问:
A mapping that converts a line segment to an arc
I'm looking for a mapping $f:\mathbb{R}^2\to\mathbb{R}^2$ that converts a line segment $AA'$ with two end points $A=(x_A,y_A)$ and $A=(x_{A'},y_{A'})$ to an arc $BB'$ with radius $r$ centered at $O$, with end points $B=(x_B,y_B)$ and $B'=(x_{B'},y_{B'})$.

This mapping should map $A$ to $B$ and $A'$ to $B'$. Does anyone know a closed-form relation for such mapping?

I know that mappings such as $w=1/z$ in complex plain can convert lines to circles. But I don't know how to modify this to convert a particular line segment to an arc given the endpoints and radius.

答:
Treat everything as complex numbers, let $P$ be a point on the line segment $AA'$. The map $$P \mapsto \frac{1}{A'-A}\left( P  - \frac{A+A'}{2}\right)$$
sends the line segment $AA'$ to the line segment joining $-\frac12$ to $\frac12$. Multiply $B'-B$ will rotate and scale the line segment to a parallel copy of $BB'$. After another translation, the map

$$P \mapsto P' \stackrel{def}{=} \frac{B+B'}{2} + \frac{B'-B}{A'-A}\left( P - \frac{A+A'}{2}\right)$$
sends line segment $AA'$ to line segment $BB'$.

To send line segment $BB'$ to an arc joining them, you first figure out the mid point $M$ of the circular arc $BB'$ and let $N = 2O - M$ be the antipodal point of $M$ on the circle holding the arc.
$$N = O - \frac{2r}{\sqrt{4r^2 - |B-B'|^2}} \left(\frac{B+B'}{2} - O\right)$$

If you perform a circle inversion with respect to $N$ and radius $R = |B - N| = |B'-N|$, the point $P'$ will get mapped to a point $P''$ on the circular arc joining $BB'$.

$$P' \mapsto P'' \stackrel{def}{=} N + \frac{R^2}{\bar{P}' - \bar{N}}$$
Please note that in denominator of above expression, you need to take complex conjugation for $P'$ and $N$.


问:
Weak law of large numbers - problems
I am currently studying the weak law of large numbers and I have understood the concept behind it. If we take a sample that is enoughly big, the mean of this sample will converge to the sequence of random variables X1,...,Xn that would be very convenient because most of the time. The sequence of random variables called Population is too big to be observed.

My problem is : I cannot find a proper example, problem that uses this theorem. All I found is the proof of it which is not required at this time of my course because my professor has not introduced the Chebyshev inequality...

Could you post here small problemss that I could work on to grasp and practice the mechanic behind the concept ?

Thank you very much

答:
Really great you are seeking to play around with the concepts! I have a couple of examples to show you here!  You will need Chebsky's inequality: If $E(X) = \mu$ and Var($X) = \sigma^2$ then $P(|X-\mu|>\epsilon) < \frac{\sigma^2}{\epsilon^2}$

$\textbf{My fisherman father:} $

My father loves to fish (he really does by the way!) each day he goes to Monk lake and if its sunny he fishes for 9 hours. If it is raining  he fishes instead for 6 hours. The probability he catches one fish in any given hour, independent of others, is $\frac{1}{3} $ and the probability he doesn't is $\frac{2}{3} $. Lets say it rains half of the time and is sunny the other half. Lets let the variables $X_W, X_M, X_Y, X_D $ denote respectively the average number of fish he catches per day in a week, month, year and decade of fishing. (i.e $X_Y$ = $\frac{1}{365} \cdot$ #Fish he caught that year. )
Can you tell me:

The Expected value of each of the $X_i$ call this $\mu_i$ for $i = W,M,Y,D $

The Variance for each $X_i$ call this $\sigma^2_i$

An upper bound  (using Markov's/Chebyshev's inequality) of $P( |X_i - \mu_i|>0.01)$ i.e the probability that $X_i$ differs from $\mu_i$ by at least $0.01$

$\textbf{Terrible tennis players:} $

I am bad at tennis, very bad. Let us say each time I play my girlfriend I only have a 20% chance of winning each match (believe me this is a generous estimation.) She agrees to the following bet: Each time we play a match, if I win I get £10 and each time and if I loose I have to pay her £3. Let us say we have played $1000$ games.

What was the expected average amount of money it cost me to play each game. i.e what was the expected average I lost per game?

What is the variance of this random number?

Again, using Markov's/Chebyshev's inequality, what is and upper bound for the probability that I either made money, or lost more than £400?

Now do these questions if we played 10000, 100000 and 1 million games.

$\textbf{Farmer Groggstern and his wild cabbage} $

Farmer Groggstern grows cabbage, a lot of it. The weight of each cabbage he grows is perfectly modelled  by a normal distribution  of Mean $2$kg and Variance $0.5^2$. He decides to harvest 500 cabbages to make soup for his entire village, coincidently of population 500 people. He turns all the cabbage into soup with entirely efficient mass transfer and adds no extra ingredients, yuck! (i.e 1kg of cabbage makes 1kg of soup) he then divides all the soup he has into 500 equal portions.

What is the expected amount of soup each person gets?

What is the variance of this random number?

What is an upper bound for the probability everyone gets 1.9 - 2.1 Kg of soup

What is an upper bound for the probability everyone gets 1.8 - 2.2 Kg of soup

What is an range for which I am 99.99% confident that everyones soup mass is within
\

Now do these questions if he harvests 1000, 2000 and 5000 cabbages for respectively 1000, 2000 and 5000 people

I hope these helped you ! If you want answers to any of these questions please comment below!  Oskar :)

ContentLicense: CC BY-SA 4.0

Title: Weak law of large numbers - problems


问:
Surprisingly Countable or Uncountable Sets
By Cantor, we know that there are different cardinalities of infinite sets. $\mathbb{N}$ is countable, $\mathbb{Z}$ is countable and - rather surprisingly - $\mathbb{Q}$ is countable, too. And even more surprisingly, $\mathbb{R}$ is uncountable.

I sometimes show Cantor's arguments in high school (where I teach) and they and their consequences are - indeed and expectedly - very surprising to learn about.

What are some other infinite sets which are surprisingly countable or uncountable?

答:
Cantor set, the set of points in $[0,1]$ whose decimal expansion has no prefixed number (for example, those nombers such that theres is no 5 in the decimal expansion).

All of them are, surprisingly (at least the first time one see them) uncountable. And the Cantor Diagonal Argument is key in proving this.


问:
Integral of test function is a test function? (Application in tensor theory of distributions)
Let $f:\mathbb{R}^m\to\mathbb{C}$ and $g:\mathbb{R}^n\to\mathbb{C}$, then we can define the tensor product $f\otimes g$ of $f$ and $g$ as the function $\mathbb{R}^{m+n}\to\mathbb{C}$ for which $f\otimes g(x,y)=f(x)g(y)$.

If $f$ and $g$ are locally integrable on resp. $\mathbb{R}^m$ and $\mathbb{R}^n$, then so is $f\otimes g$ on $\mathbb{R}^{m+n}$. We can therefore consider the regular distribution $T_f$,$T_g$ and $T_{f\otimes g}$ associated with each of these functions. The latter one is defined as $$ \langle T_{f\otimes g},\phi(x,y)\rangle = \int_{\mathbb{R}^{m+n}} f\otimes g(x,y)\phi(x,y)dxdy = \int_{\mathbb{R}^m} f(x) \left(\int_{\mathbb{R}^n} g(y)\phi(x,y)dy\right) dx, \text{ for } \phi\in \mathcal{D}(\mathbb{R}^{m+n}).$$ For this last integral to be well-defined, we would need that $\psi(x) := \int_{\mathbb{R}^n} g(y)\phi(x,y)dy$ is a test function over $\mathbb{R}^m$, i.e. $\psi \in \mathcal{D}(\mathbb{R}^m)$. I don't see how I could show this rigorously. Does continuity of $\psi$ follow directly? Can I be sure that the integral will be continuous? As for the support, is taking the intersection of $[\phi]$ with a compact $K_1\subset \mathbb{R}^m$ for which $[\phi]\subset K_1\times K_2$ (and $K_2\subset \mathbb{R}^n$) enough?

Thanks in advance.

答:
David already gave a nice self-contained answer, but I would like to add that this is just a special case of Fubini's Theorem for Schwartz distributions. See

https://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/351028#351028

There I discussed the temperate case but the same holds with the change of spaces
$$
\mathscr{S}\rightarrow \mathscr{D}=C_{\rm c}^{\infty},\
\mathscr{S}'\rightarrow \mathscr{D}',\
\mathscr{O}_{\rm M}\rightarrow \mathscr{E}=C^{\infty}.
$$


问:
Why are circles of opposite directions not homotopic to each other?
I’m learning about homotopy and the fundamental groups, and I’m having a hard time visualizing the transformation. In particular, I can’t see when there is no homotopy between loops.

An example from my class is the following picture (the loops are all confined to $S^1$. I draw them that way just to make clear what paths are being travelled).

I can see how the first loop could be deformed into the second one by doing less and less of the “repeated” portion.

Why can’t we have something similar, for example as shown below, to deform a loop into another loop of the opposite direction? Is the condition of continuity violated somehow?

Apologize for the bad quality of the pics.

=======

Edit: I’m actually thinking of the fundamental group of the circle $S^1$. My understanding is that the group is comprised of the equivalence classes of loops with a base point $x_0$ (here I take it to be $(1,0)$).
In what sense is a loop goes a full circle from $(1,0)$ counterclockwise not the same as a loop goes a full circle from $(1,0)$ clockwise? I don’t see how the procedure in the second picture does not describe a homotopy between the two.

答:
Two curves being homotopoc is context-dependent. If you are told that two curve homotopic, you should ask yourself “homotopic where”?

Your second diagram explains why the first and the second circles are homotopic on $\Bbb C$. Yes, they are! But what's important here is that they are not homotopic on $\Bbb C\setminus\{0\}$. Note that when you started to change your circle, you wer unable to avoid passing through $0$.


问:
$\mathbb R^{\omega}$ in the box topology not paracompact?
I know that we can't prove that $\mathbb R^{\omega}$ in the box topology is normal, so we can't say for sure that it is paracompact(it is Hausdorff, and every paracompact Hausdorff space is normal). Then can we prove that it is 'not' paracompact?

答: In The box product of countably many compact metric spaces, Gen. Top. Appl. $2$ ($1971$), pp. $293$-$298$, Mary Ellen Rudin showed that the continuum hypothesis implies that the box product of countably many $\sigma$-compact, locally compact metric spaces is paracompact. In particular, then it implies that the box product of countably many copies of $\Bbb R$ is paracompact, so we cannot prove in $\mathsf{ZFC}$ that this box product is not paracompact.


问:
A finite dimensional extension of a field is direct product of local rings, the Fulton way.
Theorem. If $k$ is a field, $R$ a ring containing $k$ as a subring and $R$ finite dimensional over $k$, then $R$ is isomorphic to a direct product of local rings.

I know the usual proof using the Chinese remainder theorem, but I want to understand the proof by Fulton in his 'Algebraic curves', only valid for algebraically closed $k$. Fulton shows in an entertaining way that, if $k$ is algebraically closed and if $I$ is an ideal in $k[X_1, \ldots, X_n]$ with finite vanishing set $V(I)=\left \{  P \in k^n \mid F(P)=0,
 \forall F\in I \right \}$ then $k[X_1, \ldots, X_n]/I$ is isomorphic to a direct product of local rings. Then he claims that the above Theorem can be deduced. I know how to get $R$ as a quotient $k[X_1, \ldots, X_n]/I$, using the fact that $R$ is finite dimensional over $k$, but I don't know how to assure that $I$ has a finite vanishing set. Any ideas?

答:
I worked out the excellent but succinct comment of reuns.

By hypothesis, $k$ is an algebraically closed field, $R$ is a ring with
$k \subset R$, and $R$ is a vector space over $k$ of finite dimension $n$,
with basis $\{ r_1, \ldots, r_n \}$. Since $R$ is a vector space,
$r_i \cdot r_j = \sum_{ k = 1 }^{ n } a_{i,j,k} \cdot r_k$ with unique
coefficients $a_{i,j,k} \in K$. That is,
$R \simeq k[X_1, \ldots, X_n] \mathbin{/} I$, where $I$ is generated by the
corresponding relations
$X_i \cdot X_j - \sum_{ k = 1 }^{ n } a_{i,j,k} \cdot X_k$.

Also, since $R$ is a vector space over $k$ of dimension $n$, for each $i$, the
elements of $\{ 1, r_i, \ldots, r_i^n \}$ are linearly dependent, such that a
unique monic polynomial $f_i(X_i)$ of minimal degree (at most $n$) exists with
$f_i(r_i) = 0$. These minimal polynomials can be naturally embedded in
$k[X_1, \ldots, X_n]$, and so $f_i \in I$ (because $I$ comprises, by
construction, all relations pertaining to the basis elements $r_i$).

For each $P = (u_1, \ldots, u_n) \in V(I)$, $φ(X_i) = u_i$ defines a unique
ring homomorphism $\phi$ from $k[X_1, \ldots, X_n]$ to $k$. Also, for each
$g \in I$: $\phi(g(X_1, \ldots, X_n)) = g(\phi(X_1), \ldots, \phi(X_n)) =
g(u_1, \ldots, u_n) = 0$ (since $P \in V(I)$ and $g \in I$). Thus,
$I \subset \ker(\phi)$, such that $\phi$ induces an homomorphism from
$k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$.

