问:
Shortcut solution for $f''(t)+c^2f(t)=0$?
I am trying to prove that the general solution to $f''(t)+c^2f(t)=0$ is $f(t) = acos( ct) + bsin (ct)$

Could you please tell me if it is possible to solve it this way?
$$
\frac{d}{dt}(\frac{df}{dt})=-c^2f(t)
$$

$$
df* \frac{d}{dt}(\frac{df}{dt})=-c^2f(t)*df
$$
$$
d(\frac{df}{dt})^2=-c^2 f(t)df
$$
If not for the constant C resulting from integration, this would lead to the desired solution, but I think I am not doing it right somehow.

Thanks in advance for any suggestions.

答:
HINT

Here it is another way to approach it
\begin{align*}
y'' + c^{2}y = 0 & \Longleftrightarrow y''y' + c^{2}yy' = 0\\\\
& \Longleftrightarrow  (y')^{2} + c^{2}y^{2} = k^{2}\\\\
& \Longleftrightarrow y' = \pm\sqrt{k^{2} - c^{2}y^{2}}\\\\
& \Longleftrightarrow \frac{y'}{\sqrt{k^{2} - c^{2}y^{2}}} = \pm1\\\\
\end{align*}

Can you take it from here?


问:
Understanding Slope Better
Recently, I have realized how much I have taken for granted in understanding the slope of a line and the slope of a curve in general. With that said, I wanted to clear up my understanding to make sure I am on the right track.

After reading this question: "Understanding the slope of a line as a rate of change", I see now how the slope of the line $y = f(x): = mx + b$ (where $f: \mathbb{R} \to \mathbb{R}$) is the rate of change of $y$ with respect to $x$. However, where I get a little confused is this: since the value $y$ of the function changes by an amount $\Delta y$ for every $\Delta  x$ change in $x$, therefore $y$ changes by an amount $ m =\frac{\Delta y}{\Delta x}$. So, if we were to study the line $y = \frac{2}{3}x$, does that mean the slope of the line can be interpreted as "two units of $y$ every three units of $x$"? With more relatable units, if $y$ is measured in meters and $x$ is measured in seconds, would the rate of change of a particle traveling along this line be read as "two meters every three seconds"?

This seems right, however in general for any differentiable curve $g: \mathbb{R} \to \mathbb{R}$ that may not be a line i've seen for example that $g'(3) = \frac{2}{5} $ (given in meters/seconds) is read as "two-fifths meters per second"; simply because the derivative is the rate of change at a point (or at an instant).

Is my understanding correct at all?

答:
The derivative at a point  $x_0\in\mathbb{R}$ of a differentiable function $f:\mathbb{R}\to\mathbb{R}$ is a local linearization of the function. The derivative $f'(x_0)$ is associated with  a linear transformation:
$$
h\mapsto f'(x_0)h\tag{1}
$$

Depending on whether one is talking about the derivative $f'(x_0)$, for the linear transformation (1), one has all sorts of "interpretations".

Geometrically, the graph of the linear function (1) is the tangent line to the graph of the function $f$. The "slope" $m=f'(x_0)$ of the tangent line $y=f'(x_0)(x-x_0)$ says that whenever $x$ increases $h$ units at $x_0$, i.e., from $x_0$ to $x_0+h$, the dependent variable $y$ increases $mh$ units.

If one interprets the function $f$ as a one-dimension motion of a particle (so that the variable $x$ is interpreted as time), then the sign of $m$ tells which direction ($+1$ for right and $-1$ for left) the particle is traveling, and the absolute value $|m|$ says how fast the particle travels at the moment of $x=x_0$.

ContentLicense: CC BY-SA 4.0

Title: Understanding Slope Better


问:
Resources, references, or examples for logics with finitely many sentences
Are there any interesting logics with only finitely many sentences total? Is there a reason why potential logics with this property would be trivial or badly behaved?

I'm looking for a reference to read more or an example of such a logic that's interesting as a mathematical object.

I was recently looking at some examples of interesting logics that are fragments of better-known logics, such as two-variable logic and a list of intermediate logics that are interesting logics between intuitionistic propositional logic and classical propositional logic.

One thing that I haven't come across yet, though, is a modern logic (possibly described as a fragment of another logic) with structural constraints on well-formed formulas that are so strong that only finitely many sentences are possible or only finitely many sentences up to renaming of variables are possible. Traditional term logic, even with the addition of non for modifying predicates, has finitely many sentences up to renaming of variables.

I can think of a few weird properties of such finite-number-of-sentences logics off the top of my head like the fact that ordinary introduction rules like $A, B \vdash A \land B$ will sometimes fail if the conclusion is too big to be well-formed.

NOTE: removed reference to $\alpha$-equivalence, replaced with up to renaming of variables which is more clear.

答:
All automated (computer based) theorem provers involve some logic with a finite number of sentences.

I would consider some of those automated theorem provers of interest, since they have gotten used to solve open problems in mathematics and logic, such as when the Robbins problem got solved.  Some others might not have solved any open problem, but have the ability to solve open problems with human assistance (or one might think of the computer as the human's assistant).  Also, model finders with only a finite number of possible sentences, I would think of interest, since they save people time, as computers work faster than humans in some situations.


答:
$$f''(t)+c^2f(t)=0$$
Maybe you wanted to do this ?
$$\dfrac {df'}{dt}=-c^2 f(t)$$
$$\dfrac {df'}{df}\dfrac {df}{dt}=-c^2 f(t)$$
$${df'}\dfrac {df}{dt}=-c^2 f(t)df(t)$$
$$f'{df'}=-c^2 f(t)df(t)$$
Now you can integrate.
$$f'^2=-c^2 f^2+k$$
This is separable.

Title: Shortcut solution for $f''(t)+c^2f(t)=0$?

Body:
I am trying to prove that the general solution to $f''(t)+c^2f(t)=0$ is $f(t) = acos( ct) + bsin (ct)$

Could you please tell me if it is possible to solve it this way?
$$
\frac{d}{dt}(\frac{df}{dt})=-c^2f(t)
$$

$$
df* \frac{d}{dt}(\frac{df}{dt})=-c^2f(t)*df
$$
$$
d(\frac{df}{dt})^2=-c^2 f(t)df
$$
If not for the constant C resulting from integration, this would lead to the desired solution, but I think I am not doing it right somehow.

Thanks in advance for any suggestions.


问:
Find the determinant of this matrix
$k_1,k_2,...,k_n$ are non-negative integers.
Let $M$ be an $n\times n$ matrix with entries:
$a_{i,1} = t^{k_i}, a_{i,j+1} = \frac{da_{i,j}}{dt}$ where $a_{i,j}$ is the element of $i$-th row and $j$-th column.

Prove that there exist $C$ and $r$ such that
$\det(M) = Ct^r$

I was able to reduce it to:

$\det(M) = t^{(k_1+...+k_n) - {n(n-1)\over 2}}$
$\cdot \det\begin{pmatrix}
 1& k_{1} & k_{1}(k_{1}-1)  & ... & {k_1!\over (k_1-n+1)!}\\
 1& k_{2} & k_{2}(k_{2}-1) & \vdots & \vdots\\
 \vdots& \vdots & \vdots &  \ddots \\
 1& k_{n} & k_{3} & ... & {k_n!\over (k_n-n+1)!}
\end{pmatrix}$

I have  already determined $r$, the problem is to calculate  determinant of this thing, any tips ?

答:
$\displaystyle  C=\pm \prod_{1\le i<j\le n} (k_j-k_i)$ .

I hope you know the Vandermonde Determinant and how to compute it.
If $k_i=k_j$ for some $i\neq j$  or $max\{k_1,k_2,...,k_n\}\le n-2$ then $det(M)=0$.

Title: Find the determinant of this matrix


问:
How to resize a rectangle to a specific area while maintaining the aspect ratio
Let's just say I have a value like 46,080.  I want to resize a rectangular image to have this area while maintaining its aspect ratio. What are some ways I can do that?

