答:
$$\frac 1{\sqrt{x^2+4}}=\frac 14\frac{(x^2+4)-x^2}{\sqrt{x^2+4}}=\frac 14\left(\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}\right)$$

$\int\sqrt{x^2+a^2}$ can be computed using IBP without any trig sub.

For the second part, consider the sub $t:=\sqrt{x^2+4}$ so that $t^2=x^2+4$ and $t\,\mathrm dt=x\,\mathrm dx$

$$\int\frac{x^2}{\sqrt{x^2+4}}\,\mathrm dx=\int\frac{t^2-4}{t}\frac t{\sqrt{t^2-4}}\,\mathrm dt=\int\sqrt{t^2-4}\,\mathrm dt$$

which is again computable using IBP without any trig sub.


问:
Probability interpretation
Supose we have a mouse on a labyrinth and we associate $p$ as the probability of the mouse escaping this labyrinth this turn, and $(1-p)$ for the probability of him not escaping that turn. Call $X$ the random variable associated with the number of turns that the mouse takes in order to escape.
If I want him to take, at best, $4$ turns to escape the labyrinth then I would consider the possibilities of:

$$P(X \leq 4)= P(X = 4)+P(X = 3)+P(X = 2)+P(X = 1)$$

After some calculations I arrive at the probability:

$$ P(X \leq 4)= 1-(1-p)^4 $$

On the other hand:

$$ P(X \leq 4)=1-P(X > 4)$$

Which leads to:

$$ P(X > 4) = (1-p)^4$$

This gives us the probability of him failling more then $4$ times, given that it is the sum:

$$ P(X > 4) = P(X=5)+P(X = 6)+...=p(1-p)^4+p(1-p)^5+...=p \sum_{i=4}^n(1-p)^{i}=(1-p)^4$$

Can someone give me a intuitive interpretation for this? It seems a little odd that the summation would tend to $(1-p)^4$ when starting at $4$.

答:
You have $P(X>4)=P(X=5)+...+P(X=n)+...=p\cdot\sum_{i=4}^{\infty}(1-p)^i$

This is a geometric series $a_1=(1-p)^4$ and $q=1-p$

By Geometric series formula, we have when $q<1$:

$lim_{n->\infty}\sum_{i=1}^{n}a_1 \cdot q^i = \frac{a_1}{1-q}=\frac{(1-p)^4}{1-(1-p)} = \frac{(1-p)^4}{p}$

But we had the multiplication with $p$ outside so we get $(1-p)^4$

It will work with any Geometric distribution, and for every value.

If you look for intuition, $(1-p)^4$ is the chance to "fail" in the first 4 times, so it leads to $P(X>4)$

ContentLicense: CC BY-SA 4.0

Title: Probability interpretation


问:
Strategy to win a dice-rolling game
I want to find a strategy to maximize the probability of winning the following game:

Suppose that we have two players, with each player rolls a six-sided dice. They cannot see
their own outcome but they can see the outcome of the other player, and no other information is passed between the two players. They must guess their own dice, and if they are both correct, they win. If either is wrong, they lose.

The maximum strategy I can think of is to either guess exactly the outcome of the other player, or guess an outcome that, summed by the outcome by the other player, will equal to 7. Both of them will guarantee an probability of $\frac{1}{6}$.

I just wonder whether it could be better than $\frac{1}{6}$. If so, what could it be; if not, then how to prove the above strategies are the best.

答: Let $A$ be the event of the first person guessing correctly and $B$ be the event of the second person guessing correctly.  No matter what strategy they use, it seems clear that $P(A)$ and $P(B)$ cannot be greater than $\frac16$, since they have no information that would inform their guess.  The only thing that they have control over is coordinating a strategy such that $P(B\mid A)=1$ so that the probability of one being right is the same as the probability of both being right.  As either of the strategies you mention achieve that, it seems evident to me that it is ideal.

Title: Strategy to win a dice-rolling game


问:
Definition of open sets
Open ball, centered at $$ x \in U,$$ with radius r is defined as
$$ B_r \left( x\right)=\left\{x' \in U \mid \mathrm{d} \left(x, x' \right)<r \right\}.$$

Definition of an open set:
$$ U \ \mathrm{is \ open.} \Leftrightarrow \forall x\in U: \exists r>0: B_r(x)\subset U$$

My question is: don't all points of U anyway satisfy the latter condition? Since $$ x \in U, $$ the open ball is a collection of elements in U and thus a subset of U.

答:
Your definitions are a bit off. The setting is $(X,d)$, a metric space, and the open ball $B_r(x)$ ofr $x \in X$ and $r>0$ is defined as

$$B_r(x)=\{x' \in X\mid d(x,x,') < r\}$$

(not $x \in U$).

Then a subset $U \subseteq X$ is open iff

$$\forall x \in U: \exists r>0: B_r(x) \subseteq U$$

And this defines a topology $\mathcal{T}_d$ on $X$.

You take everything relative to $U$, so as if we're working only in $(U,d)$ and then it's indeed not surprising that $U$ is then open, as the whole set is always in any topology...

Title: Definition of open sets


问:
What is the percentage of these numbers?
Consider the numbers $N$ of the form:

$2580\cdot x + 2236\cdot y$, with $x$ and $y$ nonnegative integers.

I have to calculate the density of these numbers below $10^5$.

I mean suppose $\rho (10^5)$ is the number of $N$'s below $10^5$.

Which is the value of  $\frac{\rho(10^5)}{10^5}$?

Is the brute force approach useful with a calculator?

$\gcd(2580,2236)=172$

Does it mean that roughly $1/172$ of nonnegative integers below $10^5$ has that form?

Now a question a bit harder:

what percentage of the $N$'s are also congruent to $10^m\pmod {41}$? Is this equivalent to ask what percentage of the multiples of $172$ are congruent to $10^m\pmod {41}$, for some positive $m$?

Body:
The second question

There are $23$ numbers of the form $2580 x + 2236 y,\;x,y\in\mathbb{N};\;x\ge 0,y\ge 0$ less than $10^5$ congruent to $10\bmod 41$, while the  total is $499$.

Percentage is $23/499=4.61\%$

$$\small\{12556,19608,26660,33712,40764,47816,54868,61920,61920,68972,68972,76024,76024,\\\small76024,83076,83076,83076,90128,90128,90128,97180,97180,97180\}$$
whose remainder when divided by $41$ is $10$.


问:
Why do rectangular matrices have no inverse that can solve Ax = v?
We can reverse a linear transformation using an inverse matrix to find the input vector used in the original linear transformation:

$$A^{-1}A\vec{x} = A^{-1}\vec{v}$$

Let's say we have a 2D vector that we transform into a 1D "vector":
$$\begin{pmatrix}
a & b
\end{pmatrix}*\begin{pmatrix}
x  \\
y
\end{pmatrix} = \begin{pmatrix}
ax + by
\end{pmatrix}       $$

Isn't there some matrix that can then convert that 1D vector back into the original 2D vector? Something like this:

$$\begin{pmatrix}
c  \\
d
\end{pmatrix}*\begin{pmatrix}
ax + by
\end{pmatrix}  = \begin{pmatrix}
x  \\
y
\end{pmatrix}      $$

My TA says that the original transformation does not have an inverse because a single input vector would map to multiple  output vectors. But I'm not seeing those multiple outputs in my algebra.

答:
When it exists, the solution of a linear system
$$\eqalign{
AX &= B \\
}$$
can be written using the Moore-Penrose inverse $A^+$ and an arbitrary matrix $Y$ as
$$\eqalign{
X &= A^+B + (I_n-A^+A)Y \\
}$$
When the dimensions of the matrices $(A,B,X,Y)$ are
$\Big((1\times 2),\,(1\times 1),\,(2\times 1),\,(2\times1)\Big)$ you'd rewrite this in terms of vectors and scalars $(a,\beta,x,y)$ as
$$\eqalign{a^Tx = \beta}$$
Since the pseudoinverse of a vector has a closed-form $\Big(a^+ = \frac{a^T}{a^Ta}\Big)$ so does the solution
$$\eqalign{
x = \beta\left(\frac{a}{a^Ta}\right) + \left(I_2-\left(\frac{aa^T}{a^Ta}\right)\right)y \\
}$$
Which is great -- except that there are an infinite number of possible
$y$ vectors, and so there are an infinite number of solutions.


问:
Geometric distribution, rolling a dice
We roll a fair dice until we get a number $3$. We win as many dollars as is a number of times we didn't get a $3$. If we get $3$ in a first throw, we loose $y$ dollars. What is the maximum value for $y$ so that this game is profitable for us?

I don't really know where to start. Of course I'd like to compute some kind of expectancy, but how to include this information about a possible outcome of a first roll?

答:
Let $p = \frac{5}{6}$ and let $q = (1-p)$.

First, preliminary results:

$S = \sum_{n=0}^\infty p^n = \frac{1}{1-p} = 6.$

$T = \sum_{n=1}^\infty np^{n-1} =
1 + 2p + 3p^2 + \cdots$

$= S \times (1 + p + p^2 + \cdots) = S^2 = 36.$

There are two possibilities:

1st roll is $3$, which results in $-y$.  This case has an expectation of $-qy$.

First roll is not $3$, which means that you win money.

This scenario has a probability of occuring of $p$.

Once it is given that the first roll is not $3$, the chance of you winning exactly $n$ is given by $qp^{(n-1)}.$  Therefore, assuming that your first roll is not $3$, your expected profit is

$q \times (1 + 2p + 3p^2 + \cdots) = qT.$

Therefore, your positive expectation is represented by
$p \times qT = 5.$

Therefore, your overall expectation is

$$5 - qy$$.

Therefore, for the game to be break-even, $y = 30.$

Title: Geometric distribution, rolling a dice


问:
A robust convex optimization problem
Consider a function $ f: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}  $ such that $\forall x \in \mathbb{R}^n$ the map $f(x,\cdot)$ is convex, and $\forall y\in \mathbb{R}^m$ the map $f(\cdot,y)$ is convex as well. We assume that $m<n$.

Given $y^1, ..., y^N \in \mathbb{R}^m$, and $c \in \mathbb{R}^n$, consider the following convex optimization problem.

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y^i) \leq 0 \ \ \forall i \in [1,N] $$

Let $x^\star$ be its optimizer, supposed to be unique.

Notice that the above optimization problem is equivalent to the following one.

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y) \leq 0 \ \ \forall y \in \text{conv}( \{y^1, ..., y^N\} ) $$

We now say that $f(x,y^k) \leq 0$ is a "true" constraint if it "really matters", i.e.
any optimizer of the problem

$$ \min_{ x \in \mathbb{R}^n } c^\top x \ \ \ \text{ sub. to: } \ f(x,y^i) \leq 0 \ \ \forall i \in ([1,N]\setminus\{k\}), $$

which we denote by $x_{ \neg k }^\star$, is such that $c^\top x_{ \neg k }^\star < c^\top x^\star$.

What is the number of true constraints?

答:
What are you calling a "true" constraint?
Suppose you want to minimize $c(x)=x^2$ under some sets of constraints:

$x\le1$ and $x\le2$, optimum $x^*=0$ [$f(x,y)=x-y, y_1=1, y_2=2$]. The second constraint is redundant, it can be eliminated because its a consequence of the first ($y_1 \lt y_2$).

$x\le1$, optimum $x^*=0$. [$f(x,y)=x-y, y_1=1$]. This constraint is inactive, it in not eliminated because of other constraints but because it is true near the optimum ($x^*\lt y_1$).

$x\lt-1$, optimum $x^*=-1$. [$f(x,y)=x-y, y_1=-1$]. This constraint is active, you cannot remove it without changing the optimum.

Type 3 is clearly a "true" constraint and type 1 clearly not. But what about type 2?

ContentLicense: CC BY-SA 3.0

Title: A robust convex optimization problem


答:
Let $r\neq 1$. Then
\begin{align*}
\sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r},
\end{align*}
which for $r \in (0,1)$ tends to $\frac{1}{1-r}$ as $n$ tends to infinity.

Thus, if $p\in(0,1)$, then
\begin{align*}
p \sum_{k=4}^\infty (1-p)^k &= \left( p \sum_{k=0}^\infty (1-p)^k \right) - \left(p \sum_{k=0}^{4-1} (1-p)^k\right) \\
&= p\frac{1}{1-(1-p)} - p\frac{1-(1-p)^4}{1-(1-p)} \\
&= 1 - \left(1 - (1-p)^4\right) \\
&= (1-p)^4.
\end{align*}

(In the above, the LHS of course corresponds to $\mathbb{P}(X>4) = \sum_{k=5}^\infty \mathbb{P}(X = k)$.)

Alternatively, one can re-write the LHS according to
\begin{align*}
p \sum_{k=4}^\infty (1-p)^k
&=
p \sum_{k=0}^\infty (1-p)^{k+4} \\
&=
(1-p)^4 p \sum_{k=0}^\infty (1-p)^k,
\end{align*}
when the result is established by noting that the last term equals $\frac{1}{1-(1-p)} = \frac{1}{p}$.

Title: Probability interpretation

Body:
Supose we have a mouse on a labyrinth and we associate $p$ as the probability of the mouse escaping this labyrinth this turn, and $(1-p)$ for the probability of him not escaping that turn. Call $X$ the random variable associated with the number of turns that the mouse takes in order to escape.
If I want him to take, at best, $4$ turns to escape the labyrinth then I would consider the possibilities of:

$$P(X \leq 4)= P(X = 4)+P(X = 3)+P(X = 2)+P(X = 1)$$

After some calculations I arrive at the probability:

$$ P(X \leq 4)= 1-(1-p)^4 $$

On the other hand:

$$ P(X \leq 4)=1-P(X > 4)$$

Which leads to:

$$ P(X > 4) = (1-p)^4$$

This gives us the probability of him failling more then $4$ times, given that it is the sum:

$$ P(X > 4) = P(X=5)+P(X = 6)+...=p(1-p)^4+p(1-p)^5+...=p \sum_{i=4}^n(1-p)^{i}=(1-p)^4$$

Can someone give me a intuitive interpretation for this? It seems a little odd that the summation would tend to $(1-p)^4$ when starting at $4$.


