问:
A non-abelian group of order $6$ is isomorphic to $\mathcal{S}_3$.
Let $G$ be a group of order $6$ and $H$ be a non-normal subgroup of $G$ such that $|H|=2$.

Consider Action of $G$ on set of left cosets of $H$ by left multiplication. Let $\{g_iH :1\leq i\leq 3\}$ be cosets of $H$ in $G$. Then the maps : $\phi : G\rightarrow \mathcal{S}(G/H)$, $g\mapsto \sigma_g$ (such that $\sigma_g: G/H\to G/H$, $g_iH\mapsto gg_iH$) be a morphism of group of kernel
$$
\ker (\phi)=\bigcap_{i=1}^{3} g_iHg_i^{-1}
$$

Problem I want to show that $\ker (\phi)=\{0\}$.

Any ideas are welcome.

答: Hint: You know that if $H$ isn't normal, then there is more than one conjugate subgroup. This means that for at least one $i$, $g_iHg_i^{-1} \neq H$. How many conjugate subgroups will you then have given that |H| = 2 and what will $\bigcap_{i=1}^{3} g_iHg_i^{-1}$ be?


答:
Since $\lvert G\rvert=2p$ for prime $p=3$, we can use the following theorem:

Theorem 7.3: Let $G$ be a group of order $2p$, where $p$ is a prime greater than $2$. Then $G$ is isomorphic to $\Bbb Z_{2p}$ or $D_p.$

(Here $D_p$ is the dihedral group of order $2p$.)

The theorem is fairly standard; I quoted it from page 150 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)", but you're likely to find it in any decent textbook on group theory or abstract algebra.

Note that $S_3\cong D_3$.

Since $G$ is nonabelian (as it has a non normal subgroup), then, we have $G\cong S_3$.

ContentLicense: CC BY-SA 4.0

Title: A non-abelian group of order $6$ is isomorphic to $\mathcal{S}_3$.

Body:
Let $G$ be a group of order $6$ and $H$ be a non-normal subgroup of $G$ such that $|H|=2$.

Consider Action of $G$ on set of left cosets of $H$ by left multiplication. Let $\{g_iH :1\leq i\leq 3\}$ be cosets of $H$ in $G$. Then the maps : $\phi : G\rightarrow \mathcal{S}(G/H)$, $g\mapsto \sigma_g$ (such that $\sigma_g: G/H\to G/H$, $g_iH\mapsto gg_iH$) be a morphism of group of kernel
$$
\ker (\phi)=\bigcap_{i=1}^{3} g_iHg_i^{-1}
$$

Problem I want to show that $\ker (\phi)=\{0\}$.

Any ideas are welcome.


问:
How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms?
So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were doing problems, we came about one that I have no idea how to solve. It seems to be a straight forward "here's an equation, solve for $x$." Just one problem, I ended up not being able to solve for $x$, making me feel embarrassed since I was trying to help her and I am supposed to be the one who is "good at math." Anyway, it would be a grreat help if anyone can help me out on this. The equation is....

$$4\sqrt{x-3} - \sqrt{6x-17} = 3$$

The answer comes out to be $x = 7$ because I did it on my calculator. I first tried to just square each side to get rid of the square roots. But once I "FOILed" the left side, there were still square roots and things didn't look good for me. By the way, I am sorry I don't know how to format this equation correctly.

答:
Do not square!

Move one root over equality sign.

A strictly increasing and a strictly decreasing functions can cross at no more than 1 point. It is easy to prove: 2nd crossing point will violate condition of functions being strictly increasing / decreasing.

Here is my answer to same question.

ContentLicense: CC BY-SA 3.0


问:
Convergence of series $\sum{(a_n+k)}$ where k is a positive number and $a_n$ is a positive sequence
To prove that the series $\sum{(a_n+k)}$ converges or diverges according as the series $\sum{a_n}$ converges or diverges, where $a_n$ is sequence of positive terms and k is any positive real number.

According to me a series $\sum{a_n}$ of positive terms such that $\\lim_{n\rightarrow \infty} a_n \neq 0$ diverges.
So, if $\sum a_n$ converges then $\\lim_{n\rightarrow \infty} a_n = 0$. If we consider
$$\lim_{n\rightarrow \infty} (a_n +k) =\lim_{n\rightarrow\infty} a_n+\lim_{n\rightarrow\infty}k\\
=\lim_{n\rightarrow\infty} a_n+ k\\
= 0+k\neq 0$$
Thus, series $\sum (a_n+k)$ diverges. Which is opposite of what I am supposed to prove.

Please point out mistakes if any and provide the proof of question.

答:
You've got the right idea. The result is false: $\sum a_n+k$ diverges regardless of whether $\sum a_n$ converges or diverges, given that $k>0$ and each $a_n>0.$

This is because $a_n+k>k>0$ for all $n,$ so it is impossible for $a_n\to 0$ as $n\to\infty.$

Perhaps a more straightforward way to see the divergence is by direct comparison to the series $$\sum_{n=1}^\infty k,$$ which clearly diverges to $+\infty$ for any $k>0.$


问:
$ \sum_{i=1}^{10} x_i =5 $, and $ \sum_{i=1}^{10} x_i^2 =6.1 $ ; Find the largest value of these ten numbers

Ten non-negative numbers are such that  $ \sum_{i=1}^{10} x_i =5 $,
and $ \sum_{i=1}^{10} x_i^2 =6.1 $. What is the greatest value of the
largest of these numbers?

It presents a lot of confusion to me, but I think the largest number maybe a integer. Please help me with procedure.

Source : Moscow Institute of Physics and Technology admission assessment for computer Science

答:
A General Problem

Suppose
$$
\frac{\sum_{i = 1}^n x_i}{n} = a
$$
and
$$
\frac{\sum_{i = 1}^n x_i^2}{n} = b^2
$$
Set $y_i = x_i - a$. Then (as you can check)
$$
\frac{\sum_{i = 1}^n y_i}{n} = 0
$$
So
$$
b^2 = \frac{\sum_{i = 1}^n (y_i + a)^2}{n} = \frac{\sum_{i = 1}^n  y_i^2 + 2ay_i + a^2}{n} = \frac{\sum_{i = 1}^n y_i^2}{n} + 2a \frac{\sum_{i = 1}^n y_i}{n} + \frac{a^2}{n} \sum_{i = 1}^n 1
$$
Setting
$$
c^2 = \frac{\sum_{i = 1}^n y_i^2}{n}
$$
We obtain
$$
b^2 = c^2 + 2a(0) + a^2
$$
So
$$
c^2 = b^2 - a^2
$$
Observe that such a sequence $y_i$ achieves its largest possible element precisely when $x_i$ achieves its largest possible element.

The upshot of all of this is that we've reduced the problem to finding the largest possible value of a sequence of numbers $y_i$ with a mean of $0$ (i.e. $\frac{\sum_{i =1}^n y_i}{n} = 0$) and variance of $c^2$ (i.e. $\frac{\sum_{i = 1}^n y_i^2}{n} = c^2$).

Intuitively, it's clear that the maximum should occur when $y_1 = \ldots = y_{n -1} = y$. Applying the mean and variance condition and performing a straightforward calculation yields $y_n = \sqrt{n - 1} c$. It remains to check that this is the best we can do.

Suppose for contradiction $y_n > \sqrt{n - 1} c$. Then
$$
\sum_{i = 1}^{n - 1} y_i = -y_n
$$
and
$$
\sum_{i = 1}^{n - 1} y_i^2 = nc^2 - y_n^2
$$
Thus
$$
\frac{\sum_{i =1}^{n - 1} y_i}{n -1} = -\frac{y_n}{n - 1}
$$
and
$$
\frac{\sum_{i = 1}^{n - 1} y_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1}
$$
Set $z_i = y_i + \frac{y_n}{n - 1}$. Then
$$
0 \leq \frac{\sum_{i = 1}^{n - 1} z_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1} - \left(\frac{y_n}{n - 1}\right)^2 < \frac{nc^2 - (n - 1)c^2}{n -1} - \frac{(n - 1)c^2}{(n - 1)^2} = 0
$$
which is absurd. Thus $y_n = \sqrt{n - 1} c$ is the best we can do.

In terms of $x_n$, $a$, and $b$ this is
$$
x_n = a + \sqrt{(n - 1)(b^2 - a^2)}
$$

Your Problem

In your problem $a = .5$ and $b^2 = .61$. Plugging in, you find that
$$
x_n = 2.3
$$
is the best you can do.


问:
Why can I change the sign of this expression?
I have an equation; $2\vec{k}\cdot\vec{G}=-G^2, \vec{G} = u_1\vec{b_2} + u_1\vec{b_2} + u_3\vec{b_3}$ where $u_1, u_2, u_3$ are integer.

I saw the statement that the equation $2\vec{k} \cdot \vec{G} = -G^2$ is equal to $2\vec{k} \cdot \vec{G} = G^2$ because $-\vec{G}$ is also $\vec{G}$.

But I can't understand why they are same expression.

I need more explanation.

答:
From the material:

When the scattering vector ${\bf \Delta k}$ is equal to some particular reciprocal lattice vector ${\bf G}$, $${\bf \Delta k} = {\bf G} \tag{21}$$ [...]
From (21) we found ${\bf \Delta k} = {\bf G}$ or ${\bf k} + {\bf G} = {\bf k'}$ so that the diffraction condition is written as $({\bf k} + {\bf G})^2 = k^2$ or $$2{\bf k}\cdot{\bf G} + G^2 = 0 \tag{22}$$ [...]
When ${\bf G}$ is a reciprocal lattice vector, so is $-{\bf G}$ and with this substitution we write $$2{\bf k}\cdot{\bf G} = G^2 \tag{23}$$

We first assume that ${\bf \Delta k}$ is a reciprocal lattice vector and we denote it by ${\bf G}$, but then, since ${\bf -G}$ is a reciprocal lattice vector as well, we change the notation to ${\bf \Delta k} = -{\bf G}$. Therefore, replacing ${\bf G}$ with ${\bf -G}$ in $(22)$, we arrive at $(23)$.

You can think about it like this: we first denote ${\bf \Delta k} =: {\bf G}$ and derive $(22)$ and say ${\bf H} := -{\bf G}$, replace ${\bf G}$ in $(22)$ with ${\bf -H}$, then rename ${\bf H}$ to ${\bf G}$ so that in $(23)$, ${\bf G}$ is not ${\bf \Delta k}$ anymore, but $-{\bf \Delta k}$.


问:
Let $p(z) = z^6 + 9 z^4 + z^3 + 2z + 4$ show that the two zeros of $p(z)$ lying outside the unit circle satisfy $\{ |z \pm 3i| < 1/10$
Let $p(z) = z^6 + 9 z^4 + z^3 + 2z + 4$, I want to show that the two zeros of $p(z)$ lying outside the unit circle satisfy $ |z \pm 3i| < 1/10$.

By the Rouche's theorem, $p(z)$ has four zeros contained in the unit circle. Therefore, the other two zeros must lie outside of the unit circle.

Here is my attempt:
let $f(z) = z^6 + 9z^4$ and $g(z) = z^3 + 2z + 4$.
Now $f(z)$ clearly has two zeros at $z = 3i$ and $z = - 3i$ in the domain $D = \{ |z + 3i| < 1/10\}$.
If I can show the $|g(z)| < |f(z)|$ on $\partial D$, then I can apply the Rouche's theorem.
So that $p(z) = z^6 + 9 z^4 + z^3 + 2z + 4$ has same number of zeros with $f(z) = z^6 + 9z^4$ on the domain $D = \{ |z + 3i| < 1/10\}$.
However, seems like I can't prove the inequality $|h(z)| < |f(z)|$ required by the Rouche's theorem.

Can anyone provide me some help please?

Body:
If $|z+3i|=0.1$ then $$|f(z)| =|z|^4 |z-3i ||z+3i| =0.1 |z-3i| |z|^4 \geq 0.1\cdot 5.9 \cdot (2.9)^4 >41$$
and
$$|h(z)|=|z^3 +2z +4|\geq |z|^3 +2|z| +4 =(3,1)^3 +6.2 +4 <40$$
So $$|f(z)|> |h(z)|$$ on $|z+3i|=0.1 $


问:
Understanding the "Guess and Check" definition of $NP$.
Aside from the normal complexity definition of $NP:= \bigcup_{k \geq 1} \ NTIME$, we have an alternative definition ("Guess and Check") of NP

A language $L \subseteq \Sigma^*$ is in $NP$ , if there exist a
polynomial $p: \mathbb{N} \to \mathbb{N}$ and a deterministic Turing
Machine $M$, so that for every $x \in \Sigma^*$ we have:

$$x \in L \iff \exists u \in \Sigma^{p(\vert x \vert)} : \langle x,u \rangle \in
T(M)$$

If $\langle x,u \rangle \in T(M)$ then $u$ is considered as a certificate of
$x$.

So essentially we use the string $u$ to help us determine a solution but I am not sure if I understand this definition properly.

答:
For the sake of simplicity, I am going to be a little bit informal. This is according to a nice chapter in "Graph Theory" by Bondy and Murty.

We define a decision problem as a question whose answer is either "yes" or "no". A decision problem belongs to $\mathcal{P}$ if every instance of the problem can be solve in polynomial time by some algorithm. It belongs to $\mathcal{NP}$ if, given any instance of the problem whose answer is "yes", there is a certificate validating this fact which can be checked in polynomial time; such a certificate is said to be succinct. (If the answer is "no" and there is such a certificate which can be validated in polynomial time then the problem is in $co-\mathcal{NP}$).

