问:
Covariance between coordinate x and distance in a circle.
Let $\Omega$ be a circle of radius $r$ and let $Z$ be the random variable that measures the distance from the center of the circle to a point. Let $X$ be the random variable that given $(x,y)$ associates the value of $x$. Calculate $cov(X,Z)$.

I'm having trouble thinking in a way to relate $X$ and $Z$ in order to find $E[XY]$. What limits should I consider when integrating? What I have done so far is define $p_{X,Y}(x,y)$ as:
$$
p_{X,Y}(x,y) = \frac{1}{\pi r^2},\;\text{for } x^2+y^2 ≤ r^2
$$
and
$$
p_{X,Y}(x,y) = 0\;\text{otherwise}
$$

Because I need to find $E[X]$, $E[Z]$ and $E[XZ]$ I defined first $p_{Z}$ as:
$$
p_{Z}(t) = \frac{2x}{r^2},\;\text{for } 0<t≤r
$$
Which resulted in: $E[Z]=\frac{2}{3}r$.

Then with $p_{X,Y}$ I integrated with the limits where $x$ is inside the circle to find $p_{X}$ as follows:
$$
p_X(x) = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}p_{X,Y}dy = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\frac{1}{\pi r^2}dy = \frac{1}{\pi r^2}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}dy = \frac{1}{\pi r^2}\left(\sqrt{r^2-x^2} + \sqrt{r^2-x^2}\right)= \frac{2}{\pi r^2}\sqrt{r^2-x^2}
$$

This left me with:
$$
E[X] = \int_{-r}^{r}xp_{X}(x)dx = \int_{-r}^{r}x\cdot\frac{2}{\pi r^2}\sqrt{r^2-x^2}dx = 0
$$

My main problem comes when finding $E[XZ]$. How should I define $p_{X,Z}$? I think I should be doing the double integral of $p_{X,Z}$, but how? I guess it is with establishing it as follows:
$$
\int_{-r}^{r}\int_{\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}xtp_{X,Z}dxdt
$$
I don't know if that's the way to go, because I've seen the limits of $p_{Z}$ defined from $0$ to $r$ instead of $-r$ to $r$. Any help in defining $p_{X,Z}$, the limits of these integrals or in telling me what should I do is appreciated.

答:
Let $(Z,\Phi)$ be the polar analogous of $(X,Y)$. Then $\Phi\sim U(0,2\pi)$, $Z\sim \dfrac{2z}{R^2}$, and $\Phi \perp\!\!\!\perp Z$, so

$$E\{XZ\} = E\{Z^2\cos(\Phi)\} = E\{Z^2\}E\{\cos(\Phi)\} = 0$$

because $E\{\cos(\Phi)\}=0$.

Also $E\{X\}=0$, so $cov(X,Z)=0$


问:
proof of the Krull-Akizuki theorem (Matsumura)
This set of questions refers to the proof of the Krull-Akizuki theorem given in Matsumura's Commutative Ring Theory, pages 84-85. For those who don't have the text, i will provide the details.

The Krull-Akizuki theorem reads as follows: Let $A$ be a one-dimensional Noetherian integral domain with field of fractions $K$, let $L$ be a finite algebraic extension of $K$, and $B$ a ring with $A \subset B \subset L$; then $B$ is a Noetherian ring of dimension at most 1, and if $J$ is a non-zero ideal of $B$ then $B/J$ is an $A$-module of finite length.

The proof starts by proving the following Lemma: Let $A$ and $K$ be as in the theorem, and let $M$ be a torsion-free $A$-module of finite rank $r$. Then for nonzero $a \in A$ we have $l(M/aM) \le r \cdot l(A/aA)$, where $l(\cdot)$ is the length function.

Proof of Lemma: First assume that $M$ is finitely generated. Choose elements $\xi_1,\cdots,\xi_r \in M$ linearly independent over $A$ and set $E=\sum A\xi_i$; then for any $\eta \in M$ there exists $t \in A$ with $t\neq0$ such that $t \eta \in E$...

...If we set $C=M/E$ then from the assumption on $M$ we see that $C$ is also finitely generated, so that $t C=0$ for suitable nonzero $t \in A$...we can find $C=C_0 \supset C_1 \supset \cdots \supset C_m=0$, such that $C_i/C_{i+1} \cong A/p_i$ with $p_i \in \operatorname{Spec}(A)$. Now $t \in p_i$, and since $A$ is one-dimensional each $p_i$ is maximal, so that $l(C)=m < \infty$...

Question 1: What exactly is the role of $t$ in showing that $l(C) < \infty$? It seems to me that since $A$ Noetherian and $C$ is finitely generated by construction, we obtain directly the chain $C=C_0 \supset \cdots \supset C_m=0$, with $C_i/C_{i+1} \cong A/p_i, p_i \in \operatorname{Spec}(A)$ (this is essentially theorem 6.4). From this chain and the fact that each $A/p_i$ is a field, we get that the length of $C$ is finite. What am i missing?

Question 2:Further down in the proof of the theorem, the proof shows the existence of an element $a_0 \in J \cap A$. How do we deduce from this that $l(B/J) \le l(B/a_0B)$? If $B,J$ are finite, then i can see that $l(B/J)=l(B)-l(J) \le l(B) - l(a_0B)=l(B/a_0B)$. But we don't know a priori this finiteness condition.

答:
Q1. Think about extreme case $C=\mathbb{Z}^2$, $A=\mathbb{Z}$. In this case, you won't have such finite sequence of modules. The existence of such $t$ means $C$ is a torsion module, and being a torsion module ensures the existence of such finite sequence. </p>

<p>Q2. Finiteness of $B$, $J$ cannot be assumed. I think this is just because $a_0 B \subset J$. </p>

ContentLicense: CC BY-SA 4.0

Body:
Q1. Think about extreme case $C=\mathbb{Z}^2$, $A=\mathbb{Z}$. In this case, you won't have such finite sequence of modules. The existence of such $t$ means $C$ is a torsion module, and being a torsion module ensures the existence of such finite sequence.

Q2. Finiteness of $B$, $J$ cannot be assumed. I think this is just because $a_0 B \subset J$.

ContentLicense: CC BY-SA 3.0


问:
How to prove the product rule in multivariable without chain rule?
It's easy to prove in one variable functions $f:I\subset \Bbb R \to \Bbb R$ but I tried in multivariable and I'm stuck.

Let $A$ be an open set in $\Bbb R^n$ and $a\in A$.
If $f,g:A\to \Bbb R$ are differentiable in $a$, then $f·g$ is differentiable in $a$ and
$D(f·g)(a)=g(a)·D(f)(a)+f(a)·D(g)(a)$

$f,g$ differentiable in $a$: $\lim\limits_{x\to a} \frac{f(x)-f(a)-Df(a)(x-a)}{||x-a||}=0$ and $\lim\limits_{x\to a} \frac{g(x)-g(a)-Dg(a)(x-a)}{||x-a||}=0$.

I started this way trying to prove it similar to one variable case:

$\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(a)-g(a)·Df(a)(x-a)-f(a)·Dg(a)(x-a)}{||x-a||}=\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(x)+f(a)g(x)-f(a)g(a)-g(a)·Df(a)(x-a)-f(a)·Dg(a)(x-a)}{||x-a||}=\lim\limits_{x\to a} f(a)\frac{g(x)-g(a)-Dg(a)(x-a)}{||x-a||}+\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(x)-g(a)·Df(a)(x-a)}{||x-a||}.$

In this last step, I know the first limit is zero but I don't know what to do with the other one.

答:
Everything is easier if you're fluent with big/little $o$ notation:

\begin{align*}
fg(a+h) &= f(a+h)\cdot g(a+h) \\
&= [f(a)+h\cdot D(f)(a)+o(\|h\|)]\cdot[g(a)+h\cdot D(g)(a)+o(\|h\|)] \\
&=f(a)\cdot g(a)+h[g(a)\cdot D(f)(a)+f(a)\cdot D(g)(a)]+o(\|h\|),
\end{align*}
which implies $f\cdot g$ is differentiable at $a$, and $D(f\cdot g)(a)=g(a)\cdot D(f)(a)+f(a)\cdot D(g)(a)$.


问:
What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu,\Sigma)$

What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$

My attempt is:

We know that distribution of $\sum^n_{i=1}\frac{1}{n}X_i\sim N_p(\mu,\frac{1}{n}\Sigma)$

So, $\mathbb{E}[X_i-\frac{1}{n}\sum^n_{i=1}X_i]=0$

And variance is $\text{Var}(X_i-\frac{1}{n}\sum^n_{i=1}X_i)=\Sigma-\frac{1}{n}\Sigma=\frac{n-1}{n}\Sigma$

Would that be correct?

edit:

suppose the first $X_i$ in $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ is $X_j$, $j\neq i$,so $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ becomes $X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j$

So,

$$\text{Var}(X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j)=\text{Var}(-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i+\frac{n-1}{n}X_j)=\frac{n-1}{n^2}\Sigma+\frac{(n-1)^2}{n^2}\Sigma=\frac{n-1}{n}\Sigma$$

答:
Think about a simple case, $X_1 - (X_1 + X_2 + X_3)/3$. What you have is
$Y := \frac{2}{3}X_1 - \frac{1}{3}X_2 - \frac{1}{3}X_3$, a weighted sum of $N_p(\mu, \Sigma)$ i.i.d and you know $Y$ will be normally distributed with
\begin{align}
E\,Y &= (2/3)\mu - (1/3)\mu - (1/3)\mu \\
&= 0, \\
\text{cov}(Y, Y) &= (2/3)^2\Sigma + (1/3)^2\Sigma + (1/3)^2\Sigma \\
&= (2/3)\Sigma
\end{align}


问:
Solving a Quartic Function
A user named 'Uzdawi' from another post asked a question about how to solve the quartic function of

One of the responses included an answer from the user 'Peđa Terzić', which is as follows:

Could anyone kindly elaborate on the method used to get to the answer please.

And how the same method could be used to solve the following function:

If not the same method used by Peđa Terzić, how else could you approach to solving a quartic function?

Thank you for your time.

答:
In a case like this, I would pretend I was a high-school student and use the Rational Root Theorem, which says, here, that the only rational roots (if any) will be from the set $\{\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm25,\pm50,\pm100\}$.

Starting systematically, you see that $\pm1$ aren’t roots, but $+2$ is. Then divide by $x-2$ to try to find the roots of $x^3-8x^2+5x+50$.

The RRT tells you to check $\{\pm1,\pm2,\pm5,\pm10,\pm25,\pm50\}$, and you already know that $\pm1$ are no good. Trying $+2$ is no good, but you try $-2$ and see that that’s a root too. And the quotient $(x^3-8x^2+5x+50)/(x+2)$ is $x^2-10x+25=(x-5)^2$, so that you’re done.

No ingenuity.

(In this very special case)

EDIT (addendum):

I notice that I originally omitted the possibilities $\pm20$ from the original list. I have included them now, but I might as well take the opportunity to adduce additional information:

If you know the magic of the Newton Polygon, you see that if you draw the two diagrams for the primes $2$ and $5$ and the original polynomial, both have their only vertices at $(0,2)$, $(2,0)$, and $(4,0)$. The only slopes are $-1$ and $0$. This means that a root can be only singly divisible by  $2$ or $5$. In other words, the only possibilities are from $\{\pm1,\pm2,\pm5,\pm10\}$. You’d still need to do the computational work I did above, though.

Title: Solving a Quartic Function


问:
¿Can you help me with axiom of regularity?
I understand that if we define a set such that $A=\{A\}$ we have then a contradiction because of regularity axiom, thus, a set can not be a member of itself. My question is; how can it be so, if I define a set $A=\{A,b\}$ ($b$ disjoint from $A$)?

Thanks.

答:
Given $A\in A$, regardless or not $A=\{A\}$, we can define $\{A\}$ in one of many ways:

By the axiom of pairing, if one takes it as an axiom of $\sf ZF$, the pair $\{A,A\}$ exists, and using extensionality we know this is really just $\{A\}$.

By the axiom of power set we know that $\mathcal P(A)$ exists, and by the axiom of separation, if one takes it - or as a consequence of replacement instead, one can define $\{A\}$ as the set of maximal elements of $\mathcal P(A)$ when ordered by $\subseteq$.

One can make a direct appeal to replacement, and define the function from the singleton $\{\varnothing\}$ (which exists due to power set and empty set axioms), and make it so the image is exactly $\{A\}$.

One can use a Russell-like trick and another appeal to separation, and require $\{x\in A\mid A\in x\}$, so $A$ is a member of this set, and it is sufficient to contradict the axiom of regularity.


问:
Convergence in $W^{1,\infty}(\Omega)$
I was reading some notes on Sobolev spaces and came across this statement.

"The Meyers-Serrin theorem 1.18 is false for $p = \infty$. Indeed, if $u_i \in C^\infty(\Omega)\cap
W^{1,∞}(\Omega)$ such that $u_i \rightarrow u$ in $W^{1,∞}(\Omega)$, then $u ∈ C^1(\Omega)$ (exercise)."

I thought about this for a bit and I feel like I think I have a counter-example. Let $u$ be zero on all of $\mathbb{R}$ except $u(0) =1$. Then $u \in W^{1,\infty}(\Omega)$ and the weak derivative of $u$ is the zero function. Let he sequence $u_i$ just be the constant sequence that is just the 0 function. Then  it we have that $\text{ess sup}|u_i - u| = 0$ and $\text{ess sup}|Du_i - Du| = 0$  for all $i$ and so $u_i \rightarrow u$ in $W^{1,\infty}(\Omega)$. But $u$ is clearly not continuous so $u \notin C^1(\mathbb{R})$.

Is my reasoning correct, or is there something I'm missing?

答: Here u is continuous means, up to zero measure, u is continuous. In other words, we can alter values of u on zero measure set, since they represent the same thing in Sobolev space.</p>

Title: Convergence in $W^{1,\infty}(\Omega)$


问:
find an unbiased estimator of $\sigma$ based on $\sum X_i^2$
Let $X_1,...X_n$ be iid drawn from $N(\sigma,\sigma^2)$ where $\sigma\gt0$ is the unknown parameter. Find an unbiased estimator of $\sigma$ based on $\sum X_i$, find an unbiased estimator of $\sigma$ based on $\sum X_i^2$, and use these results to determine the minimal sufficient statistic $T=(\sum X_i, \sum X_i^2)$ is complete?

