问:
Show that there are $36$ $5$-Sylow subgroups

Given the symmetric group $S_6$, we consider a $5$-Sylow subgroup. How can one show that this subgroup is isomorphic to $\mathbb{Z}/ 5 \mathbb{Z}$? I have to show that there exist $36$ such subgroups. How can I do that?

What I know:

If $G$ is a finite group, $|G|=p^n \cdot m $, where $p$ is prime, $n\ge 1$ and $(p,m)=1$, then a Sylow $p$-group in $G$ is a subgroup $H\subset G$, where $|H|=p^n$

The number of distinct Sylow-$p$-groups in $G$ is congruent to $1\mod p$

The number of distinct Sylow-$p$-groups in $G$ is a divisor of $m$.

答:
By Lagrange's theorem, the order of any subgroup divides the order of the group. Since $|S_6|=6!$ contains only one factor of $5$, and a Sylow 5-group must have order a power of $5$, the $5$-group must have order $5$. Since any group of prime order is cyclic, we see the $5$-group is $\mathbb Z/ 5\mathbb Z$.

So the subgroup must be generated by one element, a $5$-cycle. And since all $5$-cycles are conjugate, and all Sylow $5$-groups are conjugate, the problem reduces to counting the number of $5$-cycles in $S_6$, and then dividing by $4$ to account for the fact that there are $4$ in each subgroup.


问:
derivate is based on addition, is there a muliplication analogon?
like

$$
\operatorname{f^o}(x) = \lim_{h\to 1} \frac{f(x*h)}{f(x)}
$$

$$
\operatorname{f}(x)=e^x
$$

$$
\operatorname{f^o}(x) = \lim_{h \to 1} e^{x*h}/e^{x} = \lim_{h\to 1} e^{x*h-x}=e^0=1
$$

does it have a name and some theories behind it?
if not why is this alt-derivative not used?

it would be nice if someone can point me in the right direction since I'm unsure what to search for without a name.

the application is if this could be used on a special machine learning approach instead of a gradient descent based on normal derives.

I'm only interested in purely positive functions unequal zero to apply this on

edit the correct formular:
from @B.Martin book reference in the comments:

https://books.google.com.au/books?id=_Eg7p3UXJjUC&printsec=frontcover&dq=Bigeometric+Calculus:+A+System+With+a+Scale-free+Derivative%E2%80%8E&hl=en&sa=X&ved=2ahUKEwiQ4eek-_buAhVIwjgGHc_wCWYQ6AEwAHoECAMQAw#v=onepage&q=Bigeometric%20Calculus%3A%20A%20System%20With%20a%20Scale-free%20Derivative%E2%80%8E&f=false

$$
lim_{x\to a} \frac{\operatorname{f}(x)}{\operatorname{f}(a)}^{\frac{1}{ln(x)-ln(a)}}
$$

答: wrong approach, thanks for the comments was stuck on my idea there :)</p>

ContentLicense: CC BY-SA 4.0


问:
Give a counterexample showing that there exist two different matrices $X$ and $Y$ such that $X\cdot A=I_n=Y\cdot A$
If $X$ and $Y$ two matrix of order $n\times m$ such that $X\cdot A=I_n=Y\cdot A$ for some matrix $A$ of order $m\times n$ then it must be $X=Y$? I am sure not but unfortunately I did not find any counterexample: in particular if $X\cdot A=I_n$ then the linear function $f:\Bbb K^n\rightarrow\Bbb K^m$ generated by $A$ is injective and thus $A$ has rank $n$ so that if $\phi:\Bbb K^n\rightarrow\Bbb K^m$ and $\psi:\Bbb K^n\rightarrow\Bbb K^m$ are the linear functions generated by $X$ and $Y$ then
$$
\phi(y)=\psi(y)
$$
for any $y\in f[\Bbb K^n]$ but if $f[\Bbb K^n]\neq\Bbb K^m$ then could be exist $y\in\Bbb K^m$ such that
$$
\phi(y)\neq\psi(y)
$$
but I am not able to make a counterexample. Could be that $X$ and $Y$ are equal? So could someone help me, please?

答:
So let be $A:=\begin{pmatrix}1\\2\end{pmatrix}$ and we suppose that $B:=\begin{pmatrix}u&v\end{pmatrix}$ is such that $B\cdot A=I$ so that
$$
1=u+2vB=u\cdot 1+v\cdot2=B\cdot A
$$
and thus taking $u=1$ and $v=0$ for $X$ and $u=3$ and $v=-1$ for $Y$ then $X\neq Y$ and $X\cdot A=I=Y\cdot B$ as it is easy verify.


问:
What do we actually rotate with rotational matrices
We just learned about rotational matrices in a more theoretical context, but we only "glanced" at them.

My question: What exactly can a matrix rotate (vectors, other matrices, etc...?) and I still don't really get it how it actually rotates another object. Can somebody please give me a simple explanation (it can also be rigorous).

答:
Here' s a concrete example:

The rotation matrix
$$R_\phi := \begin{pmatrix}\cos(\phi) & -\sin(\phi) \\ \sin(\phi) & \cos(\phi) \end{pmatrix} $$
applied to the vector $(r\cos(\theta), r\sin(\theta))^T$ yields
$$\begin{pmatrix}\cos(\phi) & -\sin(\phi) \\ \sin(\phi) & \cos(\phi) \end{pmatrix} * \begin{pmatrix} r\cos(\theta) \\ r\sin(\theta) \end{pmatrix} = \begin{pmatrix} r\cos(\phi)\cos(\theta) - r\sin(\phi)\sin(\theta) \\ r\sin(\phi)\cos(\theta) + r\cos(\phi)\sin(\theta) \end{pmatrix} = \begin{pmatrix} r\cos(\phi + \theta) \\ r\sin(\phi + \theta) \end{pmatrix}.$$
That is, the original vector has been rotated counterclockwise through an angle of $\phi$.


问:
Flawed proof by induction that $a^{n-1}=1$
Donald Knuth "The Art of Computer Programming" Section 1.2.1 has an exercise No.2, the solution to which I don't understand. Exercise:

There must be something wrong with this proof. What is it? Theorem. Let $a$ be any positive number. For all positive integers $n$ we have $a^{n-1}=1$. Proof. If  $n = 1, a^{n-1}=a^{1-1}=a^0=1$. And by induction, assuming that the theorem is true for $1, 2, ..., n$, we have
$a^{(n+1)-1}=a^n=\frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times1}{1}=1$; so the theorem is true for $n+1$ as well.

As far as I understand the proof by induction, the solution should be approximately following:
$a^{1-1}=1$
Assume that $a^{n-1}=1$;
Therefore $a^{(n+1)-1}=1\times a$;
So $a^{(n+1)-1}$ does not equal $a^{n-1}$ and does not equal $1$, unless $a=1$

The solution to exercise reads:

The theorem has not been proved for $n=2$; in the second part of the proof, take $n=1$; we assume there that $a^{-1}=1$. If this condition is true (so that $a=1$), the theorem is indeed valid.

I want to compare it to another, correct, proof by induction in the example from here:

Adding up Odd Numbers
$1+3+5+...+(2n−1)=n^2$

Show it is true for $n=1$

$1=1^2$ is True

Assume it is true for $n=k$

$1+3+5+...+(2k−1) = k^2$ is True
(An assumption!)

Now, prove it is true for $(k+1)$

$1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)^2$
We know that $1 + 3 + 5 + ... + (2k−1) = k^2$ (the assumption above),
so we can do a replacement for all but the last term:
$k^2 + (2(k+1)−1) = (k+1)^2$
Now expand all terms:
$k^2 + 2k + 2 − 1 = k^2 + 2k+1$
And simplify:
$k^2 + 2k + 1 = k^2 + 2k + 1$
They are the same! So it is true.
So: $1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)^2$ is True

Returning to the first example, I didn't quite understand the meaning of the phrase

The theorem has not been proved for $n=2$;

In the second (correct) proof there is no proving that it holds for $n=2$, despite covering the same topic. And Knuth himself has the proof of the theorem $1+3+5+...+(2k−1) = k^2$ similar to the one above, where he too doesn't prove it for $k=2$. So what did he mean by this phrase? Does he mean that there already is an exception in the rule stated when we take arbitrary $n$, for example $n=2$, so therefore it wouldn't hold for all examples anyway? But taking arbitrary numbers and putting it into equation instead of $n$ doesn't seems to be an effective and systematic way to debunk such proofs. How would you explain what is wrong with the exercise?

答:
The crux of complete induction is that we are trying to show that given the statement is true for $\{1,\ldots n\}$, it must be true for $n+1$. For the question at hand, this means that we are allowed to assume that $a^{k-1}=1$ if $k \in \{1,\ldots n\}$. Hence, we are allowed to assume that $a^0=1,\ldots,a^{n-1}=1$.

For $n=2$, this means that we are only allowed to assume that the statement is true for $n=1$ (i.e. $a^0=1)$. Now, we want to show that the statement is true for $n+1=2$; that is, we want to show $a^{1}=1$. Already alarm bells should be ringing in your mind! The inductive step is
$$
a^{(n+1)-1}=a^n=\frac{a^{n-1}\cdot a^{n-1}}{a^{n-2}} = \frac{1 \cdot 1}{1}=1\, .
$$
Substituting $n+1=2$ into the above equation, it reads
$$
a^{2-1}=a^1=\frac{a^{0}\cdot a^{0}}{a^{1-2}}=\frac{1}{a^{-1}}\color{red}{=}1 \, .
$$
The last equality does not hold:
$$
\frac{1}{a^{-1}}=a \neq 1 \, .
$$
Induction fails because we are relying on the fact that the statement being true for $n-2$. When $n=1$, this amounts to assuming that $a^{-1}=1$, which is clearly not true for all $a$. We are only allowed to assume that the statement is true for $\{1,\ldots,n\}$, and $n-2$ might not be a member of the previous set, as we have just seen. The falseness of the proof boils down to the fact that we are assuming more than we should. For induction proofs that require us to know that a statement is true for $\{1,\ldots,n\} \wedge \{n-2\}$, we need two base cases. Once we have established that the statement is true for $1$, and $2$, then things are the same as before, since $n-2 \in \{1,\ldots,n\}$.

The later examples you give are much simpler cases of induction. The type of induction used previous is called 'complete induction', where you assume that a statement is true for $\{1,\ldots,n\}$ and show it for $n+1$. To show that the sum of the first $n$ odd numbers is $n^2$, we only need to assume that it is true for $n$, and $n+1$. There is less room for error with this weaker of form of induction. Once we have established that the statement is true for $1$, it must be true for $2$, $3$, $4$, etc.

Finally, the word 'assumption' is, in my view, somewhat confusing in the context of proof by induction. Really, we are just trying to establish two things:

The statement is true for $1$.

If the statement is true for $n$, then it must be true for $n+1$. (Or, in the case of complete induction, if the statement is true for $\{1,\ldots,n\}$, then it must be true for $n+1$.)

Then, it follows that the statement must be true for $2$, $3$, $4$, etc. Of course, in order to show that $A$ implies $B$, we have to assume that $A$ is true in the first place. But that's kind of obvious, right? I don't know why people make a meal out of the word 'assumption'. Essentially, we are trying to show that given a statement is true for $n$, then it must be true for $n+1$. And, since it is true for $1$, it must be true for $2$, $3$, $4$, etc. That's how I like to think about induction; it's clearer that way in my mind.


问:
Surjective etale morphism from an open subscheme.
I'm looking for an example of a surjective etale morphism from an open subscheme of $\mathbb{A}^d$ (which is not the whole $\mathbb{A}^d$) to $\mathbb{A}^d$. The base field is a finite field or its algebraic closure. I am just wondering whether something like this is possible.

答: If you're particularly interested in positive characteristic, you may also be interested in Proposition 5.2 of Piotr Achinger's "Wild ramification and $K(\pi,1)$ spaces" (link): Let $k$ be a field of characteristic $p > 0$, let $U$ be an affine scheme of finite type over $k$, let $g : U \to \mathbb{A}_{k}^{n}$ be an etale map. Then there is a finite etale map $f : U \to \mathbb{A}_{k}^{n}$.


问:
On what interval is $f(x)=x-\tan(x)$ increasing or decreasing?
My attempt at the solution is:
$$
f'(x) = 1- \frac{1}{\cos^2(x)} $$
We check when $f'(x)=0$ to determine when the derivative is zero.
$$
1- \frac{1}{\cos^2(x)} = 0 \iff 1 = \frac{1}{\cos^2(x)} \iff \cos^2(x)=1
$$
From that we can see that
$$
x=n\pi, n \in \mathbb Z.
$$
I don't understand how to continue. Normally we use a "signs table" in this situation when the given function is a regular one and not trigonometric.

答:
The function is defined over the intervals of the form $(-\pi/2+k\pi,\pi/2+k\pi)$, for $k$ any integer.

For simplicity, let $a_k=-\pi/2+k\pi$ and $b_k=\pi/2+k\pi$. From the known behavior of the tangent function, we can see that
$$
\lim_{x\downarrow{a_k}}(x-\tan x)=\infty,\qquad
\lim_{x\uparrow{b_k}}(x-\tan x)=-\infty
$$
(limit from the right and from the left, respectively).

