问:
Cardinality of the set of all subsets of $E$ equipotent to $E$
I'm trying to prove the following statement (an exercise in Bourbaki's Set Theory):

If $E$ is an infinite set, the set of subsets of $E$ which are equipotent to $E$ is equipotent to $\mathfrak{P}(E)$.

As a hint, there is a reference to a proposition of the book, which reads:

Every infinite set $X$ has a partition $(X_\iota)_{\iota\in I}$ formed of countably infinite sets, the index set $I$ being equipotent to $X$.

I don't have any idea how that proposition might help.

If $E$ is countable, then a subset of $E$ is equipotent to $E$ iff it is infinite. But the set of all finite subsets of $E$ is equipotent to $E$. So its complement in $\mathfrak{P}(E)$ has to be equipotent to $\mathfrak{P}(E)$ by Cantor's theorem. Hence the statement is true if $E$ is countable. Unfortunately, I don't see a way to generalize this argument to uncountable $E$.

I'd be glad for a small hint to get me going.

答:
Using the axiom of choice, every infinite set $X$ can be divided into two disjoint sets $X_0\sqcup X_1$, both of which are equinumerous with $X$. (Just well-order $X$, and take every other point in the enumeration.)

Now, consider all sets of the form $X_0\cup A$ for any $A\subset X_1$. There are $2^X$ many such $A$ and hence $2^X$ many such sets, and each is equinumerous with the original set $X$. So we've got $2^X$ many sets as desired, and there cannot be more than this, so this is the precise number.

Incidently, the stated answer to this question does in fact depend on the axiom of choice, since it is known to be consistent with $ZF+\neg AC$ that there are infinite Dedekind finite sets, and these are not equinumerous with any proper subsets of themselves. So for such an infinite set $X$, there would be only one subset to which it is equinumerous.


问:
Given two positive numbers $x,y.$ Prove that $\left|x\ln x-y\ln y\right|\leq\sqrt[3]{\left|x^{2}-y^{2}\right|}+\sqrt[3]{\left|x^{2}-y^{2}\right|^{2}}$

Problem 1. Given two poistive numbers $x, y.$ Prove that
$$\left | x\ln x- y\ln y \right |\leq\sqrt[3]{\left | x^{2}- y^{2} \right |}+ \sqrt[3]{\left | x^{2}- y^{2} \right |^{2}}$$

There exists a solution using the integral to solve, without loss of generality, assume that $x\geq y\Rightarrow t= x^{2}- y^{2}\geq 0,$ let $b= y^{2},$ we need to prove the following integral inequality in 2 cases : $b\geq\frac{1}{e^{2}}$ either $b+ t\leq\frac{1}{e^{2}}$
$$\left | \int_{b}^{b+ t}\frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x \right |\leq\sqrt[3]{t}+ \sqrt[3]{t^{2}}$$
Case 1. $b\geq\frac{1}{e^{2}}$
$$\int_{b}^{b+ t}\frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x\leq\sqrt[3]{t}+ \sqrt[3]{t^{2}}\Leftrightarrow\int_{0}^{t}\frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\leq\int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}{\rm d}x$$
Substitute $g\left ( b \right )= \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}\Rightarrow {g}'\left ( b \right )= -\frac{\ln\left ( x+ b \right )}{8\left ( x+ b \right )^{\frac{3}{2}}}$

If $t\geq 1- \frac{1}{e^{2}}$
$$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x= \int_{0}^{1- \frac{1}{e^{2}}}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x+$$
$$+ \int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{1- \frac{1}{e^{2}}}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x+$$
$$\int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x= \int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x$$
$$+ \frac{1}{2}\sqrt{b}\ln b+ \left ( 1- e^{-2} \right )^{-\frac{1}{3}}+ \left ( 1- e^{-2} \right )^{-\frac{2}{3}}- \frac{1}{2}\sqrt{b+ 1- e^{-2}}\ln\left ( b+ 1- e^{-2} \right )$$
Let $f\left ( x \right )= \frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}},$ since $f\left ( x \right )\leq 0\,{\rm as}\,36.29678351< x< 1161.52586,$
$$f\left ( x \right )\geq 0\,{\rm as}\,1- e^{-2}< x< 36.29678351, 1161.52586\leq x$$
$$\Rightarrow H= \frac{1}{2}\sqrt{b}\ln b+ \left ( 1- e^{-2} \right )^{-\frac{1}{3}}+ \left ( 1- e^{-2} \right )^{-\frac{2}{3}}- \frac{1}{2}\sqrt{b+ 1- e^{-2}}\ln\left ( b+ 1- e^{-2} \right )+$$
$$+ \int_{1- \frac{1}{e^{2}}}^{1161.52586}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x\geq 1.431081083- 1.123709381> 0$$
If $t\leq 1- e^{-2}$
$$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{t}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{1}{2}{\rm d}x\geq 0$$
Case 2. $b+ t\leq\frac{1}{e^{2}}$
$$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}+ \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}+ \frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x\geq$$
$$\geq -\frac{1}{2}\sqrt{t}\ln t+ t^{\frac{1}{3}}+ t^{\frac{2}{3}}\geq 0$$

Problem 2. Given two real numbers $x, y.$ Prove that
$$\left | \sin x- \sin y \right |\leq\left | x- y \right |$$

I'm having trouble using the way that we used in problem 1, I don't know how to devide it into some cases. On the other hand does there exist a nicer solution for problem 1 ; and another problem(s) as same as these ones as above, thank you.

ContentLicense: CC BY-SA 4.0

Body:
Alternative solution:

First we give some auxiliary results (Facts 1-2). The proofs are given at the end.

Fact 1: Let $t > 1$ and $u > 0$. Then
$$(t^2-1)^{1/3}u^{-1} + (t^2 - 1)^{2/3}u - t\ln t - 3(t-1)\ln u \ge 0.$$

Fact 2: Let $t > 1$ and $u > 0$. Then
$$(t^2-1)^{1/3}u^{-1} + (t^2 - 1)^{2/3}u + t\ln t + 3(t-1)\ln u \ge 0.$$

Now, WLOG, assume $x > y$.

Let $t = \frac{x}{y}$. Then $t > 1$. The desired inequality is written as
$$|t\ln t + (t-1)\ln y | \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}. \tag{1}$$
It suffices to prove that
$$t\ln t + (t-1)\ln y  \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}, \tag{2}$$
and
$$- t\ln t - (t-1)\ln y  \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}. \tag{3}$$
By Facts 1-2, (2) and (3) are both true.

We are done.

$\phantom{2}$

Proof of Fact 1: Denote LHS by $f(u)$. We have
\begin{align}
f'(u) &= - (t^2-1)^{1/3}u^{-2} + (t^2 - 1)^{2/3} - 3(t-1)u^{-1}\\
&= \frac{1}{u^2}(u - u_0)\Big[(t^2-1)^{2/3}u + (t^2-1)^{1/3} u_0^{-1}\Big]
\end{align}
where $u_0 = \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2(t^2 - 1)^{2/3}}$.

Thus $f'(u_0) = 0$, $f'(u) < 0$ on $(0, u_0)$, and $f'(u) > 0$ on $(u_0, \infty)$.
Thus, $f(u) \ge f(u_0)$ for all $u > 0$.

It suffices to prove $f(u_0) \ge 0$ that is
\begin{align}
&\frac{2(t^2 - 1)}{3(t-1) + \sqrt{(t - 1)(13t - 5)}} + \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2}\\
 &\quad - t\ln t - 3(t-1)\ln \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2(t^2 - 1)^{2/3}} \ge 0.
\end{align}
The proof is not difficult (Hint: Let $\frac{13t-5}{t-1} = w^2$.). We are done.

$\phantom{2}$

Proof of Fact 2: Denote LHS by $g(u)$. We have
\begin{align}
g'(u) &= - (t^2-1)^{1/3}u^{-2} + (t^2 - 1)^{2/3} + 3(t-1)u^{-1}\\
&= \frac{1}{u^2}(u - u_1)\Big[(t^2 - 1)^{2/3} u + \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2}\Big]
\end{align}
where $u_1 = \frac{2(t^2 - 1)^{1/3}}{3(t-1) + \sqrt{(t - 1)(13t - 5)}}$.

Thus $g'(u_1) = 0$, $g'(u) < 0$ on $(0, u_1)$, and $g'(u) > 0$ on $(u_1, \infty)$.
Thus, $g(u) \ge g(u_1)$ for all $u > 0$.

It suffices to prove $g(u_1) \ge 0$ that is
\begin{align}
& \frac{3(t-1) + \sqrt{(t-1)(13t-5)}}{2} + \frac{2(t^2 - 1)}{3(t-1) + \sqrt{(t-1)(13t-5)}}\\
&\quad + t\ln t + 3(t-1)\ln \frac{2(t^2 - 1)^{1/3}}{3(t-1) + \sqrt{(t - 1)(13t - 5)}} \ge 0.
\end{align}
The proof is not difficult (Hint: Let $\frac{13t-5}{t-1} = w^2$.). We are done.


问:
Discrete Math Quick Regular Expression question!
For a regular expression (a | b)*, can I write this string: aaaaaaabbbbbbbbbbbabababbb. I was thinking this means that when I choose a, I can have multiple a's and then I can choose b too after choosing multiple a's. Or is it that once I chose a, I cannot go back and choose b after finishing with a?</p>

答:
$(a|b)*$ means $a$ or $b$ zero or more times. You can only pick $a$ or $b$ at any given iteration

so $bbbbb, aaaaa, ab,$ null, $abaab$ are all correct.

ContentLicense: CC BY-SA 3.0


问:
Finding the fifth roots of unity using DeMoivre's theorem. I'm getting $-1$ as a root twice.
$$X^5 +1 = 0$$
I need to find all the five fifth roots of unity.

My method :

$$\begin{align}X &=(-1)^{1/5} \\
&= \left(\;\cos(2\pi k+\pi)+\sin(2\pi k+\pi)\;\right)^{\;1/5}
\end{align}$$
where $k = 0,1,2,3,4$. Then by using De Moivre's Theorem, I further solved it.

But I'm getting stuck at $k=2$, as I'm again getting $X=-1$ at both $k=2$ and $k=0$.

I am not able to understand as where I'm going wrong. If anyone could provide me with a detailed answer it would be very helpful for me

答:
As you suggest, by de Moivre the roots are
$$z_k =\cos \frac{\pi + 2k\pi}{5} + i\sin \frac{\pi + 2k\pi}{5}$$

$z_0 = \cos\dfrac{\pi}{5} + i\sin\dfrac{\pi}{5}\neq -1$

$z_1 = \cos\dfrac{3\pi}{5} + i\sin\dfrac{3\pi}{5}$

$z_2 = \cos\dfrac{5\pi}{5} + i\sin\dfrac{5\pi}{5} = -1$

etc.; in particular, $z_0\neq z_2$. Perhaps you forgot to divide by $5$?


Title: Finding the fifth roots of unity using DeMoivre's theorem. I'm getting $-1$ as a root twice.

Body:
$$X^5 +1 = 0$$
I need to find all the five fifth roots of unity.

My method :

$$\begin{align}X &=(-1)^{1/5} \\
&= \left(\;\cos(2\pi k+\pi)+\sin(2\pi k+\pi)\;\right)^{\;1/5}
\end{align}$$
where $k = 0,1,2,3,4$. Then by using De Moivre's Theorem, I further solved it.

But I'm getting stuck at $k=2$, as I'm again getting $X=-1$ at both $k=2$ and $k=0$.

I am not able to understand as where I'm going wrong. If anyone could provide me with a detailed answer it would be very helpful for me

Body:
$$X^5 = -1 = e^{(2k+1)\pi i}$$

$$X = e^{\frac{(2k+1)}{5}\pi i}, \quad (k=0,1,...,4)$$

$$X \in \{ e^{\pi i /5}, e^{3\pi i /5}, e^{5\pi i /5}, e^{7\pi i /5}, e^{9\pi i /5} \}$$

As you wrote, apply De Moivre's theorem.


问:
Prove that the angle given by the rotation around the cross product axis is the argmin of all possible rotations
As we know, for any vector $v$ that we want to rotate, the rotated vector $v_{rot}$ about some axis $u$ is equivalent to the sum of the parallel component of $v$ and $u$ plus the planarly-rotated perpendicular component of $v$, ala Rodriguez formula: https://people.eecs.berkeley.edu/~ug/slide/pipeline/assignments/as5/rotation.html

$v_{rot} = v_{||} + v_{\perp rot}$

Given two initial specified vectors $v$ and $v_{rot}$, we can always generate some axis $u$ by taking the cross product $\text{cross} (v, v_{rot})$ where $v$ is the initial vector.

Let's call this $u_{cross}$. $u_{cross}$ is always associated with a specific angle, $\theta_{cross}$, which can be calculated using this: https://stackoverflow.com/questions/5188561/signed-angle-between-two-3d-vectors-with-same-origin-within-the-same-plane

How do we prove that $\theta_{cross}$ is the smallest rotation angle across all $u$?

答:
The shortest possible rotation angle will have no component or rotation in the direction of either vector, because any rotation about $v$ or $v_{rot}$ results in the same vector $v$ or $v_{rot}$.  Therefore, if given a rotation with some component in the direction of $v$ or $v_{rot}$, then there will be a shorter one, without this component.

Between $v$ and $v_{rot}$, the only vector that has this property is the one orthogonal to both $v$ and $v_{rot}$, and is given by the cross product. (I guess its opposite is also orthogonal, so you could use that one too, but the rotation angle about this vector has the same magnitude even if it is negative).

There is the interesting case is when $v = \pm v_{rot}$, where there are an infinite number of "shortest axes of rotation" all in the plane orthogonal to $v$ since there are an infinite number of orthogonal vectors.


