问:
Coercivity type condition $\langle x - y , \mu(x) - \mu(y) \rangle _ {\mathbb R ^d } \leq || x-y || ^2_{\mathbb R ^d } $ implies Lipschitz
Let's define the coercivity type condition as :

$$ \forall x,y \in \mathbb R^d : \langle x - y , f(x) - f(y) \rangle _ {\mathbb R ^d } \leq  || x-y || ^2_{\mathbb R ^d } $$

It seems to me that this condition is stronger than locally Lipschitz continuous for $f$. However, I have no clue how to prove it.

If it is stronger than local. Lipschitz, what about Lipschitz continuous?

答:
A conclusion from my comment in the general case $\mathbb{R}^d$. Let $g$ be any decreasing function from $\mathbb{R}$ to $\mathbb{R}$, and define $f : \mathbb{R}^d \mapsto \mathbb{R}^d$ by $f(x) = \big(g(x_i)\big)_{1\le i\le d}$.

Then for any $x,y \in \mathbb{R}^d$, $\langle x - y , f(x) - f(y) \rangle _ {\mathbb R ^d } = \sum \limits_{i=1}^d (x_i-y_i)(g(x_i)-g(y_i)) \le 0 \le ||x-y||_{\mathbb{R}^d}^2$.

Yet if $f$ were locally Lipschitz, then so would its first coordinate $g$ be.

Proof that there exists a decreasing real function $g$ that is nowhere locally Lipschitz (i.e. not Lipschitz on any interval): with $(q_n)_{n \in \mathbb{N}}$ an enumeration of the rational numbers, let $$g(x) = - \sum \limits_{k=1}^{+\infty} \frac{\sqrt[3]{x-q_k}}{2^k (1+\sqrt[3]{|q_k|})}.$$

This function is well defined because for $x \in [-A,A]$, the general term is bounded in absolute value by $\frac{\sqrt[3]{A}+\sqrt[3]{|q_k|}}{2^k (1+\sqrt[3]{|q_k|}} \le \frac{\sqrt[3]{A}+1}{2^k}$, so the sum converges uniformly on $[-A,A]$.

Ad absurdum, assume that $g$ is Lipschitz in some open interval $I$, and let $k \in \mathbb{N}$ such that $q_k \in I$. Since $g$ is defined as a sum of decreasing functions, $$\Big|-\frac{\sqrt[3]{x-q_k}}{2^k(1+\sqrt[3]{|q_k|}} - 0 \Big| \le |g(x)-g(q_k)| \le C|x-q_k|$$ for $x$ in $I$, which is a neighborhood of $q_k$. We conclude that $x \mapsto \sqrt[3]{x-q_k}$ is Lipschitz at $q_k$, which is obviously false.


问:
Sup and inf of a sequence
I have to find $\inf$ and $\sup$ of the given sequence:
$$ a_n := \left(1+ \frac{\cos(n\pi)}{n}\right)^n + \sin\left(\frac{n\pi}{2}\right)$$
First thought that come to my mind was to consider 4 cases dependinig on trigonometric functions. For $\sup$, 2 cases, when $\sin = 1$ or $\cos = 1$, and the same for $\inf$.

Is that correct or I'm completely wrong?

答:
We assume $n$ must be a positive integer.

We will split into cases depending on the remainder of $n$ when divided by $4$. We will use the fact that $f(x) = \big(1 + \frac{m}{x}\big)^{x}$ is increasing on $x\in (1,\infty)$ for $m \geq -1$, which can be proven using basic calculus. We will also use $\displaystyle\lim_{x\to\infty}f(x) = e^{m}$.

If $n = 4k$ for some integer $k$:

$$a_{4k} = \bigg(1 + \frac{\cos(4k\pi)}{4k}\bigg)^{4k} + \sin(2\pi k)=\bigg(1 + \frac{1}{n}\bigg)^{n}$$

This sequence is increasing, so its infimum is $a_{4} =\frac{625}{256}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k} = e$.

If $n = 4k + 1$:

$$a_{4k + 1} = \bigg(1 + \frac{\cos(4\pi k + \pi)}{4k + 1}\bigg)^{4k + 1} + \sin\bigg(2\pi k + \frac{\pi}{2}\bigg) = \bigg(1-\frac{1}{n}\bigg)^{n} + 1$$

This sequence is increasing, so its infimum is $a_{1} = 1$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 1} = \frac{1}{e} + 1$.

If $n = 4k + 2$:

$$a_{4k + 2} = \bigg(1 + \frac{\cos(4k\pi + 2\pi)}{4k + 2}\bigg)^{4k + 2} + \sin(2\pi k + \pi)=\bigg(1 + \frac{1}{n}\bigg)^{n}$$

This sequence is increasing, so its infimum is $a_{2} =\frac{9}{4}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 2} = e$.

If $n = 4k + 3$:

$$a_{4k + 3} = \bigg(1 + \frac{\cos(4\pi k + 3\pi)}{4k + 3}\bigg)^{4k + 3} + \sin\bigg(2\pi k + \frac{3\pi}{2}\bigg) = \bigg(1-\frac{1}{n}\bigg)^{n} - 1$$

This sequence is increasing, so its infimum is $a_{3} = -\frac{19}{27}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 3} = \frac{1}{e} - 1$.

Then, $\inf a_{n}$ is the minimum of the infima of the subsequences making up $a_{n}$, Likewise, $\sup a_{n}$ is the maximum of the suprema of these subsequences. Thus, $\boxed{\inf a_{n} = -\frac{19}{27}\text{ and }\sup a_{n} = e.}$

ContentLicense: CC BY-SA 4.0

Title: Sup and inf of a sequence


问:
Prove that $\left| \dfrac{an^k + f(n)}{bn^l+g(n)} \right|\to \infty$
Suppose $k,l \in \Bbb{N}$, $a,b \in \Bbb{R^*}$ (non-zero reals), and consider the functions $f:\Bbb{N} \to \Bbb{N}, g:\Bbb{N} \to \Bbb{N}$, such that $\dfrac{f(n)}{n^k} \to 0, \dfrac{g(n)}{n^l}\to0$.

Prove that if $k>l$, then $\left| \dfrac{an^k + f(n)}{bn^l+g(n)} \right|\to \infty$

I'm stucked at this problem. Please if possible also tell if my attempt is at least making sense so far.

My attempt:

Let $M > 0$

There are $N_1>0$ st. $n \ge N_1 \implies \left|\dfrac{f(n)}{n^k} \right|< M$ and $N_2>0$ st. $\left| \dfrac{g(n)}{n^l}\right|<M$. Let $N = \max\{M_1,M_2\}$

$M > \left|\dfrac{g(n)}{n^l}\right| \implies Mn^l > g(n) \implies Mn^l + bn^l > g(n) + bn^l \implies \dfrac{1}{bn^l + g(n)} > \dfrac {1}{Mn^l + bn^l} $

$k > l \implies n^k > n^l \implies an^k > an^l \implies an^k + f(n) > an^l + f(n)$. (I considered $a>0$ just to see if I got something)

Then, for all $n \ge N$, $\left| \dfrac{an^k + f(n)}{bn^l + g(n)} \right| \ge \dfrac{an^l + f(n)}{Mn^l + bn^l} = \dfrac{a}{M+b}+\dfrac{1}{M+b}\dfrac{f(n)}{n^l}$

And if I could guarantee $\dfrac{f(n)}{n^l} \to 0$ I would have done it, right?

答:
Easier approach:

AS $k-l>0$, we have that:
$$\left| \dfrac{an^k + f(n)}{bn^l+g(n)} \right| = \left| \dfrac{an^k + f(n)}{bn^l+g(n)} \right| \cdot \frac{1/n^k}{1/n^k} = \left|\frac{a+f(n)/n^k}{b/n^{k-l}+g(n)/n^{k}}\right| =$$
$$= \left|\frac{a+f(n)/n^k}{(b+g(n)/n^l)/n^{k-l}}\right| = n^{k-l}\left|\frac{a+f(n)/n^k}{b+g(n)/n^l}\right| \to +\infty \left|\frac{a+0}{b+0}\right|=+\infty$$


问:
How do I find the second order derivative of a function on $\mathbb{R}^n$
I am studying systems of differential equations, i.e., $$f(\vec{x})=\dot{\vec{x}} = A\vec{x}$$ for some matrix $A_{n\times n}$, a vector valued function $x(t)=x $ and a function $f:\mathbb{R}^n \to \mathbb{R}^n$. Now, if $f$ is differentiable, then its derivative is given as $$Df= \left[\frac{\partial{f_i}}{\partial{x_j}}\right]$$ which is an $n\times n$ matrix. Now, I am studying from Perko's DE's and Dynamical Systems and the second order derivative of $f$ is given exactly as follows. Assuming $f:E \subseteq\mathbb{R}^n\to \mathbb{R}^n$, $$D^2f(x_0)(x,y) = \displaystyle\sum_{j_1,j_2=1}^{n} \frac{\partial^2{f(x_0)}}{\partial{x_{j_1}\partial{x_{j_2}}}}x_{j_1}y_{j_2}$$
for some $(x,y) \in  E \times E$.
Now, I want to the find $D^2f(x_0)$ for $$f(x) = \left(\begin{matrix} x_1+x_1x_2^2 \\ -x_2+x_2^2+x_1^2 \end{matrix}\right)$$
and $x_0 = (0,1)$ but I do not know how to use the formula. Is the sum gives a column or row of the desired derivative matrix? How do I proceed for this question and for higher order derivatives?

答:
For a real-valued function $\mathbb{R}^n \to \mathbb{R}$,
$$\displaystyle\sum_{j_1,j_2=1}^{n} \frac{\partial^2{f(x_0)}}{\partial{x_{j_1}\partial{x_{j_2}}}}x_{j_1}y_{j_2}$$
is the same as:
$$\sum_{i=1}^n \sum_{j=1}^n H_{ij}x_i y_j$$
which is just $x^THy$ where $H$ is the Hessian of $f$ at $x_0$. For a vector-valued function is gets a bit tricky when it comes to interpreting $\frac{\partial^2{f(x_0)}}{\partial{x_{j_1}\partial{x_{j_2}}}}$. It is ambiguous, because you might think this is a matrix. It is in fact a vector where the i-th component is the second derivative (to $x_{j_1}$ and $x_{j_2}$) of the i-th component of your objective. An easy way to compute the full expression is by writing the Hessian of the coordinates of your function:
$$H_1 = \begin{pmatrix}0 & 2x_2 \\ 2x_2 & 2x_1\end{pmatrix}$$
$$H_2 = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix},$$
and writing $D^2f(x_0)(x,y)$ as a vector in $\mathbb{R}^2$:
$$\begin{pmatrix}x^T H_1 y \\ x^T H_2 y\end{pmatrix}.$$


问:
Solving a separable ODE explicitly
I am stuck at solving the following separable ODE:
$$f(x)=\sqrt{1+[f'(x)]^2},$$
with the condition that $f:[0,\infty)\longrightarrow [0,\infty)$.

I first note that $f(x)\geq 1$ and that the constant function $f(x)=1$ is a solution.

By squaring both sides and separating $f'(x)$, I get that
$$[f(x)]^2=1+[f'(x)]^2 \iff \\ f'(x)=\pm\sqrt{[f(x)]^2-1}$$

Now, for $f(x)\neq 1$, using the fact that $f'(x)=\frac{df}{dx}$ and abusing the notaion $f=f(x)$, I separate the differentials as follows:
$$\frac{df}{dx}=\pm\sqrt{f^2-1} \iff\\ \frac{df}{\sqrt{f^2-1}}=\pm dx$$

Integrating both sides gives
$$\int\frac{df}{\sqrt{f^2-1}}=\int \pm dx \implies ln(f+\sqrt{f^2-1})=\pm x+C$$

We can collect the R.H.S.
\begin{cases}
x+C_1 & (1)\\
-x+C_2 & (2)
\end{cases}

For the left integral, I used the standard integral that is provided in my textbook, namely that
$$\int\frac{dx}{\sqrt{x^2+\alpha}}=\\ ln\left|x+\sqrt{x^2+\alpha}\right|+C$$

I know that I can get rid of the logarithm in the L.H.S., using $(1)$ from above,
$$ln(f+\sqrt{f^2-1})=x+C_1\iff \\ f+\sqrt{f^2-1}=e^{x+C_1}$$
Note that i have dropped the absolute value sign, due to the given condition which constraints $f=f(x)$ to the positive real numbers.

But this is where I am stuck. What am I supposed to do? I am failing to factorize $f$. I know that we can have implicit answers for separable ODEs, but I am supposed to satisfy the condition that $f:[0,\infty)\longrightarrow [0,\infty)$, i.e., I need to find all such functions $f$.

To reiterate: how do I continue to find the explicit answer to the given separable ODE?

P.S. This is my very first time posting a question, although I have read a lot of answers. Also, english is not my first language, please excuse any grammatical errors.

答:
$$
1=f(x)^2-f'(x)^2
$$
is a hyperbole equation, so you can parametrize it in hyperbolic functions, $$f(x)=\cosh(u(x)), ~~f'(x)=\sinh(u(x)).$$ In consequence of the first part, also $$f'(x)=\sinh(u(x))u'(x),$$ so that, apart from the constant solution, $u'(x)=\pm 1$. Both signs lead to the same solution family $$f(x)=\cosh(x+C).$$

Title: Solving a separable ODE explicitly


问:
Non-associative version of a group satisfying these identities: $(xy)y^{-1}=y^{-1}(yx)=x$
The following identities are a consequence of the group axioms.
$$(xy)y^{-1}=x,\quad y^{-1}(yx)=x$$
Notice we haven't mentioned an identity element, and that the above identities make sense even in the absence of associativity.

Is there any interest in the generalization of a group where these identities are taken as the sole axioms? The data would consist of

an underlying set

a law of composition

an inversion operator

Remark. These axioms allow us to solve simple equations. For example, in the additive group of integers we can argue as follows.$$x+3=4$$ $$\Rightarrow (x+3)+(-3)=4+(-3)$$ $$\Rightarrow x = 4+(-3)$$

答:
If for all $x,y \in G$, one has $(xy)y^{-1}=y^{-1}(yx)=x$, then $G$ is a special type of quasi-group that I'll call a SIP-quasigroup (“symmetric inverse property quasigroup”). However, there are quasi-groups without such inverse functions; I give an infinite family of examples due to Belousov that do have an inverse property (but not a symmetric one).