The derivative is
$$
f'(x)=1-(1+\tan^2x)=-\tan^2x
$$
which is everywhere negative, except at the points $k\pi$, so it just vanishes once in each interval $(a_k,b_k)$. Due to this, the function is strictly decreasing over each interval $(a_k,b_k)$.

You can see it from the “table of signs": the derivative is negative over $(a_k,k\pi)$ and also over $(k\pi,b_k)$, so the function is strictly decreasing over $(a_k,k\pi]$ and also over $[k\pi,b_k)$, hence strictly decreasing in the whole interval.

Beware, though, that you cannot state that the function is everywhere decreasing, because it obviously isn't.

Note that the second derivative is $f''(x)=-2\tan x(1+\tan^2x)$, so the points where the derivative vanish are inflection points.


问:
Need help understanding proof to their being a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$

Let $p_k ( k \geq 1)$ be an enumeration of all the positive primes with $p_1 = 2$ and $p_k < p_{k+1}$ for all $k \geq 1$. Prove that if $n$ is sufficiently large, then there is a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$.

HINT: Let $m$ be the largest integer with $p_m \leq n$. Consider $\sum_{k=1}^m (p_{k+1} - p_k)$ and apply the Prime Number Theorem to $\pi(x)$.

$\pi(x)$ is the prime counting function.

I didn't have a clue how to do this so I looked in the answers and this is what it says:

If $n$ is sufficently large given $n$, we have $\pi(n) \leq \frac{5n}{4 \ln(n)}$. If $m$ is the largest integer with $p_m \leq n$, then $\pi(n) = m$. If $p_{k+1} - p_k \leq \frac{1}{2} \ln(n)$ for all $k \leq m$ then

$$n \leq p_{m+1} - 1 \leq 1 + \sum_{i = 1}^m(p_{k+1} - p_k) \leq 1 + \frac{1}{2} \ln(n) \cdot m \leq 1 + \frac{1}{2} \cdot \ln(n) \cdot \frac{5}{4} \cdot \frac{n}{\ln(n)} = 1 + \frac{5n}{8}.$$

This gives a contradiction if $\frac{3n}{8} > 1$, in paarticular, for $n \geq 3$.

Now, I kind of understand most of everything that's going on, but one this that's making it very hard is the very first line. Where does the $\frac{5}{4}$ come from? I know they've re-aranged the prime number theorem to get it in this form, but I don't get where that number comes from?

Can someone tell me?

答:
Recall that by the Prime Number Theorem, we have
$$\lim_{n\to\infty} \frac{\pi(n)}{\frac{n}{\log n}}=1.$$
Thus by the definition of limit, we have that for any $\epsilon\gt 1$, there is an $N$ such that if $n\gt N$ ("if $n$ is large enough") then
$$\frac{\pi(n)}{\frac{n}{\log n}}\lt 1+\epsilon.$$
The prover picked $\epsilon=\frac{1}{4}$, because that is plenty good enough to push the proof through, as explained by leshik.

Remark: We do not really need the Prime Number Theorem to prove the result. A half century before the first proofs of PNT, Chebyshev had shown by "elementary" means that for large enough $n$, the ratio $\pi(n)/(n/\log n)$ is less than $1.12$.

ContentLicense: CC BY-SA 3.0


问:
How to evaluate cumulative distribution function w/ sigma of given SRS?
I have been given a set of 10 positive values that make up an SRS (Simple Random Sample) of a population, which has a normal distribution.

I then am asked to evaluate $a = \sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2}$ and provide the distribution. Since it is a sum, I assumed a would simply be a number.

Finally, I am asked to evaluate the following CDF: P ($\sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2} \le a$).

How should I interpret this? If the values in the SRS are all positive, then how could any subset less than the size of $n$ have the value of $a$? Wouldn't the probability of this just be $1$?

答: Since $\dfrac{x_i-\mu}{\sigma}\sim\operatorname N(0,1)$ and $x_1,\ldots,x_n$ are mutually independent, we have $ \displaystyle \sum_{i=1}^n \left( \frac{x_i-\mu} \sigma \right)^2 \sim \chi^2_n.$


问:
Prove that if sequence does not tend to infinity, then it has a bounded subsequence?
I need to prove that that if sequence does not tend to infinity, then it has a bounded subsequence. I have tried to divide my problem into 2 subcases. 1) when my sequence is bounded, by Bolzano-Weierstrass I can prove that it has a bounded subsequence. 2) when my sequence is unbounded, it also has 2 subcases. I do contradiction and assume that there is no subsequence which is bounded. Either my subseqences tend to infinity or they do not have a limit. I'm stuck at this point. What would you suggest to do next?
enter image description here</p>

答:
In my opinion it is useless to divide the problem in these two subcases because if sequence is bounded obviously it has bounded subsequence (it itself is its own sub-sequence).

I suggest a solution: let's assume that there is sequence $(a_n)$ which does not tend to infinity. It means that the following is not true:

$$\forall N > 0\, \exists n: \forall k > n \, |a_k| > N.$$

The opposite of the logical expression is
$$\exists N > 0\, \forall n: \exists k > n \, |a_k| \leq N.$$
Let's construct bounded subsequence. We will fix the $N$ from expression and for the first element one can take $a_{k_1}$ such that $k_1 > 1,|a_{k_1}| \leq N$ (indeed it exists because one can put 1 instead of $n$ in expression $\exists N > 0\, \forall n: \exists k > n \, |a_k| \leq N$). For the next term we will get $a_{k_2}$ such that $k_2>k_1,a_{k_2} \leq N$. Continuing the process we will have bounded subsequence.


问:
Constructing a Multivariate Probability Distribution Formula
Suppose there are three individuals playing a simple game, Player $1$, Player $2$, and Player $3$.

In each round of the game, one of the players is uniformly randomly selected to receive the point for that round.  The game ends when any player achieves a score of $10$ points.

How can a multivariate probability distribution formula be constructed that would express the probability of each possible combination of the total number of rounds a game will contain, and an arbitrarily chosen player's (i.e., Player $1$'s) score at the end of the game?  How does this formula change if we introduce the assumption that the arbitrarily chosen player did not win the game?

I started by noting that the chosen player has a $\frac{1}{3}$ probability of winning each round.  I think that for a fixed number of rounds $n$, the formula would be the binomial distribution ${n \choose a} (\frac{1}{3})^a (\frac{2}{3})^{1 - a}$, where by plugging a possible point score into $a$ one obtains the probability of the chosen player attaining that exact point score, but I'm not sure if this is right or where to go from here.

Body:
To calculate the probability of the game ending on a particular score $k-j-10$, you need the score to reach $k-j-9$ in any order of scoring, and then the particular player to have the final score.

As saulsplatz has said in effect, the probability that the final score is $k - j - 10$ is $$\frac{(k+j+9)!}{k!\,j!\, 9!}\left(\frac13\right)^{k+j+10}$$ with $j$ and $k < 10$, though note that $j - k -10$ is also possible $($if $k \not=j)$ with the same probability, as are $k - 10 -j$ and $j-10-k$ and $10 -k-j$ and $10-j-k$, so the overall probability of that pattern is $6$ times that expression if $k \not=j$ and $3$ times that expression if $k=j$.

So the probabilities for individual scores are (with a total of $\frac13$ since we have assumed a particular player wins)

    j   0       1       2       3       4       5       6       7       8       9
k
0       0.00002 0.00006 0.00010 0.00014 0.00015 0.00014 0.00012 0.00009 0.00006 0.00004
1       0.00006 0.00021 0.00041 0.00060 0.00070 0.00070 0.00062 0.00050 0.00038 0.00026
2       0.00010 0.00041 0.00090 0.00140 0.00174 0.00186 0.00176 0.00151 0.00119 0.00088
3       0.00014 0.00060 0.00140 0.00233 0.00310 0.00351 0.00351 0.00318 0.00265 0.00206
4       0.00015 0.00070 0.00174 0.00310 0.00439 0.00527 0.00556 0.00530 0.00464 0.00378
5       0.00014 0.00070 0.00186 0.00351 0.00527 0.00668 0.00742 0.00742 0.00680 0.00579
6       0.00012 0.00062 0.00176 0.00351 0.00556 0.00742 0.00865 0.00907 0.00869 0.00772
7       0.00009 0.00050 0.00151 0.00318 0.00530 0.00742 0.00907 0.00993 0.00993 0.00919
8       0.00006 0.00038 0.00119 0.00265 0.00464 0.00680 0.00869 0.00993 0.01034 0.00996
9       0.00004 0.00026 0.00088 0.00206 0.00378 0.00579 0.00772 0.00919 0.00996 0.00996

and the probability of a pattern is (with a total of $1$, since we do not care who wins or comes second)

    j   0       1       2       3       4       5       6       7       8       9
k
0       0.00005
1       0.00034 0.00062
2       0.00062 0.00248 0.00269
3       0.00083 0.00359 0.00837 0.00698
4       0.00090 0.00419 0.01046 0.01860 0.01318
5       0.00084 0.00419 0.01116 0.02108 0.03163 0.02003
6       0.00070 0.00372 0.01054 0.02108 0.03338 0.04451 0.02596
7       0.00053 0.00301 0.00904 0.01908 0.03179 0.04451 0.05440 0.02979
8       0.00038 0.00226 0.00715 0.01590 0.02782 0.04080 0.05213 0.05958 0.03103
9       0.00025 0.00159 0.00530 0.01236 0.02267 0.03476 0.04634 0.05517 0.05977 0.02988

and the probability that a game lasts a particular number of rounds is $($adding up to $1)$

10  0.00005
11  0.00034
12  0.00124
13  0.00331
14  0.00718
15  0.01339
16  0.02232
17  0.03402
18  0.04819
19  0.06425
20  0.08123
21  0.09750
22  0.11066
23  0.11787
24  0.11668
25  0.10592
26  0.08620
27  0.05977
28  0.02988

This means that

the equally most likely individual final scores are $10-8-8$ or $8-10-8$ or $8-10-8$, each with a probability of about $0.01034$,

the most likely final pattern is $10-9-8$ with a combined probability of about $0.05977$, and

the most likely number of rounds in the game is $23$ with a combined probability of $0.11787$.  The expected number of rounds is about $22.34469$ and the median also $23$, all rather less than the $27$ or $28$ involved in the most likely scores.

The expected final score per player is about $7.44823$, also rather less than the most likely scores might suggest.


问:
$g(x) = |f(x)|$ is differentiable at $x_0$ iff $f'(x_0) = 0$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $x_0$, and $f(x_0) = 0$. Show that $g(x) = |f(x)|$ is differentiable at $x_0$ iff $f'(x_0) = 0$.

Given that $f$ is differentiable at $x_0$ then, for every $\varepsilon >0$ there exist $\delta>0$ such that if
$$
0 < |x-x_0| < \delta \implies |\frac{f(x) - f(x_0)}{x-x_0} - L| < \varepsilon
$$

if $f'(x_0) = 0$ then

$$0 < |x-x_0| < \delta \implies |\frac{f(x) - f(x_0)}{x-x_0} - 0| < \varepsilon$$
$$ \implies |\frac{f(x)-f(x_0)}{x-x_0}| < \varepsilon$$

By the reverse triangle inequality we have that
$$|\frac{|f(x)| - |f(x_0)|}{x-x_0}| \leq |\frac{f(x)-f(x_0)}{x-x_0}| < \varepsilon$$

Then, $|x - x_0| < \delta$ implies that
$$|\frac{g(x)-g(x_0)}{x-x_0} - 0 | < \varepsilon$$

Therefore, $g$ is differentiable at $x_0$.

I'm having troubles when the hypothesis is that $g(x)$ is differentiable at $x_0$, I would like to prove that $g'(x_0) = 0$ (I'm trying to prove it without using the fact that $g$ has a minimum at $x_0$). If I could prove that then I would have:

If $g$ is differentiable at $x_0$ and $g'(x_0) = 0$, then for every $\varepsilon > 0$ there exist $\delta > 0$ such that if
$$0< |x - x_0|<\delta \implies |\frac{g(x) - g(x_0)}{x-x_0} -0| < \varepsilon$$
$$ \implies |\frac{|f(x)|}{x-x_0}| < \varepsilon$$
$$ \implies |\frac{f(x)-f(x_0)}{x - x_0} - 0| < \varepsilon$$

Therefore $f'(x_0) = 0$.

答:
$ f $ is differentiable at $ x_0 \implies$

$$\lim_{x\to x_0}\frac{f(x)}{x-x_0}=f'(x_0)\implies$$

$$\lim_{x\to x_0^+}\frac{g(x)}{x-x_0}=|f'(x_0)|$$
and
$$\lim_{x\to x_0^-}\frac{g(x)}{x-x_0}=-|f'(x_0)|$$

So,
$$g\text{ differentiable at } x_0\implies$$
$$|f'(x_0)|=-|f'(x_0)|\implies f'(x_0)=0$$

The converse is much easier.


问:
How many simultaneous independent events needed to have a certain probability of getting at least X specific outcomes
Each cup of coffee I purchase has random 60% probability of having my name spelled correctly on the cup. If I want to have an 80% chance of having at least $x$ correctly spelled coffee cups for my collection, how many coffees do I need to purchase?

I can afford an infinite number of coffees from my local coffee shop, but I can only make one order (all the cups will arrive at the same time).

