问:
Solutions of $x^2 = x + 1$ in $\mathbb Z/n\mathbb Z$
In a stage preceding the national stage of the math olympiad in Romania ( which was held online because of the pandemic ) the following question was asked:

If $A$ is the set of natural numbers $n$ for which $x^2=x+1$ has only a solution in $\mathbb Z/n\mathbb Z$ then:

A) $A$ has no elements

B) $A$ has only one element

C) The elements of $A$ are even

D) $A$ has infinitely many elements

E) ( I can't remember it ).

The oficial answer was that $A$ has only one element but when I wrote a bit of code I found $5$, $46985$, $47209$ and more to be in $A$.

Can you help me characterise the numbers in $A$ a little bit more?

Edit:

答:
If $n=p_1^{r_1}\ldots p_m^{r_m}$ then by Chinese Remainder theorem, the solution is unique $\pmod n\iff$ unique $\pmod{p_i^{r_i}}$ for every $i$. This in particular implies the solution is unique $\pmod{p_i}$.

First, we do the case when $n=p$ is prime. Obviously $p=2$ doesn't work. For other cases, we are now in a field with characteristic not $2$, so we can use the quadratic formula (being in a field guarantees we can factor into two linear factors; characteristic not 2 guarantees we can divide by 2). If there were to be a solution, the discriminant $\Delta=5$ must be a square. For it to be unique, $\frac{1+\sqrt5}2=\frac{1-\sqrt5}2$ (for a chosen square root of $5$), or equivalently $\sqrt5=0$. Hence $5=0$, and $p=5$. We check that $5$ indeed works.

Back to the original case: the prime case implies $n=5^m$ for some $m\geq1$ (since it can have no prime factor other than 5). But when $n=25$ there is already no solution (and if $m\geq 2$, a solution $\pmod{5^m}$ would give rise to a solution $\pmod{25}$, a contradiction) so $n=5$.


问:
Why is the decimal abacus more popular than the binary one?
An abacus is basically a device with beads in rows, each for 1 digit. A decimal abacus has 4 lower beads (signifying 1 unit) and an upper one (signifying 5 units). So 82 would be 3 beads plus the 5-bead on the tens row, and 2 beads on the ones. A decimal abacus just has 1 bead per row for each digit.

There are many shortcuts that make calculation on an abacus much faster for an experienced user. Is it that such shortcuts are limited on a binary one? It would be much quicker for a person to perform binary operations (even on more digits) than decimal ones, after practice. Converting decimal to binary numbers and back should also be easy on the abacus, but it could be the reason why it is not preferred.

If a binary abacus becomes popular, one could also create an octal one, with 1 4-bead and 4 1-beads, or even a hexadecimal one, with 3 4-beads and 4 1-beads.

答:
Another thing to consider. An abacus does not care how many beads you move.
Decimal abacus handy and need to do base 5 math for some reason, use it like when you do decimal math only move the carries over to the left at 5 and above instead of 10 and above.

This is why you can do hexadecimal math on a Chinese abacus since it has 5 ones counters and 2 fives counters.

Need binary? Just move fewer beads. No need to have a separate machine.

So Why is the decimal abacus more popular than the binary one? Couple of reasons.

More people use decimal numbers on a day to day basis

They can do binary operations on the same machine.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Does the converse of uniform continuity -> Preservance of Cauchy sequences hold?

We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.

Now, I would like to ask the following question:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.

How do we go about doing it?

答:
You can also prove it by contradiction.

Suppose that $f$ is not uniformly continuous. Then there exists an $\epsilon >0$ so that for each $\delta>0$ there exists $x,y \in (0,1)$ with $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$.

For each $n$ pick $x_n, y_n$ so that $|x_n-y_n| < \frac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \epsilon$.

Pick $x_{k_n}$ a Cauchy subsequence of $x_n$ and $y_{l_n}$ a Cauchy subsequence of $y_{k_n}$.

Then the alternaticng sequence $x_{l_1}, y_{l_1}, x_{l_2}, y_{l_2},..., x_{l_n}, y_{ln}, ...$ is Cauchy but

$$\left| f(x_{l_n}) - f(y_{l_n}) \right| \geq \epsilon \,.$$


问:
Self-Paced Graduate Math Courses for Independent Study
Does anyone know of any graduate math courses that are self-paced, for independent study?

I am a high school math teacher at a charter school in Texas. While I am quite happy with where I am right now, but my goal is to earn at least 18 graduate credits on math subjects so that I can teach higher-level math, and be certified as dual-credit math teacher. (HS class where students earn high school as well as college credits at the same time.) I am aware of many online graduate classes offered by some respectable universities, but all of them are semester-based, which may not be very feasible since my full-time teaching is extremely demanding, not to mention that math is anything but a casual subject.

I welcome any suggestions even for programs from outside of US, as long as they are accredited and conducted in English. (For example, I was told that the college system in the Philippine is an exact "copy cat" of the US.) For your information, I am quite comfortable studying independently, in fact, I took lots of prerequisite math classes successfully under this study mode. By the way, last year I took GRE for this purpose, my verbal + quantitative score is a decent 1200 under old scoring scale.

Thank you very much for your time and help.

答: I would sincerely stay away from online programs. Period. Doing one online class for your own benefit is fine, but an entire degree program? No way! I have served on several search committees for new hirees and believe me, an online degree on the resume is NOT a good thing. This is what I did. When I got the opportunity to teach higher level math classes, I went to an accredited intitution that offers all graduate classes in the evening and in summer. This was not conflicting with my work schedule, just have to allocate time for study with respect to your duties at home, if any. It took me about 5 years (parttime of course) Your employer may even fincancially sponsor you up to some amount. Aren't there any universities somewhere where you live? You want to study abroad? How about your job then?</p>


问:
Sampling a random $m$-dimensional subspace of $\mathbb{R}^n$ uniformly
Let $G_{n,m}$ be the Grassmanian manifold, the set of $m$-dimensional linear subspaces of $\mathbb{R}^n$.

I read in the book of Verschynin that one can sample a random element $E \in G_{n,m}$ with uniform distribution in $G_{n,m}$ by computing the column span of a random $n \times m$ Gaussian random matrix with i.i.d. $N(0,1)$ entries (meanzero with unit variance).

Can anyone give me an hint of how to prove rigorously that this object has uniform distribution in  $G_{n,m})$? Where does gaussianity play a role?

答:
The gaussianity is important because the Gaussian Orthogonal Ensemble is orthogonally invariant (which is the usual reason for Gaussian variables). The method comes from the following remark of G. W. Stewart (see the reference below):

Stewart, G. W., The efficient generation of random orthogonal matrices with an application to condition estimators. (With mircofiche section), SIAM J. Numer. Anal. 17, 403-409 (1980). ZBL0443.65027.

To generate a random orthogonal matrix uniformly, generate an i.i.d. Gaussian matrix, $G,$ compute the QR decomposition $G = QR,$ then $Q$ is the random orthogonal matrix you seek. Extension of this to arbitrary grassmannians should be easy.


问:
How do you solve this recurrence relation with divides?
I am trying to find all sequences $(a_n)$ of positive integers with the following property: for all $m$ and $n$, $a_ma_n$ divides $a_{m+n}$.  I’m not sure whether this counts as a recurrence relation or not, given the presence of “divides”.

Two sequences that satisfy it are $b^n$ for any positive integer $b$, and $n!$.  But how would I find the general solution?

答:
For any prime $p$ and $k\in \mathbb N^+$, let $\nu_p(k)$ be the largest power of $p$ dividing $k$. The divisibility condition implies $\nu_p(a_{n+m})\ge \nu_p(a_n)+\nu_p(a_m)$, so that the sequence $\nu_p(a_n)$ is super-additive in $n$ for all $p$. Clearly, this is sufficient as well.

This gives a full characterization of all sequences $a_n$ which work; they are defined by a collection of super-additive sequences $\nu_p(a_n)$ for each prime $p$, with the only constraint between the sequences being that for each $n\in\mathbb N$, $\nu_p(a_n)$ must be nonzero for only finitely many $p$.

There are several infinite classes of examples here.


问:
Percentage Problem
Trying to help a student with this, but I cannot see how I am doing it wrong. Here is the problem and the answer from the book.

Q. Tina invested $30,000$ in a stock. In the first year, the
stock increased in value by $10\%$. In the second year,
the stock decreased in value by $20\%$. What percent-
age gain is required in the third year for Tina’s stock
to return to its original value? (Round to the nearest
tenth of a percent.)

A. $13.6\%$

My work:
The first year as $30000*10\%=3000$,  $30000+3000=33000$

Second year as $33000*-20\%=-6600$, $33000-6600=26400$

Third year as $26400/30000=0.88,$ $1-0.88=.12=12\%$

By steps, please, what am I doing wrong?

答:
Your steps were fine until you calculated the third-year percentage. It looks like you calculated the percentage growth relative to the original amount ($\$30,000$) rather than relative to the amount after the second year ($\$26,400$).

You can think of this like an algebraic equation:

$$\begin{align*}
26 \ 400 \times  \left(1 + \frac{x}{100} \right) &= 30 \ 000\\[5pt]
1 + \frac{x}{100} &= \frac{30 \ 000}{26 \ 400}\\[5pt]
\frac{x}{100} &= \frac{3 \ 600}{26 \ 400} \\[5pt]
x &= \frac{3 \ 600}{264} \approx 13.6 \\[5pt]
\end{align*}$$

In short, the amount Tina needs to make up - $\$3,600$ - is $12\%$ of $\$30,000$ as you have calculated, but that is not the value the question is looking for. The question is looking for the amount that Tina needs to gain based on what she has after the second year. $\$3,600$ is $13.6\%$ of $\$26,400$, which is why it is the answer.

Title: Percentage Problem


问:
probability distribution Markov chain

Given this transition matrix of a Markov chain the question is:
What is the probability distribution for the length of a continuous period of being in state A? What is the average length of this period?

I have calculated the stationary probabilities but otherwise I am not really sure what this question is asking. Any input would be much appreciated.

答:
The stationary probabilities are unnecessary work here.

The transition matrix tells us that from state $A$, we go to state $H$ with probability $0.5$, and stay in state $A$ with probability $0.5$. So:

If we stay in $A$ for $1$ step, then as soon as we get to $A$, we immediately transition $A \to H$. What is the probability that this happens?

If we stay in $A$ for $2$ steps, then once we get to $A$, the next two steps must be $A \to A \to H$. What is the probability that this happens?

In general, if we stay in $A$ for $k$ steps, that means that after getting to $A$, the next $k$ steps must be $A \to A \to \dots \to A \to H$ (with $k-1$ transitions $A \to A$). What is the probability that this happens?

These probabilities tell us the probability distribution; from there, you can compute its expected value in the usual way.

Title: probability distribution Markov chain


问:
Is there a function continuous on $R$, such that $f(x)\in\mathbb{Q}$ iff $f(x+1)\in\mathbb{R\setminus Q}$?
I am given the following question:

Is there a function $f:\mathbb{R}\rightarrow\mathbb{R}$, continuous on $R$, such that $f(x)\in\mathbb{Q}$ iff $f(x+1)\in \mathbb{R \setminus Q}$?

I think it cannot exist because the intermediate value property would pop out numbers from $\mathbb{R}$ until I would have two rationals between two irrational numbers instead of one. I thought about having two sequences that converge at the same $x$ but I don't know how to go further with this ideea, as I don`t know how to give a rigurous formalisation of that ideea of obtaining that two numbers.

I thought about Dirichlet's function and it can be ''connected'' to our function, but I do not see how to link continuity to a property of Dirichlet's function so to obtain a result.

Any help would be warmly welcomed.

答: So, as reuns( it's the name of an account) has observed and told me: the basic ideea is to observe that if $f(x)\in\mathbb{Q}$ and $f(x+1)\in\Bbb R\backslash\Bbb Q$ for every $x\in\mathbb{R}$, then $f(x)+f(x+1)\in\Bbb R\backslash\Bbb Q$, and because the finite sum of continuous function is a continuos function and this sum is not continuous, we get the answer to the question, that a function with that property does not exist.


答:
The mistake appears to be in the last percantage calculation. You set
$$100\%=30'000$$
$$88\%=26'400,$$
so you got $12\%$. The order is incorrect. You should have set
$$100\%=26'400$$
$$113.\dot{6}\dot{3}\%=30'000,$$
So the difference is rounded $13.6\%$.

Title: Percentage Problem

Body:
Trying to help a student with this, but I cannot see how I am doing it wrong. Here is the problem and the answer from the book.

Q. Tina invested $30,000$ in a stock. In the first year, the
stock increased in value by $10\%$. In the second year,
the stock decreased in value by $20\%$. What percent-
age gain is required in the third year for Tina’s stock
to return to its original value? (Round to the nearest
tenth of a percent.)

A. $13.6\%$

My work:
The first year as $30000*10\%=3000$,  $30000+3000=33000$

Second year as $33000*-20\%=-6600$, $33000-6600=26400$

Third year as $26400/30000=0.88,$ $1-0.88=.12=12\%$

By steps, please, what am I doing wrong?


