Title: Rewriting function

Body:
Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.

This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$

However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.

EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.

Body:
Third binomial formula: $(a+b)(a-b)=a^2-b^2$
$$\frac{2}{1+\sqrt{x+1}}=\frac{2(1-\sqrt{x+1})}{(1+\sqrt{x+1})(1-\sqrt{x+1})}=\frac{-2 + 2 \sqrt{x+1}}{1^2-(\sqrt{x+1})^2}=\frac{-2 + 2 \sqrt{x+1}}{x}$$


问:
How one vector space can have more than one basis?
What I know is that a  basis contains the maximum number of linearly independent vectors in the vector space that spans the vector space and the # of basis vectors is the dimension of the vector space, if a vector space have two basis then wouldn't  the dimension of vector space change?</p>

答:
Lets consider some practical situations where a chance of basis doesn't change the dimension. Using $\mathbb{R}^3$ to model the usual three dimensional space. This happens in practice when we change units of measure, say from meter to centimeters. Something that was at the point $(1,0,0)$ when measured in meters is now at the point $(100,0,0)$ in centimeters and likewise for the other axes. Now the units we measured in are arbitrary and don't change the dimension of the space even though they change the coordinates we measured them in.

Another arbitrary choice is what direction the axes point in. Someone south of the equator might orient up to be south while someone on the north side might choose north for up. They didn't change the dimension of the space by choosing a different direction to measure in. In fact we don't even have to have the axes at right angles to each other. Say for example we tilt the canonical basis vectors a small distance towards their center of mass. While the basis is now skew we can still map out any point in three dimensions using these frame.

So while we change the basis vectors we don't change the space in which they exist and so the dimension stays the same. The dimension of the space tells us that we need so many vectors to uniquely identify each point with a unique coordinate. If we don't have enough we can't get to every point and if we have too many some of them are redundant. The dimension formalizes this geometric notion algebraically so that it can be calculated in other spaces that might not behave in the same way as $\mathbb{R}^3$ but still give us so me idea about how many coordinates we need to label every point in the space.

ContentLicense: CC BY-SA 4.0


问:
Collecting cards
The following problem is from Princeton University Mathematics competition.

Kelvin and Quinn are collecting trading cards; there are 6 distinct cards that could appear
in a pack. Each pack contains exactly one card, and each card is equally likely. Kelvin buys
packs until he has at least one copy of every card, then he stops buying packs. If Quinn is
missing exactly one card, the probability that Kelvin has at least two copies of the card Quinn
is missing is expressible as $\frac{m}{n}$ for coprime positive integers $m,n$. Determine $m + n.$

Assuming that Kelvin stops after getting $N$ cards, in which he has exactly one copy of the card, say A,  Quinn is missing, the computation of the probability of this event is a bit messy. The official solution is given below and appears incomplete:

However, we also have the probabilities for each of the new cards that appear. This
is $\frac{5}{6} \cdot \frac{4}{6} \cdots \frac{1}{6} \cdot \frac{1}{6}$ since we are fixing when A appears so we have two copies of $\frac{1}{6}$, one for A
and one for the last card that isn't A, Thus, in total, the probability is $\frac{6^5}{5!\cdot (6-n)} \cdot \frac{5!}{6^6} = \frac{1}{(6-n)\cdot 6}$

We now need to sum this over all possible n, giving us
$$ \sum_{n=0}^6 \frac{1}{6(n-6)} = \frac{49}{120} $$
Since we computed the complement, the probability we want is $1 - \frac{49}{120} = \frac{71}{120}$ and $m+n = 191$.

Some text appears to be missing at the beginning of the official solution.

答:
I answered it here: Probability of exactly 1 given everything appears. Here is a solution:

"Consider pulling cards until you pull  twice. Note that the order of  is equally likely to be the first, second, ..., sixth. Now, condition on the order of , let it be the th unique card. Let us assume that your position  is not equal to 6. Now,  is again equally likely to be the last card among the remaining (6−) cards, giving a probability of  being last 1/(6−+1).

For  to be drawn once in your scenario,  had to be the last card in the first or the second draw (do you see why?), giving the final answer (1/6)+(1/6)∑5=11/(6−+1), which matches the formula they gave. I believe they used a different counting argument, but since their solution is incomplete, and this one is simpler, you might as well use this.

As an important side note, I think the better way to think about this problem to understand all the symmetries and independences is by considering a Poisson splitting (that is how I solved it). It is a hard problem without Poisson processes for sure."

Title: Collecting cards


问:
Converting rectangular form complex numbers to polar form in different quadrants
I know how to convert a complex number from rectangular form to polar form, but I can't understand when should I add or subtract $\pi$ from/to arctan in different quadrants.

Can some one explain this for all quadrants? (maybe except for I)

答:
Normally, I'm not supposed to answer, because the OP hasn't shown any work. But this is a different situation, so if I get downvoted, okay.  The way that I was taught ("An Introduction To Complex Function Theory" : Bruce Palka) you are supposed to use the sine and cosine functions.

Let $z = (x + iy)$, where $x$ and $y$ are not both $= 0$.

Let $r = +\sqrt{x^2 + y^2}$.

Then, within a modulus of $(2\pi)$, there will be a unique angle $\theta$, such that

$\cos(\theta) = \frac{x}{r}$ and $\sin(\theta) = \frac{y}{r}.$

Then, $(x + iy) = z = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$

Therefore, you never have to concern yourself with the tangent function, or wondering which quadrant $\theta$ is in.  The quadrant is determined by whether (for example) $\frac{x}{r}$ and/or $\frac{y}{r}$ is positive.

Addendum

Responding to the OP's comment/question.
$z = -3 + i\sqrt{3}$.

$r^2 = (-3)^2 + \left(\sqrt{3}\right)^2 = 9 + 3 = 12 \implies $

$r = 2\sqrt{3}.$

Therefore, you are looking for $\theta$ such that

$\cos(\theta) = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$ and

$\sin(\theta) = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$.

Looking at the sign of $\cos(\theta)$ and $\sin(\theta)$, you immediately deduce that $\theta$ is in the 2nd quadrant.  Then, for this particular problem, you can simply draw the unit circle, identify the corresponding angle in the first quadrant, and then infer the value of $\theta$ in the 2nd quadrant.

For example, you know that $\cos(\pi/6) = \frac{\sqrt{3}}{2}.$

Therefore, you can infer, by symmetry, that
$\theta$ must be $(\pi - \pi/6) = (5\pi/6)$.

You could have instead focused on the fact that $\sin(\pi/6) = (1/2)$, and reached the same inference.  Using symmetry around a drawing of the unit circle is a good way of stretching your intuition to easily handle situations like this.

Anyway, once you determine that $(\theta = 5\pi/6)$ and
$(r = 2\sqrt{3})$, you know that

$-3 + i\sqrt{3} = 2\sqrt{3}[\cos(5\pi/6) + i\sin(5\pi/6)].$

For a more systematic way of handling situations like this,
first see this wikipedia article.

Then, you could (for example) use the appropriate formula to deduce that $\sin(\pi - \theta) = \sin(\theta).$

Then, you can easily attack a situation like this sytematically by first learning (not memorizing) the sine and cosine functions for such special angles as $\{0, \pi/6, \pi/4, \pi/3, \pi/2\}$, and the corresponding angles in each of the other 4 quadrants.

Actually, I used the verb learn to indicate that once you've got the 1st quadrant down, you should be able to infer the trig functions that correspond to the  special angles in the other quadrants, simply by looking at the unit circle, and making the intuitive leaps, around ideas involving symmetry.


问:
Rewriting function
Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.

This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$

However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.

EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.

答:
You can start with $f(x)$ by doing the following:

$$
\begin{alignat}{1}
f(x) &= \frac{-2 + 2 \sqrt{x+1}}{x} \cdot \frac{(-2-2\sqrt{x+1})}{(-2-2\sqrt{x+1})} = \frac{4-4(x+1)}{-2x-2x\sqrt{x+1}}=\frac{-4x}{-2x\cdot(1+\sqrt{x+1})}
\\ &= \left(\frac{-2x}{-2x}\right)\cdot \left( \frac{2}{1+\sqrt{x+1}}  \right) = \frac{2}{1+\sqrt{x+1}}=g(x)
\end{alignat}
$$

But what's the idea behind it?

When solving this type of problem, look at the structure of the functions you are working with. For the functions in your question:

$$
\newcommand{\altfrac}{\genfrac{}{}{0pt}{}}
\begin{alignat}{1}
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
&\altfrac{\leftarrow \text{Square root here}~~~~~}{\leftarrow \text{No square root here}}
\\[8pt]
g(x) = \frac{2}{1+\sqrt{x+1}}
&\altfrac{\leftarrow \text{No square root here}}{\leftarrow \text{Square root here}~~~~~}
\end{alignat}
$$

Going from $f(x)$ to $g(x)$, you need to make the square root appear in the denominator, so a good start would be to rationalize the numerator of $f(x)$.

Similary, from $g(x)$ to $f(x)$, you need to make the square root appear in the numerator, so a good start would be to rationalize the denominator of $g(x)$, as done in @vitamin d answer.

Title: Rewriting function


问:
Functions $f$ having the intermediate value property such that $(f(x))^2=x^2$ for all $x\in \mathbb{R}$
It seems "obvious" that there are only four such functions, namely $x,-x,|x|,-|x|$, but I cannot see the rigorous justification for that.

答: Any other function must have two points $x,y$ with $f(x)=x$, $f(y)=-y$, and $xy\gt 0$. Now apply the intermediate value property to find a point $t$ between $x$ and $y$ with $f(t)=0$; since $t$ can't be $0$ ($x$ and $y$ are both on the same side of $0$), that's your contradiction.


问:
Bernoulli Measure and Joint PDF: Are these Wiki definitions correct?
I have a question on the Bernoulli Measure and Joint PDF.  Using the Wiki definitions my results are not consistent.  I first present my understanding with definitions and then ask questions.

Wiki defines the Bernoulli Measure as:

a)
$$
P(X_1 = x_1, X_2 = x_2, ..., X_n = x_n) = p^k(1-p)^{n-k}
$$

where $k$ is # of successes in $n$ trials.  Question this omits the binomial coefficient.  Is it correct?

The above measure is a Joint PDF which Wiki defines as follows:

b)
$$
\rho_{X_1,...,X_n}(x_1, ..., x_n) = P(X_1 = x_1 \land ... \land X_n = x_n) = P(X_1 = x_1) P(X_2 = x_2 | X_1 = x_1) P(X_3 = x_3 | X_1 = x_1, X_2 = x_2) ... P(X_n = x_n | X_1 = x_1, X_2 = X_2, ..., X_{n-1} = x_{n-1})
$$

The RHS above is from application of the chain rule of conditional probability.  Using the following equality $(X = z) = \{s \in \Omega : X(s) = z\}$ it's my understanding this equals:

c)
$$
\rho_{X_1, ..., X_n}(a_1, ... , a_n) = P(X_1 = a_1, ... , X_n = a_n) = P(\{s \in \Omega : X_1(s) = a_1 \land ... \land X_n(s) = a_n\}) = P(\{s \in \Omega : X_1(s) = a_1\} \cap ... \cap \{s \in \Omega : X_n(s) = a_n\}
$$

Question is form (c) correct?

I have doubts on (c).  For example, if there is a Bernoulli Measure with sample space, probabilities and random variable:

d)
$$
X_i : \Omega \rightarrow \mathbb{R}
$$
$$
\Omega = \{\epsilon_0, \epsilon_1\}
$$
$$
X_i(\epsilon_0) = 1, X_i(\epsilon_1) = 0
$$
$$
P(\epsilon_0) = P(X_i = 1) = p, P(\epsilon_1) = P(X_i = 0) = 1 - p
$$

Now considering two steps in the measure:

e)
$$
P(X_1 = 0, X_2 = 1) = p^1 (1 - p)^{2 - 1} = p (1 - p)
$$

but when substituting events from the sample space there is an inconsistency between above and what follows.  This makes me wonder if my definition of the JPDF or Bernoulli measure is incorrect:

f)
$$
P(X_1 = 0, X_2 = 1) = P(\{\epsilon_0\} \cap \{\epsilon_1\}) = P(\{\}) = 0
$$

答:
"This makes me wonder if my definition of the JPDF or Bernoulli measure is incorrect": They are both correct, except you are talking about joint PMFs, not joint PDFs. In fact, in the Wiki link you shared, the joint PMF is already defined as $$p_{X_!,\dots,X_n}(x_1,\dots,x_n)=P(X_1=x_1\cap\cdots\cap X_n=x_n)$$
in Eq.4, which is exactly the same as (c) (maybe you missed this definition, it's right above equation (b), which you wrote above, in the link). (b) is just an equivalent form and you will see that form (c) is much more common than (b).