I've seen this question which feels related but I'm not sure how to extend that to desiring a certain number of successes.

答:
You can basically use the same technique as in the linked post. If you purchase n cups of coffee, the number of cups with your name correctly spelled on it has binomial distribution $X\sim \mathscr B(n, .6)$. We are interested in the probability that the number of cups purchased with your name correctly spelled on it is at least x, that is, $P(X\ge x)$. We know for a binomial distribution this is $\sum_{k=x}^{n}P(X=k)$. And we want this quantity to be greater than $80\%$, that is, getting at least x cups with your name correctly spelled with a probability of at least $80\%$. You simply find the smallest n such that $\sum_{k=x}^n{n\choose k}.6^k.4^{n-k}\ge .8$. This can be done using the pbinom command. If, for example, you wanted to find the number of cups you should purchase to have at least 80% chance of getting 5 cups with your name spelled right, you would find

```<code>> print(pbinom(4, 9, .6, lower.tail=FALSE))
[1] 0.7334323
</code>```

but

```<code>> print(pbinom(4, 10, .6, lower.tail=FALSE))
[1] 0.8337614
</code>```

therefore you need to buy 10 cups to have the probability of getting at least 5 cups with your name spelled right to be at least 80%. More generally, you can use the following code to find the number of purchases needed

```<code>cupsrightname=5
for (k in 1:100) {
  w=pbinom(cupsrightname-1, k, .6, lower.tail=FALSE)
  if (w>=.8) {
    print(k)
    break
  }
}
</code>```

-out

```<code>[1] 10
</code>```

Body:
You can basically use the same technique as in the linked post. If you purchase n cups of coffee, the number of cups with your name correctly spelled on it has binomial distribution $X\sim \mathscr B(n, .6)$. We are interested in the probability that the number of cups purchased with your name correctly spelled on it is at least x, that is, $P(X\ge x)$. We know for a binomial distribution this is $\sum_{k=x}^{n}P(X=k)$. And we want this quantity to be greater than $80\%$, that is, getting at least x cups with your name correctly spelled with a probability of at least $80\%$. You simply find the smallest n such that $\sum_{k=x}^n{n\choose k}.6^k.4^{n-k}\ge .8$. This can be done using the pbinom command. If, for example, you wanted to find the number of cups you should purchase to have at least 80% chance of getting 5 cups with your name spelled right, you would find

> print(pbinom(4, 9, .6, lower.tail=FALSE))
[1] 0.7334323

but

> print(pbinom(4, 10, .6, lower.tail=FALSE))
[1] 0.8337614

therefore you need to buy 10 cups to have the probability of getting at least 5 cups with your name spelled right to be at least 80%. More generally, you can use the following code to find the number of purchases needed

cupsrightname=5
for (k in 1:100) {
  w=pbinom(cupsrightname-1, k, .6, lower.tail=FALSE)
  if (w>=.8) {
    print(k)
    break
  }
}

-out

[1] 10


问:
Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$.

Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.

For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le  1
\end{align}

For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.

My questions are:

Does someone have a hint or a proof?

Can someone give me some advice on how to handle such problems during an exam?

Thanks :)

答:
I would like to solve a more general problem than this problem and that problem.

First, we find all local maxima of the function $f(x) =\prod_{k=0}^n |x-k|$ in the interval $0 \le x \le n$.

For each interval $[p,p+1)$ with $p\in \{0,...,n-1\}$ we prove that there exists 1 local maximum.

We guess the value of $v_0,v_1,...,v_n$ such that there exists $x_p \in [p,p+1)$ satisfying two following conditions
$$v_0x = v_1(x-1) =...=v_n(x-k) \tag{1}$$
$$v_0+v_1+...+v_n=0\tag{2}$$
Solve these 2 conditions, we can have a solution (attention, this is not a unique solution, but we need only 1) as follows
$$v_k=\frac{1}{x_p-k}$$
with $x_p \in [p,p+1)$ satisfying this equation
$$\sum_{0 \le k \le n}\frac{1}{x_p-k} = 0 \tag{3}$$
Note: For information, it's easy to prove that the equation (3) has exactly one root in each interval $[p,p+1)$ where $p\in \{0,...,n-1\}$.

Now, we apply the AM-GM inequality for $v_k(x-k)$ with $0 \le k \le n$ (we notice that for $k >p$, $v_k$ is negative, then the terms $v_k(x-k)$ are all positive).

\begin{align}
f(x) &=\prod_{0 \le k \le n}|x-k| \\
 &= \frac{1}{\prod_{0 \le k \le n}|v_k|} \prod_{0 \le k \le n}v_k(x-k) \\
 &\le \frac{1}{\prod_{0 \le k \le n}|v_k|} \left(\frac{\sum_{0 \le k \le n}v_k(x-k)}{n+1}   \right)^{n+1} =  \frac{1}{\prod_{0 \le k \le n}|v_k|k} \left(\frac{-\sum_{0 \le k \le n}kv_k}{n+1}   \right)^{n+1}
\end{align}

We note that,;from (1) and (3), we have
\begin{align}
-\sum_{0 \le k \le n}kv_k &=  -\sum_{0 \le k \le n}\frac{k}{x_p-k} \\
&= n+1 - x_p\sum_{0 \le k \le n}\frac{1}{x_p-k} \\
&= n+1
\end{align}

Hence,
$$f(x) \le \frac{1}{\prod_{0 \le k \le n}|v_k|} \left(\frac{-\sum_{0 \le k \le n}kv_k}{n+1}   \right)^{n+1} = \prod_{0 \le k \le n}|x_p-k|\left(\frac{n+1}{n+1}   \right)^{n+1} = \prod_{0 \le k \le n}|x_p-k|$$

The local maximum in $x \in [p,p+1)$ occurs when $x = x_p$ where $x_p \in [p,p+1)$ is the solution of (3).

Second, we prove that $f(x_0) \ge f(x_1) \ge ...\ge f(x_{[\frac{n}{2}]})$ and $f(x)$ is a symmetric function at $x = \frac{n}{2}$ (in other words, $f(x) = f(n-x)$).

It's easy to prove the second statement about the symmetry.
For the first statement, for $x \in [p,p+1)$ with $p\in \{0,...,[\frac{n}{2}]-1\}$, we have

$$ f(x+1) = \prod_{0 \le k \le n}|(x+1)-k| = \frac{x+1}{|x-n|}\prod_{0 \le k \le n}|x-k| = \frac{x+1}{|x-n|} f(x)$$
As $x \in [p,p+1)$ with $p\in [0,...,([\frac{n}{2}]-1)]$, we have then $x< \frac{n+1}{2}$, so $\frac{x+1}{|x-n|} <1$ or $f(x+1) < f(x)$. We deduce that
$$\max_{x \in [p+1,p+2)}f(x) = \max_{x \in [p,p+1)}f(x+1) < \max_{x \in [p,p+1)}f(x)$$
or for all $p\in \{0,...,[\frac{n}{2}]-1\}$ $$f(x_{p})>f(x_{p+1}) \tag{4}$$

Finally, we solve your problem, which corresponds to the case: find the maximum of $f(x)$ in the interval $ 1 \le x \le n-1$  (another particular case is where $0 \le x \le n$ was solved here ).

From (4), we have the function $f(x)$ attains its maximum value $f(x_1)$ in $x \in [1,n-1]$ when $x=x_1$ where $x_1 \in (1,2)$ is the root of the equation (3).

$$f(x) \le f(x_1) = \prod_{0 \le k \le n}|x_1-k| = x_1(x_1-1)(2-x_1) \prod_{3 \le k \le n}(k-x_1) \le \left(\frac{x_1 + (x_1-1) + 2(2-x_1)}{3}  \right)^3 \frac{1}{2}\prod_{3 \le k \le n}(k-1) = \frac{(n-1)!}{2}$$

Because $x_1$ doesn't satisfy the condtion of AM-GM inequality ($x_1 = x_1-1 = 2(2-x_1)$), we conclude that $$f(x) < \frac{(n-1)!}{2}$$


问:
Maximum capacity of box without using Calculus?
Lets use a square cardboard paper (side $s$) to make an open box. Remove four small squares from each corner and fold to make a rectangular box of height $x$.

To compute maximum possible volume, we can use AM-GM :

$$ \sqrt[3]{(s-2x)(s-2x)\color{blue}{4}x} \le \dfrac{(s-2x)+(s-2x)+\color{blue}{4}x}{3} $$

$$ \Rightarrow V \le \dfrac{2}{27}s^3$$

Note : This matches the result obtained from calculus.

Next take a rectangular paper ($L \times B$) where $L > B$.

But AM-GM won't apply directly this time, since $L-2x > B-2x$.

This can be easily done by calculus. But I want to know if a valid solution using inequalities or other pre-Calculus methods exists?

I tried :

Let $(L-2x)=\lambda (B-2x)$ for maximum capacity. Then AM-GM,

$$ \sqrt[3]{(L-2x)\lambda(B-2x)(2+2\lambda)x} \le \dfrac{(L-2x)+\lambda(B-2x)+(2+2\lambda)x}{3} $$

$$ \Rightarrow 2\lambda(1+\lambda)V \le \frac{1}{27}(L+\lambda B)^3$$

where $$ (L-2x)=\lambda(B-2x)=(2+2\lambda)x $$

gives a quadratic in $\lambda$ : $$\dfrac{\lambda (\lambda+2)}{2\lambda+1} = \dfrac{L}{B} = r$$

One obtains

$$ V \le  \dfrac{(r+\lambda)^3}{\lambda (\lambda+1)} \dfrac{B^3}{54} $$

Is this correct? If it's incorrect, can this solution be improved? Or, does a different solution using pre-Calculus methods exist?

Thank you for your time!

答:
How to determine the height $x$ of the box maximising the box volume $V$ by a method which requires neither calculus nor inequality techniques.

Length $L$ and breadth $B$ are given and it is assumed that $\,0<B\leqslant L$. The volume of the box is
$$\begin{align}
V(x)\; & =\; x(B-2x)(L-2x)
 \;=\; 4x\left(x-\frac B2\right)\left(x-\frac L2\right) \\[1.5ex]
& =\; 4x^3 -2(B+L)\,x^2 +BL\,x\tag{1}
\end{align}$$
This cubic function is strictly positive if $\,0<x<\frac B2$ or $\frac L2<x$,
and if $x\leqslant 0$ or $\,\frac B2\leqslant x\leqslant\frac L2$
then $V(x)\leqslant 0\,$.

Note that $2(B+L)=P\,$ is the perimeter and $BL=A$ is the area of the given rectangular paper.

Now the idea is to shift the function $V$ parallel to the $y$-axis such that one extremum becomes a second order zero, i.e., the function graph touches the $x$-axis without changing sign. This preserves the $x$-coordinates of the extrema whereas the number of parameters needed to characterise the shifted function is reduced by one.

Let $x_e$ be the $x$-coordinate of one extremum, let $V_e=V(x_e)$, and let $u$ be an unknown. Then
$$\begin{align}V_\text{shifted}(x) \,=\, V(x)-V_e\: & \stackrel{!}{=}\:
4(x-u)(x-x_e)^2 \\[1ex]
& =\: 4x^3 -(4u+8x_e)\,x^2 +\big(8ux_e +4x_e^2\big)\,x - 4ux_e^2\tag{2}
\end{align}$$
Compare $(2)$ and $(1)$, then uniqueness of the coefficients implies
$\,4u=\dfrac{V_e}{x_e^2},$ and, exploiting this identity for $u$ straight away, also
$$\begin{align}2\,\frac{V_e}{x_e} +4x_e^2 & \;=\; BL\tag{3}\\[1.5ex]
\frac{V_e}{x_e^2} +8x_e & \;=\; 2(B+L)\,.\tag{4}\end{align}$$
Performing $\,(4)\cdot2x_e-(3)\,$ gives
$$x_e^2 -\frac{B+L}{3}x_e +\frac{BL}{12} \;=\; 0\tag{5}$$
with the solutions
$$x_e\;=\;\frac{B+L\mp\sqrt{BL+(L-B)^2}}{6}\,.$$
Clearly the smaller value belongs to the seeked maximum.

The larger value is the local minimum of $V\,$ between $\,\frac B2$ and $\,\frac L2$.

Cross-checking the equality case $L=B=s\,$ yields $\,x_e=\frac16s$, in agreement with the OP where this value corresponds to equality in AM-GM.

Rewriting the result in terms of $P=2(L+B)$ and $A=LB$ reads
$$x_e=\frac{P\mp \sqrt{P^2-12A}}{12}\,.$$

Finally, $V_e$ is made a bit more explicit: Performing $\,(3)\cdot 2-(4)\cdot x_e\,$ gives
$$V_e \;=\;\frac13\big[2BL\,x_e -2(B+L)\,x_e^2\big]
 \;=\; \frac1{18}BL(B+L) -\frac29\big(BL+(L-B)^2\big)\,x_e$$
where $(5)$ has been used to replace $\,x_e^2\,$ by $\,\frac{B+L}{3}x_e -\frac{BL}{12}\,$.

Cross-checking the equality case $L=B=s\,$ gives $\,V_e=\dfrac2{27}s^3\,$ as desired.


