问:
Proof that all extensions of $\mathbb{Q}_p$ are of the form $\mathbb{Q}_p[\sqrt[n]{a}]$
I'm going over some notes that claim that any extension of $\mathbb{Q}_p$ of degree $n$ has the form $\mathbb{Q}_p[\sqrt[n]{a}]$ for some $a\in\mathbb{Z}_p$. It references Proposition III.12 from Serre's "Local Fields", which states, more or less, that if $F$ is an extension of $\mathbb{Q}_p$ of degree $n$, then there is a $\mathbb{Q}_p$-basis of $F$ given by $1,x,x^2,\dots,x^{n-1}$ for some $x$.

What I don't understand is how to show that this can be used to produce an $n$th root. All we know is that there is a monic polynomial of degree $n$ such that $f(x)=0$. Does that really give us some $n$th root of an element of $\mathbb{Z}_p$?

答:
If you track the reference to Serre's book, you'll see the actual result cited from there is that for every finite extension $K/\mathbf Q_p$ the ring of integers of $K$ has a power basis: $\mathcal O_K = \mathbf Z_p[\alpha]$ for some $\alpha$.  That does not mean $K = \mathbf Q_p(\sqrt[n]{a})$ and it is stronger than your "more or less" comment that $K$ has a power basis over $\mathbf Q_p$!  There was some kind of confusion between a $\mathbf Z_p$-basis of the form $\{1,\alpha,\ldots,\alpha^n\}$ and a $\mathbf Q_p$-basis of $K$ consisting of powers of the specific kind of number $\sqrt[n]{a}$.

That the ring of integers of every $p$-adic field has a power basis over $\mathbf Z_p$ is a bit surprising from the viewpoint of number fields, where the ring of integers always has a $\mathbf Z$-basis but need not have a power basis over $\mathbf Z$, a standard example being the cubic field $\mathbf Q(\gamma)$ where $\gamma$ is a root of $x^3-x^2-2x-8$.


问:
How can I prove that $\frac{a}{2} + \frac{x}{a} \geq \sqrt{2x}$
How can I prove the inequality

$$
\frac{a}{2} + \frac{x}{a} \geq \sqrt{2x}
$$
for positive $a$ and $x$.

It seems simple, but I don't see a way to prove it.

ContentLicense: CC BY-SA 4.0

Body: $$\left(\sqrt{\frac{a}{2}}-\sqrt{\frac{x}{a}}\right)^2\geq0.$$


问:
Integral of rational functions.
I want to evaluate this integral:

$$\int{\frac{ax+b}{(x^2+2px+q)^n}}dx$$

The book only says to integrate by parts $\int{\dfrac{1}{(x^2+2px+q)^{n-1}}dx}$,
for simplicity if $n = 2$ I get:
$\int{\dfrac{1}{(x^2+2px+q)}dx}=\dfrac{x}{x^2+2px+q}+\int{\dfrac{2x^2+2px}{(x^2+2px+q)^2}}dx$.

Now I don't know what to do.

答:
1.for the integrand $\dfrac{1}{x^2+2px+q}$ complete the square to get

$\frac{1}{\dfrac{1}{4}(4q-4p)^2+(p+x)^2}$

2.then substitute $u=p+x$

3.factor out $\dfrac{1}{4}(4q-4p)^2$

The integral reduced to simple trigonometric function

ContentLicense: CC BY-SA 3.0

Title: Integral of rational functions.


问:
Approximating a function by Taylor polynomial
The introduction to Taylor polynomial and remainder is given in various calculus texts with different flavors. Still, I wanted to know one more thing by comparing two things about it.

For this, consider non-constant, non-polynomial function, which is infinitely many times differentiable in an open interval $I$.

Fix $a\in I$.

Let $P(x)=f(a) + \frac{f'(a)}{1!}(x-a) + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k$ be the $k$-th degree Taylor polynomial of $f$ around $a$.

(1) If we fix degree of polynomial and approach to $a$ (i.e. if we take $x$ close to $a$) then the error between $f(x)$ and $P(x)$ goes to $0$ much faster compared to $(x-a)^{k+1}$.

This will give a polynomial approximation for $f(x)$ of degree $k$ in suitably small interval around $a$.

(2) Fix domain of interval $(a-\delta,a+\delta)$, and vary the degree of the polynomial $P(x)$; we get better and better approximation to to $f(x)$ in the domain $(a-\delta,a+\delta)$.

Question: Practically, it may depend on the function $f$, whether to come close to $a$, or take large degree of Taylor polynomial for approximation of function $f(x)$ by a polynomial. Can one illustrate me by some example(s) where one situation is better than the other?

答:
In most practical situations, the two points go hand in hand. For an interval of choice you typically determine the degree to fix and work with that. On the other hand, if estimating higher order derivatives is not possible or practical then you fix an interval within which the contribution from higher order derivatives can be ignored because the  possible error is within a specified budget.

A practical example of this from Options pricing. If you don't know what an option is think of it is a financial instrument whose price ($f$) is given as a function of a underlying asset's price ($s$), that asset's volatility ($\sigma$), time to maturity of the option ($t$), some type of risk free interest rate ($r$), and many other variables depending upon the type of option (including some properties of the option which may remain fixed). This function may not have a closed form expression in many cases.

If you want to make a quick estimation of an option's price in response to a small change in the underlying stock's price ($s$) you can use the delta-gamma approximation. This is nothing but the Taylor's approximation up to second order with:

$$
\delta = \frac{\partial{f}}{\partial{s}} \quad , \quad \gamma = \frac{\partial^2f}{\partial s^2}
$$

These approximations are only valid in a small interval because the derivatives themselves change with changes in $s$.

If you are short on computing resources, you may want to use the delta-gamma approximation while the asset's price has not deviated materially from the point at which you last calculated your deltas, gammas, and the original prices. If you are outside that range you can then calculate the derivatives and the prices using a full (expensive) revaluation. The trade off between accuracy and computing power will be driven by your tolerance for discrepancies.


答:
Since everyone here is positive, compare the squares:
$$\Bigl(\frac{a}{2} + \frac{x}{a}\Bigr)^2=\frac{a^2}{4} + \frac{x^2}{a^2}+ x \geq  2x \iff \frac{a^2}{4} + \frac{x^2}{a^2}- x=\Bigl(\frac{a}{2}-\frac{x}{a}\Bigr)^2\ge 0. $$

Title: How can I prove that $\frac{a}{2} + \frac{x}{a} \geq \sqrt{2x}$

Body:
How can I prove the inequality

$$
\frac{a}{2} + \frac{x}{a} \geq \sqrt{2x}
$$
for positive $a$ and $x$.

It seems simple, but I don't see a way to prove it.


问:
How do we apply chain rule to a function of two variables, which are also functions.
Let $f$ and $g$ be differentiable functions.

$$
\frac{\partial}{\partial z}(g(f(x+y),f(z))
$$
How do we apply the chain rule here correctly? Do we just treat $f(x+y)$ and $f(z)$ as some variables $u$ and $v$?

答: Yes.  Think of $g = g(u,v)$.  Then, you have $$\frac{\partial}{\partial z}(g(f(x+y),f(z)) = \frac{\partial g}{\partial v}(f(x+y),f(z))f'(z).$$


问:
A sum involving the Mobius function and a product over prime factors
This is a follow-up question to: What is known about partial sums involving squares of the totient function?

Using the hint given by Daniel Fischer, I was able to express the sum posed in the question in the following way:

$$\displaystyle\sum_{r \leq x} \Big( \frac{\varphi(r)}{r} \Big)^2 = x \displaystyle\sum_{r \leq x} \frac{\mu(r)}{r^2} \displaystyle\prod_{p | r} \Big(2 - \frac{1}{p} \Big) = x \displaystyle\sum_{r \leq x} \frac{\mu(r)}{r^2} \displaystyle\sum_{d | r} \frac{\mu(d)}{d} 2^{\omega(r/d)}$$

I am interested in the asymptotic behaviour of this sum, and unfortunately, I am stuck once again. Numerical experimentation suggests that:

$$\displaystyle\sum_{r \leq x} \Big( \frac{\varphi(r)}{r} \Big)^2 = \frac{6}{\sqrt{2} \pi^2} x + O(\text{something})$$

Interestingly enough, we know that:

$$\displaystyle\sum_{n = 1}^{\infty} \frac{\mu(n)}{n^2} = \frac{6}{\pi^2}$$

so I'm inclined to believe that the first sum separates nicely, when we split it into a sum from $1$ to $\infty$, minus the sum for all $r > x$.

I am only familiar with very basic analytic number theory, I don't know of any tricks to use to try to compute the product of primes in the first equation, or do something else that would allow for a clean solution. Any hints would be much appreciated!

Body:
$$\sum_{n=1}^\infty (\varphi(n)/n)^2 n^{-s}= F(s)\zeta(s)$$ where $$F(s)=\sum_{m\ge 1} a_m m^{-s}= \prod_p (1-p^{-s})(1+\sum_{k\ge 1} \frac{(p-1)^2}{p^2}p^{-sk})$$ $$= \prod_p (1+O(p^{-1})p^{-s}+\sum_{k\ge 2} O(1)p^{-2s})$$ converges absolutely for $\Re(s)>1/2$. Thus $$\sum_{n\le x} (\varphi(n)/n)^2 = \sum_{m\le 1} a_m \lfloor x/m\rfloor = x \sum_{m\le 1} a_m/m+ O(\sum_{m\le x} |a_m|)=xF(1)+O(x^{1/2+\epsilon})$$
I don't think
$$F(1)=\prod_p (1-p^{-1})(1+\frac{p-1}{p^2})$$ has a closed-form.


问:
Can there be a fragile family on $\mathcal{P}(\omega)$?
By a fragile family on $\mathcal{P}(\omega)$ (this is a made-up term), I mean a countable family $F\subseteq\mathcal{P}(\omega)$ of pairwise distinct infinite sets such that for any $n\in\omega$, the family $F':=\{X_i\smallsetminus\{n\}\mid X_i\in F\}$ is no longer pairwise distinct. That is, for some elements $X_i\neq X_j$ in $F$, $X_i\smallsetminus\{n\}=X_j\smallsetminus\{n\}$. The family is fragile, because once you take away anything from a member, two members will clash. Can there be a fragile family on $\mathcal{P}(\omega)$?

I wonder if this is a special instance of some more general theorem, so I am also curious about various generalizations, for example for families on $\mathcal{P}(\kappa)$ for uncountable $\kappa$, or varying the size of $F$, or instead of removing a singleton, we remove an infinite subset from each element in the family.

答:
Yes, there are lots of these. In my opinion the simplest is the set of all finite subsets of $\omega$: given $n$, consider $\{n, n+1\}$ versus $\{n+1\}$.

Demanding that the sets in question be infinite doesn't change the picture: consider e.g. the set of $X\subseteq\omega$ such that $X$ differs only finitely much from the set of even numbers. For each $n$, let $E=\{2k: k\in\omega\}$ and $E_n=E\Delta \{n\}$. Then $E\not=E_n$ but $E\setminus\{n\}=E_n\setminus\{n\}$.

I do not at the moment see a way to modify the notion of fragility to avoid this sort of example.


问:
Proof of the fact that the closure of a set always contains its supremum and an open set cannot contain its supremum
I would like to ask, if my proof checks out and is completely sound.

Exercise 3.2.4 from Stephen Abbot's Understanding Analysis.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Let $A$ be non-empty subset of $\mathbf{R}$ and bounded above, so that $s = \sup A$ exists. Let $\bar{A} = A \cup L$ be the closure of $A$.

(a) Show that $s \in \bar{A}$.

(b) Can an open set contain its supremum?

My Attempt.

(a) $\bar{A} = A \cup L$ is the closure of $A$ and contains the limits points of $A$. We proceed by contradiction. Assume that $s \notin \bar{A}$ and is not a limit point of $A$.

Since, $s$ is the supremum for $A$, looking at the definition of least upper bound, it must satisfy two properties: (i) $s$ is an upper bound for $A$. (ii) Given any small arbitrary, but fixed positive real $\epsilon > 0$, $(s - \epsilon)$ should not be an upper bound for $a$.

From (ii), it follows that, given any $\epsilon > 0$, there exists $t \in A$, such that $s - \epsilon < t$. Thus,
\begin{align*}
 \absval{t - s} < \epsilon
\end{align*}

Thus, every $\epsilon$-neighbourhood of $s$, $V_\epsilon(s)$ intersects $A$ in points other than $s$. So, $s$ is the limit point of $A$. Therefore, $s \in \bar{A}$, which contradicts our initial assumption. Hence, our initial assumption is false.