问:
Find the set of points at which the series is holomorphic?
Let $f$ be defined on $\mathbb{C}$. Find the set of points at which $f$ is holomorphic, where
\begin{align*}
&f(z)=\displaystyle\sum_{n\geq0}^{\infty} \frac{z^{2n}}{(5n)^n} \mbox{ for every $z$ for which} \sum_{n\geq0}^{\infty} \frac{z^{2n}}{(5n)^n} \mbox{ converges,} \\& f(z)=0 \mbox{ for every $z$ for which} \sum_{n\geq0}^{\infty} \frac{z^{2n}}{(5n)^n} \mbox{ does not converge.}
\end{align*}
I know that $f$ is holomorphic everywhere if the Cauchy-Riemann equations hold, but how do I apply that to a series? I tried setting $z=x+iy$ and then calculating the partial derivatives, but that seems to lead to nowhere in particular. Would appreciate any help I can get thanks.

答:
The sum of the series $g(z)=\displaystyle\sum_{n\geq0}^{\infty} \frac{z^{n}}{(5n)^n}$ is an entire function by Hadamard's formula since
$$\lim a_n^{1/n}=\lim\frac1{5n}=0$$
Hence $f(z)=g(z^2)$ is also an entire function.


问:
If $X$ is a normed vector space and $U\subseteq X$, how do we show $(U^\perp)_\perp=\overline U$?
Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $X$ be a normed $\mathbb K$-vector space. It is easy to see that $$U^\perp:=\{\varphi\in X':\langle x,\varphi\rangle=0\text{ for all }x\in U\}$$ is a closed subspace of $X'$ for all $U\subseteq X$ and $$V_\perp:=\{x\in X:\langle x,\varphi\rangle=0\text{ for all }\varphi\in V\}$$ is a closed subspace of $X$ for all $V\subseteq X'$.

I would like to show

If $U\subseteq X$ and $V:=U^\perp$, then $\overline U=V_\perp$.

If $V\subseteq X'$ and $U:=V_\perp$, then $\overline V=U^\perp$.

In 1. (resp. 2.) it is easy to show that $U\subseteq V_\perp$ (resp. $V\subseteq U^\perp$) and hence $\overline U\subseteq V_\perp$ (resp. $\overline V\subseteq U^\perp$).

Now I know the proof of the analogue result in the Hilbert space case, but this proof relies on the projection theorem and hence doesn't easily generalize to the present situation.

So, how do we show $V_\perp\subseteq\overline U$ (resp. $U^\perp\subseteq\overline V$)?

答:
Suppose that there exist  $ x \in  (U^\bot)_\bot:=  \{ x \in X :  f(x) =0, \text { for all } f \in U^\bot  \}   $ such that    $ x \notin \overline U $. Evidently,  $ x  $ is not zero. From a known Corollary of Hahn-Banach thm, we can find  $ f \in X^* $ such that      $ f(x)=d(x,\overline U)>0  $ and  $ f|_{\overline U}=0 $.
Since   $ f (U) = \{0\} $ we have that   $ f \in U^\bot $. Moreover, since   $ x \in  (U^\bot)_\bot  $, we get that     $ f(x) =0 $ which is impossible.

We now show that $(V_\bot)^\bot := \{ f  \in X^* : f ( V_\bot ) = \{0\}\} = \overline{V}^{w^*} $ (weak-star closure). It is easy to see that  $ \overline {V}^{w^*} \subset (V_\bot)^\bot$. Suppose that there exist  $ x_0^*    \in  (V_\bot)^\bot  \setminus \overline{V}^{w^*} $. From the strong separation thm, we can find  $ x^{**}: (X^*,w^*) \to \mathbb R \ $  linear and continuous such that

$$ \sup_{ v^*  \in \overline{V}^{w^*}} x^{**} (v^*)<x^{**}(x_0^*) . $$

Since  $ x^{**} $ is  $ w^*  $-continuous, we can find $  x\in X$ such that $ x^{**}(x^*)=x^*(x), $ for all  $ x^* \in X^* $. So,

$$ \sup_{v^*  \in \overline{V}^{w^*}} v^*(x)< x^*_0(x) . $$
This leads to a contradiction:

If   $ x\in  V_\bot, $ since  $ \overline {V}^{w^*} \subset  ( V_\bot )^\bot  $, it must be true that     $ 0 = \sup \{ v^*(x) \colon v^*  \in \overline{V}^{w^*} \}< x^*_0(x)=0 $  which is impossible.

If  $ x \notin V_\bot $, then  $ v_0^*(x)\neq 0  \text{ for some  } v_0^*\in V $.  Hence, (since  $\frac{x^*_0(x)}{v_0^*(x)} v_0^*(x) \in \overline {V}^{w^*} $)

$$ \frac{x^*_0(x)}{v_0^*(x)} v_0^*(x)  \leq \sup \{ v^*(x) \colon v^*  \in \overline{V}^{w^*} \}< x^*_0(x)   $$
which is impossible


问:
Solve sixth degree complex polynomial
I am trying to solve this equation
$$ \lambda^6+2\lambda^4+\lambda^2-s^2=0, $$
where $s=1+i\rho$.

with some manipulations I can get
$$ (\lambda^3+\lambda -1-i\rho)(\lambda^3+\lambda+1+i\rho)=0 .$$

So $\lambda^3+\lambda -1-i\rho=0$ or $\lambda^3-\lambda +1+i\rho=0$. But I can't get the roots of each polynomial.

I accept any suggestion, hint or book to read.

Another option is to properly define $s=s(\rho)$ such that the polynomial has "pretty" roots, but I can't think in other definitios for $s$.

答:
$$ \lambda^6+2\lambda^4+\lambda^2-s^2=0
\quad=(λ^3 + λ - s )(λ^3 + λ + s)$$

For each cubic, a single known-real root can be obtained from the cubic equation.

$$x=\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
+\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$

where $\quad a=1\quad b=0,\quad c=1\quad d=\pm1$

There is also a Trigonometric solution for three real roots

Title: Solve sixth degree complex polynomial


问:
Quantifiers and set theory
Given the following statement,

$$\forall x \, (x \in S)$$

When we say "for every x in S", are we saying that every possible element is in S -- and there are no other elements, except x?

答:
What you wrote amounts to

Every  $x$ in our domain of discourse is an element of $S$

i.e., $S$ is a "set of everything".

Title: Quantifiers and set theory


问:
Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$

Given $x,y,z$ are  positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$

My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}  \right)$$
$$x^2+1 \geq 2x \Rightarrow \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}\right)\leq \frac{1}{2}\left(x^2y+y^2z+z^2x\right)=\frac{3}{2}$$
$$P \geq x^3y+y^3z+z^3x - \frac{3}{2}$$
I was stuck when I was finding the minium of $x^3y+y^3z+z^3x$

Help me, thank you so much

答:
There's no minimum of $P$ at all. Consider an intersection of $x^2y+y^2z+z^2x = 3$ and $y = x^{-4}$, i.e. $x^{-2} + x^{-8}z + z^2x = 3$ or $x^9 z^2 + z + x^6 - 3x^8 = 0$. We can find from here that there exists a positive solution
$$
z = \dfrac{-1 + \sqrt{1+12x^{17}-4x^{15}}}{2x^9}.
$$
at least for $x$ big enough. Moreover,
$$
\lim_{x \to +\infty}z\sqrt{x} = \sqrt{3}.
$$
Hence
$$
\lim_{x \to +\infty}\left[ \dfrac{x^5 y}{x^2+1} + \dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\left[\dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\dfrac{z^5 x}{z^2+1} = \lim_{x \to +\infty}\dfrac{3^{\frac{5}{2}}x^{-\frac{5}{2}} x}{3 x^{-1}+1} = 0.
$$
So $P$ does not have a minimal value on the set $(x,y,z) \in (0, +\infty) \times (0, +\infty) \times (0, +\infty)$ (because $P$ is obviously positive on it).


答:
Your question doesn't specify what $U$ is.

To your question: it reads like it can be restated, "Aren't all points of $U$ interior points of $U$?"  This need not be the case, depending on what $U$ is.

Consider the real line as a metric space with the metric induced by the modulus:
$$
d(a, b) = |a - b|.
$$
Now let $U = [0,1]$.  As you can see, not all points of $U$ are interior to $U$: the endpoints $0, 1$ are not.

Finally, let me suggest a definition of "open set" that relies on prose rather than on notation, hence is probabily easier to read and process:

Definition:  In a metric space, the open sets are exactly those sets that are unions of open balls.

In the above example, $U = [0, 1]$ is not open.

Title: Definition of open sets

Body:
Open ball, centered at $$ x \in U,$$ with radius r is defined as
$$ B_r \left( x\right)=\left\{x' \in U \mid \mathrm{d} \left(x, x' \right)<r \right\}.$$

Definition of an open set:
$$ U \ \mathrm{is \ open.} \Leftrightarrow \forall x\in U: \exists r>0: B_r(x)\subset U$$

My question is: don't all points of U anyway satisfy the latter condition? Since $$ x \in U, $$ the open ball is a collection of elements in U and thus a subset of U.


问:
Binary operations of addition and multiplication.
Prove that the set $\mathbb{R}$ of all real numbers is an abelian group under the binary operation of addition (+) and a semi-group under multiplication ($\cdot$).

Prove that $\mathbb{R}^*$, the subset of non-zero $x \in \mathbb{R}$, forms an abelian group under multiplication.

答:
The point is, that the definition of semigroup, group, Abelian group are all originated from the addition and multiplication of numbers, primarily.

So to speak, this must be a trivial exercise. You have to agree with the following:

If $u,v\in\Bbb R$, then both symbols $u+v$ and $u\cdot v$ are uniquely defined in this context, and they are still elements of $\Bbb R$.

For any elements $u,v,w\in\Bbb R$, both the equations
$$u+(v+w)=(u+v)+w, \quad\quad u\cdot(v\cdot w)=(u\cdot v)\cdot w$$
hold (associativity - it basically means we can live without parenthesis if using only $+$ or only $\cdot$ operation).

Abelianness: For any elements $u,v\in\Bbb R$, we have both
$$u+v=v+u,\quad\quad u\cdot v=v\cdot u\,.$$
This is called commutativity, and together with associativity means that we can arbitrarily permute the members of a sum/product, and will still get the same result.

For $\Bbb R^*$ being group, we need that the product of elements of $\Bbb R^*$ stays within $\Bbb R^*$, and that all $u\in\Bbb R^*$ has an inverse with respect to the given operation, which is now the multiplication. So, it speaks about that we can form $1/u$ whenever $u\in\Bbb R^*$.


问:
maximum values of the Poisson distribution
good day
Can anybody help me? thanks

let $X\sim  \operatorname{Poisson}(\lambda)$

1.- For what $\lambda$ values the value of $P (X = i), i≥0$ is maximum

2.- Prove that: $E (X^n) = \lambda E[(X+1)^{n-1} ]$ and use this result to calculate the value of $E (X^3)$

答:
Hint: (1) Do you know the formula for $\mathbb P(X=i)$?  Can you differentiate it with respect to $\lambda$?

(2) write the expectations using the formula for $\mathbb P(X=i)$ and compare the term for $i$ on one side to the term for $i-1$ on the other.


答:
Typically this is written more simply as
$$\forall x\in S$$
On its own, the expression is meaningless. It only has meaning when coupled with other statements. For example, consider the following:
$$|x|\geq x~~,\forall x\in\Bbb{R}$$
In words, this means "For any real number - $5$, $\log(\gamma)$, $\sqrt{\pi}$, etc,  its absolute value is greater than or equal to the number itself." The important point is the fact we used $x$ is irrelevant - the statement has exactly the same meaning if we write
$$|\text{banana}|\geq \text{banana}~~,\forall ~\text{banana}\in\Bbb{R}$$
In other words, the statement says nothing about "$x$" itself. $x$ is simply a placeholder that we use to represent "something contained in the set $S$".

Title: Quantifiers and set theory

Body:
Given the following statement,

$$\forall x \, (x \in S)$$

When we say "for every x in S", are we saying that every possible element is in S -- and there are no other elements, except x?


问:
Difference between the definition of the dot product in infinite and finite dimensions
I know that in $\mathbf{\mathbb{R}}^n$ the definition of the dot (or scalar) product is the following:

$x.y=x^{\mathrm{T}}y$, with ''T" denoting the transpose of the vector x.

How does this definition change when working in infinite space, e.g:

$\int_{\Omega}\nabla u \cdot \nabla v \, d\Omega = 0,\, \forall v\in H^{1}_{0}\left(\Omega\right)$

am I allowed to write:

$\int_{\Omega}\left(\nabla u\right)^{\mathrm{T}}\, \nabla v\,d\Omega = 0,\, \forall v\in H^{1}_{0}\left(\Omega\right)$ ??

Thanks in advance for your answers

答:
I basically found my answer from this question, where they clarified the ambiguity I had with the definition of the inner product. In fact, when working in an infinite dimension space the inner product (a general nomenclature and definitions for different spaces) is the one used and as the dimension becomes finite its definition becomes the dot product (or you can still call it an inner product). So basically, the dot product is the definition of the inner product in a finite dimension space. Me moving from integral 1 to the second integral defined in a discretized space is completely fine as long as I remind the definition of the inner product.

I confirmed this by the comments I've gotten here.

Hope I was clear enough :)


问:
what are the arrows in between the objects $M \otimes_U N$ in the category $\mathscr{M} \otimes_U N$?
I am trying to understand a single object category. May be we can call it as monoid category.

Suppose $R$ be a commutative ring with identity and $A$ is a $R$-algebra.

$(1)$ Can we think of $A$ as a category with a single object $A$ itself and with the identity arrow
?

For better sense we can talk about monoidal category with morphisms $A \otimes A \to A$ and identity map $id \to A$  with a commutative diagram.