For each homomorphism $\phi$ from $k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$,
for each $g \in I$:
$g(\phi(X_1), \ldots, \phi(X_n)) = \phi(g(X_1, \ldots, X_n)) = \phi(0) = 0$,
that is, $(\phi(X_1), \ldots, \phi(X_n)) \in V(I)$. In particular, since
$f_i \in I$, $f_i(\phi(X_i)) = 0$, and thus $\phi(X_i)$ can assume only
finitely many values (because each such value must be a root of $f_i$), such
that only finitely many such homomorphisms exist.

Hence, there is a one-to-one correspondence between $V(I)$ and the
homomorphisms from $k[X_1, \ldots, X_n] \mathbin{/} I$ to $k$, and since there
are finitely many such homomorphisms, $V(I)$ is a finite set. Furthermore,
$V(I)$ is not empty, because otherwise $I = k[X_1, \ldots, X_n]$ (by the
Nullstellensatz, since $k$ is algebraically closed) and thus $R$ is the
zero-ring, which is excluded by the hypothesis $k \subset R$. Therefore, the
theorem from Fulton's book mentioned in the question applies, such that
$k[X_1, \ldots, X_n] \mathbin{/} I$ (isomorphic to $R$) is direct product of
local rings.


问:
Solve a cubic polynomial (given one root is four times a second root)?
So, I've been stuck on a question for a long time now:

"Solve the equation $10x^3 + 23x^2 + 5x - 2 = 0$ given that one root is four times a second root."

How would you go about solving this?
Any help would be greatly appreciated.

答:
Another (more brutal) method:

Let $x$ a root of $P(x)=10x^3+23x^2+5x-2$ such that $4x$ is also a root of $P$. Then, substituting $x$ by $4x$ in $P(x)$ implies that $x$ is a root of $640x^3+368x^2+20x-2$. Substracting $P$, we obtain that $x$ is a root of $630x^3+345x^2+15x=15x(42x^2+23x+1)$. Since $x$ is not zero, $x$ is a root of $42x^2+23x+1$. The discriminant is $19^2$, so $x$ is $\frac{-1}{2}$ or $\frac{-1}{21}$ (and $4x$ is $-2$ or $\frac{-4}{21}$). We check easily that $-2$ is a root of $P$, so we can factor $P$ as $P(x)=10(x-\frac{1}{5})(x+\frac{1}{2})(x+2)$.

ContentLicense: CC BY-SA 3.0


问:
Increasing the tail of exponential function
Suppose I generate a number of random number according to the exponential distribution and convert them into probability values in R.

```<code>prob <-  data.frame(round(rexp(2000 ,0.5) , digits = 4))
colnames(prob) <- "V"
prob$V <- prob$V / max(prob$V)
hist(prob$V , prob= TRUE, breaks =30, main=" ", xlab= "Probability Vals", col="red")
lines(density(prob$V, adjust=2), lty="dotted", col="darkblue", lwd=2)
</code>```

When I plot the distribution of the probability values, here how it looks;

figure

I would like to obtain something like

second figure

I have found something called exponentially modified Gaussian distribution and select the parameters as $\mu= -3, \sigma =1, \lambda =0.25$ as shown at this wiki page. However, I am not getting the desired output. The R code is

```<code>prob <-  data.frame(round(rexp(2000 ,0.25)+ dnorm(2000, mean = -3, sd = 1, log = FALSE)   , digits = 4))
</code>```

I was wondering if someone help me out.

答:
I gather you want an exponential-like distribution with a constant tail for $x$ which is zero away from the interval $0<x<1$. Such a PDF can be written

$$ P(x) = A \ \text{max}\big[C,e^{-\lambda x}\big]\big[\theta(x)-\theta(x-1)\big]$$

where $C<1$ and $0< 1/\lambda < 1$. $C$ is your constant tail, and the turnover from exponential-like to constant happens at $x = -\lambda^{-1}\log C$ (a positive number between 0 and 1). The normalization constant $A$ is determined by

$$1 = \int_0^1 P(x) dx = A \Big[ \int_0^{-\lambda^{-1}\log C}  e^{-\lambda x} dx  + \int_{-\lambda^{-1}\log C}^{1} C dx\Big],$$
and this gives
$$ A = -\frac{\lambda}{C(1+\log C)}$$
so

$$ P(x) = -\frac{\lambda}{C(1+\log C)} \ \text{max}\big[C,e^{-\lambda x}\big]\big[\theta(x)-\theta(x-1)\big].$$

If you want to draw random variables from this distribution, probably the best way is to sample from an exponential distribution and throw away values that don't work (A Monte Carlo method).


问:
Equivalence of gluability axiom for sheaves
Suppose $F$ is a sheaf on a topological space X. Suppose we have an open set $U\subseteq X$. For simplicity, suppose we have an open cover $U=U_1\cup U_2$. The gluability axiom states that given $f_1\in F(U_1)$ and $f_2\in F(U_2)$ such that $f_1|_{U_1\cap U_2}=f_2|_{U_1\cap U_2}$, then there is some $f\in F(U)$ such that $f|_{U_1}=f_1$ and $f|_{U_2}=f_2$.

I am trying to understand why this statement of the gluability axiom is equivalent to the following sequence being an equalizer: $F(U)\rightarrow F(U_1)\times F(U_2){{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} F(U_1\cap U_2)\times F(U_1\cap U_2)$. So to show that this implies the above axiom, suppose we have $f_1\in F(U_1)$ and $f_2\in F(U_2)$. But don't we needs to know the first map is surjective? What am I missing?

答:
Why do you think the first map needs to be surjective? This is easily seen to be false by example: consider holomorphic functions on $\mathbb C$. The natural map $\mathcal O_{\mathbb C}(\mathbb C) \to \mathcal O_{\mathbb C}(\mathbb C \setminus \left\{0\right\})$ is not surjective since $f(z) = 1/z$ is not in its image, so pairing with the restriction map to any other open set $U \not \ni 0$ you get a map of this type which is not surjective.

The hypothesis (existence of two sections that agree on $U \cap V$) is that we have a pair $(f_u,f_v)$ in the middle spot which equalizes the double arrows (i.e. lies in the difference kernel of the two arrows, if I'm getting my terminology correct). Now the sheaf axiom is equivalent to exactness, i.e. the difference kernel of the double arrows is equal to the image of the first arrow. In other words, the conclusion (existence of a section that glues $f_u$ and $f_v$) is that we can lift the pair to a section on $X$, i.e. the pair is in the image of the first map.


问:
Generating functions for combinatorics
I have no formal education in generating functions, but, based on another question, I have seen that they can be powerful for combinatorics.

Are there a few general principles for using generating functions to construct counting arguments (e.g. counting the number of subsets or number of total orderings) that can be remixed to count different quantities?

If the answer is not amenable to this forum, where can I learn more?

I've been trying to read Concrete Mathematics (Graham, Knuth, Patashnik), but the topic spans several chapters and over 100 pages. At the moment, I just have patience to learn the gist of the combinatorics applications of generating functions. Maybe later I'll go back and get an extensive education from GKP.

答:
To add to the other answers. The utility of generating functions goes further than counting-combinatorics. They are a basic tools for dealing with discrete functions, in particular with linear difference equations - and these frequently appear, typically as recursions, when solving many counting problems, or when dealing with discrete probabilities, etc.

A note for those who come from engineering: generating functions are very connected to the Z transform (an elemental tool in discrete Signal Processing) -arguably they are exactly the same thing, just different terminology - but the math and the concept is the same. And, to add further connections (perhaps further confusion?),  the Z transform can be thought as the discrete version of the Laplace transform - and also as a generalization of the discrete Fourier transform.

Title: Generating functions for combinatorics


问:
Left shift and right shift calculation ..
i need to calculate left and right shift in java. in java i am able to code this. but i like to know how it calculate . Can any body here enplane me with Example.
 I am using below logic, plz check it's correct or not.

4>>2 ans is 1

Logic
4>>2
0100>>2
0001 = 1

答:
If I correctly understand what you're doing, $$n>>m=\frac{n}{2^m}$$  and $$n<<m=n\cdot 2^m.$$ This is similar to the way that moving a decimal point one place to the left (right) corresponds to division (multiplication) by $10$.

For example, then: $$16>>1=8\\16>>2=4\\16>>3=2,$$ and so on; or the other way, $$5<<1=10\\5<<2=20\\5<<3=40,$$ and so on.


问:
How can I know if a Theorem is provable inside a logic system? (Pure Positive Implicational Propositional Logic)
Im using the pure implicational logic with  detachment, substituition and the following two axioms:

$(A) \Rightarrow [(B) \Rightarrow (A)]$

$\{(A) \Rightarrow [(B) \Rightarrow (C)]\} \Rightarrow \{[(A) \Rightarrow (B)] \Rightarrow [(A) \Rightarrow (C)]\}$

Im going through the book Logic, Mathematics, and Computer Science from Yves Nievergelt.

After presenting a collection of proofs in and about pure implicational logic, the author give a list of formulae to the reader show if it is a Theorem in pure implicational logic, but he states that some formulae with implications cant be proved inside this system.

An example is the Peirce Law $\{[(P)\Rightarrow(Q)]\Rightarrow (P)\} \Rightarrow (P)$, I tried very hard to derive this from the Axioms and using all the presented theorems, later I just did a search and the book itself says its not possible inside the implicational calculus.

Then I got stucked in the next formulae which is $[(P) \Rightarrow (R)] \Rightarrow [\{[(P) \Rightarrow(Q)] \Rightarrow (R)\} \Rightarrow (R)]$, I have no idea how can I deduce if its provable with the given axioms or no, then its becoming very frustrating, as Im just in the first collection of exercises it is somekind boring to skip.

I want to know if there is a way to determine if a given well-formed formula is unprovable inside the implicational logic, and as a newcomer to logics I wonder if its a common pratice in other books to give hidden unprovable formulae as exercises.

The author commented a preliminary version of Deduction Theorem, but even with this some formulae still unprovable for example the Peirce Law again. What I have been doing is to derive the proved Theorems in the book(the most basic ones) for myself, then I try to apply substituitions in a way that similar sub-formulas to one im trying to prove appears, so I can try to reduce to the goal.

All tips on learning logic are welcome.

答:
Remember that in the end, formal logic is just that: a formal, i.e. purely syntactical system of symbols, where rules like the ones you state merely say: "If you have a statement that looks that [this string of symbols], then you write down [this other string of symbols]"

The point is: the system itself has no understanding what the symbols are supposed to mean ... and in fact we could interpret the symbols completely differently from what they were intended to capture.

Indeed, suppose that the expressions in out system are not assigned one of two truth-values, but instead one of three values: $1$, $2$, or $3$.  Like truth-functional logic, however, we will assume that the value of larger expressions is a function of the values of smaller expressions when combined using the $\to$ operator, and it all works in accordance to the table below (which is not no longer a truth-table ... but, for lack of a better word, a "$1,2,3$-table" ... anyway, it just specific how $\to$ works as a function):

\begin{array}{cc|c}
P&Q&P \to Q\\
\hline
1&1&1\\
1&2&2\\
1&3&3\\
2&1&1\\
2&2&1\\
2&3&3\\
3&1&1\\
3&2&1\\
3&3&1\\
\end{array}

Now:  For yourself, please work out the "$1,2,3$-tables" of $A \to (B \to A)$ and $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$  (yes, for that second one you'll get $27$ rows!)

You should find that in all rows, both statements evaluate to $1$, without exception.  We could therefore call them "$1$-tautologies"

Next, consider your only rule of inference, Modus Ponens. Observe from the table above that when $P$ is $1$, and $P \to Q$ is $1$, then the value of $Q$ has to be $1$ as well. This means that if you apply Modus Ponens to two $1$-tautologies, the result will be a $1$-tautology as well.

OK, but now finally observe that when $P$ has the value of $2$, and $Q$ the value of $3$, then $((P \to Q) \to P) \to P$ evaluates to $((2 \to 3) \to 2) \to 2 = (3 \to 2) \to 2 = 1 \to 2 = 2$

So, $((P \to Q) \to P) \to P$ is not a $1$-tautology!

What this all means is that in your system of logic, when what you are doing is interpreted from this persepctive, you can only write down $1$-tautologies using your axioms, and infer further $1$-tautologies from that, but you'll never be able to to derive $((P \to Q) \to P) \to P$, since that is not a $1$-tautology. So that's why it is not provable.

Now, how did I come up with this non-standard interpretation?  Well, I just use a little computer program that systematically searches for these.  You can try and find them by hand, of course, and sometimes you'll be able to find them, but there are also times where the only way to demonstrate that some specific sentence is not provable from a set of axioms and inference rules, is to use interpretations where the expressions can take on one of four possible values, or even more.

Now, what I actually don't know is if there is a decision procedure for such provability. That is, while my computer program is able to find these 'counter-models' for certain specific cases, I know for a fact that my program will sometimes stop without giving any answer. So: is there an algorithm that will always be able to tell whether some sentence is provable from some finite set of axioms and finite inference rules or not?  I expect there is not, for you can presumably reduce that to the Halting Problem... but I am not sure.


答: The problem is steps $1\to 2$ and $4\to 5$. Perhaps you could imagine this problem as having a rubber band wrapped once around a metal bar (attached to a wall, say, like a ladder rung). Is there any way to get the rubber band off the bar? Or to wrap it around the bar but in the opposite direction?

