R(n+x)}{2 \pi i e^{2 \pi i x}}$$
But I can't see where does the $2\pi i$ in the denominator come from. What's the process there? By the way, $R(z)$ is defined as a rational function, any, but I don't believe that influences the result.

答: Use L'hopital's rule.  We have, after taking derivatives, $\dfrac {R(z)+(z-(n+x))R'(z)}{2\pi i e^{2\pi iz}}\to\dfrac{R(n+x)}{2\pi ie^{2\pi ix}}$.


答:
hint

For the first, we have

$$(\forall k\ge 1)\; 1+k^2\ge k^2$$
and
$$\frac{1}{1+k^2}\le \frac{1}{k^2}$$

So, $$\frac{\sqrt{k}}{1+k^2}\le \frac{1}{k^{\frac 32}}$$

For the second,

$$(\forall k\ge 8) \;\; k-\pi\ge \frac k2$$

and
$$\frac{1}{(k-\pi)^2}\le \frac{4}{k^2}$$

Title: Using the comparison test to determine the convergence of these series

Body:
I'm given the following problem:

$\sum_{k=1}^\infty \frac{\sqrt{k}}{1+k^2}$

$\sum_{k=1}^\infty \frac{1}{(k-\pi)^2}$

I'm asked to solve these using the comparison test, however I'm unable to determine a suitable comparison. How does one choose such a series? Is there a technique that can be used, in general?


问:
Show that the rectangular box of maximum volume inscribed in a sphere of radius $r$ is a cube
Reading an old calculus book I have found this interesting exercise:

"Show that the rectangular box of maximum volume inscribed in a sphere of radius $r$ is a cube"

I solved this using Lagrange. But I think this can be solved using partial derivatives. However, I'm not entirely sure about this. So I would like to know if this is possible.

答:
Sphere is given by $x^2 + y^2 + z^2 = r^2$.

For the rectangular box with center at the origin,

$V = 8 xyz = 8xy \sqrt{r^2 - x^2 - y^2}$

$\frac{\partial V}{\partial x} = 8y \sqrt{r^2 - x^2 - y^2}  - \frac{8x^2y}{\sqrt{r^2 - x^2 - y^2}}$

To find critical points, we equate it to zero.

As $\sqrt{r^2 - x^2 - y^2} \ne 0$ in a sphere, we get $8y (r^2 - x^2 - y^2)  - 8x^2y = 0$

Again as $y \ne 0$, $2x^2 + y^2 = r^2$.

Similarly from the other partial derivative (wrt $y$), we get $2y^2 + x^2 = r^2$

Solving them for $x, y$ and then plugging into equation of sphere to find $z$, we will notice they are all equal and of length $\frac{2r}{\sqrt3}$.


问:
$\mathbb{Q}/\mathbb{Z}$ has cyclic subgroup of every positive integer $n$?
I would like to know whether $(\mathbb{Q}/\mathbb{Z},+)$ has

$1$. Cyclic subgroup of every positive integer $n$?

$2$. Yes, unique one.

$3$. Yes, but not necessarily unique one.

$4$. Does not have cyclic subgroup of every positive integer $n$.

What I know about the given group is infinite group but every element has finite order. Please help how to proceed.

答: HINT. The order of an element $p/q \in \mathbb{Q}/\mathbb{Z}$ is the least $n$ such that $\underbrace{p/q+\cdots+p/q}_{n\text{ times}}$ is an integer.


Title: Can we assume that this separation is formed of connected sets?

Body: Suppose $Y$ is a subset of $\mathbb{R}^n$ and we want to prove that $Y$ is connected. By contradiction we assume this is not the case, and so there are sets $A$ and $B$ such that $Y=A\cup B$ and the closure of each set does not interset the other set. My question is: can we ASSUME that $A$ and $B$ are connected?

Body: No, certainly not. Take $n=1,$ $Y=\{1,2,3,4\}$ and $A= \{1,2\}, B= \{3,4\}.$


问:
The James Space
Let us define $\tilde{J}$ to be the space of all sequences $\xi=(\xi(n))_{n=1}^{\infty}$ of real numbers with finite square variation; that is, $\xi\in\tilde{J} $ if and only if there is a constant M such that for every choice of integers $((p)_{j})_{j=0}^{n}$ with $1\leq p_{0}<p_{1}<\dots<p_{n} $ we have:
$$\sum_{j=1}^{n}(\xi(p_{j})-\xi(p_{j-1}))^{2}\leq M^{2}$$

Can someone tell why is this:
If $\xi \in \tilde{J},$ then $\lim_{n\to\infty}\xi(n)$ exists and $\lim_{n\to\infty}\xi(n)=0$?