Actually I am trying to understand the product between a category of modules over a commutative ring and a module over another commutative ring. More specifically, let $U$ and $V$ be two commutative rings with  $U \subset V$. Let $\mathscr{M}$ be the category of $V$-modules and $N$ be a $U$-algebra.

Now I want to define the product $\mathscr{M} \otimes_U N$  or just $\mathscr{M} \times N$.  I have looked the definition $4.2.20$ in The Stack Project.

I understand that the objects of the new category $\mathscr{M} \otimes_U N$ will be of the form $M \otimes_U N$ or just $M \times N$ for $M \in \mathscr{M}$.

$(2)$ But what are the arrows in between the objects $M \otimes_U N$ in $\mathscr{M} \otimes_U N$ ?

If the question $(1)$ is true, then I think the arrows  or morphisms in the new category $\mathscr{M} \otimes_U N$ are just $f \otimes 1$, where $f$ is the arrow in the category $\mathscr{M}$ and $1$ denote the identity arrow in the single object category $N$.

Any comment on question $(1)$ and question $(2)$

答:
Let me answer these questions in order.

First, let $R$ be any ring with identity. In general we can regard $R$ as a category with one object, $\mathscr{R}$, defined as follows:

Objects: A singleton set $\lbrace \ast \rbrace$.

Morphisms: Define the hom-set $\mathscr{R}(\ast,\ast):= R$.

Composition: For any $r,s \in R$, define $r \circ s := rs$, i.e., the composition of endomorphisms of $\ast$ is given by multiplication in the ring $R$.

Identities: The identity on $\ast$ is $1_R$.

As you anticipated, this does become a monoidal category: $\mathscr{R}$ has monoidal functor $\otimes:\mathscr{R} \times \mathscr{R} \to \mathscr{R}$ given by sending $(\ast,\ast) \mapsto \ast$ and sending a pair of morphisms $(r,s) \mapsto r \circ s = rs$.  This actually gives a functor $\mathbf{Ring} \to \mathbf{MonCat}$, from the $1$-category of rings to the $1$-category of monoidal categories with monoidal functors between them.

Now assume that $R$ is commutative. If $A$ is a unital $R$-algebra, then following the process above we exactly get that $A$ may be represented by a one-object category $\mathscr{A}$ (just proceed mutatis mutandis as to how we built $\mathscr{R}$). Things get really annoying if we have non-unital algebras, but we can always use the unitization functor $R\mathbf{Alg} \to R\mathbf{Alg}_1$, tensor with $R$ appropriately, and then make sure to keep track of augmentation ideals appropriately. Because of this, I'm going to focus on unital $R$-algebras.

An interesting and important observation of this unital ring <--> $\mathbf{Ab}$-enriched monoidal category with one object is that an action of the category $\mathscr{R}$ is on $\mathscr{A}$ which respects all the relevant monoidal structures (an actegorical structure) is equivalent to making sure that $\mathscr{R}$ acts appropriately on the set of endomorphisms of the identity functor, $Z(\mathscr{A}) = [id_{\mathscr{A}},id_{\mathscr{A}}]$. Because the category $\mathscr{A}$ is enriched over the category of Abelian groups, $\mathbf{Ab}$, and this action is bilinear with respect to composition in $\mathscr{A}$, it isn't too rough to show that $Z(\mathscr{A})$ is a (commutative) unital ring. However, $Z(\mathscr{A}) \cong Z(A)$ as rings, so such an action corresponds to a ring morphism $R \to A$ which factors through $Z(A)$, i.e., this corresponds to how we witness $A$ as an $R$-algebra.

In this way we answer your question (1) affirmatively: Up to some isomorphism and unitization issues that I brushed under the rug, we can indeed regard an algebra $A$ as a one-object category (equipped with an appropriate action of the one-object category of $R$, of course).

Your second question is a little more involved. Following your notation, fix two commutative unital rings $U, V$ with an inclusion ring morphism $U \hookrightarrow V$. Let $V\mathbf{Mod}$ be the category of $V$-modules and let $N$ be a $U$-algebra. As far as I have seen, the category $V\mathbf{Mod} \otimes_U N$ is defined as follows (although your mileage may vary based on context of where you have seen this category):

Objects: $U$-modules $M \otimes_U N$, where we regard $M$ as a $U$-module via restriction of scalars.

Morphisms: $U$-linear morphisms $f \otimes g:M \otimes_U N \to M^{\prime} \otimes_U N$ where $f \in V\mathbf{Mod}(M,M^{\prime})$ and $g \in U\mathbf{Mod}(N,N)$.

Composition: Assume that $f \otimes g:M \otimes_U N \to M^{\prime} \otimes_U N$ and $h \otimes k:M^{\prime} \otimes_U N \to M^{\prime\prime} \otimes_U N$ are morphisms in $V\mathbf{Mod} \otimes_U N$. The composite
$$
(h \otimes k) \circ (f \otimes g) := (h \circ f) \otimes (k \circ g):M \otimes_U N \to M^{\prime\prime} \otimes_U N.
$$

Identities: For any object $M \otimes_U N$ in $V\mathbf{Mod} \otimes_U N$, the identity morphism $id_{M \otimes_U N}$ is $id_M \otimes id_N$.

Now if you write $\mathscr{N}$ as a single object category, there is a connection between
$$
V\mathbf{Mod} \otimes_U N
$$
and
$$
V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N},
$$
where this second category has some horrible abuse of notation going on. By the second category, I mean the category with objects $(M,\ast)$, for a $V$-module $M$ regarded as a $U$-module, and $\ast$ the object of $\mathscr{N}$ (you think of this being the object $M \otimes_U N$). A morphism in $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N}$ corresponds to a pair $(f,g)$ where $f$ is a morphism in $V\mathbf{Mod}$ and $g$ is a $U$-module endomorphism of $N$ induced by multiplication by an element of $N$ (so by, for instance, the left-multiplication-by $n \in N$ map). Composition is given component-wise, and the identities are $(id_M,id_{\ast}) = (id_M,1_N)$.

Generally there is an inclusion $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N} \to V\mathbf{Mod} \otimes_U N$, but it need not be an isomorphism unless $U\mathbf{Mod}(N,N) \cong N$. It is also not true that $V\mathbf{Mod} \times \mathscr{N}$ is equivalent to $V\mathbf{Mod} \otimes_U N$ or $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N}$. The issue here is that in the product category $V\mathbf{Mod} \times \mathscr{N}$, there is no restriction of scalars to $U$ business. Objects are just $(M,\ast)$ where $M$ is a $V$-module. You still need to somehow be able to collapse things via a restriction of scalars argument, and in general you can have two non-isomorphic $V$-modules that become isomorphic after restricting scalars to a smaller ring $U$.

For an example of this second phenomenon, take $U = \mathbb{Q}, V = \mathbb{Q}[x]$, and let the map $U \to V$ the standard embedding. Now note that a $\mathbb{Q}[x]$-module is equivalent to a $\mathbb{Q}$-vector space $V$ equipped with a $\mathbb{Q}$-linear endomorphism $T:V \to V$.  Choose the $\mathbb{Q}[x]$-modules as follows: Let $M = (\mathbb{Q}^2,T)$ where $T$ is a linear transformation that acts via the matrix
$$
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}
$$
in the standard basis for $\mathbb{Q}^2$ and let $M^{\prime}$ be given by the underlying vector space $\mathbb{Q}^2$ with linear transformation $T^{\prime}$ acting via the matrix
$$
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}
$$
in the standard basis of $\mathbb{Q}^2$. Then it is routine to see that in $\mathbb{Q}[x]\mathbf{Vect}$, $M \not\cong M^{\prime}$ (one linear transformation is the identity while the other is nilpotent). However, restricting scalars from $\mathbb{Q}[x]$ to $\mathbb{Q}$ gives that
$$
\operatorname{Res}(M) = \operatorname{Res}(\mathbb{Q}^2,T) = \mathbb{Q}^2 = \operatorname{Res}(\mathbb{Q}^2,T^{\prime}) = \operatorname{Res}(M^{\prime}).
$$
which shows the phenomenon.

As a final comment to this alreadly way too long answer, in order to get a category with where objects are pairs $(M,\ast)$ and where morphisms are $(f,id_{\ast})$ for a one-object category $\mathscr{N}$ and any category $\mathscr{M}$ with $M \in \mathscr{M}_0$ and $f \in \mathscr{M}_1$, then you need the category
$$
\mathscr{M} \times \underline{1},
$$
where $\underline{1}$ is the category with a single object and single arrow. Unsurprisingly, this category is canonically isomorphic to $\mathscr{M}$.


问:
Associated prime ideals of $D[[X]]$
Let $D$ be an integral domain.

A prime ideal $P$ of $D$ is called an associated prime of a principal ideal $aD$ of $D$ if $P$ is minimal over $(aD:bD)$ for some $b\in D\setminus aD$. For brevity, we call $P$ an associated prime of $D$.

My question is:

If $P$ is an associated prime of $D$, is $P[[X]]$ also an associated prime of $D[[X]]$?

答: Assume that $P$ is an associated prime of $D$. Then there $P$ is minimal over $(aD:bD)$ for some elements $a$ and $b$ of $D$. ''If  $D[[X]]$ is a flat $D-$module, by a well-known result of N. Bourbaki, $(aD:bD)D[[X]]=(aD[[X]]:bD[[X]])$''. Now, is $P[[X]]$ minimal over $(aD[[X]]:bD[[X]])$?

Title: Associated prime ideals of $D[[X]]$


问:
Compact subsets of function spaces, geometry
The subset is called compact when every open cover contains a finite subcover. In Euclidean spaces, it is easy to visualize this by imagining some open ball that contains this set, thinking about the subset of this ball that contains the set and divising this subset into the finite collection of sets. Is there a way of some kind of visualization in the case of infinite-dimensional function spaces?</p>

答:
If your metric space is complete you have that a set is totally bounded if and only if it is relatively compact.

A set a totally bounded if for every $\epsilon > 0$ there is a finite $\epsilon$-net.
A set is relatively compact if the closure is compact.

This somehow can bring your intuition from the finite dimensional case to the infinite dimensional case, when your space is complete.


问:
Smallest hexagonal lattice with no more than one point per unit Cartesian cell, as a function of rotation angle?
This answer to How to randomly but evenly distribute nodes on a plane introduces Bridson's Algorithm for Poisson-disc sampling (original paper: Fast Poisson Disk Sampling in Arbitrary Dimensions (also here) by Robert Bridson, University of British Columbia) Also see this and animation and explanation here.

I'm seeding the initial field with some regions with points on periodic lattices which will be both square and hexagonal, and of arbitrary rotation.

Bridson's method is characterized by a minimum distance $r$ that new points can be generated

For the algorithm to go fast, one chooses a square grid of side $r/\sqrt{2}$ which is small enough that there will always be either zero or one point in each cell. This way when a new point is introduced:

it must be introduced in a currently empty cell

only eight adjacent cells must be checked to evaluate the minimum distance to all existing points.

When I seed the space with preexisting array regions, I need to make sure that they never put more than one point in any cell.

Question: What are the constraints on lattice spacing $a$ and rotation angle $\theta$ for a hexagonal lattice with unit vectors:

$$\mathbf{a_1} =  a\cos \left(\theta \right) \ \mathbf{\hat{x}} + a\sin \left(\theta \right) \ \mathbf{\hat{y}}$$

$$\mathbf{a_2} =  a\cos \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{x}} + a\sin \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{y}}$$

such that there will never be more than one point in a unit square cartesian grid with any origin? (In other words, no matter how you translate the origin of the hexagonal lattice with respect to the cartesian grid.)

Rough example of application, for background only. Left: real space, right: reciprocal space (Fourier transform).

答:
There are three regular tilings of the Euclidean plane, and they correspond to regular triangular lattice, regular square lattice, and regular hexagonal lattice.

In order for a lattice to be able to describe any point configuration with at most one point per cell, with pairwise distances greater than $r_\min$, the radius of the lattice cell circumcircle must be at most $2 r_\min$.  In other words, the maximum distance between lattice cell vertices, or "cell diameter", must be $r_\min$ or less.  Otherwise, a single lattice cell could contain two points (near vertices of the cell) with their distance greater than $r_\min$.

The properties of the three lattices with the same "cell diameter" (cell circumcircle diameter $d = r_\min$) are
$$\begin{array}{l|ll|ll}
\text{Lattice} & \text{Side length} & ~ & \text{Cell area} & ~ \\
\hline
\text{Triangular} & a_△ = r_\min & \approx 1.0000 \, r_\min & A_△ = \frac{\sqrt{3}}{4} r_\min^2 & \approx 0.433 \, r_\min^2 \\
\text{Square} & a_□ = \frac{r_\min}{\sqrt{2}} & \approx 0.7071 \, r_\min & A_□ = \frac{1}{2} r_\min^2 & \approx 0.500 \, r_\min^2 \\
\text{Hexagonal} & a_⬡ = \frac{r_\min}{2} & \approx 0.5000 \, r_\min & A_⬡ = \frac{3 \sqrt{3}}{8} r_\min^2 & \approx 0.650 \, r_\min^2 \\
\end{array}$$
Maximum density of points is reached with a hexagonal lattice with $a = 2/\sqrt{3} \; r_\min$, where each cell has area $A_\min = \sqrt{3}/2 \, r_\min^2 \approx 0.866 \, r_\min^2$.  From this we can estimate that for triangular lattices, at most half the cells will be populated with points; for square lattices, $1/\sqrt{3} \approx 0.5/0.866 = 58\%$ of cells will be populated; and for hexagonal lattices, $3/4 = 75\%$ of cells will be populated.

The following illustration shows these lattices, with shaded areas showing the region where existing points can limit where/whether the center white cell can contain a new point or not.

Without subdividing the cell further, one needs to examine up to 24 triangular cells, 20 square cells, or 18 hexagonal cells.

If a larger side length $a$ is used for a lattice, some valid point configurations cannot be described using only one point per lattice cell.  (That is, with larger $a$, not all valid point sets can be generated when limiting to at most a single point per cell.)