For the Bernoulli variables, the problem lies in your sample space $\Omega$ and the probability space in general. The probability space $(\Omega,\mathcal F:=2^\Omega,P)$ defined in your question makes sense only if you consider a single Bernoulli trial. If we have a sequence $\{X_i\}_{i=1,\dots,n}$ of $n$ i.i.d. Bernoulli trials, the sample space, say $\Omega'$, should be the product space $\Omega'=\{\epsilon_0,\epsilon_1\}^n$. In our case, $n=2$ and thus, $\Omega'=\{\epsilon_0,\epsilon_1\}\times \{\epsilon_0,\epsilon_1\}=\{(\epsilon_0,\epsilon_0),(\epsilon_0,\epsilon_1),(\epsilon_1,\epsilon_0),(\epsilon_1,\epsilon_1)\}.$
So, the probability space is the triple $(\Omega',\mathcal F',P'),$ where  $\mathcal F':=\mathcal F\otimes \mathcal F$ (tensor product) and $P':=P\times P$ is the product measure.

Considering this probability space, we have \begin{align}
P'(X_1 = 0, X_2 = 1)&=P'(\{(\epsilon_1,\epsilon_0),(\epsilon_1,\epsilon_1)\}\cap \{(\epsilon_0,\epsilon_0),(\epsilon_1,\epsilon_0)\})\\ &=P'(\{(\epsilon_1,\epsilon_0)\})\\&=P(\epsilon_1)P(\epsilon_0)=(1-p)p
\end{align}
as required.


问:
Expressing “If $8$ does not divide $x^2−1$, then $x$ is even.” in formal language and negate it.
I am trying to solve the question 4.18) on the book Reading, Writing, and Proving: A Closer Look at Mathematics 2nd ed. on page 44.

The following is the question.

For an integer $x$ consider the statement: “If $8$ does not divide $x^2−1$, then $x$ is even.”

(a) State an appropriate universe for $x$.

(b) Write the statement in symbols.

(c) Negate the statement.

The following is my attempted solution.

a) An appropriate universe for $x$ is $\mathbb{R}$.

b) $\forall x,((x\in \mathbb{Z} \wedge \forall y,(y \in \mathbb{Z} \rightarrow y\ne \frac{x^2 - 1}{8})) \rightarrow \exists m,(m\in \mathbb{Z} \wedge x = 2m)).$

c) $\exists x,((x \in \mathbb{Z} \wedge  \forall y,(y \in \mathbb{Z} \rightarrow y\ne \frac{x^2 - 1}{8}) \wedge  \forall m,(m \in \mathbb{Z} \rightarrow x \ne 2m)).$ The statement can be translated in plain English language as "There exists a real number $x$ such that $x$ is an integer, $8$ does not divide $x^2 -1$, and $x$ is not even."

I need help verifying my answer.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 44). New York: Springer.

答:
As lulu noted, for a) you could nominate $\Bbb Z$, which would make the statement true, whereas your choice makes its negation true instead. Having said that, whether the statement is true or false isn't the point of the exercise.

In a universe $U$ we can write $\forall x(8\nmid x^2-1\to2|x)$ for (b); we need not mention membership of $U$. Even if we did, we'd expand $\forall x(\phi(x))$ as $\forall x\in U(\phi(x))$ or $\forall x(x\in U\to\phi(x))$, no commas needed. These $U$-mentioning approaches have two related disadvantages: they require $U$ to be an object that can appear in this formalism, and if we're making $x\in U$ an antecedent it raises the question of what $\forall$ runs over.

With my recommended concision, the negation can be written in (c) as $\exists x(8|x^2-1\land 2\nmid x)$.


答:
a. "$x$ is an integer": $x\in \Bbb Z$

b. Split by high-level/low-level construct on divisibility:

(1) "$8$ does not divide $x^2-1$": $8\not\mid (x^2-1)\quad$ OR $\quad \not\exists k\in\Bbb Z:8k=x^2-1 \quad$ OR $ \quad \forall k\in\Bbb Z:8k\neq x^2-1$

(2) "if... then": $\implies$

(3) "$x$ is even": $2\mid x \quad$ OR $\quad  \exists j\in\Bbb Z:2j=x$

The use of division in an integer universe is a little suspect, hence the use of multiplication above.

The claim is that this is always true, so $\forall x\in \Bbb Z:(\forall k\in\Bbb Z:8k\neq x^2-1) \implies (\exists j\in\Bbb Z:2j=x)$

c. The claim is in fact true, so negation will produce a false claim; you seem to have made a valid negation of your preceding statement, although there are several ways to do this.

Note there are variations on symbols and preferred usage, which changes over time also, so don't be surprised to see some slightly different answers both here and elsewhere.

Title: Expressing “If $8$ does not divide $x^2−1$, then $x$ is even.” in formal language and negate it.

Body:
I am trying to solve the question 4.18) on the book Reading, Writing, and Proving: A Closer Look at Mathematics 2nd ed. on page 44.

The following is the question.

For an integer $x$ consider the statement: “If $8$ does not divide $x^2−1$, then $x$ is even.”

(a) State an appropriate universe for $x$.

(b) Write the statement in symbols.

(c) Negate the statement.

The following is my attempted solution.

a) An appropriate universe for $x$ is $\mathbb{R}$.

b) $\forall x,((x\in \mathbb{Z} \wedge \forall y,(y \in \mathbb{Z} \rightarrow y\ne \frac{x^2 - 1}{8})) \rightarrow \exists m,(m\in \mathbb{Z} \wedge x = 2m)).$

c) $\exists x,((x \in \mathbb{Z} \wedge  \forall y,(y \in \mathbb{Z} \rightarrow y\ne \frac{x^2 - 1}{8}) \wedge  \forall m,(m \in \mathbb{Z} \rightarrow x \ne 2m)).$ The statement can be translated in plain English language as "There exists a real number $x$ such that $x$ is an integer, $8$ does not divide $x^2 -1$, and $x$ is not even."

I need help verifying my answer.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 44). New York: Springer.


问:
What can be said about holomorphy and convergence radius of $\sum_{n=0}^{\infty}a_{n+k}z^n$
Consider a function f(z) holomorphic in some neighbourhood of z = 0 expands at zero into
power series $\sum_{n=0}^{\infty}a_nz^n$ with convergence radius R. Fix a natural number k. What can be said about holomorphy and convergence radius of $\sum_{n=0}^{\infty}a_{n+k}z^n$?

I know that according to Cauchy-Hadamard theorem $lim_{n\to \infty}\sqrt[n]{|a_n|}=\frac{1}{R}$, hence I want to find what is the $lim_{n\to \infty}\sqrt[n]{|a_{n+k}|}$. I think it is a subsequence then it is the same limit but I'm not sure. Can you help me?

Thank you in advance for your help!

答:
Notice that
\begin{aligned}
|a_{n+k}|^{1/n}&=\exp\Big(\frac{1}{n}\log(|a_{n+k}|)\Big)\\
&=\exp\Big(\frac{n+k}{n}\frac{1}{n+k}\log(|a_{n+k}|)\Big)\\
&=\exp\Big(\frac{n+k}{n}\log\big(|a_{n+k}|^{1/(n+k)}\big)\Big)
\end{aligned}

From where it is easy to check that
$$\limsup_n\sqrt[n]{|a_{n+k}|}=\limsup_n\sqrt[n]{|a_n|}$$
That is, the series you mention in the OP have the same radius of convergence, which is not surprising as some of the comments explain.


问:
order of magnitude analysis
Could anyone explain how to keep track of the error terms when solving an integral approximately? For example consider to evaluate the integral $\int_0^{\pi/2}\frac{\cos^2xdx}{x^2+\epsilon^2}$ as $\epsilon \rightarrow0$ correct to two terms I would split the range at $0\ll\delta\ll\pi/2$ and evaluate $\int_0^\delta$ and $\int_\delta^{\pi/2}$ separately.

In the small region I would rescale by $x=\epsilon u$ to get:

$${1\over \epsilon}\int_0^{\delta/\epsilon}\frac{\cos^2(\epsilon u)}{u^2+1}du={1\over \epsilon}\int_0^{\delta/\epsilon}\frac{1}{u^2+1}(1-4\epsilon^4u^2)dx={1\over \epsilon}\int_0^{\delta/\epsilon}\frac{1}{u^2+1}dx+O(what)$$

What is the order of the terms left out?

答: If I remember correctly O details items of order something and smaller. In your case you are neglecting terms of order $\epsilon^2$ and smaller since this is the next smallest term in the squared cosine expansion ( I want to mention I think you need to change the power on the  $\epsilon$ so it matches the power of u)</p>

ContentLicense: CC BY-SA 3.0

Title: order of magnitude analysis


问:
Find minimum volume bounded by three planes and another plane tangent to an ellipsoid.
I am having trouble with a problem that I found in a multivariate calculus book. The problem is :

Let $a$, $b$, and $c$ be nonzero real numbers. Find the minimum volume bounded by the planes given by $x=0$, $y=0$, $z=0$, and a plane that is tangent to the ellipsoid given
by $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$.

I suspect that solving this problem would involve the use of the Lagrange multiplier method.

In my solution so far I start with defining a function :
\begin{equation}
h(x,y,z) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} - 1
\end{equation}
So that the ellipse (call it $E$) is given by :
\begin{equation}
E = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; h(x,y,z) = 0 \}
\end{equation}
We have the partial derivatives of $h$ as :
\begin{align}
h_{x}(x,y,z) & = \frac{2x}{a^{2}}\\
h_{y}(x,y,z) & = \frac{2y}{b^{2}}\\
h_{z}(x,y,z) & = \frac{2z}{c^{2}}
\end{align}
Now let $(x_{0},y_{0},z_{0}) \in E$. The tangent plane to E at $(x_{0},y_{0},z_{0})$ can be defined implicitly using a function $T$ :
\begin{align}
T_{x_{0},y_{0},z_{0}}(x,y,z) & = h_{x}(x_{0},y_{0},z_{0})(x-x_{0}) + h_{y}(x_{0},y_{0},z_{0})(y-y_{0}) + h_{z}(x_{0},y_{0},z_{0})(z-z_{0}) \\
  & = \frac{2x_{0}(x-x_{0})}{a^{2}} + \frac{2y_{0}(y-y_{0})}{b^{2}} + \frac{2z_{0}(z-z_{0})}{c^{2}}\\
\end{align}
Where the tangent plane (say $Q(x_{0},y_{0},z_{0})$) is :
\begin{equation}
Q(x_{0},y_{0},z_{0}) = \{(x,y,z) \in \mathbb{R}^{3} \; : \; T_{x_{0},y_{0},z_{0}}(x,y,z) =0 \}
\end{equation}
So I suspect the point of the problem is to find the $(x_{0},y_{0},z_{0}) \in E$ so that
the volume bounded by the planes defined by $Q(x_{0},y_{0},z_{0})$ , $x = 0$, $y=0$, and $z = 0$ is minimized. The problem that I have at this point is how to get an expression for this volume.

Can anyone help with this ?

答:
Hint.

With the help of Lagrange multipliers

Calling $\Pi\to n_1 x+n_1 y+n_3 z = 1$ we have the lagrangian

$$
L=\frac 16 \frac{1}{n_1n_2n_3} + \lambda(n_1 x+n_1 y+n_3 z - 1)+\mu\left(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} - 1\right)
$$

and the stationary points are the solutions for

$$
\nabla L = 0 = \cases{
\begin{array}{l}
 \lambda  x-\frac 16\frac{1}{n_1^2 n_2 n_3} \\
 \lambda  y-\frac 16 \frac{1}{n_1 n_2^2n_3} \\
 \lambda  z-\frac 16 \frac{1}{n_1 n_2 n_3^2} \\
 \frac{2 \mu  x}{a^2}+\lambda  n_1 \\
 \frac{2 \mu  y}{b^2}+\lambda  n_2\\
 \frac{2 \mu  z}{c^2}+\lambda  n_3 \\
 n_1 x+n_2 y+n_3 z-1 \\
 \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \\
\end{array}
}
$$

now as $\lambda, \mu$ are generic multipliers, making $\lambda_0 = 6\lambda n_1n_2n_3,\ \ \mu_0 = 6\mu n_1n_2n_3$ we follow with

$$
\nabla L = 0 = \cases{
\begin{array}{l}
 \lambda_0  x-\frac{1}{n_1} \\
 \lambda_0  y-\frac{1}{n_2} \\
 \lambda_0  z-\frac{1}{n_3} \\
 \frac{2 \mu_0  x}{a^2}+\lambda_0  n_1 \\
 \frac{2 \mu_0  y}{b^2}+\lambda_0  n_2\\
 \frac{2 \mu_0  z}{c^2}+\lambda_0  n_3 \\
 n_1 x+n_2 y+n_3 z-1 \\
 \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \\
\end{array}
}
$$

etc.


