问:
If we want to use proof by contradiction, shouldn't we first proof that a statement is either true or false?
As an example how this could go wrong "This statement is false"
We could proof this statement by contradiction as follows

Assume the statement is false $\implies$ The statement is false $\implies$ the statement is true. Which is a contradiction, hence the statement must be true.

Obviously, the statement being true leads to equally contradictory results, but in mathematics we I have never seen anyone continue to check also what happends if we assume the statement is true. And if we did, it would not be humanly possible to check all possible consequences of a statement being true.

Is there a mathematical axiom that claims that all statements are either true or false, and how can we be sure that this axiom itself does not lead to contradictions?

答:
Is there a mathematical axiom that claims that all statements are either true or false?

Yes, it is called Law of excluded middle.
This is always used for the vast majority of mathematics.

and how can we be sure that this axiom itself does not lead to contradictions?

We can't (provided we start with sufficiently strong axioms such as Peano arithmetic or ZFC).
This is known as Goedel's second incompleteness theorem.

"This statement is false"

This is a self-referential statement. While this is possible in the english language,
i am not aware of any formulation in classical mathematics of this statement. In my (and probably many other's) opinion this is therefore not a valid example to show that a mathematical statement can be neither true nor false.


问:
Rotate Quaternion A by 180 degrees
Suppose you have an arbitrary quaternion - call it A - how do you rotate it by 180 degrees?

Is there a way to do this without convert to angle-axis representation, i.e., keep it within the quaternion algebra?

答: I don't know what you mean with rotate a quaternion (which actually represents a rotation). But I guess you mean concatenate two quternions with one being a 180 degree rotation about some axis. In this case you can just use the quternion multipication for concatenating two rotations (There is rarely a case where you need to convert them to axis-angle representation). The quaternion for a 180 degree rotation about axis (x,y,z) is just (0,x,y,z).</p>

OwnerDisplayName: ina

ContentLicense: CC BY-SA 3.0

Title: Rotate Quaternion A by 180 degrees

OwnerDisplayName: Christian Rau


问:
Matrix inverse of the sum of two matrices
I know that for any given matrix $U\in\mathbb{R}^{3\times 3}$ the following matrix
$$ B = cI_3 + dU^TU $$ is invertible, where $I_3$ is the three by three identity matrix and $c,d>0$ are constants coming from the modelled physical system. Precisely, in my case I have that $U=ST$ where $S,T\in\mathbb{R}^{3\times 3}$ are both skew-symmetric matrices.
I even know that if the matrix $U^TU$ has rank 1, then there is an iterative way to compute this inverse.

Is there some reference or some formula I can see which applies to explicit the inversion of the matrix $B$? If it may help, I know that the matrix $U$ has rank 2 and it is singular.

答:
If it may help someone, I have inverted it with the Cayley-Hamilton Theorem, since the eigenvalues in my case are pretty easy to compute.

So, I get an expression similar to this one
$$ B^{-1} = c_1B^2 + c_2B + c_3I_3 $$
which is something quite simple to study.

ContentLicense: CC BY-SA 4.0


问:
Why is $\delta x$ tending to be zero in dy/dx and not $\delta y$ tending to be 0?
Why is  $\Delta x $ and not $\Delta y $ written to be tending towards 0 in case when we give limits to form dy/dx.

, $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}$$

Can we solve this equation like we solve when there is a = sign and we take denominator towards the other numerator?
Like $$\frac{3}{4}=\frac{6}{8}$$

So we get 24=24.

, $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}$$

答:
The derivative of a function is often described as the instantaneous rate of change. I will explain this by way of analogy. Say it takes you $10$ seconds to drive $100$m in a car. Your average speed is $10$m/s, but it is possible that during those $10$ seconds, sometimes you were going faster, and other times you were going slower. Even in the space of $0.1$ seconds, perhaps your speed changed a little bit during that time. So rather than trying to calculate one's speed over a specific time interval, we instead try to make rigorous the notion of what your speed is at one particular point. We do this by calculating the limit
$$
\frac{ds}{dt}=\lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t} \, .
$$
As $t$ gets closer and closer to $0$, there is less and less room for your speed to be changing during that time period. This can be visualised by zooming in on the velocity-time graph. It looks almost linear, meaning that, for example, in the space of $0.000001$ seconds, your speed—the rate of change of distance—is pretty much constant.

Because we are considering a limit, it is not possible to 'solve' the above equation in the way you would a fraction. This is because we are not considering any one value of $t$, but rather what happens at $t$ approaches $0$.


问:
“Proof” that zero is equal to one by infinitely subtracting numbers
Recently, I came across a “proof” that $0=1$. Here is how it goes:

Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. Then, we get $1-x=0-x$. So, $1-x+x=0-x+x$. Hence, $1+0=0+0$ and so $1=0$.

I could not figure out what went wrong in this proof. The result is clearly not true but the proof seems to be true. I then asked a few people and they all could not figure out what went wrong. Can someone come please help me to identify what went wrong? Thank you.

答:
Leaving questions of convergence aside, note that subtraction is not associative.

With just a three term expression of the same type:

$(1-1)-1=-1$, and

$1-(1-1)=1$

Did I just prove $1 = -1$???


问:
Solving two algebraic equations
I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):

$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$  and I need to show that $$\frac{l}{e} = \sqrt{Mr}\left(1 - \frac{2M}{r}\right)^{-1}$$

I've tried rearranging the second eqation to get $l^2$ on its own and then dividing this by $e^2$ from equation 1 to get an expression for $l^2/e^2$ but this is quite messy and I can't see a clear way to simplify it. Can anyone see a straightforward way of doing this?

答:
I concur with chenbai about the "12" being a "4" (I think it's a "2" inside the squared ratio).  Here's how I worked this out.

Rearrange the second equation to deal with the radical:

$$2Mr \ = \ l^2 \ - \ l^2 \cdot \sqrt{1 - (\frac{2M}{l})^2} \ \Rightarrow  \ l^2 \cdot \sqrt{1 - (\frac{2M}{l})^2} \ = \ l^2 \ - \ 2Mr $$

$$\Rightarrow \ l^4 \cdot (1 - \frac{4M^2}{l^2}) \ = \ l^4 \ - \ 4Mrl^2 \ + \ 4M^2r^2 $$

$$ \Rightarrow \  4Mrl^2 \ = \ 4M^2r^2 \ + \ 4M^2l^2 \ = \ 4M^2 \cdot (r^2 + l^2) \ ,  \ \ \ [1] $$

after squaring to eliminate the radical and making some further algebraic rearrangement.

Now,

$$\frac{l^2}{e^2} \ = \   \frac{4M^2 \cdot (r^2 + l^2)}{4Mr} \ \cdot \ (1 - \frac{2M}{r})^{-1} \ \cdot \ (\frac{r^2}{r^2 + l^2}) \ \ , $$

[using equation [1] and the equation for $e^2$ ]

$$\Rightarrow \ \frac{l^2}{e^2} \ = \   Mr \cdot \ (1 - \frac{2M}{r})^{-1}  \ \Rightarrow \ \frac{l}{e} \ =  \ \sqrt{Mr \cdot \ (1 - \frac{2M}{r})^{-1}} .  $$

OK, so I was partly wrong: the "Mr" is under the radical...

Title: Solving two algebraic equations


问:
What is $\det(A + I)$ when $AA^t = I$ and $\det(A) < 0$?
Notation:

$A^t$: transpose of matrix $A$.

$\det A$: determinant of matrix $A$.

$I$: the identity matrix.

We know:

$A \in M_{n\times n}(\mathbb{R}), \\
AA^t = I, \ \mathrm{and \ det}(A) < 0$

We want:  $\det(A + I)$

From $\det A = \det A^t$, $\det A^{-1} = \frac{1}{\det A}$ and $\det A < 0$ we can deduce: $\det A = -1$.  I tried calculating

$$\det(A + I) = \det(A(I + A^t)) = \det(A) \det(I + A^t) = - \det(I + A^t)$$

but that didn't lead to anything. I tried using the fact that $A + A^t$ is symmetric and $A - A^t$ is skew-symmetric but I couldn't achieve anything with that either. I thought about interpreting it via a linear transformation but couldn't come up with anything.

I learned that matrices, where $AA^t = I$, are called orthogonal but that's all I know about orthogonal matrices so I'm hoping to find a solution that doesn't get too deep into orthogonality.

Hints would be appreciated. Thank you in advance!

答:
Let $D=\det(A+I)$. You have
$$D=\det(A)\det(I+A^{-1})=\det(A)\det(I+A^{T})=\det(A)\det(I+A)=\det(A)D$$

So $D=0$ since $\det(A) \neq 1$.


问:
Show that $p^*(n)=n^{-2/3}$ is the threshold function
Show that $p^*(n)=n^{-2/3}$ is the threshold function for the property $G(n,p)$ contains at least $n/6$ pairwise vertex disjoint triangles.
I'm considering the probability that some set of $n/2$ vertices is triangle free, but still stuck.

答:
Below is my solution when I met the problem in a homework. And the main reference is

Alon, Noga, and Raphael Yuster. "Threshold functions for H-factors." Combinatorics, Probability and Computing 2.2 (1993): 137-144.

We claim that $p = n^{-2/3}$ is a threshold function for the given property. WLOG, we assume that $6$ divides $n$.

Lemma 2.1:
There exists a positive constant $c$ such that almost surely $G(n, cn^{-2/3})$ contains less than $n/6$ pairwise vertex disjoint triangles.

Proof: Let $\{A_i : i \in I\}$ denote the set of all distinct labeled copies of $K_3$ in the complete labeled graph on $n$ vertices. Let $B_i$ be the event that $A_i \subset G(n, p)$, and let $X_i$ be the indicator random variable for $B_i$. Let $X = \sum_{i \in I} X_i$ be the number of distinct copies of $K_3$ in $G(n, p)$ (note that here we do not need the copies to be vertex disjoint). Suffice it to show that $X < n/6$ with high probability. It's easy to check that
$$E[X] = \binom{n}{3} p^3 \leq c^3 n/6,$$
which can be upper bounded by $n/12$ with sufficiently small $c$, and implies that $E[X] = \Theta(n)$. For two copies $A_i$ and $A_j$ we say $i \sim j$ if they share one edge, and over all ordered pairs, let
$$\Delta = \sum_{i \sim j} \Pr[B_i \wedge B_j].$$
As $Var[X] \leq E[X] + \Delta$, it remains to show that $\Delta = o(E^2[X])$ before using Chebyshev's Inequality to complete the proof. Actually,
$$\Delta = \binom{n}{5} 5! p^5 \leq c^5 n^{5/3} = o(E^2[X]),$$
with sufficiently large $n$, completing the proof.

Lemma2.2: There exists a positive constant $C$ such that almost surely $G(n, Cn^{-2/3})$ contains at least $n/6$ pairwise vertex disjoint triangles.

Proof: Consider random graph $G(n/2, Cn^{-2/3})$, with the same notations used in the proof of Lemma $2.1$ (except that the number of vertices is $n/2$ now), we have
$$\mu = E[X] \geq C^3n/16$$
and
$$\Delta \leq C^5 n^{5/3}/32,$$
with sufficiently large $C$ and $n$.
As $\Delta > \mu$ when $C$ and $n$ are sufficiently large, by the extended Janson inequality, we have
$$\Pr[X=0] = \Pr[\bigwedge_{i \in I} \overline{B_i}] \leq e^{-\mu^2/2\Delta} \leq e^{-100n^{1/3}}.$$
Therefore, we conclude that almost surely $G(m, Cn^{-2/3})$ contains a triangle if $m \geq n/2$. From a random graph $G(n, p)$, with probability as least $(1 - e^{-100n^{1/3}})^{n/6}$, we can repeatedly do for $i \in [n/6]$,

[1] Find a triangle in the current graph on the remaining $n - 3(i-1)$ vertices

[2] Remove the $3$ vertices and all edges incident to them, now we essentially have $G(n-3i, Cn^{-2/3})$

Finally, we have $n/6$ pairwise vertex disjoint triangles with high probability, completing the proof of both the lemma and the whole problem.


问:
Recognize a circle in the complex plane
I have this equation: $z \in \mathbb{C}; \frac{1}{z} - \frac{1}{\bar{z}} = -i$

and transformed it into $2b = a^2 + b^2$ with $a=Re(z)$ and $b=Im(z)$

The original equation has a circle around $(0,1)$ with radius 1 as solution set. How can I read this from my equation?

答: Just complete squares: $$a^2+b^2-2b=a^2+(b^2-2b+1)-1=0\to a^2+(b-1)^2=1,$$ circle of radius $1$ centered at $(0,1)$.

Title: Recognize a circle in the complex plane


问:
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

答:
It seems the following.

We can easily transform the inequality to the form $(a+b+c+d-1)^2\le 4abcd+4$.

Now we estimate the infimum of a function $f(a,b,c,d)=4abcd+4-(a+b+c+d-1)^2$  on the set
$X=\{(a,b,c,d):a,b,c,d\ge 0$ and $a^2+b^2+c^2+d^2=3\}.$ Since the set $X$ is compact, the function $f$ attains its minimum on $X$ in some point $x=(a,b,c,d)\in X$.
Suppose that $a+b+c+d=M$. Then the point $x$ is also a point of an conditional extremum
of the function $f_1(a,b,c,d)=abcd$ with the constraints $g_1(a,b,c,d)=a^2+b^2+c^2+d^2-3=0$ and $g_2(a,b,c,d)=a+b+c+d–M=0$. The Lagrange Theorem implies that one of the following cases hold.