问:
Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$
I was solving a previous exam paper and there I got stuck on the following problem:

Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which

$|f(z)|=1$,

$|f(z)|\le 1$.

Any idea about how to tackle this problem will be highly appreciated. Thanks for your time.

答:
Let $z=r e^{i\theta}$, where $r\ge 0$ and $\theta\in \mathbb{R}$.
Then
$$\begin{eqnarray*}
|f(r,\theta)|
&=& \exp\left(\frac{r^2-1}{r^2-2r\cos\theta+1}\right) \\
&=& \exp\left(\frac{r^2-1}{(r-1)^2+2r(1-\cos\theta)}\right).
\end{eqnarray*}$$

I. If $\cos\theta = 1$, $|f| = \exp\left(\frac{r+1}{r-1}\right)$.
In this case $|f|\ne 1$ for any $r$.
However, $|f|<1$ for $r<1$.

II. If $\cos\theta\ne 1$, $(r-1)^2+2r(1-\cos\theta) > 0$.
Therefore, $|f| = 1$ for $r=1$ and $|f|<1$ for $r<1$.

We find, $|f| = 1$ for all
$z = e^{i\theta}$,
where $\theta \ne 2\pi n$ and $n\in\mathbb{Z}$.
This is the unit circle with $(1,0)$ deleted.

In addition, $|f|\le 1$ for all
$z = r e^{i\theta}$
such that
(1) $r=1$ and $\theta\ne 2\pi n$ or
(2) $r<1$ and $\theta\in\mathbb{R}$.
This is the unit circle and its interior with $(1,0)$ deleted.

The results agree with those of @JonClaus, who used Cartesian coordinates.

Here is a three dimensional plot of $|f|$ above the complex plane.


问:
Distribution of multiplied variable into min() function
Is distributing a variable over a $min$ call ($i \times min({1  \over i}, {1 \over j}) = min({i \over i}, {i \over j})$ correct? Intuitively it is, but I can find no difinitive answer.

PS - if anyone is curious/kind enough to read the context, I include it as a comment.

答:
Yes, it is correct.

This is due to $i$ is non-negative.

If $b \le c$ and $a \ge 0$, then we have $ab \le ac$.


问:
Insensitivity of convergence in distribution
I'm proving:

If $X_n \xrightarrow{d} X$ and $\mathop{{}\mathbb{P}}\{X_n \neq Y_n\}\to 0$, then $Y_n \xrightarrow{d} X$.

My idea is to use almost sure representation of convergence in distribution, that is

$X_n \xrightarrow{d} X$ iff their corresponding quantile $q_n \to q$ almost surely.

where we treat $q$ as random variable in $((0,1),\mathcal{B}_{(0,1)},\lambda)$, where $\lambda$ is Lebesgue measure.

But I still have no idea of whether $\lambda\{q_n \neq q_n'\}\to 0$, where $q'_n$ is corresponding quantile of $Y_n$

答:
$P(X_n \leq x)\leq P(X_n \neq Y_n)+P(Y_n \leq x)$ and

$P(Y_n \leq x)\leq P(X_n \neq Y_n)+P(X_n \leq x)$. The result follows from this.

Some details: Using the second inequality we get $\lim \sup P(Y_n \leq x) \leq \lim \sup P(X_n \leq x)=P(X\leq x)$ at continuity points of $F_X$. Similarly, the first inequality gives $\lim \inf P(Y_n \leq x) \geq \lim \inf P(X_n \leq x) =P(X\leq x)$ at continuity  points of $F_X$. So $P(Y_n \leq x) \to P(X\leq x)$ for $x$ in a dense set which implies that $Y_n \to X$ is distribution.


问:
Find the plane containing line $\frac{x-3}{2} = \frac{y+2}{2} =\frac{z-1}{3}$ and also contains its projection on the plane $2x+3y-z=5$
I am not really able to understand the question. I didn’t get the part ‘contains the projection of the line on the plane....’

How can a plane contain the projection of a line on another plane? Wouldn’t that just mean that required plane is coincident with $2x+3y-z=5$?

I know there is already such a question on MSE but I am not convinced with that answer.

答:
The first vector you provide, $2i+2j+3k$, is the direction of the line $\frac{x-3}{2}=\frac{y+2}{2}=\frac{z-1}{3}$. Surely the plane containing the line should contain this.

The second vector is the "direction" of the plane $2x+3y-z=5$, which is perpendicular to it. And if you can draw a picture on the draft, you know that the plane we want should contain this, since the projected line, the original line, and the perpendicular vector $2i+3j-k$ should be on the same plane.

Finally you need the plane to contain $(3,-2,1)$ which is on the original line. One point and two different vectors is now sufficient to determine a plane--

$a(x-3)+b(y+2)+c(z-1)=0$ is what you want. Here $(a,b,c)\perp(2,2,3)$ and $(a,b,c)\perp(2,3,-1)$ as we discussed above. You can then either solve the equation directly or use outer product.


问:
Description of all morphism of $f: \Bbb A^1_k\to E$

let $\lambda\in k-\{ 0,1 \}$and consider the affine curve $V( Y^2-X(X-1)(X-\lambda)\subset \Bbb A^2_k )$,give a description of all morphisms $f: \Bbb A^1_k\to E$

i am stack after $\lambda_1\to Y^2-X(X-1)(X-\lambda_1) $,

$\lambda_2\to Y^2-X(X-1)(X-\lambda_2) $

$\lambda_1+\lambda_2\to 2Y^2-X(X-1)(X-\lambda_1-\lambda_2) $

and i have no idea what to do next. And i don't know whether i am correct or not.Can anyone help me?

答: Giving a morphism $f:\mathbb A^1\to E$ is equivalent to giving a homomorphism of coordinate algebras in inverse direction $f^\sharp:k[X,Y]/(Y^2-X(X-1)(X-\lambda))\to k[T]$. Suppose that $X$ maps to $f$ and $Y$ maps to $g$, then $g^2=f(f-1)(f-\lambda)$. The ring $k[T]$ is factorial and polynomials $f,f-1,f-\lambda$ have no common irreducible factors, hence they all should be squares. Let $f=a^2$ and $f-1=b^2$, then $1=a^2-b^2=(a-b)(a+b)$. So $a-b$ and $a+b$ are constants, hence $f$ and $g$ are constants. So, the morphisms that you want to describe are given by maps of all $\mathbb A^1$ to a point of $E$.


问:
Questions Regarding Linear Regression
Are the slope and intercept of a simple linear regression model always normally distributed?  Is there ever a difference between the distribution of the estimated slope and intercept and the actual ones?  I have only just begun learning about the subject but I am still not clear on the details.  A final question: is the least squares method the same as linear regression in that it gives information like the $R^2$?  Thanks!</p>

答:
Are the slope and intercept of a simple linear regression model always normally distributed?

They are if the residuals are independent and normally distributed with common variance. More precise:
The linear model is given by $\vec{y} = X\vec{\beta} + \vec{\varepsilon}$, where the design matrix $X$ is $n\times(k+1)$ with $\mathrm{rk}(X)=k+1 <n$ and $\vec{\beta}$ is a vector which contains the intercept and the slopes. Additionally we assume that $\vec\varepsilon \sim N(0, \sigma^2 I)$, that is, we assume that $\vec{y} \sim \mathrm{N}_n(X\vec{\beta}, \sigma^2 I)$.
Then we have $\hat{\vec{\beta}}_{ML} \sim \mathrm{N}_{k+1}(\vec{\beta}, \sigma^2(X^T X)^{-1})$, i.e. the maximum likelihood estimator of the intercept and the slopes is multivariate normal.

Is there ever a difference between the distribution of the estimated slope and intercept and the actual ones?

The actual ones are not random, so I don't understand your question here.

Is the least squares method the same as linear regression?

The least squares method is a method that we apply to estimate the parameters, like the intercept and the slope. Quote from http://en.wikipedia.org/wiki/Linear_regression: Linear regression models are often fitted using the least squares approach, but they may also be fitted in other ways, such as by minimizing the "lack of fit" in some other norm (as with least absolute deviations regression), or by minimizing a penalized version of the least squares loss function as in ridge regression. Conversely, the least squares approach can be used to fit models that are not linear models. Thus, although the terms "least squares" and "linear model" are closely linked, they are not synonymous.

Title: Questions Regarding Linear Regression


问:
MLE for NB (Negative Binomial) point process.
I have a bunch of machines. These machines fail from time to time, stay down for a while, recover and then commence running.

First, I wanted to model the failures as a Poisson point process. This assumes the machines are homogeneous in terms of their failure rates. In this model, the estimation becomes quite easy. You count the total failures across all the machines, $n$ and the total time the machines collectively ran, $t$. Then, the maximum likelihood estimate for the failure rate of the machines becomes:

$$\lambda = \frac{n}{t}$$

So far so good. But now I want to get rid of the assumption that the machines are homogeneous. So, there is some distribution from which their failure rates are drawn. A Gamma distribution is a natural choice for this distribution since its the conjugate prior. And this is covered in section 5.4.3 of the book, Introduction to probability models by Ross (10th ed). The overall point process becomes NBD or the negative binomial (the number of events in any interval of time is negative binomial instead of Poisson). Now, how do I take my data and fit the parameters of this Negative binomial point process?

I assume getting the total failures and total machine run time per machine will probably serve as a pre-processing step (just as the total failures and total machine time across all machines were used in the Poisson case). Not sure how to make progress from there.

Including relevant equations from the book:

The gamma distribution from which the rate, $\lambda$ of the Poisson is drawn has a PDF:

$$g(\lambda) = \theta e^{-\theta \lambda} \frac{(\theta \lambda)^{m-1}}{(m-1)!}$$

And this results in the PMF of the number of events:

$$P(N(t)=n) = {n+m-1 \choose n}\left(\frac{\theta}{t+\theta}\right)^{m} \left(\frac{t}{t+\theta}\right)^n$$

答: If for a machine you observe over a span of time k failures, the number of failures in a one-hour span can be modeled by $X\sim \mathscr {NegBinom}\left(t, \frac{t}{k+t}\right)$. As $t\rightarrow \infty$, $X$ will approach the distribution $\mathscr {Pois}\left(k\right)$. For example, if in a one hour span of time you observe $k=5$ failures from a particular machine, you can create the random variable $X\sim \mathscr{NB}\left(3600, \frac{3600}{3605}\right)$ to represent the number of failures observed in a one-hour period. For each of the machines, you will have a negative binomial random with a potentially different parameter of $p=\frac{3600}{k+3600}$, so you will have to figure out what the sum of negative binomial random variables with different probability of success turns out to be.


问:
Recap of Riemann's original proof of the dimension of Moduli space $M_g$
I am trying to understand some single steps in this
recap  of Bernhard Riemann's original proof by Alexandre Eremenko
that the moduli space $M_g$
of compact Riemann surfaces with genus $g \ge 2$ has dimension
$3g-3$ (or in original context the $3g-3$ 'degrees of freedom' are
what Riemann called 'moduli')

Riemann combines what is called Riemann-Roch and Riemann-Hurwitz nowadays.
He considers the dimension of the space of holomorphic maps of degree $d$
from the Riemann surface of genus $g$ to the sphere. He computes this dimension
in two ways. By Riemann-Roch this dimension is
$2d-g+1$, for a fixed Riemann surface. (Indeed, Riemann-Roch says that
the dimension of the space of such functions with $d$ poles fixed
is $d-g+1$ (when $d\geq 2g-1$ which we may assume), but these poles can
be moved, so one has to add $d$ parameters).

On the other hand, such a function has $2(d+g-1)$ critical points by
Riemann-Hurwitz. Generically, the critical values are distinct, and can
be arbitrarily assigned, and this gives the dimension of the set of
all such maps on all Riemann surfaces
of genus $g$.

So the space of all Riemann surfaces of genus $g$ must be of dimension
$$2(d+g-1)-(2d-g+1)=3g-3.$$

There are essentially two things I struggle with.

Point 1: Why the dimension of the space of holomorphic maps
of degree $d$ from the Riemann surface $S$ of genus $g$ to the sphere
equals $2d-g+1$?

As far as can follow we are going to apply RR. Let $h$ any
holomorphic map $S \to \mathbb{P}^1$ of degree $d$ which we regard as
meromorphic function $h$ which behaves compatible with degree $d$ in
it's poles and zeros, set $(h) :=D$ and apply RR to it.
RR tells $l(D)= d -g+1$ where $l(D)$ is the $C$-dimension of
meromorphic functions $f$ on $S$
such that all the coefficients of the divisors $(f) + D$ are non-negative.

Eremenko argues then that the additional $d$ dimensions
we obtain because 'the poles of can be moved, so one has to
add $d$ parameters.'

I not understand why moving of the $d$ poles gives additionally exaxt
$d$ parameters.

Point 2: Why the difference

$$ 2(d+g-1)-(2d-g+1) $$

between the number of critical points of these moromorphic functions
(that's Riemann-Hurwitz-Thm) and the counted dimension of
the space of holomorphic maps of degree $d$ from point 1 gives exactly
the dimension of the moduli?

答:
To specify a degree $d$ map $f: X \to \mathbb{P}^1$, one first chooses an effective divisor $D$ of degree $d$ to specify the poles of $f$. Since $D$ has support consisting of $d$ points, this amounts to $d$ choices. So this by this "the poles can be moved" comment, he means that choosing a different divisor $D'$ will result in a different map with poles given by the support of $D'$ instead. Then one computes $\ell(D)$ as you have described above to get the dimension of the space of degree $d$ maps $f: X \to \mathbb{P}^1$ with $\renewcommand{\div}{\operatorname{div}} \div_\infty(f) = D$.