答:
Assume that $\xi(n)$ is not a Cauchy sequence. Then there is an $\epsilon_0>0$ such that
for every $n$
$$\tag{1}(\xi(k)-\xi(m))^2\geq \epsilon_0^2$$
for some $k,m\geq n$. Now, fix $N\geq 1$. Using $(1)$ recursively, we find pairs of indices $(k_j,m_j)$ for $j=1,...,N$ with the following ordering
$$k_1<m_1<k_2<m_2<...<k_N<m_N$$
and $$\tag{2}(\xi(k_j)-\xi(m_j))^2\geq \epsilon_0^2$$ for every $j=1,...,N$ (check that this can be done).Now let $$p_1=k_1,p_2=m_1,...,p_{2N-1}=k_N,p_{2N}=m_N$$
Then this choice of integers has the orderding $p_1<p_2<...<p_{2N-1}<p_{2N}$ and by $(2)$
$$\sum_{j=1}^{2N-1}(\xi(p_j)-\xi(p_{j+1}))^2\geq \sum_{k=1}^{N}(\xi(k_j)-\xi(m_j))^2\geq N\epsilon_0^2$$
Since, N was arbitrary we conclude that $\xi(n)$ cant have a finite square variation.

Title: The James Space


问:
How to create my own table for student's t distribution?
I'm trying to compute the values of a student's t distribution, but I don't get how is it calculated. I need something like

$$
P(T<t)=t_{n,\alpha}
$$

but im not sure where does the parameter $\alpha$ enters in this coputation, $\alpha$ is the parameter for the confidence level $1-\alpha$ and $n$ is the number of degrees of freedom.

答:
As to why you may experience some difficulties computing the values for the  Student's t-distribution, some background comments from Wikipedia on the topic of the quantile function (inverse of the CDF), to quote:

The evaluation of quantile functions often involves numerical methods...Quantile functions may also be characterized as solutions of non-linear ordinary and partial differential equations. The ordinary differential equations for the cases of the normal, Student, beta and gamma distributions have been given and solved.[4]

And further:

Student's t-distribution
(further information: Student's t-distribution)

This has historically been one of the more intractable cases, as the presence of a parameter, ν, the degrees of freedom, makes the use of rational and other approximations awkward. Simple formulas exist when the ν = 1, 2, 4 and the problem may be reduced to the solution of a polynomial when ν is even. In other cases the quantile functions may be developed as power series.[8]

My advice, get good technical support as this has been described as 'one of the more intractable cases', but with special exceptions.


问:
Nowhere-Dense-Sets Clarified
If A is a nowhere dense set, it means that "A closure" has an empty interior.
Canwe also say that "A" too has an empty interior?

I believe Yes, we can say so.

Because "A closure" is just the union of A and it's limit points. The set of limit points are always closed(since they are not open). Now when we talk about "A Closure's interior", we refer to A's interior. And so if we say A closure has an empty interior, it means "A Closure" sans the closed subset of A (which is the set of it's limit points) has an empty interior. In other words, A has an empty interior? Am I Right?

答:
If $A$ is nowhere dense, then, as $\operatorname{int}(A) \subseteq \operatorname{int}(\overline{A}) = \emptyset$, $A$ has empty interior.

But e.g. $([0,1] \cap \Bbb Q)$ has empty interior in $\Bbb R$, but is not nowhere dense, so it's quite a bit stronger than having empty interior.

Title: Nowhere-Dense-Sets Clarified


问:
Why are there no cycles of length 2?

A graph $G$ is an ordered pair $(V,E)$ of disjoint sets where
$$E\subset V^{\underline{2}}:=\left\{\{x,y\}\ |\ x,y\in V\land x\ne
 y\right\}.$$

Let $G=(V,E)$ be a graph. A path $P$ in $G$ is an ordered list of
elements of $V$ $$P=(x_0,x_1,\ldots x_n)$$ for some $1\leq n$ such
that $\{x_{i-1},x_i\}\in E$ for all $1\leq i\leq n$. We call $n$ the
length of the path $P$.