(b) An open set cannot contain its supremum. We proceed by contradiction. Let $O$ be an open set. Assume that $s \in O$.

Since $O$ is an open set, for all points $x$ belonging to $O$, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$ that is contained in $O$. In particular, $V_\epsilon(s) \subseteq O$. So, if
\begin{align*}
 s - \epsilon < t < s + \epsilon
\end{align*}

then $t \in O$, for some $\epsilon$. But, that implies, for some $\epsilon > 0$, we must have
\begin{align*}
 s < t < s + \epsilon
\end{align*}

$t \in O$. So, $s$ is not an upper bound for $O$. This is a contradiction. Our initial assumption must be false. $s \notin O$.

答:
The argument for (a) isn’t quite correct, because $s$ need not actually be a limit point of $A$. For instance, let $A=(0,1)\cup\{2\}$; then $s=2$, and for any positive $\epsilon\le 1$ the open interval $(s-\epsilon,s)$ is disjoint from $A$. And as a minor point, you don’t need to argue by contradiction.

If $s\in A$, then certainly $s\in\operatorname{cl}A$, so suppose that $s\notin A$. Let $\epsilon>0$; then $s-\epsilon< s$, so $s-\epsilon$ is not an upper bound for $A$, and therefore $A\cap(s-\epsilon,s]\ne\varnothing$. Moreover, $s\notin A$, so $A\cap(s-\epsilon,s)\ne\varnothing$. Thus, for each $\epsilon>0$ there is an $a\in A$ such that $|a-s|<\epsilon$, so $s$ is a limit point of $A$, and therefore $s\in\operatorname{cl}A$.

(Note that while there is absolutely nothing wrong with including extra detail, and it can be a good idea when you’re still learning, it’s really not necessary to say more by way of justifying the various steps than I did above.)

The argument for (b) is fine.


答: Take any infinite $A\subseteq\omega$ and let $F=\{A\}\cup\{A\mathbin{\triangle}\{n\}:n\in\omega\}$.  More generally, for each $n$, you could pick a pair of sets to be in $F$ which differ only on $n$, and the resulting $F$ will be countable.

Body:
By a fragile family on $\mathcal{P}(\omega)$ (this is a made-up term), I mean a countable family $F\subseteq\mathcal{P}(\omega)$ of pairwise distinct infinite sets such that for any $n\in\omega$, the family $F':=\{X_i\smallsetminus\{n\}\mid X_i\in F\}$ is no longer pairwise distinct. That is, for some elements $X_i\neq X_j$ in $F$, $X_i\smallsetminus\{n\}=X_j\smallsetminus\{n\}$. The family is fragile, because once you take away anything from a member, two members will clash. Can there be a fragile family on $\mathcal{P}(\omega)$?

I wonder if this is a special instance of some more general theorem, so I am also curious about various generalizations, for example for families on $\mathcal{P}(\kappa)$ for uncountable $\kappa$, or varying the size of $F$, or instead of removing a singleton, we remove an infinite subset from each element in the family.


问:
Summation of a finite series
Let

$$f(n) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1} \forall \ n \ \epsilon \ \mathbb{N} $$

If it cannot be summed , are there any approximations to the series ?

答: $$f(n)\approx \ln(2^n-1)+\gamma\approx n\ln 2+\gamma$$</p>

Title: Summation of a finite series


问:
$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty $
Let $a \in \mathbb R $ and $f,g:]a$,+∞[$\to \mathbb R$ two differentiable functions such that $g'$ is never equal to $0$. I need to prove that $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty $

I tried to use mean value theorem but it does not lead me anywhere. I am also not very familiar with proving things with $\neq$...

I am not allowed to use L'Hopital.

答:
Suppose $f'(x)/g'(x) \to -\infty.$ Then there exists $b>a$ such that $f'(x)/g'(x) <0$ for $x\in (a,b).$ But for $x$ close to $a,$

$$\frac{f(x)-f(b)}{g(x)-g(b)}=\frac{f'(c_x)}{g'(c_x)}$$ by Cauchy's MVT. Note the left side is positive for such $x$ (because $f,g\to +\infty$) while the right side is negative. This contradiction shows $f'(x)/g'(x) \to -\infty$ cannot happen.


问:
Monotonicity of Riemannian Logarithm
Let $(X,g)$ be a complete, connected, Riemannian manifold of non-positive curvature.  Let $(z_n)_{n=1}^{\infty}$ be a sequence in $X$ converging to some $z \in X$ such that
$$
d_g(z,z_{n+1})\leq d_g(z,z_n),
$$
for every $n\geq 1$; where $d_g$ is the intrinsic metric induced by the riemannian metric $g$.  By the Cartan-Hadamard theorem, we know that in this case each $\exp^{-1}_x$ is globally well-defined on $X$.

Is it then true, that the sequence of functions
$$
\begin{aligned}
F_n:X^2 &\rightarrow [0,\infty)\\
(x,y) & \mapsto \|\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)\|,
\end{aligned}
$$
converges to $(x,y)\mapsto \|\exp_{z}^{-1}(y)-\exp_{z}^{-1}(x)\|$ uniformly and monotonically (ie $F_{n+1}(x,y)\leq F_{n}(x,y)$ for each $x,y \in X$?)

答:
This is not true. Assume $X$ is the hyperbolic plane (with constant curvature $-1$). Fix any $n$, let $w=d(z_n, z)$. Extend the geodesic from $z_n$ to $z$ by distance $b$ and reach a point $x$; and at $z$ draw a geodesic of length $b$ that is perpendicular to the first geodesic, reach a point $y$. Here $b$ is a large number depending on $w$. Write  $a=d(z_n, y)$. So
$$
|\exp_z^{-1}(y)-\exp_z^{-1}(x)|=\sqrt 2 b.
$$
Now by the hyperbolic law of cosine,
$$
\cosh a=\cosh b \cdot \cosh w.
$$
Now $a, b$ are both big, $\cosh b=\frac{e^{b}}{2}(1+e^{-2b})$, so $\log \cosh b=b-\log 2+O(e^{-2b})$. $w$ is small, so $\log\cosh w=\log(1+\frac{w^2}{2}+...)
=\frac{w^2}{2}+O(w^4)$. So
$$
a-\log 2+O(e^{-2b})=b-\log 2+O(e^{-2b})+\frac {w^2}{2}+O(w^4),
$$
i.e. $a=b+\frac{w^2}{2}+O(e^{-2b})+O(w^4)$. We can take $b$ so big that we have $e^{-2b}\ll w^4$, so
$$
a=b+\frac{w^2}{2}+O(w^4).
$$
Apply hyperbolic law of cosine to the same thin triangle $z_nzy$ we get
$$
\cos\theta=\frac{\cosh a\cosh w-\cosh b}{\sinh a\sinh w}
=\frac{\cosh b\cosh w\cosh w-\cosh b}{\sinh a\sinh w}
=\frac{\cosh b\sinh^2 w}{\sinh a\sinh w}=\frac{\cosh b\sinh w}{\sinh a}.
$$
From $a=b+\frac{w^2}{2}+O(w^4)$ and $b$ is sufficiently big, this implies
$$
\cos\theta=w+O(w^2).
$$
This is the key estimate, saying that $\theta$ differs from the right angle $\pi/2$ by $O(w)$, much bigger than the Euclidean case $O(w/a)$.

So, finally,
$$
\begin{aligned}
|\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)|=&\sqrt{(b+w)^2+a^2-2(b+w)a\cos\theta}\\
=&\sqrt{b^2+w^2+2bw+b^2+b\cdot O(w^2)-2(b+w)(b+O(w^2))(w+O(w^2))}\\
=& \sqrt{2b^2+w^2+2bw+b\cdot O(w^2)-2b^2w+b^2O(w^2)}\\
=& \sqrt{2b^2+2bw-2b^2w+b^2O(w^2)}\\
=& \sqrt 2 b\sqrt{1+\frac wb-w+O(w^2)}\\
=& \sqrt 2 b\Big(1-\frac w2 +\frac {w}{2b} +O(w^2)\Big);
\end{aligned}
$$
this differs from the limit $\sqrt 2 b$ by at least $\frac{\sqrt 2 b w}{3}>1$, if $b$
is sufficiently big.

So the converge is not uniform. And there is no reason to think monotone (try travel backwards in a direction from $z$ to $z_n$...)

Title: Monotonicity of Riemannian Logarithm


问:
A question on limits using the epsilon-delta definition
I am trying to solve this problem but I keep getting stuck. This question does not seem to be particularly hard.

The question is stated below:

$
|f(x)| < \frac{1}{100}
$ if $0 < x < \delta$

Find $\delta$ if $f(x) = x\sin(x)$

Please do not tell me the answer directly but help me understand it and point me in the right direction. That way I will learn more.

Thank you for helping and stay safe!

答: Hint: Think of what happens as you increase $x$. Up until what value of $x$ can you safely say that $|x \sin(x)|$ will always be less than 0.01?


答:
You may notice that $|\sin x|\le1$ and so $|f(x)|=|x\sin x|=|x||\sin x|\le|x|$.

Since $x>0,x=|x|$ and we have $|x|<\delta$. Thus $|f(x)|<\delta$ whenever $0<x<\delta$.

Can we select an appropriate positive $\delta$ to make $|f(x)|<1/100$ true?

Title: A question on limits using the epsilon-delta definition

Body:
I am trying to solve this problem but I keep getting stuck. This question does not seem to be particularly hard.

The question is stated below:

$
|f(x)| < \frac{1}{100}
$ if $0 < x < \delta$

Find $\delta$ if $f(x) = x\sin(x)$

Please do not tell me the answer directly but help me understand it and point me in the right direction. That way I will learn more.

Thank you for helping and stay safe!


问:
Proving two Topologies to be equal
Let $(X,\mathcal T)$ and $(Y,\mathcal V)$ be topological spaces, let $f$ be a function that maps $X$ onto $Y$, and let $\mathcal U$ be the quotient topology on $Y$ induced by $f$. Proved that if $f$ is continuous and  closed then $\mathcal U= \mathcal V$.

All I can deduce with the information given in the question is that $f$ is quotient map.

Can you please tell how should I proceed with this problem. I am not good in problems related to quotient topology and I am still learning.Just telling the main elements of proof will be sufficient.I will fill details myself.

Thanks!

答:
Let $C$ be closed for $\mathcal{U}$, the quotient topology. This implies that $f^{-1}[C]$ is closed for $\mathcal{T}$, because of the definition of the quotient topology. As we assume that $f: (X,\mathcal{U}) \to (Y, \mathcal V)$ is closed (and continuous) we see that $C= f[f^{-1}[C]$ (equality holds as $f$ is onto) is closed for $\mathcal{V}$ as the image under $f$ of $f^{-1}[C]$. So $\mathcal{U} \subseteq \mathcal{V}$.

Let $C$ be closed for $\mathcal{V}$. As we assume that $f: (X,\mathcal{U}) \to (Y, \mathcal V)$ is continuous, we know that $f^{-1}[C]$ is closed for $\mathcal T$, and so as again by definition of the qutient topology $\mathcal U$ we conclude that $C$ is closed for $\mathcal U$. So $\mathcal{V} \subseteq \mathcal{U}$, and we have equality of topologies.

Title: Proving two Topologies to be equal


问:
Lower bound for $R(5, 5)$ cyclic coloring
I want a $K_5$-free $2$-coloring of $K_{42}$ to prove $R(5, 5) > 42$. The only construction I could find is from Exoo, G. (1989). A lower bound for R (5, 5). Journal of graph theory, 13(1), 97-98:

A cyclic coloring is one in which the vertices of $K_n$ are identified with the integers from $0$ to $n - 1$, and in which the color of an edge depends only on the numerical differences between its end-vertices. These differences are often called chord lengths. A cyclic coloring can be specified by giving the chord lengths (from $\mathbf{1}$ to $\mathbf{[n/2]}$) for each color.

Q1: Don't we need the chord lengths from $1$ to $n - 1$? How else will we know what color to assign to the edge $0 \leftrightarrow n - 1$, which has chord length $n - 1$?

