答:
The $(F_k)$ are a sequence of identically distributed random variables, but they not identical.  Thus $\mathsf{Var}(\sum_{k=1}^r F_k)$ does not equal $\mathsf {Var}(k F)$.

So, once you have successfully argued that they are also independent, then you may use:$$\begin{align}\mathsf{Var}(\sum_{k=1}^n F_k)&=\sum_{k=1}^n \mathsf{Var}(F_k)\\[1ex]&=n\,\mathsf{Var}(F)\end{align}$$

And how would I prove that the Fi are independent?

Independently from the count for failures and successes that have preceded any success, the trials after that will still form a sequence of independent Bernoulli random variables with the identical success rate.

Title: Number of failures until first r^th success - Expectation, Variance, Probability

Body:
I was given the following problem:

Given:

For the r.v $\displaystyle F=number\ of\ failures\ before\ first\ success$.

$\displaystyle E[ F] \ =\ \frac{1-p}{p}$ and $\displaystyle Var[ F] \ =\ \frac{1-p}{p^{2}}$.

Let's now define $\displaystyle X=number\ of\ failure\ before\ r^{th} \ success$, calculate the $\displaystyle E[ X] ,\ Var[ X]$.

My solution was:

Let's notice that if $\displaystyle X=the\ number\ of\ failures\ till\ r^{th} \ success$, then we can define it using the previous question. We have $\displaystyle r$ successes, each success has an $\displaystyle i$ number of falilures before it, for $\displaystyle 0\leqslant i\leqslant k$.

So we can define:

$\displaystyle X\ =F_{1} +F_{2} +F_{3} +...+F_{r}$.

Where each $\displaystyle F_{i} =the\ number\ of\ failures\ after\ the\ ( i-1)^{th} \ success\ and\ before\ the\ i^{th} \ success$.

For example:

$\displaystyle F_{1}$ will be the number of failures before the $\displaystyle 1^{st}$ success.

$\displaystyle F_{2}$ will be the number of failure between the $\displaystyle 1^{st}$ and $\displaystyle 2^{nd}$ success.

$\displaystyle F_{3}$ will be the number of failure between the $\displaystyle 2^{nd}$ and $\displaystyle 3^{rd}$ success.

etc.

Note that each $\displaystyle F_{i}$ is simply the number of failures before some first success, meaning, it is exactly the random variable $\displaystyle F$ we were given.

Expectation:

$\displaystyle E[ X] =E[ F_{1} +F_{2} +F_{3} +...+F_{r}] =E[ F_{1}] +[ F_{2}] +[ F_{3}] +...+[ F_{r}] =r\cdotp \frac{1-p}{p} .$

Variance:

As per the $\displaystyle Var[ X]$, I'm having some trouble calculating it.

I can't use linearity with Variance, and I fail to understand wether I can simply say that:

$\displaystyle Var[ X] \ =\ Var[ rF] =r^{2} Var[ F]$.

Or, I should prove that each $\displaystyle F_{1} +F_{2} +F_{3} +...+F_{r}$ is indepentent, and so deduct that:

$\displaystyle Var[ X] \ =rVar[ F]$.

Which one of the two is correct? And how would I prove that the $\displaystyle F_{i}$ are independet?


问:
How does the increase in overall number of events affect the peak (events/time)?
I have a (simple?) question that I hope someone will find interesting enough to help me out with.

A web site has a given number of subscribers who generate a certain amount of traffic on the web server. Hits (web requests) are distributed throughout the day and result in a peak load of x hits/second.

Question: should the number of users increase (let's say double) and follow the same usage pattern, how would the peak load (hits/second) be affected?

I strongly doubt that the answer is 2x hits/second and realise that more information might be needed, to start from the actual distribution of the hits and the average total hits per day.

Anyone intrigued and kind enough to lend a hand?

Thanks,
Paolo

答:
It does indeed depend on the distribution and number of hits.

For example, suppose everyone always logged on at the same moment. Then the peak doubles.

Now suppose that the distribution is completely random throughout the day, and that you go from 1 user to 2 users. The chance that they log on at the same time is 1/86400. Therefore the peak will almost certainly stay at 1.

Basically, if the distribution is 'filled out' by the current usage, it will go up by 2, if it's sparse, it will stay the same. In between, it will vary, but on average increase by a factor between 1 and 2.

ContentLicense: CC BY-SA 3.0


问:
How to write $\sigma(C)$?
Suppose you are given a collection $C$ of subsets of a set $\Omega$. You can always complete this set to a $\sigma$-algebra as follows. Let $F(C)$ denote all $\sigma$-algebra of $\Omega$ that contains C; the $\sigma$-algebra generated by C is

$$\sigma(C) =\bigcap_{G\in F(C)}G$$

Suppose $\Omega =\{1,2,3,4,5,6\}$ and $C=\{\{1,2\},\{2,3\}\}$

Then how can I write $\sigma(C)$?

As I guess, firstly I found its atoms. But I cannot proceed this question

My answer is $\sigma(C)=\{\emptyset, 1, 2, 3, \{1,2\}, \{2,3\}\}$

ContentLicense: CC BY-SA 4.0

Title: How to write $\sigma(C)$?

Body:
Yes, firstly find the atoms of $\sigma(C)$. There are FOUR atoms: $\{1\}$, $\{2\}$, $\{3\}$, $\{4,5,6\}$. So, we know that $\sigma(C)$ must have $2^4$ sets. In fact:

$\sigma(C)=\{\emptyset, \{1\}, \{2\}, \{3\}, \{4,5,6\}, \{1,2\}, \{1,3\}, \{1,4,5,6\},\{2,3\}, \{2,4,5,6\}, \{3,4,5,6\}, \{1,2,3\},    \{1,2,4,5,6\}, \{1,3,4,5,6\}, \{2,3,4,5,6\}, \{1,2,3,4,5,6\}\}$.


问:
Applications of the orbit stabiliser theorem
I'm trying to teach the orbit-stabiliser theorem.

One nice application of the theorem is counting the number of symmetries of different platonic solids and regular n-gons (you count the orbit and stabiliser of some feature, e.g. a particular face).  Another is counting the number of distinct 'words' that you can get my permuting the letters of a given word, e.g. BANANAS (you want the orbit of the action of $S_n$ on the letters, which you find by dividing $n!$ by the size of the stabiliser.  In this case, the stabiliser has size $3!\cdot 2!$ - you can permute the As and the Ns separately.).

I'd like a couple of other examples of applications to demonstrate that the (finite) orbit-stabiliser theorem is useful.  Ideally, they would be of obvious interest outside of group theory and wouldn't require developing much theory to understand them.  (The two examples above meet both these criteria in my mind.)

Does anyone have any good examples?

答:
The definition of Schreier generators (without the proof that they generate the subgroup) is quite easy and they provide natural solving moves for puzzles such as Rubik's cube, see e.g. page 98 in
https://www.math.colostate.edu/~hulpke/CGT/howtogap.pdf</p>


问:
Elegant solution to $e^{xe^{ix}} = e^{ix}, x \in \mathbb{R}$
I recently had this problem on an exam and I was able to find the following solution, but it seemed really ugly and I'm wondering if there is a more elegant approach to this problem.

Find all $x\in \mathbb{R}$ satisfying $e^{xe^{ix}} = e^{ix}$.
My solution was as follows (note that I use $\log$ for the complex valued natural log and $\ln$ for the real valued natural log):
Fix some $x \in \mathbb{R}$ such that $e^{xe^{ix}} = e^{ix}$ holds. First we take the $\log$ of both sides, yielding
\begin{align*}
    \log(e^{xe^{ix}}) &= \log(e^{ix})\\
    \ln|e^{xe^{ix}}| + i \arg (e^{xe^{ix}}) &= \ln|e^{ix}| + i \arg(e^{ix})\text{ by definition of } \log\\
    \ln|e^{x(\cos(x)+i\sin(x))}| + i \arg (e^{xe^{ix}}) &= \ln(1)+i \arg(e^{ix}) \text{ by Euler's formula}\\
    \ln(|e^{x\cos(x)}e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &= 0+i \arg(e^{ix})\\
    \ln(|e^{x\cos(x)}||e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\text{ since $|\cdot|$ is multiplicative}\\
\end{align*}
Note that since $x\in \mathbb{R}$, $x\sin(x), x \cos(x) \in \mathbb{R}$ as well, so $|e^{ix\sin(x)}| = 1$ and $e^{x\cos(x)} \in \mathbb{R}$. Thus we have:
\begin{align*}
    \ln(|e^{x\cos(x)}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
    x\cos(x) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
\end{align*}
Matching real and imaginary parts, we get that
\begin{align*}
        x\cos(x) &= 0\\
        \arg(e^{xe^{ix}}) &= \arg(e^{ix})
\end{align*}
From the left equality, we get $x \in \{0\} \cup \{\frac{\pi}{2}+n\pi| n \in \mathbb{Z}\}$. Clearly $x = 0$ is a valid solution for the second equality, so it suffices to check which values of $n$ satisfy the second equality. We see that
\begin{align*}
    \arg(e^{(\pi/2 +n \pi)e^{i(\pi/2 +n \pi)}}) &= \arg(e^{i(\pi/2 +n \pi)})\\
    \arg(e^{(\pi/2 +n \pi)(-1)^ni})&= \frac{\pi}{2} + n \pi \\
    (-1)^n(\pi/2 +n \pi) &=\frac{\pi}{2} + n \pi
\end{align*}
Which holds when $n$ is even, so the full solution set is $x \in \{0\}\cup \{\frac{\pi}{2} + 2m \pi| m \in \mathbb{Z}\}$.

答: For $r_1,r_2,\theta_1,\theta_2 \in \mathbb R$ we have $r_1e^{i\theta_1} = r_2e^{i\theta_2}$ if and only if $r_1=r_2$ and $\theta_1-\theta_2 \in 2\pi \mathbb Z := \{2\pi k : k \in \mathbb Z\}$. Thus $e^{-\sin x}e^{i \cos x} = e^{ie^{ix}} = e^{ix}$ implies that $e^{- \sin x} = 1$ and $\cos x - x \in 2\pi \mathbb Z$. Now, from $e^{- \sin x} = 1$ we get $x = k\pi$ for some $k \in \mathbb Z$ and from $$(-1)^k-k\pi = \cos x - x \in 2\pi \mathbb Z$$ we get...


问:
What is the probability of choosing 15 correct bags out of 40 with 8 fake ones?
In a certain television program there are 40 bags, 15 to be chosen if the participant answers correctly (but suppose he gets them all right) out of those 40 there are 8 bags that will remove the high prizes.
So what is the chance that I will choose 15 correct bags?

I thought that way:

P = probability of the correct bag

P(first bag) = 32/40, P(second bag) = 31/39 and so on...

In this logic the answer would be:

(32!/(32-15)!)/(40!/(40-15)!) = 0.01406383759 or 1.4%

But I have a feeling that this is wrong because I learned at school that when the order doesn't matter we use the combination formula, I hope someone will answer me if I'm correct :)

答:
You are correct in a sense. We have to formulas to consider and compare:
$$
nPr=\frac{n!}{(n-r)!}\text{ and }nCr=\frac{n!}{r!(n-r)!}
$$
but when you divide two of those using different values of $n$ but the same value of $r$, the two results will be equal whether using combinations (nCr) or permutations (nPr).

I will give you two reasons for this:

Suppose we give the good bags an ID 1 through 32 and the bad ones 33 through 40. Then you could consider a successful outcome either as an ordered sequence from the 32 good bags, or just some subset of the 32 bags without considering their order. For the ordered case, to compute the probability of success, you would then have to divide by ALL ordered sequences from the 40 bags. Hence $32P15/40P15$. But if you only considered groups of $15$ without considering order, it should be $32C15/40C15$.

The factor $r!$ cancels out when dividing the combinations formulas with the same $r$.


答:
$e^{xe^{ix}}=e^{ix} $ if and only if: $\ e^{xe^{ix}-ix}=1\ $.

$e^{xe^{ix}}=e^{ix} $ if and only if: $\ e^{x\cos x} e^{i(x\sin (x)-x)}=1$

$e^{xe^{ix}}=e^{ix} $ if and only if: ($\ x\cos(x)=0 \ $) and ($\ x\sin(x)-x=2k\pi \ $ with $k\in\mathbb Z$)

...

Title: Elegant solution to $e^{xe^{ix}} = e^{ix}, x \in \mathbb{R}$

Body:
I recently had this problem on an exam and I was able to find the following solution, but it seemed really ugly and I'm wondering if there is a more elegant approach to this problem.

Find all $x\in \mathbb{R}$ satisfying $e^{xe^{ix}} = e^{ix}$.
My solution was as follows (note that I use $\log$ for the complex valued natural log and $\ln$ for the real valued natural log):
Fix some $x \in \mathbb{R}$ such that $e^{xe^{ix}} = e^{ix}$ holds. First we take the $\log$ of both sides, yielding
\begin{align*}
    \log(e^{xe^{ix}}) &= \log(e^{ix})\\
    \ln|e^{xe^{ix}}| + i \arg (e^{xe^{ix}}) &= \ln|e^{ix}| + i \arg(e^{ix})\text{ by definition of } \log\\
    \ln|e^{x(\cos(x)+i\sin(x))}| + i \arg (e^{xe^{ix}}) &= \ln(1)+i \arg(e^{ix}) \text{ by Euler's formula}\\
    \ln(|e^{x\cos(x)}e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &= 0+i \arg(e^{ix})\\
    \ln(|e^{x\cos(x)}||e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\text{ since $|\cdot|$ is multiplicative}\\
\end{align*}
Note that since $x\in \mathbb{R}$, $x\sin(x), x \cos(x) \in \mathbb{R}$ as well, so $|e^{ix\sin(x)}| = 1$ and $e^{x\cos(x)} \in \mathbb{R}$. Thus we have:
\begin{align*}
    \ln(|e^{x\cos(x)}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
    x\cos(x) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
\end{align*}
Matching real and imaginary parts, we get that
\begin{align*}
        x\cos(x) &= 0\\
        \arg(e^{xe^{ix}}) &= \arg(e^{ix})
\end{align*}
From the left equality, we get $x \in \{0\} \cup \{\frac{\pi}{2}+n\pi| n \in \mathbb{Z}\}$. Clearly $x = 0$ is a valid solution for the second equality, so it suffices to check which values of $n$ satisfy the second equality. We see that
\begin{align*}
    \arg(e^{(\pi/2 +n \pi)e^{i(\pi/2 +n \pi)}}) &= \arg(e^{i(\pi/2 +n \pi)})\\
    \arg(e^{(\pi/2 +n \pi)(-1)^ni})&= \frac{\pi}{2} + n \pi \\
    (-1)^n(\pi/2 +n \pi) &=\frac{\pi}{2} + n \pi
\end{align*}
Which holds when $n$ is even, so the full solution set is $x \in \{0\}\cup \{\frac{\pi}{2} + 2m \pi| m \in \mathbb{Z}\}$.