问:
Distributional derivative of $\left|\cos|\pi x|\right|$
I am new to distributions and I have a problem finding the distributional derivative for the following function: $$\left|\cos|\pi x|\right|$$

I would be grateful if you could help to find the distributional derivative for this case. Please find the attached figure with this link.
Problem

I have tried up to some extent and I have attached a figure of that one here too. However, I could not find a way to extend that solution. Please find that figure with this link.attempt_one

答:
Since $f(x)=|\cos|\pi x||=|\cos \pi x|$ is continuous and piecewise $C^1$ the distributional derivative is just the ordinary pointwise derivative where it's defined:
$$f'(x) = \pi (-\sin \pi x) \operatorname{sign}(\cos\pi x).$$

The derivative is only piecewise continuous, piecewise $C^1$, and has steps from $-\pi$ to $+\pi$ at $x=\pm\frac12,\pm\frac32,\pm\frac52,\ldots$ It's derivative is therefore the pointwise derivative plus $-2\pi\delta(x-x_i)$ where $x_i=\pm\frac12,\pm\frac32,\pm\frac52,\ldots$:
$$f''(x) = -\pi^2|\cos\pi x| + \sum_{k\in\mathbb{Z}}\delta(x-\frac{2k+1}{2}).$$


问:
A logic question about Karl's favorite brownie recipe
I have been trying to solve the following problem:

Karl’s favorite brownie recipe uses semisweet chocolate, very little
flour, and less than 1/4 cup sugar. He has four recipes: one French, one Swiss,
one German, and one American. Each of the four has at least two of the qualities
Karl wants in a brownie recipe. Exactly three use very little flour, exactly three use
semisweet chocolate, and exactly three use less than 1/4 cup sugar.
The Swiss and the German recipes use different kinds of chocolate. The American
and the German recipes use the same amount of flour, but different kinds of
chocolate. The French and the American recipes use the same amount of flour. The
German and American recipes do not both use less than 1/4 cup sugar.
Karl is very excited because one of these is his favorite recipe. Which one is it?

The following is my attempted solution:

From three of the statements given in the problem: "The Swiss and the German recipes use different kinds of chocolate.", "The American
and the German recipes use the same amount of flour, but different kinds of chocolate.", and "Exactly three  recipes use semisweet chocolate.",
one can deduce that the German recipe does not use the semisweet chocolate because the German recipe uses different kind of chocolate from both
the Swiss and the American recipes. Because each of the four recipes has at least two of the qualities Karl wants in a brownie recipe, it follows that the German  recipe use
both very little flour and less than 1/4 cup sugar.

From three of the statements given in the problem: "The American and the German recipes use the same amount of flour, but different kinds of chocolate.", "The French and the American recipes use the same amount of flour.", and  "Exactly three recipes use  very little flour.", one can deduce that the Swiss recipe does not use very little flour.

Because both the German and the Swiss recipe have only two of the qualities Karl wants for a brownie recipe, they both are not Karl's favorite recipe.

The statement "The German and American recipes do not both use less than 1/4 cup sugar." means the German recipe or the American recipe and not both uses less than 1/4 cup sugar.
Because the German recipe uses both very little flour and less than 1/4 cup sugar, the American recipe does not use less than 1/4 cup sugar. It follows that the American recipe is not Karl's favorite.

Therefore, Karl's favorite recipe is the French recipe.

My questions are whether the statement "The German and American recipes do not both use less than 1/4 cup sugar." means the German recipe or the American recipe and not both uses less than 1/4 cup sugar and whether the statement "The German and American recipes do not both use less than 1/4 cup sugar." is equivalent to the statement "Both the German and the American recipes do not use less than 1/4 cup sugar." or not?

The question is from the book: Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 31). New York: Springer.

答:
On its own,

"The German and American recipes do not both use less than 1/4 cup sugar."

means that one or two of them use too much sugar.

But you are also told "exactly three use less than 1/4 cup sugar" so in this question only one can use too much sugar.


问:
How to prove $1 + 3 + 3^2 + ... + 3^{n-1} = (3^n - 1)/2$ by mathematical induction?

$1 + 3 + 3^2 + ... + 3^{n-1} = \dfrac{3^n - 1}2$

I am stuck at

$\dfrac{3^k - 1}2 + 3^k$

and I'm not sure if I am right or not.

Body:
Assume
$$S=1+3+3^2+3^3+...+3^{n-1}$$
in this case
$$ 3\cdot S=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{n-2}+3\cdot3^{n-1}$$
$$ 3\cdot S=3+3^2+3^3+...+3^{n-1}+3^{n}$$
Now subtracting $S$ from $3\cdot S$ we have:
$$3\cdot S-S=-1+3-3+3^2-3^2+3^3-3^3+...+3^{n-1}-3^{n-1}+3^n$$
it is easy to see that the result will be
$$2\cdot S=3^n-1$$
so
$$S=\frac{3^n-1}{2}$$
Replacing $S$ with the original sum we have:
$$1+3+3^2+3^3+...+3^{n-1}=\frac{3^n-1}{2}$$
Q.E.D


问:
Proving $ \prod_{k=0}^{n-1} (n^2-k^2) = \frac{(2n)!}{2}$
I'm probably having a brain-fart but I can't figure out why this identity holds:

$$
\prod_{k=0}^{n-1} (n^2-k^2) = \frac{(2n)!}{2}
$$

I tried using various formulae involving $\binom{2n}{n}$, without success. Any ideas?

Bonus if it helps visualise why the factors on the RHS, involving squares, would multiply to get the RHS, which does not contain any square terms.

答:
A variation: We obtain
\begin{align*}
\color{blue}{\prod_{k=0}^{n-1}\left(n^2-k^2\right)}
&=\prod_{k=0}^{n-1}(n-k)\prod_{k=0}^{n-1}(n+k)\\
&=\prod_{k=0}^{n-1}(k+1)\cdot n\cdot\prod_{k=1}^{n-1}(n+k)\tag{1}\\
&=n!\cdot n\cdot \frac{(2n-1)!}{n!}\\
&\,\,\color{blue}{=\frac{1}{2}(2n)!}
\end{align*}
and the claim follows.

In (1) we change the order of multiplication in the left-hand product $k\to n-1-k$ and split the factor $n$ from the right-hand product


问:
Baire category theorem dense subsets
I'm reading John B. Conway's book on point-set topology and there the Baire category theorem is stated as follows:

If $(X,d)$ is a complete metric space and $\{U_n\}$ is a sequence of open subsets of $X$ each of which is dense, then $\bigcap_{n=1}^\infty U_n$ is dense.

Does this mean that the subsets are dense in $X$ or in some subset of
$X$ and why is their intersection always nonempty? For example
$\mathbb Q\cap\mathbb R\setminus\{\mathbb Q\}=\emptyset$ and they are
both dense in $\mathbb R$?

Of course this is not an infinite sequence
of subsets but to me it feels like pointing to the direction that the
intersection might be empty. The proof uses the fact that the
intersection is nonempty.

答:
As noted in comments about, $\mathbb Q$ and $\mathbb R\setminus \mathbb Q$ are dense, but they are not open.

It is easier to characterize the complement $V$ of open dense subset of $\mathbb R.$

$V\subseteq\mathbb R$ is closed and contains no open interval if and only if $\mathbb R\setminus V$ is open and dense.

Examples of such $V:$

finite sets

the Cantor set

$\mathbb Z,$ or any discrete subspace

$\{0\}\cup\left\{\frac1n\mid n\in\mathbb Z^+\right\}$

Such sets tend to be “small” in comparison  with the whole set. These examples are all measure zero, for example. The general case might not have a measure, but the theorem essentially gives another way of saying such sets are “small.”

Another way to state the theorem is if $\{V_i\}$ is a countable set of closed sets, none of which contains an open ball, the their union does not contain an open ball.

Title: Baire category theorem dense subsets


问:
name of theorem for infinite order polynomial limit for small x
is there a name for this property?

$\mathop{\operatorname{lim}}_{x\to 0} \sum_{n=1}^\infty x^n = \frac{x}{1-x}$

I have seen it in a derivation but want to know where it comes from.

答:
You can show that

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

For $|x|<1$ you can take the limit $N\rightarrow \infty$ and you obtain the result. It's called geometric series.


问:
Complex Integration and trigonometric functions
I am just begining with complex integration. Please help with this.
Evaluate $\int_{-i\pi}^{i\pi}\cos z dz$ where $z$ is a complex number.

This is my effort,
I know that $\int_{C}{f(z)dz} = \int_a^b f(z(t))z'(t)dt$ where C:[a,b] is a complex curve and $a \leq t \leq b$

Step one would be to determine z(t) from the path of integration and this is the toughest part for me.

Once I have obtaine z(t) I hope the rest would be easy

答:
Thank you guys just figured it out

From
$\int_{C} f(z)dz = \int_a^b f(z(t))z'(t)dt$

But
$z(t) = p + t(q - p)$ for $t \in [0,1]$

In our case
$p = -i\Pi$ and $q = i\Pi$

Then
$z(t) = -i\Pi + t(i\Pi + i\Pi)$
$z(t) = i\Pi (2t - 1)$
and
$z'(t) = 2i\Pi$

Also
$f(z(t)) = cos(i\Pi (2t - 1))$

Therefore

$\int_{-i\Pi}^{i\Pi} coszdz = \int_0^1 cos(i\Pi(2t - 1)) 2i\Pi dt$

$= 2i\Pi \int_0^1 cos(i\Pi(2t - 1)) dt$

Let $w = i\Pi(2t - 1)$

then
$dt = \frac{dw}{2i\Pi}$

$= \int_0^1 coswdw$
$= sinw$ from 0 to 1

$= sin(i\Pi(2t - 1))$ from 0 to 1
$= sin(i\Pi) - sin(-i\Pi)$

Therefore

$\int_{-i\Pi}^{i\Pi} coszdz = 2sin(i\Pi)$


问:
$\textit{Why}$ the cases in 4 $\nmid n^2 - 3?$ (bases cases in modular induction)
This question has already been answered here. My question is, when we suppose the hypothesis is not true (proof by contradiction), namely that $4 \mid (n^2 - 3)$, we get $n^2 - 3 = 4k$ for some integer $k$ and then we have two cases. But where does the two cases come from? In addition, why does putting them into two cases work to prove the statment? My thought process would be to try to prove this statement by contradiction and somehow figure out a way to produce a contradiction similar to how $\sqrt{2}$ is proved by contradiction. Any help is great!

Let me know if I need to clarify more!

答:
If you do a proof by contradiction it might seem like you don't need two cases (but you actually do).

If $n^2 - 3 = 4k$ then $4k$ is even so $n^2=4k + 3$ is odd.  So $n$ is odd.

SO $n = 2m+1$ for some $m$ and $n^2 = 4m^2 + 4m + 1= 4(m^2 + m) + 1$.

But $n^2 = 4k + 3$ so $4(m^2 +m) + 1 = 4k +3$.

So $4(m^2+m) - 4k = 2$.

So $4(m^2 + m-k) = 2$ and $2(m^2 + m -k) = 1$.  But $1$ is odd and $2(m^2 + m-k)$ is even.  A contradiction.

In this case that case that $n$ might be even was tacitly dismissed as an impossibility.

But tacitly dismissing as an impossibility is a contradiction.  It's just a case that needn't be considered as it was directly proven to be false.

....

And you don't need just two cases. It's that two cases are sufficient.

I could have done it with $10$ cases.

Let $n = 10k + i$ where $i=0,1,2,3,...9$.

And $n^2 - 3 = 4(25k^2 + 5ki) + (i^2 -3)$.

Need to show $4\not\mid i^2 -3$.  And considering the the $10$ cases we see

$i^2 -3 = -3, -2, 1, 6, 13,22, 33,46,61,78$ are none of them divisible by $4$.


问:
Mean curvature vector
On a Riemannain manifold $M$ with a submanifold $S$, we can define the second fundamental form $H$ to be $-H(X, Y) = (\nabla_{X} Y)^{N}$ the component of $(\nabla_{X} Y)^{N}$ normal to $S$. Then mean curvature vector is defined to be the trace of the second fundamental form. I am confused as to what does it mean for a vector to be the trace of the second fundamental form.

Usually contractions/trace will yield a real-valued function on the manifold, but here we want a vector.

答:
The definition of the trace is essentially the same. You simply define the trace of  to be
$$ \operatorname{tr} = \sum_{k=1}^n H(e_k,e_k) $$
where $(_1,…,_)$ is an orthonormal basis of $_$. Or, if you have coordinate vector fields $\partial_1,\dots,\partial_$ and dual 1-forms $^1,\dots,^$ on  and the metric is $g_{ij}dx^i\,dx^j$, then
$$
\operatorname{tr} = g^{ij} H(\partial_i,\partial_j) $$
If a basis of the normal bundle is $\nu_1,\dots,\nu_{n-m}$, then you can write $$=\nu_\mu H^\mu_{ij}dx^idx^j$$ and its trace as $$^{ij}^\mu_{ij}\nu_\mu$$

Title: Mean curvature vector


问:
Understanding basic concept of prime numbers
My textbook provides a theorem but I cannot understand the structure of the sentence being used. Could someone please help me understand the meaning of this theorem?

A natural number $n>1$ is prime if and only if for all primes $p\leq \sqrt{n}$, $p$ does not divide $n$.