The coloring of $K_{42}$ starts with a coloring of $K_{43}$. Assign the chord lengths $1, 2, 7, 10, 12, 13, 14, 16, 18, 20, 21$ to red and $3, 4, 5, 6, 8, 9, 11, 15, 17, 19$ to blue. The resulting $K_{43}$ has $0$ blue $K_5$s and $43$ red $K_5$s, which have vertices of the form
$$\{i, i + 1, i + 2, i + 22, i + 23 \} \quad \text{for } i = 0, 1, \dots, 42$$

Q2: Does this use congruence classes mod $43$? When $i = 41$, the vertices would be $\{ 41, 42, 0, 20, 21 \}$?

The next step is to delete vertex $0$, producing a colored $K_{42}$ with only $43 - 5 = 38$ red $K_5$s. Finally, change the following edges from red to blue: $11 \leftrightarrow 32$ and $j \leftrightarrow j + 1$ for $j = 4, 5, 6, 7, 13, 14, 15, 16, 23, 24, 30, 33, 39, 40, 41$. One can verify with a computer that this coloring of $K_{42}$ is $K_5$-free.

I really want to see a picture of this coloring, but I can't find one.

答:
Don't we need the chord lengths from $1$ to $n−1$? How else will we know what color to assign to the edge $0↔n−1$, which has chord length $n−1$?

This edge has length $n-1$ if we go around the circle "the long way", going $0, 1, 2, 3, \dots, n-1$. It has length only $1$ if we go around the circle the other way. So we actually assign length $1$ to such a chord.

There's two ways to think about these lengths:

The chord lengths are distances in the cycle graph $C_n$ (where the vertices are also labeled $0, 1, 2, \dots, n-1$, and which contains edges $i \leftrightarrow i+1$ and $n-1 \leftrightarrow 0$).

To find the chord length of the edge $i \leftrightarrow j$, take $i - j \bmod n$, but instead of using the residues $0, 1, \dots, n-1$ as usual, use the residues $-\lfloor \frac n2\rfloor, \dots, \lceil \frac n2 \rceil$. Then, take the absolute value.

Does this use congruence classes mod $43$? When $i=41$, the vertices would be $\{41,42,0,20,21\}$?

Yes. Because the chord lengths are computed modulo $43$, the cliques also have the same structure.

I really want to see a picture of this coloring, but I can't find one.

It's hard to think of a very enlightening way to draw the coloring. Here's a picture that might not be very helpful:

I've put the points in order around the circle, with the missing vertex $0$ at the top, and the rest going $1, 2, 3, \dots, 42$ in order clockwise. The edges that have been recolored from red to blue are shown darker and thicker than the rest.

Here is the graph6 encoding for the graph of the red edges:

```<code>>>graph6<<izGOSHBG[`hBhBs`\GRhAmcH\G\\GYmcMjhBt\G^Ts`}jhA]jhAnTs`jt\G\]jhBTymcMjt\G]jt\GVTimcLt]jhBmjt\GQynTs`dt]jhBdt]jhBQynTs`cmjt\GWdt]jhAAVTymcICmjt\GO
</code>```

For example, we might use this to verify the lower bound in Mathematica as follows:

```<code>graph = ImportString[
  ">>graph6<<izGOSHBG[`hBhBs`\\GRhAmcH\\G\\\\GYmcMjhBt\\G^Ts`}jhA]\
jhAnTs`jt\\G\\]jhBTymcMjt\\G]jt\\GVTimcLt]jhBmjt\\GQynTs`dt]jhBdt]\
jhBQynTs`cmjt\\GWdt]jhAAVTymcICmjt\\GO", "Graph6"];
VertexCount[graph]
FindClique[graph]
FindClique[GraphComplement[graph]]
</code>```

The expected output is 42 for the vertex count, a $4$-vertex clique {{1, 2, 3, 15}} for a largest clique in the graph, and a $4$-vertex clique {{1, 4, 5, 9}} for a largest clique in the complement.

Body:
Don't we need the chord lengths from $1$ to $n−1$? How else will we know what color to assign to the edge $0↔n−1$, which has chord length $n−1$?

This edge has length $n-1$ if we go around the circle "the long way", going $0, 1, 2, 3, \dots, n-1$. It has length only $1$ if we go around the circle the other way. So we actually assign length $1$ to such a chord.

There's two ways to think about these lengths:

The chord lengths are distances in the cycle graph $C_n$ (where the vertices are also labeled $0, 1, 2, \dots, n-1$, and which contains edges $i \leftrightarrow i+1$ and $n-1 \leftrightarrow 0$).

To find the chord length of the edge $i \leftrightarrow j$, take $i - j \bmod n$, but instead of using the residues $0, 1, \dots, n-1$ as usual, use the residues $-\lfloor \frac n2\rfloor, \dots, \lceil \frac n2 \rceil$. Then, take the absolute value.

Does this use congruence classes mod $43$? When $i=41$, the vertices would be $\{41,42,0,20,21\}$?

Yes. Because the chord lengths are computed modulo $43$, the cliques also have the same structure.

I really want to see a picture of this coloring, but I can't find one.

It's hard to think of a very enlightening way to draw the coloring. Here's a picture that might not be very helpful:

I've put the points in order around the circle, with the missing vertex $0$ at the top, and the rest going $1, 2, 3, \dots, 42$ in order clockwise. The edges that have been recolored from red to blue are shown darker and thicker than the rest.

Here is the graph6 encoding for the graph of the red edges:

>>graph6<<izGOSHBG[`hBhBs`\GRhAmcH\G\\GYmcMjhBt\G^Ts`}jhA]jhAnTs`jt\G\]jhBTymcMjt\G]jt\GVTimcLt]jhBmjt\GQynTs`dt]jhBdt]jhBQynTs`cmjt\GWdt]jhAAVTymcICmjt\GO

For example, we might use this to verify the lower bound in Mathematica as follows:

graph = ImportString[
  ">>graph6<<izGOSHBG[`hBhBs`\\GRhAmcH\\G\\\\GYmcMjhBt\\G^Ts`}jhA]\
jhAnTs`jt\\G\\]jhBTymcMjt\\G]jt\\GVTimcLt]jhBmjt\\GQynTs`dt]jhBdt]\
jhBQynTs`cmjt\\GWdt]jhAAVTymcICmjt\\GO", "Graph6"];
VertexCount[graph]
FindClique[graph]
FindClique[GraphComplement[graph]]

The expected output is 42 for the vertex count, a $4$-vertex clique {{1, 2, 3, 15}} for a largest clique in the graph, and a $4$-vertex clique {{1, 4, 5, 9}} for a largest clique in the complement.


问:
Given a set A in which all elements are of the form: $x+y \sqrt3$; $x,y \in \mathbb{Q}$. What structure can you define...
Given a set $A$ in which all elements are of the form: $x+y \sqrt3$, $x,y \in \mathbb{Q}$.

What algebraic structure can you define with operations of addition and multiplication?

I am stuck figuring how to interpret  $x+y \sqrt3$ and use it to prove  the properties of algebraic structures... so any help would be sincerely appreciate.

答:
For ease, I will write $r=\sqrt{3}$, so $r^2=3$.

Addition:

It is closed: $$(x+yr)+(a+br) = (x+a)+(y+b)r$$ since $x+a$ and $y+b$ are again rational.

$0$ is neutral element.

$-x-yr$ is opposite to $x+yr$

Clearly it is associative and commutative so it is Abel group for addition.

Multiplication:

It is closed: $$(x+yr)\cdot (a+br) = (xa+3yb)+(xb+ya)r$$ since $xa+3by$ and $xb+ya$ are again rational.

$1$ is neutral element for multiplication.

Inverse to $x+yr\ne 0$ is $${x-ry\over x^2+3y^2}$$

Clearly it is associative and commutative.

And multiplication is distributive over addition...

So it is a field.

Notice that you can replace $\sqrt{3}$ with any $\sqrt{q}$ where $q$ is rational number which is not a square of rational number.


问:
Physical representation of volume to surface area
I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.

My question is: what is the physical representation of length in this ratio?

答:
The volume of a sphere is $V = \frac{4}{3} \pi r^3$ and the surface area is $A = 4 \pi r^2$. Thus, their ratio is given by:

$$\frac{V}{A} = \frac{r}{3}$$.

Thus, 3.3 cm is one-third of the radius of the sphere.


问:
Prove that if the altitude and median of a triangle form equal angles with sides then the triangle is right.
Problem statement:

Prove that if the altitude and median drawn from the same vertex of a nonisosceles triangle lie inside the triangle and form equal angles with its sides, then this is a right triangle.

After many attempts, I came up with this: let $ABC$ be the triangle where $CH$ is the altitude, $CM$ is the median, and name the angles $ACH = MCB = \theta$, $CAH = \alpha$, $CBM = \beta$ and $HCM = \gamma$. Using the sine theorem in the triangle $CMB$ we get $$\frac{MB}{\sin(\theta)} = \frac{MC}{\sin(\beta)},$$ while using the sine theorem in the triangle $ACM$ we get $$\frac{MC}{\sin(\alpha)} = \frac{MA}{\sin(\theta + \gamma)}.$$ Combining these equations I found $$\sin(\alpha) \sin(\theta) = \sin(\beta) \sin(\theta + \gamma).$$ Applying the identity $$\sin(a)\sin(b) = \frac{\cos(a-b) - \cos(a+b)}{2}$$ I concluded $$\cos(\alpha - \theta) = \cos(\theta + \gamma - \beta).$$ Since all angles are within $[0, \pi]$ range I concluded $$\alpha - \theta = \theta + \gamma - \beta,$$ thus $2 \theta + \gamma = \alpha + \beta$. From the original triangle we know $$2 \theta + \gamma + \alpha + \beta = \pi,$$ and combining the equations proves the triangle is right.

Is this correct? Is there a synthetic way to do it?

答:
Trigonometric Approach

We are given that $\angle MAC=\angle BAP$ and since $\triangle BPA$ is a right triangle, $\angle BAP=\frac\pi2-B$. Therefore,
$$
\angle MAC=\frac\pi2-B\tag1
$$
and, since $\angle MAC+\angle BAM=A$, we have
$$
\angle BAM=A+B-\frac\pi2\tag2
$$
Since $M$ is the bisector of $\overline{BC}$, the areas of $\triangle BAM$ and $\triangle MAC$ are equal. Thus,
$$
\begin{align}
\overbrace{\frac12cm\sin\left(A+B-\frac\pi2\right)}^{\left|\triangle BAM\right|}
&=\overbrace{\frac12bm\sin\left(\frac\pi2-B\right)}^{\left|\triangle MAC\right|}\tag{3a}\\
c\cos(C)&=b\cos(B)\tag{3b}\\[3pt]
c\sin(B)&=b\sin(C)\tag{3c}\\[3pt]
\sin(2B)&=\sin(2C)\tag{3d}
\end{align}
$$
Explanation:

$\text{(3b)}$: apply $A+B-\frac\pi2=\frac\pi2-C$ and multiply by $\frac2m$

$\text{(3c)}$: Law of Sines

$\text{(3d)}$: cross multiply $\text{(3b)}$ and$\text{(3c)}$ and multiply by $\frac2{bc}$

Since $\triangle ABC$ is not isosceles, $\text{(3d)}$ implies that $2B=\pi-2C$; which, since $A=\pi-B-C$, means
$$
\bbox[5px,border:2px solid #C0A000]{A=\frac\pi2}\tag4
$$

Geometric Approach

This is an approach similar, but slightly different, to that in Mick's answer.

Start with the diagram above. We are given that $\angle MAC$ is equal to $\angle BAP$, which is $\frac\pi2-B$ since $\angle APB$ is a right angle.

Add $N$ and $R$, the midpoints of $\overline{AC}$ and $\overline{AB}$ respectively. Then add the segments from $R$ to $N$, $M$, and $P$.

Add the circle with $\overline{AM}$ as its diameter. Since $\angle APM$ is a right angle, $P$ sits on this circle. Since $\overline{RN}$ is parallel to $\overline{BC}$, $\overline{RN}$ is also perpendicular to $\overline{AP}$.

Since $\triangle ARN$ is similar to $\triangle ABC$ with a linear ratio of $\frac12$, $Q$, the intersection of $\overline{AP}$ and $\overline{RN}$, bisects $\overline{AP}$. Thus, $\angle RPQ$ is also $\frac\pi2-B$.

Since $\overline{RM}$ is parallel to $\overline{AC}$, $\angle RMA$ is equal to $\angle MAC$, which equals $\frac\pi2-B$.