It's easy to find an unbiased estimator based on $\sum X_i$. $E\left(\frac {\sum X_i} n\right)=\sigma$, so $\bar X$ is an unbiased estimator of $\sigma$. But $E(X_i^2)=var(X_i)+(E(X_i))^2=2\sigma^2$, so it's hard to get $E(\sum X_i^2)$ to be $\sigma$. $E(\frac {\sum X_i^2)} {2n})=\sigma^2$, but how can you spit out a $\sigma$?

Neither $\sqrt {\frac {\sum X_i^2)} {2n}}$ nor $\frac n 2\cdot \frac {\sum X_i^2}{\sum X_i}$ seem to work, because you can't put the expectation into the square root, and I don't think you can split an expectation across a numerator and denominator. Or can you? Am I even allowed to include $\sum X_i$ here?

Help

答:
Denote $Y_n = \frac{1}{2n} \sum_{i=1}^n X_i^2$ with $X_i$ iid and follows the normal distribution $N(\sigma,\sigma^2)$, We will contruct an unbiased estimator with the form $\theta(Y_n,n) = \sqrt{Y_n}g(n)$

The estimator $\theta(Y_n,n)$ must satisfy the 2 following conditions for all $n$
$$E(\theta(Y_n,n))= \sigma \tag{1}$$
and
$$\theta(Y_n,n) \xrightarrow{n \rightarrow +\infty} \sigma \tag{2} $$

For the second condition (2), we know already that $\sqrt{Y_n}\rightarrow \sigma$, so it suffices to construct $g(n)$ such that $g(n)\xrightarrow{n \rightarrow +\infty} 1 \tag{3}$

For the first condition (1), we have $Y_n = \frac{\sigma^2}{2n} \sum_{i=1}^n (\frac{X_i}{\sigma})^2 =\frac{\sigma^2}{2n} Z_n$ with $Z_n$ follows the noncentral chi-squared distribution $\chi^2_{n,n}$ of $n$ degrees of freedom and non-centrality parameter $\lambda = n$. The condition (I) is so equivalent to
$$E(\sigma\frac{1}{\sqrt{2n}}\sqrt{Z_n}g(n))= \sigma$$
or
$$g(n)=\frac{\sqrt{2n}}{E(\sqrt{Z_n})}=\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})} \tag{4}$$
For information, $E(\sqrt{\chi^2_{n,n}})$ has a semi-analytical form $$E(\sqrt{\chi^2_{n,n}})=\sqrt{2}\Gamma(\frac{1+n}{2}) \text{Hypergeometric1F1Regularized}(-\frac{1}{2},\frac{n}{2}, -\frac{n}{2}) \tag{5}$$

Now, it suffices to check whether $g(n)$ in (4) satisfies the condition (3). I did check with mathematica and effectively, we have
$$\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})}\xrightarrow{n \rightarrow +\infty} 1 $$
(for information, you can also check this by using Wolfram Alpha by typing the formula (5), for example with $n=1000$ here).

Conclusion: One of the unbiased estimators of $\sigma$ from $\sum_{i=1}^n X_i^2$ is
$$\frac{\sqrt{2n}}{E(\sqrt{\chi^2_{n,n}})} \sqrt{\frac{1}{2n}\sum_{i=1}^n X_i^2} $$


问:
2nd order nonlinear ODE
I have to solve the following ODE
$$
  \frac{y''}{y'^2} + y'e^y = 0
$$
by finding a function $F(y',y)$ such that its derivative is equal to $\frac{y''}{y'^2} + y'e^y$, then using initial conditions $y(0) = 0$ and $y'(0) = 1$.

What I have done so far is this. Using the multivariable chain rule we can deduce the following:
$$
  \frac{d}{dx} \left(F(y',y)\right) = \underbrace{\frac{d}{dx} y'(x)}_{=y''} \cdot \frac{\partial F(y',y)}{\partial y'} + \underbrace{\frac{d}{dx} y(x)}_{=y'} \cdot \frac{\partial F(y',y)}{\partial y}
$$
So by simple comparison we need that $\frac{\partial F(y',y)}{\partial y'} = \frac{1}{y'^2}$ and $\frac{\partial F(y',y)}{\partial y} = e^y$.

Then by integrating both of these equations with respect to either variable $y'$ and $y$ we get
$$
\begin{align*}
  &F(y',y) = \int e^y \, dy = e^y + c(y') \\
  &F(y',y) = \int \frac{1}{y'^2} \, dy' = -\frac{1}{y'} + c(y)
\end{align*}
$$
Thus an appropriate function would be $F(y',y) = e^y - \frac{1}{y'} + C$.
So we have the relationship
$$
  \frac{d}{dx} \left(e^y - \frac{1}{y'} + c\right) = 0
$$
Integrating both sides yields
$$
  e^y - \frac{1}{y'} + c_1 = c_2
$$
Now if $c_1$ and $c_2$ were zero (or identical), this would be easy to solve using separation but as this is nonlinear and inhomogeneous, I have no idea.

答:
Continuing from your last line we've gotten to
$$\frac{1}{y'}=e^y+c_1$$
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{e^y +c_1}$$
$$\int\mathrm{d}x=\int e^y+c_1 ~\mathrm{d}y$$
$$x=e^y+c_1y+c_2$$
This is an implicit solution. Unfortunately I don't think there is a closed form for $y(x)$.

We can apply initial conditions as follows: We know that $y=0$ when $x=0$ meaning
$$0=e^0+c_1\cdot 0+c_2\implies 0=1+c_2\implies c_2=-1$$
So our equation is $x=e^y +c_1 y-1$. Taking derivatives,
$$\mathrm{D}_x\{x\}=\mathrm{D}_x\{e^y+c_1y-1\}$$
$$1=y' e^y+c_1y'$$
We are told $y'(0)=1$. So,
$$1=y'(0)e^{y(0)}+c_1 y'(0)$$
But we know $y(0)$ and $y'(0)$.
$$1=1\cdot e^0+c_1\cdot 1\implies c_1=0$$
So our solution considering ICs is
$$x=e^y-1\implies e^y=x+1\implies y=\log(x+1)$$

Title: 2nd order nonlinear ODE


问:
The fibers of the quotient map $q:\Bbb A^{n+1}\setminus\{0\}\to \Bbb P^{n}$ are irreducible.
Let $k$ be an algebraically closed field, let $\Bbb A^{n+1}$ be the
$(n+1)$-dimensional affine space, let $\Bbb P^{n}$ be the $n$-dimensional projective space, and let $q:\Bbb A^{n+1}\setminus\{0\}\to\Bbb P^{n}$ be the quotient map, that is $q(a)=k^{\times}\cdot a$.

I want to show that each preimage $q^{-1}\{P\}\subseteq\Bbb A^{n+1}\setminus\{0\}$, for $P\in\Bbb P^{n}$, is irreducible. That is,
if $U,V\subseteq\Bbb A^{n+1}\setminus\{0\}$ are open subsets such that
$U\cap q^{-1}\{P\}\neq\emptyset$ and $V\cap q^{-1}\{P\}\neq\emptyset$
then $(U\cap V)\cap q^{-1}\{P\}\neq\emptyset$. We can assume that $U$ and  $V$ are basic opens and so assume that $U=D(f)$ and $V=D(g)$ for some $f,g\in k[x_{0},x_{1},\dots,x_{n}]$. Suppose that $P=q(a)$,
for some $a\in\Bbb A^{n+1}\setminus\{0\}$. Then, the above assumptions
imply that there exist $\lambda,\mu\in k^{\times}$ such that $\lambda \cdot a\in D(f)$ and $\mu\cdot a\in D(g)$, or equivalently that $f(\lambda a)\neq 0$ and $g(\mu\cdot a)\neq 0$. Now, to prove that $D(f)\cap D(g)\cap q^{-1}\{P\}=D(fg)\cap q^{-1}\{P\}\neq \emptyset$ I must find an element $\nu\in k^{\times}$ such that $\nu\cdot a\in D(fg)$ ($\iff (fg)(\nu\cdot a)\neq 0)$.

I have no clue on how to find this element $\nu$. I have found a counterexample to the choice $\nu=\lambda\cdot \mu$.

答:
I'm quite rusty when it comes to algebraic geometry but here's an idea:

The fiber over $p = [1 \colon a_1 \colon \cdots \colon a_n]$ consists of ponts $(x_1, \ldots, x_n)$ such that $(x_0, x_1,\ldots,x_n) = (\mu,\mu a_1,\ldots, \mu a_n)$ for some nonzero $\mu$. Equivalently, we want $x_0 \neq 0$ and

$$
x_j =x_0a_j
$$

for each positive $j$, i.e. $x_j-x_0a_j = 0$ for all $j$.

Hence the preimage is

$$
q^{-1}(p) = V(x_j-x_0a_j : j > 0) \cap D_{x_0}.
$$

The map $x \in \Bbb A^1 \mapsto (x,x a_1\ldots,x a_n) \in V(x_j-x_0a_j : j > 0)$ shoulld correstrict to an isomorphism

$$
\Bbb A^1 \setminus \{0\} \to V(x_j-x_0a_j : j > 0) \cap D_{x_0}.
$$

So we're left with showing that $\Bbb A^1 \setminus \{0\} \simeq V(xy-1) \subset \Bbb A^2$ is irreducible, but indeed looking at its coordinate ring we see that

$$
k[x,y]/(xy-1) \simeq k[t,t^{-1}]
$$

is a domain.


问:
Exact sequence stays exact after tensoring if right Module is free (or projective)
Let $f: R \to A$ be a ring morphism ($R, A$ a commutative rings) and

$$ 0 \to M_1 \to M_2 \to M_3 \to 0 $$

a ex. sequence of $R$ modules. Assume, that $M_3 = R^n$ is free (or
say projective). I want to show that of we tensor this sequence
with $A$ the sequence we obtain is still exact. That's always true that
the part

$$ M_1 \otimes A \to M_2 \otimes A  \to M_3 \otimes A \to 0 $$

is exact. What we have to show is that the left induced map
$M_1 \otimes A \to M_2 \otimes A$ is injective. Since
$M_3= R^n$, after tesnor $ M_3 \otimes A= A^n$ is still free and
we have a splitting $id_{A^n} = A^n \to M_2 \otimes A \to A^n $. But
I not know why this shows that the left map $M_1 \otimes A \to M_2 \otimes A$
is injective.

答:
Denote by $\alpha$ the injection $M_1\to M_2$ and by $\beta$ the surjection $M_2\to M_3$. Since $\beta$ is surjective, projectivity of $M_3$ gives us a map $\gamma:M_3\to M_2$ making the diagram below commute.

$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
   \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& M_2\\
& \diaguparrow{\exists \gamma} @VV\beta V \\
M_3 @>>\text{id}_{M_3}> M_3
\end{CD}$$
In particular, the map $\beta$ is split, and so the splitting lemma gives us $\delta:M_2\to M_1$ a retraction of $\alpha$: ie, a map such that $\delta\circ\alpha=\text{id}_{M_1}$. Then the map $\delta\otimes\text{id}_A:M_2\otimes A\to M_1\otimes A$ is a left inverse to $\alpha\otimes\text{id}_A$, whence $\alpha\otimes\text{id}_A$ is injective, as desired.


问:
Markov chain on an $n$-dimensional hypercube?
I am looking at the following Cambridge Tripos question, which has left me rather stumped for quite a while.

A particle moves on the $2^n$ vertices of the hypercube $\{0,1\}^n$. At each step the particle is equally likely to move to each of the $n$ adjacent vertices, independently of its past motion. The initial vertex occupied by the particle is $(0,0,...,0)$. Calculate the expected number of steps until the particle

i) first returns to $0,0,...,0)$;

ii) first visits $(0,0,...,1)$;

iii) first visits $(0,0,...,1,1)$.

I have seen multiple answers on here regarding the 3D case, as well as a general $D$-dimensional case presented here, but I am quite stuck on generalising to other vertices, including the three presented here. Would be lovely to see some guidance here, thanks!

答:
2018 Part IB Paper 2 Q20H if anybody is looking.  Also asked in 1991.  The introductory part to the question was

For a finite irreducible Markov chain, what is the relationship
between the invariant probability distribution and the mean recurrence
times of states?

which can be applied here to give the answer for (i) of $2^n$ given the symmetry of the hypercube

If we call the expectation when starting zero steps away from the end point $E_0=2^n$, when starting one step away $E_1$, and starting two steps away $E_2$ then:

for (ii) we can say $E_0=1+\frac{n}{n}E_1$ as you start by stepping one step from the start, making $E_1 = 2^{n}-1$

similarly for (iii) we can say $E_1=1+\frac1n0 +\frac{n-1}{n}E_2$ as you start by stepping one step either to the end point or (more likely) to somewhere two steps away  $E_2=\frac{n}{n-1}(2^n-2)$

The pattern continues: in general for $1 \lt k \lt n$ you have $E_k=1+\frac kn E_{k-1} +\frac{n-1}{n} E_{k+1}$ so can find $E_{k+1}$ from $E_{k}$ and $E_{k-1}$.  And finally for the diametrically opposite corner you will have $E_n=1+E_{n-1}$.


答:
For (i), note that the Markov chain $(X_n)$ is clearly irreducible, so since the state space is finite, it is positive recurrent. So there is a stationary distribution, which by symmetry is just $\pi(v)=1/2^n$ for all vertices $v$. So the expected first return time is the inverse of this, i.e. $2^n$.

Now let $T_k$ be the mean time to hit the origin from a vertex of distance $k$ away, with $T_0=2^n$ as above.

For (ii), we want $T_1$. But clearly $T_0=1+T_1\implies T_1=2^n-1$.

For (iii), we want to find $T_2$. Just note that after two steps, either we are back at the origin (with probability $1/n$) , or we end up a distance $2$ away. So
$$T_0=2+\left(1-\frac{1}{n}\right)\cdot T_2\implies T_2=\frac{n\left(2^n-2\right)}{n-1}$$

Title: Markov chain on an $n$-dimensional hypercube?

Body:
I am looking at the following Cambridge Tripos question, which has left me rather stumped for quite a while.

A particle moves on the $2^n$ vertices of the hypercube $\{0,1\}^n$. At each step the particle is equally likely to move to each of the $n$ adjacent vertices, independently of its past motion. The initial vertex occupied by the particle is $(0,0,...,0)$. Calculate the expected number of steps until the particle

i) first returns to $0,0,...,0)$;

ii) first visits $(0,0,...,1)$;

iii) first visits $(0,0,...,1,1)$.

