问:
How do I solve this system of four equations? Can it be done using Gaussian elimination?
I have the following system of four equations:

$$\begin{align} &x_1 = \dfrac{1}{2} x_2 + \dfrac{1}{4} x_3 \\ &x_2 = \dfrac{1}{3} x_1 + \dfrac{1}{2} x_4 \\ &x_3 = \dfrac{2}{3} x_1 + \dfrac{1}{2} x_4 \\ &x_1 + x_2 + x_3 + x_4 = 1 \end{align}$$

How do I solve this system to get a unique solution for $\mathbf{x}$? I tried doing it by hand by substituting the first equation into the second, and then the third into the second, and then everything into the last one, but that doesn't work. I could use an online calculator, but I'd like to know what I'm doing wrong and how it is done by hand. Is there a more systematic/orderly way to do this, such as with Gaussian elimination / row reduction?

答:
If you want to use substitution.

Solve the first equation for $x_3$; plug it. Solve the second equation for $x_4$; plug it. Solve the thirs equation for $x_2$.

At this point, $x_1,x_2,x_3$ as expressed as functions of $x_1$. Plug in the last equation and solve for $x_1$ and go backward to get the other.


问:
Can you write this operation in matrix notation?
Consider the matrix $A$ and column vector $b$ where,

$A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ and  $b = \begin{bmatrix}1\\3\end{bmatrix}$

In the R statistics software, the the code A*b performs element-wise multiplication of the columns in $A$ with the column vector $b$ returning a matrix with the same dimensions as $A$, i.e.

$A*b = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} * \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}1\times1 & 2\times1\\3\times3 & 4\times3\end{bmatrix}=\begin{bmatrix}1 & 2\\9 & 12\end{bmatrix}$

Is there a name for this operation? Can this operation be expressed using common matrix notation?

答:
Let us call $\text{Diag}(b)$ the diagonal matrix, the diagonal of which is $b$

$$ \text{Diag} (b) = \begin{bmatrix} 1 & 0\\0 & 3\end{bmatrix} $$

Then

$$ \text{Diag} (b) \times A = \begin{bmatrix} 1 & 2\\9 & 12\end{bmatrix} $$

ContentLicense: CC BY-SA 4.0


问:
Conditional density of X given if X is atleast t
Let $X$ be a positive random variable with probability density funcion $f_X(x)$ and $Y := 1_{[X \ge t]}$ for a fixed $t$.

I am interested in the conditional density $f_{X|Y}(x|y)$.

Using Bayes' Theorem:

$f_{X|Y}(x|Y=y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{f_{Y|X}(y|x)f_X(x)}{f_Y(y)} = \begin{cases}\frac{f_{Y|X}(0|x)f_X(x)}{f_Y(0)}  = \frac{1_{[x < t]}f_X(x)}{p(Y = 0)} = \frac{1_{[x < t]}f_X(x)}{\int_{0}^t f_X(x) dx} &\text{when }Y = 0\\
\frac{f_{Y|X}(1|x)f_X(x)}{f_Y(1)} = \frac{1_{[x \ge t]}f_X(x)}{p(Y = 1)} = \frac{1_{[x \ge t]}f_X(x)}{1-\int_{0}^t f_X(x) dx}&\text{when }Y = 1
\end{cases} $

Considering the discrete case this makes sense. But using $f_Y(y) = p(Y = y)$ feels very illegal. Is this correct?

答:
Random variable $Y$ has no PDF so the use of $f_Y(y)$ is actually illegal.

Personally I would go for:

$P\left(X\leq x\mid Y=0\right)=\frac{P\left(X\leq x,Y=0\right)}{P\left(Y=0\right)}=\begin{cases}
\frac{F_{X}\left(x\right)}{F_{X}\left(t\right)} & \text{if }x<t\\
1 & \text{otherwise}
\end{cases}$

$P\left(X\leq x\mid Y=1\right)=\frac{P\left(X\leq x,Y=1\right)}{P\left(Y=1\right)}=\begin{cases}
\frac{F_{X}\left(x\right)-F_{X}\left(t\right)}{1-F_{X}\left(t\right)} & \text{if }x\geq t\\
0 & \text{otherwise}
\end{cases}$

Then the corresponding PDF's can be described as:

$f_{X|Y=0}\left(x\right)=\begin{cases}
\frac{f_{X}\left(x\right)}{F_{X}\left(t\right)} & \text{if }x<t\\
1 & \text{otherwise}
\end{cases}$

$f_{X|Y=1}\left(x\right)=\begin{cases}
\frac{f_{X}\left(x\right)}{1-F_{X}\left(t\right)} & \text{if }x\geq t\\
0 & \text{otherwise}
\end{cases}$

This agrees with your end result.


问:
How to find the number of combinations of $x,y,z$ that satisfy an equation
The equation is $x + y + z = 18$,
with the constraints:

$2 \leq x \leq 8$

$1 \leq y \leq 7$

$1 \leq z \leq 6 $

First I solved it so the lower bounds where zero, by substituting the following:

$x' = x - 2$, $y' = y - 1$, $z'= z - 1$.

Which gave me the combination of $16,2 = 120$, now I have to subtract the number of combination of $x,y,z$ that are over, so $x'\geq7$, $y'\geq7$, or $z'\geq6$. The problem is when I plug those into the equation I end up with a negative answer and do not know how to deal with the problem.

答:
Since $x+y \leq 8+7 = 15$ we must have $z \geq 3$ while $x+y \geq 2+1 = 3$ tells us that $z \leq 15$

Do something similar to establish bounds for $x$ and $y$ and compare to the originally set bounds.

ContentLicense: CC BY-SA 3.0


问:
Example of 2 Inverse function
Im trying to find 2 functions $f,g$ such that $f $ and $ g $ both not linear, and $ f,g:\mathbb{R}\to\mathbb{R} $ 1-1 and onto, such that $ f\left(g\left(x\right)\right)=x,\thinspace\thinspace\:g\left(f\left(x\right)\right)=x $.

(f is the inverse function of g).

There exists such functions?

Title: Example of 2 Inverse function

Body: How about $f(x)=x^3$, $g(x)=\sqrt[3]{x}$


问:
$\mathsf{ZF}$ defines a transitive class $M \models \mathsf{ZFC} + \neg\mathsf{CH}$ implies inconsistent (Kunen's Set Theory, Remark II.6.32)
This is Remark II.6.32 from Kunen's Set Theory, which puzzles me:

Suppose that working in $\mathsf{ZF}$ (or even $\mathsf{ZFC} + V = L$), one could define a transitive proper class $M$ and prove that it is a model for $\mathsf{ZFC} + \neg\mathsf{CH}$ (or even just $\mathsf{ZF} + V \neq L$). Then $\mathsf{ZF}$ is inconsistent.

I understand the proof they provided, which uses the absoluteness of rank and $L(\delta)$. What I'm confused about is that the statement itself appears to very simply contradict Godel's second incompleteness theorem - if $M \models \mathsf{ZFC} + \neg\mathsf{CH}$, then certainly $M \models \mathsf{ZF}$, so $\mathsf{ZF} \vdash \mathrm{Con}(\mathsf{ZF})$, a contradiction.

Is there a mistake in my reasoning? Thanks in advance.

答:
The existence of a proper class which we can prove each relativized ZF axiom does not prove the consistency of ZF. To see the absurdity of this, observe that $V$ is a proper class for which we can prove each relativized axiom... since each relativized axiom is just an axiom we take. So if that sufficed to prove consistency of ZF, we would be in violation of Godel's theorem from the get-go.

In order to prove the consistency of ZF within ZF, we would need to find a model that we could internally prove that every axiom holds. In other words we can prove the internally quantified statement "for every axiom $\ulcorner\phi\urcorner$ of ZF, $M\models \ulcorner\phi\urcorner$." (I use "Quine corners" to emphasize this is an internal code for the formula as a set.) This is different from the sense in this theorem, which is to say for every axiom $\phi$ of ZF we can prove "$\phi^M$."


问:
Find a closed formula for $\sum_{n\geq 0} f(n)x^n.$
Given that $f(n) =f(n-1) + 2f(n-2)$ for $n \geq 2$, with $f(1) = 1$ and $f(2) = 3$ find a closed formula for $$\sum_{n\geq 0} f(n)x^n.$$

Here I try to do the following $ \sum_{n=0}^{\infty}f(n)x^{n} = \sum_{n=1}^{\infty}f(n-1)x^{n} + 2\sum_{n=2}^{\infty}f(n-2)x^{n}$ but when I simplify and re-index I get $$1=x+2x^2,$$ which is not correct. Can someone help me?

答:
Let $F(x)$ denote the generating function for $f(n)$, i.e.

$$F(x)=\sum_{n\ge1}f(n)x^n$$

Multiply both sides of the recurrence by $x^n$ and sum both sides starting from $n=3$:

$$\sum_{n\ge3}f(n)x^n=\sum_{n\ge3}f(n-1)x^n+2\sum_{n\ge3}f(n-2)x^n$$

Now manipulate this as need to get everything in terms of $F(x)$:

$$\begin{align}
\sum_{n\ge3}f(n)x^n&=\sum_{n\ge3}f(n-1)x^n+2\sum_{n\ge3}f(n-2)x^n\\[1ex]
\sum_{n\ge1}f(n)x^n-f(1)x-f(2)x^2&=\left(x\sum_{n\ge2}f(n-1)x^{n-1}-f(1)x^2\right)+2x^2\sum_{n\ge3}f(n-2)x^{n-2}\\[1ex]
F(x)-f(1)x-f(2)x^2&=x\sum_{n\ge1}f(n)x^n-f(1)x^2+2x^2\sum_{n\ge1}f(n)x^n\\[1ex]
F(x)-x-3x^2&=xF(x)-x^2+2x^2F(x)
\end{align}$$

which is easily solved for $F(x)$.


问:
The functor $U\circ C:\mathbf{Top}^{op}\to \mathbf{Set}$ is representable
Consider the functor $C:\mathbf{Top}^{op}\to \mathbf{Ring}$ that sends an object $X$ to a continuous function $X\to \mathbb R$. Consider also the composition $U\circ C$ where $U:\mathbf{Ring}\to\mathbf{Set}$ is a forgetful functor. This composition should be representable, i.e., should be isomorphic to $\mathbf{Top}(-,Y)$ (or $H_Y$) for some topological space $Y$.

To find this $Y$, I expanded what the above means, and the expanded version says that there should be a bijection $$\{\text{functions } X\to \mathbb R \} \leftrightarrow \{\text{continuous functions } X\to Y \}.$$

So it's natural to take $Y$ to be $\mathbb R$. Then the correspondence above is supposed to send a function $f:X\to\mathbb R$ to the same function $f$ that's regarded as continuous map. But why does this map $f$ have to be continous? It looks like this argument implies that any function $X\to \mathbb R$ has to be continuous, which doesn't make sense.

答:
Your functor $C$ sends an object $X$ to the ring of continuous functions $X \to \mathbb R$. Then $U\circ C$ sends $X$ to the set of continuous functions $X\to\mathbb R$. Therefore, the left hand side should be
$$
\{\textrm{continuous functions }X\to\mathbb R\}\,.
$$


问:
Showing that the linear transformation of an elementary/Jordan measurable set is Jordan measurable
This question involves Exercise 1.1.11 in Tao's Introduction to Measure Theory, which I'm currently self-studying as a senior undergraduate.

For (1), Tao states that previous exercises show that $L(E)$ is Jordan measurable if $E$ is an elementary set, but I can't see how to apply them. I'm fine with the other part of (1).

For (2), I also don't really know how to start this. I feel like this will be clearer once I understand how to show that $L(E)$ is Jordan measurable for elementary $E$.

For (3), I'm guessing that I need to show that $D = |\det L|$ for elementary $L$ first and then reduce $L$ to a product of elementary matrices (and potentially an 'extra' matrix $A$ that can't be reduced further), but I'm not sure how to proceed after this.

答:
You can assume $L$ is invertible, otherwise $L(E)$ can be upper-bounded by an elementary set with measure $0$.

For (1), to show $L(E)$ is Jordan measurable, you only need to show that, given a box $B$, $L(B)$ is a compact convex polytope (exercise 1.1.9).

For (2), $E$ can be approximated by elementary sets $A$ and $B$ from within and from without, then you can find elementary sets $A' \subset L(A)$ and $L(B) \subset B'$ that can approximate $L(E)$.

For (3), my understandings of the "alternatively" part of the hint in the book is that you can use polar decomposition to get $L = PU$, where $P$ is a SPD matrix, which can be factorized into $O^T\Lambda O$, where $\Lambda$ is diagonal, $O$ and $U$ are orthogonal. You can use unit ball to show that the constant $D$ for $O$ or $U$ is $1$, and use box to show that $D$ for $\Lambda$ is $|\det(\Lambda)|$, so $D$ for $L$ is $|\det(L)|$


Title: How to find the number of combinations of $x,y,z$ that satisfy an equation

Body:
The equation is $x + y + z = 18$,
with the constraints:

$2 \leq x \leq 8$

$1 \leq y \leq 7$

$1 \leq z \leq 6 $

First I solved it so the lower bounds where zero, by substituting the following:

$x' = x - 2$, $y' = y - 1$, $z'= z - 1$.

