问:
Extension Theorem for the Sobolev Space $W^{1, \infty}(U)$
I am trying to find a way of extending functions in the Sobolev Space $W^{1, \infty}(U)$ to $W^{1, \infty}(\mathbb{R}^n)$ where $U\subset\mathbb{R}^{n}$ is open such that $U\subset\subset V$ for $V\subset\mathbb{R}^{n}$ also open and bounded. Furthermore, assume $\partial U$ is $C^1$.</p>

答:
I am not familiar with the Whitney extension theorem, etc. but maybe the following proof will do?

Fix $x^0\in\partial U$. Assume for now that for some ball $B=B(x^0, r)$ about $x^0$, the boundary is straightened, such that $B\cap U\subset \mathbb{R}_{+}^n$. Let $B^+=\{x\in B\ \mid \ x_n\geq 0\}$ and $B^-=\{x\in B\ \mid \ x_n\leq 0\}$. Define
\begin{equation}
\bar{u}(x)\equiv\left\{
\begin{array}{l l}
u(x', x_n)&\quad x\in B^+\\
u(x', -x_n)&\quad x\in B^-.
\end{array}\right.
\end{equation}
Let $u^+=\bar{u}|_{B^+}$ and $u^-=\bar{u}|_{B^-}$. On $\{x_n=0\}$ we observe that $u^+=u^-$. Since $u\in L_{\text{loc}}^1(U)$ we can deduce that $\bar{u}\in L_{\text{loc}}^1(B)$. We now claim that the weak derivative of $\bar{u}$ is
\begin{equation}%\label{eq: dve}
D\bar{u}(x)=\left\{
\begin{array}{l l}
Du(x', x_n)&\quad x\in B^+\\
D_{x'}u(x', -x_n)&\quad x\in B^-\\
-D_{x_n}u(x', -x_n)&\quad x\in B^-.
\end{array}\right.
\end{equation}
We prove it as follows. Let $\varphi\in C_{c}^{\infty}(B)$ and assume for now that $u\in C^1(\overline{B^+\setminus\mathbb{R}^{n-1}})$ then
\begin{align*}
\int_{B}\bar{u}\varphi_{x_n}\mathrm{d}x &=\int_{B^{+}}\bar{u}\varphi_{x_n}\mathrm{d}x+\int_{B^-}\bar{u}\varphi_{x_n}\mathrm{d}x\\
&=\int_{B^{+}}u\varphi_{x_n}\mathrm{d}x-\int_{B^+}u(y', y_n)\varphi_{y_n}(y', -y_n)\mathrm{d}y\\
&=-\int_{\{x_n=0\}}u\varphi\mathrm{d}S_x-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x+\int_{\{y_n=0\}}u\varphi\mathrm{d}S_y-\int_{B^+}-u_{y_n}(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\
&=-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^+}-u_{y_n}(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\
&=-\int_{B^+}u_{x_n}(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-u_{x_n}(x', -x_n)\varphi(x', x_n)\mathrm{d}x
\end{align*}
and using the same technique for $i\neq n$ we obtain for every multi index $\alpha$ such that $\vert\alpha\vert=1$
\begin{equation*}
\int_{B}\bar{u}D^{\alpha}\varphi\mathrm{d}x=\left\{
\begin{array}{l}
-\int_{B^+}D_{x'}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}D_{x'}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x\\
-\int_{B^+}D_{x_n}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-D_{x_n}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x.
\end{array}\right.
\end{equation*}

Let $C^+\equiv B^+\setminus\{x_n=0\}$. Since $u\in W^{1, \infty}(C^+)$ is locally integrable in $C^+$, we can define its $\varepsilon$ mollification in $C_{\varepsilon}^+$ and we also know that:
\begin{equation*}
D^{\alpha}u^{\varepsilon}(x)\equiv \eta_{\varepsilon}\ast D^{\alpha}u(x)\quad x\in C_{\varepsilon}^+, \ \vert\alpha\vert\leq1.
\end{equation*}
Here, we define the cutoff function $\eta:\mathbb{R}^n\rightarrow\mathbb{R}$ as
\begin{equation*}
\eta(x)\equiv\left\{
\begin{array}{l l}
C\exp\left[\frac{-1}{1-\|x\|^2}\right] &\quad \|x\|<1\\
0&\quad \|x\|\geq 1
\end{array}\right.
\end{equation*}
where $C$ is chosen such that $\int_{\mathbb{R}^n}\eta\mathrm{d}x=1$ and then we define $\eta_{\varepsilon}(x)\equiv\frac{1}{\varepsilon^n}\eta\left(\frac{x}{\varepsilon}\right)$.

Define $v_{\varepsilon}: C^{+}\rightarrow\mathbb{R}$ as
\begin{equation*}
v_{\varepsilon}(x)\equiv\left\{
\begin{array}{l l}
u^{\varepsilon}(x) &\quad x\in C_{\varepsilon}^{+}\\
0&\quad x\in C^{+}\setminus C_{\varepsilon}^{+}.
\end{array}\right.
\end{equation*}
Now, because $u^{\varepsilon}\in C^{\infty}(D)$ for every open bounded set $D\subset C^+$, by repeated application of the mean value theorem, the derivatives of all orders are bounded on such $D$. Therefore $D^{\alpha}u$ is Lipschitz continuous on $D$ for all $\vert\alpha\vert\leq 1$. In particular, $D^{\alpha}u$ is uniformly continuous on open bounded subsets of $C^{+}$ for all $\vert\alpha\vert\leq 1$. So we can say that $v^{\varepsilon}\in C^1(\overline{C_{\varepsilon}^{+}})$.
Now for $\vert\alpha\vert=1$ we claim that the weak derivative of $v_{\varepsilon}$ exists and is given by:
\begin{equation}
D^{\alpha}v_{\varepsilon}(x)\equiv\left\{
\begin{array}{l l}
D^{\alpha}u^{\varepsilon}(x) &\quad x\in C_{\varepsilon}^{+}\\
0&\quad x\in C^{+}\setminus C_{\varepsilon}^{+}.
\end{array}\right.
\end{equation}
We check it as follows, let $\vert\alpha\vert=1$ and let $\varphi\in C_{c}^{\infty}(C^+)$, then
\begin{align*}
\int_{C^+}v_{\varepsilon}\varphi_{x_i}\mathrm{d}x &=\int_{C_{\varepsilon}^{+}}v_{\varepsilon}\varphi_{x_i}\mathrm{d}x\\
&=-\int_{C_{\varepsilon}^+}(v_{\varepsilon})_{x_i}\varphi\mathrm{d}x+0\\
&=-\int_{C_{\varepsilon}^+}u^{\varepsilon}_{x_i}\varphi\mathrm{d}x
\end{align*}
for $i=1, \dots, n$ whence we conclude that the weak derivative exists. Since $\|D^{\alpha}v_{\varepsilon}\|_{L^{\infty}(C^+)}\leq\|D^{\alpha}u\|_{L^{\infty}(C^+)}\infty$ for all $\vert\alpha\vert\leq 1$ we conclude that $v_{\varepsilon}\in W^{1, \infty}(C^+)$.
Moreover as $\varepsilon\rightarrow0$
\begin{equation*}
D^{\alpha}v^{\varepsilon}\rightarrow D^{\alpha}u\quad\text{a.e. in } C^+, \ \vert\alpha\vert\leq 1.
\end{equation*}
We can write
\begin{equation*}
C^+\equiv\bigcup_{i=1}^{\infty}C_{1/i}^{+}
\end{equation*}
and $D^{\alpha}v_{i}: C^{+}\rightarrow\mathbb{R}$
for each $\vert\alpha\vert\leq 1$
\begin{equation*}
D^{\alpha}v_{i}(x)\equiv\left\{
\begin{array}{l l}
D^{\alpha}v^{\frac{1}{i}}(x) &\quad x\in C_{1/i}^{+}\\
0&\quad x\in C^+\setminus C_{1/i}^{+}.
\end{array}\right.
\end{equation*}
Since $D^{\alpha}v_i$ is measurable and uniformly bounded almost everywhere in $C^+$ by $\|D^{\alpha}v\|_{\infty}$ for each $\vert\alpha\vert\leq 1$, we deduce by the dominated convergence theorem that:
\begin{equation*}
\lim_{i\rightarrow \infty}\int_{C^+}D^{\alpha}v_i\mathrm{d}x=\int_{C^+}D^{\alpha}u\mathrm{d}x\quad \vert\alpha\vert\leq 1.
\end{equation*}
Since $\mathcal{L}^{n}(\mathbb{R}^{n-1})=0$ we can define $D^{\alpha}v_i$ arbitrarily there and write
\begin{equation*}
\lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}v_i\mathrm{d}x=\int_{B^+}D^{\alpha}u\mathrm{d}x \quad\vert\alpha\vert\leq 1.
\end{equation*}
Now let  $\bar{u}$ be as defined initially, $\alpha=(0, \dots, 1)$ and $\varphi\in C_{c}^{\infty}(B)$, then %replacing $D^{\alpha}v_i$ with $D^{\alpha}u_i$ we have
\begin{align*}
\int_{B}\bar{u}D^{\alpha}\varphi\mathrm{d}x &=\int_{B^{+}}\bar{u}D^{\alpha}\varphi\mathrm{d}x+\int_{B^-}\bar{u}D^{\alpha}\varphi\mathrm{d}x\\
&=\int_{B^{+}}uD^{\alpha}\varphi\mathrm{d}x-\int_{B^+}u(y', y_n)D^{\alpha}\varphi_{y_n}(y', -y_n)\mathrm{d}y\\
&=\lim_{i\rightarrow \infty}\int_{B^{+}}v_iD^{\alpha}\varphi\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}v_i(y', y_n)D^{\alpha}\varphi(y', -y_n)\mathrm{d}y\\
&=-\lim_{i\rightarrow \infty}\int_{B^+}D^{\alpha}v_i(x', x_n)\varphi(x', x_n)\mathrm{d}x-\lim_{i\rightarrow \infty}\int_{B^+}-D^{\alpha}v_i(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\
&=-\int_{B^+}D^{\alpha}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^+}-D^{\alpha}u(y', y_n)\varphi(y', -y_n)\mathrm{d}y\\
&=-\int_{B^+}D^{\alpha}u(x', x_n)\varphi(x', x_n)\mathrm{d}x-\int_{B^-}-D^{\alpha}u(x', -x_n)\varphi(x', x_n)\mathrm{d}x.
\end{align*}
We obtain the corresponding result for any other multi index $\alpha$ such that $\vert\alpha\vert=1$. So we  conclude that the weak derivative of $\bar{u}$ is as claimed.

Furthermore, we have that  for $\vert \alpha\vert\leq 1$:
\begin{equation*}
\vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B)}\leq\vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B^+)}+\vert\vert D^{\alpha}\bar{u}\vert\vert_{L^{\infty}(B^-)}=2\vert\vert D^{\alpha}u\vert\vert_{L^{\infty}(B^+)}
\end{equation*}
and hence
\begin{equation*}
\vert\vert \bar{u}\vert\vert_{W^{1, \infty}(B)}\leq 2\vert\vert u\vert\vert_{W^{1, \infty}(B^+)}.
\end{equation*}
If the boundary is not already straightened out we can do so under the map $\Phi: W\rightarrow B$ and unstraighten it under $\Psi: B\rightarrow W$ as defined in Evans, where $\Psi (B)=W$. Writing $\Phi(x)=y$ and $\Psi(y)=x$ we define $u'(y)\equiv u\circ\Psi (y)$ . Then we have:
\begin{equation*}
\vert\vert \bar{u'}\vert\vert_{W^{1, \infty}(B)}\leq 2\vert\vert u'\vert\vert_{W^{1, \infty}(B^+)}
\end{equation*}
and converting back to original coordinates we have
\begin{equation}
\vert\vert \bar{u}\vert\vert_{W^{1, \infty}(W)}\leq 2\vert\vert u\vert\vert_{W^{1, \infty}(U)}
\end{equation}
We know that $\overline{U}\subset\mathbb{R}^n$ is paracompact so it admits locally finite partitions of unity subordinate to any countable cover of open sets. Cover the boundary $\partial U$ with the cover $\{W_x\}_{x\in\partial U}$ such that each open set $W_x$ is a ball about $x$ and we obtain inequality as above for each $W_x$. The boundary is compact so we can pass to a finite subcover $\{W_i\}_{i=1}^{N}$ and have the corresponding extensions $\bar{u}_i$. Choose any $W_0\subset\subset U$ such that $U\subset\bigcup_{i=0}^{N} W_i$ and define $\bar{u}_0\equiv u$ in $W_0$.

Let $\{\zeta_i\}_{i=1}^{N}$ be a locally finite smooth partition of unity subordinate to the cover $\{W_i\}_{i=0}^{N}$. Define $\bar{u}\equiv\sum_{i=0}^{N}\zeta_i\bar{u}_i$. Then we obtain the bound:
\begin{align*}
\vert\vert \bar{u}\vert\vert_{W^{1, \infty}(\mathbb{R}^n)}&=\left\| \sum_{i=0}^{N}\zeta_i\bar{u}_i\right\|_{W^{1, \infty}(\mathbb{R}^n)}\\
&\leq\sum_{i=0}^{N}\|\zeta_i\bar{u}_i\|_{W^{1, \infty}(W_i)}\\
&\leq\sum_{i=0}^{N}\|\bar{u}_i\|_{W^{1, \infty}(W_i)}\\
&\leq \sum_{i=0}^{N}2\|u\|_{W^{1, \infty}(U)}\\
&\leq C\|u\|_{W^{1, \infty}(U)}.
\end{align*}

Furthermore we can arrange for the support of $\bar{u}$ to lie within some $V\supset\supset U$. Define $Eu\equiv\bar{u}$. Then we can see that the map $E: W^{1, \infty}(U)\rightarrow W^{1, \infty}(\mathbb{R}^n)$ is a linear operator such that

$Eu=u$ almost everywhere in $U$.

$Eu$ has support within $V\supset\supset U$

$\vert\vert Eu\vert\vert_{W^{1, \infty}(\mathbb{R}^n)}\leq C\|u\|_{W^{1, \infty}(U)}$ for $C=2N$.


问:
find and prove probability of P({D1,D2,D3,...}=N)
Let $D_1, D_2,\cdots$ be independent with $D_n\sim Unif\{1,2,\cdots,n\}$. Find and prove tthe probability of $P(\{D_1,D_2,\cdots\}=\mathbb{N})$. I am a bit lost as to how to prove this. My intuition tells it should be 1. Hope someone could help! Thanks!

答:
UPDATED ANSWER

Thanks to @Crostul (who has revised the answer below) and @TippingOctopus.

Hint: Consider the probability $P(k\notin\{D_1,D_2,\dots,D_n\})$ where $k<n$.