I have seen multiple answers on here regarding the 3D case, as well as a general $D$-dimensional case presented here, but I am quite stuck on generalising to other vertices, including the three presented here. Would be lovely to see some guidance here, thanks!


问:
Computing $N_{S_n}(S_m)$ with $m<n.$
Let $m<n$ be positive integers. Calculate $N_{S_n}(S_m)$. In particular, find when $N_{S_n}(S_m)=S_m.$

Hi. I don't know how to do this exercise. Some hint? only i have this:

Let $\sigma\in N_{S_n}(S_m)$ then $\sigma \tau\sigma^{-1}\in S_m$ for all $\tau\in S_m$. Now, I thought to use that $S_m=\langle (1\ 2),\ldots, (m-1\, m)\rangle$ but I don't see that it is very useful

答:
Here are a couple of observations to help point you in the right direction.

It’s definitely the case that $N_{S_n}(S_m)\subsetneqq S_m$ when $n\ge m+2$: consider the transposition $(m+1\;m+2)$.

Now suppose that $n=m+1$ and $m>1$. If $\sigma\in S_n$ does not fix $n$, let $k=\sigma(n)$. There is a $\tau\in S_m$ such that $\tau(k)\ne k$, and hence $(\sigma^{-1}\tau\sigma)(n)=\sigma^{-1}(k)\ne n$, so that $\sigma^{-1}\tau\sigma\notin S_m$.

Title: Computing $N_{S_n}(S_m)$ with $m<n.$


问:
Let $\{X_n\}_{n\ge1}$ a sequence of independent random variables with same distribution function $F$.
Let $\{X_n\}_{n\ge1}$ a sequence of independent random variables with same distribution function $F$. Let $x\in\mathbb{R}$ a fixed real number and $Z_i=1_{\{X_i\le x\}}$.

Show that, random variables $Z_i$ are independent and identically distributed and therefore $$F_n(x):=\dfrac{1}{n}\sum_{i=1}^nZ_i\to F(x)$$ almost surely.

Show that, for each $a<b$ $$\dfrac{1}{n}\big|\{i\in\{1,2,\ldots,n\}\mid a< X_i\le b\}\big|\to F(b)-F(a)$$ almost surely.

I have this problem, but I don't have a very clear idea of how to solve it.

答:
Hint for 1) we know that the $Z_i$ have finite moments (actually they are Bernoulli(p) with $p=P(Z_i\leq x)$) so appeal to SLLN.

$$E[Z_i] = p = P(Z_i\leq x)= F(x)$$

Since $Z_i$ are a sequence of iid random variables their sample mean will converge almost surely to their expected value.

Hint for 2) You can recast it in terms of $Z_i(a),Z_i(b)$ and use SLLN


问:
Showing $\lim_{n \to +\infty} \log(n!)/(n\log n) = 1$ without using Stirling approximation
As a passage of a bigger limit I have to show that $$ \lim_{ n \to \infty } \frac{\log(n!)}{n\log(n)} = 1. $$
I think it could be done using Stirling approximation, but I'm wondering if there's a way without that formula.</p>

答:
Using
$$\left( \frac{n}{e}\right)^n \lt n! \lt e \left( \frac{n}{2}\right)^n$$
we have
$$n \log \frac{n}{e} \lt \log n! \lt \log e+ n \log \frac{n}{2}$$

Proof:

For left side first step of induction is clear. Then

$$(n+1)!=n!(n+1) \gt \left( \frac{n}{e}\right)^n (n+1) =$$
$$=\left( \frac{n+1}{e}\right)^{n+1} \frac{(n+1)\left( \frac{n}{e}\right)^n}{\left( \frac{n+1}{e}\right)^{n+1}} \gt \left( \frac{n+1}{e}\right)^{n+1}$$

because $(n+1)\left( \frac{n}{e}\right)^n \left( \frac{n+1}{e}\right)^{-n-1}\gt 1$ is equivalent $\left(1+ \frac{1}{n}\right)^{n} \lt e$.

For right side

$$n! \lt \left(\frac{n+1}{2}\right)^{n} = e\left(\frac{n}{2}\right)^{n} \frac{\left(\frac{n+1}{2}\right)^{n}}{e\left(\frac{n}{2}\right)^{n}} = $$
$$=e\left(\frac{n}{2}\right)^{n} \frac{\left(1+ \frac{1}{n}\right)^{n}}{e} \lt e\left(\frac{n}{2}\right)^{n}$$


答:
if we have $f(x) > 0$ and $f'(x) > 0,$  then
$$ \int_{a-1}^{b} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a}^{b+1} \;  f(x) \; dx   $$

$a=2; b= n ; f: log(n!)$
$$ \int_{1}^{n} \; \log x \;  dx \;  < \; \log n! \; = \; \sum_{j=2}^n \;  \log j \; < \; \int_{2}^{n+1} \;  \log x \; dx   $$
An antiderivative of $\log x$ is $x \log x - x$

$$ 1 - \frac{n-1}{n \log n} < \frac{\log n!}{n \log n} < \frac{(n+1) \log (n+1)}{n \log n} - \frac{n + 2 \log 2 -1}{n \log n}  $$
Some terms evidently go to zero with increasing $n.$

$$  \frac{(n+1) \log (n+1)}{n \log n} = \frac{(n+1) }{n } \frac{ \log (n+1)}{ \log n}   $$ goes to $1$

Title: Showing $\lim_{n \to +\infty} \log(n!)/(n\log n) = 1$ without using Stirling approximation

Body:
As a passage of a bigger limit I have to show that $$ \lim_{ n \to \infty } \frac{\log(n!)}{n\log(n)} = 1. $$
I think it could be done using Stirling approximation, but I'm wondering if there's a way without that formula.</p>


问:
$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series
Find limit without using L'Hospital or Taylor's series:

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = $$

$$= \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{x^5} \cdot \frac{x^2}{5^{x^{2}} - 1} \cdot \frac{ x^3}{\tan^3(3x)}$$

$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$

$\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$

$\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$

$\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$

We get:

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = 1 * 2 * \frac{1}{\ln5} * \frac{1}{27} = \frac{2}{27 \ln5}$$

All credits to @Bernard. Thank you all, question closed.

答:
Just a slight variant on Bernard's answer: rewrite the fraction as

$${\ln(1+\sin(2x^5))\over\sin(2x^5)}\cdot{\sin(2x^5)\over2x^5}\cdot\left(3x\over\sin(3x)\right)^3\cdot{x^2\ln5\over e^{x^2\ln5}-1}\cdot{2\cos^3(3x)\over3^3\ln5}$$

then use the familiar limits

$$\lim_{u\to0}{\ln(1+u)\over u}=\lim_{u\to0}{\sin u\over u}=\lim_{u\to0}{e^u-1\over u}=1$$

to get the limit

$$1\cdot1\cdot1^3\cdot1\cdot{2\cdot1^3\over27\ln5}={2\over27\ln5}$$


问:
How to calculate the Rotation Matrix from other known values?
Dear Math Stackexchange community, how can I compute $R$ rotation matrix in this equation ($P=Rp+t$)? (Please see the figure. More detailed explanation is under the figure.)

The equation is for the transformation of $p=p_1,p_2,p_3$ points. $p$ is in Oxyz (local) reference frame.     $p$ consists of 3 points of $p_1,p_2,p_3$. And $p_1,p_2,p_3$ each has 3 coordinates of $x,y,z$. Therefore p is 3x3 matrix. $p=\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}$

$t$ is the distance between Oxyz(local) and OXYZ(global) reference frame.  $t=\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$. To be able to do matrix operations, we can take $t$ as $t=\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

$P$ is the coordinates of p_1, p_2, and p_3 in OXYZ global reference frame. $P$ is a 3x3 matrix.
$P=\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix}$

$R$ is the (orthogonal) rotation matrix.
$R=\begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

Therefore, all in all:
$P=Rp+t$

$\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix} = \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}.\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}+\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

We can derive the following form of the equation of  $P=Rp+t$

$(P-t)=R.p$

$(P-t).p^{-1}=R$

$p$ is an arbitrary matrix, and not always be taken of its inverse. I can't take the inverse of $p$, if last raw of $p$ is zero.

Let's look at an example to make it clear:
$P=Rp+t$

$p_1=(1,1,0)$ , $p_2=(2,4,0)$ , $p_3=(5,10,0)$ is the coordinates of three points. So in matrix form:
$p=\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}$

$t=\begin{bmatrix} 0 \\ 0 \\ 50 \end{bmatrix}$

$\alpha=10^\circ $ (rotation around the x axis)

$\beta=15^\circ $ (rotation around the y axis)

$\gamma=20^\circ $ (rotation around the z axis)

$R_x=\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha) \end{bmatrix}$

$R_y=\begin{bmatrix} cos(\beta) & 0 & sin(\beta) \\ 0 & 1 & 0 \\ -sin(\beta) & 0 & cos(\beta) \end{bmatrix}$

$R_x=\begin{bmatrix} cos(\gamma) & -sin(\gamma) & 0 \\ sin(\gamma) & cos(\gamma) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$R=R_z.R_y.R_x$

Therefore, for $\alpha=10^\circ $, $\beta=15^\circ $, $\gamma=20^\circ $, the rotation matrix is :

$R=\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$

$P=R.p+t$

$P=$$\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$$.\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$  $+\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

$P=\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}$

$p_1=(0.6131,1.2712,49.90)$,$p_2=(0.6370,4.4239,50.1533)$,$p_3=(1.5925,11.0597,50.3832)$.

Now let's take opposite example with the same values to find $R$, with the values of $P,p,t$ as known.

$P=R.p+t$

$\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}=R.$ $\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}+$ $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

Could you find $R$ here in the above equation?

To repeat the formulation:

$P=Rp+t$

$(P-t)=Rp$

$(P-t).p^{-1}=R.p.p^{-1}$

$(P-t).p^{-1}=R$

For the inverse of $p$ (where $p$ is $\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$ ), I can't take the inverse of $p$.

Here I am stuck. It is interesting for me, because for the condition that $R,p,t$ is known, $P$ can be calculated(9 unknown elements in 3x3 matrix of $P$). However, for the condition that $P,p,t$ is known, I am having problem to compute $R$ (9 unknown elements in 3x3 matrix of $R$). I should be able to compute the equivalent $R$ matrix for that inverse problem.

Do you have any idea about how to find $R$ rotation matrix? Any suggestion would be  appreciated.

Thanks in advance.

答:
Matrices can do a lot more than rotation. I think this may be part of what is making the problem difficult. Indeed to completely determine a general linear transformation in $\mathbb R^3,$ which is what a general real $3\times3$ matrix encodes,
you need to know what happens to at least three linearly independent vectors.
In this more general case, your $p$ matrix must be invertible in order for there to be a unique solution.

Limiting the scope of the transformation to a rotation changes the problem dramatically. You go from $9$ degrees of freedom in a general linear transformation in $\mathbb R^3$ (one degree of freedom for each entry in a $3\times3$ matrix) to just $3$ degrees of freedom.

Because of the limited degrees of freedom of a rotation, you do not need as much information. It is sufficient to know what happens to just two vectors that are not linearly dependent. (That is, they cannot be collinear.) Even that will give you a lot of redundant information, because we already know that the lengths of the vectors will not change and the angle between them will not change.

I'm not sure how to exploit this extra knowledge in an algorithm that relies purely on matrix arithmetic. But there are certainly ways to extract the necessary information from the two matrices
$$
\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\
 P_{1y} & P_{2y} & P_{3y} \\
 P_{1z} & P_{2z} & P_{3z} \end{bmatrix}
- \begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\
 p_{1y} & p_{2y} & p_{3y} \\
 p_{1z} & p_{2z} & p_{3z} \end{bmatrix}.$$
In fact you should only need the first two columns of each matrix.
The answer to Calculating rotation for a pair of unit vectors given initial and final states uses the two "before rotation" vectors to generate an orthogonal basis, and likewise with the two "after rotation" vectors,
and computes the rotation from one basis to the other
(for which the matrix inversion method will work, since the three basis vectors are linearly independent).
Or you could consider G Cab's answer, which starts with just a little three-dimensional vector arithmetic to set up a matrix-arithmetic solution.


问:
need help simplifying this radical $\sqrt[35]{128y^{42}}$
I am trying to figure out how to get to the solution below but have having difficulty. Can someone explain how to get to the solution.

$$\sqrt[35]{128y^{42}}$$

This is the answer but I can't figure out how to get it.

$$2^{\frac15}y^{\frac65}$$

Body:
Note that $((2y^6)^7)^{1/35}=(2y^6)^{1/5}$.

Now can you figure it out?