Which gave me the combination of $16,2 = 120$, now I have to subtract the number of combination of $x,y,z$ that are over, so $x'\geq7$, $y'\geq7$, or $z'\geq6$. The problem is when I plug those into the equation I end up with a negative answer and do not know how to deal with the problem.

Body:
Brute force yields the following 10 combinations:
$$
(8,4,6),(7,5,6),(6,6,6),(5,7,6), \\
(8,5,5),(7,6,5),(6,7,5), \\
(8,6,4),(7,7,4),\\
(8,7,3) \\
$$


问:
Strang: Simple Calculus Question
I'm looking at Gilbert Strang's introductory calculus textbook "Calculus".
https://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf.

In chapter 1, fig 1.5, Strang illustrates what happens when you alter the function f(t) = 2t + 1. He shows the following equations along with two similar graphs, the second shifted downward.

f(t) = 2t + 1

f(t) - 2 = 2t - 1

I realize that taking the first equation and subtracting 2 from the RHS alone will nudge the graph down 2 tics, but he's done it to both sides. Shouldn't this result in the same function, and the same graph? My only guess is that maybe he forgot to  relabel the axes?

Screenshot: https://i.stack.imgur.com/HOtXF.png

答:
If you rename

$$g(t) := f(t) -2$$

then you have the function $g(t) = 2t - 1$ which is what he illustrated.

If he would write $f(t) = 2t - 1$ then it would be ambiguous because you would label two different objects with the same name. This could lead to the wrong perception that

$$2t - 1 = 2t + 1$$

i.e.

$$-1 = +1$$

which is obviously not true.

Title: Strang: Simple Calculus Question


问:
Test convergence/ divergence of $\sum _{n=1}^{\infty }\:2^{-n-\left(-1\right)^n}$
I have no hint how to approach this problem.
Please Help!</p>

答: For every $n\in\mathbb{N}$ we have $2^{-n-(-1)^n}=2^{-(n+(-1)^n)}=\big(\frac{1}{2}\big)^{n+(-1)^n}=\big(\frac{1}{2}\big)^n\big(\frac{1}{2}\big)^{(-1)^n}\leq\big(\frac{1}{2}\big)^n2$. With this observation and a criteria for series convergent, you can argue and conclude that the series is convergent.


问:
Define a new operation $*$ on $G$ by $a * b = ac^{-1} b$
Let $G$ be a group and let $c$ be a fixed element of $G$.

Define a new operation $*$ on $G$ by

$$a * b = ac^{-1} b$$

I already show the associativity: for $a,b,d\in G$,
$$\begin{align*}
a * (b*d)
&= a*(bc^{-1} d)\\
&=ac^{-1}(bc^{-1}d)\\
&=(ac^{-1})(bc^{-1}d).\\
(a * b)*d&=(ac^{-1} b)*d\\
&=(ac^{-1}b)c^{-1}d\\
&=(ac^{-1})bc^{-1}d\\
&=(ac^{-1})(bc^{-1}d)\\
&=a * (b*d)
\end{align*}$$
and find the identity is $c$, how do I find the inverse?

答:
Hint: solve the equation $ac^{-1}x = c$ for $x$ and then check that also gives a solution to $xc^{-1}a = c$.

 The solution is $x = ca^{-1}c$

Aside: I now see how this arises. If you define $f : G\to G$ by $f(x) = c^{-1}x$, then $f$ is a bijection and you get a new group operation on $G$ by using $f$ to relabel the elements of $G$. Thus:
$$
a * b = f^{-1}(f(a)f(b)) = c((c^{-1}a)(c^{-1}b)) = ac^{-1}b
$$


答:
Yes, when you write

$$f(t)-2=(2t+1)-2,$$ $f$ is still the same function.

But the author shows

$$g(t):=f(t)-2=2t-1$$ next to $f(t)$.

Title: Strang: Simple Calculus Question

Body:
I'm looking at Gilbert Strang's introductory calculus textbook "Calculus".
https://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf.

In chapter 1, fig 1.5, Strang illustrates what happens when you alter the function f(t) = 2t + 1. He shows the following equations along with two similar graphs, the second shifted downward.

f(t) = 2t + 1

f(t) - 2 = 2t - 1

I realize that taking the first equation and subtracting 2 from the RHS alone will nudge the graph down 2 tics, but he's done it to both sides. Shouldn't this result in the same function, and the same graph? My only guess is that maybe he forgot to  relabel the axes?

Screenshot: https://i.stack.imgur.com/HOtXF.png


答: Hint: as $x' = x-2$, it follows that $x' \leq 6$ rather than $x' \leq 8$. Similar for $y'$ and $z'$.</p>


问:
Resolvent for Neumann Laplacian operator
I want to find the resolvent for the Laplacian with Neumann boundary conditions:
$$D(A)=\{f\in W^{2,1}(0,1):f'(0)=f'(1)=0\},\quad f\mapsto Af:=f''. $$

I know that $0$ is an eigenvalue with eigenspace spanned by the constant one function. I need to compute the resolvent $R(\lambda, A):=(\lambda-A)^{-1}$ for $\lambda<0$ which are in the resolvent set of $A$.

My try 1: Let $f\in L^1(0,1)$ and $R(\lambda,A)f=g$. Then $f=(\lambda-A)g=\lambda g - g''$. Therefore $$
\begin{align*}\int_0^1 f^2(y)\, dy &= \lambda \int_0^1g(y)f(y)dy-\int_0^1f(y)g''(y)\, dy\\
&= \lambda \int_0^1g(y)f(y)dy+\int_0^1 f'(y)g'(y)\, dy\\
&=\lambda \int_0^1g(y)f(y)dy+ f'(1)g(1)-f'(0)g(0)-\int f''(y)g(y)\, dy.
\end{align*}$$

However, I'm not sure how to proceed from here. Any reference where the resolvent is explicitly computed would also be sufficient.

Edit:

My try 2: I tried to solve by variation of parameters and would appreciate if someone verifies it:

Let $\lambda\in (-\infty,0)\cap \rho(A)$ and $\mu=-\lambda$. The solution of the auxiliary equation is $g_c(x) = A \cos\sqrt{\mu}x + B\sin \sqrt{\mu}x$ and the Wronskian is $\sqrt{\mu}$. By variation of parameters: $$\sqrt{\mu}g_p(x)= \cos \sqrt{\mu} x\int_0^x \sin \sqrt{\mu} t f(t)\, dt-\sin \sqrt{\mu} x \int_0X\cos \sqrt{\mu} t f(t)\, dt.$$
So the solution is $$g(x)=A \cos\sqrt{\mu}x + B\sin \sqrt{\mu}x+\frac{1}{\sqrt \mu}\left(\cos \sqrt{\mu} x\int_0^x \sin \sqrt{\mu} t f(t)\, dt-\sin \sqrt{\mu} x \int_0^x\cos \sqrt{\mu} t f(t)\, dt \right).$$
Differentiating:
$$g'(x)=-\sin \sqrt{\mu} x\left(A \sqrt{\mu} + \int_0^x \sin \sqrt{\mu} t f(t)\, dt\right)+\cos\sqrt{\mu} x\left(B \sqrt{\mu} -\int_0^x \cos\sqrt{\mu} tf(t)\, dt\right).$$

Using $g'(0)=0$, we get $B=0$ and using $g'(1)=0$, we get

$$A=-\frac{1}{\sqrt{\mu}} \int_0^1 \sin \sqrt{\mu} t f(t)\, dt.$$

Thus $$R(\lambda,A)f=\frac{1}{\sqrt{\mu}}\left(\cos \sqrt{\mu} x\int_x^1 \sin \sqrt{\mu} f(t)\, dt -\sin \sqrt{\mu} x\int_0^x \cos \sqrt{\mu} t f(t)\, dt\right).$$

Is this correct?

答:
Essentially what you are asking is how to solve the inhomogenous ODE
$$g''-\lambda g=f$$
subject to Neumann boundary conditions. For this I suggest you use variation of parameters, which will eventually give you an expression for $g$ in terms of $f$ and the solutions of the homogenous equation $g''-\lambda g=0$. Here are two sources for this technique, the first more applied the second with some proofs. You may also consult an ODE textbook.


问:
The Etale Fundamental Group of $\mathbb{A}_{\mathbb{F}_q}$ and $\mathbb{G}_{m,\mathbb{F}_q}$
I am learning the etale fundamental group of a scheme. And I am hugely confused by the etale fundamental group of the additive  and multiplicative group scheme $\mathbb{G}_{a}$ and $\mathbb{G}_{m}$ over a finite field $\mathbb{F}_q$.

I know that for an algebraic variety $X$ over a general field $k$, its etale fundamental group suits the following exact sequence  $1 \rightarrow \pi_1(X_{k^s}) \rightarrow  \pi_1(X) \rightarrow Gal(k^s/k) \rightarrow 1$. So what I want can be captured more or less by $\pi_1(\mathbb{G}_{a,\bar{F_q}})$ and $\pi_1(\mathbb{G}_{m,\bar{F_q}})$.

But the reduced question seems complicated too, because of the characteristic is not $0$. When characteristic $0$ everything is known and can be found, and the result is that $\pi_1(\mathbb{G}_{a,\bar{k}})=1 $ and $\pi_1(\mathbb{G}_{m,\bar{k}})=\mathbb{\widehat{Z}}$, by Riemann-Hurwitz formula or something else. But in characteristic $p$, for $\mathbb{G}_{m}$, at least you don't have the $\mathbb{Z}_p$ part of $\mathbb{\widehat{Z}}$, for the multiplication by $p$ map $\mathbb{G}_m \rightarrow \mathbb{G}_m$ is not unraminfied. And further, the etale fundamental group of $\mathbb{G}_{m}$ seems to be larger than $\prod_{l\neq p}{\mathbb{Z}_l}$, for some kind of higher ramification theory which I don't know.

I also heard the existence of Artin-Schreier extension, but totally don't understand. Can someone give me any tips on that? Or any information about the fundamental group of  $\mathbb{G}_{a}$ and $\mathbb{G}_{m}$ over finite field $\mathbb{F}_q$?

答:
Fundamental group in characteristic $p$ consists of a tame part and a wild part. the tame part is computed by the finite etale covers with prime to $p$ degree and basically behave like the situation in characteristic 0. in the case you want the tame part of the fundamental group of $G_a$ is $1$ and for $G_m$ it is $\Pi_{l\not =p}\mathbb{Z}_l$.

there is also a wild part. for this part like the usual algebraic topology it is easier to understand first cohomology. for this write the long exact sequence obtained by evaluating the short exact sequence $$0\to F_p\to O_{G_a}\to O_{G_a}\to 0$$
on $G_a$ where the last map is $Fr-id$, you get
$$0\to F_p\to \bar{F_p}[t]\to \bar{F_p}[t]\to H^1(G_a,F_p)\to H^1(G_a,O_{G_a})=0$$
the last equality is true because $G_a$ is affine and hence cohomology of coherent sheafs are  zero.

in summary first cohomology of $G_a$ is the cokernel of $Fr-id$. more explicitly if you want an etale cover of degree $p$ you have to choose a polynomial $f(t)\in k(t)$ and then add the root of $x^p-x-f(t)=0$ to $k[t]$. $H^1(G_a,F_p)$ is already a huge group, by a similar trick you can find all the etale covers of degree $p^n$.


问:
Show that $U = int(cl(U))$
Let $U \subset X$ and $V := X  \backslash  U$. Show that $U = int(cl(U))$ only if, $V = cl(int(V))$.
I found this: $cl(U)=U \cup int(cl(U))$, but i don't know if it can be used here. Is it true that $U \subset int(cl(U)) $? If it is then $U \cup int(cl(U)) = int(cl(U))$ ?. And also i found that $cl(U) = U $ if $U$ is open, but how to check if this set is open?

答:
This is equivalent to showing that if $U = int(cl(U))$ then $V = cl(int(V))$.

Let's first recall 2 properties (which is easy to prove if $X$ is a metric space but also true for topological space):

$cl(A)^c = int(A^c)$ and 2. $int(A)^c = cl(A^c)$ (which is just applying 1. on $A^c$)

Suppose $U = int(cl(U))$. Then $V = U^c = int(cl(U))^c = cl(cl(U)^c)$, by applying 2 on $cl(U)$.

But we also have $cl(U)^c = int(U^c)$ (applying 1) so then $V = cl(int(U^c)) = cl(int(V))$

I am however not sure if you can use $1$ and $2$.

Title: Show that $U = int(cl(U))$


问:
Example of a symbolic or a discrete dynamical system where $NW(f) \not\subset \overline{R(f)}$?
In this question there is an example of a continuous dynamical system with $NW(f) \not\subset \overline{R(f)}$. The definitions I am working with are exactly same as that question. I want to find a symbolic or a discrete dynamical system with the mentioned property. I would really appreciate a not-so-involved example, so that I can gain an intuition to distinguish non-wandering points from points in the closure of the set of all recurrent points.

答: One example is the orbit closure of ${}^\infty 0 1 0 1 0 0 1 0 0 0 1 \cdots 1 0^n 1 0^{n+1} 1 \cdots$ with respect to the shift map. The left tail is just repeating 0s, and on the right are 1s with increasing gaps. The only recurrent point is the all-0 configuration, but ${}^\infty 0 1 0^\infty$ is nonwandering.