Now, using the formal definition that you have provided. Our turing machine $M$ accepts every word $x \in L$ if and only if there is a word $u$ of length bounded from above by $p(|x|)$ and $M$ accepts the pair $<x,u>$. The point is that $M$ accepts $x$ using $u$ in polynomially many steps, since $u$ has size at most $p(|x|)$.

For example, solving a hereditary property based problem on a simple, undirected graph $G$ (which is a property which must hold for each induced subgraph), we have to check each induced subgraph $H$ of $G$. This boils down to enumerate all subsets of the vertex set $V$ and if $|V|=n$ there are $2^n$ possible subsets. But if a solution is provided, it could be essentially faster to check if this solution is correct.

Furtheremore, the term "Guess and check" is often used in relationship with non-deterministic Turing machines. We guess a solution of our problem instance (which is generated in a non-deterministic way) and then verify its solution in polynomially many steps by some deterministic Turing machine.

In other words: $L$ is in $\mathcal{NP}$ if $x \in L$ can be verified by an algorithm encoded by $u$ whereby $u$ has length at most $p(|x|)$, and this holds for all words of $L$.

Example: Take an instance of the SAT problem. Is $(x_1 \land \lnot x_2) \lor (x_1 \land x_3) \lor (\lnot x_2 \land x_3)$ satisfiable? Now, perform the following steps:

Guess $n$-bit vector,

Verify the guessed vector, simply by calculating the formula of the given instance.

SAT is obviously in $\mathcal{NP}$.


问:
Find the unitary operator responsible for finite translation in time.
Show that if the Hamiltonian depends on time and $[H_{t1}, H_{t2}] = 0$, the time development operator is given by:

$$U(t) = exp [ \frac{-i}{\hbar} \int_{0}^{t} dt' H(t')]$$

I am having trouble to prove it:

Since $U(dt) = 1 - \frac{iHdt}{\hbar}$,

$U(t) = U(dt1)U(dt2)U(dt3)U(dt4)U(dt5)...$

$U(t) = 1 - \sum_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$

The sum can be interpreted as an integral

$U(t) = 1 - \int_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$

$U(t) = 1 + \frac{-i}{\hbar} \int_{i = 0}^{N}{H(t') dt'}$

But this is wrong, where is the error (i really think that the error is going from the sum to the integral, but i don't know how to fix it)?

答:
The problem is when you wrote

$$U(t) = 1 - \sum_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$$

Instead you should have ($\hbar=1$)

$$ U(t)=\prod_j(1-iH_jdt_j)$$

Which is useful if $H$ is time independent. Since here $H=H(t)$, we try a different approach. Since you know the infinitesimal form

$$ U(dt)=1-iHdt$$

You can find the equation obeyed by $U(t)$ as follows

$$ U(t+dt)=U(dt)U(t)=(1-iHdt)U(t)$$

$$ U(t+dt)-U(t)=-iHdtU(t)$$

$$ \frac{U(t+dt)-U(t)}{dt}=-iHU(t)$$

The left is simply a derivative when $dt \rightarrow0$

$$ \partial_t U(t)=-iH(t)U(t)$$

Since we are told that $[H(t_1),\ H(t_2)]=0$, everything in the above differential equation commutes, and we may integrate directly. Separating 'variables', we have

$$ U^{-1}\partial_tU=-iH$$

The left hand side is a logarithmic derivative, so integrating is simple

$$ \ln(U(t))=-i\int^t_0 dt' \ H(t')$$

Where we have used $U(0)=1$. Exponentiating both sides yields the answer

$$ U(t)=\exp\left( -i\int_0^t dt' \ H(t')\right) \tag*{$\blacksquare$}$$


问:
Formula for angle in triangle
Given a triangle $ABC$ with edges $a$, $b$, and $c$ and corresponding angles
$\alpha$, $\beta$ and $\gamma$. Assume, the values of $a$, $b$ and $\gamma$ are given. Does there exist a direct theorem (like the law of cosines, the law of sines or the law of tangents) which describes the relationship between these parameters and $\alpha$?

Obviously, one can use the law of cosine to calculate $c$
and then apple the law of sine to solve the problem.
But does there exist a specific formula for the relationship which can be applied without a step in between? Maybe, someone could give me some advice. I have already read the entire article on wikipedia about trigonometry. Thanks in advance!

答:
As you already know by cosine rule we can find $c$ as
\begin{align}
c&=\sqrt{a^2+b^2-2ab\cos\gamma}
.
\end{align}

However, it's not a good idea to use the sine rule to find $\alpha$,
since in general, $\alpha$ could be obtuse.
It's better to use known identity

\begin{align}
b&=a\cos\gamma+c\cos\alpha
\end{align}

to get the final expression for $\alpha$
in terms of $a,\ b$, and $\gamma$:

\begin{align}
\alpha&=\arccos\left(
\frac{b-a\cos\gamma}{\sqrt{a^2+b^2-2ab\cos\gamma}}
\right)
.
\end{align}

Title: Formula for angle in triangle


问:
Collecting coupons which arrive as a Poisson process
I'm trying to answer the following question:

"A person collects coupons one at a time, at jump times of a Poisson process $(N_t)_{t\geq 0}$ of rate $\lambda$. There are m types of coupons, and each time a coupon of type j is obtained with probability $p_j$, independently of the previously collected coupons and independently of the Poisson process. Let T be the time the complete set is obtained.

Show that $\mathbb{P}(T<t)=\prod_{j=1}^m(1-e^{-p_j\lambda t})$. let L be the number of coupons collected by time T. Show that $\mathbb{E}L=\lambda \mathbb{E}T$."

I'm able to find the distribution of T, by splitting the process into independent subprocesses of rate $\lambda p_i$, but am not sure where to go for the second part. It seems that explicit computation $\mathbb{E}L$ or $\mathbb{E}[L|T]$ is infeasible. Is it possible to use the fact that $T$ is a stopping time here and perhaps the strong Markov property?

Thank you.

答:
Hint: Try showing ${\rm E}[T]=(1/\lambda){\rm E}[L]$, noting that $1/\lambda$ is the mean of the waiting time...

It is also interesting to note the following.

Let's show using the formula for ${\rm P}(T \leq t)$ that ${\rm E}[L] = \lambda {\rm E}[T]$ in the case when all the $p_j$ are equal, that is $p_j = 1/m$, $j=1,\ldots,m$.

On the one hand, we have
$$
{\rm E}[L] = m\bigg(\frac{1}{1} + \frac{1}{2} +  \cdots  + \frac{1}{m}\bigg) = mH_m ,
$$
where $H_m$ is the $m$th harmonic number; see the section Calculating the expectation in Wikipedia's article on the Coupon collector's problem.
On the other hand,
$$
{\rm E}[T] = \int_0^\infty  {t\frac{d}{{dt}}P(T \le t)\,dt}  = \int_0^\infty  {tm(1 - e^{ - \lambda t/m} )^{m - 1} e^{ - \lambda t/m} \frac{\lambda }{m}\,dt} .
$$
Put $c=m/\lambda$ and $x=t$. Then,
$$
{\rm E}[T] = \int_0^\infty  {x\frac{m}{c}e^{ - x/c} (1 - e^{ - x/c} )^{m - 1}\, dx} .
$$
We know from Integral yielding part of a harmonic series that the last integral evaluates to $c\sum\nolimits_{k = 1}^m {(1/k)}  = cH_m $ (which, in view of that thread, is not an easy result). Hence,
$$
\lambda {\rm E}[T] = \lambda cH_m = mH_m,
$$
and finally
$$
{\rm E}[L] = \lambda {\rm E}[T].
$$


问:
Find all integer solutions for $x^2+y^2<a$
How do I find all integer solutions for the inequality $x^2+y^2<a$, where $a$ is a known integer? I can try pairs of numbers to find solutions one by one, but what's a generalised, systematic way to find all solutions?

答:
Going from $=$ t0 $<$ complicates things but here is one approach. A general "formula" is
$\quad (A=0\implies  B\le\lfloor\sqrt{a}\rfloor-1)$
$\qquad (B=0\implies  A\le\lfloor\sqrt{a}\rfloor-1) $

For $A^2+B^2=C^2\quad A\space $ takes the form $\space (2n+1),\space $ B takes the form $\space 4n,\space $ and $\space C\space $ takes the form $\space 4n+1\space $ where $\space n\in\mathbb{N}\quad$ but we will not use this form.

We can derive functions from Euclid's formula that will help find our solutions. The formula is
$$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$

But our modified form will be
$$A=m^2-n^2\qquad B=2mn\qquad \sqrt{a}=C>m^2+n^2$$
Solving for $\space n \space$ we get
$\qquad n<\sqrt{C-m^2}\quad\text{for}\quad  m\le\sqrt{C-1}$

We can now try values of $m$ within within the limits to find which, if any yield integers. For example
$$C=65\implies m \le \sqrt {65-1}=8$$

If we now plug $(m,n)$ back into the AB fuctions of Euclid's formula, we get the largest combinations of AB with squares that sum up to $\space a=C^2\space$ and we can know that anything less of either A or B will satisfy the equation. Here is the combination of $(m,n,A,B)$ that shows the largest $A$-value that qualifies: $\quad (8,1,63,16)\quad$

$$\text{So, for}\quad a=65^2,\\
A\{63,62,61,\cdots\}\implies B\in\{15,14,13,\cdots\}\\
\land\quad\\
A\{16,15,14,\cdots\}\implies B\in\{62,61,60,\cdots\}$$

Note: $n=1$ always yields the largest $A$-value for a given $m$-value and $n=m-1$ yields the larges $B$-value. It is simply easier to reverse AB for the high-limits here.


问:
Picture for the Straddle Lemma, in Bartle's Introduction to Real Analysis (2011 4 ed), 6.1.17, p 171?
Can anyone please picture the Saddle Lemma in Introduction To Real Analysis (2011 4 ed) by Robert G. Bartle & Donald R. Sherbert, Section 6.1, Exercise 17, p 171? I'm not asking proofs.

Let $f \colon I \to \mathbb{R}$ be differentiable at $c \in I$. Establish the Straddle Lemma. Give $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $u, v \in I$ satisfy $c-\delta(\varepsilon)<u\leq c \leq v < c+\delta(\varepsilon)$, then we have $\left\lvert f(v) - f(u) - (v-u)f^\prime(c) \right\rvert \leq \varepsilon (v-u)$.

[Hint: The $\delta(\varepsilon)$ is given by Definition 6.1.1. Subtract and add the term $f(c) - c f^\prime(c)$ on the left side and use the Triangle Inequality.]

答:
I'll try to explain it for the case $u\neq v$. The case where $u=v$ yields nothing fruitful.

I believe its geometric interpretation is: at $c$ where $f$ is differentiable, if you confine yourself to a small enough neighborhood of $c$ (which is dependent on $\varepsilon$, your "error tolerance") and pick any two $u$, $v$ in that neighborhood (subject to your conditions of course), the slope of the secant joining the points $(u,f(u))$ and $(v,f(v))$ won't be too different to the slope of tangent at $c$, i.e. the difference is at most $\varepsilon$, your error tolerance.

Here's your picture:

I've exaggerated the size of $\delta$ to make the drawing less cluttered. As you can see, the angle between the blue line (tangent) and the orange line (secant) is quite small, which translate to a small difference in slope. I hope this helps!


问:
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$
let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$

maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks

答:
Hint:

It suffices to prove that
$$\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \ge \left(\frac{x^{2n+1} + y^{2n+1}}{2}\right)^{\frac{1}{2n+1}}.$$
Since this inequality is symmetric and homogeneous, assume $y = 1$ and $x \ge 1$. It suffices to prove that
$$\ln (x^{n+1} + 1) - \ln (x^n + 1) \ge \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2}.$$
Take derivative.

Update

Let $f(x) = \mathrm{LHS} - \mathrm{RHS}$. We have
$$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$


问:
Let $F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$. Find $F'\left(x\right)$
Let $$F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$$ Find $F'\left(x\right)$

Had the lower limit been a function of $x$ I would have easily calculated. I am not very familiar with the Newton-Leibniz rule in two variables.  I do not know which formula to use.

答:
You can use the most general one. The below is from Wikipedia.

So actually because the lower limit of the integral is a function of $t$ and because you are differentiating with respect to $x$, you can treat $t^2$ as a constant. And you can change the integrating variable, and for the purpose of preventing confusion let's change it to $t'$. Notice that the $t$ inside the integral is independent of the $t^2$ in the lower limit.

$$F_t(x) = \int_{a(t)}^{b(x)} g(x, t') \ dt'$$

$$\frac{\partial F_t(x)}{\partial x} = g(x, b(x)) \times \frac{\partial b}{\partial x} \ + \ g(x, a(t)) \times \frac{\partial a(t)}{\partial x} + \int_{a(t)}^{b(x)} \frac{\partial g(x, t')}{\partial x} dt'$$

If we plug in everything we get

$$f_t(x) = \frac{3x^2}{\sqrt{x^2 + x^6}} - \ 0 \ -\frac{1}{2} \int_{t^2}^{x^3} (t'^4 + x^2)^{-3/2} dt'$$

I couldn't integrate the last term, so you may try it out.

What's important here is to think about the intuition behind the formula - think of it as a proper generalisation and combining of the two formulae:

$$\frac{d}{dx} \int_a^x f(t) dt = f(x) \Rightarrow \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) \times \frac{db}{dx} - f(a(x)) \times \frac{da}{dx}$$ (which is a simple use of chain rule)

$$\frac{d}{dx} \int_a^b f(x, t) \ dt = \int_a^b \frac{\partial f(x, t)}{\partial x} \ dt$$


问:
Would Nash equilibrium in a game when only pure strategies are allowed be also a Nash equilibrium in a game when mixed strategies are allowed as well?
Would Nash equilibrium in a game when only pure strategies are allowed be also a Nash equilibrium in a game when mixed strategies are allowed as well?</p>

答:
Yes.