Title: Why are circles of opposite directions not homotopic to each other?

Body:
I’m learning about homotopy and the fundamental groups, and I’m having a hard time visualizing the transformation. In particular, I can’t see when there is no homotopy between loops.

An example from my class is the following picture (the loops are all confined to $S^1$. I draw them that way just to make clear what paths are being travelled).

I can see how the first loop could be deformed into the second one by doing less and less of the “repeated” portion.

Why can’t we have something similar, for example as shown below, to deform a loop into another loop of the opposite direction? Is the condition of continuity violated somehow?

Apologize for the bad quality of the pics.

=======

Edit: I’m actually thinking of the fundamental group of the circle $S^1$. My understanding is that the group is comprised of the equivalence classes of loops with a base point $x_0$ (here I take it to be $(1,0)$).
In what sense is a loop goes a full circle from $(1,0)$ counterclockwise not the same as a loop goes a full circle from $(1,0)$ clockwise? I don’t see how the procedure in the second picture does not describe a homotopy between the two.


问:
Connection between $p$-adic analytic groups and linear groups
I trying to prove that

a group $G$ is $p$-adic analytic if and only if is linear (as a closed subgroup of $\mathrm{GL}_n(\Bbb{Q}_p)$).

I'm not sure if the result is valid or if is necessary some additional hypothesis. If anyone knows some reference for a proof or a counterexample it would be a great help.

答:
The $p$-adic metaplectic group gives a counterexample; see here.
https://en.wikipedia.org/wiki/Metaplectic_group


问:
Scale free property of Pareto distribution
I am trying to show that the Pareto distribution is scale free, defined as:
p(bx) = g(b)p(x)
I get to this stage:
x dp/dx = p'(1)/p(1) p(x)
I have a rough solution which simply jumps to the next step:
ln(p(x)) = p(1)/p'(1) ln(x) + ln(p(1))
Can anyone explain?

答:
From the given equation, we have:

$$x \frac{\mathrm dp}{\mathrm dx} = \frac{p'(1)}{p(1)}p(x) \iff \frac{p'(x)}{p(x)} = \frac{p'(1)}{p(1)}\cdot \frac1x$$

and by the chain rule $\dfrac{\mathrm d(\log p(x))}{\mathrm dx} = \dfrac{p'(x)}{p(x)}$.

Integrating both sides with respect to $x$ gives the desired expression

$$\log p(x) = \dfrac{p'(1)}{p(1)}\log x+\log p(1)$$

because at $x = 1$ both expressions agree (thus verifying the choice of the constant of integration).


问:
Transportation-information inequalities (concentration inequalities)
The following is one example of a transportation-information inequality,  which show connections between optimal transport theory and information theory:

$$W_1(\nu,\mu) \leq \left[ 2\sigma^2 D_{KL}(\nu\Vert\mu) \right]^\frac{1}{2} $$

$W_1$ is the Wasserstein distance found in optimal transport theory, and $D_{KL}$ is the Kullback-Leibler (KL) divergence found in information theory. (Source)

What other transportation-information inequalities are out there?

答:
There are several other Transportation Inequalities. I recommend looking this paper: Transport Inequalities. A Survey.

The authors not only show different transport inequalities (e.g. Talagrand’s transport inequality for the Gaussian measure), but also prove them.


问:
Is this matrix algebra move correct?
I’ve started a linear algebra class and I have no idea if this equation is correct, suppose $A$ and $B$ are square matrices with $n$ rows, and suppose $x$ is a vector with $n$ elements. Given:

If it is true that $Ax=Bx$, is it always the case that
$A=B?$

Could you explain why this is True or False?

答:
If $ Ax= Bx $ for every vector $x$ then they are the same, since $A e_i = Be_i$

the i'th columns are equal for every i, where $e_i$ is the standard basis in $\mathbb{R}^n$

If only for a single value of $x$, $Ax=Bx$, then they are not necessarily equal. Take
$ A = \begin {bmatrix} 1 -1 \\ 1 -1 \end{bmatrix}$, $B = 2 * A$, and $x = [1, 1]^t$,
Then $Ax=Bx=[0,0]$.

Title: Is this matrix algebra move correct?


问:
Expected value of tossing a coin thrice
A coin with probability 0.6 of coming up Heads is tossed three times. Let X be the number of Heads observed. Find E(X^2) (expected value of X^2)

I understand what I need to do but I'm a bit thrown off by the E(X^2) Does this mean the following is correct:

```<code>             E(X^2) = (1*6/10)^2 + (2*9/25)^2 + (3*27/125)^2 = 1.30
</code>```

or is it:

```<code>             E(X^2) = ((1*6/10) + (2*9/25) + (3*27/125))^2 = 3.87
</code>```

Title: Expected value of tossing a coin thrice

Body:
(I'm not sure where your specific numbers are coming from, but I'm getting that $P(0)=0.064$, $P(1)=0.288$, $P(2)=0.432$, and $P(3)=0.216$.)

So then using
$$
E[X^{2}] = \sum_{x} x^{2} P(x),
$$
we can see that
$$
\begin{aligned}
E[X^{2}] &= (0^{2})(0.064) + (1^{2})(0.288) + (2^{2})(0.432) + (3^{2})(0.216) \\
&= 0.288 + 1.728 + 1.944 \\
&= 3.96
\end{aligned}
$$


问:
Multiplicity of integer partitions in iterative process
Let $(M_k)_{k\geq0}$ be a sequence of multisets. The multiset $M_0=\{[\:]\}$ has only one element, which is an empty sequence. For positive $k$, $M_k$ is a multiset of sequences of integers sorted in non-increasing order. For each sequence $s=\{s_1,s_2,\dots\,s_j\}\in M_{k-1}$, the multiset $M_k$ includes the sequences $\{s_1+1, s_2, \dots, s_j\}$, $\{s_1, s_2+1, \dots, s_j\}$, $\dots$, $\{s_1, s_2, \dots, s_j+1\}$, and $\{s_1, s_2, \dots, s_j, 1\}$ that one can obtain by adding 1 to one of the elements of $s$ or appending a new 1 at its end. If this process generates integer sequences that are not in non-increasing order, reorder the sequence.

Every element of $M_k$ is an integer partition of $k$. I am interested in the multiplicity $m$ of each integer partition in $M_k$, preferably without drawing out the entire tree for the generating process. Is there a way to express $m$ as a function of $s_1, s_2, \dots$?

It seems to me that this problem should be related to multinomial coefficients and/or some multinomial version of Pascal's triangle. But so far, I haven't been able to make that connection.

The example below shows the tree for the generating process of integer partitions and their resulting multiplicities $m$ up to $k=5$.

答:
With a few corrections in the last column (which would make the graphics messier), this looks like OEIS A036040, A080575, or A178867 which all seem to be related to Bell polynomials and vary by the order within rows (your columns).

This makes me think of Thomas Brylawski's "The lattice of integer partitions" (Disc. Math. 6 (1973) 201--219) where he developed a dominance ordering on partitions that's related to this operation.  This arose later in theoretical physics as the "sandpile model" which I've studied in a little field that I call partition dynamics.

The corrections: [4,1] can also come from [3,1] which gives it $m = 5$, and [3,1,1] can also come from [2,1,1] which gives it $m=10$.


问:
Are these two asymptotic notations equal?
I see two notations and try to understand these are the same or not? (from two views)

$O( n * (log m) * (log n))$ is equal to $O( n * (log m + log n))$ from logarithm property and from asymptotic order?

答:
No: the logarithm property says that $\log m+\log n=\log(mn)$, not $(\log m)(\log n)$. The function $n(\log m)(\log n)$ is not $O(n(\log m+\log n))$ (unless either $m$ or $n$ is bounded).

So $O(k(\log n+\log k)) = O(k\log(nk))$, although that statement has nothing to do with the $O$ notation, but is simply a fact about logarithm functions.


答:
Suppose that $M_k=B_{k-1}$ for $k\le n$. Then

$$\begin{align*}
M_{n+1}&=\sum_{i=1}^nM_iM_{n+1-i}\\
&\overset{(1)}=\sum_{i=0}^{n-1}M_{i+1}M_{n-i}\\
&\overset{(2)}=\sum_{i=0}^{n-1}B_iB_{n-1-i}\\
&=B_n\,,
\end{align*}$$

and since $M_1=B_0$ and $M_2=B_1$, the result follows by induction. Here $(1)$ is just an index shift; if it’s not clear, let $j=i-1$, rewrite the original sum in terms of $j$, and then rename $j$ to $i$. $(2)$ uses the induction hypothesis.

Title: Given $B_n = \sum_{i=0}^{n-1} (B_i B_{n-i-1})$ and $M_n = \sum_{i=1}^{n-1} (M_i M_{n-i})$, how can I show $B_{n-1} = M_n$?

Body:
I have two recursions:

$$\begin{align}
B_n &= \sum_{i=0}^{n-1} (B_i B_{n-i-1}) \\
B_1 &= 1 \\
B_0 &= 1
\end{align}$$

And:

$$\begin{align}
M_n &= \sum_{i=1}^{n-1} (M_i M_{n-i}) \\
M_2 &= 1 \\
M_1 &= 1
\end{align}$$

How can I show $B_{n-1} = M_n$?

I've found:

$$\begin{align}
B_{n-1} &= \sum_{i=0}^{(n-1)-1} (B_i B_{(n-1)-i-1}) \\
&= \sum_{i=0}^{n-2} (B_i B_{n-i-2})
\end{align}$$

Then, if I assume $M_k = B_{k-1}$ via induction, then I have:

$$\begin{align}
M_n &= \sum_{k=1}^{n-1} (M_k M_{n-k}) \\
&= \sum_{k=1}^{n-1} (B_{k-1} B_{n-k-1}) \\
&= \sum_{k=1-1}^{(n-1)-1} (B_{(k+1)-1} B_{(n-1)-(k+1)-1}) \\
&= \sum_{k=0}^{n-2} (B_k B_{n-1-k})
\end{align}$$

But I am not sure if this is kosher or how to proceed from here.


问:
Question on orbit maps
I have a group $G$ acting on a set $X$.

Let $U$ be any subset of $X$.

Let $G(x)=\{gx\mid g \in G\}$

This is basically orbit of $x$ under the action

Now I define $q$ : $X \rightarrow \frac{X}{G}$ as

$q(x) = G(x)$

Then how can I  show that $q^{-1}(q(U))$ is union of open sets $gU$ where $g \in G$?

I am proceeding like this

$$q^{-1}(q(U)) =\{x \in X:q(x) \in q(U)\}$$

$$q^{-1}(q(U))=\{x \in X:G(x) = G(y) \text { for some y in U}\}$$

From here, how can I claim the desired result?

答: Two orbits $G(x),G(y)$ are equal, if and only if, $x=gy,\, \text{for some } g\in G$. Hence, you have $$\{x\in X: G(x)=G(y), \text{ some } y\in U\}=\{x\in X: x=gy, \text{ some }g\in G,\, y\in U\}=\bigcup_{g\in G}gU.$$

Title: Question on orbit maps


问:
Mobile families of sets and pure subsets
Exercise

This is Bourbaki Theory of Sets Chapter 3 Section 4 Exercise 11, English version:

Let $A$ be a set and let $\mathcal{R}$ be a subset of the set $\mathcal{F}(A)$ of finite subsets of $A$. $\mathcal{R}$ is said to be mobile if it satisfies the following condition:

(MO) If $X$, $Y$ are two distinct elements of $\mathcal{R}$ and if $z\in X\cap Y$, then there exists $Z\subseteq X\cap Y$ belonging to $\mathcal{R}$ such that $z\notin Z$.

A subset $P$ of $A$ is then said to be pure if it contains no set belonging to $\mathcal{R}$.

a) Show that every pure subset of $A$ is contained in a maximal pure subset of $A$.

b) Let $M$ ba a maximal pure subset of $A$. Show that for each $x\in X\setminus M$ there exists a unique finite subset $E_{M}(x)$ of $M$ such that $E_{M}(x)\cup\{x\}\in\mathcal{R}$. Moreover, if $y\in E_M(x)$, the set $(M\cup\{x\})\setminus\{y\}$ is a maximal pure subset of $A$.

c) Let $M$, $N$ be two maximal pure subsets of $A$, such that $N\setminus M$ is finite. Show that $|M|=|N|$.

d) Let $M$, $N$ be two maximal pure subsets of $A$, and put $N'=N\setminus M$, $M'=M\setminus N$. Show that:

$$M'\subseteq\bigcup_{x\in N'}E_M(x).$$

Deduce that $|M|=|N|$.

Question

I was able to do all this exercise, but, in the French version, the exercise assumes only a weaker condition:

(MO') If $X$, $Y$ are two distinct elements of $\mathcal{R}$ and if $z\in X\cap Y$, then there exists $Z\subseteq X\cup Y$ belonging to $\mathcal{R}$ such that $z\notin Z$

instead of (MO), and was able to do items (a) to (c), but I do not know how to do item (d).