One of the numbers $a,b,c,d$ is equal to $0$. Then $0\le a+b+c+d\le 3\sqrt{\frac{a^2+b^2+c^2+d^2}3}=3$ and $(a+b+c+d-1)^2\le 4$ and $f(a,b,c,d)\ge 0$.

$\operatorname{rank}J(x)<2$, where $J$ is the Jacoby matrix $||\frac{\partial g_i(x)}{\partial x_j}||$. Since $J(a,b,c,d)=\left(\begin{matrix}2a & 2b & 2c & 2d\\ 1 & 1 & 1 &1\end{matrix}\right)$, we see that $a=b=c=d$. Then $a=b=c=d=\sqrt{3}/2$ and $f(a,b,c,d)=4\sqrt{3}-27/4>0$.

There exist numbers $\lambda_1$ and $\lambda_2$ such that
$(f_1+\lambda_1g_1+\lambda_2g_2)’(a,b,c,d)=0$. Then we have a system of equations.

$\cases{
bcd+2\lambda_1a+\lambda_2=0\\
acd+2\lambda_1b+\lambda_2=0\\
abd+2\lambda_1c+\lambda_2=0\\
abc+2\lambda_1d+\lambda_2=0\\
}$

Hence all numbers $a,b,c,d$ are the roots of the square equation $abcd+2\lambda_1t^2+\lambda_2t=0$. Since the inequality is symmetric, without loss of generality we may assume that one of the following cases hold:

3.1. $a=b$, $c=d$. Then we have $a^2+c^2=3/2$ and $f(a,c)=4a^2c^2+4-(2a+2c-1)^2$.  To find the minimal value of $f$, we again use  Lagrange Theorem. The cases $a=0$ or $c=0$ were considered above. The Jacoby matrix in this case is equal to $(2a$ $2c)$ and have nonzero rank, provided $a\not=0$  or $c\not=0$. So it remains to consider the case when there exists a number $\lambda$ such that

$(f+\lambda(a^2+c^2))’(a,c)=0$. Then we have a system of equations.

$$\cases{
8ac^2–4(2a+2c-1)+2\lambda a=0\\
8a^2c–4(2a+2c-1)+2\lambda c=0\\
}$$

If $a\not=0$ and $c\not=0$, then we obtain that $(2ac^2+2a^2c-2a-2c+1)(c-a)=0.$

Since $a=c$ implies $a=b=c=d$ and this case was considered above, without loss of generality we may assume that $2ac^2+2a^2c-2a-2c+1=0 $. Put $t=a+c$. Then $2ac=t^2-3/2$ and we have $0=2t^3-7t+2=(t+2)(2t^2-4t+1)$. Since $t\ge 0$, we have $t=1\pm\sqrt{2}/2$, $2ac=t^2-3/2=\mp\sqrt{2}$, and $f(a,c)=3\mp 2\sqrt{2}>0$.

3.2. $b=c=d$. Then we have $a^2+3b^2=3$ and $f(a,b)=4ab^3+4-(a+3b-1)^2$.  To find the minimal value of $f$, we again use  Lagrange Theorem. The cases $a=0$ or $b=0$ were considered above. The Jacoby matrix in this case is equal to $(a$ $3b)$ and have nonzero rank, provided $a\not=0$  or $b\not=0$. So it remains to consider the case when there exists a number $\lambda$ such that

$(f+\lambda(a^2+3b^2))’(a,b)=0$. Then we have a system of equations.

$$\cases{
4b^3–2(a+3b-1)+2\lambda a=0\\
12ab^2–4(a+3b-1)+6\lambda b=0\\
}$$

If $a\not=0$ and $b\not=0$, then we obtain that $(2ab^2+2b^3-a-3b+1)(b-a)=0.$

Since $a=b$ implies $a=b=c=d$ and this case was considered above, without loss of generality we may assume that $2ab^2+2b^3-a-3b+1=0$. Then $1-2b^2\not=0$ and
$a=(2b^3-3b+1)/(1-2b^2)=\sqrt{3-3b^2}$. If $b=1$, then $a=0$ and this case was considered above. If $b\not=1$, then $(2b^2+2b–1)/(1-2b^2)=\sqrt{(3+3b)/(1-b)}$ and $8b^5+8b^4–10b^3–8b^2+4b+1=0$. Mathcad calculated that this equation has two not real roots, one negative root, and roots $b_1\sim 0.60442635232610682922$  and $b_2\sim 0.82451890455979510644$. In the first case we obtain $f(a,b)\sim 0.409>0$, in the second case we obtain $f(a,b)\sim 0.177>0$.

From the above we can see that the initial inequality becomes an equality iff one of the numbers $a,b,c,d$ if equal to $0$ and the others are equal to $1$.


答:
Let $x=1−1−1−1−1−1−1-\cdots.$

Since $1−1=0$

$x=0-1-1-1-1-1-1-\cdots$.

Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get

--> $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. <--

Here is the error. Having a minus before the brackets negates everything inside.

So it actually becomes:

$$x = 0 - (1 + 1 + \cdots)$$

And I don't think it's a valid math operation to cross out $\infty$ on both sides, as $\infty$ is just a placeholder for very large number (not a concrete large number, so $\infty_{left} \ne \infty_{right}$).

Title: “Proof” that zero is equal to one by infinitely subtracting numbers

Body:
Recently, I came across a “proof” that $0=1$. Here is how it goes:

Let $x = 1-1-1-1-1-1-1-\cdots$. Since $1-1=0$, $x=0-1-1-1-1-1-1-\cdots$. Now, we bracket the $1-1-1-1-1-1-\cdots$ on both sides and we get $x=1-(1-1-1-1-1-1\cdots)=0-(1-1-1-1-1-1-\cdots)$. Then, we get $1-x=0-x$. So, $1-x+x=0-x+x$. Hence, $1+0=0+0$ and so $1=0$.

I could not figure out what went wrong in this proof. The result is clearly not true but the proof seems to be true. I then asked a few people and they all could not figure out what went wrong. Can someone come please help me to identify what went wrong? Thank you.


答:
Cirrcle whose center is at $(-g,-f)$ with radius $\sqrt{f^2+g^2-c}$ can be expressed as
$$z\bar z+z_0\bar z+\bar z_0 z+c=0$$ is one form of equation of circle, whose center is at $z=-z_0=-g-if$ and radius is $r=\sqrt{z_0 \bar z_0 -c}$ In rhe given question
$$\frac{1}{z}-\frac{1}{\bar z}=-i \implies z\bar z-i\bar z+iz=0 \implies z_0=-i$$
So the center of this circle is at $z=-z_0=i=0+i$ or $(0,1)$, its radius is $\sqrt{-i.i}=1$

Title: Recognize a circle in the complex plane

Body:
I have this equation: $z \in \mathbb{C}; \frac{1}{z} - \frac{1}{\bar{z}} = -i$

and transformed it into $2b = a^2 + b^2$ with $a=Re(z)$ and $b=Im(z)$

The original equation has a circle around $(0,1)$ with radius 1 as solution set. How can I read this from my equation?


问:
Show that when $p$ is a principal type, it must contain the formula that implies it
In my model theory lecture course, a partial type $p$ is defined to be principal when there is some $ϕ$ such that $T \models ∃vϕ(v)$ (for any n-place variable $V$) and for any $ψ ∈ p$, $T \models ∀v(ϕ(v) → ψ(v))$.

[and we define a partial type of a theory $T$ in variables $x$ to be a satisfiable set $P$ of
formulas in the variables $x$, containing $T$ and closed under logical deduction.]

I'm asked to show that when $p$ is a principal (complete) type, the formula $ϕ$ which features in the definition above must be in $p$.

It's clear that since $T \models ∃vϕ(v)$, and $p$ must be closed under logical deduction, that $∃vϕ(v) \in p$ , but I don't see how it follows that $ϕ(v) \in p$, which is what I'm being asked to prove.

Any suggestions you could give me would very much be appreciated.

答: As pointed out in the comment this question only makes sense for complete types. So let $p$ be a principal type, with principal formula $φ \in F_n$. Since p is complete, $φ \not\in p \implies ¬φ \in p$. Add new constant symbols $c = (c_1, c_2, ..., c_n)$. The principality condition gives us that $T \cup \{φ(c)\} \models ψ(c)$ for all $ψ(x) \in p$, and so $T \cup \{φ(c)\} \models ¬φ(c)$. It follows that $T \models ¬φ(c)$, and therefore $T \models ¬\exists x φ(x)$. This contradicts the fact that $φ$ is realised in all models of T, i.e. $Τ \models \exists x φ(x)$. Thus $φ \in p$.


问:
Integral $\int\limits^{\infty}_0\frac{\tan^{-1}t }{(1+t)^{n+1}} dt$
I'm having a good amount of trouble evaluating this: $$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}},\ n>0$$
Here are some methods I've tried:
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\frac1n\int\limits^{\infty}_0\frac{dt}{(1+t^2)(1+t)^{n}}$$
using integration by parts. I then tried more integration by parts, residue theorem, and expanding into a power series but failed. I did however use partial fractions for $n=2$ to get $1/4$.
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\frac{\pi}{2n}-\int\limits^{\infty}_0\frac1{(1+t)^{n+1}}\int\limits^{\infty}_0\frac{\sin(x)}xe^{-xt}dxdt=\frac{\pi}{2n}-\int\limits^{\infty}_0\frac{\sin(x)}xE_{n+1}(x)e^{-x}dx$$
using the Laplace Transform of $\text{sinc}(x)$ and the $E_n$-function.
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\int\limits^1_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}+\int\limits^1_0\frac{\cot^{-1}(t)t^{n-1}dt}{(1+t)^{n+1}}=\int\limits^1_0\frac{\tan^{-1}(t)\left(1-t^{n-1}\right)dt}{(1+t)^{n+1}}+\frac{\pi}{2^{n+1}n}$$
This one I felt the best about and its also where I got that at $n=1$ the integral is $\pi/4$, but I was unable to go further.

Update: I had a couple more attempts, one of which I posted as an answer, after Claude Leibovici's idea reminded me you can do partial fractions on $\frac1{(1+x^2)(1+x)^n}$.

Notice that if we write
$$\frac1{(1+x^2)(1+x)^n}=\frac{1+x}{1+x^2}-\frac{a_0+a_1x+\dots+a_{m-1}x^{m-1}}{(1+x)^n}$$
then the coefficients $a_k$ follow the pattern
$$a_0=0,\ a_1=C_1^{n+1},\ a_2=C_2^{n+1}-a_0,\ a_3=C_3^{n+1}-a_1,\ a_4=C_4^{n+1}-a_2\dots$$
The only problem is that this sequence is always infinite and the power series does not converge on all of $[0,\infty)$, so I believe the coefficients of $1$ and $x$ in the numerator of $\frac{1+x}{1+x^2}$ could be changed to avoid this.

答:
This is not an answer but it is too long for comments.

For the computation of
$$I_n=\int\limits^{\infty}_0\frac{dt}{(1+t^2)(1+t)^{n+1}}$$ it is amazing that a CAS gives a solution in terms of a generalized hypergeometric function which works very fine ... except when $n$ is an integer !

What I think is that writing
$$(1+t^2)(1+t)^{n+1}=(t+i)(t-i)(1+t)^{n+1}$$ and using partial fraction could be a solution. For example, for $n=3$, the integrand is
$$-\frac{1+i}{8(t+i)}-\frac{1-i}{8(t-i)}+\frac{1}{4 (t+1)}+\frac{1}{2
   (t+1)^2}+\frac{1}{2 (t+1)^3}$$ and
$$\int \Big[\frac{1+i}{8(t+i)}+\frac{1-i}{8(t-i)}\Big]\,dt=\frac{1}{8} \log \left(t^2+1\right)+\frac{1}{4} \tan ^{-1}(t)$$ For $n=4$ , the integrand is
$$\frac{i}{8 (t-i)}-\frac{i}{8 (t+i)}+\frac{1}{4 (t+1)^2}+\frac{1}{2 (t+1)^3}+\frac{1}{2   (t+1)^4}$$
$$\int \Big[\frac{i}{8 (t-i)}-\frac{i}{8 (t+i)}\Big]\,dt=-\frac{1}{4} \tan ^{-1}(t)$$ and, obviously,  the coefficients of the terms $\frac{1}{ t\pm i}$ are complex numbers if $n$ is odd and pure imaginary numbers if $n$ is even.

Probably, the two cases could be separatly studied.

All these integrals are in the form $I_n=a_n+b_n\pi$ but the $b_n$'s are all zero for $n=4k+2$


问:
Spivak problem on Schwarz inequality
I have a question regarding problem 19 in the 3rd Ed. of Spivak's Calculus. Specifically, part (a). The question concerns the Schwarz inequality:
$$
x_1y_1 + x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2} \ .
$$
It says to prove that if $x_1=\lambda y_1$ and $x_2 = \lambda y_2$ for some number $\lambda$, then equality holds in the Schwarz inequality.

Substituting the given values for $x_1$ and $x_2$ we have
$$
\lambda (y_1^2+ y_2^2) \leq |\lambda|(y_1^2+y_2^2) \ .
$$
It appears to me that equality can only hold if $\lambda \geq 0$. Can someone explain to me how equality holds for any given $\lambda$?