The event that the set $\mathscr{D} = \{D_1,D_2\dots\}$ contains $\mathbb{N}$ is similar to asking whether every element of $\mathbb{N}$ belongs to this set $\mathscr{D}$. The probability that a number $k$ does not belong to the set $\{D_1,D_2\dots,D_n\}$ where $k<n$ is given by
$$ P(k\notin\{D_1,D_2,\dots,D_n\}) = P(k\notin\{D_{k},D_{k+1},\dots,D_n\}) = \frac{k-1}{k}\cdot\frac{k}{k+1}\cdot\cdot\cdot\frac{n-1}{n} = \frac{k-1}{n}.$$

Now fix $k$. As $n$ tends to $\infty$, the probability of this number $k$ belonging to the set $\mathscr{D}$ is
$$ P(k\in\{D_1,D_2,\dots\}) = 1 - P(k\notin\{D_1,D_2,\dots\}) = \lim_{n\rightarrow\infty} 1-\frac{k-1}{n} = 1.$$

Hence, $P(N\subseteq \mathscr{D}) = 1$ which is your answer.

OLD ANSWER

Hint: Let $S_n = \{1,2,\dots,n\}$. Consider the following probabilities:

(i) $P(\{D_1,\dots,D_n\} = S_n) =\ ??$

(ii) $P(\{D_1,\dots,D_{n-1}\} = S_n) =\ ??$

Solution. The probability of (i) above is $1/n!$. Taking limit as $n$ tends to infinity, we get the probability as $0$. Thus,
$$ P(\{D_1,D_2\dots\} = \mathbb{N}) = \lim_{n\rightarrow\infty} P(\{D_1,\dots,D_n\} = S_n) = \lim_{n\rightarrow\infty} 1/n! = 0.$$ (The probability of (ii) above is $0$ because there is no way of that $n$ occurs above.)

Hope this helps. Do correct me if I am wrong.

ContentLicense: CC BY-SA 4.0


问:
Probability of a kill or receive
I am writing a paper for math class and have run into an issue.
I am trying to determine the probability of getting $9$ (or more) kills out of $21$ hits in a volleyball game in a certain spot by using a binomial distribution function.

For $p$ I am using a probability determined from data I collected ($\frac{45}{66}$ is the probability I found of getting a kill in zone $1$).

There are 6 zones and I have observed the probability of getting a kill in each of the 6 zones. For x trials, I am using 21 and n successes (kill). When I plug in the numbers I am getting crazy answers, for example, getting at least 9 kills out of 21 attempts in zone one using a probability of a kill at $\frac{45}{66}$, that probably is coming up as $99$%... while some of the other zones where the probability is $.158$ probability of getting $9$ or more kills out of $21$ attempts come up as $2$%?

This can’t be right, can it? My thinking is that a binomial distribution function would work as there are only two options (kill or receive) and I have the observed probabilities for $p$. Maybe I need to rethink my strategy? Help would be appreciated.

答:
If $p$ is the probability of a kill and $q = (1-p)$ and if each bernoulli trial [kill or no kill] is independent of the others, then the probability of getting exactly $k$ kills in $n$ trials is

$$\binom{n}{k}p^k q^{(n-k)}.$$

This means that if you want 9 or more kills out of 21, you want

$$\sum_{k=9}^{21} \binom{21}{k} p^k q^{(n-k)}
~~\text{where}~~ p = \frac{45}{66}.$$

Title: Probability of a kill or receive


问:
cyclical points on a conic section
$P_1, P_2, P_3, P_4$ are four arbitrary points on $xy = 1$. $P_1P_4$ intersects $P_2P_3$ at $D_1$, and similarly define $D_2, D_3$. Prove that $O, D_1, D_2, D_3$ are cyclical, where $O$ is the origin.

I have never seen a geometry problem asking to prove cyclical points on a conic section. So frankly not sure where to start. I suspect bashing coordinates would work, but that's quite a bit of efforts (and I am not sure how to go from coordinates to proving points are cyclical).

Is there some shortcut I could use by some geometric properties of $xy=1$?

答:
Here is an outline of a proof that still uses coordinates,
but it is high-level enough to make the core of the proof an easy
rearrangement without lengthy calculation.

For brevity, I rename the triangle $D_1D_2D_3$ to $ABC$.
Its side lengths are denoted $a,b,c$.
The rectangular hyperbola is named $K$ here.

I use barycentric coordinates with reference triangle $ABC$.
I also presume a projective geometry context; that is, I will refer to points at infinity and the line at infinity. For the latter, I use the symbol $L_\infty$ and remark that in barycentric coordinates,

$L_\infty$ has coefficient tuple $[1:1:1]$;

the incenter of $ABC$ is at $(a:b:c)$;
flipping the sign of one of these coordinates yields an excenter;

the points $(u:v:w)$ on the circumcircle of $ABC$ are those fulfilling
$a^2 v w + b^2 w u + c^2 u v = 0$.

Notation: I use round parentheses for point coordinates and square brackets for
coefficients of lines and curves.
Both kinds are occasionally represented in matrix form without the $:$ separator;
in such cases, I use $\cong$ to indicate linear dependence between two such matrices.

Some of the reasoning steps below may be regarded as reminders or observations
left as exercises.

Given $P_1=(u_1:v_1:w_1)$, then (a permutation of)
$P_2,P_3,P_4$ can be given as
$P_2=(-u_1:v_1:w_1)$, $P_3=(u_1:-v_1:w_1)$, $P_4=(u_1:v_1:-w_1)$.
This has nothing to do with $K$; it follows from the construction of $ABC$.

In barycentric coordinates, $K$ has its homogeneous equation in diagonal
form $K(u,v,w) = Ru^2 + Sv^2 + Tw^2 = 0$ with nonzero coefficients $R,S,T$.
(Nonzero because $K$ is not degenerate.)
This diagonality follows from item (1) and $K(P_i) = 0$ or alternatively
from the fact that the reference triangle $ABC$ is, by construction,
self-polar
with respect to $K$; thus the pole $A=(1:0:0)$ has associated polar
$\overline{BC}=[1:0:0]$ etc.

The center $O=(u_0:v_0:w_0)$ of $K$ is the pole of the line at infinity,
thus:

$$\begin{pmatrix}u_0\\v_0\\w_0\end{pmatrix}
\cong   \begin{bmatrix}
            R & 0 & 0
    \\      0 & S & 0
    \\      0 & 0 & T
    \end{bmatrix}^{-1}
\begin{bmatrix}1\\1\\1\end{bmatrix}
= \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix}$$

As a hyperbola, $K$ has two distinct points at infinity, call them
$E_1=(\hat{u}_1:\hat{v}_1:\hat{w}_1)$ and $E_2=(\hat{u}_2:\hat{v}_2:\hat{w}_2)$.
Thus $\overline{E_1E_2}=L_\infty$, which in barycentric coordinates
means

$$\begin{pmatrix}\hat{u}_1\\\hat{v}_1\\\hat{w}_1\end{pmatrix}
\times\begin{pmatrix}\hat{u}_2\\\hat{v}_2\\\hat{w}_2\end{pmatrix}
= \begin{bmatrix}\hat{v}_1\hat{w}_2-\hat{w}_1\hat{v}_2
\\\hat{w}_1\hat{u}_2-\hat{u}_1\hat{w}_2
\\\hat{u}_1\hat{v}_2-\hat{v}_1\hat{u}_2\end{bmatrix}
\cong \begin{bmatrix}1\\1\\1\end{bmatrix}$$

Since both $E_i$ lie on $K$, we must have

$$\begin{aligned}
\begin{bmatrix}R\\S\\T\end{bmatrix}
&\cong \begin{pmatrix}\hat{u}_1^2\\\hat{v}_1^2\\\hat{w}_1^2\end{pmatrix}
\times\begin{pmatrix}\hat{u}_2^2\\\hat{v}_2^2\\\hat{w}_2^2\end{pmatrix}
= \begin{bmatrix}\hat{v}_1^2\hat{w}_2^2-\hat{w}_1^2\hat{v}_2^2
\\\hat{w}_1^2\hat{u}_2^2-\hat{u}_1^2\hat{w}_2^2
\\\hat{u}_1^2\hat{v}_2^2-\hat{v}_1^2\hat{u}_2^2\end{bmatrix}
\\&= \begin{bmatrix}(\hat{v}_1\hat{w}_2-\hat{w}_1\hat{v}_2)(\hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2)
\\(\hat{w}_1\hat{u}_2-\hat{u}_1\hat{w}_2)(\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2)
\\(\hat{u}_1\hat{v}_2-\hat{v}_1\hat{u}_2)(\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2)\end{bmatrix}
\\&\cong \begin{bmatrix}\hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2
\\\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2
\\\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2\end{bmatrix}
\text{ using item (4)}
\end{aligned}$$

Since points at infinity represent slopes of lines, it makes sense to ask
whether two given points at infinity are orthogonal. This can be answered
if the equation of a circle is given: $E_1,E_2$ (at infinity) are
orthogonal if and only if $E_1$ lies on the polar of $E_2$ with respect to
the circle.
Concretely, given the symmetric bilinear form $U$ of any circle,
then $E_1,E_2$ are orthogonal if and only if $U(E_1,E_2) = 0$.

$K$ is a rectangular hyperbola, therefore $E_1,E_2$ must be orthogonal. Using item (6) with the symmetric bilinear form of the circumcircle $U$ of the reference triangle, we have

$$\begin{align}
    0 &= U(E_1,E_2) =
    \begin{pmatrix}\hat{u}_1\\\hat{v}_1\\\hat{w}_1\end{pmatrix}^\top
    \begin{bmatrix}
            0 & c^2 & b^2
    \\      c^2 & 0 & a^2
    \\      b^2 & a^2 & 0
    \end{bmatrix}
    \begin{pmatrix}\hat{u}_2\\\hat{v}_2\\\hat{w}_2\end{pmatrix}
\\  &= \begin{pmatrix}a\\b\\c\end{pmatrix}^\top
    \begin{bmatrix}
            \hat{v}_1\hat{w}_2+\hat{w}_1\hat{v}_2\hspace{-1em} & 0 & 0\quad
    \\      \quad 0 & \hspace{-1em}\hat{w}_1\hat{u}_2+\hat{u}_1\hat{w}_2\hspace{-1em} & 0\quad
    \\      \quad 0 & 0 & \hspace{-1em}\hat{u}_1\hat{v}_2+\hat{v}_1\hat{u}_2
    \end{bmatrix}
    \begin{pmatrix}a\\b\\c\end{pmatrix}
\\  &\cong \begin{pmatrix}a\\b\\c\end{pmatrix}^\top
    \begin{bmatrix}
            R & 0 & 0
    \\      0 & S & 0
    \\      0 & 0 & T
    \end{bmatrix}
    \begin{pmatrix}a\\b\\c\end{pmatrix} \tag{I}
\\  &= \frac{RST}{2}
    \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix}^\top
    \begin{bmatrix}
            0 & c^2 & b^2
    \\      c^2 & 0 & a^2
    \\      b^2 & a^2 & 0
    \end{bmatrix}
    \begin{pmatrix}R^{-1}\\S^{-1}\\T^{-1}\end{pmatrix}  \tag{II}
\end{align}$$

Now (II) tells us that the center $O$ of $K$ lies on the circumcircle of $ABC$.
Furthermore, (I) tells us that $K$ contains the incenter $(a:b:c)$ of $ABC$,
and we can generalize this to $K$ containing $(\pm a:\pm b:\pm c)$
which are the in- and excenters of $ABC$.
(This is fitting: The in- and excenters form an orthocentric system.)

Title: cyclical points on a conic section


问:
Question about a certain ideal
I have a question regaridngs the following ideal:

$$I_a:=\lbrace af+Xg:f,g \in R[X]\rbrace\subseteq R[X]$$

If $a\in R^{\times}$ then $I_a$ is a principal ideal.

Now my question.

In my textbook the definition of a principal ideal was:

$$(a)=\lbrace ab:b\in R \rbrace$$

Suppose $I_a$ is a principal ideal, does that mean $I_a=\lbrace ab: b \in R[X]\rbrace$?

thanks for the answers :)

答: To remove this question from the unanswered queue and to provide a more comprehensive solution, principal ideals are those generated by a single element, so if $I$ is a principal ideal of a ring $R$, then $I = (r)$ for some $r \in R$ (there may be more than one possible generator). So it doesn't matter what ring $R$ you're working in, you can always prove that a given ideal $I$ is principal by showing that $I = (r) = \{sr|s\in R\}$ for some $r \in R$.

Title: Question about a certain ideal


问:
What is the right notion of "group product" here?
Let $U$ denote a set of $N$ points $x_1, \ldots, x_N$ that live on the $(d-1)$-dimensional sphere. The set $U$ admits the following group actions:

The permutation group $S_N$ acts on $U$ by changing the labels of the $x_i$

The orthogonal group $\mathcal O_d$ acts on $U$ by rotating each $x_i$ individually.

My question is: how can I construct a group $G$ that has a copy of $S_N$ and $\mathcal O_d$ as subgroups, such that the action of $G$ on $U$ is consistent with the individual actions of $S_N$ and $\mathcal O_d$ above?

I assume that $G$ should be some kind of product of $S_N$  and $\mathcal O_d$, but I am not sure which kind is appropriate here (semi-product? wreath product?). Any help is very much appreciated!

答:
If you mean that an element of $\mathcal{O}_d$ acts by simultaneously rotating every $x_i$, then you just want the direct product $S_N\times\mathcal{O}_d$.  This works because the actions of $S_N$ and $\mathcal{O}_d$ commute with each other: permuting the points and then rotating all of them is the same as rotating all of them and then permuting them.

Alternatively, if you want $\mathcal{O}_d$ to be able to act by rotating just one point while leaving the others fixed (so you really have $N$ different actions, once for each point), this is exactly what the wreath product is for.  Specifically, you would want the wreath product of $\mathcal{O}_d$ by $S_N$ using the permutation action of $S_N$ on the set $\{1,\dots,N\}$.  Explicitly, this group can be described as a semidirect product $\mathcal{O}_d^N \rtimes S_N$ where $S_N$ acts on $\mathcal{O}_d^N$ by permuting the coordinates.


问:
When a restriction of a function is bijective, then is the original function bijective?
Given a function $f : A \to B$ and $C \subseteq A$, when $f|_C$ is bijective, then is $f$ bijective?

My initial assumption would be yes, since the domain of the restriction is a subset of the original one, meaning it should be a portion of the original's domain, and if the restriction of the function is bijective, then surely the original one is bijective. However, I am not certain about this, and I would appreciate if somebody helps in clearing this up. Our professor never discussed this so I am not sure if I missed some points or not.

答:
Not necessarily, you can lose injectivity.