答:
They are two different ways to calculate the same probability.

$$\frac{\binom{32}{15}}{\binom{40}{15}} = \frac{\frac{32\times\dots\times18}{15!}}{\frac{40\times\dots\times26}{15!}} = \frac{32\times\dots\times18}{40\times\dots\times26} = \frac{32}{40} \times \frac{31}{39} \times \dots \times \frac{18}{26}$$

LHS : number of correct combinations divided by number of possible combinations

RHS : your thinking

Note that the $15!$ on both the denominator and the numerator cancel each other.

Title: What is the probability of choosing 15 correct bags out of 40 with 8 fake ones?

Body:
In a certain television program there are 40 bags, 15 to be chosen if the participant answers correctly (but suppose he gets them all right) out of those 40 there are 8 bags that will remove the high prizes.
So what is the chance that I will choose 15 correct bags?

I thought that way:

P = probability of the correct bag

P(first bag) = 32/40, P(second bag) = 31/39 and so on...

In this logic the answer would be:

(32!/(32-15)!)/(40!/(40-15)!) = 0.01406383759 or 1.4%

But I have a feeling that this is wrong because I learned at school that when the order doesn't matter we use the combination formula, I hope someone will answer me if I'm correct :)


问:
Showing that $\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$ diverges
I know that the series diverge, I'm just having hard time showing it.

$$\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$$

Thanks in advance

答: Hint:  Use the comparison test with 1/log(n)</p>


问:
How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$

答:
Just a little variation, but it may be nice as an alternative.

Recall that
$$(1+\alpha(x))^\frac1{\alpha(x)}\to e$$
whenever $\alpha(x) \to 0$.

Replace $t= x - a$, to get
\begin{eqnarray}\mathcal L &=&\lim_{t \to 0} \left[\frac{\sin(t+a)}{\sin a}\right]^\frac1t=\\
&=&\lim_{t\to 0}\left(\frac{\sin t \cos a + \cos t \sin a}{\sin a}\right)^{\frac1t}=\\
&=&\lim_{t\to 0}\left(\sin t \cot a + \cos t\right)^{\frac1t}=\\
&=&\lim_{t\to 0}\left[1+(\cos t +\sin t \cot a - 1)\right]^{\frac1t}=\\
&=&\lim_{t\to 0}\left\{\left[1+(\cos t +\sin t \cot a - 1)\right]^{\frac1{\cos t +\sin t \cot a -1}}\right\}^{\frac{\cos t + \sin t \cot a -1}{t}}=\\
&=&\lim_{t\to 0}e^{\frac{\cos t -1}t +\cot a \frac{\sin t}t}=e^{0+\cot a}=e^{\cot a}.
\end{eqnarray}


答:
The function $$f(z) : \mathbb C \to \mathbb C \\ f(z) = e^z$$ is periodic with period $2\pi i$, consequently, $g(z) = f(iz)$ is periodic with period $2\pi$.  The condition $$e^{ie^{ix}} = e^{ix} \iff g(g(x)) = g(x) \tag{1}$$ therfore requires $$e^{ix} = g(x) = x + 2\pi k \tag{2}$$ for some integer $k$, in the sense that any solution to $(1)$ is a solution to $(2)$ and vice versa.  if $x \in \mathbb R$, we must then have $e^{ix} = \cos x + i \sin x = x + 2\pi k$ so $\sin x = 0$, or $x \in \mathbb \pi m$ for some integer $m$.  But this would imply $g(\pi m) = (-1)^m$, and $g(g(\pi m)) = e^{\pm i}$, which has a nonzero imaginary part, thus there is no solution.

We can also plot the locus of $$(\Re(g(g(x))-g(x)), \Im(g(g(x)) - g(x)))$$ for $x \in [0, 2\pi)$:

This shows that this function has no zeros for $x \in \mathbb R$.

Since the question was edited to fix an error, the solution is radically different.

Let $x = \theta + 2\pi k$ for some $\theta \in [0,2\pi)$ and $k \in \mathbb Z$.  Then $g(x) = e^{ix} = \cos \theta + i \sin \theta$, and $$\begin{align}
h(x) &= e^{x g(x)} - g(x) \\
&= e^{x \cos \theta} e^{i x \sin \theta} - \cos \theta - i \sin \theta \\
&= e^{x \cos \theta} \cos (x \sin \theta) - \cos \theta + i \left( e^{x \cos \theta} \sin (x \sin \theta) - \sin \theta \right).
\end{align}$$
As we require both real and imaginary parts to be zero, this implies $$\cos \theta = e^{x \cos \theta} \cos (x \sin \theta) \\ \sin \theta = e^{x \cos \theta} \sin (x \sin \theta).$$  It follows from taking the sum of the squares that $$e^{2x \cos \theta} = 1.$$  Since the exponent is real-valued, we conclude $2x \cos \theta = 0$, which is true if and only if $x = 0$ or $\theta \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.  In the first case, we obtain $$h(0) = 0,$$ and in the second, we find $x = 3\pi/2$ to be extraneous.  Just to check, we compute $$g(\pi/2 + 2\pi k) = e^{i \pi/2} = i,$$ hence $$h(\pi/2 + 2\pi k) = e^{(\pi/2 + 2\pi k)i} - i = e^{i \pi/2} - i = 0,$$ so the complete solution set is $$x \in \{0, \pi/2 + 2\pi k\}$$ for $k \in \mathbb Z$.


问:
Existence of epimorphism from a Ring to a Module
So the problem is the following, given a conmutative ring $R$ and a free $R$-module let's say $M$, such that $\dim(M)=2$.

Prove there is no epimorphism from $R$ to $M$

I've been struggling for a while to solve this problem. I believe the solution has to with what elements in the ring are sent to the basis of $M$ but I literally have no clue of where to start.

Any help would be appreciated

答: Let $\mathfrak{m}$ be a maximal ideal in $R$. Tensoring is right-exact, so if we take a surjection $R \twoheadrightarrow R^2$, we can tensor with $R/\mathfrak{m}$ to obtain a surjection $R \otimes_R R/\mathfrak{m} \cong R/\mathfrak{m} \to R^2 \otimes_R R/\mathfrak{m} \cong (R/\mathfrak{m})^2$. But $R/\mathfrak{m}=:k$ is a field and it's well-known from linear algebra that there is no $k$-linear surjection $k \to k^2$ for a field $k$.


问:
Tricks for Constructing Hilbert-Style Proofs
Several times in my studies, I've come across Hilbert-style proof systems for various systems of logic, and when an author says, "Theorem: $\varphi$ is provable in system $\cal H$," or "Theorem: the following axiomatizations of $\cal H$ are equivalent: ...," I usually just take the author's word as an oracle instead of actually trying to construct a Hilbert-style proof (can you blame me?). However, I would like to change this habit and least be in a position where I could check these claims in principle.

On the few occasions where I actually did try to construct a (not-so-short) Hilbert-style proof from scratch, I found it easier to first construct a proof in the corresponding natural deduction system, show that the natural deduction system and the Hilbert system were equivalent, and then try to deconstruct the natural deduction system into a Hilbert-style proof (in the style of Anderson and Belnap). The problem with that (apart from being tortuous) is that I would need the natural deduction system first, and it's not always obvious to me how to construct the natural deduction system given the axioms (sometimes it's not so bad; it's easy to see, for instance, that $(A \rightarrow (A \rightarrow B)) \rightarrow (A \rightarrow B)$ corresponds to contraction; but it's not always that easy...).

So I'm wondering: are there "standard tricks" for constructing Hilbert-style proofs floating around out there? Or are there tricks for constructing a corresponding natural deduction system given a set of Hilbert axioms? Or is it better to just accept proof-by-oracle?

答:
The standard method to construct a Hilbert Style proof from a Natural Deduction proof is so called Bracket Abstraction. It appeared for example in Curry and Feys, Combinatory Logic, 1958. You can extract a Lambda Expression from a Natural Deduction proof for example:

```<code>     -------- (id)
      p |- p
   ------------ (->I)
    p |- q -> p
   ---------------- (->I)
   |- p -> q -> p
</code>```

The (id) corresponds to a variable in a Lambda Expression and the (->) corresponds to abstraction in a Lambda Expression. The above proof has the Lambda Expression λx.λy.x. For more details see the Curry-Howard isomorphism.

Bracket Abstraction can turn a Lambda Expression into an SKI Expression. And an SKI Expression corresponds to a Hilbert Style proof. But Bracket Abstraction can be difficult. One attempt might lead to S(KK)I which is this odd proof:

```<code>1. ax-2: (p -> p -> q -> p) -> (p -> p) -> p -> q -> p
2. ax-1: (p -> q -> p) -> p -> p -> q -> p
3. ax-1: p -> q -> p
4. ax-mp(2, 3): p -> p -> q -> p
5. ax-mp(1, 4): (p -> p) -> p -> q -> p
6. ax-0: p -> p
7. ax-mp(5, 6): p -> q -> p
</code>```

If a little bit more is invested into the Bracket Abstraction the result might be the combinator K, and thus telling us that the Natural Deduction proof is already a Hilbert Style axiom in a certain axiomsystem.

Braket Abstraction can be put on the computer like here.

Body:
The standard method to construct a Hilbert Style proof from a Natural Deduction proof is so called Bracket Abstraction. It appeared for example in Curry and Feys, Combinatory Logic, 1958. You can extract a Lambda Expression from a Natural Deduction proof for example:

     -------- (id)
      p |- p
   ------------ (->I)
    p |- q -> p
   ---------------- (->I)
   |- p -> q -> p

The (id) corresponds to a variable in a Lambda Expression and the (->) corresponds to abstraction in a Lambda Expression. The above proof has the Lambda Expression λx.λy.x. For more details see the Curry-Howard isomorphism.

Bracket Abstraction can turn a Lambda Expression into an SKI Expression. And an SKI Expression corresponds to a Hilbert Style proof. But Bracket Abstraction can be difficult. One attempt might lead to S(KK)I which is this odd proof:

1. ax-2: (p -> p -> q -> p) -> (p -> p) -> p -> q -> p
2. ax-1: (p -> q -> p) -> p -> p -> q -> p
3. ax-1: p -> q -> p
4. ax-mp(2, 3): p -> p -> q -> p
5. ax-mp(1, 4): (p -> p) -> p -> q -> p
6. ax-0: p -> p
7. ax-mp(5, 6): p -> q -> p

If a little bit more is invested into the Bracket Abstraction the result might be the combinator K, and thus telling us that the Natural Deduction proof is already a Hilbert Style axiom in a certain axiomsystem.

Braket Abstraction can be put on the computer like here.


问:
Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$
Let a,b and c be the three sides of a triangle. Show that $\dfrac 1a+\dfrac 1b+\dfrac 1c\le \dfrac 1{a+b-c}+\dfrac 1{a-b+c}+\dfrac 1{-a+b+c}.$ When does equality occur?

I have used the Ravi-transformation and obtained

$\frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{(x+y)+(y+z)-(x+z)} + \frac{1}{(x+y)-(y+z)+(x+z)} \\ +  \frac{1}{-(x+y)+(y+z)+(x+z)} \Longleftrightarrow  \frac{1}{x+y} + \frac{1}{x+z} + \frac{1}{y+z} \leq \frac{1}{2x} + \frac{1}{2y} + \frac{1}{2z}.$

But I get stuck here. Does anyone have an idea to proceed?

答:
If I understand correctly you have solved the problem already!

Take:
$$ x = \frac{a+b-c}{2} $$
$$ y = \frac{a-b+c}{2} $$
$$ z = \frac{-a+b+c}{2} $$
And note that $x+y = a$ , $x+z = b$ , and $y+z = c$.

The last inequality that you reached is what the problem is asking to prove.


问:
Smooth maps on a manifold lie group
$$
\operatorname{GL}_n(\mathbb R) = \{ A \in M_{n\times n} | \det A \ne 0 \} \\
\begin{align}
&n = 1, \operatorname{GL}_n(\mathbb R) = \mathbb R - \{0\} \\
&n = 2, \operatorname{GL}_n(\mathbb R) = \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\Bigg| ad-bc \ne 0\right\}
\end{align}
$$

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a group.

$AB$ is invertible if $A$ and $B$ are invertible.

$A(BC)=(AB)C$

$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$A^{-1}$ is invertible if $A$ is invertible.

$$(\operatorname{GL}_n(\mathbb R) := \det{}^{-1}(\{0\}))$$
$\det{}^{-1}(\{0\})$ is open in $M_{n\times n}(\mathbb R)$.

$\det : M_{n\times n}(\mathbb R) \to \mathbb R$ is continuous, why?

$\dim \operatorname{GL}_n(\mathbb R) = n^2 - 1$, why?

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a Lie group if:

$$
\mu : G\times G \to G \\
\mu(A,B) = A\cdot B \:\text{ is smooth} \\
I(A) = A^{-1} \:\text{ is smooth}
$$

How can I show this?