If you assume an identity, then you get exactly the IP-loops (“inverse property loops”) which are reasonably well studied.

All SIP-quasigroups are quasigroups

Lemma: ${(x^{-1})}^{-1}=x$ and $(xy^{-1})y = y(y^{-1}x) = x$

Proof: [by user1] $$\begin{array}{rl}
(x^{-1})^{-1}
&=\left((x^{-1})^{-1}x\right)x^{-1} \\
&=\left((x^{-1})^{-1}\left(x^{-1}(xx)\right)\right)x^{-1} \\
&=(xx)x^{-1} \\
&=x
\end{array}$$
Then substituting $y^{-1}$ instead of $y$ into the defining axioms one gets $x = (xy^{-1})(y^{-1})^{-1} = (xy^{-1}) y$ and similarly on the other side. $\square$

Proposition: $G$ is a quasi-group.

Proof: [by Berci] We need to show $ax=b$ and $ya=b$ have unique solutions $x,y \in G$. However, the defining axioms of a SIP-quasigroup give that $x=a^{-1}(ax) = a^{-1}(b)$ and $y=(ya)a^{-1} = (b)a^{-1}$, so those are the only possible solutions. Since $a(a^{-1}b) = b$ and $(ba^{-1})a=b$ by the lemma, these are in fact solutions, and $G$ is a quasi-group. $\square$

A special type of quasigroup are the IP-quasigroup (“inverse property quasigroups”). These are the quasigroups $G$ with functions $\lambda:G \to G$ and $\rho:G \to G$ such that $x^\lambda (xy) = (yx) {}^\rho x = y$. A loop with both those properties is called an IP-loop, and every IP-loop satisfies $x^\lambda = {}^\rho x$.

Proposition: Every SIP-quasigroup is an IP-quasigroup.

Proof: Simply take $x^\lambda = {}^\rho x = x^{-1}$.

But not all quasigroups are SIP-quasigroups

We give a family of examples of IP-quasigroups that are not SIP-quasigroups.

Let $A$ be an abelian group that has an element of order not 1 or 2, and define a multiplication on $G=A \times A$ by $(a_1,b_1) \cdot (a_2,b_2) = (a_1 + a_2, b_2-b_1)$. $G$ is a quasi-group (but not a loop) with both left and right inverse properties: define $x^\lambda = (-a,-b)$ and ${}^\rho x =(-a,b)$ so that $x^\lambda (xy) = (yx) {}^\rho x = y$.

Verifications:

$G$ is a quasi-group.

Proof: Let $a,b,c,d \in A$ and we try to uniquely solve $(a,b) \cdot(x,y) = (c,d)$. The left hand side is $(a+x,y-b)$ so the unique solution is $x=c-a$ and $y=b+d$. Hence $G$ is a quasi-group. $\square$

$G$ is not a loop.

Proof: Suppose by way of contradiction that $(x,y)$ is an identity. Then $(a,b) \cdot (x,y) = (a,b)$ implies $x=0$, $y=2b$, but this depends on $b$, so $(x,y)$ is not an identity. $\square$

$G$ satisfies the left and right inverse properties, so is an IP-quasigroup.

Proof: These are just calculations: $$(-a,-b) \cdot ( (a,b) \cdot (c,d) ) = (-a,-b) \cdot (a+c,d-b) = (-a+a+c,d-b-(-b)) = (c,d)$$ and $$((c,d) \cdot (a,b)) \cdot(-a,b) = (c+a,b-d) \cdot (-a,b) = (c+a-a,b-(b-d)) =(c,d). \square$$

$G$ does not satisfy the original questions axioms for any choice of $a^{-1}$.

Proof: The left and right inverses were uniquely determined. Set $(a,b)^{-1} = (x,y)$. Then $(a,b)^{-1} ( (a,b) \cdot (c,d) ) = (x,y) \cdot (a+c,d-b) = (x+a+c,d-b-y)$ so that $x=-a$ and $y=-b$. However, $((c,d) \cdot(a,b))\cdot(x,y) = (c+a,b-d) \cdot (x,y) = (c+a+x,y-(b-d))$ so that $x=-a$ and $y=b$. Unless $b=-b$, these differ. $\square$

$G$ is an IP-quasigroup that is not a SIP-quasigroup.

Proof: Since the left and right inverses are uniquely defined, the only way for $x^{-1}$ to exist is if the two inverse agree, but they disagree on any element $(a,b)$ with $b$ not of order 1 or 2. $\square$

A loop is a SIP-quasigroup iff IP-loop

If we assume the existence of an identity, then the SIP axioms become equivalent to the IP-loop axioms.

Proposition: For loops $G$, the following are equivalent:
* $G$ is a SIP-quasigroup
* $G$ is an IP-quasigroup
* $G$ is an IP-loop

Proof: This is simply because in a loop $x^\lambda = {}^\rho x$ so that inverses are automatically symmetric. $\square$

SIP-quasigroup that is not a loop

While the axioms for a SIP-quasigroup with identity are equivalent to the standard notion of "IP-loop" or "loop with both left and right inverse properties", a SIP-quasigroup need not have an identity. For example:

$$\begin{array}{c|ccc}
\times & 1 & 2 & 3 \\ \hline
1 & 1 & 3 & 2 \\
2 & 3 & 2 & 1 \\
3 & 2 & 1 & 3
\end{array}$$

In this quasi-group we have $x^2 = x$ and $xy=yx$ and $x^{-1}=x$ and $xy=z$ as long as $x,y,z$ are distinct (a so-called Steiner quasigroup). Hence it satisfies the original post's axioms, but has no identity.

Frequency of SIP-quasigroups

The multiplication table of a quasigroup is “Latin square” and if we sample uniformly at random from Latin squares we get one (amongst many) notion of frequency of quasigroups. Here are the observed frequencies after 10000 samples at each order:

$\begin{array}{c|rrrrrrrrrr}
                &   1   &   2   &   3   &   4   &   5   &   6   &   7   &   8   &   9   &   10   \\ \hline
Group           & 100\% & 100\% &  25\% &   2\% &       &       &       &       &       &        \\
SIP-quasigroup  &       &       &  25\% &   4\% &       &       &       &       &       &        \\
IP-quasigroup   &       &       &  50\% &   8\% &       &       &       &       &       &        \\
RIP-quasigroup  &       &       &       &  44\% &  10\% &       &       &       &       &        \\
quasigroup      &       &       &       &  42\% &  90\% & 100\% & 100\% & 100\%   &   100\%   &  100\% \\
\end{array}$

A quasigroup is counted only in the first line it qualifies, so that “SIP-quasigroup” here is short for “SIP-quasigroup that is not a group”. “RIP-quasigroup” normally stands for quasigroup with right inverse property, but I use it as shorthand for “quasigroup with right inverse property or left inverse property, but not both.” I don't report percentages less than 0.5%. In particular, loops that aren't groups never broke the 0.5% barrier.

Example quasigroups

Wikipedia has several examples of quasigroups, and it might be nice to know which ones are SIP-quasigroups. I consider groups under division, Steiner quasigroups, and Moufang loops.

Proposition: A group under division is a RIP-quasigroup. It is an IP-quasigroup iff it is a SIP-quasigroup iff the group is abelian. It is a loop iff  it is an IP-loop iff the group is an elementary abelian 2-group.

Proof: Define $g\ast h = g h^{-1}$ then $a \ast x = b$ has unique solution $x=b^{-1}a$ and $y \ast a = b$ has unique solution $y=ba$, so $(G,\ast)$ is a quasigroup. Suppose that $x$ is an identity, then $g = x\ast g = xg^{-1}$ implies $x=g^2$ must be constant (and so equal to the identity of the group). If $x \ast (g \ast h) = h$, then solving for $x$ one gets $x=hgh^{-1}$ so that $(G,\ast)$ is a LIP-quasigroup iff $G$ is abelian, and then $x^\lambda = x$. If $(h \ast g ) \ast y = h$, then solving for $y$ one gets $y=g$ so that ${}^\rho g =g$ is always a right inverse. $\square$

Proposition: A Steiner quasigroup is a commutative, idempotent SIP-quasigroup that is not a loop.

Proof: If $x \neq y$, then there is a unique Steiner triple $\{x,y,z\}$ and $xy=yx = z$, so we get that $x(xy) = (yx)x = y$ since $xz=zx=y$. If $x=y$ then $x(xx) = xx = x$ since Steiner quasigroups are idempotent. Note that for any $x$, $xx=x$, but that there is a $y$ with $xy=z$ and $x,y,z$ distinct, so $x$ cannot be the identity. $\square$

Proposition: A Moufang loop is an IP-loop, so also a SIP-quasigroup.

Indeed Moufang loops are exactly the loops that remain IP-loops even after scrambling the rows, columns, and labels.

ContentLicense: CC BY-SA 3.0


问:
Show that $C$ is a convex subset of $\mathbb R^4$
So I'm given the following information:
Let $f:\mathbb R^4\rightarrow \mathbb R$ be given by

$$f(x_1,x_2,x_3,x_4)=(x_1-x_3)^2+(x_2-x_4)^2$$

and $C\subseteq \mathbb R^4$ by

$$C=\{(x_1,x_2,x_3,x_4)\in \mathbb R^4 \mid x_1^2+(x_2-2)^2\leq 1,\ x_3-x_4\geq 0\}$$

I'm asked to show that $C$ is a convex subset of $\mathbb R^4$.

I have the following definition:

A convex subset $C\subseteq \mathbb R^d$ is a subset that contains the line segment between any two of its points $x, y \in C$, i.e.,
$$(1-t)x+ty\in C$$
for every number $t$ with $0\leq t\leq 1$.

I'm just not sure how to put these things together, anything to help me get started is greatly appreciated.

答:
HINT.-$V=(x,y,z,w)\in C\iff x^2+y^2-2y\le-3$ and $z\le w$.

Let $V_1$ and $V_2$ be in $C$; we need to show that $V_3=(1-t)V_1+tV_2=((1-t)x_1+tx_2,(1-t)y_2+ty_2,(1-t)z_1+tz_2,(1-t)w_1+tw_2)\in C$ and this is verify if and only if
$$\begin{cases}((1-t)x_1+tx_2)^2+((1-t)y_2+ty_2)^2-2(t(1-t)y_2+ty_2)\le -3\\(1-t)z_1+tz_2\le(1-t)w_1+tw_2 \end{cases}$$It is the only way to proceed. What remains  is a little bit of algebraic manipulation using the fact that $0\le t\le1$ and
$$ x_1^2+y_1^2-2y\le-3,\space\space z_1\le w_1\\x_2^2+y_2^2-2y\le-3,\space\space z_2\le w_2$$


问:
Name for vector without sign/sense? (Just module and direction)
If vectors are the equivalence class of oriented segments, what is the name for the equivalence class of unoriented segments?

These "unsigned vectors" would have a module (or normal value) and a direction but no sense/sign.

An example of a use for such entities is the cross product: if we don't know the orientation/headiness of the space, we can "unify" the two possible answers as a single "unsigned vector".

答:
I do not know of any special name for this concept.
In the context of Graph theory
there is a clear distinction between directed and
undirected graphs. The edges are sometimes known as
"lines". Thus, in graph theory, both undirected lines
and directed lines appear on an equal footing. Then,
depending on context, a "line" is assumed to be either directed or undirected.

In the context of projective space the
fundamental objects are "points" which are defined
to be equivalence classes of vectors that are said
to be related if they are nonzero scalar multiples
of each other. In your case the relation is much
more restricted and only a vector and its negative
are related.

Better is the context of convex sets in Euclidean
space. Any two distinct points determine the set of
convex linear combinations of the two points. This
is the line segment with the two given points as the
end points of the segment. The displacement from one
of the endpoints to the other is a vector in the
vector space associated to the Euclidean space. Of
course, the displacement going the other way is the
negative of the first vector.

Despite this, I don't know of any special name for
the equivalence class of a vector and its negative. However, in the specific case
that you mention of the cross product of
two vectors, there is the concept of
pseudovector. The Wikipedia article states:

In three dimensions, a pseudovector is associated with the curl of a polar vector or with the cross product of two polar vectors

It also states:

A number of quantities in physics behave as pseudovectors rather than polar vectors, including magnetic field and angular velocity. In mathematics, pseudovectors are equivalent to three-dimensional bivectors, from which the transformation rules of pseudovectors can be derived.

Thus, the reason for confusion about the
status of the cross product is that the
result is better thought of as a bivector
and not as an actual ordinary or polar
vector.


问:
transformation for my self
I am trying to prove this equation of integral transformation with Borel measure. But so far have no idea. Can you help me how to start.
Suppose $(X, A, \mu)$ is a $\sigma$-finite measure space and $f: X\to [0,\infty]$ an $A$-measurable function. Let $\nu$ be a Borel measure on $[0,\infty)$ such that $\nu(K)<\infty$ for any compact set $K$ in $[0,\infty)$. Show that

$$\int_X \nu([ 0 ,f(x))\;\mu(dx) = \int_0^\infty\mu(\{\ x\in X \ | f(x)>t\})\;\nu(dt).$$

答:
This is often called the layer-cake representation. Compute

\begin{align*}
\int_{X} \nu([0,f(x))\, \mu(dx) &= \int_{X} \int_{0}^{\infty} \mathbb{1}_{[0, f(x))}\, \nu(dt)\, \mu(dx) \\
&\overset{(i)}{=} \int_{0}^{\infty} \int_{X} \mathbb{1}_{[0,f(x))}\, \mu(dx)\, \nu(dt) \\
&\overset{(ii)}{=} \int_{0}^{\infty} \int_{X} \mathbb{1}_{\{x\, |\, f(x) > t \}}\, \mu(dx)\, \nu(dt) \\
&= \int_{0}^{\infty} \mu(\{x \, | \, f(x)>t \})\, \nu(dt).
\end{align*}

Step $(i)$ follows from Tonelli's (Fubini-Tonelli)  theorem (it applies as (a) $\mathbb{1}_{[0,f(x))}$ is measurable and non-negative and (b) $\mu \otimes \nu$ is $\sigma$-finite), while step $(ii)$ follows since $\mathbb{1}_{\{x\, |\, f(x) > t \}} = \mathbb{1}_{[0,f(x))}$.