For example, $x^2$ is bijective on $[0,1]$ but not on $[-1,1].$


问:
Decomposition of wave number
Wave number can be easily obtained in one dimensional situation.
$$k=\omega/c$$
In the book, Fundamentals of Acoustics, the method of separation of variables is applied to solve two dimensional wave equation. It follows that wave number is also decomposed.
$$k_x^{2}+k_z^{2}=k^{2}$$
$k_x$ and $k_z$ are determined by the boundary conditions.

But in one dimensional wave equation, it seems that we don't need additional conditions to get wave number. Given a specific frequency $f$ and the velocity $c$ of wave is determined by medium. Then we can get  $k$.

I wonder how to determine the value of the decomposed wave number if there are no boundary conditions, for example, an infinite area. Any help would be greatly appreciated.

答:
I got the explanation in the book [1].

$k_x$ is called the trace wavenumber in the $x$ direction. $\lambda_x=2\pi/k_x$ is called the trace wavelength in the $x$ direction. $c_x=\omega/k_x$ is called the phase speed of the trace wave along the $x$ axis.

Since $k$ is a constant, the trace wavenumbers are not independent of one another. Importantly, there is no restriction on the value of $k_x$ and $k_z$; they can extend over all real numbers from $-\infty$ to $+\infty$.

If the wave propagates in the direction given by $\theta$, the angle between the direction of propagation and the Z-axis, the trace wavenumbers follow that:
$$k_x =k sin\theta$$
$$k_z =k cos\theta$$

For more imformation, you can read the book.

[1] Earl G. Williams, Fourier Acoustics, Academic Press, 1999, https://doi.org/10.1016/B978-012753960-7/50009-7

Title: Decomposition of wave number


问:
On convergence of problematic series.
Determine if the following series is converges or not

$$\sum_{n=2}^{\infty}\frac{1}{\ln^{\ln(\ln n)} (n)}$$

答:
Do it by comparison with $\sum_a^\infty \frac{1}{n}$. We want to show that ultimately, $(\ln n)^{\ln\ln n}\lt n$. Equivalently, we want to show that ultimately, $(\ln\ln n)^2\lt \ln n$.

This can be done in various ways, like L'Hospital's Rule.

Or else we can let $\ln\ln n=t$. So we want to show that for $t$ large, we have $t^2\lt e^t$. This is easy, for positive $t$, by the power series expansion, $e^t\gt \frac{t^3}{3!}$.

ContentLicense: CC BY-SA 3.0

Title: On convergence of problematic series.


答:
And if the restriction of the function is bijective, then surely the original one is bijective.

While the other answer already gave an example showing you are wrong, let me explain more intuitively why your logic is faulty.

I have 5 pairs of shoes. Let's take a look at the following statement:

"All my shoes are brown".

Now, I also noticed that if I restrict the collection of my shoes to just my everyday shoes, my dancing shoes and my hiking shoes, then the statement above is true.

Can I, from what I wrote above, conclude that the statement is true in general?

Of course not. My hiking shoes are blue.

Title: When a restriction of a function is bijective, then is the original function bijective?

Body:
Given a function $f : A \to B$ and $C \subseteq A$, when $f|_C$ is bijective, then is $f$ bijective?

My initial assumption would be yes, since the domain of the restriction is a subset of the original one, meaning it should be a portion of the original's domain, and if the restriction of the function is bijective, then surely the original one is bijective. However, I am not certain about this, and I would appreciate if somebody helps in clearing this up. Our professor never discussed this so I am not sure if I missed some points or not.


问:
Show that two segments are the same measure.
On side $AB$ and $AC$ on the outside of any triangle $ABC$ there are built equilateral triangles $ABD$ and $ACE$. Let $F$ be point such that $\sphericalangle CBF=\sphericalangle FCB=30° $. Prove that $|DF|=|FE|$.

---
Notice that triangles $DAC$ and $BAE$ are congruent, so $|DC|=|BE|$ and $\sphericalangle AEB=\sphericalangle ACD$. From this we have: $$\sphericalangle BGC=180°-\sphericalangle EGC=180°-(180°-(60°-\sphericalangle AEB)-(60°+\sphericalangle ACD))=120°$$
So, $$\sphericalangle EBF + 30°+(30°-\sphericalangle FCD)=180°-120°=60° \Rightarrow \sphericalangle EBF=\sphericalangle FCD$$
Since $|FB|=|FC|$,  triangles $DFC$ and $BEF$ are congruent so $|DF|=|EF|$.

This proof seems fine, but it only works when point $F$ is inside triangle $ABC$. Are there any other more universal ways to prove this statement using congruency of triangles?

答:
I. The proof by congruent triangles also works if $G$ is within but $F$ is outside triangle $ABC$.

We still have $\angle BGC=120^o$. And since, as OP calculates$$\angle EBF+30^o+(30^o-\angle FCD)=\angle EGC=180^o-120^o=60^o$$then$$\angle EBF+60^o-\angle FCD=60^o$$making$$\angle EBF-\angle FCD=0$$and therefore$$\angle EBF=\angle FCD$$and$$DF=FE$$

II. Further, if $F$ is within triangle $ABC$ but $G$ is outside, as in the next figure

we have$$\angle EBC=\angle EGC+\angle BCG$$i.e.$$\angle EBF+30^o=60^o+\angle FCD-30^o$$or$$\angle EBF+60^o=60^o+\angle FCD$$so that again$$\angle EBF=\angle FCD$$and$$DF=FE$$

III. Lastly, even if $F$ and $G$ are both outside triangle $ABC$

again we get$$\angle EBC=\angle EGC+\angle BCG$$i.e.$$\angle EBF+30^o=60^o+\angle FCD-30^o$$or$$\angle EBF+60^o=60^o+\angle FCD$$so that again$$\angle EBF=\angle FCD$$and$$DF=FE$$

Hence it seems the proof by congruent triangles that $DF=FE$ applies generally.


问:
Find a polynomial $P$ of the smallest degree such that $\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$
Find a polynomial P of the smallest degree such that $$\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$$

I am ususally solve this kind of "best approximation" problems by taking using $GS$ but I think this time it should be solved by Chebyshev polynomials since that was the topic of the lecture, I am not sure this is the idea here but I would like for some help here (Im not even sure if I can tag it under "chebyshev polynomials"

答:
First, let us try $\mathrm{deg} P(x) \le 2$.
Let $P(x) = ax^2 + bx + c$.
We have $$\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x =
\frac{2}{5}a^2 + \frac{4}{3}ac + \frac{2}{3}b^2 + 2c^2 - \frac{12}{11}a - \frac{12}{5} c + \frac{6}{7}.$$
It is easy to minimize a quadratic form. The best $P(x)$
is given by $P(x) = \frac{9}{11}x^2 + \frac{18}{55}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 384/21175 \approx 0.01813 > \frac{1}{100}$.

Similarly, we try $\mathrm{deg} P(x) \le 3$. The best $P(x)$ is the same as that of $\mathrm{deg} P(x) \le 2$.

Then we try $\mathrm{deg} P(x) \le 4$.
The best $P(x)$ is given by
$P(x) = -\frac{189}{187}x^4 + \frac{315}{187}x^2 + \frac{45}{187}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 1536/244783 \approx 0.006275 < \frac{1}{100}$.

Thus, the smallest degree is $4$.

Remark: How to find the best $P(x)$ if $\mathrm{deg} P(x) \le 2n$?
Let $P(x) = a_{2n}x^{2n} + \cdots + a_0$. We set
$$\frac{\partial }{\partial a_k}
\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 0, \ k=0, 1, \cdots, 2n.$$


问:
For what $n$, is $S_{n}$ a homomorphic image of $S_{n+1}$?
Let $S_n$ be the symmetric group of $n$-elements.

If there is a surjective homomorphism from $S_{n+1}$ to $S_{n}$, then there must be a normal subgroup of order $n+1$ of $S_{n+1}$ given by the kernel of the homomorphism. If $n+1$ is prime, this subgroup is cyclic. It must be generated by a cycle of length $n+1$.

But the subgroup cannot be a normal subgroup in general. Thus for some $n$, there isn't such a surjective homomorphism. Hence the question:

For what values of $n$, does there exists a surjective homomorphism $S_{n+1} \to S_n$?

答:
Let's let $\phi\colon S_{n+1}\to S_n$ be a surjective homomorphism. Then, as you said the kernel of $\phi$ must be a normal subgroup of $S_{n+1}$, and $|\text{ker }\phi|=n+1$.

Luckily, the normal subgroups of the symmetric groups are well-understood. Here is a proof that the only normal subgroup of $S_n$ is $A_n$, when $n\geq 5$. (That proof relies on the knowledge that the alternating groups $A_n$ are simple for $n\neq 4$, a proof of which can be found here.) Clearly, for $n+1\geq 5$, the alternating group $A_{n+1}\leq S_{n+1}$ will be far too large to be our kernel; therefore, we only have to check the cases $n=1,2,3$.

$n=1$ is the easiest; $S_1$ is trivial, so the trivial homomorphism $\phi_{id}\colon S_2\to S_1$ works.

For $n=2$, we can easily see that the sign homomorphism, which sends even homomorphisms to the identity and odd homomorphisms to the transpositions $(1\ 2)$, will do nicely as our homomorphism $\phi\colon S_3\to S_2$.

$n=3$ is far more interesting; $A_4$ is actually non-simple, and $S_4$ has a normal subgroup $H$ that is isomorphic to the Klein $4$-group. This subgroup, it turns out, is perfect for our kernel. A description of this homomorphism $\phi:S_4\to S_3$ is left as an exercise to the reader :-)


问:
If $x_{n+1}=x_n+x_n^{-m}, x_1>0, m \ge 2 $ then $x_{n}^{m+1} = (m+1)(n+\ln(n))+O_m(1) $
This is a generalization of
Divergence of a sequence proof

Show that
if
$x_{n+1}=x_n+\dfrac{1}{x_n^m},
x_1>0, m \ge 2
$
then
$x_{n}^{m+1}
= (m+1)(n+\ln(n))+O_m(1)
$.

(Note:
The notation $O_m(...)$
means that the
constant implied by the
big-oh
depends on $m$.)

I can show that
$x_{n}^{m+1}
= (m+1)n+O_m(\ln(n))
$
for integer $m$,
and I am pretty sure that
in this case
I can show that
$x_{n}^{m+1}
= (m+1)(n+\ln(n))+O_m(1)
$.

I don't have an
explicit expression for the
$O_m(1)$
and I don't have a proof
for real $m \ge 2$.

答:
The coefficient of $\log n$ turns out to be $\color{blue}{m/2}$, not $m+1$. Also, the "$O_m(1)$" depends on $x_1$ too.

We assume $m>-1$, and use the binomial series (BS) and the Stolz–Cesàro theorem (SCT).

Let $y_n=x_n^{m+1}$; then $y_{n+1}=y_n(1+1/y_n)^{m+1}$. Clearly $y_n\to\infty$ [as $n\to\infty$]. By (BS), $$y_{n+1}=y_n\big(1+(m+1)/y_n+o(1/y_n)\big)\implies y_{n+1}-y_n\to m+1,$$ and (SCT) gives $y_n/n\to m+1$ (as we already know from the OP).

Now we let $y_n=(m+1)n+z_n$ with $z_n=o(n)$, and we put this into the next-order (BS) $$y_{n+1}=y_n\left(1+\frac{m+1}{y_n}+\frac{m(m+1)}{2y_n^2}+o(y_n^{-2})\right),$$ which results in $$z_{n+1}-z_n=\frac{m(m+1)}{2(m+1)n+2z_n}+o(n^{-1})=\frac{m}{2n}+o(n^{-1}),$$ and this time (SCT) yields $z_n/\log n\to m/2$ (as claimed at the beginning).

Similarly, if we let $z_n=(m/2)\log n+t_n$, we get $t_{n+1}-t_n=O(n^{-2}\log n)$, hence $$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\left(x_n^{m+1}-(m+1)n-\frac{m}{2}\log n\right)$$ exists (and is finite; but I wouldn't expect it to have a closed form).


问:
Integral tends to infty
How can we prove that $\lim\limits_{\varepsilon\searrow 0} \dfrac{1}{\varepsilon} \displaystyle\int_0^1 \left [\dfrac{1}{4}-\dfrac{1}{\pi^2}\arctan^2\left (\dfrac{x}{\varepsilon}\right )\right ]^2 dx=+\infty$?

I came across this integral in studying the Heaviside aproximation function. I used the following sharp estimate:

$\arctan(x)<\dfrac{\pi^2 x}{4+\sqrt{32+(2\pi x)^2}}, \forall\ x>0$, but the limit goes to $0$ (I computed it exactly with wolfram). This estimation and many others can be found here:

https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-018-1734-7

The integral in the statement is simulated here:

https://www.wolframalpha.com/input/?i=int_0%5E1+1000000000+%281%2F4+-+%281%2Fpi%5E2%29+%28arctan%2810000000000+x%29%29%29%5E2+dx

But I wonder if it can be proved that it really tends to infinity. Maybe there are some singularities and the program gives some big errors.

Title: Integral tends to infty

Body:
We can approach the value given by @Sangchul Lee in comments.

Let
$$f(t)= \left[\frac{1}{4} - \frac{1}{\pi^2} \arctan^2 \bigl( t^2 \bigr) \right]^2 $$ and write for example
$$\int_0^\infty f(t)\,dt=\int_0^1 f(t)\,dt+\int_1^\infty f(t)\,dt$$ For the first integral, using
$$f(t)=\frac{1}{16}-\frac{t^4}{2 \pi ^2}+\frac{\left(3+\pi ^2\right) t^8}{3 \pi
   ^4}-\frac{\left(120+23 \pi ^2\right) t^{12}}{90 \pi ^4}+\frac{22 \left(7+\pi
   ^2\right) t^{16}}{105 \pi ^4}-\frac{\left(14360+1689 \pi ^2\right) t^{20}}{9450
   \pi ^4}+\frac{\left(239327+24405 \pi ^2\right) t^{24}}{155925 \pi
   ^4}+O\left(t^{28}\right)$$
$$\color{blue}{\int_0^1 f(t)\,dt \sim \frac{1}{16}-\frac{58331177}{804053250 \pi ^2}+\frac{15323818}{182739375 \pi ^4}}$$
For the second integral, using
$$f(t)=\frac{1}{\pi ^2 t^4}-\frac{2}{\pi ^3 t^6}+\frac{3-2 \pi ^2}{3 \pi ^4 t^8}+\frac{2}{\pi ^3 t^{10}}+\frac{23 \pi ^2-60}{45 \pi ^4
   t^{12}}-\frac{28}{15 \pi ^3 t^{14}}-\frac{22 \left(2 \pi ^2-7\right)}{105 \pi ^4
   t^{16}}+\frac{1636}{945 \pi ^3 t^{18}}+\frac{1689 \pi ^2-7180}{4725 \pi ^4
   t^{20}}-\frac{282}{175 \pi ^3 t^{22}}+\frac{239327-48810 \pi ^2}{155925 \pi ^4
   t^{24}}+\frac{26126}{17325 \pi ^3
   t^{26}}+O\left(\frac{1}{t^{28}}\right)$$
$$\color{blue}{\int_1^\infty f(t)\,dt \sim \frac{5946893}{22713075 \pi ^2}-\frac{474283184}{2010133125 \pi
   ^3}+\frac{1446994}{13627845 \pi ^4}}$$

So
$$\color{red}{\int_0^\infty f(t)\,dt \sim \frac{1}{16}+\frac{22169236787}{117123756750 \pi ^2}-\frac{474283184}{2010133125 \pi ^3}+\frac{500797244128}{2635284526875 \pi ^4}}$$ which is $0.0760194$.