We want to compute the dimension of $\mathcal{M}_g$, the moduli space of curves of genus $g$, but what we've really done is computed the dimension of the moduli space of pairs $(X,f)$ where $X$ is a curve of genus $g$ and $f$ is a degree $d$ map $X \to \mathbb{P}^1$. This related moduli space is called a Hurwitz space, denoted $\DeclareMathOperator{\Hur}{Hur} \Hur_{d,g}$. In order to remove this choice of a map $f: X \to \mathbb{P}^1$, we subtract off the dimension of the space of degree $d$ maps $X \to \mathbb{P}^1$ which was computed above.

You can interpret this using the projection
\begin{align*}
\pi: \Hur_{d,g} &\to \mathcal{M}_g\\
(X,f) &\mapsto X \, .
\end{align*}
We've computed the dimension of $\Hur_{d,g}$ and, for a given $X$, the dimension of the fiber $\pi^{-1}(X)$. Assuming $\pi$ is sufficiently "nice", we should have
\begin{align*}
\dim(\mathcal{M}_g) = \dim(\Hur_{d,g}) - \dim(\pi^{-1}(X)) = 2d + 2g - 2 - (2d - g + 1) = 3g-3 \, ,
\end{align*}
as desired.

This is covered in greater detail in chapter 2, section 3 (p. 255) of Principles Of Algebraic Geometry by Griffiths and Harris. Jarod Alper also recently covered this in a lecture for a course he's teaching, which you can watch here. (See here for the dimension calculation you've asked about.)


问:
Maximizing a combinatorial identity
So here is my problem. If we have a vector $\textbf{x}=(x_1,...,x_n)$ where $x_j \in \mathbb{N}$ for each $j \in \{1,...,n\}$, then is there a way to maximize the value of the following combinatorial identity:

$\sum_{j=1}^{n}\binom{2x_j}{2}$?

The problem here is that we do not know anything about the vector $\textbf{x}$ except for the fact that the entires are all natural numbers and that the constraint is $\sum_{j=1}^{n}x_{j}=k$ where $k\in \mathbb{N}$ is fixed.

答: You want one component of $\bf x$ to have value $k$ and all the rest to be $0$.  Because the term ${2x_j \choose 2}=x_j(2x_j-1)$ you will increase the sum every time you move $1$ to the largest component from any other component.  This stops when one component equals $k$.

Title: Maximizing a combinatorial identity


问:
I know how to check that the dot product is a scalar. But how do I check that that a function like temperature is a scalar?
I want temperature to be the same at the same point in $(x,y)$ and polar $(r,\theta)$. But how is that enforced that mathematically?

答: If you have a function that calculates the temperature at a point $(x, y)$, you can also calculate the temperature at the point in polar coordinates $(r, \theta)$ by using the conversion $(r\cos \theta, r\sin\theta)$. Then you can check what the temperature is at $(r\cos \theta,r\sin\theta)$ and see if it is the same as the temperature at point $(x,y)$. If it isn't, you can set the temperature at the point $(r\cos\theta, r\sin\theta)$ to be the temperature at $(x, y)$. Or you can try to find a new point $(r\cos\theta, r\sin\theta)$ that has the same temperature as at $(x, y)$, depending on what you mean.


问:
Application of closed operator theory.
Use the closed operator theory to show the following result:

Let $(f_n)$ be a sequence in $C^1[0,1]$ such that the function series $\sum_{n=1}^{\infty} f_n$ converges uniformly for a function $f$ in the interval $[0,1].$ If the series of functions $\sum_{n=1}^{\infty} f'_n$ converges uniformly in the interval $[0,1],$ show that $f \in C^1[0,1]$ and that
$$f'(t)=\sum_{n=1}^{\infty} f_{n}'(t), \quad \forall t \in[0,1].$$

Can I use the following Lemma to solve this exercise? If so, how do I define T?

Lemma (Closed operator). (Kreyszig's book) Let $T: X \to Y$ be a bounded linear operator with domain $D(T) \subset X,$ where X and Y are normed spaces. If $D(T)$ is a closed subset of $X,$ then $T$ is closed.

答:
I suppose you are using the usual norn $\|f\|_{\infty}+\|f'\|_{\infty}$ on $C^{1}[0,1]$.

Hint:Use the sequence $( \sum\limits_{k=1}^{n} f_k,\sum\limits_{k=1}^{n} f_k')$ in the graph of $T$. The graph is closed, by the Lemma.

Title: Application of closed operator theory.


问:
How to compare two rotations represented by axis-angle rotation vectors?
I currently have two rotation vectors r1 and r2 using axis-angle representations, i.e. they are 3d vectors, their norms are the rotation angles and the normalized unit vectors are rotation axes respectively.

I am confused how to compare r1 and r2. In an optimization process of a regression system, if r1 is the ground truth, and r2 is the current result produced by the system, how do I define a proper loss to encourage the system to converge to r1? Can I just use L1/L2 loss between r1 and r2?

Thanks!

答:
These vectors are actually both the Lie Algebra associated with $SO(3)$, and this is actually a vector space, so assuming that your coordinate frames line up, you can totally use something like $L_2$ loss between between them as a cost function.

The trick here is knowing what the coordinate frames are.

To convert from the Lie Algebra to the rotation matrix, we can use the matrix exponential ($SO(3)$ has a shortcut, where you can just use the rodriguez formula), but in general it's

$R = \exp\left(\lfloor r \rfloor_\times\right)$

where $\lfloor r \rfloor_\times$ is the skew-symmetric matrix formed from $r$.

Disclaimer: There are various conventions here that can be super confusing, and my conventions may not agree with yours, so you may need to play around with my equations below to make your particular implementation work.

Assuming that you define $R_1 = \exp\left(\lfloor r_1 \rfloor_\times\right)$ and $R_2 = \exp\left(\lfloor r_2 \rfloor_\times\right)$.  If $R_1$ and $R_2$ are defined in the same frame  (say, they are both "world to whatever"), then the $L_2$ norm will give you proper convergence since both $r_1$ and $r_2$ will be defined in the same coordinate frame.  If, instead they are something like "estimate to world" and "truth to world", then you need to do $J = \lVert R_1^\top r_1 - R_2^\top r_2\rVert$ because you'll want to do the subtraction in a common frame.  If for some crazy reason you have a mix of the two, then make the appropriate coordinate frame swaps.

One additional trick is that you probably want to make sure that the norm of these vectors has any extra $2\pi$ factors removed, but otherwise I do this kind of thing all the time and it totally works.  In my experience, I usually don't have axis-angle parameterizations.  Instead I have the unit quaternion or rotation matrix, so I tend to use the following formula as a cost function in bundle-adjustment or other optimization problems over rotation manifolds.

$$
J = \lVert\log(R_1 R_2^\top)\rVert
$$

where

$$
\log(R) = \frac{\theta}{2 \sin \theta}(R-R^\top)
$$

and the rotation angle

$$
\theta = \arccos\left(\frac{\mathrm{tr}(R) - 1}{2}\right).
$$


问:
What is is it meant by "analytically derive the expected value of an estimator?" in statistics?
Good evening, everyone. I am new to statistics, so I wanted to see if I could get guidance with this problem:

Consider a variable Y which follows a distribution with mean µ and variance σ2 in the population.
Suppose we take a random sample of Y of size n, and propose to estimate the mean of Y from this
dataset using two potential new estimators, the Addition Estimator and the First-ten Estimator, which
are defined as follows:

The Addition Estimator: Take the sample mean, and add on $\frac{20}{n}$

The First-ten Estimator: Select only the first ten observations in the sample, and take the sample
mean of these observations

For the sake of simplicity, we will assume that we always take samples containing more than ten obser-
vations. Remember that the sample mean can be calculated in the following manner:  $µ = ∑_n \frac{Yn}{n} $

Analytically derive the expected value of both estimators. Based on these results, is each estimator
biased or unbiased? Using your knowledge of limits, intuitively explain what you think will happen
to the bias (that is, the degree to which E[θ] deviates from θ) as n approaches infinity

I am not understanding what it means to "analitically derive de expected value" if someone could explain me what it means and how to do it, I would be very thankful.

Thank you again!

答: Analytically derive the expected value of the estimator means to take the expected value. Here is the addition estimator: $\hat a=\frac{\sum_{i=1}^n Y_i}{n}-\frac{20}{n}$ and here is the first-ten estimator $\hat b=\frac{\sum_{i=1}^{10} Y_i}{10}$. The question wants you to apply the operation $\mathbb E$ to the addition and first-ten estimators. What is $\mathbb E(\hat a)$ and $\mathbb E (\hat b)$? For the first one, it is $\mathbb E\left(\frac{\sum_{i=1}^n Y_i}{n}-\frac{20}{n}\right)=\frac{\sum_{i=1}^n \mathbb E(Y_i)}{n}-\frac{20}{n}=\frac{n\mu}{n}-\frac{20}{n}=\mu-\frac{20}{n}$. This is biased because we do not expect in the long run, if we take the addition estimators for a lot of samples, that the average estimate will be $\mu$. It will be smaller than $\mu$. But the bias will decrease as $n\rightarrow \infty$ because as n gets larger, 20/n gets smaller. If we take samples of size 10, there will be a larger bias $\mu-\left(\mu-\frac{20}{10}\right)=2$ than if we take samples of size 1000, with bias of $\frac{20}{1000}=.2$. As n goes to infinity, the bias will go to zero. Now try to answer these questions for the first-ten estimator. (1) What does $\mathbb E(\hat b)$ simplify to? (2) Is the estimator biased? (Is the expected value $\mu$?) (3) Does the bias change as $n\rightarrow \infty$, or does it stay the same?


问:
Solve Limit Tend To $\pi$, From Length Of Circle
Computing the length of a semicircle of radius 1. It's divided in $2^x$ equal arcs, I find the following formula for the total length of the chords created:

$$L=2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

When $x \to \infty$ we see, geometrically, that this expression tends to be  the length of the circle:

$$2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

My goal is to define $\pi$ as the value of that expression when $x \to \infty$ (the limit). But how can I rigorously prove that this limit exists and that the expression tends to be the length of the circle? In other words, how can I write rigorously the geometric ideas?

Note: The number  $\frac{45}{2^{x-1}}$ is in degrees.

答:
Converted to radians, we have
$$L_x=2^{x+1} \sin \left(\frac{\pi }{2^{x+1}}\right)$$ Use the double inequality
$$t-\frac{t^3}{6} \leq \sin(t) \leq t$$ which makes
$$\pi -\frac{\pi ^3}{24 \times 4^x}\leq L_x \leq \pi$$ So, if you need $k$ significant figures
$$x_{(k)}= -\frac{\log \left(\frac{3\ 10^{-k}}{\pi ^2}\right)}{2 \log (2)}-\frac{3}{2}$$ For $k=10$, then $x=16$ and computing
$$L_{16}=131072 \sin \left(\frac{\pi }{131072}\right)=\color{red}{3.141592653}29$$


问:
Show that $f$ is not continuous at (0,0) and is bounded on $\mathbb{R}^2$ but the restritions to straight line in $\mathbb{r}^2$ is cont.

if $f(x,y)=\frac{xy^2}{x^2+y^4}$ and $f(0,0)=0$ show that $f$ is not continuous at (0,0) and is bounded on $\mathbb{R}^2$ but the restritions to straight line in $\mathbb{r}^2$ is cont.

My attempt:
By changing the numerator and denominator a bit we get the function of the form $f(x,y)=\frac{z}{z^2+1}$ where $z=\frac{x}{y^2}$ and then I take two cases where $\lim_{z \to 0}\frac{z}{z^2+1} \to 0$ and  $\lim_{z \to \infty}\frac{z}{z^2+1} \to 1$ this is how I have shown that the function is bounded.

Now the next thing that I have to show is that the function is discontinuous at $(0,0)$ so we consider the line $y=x$ from here I have done that $f(x,y=x)=\frac{x^3}{x^2+x^4}$.Now,if  $lim_{x \to 0}\frac{x^3}{x^2+x^4} \to 0$ .I have got no intition as to why it would not be continuous at$(0,0)$.To prove continuity at straightline I have taken $y=mx+c$ and I have used the same tactic.Any hint to rectify my mistake would be highly appreciated.

答:
It's clear that $f$ is continuous at all points of $\mathbb{R}^2$ except possibly at $(0,0)$.

It follows that the restriction of $f$ to any line not passing through $(0,0)$ is continuous.

Next consider the restriction of $f$ to a line $L$ passing through $(0,0)$.

If $L$ is the $y$-axis, then $x=0$, so $f$ is identically $0$ on $L{\setminus}\{(0,0)\}$, hence, since we also have $f(0,0)=0$, it follows that the restriction of $f$ to $L$ is continuous.

If $L$ passes through $(0,0)$ but is not the $y$-axis, then $y=mx$ for some fixed constant $m\in\mathbb{R}$. To verify that $f$ restricted to $L$ is continuous, it suffices
to show that
$$
\lim_{x\to 0}\frac{x(mx)^2}{x^2+(mx)^4}=0
$$
You should have no trouble with that.

Finally, to verify that $f$ is not continuous at $(0,0)$, note that if $(x,y)$ approaches $(0,0)$ along the curve $x=y^2$, the limit reduces to
$$
\lim_{y\to 0}\frac{y^4}{2y^4}
$$
which is clearly equal to ${\large{\frac{1}{2}}}$, hence is not equal to $f(0,0)$.


问:
Finding the fifth roots of unity using DeMoivre's theorem. I'm getting $-1$ as a root twice.
$$X^5 +1 = 0$$
I need to find all the five fifth roots of unity.

My method :

$$\begin{align}X &=(-1)^{1/5} \\
&= \left(\;\cos(2\pi k+\pi)+\sin(2\pi k+\pi)\;\right)^{\;1/5}
\end{align}$$
where $k = 0,1,2,3,4$. Then by using De Moivre's Theorem, I further solved it.

But I'm getting stuck at $k=2$, as I'm again getting $X=-1$ at both $k=2$ and $k=0$.