NOTE: Sometimes you may see the notation $\mathbb{1}_{L(f;t)(x)} := \mathbb{1}_{\{x\, |\, f(x) > t \}}$ instead (cf.Layer cake representation)

Let us know in the comments if there are any questions.

Title: transformation for my self


问:
Prime decomposition in ring extensions
Let $R\subseteq R'$ be Dedekind domains, let $\mathfrak{p}$ be a nonzero prime ideal of $R$. Then $\mathfrak{p}R'$ is an ideal of $R'$ and it has a factorization
$$\mathfrak{p}R'=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_g^{e_g}$$
in which $\mathfrak{P}_1,\ldots\mathfrak{P}_g$ are distinct prime ideals of $R'$ and $e_1,\ldots,e_g$ are positive integers.

At this point, my textbook says something obscure (in my opinion). It says: note thet $\mathfrak{P_i}\cap R=\mathfrak{p}$ for each $i$. $\textbf{Thus}$  the integer $e_i$ is completely determined by $\mathfrak{P}_i$.

I can't understand the sentence in itself: what does it mean that the exponent is completely determined by the ideal?? Secondly, how the previous argument should show this?

Any suggestion/comment/help would be appreciated

答:
The point of the sentence is that given $\mathfrak P$, you can determine $\mathfrak p$ (in other words, there is a unique prime ideal of $R$ that contains $\mathfrak P$ in its factorization in $R'$, something that may not be obvious a priori, but which
follows from the formula $\mathfrak p = R \cap \mathfrak P$).  So given $\mathfrak P$, you can determine $\mathfrak p$, and hence (by factoring $\mathfrak p$ in $R'$) determine $e$.  Thus $e$ is an invariant of $\mathfrak P$
and the extension $R'/R$ alone, whereas a priori it is an invariant of the
triple $(R'/R, \mathfrak p, \mathfrak P)$. It is called the ramification index of $\mathfrak P$.