Edit

After the edit of the question, the problem is much simpler with
$$f(t)= \left[\frac{1}{4} - \frac{1}{\pi^2} \arctan^2 \bigl( t \bigr) \right]^2 $$ since there is an antiderivative.

Expanding the square,we face two integrals
$$I_1=\int\arctan^2(t)\,dt \qquad \text{and} \qquad I_2=\int\arctan^4(t)\,dt$$ Making $t=\tan(u)$
$$I_1=\int u^2 \sec ^2(u)\,du \qquad \text{and} \qquad I_2=\int u^4 \sec ^2(u)\,du $$ which can express in terms of the polylogarithm function.

Recombining the antiderivatives and using the bounds
$$\int_0^\infty \left[\frac{1}{4} - \frac{1}{\pi^2} \arctan^2 \bigl( t \bigr) \right]^2 =\frac{3 }{\pi ^3}\zeta (3)$$

Using, as before, series expansions
$$\color{blue}{\int_0^1 f(t)\,dt \sim \frac{1}{16}+\frac{40832183318753}{559056788915625 \pi
   ^4}-\frac{56140585778267}{452214824811750 \pi ^2}}$$
$$\color{blue}{\int_1^\infty f(t)\,dt \sim \frac{1112361876473779}{7102091799928125 \pi
   ^4}-\frac{478852611552730453}{679000559454842625 \pi
   ^3}+\frac{2370558288062177347}{2799586611272075625 \pi ^2}}$$

This gives $\sim 0.11533$ while $\frac{3 }{\pi ^3}\zeta (3)\sim 0.11630$.


问:
Find the probability that there are at least three teachers in the selection.
From $8$ teachers and $7$ students, $4$ people are randomly selected.

Find the probability that there are at least three teachers in the selection.

With this I knew I had to get the total of options.  So $\binom{15}{4}$.
Then I thought it would be $\binom{7}{3}$ showing the number of options $3$ teachers can be selected.
Then the number of options for a single option, so $\binom{15}{1}$, as the fourth selection could be a teacher or student.

The actual answer is $\frac{22}{65}$, just not sure what I'm doing wrong.

答:
You can do so by dividing them into cases:

Case I: Three teachers and one student

You can choose this in $\binom{8}{3} \times \binom {7}{1}$ ways.

So the probability for case I is given by $\frac{56 \times 7}{1365}=\frac{392}{1365}$

Case II: Four teachers and no student

You can choose this in $\binom{8}{4}$ ways.

So the probability for case II is $\frac{70}{1365}$

The final probability will be found by adding the two cases.


问:
Application of Parseval's identity to a vectorial series
Let $H$ be a Hilbert space and let $\{e_n, n\in \mathbb N\}$ be an orthonormal sequence in $H$.
Determine whether these series converge in $H$:

$\sum \frac{e_n}{n}$

$\sum\frac{e_n}{\sqrt{n}}$

The Parseval's identity tells us  that for every $x \in H$,
$$
\displaystyle \sum _m|\langle x,e_m\rangle |^2=\|x\|^2.
$$
Let $x=\displaystyle\sum_n \frac{e_n}{n}$, then $|\langle x,e_m\rangle|^2=\displaystyle\sum_n \frac{1}{n^2}\quad \forall m \in \mathbb N$. Which is a convergent Riemann series. In other hand we have
$$
\|x\|^2= \displaystyle\sum_n |\langle \frac{e_n}{n}, \frac{e_n}{n}\rangle|^2=\displaystyle\sum_n \frac{1}{n^2}
$$

and there I got confused.

Is there a merciful soul that can help me?
Thank's

答:
$\sum a_ne_n$ converges if and only if $\sum|a_n|^{2} <\infty$. Hence the first series is convergent and the second one is not.

Proof: Recall that Hilbert spaces are complete so a series in $H$ converges if and only if the partial sums form a  Cauchy sequence. $\|\sum\limits_{k=n}^{m} a_ne_n\|^{2}=\sum\limits_{k=n}^{m}|a_n|^{2}$ and a series $\sum b_n$ of non-negative numbers converges if and only if $\sum\limits_{k=n}^{m} b_k \to 0$ as $ m>n \to \infty$.


答:
This probability follows a hyper-geometric distribution since you are picking the teachers and students from a small population with no replacement.

In general and as indicated by Wikipedia, the distribution takes the following form:

$\frac{{K \choose k} {N-K \choose n-k}}{N \choose n} $

Where:

$N$ is the population size,

$K$ is the number of success states in the population,

$n$ is the number of draws (i.e. quantity drawn in each trial),

$k$ is the number of observed successes

The probability that AT LEAST three teachers will be selected is equal to the probability that exactly three teachers will be selected + the probability that exactly four teachers will be selected so we get the following:

$\frac{\binom{8}{3}\cdot \binom{7}{1}}{\binom{15}{4}}$ for the case of exactly three teachers, the numerator gives us the number of ways we can choose the teachers in your case, we are choosing three teachers out of eight where order doesn't matter so that is $\binom{8}{3}$ and for each of these choices we need to choose one student which is why we multiply it by the total number of ways of choosing one student and that is $\binom{7}{1}$, moreover we divide this by the total number of ways we can choose any four people from what we have to get the probability  and that would be $\binom{15}{4}$

$\frac{\binom{8}{4}\cdot \binom{7}{0}}{\binom{15}{4}}$ for the case of exactly four teachers, the explanation is similar to the previous case

Then you add both together since these are mutual exclusive events.

Body:
From $8$ teachers and $7$ students, $4$ people are randomly selected.

Find the probability that there are at least three teachers in the selection.

With this I knew I had to get the total of options.  So $\binom{15}{4}$.
Then I thought it would be $\binom{7}{3}$ showing the number of options $3$ teachers can be selected.
Then the number of options for a single option, so $\binom{15}{1}$, as the fourth selection could be a teacher or student.

The actual answer is $\frac{22}{65}$, just not sure what I'm doing wrong.


问:
Maximum of d(12-d)
I'm a little confused on a quite simple quadratic problem. I need to calculate the maximum of $d(12-d)$ using basic quadratics. The answer is $6$ as can also be shown by $f'(x)= -2d +12$, however this is algebra not calculus. Using the squares method I get twelve, I think I may be using the incorrect method. Could someone set me on the right track?

$d(12-d)$

$= -d^2 + 12d$

$-d^2 + 12/(-2)d + 36 = 36 $

$(d-6)^2 = 36$

$d = 12  or  0$ <- I understand these are the roots, am I not meant to calculate them?

The suggested answer is: $-(d-6)^2 + 36$, but I have no idea where it came from.

Title: Maximum of d(12-d)

Body:
$$d(12-d)=12d-d^2=-(d^2-2\cdot6\cdot d+6^2)+6^2=36-(d-6)^2$$

Now for real $d, (d-6)^2\ge 0$

$\implies d(12-d)\le 36$


问:
Producing a matrix not similar to its transpose
I know that a matrix is similar to its transpose, if the minimal polynomial over the field splits
Now how do I produce a matrix over $\mathbb{R}$ that does not satisfy given property.

Can you give me a hint ?

答:
A (square) matrix is always similar to its transponse. One reason (which is not immediate to prove) is that $A,B\in k^{n\times n}$ are similar if and only if they are similar in $\overline k^{n\times n}$: in other words, for two matrices $A,B\in k^{n\times n}$ there is an invertible matrix $P\in k^{n\times n}$ such that $B=PAP^{-1}$ if and only if there is a matrix $Q\in\overline k^{n\times n}$ such that $B=QAQ^{-1}$. In $\overline k^{n\times n}$ all matrices are triangulable, and therefore the result follows.

An intrinsic formulation of this, but not a substantially easier one to prove, is that two matrices $A,B\in k^{n\times n}$ are similar if and only if $\dim\ker (p(A))^s=\dim \ker (p(B))^s$ for all $n\ge1$ and for all irreducible polynomials $p\in k[X]$. With this formulation, you would use identities $\dim\ker M^T=\dim \ker M$ and $p(M^T)=(p(M))^T$.


问:
Morphism of algebraic function fields
Let $k$ be a field and $T$ be an indeterminate. Consider $y\in\overline{k(T)}$. So, $y$ is a root of an irreducible polynomial $P(X)\in k(T)[X]$: $P(X)=\sum_{i=0}^na_i(T)X^i$ where the $a_i$'s are in $k(T)$. Each element $z\in k(t)(y)$ can be uniquely written as:
$z=\sum_{i=0}^{n-1}\gamma_i(T)y^i$ with the $\gamma_i$'s are in $k(T)$. Denote by $\beta$ a root (in $\overline{k(T)}$) of the polynomial $Q(X)=\sum_{i=0}^na_i(h(T))X^i$, where $h(T)\in k(T)$. Define the application $\sigma:k(T)(y)\to k(T)(\beta)$ by $\sigma(z)=\sum_{i=0}^{n-1}\gamma_i(h(T))\beta^i$. I want to prove that $\sigma$ is a $k$-morphism of fields.

Any clue to prove that would be welcome.

答:
Let $h(T)\in k(T),\not \in k$ and the field isomorphism $$\sigma : F=k(T)\to \sigma(F)=k(h(T)),\qquad T\to h(T)$$

Then it is completely obvious that $$F[X]/(\sum_{j=0}^n a_j X^j) \to \sigma(F)[X]/(\sum_{j=0}^n \sigma(a_j) X^j), \qquad \sum_{m=0}^{n-1} f_m X^m \to \sum_{m=0}^{n-1} \sigma(f_m) X^m$$
is a ring isomorphism.

No need that $\sum_{j=0}^n a_j X^j$ is irreducible in $F[X]$.

Title: Morphism of algebraic function fields


问:
Is $\left(K, - \right)$ a group, semigroup, or monoid?
My professor asked us to determine whether $\left(K, - \right)$ for $K$ is a set of integers is a group, semigroup, or monoid. Using my limited knowledge in the group theory here's 4 condition that I need to check

For $a, b \in K$, $a-b$ is a uniquely defined element of $K$

It satisfies associativity

It has an identity element $e$

For each $a \in K, \exists a^{-1} \in K$ then $a - a^{-1} = e$ and $a^{-1} - a = e$

I found that $\left(K, - \right)$ only satisfy (1). My question is $\left(K, - \right)$ a group, semigroup, or monoid? or neither?

Thank you

答:
You're right in that only condition $1$ is satisfied.

As for determining what structure it is, a table on Wikipedia gives a list of various structures and the necessary properties for each:

Within these definitions, $(K,-)$ is a magma. It is not a group or semigroup or monoid, since we don't have associativity in particular, among the other axioms they require.

Now, interestingly, there is a sense in which you can say structures satisfy invertibility without an identity. Namely, as for quasigroups, $(S,\ast)$ is invertible if and only if

$$(\forall a,b \in S)(\exists ! \; x,y \in S)(a \ast x = b \text{ and } y \ast a = b)$$

In the case of $(S,\ast) = (K,-)$, then this happens to hold: $x = a-b$ and $y = b-a$, and uniqueness is obvious. Within this definition, then you could go so far as to claim $(K,-)$ is a quasigroup.


问:
Choose whether or not the given sets are equinumerous to $\mathbb{N}^\mathbb{N}$.
Which of the following sets are equinumerous with $\mathbb{N}^\mathbb{N}$?

(i) $\mathbb{N} $

(ii) $\mathbb{R}$

(iii) $2^\mathbb{R}$

(iv) $\mathbb{N} \times \mathbb{R}$

(v) $\mathbb{R}^\mathbb{R}$

(vi) $\mathbb{R}^\mathbb{N}$

(vii) The set of permutations of $\mathbb{N}$

It seems that only choice (i) is equinumerous with $\mathbb{N}^\mathbb{N}$ but I am not sure. Could anyone help explain?

答:
For the proofs I'll use $|F|$ as the cardinal of $F$ and $\mathbb N=\{1,2,3,..\}$

(i) No. In general $|2^F|>|F|$ for any set $F$.

(ii) Yes. For $\mathbb R$ define $\mathbb R_n$ as $[n-1,n)$ and for a set of the form $[a,b)$ define $[a,b)_n$ as $\left[b-\frac{b-a}{2^{n-1}},b-\frac{b-a}{2^n}\right)$, notice that this is an ordered partition of $[a,b)$, now we define $\mathbb R_{(n_1,n_2)}$ as $(\mathbb R_{n_1})_{n_2}$ and inductively $\mathbb R_{(n_1,...,n_k)}$ as $(\mathbb R_{(n_1,...,n_{k-1})})_{n_k}$ for every $r=(n_1,n_2,...)\in\mathbb N^\mathbb N$ define $b_{r,k}=(n_1,...,n_k)$ and $f:\mathbb N^\mathbb N\to\mathbb R/f(r)=\bigcap_{k\in\mathbb N}\mathbb R_{b_{r,k}}$ (notice that there always exist a point and only one point in $\bigcap_{k\in\mathbb N}\mathbb R_{b_{r,k}}$). $f$ is clearly inyective, so $|\mathbb N^\mathbb N|\leq|\mathbb R|$. Now $2^{\mathbb N}\subset\mathbb N^\mathbb N$ so $|2^\mathbb N|\leq|\mathbb N^\mathbb N|$, define $f:2^\mathbb N\to\mathbb [0,1]$ as $f((n_i)_{i\in\mathbb n})=\sum_{i\in\mathbb N}\frac{n_i}{2^i}$ (where $n_i\in\{0,1\}$), this function is surjective (binary representation) so $|\mathbb N^\mathbb N|\leq|\mathbb R|=|[0,1]|\leq|2^\mathbb N|\leq|\mathbb N^\mathbb N|$ so $|\mathbb N^\mathbb N|=| 2^\mathbb N|=|\mathbb R|$.

(iii) No. $|2^\mathbb R|>|\mathbb R|=|\mathbb N^\mathbb N|.$

(iv) Yes. Clearly $|\mathbb R|\leq|\mathbb R\times\mathbb N|$, now remember $|[0,1)|=\mathbb R$ so $|[0,1)\times\mathbb N|=|\mathbb R\times\mathbb N|$. Take $f:[0,1)\times \mathbb N\to\mathbb R/f(x,n)=x+n$, this is clearly inyective, so $|\mathbb N^\mathbb N|=|\mathbb R|=|\mathbb R\times\mathbb N|$.