I am not able to understand as where I'm going wrong. If anyone could provide me with a detailed answer it would be very helpful for me

答:
Abbreviation: $Cis(x)=\cos x +i\sin x.$ Not very common, but I like it.

"Unity" is $1.$ If $X$ is a $5$th root of unity then $X^5-1=0,$ not $X^5+1=0.$

If $t\in \Bbb R$ then $[Cis(t)=1 \iff t/2\pi\in\Bbb Z].$ Therefore if $t\in\Bbb R$ and $X=Cis(t),$ we have $$[ X^5=1] \iff [ Cis(5t)=1] \iff$$ $$\iff [5t/2\pi\in \Bbb Z] \iff$$ $$\iff [\exists n\in \Bbb Z\,(t=2\pi n/5)\,] \iff$$ $$\iff [X\in S=\{Cis(2\pi n/5): n\in \{0,1,2,3,4\}\,\}\,]$$  because if $n\in\Bbb Z$ and $n\not\in \{0,1,2,3,4\}$ then $Cis (2\pi n/5)\in S$ anyway.

So $S$ is a set of $5$  solutions to $X^5-1=0$. And $\Bbb C$ is a $field$. In a field, a polynomial equation in one variable, of degree $m,$ with $m\in\Bbb Z^+$, cannot have more than $m$ solutions. So $S$ is the set of $all$ the $5$th roots of unity.

You can also observe that if $X\in \Bbb C$ but  $X\ne Cis(t)$ for any $t\in\Bbb R$ then $\exists r,t\in \Bbb R\,(|r|\ne 1\land X=r\cdot  Cis(t)\,),$ but then $|X^5|=|r|^5\ne 1,$ so $X^5\ne 1.$


问:
There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.
There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.

Which measure of average would be the most meaningful – mean, median, mid-range, or mode? Do we at all need any measure of average?

can anyone please help me ?

答:
If you're actually running a class

The median would be most helpful to understand. Let's suppose you have homework assignments (30%), quizzes (5%), midterms (25%), and finals (40%) for students and you weighted those to come up with a score for each student $S$ between 0 and 100. Now you sort the scores and find the scores that are at the 225th and 226th. These are the medians. Or you could take the average of them which is common. That is the median score. It is the 50th percentile score. Anyone who scored above the median is in the top 50th percentile. Now do the same thing for the 80th percentile. You have 450 students: $450*.8=360$ so take the average of the 360th and 361th score to find the 80th percentile score. Anyone who scored above that score is in the top 80th percentile of the class.


问:
Induction Problem Help, writing equation
There are $n$ piles of marbles and that every pile has a different number of marbles. We want to prove that the sum of the number of marbles in all the piles is greater than or equal to $\frac{n(n + 1)}{2}$, without knowing anything beyond the fact that each pile has a different number of marbles. Use induction to prove this fact for any positive integer $n$.

I'm having trouble writing the initial equation to solve and getting started. $\sum_{i=1}^{n}$ $\geq \frac{n(n + 1)}{2}$ is what I have so far where the base case is n = 1, but should I say have a variable for the number of piles so that it's written like $\sum_{i=1}^{p}~$ p $~\geq \frac{n(n + 1)}{2}$ where p is the sum of the piles and then the base case would be p = 1?

答:
Base case : (Assuming that each pile has non zero number of marbles) Clearly the claim is true for $ 1 $ pile .

Induction Hypothesis : Suppose that the sum of number of marbles in each pile is greater than $ \frac{n(n+1)}{2} $ for some arbitrary $n$ piles.

Induction step : Now suppose that there are $n + 1$ piles. Obviously the last pile has to contain at least $ n + 1 $ marbles as the number of marbles in each pile is distinct. Let the number of marbles in last pile be $p$. We have that $ p \geq n+1$ . Hence $$ \frac{n(n+1)}{2} + p \geq \frac{n(n+1)}{2} + n + 1$$
As there are at least $\frac{n(n+1)}{2}$ marbles when there are $n$ piles from our hypothesis, we are done.


问:
How to get the density function of the inverse Burr distribution
I have to find the density function of the inverse Burr distribution with $Y = 1/X $ but by plugging it in the CDF of $F(y)$ and then deriving it to get $f(y)$, I don't get the supposed answer which is:

I'm confused since I usually do this method and it works. How can I find its inverse density function? Thanks in advance for your help!

答:
Because $Y = g(X) = 1/X$ is a monotone transformation, we use the formula
$$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|,$$ where $$f_X(x) = \frac{\alpha \gamma (x/\theta)^\gamma}{x(1+(x/\theta)^\gamma)^{\alpha+1}}.$$  This yields $$\begin{align}
f_Y(y) &= f_X(1/y) \left|\frac{d}{dy}\left[y^{-1}\right]\right| \\
&= \frac{\alpha \gamma (y\theta)^{-\gamma}}{(1/y)(1 + (y \theta)^{-\gamma})^{\alpha + 1}}\cdot \frac{1}{y^2} \\
&= \frac{\alpha \gamma (y \theta)^{-\gamma}}{y(1 + (y\theta)^\gamma)^{\alpha+1} (y\theta)^{-\gamma(\alpha+1)}} \\
&= \frac{\alpha \gamma (y \theta)^{\alpha \gamma}}{y(1+(y\theta)^\gamma)^{\alpha+1}}. \tag{1}
\end{align}$$
Note that the chosen parametrization for the inverse Burr is a bit confusing because we have $y\theta$ instead of $y/\theta$.  In other words, the density
$$f_X(x) = \frac{\tau \gamma (x/\theta)^{\tau \gamma}}{x (1 + (x/\theta)^\gamma)^{\tau+1}} \tag{2}$$ is using a scale parametrization, whereas in the density we derived in $(1)$ is a rate parametrization.  This is perhaps the source of your confusion:  they are not the same $\theta$.


问:
How do I prove that $161616 \dots 16$ ($100$ digits) cannot be written as a perfect square?
I did:
$$
16 \cdot 101010 \dots 1 \quad (99 \text{ digits})
$$

It's clear that $16 = 2^4$, satisfying the condition that the prime has to have an even power. But what about $101010 \dots 1$?

Also, I tried $16 \pmod 4$ but it's congruent to $0$.

(???)

答: Hint: Consider the $161616\ldots 16$ ($100$ digits) number (or, since the $16$ factor is a perfect square, the $101010\ldots 1$ ($99$ digits) number) in modulo $9$, using that the sum of digits has the same remainder as the value when divided by $9$, to get that it's not one of the possible congruences of $\{0,1,4,7\}$ modulo $9$ of perfect squares.


问:
How to find general equation for surface of revolution?
If a curve (let's say y=f(x)) revolve around an axis (let's say x-axis), then it would form a surface of revolution.

So is there any way to find the general equation for that surface of revolution?

答:
Yes, there is. Suppose your function is defined on $[a,b]$ for some $0\le a<b$ and the graph rotates around the $y$-axis. Then the point $(x,y)$ on your graph rotates with radius $x$. A parametrization of your surface (adding a third axis $z$ perpendicular to $x$ and $y$) is then
$$
x=t\cos\theta,\quad y=f(t),\quad z=t\sin\theta
$$
where $t\in[a,b]$ and $\theta\in[0,2\pi)$.

You can easily adapt it for the case it rotates around the $x$ axis.


问:
Using $z = x + yi$ does not reveal all solutions
The question is to find all solutions in $\mathbb{C}$ of the equation: $$z^2 - (i+1)z + i = 0$$

After expanding with $z = x + yi$, I get $(x^2 - y^2+y-x) + (2xy -x-y+1)i = 0$.

Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.

Now, we plug that into $(2xy -x-y+1) = 0$ .

Since $x,y \in \mathbb{R}$,  this only gives the solution $x=  0, y = 1$.

However, there is another solution which is $z=1$. Why this solution does not appear with this method?

答:
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.

That is your mistake here.  $0=x^2-y^2+y-x=(x-y)(x+y-1)$ so $x=y$ or $x=1-y$, not $y-1$.

Then substituting into the second equation $2xy-x-y+1=0$ gives $(x,y)=(0,1)$ or $(1,0)$.


问:
Upper and lower integration inequality
I would like to learn how to prove that the following inequality holds.

Let $F$ be a bounded function on an interval $[a,b]$, so that there exists $B\geq 0$ such that $|f(x)| \leq B$ for every $x\in [a,b]$.

Show that
$[
U(f^2,P) -L(f^2,P) \leq 2B [ U(f,P) -L(f,P) ]
]$
for all partitions $P$ of $[a,b]$.

答:
Notation. For any finction $\varphi:[a,b]\to\mathbb{R}$ we denote
$$
M_{[\alpha,\beta]}(\varphi)=\sup\limits_{x\in[\alpha,\beta]}\varphi(x)\qquad
m_{[\alpha,\beta]}(\varphi)=\inf\limits_{x\in[\alpha,\beta]}\varphi(x)
$$
where $[\alpha,\beta]\subset[a,b]$. For a given partition $P=\{\Delta_1,\ldots,\Delta_n\}$ of $[a,b]$ we denote
$$
U(f,P)=\sum\limits_{i=1}^n M_{\Delta_i}(f)|\Delta_i|\qquad
L(f,P)=\sum\limits_{i=1}^n m_{\Delta_i}(f)|\Delta_i|
$$

Proposition. Let $f:[a,b]\to\mathbb{R}$ be a real valued bounded function such that $|f(x)|\leq B$ for all $x\in[a,b]$. Let $P$ be partition of $[a,b]$, then
  $$
U(f^2,P)-L(f^2,P)\leq 2B(U(f,P)-L(f,P))
$$

Proof For each $i\in\{1,\ldots,n\}$ we have
$$
\begin{align}
M_{\Delta_i}(f^2)-m_{\Delta_i}(f^2)
&=M_{\Delta_i}(|f|^2)-m_{\Delta_i}(|f|^2)\\
&=M_{\Delta_i}(|f|)^2-m_{\Delta_i}(|f|)^2\\
&=(M_{\Delta_i}(|f|)+m_{\Delta_i}(|f|))(M_{\Delta_i}(|f|)-m_{\Delta_i}(|f|))\\
&\leq 2B(M_{\Delta_i}(|f|)-m_{\Delta_i}(|f|))\\
&\leq 2B(M_{\Delta_i}(f)-m_{\Delta_i}(f))\\
\end{align}
$$
Hence
$$
\begin{align}
U(f^2,P)-L(f^2,P)
&=\sum\limits_{i=1}^n(M_{\Delta_i}(f^2)-m_{\Delta_i}(f^2))|\Delta_i|\\
&\leq\sum\limits_{i=1}^n 2B(M_{\Delta_i}(f)-m_{\Delta_i}(f))|\Delta_i|\\
&2B(U(f,P)-L(f,P))
\end{align}
$$

Remark. This proof used several standard properties of $M_\Delta$ and $m_\Delta$ functions. These properties are proved below.

Lemma. Let $\varphi:[a,b]\to\mathbb{R}$ be a real valued bounded function such that $|\varphi(x)|\leq K$ for all $x\in[a,b]$. Let $\Delta\subset[a,b]$ then

$M_\Delta(\varphi)-m_\Delta(\varphi)=\sup_{x,y\in\Delta}|\varphi(x)-\varphi(y)|\tag{1}$

$M_\Delta(|\varphi|)-m_\Delta(|\varphi|)\leq M_\Delta(\varphi)-m_\Delta(\varphi)\tag{2}$

$M_\Delta(|\varphi|)\leq K\qquad m_\Delta(|\varphi|)\leq K\tag{3}$

Proof. 1) For all $x,y\in\Delta$ we have $m_\Delta(\varphi)\leq \varphi(x)\leq M_\Delta(\varphi)$ and $m_\Delta(\varphi)\leq \varphi(y)\leq M_\Delta(\varphi)$, so
$$
|f(x)-f(y)|\leq M_\Delta(\varphi)-m_\Delta(\varphi)\tag{*}
$$
Fix $\varepsilon>0$, then there exist $\tilde{x}\in\Delta$, $\tilde{y}\in\Delta$ such that
$M_\Delta(\varphi)<\varphi(\tilde{x})+\varepsilon/2$ and $m_\Delta(\varphi)>\varphi(\tilde{y})-\varepsilon/2$. Hence
$$
M_\Delta(\varphi)-m_\Delta(\varphi)<\varphi(\tilde{x})-\varphi(\tilde{y})+\varepsilon\tag{**}
$$
Since $\varepsilon>0$ is arbtrary $(\,^{*})$ and $(\,^{**})$ implies
$$
M_\Delta(\varphi)-m_\Delta(\varphi)=\sup\limits_{x,y\in\Delta}|\varphi(x)-\varphi(y)|
$$

2) For all $x,y\in\Delta$ we have $||\varphi(x)|-|\varphi(y)||\leq|\varphi(x)-\varphi(y)|$, hence using $(1)$ we get
$$
M_\Delta(|\varphi|)-m_\Delta(|\varphi|)=\sup\limits_{x,y\in\Delta}||\varphi(x)|-|\varphi(y)||\leq\sup\limits_{x,y\in\Delta}|\varphi(x)-\varphi(y)|=M_\Delta(\varphi)-m_\Delta(\varphi)
$$

3) Since $|\varphi(x)|\leq K$ for all $x\in\Delta$, then
$$
m_\Delta(|\varphi|)\leq M_\Delta(|\varphi|)\leq K
$$

Lemma. Let $\varphi:[a,b]\to[c,d]$ be a real valued function. Let $\psi:[c,d]\to\mathbb{R}$ be continuous non decreasing function. Then for any interval $\Delta\subset[a,b]$ we have

   $M_\Delta(\psi\circ \varphi)=\psi(M_\Delta(\varphi))\qquad m_\Delta(\psi\circ\varphi)=\psi(m_\Delta(\varphi))\tag{4}$

Proof. For all $x\in\Delta$ we have $\varphi(x)\leq M_\Delta(\varphi)$. Since $\psi$ is a non decreasing function, then for all $x\in\Delta$ we get $\psi(\varphi(x))\leq\psi(M_\Delta(\varphi))$. This implies
$$
M_\Delta(\psi\circ\varphi)\leq \psi(M_\Delta(\varphi))\tag{***}
$$
Fix $\varepsilon>0$. Since $\psi$ is continuous and non decreasing there exist $\delta>0$ such that $\psi(M_\Delta(\varphi))-\psi(t)<\varepsilon$ for all $M_\Delta(\varphi)-\delta<t<M_\Delta(\varphi)$. Moreover from definition of supremum we can find $\tilde{x}\in\Delta$ such that $M_\Delta(\varphi)-\delta<\varphi(\tilde{x)}<M_\Delta(\varphi)$. Now we set $t=\varphi(x)$ to get
$$
\psi(M_\Delta(\varphi))<\psi(\varphi(\tilde{x}))+\varepsilon\tag{****}
$$
Since $\varepsilon>0$ is arbitrary from $(\,^{***})$ and $(\,^{****})$ we get
$$
M_\Delta(\psi\circ \varphi)=\psi(M_\Delta(\varphi))
$$
The proof of the other equality is similar.