答: That is a typo. You need $\lambda\ge 0$.</p>

Title: Spivak problem on Schwarz inequality


问:
Let $G \le S_n$ ,show that if $G$ has an odd permutation then half of the elements in $G$ are even permutation and the other half are odd permutation.
Assume $G \le S_n$ ,show that if $G$ has an odd permutation then half of the elements in $G$ are even permutation and the other half are odd permutation.   Moreover conclude that $G$ has a subgroup of index $2$.  I've seen the following related proof:

Define a map  $\text{sgn}:S_n→\left\{−1,1\right\}$ which is given by:

$$  \begin{align} \text{sgn(σ)} = \left\{ \begin{array}{cc}                  1 & \hspace{5mm} \text{if σ is a product of an even number of >transpositions} \\                 -1 & \hspace{5mm} \text{if σ is a product of an odd number of transpositions} \\                 \end{array} \right. \end{align}$$

Then the kernel of the mapping would be the alternating group $A_n$,by the isomorphism theorem $S_n/A_n≅\left\{−1,1\right\}$.

Now by the Lagrange's  theorem  it  follows  that  $|A_n|= \frac{|S_n|}{2}.$

This  means  that  exactly  half  of  the  permutations  in    Sn    are  even  and  so  exactly  half  are  also  odd.

But  I  don't  see  how  to  conclude  that  for    $G \le  S_n$  with  the  given  assumptions  the  claim  that  half  of the  elements  in  $G$  are  even  permutation  and  the  other  half  are  odd  permutation  does  hold. Maybe after  this  it  looks  straight  forward  but  can  someone $2$? show  me why $G$ has a subgroup of index

答: Alternatively, if $\tau$ is an odd permutation in $G$, then the map $\sigma \mapsto \tau\sigma$ is a bijection $G \to G$ and it maps even permutations to odd ones and vice-versa.


问:
Exercise 2.7.5 of the probabilistic method
Let $H$ be a graph, and let $n > |V(H)|$ be an integer.Suppose there is a graph on $n$ vertices and $t$ edges containing no copy of $H$, and suppose that $tk>n^2\log_en$. Show that there is a coloring of the edges of the complete graph on $n$ vertices by $k$ colors with no monochromatic copy of $H$.

答:
Below is my solution when I met the problem in my homework before.

When $|V(H)| \leq 1$, it's not so meaningful, so we only need to care about the cases when $|V(H)| \geq 2$, and then $n \geq |V(H)| + 1 \geq 3$.

Let $G$ be such a graph on $n$ vertices and $t$ edges containing no copy of $H$. As we supposed that $tk > n^2 \log_e n > \binom{n}{2}$, maybe we can find some way to paste $k$ monochromatic copies of $G$ onto $K_n$ (some edges may be colored more than once, among the colors they are colored in, choosing any one can make the conclusion hold), where each edge is colored and no copy of $H$ exists in the graph.

So, let's randomly paste $k$ monochromatic copies of $G$ onto $K_n$, mathematically speaking, let $V(G) = [n]$ and $V(K_n) = \{v_i\}_{i \in [n]}$, for each color, we randomly pick a permutation of $[n]$, $\sigma: [n] \rightarrow \sigma([n])$, then for all $(i, j) \in E(V(G))$, we color the edge $(v_{\sigma(i)}, v_{\sigma(j)})$ in $K_n$ in that color (if it's already colored, then we simply recolor it). Let $X$ be the random variable of edges not colored, which is naturally the summation of $X_e$, the indicator random variable of $e$ being not colored, for all $e \in E(K_n)$, thus clearly,
$$E[X_e] = (1 - \frac{2t(n-2)!}{n!})^k = (1 - \frac{2t}{n(n-1)})^k.$$
Therefore,
\begin{align*}
    E[X] = & \sum_e E[X_e] \\
          = & \sum_e (1 - \frac{2t}{n(n-1)})^k \\
          = & \binom{n}{2} (1 - \frac{2t}{n(n-1)})^k \\
       \leq & \frac{n(n-1)}{2} e^{-\frac{2tk}{n(n-1)}} \\
          < & \frac{n^2}{2} e^{-2\log_e n} \\
          = & \frac{1}{2},
\end{align*}
as $X$ is an integer, there is a coloring by pasting $k$ monochromatic copies of $G$ with $X = 0$, i.e., each edge is colored, completing the proof.


问:
Why isn't the Euler characteristic equal to 2 for this polyhedron?
I've managed to confuse myself on a topic I thought I understood.

To set up, if a surface $M$ is a compact Riemannian manifold with boundary $\partial M$, then the Gauss-Bonnet theorem states $$\int_{M}KdS+\int_{\partial M}kds=2\pi \chi(M)$$ where $K$ is the Gaussian curvature of $M$, $k$ is the geodesic curvature of $\partial M$, and $\chi(M)$ is the Euler characteristic of the surface. Now, if our surface was $\textit{almost}$ smooth everywhere, except at some finite set of points $\left\{ x_{k}\right\}_{k\le n}\subset M$ where there were vertex singularities with angular defects $\left\{ \theta_{k}\right\}_{k\le n}$ at these points, then the Gauss-Bonnet theorem generalizes to $$\int_{M}KdS+\int_{\partial M}kds+\sum_{k\le n}\theta_{k}=2\pi \chi(M).$$ This is intuitively very pleasing, because it's essentially the statement that a surface's curvature, when viewed as a measure, can either live on the subset of the surface that's Riemannian, the boundary, or at vertices. This generalizes both Gauss-Bonnet for smooth surfaces and Descarte's theorem for the total angular defect of polyhedra.

In particular, if $M$ is a polyhedron with $n$ vertices, then $K=k=0$, and the theorem reduces to Descarte's theorem $$\boxed{\sum_{k\le n}\theta_{k}=2\pi \chi(M)}$$ This formula works beautifully for convex or toroidal polyhedra. But for some nonconvex polyhedra, I have no idea what's going on and am completely lost. To illustrate, I'll focus my attention particularly on two examples.

First, the $\textit{pentakis dodecahedron}$ is simply a regular dodecahedral surface with (short) pentagonal pyramids attached to each face. From reading its Wiki page, its Euler characteristic is $2$. No surprises here, since it has no holes or handles.

Then we come to the $\textit{small stellated dodecahedron}$, which is very similar, but with pentagonal pyramids substantially taller. It's Wiki page, Wolfram page, and this source all state, quite strangely, that it has twelve pentagrams as faces with thirty edges and only twelve vertices (which I would never even consider being the case). Its Euler characteristic then drops to $-6$. The same peculiar counting method seems to be implemented for several other nonconvex polyhedra.

I tested this by counting the small stellated dodecahedron's total angular defect, as shown here. Since each triangular face is isosceles with one angle of $36^{\circ}$ at the top and two angles of $72^{\circ}$ at the base, this is straightforward. There are twelve vertices (the pyramidal vertices) with a positive defect of $$360^{\circ}-5\cdot36^{\circ}=180^{\circ}$$ and twenty vertices (the original dodecahedron's vertices) with a defect of $$360^{\circ}-6\cdot72^{\circ}={-72^{\circ}}.$$ The total angular defect then comes to $$12\cdot 180^{\circ}-20\cdot72^{\circ}=720^{\circ}.$$ This is exactly what we would expect given that the small stellated dodecahedron has no holes or handles. This would also suggest that it has the same topology as a sphere and has an Euler characteristic of $2$. Moreover, whenever I count the vertices, edges, and faces, I get $V-E+F=32-90+60=2,$ which I'd expect anyways.

My question is, how could this $\textit{not}$ be the correct answer? My background is in analysis and differential geometry, not topology. If I'm wrong, I have no idea how. I also doubt all these sources are simultaneously wrong. But nowhere in the Gauss-Bonnet/Descarte theorem is convexity assumed. Moreover, it sounds preposterous from an analysis perspective that increasing the heights of the pentagonal pyramids to a certain point suddenly causes the total curvature to discontinuously become negative.

The sources for the stellated dodecahedron all state that this polyhedron confused early topologists. If this is just a funny way of counting faces, edges, and vertices, then who decided this was the $\text{"correct"}$ way to count them, and which way is $\text{"correct"}$ for the Gauss-Bonnet/Descarte theorem?

答:
Here is an anlogue from a lower dimension:

For polygons (aka. polygonal lines), the analogue of angular defect are the interior angles.
For a closed non-self-intersecting polygonal line with $n$ corners, the sum of these angles is $\pi(n-2)$.

But the pentagram (like the small stellated dodecahedron) is self-intersecting, and so this rule does not apply.
Note that for the pentagram these intersections are not vertices. If you consider them as vertices you actually study the star on the right.

Still, I find it strange to state that the Euler characteristic is negative, since topologically the pentagram is a circle (and the small stellated dodecahedron is a sphere).


问:
Prove that any group $G$ with $|G|=588$ is solvable
I'm stuck trying to solve this problem from my abstract algebra course:

Prove that every group of order $588$ is solvable (If you assume that all groups of certain order are solvable, you must prove it too).

First I noticed that $588=2^2\cdot 3\cdot 7^2$, so I guess the prove will use Sylow $p$-subgroups, but I don't find right what I need to do. Any help or hint will be appreciated.

Thanks in advance.

答:
You have seen, using Sylow's theorems, that there is only one Sylow $7$ subgroup, calling it $P_7$.

Then $G/P_7$ has order $12$.

But any group of order $p^2q$ is solvable.

Hence $G/P_7$ is solvable.

Then since $P_7$ is solvable, $G$ is solvable.


问:
The probabilistic method exercise 2.7.4
This is exercise 2.7.2 of the book "the probabilistic method".

Suppose $p>n>10m^2$, with $p$ prime, and let $0<a_1<a_2<...<a_m<p$ be integers. Prove that there is an integer $x$, $0<x<p$, for which the $m$ numbers $(xa_i(\mod p))\mod n$, $1 \le i \le m$ are pairwise distinct.

And I think this problem in this way. Let $\chi_x$ be the indicator for the event that the m numbers $(xa_i(\mod p))\mod n$ are pairwise distinct. Then $P(\chi_x) = \frac{n\choose m}{n^m}$, The the expected number of $(xa_i(\mod p))\mod n$ are pairwise distinct is $\sum E[\chi_x] = (p-1)\frac{n\choose m}{n^m} > \frac{(\frac{n}{m})^m}{n^{m-1}} = \frac{n}{m^m}$. Obviously, this way can not get the result that the expected number is greater than 1. Could anyone help about this problem?

答:
Below is my solution when I met the problem in homework before.

Choose integer $x$ uniformly and randomly in $[p-1]$, clearly, with the assumptions, for each $i$, $(xa_i\ \mathrm{mod}\ p)$ is distinct in $[p-1]$. Let $X$ be the random variable denoting the number of ordered pairs $(i, j)$, where $i, j \in [m]$ and $i \neq j$, which is naturally the sum of indicator random variables $X_{ij}$ for all possible $(i, j)$, where $X_{ij}$ indicates $(xa_i\ \mathrm{mod}\ p)\ \mathrm{mod}\ n = (xa_j\ \mathrm{mod}\ p)\ \mathrm{mod}\ n$. Thus,
\begin{align*}
    E[X] & = \sum_{i \neq j} X_{ij} \\
          & \leq \sum_{i} \sum_{j \neq i} \frac{\lfloor \frac{p-1}{n} \rfloor}{p-1} \\
          & \leq \sum_{i} \sum_{j \neq i} \frac{1}{n} \\
          & = \frac{m(m-1)}{n} \\
          & < \frac{m(m-1)}{10m^2} \\
          & < \frac{1}{10},
\end{align*}

which implies for some $x$, $X = 0$, i.e., $(xa_i\ \mathrm{mod}\ p)\ \mathrm{mod}\ n, 1 \leq i \leq m$ are pairwise distinct, completing the proof.

Title: The probabilistic method exercise 2.7.4


问:
Is Binary Matrix Factorization unique under given matrix sizes
I have a binary data matrix $C_{m \times n}$. I know that using Boolean algebra (logical or for summation and logical and for multiplication), this matrix can be factorized into two binary matrices $A_{m \times k}$ and $B_{k \times n}$, such that $$AB=C$$

My question is whether this factorization unique, i.e., is there any alternative $A$ and $B$ matrices that result in $C$ matrix, is the $k$ is given. We can also assume that $k$ is minimal, which means there isn't any $k' < k$ that supports this factorization.

答:
Here are two counterexamples for $k=m=n=2$:

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

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

In fact even for $k=m=n=1$, you have already the counter-example:

$$0=0 \times 1=1 \times 0=0 \times 0$$

There is a very direct combinatorial argument: pigeonhole principle for the existence of counter-examples that could work for any dimension, but that I give for example in the case of $3 \times 3$ matrices:

There are $2^9=512$ possible binary matrices: therefore, one can write $2^{18}=262144$ products  $AB$ (the "pigeons"). As in front of them, there are only $512$ possible results (the "holes"), necessarily there is a hole with (at least) 2 pigeons, i.e., a matrix $C$ with $C=A_1B_1=A_2B_2$ in two different ways...

Here is a paper giving a standard form for this type of matrices (but using rather sophisticated algebra).


问:
Symbolizing only and all other in first-order logic?
I'm having trouble symbolizing something like "Only cats play with all other cats" or "Only cats play with other cats" in first-order logic. I understand how to symbolize "only" and how to symbolize "other" but I'm not sure how to combine the two in this situation. Any help is appreciated!</p>

答:
Let $C(x)$ denote "x is a cat" and $P(x,y)$ denote "x plays with y". Then:

-"Only cats play with all other cats" translates to:
$\forall x (\forall y ((x \neq y \land C(y)) \implies P(x,y)) \implies C(x))$

This says that given x, if for all different y which are cats, x plays with y, then x is a cat. Equivalently, if something plays with all other things that are cats, then that something is a cat.