问:
Find $P(2 \leq |X| + |Y| \leq 3)$ where $X, Y \sim N(0, 4)$ are independent and normally distributed
Suppose that $X, Y \sim N(0, 4)$ are independent and normally distributed random variables.

My question is how to find $P(2 \leq |X| + |Y| \leq 3)$?

Probably, the first step is to try to solve a more general task and find the distribution of $Z := |X| + |Y|$, i.e. find $F_Z(t) = P(Z \leq t) = P(|X| + |Y| \leq t)$ for $t > 0$.

The straightforward way would be to compute the integral:

$$\int\limits_{|x| + |y| \leq t} \dfrac{1}{8\pi} e^{-(x^2 + y^2)/4}dxdy$$

This task seems to be non-trivial because the resulting integrals end up being hard to calculate. Assuming the function $\Phi(t) = \int\limits_{-\infty}^{t} \dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$ is known, the integral above requires computing something like $\int\limits_{0}^{t}e^{-x^2/4}\Phi(x) dx$ which I'm not sure how to compute with simple calculus tools that I know.

Maybe it is easier than I think or maybe there is just a different better way of solving this.

Any help would be greatly appreciated.

答:
As asked for in a comment I'll try to elaborate on the "circular symmetry" argument given in this answer.

Basically we are given two normally distributed, independent random variables $X,Y \sim N(0,4)$ and want to find the distribution function of $|X|+|Y|$. I.e. for $a>0$ we want to calculate
$$F_{|X|+|Y|}(a) = P(|X|+|Y|\leq a) $$
This means that we want to find the probability of $(X,Y)$ to lie within the set
$$A = \{(x,y)\in\mathbb{R}^2 \mid |x|+|y|\leq a\} .$$
Geometrically $A$ is a square with corners at $(-a, 0)$, $(0, -a)$, $(a, 0)$ and $(0, a)$. For example, with $a=10$:

Now, because $X$ and $Y$ are independent, the random vector $(X,Y)$ is jointly normal-distributed with covariance matrix $\begin{pmatrix}4&0\\0&4\end{pmatrix}$.
This means that it is rotationally invariant: for every angle $\alpha$, the rotated vector
$$\begin{pmatrix}
\cos \alpha& -\sin \alpha \\
\sin \alpha & \cos \alpha \\
\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$
has the same distribution as the vector $(X,Y)$ itself. (This can be checked using how a multivariate normal distribution changes under a linear transformation; note that the rotation matrix is orthogonal.) The symmetry can also be seen geometrically by plotting the density:

Hence, denoting the rotation by $\alpha$ as $R_\alpha$,
$$P((X,Y) \in A) = P(R_{-\alpha}(X,Y) \in A) = P((X,Y) \in R_\alpha(A)) .$$

We can thus equivalently compute the probability of $(X,Y)$ lying in the rotated square $A'$. Now choose $\alpha=45^\circ$; rotating the square by $45^\circ$ yields a square with axis parallel sides (the side lengths remains the same: $\sqrt{2}a$)

And now
$$P((X,Y) \in A') = P\left(|X|\leq \frac{\sqrt{2}a}{2}, |Y|\leq \frac{\sqrt{2}a}{2}\right) = \left(F_{N(0,4)}\left(\frac{\sqrt{2}a}{2}\right) - F_{N(0,4)}\left(-\frac{\sqrt{2}a}{2}\right)\right)^2 .$$


问:
derivation of equation of a hyperbola from the conic section
This is a method I tried of working out the cartesian form of a hyperbola using its definition as the cross section of a vertical cone?

I started by noting that the width of the hyperbola is the sine of a circle at some point within the cone. The radius of the circle is proportional to the height of the cone, which is the length $y$ down the hyperbola plus some constant $a$. Therefore the radius of the circle is $k(y+a)$.

The centre of the hyperbola is always the same distance from the centre of the cone at a given height $y$, at $ka$. So cosine of the angle between the middle and edge of the hyperbola at some height $y$ is $\frac{ka}{k(y+a)} = \frac{1}{1+\frac{y}{a}}$. So the width of the hyperbola $x$ at height $y$ is $x =k(y+a)\sqrt{1-\frac{1}{(1+\frac{y}{a})^2}}$ by relating the sines to cosines. The resulting equation is $x^2 = k^2(y+a)^2(1-\frac{1}{(1+\frac{y}{a})^2})$,

This simplifies to $x^2 = k^2(y^2 + 2ya)$. Is this a hyperbolic curve. How do I rearrange it to the standard form?

I understand that I'm not being very clear but I can't find a software on which to draw the problem properly. Anyway, I hope that the line of reasoning has come across.

答:
$-\frac{x^2}{k^2} + (y+a)^2 = a^2$

$-\frac{x^2}{a^2k^2} + \frac{(y+a)^2}{a^2} = 1$

This is the equation of a hyperbola on the vertical transverse axis with vertex $a$, asymptote, $y=\frac{1}{k}x$ and translated down by $a$, indeed a correct parametrisation for the curve of the hyperbola


问:
We have triangle $\triangle ABC$, with $AD\perp BC$, $\angle ABH=20^o$, $\angle HBC=40^o$ and $\angle HCB=30^o$. Find the value of angle $\angle HAC$
We have triangle $\triangle ABC$, with $AD\perp BC$, $\angle ABH=20^o$, $\angle HBC=40^o$ and $\angle HCB=30^o$. Find the value of angle $\angle HAC$

We have that $\angle HAB=30^o$, $\angle BAD=30^o$, $\angle BHD=50^o$, $\angle DHC=60^o$, $\angle AHC=120^o$. After I drew it out accurately, I worked out that the answer is $20^o$. But I can't work out how to show it. Could you please explain to me how to solve this question?

答:
As we discussed in the comments, $H$ is the orthocenter and then simple angle chasing gives $\angle HAC = 40^0$.

Alternatively, by Trigonometric form of Ceva's theorem, we have

$ \sin (\angle HBC) \ \sin (\angle HCA) \ \sin (\angle HAB) = \sin (\angle ABH) \ \sin (\angle HCB) \ \sin (\angle HAC)$

$\sin 40^0 \sin (60^0-x) \sin 30^0 = \sin 20^0 \sin 30^0 \sin x \ $ ($\angle HAC = x)$

$\sin 40^0 \sin (60^0-x) = \sin 20^0 \sin x$

$\cos (20^0 - x) - \cos(100^0 - x) = \cos (20^0-x) - \cos (20^0+x)$

$\implies x = 40^0$


问:
Posterior distribution (when prior is normal)
Suppose
\begin{align*}
a &\sim \mathcal{N}(m_0,v_1)\\
\theta \mid a &\sim \mathcal{N}(a,v_2)\\
x \mid \theta &\sim \mathcal{N}(\theta,v_3),
\end{align*}
where $m_0, v_1, v_2$ and $v_3$ are known. After oberving $x_1,...,x_n$, what is the posterior mean and variance of $a$ and $\theta$?

Could I mimic the procedure when we only have $x \mid \theta$ and a prior of $\theta$, i.e.,
$$
f(a \mid X) \propto f(X \mid a) \cdot f(a)
$$
And I was wondering what to do next.

答:
First, let's do a recap on the simpler problem. For $a \sim \mathcal{N}(m_0,v_1)$ and $x \mid a \sim \mathcal{N}(a, v_2)$, we have
$$f(a) = \frac{1}{\sqrt{2\pi v_1}}\exp\left(-\frac{(a-m_0)^2}{2v_1}\right)$$
$$f(\boldsymbol{x} \mid a) = \prod_{i=1}^n f(x_i \mid a) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi v_2}}\exp\left(-\frac{(x_i-a)^2}{2v_2}\right)$$
Then we can write
\begin{align*}
f(a \mid \boldsymbol{x}) &\propto f(a, \boldsymbol{x})\\
&= f(a) \cdot f(\boldsymbol{x} \mid a)\\
&\propto \exp\left(-\frac{(a-m_0)^2}{2v_1} - \sum_{i=1}^n \frac{(x_i-a)^2}{2v_2}\right)\\
&= \exp\left(-\frac{v_2(a-m_0)^2 + \sum_{i=1}^n v_1(x_i-a)^2}{2v_1v_2}\right)\\
&\propto \exp\left(-\frac{(v_2+nv_1)a^2 - 2(v_2m_0 + v_1n\bar{x})a}{2v_1v_2}\right)\\
&\propto \exp\left(-\frac{\left(a-\frac{v_2m_0+nv_1\bar{x}}{v_2+nv_1}\right)^2}{2\frac{v_1v_2}{v_2+nv_1}}\right).
\end{align*}
Thus, the posterior distribution of $a$ after observing $x_1,\dots,x_n$ is
$$a \mid \boldsymbol{x} \sim \mathcal{N}\left(\frac{v_2m_0+nv_1\bar{x}}{v_2+nv_1},\frac{v_1v_2}{v_2+nv_1}\right).$$
Note that the posterior mean of $a$ is a weighted average of the prior mean ($m_0$) and the sample mean ($\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$). The posterior variance of $a$ is the harmonic mean of $v_1$ and $v_2/n$. Here the main technique is to "complete the squares".

Now if we add on additional layer to the problem, same procedure still applies - we need to "complete the quadratic form" in the context of this multi-dimensional problem.
Specifically, for $a \sim \mathcal{N}(m_0,v_1)$, $\theta \mid a \sim \mathcal{N}(a, v_2)$, and $x \mid \theta \sim \mathcal{N}(\theta, v_3)$, we have
$$f(a) = \frac{1}{\sqrt{2\pi v_1}}\exp\left(-\frac{(a-m_0)^2}{2v_1}\right)$$
$$f(\theta \mid a) = \frac{1}{\sqrt{2\pi v_2}}\exp\left(-\frac{(\theta-a)^2}{2v_2}\right)$$
$$f(\boldsymbol{x} \mid \theta) = \prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi v_3}}\exp\left(-\frac{(x_i-\theta)^2}{2v_3}\right)$$
Then we can write
\begin{align*}
f(a, \theta \mid \boldsymbol{x}) &\propto f(a, \theta, \boldsymbol{x})\\
&= f(a) \cdot f(\theta \mid a) \cdot f(\boldsymbol{x} \mid \theta)\\
&\propto \exp\left(-\frac{(a-m_0)^2}{2v_1} - \frac{(\theta-a)^2}{2v_2} - \sum_{i=1}^n \frac{(x_i-\theta)^2}{2v_3}\right)\\
&\propto \exp\left(-\frac{v_2v_3(a^2 - 2m_0 a) + v_1v_3(a^2 - 2a\theta + \theta^2) + v_1v_2(n\theta^2 - 2n\bar{x}\theta)}{2v_1v_2v_3}\right)\\
&= \exp\left(-\frac{(v_2v_3 + v_1v_3)a^2 - 2v_2v_3m_0a + (nv_1v_2 + v_1v_3)\theta^2 - 2v_1v_2 n\bar{x}\theta - 2v_1v_3 \theta a \ }{2v_1v_2v_3}\right),
\end{align*}
where the exponent is a quadratic form of $(a, \theta)$, indicating that the (joint) posterior distribution of $(a, \theta)$ is $\mathcal{N}_2(\mu_a, \mu_\theta, \sigma_a^2, \sigma_\theta^2, \rho)$. Recall the kernel of a bivariate normal distribution is given by
$$\exp\left(-\frac{1}{2(1-\rho^2)}\left[\left(\frac{a-\mu_a}{\sigma_a}\right)^2 - 2\rho\left(\frac{a-\mu_a}{\sigma_a}\right)\left(\frac{\theta-\mu_\theta}{\sigma_\theta}\right) + \left(\frac{\theta-\mu_\theta}{\sigma_\theta}\right)^2\right]\right).$$
By matching the corresponding coefficients (i.e. "completing the quadratic form"), we have the following equation system:
$$\begin{cases}\frac{1}{(1-\rho^2)\sigma_a^2} &= \frac{v_2v_3 + v_1v_3}{v_1v_2v_3}\\ \frac{1}{(1-\rho^2)\sigma_\theta^2} &= \frac{nv_1v_2 + v_1v_3}{v_1v_2v_3}\\ \frac{\rho}{(1-\rho^2)\sigma_a\sigma_\theta} &= \frac{v_1v_3}{v_1v_2v_3}\\ \frac{-2}{(1-\rho^2)\sigma_a^2}\mu_a + \frac{2\rho}{(1-\rho^2)\sigma_a\sigma_\theta}\mu_\theta &= \frac{-2v_2v_3m_0}{v_1v_2v_3}\\ \frac{-2}{(1-\rho^2)\sigma_\theta^2}\mu_\theta + \frac{2\rho}{(1-\rho^2)\sigma_a\sigma_\theta}\mu_a &= \frac{-2v_1v_2n\bar{x}}{v_1v_2v_3}\end{cases}$$
Solving these equations for $(\mu_a, \mu_\theta, \sigma_a^2, \sigma_\theta^2, \rho)$, we have
$$\begin{cases}\mu_a &= \frac{(nv_2 + v_3)m_0 + nv_1\bar{x}}{nv_1 + nv_2 + v_3}\\ \mu_\theta &= \frac{v_3m_0 + (nv_1 + nv_2)\bar{x}}{nv_1 + nv_2 + v_3}\\ \sigma_a^2 &= \frac{nv_1v_2 + v_1v_3}{nv_1 + nv_2 + v_3}\\ \sigma_\theta^2 &= \frac{v_2v_3 + v_1v_3}{nv_1 + nv_2 + v_3}\\ \rho &= \frac{v_1v_3}{\sqrt{v_2v_3 + v_1v_3}\sqrt{nv_1v_2 + v_1v_3}}\end{cases}$$
Thus, the (joint) posterior distribution of $(a, \theta)$ after observing $x_1,\dots,x_n$ is
$$(a, \theta) \mid \boldsymbol{x} \sim \mathcal{N}_2(\mu_a, \mu_\theta, \sigma_a^2, \sigma_\theta^2, \rho),$$
with parameters specified above.