Title: Prime decomposition in ring extensions


问:
Is $s_n$ a multiple of $2^n$ for all $n \geq 0$?
Let's define:
$$a_{0,0} = 1$$
$$\forall k \ne 0, a_{0,k} = 0$$
$$\forall n \geq 0, \forall k, a_{n+1,k} = -(4n-2k+3)a_{n,k-1} + (2k+1)a_{n,k}$$

```<code>    |   0    1    2    3    4  ...
----+-----------------------------
  0 |   1
  1 |   1   -1
  2 |   1   -8    3
  3 |   1  -33   71  -15
  4 |   1 -112  718 -744  105
...   ...  ...  ...  ...  ... ...
</code>```

Up to the alternated sign, this is OEIS' sequence number A214406.
Now let's define:
$$s_n = \sum_{k=0}^n a_{n,k}$$
$$ s_0 = 1, s_1 = 0, s_2 = -4, s_3 = 24, s_4=-32, \dots $$

It appears that for all $n$, $s_n$ is a multiple of $2^n$. Is it provable?

Context/own efforts: so far, I was able to prove connections with the sequence of functions $(R_n)_{n \in \mathbb{N}}$ defined by:

$$ R_0(x)=1 $$
$$ R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2} \right)' $$

The connection is:

$$ s_n = 2^n R_n(1) $$

We even have:

$$ R_n(x) = 2^n \frac{P_n(x^2)}{(1+x^2)^{2n}} $$

where
$$ P_n(x) = \sum_{k=0}^n a_{n,k} x^k $$

I don't know if it helps, but this also means that the exponential generating function

$$ \sum_{n \geq 1} \frac{s_{n-1}}{2^{n-1}} \frac{x^n}{n!} $$

is the series reversion of

$$ \frac{\ln(x+1)}{2} + \frac{x}{2} + \frac{x^2}{4} = 0 + 1x + 0 x^2
+ \frac{1}{6}x^3 - \frac{1}{8}x^4
+ \frac{1}{10}x^5 - \frac{1}{12}x^6
\dots $$

I also established that

$$ s_{n+1} = -4 \sum_{k=0}^n (n-k) a_{n,k} $$

but this appears just to push the problem forward, of proving that $ \sum_{k=0}^n k a_{n,k} $ is a multiple of $2^{n-1}$ ...

Body:
I found a proof. Its sketch is:

Let
$$ f(x) = \frac{x}{1+x^2} $$

Then with $m=\lfloor n/2 \rfloor$ and $r = n \mod 2$, the following can be established by induction:

$$ f^{(n)}(x) = (-1)^m (n!) \left ( \sum_{k=0}^{m+r} (-1)^k \binom{n+1}{2k+1-r} x^{2k + 1 - r} \right) \frac{1}{ (1+x^2)^{n+1}} $$

Then one evaluates at 1, and observes, and proves, that:

$$ 2^{m+1} \frac{f^{(n)}(1)}{n!} $$

is indeed very particular:

$$ 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 1, \dots $$

In other terms,
$$
 2 f^{(n)}(1) = \frac{n!}{2^{\lfloor n/2 \rfloor}} \times
   \begin{array}{|rl|}
   +1, & n \in 8 \mathbb{N} + \{ 0, 3, 6 \} \\
    0, & n \in 8 \mathbb{N} + \{ 1, 5 \} \\
   -1, & n \in 8 \mathbb{N} + \{ 2, 4, 7 \} \\
   \end{array}
$$
For even shorter notations, define
$$ g = 2f $$
$$ g_n = g^{(n)} $$

Then
$$
 \bbox[white, border: 2px solid black]{
   g_n(1) = \frac{n!}{2^{\lfloor n/2 \rfloor}} \times
   \begin{array}{|rl|}
   +1, & n \in 8 \mathbb{N} + \{ 0, 3, 6 \} \\
    0, & n \in 8 \mathbb{N} + \{ 1, 5 \} \\
   -1, & n \in 8 \mathbb{N} + \{ 2, 4, 7 \} \\
   \end{array}
  }
  $$

Now,

$$
   \begin{array}{lcl|lcll}
   R_0 & = & 1 & R_0(1) & = & 1 & = 1 \\
   R_1 & = & \left( R_0 g \right)' = \left( g_0 \right)'     = g_1             & R_1(1) & = & \frac{1!}{2^{\lfloor 1/2 \rfloor}} \times 0 & = 0 \\
   R_2 & = & \left( R_1 g \right)' = \left( g_1 g_0 \right)' = g_2 g_0 + g_1^2 &
    R_2(1) & = & ( \frac{2!}{2^{\lfloor 2/2 \rfloor}} \times -1)(\frac{0!}{2^{\lfloor 0/2 \rfloor}} \times +1) + (\frac{1!}{2^{\lfloor 1/2 \rfloor}} \times 0)^2 = (-1)(+1) - (0)^2 & = -1 \\
   R_3 & = & \left( R_2 g \right)' = \left( (g_2 g_0 + g_1^2) g_0 \right)' = \left( g_2 g_0^2 + g_1^2 g_0 \right)' = g_3 g_0^2 + 4 g_2 g_1 g_0 + g_1^3 &
    R_3(1) & = &
    ( \frac{3!}{2^{\lfloor 3/2 \rfloor}} \times +1)(\frac{0!}{2^{\lfloor 0/2 \rfloor}} \times +1)^2
    + 4 \times ( \frac{2!}{2^{\lfloor 2/2 \rfloor}} \times -1)(\frac{1!}{2^{\lfloor 1/2 \rfloor}} \times 0)(\frac{0!}{2^{\lfloor 0/2 \rfloor}} \times +1)
    + (\frac{1!}{2^{\lfloor 1/2 \rfloor}} \times 0)^3
    = 3 \times (+1)^2 + 4 \times (-1)(0)(+1) + (0)^3 & = 3 \\
   \dots \\
   \end{array}
$$

Clearly, behind the scene are the partitions of $n$ and their coefficients listed in OEIS sequence number A145271.
The following formula holds,
$$
 \bbox[white, border: 2px solid black]{ R_n(1) = \sum_{P \mbox{ partition of } n} \mbox{A145271}(n, P) \times \prod_{p \mbox{ part of } P} g_{\mbox{size}(p)}(1) }
$$

with the nice property that it's not worth computing the contributions of partitions that contain a part equal to 1, or 5, or a multiple of 4 plus 1 (they are 0).

Finally, as a sum of products of integers, $R_n(1)$ is an integer; and $s_n$ is thus a multiple of $2^n$.


问:
Weak convergence of measures and Regularized optimal transport
Let W_p be the Wasserstein metric on the space of probability measures on $\mathcal{X}$, a compact subset of $\mathbb{R}^n$. I know that $\mu_k \rightarrow \mu$ (under weak convergence of measures) iff $W_p(\mu_k, \mu) \rightarrow 0$. Now instead of working with $W_p$, I am using the entropic regularized version of $W_p$, call it $W_{p, \lambda}$ as introduced in https://papers.nips.cc/paper/2013/file/af21d0c97db2e27e13572cbf59eb343d-Paper.pdf. Due to some pathological properties of $W_{p, \lambda}$, let us assume that the sequence $\mu_k$ is not constant and also the number of samples drawn from our measures are arbitrarily large. Then my question is: Is it still true that
$\mu_k \rightarrow \mu$ iff $W_{p, \lambda}(\mu_k, \mu) \rightarrow 0$

答:
So I’ve found the answer. [This article by Feydey et al][1] proves that for a compact metric space $X$, and a Lipschitz cost $c$. Then the Sinkhorn Divergence does metrize the weak convergence, i.e $\tilde W_{p,\epsilon}(\mu_n,\mu)\to 0 \iff \mu_n \rightharpoonup \mu$.

Just note that the Sinkhorn divergence is an unbiased version of the regularized cost, where:

$$
\tilde W_{p,\epsilon}(\alpha, \beta)^p =
2W_{p,\epsilon}(\alpha,\beta) -W_{p,\epsilon}(\alpha,\alpha)-W_{p,\epsilon}(\beta,\beta)
$$
[1]: https://hal.archives-ouvertes.fr/hal-01898858/document


问:
Prove that $n\log(n) = O(2^{\log^2(n)})$
I have to prove that $n\log(n) = O(2^{\log^2(n)})$, where $\log(x) = \log_2x$. I tried saying that: $2^{\log^2(n)} = 2^ {\log(n)\log(n)}= 2^{\log(n)^{\log(n)}} = n^{log(n)}$. I also know that $n\log(n) = \log(n^{n})$.I thought of just saying that it's true because the first function is logarithmic, but I am not sure this is correct. I also tried with the fact that $n\log(n) = O(2^n)$, but proving that $2^{\log(n)} = O(2^n)$ doesn't help. Any recommendations would be appreciated.

答: Note that $\log n + \log \log n \le 2 \log n \le (\log n)^2$ provided that $n \ge 4$ in which case $$n \log n = 2^{\log n + \log \log n} \le 2^{(\log n)^2}.$$


答:
Actually, $n\log n=o(2^{\log^2 n})$ (little oh), which is a stronger statement.
To see the, consider the functions $f(x)=x\log x$ and $g(x)=2^{\log^2 x}$.
Clearly both functions tend to infinity as $x\to\infty$. Using L'hopital's rule,

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\frac{
\frac{1+\ln x}{\ln 2}
}
{\frac{2\exp(\frac{\ln^2 x}{\ln 2})\ln x}{x\ln 2}}=\frac{x\ln 2(1+\ln x)}{2\ln 2\exp(\frac{\ln^2 x}{\ln 2})\ln x}$$
Since for sufficiently large $x$ we have $\ln x>2\ln 2$, we deduce that
for sufficiently large $x$ $\exp(\frac{\ln^2 x}{\ln 2})>\exp(2\ln x)=x^2$, and so the r.h.s of the displayed equation above is bounded above by
$$\frac{x\ln 2(1+\ln x)}{2\ln 2\cdot x^2\ln x}=O(\frac{1}{x})$$

Therefore $f(x)=o(g(x))$ as $x\to\infty$, and in particular, $n\log n=o(2^{\log^2 n})$ as $n\to\infty$.

Remark. The argument above shows that actually a much stronger statement is true, namely for every positive integer $k$,
$$\frac{f(x)}{g(x)}=o(x^{-k})\quad (x\to\infty)$$
Therefore, the ratio
$$\frac{n\log n}{2^{\log^2 n}}$$
tends to zero faster than any power of $\frac{1}{n}$ as $n\to\infty$.
In other words, for every positive integer $k$,

$$\lim_{n\to\infty}\frac{n^k\log n}{2^{\log^2 n}}$$

This of course answers the OP's question, although its  primary purpose is to show that the OP's question, is in a sense, "not the right question".

Title: Prove that $n\log(n) = O(2^{\log^2(n)})$

Body: I have to prove that $n\log(n) = O(2^{\log^2(n)})$, where $\log(x) = \log_2x$. I tried saying that: $2^{\log^2(n)} = 2^ {\log(n)\log(n)}= 2^{\log(n)^{\log(n)}} = n^{log(n)}$. I also know that $n\log(n) = \log(n^{n})$.I thought of just saying that it's true because the first function is logarithmic, but I am not sure this is correct. I also tried with the fact that $n\log(n) = O(2^n)$, but proving that $2^{\log(n)} = O(2^n)$ doesn't help. Any recommendations would be appreciated.


问:
Prove that a Collection of subsets of $E$ is a $\sigma$-algebra
Prove that:$\;$ $\mathscr C$ is a $\sigma$-algebra,with$\;$$E$ $\;$ is a infinite set.
$$\mathscr C\equiv\{A\subset\mathbb E\,|\,A\text{ is countable or }A^c\text{ is countable}\}.$$

答:
We check the three conditions in the definition one by one.

First, $\mathbb{E} \in \mathscr{C}$ because $\mathbb{E}^c = \emptyset$ is countable.

Second, we verify that $\mathscr{C}$ is closed under complementation. Fix $A \in \mathscr{C}$. Either $A$ is countable or $A^c$ is countable. We need to prove that $A^c \in \mathscr{C}$. If $A$ is countable, then $A^c \in \mathscr{C}$ by virtue of $A^c$ having a countable complement. If $A^c$ is countable, then $A^c \in \mathscr{C}$ by virtue of $A^c$ being countable.

Finally, we verify that $\mathscr{C}$ is closed under countable unions. Fix $A_i \in \mathscr{C}$ for $i = 0, 1, 2, \dots$ and let $B = \cup_{i=0}^\infty A_i$. We need to prove that $B \in \mathscr{C}$. Either all $A_i$ are countable or there is $A_k$ whose complement is countable. In the former case, $B$ is countable as a countable union of countable sets. In the latter case, $B$ has countable complement, because $B^c \subset A_k^c$.


问:
Limits of indefinite Integrals
My professor went through this very quickly and I didn't quite understood. This appeared on exercises without any context and my textbook/class notes do not talk about these limits - but I know it is related to the Fundamental Theorem of Calculus and L'Hopital rule.

How does one solve this?

$$\lim_{x \rightarrow \infty} \frac{\int_0^x\frac{\sin(t)dt}{t}-x}{x^3}$$

My guess is: I should integrate in terms of $t$ to get $F(x) - F(0)$. Then I should use L'Hopital rule in order to solve the limit. Is this right?

If so: what does it mean geometrically? What if I had a function  (ex. $x^2$ on the boundaries of the integral - should I use the composition of functions? And what if the function is undefined there? In this case we have it on $t=0$.

答:
HINT

The idea is to show that as $x \to \infty$, the numerator diverges, hence, when you look at the limit of the whole fraction, L'Hospital's rule applies. To take the derivative of the numerator, you will need the Fundamental Theorem of Calculus, i.e.
$$
\frac{d}{dx} \left[\int_a^{x} f(t) dt \right] = f(x).
$$

Can you finish this?

Title: Limits of indefinite Integrals


问:
Why can we factor out $X$ in $E(E(XY∣X))$?
Using the law of total expectation, $E(XY)=E(E(XY∣X))=E(XE(Y∣X))$

I understand the first equality is part of the definition, but why are we able to treat $X$ as a constant in the second equality and factor it out?

答:
Because $X$ is constant with respect to $X$.

The idea of $\mathbb{E}(Y|X)$ is the probability once $X$ is known, so the result is a function of $X$. Just like $\mathbb{E}(aY|X) = a\mathbb{E}(Y|X)$ when $a$ is constant wrt $X$, so too $\mathbb{E}(XY|X) = X\mathbb{E}(Y|X)$ since once $X$ is known, it is constant wrt itself.


问:
Prove that there exists $A \in F$ (σ-algebra) such that $P(A) = a$
The question

Let $\left( \Omega,F,P \right)$ be a probability space.

A set $E\in F $ is called  K-set iff

$P(E)>0$ and $B \subseteq E$ with $B \in F$ $\Rightarrow P(B) = P(E)$ or $P(B) = 0$

Note: I don't know how the above term is translated in English. I looked it up on Google but found nothing so I will name it for the sake of this post as K-set. If anyone knows the name of this term please tell me.

Suppose that the probability measure $P$ has no $K$-sets. Prove that for every $a \in [0,1]$ there exists $A \in F$ such that $P(A) = a$

Attempt

So $P$ has not $K$-sets. From this I understand that

$\forall A \in F$ with $P(A)>0$ and $\forall B \subseteq A$ with $B \in F$ we have $P(B) \neq P(A)$ and $P(B)>0$

For $a=0$, $P(\emptyset ) = 0$ and $\emptyset \in F$ since $F$ is $\sigma$-algebra.

For $a=1$, $P(\Omega) = 1$ and $\Omega \in F$ since $F$ is $\sigma$-algebra.

For $a \in (0,1)$ I thought of the idea that we can find a sequence $\{B_{n} \}$ of sets in $A$ such that $B_{n+1} \subset B_{n} \forall n $

I also tried using the Borel-Cantelli Theorem and other Propositions involving $\lim \inf B_n $ and $\lim \sup B_n $ but couldn't get to the desired result. I know that $P(B_n)$ will get smaller and smaller as $n$ gets bigger. But how do I know that $\forall a \in (0,1)$ I can find such $B_n$ such that $P(B_n) =a$ ? If this is the approach anyway.

答:
In the following, unless otherwise specified, all sets under
consideration are measurable. That is, if we say "let $A\subseteq\Omega$"
we actually mean $A\in\mathcal{F}$.

Lemma 1:
Assume that $(\Omega, \mathcal{F}, P)$ does not have $K$-set, then we have: For any $A\subseteq\Omega$ with
$P(A)>0$ and $\varepsilon>0$, there exists $B\subseteq A$ such
that $0<P(B)<\varepsilon$.

Proof of Lemma 1: Let $A\subseteq\Omega$ with $P(A)>0$ and $\varepsilon>0$
be given. Choose $B_{0}\subseteq A$ such that $0<P(B_{0})<P(A)$.
Define $A_{1}=\begin{cases}
B_{0}, & \mbox{if }P(B_{0})\leq\frac{1}{2}P(A)\\
A\setminus B_{0}, & \mbox{otherwise}
\end{cases}.$ Note that $0<P(A_{1})\leq\frac{1}{2}P(A).$ Suppose that $A\supseteq A_{1}\supseteq\ldots\supseteq A_{n}$
have been chosen. Choose $B_{n}\subseteq A_{n}$ such that $0<P(B_{n})<P(A_{n}).$
Define $A_{n+1}=\begin{cases}
B_{n}, & \mbox{if }P(B_{n})\leq\frac{1}{2}P(A_{n})\\
A_{n}\setminus B_{n}, & \mbox{otherwise}
\end{cases}.$ Clearly $0<P(A_{n+1})\leq\frac{1}{2}P(A_{n})$ and hence $P(A_{n})\rightarrow0.$
In particular, for sufficiently large $n$, $0<P(A_{n})<\varepsilon$.

We go back to your question. Let $a\in[0,1]$ be given. If $a=0$
or $a=1$, we may choose $A=\emptyset$ or $A=\Omega$ respectively.
Suppose that $a\in(0,1).$ Define an equivalence relation $\sim$ on