-"Only cats play with other cats" translates to:
$\forall x (\exists y (x \neq y \land C(y) \land P(x,y)) \implies C(x))$

Again, this says that given x, if there exists some y different from x, which is a cat, and s.t. x plays with y, then x is a cat. Equivalently, if something plays with something different which is a cat, then that something is a cat itself.


问:
Is this map an embedding?
can someone give me an idea why the map $F:M\rightarrow M\times \mathbb{R}$ s.t. $F(p)=(p,f(p))$ where $f:M\rightarrow \mathbb{R}$ is smooth, is an embedding?, a map  s. t. $dF_p$ is injective and $F$ is homeo onto his image. I can not see anything using the differencial of $F$.

答:
$F$ is smooth, since it is the (cartesian) product of smooth maps

$dF_p$ is injective, since
$$dF_p=\begin{pmatrix}I_n&*\end{pmatrix}^t$$ so $rk(dF)=n$, and $dF$ is injective. This implies that $F$ is a smooth immersion

$F$ is injective (trivially, since $(p,f(p))=(q,f(q))\leftrightarrow p=q$).

$F: M\to F(M)$ is open, since for any open $A\subset M$, $F(A)=\left(A\times \mathbb{R}\right)\cap F(M)$.

Since every bijective open map is an homeomorphis, we have that $F$ is an embedding.

Title: Is this map an embedding?


问:
what $H^1_q$ is?
I don't know the meaning of the notation $H^1_q$ (a Sobolev space)?
Where I can find the exact definition of that notation?
Thank you in advance.

答:
A good resource for all things Sobolev-space related is Sobolev Spaces by R. A. Adams and J. J. F. Fournier, $2^\mathrm{nd}$ edition.  Page $60$ there provides the definition that you're looking for:

$$H^m_q = H^{m,q}(\Omega) \equiv \mbox{the completion of } \{u\in C^m(\Omega) \mid \|u\|_{m,q} \lt \infty \}  $$
where the completion is with respect to the norm $\|u\|_{m,q}$.

In fact, as was proven in a paper by Myers and Serrin with the famously short title H=W , the spaces $H^{m,q}$ and $W^{m,q}$ are the same so most of the time a choice between $H$ and $W$ will be entirely stylistic.

Title: what $H^1_q$ is?


答:
From what I recall, the actual statement of the inequality is
$$
\left|\langle x,y \rangle\right|\leq \lVert x\rVert\cdot\lVert y\rVert
$$
or, with your notations
$$
\left|x_1y_1 + x_2y_2\right|\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}
$$
so the book is right (you'll have $|\lambda|$ on both sides). Note that this is a stronger statement than what you made, since we always have $x_1y_1 + x_2y_2\leq \left|x_1y_1 + x_2y_2\right|$.</p>

Title: Spivak problem on Schwarz inequality

Body:
I have a question regarding problem 19 in the 3rd Ed. of Spivak's Calculus. Specifically, part (a). The question concerns the Schwarz inequality:
$$
x_1y_1 + x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2} \ .
$$
It says to prove that if $x_1=\lambda y_1$ and $x_2 = \lambda y_2$ for some number $\lambda$, then equality holds in the Schwarz inequality.

Substituting the given values for $x_1$ and $x_2$ we have
$$
\lambda (y_1^2+ y_2^2) \leq |\lambda|(y_1^2+y_2^2) \ .
$$
It appears to me that equality can only hold if $\lambda \geq 0$. Can someone explain to me how equality holds for any given $\lambda$?


问:
How to guessing $y_p$ of $\frac{d^2 y}{dx^2}-y=\frac{2}{1+e^x}.$
Solve the ODE
$$\dfrac{d^2 y}{dx^2}-y=\dfrac{2}{1+e^x}.$$

I have tried to solve using undetermined coefficients.

Let $y=y_h+y_p$.

I can find the homogeneous solution, $y_h$ is $$y_h=c_1e^x+c_2e^{-x}.$$

Now I'm confused to determine $y_p$. I don't know guessing $y_p$ of $\dfrac{2}{1+e^x}$.

Any hint to determine $y_p$?

答:
The right side does not have the format for the method of undetermined parameters. For that it would have to be a sum of terms $q_k(x)e^{\lambda_k x}$ where $q_i$ are polynomials and $λ_k$ may be complex (along with the coefficients of $q_i$, then the conjugate of the term will also be in the sum for a real-valued right side).

By a slight abuse of the method one could expand the given right side into a series that has terms of that format, for $x>0$ this would be
$$
\frac{2e^{-x}}{1+e^{-x}}=2\sum_{k=1}^\infty e^{-kx}.
$$
Then each term $2e^{-kx}$ contributes a term $a_1xe^{-x}$ and $a_ke^{-kx}$ for $k\ge 2$ where
$$
a_1=-1,~~a_k=\frac2{k^2-1}\text{ for }k\ge 2
$$
A similar approach works for $x<0$ where $|e^x|<1$.


问:
if $f$ and $g$ are smoothly homotopic, then they have some common regular value
I was reading Milnor's differential topology book. in page 24 when talking about homotopy class and mod 2 degree, there is an claim:

Let $f,g:M\to N$ as smooth map between two smooth manifolds (where $M$
is compact and $N$ is connected), we assume $f$ is smoothly homotopic to
$g$, then there exists a regular value for both maps $f$ and $g$.

I can prove the following result,that is given map $f$ if it has a regular value $p$,then exist a map in the homotopy class that can be any regular value.

That is we know given $p$ there exsit a diffeomorphism $F$ which can change $p\in N$ to any $q\in N$ ,and $F$ smooth isotopy to identity maps(which is given by homogeneity lemma).

hence $f\cong f$(f smooth homotopy to itself) we also have $F\cong id$ hence $f\cong F\circ f$ now the RHS $F \circ f$  has regular value $q$

I don't know how to use Sard's theorem here to prove given $f,g$ exist common regular value.

答:
This is true for any two $f,g$. You do not need the fact that they are homotopic.

Let $M' = M + M$ be the disjoint union of two copies of $M$. Then $f, g$ induces a smooth map $\phi : M' \to N$ by taking $f$ on the first summand and $g$ on the second.

The set of regular values of $\phi$ is non-empty (it is dense in $N$). Any regular value of $\phi$ is a common regular value of $f,g$.


问:
Determining all the $1$-types of a given theory
I'm trying to solve an exercise in my lecture notes. I'm given the following theory of infinitely many disjoint infinite unary predicates:

Let $L_n$ be the language with $n$ unary predicate symbols $P_1, . . . , P_n$. Let $T_n$ be the theory asserting that each $P_i$ is infinite, the $P_i$ are disjoint, and there are infinitely many elements not in any $P_i$, for $i \leq n$. Finally, let $T = \bigcup_n T_n$.

After having proved that both $T_n$ and $T$ are complete, I'm then asked to determine all the $1$-types of $T$, show that
there's a unique 1-type $p_n$ including $P_n(x)$, and a unique type $q$ that is not
one of the $p_n$.

I have no idea how I'd go about determining all the $1$-types, and while I can certainly find a $1$-type $p_n$ including $P_n(x)$ (by setting $p_n = $ tp$_A(a) = \lbrace ϕ(x) : A \models ϕ(a)\rbrace $, where $A$ is some model with $a\in A$ such that $a\in P_n^A$), but I'm not sure how I'd show it's unique.

I know that if If $π : A → B$ is an isomorphism, $a ∈ A$, $b ∈ B$ and $π(a)= b$ that tp$_A(a) = $tp$_B(b)$, so clearly there's only one $p_n$ obtained by taking a set like tp$_A(a) = \lbrace ϕ(x) : A \models ϕ(a)\rbrace $, but I don't know how to show there are no other $1$-types containing $p_n$ (perhaps this will become clearer once I know what all the $1$-types are). I also have no idea how I'd go about finding the $q$ I'm asked for.

I'd really appreciate any help you could offer.

答:
First, you should know that in general determining all the 1-types of a given theory is difficult and that there is no automatic recipe to do it.

One particularly useful way to simplify the task is to show that the given theory admits quantifier elimination, that is, that every first-order formula $\phi(x)$ is logically equivalent modulo $T$ to a quantifier-free formula $\psi(x)$. Then understanding a 1-type amounts to understanding the possible quantifier-free formulas that it contains.

Proving quantifier elimination can be done "by hand", by exhibiting a quantifier-free formula for every existential positive formula (i.e., a first-order formula of the form $\exists \bar y(\bigwedge ...)$ where inside the conjunction only atomic formulas appear). Perhaps you can already see how to do this for your example.

Another way to do it is to use properties of the models of $T$ (which is particularly useful when $T$ is $\omega$-categorical for example, since it only has one countable model). For example, if $T$ is $\omega$-categorical, it has quantifier elimination iff its unique countable model is ultrahomogeneous (i.e., every partial isomorphism between finitely generated substructures extends to an automorphism of the whole model). You can for example check that the unique countable model of every $T_n$ is indeed ultrahomogeneous. Even if $T$ is not $\omega$-categorical, this is enough for your purposes: every fo-formula only involves finitely many of the predicates $P_i$, therefore by quantifier-elimination for $T_n$, it is equivalent modulo $T_n$ to a quantifier-free formula. This equivalence is still true modulo $T$, therefore $T$ has quantifier elimination.


问:
Find subgroup of $S_5$ isomorphic to the quaternion $Q$
I'm trying to solve this problem from my abstract algebra course:

Find a subgroup of $S_5$ (Symmetric group of order 5) isomorphic to the quaternion group $Q$.

I started writting down the elements of $Q$ to start trying some examples and see if the quaternion properties where verified:
$$Q=\{\pm 1,\pm i,\pm j,\pm j\}.$$
Obviously the element $1$ is $(1)(2)(3)(4)(5)$ in $S_5$.

Then I tried with:
$$i=(1234)(5)\ \ , \ \ -i=(1432)(5).$$

Both verify they have order $5$, and from them I get $-1=(13)(24)(5)$. Now, I'm stuck, since I think I pick wrong elements because I can't find good elements for $\pm j$ and $\pm k$. What is the easiest way to do this kind of problem where you're asked to find a subgroup isomorphic to a certain group?

Any help will be appreciated.

答:
This won't be possible. The smallest faithful permutation action of $Q_8$ is the regular one. That is, the smallest symmetric group containing $Q_8$ as a subgroup is $S_8$.

To see this, note if there is a subgroup $H\subseteq S_n$ isomorphic to $Q_8$, then $X=\{1,\cdots,n\}$ carries a group action from $H$. By the orbit-stabilizer theorem, if this action is transitive must be equivalent to an action on a coset space $H/K$, which is equivalent to $Q_8$ acting on $Q_8/N$ for some subgroup $N\le Q_8$. But every subgroup of $Q_8$ is normal, by inspection, so such a group actions would be unfaithful unless the kernel $N$ is trivial. If $X$ is not transitive, then it's a union of nonregular orbits, but since every proper subgroup of $Q_8$ contains the central element $-1$, we know $-1$ must be in the kernel of this action, so again it's not faithful.

Alternatively, we can find $2$-Sylow subgroups of $S_n$ and compare with $Q_8$. After all, if $S_n$ contained an isomorphic copy of $Q_8$, then it'd have to be contained in a $2$-Sylow of size at least $2^3$. For $n=4$ and hence also $n=5$ the $2$-Sylow is the dihedral group $D_8$ of order $2^3$ which is not isomorphic to $Q_8$. Indeed for $n=6$ and hence also $n=7$ the $2$-Sylow is $D_8\times C_2$ (contained in $S_4\times S_2$) which does not have a pair of noncommuting involutions generating a copy of $Q_8$.

Thus $S_8$ is the smallest symmetric group containing $Q_8$ (which is the permutation representation afforded by the left regular action, as used in Cayley's theorem).


问:
Relationship between the vectors
I have the following equations-

$\textbf{a}.\textbf{b}=0$

$\textbf{a}.\textbf{c}=0$

where $\textbf{a},\textbf{b}$ and $\textbf{c}$ are vectors. What are the possible relationships between $\textbf{b}$ and $\textbf{c}$?

I can think of the following three-

$\textbf{b}$ and $\textbf{c}$ are parallel

$\textbf{b}$ and $\textbf{c}$ are perpendicular (which is true in case of the basis vectors in 3D) and

$\textbf{a}$ is a zero vector.

$\textbf{b}$ and $\textbf{c}$ are vectors that are linear combinations of vectors given in 1 or 2 (this statement might need correction).

Are there any other possible relationships between the $\textbf{b}$ and $\textbf{c}$

答: $\vec a. \vec b=0=\vec a. \vec c \implies \vec a =k(\vec b \times \vec c).$

Title: Relationship between the vectors


问:
What's wrong with my proof ??
This question was on my exam but apparently my answer was wrong as I only got half of the point:
Suppose $p:X \to Y$ is an open map, prove if $A$ is open in $X$ then $q:A \to p(A)$ obtained by restricting $p$ is also an open map.

My proof: Let $U$ be open in $A$. We have that $q(U) = p(U)$ is open in $Y$, hence it's open in $p(A)$. So q is an open map.

This sounded like a quick excercise but apparently my proof needs improvement. What's wrong about my proof?