"for all primes $p\leq \sqrt{n}$, $p$ does not divide $n$". This is the part I don't understand. The premise is that $p$ is less than the square root of $n$ then the conclusion is that $p$ does not divide $n$. So for $p$ to divide $n$ it must be greater than the square root of $n$? I'm really confused as to what information and meaning I'm supposed to see in this theorem.

答:
A positive integer $n\gt 1$ is not prime if and only if there exist positive integers $a$ and $b$, $1\lt a\lt n$, $1\lt b\lt n$, with $n=ab$.

But now notice that we cannot have both $a$ and $b$ greater than $\sqrt{n}$. Why? Because if $a\gt\sqrt{n}$ and $b\gt\sqrt{n}$, then $ab\gt\sqrt{n}\sqrt{n}=n$. But $ab$ is supposed to be equal to $n$.  And we cannot have both be strictly less than $\sqrt{n}$, because then $ab\lt\sqrt{n}\sqrt{n}=n$, again a problem.

In other words, whenever you express $n$ as a product of two positive integers, you must have one less than or equal to $\sqrt{n}$, and the other one be greater than or equal to $\sqrt{n}$.

So if you are looking for proper factors of $n$ (positive integers $a$ such that $1\lt a\lt n$ and $a$ divides $n$), you only have to look for them up to $\sqrt{n}$:  if you don't find any between $2$ and $\sqrt{n}$, then there are none to find.

Of course, because any such factor $a$ must also be the multiple of a prime $p$, $p\leq a$, in fact you only need to check the primes that are between $2$ and $\sqrt{n}$ to see whether $n$ has a proper factor. If $n$ has a proper factor, then there will be a prime between $2$ and $\sqrt{n}$ that divides $n$. If no  prime between $2$ and $\sqrt{n}$  divides $n$, then $n$ has no proper factors, so $n$ will necessarily be a prime.

That's what (one of the implications in) the theorem is telling you.

You are also misidentifying the premise and conclusion. That's because the premise is itself an implication.

The premise is:

If $p$ is a prime and $p\leq\sqrt{n}$ then $p$ does not divide $n$.

The conclusion is:

$n$ is a prime.


答:
Let $n$ be a natural number.

Suppose $n > 1$.

n is prime if

for all positive integers $p$

where $p$ is prime

and $p \leq \sqrt{n}$,

$p$ does not divide $n$

and only if

for all positive integers $p$

where $p$ is prime

and $p \leq \sqrt{n}$,

$p$ does not divide $n$

For instance, $17$ is prime if (and only if) all the primes in the interval $[1,\sqrt{17}]$ do not divide $17$.  The primes in the interval $[1, \sqrt{17}]$ are $2$ and $3$.  Since $2 \not\mid 17$ and $3 \not\mid 17$, $17$ is prime.

Title: Understanding basic concept of prime numbers

Body:
My textbook provides a theorem but I cannot understand the structure of the sentence being used. Could someone please help me understand the meaning of this theorem?

A natural number $n>1$ is prime if and only if for all primes $p\leq \sqrt{n}$, $p$ does not divide $n$.

"for all primes $p\leq \sqrt{n}$, $p$ does not divide $n$". This is the part I don't understand. The premise is that $p$ is less than the square root of $n$ then the conclusion is that $p$ does not divide $n$. So for $p$ to divide $n$ it must be greater than the square root of $n$? I'm really confused as to what information and meaning I'm supposed to see in this theorem.


答:
It says that, in order to see if $n$ is prime, we don't have to test divide $n$ by every prime less than $n$.  It's enough to check all primes $\leq\sqrt n$.  For example, if we want to know if $101$ is prime, it's enough to check that it's not divisible by $2,3,5,\text{ or }7$, because the next prime, $11$ is $>\sqrt{101}$.

This is because if $n=ab$, it's impossible that both $a>\sqrt n$ and $b>\sqrt n$.


问:
Beginner conditional probablity calculation
You're selling a production machine P and offer both a maintenance contract as well as an insurance against possible failure.

You know that 45% of customers do not arrange a maintenance contract and 65% do not arrange an insurance.
25% do neither.

a) Whats the probablity of a customer, who didn't arrange an insurance, also not arranging a maintenance contract? (Solution: 38,46%)

b) Whats the probablity of a customer arranging both? (Solution: 15%)

c) Whats the probablity of a customer, who did arrange a maintenance contract, also arranging an insurance? (Solution 27.27%)

I have no idea how to arrive at these solutions.

On the first question: With "$M$ = No Maintance" and "$I$ = No Insurance" i thought about $$P(M|I)=\frac{0.45}{0.75 *0.65}$$ but that doesnt work.

答:
For the first question:

$P(M \cap I)=0.25$ so you can find the probability $P(M|I)$

For the second question:

You need to find $P(M^c \cap I^c)$

Note that $$P(M^c \cap I^c)=1-P(M \cup I)=1-[P(M)+P(I)-P(M \cap I)]$$

For the third question:

You need to find $$P(I^c|M^c)=\frac{P(M^c \cap I^c)}{P(M^c)}$$

which can be found using the result of the second question.


问:
The non-existence of a logical formula

Let $\tau = \{+, Zero \}$ be a vocabulary with binary function symbol $+$ and a constant symbol $Zero$. We define the strucutre $S = \{\mathbb{R}, +^S, Zero^S \}$, where $+^S$ and $Zero^S$ correspond to the normal addition of numbers and the constant $0$.

Prove that there is no formula $\varphi (x, y) \in FO [\tau]$ such that $\varphi (S) = \{(a, b) | a, b \in \mathbb{R}\ and\ a <b\}.$

Here's what I understand: No formula $\varphi$ can exist with two free variables $x, y$ such that the set of all $x, y$ tuples that would evaluate $\varphi$ to true is equal to $\{(x, y) | x, y \in \mathbb{R}\ and\ x<y\}$.

A reasonable way to prove this non-existence seems to me to be a proof by contradiction, but I don't see what follows. How should we pick or "generalize" $\varphi$ to fit our proof? If a proof by contradiction doesn't work, is there a theorem that we could use to prove this? I haven't been able to find any. All hints are appreciated.

答:
Welcome to MSE!

Here's a hint:

For every formula $\varphi$, the set $\varphi(S)$ is necessarily fixed by any automorphism of $S$ (do you see why?).

(Recall an automorphism of $S$ is a bijection $\mathbb{R} \to \mathbb{R}$ which preserves $0$ and $+$)

So to show $\{ (a,b) ~|~ a < b \}$ is not $\varphi(S)$ for any $\varphi$, we can show that some automorphism doesn't fix it!
With that in mind, can you find an automorphism $\sigma$ of $S$ and some points $a$ and $b$ so that

$a < b$

$\sigma a \not < \sigma b$?

This would give you the contradiction that you're looking for.

I hope this helps ^_^

Title: The non-existence of a logical formula


问:
Prove that $\forall a \in \mathbb{R} \ , \mathbb{E}\left\{|\boldsymbol{X}-\boldsymbol{a}|^{p}\right\}<\infty$ with conditions.
Let $ p\geqslant1 $ and b be a real number. Having $\mathbb{E}\left\{|\boldsymbol{X}-\boldsymbol{b}|^{p}\right\}<\infty$, I would like to prove that
$$\forall a \in \mathbb{R} \ \ \ , \mathbb{E}\left\{|\boldsymbol{X}-\boldsymbol{a}|^{p}\right\}<\infty$$

I thought of using the moment generating function. But I can't seem to see a path to the solution with that idea.

Any suggestions??

Body:
$HINT$

$$|X-a|^p\leq (|X-b|+|b-a|)^p$$ $$ \leq2^p\max\{|X-b|^p,|b-a|^p\} \leq 2^p(|X-b|^p+|b-a|^p)$$


问:
3D line integral question
Let $C$ be the curve of intersection of the cylinder $x^2 + y^2 = 1$ and the surface $z = xy$, oriented counterclockwise around the cylinder. Compute the integral $\int_C y\,dx + z\,dy + x\,dz$.</p>

答:
To do line integrals over curves in any dimension, we must first parametrize the curve with some parameter and then express the integrand and differentials in that parameter. Basically perform a variable substitution. This reduces the integral to a one dimensional integral.

The curve C
  lies on the cylinder with unit radius described by the equation, $x^{2}+y^{2}=1$
 . This cylindrical geometry calls for use of the polar coordinates $\left(r,\theta,z\right)$
 .
$$
x = r\cos\theta,
y = r\sin\theta,
z = z
$$
 The cylinder equation tells us how x
  and y
  relate to each other and the surface equation tells us how z
  is related to x,y
 . In terms of the polar coordinates we can now write the curve C
  as the vector parametrized by the parameter $\theta$
  [which is conveniently the polar coordinate of azimuthal angle].
$$
\vec{C}=\left(\cos\theta,\sin\theta,\cos\theta\sin\theta\right)
$$

,r=1
  on the cylinder.

Now, the differentials will be changed to:
$$
dx = d\left(\cos\theta\right)=-\sin\theta d\theta,
dy = d\left(\sin\theta\right)=\cos\theta d\theta,
dz = d\left(\cos\theta\sin\theta\right)=-\sin^{2}\theta d\theta+\cos^{2}\theta d\theta
$$

The integral now becomes:
$$
\int_{0}^{2\pi}-\sin^{2}\theta d\theta+\cos^{2}\theta\sin\theta d\theta+\cos\theta\left(\cos^{2}\theta-\sin^{2}\theta\right)d\theta =
$$
$$
-\frac{1}{2}\int_{0}^{2\pi}\left(1-\cos2\theta\right)d\theta-\int_{0}^{2\pi}\cos^{2}\theta d\left(\cos\theta\right)+\int_{0}^{2\pi}\left(1-2\sin^{2}\theta\right)d\left(\sin\theta\right)
$$

Where we have used the trig identities:
$$ \sin^{2}\theta = \frac{1-\cos2\theta}{2},
\cos^{2}\theta+\sin^{2}\theta = 1
$$

and the parameter $\theta$
  runs through a full $2\pi$
  rotation.

Title: 3D line integral question


答:
For a number $n>2$, if there is no prime $p$, $1 < p \leq{\sqrt{n}}$ s.t. $p | n$, then there can't be any prime $p'$,  $n > p'>\sqrt{n}$ that divides $n$. (Let's not consider the trivial divisors 1 and $n$)

Let's prove the above statement by contradiction.

Let's assume, to the contrary that there exists such a prime $p'>\sqrt{n}$, s.t. $p' | n$. It implies there exists another number $q$ s.t. $p'.q=n$, i.e., $q=\frac{n}{p'}<\sqrt{n}$.

Now, if $q$ is a prime then it's readily a contradiction, since it divides $n$, contrary to the assumption we started with.

If $q$ is not a prime then it can be written as product of smaller primes $q_1q_2...$ by the prime number theorem. All of them divides $n$ and are smaller than $q$, hence smaller than $\sqrt{n}$, leading to contradiction again.

Hence, such a $p'$ can't exist, which implies there is no prime $1<p<n$ that divides $n$, which implies $n$ is prime and it completes the proof.


问:
Discrepancy in Proof of π's Transcendence
Consider the following:

"From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{iπ}$ is transcendental.

However, from Euler's Identity:
$$e^{iπ} = −1$$

which is the root of $h(z) = z+1$

and so is algebraic. This contradicts the conclusion that $e^{iπ}$ is transcendental. Hence by Proof by Contradiction it must follow that $π$ is transcendental."

Monic, minimal polynomial expressions are of degree two. When no such polynomial exists for an irrational number it is deemed transcendental. Note that $e^{iπ} = −1$ is the root of a linear equation: $h(z) = z+1$. How is this possible? Logically, $e^{iπ} = −1$ should be the root of a quadratic equation in order for the contradiction to be warranted. Doesn't this call into question the proof of $π$ transcendence?

答: Not really. It is not true that all minimal monic polynomials are of degree $2$. Precisely, $x+1$, the minimal polynomial of $-1$, has degree $1$. There are minimal polynomials of any degree.


问:
What do we actually rotate with rotational matrices
We just learned about rotational matrices in a more theoretical context, but we only "glanced" at them.

My question: What exactly can a matrix rotate (vectors, other matrices, etc...?) and I still don't really get it how it actually rotates another object. Can somebody please give me a simple explanation (it can also be rigorous).

答:
To begin with, here is a bit of background about how you should think about linear transformations in general. Every $2 \times 2$ matrix has two pieces of information encoded in it: where the $\underline{i}$ vector lands, and where the $\underline{j}$ vector lands. Thus, in the below matrix, we know that $\underline{i}=(1,0)$ is mapped to $(a,b)$, and $\underline{j}=(0,1)$ is mapped to $(c,d)$.
\begin{bmatrix}
a & c \\
b & d \\
\end{bmatrix}
Linear transformations are essentially transformations that can be described entirely by where the $\underline{i}$ and $\underline{j}$ vectors land. If you multiply the above matrix with a $2 \times 1$ matrix, then you get
$$
\begin{bmatrix}
a & c \\
b & d \\
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
x
\begin{bmatrix}
a \\
b
\end{bmatrix}
+
y
\begin{bmatrix}
c \\
d
\end{bmatrix}
=
\begin{bmatrix}
ax + cy \\
bx + dy
\end{bmatrix}
$$
Since the $\underline{i}$ vector is mapped to $(a,b)$, let $\underline{i_{\text{new}}}=(a,b)$. Similarly, let $\underline{j_{\text{new}}}=(c,d)$. What I'm trying to emphasise here is that when you do any linear transformation, you move the basis vectors $\underline{i}$ and $\underline{j}$  to $\underline{i_{\text{new}}}$ and $\underline{j_{\text{new}}}$. The point $(x,y)=x\underline{i}+y\underline{j}$ is mapped to $x\underline{i_{\text{new}}} + y\underline{j_{\text{new}}}$, hence why each linear transformation can be described entirely by where the $\underline{i}$ and $\underline{j}$ vectors land. Notice how $(x,y)$, which is a linear combination of the $\underline{i}$ and $\underline{j}$ vectors, is mapped to the same linear combination of $\underline{i_{\text{new}}}$ and $\underline{j_{\text{new}}}$.

