问:
Strategy for second-price auction with multiples
Suppose there is an auction website.  The rules are that the winner pays second-highest bid, and if a bid is made with less than 3 minutes left on the timer, the timer is reset to 3 minutes.

A seller is auctioning a large collection of items as many individual listings, but they all have the same closing time.  (Until and unless the 3-minute-bid rule is triggered.)  There are multiples of some items.  Suppose you want to buy one of the multiples of a given item, but only one.  Is there a game-theoretically sound way to guarantee that either you win exactly one item and pay no more than your max bid, OR that all items sell to other parties for more than your max bid?

In a single-item auction, you can meet the criteria simply by placing your max bid.  (I've read assertions that you should place your bid right before the end, to avoid emotional bidders who will exceed their max bid just to outbid you, but with the 3-minute-bid rule, that may not be applicable, and maybe isn't a cleanly game-theoretic consideration, anyway.)  However, in a duplicate-item scenario, that can fail in several ways:

If you put your max bid on multiple listings, you may end up accidentally winning multiple listings and exceeding your budget.

If you don't, then you might instead arbitrarily pick one item to go after.  Let's say for sake of example that all the items have one bidder on them, including you, and so all the items are tied in price.  There are a few last minute bids (say, by bidders not carefully watching for the cheapest copy, or perhaps there's some other combination of events that could make the resulting actions optimal for each involved agent), pushing your listing's timeout past when the other listings close.  You can't switch to a cheaper listing yet, because you might still win this one.  Then, say, a latecomer outbids you.  One way or another, you may find yourself outbid, and the other listings (which sold for cheaper than your max) are already closed.

There may be some tactic where you jump between listings, incrementally raising your bid so that you're only ever highest-bid on one listing at a time, and reducing (eliminating?) the chance you'll get stuck waiting for a single listing while another sells for cheaper.  Perhaps this allows you to keep listings open long enough to guarantee an acceptable outcome?  In any case, it is not clear to me that there exists such a strategy that guarantees you don't end up in one of the undesirable results.

(This question is not purely academic, haha, but I'm interested in the game-theoretic solution.)

The best I've found on google is this: http://answers.google.com/answers/threadview/id/503078.html
But it seems like it's addressing sequentially-sold listings, and so isn't quite applicable.

To reiterate:
Is there a game-theoretically sound way to guarantee that either you win exactly one listing (of multiple duplicates) and pay no more than your max bid, OR that all listings sell to other parties for more than your max bid?

答:
It can't be done.

You certainly can't guarantee that you'll win even a single listing, as there's always the possibility that the sale price is higher than your max bid on every listing.

You also can't ever bid on more than one listing at a time, or else you run the risk that you're topped by exactly one bid in both. If both of those first-price bids are higher than your max bid, there's nothing you can do, and you'll win both listings.

In order to prevent any listing from being sold for less than your max bid, you must bid on any listing that is about to be sold for less than your max bid. This can happen for multiple listings simultaneously, which would force you to submit bids on multiple listings. This violates #2, as you can't bid on multiple listings and not risk winning more than one of them.

Since this is a second-price auction, submitting your max bid can only make somebody else pay less than your max bid. There's absolutely nothing you can do to raise the final price above your max bid amount, unless you bid more than your max bid (which is a contradiction).

You can't guarantee that you win even a single listing. You also can't guarantee that you'll win any listing that does sell for less than your maximum bid, unless you risk the possibility of winning multiple listings. Finally, there's nothing at all you can do to force others to pay more than your maximum bid, since your max bid can only force them to pay less than that in a second-price auction.

So, there is no way to guarantee that you win any listing for less than your max bid, and also no way to guarantee that any listing sells for more than your max bid. It's entirely possible that every single listing sells for less than your max bid, and that you win none of them.


问:
Flipping an unfair coin 5 times
I'm flipping an unfair coin 5 times. The probability of getting heads is $\frac{2}{3}$ and the probability of getting tails is $\frac{1}{3}$. What's the probability that at least 3 of the coins end up being heads?

My working:

If I flip 5 times, there are $2^5 = 32$ possible outcomes.

I know that there are $16$ outcomes where there are at least 3 heads (1 with 5H, 5 with 4H, and 10 with 3H).

The probability of getting 0H is $\left(\frac{1}{3}\right)^5$. Can I extend this logic for the probability of getting 1H and 2H and simply sum those numbers and subtract from 1?

答:
A coin flip follows a binomial distribution.
\begin{align*}
P(H\geq3)&=P(H=3)+P(H=4)+P(H=5) \\
&=\binom{5}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2+\binom{5}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^1+\binom{5}{5}\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^0\\
&=\frac{64}{81}
\end{align*}

ContentLicense: CC BY-SA 4.0

Title: Flipping an unfair coin 5 times


问:
How to pass minimum count of apples so that every one around the table has same amount of apples?
With three ($A_1,A_2 and A_3)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=4 $ apples,to ensure each one has equal amount of 3 apples,  A3 just  gives A1 one apple A3 $\to$ A1 =1. In this case, the minimum count of ample exchanged is 1.

With four ($A_1,A_2,A_3 and A_4)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=6,a_3=9 $ apples, to ensure each one has equal amount of 5 apples,  A1 $\to$  A2 =0, A2 $\to$ A3 =-2,  A3 $\to$ A4 ==-1, A4 $\to$ A1 =3. In this case, the minimum count of apples exchanged is 0+2+1+3=6.