Note that the posterior means of both $a$ and $\theta$ are weighted averages of the prior mean ($m_0$) and the sample mean ($\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$), with different weights. $\rho > 0$ is the posterior correlation coefficient between $a$ and $\theta$.


问:
Textbook's proof of Galois correspondence is circular
I am reading Galois Theory 4th edition by Ian Stewart and the proof of one part of the Galois correspondence is circular.

$L : K$ is a finite normal field extension inside $\mathbb{C}$ with Galois group $G, \mathscr{F}$ is the set of intermediate fields, and $\mathscr{G}$ is the set of subgroups of $G.$ For $M \in \mathscr{F}, M^*$ is the group of $M$-automorphisms of $L$ and for $H \in \mathscr{G}, H^†$ is the fixed field of $H.$

(2) The maps $∗$ and $†$ are mutual inverses, and set up an order-reversing one-to one correspondence between $\mathscr{F}$ and $\mathscr{G}.$

We need to show that $M^{*†} = M, H^{†*} = H,$ which the book proceeds to do as follows:

Suppose that $M$ is an intermediate field, and let $[L : M] = d.$ Then $|M^∗| = d$ by Theorem 10.5. On the other hand, if $H$ is a subgroup of $G$ of order $d,$ then $[L : H^†] = d$ by Corollary 11.11. Hence the composite operators $∗†$ and $†∗$ preserve $[L : M]$ and $|H|$ respectively. From their definitions, $M \subseteq M^{∗†}$ and $H \subseteq H^{†∗}.$ Therefore these inclusions are equalities.

Theorem 10.5: Let $G$ be a finite subgroup of the group of automorphisms of a field $K,$ and let $K_0$ be the fixed field of $G.$ Then $[K : K_0] = |G|.$

Corollary 11.11. If $L : K$ is a finite normal extension inside $\mathbb{C},$ then there are precisely $[L : K]$ distinct $K$-automorphisms of $L.$ That is, $|\Gamma(L : K)| = [L : K]$

If $M$ is the fixed field of $M^*,$ then Theorem 10.5 indeed gives us $|M^*| = [L : M] = d.$ However, this is equivalent to $M^{*†} = M,$ which is what we are trying to show.

Supposing that $[L:H^†]$ is a normal extension, Corollary 11.11 gives us $[L:H^†] = |\Gamma(L : H^†)| = |H^{†*}|,$ but we want this to equal $|H| = d$ and hence need to show $H = H^{†*}.$

As it stands, all we really know is that $|M^*| \le d$ and $[L:H^†] \ge d.$

答:
We start with a field $L \leq M \leq K$, and we want to show that $M^{\star\dagger} = M$. Suppose that $[L:M] = d$.

The first thing we do is apply Corollary 11.1, this tells us that $|M^{\star}| = d$ (the book says they are applying Theorem 10.5 here, but I believe this is a mistake).

Then we apply Theorem 10.5 (again the book mistakenly here says to apply Corollary 11.1) to the group $M^{\star}$, which lets us conclude that
$$[L:M^{\star\dagger}] = |M^{\star}| = d = [L:M].$$

We tie this together by remembering how the $\star$ and $\dagger$ operations are defined; $M^{*}$ is a group that fixes everything in $M$, and $M^{\star\dagger}$ is a field $L \leq M^{\star\dagger} \leq K$ that contains precisely the elements fixed by every element of $M^{\star}$, so certainly $M \leq M^{\star\dagger}$. Since they are field extensions of $L$ with the same index, they must be equal.


问:
Let $f \in \mathbb Q[x]$ be a nonconstant polynomial. Is $\mathbb Q[x]$ a finitely generated $\mathbb Q[f]$-module?
I'm trying to understand what this Lemma is saying in the case $r=1$: (from Eisenbud's Commutative Algebra)

If $r=1$, it seems to me to be saying that $T$ is a finitely generated module over $k[f]$.

Is this true?

答:
Yes, because $T$ is a $k[f]$-algebra generated by the single element $x_1$, which is integral over $k[f]$: specifically, if $f=a_0+\cdots+a_dx_1^d$, then $x_1$ is a root of $$X^d+\frac{a_{d-1}}{a_d}X^{d-1}+\cdots+ \frac{a_0-f}{a_d}$$

which is a monic polynomial of $(k[f])[X]$.


问:
On an identity between fractions, related to a determinant by K. Rohn
The following formula for a determinant is attributed to Karl Rohn (e.g. in Bruijn, de, N. G. (1955). On some multiple integrals involving determinants. Journal of the Indian Mathematical Society. New Series, 19, 133-151.):

$$
\det \left(\left(\frac{x_i-x_j}{x_i+x_j} \right)_{i, j=1, \ldots, n}\right) = \prod_{1 \le i < j \le n}\left(\frac{x_i-x_j}{x_i+x_j} \right)^2 \, .
$$

In a recent answer I provided a proof of that formula via induction. A crucial role in the induction step played the following identity:
$$ \tag{1}
\frac{a-b}{a+b}\cdot\frac{c-d}{c+d} + \frac{a-c}{a+c}\cdot\frac{d-b}{d+b} + \frac{a-d}{a+d}\cdot\frac{b-c}{b+c}\\ =
\frac{a-b}{a+b}\cdot\frac{a-c}{a+c}\cdot\frac{a-d}{a+d}\cdot\frac{b-c}{b+c}\cdot\frac{b-d}{b+d}\cdot\frac{c-d}{c+d} \, ,
$$
which came a bit surprising. $(1)$ holds for all real (or complex) numbers as long as it is well-defined, i.e. no denominator is zero. I verified it with a computer algebra system (Maxima).

One could also argue as follows:

For fixed $b, c, d$ are both sides of $(1)$ rational functions in $a$, of degree at most $3$.

These rational functions both have zeros at $a=b, c, d$ and poles at $a=-b, -c, -d$.

It follows that these rational functions differ only by a constant factor.

Therefore it suffices to show that $(1)$ holds for $a \to \infty$, i.e. that
$$ \tag{2}
\frac{c-d}{c+d} + \frac{d-b}{d+b} + \frac{b-c}{b+c} =
\frac{b-c}{b+c}\cdot\frac{b-d}{b+d}\cdot\frac{c-d}{c+d}\, .
$$
And that is a known identity, see for example Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$, where one can find some elegant proofs of $(2)$.

One could also repeat the argument:

For fixed $c, d$ are both sides of $(2)$ rational functions in $b$, of degree at most $2$.

These rational functions both have zeros at $b=c, d$ and poles at $b=-c, -d$.

It follows that these rational functions differ only by a constant factor.

Therefore it suffices to show that $(2)$ holds for $b \to \infty$, i.e. that
$$ \tag{3}
 \frac{c-d}{c+d}  = \frac{c-d}{c+d} \, ,
$$
and that is obviously true.

So $(1)$ implies $(2)$ and that implies $(3)$. On the other hand, $(3)$ can be used to prove $(2)$, and that can be used to prove $(1)$.

What I am asking for is:

Other proofs of the initial identity $(1)$.

If possible, some deeper insight how equations $(1)$, $(2)$, and $(3)$ are related.

If possible, further generalizations: What would the corresponding identity be for five numbers (if there is one)?

答:
Alternative proof: Let $z_1=a,z_2=b$, etc.
Put the LHS over a common denominator so it reads
$$\mathrm{LHS} = \frac{f(z)}{\prod_{i<j}(z_i + z_j)}$$ and note that its numerator $f$ must have degree six. But the LHS is totally antisymmetric in $z_i$, and the denominator is totally symmetric, and as such the numerator $f$ is antisymmetric and vanishes when any pair of $z$s coincide. Thus it must have a factor of a particular degree six polynomial:
$$\prod_{i<j}(z_i - z_j) \mid f$$
But $f$ is degree six. Thus the LHS and RHS are identical up to an overall constant, which is easily checked e.g. by computing a single coefficient (e.g. that of $a^3b^2c^1d^0$) or evaluating at one suitable point.

Connection: This makes clear the connection between the identities you offer: the LHS is a totally antisymmetric function of the variables involved in each case, with the common denominator totally symmetric; hence the numerator has a factor of $\prod_{i<j}(z_i - z_j)$ and is of the same degree as this product (the so-called Vandermonde determinant).

Generalization: Off the top of my head, I can't think of a simple generalization to odd numbers of variables greater than 3 [Edit: See below.] -- the most obvious generalization is to Schur's Pfaffian identity (1911), for polynomials in $2n$ variables, relying on a Pfaffian-type antisymmetrization. Namely, let $P$ be the set of all $(2n)!/(2^n n!)$ distinct pairings of $z$s, so e.g.
$$P=\{((1,2),(3,4)), ((1,3),(2,4)), ((1,4),(2,3))\}$$
for $2n=4$ variables. Then
$$
\sum_{p\in P} \sigma(p) \prod_{i=1}^{n} \frac{z_{p_{i,1}} - z_{p_{i,2}}}{z_{p_{i,1}} + z_{p_{i,2}}} = \prod_{i<j}\frac{z_i-z_j}{z_i+z_j}
$$
where $\sigma(p)$ is the signature of the permutation $(p_{1,1},p_{1,2},\ldots,p_{n,1},p_{n,2})$ of $(1,\ldots,2n)$.

Reference: This identity is more commonly expressed as
$$
\mathrm{Pf}\left(\frac{z_j - z_i}{z_j + z_i}\right)_{i,j} = \prod_{i<j} \frac{z_j - z_i}{z_j + z_i}
$$
but I've rewritten it to be more clearly in the spirit of your question.

This, and similar identities, are given in e.g. Generalizations of Cauchy’s Determinant Identity and Schur’s Pfaffian Identity (Soichi Okada).

Finally, note that squaring this Pfaffian identity immediately gives the result in the linked question.

Odd Numbers of Variables (Edit):  As Martin points out in the comments, the technique of taking $z_{2n} \to \infty$ gives an identity for $2n-1$ variables from the one above. This identity has the same number of terms in the sum on the LHS as the parent identity, but one fewer term in each product there, and $2n-1$ fewer terms in the product on the RHS. Thus there is a five variable identity with 15 terms like $\frac{a-b}{a+b}\frac{c-d}{c+d}$ with various signs on the LHS, and the RHS being the ratio of $(a \mp b)(a \mp c)\cdots(d \mp e)$.


问:
Asymptotic location of the turning points of $\ddot x = -x-{\dot x}^3$
Fix a solution $x(t)$ of the following ODE: $$\ddot x=-x-\dot {x}^3.$$

Does the energy $E(t) := x^2 + \dot x^2 $ decrease to zero as $t \to \infty$?

Denote by $x_1,x_2,\ldots $ the positive values of $x(t)$ for which $\dot x(t)=0$, ordered by time. Prove that the following limit exists: $$\lim_{n \to \infty} \frac{x_n-x_{n+1}}{x_n^3}.$$

Some Remarks

To prove 1 it is enough to prove that $\dot x \to 0$, since then $x\to c$ and the ODE forces $c=0$.

If we had $\dot x \not \to 0$ then $\dot x$ would have been bounded away from $0$ "a lot", so $\dot E=-\dot x ^4$ would decrease at a rate bounded from below, hence tend to zero. This is not yet a proof because $\dot x$ might only have spikes away from zero. Perhaps this will make its second derivative large and lead to a contradiction with the ODE?

In the linearized equation the sequence $(x_n)$ is constant, and for $t\gg 0$ we expect the system to be very close to its linearization. The limit quantifies this.

I cross-posted from the physics site.

答:
The answer to question 1 is yes.

