问:
Proving expression with binomial coefficients is integer
How do we prove that $\frac{\binom{n+1}{m+1}\binom{n}{m-1}}{m}$ for $n\ge m\ge1$ is an integer?

Both coefficients in the numerator are integers and the interval from $n+1$ to $n-m+2$ covered by the numerators of the binomial coefficients has length $m$. However, that doesn't guarantee the whole thing is an integer because those factors could be canceled by the denominators of the binomial coefficients that are not relatively prime to m.

If $m$ is prime then it's more obviously an integer because if the $m$ in the denominator of $\binom{n+1}{m+1}$ divides $n+1$ then the $m$ in the denominator of the whole expression cancels $n-m+2$, and otherwise, there are two multiples in the numerator. But how do we work with non-primes?

I also tried a counting argument but couldn't get anywhere with that.

答:
Use
$$
\frac{\binom{n+1}{m+1}\cdot\binom{n}{m-1}}{m}=\frac{\frac{n+1}{m+1}\binom{n}{m}\cdot\frac{m}{n+1}\binom{n+1}{m}}{m}
$$
and the fact that $m$ and $m+1$ are relatively prime.

Added: One can, in retrospect, devise a combinatorial interpretation. Define two sets:

$A$ is the set of pairs of words, the first word of the pair an $(n+1)$-letter word with $n-m$ $X$s and $m+1$ $Y$s, one of the $Y$s underlined, and the second word of the pair an $n$-letter word with $n-m+1$ $x$s and $m-1$ $y$s;

$B$ is the set of pairs of words, the first word of the pair an $n$-letter word with $n-m$ $X$s and $m$ $Y$s, and the second word of the pair an $(n+1)$-letter word with $n-m+1$ $x$s and $m$ $y$s, one of the $y$s underlined.

There is a bijective map from $A$ to $B$ that deletes the underlined $Y$ from the first word and inserts an underlined $y$ at the corresponding position in the second word. So $A$ and $B$ are of equal size, which gives combinatorial meaning to
$$
(m+1)\binom{n+1}{m+1}\binom{n}{m-1}=m\binom{n}{m}\binom{n+1}{m}.
$$
Because $m$ and $m+1$ have no common factor, both sides are divisible by $m(m+1)$.


问:
is it necessary for a function to be defined everywhere in it's domain to be onto or one-to-one mapping?
[this is question. I want to know about iv.] I want to know that without being defined everywhere ,can a  mapping be onto and one-to-one?
In iv - D has four elements and B has three elements, while in question only three elements are used. So it cannot be function as per definition. Then how it will onto or one to one without function? Is it possible?
1</p>

答: One- one and onto is merely a type of function . So yes the function needs to be defined everywhere in it's domain for it to be one - one or onto. The question you are referring to  will therefore will be neither defined in domain nor one- one or onto</p>

ContentLicense: CC BY-SA 4.0


问:
$\sigma$-algebra + measure implies premeasure

Definition: By a measure $\mu$ on a measurable space
$\left(X, \mathcal{M}\right)$ we mean an extended real-valued
nonnegative set function $\mu\colon\mathcal{M}\to[0, \infty]$ for which $\mu(\emptyset) = 0$ and which is countably additive.

Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0, \infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.

Question: Show that if a set function on a $\sigma$-algebra is a measure then it's a premeasure.

Let $\mu\colon\mathcal{M}\to[0,\infty]$ be a measure on a $\sigma$-algebra $\mathcal{M}$.

Firstly, as $\mu$ is a measure, by definition it follows that $\mu(\emptyset)=0$.

Finite additivity immediately follows from countable additivity by setting $E_k = \emptyset$, so that
$μ(E_k) = 0$, for $k > n$.

For countable monotonicity, let $\{E_k\}_{k=1}^{\infty}$ be any countable collection of measurable that covers a measurable set $E$. Define $F_1 = E_1$ and then define

$$F_n = E_k\setminus\left[\bigcup_{i=1}^{k-1}E_i\right]\text{ for } k\geq 2.$$

Immediately, it's seen that

$$\{F_k\}_{k=1}^{\infty}\text{ is disjoint, } \bigcup_{k=1}^{\infty}F_k = \bigcup_{k=1}^{\infty}E_k\text{ and } F_k\subseteq E_k\text{ for every } k.$$

From countable additivity and monotonicity of $\mu$ being measure,

$$\mu(E)\leq\mu\left(\bigcup_{k=1}^{\infty}E_k\right)=\mu\left(\bigcup_{k=1}^{\infty}F_k\right) = \sum_{k=1}^{\infty}\mu(F_k)\leq\sum_{k=1}^{\infty}\mu(E_k).$$

Is this correct? Also, I'm not sure the fact that $\mathcal{M}$ is a $\sigma$-algebra comes in?

答:
Your proof is correct. It can be made a little shorter.

Definition: By a measure $\mu$ on a measurable space
$\left(X, \mathcal{M}\right)$ we mean an extended real-valued
nonnegative set function $\mu\colon\mathcal{M}\to[0, \infty]$ for which $\mu(\emptyset) = 0$ and which is countably additive.

Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0, \infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.

Question: Show that if a set function on a $\sigma$-algebra is a measure then it's a premeasure.

Proof: Let $\mu\colon\mathcal{M}\to[0,\infty]$ be a measure on a $\sigma$-algebra $\mathcal{M}$.

Firstly, as $\mu$ is a measure, by definition it follows that $\mu(\emptyset)=0$.

Finite additivity immediately follows from countable additivity by setting $E_k = \emptyset$, so that
$μ(E_k) = 0$, for $k > n$.

For countable monotonicity, let $\{E_k\}_{k=1}^{\infty}$ be any countable collection of measurable that covers a measurable set $E$.

Since $\mathcal{M}$ is a $\sigma$-algebra, we have that $\bigcup_{k=1}^{\infty}E_k \in \mathcal{M}$. Since $\mu$ is a measure defined on $\mathcal{M}$, we have
$$\mu(E)\leq\mu\left(\bigcup_{k=1}^{\infty}E_k\right)\leq\sum_{k=1}^{\infty}\mu(E_k)$$
(you don't need to go through the sets $F_n$).


问:
What does abbreviation Ber(x) stand for?
I am trying to understand the protocol in the following picture for shuffled differential privacy from this paper: Distributed Differential Privacy via Shuffling.

Here, the authors use an abbreviation Ber(...) that I did not came accross yet. From my knowledge it does not mean the bit error rate. I come from a computer science background and did not yet get into too much statistics, so the abbreviation might be clear in general. Does anybody know, what they mean ?

答: In the context of probability and statistics, Ber could be short for Bernoulli.</p>


问:
Besse's Einstein manifolds typo: Euler characteristics of complex projective space is $m$
In

Besse, Arthur L., Einstein manifolds, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge, Bd. 10. Berlin etc.: Springer-Verlag. XII, 510 p. DM 198.00 (1987). ZBL0613.53001.

On page 161 there is a suspicious equality $\chi(\mathbb{C}\mathrm{P}^m)=m$. There is nothing in the errata list on last page of the book. Is there something here that I missed it or it is a typo and it should be $\chi(\mathbb{C}\mathrm{P}^m)=m+1$?

答: This is a typo. All other formulas on the page that follow are for the unreduced Euler characteristic where $\chi(\Bbb{CP}^m) = m+1$.


问:
Confusion regarding limiting variances (Casella, Statistical Inference, 2nd edition, example 10.1.8)
In the book Statistical Inference (George Casella 2nd ed.), page 470, there is an example:

$\bar{X}_n$ is the mean of $n$ iid observations, and E$X=\mu$, $\operatorname{Var}X=\sigma^2$. "If we take $T_n=1/\bar{X}_n$, we find that the variance is $\operatorname{Var}T_n=\infty$, so the limit of the variances is infinity." Why is the limit of the variance infinity? I can go as far as

$$\operatorname{Var}\frac 1{\bar{X}_n}=\operatorname{E}_{\bar{X}_n}\left[\left(\frac1{\bar{X}_n}-\operatorname{E} \frac 1{\bar X_n} \right)^2 \mid \mu\right] $$

what's next? I know that $\lim_{n\to\infty}\operatorname{Var}{\bar X_n}=0$. However, $\lim_{n\to\infty}1/\operatorname{Var}{\bar X_n}\not=\lim_{n\to\infty}\operatorname{Var}1/\bar X_n$. How can we show the variance approaches to infinity for sufficiently large $n$?

Thanks!

答:
There are many issues with Example 10.1.8 in Casella's book which ultimately lead to a lot of confusion for me when I was working through it a couple years ago. I have seen other posts on MSE that express similar confusions and so this is my attempt to make sense out of the words and equations Casella gives in the example. While I am beginning to see the bigger picture of what's going on here, I have had, and continue to have, trouble explaining it concisely. My hope is someone will read my comments and be able to piece them together into a more concise and clear answer. With that said, let me address each issue in this example separately and hopefully bring at least a little clarity to the situation.

Issue #1 (typo):

First of all, as pointed out in this question there is a typo in the first paragrpah of the example. The correct sentence should read:

...if we take $T_n=\bar X_n$, then $\lim n\mathsf{Var}\bar X_n=\sigma^2$ is the limiting variance of $T_n$.

Issue #2 ($\mathsf{Var}T_n\neq\infty$):

Now in the second paragraph we read:

If we now take $T_n=1/\bar X_n$, we find that the variance is $\mathsf{Var}T_n=\infty$, so the limit of the variances is infinity.

This claim was my first real source of confusion. You see, the variance is not infinite but rather undefined. For an explanation why, see my answer here. In fact, using similar arguments as in the linked answer we can properly conclude that all the even moments of $T_n=1/\bar X_n$ are infinite while all the odd moments are undefined, that is, $\mathsf ET_n^{2m+2}=\infty$ and $\mathsf ET_n^{2m+1}=\text{d.n.e}$ for $m=1,2,\dots$ Since $\mathsf{Var}T_n=\mathsf{E}T_n^2-(\mathsf ET_n)^2$ and $\mathsf ET_n$ is undefined it follows that $\mathsf{Var}T_n$ must also be undefined.

Issue #3 (use of $\approx$):

This issue still has me struggling to provide a clear answer. Let me explain what I have found out so far. The issue of confusion here centers around the statements:
$$
\begin{aligned}
\mathsf E\bar X_n^{-1} &\approx\mu^{-1}\\
\mathsf{Var}\bar X_n^{-1} &\approx\mu^{-4}\sigma^2/n.\\
\end{aligned}
$$
As we have shown in Issue #2 above, the mean and variance of $\bar X_n^{-1}$ are undefined so what exactly are we approximating? Let me first say that the approximation sign does not mean anything as there really isn't a definition of what "$\approx$" means. Here, Casella deserves a little slack because in the sentence preceding these claims he states

...the "approximate" mean and variance of $1/\bar X_n$ are...

So why the quotation marks around approximate? Because he thought the mean and variance were infinite and thus cannot be approximated by anything. We now know that the situation is even stranger than that because the moments do not even exist...let alone be infinite. So what exactly was Casella trying to say by these statements? Let's first take a small detour:

Small (not really) detour (we can reinterpret $\mathsf E\bar X_n^{-1}$ and $\mathsf{Var}\bar X_n^{-1}$ so that they have well-defined values):

This should strike you as a strange claim to make as we just stated that both $\mathsf E\bar X_n^{-1}$ and $\mathsf{Var}\bar X_n^{-1}$ are undefined. However, if we reinterpret the integral representation of $\mathsf E\bar X_n^{-1}$ using the Cauchy principal value we can assign a well-defined value to $\mathsf E\bar X_n^{-1}$ and subsequently $\mathsf{Var}\bar X_n^{-1}$. By the Cauchy principal value interpretation of $\mathsf E\bar X_n^{-1}$ we find
$$
\mathsf E\bar X_n^{-1}\overset{\text{p.v.}}{=}\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon}+\int_\epsilon^\infty\right)\frac{1}{x}f_{\bar X_n}(x)\,\mathrm dx=\frac{\sqrt{2n}}{\sigma}\mathcal D\left(\tfrac{\sqrt n\mu}{\sqrt 2\sigma}\right),
$$
where $\mathcal D(z):=e^{-z^2}\int_0^z e^{t^2}\,\mathrm dt$ is the Dawson integral. Note that the "$=$" sign above has been replaced with "$\overset{\text{p.v.}}{=}$" to remind us that this is not a standard equality. Why is this reinterpretation useful? Well, there is a generalized central limit theorem which states that
$$
\frac{1}{m}\sum_{k=1}^m\frac{1}{\bar X_{n,k}}\overset{d}{\to}\operatorname{Cauchy}(\mathsf E\bar X_n^{-1},\pi f_{\bar X_n}(0)).
$$
So we see that the principal value of $\mathsf E\bar X_n^{-1}$ tells us something about the center of the limiting distribution for suitably large sample means involving $\bar X_n^{-1}$ and thus gives us a useful measure of center or "mean". To see this in action let's set $\mu=\sigma=n=1$ so that $\bar X_n\sim\mathcal N(1,1)$ and define
$$
\bar Z_m=\frac{1}{m}\sum_{k=1}^{m}\frac{1}{\bar X_{n,k}}
$$
to be the sample average of $m$ i.i.d. observations of $\bar X_n^{-1}$. I generated $10^6$ values of $\bar Z_{10^6}$ in Mathematica and plotted it in a histogram (below). Overlaid on the histrogram is the density function $\operatorname{Cauchy}(\sqrt 2\mathcal D(1/\sqrt 2),\pi f_{\bar X_n}(0))$ which represents the limiting distribution of $\bar Z_m$ as $m\to\infty$:

So again se see that the principal value of $\mathsf E\bar X_n^{-1}$ provides some useful measure of center even though strictly speaking the mean does not exist. Furthermore, by using the Cauchy principal value interpretation of $\mathsf E\bar X_n^{-1}$ we can also define the variance as
$$
\mathsf{Var}\bar X_n^{-1}\overset{\text{p.v.}}{=}\mathsf E\bar X_n^{-2}-(\mathsf E\bar X_n^{-1})^2=\infty-\left(\frac{\sqrt{2n}}{\sigma}\mathcal D\left(\tfrac{\sqrt n\mu}{\sqrt 2\sigma}\right)\right)^2=\infty.
$$
So even though the result for the variance is not finite we can still claim that it is well-defined and thus the equality holds. Bringing both results together we have through the Cauchy principal value interpretation:
$$
\begin{aligned}
\mathsf E\bar X_n^{-1} &\overset{\text{p.v.}}{=} \frac{\sqrt{2n}}{\sigma}\mathcal D\left(\tfrac{\sqrt n\mu}{\sqrt 2\sigma}\right)\\
\mathsf{Var}\bar X_n^{-1} &\overset{\text{p.v.}}{=} \infty.
\end{aligned}
$$

This is starting to look like an awfully long detour but there is a reason. Going back to our expression for $\mathsf E\bar X_n^{-1}$ we can use the asymptotic expansion for the Dawson integral see equation $(9)$ here to deduce as $n\to\infty$:
$$
\tag{1}
\mathsf E\bar X_n^{-1}\sim\frac{\sqrt{2n}}{\sigma}\left(\frac{\sigma}{\sqrt{2n}\mu}+\frac{\sigma^3}{n\sqrt n\mu^3}+\mathcal O(n^{-5/2})\right)=\frac{1}{\mu}+\frac{\sigma^2}{n\mu^3}+\mathcal O(n^{-2}).
$$
Look familiar? If we drop all the higher order terms we find for large $n$:
$$
\mathsf E\bar X_n^{-1}\approx\frac{1}{\mu},
$$
which is precisely the "approximation" given by Casella in Example 10.1.8 and is the same result obtained by the (first-order) delta method. This observation is a big clue to what Casella meant by the approximations in the example.

Small detour 2 (use of $\delta$-method for approximating moments of $\bar X_n^{-1}$):

The main idea driving the delta method is that we can approximate the moments of $g(\bar X_n)$ for some function $g$ by considering the moments of the corresponding Taylor polynomial of $g$ (to some order) centered about the mean of $\bar X_n$. If the density of $\bar X_n$ is "concentrated" in the area where $g$ and its Taylor polynomial approximation show good agreement, then we can obtain a useful approximation of the moments for $g(\bar X_n)$. In our case, we consider the function $g(x)=x^{-1}$ and expand $g$ about the mean of $\bar X_n$ (i.e. $\mu\neq 0$) to get
$$
g(x)=\frac{1}{\mu}-\frac{x-\mu}{\mu^2}+\frac{(x-\mu)^2}{\mu^3}+\mathcal O((x-\mu )^3).
$$
If we define $g_2(x)=\frac{1}{\mu}-\frac{x-\mu}{\mu^2}+\frac{(x-\mu)^2}{\mu^3}$ as the second-order Taylor polynomial approximation to $g(x)$ then we can easily find that
$$
\begin{aligned}
\mathsf Eg_2(\bar X_n) &=\frac{1}{\mu}+\frac{\sigma^2}{n\mu^3}\\
\mathsf{Var}g_2(\bar X_n) &=\frac{\sigma^2}{\mu^4 n}+2\frac{\sigma^4}{n^2\mu^6}\frac{1}{\mu^3}.
\end{aligned}
$$
By dropping the higher order terms we obtain the "approximations" given by Casella. Furthermore, the expression for $\mathsf Eg_2(\bar X_n)$ is equivalent to the first two terms of the asymptotic expansion of the principal valued expression for $\mathsf E\bar X_n^{-1}$ given by $(1)$ above. Why? Furthermore, how come the $\delta$-method approach gives finite and well-defined values for moments that do not exist?

Conclusion:

The answer to the questions stated at the end of our second detour has to do with how both approaches remove singularities from the equations representing the moments. In the case of the Cauchy principal-value interpretation of $\mathsf E\bar X_n^{-1}$ we took a limit that was symmetric about the pole in the integrand which effectively canceled out the infinite components that caused the moment to become undefined. Note that this approach is in some sense "exact" in that did not make any restrictive assumptions about the density of $\bar X_n$ residing in some specified interval with high probability (i.e. does not require $n$ to be large). Likewise, the $\delta$-method also removed singularities from the problem by replacing $g(x)=1/x$ with the approximate function $g_2(x)=\frac{1}{\mu}-\frac{x-\mu}{\mu^2}+\frac{(x-\mu)^2}{\mu^3}$ which no longer has any singularities on the support of $\bar X_n$. However, unlike the Cauchy principal-value approach, the use of the Taylor series centered at $\mu$ accompanied with the fact that this Taylor series has a radius of convergence equal to $\mu$ requires that $\mathsf P(\bar X_n\in(\mu-\epsilon,\mu+\epsilon)\approx 1$, which is to say that $\bar X_n$ is observed with high probability near its mean. Since this only happens when $n$ is large we see that the first moment derived from both approaches agree as $n$ becomes large.

So what exactly to the $\approx$ signs in Casella's "approximations" mean? Well, assuming $\mu\neq 0$, as $n$ becomes very large, the density of $\bar X_n$ becomes heavily concentrated about its mean (again, by heavily concentrated we mean $\bar X_n$ is observed to be very close to $\mu$ with high probability). Simultaneously, as we continue to increase $n$ the probability that $\bar X_n$ is in the neighborhood of zero becomes vanishingly small and so the sample mean and sample variance of many $\bar X_n^{-1}$'s begins to become quasi well-behaved and settle down around particular values which correspond to the quantities derived about via the Cauchy principal value and $\delta$ method approximations. At the end of the day these approximations are not really approximating anything but rather are assigning finite values to undefined quantities, which under the right interpretation become useful when $n$ is large.


问:
Prove that $u^m \equiv -1\pmod p \iff p \equiv 1\pmod 4.$

Let $p$ be a prime number with $p \neq 2.$ Let $q = \frac {p-1} {2}$ and $u$ be an element of order $q$ modulo $p.$ Then show that there exists an integer $m$ such that $u^m \equiv -1 \pmod p$ if and only if $p \equiv 1\pmod 4.$

I tried it but failed to conclude. What I did is as follows. $$u^m \equiv -1\pmod p \implies u^{2m} \equiv 1\pmod p.$$ Hence $q\mid 2m.$ Which is same as saying $(p-1)\mid 4m.$ Now I got stuck. How do I proceed now? Any help will be warmly appreciated.

Thanks for your time.

答:
It is tagged group-theory, so let's sketch a simple conceptual proof using that

$\qquad -1 \in G\!=\!\langle u\rangle\!\iff\! G$ has elt of order $2$ $\!\iff\! 2\mid \#G =\frac{p-1}2\!$ $\iff\! 4\mid p\!-\!1$

Remark $ $ The only nontrivial part is the middle equivalence, i.e. that a cyclic group $G$ has a subgroup of order $k\!\iff\! k\mid \#G,\,$ provable very simply  using nothing deeper than the integer division algorithm. The proof in Whatsup's answer is just a special case of such when $\,k=2,\,$ and that method works more generally for $\,k = p\,$ any prime, i.e.

Theorem $ $ A cyclic group $\,G=\langle u\rangle$ has an elt of prime order $\,p\iff p\mid  q:=\#G $

Proof $\ \ (\Leftarrow)\ \ \ p\mid q \Rightarrow  (u^{q/p})^p\! = 1,\,$ and $\, u^{q/p}\!\neq 1\,$ by $\,o(u)=q,\,$ thus $\,o(u^{q/p}) = p$.

$(\Rightarrow)\ \ \ o(u^k)\!=\!p\Rightarrow u^{kp}\!=1\,$ thus $\, q\mid pk,\,$ but $\,p\nmid q\Rightarrow q\mid k \Rightarrow u^k\!=1,\,$ contra $\,o(u^k)\!=\!p$.


问:
$SU(2)$ and the $3$-sphere, with $SO(2)$ a longitude
I'm currently studying the connection between $SU(2)$ and the $3$-sphere. A longitude on the sphere is of the form $QTQ^*$ where $Q \in SU(2)$ and
$T = \{D_\lambda : \lambda \bar{\lambda} = 1 \} $, where $T$ is the group of diagonal matrices with entries: a complex number and its complex conjugate, such that determinant is $1$.

My understanding is that the subgroup $T$ of $SU(2)$ allows us to see explicitly the possible eigenvalues of our sets of transformations $QTQ^*$.

A proposition in my lecture notes says that $SO(2)$, a subgroup of $SU(2)$, can be identified with a longitude on the $3$-sphere.

In trying to make this more concrete I have written $QTQ^*$ and studied the entries more closely.

I find that for $QTQ^*$ to be a matrix of $SO(2)$ having entries contained in the reals, I need the following equations to hold:

$x_1^2+x_2^2-x_3^2-x_4^2= 0$

$x_1x_3 + x_2x_4 = 0$

and for the identification of $Q$ with some point on $S_3$:

$x_1^2+x_2^2+x_3^2+x_4^2= 1$

The problem is, these three equations seem to satisfy multiple solutions.

Then I can construct many different $Q$ to yield $QTQ^* \in SO(2)$ and the statement that any one $Q$ represents a longitude and $SO(2)$ is one such longitude does not make sense to me. I would have liked to find that only one such $Q$ is possible.

Is this the right way to think about this question? Are there more fruitful ways to perform the identification? If so, what is the central flaw of my reasoning? Thanks in advance.

答:
Any element $q \in SU(2)$ defines a longitude $q^*Tq$ in $S^3$, but for any $t \in T \cong SO(2) \subset SU(2)$, we have
$$(tq)^* T (tq) = q^* (t^* T t) q = q^* T q ,$$ so $t q \in SU(2)$ defines the same great circle that $q$ does, so there is at least a $1$-parameter family's worth of $q \in SU(2)$ that determine the same great circle.

Indeed, for all $\theta$,
$$\pmatrix{x_1&x_2\\x_3&x_4} = \frac{1}{\sqrt{2}} \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta}$$
is a solution to the quadratic system in $x_1, x_2, x_3, x_4$ given in the question statement.

In fact, direct computation shows that $SU(2)$ acts transitively on the space of $\mathcal L$ longitudes, and that the stabilizer of the longitude $T$ is $T \cong SO(2)$ itself. So we may identify $\mathcal L$ with the (right) homogeneous space $SO(2) \backslash SU(2)$, i.e., the longitudes with the right cosets $T \cdot q = q^* T q$, which in particular are conjugate copies of $SO(2)$.


问:
$L^q([\mathbb{R}])\nsubseteq L^p([\mathbb{R}])$ and $L^p([\mathbb{R}])\nsubseteq L^q([\mathbb{R}])$
Let $1\leq p<q\leq \infty$ I have to how that

$L^q([\mathbb{R}])\nsubseteq L^p([\mathbb{R}])$ and $L^p([\mathbb{R}])\nsubseteq L^q([\mathbb{R}]) $

I was trying to come up with counterexamples for the inclusion but unfortunately, I am stuck! I would appreciate any kind of help!

答:
Choose $k \in (\dfrac{1}{q}, \dfrac{1}{p})$
and we can define 2 functions $$f_{p} = \dfrac{\chi_{(0,1]}(x)}{x^{-k}}$$
$$f_q = \dfrac{\theta(x-1)}{x^{-k}}$$
Where $\chi(x)$ - is a charachteristic function and $\theta(x-1)$ is heaviside function. It's easy to see, that $f_{q} \in L_{q}(\mathbb R), f_{q} \notin L_{p}(\mathbb R)$ and $f_{p} \in L_{p}(\mathbb R), f_{p} \notin L_{p}(\mathbb R)$


问:
A confusion on $\Omega$ and $\Sigma$ functors
We know that for any pointed topological space there are two functors
$$
\Omega:\left\{\text{pointed topological spaces}\right\}\longrightarrow \left\{\text{H-groups}\right\}
$$
and
$$
\Sigma:\left\{\text{pointed topological spaces}\right\}\longrightarrow \left\{\text{H-cogroups}\right\}
$$
which are adjoint functors for each other.
Moreover, we have
$$
[X,\Omega^{n}Y]\simeq[\Sigma X,\Omega^{n-1}Y]=[\Sigma X,\Omega(\Omega^{n-2}Y)]
$$
where $Y$ is a pointed topological space with base point $y_0$. What confused me is, why is $[\Sigma X,\Omega(\Omega^{n-2}Y)]$ an abelian group? If it is, then we can conclude that all homotopy groups with $n>1$ are abelian.
I would appreciate it if you could help me!

答:
For each pointed space $Z$ we have the following five maps:

Diagonal map $\Delta : Z \to Z \times Z$

Folding map $\phi : Z \vee Z \to Z$

Inclusion map $\iota : Z \vee Z \to Z \times Z$

Coordinate flip map on product: $T : Z \times Z \to Z \times Z$

Coordinate flip map on wedge : $T' : Z \vee Z \to Z \vee Z$

Note that $T \circ \Delta = \Delta$, $\phi \circ T' = \phi$ and $T \circ \iota
 = \iota \circ T'$.

An $H$-group $G$ has a multiplication map $\mu : G \times G \to G$ such that $[\mu] \circ [\iota] = [\phi]$ (here $[-]$ denotes pointed homotopy class). This reflects the fact that multiplication has a two-sided identity. Here we do not care about associativity and existence of inverses.

Dually, an $H$-cogroup $K$ has a comultiplication map $\kappa : K \to K \vee K$ such that $[\iota] \circ [\kappa] = [\Delta]$.