问:
application on Lebesgue Dominated Convergence Theorem
If $\{f_n\}$ is a sequence of measurable functions on the measurable set $E$ such that $f_n \rightarrow f$ converges pointwise almost everywhere on $E$ where $|f| \in L(E)$ "i.e. $|f|$ is Lebesgue integrable on $E$, say, $\int_{E} |f| = a$ and $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)|= b$.

I need to show that $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)-f(x)|$ exists and to find its value.
I can see that $|f_n(x)-f(x)| \leq  |f_n(x)|+|f(x)|$ so passing the integral over $E$ then taking the limit as $n \rightarrow \infty$ proves that the limit of the integral exists, but I am still not sure how to find the limit value, my guess is to apply the Lebesgue Dominated  Convergence Theorem or the general Lebesgue Dominated  Convergence Theorem.

答:
It seems like this, I am not sure if it is correct.

Please tell me where I was wrong.

Since $\lim_{n}\int_E|f_n(x)|=b \lt ∞ $, $ {\exists}$ $N$ such that ${\forall} n \gt N$$f_n\in L(E)$.

Since $f_n \rightarrow f a.e.$, we have $f_n - f \rightarrow 0 a.e.$.

And $|f_n - f|$ domined by $|f_n| + |f|$, by Lebesgue Dominated Convergence Theorem $\int_E|f_n - f| \rightarrow \int_E 0 =0$.


问:
Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$
Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?

答:
Tedious (and therefore somewhat inferior) alternative approach that focuses on the minimum difference between distinct cubes.

Prove that the following is impossible:

$a,b \in \Bbb{Z}$

$a \neq 0 \neq b.$

$a^3 + b^3 = 1.$

For $c,k \in \Bbb{Z},~$ let $f(c,k) = (3c^2k + 3ck^2 + k^3).$

$\underline{\text{Lemma 1}}$

$(c,k \in \Bbb{Z^+}) ~\implies ~f(c,k) \neq 1.$

Proof :

$f(1,1) = 7.$

$(c \geq 1) ~\implies~ (c^2 \geq 1).$

$(k \geq 1) ~\implies~ (k^2 \geq 1 ~~~\text{and}~~~ k^3 \geq 1).$

Therefore,
$~~(c \geq 1 ~~~\text{and}~~~ k \geq 1) ~\implies~ f(c,k) \geq f(1,1).$

$\underline{\text{Lemma 2}}$

$(c,k \in \Bbb{Z^+}) ~\implies ~ \{[(c+k)^3 - c^3] \neq 1\}.$

Proof :

$[(c+k)^3 - c^3] = f(c,k)$.

Invoke Lemma 1.

If $a,b$ both positive, then the minimum value of $a^3 + b^3$ is $2$.

If $a,b$ both negative, then the maximum value of $a^3 + b^3$ is $-2$.

Therefore, $~~~~a \in \Bbb{Z^+},~ b \in \Bbb{Z^-},~~~~$ or vice versa.

Since the equation in dispute, $1 = a^3 + b^3$, is symmetric with respect to $a$ and $b$,

without loss of generality,
$~~a \in \Bbb{Z^+},~ b \in \Bbb{Z^-}$

Let $c = -b \implies$

$c \in \Bbb{Z^+}~~$ and

$1 = a^3 + b^3 = a^3 - c^3$.

This implies that $a > c.$

Therefore, $~~~\exists k \in \Bbb{Z^+}~~~$ such that $~~~a = (c + k)$.

Invoke Lemma 2.


Body:
Hint: Your answer will be equal to the coefficient of $x^{18}$ in $$(x^{2}+x^{3}+\cdots +x^{8})(x+x^{2}+\cdots +x^{7})(x+x^{2}+\cdots +x^{6})$$

Which is equal to the coefficient of $x^{14}$ in $$(1+x+\cdots +x^{6})(1+x+\cdots +x^{6})(1+x+x^{2}+\cdots x^{5})=(1-x^{7})^{2}(1-x^{6})(1-x)^{-3}$$


问:
Proof verification for matrix norm
I am studying matrix norms from here. We are defining an operator norm on a matrix as $$\|A\|=\sup_{\substack{x\in\mathbb{C}^n}\\\|x\|=1} \|Ax\|,$$
where $\mathbb{C}$ is the set of complex numbers and vector norm is considered for $\|x\|=1$. In Proposition 4.8, it is stated that if $B$ is a matrix such that $\|B\|<1$, then $I+B$ is invertible ($I$ is an identity matrix) and $\|(I+B)^{-1}\|\le \frac{1}{1-\|B\|}$.

I tried out the proof myself as it is not stated. We can write:

$$(I+B)^{-1}=\sum_{t=0}^{\infty}(-1)^tB^t=I+\sum_{t=1}^{\infty}(-1)^tB^t=I-B[I-B+B^2-B^3+...]=I-B(I+B)^{-1}$$
$$\Rightarrow \|(I+B)^{-1}\|=\|I-B(I+B)^{-1}\|\le \|I\|+\|B\|\|(I+B)^{-1}\|=1+\|B\|\|(I+B)^{-1}\|$$
$$\Rightarrow \|(I+B)^{-1}\|\le \frac{1}{1-\|B\|}.$$ The inequality in the second line follows from sub-multiplicative property of operator norm. Can someone please let me know if this proof is fine or if there is a better proof, where can I find one. Thank you.

答:
All the ideas you need are in your proof, they're just not structured in a convincing way.

I suggest that you proceed as follows: argue that because $\|B\| < 1$, the series $\sum_{t \geq 0} (-1)^tB^{t}$ converges. Then, argue that the matrix $M = \sum_{t \geq 0} (-1)^tB^{t}$ satisfies $(I + B)M = M(I + B) = I$, which allows us to conclude that $M = (I + B)^{-1}$.

Title: Proof verification for matrix norm


问:
Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots?
Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots and has only integral parameters?

Since it has two different roots the it must be of degree at least $2$.

If it is of degree $2$, then it is of the form $(x-\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})$, but this type of polynomial has parameters which are not integers.

If it is of degree $4$ then we have the polynomial $(x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})(x+\frac{\sqrt{2}}{3})$.

It remains to examine the case where it is of degree $3$. This is where I got stuck. I don't know how to examine that case. Could you please explain to me how to solve this question?

答:
The polynomial with integer coefficients with least degree that has $\frac{\sqrt{3}}{2}$ as a root is $4 x^2 - 3$.

Every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ as a root is a rational polynomial multiple of $4 x^2 - 3$.

Likewise, for $\frac{\sqrt{2}}{3}$, the polynomial is $9 x^2 - 2$.

Since these two polynomials are coprime, every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{2}}{3}$ as roots is a multiple of $(4 x^2 - 3)(9 x^2 - 2)$.


答: Hint : Let $f$ be a polynomial with integer coefficients. Then if $a+\sqrt b$ is a root of $f$, where $a, b \in \Bbb Q, \sqrt b \notin \Bbb Q$, then $a-\sqrt b$ is also a root of $f$.

Title: Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots?

Body:
Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots and has only integral parameters?

Since it has two different roots the it must be of degree at least $2$.

If it is of degree $2$, then it is of the form $(x-\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})$, but this type of polynomial has parameters which are not integers.

If it is of degree $4$ then we have the polynomial $(x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})(x+\frac{\sqrt{2}}{3})$.

It remains to examine the case where it is of degree $3$. This is where I got stuck. I don't know how to examine that case. Could you please explain to me how to solve this question?


问:
Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs

Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs as $P_1P_2,P_2P_3....P_{m-1}P_m, P_mP_1$ and any player who wins two consecutive games (ie. k and (k+1)th game) wins the match. If the chance that the match is won at the rth game is $k$, then find relation between $k$ and $r$

I haven’t really understood the question or what it wants to do, so I don’t know how to begin. Can anyone explain the question and give a starter hint?

答:
@Aditya: An attempt to simplify the presentation. We can forget about m, work out for $r=5$, and then generalize

The last two games have to be $\fbox{LW}\fbox{WL}$

and the first $(5-2)$ can either be $\fbox{WL}\fbox{WL}\fbox{WL}$ or

any  of the $3\;\fbox{WL}$ can be replaced by $\fbox{LW}$ with the caveat that once a $\fbox{LW}$ has been inserted, all subsequent ones also must change to $\fbox{LW}$

This is the rationale of $(r-2+1)$ valid strings

For this example, $Pr = \dfrac{5-2+1}{2^5}$

Putting in the general symbols, we get $Pr = \dfrac{r-1}{2^r}$


问:
The base of an isosceles triangle lies in a plane
The vertex $B$ of an isosceles $\triangle ABC$ $(AB=BC)$ does NOT lie on the plane $\alpha$, but the base $AC=18$ of the triangle lies in the plane. The angle between $CB$ and $\alpha$ is $60^\circ$. Find the angle $\measuredangle(BM,\alpha)$ where $BM$ is the median of the triangle and $AB=15$.

Is the given angle $\measuredangle ACB=60^\circ$? Because the orthogonal projection of $BC$ is the line $MC$ or not? Is the orthogonal projection of the point $B$ the point $M$ in $\alpha$ and why? I am not sure I understand which angle I am supposed to find. We can use the fact that $BM\perp AC$ because the triangle is isosceles, but what else? Thank you in advance!

答:
Imagine isosceles $\triangle ABC$ as being vertical on the plane $\alpha$, $AC$ contained in $\alpha$. $\triangle BMC$ is a $3-4-5$ triangle right-angled at $M$.

Now tilt the triangle, keeping $AC$ fixed, so $B$ moves closer to the plane. Call the projection of $B$, $P$. As $ABC$ is isosceles, $P$ will move on the perpendicular bisector of $AC$, away from $M$. Clearly in triangle $BPC$, right-angled at $P$, $\angle BCP$ is the angle between $BC$ and $\alpha$.

Initially in the vertical position, $P$ was at $M$ and $\angle BCP \approx 53.1^{\circ}$. As $B$ moves closer to the plane, this angle will reduce and go to zero when $B$ finally lies in the plane.

Thus the data $\angle BCP= 60^{\circ}$ is false; it is not compatible with $BC=15, AC=18$. $\angle BCP\le \arcsin (4/5)\approx 53.1^{\circ}$ always.


问:
Is there a $\tau$ topology for which $B$ is the base?
Please check if this is correct. $X = \{a,b,c,d,e \}$

$B = \{\{a\},\{b\},\{a,c\}\{a,b,d\}\}$ — no because set $\{a,b,c,d\}$ can't be created.

$B = \{\{a\},\{b\},\{a,c\},\{b,d\},\{b,e\}\}$ — no because set $\{a,b,c,d,e\}$ cant be created.

$B = \{\{a\},\{b\},\{e\},\{a,c\},\{b,c,d\}\}$ — yes, and topology is: $\tau = \{\{\emptyset\},\{a\},\{b\},\{c\},\{d\},\{e\},\{a,b,c\},\{a,b,c,d\},\{a,b,c,d,e\}\}$

答:
That is not correct. It is not a base because $X$ is not an union of elements of that set.

That is not correct. That set is a base of the topology$$\bigl\{\emptyset,\{a\},\{b\},\{a,c\},\{b,d\},\{b,e\},\{a,b\},\{a,b,e\},\{a,b,d\},\{a,b,c\},\{a,b,c,d\},\{a,b,c,e\},X\bigr\}.$$

It is also not correct. It is not a base because $\{a,c\}\cap\{b,c,d\}=\{c\}$ and no element of $B$ to which $c$ belongs is a subset of $\{c\}$.


问:
Define the linear transformation $T: \mathbb R^3 \to \mathbb R^2$
Define the linear transformation $\Bbb R^3\longrightarrow \Bbb R^2$ via,

$T\begin{bmatrix}x\\y\\z\end{bmatrix}\mapsto\begin{bmatrix}y+z\\y-z\end{bmatrix}.$

I'm just a bit confused on the question. When it asks to define the linear transformation, does this just mean proving the two basic rules that $T(u+v) = T(u) + T(v)$ and  $T(cu) = cT(u)$? Finding the standard matrix? Or is there something bigger to this that I am missing? Thanks in advance.

The second part asks to find all vectors which map to 0.

答:
When mathematical texts say

Define the linear transformation $\Bbb R^3\to \Bbb R^2$ via
$$
T\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}y+z\\y-z\end{bmatrix}
$$

what they are really saying is that they establish a linear transformation from $\Bbb R^3$ to $\Bbb R^2$, give it the name $T$, and its behavior is described by the given formula.

It is not, on its own, a request to you as a reader to do anything other than just remember the given information for future reference. It is not an exercise where they ask you to show that it is linear, or anything like that. However, that kind of request may come in the next sentence, depending on what kind of text you're reading.

That being said, being actively engaged in whatever text you're reading, and check that it is linear like they claim is a good habit to get into. Even though the text itself isn't asking you to do that. Eventually, will have done such checks so many times that you see it's linear as soon as you've read the symbols on the page. At least for a simple example such as this.