Question : With n people $A_i$(i from 1 to n) having  $a_i$ apples initially, what  the minimum count of apples exchanged needed so that each will have  same amount of apples in the end? (please note the question is about the minimum count of apples exchanged, not the minimum count of apple exchanges: we are counting the apples, not the exchanges.

Clarification:   The exchange can only happen between 2 neighbors. For any person  $A_i$, you can  give away $x_i$ to $A_{i+1}$ and take in $x_{i-1}$ from $ A_{i-1}$. $x_{i-1}$ and $x_{i+1}$ can be any integers (including 0).

答:
We assume that the total number of apples is divisible by the number of people. Say there are k apples per person.

I think your question isn't that through very well. It seems the only transfer that is possible is from lower indexed to higher indexed persons, so if a1 < k, or a1 + a2 < 2k, or a1 + a2 + a3 < 3k etc. there is no solution.

So I assume instead that we are going around in a circle, and person n can give to person 1. There is no "exchange", only passing apples. Any person with more than k apples needs to pass all but k apples on, as does any person who received apples putting their number over k.

We will start with one person j. If aj > k then we pass aj - k apples to person j+1 (or to 1 if j = n) and count 1 passing, then do the same for a(j+1), a(j+2) etc. If we encounter anyone with fewer than n apples then person j was not suitable.

We try this for j = 1, 2, ..., n and pick the j that leads to the smallest number of passings.

If everyone can give apples to their left or right neighbour or both then the problem gets really interesting.


问:
$B = \{w \mid w = xbaby,\text{ where }x, y \in \Sigma^*\}, \Sigma = \{a, b\}$ - DFA
I am having some trouble with this problem. I believe that the language that $B$ recognizes must included a substring "$bab$" in-order to ever hit an acceptance state. Below is a screenshot of my current progress. I was hoping for a hint into the right direction as I am not really sure how to procced.

答:
First off, recall that for a DFA each node has to have exactly one edge per symbol in your alphabet. Thus your first node cannot have two edges labelled B and the same goes for the second.

Indeed the language is as you describe given by all strings which contain the substing "bab". To draw the automata, note that (I label the node so that their name represents the substring I am currently reading.)

Node "0": start with a node. if you read an A you stay there, if you read B then you head to the node "B"

Node "B": if you read a B you stay here, if you read A you advance to the node "BA"

Node "BA": if you read a B you go to the node "BAB", if you read A you have to discard the current substring and go back to node "0"

Node "BAB": this is a dead end. Once you are in this accepting state anything you read will remain in this node.


答:
The automaton that you’ve diagrammed is not a DFA: for instance, you have two different $b$ transitions out of the initial state, something that is not possible for a deterministic finite automaton.

If $q_0$ is the initial state, you want the automaton to loop at $q_0$ as long as it is reading $a$s, not $b$s; a $b$ should send it to state $q_1$. We’ll use $q_1$ to mean I've just read a $b$, so the DFA should loop at $q_1$ as long as it continues to read $b$s. If it eventually reads an $a$ while in $q_1$, we’ll send it to $q_2$ and use that state to mean that it has just read $ba$. If it gets an $a$ at this point, that $ba$ is wasted, and in effect it’s starting over, so an $a$ should send it back to $q_0$. A $b$, on the other hand, completes the $bab$ subsequence, so the word is now known to be in $B$ and should be accepted; a $b$ should therefore send the DFA to an acceptor state $q_3$ that is a ‘sink’, meaning that every input at that point just loops the DFA at $q_3$.

Title: $B = \{w \mid w = xbaby,\text{ where }x, y \in \Sigma^*\}, \Sigma = \{a, b\}$ - DFA

Body: I am having some trouble with this problem. I believe that the language that $B$ recognizes must included a substring "$bab$" in-order to ever hit an acceptance state. Below is a screenshot of my current progress. I was hoping for a hint into the right direction as I am not really sure how to procced.


答:
Hint: You need to include more states with $g_0\overset b\to g_1\overset a\to g_2\overset b\to g_3$ and any time the substring would 'go wrong' just send back to $g_0$.

Also, you need to allow anything on the end state.


问:
Parametrizing a curve between surfaces with boundary values
I'm evaluating a line integral of the function $T= x^2 + 4xy + 2yz^3$ from $a = (0,0,0)$ to $b=(1,1,1)$ on the path $z = x^2$, and $y = x$ without using the fundamental theorem.

My question is how to factor in the boundaries of the integral when I parameterize the path in terms of $t$

So far I have:

Let $x = t$

so $r= \langle t,t,t^2\rangle$ and $dr = \langle 1,1,2t\rangle$

how do I factor in the boundaries $a=(0,0,0)$ and $b=(1,1,1)$ for my integral? After I have the boundaries, solving the line integral is not a problem

Thank you

答:
The path must be contained in the curve
$$\gamma(t)=(t,t,t^2), \, t \in \mathbb{R}$$
It is easily seen that $\gamma$ is a regular curve and, moreover, that $a=\gamma(0)$ and $b=\gamma(1)$.

We know that $\gamma'(t)=(1,1,2t) \Rightarrow ||\gamma'(t)||=\sqrt{2+4t^2}$. So, the required integral can be written as follows
$$\int_0^1 T(x,y,z) \sqrt{2+4t^2} \, \text{d} t$$

Remark: I suppose that you did mean $T$ is a scalar field, not a vector field.


问:
The consistency of "Ord is Mahlo" and inaccessible correct cardinal
In Cantor's Attic, it is stated that the theories 1 and 2 are equiconsistent.