答: You are assuming that $p(U)$ is open. Why is that? The set $U$ is an open subset of $A$, but you should deduce from this that therefore $U$ is an open subset of $X$ (and this is where the fact that $A$ is open is used; it is not true in general). Then, yes, you can deduce that $p(U)$ is indeed an open subset of $Y$.

Title: What's wrong with my proof ??


答: To be certain why you didn't get full credit, you should ask whomever graded the exercise. The most likely reason seems that you conclude that $p(U)$ is open in $p(A)$, without arguing why $U$ is open in $X$. Also, the codomain of $q$ is $Y$, not $p(A)$.

Title: What's wrong with my proof ??

Body:
This question was on my exam but apparently my answer was wrong as I only got half of the point:
Suppose $p:X \to Y$ is an open map, prove if $A$ is open in $X$ then $q:A \to p(A)$ obtained by restricting $p$ is also an open map.

My proof: Let $U$ be open in $A$. We have that $q(U) = p(U)$ is open in $Y$, hence it's open in $p(A)$. So q is an open map.

This sounded like a quick excercise but apparently my proof needs improvement. What's wrong about my proof?


问:
Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$
Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$??

答:
That not true if $R$ is a field. Did you have anything else on your mind?

I think it is also not true for local artinian rings (say $\mathbb{Z}/p^n$ ).

It will be true for semisimple rings with two different classes of irreducible modules.


答:
Let $R=\mathbb{Z}$, $M=\mathbb{Q}$, $N=R=\mathbb{Z}$. Let $\phi\in \mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$. We first show that $\phi(1)=0$.

Suppose that $\phi(1)\neq 0$. Note that $\phi(1)\in\mathbb{Z}$, and so by the Fundamental Theorem of Arithmetic, we can write
\begin{equation*}
\phi(1)=\prod_{i=1}^{m}p_{i}^{\alpha_{i}}
\end{equation*}
Let $q$ be any prime such that $q\neq p_{i}$ for all $i$. Then since $\phi$ is a $\mathbb{Z}$-module homomorphism,
\begin{equation*}
\phi(1)=\phi(q/q)=q\phi(1/q)
\end{equation*}
so that $q|\phi(1)$, a contradiction. This forces $\phi(1)=0$.

Next we show that $\phi(a)=0$ for all $a\in\mathbb{Q}$. Let $a\in\mathbb{Q}$. Then writing $a=r/s$ for some $r,s\in\mathbb{Z}^{+}$ with $\gcd(r,s)=1$,
\begin{equation*}
\phi(a)=\phi(r/s)=r\phi(1/s).
\end{equation*}
It suffices to show that $\phi(1/s)=0$. We have
\begin{equation*}
\phi(1)=s\phi(1/s)
\end{equation*}
and since $s\neq 0$, $\phi(1/s)=0$. Thus, $\phi(a)=0$ for all $a\in\mathbb{Q}$, which implies that $\phi=0$. Thus, $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$.

Body: Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. Prove that exits two $R-$modules $M,N$ different $0$ such that $Hom(M,N)=0$??


问:
Information theory reference: comparison between Mackay to Thomas and Cover
I'm a computational neuroscience student with a background in mathematics. I want to learn information theory over the summer. I am interested in its applications to neuroscience, machine learning, statistics, etc. I cannot decide between Elements of Information Theory by Cover and Thomas and Information Theory, Inference, and Learning Algorithms by David Mackay. I read recommendations saying they are both great. Also I read that while being more theoretically superficial on information theory, Mackay's book is richer in useful applications, e.g.  bayesian inference.

Has anyone read them both? What would you recommend?

Thanks.

答:
They are both great books, but completely different approaches.  MacKay's book is full of exercises, visual explanations... it tries to convince you of the intuition of concepts first; Cover and Thomas is less visual, less intuitive,  more "mathematical".

Personally, I found the narrative of Cover and Thomas hard to grasp, while MacKay is more well organized and explains to you not only the concepts but why we need them. Conversely, for some reason, I kept going back to Cover and Thomas to understand a specific concept or proof.


问:
Clifford Algebra Multiplication Intuition
I discovered Clifford Algebra recently and I am deeply impressed with its explanatory power and geometrical intuitiveness. I've been playing with the GAViewer

Home Page: http://www.science.uva.nl/research/ias/ga/viewer/content_viewer.html

Download Page: http://www.science.uva.nl/research/ias/ga/viewer/gaviewer_download.html

(highly recommended!) and I'm trying to understand Clifford multiplication of vectors intuitively. If the two vectors are parallel (e.g. they are both multiples of e1) then multiplication scales their lengths, just as scalar multiplication scales their magnitudes. This provides a beautiful geometrical analogy to multiplication, revealing it to be an analog scaling function. If the vectors are not colinear, then the product becomes a wedge product, an oriented surface with a twist that rotates in the direction from a to b in the case of c = a*b.

You can see the effect of this product by multiplying the bivector with another vector, for example d = bc, or d = b(a*b), and you will see that the bivector scales b by the ratio of a to b, and rotates b by the angle from a to b. That is a beautiful and powerful notion that accounts for many of the extraordinary invariances in Clifford algebra especially in the projective and conformal models.

Here's my problem: Multiplication is an expansive operation, creating a multiplicity of its multiplicands, this is demonstrated quite literally by the bivector c = a*b which can be pictured as sweeping vector a through b, or vice-versa, and that results in a grade increase. The product of b and the bivector c however results in a grade reduction; vector * bivector = vector. What gives?

I understand this is only true because a, b, & c all lie in the same plane, whereas multiplying by a vector out of the plane would produce a volumetric trivector. But it is still a case of multiplication producing a grade reduction. I can't picture the "sweep" to explain this operation. Can anyone help my intuition? Is the second multiplication effectively a division, as is scalar multiplication by a fractional value? Or is this an effect of the "squares-to-one" property of Clifford algebra whereby a*a = 1?

I can picture the sweep of, for example, of ab/b as vector a swept along vector b, then swept back again collapsing the surface like a venetian blind and resulting back in a. I can picture ab/a as the same sweep-and-collapse, but this time the rotational component is reversed (ab = -ba) so the collapse leaves the vector rotated by the angle a*b. But how are we to imagine a vector times a bivector within the plane?

((I happen to believe that "seeing the picture" is the most important part of mathematical understanding, and Clifford algebra is particularly revealing of the geometrical spatial principles hidden within algebra.))

答:
When you say $c = ab$ is a bivector, you're taking $a \perp b$ implicitly.  The geometric product will in general have both terms:  $ab = a \cdot b + a \wedge b$, remember?  A scalar and a bivector.

It is, admittedly, pretty hard to visualize what that is geometrically.  Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and bivectors strains thinking (and I know; I tried to explain this at a talk and people repeatedly insisted it was nonsense, but how is adding real and imaginary numbers any more sensible?).

Beyond that, it is possible to think about the generalized wedge and dot products separately and to imagine a geometric interpretation.  The wedge product of a vector and a bivector produces the trivector with magnitude of the parallelepiped, as you know.  The dot product produces the vector in the plane that is orthogonal to the first vector.

You can think of the geometric product, then, as describing a decomposition: given a bivector $B$ and a vector $v$, $v \wedge B = v_\perp B$ extracts the part of the vector orthogonal to the surface and builds a volume from that.  $v \cdot B = v_\parallel B$ extracts the part of the vector that lies in the plane and gives the orthogonal vector in the plane that could be used to reconstruct $B$ with.

You talk about sweeping vectors over each other to form the wedge product, and that's a sensible way of looking at things, but what image do we need to imagine the dot product of a vector and a bivector?  I can picture taking the bivector like a flat disc of clay, putting my hands on it, and squeezing it along the direction of $v_\parallel$.  That's the best I can conceive of it.  That's what the dot product is, after all--a reduction operation, one that forms vectors from planes and planes from volumes.

Thinking of the dot product as "division" does seem like a sensible interpretation.


答:
Your reasoning is quite in the right direction.

To show that these types are unique, first note that if $A$ is any model of $T$, then elements of $A$ which are in the same disjoint part realise the same type. Indeed, if $a,b \in P_n^A$ then we may define $σ: Α \to A$ by $σ(a) = b, σ(b)=a, σ(x)=x$ otherwise. It is easy to check that this is an automorphism of $A$, and automorphisms fix types so $tp_A(a) = tp_A(b)$. We can see that any type of this form contains $P_n(x)$, as $A \models P_n(a)$.

So, let p,q be two types that include $P_n(x)$. We may find $M \models T$ which realises both p and q, i.e. $\exists m_1, m_2 \in M$ with $p = tp_M(m_1), q = tp_M(m_2)$. Since both types contain $P_n(x)$, we see that both $m_1$ and $m_2$ lie in $P_n^M$, and so using the above we have $ p =  tp_M(m_1) = tp_M(m_2) = q$. Hence the type containing $P_n(x)$ is unique. (From there, it is relatively straightforward to show that this unique type $p_n$ is also principal, with principal formula $P_n(x)$)

Now if $A \models T$ with $A' = A \setminus \bigcup_{n} P_n^A$ non-empty, then any element $c \in A' $ defines a type different from the $p_n$, since $A \not\models P_n(c)$ for any n, i.e. $P_n(x) \not\in tp_A(c)$. By using automorphisms which swap two elements of $A'$, and a similar argument to the above, we can show that this type is unique, and furthermore it's non principal as it's not realised in all models of T (some models have $A'$ empty, for example $M = ω \times ω$ with $P_n^M = ω \times \{n\}$).

Title: Determining all the $1$-types of a given theory

Body:
I'm trying to solve an exercise in my lecture notes. I'm given the following theory of infinitely many disjoint infinite unary predicates:

Let $L_n$ be the language with $n$ unary predicate symbols $P_1, . . . , P_n$. Let $T_n$ be the theory asserting that each $P_i$ is infinite, the $P_i$ are disjoint, and there are infinitely many elements not in any $P_i$, for $i \leq n$. Finally, let $T = \bigcup_n T_n$.

After having proved that both $T_n$ and $T$ are complete, I'm then asked to determine all the $1$-types of $T$, show that
there's a unique 1-type $p_n$ including $P_n(x)$, and a unique type $q$ that is not
one of the $p_n$.

I have no idea how I'd go about determining all the $1$-types, and while I can certainly find a $1$-type $p_n$ including $P_n(x)$ (by setting $p_n = $ tp$_A(a) = \lbrace ϕ(x) : A \models ϕ(a)\rbrace $, where $A$ is some model with $a\in A$ such that $a\in P_n^A$), but I'm not sure how I'd show it's unique.

I know that if If $π : A → B$ is an isomorphism, $a ∈ A$, $b ∈ B$ and $π(a)= b$ that tp$_A(a) = $tp$_B(b)$, so clearly there's only one $p_n$ obtained by taking a set like tp$_A(a) = \lbrace ϕ(x) : A \models ϕ(a)\rbrace $, but I don't know how to show there are no other $1$-types containing $p_n$ (perhaps this will become clearer once I know what all the $1$-types are). I also have no idea how I'd go about finding the $q$ I'm asked for.

I'd really appreciate any help you could offer.


问:
Find all the irreducible representations of the group
I want to find all irreducible representations of $\{
     \left[ {\begin{array}{cc}
   a & b \\
   0 & 1 \\
  \end{array} } \right];a\in\mathbb{F}^\times_p,b\in\mathbb{F}_p
\}$. My approach was to use the fact that this group can be expressed as semi-direct product, but it gets complicated and I don't know how to proceed after that. Is there any simpler way of solving this? Can I somehow use the character of this matrices?

答:
There is a surjective map from the group to $\mathbb{F}_p^{\times}$.  This latter group is cyclic of order $p-1$, so it has $p-1$ characters ($1$-dimensional). Composed with the projection gives us $p-1$ $1$-dimensional representations of $G$.

Now, the order of the group is $p(p-1)$. The formula for the sum of squares of dimensions suggests there should be another $p-1$ irreducible representation.

Note that $G$ acts naturally on $\mathbb{F}_p$, so on the complex valued functions on $\mathbb{F}_p$.  This representation has a $1$ dimensional subrepresentation, the space of constant functions. There is a complement, the space of functions with sum of values $0$. This will be the missing irreducible representation.


问:
Proving the fundamental theorem of calculus.
I have tried proving the fundamental theorem of calculus:

Assuming: f,g,F are continuous,differentiable in the appropriate range(s)

By definition,
$$ \int _{x=a}^{x=b}g(x)dx=\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{b-a}{n}g(a+ r\dfrac{b-a}{n})\tag{E1}$$
Which can be better expressed as:
$$\int _{v=v_{1}}^{v=v_{2}}g(v)dv=\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{v_{2}-v_{1}}{n}g(v_{1}+ r\dfrac{v_{2}-v_{1}}{n})\tag{E2}$$

where $v$ is some quantity.

Now consider the integral:

$I= \int _{a}^{b}f(t)dt$.

Suppose there exists a function $F(t)$ such that $F'(t)=f(t)$.Thus,

$I= \int _{a}^{b}f(t)dt= \int _{a}^{b}\dfrac{dF(t)}{dt}dt=\int _{t=a}^{t=b}dF(t)$, which is equivalent to:
$$ \int _{F(t)=F(a)}^{F(t)=F(b)} dF(t)$$

I believe that saying $F(t)=F(a)$ is just an equivalent statement to $t=a$ think
Using $E2$,this becomes,(since $v=F(t)$, and $g(v)=1$) :

$$\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{F(b)-F(a)}{n}$$

Which is just $F(b)-F(a).$

(This is of course intuitively obvious: adding up small changes $dF$ will simply be the overall change in $F$, viz: $F(b)-F(a)$, however I wished to avoid this line of reasoning since this will entail discussions on "rigorously" treating/defining differentials)

If we consider a function $I(x)=\int _{a}^{x}f(t)dt$, then by the same reasoning, we will have:

$I(x)=F(x)-F(a)$.