$\mathcal{F}$ by $A\sim B$ iff $P(A\Delta B)=0$, where $A\Delta B=(A\setminus B)\cup(B\setminus A)$.
Intuitively, $A\sim B$ means that $A=B$ modulo a $P$-null set.
If $A\in\mathcal{F}$, we denote the equivalence class containing $A$
by $\tilde{A}$. Note that if $B\in\tilde{A}$, then $P(B)=P(A)$,
so it is well-defined to specify $P(\tilde{A})=P(A)$. Define a binary
relation on $\mathcal{F}/\sim$ by $\tilde{A}\preceq\tilde{B}$ iff
$P(A\setminus B)=0$. It is routine to verify that $\preceq$ is well-defined
(i.e., if $A_{1},A_{2}\in\tilde{A}$ and $B_{1},B_{2}\in\tilde{B}$,
then $P(A_{1}\setminus B_{1})=0$ iff $P(A_{2}\setminus B_{2})=0$).
Moreover, $\preceq$ is a partial ordering on $\mathcal{F}/\sim$.
Intuitively, $\tilde{A}\preceq \tilde{B}$ means that $A\subseteq B$ modulo a $P$-null set.

Let $\mathcal{C}=\{\tilde{A}\in\mathcal{F}/\sim\,\,\,\mid\,\,\, P(\tilde{A})\leq a\}$.
Clearly $\mathcal{C}$ is non-empty. We go to show that $\mathcal{C}$
contains a $\preceq$-maximal element. Let $\mathcal{C}_{1}=\{\tilde{A}_{i}\mid i\in I\}\subseteq\mathcal{C}$
be a chain, where $I$ is an index set. Let $b=\sup_{i\in I}P(\tilde{A}_{i})$.
Note that $b\leq a$. For each $n\in\mathbb{N}$, there exists $i_{n}\in I$
such that $P(\tilde{A}_{i_{n}})>b-\frac{1}{n}.$ Define $B_{n}=\cup_{k=1}^{n}A_{i_{k}}$.
Since $\mathcal{C}_{1}$ is a chain, we can prove by induction that
$B_{n}\sim A_{i_{k}}$ for some $k\in\{1,\ldots,n\}$. Therefore $b\geq P(B_{n})\geq P(A_{i_{n}})>b-\frac{1}{n}$. Let $B=\cup_{n}B_{n}$.
By continuity of measure, we have $P(B)=b\leq a$, so $\tilde{B}\in\mathcal{C}$.
We assert that $\tilde{B}$ is an upper bound of the chain $\mathcal{C}_{1}$.
Prove by contradiction. Suppose the contrary that there exists $j\in I$
such that $P(A_{j}\setminus B)>0$. Choose $n\in\mathbb{N}$ such
that $P(A_{j}\setminus B)>\frac{1}{n}$. Since $\mathcal{C}_{1}$
is a chain, we have $\tilde{A}_{j}\preceq \tilde{A}_{i_{n}}$ or $\tilde{A}_{i_{n}}\preceq \tilde{A}_{j}$.
If $\tilde{A}_{j}\preceq \tilde{A}_{i_{n}}$, then $P(A_{j}\setminus B)\leq P(A_{i_{n}}\setminus B)=0$
which is a contradiction. If $\tilde{A}_{i_{n}}\preceq \tilde{A}_{j}$, we have
\begin{eqnarray*}
b & \geq & P(A_{j})\\
 & = & P(A_{j}\setminus B)+P(A_{j}\cap B)\\
 & > & \frac{1}{n}+P(A_{i_{n}}\cap B)\\
 & = & \frac{1}{n}+P(A_{i_{n}})\\
 & > & \frac{1}{n}+(b-\frac{1}{n})
\end{eqnarray*}
which is also a contradiction. By Zorn's lemma, $\mathcal{C}$ contains
a maximal element $\tilde{A}$.

We go to show that $P(\tilde{A})=a$ by contradiction, then it will follow that $P(A)=a$. Suppose the
contrary that $P(\tilde{A})<a$. Choose $\varepsilon>0$ such that
$P(\tilde{A})+\varepsilon<a$. Clearly $P(A^{c})>0$. By Lemma 1,
we can choose $B\subseteq A^{c}$ such that $0<P(B)<\varepsilon$.
Now $P(\widetilde{A\cup B})=P(A)+P(B)<P(A)+\varepsilon<a$. Therefore
$\widetilde{A\cup B}\in\mathcal{C}$. Clearly $\tilde{A}\preceq\widetilde{A\cup B}$,
so $\tilde{A}=\widetilde{A\cup B}$ by maximality of $\tilde{A}$.
However, $P(A\Delta(A\cup B))=P(B)>0$, contradicting to $A\sim A\cup B$.


问:
Find the eigenvectors and eigenvalues of a linear transformation
Given the linear transformation $T(at^2+bt+c)=(2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$, how can I find the eigenvectors and eigenvalues? I know that general idea is to solve $T(v)=\lambda v$. From there, I get:
$$\lambda at^2+ \lambda bt+ \lambda c = (2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$$

This implies,

$$2a - c = \lambda a$$
$$2a + b - 2c = \lambda b$$
$$-a + 2c = \lambda c$$

This is where I get stuck. How can I solved this system? Any help is greatly appreciated!

答: That system is equivalent to the system$$\left\{\begin{array}{l}(2-\lambda)a-c=0\\2a+(1-\lambda)b-2c=0\\-a+(2-\lambda)c=0,\end{array}\right.$$which has non-trivial solutions if abd only if$$\begin{vmatrix}2-\lambda&0&-1\\2&1-\lambda&-2\\-1&0&2-\lambda\end{vmatrix}=0.\tag1$$But the LHS of $(1)$ is$$(1-\lambda)\begin{vmatrix}2-\lambda&-1\\-1&2-\lambda\end{vmatrix}=(1-\lambda)\bigl((2-\lambda)^2-1\bigr)=-(\lambda-1)^2(\lambda-3).$$Can you take it from here?


答:
Note that you are looking for non-zero solutions to $T(v) = \lambda v$. Rewrite your system as
$$
(2 - \lambda )a + 0 b - c = 0\\
2a + (1 - \lambda) b - 2c = 0\\
-a + 0b + (2 - \lambda c) = 0
$$
That is, we are looking for non-zero solutions to the equation
$$
\pmatrix{2 - \lambda & 0 & -1\\ 2 & 1-\lambda & -2\\-1 & 0 & 2 - \lambda} \pmatrix{a\\b\\c} = \pmatrix{0\\0\\0}.
$$
However, note that for a square matrix $M$, the equation $Mx = 0$ will have a non-zero solution if and only if $\det(M) = 0$. The determinant of $M$ is a polynomial on the variable $\lambda$. If you compute and expand correctly, you should get
$$
\det(M) = 0 \implies \\
-λ^3 + 5 λ^2 - 7 λ + 3 = 0 \implies\\
(\lambda - 3)(\lambda - 1)^2 = 0.
$$
The zeros of the polynomial are $1$ and $3$, which means that those are the eigenvalues of the transformation.

Now that you have the eigenvalues, you can plug these values of $\lambda$ back in and solve for the eigenvectors.

Title: Find the eigenvectors and eigenvalues of a linear transformation

Body:
Given the linear transformation $T(at^2+bt+c)=(2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$, how can I find the eigenvectors and eigenvalues? I know that general idea is to solve $T(v)=\lambda v$. From there, I get:
$$\lambda at^2+ \lambda bt+ \lambda c = (2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$$

This implies,

$$2a - c = \lambda a$$
$$2a + b - 2c = \lambda b$$
$$-a + 2c = \lambda c$$

This is where I get stuck. How can I solved this system? Any help is greatly appreciated!


答:
Factoring out $X$ corresponds to the following manipulation in the case where $(X,Y)$ have a joint density:

$$
\begin{split}
E(XY)
 &= \iint xy f(x,y)dy dx \\
 &= \int_x \left(\int_y xy f(y\mid x)f(x)dy\right)dx \\
 &= \int_x x\left(\int_y y f(y\mid x)dy\right)f(x)dx
\end{split}
$$

Title: Why can we factor out $X$ in $E(E(XY∣X))$?

Body:
Using the law of total expectation, $E(XY)=E(E(XY∣X))=E(XE(Y∣X))$

I understand the first equality is part of the definition, but why are we able to treat $X$ as a constant in the second equality and factor it out?


问:
Finding Integer Solutions to $x^2-y^2-n=0$
I am trying to find the integer solutions to the equation $x^2-y^2-n=0$.

Effectively, I am trying to find when the difference of two perfect squares is $n$

I have tried using a modified Newton's method to find the nearest point where the equation reaches parity; however, it ends up finding non-integer solutions. I am wondering if there is a way to factor/find all the integer solutions to this equation.

For my algorithm, taking the square root is prohibitively costly and I am trying to avoid using it.

Thanks for the help!

答:
Given $x^2-y^2-n=0\implies y^2+n=x^2\quad$ if $n$ is a perfect square, we have the equation for the Pythagorean theorem where $A^2+B^2=C^2$. The B-value can be any multiple of $4$ so $n=(4z)^2, z\in\mathbb{N}$. There are infinite solutions. Finding them requires a square root but all values of $\sqrt{n}$ are known beforehand so there are always values of $x,y$ to be found in a finite search.

Let's begin with a form of Euclid's formula $F(m,k)$ where
$\quad A=m^2-k^2\quad B=2mk\quad C=m^2+k^2.\quad$ We can find $m,k$ by solving $B$ for $k$ and test a finite number of m-values to see which yield integers.

\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\ge 1$
$$B=44\implies\quad \bigg\lfloor \frac{1+\sqrt{88+1}}{2}\bigg\rfloor =5 \le m \le \frac{44}{2}=22\\\land \quad m\in\{11,22\}\implies k\in\{2,1\}$$

$$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)$$

Alternatively, you can just find a list of triples and translate them to fit your equation. For instance, here are examples of triples $A,B,C$ translated to $y,n,x$.

$3,4,5\rightarrow 3,16,5\quad 15,8,17\rightarrow 15,64,17\quad 5,12,13\rightarrow 5,144,13\quad 35,16,37\rightarrow 35,256,37$


问:
Significance of Equal Angle Triangle Center
While playing around with triangle centers and came across one I did not know, the center where each triangle corner heading is equally spaced ($120^\circ$ spacing).

Does this specific triangle center have a name, and does it have any significance?

答:
This is the first isogonic centre (Kimberling $X_{13}$), which in a triangle whose angles are all less than $120^\circ$ is also the Fermat point, i.e. minimises the sum of distances to the three vertices or solves the Steiner problem for the triangle.

For a triangle with an angle at least $120^\circ$, the Fermat point is exactly on the corresponding vertex.


问:
Finding the probability that at most n events take place on any interval during a given time period
I am interested in the general case, but let us start with a smaller example. Suppose that cars arrive to a street with intensity $\lambda$ per minute. We would like to know the probability that at least two cars have arrived on the street during any five minute interval in the next hour. How can we find this?

My initial thought was to use complementary event: 1 - the probability that at most one car is on the street in any five minute period during the next hour. Hence my reasoning was something like $1 - \int_0^{55}\mathbb{P}(N(s + 5) - N(s) \leq 1)ds - \int_{55}^{60}\mathbb{P}(N(60) - N(55 + s) \leq 1)ds$.

But I quickly realized that finding this probability might not be that easy, since the intervals we are considering overlap, namely $N(5)$ and $N(5 + s)$ overlap except for the infinitesimal point $s$. So then, is the correct way to integrate just $\mathbb{P}(N(s) \leq 1$ over the region? Moreover, I think that my line of reasoning is missing something critical, since I do not see a reason, why the summation would not end up being negative.

答:
Generating Function Approach

Let's compute the probability that no $5$ minute span has more than $1$ car entry.

Poisson with $\bf{1}$ Minute Resolution

Poisson says the probability that $0$ cars enter the street in a given minute is $e^{-\lambda}$, and the probability that $1$ car enters the street in a given minute is $\lambda e^{-\lambda}$.

Let a minute with one car be represented by '${+}$', which has a probability of $\lambda e^{-\lambda}$. Let a minute with no cars be represented by '${-}$', which has a probability of $e^{-\lambda}$.

The $60$ minute span can be uniquely built from atoms of '${-}$' and '${+}{-}{-}{-}{-}$' followed by an optional terminal atom of '$\text{+}$', '$\text{+}{-}$', '$\text{+}{-}{-}$', '$\text{+}{-}{-}{-}$'

This gives a generating function of
$$
f_\lambda(x)=\frac{1+\overbrace{\ \ \lambda e^{-\lambda}x\ \ }^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}}{1-\underbrace{\ \ \ e^{-\lambda}x\ \ \ }_{-}-\underbrace{\lambda e^{-5\lambda}x^5}_{+----}}\tag1
$$
The probability that in no $5$ minute span of the hour there is more than $1$ car is
$$
\begin{align}
\left[x^{60}\right]f_\lambda(x)
&=e^{-60\lambda}\left(1+60\lambda+1540\lambda^2+22100\lambda^3+194580\lambda^4\right.\\
&\left.+1086008\lambda^5+3838380\lambda^6+8347680\lambda^7+10518300\lambda^8\right.\\[2pt]
&\left.+6906900\lambda^9+1961256\lambda^{10}+167960\lambda^{11}+1820\lambda^{12}\right)
\end{align}\tag2
$$
Computing the complementary probability, we get the probability that two or more cars will enter in a single $5$ minute period during an hour:

The Generating Function

In order to prevent any $5$ minute span from having more than $1$ car entry, above, we break down the $60$ minutes into $1$ minute atoms:

The $60$ minute span can be uniquely built from atoms of '${-}$' and '${+}{-}{-}{-}{-}$' followed by an optional terminal atom of '$\text{+}$', '$\text{+}{-}$', '$\text{+}{-}{-}$', '$\text{+}{-}{-}{-}$'

and we compute the probabilities for each of these atoms using the Poisson Distribution:

Poisson says the probability that $0$ cars enter the street in a given minute is $e^{-\lambda}$, and the probability that $1$ car enters the street in a given minute is $\lambda e^{-\lambda}$.

In the generating function, we represent an '${+}$' by $\lambda e^{-\lambda}x$ and an '${-}$' by $e^{-\lambda}x$. Thus, the power of $x$ counts the minutes and the coefficients accumulate the probabilities. Thus, for each of these atoms, we include a factor of
$$
\overbrace{\phantom{i^5}e^{-\lambda}x\phantom{i^5}}^{-}+\overbrace{\lambda e^{-5\lambda}x^5}^{+----}\tag3
$$

To account for any number of the standard atoms ('${+}$' and '${+}{-}{-}{-}{-}$'), we total the product of all powers of the standard atom:
$$
\sum_{k=0}^\infty\left(e^{-\lambda}x+\lambda e^{-5\lambda}x^5\right)^k=\frac1{1-e^{-\lambda}x-\lambda e^{-5\lambda}x^5}\tag4
$$
Since products of the standard atoms of length $4$ or more will end in at least $4$ '${-}$'s, we add optional atoms of '${+}$', '${+}{-}$', '${+}{-}{-}$', or '${+}{-}{-}{-}$'. This is included in the generating function as a factor of
$$
\overbrace{\phantom{iw^2}1\phantom{iw^2}}^\text{none}+\overbrace{\phantom{i^2}\lambda e^{-\lambda}x\phantom{i^2}}^{+}+\overbrace{\lambda e^{-2\lambda}x^2}^{+-}+\overbrace{\lambda e^{-3\lambda}x^3}^{+--}+\overbrace{\lambda e^{-4\lambda}x^4}^{+---}\tag5
$$
Thus, we arrive at the generating function, $f_\lambda(x)$, given in $(1)$. To get the probability for a span of $60$ minutes, we look at the coefficient of $x^{60}$: $\left[x^{60}\right]f_\lambda(x)$.

Poisson with $\boldsymbol{\frac1n}$ Minute Resolution

We can handle finer resolutions using atoms of '${-}$' and '${+}\overbrace{{-}{-}\cdots{-}}^{5n-1}$'. To simplify notation, we will write $e^{-\lambda/n}x=u$.
$$
\begin{align}
g_{\lambda,n}(x)
&=\frac{1+\frac\lambda{n}e^{-\lambda/n}x+\frac\lambda{n}e^{-2\lambda/n}x^2+\cdots+\frac\lambda{n}e^{-(5n-1)\lambda/n}x^{5n-1}}{1-e^{-\lambda/n}x-\frac\lambda{n}e^{-5\lambda}x^{5n}}\tag{6a}\\
&=\frac{1+\frac\lambda{n}u\,\frac{1-u^{5n-1}}{1-u}}{1-u-\frac\lambda{n}u^{5n}}\tag{6b}\\
&=\frac{\left(1-u-\frac\lambda{n}u^{5n}\right)+\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)\left(1-u\right)}\tag{6c}\\
&=\frac1{1-u}+\frac{\frac\lambda{n}u}{\left(1-u-\frac\lambda{n}u^{5n}\right)(1-u)}\tag{6d}\\