Post edit: Now that you have added the actual exercise to your question, we can be a bit more explicit. This is what they want to use $T$ for. This is the point where you are actually asked to actively do something. In this case, take the $T$ as described, and find out which elements of $\Bbb R^3$ are mapped to the origin in $\Bbb R^2$. Said differently, find all $x,y,z$ that satisfy the equations
$$
\cases{y+z=0\\y-z=0}
$$
We have three unknowns and two equations, so there ought to be many solutions.


问:
The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^o$, what is $\angle E$ equal to?
The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^\circ$, what is $\angle E$ equal to?

I tried to solve this question as follows:

$a=\angle EBC$ and $b=\angle BCA$

$2a+b=110^{\circ}$

Also $\angle ACE= 90^\circ-\frac{b}{2}$

This is as far as I got. I don't know how to work out what $\angle E$ is equal to. Could you please explain to me how to solve this question?

答:
Locate incenter $I$ (on $BE$) and connect $I$ with $C$. It is known $\angle BIC = 90+A/2$.

Also $\angle BIC$ is the exterior angle to $\triangle ICE$ where $\angle ICE=90^{\circ}$. (Why?) Clearly by exterior-angle property applied to $\triangle ICE$, $\angle E=A/2$.


问:
Does the Mandelbrot Set with a power of -1 produce a Farey Sequence, and if so, can this be connected to the Riemann Hypothesis?
I'm by no means an expert in Farey Sequences nor understanding the Riemann Hypothesis - but I do know there is some relationship between them, as alluded to in this question. As such, if the Mandelbrot Set with a -1 power (see image below) produces a Farey Sequence along the imaginary axis, could this be used to help solve/prove the non-trivial zeros for the Riemann Hypothesis? Has something like this already been attempted?

Note that the radii of the following circles will be reduced if either the bailout/threshold or the maximum iterations are increased, I don't know if they converge or not. The Escape-Time algorithm is used, so more detail might be revealed with a different algorithm.

Edit: On the subject, I noticed that as the Mandelbrot's power approaches zero, what would correspond to the first elements to the Farey Sequence move slower to the left:

I'm pretty sure that this is the Farey Sequence at this point, but all I have to show for it is just by looking at it. I wonder if algebraically solving for the boundary of this fractal will yield interesting insight...

答: Here is pixels $c$ coloured by iteration index $p$ where $|z_p|$ is minimized under iterations of $z_{n+1} = \frac{1}{z_n} + c$.  The right hand side is clamped to $\Re(c) = 0.01$ to make the structure clearer.  The Farey sequence seems to be visible (but this is no mathematical proof).


问:
Question about a proof of Second Derivatve Test

Partial Statement: Let $f$ be defined in a neighborhood of $c$ with $f'(c) = 0$. Then $f''(c) > 0 \implies f(c)$ is a local minimum of $f$

Proof: Suppose  $f$ is defined in a neighborhood of $c$ with $f'(c) = 0, \ f''(c) > 0 $. $\color{red}{\text{It can be shown}}$ there's punctured neighborhood of $N = (c - r, c) \cup (c, c + s)$ of $c$ s.t. $\displaystyle{\frac{f'(x) - f'(c)}{x - c} > 0}$ for $x \in N$ meaning $f'(x) < 0$ for $x \in (c - r, c)$ and $f'(x) > 0$ for $x \in (c, c + s)$. It follows $c$ is a local minimum as $f$ dips towards $c$ from the left and grows back up on the right of $c$.

My question is about the $\color{red}{\text{part in red of the quote above}}$. How does that follow? Here below is my attempt. If incorrect, what would be a better way to show that? Thanks.

By definition, $f$ is differentiable at $c$ if $\displaystyle{\forall\varepsilon > 0, \exists \delta>0 \ \left|\frac{f(x) - f(c)}{x - c} - f'(c)\right| < \varepsilon}$ for any $x$ with $0<|x - c| < \delta.$ Then $\displaystyle{\frac{f(x) - f(c)}{x - c} = f'(c)} > 0$. Applying the definition above one more time we get $\displaystyle{\frac{f'(x) - f'(c)}{x - c} = f''(c)} > 0$.

答:
Per the part in red:

We have $f''(c) > 0$, or

$$\lim_{x\to c} \frac{f'(x) - f'(c)}{x-c} = f''(c) > 0$$

Let's use $\frac{f''(c)}{2}$ as our $\varepsilon$. There is some $\delta$ such that for all $x$ if

$$0 < |x - c| < \delta \implies \left|\frac{f'(x) - f'(c)}{x-c} - f''(c)\right| < \frac{f''(c)}{2}$$
or
$$-\frac{f''(c)}{2} < \frac{f'(x) - f'(c)}{x-c} - f''(c)< \frac{f''(c)}{2} $$
$$\frac{f'(x) - f'(c)}{x-c} > \frac{f''(c)}{2} > 0$$

In order for the LHS to be $> 0$, for $x>c$ we must have $f'(x) > f'(c) = 0$.

Similarly, for $x < c$ we have $f'(x) < 0$.

Thus, $f$ is decreasing on some interval to the left of $c$ and increasing to the right, and so, $c$ is a local minimum.


问:
How do I know if a differential equation leads to chaos or catastrophe?
When I use catastrophe here, I mean a system exhibiting a finite number of bifurcations and by chaos, I mean a system exhibiting a (very) large number of bifurcations.

I do know that catastrophe theory is based on Thom's theorem and chaos theory on qualitative analysis but I can't get over the fact that they are 2 different theories. They seem so similar in terms of bifurcations.

So, which theory do I use before-hand to know if a differential equation leads to chaos or catastrophe and furthermore, can you please explain the exact difference?

答:
When you apply catastrophe theory, you basically make Taylor expansions of your equations close to a bifurcation point and study how the parameters affect the behaviour. Then, according to the mathematical structure of your expansions, called "germs", you might reach conclusions about the behaviour of your system close to the bifurcation points.

In a more general sense, as explained in Robert Gilmore's excellent book, catastrophe theory is a "program", in the sense that it attempts to study how the qualitative nature of the solutions of equations depends on the parameters. Although in its more general form catastrophe theory deals with any discontinuity, there are in practice insurmountable difficulties with the most general equations. In practice, what has been done to a some extent successfully is to study the discontinuities of simple gradient dynamical systems. One aspect that might explain why catastrophe theory fell into almost oblivion is the silly turmoil about it in the 70's. Another, and more important, is that catastrophe theory does not suit itself very adequately to quantitative prediction and calculation, like applied dynamical systems, although it can be used for that.

Without claiming any expertise, I have been a fan of this branch of mathematics since about 2005, literally devouring books and articles. Arcane as it seems, catastrophe theory is an illuminating and very deep set of ideas, very closely related to topology, chaos and dynamical systems in general. I would recommend the above mentioned book to anyone interested in such things. Interested people might want to look to the above mentioned book, and also Poston and Stewart's, that has an excellent and gentle introduction. Of course, there is the very hard to digest original: Structural stability and morphogenesis. Vladimir Arnold wrote a very good book on the subject, which might appeal better to modern mathematicians. Enjoy!


问:
How to convert degrees minutes seconds to hour minutes seconds (and viceversa)?
I'm reading the book "Practical Astronomy with your Calculator or Spreadsheet 4th Edition" and I want to implement formula 22: Converting between angles expressed in degrees and angles expressed in hours. It says that:

To convert between angles expressed in decimal hours and angles
  expressed in decimal degrees, simply multiply or divide by 15.

So, I have to convert a angle in degrees minutes and seconds into decimal degrees and then multiply it 15 (or divide it by 15?).

When do I have to multiply and when do I have to divide by 15?

Or

How to convert degrees minutes seconds to hour minutes seconds (and viceversa)?

答: Nope!  360 degrees is 24 hours.  360 divided by 24 is 15.  So 15 degrees is 1 hour of right ascension.</p>


问:
Show that the graph of continuous function is homeomorphic to $\mathbb{R}$

Let $f : \Bbb{R} \to \Bbb{R}$ be continuous and let $\Gamma =\{(x,f(x) \mid x \in \Bbb{R} \}$ so $\Gamma \subset \Bbb{R}^2$. Show that $\Gamma$ is homeomorphic to $\mathbb{R}$.

Define $h: \mathbb{R} \to \Gamma, h(x)=(x,f(x))$. This map is certainly continuous. What I would need to show is that it’s bijective and has continuous inverse. For the inverse I considered $h^{-1} : \Gamma \to \mathbb{R}, h^{-1}(x)=(x, f^{-1}(x))$, but I don’t have that $f$ necessarily would have an inverse? How can I find the inverse for this map?

答:
Well, you have the co-domain wrong in your putative inverse. You want $h^{-1}$ to take an ordered pair and give you a real number, but you have done the reverse.

Hint: consider $\pi(x, y) = x$.


答:
As you found out, $\angle ACE = \frac{\angle A + \angle B}{2}$

So, $\angle BCE = \frac{\angle A + \angle B}{2} + \angle C$

$\angle CBE = \frac{\angle B}{2}$

So $\angle E = 180^0 - (\frac{\angle B}{2} + \frac{\angle A + \angle B}{2} + \angle C) = ?$

Title: The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^o$, what is $\angle E$ equal to?

Body:
The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^\circ$, what is $\angle E$ equal to?

I tried to solve this question as follows:

$a=\angle EBC$ and $b=\angle BCA$

$2a+b=110^{\circ}$

Also $\angle ACE= 90^\circ-\frac{b}{2}$

This is as far as I got. I don't know how to work out what $\angle E$ is equal to. Could you please explain to me how to solve this question?


问:
Questions about godel's first incompleteness theorem
I'd rather not get into the formal proof of godel's first incompleteness theorem. But I have 2 general questions. Looking at the statement from wikipedia:

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#First_incompleteness_theorem

"Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true,[1] but not provable in the theory (Kleene 1967, p. 250)."

My first question is this... is "any consistent, effectively generated formal theory" something that can be defined and referred to? This seems problematic to me... because then we'd have self-reference wouldn't we... what about the theory from which the theorem is being stated? Meaning isn't the godel incompleteness theorem stated and proved from within a theory capable of expressing elementary arithmetic... and the theorem is referring to the theory within which it is expressed... isn't this a problem.

My second question is related to the first... if I prove godel's incompleteness theorem... it seems I cannot prove it for the very theory within which I state the theorem. Because if I did that, the so-called Godel statement would be proved true within the theory within which I'm operating... which is supposed to be impossible. So it seems to me Godel's incompleteness theorem itself is unprovable except with regards to theories within the theory I'm operating under.

Appreciate any clarification.

答:
For simplicity, I'll treat Rosser's improvement of Godel's original theorem below. In particular, "Godel sentence" should really be "Godel-Rosser sentence." This survey of Beklemishev is also a great source, once you understand the basic distinction between Godel's theorem as a global result about all appropriate theories and each particular instance of Godel's theorem.

is "any consistent, effectively generated formal theory" something that can be defined and referred to?

Yes. In particular, "effectively generated" may sound like an informal term, but it's not; more common equivalent phrasings are "computably axiomatizable" and "recursively axiomatizable," and we can talk about computable sets of sentences via Godel numbering and the $T$-predicate. (That's ahistorical, since the $T$-predicate postdates Godel's argument, but from a modern perspective this is the right way to think about things.)

because then we'd have self-reference wouldn't we

Part of the takeaway of Godel is that self-reference isn't inherently problematic. The diagonal lemma (which again was only proved in full generality post-Godel, but whatever) says exactly that a certain amount of self-reference is in fact unavoidable. Not all sets of sentences in the language of arithmetic are definable in the language of arithmetic (see Tarski for a particularly interesting example), but many are - including all computably axiomatizable theories.

isn't the godel incompleteness theorem stated and proved from within a theory capable of expressing elementary arithmetic... and the theorem is referring to the theory within which it is expressed...

Yes. First-order Peano arithmetic, $\mathsf{PA}$, is strong enough to prove Godel's incompleteness theorem (and is in fact massive overkill for this task).

isn't this a problem.

No.

if I prove godel's incompleteness theorem... it seems I cannot prove it for the very theory within which I state the theorem. Because if I did that, the so-called Godel statement would be proved true within the theory within which I'm operating... which is supposed to be impossible.

Godel's incompleteness theorem is the statement "Every computably axiomatizable theory which is consistent fails to decide its Godel sentence." That highlighted bit is fundamental. For example, $\mathsf{PA}$ can prove that statement and so consequently can prove (since it knows it is computably axiomatizable) "If $\mathsf{PA}$ is consistent then $\mathsf{PA}$ doesn't decide its Godel sentence." However, we can't go from that to $\mathsf{PA}$ deciding its Godel sentence without first proving inside $\mathsf{PA}$ that $\mathsf{PA}$ is consistent. In fact, this is how we get the second incompleteness theorem (which similarly is provable, in its general conditional form, in $\mathsf{PA}$)!


答: The inverse mapping would simply be the projection of $(x,f(x)) \in \Gamma$ onto the $x$ axis since that's the reverse of your mapping, which "lifts" the $x$ axis to $(x,f(x))$.

Title: Show that the graph of continuous function is homeomorphic to $\mathbb{R}$

Body:
Let $f : \Bbb{R} \to \Bbb{R}$ be continuous and let $\Gamma =\{(x,f(x) \mid x \in \Bbb{R} \}$ so $\Gamma \subset \Bbb{R}^2$. Show that $\Gamma$ is homeomorphic to $\mathbb{R}$.