The locus of points at which $\overline{AR}$ subtends an angle of $\frac\pi2-B$ is the circle which contains $A$, $R$, $P$, and $M$. This is the same circle that was added above since it contains $A$, $P$, and $M$. Thus, $\angle ARM$ is a right angle, and since $\overline{RM}$ is parallel to $\overline{AC}$, so is $A$.


问:
is this Integral equal?
If:
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$
$$z\Gamma(z)=\Gamma(z+1)$$
$$\Gamma(z)=\mathcal \{Me^{-x}\}(z)$$
$$\Gamma(z)=\lim_{x\to\infty}(-x^{z-1})-(-0^ze^{-0})+z\int_0^\infty x^{z-1}e^{-x}dx$$
$$z!=\prod_{n=1}^\infty\Biggl[{1\over{1+{z\over{n}}}}\biggl(1+{1\over n}\biggl)^z\Biggl]\cdot$$
Where $\mathfrak R(z)>0$ and

$$\Gamma(z+1)=z!$$
Where $\mathfrak R(z)>0$
Then is

$$z! = \int_0^\infty x+1^{z-1}e^{-x+1}dx?$$
This is what I did:

$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$
$$\Gamma(z+1)=z!$$
Then +1 and -1 cancel. Is this right?

Link to gamma function: https://en.wikipedia.org/wiki/Gamma_function

答: You mean $\Gamma(\color{blue}{z})=\int_0^\infty x^{z-1}e^{-x}dx$; $x$ is integrated over, so doesn't appear on the left-hand side. So $z!=\Gamma(z+1)=\int_0^\infty x^ze^{-x}dx$.

Title: is this Integral equal?


问:
Unable to prove the transitive relation for path connected sets

Let (X,T) be a topological space and define a relation ~ on X by x~y provided that there is a pathwise connected subset A of X such that x,y $\in A$. Prove that ~ is an equivalence relation on X.

I tried by assuming f(0)=x and f(1)=y and f'(0)=y and f'(1)= z. But I need to find a function, f* such that f*(0)=x and f*(1)=y.

But unable to do so. Can you please help in constructiong such a funtion.

答:
If $f,f':[0,1] \to X$ are the equired paths define their composition

$$f \ast f'(t):=\begin{cases}
f(2t) & t \in [0,\frac12]\\
f'(2t-1) & t \in [\frac12,1]
\end{cases}$$

for $t \in [0,1]$. This is well-defined as $(f \ast f')(\frac12)$ is defined both as $f(1) = y$ and $f'(0)=y$. Both domain sets are closed, so by the pasting lemma we have that this function is continuous whenever $f$ and $f'$ are. So it's a continuous path from $x$ to $z$ as required.

For symmetry we use $f'(t)= f(1-t)$ when $f$ is a path from $x$ to $y$ to make one from $y$ to $x$, and a constant map suffices to show reflexivity.


问:
Finding a strong enough solution to a specific PDE problem.
Let $U\subset \mathbb{R}^n$ with smooth boundary $\partial U$. And consider the expression

$$\Delta u = f.$$

$$\text{+"convenient boundary conditions"}$$

In my specific case $f\in H^2_0$. Under what (boundary) conditions is it possible to conclude a weak solution which is also in $H^2_0$ of this problem?

The problem is mainly concerned with the subscript zero in $H^2_0$. For instance if I introduce the boundary condition

$$u=0 \text{ on } \partial U\ \text{ (trace-wise),}$$

then for any $f\in L^2$, there exists a weak solution in $H^1_0$. I can regularize the solution to be also in $H^2$, hence it is in $H^1_0\cap H^2$. (I use Evans for referrences concerning to Regularity theorems.) But as far as I know $H^1_0\cap H^2$ is a strictly bigger class of functions then $H^2_0$.

Help is very much appreciated.

答: I think that you can't solve this equation in $H^2_0$. If a function $u$ is in $H^2_0$ then you have $u=0$ and $\nabla u=0$ at the boundary and these are too much boundary conditions for this equation. Let me think how one can prove the nonexistence.</p>


问:
On deformation of linear combination of linearly independent vectors
Suppose we have a linearly independent set of vectors $\{v_i\}_{i=1}^n \subset H$ where $H$ is a Hilbert space. Let
$$
Q_1 = \sum_{i=1}^n v_i
$$
Consider $\{q_i\}_{i=1}^n \subset \mathbb{C}$ with $|q_i| < 1$ for any $i$ and set
$$
Q_2 = \sum_{i=1}^n q_i v_i
$$
Is it true, that $\|Q_2\| \le \|Q_1\|$?

I thought that it could be prooved somehow by Smith orthogonalization method, but I haven't figured out how. Thank you!

Body:
Consider $v_1 = (3,0)$ and $v_2 = (-3,1)$. Then $v_1$ and $v_2$ are linearly independent. Now take $q_1 = \frac{2}{3}$ and $q_2 = 0$.

Then $Q_1 = (0,1)$ and $Q_2 = (2,0)$. Hence,
$$||Q_1|| = 1 < 2 = ||Q_2||.$$


问:
Disjoint axis-aligned rectangles in the plane
Let $A$ be some set of axis-aligned rectangles in the plane, each pair of which has empty intersection. Prove that $A$ is a countable set.

(An axis-aligned rectangle is a set of the form

$$M = {\{\langle x,y \rangle \in \mathbb{R^2} | a \leq x \leq b , c \leq y \leq d}\}$$
for $a,b,c,d$ such that $a < b$ and $ c < d$.)

Attempt:

I tried using the density of the $\mathbb{Q}$ in $(\mathbb{R},\leq)$, but without any success.

答: You're on the right track. See if you can show that every axis-aligned rectangle contains a rational point (a point $(x,y)$ with $x,y\in\mathbb Q$). From this, using that the rationals are countable, why does the problem follow?


答:
Hint:

$|\sin x|\le |x|$ for all $x$, so $|f(x)|\le\cdots$


答:
Depending on your definition of open and closed sets in $\mathbb{R}$, and depending on what previous theorems you have been given:

A boundary point $x$ for a non-empty set $A$ is a point such that in any open interval around $x$, no matter how small, there will be at least one point in the interval that is in $A$ and one point in the interval that is not in $A$.

Given any non-empty set $A$ that is bounded above, the supremum (i.e. least upper bound) of $A$ is a boundary point of $A$.

Any non-empty set that contains one of its boundary points can not be an open set.

Any non-empty closed set must contain all of its boundary points.  This last assertion is a consequence of defining a non-empty set as closed if and only if its complement is open.

Title: Proof of the fact that the closure of a set always contains its supremum and an open set cannot contain its supremum

Body:
I would like to ask, if my proof checks out and is completely sound.

Exercise 3.2.4 from Stephen Abbot's Understanding Analysis.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Let $A$ be non-empty subset of $\mathbf{R}$ and bounded above, so that $s = \sup A$ exists. Let $\bar{A} = A \cup L$ be the closure of $A$.

(a) Show that $s \in \bar{A}$.

(b) Can an open set contain its supremum?

My Attempt.

(a) $\bar{A} = A \cup L$ is the closure of $A$ and contains the limits points of $A$. We proceed by contradiction. Assume that $s \notin \bar{A}$ and is not a limit point of $A$.

Since, $s$ is the supremum for $A$, looking at the definition of least upper bound, it must satisfy two properties: (i) $s$ is an upper bound for $A$. (ii) Given any small arbitrary, but fixed positive real $\epsilon > 0$, $(s - \epsilon)$ should not be an upper bound for $a$.

From (ii), it follows that, given any $\epsilon > 0$, there exists $t \in A$, such that $s - \epsilon < t$. Thus,
\begin{align*}
 \absval{t - s} < \epsilon
\end{align*}

Thus, every $\epsilon$-neighbourhood of $s$, $V_\epsilon(s)$ intersects $A$ in points other than $s$. So, $s$ is the limit point of $A$. Therefore, $s \in \bar{A}$, which contradicts our initial assumption. Hence, our initial assumption is false.

(b) An open set cannot contain its supremum. We proceed by contradiction. Let $O$ be an open set. Assume that $s \in O$.

Since $O$ is an open set, for all points $x$ belonging to $O$, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$ that is contained in $O$. In particular, $V_\epsilon(s) \subseteq O$. So, if
\begin{align*}
 s - \epsilon < t < s + \epsilon
\end{align*}

then $t \in O$, for some $\epsilon$. But, that implies, for some $\epsilon > 0$, we must have
\begin{align*}
 s < t < s + \epsilon
\end{align*}

$t \in O$. So, $s$ is not an upper bound for $O$. This is a contradiction. Our initial assumption must be false. $s \notin O$.


Title: On deformation of linear combination of linearly independent vectors

Body:
Suppose we have a linearly independent set of vectors $\{v_i\}_{i=1}^n \subset H$ where $H$ is a Hilbert space. Let
$$
Q_1 = \sum_{i=1}^n v_i
$$
Consider $\{q_i\}_{i=1}^n \subset \mathbb{C}$ with $|q_i| < 1$ for any $i$ and set
$$
Q_2 = \sum_{i=1}^n q_i v_i
$$
Is it true, that $\|Q_2\| \le \|Q_1\|$?

I thought that it could be prooved somehow by Smith orthogonalization method, but I haven't figured out how. Thank you!

Body:
Of course not true.

Take $v_{1} = <a,0>$ and $v_{2} = <-a,b>$ where b != 0.
$$||Q_{1}|| = |b| $$

But, if $q_{1} = 1$ and $q_{2} = 0$ then
$$||Q_{2}|| = |a|$$


问:
Inequality proof, power function, 'sum of the power and power of the sum'
I got a question regarding the following two relations:

$$0<c<b<a, 0<\alpha<1,$$

$$(1-\alpha)\log\frac{3}{2}\text{ and  }\log(a^\alpha+b^\alpha+c^\alpha)-\alpha \log(a+b+c)-\left[\log(a^\alpha+b^\alpha)-\alpha\log(a+b)\right]$$

I checked a few numbers, and made the figure using Mathematica, which shows the former is larger than the second one.

I wonder if you guys have ideas on how to prove it analytically.

Thanks a lot!

答:
Your inequality doesn't hold. Take $\alpha = 1/2, a = 1, b=.75, c=.749$.

$LHS=0.202732554, RHS = 0.202892973 > LHS$.

On the other hand if $a=1, b=.9, c=.8$ then $LHS > RHS$ so the other direction doesn't hold either.

BTW: if you fix $a$ and $b$ you can prove the derivative of RHS w.r.t. $c$ is positive, so $RHS$ takes its maximum when $c=b$. After you replace $c$ with $b$ in RHS however, RHS as a function of $b$ is not monotone (if $a=b=c$ you have equality).


问:
In Cantor's Theorem, can the diagonal set D be empty?
Cantor's Theorem:

For any set $A, |P(A)| > |A|$. That is, the power set of any set $A$ has a strictly greater cardinality than set $A$.

Proof of not surjection:

Suppose that there is an onto $f : A → P(A)$. Let $D$ be defined as follows:

$D = \{x ∈ A\ |\ x \notin f(x)\}$.

Note that, $D ⊆ A$, hence $D ∈ P(A)$.
Since $f$ is onto, then there is an $x ∈ A$ such that $f(x) = D$. Now either $x ∈ D$ or $x \notin D$. If
$x ∈ D$, then $x \notin f(x)$ by definition of $D$. But $f(x) = D$, so $x \notin D$, a contradiction. If $x \notin D$, then $x ∈ f(x)$. But $f(x) = D$, so $x ∈ D$, a contradiction. In any case, a contradiction arises, hence the assumption that there is an $x ∈ A$ such that $f(x) = D$ is false. Hence $f$ cannot
be onto.

My Question:

What if $f(x)=\{x\}$? Then wouldn't $D$ be empty? Please tell me why this won't work.

答:
Whether $D$ is empty or not is irrelevant. All we need is $D \in P(A)$ (which is also true if it were empty) and so some $x \in A$ exists such that $f(x)=D$. This is all we use. If $D$ were empty, $x \notin D$ obviously and then we get a contradiction because we'd then conclude (as in the argument you have) that $x \in D$, and we still have the same solid contradiction ($x \in D$ and $x \notin D$ both would hold).

It makes no difference.


问:
$f'\rightarrow 0$ , if $f(n)\rightarrow l (n\in\mathbb{Z})$, then $f(x)\rightarrow l$?

Let $f:(0,\infty)\rightarrow \mathbb{R}$ be differentiable and $f'(t)\rightarrow 0$ as $t\rightarrow \infty$.