$\mathsf{ZFC}$ + $\delta$ is inaccessible and $V_\delta\prec V$ (the latter expressed as a scheme in the language $\{\delta,\in\}$)

$\mathsf{ZFC}$ + Ord is Mahlo

I have a question about the proof on the website and a question of why another proof doesn't just prove that 1 is strictly stronger in consistency than 2. These two questions are related so I am asking them in one post.

The proof on the website states

if $\delta$ is an inaccessible correct cardinal, then since $V_\delta\prec V$ it follows that any class club definable in $V$ with parameters below $\delta$ will be unbounded in $\delta$ and hence contain $\delta$ as an element and consequently contain an inaccessible cardinal.

My first question: I take it that this proof is meant to show that theory 1 implies theory 2, and hence Con(1) implies Con(2). I don't understand why "Ord is Mahlo" is implied. I can only see that this shows that 1 implies "every club $C\subseteq\mathsf{Ord}$ definable from parameters in $V_\delta$ contains an inaccessible". But "Ord is Mahlo" states (schematically) "every club $C\subseteq\mathsf{Ord}$ definable from parameters in $V$ contains an inaccessible". So I don't see how this latter stronger statement is proved in the proof.

My second question: why doesn't the following proof show that Theory 1 proves the consistency of Theory 2? More specifically, in Theory 1, we show that $V_\delta\vDash $ "Ord is Mahlo"

The proof: Suppose $V_\delta\vDash \varphi$ defines a club. Then by elementarity, $V\vDash \varphi$ defines a club (and this club is unbounded below $\delta$). So $V$ thinks that this club contains an inaccessible cardinal. Reflecting this down to $V_\delta$, we see that $V_\delta\vDash $ the club defined by $\varphi$ contains an inaccessible cardinal. Hence $V_\delta\vDash $ "Ord is Mahlo". And so Theory 1 proves the consistency of Theory 2.

What went wrong with this proof? Thank you!

答:
Re: your second question, there's a scheme issue going on.

For each formula $\varphi$, Theory $1$ can prove that $V_\delta$ satisfies the "$\mathsf{Ord}$ is Mahlo" instance corresponding to $\varphi$. But this does not mean, however, that Theory $1$ can prove

for every $\varphi$, $V_\delta$ satisfies the "$\mathsf{Ord}$ is Mahlo" instance corresponding to $\varphi$.

So Theory $1$ stops short of proving that $V_\delta$ actually satisfies (all of) "$\mathsf{Ord}$ is Mahlo."

It may be easier to consider a large cardinal free example. Working in the expanded language $\{\in,\delta\}$, let $S$ be the $\{\in\}$-theory $\mathsf{ZFC}$ together with a sentence saying $V_\delta\models\varphi$ for each $\varphi\in\mathsf{ZFC}$. Then $S$ is a conservative extension of $\mathsf{ZFC}$, even though at a glance it may look like $S$ proves $V_\delta\models\mathsf{ZFC}$.

What about your first question?

Well, contrary to a previous edit of this answer - I worked too fast - we do indeed have that Theory $1$ outright implies Theory $2$ (not merely each axiom in the relativization of Theory $2$ to $V_\delta$). This is a cute application of elementarity. The parameter-free version is easy to check. Working in Theory $1$, suppose $\varphi$ defines a club in $V$. By elementarity, $\varphi$ defines a club in $V_\delta$ and so $\delta\in \varphi^V$. So in $V$ we have that $\varphi$ contains an inaccessible, namely $\delta$ itself.

Now on the face of it this breaks down when we fold in parameters: what if $\varphi$ involves a parameter in $V\setminus V_\delta$? Here we work a little more coarsely: rather than reflecting single-sentence instances, we reflect parameterized instances (we reflect the statement "There is no tuple of parameters turning $\varphi$ into a failure of "$\mathsf{Ord}$ is Mahlo"). So in fact Theory $1$ proves Theory $2$.

Reflecting to $V_\delta$ we get that Theory $1$ proves each axiom of Theory $2$ relativized to $V_\delta$ - however, Theory $1$ does not prove "$V_\delta$ satisfies Theory $2$" per the above.


问:
What does proving the Riemann Hypothesis accomplish?
I've recently been reading about the Millenium Prize problems, specifically the Riemann Hypothesis. I'm not near qualified to even grasp the entire problem, but seeing the hypothesis and the other problems I can't help wonder: what is the practical use of solving it?

Many researchers have spent a lot of time on it, trying to prove it, but why is it so important to solve the problem?

I've tried relating the situation to problems in my field. For instance, solving the $P \ vs. NP$ problem has important implications should it be shown to be either $P = NP$ or $P \neq NP$: for instance, cryptographic algorithms, but it's hard to say WHY the Riemann Hypothesis, or other problems, are so important.

Perhaps a partial answer could be made by seeing which solutions proof of the Poincaré Conjecture has lead to.

答:
The Riemann hypothesis is a conjecture about the Riemann zeta function
$$\zeta(s)=\sum_{n=1}^{\infty}\dfrac{1}{n^s}$$
This is a function $\mathbb{C} \rightarrow \mathbb{C}$. With the definition I have provided the zeta function is only defined for $\Re(s)\gt1$. With some complex analysis you can proof that there is a continuous (actually holomorphic if you know what it means) extension of the function so that it is defined in whole $\mathbb{C}$. The Riemann zeta function has some trivial zero points like $-2,-4,-6.$ The hypothesis says that the other zero points lie on the critical line $\Re(s)=\dfrac12$. This hypothesis had many application in analysis and number theory. The first proof of the prime number theorem used this conjecture.