问:
What does "does not cancel on the interval" mean?
I have a question which asks me to consider the system
$$f(x)=a+bx+\frac{P(x)}{Q(x)},\qquad x\in I,$$
where $P$ and $Q$ are polynomials, and stipulates "$Q(x)$ does not cancel on the interval $I$ ".

What does this mean?

Does this mean $Q(x)\neq0$ on this interval, or that $Q(x)\neq P(x)$ on this interval, or perhaps something else? I have tried Googling the expression but couldn't find anything, and I haven't came across it before.

答:
It might mean that whenever $Q(x)=0$, then $P(x)\ne 0$. And the other way around, $P(x)=0\implies Q(x)=0$, but that is probably the less interesting case.

As $P,Q$ are polynomials, the usual use of "cancel" is that a common factor can be removed, here that could be interpreted the way that $P,Q$ do not have a common factor that has roots inside that interval.

But it might also be an unfortunate word choice and the condition to express was that "$Q$ does not vanish on $I$", which indeed then means that there are no roots of $Q$ in $I$.


问:
Determine the probability that no guest will receive the proper hat.
Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.

My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.

I think the answer should be something like:

$\frac{}{4!}$

where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.

How should I think about this problem instead?

So the book gives the answer 3/8 and myclassmate said the answer is

$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$

But I don't understand why this is?

答:
Pr

n
i=1
Ai

=
n
i=1
Pr(Ai) −

i<j
Pr(Ai
∩ Aj ) +

i<j<k
Pr(Ai
∩ Aj
∩ Ak)
−

i<j<k<l
Pr(Ai
∩ Aj
∩ Ak
∩ Al) + . . .

(−1)n+1 Pr(A1 ∩ A2 ∩ . . . ∩ An). We'll use this probability union formula for the solution. Let's assume that a first man will recieve a proper hat  and the probability for this event is P(A1) = 1/n and as there are 4 cases like thet the sum is n*(1/n)=1

Then the probability that first or second men will receive proper hat is P(A1 or A2)=1/n*(n-1) and as there are Cn,2 ways to place these two men the sum will be Cn,21(n(n-1))

Then the probability that any three men will receive a proper hat will be P(A1 or A2 or A3) = 1/n(n-1)(n-2) and the sum will be Cn,3*1/n(n-1)(n-2)

The sum P(A1 or A2 or A3 or A4) = 1/n!
Let's compute:
1 - 1/2! + 1/3! - 1/4! = 15/24 This is when each man will receive a proper hat and the probability that no man will receive a proper hat is 1-15/24 = 9/24 = 3/8


问:
Slutsky's Theorem XnYn-CX

My Question is How they got step c and Why d can conclude e step.

答:
For step (c), let $E_1$ denote the event $\{x < X_n \leq y\}$, and let $E_2$ denote the event $\{\left| Y_n \right| < \varepsilon / M\}$. Then we have
\begin{align*}
\mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) &= \mathbb{P}(E_1 \cap E_2)\\
&= \mathbb{P}(E_1) + \mathbb{P}(E_2) - \mathbb{P}(E_1 \cup E_2)\\
&= [1 - \mathbb{P}(E_1^c)] + [1 - \mathbb{P}(E_2^c)] - \mathbb{P}(E_1 \cup E_2)\\
&= 1 - \mathbb{P}(E_1^c) - \mathbb{P}(E_2^c) + [1 - \mathbb{P}(E_1 \cup E_2)]\\
&\geq 1 - \mathbb{P}(E_1^c) - \mathbb{P}(E_2^c)\\
&= 1 - \mathbb{P}(\{x < X_n \leq y\}^c) - \mathbb{P}(\{\left| Y_n \right| < \varepsilon / M\}^c)\\
&= 1 - \mathbb{P}(\{x < X_n \leq y\}^c) - \mathbb{P}(\left| Y_n \right| \geq \varepsilon / M),
\end{align*}
where $"\geq"$ comes from the fact that the probability of any event ($E_1 \cup E_2$ in this case) is less or equal to 1.

For step (e), given that we have concluded the result from step (d), i.e.
$$\lim_{n\rightarrow \infty} \mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) > 1 - \xi.$$
Together with the result from step (b) and the monotonicity of limit, we have
$$1 \geq \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| < \varepsilon) \geq \lim_{n\rightarrow \infty}\mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) > 1 - \xi.$$
Since this holds for any $\xi > 0$, using the Squeeze Theorem, we conclude that
$$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| < \varepsilon) = 1,$$
which then implies that
$$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) = 0.$$
Lastly, note that $\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) > \mathbb{P}(\left| X_nY_n \right| > \varepsilon)$. Using the monotonicity of limit, we have
$$0 = \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) \geq \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| > \varepsilon) \geq 0.$$
Using the Squeeze Theorem once again, we can conclude that
$$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| > \varepsilon) = 0.$$

Remark: The last part of proof gives a rigorous justification for the fact that the inequality sign is strict (i.e. $">"$ instead of $"\geq"$) from the result in part (e). However, this would not affect the remaining proof for the Slutsky's Theorem.

Title: Slutsky's Theorem XnYn-CX


问:
How many 4-element DNA sequences containg exactly three of four bases A,T,C,G?
```<code>How many 4-element DNA sequences containg exactly three of four bases A,T,C,G?

Hello, my answer is c(4,3)*c(3,1)*(4!/2!) because

by saying c(4,3), I choose three of four bases, e.g. ATC_
by saying c(3,1), I choose extra one of chosen three bases, e.g. ATCT
by saying 4!/2!, I permutated them, i.e., there is 4!/2! possible permutations of ATCT.
</code>```

So, I found the answer as 144, but there is a solution on the internet like this:

```<code>The objective is to find the 4 element DNA sequence that contain exactly three of the four bases A, C, G, and T.
Three of the four bases, but not with base A.
There are 3 ways for first base as it cannot be A.
There are 3 ways for second base as it cannot be A.
There are 2 ways for third base as it cannot be A and nor first or second base.
There are 1 way for fourth base as it cannot be A and nor first or second or third base.
By product rule, 3.3.2.1 =18.
Exactly three of four bases.
As there are 4 possible bases that are excluded and there are 18 DNA sequences without corresponding base but with three of the four bases.
By product rule, 4.18 = 72.
Therefore, there are 72 possible way for 4 DNA bases.
</code>```

I wonder which answer is correct 144 or 72? and if my solution with the answer 144 is correct, then why is the solution on the internet with answer 72 incorrect? or if my solution is incorrect, then why?

答:
There are four ways to exclude a base.
There are three choices for which of the remaining bases shows up twice.
There are $4!/2$ ways to arrange $4$ items when $2$ of them are identical.
Given that, there are $4 \cdot 3 \cdot \frac{24}{2} = 144$ ways.

Check: let's do the cases removing G. There should be $36$ of them.
Let's say the A is doubled. There should be $12$ of those:

With AA together:
AACT

AATC

CAAT

TAAC

CTAA

TCAA

With AA separated:

CATA

TACA

ACTA

ATCA

ACAT

ATAC


问:
The derivative of $\int_0^xxf(t)dt$ and why we cannot treat $xf$ as continuous.
This comes from Calculus by Michael Spivak, 3rd ed., chapter 14, Problem 11.

Find $F'(x)$ if $F(x) = \int_0^x xf(t)dt$. The answer is not $xf(x)$; you should perform an obvious manipulation on the integral before trying to find $F'$.)

Here, Spivak fails to include the assumption that $f$ is continuous.

With this assumption, the answer is $\int_0^xf(t)dt + xf(x)$, and is solved here, here, and here.

I understand the solution. I was able to do it myself. It makes sense to me. I also solved it using the $\lim_{h\to 0}$ approach, and that worked too.

Yet, naively, it seems like the answer should be $xf(x)$. It seems like since $x$ "behaves like a constant" in that integration, and because $f$ is continuous, then $xf$ must also be continuous, and so $F'(c) = cf(c)$.

I'm hoping to build better intuition about why this is wrong.

I was thinking perhaps one way of looking at it is that $xf$ cannot be assumed to be continuous because we are free to change the value of $x$ somewhat arbitrarily. In fact, at any fixed point $c$, $xf(c)$ varies with $x$, so certainly we cannot guarantee anything about the value of $xf(t)$ simply by restricting $t$ to certain values.

Does this make any sense? I'm sorry to use such sloppy reasoning/language/arguments.

There's a good chance I just don't yet have the tools to fully "get it" yet, but I appreciate any insight you might be able to share.

Edit

Ian's answer helps explain something that confused me in Spivak's solution to a later problem.

The next two problems are:

Prove that if $f$ is continuous, then
$$\int_0^xf(u)(x-u)du = \int_0^x\left(\int_0^uf(t)dt\right)du.$$
Hint: Differentiate both sides, making use of Problem 11.

Use Problem 12 to prove that

$$\int_0^xf(u)(x-u)^2du = 2\int_0^x\left(\int_0^{u_2}\left(\int_0^{u_1}f(t)dt\right)du_1\right)du_2.$$

These each require fairly straightforward calculations, the details of which we can skip here. However, in the published Answer Book solution for the latter, problem 13, Spivak does something that initially I found baffling. He begins:

Applying Problem 12 to $g(u) = f(u)(x-u)$, we obtain
$$\int_0^xf(u)(x-u)^2du =\int_0^x[f(u)(x-u)](x-u)du$$
$$=\int_0^x\left(\int_0^uf(t)(x-t)dt\right)du$$

This substitution is misleading. The result he's using from problem 12. applies only to functions that do not depend on $x$, while Spivak's $g$ here clearly does depend on $x$.

I think he gets away with it here only because $\frac{\partial}{\partial x}[f(u)(x-u)] = f(u).$

答:
In this particular case, you can solve the problem as

$$\frac{d}{dx} \int_0^x x f(t) dt = \frac{d}{dx} \left ( x \int_0^x f(t) dt \right ) = \int_0^x f(t) dt + x f(x)$$

by the product rule together with FTC.

More generally you have the Leibniz rule:

$$\frac{d}{dx} \left ( \int_{a(x)}^{b(x)} f(t,x) dt \right ) = f(b(x),x) b'(x) - f(a(x),x) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(t,x) dt.$$

The Leibniz rule captures the dependence of the integral on $x$ through both the limits of integration and the integrand. In your problem, giving the answer $x f(x)$ would only encapsulate the dependence on the limits of integration.


问:
classifying space of the normalizer of maximal torus and borel construction
I want to show that $B(NT) \simeq (BT)_{hW}$, where $NT$ is the normalizer of the maximal torus, $W$ is the Weyl group (that is $W = NT/T$) and $(BT)_{hW}$ is the homotopy orbit/Borel construction, $(BT)_{hW} = EW \times_W BT$.

What I did: B(NT) can be modeled by $E(NT) \times_{NT} E(NT/T)$, since $E(NT/T)$ is contractible. Note that $E(NT)/T \simeq BT$, as $T$ is a normal subgroup. Hence, $E(NT) \times_{NT} E(NT/T) \simeq E(NT) \times_{NT/T} E(NT/T) \simeq BT \times_W E(W)$. This last step, namely $E(NT) \times_{NT} E(NT/T) \simeq E(NT) \times_{NT/T} E(NT/T)$, should be correct but I am having difficults to make it precise in order to really convince myself. How could one explain it?

Is there another way of showing this? Maybe using the fact that from the exact sequence $1 \to T \to NT \to W \to 1$ there is a fibration sequence $BT \to BNT \to BW$?

Many thanks!

答: This is true for any normal (closed) subgroup $N$ of a (topological) group $G$. In general, if $G$ acts freely on $X$, then $X/G\simeq X_{hG}$ (see equivariant cohomology in case of free actions (basic question)). We can take then take $(EG)/N$ as a model for $BN$ . The residual action of $G/N$ on $(EG)/N$ is free, therefore we obtain $BN_{hG/N}\simeq (EG/N)/(G/N)\cong EG/G = BG$


问:
Nested little $o$ notation
Is it possible to "nest" little $o$ notations?

For example, in showing the chain rule for a function $f(u(x))$ (the example is taken from here, if we define $k = h\times[ u'(x) + o(h) ]$, for $h\in\mathbb{R}$ can we describe
$$
o(k) = o(h \times [o(h) + u'(x)]) = o(h\times u'(x))
$$
or does this not make sense?

It seems that in this case, since $o(h)$ is dominated by $h$, it'd be fine here, but not sure about this or about nesting generally.

答: This idea is workable; let's spell out how it works. I'll use different notation so we don't have to remember which variable's small variation is $h$, which is $k$ etc. We have$$\delta u\in u^\prime(x)\delta x+o(\delta x),\,\delta f\in f^\prime(u)\delta x+o(\delta u),$$where each ${}^\prime$ has an obvious-in-context meaning. So$$\delta u\in f^\prime(u)(u^\prime(x)\delta x+o(\delta x))+o(u^\prime(x)\delta x+o(\delta x)).$$Even easier to read is$$\delta u\in(u^\prime+o(1))\delta x,\,\delta f\in(f^\prime+o(1))\delta u=(f^\prime+o(1))(u^\prime+o(1))\delta x.$$The $\delta x$ coefficient is $f^\prime u^\prime+\underbrace{o(f^\prime)+o(u^\prime)+o(1)}_{o(1)}$.