If $G$ is an  $H$-group and $K$ an $H$-cogroup, then the set $[K,G]$ of pointed homotopy classes gets two multiplications, one induced from $\mu$ (call it $\cdot$) and one induced from $\kappa$ (call it $*$). Both multiplications agree. To see this, look at the following diagram:

$\require{AMScd}$
\begin{CD}
K  @>{\kappa}>> K \vee K  @>{f \vee g}>> G \vee G @>{\phi}>> G \\
@V{id_K}VV @V{\iota}VV @V{\iota}VV @V{id_G}VV \\
K @>{\Delta}>> K \times K @>{f \times g}>> G \times G @>{\mu}>> G
\end{CD}
It commutes up to homotopy, the bottom row represents $[f] \cdot [g]$ and top row represents $[f] * [g]$.

Let us now check that multiplication in $[K,G]$ is commutative:
$$[g] \cdot [f] = [\mu \circ (g \times f)\circ \Delta] = [\mu \circ(g \times f)\circ T\circ \Delta] = [\mu \circ T \circ (f \times g)\circ \Delta] \\= [\mu \circ T \circ \iota \circ (f \vee g)\circ \kappa] = [\mu \circ \iota \circ T' \circ (f \vee g)\circ\kappa] = [\phi \circ T' \circ (f \vee g)\circ\kappa] \\= [\phi \circ (f \vee g)\circ\kappa] = [f] * [g] = [f] \cdot [g].$$

Intuitively, the reason for commutativity is that the coordinate flip is "absorbed" both by $\Delta$ and $\phi$.

This shows in particular that $[\Sigma X, \Omega Y]$ is always an abelian group.


问:
$G$ is a finite subgroup of $\Gamma(L:K), [L:K] = pq,$ and $[G^{\dagger} : K] > 1,$ prove that $G$ is cyclic.

If $G$ is a finite subgroup of $\Gamma(L:K), [L:K] = pq,$ and $[G^{\dagger} : K] > 1$ where $p, q$ are distinct primes, prove that $G$ is cyclic.

This was a problem on one of my exams and I proceeded to try the following: $[G^{\dagger} : K] = p, q, pq$ by the Tower Law, so $|G| = |\Gamma(L:K)|/|G^{\dagger} : K] = pq/|G^{\dagger} : K] = 1, p, q,$ and from here $G$ is cyclic by Lagrange's Theorem or otherwise.

However, this assumes $|\Gamma(L:K)| = [L:K]$ and $G^{\dagger *} = G,$ which requires normality of $L:K.$ Without normality, you can still say $[L:G^{\dagger}] = 1, p, q,$ and that $G^{\dagger *} \subseteq \Gamma(L:G^{\dagger}).$ Under the assumption of normality, $|\Gamma(L:G^{\dagger})| = [L : G^{\dagger}]$ and we may finish as before since $G \subseteq G^{\dagger *}.$

However, not every extension of prime degree is normal (consider $\mathbb{Q}(2^{1/3}) : \mathbb{Q}$ for example), so you can't prove $L:G^{\dagger}$ is normal, and I didn't notice the mistake in time.

There's a similar problem with $[L:K]$ being merely prime, but there's no way I'd have come up with the solution there and generalized it within the time limit of the exam.

Note: For a subfield $H, H^*$ is the group of $H$-automorphisms of $L$ and for a subgroup $G, G^{\dagger}$ is the fixed field of $G.$ $\Gamma(L:K)$ is the group of all $K$-automorphisms of $L.$ Forget Galois, you do not know Galois.

答: Be careful, in your note you mentioned that Aut(L/K) the Galois group of L over K; it is when the extension is Galois (normal and separable), but not always. By defining Aut(L/K) as the Galois group, you're automatically assuming L/K is Galois and you're done, as you said.</p>


问:
Showing $f_*\mathcal{O}_Y(E) = \mathcal{O}_X$ if and only if $E$ is effective.
This arises from lemma 7.11 and remark 7.12 from the book ‘Higher-Dimensional Algebraic Geometry’ by Debarre. Essentially, lemma 7.11 states the following:

“Let $f: Y\rightarrow X$ be a proper birational morphism between smooth varieties. Then for any effective divisor $E$ on $Y$ contained in the exceptional locus of $f$, one has $f_*\mathcal{O}(Y)(E) \cong \mathcal{O}_X$.”

Now suppose $F$ is any exceptional divisor of $f$. We write $F = F_1 -F_2$ as a difference of effective divisors, where $F_1$ and $F_2$ have no common components. Certainly, $f_*\mathcal{O}_Y(F) = f_*\mathcal{O}_Y(F_1-F_2)$ injects into the sheaf $f_*\mathcal{O}_Y(F_1) \cong \mathcal{O}_X$.

The author then states that ‘it is therefore a sheaf of ideals that defines a subscheme of $X$ supported on $f(F_2)$. It follows $f_*\mathcal{O}_Y(F) \cong \mathcal{O}_X$ iff $F$ is effective.’

My understanding is that if $f_* \mathcal{O}_Y(F) \cong \mathcal{O}_X$. Then $F_2$ must be zero necessarily. How do we conclude that in this case $F_2 =0$? Any help given would be greatly appreciated!

答:
If $f_*\mathcal{O}_Y(F) \cong \mathcal{O}_X$ then by pullback-pushforward adjunction there is a morphism
$$
\mathcal{O}_Y \cong f^*\mathcal{O}_X \to \mathcal{O}_Y(F).
$$
Moreover, on the locus where $f$ is an isomorphism, this morphism is the identity. This shows that $F$ is an effective divisor supported on the exceptional locus of $f$.


问:
How to find focus and directrix of $4y = x^2 -16$
How would I go about finding the focus and directrix of $4y = x^2 -16$ $(1)$

I did the $SP^2 = QP^2$ and got $(y+b)^2 = (x-a)^2 + (y-b)^2$.

This gave me: $4by = x^2 -2ax +a^2$. Comparing this to $(1)$. I get $a = 0$ but I don't know what to do to get $b$. Can someone advise?

Edit: my workings:

答:
Use definition of parabola as stated in comment:

$\sqrt{[(y-b)-a]^2+x^2}=(y-b)+a$

$(y-b)^2-2a(y-b)+a^2+x^2=(y-b)^2+2a(y-b)+a^2$

Which gives:

$4a(y-b)=x^2$

Comparing with given equation $x^2=4y+16$ we get:

$b=-4$ and $a=1$, where b is value of translation in y direction. The coordinates of vertex is (0, -4), therefore:

Equation of directrix is L: $y=-1-4=-5$.

Coordinate of focus is: $F(0, -3)$.


问:
Existence of convergent sequence
IF $U=\{u\in R^n : ||u||\leq 1\}$, how do I know that there exists a sequence $\{u_k\}$ such that $\lim u_k=0$?</p>

答: Take $u_{k} = 0 \quad \forall\; k$.</p>

ContentLicense: CC BY-SA 3.0

Title: Existence of convergent sequence

Body: Take $u_{k} = 0 \quad \forall\; k$.


问:
Understanding proof of Collins' Weingarten calculus formula
I am trying to understand Collins' paper on Weingarten calculus. You can find it here: click!

On page 7, between the formulas $(2.6)$ and $(2.7)$, he claims that the action $\pi^{\otimes q} \otimes \Psi$ is on $M_d(\mathbb{C})^{\otimes q}$, but, as far as I understand, since both $\pi^{\otimes q}$ and $\Psi$ are on  $M_d(\mathbb{C})^{\otimes q}$, the tensor product action should be on $M_d(\mathbb{C})^{\otimes q} \otimes M_d(\mathbb{C})^{\otimes q}$, right? However, the minimal invariant projections $I_\lambda$ are multiplicated by a $\tilde{B}$, which is an element of $M_d(\mathbb{C})^{\otimes q}$, confirming that the afore-mentioned action should be on $M_d(\mathbb{C})^{\otimes q}$. Am I missing something?

Moreover, he claims that we know, from representation theory, that an algebra is abelian and that the minimal invariant projections are indexed by partitions of $q$. Could you provide me with a reference on that? I would be very interested by a reference carrying out concrete examples for small $q$.

答:
I mean, sure, technically putting a $\otimes$ symbol between $\pi^{\otimes q}$ and $\Psi$ ought to imply an action on the tensor product $M_d(\mathbb{C})^{\otimes q}\otimes M_d(\mathbb{Q})^{\otimes q}$, but that's not what they mean. Since the action of $\Psi$ normalizes the action of $\mathbb{U}_d^q$ (adopting their notation of $\mathbb{U}_d$), so there is an action of the semidirect product $\mathbb{U}_d^q\rtimes S_q$, which is known as the wreath product $\mathbb{U}_d\wr S_d$, on $M_d(\mathbb{C})$. Maybe you want to say $\pi^{\otimes q}\cdot\Psi$ instead of $\pi^{\otimes q}\otimes\Psi$.

The keyword to start searching is "Schur-Weyl duality."


问:
Essential spectrum of operators whose resolvent difference is compact
Suppose that $T,S$ are densely defined, closed (unbounded) operators on a separable Hilbert space such that there exists $z \in \mathbb C$ in the intersection of resolvent sets of $T$ and $S$ for which $(T-z)^{-1}-(S-z)^{-1}$ is compact. Does it follow that $T$ and $S$ have the same essential spectrum?

Remark: in the application I have in mind $S-T$ is unbounded and domains of $S$ and $T$ are not necessarily equal.

答:
It has been shown in J. J. Buoni ''An Essential Spectra Mapping Theorem'', J. Math. Anal. Appl. 56 (1976) that
$$ \sigma_{\mathrm{ess}}(S) = \{ z' \in \mathbb C \, | \, \exists w \in \sigma_{\mathrm{ess}}((S-z)^{-1}) \ \  (z'-z)w =1 \}. $$
The same is true with $T$ replaced by $S$. Since $(T-z)^{-1}-(S-z)^{-1}$ is compact, $\sigma_{\mathrm{ess}}((T-z)^{-1}) = \sigma_{\mathrm{ess}}((S-z)^{-1})$, so the above formula entails $\sigma_{\mathrm{ess}}(S) = \sigma_{\mathrm{ess}}(T)$.


问:
Limit derivation of generalized mean, $-\infty$ case yields contradiction with ones as input
Consider the following function (a CES utility function or weighted power mean):

$$u_\rho(\vec x) = \left(\sum_{i=1}^n a_i x_i^\rho\right)^{1/\rho}$$

for parameters $a_i \geq 0$, $\sum a_i =1$, and $\rho\leq1$.

It can be shown (see the Wikipedia link above or this Economics SE question) that taking the limit as $\rho\to -\infty$ yields

$$u_{-\infty}(\vec x) = \min_i a_i x_i$$

Consider the case where $x = \vec 1$. Clearly

$$u_\rho(\vec 1) = \left(\sum_{i=1}^n a_i \right)^{1/\rho} = 1^{1/\rho} = 1, \quad\forall \rho$$

But the limit above says that $$u_{-\infty}(\vec 1) = \min_i a_i \neq1$$
(unless $n=1$).

What explains the contradiction?

The formulas above can also be found in the following reference at p. 139:

Nisan, Noam, Tim Roughgarden, Èva Tardos, and Vijay V. Vazirani, eds. 2007. Algorithmic Game Theory. Cambridge University Press.

Here is a graph that may shed some light (or perhaps just add confusion):

Due to numerical errors, the function evaluates to zero for $\rho < -900$ or so, but it seems from the graph that the limit (blue) is approaching a value between $0.45$ and $0.50$, not the lower value of $0.11$ predicted (red).

```<code>import numpy as np
import matplotlib.pyplot as plt

np.random.seed(987)

def ces(a, x, rho):
    return sum(a * x**rho)**(1/rho)

def leontief(a, x):       # Limit of CES as rho -> -inf
    return min(a * x)

a = np.random.rand(3)
a /= sum(a)
x = np.random.rand(3)

rho = list(range(-1000, 2))
rho.remove(0)
plt.figure(figsize=(10, 7), facecolor="white")
plt.plot(rho, [ces(a, x, r) for r in rho], label="CES")
plt.xlabel(r"<span class="math-container">$ρ$</span>"), plt.ylabel(r"<span class="math-container">$u_ρ(x)$</span>")
plt.axhline(leontief(a, x), color="crimson", label="Leontief")
plt.legend()

print(a*x)
# [0.11114134 0.19854696 0.48172427]
</code>```

答:
As the example shows the limit expression is not correct. The true one should be

$$u_{-\infty}(\vec x) = \min_i x_i$$

(this requires that all $a_i>0$; but $a_i=0$ doesn't really make sense because then $u_\rho(\vec x)$ wouldn't depend on $x_i$ at all).

As the name mean suggests, $u_\rho(\vec x)$ lies always inside the interval $[\min_i x_i, \max_i x_i]$.

The Wikipedia link states that the ordinary (non-weighted) power mean converges to $\min_i x_i$ as $\rho \to -\infty$:
$$\lim_{\rho\to-\infty} \left(\frac{x_1^\rho+\dots+x_n^\rho}{n}\right)^{1/\rho} = \min_i x_i$$

How does this relate to the weighted power mean?

An interesting observation is that if the weights $a_i$ are rational numbers, then the weighted power mean can be written as an ordinary power mean. Example: Suppose $a_1 = \frac{1}{6}$, $a_2 = \frac{2}{6}$, $a_3 = \frac{3}{6}$.
Then
$$(a_1x_1^\rho + a_2x_2^\rho + a_3x_3^\rho)^{1/\rho} = \left(\frac{x_1^\rho + 2x_2^\rho + 3x_3^\rho}{6}\right)^{1/\rho} = \left(\frac{x_1^\rho + x_2^\rho + x_2^\rho + x_3^\rho + x_3^\rho + x_3^\rho}{6}\right)^{1/\rho}  $$
so the weighted power mean of $x_1,x_2,x_3$ with weights $a_1,a_2,a_3$ is the same as the ordinary power mean of $x_1,x_2,x_2,x_3,x_3,x_3$. (Essentially: giving more weight to variables is simulated by repeating them.)

Using this it follows that the weighted power mean also converges to $\min_i x_i$ as $\rho \to -\infty$.

Note: This method only works for rational weights $a_i$ but since real numbers can be approximated arbitrarily well by rational numbers it should be intutive that the result also holds for real weights.


问:
Find polynomials $M_1(x)$ and $M_2(x)$ such that $(x+1)^2M_1(x) + (x^2 + x + 1)M_2(x) = 1$
I have been trying to solve this with no success. Could you suggest me a solution?</p>

答:
Back to the basics! Remember how you do it for integers? For example, how would you find $x$, $y$ such that $23x+47y=1$? That's right, Euclid's algorithm.

Let's try to calculate the GCD of $(x+1)^2$ and $x^2+x+1$.

$$
(x+1)^2 - (x^2+x+1) = x\\
(x^2+x+1) - x(x+1) = 1
$$
Hence,
$$(x^2+x+1) - ((x+1)^2 - (x^2+x+1))(x+1) = 1$$
Therefore,
$$- (x+1)(x+1)^2 + (x+2)(x^2+x+1) = 1$$


问:
Permutations of Symmetric Group of Order 3
Find an example, in the group $S_3$ of permutations of $\{1,2,3\}$, of elements $x,y\in S_3$ for which $x^2 = e = y^2$ but for which $(xy)^4$ $\not=$ e.</p>

答:
However, you have to solve the problem theoretically; I 'd like to do it for you by using GAP. Indeed, after solving such this computational problem by using theorems and facts, it is a fun doing that by machine. Here you are:</p>