Let $G=(V,E)$ be a graph. A cycle is a path $(x_0,x_1,\ldots,x_n)$
such that $x_i\ne x_j$ for all $i\ne j\in[0,n-1]$ and $x_0=x_n$. Note
that a cycle has length at least 3.

Why do these definitions imply that a cycle must have a length of at least 3? For example, let $V:=\{a,b\}$ where $a\ne b$. Write $x_0:=a$, $x_1:=b$ and $x_2:=a$. Then $(x_0,x_1,x_2)$ is a path of length 2 which satisfies the definition of cycle.

答: This is not stated as it is written here, but it is conventional in most definitions of a path that each vertex in the path is distinct (except perhaps the end points if it forms a cycle). This also implies that each edge is distinct, and why it is noted that a cycle must have length at least 3. The way this definition of a path is written is more consistent with the definition of a walk, where revisiting a vertex is allowed.

Title: Why are there no cycles of length 2?


问:
Show that a prime p divides the sum of polynomials with integer coefficients of degree less than p - 1
Here's the full question: Let $F\in\mathbb Z[x]$ be a polynomial of degree less than $p-1$, show that
$$
p | F(0) + F(1) + ... + F(p-1)
$$

I don't really know how I should get started with the problem, I've thought about Fermat's little theorem but since $F$ is of degree less than $p-1$, it doesn't really apply. Any hints on what theorems I should consider or approach to try would be great, thanks!

答:
Claim 1: For  a fixed $ 1 \leq k \leq p-2$,  $p \mid \sum_{i=1}^{p-1} i^k$.

This almost corresponds to $ F(x) = x^k$ in the question.

Proof: Work mod $p$. Let $ g$ be a primitive root modulo $p$.

Then, $ \{ 1, 2, \ldots, p-1 \} = \{ g^0, g^1, \ldots g^{p-1}\}$.

So $ \sum_{i=1}^{p-1} i^k = \sum_{j=1}^{p-1} g^{jk} = \frac{ g^{(p-1)k } - 1 } { g^ k  - 1 }$ .

Note that $ p \not \mid g^k - 1$ and $ p \mid g^{(p-1)k } - 1 $, hence $p$ divides the fraction.

Claim 2: For  a fixed $ 0 \leq k \leq p-2$,  $p \mid \sum_{i=0}^{p-1} i^k$.

This corresponds to $ F(x) = x^k$ in the question.

Proof: For $ k = 0$, the sum is $p$, so the claim is true.

For $k>0$, this is (almost) the previous claim.

Note: For $ k = p - 1$, the statement isn't true since the sum is $ \equiv p-1 \pmod{p}$. Which of the above steps fail?

Corollary: The question is proved.

$F(x) = \sum_{k=0}^{p-2} a_k x^k,$ and the question is true for each $a_k x^k$, so it's true for the sum.

Note: We can prove Claim 1 without the machinery of primitive roots.

EG Using Method of Differences to show that for prime $n$, $ n \mid \sum_{i=1}^{n} i^k$ because the denominators of the coefficients $\mid (p-1)!$ and the numerator is a multiple of $p$.

Or appeal to Faulhaber's formula.


问:
Area of a region in first quadrant
Find the area of the region bounded by  paraboloid $z = x^2 + y^2 $ lies below $z = 4$  and in the first octant.

Where I am going wrong? What is the correct area?

My work:

$A = \int \int_D \sqrt{ \left (\dfrac{\partial z}{\partial x} \right ) +  \left (\dfrac{\partial z}{\partial y} \right ) + 1} \ dA $

$z = x^2+y^2$

$z = 4$

$x^2+y^2 =4$

$\left (\dfrac{\partial z}{\partial x} \right )^2 +  \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 4(x^2+y^2) +1$

$\left (\dfrac{\partial z}{\partial x} \right )^2 +  \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 4(4) +1$

$\left (\dfrac{\partial z}{\partial x} \right )^2 +  \left (\dfrac{\partial z}{\partial y} \right ) ^2+ 1 = 17$

The area is,

$A = \sqrt{17} \int \int_D dA$

$A = \sqrt{17} $(Area of circle with radius 2)

$A = 4 \sqrt{17} \pi$

Is this area correct(in first quadrant). If this is wrong kindly show me the direction where I am going worng.