If $f(n)\rightarrow l$ as $n\rightarrow \infty (n\in \mathbb{Z})$, then $f(t)\rightarrow l$ as $t\rightarrow \infty$ ($t\in\mathbb{R})$.

I am currently stuck at this problem.

How to prove this? I am having difficulties relating the limit for integers and the limit for reals when $f'\rightarrow 0$.

Thanks in advance!

答:
Let $\lfloor t \rfloor$ denote the largest integer smaller than $t$. In general, we have
$$|f(t) - l| <|f(t) - f(\lfloor t \rfloor)| +  |f(\lfloor t \rfloor) - l|.$$

Because $f(n) \to l$, the second term can be made small if $t$ is large enough.

The first term can be made small for large $t$ by the mean value theorem. It is equal to $$|f(t) - f(\lfloor t \rfloor)| = |t-\lfloor t \rfloor||f'(c)| \le |f'(c)|$$ for some $c$ between $t$ and $\lfloor t \rfloor$. When $t$ is large, $|f'(c)|$ is small because of the convergence of the derivative to zero. [The intuition here is that $f$ gets flatter and flatter as $t \to \infty$, so the function cannot vary very much between two integers.]


问:
Diagonalizability and Topology
In an exchange with one of my math professors, I was told that diagonalizability can differ depending on whether one is dealing with the field of real numbers or the complex numbers.

I learned in Real Analysis last term that concepts like "compactness" similarly vary over different fields.

My question given the above is simply this - is diagonalizability of matrices  (or their underlying linear systems) also a topological concept, or related to topology in some way?  I have not taken topology (only linear algebra) so I apologize in advance if this is an obvious/stupid question.

答:
Well, sort of. In order for a square matrix $M$ to be diagonalizable over a field $F$ its eigenvalues need to exist over $F$; equivalently, its characteristic polynomial needs to have all of its roots in $F$. $M$ can have any (monic) characteristic polynomial, thanks to the existence of companion matrices. Over $\mathbb{R}$ there are matrices such as rotation matrices

$$R_{\theta} = \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]$$

which don't have real eigenvalues (for most values of $\theta$), or equivalently whose characteristic polynomials don't have roots over $\mathbb{R}$. However, by the fundamental theorem of algebra every polynomial with real or complex coefficients has all of its roots over $\mathbb{C}$, so this difficulty disappears over $\mathbb{C}$ (e.g. all rotation matrices are diagonalizable).

The connection to topology is that many proofs of the fundamental theorem of algebra make use of topological ideas, and some of them explain why FTA is true for $\mathbb{C}$ but not $\mathbb{R}$ in terms of topological differences between them, the most basic one being that $\mathbb{R}$ is $1$-dimensional but $\mathbb{C}$ is $2$-dimensional. One such proof involves applying the Lefschetz fixed point theorem to the complex projective spaces $\mathbb{CP}^n$; if you try to apply the proof to the real projective spaces $\mathbb{RP}^n$ it will succeed in odd dimensions (corresponding to the fact that every polynomial of odd degree over $\mathbb{R}$ has a real root) but fail in even dimensions (a polynomial of even degree over $\mathbb{R}$ can have no real roots).

Title: Diagonalizability and Topology


问:
Understanding the notation of uniform norm
I'm currently struggling my way through Real Analysis and I don't know how to "read" the notation for a uniform norm.

$\left\lVert f\right\rVert_{\infty}:=\sup_{x \in X}|f(x)|$

Could someone please break down the meaning of each symbol and the information it is trying to tell me? I'll outline were I'm specifically confused,

For $\left\lVert f\right\rVert_{\infty}$, the double verticle bars represents the norm or "length" of vector within a Euclidean Space. So we would have the length of the function $f$. But what is the meaning and purpose of the subscript infinity? What information is this suppose to tell me?

For $\sup_{x \in X}|f(x)|$, $\sup$ is the lowest upper bound of some value. So we have the lowest upper bound for the absolute value of the function $f(x)$. But what is the meaning of the subscript ${x \in X}$?

The final question is how both these statements are equal but I first need to know the purpose of the notation to answer that.

答:
$\sup_{x\in X}|f(x)|$ is the lowest upper bound of the set
\begin{equation}
\{|f(x)|, x\in X\}\subset {\mathbb R}
\end{equation}
The $\infty$ in the norm notation comes from the fact that for a collection of numbers $f_1\cdots f_n$ one has
\begin{equation}
\max_i |f_i| = \lim_{p\to \infty} \left(\sum_i |f_i|^p\right)^{1/p}
\end{equation}
Its purpose is to differentiate this particular norm from other function norms that could be defined on $X$. It is a cultural element.


问:
Degree One Map induces Surjections on Homology
Is the following statement true:

If $f:M\to N$ is a degree one map of compact closed manifolds, then $f$ induces surjections $f^*:H_q(M)\to H_q(N)$.

I found this claimed on http://topospaces.subwiki.org/wiki/Degree_one_map . The $H_1$ case is clear to me, but I don't see how to generalize this.

答:
By Poincare duality every class $x \in H_p(M;R)$ is of the form $[M] \frown \psi$, for $\psi \in H^{n-p}(M; R)$.

Now if $y \in H^p(N; R)$, then we know $y = \psi_y \frown [N]$.

Now generally for a homology class $x \in H_*(M;R)$ and $\psi \in H^*(N;R)$ we have that $f_*(x) \frown \psi = f_*(x \frown f^*\psi)$. Applying this to $x = [M]$ and $y = \psi_y$ we find that

$$y = [N] \frown \psi_y = f_*([M] \frown f^* \psi_y),$$ and in particular $y$ is in the image of $[M] \frown f^* \psi_y$.

Therefore, the induced map on homology of a degree-1 map is necessarily injective.


问:
sum of increasing and decreasing function has at most one maximum
Let $f(x) \geq 0$ and $g(x) \geq 0$, both be strictly increasing and concave functions in $x>0$.

Can we show that $h(x) = f(x) - g(x) $ has at most one mode (i.e., global maximum) by the first assumptions?

Update: I added the condition on concavity for $f$ and $g$.

Thanks in advance.

答:
This statement is untrue. Let $f(x)=x+\sin x+1$, $g(x)=x$. Both functions are increasing, but $h(x)=\sin x+1$ has infinitely many global maxima.

As noted in the comments, there is an even simpler counter-example: $f(x)=g(x)=x$.


问:
Linear control theory and transition matrix problem
So I want to prove the following statement:

Consider an $n$-dimensional time-varying system $\dot{x} = A(t)x$, where $A(t)$ is continuous. If $A^{\intercal}(t)=-A(t)$ $\forall t\in\mathbb{R}$, then $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$, where $\Phi(t,s)$ is the state transition matrix.

So first of all, I know the following:

\begin{align}
\frac{\partial}{\partial s}\Phi(t,s) = -\Phi(t,s)A(s) \iff \frac{\partial}{\partial s}\Phi^{\intercal}(t,s) = -A^{\intercal}(s)\Phi^{\intercal}(t,s) = A(s)\Phi^{\intercal}(t,s).
\end{align}

So how do I go from here to show $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$? It feels like I am on the track, but I am stuck.

Edit

If you are unfamiliar with the transition matrix, this appears in the solution to system of differential equations:

$x(t)=\Phi(t,s)x(s)$

where $x(s)$ is the initial state.

答:
Hint: It suffices to note that $\|\Phi(t,s)x\| = \|x\|$ for every initial state vector $x$. Try to show that
$$
\frac {\partial }{\partial t} \|\Phi(t,s) x\|^2 = 0.
$$

In detail: a matrix $M$ satisfies $M^\top = M^{-1}$ if and only if it satisfies $\|Mx\| = \|x\|$ for all vectors $x$. So, it suffices to show that we have $\|\Phi(t,s)x\| = \|x\|$ for every possible $x$.

Note that for any $x$, $x(t) = \Phi(t,s)x$ satisfies the differential equation $\dot x(t) = A(t) x$. It follows that
\begin{align}
\frac \partial{\partial t} \|\Phi(t,s)x\|^2 &=
\frac \partial{\partial t} [x(t)^\top x(t)] = \dot x^\top(t) x(t) + x^\top(t) \dot x(t)
\\ & =
[A(t) x(t)]^\top x(t) + x^\top(t)[A(t) x(t)]
\\ & =
x^\top(t)A^\top(t)x(t) + x^\top(t)A(t)x(t)
\\ & = x^\top(t)[-A(t)]x(t) + x^\top A(t) x(t) = 0.
\end{align}
Thus, it must hold that $\|\Phi(t,s)x\|$ is constant. It follows that
$$
\|\Phi(t,s) x\| = \|\Phi(s,s)x\| = \|Ix\| = \|x\|.
$$
That is, for all $t$, the matrix $M = \Phi(t,s)$ satisfies $\|Mx\| = \|x\|$ for all $x$. Thus, for all $t$, the matrix $M = \Phi(t,s)$ satisfies $M^\top = M^{-1}$, which is what we wanted.


问:
Summation of a finite series
Let

$$f(n) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1} \forall \ n \ \epsilon \ \mathbb{N} $$

If it cannot be summed , are there any approximations to the series ?

答:
$f(n)=H_{2^n-1}$, the $(2^n-1)$-st harmonic number. There is no closed form, but link gives the excellent approximation

$$H_n\approx\ln n+\gamma+\frac1{2n}+\sum_{k\ge 1}\frac{B_{2k}}{2kn^{2k}}=\ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\ldots\;,$$

where $\gamma\approx 0.5772156649$ is the Euler–Mascheroni constant, and the $B_{2k}$ are Bernoulli numbers.

Title: Summation of a finite series


答:
"What if $f=g$? Both are strictly increasing.In particular $f=g=x$."

As I said in the comments $f=g$ is a sort of trivial example.

If you want a non trivial one , let $f=x+1,g=x$.

Why am I even writing an answer? Someone already answered, may be because of reputation points. Sorry, if I sound greedy. Its 3:45AM here , I should sleep.

Title: sum of increasing and decreasing function has at most one maximum

Body:
Let $f(x) \geq 0$ and $g(x) \geq 0$, both be strictly increasing and concave functions in $x>0$.

Can we show that $h(x) = f(x) - g(x) $ has at most one mode (i.e., global maximum) by the first assumptions?

Update: I added the condition on concavity for $f$ and $g$.

Thanks in advance.


答: Sometimes a figure is worth 1000 words:

Body: Sometimes a figure is worth 1000 words:


问:
Lipschitz Equivalents
How can I show that this proposition. Can you help me, please.

In this set, give a set $X$ different from null and $c$ and $d$ measures so that $c$ and $d$ aren't Lipschitz equivalents. but $\tau_c= \tau_d$.

答: On $\Bbb R$, $d(x,y)=|x-y|$ and $d'(x,y)=\min(|x-y|,1)$ are equivalent metrics in that $\tau_d =  \tau_{d'}$ but they are not Lipschitz equivalent.

Title: Lipschitz Equivalents


问:
How many rational elements are in the multinomial? $(\sqrt[3]{x}+\sqrt[5]{x}+2\sqrt{x})^5$

How many rational elements are in the multinomial?
$(\sqrt[3]{x}+\sqrt[5]{x}+2\sqrt{x})^5$

According to the multinomial theorem, $\sum_{i+j+k=5}^{}\frac{5!}{i!j!k!}2^k\ x^{\frac{1}{3}i+\frac{1}{5}j+\frac{1}{2}k}$

So I think it's rational when $k \ \in  \  \mathbb{N}$ and $(\frac{1}{3}i+\frac{1}{5}j+\frac{1}{2}k)\ \in  \  \mathbb{N}$ and $i+j+k=5$

Which satisfy only when $i=3,j=0,k=2$ and $j=5,i=k=0$ so if's correct then are only 2 rational elements in the sum iteration?

Additional question

Needing to find how many elements are in the multinomial, so what I figured is finding in how many for the equations $a + b + c= 5$ to be formed? which is $D(3,5) = C(3+5-1,5)$ but I am not sure about it

答:
I agree with the analysis in the first part of your query.  There is one ambiguity:

In the case of $i=3,k=2$, this can be regarded as one element or the sum of $\binom{5}{3} = 10$ elements, depending on your interpretation of the word element.

Depending on your interpretation of the word element, the second part of your query can be regarded as a Stars and Bars problem, where you want the number of non-negative integer solutions to

$$i + j + k = 5.$$

The enumeration is

$$\binom{5 + [3-1]}{[3-1]} = \binom{7}{2} = 21.$$

This agrees with the number that you (the OP) computed.