```<code>gap> S3:=SymmetricGroup(3);;
gap> o2:=Filtered(S3,x->Order(x)=2);;
gap> eo2:=Elements(o2);
                        [(2,3),     (1,2),     (1,3)]
gap> for i in [1..3] do
         for j in [1..3] do
             if (eo2[i]*eo2[j])^4 <> () then Print(eo2[i],"    ",eo2[j],"\n"); fi;
         od;
     od;

                           (2,3)            (1,2)
                           (2,3)            (1,3)
                           (1,2)            (2,3)
                           (1,2)            (1,3)
                           (1,3)            (2,3)
                           (1,3)            (1,2)
</code>```

Body:
However, you have to solve the problem theoretically; I 'd like to do it for you by using GAP. Indeed, after solving such this computational problem by using theorems and facts, it is a fun doing that by machine. Here you are:

gap> S3:=SymmetricGroup(3);;
gap> o2:=Filtered(S3,x->Order(x)=2);;
gap> eo2:=Elements(o2);
                        [(2,3),     (1,2),     (1,3)]
gap> for i in [1..3] do
         for j in [1..3] do
             if (eo2[i]*eo2[j])^4 <> () then Print(eo2[i],"    ",eo2[j],"\n"); fi;
         od;
     od;

                           (2,3)            (1,2)
                           (2,3)            (1,3)
                           (1,2)            (2,3)
                           (1,2)            (1,3)
                           (1,3)            (2,3)
                           (1,3)            (1,2)


问:
Hilbert And Banach dual spaces
Are, and in general, all Hilbert spaces identified to their dual ?, i.e $H'\equiv H$

What about Banach spaces? And can someone please state the weak convergence in each?

答:
A vector space (a fortiori, Hilbert space, Banach space) is isomorphic to its dual if, and only if, it is finite dimensional. For any infinite dimensional vector space, the dual space is always "larger" than the predual, in the precise technical sense that there exists an injection from the predual into the dual, but never a bijection.

Additionally, the isomorphism between a finite dimensional vector space and its dual is dependent on a choice of vector basis. It is not a "natural" isomorphism, in the language of category theory. However, the isomorphism between a vector space and its double dual (the dual of the dual) IS natural, in that it can be defined without a choice of basis.

Studying this phenomenon is anecdotally what pushed Saunders MacLane and Samuel Eilenberg to invent category theory, by the way.

Edit: I was mistaken in the above, due to an ambiguity on the notion of "dimension" as used by an algebraist, and as used by an analyst; and have provided in the comments below a document which clarifies my point, and explains my mistake in your specific context.

Title: Hilbert And Banach dual spaces


问:
Is $x^x$ differentiable when $x<0$?
The function $f(x)=x^x$ is only defined for negative values of $x$ if $x$ is a rational number $a/b$, where $b$ is odd. Initially, I thought this meant that the limit
$$
f'(a)=\lim_{h\to0}\frac{(a+h)^{a+h}-a^a}{h}
$$
cannot exist if $a$ is a negative number, since every neighbourhood of $a$ would contain values for which $x^x$ does not make any sense at all. However, this post—about $\lim_{x\to0}\frac{\sin(1/x)}{\sin(1/x)}$—suggests that provided every neighbourhood of $a$ contains values for which $x^x$ is defined, there is no issue with taking limits at the point.

So if $a$ is negative number that is in the domain of $f$, then does $f'(a)$ exist? And if so, is there a general formula for $f'(a)$?

答:
Differentiability is typically only defined for functions defined on an open subset of $\Bbb R$. So although you could make the case that the limit exists because the domain of the function is
$\{x\in\Bbb R: x>0 \text{ or }x=a/b\text{ for odd }b\}$, this is not enough to make the function differentiable at any $x<0$.

Title: Is $x^x$ differentiable when $x<0$?


问:
Show that $g(s):=(\beta -\alpha)^{-1}\int _{\alpha }^{\beta }f(\frac{x}{s})dx$ is continuous
Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x+1)=f(x), \forall x\in \mathbb{R}$. Given $\alpha < \beta$, define the function $g:\mathbb{R} \to \mathbb{R}$ as $g(s):=(\beta -\alpha)^{-1}\int _{\alpha }^{\beta }f\left(\frac{x}{s}\right)dx$ if $s \ne 0$ and $g(0)=\int_0^1f(x)dx$. Show that the function $g$ is continuous.

I know that since $f(x+1)=f(x), \forall x\in \mathbb{R}$, $f$ is periodic. Using this fact how do I show that $\int _{\alpha }^{\beta }f\left(\frac{x}{s}\right)dx$ is continuous on $(0,1]$?

答:
Since $f$ is periodic and continuous, it is uniformly continuous on $\mathbb{R}$. Suppose for any given $\varepsilon>0$ we have
$$
|f(t)-f(r)|<\varepsilon\quad\textrm{if   }|r-t|<\delta.
$$
Given $s_0>0$, we can pick $s$ close enough to $s_0$ in such a way that
$$
|x||1/s-1/s_0|<\delta
$$
for all $x\in[\alpha,\beta]$ (for that it is enough to choose $\max\{|\alpha|,|\beta|\}|1/s-1/s_0|<\delta$). With that choice
$$
|f(x/s)-f(x/s_0)|<\varepsilon
$$
for all $x\in[\alpha,\beta]$. Then $|g(s)-g(s_0)|<\varepsilon$ as well.


答: Turns out we have $|G| = [L:G^{\dagger}] = 1, p, q$ without any assumption of normality, and the rest is easy.

Title: $G$ is a finite subgroup of $\Gamma(L:K), [L:K] = pq,$ and $[G^{\dagger} : K] > 1,$ prove that $G$ is cyclic.

Body:
If $G$ is a finite subgroup of $\Gamma(L:K), [L:K] = pq,$ and $[G^{\dagger} : K] > 1$ where $p, q$ are distinct primes, prove that $G$ is cyclic.

This was a problem on one of my exams and I proceeded to try the following: $[G^{\dagger} : K] = p, q, pq$ by the Tower Law, so $|G| = |\Gamma(L:K)|/|G^{\dagger} : K] = pq/|G^{\dagger} : K] = 1, p, q,$ and from here $G$ is cyclic by Lagrange's Theorem or otherwise.

However, this assumes $|\Gamma(L:K)| = [L:K]$ and $G^{\dagger *} = G,$ which requires normality of $L:K.$ Without normality, you can still say $[L:G^{\dagger}] = 1, p, q,$ and that $G^{\dagger *} \subseteq \Gamma(L:G^{\dagger}).$ Under the assumption of normality, $|\Gamma(L:G^{\dagger})| = [L : G^{\dagger}]$ and we may finish as before since $G \subseteq G^{\dagger *}.$

However, not every extension of prime degree is normal (consider $\mathbb{Q}(2^{1/3}) : \mathbb{Q}$ for example), so you can't prove $L:G^{\dagger}$ is normal, and I didn't notice the mistake in time.

There's a similar problem with $[L:K]$ being merely prime, but there's no way I'd have come up with the solution there and generalized it within the time limit of the exam.

Note: For a subfield $H, H^*$ is the group of $H$-automorphisms of $L$ and for a subgroup $G, G^{\dagger}$ is the fixed field of $G.$ $\Gamma(L:K)$ is the group of all $K$-automorphisms of $L.$ Forget Galois, you do not know Galois.


问:
Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space.

Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space. Here $d_E$ is just the restriction of $d_2$ to $E$.

I'm trying to show this using the preimage definition. I have that $$E=\{(x,y)\mid f(x,y)=0\}$$ where $f(x,y)=y-x^2$. Note that $f$ is continous so if $E$ is the preimage of a closed set then $E$ is closed and since $\mathbb{R}^2$ is complete and $E \subset \mathbb{R}^n$ this would imply that $E$ is also complete. I'm not sure I entirely understand the concept of preimage. It's quite clear that it's just the set of elements that map to the image of $f$, but how do I find it? I'm tempted so say that $$f^{-1}(\{0\}) = E$$ but this would mean that only $0$ gets mapped to $0$ which isn't true since if $x=y=1$ then $f(x,y)=0$ so the preimage would be $f^{-1}(\{0,1\})$?

答: We define $F(x,y)=x^2-y$. So $F^{-1}(\{0\})=\{(x,y):x^2=y\}=E$, and since $F$ is continuous as you mentioned, and ${0}$ is closed in $\Bbb{R}$ we get that $E$ is closed in $\Bbb{R^2}$ (as the preimage of a closed set under a continuous function). So $E\subseteq\Bbb{R^2}$ is a closed subset in a complete metric space and therefore $(E,d_E)$ is complete.


答:
About your concern regarding the preimage notation:
$$E = f^{-1}(\{0\})$$
is really just another way of writing the definition of $E$:
$$(x,y) \in f^{-1}(\{0\}) \iff f(x,y)\in \{0\} \iff f(x,y) = 0.$$

Body:
Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space. Here $d_E$ is just the restriction of $d_2$ to $E$.

I'm trying to show this using the preimage definition. I have that $$E=\{(x,y)\mid f(x,y)=0\}$$ where $f(x,y)=y-x^2$. Note that $f$ is continous so if $E$ is the preimage of a closed set then $E$ is closed and since $\mathbb{R}^2$ is complete and $E \subset \mathbb{R}^n$ this would imply that $E$ is also complete. I'm not sure I entirely understand the concept of preimage. It's quite clear that it's just the set of elements that map to the image of $f$, but how do I find it? I'm tempted so say that $$f^{-1}(\{0\}) = E$$ but this would mean that only $0$ gets mapped to $0$ which isn't true since if $x=y=1$ then $f(x,y)=0$ so the preimage would be $f^{-1}(\{0,1\})$?


答:
Given any function $g:X\to Y$ and $A\subseteq Y$ we have that

$$g^{-1}(A)=\{x\in X\ |\ g(x)\in A\}$$

This is simply the definition.

And so in your case $f^{-1}(\{0\})=E$ pretty much by the definition.

but this would mean that only 0 gets mapped to 0

That doesn't even make sense, since $0\not\in\mathbb{R}^2$ and $f^{-1}(\{0\})\subseteq\mathbb{R}^2$. Note that $(x,y)\in f^{-1}(\{0\})$ means that $f(x,y)=0$. Its equivalent.

if $x=y=1$ then $f(x,y)=0$

Yes, and thus for $x=y=1$ we have that $(x,y)=(1,1)$ is an element of $f^{-1}(\{0\})$. And at the same time an element of $E$.

Title: Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space.


问:
Need help understanding function notation
I just finished my test and I have no idea what one of the questions was. No academic dishonesty here, I already got graded and I got 75%, but I am just stuck on that exercise and have no idea how to understand the function in the question.

It was supposed to be an integration of a function, and it looked like this:

$$f: \mathbb{R}^2 \rightarrow \mathbb{R}; (x,y) \rightarrow x$$

I dont even remember the teacher using this notation, any help is appreciated, thanks!

答: $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ means that $f$ is a mapping from 2d space over real number to 1d space over real number, e.g. an ordered pair (a 2-tuple) of real numbers $(x,y)$ to a single real number. In your example it maps the tuple to its first component, as described by $(x,y) \rightarrow x$.


问:
What Does the Antiderivative of This Multivariable Function Mean?
In this video, the instructor uses Undetermined Coefficients (or at least what would be called Undetermined Coefficients in the ODE world) to find a particular solution to the PDE $U_t + cU_x = f(x + ct)$, where $c$ is a parameter present in the original Wave Equation.

The guess used is $U = aF(x + ct)$, where $a$ is the constant to be determined, and $F$ is said to be the antiderivative of $f$.  However, I don't recall from MV Calc any single notion of an antiderivative of a multivariable function, only antiderivatives taken with respect to one of the function's inputs.  The closest thing I can think of would be a potential function, where one begins with a vector field and pieces together a single function whose gradient is that vector field, but I don't think that matches what's going on here.