Here is an example.

```<code>Desired Area = 46,080

image width = 231
image height = 228
image area = 52,668
</code>```

Decrease the dimensions of the image so it has an area of 46,080 while keeping its aspect ratio.

答:
Let the desired area be $A$, and let the current dimensions be $x$ and $y$. Then, the current area is $xy$, so we must scale the area by a factor of $\displaystyle\frac{A}{xy}$.

Now, because are is proportional to the square of side length, side length is proportional to the square root of area, so we must scale $x$ and $y$ by $\displaystyle\sqrt{\frac{A}{xy}}$.

Thus, the new dimensions are $\displaystyle x\cdot \sqrt{\frac{A}{xy}}\text{ by }y\cdot\sqrt{\frac{A}{xy}}$, equivalent to $\boxed{\sqrt{\frac{Ax}{y}}\text{ by }\sqrt{\frac{Ay}{x}}.}$

For your case, just plug in $A = 46080, x = 231, y = 228$. You may want to round off the values you find.


问:
System of differential equations: Inverse matrix of a fundamental matrix
I'm trying to show:

Let $A:[0,\infty[\to \mathcal{M}(n,\mathbb{R})$ a function and suppose that all solutions of the system of differential equations:
$$\vec{x'}(t)=A(t)\vec{x}(t) \ \ \ (\star)$$
are bounded when $t\geq 0$. If $X(t)$ is a fundamental matrix (a matrix of fundamental solutions) of the system $(\star)$ show that $X^{-1}(t)$ is bounded for $t\geq 0$ if and only if the function $t\to \int^t_0 \operatorname{tr}A(s)ds$ bounded below.

I know that for the system $(\star)$, a fundamental matrix is $e^{tA}=e^{tPJP^{-1}}=Pe^{tJ}P^{-1}$ where $J$ is the Jordan form of $A$, but I have no idea how to prove the exercise.

Thanks for your help.

答:
The determinant $f(t)=\mathrm{det}\,X$ of the fundamental matrix satisfies the equation
$$(\ln f)'=(\ln\mathrm{det}\,X)'=\left(\mathrm{Tr}\ln X\right)'=\mathrm{Tr}\left(X'X^{-1}\right)=\mathrm{Tr}\,A,$$
so that $\displaystyle f(t)=\mathrm{const}\cdot\exp\int^t\mathrm{Tr}\,A(t)\,dt$. The statement for $X^{-1}$ follows from this relation and boundedness of $X$.</p>

ContentLicense: CC BY-SA 3.0


问:
Linear transformations with respect to a basis
Problem: I have a question which has a linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $T(x,y,z)=(0,x,y)$. With an ordered basis $F = \{(1,1,1)^T, (1,1,0)^T, (0,1,1)^T\}$ and $E$ the standard basis.

Part of the solution says that $_{F}T_{E}=\begin{pmatrix}0 & 0 & 0\\1 & 1 &0 \\1 & 1 & 1\end{pmatrix}$.

Attempt: However I thought that we know the matrix which takes the standard basis vectors to the their mapping under $T$ is $_{E}T_{E}=\begin{pmatrix}0 & 0 & 0\\1 & 0 &0 \\0 & 1 & 0\end{pmatrix}$. So the $F$ basis vectors must go to $_{E}T_{E}(1,1,1)^T= (0, 1, 1)^T,\hspace{2mm}_{E}T_{E}(1,1,0)^T= (0, 1, 1)^T,\hspace{2mm}_{E}T_{E}(0,1,1)^T= (0, 0, 1)^T$ which are the images represented in the $E$ basis, of the mappings under $T$. So the matrix which takes the $F$ basis vectors w.r.t. $F$ to the image of the $F$ basis vectors w.r.t. $E$ under the transformation (i.e. $_{E} T _{F}$) would be $_{E}T_{F}=\begin{pmatrix}0 & 0 & 0\\1 & 1 &0 \\1 & 1 & 1\end{pmatrix}$.

Why is what I've done the opposite of the solution?

答:
Consider the coordinate mappings:

$X_F:\mathbb{R}^3\to\mathbb{R}^3$
(sends a vector to its coordinates in the basis F)
and
$X_E:\mathbb{R}^3\to\mathbb{R}^3$
(sends a vector to its coordinates in the basis E).

We note that (since E is the elementary basis and our vector space is just $\mathbb{R}^3$ already) $X_E^{-1}$ is just the identity map $1_{\mathbb{R}^3}$. Write E and F respectively as follows:
$\{E^1,E^2,E^3\}$.
and
$\{F^1,F^2,F^3\}$.

Note that multiplying a 3x3 matrix $M\in M_3(\mathbb{R})$ by $E^j$ will just give us the $j^{th}$ column of M. So we compute $_ET_F$ as follows:

$_ET_FE^j = X_F(T(X_E^{-1}(E^j))) = X_F(T(E^j))$.

Hence:

$_ET_FE^1 = X_F(T(E^1)) = X_F((0,1,0)^T) = (1,1,0)^T$
$_ET_FE^2 = X_F(T(E^2)) = X_F((0,0,1)^T) = (0,1,1)^T$
$_ET_FE^3 = X_F(T(E^3)) = X_F((0,0,0)^T) = X_F(0) = (0,0,0)^T$
so that

$_ET_F = \begin{pmatrix} 1 &0 &0\\ 1 &1 &0\\ 0 &1 &0\end{pmatrix}$.


问:
What is the difference between a partition and a subinterval?
For example the norm of a partition is the widest subinterval</p>

答:
Here is a formal definition of partitions of intervals:

In mathematics, a partition of an interval $[a, b]$ on the real line is a finite sequence $x_0, x_1, x_2, ..., x_n$ of real numbers such that
$$
a = x_0 < x_1 < x_2 < ... < x_n = b.
$$

Such partition gives you $n$ "subintervals" of $[a,b]$,

$$
[x_{k-1},x_{k}],\quad k=1,\cdots,n\tag{1}
$$

The norm (or mesh) of the partition
$$
x_0 < x_1 < x_2 < ... < x_n
$$
is the length of the longest of the subintervals in (1):
$$
\max\{|x_i − x_{i−1}| : i = 1, ... , n \}.
$$

To see a simple example for the interval $[0,10]$,
take the partition:
$$
0<1<6<8.5<10
$$
This partition gives you four subintervels:
$$
[0,1],\quad [1,6],\quad [6,8.5],\quad [8.5,10]
$$
and the norm of this partition is $5$.


问:
Solution to Cauchy problem via method of characteristics
I need to solve the following Cauchy problem:
$$\left\{\begin{matrix}xu_x-yu_y = u\\
u(x,0) = h(x)\end{matrix}\right.$$

What we know so far is the method of characteristics, so here we go:

Define $X(s,t) = (x(s,t),y(s,t),z(s,t))$, we want then to solve the problem
$$\left\{\begin{matrix}\frac{dX(s,t)}{dt} &= (x(s,t),-y(s,t),z(s,t))\\
X(s,0) &= (s,0,h(s))\end{matrix}\right.$$

if the functions $x(s,t),y(s,t)$ are invertible, we will find the solution $u(x,y) = z(s(x,y),t(x,y))$ which satisfies our problem. We then have:

$\frac{dx(s,t)}{dt} = x(s,t) \implies x(s,t) = e^tk_1(s)\, \mbox{ given that }x(s,0) = s \implies x(s,t) = e^ts$
$\frac{dy(s,t)}{dt} = -y(s,t) \implies y(s,t) = e^{-t}k_2(s)\, \mbox{ given that }y(s,0) = 0 \implies y(s,t) = 0\, \forall s,t$${\color{red}{\mbox{  Here's the problem}}}$
$\frac{dz(s,t)}{dt} = z(s,t) \implies z(s,t) = e^tk_3(s)\, \mbox{ given that }z(s,0) = h(s) \implies z(s,t) = e^th(s)$

As the bondary values are given with $y=0$, the field vector defined by this PDE has component $0$ in the $y$ direction, so any solutions passing by the hiperplane $y=0$ must stay there. So I'm actually solving the ODE

$$\left\{\begin{matrix}xu_x = u\\
u(x) = h(x)\end{matrix}\right.$$
which we can easily solve by $\frac{u_x}{u} = \frac{1}{x} \implies \ln(u(x)) = \ln(x)+C \implies u(x) = C_2x.$

Therefore our problem only has solutions if the function $h(x) = Kx$ for some constant $K$, and that solution would be $u(x,y) = Kx \;\forall y$.

Is this correct?

答:
I would say the problem is ill-posed.  The first difficulty is you didn't say where the equation holds.  Is $x u_{x} - yu_{y} = u$ supposed to hold in $\{(x,y) \in \mathbb{R}^{2} \, \mid \, y > 0\}$?  Where is the "action" taking place?

The real difficulty is the boundary $y = 0$ is "characteristic."  As you noticed, any characteristic that starts on this line stays there for all time.  Hence the information about $u$ on the line doesn't propagate anywhere --- so it tells me nothing about $u$ elsewhere in the domain.  (Hence part of the reason why you should specify where the equation holds...)

It could be an interesting exercise to try to find a function $h$ for which there's more than one solution to the PDE.  However, again, you need to specify where the equation holds.


问:
logic question inferences
Here are two inferences A and B, one of which is valid, but the other not valid.

Inference A

Every human is mortal.

Socrates is human.

Therefore Socrates is mortal.

Inference B

Every hero likes women.

Taro likes women.

Therefore Taro is a hero.

Which one is correct, and in what sense it is correct ?

Can somebody give the hints to solve it ?

答:
Draw a circle. Draw another circle, inside the first. Draw a point inside one of the circles and see if being inside one the circles implies being inside the other.

The difference is like between necessary and sufficient conditions. Being a human is sufficient to be mortal. Being mortal is necessary but not sufficient to be human. Similarly, being a lover is necessary for being a hero but not sufficient.

In probability, this fallacy is know as Confusion of inverse

Title: logic question inferences


问:
How do I find the tangent line for these two points?
I need to find the slope of the tangent line to the circle, but I forgot how to find the tangent line for each of the points (0.8, -0.6) and (0.8, 0.6)

答:
You can find the slope directly with the derivative like Joshua suggests.

Another way is to note that the tangent is perpendicular to the radius:

Find the center of the circle.

Calculate the slope of the radius to each point.

From these, calculate the slope of the perpendicular.

Can you follow this and take it home from here?


问:
Which number is different from the others? $526, 793, 382, 764, 329, 894, 932, 397, 478, 489, 652$
So, in a circle we have numbers $526, 793, 382, 764, 329, 894, 932, 397, 478, 489, 652$ (arranged like in the photo). The question is to find one number which is different from others.

We are given 5 options to be choose from which are :"$526, 329, 382, 793, 932$"

I dunno if the arrangement have something to do with the number or not. For the pattern I have found so far is that some numbers is the same with different arrangement: $(526, 652), (793, 397), (329, 932) , (894, 489)$. That leaves $764, 478, 382$ and $(764, 478)$ have two numbers which are the same so that leaves $382$ so is $382$ the answer of the problem?

答: That 764 looks more like a 784 -- compare what the other sixes look like. And 784 is a rearrangement of 478. So there is even more evidence that 382 is the 'odd' man out.</p>


答:
This can be easily solved using implicit differentiation.

The equation of the unit circle is $x^2+y^2=1$; take the derivative of both sides with respect to $x$, we get $\frac{\text{d}}{\text{d}x}(x^2+y^2)=\frac{\text{d}}{\text{d}x}1$, which gives $2x+2y\frac{\text{d}y}{\text{d}x}=0$ from chain rule. Rearranging gives $\frac{\text{d}y}{\text{d}x}=-\frac x y$. Substitute in those coordinates, we get $m_1=-\frac{4}{3}, m_2=\frac{4}{3}$.

However, since you posted this question in the geometry tag, I will provide an alternative approach, using only coordinate geometry.

The equation of the line passing through the origin $O(0,0)$ and a point on the circle $A(x,y)$ has a slope $\frac{y}{x}$. From the property of circles, we know that the tangent is perpendicular to the radius, which means the product of their slopes is $-1$. Suppose the slope of the tangent is $m$, we get $m\frac{y}{x}=-1\implies m=-\frac x y$. The result is the same as in the calculus method.

Title: How do I find the tangent line for these two points?

Body: I need to find the slope of the tangent line to the circle, but I forgot how to find the tangent line for each of the points (0.8, -0.6) and (0.8, 0.6)


Title: How to resize a rectangle to a specific area while maintaining the aspect ratio

Body:
Let's just say I have a value like 46,080.  I want to resize a rectangular image to have this area while maintaining its aspect ratio. What are some ways I can do that?

Here is an example.

```<code>Desired Area = 46,080

image width = 231
image height = 228
image area = 52,668
</code>```

Decrease the dimensions of the image so it has an area of 46,080 while keeping its aspect ratio.

Body:
The aspect ratio  $$\frac{231}{228}=1.013157895$$
The area ratio $$\frac{52668}{46080}=0.874914559$$
The area is gets a percentage from both sides so the contribution would be the square root of the area ratio.
$$\sqrt{0.874914559}=0.935368676$$
$$231\times 0.935368676=216.070164\\
228\times 0.935368676=213.264058$$

$$\frac{216.070164}{213.264058}=1.013157895$$

$$216.070164\times 213.264058$=46080$$


答:
As far as "shortcuts" go: since this is a 2nd order autonomous equation, its solution set has dimension $2$. So if you can demonstrate two solutions, one not a scalar multiple of the other, then it's automatic that all solutions are linear combinations of those two.

The nature of this equation screams $\sin(cx)$ and $\cos(cx)$ if you are familiar with what happens during differentiation of trig functions and the chain rule.

So that would be one shortcut here to reach the solution set.


问:
About the covering spaces of comb space.
I am working in this exercise:

Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the four sides
  of the square $[0,1] \times [0,1]$ together with the segments of the
  vertical lines $x = 1/2, 1/3, 1/4,\dots$ inside the square.  Show that
  for every covering space $\tilde{X} \to X$ there is some neighborhood
  of the left edge of X that lifts homeomorphically to $\tilde{X}$.
  Deduce that $X$ as no simply-connected covering space.

I just don't know why the argument here and here about pasting the homeomorphisms works.

The argument goes like this. We start by taking a finite cover of the left edge composed of evenly covered open sets. Next we take, from the open cover, an evenly covered neighborhood $U_1$ of $0$ and some $V_1$ in the inverse image such that $V_1\cong U_1$ by the covering map $p$.

Now we take a second open set $U_2$ in the cover which intersects both $U_1$ and its complement. We take some point $e_1$ in $V_1$ such that $p(e_1)\in U_1\cap U_2$ and then, observing that
$$p^{-1}(U_2)=\bigsqcup_{i\in I}W_i, W_i\cong_p U_2$$
we take $V_2=W_i$ such that the image of $e_1$ is in $U_1\cap U_2$.

Then $p|_{V_j},j=1,2$ is a homeomorphism, which can be glued to a surjective continuous function $p:V_1\cup V_2\to U_1\cup U_2$. To see it's an homeomorphism it's enough to show it's 1-1.

If I suppose $p(x)=p(y)$ with $x\in U_1,y\in U_2$ then $p(x)\in V_1\cap V_2$. But I don't see how to arrive to a contradiction here, or to $x=y$.

答:
I was once stuck here, same as you (were). Maybe it's about the perspective. We do say "evenly covered neighborhood of $x$", but in fact this property of "being evenly covered" is not subject to the choice of $x$.

Following your notation, $A:=U_1\cap U_2$ is evenly covered, being the intersection of two evenly covered open sets. Now that we have $V_1\subset \tilde{X}$, a homeomorphic copy of $U_1$, there is some $\tilde{A}\subset V_1$ such that $p|_{\tilde{A}}:\tilde{A}\to A$ is a homeomorphism. Then there is a homeomorphic copy $V_2\subset \tilde{X}$ of $U_2$ that contains $\tilde{A}$, so $A\subset p(V_1\cap V_2)$. Moreover, $p(V_1\cap V_2)\subset p(V_1)\cap p(V_2)=U_1\cap U_2=A$, so $p|_{V_1\cap V_2}$ is a homeomorphism to $A$. This implies we can define the inverse of $p|_{V_1\cup V_2}$ by simply combining $p|_{V_1}^{-1}$ and $p|_{V_2}^{-1}$, which shows $p|_{V_1\cup V_2}$ is a homeomorphism.


问:
Functions with mutually orthogonal gradients at every point
I am not completely sure if the question is too broad, but here goes. If we are given a function $f:\mathbb{R}^2 \to \mathbb{R}$, is there any way to find other non-constant functions $g:\mathbb{R}^2 \to \mathbb{R}$ such that at every point their gradients are orthogonal? In other words, given $f$ can we find a non-constant $g$ such that $\nabla f\cdot \nabla{g}=0$ for every $(x,y) \in \mathbb{R}^2$? For given $f$ I am interested in a criterion if such functions $g$ exist, and if so how would be a way to find them.

As an example, if $f(x,y)=x$, then $g(x,y)=y$ satisfies that property. If $f(x,y)=x^2+y^2$? It would have to be a solution of the PDE $xf_x+yf_y=0$, which unfortunately I don't remember my PDEs now to try to solve.

The broader question is essentially if we can say anything about the following PDE for a given $f$.

$$f_xg_x+f_yg_y=0$$

Assume any nice conditions you may want about $f$, such as $f\in C^{\infty}$ etc. If the topic is too broad for an answer, I would at least appreciate any pointers for a direction to read further about it. This also relates to products of harmonic functions, specifically a product of two harmonic functions is also harmonic if the above condition on the gradients is satisfied.

答:
This is not at all my area but some scattered thoughts. The closest keyword I've been able to find so far is orthogonal trajectories; maybe these are also called orthogonal families of curves.

I think it is possible to understand solutions using the method of characteristics, but I'm not very familiar with it. For $f(x, y) = x^2 + y^2$ the method gives that we want to solve the Lagrange-Charpit equations

$$\frac{dx}{dt} = x, \frac{dy}{dt} = y, \frac{dz}{dt} = 0$$

which gives $x(t) = x(0) e^t, y(t) = y(0) e^t, z(t) = z(0)$. The method of characteristics gives that the graph of $z = g(x, y)$ must be traced out by these curves, so $z(0) = g(x(0) e^t, y(0) e^t)$. So $g$ must be constant on every ray through the origin; in polar coordinates this gives that $g$ must be a function of the angle $\theta$ only. The solution $g(x, y) = \frac{y}{x}$, which can be found by looking for a formal Laurent series solution, is $\tan \theta$. We also have solutions

$$g(x, y) = \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}$$
$$g(x, y) = \cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$$

disregarding their singularities at the origin. Geometrically the point is that $x g_x + y g_y = 0$ says that the gradient of $g$ at any point always points orthogonal to the outward radial direction, or equivalently always points in the $\theta$ direction, which in retrospect is clear. It follows that $\lim_{(x, y) \to (0, 0)} g(x, y)$ does not exist unless $g$ is constant (since if $g$ is nonconstant then it is nonconstant  with the same values on every circle), so no nonconstant solution to $x g_x + y g_y = 0$ is defined on all of $\mathbb{R}^2$.