Title: Upper and lower integration inequality


答:
See @user10354138's answer for your question, but I just wanted to add: Another way you can do this is just using the straight-up quadratic equation on $z^2−(i+1)z+i=0$. Namely,
$$z = \frac{i+1 \pm \sqrt{(1+i)^2-4i}}{2} = \frac{i+1 \pm (-1+i)}{2}$$
Therefore, $z = i$ or $z=1$, which are the two answers you got with your method.

Title: Using $z = x + yi$ does not reveal all solutions

Body:
The question is to find all solutions in $\mathbb{C}$ of the equation: $$z^2 - (i+1)z + i = 0$$

After expanding with $z = x + yi$, I get $(x^2 - y^2+y-x) + (2xy -x-y+1)i = 0$.

Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.

Now, we plug that into $(2xy -x-y+1) = 0$ .

Since $x,y \in \mathbb{R}$,  this only gives the solution $x=  0, y = 1$.

However, there is another solution which is $z=1$. Why this solution does not appear with this method?


答: Hint: If $1616\dots 16$ ($100$ digits) is a perfect square, so is $101010\dots 101$ ($99$ digits), since $16$ is a perfect square. Now consider the latter number: is it $\equiv 1 \pmod 8$?

Title: How do I prove that $161616 \dots 16$ ($100$ digits) cannot be written as a perfect square?

Body:
I did:
$$
16 \cdot 101010 \dots 1 \quad (99 \text{ digits})
$$

It's clear that $16 = 2^4$, satisfying the condition that the prime has to have an even power. But what about $101010 \dots 1$?

Also, I tried $16 \pmod 4$ but it's congruent to $0$.

(???)


问:
P is a set containing n elements. A subset 'A' of P is selected at random.
... The set P is reconstructed by replacing its elements. A subset 'B' of P is again selected at random.
What is the number of ways of selecting the subsets such that $A∩B=\phi$  And what is the number of unordered pairs $(A, B)$ such that $A∩B=\phi$?

So I figured out that if we select r elements out of the n and assign them to A, we will have $n_{C_r}$ ways. To select elements for B such that it has no elements in common with A, we will select the subsets of P with $n-r$ elements. So that gives us total number of ways $= \sum_{0}^{n}{n \choose r}\cdot(2^{n-r})$ which is the expansion for $(2+1)^n =3^n$.

I'm stuck at finding the number of unordered pairs. Can someone please help me with the approach I'm supposed to use for this?

答:
The number of ways of selecting the subsets $A$ and $B$ such that $A \cap B = \varnothing$ is the number of ways of dividing the elements of $P$ into three buckets, which we can label $A$, $B$, and $N$ (for Neither).  Each element of $P$ ends up in one of $3$ possible buckets and each different division of elements into buckets yields a different ordered pair, so the number of choices for ordered pairs of $A$ and $B$ is $3^n$.  That saves you the trouble of computing the sum.  Your analysis, though, is also correct.

Since every ordered pair $(A, B)$ except for $(\varnothing, \varnothing)$ can be matched with the corresponding ordered pair $(B, A)$, there are $1+ \frac{3^n-1}{2}$ unordered pairs.


问:
Four dimensional field over complex numbers
A guy in Facebook claims he's come up with an algebraic field extension to the complex plane. He's defined the unit multiplications as $i^2=-1$, $j^2=i$ and $k^2=-i$. This implies that $ij=ji=k$, $ik=ki=-j$ and $jk=kj=-1$.

Furthermore complex numbers are seemingly a subset of this space, since each complex number $z = a + bi$ can be represented as $a + bi + 0j + 0k$.

So far I've checked that both addition and multiplication in this space are indeed associative and commutative, the addition and multiplication are distributive, and that $0$ and $1$ are the additive and multiplicative inverses respectively. I've yet to find an example of a non-inversible element.

Could someone help explain what's going on?

答: If $j^2=i$, then $j$ is a squareroot of $i$. This implies that $j\in \mathbb{C}$ as $\mathbb{C}$ is algebraically closed. Furthermore, if $ij=k$, then $k\in \mathbb{C}$ as well. This is not an extension, there are no elements being added.


问:
On Perron's formula with remainder
I am following Tenenbaum's Introduction to Analytic and Probabilistic Number Theory to prove Perron's formula with remainder. Particularly, in section II.2.1 the book states that

$$
\left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x}
$$

where $k>\kappa>0$ and $x>1$, but when I perform integration by parts, I get

$$
\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds={x^\kappa\over\log x}\left({x^{iT}\over\kappa+iT}-{x^{-k+iT}\over\kappa-k+iT}\right)+{1\over\log x}\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s^2}\mathrm ds
$$

Now taking absolute value on both side, the equation becomes

$$
\left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x}+{1\over|\kappa-k+iT|}+{1\over\log x}\int_{\kappa-k+iT}^{\kappa+iT}{\mathrm ds\over|s|^2}
$$

Apparently, my bound is much more loose than that of the book, and I wonder what I can do more to make the bounds more tight.

答:
Using Greg Martin's comment, I get that

$$
\begin{aligned}
\left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|
&=\left|\int_{\kappa-k}^\kappa{x^{t+iT}\over t+iT}\mathrm dt\right| \\
&\le\int_{\kappa-k}^\kappa{x^t\over|t+iT|}\mathrm dt
\end{aligned}
$$

By the property of right triangle, we see that $|t+iT|\ge T$, so

$$
\left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le\frac1T\left|\int_{\kappa-k}^\kappa x^t\mathrm dt\right|\le{x^\kappa-x^{\kappa-k}\over T|\log x|}
$$

Since $x>1$ and $\kappa>k$, we have $x^{\kappa-k}<x^\kappa$. Consequently, we obtain Tenenbaum's bound that

$$
\left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x}
$$

Title: On Perron's formula with remainder


问:
Can there be a generalization for n in the equation $2^x = a-n$??
This might seem a vague question, but let me give you its background before going on.
I have obtained an equation where a=2mn+m+n, where all a,n and m are natural numbers and neither 'm' or 'n' or both of them can be equal to 0 (0 is not a natural number, but a whole number).

Not all values of a i.e. the natural numbers satisfy 2mn+m+n and these are the numbers that I need to find (i.e. the values for 'a' which don't satisfy $2mn+m+m$).

For this, I see that 2mn+m+n can be written as $(2mn+m) + n = (2n+1)m + n$.
All the numbers can be represented as (2n+1)m except for $2^x$.
This is because, in $(2n+1)m$, $2n+1$ is an odd number, i.e. all the odd numbers can written as such.
Similarly, only those even numbers whose prime factorization consists purely of even numbers i.e. 2, CANNOT be written in this form i.e. $2^x$ numbers.

We have some values for 'a' which are not equal to 2mn+m+n and we're looking to find those.
Similarly, we see that (2n+1)m also CANNOT be written as $2^x$.

Thus, $a ≠ 2mn+m+n$, $2^x ≠ (2n+1)m$  i.e. $2mn+m$.

Therefore, $2^x+n≠ 2mn+m+n$ (the inequality is maintained here),
$2^x+n ≠ a$

Now, a is not alone but, itself is a part of the equation $2a+1$, where 'a' CAN be represented as
2mn+m+n, i.e. '2a+1' is an odd composite number, as in $2a+1$, a can be written as 2mn+m+n, how??
$(2n+1)(2m+1) = 4mn+2m+2n+1 = 2(2mn+m+n)+1$, while the values for a which don't satisfy $2mn+m+n$, $2a+1$ is a
prime number.

Therefore, $a ≠ 2^x+n$. Similarly, $2a+1 ≠ 2(2^x+n)+1$ (again, inequality is maintained).

$2^{x+1} + 2n+1≠2a+1$.
Thus, $2^{x+1} ≠ 2a-2n$,

$2^{x+1} ≠ 2(a-n)$,

$2^x≠a-n$, where 'a' can be represented as 2mn+m+n.

i.e. $2^x≠2mn+m+n-n$,

giving us,
$2^x≠2mn+m$,
$2^x≠(2n+1)m$

And, where 'a' CANNOT be represented as $2mn+m+n$, (i.e. where $2a+1$ is prime)

$2^x=a-n$

$2^x$ is given, and there is a pre requisite condition for $a$.
So, my question -
Is there any way to generalize the values for $n$??

答: An easier way to do this would be to consider the equivalent equation $$2a+1=(2m+1)(2n+1).$$ It means that the odd number $2a+1$ can't have any non-trivial (i.e. different from $1$ or itself) odd divisors, meaning it must be an odd prime number.


问:
Weak convergence in Sobolev space $W_0^{1, 2}(\Omega):$ how it works?
Let $\Omega$ be an open bounded domain in $\mathbb{R}^N$ and let $J:W_0^{1, 2}(\Omega)\setminus\{0\}\to\mathbb{R}$ be a $\mathcal{C}^1$ functional. Let $(u_n)_n$ be a sequence such that
$$ u_n\to u \mbox{ in } L^2(\Omega), \quad u_n\rightharpoonup u \mbox{ in } \ W_0^{1, 2}(\Omega)\quad\mbox{ and }\quad J^{\prime}(u_n)\to 0 \mbox{ in } W^{-1, 2}(\Omega).$$

My question is: under these hypotheses, it is true that
$$J^{\prime}(u_n)(u_n -u)\to 0\quad\mbox{ and }\quad J^{\prime}(u)(u_n -u)\to 0?$$

If yes, could anyone please explain me why?

I guess it follows from the assumptions that $J^{\prime}(u_n)\to 0 \mbox{ in } W^{-1, 2}(\Omega)$ and $u_n\rightharpoonup u \mbox{ in } \ W_0^{1, 2}(\Omega)$, but I don't know how to use these informations.

Could anyone please help? Thank you in advance!

答:
If $x_n\to x$ in $X$ and $f_n\rightharpoonup f$ in $X^*$ then $f_n(x_n)\to f(x)$. To prove this, use
$$
f_n(x_n)- f(x) = f_n(x_n-x) + (f_n-f)(x)
$$


答:
Your Facebook guy seems to have stumbled upon some interesting identities among $8$-th roots of unity. To wit,
$$
j = e^{\pi i/4} = \tfrac{\sqrt{2}}{2} (1+i)
\qquad \text{and} \qquad
k = e^{3\pi i/4} = \tfrac{\sqrt{2}}{2} (-1+i)
$$
satisfy all of your identities within $\mathbb{C}$. (Notice that $i = j^2$ and $k = j^3$.) This had to be the case since $\mathbb{C}$ is algebraically closed. This means that any $\mathbb{R}$-linear combination of $1, i, j, $ and $k$ can actually be expressed just in the span of $1, i$.

Note: the pair $-j, -k$ would also work.

Title: Four dimensional field over complex numbers

Body:
A guy in Facebook claims he's come up with an algebraic field extension to the complex plane. He's defined the unit multiplications as $i^2=-1$, $j^2=i$ and $k^2=-i$. This implies that $ij=ji=k$, $ik=ki=-j$ and $jk=kj=-1$.

Furthermore complex numbers are seemingly a subset of this space, since each complex number $z = a + bi$ can be represented as $a + bi + 0j + 0k$.

So far I've checked that both addition and multiplication in this space are indeed associative and commutative, the addition and multiplication are distributive, and that $0$ and $1$ are the additive and multiplicative inverses respectively. I've yet to find an example of a non-inversible element.

Could someone help explain what's going on?


问:
$p_{n}$ be the probability that $C+D$ is perfect square. Compute $\lim\limits_{n \to \infty}\left(\sqrt{n} p_{n}\right)$

Assume $C$ and $D$ are randomly chosen from $\{1,2, \cdots, n\}$. Let $p_{n}$ be the probability that $C+D$ is perfect square. Compute $$\lim\limits _{n \to \infty}\left(\sqrt{n} p_{n}\right)$$

My Approach: My assumptions were that $C,D$ can be equal and the order of $C,D$ are not considered [I don't know I assumed correct, but in the question it was not clear.]

The number of ways a number $k$ can be partitioned into $2$ parts is equal to $\Big\lfloor\frac{k}{2}\Big\rfloor$
Now with the numbers $\{1,2, \cdots, n\}$ all the numbers from $2$ to $2 n$ can be made by choosing $2$ numbers and adding them up. So number of perfect squares in this region are $\big\lfloor\sqrt{2 n}\big\rfloor-1$ ($-1$ because we cannot make $1$). So the probability of choosing $C, D$ such that $C+D$ is a perfect square is
$$p_n=
\frac{\displaystyle{\sum_{k=2}^{\big\lfloor\sqrt{2 n}\big\rfloor} \Bigg\lfloor\frac{k^{2}}{2}\Bigg\rfloor}}{n^{2}}
$$

Now I cant progress further.