Title: need help simplifying this radical $\sqrt[35]{128y^{42}}$

Body:
I am trying to figure out how to get to the solution below but have having difficulty. Can someone explain how to get to the solution.

$$\sqrt[35]{128y^{42}}$$

This is the answer but I can't figure out how to get it.

$$2^{\frac15}y^{\frac65}$$

Body:
$$\sqrt[35]{128y^{42}}$$
$$=(128y^{42})^{\frac{1}{35}}$$
$$=(2^7 y^{7 \times6})^{\frac{1}{5 \times 7}}$$
$$=2^{\frac{7}{7 \times5}} y^{\frac{7 \times6}{7 \times5}}$$
$$=2^{\frac{1}{5}}y^{\frac{6}{5}}$$


答:
You have
$$S=X_i - \frac{1}{n}\sum_{j=1}^nX_j = (\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T \mathbf{X}$$
where $\mathbf{e}_i\in \mathbb{R}^n$ the i-th standard basis vector, $\mathbf{1}_n \in \mathbb{R}^n$ the all ones vector  and $\mathbf{X}$ the vector of $X_i$ which follows the multivariate normal distribution  $N_n(\mu \mathbf{1}_n, \Sigma.\mathbb{I}_n)$  ($\mathbb{I}_n \in \mathbb{R}^{n\times n}$ the identity matrix)

The term $S$ is so equal in distribution to
$$
\begin{align}
S &=(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T \mathbf{X} \\
&=(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T N_n(\mu \mathbf{1}_n, \Sigma.\mathbb{I}_n) \\
&=N(\mu(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T \mathbf{1}_n,\Sigma(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n))
\end{align}
$$

In other words, $S$ follows a univariate normal distribution $N(0,(1-\frac{1}{n})\Sigma)$ of mean
$$\mu(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T\mathbf{1}_n=\mu(1-\frac{1}{n} -(n-1)\frac{1}{n})=0$$
and variance
$$\Sigma.(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n)^T(\mathbf{e}_i - \frac{1}{n}\mathbf{1}_n) =\Sigma.((1-\frac{1}{n})^2+(n-1)\frac{1}{n^2}) =(1-\frac{1}{n})\Sigma $$

Body:
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu,\Sigma)$

What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$

My attempt is:

We know that distribution of $\sum^n_{i=1}\frac{1}{n}X_i\sim N_p(\mu,\frac{1}{n}\Sigma)$

So, $\mathbb{E}[X_i-\frac{1}{n}\sum^n_{i=1}X_i]=0$

And variance is $\text{Var}(X_i-\frac{1}{n}\sum^n_{i=1}X_i)=\Sigma-\frac{1}{n}\Sigma=\frac{n-1}{n}\Sigma$

Would that be correct?

edit:

suppose the first $X_i$ in $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ is $X_j$, $j\neq i$,so $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ becomes $X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j$

So,

$$\text{Var}(X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j)=\text{Var}(-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i+\frac{n-1}{n}X_j)=\frac{n-1}{n^2}\Sigma+\frac{(n-1)^2}{n^2}\Sigma=\frac{n-1}{n}\Sigma$$


Body:
$$\sqrt[35]{128 y^{42}} \iff {(128 y^{42})}^{\frac{1}{35}}$$
$$\iff {(128)}^{\frac{1}{35}} \times y^{\frac{42}{35}}$$
This simplifies to$${(2)}^{\frac{7}{35}} \times y^{\frac{6}{5}}$$
Finally, you get $$2^{\frac{1}{5}} y^{\frac{6}{5}}$$ as the answer.
$$ \therefore \sqrt[35]{128 y^{42}} = 2^{\frac{1}{5}} y^{\frac{6}{5}}$$


问:
About restriction of linear map
Let $A$ be a linear map from the vector space $X$ to $Y$ and $T$ be a subspace of $X$ . I want to understand what is the meaning of saying that the restriction $A_{|T}: \to A(T)$ is invertible. Could anybody explain me what is this restriction map?

Thanks for the help.

答:
One way of thinking about it is that if $t$ and $t'$ are in $T$, then $A(t)=A(t')$ implies that $t=t'$.

The map $A_{|T}$ is only defined on $T$, but everywhere on $T$, it goes the same thing as $A$. If $t\in T$, then $A_{|T}(t)=A(t)$.

Here is a familiar example, though not in a linear algebra setting. Let $f(x)=x^2$, for the set $A$ of all real numbers. Then $f(x)$ is not invertible, because for example (-2)^2=2^2$.

But if we restrict $x$ to the set $T$ of non-negative reals, the restriction of $f$ to $T$, which we could call $f_{|T}$, is invertible. If for example we know that  $f_{|T}(x)=9$, then we know that $x=3$.

Title: About restriction of linear map


问:
Prove an inequality involving the absolute value of the spatial derivative of the solution to the heat equation.
Given the initial value problem $$u_t=ku_{xx}$$ $x \in \mathbb{R}$, $t>0$ $$u(x,0)=f(x)$$ $x \in \mathbb{R}$ where $f$ is some square integrable function in $\mathbb{R}$, prove there exist constant $c$ such that $$|u_x| \leq c t^{-3/4} \| f(x) \|$$

My attempt, consider the solution to the heat equation $$u(x,t)=\frac{1}{(4k \pi t)^{1/2}} \int_{- \infty }^{\infty } e^{-\frac{(x-y)^2}{4kt}} f(y)dy$$
If I differentiate it I get $$u_x=\frac{1}{(4k \pi t)^{1/2}} \int_{- \infty }^{\infty } e^{-\frac{(x-y)^2}{4kt}} \left( \frac{-2(x-y)}{4kt} \right) f(y)dy$$
Now $$|u_x|=\left|\frac{1}{(4k \pi t)^{1/2}} \int_{- \infty }^{\infty } e^{-\frac{(x-y)^2}{4kt}} \left( \frac{-2(x-y)}{4kt} \right) f(y)dy \right| \leq$$
$$\left( \frac{1}{(4k \pi t)^{1/2}} \int_{- \infty }^{\infty } e^{-\frac{(x-y)^2}{2kt}} \left( \frac{-2(-1)(x-y)}{2kt} \right) \frac{1}{4kt} \right)^{1/2} \| f(x) \|$$
I compute the integral using the error function by letting $z=\frac{-(x-y)^2}{2kt}$ I get that $$|u_x| \leq c t^{-3/2} \| f(x) \|$$ for $$c=\frac{1}{4k (4k \pi )^{1/2}} \left( \frac{\sqrt{ \pi } }{2} \right)^{1/2}$$
But this is wrong since I got a $t^{-3/2}$ factor instead of the desired $t^{-3/4}$.
I spend a lot of time trying to figure out my mistake but couldn't find it.
Can you help me spot my mistake?

答:
You're on the right track.
\begin{equation}
u_x(x, t) = -\frac{1}{(4\pi kt)^{1/2}} \int_{-\infty}^{\infty}\frac{1}{2kt}(x - y)
e^{-(x - y)^2/4kt}\, f(y)\, dy.
\end{equation}
By the Cauchy-Schwarz inequality
\begin{align}
\vert u_x(x, t) \vert
&= \frac{1}{(4\pi kt)^{1/2}} \frac{1}{2kt} \left| \int_{-\infty}^{\infty}(x - y)
e^{-(x - y)^2/4kt}\, f(y)\, dy \right| \\
&\leq \frac{1}{(4\pi kt)^{1/2}} \frac{1}{2kt} \left( \int_{-\infty}^{\infty}(x - y)^2 e^{-(x - y)^2/2kt}\, dy \right)^{1/2} \left( \int_{-\infty}^\infty f^2(y) \, dy\right)^{1/2} \\
&= \frac{1}{(4\pi kt)^{1/2}} \frac{1}{2kt} \left( \int_{-\infty}^\infty w^2 e^{-w^2/2kt}\, dw\right)^{1/2}\left( \int_{-\infty}^\infty f^2(y) \, dy\right)^{1/2}.
\end{align}
In the integral over $w$ make the change of variable $v = w/\sqrt{kt}$. Then you're left with
\begin{equation}
\int_{-\infty}^\infty w^2 e^{-w^2/2kt}\, dw = (kt)^{3/2}\int_{-\infty}^\infty v^2 e^{-v^2/2}\, dv.
\end{equation}
Keeping in mind you need the square root of this last relation, start collecting up the powers of $t$ to see that $\vert u_x(x, t) \leq C t^{-3/4} \vert f \vert_{L^2}$.

Revisit how you applied Cauchy-Schwarz. If nothing else you ought to have ended up with a $(x - y)^2$ in the mix.


问:
$\varphi(d)$ elements of order $d$ in $\mathbb{U}_n$ (defined in the question)

$$\mathbb{U}_{n\in\mathbb N^*}=\left\{\exp\left(i\frac{2\pi k}{n}\right),\,0\leq k\leq n-1\right\}$$
$$\varphi(d)=\text{card}\{k\in[[1,d]],\,\text{gcd}(k,d)=1\}$$
If $z\in\mathbb U_n$, we call order of $z$ the smallest integer $d\geq1$ such that $z^d=1$.

Let $d\in[[1,n-1]]$ such that $d|n$ (say $n=dc$ for some $c\in\mathbb N$).

Show that there are exactly $\varphi(d)$ elements of order $d$ in $\mathbb{U}_n$.

I'm going to use $a\wedge b\,$ for the gcd.

$[[a,b]]$ is an integer interval.

Let $z\in\mathbb U_n$ and let it be of order $d$, then
$$z^d=\exp\left[\left(i\frac{2\pi k}{n}\right)d\right]=\exp\left(i\frac{2\pi k}{c}\right)=1$$
Since $z$ is of order $d$, $k=k'c=k'\frac nd$ for some $k'\in\mathbb N$ (or else $c$ doesn't divide $k$).

Claim: For the above to hold, we require that $k'\wedge d=1$.

Proof: Suppose that $k'\wedge d>1$, then $z=\exp\left(i\frac{2\pi k}{n}\right)=\exp\left(i\frac{2\pi k'}{d}\right)$.

Because $k'\wedge d>1$, we get $z=\exp\left(i\frac{2\pi a}{b}\right)$ where $b<d$, thus the order of $z$ is at worst $b$, and could be less if $a\wedge b>1$, which goes against the first hypothesis that the order of $z$ is $d$.

In the above, I have put $k'=a(k'\wedge d)$ and $d=b(k'\wedge d)$ for some $a,\,b\in\mathbb N$.

There are $\varphi(d)$ of such $k'$, therefore:

there are exactly $\varphi(d)$ elements of order $d$ in $\mathbb{U}_n$

If anything is amiss, then please let me know. Thank you for your time!

答:
In short, this is true because $U_n\cong\Bbb Z_n$.

And a basic fact about cyclic groups is that $|g|=d\implies |g^k|=\dfrac d{(d,k)}$.  It follows by taking a generator that there's a unique cyclic subgroup of order $d$ for every $d$ dividing $n$, and that there are $\varphi(d)$ generators for that subgroup.


答:
Just a thought:

Let $w(x)$ denote the weight of $x\in\{0,1\}^n$, i.e., $w(x)=\sum x_j$. There is no one-step path between two $x$'s with the same weight; taking each step, the weight changes by $+1$ or $-1$. Take the weight as the state, so, there exist $n+1$ states associated with the weights. Let $p_{i}^+$ denote the probability of transition from state $i$ to state $i+1$, and $p_i^-$ denote the transition from state $i$ to $i-1$, then $p_i^-=\frac{i}{n}$, and $p_{i}^+=1-\frac{i}{n}$.

At each state, due to the symmetry of hyper-cube, members are equiprobable, i.e., the probability of visiting a specific member of $i^{th}$ state, given you are at the $i^{th}$ state, is$\dfrac{1}{{n\choose i}}$.

I suspect that it builds a simple Markov model to answer the question.


问:
Prove that $\lim_{n\to\infty} \left(\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right) =2$

Prove that $$\left(\displaystyle\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right)_{n \in \mathbb{N}}\to 2$$

My idea was to find some $N$ to do the definition job of convergence.

\begin{align}\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right |&=\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+8n-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+2}{4n - 1} \right |\\
&\leq \left |\dfrac{\sqrt{n}+2}{n}\right |\\
&\leq \left |\dfrac{\sqrt{n}}{n}\right |+ \left |\dfrac{2}{n}\right |\\
&\leq \dfrac{1}{\sqrt{n}}+2\\
&\leq 3
\end{align}

so I concluded that the sequence is bounded, so my idea was trying to show that this sequence is monotonous, so I could conclude that it converges. Taking $n = 4k$ I get a subsequence that converges to $2$, so if I knew that the sequence is convergent, the problem would be over.

答:
With what you've done, you don't need to show the sequence is monotonous. To show this converges, we want to say that for all $\varepsilon > 0$ there exists an $N$ such that for all $n>N$
$$\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right | < \varepsilon$$
And if we stop before your last step, we get
$$\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right | \leq \frac{1}{\sqrt{n}}+\frac{2}{n}$$
Then, pick $N$. one choice might be $N=\lceil 4/\varepsilon^2\rceil$, in which case (for all $\varepsilon <1$)
$$\left |\dfrac{\sqrt{N}(1 + 8\sqrt{N})}{4n - 1}-2 \right | \leq \frac{1}{\sqrt{N}}+\frac{2}{N} \leq \frac{\varepsilon}{2}+\frac{\varepsilon^2}{2}< \varepsilon$$
And since $1/\sqrt{n}+2/n$ is decreasing in $n$, for any fixed $\varepsilon>0$, we have chosen an $N$ such that for all $n>N$
$$\left |\dfrac{\sqrt{N}(1 + 8\sqrt{N})}{4n - 1}-2 \right | < \varepsilon$$
And that is precisely the definition of a convergent sequence.


答:
This applies to a function between sets in general, not only vector spaces.

Given $f:X \rightarrow Y$ and a subset $T \subset X$, the restriction $f_{|T}:T \rightarrow Y$ is by definition $f_{|T}(x) = f(x)$. That is, the function is the same, except that its domain is restricted.

If $f$ is a linear map (or any kind of homomorphism), its restriction will also be one by inheritance.

Title: About restriction of linear map

Body:
Let $A$ be a linear map from the vector space $X$ to $Y$ and $T$ be a subspace of $X$ . I want to understand what is the meaning of saying that the restriction $A_{|T}: \to A(T)$ is invertible. Could anybody explain me what is this restriction map?

Thanks for the help.


问:
Prove that $\displaystyle{\lim_{x \to a}} \sin x = \sin a$ using epsilon-delta definition
I've been struggling with this question. I don't know how to get rid of the sine function after getting to
$$|\sin x - \sin a| \leq 2 \left| \sin \dfrac{x - a}{2} \right|$$
Half angle formula isn't really useful imo, not sure what to do from here.

答:
Thinking about an identity was a good idea. However, use $\sin(\alpha)-\sin(\beta)=2\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})$.

We want to chek that $\forall \epsilon>0$, $\exists$ $\delta>0$ such that:

$|x-c|<\delta \implies |\sin(x)-\sin(c)|<\epsilon$.

Let $\epsilon=\delta$, then $|\sin(x)-\sin(c)|=2\cos(\frac{x+c}{2})\sin(\frac{x-c}{2})\leq2|\frac{x-c}{2}|=|x-c|<\delta=\epsilon$.

We can justify this last step with the following:

$|\cos(x)|\leq 1 ~\forall x\in\mathbb{R}$

$|\sin(x)|<|x| ~\forall x\in\mathbb{R}$


问:
Hölder estimation for weak solution of Laplace equation
Let $B_{R_0}(x_0)=\{y\in \mathbb R^n :  |y-x_0|\le R_0\}$, $u\in H^1_{loc}(\mathbb R^n)$. $\alpha\in (0,1)$.

If for any $0<R\le R_0$, we have
$$
\sup_{B_{R}~(x_0)} u -\inf_{B_{R}~(x_0)} u \le C_1 R^\alpha
$$
where $C_1$ is a positive constant, then how to show there is positive constant $C_2>0$ and $\theta\in (0,1)$   such that
$$
[u]_{\alpha, B_{\theta R_{~0}}(x_0)} \le C_2
$$
where $[\cdot]_{\alpha, B_{\theta R_{~0}}(x_0)} $ is Holder seminorm.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For makeing the problem more complete, I add the proof which the problem origin from. In fact, I have two problem in this proof. Firstly, how to get (1) by Lemma 5.1.5 ?  Second, why (1) means (2) ?