In general the Lagrange-Charpit equations are

$$\frac{dx}{dt} = f_x, \frac{dy}{dt} = f_y, \frac{dz}{dt} = 0$$

and unfortunately I don't know how to say anything about solutions at this level of generality. $g$ is again determined by the condition that $z(0) = g(x(t), y(t))$, so $(x(t), y(t))$ parameterizes the level curves of $g$.

Here is at least another interesting example. If we take $f(x, y) = xy$ then $\frac{dx}{dt} = y, \frac{dy}{dt} = x$ which has solutions

$$(x, y) = a(\cosh t, \sinh t) + b(\sinh t, \cosh t)$$

for constants $a, b$. These curves trace out hyperbolas $x^2 - y^2 = a^2 - b^2$, so the condition is that $g$ must be constant on all such hyperbolas. So $g$ must be a function of $x^2 - y^2$, and indeed if $g(x, y) = x^2 - y^2$ then $(g_x, g_y) = (2x, -2y)$.

These examples suggest the following general fact, which is that if $(f, g)$ is a pair of functions satisfying $f_x g_x + f_y g_y = 0$ then the same is true of the pair of functions $(h_1(f), h_2(g))$, for any functions $h_1, h_2$, since the gradients at any point are scalar multiples of the original gradients. So the "invariant content" of the solutions really is the families of orthogonal level curves of $f$ and $g$, rather than any particular function with those level curves, unless we ask for more hypotheses such as harmonicity as you mention.


问:
What is the value of $-\log\left(\log\left(\frac{2}{\pi^2}\right)+2\log(\pi)-\log(2)\right)$?
I'm struggling with this one. Assuming the natural log, Mathematica gives the following result:
$$-\log\left(\log\left(\frac{2}{\pi^2}\right)+2\log(\pi)-\log(2)\right)=\infty$$
But working it out myself I see that:
$$\log\left(\frac{2}{\pi^2}\right)+2\log(\pi)-\log(2) = 0$$
Which simplifies to:$$-\log(0)$$
Is this undefined? Is the original equation malformed?

Usually, I can make a mental picture to understand this stuff but I'm struggling here. I sense a fracturing, perhaps akin to what happens in rocks during an earthquake, or when quaternions go rogue in an otherwise decent subroutine.

Curiously, WolframAlpha gives the following transcendental result:
$$-\log\left(\log\left(\frac{2}{\pi^2}\right)+2\log(\pi)-\log(2)\right)=50\log(2)$$

I feel like I'm missing something really obvious but I can't quite sniff it out.

答:
Calculate the middle part first. Apparently $\log(\frac{2}{\pi^2})+2\log(\pi)-\log(2)=\log(2)-\log(2)=0$ from logarithmic identities.

The final result, $-\log(0)$, is undefined since $0$ is a singularity point for the logarithmic function. We may nevertheless interpret the Mathematica result as a limit, as $\displaystyle\lim_{x\rightarrow0}-\log(x)=+\infty$.


问:
How to understand the convention on describing the "position" of mathematical objects
When studying algebra, I find that it seems like a convention that we use positional words like on, under, over etc. to describe relations between things. For example:

$V$ is a vector space over $F$.

$G$ is a group under multiplication.

the operation on $G$ is defined by the usual addition.

Positional words are also used in other situations, for example we say:

Analysis on complex manifolds.

My observation is that such convention prefer to put structures/objects/operations on top of the (in some sense "more basic") things what define them.

To me, they are somewhat natural to say. But as I'm not a naive speaker, they are still pretty vague, for example, why we choose to say vector space over field and operation on set but not the other way around.

So, may I ask for the precise reasons why we "visualize" them as such?

答:
Although each preposition has a core meaning or set of (generally closely related) core meanings, prepositional usage is highly idiomatic, and not just in English: this is true in general. Once you get away from those core meanings, choice of preposition is largely a matter of idiom.

For instance, there is no outstandingly obvious choice of preposition to express the relationship between a group and its operation; the use of under, as in a group under multiplication, is simply idiomatic. It makes some intuitive sense if we think of the operation as something imposed on the underlying set, but one could make cases for other ways of thinking about the relationship.

Some of these idioms are already present in non-mathematical language. For instance, it is entirely idiomatic to speak of operating on or performing operations on things in general; to speak of addition, say, as an operation on the real numbers is just an instance of this existing idiom. In other cases a mathematical prepositional usage may not have clear non-mathematical parallels, and there may be no way to tell how it got started; presumably someone started using it, others picked it up, and it eventually became a standard idiom.


答:
As in many languages, prepositions in English are a complete mess for non-native speakers. Any given preposition can have many different meanings (and the groupings of these meanings varies from language to language).

So my best answer to your question is that, while "over", "under", and "on" can all refer to positional relationships, they don't necessarily refer to positional relationships. Here are a few examples of many different usages of "over", "under", and "on". To me, the following example sentences from those links come closer to how these prepositions are used in mathematics than interpreting them as positional:

"There was a lot of discussion over who should get the job." (connected with)

"Children will educate themselves under [the] right conditions." (an aspect of)

"We lived on an estate." (in a setting)

Title: How to understand the convention on describing the "position" of mathematical objects

Body:
When studying algebra, I find that it seems like a convention that we use positional words like on, under, over etc. to describe relations between things. For example:

$V$ is a vector space over $F$.

$G$ is a group under multiplication.

the operation on $G$ is defined by the usual addition.

Positional words are also used in other situations, for example we say:

Analysis on complex manifolds.

My observation is that such convention prefer to put structures/objects/operations on top of the (in some sense "more basic") things what define them.

To me, they are somewhat natural to say. But as I'm not a naive speaker, they are still pretty vague, for example, why we choose to say vector space over field and operation on set but not the other way around.

So, may I ask for the precise reasons why we "visualize" them as such?


问:
How to Solve a Composite Function
If $f(x)=1+2x$ and $g(f(x))=x+3$, what is the value of $g(5)$? Help would be appreciated on how to solve this problem as there has been debate over the right answer ($5$ or $8$).

答: Since we've been given the definition of $g(f(x))$, we may compute the $x$ values for which $f(x)=5$. That's $1+2x=5\implies x=2$, so $g(5)=2+3=5$ should be correct.

Title: How to Solve a Composite Function


问:
How to find the perimeter of a shaded region when one side isn't given?
The problem is as follows:

Use the figure from below to find the orange shaded regions.

$\begin{array}{ll}
1.&16(4+\pi)\,\textrm{in}\\
2.&8(12+\pi)\,\textrm{in}\\
3.&4(16+\pi)\,\textrm{in}\\
4.&12(8+\pi)\,\textrm{in}\\
\end{array}$

I'm confused exactly how to approach this problem?. The thing is how to get that side for the little square which is on the top left?.

Because the side has some displacement or off set from the top makes it difficult to get the radius.

I've attempted to do the following:

Assuming $a$ is the size of one side on the little square:

Where $P$ is perimeter:

$P=(8+4\sqrt{2}-a)\sqrt{2}\cdot \pi+4(8+4\sqrt{2}-a)+2(4)(a)$

But this expression is not possible to simplify. I mean the term $a$ doesn't cancel.

What can it be done here?. Can someone help me please?. Is there any trick here or what?.

答:
Hint: if $r$ is the radius, the side of the large square is $r\sqrt 2$ and the side of the small square is $r-\frac{r\sqrt 2}{2}$.
Thus, $r+\frac{r\sqrt 2}{2}=8+4\sqrt 2$, $r=8$


答:
There are problably multiple answers to this as there isn't a clear criterium for being different. However, for what it's worth, $397$ is the only prime number in the image.

EDIT: Just realized that $397$ is not one of the options. I leave the answer here for oddity's sake unless someone wants me to delete it.

Title: Which number is different from the others? $526, 793, 382, 764, 329, 894, 932, 397, 478, 489, 652$

Body:
So, in a circle we have numbers $526, 793, 382, 764, 329, 894, 932, 397, 478, 489, 652$ (arranged like in the photo). The question is to find one number which is different from others.

We are given 5 options to be choose from which are :"$526, 329, 382, 793, 932$"

I dunno if the arrangement have something to do with the number or not. For the pattern I have found so far is that some numbers is the same with different arrangement: $(526, 652), (793, 397), (329, 932) , (894, 489)$. That leaves $764, 478, 382$ and $(764, 478)$ have two numbers which are the same so that leaves $382$ so is $382$ the answer of the problem?


问:
Some misunderstandings while reading the proof of Berry-Esseen Theorem
I'm reading the proof of the Berry-Esseen Theorem from Varadhan's notes and I'm having trouble with a couple of steps.
The proof starts at the top of page 97.

Confusion 1: On page 98 while estimating $\lambda[a,\infty)-\mu[a,\infty)$, there is the assumption that $\mu$ can be represented as $fdm$ with $m$ being the Lebesgue measure and $f$ being a bounded function -

Therefore, if we assume that $\mu$ has a density bounded by $C$.

I am ok with this since these computations are applied to the case when $\mu$ is the Gaussian measure.
However, in the next line they seem to be assuming the same thing about $\lambda$ -

Since we get a similar bound in the other direction as well,

$\lambda$ is intended to be the pushforward-measure for $\frac{X_1 + \dots + X_n}{\sqrt{n}}$ where $X_i$ are iid with mean $0$ variance $1$ and with $(2+\alpha)$ moment finite.

Is the bounded density assumption on $\lambda$ being used? If so, how is it valid?

Confusion 2: At the bottom of page 98, they have the expansion for the c.f. $\hat{\lambda}_n$ as
$$
\hat{\lambda_n}(t) = \left(1 - \frac{t^2}{2n} + O\left(\frac{1}{n^{1+\alpha/2}}\right)\right)^n
$$

How do I get the rate of convergence of this to $\exp(-\frac{t^2}{2}$) as
$$
\left| \hat{\lambda}_n(t) - e^{-t^2/2}\right| \leq C\frac{|t|^{\alpha+2}}{n^\alpha} \ \ \ \text{  for }\ \ \ |t| < n^{\frac{\alpha}{2+\alpha}}?
$$
In the central limit theorem, I think one can justify taking logs and appealing to L'Hopital's rule to show convergence.

How do I effectivise the argument in this case?

Thanks for reading.

答:
The bounded density assumption on $\mu$ suffices. The "similar bound" Varadhan is referring to is
$$\mu[a,\infty)-\lambda[a,\infty) \le 2Ch+\mu[a+2h,\infty)-\lambda[a,\infty) \le \int f_{a+h,h} \, d(\mu-\lambda) \,.$$

It is important that in the first line of (3.37) two suprema are compared- the inequality need not hold for each $a$.

Regarding your confusion 2: You are missing a factor $|t|^{2+\alpha}$ inside the $O( \cdot)$.
After taking logs and using the Taylor expansion of $\log$, I get the bound $\frac{|t|^{2+\alpha}}{n^{\alpha/2}}$ so the power in the denominator is different than in the notes.  Correcting this will lead to a weaker final estimate (which is not sharp anyway, see [1])

[1] Katz, Melvin L. Note on the Berry-Esseen Theorem. Ann. Math. Statist. 34 (1963), no. 3, 1107--1108.   https://projecteuclid.org/download/pdf_1/euclid.aoms/1177704037


问:
Spline terminology: Can a spline be represented by a Bézier curve?
In this question it was stated that a B-spline is just a representation of a Bézier curve. I wonder if this also holds true for Bézier curves. Splines are defined piecewise by polynomials and so are Bézier curves. Is it correct to say that Bézier curves are splines? Is the term spline like a umbrella term for B-splines, Bézier curves, composite Bézier curves, NURBS etc.? Are they all splines at the end? Or are they all just representatives?</p>

答:
A spline is a piecewise continuous curve. Each piece (or segment) of the spline could be represented by polynomial functions, trigonometric functions, exponential functions,...., etc with polynomial function being the most popular choice.

A B-spline curve (sometimes referred as B-splines for simplification) is a polynomial spline and each segment of a B-spline curve is geometrically equivalent to a Bezier curve (of the same degree). A NURBS curve is basically a rational B-spline curve where each piece is a rational polynomial function.

A Bezier curve is also a polynomial curve and we can analytically convert a polynomial curve into a Bezier curve or a Bezier curve into a polynomial curve. However, a Bezier curve is not really a spline and it would be confusing or even inappropriate to use terms like "Bezier spline".


问:
Why is the zero module not the direct limit of every system?
Let $M_i$ be $A$-modules. Let $\mathbf{M}=(M_i,\mu_{ij})$ be a directed system over $I$ where $I$ is a directed set. I'm failing to see an obvious fact which tells me why the following statement is not true in general:
$$ \lim_{\rightarrow} M_i = 0 $$

答:
A reasonable way to prove something like this is to come up with a counterexample.

Let's consider a simple directed set, such as a chain: $M_1 \to M_2 \to \dots$.