Define $h: \mathbb{R} \to \Gamma, h(x)=(x,f(x))$. This map is certainly continuous. What I would need to show is that it’s bijective and has continuous inverse. For the inverse I considered $h^{-1} : \Gamma \to \mathbb{R}, h^{-1}(x)=(x, f^{-1}(x))$, but I don’t have that $f$ necessarily would have an inverse? How can I find the inverse for this map?


问:
How to find the number of combinations of $x,y,z$ that satisfy an equation
The equation is $x + y + z = 18$,
with the constraints:

$2 \leq x \leq 8$

$1 \leq y \leq 7$

$1 \leq z \leq 6 $

First I solved it so the lower bounds where zero, by substituting the following:

$x' = x - 2$, $y' = y - 1$, $z'= z - 1$.

Which gave me the combination of $16,2 = 120$, now I have to subtract the number of combination of $x,y,z$ that are over, so $x'\geq7$, $y'\geq7$, or $z'\geq6$. The problem is when I plug those into the equation I end up with a negative answer and do not know how to deal with the problem.

答:
As you say, you need to count the solutions in non-negative integers to $x'+y'+z'=14$ that violate at least one of the constraints $x'\le 6$, $y'\le 6$, and $z'\le 5$. This is an inclusion-exclusion argument.

First count the solutions with $x'\ge 7$. That’s the number of solutions in non-negative integers to $x''+y'+z'=7$, where $x''=x'-7$. You know how to compute this: it’s $$\binom{7+3-1}{3-1}=\binom92=36\;.$$

Similarly, you find that there are $\dbinom92=36$ solutions with $y'\ge 7$ and $\dbinom{10}2=45$ with $z'\ge 5$. These all have to be subtracted from $120$, leaving $120-36-36-45=3$ valid solutions.

However, some invalid solutions were removed twice, because they violated two constraints; these must be added back in once.

How many solutions to $x'+y'+z'=14$ have $x'\ge 7$ and $y'\ge 7$?

How many have $x'\ge 7$ and $z'\ge 6$?

How many have $y'\ge 7$ and $z'\ge 6$?

All of those must be added back in. If there were any solutions that violated all three constraints, they would now have been counted once in the $120$, subtracted $3$ times in the initial correction, and added back in $3$ times in the second correction, so they would be included in the total and would need to be subtracted. However, there are none, so no further correction is needed.


问:
derivative formula $\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = (n-1)\mathbf{a}$
Assume $\mathbf{r}=\mathbf{x}−\mathbf{x}′$ is the position vector in $\mathbb{R}^n$, for constant $\mathbf{a}$, we have

$$\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = (n-1)\mathbf{a}.$$

This comes from fig.6 in Tutorial on Geometric Calculus by David Hestenes. Can anyone help me to derive it? I thought that: $\nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = \epsilon^{ijk}\partial_i a_j x_k = 0$.

答:
I would think it is still applicable for traditional vector analysis approach. With $\mathbf{r}\in\mathbb{R}^n$, then
\begin{align}
[\nabla\times(\mathbf{a}\times\mathbf{r})]&=\varepsilon_{ijk}\partial_j[\mathbf{a}\times\mathbf{r}]_k=\varepsilon_{ijk}\partial_j(\varepsilon_{klm}a_{\ell}r_m)\\
&=\varepsilon_{ijk}\varepsilon_{klm}(r_m\partial_ja_{\ell}+a_{\ell}\partial_jr_m)\\
&=(\delta_i^{\ell}\delta_j^m-\delta_i^m\delta_j^{\ell})(r_m\partial_ja_{\ell}+a_{\ell}\partial_jr_m)\\
&=r_j\partial_ja+a\partial_jr_j-r\partial_ja_j-a_j\partial_j{r}\\&=(\mathbf{r}\cdot\nabla)\mathbf{a}+\mathbf{a}(\nabla\cdot\mathbf{r})-\mathbf{r}(\nabla\cdot\mathbf{a})-(\mathbf{a}\cdot\nabla)\mathbf{r}\\
&=\mathbf{a}(\nabla\cdot\mathbf{r})-(\mathbf{a}\cdot\nabla)\mathbf{r}\\
&=n\mathbf{a}-\mathbf{a}\\
&=(n-1)\mathbf{a}
\end{align}
(where I have mixed lower and upper indices).


问:
What is the smallest number of tosses for all outcomes?
What is the smallest number of tosses that need to be done to get all the possible outcomes $\{1, 2, 3, 4, 5,6\}$ of a right dice with a reliability of $0.99$?

My attempt is to use the Bernoulli scheme and the Moivre–Laplace theorem:
$$P\left(\left|\frac{\mu}{n}-p\right|<\delta\right)=2\Phi\left(\delta \sqrt{\frac{n}{p \cdot q}}\right)=0.99$$ but I don't know how to define $\delta$ here.

答:
you want to calculate the probability that 6 events all happen. This can be done with inclusion exclusion.