Which implies, $I'(x)=F'(x)=f(x)$

According to me, This completes the proof of both parts: part 1 and the evaluation theorem also.

However, this, in my view is different from the proof given in Thomas'-calculus (or just any standard textbook), since it does not make use of the Mean value theorem anywhere.

Also, this proof seems to be significantly shorter.

All of this makes me believe there is something wrong, or "non-rigorous" about my proof. My guess is the cancellation of $dt$ with $dt$. If anything else, I will be glad if someone points it out.

答:
You have not stated precisely the parts of FTC you want to prove so I'll need to guess.  In Wikipedia's page on the FTC, there are two parts. I'll refer to them as the first and second part:

"First part" Let $f$ be a continuous real-valued function defined on a closed interval $[a, b] .$ Let $F$ be the function defined, for all $x$ in $[a, b],$ by
$$
F(x)=\int_{a}^{x} f(t) d t
$$
Then $F$ is uniformly continuous on $[a, b]$ and differentiable on the open interval $(a, b),$ and
$$
F^{\prime}(x)=f(x)
$$
for all $x$ in $(a, b)$.

"Second part" Let $f$ be a real-valued function on a closed interval $[a, b]$ and $F$ an antiderivative of $f$ in $[a, b]$ :
$$
F^{\prime}(x)=f(x)
$$
If $f$ is Riemann integrable on $[a, b]$ then
$$
\int_{a}^{b} f(x) d x=F(b)-F(a)
$$

In the first half of your Question, you want to end up with the result $\int_a^b f(x) dx = F(b)-F(a)$. So this means you are trying to prove something like the second part.

The "cancellation of $dt$"  can actually be interpreted (while staying in the realm of e.g. Riemann integration, or for nice functions, the definition of an integral that you gave) as a theorem called change of variables ($F$ is the new variable), and usually comes after the proof of the FTC and so its use here would be circular. Therefore you should avoid this, but now you are stuck with $\int_a^b \frac{dF}{dt}(t) dt$, and when you apply (E1), its not clear to me how to obtain $\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{F(b)-F(a)}{n}$.

(In a chatroom, it was confirmed that OP was thinking of $dF(t)$ not as a Stieltjes type object, but rather as $dF$ where $F$ is a new variable; hence the alleged applicability of E1.)

For the second half of your Question, it seems you have also assumed the existence of an antiderivative $F$. In fact, the existence of an antiderivative is part of the FTC. Actually, the first part of the FTC shows that $I$ is differentiable, and is an antiderivative, i.e. $I'=f$. These two things are proven at the same time in Wikipedia.

So I believe what you have shown is: assuming

$(A)$ the second part of the FTC (this is the 'by the same reasoning' bit),

$(B)$ that $I$ is differentiable, and

$(C)$ that there is an antiderivative $F$,

then in fact $I(b)=F(b)-F(a)$ (and hence $I'=F'=f$). This is true:

$$I(b) \overset{\text{defn of }I}= \int_a^b f \overset{(C)}= \int_a^b F' \overset{(A)}= F(b)-F(a) \overset{(B)}\implies I'=F'=f.$$
But it is much less than the full FTC, which does not assume $(A)$, $(B)$ and $(C)$.

(See also the Corollary in wikipedia, which also has a nice short proof, but it requires the first part of the FTC; under your regularity assumptions, its the same as the second part of the FTC, so you can skip that proof.)


问:
How to deduce $\forall x \forall y \exists z(Gxz \land Gzy)$
The task is to deduce this:

$\forall x \forall y \exists z(Gxz \land Gzy)$

From this:

$ \forall x (Gxx) $

And this:

$ \forall x \forall y ( x \neq y  \to \exists z(Gxz \land Gzy) $

Using this instance of the Law of Indiscernibility of Identicals:

$ a = b \to (Gaa \equiv Gab) $

I tried to proceed by reductio, assuming as an additional premise this:

$ \neg \forall x \forall y \exists z (Gxz \land Gzy) $

And from this premise (and its instances), the instance of the Law, and the second premise above (and its instances), I was able to deduce this:

$\exists z (Gaz \land Gzb) $

Which is an instance of what I trying to prove but I won't be able to generalize appropriately to reach my conclusion, nor does it follow from the correct premises: I don't know how to use the first premise $ \forall x (Gxx) $ effectively, and further my reductio is not doing what it should be.

Thanks for any help. I really appreciate it!

This is from Goldfarb's Deductive Logic, IV5b.

答:
There are two cases $x = y$ and $x \not= y$.

if $x = y$, then $\exists z (G x z \land G z y)$ is $\exists z (G x z \land G z x)$ which holds with $z = x$ because $\forall x (G x x)$.

If $x \not= y$, then $\exists z (G x z \land G z y)$ follows from $\forall x \forall y (x \not= y \implies \exists z (G x z \land G z y))$.

In both cases, $\exists z (G x z \land G z y)$ follows as required.


问:
Convex set with empty interior lies in a proper affine subspace - infinite dimension
Given an infinite dimensional normed vector space $X$, $S$ is a convex subset of $X$ with empty interior. Is it true that $S$ lies in a proper affine subspace of $X$?

There is a similar question but it only covers the finite dimensional case.

答: No. Take $X=L^2(0,1)$ and $S=\{u\in X:\ u\ge 0 \ a.e.\}$. Then $S$ has no interior points, but $X=S-S$. Every $L^2$-function can be written as the difference of two non-negative $L^2$-functions.


问:
Why is $\{n=4r+1,r = {n-1\over 2}\}\subset \mathbb{P}$ true under these conditions?
Let $p=p_k$, $q=p_{k+1}$ and $r=p_{k+2}$, where $p_m$ denotes the $m$th prime.

I conjecture that whenever $n$ is prime, where $n$ is defined as follows:
$$n = 1+\left(\left\lfloor{p\over q}\right\rfloor+r\right)
\left\lfloor{(p+r)(q^2+pr)\over(pqr)}\right\rfloor$$
then:
$$ \frac{n-1}{4} \qquad\text{and}\qquad 4r+1$$ are both prime.

Although I've tried vigorously, I have no explanation as to why $n=4r+1,r = {n-1\over 2}$, although I heavily suspect that the solution rests on twin primes.

Alternative form:

$$n = 1+ z \left \lfloor{x\over y}+{z\over y}+{y\over z}+{y\over x}\right\rfloor+\left \lfloor{x\over y}\right\rfloor \left \lfloor{x\over y}+{z\over y}+{y\over z}+{y\over x}\right\rfloor$$

Example:

Terms $1\to 5$, where $t_m$ denotes the $m$th number $k$ which satisfies the conditions described in this question:

$$t_1 = 29 = 1+\left(\left\lfloor{3\over 5}\right\rfloor+7\right)\left\lfloor{(3+7)(5^2+3\cdot 7)\over(3\cdot 5\cdot 7)}\right\rfloor$$
$$\frac{29-1}{4}  = 7,\qquad 4\cdot 7+1 = 29$$

$$t_2 = 53 = 1+\left(\left\lfloor{7\over 11}\right\rfloor+13\right)\left\lfloor{(7+13)(11^2+7\cdot 13)\over(7\cdot 11\cdot 13)}\right\rfloor$$
$$\frac{53-1}{4}  = 13,\qquad 4\cdot 13+1 = 53$$

$$t_3 = 149 = 1+\left(\left\lfloor{29\over 31}\right\rfloor+37\right)\left\lfloor{(29+37)(31^2+29\cdot 37)\over(29\cdot 31\cdot 37)}\right\rfloor$$
$$\frac{149-1}{4}  = 37,\qquad 4\cdot 37+1 = 149$$

$$t_4 = 173 = 1+\left(\left\lfloor{37\over 41}\right\rfloor+43\right)\left\lfloor{(37+43)(41^2+37\cdot 43)\over(37\cdot 41\cdot 43)}\right\rfloor$$
$$\frac{173-1}{4}  = 43,\qquad 4\cdot 43+1 = 173$$

$$t_5 = 269 = 1+\left(\left\lfloor{59\over 61}\right\rfloor+67\right)\left\lfloor{(59+67)(61^2+59\cdot 67)\over(59\cdot 61\cdot 67)}\right\rfloor$$
$$\frac{269-1}{4}  = 67,\qquad 4\cdot 67+1 = 269$$

答:
Note that $\lfloor \frac{p}{q} \rfloor=0$, since $0<\frac{p}{q}<1$. Thus $n=1+r\lfloor \frac{(p+r)(q^2+pr)}{pqr} \rfloor=1+r \lfloor \frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p} \rfloor$.

Note that for $1<x<2$, we have $2<x+\frac{1}{x}<\frac{5}{2}$, since $x+\frac{1}{x}>2 \Leftrightarrow x^2+1>2 \Leftrightarrow (x-1)^2>0$ and $x+\frac{1}{x}<\frac{5}{2} \Leftrightarrow 2x^2+2<5x \Leftrightarrow (2x-1)(x-2)<0$.

We clearly have $p<q<r$, and by Bertrand's postulate $q<2p, r<2q$, so $1<\frac{r}{q}, \frac{q}{p}<2$, so by above we have $4<\frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p}<5$, so $n=1+r \lfloor \frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p} \rfloor=1+4r$.

It is now straightforward to see that when $n$ is prime, $\frac{n-1}{4}=r$ and $4r+1=n$ are prime.


答:
Let $A$ be open in $X$;

Then any open set $O$ in $A$ can be written as

$O=U \cap A$, where $U$ is open in $X;$

$O$ as the intersection of $U$ and $A$, which are both open in $X$, gives and open set $O$ in $X$.

Hence $f(O)$ is open in $X$.

Title: What's wrong with my proof ??


问:
How to prove this using Combinatorial proof?
Here is the problem:

$\displaystyle\sum_{k=0}^{p}\binom{p+q-k}{q}\binom{r+k}{r}\ =  \binom{p+q+r+1}{p}$

How to prove this using combinatorics? I don't want to use algebra or something. My idea was that since we are not explicitly choosing $k$, we probably shouldn't choose from a subset which can be added to $r$ or subtracted from $p+q$,instead we can imagine a bit string of probably p+q+r+1 length with $r+1$ ones and $p+q$ zeroes and consider counting on basis of occurrence of last one. But it isn't working out to be correct. Can someone please help

答:
Imagine choosing $q+r+1$ numbers from $1,...,p+q+r+1$ such that the $r+1$ - th smallest number is $r+k+1$.

Choose $r$ numbers from $1,...,r+k$, $\binom{r+k}{r}$

$r+k+1$ is already determined as the $r+1$ - th number

Then choose remaining $q$ numbers from $r+k+2,...,p+q+r+1$, $\binom{p+q-k}{q}$

In total, there are $\binom{p+q-k}{q}\binom{r+k}{r}$ possibilities. Now what if we sum for all possible values of $k$? We get the number of all possible $q+r+1$ combination from $1,...,p+q+r+1$. Isn’t that $\binom{p+q+r+1}{q+r+1}=\binom{p+q+r+1}{p}$?


问:
Cardinality of finite sequences of infinite set
I want to prove that if $A$ is a infinite set, then $|Fin(A)|=|FS(A)|=|A|$, where $Fin(A)$ is the set of all finite subsets of $A$ and $|FS(A)|$ is the set of all finite sequences. Firstly, to prove $|Fin(A)|=|A|$,

$$|Fin(A)|=|\bigcup_{n < \omega}[A]^{n}|=|\bigcup_{n < \omega}A|=\sum_{n <\omega}|A|=|A|\aleph_0=|A|$$
where in the second equality i use that $|[A]^{n}|=|A^{n}|=|A|$ (i'm also using that $|A \times A|=|A|$, thanks axiom of choice).

For the second, I would to use a similar argument and write $FS(A)=\bigcup_{n < \omega}A^{n}$, but i'm not truly sure about that equality.

答:
It appears to me that you are asking about the case of finite sequences as you worked out the finite subsets yourself. There are also other questions that think about finite sets: The cardinality of the set of all finite subsets of an infinite set. Here is my proof of your theorem.

If $Card(A) = \kappa$ is an infinite cardinal  then $Card(A \times A) = Card(A^{< \omega}) = \kappa$.

Proof

$Card(A \times A) = Card(A) \cdot Card(A) = \kappa \cdot \kappa = \max(\kappa,\kappa) = \kappa$.

Clearly, by finite induction, it follows that $Card(A^n) = \kappa$ for all $n < \omega$.

Now, observe that $A^{< \omega} = \bigcup_{n < \omega} A^n$ is bijective with $\mathbb{N} \times A$ or equivalently with $\mathbb{N} \times \kappa$. So: $Card(A^{< \omega}) = Card(\mathbb{N} \times \kappa) = \mathbb{N} \cdot A = \max(\mathbb{N}, \kappa) = \kappa$.

The bijection seems easy, choose $x \in \bigcup_{n < \omega} A^n$ then $x \in A^i$ for a unique $i$ (indeed, this union must be disjoint since each component has a different length). Then assign $x$ to $(i,x)$. The inverse would assign $(i,x)$ to the element $x$ of $A^i$.