The problem at hand is to find $a_x$ for one lattice, given a larger than minimum other lattice $a_y$ and the relative rotation $\theta$ between the two.  Because of the above, I believe this is the wrong way to try to fix the underlying problem.

What I would suggest, is to use the desired lattices (with arbitrary spacings, as long as they are not less than the ones stated above wrt. the minimum pairwise distance $r_\min$), but additionally verify and add each point to a minimal square lattice (which is computationally the simplest to implement, and likely most efficient on current hardware).

That is, an arbitrary lattices can be used to generate the points, but only when the minimal square lattice also allows each new point.  While the minimal square lattice has smaller cells than the lattices used to generate points, it allows all possible point configurations (with pairwise distances greater than $r_\min$), and is therefore well suited for verification (that a point to be added does not break the minimum pairwise distance rule).


问:
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution?
The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:

$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)

答:
Knowing the answer, as i mentioned it in the comments, the substitution in
the answer of J.G. has best chances to work, and it works in a line.

Alternatively...
Starting from the last expression in the OP, we may use the Euler substitution
$$
t = \sqrt{\frac{u+4}u}=\sqrt{1+ \frac 4u}\ .
$$
Then we have formally successively $t^2=1+\frac 4u$, $t^2-1=\frac 4u$,
$u=\frac 4{t^2-1}$,
$du=-\frac{8t}{(t^2-1)^2}\; dt$.

The substitution from $x$ to $t$ is thus $\color{blue}{t=\frac{\sqrt{x^2+4}}x}$. (Assuming $x>0$.)
Then with the used substitutions
$$
\begin{aligned}
\int\frac{dx}{\sqrt{x^2+4}}
&=\frac12\int\frac{du}{\sqrt{u(u+4)}}
=\frac12\int\frac{du}{u\sqrt{\frac{u+4}u}}\\
&=\frac12\int\frac{-\frac{8t}{(t^2-1)^2}\; dt}{\frac 4{t^2-1}\cdot t}
=-\int\frac{dt}{t^2-1}\\
&=
\frac 12\log\frac{t+1}{t-1}+C
=
\frac12
\log
\frac
{\color{blue}{\frac{\sqrt{x^2+4}}x}+1}
{\color{blue}{\frac{\sqrt{x^2+4}}x}-1}+C
\\
&=\log(\sqrt{x^2+4}+x)+C'\ .
\end{aligned}
$$
(At the last point we note that $(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)$ is a constant.)


问:
Multiplying power formal series when reindexing is required?
Normally, when I multiply the two following series:

$$\sum_{n\ge0}a_nx^n$$

$$\sum_{n\ge0}b_nx^n$$

I have:

$$\sum_{n\ge0}(\sum_{i=0}^na_ib_{n-i})x^n$$

Which is nice and intuitive. However, what if the exponents of x, or the indices don't match up? For example, what if I wanted to multiply these two:

$$\sum_{n\ge0}a_nx^{100n}$$

$$\sum_{n\ge0}b_nx^n$$

If I were to try to do this I'd reindex the second series to:

$$\sum_{\frac{n}{100}\ge0}b_nx^{100n}$$

But then the indices wouldn't line up. Am I missing something?

答:
We can nicely see the similarities when adding an intermediate step.

On the one hand we have
\begin{align*}
\sum_{k=0}^\infty a_kx^k\sum_{l=0}^\infty b_lx^l
&=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n
=\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{n}} a_kb_{n-k} x^n\\
\end{align*}

On the other hand we have
\begin{align*}
\sum_{k=0}^\infty a_kx^{100k}\sum_{l=0}^\infty b_lx^l
&=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{100k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n
=\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{\left\lfloor\frac{n}{100}\right\rfloor}}a_kb_{n-100k} x^n\\
\end{align*}

Comment:

In the middle sums we have rearranged the terms according to increasing powers of $x^n$. Consequently the conditions are $k+l=n$ resp. $100k+l=n$ besides $k,l\geq 0$.

In the right-hand sums we eliminate $l$ by substituting $l=n-k$ resp. $l=n-100k$. Consequently is the upper limit of the inner sum $n$ resp. $\left\lfloor\frac{n}{100}\right\rfloor$.


问:
Strong operator convergence of unilateral shift.
Giving an infinite dimensional separable Hilbert space $X$, with an o.n. basis $(e_n)$, we have the standard left and right shift operators $L$ and $R$. I am trying to study the convergence (in the strong operator topology) of $R^nL^n$ and $L^nR^n$. For the former I suspect that it converges to 0 ($L^n$ deletes the first $n$ entries and the $R^n$ puts 0's in their place). For the latter $L^nR^n$ leaves nothing changed so i assume the sequence converges to the identity map. However I do not know how to formally prove this? For starters $R^n$ does not converge in the strong topology (while $L^n does) so I cannot use convergence of compositions.

答:
Since $LR=I$, you immediately get that $L^nR^n=I$ for all $n$, no limits required. Formally, this is shown by induction: if $L^nR^n=I$, then $L^{n+1}R^{n+1}=L^nLRR^n=L^nR^n=I$.

The other way around, the composition $R^nL^n$ equals the orthogonal projection onto $\overline{\operatorname{span}}\{e_n,e_{n+1},\ldots\}$. So indeed it goes to zero sot.


问:
Percentage question
Certain alloy contains 20% of copper and 5% of tin. How much of copper and how much tin must be added to 100kg of this alloy to get another with 30% of copper and 10% of tin?

I can't figure how to start it

答:
My list of leading questions as used in the comment section to the question:

 How much copper do the 100kg of alloy contain?

.

 Now, if you add x kg of copper and y kg of tin, how much copper does the result contain?

.

 If you add x kg of copper and y kg of tin, how much kg does the result have?

.

 So, what is the percentage of copper in the result? This should give you one equation. You get a similar equation for the tin. Then you solve the system of two equations.

Title: Percentage question


问:
Constructing continuous function using locally finite basis
(Munkres) Let $\mathscr{A}$ be a collection of subsets of $X$ such that $X=$ $\bigcup_{A \in \mathscr{A}} A,$ and let $f: X \rightarrow Y$ be such that $\left.f\right|_{A}$ is continuous, for every $A \in \mathscr{A}$.

We say $\mathscr{A}$ is locally finite if for every $x \in X,$ there is a neighborhood $U \subset X$ of $x$ such that $\{A \in \mathscr{A} \mid U \cap A \neq \emptyset\}$ is finite. Show that if $\mathscr{A}$ is locally finite and if each $A \in \mathscr{A}$ is closed in $X,$ then $f$ is continuous. Give an example to show that the result may fail if $\mathscr{A}$ is not locally finite.

I'm new to topology. My attempt so far:

$\forall\;V\subseteq Y$ such that $V$ is open in $Y$, $f^{-1}(V)\cap A$ is open in $A$ for every $A \in \mathscr{A}$. Another point to note is, if $x \in int\;A$ then $f$ is continuous at $x$, therefore we only need to prove continuity for points on the boundaries of sets in $\mathscr{A}$ (even though I'm confused about the fact that in analysis we talk about continuity at a point but in topology we only talk about continuity of the function as a whole, defined by pre-images of open sets being open). Also, for some finite subcollection $\{A_i\}_i$ of $\mathscr{A}$, $x \in \cap_i A_i$ and $x \notin A$ for all other $A \in \mathscr{A}$. For any sequence $x_n \rightarrow x$, $\exists N$ s.t. for all $n > N$, $x_n \in A_i$ for some $i$ (we would not be able to say this without local finiteness). Therefore by continuity of $f|_A$, $f(x_n) \rightarrow f(x)$.

I did the sequential proof based off my analysis courses where I have dealt with only metric spaces. I'm not sure if the sequential proof is correct in the context of topology.

答:
The key is that a locally finite collection $\mathcal{A}$ of sets is so-called closure-preserving in the sense that $\overline{\bigcup \mathcal{A}} = \bigcup \{\overline{A}\mid A \in \mathcal{A}\}$.

Then if $\mathcal{A}$ is a locally finite family of closed sets and $f\restriction_A$ is continuous for every $A \in \mathcal{A}$, then $f$ is indeed continuous: let $F \subseteq Y$ be closed. By the assumptions $(f\restriction_A)^{-1}[F] = f^{-1}[F] \cap A$ is closed in $A$ for every $A \in \mathcal{A}$, and as all such $A$ are themselves closed in $X$, $f^{-1}[F] \cap A$ is closed in $X$ as well, for all $A \in \mathcal{A}$. It's trivial that the family $\{f^{-1}\cap A \mid A \in \mathcal{A}\}$ is also locally finite (a "shrinking" of $\mathcal{A}$) and so by closure preservingness:

$$\overline{f^{-1}[F]} = \overline{\bigcup \{ A \cap f^{-1}[F]\mid A \in \mathcal{A}\}} = \bigcup \{A \cap f^{-1}[F]\mid A \in \mathcal{A}\}  =f^{-1}[F]$$

so that $f^{-1}[F]$ is closed and so $f$ is continuous.

Finally to show the closure prervingness of the locally finite $\mathcal{A}$, we note that for every fixed $A \in \mathcal{A}$, we have that $\overline{A} \subseteq \overline{\bigcup \mathcal{A}}$ by mononicity, and so also

$\bigcup \{\overline{A}\mid A \in \mathcal{A}\} \subseteq \overline{\bigcup \mathcal{A}}$, is then clear. For the reverse, we pick $x \in \overline{\bigcup \mathcal{A}}$ and we pick an open neighbourhood $O_x$ such that $\mathcal{A}(x):=\{A \in \mathcal{A}: A \cap O_x \neq \emptyset\}$ is finite. It's clear that $x \notin \overline{A}$ for any $A \notin \mathcal{A}(x)$. So if $x \notin A$ for every $A \in \mathcal{A}(x)$, we could pick a neighbourhood $O_A$ of $x$ that each $O_A \cap A = \emptyset$, for each $A \in \mathcal{A}(x)$. But then the open neighbourhood (finite intersection) $O_x \cap \bigcap_{A \in \mathcal{A}(x)} O_A$ of $x$ would be disjoint from $\bigcup \mathcal{A}$ altogether, which is a contradiction. So in fact, $x$ must be in one of the set $\overline{A}$ for some $A \in \mathcal{A}(x)$, so in $\bigcup \{\overline{A}\mid A \in \mathcal{A}\}$ as well ,finishing the proof of the inclusion.


问:
What's Special about Rowland’s Prime-Generating Sequence?
Recently I asked this question and quickly got back some excellent responses.
I asked the question because I came across a paper by Eric Rowland called "A Natural Prime-Generating Recurrence". The main results on page 4. It appears there was much fanfare in academic news when the paper was released which turns out to be basically the question I asked yesterday except, instead of subtracting the lpf, he added it causing his function to climb indefinitely, whereas mine is decreasing.

So basically Rowland's iterations is
$$f(n) = n + lpf – 1$$ starting at 5 (where lpf = least prime factor of $n$) which yields:
$$5→9→11→21→23→45→47→93→95→99→101→201→203→209…$$

The sequence above is exactly the sequence Roweland has in column 2 on page 4 of his article without the duplicates.

Am I missing why Roweland's sequence is a significant discovery?

答:
This is OEIS sequence A190894, which is not Rowland's sequence A106108 but an auxiliary sequence used in proving
the properties of Rowland's sequence. Rowland's sequence is
$$ 7, 8, 9, 10, 15, 18, 19, 20, 21, 22, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 69, \ldots $$
which has the recurrence $a(n) = a(n-1) + \gcd(n, a(n-1))$ with $a(1) = 7$,
and the property that the difference between any two terms is either $1$ or a prime.


问:
Find all linear functionals $X \to \mathbb{R}$ such that $|\varphi(x)\| \leq \|x\|^2$

Let $X$ be a normed space. Find all linear functionals $X \to \mathbb{R}$ such that $|\varphi(x)\| \leq \|x\|^2$

This was an exam task(easy one, judging by the amount of points possible to obtain from it) I was unable to do.

I tried taking supremum:

$$
|\varphi(x)\| \leq \|x\|^2 \implies \\
\sup_{\|x\|=1}| \varphi(x)| \leq 1^2 \implies \\
\|\varphi\| \leq 1
$$
so all bounded functionals with norm lesser equal to 1 but it was wrong.

How does one solve this task?

答:
Hint Let $a \in X$.

Then for all $y \in \mathbb R \backslash \{ 0\}$ you have
$$
|y|\cdot |\phi(a)|=|\phi(ay)| \leq \| ay \|^2 \leq |y|^2 \|a\|^2
$$

Divide by $|y|$ and set $|y| \to 0$.


问:
Identity for $e$ in terms of the Fibonacci sequence.
The following identity appears in Martin Gardner's paper, "Dr. Matrix on Little Known Fibonacci Curiosities:

$$e = \frac{1 + 1 + \frac{2}{2!} + \frac{3}{3!} + \frac{5}{4!} + \frac{8}{5!} + \frac{13}{6!} + \frac{21}{7!} + \frac{34}{8!} + \frac{55}{9!} + \cdots}{1 + 0 + \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} + \frac{3}{5!} + \frac{5}{6!} + \frac{8}{7!} + \frac{13}{8!} + \frac{21}{9!} + \cdots} $$

How can we prove this?