Attempt

There is my solution to items (a) to (c) assuming only (MO') and item (d) assuming (MO).

a) Straightforward application of Zorn's lemma. In fact, if $\mathcal{A}$ is the set of all pure subsets of $A$ containing a given pure subset $P$, then for every totally ordered subset $\mathcal{C}$ of $\mathcal{A}$, if $\bigcup\mathcal{C}$ is not pure, then it contains an $R\in\mathcal{R}$, but $R$ is finite, say, $R=\{a_1,\dots,a_n\}$, so for $i=1,\dots,n$ there is a $Q_i\in\mathcal{C}$ such that $a_i\in Q_i$, then there is a $Q\in\mathcal{C}$ such that $Q_1,\dots,Q_n\subseteq Q$, so $R\subseteq Q$, contradicting the purity of $Q$; therefore $\bigcup\mathcal{C}\in\mathcal{P}$.

b) Because of maximality of $M$, there is an $E\in\mathcal{F}(M)$ such that $E\cup\{x\}\in\mathcal{R}$.

If $F\in\mathcal{F}(M)$, $F\neq E$ and $F\cup\{x\}\in\mathcal{R}$, then $E\cup\{x\}\neq F\cup\{x\}$ and $x\in\left(E\cup\{x\}\right)\cap\left(F\cup\{x\}\right)$, so by (MO') the set $\left(\left(E\cup\{x\}\right)\cup\left(F\cup\{x\}\right)\right)\setminus\{x\}$ is not pure, but:

$$\left(\left(E\cup\{x\}\right)\cup\left(F\cup\{x\}\right)\right)\setminus\{x\}\subseteq E\cup F\subseteq M,$$

so $M$ will not be pure, a contradiction.

Let $y\in E$. If $R\subseteq(M\cup\{x\})\setminus\{y\}$ and $R\in\mathcal{R}$, then $R\subseteq M\cup\{x\}$, so, by purity of $M$, $R=G\cup\{x\}$ for some $G\in\mathcal{F}(M)$, and by (a) we have $G=E$, but $y\notin G$ and $y\in E$, a contradiction. Therefore $(M\cup\{x\})\setminus\{y\}$ is pure.

Let $z\in A\setminus((M\cup\{x\})\setminus\{y\})$, then $z=y\text{ or }(z\notin M\text{ and }z\neq x)$.

For the case $z=y$, we have $((M\cup\{x\})\setminus\{y\})\cup\{y\}=M\cup\{x\}$, that is not pure.

For the case $z\neq y$, then $z\notin M$ and $z\neq x$, so there is a $H\in\mathcal{F}(M)$ such that $H\cup\{z\}\in\mathcal{R}$, so:

If $y\notin H\cup\{z\}$, then:

$$H\cup\{z\}\subseteq (M\cup\{z\})\setminus\{y\}\subseteq((M\cup\{x\})\setminus\{y\})\cup\{z\}.$$

If $y\in S$, then $y\in(E\cup\{x\})\cap(H\cup\{z\})$, so $((E\cup\{x\})\cup(H\cup\{z\}))\setminus\{y\}$ is not pure, and:

$$((E\cup\{x\})\cup(H\cup\{z\}))\setminus\{y\}\subseteq((M\cup\{x\})\setminus\{y\})\cup\{z\}.$$

Therefore $((M\cup\{x\})\setminus\{y\})\cup\{z\}$ is not pure. So $(M\cup\{x\})\setminus\{y\}$ is maximal.

c) Induction on $|N\setminus M|$.

If $|N\setminus M|=0$, then $N\subseteq M$, so by maximality of $N$ we have $N=M$, so $|M|=|N|$.

If $|N\setminus M|>0$, then there is a $m\in M\setminus N$ and there is an $n\in E_N(m)$, so by item (b) the set $N'=(N\cup\{m\})\setminus\{n\}$ is pure maximal and $|N'\setminus M|<|N\setminus M|$, so by induction hypothesis we have $|M|=|N'|$, but $|N'|=|N|$, so $|M|=|N|$.

d) If we assume (MO), then for $m\in M'$ we have $E_N(m)\cup\{m\}\in\mathcal{R}$, so $E_N(m)\cup\{m\}\nsubseteq M$, but $m\in M$, so $E_N(m)\nsubseteq M$, so there is an $x\in E_N(m)$ such that $x\notin M$, so $x\in N'$, and $E_M(x)\cup\{x\}\in\mathcal{R}$, therefore:

$$x\in(E_M(x)\cup\{x\})\cap(E_N(m)\cup\{m\}),$$

so we have two cases:

If $E_M(x)\cup\{x\}=E_N(m)\cup\{m\}$, then $x\in N'$ and $m\in E_M(x)$.

If $E_M(x)\cup\{x\}\neq E_N(m)\cup\{m\},$ then by (MO) the set $((E_M(x)\cup\{x\})\cap(E_N(m)\cup\{m\}))\setminus\{x\}$ is not pure, but it is contained in $M$, a contradiction.

Finally, by virtue of (c), we are reduced to the case where $M'$ and $N'$ are infinite, so:

$$|M'|\leq|\bigcup_{x\in N'}E_M(x)|\leq\sum_{x\in N'}|E_M(x)|\leq\sum_{x\in N'}\aleph_0=|N'|\aleph_0=|N'|,$$

and analogously $|N'|\leq |M'|$, so $|M'|=|N'|$, and we conclude that $|M|=|N|$.

答:
Let me start with some context. Minimal sets in $\mathcal{R}$ are exactly the circuits of a (finitary) matroid on $A$, i.e. minimal dependent sets. A matroid is a notion that generalizes linear dependence in a vector space. In particular, pure maximal sets are exactly the bases of said matroid. The invariance of their cardinal is thus a well known fact that generalizes the invariance of the cardinal of bases in vector spaces, cf. Rado, R. 1949. « Axiomatic Treatment of Rank in Infinite Sets ». Canadian Journal of Mathematics 1 (4): 337‑43. https://doi.org/10.4153/CJM-1949-031-1.

However, that does not answer the first part of question d. Knowing that we are looking at a matroid also simplifies that question. Let $Y = \bigcup_{x\in N'} E_M(x)$. Then $Y \subset M$ generates N'. It follows that $Y \cup (M\cap N)$ generates $N$ and hence is a subset of $M$ which is itself a basis. So it is equal to $M$ and we do have $M' \subset Y$.

That being said, one can prove the inclusion directly. A complete solution (in French) to the exercice can be found here: https://www.bourbaki.fr/TEXTES/E-III-4-11.pdf. Let me translate the argument for the proof of $M' \subset \bigcup_{x\in N'} E_M(x)$.

Fix $y\in M'$. Let us prove that there exists $x \in E_N(y)\setminus M$ such that $y\in E_M(x)$. We proceed by induction on the cardinal of $E_N(y)\setminus M$, which is non-empty, by purity of $M$. Pick some $x \in E_N(y)\setminus M$. We also have that $E_M(x)\setminus N$ is non-empty and pick $y'$ in it. If $y = y'$ we are done. Otherwise, consider $M'' = (M\cup\{x\})\setminus \{y\}$ which is maximal pure. Since $E_{N}(y)\cap M'' \subset (E_N(y)\cap M)\setminus \{x\}$, by induction, we find $x'\in E_N(y)\cap M''$ such that $y\in E_{M''}(x)$.

If $y' \not\in E_{M}(x')$, then $E_M(x') \subset M''$ and hence $E_M(x') = E_{M''}(x')$ contains $y$. Otherwise, by (MO'), there exists some $Z\subseteq E_M(x)\cup\{x\}\cup E_M(x')\cup\{x'\}$ with $y'\not\in Z$. Then $Z\subset M''\cup\{x'\}$ and hence is equal to $E_{M''}(x')$; a set that contains $y$. It follows that either $E_M(x)$ or $E_M(x')$ contains $y$.


问:
Different linear programming versions of optimal transport
What is the difference between these two different versions of the linear programming optimization set-up for optimal transport (OT)?  how to reconcile them mathematically to show that they are equivalent?

OT Linear Programming #1

(Source)
\begin{align*}
\min_{\boldsymbol\Gamma} \quad & \langle\mathbf{C},\boldsymbol\Gamma  \rangle = \sum_{ij}\mathbf{C}_{ij}\boldsymbol\Gamma_{ij} \\
\mathrm{s. t.}  \quad  & \boldsymbol\Gamma \mathbf{1} = \mathbf{\alpha} \\
&\boldsymbol\Gamma^{\top}\mathbf{1} = \mathbf{\beta}
\end{align*}

$\boldsymbol C$ is the distance matrix, which contains elements$\Vert x - y \Vert$,

$\boldsymbol \Gamma$ is the transport matrix, which contains elements $\gamma(x,y)$,

$\alpha$ and $\beta$ are the source and target probability distributions, respectively, of variables $X$ and $Y$.

OT Linear Programming #2

(Source)
$$
\begin{array}{rrcl}
\min_{\mathbf{x}} \ & \mathbf{c}^T \mathbf{x} &  & \ \\
\mathrm{s.t.} \ & \mathbf{A} \mathbf{x} & = & \ \mathbf{b} \\
 \  & \mathbf{x} & \geq &\ \mathbf{0}
\end{array}$$

$\mathbf{c}$ is a vectorization of the distance matrix $\boldsymbol C$,

$\mathbf{x}$ is a vectorization of the transport matrix  $\boldsymbol\Gamma$,

while the source and target distributions are $\mathbf{b} = \begin{bmatrix} \alpha \\ \beta\end{bmatrix}$

答:
To reconcile both, note that $\bf x$ is obtained by stacking the columns of $\bf \Gamma$. For the cost, $\bf c$ is obtained by stacking the columns of the cost matrix $\bf C$.
Also, let $\mathbb I_n$ be an $n\times n$ identity matrix. Therefore, make:
$$
{\bf A} = \begin{bmatrix}
{\bf 1}_n^\top \otimes \mathbb I_m\\
\mathbb I_n \otimes {\bf 1}_m^\top
\end{bmatrix}
$$

Finally, the constraint ${\bf x}\geq 0$ is the same as the constraint that the transport plan $\bf \Gamma$ must be positive. This can be found in page 38 in the book Computational Optimal Transport, by Peyré and Cuturi. The pdf of the book is freely avaiable.


问:
$m^n$ has exactly $9$ digits, all of these digits are distinct.
I was reading the following problem in a book: $2^{29}$ has exactly $9$ digits, all of these digits are distinct, what is the missing digit?

I tried to solve it, and I did it correctly.

Suddenly, the following two questions popped into my mind:

Find the number of pairs of natural numbers $m$ and $n$ $(n \ne 1)$ such that $m^n$ has exactly $9$ digits, all of these digits are distinct.

Find the least $9$-digit number that can be expressed as $m^n$ where $m$ and $n$ are natural numbers and $n \ne 1$.

These questions may be solved using a software, but I am not sure if we can use a purely mathematical way.

Do not provide a solution. I am just asking for useful hints/formulae/techniques, then I will try to solve them.

Your help would be appreciated. THANKS!

Body:
My computer says:
$$10124^{2}=102495376$$
$$10128^{2}=102576384$$
$$10136^{2}=102738496$$
$$10214^{2}=104325796$$
$$10278^{2}=105637284$$
$$11826^{2}=139854276$$
$$12363^{2}=152843769$$
$$12543^{2}=157326849$$
$$12582^{2}=158306724$$
$$12586^{2}=158407396$$
$$13147^{2}=172843609$$
$$13268^{2}=176039824$$
$$13278^{2}=176305284$$
$$13343^{2}=178035649$$
$$13434^{2}=180472356$$
$$13545^{2}=183467025$$
$$13698^{2}=187635204$$
$$14098^{2}=198753604$$
$$14442^{2}=208571364$$
$$14676^{2}=215384976$$
$$14743^{2}=217356049$$
$$14766^{2}=218034756$$
$$15353^{2}=235714609$$
$$15681^{2}=245893761$$
$$15963^{2}=254817369$$
$$16549^{2}=273869401$$
$$16854^{2}=284057316$$
$$17252^{2}=297631504$$
$$17529^{2}=307265841$$
$$17778^{2}=316057284$$
$$17816^{2}=317409856$$
$$18072^{2}=326597184$$
$$19023^{2}=361874529$$
$$19377^{2}=375468129$$
$$19569^{2}=382945761$$
$$19629^{2}=385297641$$
$$20089^{2}=403567921$$
$$20316^{2}=412739856$$
$$20513^{2}=420783169$$
$$20754^{2}=430728516$$
$$21397^{2}=457831609$$
$$21439^{2}=459630721$$
$$21744^{2}=472801536$$
$$21801^{2}=475283601$$
$$21877^{2}=478603129$$
$$21901^{2}=479653801$$
$$22175^{2}=491730625$$
$$22456^{2}=504271936$$
$$22887^{2}=523814769$$
$$23019^{2}=529874361$$
$$23113^{2}=534210769$$
$$23178^{2}=537219684$$
$$23439^{2}=549386721$$
$$23682^{2}=560837124$$
$$23728^{2}=563017984$$
$$23889^{2}=570684321$$
$$24009^{2}=576432081$$
$$24237^{2}=587432169$$
$$24276^{2}=589324176$$
$$24441^{2}=597362481$$
$$24807^{2}=615387249$$
$$25059^{2}=627953481$$
$$25279^{2}=639027841$$
$$25572^{2}=653927184$$
$$25941^{2}=672935481$$
$$26152^{2}=683927104$$
$$26351^{2}=694375201$$
$$26409^{2}=697435281$$
$$26733^{2}=714653289$$
$$27105^{2}=734681025$$
$$27129^{2}=735982641$$
$$27209^{2}=740329681$$
$$27273^{2}=743816529$$
$$27984^{2}=783104256$$
$$28171^{2}=793605241$$
$$28256^{2}=798401536$$
$$28346^{2}=803495716$$
$$28582^{2}=816930724$$
$$28731^{2}=825470361$$
$$29034^{2}=842973156$$
$$29106^{2}=847159236$$
$$29208^{2}=853107264$$
$$30384^{2}=923187456$$
$$2^{29}=536870912$$
So there are quite a few squares and just one higher power with that property.