Now, let's consider the rotation matrix $R_{\theta}$ that represents a rotation of $\theta$ radians anticlockwise about the origin. $\cos\theta$ and $\sin\theta$ are, by definition, the $x$- and $y$-coordinates on a circle that you land on after making an angle of $\theta$ with the positive $x$-axis, meaning that
$$
\underline{i_{\text{new}}}=(\cos\theta,\sin\theta) \, .
$$
Furthermore,
$$
\underline{j_{\text{new}}}=(\cos(\theta + \pi/2),\sin(\theta + \pi/2)) \, ,
$$
as the $\underline{j}$ vector is at right angles to the $\underline{i}$ vector. Since the derivatives of $\cos\theta$ and $\sin\theta$ are, respectively, $\cos(\theta + \pi/2)$ and $\sin(\theta + \pi/2)$, the formula for $\underline{j_{\text{new}}}$ can be simplified to
$$
\underline{j_{\text{new}}}=(-\sin\theta,\cos\theta) \, .
$$
Thus, $R_{\theta}$ is given by
$$
R_{\theta} =
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
$$
Since rotation is a linear transformation, we can describe its effects by considering where the $\underline{i}$ and $\underline{j}$ vectors land. This means that we can rotate any point $(x,y)$ by $\theta$ radians about the origin by multiplying $R_{\theta}$ with the vector $x\underline{i} + y\underline{j}$:
$$
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
x
\begin{bmatrix}
\cos \theta \\
\sin \theta
\end{bmatrix}
+
y
\begin{bmatrix}
-\sin\theta \\
\cos \theta
\end{bmatrix}
=
\begin{bmatrix}
x\cos\theta - y\sin \theta \\
x\sin \theta + y\cos \theta
\end{bmatrix}
$$
For an example of when this might be useful, consider the 'negative reciprocal' rule: if a line has a gradient of $m$, then the line perpendicular to that line has a gradient of $-1/m$. This can be proven using matrices. Consider the line $l_1 $that joins the origin with the point $(x,y)$, and thus has a gradient of $y/x$. The line $l_2$, which is perpendicular to $l_1$, makes an angle of $\pi/2$ radians with $l_1$. Hence, we can map $l_1$ to $l_2$ using the rotation matrix
$$
R_{\pi/2} =
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
$$
Again, the entries of the above matrix can be worked out by considering where the basis vectors land. Now, if we perform a $\pi/2$ rotation on $(x,y)$, look at what happens:
$$
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
x
\begin{bmatrix}
0 \\
1
\end{bmatrix}
+
y
\begin{bmatrix}
-1 \\
0
\end{bmatrix}
=
\begin{bmatrix}
- y \\
x
\end{bmatrix}
$$
The point $(x,y)$ is mapped to $(-y,x)$, which has a gradient of $-x/y$. Of course, this only proves the rule when the two lines intersect at the origin, but it's easy to visualise that wherever the two lines intersect, this negative reciporocal relationship holds.

For a more sophisticated application of rotation matrices, you can use them to prove the addition formulae for $\cos$ and $\sin$:
\begin{align}
\cos(\alpha + \beta) &= \cos\alpha\cos\beta - \sin\alpha\sin\beta \\
\sin(\alpha + \beta) &= \sin\alpha\cos\beta + \cos\alpha\sin\beta \, .
\end{align}
How? Well, for any two $2 \times 2$ matrices $A$ and $B$, their product $AB$ is the linear transformation described by applying $B$, then applying $A$. The order of operations is similar to function notation, where $f \circ g$ means 'apply $g$, then apply $f$'. Clearly then, $R_{\alpha}R_{\beta} = R_{\alpha + \beta}$, since a rotation of $\beta$ radians, followed by an alpha of $\alpha$ radians, is surely the same as a rotation by $\alpha + \beta$ radians. But, if we multiply out $R_{\alpha}$ and $R_{\beta}$ the long way, we obtain formulae for these rotations written in terms of $\cos\alpha$, $\sin\alpha$, $\cos\beta$, and $\sin\beta$; equating these two leave you with the addition formulae. I'll leave this for you to work out. Please let me know how you get on, and I'll help you with any questions you have.


问:
Is p.l.(piecewise linear) map continuous?
I'm reading Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology", and I tried to show that the composition of two p.l. maps is p.l. (this is an exercise 1.6(2)) , but I couldn't. I think if any p.l. map is always continuous, then I can show that. So what I want to ask is p.l. map is always continuous? And if it doesn't hold, how can I show that the composition of two p.l. maps is p.l.?

(These are the definitions in the book:

For $a\in \mathbb{R}^m, L\subset\mathbb{R}^m$, $aL:=\{(1-t)a+tb; b\in L, t\in[0,1]\}$ is a cone if each point $x\in aL$ which is not $a$ is uniquely expressed by the form $(1-t)a+tb$ ($b\in L, t \in [0,1]$).

$P\subset\mathbb{R}^m$ is a polyhedron if each point $a\in P$ have its cone neighborhood $N=aL\subset P$ with $L$ compact.

$f:P\to Q$ ($P\subset\mathbb{R}^m,Q\subset\mathbb{R}^n$ are polyhedra) is p.l. if each point $a\in P$ have its cone neighborhood $N=aL$ ($L$:compact) such that $f((1-t)a+tb)=(1-t)f(a)+tf(b)$ for any $b\in L,t\in[0,1]$.)

答: Yes, PL maps are always assumed to be continuous. If you read page 1 of their book,  they say "a map is a continuous function." It is very common among topologists to use the word "map" only for continuous maps.</p>


问:
Need help understanding Hilbert's Nullstellensatz (weak version)
The theorem is stated in my lecture notes as follows:

Every proper ideal $\mathcal{a}$ of $k[x_1 , \dots , x_n]$ admits a point in $V(\mathcal{a};l)$ over some algebraic field extension l/k. In particular, if $k$ is algebraically closed, then $l=k$.

I am having a hard time understanding this last part.

For example, let $k = \mathbb{C}$, let $S = \{ x^2 + 1 , x^2 - 1 \} \subset \mathbb{C}[x]$, and let $a_S$ be the ideal generated by $S$. In this case, $\mathbb{C}$ is algebraically closed, so the theorem says there should be a point in $\mathbb{C}$ satisfying both $x^2 + 1$ and $x^2 - 1$. However, these polynomials have no common roots.

Where am I going wrong here?

答:
The ideal $a_S$ contains $x^2+1$ and $x^2-1$, so it contains $1=\frac{1}{2}((x^2+1)-(x^2-1))$ and hence $a_S = \mathbb{C}[x]$ is not a proper ideal, so the theorem (as stated) doesn't apply.

In this case (where the ideal is the whole ring), points in $V(\mathbb{C}[x],\mathbb{C})$ are also solutions to the equation $1=0$ (by the above expression for $1$), so there are no points: $V(\mathbb{C}[x],\mathbb{C}) = \varnothing$.


答:
Firstly, the formula as written is false. On the left hand side you take a limit as $x$ approaches $0$, yet on the right hand side you have a function of $x$. The correct formula is that for all $x\in \mathbb R$ with $|x|<1$ it holds that $$\sum_{n=1}^\infty x^n=\frac{x}{1-x}$$

which is known as the sum of a geometric series (it actually holds true for all complex numbers $z$ with $|z|<1$ and the formula fails for any number with absolute value bigger than or equal to $1$). Here is an informal argument (that can be turned completely formal) for proving it: believing (this is the informal part) that the series actually converges, say to $L$, we have that (with a bit more formal stuff lurking around) $$(1-x)L=(1-x)\cdot \sum_{n=1}^\infty x^n=\frac{x}{1-x}=\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty x^{n+1}=x$$ from which the formula follows. There are other ways of showing the formula holds, and, of course, they all require some amount of limit formalism.

One says that the function $\frac{x}{1-x}$ admits a power series representation with center $x=0$ and radius of convergence $1$.

Title: name of theorem for infinite order polynomial limit for small x

Body:
is there a name for this property?

$\mathop{\operatorname{lim}}_{x\to 0} \sum_{n=1}^\infty x^n = \frac{x}{1-x}$

I have seen it in a derivation but want to know where it comes from.


答:
As it was pointed out already, it is allowed in the theory for the minimal polynomial to be linear.

If you do not like this, you can reach a contradiction by observing that
$$
e^{ i \frac{\pi}{2}}=i
$$
is algebraic with minimal polynomial $x^2+1$. This implies that $\frac{\pi}{2}$ is transcendental, and it is easy to show that implies that $\pi$ must also be transcendental.

Title: Discrepancy in Proof of π's Transcendence

Body:
Consider the following:

"From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{iπ}$ is transcendental.

However, from Euler's Identity:
$$e^{iπ} = −1$$

which is the root of $h(z) = z+1$

and so is algebraic. This contradicts the conclusion that $e^{iπ}$ is transcendental. Hence by Proof by Contradiction it must follow that $π$ is transcendental."

Monic, minimal polynomial expressions are of degree two. When no such polynomial exists for an irrational number it is deemed transcendental. Note that $e^{iπ} = −1$ is the root of a linear equation: $h(z) = z+1$. How is this possible? Logically, $e^{iπ} = −1$ should be the root of a quadratic equation in order for the contradiction to be warranted. Doesn't this call into question the proof of $π$ transcendence?


问:
Why is $\sin(\tanh x) + \sinh(\tanh x)$ almost exactly $2\tanh x$?
I was trying to come up with some approximations for the solution to the differential equation $y'' + \operatorname{sgn}(y') + y = 0$ and noticed while I was messing around that $\sin(\tanh x) + \sinh(\tanh x)$ is almost exactly $2\tanh x$. Looking at the series expansions for $\sinh$ and $\sin$ it's not immediately obvious why this would be the case. It makes sense that high-order terms should converge to a constant value for large $x$, and that near $x=0$ they should be approximately zero. This seems to imply that for some intermediate values of $x$ around something like $x=0.5$ there should be disagreement, but this seems like a remarkably good fit for every value of $x$. Why might this be?

答:
We can ignore the $\tanh$ component momentarily and consider instead the function
$$f(x) = \frac{\sin x + \sinh x}{x}.$$  This has series expansion
$$f(x) \sim 2 + \frac{x^4}{60} + \frac{x^8}{181440} + O(x^{12}).$$  Therefore, on the open interval $x \in (-1,1)$, $f(x)$ has exceptionally small deviation from $2$, thus on the real line, $$g(x) = \frac{\sin \tanh x + \sinh \tanh x}{2 \tanh x}$$ will be very close to $1$.

In other words, it is not simply that $\sin x \approx x$ and $\sinh x \approx x$, but that these approximations are in a sense opposing, thus terms cancel out in the series expansion:

$$\sin x \sim x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + O(x^9) \\
\sinh x \sim x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + O(x^9) \\
$$
so their sum cancels up to $O(x^4)$, which after dividing by $x$, gives the first nonzero error term being $O(x^4)$.


问:
Making sense of measure-theoretic definition of random variable
I have some understanding of measure theory / real analysis, and some understanding of probability theory, but I'm having some trouble putting the two together.

According to Wikipedia:

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $(E, \mathcal{E})$ a measurable space. Then an $(E, \mathcal{E})$-valued random variable is a function $X : \Omega \to \mathcal{E}$ which is $(\mathcal{F}, \mathcal{E})$-measurable.

Now for example, let's take $X$ be a standard Gaussian random variable, $X \sim \mathcal{N}(0, 1)$.

I think $E = \mathbb{R}$ since $X$ takes values in $\mathbb{R}$.

Also, we should have $\mathcal{E} = \mathscr{B}(\mathbb{R})$ the Borel $\sigma$-field of $\mathbb{R}$.

But, what should $(\Omega, \mathcal{F}, P)$ be?

Furthermore, let's try to calculate $\mathbb{E}[X]$ the mean of $X$. By Wikipedia's definition,

$$\mathbb{E}[X] = \int_\Omega X\, dP = \int_\Omega X(\omega)\, P(d\omega).$$

This raises some questions.

How does this relate to the elementary computation:
$$\mathbb{E}[X] = \int_{-\infty}^{\infty} x\cdot f_X(x)\, dx$$
How does $f_X : \mathbb{R} \to \mathbb{R}^{\geq 0}$ relate to the measure-theoretic definition of $X$?

What is the meaning of $P(d\omega)$? $P$ is a measure so it makes sense to integrate $dP$, but what is $d\omega$?