Title: Nested little $o$ notation


问:
Integration in Banach spaces - interesting, nice and non-trivial examples needed
I am interested in $\textbf{Integration in Banach spaces}$. Here is a little motivation for my question:

Let $\left(X,\|\cdot\| \right)$ be a Banach space, $a,b \in \mathbb{R}$ with $a<b$ and $f \colon [a,b] \longrightarrow X$ a function. How can we integrate such a function?

I could already find an answer with the $\textbf{Riemann Integral for Banach space-valued functions}$ (which is quite similar to the comon Riemann Integral) and the $\textbf{Bochner Integral}$ (which is similar to the Lebesgue Integral).

But so far I only know some theoretical results about those integrals (only the basical ones) and I have not yet seen or calculated a practical example.

Now I wonder if anybody could present me different examples of such a integral. (I am looking for such nice and epical integrals we know from Complex analysis or we could calculate using an $d$-dimensional Spherical coordinate system or something similar.)

I am also looking for any kind of (nice) calculations involving Integration in Banach Spaces.
If anybody knows a rewarding (not too hard) theorem/proof involving Integration in Banach Spaces this would also interest me.

I hope you understand what I am searching for...

答:
Elaborating a bit on @QiaochuYuan's comment: for almost every purpose, we certainly want the property that $T\int f=\int Tf$, where $f$ is $V$-valued, and $T$ is a continuous linear map $V\to W$. Under very mild hypotheses on the spaces $V,W$ (locally convex, quasi-complete suffices), by Hahn-Banach the desired property already follows from requiring $\lambda\int f=\int \lambda\circ f$ for continuous linear functionals $\lambda$. (We would also want estimates on $\int f$, but/and that does/will come out of one of several constructions.) This characterization is often called a Gelfand-Pettis integral, or weak integral (not because the conclusion is weak, but because the hypothesis is weak).

The condition with linear functionals already implies (by Hahn-Banach) that if there exists any vector $v=\int f\in V$ at all, it is unique. There's no ambiguity in what it is... if it exists.

Not all uniquely characterized things exist, as it turns out, so proof of existence (under various hypotheses) is important. Constructions suffice, and may prove useful incidental properties.

This set-up already allows the Schwartz-Grothendieck vector-valued extension of Cauchy-Goursat theory of complex functions in one variable.

Many basic purposes are fulfilled by looking at Banach-space-valued functions, perhaps continuous with compact support. But in practice we often want the weaker "strong" topology on operators. And we may want test-function-valued functions (not Frechet, but strict-colimit-of-Frechet), or distribution-valued. A small amount of work does verify that all spaces $V$ are "quasi-complete" (and locally convex), so that compactly-supported, continuous, $V$-valued functions (as a useful example) have integrals. In a very positive sense, we can prove that the situation is fairly "idiot-proof", although understanding that point is considerably more complicated than simply using the ideas in blissful naivete. :)


问:
Is there a real number, that is proven to be normal in every base, whose digits can be enumerated by an algorithm?
To clarify, I mean every natural number base $b$ where $b \geq 2$. If so, what is the algorithm to generate the number (and what is the number, if it has a name)?</p>

答: Wikipedia reveals that the answer is yes, citing the paper "An example of a computable absolutely normal number" by Becher and Figueira (Theoretical Computer Science 270(1-2), 2002).


答:
The Stirling number of the second kind $\left\{n \atop k\right\}$ counts the number of partitions of an $n$-set into exactly $k$ parts, so $k! \left\{n \atop k\right\}$ is the number of surjections of an $n$-set onto a $k$-set.  Your desired count is hence
$$\binom{4}{3} 3! \left\{4 \atop 3\right\} = 4 \cdot 6 \cdot 6 = 144.$$

Body:
```<code>How many 4-element DNA sequences containg exactly three of four bases A,T,C,G?

Hello, my answer is c(4,3)*c(3,1)*(4!/2!) because

by saying c(4,3), I choose three of four bases, e.g. ATC_
by saying c(3,1), I choose extra one of chosen three bases, e.g. ATCT
by saying 4!/2!, I permutated them, i.e., there is 4!/2! possible permutations of ATCT.
</code>```

So, I found the answer as 144, but there is a solution on the internet like this:

```<code>The objective is to find the 4 element DNA sequence that contain exactly three of the four bases A, C, G, and T.
Three of the four bases, but not with base A.
There are 3 ways for first base as it cannot be A.
There are 3 ways for second base as it cannot be A.
There are 2 ways for third base as it cannot be A and nor first or second base.
There are 1 way for fourth base as it cannot be A and nor first or second or third base.
By product rule, 3.3.2.1 =18.
Exactly three of four bases.
As there are 4 possible bases that are excluded and there are 18 DNA sequences without corresponding base but with three of the four bases.
By product rule, 4.18 = 72.
Therefore, there are 72 possible way for 4 DNA bases.
</code>```

I wonder which answer is correct 144 or 72? and if my solution with the answer 144 is correct, then why is the solution on the internet with answer 72 incorrect? or if my solution is incorrect, then why?


问:
How can I prove that every point in the boundary of the unit disc in $\mathbb{R}^2$ is a extreme point?
I know that a point in the circunference can't be in a segment with the edges inside de disk, but I'm having problems at proving it formally.

We defined extreme point of a set $S$ as a point $x_0 \in S$ such that $x_0 = \alpha x + (1-\alpha)y \Longleftrightarrow x_0 = y = x$ for $x,y \in S, \alpha \in [0,1]$

I tried with a proof by contradiction assuming there exists a segment with the edges inside of the disk that contains it but I don't know what steps to take to arrive to a contradiction.

Does someone have any ideas that could help?　Thanks.

Edit: Does this lead anywhere? (I don't think so)
For any $x,y \in D, x_0 \in \partial D, || \alpha x + (1-\alpha)y|| \leq | \alpha |||x|| + |1-\alpha|||y|| = ||x_0|| = 1 \Longleftrightarrow ||x|| = ||y|| = 1$

答:
As per what you've done, we can let $x, y \in D$ and $x_0 \in \partial D$ such that $x_0 = \lambda x + (1 - \lambda)y$ for some $\lambda \in (0, 1)$. As you've noticed, this can only happen if $\|x\| = \|y\| = 1$.

Let's first consider the special case of $\lambda = \frac{1}{2}$. In this particular case, we can fall back on some circle geometry. The line segment between $x$ and $y$ is a chord in the circle $\partial D$, and $x_0$ is the midpoint. By circle geometry, I would expect the line segment from $0$ (the centre) to $x_0$ (the midpoint) should be perpendicular to $y - x$ (the chord). Let's verify:
$$x_0 \cdot (y - x) = \frac{1}{2}(y + x) \cdot (y - x) = \frac{1}{2}(y \cdot y - x \cdot x) = \frac{1}{2}(1^2 - 1^2) = 0,$$
as expected. Note also that
$$x_0 + \frac{y - x}{2} = \frac{y + x}{2} + \frac{y - x}{2} = y,$$
so by Pythagoras's theorem,
$$1 = \|y\|^2 = \|x_0\|^2 + \left\|\frac{y - x}{2}\right\|^2.$$
This implies that $\|x_0\| = 1$ if and only if $\left\|\frac{y - x}{2}\right\|^2 = 0$, i.e. $y = x$.

Now, the other cases, where $\lambda \neq \frac{1}{2}$, follow from the fact that every point in the line segment between $x$ and $y$ is either a convex combination of $x$ and $\frac{x + y}{2}$ or a convex combination of $y$ and $\frac{x + y}{2}$. As you already proved, in order for such a convex combination to be on the boundary circle, both endpoints must be on the boundary circle! But, we just proved that $\frac{x + y}{2}$ is not (unless $x = y$), and so the theorem is proven.


问:
Uniformly integrability of inverse of sample moments
I am interested in the uniform integrability of the set $\{Y_n\}_{n\ge 3}$ where

$$
Y_n = \bigg(\frac{1}{n} \sum_{i=1}^n X_i^k\bigg)^{-1},
$$
where the $X_i$'s are i.i.d observations of a continuous random variable $X$, $k\ge 1$, and $E[X] < \infty$.

Is $\{Y_n\}_{n\ge 3}$ uniformly integrable for $k = 1$ which corresponds to the inverse of the sample mean? What about $k > 1$? In particular, the inverse of the second sample momemt, i.e., $k=2$, is what I am most interested in.

If it is not uniformly integrable in general, is it uniformly integrable if we restrict the allowable $X$?

答:
Cannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI.

Note that
$$
\mathbb{E}\left(|Y_n| 1\{|Y_n|\geq M\}\right) \geq M\mathbb{P}\left(|Y_n|\geq M\right) = M\mathbb{P}\left(|\bar{X}_n|\leq \frac{1}{M}\right) \to M,
$$
as $n\to \infty$ by laws of large numbers ( $\bar{X}_n\to 0$ in probability), now you take the limit as $M\to \infty$ to obtain

$$
\liminf_{M\to \infty }\liminf_{n\to \infty }\mathbb{E}\left(|Y_n| 1\{|Y_n|\geq M\}\right)= \infty
$$

Since you are most interested in $k=2$, let us take $k$ is even. Let us show below that an "allowable" $X$ is the following: $X$ is a random variable with $\mathbb{E}(X^{-2k})<\infty$, then $\{Y_n\}$ is UI.

(Note that the conditions in 2 are sufficient but might not be necessary )

Consider the function $\frac{1}{x}$, it is convex on $(0,\infty)$, now using Jensen's inequality you get (I am using $k$ is even to make sure $X_i^k$'s are non-negative, also the inequality below is trivial when any of the $X_i$ is zero)
$$
\left(\frac{1}{n}\sum_{i=1}^n X_i^k\right)^{-1} \leq \frac{1}{n}\sum_{i=1}^n X_i^{-k} .
$$
Thus
$$
\mathbb{E}\left(\left|\left(\frac{1}{n}\sum_{i=1}^n X_i^k\right)^{-1}\right| 1\{\left|\left(\frac{1}{n}\sum_{i=1}^n X_i^{k}\right)^{-1}\right|\geq M\}\right)\\ \leq \frac{1}{n}\sum_{i=1}^n\mathbb{E}\left( X_i^{-k}1\{\left|\left(\frac{1}{n}\sum_{i=1}^n X_i^{k}\right)^{-1}\right|\geq M\}\right) \\
= \mathbb{E}\left( X_1^{-k}1\{\left|\left(\frac{1}{n}\sum_{i=1}^n X_i^{k}\right)^{-1}\right|\geq M\}\right) \leq \mathbb{E}^{1/2}\left( X_1^{-2k}\right)\mathbb{P}^{1/2}\left(\frac{1}{n}\sum_{i=1}^n X_i^{k}\leq \frac{1}{M}\right).
$$
In the last step, we use Cauchy–Schwarz inequality. Now $X_i^{-k}$'s are i.i.d random variables with finite second moment. Let us now consider two cases:

Case I:  $\mathbb{E}(X_i^{k})<\infty$.  Then $\frac{1}{n}\sum_{i=1}^n X_i^{k} \to \mathbb{E}(X^k)$ as $n\to\infty$ in probability by laws of large numbers. Moreover $\mathbb{E}(X^k)>0$ as otherwise $X=0$ with probability $1$, which will violate $\mathbb{E}(X^{-k})<\infty$. Therefore $\frac{1}{n}\sum_{i=1}^n X_i^{k} \to \mathbb{E}(X^k)>0$ and thus choosing $M$ large we can complete the proof.

Case II: $\mathbb{E}(X_i^{k})=\infty$. Here also you can show that for an arbitrary positive constant $c>0$, probability that $\frac{1}{n}\sum_{i=1}^n X_i^{k} >c$ goes to $1$ as $n \to \infty$. Therefore again choose an $M$ large enough to complete the proof.


问:
Infimium between two sets with a bijection
People, I need some help about a thing: I am studying analysis and, between my thoughts about supremum and infimum, I started to ask myself the following:

Let A, B $\neq \emptyset$ and $f:A\rightarrow B$ be a bijection. Assume that there A and B are bounded below. Show that $f(\inf(A))=\inf(B)$.

Is that sentence true? Why? I did not found a way to prove it, since I can not assume that $f$ takes the infimium from one set to the another, once $\inf(A) \notin A$, in general.

答:
I am interpreting "inferiorly limited" as "bounded below" since I haven't seen that terminology before; and this does not seem true at all.

Suppose $A = \{1,2,3\}, B=\{4,5,6\}$, then $\text{inf}(A) = min(A)= 1$, $\text{inf}(B)=min(B)=4$.

And we have the bijection $f: A\rightarrow B, f(x)=x+3$.

Giving us a contradiction.


问:
Why is there this recurring pattern in the derivative of $x\cdot e^{-k\cdot x}$?
I was doing an exercise for which I needed to take the derivative(s) of the function

$$f_k(x)=x\cdot e^{-k\cdot x}$$

Doing this a few times, I noticed some sort of recurring pattern. I concluded that one can write the $n$-th  derivative of the function as

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}((-1)^nkx-(-1)^n\cdot n)$$

or perhaps, to improve readability

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(-kx+n)$$
for odd $n$ and

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(kx-n)$$
for even $n$.