答:
The correct form of this expression seems to be:
$$e=\frac{1+1+\frac{2}{2!}+\frac{3}{3!}+\frac{5}{4!}+\frac{8}{5!}+\frac{13}{6!}+\ldots}{1-0+\frac{1}{2!}-\frac{1}{3!}+\frac{2}{4!}-\frac{3}{5!}+\frac{5}{6!}-\ldots} = \frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}}$$

There are two useful observations that can be used to prove this statement:

$\sum\limits_{k=0}^\infty \frac{x^k}{k!}=e^x$ (the definition of $e^x$)

$F_n=\frac{\varphi^n-\psi^n}{{\varphi-\psi}}$, where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$ (the Euler-Binet formula)

Now we are all set for the proof!
$$\begin{eqnarray}
\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{\varphi^{k+1}-\psi^{k+1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(\varphi\sum\limits_{k=0}^\infty \frac{\varphi^k}{k!}-\psi\sum\limits_{k=0}^\infty \frac{\psi^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\varphi e^\varphi - \psi e^\psi\right) \\
\end{eqnarray}$$

$$\begin{eqnarray}
\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{(-1)^k\varphi^{k-1}-(-1)^k\psi^{k-1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(
\frac{1}{\varphi}\sum\limits_{k=0}^\infty \frac{(-\varphi)^k}{k!} -
\frac{1}{\psi}\sum\limits_{k=0}^\infty \frac{(-\psi)^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi}\right) \\
\end{eqnarray}$$

Dividing these two yields: $$\frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}} = \frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} }$$

Since $\varphi + \psi = 1$ and $\varphi\psi = -1$, we this can be simplified as
$$\frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} } =
\frac{\varphi e^\varphi - \psi e^\psi}{ e^{-1}\left(\varphi e^{\varphi} - \psi e^{\psi}\right)} = e$$

Q.E.D.


答:
Just observe that
$$\frac{f(x)}{x^n}=a_n+a_{n-1}x^{-1}+\cdots +a_0x^{-n} $$
and the limit of the right hand side is $a_n\ne0$.

Body:
Show that $f(x)=a_{n}x^n+a_{n-1}x^{n-1}+...+a_{0}$ is $\Theta(x^n)$.

In an attempt to show the lower bound $\Omega(x^n)$, I got stuck at the following assuming $k_{1}= 1$:
$$\mid f(x) \mid=\mid a_{n}x^n+a_{n-1}x^{n-1}+...+a_{0}\mid$$
$$\ge \mid k_{1}x^n+k_{1}x^{n-1}+...+k_{1}\mid$$
$$\ge \mid k_{1}(x^n+x^{n-1}+...+1)\mid$$
$$\ge \mid k_{1}(x^n)\mid$$

Question:why $\mid a_{n}x^n+a_{n-1}x^{n-1}+...+a_{0}\mid  \ge  \mid k_{1}x^n+k_{1}x^{n-1}+...+k_{1}\mid$ as $a_{n}, a_{n-1}, ...$ could be larger numbers than $k_{1}$?


问:
if $a^{2}+b^{2}=1$ Prove that $-\sqrt{2} \le a+b \le \sqrt{2}$
$a,b \in \mathbb{R}$

My attempt:
$$a^{2}+b^{2}=1 \iff a^{2}+2ab+b^{2}=1+2ab$$
$$(a+b)^2=1+2ab \iff \mid a+b \mid =\sqrt{1+2ab} \iff a+b = \pm\sqrt{1+2ab}$$

But notice that $\sqrt{1+2ab} \in \mathbb{R} \iff 1+2ab \geq 0 \iff 2ab \geq -1$ .

let's take the case where :$2ab=-1$.
$$a^{2}+2ab+b^{2}=0 \iff a+b=0 \iff a=-b.$$

When I plug this result into the original equation, I get:
$$b=\frac{1}{\sqrt{2}}$$
And now, I don't know where to go.

答:
Here is an alternative route:

By the AM–GM inequality, we have
$$
{\frac  {a+b}2}\geq {\sqrt  {ab}}
$$
and so $a^2+b^2+2ab \ge 4ab$, which implies $2ab \le 1$. Then
$$
(a+b)^2 = a^2+b^2+2ab\le 2
$$
gives
$$
|a+b| \le \sqrt 2
$$
as required.


问:
Lie groupoid associated to an incomplete vector field
Let $M$ be a smooth manifold and $X \in \mathfrak{X}(M)$ be a vector field. If $X$ is complete, its flow defines a group action $\Bbb R \circlearrowright M$ via $t \cdot x \doteq \Phi_X(t,x)$. This gives us an action groupoid $\mathcal{G} \rightrightarrows M$, where $$\mathcal{G} = \{(t,x,y) \in \Bbb R \times M \times M \mid x = \Phi_X(t,y)\},$$with ${\sf s}(t,x,y) = y$, ${\sf t}(t,x,y) = x$, $M \hookrightarrow \Bbb R \times M \times M$ via $x \mapsto (0,x,x)$ and the operation is $(t,x,y)\cdot (s,y,z) = (t+s,x,z)$, everything standard here.

In Exercise 1.9 of Eckhard Meinrenken's notes, he wants to consider the case where $X$ is not complete, and the flow domain is just an open set $U \subseteq \Bbb R \times M$ (containing $\{0\} \times M$). The exercise asks us to show that $U$ can be turned into a Lie groupoid $U\rightrightarrows \Bbb R$ instead, apparently unrelated to the action groupoid described above.

It's completely unclear to me how to go about this. First because we're changing the manifold of units. Second because there doesn't seem to be a natural way to embed $\Bbb R \hookrightarrow U$. Third because things will go wrong with the operation above, since for $x \in M$ with $(t,x),(s,x) \in U$, one might not have $(t+s,x) \in U$ (duh).

Naively writing $$U = \bigsqcup_{x \in M} (a_x,b_x)$$and trying to define the source and target maps by relating $t$ in $(t,x)$ with $a_x$ and $b_x$ also seems bad, since we could have $a_x = -\infty$ and/or $b_x = +\infty$ even with $X$ incomplete.

Can you give me some ideas? I don't care for a full solution, I'm doing this for fun. If you tell me how to embed $\Bbb R \hookrightarrow U$ I'd say I already have a decent chance of figuring out the rest on my own. I don't know.

答:
The source map is the projection $s: U\to M$, $s(t,q)= q$, the target map is the local flow $\Phi:U\to M$ of your vector field.  The unit map sends $q\in M$ to $(0,q)$. The composition/multiplication in the groupoid is given by
$$ (t,\Phi(s,q))\circ (s, q) = \Phi(t+s, q).$$

(Note that my convention is that the source of $(s,q)$ is $q$ and the target is $\Phi(s,q)$).


答:
Actually you don't need to focus on the case $2ab=-1$. Since we know that $|a+b| = \pm \sqrt{1+2ab}$, we just need to prove that $2ab \leqslant 1$.

To do this, observe that

$$ 1-2ab = a^2+b^2-2ab = (a-b)^2 \geqslant 0 $$

Title: if $a^{2}+b^{2}=1$ Prove that $-\sqrt{2} \le a+b \le \sqrt{2}$

Body:
$a,b \in \mathbb{R}$

My attempt:
$$a^{2}+b^{2}=1 \iff a^{2}+2ab+b^{2}=1+2ab$$
$$(a+b)^2=1+2ab \iff \mid a+b \mid =\sqrt{1+2ab} \iff a+b = \pm\sqrt{1+2ab}$$

But notice that $\sqrt{1+2ab} \in \mathbb{R} \iff 1+2ab \geq 0 \iff 2ab \geq -1$ .

let's take the case where :$2ab=-1$.
$$a^{2}+2ab+b^{2}=0 \iff a+b=0 \iff a=-b.$$

When I plug this result into the original equation, I get:
$$b=\frac{1}{\sqrt{2}}$$
And now, I don't know where to go.


问:
$\int_{\mathbb{R}}|f(t)|^2dt=\int_{\mathbb{R}}|f'(t)|^2dt$ implies $f(t)=\mathbb{x}_{i}|f(t)|$
Let $f \in C^{1}(\mathbb{R},\mathbb{R}^m)$ be such that $f$ and $f'$ are square integrable and
$$\{t:f(t)=0\} \subset \{t:f'(t)=0\}$$
$$ |\{t:f(t)=0\}|=n\in \mathbb{N}$$
Prove that if $\int_{\mathbb{R}}|f'(t)|^2\, dt=\int_{\mathbb{R}}|f|'^2(t)\,dt$, then there are $n+1$ points
$$\mathbb{x}_{1},\mathbb{x}_{2},...,\mathbb{x}_{n+1}\in S^{m-1}=\{\mathbb{x}\in \mathbb{R}^m:|\mathbb{x}|=1\} $$
such that for each $t \in \mathbb{R}$, $f(t)=\mathbb{x}_{i}|f(t)|$ for some $i$.

I got this problem from Postech Math Competition.

Thanks in advance.

答:
Here's a sketch. Transform to polar coordinates. Write $f(t) = r(t) \eta(t)$, where $r \colon \mathbb{R} \to [0,\infty)$ and $\eta(t) \in S^{m-1}$. Then
$$|f|'(t) = r'(t), \quad |f'(t)|^2 = |r'(t) \eta(t) + r(t) \eta'(t)|^2 = |r'(t)|^2 + r(t)^2 |\eta'(t)|^2$$ because $\eta \perp \eta'$. Hence our condition becomes
$$\int r'^2(t) \, dt = \int r'^2(t) + r(t)^2 |\eta'(t)|^2 \, dt,$$
which implies that $r(t)\eta'(t) = 0$ for all $t$. Thus either $r(t) = 0$, in which case we are on one of the $n$ points where $f(t) = 0$, or $\eta'(t) = 0$, in which case we stay on the same line. Since there are $n+1$ segments separated by the $n$ points, there are at most $n+1$ different lines we stay on.</p>


问:
Is $A=B$ if $A=\{a,b\}$ and $B=\{a,b,∅\}$?
I know that $A∩∅=∅$. But what about $A∩\{∅\}$?

Shouldn't  $A∩\{∅\}=\{∅\}$.

If $A=\{a,b\}$ and $B=\{a,b,∅\}$ then  we can conclude $A=B$?

答:
It's really simple:
$$A\cap\{\varnothing\} = \left\{\begin{array}{ll}
\varnothing &\text{if }\varnothing\notin A,\\
\{\varnothing\} &\text{if }\varnothing \in A.
\end{array}\right.$$

If $A=\{a,b\}$ and $B=\{a,b,\varnothing\}$, then $A=B$ if and only if $a=\varnothing$ or $b=\varnothing$.

(Informally: an empty bag is not the same thing as a bag that contains nothing except for an empty bag)


答:
Hint (1): Why did you choose the case $2ab = -1$ ? Well, as you showed, this is the smallest possible value for $2ab$ . What can you say about $|a+b|$ when $2ab > -1$ ?

Hint (2) for a different method: Consider a circle of radius 1 around the center of coordinates. The equation of this circle is $x^2 + y^2 = 1$ . Now consider the set of parallel lines $x+y = c$ where $c$ can be any number. Some of these lines touch the circle. Among those lines, which one has the highest value of $c$ ?


问:
Inequalities of Two i.i.d. Random Variables

The Problem

Let $X_1$ and $X_2$ be a continuous i.i.d. random variables.  By definition, we know that...

Independence: $\mathbb P(X_1 \in A \ \cap \ X_2 \in B)=\mathbb P(X_1 \in A) \ \mathbb P(X_2 \in B)$

Identically Distributed: $\mathbb P(X_1 \le x) = \mathbb P(X_2 \le x) \ \ \forall \ \ x \in \mathbb R^1$

My hypothesis is that:
$$\mathbb P(X_1 > X_2) = 0.5$$
I have attempted to prove this by using the above definitions to form the expression:
$$\mathbb P(X_1 \ge x \ \cap \ X_2 < x)=\mathbb P(X_2 \ge x \ \cap \ X_1 < x) \ \ \forall \ \ x \in \mathbb R^1$$

My Questions

Is this a valid proof, and am I missing anything?

This seems like a fundamental property of i.i.d. random variables.  Is this already a named theorem?

答:
There is a standard trick involved here. Instead of giving you the full solution, let me point you in the right direction instead. Note that the probability you are interested in is equivalent to $\mathbb{P}(X_1 - X_2 > 0).$ Hence it would be enough to determine the cumulative distribution function of the variable $X_1 - X_2.$ For this it would be enough to first determine the CDF of $-X_2$ and then determine the CDF of the difference by explicitly computing the density of $X_1 - X_2$ in terms of the densities of $X_1$ and $-X_2$ (which you would have determined in the previous step). Is there any important Theorem that comes to mind when looking at it this way?

Please do let me know if you manage to fill in the blanks by yourself and if not, I shall do that.


答:
Using the triangle inequality and the equivalence between the $\ell^1$ and the $\ell^2$ norm, we have
$$|a+b| \leq |a|+|b| = \|(a,b)\|_1 \leq \sqrt{2}\, \|(a,b)\|_2 = \sqrt{2}\,\sqrt{a^2+b^2} = \sqrt{2}.$$


答:
Since $a^2+b^2\ge2ab$

$\bigg(a^2+b^2-2ab=\left(a+b\right)^2\ge0\bigg)$

$2ab\le1$.

So, $a^2+b^2+2ab\le2$

$(a+b)^2\le2$.

When you square root both sides, there will be two equations

$a+b\le\sqrt2$

$a+b\ge-\sqrt2$

Combining them gives

$-\sqrt2\le a+b\le\sqrt2$

I'm new to MathJax so please forgive me for weird signs.


答:
Note that the inequality you wanted to prove is equivalent to $(a+b)^2\leq 2$, so once you've obtained that $(a+b)^2=1+2ab$, the next step should be trying to prove that $1+2ab \leq 2$. This can be proven using the AM-GM inequality, which says that $2ab \leq a^2+b^2$, where the equality is taken when $a=b$. Thus $1+2ab \leq 1+a^2+b^2 = 2$.

In your attempt, you've actually reversed the logic a little bit by first declaring that $\sqrt{1+2ab}$ is real, then trying to prove the condition for its real-ness. This doesn't make sense. You can prove that $2ab\geq -1$ by rearranging $(a+b)^2\geq 0$ and using the fact that $a^2+b^2 = 1$. The equality here is taken when $a = -b$, which leads to the values $b = \pm \frac{1}{\sqrt{2}}$ that you solved for. In any case, this doesn't help you prove the original inequality, because it provides a lower-bound, not an upper bound to $(a+b)^2$.


问:
spectral radius of a unilateral forward weighted shift
Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward weighted shift. i.e. $Te_n=t_n e_{n+1}$. And $t_n = (gcd(n, 2^n))^{-1}$

Let $\rho(T)$ be the spectral radius of $T$.