What happens if you take a non-zero module $M$, and let $M_i = M$ for all $i$, and make the transition maps $\mu_i:M_i \to M_{i+1}$ all equal to the identity on $M$?

One can come up with even simpler examples - what if I took the directed set to just have one element?


问:
Finite order operators has pure point spectrum
I have to prove that the spectrum of an operator $T:H\rightarrow H$, $H$ Hilbert, s.t. $T^k=id_H$ has pure point spectrum. Now, it is obvious that $\sigma_0(T)\subseteq\{e^{i\mu2\pi/k}\}_{\mu=0}^{k-1}$ is the point spectrum of $T$ and $\sigma_0(T)\subseteq\sigma(T)$. However, I cannot find out how to prove that the two spectra coincide. I guess one has to prove that $T-\lambda{id_H}$ is invertible whenever $\lambda\notin\sigma_0(T)$. I thought I can do it in this way: $T-\lambda id_H$ is clearly injective under the assumptions on $\lambda$ (otherwise $\lambda\in\sigma_0(T)$). To prove that it's surjective, one may use the fact that $H=ker(T-\lambda id_H)\oplus Im(T-\lambda id_H)=\{0\}\oplus Im(T-\lambda id_H)=Im(T-\lambda id_H)$ (the equalities are isometric equivalences). Then, the open mapping theorem would imply the continuity of $(T-\lambda id_H)^{-1}$. Is it correct?

However, I'd like to avoid to invoke the open mapping theorem. Is there an easy way to do so?

答:
Your argument to show that $\sigma(T)\subset\{\lambda:\ \lambda^k=1\}$, minus the typos, is fine. It is part of the more general Spectral Mapping Theorem (the "polynomial version"s says that if $p(T)=0$ for some polynomial, then $\sigma(p(T))=p(\sigma(T))$).

But, it has nothing to do with the point spectrum. Your argument for the point spectrum is flawed, as far as I can tell, because you have not shown that the image of $T-\lambda I$ is closed.

I might be wrong, of course, but I don't think there is an elementary argument to show this.

One way to prove the result by using some machinery, is to use the Holomorphic Functional Calculus. Once you know that the spectrum is discrete, given $\lambda_0\in\sigma(T)$ you can construct a holomorphic function (a polynomial, actually) $p$ with $p(\lambda_0)=1$ and $p(\lambda)=0$ for $\lambda\in\sigma(T)\setminus\{\lambda_0\}$. The Functional Calculus then gives you an operator $p(T)$. As $p^2=p$ on $\sigma(T)$, we get $p(T)^2=p(T)$. As this is nonzero, take $x$ such that $p(T)x\ne0$; this is an eigenvector for $\lambda_0$, since
$$
Tp(T)x=\lambda_0\,p(T)x.
$$
This equality occurs due to the equality of functions $tp(t)=\lambda_0p(t)$ on $\sigma(T)$.


答:
Hint

Inference B

$\varphi(x)$ mean $x$ likes women

$Hero(p)$ means $p$ is a hero

1) $\forall p:Hero(p)\rightarrow \varphi(p)$

2) $\varphi(taro)$

3) $\varphi(taro)\rightarrow Hero(taro)$

So the second inference is

$(P\rightarrow Q)\rightarrow (Q\rightarrow P) $

Is this always true?

Inference A

But when we talk about socrate we have a different thing:

$\varphi(x)$ mean $x$ is mortal

$Human(p)$ means $p$ is a human

1) $\forall p:Human(p)\rightarrow \varphi(x)$

2) $Human(socrate)$

3) $Human(socrate)\rightarrow \varphi(socrate)$

that is very different.

Title: logic question inferences

Body:
Here are two inferences A and B, one of which is valid, but the other not valid.

Inference A

Every human is mortal.

Socrates is human.

Therefore Socrates is mortal.

Inference B

Every hero likes women.

Taro likes women.

Therefore Taro is a hero.

Which one is correct, and in what sense it is correct ?

Can somebody give the hints to solve it ?


问:
Polynomials with least value of n
Let $p(x)$ be a polynomial of degree $n$ with all integer coefficients . If $p (2^{1/5}- 2^{-1/5} )= 2015$ then what is
the least possible value of $n~?$

答:
The presence of fifth roots implies that $n\ge5$. Now note that $a=2^{1/5}-2^{-1/5}$ satisfies
$$a^5=2-5\cdot2^{3/5}+10\cdot2^{1/5}-10\cdot2^{-1/5}+5\cdot2^{-3/5}-\frac12$$
Continuing this way with the expansions of $a^3$ and $a$ you should work out
$$2a^5+10a^3+10a-3=0$$
$$2a^5+10a^3+10a+2012=2015$$
and thus $n=5$.

Title: Polynomials with least value of n


问:
Contradiction or Contrapositive divides proof?
I'm just working through the Book of Proof and have come up against the following question:

Prove $(n\in \mathbb{Z})\Rightarrow (4\nmid (n^{2}-3))$

It's in the section for either direct or contrapositive proof. I have a proof by contradiction that I think works:

Proof (Contradiction):

Suppose that there is some $n\in \mathbb{Z}$ where $4\mid(n^{2}-3)$. By definition, therefore, $n^{2}-3 = 4x$ where $x\in \mathbb{Z}$.

Case 1: $n$ is even. Therefore $n = 2a, a\in \mathbb{Z}$. Therefore, $(2a)^{2} - 3 = 4x \Leftrightarrow 4a^{2} - 3 = 4x$. Therefore, $2b + 1 = 2c$ where $b=2a^{2} - 2, b\in \mathbb{Z}$ and $c = 2x, c\in \mathbb{Z}$. Therefore we have a contradiction - LHS Odd, RHS even.

Case 2: $n$ is odd. Therefore $n = 2d + 1, d\in \mathbb{Z}$. Therefore, $(2d +1)^{2} - 3 = 4x \Leftrightarrow 4d^{2} + 4d - 2 = 4x \Leftrightarrow 2(2d^{2} + 2d - 1) = 2(2x)$. Dividing both sides by 2 we get $2e + 1 = 2x$ where $e=d^{2}+d-1, e\in \mathbb{Z}$. Therefore we have a contradiction - LHS Odd, RHS even.

Both cases result in a contradiction therefore $(n\in \mathbb{Z})\Rightarrow (4\nmid (n^{2}-3))$
$\blacksquare$

My question is how would I approach a contrapositive proof? Maybe I can factor the $n^{2}$ term somehow to show that it's values must be from $\mathbb{Q}$ or $\mathbb{R}$?

答:
The proof by contrapositive would be the following:

Prove $4 \mid n^{2}-3 \implies n \notin \mathbb{Z}$.

Suppose $4 \mid n^{2}-3$. Then $n^{2}-3 = 4k \implies n^{2}-3 \equiv0 \pmod 4 \implies n^{2} \equiv 3\pmod 4 $.

But this has no solution as $n^{2} \equiv 0,1 \pmod 4$ for all $n$.

This means that our initial equivalence has no solution, i.e., no integer solution. We have proved that if there is one, then it must be a non-integer.


问:
What is the difference between a partition and a subinterval?
For example the norm of a partition is the widest subinterval</p>

答: A partition of a set $S$ is a collection of subsets $S_i\subset S$ such that the union $\bigcup S_i=S,$ and each distinct $S_i$ have empty intersection. In the case of $S=[a,b]\subset \Bbb R^1$, this gives you non-overlapping subintervals of $[a,b]$ which together forms the entirety of $[a,b]$.


问:
Examples (or characterization) of conditionally complete vector lattices
Are there examples of conditionally complete vector lattices that are not subsets of measurable functions (with order induced by cone of non-negative functions)?

I ask, because there are results in this direction: every Hilbert lattice is isomorph to some $L^2$ with order induced by pointwise inequalities, every lattice on a Hilbert space with self-dual positive cone is isomorph to $L^2$.

Are there similar results that say: a conditionally complete lattice with some extra structure is isomorph to some $L^2$?

答:
There are conditionally order complete vector lattices (even conditionally order complete Banach lattices) which are not isomorphic to an $L^2$-space. For instance, for every measure space $(\Omega,\mu)$ and for $p \in [1,\infty)$ the space $L^p(\Omega,\mu)$ is conditionally order complete. If $(\Omega,\mu)$ is $\sigma$-finite, then $L^\infty(\Omega,\mu)$ is conditionally order complete, too.

If an $L^p$-space is infinite-dimensional and $p \not=2$, then it is not Banach space isomorphic to an $L^2$-space. Hence, it is not vector lattice isomorphic to an $L^2$-space, either (since each vector lattice isomorphism between two Banach lattices is automatically a Banach space isomorphism).

On the other hand, there are various representation theorems on how Archimedean vector lattices can be represented as sublattices of certain function spaces. An overview over many such representation thereoms can be found in Chapter 7 of the classical book "Riesz Spaces I" (1971) by Luxemburg and Zaanen.

In particular, every Archimedean vector lattice is a sublattice of a space of functions (but these functions might take the values $\infty$ and $-\infty$ on a "small" subset of their domain). This is the content of [op. cit., Section 44 in Chapter 7].


问:
Shortcut solution for $f''(t)+c^2f(t)=0$?
I am trying to prove that the general solution to $f''(t)+c^2f(t)=0$ is $f(t) = acos( ct) + bsin (ct)$

Could you please tell me if it is possible to solve it this way?
$$
\frac{d}{dt}(\frac{df}{dt})=-c^2f(t)
$$

$$
df* \frac{d}{dt}(\frac{df}{dt})=-c^2f(t)*df
$$
$$
d(\frac{df}{dt})^2=-c^2 f(t)df
$$
If not for the constant C resulting from integration, this would lead to the desired solution, but I think I am not doing it right somehow.

Thanks in advance for any suggestions.

答:
Another way to do it.

$$f''+c^2f=0$$ Switch variables
$$-\frac {t''}{[t']^3}+c^2f=0$$ reduction of order $p=t'$
$$-\frac {p'}{p^3}+c^2f=0\implies p=\pm\frac{1}{\sqrt{k_1-c^2 f^2}}$$
$$t+k_2=\pm \int \frac {df}{\sqrt{k_1-c^2 f^2}}$$ Just finish and inverse.


问:
Finding the number of pairs (a,b) such that gcd(a,b,n)=1
This is question 6E from the 2019 Cambridge Mathematics Tripos paper 1A (which can be accessed at https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2019/paperia_4_2019.pdf).

"Let $n\geq 2$ be an integer. Let $X = \{0,1,2,...,n-1\}$ and $Y=\{(a,b) \in X^2\mid \operatorname{gcd}(a,b,n)=1\}$.

Show that $$|Y|=n^2 \prod_{p \mid n}(1-\frac{1}{p^2})$$ where the product is over all primes $p$ dividing $n$."

The previous part of the question asks respondents to state the inclusion-exclusion principle, so I believe that this is supposed to be a hint.

My thoughts on the solution include the following:

If we have $\operatorname{gcd}(a,n)=k$ and $\operatorname{gcd}(b,n)=j$, then the condition that $\operatorname{gcd}(a,b,n)=1$ seems to imply that:

$k$ and $j$ must be relatively prime

$\operatorname{gcd}(a,b)=m$, where $\operatorname{gcd}(m,k) = \operatorname{gcd}(m,j)=1$.

In fact, these two conditions collectively seem to imply that $\operatorname{gcd}(a,b,n)=1$. But I could not get anywhere with this. I am also aware of the totient function $\phi(n)$, and I've tried to extend the proof that $$\phi(n)=\prod_{p\mid n} n(1-\frac{1}{p})$$ but was unable to. I noticed that the desired form in the question is actually equivalent to $\phi(n^2)$, but was unable to use this to generate a proof.

Any hints / solutions would be appreciated.

答:
Hint:
\begin{align*}
\#Y = \sum_{\substack{a\in X \\ b\in X \\ (a,b,n)=1}} 1 = \sum_{\substack{a\in X \\ b\in X}} \sum_{d\mid (a,b,n)} \mu(d) = \sum_{d\mid n} \mu(d) \sum_{\substack{a\in X \\ b\in X \\ d\mid a \\ d\mid b}} 1.
\end{align*}


答:
Set $k = j-i$, so replace all uses of $j$ with $k+i$.

For the lower bound, subtract $i$ from both sides of the assignment.

For the upper bound, subtract $i$.  ("... to $k+i = n$ ...".)

In the summand, simplify $k+i-i$.

This should leave a sum you know how to evaluate.

Title: How to evaluate the sum $\sum_{j=i+1}^n (j-i)$?

Body:
How would I go about solving this summation?

$$\sum_{j=i+1}^n (j-i)$$

So far I have:

Using variable substitution

let ${k = j - i}$

$$ \sum_{j=i+1}^n (j-i) =  \sum_{k=1}^{n-i} k$$

$$S = 1 + 2 + 3 + ... + (n - i)-2 + (n - i) - 1 + (n - i) $$

$$S = (n-i) + (n-i)-1 + (n-i)-2 + ... + 3 + 2 + 1$$

$$2S = (n-i+1)(n-i)$$

$$S = 1/2(((n-i+1)(n-i))) $$

$$\sum_{j=i+1}^n (j-i) = 1/2(((n-i+1)(n-i))) $$

Am I following this right?


问:
These two methods of solving the $\tan(2\theta + \frac{\pi}{4})$ ，which one is correct?
I just prefer the Method 2 because of the period, but I don’t know how could Method 1 is incorrect if it comes out with the same answer as Method 2 does.