Lemma 5.1.5:  $\omega$ is nonnegative undecreasing function on $[0, R_0]$. If exiting $0<\theta, \eta<1, 0<\gamma\le 1, K\ge 0$ such that
$$
\omega(\theta R) \le \eta \omega(R)+KR^\gamma, ~~~~~~0<R\le R_0,
$$
then, there are constants $\alpha\in (0, \gamma), C>0$ depending on $\theta,\eta,\gamma$ such that
$$
\omega(R)\le C(\frac{R}{R_0})^\alpha(\omega(R_0)+ KR_0^\gamma), ~~~~~~0<R\le R_0.
$$

Theorem 5.1.6: $u\in H_{loc}^1(R^n)$ is bounded weak solution of
$$
-\Delta u=0
$$
then, there is $\alpha\in (0,1)$ such that for any bounded domain $\Omega\subset R^n$, there is
$$
[u]_{~\alpha;~\Omega} \le C
$$
where $C$ depending on $n, \Omega$.

Proof of Theorem 5.1.6:   For any fixed $x^0\in R^n$，$R>0$, let
$$
m(R)=\inf_{B_R} u,~~~ M(R)=\sup_{B_R} u
$$
where $B_R=B_R(x^0)$ is ball. Let
$$
v(x)= u(x)- m(R),~~~ \omega(x)= M(R)-u(x).
$$
Then $v,\omega\in H^1_{loc}(R^n)$  are nonnegative bounded. And in the sense of weak, we have
$$
-\Delta v = -\Delta \omega =0.
$$
By the Harnack inequation, we have
$$
\sup_{B_{R/2}} v\le C\inf_{B_{R/2}} v,~~~~~~\sup_{B_{R/2}} \omega\le C\inf_{B_{R/2}} \omega
$$
Namely,
$$
M(R/2)-m(R)\le C(m(R/2)-m(R))\\
M(R)-m(R/2)\le C(M(R)-M(R/2))
$$
we can assume $C>1$. Then, by the above two inequation, we have
$$
M(R/2)-m(R/2)\le \frac{C-1}{C+1}(M(R)-m(R))
$$
Letting $f(R)=M(R)-m(R), \eta=\frac{C-1}{C+1}$, $f$ is nonnegative nondecreasing, and
$$
f(R/2)\le \eta f(R).
$$
By  Lemma 5.1.5, there is $\alpha\in(0,1)$ such that
$$
f(R)\le CR^\alpha        \tag{1}
$$
namely
$$
[u]_{\alpha; B_{R}(x^0)} \le C    \tag{2}
$$
Finally, just using finite covering for $\overline \Omega$.

答:
There is a bit of an abuse of notation in the text you quoted. They're proving that for all $x_0\in \mathbb{R}^n$ and all $R>0$ it holds,
$$
f(R/2)\leq \theta f(R),
$$
where $f(R)= M(R)-m(R)$ and $\eta=(C-1)/(C+1)$. Now fix any $R_0>0$ and apply the Lemma with $f=\omega$, $\theta=1/2$, and $K=0$ to obtain
$$
f(R)\leq C\left( \dfrac{R}{R_0}\right)^\alpha f(R_0), \qquad 0<R<R_0,
$$
for some $\alpha>0$.

From here tha Hölder continuity follows: Fix $x_0 \in \mathbb{R}^n$ and $R_0>0$. Let $x,y\in B_{R_0/2}(x_0)$. Without loss of generality $u(x)>u(y)$ and we have two cases.

Case 1. $|x-y|>R_0/4$. Here we simply estimate
$$
\begin{split}
|u(x)-u(y)| &= u(x)-u(y)\leq \sup_{B_{R_0/2}(x_0)} u - \inf_{B_{R_0/2}(x_0)}u = f(R_0/2) \\ &\leq C\left(\dfrac{R_0/2}{R_0}\right)^\alpha f(R_0) \\  & \leq \dfrac{Cf(R_0)}{R_0^\alpha} |x-y|^\alpha.
\end{split}
$$
This gives the result in this case with $C_2= Cf(R_0)/R_0^\alpha$.

Case 2. $|x-y|\leq R_0/4$. Here note that, if $s=|x-y|$, then $B_s(x)\subset B_{R_0}(x_0)$ and so, estimating as before,
$$
|u(x)-u(y)| \leq \sup_{B_s(x)} u - \inf_{B_s(x)} u = f(s, x) \leq C\left( \dfrac{s}{R_0/2}\right)^\alpha f(R_0/2, x),
$$
where $f(s, x)$ is the same as $f$ but with the balls now centered at $x$ instead of $x_0$ (this is why it's important that we proved the estimate for all $x_0$, in particular it holds for $x$). Since $f(R_0/2, x)\leq f(R_0)$ we arrive at
$$
|u(x)-u(y)|\leq \dfrac{2^\alpha C f(R_0)}{R_0^\alpha} |x-y|^\alpha.
$$
This is the claim with $C_2= 2^\alpha Cf(R_0)/R_0^\alpha$.

Remark. Notice that the appearance of $f(R_0)/R_0^\alpha$ is natural, simply because of the way the quantity $[u]_\alpha$ behaves under the scaling $u(x)\mapsto u(\lambda x)$.


问:
Help with $\cos(z)=3/2$
We were doing solutions to trigonometric equations and pre-calculus and I stumbled across a problem that had no real solutions but I inquired about the complex ones. Please tell me if I have made any mistakes, my own answer differs from that of Wolframalpha's.

$$\cos(z)=\frac32 \\
\frac{e^{iz}+e^{-iz}}{2}=\frac32 \\
e^{iz}+e^{-iz}=3 \\
e^{2iz}-3e^{iz}+1=0 \\
Z=e^{iz} \\
Z^2-3Z+1=0 \\
Z=\frac{3\pm\sqrt{3^2-4(1)(1)}}{2} \\
e^{iz}=\frac{3\pm\sqrt{5}}{2} \\
z=-i\ln(3\pm\sqrt 5)+i\ln(2)$$

答:
More directly,
$$\cos z = \cos(z-2\pi n)= \cosh (-i (z+2\pi n))=\frac32
$$
which leads to $-i (z-2\pi n)=\pm \cosh^{-1} \frac32$ and the solutions
$$z= 2\pi n \pm i \cosh^{-1} \frac32 =  2\pi n  \pm 2 i\ln \frac{\sqrt5+1}2
$$

Title: Help with $\cos(z)=3/2$


问:
Base Conversion: Question on common algorithm using Euclidean division
I can see that this algorithm converts bases, however have questions on whats actually going on.  The algorithm is presented, then questions follow.

Algorithm

To convert a number $X$ to base $B$.

$k = 1$

$X \bmod B = R_k$,
$Q = \frac {X - R_k} B$

where $\bmod$ is defined:
$q \in \mathbb{Z}, r = X - Bq, |r| < |B|$

$(Q = 0) \implies$ Goto #3

$(Q \ne 0) \implies (X = Q) \land (k = k + 1) \land $ Goto #1

Converted string is $R_n, ..., R_1$

What is happening mathematically that $R_k$, the remainders, correspond to the converted numbers digits?  What significance is there to dividing by the new quotient $X = Q$ each iteration?

I have seen this algorithm a few times, however have yet to see a detailed mathematical explanation on why it works.

Body:
${a_ma_{m-1}a_{m-1}a_{m-3}....a_3a_2a_1a_0}_b = a_mb^m + a_{m-1}b^{m-1} + a_{m-2}b^{m-2}+a_{m-3}b^{m-3}+......+a_3b^3 + a_2b^2 + a_1b + a_0=$

$b(a_mb^{m-1} + a_{m-1}b^{m-2}+ a_{m-2}b^{m-3}+........ + a_3b^2 + a_2b + a_1) + a_0 =$

$b^(b(a_mb^{m-2} + a_{m-1}b^{m-3} + a_{m-3}b^{m-4}+.......+ a_3b + a_2) + a_1) + a_0 = $

$b(b(b(a_mb^{m-1} + a_{m-1}b^{m-4} + a_{m-3}b^{m-5}+.......+ a_3) + a_2) + a_1) + a_0 =$

$.....$

$b(b(....(a_mb^2 + a_{m-1}b + a_{m-2}) + a_{m-3} + ....... + a_3) + a_2) + a_1) + a_0=$

$b(b(b(....(a_mb + a_{m-1}) + a_{m-2}) + a_{m-3} + ....... + a_3) + a_2) + a_1) + a_0=$

So each of the $a_k$ digit is the result of taking a remainder when dividing by $b$.


问:
What are some tips for finding the infimum and supremum when bounds are not given explicitly?
I have homework regarding finding the infimum and supremum of a set, but the examples from class and examples I can find online all describe the elements of the set as being bounded between values. What are some tips for finding the infimum and supremum when bounds are not given explicitly? For example, this is my homework problem:

Find the infimum and supremum of A={ $\frac{m+2n}{2m+n}$ : m,n $\in$ $\mathbb{N}$]. Since m and n are natural numbers, I know they have to be positive whole numbers (greater than zero), but I can't figure out where to go from there in order to find the inf and sup.

Any advice helps! Thank you :)

答:
Play with the expression, to write it in a simpler form. For example, divide top and bottom by by $n$, to obtain the equivalent expression
$$
\frac{(\frac mn) +2}{2(\frac mn) +1}$$
This tells you that the expression is a function of the ratio $m/n$. Now you don't need to worry about varying $m$ and $n$ separately.

If you now write
$$f(x):=\frac{x+2}{2x+1}$$
then your set $A$ is the set of all $f(x)$ where $x$ is positive and rational.

Since this is a question about inf and sup for set $A$, you are looking for the global min and max for $f(x)$ as $x$ varies over positive rational numbers. This should remind you of using calculus.

The twist here is calculus will optimize $f(x)$ over all real $x>0$. You can still use calculus to determine the largest and smallest possible values for $f(x)$ for real $x>0$, but you need to show that these largest/smallest values can be also approached through rational $x$.


问:
Experiment and Sample Space
" From a bin of 20 apples, a sample of 4 is chosen. The number of rotten apples in the sample is counted. What is the sample space ? "

The above question seems very simple but I am confused about finding the actual Experiment and the Sample Space involved in this question.

This is what I know :

Experiment : Any activity that leads to an observable outcome.

Sample Space : The set of all possible outcomes of an experiment.

Event : A subset(desired outcomes) of the sample space.

But, I am not sure if the sample space in the given question consists of just 4 elements( a sample) or it is a set of all possible 4-element samples from the 20 apples.

Also, What exactly is the 'Experiment' in the question ? Is it picking ' a sample of 4 apples ' ?

The only thing I am certain about is the Event. Here, the event is ' Counting the number of rotten apples in a sample '.

Could someone please explain to me about the exact 'Experiment' and 'Sample Space' of the given question ?

答:
You are correct about the “experiment” — it is drawing 4 apples (without replacement I assume).

Given this experiment, the sample space is the set of possible outcomes.

E.g., $GGGG, RGGG, RRGG, ...$

Title: Experiment and Sample Space


问:
What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu,\Sigma)$

What would be the distribution of $(X_i-\frac{1}{n}\sum^n_{i=1}X_i)?$

My attempt is:

We know that distribution of $\sum^n_{i=1}\frac{1}{n}X_i\sim N_p(\mu,\frac{1}{n}\Sigma)$

So, $\mathbb{E}[X_i-\frac{1}{n}\sum^n_{i=1}X_i]=0$

And variance is $\text{Var}(X_i-\frac{1}{n}\sum^n_{i=1}X_i)=\Sigma-\frac{1}{n}\Sigma=\frac{n-1}{n}\Sigma$

Would that be correct?

edit:

suppose the first $X_i$ in $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ is $X_j$, $j\neq i$,so $X_i-\frac{1}{n}\sum^n_{i=1}X_i$ becomes $X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j$

So,

$$\text{Var}(X_j-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i-\frac{1}{n}X_j)=\text{Var}(-\frac{1}{n}\sum^n_{i=1,i\neq j}X_i+\frac{n-1}{n}X_j)=\frac{n-1}{n^2}\Sigma+\frac{(n-1)^2}{n^2}\Sigma=\frac{n-1}{n}\Sigma$$

答:
And variance is $\operatorname{Var}(X_i-\frac{1}{n} \sum^n_{i=1} X_i)=\Sigma-\frac{1}{n}\Sigma=\frac{n-1}{n}\Sigma$

Would that be correct?

Be careful: The difference is between random variables that are correlated. You could say the variance is the sum of the two variances minus the covariances:
\begin{align}
& \operatorname{var}\left( X_i - \frac 1 n \sum_{j=1}^n X_j \right) \\[8pt]
= {} & \operatorname{var}(X_i) + \frac 1 {n^2} \operatorname{var}\left( \sum_{j=1}^n X_j \right) \\[8pt]
& \qquad {} - \operatorname{cov}\left( X_i,\,\, \frac 1 n \sum_{j=1}^n X_j \right) - \operatorname{cov}\left( \frac 1 n \sum_{j=1}^n X_j , \,\, X_i \right).
\end{align}
The covariance between random vectors $U\in\mathbb R^{k\times1},\,V\in\mathbb R^{\ell\times1}$ with respective expectations $\mu,\nu$ is
$$
\operatorname{cov}(U,V) = \operatorname E\Big( (U-\mu)(V-\nu)^\top \Big) \in \mathbb R^{k\times\ell}.
$$
Corollary:
$$
\operatorname{cov}(V,U) = \Big( \operatorname{cov}(U,V)\Big)^\top.
$$

I would argue as follows. We have
$$
X_1,\ldots,X_n \sim \text{i.i.d.} \operatorname N_p(\mu,\Sigma)
$$
We seek the distribution of
$$
X_i-\frac 1 n \sum^n_{j=1} X_j.
$$
(Notice that on the line above I distinguish between $i$ and $j.$)
\begin{align}
& X_i-\frac 1 n \sum^n_{j=1} X_j \\[8pt]
= {} & {-\frac{X_1} n} - \frac{X_2} n - \cdots - \frac{X_{i-1}} n + \left( 1 - \frac 1 n \right) X_i \\[8pt]
& {} \qquad {} - \frac{X_{i+1}} n - \cdots - \frac{X_n} n.
\end{align}
The terms in this sum are independent, so the variance is
\begin{align}
& \frac\Sigma{n^2} + \cdots + \frac\Sigma{n^2} + \left( 1 - \frac 1 n \right)^2 \Sigma + \frac\Sigma{n^2} + \cdots + \frac\Sigma{n^2} \\[10pt]
= {} & \left( \frac{n-1}{n^2} + \frac{(n-1)^2}{n^2} \right) \Sigma = \frac{n-1} n \Sigma.
\end{align}
$$
\text{So it's } \operatorname N_p\left(0, \frac{n-1} n \Sigma\right).
$$


问:
Looking for an alternative textbook for Gamelin's Complex Analysis
I am currently using Gamelin's textbook for studying Complex Analysis. (I just finished first seven chapter, I'm still a beginner on this subject.)