问:
Confusing notation and brackets
I am curious about how to differ between brackets for function calls like $f(x)$ and brackets for multiplication $x(y+1)$. I think there is the possibility to get confused by this, because the first example shows a function call, and the second a multiplication. Does there exist some general style guide for something like this? This is extremly confusing if I have a equation like $g(a,b) = \frac{2f(a)}{h(b+1)}$, where $h$ is not a function, but $f$ is a function.

答:
I think the convention is just to remember which is which. I can see how your example may be confusing. I will give you two strategies for making it easier to distinguish which is which:

Name your variables consistently and intuitively! Your example is en excellent illustration of a poor choice of variable names. Since $f$ and $g$ are functions, $h$ is a poor choice for a variable which is not a function. Choose variables that are alphabetically contiguous for things "of the same type". What I mean by "the same type" will vary between situations. Bottom line: $h$ should get another name in your example.

If there are many brackets in an expression, you are allowed to use other brackets than $()$, like for example $[]$. In expressions involving nested brackets, some people choose $[]$ for the outer brackets and $()$ for the inner brackets. However, you should not substitute $[]$ for $()$ in function calls because some authors use this to distinuish between a function value, like $f(x)$, and the image of a set under some function, like $f[X]$.

Title: Confusing notation and brackets


问:
Is uncountable subset of separable space separable?
I have to prove that any uncountable $B\subseteq \mathbb{R}$, where $(\mathbb{R},\epsilon^1)$ is euclidean topology and topology on B is relative, is separable. And I know it's true because every subset of separable metric space is separable. </p>

<p>But what if we are given separable space $(X,\tau)$, $X$ uncountable, and $A \subseteq X$ uncountable subset with relative topology. Is $(A,\tau_A)$ separable and if it is, how to prove it? </p>

答:
Another example is this；

Let $ \omega$ denote infinite countable cardinal and $\mathcal E \subset [\omega]^\omega$ is a maximal almost disjoint family, where $[\omega]^\omega = \{ A \subset \omega: |A| = \omega \}$. Let $\Psi(\mathcal E)$ denote the topological space whose point-set is $\omega \cup \mathcal E$, with the topology generated by isolating each $\alpha \in \omega$, and the basic nbhds about $E \in \mathcal E$ are all sets of the form $\{E\}\cup (E\setminus F)$, where $F \in [E]^{< \omega}$. This we can called  $\Psi$ space.

Claim: It is separable. However the subspace $\mathcal E$ of $\Psi$ is uncountable closed discrete, and hence it is not separeblae.


问:
In unitary space $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$.

Prove that in unitary space if $\|x_n\| \to \|x\|$ and $\langle x_n, x\rangle \to \|x\|^2$ implies $x_n \to x$.

Now, so far I have just written

$\|x_n\| \to \|x\| \implies \lim_\limits{n\to \infty} |\|x_n\|-\|x\|| = 0$

and

$\langle x_n, x\rangle \to \|x\|^2 \implies \langle x_n, x\rangle \to \langle x, x\rangle \implies \lim_\limits{n\to \infty}  | \langle x_n, x\rangle - \langle x, x\rangle | = \lim_\limits{n\to \infty} | \langle x_n - x, x\rangle |$

I am however, to show that $\|x_n - x \| \to 0$.

How can I do that?

答:
$$\|x_n-x_0\|^2 = | \langle x_n-x_0, x_n-x_0\rangle| = | \langle x_n, x_n-x_0\rangle| - \langle x_0, x_n-x_0\rangle| \leq | \langle x_n, x_n-x_0\rangle | + | \langle x_0, x_n-x_0\rangle|$$
Now because $\langle x_n, x_0\rangle \to \|x_0\|^2 \implies \lim_\limits{n\to \infty} | \langle x_n - x_0, x_0\rangle | = 0$ then the second term:
$$
| \langle x_0, x_n-x_0\rangle| = \overline{| \langle x_n-x_0, x_0\rangle|} \to 0
$$

The first term:
$$
| \langle x_n, x_n-x_0\rangle | = \overline{| \langle x_n-x_0, x_n\rangle |} = \overline{| \langle x_n, x_n\rangle -  \langle x_0, x_n\rangle |} = \overline{|\|x_n\|^2 - \overline{\langle x_n, x_0\rangle}|} = \overline{|\|x_n\|^2 - \|x_0\|^2} \to 0
$$
because of $\|x_n\| \to \|x_0\| \implies \|x_n\|^2 \to \|x_0\|^2 =\implies \lim_\limits{n\to \infty} |\|x_n\|^2-\|x_0\|^2| = 0$

Finally
$$
\|x_n-x_0\|^2 \to 0
$$


问:
Inequality with powers
I would like to prove that for $0<a<1$ and $0<b<1$:

$a^{1-p}*b^p+(1-a)^{1-p} (1-b)^p<1~~~if~~~0<p<1$

and

$a^{1-p}*b^p+(1-a)^{1-p} (1-b)^p>1~~~if~~~p>1$

Any help is welcome.

I was trying to approach the problem from different angles:

I tried to assume that $a>b$ and increase the formula for $p<1$, by changing b to a in the first summand and a to b in the seconnd, getting:

$a^{1-p}*a^p+(1-b)^{1-p} (1-b)^p=a+1-b>1$, which is, evidentelly, not the result I would like to obtain.

I tried to make a very similar trick with moving $a^{1-p}*b^p$ out and substituting a with b, assuming that a>b:

$a^{1-p}*b^p(1+(1-1/a)^{1-p} (1-1/b)^p)1$ - overapproximation, evidently.

Title: Inequality with powers

Body:
Sorry that I made very stupid mistakes.

Let
$$f(p) = a^{1-p}b^p+(1-a)^{1-p}(1-b)^p$$
$$f'(p)=a \ln(b/a) (b/a)^p + (1-a) \ln[(1-b)/(1-a)] [(1-b)/(1-a)]^{p}$$
$$f''(p) = a \ln^2(b/a) (b/a)^p + (1-a) \ln^2[(1-b)/(1-a)] [(1-b)/(1-a)]^{p} > 0$$
and
$$f(0) = f(1) = 1$$


问:
Hash collision probability
I am trying to show that the probability of a hash collision with a simple uniform 32-bit hash function is at least 50% if the number of keys is at least 77164. I have figured out how to plot a graph on python and then read off the values and percentages there, but I can't seem to figure out a formal proof. I've played around a bit and have come to this formula

$$1-e^{-k(k-1)/(2N)}.$$

That is what I used in python and I used a loop to sub in different values for $k$. Does anyone know how I can use this to solve it by hand since I don't think my proof via python would be valid?

答:
First of all, the formula is approximate, see the below part for the approxiamation;

$$ p(n;H) \approx 1 - e^{-n(n-1)/(2H)} \approx 1-e^{-n^2/(2H)} \tag{1}\label{1}$$

This is the probability of uniform randomly selecting $n$ values from $H$ values with repetitions.

What I understand that you need a graph that shows the percentages like $10\%, 20\%, 30\%, 40\%,$, etc. For this, you need to invert the above expression \ref{1}.

$$n(p;H) \approx \sqrt{2H\ln\frac{1}{1-p}}$$

Simply plug your required probability to get the required selection.

$$n(0.5;H) \approx 1.774 \sqrt{H}$$

If we plug $H=2^{32}$ on Wolfram Alpha, see here, then

$$n(0.5;H) \approx \sqrt{2\cdot 2^{32} \ln\frac{1}{1-(1/2)}} = 77162.74324...$$

I can't seem to figure out a formal proof.

The formal proof comes with the birthday pardox, we will look at the only hash part, where the hash function has output space $H$. Let  \bar p(n) is the probability of no two hash value collide is then $$p(n) = 1 - \bar p(n) \tag{2}\label{2} $$ is the probability of at least two hash values are colliding.

$$\bar p(n) = \left(\frac{H-1}{H}\right) \cdots \left(\frac{H-k}{H}\right) = \prod_{k=1}^{n-1}\left(1-\frac{k}{H}\right) \tag{3}\label{3}$$ where $p(n)$ is the probability of at least two hash is colliding.

The approximation starts with $e^x \approx 1 +x $ for $|x| \ll 1$

Set $x = - \frac{a}{H}$ then $e^{-a/H} \approx 1 - \frac{a}{H}$. Now use this formula to replace each of the product in equation \ref{3}

\begin{align}
\bar p(n) & \approx  e^{-1/H} \cdot e^{-2/H} \cdots e^{-(n-1)/H}\\
           & \approx  e^{-\left( 1+2+\cdots + (n-1)/H\right)}\\
           & \approx  e^{-n(n-1)/(2H)}
\end{align}

Now use \ref{3} with the above result $$ p(n) = 1 - \bar p(n) \approx 1 -e^{-n(n-1)/(2H) $$ We can now parametrize as

$$p(n) \approx 1 - e^{-n(n-1)/(2H)} \approx 1-e^{-n^2/(2H)} $$

Note: I've used the $n$ instead of $k$.

Title: Hash collision probability


问:
proof that a stable matching in the roommate's problem is not guaranteed when there are even numbers
I came up with the following:

I don't know if I have proven this appropriately. A friend of mine told me I had to match C with D and from there prove that there is a blocking pair. I'm not sure which way is the correct way.

答:
The approach using Irving's algorithm should be fine to prove no stable matching exists. If during phase one of the algorithm any participant is eventually rejected by all other participants, this indicates that no stable matching is possible. For the rankings in your problem, by Irving's algorithm we have

A proposes to B, B accepts.

B proposes to C, C accepts.

C proposes to A, A accepts.

D proposes to A, A rejects.

D next proposes to B, B rejects.

D finally proposes to C, C rejects.

D has been rejected by all other participants and no stable matching is possible.

We can also reason about it by saying in any solution A, B, or C must be paired with D. For anyone who is with D, another person will have rated them highest, and D’s partner will prefer this other person over D. Also note that D is last on every persons preferences.


问:
Finding the order of a pole of a complex function
So many solutions on here seem to use the following result: if $f(z)$ has a pole at $z_0$, and\ $\lim_{z \to z_0}(z-z_0)^{m}f(z)$ is finite and non-zero (or possibly just finite?), then $z_0$ is a pole of order $m$. How is this result proved?

答:
Assume that $\lim_{z \rightarrow z_0} (z-z_0)^m f(z) < \infty$. This means that $(z-z_0)^m f(z)$ has at most (or worst if you will) a removable singularity at $z_0$. Make sure you understand why this is or prove it!

Consider tha Laurent series of $f$ around $z_0$:

$$f(z) = \sum_{n = -\infty}^{\infty} a_n (z-z_0)^n$$

$$(z-z_0)^mf(z)  = (z-z_0)^m \sum_{n = -\infty}^{\infty} a_n (z-z_0)^n = \sum_{n = -\infty}^{\infty} a_n (z-z_0)^{n+m}$$

Since, as we commented earlier, this function has at most a removable singularity at $z_0$, $a_n = 0$ for all $n+m < 0 \implies n < -m$, i.e.

$$(z-z_0)^m f(z) = \sum_{n=-m}^{\infty} a_n(z-z_0)^{n+m}.$$

We divide by $(z-z_0)^m$ and get

$$f(z) = \sum_{n=-m}^{\infty} a_n (z-z_0)^n,$$

which shows that $z_0$ is a pole of order $m$.


问:
How to sketch the region described by transformed random variables U and V
I have the following question, for which the first part of (a), I had no trouble with, however, when it comes to sketching the region described (highlighted in yellow) - problem is in the link further below. I'm a bit lost on how to even approach this.

for context and to know that I haven't just run here, I have watched several videos which I thought would help on sketching the range. However, none have worked, and this is not covered in any of my lectures, which makes me assume this is considered a pre-requisite, which unfortunately I don't have in my background or don't recall.

I have spent over a day looking for a way on how to tackle this and the most relevant content I found is the Youtube video below, however, this was a F_x(x), rather than a joint pdf, and going from one to the other doesn't seem like something I should take for granted or assume is too easy - also, I can't be 100% certain this is the correct track, but I think so.

video I tried:
https://www.youtube.com/watch?v=Vf7ER2GSenI

Could someone please advise how to approach this problem...

THE MATH PROBLEM

Thank you very much for the help!
It is much appreciated.

答:
Set

$$\begin{cases}
u=xy   \\
v=\frac{x}{y}
\end{cases}\rightarrow\begin{cases}
x=\sqrt{uv}   \\
y=\sqrt{\frac{u}{v} }
\end{cases}$$

it is self evident that it must be also

$$\begin{cases}
1\leq uv<+\infty ,& (1)  \\
1\leq \frac{u}{v} <+\infty,&(2)
\end{cases}$$

Plotting this region in the 1st Quadrant of Cartesian Axes you get the requested solution

In fact (1)is satisfied under the condition $u\geq \frac{1}{v}$ while (2) is satisfied when $u\geq v$

When $v\in (0;1)$ the first condition (1) is stronger, when $v>1$ is the second one (2) to be the strongest


问:
In an integral, why does logarithmic function of an exponential completely drop out?
How to show that

$$-\int_{-\infty}^{\infty}  e^{-(x-\mu)^2 / 2\sigma^2} \ln\left[  e^{-(x-\mu)^2 / 2\sigma^2} \right] \mathrm{d} x$$

equals this:

$$-\int_{-\infty}^{\infty}  e^{-(x-\mu)^2 / 2\sigma^2}  \mathrm{d} x$$

If the exponential function appears twice in the first formula as its non-logged self multiplied by its logged self, why does the first formula reduce to an integral of only the non-logged exponential, while the logarithm completely drops out?