We get $\sum\limits_{i=0}^6 \binom{6}{i} (-1)^i (\frac{i}{6})^N$ which we can approximate from above as $1-6(5/6)^N$ so we are going to require $6(\frac{5}{6})^N < 1/100$. we can solve for $N$ and we get $N=\frac{-log(600)}{log(5)-log(6)}$ which is approximately $35.086$, so if I had to guess I would say $35$ is the first value, although apparently I got it wrong and it's $36$, but to be fair with $35$ you get approximately $0.9898$. Unless there is something wrong with my code (which I include next).

```<code>#include<bits/stdc++.h>
using namespace std;

const int MAX = 15;
int bino[MAX][MAX];

long double pot(long double b,int e){
        long double res = 1;
        while(e){
                if( e%2) res = res*b;
                b = b*b;
                e/=2;
        }
        return res;
}

int main(){
        int n;
        cin >> n;
        for(int i=0;i<MAX;i++){
                bino[i][0] = 1;
        }
        for(int i=1;i<MAX;i++){
                for(int j=1;j<MAX;j++){
                        bino[i][j] = bino[i-1][j] + bino[i-1][j-1];
                }
        }
        long double res = 0;
        for(int i = 0;i<=6;i++){
                long double summand = 1;
                summand = summand*bino[6][i];
                summand = summand*pot(-1,i);
                summand = summand*pot((1.0/6)*i,n);
                res += summand;
        }
        cout << res << endl;

}
</code>```

Body:
you want to calculate the probability that 6 events all happen. This can be done with inclusion exclusion.

We get $\sum\limits_{i=0}^6 \binom{6}{i} (-1)^i (\frac{i}{6})^N$ which we can approximate from above as $1-6(5/6)^N$ so we are going to require $6(\frac{5}{6})^N < 1/100$. we can solve for $N$ and we get $N=\frac{-log(600)}{log(5)-log(6)}$ which is approximately $35.086$, so if I had to guess I would say $35$ is the first value, although apparently I got it wrong and it's $36$, but to be fair with $35$ you get approximately $0.9898$. Unless there is something wrong with my code (which I include next).

#include<bits/stdc++.h>
using namespace std;

const int MAX = 15;
int bino[MAX][MAX];

long double pot(long double b,int e){
        long double res = 1;
        while(e){
                if( e%2) res = res*b;
                b = b*b;
                e/=2;
        }
        return res;
}

int main(){
        int n;
        cin >> n;
        for(int i=0;i<MAX;i++){
                bino[i][0] = 1;
        }
        for(int i=1;i<MAX;i++){
                for(int j=1;j<MAX;j++){
                        bino[i][j] = bino[i-1][j] + bino[i-1][j-1];
                }
        }
        long double res = 0;
        for(int i = 0;i<=6;i++){
                long double summand = 1;
                summand = summand*bino[6][i];
                summand = summand*pot(-1,i);
                summand = summand*pot((1.0/6)*i,n);
                res += summand;
        }
        cout << res << endl;

}


问:
There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each.

There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each. This means there were about $11^{107}$ possible seating plans.

I read that in a book but didn't really understand the solution. I mean the solution doesn't consider the fact that each table can have a max of $10$ people and it actually doesn't care about the capacity of the table. Can anyone explain it to me?

答: The simplest way to read the problem is to consider all the seats distinct, so there are $110$ of them.  In that case the first person has $110$ choices of seat, the next has $109$, the next has $108$ and so on.  This gives $\frac {110!}{3!}$ possible seating plans.  As in my comment, you may want to consider rotating people around a table the same seating plan.  Each table has $11$ rotations, so you should divide by $11^{10}$.  Maybe you want to consider swapping groups of people among the tables, so the tables are indistinguishable.  Then you should divide by $11!$.  You need to specify what makes a seating plan unique.


问:
Eigenvalues of block matrix comprising diagonal matrices
This seems simple, but I can't seem to solve it.
Let
$$A = \begin{pmatrix} D_{11} & \dots & D_{1n} \\ D_{21} & \dots & D_{2n} \\ \vdots & \vdots & \vdots\\ D_{n1} & \dots & D_{nn} \end{pmatrix}$$
be a block matrix such that
(1) $A$ is symmetric
(2) each $D_{ij}$ is diagonal.
The goal is to find the eigenspectrum of $A$.

Special case) If $D_{ij} = c_{ij}I$, would replacing each block $D_{ij}$ by the scalar $c_{ij}$ give the same eigenspectrum?

General) Without this additional assumption, can we compute the eigenspectrum of $A$ using a surrogate matrix $\tilde{A}$ with block entries $A_{ij}$ replaced by a characterization of $D_{ij}$? If so, what would be this characterization?

I have seen this post where we can reorder the rows/columns to preserve eigenspectrum and reduce computation, but I think this question is a bit more specific and hopefully there exists a more analytical answer.

答:
For the general case, suppose that the diagonal matrices have size $m \times m$, and let $d_{ijk}$ denote the $k$th diagonal entry of the diagonal matrix $D_{ij}$. Your matrix can be written in the form
$$
A = \sum_{i,j = 1}^n \sum_{k = 1}^m d_{ijk} \cdot E_{ij}^{(n)} \otimes E_{kk}^{(m)}
$$
where $\otimes$ denotes a Kronecker product and $E_{ij}^{(n)}$ is the $n \times n$ matrix with a $1$ as its $i,j$ entry and zeros elsewhere. This matrix is similar to the matrix
$$
B = \sum_{i,j = 1}^n \sum_{k = 1}^m d_{ijk} \cdot E_{kk}^{(m)} \otimes E_{ij}^{(n)},
$$
which has the block-diagonal form
$$
B = \pmatrix{B_1\\ & \ddots \\ && B_n}
$$
where the $i,j$ entry of $B_k$ is $d_{ijk}$.

In other words, the spectrum of $A$ is the combined spectrum of each of the $n \times n$ matrices $B_1,\dots,B_m$.


问:
Combinatorics proof counting
The question is counting the number $b_{p,q}$ of binary strings with no consecutive $1$’s, with a $0$ at each end. With q 1’s and p 0’s.

How do I prove $b_n = b_{n-1} + b_{n-2}$ is equivalent to saying $b_{p,q}= b_{p-1,q-1}+ b_{p-1,q}$  if $n = p+q$. where $n$ is the number of digits available

Any help would be appreciated!

Title: Combinatorics proof counting

Body:
Here is why $b_{p,q}=b_{p-1,q}+b_{p-1,q-1}$ implies $b_n=b_{n-1}+b_{n-2}$:
$$\begin{aligned}
b_n
&= \sum_{p} b_{p,n-p}
\\
&= \sum_{p} b_{p-1,n-p}+b_{p-1,n-p-1}
\\
&= \sum_{p} b_{p-1,(n-1)-(p-1)}+\sum_pb_{p-1,(n-2)-(p-1)}
\\
&= \sum_{p} b_{p,(n-1)-p}+\sum_pb_{p,(n-2)-p}
\\
&= b_{n-1}+b_{n-2}
\end{aligned}$$


答: The probability that at least one number does not appear after n tosses is $Q=6(\frac{5}{6})^n-15(\frac{4}{6})^n+20(\frac{3}{6})^n-15(\frac{2}{6})^n+6(\frac{1}{6})^n$, so the desire probability is $P=1-Q$.  Solve for $n$ with $Q=0.01$

Title: What is the smallest number of tosses for all outcomes?

Body:
What is the smallest number of tosses that need to be done to get all the possible outcomes $\{1, 2, 3, 4, 5,6\}$ of a right dice with a reliability of $0.99$?

My attempt is to use the Bernoulli scheme and the Moivre–Laplace theorem:
$$P\left(\left|\frac{\mu}{n}-p\right|<\delta\right)=2\Phi\left(\delta \sqrt{\frac{n}{p \cdot q}}\right)=0.99$$ but I don't know how to define $\delta$ here.


问:
Show that $U = int(cl(U))$
Let $U \subset X$ and $V := X  \backslash  U$. Show that $U = int(cl(U))$ only if, $V = cl(int(V))$.
I found this: $cl(U)=U \cup int(cl(U))$, but i don't know if it can be used here. Is it true that $U \subset int(cl(U)) $? If it is then $U \cup int(cl(U)) = int(cl(U))$ ?. And also i found that $cl(U) = U $ if $U$ is open, but how to check if this set is open?

答:
Let $U \subseteq X$. Define the interior of $U$ to be the largest open set contained in $U$ (the existence of such is guaranteed by taking the union over all open sets contained in $U$, and this collection is nonempty since the empty set is in there). Denote this by $U^{\mathrm{o}}$. Define the closure of $U$ to be the smallest closed set containing $U$. Denote this by $\overline{U}$ (the existence of such is guaranteed by taking the intersection over all closed sets containing $U$, and this collection is nonempty since $X$ is in there).

Example: Let $U = (0,1]$. Then $\overline{U} = [0,1]$, $U^\mathrm{o} = (0,1)$, $\overline{(U^\mathrm{o})} = [0,1],$ $(\overline{U})^\mathrm{o} = (0,1).$

Claim 1: If $A \subseteq B \subseteq X$, then $B^c \subseteq A^c$, where we define $A^c = X \setminus A$.

Proof: Let $x \in B^c$, then $x \notin B$ per definition. Since $A \subseteq B$, $x \notin B$ implies $x \notin A$, so $x \in A^c$.

Claim 2: Let $U \subseteq X$ be a set. Then $(\overline{U})^c = (U^c)^\mathrm{o}$.

Proof: For notational convenience, let $V = U^c$. We note $U \subseteq \overline{U}$, so $(\overline{U})^c \subseteq V = U^c$ and $(\overline{U})^c$ is open. By maximality, we get $(\overline{U})^c \subseteq V^\mathrm{o} \subseteq V$. Taking complements again, we have $V^c = U \subseteq (V^\mathrm{o})^c$. By minimality of the closure, we get $U \subseteq \overline{U} \subseteq (V^\mathrm{o})^c$. Taking complements again, we have $V^\mathrm{o} \subseteq (\overline{U})^c$. This shows $V^\mathrm{o} = (\overline{U})^c$.

Claim 3: Let $U \subseteq X$ be a set. Then $(U^\mathrm{o})^c = \overline{(U^c)}$.

Proof: Exercise. It's the same kind of argument as in Claim 2.

Claim 4: Let $U \subseteq X$, $V = U^c$. If $U = (\overline{U})^\mathrm{o}$, then $V = \overline{(V^\mathrm{o})}$.

Proof: $V = U^c = ((\overline{U})^\mathrm{o})^c$. By Claim 3, we have $((\overline{U})^\mathrm{o})^c = \overline{((\overline{U})^c)}$. By Claim 2, we have $\overline{((\overline{U})^c)} = \overline{((U^c)^\mathrm{o})} = \overline{(V^\mathrm{o})}$.

Claim 5: Let $U \subseteq X$, $V = U^c$. If $V = \overline{(V^\mathrm{o})}$, then $U = (\overline{U})^\mathrm{o}$.

Proof: Exercise. It's the same argument as in Claim 4.

Title: Show that $U = int(cl(U))$


问:
$f\in C[a, b] $ Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac1{f(x)} \mathrm{dx} \ge 2\sqrt{\frac mM} (b-a) $
Let $f\in C[a, b] $. Assume that $m> 0$ and $M$ are minimum and maximum values of $f$ over the interval $[a, b]$.Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac 1{f(x)} \mathrm {dx} \ge 2\sqrt{\frac mM} (b-a) $
$\bf{Try} :$
If we can say $\int_{a}^{b} \frac {1} {f(x)} \mathrm dx =\frac {1}{\int_{a}^{b} f(x)} \mathrm dx$ $\bf (\text{I don't know the proof of this claim} )$ then taking $\int_{a} ^{b} f(x) \mathrm dx =y$ we can apply A.M.-G.M. inequality to get

$\frac yM +\frac my \geq 2\sqrt{\frac mM} $.

$\bf \text{But how the factor (b-a) will come in the right hand side?} $
I want to know the answer of both the questions. A alternative way of solving the problem will be interesting to learn if exists.

答:
$\int_{a}^{b} \frac {1} {f(x)} \mathrm dx =\frac {1}{\int_{a}^{b} f(x)} \mathrm dx$ does not hold in general – try it with $f(x) = x$ on $[a,b]=[1, 2]$.

But you are on the right track with the AM-GM inequality. If you apply it to the integrands, not to the integrals, then everything figures out nicely:
$$
\frac{1}{M}\int_{a}^{b} \, f(x) dx +m\int_{a}^{b} \frac 1{f(x)} \, dx
=  \int_a^b \left( \frac{f(x)}{M}+ \frac{m}{f(x)}\right) \, dx \\
\ge \int_a^b 2 \sqrt \frac mM \, dx = 2 \sqrt \frac mM (b-a) \, .
$$

This inequality holds as long as $m$, $M$ and $f(x)$ are strictly positive . The fact that $m$ is the minimum and $M$ the maximum of $f$ is not used.

Equality holds if $\frac{f(x)}{M}= \frac{m}{f(x)}$ on $[a, b]$, that is if $f$ is constant.


问:
Witt Vector Motivation for Teichmüller representatives
I'm quite comfortable with the motivation for Teichmüller representatives given in this MO post, but I'm curious about something said in the wikipedia article:

The elements of $\mathbb{Z}_p$ can be expanded as (formal) power series in $p$:

$$a_0 + a_1 p^1 + a_2 p^2 + \cdots,$$

where $a_i$ are usually taken from the integer interval $[0,p-1]=\{0, 1, \ldots, p-1\}.$ Of course, this power series usually will not converge in $\mathbb{R}$ using the standard metric on the reals, but it will converge in $\mathbb{Z}_p,$ with the $p$-adic metric. We will sketch a method of defining ring operations for such power series.

Letting $a+b$ be denoted by $c$, one might consider the following definition for addition:

$$\begin{align}
c_0 &\equiv a_0+b_0 && \bmod p \\
c_0+c_1 p &\equiv (a_0+b_0) + (a_1 +b_1)p && \bmod p^2 \\
c_0+c_1 p+c_2 p^2 &\equiv (a_0+b_0) + (a_1 +b_1)p+(a_2+b_2) p^2 &&\bmod p^3
\end{align}$$

and one could make a similar definition for multiplication. However, this is not a closed formula, since the new coefficients are not in the allowed set $[0,p-1]$.

I've bolded the last line above to emphasize it. This example is used to motivate choosing a new set of representatives, but I don't see the problem. If I carry out the process of solving those equations for the sum $c=a+b$ with, say, $p=7, a = 55 = 5\cdot 7 + 5$ and $b = 46 = 4\cdot 7 + 6$, I obtain values of $c_0 = 4, c_1 = 3, c_2=1$ which are within the allowed set $\{0, \ldots, 6\}$. I get the same thing if I use addition with carrying. I don't see how any other example can fail to be closed in the sense above. What am I doing wrong or misunderstanding here?

答:
The equations given there for $c_i$ are more naive than you think: in your example, they say that $(5 \cdot 7 + 5) + (4 \cdot 7 + 6) = 9\cdot 7 + 11$.  Converting this into $1\cdot 7^2 + 3 \cdot 7 + 4$ requires carrying, which is a finicky piecewise-defined operation that we want to avoid.

Perhaps this will better explain how non-algebraic the carrying operation is: given two two-digit numbers $a_1 a_0$ and $b_1 b_0$ written in base $p$, you know how to express their sum $c_2 c_1 c_0$ in base $p$ by carrying. This process involves working in $\mathbb Z$ and checking various inequalities, e.g. whether $a_0 + b_0 \geq p$. Can you find a way to calculate the digits $c_0, c_1, c_2 \in \mathbb F_p$ in terms of the digits $a_0, a_1, b_0, b_1 \in \mathbb F_p$ without making any reference to $\mathbb Z$?

The boldfaced sentence seems to be mixing up two senses of the word "closed".  On the one hand, base-$p$ expansions are not closed under the operation of naive addition (without carrying), since for example $9 \cdot 7 + 11$ is not a valid base-$p$ expansion.  (This is of course just a reflection of the fact that the subset $\{0, \dots, p-1\} \subset \mathbb Z$ is not closed under addition.)  On the other hand, it is difficult to give an explicit closed formula for the operation of addition with carrying (as opposed to an algorithm for computing it).