问:
PDE variables separation problem
I suspect I'll be minused for that but I don't know what to do anymore.

Here's the thing:
$$\begin{cases} u_{tt} = u_{xx} & t \geq 0, x\in (0,\pi) \\ u(0,x)=x, u_t(0,x)=1  & x\in[0,\pi] \\ u_x(t,0) = u_x(t,\pi)=0 & t\geq 0 \end{cases}$$

I'm told to solve this by variables separation.
It differs from other examples because I have first derivatives with respect to $x$.

What I get form it, using typical algorithm, is
$$u(t,x) = \sum_{n=1}^{\infty} \cos{nx} (C_n \cos{nt}+D_n\sin{nt}) $$
Applying the condition
$$u(0,x) = x = \sum_{n=1}^{\infty} C_n\cos{nx}$$
and here I'm stuck because I can't make a cosine series here, because cosine series for $x$ has non-zero $a_0 = \frac{\pi}{2}$.

Should I change some variable or substitute other function? Some advices here were to use Fourier Transform but I have to do it by variables separation.

答:
If you follow the separation of variables procedure
$$
                    u(t,x)=X(x)T(t) \\
                  \frac{X''}{X}=\lambda=\frac{T''}{T}
$$
and impose $X'(0)=X'(\pi)=0$, then you end up with solutions
$$
                       X=1,\cos(x),\cos(2x),\cos(3x),\cdots.
$$
That is
$$
                  X_0 = 1,X_1=\cos(x),X_2=\cos(2x),\cdots.
$$
The corresponding $T_0$ solution satisfies $T_0(t)=A_0+B_0 t$. You overlooked this term because it has only a constant solution in $x$. The full general solution is
$$
        u(t,x) = (A_0+B_0t)+\sum_{n=1}^{\infty}(A_n\cos(nt)+B_n\sin(nt))\cos(nx).
$$
The constants are determined by
$$
       x=u(0,x)=A_0+\sum_{n=1}^{\infty}A_n\cos(nx) \\
       1=u_t(0,x)=B_0+\sum_{n=1}^{\infty}nB_n\cos(nx).
$$
$B_0=1$ and $B_n=0$ for $n=1,2,3,\cdots$.

Title: PDE variables separation problem


问:
Probability of a four letter word from a sequence of n random letters
I know this is easy, but my high school maths has failed me.

Question: I generate an 8 letter random string. What is the probability that within these 8 letters I will find a particular 4 letter word?

Each letter is A-Z. Repeats are allowed. What are the chances my string will contain the word "ABCD" for example?

EDIT: To clarify.. I do care about ordering. I want to know if "ABCD" occurs within my randomly generated string. But I don't care if "ABDC" occurs. In other words I want to know the probability of A followed by B followed by C followed by D occurring.

If it helps to understand why I am asking this. Our software generates 8 letter random strings which are used for logins to a web site. Very occasionally, these 8 letter random strings contain swear words. I want to know how to calculate the probability of a particular swear word occurring in a randomly generated 8 letter string.

答:
The answer is 2284879/208827064576 if no letters are repeated. This is roughly $5\times 10^{-12}$ less than Marc's answer 5/456976. It is also yybtcbk's answer.