Tell me if I am correct. Also are my assumptions correct ?

Edit: The answer of the limit is $\frac{4(\sqrt{2}-1)}{3}$

答: We have $1^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} \sim n^3/3.$ Now drop the floors and include $k=1$ in the sum to get $\sqrt{n} p_n \sim \sqrt{n} \cdot (\sqrt{2n})^3/(3n^2) = 2\sqrt{2}/3.$ You can go back and clean this up.


问:
$z^5 = -32$ Where I am doing wrong?
I believe my steps are correct but not sure why I am not getting the right answer. Following is my solution.

\begin{align*}
z^5 & = (-2)^5 \cdot (1 + 0\cdot i)\\
 & = (-2)^5 (\cos(2\pi k) + i\sin(2\pi k))  & k \in \mathbb{Z}\\
 & = (-2)^5 e^{2\pi ki} &
\end{align*}

Hence,
$$z = -2e^{2i\pi k/5}, k = 0,1,2,3,4$$

However, the answer is $z = -2, e^{\pm i\pi/5}, e^{\pm 3i\pi/5}$ (see below).

Edit: I still don't understand why my solution is wrong. Let $z = -2e^{\frac{2i\pi}{5}}$. Then
\begin{align*}
z^5 & = -32e^{2\pi i}\\
& = -32 (\cos 2\pi + i\sin 2\pi)\\
& = -32 \cdot (1 + 0\cdot i)\\
&= -32
\end{align*}

答:
That is not the actual trigonometric form of $-32$. The radius needs to be a positive number. You also don't need to account for every possible $k$ among the angles.

You should write $z^5 = 2^5 (\cos(\pi) + i \sin \pi)$ and that should lead you to the correct solutions.

Title: $z^5 = -32$ Where I am doing wrong?


问:
Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable.

Let $R=k[x,y]$ be a polynomial ring ($k$, of course, is a field). Show that
  $R/(xy-1)$ is not isomorphic to a polynomial ring in one variable.

答:
We have $k[x,\frac{1}{x}]$ as a ring. Form a polynomial ring $k[x,\frac{1}{x}][y]$ with y as the variable. Construct an evaluation map $\phi$ $$\phi:k[x,\frac{1}{x}][y] \to k[x,\frac{1}{x}], \text{with }x\to x, \frac{1}{x} \to \frac{1}{x}, y \to \frac{1}{x}$$ then $\phi$ is obviously a ring morphism. Now by factor theorem, if $f\in k[x,\frac{1}{x}][y], f\in \ker\phi$, then $f(1/x) = 0$, which gives $(y-1/x)|f$. Note that 1/x is a taken as a constant here.

Now if $f\in k[x][y] \subset k[x,\frac{1}{x}][y] $, then $f = g(y-1/x), g \in k[x,\frac{1}{x}][y]$. Now you only need to show that g is in $k[x][y]$ and then we've shown $k[x][y]/(xy-1) \cong k[x,1/x]$, now check for the accepted answer by user26857.


问:
Is $\{ \text{Lebesgue points} \} \subset \{x: |f(x)| \leq \|f\|_\infty\}$?
Let $f$ be a locally Lebesgue integrable function $f$ on $\Bbb R^n$. Let $L_f$ be the set of Lebesgue points  of $f$. Then $(L_f)^c$ has Lebesgue measure zero.

By definition of the $L^\infty$ norm, if $S$ is the set defined by
$$S=\{x: |f(x)| \leq \|f\|_\infty\},$$
then $S^c$ has Lebesgue measure zero, that is $|f(x)|\leq \|f\|_\infty$ a.e.

My question: Is $L_f = S$? Or at least is the inclusion $L_f \subset S$ true?

I know this question might look strange, but I'm reading an article and I think they're using this fact, but I'm not sure about that and I don't know how to prove it, maybe there's something about Lebesgue points that I'm ignoring.

答:
(If $\|f\|_\infty$ means $\|f\|_{L^\infty(\mathbb R^n)}$ then its trivially true for some $f\in L^1_{loc}$ since this would be just a weird way to write $\{x:f(x)\le \infty\}$.)

If $x$ is a Lebesgue point, then by Lebesgue Differentiation Theorem,
$$ |f(x)| = \lim_{r\to 0^+}\frac1{|B_r|}\int_{B_r(x)}|f(y)|dy \le \lim_{r\to 0} \|f\|_{L^\infty(B_r(x))} \le \|f\|_{L^\infty(\mathbb R^n)},$$
proving that $L_f \subset S$.

The opposite inclusion is false: let $f$ be the following particular representative of $1\in L^1_{\text{loc}}$,
$$ f(x)= \mathbf 1_{x\neq 0} .$$ Then it is easy to see that $x=0$ is not a Lebesgue point, but $0=f(0) \le \|f\|_{L^\infty} = 1$. So $0\in S \setminus L_f$.


问:
How to integrate this modified Bessel function of second kind
I am getting stuck in integrating the below equation that includes modified Bessel function of second kind:

$$\int_0^{\frac{y}{\eta}}K_0\left(\frac{|x|}{\sqrt{\sigma_a^2\sigma_b^2}}\right)\mathrm{d}x$$

Any help in this regard is highly appreciated.

答:
Make the appropriate change of variable and
$$\int K_0(t)\,dt=\frac{\pi}{2} \,  t \,(\pmb{L}_{-1}(t) K_0(t)+\pmb{L}_0(t) K_1(t))$$ where appear  the modified Struve functions.

Learn about them since they appear in most integrals of Bessel functions.


问:
A question about the derivative of trace involving Hadamard product
Assume that $X\in\mathbb{R}^{n\times n}\geq 0$, $Y\in\mathbb{R}^{n\times k}\geq 0$ and $Z\in\mathbb{R}^{k\times n}\geq 0$. Let us define the function $f(Y,Z)$ as follows:
$$f(Y,Z)=\Vert X-YZ\Vert_W^2:=\mathrm{trace}\left((X-YZ)^T(W\circ(X-YZ))\right),$$
where $\circ$ is the Hadamard product and $W\in\mathbb{R}^{n\times n}\geq 0$ is symmetric.

It can be seen that
\begin{align*}
f(Y,Z)&=\mathrm{trace}\left((X^T-Z^TY^T)(W\circ(X-YZ))\right)\\
&=\mathrm{trace}(X^T(W\circ X))-\mathrm{trace}(X^T(W\circ (YZ)))-\mathrm{trace}(Z^TY^T(W\circ X))
+\mathrm{trace}(Z^TY^T(W\circ (YZ))).
\end{align*}
Now, according to the fact that $\mathrm{trace}(AB)=\mathrm{trace}(BA)$, and $\mathrm{trace}(A^T(B\circ C)=trace((A\circ B)^TC)$, it follows that
$$f(Y,Z)=\mathrm{trace}(X^T(W\circ X))-2\mathrm{trace}(Z^TY^T(W\circ X))
+\mathrm{trace}(Z^TY^T(W\circ (YZ))).$$

I have a question about the derivative of $f$ with respect to $Y$ or $Z$. To this end, we have
$$\frac{\partial \mathrm{trace}(Z^TY^T(W\circ X))}{\partial Y}=(W\circ X)Z^T,\quad and\quad \frac{\partial \mathrm{trace}(Z^TY^T(W\circ X))}{\partial Z}=Y^T(W\circ X).$$

What can it be said about the following items?
$$\frac{\partial \mathrm{trace}(Z^TY^T(W\circ (YZ)))}{\partial Y},\quad and\quad\frac{\partial \mathrm{trace}(Z^TY^T(W\circ (YZ)))}{\partial Z}.$$

I would be appreciate if someone can help me.

答:
If we define $\phi=\mathrm{trace}(Z^TY^T(W\circ (YZ)))$ and $A=YZ$, it can be observed that
$$\phi=\mathrm{trace}(Z^TY^T(W\circ (YZ)))=\mathrm{trace}(A^T(W\circ A)).$$
For more convenience, let's define
$\mathrm{trace}(C^TD)=C:D$. Hence,
\begin{align*}
\phi=W\circ A:A,
\end{align*}
and we can conclude that
\begin{align*}
d\phi&=2W\circ A:dA\\
&=2W\circ (YZ):d(YZ)\\
&=2W\circ (YZ):(dY\,\,Z+Y\,\,dZ)\\
&=2(W\circ (YZ))Z^T:dY+2Y^T(W\circ (UV)):dZ.
\end{align*}
As a consequence, we get
\begin{align*}
&\frac{\partial\phi}{\partial Y}=2(W\circ (YZ))Z^T,\\
&\frac{\partial\phi}{\partial Z}=2Y^T(W\circ (UV)).
\end{align*}


问:
Can someone explain how to get the minimal polynomial for this matrix?
This is example 8.45 from Sheldon Axler's Linear Algebra Done Right. $T$
is the operator on $\mathbb{C}^5$ whose matrix with respect to the standard basis is:\begin{matrix}
0 & 0 & 0 & 0 & -3\\
1 & 0 & 0 & 0 & 6\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0
\end{matrix}
Find the minimal polynomial of $T$.
The solution is supposed to be $z^5-6z+3$ but I don't understand the steps to get to this.

答:
If you know about companion matrix this is obvious.

If not, you can compute the characteristic polynomial of $A$ and get $\chi_A(x)=x^5-6x+3$.  Since $A$ has no repeated eigenvalues, a fact you can easily check with the roots of $\chi'_A(x)=5x^4-6$ (or use Eisenstein criterion), the minimal polynomial coincide with characteristic polynomial.

Or you can show by direct computation that $T^4,T^3,T^2,T,I$ are linearly independent (look at where $e_1$ got sent to).  Hence the minimal polynomial must be a quintic, etc.


答:
I don't have reputation to comment, Hence, suggesting here.

if a positive integer n is a perfect square then n can not be written in form of 4k+3, for k is an integer.

Square of an integer leaves reminder 0 or 1 upon division by 4.

More importantly , the square of any integer is either of the form 3k or 3k+1. In this case n= 16....16 (100 digit) , leaves 2 as remainder when divided by 3.  ${n = 2 (mod 3)}$, hence it doesn't seem a perfect square.

This means no square can be written in form of ${3k+2}$, why? proof... By Division Algorithm every integer can be written in form of ${3k, 3k+1}$ or ${3k+2}$. if ${n=3k}$ then ${n^2}$ also of this form. if ${n=3k+1}$ then ${n^2=3K+1}$, where ${K=3k^2+2k}$ But if ${n=3k+2}$ then ${n^2=3K+1}$ where ${K=3k^2+4k+1}$, we've ${n}$ in form of ${3k+2}$ ..so, it can't be perfect square !


问:
Pearson coefficient may not be bounded by 1
I am reading this paper. On the paragraph below Theorem 2.2, it says

However, without
the normality assumption, Pearson coefficient, ρ, may be problematic. Indeed,
as shown in Frechet (1957), ρ may not be bounded by 1 in absolute value and
the bounds differ for different distributions

I find difficulty to understand this statement. I am unable to read the Frechet's paper because it is not in English.

Frechet, M. (1957). Les tableaux de correlations dont les marges sont donnees, Annales
de l’Universite de Lyon, Sciences Mathematiques at Astronomie, Serie A 4, 13-31.

I thought the Crachy Schawarz inequality would guarantee that Pearson Coefficient lies in [-1,1]. Why does the Pearson Coefficient may not bounded by 1?

答: There is a bit of ambiguous terminology in that paper. The Pearson coefficient is still guaranteed to be in $[-1, 1]$, but what the authors mean is that the bounds could be tighter for certain distributions. The example here is that the correlation between $A = e^X$ and $B = e^Y$ is bounded within $[-1/3, 1]$, where $X, Y \sim N(0,1)$.


问:
taking the power of upper triangular matrices
When a matrix $A$ is diagonalizble, i.e. $A = P \Lambda P^{-1}$, then it is very easy to compute $A^i$: it is just $P \Lambda^i P^{-1}$ where $\Lambda^i$ is the diagonal matrices with the eigenvalues exponentiated.

I have an upper triangular matrix where the diagonal is all 0s (this means that all the eigenvalues are 0). I would expect somehow to be able to do something similar to the exponentiation of $A$ above, but that's not possible because such upper triangular matrix is not diagonalizible. Is there another efficient way of computing the power of an upper triangular matrix with all 0s on the diagonal?

答: Every square matrix has a unique Jordan decomposition up to permutations of nonsingular matrix P. Let T be your upper triangular matrix where diagonal elements are all zero. Then $T=P^{-1}JP$. $T^n=P^{-1}J^n P$.


问:
Algebraic argument for why any $A_5$ in $S_6$ can be extended to an $S_5$ in $S_6$
It is well known that $S_5$ is a subgroup of $S_6$ in a way that acts transitively on $6$ points, a surprising fact related to the outer automorphism of $S_6$. One can see this using the icosahedron: its rotational symmetry group is isomorphic to $A_5$, and acts transitively on the $6$ axes of the icosahedron that go through opposite vertices. One can show that this exotic $A_5$ in $S_6$ extends to an $S_5$ in $S_6$ (see this excellent page by John Baez), using some geometric observations.

Is there a purely algebraic way to see that any $A_5$ including into $S_6$ extends to an inclusion of $S_5$ in $S_6$, that avoids working directly with the icosahedron? If such an argument were slick enough, I might prefer it to the way Baez gives (not that I don't love the icosahedron!)

答:
$A_5 \cong {\rm PSL}(2,5) < {\rm PGL}(2,5) \cong S_5$.

I suppose that doesn't show that the $A_5$ that is isomorphic to ${\rm PSL}(2,5)$ is the same as (i.e. conjugate to in $S_5$) the $A_5$ that is the symmetry group of the icosahedron, but your could not really prove that without thinking about the icosahedron!