The ODE can be written as a system $\dot x = y$, $\dot y = -x - y^3$, and with $E=\frac12(x^2+y^2)$ we have $\dot E = - y^4 \le 0$, so $E$ is a weak Lyapunov function. The set where $\dot E=0$ is the $x$-axis, and it contains no complete trajectories except the origin. (Indeed, the vector field $(\dot x,\dot y)$ equals $(0,-x)$ when $y=0$, so the other trajectories intersect the $x$-axis orthogonally). Therefore LaSalle's theorem says that the origin is an asymptotically stable equilibrium; in fact, it's even globally asymptotically stable, since $E \to \infty$ as $\sqrt{x^2+y^2} \to \infty$. That is, all trajectories converge to the origin as $t \to +\infty$.


问:
L2 norm inequality respect to convolution on $S^1$
I am trying to show if $f,g\in C^0(S^1)$, then $$||f\ast g||^2_{L^2(S^1)}\leq ||f\ast f||_{L^2(S^1)}||g\ast g||_{L^2(S^1)}.$$ But nothing comes out in mind, any ideas?

答:
By Young's convolution inequality we know that if $f\in L^2(S^1)$ and $g\in L^2(S^1)$ then
$$
||f*g||_{\infty}\leq||f||_2||g||_2
$$

and
$$
||f*g||_{2}\leq||f*g||_{\infty}\
$$
since $m(S^1)=1$. This tells us that the function $f*g\in L^2$ and we can use the Parseval's identity which gives us

$$
||f*g||^2_{2}=\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*g}(n)|^2.
$$

Next we use the fact that if $f,g\in L^1$ then $\widehat{f*g}(n)=\hat{f}(n)\hat{g}(n)$ to get
$$
\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*g}(n)|^2=\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)\hat{g}(n)|^2=\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2|\hat{g}(n)|^2.
$$
Now we use the Cauchy-Schwartz inequallity

$$
\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2|\hat{g}(n)|^2\leq\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^4}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\hat{g}(n)|^4}=\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*f}(n)|^2}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{g*g}(n)|^2}.
$$

And lastly we use the Parseval's identity again to get.$$
\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{f*f}(n)|^2}\sqrt{\displaystyle\sum_{n=-\infty}^{\infty}|\widehat{g*g}(n)|^2}=||f*f||_2||g*g||_2.
$$


问:
Prove that the point-slope form of a linear equation does not depend on the point
I am stucked on the following challenge:
"If the line determined by two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ is not vertical, and therefore has slope $(y_2-y_1)/(x_2-x_1)$, show that the point-slope form of its equation
is the same regardless of which point is used as the given
point."
Okay, we can separate $(x_0, y_0)$ from the form to get:
$$y(x_2-x_1)-x(y_2-y_1) = y_0(x_2-x_1)-x_0(y_2-y_1)$$
But how exclude this point $(x_0,y_0)$ and leave only $x, y, x_1, y_1, x_2, y_2$ in the equation?
UPDATE: There is a solution for this challenge:
$$(y_1-y_2)x+(x_2-x_1)y=x_2y_1-x_1y_2$$
From the answer I found that
$$y_2(x-x_1)-y_1(x-x_2)=y(x_2-x_1)$$
... but why this is true?

答:
Thanks to saulspatz, the solution is to simply show, that whether we are using $(x_1, y_1)$ or $(x_2, y_2)$ as the given point, the equation does not change. So both equations:
$$y-y_1=m(x-x_1)$$
$$y-y_2=m(x-x_2)$$
reduce to the
$$(y_1−y_2)x+(x_2−x_1)y=x_2y_1−x_1y_2$$


问:
Why does $\operatorname{Log}(e^{3+4i})=3+(4-2\pi)i$?
I believe I am more used to the concept in the Real Numbers, but I am not sure why the answer here includes $-2\pi$. I'm assuming it has something to do with where $z=3+4i$ falls in the complex plane? Any help understanding this would be appreciated.

答:
The complex logarithm $\operatorname{Log}$ is a function. Which means for any given input, it can only give a single output. However, each non-zero complex number has infinitely many logarithms. So to make a function we need to make some kind of choice. One standard choice is to always give an output with imaginary part in the interval $(-\pi,\pi]$.

One logarithm of $e^{3+4i}$ is obviously $3+4i$. However, it has a slightly too large imaginary part for our taste. So we move down one step to $3+(4-2\pi)i$ to get a logarithm in the right range. That's how $\operatorname{Log}$ works.


问:
How do I simplify $\log (1/\sqrt{1000})$?
How do I simplify $\log \left(\displaystyle\frac{1}{\sqrt{1000}}\right)$?

What I have done so far:

1) Used the difference property of logarithms
$$\log \left(\displaystyle\frac{1}{\sqrt{1000}}\right) = \log(1) - \log(\sqrt{1000}) $$

2) Used the exponent rule for logarithm

$$\log (1) - \frac{1}{2}\log (1000) $$

I'm stuck at this point. Can someone explain why and what I must do to solve this equation?

答:
$\log_{10} \left( \displaystyle \frac{1}{\sqrt{1000}} \right) = \log_{10} \left( \displaystyle \frac{1}{\sqrt{10^3}} \right) = \log_{10} \left( \displaystyle \frac{1}{{10^\frac{3}{{2}}}} \right) = \log_{10} \left( \displaystyle {{10^\frac{-3}{{2}}}} \right) = \displaystyle {{\frac{-3}{{2}}}}$

Sorry I accidently erased an answer. Please restore. I submit mine as another.


问:
If $A,B$ are positive definite matrix then I need to prove so is $ABA^*$
If $A,B$ are positive definite matrix then I need to prove so is $ABA^*$, here is what I have done

$$x^*ABA^*x=(A^*x)^*B(A^*x)=y^*By>0$$, is it okay? $y=A^*x$

答: Yes, that's almost ok. You should mention that $y = A^*x$ cannot be $0$ for $x \ne 0$, as $A$ is invertible. Moreover your computation shows that $A > 0$ isn't necessary, $A$ invertible suffices for $ABA^*$ being positive definite for positive definite $B$.</p>


答:
I don't think it would be:

$3+4i=3(4-2\pi)i$

$3+4i=3+4i-2\pi i$

$0\neq -2\pi i$

Title: Why does $\operatorname{Log}(e^{3+4i})=3+(4-2\pi)i$?

Body: I believe I am more used to the concept in the Real Numbers, but I am not sure why the answer here includes $-2\pi$. I'm assuming it has something to do with where $z=3+4i$ falls in the complex plane? Any help understanding this would be appreciated.


问:
How do I make an exponential function that doubles every fifth integer.
I don't understand exponential functions too well, so if you know the answer, please also show me the process I would use to come up with it.

