问:
How to minimise the area of the folded part of a piece of paper when it touches the other side
I am studying maths purely out of interest and have come across this question in my text book:

A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on CD. If AB = a and AD = b, where $a \ge\frac{2b}{\sqrt3}$, show that the least possible area of the triangle APQ is obtained when the angle AQP is equal to $\frac{\pi}{3}$. What is the significance of the condition $a \ge\frac{2b}{\sqrt3}$?

I realise I need to get an equation for the area of the fold, which I can then differentiate. Even then, I am not sure how to relate this to angle AQP.I have looked at solutions on the internet but they tend to look at the length of the crease, not the area of the fold.

I have taken A in my diagram to represent its position before the fold. I could be mistaken.

This is how I visualise it:

I have said:

$x^2 = m^2 + (b - x)^2 = m^2 + b^2 - 2bx +x^2 \implies m^2 = 2b(x - \frac{b}{2})$

$L^2 = (L - m)^2 + b^2 \implies L = \frac{m^2 + b^2}{2m}$

$y^2 = x^2 + L^2 = x^2 + \frac{(m^2 + b^2)^2}{4m^2} = x^2 + \frac{(2bx - b^2 + b^2)^2}{4m^2} = x^2 + \frac{4b^2x^2}{8b(x - \frac{b}{2})} = \frac{bx^2}{2x - b}$

After this I am not sure how to proceed.

答:
If $\angle AQP = \theta$, $\angle APQ = 90^0 - \theta$.

Draw a perp from $A_1$ to line segment $AP$ and say it is point $A_2$ on line segement $AP$.

Then $A_1A_2 = b$ and $\angle A_2PA_1 = 2 \ \angle APQ = 180^0 - 2\theta \ $.

So $AP = L = \frac{A_1 A_2}{\sin(180^0 - 2 \theta)} = \frac{b}{\sin (180^0 - 2 \theta)} = \frac{b}{\sin 2 \theta} \ $ (in $\triangle A_1PA_2$)

$AQ = x = L \cot \theta = \frac{b}{2 \sin^2 \theta}$

Area of $\triangle APQ = \frac{1}{2} \ L \ x = \frac{b^2}{8 \sin^3 \theta \cos \theta}$

So to minimize area of $\triangle APQ$, we need to maximize $f(\theta) = \sin ^3 \theta \ \cos \theta$.

$f'(\theta) = 3 \sin^2 \theta \cos^2 \theta - \sin^4 \theta = 0$

$\implies \sin \theta = 0, \tan \theta = \pm \sqrt3$. This leads to $\theta = \frac{\pi}{3}$.

Which also tells that $L = \frac{b}{\sin 2 \theta} = \frac{2b}{\sqrt3}$

Also note that for us to be able to fold such that we obtain minimum area,

$a \geq L =  \frac{2b}{\sqrt3}$.


问:
Can one estimate the uncertainty over the estimated outcome probabilities of an event?
Say I'm making the following bet: I flip a coin. If it's heads, I win \$2; if it's tails, I lose \$1. The coin is not unbiased in general: there is some probability $p\in[0,1]$ that it's going to be heads.

If I know $p$, I can estimate my average return over such a bet, and decide whether it's a good idea to wage it.
Suppose I have some information that would allow me to make an estimate over what $p$ is.
However, I am not certain whether my analysis of such information is accurate.

Is there a way to formalise this sort of situation? That is, if I estimate the outcome probabilities of some event, is there a way to meaningfully take into account how (not) confident I am in my own analysis?

For example, say the real probability, as estimated by someone else by tossing the coin many times, is $p=0.1$. Having no information about $p$, Alice's (who is deciding whether the bet is profitable) Bayesian prior on the probability should be $1/2$, and thus she would conclude that the bet is profitable on average (the win reward being larger than the loss cost). Similarly, she might not be very good at evaluating the available information and wrongly estimate $p$ to be $1/2$, leading to the same conclusion.
Such conclusion would be wrong, and she would end up losing money, on average.

Now Bob, who is more careful, knows he is not that great at evaluating the available information, and wants to take this into account for his own estimate.
Is there a general way to do this sort of thing?

Of course, if the bet is repeated multiple times, one can update their prior using the observed outcomes. Here I'm however considering the "first-shot" scenario, that is, what one would estimate before tossing any coin.

答: You can assume a probability distribution for $p$. At fixed $p$, you know how to compute the expected outcome: then you have to average the expected outcome over $P(p)$ to decide if the bet is worth the risk.

ContentLicense: CC BY-SA 4.0


问:
G is a graph with n points.If it exists a number c > 0,,makes that g(G) ≥ cn.(g(G) is the girth of the G).How to prove that e(G) ≤ n + o(n)?

Suppose $G$ is a graph with $n$ points. If it exists a number $c > 0$, makes that $g(G) ≥ cn\cdot\left(g(G) \text { is the girth of the G}\right)$. How to prove that $e(G) ≤ n + o(n)$?

I want to use mathematical induction to solve this question, but there is no good idea of using mathematical induction. I want to change this question to a normal graph that only has cyclic with length $[cn]$, and try to think the situation has the most edges, but I failed.

答: I have solve this question by myself.Just think about c=1/2^n(n=0,1,2,3...),and for graph G, divide points of G to two part and use mathematical induction to n will solve this question.</p>


问:
Prove, preferably with elementary methods, that $\int_0^\infty x^5 e^{-x^4}~dx=\frac{\sqrt\pi}{8}$
In the final question of the MIT Integration Bee $2020$ qualifier exam, the participants were asked to evaluate
$$\int_0^\infty x^5 e^{-x^4}~dx$$
and in the solutions they give the answer as $$\frac{\sqrt\pi}{8}$$
Most, if not all the other questions can be completed with the use of elementary methods and functions, but I could not solve this question. The integrand is strongly suggestive of the Gamma function and the Gaussian integral, but I am not competent with these yet, so I couldn't employ them to evaluate the integral.

So, how would I solve this integral? If you must resort to non-elementary functions and results and techniques, then so be it.

Thank you for your help.

答:
$$\int_0^\infty x^5 e^{-x^4}~dx$$
Let $x^2= u$.

$$\frac12\int_0^\infty u^2 e^{-u^2}~du$$

$$\frac12\int_0^\infty u \cdot u e^{-u^2}~du$$

Now use integration by parts and use that $\int_0^\infty e^{-u^2}~du=\frac{\sqrt\pi}{2}$


问:
Is the following matrix invertible or not? Provide condition if needed.

The matrix is $n\times n$

The nth row is $[n^1,n^2,n^3....n^n]$

I tried the first few examples:
$\begin{pmatrix}1\end{pmatrix}$ ,$\begin{pmatrix}1 & 1\\ 2 & 4\end{pmatrix}$,$\begin{pmatrix}1 & 1 & 1\\ 2 & 4 & 8\\3 & 9 & 27\end{pmatrix}$. They are obviously invertible.

But afterwards I got confused. Do I try to prove that the determinant keeps increasing, or should I try proving by induction?

答:
Given $1,...,n$, all of the polynomials of degree $n$ that has this numbers as roots are:

$p(x)=k(x-1)(x-2)...(x-n)=k(\tilde{a}_0+\tilde{a}_1x+...+\tilde{a}_{n-1}x^{n-1}+\tilde{a}_n x^n), k\neq 0$

So we are saying that the linear system
$$\begin{cases} a_0+ja_1+j^2a_2+...+j^na_n=0\end{cases} \forall j\in \{1,...,n\}$$
Has this set of solutions $\mathcal{S}=\{(k\tilde{a}_0,..,k\tilde{a}_n)|k\in \mathbb{R}\}$. Notice that if we fix the value of the first component of the solution (for example $a_0=1$), then the system has a unique solution, because $k\tilde{a}_0=1 \Rightarrow k=\tilde{a}_0^{-1}$(notice that $|\tilde{a}_0|=n!\neq0$ ), so the unique solution would be $(1,\tilde{a}_0^{-1}\tilde{a}_2,..,\tilde{a}_0^{-1}\tilde{a}_n)$. We are saying that this system:
$$\begin{cases} 1+ja_1+j^2a_2+...+j^na_n=0\end{cases} \forall j\in \{1,...,n\}$$
Has unique solution. We can easily rewrite the system as:
$$\begin{cases} ja_1+j^2a_2+...+j^na_n=-1\end{cases} \forall j\in \{1,...,n\}$$
So we are saying that this square system has unique solution, that is to say that its coefficients matrix is non singular. And the coefficients matrix of this system is exactly the matrix of your problem.


问:
$f_n\xrightarrow{L^p}f \implies f_n'\xrightarrow{L^p}f'$?
Let $(f_n)_{n\ge 1}$ be a sequence of function that is differentiable. such that $f_n\xrightarrow{L^p}f$ converges. I was wondering if $f_n'\xrightarrow{L^p}f'$. So far I could only find this result but it doesn't really apply here I think.

Body:
No.

Example: $p=1$, $f_n$ and $f$ in $L^1[0,1]$ let be given by

$$ f_n(x) =x^n , \quad f(x)=0.$$

Then

$$||f_n-f||_1 = \frac{1}{n+1} \to 0,$$

as $ n \to \infty,$ but

$$||f_n'-f'||_1 = 1$$

for all $n$.


答:
I think what you're describing is called ambiguity. When we know that something is random, and we know the distribution that is called uncertainty. When we know that something is random, but we do not know the distribution that is called ambiguity.

There are several ways how we can deal (make decisions) with ambiguity. The simplest approach is Bayesian, i.e. we assume that there is some subjective distribution of the parameter. Some other approaches to decision-making under ambiguity which I am aware of are maximin and minimax regret. Google "decision making under ambiguity" for more details.

Let me illustrate the latter two criteria. Suppose we know ex-ante that $p\in[\underline{p}, \bar{p}], 0\le \underline{p} \le \bar{p} \le 1$. Also, suppose that we want to choose $\delta\in[0, 1]$ which is the probability with which we play the game you've described. Then maximin criterion looks as follows: $$\delta^{MM} \in arg \underset{\delta \in [0, 1]}{\max} \underset{p \in [\underline{p}, \bar{p}]}{\min} \delta(3p-1)$$
the solution would be $$\delta^{MM} = \begin{cases}
0, \underline{p} < \frac{1}{3}\\
1, \underline{p} \ge \frac{1}{3}
\end{cases}$$
Minimax regret criterion would be:
$$\delta^{MMR} \in arg \underset{\delta\in[0, 1]}{\min} \underset{p \in [\underline{p}, \bar{p}]}{\max} \underbrace{\underset{\delta'\in[0, 1]}{\max} (3p-1)(\delta'-\delta)}_{\text{Regret}}$$
and the solution is $$\delta^{MMR} = \begin{cases}
0,\ \  &\underline{p} \le \bar{p} \le \frac{1}{3}\\
\frac{3\bar{p} - 1}{3(\bar{p} - \underline{p})}, &\underline{p} \le \frac{1}{3} < \bar{p}\\
1, & \frac{1}{3} \le \underline{p}<\bar{p}
\end{cases}$$

Title: Can one estimate the uncertainty over the estimated outcome probabilities of an event?

Body:
Say I'm making the following bet: I flip a coin. If it's heads, I win \$2; if it's tails, I lose \$1. The coin is not unbiased in general: there is some probability $p\in[0,1]$ that it's going to be heads.

If I know $p$, I can estimate my average return over such a bet, and decide whether it's a good idea to wage it.
Suppose I have some information that would allow me to make an estimate over what $p$ is.
However, I am not certain whether my analysis of such information is accurate.

Is there a way to formalise this sort of situation? That is, if I estimate the outcome probabilities of some event, is there a way to meaningfully take into account how (not) confident I am in my own analysis?

For example, say the real probability, as estimated by someone else by tossing the coin many times, is $p=0.1$. Having no information about $p$, Alice's (who is deciding whether the bet is profitable) Bayesian prior on the probability should be $1/2$, and thus she would conclude that the bet is profitable on average (the win reward being larger than the loss cost). Similarly, she might not be very good at evaluating the available information and wrongly estimate $p$ to be $1/2$, leading to the same conclusion.
Such conclusion would be wrong, and she would end up losing money, on average.