While I know how to compute each derivative, I do not understand why the derivative follows this pattern-like structure. Is there any intuitive idea behind this?

Clarification: By "intuitive", I mean anything that does not purely rely on blindly computing each integral. A very beautiful way to represent derivatives visually is shown in this 3blue1brown video; I don't know if anything like this is possible to explain the above described pattern. Basically, I am looking for something which actually explains why there is a pattern instead of just noticing it.

Note: This is not the actual homework assignment, it just came up when doing it.

答:
Following the hint of @HansLundmark is quite convenient.

The general Leibniz rule gives us
\begin{align*}
\color{blue}{\frac{\partial^n}{\partial x^n}}\color{blue}{\left(x e^{-k\cdot x}\right)}
&=\sum_{j=0}^n\binom{n}{j}\left(\frac{\partial^j}{\partial x^j} x\right)\left(\frac{\partial^{n-j}}{\partial x^{n-j}} e^{-k x}\right)\tag{1}\\
&=\binom{n}{0}x\left(\frac{\partial^{n}}{\partial x^{n}}e^{-k x}\right)+\binom{n}{1}\left(\frac{\partial^{n-1}}{\partial x^{n-1}}e^{-k x}\right)\tag{2}\\
&=x\left(-k\right)^ne^{-k x}+n\left(-k\right)^{n-1}e^{-k x}\\
&\,\,\color{blue}{=(-1)^nk^{n-1}e^{-kx}\left(kx-n\right)}
\end{align*}
Here we do not have to look for a common pattern, since the Leibniz rule gives us the wanted expression right at hand.

Comment:

In (1) we get by the general Leibniz rule more or less the final expression without the need to calculate derivatives iteratively.

In (2) we use the nice fact that $\left(\frac{\partial^j}{\partial x^j} x\right)=0$ if $j>1$ so that only two summands give a contribution.


问:
Laurent Series Expansion Solution Verification
Write the Laurent Series of
$$ \frac{1}{z(1-z)^2}$$

Now the singularities are $z=0$ and $z=1$.

My attempt:

For $z=0$ case
$$= \frac{1}{z}.(1-z)^{-2}$$
$$=\frac{1}{z}.[1+2z+3z^2+...]$$

For $z=1$ case
$$=\frac{1}{(1-z)^2}.\frac{1}{(1+z-1)}$$
$$=\frac{1}{(1-z)^2}.(1+(z-1))^{-1}$$
$$=\frac{1}{(1-z)^2}.[1+(z-1)+(z-1)^2+(z-1)^3+...]$$

I used the $(1+x)^{-n}$ expansion in the last line.

Am I doing it right? Is there a better approach?

答:
Here's an approach that uses partial fraction decomposition:
\begin{align}
\frac{1}{z(1-z)^2}
&= \frac{1}{z} + \frac{1}{1-z} + \frac{1}{(1-z)^2} \\
&= \frac{1}{z} + \sum_{n \ge 0} z^n + \sum_{n \ge 0} \binom{n+1}{1} z^n \\
&= \frac{1}{z} + \sum_{n \ge 0} (n+2) z^n
\end{align}


问:
Famous Problems Where We Only Know the Elementary
Define a graph with vertex set $\mathbb{R}^2$ and connect two vertices if they are unit distance apart. The famous Hadwiger-Nelson problem is to determine the chromatic number $\chi$ of this graph. For the problem as stated (there are many variations), the best known bounds are $4 \leq \chi \leq 7$. The lower bound comes from the existence of a clever subgraph on just seven vertices that is readily seen to require four colors. The upper bound comes from a straightforward tiling of the plane by monochromatic hexagons that admits a proper $7$-coloring.

I like to show this problem to young students because they are always fascinated by the fact that we can cover "everything" that is known about it in just a few minutes. What are some other problems of this type?

What are some famous problems for which the best known results are fairly obvious or elementary?

Update: Aubrey de Grey has recently improved the lower bound to 5 using an elaborate subgraph, so we now know slightly more than the elementary.

答:
Is $\pi$ a normal number?

All we know is that rational numbers are not normal, and that $\pi$ is irrational.

You can just as well replace $\pi$ by many other constants, and we don't know much about any of them.


问:
Prove that these are equivalence relations.

1) $\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : 2|x \ \land \ 2|y \}$

$$\text{reflexive: } (2|x \land 2|y)\mathbf{R}(2|y \land 2|x)  \ \textbf{[True]}\\ \text{symmetric: } (2|x_1 \land 2|y_1) \mathbf{R}(2|x_2 \land 2|y_2) \implies (2|x_2 \land 2|y_2) \mathbf{R}(2|x_1 \land 2|y_1)  \ \textbf{[True]}\\ \text{transitive: } (2|x_1 \land 2|y_1) \mathbf{R}(2|x_2 \land 2|y_2), (2|x_2 \land 2 | x_2)\mathbf{R}(2|x_3 \land 2|y_3) \implies (2 |x_1 \land 2|y_1)\mathbf{R}(2|x_3 \land 2|y_3) \ \textbf{[True]}\ $$

2) $\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : x|y \ \lor \ y|x \}$

$$\text{reflexive: } (x|x \lor y|y) \ \textbf{[True]} \\ \text{symmetric: } (x|y \lor y|x) \implies (y|x \lor x|y) \ \textbf{[True]} \\ \text{transitive: } (x|y \ \lor \ y|x) \ , \ (y|z \ \lor z|y) \implies (x|z \lor z|x) \ \textbf{[True]} $$

are my proofs right?

答:
Actually, it's only 2(symmetric) and 2(transitive) you got right. For 2-reflexive, you should prove that $xRx$, which is equivalent to proving that $(x|x \vee x|x)$ (note there's no $y$).

More importantly, your notation in 1) doens't make sense. It doens't make sense at all to write $(2|\wedge 2|)R(2|\wedge2|)$, cause you have to have just a natural number on either side of $R$. Remember, it's the numbers that are related with $R$, not the conditions. Take 1(symmetric) for example. You need to show that
$$
xRy \implies yRx
$$
and the definition of $R$ says that $xRy$ means that $2|x\wedge 2|y$, so you need to show that
$$
2|x \wedge 2|y \implies 2|y \wedge 2|x
$$
Makes sense?

You should also think about what you're actually proving - I claim that the $R$ in 1) is not an equivalence relation at all. Can you see which point fails?


问:
Show that the relation R on set A is transitive if and only if $R^{n} \subseteq R$ for all positive integers n
I tried to prove this using induction but i get stuck at a point and the proof fails. I dont know what am I doing wrong! Can someone help me prove this? I have tried all I possibly could</p>

答:
Since this is an if and only if theorem, there are actually two implications to prove:

if $R^n\subseteq R$ for all $n\in\Bbb Z^+$, then $R$ is transitive; and

if $R$ is transitive, then $R^n\subseteq R$ for all $n\in\Bbb Z^+$.

The first of these does not require induction. The easiest way to prove it is to prove the contrapositive: if $R$ is not transitive, then there is a positive integer $n$ such that $R^n\nsubseteq R$. Suppose that $R$ is not transitive; then there are $a,b,c\in A$ such that $\langle a,b\rangle$ and $\langle b,c\rangle\in R$, but $\langle a,c\rangle\notin R$. By definition, however, $\langle a,c\rangle$ is in $R^2$, so $\langle a,c\rangle\in R^2\setminus R$, and therefore $R^2\nsubseteq R$. This shows that if $R^2\subseteq R$, then $R$ is transitive, so of course if $R^n\subseteq R$ for all $n\in\Bbb Z^+$, then $R$ is transitive.

For the other direction we have to prove that if $R$ is transitive, then $R^n\subseteq R$ for each $n\in\Bbb Z^+$. We can do this by induction on $n$. Of course $R^1=R\subseteq R$; that gets the induction started. Now for the induction step suppose that $R^n\subseteq R$ for some $n\in\Bbb Z^+$; we need to show that $R^{n+1}\subseteq R$. Let $\langle x,y\rangle\in R^{n+1}$; then by definition there is an $a\in A$ such that $\langle x,a\rangle\in R^n$ and $\langle a,y\rangle\in R$. Can you see now how to use the induction hypothesis and the fact that $R$ is transitive to show that $\langle x,y\rangle\in R$ and hence that $R^{n+1}\subseteq R$?


问:
What is the particular solution for the PDE?
Consider the PDE:
$$
u_t(x,t) = ku_{xx}(x,t)+F(x,t),
$$
$$
u(0,t)=0, \hspace{0.2cm} u(L,t)=0
$$
$$
u(x,0)=\varphi(x)
$$
With $F(x,t)$ and $\frac{\partial F}{\partial x}$ continuous. And $F(0,t)=F(L,t)=0, \forall t>0$.

I know that the solution of the homogeneous system is:
$$
u(x,t)=\sum_{n=1}^{\infty}c_n(t)\phi_n(x)=\sum_{n=1}^{\infty}c_ne^{k\lambda_nt}\phi_n(x)
$$
With
$$
\lambda_n=-\left(\frac{n\pi}{L}\right)^2, \hspace{0.2cm} \phi_n(x)=\sin\frac{n\pi x}{L}
$$
$$
c_n = \frac{2}{L}\int_{0}^{L}\varphi(x)\sin\frac{n\pi x}{L}dx
$$

Now I am looking for the particular solution $v(x,t)$ for the inhomogeneous system that will allow me to find the  general solution:
$$
v_t(x,t)=kv_{xx}(x,t)+F(x,t),\hspace{0.22cm} (3)
$$
$$
v(0,t)=0, \hspace{0.2cm} v(L,t)=0
$$
$$
v(x,0)=0
$$

答:
I don't see a real benefit of doing it in 2 seperate steps:

You start exactly the way you did with the following Ansatz which satisfies the boundary conditions:

$$
u(x,t)=\sum_{n=1}^{\infty}c_n(t)\phi_n(x) \mbox{ , with }
\phi_n(x)=\sin\frac{n\pi x}{L}
$$
Now you plug it into the equation and get
$$
\sum_{n=1}^{\infty}c'_n(t)\phi_n(x)=-k\sum_{n=1}^{\infty}c_n(t)\phi_n(x)\left(\frac{n\pi}{L}\right)^2+F(t,x)\\
\Rightarrow \int_0^L\sum_{n=1}^{\infty}c'_n(t)\phi_n(x)\phi_j(x)dx=-\int_0^Lk\sum_{n=1}^{\infty}c_n(t)\phi_n(x)\phi_j(x)\left(\frac{n\pi}{L}\right)^2dx+\int_0^LF(t,x)\phi_j(x)dx\\
$$
Where I have multiplied with $\phi_j(x)$ and integrated over the domain. Now since the $\phi_n$ are orthogonal we end up with
$$
\frac{L}{2}c'_j(t)=-kc_j(t)\frac{j^2\pi^2}{2L}+\int_0^LF(t,x)\phi_j(x)dx
$$
Now you have to solve this ODE for the $c_j$'s and finally plug it in the inital condition.


问:
Converting into alternate e-Form
I was checking my results of an exercise online. The Maximum of my function was (1): $$\frac{4^{\frac{-1}{\log(2)}}}{\log^2(2)}$$

Now I wolframalpha tells me the alternate form is (2): $$\frac{1}{e^2\log^2(2)}$$

I feel like this form could be useful for future problems sometimes, but I absolutely have no clue how to get there. Could someone help me converting the maximum (1) into the alternate form (2)?

Title: Converting into alternate e-Form

Body:
Assuming  $\frac{4^{\frac{-1}{\ln2}}}{\ln^2(2)}$ = $\frac{1}{e^2\ln^22}$

$\frac{4^{\frac{-1}{\ln2}}}{1}$ = $\frac{1}{e^2}$

${4^{\frac{1}{\ln2}}}$ = ${e^2}$

$\frac{\ln4}{\ln2} $=2$\ln e$

$e$= $\frac{\ln 4}{2\ln2}$  =1

Reverse the above steps, we can prove  from (1) to (2).


答:
Firstly we have $4=2^2$. So we get $$\frac{(2^2)^{\frac{-1}{\log(2)}}}{\log^2(2)}=\frac{2^{\frac{-2}{\log(2)}}}{\log^2(2)}$$

I´ve applied the rule $\left(\color{blue}a^{\color{red}b}\right)^{\color{green}c}=\color{blue}a^{\color{red}b\cdot \color{green}c}$. Furthermore it is  $2=e^{\log(2)}$.

$$=\frac{\left(e^{\log(2)}\right)^{\frac{-2}{\log(2)}}}{\log^2(2)}$$

Applying the rule above again and then cancelling $\log(2)$.

$$=\frac{e^{\log(2)\cdot {\frac{-2}{\log(2)}}}}{\log^2(2)}=\frac{e^{-2}}{\log^2(2)}=\frac{1}{e^2\cdot \log^2(2)}$$

Title: Converting into alternate e-Form

Body:
I was checking my results of an exercise online. The Maximum of my function was (1): $$\frac{4^{\frac{-1}{\log(2)}}}{\log^2(2)}$$

Now I wolframalpha tells me the alternate form is (2): $$\frac{1}{e^2\log^2(2)}$$

I feel like this form could be useful for future problems sometimes, but I absolutely have no clue how to get there. Could someone help me converting the maximum (1) into the alternate form (2)?