答:
You can use the following derivations:

$\int X dP = \int t  dPX^{-1}(dt) = \int t dF = \int t f_{X} dx$ where F is the distribution function for $X$ and $PX^{-1}$ is the pull back measure given by $PX^{-1}(B)=P(X^{-1}(B))$.


问:
Does abelianization change the rank of $\pi_1(X)$?
I am looking at the relation between the first homology group $H_1(X)$ and the fundamental group $\pi_1(X)$. Given a path-connected space $X$, I have some grasp of what $\pi_1(X)$ should be intuitively (as far as I am aware, it can be informally be thought of as the number of 'holes' in the space). Thus, it would be a very easy to deduce $H_1(X)$ if I knew that taking the abelianization $\pi_1(X)_{ab}$ does not change the rank of the fundamental group.

As an example, $S^1 \vee S^1$ has two 'holes', and so intuitively one might think that $\pi_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}$. It turns out also that $H_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}$, so they have the same rank.

My question is, does this always hold? Is there some example where the rank of $H_1(X)$ and $\pi_1(X)$ are different?

答: If $G$ is a group, then $G$ and $G_{ab}$ need not be of the same rank. For example, $A_5$ has rank 2, while its abelianization is the trivial group, with rank 0. Every group is the fundamental group of some path-connected space, so this shows there is a space $X$ such that $H_1(X)$ and $\pi_1(X)$ have different ranks.</p>


问:
the proof of faithfully flat module is faithful
I know faithful and flat module is not always faithfully flat.
For example $\mathbb Q \otimes_{\mathbb Z} \mathbb Z/2\mathbb Z=0$ implies $\mathbb Q$ is not faithfully flat as $\mathbb Z$-module unless it is faithful and flat as $\mathbb Z$-module.

But is the converse true?

I want to know the proof of

faithfully flat module is faithful and flat（flatness is obvious）

答:
Consider the annihilator $I$ of a flat module $M$ and the commutative diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> I\otimes_R M @>>> R\otimes_R M @>>> R/I\otimes_R M @>>> 0 \\
@. @VfVV @VgVV @VhVV \\
0 @>>> IM @>>> M @>>> M/IM @>>> 0
\end{CD}
$$
The obvious morphisms $g$ and $h$ are isomorphisms even if $M$ is not flat; if $M$ is flat, also $f$ is an isomorphism. Since $IM=0$ by definition, we have $I\otimes_RM=0$ and, if $M$ is also faithfully flat, this implies $I=0$.