To solve variants of this problem you may use this python code.

```<code>######## compile word to dfa

alphabet = 'abcdefghijklmnopqrstuvwxyz'

def bitstoint(bs):
    n = 0
    for b in reversed(bs):
        n = 2*n + b
    return n

def compile_word_to_dfa_that_matches_it_as_substring(word):
    """
    The state of the automaton must tell, for each i, whether the
    input so far ends in word[:i].

    If the input so far ends in word[:i], and the next character is
    indeed word[i], then the input so far ends in word[:i+1];
    otherwise it doesn't.

    As an exception to this rule, if the input so far ends in word,
    then this will forever be the state of the automaton.
    """
    assert set(word) <= set(alphabet)
    N = 2**(len(word) + 1)  # all states are 0 <= s < N; (s & (1 << i)) is valid for 0 <= i <= len(word)
    def transition(s, c):
        if s == N-1:
            return N-1  # accepting state
        if s & (1 << (len(word)-1)) and word[-1] == c:
            return N-1
        return 1 + 2*bitstoint([
            int(s & (1 << i) and word[i] == c)
            for i in range(len(word))
        ])

    transition_matrices = {
        c: [[int(transition(i, c) == j) for j in range(N)] for i in range(N)]
        for c in alphabet
    }
    initial_state = 1
    final_state = N-1
    return (transition_matrices, initial_state, final_state)

######## (optional: optimize the dfa)

######## probability calculation

import numpy as np

def calculate_acceptance_probability(dfa, q, t):
    transition_matrices, initial_state, final_state = dfa
    Aq = sum([np.array(transition_matrices[c])*q[c] for c in alphabet]).transpose()
    N = Aq.shape[0]
    v = np.array([int(i == initial_state) for i in range(N)])
    for _ in range(t):
        v = np.dot(Aq, v)
    return v[final_state]

######## demo

word = 'joel'  # the 4 letter word
t = 8          # the total number of letters

#number_type = float
import fractions
number_type = fractions.Fraction

dfa = compile_word_to_dfa_that_matches_it_as_substring(word)
q = {c: number_type(1)/len(alphabet) for c in alphabet}
answer = calculate_acceptance_probability(dfa, q, t)

print(f"The accepted stackexchange answer: {number_type(5)/26**4}")
print(f"The exact answer with automaton: {answer}")

""" demo output

box% python -i main.py
The accepted stackexchange answer: 1.094149364518049e-05
The exact answer with automaton: 1.094148885652915e-05
>>>

box% python -i main.py
The accepted stackexchange answer: 5/456976
The exact answer with automaton: 2284879/208827064576
>>>

"""
</code>```

Body:
The answer is 2284879/208827064576 if no letters are repeated. This is roughly $5\times 10^{-12}$ less than Marc's answer 5/456976. It is also yybtcbk's answer.

To solve variants of this problem you may use this python code.

######## compile word to dfa

alphabet = 'abcdefghijklmnopqrstuvwxyz'

def bitstoint(bs):
    n = 0
    for b in reversed(bs):
        n = 2*n + b
    return n

def compile_word_to_dfa_that_matches_it_as_substring(word):
    """
    The state of the automaton must tell, for each i, whether the
    input so far ends in word[:i].

    If the input so far ends in word[:i], and the next character is
    indeed word[i], then the input so far ends in word[:i+1];
    otherwise it doesn't.

    As an exception to this rule, if the input so far ends in word,
    then this will forever be the state of the automaton.
    """
    assert set(word) <= set(alphabet)
    N = 2**(len(word) + 1)  # all states are 0 <= s < N; (s & (1 << i)) is valid for 0 <= i <= len(word)
    def transition(s, c):
        if s == N-1:
            return N-1  # accepting state
        if s & (1 << (len(word)-1)) and word[-1] == c:
            return N-1
        return 1 + 2*bitstoint([
            int(s & (1 << i) and word[i] == c)
            for i in range(len(word))
        ])

    transition_matrices = {
        c: [[int(transition(i, c) == j) for j in range(N)] for i in range(N)]
        for c in alphabet
    }
    initial_state = 1
    final_state = N-1
    return (transition_matrices, initial_state, final_state)

######## (optional: optimize the dfa)

######## probability calculation

import numpy as np

def calculate_acceptance_probability(dfa, q, t):
    transition_matrices, initial_state, final_state = dfa
    Aq = sum([np.array(transition_matrices[c])*q[c] for c in alphabet]).transpose()
    N = Aq.shape[0]
    v = np.array([int(i == initial_state) for i in range(N)])
    for _ in range(t):
        v = np.dot(Aq, v)
    return v[final_state]

######## demo

word = 'joel'  # the 4 letter word
t = 8          # the total number of letters

#number_type = float
import fractions
number_type = fractions.Fraction

dfa = compile_word_to_dfa_that_matches_it_as_substring(word)
q = {c: number_type(1)/len(alphabet) for c in alphabet}
answer = calculate_acceptance_probability(dfa, q, t)

print(f"The accepted stackexchange answer: {number_type(5)/26**4}")
print(f"The exact answer with automaton: {answer}")

""" demo output

box% python -i main.py
The accepted stackexchange answer: 1.094149364518049e-05
The exact answer with automaton: 1.094148885652915e-05
>>>

box% python -i main.py
The accepted stackexchange answer: 5/456976
The exact answer with automaton: 2284879/208827064576
>>>

"""


问:
Finding Transformation Matrices Along The Line $y = 2x$
It seems I have fallen behind in my Methods of Theoretical Physics class, and am asking here as I need explanations and resources to learn. Let's say we are given the following:

"Find the transformation matrix, $N$, which stretches vectors by a factor of 4 along the line $y = 2x$."

How would you go about this? If this just means "the matrix that stretches the y-axis by a factor of 4 and rotates such that $y' = 2x$", then I think I can do that, but would rather the help to be sure.

答:
Choose a basis for the coorinate system.

$e_1 = \begin {bmatrix} 1\\2\end{bmatrix}, e_2=\begin{bmatrix}-2\\1 \end{bmatrix}$

That is one vector in the direction of $y=2x$ and one orthogonal.

$B = \begin{bmatrix} 1&-2\\2&1\end{bmatrix}$ will transform a vector in our basis to the standard basis.

$B^{-1}$ will then transform a vector in the standard basis to our basis.

$\begin{bmatrix} 4\\&1\end{bmatrix}$ will transform a vector in our basis as desired.

$B\begin{bmatrix} 4\\&1\end{bmatrix}B^{-1}$ will take a vector in the standard basis, transform it to our adjusted frame of reference, dilate in one direction, leaving the other unchanged, and transform back to the standard basis.

$\begin{bmatrix} 1&-2\\2&1\end{bmatrix}\begin{bmatrix} 4\\&1\end{bmatrix}\begin{bmatrix} \frac15&\frac 25\\-\frac25&\frac 15\end{bmatrix}= \begin{bmatrix}\frac 85&\frac 65\\\frac 65&\frac {17}5\end{bmatrix}$

OwnerDisplayName: TheAkashain


问:
Distribution of the Sum of Uniform Random Variable and Another Random Variable
This could have been asked somewhere else, but I couldn't find it anywhere. Assume $X$ and $Y$ are random variables which take values in $\mathbb{Z}_n$ - i.e., $\{0, 1, 2, ..., n-1\}$. Also assume that $X$ is uniformly distributed. What is the distribution $X+Y$?

My approach: Let $Z$ be a random variable defined by $X+Y$. Then, since the maximum value that $X$ and $Y$ can take is $n-1$, $Z$ can take a maximum number of $2(n-1) = 2n -2$, which means that $Z$ takes values in $\mathbb{Z}_{2n-1}$. Now to get the probability mass function of $Z$, we have that
\begin{align*}
P(Z = z) &= P(X+Y = z)\\
&= P(X= x, Y = z -x) \\
&= P(X=x) P(Y = z-x) \\
&= \frac{1}{n} P(Y = z - x)
\end{align*}
which shows that $Z$ has the same distribution as $Y$ and it takes values in $\mathbb{Z}_{2n-1}$. Is my logic correct? If not, how else can I find the distribution of $X+Y$? I would appreciate any help on this!

答:
Your own calculations are almost correct however as suggested in my comment you fail to consider all values that $X$ can take. The formula for the probability mass function of $Z=X+Y$ is therefore

$$P(Z=z) = P(X+Y=z) = \frac{1}{n} \sum_{x=0}^{n-1}P(Y=z-x),$$
It should however be noted, that some of the terms on the right hand side will probably be $0$, since $P(Y = z-x)=0$, when $z-x \notin \{0,\dots,n-1\}$ or equivalently, when $z \notin \{x ,\dots,x+n-1\}$.

Let us consider a concrete example, where $X$ and $Y$ are both uniform on $\{0,1,2\}$, then by our derived formula
\begin{align*}
P(X+Y = 0) &= \frac{1}{3}(P(Y=0)+P(Y=-1)+P(Y=-2)) \\ &= \frac{1}{3}P(Y=0) =\frac19
\end{align*}
and
\begin{align*}
P(X+Y = 1) &= \frac{1}{3}(P(Y=1)+P(Y=0)+P(Y=-1)) \\ &= \frac{1}{3}(P(Y=1)+P(Y=0)) =\frac29
\end{align*}
and by continuing the calculations you should get a distribution, which looks like this


问:
Matrix Determinant Formula

Problem. Let $A$ be a non-singular $n\times n$ matrix and let $\Gamma=[\Gamma_1\quad\Gamma_2]$ be an $n\times n$ orthogonal matrix where $\Gamma_1$ is $n\times n_1$, $\Gamma_2$ is $n\times n_2$ and $n=n_1+n_2$. Show that $$\det(\Gamma_1^TA\Gamma_1)=\det(A)\det(\Gamma_2^TA^{-1}\Gamma_2).$$

My Attempts. Here we make use of the property of orthogonal matrix:
\begin{align}
\det(A)=\det(\Gamma^TA\Gamma)=\det\left(\begin{bmatrix}
\Gamma_1^T \\ \Gamma_2^T
\end{bmatrix}A\begin{bmatrix}
\Gamma_1 & \Gamma_2
\end{bmatrix}\right)=\det\left(\begin{bmatrix}
\Gamma_1^TA\Gamma_1 & \Gamma_1^TA\Gamma_2 \\ \Gamma_2^TA\Gamma_1 & \Gamma_2^TA\Gamma_2
\end{bmatrix}\right).
\end{align}
Since $A$ is non-singular, $\Gamma_1^TA\Gamma_1$ is also non-singular. Thus,
\begin{align}
\det(A)=\det(\Gamma_1^TA\Gamma_1)\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right).
\end{align}
If the formula we want to prove is true, we would have
\begin{align}
1&=\det(\Gamma_2^TA^{-1}\Gamma_2)\cdot\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right) \\
&=\det\left(\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right).
\end{align}
Nonetheless, I have no idea how to simplify the terms in the parenthesis because I only have $\Gamma_1\Gamma_1^T+\Gamma_2\Gamma_2^T=I$. Hope anyone has good suggestions.

答:
To simplify the terms, we indeed have to use the formula $\Gamma_2\Gamma_2^T=I-\Gamma_1\Gamma_1^T$. Here
\begin{align*}
\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2&=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_2 \\
&=\Gamma_2^TA^{-1}A\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2 \\
&=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2
\end{align*}
and
\begin{align*}
&\quad\ \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\Gamma_2^TA^{-1}(I-\Gamma_1\Gamma_1^T)A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\Gamma_2^TA^{-1}A\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=\color{red}{\Gamma_2^T\Gamma_1}(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_1\color{red}{\Gamma_1^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}}\Gamma_1^TA\Gamma_2 \\
&=0-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2=-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2.
\end{align*}
Thus,
\begin{align*}
&\quad\ \Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2 \\
&=I-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1A\Gamma_2-(-\Gamma_2^TA^{-1}\Gamma_1\Gamma_1^TA\Gamma_2)=I.
\end{align*}
The formula thus follows.

P.S. If anyone has simpler solutions, please share it with me!! This calculation is too complicated.

Title: Matrix Determinant Formula


问:
Finding the limit to the power of a sequence.
Let $a_n$ be a sequence such that n $\in{N}$ and $a_n$ $\ne 0$. Assume that $\lim_{n\to \infty}a_n=0$.   Show that $\lim_{n\to \infty}(1+a_n)^{1/a_n}=e$.

Really having a tough time with this one. Can anyone help? This question is in the subsequential limit problem section.

答:
Hint: Rewrite
$$(1+a_n)^{1/a_n}=e^{\frac{1}{a_n}\ln(1+a_n)}$$
and compute  the limit of the exponent using Taylor polynomials or the definition of the derivative of $\ln$.


问:
If I have 4 determinants for 4 parts of one matrix, how can I compute the determinant for the matrix as a whole?

M=[AB 0C] (A and B build the first row, 0 and C build the second row.)

The determinant of Matrix A is 2. The determinant of Matrix B is 3. The determinant of Matrix C is 5.
What's the determinant of the Matrix M
Here is the exercise that I am referring to

Can I just use the 4 determinants to build a 2*2 matrix and then calculate "|23 05|(2 and 3 sit in the first row, 0 and 5 sit in the second row)"?

答:
What you're looking for is the determinant of a block matrix.

What you describe (constructing a matrix of determinants) would yield the correct result, but would not be a valid procedure.
Namely, it's true that
$$
\mathrm{det}\bigg(\begin{array}{cc}
A & B \\ 0 & C
\end{array} \bigg) = \mathrm{det}(A) \cdot \mathrm{det}(C).
$$
In general, the formulas for determinants are ugly, but I would recommend reading the Wikipedia article linked above to learn more about them.


问:
Every $R$-linear map of $S^{-1} R$ Modules is automatically a $R_S$-linear map
I am currently learning a little bit about localization in rings. I am stuck with this simple looking question:

"Every $R$-linear map of $S^{-1} R$ Modules is automatically a $S^{-1} R$-linear map"

Here, $R$ is a commutative and unitary ring and $S\subseteq R$ is a multiplicative monoid of $R$. But what can I try now? Applying the basic definition:

Let $\phi : M \rightarrow N$, with $M,N$ being $S^{-1} R$ Modules, $\phi$ is  $R$-linear, so:
$$ \phi(\lambda m + n) = \lambda \phi(m) + \phi(n) \hspace{3mm}\forall m \in M, n \in N, \lambda \in R $$

But what can I do now? I have to show, that its also equal for $\lambda \in R_S$, but how can I do that? Can someone give me a tip?

答:
Let $\phi$ as in your problem; for convenience we denote $(1\big/s)m$ by $m\big/s$ for any $s\in S$, $m\in M$. Because $\phi$ is $R$-linear, it suffices to show that $\phi(m\big/s)=\phi(m)\big/s$ for any $m\in M, s\in S$. (Why? Explanation below, but think it through yourself first.)

 Suppose the above equality holds. Since $\phi$ is $R$-linear it is in particular additive, so to show $S^{-1}R$-linearity we need only show that $\phi(\lambda m)=\lambda\phi(m)$ for any $m\in M,\lambda\in S^{-1}R$. But every element of $S^{-1}R$ is of the form $r\big/s$ for some $r\in R,s\in S$, and we have $$\phi((r\big/s)m)=\phi(rm\big/s)=\phi(rm)\big/s=r\phi(m)\big/s=(r\big/s)\phi(m)$$ where the second equality holds by the equation above and the third equality holds by $R$-linearity.

Now, by $R$-linearity we have that $s\phi(m\big/s)=\phi(sm\big/s)=\phi(m)$, and so multiplying both sides by $1\big/s$ gives the desired equality.


问:
Using modular arithmetic in exponent