(v) No. $|\mathbb R^\mathbb R|\geq|2^\mathbb R|>|\mathbb N^\mathbb N|$.

(vi) Yes. Clearly $|\mathbb R|\leq|\mathbb R^\mathbb N|$. Take $[0,1)$ and for each $x\in[0,1)$ take its unique binary representation $x=0,x_1x_2...$ such that $x_n\in\{0,1\}$ and there isn't an $N$ such that $x_n=1$ for all $n\geq N$. Now define $y_n^k=x_{{k+n\choose 2}-(n-1)}$, $y_n=0,y_n^1y_n^2...$ and $f:[0,1)\to\mathbb [0,1]^\mathbb N/f(0,x_1x_2,...)=(y_1,y_2,...)$. To help you visualize what I'm doing here: $$0,x_1\color{blue}{x_2}x_3\color{red}{x_4}\color{blue}{x_5}x_6\color{yellow}{x_7}\color{red}{x_8}\color{blue}{x_9}x_{10}...\to(y_1,\color{blue}{y_2},\color{red}{y_3},\color{yellow}{y_4},...)$$
Notice that $[0,1)^\mathbb N\subset f([0,1))$ so $|\mathbb R^\mathbb N|=|[0,1)^\mathbb N|\leq|[0,1)|=|\mathbb R|=|\mathbb N^\mathbb N|$.

(vii) Yes. The set of permutations of $\mathbb N$ ($\mathbb N!$) is a subset of $\mathbb N^\mathbb N$ so $|\mathbb N!|\leq|\mathbb N^\mathbb N|$. Now take $f:\mathbb N\to\{0,1\}$ and define $g_f:\mathbb N\to\mathbb N$ as follows: if $f(n)=0$ then $g_f(2n-1)=2n-1$ and $g_f(2n)=2n$, and if $f(n)=0$ then $g_f(2n-1)=2n$ and $g_f(2n)=2n-1$, the map $f\to g_f$ is inyective and $g_f\in \mathbb N!$, so $|2^\mathbb N|\leq|\mathbb N!|\leq|\mathbb N^\mathbb N|$ which implies $|\mathbb N^\mathbb N|=|\mathbb N!|$.


问:
Polar equation to cartesian equation help
Since this got closed
Convert Polar equation to Cartesian coordinates

So I ended up doing some work and I don't know if I'm headed in the right direction, or maybe if I solved it correctly.

problem:
$$r\cos(\theta) = 5r^2\sin(2\theta)$$

Steps taken:

$$\begin{align}x&=10r^2\sin(\theta)\cos(\theta)\\x&=10r\sin(\theta)r\cos(\theta)\\x&=10yx\\y&=1/10\end{align}$$

Assuming that I'm correct with how to handle converting polar to cartesian, this would be just a horizontal line. I'm confused because I have to give cartesian coordinates, and also describe the curve.

答: $x(1-10y)=0$ describes the curve. So it's two straight lines $x=0$ and $y=\frac{1}{10}$


问:
if $f$ is lebesgue integrable and additive then $f$ is continuous
How can I prove that if $f$ is Lebesgue integrable in every interval and additive then $f$ is continuous.

答:
Suppose that $f$ is locally Lebesgue integable and solves the Cauchy functional equation

$$ \forall x, y \in \mathbb{R} \ : \quad f(x+y) = f(x) + f(y). $$

Then for each given $x$, we have $ f(x) = f(x+t) - f(t) $ for all $t$, and so, integrating both sides for $0 \leq t \leq 1$ gives

$$ f(x) = \int_{x}^{x+1} f(t) \, \mathrm{d}t - \int_{0}^{1} f(t) \, \mathrm{d}t. $$

Now by the absolute continuity of antiderivative, the right-hand side is continuous, and therefore $f$ is continuous.


问:
Growth Rate of Integral of Harmonic functions
This problem comes from a statement in the book $\textit{Elliptic Partial Differential Equations: Second Edition}$ written by Han and Lin. In its chapter 1, there's Lemma 1.41 (in the page 22) as follows.

Suppose $\{a_{ij}\}$ is a constant positive definite matrix such that for all $\xi\in\mathbb R^n,$
$$\lambda|\xi|^2\le a_{ij}\xi_i\xi_j\le \Lambda|\xi|^2$$
for some constant $0<\lambda\le\Lambda.$ Suppose $u\in C^1(B_1)$ satisfies
$$\int_{B_1}a_{ij}D_iuD_j\varphi=0$$
for all $\varphi\in C^1_0(B_1),$ then there exists $c>0$ such that for all $\rho\in(0,1],$
$$\int_{B_\rho}|u|^2\le c\rho^n\int_{B_1}|u|^2.$$

In the statement above, $B_\rho\subseteq\mathbb R^n$ is the open ball centered at the origin with radius $\rho.$

My question is the first step in its proof. It says that we only need to treat the case $\rho\in(0,\frac 12]$ since the inequality is trivial for $\rho\in(\frac 12,1].$ I could not understand why it is trivial in the latter case.

Any comment is appreciated!

答:
Perhaps the 1/2 is a red herring; it could be any constant bigger than 0. The point is that trivially, merely because $B_\rho\subset B_1$,
$$ \int_{B_\rho} |u|^2 \le \int_{B_1} |u|^2. $$
The important part of the lemma is the decay rate as $\rho\to 0$: in this regime, $c\rho^n$ is an improvement over the above constant 1.


问:
How to calculate the divergence by the definition
I am trying to find the divergence of this field: $E = \frac{q(1-\sqrt{\alpha r})}{4 \pi \epsilon r^2} \hat r$

I already found the surface integral, that is, $\int EdS = \frac{q(1-\sqrt{\alpha r_{1}})}{\epsilon}$

I am having trouble to evaluate how to calculate the divergence by the definition now. (It has to be by definition

答: If $\vec{E}=f(r)\hat{r}$ then integration over the radius-$r_1$ circle centred on the origin gives$$\int\vec{E}\cdot d\vec{S}=\int f(r)\hat{r}\cdot(dS)\hat{r}=\int f(r)dS=\int f(r_1)dS=f(r_1)\int dS=4\pi r_1^2f(r_1).$$


问:
Percentage profit/loss
By selling two tables for 1200 each, a man gains 10% on one and loses 10% on the other. What is his gain/loss percent in the whole transaction?

I calculated cost price for table one by equating selling price to 1.1 cost price.and got the cost price as 1,091(approx) and for table 2 I equated selling price to .9 cost price which comes out to be 1,333 (approx). Total cost price becomes 2024. There is a loss of 24. Now if I calculate loss on cost price the answer is 0.99 which is close to the answer given in the book which 1% loss.

Is there some other easier way to solve this?

答:
As said before, there is no easier way to solve this. However, you can avoid any rounding errors by calculating at the very last step, instead of using approximations.

Calculate $\frac{1200}{1.1} + \frac{1200}{0.9}$ directly, as to get the original price, you undo the $1.1$ and $0.9$ multipliers on the tables.

Then the percentage change is:

$$\frac{\text{new} - \text{old}}{\text{old}} = \frac{2400-(1200/1.1 + 1200/0.9 )}{1200/1.1 + 1200/0.9} \times 100 \% \approx -1.01 \%$$

or approximately a $1 \%$ decrease.

Title: Percentage profit/loss


问:
Prove for all $t\in \mathbb{N}$ there are natural $t$ consecutive number such that every one of them are not power of primes
Prove for all $t\in \mathbb{N}$ there are natural $t$ consecutive number such that every one of them are not power of primes.

$q$ will be Power of primes if $\exists p $ prime such that $q=p^t$ , $q,p,t \in \mathbb{N}$

Hint (1): Chinese remainder theorem

Hint (2): Solve. $n\equiv−1\pmod{2\times3}$ and  $n\equiv−2\pmod{5\times7}$

Attempt:
Solve the hint (2)

Chinese remainder theorem

$$M= 2\cdot 3 \cdot 5 \cdot 7 ...\cdot k...$$
$$M_1= 5\cdot 7$$
$$M_2 = 2\cdot 3$$
$$M_n = p_{k-1}\cdot p_k$$
$m_i$ , $M_i$ are coprime for all $i$ , Let $y_i$ be the inverse of $M_i$ mod $m_i$ .
$$n=\sum^{k}_{i=1}a_i\cdot M_i\cdot y_i$$

$$n=(-1)\cdot5\cdot7\cdot5+(-2)\cdot2\cdot3\cdot6=-247$$

$$n\equiv-247\pmod{2\cdot3\cdot5\cdot7}$$
$$n\equiv 173\pmod{2\cdot3\cdot5\cdot7}$$

答:
As lulu's question comments indicate, one way to solve this is to use modular equations and the Chinese remainder theorem. You don't actually need to solve the specific given example, apart from it helping you to understand the concept. As for solving it, though, a somewhat easier example to use instead is

$$n \equiv 0 \pmod{2 \times 3} \tag{1}\label{eq1A}$$

$$n \equiv -1 \pmod{5 \times 7} \tag{2}\label{eq2A}$$

The first one shows $n$ is a multiple of $6$. With the second one, a fairly easy way to solve it is to note the first positive value is $34 \equiv 4 \pmod{6}$, and you want to add an integral multiple of $35 \equiv -1 \pmod{6}$ to get $n$, with $6 \mid n$. This multiple is $4$, so the result is $n = 34 + 4(35) = 174$. Now, $n = (2)(3)(29)$ and $n + 1 = (5^2)(7)$. In particular, there's at least $2$ prime factors for both $n$ (i.e., $2$ and $3$) and $n + 1$ (i.e., $5$ and $7$).

For the general case, I believe it's a bit easier to start at $-1$ instead of $0$, as lulu suggested. In that case, there would be $t$ equations, where if $p_i$ is the $i$'th prime, then the $j$'th modulus equation, for $1 \le j \le t$, would be

$$n \equiv -j \pmod{p_{2j-1} \times p_{2j}} \implies n + j \equiv 0 \pmod{p_{2j-1} \times p_{2j}} \tag{3}\label{eq3A}$$

Since all of the moduli are coprime, the Chinese remainder theorem guarantees there exists a unique solution modulo the product of these moduli. Also, $n + j$ has prime factors of $p_{2j-1}$ and $p_{2j}$, so it can't be a power of a prime.

For another method, Piquito's question comment suggested using $n!$. A fairly simple way to use this idea is to consider, for $2 \le k \le t + 1$,

$$n_k = (2t + 2)! + k \tag{4}\label{eq4A}$$

Note $k \mid (2t + 2)!$ so, if $k$ has $2$ or more distinct prime factors, then those prime factors will divide $n_k$, so it can't be a power of a prime. If, instead, $k = p^{m}$ for some prime $p$ and integer $m \ge 1$, then we get

$$n_k = p^{m}\left(\frac{(2t + 2)!}{p^{m}} + 1\right) \tag{5}\label{eq5A}$$

For simpler algebra, let

$$f = \frac{(2t + 2)!}{p^{m}} \tag{6}\label{eq6A}$$

Since $2k = 2p^{m} \le 2(t + 1)$, then it's a separate factor from $k$ of $(2t + 2)!$. This means $p^{m} \mid f$, so $p \not\mid f + 1$. Since $f + 1 \gt 1$, it must have prime factors other than $p$, so $n_k$ is also not a prime power in this case either.


问:
Variance of number of heads
Given two coins with chance of heads $p=0.6$ for a coin $X$, and $p=0.7$ for a coin $Y$. Pick one coin uniformly at random. Toss the coin 10 times. Find the variance $Z$ of the number of heads. I am thinking of just using the definition of variance.

$E[Z]^2 = \sum_{n=0}^{10}\frac{n}{2}\left(\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{6}{10}\right)^{n}\left(\frac{4}{10}\right)^{10-n}+\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{7}{10}\right)^{n}\left(\frac{3}{10}\right)^{10-n}\right) = 42.25$
$E[Z^2] = \sum_{n=0}^{10}\frac{n^{2}}{2}\left(\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{6}{10}\right)^{n}\left(\frac{4}{10}\right)^{10-n}+\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{7}{10}\right)^{n}\left(\frac{3}{10}\right)^{10-n}\right) = 44.75$

$Var[Z] = 2.5$

But is there a faster, cleaner way without all this computation?

答:
your calculations are correct!. To use a shortcut, observe that

Z probability function, using total probability theorem, is the following

$$\mathbb{P}[Z=z]=\frac{1}{2}\mathbb{P}[Z=z|p=0.6]+\frac{1}{2}\mathbb{P}[Z=z|p=0.7]$$

Now observe that the second simple moment for a binomial $B(n;p)=np(1-p)+(np)^2$

Thus the variance is

$$\mathbb{V}[Z]=\mathbb{E}[Z^2]-\mathbb{E}^2[Z]=0.5[(2.4+6^2)+(2.1+7^2)]-[0.5(6+7)]^2=2.5$$

Title: Variance of number of heads


问:
Proving if a function from $\mathbb R^2$ to $ \mathbb R^2$ is convex?
Say $f(x,y)$ = [ $x + y $$\quad$ $ 2x + 7y$ ] , how do I prove that this function is convex?

I know in order to prove that a function from $R^2$ --> $R^1$ is convex then the Hessian of that function has to be pos-semi-def. So is it then sufficient to say that if,

$f(x,y) = x + y$ $\quad$ is convex

and

$f(x,y) = 2x+7y$ $\quad$ is convex,

then

$f(x,y)$ = [$x+y$ $\quad$ $2x+7y$] $\quad$ is also convex?

My intuition says yes, but I haven't been able to find anything supporting this online.

答: By definition, a function $f: \mathbf{X}\to\bar{\mathbb{R}}$ is convex if $\text{epi}f$ is a convex set. Therefore convexity is a definition which applies to real-valued functions only. Vector-valued functions simply don't posses the property of being be convex or not.


问:
A conjecture on a product of a positive-definite matrix and a vector
let $A$, $B$ positive-definite matrices,  and $v$ a vector. $A,B$ don't necessarily commute. Let $w:=(A+B)^{-1}Av$. Is it always the case that $|w_i| \le |v_i|$?

Body:
No. When
$$
A=\pmatrix{2&-1\\ -1&1},\ B=\pmatrix{1&-2\\ -2&5},
$$
\begin{aligned}
(A+B)^{-1}A
&=(I+A^{-1}B)^{-1}\\
&=\left[I+\pmatrix{1&1\\ 1&2}\pmatrix{1&-2\\ -2&5}\right]^{-1}\\
&=\pmatrix{0&3\\ -3&9}^{-1}\\
&=\pmatrix{1&-\tfrac13\\ \tfrac13&0}.
\end{aligned}
Therefore the Euclidean norm of the first column of $(A+B)^{-1}A$ is greater than $1$.