This is definitely a nice book, it covers everything I need, its every comprehensive.

However, Gamelin's writing style is really killing me.

For example, how he uses the language of differential form to prove Cauchy's theorem,
how he talks about homotopy without using the word homotopy.

There are also a lot of minor problems in the text.
such as he defines analyticity for a function on a set, but later he uses analyticity for a function on a point.

So I want to find a textbook that is about the level of Gamelin, could be deeper, on Complex Analysis, that is more rigorous or formal than Gamelin's book.
(I really enjoy the style of Rudin, Munkres, and Hungerford)
(Also I have enough background from analysis and topology to discuss things like forms and homotopy.)

I heard that Ahlfors is a great book, but it is also really old. I'm worried about the notation maybe a little bit outdated.

答:
Serge Lang wrote more books than just about any one I know of.  He was pretty good at it.  His Algebra is very highly regarded, for one example.

Professor Hung-Hsi Wu chose Lang's Complex Analysis as the text when I took $185$ up at Berkeley.  I believe it was a very good choice, and enjoyed it alot. I dare say it's a beautiful treatment of a beautiful subject (but then so was Wu's teaching).  Maybe have a look.

Btw I sat in on Gamelin's Complex Analysis course as a grad student at UCLA, and, while I may be mistaken, it seems to me that he used Ahlfors' as the text, rather than his own book, for what it's worth.


问:
How determine the distribution of $S_{n-1}^{2}$?
If $ X_1,\dots,X_n $ is a random sample with $X_i \sim N(\mu,\sigma^2) $, how determinate the distribution of $S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 $?

I tried to apply the method of moments as follow:

$S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 = \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 $

$$ M_{S_{n-1}^2 }(t) = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 ) } ] =  E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{t(- \frac{n}{n-1} \bar{X}^2 ) } ] = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] $$
$$= E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] = \prod_{i=1}^n E[ e^{t ( \frac{1}{n-1} X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ]$$

But in this point I'm lose. Should I apply the following?

$$ X_i \sim N( \mu,\sigma^2 ) \Rightarrow X_i = \sigma Z + \mu  $$
$$ Y = (\sigma Z+ \mu )^2 \Rightarrow M_Y(t) = \frac{1}{\sqrt{1-2t\sigma^2}} \exp\left(  \dfrac{\mu^2t}{1-2t\sigma^2}\right) , \ \text{if } \ 2\operatorname{Re}(t\sigma^2) < 1$$

Or isn't this the simplest way to determinate $S_{n-1}^2$?

答:
The term $S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 $ is called sample variance and its distribution is
$$S_{n-1}^2 = \frac{\sigma^2}{n-1} \chi^2(n-1)$$
where $\chi^2(n-1)$ the chi squared distribution of $n-1$ degrees of freedom. One of the proofs can be found here: https://online.stat.psu.edu/stat414/lesson/26/26.3


答: Use the mean value theorem.  That is, $|\sin(x) - \sin(a)| \leq |\cos(\psi)||x-a|$ for some $\psi \in (x, a)$.  I will let you finish the proof.

Title: Prove that $\displaystyle{\lim_{x \to a}} \sin x = \sin a$ using epsilon-delta definition

Body:
I've been struggling with this question. I don't know how to get rid of the sine function after getting to
$$|\sin x - \sin a| \leq 2 \left| \sin \dfrac{x - a}{2} \right|$$
Half angle formula isn't really useful imo, not sure what to do from here.


答: Ahlfors book may be old but it is one of the best books on the subject and I highly recommend that you struggle through it.  If you are looking for a more modern book then I suggest Robert Greene's Function Theory of One Complex Variable.</p>

Title: Looking for an alternative textbook for Gamelin's Complex Analysis

Body:
I am currently using Gamelin's textbook for studying Complex Analysis. (I just finished first seven chapter, I'm still a beginner on this subject.)

This is definitely a nice book, it covers everything I need, its every comprehensive.

However, Gamelin's writing style is really killing me.

For example, how he uses the language of differential form to prove Cauchy's theorem,
how he talks about homotopy without using the word homotopy.

There are also a lot of minor problems in the text.
such as he defines analyticity for a function on a set, but later he uses analyticity for a function on a point.

So I want to find a textbook that is about the level of Gamelin, could be deeper, on Complex Analysis, that is more rigorous or formal than Gamelin's book.
(I really enjoy the style of Rudin, Munkres, and Hungerford)
(Also I have enough background from analysis and topology to discuss things like forms and homotopy.)

I heard that Ahlfors is a great book, but it is also really old. I'm worried about the notation maybe a little bit outdated.


答:
When we restrict a function to a subset of the domain it means we are interested now in what the function is doing to that subset. We don't care what $A(x)$ is when $x\notin T$.

To say that it is invertible is to say that even if $A$ itself is not invertible as a function, when we only consider what $A$ does to the elements of $T$, we can invert the function.

Namely, there exists $B\colon A(T)\to T$ such that $B(A(t))=t$ for every $t\in T$. This condition may be falsified when we consider $A$'s action on $X$ itself, though, but this is not what we came for.

Title: About restriction of linear map


问:
Find $\limsup_{n \rightarrow \infty}A_n$ and $\liminf_{n \rightarrow \infty}A_n$
I am new to probability theory and doing some exercise behind a textbook "A Probability Path" written by Resnick.

I have a sense that the $\limsup A_n=(0,1]$ and $\liminf A_n=0$. I am not sure whether I am right or not. But I hope someone can show me some idea.

答:
$\limsup_n A_n$ is the collection of elements that are in $A_n$ infinitely often.

$\liminf_n A_n$ is the collection of elements that are in $A_n$ for all but a finite number of $n$.

It is clear that $\liminf_n A_n \subset \limsup_n A_n$.

You are given $a_n >0, b_n >1$ and $a_n \to 0, b_n \to 1$.

Hence if $x \in (0,1]$ we see that $x \in A_n$ for $n$ greater than some $N$ (dependent on $x$, of course). Hence $x \in \liminf_n A_n$.

Note that if $y \le 0$ then $y \notin A_n$ for any $n$. Also, if $y >1$ then after some $N$ we have $b_n < y$. Hence $\liminf_n A_n = \limsup_n A_n = (0,1]$.


问:
check if a function is holomorphic
Let $f:\mathbb{C} → \mathbb{C}$ be defined as

$$f(z)= \begin{cases}\frac{x^3y(y − ix)}{x^6 +y^2}&\text{ if }z = x + iy= 0\\ 0 &\text{ if }z=x+iy=0\end{cases}$$

And holomorph at $z = 0$

I tried to apply the definition of holomorphic application, which is as follows:

$\displaystyle \lim_{(a, b) \to (0,0)} \frac{f (z_0 + h) - f (z_0)}{ h}$

Defining:

$f(x, y) = (xy^2 /x^4 -1, -x^2 y / x^4 -1)$

$f(x + a, y + b) = ((x + a) (y + b)^2 / (x + a)^4 -1, - (x + a)^2. (y + b) / (x + a)^4 -1)$

$f (z_0) = f (x, y) = (xy^2 / x^4 -1, -x^2 y / x^4 -1)$

$ h = (a, b)$

My question is how to solve this limit ...

My question is to define the limit of homologous functions or Verify if it is differentiable from a real point of view and use the cauchy rieman equations.

Thank you!

答: Write $f = u + iv$ and check that the Cauchy Riemann equations hold.

Title: check if a function is holomorphic


问:
Properties of Euler's totient function
I'm having some trouble understanding the following proposition:

Let $\phi$ be the Euler's totient function. If $p$ is a prime number and $n,m \in \mathbb N$, then:

$\phi(p)=p-1$

$\phi(p^n) = p^n - p^{n - 1}$

If $\gcd(m,n)  = 1$, then $\phi(mn)=\phi(m)\phi(n)$

I can understand and prove proposition number (1), but as for (2) and (3) I'm not understanding intuitively what they are saying and why they are true. Can someone explain to me the intuition behind proposition (2) and (3) and show me the proof?  Thanks

答:
Let $(a,b)$ denote the gcd of $a$ and $b$, then

$$
\varphi(n)\triangleq\sum_{k\le n,(k,n)=1}1=\sum_{k\le n}\left\lfloor1\over(k,n)\right\rfloor
$$

where $k$ runs from 1 to $n$. By the property that

$$
\sum_{d|n}\mu(d)=\left\lfloor\frac1n\right\rfloor
$$

we obtain

$$
\begin{aligned}
\varphi(n)
&=\sum_{k\le n}\sum_{d|k,d|n}\mu(d)=\sum_{d|n}\mu(d)\sum_{k=jd\le n}1 \\
&=\sum_{d|n}\mu(d)\sum_{j\le n/d}1=\sum_{d|n}\mu(d)\cdot\frac nd \\
\end{aligned}
$$

Because the $n$ and Mobius function $\mu(n)$ are multiplicative, we deduce that $\varphi(n)$ is multiplicative as well due to the property of Dirichlet convolution.

For prime power $p^n$ ($n\ge1$), we have

$$
\varphi(p^n)=\sum_{k=0}^n\mu(p^k)p^{n-k}=\mu(1)p^n+\mu(p)p^{n-1}=p^n-p^{n-1}
$$

Title: Properties of Euler's totient function


答:
Here's a different way from "NN2"'s answer.
$$
\left[ \begin{array}{c} X_1 \\ \vdots \\ X_n \end{array} \right] \sim \operatorname N_n\left( \left[ \begin{array}c \mu \\ \vdots \\ \mu  \end{array} \right], \, \sigma^2 I_n \right).
$$
The mapping
$$
\left[ \begin{array}{c} X_1 \\ \vdots \\ X_n \end{array} \right] \mapsto \left[ \begin{array}{c} X_1 - \overline X \\ \vdots \\ X_n -\overline X \end{array} \right]
$$
is an orthogonal projection onto an $(n-1)$-dimensional subspace and it takes $\left[ \begin{array}c \mu \\ \vdots \\ \mu  \end{array} \right]$ to $\left[ \begin{array}c 0 \\ \vdots \\ 0 \end{array} \right].$

The distribution of the vector above is spherically symmetric about the mean, i.e. the density is constant on spheres centered at the vector that is the expected value.

Consequently, the distribution of the orthogonal projection of our random vector onto that $(n-1)$-dimensional subspace is spherically symmetric. We can express in terms of an orthonormal basis of that $(n-1)$-dimensional subspace, as $(U_1,\ldots,U_{n-1})^\top.$

So now we have
$$
(X_1-\overline X)^2 + \cdots + (X_n - \overline X)^2 = U_1^2 + \cdots + U_{n-1}^2
$$
and $U_1,\ldots,U_n \sim\text{i.i.d.}\operatorname N(0,\sigma^2).$ Hence
$$
\frac{U_1^2 + \cdots + U_{n-1}^2}{\sigma^2} \sim \chi^2_{n-1}.
$$

Title: How determine the distribution of $S_{n-1}^{2}$?

Body:
If $ X_1,\dots,X_n $ is a random sample with $X_i \sim N(\mu,\sigma^2) $, how determinate the distribution of $S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 $?

I tried to apply the method of moments as follow:

$S_{n-1}^2 = \frac{1}{n-1} \sum \limits_{i=1}^n (X_i - \bar{X})^2 = \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 $

$$ M_{S_{n-1}^2 }(t) = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2 - \frac{n}{n-1} \bar{X}^2 ) } ] =  E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{t(- \frac{n}{n-1} \bar{X}^2 ) } ] = E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)} e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] $$
$$= E[ e^{t ( \frac{1}{n-1} \sum \limits_{i=1}^n X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ] = \prod_{i=1}^n E[ e^{t ( \frac{1}{n-1} X_i^2)}] E[e^{(-t \frac{n}{n-1} \bar{X}^2 ) } ]$$

But in this point I'm lose. Should I apply the following?

$$ X_i \sim N( \mu,\sigma^2 ) \Rightarrow X_i = \sigma Z + \mu  $$
$$ Y = (\sigma Z+ \mu )^2 \Rightarrow M_Y(t) = \frac{1}{\sqrt{1-2t\sigma^2}} \exp\left(  \dfrac{\mu^2t}{1-2t\sigma^2}\right) , \ \text{if } \ 2\operatorname{Re}(t\sigma^2) < 1$$

Or isn't this the simplest way to determinate $S_{n-1}^2$?


问:
Total area of squares.
We have a square whose length is $1$ unit. Every time we rotate by $\theta$ and scale the square such as you see in the image. Does the total area of squares converge if $\theta $ goes to $0$?

答:
Consider a square of side $a$. If the new square rotated by $\theta$ has side of length $b$, we then have
$$b \cos(\theta) + b \sin(\theta) = a \implies b = \dfrac{a}{\sqrt2 \sin(\theta+\pi/4)}$$
Hence, the $n^{th}$ square will have a side of length $$\dfrac{a}{2^{n/2}\sin^n(\theta+\pi/4)}$$ and thereby an area of
$$\dfrac{a^2}{2^n \sin^{2n}(\theta+\pi/4)}$$
Hence, we want to know for what $\theta$
$$\sum_{n=0}^{\infty} \dfrac{\csc^{2n}(\theta+\pi/4)}{2^n}$$ converges.
Can you finish it off from here?</p>

Title: Total area of squares.


问:
Find all integer pairs $(a,b)$ such that all roots of $x^3+ax^2+bx+10$ are integers.
Find all integer pairs $(a,b)$ such that all roots of $f(x)=x^3+ax^2+bx+10$ are integers.

Let the roots of $f$ are $p,q,r$.
Then, by Vietta, $p+q+r=-a, pq+pr+qr=b$, and $pqr=-10$. It means that all roots of $f$ are the multiple of the factor of $10$.

Whats next? What if $c$ in $f(x)=x^3+ax^2+bx+c$ is a big number, for example $200$?