问:
Prove that $F$ is non-singular and find the orthogonal projection of $ v$
Let be the follow bilinear form on $\mathbb{R}^4$ given by the matrix:

\begin{pmatrix}
    1 & 1 & 1 & 1 \\
    1 & 0 & 0 & 0 \\
    1 & 0 & 1 & 1 \\
    1 & 0 & 1 & 0
  \end{pmatrix}

Consider the subspace $F$, generated by $ w_1 = (1, 1, 0, 1) $ and $ w_2 = (0, −1, 1, 1)$. Prove that $F$ is non-singular and find a basis for $F^{\perp}$. Find the orthogonal projection of $ v = (1, 0, −1, 1) $ over $F$.

Is there an easier way to show that $F$ is non-singular? My attemp:

$F$ non-singular if $rad(F) = \{ {0_{\mathbb{R}^4}} \} $.

We calculate $F^{\perp}$ = $\langle (-1/2,1,1/4,0),(0,0,-1/2,1) \rangle $

$ rad(F) = F \cap  F^{\perp} = \{ {0_{\mathbb{R}^4}} \}  $

Then, I have several problems to find the projection of $v$ over $F$. I know that I need an orthogonal basis for $F$, let be $W=(w_1, w_2)$, and then we can do something like:

$ p^r_F (v) = \frac{\langle v,w_1 \rangle}{\langle w_1,w_1 \rangle } w_1 + \frac{\langle v,w_2 \rangle}{\langle w_2,w_2 \rangle } w_2 $

But, well, I'm not sure at all about that last step... Any help for that questions?? Thanks!!

Body:
Let be $W_F = (w_1,w_2)$ an orthogonal basis on $F$. For instance,
$ w_1 = (1,1,0,1) $
$ w_2 = (-2,7,5,3) $.

Then,

$ M ( \langle \ ,\ \rangle , W_F ) = \begin{pmatrix}
    5 & 0 \\
    0 & 55 \\
  \end{pmatrix} $

So we finally find,

$ p^r_F (v) = \frac{2}{5} w_1 + \frac{-4}{55} w_2 = (\frac{6}{11},\frac{10}{11},\frac{-4}{11},\frac{2}{11}) $


问:
Do overlapping left and right coset spaces have a name?
I have been studying they symmetries of the square, D₄. I would like to use the subgroups that are not normal to make a coset space, but since they are not normal, the right coset space is different than the left coset space. However, I did notice that the left and right coset space of pairs of subgroups in the same conjugacy classes overlap. Is there a name for this whole system?

This picture is one of the two conjugacy systems. There are four overlapping coset spaces in the form of the two rows and columns.

答:
Contrary to the comment, these are double cosets.  They are not all of the double cosets. They are simply the double cosets that are also cosets.  Usually double cosets, being unions of cosets, are larger than cosets, but when the a left coset $H_1g$ of $H_1$ coincides with a right coset $gH_2$ of $H_2$, then $H_1gH_2$ coincides with both of them.  This happens precisely when $H_1$ and $H_2$ are conjugate, and the number of such cosets is $[G:H_1]/|Orb| = [N_G(H_1):H_1] = [N_G(H_2):H_2]$, where $|Orb|$ is size of the orbit of $H_1$ under conjugation.

In this case $[N_G(H_1):H_1]=2$ since $N_G(H_1)$ is generated by $H_1$ and the center (the 180 degree rotation), so there are two coset-sized double cosets in $H_1\backslash G/H_2$ and similarly for $H_2\backslash G/H_!$.  In each case the other double coset has the other $4$ elements of $G$, for a total of $3$ double cosets in each case.

Specifically, if I am able to read your picture properly (the fonts in it are a bit blurry), $H_1\backslash G/H_2$ consists of the double cosets $H_1eH_2=\{e,rf,rr,rrrf\}$, $H_1rH_2=\{r,rrf\}$, and $H_1fH_2=\{f,rrr\}$.  Note that the double cosets with two elements are precisely the sets that are simultaneously left cosets of $H_2$ and right cosets of $H_1$.

I'll leave $H_2\backslash G/H_1$ as an exercise so you can validate your understanding.


问:
How to compute change of basis?
Let $v_1=(2,2,1,3)$ and $v_2=(0,1,1,1)$ be two vectors in $\mathbb{R}^4$, and let $V$ be the subspace which is the span of these two vectors.

According to what I am working on, there exists a basis for $V$ of the form $w_1=(1,a,0,b)$ an $w_2=(0,x,1,y)$ with integer $a$,$b$,$x$, and $y$.

I have tried figuring out the basis in a sort of ad hoc way. But there is a systematic way to compute such a basis?

答:
You need to build $w_1$ out of $v_1$ and $v_2$. To get the first component, $v_2$ is useless, so you have to take 1/2 of $v_1$. That gives you (1,1,1/2,3/2).  To make the third component zero, you have to subtract enough copies of $v_2$ to get rid of the 1/2.  Since $v_2$ has a 1 in that place, you subtract $\frac{1}{2}v_1$ to get $$(1,1,1/2,3/2) - (0, 1/2, 1/2, 1/2) = (1, 1/2, 0, 1) = w_1$$
You find the other vector similarly.

Title: How to compute change of basis?


问:
Are there mini-mandelbrots inside the julia set?
I've seen a julia set zoom but it is not nearly as interesting as a mandelbrot zoom. I also have not seen corresponding julia sets for zooms in the mandelbrot deeper than the original image. I'm wondering if it is possible to locate some form of a mini mandelbrot inside the julia set.</p>

答:
This answer has two parts: a partial negative and a partial positive.

Why Julia sets are typically much simpler than the Mandelbrot set

Julia sets of rational functions can be computed using an inverse iteration technique that shows them to be something close to self-similar.  This helps explain the extreme regularity displayed when zooming into most Julia sets.  For example, here we zoom in to the Julia set for $f(z)=z^2-1$ increasing the magnification by a factor of the Golden ratio with each step.

Unlike this, the Mandelbrot set gets more complicated with each step in magnification, rather than staying at the same basic level of complexity.

Julia sets that contain part of a Mandelbrot set

Nonetheless, these inverse images are not strict self-similarities (just not so far off for many functions) so more complicated behavior is possible.  In fact, the Julia sets of some rational functions do indeed contain some Mandelbrot type behavior.  Here's the Julia set of

$$f(z) = \frac{z^3-z}{1+4 z-z^2}$$

Note the appearance of something that looks like the period two bulb of the Mandelbrot set.  I believe that there is, in fact, a genuine correspondence here but that the true nature is not yet fully understood.  It appears to be related to the fact that the period two bulb is missing from the parameter space for a related family of functions:

$$f_a(z) = \frac{z^3-z}{1+az-z^2}$$

The study of this family is ongoing by Jane Hawkins and her students at UNC Chapel Hill and you can find more in her papers, particularly the ones that have the term "rational maps" in the title.


问:
A Theorem of Dimension of Fiber in Qing Liu (Theorem 4.3.12)
I'm reading Qing Liu. I feel confused about the proof of theorem 4.3.12 and corollary 3.14.

Theorem 4.3.12. Let $f$ be a morphism of locally Noetherian schemes. Let $x\in X$ and $y= f(x)$. Then $\dim (\mathcal{O}_{X_y,x})\geq \dim (\mathcal{O}_{X,x})-\dim (\mathcal{O}_{Y,y})$. If, moreover, $f$ is flat, then we have equality.

My question:

1.(red sentences) How do we know $t$ is not invertible in $B=\mathcal O_{X,x}$?

2.(green sentence) After base change with respect to $Y'$, $X'\rightarrow X$ is a closed immersion. But why $x$ is in $X' = X\times_YY'$?

Corollary 4.3.14. Let $f:X\rightarrow Y$ be a flat subjective morphism of algebraic varieties. We suppose that $Y$ is irreducible and that $X$ is equidimensional. Then for every $y\in Y$, the fiber $X_y$ is equidimensional, and we have $\dim(X_y) = \dim(X)-\dim(Y)$

My Question: Qing Liu defines algebraic variety as Noetherian scheme of finite type over $k$. That means $X_i$ may not be reduced. But Proposition 2.5.23(a) only works for domains. In this case how to get the highlight equality?

Theorem 2.5.15 Let $(A,\frak m)$ be a Noetherian local ring, $f\in \frak m$. Then we have $\dim(A/fA)\geq \dim(A)-1$. Moreover, equality holds if $f$ is not contained in any minimal prime ideal of $A$.

Theorem 2.5.23(a) Let $A$ be a finitely generated integral domain over $k$. Let $p$ be a prime ideal of $A$, we have $\operatorname{ht}(p) + \dim(A/p) = \dim(A)$.

答:
$A\to \mathcal{O}_{X,x}$ is a local homomorphism of local rings, so it sends $\mathfrak{m}_A$ to $\mathfrak{m}_x$. $t$ is in the first by assumption, so it's in the second, and the maximal ideal of a local ring is precisely the non-invertible elements.

Since $Y'\to Y$ hits $y$ by construction, there are maps from $\operatorname{Spec} k(x)$ to $Y'$ and $X$ which agree after composing with the map to $Y$. So there's a map $\operatorname{Spec} k(x)\to X'$ which after composing with $X'\to X$ agrees with the map $\operatorname{Spec} k(x)\to X$ which picks out $x$. As $X'\to X$ is a closed immerison, this means $x\in X'$.

Dimension is a purely topological invariant insensitive to the scheme structure. So you can replace your scheme by its reduction.


问:
Compute $\lim\limits_{n\to\infty}n^2\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right)$

A problem I saw recently is to compute $$\lim_{n\to\infty} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)$$

I thought it would be fun to give this a go but it completely stumped me. I am not aware of any of the standard techniques used to compute $\infty\times (e-e)$ - type limits, bar possibly L'Hôpital's Rule. But I doubt that would be a good idea here, at best it would be ugly. I would appreciate solutions using any methods, the more elegant the better of course. Though I welcome any relevant hints as well if full solutions are discouraged.

答:
Using only the Binomial Theorem and Geometric Series Formula:
$$
\begin{align}
&n^2\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right)\tag1\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}-\frac{n}{n+1}\right)\tag2\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\left(1-\frac1{(n+1)^2}\right)^{n+1}-\left(1-\frac1{n+1}\right)\right)\tag3\\
&=n^2\left(1+\frac1n\right)^{n+1}\left({\scriptsize\left(\color{#C00}{1-\frac1{n+1}}+\color{#090}{\frac{n}{2(n+1)^3}}+\color{#00F}{\sum_{k=3}^{n+1}\binom{n+1}{k}\frac{(-1)^k}{(n+1)^{2k}}}\right)-\left(\color{#C00}{1-\frac1{n+1}}\right)}\right)\tag4\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\color{#090}{\frac{n}{2(n+1)^3}}+\color{#00F}{O\!\left(\frac1{n^3}\right)}\right)\tag5\\
&=\left(1+\frac1n\right)^{n+1}\left(\frac{n^3}{2(n+1)^3}+O\!\left(\frac1n\right)\right)\tag6\\[6pt]
&\to\frac e2\tag7
\end{align}
$$
Explanation:

$(1)$: quantity of which to find the limit as $n\to\infty$

$(2)$: pull out a factor from both terms in the difference

$(3)$: rewrite both minuend and subtrahend as a difference from $1$

$(4)$: apply the binomial theorem

$(5)$: cancel the red terms and note that

$\phantom{\text{(5):}}$ $\sum\limits_{k=3}^{n+1}\binom{n+1}{k}\frac1{(n+1)^{2k}}\le\sum\limits_{k=3}^{n+1}\frac1{k!}\frac1{(n+1)^k}\le\sum\limits_{k=3}^\infty\frac1{6\cdot4^{k-3}}\frac1{(n+1)^3}=\frac2{9(n+1)^3}$

$(6)$: distribute $n^2$ over the difference

$(7)$: limit of a product is the product of the limits


问:
Induction and well ordering principle
Induction stays that is we have $P(0)$ is true and $\forall n \in \mathbb{N},\ P(n) \implies P(n+1)$ then $P(n)$ is true for every $n \in \mathbb{N}$.

I found a proof that Induction conclusion is true  if we suppose that the well ordering principle is true. And the proof goes like this : Suppose that the induction conclusion is false. Meaning that the set $\{ n,\ P(n) \text{ is false } \}$ is non empty. By the well ordering principle, it contains a least element $n_0$ . $n_0 >0$ because $P(0)$ is true. Then $P(n_0 - 1)$ must be true otherwise it contradicts the definition of $n_0$ but using the second point of the induction, we must also have $P(n_0)$ true, hence the contradiction.

I have three questions :

Is this proof correct (there is no hidden circularity in the reasoning)

Is there an implication the other way around ?

Which principle can be first implied from Peano axioms ? Induction, or Well ordering principle ?

Thank you for your answer

答:
It's correct, you've basically proved principle of weak induction assuming the well ordering principle.