问:
Cauchy Problem For Inhomogeneous Transport Equation
Here is a problem:

$\left\{\begin{array}{l}
2 u_{t}-u_{x}=f(t, x), \quad t>0 \\
\left.u\right|_{t=0}=\varphi(x)
\end{array}\right.$

My solution

$\frac{d t}{2}=\frac{d x}{-1}=\frac{d y}{f(x, c)}$

Find the first characteristic:

$\frac{d t}{2}=\frac{d x}{-1} \Rightarrow C_{1} = x+\frac{1}{2} t$

And the second one:

$\frac{d t}{2}=\frac{d u}{f(x, y)} \Rightarrow \frac{d u}{d t}=f\left(t, C_{1}-\frac{1}{2} t\right)$

And solution is a linear combination of them:

$u(x, t)=\varphi\left(C_{1}-\frac{1}{2} t\right)+\int_{0}^{t} f\left(s, C_{1}-\frac{1}{2} s\right) d s$

Am I right? If not, how to solve it correctly?

答:
I agree with your result :

$$u(x, t)=\varphi\left(C_{1}-\frac{1}{2} t\right)+\int_{0}^{t} f\left(s, C_{1}-\frac{1}{2} s\right) d s$$
But I think that is not completely finished because in the final form of result $C_1$ depends on $x$ and $t$ :
$$u(x, t)=\varphi\left(x\right)+\int_{0}^{t} f\left(s, x+\frac{t-s}{2} \right) d s$$


问:
Simplify Double Integral Joint Density
$f(x, y) = Axe^{-x(y+1)^2}$ for $x > 0$ and $y > 0$, $0$ otherwise. To solve for $A$, I have:
$$1=\int_{0}^{\infty}\int_{0}^{\infty}Axe^{-x\left(y+1\right)^{2}}dxdy$$ But I do not know how to simplify this integral.

答: Let $t >0$ Integrating by parts we get $\int_0^{\infty} xe^{-tx}dx=-\frac 1 t xe^{-tx}|_0^{\infty} +\frac1  t\int_0^{\infty} e^{-tx} dx=\frac1  {t^{2}}$. Hence, taking $t=(y+1)^{2}$ we get  $1=A\int_0^{\infty} \frac 1 {({y+1})^{4}} dy$. Can you continue?

Title: Simplify Double Integral Joint Density


Title: Simplify Double Integral Joint Density

Body:
$f(x, y) = Axe^{-x(y+1)^2}$ for $x > 0$ and $y > 0$, $0$ otherwise. To solve for $A$, I have:
$$1=\int_{0}^{\infty}\int_{0}^{\infty}Axe^{-x\left(y+1\right)^{2}}dxdy$$ But I do not know how to simplify this integral.

Body:
you can rewrite the integral in the following way

$$\underbrace{\int_0^{\infty}[x(y+1)^2]e^{-x(y+1)^2}d[x(y+1)^2]}_{\int_0^{\infty}ue^{-u}du=1}\int_0^{\infty}\frac{1}{(y+1)^4}dy=\int_0^{\infty}\frac{1}{(y+1)^4}dy=$$

$$=-\frac{(y+1)^{-3}}{3}\Bigg|_{y=0}^{y=\infty}=\frac{1}{3}$$

Thus $$A=3$$


问:
Proof: Sum of dimension of orthogonal complement and vector subspace

Let $V$ be a finite dimensional real vector space with inner product $\langle \, , \rangle$ and let $W$ be a subspace of $V$. The orthogonal complement of $W$ is defined as
  $$
W^\perp= \left\{ v \in V \,:\, \langle v,w \rangle = 0 \text{ for all } w \in W \right\}.
$$
  Prove the following: $\dim W + \dim W^\perp= \dim V$.

I'm not sure how to find the relationship between number of basis vectors in $W$ and $W^\perp$.

答:
There is something missing in @Solumilkyu 's answer.

In order to show that $W$ and $W^\perp$ have bases, we must show that $W, W^\perp \neq \{\vec{0}\}$, which cannot always be satisfied.

In fact, if $W=V$, then from $W \cap W^\perp=\{\vec{0}\}$ (To prove this, let $\vec{x} \in W \cap W^\perp$. Then $<\vec{x},\vec{x}>=0 \to \vec{x}=\vec{0}$) we can conclude that $W^\perp=\{\vec{0}\}$. Then $\dim W+\dim W^\perp=\dim W$. If $W\neq V$, then from the fact that $v \in V$ implies that $v=v_1+v_2$ for $v_1 \in W \wedge v_2 \in W^\perp$, we can conclude that $W^\perp \neq \{\vec{0}\}$ just by setting $v \in V \setminus W \wedge v \neq \{\vec{0}\}$ and $v_1 = \vec{0}$.

On the other hand, if $W = \{\vec{0}\}$, then obviously $W^\perp = V$.


问:
Can we deduce strictly increasing on a sub-interval for such a function?
Assume that we have a function $f : [0, 1] \rightarrow [0, 1]$ which is differentiable, $f(0) = 0$ and $f(1)=1$. Can we deduce that there exists an open sub-interval in $[0, 1]$ in which $f$ is strictly increasing over that?

Note that here, continuity of $f^{\prime}$ is not mentioned.

答:
The answer is no, a counterexample (and discussion/history around it) can be found in

K. C. Ciesielski - "Monsters in Calculus", page 4.

The basic idea is that if one picks the Pompeiu function $h$, there exists a dense set $G$ of translates such that for each $t \in G$ the zeroes of the derivatives of $h(x)$ and $h(x-t)$ do not coincide. Then, it is easy to conclude that the derivative of $h(x)-h(t-x)$ is positive on a dense subset and negative on another dense subset of $\mathbb R$.

Edit simplified the next

Let $g(x)=h(x)-h(t-x)$. Then, since $g$ is not constant, there exists some $a <b$ such that $g(a) \neq g(b)$.

Then for the right values of $\alpha, \beta$
$$f(x)= \alpha g( a+ (b-a)x) +\beta$$
satisfies $f(0)=0$ and $f(1)=1$.


问:
On convergence of problematic series.
Determine if the following series is converges or not

$$\sum_{n=2}^{\infty}\frac{1}{\ln^{\ln(\ln n)} (n)}$$

答:
Note that $$\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)=\ln1-\ln(\ln^{\ln(\ln n)}(n))=-\ln(\ln n)\ln(\ln n)=-\left[\ln(\ln n)\right]^2$$
Therefore
$$\sum_{k=2}^\infty\frac{1}{\ln^{\ln(\ln n)}(n)}=\sum_{k=2}^\infty e^{\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)}=\sum_{k=2}^\infty e^{-\left[\ln(\ln n)\right]^2}$$
It's not too hard to show that $\left[\ln(\ln n)\right]^2<\ln n$ for sufficiently large $n$. Therefore, for sufficiently large $n$, we have $e^{-\left[\ln(\ln n)\right]^2}>e^{-\ln n}=\frac{1}{n}$, so the series diverges (by comparison with the harmonic series).</p>

Title: On convergence of problematic series.


问:
Why is the exponent when working in $L^p$ spaces often dropped?
Formally, a function belong $L^p$ if $[\int |f|^p d\mu]^{1/p} < \infty$, but it seems quite common in proofs to drop the exponent? Why?

答:
In pure math, usually we don't care about the exact value of a non-negative real number. Rather, we care about its "categories'':

it is zero

it can be arbitrarily small

it is a constant

it can be arbitrarily large

it is $\infty$

If we drop the exponent, then the math expression is simplified while these "categories'' are still preserved. On the other hand, if we care about the exact value (e.g. for proving a tight inequality), then the exponent may need to be kept, because it serves as a normalization. E.g. Consider the difference between variance and standard deviation in probability theory.


问:
Enumerate the $m^n$ indexes of $m$ points along each axis of an $n$-dimensional space
This is a little programming problem where I need to enumerate, as $i=1,2,3,\ldots,m^n$, the sequences of coordinates
$$\begin{array}{rc}
 i=1\ldots m: & (1,1,...,1,1), (1,1,...,1,2), ..., (1,1,....,1,m)\\
 i=m+1\ldots 2m: & (1,1,...,2,1), (1,1,...,2,2), ..., (1,1,....,2,m)\\
      & \vdots\\
i=m^n-m+1\ldots m^n: & (m,m,...,m,1), (m,m,...,m,2), ..., (m,m,....,m,m)
 \end{array}$$
where each sequence has $n$ numbers.

So I need a programmable (preferably easily/straightforwardly programmable:) function $f:\mathbb N\to\mathbb N^n$ that generates these sequences in the illustrated order as $i=1,\ldots,m^n$.
(Looks like maybe some recursion of the inverse pairing function, https://en.wikipedia.org/wiki/Pairing_function, but I can't see how to do that. And hopefully there's an easier way.)

      E d i t
------------------

In case it's of any interest, below's the very short-and-easy C function
implementing @Haran's answer. Note that since this is C, my needed indices actually go from $0$ to $m-1$ rather than the $1$ to $m$ that I illustrated in the question above...

```<code>/*===========================================================================
 * Function:    clxcoord ( i, ndim, npts )
 * Purpose:     enumerates coordinates (x_1,x_2,...,x_ndim)
 *              from (0,0,...,0) to (npts-1,npts-1,...,npts-1)
 *              corresponding to i=0...npts^ndim-1
 * --------------------------------------------------------------------------
 * Arguments:   i (I)           int containing 0...npts^ndim-1
 *              ndim (I)        int containing number of dimensions
 *              npts (I)        int containing number of pts along each axis
 * Returns:     (int *)         i^th sequence of coords (x_1,x_2,...,x_ndim)
 *                              or NULL for any error
 * --------------------------------------------------------------------------
 * Notes:     o
 *======================================================================== */
/* --- entry point --- */
int *clxcoord ( int i, int ndim, int npts ) {
  static int xbuff[16][99];
  static int ibuff = 0;
  int    *xcoord = xbuff[ibuff];
  int    idim = 0;
  if ( ++ibuff >= 16 ) ibuff=0;
  for ( idim=0; idim<ndim; idim++ ) {
    xcoord[(ndim-1)-idim] = i%npts;
    i /= npts; }
  return ( xcoord );
  } /* --- end-of-function clxcoord() --- */
</code>```

And here's the short snippet testing the function, followed by its output...

```<code>  if ( testnum == 3 ) {
    int *xcoord=NULL, *clxcoord();
    int ndim=3, maxcoord=4;
    int icoord=0,i=0, ncoords=(int)(pow((double)maxcoord,(double)ndim)+0.5);
    for ( icoord=0; icoord<ncoords; icoord++ ) {
      xcoord = clxcoord(icoord,ndim,maxcoord);
      printf(" icoord=%04d: (",icoord);
      for ( i=0; i<ndim; i++ )
        printf("%d%s",xcoord[i],(i<ndim-1?",":")\n")); }
    } /* --- end-of-if(testnum==3) --- */
 icoord=0000: (0,0,0)
 icoord=0001: (0,0,1)
 icoord=0002: (0,0,2)
 icoord=0003: (0,0,3)
 icoord=0004: (0,1,0)
 icoord=0005: (0,1,1)
 icoord=0006: (0,1,2)
 icoord=0007: (0,1,3)
 icoord=0008: (0,2,0)
 icoord=0009: (0,2,1)
 icoord=0010: (0,2,2)
 icoord=0011: (0,2,3)
 icoord=0012: (0,3,0)
 icoord=0013: (0,3,1)
 icoord=0014: (0,3,2)
 icoord=0015: (0,3,3)
 icoord=0016: (1,0,0)
 icoord=0017: (1,0,1)
...
 icoord=0047: (2,3,3)
 icoord=0048: (3,0,0)
 icoord=0049: (3,0,1)
 icoord=0050: (3,0,2)
 icoord=0051: (3,0,3)
 icoord=0052: (3,1,0)
 icoord=0053: (3,1,1)
 icoord=0054: (3,1,2)
 icoord=0055: (3,1,3)
 icoord=0056: (3,2,0)
 icoord=0057: (3,2,1)
 icoord=0058: (3,2,2)
 icoord=0059: (3,2,3)
 icoord=0060: (3,3,0)
 icoord=0061: (3,3,1)
 icoord=0062: (3,3,2)
 icoord=0063: (3,3,3)
</code>```

Now, ... if I can only get @Haran to do the rest of my work for me :)

答: Simply convert $i-1$ to base $m$. Then, take convert all the digits (with leading $0s$) into a sequence of length $n$. Convert digits such as $A$ into $10$, $B$ into $11$ and so on. Finally, add $1$ to all the values in the sequence. This gives you the required.


问:
Formal power series and the quadratic formula
I was recently reviewing a derivation of getting a formula for Catalan numbers in closed form. I understand the mechanics of the derivation, with the exception of one significant step:

If $f(x) = c_0 + c_1x +c_2x^2 + \cdots $ is the generating series for the Catalan numbers $c_n, n \geq 0$, then it satisfies the relation $xf(x)^2 - f(x) + 1 = 0$. But the next step of the proof is to solve for $f(x)$ by using the quadratic formula, which gives $f(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}$ (and then we take the 'minus' since the 'plus' doesn't work). My question is regarding the legality of applying the quadratic formula (qf) here. We initially learn the qf in elementary algebra, which is in the context of the $\textit{field}$ of real numbers. In particular, division and square roots are permitted. Here we are working with the $\textit{ring}$ of formal power series. Specifically, not all fps have multiplicative inverses, and indeed the fps $2x$ is not invertible. I see that the idea is that the numerator (after simplification) can be seen to have a factor of $x$, and so the idea is that we simply want to 'cancel' these two factors of $x$, but this strikes me as non-rigorous. I appreciate that we can prove we get the correct answer a posteriori (since it satisfies the recurrence relation that the Catalan numbers need to satisfy), but the actual method of finding the answer seems to be informal symbol manipulation based on our intuition of the qf in a field. So I suppose my specific question is:

Why are we allowed to divide by $x$ here when $x$ is not invertible in the ring of fps?

My guess is that it is allowed due to analytic convergence of the series on some positive radius of convergence, but I am really looking for some combinatorial/algebraic reasoning instead, which could potentially generalize to other situations. Does such reasoning exist?

If anyone could explain why performing the qf here is justifiable from a combinatorial/algebraic perspective, I would be very appreciative.

答:
If $f$ has positive radius of convergence $R$, for any fixed $x\in(0,\,R)$ we know $f(x)$ is finite, and calculable by the method you've seen. (Note we no longer need to work in a ring of polynomials over a variable.) In particular, continuity at $x=0$ ensures $f(x)=\frac{1-\sqrt{1-4x}}{2x}$ for all such $x$.

So the hard question is how we know $R>0$; indeed, the convergence of $f$ for some $x$ of interest is necessary to prove $xf^2-f+1=0$. Alternatively, we can show $\frac{1-\sqrt{1-4x}}{2x}$ generates $\tfrac{1}{n+1}\binom{2n}{n}$ and this satisfies the initial conditions and recursion relation; the worst you can say about this reasoning is that it "got lucky".