答:
If $D$ denotes the surface of the given solid, then the area of $D$ over the paraboloid is

$$\begin{align}
\iint_S \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}\,\mathrm dA&=\iint_S\sqrt{1+4(x^2+y^2)}\,\mathrm dx\,\mathrm dy\\[1ex]
&=\int_0^{\frac\pi2}\int_0^2r\sqrt{1+4r^2}\,\mathrm dr\,\mathrm d\theta
\end{align}$$

where $x=r\cos\theta$ and $y=r\sin\theta$. You would then get an area of $\dfrac{\left(17^{\frac32}-1\right)\pi}{24}$.

Title: Area of a region in first quadrant


问:
Eigenspace and Generalized Eigenspace on polynomial map
Consider the linear operator $T$ on $P(\mathbb{R})$ defined by $T:p(x) \mapsto p(x)+xp''(x)$.

I need to a) check that 1 is an eigenvalue of $T$ and compute the eigenspace $E_1$ and b) find the generalized eigenspace.

My attempt: a) Our vector space is any polynomial and our field is the reals. We need to show that $T(p(x)) = \lambda p(x)$. When $\lambda = 1$ we get that $T(p(x)) = p(x)$ which implies $p(x) + xp''(x) = p(x)$ and thus $xp''(x) = 0$. Hence $1$ is an eigenvalue for any polynomial of degree 1 or 0. And so $E_1 = \{ a + bx | a,b \in \mathbb{R}\}$.

b) Now is where I'm really confused. The generalized eigenspace will be the set {$v ∈ V$ | $(T − λI)^mv = 0$, m > 0 }. The issue is I have no idea what $(T − I)^mv = 0)$ is supposed to mean for this map.

Any help here would be great. Thanks

答:
I'll answer the only question in the question :)

The issue is I have no idea what $(T − I)^mv = 0)$ is supposed to mean for this map

$$
\because (T − I)p(x) = xp''(x)\\
\therefore (T − I)^2p(x) = x\frac{d^2}{dx^2}xp''(x)
$$

so on and so forth.


问:
Union of countable set with infinite set
Suppose $A$ is infinite and $B$ is countable. I wish to show that $|A \cup B| = |A|$. Clearly $|A \cup B|\geq |A|$. If $B$ is finite, then the rightwards 'shift' function is injective from $|A \cup B| \to |A|$ and hence $|A \cup B| \leq |A|$. (e.g. If $B = \{1, 2\}$ and $A = \{3,4, 5, \dots\}$, let $f(n) = n +2$). But what can  I do if $B$ is countably infinite?

答:
I will assume that $A$ and $B$ are disjoint sets.

If $A$ is infinite, then it has a countable subset $\{a_1,a_2,a_3,\ldots\}$. And $B$ can be written as $\{b_1,b_2,b_3,\ldots\}$. So, consider the map$$\begin{array}{ccc}A\cup B&\longrightarrow&A\\x&\mapsto&\begin{cases}x&\text{ if }x\in A\setminus\{a_1,a_2,a_3,\ldots\}\\a_{2n}&\text{ if }x=a_n\\a_{2n-1}&\text{ if }x=b_n.\end{cases}\end{array}$$


问:
Defining Functors when there are Choices Involved
I'm a little confused about how we formally define functors in certain instances, namely when there is no canonical choice for precisely which object of the target category we send each object in the source category to. This sounds a little vague, so I'll go by way of examples.

The functor $\operatorname{Hom}(\bullet,B)$ from the category of abelian groups to itself. I have no issue here; Given an abelian group $A$, there is a well defined set $\operatorname{Hom}(A,B)$ which is given an abelian group structure.

The functor $\bullet \otimes B$. This gives me a little more trouble, since there are many equivalent ways of defining the tensor product $A \otimes B$. Formally a functor has to send $A$ to a precise abelian group, not a class of abelian groups. Do we just make an arbitrary choice for every single possible $A$ and call it a day? This doesn't really sit right with me, but since there's a specific way the tensor product is usually defined (as a quotient of a free abelian group), this isn't that big of a deal to me.

The functors Ext and Tor. These pose a real problem for me, since the exact group the functors assign to a pair of abelian groups depends on the choice of free (or injective for the case of Ext) resolution we go with. Sure there are canonical equivalences between all the possible resultant groups, but there's no canonical way of choosing each one. If a functor is supposed to assign a single target object to each source object, how do we go about making that choice?

I was initially thinking that the axiom of choice could let us pick an arbitrary target object for each source object, but in general the class of objects isn't even a set, so the axiom of choice can only be used for small categories.