Now Bob, who is more careful, knows he is not that great at evaluating the available information, and wants to take this into account for his own estimate.
Is there a general way to do this sort of thing?

Of course, if the bet is repeated multiple times, one can update their prior using the observed outcomes. Here I'm however considering the "first-shot" scenario, that is, what one would estimate before tossing any coin.


Title: Prove, preferably with elementary methods, that $\int_0^\infty x^5 e^{-x^4}~dx=\frac{\sqrt\pi}{8}$

Body:
In the final question of the MIT Integration Bee $2020$ qualifier exam, the participants were asked to evaluate
$$\int_0^\infty x^5 e^{-x^4}~dx$$
and in the solutions they give the answer as $$\frac{\sqrt\pi}{8}$$
Most, if not all the other questions can be completed with the use of elementary methods and functions, but I could not solve this question. The integrand is strongly suggestive of the Gamma function and the Gaussian integral, but I am not competent with these yet, so I couldn't employ them to evaluate the integral.

So, how would I solve this integral? If you must resort to non-elementary functions and results and techniques, then so be it.

Thank you for your help.

Body: $$\frac12\int x^4e^{-x^4}dx^2=\frac12\int y^2e^{-y^2}dy=\frac12\int y\,ye^{-y^2}dy=-\frac14ye^{-y^2}+\frac14\int e^{-y^2}dy.$$


问:
If $L/F$ and $F/E$ are separable, then $L/E$ is separable
Well, I'm stuck trying to prove the following about separable extensions.

If $L/E$ is a extension (not necessarily finite) such that $L/F$ and $F/E$ are both separable, then $L/E$ is also separable.

My problem is I need to prove this without using the primitive element theorem, facts about degree of separability-inseparability or the fundamental theorem of Galois Theory because this appears as an exercise before those results.

My idea was to prove that $L/F$ and $F/E$ are Galois extensions and then show $E=L^{G(L/E)}$. So, for a well-known theorem $L/E$ is a Galois extension and in particular, $L/E$ is separable, but I can't prove that $L/F$ is normal. Is it correct if I take $L$ to be the normal closure of $F$? Thanks in advance.

答:
Lemma 1. Let $E/k$ be separable, and let $\sigma: k\rightarrow k^a$. Then $\sigma$ extends to an embedding of $E$ in $k^a$, and there are $[E:k]$ distinct extensions of $\sigma$. The number of such distinct monomorphism is $\lt [E:k]$ if $E/k$ is not separable.

Proof. We will prove the lemma by induction. Let $\alpha\in E$, then the minimal polynomial $\text{Irr}(\alpha,k)$ has no multiple roots. Any embedding of $k(\alpha)$ sends $\alpha$ to a root of $\text{Irr}(\alpha,k)$, so there are $\text{deg}(\text{Irr}(\alpha,k))=[k(\alpha):k]$ distinct extensions. For any $\beta\in k(\alpha)$, since $\text{Irr}(\beta,k(\alpha))\mid \text{Irr}(\beta,k)$, it follows from our assumption that $\text{Irr}(\beta,k(\alpha))$ is separable. By induction hypothesis, the number of distinct extensions of a given embedding of $k(\alpha)$ on $E$ is $[E:k(\alpha)]$. Hence the number of distinct extensions of $\sigma$ on $E$ is $[E:k(\alpha)]$. The second assertion is proven by essentially the same proof.

Lemma 2. Let $k\subset k(a_1)\subset k(a_1,a_2)\subset\cdots\subset k(a_1,\cdots,a_n)$ where $a_i$ is separable over $k(a_1,\cdots, a_{i-1})=E_{i-1}$, and $E=k(a_1,\cdots,a_n)$. Then the identity on $k$ extends to $[E:k]$ distinct monomorphism $\sigma: E/k\hookrightarrow k^a/k$.

Proof. By induction.

Proposition. Let $k\subset k(a_1)\subset k(a_1,a_2)\subset\cdots\subset k(a_1,\cdots,a_n)$ where $a_i$ is separable over $k(a_1,\cdots, a_{i-1})=E_{i-1}$, and $E=k(a_1,\cdots,a_n)$. Then $E/k$ is separable.

Proof. By Lemma $2$, the identity on $k$ extends to $[E:k]$ distinct monomorphism $\sigma: E/k\hookrightarrow k^a/k$.  By Lemma $1$, this implies that $E/k$ is separable.

Theorem. Let $F/k$ and $E/F$ be finite separable. Then $E/k$ is separable.

Proof. By our previous proposition, $F(a)$ is separable over $k$ for each $a\in E$. This shows that $E/k$ is separable.


答:
Let $x=\sqrt t$ to make
$$I=\int x^5 e^{-x^4}\,dx=\frac{1}{2}\int t^2\,e^{-t^2} \,dt$$ One integration by parts to get
$$I=\frac{\sqrt{\pi } }{8} \text{erf}(t)-\frac{1}{4}t \, e^{-t^2} $$ and then the result for the definite integral.


问:
compute $\int_\gamma (\sin xy+xy\cos xy+2x )\ dx\ +(x^2\cos xy+e^y) \ dy=e-1$
The question is :
$$ F=(\sin xy+xy\cos xy+2x, \ x^2\cos xy+e^y)$$
Find a curve $\gamma$ such that $ \int_\gamma F=e-1$

I have tried to solve it like this:
$$ \text{The field is conservative,}\quad F=\nabla f, \quad f=x\sin xy+e^y+x^2+C $$
I couldn't solve it fully I've just found the potential $f$

Any suggestion would be great, thanks

答: Once you have that the field is conservative, you just need to find two points where the underlying scalar field assumes values of $e$ and $1$. Example points are $(x,y)=(0,1)$ and $(0,0)$. Thus $\gamma$ can be the line segment from $(0,1)$ to $(0,0)$.


问:
Simple numerical methods for calculating the digits of $\pi$
Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?</p>

答:
Check out the wiki page on approximations to pi, http://en.wikipedia.org/wiki/Numerical_approximations_of_%CF%80#20th_century

I really love Ramanujan's method, which I recall converges extremely rapidly.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 2.5


问:
Is mathematics one big tautology?
Is mathematics one big tautology? Let me put the question in clearer terms:

Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work!

If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic?

I'm trying to figure out the true nature of what's going on here.

答:
There are a few things to mention here; without checking what others have already said, I'll quickly sum them up.

First, there are more than only one set of axioms in Mathematics. Indeed, the beauty, and diversity, of mathematics arises from the diversity, and beauty, of the different axiomatic systems that have been introduced thus far. Certainly, it is not a novel act to introduce such a system, but it is so if the system is itself novel or interesting. Since, axioms are supposed to be, by definition "fundamental," introducing two sets of logically equivalent axioms (which differ, in the end, only in presentation) will not result in two (wholly, perhaps I should clarify this more) distinct theories. On a related note, it may very well be difficult to establish the logical equivalence of two sets of axioms. For more on the discussion of axiomatic systems, I recommend you study the development of and current set theory and formal logic.

Secondly, you must understand the difference between logical equivalence and identity. For example, the statement $P \implies Q$ is logically equivalent to $ \neg  P \lor Q$, however, they are not the same statement. Indeed, both statements say the same thing, but they differ, as mentioned prior, in presentation; you may use one over the other, however, you may not be able to derive the same conclusions depending on your choice. For more on the difference between the two, see this and this.

Thirdly, and perhaps the most obvious to mention here, there are certain propositions in Mathematics that can neither be proved nor disproved. Indeed, this is Godel's theorem.

To answer your question about the "creation" of new and novel truths, theorems can simply be interpreted as new ways of looking at combinations of axioms. For example, $${12 \over \pi^2}\sum_{k = 1}^\infty {1\over{k^2}} = 2 \tag{1}$$ is just another way (I would say new, but that is not really the case) of saying $$2 = 2 \tag{2}.$$ And, certainly, there is nothing novel about saying $(1)$ in this light. But! The point of the exercise is to draw relations that are not immediately obvious, and relations such as they may be useful in drawing other relations and equivalences. It is after all, as you said, a matter of drawing a string of equivalences, but again, keep in mind we are not limited to one set of axioms.

Title: Is mathematics one big tautology?


答: You just need to find 2 points $(x_1,y_1)$, $(x_2,y_2)$ such that $f(x_2,y_2)-f(x_1,y_1) = e-1$. After you do this, just choose any $\gamma$ with those endpoints.

Title: compute $\int_\gamma (\sin xy+xy\cos xy+2x )\ dx\ +(x^2\cos xy+e^y) \ dy=e-1$

Body:
The question is :
$$ F=(\sin xy+xy\cos xy+2x, \ x^2\cos xy+e^y)$$
Find a curve $\gamma$ such that $ \int_\gamma F=e-1$

I have tried to solve it like this:
$$ \text{The field is conservative,}\quad F=\nabla f, \quad f=x\sin xy+e^y+x^2+C $$
I couldn't solve it fully I've just found the potential $f$

Any suggestion would be great, thanks


Body:
Let $ \gamma(t):=t(0,1)$ for $t \in [0,1].$ Then

$$\gamma(0)=(0,0)$$

and

$$ \gamma(1)=(0,1).$$

Then

$$\int_\gamma F=f( \gamma(1))-f(\gamma(0))=e-1.$$


问:
Calculating gradf in spherical polar basis?
If $f(\mathbf{r})=\vert\mathbf{r}\vert^4$

How would you calculate $\operatorname{grad} f$ in spherical polar basis?

答:
Hint:

Take advantage of the fact that $|\vec r|=r$; and $g'(r)=4r^3$ for $g(r)=r^4$.

For the whole story, you can read: https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates


问:
for F ⊂ K ⊂ E if E/F ve E/K Galois gextensions then K/F is also Galois extension.
Consider $E,K,F$ being fields such that $F \subset  K \subset  E$. I'm trying to find $E,K,F$ s.t. $E/K$ and $E/F$ are Galois, but $K/F$ is not. Could someone help to provide such fields ?

答:
Hint

Take $$E=\mathbb Q(\sqrt[3]2,i),\quad K=\mathbb Q(\sqrt[3]2)\quad \text{and}\quad F=\mathbb Q.$$


答:
As you already found $f (x, y) = x\sin xy+e^y+x^2 + C$

Given the line integral is $e - 1$, it is easy to see that we can assign end point as $y = 1$ and starting point as $y = 0$.

Now to make sure other terms are zero, we can assign $x = 0$ for both the start and end points.

So starting points and end points are  $A(0,0)$ and $B(0,1)$ respectively.

Now given that the line integral is path independent for the given vector field, it could be any smooth curve from point $A$ to $B$ - for example, a straight line or say, a semicircle of radius $\frac{1}{2}$ with center at $(0, \frac{1}{2})$.


问:
Intuitive explanation for orthogonality relations between matrix elements of irreps
It is well known that the matrix elements of complex irreducible representations of a finite group $G$ are orthogonal, i.e.
$$
\langle T_{ij}^\lambda, T_{kl}^{\rho} \rangle = \frac{1}{|G|} \sum_{g\in G} T_{ij}^\lambda(g) T_{kl}^\rho(g)^* = \frac{\delta_{ik} \delta_{jl} \delta_{\lambda \rho}}{d_\lambda},
$$
where $T_{ij}^\lambda$ is the $(i,j)$ matrix element of irreducible representation (irrep) indexed by $\lambda$, $d_\lambda$ is the dimension of this irrep, and $\delta_{\lambda \rho}$ means that $\lambda$ and $\rho$ correspond to an equivalent irrep.

The above can be strengthened, because in fact the matrix elements of irreps form a basis for the space of functions on $G$.

As a consequence, we can expand a function on $G$ as a linear combination of matrix elements, giving us some kind of generalization of classical Fourier theory. I want to explain this idea to an audience that knows classical Fourier theory but hardly anything about representations. My goal is to keep it high level and intuitive. I have looked at some proofs of the statement above, but they all seem fairly long, reliant on other results, and it's hard to distill the "essence" of why the above must be true. I would appreciate a short and intuitive explanation of the result above, that conveys the gist to an audience who does not yet know much representation theory.