Later in the problem, he seems to take $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$ separately, in both cases yielding $f$, but it seems to me that baring certain exceptional cases, it is not possible to find a function whose partial with respect to each of its variables yield a single, given function.  What is going on here?

答:
If $f$ is a function of a single variable and $s$ is a function of two variables $x$ and $t$ the composition $f \circ s$ is also a function of $x$ and $t$.  This is often written in a convenient but slightly imprecise form as $$f(x,t) = f(s(x,t)).$$  In your case $f$ is a function of a single variable and its antiderivative in the usual sense is $F$. The function $s$ is defined as $s(x,t) = x + ct$.

Using the same notation you may write $$F(x,t) = F(s(x,t)) = F(x + ct).$$ The chain rule states  $$\frac{\partial F}{\partial x} (x,t) = F'(s(x,t))\frac{\partial s}{\partial x}(x,t) = f(s(x,t))$$
and similarly
$$\frac{\partial F}{\partial t} (x,t) = F'(s(x,t))\frac{\partial s}{\partial t}(x,t) = c f(s(x,t))$$
giving you $$c \frac{\partial F}{\partial x}(x,t) = \frac{\partial F}{\partial t}(x,t).$$


答: That's true that $f$ is continuous, and $E = f^{-1}(0)$, and since point zero is closed in $\mathbb R$ then by continuity of $f$ set $E$ is closed.You misunderstand the symbol $f^{-1}$, it's not and inverse map, we define the preimage of set $X \subset \mathbb R$ the same way$$f^{-1}(X) = \{(x, y) \in \mathbb R | f(x,y) \in X\}$$


问:
Find $\int_{\frac{1}{e}}^{e}|\ln x| dx$
How to calculate this integral?

$$\int_{\frac{1}{e}}^{e}|\ln x| dx$$

It is an obvious integral if the function is under the integral without modulus. The whole catch is in the module.

答:
You can use that

$$
\int_{1/e}^{e}\left|\ln\left(x\right)\right|\text{d}x=\int_{1/e}^{1}\left|\ln\left(x\right)\right|\text{d}x+\int_{1}^{e}\left|\ln\left(x\right)\right|\text{d}x
$$
Can you now discuss the value of $\left|\ln\left(x\right)\right|$ in each interval ?

Title: Find $\int_{\frac{1}{e}}^{e}|\ln x| dx$


问:
Given a coin that produces the probability $\lambda$, what is a constructive way to produce $f(\lambda)$, where $f$ is continuous non-Hölder?
Given a coin of "bias" $\lambda$, sample the probability $f(\lambda)$. This is the Bernoulli factory problem, and it can be solved only for certain functions $f$. (For example, flipping the coin twice and taking heads only if exactly one coin shows heads, we can simulate the probability $\lambda(1-\lambda)$.)

Usually, the Bernoulli factory problem can be solved for $f$ if and only if—

$f$ is continuous in $[0, 1]$, and

$f$ is identically $0$, identically $1$, or polynomially bounded away from $0$ and $1$ (roughly speaking, the function takes on only values in $ [0, 1]$ and its graph doesn't touch $0$ or $1$ except at the points $0$ and/or $1$)

(Keane and O'Brien 1994).

This question is a continuation of another question of mine, but this time we focus on the case when $f(\lambda)$ is a continuous non-Hölder function, which roughly means a continuous function with an exponentially steep slope (steeper than any $n$th-root).

The question just linked to seeks ways to compute polynomials that converge from above and below to $f$ in a manner that solves the Bernoulli factory problem for $f$. These polynomials form an approximation scheme for $f$. See that question for a formal statement of such approximation schemes.

There are two algorithms (one by Thomas and Blanchet $2012$, another by Łatuszyński et al.) to simulate a function $f$ via an approximation scheme. Roughly speaking, the algorithms work as follows:

Generate U, a uniform random number in $[0, 1]$.

Flip the input coin (with "bias" $\lambda$), then build an upper and lower bound for $f(\lambda)$, based on the outcomes of the flips so far. In this case, these bounds come from two degree-$n$ polynomials that approach $f$ as $n$ gets large, where $n$ is the number of coin flips so far in the algorithm.

If U is less than or equal to the lower bound, return 1. If U is greater than the upper bound, return $0$. Otherwise, go to step $2$.

The result of the algorithm is $1$ with probability exactly equal to $f(\lambda)$, or $0$ otherwise.

In general, the rate at which a function can be simulated by these algorithms depends on the smoothness of $f$. It roughly corresponds to the rate of convergence of an approximation scheme for $f$. I list approximation schemes for different kinds of functions $f$, which work with the two algorithms described above.

For example, roughly speaking, a $C^0$ continuous function $f$ can be simulated at the rate $O(n^{\alpha})$ only if $f$ is $\alpha$-Hölder continuous (Holtz et al. 2011). This seems to imply that a function not meeting a Hölder condition (a non-Hölder function) can achieve, at best, a simulation rate of $O(1)$, so that the polynomials won't necessarily converge uniformly for all $\lambda \in [0, 1]$ and for all non-Hölder functions, so that the algorithms above won't terminate in all cases.

<s>And this suggests to me that we need to add the following extra condition to the Bernoulli factory problem, since it appears that no approximation scheme exists for non-Hölder functions:

$f$ is $\alpha$-Hölder continuous in its domain for some $\alpha > 0$.

Is this true? Is there really no algorithm to sample the probability $f(\lambda)$ for a non-Hölder function $f$, given sample access to the probability $\lambda$?</s>

This suggests that a different approach may be needed for Hölder functions. Thus: Is there a constructive method to approximate non-Hölder functions with polynomials for the Bernoulli factory problem (in the form of a formula to compute their coefficients), so that we can sample the probability $f(\lambda)$ for a non-Hölder function $f$, given sample access to the probability $\lambda$?

Note: An example of a non-Hölder function is the following, with a "cusp" at $0$:

$1/10$ if $\lambda = 0$, and

$-1/(2*\ln(\lambda/2))+1/10$ otherwise.

Another example is the following, where the cusp is at $3/10$:

$1/10$ if $\lambda = 3/10$, and

$-1/(2*\ln(\frac{|\lambda-3/10|}{2}))+1/10$ otherwise.

A third example is the following monotone function:

$3/10$ if $\lambda = 3/10$,

$1/(2*\ln(\frac{3/10-\lambda}{2}))+3/10$ if $\lambda < 3/10$, and

$-1/(2*\ln(\frac{\lambda-3/10}{2}))+3/10$ otherwise.

Roughly speaking:

A non-Hölder function is continuous but has an exponentially steep slope.

If a continuous function has no vertical slope, the function is 1-Hölder continuous.

REFERENCES:

Thomas, A.C., Blanchet, J., "A Practical Implementation of the Bernoulli Factory", arXiv:1106.2508v3 [stat.AP], 2012.

Łatuszyński, K., Kosmidis, I., Papaspiliopoulos, O., Roberts, G.O., "Simulating events of unknown probabilities via reverse time martingales", arXiv:0907.4018v2 [stat.CO], $2009/2011$.

Keane, M. S., and O'Brien, G. L., "A Bernoulli factory", ACM Transactions on Modeling and Computer Simulation 4(2), 1994.

Holtz, O., Nazarov, F., Peres, Y., "New Coins from Old, Smoothly", Constructive Approximation $33$ ($2011$).

Update:

I have become aware of so-called Dini-continuous functions, which are a superset of Hölder continuous functions.

Essentially, a Dini-continuous function has a slope no "steeper" than that of $\phi(\lambda)$ for some increasing non-negative function $\phi$ with a finite integral of $\phi(\lambda)/\lambda$ over $[0, 1]$.

However, I haven't found any paper yet on the rate at which polynomials can approximate a Dini-continuous (but not Hölder-continuous) function (let alone in a manner that solves the Bernoulli factory problem).

Moreover, the following function, adapted from this question, is apparently not even Dini continuous:

$1/10 + (1/(1+|\ln(\lambda)|))/2$ if $\lambda>0$, and

$1/10$ otherwise.

答:
I have worked on this and found the following result, which shows that any continuous function meeting the two conditions in my question admits an approximation scheme with polynomials (in a manner that solves the Bernoulli factory problem). This includes continuous functions that are not Hölder continuous. The method of proving the result goes back to Nacu and Peres (2005).

Result: Let $f(\lambda)$ be a continuous function that maps [0, 1] to [0, 1], and let $X$ be a hypergeometric($2n$, $k$, $n$) random variable. Let $\omega(x)$ be a modulus of continuity of $f$ (a non-negative and nondecreasing function in the interval [0, 1], for which $|f(x)-f(y)|\le\omega(|x-y|)$ for all $x$ in [0, 1] and all $y$ in [0, 1]). If $\omega$ is concave on [0, 1], then the expression—$$|\mathbb{E}[f(X/n)]-f(k/(2n))|, $$is bounded from above by—

$\omega(\sqrt{\frac{1}{8n-4}})$, _for all n≥1 that are integer powers of 2, _

$\omega(\sqrt{\frac{1}{7n}})$, for all n≥4 that are integer powers of 2, and

$\omega(\sqrt{\frac{1}{2n}})$, for all n≥1 that are integer powers of 2.

Proof: ω is assumed to be non-negative because absolute values are non-negative. To prove the first and second bounds: abs(E[f(X/n)] − f(k/(2 * n))) ≤ E[abs(f(X/n) − f(k/(2 * n))] ≤ E[ω(abs(X/n − k/(2 * n))] ≤ ω(E[abs(X/n − k/(2 * n))]) (by Jensen's inequality and because ω is concave) ≤ ω(sqrt(E[abs(X/n − k/(2 * n))]2)) = ω(sqrt(Var[X/n])) = ω(sqrt((k*(2 * n − k)/(4*(2 * n −1)*n2)))) ≤ ω(sqrt((n2/(4*(2 * n −1)*n2)))) = ω(sqrt((1/(8*n −4)))) = ρ, and for all n ≥4 that are integer powers of 2, ρ ≤ ω(sqrt(1/(7*n))). To prove the third bound: abs(E[f(X/n)] − f(k/(2 * n))) ≤ ω(sqrt(Var[X/n])) ≤ ω(sqrt(1/(2*n))). □

The only technical barrier, though, is to solve the functional equation—$$\delta(n) = \delta(2n) + \kappa(n), $$ where $\kappa(n)$ is one of the bounds proved above, such as $\omega\left(\sqrt{\frac{1}{8n-4}}\right)$.

In general, the solution to this equation is—$$\delta(2^m) = \sum_{k=m}^\infty \kappa(2^k), $$ where $n = 2^m$ and $m\ge0$ are integers, provided the sum converges.

In this case, the third bound has a trivial solution when $\omega(x)$ is of the form $cx^\alpha$, but not in general.

Now, we approximate $f$ with polynomials in Bernstein form of power-of-two degrees. These are two sequences of polynomials that converge to $f$ from above and below, such that their coefficients "increase" and "decrease" just as the polynomials themselves do. In general, for all $n\ge1$ that are integer powers of 2:

The $k$th Bernstein coefficient for the upper polynomial of $n$th degree is $f(k/n) + \delta(n)$.

The $k$th Bernstein coefficient for the lower polynomial of $n$th degree is $f(k/n) - \delta(n)$.

Thus, for the polynomials of degree $n$, $\delta(n)$ is the amount by which to shift the approximating polynomials upward and downward.

Remark:

Theorem 26 of Nacu and Peres (2005) and the proof of Keane and O'Brien (1994) give general ways to simulate continuous factory functions $f(\lambda)$ on the interval $[0, 1]$.  The former is limited to functions that are bounded away from 0 and 1, while the latter is not.  However, both methods don't provide formulas of the form $f(k/n) \pm \delta(n)$ that work for a whole class of factory functions (such as formulas of the kind given earlier in this answer).  For this and other reasons, given below, both methods are impractical:

Before a given function $f$ can be simulated, the methods require computing the necessary degree of approximation (finding $k_a$ or $s_i$ for each polynomial $a$ or $i$, respectively).  This work has to be repeated for each function $f$ to be simulated.

Computing the degree of approximation involves, among other things, checking whether the approximating polynomial is "close enough" to $f$, which can require a symbolic maximization.

This computation gets more and more time-intensive with increasing degree.

For a given $f$, it's not guaranteed whether the $k_a$'s (or $s_i$'s) will show a pattern or keep that pattern "forever" — especially since only a finite number of approximation degrees can be computed with these methods.

REFERENCES:

Nacu, Şerban, and Yuval Peres. "Fast simulation of new coins from old", The Annals of Applied Probability 15, no. 1A (2005): 93-115.


问:
How to compute complete elliptic integral of the first kind in explicit form using elementary functions?
How to compute complete elliptic integral of the first kind in explicit form using elementary functions?
If it is not possible to compute complete elliptic integral of the first kind in explicit way, then is it possible to define the result? How would you define solution?</p>

答:
If it is not possible to compute complete elliptic integral of the first kind in explicit way, then is it possible to define the result? How would you define solution?

There is a general definition of the integral $\int_a^b f(x)\,dx$ which works for any continuous function $f:[a,b]\to\mathbb R$ (and also for some functions that are not continuous). The integrals such as
$$\int_0^{\pi/2} \frac{1}{\sqrt{2\cos^2\theta+3\sin^2\theta}}\,d\theta$$
and
$$\int_0^{\pi/2} \sqrt{2\cos^2 (e^\theta)+3\sin^2 \sqrt{\theta}}\,d\theta$$
are perfectly well-defined, regardless of our ability to write down their value in any explicit way. For many integrals there is no explicit formula for the result, which usually calls for some sort of approximation.