Added: You can avoid thinking about the icosahedron by proving directly that, up to conjugacy in $S_6$, $A_6$ has a unique transitive subgroup isomorphic to $A_5$. Here is a sketch proof.

$A_5$ is generated by $x,y$ with $x$, $y$, and $xy$ having order $2$, $3$ and $5$, respectively.

In the image in $A_6$, we can assume that $x=(1,2)(3,4)$. For $\langle x,y \rangle$ to be transitive, $y$ must consist of two $3$-cycles.

The essentially different possibilities for $y$ (by which I mean up to conjugation in $S_6$) are

(i) $y=(1,3,5)(2,4,6)$, $xy$ has order $3$;

(ii) $y=(1,3,5)(2,6,4)$, $xy$ has order $4$;

(iii) $y=(1,2,3)(4,5,6)$, $xy$ has order $5$.


问:
If a cricketer has a $37$% chance of hitting a six, what is the probability of him hitting a six in 2 deliveries?

If a cricketer has a $37$% chance of hitting a six, what is the
probability of him hitting a six in 2 deliveries?

Our intuition says the probability will increase. So it will be $2(37/100)$. But it doesn't, seem right. Also, it won't decrease like $(37/100)^2$.

What is the correct answer? My friend says he calculated the answer to be $55.5$%. Is that correct?

答:
You have to add (considering you want to compute the probability of him hitting a six at least one time in two deliveries):

The chances of him hitting a six first and then NOT hitting a six.

The chances of him NOT hitting a six first and then hitting a six.

The chances of him hitting a six in both deliveries.

Knowing that the chances of him hitting a six are $\frac{37}{100}$, we conclude that the chances of him not hitting a six are $\frac{63}{100}$, so this gives us the following result:
$$\frac{37}{100}\cdot\frac{63}{100} + \frac{63}{100}\cdot\frac{37}{100}+\frac{37}{100}\cdot\frac{37}{100}=\boxed{\frac{6031}{10000}},$$
so we can conclude that the probability of him hitting at least a six is $\frac{6031}{10000}$, to be said $60.31$%.

Another way faster to do this is to just calculate the probability  of him NOT hitting a six both times (since it's the only case where the cricketer does not hit any six), and subtract that number to $1$. This would give us the same result:
$$1-\frac{63}{100}\cdot\frac{63}{100}=\frac{6031}{10000}.$$


答:
This is not an answer but an answer to some comments which is simply too long to be a comment itself:

Intuitively this construction feels like a four dimensional space, but it's not. Is there some graphic or layman's explanation that would make it more clear what is actually happening in this construction?

I do not have a graphical explanation. The crucial point indeed is that the space is not four dimensional. If you are familiar with group presentations, a very similar issue occurs. Suppose you have a group generated by a bunch of elements ($a,b,c$) and you are giving a bunch of relations in between those elements ($a^2b=cb, \dots$). It is possible that you have so many relations that one does not really need all generators (for example $a^2b=cb$ implies that $a^2=c$ and thus $c$ is redundant). In this you at first might be tempted to think that this group needed three generators, but actually less suffice.

Your facebook guy did something similar. At first glance it looks like you need 4 independent vectors te express everything, but these vectors are really not independent. That's the crux of the current story.

Or alternatively, what would be the implications if we were to suppose that $z^2=i$ had more than two solutions?

If $z^2-i=0$ has more than two solutions, you cannot be working in a field. Indeed, you can easily prove that any polynomial $P(X)\in K[X]$ of degree $n$ over a field $K$ cannot have more than $n$ roots. The proof is straightforward as soon as you have the division algorithm for polynomials over a field.

Edit: I'll include a proof of the latter statement. Let $P(X)\in K[X]$ be a polynomial. We can write $$P(X)=\sum_{i=0}^na_iX^i$$ with each $a_i\in K$ and $a_n\neq 0$. First note that we may assume that $a_n=1$ by multiplying the polynomial by $a_n^{-1}$ (here I use that $K$ is a field). Furthermore, doing so does not change the roots of $P(X)$ (that's a claim that you can prove as an exercise).

Now I claim the following: If $c\in K$ is a root of $P(X)$ (i.e. $P(c)=0$), then $$P(X)=(X-c)Q(X)$$ for some polynomial $Q(X)\in K[X]$ with $\deg(Q(X))=\deg(P(X))-1$. Indeed, by the division algorithm for polynomials, one can write $$P(X)=(X-c)Q(X)+R(X)$$ where $Q(X),R(X)\in K[X]$ and $\deg(Q(X))\leq \deg(P(X))$ and $\deg(R(X))<\deg(P(X))=1$. By assumption, $P(c)=(c-c)Q(c)+R(c)=R(c)=0$. On the other hand, $\deg(R(X))=0$ and thus $R(X)=0$ as it is a constant polynomial. Thus $P(X)=(X-c)Q(X)$ as required.

Hence we now showed that if $c$ is a root of $P(X)$, then we find that $P(X)=(X-c)Q(X)$ with $\deg(Q(X))=\deg(P(X))-1$. Now assume that $P(X)$ has more roots than its degree. Iterating the above procedure, you find that $P(X)=(X-c_1)(X-c_2)\dots (X-c_m)Q(X)$ with $m>\deg(P(X))$, but then $\deg(P(X))\geq m$, a contradiction!


问:
"A Simple Proof of Zorn's Lemma" article. (explanation of a step)
There is a short proof of Zorn's lemma by J.W.Lewin:
A Simple Proof of Zorn's Lemma.pdf

Summary:

I don't understand this step:
"Therefore if z is the least member of
$A\setminus P(B, y)$, we have $P(A, z) = P(B, y)$."

I'm interested in both  inclusions from left to right and from right to left.

A subset A of X is conforming if the following two conditions hold:

The order $<$ is a well order of the set A.

For every element $x\in A$, we have $x = f(P(A, x))$.

Theorem
 If A and B are conforming subsets of X and $A\neq B$, then one of these two sets is an initial segment of the other.

Proof We may assume that $A\setminus B\neq\emptyset$. Define x to be the least member of
 $A\setminus B$. Then $P(A, x) \subseteq B$. We claim that $P(A, x) = B$. To obtain a contradiction,
 assume that $B\setminus P(A, x)\neq\emptyset$, and define y to be the least member of
 $B\setminus P(A, x)$. Given any element $u\in P(B, y)$ and any element $v\in A$ such that
 $v < u$, it is clear that $v \in P(B, y)$. Therefore if z is the least member of
 $A\setminus P(B, y)$, we have $P(A, z) = P(B, y)$. (???)

答:
Actually this proof is not correct!

Suppose that X is a partial order under ≤. Let Low(A) and Up(A) be respectively the sets of all strict lower bounds and all strict upper bounds of a subset A of X; and let P(A, x) be all strict predecessors of x in A, i. e. A ∩ Low({x}).

Let g a choice function that assigns to every non-empty set one of its elements; and let  f be a function that assigns to every chain C in X the strict upper bound f(C) = g(Up(C)).

Now suppose that x and y ∈ X are incomparable (not x ≤ y and not y ≤ x) and that A = {x} and B = {y} and that g(the union of A and Up(A)) = x and g(the union of B and Up(B)) = y. So x = f(P(A,x)) and y = f(P(B,y)).

Now the sets A = {x} and B = {y} are conforming as defined in Lewin’s paper, because both are well-ordering and the second condition holds for both. But we supposed that x and y are incomparable; so are the sets A = {x} and B = {y}. So the whole lemma in the bottom of the first page of the paper is incorrect.


问:
Integrability of a function (Darboux)
Question:

Let $f: [a,b] \to \mathbb{R}$ and assume $0 \leq f(x) \leq B$ for  $x\in [a,b]$.   Show that
  $$U(f^2,P) -L(f^2,P) \leq 2B\ ( U(f,P) - L(f,P) )$$
  for all partitions $P$ of $[a,b]$.

Show that if $f$ is integrable on $[a,b]$, then so is $f^2$ [no positivity assumption here].

I found this hint buried in the latex file my prof gave me.

(Hint: $f(x)^2 -f(y)^2 = (f(x) + f(y)) \cdot (f(x) -f(y))$.)

It really wasn't easy to put it to good use...

$$\begin{align}
U(f^2,P) -L(f^2,P) &= (U(f,p) + L(f,p)) \cdot (U(f,p) - L(f,p)) \\
&\leq (|U(f,p)| + |L(f,p)|) \cdot (U(f,p) - L(f,p))
\end{align}$$

We know $|U(f,p)|\geq |L(f,p)|$, and since $f(x) \leq B$, $|f(x)| \leq B$ for all $x \in [a,b]$ where $B \in \mathbb {R}$.

Knowing that  $|L(f,p)| \leq |U(f,p)|\leq |f(x)| \leq B$, we get:

$$(B + B) \cdot (U(f,p) - L(f,p)) = 2B\ (U(f,P) - L(f,P))$$

Hence $U(f^2,P)-L(f^2,P) \leq 2B\ (U(f,P) - L(f,P))$.

Anyone agree with me?

Part 2)

Theorem: Let $f:[a,b] \to [c,d]$ be Daraboux-integrable and $g:[c,d] \to \mathbb {R}$ be continuous. Then the composition $g \circ f$ is Daraboux-integrable.

Let $g=x^2$, so that $g \circ f=f^2$. Since $g$ is continous on $[c,d]$ for all $c,d \in \mathbb {R}$, $f^{2}$ is integrable as long as $f$ is integrable on $[a,b]$ by the theorem above.

答:
$\color{red} \cdots$ $U(f,p)$ generally refers to the "upper sum," not a step function.  Instead, define two step functions $s$ and $t$ over a partition $P$ such that $s(x) \le f(x) \le t(x)$.  Let $s$ be maximal while still requiring $s(x) \le f(x)$ and let $t$ be minimal.  Then $s^2$ and $t^2$ are step functions corresponding to $f^2$.

Your comment that "hence $U(f,p) \le B$" threw me off a bit.  You can simply conclude that $t(x) \le B$ since $f(x) \le B$ and $t$ is a step function taking values on $f$.

And I would recommend turning the inequality $[B+B]( U(f,p) - L(f,p) ) \le 2B ( U(f,p) - L(f,p) )$ into an equality.

Both parts give correct proofs (if we replace the instances of $U(f,p)$ and $L(f,p)$ appropriately with $s$ and $t$).  However, it sounds like your professor may want you to prove the second part directly from the first.  If that is the case, simply pay attention to the fact that $U(f,p) - L(f,p) \to 0$ as the partitions become finer, while $2B$ is a fixed constant.  $\color{red} *$ But be careful, because you have to drop the positivity assumption for this part.

There are a few lines in there which are unnecessary.  For instance, you write:

$U(f,p)|≥|L(f,p)|$ since $f(x)≤B |f(x)|≤B$ for all $x∈[a,b]$ where $B∈\mathbb R$

While all these statements are true, and you should definitely mention $U(f,p) \ge L(f,p)$, the second part is both unnecessary and trivial and makes the proof a bit messy (since it makes it sound like the first statement follows from the second).  So I would recommend getting rid of the part beyond "since."

You also write:

$|L(f,p)|≤|U(f,p)|≤|f(x)|≤B$

By using the absolute value signs, do you mean to be taking the maximum value of the functions?

Title: Integrability of a function (Darboux)


答:
$C$ and $D$ are chosen independently, so the number of pairs $(C,D)\in[n]^2$ with $C+D=k^2$ is
$$
\begin{cases}
k^2-1 &\text{if }2\leq k\leq \sqrt{n+1}\\
2n+1-k^2& \text{if }\sqrt{n+1}<k\leq\sqrt{2n}\\
0 & \text{otherwise}
\end{cases}
$$
So
\begin{align*}
\sqrt{n}p_n&=\frac{\displaystyle
\sum_{k=2}^{\lfloor\sqrt{n+1}\rfloor} (k^2-1)+\sum_{k=\lfloor\sqrt{n+1}+1\rfloor}^{\lfloor\sqrt{2n}\rfloor}(2n+1-k^2)}{n^{3/2}}\\
&\sim n^{-3/2}
\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} k^2+n^{-3/2}\sum_{k=\lfloor\sqrt{n}\rfloor}^{\lfloor\sqrt{2n}\rfloor}(2n-k^2)\\
&\sim\int_0^1 x^2\,\mathrm{d}x+\int_1^{\sqrt{2}}(2-x^2)\,\mathrm{d}x=\frac43(\sqrt{2}-1)
\end{align*}

Title: $p_{n}$ be the probability that $C+D$ is perfect square. Compute $\lim\limits_{n \to \infty}\left(\sqrt{n} p_{n}\right)$

Body:
Assume $C$ and $D$ are randomly chosen from $\{1,2, \cdots, n\}$. Let $p_{n}$ be the probability that $C+D$ is perfect square. Compute $$\lim\limits _{n \to \infty}\left(\sqrt{n} p_{n}\right)$$

My Approach: My assumptions were that $C,D$ can be equal and the order of $C,D$ are not considered [I don't know I assumed correct, but in the question it was not clear.]

The number of ways a number $k$ can be partitioned into $2$ parts is equal to $\Big\lfloor\frac{k}{2}\Big\rfloor$
Now with the numbers $\{1,2, \cdots, n\}$ all the numbers from $2$ to $2 n$ can be made by choosing $2$ numbers and adding them up. So number of perfect squares in this region are $\big\lfloor\sqrt{2 n}\big\rfloor-1$ ($-1$ because we cannot make $1$). So the probability of choosing $C, D$ such that $C+D$ is a perfect square is
$$p_n=
\frac{\displaystyle{\sum_{k=2}^{\big\lfloor\sqrt{2 n}\big\rfloor} \Bigg\lfloor\frac{k^{2}}{2}\Bigg\rfloor}}{n^{2}}
$$

Now I cant progress further.