答:
There are two important underlying facts:

If you take the action of $G$ on $End(V_{\lambda}) $, this decomposes as $d_{\lambda}$ copies of $V$. You can think that $g$ acts on each column separately.

The function

$$ A, B \mapsto \frac{1}{|G|} \sum_{g \in G} tr(gA) tr(gB)^*$$

gives an invariant pairing
$$ End(V_{\lambda}) \otimes_{\mathbb{C}} End(V_{\mu} ) \to \mathbb{C}$$

Given these two facts, you can see that for $\lambda \neq \mu$ this must be zero, because there is no invariant pairing between two different irreducible representations. In the case $\lambda = \mu$, you know that when restricted to a single colimn in both components it must be a multiple scalar of the unique invariant product $V_{\lambda} \otimes V_{\lambda} \to \mathbb{C} $, which is the standard hermitian product in some basis (the one in which $G$ acts in a unitary way). Here I think you can put under the carpet the computations showing the scalar is exactly 1, and conclude that

$$ \frac{1}{|G|} \sum_{g \in G} tr(gA) tr(gB) = A B^t $$


问:
showing that the Euler's number is irrational
Our teacher wants us to do the following:
Suppose that e is rational i. e $e=\frac{a}{b}$ where $a,b\in\mathbb{N}$. Choose $n\in\mathbb{N}$ such that $n>b$ and $n>3$. Use the following inequality

$0<e-(1+\frac{1}{1!}+\frac{1}{2!}+....+\frac{1}{n!})<\frac{3}{(n+1)!}$ and replace e with $\frac{a}{b}$. Then multiply both sides of the inequality with $n!$ to see that the inequality leads to existence of an integer N that satisfies $0<N<\frac{3}{4}$ which leads to a contradiction.

I have tried to to it but I am not getting there...
I choose $n=4$ and $b=3$ and
$0<\frac{a}{b}-1-1-\frac{1}{2}-\frac{1}{6}-\frac{1}{24}<\frac{3}{5!}$

$0<8a-\frac{17}{24}<\frac{3}{5}$

$0<8a<\frac{157}{120}$

$0<a<\frac{157}{960}$.....

How should I do it???

答:
By assumption, $n\ge b$. Therefore, it is possibly to bring the fraction $\frac ab$ to the denominator $n!$ as $n!$ is a multiple of $b$ (which is one of the factors in $1\cdot2\cdots n$). Likewise, all the other fractions can be braought to denominator $n!$ and hence the sum of all these fractions can be written with denominator $n!$. After multiplication with $n!$, only an integer - the numerator - remains.

Specifically, if you multiply
$$ 0<\frac a3-1-1-\frac12-\frac16-\frac1{24}<\frac3{120}$$
with $4!=24$, you should get
$$0<8a-24-24-12-4-1<\frac35. $$
Whatever $a$ is, it is an integer, hence so is the expression $8a-24-24-12-4-1$. But there is no integer strictly between $0$ and $\frac 35$.

Of cours, it is not suffucuent to consider this for just a specifically chosen $b=3$ (and $n=4$) because we don't know what the correct denomianator in a fraction for $e$ might be. It is quite clear that $b=3$ is a bad choice already because $2.\bar 6<2.7128\ldots<3$ rules that out. The argument above works for any integer $b$.


问:
Real life examples for eigenvalues / eigenvectors
There are already good answers about importance of eigenvalues / eigenvectors, such as this question and some others, as well as this Wikipedia article.

I know the theory and these examples, but now in order to do my best to prepare a course I'm teaching, I'm looking for ideas about good real life examples of usage of these concepts.

Do you know some good simple real-life examples (in economics or data analysis or anything else), in which  the usage of eigen values/vectors is a crucial tool?

答:
Eigenvectors are axes, eigenvalues are distances along these axes

Eigenvectors are the base of dimensional reduction techniques like PCA (principal component analysis), extremely useful in situations where we want to reduce the number of dimensions to a more practical one.

Concrete example: We want to find similar pictures in a large set and show the relationships in 2D (we don't know what are the similarity criteria, not either their number):

The result above is obtained using a non guided simple dimension reduction technique. Probably it isn't very useful, but it illustrates the creation of partial clusters, e.g. the direction the person is looking to, or the color of the skin, or whether the person opens the mouth, or who the person is. While there are plenty of criteria measured, we are able to use a 2D system to reveal their effects combined, this is actually the expected benefit. The algorithm here found two super abstract criteria which are eigenvectors, and returned the corresponding pair of eigenvalues for each picture, used as individual coordinates to arrange the set.

Face features as eigenvector: Eigenface

Using eigenvectors is a base technique in face recognition where we want to associate a name to a person picture. The eigenvectors in this case are eigenfaces.

Imagine we got black and white images of 47x62 pixels which can have some gray attribute, we actually have data with a value in 1348 dimensions:

Not all pixels are critical in a given image. We want to reduce the number of dimensions to the "useful" ones, without losing the main features of the images. Say we want to move from 1348 to 150 dimensions. This is the usual way of pre-processing images before doing some image classification, like face recognition, in order to decrease CPU workload.

The reduction is done by finding eigenvectors of the input images, these eigenvectors can be seen as basis images, from which the complete (actually nearly complete) images can be reconstructed.

Below are the 32 first eigenvectors, out of 150 which were computed by some PCA, in order of usefulness, that is they are dimensions along which original images have the highest variance. These images are somehow like the major harmonics of a sound (obtained using a Fourier transform):

Note the three first eigenvectors are luminosity related, the main variance in images not shot in controlled conditions.

To reconstruct the images from these eigenvectors, we only need to know the associated eigenvalues. The information to store and/or process for each image is now a vector of 150 eigenvalues, instead of the original vector of 1348 pixel values. A large gain. Still not much information has been lost. Here is a subset of original images, and images reconstructed using the eigenvectors shown above:

The number of eigenvectors used (and their choice) determines how much variance from the pixels is lost. This number is labeled below as "number of components", from the name of the reduction technique used: The principal component analysis (PCA):

For this graph, we see:

With the 32 eigenvectors shown above, we can reconstruct images and get back more than 80% of the original pictures.

With 150 eigenvectors, more than 95% of the information can be retrieved.

This is somehow similar to image compression, but much more too, because the eigenvalues can be used directly to make a relation between the image and the name of the person, we just need to train the classifier with a set of known images associated with the correct category (here the name to associate).

(I took my examples from this excellent book on machine learning. Faces pictures are from the LFW dataset)


问:
Polynomial Equations for the Rank of a Power of a Matrix
If I have some $n \times n$ matrix $X$ (in my case, I happen to know that X is nilpotent and in Jordan normal form), how can I write the condition that $\text{rank}(X^r)= k$ as a polynomial equation (to represent a set of matrices as a variety) in terms of the entries of the matrix? I would be interested in a Sage program that does this, or writing my own program.

EDIT: I know that $\text{rank}(X) \leq k$ if and only if the determinant of every $(k+1) \times (k+1)$ minor of $X$ is zero, but I'm not sure how to get a set of polynomial equations which will give me exact equality.

答: The set of matrices with rank exactly $k$ is the open subset of those of rank at most $k$ defined by the non-vanishing of at least one $k$ by $k$ minor---thus, it is not naturally realized as a closed subvariety of the space of all matrices. You can use the usual trick to identify this with an affine variety if you wish: the point is that the set of non-zero elements $x$ of a field $F$ is isomorphic to the affine variety $xy=1$ in $F^2$.


问:
$ad + cb = ac + bd$ s.t. $a < b$ and $c < d$
Is it possible to find numbers $a, b, c, d > 0$ such $a < b$ and $c < d$ and

$$
ad + cb = ac + bd
$$
?

Attempt based on comment:

Assume it is true, then
$$
ad + cb - ac - bd = a(d-c) + b(c-d) = 0
$$
Next,
$$
a(d-c) = b(d-c).
$$
Next, since $d\neq c$, we have
$$
a=b,
$$
which contradicts to the original assumption $a < b$.

答:
$ad + cb = ac + bd$ is equivalent to $(a-b)(d-c) = 0$.  So we can have $c=d$ which contradicts the assumption $c<d$, or $a=b$ which contradicts the assumption $a<b$.

Hence, it is not possible to find numbers $a, b, c, d > 0$ such $a < b$ and $c < d$ and
$
ad + cb = ac + bd
$.


问:
Ballistic Motion in Space with One Gravitating Body.
I had an interesting math problem presented to me some time ago by a friend (he stated it in non-mathematical terms). At what angle would you launch a projectile from a spaceship/satellite such that it left that object and went on to hit another orbiting object? Then as a supplemental question he asked at what angle would you launch that projectile to hit the other orbiting object in the least amount of time?

I assumed that the objects were only acted upon by a single spherically symmetric mass distribution such that I could treat it as a one-body problem for each object. Further, I assumed it all took place in the plane with a polar coordinate system so that I ended up with this simple system of nonlinear autonomous ODE's,

$$ \begin{bmatrix} \frac{d \theta}{dt}  \\ \frac{dv}{dt}\\ \frac{dr}{dt} \end{bmatrix} =  \begin{bmatrix} \frac{h}{r^{2}} \\ \frac{h^{2}}{r^{3}} -\frac{\mu}{r^{2}} \\ v \end{bmatrix}.$$

Where the inital conditions for the projectile would be $\{ \theta_{i} , r_{i}, v_{\beta}\cos(\phi - \theta_{i}) + v_{i}\}$ with an $h_{\phi} = r_{i} (v_{\beta}\sin(\phi - \theta_{i}) + v_{\theta i})$ and the target objects' initial conditions are $\{ \theta_{i}^{'} , r_{i}^{'}, v_{i}^{'}\}$ with $h'= r^{'}_{i} v_{\theta i}^{'}$. Where $\phi$ is the launch angle from the polar axis while $v_{\beta}$ is the magnitude of the projectiles velocity (which I assume to not change, only its launch direction) and $\mu$ is a constant relating to the gravitational field strength of the attracting object. Below is a picture depicting the general initial and final conditions.

After all these preliminaries, i'm basically asking if there is a variational calculus or other simpler solution to solving this problem as perhaps a boundary problem of sorts mixed with an initial ODE problem. That is, aside from computationally pouring on through thousands of trajectories with minutely differing $\phi$'s then numerically guessing at the appropriate approximate launch angle or angles that solve my question(s).

Which. . . isn't what I exactly want to do and I desire to know if there is an equation, a single or system of ODE's, that I could solve themselves for this launch angle which gives a least time of travel or gives launch angles that would lead to a hit (irrespective of the time of travel). If you can help in anyway, this would be most appreciated. I'm a sophomore College student with little knowledge of solving robustly ODE's or even programming solvers for them.

答:
Another approach is to assume all trajectories lie the same plane, and use Kepler's laws to turn this into a geometry problem. While it may be exactly solvable in some specific cases, after an initial dive it looks like this approach leads to a system of equations that can only be solved numerically. Still, that might be far preferable to numerically solving some ODEs.

Without getting into the messy algebra, the main tools of this approach would be that

All the orbits are conic sections. In particular, the target is traveling in an ellipse (as is the launcher, less importantly), and the projectile has discrete possibilities of traveling in an ellipse or hyperbola (or, as a limiting case, a parabola). It seems like assuming the target is traveling in a circle rather than a general ellipse is a valid simplifying assumption, but doesn't make much of a difference for numerical solutions.

Kepler's second law states that equal areas are swept by orbits in equal times. With not too much work, you can relate the area between the lines OA from the origin to the initial point A, OB, and a conic section C from A to B (call that TC). This could be generalized if the objects are allowed to complete one or more orbits before colliding, but that's quite likely to not be the least-time solution.