You can prove the well ordering principle assuming the principle of weak induction.
As the induction step, you are given that all non-empty subsets of $\mathbb{N}$ of size $k - 1$ have a least element, and in this step you must show that all non-empty subsets of $\mathbb{N}$ of size $k$ have a least element. This is trivial to show.

None of those two are implied by Peano's axioms. These are axioms (although equivalent) on their own.

Title: Induction and well ordering principle


问:
Complex numbers in 3 dimensions
There was a question about this topic a while back, see here. It is known that there is no great way to design 3-dim complex numbers: any representation will have a number of drawbacks, such as impossibility to define an inverse, non-commuting products, or a solution that is trivial and of no value. I was first interested in this topic by looking at bicomplex numbers. Though being 4-dim and similar to quaternions, they do commute. But they don't always have an inverse.

The 3-dim complex numbers that I propose here commute, and many have an inverse. I don't know if they are associative and if they accept a matrix representation like complex numbers, quaternions, or bicomplex numbers. I am interested in an answer to these questions.

I define these 3-dim complex numbers as follows. Let $z_1 = a_1 + b_1 i + c_1 j$ and $z_2 = a_2 + b_2 i + c_2 j$. Then the product is defined as

$$z_1 * z_2 = (a_1a_2-b_1b_2+c_1c_2) + (b_1a_2 +a_1b_2) i + (c_1b_2+b_1c_2) j.$$

The first issue is that it is not compatible with the scalar product:

$$(a_1 + b_1 i + c_1 j) * a_2 = a_1 a_2 + b_1 a_2 i \neq (a_1 + b_1 i + c_1 j) \cdot a_2.$$

Note that $i^2=-1, ij = j$ and $j^2=1$. When the inverse exists, it is given by the following formula:

$$(a + bi + cj)^{-1} = \frac{a}{a^2+b^2+c^2} - \frac{bi}{a^2+b^2+c^2}+\frac{cj}{a^2+b^2+c^2}.$$

I used Mathematica to find the inverse, see here.

Update

Another interesting set of 3-dim complex numbers can be found here. The multiplication is associate and commutative, but again, not all of these numbers have an inverse. In fact, fewer than in my own set.

答:
There is another construction of 3D complex numbers that has associative and commutative multiplication: 3dcomplexnumbers.net. This construction, discovered in parallel by Shlomo Jacobi and Reinko Venema (who is behind the website 3dcomplexnumbers), is easier to construct, but hard to find the inverse.

The set constructions, sometimes denominated as $\mathbb R^3$, is $\mathbb J^-_3\circ \mathbb R$, where we have the imaginary unity $j$ where $j^3 = - 1$.

Every 3D number can be represented as $X = (a, b, c) = a + bj + cj^2$.

$$
\begin{align}
X Y &= (a, b, c)(d, e, f)\\
&= a(d, e, f) + bj(d, e, f) + cj^2(d, e, f)\\
&= a(d, e, f) + b(- f, d, e) + c(- e, - f, d)\\
&= (ad, ae, af) + (- bf, bd, be) + (- ce, - cf, cd)\\
&= (ad - bf - ce) + (ae + bd - cf)j + (af + be + cd)j^2
\end{align}
$$

There's also the matrix representation of these numbers (seeing $X$ as a column vector):

$$
M(X) = \left[\begin{matrix}X & jX & j^2 X\end{matrix}\right] = \left[\begin{matrix}
a & - c & - b\\
b & a & - c\\
c & b & a
\end{matrix}\right]
$$

Also, $XY$ (or $X^n Y^m \;\;\forall n, m\in\mathbb R$) can be calculated by the matrix multiplication (and if you diagonalize the matrix, you can even calc $\exp$ or $\ln$ of any $X$).

Note: the module is not necessarily conservative over multiplication. But the determinant always is.

There are two important numbers to remark:

The number $\alpha$, where $n\alpha = \sum_{i = 0}^{n - 1} (j^n)^2$ (or $3\alpha = j^0 + j^2 + j^4 = 1 + j^2 - j = 1 - j + j^2$).

$\alpha^n = \alpha\;\forall n\in\mathbb N^*$.

$\alpha$ is non-invertible.

$X\alpha = (X\cdot \alpha) \alpha =\;<X, \alpha> \alpha$ (multiplying by $\alpha$ projects the number in the $\alpha$ line).

The number $\tau := \ln{j}$ (or $j^2$, if $\det M(j) = - 1$)

$\tau$ in the complex numbers ($\mathbb C$) is $\pi i$.

$\tau$ and $\tau^2$ are non-invertible.

$\tau^3 = - k \tau$, in a similar way to $i^3 = - 1 i$.

In 3D, $\tau = {2\pi\over 3\sqrt 3} (j - j^2)$.

The set of non-invertible numbers are the plane $x - y + z = 0$ and the (orthogonal to this plane) line between $0$ and the number $\alpha$. In 4 dimensions constructed this way, the non-invertibles are just two lines.

There's even the exponencial circle $e^{\tau t}$, with center in $\alpha$ and passing by $1$, $- j$, $j^2$...

And finally, as I gave you mostly general rules, it's possible to develop many other nd complex structures. The first site I mention shows it for 3, 4, 5  and 7 dimensions, and the use of $\mathbb C$ as scalars in 3 dimensions.

Final note: in odd dimensions ($\mathbb J^-_n$ where $n = 2m + 1$), there's no solution to $j^n = - 1$.

Title: Complex numbers in 3 dimensions


问:
Continuity of the product of two $L^p$ functions
If two functions are in $L^2$, is their product continuous? I am not sure how to connect functions in $L^p$ spaces with continuity?

答:
No, functions in $L^2$ are not necessarily continuous, and their product is not necessarily continuous. The connection with continuous functions is that the space of continuous functions is dense in $L^2$, so that for any $f \in L^2$ and $\epsilon > 0$, there is a continuous function $h$ and a function $g \in L^2$ with small norm $||g||_2 \leq \epsilon$ such that $f = h + g$.

Thus, there is a continuous functions arbitrarily close to any $L^2$ function, close being in the $L^2$ norm.


问:
A loss and gain problem
This is a very simple but confusing puzzle.

A customer buys goods worth $200$ rupees from a shop. The shopkeeper selling these goods makes zero profit from this purchase.

The lady gives him a $1000$ rupee note.

The shopkeeper has no change, so he goes next door to another shopkeeper to get change for the $1000$ rupee note. He keeps $200$ for himself and returns $800$ to the customer.

Later, the second shopkeeper from next door comes back with the $1000$ rupee note with a stamp on it saying "counterfeit" and takes his money back.

How much loss does the first shopkeeper face?

Title: A loss and gain problem

Body:
i) Assuming that the second shop keeper $S_2$ is not doing something fake (saying the bill is fake):

$L=-0+800+x=800+x$

$S_1=+0 -0+1000-800-x+0-1000=-800-x$

$S_2=+0-1000-0+1000$

So shop keeper looses 1000, where $x=200$ and $0$ is the fake bill.

ii) If the the second shop keeper $S_2$ is doing something fake:

$L=-1000+800+x=-200+x=0$

$S_1=+1000 -1000+1000-800-x+0-1000=-800-x=-1000$

$S_2=+1000-1000-0+1000=1000$

In both cases $S_1$ looses $1000$.


答:
Not necessarily. By saying a function is in $L^2(\Omega)$ you are just asking that the function $f$ is square integrable in $\Omega$.

What you can say is that, if $ 1\le p < \infty$ then you can be sure you can find a continuous function that is close as much as you want ($\epsilon$-close) to a continuous function. If you want to check it more, just look for continuous functions with compact supports are dense in L^p

Title: Continuity of the product of two $L^p$ functions

Body: If two functions are in $L^2$, is their product continuous? I am not sure how to connect functions in $L^p$ spaces with continuity?


问:
Determine whether the given statement for a Cauchy problem of a partial differential equation is true or false?

Problem: Let $a$ be a fixed real constant. Consider the first order partial differential equation $~u_t+au_x=0, ~~x\in\mathbb R,~ t>0~$ with the initial data $~u(x,0)=u_0(x),~~x\in\mathbb R~$ where $~u_0~$ is a continuously differentiable function. Then determine whether the following statement is true or false:

If $~u_0~$ vanishes outside a compact set then for each fixed $~T>0~$ there exists a compact set $~K_T\subset \mathbb R~$ such that $~u(x,T)~$ vanishes for $~x\notin K_T~.$

My work: By Lagrange's subsidiary equation $$\dfrac{dx}{a}=\dfrac{dt}{1}=\dfrac{du}{0}$$
Which implies, $~~u=c_1~$ and $~x-at=c_2~$ where $~c_1~$ and $~c_2~$ are arbitrary independent constants.
 Therefore the solution of the given PDE is $~u(x,t)=\phi(x-at),~$ where $~\phi(x)~$ is an arbitrary function.

The given initial data, $~u(x,0)=u_0(x)\implies u_0(x)=\phi(x)~.$

Hence the solution of the given PDE is $~u(x,t)= u_0(x-at)~.$

From here I am totally in dark. I have no idea how to proceed further. Please help to solve the problem.

Note: If $~u_0(x)~$ is a bounded function, then $~u_0(x-at)~$ is also bounded function and hence $~u(x,t)~$ is bounded.

答:
You have already solved the problem. Lets say $[L,U]\in\mathbb{R}$ is the compact set outside of which $u_0$ vanishes. Since you have the given solution $u(x,t)=u_0(x-at)$ you just have to shift the compact set accordingly $K_T=[L+aT,U+aT]$.

Now lets try it out. Lets assume $x<L+aT\Rightarrow x-aT<L\Rightarrow u(x-aT,T)=u_0(x-aT)=0$ Thus, for $x<L+aT$, the function $u(x,T)$ vanishes. (The same holds true for the upper limit)

The intuition behind it is that these (transport) equations have a finite (in this case even constant) speed of propagation. Meaning information can only move with a certain velocity. This is not necessarily true for all PDEs. For example the heat equation features an infinite speed of propagation meaning that initial data on a compact set will spread to the whole domain as soon as the clock starts.


答: The only minor  gap if at all is that you assume that every natural number $\ne0$ has a predecessor. In fact this property is what distinguishes the naturals from other well-orders infinite sets.

Title: Induction and well ordering principle

Body:
Induction stays that is we have $P(0)$ is true and $\forall n \in \mathbb{N},\ P(n) \implies P(n+1)$ then $P(n)$ is true for every $n \in \mathbb{N}$.

I found a proof that Induction conclusion is true  if we suppose that the well ordering principle is true. And the proof goes like this : Suppose that the induction conclusion is false. Meaning that the set $\{ n,\ P(n) \text{ is false } \}$ is non empty. By the well ordering principle, it contains a least element $n_0$ . $n_0 >0$ because $P(0)$ is true. Then $P(n_0 - 1)$ must be true otherwise it contradicts the definition of $n_0$ but using the second point of the induction, we must also have $P(n_0)$ true, hence the contradiction.

I have three questions :

Is this proof correct (there is no hidden circularity in the reasoning)

Is there an implication the other way around ?

Which principle can be first implied from Peano axioms ? Induction, or Well ordering principle ?

Thank you for your answer


答:
No. There exist discontinuous square-integrable functions, e.g. the indicator function of a set of finite measure, e.g. $\mathbb{1}_{[0, 1]}$. Since $\mathbb{1}_{[0, 1]} \cdot \mathbb{1}_{[0, 1]} = \mathbb{1}_{[0, 1]}$ we see that a product of two square-integrable functions may be discontinuous.

There is a subtlety here. The space $L^2(S, \mu)$ on $S$ relative to measure $\mu$ does not strictly speaking consist of functions. Instead, it is defined as the quotient

$$
L^2(S, \mu) = \mathcal{L}^2(S, \mu) / \mathcal{N}
$$

where $\mathcal{L}^2(S, \mu)$ is the set of functions on $S$ square-integrable relative to measure $\mu$ and $\mathcal{N}$ is the kernel of the $L^2$ norm. In particular, we identify two square-integrable functions $f$ and $g$ whenever $f(x) = g(x)$ almost everywhere. Thus, the elements of $L^2$ are strictly speaking equivalence classes of functions. A given equivalence class may simultaneously have continuous and discontinuous representatives. For example, the equivalence class of the zero function contains continuous $f(x)=0$ and discontinuous $g(x)=\mathbb{1}_{\{0\}}$.


问:
(Simple) Examples on Non-Commutative Rings
Looks like it is easier to find example of commutative rings rather than
non-commutative rings. Prabably the easiest examples of the former are
$\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various
literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non-commutative kind) are not that easy.
One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.