I'm trying to figure out how to make an exponential function that doubles every five whole integers. That would mean that:

```<code>|input|output|
|:---:|:----:|
|f(1) |1     |
|f(6) |2     |
|f(11)|4     |
|f(16)|8     |
</code>```

et cetera.

I know some people have suggested using a best-fit application, but I don't think those things are too accurate. Is there a tried and true process to creating a function that matches these outputs?

答:
You want an exponential function. This means it is of the form
$$
f(n)=a\cdot b^n
$$
for some numbers $a,b$. You also want $f(1)=1,f(6)=2$. That's enough to solve your problem:
$$
\cases{a\cdot b^1=1\\
a\cdot b^6=2}
$$
Dividing equation two by equation one, we get $b^5=2$. This is exactly the equation fifth roots are made to solve: $b=\sqrt[5]2$. Inserting this into equation one, we get $a=\frac{1}{\sqrt[5]{2}}$. Thus our function must be
$$
f(n)=\frac{\sqrt[5]2^n}{\sqrt[5]2}
$$


问:
Type of singularities of $\frac{z}{e^z-1}$
I don't really understand how one can find the type of singularities for a given function. Say if
$$f(z) = \frac{z}{e^z-1}$$
then I know that the singularities are at $z = 2n\pi i$

However, how do I find the type? If I try to write out Laurent series for this by using
$$e^z = 1 + z + z^2/2! + z^3/3! + \cdots $$, then I got a summation in the denominator, which I don't know how to rearrange as a Laurent series.

Can anyone please give me a hint?

答:
On $z=0$ you can remove the singularity.

For the other use that a singularity at $z_0$ is a pole of order $k$ when $f(z)\cdot (z-z_0)^k$ does have a removable singularity at $z_0$.


问:
Number of zeros elementary functions
While studying functions and zeros of functions i came across the following problem:

Problem:

Knowing that $1,09<\ln 3<1,1,$ determine the number of zeros of $J: \mathbb{R} \rightarrow \mathbb{R}$ given by $J(x)=3^{x}-1-2 x^{2}$

Attempt:

One quickly notices that $x = 0$ satisfies the expression given in the problem. Now, one can draw the functions $3^{x}$ and $2x^2 + 1$ and notice that for $x < 0$ is not possible  that $3^x = 1 + 2x^2$

Therefore, we should investigate what happens when $x > 0$. For that, i tried solving $$
g^{\prime}(x)>h^{\prime}(x)
$$

where $g(x) = 3^x$ and $h(x) = 1 + 2x^2$. The ideia behind solving the above inequality comes from the fact that if $g^{\prime}(x)>h^{\prime}(x)$ for every $x > 0$, then it would not possible for $g(x) = h(x)$ for $x > 0$ (Just intuition, i dont know if that holds in general)

However, $$
g^{\prime}(x)>h^{\prime}(x) \Rightarrow \ln (3) \cdot 3^{x}>4 x+1
$$

It looks like we are on the right track, since the term $\ln(3)$ just appeared. Now, by observing that $$
1,1 \cdot 3^{x}> ln(3) \cdot 3^x> 4 x+1>4 x
$$

we see that $$
\frac{1,1 \cdot 3^{x}}{x}>4
$$ which i dont know how to solve.

Can someone help?
Thanks in advance, Lucas

答:
In fact, it is easy to verify that $x = 0, 1, 2$ are all zeros of $J$.

We can show that these are the only zeros of $J$, in the following order:

Show that $J''$ is strictly increasing, and deduce that $J''$ has exactly one zero, which lies in the interval $(0, 2)$. We call this zero $z$.

Show that $J'$ is strictly decreasing on the interval $(-\infty, z)$ and strictly increasing on the interval $(z, \infty)$. Deduce that $J'$ has exactly two zeros, which lie in the intervals $(0, 1)$ and $(1, 2)$, respectively. We call them $z_1$ and $z_2$.

Show that $J$ is strictly increasing on the intervals $(-\infty, z_1)$ and $(z_2, \infty)$ and is strictly decreasing on the interval $(z_1, z_2)$. Thus $J$ has at most three zeros, which we have all found.

Title: Number of zeros elementary functions


问:
Constructing self-complementary graphs
How does one go about systematically constructing a self-complementary graph, on say 8 vertices?

[Added: Maybe everyone else knows this already, but I had to look up my guess to be sure it was correct: a self-complementary graph is a simple graph which is isomorphic to its complement.  --PLC]

答:
Here's a nice little algorithm for constructing a self-complementary graph from a self-complementary graph $H$ with $4k$ or $4k+1$ vertices, $k = 1, 2, ...$ (e.g., from a self-complementary graph with $4$ vertices, one can construct a self-complementary graph with $8$ vertices; from $5$ vertices, construct one with $9$ vertices).

See this PDF on constructing self-complementary graphs.

Title: Constructing self-complementary graphs


问:
The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$
There is an exercise on $\sin x$. How could I see that for any $0<x< \frac \pi 2$, $\frac 2 \pi x \le \sin x\le x$?

Thanks for your help.

答: $\sin(x) -x$ is a decreasing function on $(0,\pi/2)$(look at the derivative) and is equal to zero at zero. Can you fill in the details now?</p>


问:
Convergence of $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}$
If $\sum_{n=1}^{\infty}\frac{a_n}{n}=a$ converges to a finite number with $a_n \geq 0$ for all $n \geq 1$ does this also imply that $$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}=0$$ I tried using $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{n}$ as a bound so that I can use comparison test but I can't prove that it converges either. How would I go about proving this statement?

答:
Note that for any $\epsilon > 0$, there exists $N$ such that

$$0 \leqslant \sum_{i=1}^n \frac{a_i}{i+n} = \sum_{i=1}^N \frac{a_i}{i+n}+ \sum_{i=N+1}^n \frac{a_i}{i+n} \leqslant  \sum_{i=1}^N \frac{a_i}{i+n}+ \sum_{i=N+1}^n \frac{a_i}{i} \\ <  \sum_{i=1}^N \frac{a_i}{i+n} + \frac{\epsilon}{2}$$

Try to finish from here.


问:
one substitution in Chern's intrinsic proof of Gauss-Bonnet-Chern theorem
In Chern's proof for Gauss-Bonnet-Chern theorem, he claims that
$$
\varepsilon_{i}u_{i_1}u_j\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=P_k+2(p-k-1)\Sigma_k
$$
where
$$
P_k=\varepsilon_{i}u_{i_1}^2\Omega_{i_1i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}
$$
and
$$
\Sigma_k=\varepsilon_{i}u_{i_1}u_{i_3}\Omega_{i_3i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}
$$

By direct computations:

$j=i_1$: we get $P_k$.

$j=i_3,\dots,i_{2p-2k}$: we get $\Sigma_k$.

My question is: why
$$
\sum_{j=i_{2p-2k+1}}^{2p}\varepsilon_{i}u_{i_1}u_j\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=0?
$$
It seems that
$$
\varepsilon_{i}u_{i_1}u_{j}\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}\\=\varepsilon_{i}u_{i_1}u_{i_{2p-2k+1}}\Omega_{i_{2p-2k+1}i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}
$$
for all $j\in\{2p-2k+2,\dots,2p\}$, then why
$$
\varepsilon_{i}u_{i_1}u_{i_{2p-2k+1}}\Omega_{i_{2p-2k+1}i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=0?
$$

Any help would be appreciated.

The link for Chern's original proof is:
https://www.maths.ed.ac.uk/~v1ranick/papers/chern7.pdf

答:
Here's all that is going on. It is the skew symmetry in the $\epsilon_i$ symbol. Because $\theta_{i_3}\wedge\dots\wedge\theta_{i_{2p-2k}}$ is a form of even degree, we have for $\mu$ odd in $2p-2k+1,\dots,2p-1,2p$,
$$\Omega_{i_\mu i_2}\wedge (\dots) \wedge \Omega_{i_\mu i_{\mu+1}} = \Omega_{i_\mu i_{\mu+1}}\wedge (\dots) \wedge \Omega_{i_\mu i_2},$$
but switching $2$ and $\mu+1$ introduces a factor of $-1$ in the $\epsilon_i$. Thus, the sum (for $j=i_\mu$ fixed) is zero.
And if $\mu$ is even, first write $\Omega_{i_\mu i_2}=-\Omega_{i_2i_\mu}$ and make the same argument with $2$ and $\mu-1$.


问:
How to figure out equilibrium and isoclines
I have this question :

For the dynamical system
$$\begin{align*}
\dot{x}&= (2-x-y)x \\[5pt]
\dot{y}&= (1-3x-4y)y\\[5pt]
\end{align*}$$

Find all the equilibrium points of the system. Determine the horizontal and vertical isoclines of the system.

So usually I go about just solving them as to set $f(x) =$ RHS and then solve for eq. pts but I'm not sure how to do this. I plugged into a graphing website but would want an explanation / method on how to do this

Thank you:)

答:
To solve this kind of problems, what you usually have to do is start finding the nullclines, i.e. the subsets of $\mathbb{R}^2$ where either $f(x,y) = \dot{x}=0$ or $g(x,y)=\dot{y}=0$. Then, to find the equilibria, if any, you just have to find the intersections between these two sets. Indeed, an equilibrium point is a point where both $\dot{x}=\dot{y}=0$.

In your case, as you've mentioned, you can set $\dot{x} = f(x,y) = (2-x-y)x$ and $\dot{y} = g(x,y) = (1-3x-4y)y$.

We now start finding the zeros of $f(x,y)$, i.e. the points where the vector field is vertical. To do so we first put the parenthesis equal zero and then the remaining part.

So $f(x,y)=0$ if either $2-x-y=0$ or $x=0$. The first set of points  defines the line $y = 2-x$.

Then, you look where the vector field is horizontal, i.e. where $g(x,y)=0$. Here the reasoning is the same, so you get that either $y=-3/4x+1/4$ or $y=0$.

So now that you have found the isoclines (or nullclines to be precise), you can find their intersections to find the equilibria.

We start fixing $x=0$, which makes $f=0$. We see that the only way to have that even $g(0,y)=0$, is to set either $y=0$, getting the first equilibrium $P_1=(0,0)$, or $y=1/4$ so that you get the point $P_2=(0,1/4)$ which lives in the line where $g$ vanishes.

Then we choose the other set where we have seen that $f$ vanishes, i.e. the line $y=2-x$ and see if for some points in the set $\{(x,2-x):x\in\mathbb{R}\}$ the function $g$ vanishes, i.e. if it has non trivial intersection with the points we have found before.

We see that if $x=0$, we get $y=2$ and hence this choice makes both $f=g=0$, and gives the third equilibrium point $P_3=(0,2)$. To conclude, we just need the intersection between the two lines $y=2-x$ and $y=-3/4x+1/4$ which stands at the forth and last equilibrium $P_4 = (7,-5)$.


问:
Why is the possibility relation $\mathcal K_i$ an equivalence relation on $S$?
I have just begun reading about Epistemic Logic from Reasoning About Knowledge, Fagin et. al. $\mathcal K_i$ is defined as the possibility relation between two worlds, and the author says that for the most part - one would want $\mathcal K_i$ to be an equivalence relation. I do not understand the author's intuition.

Here is some setup to give you background - there are $n$ agents, $\{1,2,...,n\}$. As usual, $\Phi$ denotes a set of propositional variables, and $S$ is a non-empty set of states/possible worlds for the Kripke structure. $\pi$ is a valuation, which associates each state in $S$ with a map as follows:
$$\pi(s): \Phi\to\{T,F\}$$
Then, the author says the following about $\mathcal K_i$:

The binary relation $\mathcal K_i$ is intended to capture the possibility relation according to agent $i$: $(s, t) \in \mathcal K_i$ if agent $i$ considers world $t$ possible, given his information in world $s$. We think of $\mathcal K_i$ as a possibility relation, since it defines what worlds agent $i$ considers possible in any given world. Throughout most of the book (in particular, in this chapter), we further require that $\mathcal K_i$ be an equivalence relation on $S$. We take $\mathcal K_i$ to be an equivalence relation since we want to capture the intuition that agent $i$ considers $t$ possible in world $s$ if in both $s$ and $t$ agent $i$ has the same information about the world, that is, the two worlds are indistinguishable to the agent. Making $\mathcal K_i$ an equivalence relation seems natural and turns out to be the appropriate choice for many applications.

I find it difficult to understand what intuition the authors are trying to capture with such a definition. Could someone help me better understand the motivation and reason behind such an assumption? Thanks a lot!

答:
There are several motivations behind taking possibility relations to be equivalence relations. Some are logical, some have to do with applications of epistemic logic.

From a logical point of view, using equivalence models (i.e. modal models with equivalence relations) allows reducing modal models to a very simple form, at least in the monomodal case (just one possibility relation): you can equate such models with sets of classical propositional models (sets of atoms). Correspondingly, the satisfiability problem for formulas in equivalence models is extremely simple: it is NP-complete and so does not add any complexity to the case of classical propositional logic.

From a pragmatic point of view using equivalence models is a natural way of formalizing distributed systems, which is a core area of applying epistemic logic to computer science. Fagin et al. provide such applications in the very book you are citing.


答: Consider the function $y=\frac{sin(x)}{x}$. Use calculus techniques to find its range and hence deduce the desired inequality (once you've specified the domain for the inequality).</p>

Title: The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$

Body:
There is an exercise on $\sin x$. How could I see that for any $0<x< \frac \pi 2$, $\frac 2 \pi x \le \sin x\le x$?

Thanks for your help.


问:
Calculate volume of intersection two non-aligned cuboids
I'm on the lookout to finding an algorithm that calculates the volume of the intersection between two cuboids. Most answers so far assume all the axes of the two cuboids in question are nicely aligned. Not the case in my problem. See figure below

Finding out what corners of the two cuboid are in each others space is not that difficult. Also determining some boundary points on the cuboids faces shouldn't be problematic. The hard part where I get lost is to determine what kind of shape the intersection actually is and how to calculate it's volume. I think it's safe to assume the intersection will be convex. Any suggestions?

答:
Hints:

If $C_1$ and $C_2$ denote the two cuboids, you can assume for all the calculations that $C_1$ is a cube with unit sidelengths (with, at the very end, a multiplication by the 3 initial dimensions of $C_1$, in order to get the real volume).