Consider a simultaneous move game. Denote the payoff function by $\pi$, the strategy of player $i$ by $s_i$, and the strategy profile of other players by $s_{-i}$. The Nash equilibrium if only pure strategies are allowed implies
$$\pi_i(s_i,s_{-i})\ge \pi_i(\hat{s}_i,s_{-i}) \forall i, \forall \hat{s}_i\neq s_i,$$
that is, $s_i$ is a best response to the profile $s_{-i}$ of all other players, and this holds for all players, by the definition of a Nash equilibrium.

Now suppose you allowed mixing (but do not add any new pure strategies to the strategy set). Consider again player $i$. Hold the strategies of all other players $-i$ fixed. Is $s_i$ - the pure strategy - still a best response in this case, and hence the above strategy profile a Nash equilibrium?

Yes, because the payoff of playing say strategy $s'$ with probability $p$ and playing $s''$ with $1-p$ is
$$p\pi_i(s',s_{-i})+(1-p)\pi_i(s'',s_{-i}),$$
i.e., it is a linear combination of the payoffs from the pure strategies. Hence,
$$\pi_i(s_i,s_{-i})\ge \pi_i(\hat{s}_i,s_{-i}) \forall i, \forall \hat{s}_i\neq s_i \\ \implies \\\pi_i(s_i,s_{-i})\ge p\pi_i(s',s_{-i})+(1-p)\pi_i(s'',s_{-i}) \forall s',s'', \forall i,$$
i.e., the payoff from playing the pure strategy (given that all other players stick with their pure strategies $s_{-i}$) is at least as large as the expected payoff from playing any mixed strategy, i.e., it remains a best response.

This is because it was a best response to begin with when only allowing pure strategies, and a mixed strategy can never do better than any pure strategy it mixes over, because the expected payoff from mixing is a linear combination of the payoffs of the pure strategies.

Final note: It may still be that additional mixed strategy equilibria exist, but the pure strategy equilibria survive in any case.


问:
What is the real dimension of a manifold?
This is just hypothetic: If I'm given a function $f(x,y,z)$ like:

$f(x,y,z) = x^2 + y^2 -z^2 +c$

The theory proclaims that the dimension is: $\underbrace{\texttt{number of columns}}_{N} - \underbrace{\texttt{number of rows}}_{r}\:$  of $D(f)$. In this case:

$D(f) = [2x \quad 2y \quad 2z]$, hence $\texttt{dim}  = 3-1 = 2$

This correspond with the plot: For $c = -1$ I'm getting a hyperboloid.