问:
Defining $A \in \mathcal{P}(\Bbb N \times\Bbb N)$ such that it is not any member of a countable subset $M \subseteq \mathcal{P}(\Bbb N \times \Bbb N)$
$
\newcommand{\N}{\mathbb{N}}
\newcommand{\P}{\mathcal{P}(\N \times \N)}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\i}{^{(i)}}
\newcommand{\x}{^{[x]}}
\newcommand{\y}{^{[y]}}
\newcommand{\I}{\mathcal{I}}
\newcommand{\J}{\mathcal{J}}
$
The Problem: Define $M := \set{A_i}_{i \in \N} \subseteq \P$ to be a countable susbset of $\P$. How would we define some $A \in \P$ such that $A \ne A_i$ for every $i \in \N$?

Context: The context for the problem is this unanswered MSE question from $2018$ that has not been answered yet. I've been trying to answer it for a few days, but I'm not certain of the approach I'm taking. I would like a sort of constructivist approach to the problem, in the sense of an explicit or direct way to define the set $A$ in question (as opposed to, say, just handwaving things away by saying a certain element of $A$ is just in a complement of some $A_i$).

The original problem also specifies the condition $(0,0) \not \in A$, though it's not clear if this is an important restriction in my eyes. (Of course this also means we're adopting the convention $\N := \set{0,1,2,3,\cdots}$.)

My Approach So Far: My goal is sort of in the vein of a diagonalization argument. Since $M$ is countable and $A,A_i$ are at most countable themselves (as $|\N \times \N| = \aleph_0$ per, e.g. here), we may write:

$$M = \set{A_i}_{i=0}^\infty \qquad A = \set{a\i}_{i \in \I} \qquad A_i = \set{a_{i,j}}_{j \in \J}$$

as a means of indexing the members of each set. $\I$ and $\J$ are index sets such that $|\I|,|\J| \le \aleph_0$. We may likewise order the elements of each. Finally, define $a_{i,j} \x$ to be the first coordinate of $a_{i,j}$, and $a_{i,j} \y$ the second. That is,

$$a_{i,j} = \left( a_{i,j} \x \; , \; a_{i,j} \y \right)$$

Then, to ensure $A \ne A_i$, I want to ensure $A$ contains an element different from each $A_i$. We can break this into cases for every possible $A_i$:

Suppose $A_i = \varnothing$. Then let $a\i :=  (1,1)$.

Suppose $A_i$ is finite and nonempty. Then take
$$a^{(i)} := \left( 1+ \max_{0 \le j \le |A_i|} a_{i,j} \x \; , \; 1+ \max_{0 \le j \le |A_i|} a_{i,j} \y \right)$$
In short: $a \i$ is $1$ bigger (or more) in each component than any of $A_i$'s elements.

Suppose $A_i$ is infinite, but not equal to $\N \times \N$. Then ... ?

Suppose $A_i = \N \times \N$. Then ... ?

I'm pretty confident in my first three cases, but handling the cases of $A_i$ infinite are proving to be sticking points for my approach - in particular the case of $A_i = \N \times \N$, I'm really not sure how to handle that one. I guess we could just say $a\i \in A_i^c$, but that feels a little hand-wavey (especially for, again, $A_i = \N \times \N$), and I'm more interested in defining the elements of $A$ directly in some fashion, if at all possible.

Someone brought up, in the comments of the original question, about using the axiom of choice and ZFC set theory. If strictly necessary, I'm fine with the use of axiom of choice or alternative set theories (e.g. NBG), though the less we have to invoke those, the better in my opinion, since I'm not particularly well-versed in these sorts of decisions and their ramifications.

Anyhow - does anyone have some bright ideas as to how I might be able to finish this approach? And how does the restriction of $(0,0) \not \in A$ affect the original problem?

答:
It’s possible to do something along the lines that you have in mind, but you can also (and much more easily) do it all at once: let

$$A=\{\langle n,n\rangle\in\Bbb N\times\Bbb N:\langle n,n\rangle\notin A_n\}\,.$$

For each $n\in\Bbb N$ the sets $A$ and $A_n$ disagree on $\langle n,n\rangle$: one contains it, and the other doesn’t.

If you want to ensure that $\langle 0,0\rangle\notin A$, change the definition: let

$$A=\{\langle 1,n\rangle\in\Bbb N\times\Bbb N:\langle 1,n\rangle\notin A_n\}\,.$$


问:
Finding roots of an equation dealing complex numbers
For the equation
$z^6
 + 6z + 20 = 0,$ z is a complex number

Putting z=x+iy is tiresome . Moreover my question demands to find the roots lying in each quadrant . Is there an easy way out?

Edit.

The actual statement is
The number of roots in 1st,2nd,3rd ,4th quadrant are

答:
Set $z=a+ib\;$ and expand
$$(a+ib)^6 + 6(a+ib) + 20 = 0 \tag{1}$$

Equating real and imaginary parts to $0$ we obtain the system
$$\begin{aligned}a^6-15a^4b^2+15a^2b^4-b^6+6a+20&=0 \quad\quad\quad\quad (*)\\
6a^5b-20a^3b^3+6ab^5+6b&=0 \quad\quad\quad\quad (**)\end{aligned}$$
Recall that the system has six solutions $(a,b)\in \mathbb{R}\times\mathbb{R}.$ Clearly $ab\neq0.$

Take the equation $(**)$ and divide it by $2b.$ We obtain $$3a^5-10a^3b^2+3ab^4+3=0 \tag{2}$$ Let us consider $(2)$ as equation in variable $a.$

Descartes rule of signs says that $(2)$ has $2$ positive and $3$ negative roots $a$ (more precisely, two $a$'s have positive real part and three have negative real part, but we know that all $a$'s are real).

Recall that all solutions of $(1)$ are in pairs  $(a_1,\pm b_1), (a_2,\pm b_2),(a_3, \pm b_3).$

Only one of $a$'s is missing in $(**)$, thus we can conclude that $(**)$ has one positive double root, one negative double root and one simple root $a.$

Therefore, one solution of the original equation $(1)$ lies in the first quadrant, two in the second, two in the third and one in the fourth.

To verify our result, let us apply Descartes rule to $(*):$

if $\;20-b^6<0$ then $(*)$ has $3$ positive and $3$ negative roots $a_i,$ which is impossible

if $\;20-b^6=0$ then $(*)$ has $2$ positive and $3$ negative roots $a_i$

if $\;20-b^6>0$ then $(*)$ has $2$ positive and $4$ negative roots $a_i$

Our conclusion is justified. As a by-result, necessarilly $\;20-b^6\geq 0.$


问:
How do we apply chain rule to a function of two variables, which are also functions.
Let $f$ and $g$ be differentiable functions.

$$
\frac{\partial}{\partial z}(g(f(x+y),f(z))
$$
How do we apply the chain rule here correctly? Do we just treat $f(x+y)$ and $f(z)$ as some variables $u$ and $v$?

答:
We consider:
\begin{align*}
g=g(u,v)\qquad\qquad &u=u(x,y,z)\\
&v=v(x,y,z)
\end{align*}
In the current situation we have:
\begin{align*}
&u(x,y,z)=f(x+y)\qquad\qquad v(x,y,z)=f(z)\\
&\qquad\ \,\frac{\partial u}{\partial z}=0\qquad\qquad\qquad\qquad\quad\ \  \frac{\partial v}{\partial z}=f^\prime(z)
\end{align*}

Application of the chain rule gives
\begin{align*}
\frac{\partial}{\partial z}g(u(x,y,z),v(x,y,z))&=
\frac{\partial g}{\partial u}\,\frac{\partial u}{\partial z}
+\frac{\partial g}{\partial v}\,\frac{\partial v}{\partial z}\\
\color{blue}{\frac{\partial}{\partial z}g(f(x+y),f(z))}&\color{blue}{=}
\frac{\partial g}{\partial u}\, \cdot 0
+\color{blue}{\frac{\partial g}{\partial v}\,f^{\prime}(z)}
\end{align*}
in accordance with the already given answer.


问:
What (if anything) can we infer about a first order theory from its category of models?
Does the structure of the category of models of a given first order theory where the arrows are embeddings tell us anything interesting about the theory itself?

Two models $M$ and $M'$ are isomorphic if and only if there exists an embedding $f : M \to M'$ whose inverse $f^{(-1)}$ is an embedding from $M'$ to $M$. Also, embeddings compose. Thus, we have a category.

As a trivial observation, we can rehash the Löwenheim–Skolem theorem in categorical language, if we take some effort to remove the finite models ahead of time.

Picking a satisfiable theory $T$ with no finite models produces a category $C$. If we take $C$ and produce a new category $D$ that identifies all the arrows in every homset in $C$, we get a category that is isomorphic to the category of infinite cardinals, where the homset $\text{Hom}(\kappa, \lambda)$ is a singleton if and only if $\kappa \le \lambda$ and $\text{Hom}(\kappa, \lambda)$ is empty otherwise.

The above statement is a consequence of the Löwenheim–Skolem theorem; it might be equivalent, though, if "cardinal gaps" are always detectable by the structure of the category.

答:
There is a very interesting result by Campion, Cousins, and Ye in the context of elementary embeddings. In particular, the Lascar group of a first order theory $T$ is isomorphic to the fundamental group of the classifying space of $Mod(T)$ (where, $Mod(T)$ is the category of models of $T$ with elementary embeddings).

Here is a link to the paper.


问:
Quotient space of vector spaces involving continuous functions.
I'm stuck on the follow question:

Let $\mathcal{C}([a,b],\mathbb{R})$ the vector space of the continuous functions $f:[a,b] \to \mathbb{R}$ and let $S$ the subspace of the constant functions of $[a,b]$. Show that the quotient space $\mathcal{C}([a,b],\mathbb{R})/S$ is isomorphic to $W$, where $W$ is the subspace of the continuous functions $g:[a,b] \to \mathbb{R}$ s.t $g(a)=0$.

Usually, this kind of questions is not so hard. I just need some sobrejective linear function $\phi: \mathcal{C}([a,b],\mathbb{R}) \to W$, where $\ker \phi = S$. However, I didn't see this function $\phi$, because this quotient space is kind different. I guess, it is no so intuitive, so do you have some hint for me?

答:
For any $\;f\in \mathcal C(a,b]\;$, define the scalar $\;k(f,a):= f(a)\;$ , and then define

$$\phi: \mathcal C[a,b]\to W\;,\;\;\phi(f):= f(x)-k(f,a)$$

Check stuff...and there you go.


答: Hint: Try $\phi(f)=x\mapsto f(x)-f(a)$.

Title: Quotient space of vector spaces involving continuous functions.

Body:
I'm stuck on the follow question:

Let $\mathcal{C}([a,b],\mathbb{R})$ the vector space of the continuous functions $f:[a,b] \to \mathbb{R}$ and let $S$ the subspace of the constant functions of $[a,b]$. Show that the quotient space $\mathcal{C}([a,b],\mathbb{R})/S$ is isomorphic to $W$, where $W$ is the subspace of the continuous functions $g:[a,b] \to \mathbb{R}$ s.t $g(a)=0$.

Usually, this kind of questions is not so hard. I just need some sobrejective linear function $\phi: \mathcal{C}([a,b],\mathbb{R}) \to W$, where $\ker \phi = S$. However, I didn't see this function $\phi$, because this quotient space is kind different. I guess, it is no so intuitive, so do you have some hint for me?

Body: Hint: Try $\phi(f)=x\mapsto f(x)-f(a)$.


问:
Prove that a wff built up only with $\lnot$ and $\leftrightarrow$ is a tautology iff $\lnot$ and each statement letter occur an even number of times.
First of all, I know how to prove this theorem below:

A statement form that contains $\leftrightarrow$ as its only connectives is a tautology iff  each statement letter occurs an even number of times.

To prove the theorem, I first derived this corollary:

A statement built up using only $\leftrightarrow$ is true iff there are an even number of false terms.

But now, I have to prove the theorem below:

Proving that a statement form that contains $\lnot$ and $\leftrightarrow$ as its only connectives is a tautology iff $\lnot$ and each statement letter occur an even number of times.

The bold texts are the only differences comparing it to the previous question that I could solve. I tried modifying the corollary to help me prove the theorem but it didn't help me, or modifying the corollary made it impossible to prove. Is there anything else I could do?