What should I write down if I get an answer which does not in the domain that the question require? $\theta = \frac{11\pi}{24}$ (unqualified)?

答:
Notice that $\tan(2\theta+\pi/4)$ is a periodic function, with period equal to $\pi/2$. So if you have a solution outside the domain, you can add (or subtract) multiples of $\pi/2$ to get it in the desired range.

So if $\theta=-\pi/24$ is a solution outside the range, add $\pi/2$ or $2\pi/2=\pi$ to get it in the $[0,\pi]$ range.


问:
Energy functional in Poisson's equation: what physical interpretation?
Let's consider this boundary-value problem:

$$\begin{cases} -\Delta V = \rho & \rm{in}\ \Omega
\\
V=0 & \rm{on}\ \partial \Omega \end{cases}.$$

We know that this problem has a variational formulation: its solutions are critical points of the energy functional

$$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx,$$

(cfr. Evans Partial differential equations 2nd ed., §2.2.5).

Now let us look at one (of the many) possible physical interpretations of this problem, the electrostatic one. If units are chosen so that $\varepsilon_0=1$, this equation describes the electric potential $V$ in a region $\Omega$ filled with a charge distribution $\rho$ and whose boundary is grounded.

It would be nice if the energy functional introduced above coincided with the total energy of this physical system. Unfortunately, it seems to me not to be so. In fact, as I read in Feynman's Lectures on physics, vol.II, §8-5, the total energy of this system consists of two additive parts: one is

$$\frac{1}{2}\int_{\Omega} \lvert \nabla V \rvert^2\, dx$$

and is due to the electric field; the other is

$$\frac{1}{2}\int_{\Omega} V\rho\, dx$$

and is due to the charge distribution.

This induces me to think that the
  energy functional should be

$$U[u]=\int_{\Omega} \left(\frac{1}{2}\lvert \nabla u \rvert^2 + \frac{1}{2}u\rho \right)\, dx,$$
  instead of the correct
  $$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx.$$
  Why am I wrong?

答:
In the Lagrangian formulation of classical mechanics, the functional that we extremise to derive the Euler-Lagrange equation of motion is the Lagrangian, defined as

$$L[u] = T[u] - V[u]$$

where $T[u]$ is the kinetic part and $V[u]$ is the potential part. Contrast this with the Hamiltonian $H=T+V$ from which we derive Hamilton's equations of motion (these are equivalent to the Euler-Lagrange equations, and the two are related by Legendre transform). That should explain the minus sign discrepancy.

I am not sure about the factor of 1/2. Is it possible that we are simply defining $\rho_{\mathrm{Feynman}} = 2\rho_{\mathrm{Evans}}$?


问:
System of two simple modular equations
$$x \equiv -7 \mod 13$$
$$x \equiv 39 \mod 15$$

I need to find the smallest x for which these equations can be solved. I've been always doing this using Chinese Reminder Theorem, but it seems that it doesn't work here, I'm not sure why though. Let's see:

$N = 13*15 = 195$

$N_{1} = \frac{195}{13} = 13$

$N_{2} = \frac{195}{15} = 15$

$GCD(13, 13) = 1*13 + 0*13$

$GCD(15,15) = 1*15 + 0*15$

$$x = 1 * (-7) * 13 + 1 * 39 * 15 = 480$$

The answer however is supposed to be $x = 84$...What's wrong?

答:
First, you can check your answer by taking it mod $13$ and $15$ (it's not right).  Second, there are several formulas for applying the CRT; however the method you're using doesn't agree with any of them.  It's difficult to find the place where your method differs from the one you were given, without knowing the latter.

The most common formula to solve $x\equiv x_1\pmod{a_1}, x\equiv x_2\pmod{a_2}$ is $$x\equiv x_2a_1 (a_1)^{-1} + x_1 a_2(a_2)^{-1}$$

Where $(a_1)^{-1}$ is computed modulo $a_2$, and vice versa.  In this case, $a_1=13, a_2=15$.  We have $(a_1)^{-1}=7, (a_2)^{-1}=7$ by coincidence, so $x=(39)13(7)+(-7)15(7)=2814$, which is equivalent to $84$, modulo $13\times 15=195$.

Title: System of two simple modular equations


答:
The general idea may help you to choose.
$$\tan\left(2\theta+\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{6}\right)\\
\implies 2\theta+\frac{\pi}{4}=k\pi+\frac{\pi}{6}\\\implies
\theta+\frac{\pi}{8}=k\frac{\pi}{2}+\frac{\pi}{12}\\\implies
\theta=k\frac{\pi}{2}-\frac{\pi}{24}\\\implies0\leq \theta \leq \pi\\\implies
0\leq k\frac{\pi}{2}-\frac{\pi}{24} \leq \pi\\\implies k=1,2.$$

Title: These two methods of solving the $\tan(2\theta + \frac{\pi}{4})$ ，which one is correct?

Body:
I just prefer the Method 2 because of the period, but I don’t know how could Method 1 is incorrect if it comes out with the same answer as Method 2 does.

What should I write down if I get an answer which does not in the domain that the question require? $\theta = \frac{11\pi}{24}$ (unqualified)?


问:
Singular value decomposition for unitary matrices
I know the title is strange, but there are many instances in Quantum Information in which one is interested, not in diagonalizing a unitary matrix, but instead in finding its singular value decomposition (SVD from now, the theorem of existence is Cor. 2.4 in Nielsen-Chuang). I was checking how to actually compute this decomposition in here, but the example is based in the usual rectangular matrix $A$ with real entries.

In order to compute $A = USV$, one needs to compute $W = AA^T$ ($T$ is the transpose) and the eigenvectors of $W$ form the columns of $U$; likewise the eigenvectors of $\tilde{W}=A^TA$  form the columns of $V$. Finally $S$ is a diagonal matrix in which the nonzero entries are the squared root of the eigenvalues of $W$.

My question is: how this would be modified if $A$ is unitary. Obviously $AA^\dagger=\mathbb{1}$, so it seems still needed that $W = AA^T$ (is this correct?). What about the diagonal values of $S$? Shouldn't carry a complex conjugation aside from the square root?

答:
In the complex setting, a singular value decomposition is given by a factorization $A = US V^\dagger$ where $U$ and $V$ are unitary matrices of the appropriate sizes and $S$ is a diagonal matrix with nonnegative entries. (Here, $(\cdot)^\dagger$ is the adjoint/conjugate transpose.) If $A$ is unitary, then one singular value decomposition is attained by setting $U = A$ and $S = V = I$ (for $I$ the identity matrix). Another is given by $U = S = I$ and $V = A^\dagger$. Indeed, for any unitary matrix $U$, setting $S = I$ and $V = A^\dagger U$ gives a singular value decomposition of $A$.

The key observation is that all of $A$'s singular values are equal to $1$ (this follows by the fact that $A^\dagger A = AA^\dagger =I$ which has all ones for eigenvalues). The singular vectors are not unique: the singular subspaces associated with a single singular value are unique. As such, for unitary $A$, there is one singular subspace consisting of all vectors, and a singular value decomposition can be constructed by any unitary matrix $U$ whose columns will form a basis for this space.


问:
Proving the curl is independent of the system
Q: Prove that the differential operators of the divergence and the curl are independent of the system

My lecturer provided the proof for the divergence using index notation, but I wasn't sure how to prove the curl; could someone clarify whether my proof is correct?

Divergence:

$$ \begin{align}
\nabla' \cdot \textbf{A}' &= \frac{\partial A'_i}{\partial x'_i} \\
&= \frac{\partial x_j}{\partial x'_i} \frac{\partial}{\partial x_j} L_{ik} A_k \\
&= L_{ij} L_{ik} \frac{\partial A_k}{\partial x_j} \\
&= \delta_{jk} \frac{\partial A_k}{\partial x_j} \\
&= \frac{\partial A_j}{\partial x_j} \\
&= \nabla \cdot \textbf{A}
\end{align} $$

Therefore proven.

My proof for the application of this similarly to the curl is as follows:

$$ \begin{align}
\nabla' \times \textbf{A}' &= \epsilon_{ijk} \frac{\partial A'_k}{\partial x'_j} \\
&= \epsilon_{ijk} \frac{\partial x_l}{\partial x'_j} \frac{\partial}{\partial x_l} L_{mi} A'_m \\
&= \epsilon_{ijk} L_{jm} L_{jl} \frac{\partial A_m}{\partial x_l} \\
&= \epsilon_{ijk} \delta_{ml} \frac{\partial A_m}{\partial x_l} \\
&= \epsilon_{ijk} \frac{\partial A_l}{\partial x_l} \\
&= \nabla \times \textbf{A}
\end{align} $$

Is this a correct proof? Have I made any mistakes and have basically just fudged it? Or am I entirely correct? Any help appreciated!

答:
There are some minor mistakes.  The first mistake is the following.  You substitute $A'_k$ with $L_{mi} A_m$.  This is incorrect for two reasons.  The first is that it has to depend on $k$, since $A'_k$ does.  The second is that you expressed the matrix multiplication part incorrectly (i.e., you got rows and columns reversed).

More importantly, the fact that you "basically just fudged it" indicates that you do not understand what you are doing.  I am not trying to place blame here.  (I have "fudged" my way through many concepts the first time I encountered them.)  But my point is this: Even if your proof is correct, it is useless to you and your education, because you do not understand why it is correct.

I would recommend two steps.  First, make sure you actually understand all of the following five facts.  Next, go through the proof of the divergence again and see if you can justify each step using one of these five facts.  Then you will be ready to correct your proof of the curl.

$x' = L x$

$A' = L A$

When multiplying a matrix $L$ and a vector $x$, you get (using Einstein notation) $(Lx)_i = L_{ij} x_j$.

An orthogonal matrix satisfies $L^T L = I$.  (Try writing this as a sum over matrix elements.)

If $x'$ is a function of $x$, then we can write
$$ \frac{\partial}{\partial x'_i} = \frac{\partial x_j}{\partial x'_i} \frac{\partial}{\partial x_j} $$


问:
How to get the angle formed by an hexagon which has two lines on it?
The problem is as follows:

In a regular hexagon $ABCDEF$ a point $G$ is located on $CD$ such that
$AG$ and $BE$ intersect at point $R$. If $\angle AGD = 110^\circ$.
Find $\angle BCR$.

The alternatives given in my book are:

$\begin{array}{ll}
1.&40^\circ\\
2.&35^\circ\\
3.&50^\circ\\
4.&45^\circ\\
\end{array}$

I'm stuck on this problem. However since this problem does not include a graph. I did my best effort as an interpretation of what has been mentioned.

And I believe the problem is indicating this figure:

But what to do from here?. I cannot spot which sort of theorem or identity relying on euclidean geometry approach could be used to solve this problem?.

Can someone help me here?. I've attempted to look for congruence or maybe construction but I just can't spot else. Please include a drawing in your answer. Please help.

答:
The answer is 50 degrees.

First, construct a straight line from C to R. Then, label the point at which this meets AF as $G'$.

Recall that in a regular hexagon, the internal angles are $120^\circ$. This implies that $\angle AFE=120^\circ$, and $\angle BEF=60^\circ$.

Also, note that the lines CG' and AG are "reflections" of each other across the line BE. Thus, $\angle CG'F=\angle AGD=110^\circ$ by symmetry. Note that G'REF is a quadrilateral, which has an internal angle sum of $360 ^\circ$. Hence, we conclude that $\angle G'RE=360^\circ-120^\circ-60^\circ=70^\circ$. This of course means that $\angle BRC=\angle G'RE=70^\circ$, due to vertically opposite angles being equal.  Note that $\angle RBC =60 ^\circ$. From these, we can conclude that $\angle BCR=50^\circ$ since the angles of the triangle BCR add up to $180 ^\circ$.


问:
Trace identity of product of matrices
I have to express the trace of a product of matrices A and B in terms of traces of individual matrices. I know that $\operatorname{tr}(AB) \neq \operatorname{tr}(A)\operatorname{tr}(B)$, and for the tensor product $\operatorname{tr}(A \otimes B) = \operatorname{tr}(A) \operatorname{tr}(B)$. However, is there a nice function $f(\cdot)$ such that $\operatorname{tr}(AB) = f(\operatorname{tr}(A), \operatorname{tr}(B))$?

Any help will be greatly appreciated.

Regards,
H.A.S.

答:
I don’t follow. The mere fact that in dimension 1 we have a product doesn’t imply that a general function would be just a product. Maybe there is a function that in one dimension is just multiplication but in other dimensions is something else.
Granted the intuition is that such a function Durant exist. But a proof of non existence must be more robust than this. The following argument is not sound:

For dimension one tr(AB)=tr(A)tr(B)

For a dimension rather than 1, i.e 2 the formula above fails
Hence no formula Can exist.

Without having any assumption on  the supposed function it’s not possible to prove it can’t exist.