I want to show that general special linear group is Lie group. I could not show the my last step. How can I show that $AB$ and $A^{-1}$ are smooth? please help me I want to learn this. Thanks

Here the my handwritten notes http://i.stack.imgur.com/tkoMy.jpg

答:
Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$. </p>

<p>The other bit is similar, if you know what crammer's rule is. </p>

Title: Smooth maps on a manifold lie group

Body:
Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$.

The other bit is similar, if you know what crammer's rule is.


问:
Tests for normal distribution
Given some data, is there any test to determine if the data fits a normal distribution ( the mean and variance are not mentioned )</p>

答:
The Shapiro-Wilk test is a formal test of normality-in-general for a random sampled data - without reference to specific numerical values of the mean $\mu$ or variance $\sigma^2.$

For example, vector x has observations from a nomral
population and vector y has observations from an exponential population.

```<code>set.seed(2020)
x = rnorm(100, 1, 1)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 -2.039   0.438   1.120   1.109   1.739   4.202
shapiro.test(x)

        Shapiro-Wilk normality test

data:  x
W = 0.98906, p-value = 0.5895
</code>```

The null hypothesis that the population is normal is not rejected at the 5% level: P-value = $0.59 > 0.05 = 5\%.$

```<code>y = rexp(100, 1)
summary(y)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.008152 0.329992 0.810329 1.175030 1.679118 4.713232
shapiro.test(y)

        Shapiro-Wilk normality test

data:  y
 W = 0.85789, p-value = 2.346e-08
</code>```

The null hypothesis that the population is normal is rejected at the 5% level: P-value $\approx 0 < 0.05 = 5\%.$

Also, as suggested in the Comment by @MatthewPilling one can look at a normal probability plot (Q-Q plot), in which normal data should roughly follow a straight line. (One should not be too fussy about fit to a straight line for the lower and highest values in the sample.

Here are Q-Q plots of vectors x (left panel) and y.

```<code>par(mfrow=c(1,2))
 qqnorm(x); qqline(x, col="darkgreen")
 qqnorm(y); qqline(y, col="darkgreen")
par(mfrow=c(1,1))
</code>```

With samples as large as $n = 100$ is is usually
possible to distinguish samples from normal populations for samples from non-normal populations. But for smaller
samples, the distinction may not be so clear.

Here is a repetition of the above, but with samples of
size $n = 15.$

```<code>set.seed(1213)
x = rnorm(15, 1, 1)
shapiro.test(x)<span class="math-container">$p.val
[1] 0.02057266
y = rexp(15, 1)
shapiro.test(y)$</span>p.val
[1] 0.002650913
</code>```

The Shapiro-Wilk test mistakenly rejects normality of x at the
5% level (moderately close call), correctly rejects
normality of y with a very small P-value.

Normal probability plots: We might be willing to excuse
the ragged plot at left for x as "perhaps roughly linear", but the plot at right 'y' is clearly not linear.

Boxplots show some left-skewness for x (at left), but three high high outliers out of $n=15$ for y.