In order to give an anwer to your question a would like to refer to this website, where you can find tons of applications of the Riemann hypothesis.

ContentLicense: CC BY-SA 3.0


问:
$\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.
Suppose $X_n \rightarrow X$ a.s. as $n \rightarrow \infty$ and that $X_n \geq0$ for all $n, \omega$. Furthermore, $\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.

I know that a sequence of random variables converges almost surely to $X$ if for every $\epsilon > 0$ we have
\begin{equation}
P(\lim_{n \rightarrow \infty}|X_n - X|< \epsilon) = 1
\end{equation}

But I'm not sure how to proceed. Some hints would be greatly appreciated.

答:
Another (similar) approach is as follows.

For every $\epsilon > 0$, choose $K(\epsilon)$ so that $E[|X_n|1_{X_n\geq K(\epsilon)}]<\epsilon$ and $E[|X|1_{X\geq K(\epsilon)}] < \epsilon$ (this is simply applying the definition of UI to $\{ X_n \}_n \cup \{X\}$.
Compute

$$
\begin{align}
E|X_n-X| \leq
& E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\
+ &E[|X_n|1_{X_n > K(\epsilon), X \leq K(\epsilon)}] \\
 +& E[|X_n|1_{X_n \leq K(\epsilon), X > K(\epsilon)}] \\
+& E[|X|1_{X_n > K(\epsilon), X \leq K(\epsilon)}]\\
+& E[|X|1_{X_n \leq K(\epsilon), X > K(\epsilon)}]\\
\leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\
 + &  E[|X_n|1_{X_n > K(\epsilon)}] \\
+&E[|X|1_{X > K(\epsilon)}] \\
+&   E[|X_n|1_{X_n > K(\epsilon)}] \\
+&E[|X|1_{X > K(\epsilon)}] \\
\leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}]   + 4\epsilon.
\end{align}
$$

Sending $n \to \infty$ we find
$$
\limsup_n E|X_n-X| \leq 4\epsilon.
$$
Now send $\epsilon \to 0$.


问:
Identity on the sum $\sum_{i=1}^{n-2}{i-1 \choose k}{n-2-i \choose k-1}={n-2 \choose 2k}$
I need to evaluate the first summation for $k\geq1$ and $n\geq2$.
Some computer calculations allow the identification with the expression ${n-2 \choose 2k}$
above but, how can it be derived? Are there different methods or any
combinatorial interpretation?

答:
Trying to verify that

$$\sum_{q=1}^n {q-1\choose k} {n-q\choose k-1} =
{n\choose 2k}$$

using that $k\ge 1$ and $n\ge 1$ we get
$q-1\ge k$ and $q\le n-k+1$ and may re-write as

$$\sum_{q=k+1}^{n-k+1} {q-1\choose k} {n-q\choose k-1}
= \sum_{q=0}^{n-2k} {q+k\choose k} {n-(k+1)-q\choose k-1}
\\ = \sum_{q=0}^{n-2k} {q+k\choose k} {n-k-1-q\choose n-2k-q}
= [z^{n-2k}] (1+z)^{n-k-1}
\sum_{q=0}^{n-2k} {q+k\choose k} \frac{z^q}{(1+z)^q}.$$

Now the coefficient extractor enforces the upper limit of the sum
and we may extend $q$ to infinity, getting

$$[z^{n-2k}] (1+z)^{n-k-1}
\frac{1}{(1-z/(1+z))^{k+1}}
= [z^{n-2k}] (1+z)^{n-k-1} (1+z)^{k+1}
\\ = [z^{n-2k}] (1+z)^n = {n\choose n-2k} = {n\choose 2k}.$$

This is the claim.


问:
Research Paper and Affine Subspace
I was reading a research paper titled Purity and Reid's Theorem by A.Blass and J.Irwin and i have the problem with the explanation of the proof of the first theorem, that is theorem 1.1.
In the proof of the first theorem, G is a torsion-free abelian group of infinite rank $\kappa$. At the end of the proof of theorem 1.1, the author says that :

Equivalently, $e(\alpha)$ should not be in the affine subspace of $\bar{G}$ spanned by $(f_{\beta}-g_{\alpha}, \beta < \alpha ) \cup (g_{\alpha}) $where $\bar{G}$ is the divisible hull of $G$.

Question:

What does it mean by affine subspace of $\bar{G}$ spanned by $(f_{\beta}-g_{\alpha}, \beta < \alpha ) \cup (g_{\alpha})$?

The research paper can be obtained from the web.

答:
Classically, an affine space is a torsor for a vector space; that is, it's a set equipped with a free transitive action of a vector space. Intuitively, it's a vector space that doesn't know where its origin is. For example, any translation of a subspace of a vector space is an affine space.

Affine spaces don't have a notion of addition of vectors, but they do have a well-defined notion of affine combination of points, which you can intuitively think of as a weighted average. The affine subspace spanned by a collection of points in an affine space is the affine space of affine linear combinations of those points.

In this case I guess "affine space" means a torsor for an abelian group. The definitions above generalize readily to modules over a commutative ring.

Title: Research Paper and Affine Subspace


答:
The Millennium problems are not necessarily problems whose solution will lead to curing cancer. These are problems in mathematics and were chosen for their importance in mathematics rather for their potential in applications.