答:
You're exactly right that, to turn certain constructions into functors, we need an axiom of global choice, that is, an axiom of choice for classes: a product of and class of nonempty classes is nonempty. For a simple example, given an arbitrary large category $\mathcal C$ which admits products and a fixed object $c$ of $\mathcal C$, the axiom of global choice is needed to construct a functor $c\times (-):\mathcal C\to \mathcal C$ assigning to every $c'\in \mathcal C$ a product $c\times c'$.

In the most common foundations used in category theory, this is no problem. Typically one might use Grothendieck universes, in which every large category becomes small in an appropriate universe, and then this global axiom of choice reduces to the usual one together with a large cardinal axiom. Alternatively one might use NBG set theory, which directly axiomatizes both sets and classes and implies no new statements about sets beyond ZFC. There we explicitly assume there exists a global choice function, namely a function from the class of all nonempty classes to itself choosing an element of every nonempty class.

In a more constructive paradigm, it's more appropriate to modify the definitions so that, for instance, for a category to admit products means for it to come equipped with chosen product functors. Alternatively, one can consider anafunctors, which are generalizations of functors which are intended to precisely model the notion of a functor that's well defined only up to canonical isomorphism.

In practice, no matter the approach taken, there is never any serious gap between an anafunctor and a functor, and most mathematicians pay no notice whatsoever to issues like those you describe, even in the case of derived functors and other elaborate constructions involving numerous choices. As long as the choices are unique up to canonical isomorphism, they are almost never distinguished from choices that are literally unique. A final foundational framework in which this practical perspective is realized most closely is in homotopy type theory, where the notions of equality and of canonical isomorphism become precisely the same. This is, to my mind, the correct attitude toward such issues.


问:
Creating a version of conical helix (parametric) about the border of a paraboloid.
Im just getting to know a little more about helix's and i thought it was cool to visualize these helix's in 3D as regular helix's that extend uniformly across the z axis but what was far more interesting was the visualizations of conical as well as spherical spirals. They usually operate by making the radius of the spiral a function of Phi but naturally after examining this i began to wonder what a spiral about the edge of a paraboloid would look like and how you would represent this parametrically. I should start by asking if these equations exist somewhere where i could see them and if not what would they be.

In other-words for a paraboloid with the equations

```<code>x(θ,φ) = cos(φ) * θ
y(θ,φ) = sin(φ) * θ
z(θ,φ) = θ^2
</code>```

what are the equations for a spiral whos radius is about the border of the paraboloid

答:
You can take a whole family of spirals $S_k$ like this, for a given fixed $n$, (on the figure below, there are $n=15$ spirals):

$$\begin{cases}x&=&\cos(t+2\pi k/n) t\\y&=&\sin(t+2 \pi k/n)t\\z&=&t^2\end{cases}$$

(check that $x^2+y^2=z$, proving that the different spirals are on the paraboloid).

The spirals have different colors, from $k=0$ (pure red) to $k=n-1$ (pure blue).

or, in a lateral view:

Matlab code for the figure:

```<code>clear all;close all;hold on;axis([-5,5,-5,5,0,25])
[X,Y]=meshgrid(-4:0.01:4);
Z=min(X.^2+Y.^2,26);
surf(X,Y,Z,'edgecolor','none');alpha(0.4);view([14,70])
n=15;t=0:0.005:4.;
for k=0:n-1
    T=t+2*pi*k/n;
    plot3(cos(T).*t,sin(T).*t,t.^2,'k','color',[1-k/n,0,k/n]);
end;
</code>```

Body:
You can take a whole family of spirals $S_k$ like this, for a given fixed $n$, (on the figure below, there are $n=15$ spirals):

$$\begin{cases}x&=&\cos(t+2\pi k/n) t\\y&=&\sin(t+2 \pi k/n)t\\z&=&t^2\end{cases}$$

(check that $x^2+y^2=z$, proving that the different spirals are on the paraboloid).