&=\frac1{1-u}+\left(\frac1{1-u-\frac\lambda{n}u^{5n}}-\frac1{1-u}\right)\frac1{u^{5n-1}}\tag{6e}
\end{align}
$$
Since all coefficients of $\frac1{1-u}$ are $1$, we get
$$
\begin{align}
\left[x^{60n}\right]g_{\lambda,n}(x)
&=e^{-60\lambda}\!\left[u^{60n}\right]g_{\lambda,n}(x)\tag{7a}\\[9pt]
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\frac1{1-u-\frac\lambda{n}u^{5n}}\tag{7b}\\
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\left(u+\frac\lambda{n}u^{5n}\right)^k\tag{7c}\\
&=e^{-60\lambda}\!\left[u^{65n-1}\right]\sum_{k=0}^\infty\sum_{j=0}^k\binom{k}{j}u^{k-j}\left(\frac\lambda{n}\right)^ju^{5nj}\tag{7d}\\
&=e^{-60\lambda}\sum_{j=0}^{12}\binom{65n-1-(5n-1)j}{j}\left(\frac\lambda{n}\right)^j\tag{7e}
\end{align}
$$
For $n=1$, $\text{(7e)}$ agrees with $(2)$.

Infinite Resolution

The limit as $n\to\infty$ of $\text{(7e)}$ is
$$
\begin{align}
p(\lambda)
&=e^{-60\lambda}\sum_{j=0}^{12}\frac{(65-5j)^j}{j!}\lambda^j\tag{8a}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{8b}
\end{align}
$$

At $\frac14$ car per minute, the probability is $99.91\%$ that two or more cars will enter in a single $5$ minute period during an hour.


问:
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$
$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

Can we write it as following

$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$

Let's see what happens:-

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$

In the same way for further terms, we will get $0$

Let's also confirm for general term

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$

So the whole expression $E$ will be zero

But actual answer is $1$

Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$

Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?

I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.

But I got the following counter thought:-

$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$

As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$

$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$

$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$

Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.

But surprisingly $\dfrac{1}{3}$ is the correct answer.

I am feeling very confused in these two things. Please help me.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 1}^{n}{n \over n^{2} + k}} =
n\sum_{k = 0}^{n - 1}{1 \over k + 1 + n^{2}}
\\[5mm] = &\
n\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + n^{2}} -
{1 \over k + n + 1 + n^{2}}} = n\pars{H_{n^{2} + n} - H_{n^{2}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
n\braces{\bracks{\ln\pars{n^{2} + n} + \gamma +
{1 \over 2n^{2} + 2n}} -
\bracks{\ln\pars{n^{2}} + \gamma + {1 \over 2n^{2}}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
n\ln\pars{1 + {1 \over n}}
\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\bbx{1} \\ &
\end{align}
$\ds{H_{z}}$ is a Harmonic Number and $\ds{\gamma}$ is the Euler-Mascheroni Constant.


问:
Bijection example given set of integers to natural numbers with condition.
State an example of a bijection $f:\mathbb{Z}\to\{ 2n \space|\space n\in\mathbb{N} \}$.

I know for a bijection the two sets must have the same cardinality. An example I came up with for this condition is:
$$
f(n)=2n^2, f:\{ ...,-3,-2,-1,0,1,2,3,... \}\to\{ 2,4,6,8,... \}
$$

But I'm not sure if this is valid or if I am missing something?

答: What about $f(n) =4(n+1)$ for $n\ge 0$ and $f(n)=-4n-2$ for $n \lt 0$?


问:
Find the sum $\sum_{k=0}^{N-1} (-1)^k \omega_m^k$
Let $\omega = exp(\frac{2 \pi i}{N})$ be an $Nth$ root of unity where $N$ is even, and further define $\omega_m = \omega^m$. Then I want to compute the value of the sum
$$ \sum_{k=0}^{N-1} (-1)^k \omega_m^k $$

So far I have done,
\begin{align*}
\sum_{k=0}^{N-1} (-1)^k \omega_m^k & = 2 \sum_{k=0}^{\frac{N}{2}} \omega_m^{2k} - \sum_{k=0}^{N-1} \omega_m^k \\
&= 2 \sum_{k=0}^{\frac{N}{2}} \omega_m^{2k} - \begin{cases} 0 & m \equiv 0 (\mod N) \\ N & m \not\equiv 0 (\mod N) \end{cases} \\
\end{align*}
So it suffices to find the first sum, which is where I am stuck. I can't think of a way to get a nice closed formula for the sum. Does anyone have any ideas on how to proceed, thanks in advance.

Body: Just use$$\sum_k(-\omega_m)^k=\frac{1-(-\omega_m)^N}{1+\omega_m}=\frac{1+(-1)^{N+1}}{1+\exp\frac{2m\pi i}{N}}.$$


问:
Show that the eigenvalues of this positive semi-definite matrix converge
Let $(H_n)$ and $(\Lambda_n)$ be sequences of $n\times k$ matrices respectively (growing number of rows), where $k\in\mathbb{N}$ is fixed. We make the following two assumptions:

$H'_nH_n \to I_k$ as $n\to \infty$.

$\Lambda'_n\Lambda_n \to \Sigma$ as $n\to \infty$, where $\Sigma$ is  a positive definite matrix with distinct eigenvalues.

Here convergence is with respect to the Frobenius norm, which I denote $||\cdot||$.

I want to show that the largest $k$ eigenvalues of the matrix $H_n \Lambda'_n \Lambda_n H'_n$ converge to those of $\Sigma$ (the rank of $H_n \Lambda'_n \Lambda_n H'_n$ is at most $k$, so all other eigenvalues must be zero).

Any ideas on how to proceed is greatly appreciated.

答:
This one is surprisingly easy, as long as the $k$ leading eigenvalues are non-zero.

It is well known that the eigenvalues of an $n \times n$ matrix depend continuously on the entries of that matrix. With that said, we have
$$
\vec \lambda^{\downarrow k}(H_n \Lambda'_n \Lambda_n H_n') = \vec \lambda^{\downarrow}(\Lambda_n' \Lambda_nH_n'H_n)
$$
as a consequence of the fact that $H_n \Lambda'_n \Lambda_n H_n'$ is positive semidefinite and the fact that $AB$ and $BA$ have the same non-zero eigenvalues (for any matrices $AB$ such that both $AB$ and $BA$ are square).

It follows that
$$
\lim_{n \to \infty}\vec \lambda^{\downarrow k}(H_n \Lambda'_n \Lambda_n H_n') = \\
\lim_{n \to \infty}\vec \lambda^{\downarrow}(\Lambda_n' \Lambda_nH_n'H_n) = \\
\vec \lambda^{\downarrow}\left(\lim_{n \to \infty} \Lambda_n'\Lambda_n H_n' H_n \right) = \\
\vec \lambda^{\downarrow}\left([\lim_{n \to \infty} \Lambda_n'\Lambda_n] [\lim_{n \to \infty}H_n' H_n] \right) = \\
\vec \lambda^{\downarrow}(\Sigma I_k) = \vec \lambda^{\downarrow}(\Sigma),
$$
which is what we wanted.


问:
To Show that, $f$ cannot be expressed as the product of two polynomials with integer coefficients and degree $>1.$
$\textbf{Problem}$- Let, $f(x)=x^n+ax+p$, a $\in \mathbb{Z}.$ $p$ is a prime. Also, $p>|a|+1$. Show that, $f$ cannot be expressed as the product of two polynomials with integer coefficients and degree $>1.$

For this problem, which is a classical polynomial problem, I have been trying to apply the Vieta s theorem and irreducible criterion,but I can't structure my proof. A little bit rigorous proof is appreciable.
Thank you.

This problem has been lying unanswered... Kindly help ...I don't know why this problem is lying unanswered... Is there is any typo problem in the problem?

答:
If you could factor the polynomial it must be in the form.  :

$$ (x^{bb}+... + b_1x + b_0)  (x^{cc} + ... + c_1x+c_0)$$

$$bb+cc = n$$
$$ b_0  c_0 = p$$
$$ b_1c_0 + b_0 c_1 = a$$

$$ p> |b_1c_0 + b_0 c_1|+1$$

Since p is prime you must iterate through the 4 possibilities for $b_0$ and $c_0$ :
{1,p}, {-1,-p}, {p,1}, {-p,-1}.

Plug these 4 possibilities into the inequality above and you'll see that the inequality cannot be satisfied with integer values for the $b_1$ and $c_1$.  Picking $b_1$ and $c_1$ from {1,-1} gets you  $|b_1c_0 + b_0c_1|$ only as small as $p-1$.


问:
Plot of $f'(x)$ against $f(x)$
What does the plot of $f'(x)$ against $f(x)$ represent?

答: It represents the derivative in terms of the original function. Without context, that's as much as I can answer.</p>

OwnerDisplayName: GitGoodCodes

Title: Plot of $f'(x)$ against $f(x)$


问:
Is the existential quantifier commutative in intuitionistic higher order logic?
In higher order logic, one sometimes defines the existential quantifier by
$$(\exists x. \phi(x)) := (\forall \rho.(\forall x.\phi(x)\to\rho)\to\rho).$$
I am having trouble proving that
$$(\exists x.\exists y.\psi(x,y)) \to (\exists y.\exists x.\psi(x,y)).$$

Is this even true? If not, what would be a counter example?

答:
The trick is to unfold and instantiate both instances of $\exists$ in the hypothesis with the right hand side as follows:

Let $\rho_0 := \exists y.\exists x.\psi(x,y)$, then the following is a derivation tree (read from bottom to top):

\begin{align*}
(\exists x.\exists y.\psi(x,y))&\vdash \rho_0\\
((\forall x.(\exists y.\psi(x,y)) \to \rho_0)\to \rho_0) &\vdash \rho_0\\
&\vdash(\forall x.(\exists y.\psi(x,y)) \to \rho_0)\\
&\vdash_x (\exists y.\psi(x,y))\to\rho_0\\
\exists y.\psi(x,y)&\vdash_x\rho_0\\
(\forall y.(\psi(x,y) \to\rho_0)\to\rho_0)&\vdash_x \rho_0\\
&\vdash_x \forall y.(\psi(x,y)\to\rho_0)\\
&\vdash_{x,y}\psi(x,y) \to(\exists y.\exists x.\psi(x,y))\\
\psi(x,y)&\vdash_{x,y}\psi(x,y),
\end{align*}

Where the last step follows from the general observation that (again, from bottom to top)
\begin{align*}
\phi(x)&\vdash_{x}\exists x.\phi(x)\\
\phi(x),(\forall x.\phi(x)\to\rho) &\vdash_{x,\rho} \rho\\
\phi(x),\phi(x)\to\rho &\vdash_{x,\rho}\rho\\
\rho&\vdash_\rho\rho.
\end{align*}


答:
We have

$$x^2-y^2-n=0\iff(x+y)(x-y)=n$$

so if you are looking for integer solutions, all you need to do is factorize $n$ as $ab$, let $x+y=a$ and $x-y=b$, obtaining $x=(a+b)/2$ and $y=(a-b)/2$, and then observe that the factorization $n=ab$ must have factors of the same parity (in order for $x$ and $y$ to be integers). This parity condition can always be met if $n$ is odd or a multiple of $4$, but not if $n$ is twice an odd number.

When $n$ is odd, there is always the factorization $n\cdot1$, giving $x=(n+1)/2$ and $y=(n-1)/2$. When $n=4m$, there is always the factorization $(2m)\cdot2$, giving $x=m+1$, $y=m-1$. When $n$ (or $m$) is prime, that's all you get (up to sign). But in composite cases, you get additional solutions.

Remark: I finally read the comments beneath the OP and saw that this approach was already described there. To reiterate a point made there, solving $x^2-y^2-n=0$ is essentially equivalent to factoring $n$.  For odd integers $n$, it is known as Fermat's factorization method, which is the starting point for the more powerful Quadratic sieve approach to factoring.

Title: Finding Integer Solutions to $x^2-y^2-n=0$

Body:
I am trying to find the integer solutions to the equation $x^2-y^2-n=0$.

Effectively, I am trying to find when the difference of two perfect squares is $n$

I have tried using a modified Newton's method to find the nearest point where the equation reaches parity; however, it ends up finding non-integer solutions. I am wondering if there is a way to factor/find all the integer solutions to this equation.

For my algorithm, taking the square root is prohibitively costly and I am trying to avoid using it.

Thanks for the help!


问:
Simplest Non-Abelian Example of Two Isomorphic Direct Products $A\times B$ and $A \times C$ for which $B$ is not isomorphic to $C$
Suppose we have two groups $ A \times B$ and $A \times C$ which are isomorphic to each other, but $B$ is not isomorphic to $C$. Can you find non-abelian groups $A,B$, and $C$ satisfying these criteria? The simpler the example, the better, but I am also curious about more contrived examples, as well.

I failed to come up with any non-abelian examples and could only come up with abelian ones. Can you come up with any?

答: Pick any nonabelian group $G$ with $G\not\cong G\times G$. Then take $A=\prod_{i\in\omega} G$, $B=G$, $C=G\times G$.


问:
A sufficient condition to determine two weights are equal

The above sreenshot is from Takesaki's book (Vol 2, Chapter VIII).

In the proof of Proposition 3.15. I met with some questions.

Question1: If we let $g(x)=\varphi(axb), h(x)=\psi(axb)$, we have $g(x)=h(x)$ for all $x\in m_0$. $m_0$ is not dense in $M$, how to conclude $g(x)=h(x)$ for all $x\in M$.

Question 2: why can we construct an increasing net in $m_0^{+}$ converging $\sigma$-strongly to 1. I think we can construct a net in $m_0$ which is $\sigma$-weakly covergent to 1 since 1 is in $m_{\varphi}$ and $m_0$ is a $\sigma$-weakly dense $*$-subalgebra of  $m_{\varphi}$

答:
It doesn't say it in the statement, but if you read the proof he is assuming that both weights are normal. That answers 1.

For 2, one way to argue is to instead the unital C$^*$-algebra $\overline{\mathfrak m_0}$. Let $\{a_j\}$ be an approximate unit. It is well-known that $a_j\nearrow1$ wot, which agrees with the $\sigma$-weakly convergence since the net is bounded. Working in the norm closure of $\mathfrak m_0$ does not hamper the proof, since everything is norm-continuous.


问:
The importance and point in the minor axis of a hyperbola
The minor axis of an ellipse can be easily determined , by finding out the smaller axis among the enclosed figure..however the determination of the minor axis of a hyperbola is rather confusing for me.

The both axis of a hyperbola on the transverse axis and the conjugate axis.
The 'a' is the point at which the hyperbola intersects the transverse axis. How then will the 'b' be figured when the hyperbola doesn't actually intersect the conjugate axis ?

答: Given a hyperbola, one can define its conjugate hyperbola as having the same asymptotes and the same focal distance, but with foci rotated by 90° about the centre. The conjugate axis is then the transverse axis of the conjugate hyperbola.


答: In general, the only thing you can tell from a velocity-displacement graph is what the velocity is given the displacement, and vice-versa (although if the graph isn't one-to-one, then we may have various different values of displacement for a given value of velocity).</p>

Title: Plot of $f'(x)$ against $f(x)$

Body: What does the plot of $f'(x)$ against $f(x)$ represent?


问:
Describe the open balls of the metric $d_v$ given by $d_v(p,q)= 1$ or $d_v=|p_2-q_2|$
For points $p=(p_1,p_2)$ and $q=(q_1,q_2)$ in $\mathbb{R}^2$ define
$$d_v(p,q)=\begin{cases} 1\, \text{if $p_1 \neq q_1$ or $|p_2-q_2| \geq 1$} \\