答:
To solve the problem, observe that, for every formulas $\varphi$ and $\psi$, the formulas
$$\lnot (\varphi \leftrightarrow \psi) \qquad\qquad \lnot \varphi \leftrightarrow \psi \qquad\qquad \varphi \leftrightarrow \lnot \psi
$$
are logically equivalent. It means that in the fragment $\{\lnot, \leftrightarrow\}$ of propositional logic, the truth value of a formula depends on the number of occurrences of $\lnot$, but not on their position. And since the
truth values are only two, not even the number of occurrences of $\lnot$ really matters, but only the parity of such a number.

This property gives an intuition about the way to generalize the corollary from the fragment $\{\leftrightarrow\}$ to the fragment $\{\lnot, \leftrightarrow\}$. The generalization is the lemma below.

Lemma. Let $\varphi$ whose only connectives are $\lnot$ and $\leftrightarrow$, and let $v$ be a valuation.
Then, $v$ makes $\varphi$ true if and only if the sum of
the number of occurrences of $\lnot$ in $\varphi$ and the number of occurrences of statement letters in $\varphi$ made false by $v$ is even.

Proof. By structural induction on the formula $\varphi$. Let us denote by $\#^\lnot_\varphi$ the number of occurrences of $\lnot$ in $\varphi$, and by $\bot^v_\varphi$ the number of occurrences of statement letters in $\varphi$ made false by $v$. We want to prove that $v$ makes $\varphi$ true if and only if $\#^\lnot_\varphi + \bot^v_\varphi$ is even.

Base case. The formula $\varphi$ is a statement letter, hence $\#^\lnot_\varphi = 0$ and either $\bot^v_\varphi = 1$ (if $v$ makes $\varphi$ false) or $\bot^v_\varphi = 0$ (if $v$ makes $\varphi$ true). Clearly, $\varphi$ is true if and only if $\bot^v_\varphi = \#^\lnot_\varphi + \bot^v_\varphi$ is even.

Inductive step. Since $\lnot$ and $\leftrightarrow$ are the only connectives used in the formula $\varphi$, there are only two cases.

$\varphi = \lnot \psi$. By induction hypothesis, $v$ makes $\psi$ true if and only if $\#^\lnot_\psi + \bot^v_\psi$ is even.
Clearly, $\#^\lnot_\varphi = \#^\lnot_\psi + 1$ and $\bot^v_\varphi = \bot^v_\psi$. Therefore, $v$ makes $\varphi$ true if and only if $v$ makes $\psi$ false (since $\varphi$ is the negation of $\psi$) if and only if $\#^\lnot_\psi + \bot^v_\psi$ is odd if and only if $\#^\lnot_\varphi + \bot^v_\varphi$ is even.

$\varphi = \psi_1 \leftrightarrow \psi_2$. By induction hypothesis, for each $i \in \{1,2\}$, $v$ makes $\psi_i$ true if and only if $\#^\lnot_{\psi_i} + \bot^v_{\psi_i}$ is even. Clearly,  $\#^\lnot_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}$ and $\bot^v_\varphi = \bot^v_{\psi_1} + \bot^v_{\psi_2}$.
Cases:

$v$ makes both $\psi_1$ and $\psi_2$ true; then, $v$ makes $\varphi$ true (according to the truth table of $\leftrightarrow$) and $\#^\lnot_\varphi + \bot^v_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}  + \bot^v_{\psi_1} + \bot^v_{\psi_2} = (\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}) + (\#^\lnot_{\psi_2} + \bot^v_{\psi_2})$ is even because so are $\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}$ and $\#^\lnot_{\psi_2} +  \bot^v_{\psi_2}$.

$v$ makes both $\psi_1$ and $\psi_2$ false; then, $v$ makes $\varphi$ true (according to the truth table of $\leftrightarrow$) and $\#^\lnot_\varphi + \bot^v_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}  + \bot^v_{\psi_1} + \bot^v_{\psi_2} = (\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}) + (\#^\lnot_{\psi_2} + \bot^v_{\psi_2})$ is even because both $\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}$ and $\#^\lnot_{\psi_2} +  \bot^v_{\psi_2}$ are odd.

$v$ makes $\psi_1$ true and $\psi_2$ false; then, $v$ makes $\varphi$ false (according to the truth table of $\leftrightarrow$) and $\#^\lnot_\varphi + \bot^v_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}  + \bot^v_{\psi_1} + \bot^v_{\psi_2} = (\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}) + (\#^\lnot_{\psi_2} + \bot^v_{\psi_2})$ is odd because $\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}$ is even and $\#^\lnot_{\psi_2} +  \bot^v_{\psi_2}$ is odd.

$v$ makes $\psi_1$ false and $\psi_2$ true; then, $v$ makes $\varphi$ false (according to the truth table of $\leftrightarrow$) and $\#^\lnot_\varphi + \bot^v_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}  + \bot^v_{\psi_1} + \bot^v_{\psi_2} = (\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}) + (\#^\lnot_{\psi_2} + \bot^v_{\psi_2})$ is odd because $\#^\lnot_{\psi_1} +  \bot^v_{\psi_1}$ is odd and $\#^\lnot_{\psi_2} +  \bot^v_{\psi_2}$ is even. $\qquad\qquad\qquad\square$

From the lemma above, it easily follows the syntactic carachterization of tatulogies in the fragment $\{\lnot, \leftrightarrow\}$: a formula $\varphi$ whose only connectives are $\lnot$ and $\leftrightarrow$ is a tautology if and only if $\lnot$ and each statement letter occur an even number of times in $\varphi$.

Proof.

\begin{align} \varphi \text{ is a tautology } &\iff \text{ every valuation } v \text{ makes } \varphi \text{ true }
\\
&\overset{\text{by lemma}}{\iff} \text{ for every valuation $v$, the sum of the number of}
\\
&\qquad \quad \text{occurrences of $\lnot$ in $\varphi$ and the number of occurrences}
\\
&\qquad \quad \text{of statement letters in $\varphi$ made false by $v$ is even }
\\
&\iff \lnot \text{ and each statement letter occur an even number of times in $\varphi$.}
\end{align}
The last equivalence holds because the number of occurrences of $\lnot$ in $\varphi$ does not depend on $v$, and because of the remark below.
In particular, the remark  below excludes the possibility that the sum of the number of occurrences of $\lnot$ in $\varphi$ and the number of occurrences of statement letters in $\varphi$ made false by $v$ is even because the two numbers are odd.
$\qquad\qquad\qquad \qquad\qquad \square$

Remark. Let $\varphi$ be a formula and $p$ a statement letter. Let $v$ be a valuation and $\overline{v}$ be a valuaiton with the same values as $v$ for every statement letter but $p$. The number of occurrences of statement letters in $\varphi$ made false by $v$ and the number of occurrences of statement letters in $\varphi$ made false by $\overline{v}$ have the same parity if and only if the number of occurrences of $p$ in $\varphi$ is even.

Thanks to the lemma above, in the fragment $\{\lnot, \leftrightarrow\}$ you can provide not only a syntactic carachterization of tautologies but also syntactic carachterizations of contradictions (a formula that is always false, in every row of its truth table) and of fomulas that are neither tautologies nor contradictoins.
Indeed, let $\varphi$ be a formula whose only connectives are $\lnot$ and $\leftrightarrow$.

$\varphi$ is a contradiction if and only if $\lnot$ occur an odd number of times in $\varphi$, and each statement letter occur an even number of times in $\varphi$;

$\varphi$ is neither a tautology nor a contradiction if and only if some statement letter occur an even number of times in $\varphi$.


问:
On applying the mean value theorem for multiple functions
Is there some rule about multiple functions(all continuous and differentiable in some specific interval) having their average slopes (as obtained using LMVT) be equal to the slope at the same point? As in, are the c-values of all functions in a specific interval the same?

Multiple problems I've come across seem to rely on this principle being true. If it is, how is it proved?

答:
Consider $f(x)=\sin(x)$ and $g(x)=\cos(x)$ on $[0,2 \pi]$. You can actually find out the points where the slope is equal to the average slope and convince yourself that there is no common point for all.

There is one result in this direction, which looks like a generalisation: The Cauchy's mean value theorem. It says that if  if the functions $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then, there exists some $c \in (a,b)$ such that
$$
f'(c) (g(b)-g(a))=g'(c)(f(b)-f(a))
$$

When, none of the terms above is zero, this relation can be written as
$$
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}
$$
which is similar but not the samea s what you want.


答: $\mathcal{C}([a,b],\mathbb{R})/S$ defined with equivalence $\left(x(t) \sim y(t)\right) \Leftrightarrow \left( x(t) - y(t) = \mathrm{const}\right)$, so $\mathcal{C}([a,b],\mathbb{R})/S$ consists of classes of functions, whose difference is constant (something like in case of antiderivatives). You may think of each class as set of functions $x(t) + C$ for fixed $x(t) \in \mathcal{C}([a,b],\mathbb{R})$. So, in each class you can find function that is zero in $t = a$ and vice versa, for each function $f(t) \in \mathcal{C}([a,b],\mathbb{R})$ with $f(a) = 0$ there is a class of equivalent functions if form of $f(t) + C$. So, this mapping is isomorphism you are looking for.


问:
RSA: What message will Alice receive?
In RSA, Alice chooses $p=47$, $q=57$, public key ($n=2679$, $e=11$). When Bob sends the message $m=3$, what is the message that Alice will read?</p>

答:
It's not entirely clear what the question is asking, especially because (as noted in the comments), Alice has made a mistake. I suspect that the point of the exercise is to drive home the relevance of picking $p$ and $q$ which are primes (or, more precisely, of correctly computing $\varphi(n)$), and that we're assuming "academic" RSA rather than "real-world" RSA.

Bob will send $3^{11} \pmod{2679}$ which is $333$.

If Alice computes $d$ such that $11d = 1\pmod{(p-1)(q-1)}$, then what is $333^d \pmod{2679}$? (Hint: it isn't $333$. I suspect that this is the answer you're looking for).

On the other hand, if Alice was deliberately using a three-prime RSA (with the horrifically small $3$ as one of them), she will compute $d$ such that $11d = 1 \pmod{\textrm{what?}}$ and get $333^d = 3 \pmod{2679}$.

Title: RSA: What message will Alice receive?


问:
Minimum integer $m$ such that $\lim_{(x,y)\to(0,0)} \frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\in \mathbb{R}$
Find the minimum $m\in\mathbb{N}$ such that
$$\lim_{(x,y)\to(0,0)} \frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\in \mathbb{R}$$
My attempt: lets consider the modulus of the limit, so it is
$$0 \leq \lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}\leq\lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{\sqrt{x^2}}=\lim_{(x,y)\to(0,0)} \frac{|x|^{\frac{m}{3}}|x-y|}{|x|}$$
$$=\lim_{(x,y)\to(0,0)} |x|^{\frac{m}{3}-1}|x-y|=\lim_{(x,y)\to(0,0)} |x|^{\frac{m-3}{3}}|x-y|$$
Since $|x-y| \to 0$ as $(x,y) \to (0,0)$, the only problem can occur when $|x|^{\frac{m-3}{3}}$ goes to the denominator as $m$ varies; so it must be $m-3 \geq 0 \iff m \geq 3$; so if $m \geq 3$ the limit is finite and it is $0$. Is this correct?

Another question: if I try to solve this with polar coordinates I find another value, so one way must be wrong. Let $x=r \cos t$ and $y=r \sin t$, with $r \geq 0$ and $0 \leq t <2\pi$, considering the modulus of the limit it is
$$0 \leq \lim_{r \to 0^+} \frac{|r \cos t|^{\frac{m}{3}}|r\cos t-r \sin t|}{r}0 = \lim_{r \to 0^+} \frac{r^{\frac{m}{3}}|\cos^{\frac{m}{3}} (t)|r |\cos t- \sin t|}{r}=\lim_{r \to 0^+} r^{\frac{m}{3}}|\cos^{\frac{m}{3}}t||\cos t- \sin t|$$
$$\leq \lim_{r \to 0^+} r^{\frac{m}{3}}|\cos t|(|\cos t|+|\sin t|)\leq \lim_{r \to 0^+} r^{\frac{m}{3}}\cdot1\cdot(1+1)=2\lim_{r \to 0^+} r^{\frac{m}{3}}$$
And this limit is finite if $\frac{m}{3} \geq 0 \iff m \geq 0$; where is my mistake? I suspect that it is when I suppose that $|x|^{\frac{m-3}{3}}$ can go to the denominator, because the estimation doesn't give informations for $\frac{m-3}{3} \leq 0$ so I must study what happens for $\frac{x^{\frac{m}{3}}|x-y|}{\sqrt{x^2+y^2}}$ for $m\in\{0,1,2\}$. Thanks.