Show $\rho(T)>0$.

Things I know:

I know that $\rho(T)= \sup \{ |\lambda|: \lambda \in \sigma(T)\} = \lim_{n \rightarrow \infty} \| T^n\|^{1/n}$.

Also, I showed that $\| T\| = \sup_n |t_n|$.

I was also given the hint to find $T^{2^k}$.

So I have that

$(Te_n)^{2^k}= (t_ne_n)^{2^k} = (t_n)^{2^k} (e_n)^{2^k} =(gcd(n, 2^n))^{-1})^{2^k} (e_n)^{2^k} $.

Not sure how to proceed from here.

Thank you!

答:
Well your last equation does not make any sense at all, what is a power of a vector of a Hilbert space? Note that $T^{2^k}$ means the composition of $T$ with itself $2^k$ times.

As you said, by the Gelfand-Beurling formula we have that $\rho(T)=\lim_{n\to\infty}\|T^n\|^{1/n}$. Since this limit exists, we also have that
$$\rho(T)=\lim_{n\to\infty}\|T^{2^n}\|^{\frac{1}{2^n}}.$$

Lemma: Fix an integer $p\geq1$ and let $T:H\to H$ be any bounded operator with $Te_n=c_n\cdot e_{n+p}$ for all $n$ where $c_n\in\mathbb{C}$. Then $\|T\|=\sup_{n\geq1}|c_n|$ (so $T$ is a bounded  operator if-f $(c_n)$ is a bounded sequence). I leave this to you to prove, it's easy.

Now $T^me_k=T^{m-1}(t_ke_{k+1})=T^{m-2}t_k\cdot t_{k+1}e_{k+2}=\dots=(t_k\cdot t_{k+1}\cdots t_{k+m-1})e_{k+m}$ and this is true for all $k,m$. Therefore, for $m$ being a power of $2$ we have that
$$T^{2^n}e_k=(t_k\cdot t_{k+1}\cdots t_{k+2^n-1})e_{k+2^n}$$

 Another exercise: Show that for any $k,n\in\mathbb{N}$ we have that
$$\require{cancel}\cancel{gcd(k,2^k)\cdot gcd(k+1,2^{k+1})\cdot gcd(k+2,2^{k+2})\cdots gcd(k+2^n-1,2^{k+2^n-1})=2^{2^n-1}}$$

Using our lemma and the above equation we conclude that $$\require{cancel}\cancel{\|T^{2^n}\|=2^{1-2^n}}$$
so
$$\require{cancel}\cancel{\|T^{2^n}\|^{1/2^n}=2^\frac{1-2^n}{2^n}=2^{1/2^n}\cdot 2^{-1}\xrightarrow{n\to\infty}\frac{1}{2}}$$

and we conclude that $\require{cancel}\cancel{\rho(T)=1/2>0}$.

Edit

The striked text is wrong, the exercise is not true as stated, obviously the RHS cannot be independent of $k$. I am sorry for such a big mistake. However, the following is true:
Note that by our lemma we have that
$$\|T^{2^n}\|=\sup_{k\geq1}\frac{1}{gcd(k,2^k)\cdot gcd(k+1,2^{k+1})\cdots gcd(k+2^n-1,2^{k+2^n-1})}\geq\frac{1}{gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})}$$

We will compute the RHS of this inequality with induction. Note that in general, an integer $k$ is uniquely written as $2^m\cdot(2l+1)$, where $m,l\geq0$ and then $gcd(k,2^k)=2^m$.

Claim: For all $n\geq1$ we have that $gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})=2^{2^n-1}$.

Proof of the claim: With induction on $n$. Obviously this is true for $n=1$. Suppose that this is true for some $n$, so
$$\prod_{k=1}^{2^n}gcd(k,2^k)=gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})=2^{2^n-1}.$$
Now
$$\prod_{k=1}^{2^{n+1}}gcd(k,2^k)=\prod_{k=1}^{2^n}gcd(k,2^k)\cdot\prod_{k=2^n+1}^{2^{n+1}}gcd(k,2^k)=2^{2^n-1}\cdot\prod_{k=2^n+1}^{2^{n+1}}gcd(k,2^k)=$$
$$=2^{2^n-1}\cdot\prod_{k=1}^{2^n}gcd(k+2^n,2^{2^n+k}).$$

Now let $1\leq k\leq 2^n$. Note that if $k=2^{m_k}\cdot(2l_k+1)$ where $m_k,l_k\geq0$, then $m_k\leq n$, because $1\leq 2l_k+1$, so $2^{m_k}\leq k\leq 2^n$. Also, if $m_k=n$, then $k=2^n$. Now obviously $gcd(2^n+2^n,2^{2^n+2^n})=2^{n+1}$. On the other hand, if $m_k<n$ then $n-m_k>0$, so $2^n+k=2^n+2^{m_k}(2l_k+1)=2^{m_k}(2^{n-m_k}+2l_k+1)$, so $gcd(2^n+k,2^{2^n+k})=2^{m_k}=gcd(k,2^k)$. Therefore,
$$\prod_{k=1}^{2^n}gcd(k+2^n,2^{k+2^n})=gcd(2^n+2^n\cdot 2^{2^n+2^n})\cdot\prod_{k=1}^{2^n-1}gcd(k+2^n,2^{k+2^n})=$$
$$=2^{n+1}\cdot\prod_{k=1}^{2^n-1}gcd(k,2^k)=2^{n+1}\cdot\frac{1}{2^n}\cdot\prod_{k=1}^{2^n}gcd(k,2^k)=2^{n+1-n}\cdot2^{2^n-1}=2^{2^n}$$
so
$$\prod_{k=1}^{2^{n+1}}gcd(k,2^k)=2^{2^n-1}\cdot2^{2^n}=2^{2^n+2^n-1}=2^{2^{n+1}-1}$$
as we wanted! ${\blacksquare}$

Now using the estimate on $\|T^{2^n}\|$, we have that
$$\|T^{2^n}\|\geq\frac{1}{2^{2^n-1}}$$
so $$\rho(T)=\lim_{n\to\infty}\|T^{2^n}\|^\frac{1}{2^n}\geq\lim_{n\to\infty}\frac{1}{2^{1-\frac{1}{2^n}}}=\frac{1}{2}>0.$$

A final comment: To be honest I now think that more generally one can prove that for all $n,k\geq1$ we have that $$\prod_{j=1}^{2^n} gcd(k+j-1,2^{k+j-1})=2^{2^n+k-1}$$
and if this is true then all the inequalities in the edit actually become equalities so $\rho(T)=1/2$ indeed, but I think the above calculations were tiresome enough!


答: As $a^2 + b^2 = 1$ you can find $\theta \in [0,2\pi]$ such that $a = \cos(\theta)$ and $b = \sin(\theta)$. Now you can study the variations of the function $f : [0,2\pi] \to \mathbb{R}$ defined by $f(\theta) = \cos(\theta) + \sin(\theta)$ and you will see that for all $\theta$, $|f(\theta)| \leq \sqrt{2}$.


问:
Relating arcs and paths in topology
There seems to be slightly varying definitions in the literature for {\bf arc} and {\bf path}. So, here I am using the following: If $S$ is a space and $I$ denotes the closed unit interval, and if $f:I\to S$, then define: (A) if $f$ is continuous, then $f$ is called a {\bf path}, and (B) if $f$ is a homeomorphism, then $f(I)$ is called an {\bf arc}. I have two questions. From the literature, it seems that two statements must be true: (1) If $S$ is a metric space, then $S$ is Hausdorff. (2) If $S$ is a Hausdorff space, $X\subset S$ contains at least two points, and $f:I\to X$ is onto, then $f$ is a path if and only if $f(I)$ is an arc ($f$ itself is not the homeomorphism, but some onto function $g:I\to f(I)$ is).

Question 1: Are statements (1) and (2) true? If not, what is wrong?

Question 2: If all above is correct, then is the following true? Suppose $S$ is a metric space, $f:I\to S$ is a path, and $J=[a,b],a<b$ is a closed sub-interval of $I$. Then $f(J)$ is an arc.

答:
Yes, a metric space is Hausdorff.

And (2) is also true in this improved variant (but rather hard to show, we need theory about Peano continua and characterisations of $[0,1]$, all early 20-th century stuff, but not very generally taught anymore): If $x \neq y$ in a Hausdorff space and there exists a path $f:I \to X$ from $x$ to $y$ then we can also find an arc from $x$ to $y$. In shorter terms, a path-connected Hausdorff space is arcwise-connected as well.

Title: Relating arcs and paths in topology


问:
Rotation angle from quaternion via atan2
I need some help with understanding the rotation angle determination via atan2.
The formula which I found in this forum is following:

$\phi=2\tan^{−1}(\sin(\phi/2)/\cos(\phi/2))$

This is derived from the unit quaternion.
$a=(\cos(\phi/2),\sin(\phi/2)\vec{n})$

I understand that I get following equation by transforming the scalar part of the quaternion:
$\phi=2\cos^{-1}(a_1)$

But how do I get the atangens equation ?