答:
From the rational root theorem every rational root of every monic polynomial with integer coefficients must be integer.

So with that in mind we can say that since a and b are integers the roots p, q and r also integers.

And as you already stated pqr = -10

I believe the corresponding integer values would be (-2,5,1), (2,-5,1), (2,5,-1), (-2,-5,-1) and (10,-1,1), (10,1,-1)

This gives you 6 values of (a,b)


问:
Density of a function of random variables $Y=\cos (X)$
I'm really struggling to solve this question:
Let $X$ be a random variable with $X \sim Exp(1)~$, Find the the density of $Y=\cos (X)$.
I try to solve by dividing the domain in two parts, $(i-1)\pi$ and $i\pi$ to obtain the invert functions but after that I'm totally clueless, I don't know what to do. I appreciate any help.

答:
That is indeed the way to go.

Rather than $i$, we shall use $n \in \Bbb N_0$.

The points for $Y$ shall be a fold of all points where $X=(2n\pi+\arccos(Y))$ or $X=(2(n+1)\pi-\arccos(Y))$ where $\arccos:[-1..1]\mapsto[0..\pi]$

Noting that for $y\in[-1..1]$, we have $\dfrac{\mathrm d \arccos(y)}{\mathrm d y}=\dfrac{-1}{\surd(1-y^2)}$

Thus we shall use the Jacobian transformation:

$$\begin{align}f_{Y}(y) &=\dfrac{\mathbf 1_{y\in[-1..1]}}{\surd(1-y^2)} \sum_{n=0}^\infty \left(f_{X}(2n\pi+\arccos(y))+ f_X(2(n+1)\pi-\arccos(y))\right)\end{align}$$

From here it is a matter of substituting with the probability density function for $X$, distributing out all factors not containing $n$, and expressing resulting the Geometric Series in its closed form.


问:
2nd order nonlinear ODE
I have to solve the following ODE
$$
  \frac{y''}{y'^2} + y'e^y = 0
$$
by finding a function $F(y',y)$ such that its derivative is equal to $\frac{y''}{y'^2} + y'e^y$, then using initial conditions $y(0) = 0$ and $y'(0) = 1$.

What I have done so far is this. Using the multivariable chain rule we can deduce the following:
$$
  \frac{d}{dx} \left(F(y',y)\right) = \underbrace{\frac{d}{dx} y'(x)}_{=y''} \cdot \frac{\partial F(y',y)}{\partial y'} + \underbrace{\frac{d}{dx} y(x)}_{=y'} \cdot \frac{\partial F(y',y)}{\partial y}
$$
So by simple comparison we need that $\frac{\partial F(y',y)}{\partial y'} = \frac{1}{y'^2}$ and $\frac{\partial F(y',y)}{\partial y} = e^y$.

Then by integrating both of these equations with respect to either variable $y'$ and $y$ we get
$$
\begin{align*}
  &F(y',y) = \int e^y \, dy = e^y + c(y') \\
  &F(y',y) = \int \frac{1}{y'^2} \, dy' = -\frac{1}{y'} + c(y)
\end{align*}
$$
Thus an appropriate function would be $F(y',y) = e^y - \frac{1}{y'} + C$.
So we have the relationship
$$
  \frac{d}{dx} \left(e^y - \frac{1}{y'} + c\right) = 0
$$
Integrating both sides yields
$$
  e^y - \frac{1}{y'} + c_1 = c_2
$$
Now if $c_1$ and $c_2$ were zero (or identical), this would be easy to solve using separation but as this is nonlinear and inhomogeneous, I have no idea.

Title: 2nd order nonlinear ODE

Body:
$$\frac{y''}{y'^2} + y'e^y = 0$$
You can integrate directly:
$$-\left( \dfrac 1 {y'} \right)'+(e^y)'=0$$
$$-\dfrac 1 {y'}+e^y=C_1$$
$$-1+y'e^y=C_1y'$$
$$-x+e^y=C_1y+C_2$$


问:
How to approximate the integral $\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx$ for a constant $\mu \gg 1$
I think this integral has no exact solution, but I've found that you can approximate it for large $\mu$ as:

$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx \approx \sqrt{\frac{\pi \mu}{2}} e^{-\mu}$$

This is used a couple of times in a book I'm reading. I've checked it numerically and it works, but I wanted to get the symbolic expression.

I've tried expanding the square root:
$$\sqrt{x^2-\mu^2}=\mu\sqrt{-(1-x^2/\mu^2)}\approx i\mu \left(1-\frac{1}{2}\frac{x^2}{\mu^2}-\frac{1}{8}\frac{x^4}{\mu^4}-...\right)$$

That means that I get an imaginary solution as far as I know, because the first term is a simple logarithmic integral:

$$\int_{\mu}^{\infty} \frac{i\mu}{e^x-1}dx=-i\mu\ln{|e^{-\mu}-1|}$$

And the following terms are integrals that I don't know how to solve. So I guess that this is not the way to proceed.

Any ideas on how to do this approximation, or a more exact solution? Appreciate your help.

答:
Having been stuck with this integral, now using @Semiclassical's elegant approach, I think that we can go further.

$$I=\frac{\sqrt{2\mu}}{e^\mu }\int_0^\infty\frac{\sqrt{z+\frac {z^2}{2\mu}}}{e^z-e^{-\mu}}\,dz\tag1$$

Expanding $\sqrt{z+\frac {z^2}{2\mu}}$ as a series, we have

$$\color{blue}{I=2\sqrt{\pi}\,\,\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{3}{2}\right)\,\,\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right)}{(2\mu)^{n-\frac{1}{2}}\,\,\Gamma \left(\frac{3}{2}-n\right)\,\, \Gamma
   (n+1)} }\tag2$$

and the first term is then
$$a_0=\sqrt{\frac{\pi \mu}{2}}  \text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)=\sqrt{\frac{\pi \mu}{2}} \sum_{k=1}^\infty \frac {e^{-k\mu}}{k^{\frac 32}}$$ and then the required approximation using the first term of the last summation. Doing the same (that is to say using $\text{Li}_{n+\frac{3}{2}}\sim e^{-\mu}~~\forall n$) a better approximation would be
$$I=\sqrt{\frac{\pi\mu}{2}} e^{-\mu } \left(1+t-\frac{5 t^2}{6}+O\left(t^3\right) \right) \qquad \text{with} \qquad t=\frac{3}{8 \mu }$$

Edit

To better see the approximation, we could rewrite
$$I=\sqrt{\frac{\pi \mu}{2}} e^{-\mu } \big[e^{\mu } \,\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right) \big]\Bigg[1+\sum_{n=1} ^\infty B_n\Bigg]$$ with
$$B_n=\frac{ \Gamma \left(n+\frac{3}{2}\right)}{(2\mu)^n\,\,\Gamma   \left(\frac{3}{2}-n\right)\,\,\Gamma (n+1)}\,\,\frac{\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right)  } {\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)  }$$

$$e^{\mu } \,\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)=1+\frac{e^{-\mu }}{2 \sqrt{2}}+\frac{e^{-2 \mu }}{3 \sqrt{3}}+\cdots$$
$$\frac{\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right)  } {\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)  }=1- 2^{-n-\frac{3}{2}} \left(2^n-1\right)e^{-\mu }+\cdots$$


答:
Set $f=u + iv$, so $u=\frac{xy^2}{x^4-1}$ and $v=\frac{-x^2y}{x^4-1}$.

calculating the partial derivatives we have:

$$\frac{\partial u}{\partial x} = \frac{y^2(x^4-1)-xy^2(4x^3)}{(x^4-1)^2}$$

And

$$\frac{\partial v}{\partial y}= \frac{-2xy(x^4-1)-(-x^2y)(4x^3)}{(x^4-1)^2}$$

as the first equation is already different, it follows that f is not holomofa at 0.

Title: check if a function is holomorphic

Body:
Let $f:\mathbb{C} → \mathbb{C}$ be defined as

$$f(z)= \begin{cases}\frac{x^3y(y − ix)}{x^6 +y^2}&\text{ if }z = x + iy= 0\\ 0 &\text{ if }z=x+iy=0\end{cases}$$

And holomorph at $z = 0$

I tried to apply the definition of holomorphic application, which is as follows:

$\displaystyle \lim_{(a, b) \to (0,0)} \frac{f (z_0 + h) - f (z_0)}{ h}$

Defining:

$f(x, y) = (xy^2 /x^4 -1, -x^2 y / x^4 -1)$

$f(x + a, y + b) = ((x + a) (y + b)^2 / (x + a)^4 -1, - (x + a)^2. (y + b) / (x + a)^4 -1)$

$f (z_0) = f (x, y) = (xy^2 / x^4 -1, -x^2 y / x^4 -1)$

$ h = (a, b)$

My question is how to solve this limit ...

My question is to define the limit of homologous functions or Verify if it is differentiable from a real point of view and use the cauchy rieman equations.

Thank you!


问:
prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx$ for a continuous function $f$
Let $f\colon [0, 1] → \mathbb{R}$ be a continuous function. Given, $f(0) = 0$ and $\int_0^1 f(x)\, dx = 0$, prove that $∃ c ∈ (0, 1)$ such that $c^2f(c) = 2\int_0^cxf(x) \,dx.$

I need some help on how to use the fundamental theorem to prove this result.

My attempt: Let $F'(x)=f(x)$, the $\int_0^1 f(x)\, dx = 0$ implies that $F(1)=F(0)$. Now we take $H(x)=xf(x)$ and $\int_0^c xf(x)dx=\frac{c^2f(c)}{2}-\int_0^c\frac{x^2f'(x)}{2}dx$. This is where I'm stuck.

答:
For $x\geq 0$ fixed define $\varphi:[0,x]\longrightarrow \mathbb{R}$ by $\varphi(t)=t$. Since $f,\varphi$ are continous on $[0,x]$ and $\varphi$ is non$-$negative, then by this we can locate $\xi_x\in[0,x]$ such that $$\int_0^xf(t)\varphi(t)dt=f(\xi_x)\int_0^x\varphi(t)dt=\frac{f(\xi_x)x^2}{2}$$ Also, $f$ being continuous on $[0,1]$ with $\int_0^1f(t)dt=0$ implies there exists $a \in (0,1)$ such that $f(a)=0$. (This is easily proved by a straight$-$forward application of Rolle's Theorem to the function $x \longrightarrow \int_0^xf(t)dt$) Define $G:[0,a]\longrightarrow \mathbb{R}$ by $$G(x)=\int_0^xtf(t)dt-\frac{x^2f(x)}{2}$$ From our initial commentary we can express $G$ as $$G(x)=\frac{x^2}{2}\bigg(f(\xi_x)-f(x)\bigg)$$ Now set $m=\min\Big\{f(x):x\in [0,a]\Big\}$ and $M=\max\Big\{f(x):x\in [0,a]\Big\}$. Since $f$ is continuous on the compact set $[0,a]$ we can locate $x_1,x_2\in [0,a]$ such that $f(x_1)=m$ and $f(x_2)=M$. Now $G(x_1)\geq 0$ while $G(x_2)\leq 0$ so by $\text{IVT}$ there exists $c$ between $x_1$ and $x_2$ (inclusive) such that $G(c)=0$. We now need only prove that $c$ can be chosen to belong to the open interval $(0,1)$.

If $mM<0$, then $f(0)=0=f(a)$ implies $x_1,x_2\in (0,a)$; this means $c$ must also belong to $(0,a)\subset (0,1)$ and we're done. On the other hand, if $m=M=0$, then $f\equiv 0$ on $[0,a]$, and taking $c$ to be any real number in $(0,a)$ will work. Now if $m=0$ and $M>0$ we have $x_2 \in (0,a)$ and $G(a)\geq 0$ so by IVT find $c\in [x_2,a] \subset (0,1)$ with $G(c)=0$. A similar argument applies to the case when $m<0$ and $M=0$ completing the proof.


问:
How many possible ways are there to print 125 pages on 5 printers?
I think I understand this problem but I want to make sure I do get it thoroughly. In my mind, this problem is asking you to assign a single page to a single printer, but it doesn't really matter the order in which they are printed. I.E., we don't care if page 1 is sent to printer 1 first, or if printer one gets page 2 first and then page 1 is sent.

In my mind, then, we would simply take a page and say, how many choices do I have to print it? That would be 5 for the 1st page. Now page 2 also has 5 choices, and so on and so on, so that means the solution would be 5 x 5 x ... x 5 = 5 ^ 125.

Is this the correct answer or am I implicitly slipping in order some how and need to remove the number of permutations from this number? Thanks for your help in understanding this problem!

答: OK, so it seems that my original thinking is indeed the correct answer as if you choose to start with page 2, then go to page 3, etc., etc., until you wrap back around and finish with page 1, there are still the same number of ways to distribute all the pages to each printer as if you had started with page 1. Therefore 5 ^ 125 is the answer.


答:
Using the fact that $\overline X$ and $S_{n-1}^2$ are independent in normal distribution.

$\sum_{i}(X_i-\mu)^2=\sum_{i}(X_i-\overline X)^2+n(\overline X-\mu)^2$

$\underbrace{\dfrac{\sum_{i}(X_i-\mu)^2}{\sigma^2}}_W=\underbrace{\dfrac{\sum_{i}(X_i-\overline X)^2}{\sigma^2}}_{W_1}+\underbrace{\dfrac{n(\overline X-\mu)^2}{\sigma^2}}_{W_2}$

$W=W_1+W_2$

$W\sim \chi_{n}^2 ,\ \ \ W_2\sim\chi_{1}^2 ; \text{Let me know if you didnt get this, I will elaborate.}$

$W_2$ is a function $\overline X$ and $W_1$ is function $S_{n-1}^2$ so $W_1$ amd $W_2$ are independent.

Therefore,

$M_W{(t)}=M_{W_{1}}(t)\cdot M_{W_{2}}(t)$

$M_{W_{1}}(t)=\dfrac{M_{W}(t)}{M_{W_{2}}(t)}=\dfrac{(1-2t)^{\frac{-n}{2}}}{1-2t)^{\frac{-1}{2}}}=(1-2t)^{\frac{-(n-1)}{2}}\ \ \ ,t< \frac{1}{2}$

$W_1=\dfrac{\sum_{i}(X_i-\overline X)^2}{\sigma^2}=\dfrac{(n-1)S_{n-1}^2}{\sigma^2} \sim \chi_{n-1}^2$


问:
Approaches to solving $y''(1-x^2)-2xy'+2y=0$?
I know it could be done via power series but what other methods are there?</p>

答:
Let $y=x z$ and the equation becomes
$$2\left(1-2 x^2\right) z'-x \left(x^2-1\right) z''=0$$ Reduction of order $p=z'$ gives
$$2\left(1-2 x^2\right) p-x \left(x^2-1\right) p'=0$$ which is separable. So
$$p=\frac{c_1}{x^2 \left(1-x^2\right)}=c_1 \left(\frac{1}{x^2}+\frac{1}{2 (x+1)}-\frac{1}{2 (x-1)} \right)$$ which is simple.