Thus, the spatial coordinates of the collision will be at either an (Ellipse, Ellipse) intersection or (Hyperbola, Ellipse) intersection (or (Parabola, Ellipse), but that's a special case). Take the case where the projectile is launched into an elliptical orbit at fixed initial point A, and must hit a target traveling on a circular orbit from initial point B. The ellipse has three parameters in general (semi-major and semi-minor axes, and an angle of rotation). The total set of equations to solve is thus, schematically,

```<code>Ellipse intersects A
Ellipse intersects C
Circle intersects C
T[Ellipse](A,C) = T[Circle](B,C) = DeltaT
</code>```

These are essentially four equations (as the first has two components) with five unknowns: the two coordinates of the intersection point C, the three parameters of the ellipse. So, one parameter - e.g., one of the ellipse parameters - can be scanned over, and a solution (if any) and corresponding DeltaT found for each value. This would have to be repeated for the other cases (where the projectile travels in a hyperbola or a parabola).

Body:
Another approach is to assume all trajectories lie the same plane, and use Kepler's laws to turn this into a geometry problem. While it may be exactly solvable in some specific cases, after an initial dive it looks like this approach leads to a system of equations that can only be solved numerically. Still, that might be far preferable to numerically solving some ODEs.

Without getting into the messy algebra, the main tools of this approach would be that

All the orbits are conic sections. In particular, the target is traveling in an ellipse (as is the launcher, less importantly), and the projectile has discrete possibilities of traveling in an ellipse or hyperbola (or, as a limiting case, a parabola). It seems like assuming the target is traveling in a circle rather than a general ellipse is a valid simplifying assumption, but doesn't make much of a difference for numerical solutions.

Kepler's second law states that equal areas are swept by orbits in equal times. With not too much work, you can relate the area between the lines OA from the origin to the initial point A, OB, and a conic section C from A to B (call that TC). This could be generalized if the objects are allowed to complete one or more orbits before colliding, but that's quite likely to not be the least-time solution.

Thus, the spatial coordinates of the collision will be at either an (Ellipse, Ellipse) intersection or (Hyperbola, Ellipse) intersection (or (Parabola, Ellipse), but that's a special case). Take the case where the projectile is launched into an elliptical orbit at fixed initial point A, and must hit a target traveling on a circular orbit from initial point B. The ellipse has three parameters in general (semi-major and semi-minor axes, and an angle of rotation). The total set of equations to solve is thus, schematically,

Ellipse intersects A
Ellipse intersects C
Circle intersects C
T[Ellipse](A,C) = T[Circle](B,C) = DeltaT

These are essentially four equations (as the first has two components) with five unknowns: the two coordinates of the intersection point C, the three parameters of the ellipse. So, one parameter - e.g., one of the ellipse parameters - can be scanned over, and a solution (if any) and corresponding DeltaT found for each value. This would have to be repeated for the other cases (where the projectile travels in a hyperbola or a parabola).


问:
Mapping randomness to binary
I'm working on an algorithm, to map entropy from random events like coin flips, dice, user input, timing delays, and so on (any arbitrary random event), to binary, in a deterministic and reversible manner without introducing any bias.

For example:

Using coin flips is easy, tails <=> 0 and heads <=> 1, because every coin flip, gives you 1 bit of entropy.

Using an 8-sided dice is also easy, 1 <=> 000 to 8 <=> 111, because every roll of an 8-sided dice, gives you exactly 3 bits of entropy.

These examples are 2^N. But what about events that do not fit nicely in 2^N. For example a regular 6-sided dice. Log2(6) ~= 2.5 bits.

So a 6-sided dice, is not enough entropy to get 3 bits, but is enough for 2 bits at least, (as we have ~2.5 bits of entropy). So how do we get at least 2 bits of entropy from a 6-sided dice?

Option 1 - The mapping could be:

1 <=> 00

2 <=> 01

3 <=> 10

4 <=> 11

5 <=> ?

6 <=> ?

If we rolled a 5 or 6, we could ask the user to roll again, (or in a general sense, for a computer program), we could error out, or discard, and collect new data for entropy. This option seems a bit wasteful because we are discarding the entropy we could of used, and we are re-rolling, or re-collecting more entropy.

Option 2, could be to truncate:

1 <=> 000

2 <=> 001

3 <=> 010

4 <=> 011

5 <=> 100

6 <=> 101

However this does not work, the least significant bit is ok, it has 50% chance of a 0 or 1. But the middle bit has a 4/6 chance of '0' and 2/6 chance of '1', and the last bit is also not 50/50. One explanation for this, is we are inflating the ~2.5 bits of entropy from a 6-sided dice to 3 bits of entropy, so we will be introducing some bias. We can only use 1 bit of entropy here (the least significant bit), however, if the source of entropy does not have an even probability, we may not even be able to get 1 bit.

Even if we ask the user to roll a 6-sided dice, 1 time (~2.5 bits), 2 (~5.2), 3 (~7.8), 4 (~10.3), 5 (~12.9), ... 10 times (~25.8 bits), we never get an whole number of bits, so do we get them to discard all of the dice rolls? Surely not. We should be able to collect 25 bits (or less?) of entropy from 10 6-sided dice rolls? If so, how?

答:
Please check Shannon-Fano Encoding, also Huffman coding. The notion of entropy is inherently asymptotic; if you adopt the optimum encoding to represent $X$, for long enough sequence of $\{X_i\}_{i=1}^{n}$ (as $n\rightarrow \infty$), the length of representation tends to $nH(X)$.

In practice, additional concerns must be taken into account. Trivially, any binary sequence must be uniquely decoded. In the case of a fair dice, you may adopt the following encoding:

$1: 00, 2: 01, 3: 100, 4: 101, 5: 110, 6: 111$

which results in an average length of 2.66 bits for a long enough sequence. The overhead of $2.66-2.58=0.08$ bits is to ensure unique decoding. For example, if you want to decode:

$001110100010001101...$

it can only be decoded as:

$00, 111, 01, 00, 01, 00, 01, 101, ...$

that is:
$1, 6, 2, 1, 2, 1, 2, 4, ...$

Title: Mapping randomness to binary


问:
Infinity and Negative Infinity Logical Query
If infiniti is the highest theoretical number then what is the opposite of it? At first glance one might think 0 but that is not true. Research (aka a quick google search) has shown me that the opposite of infinity is called infinitesimal(sorry if I spelled that wrong) which is another unquantifiable number but it is basically zero point infinite zeros followed by a single one, as I understand it. Now if the definition of negative is “the opposite of a number” and we know that 0.000… infinitely, followed by a one, is the opposite of infinity, then we know that because of the transitive property then -infinity=0.0…1. This means that the highest theoretical negative number is actually a positive number. This leads me to think that there are no negative numbers and that all negative numbers are just decimal. Is my logic completely flawed or does it make sense?</p>

答:
Is my logic completely flawed or does it make sense?

Your question makes a lot of separate claims that each have different issues, so I think the best way to clear things up is to comment on some individual lines.

Clarifying the Logic

If infiniti is the highest theoretical number

"Infinity" (note the "y" at the end in English) is not just one idea, and the ideas related to infinity are not always called "number"s, and never have all the nice properties that "real numbers" do. For a survey of some related ideas, see the English Wikipedia page for Infinity or the Math StackExchange question Understanding infinity.

what is the opposite of it?

Dictionary.com has a definition for "opposite": "contrary or radically different in some respect common to both...". This can me used in a variety of different ways in mathematics. Here are two main ones:

In arithmetic, "opposite" often means the additive inverse or negative of a regular finite number. So you might hope that a given concept of infinity has something analogous, like a "negative infinity".

If you think of infinity as in some sense "very big (and positive)", then a different application of the definition would make the opposite "very small (and positive)". It is in this sense that the word "infinitesimal" may be used. For regular finite numbers, this kind of opposite might be the multiplicative inverse or reciprocal.

then we know that because of the transitive property...

This entire line of reasoning is invalid because you are treating the two ideas of "opposite" above as if they are the same thing. As HallaSurvivor alluded to in a comment, this is very similar to saying: "$-2$ is an opposite of $2$ because it's the same size but negative. And $1/2$ is an opposite of $2$ because multiplying by $1/2$ shrinks something by the same factor that multiplying by $2$ grows it. Since they're both 'opposites', $-2=1/2$."

Clarifying Infinity

As discussed above, there are at least two senses of "opposite" you might be interested in: something like "negative infinity" and something like "reciprocal infinity". What sense these do or do not make depends a lot on what meaning of infinity you're working with. Without getting into all of the possible interpretations in detail, I'll just outline a few key examples.

Beyond regular decimals?

zero point infinite zeros followed by a single one

This would not represent a real number, and does not have a standard definition. People have tried making up new number systems where this sort of thing would convey some meaning, but it's not easy and to my knowledge hasn't ever been shown to work out nicely. This matter is discussed a bit in answers to the Math StackExchange question Is it possible to create the smallest real positive number by axiome?.

Calculus $\infty$

In Calculus, the lemniscate $\infty$ is used to represent an idea like "a function or sequence gets (and stays) greater than any finite positive number". For instance, we might say "the limit of $(1,4,9,16,\ldots)$ is $\infty$" because the sequence stays above $100$ after the tenth term, stays above $1000$ after the $31^{\text{st}}$ term, etc.

Analogously, it's common to then use $-\infty$ to represent things that get and stay less than any finite negative number". For example, $(17-1,17-4,17-9,17-16,17-25,\ldots)=(16,13,8,1,-8,\ldots)$ might be said to have a limit of $-\infty$. Note that we cannot usefully think of this $-\infty$ as an additive inverse to $\infty$. For example, if we add the two sequences above term by term, we get $(17,17,\ldots)$ which stays at $17$ and does not approach $0$.

As for $1/\infty$, since the reciprocals of a sequence that gets larger and larger get smaller and smaller, some may write $1/\infty=0$. In this context, people would not generally call that an infinitesimal, just $0$.

Complex Calculus $\infty$

When doing Calculus with the complex numbers in "complex analysis", we are not limited to two directions on the number line since there is a whole complex plane to work with. In that context, we often use the symbol $\infty$ to represent things that get and stay further away from $0$ than any finite positive distance, no matter the direction.  A way of visualizing that is the Riemann sphere. In that usage, both sequences above have numbers moving away from zero, so their behavior might both be represented by $\infty$, and there is no $-\infty$ (or perhaps we would declare $-\infty=\infty$).

Similarly to the real numbers, reciprocals of complex numbers far from $0$ are close to $0$, so it would be common to write $1/\infty=0$.

Infinite Sizes

Another common use of the ideas of infinity is in giving names to the sizes of infinite sets in the study of "cardinality". But every set has at least $0$ elements, so "negative infinity" would make no sense in that sort of context. And it doesn't make sense to have between $0$ and $1$ elements, so "reciprocal infinity" wouldn't make sense either.

Abstract Settings

In more obscure mathematics, we might have a sort of "number system" where arithmetic and order work out in a fairly normal way, but there are now new numbers larger than any positive integer. Many of these are called "non-Archimedean ordered fields". In such a system, there is not just one "infinity" so we would not use the symbol $\infty$, but we could have "an infinite element $H$" and then things like $1+H>H$, $-H+H=0$, and $0<1/H<1/100$ might all make perfect sense.

In this context, "positive infinite" would mean "greater than any positive integer". And "positive infinitesimal" would mean "positive but less than the reciprocal of any positive integer". $0$ might be considered an "infinitesimal", but $1/0$ would still not make sense.


问:
What is this matrix/operator notation $\otimes$?
Some notes I'm reading say that:

$$x = \sum_{i=1}^m c_i\langle x,u_i\rangle u_i$$ for all $x\in\mathbb{R}^n$ (where $c_i > 0$ are positive numbers and $u_i$'s are unit vectors) can be equivalently written in matrix (or operator) notation as $$\sum_{i=1}^m c_i u_i\otimes u_i = I_n$$ where $I_n$ is the identity map on $\mathbb{R}^n$, and for any unit vector $u$, $u\otimes u$ is the rank-one orthogonal projection onto the span of $u$, i.e. the map $x\mapsto \langle x,u\rangle u$. The trace of this projection is $1$.

So, I haven't seen the above notation $\otimes$ earlier - and I'm a little confused what it means? Could someone help me understand?