For samples as small
as $n = 15,$ one can often only speculated about normality.

```<code>boxplot(x, y, col="skyblue2", pch=19, names=T)
</code>```

Title: Tests for normal distribution

Body:
The Shapiro-Wilk test is a formal test of normality-in-general for a random sampled data - without reference to specific numerical values of the mean $\mu$ or variance $\sigma^2.$

For example, vector x has observations from a nomral
population and vector y has observations from an exponential population.

set.seed(2020)
x = rnorm(100, 1, 1)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 -2.039   0.438   1.120   1.109   1.739   4.202
shapiro.test(x)

        Shapiro-Wilk normality test

data:  x
W = 0.98906, p-value = 0.5895

The null hypothesis that the population is normal is not rejected at the 5% level: P-value = $0.59 > 0.05 = 5\%.$

y = rexp(100, 1)
summary(y)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.008152 0.329992 0.810329 1.175030 1.679118 4.713232
shapiro.test(y)

        Shapiro-Wilk normality test

data:  y
 W = 0.85789, p-value = 2.346e-08

The null hypothesis that the population is normal is rejected at the 5% level: P-value $\approx 0 < 0.05 = 5\%.$

Also, as suggested in the Comment by @MatthewPilling one can look at a normal probability plot (Q-Q plot), in which normal data should roughly follow a straight line. (One should not be too fussy about fit to a straight line for the lower and highest values in the sample.

Here are Q-Q plots of vectors x (left panel) and y.

par(mfrow=c(1,2))
 qqnorm(x); qqline(x, col="darkgreen")
 qqnorm(y); qqline(y, col="darkgreen")
par(mfrow=c(1,1))

With samples as large as $n = 100$ is is usually
possible to distinguish samples from normal populations for samples from non-normal populations. But for smaller
samples, the distinction may not be so clear.

Here is a repetition of the above, but with samples of
size $n = 15.$

set.seed(1213)
x = rnorm(15, 1, 1)
shapiro.test(x)$p.val
[1] 0.02057266
y = rexp(15, 1)
shapiro.test(y)$p.val
[1] 0.002650913

The Shapiro-Wilk test mistakenly rejects normality of x at the
5% level (moderately close call), correctly rejects
normality of y with a very small P-value.

Normal probability plots: We might be willing to excuse
the ragged plot at left for x as "perhaps roughly linear", but the plot at right 'y' is clearly not linear.

Boxplots show some left-skewness for x (at left), but three high high outliers out of $n=15$ for y.

For samples as small
as $n = 15,$ one can often only speculated about normality.

boxplot(x, y, col="skyblue2", pch=19, names=T)


问:
Details about a Recurrence Relation problem.
I am trying to understand Recurrence Relations through the Towers of Hanoi example, and I am having trouble understanding the last step:

If $H_n$ is the number of moves it takes for n rings to be moved from the first peg to the second peg, then:

$H_n = 2H_{n−1} + 1$

$H_n = 2(2H_{n−2} + 1) + 1 = 2^2H_{n−2} + 2 + 1$

$H_n = 2^2(2H_{n−3} + 1) + 2 + 1 = 2^3H_{n−3} + 2^2 + 2 + 1 $

$\vdots$

$H_n = 2^{n−1}H_1 + 2^{n−2} + 2^{n−3} +· · ·+2 + 1$

$H_n = 2^{n−1} + 2^{n−2} +· · ·+2 + 1$

$H_n = 2^n − 1.$

I am having trouble understanding the last two steps. How does one go from this:

$H_n = 2^{n−1} + 2^{n−2} +· · ·+2 + 1$

To this:

$H_n = 2^n − 1$

?

答:
The last few lines contain errors. The line

$$H_n=2^{n-1}H_1+2^{n-2}+2^{n-3}+\ldots+2+1$$

is correct, but the next line is not: since $H_1=1$, it should be

$$H_n=2^{n-1}+2^{n-2}+2^{n-3}+\ldots+2+1\;.$$

Apparently the exponents got dropped down onto the main line of text. This is now the sum of a geometric series:

$$H_n=\sum_{k=0}^{n-1}2^k=\frac{2^n-1}{2-1}=2^n-1\;,$$

and again it appears that an exponent was accidentally dropped down to the main line.


问:
Best book for tensor products
I'm trying to improve my understanding of tensor of products of modules, my algebra background is one semester of undergrad algebra and one semester of graduate algebra, in which I struggled through the section on tensor products. In my graduate course, we used Dummit and Foote and I generally liked their exposition, but found the tensor product section very confusing, so I'm looking for other sources. I'm not interested in anything specific, i.e. not just limited to vector spaces, or $\mathbb{Z}$ modules, although I don't really care about the non-commutative case. Does anyone have any recommendations?

答:
You can read about tensor products all over the place.

The wikipedia page (https://en.wikipedia.org/wiki/Tensor_product_of_modules) is reasonably good and complete if you're looking for a free resource.

Personally, I'm partial to Atiyah and Macdonald's treatment in Chapter 2 of their Commutative Algebra. It has the benefit of being quite to the point and there are some useful and pleasant exercises.

You could also try Algebra by Lang.

Keith Conrad's notes: https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf also appear quite complete, and I've heard good things about the writing.

Title: Best book for tensor products


问:
Suppose $f(x)-f(y)=h(g(x)-g(y))$, then $h$ is linear?
Background $f,g$ are real functions, $h$ is increasing, satisfying:

$$f(x)-f(y)=h(g(x)-g(y)),$$

What function is $h$?

Motivation: At a first glance, $h$ can be any function, since $f,g$ are arbitrary functions. However, things are not that simple.

Hypothesis: $h$ must be linear.

My try: Let $y=c$, where $g(c)=0$ (Suppose this zero exist)

$$f(x)-f(c)=h(g(x)-g(c))$$
So,
$$f(x)=h(g(x)-g(c))+f(c)$$

Then, $h(g(x)-g(y))=h(g(x)-g(c))+f(c)-(h(g(y)-g(c))+f(c))$

$$h(g(x)-g(y))=h(g(x)-g(c))-h(g(y)-g(c))$$

Since $g(c)=0$,

$$h(g(x)-g(y))=h(g(x))-h(g(y))$$

By definition, $h$ is linear functional. That is, $h(x)=ax+b$.

However, I have no idea how to prove this without assuming that $g(c)=0$ for some constant $c$.

答:
You do not give precise informations about $f, g, h$, so let us assume that all are functions $\mathbb R \to \mathbb R$. They satisfy
$$f(x)-f(y)=h(g(x)-g(y)) .\tag{1}$$

Let us first observe that on both sides of $(1)$ only the differences $f(x) - f(y)$ and $g(x) - g(y)$ occur. Thus for any two numbers $c,d$ the functions $f_c(x) = f(x) + c$ and $g_d(x) = g(x) + d$ also satisfy $(1)$. This means that w.l.o.g. we may assume that $f(0) = g(0) = 0$. With $y = 0$ we get
$$f(x) = h(g(x))  \tag{2}$$
and therefore
$$h(g(x)) - h(g(y))  = h(g(x) - g(y)) \tag{3}$$
Clearly, $(3)$ is equivalent to $(1)$  - if $(3)$ is satisfied, we may define $f$ via $(2)$. The crucial condition $(3)$ only depends on the image $G  =g(\mathbb R)$. In fact, it says that that for all $a, b \in G$ we have
$$h(a) - h(b) = h(a - b) . \tag{4}$$

If we do not have restrictions on $g$, then $G$ can be any nonempty subset of $\mathbb R$ containing  $0 = g(0)$. Condition $(4)$ gives a restriction for $h$ only on the set $G^* = \{ a - b \mid a, b \in G \}$. This is a symmetric set ($G^*
 = -G^*$) containing $0$. Outside of $G^*$ we cannot say anything about $h$. But what we can say is that $h$ is an odd function on $G^*$ (take $a  = 0$). In particular, $h(0) = 0$.

Now assume that $g$ and $h$ are continuous. Then either $G = \{0\}$ (and $G^* = \{0\}$) or $G$ is an interval. In the first case $h$ can be any function with $h(0) = 0$. In the second case $G^*$ is a symmetric interval. Pick $r  \in G^*$, $r > 0$. We get $h(r) - h(r/2) = h(r - r/2) = h(r/2)$, thus $h(r/2) = h(r)/2$. Inductively we get $h(r/2^n) = h(r)/2^n$. By another induction we get
$$h(k\cdot r/2^n) = k/2^n \cdot h(r) \tag{5}$$
for $k=1,\ldots,2^n$. Note that $h((k+1)\cdot r/2^n) = h(k\cdot r/2^n  -(-r/2^n)) =  h(k\cdot r/2^n) - h(-r/2^n) = h(k\cdot r/2^n) + h(r/2^n)$. The set of all $k\cdot r/2^n$ is dense in $[0,r]$, thus $h(x) = x/r \cdot h(r)$ for $x \in [0,r]$. This means that $h$ is $\mathbb R$-linear on $[-r,r]$, $h(x) = \lambda x$. But formula $(5)$ is true for all $r' \in G^*$ with $r' > 0$. Since $r'/2^m \in [0,r]$ for some $m$, we see that $h$ is linear on $G^*$.

If $h$ is not required to be continuous, then it may have exotic behavior. Here is an example. Assume that $g$ is surjective, i.e. $G = \mathbb  R = G^*$. Let $\{b_\alpha\}_{\alpha \in A}$ be a basis of $\mathbb R$ as a vector space over $\mathbb Q$. We may assume that all $b_\alpha > 0$ (otherwise replace $b_\alpha$ by $-b_\alpha$). Pick $\alpha_0$ such that the sets $A^- =\{\alpha \in A \mid b_\alpha \le b_{\alpha_0}\}$ and $A^+ =\{\alpha \in A \mid b_\alpha > b_{\alpha_0}\}$ are nonempty. Define
$$h(b_\alpha) = \begin{cases} b_\alpha  & \alpha \le \alpha_0 \\ 2b_\alpha  & \alpha >\alpha_0 \end{cases}$$
This determines a unique $\mathbb Q$-linear map $h : \mathbb R \to \mathbb R$, thus it satisfies $(4)$. It is not $\mathbb R$-linear and not continuous. But I  must admit that I am not sure whether it is increasing.


问:
Asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$

What is the asymptotic expansion of $\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt$? As $x\to 1$.

I took the series expansion of the integrand and got $e^{\frac{1}{\log(x)}}\approx 1+\frac{1}{\log(x)}+\frac{1}{2\log(x)}+ \cdot\cdot\cdot$

and then I integrated term by term...

I got...

$\int_0^x \exp\big( \frac{1}{\log(t)} \big)~dt \approx x+li(x)+\frac{1}{2}\big(li(x)-\frac{x}{\log(x)}\big)+ \cdot\cdot\cdot$

where $li(x)$ is the logarithmic integral.

How right is this?

答:
Write $\varphi(x) = \exp(1/\log x)$ and note that $\varphi^{-1} = \varphi$. So by substituting $y = \varphi(t)$, or equivalently $t  = \varphi(y)$,

\begin{align*}
I(x)
:= \int_{0}^{x} \exp\left(\frac{1}{\log t}\right) \, \mathrm{d}y
&= \int_{1}^{\varphi(x)} y \varphi'(y) \, \mathrm{d}y \\
&= \left[ y \varphi(y) \right]_{1}^{\varphi(x)} + \int_{\varphi(x)}^{1} \varphi(y) \, \mathrm{d}y \\
&= c + x \varphi(x) - \int_{0}^{\varphi(x)} \varphi(y) \, \mathrm{d}y,
\end{align*}

where

$$c := \int_{0}^{1} \varphi(y) \, \mathrm{d}y \approx 0.27973176.$$

Using the expansion $ \varphi(y) = \sum_{n=0}^{\infty} \frac{1}{n!(\log y)^n} $ which converges uniformly for $0 < y \leq \varphi(x)$, we can perform term-wise integration, obtaining

\begin{align*}
I(x)
&= c + x \varphi(x) - \sum_{n=0}^{\infty} \int_{0}^{\varphi(x)} \frac{1}{n!(\log y)^n} \, \mathrm{d}y \\
&= c  - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \int_{0}^{\varphi(x)} \frac{\mathrm{d}y}{(\log y)^{n+1}}
\end{align*}

Now by using the integral formula

$$ \int \frac{\mathrm{d}y}{(\log y)^{n+1}}
= \frac{1}{n!} \left( \operatorname{li}(y) - \sum_{k=1}^{n} \frac{(k-1)! y}{(\log y)^k} \right), $$

where $\operatorname{li}(x) = \int_{0}^{x} \frac{\mathrm{d}t}{\log t}$ is the logarithmic integral function, we get

\begin{align*}
I(x)
&= c  - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{n!(n+1)!} \left( \operatorname{li}(\varphi(x)) - \varphi(x) \sum_{k=1}^{n} (k-1)! (\log x)^k \right) \\
&= \boxed{ c - a_0 \operatorname{li}(\varphi(x)) + \varphi(x) \sum_{k=1}^{\infty} (k-1)! a_k (\log x)^k }, \tag{*}
\end{align*}

where

$$ a_k := \sum_{n=0}^{\infty} \frac{1}{(n+k)!(n+k+1)!}. $$


问:
How many n-digit numbers can we generate when each digit can be repeated max k times?
In case each digit can be repeated 2 times, I concluded that the amount of posibilities of numbers with

3 digits: 1000 digits - [000, 111, ... 999] 10 = 990 possibilities

4 digits: 10 * 990 - [0001, 0010, 0100, ..., 999n, 99n9, 9n99] 10 * 3 * 9 = 9,630

5 digits: 10 * 9630 - [000nm, 00n0m, 0n00m, 00nm0, 0n0m0, 0nm00] 10 * 6 * 9 * 9 = 91,440

6 digits: 10 * 91440 - 10 * 10 * 9 * 9 * 9

So, for n digits, it adds
$$
 10 * (the.amount.of.n-1.posibilities) - \sum_{i=n-2}^0 i * 9^{(n-2)}
$$

In case each digit can be repeated 3 times, I concluded that the amount of posibilities of numbers with

4 digits: 4 digit: 10000 - [0000, 1111, ..., 9999] 10 = 9,990

5 digits: 10 * 9990 - [0000n, 000n0, 00n00, 0n000] 10 * 4 * 9 = 99,540

6 digits: 10 * 99540 - 10 * 10 * 9 * 9 = 987,300

7 digits: 10 * 987300 - 10 * 20 * 9 * 9 * 9 = 9,727,200

I don't see the steps 1,4,10,20 are related.

And how would you write it for an n-digit numbere where each digit can be repeated maximum k times?

答:
A partial solution can be given using exponential generating functions. We are trying to count the partitions of $n$ digits into 10 collections such that each collection has at most k elements. The number of ways of doing this is $n!$ times the $n$-th coefficient in the polynomial expansion of
$$
(1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^{k}}{k!})^{10}.
$$


问:
Find a determinant of this matrix.
I've got matrix
\begin{equation*}
A=\left(
\begin{array}{ccccccc}
a & 0 & 0 & \ldots & 0 & 0 & b \\
0 & a & 0 & \ldots & 0 & b & 0 \\
0 & 0 & a & \ldots & b & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0 & b & \ldots & a & 0 & 0 \\
0 & b & 0 & \ldots & 0 & a & 0 \\
b & 0 & 0 & \ldots & 0 & 0 & a \\
\end{array}
\right)
\end{equation*}

оf $2n$ size (number columns and rows). I need to find it's determinant using a traditional formula:

$$\det A = \sum_{\sigma \in S_{2n}}{\text{sgn}(\sigma)\prod_{i=0}^{2n}{A_{i\sigma(i)}}}$$

Where $\text{sgn}(\sigma)$ is $(-1)^{inv(\sigma)}$, and $inv(\sigma)$ is the number of inversions in $\sigma$.

It's easy to see that non-null terms in will be if and only if $\sigma(i) \in \{i, 2n - i + 1\}$.

So, the answer for me looks like: $\text{sgn}(\sigma_1)a^{k_1}b^{2n - k_1} + \text{sgn}(\sigma_2)a^{k_2}b^{2n - k_2}\ldots$. (where each $sgn(\sigma_i)$ is $\pm1$).

But I have no idea how to expand this and count all $\sigma$'s.

答: This is simply the value $(a^2-b^2)^n$

Title: Find a determinant of this matrix.

Body: This is simply the value $(a^2-b^2)^n$


问:
Complex Analysis: Picard's theorem applying to cot(z)
Why doesn't Picard's Theorem apply to $cot(z)$? Under my understanding, $cot(z)$ has an essential singularity at $\infty$, because there is no punctured neighborhood of $0$ in which $cot(1/z)$ is bounded. Since $cot(z)$ has this singularity, Picard's Thm implies it assumes all but possibly 1 complex value. However, there is no $z$ for which $cot(z) = i$, and no $z$ for which $cot(z) = -i$. This is seemingly a contradiction. Where did I go wrong?

答: The answer is in a comment by Conrad, the singularity at $\infty$ is essential but not isolated, and Picard only applies to isolated.


问:
Find formula for $\sum_{k = 1}^{n}(k^2 \cdot x^{k - 1})$.
It is almost obvious that we have to use derivative in order to find the answer. However, I couldn't see this one to be derivative of something.

It would be amazing if you share your thoughts on this problem. Also, I would be thankful if you would explain what was intuition behind your solution. Thanks!

答:
Let $$f(x)=\sum_{k=0}^nx^k,$$ so that $$
\begin{align}
f'(x)&=\sum_{k=1}^n kx^{k-1}\\
xf'(x)&=\sum_{k=1}^n kx^k\\
f'(x)+xf''(x)&=\sum_{k=1}^n k^2x^k\\
\end{align}$$

Now just evaluate $f(x)$ and fill in the details.


问:
residue of $f(z)=\sqrt[3]{z^2(1-z)}$ at $\infty$
I'm trying to find the residue of $f(z)=\sqrt[3]{z^2(1-z)}$ at $\infty$, while computing $\int^1_0  \sqrt[3]{x^2(1-x)} dx$

I know that I can compute it using $g(w) = -w^{-2} f(1/w)$ at $w = 0$ (Residue at infinity, contradiction?).

So I have  $g(w)=\frac{-1}{w^2} \sqrt[3]{\frac {1}{w^2} (1-\frac{1}{w})}=\frac{-1}{w^2} \sqrt[3]{\frac {1}{w^3} (w-1)}$.

And here I don't know what to do. I want to put $\frac{1}{w^3}$ out of the root, can one do this? And why? And if not what one can do? Could you please help me to find the residue explicitly?

答: There is no way to get the residue at $z=\infty$ for this function $f(z)=\sqrt[3]{z^2(1-z)}$. This is because $f(z)$ is not analytic for $|z|>R$ for any large $R>0$.


问:
Generalize a sum given first elements
Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as

$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$

generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)

$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$

I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.

答:
It is easy to define $f$ recursively. This is easier to see when we reverse the order of the terms:

\begin{align*}
f(0) &= 1 \\
f(1) &= \rho + 2 = \rho f(0) + 2 \\
f(2) &= \rho^2 + 2\rho + 3 = \rho f(1) + 3 \\
f(3) &= \rho^3 + 2\rho^2 + 3\rho + 4 = \rho f(2) + 4
\end{align*}

We can define $f$ as $f(0) = 1$, $f(n) = \rho f(n-1) + (n + 1)$ for $n \geq 1$.

Paras Khosla’s answer shows a different way of writing this definition.

Title: Generalize a sum given first elements


问:
How can I find the oblique asymptote of $y=\frac{x^3-2x^2+x-1}{x^2+1}$?
Does any one has any idea about how to find the oblique asymptote of this curve?

$$y=\dfrac{x^3-2x^2+x-1}{x^2+1}$$

(original problem image)

答:
$$  \left(   x^{3}  - 2 x^{2}  +  x  - 1 \right)  =  \left(   x^{2}  + 1 \right)  \cdot \color{magenta}{  \left(   x  - 2 \right) } +  \left( 1  \right)  $$
so your fraction is
$$ x-2 + \frac{1}{x^2 + 1} $$
and, for large $|x|,$   ......


问:
The convergence or divergence of two improper integrals
I want to investigate the  convergence  or divergence of the  following   improper integrals :

$$(1)\displaystyle\int_{0}^{+\infty} (-1)^{[x^2]}dx;$$
$$(2)\displaystyle\int_{-\infty}^{+\infty} \frac{dx}{\prod_{k=1}^{n}|x-a_k|^{p_k}},$$
where $a_1<a_2<\cdots<a_n.$

I have no idea about them. I would appreciate it if someone can give some suggestions and comments. Thanks.

答:
For the first integral we can use the substitution $u = x^2$

$$\int_0^\infty (-1)^{[x^2]}dx =\int_0^\infty \frac{(-1)^{[u]}}{2\sqrt{u}}du$$

$$= \sum_{n=0}^\infty \int_n^{n+1}\frac{(-1)^n}{2\sqrt{u}}du = \sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}}$$

which converges by alternating series test. The second integral requires

$$p_k < 1 \hspace{20 pt} 1\leq k \leq n$$ and $$\sum_{k=1}^n p_k > 1$$

by p-test.


问:
Applying the central limit while having the sample standard deviation
I have the following exercise:

In a research report, Richard H. Weindruch of the UCLA Medical School
claims that mice with an average life span of $32$ months will live to
be about $40$ months old when $40$% of the calories in their diet are
replaced by vitamins and protein. Is there any reason to believe that
$μ < 40$ if $64$ mice that are placed on this diet have an average life of
$38$ months with a standard deviation of $5.8$ months? Use a P-value in
your conclusion.

From my personal understanding, in applying the central limit theorem we dividethe difference of the means by the sample standard deviation which is given as $5.8$, however in the solution they did the following:

$z=\frac{38-40}{\frac{5.8}{\sqrt{64}}}$, what I don't understand is that why we divided $5.8$ by $\sqrt{64}$, shouldn't we do that iff this value was the standard deviation for the population which isn't the case here.

答:
Many comments, some confusion, maybe no resolution:

Seems to me this is a left-sided, 1-sample t test of
$H_0: \mu=40$ against $H_1: \mu < 40$ based on sample mean and standard deviation
$\bar X =38, S = 5.8,$ respectively, for $n=64$ random observations
from a normal sample. Output from a recent release
of Minitab, which accepts summarized data:

```<code>One-Sample T

Test of μ = 40 vs < 40

 N    Mean  StDev  SE Mean  95% Upper Bound      T      P
64  38.000  5.800    0.725           39.210  -2.76  0.004
</code>```

I will leave it to you to show how to obtain the $T$ statistic (following @Henry's Comments)
and to show that you reject $H_0$ at the 5% level (or even the 1% level).

It is easy to see that $\bar X = 38 < 40.$
You have evidence that it is significantly less than 40, in a statistical sense.