问:
Maximum of d(12-d)
I'm a little confused on a quite simple quadratic problem. I need to calculate the maximum of $d(12-d)$ using basic quadratics. The answer is $6$ as can also be shown by $f'(x)= -2d +12$, however this is algebra not calculus. Using the squares method I get twelve, I think I may be using the incorrect method. Could someone set me on the right track?

$d(12-d)$

$= -d^2 + 12d$

$-d^2 + 12/(-2)d + 36 = 36 $

$(d-6)^2 = 36$

$d = 12  or  0$ <- I understand these are the roots, am I not meant to calculate them?

The suggested answer is: $-(d-6)^2 + 36$, but I have no idea where it came from.

答: You can use your knowledge of parabolas. The parabola $y=x(12-x)$ opens downwards, and reaches its maximum halfway between the two points where it crosses the $x$-axis. These two points can be read off from the equation. </p>

Title: Maximum of d(12-d)


问:
If $P$ is a fifth degree polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0, 1, 2, 3, 4$ and $5$, then find $P(6)$.
So I let $P(x)=ax^5+bx^4+cx^3+dx^2+ex+f$, and obviously $f=0$. Then,

$2^5a+2^4b+2^3c+2^2d+2e={1\over 2}$

$3^5a+3^4b+3^3c+3^2d+3e={2\over 3}$

$4^5a+4^4b+4^3c+4^2d+4e={3\over 4}$

$5^5a+5^4b+5^3c+5^2d+5e={4\over 5}$

and we are looking for $6^5a+6^4b+6^3c+6^2d+6e=$

Then I got stuck...
Could anyone at least give me some hints or methods to continue?

答: Hint: Note that $Q(x) = (x+1)P(x)-x$ is a polynomial of degree at most $6$ which satisfies $Q(k) = 0$ for $k = 0,1,2,3,4,5$. Hence, $Q(x) = Cx(x-1)(x-2)(x-3)(x-4)(x-5)$ for some constant $C$. To solve for $C$, try to figure out what $Q(-1)$ must be. Then, finding $P(x)$, and thus, $P(6)$ will be easy.


问:
Internal product of rings is homomorphic image of direct product
Let $R$ be a ring (commutative with $1$) and $A,B \subseteq R$ subrings (both still having the same $1$). Then we can consider the outer direct product of the rings $A \times B$ with elementwise addition/multiplication. We can also consider the inner product
$$ AB= \{ \sum_{i=0}^n a_i b_i \; | \; n \in \mathbb{N}, a_i \in A ,b_i \in B  \} \subseteq R $$
with addition/multiplication in $R$ (which is the smallest subring of $R$ containing both $A$ and $B$).

Is the ring $AB$ a homomorphic image of the ring $A \times B$? Is it true and if so, is there a reference for ist? If it is not true, what counterexample do we have?

If $A \subseteq B$ this is the case as $\phi \colon A\times B \to AB=B,(a,b) \mapsto b$ is an epimorphism. But what about the general case? The obvious map $(a,b) \mapsto ab$ is not a ring-homomorphism and I could not find any other map which would work.

答:
Suppose that there exists $f\colon A\times B\to AB$; then $f(1,1)=1$; set $x=f(1,0)$ and $y=f(0,1)$. Then $xy=f(1,0)f(0,1)=f(0,0)=0$, but also $x+y=1$,  $x^2=x$ and $y^2=y$.

Take your favorite ring with no nontrivial idempotents (a field, for instance) and having at least two distinct subrings.


问:
Why does the Alternating Series Test require $b_n$ to be > 0?
So one of the prerequisites for applying the alternating series test is that the serie is of the form $$\sum_{n=1}^{\infty} (-1)^{n-1}b_n$$
Where $b_n$ has to be greater than 0. I was wondering why $b_n$ has to be greater than zero?

It would obviously still alternate if $b_n$ was less than zero, so why is this a condition?

Thanks in advance!

答:
There is no principal difference between series where all $b_n > 0$ and where all $b_n< 0$ from convergence point of view. Convergence of 2 such series are equivalent and sum differs, when exists, by $−1$.

Analogically, there is no difference in alternating series from convergence point of view if we consider all $b_n > 0$ and/or where all $b_n< 0$. If $b_n < 0$, then we can write $b_n = - (-b_n)$ and take minus out of summing in partial sums and, if series converged, out of infinite sum and consider convergence question for $(-b_n) > 0$: $\sum\limits_{n=1}^{\infty} (-1)^{n-1}b_n=-\sum\limits_{n=1}^{\infty} (-1)^{n-1}(-b_n)$.


问:
If a partition $(A_i)_{i\in I}$ satisfies $P(A_i)>0$ then $I$ is at most countable
I want to show that if a partition $(A_i)_{i\in I}$ satisfies $P(A_i)>0$ for all $i$ then $I$ is at most countable to avoid problems in a larger proof. I know this is true, however, I'm finding it difficult to prove/argue this seemingly obvious statement.

I tried arguing using covers of the compact $[0,1]$, which the sum $\sum_i P(A_i)$ must "fill", but that approach already assumed that the partition was finite.

Am I missing some bit of topology? Or is there another way entirely?

答:
No topology is involved here. I suppose your $P$ is a probability measure (or at least a finite measure).

Consider $I_j=\{i: P(A_i) >\frac 1j\}$ where $j$ is a positive inrteger. If there are $n$ elements in this set, say $i_1,i_2,..,i_n$ then $P(\Omega) \geq P(\bigcup_{1\leq k \leq n} A_{i_k}) = \sum\limits_{k=1}^{n} P(A_{i_k})>\frac n j $. This proves that there are at most finitely  many points in this set. Now take union over all $j$ and you will get the proof.  ($I=\bigcup_j I_j$).


问:
Integral tends to infty
How can we prove that $\lim\limits_{\varepsilon\searrow 0} \dfrac{1}{\varepsilon} \displaystyle\int_0^1 \left [\dfrac{1}{4}-\dfrac{1}{\pi^2}\arctan^2\left (\dfrac{x}{\varepsilon}\right )\right ]^2 dx=+\infty$?

I came across this integral in studying the Heaviside aproximation function. I used the following sharp estimate:

$\arctan(x)<\dfrac{\pi^2 x}{4+\sqrt{32+(2\pi x)^2}}, \forall\ x>0$, but the limit goes to $0$ (I computed it exactly with wolfram). This estimation and many others can be found here:

https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-018-1734-7

The integral in the statement is simulated here:

https://www.wolframalpha.com/input/?i=int_0%5E1+1000000000+%281%2F4+-+%281%2Fpi%5E2%29+%28arctan%2810000000000+x%29%29%29%5E2+dx

But I wonder if it can be proved that it really tends to infinity. Maybe there are some singularities and the program gives some big errors.

答:
OHH, please sorry. I made a mistake putting an extra $2$. It should be:

$$\lim\limits_{\varepsilon\searrow 0} \dfrac{1}{\varepsilon} \displaystyle\int_0^1 \left [\dfrac{1}{4}-\dfrac{1}{\pi^2}\arctan^2\left (\dfrac{x}{\varepsilon}\right )\right ]^2 dx=+\infty$$

Title: Integral tends to infty


问:
Finding the error term in the Cauchy-Schwarz inequality for integrals
This question is inspired from chapter 13 problem 39biii of Spivak's Calculus:

The Schwarz inequality (SI):

$\big[\sum_{i=1}^n x_iy_i \big]^2 \leq \big[\sum_{i=1}^n x_i^2 \big]\cdot \big[\sum_{i=1}^n y_i^2 \big]$

The Cauchy-Schwarz inequality for integrals (CSI):

$\big[\int_a^b f(x)g(x)dx\big]^2 \leq \big[\int_a^b [f(x)]^2dx\big]\cdot \big[\int_a^b [g(x)]^2dx\big]$

Problem 39biii) Prove CSI by adapting the proof of SI from problem 2-21 (This will take some imagination).

The proof of SI from problem 2-21 is as follows:

$\big[\sum_{i=1}^n x_i^2 \big]\cdot \big[\sum_{i=1}^n y_i^2 \big] = \big[\sum_{i=1}^n x_iy_i \big]^2 + \sum_{i < j} (x_iy_j-x_jy_i)^2 \geq \big[\sum_{i=1}^n x_iy_i \big]^2$

I understand this proof, but I can't see how to generalise it to prove CSI. The general idea behind this proof is to calculate the error term of SI, which is that $\sum_{i < j} (x_iy_j-x_jy_i)^2$ term. So my only idea is to calculate the error term of CSI, which (unsimplified) would be:

$E = \big[\int_a^b [f(x)]^2dx\big]\cdot \big[\int_a^b [g(x)]^2dx\big] - \big[\int_a^b f(x)g(x)dx\big]^2$

But I can't see how to simplify $E$ into a form (I'm guessing a sum of positive integrals?) which is clearly positive. Or maybe Spivak had another approach in mind?

Any ideas?

Body: Write$$\begin{align}E&=\int_a^bf^2(x)dx\int_a^bg^2(y)dy-\int_a^bf(x)g(x)dx\int_a^bf(y)g(y)dy\\&=\int_{[a,\,b]^2}(f^2(x)g^2(y)-f(x)f(y)g(x)g(y))dxdy\\&=\int_{a\le x\le y\le b}(f^2(x)g^2(y)+f^2(y)g^2(x)-2f(x)f(y)g(x)g(y))dxdy\\&=\int_{a\le x\le y\le b}(f(x)g(y)-g(x)f(y))^2dxdy.\end{align}$$


问:
$A,B \subset (X,d)$ and $A$ is open dense subset, $B$ is dense then is $A \cap B$ dense?
I am trying to solve this problem, and i think i did something, but finally i couldn't get the conclusion. The question is:

Let $(X,d)$ be a metric space and let $A,B \subset X$. If $A$ is an open dense subset, and $B$ is a dense subset, then is $A \cap B$ dense in $X$?.

Well, i think this is true. We have to show that $\overline{A \cap B}=X$. Or in other words, $B(x,r) \cap (A \cap B) \neq \emptyset$ for any $x \in X$. Since $A$ is dense in $X$, so we have $B(x,r) \cap A \neq\emptyset$. That means there is a $y \in B(x,r) \cap A$. Which means there is a $y \in A$. And since $A$ is open we have $B(y,r_{1}) \subseteq A$ for some $r_{1} > 0$. Couldn't get any further. I did try some more from here on, but couldn't get it. Any idea for proving it or giving a counter example.

答:
Let $(X,\tau)$ be a topological space. Let $A$ and $B$ be two dense subsets of $(X,\tau)$  with $A$ is an open set.

Let $G\in \tau$. Then since $A$ is dense in $(X,\tau)$, $G\cap A\neq\emptyset$.

Now since $B$ is dense in $(X,\tau)$, for open set $G\cap A$ which is a
nonempty set, $(G\cap A)\cap B\neq \emptyset$.

Hence $(A\cap B)$ is a dense set in $(X,\tau)$.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Error in my argument that $A^m=0$ implies $A^{m-1}=0$ for matrix $A$?
So nilpotent is the matrix where $A^m =0$ but $A^{m-1}$ is not equal to $0$. Well here's my logic-

$$A^{m-1} = I \cdot A^{m-1} = A^{-1} \cdot A \cdot A^{m-1} = A^{-1} \cdot A^m = A^{-1} \cdot 0 = 0$$

This gives $A^{m-1}$ always zero when $A^m=0$! Now I am sure there's some glitch in my logic so can someone point that out?
(Note that $A^{-1}$ here is inverse of matrix $A$)

Sorry for the weird symbols but I am not good at coding those math symbols.

答:
Yes, but you don't know if $A^{-1}$ actually exist. In fact it does not.

Say $$A =\pmatrix{0&1&0\\0&0&1\\0&0&0}\implies A^2 = \pmatrix{0&0&1\\0&0&0\\0&0&0}$$

and $A^3=0$. Notice that $A^{-1}$ does not exist since $\det A =0$.

Why it newer exists? If $\lambda$ is it eigenvalue and $v$ eigenvector, then we have $$Av = \lambda v\implies 0=A^{n}v =\lambda^n v \implies \lambda =0$$

So all it eigenvalues are $0$ and thus $\det A =0$ and so $A^{-1}$ does not exists.


答:
You're assuming that $A^{-1}$ exists!

You'll find that nilpotent matrices cannot have inverses. In fact, your contradiction proves this!

I hope this helps ^_^

Title: Error in my argument that $A^m=0$ implies $A^{m-1}=0$ for matrix $A$?

Body:
So nilpotent is the matrix where $A^m =0$ but $A^{m-1}$ is not equal to $0$. Well here's my logic-

$$A^{m-1} = I \cdot A^{m-1} = A^{-1} \cdot A \cdot A^{m-1} = A^{-1} \cdot A^m = A^{-1} \cdot 0 = 0$$

This gives $A^{m-1}$ always zero when $A^m=0$! Now I am sure there's some glitch in my logic so can someone point that out?
(Note that $A^{-1}$ here is inverse of matrix $A$)

Sorry for the weird symbols but I am not good at coding those math symbols.


问:
Find $x, y$ and $z$ such that $x+y+z=23$, $xy+yz+xz=144$ and $xyz=252$, where $x\gt y\gt z$.
All I can think of is that $x$, $y$ and $z$ are the roots of the multiples of polynomial $P(x)=x^3-23x^2+144x-252$, and I found the roots by rational root test.

However this method is not elegant IMO because if the roots are not rational, I have no idea how to find them anymore.

So, could anyone introduce me a general method or at least a better method in solving system equations such as $x+y+z=a$, $xy+yz+xz=b$ and $xyz=c$, where $a, b, c\in \mathbb R$?

答:
You can apply this identity:

$$(ab+bc+ca)(a+b+c)-abc=(a+b)(b+c)(c+a)$$

Putting values you get:

$(x+y)(y+z)(z+x) =3060$

Now you have to decompose 3060 and equate factors , you will have some system of equations to solve.


问:
Variance of number of heads
Given two coins with chance of heads $p=0.6$ for a coin $X$, and $p=0.7$ for a coin $Y$. Pick one coin uniformly at random. Toss the coin 10 times. Find the variance $Z$ of the number of heads. I am thinking of just using the definition of variance.

$E[Z]^2 = \sum_{n=0}^{10}\frac{n}{2}\left(\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{6}{10}\right)^{n}\left(\frac{4}{10}\right)^{10-n}+\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{7}{10}\right)^{n}\left(\frac{3}{10}\right)^{10-n}\right) = 42.25$
$E[Z^2] = \sum_{n=0}^{10}\frac{n^{2}}{2}\left(\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{6}{10}\right)^{n}\left(\frac{4}{10}\right)^{10-n}+\operatorname{nCr}\left(10,\ n\right)\cdot\left(\frac{7}{10}\right)^{n}\left(\frac{3}{10}\right)^{10-n}\right) = 44.75$

$Var[Z] = 2.5$

But is there a faster, cleaner way without all this computation?