Now to my question: If I choose $x, y$ or $z$ to be $0$ I'm still getting a $\text{3-D-shape}$. If I'm not understanding the theorem wrong the dimension has to be reduced however. Maybe it's the way I plot. I'm doing this with Matlab:

```<code>f = @(x,y,z) x.^2 + y.^2 - 0.^2-1;

fimplicit3(f)
</code>```

Her I set $z$ to $0$, but it shows me the surface of a cylinder still...

答:
The number of columns is still 3, $D(f) = [2x \quad 2y \quad 0]$. Just in this case one of the columns is 0. The function is still a function from $\mathbb{R}^3$ to $\mathbb{R}$, so $D(f)$ has 3 columns and 1 row.

Either way, you can still get strange examples like $x^2+y^2+z^2=1$, where the dimension should be 2 but the 0 set is just a point. Dimension=(number of columns)-(number of rows) not always works. Although it will work most of the times (when there are no self intersections) in points where $D(f)\neq 0$.


问:
Limitation of Henkin sematics in second order logic
I am reading Enderton's A Mathematical Introduction to Logic.

The book explains that compactness fails in second order logic by a counter example
$\Sigma = \{\neg \lambda_{\infty} , \lambda_2, \lambda_3,\lambda_4,\lambda_5, \cdots \}$ where $ \lambda_{\infty} = \exists X[\forall u \forall v \forall w (Xuv \rightarrow Xvw \rightarrow Xuw) \land \forall u \neg Xuu \land \forall u \exists x Xuv ]$
(Existence of transitive irreflexive relation on the entire set having no last element) and $ \lambda_n = \exists x_1 \exists x_2 \cdots \exists x_n(x_1 \neq x_2 \land x_1 \neq x_3...\land x_{n-1} \neq x_n) $ (There exist at least n different points)

Then the book explains many sorted logic and a 'structure' which is a reduction of second order language to first order many sorted logic by adding sorts of function and relation universe and using appropriate definitions, (I assume it is the Henkin semantics) and shows enumerability, compactness and LST theorem.

My question is about satisfiability of $\Sigma $ on the henkin semantics. To me it looks that $\Sigma$ still is a counter example at first glance. Furthermore, what makes compactness(and other good propertices) possible in Henkin semantics but not in general(usual) semantics? What is the limitation of henkin semantics compared to general semantics?

答:
In a Henkin model, the second-order quantifiers are not required to range over all possible properties/relations over the domain of objects. In other words, the domain of the second-order quantifiers can be just some of the properties/relations which could be in principle be defined over the domain of objects.

So take e.g. a Henkin model with (i) infinitely many objects in the domain of the first-order quantifiers, but where (ii) the only (two-place) relations in the domain of the second-order quantifiers are reflexive. Then $\lambda_\infty$ is false (because there doesn't exist a relation $X$ which is irreflexive etc.); so this model indeed satisfies $\Sigma$.

Think of it this way. In the Henkin semantics, we can have fewer relations in the domain of the second-order domain than in the full semantics: that's why it is easier for a second-order existential quantification to fail (easier for its negation to be satisfied) when interpreted Henkin-style.


问:
Understanding why in this case $\det A = \det A'$, and more importantly why two separate $A$ and $A'$ exist
In my linear algebra class we have learnt that each Linear Transformation $T:V \rightarrow V$ where $\dim V$ is finite has a square matrix associated with it (it is clear to that the action of matrix multiplication is equivalent to performing the linear transformation, and I know how to obtain that matrix when the transformation is provided). Similarly, we have gone on to define other notions such as the determinant of any linear transformation.

However, we have recently started the topic of Eigenvalues and Eigenvectors and we have obtained this new result:

If we have a linear transformation $\hat{A}:V\rightarrow V$ and the matrices of that linear transformation which are obtained using two separate bases say $\mathcal{B}$ and $\mathcal{B'}$ are $A$ and $A'$ respectively, then $\det A = \det A'$.

This statement is alien to me, because until now I was operating under the assumption that there is only one matrix associated to each linear transformation. This, however, implies that there is more than one and that the obtained matrix depends on the choice of basis of our vector space.

I am having difficulty wrapping my head around this concept. How is this the case? Trying out a few examples using the different bases always results in the same matrix.

Is there an explanation/example for why and how this is true?

答:
There is no such thing as the matrix of a linear map. If you have a linear map $f\colon V\longrightarrow V$ and if $B=\{v_1,v_2,\ldots,v_n\}$ is a basis of $V$, then the matrix of $f$ with respect to $B$ is defined as follows: for each $i\in\{1,2,\ldots,n\}$ you write $f(v_i)$ as $a_{1i}e_1+a_{2i}e_2+\cdots+a_{ni}e_n$ and then that matrix will be$$\begin{bmatrix}a_{11}&a_{12}&\ldots&a_{in}\\a_{21}&a_{22}&\ldots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\ldots&a_{nn}\end{bmatrix}.$$

If, for instance, $V=\Bbb R^2$ and $f(x,y)=(x+y,x-y)$, then the matrix of $f$ with respect to $B=\{(1,0),(0,1)\}$ will be$$\begin{bmatrix}1&1\\1&-1\end{bmatrix},$$since$$f(1,0)=(1,1)=1\times(1,0)+1\times(0,1)$$and$$f(0,1)=(1,-1)=1\times(1,0)+(-1)\times(0,1).$$But the matrix of $f$ with respect to $B'=\{(1,0),(1,1)\}$ is$$\begin{bmatrix}0&2\\1&0\end{bmatrix},$$since$$f(1,0)=(1,1)=0\times(1,0)+1\times(1,1)$$and$$f(1,1)=(2,0)=2\times(1,0)+0\times(1,1).$$


问:
Prove by contradiction that there is an $i \in [n]$ such that $x_i \geq 2$ if $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$
Prove this statement using a proof by contradiction:

Let $n$ be a natural number. If $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$ and $\sum_{i=1}^{n}{x_i} = n+1$ then there is an $i \in [n]$ such that $x_i \geq 2$

I'm not sure how to approach this problem with the proof by contradiction. So far, I assume that there is no $i \in [n] : x_i \geq 2$. Then, $\sum_{i=1}^{1}{x_i} < 2$. Finally, because $\sum_{i=1}^{1}{x_i} = n + 1 = 2$, I can conclude that by contradiction this is false.

I'm not entirely sure if this logic is a valid proof by contradiction or what a better proof would be. Any help is appreciated

答:
The mistake is that you are focusing only on $\sum_{i=1}^1 x_i = x_1$ to get your contradiction, but we don't know anything about this sum. The sum that we do know a lot about is $\sum_{i=1}^n x_i = x_1 + x_2 + \cdots + x_n = n+1$, so we should try and contradict this assumption.

To be clear, the reason why we don't know anything about $\sum_{i=1}^1 x_i$ is because $n$ was given and is fixed. We don't know what the value of $n$ is, nor do we have control over what value it is, just that we have it

So with proof by contradiction, we would want to try to contradict some assumption or some known fact.

As you pointed out, we assume that there is no $i \in [n]$ such that $x_i \geq 2$. What's another way to say this? Well, this is saying for every $i \in [n], \, x_i < 2$. Now we do have control of this $i$, so we know that $x_i = 0,1$, as each is a nonnegative integer.

Now if we consider the sum $\sum_{i=1}^n x_i = x_1+x_2 + \cdots + x_n$, we note that each $x_i$ could equal $0$ or it could equal $1$. Obviously, the maximal possibility is that each $x_i = 1$, so we have an inequality:
$$\sum_{i=1}^n x_i = \underbrace{x_1}_{\leq 1} + \underbrace{x_2}_{\leq 1} + \cdots + \underbrace{x_n}_{\leq 1} \,\, \leq \,\, \underbrace{1 + 1 + \cdots + 1}_{n-\text{times}} = n.$$
So the sum can be at most $n$. But this is a contradiction, as we know that $\sum_{i=1}^n x_i = n+1$.


问:
(fractals) What is the link between my bifurcation diagram of a population and the mandelbrot set?
This is my first time here as I'm investigating fractals.

I started of by writing about the similarity between a bifrucation diagram of a population, using formula $z_{n}=a*z_{n-1}*(1-z_{n-1})$.

I ended up getting the following image:bifurcation diagram

I compared this to the mendelbrot set, without seeing to much similarity at first, but in a youtube video, I saw that there 2 are very similar, when placed on top of eachother:Mendelbrot set, placed on bifurcation diagram

What could be a way of explaining the way that the periodic behaviour of both graphs are similar if not identical?
By taking the fracture of the lengths I discovered the Feigenbaum constant, 4,669... (as expected), though I was curious if there is another explanation for this behaviour.

Thank you all in advance.

Greetings!
Have a nice day!

答:
There is a simple change of coordinates between $x \to r x (1 - x)$ (Logistic Map) and $y \to y^2 + c$ (Mandelbrot set): $$c=\frac{1-(1-r)^2}{4}$$ $$r = 1 + \sqrt{1 - 4 c}$$

Reference: https://en.wikipedia.org/wiki/Talk:Logistic_map#Very_old_discussions


问:
Proof that $\mathcal{O}_X(D(f))=A(X)_f$
I am confused by the highlighted line in the following proof given by Gathmann in his set of algebraic geometry notes:

In particular this seems to be saying that if two functions $f$ and $h$ vanish at precisely the same points, then they are the same function. But surely $f(x)=x^2$ and $h(x)=x$ vanish at precisely the same points. Am I missing something here? Why can we identify $f_a$ and $h_a$ in proof?

答:
Here is a general way of looking at the argument that Gathmann is making. I'll assume we're working in affine space over an algebraically closed field.

Claim: Suppose a rational function $\varphi$ is defined on a distinguished open set $D(h)$. Then we can write $\varphi = \frac{g}{h^n}$ for some natural number $n$ and some polynomial $g$.

Proof: Since $\varphi$ is rational, we can write $\varphi = \frac{g'}{f}$, for polynomials $g'$ and $f$, and since $\varphi$ is defined on $D(h)$, $f$ does not vanish on $D(h)$. Thus, $D(h)\subseteq D(f)$, so $V(f)\subseteq V(h)$. By the Nullstellensatz, $h\in  I(V(h))\subseteq I(V(f)) = \sqrt{(f)}$. So there is some natural number $n$ and some polynomial $p$ such that  $h^n = pf$. Defining $g = pg'$, we have $\varphi = \frac{pg'}{pf} = \frac{g}{h^n}$. $\square$

If we don't actually care about the specific polynomial $h$, just the open set $D(h)$, we can go further and remove the power of $n$.

Claim: Suppose a rational function $\varphi$ is defined on a distinguished open set $U$. Then there are polynomials $g$ and $h$ such that $U = D(h)$ and $\varphi = \frac{g}{h}$.

Proof: Since $U$ is a distinguished open set, we have $U = D(h')$ for some polynomial $h'$. By the previous claim, we can write $\varphi = \frac{g}{(h')^n}$ for some natural number $n$ and some polynomial $g$. Let $h = (h')^n$. Since $D(h') = D((h')^n)$, we have $U = D(h)$, and $\varphi = \frac{g}{h}$.

Title: Proof that $\mathcal{O}_X(D(f))=A(X)_f$


问:
Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

The value of $x$ satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ is given by

$2 \cos(10°)$

$2 \cos(20°)$

$2 \cos(40°)$

$2 \cos(80°)$

Using $x=2\cos(\theta)$  we get
$2\cos(\theta)=\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}}$

$2\cos(\theta)=\sqrt{2+\sqrt{2-2\cos\left(\frac{\theta}{2}\right)}}$

$2\cos(\theta)=\sqrt{2+2\sin\left(\frac{\theta}{4}\right)}$

$4\cos^2(\theta)={2+2\sin\left(\frac{\theta}{4}\right)}$

After this step I am not able to solve it

Body:
My illustration is outlined below

$\begin{array}{l}
4{\cos ^2}\theta  = 2\left( {1 + \sin \frac{\theta }{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin {{2.5}^o}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \cos {{87.5}^o}} \right)\\
\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin {5^o}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \cos {{85}^o}} \right)\\
\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin \frac{{{{40}^o}}}{4}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin {{10}^o}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \cos {{80}^o}} \right)\\
\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin \frac{{{{80}^o}}}{4}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin {{20}^o}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \cos {{70}^o}} \right)
\end{array}$\

Hence the option is $0=40^o$


答: Your question is confusing. Do you mean this? Namely, that if $T_A$ is the matrix for $T$ with respect to the basis $A$ and if $B$ is the change of basis matrix $A\to A'$ then $T_{A'}=B^{-1}T_AB$, so $\det T_{A'}=\det T.$

Title: Understanding why in this case $\det A = \det A'$, and more importantly why two separate $A$ and $A'$ exist

Body:
In my linear algebra class we have learnt that each Linear Transformation $T:V \rightarrow V$ where $\dim V$ is finite has a square matrix associated with it (it is clear to that the action of matrix multiplication is equivalent to performing the linear transformation, and I know how to obtain that matrix when the transformation is provided). Similarly, we have gone on to define other notions such as the determinant of any linear transformation.

However, we have recently started the topic of Eigenvalues and Eigenvectors and we have obtained this new result:

If we have a linear transformation $\hat{A}:V\rightarrow V$ and the matrices of that linear transformation which are obtained using two separate bases say $\mathcal{B}$ and $\mathcal{B'}$ are $A$ and $A'$ respectively, then $\det A = \det A'$.

This statement is alien to me, because until now I was operating under the assumption that there is only one matrix associated to each linear transformation. This, however, implies that there is more than one and that the obtained matrix depends on the choice of basis of our vector space.

I am having difficulty wrapping my head around this concept. How is this the case? Trying out a few examples using the different bases always results in the same matrix.

Is there an explanation/example for why and how this is true?


答:
Let's consider the transformation $T:\Bbb R^2\to\Bbb R^2$ that doubles the $x$-coordinate and leaves the $y$-coordinate unchanged. This can be shown to be linear, and the matrix of $T$ with respect to the usual basis of $\{\vec e_1,\vec e_2\}=\{(1,0)^\dagger,(0,1)^\dagger\}$ is $\begin{bmatrix}2 & 0\\0 & 1\end{bmatrix}.$ This is because $T(\vec e_1)=2\vec e_1+0\vec e_2,$ and $T(\vec e_2)=0\vec e_1+1\vec e_2,$ which gives us the columns of the matrix.

On the other hand, we could take $\{\vec e_1,\vec e\}=\{(1,0)^\dagger,(1,1)^\dagger\}$ as a basis, instead. In that case, $T(\vec e_1)=2\vec e_1+0\vec e$ and $T(\vec e)=(2,1)^\dagger=1\vec e_1+1\vec e.$ Thus, the matrix of $T$ with respect to this basis is $\begin{bmatrix}2 & 1\\0 & 1\end{bmatrix},$ instead. However, you should clearly see that the determinant of both of these matrices is $2.$

The reason this happens is that for any two bases $B$ and $B'$ and any linear transformation $T$ with matrices $A$ and $A'$ (with respect to the bases $B$ and $B',$ respectively), there is an invertible matrix $Q$ such that $A'=Q^{-1}AQ.$ Hence, $$\begin{eqnarray}\det(A') & = & \det(Q^{-1})\det(A)\det(Q)\\ & = & \det(Q^{-1})\det(Q)\det(A)\\ & = & \det(Q^{-1}Q)\det(A)\\ & = & \det(I)\det(A)\\ & = & \det(A).\end{eqnarray}$$

If I recall correctly, the only transformations with unique associated matrices are scalar multiplication maps.


问:
Need Help understanding the philosophy behind splitting the integrals into sqrt(x)
Quite occasionally we see in number theory the below integral evaluated as such:

$$\int_{2}^{x} \frac{dt}{\log t}= \int_{2}^{\sqrt x}\frac{dt}{\log t} + \int_{\sqrt x}^{x}\frac{dt}{\log t} $$

Why the choice of $\sqrt x$ And how does it help facilitate the integral and estimating it.

答:
The first part fits inside a box of width $\sqrt x$ and height $1/\ln2$ so its integral is less than $\sqrt x/\ln 2$.

The second part fits inside a box of width $x$ and height $1/\ln\sqrt x$ so its integral is less than $2x/\ln x$.
The first part is narrow and tall, the second part is wide but short.  Their combined area is much less than the simplest estimate of $x/\ln 2$ that comes from a single box of width $x$ and height $1/\ln2$.

$\sqrt x$ was chosen to be a fairly simple compromise, where $\sqrt x=o(x)$  and $1/\ln\sqrt x=O(1/\ln x)=o(1/\ln2)$

$$\int_2^{\sqrt x}\frac{dt}{\ln t}\lt\int_2^{\sqrt x}\frac{dt}{\ln2}\\=\frac{\sqrt x -2}{\ln2}$$

$$\int_{\sqrt x}^x\frac{dt}{\ln t}\lt\int_{\sqrt x}^x\frac{dt}{\ln\sqrt x}\\ \lt\frac x{\ln\sqrt x}$$


问:
Difference between Fourier integral and Fourier transform

What is the difference between Fourier integral and Fourier transform?

I know that for Fourier integral, the function must satisfy
$\int_{-\infty}^\infty  |f(t)| dt < \infty$, but what if I have a function that satisfies this condition: what does it mean to calculate Fourier transform and Fourier integral?

Fourier Integral:
$$f(t) =  \int_0^ \infty  A(\omega)\cos( \omega t) + B( \omega )\sin( \omega t)  d \omega$$
where
$A(\omega ) =  \frac{1}{\pi}   \int_{- \infty }^\infty f(t)\cos( \omega t) dt$ and $B(\omega ) = \frac{1}{\pi}  \int_{- \infty }^ \infty  f(t) \sin( \omega t) dt$

Fourier Transform:
$$F(\omega ) =  \frac{1}{2 \pi }   \int_{- \infty }^\infty  f(t) e^{-i \omega t}  dt$$

答:
Your question is a bit ambiguous, since you don't state what you mean by fourier integral and fourier transform.

One possible source of confusion is that, while the fourier transform is indeed a linear isometry on $L^2$, the integral $$
  \int_{-\infty}^\infty f(t)e^{-i2\pi\omega t} \,dt
$$
does not converge for every $f \in L^2$. It does, however converge for every $f \in L^1 \cap L^2$, and the fourier transform on the full space $L^2$ can therefore be defined as the unique extension of the transform defined by the integral on $L^1 \cap L^2$. The result is then sometimes called the Fourier-Plancherel-Transform, but sometimes also simply the fourier transform on $L^2$.

Or you could simply be referring to the difference between the integral one uses to compute the coefficients of a fourier series and the integral used to define the fourier transform (on $L^2 \cap L^2$).


问:
Finding all natural $x$, $y$, $z$ satisfying $7^x+1=3^y+5^z$
The problem goes as follows:

Find all possible pairs of $x,y,z \in \mathbb{N}$ which satisfy the equation $7^x+1=3^y+5^z$

My first instinct was to continue by modding, but I don't think I can get anything out of it. The obvious solutions seem to be $x,y,z=1$ and $x,y,z=0$, but I am not really sure how to approach the problem.

Thanks in advance!

答:
The two solutions you found are indeed all of them. In fact, we can solve this with a straight mod bash, which was surprising to me.

If any of $x,y,z$ are $0$, then we easily get $x=y=z=0$. If $y=1$, then $7^x=5^z+2$, and we readily get that $z=1$, so $x=1$ too. So now assume $y>1$ and $x,z>0$. Let the mod bash begin.

Taking mod $3$ implies that $z$ is odd. Taking mod $4$ implies that $x$ and $y$ have the same parity. Taking mod $5$ then implies that $x\equiv y\equiv 1\pmod 4$.

Now taking mod $7$, we get that either (a) $y\equiv1\pmod{12}$ and $z\equiv1\pmod{6}$ or (b) $y\equiv5\pmod{12}$ and $z\equiv5\pmod6$.

Now taking mod $9$, in case (a) we have $x\equiv5\pmod{12}$ and in case (b) we have $x\equiv9\pmod{12}$.

Consider case (a). If $z\equiv1\pmod{12}$, then taking mod $13$ gives $11+1\equiv3+5\pmod{13}$; if $z\equiv7\pmod{12}$ we still get a contradiction $11+1\equiv8+5\pmod{13}$.

Similarly in case (b), we get a contradiction in mod $13$, considering both cases $z\equiv 5\pmod{12}$ and $z\equiv11\pmod{12}$.


问:
Every connected, open surface with the infinitely generated fundamental group is the interior of some non-compact surface with boundary
This question is based on the discussion in the comment section of my previous post. I used the fact: "Every connected, open surface with the finitely generated fundamental group is the interior of some compact surface" to prove open annulus covers any orientable non-simply-connected surface without boundary. My current question is the following:

$\textbf{Problem:}$ Is this also true: Every connected, open surface with the infinitely
generated fundamental group is the interior of some non-compact
surface with boundary.

$\textbf{Attempt:}$ I hope this is not in general true. I am trying to use $\text{Theorem 3}$ of Ian Richard's classification theorem on open surfaces. It says:

Every non-compact surface without boundary can be obtained by first removing a closed totally disconnected subset $X$ of $\Bbb S^2$, then removing the interior of a finite or countable number of non-overlapping closed discs, and after that suitably identifying the boundaries of these
discs in pairs(to produce either handle or cross cap).

Now, if $X$ has an isolated point $p$, then removing $X$ from $\Bbb S^2$ is the same as removing $X\backslash\{p\}$ from $\Bbb D:=\{z\in \Bbb C:|z|<1\}$ using one-point compactification. Now, every other operation in $\text{Theorem 3}$ are happening in a smaller open disc $\{z\in \Bbb C:|z|<\varepsilon\}$ for some $0<\varepsilon<1$. So, considering the boundary $\{z\in \Bbb C:|z|=1\}$ of $\Bbb D$ we are done in this special case.

Some examples of connected, open surface with the infinitely generated fundamental group are $\Bbb R^2\backslash\text{cantor set}$, Loch ness Monster surface, Jacob Ladder surface. Note that all infinite-type surface(with or without boundary) are homotopically equivalent to $\displaystyle\bigvee_{n\in\Bbb N}\Bbb S^1$.

$\bullet$ Is my reasoning in the $\textbf{Attempt}$ section correct?

$\bullet$ Is $\Bbb S^2\backslash \text{cantor set}$ an example of an
open surface for which the statement of $\textbf{Problem}$  false?

$\bullet$ Vague Question: Can we make some decision on the
$\textbf{Problem}$ using some kind of cohomology or other useful
end-functors?

Thanks for your interest.

答:
I assume that when you say "with boundary," you mean "with nonempty boundary." Indeed, every (connected) open surface is homeomorphic to the interior of a surface with nonempty boundary. Let me explain how to do this in the case of the complement to the standard Cantor set $C$ in $S^2=R^2\cup \{\infty\}$, $C\subset [0,1]$, $C$ contains $1$. Consider the open unit disk $D\subset R^2$ centered at $(2,0)$. Take the surface with nonempty boundary $X\subset R^2\cup \{\infty\}$ obtained by removing from $S^2-C$ the disk $D$. Then $X\setminus \partial X$ equals $S^2-(C\cup cl(D))$. I claim that the latter is homeomorphic to $S^2-C$. Indeed, consider the quotient space $Z$ of $S^2$ obtained by collapsing $cl(D)$ to a single point. It is easy to see that $Z$ homeomorphic to $S^2$. The restriction of the quotient map $q: S^2\to Z$ to $C$ is 1-1, hence, by compactness of $C$, a homeomorphism to its image. Hence, the pair $(Z,q(C))$ is homeomorphic to $(S^2,C)$. Claim follows.

The general case can be proven using the above argument in conjunction with the quoted sentence from Richards' paper. (There is some work to be done but you can probably do this yourself.)

In fact, one can prove more (using a refined version of the above argument):

Theorem. Let $S$ be an open connected surface with boundary. Then there exists a surface with boundary $F$ whose interior is homeomorphic to $S$ and such that  $\partial F$ is dense in the end-compactification $\bar{F}$ of $F$ (i.e. the closure of $\partial F$ in $\bar{F}$ contains $e(F)= \bar{F}\setminus F$).


问:
Using 6 Sided Dice to Get an Even Distribution of Numbers $1$ to $50$
Is this possible?

I have a $6$ sided dice (call it a $d6$).

I want a way of getting an even distribution of the numbers $1$ to $50$ using that dice with no or minimal rerolls.

A way with a lot of rerolls.

A. Roll d6. $1 - 3 $means use table $A$, $4 - 6$ means use table $B$.

B. Table $A$ needs you to roll $2$ dice. First Dice gives you the column and second dice gives you the row. The numbers $1$ to $25$ are in the table, with the other $11$ spaces telling you to reroll. (Table B is similar, but with numbers $26$ to $50$).

The only problem with this is there are $22$ reroll spaces.

What I want is a way to get an even distribution, but with no rerolls or very few.

答: Rolling three dice gives you a number from $1$ to $216$.  For a start, you can use the numbers $1-200$ and reroll the last $16$ but remember what the number is.  Now one more roll gives you a number $1-96$ by multiplying your remembered number by $6$, subtracting the new roll, and adding $1$.  You can use $1-50$ and reroll $46$ values.  Again remember the remainder, roll again, and you get a number $1-6\cdot 46=1-276$.  You can use $1-250$ and keep $1-26$.  Keep going.  This has to be optimal because you are using all the information you get.  I haven't computed the average number of throws.


问:
Does every finite group $G$ have a set of generators such that the sum of the orders of the generators is less than or equal to $|G|$?
Does every finite group $G$ have a set of generators such that the sum of the orders of the generators is less than or equal to $|G|$?

This is surely true but I am failing to see why.

This is easily seen to be true for the symmetric group, abelian groups, the quaternions, and it follows from their classification that simple groups have this property too.

答: We prove this by induction on $|G|$, the base case being the trivial group.  Now suppose $G$ is nontrivial and let $H\subset G$ be a maximal proper subgroup.  By the induction hypothesis, there is a set $S$ of generators for $H$ such that the sum of the orders of elements of $S$ is at most $|H|$.  By maximality of $H$, $G$ is generated by $H$ together with one more element $x\in G$.  If $x$ has order $|G|$, then it generates $G$ and we can take $\{x\}$ as our set of generators for $G$.  Otherwise, the order of $x$ and $|H|$ are both proper divisors of $|G|$, and in particular they are both at most $|G|/2$.  So their sum is at most $|G|$, and so the sum of the orders of elements of $S\cup\{x\}$ is at most $|G|$.


问:
What are the bases of a function space (Hilbert space)?
1. Motivation

I was learning about the Hilbert space and function spaces, i.e., roughly speaking, infinite-dimensional vector spaces.

Let's now think about ordinary 3D Euclidean vectors. A vector $\vec{x}$ may be given by

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} \vec{x} = \mycolv{1\\3\\2}$$

and this is equivalent to saying that $\vec{x} = 1 \hat{i} + 3 \hat{j} + 2 \hat{k}$. So even if we think of this vector as an ordered triple, $(1, 3, 2)$, then a mathematical structure behind is saying that a vector can be demonstrated in the form of a linear combination of basis vectors.

2. My understanding of function spaces

As someone studying engineering and hasn't been exposed to rigorous mathematical proofs of linear algebra, I understood, through quite an intuitive approach, why function spaces are infinite-dimensional vector spaces. I considered the inner product defined, for example, of functions $\phi: R \rightarrow C$ and $\psi: R \rightarrow C$.

$$\langle \phi | \psi \rangle = \int \phi^*(x) \psi(x) \ dx$$

Intuitively speaking this is about adding up all the differentials $\phi^*(x) \psi(x) \ dx$, which is analogous to the scalar product in Euclidean space, so I thought each and every function values corresponding to individual (though an infinite number of) elements of the domain should all be the components of vectors $\phi(x)$ and $\psi(x)$. (And thus both function spaces should be of an infinite dimension.) So, like the below:

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} |\psi(x) \rangle = \mycolv{...\\\psi(a-2\epsilon) \\ \psi(a-\epsilon) \\\psi(a)\\\psi(a + \epsilon) \\ \psi(a+2 \epsilon) \\ ...}$$

for some $a \in R$.

3. Question

It's fine up to here. But, there should at least be an infinite number of linearly independent basis vectors that make up the actual infinite-dimensional column vector, i.e., $|\psi(x)\rangle$. As if $\hat{i}, \hat{j}, \hat{k}$ corresponded respectively to the coefficients $1, 3, 2$, there should be basis vectors that correspond to each of $\psi(a), \psi(a+\epsilon), \psi(a-\epsilon), $ and so on. What are they?

答:
Basically, a vector space $V$ (also called linear space) is a set with two operations: multiplication with scalars (e.g. real or complex numbers) and addition.

A basis $B$ of the vector space $V$ is a subset such that every element $v \in V$ can be written as a unique linear combination of the elements in $B$.

In a Hilbert space the linear combination can contain an infinite number of terms, and the sum of the infinite sum is defined through the norm induced by the inner product.

The elements in a function space are functions, and so are the elements in a basis of such a space.

An example of a Hilbert space is $L^2([0,2\pi], \mathbb{C}),$ the linear space of functions $f : [0,2\pi] \to \mathbb{C}$ such that $\int_0^{2\pi} |f(x)|^2 \, dx$ is finite. The inner product is given by $\langle f, g \rangle = \int_0^{2\pi} \overline{f(x)}\,g(x)\,dx$. One basis for this space is given by $\{ e^{ikx} \mid k\in\mathbb{Z} \},$ where $e^{ikx}$ is restricted to $[0,2\pi]$.


问:
Orthogonal group derived using tensor notation $f(v)=v_av_bm^{ab}$, instead of inner product notation $f(v)=v^Tv$?
I wish to replicate the following proof using tensor notation. Let $v$ be a vector of $\mathbb{R}^n$. First I define this function:

$$
f(v)=v^Tv\\
$$

I now ask, what are the transformations that leave $f$ invariant:

$$
f(Ov)=(Ov)^T (Ov)=v^TO^TOv
$$

which is invariant iff $O^TO=I$.

Now using tensor notation, I write $f$ as follows:

$$
f(v)=v_av_bm^{ab}
$$

for a 2D vector:

$$
f(v)=v_0v_0m^{00}+v_0v_1m^{01}+v_0v_1m^{10} +v_1v_1m^{11}
$$

Then, if we assume the $m_{ab}$ is the identity matrix, we get:

$$
f(v)=v_0^2+v_1^2
$$

What happen when I inject a transformation into $f$? Is the following the correct approach?

$$
f(Ov) = (Ov)_a(Ov)_b m^{ab}
$$

How do I then eliminate $O$ such that the condition $O^TO=I$ emerges, using this kind of notation?

答: The $O$ for which $v\mapsto Ov$ preserves $v^Tmv$ (where without loss of generality $m$ is symmetric, so is diagonalisable with real eigenvalues) are not in general the orthogonal $O$, but the solutions to $O^TmO=m$. Indeed$$(Ov)_a(Ov)_bm^{ab}=O_a^{\:c}v_cO_b^{\:d}v_dm^{ab}\equiv v_cv_dm^{cd}\iff m^{cd}=O_a^{\:c}O_b^{\:d}m^{ab}=(O^TmO)^{cd}.$$


问:
Area of an equilateral triangle with the radii of the inscribed circle
Find the area of an equilateral triangle $ABC$ with the radii of the inscribed circle $r$.

 The answer given in my book is $3r^2\sqrt3$.

We know that $S=pr$ where $p$ is the semiperimeter and $r$ is the radii of the inscribed circle. How can I express $p$ with $r$? Thank you in advance!

答:
Welcome to MSE. Since $p = 3a/2$, where $a$ is the length of a side of the triangle, so it suffices to relate $r$ to $a$.

To do this, see the figure here below equation (2). We can make a right triangle with legs of length $a/2$ and $r$, with one angle equal to $30°$. You can now use trigonometry to solve for $a$ in terms of $r$.


答:
Your intuition that a function space is an infinite dimensional vector space with each point in the domain corresponding to a coordinate is correct.

The interesting function spaces come with a norm. Then a basis is a set of vectors such that every vector in the space is the limit of a unique infinite sum of scalar multiples of basis elements - think Fourier series.

The uniqueness is captures the linear independence.

These vector spaces also have infinite bases such that every element is a finite linear combination of scalar multiples of basis vectors, but those bases are not useful in analysis. See
https://en.wikipedia.org/wiki/Basis_(linear_algebra)#Analysis

Title: What are the bases of a function space (Hilbert space)?

Body:
1. Motivation

I was learning about the Hilbert space and function spaces, i.e., roughly speaking, infinite-dimensional vector spaces.

Let's now think about ordinary 3D Euclidean vectors. A vector $\vec{x}$ may be given by

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} \vec{x} = \mycolv{1\\3\\2}$$

and this is equivalent to saying that $\vec{x} = 1 \hat{i} + 3 \hat{j} + 2 \hat{k}$. So even if we think of this vector as an ordered triple, $(1, 3, 2)$, then a mathematical structure behind is saying that a vector can be demonstrated in the form of a linear combination of basis vectors.

2. My understanding of function spaces

As someone studying engineering and hasn't been exposed to rigorous mathematical proofs of linear algebra, I understood, through quite an intuitive approach, why function spaces are infinite-dimensional vector spaces. I considered the inner product defined, for example, of functions $\phi: R \rightarrow C$ and $\psi: R \rightarrow C$.

$$\langle \phi | \psi \rangle = \int \phi^*(x) \psi(x) \ dx$$

Intuitively speaking this is about adding up all the differentials $\phi^*(x) \psi(x) \ dx$, which is analogous to the scalar product in Euclidean space, so I thought each and every function values corresponding to individual (though an infinite number of) elements of the domain should all be the components of vectors $\phi(x)$ and $\psi(x)$. (And thus both function spaces should be of an infinite dimension.) So, like the below:

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} |\psi(x) \rangle = \mycolv{...\\\psi(a-2\epsilon) \\ \psi(a-\epsilon) \\\psi(a)\\\psi(a + \epsilon) \\ \psi(a+2 \epsilon) \\ ...}$$

for some $a \in R$.

3. Question

It's fine up to here. But, there should at least be an infinite number of linearly independent basis vectors that make up the actual infinite-dimensional column vector, i.e., $|\psi(x)\rangle$. As if $\hat{i}, \hat{j}, \hat{k}$ corresponded respectively to the coefficients $1, 3, 2$, there should be basis vectors that correspond to each of $\psi(a), \psi(a+\epsilon), \psi(a-\epsilon), $ and so on. What are they?


问:
On Probability Measures on $\{-1,+1\}^n$
Let's put it formally...

Let $n$ be a positive integer and $E = \{-1,+1\}^n$.
Consider the set $\mathcal{P}$ of all the probability measures $\mu$ on $E$ such that
$$
\mu ( \pi_i^{-1} \{-1\} ) = \mu ( \pi_i^{-1} \{+1\} ) = \frac12
$$
where $\pi_i : E \rightarrow \{-1,+1\}$ is the canonical projection onto the $i^\text{th}$ component of $E$, i.e. $\pi_i(x_1, \dots, x_n) = x_i$.

How can I parametrize $\mathcal{P}$?

Also, letting $\mathcal{C}$ be the set of all $n \times n$ correlation matrices, consider the map