Integrating
$$z=c_1\left(-\frac{1}{x}-\frac{1}{2} \log (1-x)+\frac{1}{2} \log (x+1)\right)+c_2$$
$$y=c_1\left(-1-\frac{1}{2} x\log (1-x)+\frac{1}{2}x \log (x+1)\right)+c_2 x$$


问:
If $f: \mathbb{R} \to \mathbb{R}$ such that $\displaystyle \lim_{x \to \pm \infty}\sqrt{|x|}(f(x)- cos(x))=1$ is continous, it is uniformly continous.
There is a continous function $f: \mathbb{R} \to \mathbb{R}$ such that: $\displaystyle \lim_{x \to \pm \infty}\sqrt{|x|}(f(x)- \cos(x))=1$. Prove that function $f$ is uniformly continous.

I know that there are proofs on this forum that if $g$ is bounded and continous, $g$ is uniformly continous, but I don't know how to apply those to my situation.

Maybe I need to show that if function if $\sqrt{|x|}(f(x)- \cos(x))$ is bounded then $f$ is bounded?

答:
Principle

This is a really lovely problem, a case of "small $x$ , big $x$". The existence of the limit pretty much decides $f(x)$ for large values of $x$, while for small values of $x$ uniform continuity follows from the fact that a continuous function on a compact(same as closed and bounded) interval  is uniformly continuous.

Another point : it is true that $f$ is bounded, but we can actually derive the stronger statement that it is uniformly continuous without requiring this fact.

Let's quickly note that if
$
\sqrt{|x|}{(f(x)-\cos x)} \to 1
$
then for large $x$, $f(x)-\cos x \approx \frac{1}{\sqrt{|x|}}$ i.e. $f(x)$ behaves like $\cos x$ for large enough $x$. Because $\cos(x)$ is uniformly continuous (this is not difficult to prove from its periodicity), we would then expect the same from $f$.

The fundamental quantity to control for uniform continuity is $|f(x)-f(y)|$. So we want to express $f(x)-f(y)$ in terms of the approximation we have obtained. We use a split-into-three inequality , often called an $\frac{\epsilon}{3}$ split. The aim of this inequality is this : $f(x)$ is close to $\cos(x)$, $f(y)$ is close to $\cos(y)$ and $\cos(x)$ is close to $\cos(y)$. It's like : showing $f$ values are close, is transferred to showing that $\cos$ values are close, which we know is true.

This bounds $f(x)-f(y)$ for large enough $x,y$ close enough, which is all we need.

The large $x$ values

Let $x,y \in \mathbb R$ and $\epsilon>0$. Note the inequality :
$$
|f(x)-f(y)| \leq |f(x)-\cos(x)|+|\cos(x) - \cos(y)| + |f(y)-\cos(y)| \tag{1}
$$

which is true for all $x,y$.

First : by uniform continuity of $\cos(x)$, there is a $\delta_1>0$ such that for all $|x'-y'|<\delta_1$ we have $|\cos(x')-\cos(y')| < \frac{\epsilon}{3}$.

Now, for the other two terms we use the limit : let's see what it means. For every $\epsilon'>0$ there is $r'>0$ large enough such that $|x|>r'$ implies $|\sqrt{|x|}(f(x)-\cos(x))-1| < \epsilon'$. That is, $\frac{1-\epsilon'}{\sqrt{|x|}}<|f(x)-\cos(x)| <\frac{1+\epsilon'}{\sqrt{|x|}}$ from the triangle inequality.

This is helpful! Let's take $\epsilon'=1$, and so we get $r'>0$ such that the above happens. In particular, $|f(x)-\cos(x)| < \frac{2}{\sqrt{|x|}}$ for all $|x|>r'$.

But now life is simple : take $r''>0$ such that $\frac{2}{\sqrt{|x|}} < \frac{\epsilon}{3}$ for all $|x|>r''$. Then if $r = \max\{r',r''\}$, $|x|>r$ implies that $|f(x)-\cos(x)| < \frac{\epsilon}{3}$.

Finally, the following is true by $(1)$ :

For all $x,y$ with $|x|,|y|>r$ and $|x-y| < \delta_1$, we have $|f(x)-f(y)| < \epsilon$.

Small values

We never used the fact that $f$ is continuous above, so where does it come into play? For small values. The following result is standard, and hopefully has been studied. If not, I will attach the required links.

A continuous function on a closed and bounded subset of $\mathbb R$ is uniformly continuous.

We apply this, to $f$ restricted to $[-r-\delta_1,r+\delta_1]$ , where $r$ is coming from the previous argument. Of course this is closed and bounded, and $f$ is continuous, hence $f$ is uniformly continuous on this interval. You'll see why I added the $\delta_1$ part in some time.

Thus, for the given $\epsilon$, we have a $\delta_2 >0$ such that for all $|x-y| < \delta_2$ and $x,y \in [-r-\delta_1,r+\delta_1]$ we have $|f(x)-f(y)|<\epsilon$.

Conclusion

To conclude, we need to patch these up. Let $\delta = \min\{\delta_1,\delta_2\}$. If $|x-y| < \delta$, there are three possibilities :

$|x|,|y|<r$. We use the "small values" discussion to conclude that $|f(x)-f(y)|<\epsilon$.

$|x|,|y|>r$. We use the "large values" discussion to conclude that $|f(x) - f(y)| < \epsilon$.

One of $|x|,|y| < r$ and other is $>r$. Say $|x|<r$. Then, because $|x-y|<\delta \leq \delta_1$, we get by the triangle inequality that $|x|,|y| < r+\delta_1$, and hence the "small values discussion" applies (can you now see why the interval was extended?) to conclude that $|f(x)-f(y)|<\epsilon$.

Principles to be kept in mind

Using limits to control large values of $f(x)$.

Using the $\frac \epsilon 3$ method to transfer closeness of $f$ to closeness of $\cos$, allowing us to use the uniform continuity of $\cos$.

Controlling small values of $x$ using a result about continuity on closed and bounded intervals.

Extending the interval by a $\delta_1$ factor to "patch" the two arguments together.

Sidenote

Please let me know if you do not know either of these facts, which should have been proved prior to the introduction of this question :

$\cos(x)$ is uniformly continuous on $\mathbb R$.

A continuous function defined on a closed and bounded interval of $\mathbb R$ is uniformly continuous.

I can direct you to sources, or we can have a conversation about why these are true.

Edit

To see that $\cos$ is uniformly continuous, we note that it has a period of $2 \pi$. Therefore, controlling it in the interval $[0,2\pi]$ is enough. But we again need to do a little "patching", so the interval is not quite $[0,2 \pi]$ for a "small" situation.

More precisely : note that $\cos$ is continuous and $[0,4\pi]$ is uniformly bounded, therefore $\cos$ is uniformly continuous on $[0,4\pi]$. Now, let $\epsilon>0$. There exists a $\delta>0$ such that for $x,y \in [0,4 \pi]$ we have $|x-y|<\delta$ implies $|\cos(x) - \cos(y)| < \epsilon$.

Now, let $x,y$ be arbitrary. We can, by subtracting multiples of $2 \pi$, ensure that there is are $x'$ and $y'$ in $[0,4 \pi]$ such that $x'-x$ and $y'-y$ are integer multiples of $2 \pi$, so that $f(x) = f(x')$ and $f(y) = f(y')$. Now, suppose that $|x-y|<\delta$. Prove that $x',y'$ can be chosen such that $|x'-y'|<\delta$, then by uniform continuity $|f(x) - f(y)| = |f(x')-f(y')| < \epsilon$. So this $\delta$ always works and we are done.


问:
Euclid Division Lemma Proof for J < 0
I am reading George E. Andrews' Number Theory and am at the portion where he proves Euclid's Division Lemma.

The lemma states that

For any integers $k (k>0)$ and $j$, there exist unique integers $q$ and $r$ such that $0≤r<k$ and $j=qk+r.$

I am having difficulty understanding the final part of the proof, where the uniqueness of $q$ and $r$ when $j < 0$ is proven.

By this point it has been established that the lemma is true for $j>1$ because by the basis representation theorem, $j$ has a unique representation to the base $k$

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0$$
$$=k(a_sk^{s-1}+a_{s-1}k^{s-2}+...+a_1)+a_0$$
$$=kq+r$$

Where $0≤r=a_0<k$.

Supposing a second pair $q'$ and $r'$ exists, there is a representation for $q'$ to the base $k$:

$$q'=b_tk^t+b_{t-1}k^{t-1}+...+b_1k+b_0.$$

Thus,

$$j=kq'+r'$$
$$=b_tk^{t+1}+b_{t-1}k^t+...+b_1k^2+b_0k+r'.$$

But,

$$j=a_sk^s+a_{s-1}k^{s-1}+...+a_1k+a_0.$$

By the uniqueness of the representation of $j$ to base $k$, the following may be gleaned: $r'=r=a_0, \ a_{i+1}=b_i, \ \text{and} \ t=s-1$. Therefore,

$$q'=b_tk^t+...+b_1k+b_0$$
$$=a_sk^{a-s}+...+a_2k+a_1$$
$$=q.$$

Therefore the lemma is true when $j>0$. And if $j=0, \ q=r=0$ is the only possible solution where $0≤r<k$. It is left to prove that the lemma is true when $j<0$.

Here we run into my issue. He writes,

If $j<0$, then $-j>0$, and there exist unique integers $q'' \ \text{and} \ r''$ such that

$$-j=kq''+r''.$$

If $r''=0$, then $j=k(-q'')$; thus we may take $q=-q'' \ \text{and} \ r=0$.

If $r''\neq 0$, then

$$j=-kq''-r''$$
$$=k(-q''-1)+(k-r''),$$

and we may take $q=-q''-1, \ \text{and} \ r=k-r''$.

In either case, $q$ and $r$ satisfy equation $j=kq+r$. Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

I understand the reason for using the equation $-j=kq''+r''$, since it has already been established that when the equation is greater than $0$, there exists unique integers $q''$ and $r''$. This is useful because it is equivalent to $j=-kq''-r''$, so if there exists unique integers $q''$ and $r''$ for $-j$, they must also exist for $j$, and so we may conclude that the lemma is true when $j<0$, correct? But this raises the question, why doesn't the proof end there? I am missing something.

There is a reason why we treat cases $r'' \neq 0$ and $r''=0$ separately. Why is that?

Furthermore, how is the fact that in the former case $q=-q''$ and $r=r''=0$ relevant to the proof? And in the latter case, how do $q=-q''-1$ and $r=k-r''$ satisfy the equation $j=kq+r$?

Finally, the phrasing of the last sentence is confusing:

Uniqueness for negative $j$ follows from uniqueness for $-j$, which is then positive.

It seems as if "negative $j$" and "$-j$" are different things, otherwise the statement would be begging the question. But what then is "negative $j$?" The "$j$" from the equation $j=kq+r$ multiplied by $-1$? I assume "$-j$" is from the equation $-j=kq''+r''.$

I realize I have asked a great many questions and probably have not been clear enough in stating my confusions. I am very new to number theory and am learning it independently, so I am very prone to misunderstanding the content. This is the only place I can reliably go for clarification. My apologies for the length of this question, I have tried to be as specific as possible to avoid miscommunication.

答:
Consider this.  If $j < 0$ then $|j| > 0$.

So there is a $q, r$ so that $|j| = qk + r$ where $0 \le r < k$.  But that means

$j = -qk -r$.  we can't use $-r$ because obviously we do not have $0 \le -r < k$.

What we do have is $-k < -r \le 0$.  Which is not what we need.

If if $r = 0$ we are okay because $-r= r = 0$ and we do have $j = (-q)k + 0$ as an expression.  That's just fine.

But if $-k < -r < 0$ we have to do something... what can we do?

Well just add $k$.  $0 < k -r < k$

And we have $j = -qk -r=$

$-qk +(k-r) -k =$

$(-q-1)k + (k-r)$.

....

This is basically a "fence post" indexing issue about worrying "well where exactly is our starting point"

====

Consider this:  $qk + 0$ through $qk + (k-r)$ will go from $qk$ to $(q+1)k -1$ inclusively.  this is true for all $q$ positive or negative.

But if $q$ is negative this...."looks weird" we expect, if $q$ is negative for $(q+1)k - 1$ to be $(-|q|+1)k -1$ the same as $(-|q+1|)k -1$ and .... it isn't; $(-|q|+1)k -1 = (-|q-1|)k -1$

but let's look  closely:

If $q = 2$ we go from $2k$ through $3k -1$ with a typical item being $2k + r$.

If $q= 1$ we go from $k$ through $2k -1$ with a typical item being $k + r$.

If $q=0$ we go from $0$ to $k-1$ with a typical item being $ r$.

If $q=-1$ we go from $-k$ to $-1$ with a typical item being $-k +r = -(r-k)$

If $q=-2$ we go from $-2k$ to $-k-1$ with a typical item being $-2k +r=-3k- (k-r)$.

Switching the signs, but not changing the fact that we still count up and we don't switch to counting down really does a number on our intuition, doesn't it.

...

Further illustration if $k = 7$ (just for the sake of illustrion:

$q= 2:  [14+0=14,14 + 1=15; 14+2=16;14+3=17;14+4=18;14+5=19; 14+6=20]$

$q=1: [7+0=7,7 + 1=8; 7+2=9;7+3=10;7+4=11;7+5=12; 7+6=13]$

$q=0: [0+0=0,0 + 1=1; 0+2=2;0+3=3;0+4=4;0+5=5; 0+6=6]$

$q=-1: [-7 + 0=-7;-7+1=-6; -7+2=-5;-7+3=-4;-7+4=-3;-7+5 = -2;-7+6 = -1]$.

$q=-2: [-14 + 0=-14;-14+1=-13; -14+2=-12;-14+3=-11;-14+4=-10;-14+5 = -9;-14+6 = -8]$.

Are you seeing it?

Title: Euclid Division Lemma Proof for J < 0