|p_2-q_2| \, \text{if $p_1=q_1$ and $|p_2-q_2|<1$}\end{cases}$$

Describe the open balls in the metric $d_v$.

By definition of open ball we have that $B(p,\varepsilon)= \lbrace q\in \mathbb{R}^2 \mid d(p,q)< \varepsilon \rbrace$, from here we had two cases

Case I

$B(p,\varepsilon)= \lbrace q\in \mathbb{R}^2 \mid 1< \varepsilon \rbrace$ then

$p_1 \neq q_2$ or $|p_2-q_2| \geq 2$. If $p_1 \neq q_2$ then $q=(q_1,q_2)$ be in the open ball for any $U\subset \mathbb{R}^2 \setminus \lbrace x=p_1 \rbrace$.

Case II

$|p_2-q_2|\geq 1$ then $1+p_2<q_2<p_2-1$ therefore $q$ are in the open ball for $\mathbb{R} \times [1-p_2,p_2-1]$.

Finally the open balls for $d_v$ are $\lbrace p_1 \rbrace \times (p_2-1,p_2+1)$

$U\subset \mathbb{R}^2 \setminus \lbrace x=p_1 \rbrace $ or $W \subset \mathbb{R}  \times [1+p_2,p_2-1]$

答:
I’m afraid that I can make little sense of your reasoning. The nature of an $\epsilon$-ball in this metric depends entirely on whether $\epsilon\le 1$.

Let $p=\langle x,y\rangle$. Then

$$B(p,\epsilon)=\begin{cases}
\{x\}\times(y-\epsilon,y+\epsilon),&\text{if }0<\epsilon\le 1\\
\Bbb R^2,&\text{if }\epsilon>1\,.
\end{cases}$$


问:
Question about Poincare disc model D
Suppose σ is a circle centered at (h,k) such that σ∩D is a P-line.

a. Show that the equation of σ is x^2+y^2-2xh-2yk+1=0.

b. Find the P-length d(A,B), where A=(-1/2,0) and B=(0,-1/2).

And then we are asked the following question,

For a, I let the point (a,b) be the point lies on σ and intersect the disc orthogonally,
hence I have (x-h)^2 + (y-k)^2 = (a-h)^2 +(b-k)^2
and since a^2 +b^2 =1, I get x^2+y^2-2xh-2yk=1-2ah-2bh
which is a bit off from the desired answer.

For b, I tried to find where A and B intersect with the circle by plugging in the value of A and B into the equation from a. Then used the formula d(AB)=|log(AB,PQ)| to find the p-length. But I get a very strange number.

And for c , I don't know where to start with.

答:
Question a) : the equation of circle $\sigma$ is

$$(x-h)^2+(y-k)^2=r^2 \ \iff \ x^2+y^2-2hx-2ky=r^2-(h^2+k^2)\tag{1}$$

Besides, using the orthogonality property, Pythagoras gives us:

$$1+r^2=h^2+k^2 \ \iff \ r^2-(h^2+k^2)=-1\tag{2}$$

Plugging (2) in (1), we obtain the desired result.

Question b) : formula $$d(A,B)=|\operatorname{ln}(A,B;P,Q)|$$ is too difficult to apply here (I have begun the computations, obtained the equation $x^2+y^2+\tfrac52(x+y)+1=0$ of the circle containing $A$ and $B$, but afterwards, it becomes too complicated).
Fortunately, there is another formula with complex numbers $a=-1/2$ and $b=-i/2$ associated with $A$ and $B$:

$$d(A,B)=2 \operatorname{arctanh} \left|\frac{b-a}{1-\overline{a}b}\right|=2 \operatorname{arctanh}|\tfrac{10-6i}{17}|\approx 1.681$$

Please note that there is "logarithm" function hidden in the previous formula; indeed:

$$\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$$

See a proof here.

Title: Question about Poincare disc model D


问:
When has this sequence a triangular shape?
Can someone explain to me why y[n] has a triangular shape? From what I have found, there is a specific range $n_0$ where y[n] is triangular.
But right now I have no clue where to start.

$^2$ Search in a table    of  Fourier transform   properties

答:
A simple way to see this: taking the square of a Fourier transform is just multiplying its Fourier transform with itself. Now, taking the inverse transform of a product of two Fourier transforms is the same as convolving the two original functions. Here you are convolving a step function with itself. Now, think about how convolutions work and it should be evident that the result is a triangle function. But if you don't see it, just work out the convolution product of two step functions.

EDIT: Here's a detailed worked out solution. Using the definition from wikipedia for the DFT:

$$\text{DFT}(x[n])=\sum_{n=0}^{N-1}(u[n]-u[n-n_0])e^{-\frac{2\pi i}{N}k n} = \sum_{n=0}^{n_0-1}e^{-\frac{2\pi i}{N}k n} \; .$$

If you multiply this DFT with itself, you get the following sum

$$(\text{DFT}(x[n]))^2= \left(\sum_{n=0}^{n_0-1}e^{-\frac{2\pi i}{N}k n}\right)\left(\sum_{m=0}^{n_0-1}e^{-\frac{2\pi i}{N}k m}\right) = \sum_{n=0}^{n_0-1}\sum_{m=0}^{n_0-1}e^{-\frac{2\pi i}{N}k (m+n)} \; .$$

With a change of variables $t=m+n$, you can now write this as

$$(\text{DFT}(x[n]))^2 = \sum_{t=0}^{2(n_0-1)} z[t] e^{-\frac{2\pi i}{N}k t} \; .$$

where $z[t]$ is

$$z[t]=\begin{cases}1, \text{ if } t=0=0+0,\\
2, \text{ if } t=1=1+0=0+1,\\
3, \text{ if } t=2=2+0=1+1=0+2,\\
\ldots\\
2, \text{ if } t=2n_0-3=n_0-1+n-0-2=n_0-2+n_0-1,\\
1, \text{ if } t=2n_0-2=n_0-1+n_0-1,\\
\end{cases}$$

which is a triangular function. Note however that if $2(n_0-1)\geq N$, then because of periodicity, $z[t]\neq y[t]$. You'll have to add the $z[t]$'s corresponding to a same frequency to obtain the correct expression for $y[t]$, since $y[t]$ is only defined over the range $[0,N-1]$ just like $x[n]$.

So the range they are talking about in the question is $0\leq2(n_0-1)<N$.


问:
What accounts for the special relationship between Euclidean geometry and other branches of math?
Many times there are problems which are, in a sense "outside of geometry", but are nevertheless amenable to a geometric approach. For example, I may be asked to prove that the ranges of any two affine one dimensional functions must have either one, zero, or infinitely many values in common. An easy proof is to say, "One-dimensional affine functions are lines in Euclidean geometry. Such lines intersect once, no times, or infinitely many times according to Euclid's postulates."

In another case, we may be asked to prove a trigonometric identity or inequality. But instead of appealing to their analytic definitions, we may often draw a circle and some lines and use the theorems of Euclidean geometry.

The theories of linear equations and trigonometry can be developed without bringing in any axioms of Euclid. Typically both theories can be derived from the axioms of set theory. The postulates of Euclid, however, cannot be so derived.

1) Why is Euclidean geometry special in this way? Why won't we be able to use another sort of geometry (hyperbolic?) to such proofs? In other words, Euclidean geometry seems more natural than other sorts - but why?

2) Are these proofs rigorous? Should a geometric argument be just as convincing as a proof in language of the original theory?

答:
You don't need Euclidean geometry to prove that lines intersect once, no times, or infinitely many times. This can be proven in affine geometry, the point being that affine geometry has a notion of lines and parallel lines but not a notion of length. Affine geometry in turn is more or less linear algebra, which is thoroughly embedded into modern mathematics.

Once you prove that linear algebra in $\mathbb{R}^n$ (possibly as an inner product space if you need lengths or angles) models whatever part of Euclidean geometry you need, a proof using Euclidean geometry is completely rigorous. Other kinds of geometry are also useful in mathematics (e.g. hyperbolic geometry is indispensable to low-dimensional topology) but the basic objects in them (e.g. lines) are not as straightforward (!) as lines in Euclidean geometry.

I can't really make sense of the statement

The postulates of Euclid, however, cannot be so derived.

Would you say that the axioms of group theory can't be derived from the axioms of set theory? In my opinion this is a "wrong question." You can write down Euclid's axioms and ask what sorts of models they have, and you can certainly write down models of Euclid's axioms using ordinary set theory.


问:
What are some examples of incomplete nonextendible manifolds?
A connected Riemannian manifold $(M,g)$ is said to be extendible if it is isometric
to a proper open subset of a larger connected Riemannian manifold (this is the definition given in Lee's Introduction to Riemannian manifolds). I was able to show that a complete Riemannian manifold is nonextendible. I would like to find an example of incomplete Riemannian manifold which is not extendible. According to Wikipedia such manifolds exist, but no reference is given.

What are some (relatively simple) examples of connected incomplete nonextendible Riemannian manifolds ?

I tried looking at the case of the real numbers $\mathbf{R}$, without great success. More specifically I tried to find conditions on a metric on $\mathbf{R}$ to be extendible. Suppose that there is a metric $g$ on $\mathbf{R}$ and an isometric embedding $\varphi:(\mathbf{R},g)\hookrightarrow (M,g_0)$ with $M$ a connected 1-manifold and with $\varphi(\mathbf{R})\subsetneq M$. For example we can look at the case $M=\mathbf{R}$ and $\varphi(\mathbf{R})=(-\infty,0)$ with $\varphi$ increasing. Writte $g_0 = f(x)dx^2,~f>0$. Then, $\varphi$ bieng an isometry, we get

$$g = \underbrace{f\circ\varphi(x)\varphi^\prime(x)^2}_{h(x):=}dx^2.$$

Now $\varphi$ is strictly increasing and $\lim_{x\to+\infty} \varphi(x)=0$ so there is a sequence $(x_n)_n\in\mathbf R^\mathbf N$ converging to $+\infty$ such that

$$\lim_{n\to\infty}\varphi^\prime(x_n)=0.$$

Since $f$ in bounded near $0$, we get
$$\lim_{n\to\infty}h(x_n)=0.$$

Therefore if we want to construct nonextendible metric $g=h(x)dx^2$ on $\mathbf R$ "by hand", we might want to assume $h$ is bounded bellow by a positive constant $\varepsilon>0$ (so that there is no sequence $(x_n)_n\in\mathbf R^\mathbf N$ with $\lim_{n\to\infty} \vert x_n\vert = +\infty$ and $\lim_{n\to\infty}h(x_n)=0$). However I am pretty sure that these metrics are all complete (even if I don't have an actual proof of this fact), so I don't think I am gonna find an example in this way.

In dimension two, the only incomplete manifolds I can think of are punctured surfaces, but these are all extendible.

Any kind of help will be greatly appreciated.

答: For a simple example, take the surface $C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,z>0\}$ with the metric inherited from the Euclidean metric on $\mathbb{R}^3$.  This is a cone without its vertex.  To complete it as a metric space, you would just need to add in the vertex, but this completion is not a Riemannian manifold since it is not smooth at the vertex.


问:
How to formally write this statement on the existence of the limits?

The existence of the limit from one side does not entail the existence of the limit from the other side.

(Source: Schaum's Outline of Calculus, 6th Edition, by F. Ayres, E. Mendelson, p. 57)

How would you, using the precise definitions of the limits, formally write the above statement?

The precise definitions of the left-hand limit and the right-hand limits according to this source are as follows:

答:
Given a function $f:X \rightarrow Y$ and some fixed $x_0 \in X$, the existence of

$\lim_{x \rightarrow x_0^-} f(x)$

does not guarantee the existence of

$\lim_{x \rightarrow x_0^+} f(x)$

and vis versa.


答:
If we denote the statements "$\lim_{x\to a^{+}}f(x)$ exists" and "$\lim_{x\to a^{-}}f(x)$ exists" as they are, then your statement can be symbolized as follows

$$\neg\left(\lim_{x\to a^{+}}f(x)\text{ exists}\Rightarrow\lim_{x\to a^{-}}f(x)\text{ exists}\right)\wedge\neg\left(\lim_{x\to a^{-}}f(x)\text{ exists}\Rightarrow\lim_{x\to a^{+}}f(x)\text{ exists}\right)$$

Equivalently, by DeMorgan's,

$$\neg\left[\left(\lim_{x\to a^{+}}f(x)\text{ exists}\Rightarrow\lim_{x\to a^{-}}f(x)\text{ exists}\right)\vee\left(\lim_{x\to a^{-}}f(x)\text{ exists}\Rightarrow\lim_{x\to a^{+}}f(x)\text{ exists}\right)\right]$$

Edit: We can further reveal the logical structure of the statement by making use of the following definitions:

$\lim\limits_{x\to a^{\pm}}f(x)\text{ exists}\iff\left(\exists L\in\mathbb{R}\right)\left(\lim\limits_{x\to a^{\pm}}f(x)=L\right)$

$\lim\limits_{x\to a^+}f(x)=L\iff(\forall\varepsilon>0)\{(\exists\delta >0)[(\forall x\in\mathbb{R})(a<x<a+\delta\Rightarrow |f(x)-L|<\varepsilon)]\}$

$\lim\limits_{x\to a^-}f(x)=L\iff(\forall\varepsilon>0)\{(\exists\delta >0)[(\forall x\in\mathbb{R})(a-\delta<x<a\Rightarrow |f(x)-L|<\varepsilon)]\}$

Similar definitions hold for left-hand limits.

I won't present the fully expanded statement here; it's way too big and messy.

Title: How to formally write this statement on the existence of the limits?

Body:
The existence of the limit from one side does not entail the existence of the limit from the other side.

(Source: Schaum's Outline of Calculus, 6th Edition, by F. Ayres, E. Mendelson, p. 57)

How would you, using the precise definitions of the limits, formally write the above statement?

The precise definitions of the left-hand limit and the right-hand limits according to this source are as follows:


答:
Let $A_0$, $B_0$, $C_0$ be Abelian groups such that $A_0\times B_0\cong A_0\times C_0$ while $B_0\not\cong C_0$. Let $D$ be a centerless non-Abelian group, e.g., $D=S_3$. Let $A=A_0\times D$, $B=B_0\times D$, $C=C_0\times D$. Then $A$, $B$, $C$ are non-Abelian groups, and $A\times B\cong A\times C$, but $B\not\cong C$, since the center of $B$ is isomorphic to $B_0$ while the center of $C$ is isomorphic to $C_0$.

Alternatively, let $A$ be a non-Abelian group such that $A\cong A\times D$ for some non-trivial group $D$; let $B$ be a non-Abelian group such that $B\not\cong D\times B$, e.g. any finite non-Abelian group; and let $C=D\times B$. Then $A$, $B$, $C$ are non-Abelian groups, and $B\not\cong C$, but $A\times B\cong(A\times D)\times B\cong A\times(D\times B)\cong A\times C$. (Here, if $G$ is a non-Abelian group such that $G\not\cong G\times G$, then, taking $A=G^\infty$ and $D=B=G$, we get the example in Arturo Magidin's answer.)

Title: Simplest Non-Abelian Example of Two Isomorphic Direct Products $A\times B$ and $A \times C$ for which $B$ is not isomorphic to $C$

Body:
Suppose we have two groups $ A \times B$ and $A \times C$ which are isomorphic to each other, but $B$ is not isomorphic to $C$. Can you find non-abelian groups $A,B$, and $C$ satisfying these criteria? The simpler the example, the better, but I am also curious about more contrived examples, as well.