$c: \mathcal{P} \rightarrow \mathcal{C} $ that assigns to each probability measure in $\mathcal{P}$ its correlation matrix.
Is $c$ injective? Surjective?

For the simple case $n=2$, it is clear to me that we can put $\mathcal{P}$ in a bijection with $[0,\frac12]$. For $a \in [0,\frac12]$, define $\mu$ as
$$
\mu \{-1,-1\} = \mu \{+1,+1\} = a \\
\mu \{-1,+1\} = \mu \{+1,-1\} = \frac12 - a
$$
It's also clear that $c$ is bijective since the correlation is $4a-1$.

But what about the case of arbitrary $n$?

答:
The measures you are interested in are called symmetric Bernoulli distributions. The question which correlations are possible for such distributions is answered in the paper Admissible Bernoulli correlations.

To quote from the paper:

Consider a vertex of the n-dimensional cube $v\in\{0,1\}^n$. For
instance, when $n=5$, $v=(0,0,1,0,1)$ is such a vertex. Let 1 denote
the vector of all 1’s. Then for any $v\in\{0,1\}^n$, the distribution
$\text{Unif}(\{v,1−v\})$ (discrete uniform distribution over two
points: $v$ and $1−v$) has marginals that are all uniform over the
pair $\{0,1\}$. Hence all such distributions are multivariate
symmetric Bernoulli.

Any convex combination of multivariate symmetric Bernoulli
distributions will also be multivariate symmetric Bernoulli. Our main
result is that any admissible correlation structure can also be
realized as the correlation structure of such a convex combination.

They further characterize (in their Theorem 2) all possible correlations of symmetric Bernoulli distributions in terms of the $\text{CUT}_n$ polytope.


问:
Mix of first price and second price auction problem.
Suppose we have two participants. They play the game with utilities: $u_1 = v_1 - b_2$ if $b_2 \le b_1$ and zero otherwise. For the second participant $u_2 = v_2 - b_2$ if $b_2 \ge b_1$ and zero otherwise. Where $v_1, v_2$ - known non-negative values.
We want to determine does there a weakly dominant strategy for both players?

For player $1$ there is a weakly dominant strategy: $b_1 = v_1$ (it's easy to show by varying $b_1$).

For the second player, I proved that there is no weakly dominant strategy. Suppose w.d.s. $b_2^* > v_2$ then $0 > u_2(b_2^*, v_2) < u_2(v_2, v_2) = 0$ - contradiction. Now suppose that $0 < b_2^* < v_2$, then since there is a gap between $b_1^*$ and $v_2$ we can find $\epsilon_1 < \epsilon_2$ such that: $b_2^* + \epsilon_1 \in (b_2^*, v_2)$ and $b_2^* + \epsilon_2 \in (b_2^*, v_2)$. For these two points we have: $0 = u_2(b_2^*, b_2^* + \epsilon_1) < u_2(b_2^* + \epsilon_2, b_2^* + \epsilon_1)$, the last term is greater than zero, so we have contradiction. $b_2^* = 0$ is also not weakly dominant strategic, since for all points $u_2(0, b_1) = 0$ and $u_2 \ge 0$ for all $v_2 \ge 0$. So we need to check only one case: $b_2^* = v_2$. But in this case: $0 = u_2(b_2^*, b_1) < u_2(b_1 + \epsilon, b_1) = (v_2 - b_1 - \epsilon)$ for $b_1^* < v_2$ and $\epsilon: b_1^* + \epsilon < v_2$. So there is no weakly dominant strategic for second player.

Am I right? There is a problem. There is also a question: find all N.E. of form $(v_1, v_2)$, but I'm not sure whether there is at least one N.E., since there is no weakly dominant one. Any ideas?

答:
OK, suppose the equilibrium strategy of 1 is $b_1=b_1$ - you already showed there is no strictly better strategy as it is weakly dominant, so this is a good candidate. To find a Nash equilibrium, we need the best response of 2 to that strategy.

That is, we need
$$b_2=\arg \max_b 1_{\{b\ge b_1 \}}(v_2-b),$$
where $1_{\{b\ge b_1 \}}$ is the indicator function with the case where 2 has the larger bid. First, clearly $b_2>v_2$ is strictly dominated (e.g., by $b_2=-\infty$), so this cannot be an equilibrium strategy. Second, $b_2=v_2$ yields a profit of 0, again not optimal, as we should be able to get a positive payoff in some cases.

From this we can already conclude that, in equilibrium, 1 will get the item if $v_1>v_2$, since the above implies $v_1=b_1>v_2>b_2$. So we focus on the interesting case $v_2>v_1=b_1$.

And we need to clarify what happens in case of $b_1=b_2$, i.e., we need a tie breaking rule. Your example above is inconsistent, as it gives both bidders the item in case of equality (which is not feasible). So I will adopt the standard tie breaking rule that the item is given to each with 1/2 probability in case of a tie.

Moreover, we need to clarify whether the bidding grid is discrete or continuous. If discrete, let $\epsilon$ be the smallest increment.

Now, if $v_2>v_1$ and a discrete grid, then I claim the best response is $b_2=v_1+\epsilon$. Obviously, a larger bid is never better, as this is already enough to get the item, and a large bid only reduces the payoff. A smaller bid of $b_2=v_1$ expects a profit of $(v_2-v_1)/2$, since the item is obtained only with probability 1/2. This is smaller than $(v_2-v_1-\epsilon)$ (payoff from my strategy) for $\epsilon>0$ small enough (i.e., the discrete grid fine enough). Finally, an even smaller bid $b_2<v_1$ is also inferior, as the item is not obtained and the profit is 0. So, for $\epsilon>0$ small enough, this is a Nash equilibrium.

Moreover, if $v_2>v_1$ and a continuous grid, then there exists no Nash equilibrium. Why? Intuitively, the best response is just above $b_1=v_1$. However, there exists no smallest $b_2$ such that $b_2>b_1$ - you can always find a smaller one. Hence, the definition of a best response is not fulfilled and this is not a Nash equilibrium.

Final note: A Nash equilibrium can exist without weakly dominant strategies. A dominant strategy equilibrium is always a Nash equilibrium, but a Nash equilibrium need not be a dominant strategy equilibrium. Hence, finding that no weakly dominant strategy exists is not sufficient to conclude that no equilibrium exists.


问:
Counting problem, number of moves
In a game, you are standing somewhere and a move consists of either moving forward any number of steps, not moving at all, or moving $1$ step backwards. If you complete $n$ moves, find the number of sequences that end with you in your starting position and which never go behind your starting position.

If the "never go behind the starting position" condition weren't there, the answer would just be $\binom{2n-1}{n-1}$. However I'm not on sure how to proceed from here.

答: There is a bijection between walks of your type with $n$ moves, and Dyck paths. Encode a walk of your type as a sequence of $n$ integers $(a_1,\dots,a_n)$, where $a_i\in \{-1,0,1,\dots\}$ for each $i\in \{1,\dots,n\}$. To get the corresponding Dyck path, replace each $a_i$ with a sequence of $(a_i+1)$ up steps and $1$ down step. Note your path stays to the right of the origin if and only if the corresponding Dyck path stays above the $x$-axis. Of course, the number of Dyck paths is $\frac1{n+1}\binom{2n}n$.

Title: Counting problem, number of moves


问:
Vector field whose rotation is normal to a half ellipsoid
I have a (special) half ellipsoid
$$\tilde F=\left\{(x,y,z)\in\mathbb{R}\ \bigg|\ \frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}=1,\ z\geq0\right\}$$
and need a vector field $f$ whose rotation is normal to that surface, so something like
$$\begin{cases}\partial_yf_z-\partial_zf_y=\frac{x}{a^2}\\\partial_zf_x-\partial_xf_z=\frac{y}{a^2}\\\partial_xf_y-\partial_yf_x=\frac{z}{b^2}\end{cases}\qquad\forall (x,y,z)\in\tilde F $$
However, I'm not sure if something like this is even possible.

In case you're wondering why I want this: I want to determine the surface area of the (full) ellipsoid
$$F=\left\{(x,y,z)\in\mathbb{R}\ \bigg|\ \frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}=1\right\}$$
Therefore I want to use the half ellipsoid, so
$$v_2(F)=2v_2(\tilde F)$$
$\tilde F$ then is a compact subset of the submanifold $F$ with a smooth boundary (I'm not sure I translated this correctly from the german glatter Rand)
$$\partial\tilde F=\left\{(x,y,0)\in\mathbb R^3\ \bigg|\ x^2+y^2=a^2\right\}$$
So one can apply the Kelvin–Stokes theorem
$$\int_{\partial\tilde F}\langle f,d\vec s\rangle=\int_{\tilde F}\langle\mathrm{rot} f(x),\nu(x)\rangle dS(x)$$
Which looks similar to the surface I'm looking for,
$$v_2(\tilde F)=\int_{\tilde F}dS(x)$$
So I merely have to achieve that $\langle\mathrm{rot} f(x),\nu(x)\rangle=1$ (and then calculate the line integral of the boundary). As $\nu$ is the normalized normal field to the ellipsoid, I can achieve this by requiring $\mathrm{rot} f(x)=\nu(x)$ - which leads to this question.

It's kind of hard to explain (and draw) what I mean, but my idea was to modify $\nu$ outside of $\tilde F$ in a way, that the divergence is $0$, by letting the flow run in from the bottom - drawn in two dimensions:

If $\nu$ then is normed (such that $\|\nu(x)\|=1\ \forall x$) the divergence should be $0$ (right?).

答:
See this post here: for a fixed vector field $\mathbf{v}$ in a simply connected region $R$, we can find a vector field $\mathbf A$ such that $\mathbf v = \nabla \times \mathbf A$ if and only if $\nabla \cdot \mathbf v = 0$ in $R$.

Unfortunately, in this case the divergence
$$\nabla \cdot \mathbf v = (1/a^2) + (1/a^2) + (1/b^2) $$
is never zero.


问:
Question about finding the dimension of Ker of a matrix!
First year student here trying to pass the Linear Algebra exam! Found this in one of the practice quizzes! I tried to find a way to acchive the result but i cant figure it out! Can someone help me?
So the question goes like this!

Let R^(3,2) be the vector space of real matrices with 3 rows and 2 columns and R^(2,2) be the vector space of real matrices with 2 rows and 2 columns. Let us consider the linear map:

f(X)=(1 2 1, 0 0 0)*X where X∈R^(3,2).

Which of the following statements is true?
A) f is sujective;

B)dim(Ker(f))=2

C)dim(Ker(f))=4

D) f is injective

Tee correct one is C but i cant figure it out how to solve it in case something similar comes up in my exam! Can you help me?

答: For any two finite dimensional vector spaces $V$ and $W$ with a linear map $\varphi:V \rightarrow W$ the dimension of the image plus the dimension of the kernel will be the dimension of $V$. We also know that the dimension of the image can be no greater than the dimension of $W$. In this case you have a six dimensional space of $3 \times 2$ matrices mapping to the four dimensional space of $2 \times 2$ matrices. The matrix you're given will have an image of the form $$\pmatrix{x & y  \\ 0 & 0}$$ and since the image is two dimensional. This means the dimension of the kernel is $6 - 2 = 4$.


问:
Techniques of estimating integrals, asymptotic behavior of Li(x)
I need help understanding the logic and build an rigorous intuition proving that if $x \gt 2$, then:

$$\int_{2}^{x} \frac{dt}{t \log^{3} t} \lt \frac{1}{2 \log^{2}2}$$

答:
Notice that
$$\int_2^x \frac{dt}{t\ln^3(t)}<\int_2^{\infty} \frac{dt}{t\ln^3(t)}=\frac{1}{2\ln^2(2)}$$
and we are done.


问:
conditional expectation of disjoint events
Let $B_{1},B_{2},\dots,B_{n} \in \mathcal{F}$
partition of $\Omega$

and $\mathcal{P}=\sigma(B_{1},B_{2},\dots,B_{n})$

If $P(B_j)>0 , j = 1,\dots,n$.

Then show that :

$$\mathbb{E}[X|\mathcal{P}] = \sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}}$$
where :

$$\mathbb{E}[X|{B_i}] = \frac{\mathbb{E}[ X \mathbf{1}_{B_{j}}]}{P(B_{j})}$$

Hint : (all the non empty events of $\mathcal{P}$ have the form of $B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}$ )

With $\bigsqcup$ we denote the union of disjoint events.

My effort is :

\begin{align*} \mathbb{E}\left[X | B_{1} \bigsqcup B_{2} \bigsqcup,\dots \bigsqcup B_{j}\right] &= \frac{1}{P(B_{1} \bigsqcup B_{2} \bigsqcup \dots \bigsqcup B_{j})} \mathbb{E} \left[X \cdot \mathbf{1}_{B_{1} \bigsqcup B_{2} \bigsqcup,\dots \bigsqcup B_{j}} \right] \\
&=\frac{1}{   P\left( \bigsqcup_{j=1}^{n} B_{j} \right)  }\mathbb{E}\left[ X \left[ \mathbf{1}_{B_{1}} + \mathbf{1}_{B_{2}} + \dots + \mathbf{1}_{B_{j}} \right] \right] \\
&=\frac{1}{\sum_{j=1}^{n} P(B_{j})} \mathbb{E}(X\cdot \mathbf{1}_{B_{1}}) + \mathbb{E}(X \cdot \mathbf{1}_{B_{2}}) + \dots +\mathbb{E}(X \cdot \mathbf{1}_{B_{j}}) \\
&=\sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}}
\end{align*}

But why it says $B_{j1},B_{j2},\dots,B_{jk}$

Proposal :

\begin{align*} \mathbb{E}\left[X | B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}\right] &= \frac{1}{P(B_{j1} \bigsqcup B_{j2} \bigsqcup \dots \bigsqcup B_{jk})} \mathbb{E} \left[X \cdot \mathbf{1}_{B_{j1} \bigsqcup B_{j2} \bigsqcup,\dots \bigsqcup B_{jk}} \right] & \text{by definition}\\
&=\frac{1}{   P\left( \bigsqcup_{j=1}^{n} B_{j} \right)  }\mathbb{E}\left[ X \left[ \mathbf{1}_{B_{j1}} + \mathbf{1}_{B_{j2}} + \dots + \mathbf{1}_{B_{jk}} \right] \right]& \text{by mutual exclusivity} \\
&=\frac{1}{\sum_{j=1}^{n} P(B_{j})} \mathbb{E}(X\cdot \mathbf{1}_{B_{j1}}) + \mathbb{E}(X \cdot \mathbf{1}_{B_{j2}}) + \dots +\mathbb{E}(X \cdot \mathbf{1}_{B_{jk}})&\text{by linearity of expectation} \\
&=\sum_{j=1}^{n}\mathbb{E}[X| B_{j}]\mathbf{1}_{B_{j}}
\end{align*}

答:
Hint

What has to be proved it's that for all $F\in \mathcal P$ (i.e. for $F$ of the form $B_{i_1}\sqcup\cdots \sqcup B_{i_k}$), $$\mathbb E[X\boldsymbol 1_F]=\mathbb E\left[\sum_{j=1}^n \mathbb E[X|B_j]\boldsymbol 1_{B_j}\boldsymbol 1_F\right].$$

Try to adapt what you did in your original post to $\mathbb E[X\boldsymbol 1_F]$ and conclude.

Edit

We can suppose WLOG that $F=B_1\sqcup\cdots B_k$ (where $k\leq n$).

Then

\begin{align*}
\mathbb E[X\boldsymbol 1_F]&=\sum_{i=1}^k\mathbb E[X_i\boldsymbol 1_{B_i}]\\
&= \sum_{i=1}^k\mathbb P(B_i)\mathbb E[X_i\mid B_i]\\
&=\mathbb E\left[\sum_{i=1}^k\mathbb E[X\mid B_i]\boldsymbol 1_{B_i}\right]\\
&=\mathbb E\left[\sum_{i=1}^n\mathbb E[X\mid B_i]\boldsymbol 1_{B_i}\boldsymbol 1_F\right],
\end{align*}
as wished.


答:
$$\int_{2}^{x} \frac{dt}{t \log^{3} t} = \frac12\left(\frac{1}{\log^{2}2}-\frac{1}{\log^{2}x}\right)<\frac{1}{2\log^{2}2}$$
How much logic would one need for that? How much logic can one muster while seriously combining "rigorous" and "intuition"?

Title: Techniques of estimating integrals, asymptotic behavior of Li(x)

Body:
I need help understanding the logic and build an rigorous intuition proving that if $x \gt 2$, then:

$$\int_{2}^{x} \frac{dt}{t \log^{3} t} \lt \frac{1}{2 \log^{2}2}$$


问:
Why is $A \rtimes B \simeq \mathbb Z \rtimes \mathbb Z/2\mathbb Z$
Hello everyone I have a hard time trying to resolve this problem if anyone could help it would be a lot appreciated.

Let
$$f_1\colon\mathbb R\rightarrow \mathbb R,\,f_1(x)=-x,\quad f_2\colon\mathbb R\rightarrow \mathbb R,\,f_2(x) = -x +1$$

We define $D_\infty = \langle f_1,f_2 \rangle$ , the infinite dihedral group and $A = \langle f_2 \circ f_1 \rangle$ and  $B=\langle f_2\rangle$,  then we have :

$$D_\infty = A \rtimes B \simeq \mathbb Z \rtimes \mathbb Z/2\mathbb Z$$

How can we prove that statement ?

Should we prove that  $f_1 ^2=f_2 ^2=1$ , so I can use the following answer ? I also found here exactly what I want but how can I prove it using $f_1$ and $f_2$.

( How can we also prove infinite order of $\langle f_2 \circ f_1\rangle$, I thought maybe we have to compute $(f_2 \circ f_1)^k(x)$?)

答:
$B$ and $\frac{\Bbb Z}{2\Bbb Z}$ act faithfully on $A$ and $\Bbb Z$, respectively, via $A$ automorphisms and $\Bbb Z$ automorphisms. The map $$\Psi:\langle f_2\circ f_1\rangle\rtimes \langle f_2\rangle\to\Bbb Z\rtimes\frac{\Bbb Z}{2\Bbb Z}\;\;\;\;\;\;\;\;\;\Psi:((f_2\circ f_1)^j,f_2^k)\mapsto (j,\mathbf k)\;\;\;\;$$ is obviously a bijection where $j\cdot(-1)^k$ is the action of the residue class $k+2\Bbb Z$ on the integer $j$ while$$f_2^k\circ(f_2\circ f_1)^j\circ f_2^k=(f_2\circ f_1)^{j\cdot(-1)^k}$$ is the action of $f_2^k$ on $(f_2\circ f_1)^j$ for any $j,k\in\Bbb Z$. Consequently,

$$\begin{align}
\Psi(((f_2\circ f_1)^j,f_2^k)\cdot((f_2\circ f_1)^s,f_2^t))&=\Psi((f_2\circ f_1)^{j+s\cdot(-1)^k},f_2^{k+t}))\\
&=(j+s\cdot(-1)^k,k+t+2\Bbb Z)\\
&=(j,k+2\Bbb Z)\cdot(s,t+\Bbb Z)\\
&=\Psi(((f_2\circ f_1)^j,f_2^k))\cdot\Psi((f_2\circ f_1)^s,f_2^t))
\end{align}$$ $$\therefore\;\langle f_2\circ f_1\rangle\rtimes \langle f_2\rangle\approx\Bbb Z\rtimes\frac{\Bbb Z}{2\Bbb Z}$$


问:
Number of sequences in A, B, and C, where number of A's is even and number of B's is odd.
Given three letters A,B and C. Find number of words of length 11 in which A occurs even number of times and B occurs odd number of times.

( Caution: A can occur 0 number of times and there is no restriction on C )

My Try: I tried making cases but it doesn't work as there will be more than $50$ cases.

Is there any mechanism to solve this?

Even a trinomial expansion won't work as again we have to find possibilities for $n+m+l = 11$ with n even and m odd.

答:
If there were $n$ letters, the number of valid sequences would be $(3^n-(-1)^n)/4$.

To prove this, call a sequence valid if it satisfies your given conditions, and call a sequence partially valid if the number of occurrences of "A" or "B" is odd. There are two steps:

The number of partially valid sequences is $(3^n-(-1)^n)/2$. To see this, we will divide the set of $3^n$ sequences into pairs, where one sequence in each pair is valid. Namely, the mate of a given sequence is found by changing the first occurrence of "A" or "C" into the other letter, "C" or "A". This pairs up almost all sequences; the only one unpaired is the sequence of all "B"s. This is valid if and only if $n$ is odd. Therefore,