As J.M. said, the AGM method is an efficient way to compute $K(k)$. See, for example,

Landen's transformation

Arithmetic–geometric mean

Evaluation of the complete elliptic integrals by the AGM method

AGM (Wolfram)


问:
Proving expression with binomial coefficients is integer
How do we prove that $\frac{\binom{n+1}{m+1}\binom{n}{m-1}}{m}$ for $n\ge m\ge1$ is an integer?

Both coefficients in the numerator are integers and the interval from $n+1$ to $n-m+2$ covered by the numerators of the binomial coefficients has length $m$. However, that doesn't guarantee the whole thing is an integer because those factors could be canceled by the denominators of the binomial coefficients that are not relatively prime to m.

If $m$ is prime then it's more obviously an integer because if the $m$ in the denominator of $\binom{n+1}{m+1}$ divides $n+1$ then the $m$ in the denominator of the whole expression cancels $n-m+2$, and otherwise, there are two multiples in the numerator. But how do we work with non-primes?

I also tried a counting argument but couldn't get anywhere with that.

答:
Write this as
$$
\begin{aligned}
\color{blue}{\frac{1}{m}}\binom{n+1}{m+1}\color{blue}{\binom{n}{m-1}}
&=\color{blue}{\frac1{n+1}}\binom{n+1}{m+1}\color{blue}{\binom{n+1}{m}}
\\
&=\frac{m+1}{n+1}\binom{n+1}{m+1}\binom{n+1}{m}-\frac{m}{n+1}\binom{n+1}{m+1}\binom{n+1}{m}
\end{aligned}
$$
then apply the absorption identity to each summand.


问:
Intuition behind structural weakening
According to the Fitch deductive system I'm using in my logic class, the following is a valid inference:

Premise: Q

Step 1: Assume P

Step 2: Q by reiteration

Step 3: P → Q by conditional intro.

Basically it seems that, given the truth of any arbitrary atomic sentence, you can deduce that any other arbitrary atomic sentence, if also true, implies the first sentence. And it is this result that I'm having trouble wrapping my head around. Couldn't it be the case that two sentences can be true without any relation of implication holding between them? This certainly seems to be the case in informal reasoning and with facts about the world, so what's the rationale behind this being a valid step in propositional logic?

Edit: the initial title ("motivation for entailment in propositional logic") was misleading - this question is more about structural weakening than it is the material conditional or entailment per se. I made this mistake beacuse I myself wasn't clear on weakening until the question was answered.

答:
I do not know quite well the fitch style proof, but in propositional logic it is easy to check that for whatever P, it implies Q->P. If you make the right truth table, it will be clear. Even if this is not quite intuitively appealing, it depends on how the material conditional is defined:
P->(Q->P) can never be false. If it were, P should be true and the consequent false. But since P is true, Q->P can't be false. Therefore, P->(Q->P) must be true.</p>

Title: Intuition behind structural weakening


问:
Transition Probability Matrix of a Particular Markov Chain
I have come across a "MC that represents the success runs of binomial trials with success p".

So does the state of the MC represent the total number of successes for the trials?

The transition probability matrix is

$$ \begin{bmatrix}(1-p)&p&0&0&...\\(1-p)&0&p&0&...\\(1-p)&0&0&p&...\\...&...&...&...\end{bmatrix} $$

So e.g. state $n \rightarrow n+1 $ with probability p, but why does state $ n  \rightarrow 0$ with probability (1-p)? Shouldn't it be state $ n \rightarrow n$ with probability (1-p) instead?

Is there a particular name for this Markov Chain?

Thank you!

答:
This is the Markov chain for the runs of consecutive successes, not for the total number of successes.

Suppose that you are playing many rounds of a game, and you win each round with probability $p$. This Markov chain keeps track of your streak: how many wins have you had since the most recent loss? After each game, the streak changes as follows:

If you win the game, your streak increases from $n$ to $n+1$.

If you lose the game, your streak resets from $n$ to $0$.


问:
Can gradient descent on covariance of Gaussian cause variances to become negative?
I have a Gaussian distribution with likelihood

$$p(x; \mu, \Sigma) = (|2 \pi |\Sigma|)^{-1/2} \exp(-\frac{1}{2} (x - \mu)^T \Sigma^{-1} (x-mu))$$

and I've computed the gradient of the log likelihood with respect to $\Sigma$ as

$$\nabla_{\Sigma} \log p(x; \mu, \Sigma) = -\frac{1}{2}(\Sigma^{-1} - \Sigma^{-1} (x-\mu)(x-\mu)^T \Sigma^{-1})$$

I'm now trying to optimize $\Sigma$ using gradient descent but my $\Sigma$ is gaining negative diagonal entries, which doesn't make sense as the diagonal entries are variances and thus be non-negative. Which of the following is true?

a) Is this a known possibility when using gradient descent on covariances?

b) Is my derivation incorrect?

c) Is there an error with my gradient descent implementation?

答:
A friend was able to provide me with an answer. I can't rule out an error in my code, but in general, yes, directly optimizing positive (semi-)definite matrices using gradient descent can result in non-positive (semi-)definite matrices.

This article provides two solutions:

Replace the unconstrained optimization problem with a constrained optimization problem (not recommended)

Reparameterize the covariance to ensure the resulting covariance is P(S)D. For instance, define $\Sigma = A^T A$ and optimize $A$ instead.


问:
approximate mean absolute deviation using histogram
Given a sequence $A= a_1, a_2, \cdots,a_i, \cdots,a_n$, where $a_i \in \mathbb{R}$.

We can calculate the MAD(mean absolute deviation) via $G(A) = \frac{1}{n} \sum_{i=1}^n|a_i - \textbf{median}(A)|$, where $|.|$ is absolute value.

On the other hand, we can approximate the calculation of $G(A)$ using a $k$ bin histogram.

Compute a histogram of $A$ with $k$ bin and obtain $W=(w_1, \cdots, w_i, \cdots, w_k)$ and edges $E=(e_0, e_1, \cdots, e_k)$.

using $W, E$, we can compute an approximate version of $G(A)$, called $S(W, E)$.

I am wondering how can one bound the difference between $G(A)$ and  $S(W, E)$?

And apparently, $k$ is an important factor here. With $k > n$, the two would be equvilant, i probably shouldn't use the histogram approach

But normally we would choose $k<<n$, what would be the consequences?

As Ian pointed out, there are different ways to define $S(W, E)$, e.g.  1, the exact MAD of the distribution with PDF given by the histogram. Or  2, collapse each bin into its midpoint and use a discrete distribution instead.

My goal is to find an as accurate as possible approximation, so whichever strategy is accurate would be the ideal implementation. Meanwhile, I also want to have some error bound, so if whichever is easy to have an error bound is also appreciated.

答:
Say you define the approximate MAD by $\tilde{M}=\frac{1}{n} \sum_{i=1}^n |m_{j(i)}-m|$ where

$b_j$ is the midpoint of bin $j$

$j(i)$ sends $i$ to the index of the bin where $a_i$ is located

$m$ is (for simplicity) the exact median. Again for simplicity let us say $n$ is odd so $m=a_{(n+1)/2}$.

For simplicity let us also say that the median is one of the edges, say $m=e_{k^*}$. Then

Then

$$\tilde{M}=\frac{1}{n} \sum_{i=1}^{(n-1)/2} -(m_{j(i)}-m) + \frac{1}{n} \sum_{i=(n+1)/2}^n (m_{j(i)}-m).$$

Meanwhile

$$M=\frac{1}{n} \sum_{i=1}^{(n-1)/2} -(a_i-m) + \frac{1}{n} \sum_{i=(n+1)/2}^n a_i-m.$$

It follows that the difference is

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

Thus in absolute value it is less than the average bin radius. This bound is not at all guaranteed to be tight, though.


问:
Is there set with no interior in $\mathbb{R}^2$, the intersection with $\lambda$-a.e. set $ \{x\} \times \mathbb{R}$ has an interior in $\mathbb{R}$?
I am searching for a set that has no interior in $\mathbb{R}^2$ but when I intersect with the sets $\{x\} \times \mathbb{R}, x \in \mathbb{R}$ it has $\lambda$-a.e. an interior in $\mathbb{R}$, where $\lambda$ is the Lebesgue measure. The same should hold true for the sets $\mathbb{R} \times \{y\}, y \in \mathbb{R}$. My idea was to take $(q_n)_{n \in \mathbb{N}} = \mathbb{Q}^2$ and define $U = \bigcup_{n \in \mathbb{N}} B_{\frac{1}{2^n}}(q_n)$. Then I consider $A = U^C$, which has no interior in $\mathbb{R}^2$ since it contains no point of $\mathbb{Q}^2$. Now when I fix $x \in \mathbb{R}\setminus \mathbb{Q}$ does the set $C = A \cap (\{x\}\times \mathbb{R})$ have an interior?

答:
You could recursively construct a set that is dense in the plane, but intersects every line in at most two points. Its complement in the plane answers your question.

So called Mazurkiewicz set intersects every line in exactly two points, but to answer your question it is enough to construct a set dense in the plane that intersects every line in at most two points, which is easier to construct recursively in just countably many steps. Pick a countable base $\{U_n:n\ge 1\}$ for the topology of the plane, and at step $n$ pick a point $p_n\in U_n$ that does not lie on any of the (finitely many) lines that go through any two of the points that have already been picked at previous stages.

There is some literature about such sets, and I am bit lazy and only found one paper (because I wrote it) but you could find more along these lines (and, perhaps, better :). See

Non-regularity of some topologies on $\mathbb R^n$ stronger than the standard one.

Mathematica Pannonica (1994)

Volume: 5, Issue: 1, page 105-110.

https://eudml.org/doc/231430

In general this is also related to questions about joint continuity vs separate continuity, one could google that too for more papers.

Here, I remembered the term "two-point set", and googled some more links:

Can two-point sets be Borel?

https://mathoverflow.net/questions/272527/can-two-point-sets-be-borel

(there are more references if you follow the above link)

The axiom of choice and two-point sets in the plane

Arnold W. Miller

https://www.math.wisc.edu/~miller/res/two-pt.pdf

The Two Points Theorem of Mazurkiewicz,

Peter Komjath

Journal of Combinatorial Theory, Series A 99, 371–376 (2002)

doi:10.1006/jcta.2002.3278

https://www.sciencedirect.com/science/article/pii/S0097316502932784


问:
Is there a norm in which the vector space of all sequences with the induced metric is complete?
The question of knowing whether is possible put a norm in a vector space is something new to me. I liked the Ivo ideas and I have wondered a little more about a particular case.

Is there a norm in which the vector space of all sequences with the induced metric is complete? This question sounds harder to answer. The completion of a metric space is something that is usually done to achieve this result. Is the completion of the norm given by Ivo ideas isomorphic to the vector space of all sequences? If not, is there a norm in which the vector space of all sequences with the induced metric is complete?

答: To help people out, any separable Banach space have Hamel dimension of $\mathfrak{c}$, the second Beth number, as proven here or here. Since the Hamel dimension of all real sequences have the same cardinality $\mathfrak{c}$ and assuming the Continuum hypothesis, we can find a linear bijection $\phi$ from $l^{2}$ to the space of all sequences, say $\mathbb{R}^{\mathbb{N}}$. Hence, given a vector $w\in \mathbb{R}^{\mathbb{N}}$, define the norm $$\|w\|_{\mathbb{R}^{\mathbb{N}}} = \|\phi^{-1}(w)\|_{l^{2}}.$$ This norm define a  complete Banach space. To define a complete Hilbert space it's just define a inner product in the same way.


问:
Solving a non-linear optimisation problem with min type constraint.
The problem is $\min_{\boldsymbol{\mu}}f(\boldsymbol{\mu})$ given that
$\min|\mu_i-\mu_j|\geq\delta>0$, where $\boldsymbol{\mu} = (\mu_1,\ldots,\mu_n)^{\rm T}$. I am having difficulty to convert $\min|\mu_i-\mu_j|\geq\delta>0$ to a set of linear constraints so that I can get the solution (at least numerically). Any help will be appreciated. Thanks in advance.

答:
Your desired constraint
$$\min_{i<j} |\mu_i-\mu_j| \ge \delta$$
is equivalent to the set of constraints
$$|\mu_i-\mu_j| \ge \delta \quad\text{for all $i<j$} \tag1$$
You can linearize $(1)$ by introducing binary variable $z_{i,j}$ and linear big-M constraints
\begin{align}
\mu_j - \mu_i + \delta &\le Mz_{i,j} &&\text{for all $i<j$} \tag2 \\
\mu_i - \mu_j + \delta &\le M(1-z_{i,j}) &&\text{for all $i<j$} \tag3 \\
\end{align}
Constraint $(2)$ enforces $z_{i,j}=0 \implies \mu_i-\mu_j \ge \delta$.
Constraint $(3)$ enforces $z_{i,j}=1 \implies \mu_j-\mu_i \ge \delta$.


问:
Why Is $y=|x|$ A Cone?
Looking at

$$y=|x|$$

is an example of a cone that is not convex.
It is not convex because for $x_1=(1,1),x_2=(-1,1)\in \{x,|x|\}$ there is not $t\in[0,1]$ s.t $tx_1+(1-t)x_2\in \{x,|x|\}$

But why is it a cone?

答:
The locus of points where $y = |x|$ is indeed not convex. It's rays that meet at the origin.

You probably mean that the set of points where $y \ge  |x|$ is a cone.

Draw a picture.

Title: Why Is $y=|x|$ A Cone?


问:
Understanding (GIT) quotient by action on End(W)
Suppose we have a subgroup $H < \mathrm{Gl}(W)$, where $W = \mathbb{C}^n$, which acts on $\mathrm{End}(W)$ by precomposing: $a \mapsto h\circ a$, for $h\in H$.