答:
Fairly concrete examples of noncommutative rings arise in  calculus when studying differential equations using operator algebra. For example, consider the ring of linear operators generated by the derivative $\, D = d/dx\,$ and the operation of multiplication by $\,x,\,$ i.e. $\, f\mapsto xf,\,$ where $\,(L+M)f = Lf + Mf\, $ and $\,(LM)(f) = L(Mf)),\,$ i.e. multiplication is composition of operators. This  bivariate ring of differential  polynomials $\,\Bbb R\langle x,D\rangle$ is noncommutative since
$$ Dx = xD + 1\ \ \ {\rm i.e.}\ \ (Dx)(f) = D(xf) = x Df +  f = (xD + 1) f$$

There is also a discrete analog for difference equations (recurrences), consisting of polynomials in the shift operator $\,Sf(n) = f(n\!+\!1),\,$ and multiplication by $\,n,\,$ i.e. $\, f\mapsto nf.\,$ Then

$$ S n = (n\!+\!1) S\ \ \ {\rm i.e.}\ \ \ (Sn)(f(n)) = S(nf(n)) = (n\!+\!1)f(n\!+\!1) = ((n\!+\!1)S)(f(n))$$

Both of these operator algebras prove useful because we can sometimes employ noncomutative analogs of familiar polynomial arithmetic, e.g. factoring operator polynomials in order to solve differential and difference equations, e.g. solving the Fibonacci recurrence by factoring it as $\,(S-\phi)(S-\bar \phi) f_n = 0,\,$ so $\,f_n = c\phi^n + d\bar \phi^n,\,$ or the noncommutative example here
$$ (n\!-\!1)\ S^2\! - (3n\!-\!2)\ S + 2\,n\, =\, ((n\!-\!1)\, S - n)\ (S - 2)$$
Special cases go by various names, e.g  the method of characteristic polynomials, or the classic Heaviside operator calculus, etc.


问:
If $\langle a\rangle$ is a cyclic group of order $90$, write $a$ as the product of $p$-elements, for various primes $p$.
We say that $a$ is a $p$-element if the order of $a$ is $p^n$ for some integer $n\ge0$.
The primes dividing $p$ are $2,3$ and $5$.

Thus the $2$ - elements are $\{e,a^{45}\}$.

The $3$ - elements are $\{e,a^{10},a^{20},a^{30},a^{40},a^{50},a^{60},a^{70},a^{80},a^{90}\}$.

The $5$ - elements are $\{e,a^{18},a^{36},a^{54},a^{72}\}$

How can I write $a$ as product of these?

答: Since $a^{36},a^{10} $ and $a^{45} $ have orders $5,9$ and  $2$, respectively,  you can take their product.


问:
Formal logic question - why is this invalid?
I typed this into a logic calculator, and it comes out as invalid:

\begin{align*}
\forall x, & E(x) \implies I(x) \vee A(x) \\
\forall x, & A(x) \implies \neg E(x) \\
\forall x, & E(x) \implies \neg A(x)\\
\exists x, & E(x) \implies I(x)\\
\exists x, & E(x)\\
E(x)⟹I(x)
\end{align*}

Can someone help me explain why it's invalid?

https://www.umsu.de/trees/#((((((%E2%88%80y((Ex%E2%86%92(Ix%E2%88%A8Ax))))%E2%88%A7(%E2%88%80y((Ax%E2%86%92(%C2%ACEx)))))%E2%88%A7(%E2%88%80y((Ex%E2%86%92(%C2%ACAx)))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%AC(Ex%E2%86%92Ix))))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%ACEx)))))%E2%88%A7(Ex%E2%86%92Ix))

答:
Long comment

As you have written it in the logic calculator, it is a single formula (a conjunction) and not an argument.

Thus, the calculator's answer is correct: the formula is invalid.

As answered by the tool, consider a domain $D = \{ 0 \}$ where we have that $E(0), I(0)$ and $A(0)$ hold. The part $\forall x (Ex \to \lnot Ax)$ is not satisfied.


问:
Infimium between two sets with a bijection
People, I need some help about a thing: I am studying analysis and, between my thoughts about supremum and infimum, I started to ask myself the following:

Let A, B $\neq \emptyset$ and $f:A\rightarrow B$ be a bijection. Assume that there A and B are bounded below. Show that $f(\inf(A))=\inf(B)$.

Is that sentence true? Why? I did not found a way to prove it, since I can not assume that $f$ takes the infimium from one set to the another, once $\inf(A) \notin A$, in general.

答: Here's a contradiction: $A = B = [0, 1]$, and $f(x) = 1 - x$. Then $f(\inf A) = f(0) = 1 \neq \inf(B)$.


问:
The function $\arcsin(\sin(x))$
I have this question but I'm really not sure about my solution. Does this make sense?

Given the function $g(x) = \arcsin(\sin x))$, find its domain and where it is differentiable.

So I assume it's enough to say that $\sin$'s co-domain is $[-1,1]$ and so $\arcsin$'s domain, being its inverse function, is $[-1,1]$.

Now $\sin$ is differentiable on $\mathbb R$ and $\arcsin$ is differentiable on $(-1,1)$ (is it?) and so using the chain rule we can say that $g$ is differentiable at all $a \in \mathbb R$ for which $\sin(a) \neq 1$ or $\sin(a) \neq -1$. Which would be $\mathbb R $ \ {$-\pi/2, \pi/2$}.

Thanks!!

答:
Domain is $\mathbb{R}$. It's differentiable on $\mathbb{R} \setminus \{\frac{\pi}{2} + n\pi | n \in \mathbb{Z}\}$. Check this by observing that
$$x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \Rightarrow \arcsin (\sin x) = x  $$
Try to write the formula when $x$ is not in that interval.

Title: The function $\arcsin(\sin(x))$


问:
Converting into alternate e-Form
I was checking my results of an exercise online. The Maximum of my function was (1): $$\frac{4^{\frac{-1}{\log(2)}}}{\log^2(2)}$$

Now I wolframalpha tells me the alternate form is (2): $$\frac{1}{e^2\log^2(2)}$$

I feel like this form could be useful for future problems sometimes, but I absolutely have no clue how to get there. Could someone help me converting the maximum (1) into the alternate form (2)?

Title: Converting into alternate e-Form

Body:
It's a logarithmic rule-- a transformation rule, notice that $\log^2{(2)}$ appears common, so focus on $4^{\frac{-1}{\log{2}}}$
$$4^{\frac{-1}{\log{2}}} = 4^{-\frac{1}{\log{2}}}$$
$$ \frac{1}{\log_{a}{b}} = \log_{b}{a}$$
$$4^{-\frac{1}{\log{2}}} = 4^{-\log_{2}{e}}$$
$$4^{-\log_{2}{e}} = 2^{-2\log_{2}{e}}$$
$$2^{(\log_{2}{e})(-2)}$$
There's also another transformation rule
$$ a^{log_{a}{b}} = b$$
$$2^{(\log_{2}{e})(-2)} = (2^{\log_{2}{e}})^{-2}$$
$$e^{-2}$$


答:
A picture is worth a thousand words:

The slopes are $\pm 1$.  Clearly the domain is the whole real line and it is differentiable except at the corners, which come at $\frac \pi2+k\pi$ for $k$ an integer.

Title: The function $\arcsin(\sin(x))$

Body:
I have this question but I'm really not sure about my solution. Does this make sense?

Given the function $g(x) = \arcsin(\sin x))$, find its domain and where it is differentiable.

So I assume it's enough to say that $\sin$'s co-domain is $[-1,1]$ and so $\arcsin$'s domain, being its inverse function, is $[-1,1]$.

Now $\sin$ is differentiable on $\mathbb R$ and $\arcsin$ is differentiable on $(-1,1)$ (is it?) and so using the chain rule we can say that $g$ is differentiable at all $a \in \mathbb R$ for which $\sin(a) \neq 1$ or $\sin(a) \neq -1$. Which would be $\mathbb R $ \ {$-\pi/2, \pi/2$}.

Thanks!!


问:
Self-study resources for basic probability?
I am taking a Computer Science class soon that requires a solid knowledge of the basics of probability.  I've only had minimal exposure to probability in classes I've taken in the past, so I need to get up to speed quickly.  Can anyone recommend some good self-study resources (e.g. books, online classes, web sites) that I could use to teach myself the fundamentals?</p>

答: An excellent introductory probability book for self-study  is Henk Tijms, Understanding Probability (Cambridge University Press, 2nd ed., 2007). It distinguishes itself from other introductory probability texts by its emphasis on how probability works and how to use it.  </p>

ContentLicense: CC BY-SA 2.5


问:
Show that the relation R on set A is transitive if and only if $R^{n} \subseteq R$ for all positive integers n
I tried to prove this using induction but i get stuck at a point and the proof fails. I dont know what am I doing wrong! Can someone help me prove this? I have tried all I possibly could</p>

答:
Well the definition of $R^k$ is the set of all $(a,b)$ there that there exist $c_1,...,c_{k+1}$ with $a=c_1, b=c_{k+1}$ and $(c_i, c_{i+1}) \in R$ for all $i: 1\le i < k+1$..

If $R$ transitive and $a,b,c_i$ are as above, if we have $(a,c_1) \in R$ and if we have $(a,c_i)\in R; i\le k$ the we have $(c_i, c_{i+1})\in R$ and as $R$ is transitive we have $(a,c_{i+1})\in R$. So be induction we have $(a,b) \in R$ so $R^n \subset R$.

If $R^n \subset R$ for all $n$ then $R^2 \subset R$.  And if $(a,b) \in R$ and $(b,c) \in R$ then $(a,c) \in R^2\subset R$ so $(a,c) \in R$.  So $R$ is transitive.


问:
Connection between two analytic notions of "spectrum"
In Spectral Theory and Analytic Geometry over Non-Archimedean Fields Berkovich defines the spectrum of a commutative, unital Banach ring $A$ to be the set of all bounded multiplicative seminorms on $A$ with the weak topology with respect to the family of functions
$$\phi_f: || \mapsto |f|$$
In another class, I learned the definition of spectrum $\sigma(T)$ of an element of a Banach algebra to be the set of all the $\lambda$ such that $T - \lambda I$ is non-invertible. Many times I try to look something up related to Berkovich's text (such as the Gel'fand transform, characters),  and I find a concept by the same or similar name that seems similar but is definitionally different. So are the two definitions connected somehow? This section of Wikipeda seems promising, as well as the page on the Gel'fand representation, but I could use some help connecting the dots.

答:
Make note of the following theorem:

Theorem (Mazur)

If $A$ is a normed algebra over $\Bbb R$ together with a multiplicative norm (ie $\|a\cdot b\|=\|a\|\cdot\|b\|$ for all $a,b\in A$) then $A$ is isometrically isomorphic (as an $\Bbb R$-algebra) to one of the following:

$\Bbb R$

$\Bbb C$

$\mathcal H$ (the quaternions)

Now if $A$ is Banach-algebra and $\|\cdot\|_s$ is a multiplicative semi-norm, let $N=\{a\in A\mid \|a\|_s=0\}$, then $N$ is a (closed) two-sided ideal and $\|\cdot\|_s$ descends to a multiplicative norm on $A/N$. So by the theorem you've got that $A/N$ is either $\Bbb R, \Bbb C$ or $\mathcal H$. Since we are asking $A$ to be commutative $A/N$ must also be commutative and $\mathcal H$ is out of the picture. Additionally I am assuming that when you say "Banach algebra" you are talking about a complex Banach algebra, since thats the common convention, so $\Bbb R$ is also out.

Hence every multiplicative semi-norm $\|\cdot\|_s$ is of the form:

$$A\to \Bbb C \to \Bbb R_{≥0}, \qquad a\mapsto \psi(a) \mapsto |\psi(a)|$$
for some unital algebra morphism $\psi: A\to \Bbb C$, these morphisms are also called "characters".

This is the connection of the first definition to the Gel'fand transform story, this story is very well expositioned and you can find it explained in almost every introductory textbook on operator algebras. For convenience I'll summarise the connection with the spectrum $\sigma(a)$ of elements of $A$.

Denote with $\Delta$ the space of characters and give it the weak* topology, ie the topology induced by the semi-norms $|\psi|_a= |\psi(a)|$, since $\Delta$ is weak* closed in the unit ball of the dual $A'$ you get that $\Delta$ is compact. Further there is a continuous algebra morphism
$$A\to C(\Delta), \qquad a\mapsto [\psi\mapsto \psi(a)]$$
called the Gel'fand transform. You have that:

Each element $\psi\in\Delta$ is uniquely determined by the semi-norm $a\mapsto |\psi(a)|$ and the inclusion into the space of multiplicative semi-norms is a homoemorphism if you give the semi-norms the topology you have indicated, ie $\Delta$ is the spectrum in the sense of Berkovich.

$a(\Delta) \subseteq \sigma(a)$, ie the image of the spectrum under $a$ is contained in the spectrum of $a$.

The map $\Delta\mapsto \mathrm{Spec}_m(A)$, $\psi\mapsto \ker(\psi)$ is well defined and bijective, ie $\Delta$ is equal to the maximal spectrum (in the sense of algebraic geometry) - but this map won't generally be a homeomorphism since in the Banach algebra setting you have a very nice analytic topology on $\Delta$ thats not available in the general algebraic setting.

The map $A\to C(\Delta)$ is an isomorphism iff $A$ is a $C^*$-algebra.

However there are commutative Banach-algebras for which $\Delta$ is just a point.

The proofs look roughly like this:

Note that for any $\psi\in\Delta$ the sub-space $\ker(\psi)$ is an ideal and it has co-dimension $1$, the quotient $A/\ker(\psi)$ is then isomorphic to $\Bbb C$ as a one-dimensional unital $\Bbb C$ algebra, $\psi$ necessarily factors over the quotient to $A/\ker(\psi)$, but there is only one non-zero morphism from this algebra to $\Bbb C$, so $\psi$ is uniquely determined by $\ker(\psi)$. Hence for each semi-norm $\|\cdot\|_s$ there is at most one character $\psi$ with $\|a\|_s=|\psi(a)|$ for all $a$.