There are plenty of important open problems in mathematics, and the Clay Institute had to narrow it down to seven. Whatever the reasons may be, it is clear such a short list is incomplete and does not claim to be a comprehensive list of the most important problems to solve. However, each of the problems solved is extremely central, important, interesting, and hard. Some of these problems have direct consequences, for instance the Riemann hypothesis. There are many (many many) theorems in number theory that go like "if the Riemann hypothesis is true, then blah blah", so knowing it is true will immediately validate the consequences in these theorems as true.

In contrast, a solution to some of the other Millennium problems is (highly likely) not going to lead to anything dramatic. For instance, the $P$ vs. $NP$ problem. I personally doubt it is probable that $P=NP$. The reason it's an important question is not because we don't (philosophically) already know the answer, but rather that we don't have a bloody clue how to prove it. It means that there are fundamental issues in computability (which is a hell of an important subject these days) that we just don't understand. Solving $P \ne NP$ will be important not for the result but for the techniques that will be used. (Of course, in the unlikely event that $P=NP$, enormous consequences will follow. But that is about as likely as it is that the Hitchhiker's Guide to the Galaxy is based on true events.)

The Poincaré conjecture is an extremely basic problem about three-dimensional space. I think three-dimensional space is very important, so if we can't answer a very fundamental question about it, then we don't understand it well. I'm not an expert on Perelman's solution, nor the field to which it belongs, so I can't tell what consequences his techniques have for better understanding three-dimensional space, but I'm sure there are.

Title: What does proving the Riemann Hypothesis accomplish?

Body:
I've recently been reading about the Millenium Prize problems, specifically the Riemann Hypothesis. I'm not near qualified to even grasp the entire problem, but seeing the hypothesis and the other problems I can't help wonder: what is the practical use of solving it?

Many researchers have spent a lot of time on it, trying to prove it, but why is it so important to solve the problem?

I've tried relating the situation to problems in my field. For instance, solving the $P \ vs. NP$ problem has important implications should it be shown to be either $P = NP$ or $P \neq NP$: for instance, cryptographic algorithms, but it's hard to say WHY the Riemann Hypothesis, or other problems, are so important.

Perhaps a partial answer could be made by seeing which solutions proof of the Poincaré Conjecture has lead to.


问:
For integers $n,r$, let $\binom nr = \begin {cases} \binom nr & n\ge r\ge 0 \\ 0 & \text{otherwise} \end {cases}$

For integers $n,r$, let $\binom nr = \begin {cases} \binom nr & n\ge r\ge 0 \\ 0 & \text{otherwise} \end {cases}$. Find the maximum value of $k$ for which the sum $\sum_{i=0}^k \binom {10}{i} \binom{15}{k-i} +\sum_{i=0}^{k+1} \binom {12}{i} \binom {13}{k+1-i}$ exists

I haven’t really understood the question and I have no idea how to begin. Can I get an explanation on what this question actually wants?

答:
There is no maximum: that sum exists for all integers $k$, if one follows normal practice and takes $\sum_{i=0}^\ell f(i)$ to be $0$ when $\ell<0$, and it exists for all $k\ge 0$ if one takes such sums to be undefined. In fact

$$\begin{align*}
&\sum_{i=0}^k\binom{10}i\binom{15}{k-i}+\sum_{i=0}^{k+1}\binom{12}i\binom{13}{k+1-i}\\
&\qquad=\binom{25}k+\binom{25}{k+1}\\
&\qquad=\binom{26}{k+1}
\end{align*}$$

for all integers $k$, where I used the Vandermonde identity for the first step.


问:
N'th derivative with chain rule
I am trying to find a general form of the chain rule for higher derivatives, using the general Leibniz rule I got to the following formula. However it doesn't seem to work. I suspect it has something to do with the $n-1$ but I am unsure.
$$D^{\left(n\right)}\left(f\left(g\left(x\right)\right)\right)=\sum_{k=0}^{n-1} \binom {n-1}{k}D^{\left(k+1\right)}g\left(x\right)D^{(n-k)}\left(f'\left(g\left(x\right)\right)\right)$$
(Where $D^{\left(n\right)}f(x)$ is the $n$'th derivative of $f(x)$)

The way I derived this is by using $\left(f\left(g\left(x\right)\right)\right)^{\prime}=g^\prime(x)f^\prime(g(x))$ and then applying Leibniz rule on this for higher derivatives.

EDIT- I have managed to proof that the following formula works. Can it be proofed it is equivalent to Faà di Bruno's formula?.
$$\sum_{k=0}^{n-1} \binom {n-1}{k}D^{\left(n-k\right)}g\left(x\right)D^{\left(k\right)}\left(f'\left(g\left(x\right)\right)\right)$$

答:
I think you messed up something. You cannot apply the product rule as you did it.
Think about that. Take a look at small n.

What you are looking for is https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula

Title: N'th derivative with chain rule


问:
How can we calculate the side $AC$?
In the triangle $ABC$ it holds that $|AB| = 10$, for the midpoint $C'$ of the side $AB$ it holds that $|CC'| = 8$ and the interior angle at $C$ at the triangle $BCC'$ is $\frac{\pi}{6}$.

(a) What can we say about the distance between $A$ and $C$?

(b) Let $M$ be the midpoint ofthe circumcenter of the triangle. Determine all possibilities for the angle $\angle AMB$.

Using the cosine law we could calculate the side $BC$, right?

But how can we calculate the side $AC$?