We are given a distinguished linear subspace $L_0 \subset W$ and we consider the affine variety
$$X = \bigcup_{h \in H} \mathrm{End}(W, h(L_0))$$
which contains the distinguished subvariety $Y = \mathrm{End}(W, L_0)$.

Problem: I would like to prove $X //H \simeq Y//H_0$ where $H_0$ is the stabiliser of $L_0$ (for the natural action of $H$ on $W$).

I am struggling to find a formal proof of this. Intuitively it should work, since for $h\in H$ we have $h\cdot Y=\mathrm{End}(W, h(L_0))$, so we can view $X = \bigcup_{h \in H} h\cdot Y$ and $H_0$ is the stabiliser of $Y$

but I'm a bit stuck and not so familiar with the details of GIT (if it has to do with that). Any suggestion?

答: Check the composition $Y\hookrightarrow X\twoheadrightarrow X/H$ is $H_0$-equivariant, which means there is a well-defined map $Y/H_0\to X/H$. Can you show this map is one-to-one and onto?


问:
Find $\gamma$ such that $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{-5},\sqrt{2})$
I have to solve the following problem:

Find $\gamma$ such that $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{-5},\sqrt{2})$

I am a bit confused about how to do it but as $\mathbb{Q}(\sqrt{-5},\sqrt{2})$ has degree $4$,then $\mathbb{Q}(\gamma)$ must also have degree $4$. I think this means that we must be able to write:

$$a+b\gamma + c \gamma^2 + d\gamma^3=\sqrt{-5}\\a'+b'\gamma + c' \gamma^2 + d'\gamma^3=\sqrt{2}$$

And find a suitable $\gamma$, I suspect $\gamma=\sqrt{-5}\sqrt{2}$ but I still can't solve it for $a,b,c,d$.

答: Consider $\gamma = \sqrt{2} + \sqrt{-5}$. By rationalizing the denominator, it follows that $$\frac{7}{\gamma} = \sqrt{2} - \sqrt{-5}$$ and hence the field $\Bbb Q(\gamma)$ contains $\sqrt{2} - \sqrt{-5}$. Now $$\sqrt{2} = \frac{1}{2}\left(\gamma + \frac{7}{\gamma}\right) \in \Bbb Q(\gamma)$$ and so $\Bbb Q(\gamma)$ contains $\gamma - \sqrt{2}$, or $\sqrt{-5}$. Therefore $\Bbb Q(\gamma) \supset \Bbb Q(\sqrt{2}, \sqrt{-5})$. As $\gamma \in \Bbb Q(\sqrt{2}, \sqrt{-5})$, the reverse containment $\Bbb Q(\gamma) \subset \Bbb Q(\sqrt{2}, \sqrt{-5})$ follows.


问:
Can the diagonal elements of a precision matrix be 0
I have this confusion that why the diagonal elements of the precision matrix cannot be 0?

Any suggestions will be much appreciated

答: The covariance matrix $E[(X-EX)(X-EX)^T]$ is positive-semidefinite, so (assuming it's invertible at all) positive definite, and so its inverse is also positive-definite. The diagonal elements of any positive-definite matrix $M$ are always strictly positive, because they're given by $e^T M e$ where $e = e_1,...,e_n$ runs through the standard basis vectors of $R^n$.</p>


问:
Interior versus Boundary Points Clarification
Let a,b,c be real numbers such that a<c<c and suppose that U = (a,c)u(c,b) is an open interval punctured at c. Does U have interior points? Does U have boundary points?
I understand that all points in U are interior points. However, I am struggling with the boundary points.
Is it logical to state that there U has no boundary points since all points in U are interior points? Can someone please tell me if my logic is accurate?</p>

答:
Given any set $A$ in a topological space $X$ (which in your example is $\mathbb{R}$), the boundary of $A$ or the frontier of $A$ is defined by:

$\overline{A}\cap \overline{A^{com}}$, where $A^{com}$ denote the complement of $A$.

In your case $A= (a,c)\cup (c,b)$. Thus, $\overline{A}=[a,b]$ and $A^{com}=(-\infty, a]\cup [b, \infty)\cup \{c\}.$ Note that $A^{com}$ is already closed, since $A$ is open. Therefore, $\overline{A^{com}}=(-\infty, a]\cup [b, \infty)\cup \{c\}.$ Now, it is trivial to see that $\overline{A}\cap \overline{A^{com}}=\{a,b,c\}.$ Therefore, the boundary or the frontier of the set $A$ is given by $\{a,b,c\}$.

You can describe the boundary of a set in a metric space in another way too. Given any set $D$ in a metric space $X$, a point $x_0\in X$ is said to be a boundary point of $D$ if each ball containing $x_0$ contains at least one point of $D$ and at least one point of $D^{com}$. Observe that the boundary $\{a,b,c\}$ (of the set $A$) obtained in case of your example, exactly satisfies this property.


问:
Show that $P(W_t \in [-1,1] \; \forall t \geq 0) = 0$

Show that $P(W_t \in [-1,1] \; \forall t \geq 0) =  0$ where $W_t$ is a standard Wiener process on some probability space $(\Omega, \mathcal{F}, P)$.

My attempt:

It holds that $W_t = W_t - 0 = W_t - W_0 \sim N(0,t) \Longrightarrow \dfrac{W_t}{\sqrt{t}} \sim N(0,1)$. So we have

\begin{aligned}
P(W_t \in [-1,1]) &= P(-1 \leq W_t \leq 1) = P\left(-\frac{1}{\sqrt{t}} \leq \frac{W_t}{\sqrt{t}} \leq \frac{1}{\sqrt{t}}\right) \\ &= \Phi\left(\frac{1}{\sqrt{t}}\right) - \Phi\left(-\frac{1}{\sqrt{t}}\right) = \Phi\left(\frac{1}{\sqrt{t}}\right) - \left(1 - \Phi\left(\frac{1}{\sqrt{t}}\right)\right)\\
&= 2\Phi\left(\frac{1}{\sqrt{t}}\right) - 1
\end{aligned}

but this obviously does not equal $0$ for all $t\geq 0$.

Anyone can explain what I'm doing wrong? I recently started to study Stochastic processes.

答: You seem to be mixing up $$P(W_t \in [-1,1] \; \forall t \geq 0) =  0$$ with $$P(W_t \in [-1,1] ) =  0 \; \; \forall t \geq 0$$ The quantifier "$\forall t\geq 0$" is inside the probability, so you can let $t\to\infty$ in the expression you found, which does give an answer of $0$.


答:
You know that for any $s$,
$$P( W_t \in [-1, 1], \forall t) \leq P(W_s \in [-1,1]) = 2\Phi\left(\frac{1}{\sqrt{s}}\right) - 1.$$
Send $s \to \infty$!

Body:
Show that $P(W_t \in [-1,1] \; \forall t \geq 0) =  0$ where $W_t$ is a standard Wiener process on some probability space $(\Omega, \mathcal{F}, P)$.

My attempt:

It holds that $W_t = W_t - 0 = W_t - W_0 \sim N(0,t) \Longrightarrow \dfrac{W_t}{\sqrt{t}} \sim N(0,1)$. So we have

\begin{aligned}
P(W_t \in [-1,1]) &= P(-1 \leq W_t \leq 1) = P\left(-\frac{1}{\sqrt{t}} \leq \frac{W_t}{\sqrt{t}} \leq \frac{1}{\sqrt{t}}\right) \\ &= \Phi\left(\frac{1}{\sqrt{t}}\right) - \Phi\left(-\frac{1}{\sqrt{t}}\right) = \Phi\left(\frac{1}{\sqrt{t}}\right) - \left(1 - \Phi\left(\frac{1}{\sqrt{t}}\right)\right)\\
&= 2\Phi\left(\frac{1}{\sqrt{t}}\right) - 1
\end{aligned}

but this obviously does not equal $0$ for all $t\geq 0$.

Anyone can explain what I'm doing wrong? I recently started to study Stochastic processes.


问:
Solving Differential Equation $4x^2yy'=3x(3y^2+2)+2(3y^2+2)^3$
I have tried the solve following differential equation with Riccati method but couldnt do.
Can someone help how can I find the particular solution if it is?

$$4x^2yy'=3x(3y^2+2)+2(3y^2+2)^3$$

答:
Hint:
$$4x^2yy'=3x(3y^2+2)+2(3y^2+2)^3$$
Substitute  $u=3y^2+2$

Then you have a Bernoulli's DE.
$$\dfrac 23x^2u'=3xu+2u^3$$


问:
How to avoid problems in modelchecking if we don't allow any $q_f \in F$?
We get a specification $\varphi \in \mathsf{LTL}$ and we translate a real system to some NFA $A$.

The problem is to construct an NFA $A'$ with $L(A) = L(\neg \varphi)$, so we can check for $L(A') \cap L(A) = \emptyset$

Question: When translating a real system, which shouldn't end (i.e like an OS), to a NFA $(Q, \Sigma, \delta, I, F)$, we need to avoid final states $q_f \in F$. But when doing this, we get $L(A) = \emptyset$ since we never accept anything? So how can we avoid this in modelchecking?

答:
In my understanding, this problem can be resolved by making every state a final state.
This means, there are no special final states and still, words can be accepted. The steps of a path in the automaton correspond to time steps and when every state is a final one, the 'word' is accepted in every time step, meaning that is is still a valid system.</p>


问:
Is there any self complementary graph of order 5 other than C5
The concept of a self-complementary graph as I learned is, a graph that is isomorphic to its complement is called self-complement. I have some examples in mind of the self-complementary graph of order five but I am confused about, how can we find a self-complementary graph of order five other than copies of C5. Kindly give me the answer with an example.</p>