问:
World's smallest (Exception-based) proof-checking language (in Python). In what way is Coq, Lean, Isabelle "better"?
As you may already know, the codebases written for Coq, Lean, etc. are humungous.  Thus a goal in a new proof assistant might be to simplify things.  Suppose we restrict our attention not to general proof theory but to categorical contexts.   The closest concept I can think of would be a "CAS for category theory (with proofs)".

```<code>C = Cat('C')
X = C.ob('X')
i = Id(X, 'i')
j = Id(X, 'j')

with Proof(goal=Eq(i,j), given=[C, X, i, j]) as p:
    eq1 = p.equals(i, j(i))
    eq2 = p.equals(j(i), j)
    p.QED(p.transitivity(eq1, eq2))

print(p.proposition_text())
print(p.proof_text())
</code>```

The idea is that print will present precisely the following text to the user:

Prop. Let $C$ be a category, $X$ an object in $C$, and $i,j$ identity
morphisms on $X$.  Then $i = j$.

Proof.
$$i = i\circ j \tag{1}$$
$$i\circ j = j \tag{2}$$
Therefore, $i = j$ by transitivity of equality and (1), (2)  $\blacksquare$

Thus the most basic proof in Category can be given in English using classes Ob, Cat, Arrow, Id, Proof, and Eq.    The methods of the proof instance p encode how to work with and return equalities.

Thus a proof of a statement is valid if and only if we can construct the statement's goal data.

Questions.  Would you consider using such a python library as a student of category theory?  How would Coq, Lean, or Isabelle be superior on this particular proof?  The last question is so that I can make changes to the design, so that it is on par with the other proof assistants.

How does proof checking work?  Each method call on the p object will perform a run-time assert of the proposed equality against the given equalities.  Inclusion of i,j into the givens automatically includes their definitional equalities.  Should any assertion fail, the mathematician (or coder / you) will be informed in their IDE of the AssertionFailure.  So that is the proof-checking  mechanism - it makes use of the exception system within a programming language, which is in this case Python's.

If the with block is exited without a successful call to p.QED then a ProofError will be thrown (another exception).  The data of the argument to QED must equal p.goal.

Simply doing p.equals(i,j) from the start will result in a ProofError, since the reference i does not equal the reference j.  That is to say, a real proof environment is indeed formed - a.k.a. you can't cheat on a proof.

答:
This thing would be possible. I really struggle to believe it would be remotely as ergonomic as e.g. Lean, though. Phrasing your proofs as types in a strong type system gives the compiler perfect knowledge, and allows it to be extremely helpful (if you haven't experienced Agda or Idris's hole-filling, it is actually miraculous). The extremely dynamic nature of Python means you are crippled from the start. If the user promises they will stick only to using your built-ins, then sure, you could get an experience like Lean's or Coq's where the interactive proof assistant tells you all the information available, but you're embedded in Python where anything goes, so it's quite a bit of trust you're putting on the user.

Additionally, Python is based around mutable state. Your proof terms are inherently dynamic, and the fact that your language is running inside Python means you are inviting the user to knit together proof terms using mutability. I do not believe this will lead to readable or maintainable proofs, and it also makes the job of the kernel a lot harder: if I construct a proof term in function f, then I use it in proof h, then I mutate it in function g, you're going to have to track this somehow so that you can re-evaluate the construction of h. This sounds like a pretty big undertaking to me, not that I'm deeply familiar with Python's reflection APIs. An immutable language rules out this entire can of worms.


答:
Answering to " ..it can rotate also polynomial functions?"

Since $\bf R_{\varphi}$ is orthogonal , it preserves the dot product, hence the distances and the angles.

so if you apply it to the vector
$$
\left( {\matrix{ x  \cr  {f(x)}  \cr  } } \right)\quad
 or\quad \left( {\matrix{  {x(t)}  \cr {y(t)}  \cr } } \right)
$$
you get a rotated representation of the graph of $y(x)$ or of the curve $(x(t),y(t))$ which preserves the distances (lengths, derivatives,angles, integrals, ..).

Title: What do we actually rotate with rotational matrices

Body:
We just learned about rotational matrices in a more theoretical context, but we only "glanced" at them.

My question: What exactly can a matrix rotate (vectors, other matrices, etc...?) and I still don't really get it how it actually rotates another object. Can somebody please give me a simple explanation (it can also be rigorous).


问:
Convergence of the integral of $\frac{\cos(x)}{x^2}$
For $n$ in the natural numbers let

$$a_n = \int_{1}^{n} \frac{\cos(x)}{x^2} dx$$

Prove, for $m ≥ n ≥ 1$ that $|a_m - a_n| ≤ \frac{1}{n}$ and deduce $a_n$ converges.

I am totally stuck on how to even go about approaching this. All help would be very gratefully received!

答:
We only need to show $\{b_j\}$ is Cauchy, where $b_j=\int_{1}^{j} \frac{\cos x}{x^2} dx$

Proof: $b_j-b_k=\int_{1}^{j} \frac{\cos x}{x^2} dx-\int_{1}^{k} \frac{\cos x}{x^2} dx=\int_{k}^{j} \frac{\cos x}{x^2} dx$

That is,
$$|b_j-b_k|=|\int_{k}^{j} \frac{\cos x}{x^2} dx| \leq \int_{k}^{j} |\frac{\cos x}{x^2}| dx$$

(Recall that $|\int_{a}^{b}f(x) dx| \leq \int_{a}^{b}|f(x)| dx$)

We continue our above analysis,
$$\leq \int_{k}^{j}\frac{1}{x^2}  dx=-\frac{1}{x}\Biggr|_{k}^{j}= -\frac{1}{j}+\frac{1}{k}<\frac{1}{k}$$

Thus,
$$|b_j-b_k|<\frac{1}{k}<\epsilon$$,whenever $k>\frac{1}{\epsilon}$, viz., $k>N,N=[\frac{1}{\epsilon}]+1$

To summarize, $\forall \epsilon>0$, let $N=[\frac{1}{\epsilon}]+1$, then whenever $j>k>N$, we have $|b_j-b_k|<\frac{1}{k}<\epsilon$.


问:
Question about improper integral
Does $\forall \epsilon \in (0,a]: \int _{\epsilon}^{a} f(x)dx =0$ imply that $\lim_{h \rightarrow 0} \int _h^a f(x)dx=0$. I guess it is somehow the definition of it, but I need to know this exactly. </p>

答:
Hint

If we have a function $f$ s.t.
$$f(x)=0\quad \forall x\in(0,a]$$
then what's $\lim_{x\to0}f(x)$?

Title: Question about improper integral


问:
Prove that $\lim_{x\to0}f(x)g(x)$ does not exist
There are already some questions concerning problem 23 of the fifth chapter of "Calculus", but they mostly focus on part (c). I altered the question so that it contains all the information of the context. (Spivak Calculus Chapter 5 Problem 23 (b))

23 (b) Suppose that $\lim_{x\to 0} |f(x)| = \infty$. Prove that if $\lim_{x\to0}g(x)$ does not exist, then $\lim_{x\to0}f(x)g(x)$ also does not exist.

The given answer:

Clearly, if $\lim_{x\to0}f(x)g(x)$ exists, then $\lim_{x\to0}g(x)=0$.

Unfortunately this is not clear to me, so I tried to work out a proof for this problem.

Proof by contradiction:

$\lim_{x\to 0} |f(x)| = \infty$. Thus, for every $N\in\mathbb{R}_{>0}$ there is a $\delta>0$ such that,

$\textrm{for every $x$ with  } 0<|x-0|=|x|<\delta,\quad|f(x)|>N, \tag{1}$

Suppose $\lim_{x\to0}f(x)g(x)=m$. Then for every $\varepsilon>0$ there is a $\delta_1>0$ such that,

$\textrm{for every $x$ with  } 0<|x-0|=|x|<\delta_1,\quad|f(x)g(x)-m|<\varepsilon. \tag{2}$

Considering all $x$ satisfying $0<|x|<min(\delta,\delta_1)$ it follows from $(1)$ and $(2)$ that $$|f(x)g(x)|\left(\frac{1}{|f(x)|}\right)<(\varepsilon+|m|)\left(\frac{1}{N}\right),$$ which is$$|g(x)|=|g(x)-0|<\frac{\varepsilon+|m|}{N}.$$ Thus $\lim_{x\to0}g(x)=0$, a contradiction!

Does this proof work? Is it unnecessary to prove this statement?

答:
I think it's worth proving and your proof makes sense to me.

Your proof shows that if both $|f(x)| \to \infty$ and $f(x)g(x) \to m$ then we must have $g(x) \to 0$.

If $\lim_{x\to 0} g(x) \neq 0$, the other two statements cannot both be true. So, if $|f(x)| \to \infty$ but $g(x) \not\to$ 0, then the limit for their product $f(x)g(x)$ does not exist.

Spivak's "clearly" combined with not deigning to give a proof is a little obnoxious.

Here are 2 similar proofs showing if both $|f(x)| \to \infty$ and $f(x)g(x) \to m$ then $g(x) \to 0$ :

Proof 1 (by contradiction)

Show that if $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} f(x)g(x) = \ell$, then $\lim_{x \to a} g(x) = 0$.

Suppose $\lim_{x \to a} g(x) \neq 0$

Then there must exist some $\varepsilon_g > 0$ such that for all $\delta > 0$ there is some $x$ such that

$$0<|x-a| < \delta \text{ but } |g(x)| > \varepsilon_g$$

(that is, we can always find some $x$ arbitrarily close to $a$ such that $|g(x)| > \varepsilon_g$).

Because we can make $|f(x)|$ arbitrarily large, we can restrict ourselves to $x$ close enough to $a$ such that $|f(x)| > \frac{|\ell| + \varepsilon_g}{\varepsilon_g}$.

This means, for any $\delta > 0$ there is some $x$ with

$$0 < |x-a| < \delta \text{ but } |f(x)g(x)| > \frac{|\ell| + \varepsilon_g}{\varepsilon_g} \cdot \varepsilon_g=|\ell| + \varepsilon_g$$

Therefore

$$|f(x)g(x) - \ell| \geq |f(x)g(x)| - |\ell| > \varepsilon_g$$

which contradicts $\ell$ being the limit of $f(x)g(x)$.

Proof 2 (not by contradiction)

Suppose  $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} f(x)g(x) = \ell$.

For any $M > 0$ there exists some $\delta_M > 0$ such that for all $x$ if
$$0 < |x-a|< \delta_M \text{ then } |f(x)| > M$$

The limit for $f(x)g(x)$ tells us that for any $\varepsilon > 0$ there exists some $\delta_{\ell} > 0$ such that for all $x$ if

$$0 < |x-a|< \delta_{\ell} \text{ then } |f(x)g(x) - \ell| < \varepsilon$$

Now, for any $\varepsilon$ we can set $M = \frac{\varepsilon + |\ell|}{\varepsilon}$ and pick $\delta_{min} = \min(\delta_M, \delta_{\ell})$.

So, for all $x$ if

$$0 < |x-a| < \delta_{min} \text{ we have both } |f(x)g(x) - \ell| < \varepsilon$$

$$\text{and } |f(x)| > M = \frac{\varepsilon + |\ell|}{\varepsilon}$$

$$|f(x)g(x)| - |\ell| \leq |f(x)g(x) - \ell| < \varepsilon$$

$$|f(x)g(x)| < |\ell| + \varepsilon$$

$$|g(x)| < \frac{|\ell| + \varepsilon}{|f(x)|} < \frac{|\ell| + \varepsilon}{M} = \frac{|\ell| + \varepsilon}{\frac{|\ell|+\varepsilon}{\varepsilon}} = \varepsilon$$

Thus, for any $\varepsilon > 0$ there is a $\delta_{min}$ such that for all $x$ if
$$0 < |x-a| < \delta_{min} \text{ then } |g(x)| < \varepsilon$$

or
$$\lim_{x \to a} g(x) = 0$$


问:
Prove for non-negative RV , $\mathbb{E}\{\boldsymbol{X}\}$ is finite iff $\sum_{n=0}^{\infty} \mathbb{P}(X>n \epsilon)$ converges for all $\epsilon>0$
I want to know how to prove that for a non-negative random variable X,  its first moment is finite ($\mathbb{E}\left\{X\right\}<\infty$ ) if and only if
$$
\sum_{n=0}^{\infty} \mathbb{P}(X>n \epsilon) \text { converges for all } \epsilon>0
$$
I tried to upperbound and lowerbound the expectation with quantities related to $\sum_{n=0}^{\infty} \mathbb{P}(X>n \epsilon)$ but with no result.

Any ideas??

答: Note that $\sum_{n=0}^{\infty}P(X>n\epsilon) \leq \sum_{n=0}^{\infty}P(\frac{X}{\epsilon} \geq n) \leq \frac{1}{\epsilon}\Bbb{E}[X]<+\infty$


问:
The category $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian is $\sf{A}$ is abelian and $\sf{C}$ is small. Why the set-theoretic condition?
Proposition 1.4.4 in F. Borceux Handbook of Categorical Algebra II says that if $\sf{A}$ is an abelian category and $\sf{C}$ is a small category, then the category of functors $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian.

I have two questions:

Is this true without demanding $\sf{C}$ to be small? The only place where Borceux proof seem to demand this condition is where he uses that in this case limits and colimits are computed pointwise. But I think I know how to prove that products, kernels and cokernels are computed pointwise without demanding $\sf{C}$ to be small.

If the theorem is true without this condition, then why aren't limits and colimits computed pointwise in functor categories without this assumption?

答:
It is "merely" a problem of formalities.
Borceux mentions both the universe axiom and Gödel-Bernays (NBG) set theory as solutions to the problem of working with large categories, but they are not equally powerful.
Borceux's definition of "category" requires that the collection of objects is a class and that the collection of all morphisms between any two objects is a set – so he is implicitly working in NBG.
Thus, in this definition, there are two obstructions to forming the category of all functors $\mathcal{A} \to \mathcal{B}$ when $\mathcal{A}$ is not small:

The collection of all functors $\mathcal{A} \to \mathcal{B}$ is so big that it is not even a class!
If it were a class, then a functor $\mathcal{A} \to \mathcal{B}$ would be a set, since a set is defined to be a collection that is a member of some class.
But then we could use the axiom of replacement to deduce that the class of objects of $\mathcal{A}$ is a set – contradiction.

The collection of natural transformations between two functors $\mathcal{A} \to \mathcal{B}$ is also too big to be a class, let alone a set, by the same argument.

That said, these are not serious problems.
If one uses the universe axiom instead of NBG then there is a more appropriate definition of "category" that permits the formation of functor categories between any two categories.
All the basic theorems hold, so if $\mathcal{B}$ has limits of a certain shape then limits of that shape exist and are computed componentwise in the category of all functors $\mathcal{A} \to \mathcal{B}$, etc.
In particular, if $\mathcal{B}$ is an abelian category then the category of all functors $\mathcal{A} \to \mathcal{B}$ is also abelian.


问:
Intuitively understanding why the volume of a cylinder of height $h$ and radius $R$ is $\pi R^2h$
It's well known that a cylinder with height $h$ and the base radius $R$ will have a volume of

$$V = \pi R^2h$$

I'm trying to derive an intuitive understanding of it. I have two ways, one seems to be "working" (as in - it yields the correct result) and another one is failing, see below.

Method 1

If we agree that the area of a circle is known as $\pi R^2$ then we can think of a volume as a sum of all "areas of circles along the heights". So if we imagine that there is such an "object" that has a height of "one point" then its volume is equal to the area of the circle it has as a base. We then see that there will be exactly $h$ points in the height of the cylinder, so there will be exactly $h$ such "objects". Hence we can just multiply a "volume" of one "object" by $h$ which will yield the correct cylinder volume. I believe this method is just a "naive" version of a more rigorous calculus approach with defining the limit for $dx$ where $x$ will be along the height of the cylinder

Method 2

Now let's look at the cylinder as a rectangle rotated along one of its sides. If we have $h$ as the cylinder height and $R$ as a radius, then it's an $h \times R$ rectangle rotated by $2\pi$ around the side with length $h$. So the volume is the "sum" of all the "objects" with the height of "one point" and the base a rectangle. There are exactly $2\pi R$ points along the circumference of the circle that the rectangle will follow. Therefore we will have $2\pi R$ of these "objects". So we just need to multiple the "volume" of one such "object" by the number of these "objects" to get the cylinder volume. The "volume" of this "object" will be just the area of a rectangle, since it has a height of "one point". To conclude, we have $2\pi$ of these "objects", thus the volume must be $V = 2\pi R \times V_{\text{object}} = 2\pi R \times h R = 2\pi R^2h$.

But method 2 yields incorrect result, it's exactly $2$ times more than it should be. Why is that? I.e. why is that the point-based construction worked in the first case but failed in the second case? And how to make it work properly? What about rotation in general (i.e. not only a cylinder)?

答:
The problem with the second approach is that although the outer side of the rectangle travels a distance of $2\pi R$ during your rotation, the inner side travels a distance of $0$, so you're not creating as much volume as you would if you extruded the rectangle "straight out". In fact, since the distance traveled by a point on the rectangle is a linear function of its horizontal position, the average distance traveled is the just the average of the inner and outer values: $\pi R$. Multiplying this average by the area of the rectangle gives you the correct volume.

What you've discovered is that it's not safe to think of "one point" as a well-defined positive distance. The cylinder is indeed the disjoint union of a bunch of one-point-thick sheets as you've described, but in decomposing it like this you've lost information about the "relative sizes" of the different "zero" thicknesses at different places in the sheet.

Instead of these sheets, you should consider the thin wedge of the cylinder that corresponds to a very small change $d\theta$ in the rotation angle. When viewed from the top, this wedge looks like a thin triangle with base $R$ and height $R\times d\theta$, and therefore area $R^2d\theta/2$. As we sweep $\theta$ from $0$ to $2\pi$, the $d\theta$s add up to $2\pi$, so the wedge volumes $\frac{R^2h}2d\theta$ add up to $\pi R^2h$.


问:
Delta System Lemma: Kunen’s proof.
I'm trying to understand Kunen's Delta System Lemma proof (Set Theory: Introduction to Independence Proofs, Chapter II, Theorem 1.6).
I'm heaving issues on understanding the last line:

"Since $|\alpha_0^{ < k}|<\theta$ there is an $r \subset \alpha_0$ and $B\subset A_2$ with $|B|=\theta$ and $\forall x \in B(x\cap \alpha_0=r)$, whence $B$ forms a $\Delta$-system with root $r$."

I didn't understand the whole sentence: Why is there such r and such B? And why is it a Delta system?
I understood the previous parts of the proof, but I'm not understanding the ending :(

Adding more context:
We want to construct a $\Delta$ system $B\subset A$ such that $|B|=0$. At this point, We already have constructed a subset $A_2$ of $A$ such that $|A_2|=\theta$ and $x \cap y\subset\alpha_0 < \theta $ whenever $x$ and $y$ are distinct members of $A_2$.

PS: A $\Delta$-system is a set $A$ such that there exists $r$ such that $x \cap y=r$ whenever $x$ and $y$ are distinct members of $A$.

答:
At that point in the argument you know that $x\cap y\subseteq \alpha_0$ whenever $x$ and $y$ are distinct elements of $\mathscr{A}_2$. For each $x\in\mathscr{A}_2$ we know that $|x|<\kappa$, so $x\cap\alpha_0\in[\alpha_0]^{<\kappa}$. For each $z\in[\alpha_0]^{<\kappa}$ let $\mathscr{A}_2(z)=\{x\in\mathscr{A}_2:x\cap\alpha_0=z\}$. Then $$\mathscr{A}_2=\bigcup_{z\in[\alpha_0]^{<\kappa}}\mathscr{A}_2(z)\;.$$

$|\mathscr{A}_2|=\theta>|\alpha_0^{<\kappa}|$, and $\theta$ is regular, so $\theta$ is not the union of $|\alpha_0^{<\kappa}|$ properly smaller sets; thus, there must be some $r\in[\alpha_0]^{<\kappa}$ such that $|\mathscr{A}_2(r)|=\theta$.

Now take $\mathscr{B}=\mathscr{A}_2(r)$. For all $x\in\mathscr{B}$ we have $x\cap\alpha_0=r$, and $\{x\setminus\alpha_0:x\in\mathscr{B}\}$ is a pairwise disjoint family by the construction of $\mathscr{A}_2$, so $\mathscr{B}$ is a $\Delta$-system with root $r$.

Title: Delta System Lemma: Kunen’s proof.


问:
Property of Kulkarni-Nomizu product
I am reading about Weyl tensor in Introduction to Riemannian manifolds by John M. Lee. In Lema 7.22 (e), there is the next assertion.

Let $V$ be an $n$-dimensional vector space endowed with a scalar product $g$, let $h$ and $k$ be symmetric $2$-tensors on V, let T be and algebraic curvatura tensor on $V$, and let tr_g denote the trace on the first and las indices. Then $g(T,h \mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} g)=4g(tr_g T,h)$.

$\mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}}$ denotes de Kulkarni-Nomizu product.
I am a bit lost in my try of the proof, what is the meaning for the metric applied on $(0,4)$ covariant tensors?

答:
Inner products of tensors are explained in Proposition 2.40.

To get started on the proof, try the strategy I used in the proof of part (b): expand everything in terms of a basis.

Title: Property of Kulkarni-Nomizu product


答:
Well done for trying to get this intuition. The insight comes from calculus. It’s because in your 2nd method, you put too much material near the center.

A better approximation would be to use triangles (they are pointy at one end, and that point belongs near the center of the cylinder). Triangles’ areas are half that of the rectangle with same height and base, so that’s why you overestimated by a factor of two in the second method.

More rigorously, the area element in polar coordinates is $r \ dr \ d\theta$. You should let me know how much exposure to calculus before I explain further.

Title: Intuitively understanding why the volume of a cylinder of height $h$ and radius $R$ is $\pi R^2h$

Body:
It's well known that a cylinder with height $h$ and the base radius $R$ will have a volume of

$$V = \pi R^2h$$

I'm trying to derive an intuitive understanding of it. I have two ways, one seems to be "working" (as in - it yields the correct result) and another one is failing, see below.

Method 1

If we agree that the area of a circle is known as $\pi R^2$ then we can think of a volume as a sum of all "areas of circles along the heights". So if we imagine that there is such an "object" that has a height of "one point" then its volume is equal to the area of the circle it has as a base. We then see that there will be exactly $h$ points in the height of the cylinder, so there will be exactly $h$ such "objects". Hence we can just multiply a "volume" of one "object" by $h$ which will yield the correct cylinder volume. I believe this method is just a "naive" version of a more rigorous calculus approach with defining the limit for $dx$ where $x$ will be along the height of the cylinder

Method 2

Now let's look at the cylinder as a rectangle rotated along one of its sides. If we have $h$ as the cylinder height and $R$ as a radius, then it's an $h \times R$ rectangle rotated by $2\pi$ around the side with length $h$. So the volume is the "sum" of all the "objects" with the height of "one point" and the base a rectangle. There are exactly $2\pi R$ points along the circumference of the circle that the rectangle will follow. Therefore we will have $2\pi R$ of these "objects". So we just need to multiple the "volume" of one such "object" by the number of these "objects" to get the cylinder volume. The "volume" of this "object" will be just the area of a rectangle, since it has a height of "one point". To conclude, we have $2\pi$ of these "objects", thus the volume must be $V = 2\pi R \times V_{\text{object}} = 2\pi R \times h R = 2\pi R^2h$.

But method 2 yields incorrect result, it's exactly $2$ times more than it should be. Why is that? I.e. why is that the point-based construction worked in the first case but failed in the second case? And how to make it work properly? What about rotation in general (i.e. not only a cylinder)?


答: For method 2, those "objects" actually have dimensions of $\frac12\times r\times h\times d\theta$ (think triangular prism or cylindrical sector).  To calculate the volume, we can sum up the volume of these "objects" along the length of the circumference:  $$\int_0^{2\pi r}\frac12rh\,d\theta=\pi r^2h.$$