Note that $a-\psi(a)\Bbb1$ cannot be invertible because under $\psi$ its image is $0$, if it were invertible there would be a $b$ with $1=\psi(\Bbb1)=\psi(b(a-\psi(a)\Bbb1))=\psi(b)(\psi(a)-\psi(a))=0$. As such $\psi(a)\in\sigma(a)$ for all $a$.

$\ker(\psi)$ is a co-dimension $1$ ideal and hence maximal. Further if $\mathfrak m$ is maximal then every non-zero element of $A/\mathfrak m$ must be invertible (else if $a+\mathfrak m$ were not invertible the ideal generated by $a$ and $\mathfrak m$ would be proper) - by a general theorem you get that $A/\mathfrak m$ is isomorphic to $\Bbb C$ and then the quotient map $A\to A/\mathfrak m$ is a character with kernel $\mathfrak m$. Since characters are uniquely determined by their kernels (see point 1) this correspondence is then bijective.

Consult a book on C* algebras for this one.

Look at a Banach space, choose some element to be the $\Bbb1$ and let any multiplication $a\cdot b=0$ when neither $a$ or $b$ are proportional to $\Bbb 1$.


问:
Finding the closure of a set using only topological definitions
Consider the set $X = \{(-1)^n - \frac{1}{n}: n \in \mathbb{N}, n > 0\}$.

I'm trying to find the closure of the set.

I know intuitively using real analysis that the closure includes $X$ and the limit points 1 and -1. So the closure must include these as well.

But how do I show that the elements of $X$ and 1 and -1 are the only elements of the closure using topological notions of closure. i.e. the intersection of all closed sets containing $X$. Couldn't there possibly be more elements in the closure of $X$?

I also have the result that $x \in \overline{X}$ iff every open set $A$ containing $x$ is such that $A \cap X \neq \emptyset$.

答: Take $y\notin X\cup\{1,-1\}$. Since $y$ is not a accumulation point of the sequence, there is some $r>0$ such that $(y-r,y+r)$ contains only finitely many elements of the sequence. If $F$ is that finite set, let $\delta=\min\{|y-t|\mid t\in F\}$. Since $y$ does not belong to the sequence, then it does not belong to $F$, and therefore $\delta>0$. So, $\left(y-\min\{r,\delta\},y+\min\{r,\delta\}\right)$ has no element of $X\cup\{1,-1\}$. This proves that $y\notin\overline{X\cup\{1,-1\}}$.


Title: Converting into alternate e-Form

Body:
Let $x=4^{-1/\log2}$. Then
$$
\log x=-\frac{1}{\log2}\log4=-\frac{1}{\log2}\cdot 2\log2=-2
$$
which means $x=e^{-2}$.


问:
Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$
Problem:

If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then

(a) $S$ is an odd integer

(b) $S + 1$ is not divisible by $9$

(c) The integer next above $(5+2\sqrt{6})^{2n+1}$ is divisible by $10$

(d) $S-1 = \frac{t}{t-1}$

Please guide me how to proceed with this question...

答:
Everything flows from the hint by Ross Millikan.

Let $(5+2\sqrt{3})^{2n+1}=a_n+b_n\sqrt{3}$. (we can imagine computing $a_n$ and $b_n$ by expanding using the Binomial Theorem). Note that $a_n$ and $b_n$ are integers. Now imagine expanding $(5-2\sqrt{3})^{2n+1}$. We get $a_n-b_n\sqrt{3}$. It follows that
$$(5+2\sqrt{3})^{2n+1}+(5-2\sqrt{3})^{2n+1}=2a_n.\tag{$1$}$$
Define $S_n$ and $t_n$ as in the question, except we have added subscripts to make clear the dependence on $n$. By the hint of Ross Millikan, $S_n=2a_n-1$.  We can conclude immediately that $S$ is odd.

Note that
$$\begin{align}a_{n+1}+b_{n+1}\sqrt{3}&=(5+2\sqrt{3})^{2n+3}=(a_n+b_n\sqrt{3})(5+2\sqrt{3})^2\\  &= 37a_n+60b_n+(20a_n+37b_n)\sqrt{3}\end{align}.\tag{$2$}$$
Thus
$$a_{n+1}=37a_n+60b_n\qquad\text{and}\qquad b_{n+1}=20a_n +37b_n.\tag{$3$}$$
Note that $a_0=5$ and $b_0=2$. The integer just above $S_n$ is $2a_n$. This is divisible by $10$ if $n=0$. It follows by induction, using the first recurrence in $(3)$, that $10$ divides $2a_n$, that is, $S_n+1$, for all $n$.

About non-divisibility of $S_n+1=2a_n$ by $9$, we prove that $a_n$ is not divisible by $3$. Certainly $a_0=5$, so $a_0$ is not a multiple of $3$. But then by the recurrence $a_{n+1}=37a_n+60b_n$, we see that $a_1$ is not divisible by $3$, but then again by the recurrence, $a_2$ is not divisible by $3$, and so on.

The computation in (d) is left to you.


问:
Solutions of $x^2 = x + 1$ in $\mathbb Z/n\mathbb Z$
In a stage preceding the national stage of the math olympiad in Romania ( which was held online because of the pandemic ) the following question was asked:

If $A$ is the set of natural numbers $n$ for which $x^2=x+1$ has only a solution in $\mathbb Z/n\mathbb Z$ then:

A) $A$ has no elements

B) $A$ has only one element

C) The elements of $A$ are even

D) $A$ has infinitely many elements

E) ( I can't remember it ).

The oficial answer was that $A$ has only one element but when I wrote a bit of code I found $5$, $46985$, $47209$ and more to be in $A$.

Can you help me characterise the numbers in $A$ a little bit more?

Edit:

答:
If $m$ is a solution, then so is $1-m.$ (Why?)

So if $m$ is unique, then $1-m\equiv m\pmod n.$ So $n$ must be odd, and thus $$m\equiv \frac{n+1}2\pmod n.$$

So $$\frac{(n+1)^2}4\equiv \frac{n+1}2+1\pmod n$$

Multiplying by $4,$ we get:

$$(n+1)^2\equiv 2(n+1)+4\pmod n$$

Which is equivalent to:

$$0\equiv n^2\equiv 5\pmod n$$


问:
Counit of adjunction between $\mathbf{Vect}_k$ and $\mathbf{Set}$
I have some clarificational questions to this answer (definitions of $F,U,\varepsilon$ should be clear from this):

As to $\varepsilon_V$, you know that the result is an element of $V$ and $\varepsilon_V$, being an arrow in $\mathbf{Vect}_k$ is a vector space homomorphism, so, by definition, it takes the addition (and other operations) of $FU(V)$, which is a "formal" addition, and maps it to the addition of $V$, which is a "real" addition. Of course, $V$ could itself be a vector space of formal linear combinations, but the point is that $\varepsilon_V$ has no choice but to interpret the "formal" operations in $FU(V)$ as the operation so $V$ whatever they are. Formal $k$-linear combinations are not, in general, elements of $V$.

From the same answer, formal linear combinations are of the form $r_{s_1}\cdot v_{s_1}+\cdots+r_{s_n}\cdot v_{s_n}$ where $s_1,\dots,s_n\in S$ and are distinct, and $r_{s_1},\dots,r_{s_n}\in k$ (in our case, $S=U(V)$). Consider the simplest case when the formal linear combination is just $v_{s_1}$. There's no any formal addition or multiplication. Is this formal combination supposed to be mapped to $s_1$ under $\varepsilon_V$? (I believe the answer I referred to above doesn't discuss this.) Further, is $r_{s_1}\cdot v_{s_1}$ supposed to be mapped to $r_{s_1}\cdot_V {s_1}$, where $\cdot $ is a formal multiplication and $\cdot_V$ is mupltiplication $V\times k\to V$? And then more generally, I suppose $$r_{s_1}\cdot v_{s_1}+\cdots+r_{s_n}\cdot v_{s_n}\mapsto r_{s_1}\cdot_V s_1 +_V\dots+_V r_{s_n}\cdot_V s_n$$ where $+_V$ is addition on $V$ (which is, I believe, referred to as "real" addition in the quote above). Is that right?

And lastly, how do we know that $\varepsilon: FU(V)\to V$ does exactly what the answer I referred to described (and what I tried to expand on)? Is it because this is the only sensible way to define  $\varepsilon_V$ and because the counit of adjunction is unique? Or does this follow from some facts in the theory of "term algebras" mentioned in the answer?

答:
To your first question: Yes, you are right. In the answer you are referring to, the notation $v_s$ instead of $s$ was used to better distinguish the formal sum $\sum r_s v_s$ from the sum taken in $V$, $\sum r_s s$. Under the counit $\epsilon$, $v_s$ is then mapped to $s$, and the formal sum $r_{s_1}v_{s_1}+\dots +r_{s_n}v_{s_n}$ is mapped to the evaluation of this sum in $V$, which is $r_{s_1}s_1 + \dots r_{s_n}s_n$.

To the second part: Look at the more general setting of an adjunction

Let $F: C\rightarrow D$ and $G: D\rightarrow C$ be functors together with a natural isomorphism with components $D(Fc,d) \simeq C(c,Gd)$. Then,fixing $c$, one can see that the object $Fc$ represents the functor $d\mapsto C(c,Gd)$. By the Yoneda lemma, one can see that the natural isomorphism $D(Fc,-) \Rightarrow C(c,G-)$ is determined by an element of $C(c,GFc)$, which we call $\eta_c$ and which we call the component of the unit of the adjunction on $c$. In the same manner, the observation that $Gd$ represents the contravariant functor $c\mapsto D(Fc,d)$ gives an element $\epsilon_d$ of $D(FGd,d)$, which we call component of the counit at $d$.

This discussion shows that the counit is obtained by the construction of the Yoneda lemma: If we call the natural isomorphism of the adjunction $\alpha$, then $\epsilon_d$ is given by $\alpha^{-1}_{Gd,d} \text{Id}_{Gd}$, which is often called the transpose of $\text{Id}_{Gd}$.

So to find $\epsilon_V$ in this particular case, we have to find the transpose of $\text{Id}_{UV}$ under the free-forgetful adjunction of $\text{Vect}_k$. So we first need to know: if we are given a set $S$ and a vector space $V$, which morphism in $\text{Vect}_k$ does a morphism $S\rightarrow UV$ correspond to under this adjunction? The answer is the following: A map $f: S\rightarrow UV$ is send to the vector space homomorphism $\phi: FS \rightarrow V$ that sends $s\in S$ to the vector $f(s)$, but this time seen as element of $V$ instead of $UV$, and a formal sum $\sum r_{s_i} v_{s_i}$ in $FS$ to the sum in $V$ that is $\sum r_{s_i} f(s_i)$.

So, what linear map does the identity morphism in Set $\text{Id}: UV\rightarrow UV$ correspond to? Well, a formal sum in $UV$ is exactly sent to its result in $V$ under the vector space addition and scalar multiplication!

Therefore, the counit of the free-forgetful adjunction works as described in the question you linked.


问:
Test if a 6-sided die is fair. How many degrees of freedom in the $\chi^2$ distribution?
For testing if a 6-sided die is fair I learnt that we have to use a so-called $\chi^2$ test. If we pick confidence 5% (i.e. false rejection probability 5%), at the end it boils down to doing a lookup in the $\chi^2$ distribution tables to find the threshold value.

So my question is: which $\chi^2$ distribution exactly i.e. with how many degrees of freedom?

I read somewhere (here actually: https://rpg.stackexchange.com/a/70803)

that for a k-sided die I need to use the table of the $\chi^2$ distribution with $(k-1)$ degrees of freedom. Is this correct and where does that stem from? Sounds   somehow counter-intuitive to me.

I don't need a detailed deep explanation but maybe some basic explanation,

and maybe some references to learn more about this asymmetry.

So for my 6-sided die, I will have to do a lookup in the $\chi^2$ distribution tables for $\chi^2$ with $5$ degrees of freedom, correct?

答:
The reason you only have $k-1$ degrees of freedom is that if you roll a $k$-sided die $100$ times, you get $k$ different counts (the number of times you roll $1$, the number of times you roll $2$, and so on) but not all $k$ of them can be chosen separately. Once you have chosen $k-1$, I can tell you the $k^{\text{th}}$: it's just whatever is left over from $100$.

(Equivalently, if we divide through by $100$, or whatever the total number of samples was, you get $k$ proportions: but one of them can be deduced from the others, because all $k$ proportions together must add up to $1$.)

In general, the number of degrees of freedom in a $\chi^2$ test is the number of proportions that need to be specified before the others can be deduced.

So yes: if the die has $6$ sides, then the number of degrees of freedom is $5$. If you tell me that $10\%$ of your rolls were $1$s, $10\%$ were $2$s, $10\%$ were $3$s, $20\%$ were $4$s, and $20\%$ were $5$s, I can deduce that $30\%$ were $6$s: there is no freedom in specifying the last percentage.