For odd positive integers $n_1$ and $n_2$ I am giver a relation:

$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$

So I'm proving something and on some step of my prove I want to use giver relation and write something like this:

$(-1)^{(n_1-1)/2 + (n_2-1)/2 }= (-1)^{(n_1n_2-1)/2}$

Can I do this? And what do I need to claim if I want to use giver relation like this?

答:
Hint: Prove that

$$(-1)^a \equiv (-1)^b \Leftrightarrow a\equiv b \pmod{2}.$$

Title: Using modular arithmetic in exponent


Title: Using modular arithmetic in exponent

Body:
For odd positive integers $n_1$ and $n_2$ I am giver a relation:

$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$

So I'm proving something and on some step of my prove I want to use giver relation and write something like this:

$(-1)^{(n_1-1)/2 + (n_2-1)/2 }= (-1)^{(n_1n_2-1)/2}$

Can I do this? And what do I need to claim if I want to use giver relation like this?

Body:
Yes you can. From definition, for some $k\in\mathbb Z$:

$$\frac {n_1n_2-1}2+2k=\frac{n_1-1}2+\frac{n_2-1}2$$

Hence:

$$(-1)^{(n_1-1)/2+(n_2-1)/2}=(-1)^{2k+(n_1n_2-1)/2} = (-1)^{2k}(-1)^{(n_1n_2-1)/2}=(-1)^{(n_1n_2-1)/2}$$


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

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

My try

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

Any help is welcome.

答:
MacLaurin Expansion of $m^y$ is
$$m^y=1+y \log m+O\left(y^2\right)$$
Therefore
$$m^{1/x}=1+\frac{\log m}{x}+O\left(1/x^2\right)\tag{1}$$
Integrate
$$\int_1^m \left(\frac{\log m}{x}+1\right) \, dx=m+\log ^2 m-1\tag{2}$$
Thus
$$\underset{m\to \infty }{\text{lim}}\frac{\int_1^m m^{1/x} \, dx}{m}=\underset{m\to \infty }{\text{lim}}\frac{m+\log ^2 m-1}{m}=\underset{m\to \infty }{\text{lim}}\left(1+\frac{\log ^2 m}{m}-\frac{1}{m}\right)=1$$

Edit

$(1)$

$$\int_1^m \frac{1}{x^2} \, dx=1-\frac{1}{m}$$
$(2)$ becomes
$$\int_1^m \left[\left(\frac{\log m}{x}+1\right) +O\left(1/x^2\right)\right]\, dx\le m+\log ^2 m-1 +1-\frac{1}{m}=m+\log ^2 m-\frac{1}{m}$$
Thus to a greater extent $(2)$ holds


问:
Largest possible radius of a circle lying in a parabola

Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$

and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?

答:
For the circle to be entirely within the parabola means the parabola and the circle intersect at exactly one point.

However solving $(x+r)^2 + y^2 = r^2$ and $y^2 + 12x = 0$ we actually will find three (potential) points of intersection.

$y^2=-12x$ so $(x+r)^2 +y^2 =(x+r)^2 -12x = r^2$ and $x(x+2r -12) =0$ so

$x = 0$ and $y=0$,  or $x = (12-2r)$ and $y = \pm \sqrt{12(2r-12)}$.

$x=0, y=0$ is okay, but we can't have $x = (12-2r)$ and $y= \pm \sqrt{12(2r-12)}$ be two more points of intersection.

The only we to avoid those two potential points existing is if either: i) $(12-2r, \pm \sqrt{12(2r-12)}) = (0,0)$; or in other words if $r=6$; or ii) if $\sqrt{12(2r-12)}$ doesn't exist; or in other words if $12(2r-12) < 0$ or in other words if $r < 6$.

So if $r \le 6$ then only one point of intersection exists.

If $r > 6$ then three points of intersection exist.

As we figure that three points of intersection was impossible, we concluder $r < 6$>

......

Hmm, I suppose we have to justify my very first sentence "For the circle to be entirely within the parabola means the parabola and the circle intersect at exactly one point".

Note: if $r > 6$ and if $2r -12 < x < 0$ then the points of the circle $y=\pm\sqrt{r^2 - (x+r)^2}$ will be so that $y^2 < -12x$ so those points of the circle are outside the parabola.

========

Maybe a clear explanation of what algebraically what a circle being within a parabola means.

A point $(w,v)$ is "inside" the parabola $y^2 + 12x =0$ or $y = \pm \sqrt {-12x}$ if the point $(w,v)$ is left of the vertex of $y^2 + 12x = 0$; that is to say that $w < 0$, and if the point $(w,v)$ is withing the bounds of the parabola at $x = w$; that is to say that if $y^2 + 12w =0$ so $y =\pm \sqrt {-12w}$ then the value of $v$ is so that $-\sqrt{-12w} \le v \le \sqrt{-12w}$.

Now the equation of the circle is $(x+r)^2 + y^2 = r^2$ so if we have a point $(w,v)$ on the circle that point is satisfies $(w+r)^2 + v^2 = r^2$ and if the circle is inside the parabola we must have $-\sqrt{-12w} \le v \le \sqrt{-12w}$.

Solving:  $w^2 + 2wr +r^2 + v^2 = r^2$ so $v =\pm\sqrt{-w(w+2r)}$ so we need $-\sqrt{-12w} \le -\sqrt{-w(w+2r)} \le \sqrt{-w(w+2r)}\le \sqrt{-12w}$.

For all points of the circle we have $w$ between $-2r, 0$ so $w \le 0$.  If $w=0$ we have $v=0$ and the inequality holds.

If $w < 0$ then $\sqrt{-w} > 0$ and we can divide the inequality be $\sqrt{-w}$ to get

$-\sqrt{12}\le -\sqrt{w+2r} \le \sqrt{w+2r} \le \sqrt{12}$ or $0 \le w+2r \le 12$. for all $w < 0$.  That means that $2r < 12 + |w|$ for all $|w|>0$ or that $r < 6 + \epsilon$ for all $\epsilon > 0$.

Sigh... okay, this throws everyone the first time the see it but that means $r \le 6$.  (If $r > 6$ then if we set $\epsilon = r-6$ we have $\epsilon > 0$ and we get the contradiction: $r < 6 + \epsilon = 6 +(r-6) = r$.)

.....

Hmmm.... this proof is nowhere near  as clear or easy to follow as my first argument that if a circle is within a parabola that means it intersects at exactly one point.

But.... although it is geometrically obvious that that statement about intersection is true, I'm not sure we can justify it without an argument.


问:
How to find the minimum of $x+y^2+z^3$?
let $x,y,z>0$, and $x+3y+z=9$, find the minimum of

$$x+y^2+z^3$$

I think this problem is very interesting. I have found this
when $$x=\dfrac{9}{2}-\dfrac{1}{\sqrt{3}},y=\dfrac{3}{2},z=\dfrac{1}{\sqrt{3}}$$

I belive this inequality have $AM-GM$ methods,becasue I have see this same problem can use $AM-GM$ methods,and I think this methods is very very nice.

if $a,b,c>0$,and $ a+b^2+c^3=\dfrac{325}{9}$, prove that
$$a^2+b^3+c^4\ge\dfrac{2807}{27}$$

my methods: let $a=x,b=y,c=z$,and then
$$x+y^2+z^3=\dfrac{325}{9}$$
use $AM-GM$,we have
$$a^2+x^2\ge 2ax$$
$$b^3+b^3+y^3\ge 3yb^2$$
$$c^4+c^4+c^4+z^4\ge 4zc^3$$

then we have
$$a^2+b^3+c^4+x^2+\dfrac{y^3}{2}+\dfrac{z^4}{3}\ge 2ax+\dfrac{3}{2}b^2+\dfrac{4}{3}zc^3$$

答:
Since $x+y^2+a^2+z^3+b^3+b^3\ge x+2ay+3b^2z$, now let $2a=3, 3b^2=1$. You can determine $ y=a=\frac{3}{2}, z=b=\frac{1}{\sqrt{3}}$ and $x=9-3y-z$ if equality holds.

You do get an AM-GM approach.

Title: How to find the minimum of $x+y^2+z^3$?


问:
How to find the base number of of two number systems when there's a difference in the quanity of numbers?
The problem is as follows:

In two numbering systems which are of consecutive bases, there are
about $120$ three-digit numerals more than the other. Find the
smallest of these bases.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{base 11}\\
2.&\textrm{base 13}\\
3.&\textrm{base 9}\\
4.&\textrm{base 15}\\
5.&\textrm{base 14}\\
\end{array}$

This problem makes me confused. How exactly should it be assessed?. I'm assuming that the author of this problem intends to say that there are two numbers as follows:

$\overline{abc}_{(\alpha)}$ and $\overline{cde}_{(\beta)}$

Then I believe it mentions that there is a difference between $120$ numbers between these digits. Therefore in order to avoid falling into the telephone post error then I think it meant:

$\overline{abc}_{(\alpha)}-\overline{cde}_{(\beta)}=120-1=119$

This assumes that the first number is greater than the second.

But here's the catch, as it stands there are too many variables to guess or poinpoint where to start any sort of inspection which makes the problem to be a guessing game.

Can someone help me with an approapiate method and a step by step approach on how to solve this problem?. It would help me a lot that the answer would be an easy to understand approach.

答:
We have base $b$ and base $b+1$. In base $b$, there are $b^3-b^2$ three-digit numbers. Likewise, in base $b+1$ there are $(b+1)^3-(b+1)^2$, so this count exceeds that for base $b$ by
$$(b+1)^3-(b+1)^2-b^3+b^2= 3b^2+b $$
and this is given to be "about 120". If we ignore the $b$, this means that $3b^2\approx 120$, so $b^2\approx 40 $, and $b$ should be $6$ or $7$. With $b=6$, $3b^2+b=114$, and with $b=7$, $3b^2+b=154$. So it seems that $b=6$ was intended.


问:
Legendre symbols
So for odd prime $p$, I have Legendre symbol $(\frac{-1}{p}) = (-1)^{(p-1)/2}$, and let $n\geq 3$ be an odd positive integer.

I have to show that for odd positive integers $n_1$ and $n_2$, we have

$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$

And from this I have to deduce $(\frac{-1}{n}) = (-1)^{(n-1)/2}$.

So I have manage to show the first part:

$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$

$\frac{n_1n_2-n_1-n_2+1}{2} \equiv 0\pmod{2}$

$\frac{(n_1-1)(n_2-1)}{2} \equiv 0\pmod{2}$

$n_1,n_2$ odd $\implies (n_1-1),(n_2-1)$ even $\implies n_1-1=2a, n_2-1=2b; a,b\in \mathbb{Z}$

$\frac{4ab}{2} \equiv 0\pmod{2}$

$2ab \equiv 0\pmod{2}$

$0 \equiv 0\pmod{2}$

So now, I have to somehow use this to deduce $(\frac{-1}{n}) = (-1)^{(n-1)/2}$, but I have no idea what to do. Any suggestions?

答:
You already know the result when $n$ is prime, so assume that $n$ is composite and positive. Then there exist positive integers $n_1$ and $n_2$ such that $2\le n_1\le n_2<n$ and $n_1 n_2 = n.$ What determines whether a power of $-1$ is $1$ or $-1$ is the exponent modulo $2.$ By what you have already proven, $$(-1)^{\frac{n-1}{2}}=(-1)^{\frac{n_1 n_2 -1}{2}} = (-1)^{\frac{n_1-1}{2}+\frac{n_2-1}{2}}=(-1)^{\frac{n_1-1}{2}}\cdot (-1)^{\frac{n_2-1}{2}}.$$
You can use strong induction on $n,$ so we get $$(-1)^{\frac{n-1}{2}}=\left(\frac{-1}{n_1}\right)\left(\frac{-1}{n_2}\right).$$ This is equal to $\left(\frac{-1}{n}\right)$ by the complete multiplicativity of the Jacobi symbol (in either entry when the other entry is fixed), assuming that is what you are using for $\left(\frac{-1}{n}\right)$ since the Legendre symbol is defined only for primes in the lower entry.

Title: Legendre symbols


问:
Why does the homology of the this plain simplicial complex vanish?
I can't find a solution to the following exercise:

Let $K$ be a plain simplicial complex, which can be represented as a union of simplices
$\sigma_1, \dots, \sigma_n$ (and their boundary simplices), such that for some integer $d \geqslant 2$ we have

$\operatorname{dim} \sigma_i \geqslant d$ for all $i$ and

for every $1 \leqslant i < j \leqslant n$ the intersection $\sigma_i \cap \sigma_j$ has dimension at most $d − 2$.

Show that the homology of $K$ vanishes in dimension $d$ and above.

Could somebody eventually help me? If any definition is not clear, just ask me and I will give it to you.

答:
Let's write out the boundary of each $\sigma_i$:
$$\partial \sigma_i = \tau_{i,0} + \tau_{i,1} + \tau_{i,2}
$$
where $\tau_{i,j}$ ($j=0,1,2$) is a simplex of dimension given by $\text{dim} \tau_{i,j} = \text{dim}\sigma_i - 1$ (in that summation, each term might have a $+1$ or $-1$ coefficient, which I am suppressing for ease of notation).

Your hypothesis implies that if $i \ne i'$ then $\tau_{i,j} \ne \tau_{i',j'}$ (no matter what $j$ and $j'$ are).

So, now choose some dimension $\ge d$ and consider a chain of that dimension:
$$c = \sum_i {a_i} \sigma_i
$$
Taking the boundary we get
$$\partial c = \sum_i a_i (\tau_{i,0} + \tau_{i,1} + \tau_{i,2})
$$
To determine whether $c$ is a cycle, i.e. whether $\partial c = 0$, we must collect terms and see if there is any cancellation. BUT, there is no cancellation at all, because of the fact that $\tau_{i,j} \ne \tau_{i',j'}$ if $i \ne i'$. So if $c \ne 0$ then $\partial c \ne 0$.

Thus the cycle group in the chosen dimension is trivial, and so the homology group (cycles modulo boundaries) is also trivial.


问:
Question about Banach Limit
I just studied the Banach Limit $\Lambda$ in my Functional Analysis class and I am trying to answer the following question

Can one define a generalised limit for bounded functions $f : \mathbb{R} \to \mathbb{R}$?

For now, I constructed a functional $\mathcal{L}$ from the space $\mathcal{B}$ of bounded linear functions to $\mathbb{R}$ in the following way:

notice that every real sequence $(x_n)_n$ can be seen as a function $f: \mathbb{N} \to \mathbb{R}$;

notice that the set $S=\{s \text{ such that } s:\mathbb{N} \to \mathbb{R}, s \text{ bounded}\}$ forms a subspace of $\mathcal{B}$;

use Hahn-Banach Theorem to define a linear functional $\mathcal{L}$ extending $\Lambda$ from $S$ to all $\mathcal{B}$ and such that
$$
|\mathcal{L}(f)| \leq \|f\| \quad \text{ for every } f \in \mathcal{B}.
$$

Such functional is linear and normalized by construction, and positivity is pretty straightforward. I am having trouble with shift invariance. Any suggestions?

答:
I guess $S$ is a quotient of $\mathcal{B}$,  rather than a subspace,  correct?

That is, we may define a projection mapping
$\pi :\mathcal{B}\to S, $
by letting $\pi (f)$ be the restriction of $f$ to $\mathbb N$, for every $f$ in $\mathcal B$.

Choosing a Banach limit $\Lambda :S\to \mathbb R$, we  then define  $\mathcal{L}$ to be the composition
$$
  \mathcal B \quad {\buildrel \pi \over \longrightarrow} \quad S \quad {\buildrel \Lambda \over \longrightarrow} \quad \mathbb R.
  $$
The OP has already observed  that this functional has all of the required properties, except possibly for translation invariance.

Denoting by $T_y(f)$ the translation of $f$ by $y$,  namely
$$
  T_y(f)|_x =  f(x-y),
  $$
the missing property is thus
$$
  \mathcal L(T_y(f)) = \mathcal L(f).
  \tag 1
  $$
Observe however  that $\Lambda $ is invariant by integer translations, so this  easily implies that
(1) holds provided $y$ is an integer number.  In particular the correspondence
$$
  y\mapsto \mathcal L(T_y(f))
  \tag 2
  $$
is seen to be periodic with period 1.

When desired invariance properties are absent,
it is a common trick in the theory of group representations to average out.  That is,
in order to overcome the above difficulty
let us   define   $\tilde{\mathcal L}$ by
$$
  \tilde{\mathcal L}(f) = \int_0^1 \mathcal L(T_y(f))\, dy.
  $$
Given any $z$ in $\mathbb R$, we then have that
$$
  \tilde{\mathcal L}(T_z(f)) =
  \int_0^1 \mathcal L(T_{y+z}(f))\, dy =
  \int_z^{z+1} \mathcal L(T_{y}(f))\, dy =
  \int_0^1 \mathcal L(T_{y}(f))\, dy = \tilde{\mathcal L}(f),
  $$
where the penultimate  step is due to the periodicity of (2).

It is now easy to see that $\tilde{\mathcal L}$ satisfies all of the required conditions.

Title: Question about Banach Limit


问:
Find a monic polynomial over $\mathbb{Q}$ whose Galois group is $V_4$