答:
You can use the law of total variance.  Let $C$ be a random variable which equals $0$ if coin $X$ is chosen and equals $1$ if coin $Y$ is chosen.  Then $Z|C=0$ has the distribution Bin$(10,0.6)$ and $Z|C=1$ has the distribution Bin$(10,0.7)$.  So the random variable $\mathbb{E}[Z|C]$ takes the values $10 \cdot 0.6 = 6$ and $10 \cdot 0.7 = 7$ with probability $1/2$ each, and the random variable $\operatorname{var}(Z|C)$ takes the values $10 \cdot 0.6 \cdot 0.4$ and $10 \cdot 0.7 \cdot 0.3$ with probability $1/2$ each.

Therefore we can calculate
$$
\operatorname{var}(Z) = \mathbb{E}[\operatorname{var}(Z|C)] + \operatorname{var}(\mathbb{E}[Z|C]) = \frac12( 10 \cdot 0.6 \cdot 0.4 + 10 \cdot 0.7 \cdot 0.3) + \frac14 = 2.5
$$

Title: Variance of number of heads


问:
Closed form for $\sum^{\infty}_{{i=n}}ix^{i-1}$
How can I find a closed form for:

$$\sum^{\infty}_{{i=n}}ix^{i-1}$$

It looks like that's something to do with the derivative

答:
Recall the power series:

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

holds for all $|x|<1$.

Then take the derivates, you get:

$$\frac{1}{(1-x)^2}=\sum_{n\geq1}nx^{n-1}$$


答:
Since $A^{m-1} \ne 0$, there is $x \ne 0$ such that $y:=A^{m-1}x \ne 0.$ Then we get

$$Ay=A^mx=0.$$

This shows that $0$ is an eigenvalue of $A$. Therefore $A$ is not invertible.


答:
$F=\Bbb{Q}((x))$ (the formal Laurent series $\sum_{n\ge -N} a_n x^n$ with rational coefficients) is a field so it works really the same way as over $\Bbb{R}$.

You need to check that the discriminant $1-4x$ of your quadratic polynomial $xT^2-T+1\in F[T]$ has a square root in $F$, which is $$a=\sum_{k\ge 0} {1/2\choose k} (-4x)^k\in F$$

Then $$xT^2-T+1= x(T-\frac{1+a}{2x})(T-\frac{1-a}{2x})\in F[T]$$ $f\in F$ is a root of $xT^2-T+1$ implies either that $f=\frac{1+a}{2x}$ or $f=\frac{1-a}{2x}$.

Title: Formal power series and the quadratic formula

Body:
I was recently reviewing a derivation of getting a formula for Catalan numbers in closed form. I understand the mechanics of the derivation, with the exception of one significant step:

If $f(x) = c_0 + c_1x +c_2x^2 + \cdots $ is the generating series for the Catalan numbers $c_n, n \geq 0$, then it satisfies the relation $xf(x)^2 - f(x) + 1 = 0$. But the next step of the proof is to solve for $f(x)$ by using the quadratic formula, which gives $f(x) = \frac{1 \pm \sqrt{1 - 4x}}{2x}$ (and then we take the 'minus' since the 'plus' doesn't work). My question is regarding the legality of applying the quadratic formula (qf) here. We initially learn the qf in elementary algebra, which is in the context of the $\textit{field}$ of real numbers. In particular, division and square roots are permitted. Here we are working with the $\textit{ring}$ of formal power series. Specifically, not all fps have multiplicative inverses, and indeed the fps $2x$ is not invertible. I see that the idea is that the numerator (after simplification) can be seen to have a factor of $x$, and so the idea is that we simply want to 'cancel' these two factors of $x$, but this strikes me as non-rigorous. I appreciate that we can prove we get the correct answer a posteriori (since it satisfies the recurrence relation that the Catalan numbers need to satisfy), but the actual method of finding the answer seems to be informal symbol manipulation based on our intuition of the qf in a field. So I suppose my specific question is:

Why are we allowed to divide by $x$ here when $x$ is not invertible in the ring of fps?

My guess is that it is allowed due to analytic convergence of the series on some positive radius of convergence, but I am really looking for some combinatorial/algebraic reasoning instead, which could potentially generalize to other situations. Does such reasoning exist?

If anyone could explain why performing the qf here is justifiable from a combinatorial/algebraic perspective, I would be very appreciative.


问:
Empty relation on non-empty set - is it reflexive?
I have here a relation defined by a conjunction of contradictory conditions, i.e.
 $R = \{ \langle x,y \rangle \in \mathbb N ^2  |  x = y^2 \land y = -x \}$

So, at least as I understand my undergrad course so far, this produces an empty relation, since there are no pairs of natural numbers such that one is the negative of the other.

I can see why such a relation would be symmetrical and transitive, since both of these properties involve an implication.

However, I am trying to determine if this relation is reflexive or irreflexive. My professor suggested that this would be computable for any relation, but in my mind, reflexivity is defined based on the presence of at least one element in the relation's domain. And if there are no elements in the relation, then how can they be in relation with themselves?

Some help, any help, would be appreciated!!

P.S. If anybody can recommend a modern manual for this subject, I'd really appreciate it. There is no recommended reading in this course, and the classes aren't very enlightening. I work better from a book.

答:
Assuming your definition of $\Bbb N$ excludes $0$ (since some do, some don't), then you are correct in concluding that $R = \varnothing$, since both equations are satisfied only for $(x,y) \in \{(0,0),(1,-1)\}$. You can prove this algebraically, or use the fact that $\Bbb N$ contains no negatives, of course, but the graph makes it especially clear in my opinion:

The question of whether the relation $R = \varnothing$ is reflexive is slightly complicated though. In the case the relation is over a nonempty set, then it is easy to verify that it is not reflexive, since no element of that nonempty set is related to anything, especially not itself. (For instance, $(1,1) \not \in R = \varnothing \subseteq \Bbb N^2$ in your case.)

However, in the hypothetical that we're defining the relation $R$ over the empty set, then the empty relation is reflexive. Namely, this is because there exists no $x \in \varnothing$ for which $(x,x) \not \in R = \varnothing$ -- i.e. vacuous logic.

Curiously, you can also see that while relations over a nonempty set are always reflexive, irreflexive, or neither, (empty) relations over the empty set are both reflexive and irreflexive.

While not strictly related to your example, I figured it would be worth noting both cases (nonempty and empty sets) for completeness's sake. Of course, your example is simply not reflexive, since $\Bbb N^2 \ne \varnothing$.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


答:
$$\sum^\infty_{r=n}rx^{r-1}=\dfrac {d\left(\sum^\infty_{r=n}x^r\right)}{dx}$$

From here, $\sum^\infty_{r=n}x^r=\dfrac{x^n}{1-x}$ if $|x|<1$

Alternatively,

let $S_n=\sum^m_{r=n}rx^{r-1}=n\cdot x^{n-1}+(n+1)x^n+(n+2)x^{n+1}+\cdots+(m-1)x^{m-2}+mx^{m-1}$

So, multiplying by $x,$   $x\cdot S_n=\sum^\infty_{r=n}rx^r=n x^n+(n+1)x^{n+1}+(n+2)x^{n+2}+\cdots+(m-1)x^{m-1}+mx^m$

On subtraction, $$(1-x)S_n=nx^{n-1}+(x^n+x^{n+1}+\cdots+x^{m-1})-mx^m$$
$$=nx^{n-1}+x^n\frac{(1-x^{m-n})}{1-x}-mx^m=nx^{n-1}+\frac{(x^n-x^m)}{1-x}-mx^m$$

Now, if $m\to\infty$ and $|x|<1,x^m\to0$

From this, $\lim_{m\to\infty}mx^m=0$ if $|x|<1$

Title: Closed form for $\sum^{\infty}_{{i=n}}ix^{i-1}$

Body:
How can I find a closed form for:

$$\sum^{\infty}_{{i=n}}ix^{i-1}$$

It looks like that's something to do with the derivative


问:
Transitivity - $R=\{(a,b)\in \mathbb{R} \times \mathbb{R} : |a-b|<1\}$
Hello Mathematics Community,

I have the relation:

$$R=\{(a,b)\in \mathbb{R} \times \mathbb{R} : |a-b|<1\}.$$

Is it true that this relation is not transitive, because:

Let $a=0, b=0$ and $c=0$:

$$ |0-0| < 1 \Rightarrow |0-0|<1$$

is true.

But if $a=b$ or $a,b \geq1$ and $c\geq1$ therefore this is not true.
So the relation is not transitive.

sincerely,
M.Hisoka

答:
As noted by olsen5 in a comment:

The relation is transitive, but your argument is a bit haphazard. To show a relation is not transitive, a simple counterexample suffices. Observe that $(0,1/2),(1/2,1) \in R$: this is because

$$\left| 0 - \frac 1 2 \right| = \left| \frac 1 2 - 1 \right| = \frac 1 2 < 1$$

However, $(0,1) \not \in R$ because $|0-1| = 1 \not < 1$.

Thus, in the framework of the variables prescribed, when we want to show that $(a,b),(b,c) \in R$ do not necessarily imply $(a,c) \in R$, take $a = 0, b = 1/2, c = 1$.

Mostly just posting this in order to get this answer out of the unanswered queue, and posting as Community Wiki since I have nothing further to offer.


问:
Poisson process probabilities example
Let X(t) be a Poisson process with parameter $\lambda$. Calculate:
$$
\mathbb{P}(3X(2) = 2X(4) | X(1) \leq 3)
$$

Now, I tried to tweak the left side:
$$
3X(2) = 2X(4) \\
X(2) + X(2) + X(2) = X(4) + X(4) \\
X(4) + X(4) - X(3) - X(3) + X(3) + X(3) - X(2) - X(2) - X(2) = 0 \\
$$
Denoting $Y,Z$ as $X(4)-X(3), X(3)-X(2)$ respectively:
$$
2Y+2Z-X(2) = 0
$$
X(2),Y,Z are independent Poisson variables and $2Y+2Z-X(2)$ has a Poisson distribution with $3 \lambda$.

Back to main equation:
$$
\mathbb{P}(3X(2) = 2X(4) | X(1) \leq 3) = \frac{\mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))}
$$
The denominator $\mathbb{P}(X(1) \leq 3))$ is easy to calculate.

Unfortunately, $2Y+2Z-X(2)$ is not independent from $X(1)$ so I cant just do:

$$
\frac{\mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))} = \frac{\mathbb{P}(2Y+2Z-X(2)=0)\mathbb{P}( X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))} \\
= \mathbb{P}(2Y+2Z-X(2)=0) = e^{-3\lambda}
$$

Can anyone take it from here? What do I do? I thought of doing stuff like(denote $X(2)-X(1)$ as Q)
$$
\mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3) =\mathbb{P}(2Y+2Z-(X(2)-X(1))-X(1)=0, X(1) \leq 3) = \mathbb{P}(2Y+2Z-Q=X(1), X(1) \leq 3)
$$
and now I can just iterate:
$$
\mathbb{P}(2Y+2Z-Q=X(1), X(1) \leq 3)  = \mathbb{P}(2Y+2Z-Q=0), X(1)=0) +  \mathbb{P}(2Y+2Z-Q=1), X(1)=1) +  \mathbb{P}(2Y+2Z-Q=2), X(1)=2) +  \mathbb{P}(2Y+2Z-Q=3), X(1)=3)
$$
where $2Y+2Z-Q$ is just Poisson with $3\lambda$ and $X(1)$ Poisson with $\lambda$. Would that be correct?

Title: Poisson process probabilities example

Body:
Your method is correct, but using the independence of $X(t_2)-X(t_1)$ and $X(t_4)-X(t_3)$ when $t_3\ge t_2$ we can write your solution even more structured.
$$
\Pr\{3X(2)=2X(4),X(1)\le 3\}{=
\sum_{k=0}^3\Pr\{3X(2)=2X(4),X(1)=k\}
\\=
\sum_{k=0}^3\sum_{u=0}^\infty\Pr\{X(2)=2u,X(4)=3u,X(1)=k\}
\\=
\sum_{k=0}^3\sum_{u=0}^\infty\Pr\{X(2)-X(1)=2u-k,X(4)-X(2)=u,X(1)=k\}
\\=
\sum_{k=0}^3\sum_{u=0}^\infty\\\Pr\{X(2)-X(1)=2u-k\}\Pr\{X(4)-X(2)=u\}\Pr\{X(1)=k\}
\\=
\sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty\\\Pr\{X(2)-X(1)=2u-k\}\Pr\{X(4)-X(2)=u\}\Pr\{X(1)=k\}
\\=
\sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty e^{-4\lambda}\frac{\lambda^{2u-k}(2\lambda)^u\lambda^k}{(2u-k)!u!k!}
\\=
\sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty e^{-4\lambda}\frac{(2\lambda^3)^u}{(2u-k)!u!k!}
}
$$
I wonder if the last summation has a closed form.


问:
Ultrametric completion does not introduce new norms
I could not solve this problem:

Prove that for a non-Archimedian field $K$ with completion $L$, $$\left\{|x|\in\mathbb R \mid x\in K\right\} =\left\{|x|\in\mathbb R \mid x\in L\right\}$$

I considered a Cauchy sequence in $K$ with norms having limit $l$, but I could not construct an element of $K$ with norm $l$ from the sequence.

Will anyone please show how to prove it?

答:
Suppose that $\{x_n\}$ is a Cauchy sequence under the ultrametric $|\cdot|$, and consider the sequence $\{|x_n|\}$ of real numbers.

If the sequence is eventually constant, then there is nothing to do: the absolute value of the limit is equal to the absolute value of some element of $K$.

If the sequence is not eventually constant, then we can find a subsequence in which $|x_{n_k}|\neq|x_{n_{k+1}}|$ for all $k$. Since a sequence is Cauchy in an ultrametric if and only if the sequence of consecutive differences $|a_i-a_{i+1}|$ goes to $0$, and we have
$$|x_{n_k} - x_{n_{k+1}}| = \max\{|x_{n_k}|,|x_{n_{k+1}}|\}$$
(since the norms of $x_{n_k}$ and of $x_{n_{k+1}}$ are different), then we must have
$$\lim_{k\to\infty}|x_{n_k}| = 0,$$
and hence, since the original sequence is Cauchy,
$$\lim_{n\to\infty}|x_n| = 0.$$
Therefore, the absolute value of the limit is $0$, which is the absolute value of some element of $K$ as well.

In either case, the absolute value of the limit of a Cauchy sequence of elements of $K$ is always equal to the absolute value of an element of $K$, so the set of absolute values of elements of the completion is equal to the set of absolute values of elements of $K$.