Note: DF$=64-1=63$ for the $T$-statistic. P-value from R statistical software, where pt is the CDF of a t-distribution:

```<code>pt(-2.76, 63)
[1] 0.003780076
</code>```

Body:
Many comments, some confusion, maybe no resolution:

Seems to me this is a left-sided, 1-sample t test of
$H_0: \mu=40$ against $H_1: \mu < 40$ based on sample mean and standard deviation
$\bar X =38, S = 5.8,$ respectively, for $n=64$ random observations
from a normal sample. Output from a recent release
of Minitab, which accepts summarized data:

One-Sample T

Test of μ = 40 vs < 40

 N    Mean  StDev  SE Mean  95% Upper Bound      T      P
64  38.000  5.800    0.725           39.210  -2.76  0.004

I will leave it to you to show how to obtain the $T$ statistic (following @Henry's Comments)
and to show that you reject $H_0$ at the 5% level (or even the 1% level).

It is easy to see that $\bar X = 38 < 40.$
You have evidence that it is significantly less than 40, in a statistical sense.

Note: DF$=64-1=63$ for the $T$-statistic. P-value from R statistical software, where pt is the CDF of a t-distribution:

pt(-2.76, 63)
[1] 0.003780076


问:
$A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.
Problem: The square real matrix $A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

I think I've got this figured out if I can say that the characteristic polynomial has multiplicity 1 for all of its roots, but I don't know why that's true from what we're supposing here. Can anyone bridge that gap for me? Or is there an easier way to go about it that I'm not seeing?

答: It isn't true.  The characteristic and minimal polynomial of $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are both $x^2$.


答:
Polynomial division can be written as $$f(x)=d(x)q(x)+r(x),$$ where $f(x)$ is the polynomial, $d(x)$ is the divisor, $q(x)$ is the quotient, and $r(x)$ is the remainder.  Rearranging the equation, we can write $$\frac{f(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)},$$
where $\dfrac{f(x)}{d(x)}$ is now a rational function and $q(x)$ is the asymptote provided that $r(x)\neq0$.

Your polynomial division should result in the equation $$x^3-2x^2+x-1=(x^2+1)(x-2)+1\implies\frac{x^3-2x^2+x-1}{x^2+1}=\underbrace{x-2}_\text{asymptote}+\frac1{x^2+1}.$$

Title: How can I find the oblique asymptote of $y=\frac{x^3-2x^2+x-1}{x^2+1}$?

Body:
Does any one has any idea about how to find the oblique asymptote of this curve?

$$y=\dfrac{x^3-2x^2+x-1}{x^2+1}$$

(original problem image)


Body:
If

$\begin{array}\\
f(x)
&= \dfrac{x^3-2x^2+x-1}{x^2+1}\\
\text{then}\\
f(x)-x
&= \dfrac{x^3-2x^2+x-1}{x^2+1}-x\\
&= \dfrac{x^3-2x^2+x-x^3-x}{x^2+1}\\
&= \dfrac{-2x^2}{x^2+1}\\
\text{and}\\
f(x)-x+2
&= \dfrac{-2x^2}{x^2+1}+2\\
&= \dfrac{-2x^2+2x^2+2}{x^2+1}\\
&= \dfrac{2}{x^2+1}\\
\end{array}
$

so
$f(x)
=x-2+\dfrac{2}{x^2+1}
$.


答:
Note

$$\frac{x^3-2x^2+x-1}{x^2+1}= x-2 +\frac1{x^2+1}$$

which indicates that the oblique asymptote is $x-2$.


问:
Examples of elementary substructures of $(\mathbb{C},0,1,+,-, \times)$
I wonder what are some examples of elementary substructures of $(\mathbb{C},0,1,+,-, \times)$. I know that $(\mathbb{R},0,1,+,-, \times)$, $(\mathbb{Q},0,1,+,-, \times)$ are not since they don't satisfy $x^2=-1$.

Thanks!

答:
As commenters have mentioned, any elementary substructure of $\mathbb{C}$ must satisfy the same first order properties. But it is well known that $\mathsf{ACF}_0$ (the theory of Algebraically Closed Fields of characteristic $0$) is model complete. This follows, for instance, from quantifier elimination.

How is this helpful?

Any elementary substructure of $\mathbb{C}$ must, in particular, model $\mathsf{ACF}_0$, since it is clear that $\text{Th}(\mathbb{C}) \supseteq \mathsf{ACF}_0$. But by model completeness, we get the converse: any submodel $\mathfrak{M}$ of $\mathbb{C}$ that models $\mathsf{ACF}_0$ must be elementary!

So we're looking for the subrings of $\mathbb{C}$ which are themselves algebraically closed fields of characteristic $0$.

We see one obvious elementary substructure: $\overline{\mathbb{Q}}$. However, there are lots more! For a more detailed discussion about the algebraically closed subfields of $\mathbb{C}$, see here for instance.

I hope this helps ^_^


答:
As stated, this is not true. Take $A=\left(\begin{array}{cc}1&1\\
0&1\end{array}\right)$. Then minimal and characteristic polynomial of $A$ are both $(x-1)^2$ (and thus equal), but $A$ is not diagonalizable.

A criterion for diagonalizability (over a suitable extension) is that the minimal polynomial is squarefree.

Body:
Problem: The square real matrix $A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

I think I've got this figured out if I can say that the characteristic polynomial has multiplicity 1 for all of its roots, but I don't know why that's true from what we're supposing here. Can anyone bridge that gap for me? Or is there an easier way to go about it that I'm not seeing?


问:
Subrings of $\mathbb{Q}$
Let $p$ be prime. Suppose $R$ is the set of all rational numbers of the form $\frac{m}{n}$ where $m,n$ are integers and $p$ does not divide $n$.

Clearly then $R$ is a subring of $\mathbb{Q}$.

I now want to show that if $\frac{m}{n}$ belongs to any proper ideal of $R$ then $p|m$.

Can someone point me in the right direction.

Thanks

答:
You can view $R$ as the localization of the ring of integers outside $p$, meaning
$$R=\mathbb{Z}_{S}=\{\frac{m}{m}:m\in\mathbb{Z},n\in S\}$$
where $S\subseteq\mathbb{Z}$ is the multiplicative subset of $\mathbb{Z}$ given by $S:=\mathbb{Z}\setminus p\mathbb{Z}$.

Now, $\mathbb{Z}$ is a Dedekind domain (it is a PID), hence every localization of $\mathbb{Z}$ outside a prime $p$ is a DVR, meaning a PID and local (i.e only one maximal ideal). If $\frac{m}{n}$ belongs to a proper ideal of $R$, then it is contained in a maximal ideal of $R$. But $R$ is local and its unique maximal ideal is $p\mathbb{Z}_S$, which is the set of fractions $\frac{a}{b}$ with $a\in p\mathbb{Z}$ and $b\notin p\mathbb{Z}$. Thus $\frac{m}{n}$ belongs to $p\mathbb{Z}_S$, implying $m\in p\mathbb{Z}$.

Title: Subrings of $\mathbb{Q}$


问:
Rotational invariance of Bell state
In this lecture on quantum mechanics, Umesh Vazirani shows with a simple calculation that $$\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle = \frac{1}{\sqrt{2}}|uu\rangle + \frac{1}{\sqrt{2}}|u^\perp u^\perp \rangle$$

for any state $u$.  Taking away the physics notation, I translate this as

$$e_1 \otimes e_1 +  e_2 \otimes e_2 = u\otimes u + u^\perp \otimes u^\perp$$

in the the vector space $V \otimes V$, $V = \mathbb{C}^2$, where $e_1,e_2$ is the standard basis of $\mathbb{C}^2$ and $u, u^\perp$ is any another basis attained by applying a (real) rotation to $e_1, e_2$.  Is there a coordinate-free or more conceptual proof than the one Vazirani gives?  What's the geometric meaning of this vector? Can we generalize?

答:
Remark 1: In quantum mechanics, the most general transformation mapping pure states of a two-level quantum system (aka qubit) to other pure states is represented as a unitary matrix in $SU(2)$. However, as Vazirani points out in the video and you note in your question, his proof only applies to real rotations, i.e. matrices in $SO(2) \subset SU(2)$.

Proof: Let $\beta_{00} = |00\rangle + |11\rangle$ where we ignore normalization since vector norm is preserved by unitary transformations. A key property of $\beta_{00}$ which we need is that for any $A \in GL(2, \mathbb{C})$

$$
(A \otimes I) \beta_{00} = (I \otimes A^T) \beta_{00}\tag1.
$$

This can be shown easily by writing it out in the matrix notation or by following indices in the tensor notation. However, since you requested a coordinate-free or more conceptual proof, we will instead use tensor networks employing some basic properties shown in this paper. Note that the tensor network representation of $\beta_{00}$ is a simple bend (so called "cup") and $(1)$ is merely a restatement of equation $(46)$ in the paper.

Now, let $U \in SO(2)$ be the real rotation that maps $|0\rangle$ to $|u\rangle$ and $|1\rangle$ to $|u^\perp\rangle$. Applying it to both qubits and using $(1)$ we obtain

$$
(U \otimes U) \beta_{00} = (I \otimes U^TU) \beta_{00}
$$

but $U$ is orthogonal so

$$
(U \otimes U) \beta_{00} = \beta_{00}.
$$

On the other hand,

$$
(U \otimes U) \beta_{00} = |uu\rangle + |u^\perp u^\perp\rangle.
$$

Comparing the last two equations, we obtain the desired rotational invariance.

Remark 2: For a generic unitary $U$ the proof above fails, because $U^TU$ is not necessarily identity. For a concrete counterexample consider $U=S$ where

$$
S = \begin{pmatrix}
1 & 0\\
0 & i
\end{pmatrix}
$$

where we obtain

$$
(S \otimes S)\beta_{00} = |00\rangle - |11\rangle \ne \beta_{00}.
$$

Remark 3: We can generalize the above result if we replace $\beta_{00}$ with a different Bell state. In his video Vazirani mentions that $\beta_{11} = |01\rangle - |10\rangle$ is invariant under general unitary transformations. Let us see how this happens.

First, note that $\beta_{11} = (Z \otimes X)\beta_{00}$ where

$$
Z = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} \,\,\,\,
X = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
$$

From equation $(45)$ in the paper linked above, we see that $\beta_{11}$ is the Levi-Civita tensor in two dimensions. An important property of this tensor, expressed in $(12)$ in the paper, is

$$
(U \otimes U)\beta_{11} = (\det U) \beta_{11}.
$$

Thus, for $U \in SU(2)$ we have

$$
(U \otimes U)\beta_{11} = \beta_{11}
$$

so $\beta_{11}$ is invariant under arbitrary unitary transformations in $SU(2)$ applied to both qubits.

Title: Rotational invariance of Bell state


问:
Probability problem about having at least as many friends as that person who is a randomly selected friend from Facebook
This question is about a Facebook friend problem. If you select a Facebook friend at random, the probability that you have at least as many friends as that person is about 50%. Is that true or false? And why?

So mainly, the question is what is the probability that I would have at least as many friends as the randomly selected friend from FB? Could it be explained simply using probabilities (or Bayes theorem), if so how?

many thanks

答:
Alternative approach without much math.

Assume that there are two people who are going to come in contact on Facebook: Person-1 and Person-2.

Assume that Person-1 has more Facebook friends than Person-2.

Assume that nothing else is known about the two people, other than that one of them is going to reach out to the other, on Facebook.

There are three possibilities:

People with more Facebook friends are significantly more likely than people with fewer Facebook friends to initiate contact.

People with more Facebook friends are significantly less likely than people with fewer Facebook friends to initiate contact.

People with more Facebook friends are neither significantly more likely nor significantly less likely to initiate contact.

Personally, I would consider anything in the ($45$% - $55$%) range to not be significant.

Each of the above assumptions is plausible.  Each assumption will lead to a different conclusion on whether the person that you are selecting probably has more Facebook friends than you do, probably has fewer Facebook friends than you do, or neither.

To see this, consider that with you coming into contact with the person that you selected (at random), the contact could have been initiated by the other person, but it wasn't.  Similarly, some other Facebook person might randomly contact you.

Personally, my blind speculation is that assumption 1 above is more likely than the other two.  It wouldn't surprise me if upwards of $55$% of the time the person who initiates contact has more friends.


答:
We have to focus on permutations $\sigma \in S_{2n}$ such that $\sigma(i) \in\{i, 2n-i+1\}$ for all $i \in \{1, \ldots, 2n\}$.

To uniquely determine such a permutation, we have to pick $k$ numbers out of $1, 2, \ldots, n$ which will be fixed points of $\sigma$. If $i$ is one of the remaining $n-k$ numbers, then set $\sigma(i)=2n-i+1$.

The action of $\sigma$ on $n+1, \ldots, 2n$ is now uniquely determined like this:
$$\sigma(i) = \begin{cases} 2n-i+1, &\quad\text{if }i \in \sigma(\{1,\ldots, n\}),\\
i,  &\quad\text{otherwise}.\\\end{cases}$$

To find the sign of $\sigma$, notice that $\sigma$ is a composition of $n-k$ transpositions $(i,  2n-i+1)$ each having sign $-1$ so $\operatorname{sgn}(\sigma) = (-1)^{n-k}$.

Now if $i$ is a fixed point of $\sigma$ then $A_{i\sigma(i)} = a$ and otherwise $A_{i\sigma(i)} = b$. Since $\sigma$ as above has $2k$ fixed points and $2n-2k$ other points, it yields the product $a^{2k}b^{2n-k}$ in the determinant formula.