Title: Trace identity of product of matrices


答:
Neither.  Before restricting to the interval, you should have an infinite family of solutions, all separated by half(!)(footnote) the period of tangent.

\begin{align*}
\tan \left( 2\theta + \frac{\pi}{4} \right) &= \frac{1}{\sqrt{3}}  \text{,} \\
2\theta + \frac{\pi}{4}  &=  \arctan\left( \frac{1}{\sqrt{3}} \right) + \pi k, k \in \Bbb{Z} \\
&=  \frac{\pi}{6} + \pi k, k \in \Bbb{Z}  \text{,}  \\
2\theta  &=  \frac{\pi}{6} - \frac{\pi}{4} + \pi k, k \in \Bbb{Z} \\
  &=  \frac{-\pi}{12}  + \pi k, k \in \Bbb{Z}  \text{, and}  \\
\theta  &=  \frac{1}{2} \left( \frac{-\pi}{12}  + \pi k \right), k \in \Bbb{Z} \\
  &=  \frac{-\pi}{24}  + \frac{\pi}{2} k, k \in \Bbb{Z} \text{.}
\end{align*}

So, as $k$ ranges through the integers, every angle $-\pi/24+ (\pi/2)k$ yields a value for $\theta$ solving the equation.  We wish to restrict to angles in the interval $[0,\pi]$, so there is no need to consider negative $k$s further.  We make a table and for this, it is convenient to rewrite the "$\pi/2$" with denominator $24$ to match the angle: $-\pi/24 + (12\pi/24)k$.

\begin{align*}
k &= 0  &  \theta &= -\pi/24  \\
k &= 1  &  \theta &= 11\pi/24  \\
k &= 2  &  \theta &= 23\pi/24  \\
k &= 3  &  \theta &= 35\pi/24  \\
\end{align*}
Clearly, increasing $k$ further gives no more solutions in the interval.  Therefore, the solutions in the interval are $\theta = 11\pi/24$ and $\theta = 23\pi/24$.

(footnote)  Why half?  Because $\theta$ appears multiplied by $2$.  Note that $\tan(2\theta)$ has period $\pi/4$.  If you just treat the period as some afterthought to your computation, you are very likely to make the error made in your work and fail to find all the solutions.  If you carry the period through from the instant you apply inverse functions (and remember that multiplying or dividing both sides of the equation applies to both terms on the right), you don't have to make any special adjustments in your handling of the period based on what $\theta$'s coefficient is.


问:
Showing uniform convergence in probability
Suppose you want to show $sup_{x\in D}|f_n(x)|\to_p 0$, for $n\to \infty$, where $D\subset \mathbb R$ is a compact interval, $f$ is continuous depending on one or more random variables, and $\to_p$ means convergence in probability. For example, $f_n(x)=\sum_{i=1}^n(X_i-x)$ (this, however, is not the problem).

Because showing statements as the one above directly is rather difficult I was wondering if it is sufficient to show $sup_{x\in D}| Ef_n(x)|\to 0$ and $sup_{x\in D}| Var(f_n(x))|\to 0$. Where $E$ and $Var$ are the expectation and variance operator, respectively. If some of you know a good read on this I appreciate your suggestions. Thanks in advance. Cheers.

答:
It's not enough. Let $X_i$ be independent uniform $[0,1]$ random variables and $f_i(x) =  e^{i(X_i-x)^2}$.  Then both $\sup_{x\in(0,1)}E(f_n(x))$ and $\sup_{x\in(0,1)}Var(f_n(x))$ converge to zero, but $\sup_{x\in(0,1)} f_n(x) = 1$ almost surely for every $n$.

Billingsley is a classic on this sort of thing.  It's got all the examples of things like this you should watch out for. (Or you can always google "measure theory lecture notes", and get something for free.


问:
Distribution of 25 book titles between 4 different authors
Can someone please help me? I've been trying to do this for hours:

You are part of a reading group. Each member will read a book from 1 out of 4 different authors(W,X,Y,Z). In how many ways is it possible to select 25 books from theses authors so there are at least 5 books from each author and, at maximum, 6 books from author X?

obs: the titles of the 25 books don't matter, only who the author is

答:
As the titles of the books don't matter but only authors, this translates to finding number of ways to pick different number of books from authors with given restrictions. As you need at least $5$ books of each author ($20$ books across $4$ authors), you are only left with $5$ books to pick.

Take two cases,

i) $5$ books from author $X$ - So you need to pick $5$ books from $3$ remaining authors

ii) $6$ books from author $X$ - you need to pick $4$ books from $3$ remaining authors.

You can use stars and bars method to find number of ways for both cases $1$ and $2$.


问:
Finding Pythagorean triple where the squares of sides are weighted
Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:

$$a^2 + 3 b^2 = 4 c^2$$

Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how to approach this problem.

答:
For the Diophantine equation in general form.

$$aX^2+bY^2=(a+b)Z^2$$

You can record the parameterization of solutions.

$$X=bp^2-(a+b)s^2$$

$$Y=bp^2+2(a+b)ps+(a+b)s^2$$

$$Z=b(p+s)^2+as^2$$

Or in the form that is given in another answer...

$$X=bp^2-2bps-as^2$$

$$Y=bp^2+2aps-as^2$$

$$Z=bp^2+as^2$$


问:
The "almost Hausdorff measure" fails to be subadditive.
I am working with some "measure" (I think it is not a measure since it is not subadditive) $$m(E)=\inf \left\{ \sum\limits_{k=1}^\infty \operatorname{diam} F_k:E\subset\bigcup\limits_{k=1}^\infty F_k \right\}.$$ Now I want to prove it is not subadditive. So I consider two disjoint segments $I =\{x\in \mathbb{R}^d : x_i=0\text{ if }i\neq 1\text{ and }0\le x_1 \le 1\}$ and $J=I+h$ where $h\in\mathbb{R}^d$ and $|h|<\varepsilon$. Now I have to show that $m(I)=m(J)=1$ and $m(I\cup J)\le 1+\varepsilon$. First of all, I can show $m(I)\le 1$ because I can just recover it by I which has diameter $1$. But since we don't have subadditivity, how to show $m(I)\ge 1$?

To show $m(I\cup J)\le 1+\varepsilon$ I am not sure how to proceed. The way I tryed it, is to cover the set $I\cup J$ by the parallelogram generated by $e_1=(1,0,0,\ldots,0)$ and $h$ and it's greater diagonal gives the diameter of this set but I think this is not the way to go.

答: $m(\cdot)$ is subadditive- You are just checking that it is not additive on disjoint closed sets. This is easiest if you take $h=  \epsilon e_2=(0,\epsilon,0,...,0).$


问:
Homework involving unitary diagonalization
I was given as an assignment to diagonalize the following matrix:

$\left(\begin{array}{cc}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{array}\right)$

I started by finding the eigenvectors and got:

$v_{1}=\left(\begin{array}{c}
1\\
-i
\end{array}\right)$,  $v_{2}=\left(\begin{array}{c}
1\\
i
\end{array}\right)$

then I normalized the vectors and composed a unitary matrix:

$U=(v_1|v_2)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1\\
-i & i
\end{array}\right)$

The problem is in the final step:

$U^{*}AU=\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & i\\
1 & -i
\end{array}\right)\right)\left(\begin{array}{cc}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{array}\right)\left(\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1\\
-i & i
\end{array}\right)\right)$

This doesn't produce a diagonal matrix.

Is there any mistake in these stages?

Many thanks.

答:
I think it's just a matter of computational mistake:

$$P=\frac1{\sqrt 2}\begin{pmatrix}\;1&1\\\!\!-i&i\end{pmatrix}\implies P^{-1}=\frac1{\sqrt 2}\begin{pmatrix}1&\;i\\1&\!\!-i\end{pmatrix}$$

And now check that you indeed get

$$P^{-1}AP=\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}$$

as expected.


问:
Does $a_n:=\left(1+\frac1{n+c}\right)^n$ converge to $e$, for $c\in\Bbb{N}$?
In my lecture, we defined $e$ as
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n = \sum_{k=0}^\infty \frac 1{k!} = e$$

While playing around with those definitions, I stumbled onto
$$a_n := \left(1 + \frac{1}{n +c}\right)^n$$ where $c \in \mathbb{N}$. From plotting it, it seems like it converges to $e$ as well, but slower if $c$ gets larger, but I can't find a proof for it.

Is my assumption true, and if so how would I go about proving it?

Body:
$$a_n=\left(1+\frac1{n+c}\right)^n\implies \log(a_n)=n \log\left(1+\frac1{n+c}\right)$$
$$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{(n+c)^k}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{n^k}\frac 1{\left(1+\frac{c}{n}\right)^k }$$
$$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty\frac{(-c)^k-(-c-1)^k}{k}\frac 1{n^k}$$ Computing the first terms
$$\log\left(1+\frac1{n+c}\right)=\frac{1}{n}-\frac{2c+1}{2n^2}+\frac{3c^2+3c+1}{3n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(a_n)=1-\frac{2c+1}{2n}+\frac{3c^2+3c+1}{3n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=e \left(1-\frac{2 c+1}{2 n}+\frac{36 c^2+36c +11}{24
   n^2}\right)+O\left(\frac{1}{n^3}\right)$$


答:
Due to the periodicity of the function $\tan$ (just examine its graph), in general for $n\in\mathbb Z,$ $$\tan\alpha=\tan\beta\iff\alpha=n\pi+\beta.$$

So for $n\in\mathbb Z,$
\begin{equation}
\begin{aligned}
\tan\left(2\theta+\frac{\pi}4\right) &=\frac1{\sqrt3}\\
\iff\tan\left(2\theta+\frac{\pi}4\right) &=\tan\left(\tan^{-1}\left(\frac1{\sqrt3}\right)\right)\\
\iff\tan\left(2\theta+\frac{\pi}4\right) &=\tan\left(\frac{\pi}6\right)\\
\iff2\theta+\frac{\pi}4 &= n\pi+\frac{\pi}6\\
\iff\theta &= \frac{\pi}{24}\left(12n-1\right).
\end{aligned}
\end{equation}
If $\theta\in[0,\pi],\;$ then $n=1$ or $2\,$ and $\,\theta=\frac{11\pi}{24}$ or $\frac{23\pi}{24}.$


问:
Number of ways to select a subject for each day in a week.
I am attempting to solve the following question:

A student wants to make up schedule for a seven-day period during which she will study one subject each day. She is taking four subjects - Mathematics, Physics, Chemistry, and Economics. Find the number of schedules that devote at least one day to each subject.

My attempt is as follows:

On the fisrt day, I have $4$ ways, on the second day, there are $3$ ways, on the third day, there are $2$ ways, and on the fourth day, there's only $1$ way, and for the remaining $3$ days, I can select any one of the subjects

答:
Labelling the subjects $A,B,C,D$, there are $3$ possible patterns of subjects, 4-1-1-1 of a kind, eg AAAABCD, 3-2-1-1 of a kind, eg AAABBCD, and 2-2-2-1 of a kind, eg AABBCCD
We need to determine ways to select subjects to fit a pattern, and permute over the $7$ day period

$4-1-1-1\;of\;a\;kind:\;[\binom 4 1\binom3 3]\times[\frac{7!}{4!}]$

$3-2-1-1\;of\;a\;kind:\;[\binom 4 1\binom 3 1\binom2 2]\times[\frac{7!}{3!2!}]$

$2-2-2-1\;of\;a\;kind:\;[\binom4 3\binom1 1]\times[\frac{7!}{2!2!2!}]$

Compute and add up


问:
Are rings in algebraic topology generally assumed commutative?
There's not much to add to the question statement. I know that in algebraic geometry and commutative algebra, one generally works with commutative rings, but I'm unsure if the same holds in algebraic topology.</p>

答: The cohomology ring is (only) skew-commutative.</p>


答:
If you see someone take cohomology with coefficients in a ring that ring is always commutative (and usually either $\mathbb{Z}$ or a field); it doesn't really help in any way to take it to be noncommutative. On the other hand, cohomology rings themselves are graded-commutative rather than commutative on the nose, meaning the product satisfies

$$xy = (-1)^{|x| |y|} yx$$

if $x, y$ are homogeneous and $|x|, |y|$ denotes their degrees. This means they commute unless they are both odd in which case they anticommute. There are good reasons for this but it's a long story.

Noncommutative rings do occasionally show up, e.g. for various purposes one considers the group ring $\mathbb{Z}[\pi_1]$ of a fundamental group (as in L-theory), but I think it will generally be clear when this is happening.

Title: Are rings in algebraic topology generally assumed commutative?

Body: There's not much to add to the question statement. I know that in algebraic geometry and commutative algebra, one generally works with commutative rings, but I'm unsure if the same holds in algebraic topology.</p>


问:
An intuitive doubt on limits
So I have been studying iterated limits and I have a doubt.

The iterated limits of a two variable function $f(x,y) \mbox{ at } (\alpha, \beta)$ are defined as :
$$\lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ and } \lim_{y \to \beta} (\lim_{x \to \alpha} f(x,y))  $$
The thing is that the iterated limit exists only if the internal one related to it exists, so :
$$ \lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ exist iff }  (\lim_{y \to \beta} f(x,y)) \mbox{ exist independently  }  $$
same is true for the other iterated limit.

now look at this function :
$$f(x,y) = xsin(1/y)$$
the limit: $$\lim_{y \to 0} xsin(1/y) \mbox{ does not exist}$$
and that makes sense as y tending to zero causes $sin(1/y)$ to oscillate and it will be some undetermined but finite number(changing rapidly as y approaches zero) times $x$ at the end, which means we don't know the limit.

But here in reality, later $x$ also approaches zero, which means that at the end its :
$$\mbox{(some undetermined but finite number, oscillating rapidly)} \times \mbox{ (a number approaching 0)}$$
logically, no matter what undetermined value $sin(1/y)$ takes, that value will be multiplied to a number really close to zero, so the limit should be zero.
But this is not the case.