I need to find a polynomial over $\mathbb{Q}$ whose Galois group is $V_4=\langle(12)(34),(13)(24)\rangle\subset S_4$. I can find examples of such polynomials, but I wonder if one can construct such a polynomial by hand.

Can anyone offer some ideas? Thanks!

答: What about $\,f(x)=(x^2-2)(x^2-3)\in\Bbb Q[x]\,$ ? In fact, any other biquadratic of this form $\;(x^2-a)(x^2-b)\;$ works, when neither $\,a,b,ab\,$ is a square.</p>


问:
Finding a kind of matrix
Given $h > 0$, let

$$B := \begin{pmatrix} 0 & -2 \\ \frac{1}{2h} & 0 \end{pmatrix}$$

or, more generally, let

$$B := \begin{pmatrix} 0^{n\times n} & -2 I^{n\times n}  \\ \frac{1}{2h}I^{n\times n}  & 0^{n\times n} \end{pmatrix}$$

Is there an explicit form for $e^{B}$?

答:
by blocked multiplication

$B^2 = \left[\begin{matrix}(-2I_n)(\frac{1}{2h}I_n) & \mathbf{0}\\\mathbf{0} & (\frac{1}{2h}I_n)(-2I_n)\end{matrix}\right]  = \frac{-1}{h} I_{2n}$

which means that $B$ has two distinct eigenvalues given by the two distinct solutions to $x^2 = \frac{-1}{h}$.

This also tells you that $\mathbf 0 = B^2+\frac{1}{h} I_{2n} =\big(B-\lambda_1 I \big)\big(B-\lambda_2 I \big)$ is the minimal polynomial of $B$.  This implies $B$ is diagonalizable over $\mathbb C$ so that is one way to finish.  A nicer finish is below:

Treated as abstract vectors, $\Big\{I,B\Big\}$ form a basis for $B^k$ for natural numbers $k$.  Or if you prefer a more concrete approach use the vec operator which stacks one column on top of the other to get a 'big coordinate vector' with $n^2$ components and consider $\Big\{\text{vec}\big(I\big),\text{vec}\big(B\big)\Big\}$ and  $\text{vec}\big(B^k\big)$ may be written as a linear combination of those two independent vectors.

either way you approach it, let  $v_k$ be the vector associated with the kth power of $B$.  (Note: the zero'th power of $B$ is the identity matrix).

from here use the Companion matrix to model the recurrence

$C := \begin{bmatrix}
0 & -a_o\\
1 & -a_{1}
\end{bmatrix} = \begin{bmatrix}
0 & -\frac{1}{h}\\
1 & 0
\end{bmatrix}$

and working through the recurrence (with $\mathbf e_1$ being the $2\times 2$ first standard basis vector), for $k=0,1,2,3,...$

$\bigg[\begin{array}{c|c}
v_0 & v_1
\end{array}\bigg]\mathbf C^k\mathbf e_1 = \bigg[\begin{array}{c|c}
v_k & v_{k+1}
\end{array}\bigg]\mathbf e_1 = v_k$

Thus

$$
\begin{align}
&e^B\\
&=\sum_{k=0}^\infty\frac{1}{k!} B^k\\
&=\sum_{k=0}^\infty\frac{1}{k!}\big( v_k\big)\\
&=\sum_{k=0}^\infty\frac{1}{k!}\Big( \bigg[\begin{array}{c|c}
v_0 & v_1
\end{array}\bigg] C^k\mathbf e_1 \Big)\\
&=\bigg[\begin{array}{c|c}
v_0 & v_1
\end{array}\bigg]\Big(\sum_{k=0}^\infty\frac{1}{k!} C^k \Big)\mathbf e_1\\
&=\bigg[\begin{array}{c|c}
v_0 & v_1
\end{array}\bigg]\big(e^C\big)\mathbf e_1\\
&=\bigg[\begin{array}{c|c}
v_0 & v_1
\end{array}\bigg]\big(e^C\mathbf e_1\big)\\
\end{align}    $$

So you want to compute $e^C$ then 'grab' its first column.  There are various ways to do this.  As mentioned in the comments, there's a closed form approach for the matrix exponential of a $2\times 2$ matrix.  Alternatively you can directly diagonalize $C$ which is easy since you know it has distinct eigenvalues $\big\{\lambda_1,\lambda_2\big\}$ and the Vandermonde matrix diagonalizes it.  You can also exploit the known form of the solution. But the point is you have an $2n\times 2n$ matrix that is diagonalizable and has a degree 2 minimal polynomial.  So rather than directly computing the matrix exponential of a $2n\times 2n$ matrix, the problem reduces to computing the matrix exponential of a nice $2\times 2$ matrix.

Title: Finding a kind of matrix


问:
if $x_1,x_2,\dots,x_8\sim N(0,52)$, then what is the distribution of $\frac{\bar{x}}{\sqrt{52/8}}$
I am trying to find what the distribution for $x_1$, $x_2,\dots, x_8$ $\sim$ $N(0,52)$ is. This is a normal distribution question and I'm not adding these probabilities. $x_1$, $x_2,\dots, x_8$ means there are 8 i.i.d variables.

A $N(0, \frac{\bar{x}}{\sqrt{52/8}})$

B $N(0, \frac{1}{\sqrt{52/8}})$

C $N(0,1)$

D $(\bar{x},1)$

E ${\rm Unif}[0,1]$

I am not sure, but I think the answer is A because that has the mean and standard deviation of the sample.

答:
The answer is not a) Since this seems a lot like homework, I’ll just say that you need to consider what distribution does the sum of normally distributed variables follow. After that, it is just simple algebra with the properties of variance.

Sum of normals: https://en.m.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables


问:
How can I solve $\lim_{x\to a}\; \frac{\cot(x)-\cot(a)}{x-a} $ without using L'Hopital's rule?
Through some steps I got to this:

$$\lim_{x\to a}\; \frac{\frac{\cos(x)\sin(a)-\cos(a)\sin(x)}{x-a}}{\sin(x)\sin(a)}$$

The denominator is fine because if I place in $a$, I get $\sin^2(a)$. But the numerator should be $-1$ which I can't calculate without using the rule. I can't continue from here.

Thanks in advance

答:
$$\lim_{x\to a}\frac{\cot x-\cot a}{x-a}=(\cot x)'|_{x=a}=-\frac1{\sin^2 a}=-\csc^2(a)$$

Another way:

$$\frac{\cot x-\cot a}{x-a}=\frac{\frac{\cos x}{\sin x}-\frac{\cos a}{\sin a}}{x-a}=\frac{\sin a\cos x-\sin x\cos a}{(x-a)\sin x\sin a}=$$

$$\stackrel{\text{Trig. Identity}}=-\frac{\sin(a-x)}{a-x}\cdot\frac1{\sin x\sin a}\xrightarrow[x\to a]{}\stackrel{\text{known limit+continuity}}(-1)\cdot\frac1{\sin^2a}=-\csc^2(a)$$


答:
Let's define $B = \Gamma^T A \Gamma$ and take the notation:
$$
\begin{aligned}
\Gamma_{11} = \Gamma_1^TA\Gamma_1 \\
\Gamma_{12} = \Gamma_1^TA\Gamma_2 \\
\Gamma_{21} = \Gamma_1 A \Gamma_2^T \\
\Gamma_{22} = \Gamma_2^T A\Gamma_2
\end{aligned}
$$
Let's denote the Shur complement of $\Gamma_{11}$ as $B/ \Gamma_{11} =  \Gamma_{2}^{T} A \Gamma_{2}-\Gamma_{2}^{T} A \Gamma_{1}\left(\Gamma_{1}^{T} A \Gamma_{1}\right)^{-1} \Gamma_{1}^{T} A \Gamma_{2}$ which is the same as $B / \Gamma_{11} = \Gamma_{22} - \Gamma_{21} \Gamma_{11}^{-1} \Gamma_{12}$. Using this notation we get:
$$
\begin{aligned}
&\det(A) = \det(B) = \det(\Gamma_{11})\det(B / \Gamma_{11})  \\
&\det(\Gamma_1^T A \Gamma_1) = \det(A)\det((B / \Gamma_{11})^{-1})
\end{aligned}
$$
We then compute $B^{-1}$:
$$ B^{-1} =
\left(
\begin{matrix}
\Gamma_{11}^{-1} + \Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\Gamma_{21}\Gamma_{11}^{-1} & -\Gamma_{11}^{-1} \Gamma_{12}(B / \Gamma_{11})^{-1}\\
-(B / \Gamma_{11})^{-1} \Gamma_{21} \Gamma_{11}^{-1} & (B / \Gamma_{11} )^{-1}
\end{matrix}
\right)
$$

We also have

$$
B^{-1} = \Gamma^T A^{-1} \Gamma =
\left (
\begin{matrix}
\Gamma_1^T A^{-1} \Gamma_1 & \Gamma_1^T A^{-1} \Gamma_2 \\
\Gamma_1 A^{-1} \Gamma_2^T & \Gamma_2^T A^{-1} \Gamma_2
\end{matrix}
\right )
$$

By checking the dimensions we get:
$$
\Gamma_2^T A^{-1} \Gamma_2 = (B / \Gamma_{11} )^{-1}
$$

From where we get $\det(\Gamma_1^T A \Gamma_1) = \det(A)\det(\Gamma_2^T A^{-1} \Gamma_2 )$.

Title: Matrix Determinant Formula

Body:
Problem. Let $A$ be a non-singular $n\times n$ matrix and let $\Gamma=[\Gamma_1\quad\Gamma_2]$ be an $n\times n$ orthogonal matrix where $\Gamma_1$ is $n\times n_1$, $\Gamma_2$ is $n\times n_2$ and $n=n_1+n_2$. Show that $$\det(\Gamma_1^TA\Gamma_1)=\det(A)\det(\Gamma_2^TA^{-1}\Gamma_2).$$

My Attempts. Here we make use of the property of orthogonal matrix:
\begin{align}
\det(A)=\det(\Gamma^TA\Gamma)=\det\left(\begin{bmatrix}
\Gamma_1^T \\ \Gamma_2^T
\end{bmatrix}A\begin{bmatrix}
\Gamma_1 & \Gamma_2
\end{bmatrix}\right)=\det\left(\begin{bmatrix}
\Gamma_1^TA\Gamma_1 & \Gamma_1^TA\Gamma_2 \\ \Gamma_2^TA\Gamma_1 & \Gamma_2^TA\Gamma_2
\end{bmatrix}\right).
\end{align}
Since $A$ is non-singular, $\Gamma_1^TA\Gamma_1$ is also non-singular. Thus,
\begin{align}
\det(A)=\det(\Gamma_1^TA\Gamma_1)\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right).
\end{align}
If the formula we want to prove is true, we would have
\begin{align}
1&=\det(\Gamma_2^TA^{-1}\Gamma_2)\cdot\det\left(\Gamma_2^TA\Gamma_2-\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right) \\
&=\det\left(\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_2-\Gamma_2^TA^{-1}\Gamma_2\Gamma_2^TA\Gamma_1(\Gamma_1^TA\Gamma_1)^{-1}\Gamma_1^TA\Gamma_2\right).
\end{align}
Nonetheless, I have no idea how to simplify the terms in the parenthesis because I only have $\Gamma_1\Gamma_1^T+\Gamma_2\Gamma_2^T=I$. Hope anyone has good suggestions.


问:
Heat loss of a cube insulated on four sides.
I'm looking for a way to make sense of this. A cube, perfectly insulated on four sides is heated to an initial temperature of 100C. The two non-insulated sides will be the top and the bottom of the cube. We place the cube on a plate which maintains a temperature of 0C. The loss of heat through the top is given, and we are left to solve a pretty standard PDE.

Here's the issue. I'm saying the insulated sides correspond to the BCs of the $X(x)$ and $Y(y)$ functions after separation. When I go to compute the Fourier coefficients I end up with 0! The entire problem collapses. I've never encountered a problem where this happens... Am I just supposed to say the eigenvalues must be 0? I'm not sure if this physically makes sense, but would this just be equivalent to an infinite number of thin rods, implying that solution is independent of $x$ and $y$ values?

答:
Assume a unit cube $-1/2 < x < 1/2$, $-1/2 < y < -1/2$ and $0 \le z \le 1$. The heat equation is
$$
              u_t = k^2\nabla^2u
$$
The bottom is at held at $T=0$:
$$
                   u(t,x,y,0)=0.
$$
The heat dissipation on the top surface is specified:
$$
             u_{z}(t,x,y,1)=C(x,y).
$$
The four sides are insulated:
$$
           u_x(t,-1/2,y,z)=u_{x}(t,1/2,y,z)=0 \\
           u_y(t,x,-1/2,z)=u_{y}(t,x,1/2,z)=0
$$
I don't believe this has a $0$ solution.


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

答:
No, this is not the case: we can have an automorphism of a group which is not an inner automorphism but which nonetheless fixes the center pointwise - even when the center is nontrivial.

(As Mark Bennet observes, by contrast every automorphism whatsoever fixes the center in the weak sense that it restricts to a self-bijection of the center, but pointwise-fixing is more complicated. In particular, any nontrivial automorphism of an abelian group does not fix the center - which is after all the whole group - pointwise.)

For example, let $G$ be a group with trivial center which has some automorphism $\alpha$ which is not inner, and let $A$ be any nontrivial abelian group. Consider the group $G\times A$. The center of $G\times A$ is exactly the image of the map $A\rightarrow G\times A: a\mapsto (e_G, a)$ and so is nontrivial, and the automorphism defined by $$\hat{\alpha}: G\times A\rightarrow G\times A: (g,a)\mapsto (\alpha(g),a)$$ fixes that center. However, $\hat{\alpha}$ is not an inner automorphism of $G\times A$ since $\alpha$ wasn't an inner automorphism of $G$.


问:
Transportation-information inequalities (concentration inequalities)
The following is one example of a transportation-information inequality,  which show connections between optimal transport theory and information theory:

$$W_1(\nu,\mu) \leq \left[ 2\sigma^2 D_{KL}(\nu\Vert\mu) \right]^\frac{1}{2} $$

$W_1$ is the Wasserstein distance found in optimal transport theory, and $D_{KL}$ is the Kullback-Leibler (KL) divergence found in information theory. (Source)

What other transportation-information inequalities are out there?

答:
From this answer regarding books on Concentration of Measure, the following textbook gave good coverage of many transportation-information linkages, having full chapters on both the entropy approach and the transportation approach.

Concentration Inequalities: A Nonasymptotic Theory of Independence by
Stéphane Boucheron, Gábor Lugosi, and Pascal Massart, 2013.

although I could see it to be inaccessible to those new to the field since alot of the theoretical derivations of the inequalities stop short of providing the completed forms of information theoretic measures we are more familiar with. There is also a lack of actual applications, given that concentration of measure is a construct of independent variables, so other recommended textbooks on applied transportation-information would be helpful.


问:
Can power set of any set equals to $\phi$
In my topology quiz this question was asked and I couldn't do it right. So, I am posting it here.

Let A be any set . Let P(A) be the power set of A, that is the set of all subsets of A, Then how is the statement" P(A)=$\phi$ for some A" a false statement?

I think if A =$\phi$ then P(A) is also $\phi$ but in answer key it is marked as false ?

Why?

答: Assume $\exists A$ such that $\mathcal P(A)=\emptyset$. This means that there is no $B$ such that $B\subseteq A.$ But we know that $A\subseteq A$, hence there does exist a $B(=A)$ which is a contradiction. Therefore no $A$ exists such that $\mathcal P(A)=\emptyset$.


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

$2x+y=1$

$x-y=0$

I can represent these as:

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

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

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

答:
In this problem, if you represent

$2x+y=1$

$x-y=0$ as

$-y+x=0$

$y+2x=1$

then you get the matrix

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

So in general what you want to do is rearrange the variables in whichever order you prefer, then rearrange the order of the equations so that the equation with a 1 is in the position of x in the rearranged variables.

As an example if you had

$2x+y=1$

$x-y=0$

$x+y+z=0$

I can represent these as:

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

If you want to rearrange the variables so that it $xyz$ goes to $zyx$ then you would consider the equations as

$y+2x=1$

$-y+x=0$

$z+y+x=0$

To get the $x$ next to the 1, since for $zyx$, $x$ is in the third position, you want the equation with a 1 to be in the third position also, so

$-y+x=0$

$z+y+x=0$

$y+2x=1$ to get the matrix

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

Title: Rearranging elements in a matrix


问:
Non-separable compact space
Off the top of my head, I can't think of a non-separable compact space.

Can you provide a good example?

答:
Based on this question I asked recently, the one-point compactification of a non-separable metric space.

I specifically asked about the space of cadlag real-valued functions on $[0,1]$ under the supremum metric, but there are lots of other non-separable metric spaces out there, and other compactifications where the same argument applies.

The metric space is an open subset of its one-point compactification, and is not separable, so the one-point compactification is also not separable.

Title: Non-separable compact space