问:
Show the set of element in $\Bbb R^n$ which the product of each element greater than $1$ is convex
I know it might be easy for you, but I have tried to show that the following set is convex for more than three days.

$$A := \left\lbrace (a_1,\dots,a_n) \in \mathbb{R}^n : x_i \geq 0, i = 1,\dots,n, \prod_{i=1}^{n} x_i \geq 1 \right\rbrace$$

I can prove only the first condition ($x_i \geq 0$) that is $(\lambda x_i +(1-\lambda) y_i)\geq 0$ for all $i$ but I cannot address how come of

$$\prod_{i=1}^{n}(\lambda x_i +(1-\lambda) y_i) \geq 1$$

Could you hint or tell me some to let me figure out of this? Thank you in advance.

答:
You can take the logarithm of both sides for the second condition:

$$
\prod_{i=1}^n x_i \geq 1 \Leftrightarrow \sum_{i=1}^n \log (x_i) \geq 0 \Leftrightarrow
-\sum_{i=1}^n \log (x_i) \leq 0.
$$

This is the sublevel set of a convex function, which is itself a convex set.


答:
Applying AM-GM once we get
$$\frac{1}{M}\int_{a}^{b} f(x) \mathrm{d}x +m\int_{a}^{b} \frac1{f(x)} \mathrm{d}x \ge 2\sqrt{\frac{m}{M}}\sqrt{\underbrace{\int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x}_{\text{we need } (b-a)^2 \text{ here}}}$$
So, we need to show that
$$\int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x \ge  (b-a)^2$$
which is obvious according to the Cauchy-Bunyakovsky-Schwarz inequality:
$$(b-a)^2 = \left(\int_a^b 1~ \mathrm dx\right)^2 = \left(\int_a^b \sqrt{f(x)}\sqrt{\frac{1}{f(x)}}\mathrm dx\right)^2 \le \int_{a}^{b} f(x) \mathrm{d}x\int_{a}^{b} \frac1{f(x)} \mathrm{d}x$$

Title: $f\in C[a, b] $ Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac1{f(x)} \mathrm{dx} \ge 2\sqrt{\frac mM} (b-a) $

Body:
Let $f\in C[a, b] $. Assume that $m> 0$ and $M$ are minimum and maximum values of $f$ over the interval $[a, b]$.Then show that $\frac{1}{M}\int_{a}^{b} f(x) \mathrm{dx} +m\int_{a}^{b} \frac 1{f(x)} \mathrm {dx} \ge 2\sqrt{\frac mM} (b-a) $
$\bf{Try} :$
If we can say $\int_{a}^{b} \frac {1} {f(x)} \mathrm dx =\frac {1}{\int_{a}^{b} f(x)} \mathrm dx$ $\bf (\text{I don't know the proof of this claim} )$ then taking $\int_{a} ^{b} f(x) \mathrm dx =y$ we can apply A.M.-G.M. inequality to get

$\frac yM +\frac my \geq 2\sqrt{\frac mM} $.

$\bf \text{But how the factor (b-a) will come in the right hand side?} $
I want to know the answer of both the questions. A alternative way of solving the problem will be interesting to learn if exists.


问:
The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^o$, what is $\angle E$ equal to?
The angle bisectors of $\angle B$ and $\angle C_{ex}$ intersect at point $E$. If $\angle A=70^\circ$, what is $\angle E$ equal to?

I tried to solve this question as follows:

$a=\angle EBC$ and $b=\angle BCA$

$2a+b=110^{\circ}$

Also $\angle ACE= 90^\circ-\frac{b}{2}$

This is as far as I got. I don't know how to work out what $\angle E$ is equal to. Could you please explain to me how to solve this question?

答:
Suppose the measure of the angle $\angle ABC$= $2x$ and the $\angle C_{ex}$ = $2y$ .
By the exterior angle theorem, we have

$70^{\circ}$+$2x$ = $2y$

$y$ - $x$ = $35^{\circ}$

$\angle ACB$ = $180^{\circ}$-$2y$  (because both the angles add upto $180^{\circ}$)

Since $EB$ and $EC$ are the angle biectors of $\angle B$ and $\angle C$ respectively, $\angle EBC$ = $x$ and $\angle ACE$ = y.

Considering the triangle $BEC$,

$\angle E$ + $\angle EBC$ + $\angle BCE$ = $180^{\circ}$

$\angle E$ + $x$ +  $180^{\circ}$-$2y$ + $y$ = $180^{\circ}$  ( since $\angle ACB$ + $\angle ACE$ = $\angle BCE$)

$\angle E$ = $y$ - $x$ = $35^{\circ}$


问:
Is there a smooth homotopy lifting theorem?
Given a smooth map $p\colon E \to B$ of smooth manifolds with the continuous homotopy lifting property, does $p$ satisfy the smooth homotopy lifting property?

答: A trivial counter-example to this is the map $p(x)=x^3$, $p: E={\mathbb R}\to B={\mathbb R}$. It is a homeomorphism, hence, satisfies the continuous homotopy lifting property (HLP). To see that it fails to have the smooth HLP, consider the maps $f_0(x)=0, f_1(x)=x$ and the (smooth) linear homotopy $f_t$ between the two. The map $f_0$ of course lifts to the constant map $\tilde{f}_0(x)=0$ with respect to $p$. On the other hand, the map $f_1$ does not have a smooth lift (the only possible lift would be $\tilde{f}_1(x)=x^{1/3}$ which is not smooth). Hence, the smooth HLP fails for the map $p$.


问:
Is there some special algorithm to minimize polynomial functions?
I have an optimization problem of the form

$$\begin{array}{ll} \underset{\textbf{x} \in \Bbb R^4}{\text{minimize}} & f(\textbf{x})\\ \text{subject to} & \textbf{x} \in \Omega\end{array}$$

where $f$ is a degree-$8$ multivariate polynomial function,  and $\Omega$ is a $4$-dimensional box (i.e., the Cartesian product of $4$ closed intervals). I would like to solve it in the most efficient way.

答:
As mentioned in a comment above, you can optimize a polynomial over a semialgebraic set by transforming the problem into a question regarding the positivity of a polynomial over a semialgebraic set---which can be answered using a positivestellensatz. Parrillo and Lasserre independently constructed a sequence of converging SDP relaxations that can be used to address this problem. You can read about it in the article posted above (link respost: https://arxiv.org/pdf/math/0103170.pdf), here: http://www.princeton.edu/~aaa/Public/Presentations/IOS16_deKlerk.pdf, or here: https://doi.org/10.1017/CBO9781107447226.

It seems that you want to solve a specific problem. You can use one of the available packages for sum-of-squares, for example:

SOSTOOLS: http://www.cds.caltech.edu/sostools/

Gloptipoly: https://homepages.laas.fr/henrion/software/gloptipoly/

Yalmip: https://yalmip.github.io/tutorial/momentrelaxations/

That being said, I will perhaps misuse the no free lunch theorem, to say that: there is no single best algorithm for all polynomial optimization problems. Although, SDP relaxations are great or are the best in general, perhaps for specific problems you can do even better. For a degree-8 multivariate polynomial over a box, I would first try Wolfram Alpha or Mathematica. I would bet that symbolic methods work great over four dimensions.


答: You want to show that the boundary of the set is convex, and for that you need to show that the Hessian of $\prod x_i$ is convex. The $ij$-th element of the hessian is $\prod_{k\neq i, j} x_k.$ (the diagonal elements are zero). Divide through by $\prod x_k,$ to get the $ij$-the element equal to $\frac1{x_i}{x_j}.$ Let $a_k = \frac1{x_k}.$ Your problem now reduces to showing that the matrix $A$ whose $ij$-th element is $a_i a_j,$ where the $a_k$ are positive, is positive definite. But that matrix is $D(J-I)D,$ where $D$ is the diagonal matrix of $a_k$s, so the question reduces to checking that $J-I$ is positive definite. Its eigenvalues are $n-1$ and $0,$ so it is, indeed, positive (semi) definite whenever $n\geq 1.$

Title: Show the set of element in $\Bbb R^n$ which the product of each element greater than $1$ is convex

Body:
I know it might be easy for you, but I have tried to show that the following set is convex for more than three days.

$$A := \left\lbrace (a_1,\dots,a_n) \in \mathbb{R}^n : x_i \geq 0, i = 1,\dots,n, \prod_{i=1}^{n} x_i \geq 1 \right\rbrace$$

I can prove only the first condition ($x_i \geq 0$) that is $(\lambda x_i +(1-\lambda) y_i)\geq 0$ for all $i$ but I cannot address how come of

$$\prod_{i=1}^{n}(\lambda x_i +(1-\lambda) y_i) \geq 1$$

Could you hint or tell me some to let me figure out of this? Thank you in advance.


问:
Prove factor of $x^3-2$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$
By Eisenstein, $x^3-2$ is irreducible over $\mathbb{Q}$. Consider $\mathbb{Q}(\alpha)$, where $\alpha= \sqrt[3]2$. Over $\mathbb{Q}(\alpha)$, we have that $x^3-2 = (x-\alpha)h(x)$ for some $h\in \mathbb{Q}(\alpha)[x]$ of degree 2. How can I show that this $h$ is irreducible over $\mathbb{Q}(\alpha)$? Since the only possible factors of $h$ are linear, I think it suffices to show that $h$ has no roots in $\mathbb{Q}(\alpha)$:

Suppose that $h(\beta)=0$ with $\beta\in\mathbb{Q}(\alpha)$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}]=3=[\mathbb{Q}(\alpha):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]$. We know that $\beta\notin\mathbb{Q}$ (otherwise $x^3-2$ would have been reducible over $\mathbb{Q}$), so $[\mathbb{Q}(\alpha):\mathbb{Q}(\beta)]=3$. How do I get a contradiction from this?

Thanks.

答:
The function $f(x) = x^3-2$ has only one real-valued zero, namely $\sqrt[3]{2}$. (It has only one, since $f'(x) = 3x^2 > 0$). As a polynomial of degree $3$ it has two more zeros, which have to be in $\mathbb{C} \setminus \mathbb{R}$.

We can write
\begin{align}
f(x) = (x- \sqrt[3]{2})\cdot h(x)
\end{align}
where $h$ is a polynomial of degree $2$. Due to that, the function $h(x)$ is irreducible (over $\mathbb{Q}(\sqrt[3]{2})$) if it has no zeros in $\mathbb{Q}(\sqrt[3]{2})$. As mentioned, the two other zeros are in $\mathbb{C} \setminus \mathbb{R}$. But $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. Hence, the two other zeros are not contained in the field extension $\mathbb{Q}(\sqrt[3]{2})$. As a result $h(x)$ has to be irreducible.


问:
Given right-angle triangle $\triangle ABC$ $(\angle A=90^o)$, with $BC=10$ and $AC=6$. Find the length of $DE$.
Given right-angle triangle $\triangle ABC$$(\angle A=90^o)$, with $BC=10$ and $AC=6$. Circle is tangent to $BC$, goes through $A$ and intersects with $AB$ and $AC$ at the points $E$ and $D$ respectively. Find the length of $DE$.

I tried to solve this question as follows:

$AB=8$ from Pythagoras.

Since $DA\perp AE$ then $DE$ is diameter of the circle.

After drawing it out accurately, I see that $DE=4.8$, but I don't know how to work this out mathematically. Could you please explain to me how to solve this question?

答: The diagrams below shows that there is simply insufficient information to determine the length of $DE$.


问:
Why is the unitization of $M_\infty(\mathbb{C})$ not complete?
Consider the space $M_\infty(\mathbb{C})$ of infinite matrices with only finitely many nonzero entries. This is a pre-C$^*$-algebra, and we can consider its unitization $M_\infty^1(\mathbb{C})$ (the unit will be an infinite matrix with ones on the whole diagonal). From what I understand, this space is not complete (its completion is the C$^*$-algebra $\mathcal{K}$ of compact operators on a separable infinite-dimensional Hilbert space).

Why is $M_\infty^1(\mathbb{C})$ not complete? What is an example of a Cauchy sequence in this space that does not converge?

答:
Consider the matrices $x_n$ given by
$$x_n=1\oplus\frac12\oplus\cdots\oplus\frac1{2^{n-1}}\in M_n(C)$$
Then the sequence $(x_n)_{n\in\mathbb N}$ is Cauchy in $M_\infty(\mathbb C)$, with
$$\|x_n-x_m\|=\frac{1}{2^{\min\{m,n\}}},$$
but $(x_n)_{n\in\mathbb N}$ does not converge in $M_\infty(\mathbb C)$ (or its unitization), because any such limit would have $(n,n)$-entry given by $\frac{1}{2^{n-1}}$, hence has infinitely many nonzero entries.


问:
Some (maybe) basic estimate of expected values involving brownian motion
I have the following estimation in a paper I'm reading about Brownian Motion on a Torus:
$$
1+\theta\ \mathbb E_x[\tau_A] + \sum_{i=2}^\infty \left(\theta\sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]\right)^i \leq \exp\left(\theta\mathbb E_x[\tau_A] + 2\theta^2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^2\right)
$$
where $\mathbb T_2$ is a Torus, $A\subset\mathbb T_2$ closed, $x\in\mathbb T_2$ and $0<\theta<\frac 1 2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^{-1}$.

Ideas: To prove this,  I have already tried the Taylor series of $e^x$, the geometric sum, the Jensen inequality and a few other things, however nothing worked so far.

I'd be really glad if you could give me some thoughts about what else to do.

答:
Assuming you are missing a $\theta$ in the exponential, this is just
$$
1+z\leq e^{z}
$$
for $z\geq 0$.


答:
This has nothing to do with probablitities.

Set $x = \theta \mathbb E_x[\tau_A]$ and $y = \theta \sup_y \mathbb E_y[\tau_A]$. Then $0 x \leq y < \frac{1}{2}$. By the geometric series we get
$$ \sum_{i=2}^\infty y^i = \frac{1}{1-y} - 1 - y = \frac{y^2}{1-y} < 2 y^2.$$
Hence
$$ 1 + x + \sum_{i=2}^\infty y \leq 1 + x + 2y^2 \leq \exp(x+2y^2).$$
The last step uses the inequality
$$1+z \leq \exp(z), \forall z > 0,$$ which follows from the Taylor expansion
$$ \exp(z) = 1 + z + \frac{z^2}{2} + \dotsb.$$

Title: Some (maybe) basic estimate of expected values involving brownian motion

Body:
I have the following estimation in a paper I'm reading about Brownian Motion on a Torus:
$$
1+\theta\ \mathbb E_x[\tau_A] + \sum_{i=2}^\infty \left(\theta\sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]\right)^i \leq \exp\left(\theta\mathbb E_x[\tau_A] + 2\theta^2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^2\right)
$$
where $\mathbb T_2$ is a Torus, $A\subset\mathbb T_2$ closed, $x\in\mathbb T_2$ and $0<\theta<\frac 1 2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^{-1}$.

Ideas: To prove this,  I have already tried the Taylor series of $e^x$, the geometric sum, the Jensen inequality and a few other things, however nothing worked so far.

I'd be really glad if you could give me some thoughts about what else to do.


问:
Define the linear transformation $T: \mathbb R^3 \to \mathbb R^2$
Define the linear transformation $\Bbb R^3\longrightarrow \Bbb R^2$ via,

$T\begin{bmatrix}x\\y\\z\end{bmatrix}\mapsto\begin{bmatrix}y+z\\y-z\end{bmatrix}.$

I'm just a bit confused on the question. When it asks to define the linear transformation, does this just mean proving the two basic rules that $T(u+v) = T(u) + T(v)$ and  $T(cu) = cT(u)$? Finding the standard matrix? Or is there something bigger to this that I am missing? Thanks in advance.

The second part asks to find all vectors which map to 0.

答:
The first part of the question is perfectly answered by Arthur , they have already defined the linear transformation

For the second part it is all the set of points {${(k,0,0)|k \in R}$}. Since the y,z components are getting reduced to zero. You can easily verify the Rank nullity theorem here.


问:
Simplifying this expression involving complex numbers
I wish to reduce the following expression :
$$
\dfrac{\mathrm{i}\left(-4t+\sin\left(4\left(t-1\right)\right)+4\right)-\cos\left(4\left(t-1\right)\right)+1}{2{\pi}^2\left(t-1\right)^2}
$$
to get
$$