All in all, we get

$$\det A = \sum_{\sigma \in S_{2n}} \operatorname{sgn}(\sigma) \prod_{i=1}^{2n} A_{i\sigma(i)} = \sum_{k=0}^{n}{n\choose k}(-1)^{n-k} a^{2k}b^{2n-2k} = (a^2-b^2)^n.$$

Title: Find a determinant of this matrix.

Body:
I've got matrix
\begin{equation*}
A=\left(
\begin{array}{ccccccc}
a & 0 & 0 & \ldots & 0 & 0 & b \\
0 & a & 0 & \ldots & 0 & b & 0 \\
0 & 0 & a & \ldots & b & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0 & b & \ldots & a & 0 & 0 \\
0 & b & 0 & \ldots & 0 & a & 0 \\
b & 0 & 0 & \ldots & 0 & 0 & a \\
\end{array}
\right)
\end{equation*}

оf $2n$ size (number columns and rows). I need to find it's determinant using a traditional formula:

$$\det A = \sum_{\sigma \in S_{2n}}{\text{sgn}(\sigma)\prod_{i=0}^{2n}{A_{i\sigma(i)}}}$$

Where $\text{sgn}(\sigma)$ is $(-1)^{inv(\sigma)}$, and $inv(\sigma)$ is the number of inversions in $\sigma$.

It's easy to see that non-null terms in will be if and only if $\sigma(i) \in \{i, 2n - i + 1\}$.

So, the answer for me looks like: $\text{sgn}(\sigma_1)a^{k_1}b^{2n - k_1} + \text{sgn}(\sigma_2)a^{k_2}b^{2n - k_2}\ldots$. (where each $sgn(\sigma_i)$ is $\pm1$).

But I have no idea how to expand this and count all $\sigma$'s.


问:
Mistake converting $2.3151515...$ to a fraction?
So I think I made a mistake but I just want to confirm I'm getting this right:

In the example, we had the number $2.3151515...$ and were supposed to convert this number into a fraction.
It was then shown to be equal to the sequence $2.3 + \frac{15}{10^3}+\frac{15}{10^5}+\frac{15}{10^7}$ etc.
As such the common ratio is $r = \frac{1}{100}$

Thus we get the sum of the sequence equalling $(2.3)\frac{1}{1-\frac{1}{100}} = \frac{230}{99}$
The problem is, when I checked my solution $\frac{230}{99}$, it's rougly $2.32323232...$ yet the right solution would actually be $\frac{382}{165}$

Does the problem lie within the simplification of the sequence or am I doing something wrong?

Thanks in advance.

答:
The $2.3$ is not be part of your series.  The first term in the series is actually $\dfrac{15}{10^3}$ with the common ratio of $\dfrac1{10^2}$.

Your sum should be $$2.3+\frac{\frac{15}{10^3}}{1-\frac1{10^2}}=\frac{382}{165}$$ as desired.


Title: Mistake converting $2.3151515...$ to a fraction?

Body:
So I think I made a mistake but I just want to confirm I'm getting this right:

In the example, we had the number $2.3151515...$ and were supposed to convert this number into a fraction.
It was then shown to be equal to the sequence $2.3 + \frac{15}{10^3}+\frac{15}{10^5}+\frac{15}{10^7}$ etc.
As such the common ratio is $r = \frac{1}{100}$

Thus we get the sum of the sequence equalling $(2.3)\frac{1}{1-\frac{1}{100}} = \frac{230}{99}$
The problem is, when I checked my solution $\frac{230}{99}$, it's rougly $2.32323232...$ yet the right solution would actually be $\frac{382}{165}$

Does the problem lie within the simplification of the sequence or am I doing something wrong?

Thanks in advance.

Body: The correct answer is $2.3+\dfrac{\dfrac{15}{10^3}}{{1-\dfrac1{100}}}=\dfrac{382}{165}$


问:
If $(i, H, B)$ is an abstract Wiener space, then $B^*$ is dense in $H$?
Let $(i, H, B)$ be an abstract Wiener space - that is, $H$ is a Hilbert space, $||\cdot ||$ is a measurable norm on $H$, $B$ is the completion of $H$ with respect to $|| \cdot||$, $i$ is the inclusion map from $H$ to $B$.

Then we can embed $B^*$ in $H$ as follows: as any measurable norm on $H$ is weaker than the norm on $H$, the restriction of any element of $B^*$ to $H$ is an element of $H^*$ - thus, by Riesz representation theorem, we can associate this element to a unique element of $H$.

So far this is what I have (don't know if this will help):

A characterization of $B^*$ is:
$$ B^* = \left\{x \in H: \sup_{y \in H, y \neq 0} \frac{\langle x, y\rangle}{||y||} < \infty \right\}$$

(Note that $\langle\cdot, \cdot \rangle$ is the inner product on $H$ and $||\cdot||$ is the measurable norm on $H$. Also, if this quantity exists and is finite then it is the $B^*$ norm of $x$)

Apparently, $B^*$ is dense in $H$, but it is not apparent to me why this is true.

答:
Write $i^*$ for the natural map from $B^*$ into $H$.  It is indeed the adjoint of $i$ since for $f \in B^*$, $h \in H$ we have $\langle i^*f, h \rangle_H = f(ih)$.

Suppose $h_0 \in (i^*(B^*))^\perp$, so $\langle i^* f, h_0 \rangle =0$ for all $f \in B^*$.  This means $f(i h_0) = 0$ for all $f \in B^*$, so by the Hahn-Banach theorem, $i h_0 = 0$ which means $h_0 = 0$.  This implies $i^*(B^*)$ is dense in $H$.


问:
an elementary ring problem
Problem:

Let $a, b \in \mathbb{Z}$. the goal of this problem is to define the set $E$ of elements of  $\mathbb{Z}$ congrus to $a$ modulo 2 and to $b$ modulo 3 .

Show that
$$
\begin{aligned}
\varphi: \mathbb{Z} & \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \\
& x \quad \mapsto \quad(x+2 \mathbb{Z}, x+3 \mathbb{Z})
\end{aligned}
$$
is a ring homomorphism

Determine the kernel of $\varphi$.

Show that the couple $(a+2 \mathbb{Z}, b+3 \mathbb{Z}) \in \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$ admits an antecedent by $\varphi$. (we can use Bézout's identity)

What I could answer:

1.

we know that $(\mathbb{Z} / n \mathbb{Z},+, \times) $ is a commutative ring, in particular $(\mathbb{Z} / 2 \mathbb{Z},+, \times)$ and $(\mathbb{Z} / 3 \mathbb{Z},+, \times)$ are commutative rings,

we need to prove that
\begin{array}{l}
\text { - } \forall x, y \in Z, \varphi(x+y)=\varphi(x)+\varphi(y) \\
\text { - } \forall x, y \in Z, \varphi(x \times y)=\varphi(x) \times \varphi(y) \\
\text { - } \varphi(1_Z)=1_{ \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}}
\end{array}

$\varphi(x+y) =  (x+y + 2 \mathbb{Z}, x+y + 3 \mathbb{Z}) \\ \iff \varphi(x+y) =  (x+2\mathbb{Z} + y+2\mathbb{Z}, x+3 \mathbb{Z} + y+3 \mathbb{Z}) \\ \iff \varphi(x+y) =  \varphi(x) + \varphi(y)$

since in $\mathbb{Z} / 2 \mathbb{Z}$ and in $\mathbb{Z} / 3 \mathbb{Z}$ we have:
$\forall (\overline{a},\overline{b} ) \in (\mathbb{Z} / 2 \mathbb{Z})²$ or in $(\mathbb{Z} / 3 \mathbb{Z})² \\ \overline{a} + \overline{b}= \overline{a + b}$

since $\exists k, l \in \mathbb{Z}, x=a+k n \text { et } y=b+\ln $

so $x+y = a+b + (k+l) n $

Also, in the same way, we also have $\overline{a} \times \overline{b} = \overline{a \times b} $ since

$\exists k, l \in \mathbb{Z}, x=a+k n \text { et } y=b+\ln \\ x \times y=a \times b+(k b+a l+k l) n$

For the neutral element with respect to the multiplicative operation,

We have

$\varphi(1_Z) = (1,1) = 1_{ \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}}$

2.(not sure if this  is right or if I can write it better)

The kernel of $\varphi$ is the set of elements that are multiples of $3$ and $2$ at the same time so it's the set { $x \quad  | \quad x = 6 k, \forall k \in Z$ }

3. (I don't know what to prove here with Bézout's identity that states : $\exists u,v \in Z² \quad au+bv=\mathbf{1} \Leftrightarrow pgcd(a ; b)=1$)

we have that $card(\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}) \leq card(Z)$ and by definition of $\varphi$ we can see that each element of $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$ has at least one antecedent from $Z$.

Any mistake pointing , indications or suggestions are welcome, thank you so much.

答:
I think you have all the right ideas, but your phrasing is a bit awkward and hard to read in a few places. For example in the first section, you don't need to introduct $k,l,n$ and all these extra letters. E.g. for additivity, you can just say something like:
\begin{align*}
\varphi(x+y) &= ((x+y) + 2\mathbb{Z},(x+y) + 3\mathbb{Z})\\
&=((x + 2\mathbb{z}) + (y + 2\mathbb{Z}), (x + 3\mathbb{z}) + (y + 3\mathbb{Z}))\\
&=(x + 2\mathbb{z}, x + 3\mathbb{z}) + (y + 2\mathbb{z}, y + 3\mathbb{z})\\
&=\varphi(x) + \varphi(y)
\end{align*}
Which is basically what you wrote in the first few lines of your explanation. Presumably you're only working on this problem after having spent some time at some point working with rings of the form $n\mathbb{Z}$, and you don't need to go back to the essentials every time.

For 2), I think you can make the idea more precise in the following way:

Fix $x \in \ker\varphi$. Then
$$
(\bar{x},\bar{x}) = (\bar{0},\bar{0})
$$
So we have that $x \in 2\mathbb{Z}$ and $3\mathbb{Z}$, which implies $x \in 6\mathbb{Z}$.

For the third problem, fix some $(\bar{a},\bar{b}) \in \mathbb{Z}/2\mathbb{Z}, \times \mathbb{Z}/3\mathbb{Z}$. We want to find prove that there exists some $x \in \mathbb{Z}$ such that $x \equiv a \mod 2$ and $x \equiv b \mod 3$. A really simple way to do this is to let $x=b$ if $a$ and $b$ have the same parity and let $x = b+3$ if they have different parities. In either case $\varphi(x) = (a+2\mathbb{Z},b+\mathbb{Z})$ as desired.

Your argument about cardinality does not make sense to me.

Title: an elementary ring problem


问:
Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$
Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$

How would I go about rearranging this to deduce a value for x?

Thanks for the help.

答:
You have a system of two equations

$$\begin{cases}\sqrt{x^2+ay^2}-\sqrt{x^2+y^2} = z \\ \sqrt{x^2+ay^2}+\sqrt{x^2+y^2}
 = \frac{(a-1)y^2}{z}\end{cases}$$

where the second one is a rationalization of the first. Square both equations and add them

$$4x^2 + 2(a+1)y^2 = z^2 + \frac{(a-1)^2y^4}{z^2} \implies x^2 = \frac{z^2}{4} - \frac{(a+1)y^2}{2}+\frac{(a-1)^2y^4}{4z^2}$$

You can take the positive or negative square root as needed.


问:
Prove that every orthogonal matrix has a null space {0}
I am currently working on some practice problems regarding orthogonality and its properties, and one of the proofs I am trying requires that I show prove that "every orthogonal matrix has a null space {0}". The solution provided by my professor only states: $Q^{-1} = Q^{T}$ with no explanation. If anyone could offer some guidance on how to go about writing a more detailed proof the statement, I would greatly appreciate it! Thank you!

答: Note that for an orthogonal matrix $Q$, we have $Q^{-1}=Q^T$. This means $Q$ is invertible (its inverse is $Q^T$), and we know that invertible matrices have trivial kernels (i.e. their null space is $\{0\}$).


问:
Splitting of prime ideals in algebraic extensions
I'm reading Algebraic Number Theory by Jurgen Neukirch. I have some problems with some of the exercises in Section 9 of Chapter 1.

They are:

1) If $L / K$ is a Galois extension of algebraic number fields with non-cyclic
Galois group, then there are at most finitely many non-split prime ideals of $K$.

2) Let $L / K$ be a finite (not necessarily Galois) extension of algebraic number
fields and $N / K$ the normal closure of $L / K$. Show that a prime ideal $p$ of $K$ is totally
split in $L$ if and only if it is totally split in $N$.

I have worked on them for a long time but couldn't get any idea of it.

Can you please help? Thank you!

答:
The following  is a rather incomplete answer.

In the same section of Neukirch, there is the following exercise:

Exercise 3: If a prime ideal $\mathfrak{p}$ of $K$ is totally split in two separable extensions $L$ and $L'$ of $K$, then it also totally split in the compositum $LL'$.

It is clear that question 2) reduces to Exercise 3 since the Galois closure is the compositum of all "conjugates" of L. Each conjugate $L'$ is isomorphic to $L$, so $\mathfrak{p}$ splits completely in $L'$, too.

A reference for the proof of Exercise 3 is pp.40 (chapter II, section 1) of Lang's "algebraic number thoery", but it uses completion, which is yet to be introduced in Neukirch's book.

However, I had a hard time to prove Exercise 3 directly. I am also curious about the following exercise on the same page, which really puzzles me (I was hoping to reduce Ex3 to it).

Exerciese 2: For every integeral ideal $\mathfrak{a}$ of $\mathcal{O}_L$, there exists a $\theta \in \mathcal{O}_L$ such that the conductor $\mathfrak{F}=\{\alpha\in \mathcal{O}_L\mid \alpha \mathcal{O}_L \subseteq \mathcal{O}_K[\theta]\}$ is coprime to $\mathfrak{a}$ and $L=K(\theta)$.


问:
$A$ is a finitely gen. $k$-algebra. How to show that the set of closed points of $\operatorname{Spec}A$ is dense without using that $A$ is Jacobson?
Let $A$ be a finitely generated $k$-algebra.

I would like to show that the set of closed points of $\operatorname{Spec}A$ is dense, but without using the fact that the nilradical of a finitely generated $k$-algebra is the intersection of all maximal ideals.

I know that the closed points of $\operatorname{Spec}A$ are the maximal ideals of $A$.

I also know that if $\frak m$ is a maximal ideal of $A$, then $A/\frak m$ is a finite extension of $k$, by Hilbert's Nullstellensatz.