According to various references and my professor the limit :
$$ \lim_{x \to 0} (\lim_{y \to 0} xsin(1/y)) \mbox{ does not exist}$$
This seems paradoxical and counterintuitive , please share your wisdom and point where I went wrong, or if this is right.

答:
I am not sure if I fully understand your question. To define a limit (iterated in this case), or any other concept for that matter, in a well defined manner one must make sure that the values of limits make sense and are uniquely defined in ALL cases. Just because in the special case of x sin(1/y) we can argue that the limit exists does not warrant the general definition to allow for cases where the independent limit is not defined.

In other words, the definition does not imply that the limit CAN NEVER BE DEFINED unambiguously. Only that, if you need it to be defined unambiguously regardless, then you should assume $ (\lim_{y \to \beta} f(x,y))$  exist independently. As a matter of fact, you might be able to weaken this assumption (intuitively, not by much). You might be able to show that in some cases  $\lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ exist}$ even if  $(\lim_{y \to \beta} f(x,y))$  does not exist independently.

Definitions are designed to work. They are not necessarily optimal.

Title: An intuitive doubt on limits


问:
Subset of a countable set is countable
I am trying to understand the full details of a proof on countability. The exact statement is:

If $A \subset B$ where $B$ is countable, then $A$ is countable.

"Countable" here means finite or countably infinite, but the case where $A$ (or $B$, and hence $A$) are finite are rather vacuous, so let's only consider the case where $A$ and $B$ are infinite. The first line of the proof states:

"It suffices to consider the case where $B = \mathbb{N}$ and $A \subset \mathbb{N}$."

I am trying to understand why this is the case. Here is my best attempt. If $B$ is countable, there is a bijection $f: \mathbb{N} \to B$ and hence another bijection $f^{-1}: B \to \mathbb{N}$. So $f^{-1} (B)$ is countable (in fact, it's the entirety of $\mathbb{N}$), but $A \subset B$, so $f^{-1} (B) \subset f^{-1} (A) = \mathbb{N}$. If we prove that $f^{-1} (B)$ is countable, we have the opposite inclusion and a bijection $g: f^{-1} (B) \to \mathbb{N}$. Then we can "apply" $f$ to compose bijections, getting a bijection from $B$ to $\mathbb{N}$.

The last step is very shaky and I don't think it works. Any idea?

答:
The composition of two injective maps is injective.  Compose the inclusion of the subset with the injection into the naturals, to get an injection of the subset into the naturals.

(This question was asked a day or two ago, and is a dupe. Although there the OP insisted on a proof by induction, rather ridiculously.)

Title: Subset of a countable set is countable


答:
Following your approach by splitting the sum you can use $\sum_{k=1}^m k = \frac 12 m(m+1)$ twice as follows:

\begin{eqnarray*} \sum_{j=i+1}^n (j-i)
& = & \sum_{j=i+1}^n j - \sum_{j=i+1}^n i \\
& = & \underbrace{\sum_{j=i+1}^n j + \sum_{j=1}^{\color{blue}{i}} j}_{\sum_{\color{blue}{j=1}}^n j}- \sum_{j=1}^{\color{blue}{i}} j  - i\sum_{j=i+1}^n 1 \\
& = & \sum_{\color{blue}{j=1}}^n j - \sum_{j=1}^{\color{blue}{i}} j  - i\sum_{j=i+1}^n 1 \\
& = & \frac 12 n(n+1) - \frac 12 i(i+1) - i(n-i)\\
& \stackrel{some \, algebra}{=} & \frac 12(n-i)(n+1-i)
\end{eqnarray*}


答:
About your example $$x\sin \frac 1y,$$ the limit does not exist at $(0,0)$ because if you fix any value of $x$ (you may take $x=1$ or whatever), you find that the limit as $y\to 0$ is undefined. On the other hand fix a value of $y\ne 0$ (say take $y=1/π$). Then the limit as $x\to 0$ is $0.$ Since for the limit to exist it shouldn't matter how $(x,y)$ approaches the origin, it follows that the limit doesn't exist.

Your mistake is taking the limit of an undefined quantity. Yes, the quantity is bounded, but that doesn't tell us exactly what it is. So taking the outer limit is quite meaningless -- it is undefined as well. Or, in other words, does not exist.

Title: An intuitive doubt on limits

Body:
So I have been studying iterated limits and I have a doubt.

The iterated limits of a two variable function $f(x,y) \mbox{ at } (\alpha, \beta)$ are defined as :
$$\lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ and } \lim_{y \to \beta} (\lim_{x \to \alpha} f(x,y))  $$
The thing is that the iterated limit exists only if the internal one related to it exists, so :
$$ \lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ exist iff }  (\lim_{y \to \beta} f(x,y)) \mbox{ exist independently  }  $$
same is true for the other iterated limit.

now look at this function :
$$f(x,y) = xsin(1/y)$$
the limit: $$\lim_{y \to 0} xsin(1/y) \mbox{ does not exist}$$
and that makes sense as y tending to zero causes $sin(1/y)$ to oscillate and it will be some undetermined but finite number(changing rapidly as y approaches zero) times $x$ at the end, which means we don't know the limit.

But here in reality, later $x$ also approaches zero, which means that at the end its :
$$\mbox{(some undetermined but finite number, oscillating rapidly)} \times \mbox{ (a number approaching 0)}$$
logically, no matter what undetermined value $sin(1/y)$ takes, that value will be multiplied to a number really close to zero, so the limit should be zero.
But this is not the case.

According to various references and my professor the limit :
$$ \lim_{x \to 0} (\lim_{y \to 0} xsin(1/y)) \mbox{ does not exist}$$
This seems paradoxical and counterintuitive , please share your wisdom and point where I went wrong, or if this is right.


问:
How to prove to be an irrational number? Like $\sqrt{2}$, $\sqrt{3}$, or $\sum\limits_{k=1}^{\infty} \frac{1}{n^2}=\pi^2/6$
As we know $\sqrt{2},\sqrt{3}$ are irrational numbers. And I see some proofs on the net.

So I doubt that how $e,\pi$ or already known irrational numbers are proved to be irrational.

In fact, I got interested in Riemann zeta function
$$\zeta(s)=\sum_{n=0}^{\infty} \frac{1}{n^s},$$

we know $\zeta(2)=\pi^2/6$ from Euler, 1737.

One mathematician (sorry to forgot his name) proved $\zeta(3)$ to be also irrational 40 years before.

Can somebody explain how he could do with that? To understand Apéry's theorem, is it very hard?

An question raises that could one real number make up of two different irrationals (for example: $e,\pi$), $e\pi$, or others can be rational?

答:
It is known to be very difficult to find the arithmetic nature of most real numbers such as the constants that appear in mathematical analysis, e.g. $\zeta (5)=\sum_{n\geq 1}\frac{1}{n^{5}}$.

Roger Apéry proved directly that $\zeta(2)$ is irrational. And he was the first to prove that $\zeta(3)$ is irrational too. He constructed two sequences $(a_n),(b_n)$ $[1]$

$$\begin{equation*}
a_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}c_{n,k},
\qquad b_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2},\end{equation*}$$

where

$$\begin{equation*}
c_{n,k}=\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{\left( -1\right)
^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\quad k\leq n.
\end{equation*}$$

The ratio $a_n/b_n\to\zeta(3)$ and has the following properties:

$2(b_{n}\zeta (3)-a_{n})$ satisfies $\lim\sup \left\vert 2(b_{n}\zeta (3)-a_{n})\right\vert^{1/n}\le(\sqrt{2}-1)^4 $.

$b_{n}\in \mathbb{Z},2(\operatorname{lcm}(1,2,\ldots ,n))^{3}a_{n}\in
\mathbb{Z}$.

$\left\vert b_{n}\zeta (3)-a_{n}\right\vert >0$.

This is enough to prove  the irrationality of $\zeta (3)$ by contradiction. $[2]$.

There is The Tricki entry To prove that a number is irrational, show that it is almost rational that gives two examples and explains the principle of some proofs of the irrationality of numbers.

References.

$[1]$ Poorten,  Alf., A Proof that Euler Missed…, Apéry’s proof of the irrationality of $\zeta(3)$. An informal report, Math. Intelligencer 1, nº 4, 1978/79, pp. 195-203.

$[2]$ Fischler, Stéfane, Irrationalité de valeurs de zêta (d’ après Apéry, Rivoal, …), Séminaire Bourbaki 2002-2003, exposé nº 910 (nov. 2002), Astérisque 294 (2004), 27-62

$[3]$ Apéry, Roger (1979), Irrationalité de $\zeta2$ et $\zeta3$, Astérisque 61: 11–13


问:
draw a convex grid polygon with 40 sides
how do I draw a convex grid polygon with 40 sides such that it is both horizontally and vertically symmetrical and all internal angles are less than 180 and vertices lie on integer coordinates.</p>

答:
One way to think about this is to have the polygon centered at $(0,0)$ and to have each quadrant contain $10$ sides.  In order to have a convex polygon, the coordinates must follow some pattern where the slopes of consecutive sides either increase or decrease throughout the entire quadrant.

Below is an example of a set of coordinates that work for one quadrant:
$(0,55),(10,54),(19,52),(27,49),(34,45),(40,40),(45,34),(49,27),(52,19),(54,10),(55,0)$


问:
Extremely accurate fractions for square roots

Approximation

The following is a simple and amazingly accurate way to get a rational approximation to square roots:

To find $\sqrt n$, guess a fraction $p/q$ near $\sqrt n$. (So $nq^2 \approx p^2$).

Let $a = nq^2 + p^2$.

Let $b = nq^2 - p^2$.

$\displaystyle \sqrt n \approx \frac{a - b^2/(2a)}{2pq}$

For accuracy we want $|b|$ to be small, and can usually get it to be 1, without much effort.

Explanation

Start with $s_0=p/q$ and use the Babylonian Method to get a first estimate
$$s_1 = \frac{s_0 + n/s_0}{2} = \frac{p^2 + nq^2}{2pq} = \frac{a}{2pq}$$
This excess will always be an overestimate by approximately
$$\frac{(s_0 - n/s_0)^2}{2s_1} = \frac{(p^2 - nq^2)^2}{4pqa} = \frac{b^2}{4apq}$$
Subtracting this excess gives the approximation formula.

Problem

I want to get an estimate of (or a bound for) the error of approximation. My guess is it is something like $|p/q-\sqrt n|^4$.

Examples

For $\sqrt 2$ try $p/q=3/2$, $a=7$, $b=1$,  $\sqrt 2 \approx \frac{17 - 1/34}{12}$

$\qquad\color{red}{1.41421}356237310 \quad\text{exact}\\
 \qquad\color{red}{1.41421}568627451 \quad\text{approximation}$

If instead we start with a better estimate $p/q=7/5$ we get $a=99$, $b=1$ and $\sqrt 2 \approx \frac{99 - 1/198}{70}$, giving a much better accuracy:

$\qquad\color{red}{1.41421356}237310 \quad\text{exact}\\
 \qquad\color{red}{1.41421356}421356 \quad\text{approximation}$

For $\sqrt 3$ try $p/q=7/4$, $a=97$, $b=1$,  $\sqrt 3 \approx \frac{97 - 1/194}{56}$

$\qquad\color{red}{1.7320508}0756888 \quad\text{exact}\\
 \qquad\color{red}{1.7320508}1001472 \quad\text{approximation}$

For $\sqrt 5$ try $p/q=9/4$, $a=161$, $b=1$,  $\sqrt 5 \approx \frac{161 - 1/322}{72}$

$\qquad\color{red}{2.236067977}49979 \quad\text{exact}\\
 \qquad\color{red}{2.236067977}91580 \quad\text{approximation}$

For $\sqrt 7$ try $p/q=8/3$, ...

$\qquad\color{red}{2.64575131}106459 \quad\text{exact}\\
 \qquad\color{red}{2.64575131}233596 \quad\text{approximation}$

For $\sqrt 11$ try $p/q=10/3$, ...

$\qquad\color{red}{3.316624790}35540 \quad\text{exact}\\
 \qquad\color{red}{3.316624790}61977 \quad\text{approximation}$

For $\sqrt 13$ try $p/q=18/5$, ...

$\qquad\color{red}{3.60555127546}399 \quad\text{exact}\\
 \qquad\color{red}{3.60555127546}653  \quad\text{approximation}$

For $\sqrt 15$ try $p/q=31/8$, ...

$\qquad\color{red}{3.8729833462074}2 \quad\text{exact}\\
 \qquad\color{red}{3.8729833462074}5 \quad\text{approximation}$

As an example where $b \ne 1$, for this last example, if we tried $p/q = 27/7$, we would get $a=1464$, $b=6$ giving the answer $\frac{1464 - 36/2928}{378}$ which  3.872983346 3440

$\qquad\color{red}{3.872983346}20742 \quad\text{exact}\\
 \qquad\color{red}{3.872983346}34400 \quad\text{approximation}$

答:
If you start with a rational number, use Newton method to generate a more and more accurate rational number.