Tell me if I am correct. Also are my assumptions correct ?

Edit: The answer of the limit is $\frac{4(\sqrt{2}-1)}{3}$


问:
Uniform convergence of $f_n(x)=\frac1{1+nx}$ on $(0,\infty)$
Problem: Let $f_n(x)=\displaystyle\frac{1}{1+nx}$ for $x\in(0,\infty)$ and $n\in\mathbb N$.

$\textbf{a)}$ Show that for each $\delta>0$, $\{f_n\}_{n=1}^\infty$ is uniformly convergent on $[\delta,\infty)$, and determine to which function.

This is the easy part. Fix $x\in[\delta,\infty)$. Then $\delta\leq x$ implies that
$$\vert f_n(x)\vert=\left\vert\frac{1}{1+nx}\right\vert\leq\frac{1}{1+n\delta}\to0\quad\text{as }n\to\infty.$$
Since $x\in[\delta,\infty)$ was arbitrary, it follows that $\|f_n-f\|_\infty\to0$ as $n\to\infty$ where $f(x)=0$ for all $x\in[\delta,\infty).$

$\textbf{b)}$ Show that $\{f_n\}_{n=1}^\infty$ is not uniformly convergent on $(0,\infty)$.

Fix $x\in(0,\infty)$ and let $\varepsilon>0$ be given. Choose $N\in\mathbb N$ such that $N>\displaystyle\frac{1}{\varepsilon x}$. Then for all $n\geq N$ we have
$$\vert f_n(x)\vert=\frac{1}{1+nx}\leq\frac{1}{nx}<\varepsilon.$$
Therefore, $f_n\to0$ as $n\to\infty$ pointwise on $(0,\infty).$ So we assume that the convergence is uniform and try to derive a contradiction. Pick $\varepsilon=1/2.$ Then there is some $N(\varepsilon)\in\mathbb N$ such that $\vert f_n(x)\vert<\varepsilon$ for all $n\geq N(\varepsilon)$ and all $x\in(0,\infty)$. Therefore, we may choose $n$ and $x$. Pick $n=N(\varepsilon)$ and $x=1/3N(\varepsilon)$. Then $\vert f_n(x)\vert=\displaystyle\frac{1}{1+1/3}>1/2=\varepsilon$ and we have a contradiction. It follows that $\{f_n\}_{n=1}^\infty$ does not converge uniformly on $(0,\infty).$

Do you agree with my proof above? I am specifically concerned about the last argument in part (b).

Thank you in advance for any help, your time is much appreciated.

答:
A most straightforward argument for part $b)$ is to notice that
$$f_n \left( \frac{1}{n}\right) = \frac{1}{2}$$

does not tend to $0$, hence $(f_n)$ cannot converge uniformly.


答:
Following your logic, the probability would be $3\cdot(37/100)>1$ in three deliveries.

It's true that the chances of hitting a six in three deliveries are very high, but the probability cannot exceed $1$.

In this kind of problems it's easier if you compute the probability of not hitting a six in $n$ deliveries, because these are independent events and the combined probability is obtained by multiplying the probabilities of the single events.

Thus the probability the cricketer doesn't hit a six in $n$ deliveries is $(1-p)^n$, where $p$ is the probability of hitting a six.

In order to find the probability you want, just compute the complement:
$$
1-(1-p)^2=1-\dfrac{67^2}{100^2}=\frac{6031}{10000}=60.31\%
$$
In three deliveries
$$
1-(1-p)^3\approx75\%
$$

Body:
If a cricketer has a $37$% chance of hitting a six, what is the
probability of him hitting a six in 2 deliveries?

Our intuition says the probability will increase. So it will be $2(37/100)$. But it doesn't, seem right. Also, it won't decrease like $(37/100)^2$.

What is the correct answer? My friend says he calculated the answer to be $55.5$%. Is that correct?


问:
Families of function which are locally uniformly bounded but not uniformly bounded
Consider the following problem asked in conformal mapping and Riemann mapping theorem section of ponnusamy and silvermann complex variables with applications(page 394).

question : What kind of families of function are locally uniformly bounded but not uniformly bounded.

Definition of both are clear to me but still not able to solve it.

Can you please tell how should I proceed?

答:
A simple example is $f_n(z) = nz^n$ on the unit disk. $(f_n)$ is not uniformly bounded since $\sup_{|z| < 1} |f_n(z)| = n$ is not bounded.

But for each $0 \le r < 1$ is  $\sup_{|z| < r} |f_n(z)| = nr^n$ a convergent and therefore bounded sequence, so that $(f_n)$ is uniformly bounded on all disks $B_r(0)$ with $r< 1$, and therefore a on all compact subsets of the unit disk.


问:
Partial derivative with two different limits
$$z = f \left( u(t), \ v(t)\right) $$

In my lecture notes, it is said that

$$ \lim_{h\rightarrow0}\frac{f\left(u\left(t+h\right){,}\ v\left(t+h\right)\right)-f\left(u\left(t\right){,}\ v\left(t+h\right)\right)}{h}=\frac{\partial f}{\partial u}\frac{du}{dt}.$$

However I would have defined that derivative as the following limit:

$$ \lim_{h\rightarrow0}\frac{f\left(u\left(t+h\right){,}\ v\left
(t\right)\right)-f\left(u\left(t\right){,}\ v\left(t\right)\right)}{h} $$

Could someone clarify how these limits are the same?

答:
I think this is one of those rare occasions when L'Hospital's Rule is your friend.

We cannot prove this without some assumptions on $f,u,v$: I am going to assume that all the derivatives I mention exist and are continuous.

Let $$H(h):=f(u(t+h),v(t+h))-f(u(t),v(t+h)).$$

To calculate $\lim_{h\to 0}\frac{H(h)}{h}$ it suffices to calculate $\lim_{h\to 0}H'(h)$, provided this exists. That is we want to calculate the limit as $h\to 0$ of
$$
f_{u}(u(t+h),v(t+h))u'(t+h)+f_{v}(u(t+h),v(t+h))v'(t+h)\\
-f_{u}(u(t),v(t+h))\cdot 0-f_{v}(u(t),v(t+h))v'(t+h).
$$

Now we can use the (assumed) continuity and take the limit to get
$$
f_{u}(u(t),v(t))u'(t)+f_{v}(u(t),v(t))v'(t)\\
-f_{u}(u(t),v(t))\cdot 0-f_{v}(u(t),v(t))v'(t)
$$
which is just
$$
f_{u}(u(t),v(t))u'(t).
$$

The other limit also evaluates to $f_{u}(u(t),v(t))u'(t)$: either use the definition of partial derivative (if it indeed applies) or argue in the same way as I have done.


问:
Relationship of the support of a test function with the support of distribution.

Let $\phi\in D(\Omega)$ and $f\in D'(\Omega)$. If $\phi$ is $0$ in a neighbourhood of $\operatorname{supp}(f)$, then how will we prove that $\langle f, \phi
\rangle$ is also $0$?
  Will it be sufficient if $\phi$ vanishes on $\operatorname{supp}(f)$?

Definition of $\operatorname{supp}(f)$:

Let $f$ be a distribution, and $U$ an open set in $\Bbb R^n$ such that, for all test functions $\phi$ with the support of $\phi$ contained in $U$, $f(\phi) = 0$. Then $f$ is said to vanish on $U$. Hence we can define the support of $f$ as the complement of the largest open set on which $f$ vanishes.

Alternatively, we can say $\operatorname{supp}(f)$ is the subset of $\Bbb R^n$ such that $x \in \operatorname{supp}(f)$, if and only if for all neighbourhoods $\mathcal U$ of $x$, there is $\phi \in D(\mathcal U)$ such that $\langle f, \phi\rangle \neq 0$.

答: If $\phi$ is $0$ in a neighbourhood of $\textrm{supp}(f)$, then the support of $\phi$ is a closed set that does not meet $\textrm{supp}(f)$. In particular this implies that $f$ vanishes on $\textrm{supp}(\phi)$ in the sense of your definition. So $f(\phi)=0.$</p>


问:
Understanding why $B$ is generated by a finite number of monomials and how this affects $B_0$.
Here is the question I want to understand its statement:

Let $B = \mathbb C[x_1, x_2, x_3, x_4]$ be $\mathbb Z$-graded so that each $x_i$ is homogeneous and $\deg(x_1, x_2, x_3, x_4) = (1, -2, 3, -4).$ Find $B_0.$

Hint: This ring is generated by a finite number of monomials.

My thoughts:

To find $B_0$ I have to solve this equation $$e_1 -2 e_2 + 3e_3 -4 e_4  = 0.$$

But this equation has infinitely many solutions, so I am guessing that here comes the importance of the given hint but for my life I do not understand how? could someone explain this to me ,please?

Some details of why specifically the above equation:

As I know that I am going to put a $\mathbb Z$-grading on $B$ where $B = \bigoplus_{i \in \mathbb Z} B_i$ and $x_i \in B_{\deg x_i}.$So according to the degrees we have in the question we have $x_1 \in B_1, x_2 \in B_{-2}, x_3 \in B_3, x_4 \in B_{-4}.$ And because I know that $B_i$ is the vector space with basis $\{X_1^{e_1}, X_2^{e_2}, X_3^{e_3}, X_4^{e_4}\}$ where $e_1 \deg x_1 + e_2 \deg x_2 + e_3 \deg x_3 + e_4 \deg x_4 = i.$

答:
I claim $B_0$ is generated as a $\Bbb C$-algebra by the following monomials: $$1,x_1^2x_2, x_1^4x_4, x_2^3x_3^2, x_3^4x_4^3, x_1x_3x_4, x_1x_2^2x_3, x_2x_3^2x_4.$$ We'll show that given any monomial in $B_0$, it's a product of the above and a constant.

Suppose $m=x_1^ax_2^bx_3^cx_4^d$ belongs to $B_0$. This is equivalent to $a-2b+3c-4d=0$. We note that $x_1^2x_2$, $x_1^4x_4$, $x_2^3x_3^2$, and $x_3^4x_4^3$ are all in $B_0$, so if we're looking for other generators $m$ of $B_0$, it suffices to consider $m$ after factoring out as many of these monomials as possible. This gives us the following:

If $a\geq 2$ and $b\geq 1$, then we can write $m=x_1^2x_2m'$, so we may assume that either $a<2$ or $b=0$.

If $a\geq 4$ and $d\geq 1$, then we can write $m=x_1^4x_4m'$, so we may assume that either $a<4$ or $d=0$.

If $b\geq 3$ and $c\geq 2$, then we can write $m=x_2^3x_3^2m'$, so we may assume $b<3$ or $c<2$.

If $c\geq 4$ and $d\geq 3$, then we can write $m=x_3^4x_4^3m'$, so we may assume $c<4$ or $d<3$.

Now we have some case work. Suppose $m$ is a generator, and we'll go through all the combinations of the observations above.

If $b=d=0$, then $a=c=0$, and so $m=1$.

If $b=0$ and $a<4$, then we can apply our fourth observation: either $c<4$, and we find $(1,0,1,1),(2,0,2,2)$ as nontrivial solutions, or $d<3$ and we find the same solutions.

If $d=0$ and $a<2$, then we can apply the third observation: either $b<3$, and we get $(1,2,1,0)$, or $c<2$ and we find the same solution.

If we assume $a<2$ and $b,d\neq 0$, we can run through the four ways to combine observations 3 and 4:

If $b<3$ and $c<4$, then $\max(a+3c)=10$ and we have the solutions $(1,2,1,0)$, $(1,0,1,1)$, $(0,1,2,1)$, $(1,0,1,1)$, $(1,1,3,2)$, and $(1,3,3,1)$; the new only new generator we get here is $(0,1,2,1)$, as $(1,1,3,2)=(1,0,1,1)+(0,1,2,1)$ and $(1,3,3,1)=(1,0,1,1)+(0,3,2,0)$.

If $b<3$ and $d<3$, then $\min(-2b-4d)=-12$ and all the solutions we get are things we've seen before.

If $c<2$ and then $\max(a+3c)=4$ and we've seen all of those solutions before.

So our list of monomials from the start is a complete list of generators. There's probably a faster/easier way to do this, and I'll be glad to upvote someone who produces it.


问:
Why the functor from a Vector Space to its dual space can not be covariant?
Why can't covariant? I mean, a functor is made of two data. Part one: object$\rightarrow$object  . And Part two: morphism$\rightarrow$morphism . Why can't there exist a functor whose part one is like V$\rightarrow$V*(V* means the dual space) and part two is covariant?why must contravariant?
I cannot give a construction of a covariant functor. But it doesn't mean there doesn't exist one. Can we use any method to prove such a functor doesn't exist?

If you still cannot fully understand my question, please look at this picture.
An example

My question is for the last sentence "I'm not sure if there's a way to do this---the arrows simply don't line up". This is not a math proof. And I want a math proof of this fact.

Or if I am wrong. There actually exist such a functor. Then can you give a construction or proof of the existence.

Thank you!

答:
Obviously there exists a covariant functor from $Vect_{\mathbb{F}}$ to itself which takes $V$ to $V^*$. Call this functor $F$. Apply a super-strong version of the axiom of choice to pick a basis $(e^V_i)_{i\in I_V}$ for every $V$ and for given linear $f:V\to W$ given by
$$
e_i^V\mapsto \sum_{j\in I_W} f_j e^W_j,
$$
we simply define $F(f)((e_i^V)^*)=\sum_{j\in I_W} f_j (e_j^W)^*$, where $v^*$ of course denotes the dual-basis vector to $v$ given the chosen bases. Since linear maps are uniquely given by how they act on basis elements, this does give a well-defined map. It's also trivial to check that $F$ maps compositions to compositions.

Now, of course, this functor is incredibly silly and not very interesting. But it does exist.