Then, select a parameterization standard for $C_2$. If you choose coordinates of center + directions of axes + lengths, you will need as many as $3+2+2+3=10$ parameters for that (3 for the center, 2 for the unit vector of the first axis, 2 for the unit vector of the second axis, 0 for the third axis which is the cross product of the previous vectors ; finally, 3 for the length, width and height). Any other parameterization method will need as well $10$ parameters.

Use "marching cubes" method for taking into account the different cases of intersection that can occur.

Finally use, as advised by @saulspatz, a decomposition into tetrahedra whose volumes' computation is easy.


问:
Inverse of metric tensor
The following is a homework example, but I have the feeling that it is not stated correctly, and I would like to hear some opinions on that:

Let $g_{ij}$ be a symmetric non-degenerate tensor field. Now define:
$$ B_i \enspace := \enspace g_{ij} A^j \qquad \text{and} \qquad C^i \enspace := \enspace g^{ij} B_j $$
Show that
$$ C^i \enspace = \enspace A^i \tag{1}$$

Equation (1) shows that $g_{ij}$ is inverse to $g^{jk}$ -- existence of such an inverse is guaranteed by non-degeneracy.

The tensor field $g$ is in the following examples introduced as the metric. However, until now I always worked with defining $g^{jk}$ to be the inverse of $g_{ij}$. Now that I have to show it, fail to accomplish that. I have the feeling that it may not be possible from the information given.

Any ideas?

答:
until now I always worked with defining $g^{jk}$ to be the inverse of $g_{ij}$. Now that I have to show it

That's not how I read the question. I read it as asking you to use $C^i=g^{ij}g_{jk}A^k=\delta^i_kA^k=A^i$, where $g^{ij}g_{jk}=\delta^i_k$ is the definition of $g^{ij}$ used in the definition of $C^i$.

Title: Inverse of metric tensor


问:
Lie algebra of $\mathrm{SL}(2, \mathbb{C})$ using method of differential geometry
I am asking a new question because people did not understand what  I meant and closed the other question.
I know that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ are traceless matrix and I can proof it. What I am interested is proof it using the methods of differential geometry.

Let $\gamma(t)$  be a curve on a manifold $M$ and $f$ a real function. We define the vector $V$ as the operator
$V\cdot f=\frac{d}{dt}f(\gamma )$

Now let $U$ be the set
$$
U:=\left\{\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \in \mathrm{SL}(2, \mathbb{C}) \mid a \neq 0\right\}
$$
and define the map
$$
\begin{array}{l}
x: \quad U \rightarrow x(U) \subseteq \mathbb{C}^{*} \times \mathbb{C} \times \mathbb{C} \\
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \mapsto(a, b, c)
\end{array}
$$

Te Lie algebra $\mathcal{L}(G),$ of a Lie group is defined by
$$
\mathcal{L}(G):=\left\{X \in \Gamma(T G) \mid \forall g, h \in G:\left(\ell_{g}\right)_{*}\left(\left.X\right|_{h}\right)=X_{g h}\right\}
$$
which can be  proved to be isomorphic to the Lie algebra $T_{e} G$ with Lie bracket
$$
[A, B]_{T_{e} G}:=j^{-1}\left([j(A), j(B)]_{\mathcal{L}(G)}\right)
$$
induced by the Lie bracket on $\mathcal{L}(G)$ via the isomorphism $j$
$$
\left.j(A)\right|_{g}:=\left(\ell_{g}\right)_{*}(A)
$$
Any $A \in T_{\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)} \mathrm{SL}(2, \mathbb{C})$ can be written as
$$
A=\alpha\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\beta\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\gamma\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)},
$$
then we can show that $$j\left(\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{1}}-x^{2} \frac{\partial}{\partial x^{2}}+x^{3} \frac{\partial}{\partial x^{3}}$$
$$j\left(\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{2}}$$

$$j\left(\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{2} \frac{\partial}{\partial x^{1}}+\frac{1+x^2x^3}{x^1}\frac{\partial}{\partial x^{3}}$$
and so
$$
 j(A)=(\alpha x^1+\gamma x^2)\left(\frac{\partial}{\partial x^{1}}\right)+(-\alpha x^2+\beta x^1)\left(\frac{\partial}{\partial x^{2}}\right)+\left(\alpha x^3+\gamma\frac{1+x^2x^3}{x^1} \right) \left(\frac{\partial}{\partial x^{3}}\right) \tag 1
$$
On the other hand we can show that by other method  that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ consist of traceless matrices.

My question is how can we identify $(1)$ with a traceless matrix?

答:
From the formula above we can prove that

$$\left[\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=2\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$
$$\left[\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=-2\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$

$$\left[\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$

Now we define the linear map $$\rho \left(\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
1 & 0 \\
0 & -1
\end{array}\right)}=X_1$$
$$\rho \left(\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
 0 & 1 \\
0 & 0
\end{array}\right)}=X_2$$
$$\rho \left(\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right)}=X_3$$

Now $\left[X_1,X_2 \right]=2X_2 \quad$,$\left[X_1,X_3 \right]=-2X_3\quad$,$\left[X_2,X_3 \right]=X_1\quad$
Since $\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \quad$,$\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$ and $\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$ are basis of this lie algebra  $\rho$ is a representation of the lie algebra of $\mathrm{SL}(2, \mathbb{C})$.

So traceless matrix are representation of the lie algebra of $\mathrm{SL}(2, \mathbb{C})$


答:
For any natural $K$, you can write

$$0\leq\sum_{i=1}^n\frac{a_i}{i+n}= \sum_{i=1}^K\frac{a_i}{i+n}+\sum_{i=K+1}^n\frac{a_i}{i+n}\leq \sum_{i=1}^K\frac{a_i}{i+n}+\sum_{i=K+1}^n\frac{a_i}{i}.$$

As $\sum_{i=1}^\infty\frac{a_i}{i}=a<\infty$, [any tail must converge to zero][1], which means that $\lim_{n\to\infty}\sum_{i=K+1}^n\frac{a_i}{i}=0$. Thus, from the above expression,

$$0\leq\lim_{n\to\infty}\sum_{i=1}^n\frac{a_i}{i+n}\leq \lim_{n\to \infty}\sum_{i=1}^K\frac{a_i}{i+n}+\lim_{n\to \infty}\sum_{i=K+1}^n\frac{a_i}{i}=\sum_{i=1}^K\lim_{n\to\infty}\frac{a_i}{i+n}=0.$$
[1]: https://proofwiki.org/wiki/Tail_of_Convergent_Series_tends_to_Zero

Title: Convergence of $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}$

Body: If $\sum_{n=1}^{\infty}\frac{a_n}{n}=a$ converges to a finite number with $a_n \geq 0$ for all $n \geq 1$ does this also imply that $$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}=0$$ I tried using $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{n}$ as a bound so that I can use comparison test but I can't prove that it converges either. How would I go about proving this statement?


问:
What is infinity times the reciprocal of infinity?
I was talking with a friend about interesting properties of numbers and their theoretical contradictions and solutions when we came up with this. What is the answer?

So...

$x * ∞ = ∞$

and...

$\frac{1}{x}*x=1$

So what do you get when you do...

$\frac{1}{∞}*∞$? $∞$ or $1$?

答:
John Wallis, who originally introduced the symbol "$\infty$" in the 17th century, used it to denote a specific infinite number, and furthermore exploited an infinitesimal of the form $1/\infty$ in area calculations. Extended number systems that include infinitesimals were used by such giants as Leibniz, Euler, and Cauchy, and are in active use today.

Today the symbol $\infty$ is not generally used in this sense.  However, the wording of the question by the OP suggests that this is the sense he may have in mind, rather than the traditional meaning in the context of "indeterminate forms".  The OP should keep in mind that the modern use of the symbol is different from its historical use.

If $H$ is an infinite number (meaning that it is greater than every real number), then $H \times \frac{1}{H}=1$ as for ordinary real numbers.  See for example http://www.math.wisc.edu/~keisler/calc.html

The answer given by user mixedmath assumes that you are dealing with "indefinite forms", and is also correct.


问:
Baire space is homeomorphic to countably many copies of itself
On wikipedia I found that the Baire space $\mathcal{N}$ is homeomorphic to the product of a countable number of copies of itself, however, I haven't been able to find a proof.

The Baire space is defined as $\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ with the product topology, $\mathbb{N}$ being endowed with the discrete topology.

Can anyone help me, please?

答:
Well, note that $\Bbb N \times \Bbb N$ has a bijection with $\Bbb N$, call it $h$, and let $g: \Bbb N \to \Bbb N \times \Bbb N$, $g(n) = (g_1(n), g_2(n))$ be the inverse, with $h(g_1(n), g_2(n)) = n$ for all $n$ etc.

Then map $f \in (\Bbb N^{\Bbb N})^{\Bbb N}$ to $\Phi(f) \in \Bbb N^{\Bbb N}$ by the rule $$\Phi(f)(n) = f(g_1(n))(g_2(n))$$
and show it's a bijection and a homeomorphsim in the product topologies.

This is ofetn summarized as

$$\mathcal{N}=\Bbb N^{\Bbb N} \simeq \Bbb N^{\Bbb N \times \Bbb N} \simeq (\Bbb N^{\Bbb N})^{\Bbb N}=\mathcal{N}^{\Bbb N}$$

where all homeomorphisms are natural (categorical), basically coordinate shuffles.


问:
Number of branch points for a rational map with degree d is 2(d-1)?
Let $R(z)=\frac{P(z)}{Q(z)}$ where P and Q are complex polynomials. A point in $CP^1$ is said to be a branch point of $R$ if for every neighborhood around $z_0$, R is not a homeomorphism restricted the the neighborhood (in my text we consider rational functions to map into the riemann sphere).  The degree of R is the max of the degree of $P$ and $Q$.  Don't see why the number of branch points for $R$ is $2(d-1)$.  The problem points we should consider are probably where the denominator is zero or when $R$ has a critical point, otherwise we could use inverse functions theorem.  The number of critical points is just the number of zeros of $Q'P+P'Q$ but I don't see how it relates.

答:
Let $R=P/Q$ be a rational function of degree $d=\max\{\text{deg } P, \text{deg } Q\}$.

Since $R$ is continuous, your statement is equivalent to:
A point in $CP^1$ is said to be a branch point of $R$ if $R$ fails to be injective on every neighborhood of $z_0$.

Suppose $z_0$ is a branch point of $R$. Then for each neighborhood $B(z_0,n^{-1})= \{z: d(z_0,z)<n^{-1} \}$ ($n=1,2,\ldots$), we may choose two distinct points $z_n, w_n \in B(z_0,n^{-1})$ such that $R(z_n)=R(w_n)$. By the mean value theorem, for each pair of points $z_n, w_n$ ($n=1,2,\ldots$), we may find some point $c_n$ on the straight line from $z_n$ to $w_n$ such that $R'(c_n)=0$. Since $R'$ is continuous and $\lim_{n \to \infty} c_n=z_0$, we have that $R'(z_0)=0$.

Now without proof I will note that the degree of
$$R'(z)=\frac{P'(z)Q(z)-P(z)Q'(z)}{\left[Q(z)\right]^2}$$
turns out to be exactly $2(d-1)$.


问:
Corollary of convergence in distribution
Can I say that if $\xi_n \xrightarrow {d} \xi$ and $\xi_n, \xi$ are non-negative random variables with finite expected value then
$$\mathbb{E}\xi \le \lim{\inf{}_{n\to\infty}\mathbb{E}\xi_n}?$$
I have problems to disprove or prove it. May somebody can help me? I would be grateful.

答: Using Skorohod's representation theorem, there exists a probability space $(\Omega, \mathcal F, \mathbb P)$, and random variables $X_n,X$ on that space, such that $X_n\stackrel{d}= \xi_n$, $X\stackrel{d}= \xi$, and $X_n\xrightarrow{\text{a.s.}} X$. Conclude using Fatou's lemma that $$E\xi =EX\le \liminf_n EX_n=\liminf_n E\xi_n.$$


问:
Intuition behind binomial variance
Suppose that I perform a stochastic task $n$ times (like tossing a coin) and that $p$ is the probability that one of the possible outcomes occurs. If $K$ is the stochastic variable that measures how many times this outcome occurred during the whole experiment, and if all the events are mutually independent, then the probability that $K$ is equal to a specific $k$, with $0\leq k\leq n$, is
$$\mathrm{Pr}\{K=k\} = \binom n k p^k (1-p)^{n-k} =: b_n(k), $$
which makes intuitive sense to me.

Now suppose that I want to know what the average, or expected, value of $K$ is going to be: the formulae tell me that
$$\begin{split}
\langle K\rangle &= \sum_{k=0}^n k b_n(k) = \sum_{k=0}^n k \binom n k p^k (1-p)^{n-k} \\
&= \sum_{k=0}^n k \frac{n(n-1)!}{k(k-1)!(n-k)!} p p^{k-1} (1-p)^{n-k} \\
&= np \sum_{\kappa=1}^\nu \frac{\nu!}{\kappa!(\nu-\kappa)!} p^\kappa (1-p)^{\nu-\kappa} = np(p + 1- p)^{\nu} \\ &= np,
\end{split}$$
where I’ve made the substitutions $\kappa = k-1$ and $\nu = n-1$. Notwithstanding the mathematical certainty of this derivation, it also makes perfect intuitive sense to me that $np$ should be the expected value of $K$, since it is the product of the probability of the outcome times the number of trials performed: if there’s a $1/6$ chance that I roll a 5 on a fair dice, and I throw it $600$ times, then I expect to see a 5 about $100$ of those times.

If instead I want to know how I should expect the outcomes to vary around the expected value, I may compute the variance of $K$: with the same substitutions as before,
$$\begin{split}
 \mathrm{Var}[K] &= \langle K^2\rangle - \langle K \rangle^2 = \left(\sum_{k=0}^n k^2 b_n(k) \right) -n^2p^2 \\
&= \left( \sum_{k=1}^n k^2 \binom n k p^k (1-p)^{n-k}\right) -n^2p^2 \\
&= \left( np \sum_{\kappa = 0}^\nu (\kappa +1) \binom \nu \kappa p^\kappa (1-p)^{\nu-\kappa} \right)-n^2p^2 \\
&= \left(np \Big(\sum_{\kappa=0}^\nu \kappa b_\nu(\kappa) +(p + 1-p)^\nu \Big) \right) -n^2p^2 \\
&= np(\nu p + 1) - n^2 p^2 = np( np - p + 1 - np) \\ &= np(1-p).
\end{split}$$
Again, the derivation is mathematically crystalline; but why should I expect that this be the formula for the variance of $K$? Why does multiplying the expected value of $K$ times the probability that my outcome doesn’t occur give me a measure of the dispersion of $K$? In other words, how can I justify this formula for variance intuitively in a similar way as I can with the formula for the mean?