$$
\text{# partially valid sequences}=
\begin{cases}
(3^n-1)/2 & n \text{ is even}
\\
(3^n-1)/2+1 & n \text{ is odd}
\end{cases}
$$
This works out to be the same as $(3^n-(-1)^n)/2$.

Exactly half of the partially valid sequences are valid. A similar method works; given a partially valid sequence, you can change whether or not it is valid by changing the first instance of "A" or "B" to the other letter, "B" or "A".

The matching, or pairing, I described in both steps is often referred to as a sign reversing involution.

Let me finish by saying that I think the best way to do this problem is to use EGF's, as awkward suggested in a comment. When you do so, a very simple formula falls out, which motivated me to look for this combinatorial proof. That is, I did not pull this out of thin air.


问:
Let $F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$. Find $F'\left(x\right)$
Let $$F\left(x\right)=\int_{t^2}^{x^3}{\frac{\mathrm{d}y}{\sqrt{x^2+y^4}}}$$ Find $F'\left(x\right)$

Had the lower limit been a function of $x$ I would have easily calculated. I am not very familiar with the Newton-Leibniz rule in two variables.  I do not know which formula to use.

答:
A simple solution to our problem would be either removing $ x $ from the integrand, or removing it from the bounds, in general it is not always possible to remove the $ x $ from the integrand, it's always possible to remove it from the bounds though.

Since we're dealing with : $ F_{t}: x\mapsto\int_{t^{2}}^{x^{3}}{\frac{\mathrm{d}y}{\sqrt{x^{2}+y^{4}}}} $, then, for $ \left(x,t\right)\in \mathbb{R}^{2} $, using the change of variable $ \left\lbrace\begin{aligned}y&=\left(x^{3}-t^{2}\right)u+t^{2}\\ \mathrm{d}y&=\left(x^{3}-t^{2}\right)\mathrm{d}u\end{aligned}\right. $, we're left with : $$ F\left(x\right)=\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}} $$

It's now possible to use the product rule, then to switch the derivative with the integral sign in order to calculate our derivative : \begin{aligned} \small\frac{\mathrm{d}F}{\mathrm{d}x}\left(x\right)&=\small 3x^{2}\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}}+\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{\partial}{\partial x}\left(\left(x,u\right)\mapsto\frac{1}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}\right)\,\mathrm{d}u} \\ &=\small 3x^{2}\int_{0}^{1}{\frac{\mathrm{d}u}{\sqrt{x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}}}}-\left(x^{3}-t^{2}\right)\int_{0}^{1}{\frac{12u x^{2}\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{3}+2x}{2\left(x^{2}+\left(\left(x^{3}-t^{2}\right)u+t^{2}\right)^{4}\right)^{\frac{3}{2}}}\,\mathrm{d}u}\\ \frac{\mathrm{d}F}{\mathrm{d}x}\left(x\right)&=\frac{3x^{2}}{x^{3}-t^{2}}\int_{t^{2}}^{x^{3}}{\frac{\mathrm{d}y}{\sqrt{x^{2}+y^{4}}}}-\frac{1}{x^{3}-t^{2}}\int_{t^{2}}^{x^{3}}{\frac{6\left(y-t^{2}\right)x^{2}y^{3}+\left(x^{3}-t^{2}\right)}{\left(x^{2}+y^{4}\right)\sqrt{x^{2}+y^{4}}}\,\mathrm{d}y}\end{aligned}


问:
There is a plane given the equation. If a sphere is tangent to this plane, what is the equation for the sphere?(Sphere Center: ${M(4,-2.3)}$)
This is plane equation:

${-2x+y-2z+7=0}$

What is the equation for the sphere tangent to this plane? Center of the sphere = ${M(4,-2,3)}$

Sphere equation form:

${(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=r^{2}}$

If center values are placed in place:

${(x-4)^{2}+(y+2)^{2}+(z-3)^{2}=r^{2}}$

So,
Since the sphere is tangent to the plane, they intersect at a common point, and this intersection ${P(x,y,z)}$ point must satisfy both the plane equation and the sphere equation, right?

I request your help.

答:
The norm of the normal vector $(-2, 1, 2)$ is $3$. Hence the radius of the sphere is
\begin{equation}
r = |-2(4)+(-2)-2(3)+7|/3 = 9/3=3
\end{equation}
The point where they cross is $(4, -2, 3) \pm (-2, 1, -2)r/3 = (2, -1, 1)$.


答:
How can I parametrize $\mathcal{P}$?

This answer should help.

Taking the example ($n=3$), and given that here the marginals are uniform, hence $E[X_i]=0$, we have the parametrization

$$ P(x_1,x_2,x_3)=\frac{1}{2^3}\left( a_{123} x_1 x_2 x_3 +
a_{12} x_1 x_2  + a_{23} x_2 x_3  + a_{13} x_1 x_3  +
1
\right) $$

where $a_{i...k}=E[X_i ... X_k]$. In particular, here, $a_{i,j}=E[X_i X_j] = C_{i,j}$ for $i\ne j$.

(Obvioulsy $C_{i,i}=1$)

I think this answers your second question. Fixing the covariance matrix only fixes the second-order terms (just as fixing the marginals fixes the first order terms). This gives a one-to-one mapping only for the bivariate case. For $n>2$ there are additional terms, which correspond to additional degrees of freedom.
Hence, for a given covariance matrix there exist (in general) several (infinite) distributions.

Title: On Probability Measures on $\{-1,+1\}^n$

Body:
Let's put it formally...

Let $n$ be a positive integer and $E = \{-1,+1\}^n$.
Consider the set $\mathcal{P}$ of all the probability measures $\mu$ on $E$ such that
$$
\mu ( \pi_i^{-1} \{-1\} ) = \mu ( \pi_i^{-1} \{+1\} ) = \frac12
$$
where $\pi_i : E \rightarrow \{-1,+1\}$ is the canonical projection onto the $i^\text{th}$ component of $E$, i.e. $\pi_i(x_1, \dots, x_n) = x_i$.

How can I parametrize $\mathcal{P}$?

Also, letting $\mathcal{C}$ be the set of all $n \times n$ correlation matrices, consider the map
$c: \mathcal{P} \rightarrow \mathcal{C} $ that assigns to each probability measure in $\mathcal{P}$ its correlation matrix.
Is $c$ injective? Surjective?

For the simple case $n=2$, it is clear to me that we can put $\mathcal{P}$ in a bijection with $[0,\frac12]$. For $a \in [0,\frac12]$, define $\mu$ as
$$
\mu \{-1,-1\} = \mu \{+1,+1\} = a \\
\mu \{-1,+1\} = \mu \{+1,-1\} = \frac12 - a
$$
It's also clear that $c$ is bijective since the correlation is $4a-1$.

But what about the case of arbitrary $n$?


问:
How did they get an analytical solution to this second-order nonlinear PDE?
A (non-math) paper I'm working through presents the following differential equation (and solution) in a casual way:

$$
0 = \frac{(1-\gamma)^{2}}{\gamma} e^{-\rho t} \bigg[ \frac{e^{\rho t} J_{x}}{\beta} \bigg]^{\frac{\gamma}{\gamma-1}} + J_{t} + [(1-\gamma) \eta / \beta + r x] J_{x} - \frac{J_{x}^{2}}{J_{xx}} \frac{(\alpha - r)^{2}}{\alpha \sigma^{2}},
$$
subject to $J(x,T)=0$.

A solution is
$$
J(x,t) = \delta \beta^{-\gamma} e^{-\rho t} \bigg[ \frac{\delta(1-e^{-\left( \frac{\rho - \gamma \nu}{\delta}\right)(T-t)})}{\rho - \gamma \nu} \bigg]^{\delta} \bigg[ \frac{x}{\delta} + \frac{\eta}{\beta r}(1-e^{-r(T-t)}) \bigg]^{\gamma},
$$
where $\delta \equiv 1 - \gamma$ and $\nu \equiv r + (\alpha - r)^{2}/2 \delta \sigma^{2}$.

It seemed rather matter-of-fact, but I would have had no idea how to solve (what appears to be) such a complicated PDE.
I've been looking through Polyanin's Handbook of Nonlinear Partial Differential Equations, but I haven't (yet) been able to find anything.

Any ideas on how they did it?
(Or recommendations for relevant references?)

答:
The second derivative $J_{xx}$ in the denominator is unusual, but maybe
some progress is possible. The explicit time exponentials can be
removed by substituting
$$
      J(x,t) = e^{(\gamma-1)\rho t}Y(x,t).
$$
I won't keep track of the many constants, but you get an equation of
the form
$$
 c_1 Y_{x}^{\gamma/(\gamma-1)}
     +c_2Y+c_3Y_t
    +(c_4+rx)Y_x-c_5\frac{Y_x^2}{Y_{xx}}
     = 0.
$$
It is still a strange equation. But maybe it suggests powers of
something plus $x$. So try
$$
 Y(x,t) = (h(t)+x)^\gamma.
$$
We get
$$
 c_1\gamma^{\gamma/(\gamma-1)}(h+x)^\gamma
    +
 c_2(h+x)^\gamma+c_3\gamma(h+x)^{\gamma-1}h_t
 $$ $$
 +
 (c_4+rx)\gamma(h+x)^{\gamma-1}
 -c_5\frac{(\gamma(h+x)^{\gamma-1})^2}{\gamma(\gamma-1)(h+x)^{\gamma-2}}
 = 0.
$$
Factor out $(h+x)^\gamma$ to get
$$
 c_1\gamma^{\gamma/(\gamma-1)}
 +
 c_2+\gamma\frac{c_3 h_t+c_4+rx}{h+x}
  -c_5\frac{\gamma}{\gamma-1}
 = 0.
$$
It is easy to find a function $h(t)$ satisfying
$$
 c_3h_t+c_4 = rh.
$$
Then you are left with the question whether the constants
$$
 c_1\gamma^{\gamma/(\gamma-1)}
 +
 c_2+\gamma r
  -c_5\frac{\gamma}{\gamma-1}
 = 0?
$$
If so then it seems that you have a solution. I have not tried
to see whether it is the same as the one proposed in the paper, or
even check the intial value.

Of course, the equation is so unusual that we probably don't
know whether the intial value problem has a unique solution.


问:
Is it possible that some Integrals do not have solutions? What about this one?
I cannot figure out how to solve this integral with the commonly known methods of integration. The different online integral calculators that I have tried fail to solve it.

$ \int  \sin(a x) ((b-x)^2-c^2)^{-3/2} dx $

where a, b and c are constants.

Is there a solution? Thanks.

答:
Unlike derivatives, not every elementary function has an elementary antiderivative. Elementary meaning polynomials, trigonometric functions, logarithmic functions, exponential functions and their inverse functions.

One example is $$\int e^{-x^2} dx$$

which cannot be represented by any elementary function. I am pretty certain your function cannot be represented in elementary terms.


问:
Is the topology induced by the bases of a finite-dimensional real vector space equal to the topology induced by the norms?
It is well known that all norms on a $n$-dimensional real vector space induce the same topology (e.g. John Lee, Introduction to Smooth Manifolds, 2nd edition, Example 1.24). However, bases induce a "natural" topolgy, too: Let $(v_1,\ldots,v_n)$ be a basis, then we can define a topolgy by requiring that the isomorphism
\begin{align}
\Phi\colon\mathbf{R}^n&\to V\\
(x_1,\ldots,x_n)&\mapsto\sum_{i=1}^nx_iv_i
\end{align}
(and its inverse) are continuous: That is, $U\subset V$ is open if and only if $\Phi^{-1}(U)$ is open. One can show that the topology doesn't depend on the basis by using the fact that all linear maps from $\mathbf{R}^n$ to $\mathbf{R}^n$ are continuous, $L(\mathbf{R}^n,\mathbf{R}^n)\subset C(\mathbf{R}^n,\mathbf{R}^n)$.

My question is: Is the topology induced by the bases of a finite dimensional real vector space equal to (or contained in) the topology induced by the norms?

答: It's the same topology, since it is induced by the norm$$\left\|\sum_{i=1}^nx_iv_1\right\|=\sqrt{\sum_{i=1}^nx_i^{\,2}}.$$And this is true because, with respect to this norm, both $\Phi$ and its inverse are continuous.


问:
convexity of a piecewise function
How can I show that the below function is convex in $x$ and convex in $e$?  $x_1 , x_2\geq 0$ and $e$ is a random variable which has a limited expectation.

\begin{equation}
Q(x,e)=\begin{cases}
1-x_1 & 0\leq e <x_1+x_2\\
e+1-2x_1-x_2 & x_1+x_2 \leq e <1+x_2\\
2(e-x_1-x_2) & 1+x_2\leq e
\end{cases}
\end{equation}

答:
It seems like there is an error in the function definition. Take $x_1=e=2$ fixed, then the function simplifies to:
$$f(x) = Q((2,x),2) = \begin{cases}-1 & 0 \leq 2 < 2+x \\ -1-x & 2+x \leq 2 < 1+x \\ -2x & 1+x \leq 2 \end{cases}$$
The second case can never occur, and the other two cases can be simplified to:
$$f(x) = \begin{cases}-1 & 0 < x \\ -2x & x \leq 1 \end{cases}$$
How do I evaluate this function at $x=0.5$?

Title: convexity of a piecewise function


问:
Upper bound for - $\sum_{i=1}^n i {2n \choose n+i} $.
I am tryin to calculate an upper bound for the following sum:
$\sum_{i=1}^n i {2n \choose n+i} $.
I know that the limit should be $\Theta (4^n\sqrt n)$ but I am not sure how to proceed. I already tried to bound by the biggest element, but the bound was too high.

答:
Here's an outline of something that might work:

Show that $\binom{2n}{n+i}\leq \binom{2n}{n}\cdot \exp(-\Omega(i^2/n)).$ (The inequality $1-x\leq \exp(-x)$ might come in handy.)

Use Stirling's approximation to note that $\binom{2n}{n} =\Theta(4^n/\sqrt{n})$. (You can also try to prove it directly. Hint here.)

The above should reduce your problem to upper bounding a sum of the form $\sum_{i} i\cdot \exp(-\Omega(i^2/n)).$ Upper bound this by a suitable integral to get the result.


答: Yes. Let $\Phi : \mathbb R^n \to V$ be a linear isomorphism and $\lVert - \rVert$ be a norm on $\mathbb R^n$. Then $\lVert v \rVert_\phi = \lVert \Phi^{-1}(v) \rVert$ defines a norm on $V$. Clearly the norm-topology induced by $\lVert - \rVert_\phi$ is the same as the topology obtained by transferring the open sets in $\mathbb R^n$ via $\Phi$ to $V$.

Title: Is the topology induced by the bases of a finite-dimensional real vector space equal to the topology induced by the norms?

Body:
It is well known that all norms on a $n$-dimensional real vector space induce the same topology (e.g. John Lee, Introduction to Smooth Manifolds, 2nd edition, Example 1.24). However, bases induce a "natural" topolgy, too: Let $(v_1,\ldots,v_n)$ be a basis, then we can define a topolgy by requiring that the isomorphism
\begin{align}
\Phi\colon\mathbf{R}^n&\to V\\
(x_1,\ldots,x_n)&\mapsto\sum_{i=1}^nx_iv_i
\end{align}
(and its inverse) are continuous: That is, $U\subset V$ is open if and only if $\Phi^{-1}(U)$ is open. One can show that the topology doesn't depend on the basis by using the fact that all linear maps from $\mathbf{R}^n$ to $\mathbf{R}^n$ are continuous, $L(\mathbf{R}^n,\mathbf{R}^n)\subset C(\mathbf{R}^n,\mathbf{R}^n)$.

My question is: Is the topology induced by the bases of a finite dimensional real vector space equal to (or contained in) the topology induced by the norms?


问:
There are $2^{\omega(a)}$ pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$
It looks as there are $2^{\omega(a)}$ ordered pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$, where $\omega(a)$ is the number of different prime factors of $a$.

Proof?

It's also true that there are  $n^{\omega(a)}$ n-tipples $(a_1,\dots ,a_n)$ such that $a=a_1\cdots a_n$.

答:
Hint: a pair $\,(b,c)\,$ corresponds to a partition $\, A = B\cup C\, $ of the set of primary factors $\,p_i^{e_i}\,$ of $\,a,\,$ hence mapping $\,b\,$ to its corresponding subset $B$ of the partition bijects these pairs with all subsets of the set $A$ of primary factors of $a$, of which there are $2^n,\,$ since $a$ has $n$ distinct prime factors.

For example if $\,a = 2^i 3^j 5^k$ then $\,A = \{ 2^i, 3^j, 5^k\}$ and the coprime factorization corresponding to the partition $\, A = \{2^i ,3^j\}\cup \{5^k\}\,$ is $\,a = \color{#c00}{2^i 3^j}\times 5^c = \color{#c00}b\times c,\,$ where $\,b\,$ corresponds to the product of the elements in the first set, i.e. $\,\color{#c00}{b = \prod\, \{2^i,3^j\}}$

Equivalently these coprime factorizations correspond to  idempotents of $\,\Bbb Z/a \,\cong\, \Bbb Z/b \times \Bbb Z/c,\,$ elements which govern such ring factorizations. This correspondence between element vs. ring factorizations holds much more generally - as proved here if $R$ is a UFD then proper factorizations of the ring $\,R/a\,$ correspond to coprime (i.e. comaximal) factorizations of the element $\,a = b\times c,\,$ just like in the classical CRT = Chinese Remainder Theorem.


问:
A delta function conjecture: almost any function can be a delta kernel
I have been thinking about delta kernels, and I think I have come up with a surprising result:

If $f\colon\Bbb{R}\to\Bbb{R}$ is such that $\int_{-\infty}^\infty f(x)\,dx=L$ is finite and nonzero, then $g_k(x)=\frac kLf(kx)$ is a delta kernel.

Note that by "delta kernel", I mean a sequence of integrable functions $g_k(x)$ such that for any absolutely integrable function $h\colon\Bbb{R}\to\Bbb{R}$ that is continuous at $0$, $\lim_{k\to\infty}\int_{-\infty}^\infty g_k(x)h(x)\,dx=h(0)$.

I find this an interesting problem to think about, so I thought I'd post it here. I'll post the proof as an answer if no one else does (unless I've missed some extra hypotheses, in which case someone will point it out, I'm sure).

答:
There is already a theorem which resembles your result:

Suppose $d$ is a non-negative function with the property:$$\int_{-\infty}^{+\infty }d(s)\,ds=1$$Then the sequence $d_k(t):=k\cdot d(kt)$ is a Dirac sequence.

A Dirac sequence has the properties:

$d_k\geq0,\forall k$

$\int_{-\infty}^{+\infty }d_k(s)\,ds=1,\forall k$

$\forall r>0$ and $\forall \varepsilon>0,$ there exists $N\in\mathbb N$ such that $\forall k>N$ we have $\int_{\mathbb{R}\setminus[-r,r]}d_k(s)\,ds<\varepsilon$


问:
translation followed by rotation
I am kind of confused about 2D transformations. I read that rotation always has to be done about the origin. But there are questions in which you are supposed to translate first and then rotate. Translation will move the shape away from the origin. </p>

答:
I guess that OP's confusing is: why there is a different effect between translation followed by rotation and rotation followed by translation. If so, I give here simple explanation, both for rotation (around the origin) and rotation around some point.

Mathematical Explanation Translation and rotation are transformations, say $T(x)=x+b$ and $R(x)=Ax$ with $A$ is $2\times 2$ matrix; and in generally, their product
$$T(R(x))=T(Ax)=Ax+b,$$
$$R(Tx)=R(x+b)=A(x+b)=Ax+Ab$$
is NOT commutative, that is $T R\not= R T$.

Graphical Explanation For the simplest case, given a point $A$. When rotating it with angle ROT and then shifting to SHI, we get the red curve. Similarly, when shifting it to SHI and then rotating it with angle ROT, we get the blue curve. As you see, they are much different.