I would like to show that if $f \in A$ and $D(f) \ne \varnothing$, then $D(f)$ contains a maximal ideal of $A$.

I know that $A_f$ is also a finitely generated $k$-algebra as well.

答: Write $A\cong k[x_1,\cdots,x_n]/I$ as a quotient of a polynomial ring. Then $D(f)$ is the spectrum of $A_f\cong k[x_1,\cdots,x_n,z]/(I,zf-1)$, and by the assumption that $D(f)\neq\emptyset$, we have that $(I,zf-1)$ is a proper ideal of $k[x_1,\cdots,x_n,z]$ and is hence contained in a maximal ideal $\mathfrak{m}$. This gives a closed point in $D(f)$.


问:
Time Average of a function $f(\phi(t),t)$, $\phi(t)$ is only varying very slowly with time.
I am a bit stumped on a task. Namely I have the function:
\begin{align}
f(\phi(t),t) = \cos{(\omega (t-t_0))} \, \cos{\phi} + \cos^2{(\omega (t-t_0))}\,\sin{\phi}\,.
\end{align}
And my task is to calculate the time average. The only thing I know is that $\phi(t)$ is only varying very slowly with time. Personally, however, this means nothing to me. Maybe it's something I had 4 or 5 semesters ago, but it's not in my head anymore. I guess I just develop the function to $\phi$ using Taylor and then from the second order I can ignore the terms because $\frac{\text{d}^2\phi}{\text{d}t^2}$ is very small. But I am not sure. Can anyone here help me briefly?

答:
I have now gone the following way. But this is wrong, because the solution should be given at the end in quantities of $\phi(t)$.

Now, we assume that $\phi(t)$ is only varying very slowly with time. This means:
\begin{align*}
    \phi(t) &= \phi(t_0) + \dot{\phi}(t_0) (t-t_0) + \underbrace{\frac{\ddot{\phi}(t_0)}{2}}_{=\,\text{very small}} (t-t_0)^2 + \ldots \\
    &\approx \phi_0 + \omega_\phi (t-t_0)
\end{align*}
For the terms now applies:
\begin{align*}
    \langle\cos{\phi(t)}\,\cos{\left(\omega(t-t_0)\right)}\rangle &= \frac{1}{2}\left\langle\cos{\left[(\omega_\phi + \omega)t -\omega t_0 -\omega_\phi t_0 + \phi_0\right]}\right\rangle \\
    &\hspace{10pt} + \frac{1}{2}\left\langle\cos{\left[(\omega_\phi - \omega)t + \omega t_0 -\omega_\phi t_0 + \phi_0\right]}\right\rangle\\
    &= \frac{1}{2}\left\{\begin{array}{cc} 0, & \omega_\phi \ne \omega \\ \cos{\left[\omega t_0 -\omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = \omega \end{array}\right. \\
    \langle\cos^2{\left(\omega(t-t_0)\right)}\,\sin{\phi(t)}\rangle &= \frac{1}{4}\left\langle\sin{\left[(2\omega+\omega_\phi)t - 2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}\right\rangle \\
    &\hspace{10pt} -\frac{1}{4} \left\langle\sin{\left[(2\omega-\omega_\phi)t -2\omega t_0 + \omega_\phi t_0 - \phi_0\right]}\right\rangle + \frac{1}{2}\langle\sin\left[\phi(t)\right]\rangle \\
    &= \frac{1}{4}\left\{\begin{array}{cc} 0, & \omega_\phi \ne \omega \\ \sin{\left[2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = 2\omega \end{array}\right.
\end{align*}
The final solution is now:
\begin{align*}
    \langle f(\phi(t),t) \rangle = \left\{\begin{array}{ll} 0, & \omega_\phi \ne \omega, \omega_\phi \ne 2\omega \\
    \frac{1}{2}\cos{\left[\omega t_0 -\omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = \omega \\
    \frac{1}{4}\sin{\left[2\omega t_0 - \omega_\phi t_0 + \phi_0\right]}, & \omega_\phi = 2\omega \end{array}\right.
\end{align*}


答:
Bring $\sqrt{x^2+y^2}$ to the right

$\sqrt{x^2+ay^2}=z+\sqrt{x^2+y^2}$

Square both sides

$x^2+ay^2= z^2+2z\sqrt{x^2+y^2}+x^2+y^2$

Get rid of $x^2$ on both sides, group together the $y^2$ terms and bring $z^2$ to the left

$(a-1)y^2-z^2=2z\sqrt{x^2+y^2}$

Square both sides again to get rid of the $\sqrt{}$ sign on the right

$\left((a-1)y^2-z^2\right)^2=4z^2(x^2+y^2)$

...at least that’s how Stewart Calculus does it.

Title: Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$

Body:
Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$

How would I go about rearranging this to deduce a value for x?

Thanks for the help.


问:
There are at least two primes with power 1 in $k!$ for $k>11$
My question is: need to prove that $\forall \ k \geq 11, \exists$ at least 2 prime numbers with power of 1, in k! factorization. I know I missed something, and that is why I'm looking for a hint. Thanks for help!

*** I can use that $\forall \ k > 5$ there exist at least $2$ prime numbers in the next interval: $(k,2k)$.

答:
It is easily verified that the property is valid for $k\le5$.

For $k\gt5$, Ramanujan showed in 1919 that between $k$ and $2k$ there are at least two distinct primes (a refinement of Bertand's postulate). Then taking $$\left\lfloor\frac k2\right \rfloor\lt p\lt q\lt k\lt p^2\lt q^2$$ we have $p$ and $q$ as examples showing the property. In fact:
$$k!=1\cdot2\cdot3\cdots p\cdots q\cdots k$$ where all factor to the right of $p$ is coprime with $p$ and so is with $q$.


问:
Let $\tau : \Omega \to \{0,1,2...,\infty\}$ and $\{ X_n :n \geq 0\}$ be an adapted process. Is $\tau $ as topping time?
Let $\tau : \Omega \to \{0,1,2,\cdots,\infty\}$ and  $\{ X_n :n \geq 0\}$ be an adapted process. Which of the following is a better defined stopping time with respect to $\{F_n : n \geq 0 \}$?

$\tau = \inf \left\{  n\geq 0 : X_n >X_{n+1}\right\}$

$\tau = \inf \left\{  n\geq 0 : \frac{1}{n}\sum_{i=1}^{n} X_{i}<0\right\}$

If $\tau$ is stopping time, then $(\tau -1 )\vee 0 $  is a stopping time.

My guess is the last one because the minimum of two stopping times is a stopping time. I am not sure though whether 0 is well defined in this case.

I would appreciate it if you can correct me.

Thanks.

答:
You are correct that the minimum of two stopping times is a stopping time, and there is not a problem with $0$ being well-defined and a stopping time, but check whether or not $\tau - 1$ is a stopping time.

I think that the second one is is also a stopping time.  We would need to show that $\{ \tau \le n \} \in \mathcal F_n$ for all $n$, and $$ \{ \tau \le n \} = \bigcup_{m=1}^n \left\{ \frac 1m \sum_{k=1}^m X_k < 0 \right\}$$

is a finite union of sets in $\mathcal F_n$ so $\{\tau \le n \} \in \mathcal F_n$ and hence $\tau$ is a stopping time.


问:
Characterizations of diagonalizability of operators on infinite-dimensional vector spaces
I would like to understand the subset $\mathcal{B}_{\mathrm{diag}}(\mathcal{H}) \subset \mathcal{B}(\mathcal{H})$ of diagonalizable operators on some potentially infinite-dimensional Hilbert space better. Specifically, I would like to prove the technical lemma below.

Sometimes, diagonalizable means “unitarily diagonalizable”, i. e. normal with respect to the adjoint ${}^*$ of the given scalar product with $\langle \, \cdot \, , \, \cdot \, \rangle$. But I have the more general meaning in mind where you replace unitary with invertible:

Equivalent Characterizations of Diagonalizability

$H \in \mathcal{B}(\mathcal{H})$ is diagonalizable.

$H$ admits a functional calculus for bounded Borel functions $f : \mathbb{C} \longrightarrow \mathbb{C}$.

There exists a positive, bounded, invertible operator $W = W^* \in \mathcal{B}(\mathcal{H})$ with bounded inverse $W^{-1} \in \mathcal{B}(\mathcal{H})$ so that $H$ is normal with respect to the weighted scalar product $\langle \varphi , \psi \rangle_W := \langle \varphi , W \psi \rangle$.

$H = H_{\Re} + \mathrm{i} H_{\Im}$ is the sum of two bounded commuting operators that are selfadjoint with respect to some weighted scalar product $\langle \varphi , \psi \rangle_W := \langle \varphi , W \psi \rangle$.

I could add other characterizations (e. g. that $H$ is related to a multiplication operator by a similarity transform), but for my purposes, I think this list suffices. First of all, is this list of characterizations correct?

At the end of the day I would like to prove the following:

Lemma

If $\lambda \mapsto H(\lambda) \in \mathcal{B}_{\mathrm{diag}}(\mathcal{H})$ is a continuous path in the set of diagonalizable operators, then there exists a continuous positive-selfadjoint-operator-valued map $\lambda \mapsto W(\lambda) \in \mathcal{B}(\mathcal{H}) \cap \mathcal{B}(\mathcal{H})^{-1}$ so that $H(\lambda)$ is normal with respect to $\langle \, \cdot \, , \, \cdot \, \rangle_{W(\lambda)}$.

Everything except for the continuity of $\lambda \mapsto W(\lambda)$ seems clear to me. That is because the choice of $W$ is not unique — for matrices, for example, you are free to choose units of length for each of the eigenvectors and $W$ declares them to all have length $1$.

答:
For future reference, I came across volume 3 of Dunford and Schwartz's book series on linear operators, which is dedicated to spectral operators. These are a generalization of what I called diagonalizable operators; more precisely, they are in the language of Dunford & Schwartz scalar operators on a Hilbert space (cf. Chapter XV.6.4, Theorem 4).

A spectral operator $H$ is an operator on a Banach space with a projection-valued measure $\{ P(\Lambda) \}_{\Lambda \in \mathfrak{B}(\mathbb{C})}$ over the Borel sets $\mathfrak{B}(\mathbb{C})$ of the complex plane (oblique-projection-valued as there is no Hilbert space adjoint), which (a) commute with the operator (and therefore, the spectral subspaces are left invariant) and (b) where the spectrum of the restricted operator $\sigma \bigl ( H \vert_{\mathrm{ran} \, P(\Lambda)} \bigr ) \subseteq \overline{\Lambda}$ (Chapter XV.2.2, Definition 2).

The projection-valued measure of a spectral operator is unique (Chapter XV.3.5, Corollary 8).

This leads to a unique decomposition (Chapter XV.4.3, Theorem 5) of a spectral operator
\begin{align*}
H &= S + N
\end{align*}
into a scalar part (Chapter XV.4.1, Definition 1),
\begin{align*}
S &= \int_{\mathbb{C}} \lambda \, \mathrm{d} P(\lambda)
\end{align*}
and a quasi-nilpotent or residual part,
\begin{align*}
N &= H - S
.
\end{align*}
I have just read the beginning, but Dunford & Schwartz seem to develop the basic theory of spectral operators in the book over the course of almost 700 (!) pages. Their work was totally under my radar, so I hope if someone looks for “diagonalizable” in the future, they might find their book useful. Dunford also has a shorter article on that subject, which summarizes the basic definitions and properties.
\medskip

Reference: Linear Operators, Part III, Spectral Operators

Nelson Dunford, Jacob T. Schwartz, William G. Bade and Robert G. Bartle

Pure and Applied Mathematics, Volume VII

Wiley-Interscience, 1971


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答: From Jensen's inequality, the mean of the square root is less than the square root of the mean, unless the numbers are equal. This is true for more than two numbers as well.</p>


答:
Let $A\in M_{n \times m}(\mathbb{K})$ for some field $\mathbb{K}$.
Since $A$ is orthogonal then by definition $A^{t}=A^{-1}$ it implies that
$A$ is a invertible matrix.

Now consider the null space given by
$$Ker(A)=\lbrace x\in \mathbb{R^m} \mid A \cdot x=0_{n \times m}\rbrace$$
$$ Ker(A)=\lbrace x\in \mathbb{R^m} \mid x=0_{m \times n}A^{-1}\rbrace $$
$$Ker(A)=\lbrace x=(0,0,0, \cdots,0)\rbrace$$
$$Ker(A)=\lbrace 0 \rbrace$$

Title: Prove that every orthogonal matrix has a null space {0}

Body: I am currently working on some practice problems regarding orthogonality and its properties, and one of the proofs I am trying requires that I show prove that "every orthogonal matrix has a null space {0}". The solution provided by my professor only states: $Q^{-1} = Q^{T}$ with no explanation. If anyone could offer some guidance on how to go about writing a more detailed proof the statement, I would greatly appreciate it! Thank you!


问:
Theorical question about union of two regular languages
I have seen the propity of $L_1\cup L_2$ is a regular language if $L_1$ & $L_2$ are regular languages.

But it works backwards? If i have a regular language L, always exist two others regular languages that together makes L?

Can i have a demostration?

答: Yes. Let $L_0=\{w\in L:|w|\text{ is even}\}$ and $L_1=\{w\in L:|w|\text{ is odd}\}$; if $L$ is regular, so are $L_0$ and $L_1$, as may be seen by modifying a DFA for $L$ to keep track of the parity of the input, and their union is clearly $L$.


问:
The naturality of the canonical projections?
The first isomorphism theorem (for groups) states that $G/ \text{Ker} (\varphi) = \varphi(G)$.
This is called the canonical projection, but also sometimes the natural projection.
I wondered if this is because these canonical projections were actually natural transformation between appropriate functors.

答:
Here's what I tried (not a complete answer to my question)
$\require{AMScd}$