What is the meaning of rank-one orthogonal projection here, and what does the operator really do? How is this equivalent to the first condition?

答:
It's the Outer/Tensor product and defined by $u \otimes v = uv^\top$ or alternatively, you can think of it as the function

$$ (u \otimes v)(x) = \langle v, x \rangle u, $$

which is of course what you get when you multiply $uv^\top$ and $x$ together.

Note: there is also a Kronecker product which uses the same symbol and the relationship is

$$ u \otimes_{\rm out} v = u \otimes_{\rm kr} v^\top. $$

The Kronecker product of two column vectors $(a_i), (b_j)$ is a vector whose $(in + j)$-th entry is $a_ib_j$. The outer product is a matrix whose $(i,j)$-th entry is $a_ib_j$.


问:
A question on finite-order operators
I have a question: is it true that if $R\in\mathbb{C}^{n\times n}$ is s.t. $R^k=1$ for some $k>1$ (that must be the smaller one for which this happens), then there exists $x\in\mathbb{C}^n$ s.t. $R^jx\neq R^lx$ for all $j,l=0,\ldots,k-1$, $j\neq l$? I think the answer is affirmative and I guess this derives from the injectivity of $R$, which follows by its invertibility. However, I cannot write it down appropriately. I guess that if $\forall x\in\mathbb{C}^n$ there exist $j\neq l\in\{0,\ldots,k-1\}$ s.t. $R^j x=R^lx$, then setting $N=l.c.m.\{j\in\{1,\ldots,k\} : \exists y\in\mathbb{C}^n \ s.t. \ R^jy=y\}$, it should be $N=k$, otherwise $R$ would not have order $k$. By assumption the assumption, $\forall x\in\mathbb{C}^n$ there exists $j\in\mathbb{Z}$ s.t. $R^jx=x$ (we use the invertibility of $R$ here). Then, I don't know how to proceed.

答: Since $R$ is invertible (having $R^{-1}=R^{k-1}$), if for all $x$ we had $R^j(x)=R^l(x)$, it would mean $R^j=R^l$, implying $1=R^{l-j}$ which would contradict to the minimality of $k$.

Title: A question on finite-order operators


问:
How to minimise the area of the folded part of a piece of paper when it touches the other side
I am studying maths purely out of interest and have come across this question in my text book:

A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on CD. If AB = a and AD = b, where $a \ge\frac{2b}{\sqrt3}$, show that the least possible area of the triangle APQ is obtained when the angle AQP is equal to $\frac{\pi}{3}$. What is the significance of the condition $a \ge\frac{2b}{\sqrt3}$?

I realise I need to get an equation for the area of the fold, which I can then differentiate. Even then, I am not sure how to relate this to angle AQP.I have looked at solutions on the internet but they tend to look at the length of the crease, not the area of the fold.

I have taken A in my diagram to represent its position before the fold. I could be mistaken.

This is how I visualise it:

I have said:

$x^2 = m^2 + (b - x)^2 = m^2 + b^2 - 2bx +x^2 \implies m^2 = 2b(x - \frac{b}{2})$

$L^2 = (L - m)^2 + b^2 \implies L = \frac{m^2 + b^2}{2m}$

$y^2 = x^2 + L^2 = x^2 + \frac{(m^2 + b^2)^2}{4m^2} = x^2 + \frac{(2bx - b^2 + b^2)^2}{4m^2} = x^2 + \frac{4b^2x^2}{8b(x - \frac{b}{2})} = \frac{bx^2}{2x - b}$

After this I am not sure how to proceed.

答:
I think you were almost there. You don't need $y$ to compute the area, so you can omit that part of your calculation. So you have two variables, $x$ and $m$, but these are related. You've solved for $m^2$ in terms of $x$, but you can instead solve for $x$ in terms of $m^2$. Then you can minimize $A=\frac{1}{2}xL$, which is a function of the single variable $m$.

Notice that there are natural endpoints for this minimization. Start by assuming that the piece of paper is semi-infinite. That is, let $a$ go to infinity so you don't have to worry about not getting a triangle when you make the fold. You should be able to see that $m$ must be greater than $0$. (As $m$ approaches $0$, the area of the triangle approaches infinity.) On the other hand, $m$ can be at most $b$ (since $m\le x\le b$), at which point the triangle area is $\frac{1}{2}b^2$. So you are minimizing on the interval $(0,b]$.


问:
Prove that $A$ is compact $\Leftrightarrow$ $X(A)$ is compact
Let:

$I(p,q)$ - closed segment with ends $p,q\in \mathbb R^3$

$A\subset (0,1)\subset \mathbb R$

$X(A)=\bigcup \{I \Big( (x,0,\tan x),(x,x^3,0) \Big):x\in A \} \subset \mathbb R^3$ with the Euclidean metric

Prove that: subset $A$ of segment $(0,1)$ with Euclidean topology is compact $\Leftrightarrow$ $X(A)$ is compact in the Euclidean space $\mathbb R^3$

My attempts:

1) $A$ is compact $\Rightarrow$ $X(A)$ is compact

We know that closed segment $I(p,q)$ can write by definition as $I(p,q)=\big\{t\cdot p+(1-t)\cdot q\ \big|\ t\in[0,1]\big\}\subseteq\mathbb{R}^3$. Let: $v(x)=(x,0,\tan x)
$, $w(x)=(x,x^3,0)$. Of course functions $v,w$ are continuous for $x\in (0,1)$. This implies that $X(A)$ is a continuous image of $A\times [0,1]$ via $(x,t)\mapsto t\cdot v(x)+(1-t)\cdot w(x)$.
 Since $A$ is compact then $A\times [0,1]$ is compact too (by Tychonoff's theorem, the cartesian product of compact spaces is compact). So $X(A)$ compact as an image of a compact set by continuous mapping.

My questions:

Firstly, I am not sure my proof "$A$ is compact $\Rightarrow$ $X(A)$ is compact" is correct so I would be grateful if you could comment on it. Secondly, I have no idea how to prove the implication: $A$ is compact $\Leftarrow$ $X(A)$ is compact.

答:
Your proof is correct. For the converse consider the projection map $p : \mathbb R^3 \to \mathbb R, p(x_1,x_2,x_3) = x_1$. This is continuous and $p(X(A)) = A$. Note that the first coodinate of $tv(x) + (1-t)w(x) \in A(X)$ is $x$.

Also observe that your map $F: A \times [0,1] \to \mathbb R^3, F(x,t) = tv(x) +(1-t)w(x)$, is an embedding because it is clearly injective and $A \times [0,1]$ is compact. This means that $A \times [0,1] \approx X(A)$.


问:
Prove that Gauss map on M is surjective
Let $M$ be a closed, orientable, and bounded surface in $\mathbb{R}^3$.

(a) Prove that the Gauss map on $M$ is surjective.

(b) Let $K_+(p) = \max \{0, K(p)\}$.  Show that
$$
\int K_+dA \ge 4\pi.
$$
in the area of $M$.  Do not use the Gauss-Bonnet Theorem to prove this.

答: Thinking about Hilbert's proof it seems that the first touch point of tangency of a given plane to the surface must touch it at a point of non-negative Gauss curvature. For at a point of negative curvature the tangent plane is crossed by the surface - saddle point. Does this give part 2 of the question? On more thought this seems to work. The first touch point of the plane must be tangent at a point of non-negative Gauss curvature because the surface would cut the plane at more than one point at a saddle point. So the Gauss map is surjective onto the sphere just from the points of non-negative curvature. Now pull back the volume form of the sphere and use change of variables. BTW: What can one say about the topology of points of zero Gauss curvature on the surface - say if it has measure zero?</p>

Title: Prove that Gauss map on M is surjective


问:
why does the null space contain nonzero vectors if the linear system is inconsistent?
I am reading through Boyd's optimization book(section 10.1.1, page 522) and I can't understand the highlighted statement in the screenshot below. I do not understand why there are nonzero vectors in the null space if the KKT system is not solvable?

答:
If KKT system (10.4) is not solvable, then the linear map defined by the KKT (square) matrix in (10.4) is not surjective. ("Surjective" means (10.4) has a solution for every pair of $(q,b)$.)

By the rank-nullity theorem, it must not be injective.

Therefore the null space of the KKT matrix contains nonzero vectors.

May be useful to think about a simpler example.

Suppose $A$ is a $2\times 2$ matrix and $b\in\mathbb{R}^2$ is a nonzero vector.

If $Ax=b$ has no solution, i.e., "not solvable", then the linear map $x\mapsto Ax$ is not surjective since there is no pre-image for the vector $b$. Then by the rank-nullity theorem, the null space for $A$ must contain nonzero vectors.


答: Since $R$ is invertible the claim is equivalent to saying that there exists some $x\in\Bbb C^n$ such that $R^jx\ne x$ for all $1\leq j\leq k-1$. Since $R^j$ is not the identity matrix for $j=1,\dots, k-1$ the set of $x\in\Bbb C^n$ with $R^jx=x$ is a proper subspace (i.e. $\ker (R^j-I)$) of $\Bbb C^n$. If for all $x\in \Bbb C^n$ there was some $j=1,\dots,k-1$ with $R^jx=x$ then $\Bbb C^n$ would be the finite union of proper subspaces which is absurd.

Title: A question on finite-order operators

Body: I have a question: is it true that if $R\in\mathbb{C}^{n\times n}$ is s.t. $R^k=1$ for some $k>1$ (that must be the smaller one for which this happens), then there exists $x\in\mathbb{C}^n$ s.t. $R^jx\neq R^lx$ for all $j,l=0,\ldots,k-1$, $j\neq l$? I think the answer is affirmative and I guess this derives from the injectivity of $R$, which follows by its invertibility. However, I cannot write it down appropriately. I guess that if $\forall x\in\mathbb{C}^n$ there exist $j\neq l\in\{0,\ldots,k-1\}$ s.t. $R^j x=R^lx$, then setting $N=l.c.m.\{j\in\{1,\ldots,k\} : \exists y\in\mathbb{C}^n \ s.t. \ R^jy=y\}$, it should be $N=k$, otherwise $R$ would not have order $k$. By assumption the assumption, $\forall x\in\mathbb{C}^n$ there exists $j\in\mathbb{Z}$ s.t. $R^jx=x$ (we use the invertibility of $R$ here). Then, I don't know how to proceed.


问:
Decay of high momentum fluctuations
Let $\chi(k)=e^{-k^2}$ and define
$$
\Gamma(x)=\int_{\mathbb{R}^d} e^{ikx} \frac{1}{k^2}(\chi(k)^2-\chi(Lk)^2)
$$
where $L>1$. Then how would one prove that $\Gamma$ decays exponentially, i.e.,
$$
|\Gamma(x)| \le A_d e^{-|x|/L}
$$
for some constant $A_d$

答:
By the Fundamental Theorem of Calculus
$$
e^{-2k^2}-e^{-2L^2k^2}=\int_{1}^{L^2}dt\ 2k^2 e^{-2tk^2}\ .
$$
Insert that and use Fubini to get
$$
\Gamma(x)=2\int_{1}^{L^2}dt\ \int_{\mathbb{R}^d}d^dk\ e^{ikx-2tk^2}\ .
$$
Then use the formula for the Fourier transform of a Gaussian to obtain
$$
\Gamma(x)=2\left(\frac{\pi}{2}\right)^{\frac{d}{2}}\int_{1}^{L^2}
t^{-\frac{d}{2}}\ e^{-\frac{x^2}{8t}}\ .
$$
By expanding $\frac{1}{8t}(|x|-4\sqrt{t})^2\ge 0$ we get the lower bound
$$
\frac{x^2}{8t}\ge\frac{|x|}{\sqrt{t}}-2\ge \frac{|x|}{L}-2
$$
which can be used to get the bounds on the integral
$$
0\le \Gamma(x)\le A_{d,L}e^{-\frac{|x|}{L}}
$$
with
$$
A_{d,L}=2e^2\left(\frac{\pi}{2}\right)^{\frac{d}{2}}\int_{1}^{L^2}
t^{-\frac{d}{2}}\ .
$$
Finally, if $d>2$ one can get rid of the $L$ dependence in the constant $A$ because the last integral converges at $L=\infty$.

Title: Decay of high momentum fluctuations


问:
Even maps of any even degree
From Hatcher :

Show that if $n$ is odd, then there exists even continuous functions from $S^n \to S^n$ of any even degree

What I know is there exists maps of any degree from any $S^n \to S^n$ (n need not be odd). But these are not even functions necessarily. How do I fix that?

Edit: $f:S^n\to S^n$ is said to be an even map if $f(x)=f(-x),\forall x\in S^n$.

答:
This is a suggestion on how you can understand where to get such a map. It might not be the clearest explanation but I still make it an answer to provide a picture.

Take an arbitrary $2k$ and $f: S^{n} \to S^{n}$ of degree $k$. This is how the sum $f+f$ looks:

Here the blue color shows how the wedge looks "from $f$'s perspective". The definition of the sum in homotopy groups does not require that you invert all coordinates on the lower little sphere, but you can still do that with no (homotopic) change if the number of coordinates is even.
This way homotopically you will get $f+f$, which is clearly of the desired degree, and your map will be even.

Title: Even maps of any even degree


问:
Why is the universal quantifier $\forall x \in A : P(x)$ defined as $\forall x (x \in A \implies P(x))$ using an implication?
And the same goes for the existential quantifier: $\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \wedge P(x))$. Why couldn’t it be: $\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \implies P(x))$ and $\forall x \in A : P(x) \; \Leftrightarrow \; \forall x (x \in A \wedge P(x))$?</p>

答: Consider the expression $\forall x \in A : P(x) \; \Leftrightarrow \; \forall x (x \in A \wedge P(x))$. Assuming $A$ to be a proper subset of the domain of discourse, the expression will always be false, because by definition there are $x$ values in the domain of discourse which are not in $A$.</p>


问:
Three people are seated in 3 our of 5 chairs . After a break , in how many ways can they be seated so that no person occupies the same same chair
I tried doing it by first counting the total arrangements i.e. $^5p_3$ = 60 then subtracting the non desired arrangements fixing the first person at its place and arranging the other 2 person .This can done for the rest of the 2 people therefore $3*(^4P_2)$
So ans should be $^5p_3$ - $3*(^4P_2)$ = 24
I know i have subtracted some same cases twice but i am not sure how to identify those cases.
Is there any other approach also ??

答:
You ask for any other approach.

Since the numbers are small we can split the cases up according to the number of originally unoccupied chairs that are used.

No originally unoccupied chair used

Then the first person can move chair in $2$ ways and that fixes where everyone goes.

One used

We can choose the chair and the person who now sits in it in $2.3=6$ ways. The other two people either switch seats or one of them sits in the newly unoccupied seat. So each of the $6$ ways has $3$ arrangements.

Both used

There are $3.2$ arrangements in the two originally unoccupied chairs. The other person can now move chair in $2$ ways.

The total is $2+18+12=32$.


问:
Proof of Alternating Binomial Coefficient Identity relating
I am looking for a proof of the following binomial identity. I encountered it in an article on Euler’s derivation of the gamma function. Euler begins by evaluating the integral:

$$\int_0^1 x^a(1-x)^n\,dx$$

He performs a binomial expansion on the integrand and makes use of the following identity involving alternating binomial coefficients:

$$\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{1}{a+k+1} = n! \prod_{k=0}^{n}\left( \frac{1}{a+k+1} \right).$$

It is not obvious to me on inspection that this formula is true, but I have found it cited elsewhere in the mathematical literature, without proof. I have verified it by hand for the cases of n=2 and n=3, but I am unable to prove the identity in general. I am looking for a good combinatorial or inductive proof of this very interesting fact.

I am aware that this formula relates to the beta function, but I am trying to derive this relationship without reference to this topic.

Thank you for enlightening me.

答:
By the partial fractions theorem, we know
$$\frac{n!}{a(a+1) \cdots (a+n+1)} = \frac{b_0}{a} + \frac{b_1}{a+1} + \dots + \frac{b_{n+1}}{a+n+1}$$
for some set of numbers $b_0, b_1, \dots , b_{n+1}$. To find $b_k$, multiply both sides of the equation by $b+k$ and then set $a=-k$ in the resulting equation.  The result is
$$b_k = (-1)^k \binom{n}{k}$$


问:
showing that the Euler's number is irrational
Our teacher wants us to do the following:
Suppose that e is rational i. e $e=\frac{a}{b}$ where $a,b\in\mathbb{N}$. Choose $n\in\mathbb{N}$ such that $n>b$ and $n>3$. Use the following inequality

$0<e-(1+\frac{1}{1!}+\frac{1}{2!}+....+\frac{1}{n!})<\frac{3}{(n+1)!}$ and replace e with $\frac{a}{b}$. Then multiply both sides of the inequality with $n!$ to see that the inequality leads to existence of an integer N that satisfies $0<N<\frac{3}{4}$ which leads to a contradiction.

I have tried to to it but I am not getting there...
I choose $n=4$ and $b=3$ and
$0<\frac{a}{b}-1-1-\frac{1}{2}-\frac{1}{6}-\frac{1}{24}<\frac{3}{5!}$

$0<8a-\frac{17}{24}<\frac{3}{5}$

$0<8a<\frac{157}{120}$

$0<a<\frac{157}{960}$.....

How should I do it???

答:
You’re not supposed to pick a specific value of $n$, like $4$; you’re supposed to look at arbitrary integers $n>\max\{b,3\}$. Suppose that $n>b$ and $n>3$. Multiplying the given inequality by $n!$, you get

$$0<en!-n!\left(1+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}\right)<\frac{3n!}{(n+1)!}\;,$$

and if you now replace $e$ in this inequality by $\frac{a}b$, you get

$$0<\frac{an!}b-n!\left(1+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}\right)<\frac{3n!}{(n+1)!}\;.$$

Now multiply out and do some simplifying:

$$0<\frac{an!}b-\left(n!+\frac{n!}{1!}+\frac{n!}{2!}+\ldots+\frac{n!}{n!}\right)<\frac3{n+1}\;.$$

Use the fact that $n>b$ to explain why $\frac{an!}b$ is an integer. All of the fractions $\frac{n!}{k!}$ with $1\le k\le n$ are integers (why?), so the whole middle expression is an integer lying strictly between $0$ and $\frac3{n+1}$. But $n>3$, so this is impossible; why?


问:
Equilibrium Price in a Double Auction
I am trying to understand how to find the equilibrium price in a double auction.

From wikipedia:

Natural ordering

Order the buyers in decreasing order of their bid: b1≥b2≥...≥bn.

Order the sellers in increasing order of their bid: s1≤s2≤...≤sn. Let k be the largest index such that bk≥sk (the "breakeven index").
Every price in the range [max(sk,bk+1),min(bk,sk+1)] is an equilibrium
price, since both demand and supply are k. It is easier to see
this by considering the range of equilibrium prices in each of the 4
possible cases (note that by definition of k, bk+1 < sk+1)

What does it mean for the "supply and demand to be at k"?

Saying something like "the demand is 23" doesn't make any sense to me.

答:
You should have seen this when a demand curve was defined.  Saying the demand at $50$ is $23$ means if you price the object at $50$ you will sell $23$ of them.  Another point on the demand curve could be that the demand at $40$ is $29$, which would say that there are six people who will buy at $40$ but will not buy at $50$.  The demand curve is just the number that will be bought as a function of the price.

In your example, let $n=10$ and let there be one bid each of $10,20,30\ldots 100$.  The demand at $55$ is $5$ because the five highest bidders, $b_1$ through $b_5$, are willing to buy at $55$ while the other five are not willing to buy.

Now make up a set of sales offers.  I would avoid having any at an exact multiple of $10$.  It doesn't have to be $10$ of them.  Follow the process described.  Note how it finds a price range where you have the same number of buyers and sellers that will accept the price.

Title: Equilibrium Price in a Double Auction


问:
Interpreting the solution of the heat equation on the whole line
Can someone provided some insight on what is meant by "weighted average"?

Allow me to give some exposition.

So given the heat equation on the whole line

$$\begin{cases}
u_{t}-ku_{xx}=0\quad\quad -\infty <x<\infty ,\;t>0\\
u(x,0)=\phi (x)
\end{cases}
$$

we know its solution is given by

$$u(x,t)=\int_{-\infty}^{\infty}S(x-y,\;t)\phi(y)\;dy,$$

where $S(x,t)$ is the heat kernel

$$S(x,t)=\frac{1}{\sqrt{4\pi kt}}\;e^{-x^2/4kt}\quad t>0.$$

The heat kernel, as we know it, represents the evolution of temperature for some region in space. In this case, for a 1-D infinite rod.

Now, in Walter A. Strauss' An Introduction to Partial Differential Equations,
Strauss claims that the solution $u(x,t)$ is
"a kind of weighted average of the initial values around the point $x$."

He later justifies this notion by writting the solution as follows

$$\int_{-\infty}^{\infty}S(x-y,\;t)\phi(y)\;dy\simeq \sum_t S(x-y_i\;t)\phi(y_i)\Delta y_i $$

and suggests that the solutions $S(x-y_i)$ are weighted by $\phi(y_i)$ .

My interpretation of the solution (in the context of 1-D heat flow in a rod) is that once time starts, the initial temperature distribution $\phi(x)$ spreads throughout the rod and as time goes on, the temperature value at a fixed point $x$ is given/determined by the average value the temperature of nearby points.

Is my intuition on track?

Also, what exactly does Strauss mean when he says $u$ is a "weighted average"?

Any input is much appreciated.

答: The literal interpretation of this formula is as follows: The function $S(x-y_1)$ is the normal distribution shifted from $x=0$ to a specific $x$ value $= y_1$. All values of this shifted function along the $x$-axis is then multiplied by all the corresponding values along the $x$-axis of the (unshifted) initial condition function. This number is multiplied by delta $y$. All these numbers are added up. As the area below the normal function is $1$, the resulting sum becomes a weighed average, i.e. the initial condition is weighed by the corresponding values of the normal distribution located at the specific position $x=y_1$. To get the value at another $x$ position, you have to shift the normal function to that new $x$-position and repeat the calculation. The procedure may be described as finding a moving average or as a convolution.


问:
Visual Proof for Congruence Relation
Can anybody please provide a convincing visual proof that

$a ≡ b \pmod {n}$ if $a-b$ is a multiple of $n$

is equivalent to

$a ≡ b \pmod n$ if $a$ and $b$ have the same remainder modulo $n$?

Edit: I am thinking of number lines, coloured squares, circle sectors etc. The symbolic proof would be helpful too, but I want to understand this in a more concrete way. Having the connections made between the symbolic proof and a more visual one would be ideal.

答:
I suggest some bottles, that they all have the same remainder of a specific number as the below figure.

may help you to visuallize.

Title: Visual Proof for Congruence Relation


问:
primitive roots of 17
I what to show that if $a$ and $b$ are primitive roots modulo prime number $p$ then $ab$ is not primitive root modulo $p$.
I want to use a counter example.
let $a = 5$, $b = 7$ and $p=17$

$O(5)=16$ and $O(7)=16$ then $5$ and $7$ are primitive roots modulo $17$.
Now $ab= 35$. How can I continue?

答:
For any odd prime $p$ and any $a$ co-prime to $p$, we have $a^{(p-1)/2}\equiv\pm 1\bmod p$. If in addition $a$ is a primitive root, we must have $a^{(p-1)/2}\equiv -1\bmod p$.

So if $a$ and $b$ are primitive roots, what is $(ab)^{(p-1)/2}\bmod p$?

Title: primitive roots of 17


问:
A doubt on the Sobolev space $W_0^{1,p}(\Omega)$
Let $\Omega \subset \Bbb R^d$ be open and $1\leq p<\infty$. Recall that $W_0^{1,p}(\Omega)$ is the closure of $C_c^\infty(\Omega)$(smooth function with compact support in $\Omega$) in $W^{1,p}(\Omega)$ where $$W^{1,p}(\Omega)= \{u\in L^p(\Omega): \nabla u \in L^p(\Omega) \}$$
is equiped with the norm $$\|u\|^p_{W^{1,p}(\Omega)}= \|u\|^p_{L^p(\Omega)} + \|\nabla u\|^p_{L^p(\Omega)}$$

Define
$$W_\Omega= \{u\in W^{1,p}(\Bbb R^d): u =0 ~~a.e.~~ on~~ \Bbb R^d\setminus \Omega \}$$
Clearly $W_\Omega$ is closed subspace of $W^{1,p}(\Bbb R^d)$ and we have $W_0^{1,p}(\Omega)\subset W_\Omega$

Question: Do we have $W_0^{1,p}(\Omega)= W_\Omega$?

Or is there  a counter example?

答:
Another situation in which equality fails is $\Omega = (-1,1)^d \setminus (\{0\}\times (-1,1)^{d-1})$. Then,
$$W_\Omega = W_0^{1,p}( (-1,1)^d )$$
since the a.e.-equality does not see the hyperplane $\{x | x_1 = 0\}$.

However, you have (without any further assumptions on $\Omega$)
$$W_0^{1,p}(\Omega) = \{ v \in W^{1,p}(\mathbb R^d) \;|\; v = 0 \text{ q.e. on } \mathbb R^d \setminus \Omega \}.$$
Here, "q.e." means quasi-everywhere (i.e., up to subsets of capacity zero) and we use the quasi-continuous representative of $v \in W^{1,p}(\mathbb R^d)$.

This can be found in Theorem 4.5 in the book
"Nonlinear potential theory of degenerate elliptic equations"
(1993)
by
Heinonen, Kilpeläinen, Martio.


问:
Does the inverse of each element of a sample from a random variable constitute a sample from the inverse of that variable?
Imagine you have a sample of $n$ elements from a random variable $X$. If I take the inverse of every element in that sample, is it always the case that the resulting set of numbers constitute a sample from $\frac{1}{X}$ ?

Thanks in advance

答:
Yes.

The inverses remain independent.

The rest of it is just thinking through what exactly it means for a random variable to have the distribution of the random variable $\frac{1}X$.


问:
Maximum number of non-congruent triangles with side lengths integers and less than $9$

Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.

I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?

答:
Yes, a systematic approach is to count the number of triples of three integers which satisfy the triangle inequality. All that this requires is that the two shorter lengths must add up to more than the greatest length.

Longest side 8

The other sides can then be 8 and 8,8 and 7,...,8 and 1 OR 7 and 7, 7 and 6,...,7+2 OR etc.

The total number is $8+6+4+2=20$.

Longest side 7

The total number is $7+5+3+1=16$

Longest side 6,5, ...

The totals for these are $12,9,6,4,2,1$.

Grand total

We must now add these  numbers, obtaining $70$.

A generalisation

Note the numbers we added above are the squares plus twice the triangular numbers:-
$$1+4+9+16+2(1+3+6+10).$$

We can generalises this result to other bounds on the side lengths.


问:
Does a finitely complete poset with the descending chain condition have minimum?
The question is really in the title. I would like to prove that there exists the minimum for of a partially ordered set $(P,\leq)$ having the property that any totally ordered subset (chain) has minimum, and that for any two elements in $P$ (and hence for any finite subset) there exists in $P$ a lower bound for these elements.

In every counterexample that I've tried to define always comes up in the end a chain not having minumum, so I persuaded myself that this should be true, but I don't know how to prove it, since not every poset is finite union of chains... Does anyone have suggestions about that?

Thank you so much!

答:
Suppose $P$ not having a minimum. Let's define a chain not having minimum. By induction, set $x_0$ any element in $P$, since $x_0$ is no minimum, there exists $y$ such that $y\not\geq x_0$, that means either $y<x_0$ or $y$ and $x_0$ are not comparable.

In the former case set $x_1=y$, in the latter find an element $z\leq x_0,y$ (which won't be neither of the two, o/w they would be comparable) and set $x_1=z$.


问:
Hydraulic radius of a complex shape
I'm working on a project involving thermoacoustics, and one of the important parameters is known as the hydraulic radius. If you have a pipe with some odd geometry, the hydraulic radius is its cross-sectional area of flow (i.e.: of the area not occupied by the shape) divided by its wetted perimeter.

The project I'm working on uses a "stack" for a certain component of the piping system. This comprises a series of plates with some thickness t and spacing x that fit within a tube. One of the plates is bisected widthwise by a diameter of the tube. If the thickness, number of plates, and tube diameter are known, what is the hydraulic radius of this shape?

The white area inside the red circle is the cross-sectional flow area
The perimeter of that white area is the wetted perimeter
The ratio of these two values is the value in question

答:
One reasonable approximation would be assuming that every white rectangle has some width $w_i$ and height $x$. So your hydraulic radius, according to your description, should be computed as

$$\frac{\text{area}}{\text{perimeter}} =
\frac{\sum_i w_i x}{\sum_i 2w_i + 2x} \approx
\frac{x\sum_i w_i}{2\sum_i w_i} =
\frac x2$$

This approximation works if

The number of plates is sufficiently high, so that the rectangle approximation is sound

The special cases at the ends of the sequence, where shapes differ even further from the rectangular case, can be ignored as well

You have $x\ll w_i$ so that you can ignore the short sides of the rectangles

If this is not enough, then I fear you'll have to do actual computations for all the dimensions involved. I doubt you'll end up with a simple decription without such approximations.

Title: Hydraulic radius of a complex shape