\left(\frac{\sin(2(t-1))}{\pi(t-1)}\right)^{2}
$$

答:
Euler's formula helps:
$$e^{-4 i(t-1)} \,=\, \cos({-4 (t-1)}) +i \sin({-4 (t-1)}) \,=\, \cos({4 (t-1)}) -i \sin({4 (t-1)}). $$

From there, you might use well-known formulas for $\cos(2\theta)$ and $\sin(2\theta)$.


问:
Does Differential Topology or Differential Geometry play a larger role in Chaos Theory?
I'm an undergraduate on somewhat of a time constraint in school.  I have room in my remaining schedule for a semester of either Differential Geometry or Differential Topology.

I understand the topological side deals with primarily global aspects of a manifold, whereas the geometrical side associates with interesting local structures.

To be concise-  I'm interested in chaotic dynamical systems and P.D.E., so which subject would be more worthwhile, i.e. which subject's vocabulary/method would equip for further reading in the topics I am interested in?  Or am i naive in forcing myself to choose between the two topics, as similar as they are?

Thank you!

答: I took both. I feel differential geometry is more tangible and computational, while differential topology is more conceptual.</p>


问:
Confusion using Ito product rule in stochastic calculus
To calculate the Ito differential of the quantity $\frac{A}{B}$ I can use the Ito product rule which gives

$$
d (\frac{A}{B})=\frac{dA}{B}-\frac{A}{B^{2}}dB-\frac{1}{B^{2}}dA dB
$$

if I now let $B=A$, I should obtain the result

$$
d(\frac{A}{A})=d(1)=0.
$$

However, substituting into the above the second-order term may be nonzero

\begin{align}
d(\frac{A}{A})&=\frac{dA}{A}-\frac{A}{A^{2}}dA-\frac{1}{A^{2}}dA dA\\
&=-\frac{1}{A^2}dA^2.
\end{align}

Is anyone able to help me understand what I have done wrong here?

答:
As pointed out in the comments I incorrectly calculated $d(B^{-1})$ by not taking the chain rule to second order. It should be

$$
d(\frac{A}{B})=\frac{dA}{B}+A(-\frac{1}{B^2}dB+\frac{1}{B^3}dB^2)-\frac{dA dB}{B^2}.
$$

Which is zero upon substituting $A=B$, as it should be. I hope anyone seeing this ends up wasting less time on this than I did.


问:
Theorem for some structure that it cannot be axiomatized in any language?
Godel’s theorem states that there cannot be an enumerable axiomatization of arithmetic $(\mathbb N,0,1,+)$ in first order logic. By implication, there cannot be any axiomatization in any language that is complete.

But this doesn’t mean that arithmetic cannot be axiomatized at all. E.g. we have second order Peano axioms that characterizes arithmetic up to isomorphism. It just means that we cannot derive all propositions from this axiomatization (which is possible because second-order logic is not complete).

My question is: is there a different theorem, similar in spirit to Godel’s, for some mathematical structure that can be constructed explicitly, that there cannot be an axiomatization that characterizes it up to isomorphism, in any language? (I.e. not first order, or second order, or higher order, or anything else that I don’t know about).

Intuitively, I would guess there might be, because: There is only a countable amount of possible enumerable axiomatizations in all languages combined, since there is only a countable amount of Turing machines. On the other hand there is an uncountable amount of structures. So there must be some structure that cannot be axiomatized. My question is whether there is an explicit constructable structure where we know that it cannot be.

答:
A crucial point in your question is the following broad claim:

There is only a countable amount of possible enumerable axiomatizations in all languages combined, since there is only a countable amount of Turing machines.

There's an implicit restriction here to logics which are "computably describable." Of course this is necessary to hope for a positive answer, but it's not immediately clear what this should mean - especially since you explicitly want to include things like second-order logic whose semantics are absurdly complicated.

One possible idea is the following: roughly, say that a definable logic consists of a computable set of things we call sentences and a semantics given by a formula in the language of set theory. This encapsulates all the higher-order logics, computable infinitary logic, and so forth. I don't know of a single candidate for a natural logic which has countably many sentences but is not a definable logic in the above sense, so this seems reasonable.

It turns out, however, that it is consistent that every structure can be pinned down up to isomorphism by some sentence in some definable logic. This is a consequence of the existence of pointwise-definable models; the obstacle to the "obvious" proof of contradiction here is that definable logics can be arbitrarily complicated to define in terms of the Levy hierarchy so we can't definably diagonalize against them (per Tarski).

In fact, even handling second-order logic alone is quite difficult. While a quick counting argument shows that there must be a countable structure which is not pinned down up to isomorphism by any computable set of second-order sentences, there is no known candidate, and in fact it is consistent with $\mathsf{ZFC}$ that every countable structure is pinned down up to isomorphism by its full second-order theory (see the discussion here). And any attempt to develop a "Godel-flavored" analysis of second-order logic immediately runs into the problem that the entailment relation for second-order logic is not second-order definable in arithmetic (in contrast with the situation for first-order logic, which follows from the completeness theorem).

That said, there is a version of Godel's first incompleteness theorem which applies to any (set-sized) logic.

First, let me prove my statement above about the complexity of the second-order entailment relation. The syntax for $\mathsf{SOL}$ is computable, and there is a second-order sentence $\theta$ which pins down $\mathbb{N}$ up to isomorphism. If the second-order entailment relation were second-order definable over $\mathbb{N}$, the full second-order theory of $\mathbb{N}$ would be second-order definable in $\mathbb{N}$. This would give us a second-order-definable-over-$\mathbb{N}$ listing of the second-order-definable-over-$\mathbb{N}$ sets; diagonalizing against that listing would give us a contradiction. (This last sentence is really just Tarski's argument, but I think this phrasing makes it easier to think about.)

This generalizes in a precise, if messy, way to any context where we have a logic $\mathcal{L}$ and a structure $\mathbb{A}$ sufficient for developing the syntax of $\mathcal{L}$ (this generalizes $\mathbb{N}$). Then either $\mathcal{L}$-entailment is too complicated to define in $\mathbb{A}$, or no good approximation to the $\mathcal{L}$-theory of $\mathbb{A}$ (this generalizes "no consistent completion of $\mathsf{Q}$") is "nicely definable" in $\mathbb{A}$.

Incidentally, this idea leads to a proof of the usual first incompleteness theorem which is "purely model-theoretic" in a precise sense, see here (although I don't know its exact provenance). And while the added generality seems useless for the most part, I did use the more general version above to prove a small result about infinitary logics in a relatively short way.


问:
Equation manipulation tips/explanation for $2v_1w_1v_2w_2=v_1^{\ \ 2}w_2^{\ \ 2}+v_2^{\ \ 2}w_1^{\ \ 2}$ and its solution of $\mathbf{w}=a\mathbf{v}$
I was hoping I could get some equation manipulation tips for finding a particular solution to the following equality:

$2v_1w_1v_2w_2=v_1^{\ \ 2}w_2^{\ \ 2}+v_2^{\ \ 2}w_1^{\ \ 2}$

where $\mathbf{v}=\langle v_1,v_2\rangle$ and $\mathbf{w}=\langle w_1, w_2 \rangle$.

I know that the answers are:

$\mathbf{v}=0$ or $\mathbf{w}=0$ or $\mathbf{w}=a\mathbf{v}$ for some scalar $a$

I can see that all of the proposed solutions work just fine, but if I had quickly scanned this equation, I would have only concluded that equality holds when $\mathbf{v}=0$ or $\mathbf{w}=0$.

How do I reformulate the equation so that the final proposed solution ($\mathbf{w}=a\mathbf{v}$ for some scalar $a$) becomes apparent?

Here is what I have so far:

$$2v_1w_1v_2w_2=v_1w_1v_2w_2+v_1w_1v_2w_2$$

and

$$v_1^{\ \ 2}w_2^{\ \ 2}+v_2^{\ \ 2}w_1^{\ \ 2}=v_1v_1w_2w_2+w_1w_1v_2v_2$$

Equating these:

$$v_1w_1v_2w_2+v_1w_1v_2w_2=v_1v_1w_2w_2+w_1w_1v_2v_2$$

The only trick I can think of to reveal the $\mathbf{w}=a\mathbf{v}$ structure is the following:

$$\color{red}{v_1w_1v_2w_2}+\color{blue}{v_1w_1v_2w_2}=\color{red}{v_1v_1w_2w_2}+\color{blue}{w_1w_1v_2v_2}$$

$$\color{red}{v_1w_1v_2w_2}-\color{red}{v_1v_1w_2w_2}= \color{blue}{w_1w_1v_2v_2}-\color{blue}{v_1w_1v_2w_2}$$

From here, assume that the red colored (or blue colored) expressions sums to $0$. For example:

$$\color{red}{v_1w_1v_2w_2}-\color{red}{v_1v_1w_2w_2}=0$$

$$v_1w_1v_2w_2=v_1v_1w_2w_2$$

$$\text{Canceling terms and rearranging}$$

$$\frac{v_1}{v_2}=\frac{w_1}{w_2}$$

Working on the blue terms (still with the assumption that they sum to $0$) results in the same expression...so we effectively have $0=0$ and the requirements to arrive at that solution is the same for each side (i.e. $\frac{v_1}{v_2}=\frac{w_1}{w_2}$)

I interpret the expression of $\frac{v_1}{v_2}=\frac{w_1}{w_2}$ as necessarily implying that $\mathbf{w}=a\mathbf{v}$ because of how fractions works...but I am not really certain what the constraint that the sum equals zero is really saying. If someone could lend some insight on this final point, it would be appreciated!

答:
$$2v_1w_1v_2w_2=v_1^2w_2^2+v_2^2w_1^2$$

$$(v_1w_2-v_2w_1)^2=0$$

Hence we have $$v_1w_2=v_2w_1$$ which is what you obtained.

The determinant of the matrix  $\begin{bmatrix} v_1 & v_2 \\ w_1 & w_2\end{bmatrix}$ is $0$, hence the conclusion.

Note: Be careful when you divide by a number, we should check that it is non-zero.


问:
Why is the difference of two vectors parallel to the vector connecting the two vectors?
Say we have any two vectors $u$ and $v$. We want to find a vector $w$ that is parallel to the vector that goes from $u$ to $v$ (is there a standard notation for this vector?). Then I already know that such a vector can be found simply by subtracting one of the two vectors from the other. Can you give me some intuition why this works? Probably what I am not understanding is how you get the vector connecting $u$ and $v$ in the first place.

答:
While there's already an accepted answer, I'd like to add some visual intuition to the ideas. In particular, I want to address the OP's sentence:

the vector that goes from $u$ to $v$

A very important aspect of vector spaces is that all vectors "start" from the origin. There is no such thing as a vector starting from the tip of another vector. What I have pictured in orange are not vectors, there are only helpful to get an intuition for vector addition. The "real" $u$, $v$ and $w$ vectors are in black, and sure, it's easy to understand $u+w=v$ if you follow the path (black $u$, orange $w$), but that doesn't mean that the orange $w$ is a member of the vector space. Same thing with $w=v-u$, which is $w=v+(-u)$.


问:
Solving a Linear Congruence
I've been trying to solve the following linear congruence with not much success:

$19\equiv 21x\pmod{26}$

If anyone could point me to the solution I'd be grateful, thanks in advance

Title: Solving a Linear Congruence

Body: $21^{-1}\equiv5\mod26\implies x\equiv95\equiv17\mod26$


问:
Let $E$ be normed space. Show that the sets $E \setminus \{0\}$ and $A=\{x \in E \mid 1 < |x|< 2 \}$ are homeomorphic.

Let $E$ be normed space. Show that the sets $E \setminus \{0\}$ and $A=\{x \in E \mid 1 < |x|< 2 \}$ are homeomorphic.

I’m trying to learn about homeomorphisms and not entirely sure how to approach these problems. I need to find a map $\varphi:E \setminus \{0\} \to A$ thats continuous bijection with continuous inverse. If I would consider $E = \mathbb{R}^2$, then A would be the strips between $(-2,-1)$ and $(1,2)$. How does one approach these problems? Should I try to visualize this first or how should I start thinking about candidates for $\varphi$?

答:
Perhaps you should start with the simplest $E=\mathbb{R}$ case. Can you find a homeomorphism $f:\mathbb{R}\backslash\{0\}\to (-2,-1)\cup(1,2)$? Such homeomorphism can be constructed from a homeomorphism $g:(0,\infty)\to(1,2)$ by taking $-g$ on the negative part. Finally $g$ can be constructed from some trigonometric functions, which I leave as an exercise.

For general $E$ this becomes quite easy now. Because we can simply rescale each vector via our $f:\mathbb{R}\backslash\{0\}\to (-2,-1)\cup(1,2)$ homeomorphism. I.e.

$$F:E\backslash\{0\}\to A$$
$$F(v)=f(\lVert v\rVert)\cdot \frac{v}{\lVert v\rVert}$$

which simply says "take vector $v$ and change its norm from $\lVert v\rVert$ to $f(\lVert v\rVert)$". This "change norm" is done by first normalizing $v$ via $\frac{v}{\lVert v\rVert}$ and then scaling via multiplication by $f(\lVert v\rVert)$. I leave as an exercise that it is continuous.

You can also check (via tedious calculation) that the explicit inverse of $F$ is

$$G:A\to E\backslash\{0\}$$
$$G(v)= f^{-1}(\lVert v\rVert)\cdot\frac{v}{\lVert v\rVert}$$

which is automatically continuous by previous argument.


问:
Why does the expansion of the Riemann zeta function for $\zeta(3)$ to $\sum_{n=1}^{\infty}\frac{n^3}{n^6}$ not allow for substitution of the latter?
For the Riemann-zeta function $\zeta(s)$ can we represent $\zeta(3)$ as the following?
$$\zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}=\sum_{n=1}^{\infty}\frac{1}{\biggl(\frac{n^6}{n^3}\biggr)}=\sum_{n=1}^{\infty}\frac{n^3}{n^6}$$

Because $$\zeta(6)=\sum_{n=1}^{\infty}\frac{1}{n^6}=\frac{\pi^6}{945}$$

is it appropriate to write this as

$$\zeta(3)=\sum_{n=1}^{\infty}\frac{n^3}{\biggl(\frac{\pi^6}{945}\biggr)}=\sum_{n=1}^{\infty}\frac{945n^3}{\pi^6}$$

This answer just diverges, so what am I missing? If I take out the constant to yield
$$\zeta(3)=\frac{945}{\pi^6}\sum_{n=1}^{\infty}n^3$$ the answer still diverges.

答:
Unfortunately, the sum operation is not multiplicatively distributive.
$$\sum
ab\neq \sum a \cdot \sum b$$
What you can do is

$$\sum_{n\geqslant 1} \dfrac{1}{n^3}=\sum_{n=m\geqslant 1}\dfrac{m^3}{n^6}=\left[\sum_{n,m\geqslant 1}-\sum_{n\neq m\geqslant 1}\right]\dfrac{m^3}{n^6}=\left[\sum_{n,m\geqslant 1}-\sum_{n> m\geqslant 1}-\sum_{m> n\geq 1}\right]\dfrac{m^3}{n^6}$$ of which, the first part is

$$\zeta(6)\sum_{m\geqslant 1} m^3.$$