答: The problem occurs as even the term $|x-y|$ contains a term of $x$, which will offset the exponent of $|x|^{\frac{m-3}{3}}$ by $1$. If we take this into account, we would get $|x|^{\frac{m}{3}}$, and so $m\geq 0$.


问:
Proving $\int_{[0,1] \times [0,1]} 1_S$ does not exist
Let $S = \{(x,y)\in \mathbb{R}^2 | x$ rational, $0<x<1$, and write $x = p/m$ in lowest form, $y = k/m, k = 1, \cdots, m-1 \}$, i.e. $S=\bigcup_{m\in\Bbb N}\{(p/m,\,k/m)|k,\,p\in\{1,\,\cdots,\,m-1\},\,(p,\,m)=1\}$. I am trying to prove that $\int_{[0,1] \times [0,1]} 1_S$ does not exist, where $1_S(x,y) = 1$ if $(x,y) \in S$ and $1_S(x,y) = 0$ otherwise.

I thought the set of discontinuities of $1_S$ is the set $S$ itself, and since $S$ is a countable set, $1_S$ is integrable. Can someone point out where I'm wrong? Thanks!

答:
Actually, both $S$ and $[0,1]^2\setminus S$ are dense in $[0,1]^2$, therefore the set of discontinuities of $1_S$ is $[0,1]^2$.

Density of $[0,1]^2\setminus S$ is easy and you seem already persuaded of that.

For density of $S$, let $(x,y)\in (0,1)^2$. Notice that for any $\alpha\in (0,1)$ and for all $\varepsilon>0$, there is a rational number $q$ in the form $\frac{n}{2^m}$ such that $\left\lvert q-\alpha\right\rvert<\varepsilon$. Also observe that we may find one such that $n$ is odd (and therefore such that $\frac{n}{2^m}$ is reduced) and, for all $N\in\Bbb N$, we may find one such that $m>N$. Putting all these together with the appropriate care, I state that for all $\alpha\in(0,1)$, for all $\varepsilon>0$ and for all $N\in\Bbb N$ there is some rational number $q\in(0,1)$ such that $\lvert q-\alpha\rvert<\varepsilon$ and such that there is some $m>N$ such that $2^mq$ is an odd natural number. And this is rather easy to prove.

Now, let $\varepsilon>0$ and consider $q_y=\frac{2n_y+1}{2^{m_y}}$ a rational number such as above obtained for $\alpha=y$ and $N=1$. Moreover, consider $q_x=\frac{2n_x+1}{2^{m_x}}$ a rational number obtained as above with $\alpha=x$ and $N=m_y$. Notice that $$q_y=\frac{2^{m_x-m_y}(2n_y+1)}{2^{m_x}}.$$

$2^{m_x-m_y}(2n_y+1)$ is a positive natural number because $m_x>m_y$ and, moreover, $2^{m_x-m_y}(2n_y+1)<2^{m_x}$ because $0<q_y<1$. Since $\frac{2n_x+1}{2^{m_x}}$ is reduced, we have that $(q_x,q_y)\in S$. Also, $$\lVert (q_x,q_y)-(x,y)\rVert\le \lvert q_x-x\rvert+\lvert q_y-y\rvert<2\varepsilon$$

This proves that $\overline S\supseteq(0,1)^2$, and therefore $\overline S=[0,1]^2$.


问:
If $W$ and $W^{\perp}$ are subspaces of $V$, both with orthonormal basis, then $V = W \oplus W^{\perp}$
Let $W$ be a subspace of $V$, $W^{\perp}$ it's orthogonal complement in $V$, and say that both have orthonormal basis, and that $V$ is a finite dimensional inner product space, then $V = W \oplus W^{\perp}$.

I can see that $W \cap W^{\perp} = \{0\}$ but can someone show me why $\mathrm{span}\{W,W^{\perp}\}=V$?

Thanks

答:
Take $v\in V$. If $\{w_1,\ldots,w_n\}$ is an orthonormal basis of $W$, then

$$w=\langle v,w_1\rangle w_1+\langle v,w_2\rangle w_2+ \dots + \langle v,w_n\rangle w_n\in W.$$

And $v-w$ is orthogonal to each $w_k$ and therefore $v-w\in W^\perp$. So,$$v=w+(v-w)\in\operatorname{span}\{W,W^\perp\}.$$


问:
What do cones have to do with quadratics? Why is $2$ special?
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

$$\rule{5cm}{0.4pt}$$

EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

$$\rule{5cm}{0.4pt}$$

EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

$$\rule{5cm}{0.4pt}$$

EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

$$\rule{5cm}{0.4pt}$$

EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)

答:
A reason "2" is special for physics is Newton's second law, which relates force to acceleration (not velocity) and that's a second derivative.  Well, there's also the role of "2" in inverse square laws.

The reason "2" is special in geometry through quadratic forms in several variables is that quadratic forms in several variables have a few nice properties.

Every quadratic form can be diagonalized to remove all cross terms, so you can focus on the case of diagonal quadratic forms $a_1x_1^2 + \cdots  + a_nx_n^2$.  (Strictly speaking this is not true for quadratic forms over fields of characteristic $2$, but you don't get geometric intuition from characteristic $2$.)  In contrast to that, cubic forms may not be able to be diagonalized, even over $\mathbf C$.  For example, the cubic form $y^2z - x^3 + xz^2$ (whose zero set in dehomogenized form is given by the equation $y^2 = x^3 - x$) can't be diagonalized over $\mathbf C$: see my comments here

Every nonsingular quadratic form has a large group of automorphisms thanks to the construction of reflections. It's called the orthogonal group of the quadratic form. In contrast to that, the "orthogonal group" of a higher-degree homogeneous polynomial $f(\mathbf x)$ (that means the group of linear transformations $A$ preserving the polynomial: $f(A\mathbf x) = f(\mathbf x)$) is often finite, e.g., the only isometries of $x_1^n + \cdots + x_n^n$ for $n \geq 3$ are coordinate permutations and multiplying coordinates by $n$th roots of unity.

Fundamental to geometry is the concept of orthogonality, which you want to be a symmetric bilinear relation: $v \perp w$ if and only if $w \perp v$, and if $v \perp w$ and $v \perp w'$ then $v \perp (ax + a'w')$ for all scalars $a$ and $a'$.  This suggests looking at bilinear forms $B(v,w)$ on a vector space and asking when the relation $B(v,w) = 0$ (an abstract version of "$v \perp w$") is symmetric.  It turns out this happens if and only if $B$ is symmetric or alternating.  The first case is, outside of characteristic $2$, closely related to studying the quadratic form $Q(v) = B(v,v)$.


问:
Intuition of countable sets using discrete sets
I'm trying to think of a way to intuitively explain myself what a countable set means.

So far as I understand it, in very simple words, a set $S$ is countable iff you can "name" or "list" down each and every member of $S$, so in a way, you've "counted" the elements of $S$ by "naming" them.

This intuition is directly inspired from the definition that $S$ is countable iff $S = \varnothing$ or $\exists$ a surjection $f : \mathbb{N} \to S$ so $f$ is a sequence in $S$ and thus you list down the members of $S$ as the sequence $\left(f(n) \right) = \left( f(1),f(2), \ldots, f(n), \ldots \right)$

Before, I go ahead. Let me know if this "way" of thinking about countable sets is correct or does it have any flaws?

Following that intuition, would it be correct to say that $S$ is countable if $S$ is a set of discrete points. What I mean by discrete is that there exists a neighborhood $N(s)$ around each members $s$ of $S$ such that $N(s) \cap \{s \} = \{s\}$.

Now, I know that "discreteness" is a topological property and it has nothing to with countablity, but hey, I'm trying to look for an intuition so think of $S$ (in this case) as some kind of metric space or something.

The only flaw, I see with the latter intuition is that if $(S,d)$ is a discrete metric space, i.e. $d(x,y)=1$ if $ x=y$ and $0$ otherwise, then any set $S$ now becomes countable, for e.g. if $S= \mathbb{R}$ but cardinality of $\mathbb{R}$ is not the same as that of $\mathbb{N}$, so this intuition changes the cardinality itself. I mean cardinality is cardinality, the number of elements in a set shouldn't change no matter what, right? I feel like discreteness $\Rightarrow$ countability but converse doesn't hold maybe or something. So this intuition is lacking something, it would be very helpful if someone could complete or fix this intuition for me. Thanks!

Any other kinds of intuitions are also very welcomed.

答:
Before, I go ahead. Let me know if this "way" of thinking about countable sets is correct or does it have any flaws?

Yes, this is correct. It's right there in the name: a set is countable if you can count it.

Following that intuition, would it be correct to say that $S$ is countable if $S$ is a set of discrete points.

This is not quite right. The problem is that any set, arbitrarily large, can be equipped with the discrete metric, or the discrete topology.

However, what is true is that if a subspace of $\mathbb{R}^n$ equipped with the subspace topology is discrete, then it is at most countable, and this is a nice exercise. So, loosely speaking, uncountable discrete spaces are "too big to fit into Euclidean space."

Any other kinds of intuitions are also very welcomed.

I wrote this in a comment but it's worth writing in the answer: a pretty good intuition is that a set is countable if its elements can be described using "a finite amount of data." In fact you can prove that the set of finite words on any finite alphabet (an "alphabet" is just a set and a "word" is a finite list of elements of that set) is countable, which is a very nice exercise and a pretty good way of making rigorous sense of the meaning of "a finite amount of data." For example the set of English sentences, or Python programs, or $\LaTeX$ formulas, is countable.

On the other hand, speaking somewhat informally, the real numbers are uncountable because to specify an arbitrary real number you need to specify infinitely many digits, which is "an infinite amount of data." Checks out!

The uncountability of $\mathbb{R}$ has the following somewhat disturbing implication: only countably many real numbers are describable by English sentences, Python programs, or $\LaTeX$ formulas. You can check out the Wikipedia articles on computable numbers and definable numbers for more on this.


问:
Induced map on Eilenberg-MacLane space
Let $X$ be an $n-1$ connected space. Why does a map $X\rightarrow K(\pi_n(X),n)$ that induces an isomorphism on $\pi_n$ exist and what is this map?</p>

答: As stated, the statement is false.  For example, let $X$ be the Hawaiian earring.  Then $X$ is 0-connected.  But $X$ can't have a map to an Eilenberg-Maclane space that's an isomorphism on $\pi_1$, because as a compact space it would have to have compact image, which would necessarily have finitely-generated $\pi_1$ in a CW complex like an Eilenberg-Maclane space, and $X$ has infinitely generated fundamental group.  The statement is true for if $X$ is a CW-complex, as wckronholm explains.</p>

Title: Induced map on Eilenberg-MacLane space


问:
Finding a subspace of orthogonal projection
I would like some help solving this question

P is an orthogonal projection that is defined by:

$$P:\Bbb R^3\to \Bbb R^3$$
$$P(v)=1/180\begin{pmatrix}144&72&0\\72&36&0\\ 0&0&180\end{pmatrix}·v$$
How do I find subspace $W$ so that $$P=P_w, \quad ?$$
I've searched everywhere online but i can't seem to find a solution to it.

答:
Let $A$ the associated matrix.

We verify that $A^2=A$.

The null space of A is $\text{span}\left\{\begin{pmatrix}-10\\20\\0\end{pmatrix}\right\}$

The column space of A is $\text{span}\left\{\begin{pmatrix}0\\0\\1\end{pmatrix}, \begin{pmatrix}4/5\\2/5\\0\end{pmatrix}\right\}$

They are orthogonal.

P is the orthogonal projection on $W=\text{span}\left\{(0,0,1),(4/5,2/5,0)\right\}$


问:
Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd.
I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.</p>

答:
Since $x$ is odd, $x=2n+1$ for some integer $n$, and therefore $(x-2)^2=(2n-1)^2=4n^2-4n+1$. Thus,

$$\left\lfloor\frac{(x-2)^2}4\right\rfloor=\left\lfloor n^2-n+\frac14\right\rfloor=n^2-n\;.$$

And $n=\dfrac{x-1}2$, so the expression reduces to

$$\left(\frac{x-1}2\right)^2-\frac{x-1}2=\frac{x^2-4x+3}4\;.$$