答:
We have:
$$\int f(x)\ln(f(x))dx$$
$e^u=f(x)$ so: $e^udu=f'(x)dx$ or $dx=\frac{e^u}{f'(x)}du$
and so:
$$f(x)\ln(f(x))dx=ue^u\frac{e^u}{f'(x)}du$$
and now we just need to work out what $f'(x)$ is:
$$f(x)=e^{-(x-\mu)^2/2\sigma^2}$$
$$f'(x)=-\frac{x-\mu}{\sigma^2}f(x)$$

and if we take logs of our function:
$$\ln(f(x))=-(x-\mu)^2/2\sigma^2$$
so:
$$x-\mu=\sqrt{-2\sigma^2\ln(f(x))}$$
$$-\frac{x-\mu}{\sigma^2}=-\sqrt{\frac{-2\sigma^2\ln(f(x))}{\sigma^4}}=-\frac{\sqrt{-2\ln(f(x))}}{\sigma}=-\frac{\sqrt{-2u}}{\sigma}$$
giving:
$$f(x)\ln(f(x))dx=-ue^{2u}\sigma\frac{1}{\sqrt{-2u}e^u}du=\frac{i\sigma \sqrt{u}e^u}{\sqrt{2}}du$$
now just work out the limits, hope this helps!


问:
Haar measure $\tau$-additive?
I'm reading some results from Measure Theory Volume 4 by D.H. Fremlin, and I'm stuck on something.

This is pulled out of one of his lemmas (stated more generally for topological groups):

A Haar measure $\mu$ on a locally compact group $G$ is strictly positive.

The proof is quite short:

It is sufficient to show that $\mu$ is positive on non-empty open sets containing 1.

Suppose for a contradiction that $U$ is an open set containing $1$ and $\mu(U) = 0$.

Then $G = \bigcup\limits_{x\in G}xU$, and since $G$ is $\tau$-additive, $\mu(G) = 0$, a contradiction.

I assume that $\tau$-additive means that $\mu(\bigcup_{\alpha\in J}E_{\alpha}) \leq \sum\limits_{\alpha\in J}\mu(E_{\alpha})$ is this correct?

If so, then why do Haar measures have this property.  If not what does it really mean?

答:
I don't know what $\tau$-additive means, but if $G$ is a locally compact Hausdorff abelian group and $\mu$ is a left Haar measure, then $\mu(U)\neq 0$ for $U$ a non-empty open subset of $G$, and the proof is almost what you've written (except, again, I don't know what is meant by $\tau$-additive). Remember that Haar measures are non-zero by definition. They are also inner regular on open sets. So suppose $\mu(U)=0$. Then, by left invariance, $\mu(gU)=0$ for all $g\in G$. Now, if $K$ is a compact subset of $G$, then $K$ is covered by finitely many translates of $gU$ of $U$ (here we use that $U$ is non-empty). So $\mu(K)=0$ by sub-additivity. Finally, by inner regularity on opens, $\mu(G)$ is equal to the supremum of $\mu(K)$ as $K$ ranges over the compact subsets of $G$. But this supremum is $0$ by what we've just shown, contrary to the assumption that $\mu\neq 0$.

EDIT: Assuming the set $J$ in your inequality is supposed to be countable, then this is countable sub-additivity (or I guess just sub-additivity) of $\mu$, and every measure has this property. I guess now that I think about it, the way ``$\tau$-additivity" is being invoked seems to be some sort of extension of countable sub-additivity to collections of sets which might not be countable (since the union in the proof cited might not be a countable one). I don't know why this property holds for Haar measures. So I guess my answer doesn't really address the question, although the proof I've given does show that one need not resort to $\tau$-additivity, whatever it is, so I will leave it for now, but will delete it if the OP requests it.

Title: Haar measure $\tau$-additive?


问:
Relationship between the vectors
I have the following equations-

$\textbf{a}.\textbf{b}=0$

$\textbf{a}.\textbf{c}=0$

where $\textbf{a},\textbf{b}$ and $\textbf{c}$ are vectors. What are the possible relationships between $\textbf{b}$ and $\textbf{c}$?

I can think of the following three-

$\textbf{b}$ and $\textbf{c}$ are parallel

$\textbf{b}$ and $\textbf{c}$ are perpendicular (which is true in case of the basis vectors in 3D) and

$\textbf{a}$ is a zero vector.

$\textbf{b}$ and $\textbf{c}$ are vectors that are linear combinations of vectors given in 1 or 2 (this statement might need correction).

Are there any other possible relationships between the $\textbf{b}$ and $\textbf{c}$

答:
Just some notes:

The inner product of two vectors $\vec{a}$ and $\vec{b}$, such as $\vec{c} = \{c_1, \cdots , c_n \}$ e $\vec{b} = \{b_1, \cdots  , b_n \}$, we have

$$\{(c_1,c_2,\cdots, c_n ): c_i \in \mathbb{R}, \forall i = 1, 2, 3, \cdots, n \}$$
$$\{(b_1,b_2,\cdots, b_n ): b_i \in \mathbb{R}, \forall i = 1, 2, 3, \cdots, n \}$$

So,
$$\vec{c}\cdot \vec{b} = c_1 b_1 + c_2 c_2 + \cdots + c_n c_n = \sum_{i=1}^{n} c_i b_i  $$

Assuming the space $E^3$ we have $n = 3$

$$\vec{c} \cdot \vec{b} = c_1 b_1 + c_2 b_2 + c_3 b_3$$

Note: For vectors $\vec{a}, \vec{b}, \vec{c}$ I'm using the orthonormal basis $(\vec{i}, \vec{j}, \vec{k})$

It seems that you are giving some assumptions based on
$$\text{If } \vec{a} \perp \vec{b} \text{ and }  \vec{a} \perp \vec{c}, \text{ so } \vec{b} \perp \vec{c}    $$
But the statement above is not true. Consider the vectors as
$\vec{a} = (-5, 1, 1)$, $\vec{b} = (1, 2, 3)$, $\vec{c} = (2, 5, 5)$ and you will see the problem.

For
$\vec{c}\perp \vec{b} \Leftrightarrow \vec{c}\cdot \vec{b} = 0$. So, $\cos(\theta) = \dfrac{\vec{c}\cdot \vec{b}}{\Vert\vec{c}\Vert  \Vert\vec{b}\Vert}=0 \Leftrightarrow \theta = \dfrac{\pi}{2}$

But note that in the dot product you cannot cancel $\vec{a}$

$$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Leftrightarrow  \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0 \Leftrightarrow  \vec{a}(\vec{b} - \vec{c}) = 0 \Leftrightarrow  \vec{a} \perp (\vec{b} - \vec{c})   $$

Title: Relationship between the vectors


问:
Equivalence Relations lists
Let $X$ be a non-empty set and $n\in\mathbb N$. Then $X^n$ is the Cartesian product of $n$ copies of $X$. A relation can be defined on $X^n$ by
$(a_1,a_2,\dots,a_n) \sim (b_1,b_2,\dots,b_n)$
if and only if every $x\in X$ appears the same the number of times in
the first list as it does in the second.

Question:Let $X=\mathbb R^3$. For the relation above, list all elements which are related to $(0,1,2)$.

Would the answer to the question be all the possible ordered triplets, for example $(0,0,0), (0,0,1),\dots ,(1,1,1), \dots, (2,2,2), \dots, (1,0,2)$ etc.?

答:
Your suggested answer is incorrect.

Your condition states that

every $x\in X$ must appear the same number of times in the first
list as it does in the second.

So if $(a,b,c)\sim (0,1,2)$ then $(a,b,c)$ must contain one $0$, one $1$ and one $2$. This should be enough for you to figure out all the triples related to $(1,2,3)$.

Title: Equivalence Relations lists


问:
Can you simplify an equation to use explicit differentiation instead of implicit differentiation?
Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get: $$\frac{12+18y}{(1-3x)^{2}}$$
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)?
By doing that I got: $$\frac{48}{(1-3x)^{3}}$$
I'm not sure if this is a correct answer as I am new to differentiation.
I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?

Thank you.

Body:
Sure you can in this case
$$y=\frac{2 (x+1)}{1-3 x}$$
and then
$$y'=\frac{8}{(1-3 x)^2}$$
and finally
$$y''=\frac{48}{(1-3 x)^3}$$


问:
Proving that the limit of the lagrange error bound of $Ln(1-x)$ converges to 0 at $|x| \lt 1$
It is a well known fact that:
$$Ln(1-x) = \sum_{k=1}^{\infty} -\frac{x^k}{k}, \qquad \text{For } |x| \lt 1$$
It can easily be proven that the sum converges only if $|x| \lt 1$. For one to say that $Ln(1-x)$ equals this series though, it must also be proven for which $x$ it is true that the limit of the lagrange error bound as the amount of terms $n \to \infty$, is $0$.
$$ \lim_{n \to \infty} M_n *\frac{(x-a)^n}{n!}   $$
where $M_n$ is the greatest value of $f'^{n+1}$ between $a$ and $x$.

$|x|$ being  $\lt 1$ in $Ln(1-x)$ amounts to $0 \lt x \lt 2$ in $Ln(x)$. So we make a taylor-expansion of $Ln(x)$ around $a=1$, and attempt to prove that the error bound approaches $0$ when $x \in (0, 2)$. It can easily be shown that the $n+1$'th derivative of $Ln(x)$ is:
$$ \frac{n!}{x^{n+1}} $$
If $x \ge 1$, then the biggest value this takes is when $x=1$, so simply $n!$. Substituting into the lagrange error bound we get:
$$ \lim_{n \to \infty} n!*\frac{(x-1)^n}{n!} = \lim_{n \to \infty} (x-1)^n $$
This does not reach $0$ if $x-1 \ge 1$, so $x$ must be less than $2$.

If $x \lt 1$, then the biggest value the $n+1'th$ derivative takes is when $x=x$, so $n!/x^{n+1}$:
$$ \lim_{n \to \infty} \frac{n!}{x^{n+1}}*\frac{(x-1)^n}{n!} =
\lim_{n \to \infty} \frac{(-1)^n}{x}*\frac{(1-x)^n}{x^n} $$
Taking the root test, we now get:
$$\lim_{n \to \infty} \left|\frac{-1}{x^{1/x}}*\frac{1-x}{x} \right|  $$
$x^{1/x}$ approaches $1$, and the absoloute value signs from the root test remove the $-1$. Hence, for this to converge, it must be the case that:
$$\left|\frac{1-x}{x} \right| \lt 1 \to x \gt \frac{1}{2}$$
But then the interval on which the taylor expansion is valid is $(1/2, 2)$ and
not $(0,2)$. What am i doing wrong here?

答:
It looks like you may have messed up the the error term - I am not sure how you got to the conclusion $x < 2$.

Define $$e_n(x) = \log(1-x) + \sum_{r=1}^{n-1} \frac{x^r}{r}$$

so that for $|x| < 1$, the Lagrange form for the exact error is given by $e_n(x) = \frac{d^{n}}{dx^{n}} \log(1-\xi) \frac{x^n}{n!}$ for some $\xi$ satisfying either $ 0 \leqslant \xi < x$ or $x < \xi \leqslant 0$.  Now
$$
\frac{d^n}{dx^n } \log(1-x) = - \frac{(n-1)!}{(1-x)^n}
$$
so that,
$$ e_n(x) = -\frac{x^n}{n} \cdot \frac{1}{(1-\xi)^n}.$$
Now, if $\xi$ is not known, when $0 \leqslant x < 1$
$$0 \leqslant \frac{x}{1-\xi} < \frac{x}{1-x}$$
so for positive $x$ the Lagrange error can only be sure to converge to zero if $0 \leqslant x < \frac{1}{2}$.  In similar vein, for negative $x$ we can obtain
$$\frac{|x|}{1-\xi}\leqslant |x|$$ and in this case we can be sure that the Lagrange error converges to zero for any $-1 < x \leqslant 0$.  Taken together we can see that the Lagarange error will surely have limit zero for $-1 < x < \frac{1}{2}$.

This does not mean it does not converge to zeror for all $|x| < 1$, but only that the uncertainty as to the value of $\xi$ means it is not going to be easy to prove so when $x \geqslant \frac{1}{2}$.

There is another error form, which works for functions with a continuous $n$th derivative,

$$e_n(x) = \int_0^x \frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}~dt.$$

This is derived using integration by parts.

In our case, with $f(x) = \log(1-x)$, we will meet the continuous derivative requirement if we limit the interval $x\in(-1,1)$ to a closed sub-interval.  Then, concentrating only on the harder case where $x \geqslant 0 $
$$ e_n(x) = \int_0^x -\frac{1}{(1-t)^n} \cdot(x-t)^{n-1} ~ dt.$$
Substitute $u = (x-t)(1-t)$ to obtain,
\begin{align}
e_n(x) = -\int_0^x \frac{u^{n-1}}{1-u} ~ du \tag{1}\label{Eq1}
\end{align}
which incidentally is exactly the same error estimate that would be obtained (with less effort if we started with $\log(1-x)$ as the integral of the geometric progression and estimated the error from the well known summation formula.  Be that as it may,
continuing from \eqref{Eq1},
$$|e_n(x)| \leqslant \frac{1}{1-x}\int_0^x u^{n-1} ~ du = \frac{x^n}{n(1-x)}$$
which now can be seen to have limit zero as $n \to \infty$, uniformly on a closed sub-interval of $(0,1)$.

Worth noting much of this could be obtained without resort to an exact error formula simply from the ratio test and the uniform convergence of all power series on a closed sub-interval of their circle of convergence.