答:
Polar form for quaternions uses three variables, which I will denote as r, p, and n.
To find the atan of a quaternion the variable n is needed:
the quaternion a+bi+cd+dj is expressed as a+v, n = v/abs(v)

```<code>atan(q) = -(n/2) ln( (n-q)*((n+q)^-1) )
</code>```

This is essentially the formula for atan of a complex number because just as a function of a complex number can be obtained as a two-dimensional stretch, shift, or rotation, the same thing can be done with quaternions, and, most stretches, shifts, or rotations only require points on a plane, n essentially denotes a plane slicing through four dimensions, and technically treating anything like a complex number, where the complex part, instead of i, is a unit versor n, works just fine.

For example, the cos of a quaternion:

a = r(cos(p))

cos(q)

= cos(r(cos(p)+n sin(p)))

= cos(a+n*sin(p))

= cos(a) cos(n r sin(p)) - sin(a) sin(n r sin(p))

= cos(a) cosh(r sin(p)) - n sin(a) sinh(r sin(p))

Body:
Polar form for quaternions uses three variables, which I will denote as r, p, and n.
To find the atan of a quaternion the variable n is needed:
the quaternion a+bi+cd+dj is expressed as a+v, n = v/abs(v)

atan(q) = -(n/2) ln( (n-q)*((n+q)^-1) )

This is essentially the formula for atan of a complex number because just as a function of a complex number can be obtained as a two-dimensional stretch, shift, or rotation, the same thing can be done with quaternions, and, most stretches, shifts, or rotations only require points on a plane, n essentially denotes a plane slicing through four dimensions, and technically treating anything like a complex number, where the complex part, instead of i, is a unit versor n, works just fine.

For example, the cos of a quaternion:

a = r(cos(p))

cos(q)

= cos(r(cos(p)+n sin(p)))

= cos(a+n*sin(p))

= cos(a) cos(n r sin(p)) - sin(a) sin(n r sin(p))

= cos(a) cosh(r sin(p)) - n sin(a) sinh(r sin(p))


问:
How to solve this integral
Wolfram integrator is unable to solve this:

$$
 \int_0^\pi \frac{1}{\sqrt {(a-\cos(t))^2+(b-\sin(t))^2+c^2}}\,\mathrm dt
$$

Any suggestions? Thanks!

答:
Note that $(a-\cos(t))^2 + (b-\sin(t))^2 + c^2 = a^2 + b^2 + c^2 + 1 - 2(a \cos(t) + b \sin(t))$.

Also note that $a \cos (t) + b \sin(t) = \sqrt{a^2 + b^2} \sin (t + F)$.

Let $d = a^2 + b^2 + c^2 + 1$ and $e = 2\sqrt{a^2 + b^2}$. Then, our integral becomes

$\displaystyle\int_0^\pi \frac{1}{\sqrt{d - e \sin(t + F)}}\mathrm dt$, where $d,e$, and $F$ are constants. This is elliptic and has no simple solution in terms of elementary functions.

Title: How to solve this integral


问:
Is this some sort of modular form? $f(\tau)=\eta(a\tau)\eta(b\tau)$
I recently found the following. Let $a,b\in\Bbb N$ with $24|(a+b)$ and consider the function
$$f(\tau)=\eta(a\tau)\eta(b\tau).$$
We have the following symmetry relations:
$$\begin{align}
f(\tau+1)&=f(\tau),\tag1\\
f\left(-\tfrac{1}{ab\tau}\right)&=-i\tau\sqrt{ab}f(\tau),\tag2
\end{align}$$
which I will prove at the bottom of the post.

Considering that $\eta(\tau)$ is a modular form of weight $1/2$, is $f(\tau)$ some sort of modular form?

Just a word on notation. Throughout this question, I will use the notational conventions $\tau\in\Bbb H=\{z\in\Bbb C:\Im z>0\}$, $q:=e^{2i\pi \tau}$, and $\eta(\tau)=q^{1/24}\prod_{n\ge1}(1-q^n)=q^{1/24}(q;q)_\infty$.

Context: I was messing around with the $\eta$-function and I saw on Wikipedia that
$$\eta(8\tau)\eta(16\tau)=\sum_{m,n\in\Bbb Z\\ m\le|3n|}(-1)^mq^{(2m+1)^2-32n^2},$$
which I found interesting, because the $q$-expansion
$$\eta(\tau)=\sum_{n\in\Bbb Z}(-1)^nq^{(6n-1)^2/24}$$
has no integer-powers of $q$. After a little experimenting, I was able to show that the $q$-expansion for $\eta(a\tau)\eta(b\tau)$ has integer-powers of $q$ when $24|(a+b)$. So my next query was whether or not $f(\tau)$ was a modular form. Clearly the relation $(2)$ is not of the form $g(-1/\tau)=\tau^kg(\tau)$, but it is pretty similar, so I thought that maybe $f(\tau)$ was a modular form in some other sense, or for some subgroup of $\mathrm{SL}_2(\Bbb Z)$.

I also learned that $\eta$-quotients of the form $\prod_{d|n}\eta(d\tau)^{r_d}$ can be modular forms under certain restrictions on $r_d$ and $n$. Perhaps $f(\tau)$ satisfies these?

Forgive me if this question is obvious, I have only learned about this kind of thing from textbooks and Wikipedia, and I have yet to take any classes in number theory.

Proofs:

$(1)$: We can see that
$$f(\tau)=\eta(a\tau)\eta(b\tau)=q^{(a+b)/24}(q^a;q^a)_\infty (q^b;q^b)_\infty=q^k(q^a;q^a)_\infty (q^b;q^b)_\infty,$$
for some $k\in\Bbb N$. Since $a,b\in\Bbb N$, clearly each product $(q^s;q^s)_\infty$, where $s=a,b$, will contribute only terms with integer-powers of $q$ to the $q$-expansion of $f$. And since $q^m$ is invariant under the transformation $\tau\mapsto\tau+1$ for integer $m$, we have that $f(\tau+1)=f(\tau)$.

$(2)$: From the relation
$\eta(-\tfrac1{\tau})=\sqrt{-i\tau}\eta(\tau)$
it is easy to prove $(2)$.

答:
Let $a,b\in\Bbb N$ such that $a+b=24$.

Theorem: The function
$$f(\tau)=\eta(a\tau)\eta(b\tau)$$ is a cusp form of weight $1$ for the congruence subgroup $$\Gamma_1(ab)=\left\{\gamma\in\mathrm{SL}_2(\Bbb Z):\gamma\equiv\begin{pmatrix}1 & *\\ 0 & 1\end{pmatrix}\pmod{ab}\right\}.$$

Proof: Since $a+b=24$, we can write $$f(\tau)=q(q^a;q^a)_\infty(q^b;q^b)_\infty,$$
which is invariant under the map $\tau\mapsto \tau+1$, so
$$f(\tau+1)=f(\tau).$$
Then, take any $$\gamma=\begin{pmatrix}p_1 & p_2\\ p_3 & p_4\end{pmatrix}\in\Gamma_1(ab),$$
and define
$$\gamma (z):= \frac{p_1z+p_2}{p_3z+p_4},$$
as well as $$\gamma_M (z):=\frac{p_1z+Mp_2}{\frac{p_3}{M}z+p_4}$$ for any nonzero integer $M$.
We have
$$\begin{align}
f(\gamma (z))=f\left(\frac{p_1z+p_2}{p_3z+p_4}\right)&=\eta\left(a\frac{p_1z+p_2}{p_3z+p_4}\right)\eta\left(b\frac{p_1z+p_2}{p_3z+p_4}\right)\\
&=\eta\left(\frac{ap_1z+ap_2}{p_3z+p_4}\right)\eta\left(\frac{bp_1z+bp_2}{p_3z+p_4}\right)\\
&=\eta\left(\frac{p_1(az)+ap_2}{\frac{p_3}{a}(az)+p_4}\right)\eta\left(\frac{p_1(bz)+bp_2}{\frac{p_3}{b}(bz)+p_4}\right)\\
&=\eta\left(\gamma_a(az)\right)\eta\left(\gamma_b(bz)\right).
\end{align}$$
Then
$$\det\gamma_M=\det\begin{pmatrix}p_1 & Mp_2\\ \tfrac{p_3}{M} & p_4\end{pmatrix}=p_1p_4-\left(Mp_2\right)\left(\tfrac{p_3}{M}\right)=p_1p_4-p_2p_3=\det\gamma=1,$$
and since $a$ and $b$ both divide $ab$ and by definition $p_3\equiv 0\pmod{ab}$, we have that $\gamma_a,\gamma_b\in\mathrm{SL}_2(\Bbb Z)$, and thus
$$\eta(\gamma_M(Mz))=\varepsilon(\gamma_M)(\tfrac{p_3}{M}Mz+p_4)^{1/2}\eta(Mz)=(p_3z+p_4)^{1/2}\varepsilon_M\eta(Mz),$$
for $M=a,b$. Thus
$$f\left(\frac{p_1z+p_2}{p_3z+p_4}\right)=\varepsilon_a\varepsilon_b (p_3z+p_4)f(z).\tag1$$
Here, $$\varepsilon\left[\begin{pmatrix}A & B\\ C & D\end{pmatrix}\right]=\begin{cases} e^{i\pi B/12} & C=0,\, D=1\\
\exp\left[i\pi\left(\tfrac{A+D}{12C}-\tfrac14-s(D,C)\right)\right] & C>0,
\end{cases}$$
with
$$s(D,C)=\sum_{n=1}^{C-1}\frac{n}{C}\left(\frac{Dn}{C}-\left\lfloor\frac{Dn}{C}\right\rfloor-\frac12\right),$$
and of course $\varepsilon_M=\varepsilon(\gamma_M)$.

From $(1)$ we then know that $f$ is a modular form of weight $1$ for $\Gamma_1(ab)$. As @reuns said, $f$ is a cusp form, because $f^{24}(\tau)=\Delta^{24}(a\tau)\Delta^{24}(b\tau)$, where $\Delta$ is the Modular Discriminant, which is a cusp form.

QED


问:
Peer rated - relative ordering of people based on IQ
I am trying to develop a peer rated IQ network which can be implemented in the real world. To summarise the problem statement, consider an ecosystem of 100 people. The aim is to rank the 100 people in decreasing order of IQ.

Data is available in the following format. Consider a 100x100 table. Each entry A[i][j] in the table denotes what the i'th person thinks of j'th person's IQ. There is no entry for A[i][i]. Also, there may be a null entry in some A[i][j], if the two people don't know each other.

Using this data, I need to sort the 100 people in decreasing order of IQ.

[PS: There is no correct answer to the problem. I am looking for different approaches.]

答:
Some reflections...

You may know that the ranking a certain individual has done of the 100 persons is called a permutation of the set $1,2,...100$.

An important idea is to express how two "permutations" differ.

There are different ways to express a distance between two different permutations $\pi_1$ and $\pi_2$.

One of the most interesting is Cayley distance which is the minimal number of transpositions (elementary exchanges between the values at two different positions) one has to apply to $\pi_1$ in order to obtain $\pi_2$. See this.

Once you have obtained all the mutual distances between the different "rankings", you should observe ideally a cluster of coherent answers, and outliers that should hopefully be eliminated (people who are not reliable...). Afterwhile, a mean must be found to "average" the good ones...

Another track: Spearman rank correlation.

The Cayley distance can be computed using the decomposition of a permutation into a product of cycles ; see here.

An interesting text.


问:
What is the probability of getting 2 green balls out of 4 draws in a finite sample space through method of counting?
I am not understanding why trying to find the probability through counting isn't working for the following example:

Example: You have an urn with 3 Red balls and 3 Blue balls. If you draw 4 balls out of the urn, what is the probability of getting exactly 2 Red balls?

So the obvious solution is P(2 Red balls) = 3/6 * 2/5 * 3/4 * 2/3 = 0.1

However if I try to find P(2 Red balls) by counting, I get a different answer:
Out of 4 draws, you can get the following combinations:

3 Red, 1 Blue: C(4 balls, 3 Red) = 4 total combinations

2 Red, 2 Blue: C(4 balls, 2 Red) = 6 total combinations

1 Red, 3 Blue: C(4 balls, 1 Red) = 4 total combinations

So if I divide the total number of outcome combinations of having exactly 2 Red balls from 4 draws against the total number of outcome combinations from 4 draws, I get: P(2 Red balls) = 6/14 = 0.428 which is not 0.1.

What is wrong with my method of counting?

答:
There are $\binom{6}{4}=15$ ways to draw 4 balls out of the urn with 6 balls.

There are $\binom{3}{2} \times \binom{3}{2} = 9$ ways to draw two of the three red balls and two out of the three blue balls.

So the correct answer is $\frac{9}{15}=\frac{3}{5}$.

Your "obvious method" assumes you draw RRBB (or BBRR) in that fixed order. So you count too few.


问:
How to generate a random vector, guaranteed to be within the hemisphere with respect to another vector?
Given a normalized vector N, how can one generate a random direction vector that is guaranteed to be in the hemisphere with respect to N (i.e. the hemisphere where N is exactly in the middle)? The way I am currently doing this is to sample a random direction vector d and dot it with N and keep that vector if the dot product is greater than 0. This method is not guaranteed to generate a vector in the hemisphere that I am interested in as ~50% of the random vectors would have a 0 or negative dot product result.

I saw somewhere that there is a way to transform the randomly generated vector d to place it in the right hemisphere using a transformation matrix but I don't know how to do it. Can someone write a [pseudo]code for how one generate a direction vector using the transformation method?

答:
To answer your followup question from the comments: If you're in 3D, we have the convenient fact that the $z$ coordinate of a random point on the unit sphere is uniformly distributed. So if you take $z\sim U(\cos\theta,1)$, $\phi\sim U(0,2\pi)$, $r=\sqrt{1-z^2}$, then $\left(r\cos\phi,r\sin\phi, z\right)$ is a random point on the sphere within $\theta$ radians of the positive $z$-axis.

To transform this to be within $\theta$ radians of $N$, multiply it by an orthonormal matrix whose last column is $N$. (In other words, apply the transformation $(x,y,z)\mapsto(xV,yW,zN)$ where $V$ and $W$ are unit vectors that are orthogonal to $N$ and to each other.)


问:
Compact subsets of function spaces, geometry
The subset is called compact when every open cover contains a finite subcover. In Euclidean spaces, it is easy to visualize this by imagining some open ball that contains this set, thinking about the subset of this ball that contains the set and divising this subset into the finite collection of sets. Is there a way of some kind of visualization in the case of infinite-dimensional function spaces?</p>

答:
The topologies for infinite dimensional Banach spaces can be a bit tricky. One nice "measure" of their weird behavior with respect to compactness is looking at the unit ball (vectors of norm $\leq 1$). The unit ball in a Banach space is compact (with respect to the norm topology) if and only if the Banach space is finite dimensional. In some sense, there are too many open sets. One of the classical results of functional analysis is the Banach-Alaoglu theorem, which states that if $X^*$ is the dual space of some Banach space, then its unit ball is compact in the weak-$*$ topology. (As a side note, in this kind of weak topology for an infinite dimensional Banach space, the (nonempty) open sets are all unbounded. ) In my experience working with weak topologies, I think the best way of thinking about compactness is in terms of nets. More precisely, a (topological) space is compact iff every net has a convergent subnet. This is because weak convergence is easier to check than finite subcoverings. Suppose $X$ is Banach and $Y$ a family of functionals on $X$, and $X$ is given the $Y$-weak topology. Then $x_{\alpha} \rightarrow x$ weakly if $l(x_{\alpha}) \rightarrow l(x)$ in $\mathbb{C}$ for all $l\in Y$.