For example, for $\sqrt{15}$, using as you did $x_0=\frac {31}8$, you will generate numbers $x_n=\frac{a_n}{b_n}$
$$\left(
\begin{array}{cccc}
 n & a_n  & b_n & \text{decimal} \\
 0 & 31 & 8 & 3.8750000000000000000 \\
 1 & 1921 & 496 & 3.8729838709677419355 \\
 2 & 7380481 & 1905632 & 3.8729833462074524357 \\
 3 & 108942999582721 & 28128961537984 & 3.8729833462074168852
\end{array}
\right)$$

If you start with $x_0=\frac p q$ for $x^2=a$ ($a$ being an integer or a rational number), you have
$$x_1=\frac{a q}{2 p}+\frac{p}{2 q}$$
$$x_2=\frac{a p q}{a q^2+p^2}+\frac{a q}{4 p}+\frac{p}{4 q}$$
$$x_3=\frac{1}{8} \left(\frac{16 a p q \left(a q^2+p^2\right)}{a^2 q^4+6 a p^2
   q^2+p^4}+\frac{4 a p q}{a q^2+p^2}+\frac{a q}{p}+\frac{p}{q}\right)$$


问:
Criterion for a relatively shift weakly compact sequence of measures to be actually relatively weakly compact
I'm trying to understand a result given in Theorem 2.3.7 (see below) of Linde's Probability in Banach spaces: Stable and Infinite Divisible Distributions.

Let $(\mu_n)_{n\in\mathbb N}$ be a sequence of Radon probability measures on the Borel $\sigma$-algebra $\mathcal B(E)$ of a Banach space $E$ such that there is a sequence $(x_n)_{n\in\mathbb N}\subseteq E$ such that $(\mu_n\ast\delta_{x_n})_{n\in\mathbb N}$ is relatively compact (denoted by "w.r.c." below) with respect to the topology of weak convergence of measures.

If we denote the topology of compact convergence on $E'$ by $\tau_c$, we can show that the family $\{\hat\mu:\mu\in\mathcal F\}$ of characteristic functions of a bounded (in total variation norm) and tight family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is uniformly $\tau_c$-continuous.

Assume now that there is a $\delta>0$ such that $C_\delta:=\left\{\left.\hat\mu_n\right|_{V_\delta}:n\in\mathbb N\right\}$ is $\tau_c$-equicontinuous at $0$ (which is actually equivalent to being uniformly $\tau_c$-equicontinuous, which in turn is equivalent to being relatively compact with respect to the uniform topology on $C(V_\delta)$), where $V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$.

In conclusion, $C_\delta$ and $\left\{\left.\widehat{\mu_n\ast\delta_{x_n}}\right|_{V_\delta}:n\in\mathbb N\right\}$ are both $\tau_c$-equicontinuous at $0$.

This should yield the existence of $\rho>0$ and a compact $K\subseteq E$ such that $$\left|1-\hat\mu_n(\varphi)\right\|<\frac\varepsilon2\tag1$$ and $$\left|1-e^{{\rm i}\langle x_n,\:\varphi\rangle}\hat\mu_n(\varphi)\right\|<\frac\varepsilon2\tag2$$ for all $\varphi\in V_\delta$ with $p_K(\varphi):=\sup_{x\in K}|\langle x,\varphi\rangle|<\rho$ and $n\in\mathbb N$.

However, as you can see in the proof of Linde below, it seems like he is assuming $\rho=1$. That doesn't make sense to me. Why should $(1)$ and $(2)$ be smaller than $\varepsilon/2$ for an arbitrary $\varepsilon$ as long as $p_K(\varphi)\le1$?

答: If $K$ is a compact subset of $E$, then so is $RK = \{Rx \, \mid \, x \in K\}$, no matter the choice of $R > 0$.  Further, given $a \in E'$, $\sup \left\{ |\langle a,x \rangle| \, \mid \, x \in RK \right\} = R \sup \left\{ |\langle a,x' \rangle| \, \mid \, x' \in K \right\}$.  Hence, by dilation, we can replace $\rho$ by $1$.


问:
Matrices over $\mathbb{F}_p$ satisfying a Determinant and Trace Condition
Fix an odd prime $p$, and integer $n \geq 1$, an $\alpha \in \mathbb{F}_p^{\times}$, and $\delta, \tau \in \mathbb{F}_p$. Let $T = (t_{ij})$ denote an arbitrary symmetric $n \times n$ matrix mod $p$. I wish to show the sets below have the same cardinality:

The symmetric matrices mod $p$ with fixed determinant $\delta$ and trace $\tau$,

The symmetric matrices mod $p$ with fixed determinant $\delta$ and satisfying the trace condition

\begin{align*}
\alpha t_{11} + \alpha^{-1} t_{22} + t_{33} + \cdots + t_{nn} = \tau.
\end{align*}

This statement is true if we instead count all such matrices over the (at most degree $2$) extension $\mathbb{F}_p(\sqrt{\alpha})$. That is, in this setting we have the explicit bijection

\begin{align*}
T \mapsto UTU^t
\end{align*}

where

\begin{align*}
U = \textrm{diag}(\sqrt{\alpha}^{-1},\sqrt{\alpha},1,\ldots,1).
\end{align*}

答:
I'll rename your matrix to $M$ (because I don't like the conflict with $T$ meaning transpose). We'll try to construct a bijection of the form $M \mapsto AMA^T$ where $A$ is a block diagonal matrix with a top-left $2 \times 2$ block $\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ and $1$s on the diagonal otherwise, and whose determinant squares to $1$. A computation gives

$$\text{tr}(AMA^T) = (a^2 + c^2) m_{11} + (ab + cd) m_{21} + (ab + cd) m_{12} + (b^2 + d^2) m_{22} + \sum_{i=3}^n m_{ii}$$

so we're done if we can find $a, b, c, d$ such that

$$a^2 + c^2 = \alpha$$
$$b^2 + d^2 = \alpha^{-1}$$
$$ab + cd = 0$$
$$(ad - bc)^2 = 1.$$

As you remark, if $\alpha$ has a square root this is easy since we can take $b = c = 0, a = \sqrt{\alpha}, d = \sqrt{\alpha}^{-1}$. In fact it is always possible: the equation $a^2 + c^2 = \alpha$ has a solution in $\mathbb{F}_p$ for every $\alpha \in \mathbb{F}_p^{\times}$. This says equivalently that the norm map $\mathbb{F}_p[i]^{\times} \to \mathbb{F}_p^{\times}$ is surjective, which is true whether or not $\mathbb{F}_p[i]$ is a field. Then we can take $b, d$ such that

$$d + bi = \frac{1}{a + ci} = \frac{a - ci}{a^2 + c^2} = \frac{a - ci}{\alpha}$$

so $N(a + ci) = \alpha$ gives $N(d + bi) = \alpha^{-1}$, and $(a + ci)(d + bi) = (ad - bc) + (ab + cd) i = 1$, so $ab + cd = 0$ and $ad - bc = 1$ automatically, which is all the conditions we needed. Probably a more conceptual argument is possible; I guess this is some standard fact about quadratic forms.

Some more details about the claimed surjectivity of the norm map. We can split into two cases: if $p \equiv 1 \bmod 4$ then $\mathbb{F}_p[x]/(x^2+1) \cong \mathbb{F}_p \times \mathbb{F}_p$ with the isomorphism sending $a + bx$ to $(a + bi, a - bi)$ where $i^2 = -1$ in $\mathbb{F}_p$. In this case the norm map is just multiplication $(x, y) \mapsto xy$ and surjectivity is clear.

If $p \equiv 3 \bmod 4$ then $\mathbb{F}_p[i] \cong \mathbb{F}_{p^2}$ whose multiplicative group has a generator $g$ whose norm is $N(g) = g \overline{g} = g^{p+1}$. Since $g$ has order $p^2 - 1$ it follows that $N(g)$ has order $p - 1$.

(If $p = 2$ then $\mathbb{F}_p[i] \cong \mathbb{F}_p[\epsilon]/\epsilon^2$ and the surjectivity of the norm map still holds in this case but is trivial.)

A somewhat more conceptual version of the computation above is to write $\text{tr}(AMA^T) = \text{tr}(A^T A M)$ where $A^T A$ has top-left $2 \times 2$ block

$$\left[ \begin{array}{cc} a & c \\ b & d \end{array} \right] \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a^2 + c^2 & ab+cd \\ ab + cd & b^2 + d^2 \end{array} \right].$$

I suppose this makes the effect of this transformation on the diagonal entries and hence on the trace a little clearer. We want $\text{tr}(A^T A M)$ to have the effect of multiplying the first diagonal entry by $\alpha$ and the second by $\alpha^{-1}$, and $\text{tr}(A' M)$ accomplishes this where $A' = \text{diag}(\alpha, \alpha^{-1}, \dots)$, so by the nondegeneracy of the trace form we require $A^T A$ to have top-left block $\left[ \begin{array}{cc} \alpha & 0 \\ 0 & \alpha^{-1} \end{array} \right]$. The rest of the argument still isn't quite as slick as I'd like it to be though. What we've shown is that the quadratic form $\alpha x^2 + \alpha^{-1} y^2$ is equivalent to the quadratic form $x^2 + y^2$. But I'm missing a general sense of when quadratic forms over finite fields are equivalent.


问:
Logic without the Law of Identity
The wikipedia page on the laws of thought mentions the The three traditional laws: The Law of the Excluded Middle, Law of Non-contradiction and The Law of Identity. These laws are accepted and used by the vast majority of mathematicians, but not by all.

Intuitionists reject the Law of the Excluded Middle, as it opens the possibility for non-constructive proofs. As such, they may prefer to use some form of intuitionistic logic, where the law is not a tautology.

Dialethics reject the Law of Non-contradiction, on the philosophical standpoint that some statements are contradictory by nature. As such, they may prefer to use some form of paraconsistent logic, where the law is not a tautology.

What about the Law of Identity?

This question is two fold:

Is there a philosophical school that rejects the law of identity as a law of thought?

Is there system of logic in which the law of identity is not a tautology?

答:
General Semantics as founded by Alfred Korzbiski rejects the principle of identity, as there can't exist a description of empirical facts which is ever one-to-one (according to that school/discipline).  As they say, the map is not the territory, but if useful bears a similar structure.  I think some of St. Leśniewski's work also rejected the principle of identity, and instead had a concept where statement forms were only 'equiform' instead of 'identical', and if I recall correctly, I think Lukasiewicz only used the term 'equiform' in some of his later work to avoid any sort of question like "how can 'Cpp' in this position be the same as 'Cpp' in this other position, especially considered as material objects?"

Schrodinger logics (see bottom of the link) reject the principle of identity.  According to that wikipedia article, and this paper, there has existed motivation for that rejection originating in the study of quantum mechanics.

Title: Logic without the Law of Identity


问:
Vanishing of a certain Tor
I am reading about the construction of the Affine Grassmannian in Dennis Gaitsgory's seminar notes
and there are some commutative algebra facts that I am not able to figure out by myself apparently, like the following:

Let $k$ be an algebraically closed field, $A$ a finite type $k$-algebra and $A\subseteq B$ some (non finite type) extension. Let $M$ be a finitely generated $A[[t]]$-module which is flat over $A$ (actually finite free over $A$) and $t$ acts nilpotently on $M$. Then, $\operatorname{Tor}^{A[[t]]}_n(M,B[[t]])=0$ for all $n>0$.

Intuitively, since we get $B[[t]]$ from $A[[t]]$ by extending only the coefficient ring $A$ in a "free" way, flatness over $A$ of $M$ should suffice, but I can't make it into a proof. Notice that $A[[t]]\otimes _A B\ne B[[t]]$ in general. Am I missing something obvious?

Edit:

After reading it again, it seems that the only additional hypothesis I missed is that $t$ acts nilpotently on $M$ which I think does not follow from what I have written. The relevant place in the notes is the first line on page 7 and a bit before that.

答:
Here's an attempt (could be mistakes so be wary!).

We have
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]] \cong (M \otimes_{A[[t]]}^{\mathbb{L}} A[[t]]/t^n) \otimes_{A[[t]]}^{\mathbb{L}} B[[t]].
$$

By associativity this is (quasi-isomorphic) to
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]]/t^n.
$$
Since $A[[t]] \rightarrow A[[t]]/t^n$ is surjective, this is the same as
$$
M \otimes_{A[[t]]/t^n}^{\mathbb{L}} B[[t]]/t^n.$$

And this satisfies the conclusion you want because of your conditions on $M$.

Title: Vanishing of a certain Tor


问:
Prove any subgroup $A$ of $n$-elements-generated group $G$ has can be generated by $2n[G:A]$ elements
If $G$ can be generated by n elements,
$A$ is a subgroup of $G$ and index $(G:A)$ is finite,

I was required to prove: $A$ can be generated by $2n(G:A)$ elements.

There is an answer that A generally cannot be generated by only n elements when G is non-abelian, but the upper bound here is much higher.(If a group $G$ is generated by $n$ elements, can every subgroup of $G$ by generated by $\leq n$ elements?)

Thank you!

答:
Welcome to MSE!

It seems this answer does what you want. In particular, the factor of $2n$ comes from ensuring the generating set $Y$ of $G$ symmetric, and the factor of $(G:A)$ comes from taking a transversal $S$ of $G/A$.

Following that answer, if we write $[g]$ to mean the chosen representative of $Ag$ in $S$, induction on word length shows $\{ sy [sy]^{-1} ~|~ s \in S, y \in Y \}$ (which has size $2n[G:A]$) generates $A$.

I hope this helps ^_^