The spirals have different colors, from $k=0$ (pure red) to $k=n-1$ (pure blue).

or, in a lateral view:

Matlab code for the figure:

clear all;close all;hold on;axis([-5,5,-5,5,0,25])
[X,Y]=meshgrid(-4:0.01:4);
Z=min(X.^2+Y.^2,26);
surf(X,Y,Z,'edgecolor','none');alpha(0.4);view([14,70])
n=15;t=0:0.005:4.;
for k=0:n-1
    T=t+2*pi*k/n;
    plot3(cos(T).*t,sin(T).*t,t.^2,'k','color',[1-k/n,0,k/n]);
end;


问:
Find $\mathbb{P}(Y > 0 \mid X = 1)$ for bivariate normal $(X, Y)$.
A friend asked me about a (homework) question of the following form:

The random vector $(X, Y)$ is bivariate normal with means $=0$ and standard deviations $=1$, the correlation is $\rho = \frac{1}{2}$. Find $\mathbb{P}(Y > 0 \mid X = 1)$.

So I've never taken calculus-based probability, so my first instinct is to say that the answer is 0, because we're conditioning on an event of probability 0. But I feel like this is a cop-out answer, so my intuition tells me to just pretend like we're working in one dimension and do something like:

$$\mathbb{P}(Y > 0 \mid X = 1) = \frac{\int\limits_{0}^\infty f_{XY}(1,y) \; dy}{\int\limits_{-\infty}^\infty f_{XY}(1,y) \; dy} $$

where $f_{XY}$ is their joint distribution.

So I have two questions. First, is this the correct interpretation? Second: if it is, it seems to me like finding a closed-form answer is unreasonable for a calculus-based probability class, since the integrand is an exponential with a nontrivially quadratic argument, but maybe I'm missing something.

答: In general, evaluating this integral would be too difficult for a probability class, but you don't need to evaluate the integrals thanks to a lot of special properties of multivariate normal distributions. Take a look at Wiki: Multivariate normal distribution for an overview.</p>


问:
Measurability of an a.s limit
Let $(\Omega,\mathcal{F},P)$ be a probability space, $(\mathcal{F}_n)_n$ a decreasing
sequence of sub-$\sigma$-algebra of $\mathcal{F}.$ Let $(X_n)_n$ be a sequence of r.v. converging a.s to $X.$ Prove that if for all $n \in \mathbb{N}, X_n$ is $\mathcal{F}_n$-measurable then $X$ is $\bigcap_{n \in \mathbb{N}}\mathcal{F}_n$-measurable.

We need to show that for a fixed $n \in \mathbb{N},U \in B(\mathbb{R}),X^{-1}(U) \in \mathcal{F}_n.
$

Any suggestions is appreciated.

答:
Note that for all $n$
$$\{X \in U\} = \bigcap_{q \in \mathbb N} \bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}.$$

Now, $\{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$ is $\mathcal F_n$-measurable for all $m \geq n$ because $(\mathcal F_n)_n$ is decreasing. So,
$$\bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$$
is $\mathcal F_n$-measurable for all $r \geq n$. So,
$$\bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$$
is $\mathcal F_n$-measurable, and therefore so is
$$\bigcap_{q \in \mathbb N} \bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}.$$

Title: Measurability of an a.s limit


问:
Newton-Raphson termination criteria for large problems
When using Newton-Raphson to solve a system of equations of the form:
$$\mathbf{r}(\mathbf{x})=\mathbf{0}, \quad \mathbf{r}=[r_1, r_2,...,r_N], \quad \mathbf{x}=[x_1, x_2,...,x_N]$$
The termination criterion is usually based on the euclidean norm of $\mathbf{r}$:
$$ \sqrt{r_1^2+r_2^2+...+r_N^2} \le tolerance $$
As far as I understand, this criterion guarantees that the largest error for a single equation, $r_i$, will always be smaller than the desired tolerance. However, I feel like such a criterion is 'unfair' when the number of equations increases. An extreme example would be that of a system where $N=1$ (a single equation) vs a system where $N \gg 1$.

For $N=1$ we can assume that the value of $r_1$ at the point of termination will be close to the set tolerance.

For $N \gg 1$ my intuition is that each $r_i$ will be significantly smaller than the tolerance, since they are all added up to produce a value close to the tolerance. This means that more iterations are needed, although the set tolerance stays the same.

Then, my question would be: Is there a termination criterion that will provide solutions with the same order of accuracy, regardless of the number of the equations?

答: Yes. Replace the Euclidian norm with the infinity norm and terminate when $$\|\mathbf{r}\|_\infty \leq \tau$$ where $\tau$ is the tolerance.