答:
a) Apply the sine rule to $BCC’$

$$\sin B= \frac85\sin\frac\pi6=\frac45,\>\>\>\>\>\cos B = \pm\frac35
$$
Then, apply the cosine rule to $ACC’$

$$AC^2 = 8^2+ 5^2 - 2\cdot8\cdot 5\cos(B+\frac\pi6)=121\pm24\sqrt3
$$
Thus, two possibilities for the distance $AC = \sqrt{ 121\pm24\sqrt3} $.

b) Note $\angle AMB =2C$ and apply the sine rule to $ABC$
$$ {\sin C}=\frac {AB}{AC}\sin B = \frac{10}{ \sqrt{ 121\pm24\sqrt3} }\cdot \frac45= \frac8{ \sqrt{ 121\pm24\sqrt3} }$$
Thus, two possible anglular values
$$\angle AMB = 2C = 2\arcsin \frac8{\sqrt{ 121\pm24\sqrt3} }$$

Title: How can we calculate the side $AC$?


问:
What is the Krull dimension of a polynomial ring over a PID?
Recently, I proved this statement:

Proposition: Let $R$ be a PID. The prime ideals of $R[y]$ are precisely the ideals of the following form:

$(0)$,

$(f(y))$ where $f$ is an irreducible polynomial in $R[Y]$

$(p, f(y))$ where $p \in R$ is prime and $f(y)$ is irreducible in $R[y]$ and its image is irreducible in $(R/p)[y]$.

Note: Maximal ideals are of the third form and vice versa.

Using this statement I would like to show that the Krull dimension of $k[x,y]$ is 2 where $k$ is a field. However, I'm stuck at showing that there can't be a chain of prime ideals of the form $\{0\}\subsetneq (f)\subsetneq (p,g)$ i.e a chain involving every ideal of the types shown above.

Any ideas? In the case when R is a general PID would the Krull dimension be 2 or 3?

答: Every PID which is not a field has Krull dimension $1$. Also, every PID is Noetherian, and if $R$ is Noetherian whose Krull dimension is $n$, then $R[x]$ has Krull dimension $n+1$. Therefore the answer to your question is that that Krull dimension of $R[y]$ is $2$.


问:
Diameter metric space
Suppose $(A, d)$ is a metric space with a non-empty subset $E \subseteq A$. Let $S = \{d(a, b) : a, b \in E\}$. Define $\phi (E)$ be the diameter of $E$ so that $\phi (E) = \sup S$, and $\phi(E) = \infty$ if $S$ is unbounded. Prove the following:

(a) If $E \subseteq A$ is non-empty then $\phi(E) = \phi(\overline{E})$ (the diameter of $E$ is equal to that of the closure of $E$).

(b) If $\phi (E) < \epsilon$ for some $\epsilon > 0$ and $E \cap B_{\epsilon}(x) \neq \emptyset$ for some $x$, then $E \subseteq B_{2\epsilon}(x)$.

For (a) I came up with a proof by contradiction but I am curious as to whether there are other proofs. After much time spent, I still cannot figure out a proof for (b). Any assistance or alternatively proofs are appreciated.

答:
Here is a direct proof of (a).

First, since $E \subseteq \overline{E}$, we have $\{d(a,b) : a,b \in E\} \subseteq \{d(a,b) : a,b \in \overline{E}\}$, so
$$\phi(E) = \sup \{d(a,b) : a,b \in E\} \leq \sup \{d(a,b) : a,b \in \overline{E}\} = \phi(\overline{E}).$$

Next, let $a,b \in \overline{E}$ be arbitrary. There exist sequences $(x_n), (y_n)$ in $E$ such that $x_n \to a$ and $y_n \to b$ as $n \to \infty$. Now by the triangle inequality,
$$d(a,b) \leq d(a,x_n) + d(x_n,y_n) + d(y_n,b) \leq d(a,x_n) + \phi(E) + d(y_n,b)$$
for all $n$. Thus,
$$d(a,b) \leq \inf \{d(a,x_n) + \phi(E) + d(y_n,b) : n \in \mathbb{N}\} = \phi(E).$$
Since $a$ and $b$ were arbitrary, $\phi(E)$ is an upper bound for $\{d(a,b) : a,b \in \overline{E}\}$. Therefore,
$$\phi(E) \geq \sup \{d(a,b) : a,b \in \overline{E}\} = \phi(\overline{E}).$$

We have now shown $\phi(E) \leq \phi(\overline{E})$ and $\phi(E) \geq \phi(\overline{E})$, so we conclude that $\phi(E) = \phi(\overline{E})$. $\square$

Title: Diameter metric space


问:
If a sequence converges to a, prove the squares of its terms converge to a^2
Given:
$$a_n\to a$$
 That means :
For all $\epsilon^*:=\frac{\epsilon }{2|a|+1} $  we can find an N such that for all n>N $\left |a_n-a  \right |<\frac{\epsilon }{2|a|+1}$

I have to prove
$$(a_n)^2 \to a^2 $$

At first we guess
$\left |a_n+a  \right |=\left |a_n-a+a+a  \right |<\frac{\epsilon }{2|a|+1 }+2|a | $

Now we can prove

$$(a_n)^2 \to a^2 $$
$\left |(a_n)^2-   a^2 \right |=\left |(a_n-a)(a_n+a)\right |=\left |a_n-a  \right |\left |a_n+a  \right |<\frac{\epsilon }{2|a|+1}(\frac{\epsilon }{2|a|+1}+2|a |)<\epsilon^2+\epsilon$

I changed my epsilon and i think it works now??
any suggestion ?