The source I used for functional analysis is Reed, Simon's book Functional Analysis Vol I.


问:
Show $\mathbb R^{\ge 0}$ can be taken as a model for $\mathbb R^2/\sim$
Paolo Aluffi says something to the effect of what's in the title in his intro to math notes. Link below.

$\mathbb R^2/\sim$ is a quotient of $\mathbb R^2$ modulo the equivalence relation given by $(x_1, y_1) \sim (x_2, y_2) \iff x_1^2 + y_1^2 = x_2^2 + y_2^2.$ If $[(a, b)] \in \mathbb R^2/\sim$, then  $[(a, b)] = \{(x, y): x^2 + y^2 = a^2 + b^2\}$ by $\sim.$ So, $\mathbb R^2/\sim$ is a set of concentric circles centered at the origin with the radius $\sqrt{a^2 + b^2}$. Note, $[(0, r)] = \{(x, y): x^2 + y^2 = r^2\}$ is a circle centered at the origin and so $\mathbb R^2/\sim = \bigcup [(0, r)]$. Thus we want to show a bijection from $\mathbb R^{\ge 0}$ to $\bigcup [(0, r)]$. I think the identity function $\color{red}{r_{(\in \  \mathbb R^{\ge 0})} \to [(0, r)]}$ should work since identity is bijective. What if we replace $\mathbb R^{\ge 0}$ with $\mathbb R$? Since the range of $e^x$ is $(0, \infty)$ and since $e^x$ is bijective, $\color{blue}{\text{$x_{(\in \ \mathbb R)} \to [(0, e^x)]$  should(?) work with a caveat: $e^x$  misses  $r = 0$}}.$

My questions:

Does the function in red above work?

Is the part in blue above reasonable? If so,

How can we augment the function in blue above such that it doesn't miss $[(0, 0)]?$

Thanks.

Introduction to Advanced MathematicsCourse notes by Paolo Aluffi

答:
Yes, the function in red works perfectly fine and is clearly bijective. You are also correct that the image of $e^x$ is $(0,\infty)$ and not $[0,\infty)$.

Note that if $f:[0,\infty)\to\mathbb{R}$ was continuous, then $f^{-1}[\mathbb{R}]=[0,\infty)$ is open, which is not true, so you will not be able to find a 'nice' function as a bijection between $\mathbb{R}$ and $[0,\infty)$.


问:
$\mathbb{Z}[\sqrt{11}]$ is norm-euclidean
I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$

By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$

For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:

Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$

Case 1: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$

Case 2: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$

Case 3: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z $ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.

This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $ \frac{3}{2} \leq |x-a| \leq 2$

答:
I am a novice and learning to write simple proofs. I welcome corrections and suggestions. Below is a detailed workout. Most maths students will find it unnecessarily verbose. Oppenheim (1934) proved, among other proofs, that $\mathbb{Z}[\sqrt{11}]$ has a division algorithm and hence a Euclidean domain. We unpack his proof here.

Let $R = \mathbb{Z}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Z} \} $ be a quadratic integer ring. It is said to be a Euclidean domain if there are:

(a) a well defined norm $N(\alpha) \leq N(\alpha\beta)$, and

(b) a division algorithm $\alpha = q \beta +r$ such that either (i) $r=0$, or (ii) $r \ne 0$ and $N(r) < N(\beta)$

with $\alpha,\beta \in R$, $\beta \ne 0$ and some quotient and remainder $q,r \in R$.

Let $\alpha = s+t\sqrt{11} \in R$. Define norm $N: R \setminus\{0\} \rightarrow \mathbb{Z}_{\geq 0}$ by setting $\alpha \mapsto |s^2-11t^2|$, which is well defined in $R$. Proof omitted. Norm $N$ is also well defined in $\mathbb{Q}[\sqrt{11}]$. Proof omitted. There exists $\alpha/\beta \in \mathbb{Q}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Q} \} \supset R $ with a numerator and a denominator $\alpha, \beta \in R$. Proof omitted.

Suppose the nearest lattice point to point $\alpha / \beta$ is $m+n\sqrt{11} \in R$ (i.e. $m,n \in \mathbb{Z}$) with $\alpha / \beta = (m+n\sqrt{11}) + (a+b\sqrt{11})$, $a,b \in \mathbb{Q}$, $-\frac{1}{2} \leq a \leq \frac{1}{2}$ and $-\frac{1}{2} \leq b \leq \frac{1}{2}$. We will consider lattice point $(m,n) = m+n\sqrt{11}$ and five others nearby. They, $(m+x,n+y)$ with $(x,y) \in \{ (0,0), (1,0), (-1,0), (2,0), (2,1), (5,2) \} $, are candidate quotients  to a division algorithm. By considering only positive values of $a,b$ such that $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, similar arguments could be made for the three other quadrants by symmetry. With $q=(m+x,n+y)$, rewrite $\alpha = q \beta +r$ as:

\begin{align}
    \alpha / \beta &= q + r/\beta    \\
    &= q + \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)
\end{align}

Condition (b)(ii) if $r \ne 0$, $N(r) < N(\beta)$ is true $\iff$ $N(r)/N(\beta) < 1$ $\iff$ $N(r/\beta) < 1$ $\iff$ $N \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)$ $= |(x-a)^2 - 11(y-b)^2| < 1$. Proof omitted.

Proposition: A division algorithm holds for $\mathbb{Z}[\sqrt{11}]$ if both inequalities:

\begin{align}
    U(x,y): &\ (x-a)^2 - 11(y-b)^2 < 1 \ \text{and}    \\
    V(x,y): &\ 11(y-b)^2 - (x-a)^2 < 1 \ \text{are true,}
\end{align}

for all $a,b \in \mathbb{Q}$, $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, and some $x,y \in \mathbb{Z}$.

Rewrite not $U(x,y)$ and not $V(x,y)$ as $P(x,y)$ and $N(x,y)$ respectively.

\begin{align}
    P(x,y): &\ (x-a)^2 \geq 1 + 11(y-b)^2    \\
    N(x,y): &\ 11(y-b)^2 \geq 1 + (x-a)^2
\end{align}

Suppose to the contrary that the contrapositive of the above proposition is true. It asserts that a division algorithm does not hold for $\mathbb{Z}[\sqrt{11}]$. Then either $P(x,y)$ or $N(x,y)$ is true for some $a,b$ and all $x,y$.

Consider $\mathbf{P(0,0)} : (-a)^2 \geq 1 + 11(-b)^2$. It implies $a^2 \geq 1$. Take the positive root and we have $a \geq 1$ which contradicts $0 \leq a \leq \frac{1}{2}$. Since $P(0,0)$ is false, $N(0,0)$ must hold.

$N(0,0): 11b^2 \geq 1+a^2$

Consider $\mathbf{P(1,0)} : (1-a)^2 \geq 1 + 11b^2$. By $N(0,0)$,

\begin{align}
(1-a)^2 &\geq 1 + 11b^2 \geq 2+a^2    \\
(1-a)^2 &\geq 2+a^2    \\
1-2a+a^2 &\geq 2+a^2    \\
-2a &\geq 1    \\
a &\leq -\frac{1}{2}
\end{align}

which contradicts $0 \leq a \leq \frac{1}{2}$. Since $P(1,0)$ is false, $N(1,0)$ must hold.

$N(1,0): 11b^2 \geq 1 + (1-a)^2$

https://www.desmos.com/calculator/rhrza9vqpx
https://www.desmos.com/calculator/pccntyashy

Consider $\mathbf{P(-1,0)} : (-1-a)^2 \geq 1 + 11b^2$. By $N(1,0)$,

\begin{align}
    (1+a)^2 &\geq 1 + 11b^2    \\
    (1+a)^2 - 1 &\geq 11b^2 \geq 1 + (1-a)^2    \\
    1 + 2a + a^2 - 1 &\geq 1 +  1 - 2a + a^2    \\
    4a &\geq 2    \\
    a &\geq \frac{1}{2}    \\
\end{align}

Then $a = \frac{1}{2}$ since $0 \leq a \leq \frac{1}{2}$. It follows that:

\begin{align}
    (1+\frac{1}{2})^2 - 1 &\geq 11b^2 \geq 1 + (1-\frac{1}{2})^2    \\
    \frac{5}{4} &\geq 11b^2 \geq \frac{5}{4}
\end{align}

But $b = \sqrt{ \frac{5}{44} } \notin \mathbb{Q}$ contradicts $b \in \mathbb{Q}$. Since $P(-1,0)$ is false, $N(-1,0)$ must hold.

$N(-1,0): 11b^2 \geq 1 + (1+a)^2$

Consider $\mathbf{N(2,0)} : 11b^2 \geq 1 + (2-a)^2$.

\begin{align}
    11b^2 &\geq 1 + (2-a)^2 \geq \frac{13}{4} > \frac{11}{4}    \\
    b^2 &> \frac{1}{4}
\end{align}

Take the positive root and we have $b > \frac{1}{2}$. But it contradicts $0 \leq b \leq \frac{1}{2}$. Since $N(2,0)$ is false, $P(2,0)$ must hold.

$P(2,0): (2-a)^2 \geq 1 + 11b^2$

Consider $\mathbf{N(2,1)} : 11(1-b)^2 \geq (2-a)^2$. By $N(2,0)$,

\begin{align}
    11(1-b)^2 &\geq 1 + (2-a)^2 \geq 1 + 11b^2    \\
    11 - 22b + 11b^2 &\geq 1 + 11b^2    \\
    9 &\geq 22b    \\
    b &\leq \frac{9}{22}
\end{align}

By $N(-1,0)$,

\begin{align}
    11b^2 &\geq 1 + (1+a)^2 \geq 2 > \frac{81}{44}    \\
    $b^2 &> (\frac{9}{22})^2
\end{align}

Take the positive root and we have $b > \frac{9}{22}$. But it contradicts $b \leq \frac{9}{22}$. Since $N(2,1)$ is false, $P(2,1)$ must hold.

$P(2,1): (2-a)^2 \geq 1 + 11(1-b)^2$

By $P(2,1)$ and $N(-1,0)$, they imply:

\begin{align}
    (2-a)^2 &\geq 1 + 11(1-b)^2    \\
    (2-a)^2 &\geq 1 + 11 -22b + 11b^2    \\
    (2-a)^2 -12 + 22b &\geq 11b^2 \geq 1 + (1+a)^2    \\
    4 -4a + a^2 -12 + 22b &\geq 2 + 2a + a^2    \\
    22b &\geq 10 + 6a    \\
    11b &\geq 5 + 3a
\end{align}

Out of interest. Take the least possible value to the RHS; it implies $11b \geq 5$. Likewise, take the greatest possible value to the LHS; it implies $11 \geq 10 + 6a$ $\implies$ $1 \geq 6a$.

Consider $\mathbf{N(5,2)}: 11(2-b)^2 \geq 1 + (5-a)^2$. By $P(2,1)$, it implies:

\begin{align}
    11(2-b)^2 &\geq 1 + (5-a)^2    \\
    11(2-b)^2 &\geq 26 -10a +a^2    \\
    11(2-b)^2 &\geq (4 - 4a + a^2) + 22 -6a    \\
    11(2-b)^2 - 22 + 6a &\geq (2-a)^2 \geq 1 + 11(1-b)^2    \\
    22 - 44b + 11b^2 + 6a &\geq (2-a)^2 \geq 12 -22b + 11b^2    \\
    10 + 6a &\geq 22b    \\
    5 + 3a &\geq 11b
\end{align}

By $11b \geq 5 + 3a$ and $5 + 3a \geq 11b$, we have $5 + 3a = 11b$. Retrace backwards the argument leading to $5 + 3a \geq 11b$. We have $11(2-b)^2 - 22 + 6a = (2-a)^2 = 1 + 11(1-b)^2$. Take the middle and the RHS of this chain of equality and name it $P'(2,1)$. Establishing the following two equalities are not strictly necessary for this proof. Since Oppenheim (1934) mentioned them, we will include them here. Take the LHS and the middle of the above chain of equality, we have $11(2-b)^2 - 22 + 6a = (2-a)^2$ $\implies$ $11(2-b)^2 = (4 - 4a + a^2) + 22 -6a$ $\implies$ $11(2-b)^2 = 26 -10a +a^2$ $\implies$ $11(2-b)^2 = 1 + (5-a)^2$. Name it $N'(5,2)$. Retrace backwards the argument leading to $11b \geq 5 + 3a$. We have $(2-a)^2 -12 + 22b = 11b^2 = 1 + (1+a)^2$. Take the middle and the RHS of this chain of equality and name it $N'(-1,0)$. And we have:

\begin{align}
    P'(2,1)&: (2-a)^2 = 1 + 11(1-b)^2    \\
    N'(5,2)&: 11(2-b)^2 = 1 + (5-a)^2    \\
    N'(-1,0)&: 11b^2 = 1 + (1+a)^2
\end{align}

By $N'(-1,0)$ and $5 + 3a = 11b$,

\begin{align}
    11b^2 &= 1 + (1+a)^2    \\
    (11b)^2 &= 11 + 11(1+a)^2    \\
    (5+3a)^2 &= 11 + 11(1+a)^2    \\
    25 + 30a + 9a^2 &= 22 + 22a + 11a^2    \\
    2a^2 - 8a &= 3 > 0
\end{align}

It implies $a<0$ or $a > 4 > \frac{1}{2}$ but they contradict $0 \leq a \leq \frac{1}{2}$. And hence $N(5,2)$ is false, $P(5,2)$ must be true.

$P(5,2): (5-a)^2 \geq 1 + 11(2-b)^2$

Consider $\mathbf{P(5,2)}$, by $N'(-1,0)$, it implies:

\begin{align}
    (5-a)^2 &\geq 1 + 11(2-b)^2    \\
    25 - 10a + a^2 &\geq 1 + 11(2-b)^2    \\
    23 - 12a + 1 + (1 + 2a + a^2) &\geq 1 + 11(2-b)^2    \\
    23 - 12a + \big( 1 + (1 + a)^2 \big) &\geq 1 + 11(2-b)^2    \\
    23 - 12a + 11b^2 &\geq 45 - 44b + 11b^2    \\
    44b &\geq 22 +12a    \\
    22b &\geq 11 + 6a
\end{align}

Possible range of the LHS is $[0,11]$ and that of RHS is $[11,14]$. Therefore, they only possible values are $b = \frac{1}{2}$ and $a=0$. Substitute $b = \frac{1}{2}$ back into $P(5,2)$, and it asserts,

$(5-a)^2 \geq 1 + 11(2- \frac{1}{2} )^2 = \frac{103}{4} > (\frac{10}{2})^2 = 5^2$

Take the positive root (since negative root $a-5 \geq 0$ is false) and we have $5-a > 5$ $\implies$ $a < 0$. But it contradicts $a = 0$. And hence $P(5,2)$ is false.

Since both $N(5,2)$ and $P(5,2)$ are false, they contradicts the proposition that either $P(x,y)$ or $N(x,y)$ is true for some $a,b$ and all $x,y$. And hence, the proposition that a division algorithm does not hold is false. Therefore, $\mathbb{Z}[\sqrt{11}]$ has a division algorithm and hence a Euclidean domain. $\ \ \ \ \blacksquare$


问:
The mean is different from the mode and the median
I am reading a study in which the mean age of a population is 41 years. The mode and median are however both 35 years. I do not have access to the raw data. How I can interpret the difference between the mean and the mode/median?</p>

答:
Here is a very primitive example: imagine the dataset has only five entries:

34, 35, 35, 48, 53

The mode and median are 35 and the mean is 41. This basic histogram shows what you might expect: an overshoot of a small number of entries with ages that are much higher than the mode.


答:
This can be rewritten

$
\int_0^{\pi} \frac{1}{\sqrt{a^2 + b^2 + c^2 +1 - 2a \cos t - 2b \sin t}} dt
$

which has the form

$
A \int_0^{\pi} \frac{1}{\sqrt{1 + B \sin(t+\phi)}} dt
$

for appropriate constants $A$, $B$, and $\phi$. I just tried the indefinite integral

$
\int \frac{1}{\sqrt{1 + B \sin(t+\phi)}} dt
$

in Wolfram Alpha. It provided a closed form answer.

I hope this helps.

Title: How to solve this integral

Body:
Wolfram integrator is unable to solve this:

$$
 \int_0^\pi \frac{1}{\sqrt {(a-\cos(t))^2+(b-\sin(t))^2+c^2}}\,\mathrm dt
$$

Any suggestions? Thanks!


问:
a little riddle, prime numbers and modulo
Is there a prime number $k$ , such that $k$  is not $1\bmod3$ , not $2
\bmod4$ , not $3\bmod5$ , not $5
bmod7$ etc..
Meaning, $k$ is not $n\bmod(n+2)$ for any $n=1,....,100000000$ ?

答:
Instead of $100000000$ we will write $N$ to make this more general.

From Dirichlet's theorem, we can choose $p$ (your $k$) to be of the form $a(N+2)!-1$. Suppose $p \equiv m \pmod{m+2}$ for some $m \in \{1,\ldots,N\}$. Then $p+2 \equiv 0 \pmod{m+2}$, so $m+2 \mid p+2 = a(N+2)!+1$. But we also know that $m+2 \mid (N+2)!$, and that's a clear contradiction, so our $p$ satisfy the statement of the problem.