答:
For this kind of questions, I like to simply use sage with geng. Try this code

```<code>for g in graphs.nauty_geng("{0} -c".format(5)):
    if g.is_isomorphic(g.complement()):
        g.show()
</code>```

Yielding to

and

Body:
For this kind of questions, I like to simply use sage with geng. Try this code

for g in graphs.nauty_geng("{0} -c".format(5)):
    if g.is_isomorphic(g.complement()):
        g.show()

Yielding to

and


问:
How to show that the $ax+b$ group is locally compact?
Let $ax+b$ be the group of affine transformations $x\mapsto ax+b$ with $a>0$ and $b\in \mathbb{R}$.  How do you topologize this group?  As a group it is isomorphic to the set of $2$x$2$ matrices of the form \begin{bmatrix}a & b \\0 & 1\end{bmatrix}.  But unless I am greatly mistaken this set of matrices connot be an open subset of $\mathbb{R}^{4}$ since two entries are fixed.  So how do you argue that $ax+b$ is a locally compact group?

答: $a>0, b\in \mathbb{R}$ gives you a pretty obvious homeomorphism with the (open) upper halfspace. Is that locally compact?


答: The topology on this group is inherited from $\mathbb{R}^2$, and when identified in $\mathbb{R}^2$ it is an open half-plane. Definitely locally-compact.

Title: How to show that the $ax+b$ group is locally compact?

Body: Let $ax+b$ be the group of affine transformations $x\mapsto ax+b$ with $a>0$ and $b\in \mathbb{R}$.  How do you topologize this group?  As a group it is isomorphic to the set of $2$x$2$ matrices of the form \begin{bmatrix}a & b \\0 & 1\end{bmatrix}.  But unless I am greatly mistaken this set of matrices connot be an open subset of $\mathbb{R}^{4}$ since two entries are fixed.  So how do you argue that $ax+b$ is a locally compact group?


问:
Poiseuille flow: can I avoid the simplification that $u=u(y)$?
Let's consider a viscous, incompressible and irrotational fluid that flows in a canal in two dimensions. It is reasonable to assume that the velocity of the fluid has the form $\vec v = u(x,y,t) \hat x $, or in other words, we are assuming that the velocity has just the horizontal component $u(x,y,t)$.

Under these hypotheses, the incompressibility condition $\vec \nabla \cdot \vec v = 0$ gives us that $\partial_x u = 0$, i.e. $u$ depends only on $t$ and $y$.

The Navier-Stokes equations then become
$$ (1) \qquad \begin{cases}
\partial_tu = -\partial_xp + \nu \partial_{yy}u \\
-\partial_y p = 0
\end{cases}$$

At this point one usually says "suppose that the velocity does not depend on time": the first equation of the system $(1)$ simplifies to $\nu \partial_{yy}u = \partial_x p $ and one obtain a closed form of the pressure $p$ and the velocity $u$.

My question is: while the assumptions made at the beginning seem resonable with the usual context of fluid dynamics, the last hypothesis seems to be made just to derive in few steps a solution of the problem. Although it is said "a physical theory will always be wrong" because of the assumptions made to translate the real world into some mathematical equations, I ask:

Can the system $(1)$ have solutions without the simplification that $u = u(y)$?

答:
You have reduced the Navier-Stokes equations to a simple form by making a number of assumptions that may or may not be realistic or commensurate with a physical configuration.  To begin with, restricting to two-dimensional flow where there is no dependence on the spanwise coordinate $z$ nor a non-zero velocity component $w$ in that direction can only be true assuming infinite extent of the domain in the $z-$direction.  Furthermore, by assuming there is no dependence on the streamwise coordinate $x$, you are making the assumption of fully developed flow far from an entrance to the channel where the velocity field must assume a more complicated form. Nevertheless, if the physical dimensions are such that the the spanwise extent is very large and we are examining the flow far from an entrance, this can be a good approximation.

Of more immediate concern is that you have not mentioned boundary conditions.  If the channel is confined between solid stationary boundaries at $y = 0,L$, then we must impose the boundary conditions $u(0) = u(L) =0$.  A further assumption is needed about the pressure gradient.  The simplest would be that it is constant, $\frac{\partial p}{\partial x} = -G < 0$. In this case, (1) admits a steady solution obtained by first integrating twice with respect to $y$ to get

$$u(y) = C_1 +C_2y - \frac{Gy^2}{2\nu},$$
and, after applying boundary conditions, we obtain the velocity profile for steady plane Poiseuille flow:

$$u(y) = \frac{GL}{2\nu}\left( y - \frac{y^2}{L}\right)$$

Again, this is an idealization for the flow between infinite parallel plane boundaries with a constant pressure gradient.

Changing the configuration or boundary conditions by introducing time-dependency will lead to solutions of (1) with the full functional form $u(y,t)$.  An example would be when the pressure gradient is itself time-dependent, for example oscillating sinusoidallly in time.

A classic variation with a time-dependent solution is the Stokes problem. In this case the domain is of semi-infinite extent $0 \leqslant y < \infty$ and the flow is driven not by pressure but, instead, by the sinusoidal oscillation of a plate at $y=0$ requiring the boundary condition $u(0,t) = U\cos \omega t$.  In this case, the pressure gradient is taken to be zero and the closed form solution to (1) is

$$u(y,t) = Ue^{-\sqrt{\frac{\omega}{2\nu}}y}\cos \left(\omega t - \sqrt{\frac{\omega}{2\nu}}y \right)$$

A closed-form time dependent solution can even be obtained when $0 \leqslant y \leqslant L$ and there is a stationary wall at $L$ where $u(L,t) =0$ for all $t$.

Even in simple configurations like flow between infinite parallel plates,  the full Navier-Stokes equations admit extremely complex  time-dependent solutions that reflect the instability of unidirectional, laminar flow and the transition to fully developed turbulence.


问:
Show that the norms $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are both continuous.

Show that the norms  $\|\cdot\|_1 : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ and $\|\cdot\|_\infty : (\mathbb{R}^n,d_2) \to (\mathbb{R}, |\cdot|)$ are both continuous.

For the first one if $x \in \mathbb{R}^n$, then $\|x\|_1=|x_1|+|x_2|+\dots+ |x_n|$ this seems just to be a sum of multiple projections e.g $\operatorname{pr}_j: \mathbb{R}^n \to \mathbb{R}$, where $j  \in \{0, \dots n\}$. So as the sum of continuous functions this is continuous? For the second one I'm not sure how to approach. I know that if I have for example $\max\{f,g\}$ then this equals $\frac12(f+g)+\frac12|f-g|$ so do I have that $$\max\{|x_1|,|x_2|, \dots|x_n|\}= \frac1n(|x_1|+\dots|x_n|)+\frac1n||x_1|-|x_2|-\dots-|x_n||$$ and I could show that this is continuous by the fact that all $x_i$ are continous since the previous result?

答:
A easy counter example: $\max\{1,2,3\} = 3$, but
$$ \dfrac{1}{3}\big(1+2+3\big) + \dfrac{1}{3}\big| 1-2-3 \big| = 2+\dfrac{4}{3} = 3\dfrac{1}{3} \ne 3, $$

Hint:
(I guess the norm $d_2$ means $\|\cdot\|_2$.)

You can prove the following results in turn:

The finite sum of continuous functions is continuous, And all $|x_j|$ are continuous.

$\max\{a,b\}=\dfrac{1}{2}(a+b)+\dfrac{1}{2}|a-b|$.

$\max\{a,b,c\}=\max\big\{\max\{a,b\},c\big\}$.


答: Let $G$ be your group and consider the bijection$$\begin{array}{rccc}\psi\colon&(0,\infty)\times\Bbb R&\longrightarrow&G\\&(a,b)&\mapsto&\begin{bmatrix}a&b\\0&1\end{bmatrix}.\end{array}$$Then, consider the distance defined on $G$ by $d(M,N)=\bigl\|\psi^{-1}(M)-\psi^{-1}(N)\bigr\|.$


问:
How would I integrate $\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$?
$$\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$$
I have known this equals to $\frac{\pi^2}{16}$, but I cannot prove it.

Body:
Note

\begin{align}
I=&\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx\\
= &\int_0^1 \frac{\frac12\ln(1-x^2)+\ln(1+ \frac x{\sqrt{1-x^2}})}{x}\,dx=\frac12I_1+I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln(1-x^2)}{x}\,dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&= \int_0^1 \frac{\ln(1-x)}{x}\,dx+ \int_0^1 \frac{\ln(1+x)}{x}\,dx
= -\int_0^1 \frac{\ln(1+x)}{x}\,dx\\
I_2 &= \int_0^1 \frac{\ln \left(1+\frac x{\sqrt{1-x^2}}\right)}x
\overset{x=\frac t{\sqrt{1+t^2}}}=\int_0^1\frac{\ln(1+t)}{t(1+t^2)}dt+ \int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t^2)}dt}\\
&=\int_0^1 \frac{\ln(1+t)}t dt - \frac14 \int_0^1 \frac{\ln t}{1+t}dt
=\frac54 \int_0^1 \frac{\ln(1+t)}t dt
\end{align}
Substitute $I_1$ and $I_2$ into (1) to obtain

$$I= \frac34 \int_0^1 \frac{\ln(1+t)}t dt =\frac34\cdot \frac{\pi^2}{12}=\frac{\pi^2}{16}
$$
$\int_0^1 \frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$


问:
How to express exclusive intersections?
I have a function $f:\mathbb{R}^2 \rightarrow\mathbb{R}, \ \mathbf{x} \mapsto \sum_{i = 1}^{n} w_i \mathbf{1}_{A_i}(\mathbf{x})$ for $w_i \in \mathbb{R}$ and the $A_i$s are allowed to intersect. Thus for the case of $n=3$ we might have something like:

($A_1 = A, A_2 = B, A_3 = C$ here)

where at each exclusive intersection e.g. $AB = A \cap B \setminus C$, $f$ takes the value $w_A + w_B$. Now, I would like to express $f$ as the sum of indicators over non-intersecting sets (i.e. a partition of the underlying space):

$$f(\mathbf{x}) = (w_A + w_B + w_C) \mathbf{1}_{ABC}(\mathbf{x}) + (w_A + w_B) \mathbf{1}_{AB}(\mathbf{x}) + (w_A + w_C) \mathbf{1}_{AC}(\mathbf{x}) + \ldots$$

How to express this for general $n$? (In the same vein as XOR, the notion of exclusive intersection would really help here, but I'm unsure of its notation.)

答:
Let $X=\{A_1,A_2,\ldots, A_n\}$.
For each $S\in\mathcal P(X)$ denote $M(S)=\bigcup S\setminus \bigcup(\mathcal P(X)\setminus S)$ and $W_S=\sum_{A\in S}w_A$. Then
$$f(x)=\sum_{S\subseteq X} W_S\mathbf 1_{M(S)}$$ is such a sum. Note that $M(S)$ is the "exclusive intersections" of all $A_i\in S$.

Alternatively, we have a map $\phi\colon\mathbb R^2\to X$, $\mathbb x\mapsto \{A\in X\mid x\in A\}$ and can write $\phi^{-1}(S)$ isntead of $M(S)$ above.

Title: How to express exclusive intersections?


问:
Help finding an explicit solution to this ODE
I have the following ODE:

$$\dot{r}=(r-r^{3})\sin^{2}(\theta)$$

I have been trying to find an explicit solution for $r$, with little success. I have tried separating variables but then get stuck trying to make it explicit. Any suggestions? Thanks.

答:
$$\frac{dr}{d\theta}=(r-r^3)\sin^2\theta$$
Separating the variables:
$$\int\frac{1}{r-r^3}dr=\int\sin^2\theta d\theta$$
First, let's focus on the integral on the left hand side.
$$\begin{align}
\int\frac{1}{r-r^3}dr&=\int\frac{1}{r(1-r)(1+r)}dr=\int\frac{1}{r}+\frac{1}{2(1-r)}-\frac{1}{2(1+r)}dr\\
&=\ln\lvert r\rvert-\frac{1}{2}\ln\lvert1-r\rvert-\frac{1}{2}\ln\lvert1+r\rvert+C\\
&=\frac{1}{2}\ln\left\lvert\frac{r^2}{1-r^2}\right\rvert+C\\
&=\frac{1}{2}\ln\left\lvert\frac{r^2-1+1}{1-r^2}\right\rvert+C=\ln\left\lvert\frac{1}{1-r^2}-1\right\rvert+C\end{align}$$
Now for the right integral.
$$\int\sin^2\theta d\theta=\frac{1}{2}\int1+\cos 2\theta d\theta=\frac{1}{2}\theta+\frac{1}{4}\sin2\theta+D$$
Hence,
$$\ln\left\lvert\frac{1}{1-r^2}-1\right\rvert=\theta+\frac{1}{2}\sin2\theta+C_1$$
And now we can rearrange to find $r^2$, and thus find $r$.

I hope that helps. If you have any questions please don't hesitate to ask :)


问:
Prove or disprove the quasi-concavity and quasi-convexity of $f(x,y)=xa^{y-1}+b$
Consider $f(x,y)=xa^{y-1}+b$ for $(x,y)\in\mathbb{R}^2_{++}$, $a\in(0,1)$, and $b\in\mathbb{R}$. I am trying to evaluate the quasi-concavity/quasi-convexity of $f$.

My attempt: To see quasi-concavity, consider $g(x,y)=ln(x)+(y-1)ln(a)$. The Bordered Hessian for $g$ is

\begin{pmatrix}
0&\frac{1}{x}&ln(a)\\
\frac{1}{x}&-\frac{1}{x^2}& 0\\
ln(a)& 0&0
\end{pmatrix}

The determinant of the 1st order Bordered Hessian is $-\frac{1}{x^2}<0$, and the determinant of the Bordered Hessian itself is $(\frac{ln(a)}{x})^2>0$. This alternation of signs implies that $g$ is quasi-concave. Consider $h(x)=e^x+b$, which clearly is increasing. We have that $f(x,y)=h(g(x,y))$, so as $g$ is quasi-concave and $h$ an increasing transformation, $f$ is quasi-concave as well.

Where I'm stuck at is figuring out if $f(x,y)$ is quasi-convex also. I've tried different transformations to use a similar technique to the one I used showing quasi-concavity, but without much luck. Any help is appreciated.

答: The function is not quasiconvex. For quasiconvexity, the sublevel set $\left\{ (x,y) : f(x,y) \leq \alpha\right\}$ needs to be convex for any $\alpha \in \mathbb{R}$. Take $\alpha = 1+b$, then you need $\{(x,y) : x \leq c^{y-1}\}$ to be convex (with $c = 1/a > 1$). This set is clearly not convex, because $(x,y)=(1,1)$ and $(x,y)=(c^{2},3)$ are in the sublevel set, but the midpoint $(x,y)=((1+c^2)/2, 2)$ is not.


问:
Is this a valid PMF question?
Suppose that the random variable Y : Ω → R where Ω = N is given by Y(n) = n .
Find E(Y), Var(Y) if the probability mass function of Y is given by :

\begin{equation}
  p(x) = \left \{
  \begin{aligned}
    &\frac{31}{108}x^x, && \text{if}\ x=2,3 \\
    &0, && \text{otherwise}
  \end{aligned} \right.
\end{equation}

Compute the values of E(2Y + 1) and V ar(2Y + 1) as well.

How does this question work if the sum of the pmf values is over 1?
I calculated the sum as,
\begin{equation}
  P(x) = (2*\frac{31}{108}2^2) +( 3*\frac{31}{108}3^3) = 10.0463.
\end{equation}

Or am I misunderstanding something? I thought that if the sum of probabilities of a pmf function is over 1, then it cannot be a pmf for a rv.

答: The pmf of $2Y+1$ is not $2p(x)+1$, it is $p(\frac{1}{2}(x-1))$ (the relationship here is computing $F(Y)$ you get pmf $p(F^{-1}(Y))$, so when you sum over the probabilities you will get the same numbers just located in different places. What you computed was $E[2Y+1]$, not $P(x)$.

Title: Is this a valid PMF question?


问:
number of valuation ring of $\Bbb Q_p((T))$
I would like to know the number of valuation rings of  $\Bbb Q_p((T))$.
I know $\Bbb Q_p$ has $2$ valuation rings, that is,$\Bbb Q_p$ and $\Bbb Z_p$.
Every algebraic extension of $\Bbb Q_p$ has more than $2$ valuation rings because of extension theorem on valuation.But $\Bbb Q_p((T))$ is not algebraic over $\Bbb Q_p$,I am at a loss.

How many valuation rings of  $\Bbb Q_p((T))$ are there?

Thank you in advance.

答:
There is only one (non-trivial) discrete valuation on $\Bbb{Q}_p((T))$.

For all $f\in 1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]]$ the binomial series gives that $f^{1/n}\in \Bbb{Q}_p((T))$ whenever $p\nmid n$. Therefore, that $v$ is discrete implies $v(f)=0$.

Decompose
$$\Bbb{Q}_p((T))^\times = T^\Bbb{Z} p^\Bbb{Z} \langle \zeta_{p-1}\rangle \ (1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]])$$
Thus, it remains to find $v(\zeta_{p-1}),v(p)$ and $v(T)$.

Since  $(\zeta_{p-1})^{p-1}=1$ we must have $v(\zeta_{p-1})=0$.

If $v(p)\ne 0$, then $p$ is a multiple of $1$ so we must have $v(p)>0$. For $r$ large enough we have $v(p^{-r} T)<0$ so that $v(1+p^{-r} T)<0$, a contradiction.

Whence $v(p)=0$. We must have $v(T)\ne 0$ for $v$ being non-trivial.

$v(T)<0$ would contradict $v(1+T)=0$ so we must have $v(T)>0$ and hence $$\{ x\in \Bbb{Q}_p((T))^\times,v(x)\ge 0\} = \Bbb{Q}_p[[T]]-0$$