答:
I am not sure if this works as a solution, so someone should check this:

Since $a_n \rightarrow a$ we know that for any $\epsilon > 0$ there exists an $N$ such that for all $n \geq N$ we have

$$ \lvert a_n - a \rvert \leq \lvert x_n\rvert + \lvert a\rvert < \epsilon $$
$$ \lvert a_n - a \rvert^2 \leq \left( \lvert x_n\rvert + \lvert a\rvert \right)^2 < \epsilon^2 $$

It follows that

$$ \left( \lvert x_n\rvert + \lvert a\rvert \right)^2 = x_n^2 + 2x_na + a^2 < \epsilon^2 $$
$$ \lvert x_n^2 - a^2 \rvert < x_n^2 + 2x_na + a^2 < \epsilon^2$$

So if $x_n$ converges to a then $ x_n^2 $ converges to $a^2$


问:
Find the closure of the following set
Hello guys I was revising and I saw this question but didn't know what is the steps to show it

Also sorry for not being familiar with who to type math notations here

https://m.imgur.com/sUvIRci

答: Just to give you a hint, the closure of a set is the set with all of it's limit points. So for instance, the closure of the rationals $\mathbb Q$ is $\mathbb R$, since $\mathbb R$ is formed by taking the limits of $\mathbb Q$. I don't want to give you the answer, I want to motivate it. I am sorry if this didn't help.

Title: Find the closure of the following set


问:
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero
According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.

答:
Similar to Gary's answer but slightly shorter for the asymptotics.

With $x=\frac{n}{2}+\frac{1}{4}$ and $y=\frac{1}{2}$ the Beta integral becomes

$$\text{Beta}(x,y)=\int_{0}^{1} t^{x-1}(1-t)^{y-1}\,dt\\
\overset{t\to exp(-u/x)}=\frac{1}{x} \int_{0}^{\infty} e^{-u}(1-e^{-u/x})^{y-1}\,du\\
\overset{exp(-u/x)\simeq 1-\frac{u}{x}}\simeq\frac{1}{x} \int_{0}^{\infty} e^{-u}(u/x)^{y-1}\,du = x^{-y}\; \Gamma(y)=x^{-y}\sqrt{\pi}$$

whence follows that the original expression for $n \to \infty$ goes like $\frac{1}{\sqrt{x}}=\sqrt{\frac{2}{n}}$.

EDIT: The obvious (and easy to remember) substitution $t^x \to v$ gives the even shorter sequence for the leading term of the asymptotics:

$$\text{Beta}(x,y) = \frac{1}{x} \int_{0}^{1} (1-v^{\frac{1}{x}})^{y-1}\,dv=\frac{1}{x} \int_{0}^{1} (1-e^{\frac{\log(v)}{x}})^{y-1}\,dv\\
\simeq \frac{1}{x} \int_{0}^{1} (\frac{-\log(v)}{x})^{y-1}\,dv=x^{-y}\; \Gamma(y) $$
where Euler's favorite form of $\Gamma$ appears.


问:
Poisson equation, exact solution
I have to prove that the solution $u$ of the following Poisson equation
$$-\Delta u(x,y)=\sin(\pi x)\sin(\pi y)$$
for $(x,y)\in[0,1]^2$ with boundary conditions
$$u(0,y)=u(1,y)=u(x,0)=u(x,1)=0$$
can be written as $v(x)\sin(\pi y)$.

So far I've written $u(x,y)=v(x,y)\sin(\pi y)$ and tried to get $\partial_y v(x,y)=0$, but I am stuck at
$$\left((-\partial_x^2-\partial_y^2+\pi^2)v(x,y)-\sin(\pi x)\right)\sin(\pi y)=2\pi\partial_y v(x,y)\cos(\pi y)$$

Any idea?

答: You can try this solution  $u(x, y)=\frac1{2\pi^2}\sin(\pi x)\sin(\pi y)$. You can verify directly that it obeys the given Poisson equation and the boundary conditions. So $v(x)=\frac1{2\pi^2}\sin(\pi x)$. Since the Poisson equation+Dirichlet boundary conditions in a closed and finite boundary has a unique solution, we are done.

Title: Poisson equation, exact solution


问:
When does conjugation in simple field extensions coincide with the identity?
Suppose I have a field $F$ and a simple extension of $F$, $F(\alpha)$, and let's assume that $\alpha^2 \in F$. Now take the "conjugation" automorphism $\sigma:F(\alpha) \to F(\alpha)$, $\sigma(a + b\alpha) = a-b\alpha$ for $a, b \in F$. My question is, does this map $\sigma$ ever coincide with the identity map? My guess is that for this to happen, we would need $\sigma(\alpha) = -\alpha = \alpha$ somehow. So every element is its own additive inverse, so maybe something about characteristic of $F$ being $2$. Is there a more structured way to think about this?

答:
That's already it. You need that   $2=0$ to make $-\alpha=\alpha$. This is called a purely inseparable extension.
Perhaps you also count the case that already $\alpha\in F$ and the extension is not really an extension  to begin with.


问:
Recurrence Tree Method: $T(n) = 3(2n/3) + 1$ Why is the solution in $\log_{3/2}$ and not base $\log 2/3$?
I feel this is more of a question in the logarithms, but I am confused  why this tree reoccurrence

$$T(n) = 3T(2n/3)+1$$

solution is derived from $\log_{3/2}$ rather than  $\log_{2/3}$? My thinking about this problem:

$\bullet$ height of this tree is = $\log_{2/3}n$

$\bullet$ The number of nodes is $3^{nodes}= 3^{\log_{2/3}n} = n^{\log_{2/3}3}  \approx n^{-2.7}$

$\bullet $ This would make the solution $\Theta(n^{-2.7}) $, but that isn't correct, the correct solution is $\Theta(n^{2.7}) $, which would be obtained using $log_{3/2}$ .