EDIT. Up until now, I’ve received answers that are just perfectly good explanations of how to derive the formula for the variance of $K$ in ways that differ from the one presented above. That’s not what I’m asking for. The ideal answer should contain as few formulae as possible, and use simple enough words to explain not why the formula is mathematically true, but why it’s reasonable and couldn’t possibly be otherwise – something like the intuitive explanation for $\langle K\rangle = np$ that I gave above.

答:
I recently asked myself identically this question. The conclusion I came to is that the variance is a function of the mean because the mean is on a bounded interval. That is, obviously/by definition, half of the outcomes have to be above the mean. When the mean approaches 1 (or likewise, as it approaches 0), it becomes harder to have a large variance, because the outcomes that would have been required to "average out" the lower valued outcomes simply cannot exist because of the interval on which the values are defined.
Graphically, these are the distributions you get at different means (here the "means" are the 'r_obs' values defined in the legend). For the blue one (u = 0.05) to have a larger variance, since it has to stretch out symmetrically because of the particular model in question, you can see it would have to stretch past 0 to the left, which is not allowed.

Title: Intuition behind binomial variance


问:
Coin toss probability calculation

A gambler bets on coin flips. With each flip, he wins $1$ dollar with probability $p$, and loses $1$ dollar with probability $1-p$. He starts with $2$ dollar and stops when he reaches either $0$ or $5$ dollar.

What is the probability that the gambler has $\$5$ in the end?

What is the expected number of coin flips he bets on?

Ok, so basically my assumption towards these question is that:

Since $p$ and $1-p$ have been specified, its a biased coin.

The steps are $0$ to $5$, and it starts from step $2$.

Could anyone please confirm or maybe write down your answers if this is correct? or if I have left out any answers could any experts please help me with this question? Your help is much appreciated.

答:
The transition matrix for your absorbing Markov chain is

$$P=
\begin{bmatrix}
 0 & p  & 0 &0&0& 1-p \\
 1-p & 0 & p & 0 & 0&0 \\
 0 & 1-p & 0 & p & 0&0 \\
 0 & 0 & 1-p & 0 & p & 0\\
 0 &0 & 0 & 0 & 1 & 0\\
 0 &0 & 0 & 0 & 0 & 1\\
\end{bmatrix}
\begin{matrix}
\dots$1\\
\dots$2\\
\dots$3\\
\dots$4\\
\dots$5\\
\dots$0\\
\end{matrix}$$

So

$$Q=
\begin{bmatrix}
 0 & p  & 0 &0 \\
 1-p & 0 & p & 0 \\
 0 & 1-p & 0 & p \\
 0 & 0 & 1-p & 0 \\
\end{bmatrix}
$$

$$R=
\begin{bmatrix}
0 & 1-p \\
0 & 0 \\
0 & 0 \\
p & 0\\
\end{bmatrix}
$$

and

$$
\begin{align}
N&=(I-Q)^{-1}\\
&=\left(
\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{bmatrix}-
\begin{bmatrix}
 0 & p  & 0 &0 \\
 1-p & 0 & p & 0 \\
 0 & 1-p & 0 & p \\
 0 & 0 & 1-p & 0 \\
\end{bmatrix}\right)^{-1}\\
&=\left(
\begin{bmatrix}
 1 & -p  & 0 &0 \\
 p-1 & 1 & -p & 0 \\
 0 & p-1 & 1 & -p \\
 0 & 0 & p-1 & 1 \\
\end{bmatrix}
\right)^{-1}
\end{align}
$$

You can invert this yourself because I'm too tired and lazy.

The point is that $B=NR$ will give you the answer to part 1

And $t=N\mathbf{1}$ will give you the expected number of steps for each starting position - you want the second row value.

Title: Coin toss probability calculation


答:
Note that there is a point after which the sum of all terms is below $\epsilon$.

All terms before this point go to zero, obviously, as $\frac{a_i}{i+n}$ converges to zero for any $i$.

Since the $\epsilon$ is chosen arbitrarily, the tail converges to zero.

A other way to look at it, the tail of the series can be as small as desired. Hence as $n$ goes to infinity it should vanish. To prove it rigorously, let $\epsilon \in \mathbb R_+$, now for any $\epsilon$ there is $k_1$

$$\sum_{i=k}^n \frac{a_i}{i+n}<\sum_{i=k}^n \frac{a_i}{i}<\frac{\epsilon}{2}, \space \space k_1<n$$

and

$$\sum_{i=1}^k \frac{a_i}{i}<C$$

So that there is $k_2$
$$\sum_{i=1}^k \frac{a_i}{i+n}<\frac{\epsilon}{2}, \space \space \space k_2<n,$$

then choose $k_3=max(k_1,k_2)$.


问:
An easier evaluation of $\det\limits_{1\leqslant i,j\leqslant n}\left\{\frac{x_i-x_j}{x_i+x_j}\right\}$
I'm looking for an easier proof of the identity (attributed to K. F. W. Rohn)
$$R_n(\bar{x}):=\det_{1\leqslant i,j\leqslant n}\left\{\frac{x_i-x_j}{x_i+x_j}\right\}=\prod_{i<j}\left(\frac{x_i-x_j}{x_i+x_j}\right)^2,$$
where $\bar{x}:=(x_1,\ldots,x_n)$ and $n$ is even (if $n$ is odd then clearly $R_n(\bar{x})=0$).

This identity is used in the end of this answer of mine.

The proof I have (see my own answer below) obtains it from the formula for
$$D_n(\bar{x},\bar{y},\bar{z}):=\det_{1\leqslant i,j\leqslant n}\left\{\frac{x_i+z_j}{x_i+y_j}\right\}$$
found by H. J. A. Duparc; this time, $n$ is not necessarily even.

But it looks convoluted. I don't have access to Rohn's works - did he have simpler arguments?..

答:
This question was linked to that question which brings up the fact that you can prove the identity above by squaring the identity
$$\mathrm{Pf}\left(\frac{z_j - z_i}{z_j + z_i}\right)_{i,j} = \prod_{i<j} \frac{z_j - z_i}{z_j + z_i}$$
using the standard fact that the square of the Pfaffian is the determinant.

This identity in turn comes from the fact the LHS can be put over the common denominator on the RHS with a numerator which is of the same degree as that of the RHS. Then one reasons the LHS is antisymmetric in $z$ and hence the numerator of the RHS divides that of the LHS. Then checking one coefficient gives the identity.


问:
Union of countable sequence of countable sets
I realize this question has been asked and answered multiple times, but I'm still not understanding the reasoning. For reference, I've read this, this, and this.

Theorem 2.12. Let {}, =1,2,…, be a sequence of countable sets, and put

$$S = \bigcup\limits_{n=1}^{\infty}E_n$$

Then  is countable.

Proof. Let every set  be arranged in a sequence {},
=1,2,3,…, and consider the infinite array

in which the elements of  form the th row. The array contains
all elements of . As indicated by the arrows, these elements can be arranged in a
sequence

11;21,12;31,22,13;41,32,23,14;…(*)

If any two of the sets  have elements in common, these will appear
more than once in (∗). Hence there is a subset  of the set of all
positive integers such that ∼, which shows that  is at most
countable. Since 1⊂, and 1 is infinite,  is infinite, and thus
countable. ◼

I understand if there's an injection $F: S\rightarrow \mathbb{N}$, S is countable. I don't understand how utilizing the diagonals leads to such an injection. Since there's 1 element in the first diagonal, 2 in the second, ..., n in the n-th, (n - 1) in the (n + 1)-th, etc., how is it possible to say that since there are n elements at most in any diagonal the set is countable?

答:
It's hard, at least for me, to make the diagonal argument clearer, so I offer you the next. Define the following sets:

$$\begin{cases}A_0=\{1,3,5,7,...,2n-1,...\}\\{}\\
A_1=\{2,6,10,14,...,4n-2,...\}\\{}\\
A_2=\{4,12,20,28,...,8n-4,...\}\\{}\\
...................................\\{}\\
A_n=\{2^n, 2^n+2^{n+1}, 2^n+2\cdot2^{n+1},...,n2^{n+1}-2^n,...\\{}\\..........................................| \end{cases}$$

Check all the above sets are pairwise disjoint and that their union is $\;\Bbb N\;$. Now you have a nice bijection between $\;\Bbb N\;$ and that countable union $\;\{E_n\}\;$ of countable sets: first, map each $\;E_n\to A_n\;$, and then map each element of $\;E_n\to $ an element of $\;A_n\;$ ( can you see why is this possible?).


问:
Application of theorem on gluing vectors in lattice theory to $E_8$
I'm currently learning about gluing vectors in lattice theory, mainly from The (Sensual) Quadratic Form by Conway & Fung, and Sphere Packings, Lattices and Groups by Conway & Sloane. In the latter, in page 100, there is a theorem which states

Theorem 1: If a unimodular lattice $L$ is formed by gluing together two lattices $L_1$ and $L_2$ in such a way that there is no
  self-glue, i.e., if

$$L_1 = (L_1 \otimes \mathbb{R}) \cap L, ~~ L_2 = (L_2 \otimes
 \mathbb{R}) \cap L,$$

then the dual quotients $L_1^*/L_1$ and $L_2^*/L_2$ are isomorphic groups.

In working with the $E_8$ lattice, one finds rank 8 sublattices by using the extended diagram algorithm, getting the eight lattices

$$A_8,~A_1A_2A_5,~A_4^2,~D_5A_3,~E_6A_2,~E_7A_1,~A_1A_7,~D_8$$

Of these, four are products of two lattices and can be extended to $E_8$ again by including a glue vector corresponding to the simple root which was deleted in the extended diagram algorithm. Now, I find that Theorem 1 holds in the cases
$$E_7A_1, ~~~~~~~~~~~~~~~ E_7^*/E_7 \simeq A_1^*/A_1 \simeq \mathbb
Z_2,$$
$$E_6A_2, ~~~~~~~~~~~~~~~ E_6^*/E_6 \simeq A_2^*/A_2 \simeq \mathbb
Z_3,$$
$$D_5A_3, ~~~~~~~~~~~~~~~ D_5^*/D_5 \simeq A_3^*/A_3 \simeq \mathbb
Z_4.$$

However, for the case $A_1A_7$ the dual quotients are $\mathbb Z_2$ and $\mathbb Z_8$, which are obviously not isomorphic. Given that the glue vector required for extending $A_1A_7 \to E_8$ is not orthogonal to either of these sublattices, Theorem 1 should hold.

Can anyone please tell me what am I doing wrong here?

Incidentally, the lattices $A_1A_2$ and $A_5$ have isomorphic dual quotients $\mathbb Z_2 \times \mathbb Z_3$ and $\mathbb Z_6$. Does this mean anything?

答:
The reason that Theorem 1 does not hold for the sublattice $A_1\oplus A_7$ is that $A_7$ does have a self-glue in $E_8$.

Let us define the $E_8$ lattice as the set of vectors $v = (v_1,...,v_8)$ such that all $v_i$'s are simultaneously integer or half-integer, and $\sum_{i = 1}^8 v_i \in 2 \mathbb Z$, as usual.

An embedding of $A_1\oplus A_7$ into $E_8$ is given by the basis vectors

$$ A_1: (\tfrac12, \tfrac12, \tfrac12, \tfrac12, \tfrac12, \tfrac12, \tfrac12, \tfrac12) $$

and

$$A_7: (1,-1,0,...,0), (0,1,-1,0,...,0),...,(0,...,0,1,-1). $$

Now, the real span of this particular $A_7$ is the set of points $\{x = (x_1,...,x_8) | \sum_{i = 1}^8 x_i = 0\}$, which contains in particular the vector $v = (\tfrac12, \tfrac12, \tfrac12, \tfrac12, -\tfrac12, -\tfrac12, -\tfrac12, -\tfrac12)$. This vector is a self-glue for $A_7$, extending it to $E_7$, and we recover the case $A_1 \oplus E_7 \subset E_8$, which satisfies Theorem 1.

In the theory of lattices we would equivalently say that $A_7$ is not primitively embedded into $E_8$.

The lattices $A_1 \oplus A_2$ and $A_5$ can be shown to be primitively embedded in $E_8$ such that they are mutually orthogonal, hence they satisfy Theorem 1.


问:
If $a, b$ are positive in a unital $C^*$-algebra and $a+b = 0$ are then $a$ and $b$ also $0$?
The title already says most of the question (our definition of positive means self-adjoint and having a positive, real spectrum), but the reason I was wondering about this is because we just covered this topic in functional analysis and I was asking myself if this property holds in general besides just in the real numbers etc. It would seem intuitive especially when people use the notation $a \geq 0$ for positivity (even though it's a bit misleading in my opinion). Could someone give me some insight if this is true or if it has counterexamples?

答: If $a$ is positive then $\sigma(a)\subseteq [0,\infty)$.  If moreover $a=-b$, with $b$ positive, then $\sigma(a)\subseteq (-\infty,0]$, by the spectral mapping Theorem.    Thus $\sigma(a)=\{0\}$, so the spectral radius of $a$ is zero, and hence also its norm.


问:
The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$
There is an exercise on $\sin x$. How could I see that for any $0<x< \frac \pi 2$, $\frac 2 \pi x \le \sin x\le x$?

Thanks for your help.

答:
For $x \in \left[0, \frac{\pi}{2}\right]$, we have $\sin''(x) = -\sin(x) \le 0$. So the sine function is concave on $\left[0, \frac{\pi}{2}\right]$. So the inequality follows from the principle (I suggest drawing the graph to see it clearly) :

$$\textrm{secant} \le \textrm{function} \le \textrm{tangent}$$


问:
Analysis of series using ratio test

$$\sum_{n=1}^{\infty} \frac{n^2+1}{n+2}$$

$$\sum_{n=1}^{\infty} \frac{b (n!)}{n^n}$$ where $b \in \mathbb{N}$.

$$\sum_{n=1}^{\infty} \frac{1}{10^{n+(-1)^n}}$$ and yes, the entire thing is an exponent.

$$\sum_{n=1}^{\infty} (-1)^n \frac{5^n}{2^n+1}$$

Given these equations above, which of the following would the ratio test work on? The only fact I know about the ratio test is that if the limit of $|\frac{a_{n+1}}{a_{n}}|=1$, then it fails.

For the 1st, I simply applied the ratio test and got 1, so the ratio test is ineffective. For the 2nd, the $b$ actually cancels out, so it's not relevant. It ends up being something like $\frac{n^n}{(n+1)^n (n+1)}$, which is not 1.

For the 3rd, I separated the $10^{n+(-1)^n} \implies 10^n  10^{(-1)^n}$ and it ended up being a real number that is $0$? I wasn't sure about this result. The last one was just expanding and simplifying, and I got $\frac{-5}{2}$.

Overall, only the first one fails the ratio test, if I calculated it correctly. Let me know if I did anything wrong

Also, I know that I can use easier tests, like for the 4th, use the alternate series test, but this is how it is.

答:
Indeed it fails here, for the reason you gave.

Note that$$\lim_{n\to\infty}\frac{n^n}{(n+1)^n(n+1)}=\lim_{n\to\infty}\frac1{(1+1/n)^n(n+1)}=0.$$Therefore, the test does not fail in this case.

In this case, the quotient is $10^{-3}$ in half the cases and $10$ in the other half. So, the test fails in this case.

In other to apply the ratio test in the fourth case, apply it to the absolute values (that is, to the series $\sum_{n=1}^\infty\frac{5^n}{2^n+1}$). You will get $\frac52$. So, the test works (and it tells you that the series diverges).

Title: Analysis of series using ratio test