```<code>// http://asymptote.ualberta.ca/
unitsize(1cm);
import math;
import graph;
add(grid(5,5,gray));
axes("<span class="math-container">$x$</span>","<span class="math-container">$y$</span>",gray);
pair A=(2.5,1), O=(0,0);
real ROT=35;
pair SHI=(2,1);
pair M=shift(SHI)*rotate(ROT)*A;
pair N=rotate(ROT)*shift(SHI)*A;

path A1=arc(O,abs(A),degrees(A),degrees(A)+ROT);
path  A1r=relpoint(A1,1)--relpoint(A1,1)+SHI;
path AtoM=A1&A1r;
draw(AtoM,red);

pair A2=A+SHI;
path AtoN=A--A2&arc(O,abs(A2),degrees(A2),degrees(A2)+ROT);
draw(AtoN,blue);

dot("<span class="math-container">$O$</span>",O,SW);
dot("<span class="math-container">$A$</span>",A,S);
dot(Label("<span class="math-container">$M$</span>",Fill(white)),M,1.5S,red);
dot("<span class="math-container">$N$</span>",N,plain.N,blue);
</code>```

Title: translation followed by rotation

Body:
I guess that OP's confusing is: why there is a different effect between translation followed by rotation and rotation followed by translation. If so, I give here simple explanation, both for rotation (around the origin) and rotation around some point.

Mathematical Explanation Translation and rotation are transformations, say $T(x)=x+b$ and $R(x)=Ax$ with $A$ is $2\times 2$ matrix; and in generally, their product
$$T(R(x))=T(Ax)=Ax+b,$$
$$R(Tx)=R(x+b)=A(x+b)=Ax+Ab$$
is NOT commutative, that is $T R\not= R T$.

Graphical Explanation For the simplest case, given a point $A$. When rotating it with angle ROT and then shifting to SHI, we get the red curve. Similarly, when shifting it to SHI and then rotating it with angle ROT, we get the blue curve. As you see, they are much different.

// http://asymptote.ualberta.ca/
unitsize(1cm);
import math;
import graph;
add(grid(5,5,gray));
axes("$x$","$y$",gray);
pair A=(2.5,1), O=(0,0);
real ROT=35;
pair SHI=(2,1);
pair M=shift(SHI)*rotate(ROT)*A;
pair N=rotate(ROT)*shift(SHI)*A;

path A1=arc(O,abs(A),degrees(A),degrees(A)+ROT);
path  A1r=relpoint(A1,1)--relpoint(A1,1)+SHI;
path AtoM=A1&A1r;
draw(AtoM,red);

pair A2=A+SHI;
path AtoN=A--A2&arc(O,abs(A2),degrees(A2),degrees(A2)+ROT);
draw(AtoN,blue);

dot("$O$",O,SW);
dot("$A$",A,S);
dot(Label("$M$",Fill(white)),M,1.5S,red);
dot("$N$",N,plain.N,blue);