问:
show $f$ complex analytic is constant
$U\subseteq \mathbb{C}$ open and connected, $f : U \rightarrow \mathbb{C}$ complex analytic function.

Suppose $f(z) = u(x, y) + iv(x, y)$, $z = x + iy$, and $u$ and $v$ are
real-valued functions.

Suppose that $v$ only takes integer values on $U$.

Using the Cauchy–Riemann equations and the fundamental theorem of the calculus, show $f$ is constant.

I'm not really sure what is meant by "$v$ only takes integer values on $U$". Does it mean that $v(x,y)=c$ for all $x,y \in U$ or does it mean that $v(x,y)=A(x)B(y)$ such that it equals an integer for all $x,y$?

I'm also just unsure in general how to use Cauchy Riemann and FTC to show f is constant.

答:
You also need the assumption that $U$ is connected. First, note that $v$ must be continuous. Since it has integer values, it is constant. Then, using Cauchy-Riemann equations,

$$ \partial_x u = \partial_y v = 0$$

so

$$f'(x+iy) = \partial_x u (x,y) + \partial_x v (x,y) = 0$$

everywhere on $U$.

Note that connected implies path-connected. Fix $z_0 \in U$. For $z \in U$, choose a path $\sigma: [0,1] \to U$ from $z_0$ to $z$. Then,

$$f(z) - f(z_0) = \int_\sigma f'(z) dz = 0 $$

Title: show $f$ complex analytic is constant


答:
I'm going to assume $U$ is connected for this exercise.

Suppose that $f = u(x,y) + iv(x,y)$ only takes integer values, that means for every element $e$ of the set $U$ $f(e)$ is an integer.

We have that $f$ is complex analytic i.e. at all point $u_i =(x,y)$ in the set $U$ $f$ satisfies the cauchy riemann equations given as:

$$ \frac{\partial u}{\partial x} =  \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$$

So now here is our proof by contradiction. Suppose $f$ is NOT constant. That is we can find two points $u_1, u_2 \in U$ such that $f(u_1)\ne f(u_2)$ now because $U$ is connected there exists a path $P$ from $u_1$ to $u_2$ (obviously $u_1$ and $u_2$ are the two endpoints of this path). Let $u_{1,k}$ be a sequence of points in $P$ that starts with $u_{1,0} = u_1$ and has a limit point $L_1$ (still on the path) with the property that for any $\epsilon >0$ $f(L_1+\epsilon) = f(u_2)$ and $f(u_{1,k})= f(u_1)$. Similarly we can define $u_{2,k}$ starting from the other endpoint and going in the other direction.

It's clear that that at the point $L_1$ our function ceases to be continuous let alone complex differentiable. Or you could say, the Cauchy Riemann Equations are not well defined. Since it is an abrupt jump.

This contradicts one of our assumptions, and so we are forced to conclude that $f$ is not constant.

Title: show $f$ complex analytic is constant

Body:
$U\subseteq \mathbb{C}$ open and connected, $f : U \rightarrow \mathbb{C}$ complex analytic function.

Suppose $f(z) = u(x, y) + iv(x, y)$, $z = x + iy$, and $u$ and $v$ are
real-valued functions.

Suppose that $v$ only takes integer values on $U$.

Using the Cauchy–Riemann equations and the fundamental theorem of the calculus, show $f$ is constant.

I'm not really sure what is meant by "$v$ only takes integer values on $U$". Does it mean that $v(x,y)=c$ for all $x,y \in U$ or does it mean that $v(x,y)=A(x)B(y)$ such that it equals an integer for all $x,y$?

I'm also just unsure in general how to use Cauchy Riemann and FTC to show f is constant.


答:
When the numbers are small, there is nothing wrong with forgoing all elegance, taking off your shoes, and counting on your fingers and toes.

Assume that Person-1 is originally seated in Seat-1, Person-2 in Seat-2, and Person-3 in Seat-3.

There are two basic cases:

Case 1: Person-1 shifts to Seat-2 (which was already occupied)

Case 2: Person-1 shifts to Seat-4 (which was not already occupied).

Case 1: Person-1 shifts to Seat-2

You will end up multiplying the enumeration by $2$, to include the case of Person-1 shifting to Seat-3, which by symmetry, should have the same analysis.

Assuming that Person-1 shifts to Seat-2, if Person-2 shifts to Seat-3, then Seats 2 and 3 are now taken, so there are 3 seats that Person-3 can take. If Person-2 shifts to any of the other remaining seats (i.e. Seat-1, Seat-4, Seat-5), then there will only be two Seats that Person-3 can shift to.

Therefore, in Case-1, if Person-1 shifts to Seat-2, the total number of new seating arrangements is

$$(1 \times 3) + (3 \times 2) = 9.$$

As indicated at the start of Case-1 analysis, you need to multiply this by $2$, since by symmetry, you get the same analysis if Person-1 shifts to Seat-3.

Therefore, the total for Case-1 is

$$2 \times [(1 \times 3) + (3 \times 2)] = 18.$$

Case 2: Person-1 shifts to Seat-4

You will end up multiplying the enumeration by $2$, to include the case of Person-1 shifting to Seat-5, which by symmetry, should have the same analysis.

If Person-1 shifts to Seat-4, and Person-2 shifts to Seat-3, then there are 3 seats that Person-3 can shift to.  If Person-1 shifts to Seat-4 and Person-2 shifts to either Seat-1 or Seat-5, then there are only two seats that Person-3 can shift to.

Therefore, in Case-2, if Person-1 shifts to Seat-4, the number of possible new seating arrangements are

$$(1 \times 3) + (2 \times 2) = 7.$$

As in Case 1, this needs to be multiplied by $2$, since Person-1 could (similarly) shift to Seat-5.

Therefore, in Case-2, the enumeration is

$$2 \times [(1 \times 3) + (2 \times 2)] = 14.$$

Final Answer:

$$14 + 18 = 32.$$

Title: Three people are seated in 3 our of 5 chairs . After a break , in how many ways can they be seated so that no person occupies the same same chair

Body:
I tried doing it by first counting the total arrangements i.e. $^5p_3$ = 60 then subtracting the non desired arrangements fixing the first person at its place and arranging the other 2 person .This can done for the rest of the 2 people therefore $3*(^4P_2)$
So ans should be $^5p_3$ - $3*(^4P_2)$ = 24
I know i have subtracted some same cases twice but i am not sure how to identify those cases.
Is there any other approach also ??


问:
Converting a daily standard deviation to an annual one
This post discusses how to convert a daily standard deviation into an annualized one.

Standard Deviation Annualized

However, it doesn't include a proof.

I've tried to prove it myself by using a summation of Gaussians and trying to calculate the standard deviation of the set, but I'm getting stuck, and I'm sure there's an easier way.

The question- why is:

$$\sigma_{\rm annual} = \sigma_{\rm measured} \sqrt{T}$$

答:
You don't need to assume each sample point is drawn from a Gaussian distribution, or even that they have the same distribution.  It is enough for them to be pairwise uncorrelated and have the same standard deviation.  If $\  D_1, D_2,\dots,D_T\ $ are pairwise uncorrelated with means $\ d_1,d_2,\dots, d_T\ $ and standard deviations $\ \sigma_1,\sigma_2,\dots,\sigma_T\ $, then the variance of $\ \sum_\limits{i=1}^TD_i\ $ is
\begin{align}
\sigma_\text{annual}^2&=\mathbb{E}\left(\left(\sum_{i=1}^T\big(D_i-d_i\big)\right)^2\right)\\
&=\mathbb{E}\left(\sum_{i=2}^T \sum_{j=1}^{i-1} \big(D_i-d_i\big)\big(D_j-d_j\big)\right)\\
&\hspace{2em}+\mathbb{E}\left(\sum_{i=1}^T \big(D_i-d_i\big)^2\right)\\
&\hspace{2em}+\mathbb{E}\left(\sum_{i=1}^{T-1} \sum_{j=i+1}^T \big(D_i-d_i\big)\big(D_j-d_j\big)\right)\\
&=\sum_{i=1}^T \mathbb{E}\left(\big(D_i-d_i\big)^2\right)\\
&=\sum_{i=1}^T \sigma_i^2\ ,
\end{align}
the first and third terms on the right of the second equation's vanishing because $\  D_1, D_2,\dots,D_T\ $ are pairwise uncorrelated. Thus, if $\ \sigma_i=\sigma_0\ $ for all $\ i\ $, then
\begin{align}
\sigma_\text{annual}&=\sqrt{T \sigma_0^2}\\
&=\sigma_0\sqrt{T}
\end{align}


问:
How many permutations of numbers $(1,2,3, .., n)$ are there in which at least $2$ elements are in their original place?
How many permutations of numbers $$(1,2,3, .., n)$$ are there in which at least $2$ elements are in their original place?

My idea for solving:
$n! - |\text{no element stands in its original place}| - |\text{exactly one element is in its original place}|$

Is it correct?

答:
We split the problem into subproblems.

How many exist with exactly one element in the same place?

Select one element to remain in the same place.
For each selection, we find the derangement of the remaining (n-1) elements. . This tells us that there are $\frac{(n-1)!}{e}$ (rounded to the nearest integer so written $\left\lfloor \frac{(n-1)!}{e} +0.5 \right  \rfloor$ ) ways to permutate n-1 elements so none are in the same place. This gives us $n \times \left\lfloor \frac{(n-1)!}{e} +0.5 \right  \rfloor$ as we multiply by n for all the choices of fixed elements.

How many exist elements with none in the same place?

Similarly we get $\left\lfloor \frac{(n)!}{e} +0.5\right\rfloor$.

So the final answer is $n! - \left\lfloor \frac{(n)!}{e} +0.5\right\rfloor - n \times \left\lfloor \frac{(n-1)!}{e} +0.5\right\rfloor $


问:
Vickrey vs English Auction Equivalence
Wikipedia states:

When the auction involves a single item for sale and each participant
has as an independent private value for the item auctioned, the
expected payment and expected revenues of an English auction is
theoretically equivalent to that of the Vickrey auction, and both
mechanisms have weakly dominant strategies.[1] Both the Vickrey and
English auction, although very different procedurally, award the item
to the bidder with the highest value at a price equal to the value of
the second highest bidder.[2]

In an English auction, the object goes to the highest bidder who pays at the price they proposed.

In a Vickrey auction, the object goes to the highest bidder, who pays at the second-highest price.

Given this, how are they "equivalent" to each other? The only thing common between the two, per my understanding, is that the highest bidder always wins.

My guess is that, in an English auction, when the second-highest bidder eventually places a bid (say b0) corresponding to their max affordable value, the highest bidder can place another bid ever so slightly higher (say b1). The object is then sold at b1.

In a Vickrey auction, the highest bidder would pay b0.

Since b0 is almost equal to b1, I imagine that's where the equivalence comes from.

答: Because in the Vickrey auction, each bidder is bidding their value (since it is a closed auction), whereas in the English open auction, the winning bidder has bid a price just at-or-above the second-highest value (instead of bidding their value which is the highest).</p>

Title: Vickrey vs English Auction Equivalence


问:
Differential Equation Find general solution of y'' - y =cosh(x) using variation of parameters
Hello I am having some issues with the simplification of the DE, I am okay up on till $$y_p(x)=v_1(x)y_{p1}(x) + v_2(x)y_{p2}(x) $$

$$ \frac12 e^{-x}\left(\frac {-e^{2x}}4-\frac x2\right)+ \frac 12e^{x}\left(\frac {-e^{-2x}}4+\frac x2\right)$$
I cannot simplify it to get $\frac x2 \sinh(x)$ I would appreciate some clarification if possible

答:
The simplification is tricky if you aren't looking for it. You get rid of the $\tfrac14\cosh(x)$ when you combine the complementary solution with the particular solution.

Simplification from where you are:

Group like terms:: $$\tfrac14x(e^x - e^{-x}) - \tfrac18(e^x + e^{-x})$$

This simplifies to:: $$\tfrac12x\sinh(x) - \tfrac18(e^x + e^{-x})$$
You don't want to simplify the last term into $-\tfrac14\cosh(x)$ because of the following:
$$
Y = Y_c + Y_p.
\\
  Y = (A)e^{-x} + (B)e^x - \tfrac18(e^x + e^{-x}) + \tfrac12x\sinh(x).
\\
(A - \tfrac18)e^{-x} = Ce^{-x}\\ (B - \tfrac18)e^x = De^{x}.
$$
The arbitrary constant from the complimentary solution combined with the $-\tfrac18$ from the particular solution can be simplified as another arbitrary constant.


问:
Prove that the union of three subspaces of $V$ is a subspace iff one of the subspaces contains the other two.
Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two.

I can do this problem when I am working in only two subspaces of $V$ but I don't know how to do it with three.

What I tried is:
If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??).

If the union of three subspaces is a subspace.....
How do I prove that one of the subspaces must contain the other two from here?

*When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?

答:
With regard to Gina and Jeff's answers, I believe the simplification at the start can be made easier (at least in my opinion).

Suppose that subspaces $U_1,U_2,U_3$ union to form a subspace.
We have 2 cases.

Case 1: Suppose $U_i\subseteq \cup_{j\neq i}U_j$ for some $i\in \{1,2,3\}$.
WLOG, let $i=1$. Then the problem is reduced to the case of 2 subspaces $U_j, j=2,3$.
WLOG again, we have $U_2\subseteq U_3$, then $U_1\subseteq U_2 \cup U_3 = U_3$.
Hence $U_3$ contains the other 2 sets.

Case 2: $\forall i\in \{1,2,3\}, U_i\not\subseteq \cup_{j\neq i}U_j$.
Then for each $i$, there exists $\mathbf{u}_i\in U_i$ such that $u_i\notin \cup_{j\neq i}U_j$, i.e.  $u_i\notin U_j$ for $j\neq i$.

After this, we can make a similar argument to that of Gina's. For simplicity, I assume the field in concern to be the real/complex field like in Axler's book.

$\mathbf{u}_1 + \mathbf{u}_2, 2\mathbf{u}_1 + \mathbf{u}_2, \mathbf{u}_1 + 2\mathbf{u}_2$ must all be in $U_3\setminus (U_1 \cup U_2)$. (Note that all these vectors are in $\cup_{1\leq i\leq 3} U_i$.) In particular, these vectors lie in $U_3$. Hence, $\mathbf{u}_1, \mathbf{u}_2 \in U_3$, a contradiction.

Hence only case 1 is possible and we are done.


问:
Maple: How to convert Cylindrical coordinates to Cartesian coordinates?
We get some expression in Cylindrical coordinates (r, ϕ, z ) like : expr := r*z^2*sin((1/3)*ϕ) we need to convert it into Cartesian coordinates and than back to Cylindrical coordinates. How to do such thing?

So I found something like this : eval(expr, {r = sqrt(x^2+y^2), z = z,ϕ= arctan(y, x)}) but it seems incorrect, how to correct it and how make eval to convert backwords from Cartesian  to Cylindrical?

ϕ == ϕ

So I try:

```<code>R := 1;

H := h;

sigma[0] := sig0;

sigma := sigma[0]*z^2*sin((1/3)*`ϕ`);

toCar := eval(sigma, {r = sqrt(x^2+y^2), z = z, `ϕ` = arctan(y, x)});

toCyl := collect(eval(toCar, {x = r*cos(`ϕ`), y = r*sin(`ϕ`), z = z}), `ϕ`)
</code>```

It looks close to true but look:

why arctan(r*sin(ϕ), r*cos(ϕ)) is not shown as ϕ?

答: I don't speak Maple, but it looks like your eval takes you from Cartesian to cylindrical coordinates.  The inverse is $x=r \cos \phi , y=r \sin \phi, z=z$.  The Wikipedia link you have gives this, though using $\rho$ instead of $r$</p>


问:
Radical of $\mathfrak{gl}_n$
I find it intuitive enough that the radical of $\mathfrak{gl}_n\mathbb F$ is the scalar matrices, but I have trouble finding an easy, but complete proof:

Proof. Let $\mathfrak s$ denote the scalar matrices. Clearly $\mathfrak s\subset\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$. Suppose that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is generated by more than one element, so that $X\in\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$, but $X\notin\mathfrak s$. We can change basis such that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is upper-triangular. (The change of basis leaves the scalar matrices invariant.) Then $X$ is upper-triangular.

I want to conclude that there exists a $Y\in\mathfrak{gl}_n\mathbb F$ s.t. $[X,Y]$ is not upper-triangular. How can I see that $Y$ always exists, except by waving hands?

答:
If $I$ is a solvable ideal of $\mathfrak{gl}_n(\mathbb{F})$ then $I\cap \mathfrak{sl}_n(\mathbb{F})$ is a solvable ideal of $\mathfrak{sl}_n(\mathbb{F})$ which is semisimple (in fact simple) as long as the characteristic of $\mathbb{F}$ does not divide $n$, so the radical of $\mathfrak{gl}_n(\mathbb{F})$ intersects trivially with $\mathfrak{sl}_n(\mathbb{F})$, and this forces it to have dimension $1$.

On the other hand, if we take the example $\mathfrak{gl}_2(\mathbb{F}_2)$ then one can check that this is actually solvable, so in this case the radical does not just consist of the scalar matrices (the derived subalgebra is contained in $\mathfrak{sl}_2(\mathbb{F}_2)$ which contains the scalar matrices. The quotient of $\mathfrak{sl}_2(\mathbb{F}_2)$ by the scalar matrices has dimension $2$ and is thus solvable).

Title: Radical of $\mathfrak{gl}_n$


问:
Quick Volumes with Lines Question
So I was given the following prompt with the graph:

"Let $R$ be the region bounded above by the graph of $y=1-x^2$ and below by the graph of $y=x^2-1$, for $-1\le x\le1$, as shaded in the figure above. What is the volume of the solid generated when the region $R$ is revolved about the horizontal line $y=3$?"

I understand that the bounds of the integral that would solve for the volume would be from $-1$ to $1$, but I'm a bit confused about what the equation for the volume might look like here. I was thinking it might look something like the following: $\int((1-x^2+3)^2-(x^2-1+3)^2)$, but I'm not too sure. Any help would be appreciated!

答:
Bad description has caused ambiguity.  The shaded area represents the area below $(1-x^2)$ and above $(x^2 - 1)$.  I'm going to assume that it is intended that the volume of solid rotated around the line $y = 3$ will have the shaded area as its cross-sectional area.

In general, the volume will be

$$\pi \times \int_{-1}^1 \{[R(x)]^2 - [r(x)]^2\}dx$$

where $R(x)$ represents the outer radius that extends from $y = 3$ down to $y = (x^2 - 1)$ and $r(x)$ represents the inner radius that extends from $y = 3$ down to $(1 - x^2).$

Title: Quick Volumes with Lines Question