问:
symmetries of a circle and the $O_2$ group: why there are both rotations AND reflections, and not either rotations OR reflections?(geometric view)
I'm reading "Camila Jordan, David Jordan Groups - Modular Mathematics series", section 1.2 Symmetries of a circle.

In this section we consider the unit circle. A typical point on the unit circle has coordinates $(\cos\theta, \sin\theta)$ for some $\theta$. In contrast to the square, this circle has infinitely many symmetries. They are all the rotations $rot_{\theta}$ together with all the reflections $ref_{\theta}$. As for the square, these symmetries form a group, the group of symmetries of the circle, but this time the group is infinite. The neutral element is $rot_0$. Each rotation $rot_{\theta}$ has the rotation $rot_{-\theta}$ as inverse while each reflection $ref_{\theta}$ is its own inverse. This group is known as orthogonal group $O_2$. The group $O_2$ acts on the circle by rotations and reflections. Now let us consider the stabilizer of an arbitrary point $P=(\cos\theta, \sin\theta)$ on the circle. Which elements of the group $O_2$ send $P$ to itself? There are just two, namely $rot_0$ and $ref_{2\theta}$

My question is: why does this $O_2$ consist of both rotations and reflections? As if we take any arc on a circle, we can cover all the circle by rotating this arc. Why we would need reflections then?

答:
Think of the circle as being oriented, say counterclockwise. Then rotations preserve this orientation, while reflections change it. Hence there are at least two kind of symmetries...

Note however that rotations can be obtained by composing two reflections about different axes. So in some sense reflections alone do suffice.


问:
Open cover of a set of real numbers
Using the definition of the open cover, I can prove that if a function is bounded on any open set $(x-\delta(x),x+\delta(x)), x \in [a,b]$, then that function is also bounded on $[a,b]$. Using the contrapositive, we would come to the conclusion that if a function is not bounded on $[a,b]$, then it would not be  bounded on any open set $(x-\delta(x),x+\delta(x)), x \in [a,b]$. This doesn't seem to make sense for me, can someone please clarify or help me understand how this is true?

答:
The contrapositive is not what you stated but instead

if a function is not bounded on $[a,b]$, then there exists an open set $(x-\delta(x),x+\delta(x)), x \in [a,b]$ where the function is not bounded

Title: Open cover of a set of real numbers


问:
Integral of $\int^1_0 \frac{dx}{1+e^{2x}}$
I am trying to solve this integral and I need your suggestions.

I think about taking $1+e^{2x}$ and setting it as $t$, but I don't know how to continue now.

$$\int^1_0 \frac{dx}{1+e^{2x}}$$
Thanks!

Body:
$\int^{1}_{0}\frac{dx}{1+e^{2x}}=$ $ \int^{1}_{0}\frac{e^{-2x}dx}{1+e^{-2x}}=$ $-\frac{1}{2}\int^{1}_{0}\frac{(1+e^{-2x})'dx}{1+e^{-2x}}=$
$=-\frac{1}{2}ln(1+e^{-2x})|^{1}_{0}= \frac{1}{2}ln\frac{2e^{2}}{1+e^{2}}$


问:
Why do we say A given B in Bayes theorem?
I enjoyed this blog on visualizing Bayes theorem but I have a question on the wording for .

$p(A\mid B) = p(A,B)/p(B)$

In words I understand this to be "probability of A given B"

if we are "given B" why do we need to divide by p(B) ?   Surely if I am "given" something then its probability is $1$?

Why do we say it this way?

[Update]

I am seeing that p(B) = 1 (because we are "given" it )

thus p(A,B)/p(B) = p(A,B)

I am looking for the right words to articulate that we still need to divide by p(B) and that it is not 1

If we were saying "given we know the probability of B" instead of "given B" I would not be so confused.

答:
if we are "given B" why do we need to divide by p(B) ? Surely if I am "given" something then its probability is 1?

As I understood probability, it is a property not of a unique event but of a stable event producing scheme. For instance, when we toss a fair coin then a probability of a head in $1/2$. On the other hand, this scheme is missed in a known joke that a probability to meet a crocodile at the street is $1/2$.

The probability of an event provided by the scheme does not change when the event happens (or not). For instance, when we have a head of the tossed coin then the probability of this event is still $1/2$.

Thus, given an event $B$, we don’t change the probability of $B$, but, calculating a conditional probability $P(A|B)$ we do this with respect to a changed scheme, where only those its instances when $B$ happens are taken into account.