Given $G$ and $\varphi: G \to \varphi(G)$. We try to construct a functor $F: \mathbb{C} \to \mathbb{C}$, such that there is a natural transformation $\eta : \text{id} \to F$. Here $\text{id}$ is the identity functor and $\mathbb{C}$ is the category of all groups. To alleviate notation let $N=Ker(\varphi)$, also consider the following two properties a group $H$ may have:

$(1)$ There is at least some morphism $\alpha: H \to G$ such that $\alpha(H) - N \not = \emptyset$, i.e. there is some $h \in H$ such that $\alpha(h) \not \in N$

$(2)$ There is at least some morphism $\beta: G \to H$ such that $\beta(N) \not = 1$

Groups with these properties have some of the structure of their $\eta$-images determined by the naturality condition, and furthermore these are the only groups, such that there is some restriction in their $\eta$-images. To see these, take a group $K$ that does not have $(1)$ nor $(2)$. Then, naturality for a morphism $f: K \to G$ is:

\begin{CD}
H @>{f}>> G \\
@VVV @VVV \\
\eta H @>{\eta f}>> G/N;
\end{CD}

If $f$ is just the trivial homomorphism then this just forces $\eta f$ to be a trivial homomorphism so $\eta H$ is "free", meaning there's no restrictions on what $\eta H$ must be in this commutative square. To see that it is "free" in any commutative commutative square where $H$ is the domain and $G$ is the codomain, by the lack of property $1$, we have that $f(H) - N = \emptyset$, so the above commutative square looks like:

\begin{CD}
h @>{f}>> f(h) \\
@VVV @VVV \\
\eta h @>{\eta f}>> 1;
\end{CD}

Again, this forces $\eta f$ to be the trivial homomorphism so that $\eta H$ is free. The last case, where $H$ is the domain and there is a morphism to any group $K$ is similar. Similarly for when $H$ is the codomain.

This shows that for the natural transformation to exist, I only  need to check it exists for those groups satisfying $(1)$ or $(2)$. However these doubts remain:

Is there actually a well defined $\eta$ on those groups such that the naturality condition holds?

If yes to the above, is the naturality condition enough to uniquely determine the $\eta$ images of those groups?

Here are some last thoughts:

Interestingly if this $\eta$ actually exists, then every group homomorphism is natural, since for $f: A \to B$ I can first get a natural transformation such that $A \to A/\text{Ker}(f)$ is a component of $\eta$, but by the first isomorphism theorem this is just the same as $A \to B$.

The sort of construction that I tried to do reminds me of field extensions, I am not well versed in this topic but I think there's more than a vague connection.


答:
Here I propose an alternative way to solve it for the sake of curiosity.

More precisely, what about polar coordinates? If we set $x = r\cos(\theta)$ and $y = r\sin(\theta)$, one gets that
\begin{align*}
\sqrt{x^{2} +  ay^{2}} - \sqrt{x^{2}+y^{2}} = z & \Longleftrightarrow r\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - r = z\\\\
& \Longleftrightarrow r = \frac{z}{\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - 1}
\end{align*}

Consequently, one has that
\begin{align*}
y = \frac{z\sin(\theta)}{\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - 1} & \Rightarrow y\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} = y + z\sin(\theta)\\\\
& \Rightarrow y^{2}(\cos^{2}(\theta) + a\sin^{2}(\theta)) = y^{2} + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow y^{2}(1 + a\sin^{2}(\theta)) = y^{2}(1 + \sin^{2}(\theta)) + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow a\sin^{2}(\theta)y^{2} = y^{2}\sin^{2}(\theta) + 2yz\sin(\theta) + z^{2}\sin^{2}(\theta)\\\\
& \Rightarrow a\sin(\theta)y^{2} = y^{2}\sin(\theta) + 2yz + z^{2}\sin(\theta)\\\\
& \Rightarrow \sin(\theta)(ay^{2} - y^{2} - z^{2}) = 2yz\\\\
& \Rightarrow \sin(\theta) = \frac{2yz}{y^{2}(a-1) - z^{2}}
\end{align*}

Once you know $\sin(\theta)$, you do also know $\cos(\theta)$. Consequently, we obtain the value of $x$.

It is worth emphasizing that one must check for each values of $\theta$ and $a$ the above-mentioned expressions make sense.


问:
Permutations of $[n]$ with $k$ peaks
Is there a formula for the number of permutations of $[n]$ with $k$ peaks?

Here is some information on the OEIS about these numbers, but no such formula is given. I'm sure such a formula exists due to the following question posed in Miklós Bóna's book, Combinatorics of Permutations (p.40, Q.33). Unfortunately the Google book is only a preview and the required solutions are missing.

EDIT: Just to clarify, I understand that an element $a_k$ of a permutation $[n]=123\cdots n$ is called a peak when $a_{k-1}<a_k$ and $a_k>a_{k+1}$. We define $a_0=a_{n+1}=0$.

答:
You can get a recurrence relation for these numbers by defining the numbers $a_{nk}^{\pm\pm}$, where the first sign is $+$ for permutations beginning with a peak and $-$ otherwise, and the second sign likewise for ending with a peak. Thus, for instance, $a_{nk}^{++}$ counts the number of permutations of $[n]$ with $k$ peaks that begin and end with a peak. (Clearly $a_{nk}^{+-}=a_{nk}^{-+}$.) Then by considering the possible places in which you can insert $n+1$ in a permutation of $[n]$ you get

$$
a_{n+1,k}^{++}=\sum_{i=0}^n\left(\sum_{j=0}^{k-1}a_{ij}^{+-}a_{n-i,k-1-j}^{-+}+2\sum_{j=0}^ka_{ij}^{++}a_{n-i,k-j}^{-+}+\sum_{j=0}^{k+1}a_{ij}^{++}a_{n-i,k+1-j}^{++}\right)
$$

and similar relations for $a_{n+1,k}^{-+}$ and $a_{n+1,k}^{--}$, with suitable initial and boundary conditions.

Title: Permutations of $[n]$ with $k$ peaks


问:
The coefficient and asymptotic in generating function

Let $L_n$ be the set of all the paths from $(0,0)$ to $(n,0)$ such that every step is $u=(1,1)$ , $d=(1,-1)$ and $r=(2,0)$.
Notice that the path could go under the $x$ axis.

a. Write a generating function of the number of paths in $L_n$.

b. Analyze the asymptotic of $|L_n|$ for a large n.

My trial:
in part a, I wrote:

Every path = (empty path) or (r every path) or (u Schröder path d every path) or (d Schröder path u every path).

Now, denote our g.f. by $A(x)$ and Schröder g.f. by $S(x)$, and denote $x$ to count half of the length, get: $A(x)=1+xA(x)+xS(x)A(x)+xS(x)A(x)$. Thus,

$$A(x)=\frac{1}{1-x-2xS(x)}$$

Then if we substitute the formula of $S(x)=\frac{1-x-\sqrt {(1-x)^2-4x}}{2x}$ :

$$A(x)=\frac{1}{\sqrt {(1-x)^2-4x}}$$

In part b, to analyze the asymptotic of $|L_n|$, I guess I should find the coeifficient of $[x^n]$ in $A(x)$ where $A(x)=\sum_{n\geq 0} |L_n| x^{2n}$.
However if we solve denominator of $A(x) = 0$ we get

\begin{align*}
(1-x)^2-4x=0
\quad\Rightarrow\quad
x^2-6x+1=0
\end{align*}

so, $x=3 \pm 2\sqrt2$. So the g.f. will be analytic if we delete the intervals $(3+\sqrt2, \infty)$ and $(-\infty,3-\sqrt2)$.
Write:
$A(x)=\frac{1} {\sqrt{(x- r_1)(x-r_2)}}=\frac{1}{\sqrt{x-r_1}} \frac{1}{\sqrt{x-r_2}}= \frac{1}{\sqrt{1-x/r_1}} \frac{1}{\sqrt{1-x/r_2}}$ where $r_1=3+\sqrt{8}$ and $r_2=3-\sqrt{8}$.
Then, $A(x)=(\sum_{n\geq 0}  (1/r_1)^n x^n \sum_{n\geq 0} (1/r_2)^n x^n)^{1/2}= \sum_{n\geq 0} \sum_{j=0}^{n} (1/r_1)^j (1/r_3)^{n-j} x^n$ ,
here I used Caushy formula.

So, $[x^n] A(x)= \sum_{j=0}^{n} (1/r_1)^j (1/r_2)^{n-j}$

$L_n = \sum_{j=0}^{n} (r_2/r_1)^j (1/r_2)^n$.
But this was not much useful!

Edited:
Another approach, which does not use the Legendre polynomial:

*I found in oeis the sequence of this problem is A001850 and it shows that the coefficient of the series is $\sum_{k=0}^{n} {n\choose k}{n+k \choose k}$ .In addition,  wikipedia page : https://www.google.com/url?sa=t&source=web&rct=j&url=https://en.m.wikipedia.org/wiki/Delannoy_number, mentioned the coefficient and the asymptotic of the above  g.f : $\frac{1}{\sqrt{1-6x+x^2}}$  which is the g.f of the  Delannoy number, but without explanation/proof.
How they find the coefficient of this generating function, and how they analayzed its asymptotic then?

Many thanks!

答:
Part (b)
$$[x^n] \frac{1}{\sqrt{1-6x + x^2}} = [x^n] \frac{1}{\sqrt{1-2xt + x^2}}\big|_{t=3}= [x^n]\sum_{n=0}^\infty P_n(3) x^n $$
where the generating function of the Legendre polynomials has been used.  The exact answer is thus $P_n(3).$  Look up the asymptotics for the Legendre polynomials and you'll find

$$ P_n(t) \sim \frac{1}{\sqrt{2 \pi n}} \frac{ (t+\sqrt{t^2-1})^{n+1/2} }{(t^2-1)^{1/4}}$$ when $t$ is large (outside the oscillatory region.)  Putting in $t=3$ and comparing the asymptotic formula to the exact, we find that, for $n=50,$ the asymptotic formula exceeds the actual by only 0.23%.

The instinct for paying attention to where the singularities are, at $3\pm\sqrt{8},$ (there's a typo in some of your problem setup) is a good one. The penultimate equation utilizes the largest root.  Making this approach rigorous is often called 'singularity analysis.'  I've had Legendre polynomials on the brain for a while, so I didn't want to reinvent a well-known result (I think the above equation can be found in Bender and Orszag.)  Ultimately, an integral representation for the Legendre polynomials is analyzed using the well-known saddle-point technique.


问:
Solving $(\sin x-\cos x)^2+\tan x=2\sin^2x$
From this post (Solving $\frac{\cos^23t}{\tan t}+\frac{\cos^2t}{\tan3t}=0$), which is my post, I have tried solving the last equation. I have gotten a solution(s). Would like to see if I did correctly.

$$\bigl(\sin\left(x\right)-\cos\left(x\right)\bigr)^2+\tan\left(x\right)=2\sin^2\left(x\right)$$

Expanding:
$$sin^2\left(x\right)-2sin\left(x\right)\cos\left(x\right)+\cos^2\left(x\right)+tan\left(x\right)=2\sin^2\left(x\right)$$
Using $\sin^2\left(x\right)+\cos^2\left(x\right)=1$, and bringing the RHS to the LHS, the expansion becomes:
$$-2sin\left(x\right)\cos\left(x\right)-2\sin^2\left(x\right)+1+\tan\left(x\right)=0$$
Then using $\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$ and multiplying the LHS by cos(x), I get the following:
$$\frac{-2\sin\left(x\right)\cos^2\left(x\right)-2\sin^2\left(x\right)\cos\left(x\right)+\cos\left(x\right)+\sin\left(x\right)}{\cos\left(x\right)}=0$$

Now strictly dealing with the numerator, making two separate terms, one being $-2\sin\left(x\right)\cos^2\left(x\right)-2\sin^2\left(x\right)\cos\left(x\right)$ and the other being $\cos\left(x\right)+\sin\left(x\right)$, then factoring out $-2\sin(x)\cos(x)$, I get

$$\color{blue}{-2\sin\left(x\right)\cos\left(x\right)+1}\color{red}{\bigl(\cos\left(x\right)+\sin\left(x\right)\bigr)}+\color{red}{\bigl(\cos\left(x\right)+\sin\left(x\right)\bigr)}=0$$

This makes two terms
$$\biggl(\color{blue}{-2\sin\left(x\right)\cos\left(x\right)+1}\biggr)\biggl(\color{red}{\cos\left(x\right)+\sin\left(x\right)}\biggr)=0$$

For the first term, aka the red term, I got $$\tan(x)=-1$$
where $$x=\frac{3\pi}{4}+n\pi $$

and the second term, aka the blue term.
using the identity $\sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right)$
$$\sin(2x)=1$$
where $$x=\frac{\pi}{4}+n\pi$$

Thanks for reading!

答:
Alternatively

\begin{align}
& (\sin x-\cos x)^2+\tan x-2\sin^2x\\
=& (1-\sin2x )+\tan x -(1-\cos 2x) \\
=&- \frac{2\tan x}{1+\tan^2 x}+ \tan x+\frac{1-\tan^2x}{1+\tan^2 x}\\
= &\frac1{1+\tan^2x}(\tan x-1)^2(\tan x+1)=0
\end{align}
Thus, $\tan x= \pm1$, leading to the solutions
$$x =\frac\pi4 +\frac{\pi n}2$$


问:
For n even, Fermat's Last Theorem has no Twin Pythagorean Triples.
This is my first naïve attempt at stating and proving a Lemma. Is it valid? Thanks!

Definitions.

Twin Pythagorean Triple. A Pythagorean Triple (a,b,c) for which two values are consecutive integers. Some examples are $(3, 4, 5), (5, 12, 13), (7, 24, 25),...$

Fermat's Last Theorem. For any integer n > 2, the equation $a^n+b^n=c^n$ has no positive integer solutions.

Mihăilescu's Theorem. The integers $2^3$ and $3^2$ are two powers of natural numbers whose values ($8$ and $9$, respectively) are consecutive. The theorem states that this is the only case of two consecutive powers.

Rules.

Assuming FLT has not yet been proved!

Lemma.

For $n$ even, Fermat's Last Theorem has no Twin Pythagorean Triples.

Proof.

 Let $n = 2k$ $(k \in \mathbb N, k>1)$

$a^n+b^n=c^n$

can be expressed as

$(a^k)^2+(b^k)^2=(c^k)^2$

Notice the Pythagorean Triple

$(a^k,b^k,c^k)$

Assume Twin Triple, then

$(b^k-a^k = 1)$ or $(c^k-b^k = 1)$

resulting in pairs of two consecutive powers

$(a^k,b^k)$ or $(b^k,c^k)$

However, by Mihăilescu's Theorem, the only case of two consecutive powers is

$(2^3,3^2)$

Consequently

$(b^k-a^k > 1)$ or $(c^k-b^k > 1)$

Therefore, no Twin Triple exists for $k>1$. $\square$

答:
$FLT$ having already been proved it is clear that this is not possible. Then by "rule of the game" we assume that $FLT$ has not yet been proved.
$$(x^n)^2+(y^n)^2=(z^n)^2$$
By definition we want that $y^n-x^n=1$ or $z^n-y^n=1$. This is trivially impossible because
$$y^n-x^n=(y-x)P(x,y)\\z^n-y^n=(z-y)Q(z,y)$$
Do we need more explanation? I have not understood the problem?