I failed to come up with any non-abelian examples and could only come up with abelian ones. Can you come up with any?


问:
Proof of the Riemann-Lebesgue Theorem in function spaces
Let $1<p\leq\infty$ and $f\in L^{p}_{loc}(\mathbb{R})$ be a $1$-periodic function. Set $$f_{n}(x)=f(nx)$$ and $$\bar{f}:=\int\limits_{0}^{1}f(x)\, \mathrm{d}x.$$ Then prove that $f_n\rightharpoonup\bar{f}$ in $L^{p}(a,b)$ if $1<p<\infty$ and $f_{n}$ weakly* converges to $\bar{f}$ in $L^{\infty}(a,b)$ if $p=\infty$, as $n\to\infty$.

答:
$$\int_0^1f_n(x)g(x)dx=\int_0^n\frac1nf(x)g(x/n)dx=\sum_{k=0}^{n-1}\frac1n\int_0^1f(x)g((x+k)/n)dx.$$
Now observe that if $g$ is continuous then
$$\sum_{k=0}^{n-1}\frac1ng((x+k)/n))\to\int_0^1g(x)dx,$$
and apply Lebesgue's dominated convergence theorem. Then use that the continuous functions are dense in $L^q(0,1)$ if $q<\infty$ to obtain the assertion in case $(a,b)=(0,1)$ and $1<p<\infty$. The case $(a,b)=(0,N)$ follows by the periodicity of $f$, and from this the general case follows.

How to prove the case $p=\infty$ is not so clear to me since there is no simple presentation of the dual space. Are you sure that not weak$^*$ convergence in $(L^1(a,b))'$ is meant?


问:
Finding subgroups of $\mathbb Z_4 \times \mathbb Z_2$. Why can't there be more?
My solution consists of taking direct products of both $\mathbb Z_4$ and $\mathbb Z_2$ and then their direct product. Apart from that I took $\langle 2,0\rangle, \langle 2,1\rangle, \langle 1,1\rangle$ as subgroup generators and generated more subgroups by brute force.

This doesn't seem an efficient method. How do I know when I have to stop? This is a small question but in the context of bigger questions, how do I know when I have found all the subgroups.

答:
The magic words are “Goursat’s Lemma” (or “subdirect products”), but I don’t think you’ll find it less “work intensive”. It is a description of all subgroups of a direct product $G\times H$.

For general comments see here; for an example of an application see here.

Basically, if $G$ and $H$ are groups, then the subgroups of $G\times H$ correspond to the following five pieces of information:

A subgroup $A$ of $G$;

A normal subgroup $N$ of $A$;

A subgroup $B$ of $H$;

A normal subgroup $K$ of $B$;

An isomorphism $f\colon (A/N)\to (B/K)$.

The 5-tuple $(A,N,B,K,f)$ then corresponds to the subgroup
$$S(A,N,B,K,f) = \{(g,h)\in A\times B\mid f(gN)=hK\}.$$

That these are all subgroups is fairly easy to verify; Goursat’s Lemma is the proof that every subgroup is of this form.

If you understand $G$ and $H$ very well, then this may lead to a less work-intensive process; but if not, then you end up having to list all subgroups of $G$ and their quotients, and all subgroups of $H$ and their quotients, and then all possible isomorphisms between pairs of such quotients.

For the example here it is relatively straightforward, but you may find it just as work-intensive as your original method.

The possible quotients-of-subgroups of $\mathbb{Z}_2$ are $\mathbb{Z}_2$ itself (with $B=\mathbb{Z}_2$ and $K=\{0\}$), and the trivial group (with either $B=K=\mathbb{Z}_2$, or $B=K=\{0\}$). The possible quotients-of-subgroups of $\mathbb{Z}_4$ are $\mathbb{Z}_4$ (with $A=\mathbb{Z}_4$, $N=\{0\}$), $\mathbb{Z}_2$ (with $A=\mathbb{Z}_4$ and $N=\langle 2\rangle$; and with $A=\langle 2\rangle$ and $N=\{0\}$); and the trivial group (with $A=N=\mathbb{Z}_4$, $A=N=\langle 2\rangle$, and $A=N=\{0\}$). So you get the following subgroups, when we pair the $\mathbb{Z}_2$s together, and when we pair the trivial groups together:

When $A=\mathbb{Z}_4$, $N=\langle 2\rangle$, $B=\mathbb{Z}_2$, $K=\{0\}$, the subgroup consists or $(1,1)$, $(2,0)$, $(3,1)$, and $(0,0)$; i.e., $\langle (1,1)\rangle$.

When $A=\langle 2\rangle$, $N=\{0\}$, $B=\mathbb{Z}_2$, $K=\{0\}$, the subgroup consists of $(2,1)$ and $(0,0)$; i.e., $\langle (2,1)\rangle$.

When $A=N=\mathbb{Z}_4$ and $B=K=\mathbb{Z}_2$, you get $\mathbb{Z}_4\times\mathbb{Z}_2$.

When $A=N=\mathbb{Z}_4$ and $B=K=\{0\}$, you get $\mathbb{Z}_4\times\{0\}$.

When $A=N=\langle 2\rangle$ and $B=K=\mathbb{Z}_2$, you get the $\langle 2\rangle\times \mathbb{Z}_2$.

When $A=N=\langle 2\rangle$ and $B=K=\{0\}$, you get $\langle 2\rangle\times\{0\}$.

When $A=N=\{0\}$ and $B=K=\mathbb{Z}_2$ you get $\{0\}\times\mathbb{Z}_2$.

When $A=N=\{0\}$ and $B=K=\{0\}$ you get $\{0\}\times\{0\}$.


问:
How to define a function to show that infinite sets are countable
To prove cardinality for infinite sets, I know that I have to show that there is a bijection. However, I'm having some trouble defining a function for my sets.

For example: Let O be the set of all odd integers. Prove that O
has the same cardinality as 2Z, the set of all even
integers.

In this case, I know that I have to define a function which is "Define f: O to 2Z s.t f(n) = n - 1."

However, I don't understand where the n - 1 comes from

答:
The idea here is to define a function in such a way as to show that you can 'match up' pairs of elements for either set. There is nothing particularly special about the function you gave in your example, for example the map:
$$
g:O\to 2\mathbb{Z}, n \mapsto n -3
$$
Also gives a bijection between the two sets. You just need to make sure that the function is defined such that it really maps odd integers to even ones. Then you check that such a function is surjective and injective, i.e. that it is a bijection. A good way to build intuition about such function is to try modifying the examples that you've seen before (such as the map $f$ you gave in your question) and see whether they still have the same properties.

Alternatively, you could think about such maps in terms of how they map a particular element. In the example of bijections between the odd integers and the even integers, we can write any odd integer $n$ as $2k+1$ where $k$ is some integer. In this form, we can see that there is an even integer hiding inside the odd one, in a sense, suggesting a map:
$$
2k+1 \mapsto 2k
$$
But of course this is precisely the map $f$ you gave as an example.

To show that such a map, say $f$ is onto, we need to show that for every element $m$ in the range, there is an element $n$ in the domain such that $f(n)=m$. For example, to show that $f$ is onto, fix an arbitrary element $m$ in the range, $2\mathbb{Z}$. Now we want to construct an element of $O$ that maps to $m$. This is straightforward, since:
$$
f(m+1) = (m+1)-1 = m
$$
and since $m$ is even, we know that $m+1$ is odd. Thus we have shown that for any even integer $m$, there exists an odd integer $n$ such that $f(n) = m$. By definition, this means that $f$ is onto.


问:
Combinatorics ( colouring a checkerboard)
Question:

If we identify colourings related by rotations of a checkerboard, how many ways can you paint it if it has 2Tx2T fields with m colours?

My attempt:

I'm struggling to begin this. The part of the question I am confused with is the "if we identify colourings related by rotations of a checkerboard".

答:
A solution that does not use Burnside's lemma explicitly:

There are $m^{4t^2}$ possible boards, and naively we must divide this by $4$. However:

$m^{2t^2}$ boards are equivalent under a $180^\circ$ turn, so we must "boost" the count by $m^{2t^2}$ so that these boards are counted once (not $0.5$ times)

$m^{t^2}$ boards are equivalent under a $90^\circ$ turn, so we must "boost" the count by $2m^{t^2}$ so that these boards are counted once (not $0.5$ times; before the first bullet point above they are counted $0.25$ times)

Thus the final answer is, writing $z=m^{t^2}$ (the number of ways to colour a fixed $t×t$ grid with $m$ colours),
$$\frac{z^4+z^2+2z}4$$


问:
Let X be a topological space and A, B $ \subset $ X compact.....
Let X be a topological space and A, B $ \subset $ X compact.

a) Show that A $ \cup $ B is compact.

b) Show that if X is Hausdorff, then A $\cap$ B  is compact.

c) Show that the hypothesis that X is Hausdorff cannot be omitted in the previous item.

My attempt

a)
Let X be a topological space and A, B $\cup$ where A and B are compact. We will claim A $\cup$ B is compact.

Let $\{O_{\alpha}\}_{\alpha \in I}$ be an open for A $\cup$ B, so A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\alpha}$.

But A $\subset$ A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\alpha}$.

So since  A is compact, A $\subset$ $\bigcup_{i=1}^n O_{\alpha_i}$ wherer $\alpha_i \in I$

Likewise,

B $\subset$ A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\beta}$.

So since  B is compact, B $\subset$ $\bigcup_{j=1}^m O_{\beta_j}$, wherer $\beta_i \in I$.

therefore,  A $\cup$ B $ \subset \bigcup_{\alpha \in I}^n \cup  \bigcup_{\beta \in I}^m$, ando so,  A $\cup$ B is compact.

b)
Since X is $ T_2 $ and A, B $ \subset $ X compact, we have that A and B are closed.

So, let $ U_{\alpha} $ be an open cover of A $ \cap $ B

So, we have A $ \cap $ B $ \subset \bigcup_{\alpha} U_{\alpha} $

So, B $ \subset \bigcup_{\alpha} U_{\alpha} \cup$ (X\A)

Since A is closed (X∖A) is open

Therefore,

$ U_{\alpha} \cup $ X∖A is an open cover of A.

Now we have that B is compact, so it follows by definition that a finite subcover, {$U_{\alpha_1},
U_{\alpha_2},…, U_{\alpha_n }$, X∖A} of A exists.

Therefore,

A $\cap$ B  $\subset \{U_{\alpha_1}, U_{\alpha_2},…, U_{\alpha_n}\}$ and therefore A $\cap B$ is compact.

c)
If it is not $ T_2 $ we cannot assume A, B closed and in this case the demonstration does not take place.

Thanks in advance for any comments.

答:
Your arguments for (a) and (b) are basically correct, but in places you have badly misused notation. For example, in (b) you write:

So, let $U_\alpha$ be an open cover of $A\cap B$.

So, we have $A\cap B=\bigcup_\alpha U_\alpha$.

If $U_\alpha$ is an open cover of $A\cap B$, as you say in the first sentence, it is a family of open sets. In the second sentence, however, you make it clear that $U_\alpha$ is a single member of some indexed family of open sets. Thus, your first sentence should have read something like this:

So, let $\{U_\alpha:\alpha\in I\}$ be an open cover of $A\cap B$.

Later you write:

$U_\alpha\cup X\setminus A$ is an open cover of $A$

This says that a single set $U_\alpha\cup X\setminus A$ covers $A$. There are two problems here. First, the expression is ambiguous: should it be read $(U_\alpha\cup X)\setminus A$, or should it be read $U_\alpha\cup(X\setminus A)$? Use parentheses when there’s any real possibility of ambiguity. The real problem, though, is that you actually meant something completely different: you meant that $\{U_\alpha:\alpha\in I\}\cup\{X\setminus A\}$ is an open cover of $A$.

By the way, you can avoid indexing altogether. For (a), for instance, let $\mathscr{U}$ be an open cover of $A\cup B$. $A$ and $B$ are compact, and $\mathscr{U}$ is clearly an open cover of each of them, so there are a finite $\mathscr{U}_A\subseteq\mathscr{U}$ that covers $A$ and a finite $\mathscr{U}_B\subseteq\mathscr{U}$ that covers $B$. But then $\mathscr{U}_A\cup\mathscr{U}_B$ is a finite subset of $\mathscr{U}$ that covers $A\cup B$, so $A\cup B$ is compact. Similarly, in (b) you can start with an open cover $\mathscr{U}$ of $A\cap B$. Then $\mathscr{U}\cup\{X\setminus A\}$ is an open cover of the compact set $B$, there is a finite $\mathscr{U}_B\subseteq\mathscr{U}$ such that $\mathscr{U}_B\cup\{X\setminus A\}$ covers $B$. $(A\cap B)\cap(X\setminus A)=\varnothing$, so $\mathscr{U}_B$ covers $A\cap B$, which is therefore compact.

Your answer for (c) is insufficient: you’ve shown that if $X$ is not Hausdorff, the argument that you gave for (b) cannot be carried out, but that does not rule out the possibility that there is some other argument that does not require $X$ to be Hausdorff. To answer (c), you must come up with an actual counterexample. One of the nicest is a convergent sequence with two limits:

Example. Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are any two points not in $\Bbb N$. Points of $\Bbb N$ are isolated. For each finite $F\subseteq\Bbb N$ let $$B(p,F)=\{p\}\cup(\Bbb N\setminus F)$$ and $$B(q,F)=\{q\}\cup(\Bbb N\setminus F)\,;$$ the sets $B(p,F)$ form a local open nbhd base at $p$, and the sets $B(q,F)$ form a local open nbhd base at $q$. Thus, a set $U\subseteq X$ is open if and only if either $U\subseteq\Bbb N$, or $X\setminus U$ is finite. $X$ is a $T_1$ space and is almost Hausdorff: the only two points that do not have disjoint open nbhds are $p$ and $q$.

Now let $A=X\setminus\{q\}$ and $B=X\setminus\{p\}$; I leave the easy verification that $A$ and $B$ are compact to you. $A\cap B=\Bbb N$, and clear $\big\{\{n\}:n\in\Bbb N\big\}$ is an open cover of $A\cap B$ with no finite subcover.


问:
An entire function satisfying $|f(z)|\le |z| |x| $ on $\mathbb{C}$ is identically zero.
Suppose $f$ is an entire function satisfying $|f(z)|\le|z||x|$ where $x=\text{real part of }z$.
Then prove that $f(z)=0$ for every $z\in \mathbb{C}$

I am able to prove that $f(0)=0$ and $f'(0)=0$. I need to prove that all derivatives vanish at 0. How do I prove that? Help please.

答: The hypothesis implies that $|f(z)| = 0$ (and hence $f(z) = 0$) for all pure imaginary $z$.  Since the imaginary axis is a set with a limit point, it follows by analytic continuation that $f(z) = 0$ for all $z$.


问:
Commutative Ring Epimorphism on quotients
Let $R$ be a commutative ring and $I$,$J$ be ideals of $R$.

Define $f: R \rightarrow R/I \times R/J$ as $f(a) = (a+I,a+J)$.

Prove that if $f$ homomorphism is surjective, then $I+J = R$.

..

I think $\gcd([R:I],[R:J])$ must be 1.

I've found Ker$f$ = $I \cap J$ and tried first isomorphism theorem but could not go any further.

答:
If $f$ is surjective then, for any $(a+I,b+J) \in R/I\times R/J$, there exists some $c \in R$ such that $f(c) = (a+I,b+J)$, i.e.
$$
(c+I,c+J) = (a+I,b+J)
$$
Applying this $((a+1)+I,a+J)$ we get that there exists $c \in R$ such that
$$
(c+I,c+J) = ((a+1)+I,a+J)
$$
Applying the coset criterion, this implies that $a+1-c \in I$ and that $c-a \in J$, so we have that
$$
(a+1-c) + (c-a) = 1 \in I + J
$$
Thus $I$ and $J$ are relatively prime, so $I+J = R$ as desired.