I guess I have a couple of confusion: 1) In the Recurrence part of a recurrence equation the T(n/x) I always assumed would create height $\log_x n$, therefore $T(2n/3)$ would create height $\log_{2/3}n$? 2) I didn't realize the nature of log bases, if the base is a rational fraction then the inverse of the base is produces the same result, but opposite signs. 3)How should I be thinking about the fractional part of the these recurrences equations?

答:
As

$$
T\left(\left(\frac 23\right)^{\log_{\frac 23}n}\right)=3T\left(\left(\frac 23\right)^{\log_{\frac 23}\frac 23 n}\right)+1
$$

making $\mathcal{T}(\cdot) = T\left(\left(\frac 23\right)^{(\cdot)}\right)$ and $z = \log_{\frac 23}n$ we follow with the recurrence

$$
\mathcal{T}(z) =3\mathcal{T}(z+1)+1
$$

with solution

$$
\mathcal{T}(z) = 3^{1-z}c_0+\frac 12(3^{-z}-1)
$$
and going backwards with $z =  \log_{\frac 23}n$ we get

$$
T(n) = 3^{\log{\frac 23}n+1}c_0 + \frac 12\left(3^{\log{\frac 23}n}-1\right)
$$


答: $\sin (\pi/6)/5=\sin \beta  /8$, then third angle in the triangle is $\tau=180-\beta-30$, then angle $\rho=180-\tau$. We have now the second triangle, there is the angle $\rho$, and $AC'=5,CC'=8$, we can use cosine formula $(AC)^2=8^2+5^2-80*\cos \rho$.

Title: How can we calculate the side $AC$?

Body:
In the triangle $ABC$ it holds that $|AB| = 10$, for the midpoint $C'$ of the side $AB$ it holds that $|CC'| = 8$ and the interior angle at $C$ at the triangle $BCC'$ is $\frac{\pi}{6}$.

(a) What can we say about the distance between $A$ and $C$?

(b) Let $M$ be the midpoint ofthe circumcenter of the triangle. Determine all possibilities for the angle $\angle AMB$.

Using the cosine law we could calculate the side $BC$, right?

But how can we calculate the side $AC$?


问:
Block Matrices and Matrix multiplication
Suppose i have a vector space $V=\mathbb{F}^{3\times2}$ of $3\times2$ matrices and a vector space $W=\mathbb{F}^{2\times2}$ of $2\times2$ matrices over the field $\mathbb{F}$.

Any Linear transformation $T: V\to W$ would take a $3\times2$ matrix and transform it into a $2\times2$.

I tried to compare this with the fact that multiplying a $2\times3$ matrix by a $3\times2$ matrix also results in a $2\times2$ matrix.

But obviously this $2\times3$ matrix is not the matrix representation of $T$.

To better understand, i defined a standard basis for $V$ and made some algebra resulting in this fact :

http://postimg.org/image/uhh8e02fx/

Where the elements of the $6\times1$ matrix are the vectors from the standard basis of $\mathbb{F}^{3\times2}$ after being transformed by $T$.

By all of this, i found that there are at least two ways to implement matrix multiplication to result in a $2\times2$ matrix.

My question :

Is the ( $1\times6$ times $6\times1$ ) matrix multiplication the only one that is equivalent to the ($2\times3$ times $3\times2$) matrix multiplication ?

I can't find detailed information about matrix multplication, even on linear algebra books ( even wikipedia http://en.wikipedia.org/wiki/Matrix_multiplication  doesn't mention the equivalent matrix multiplications i found out empirically  ).

Is there any topic name, site, resource, book that can teach much more than the algorithm approach (row dot column ) of matrix multiplication ?

答:
Not all linear maps between matrix spaces arise from one-sided matrix multiplication.

Indeed, the space of linear maps from $F^{3\times 2}$ to $F^{2\times 2}$ is $24$-dimensional ($6\cdot 4=24$). Some of them are realized as matrix multiplication $M\mapsto AM$ where $A\in F^{2\times 3}$. However, the matrix $A$ has only $6$ entires,  so the linear spaces of all such transformations is $6$-dimensional: it is a  proper subspace of $L(F^{3\times 2},F^{2\times 2})$.

However, every linear map from $F^{3\times 2}$ to $F^{2\times 2}$ can be written as a finite sum of two-sided multiplication maps $M\mapsto AMB$. Indeed, $24$ terms suffice:
$$M\mapsto \sum_{j=1}^{24} A_j MB_j$$
where each matrix $A_j$ and $B_j$ has one nonzero entry. To see why, observe that by placing $1$ at one entry of $A$ and one entry of $B$ one can move any given entry of $M$ into any desired position in $AMB$.


问:
Is the input for a linear map $\phi_{\alpha}$ with basis $\alpha$, a coordinate vector which is relative to $\alpha$ or to the standard basis?
Suppose you have a linear map, $\phi_{\alpha}$ which represents the
transformation of space given the basis $\alpha$.

I was under the impression that for $\phi_{\alpha}(\boldsymbol{x})$,
$\boldsymbol{x}$ is relative to the standard basis, and then gets
transformed. I have seen that it is indeed the converse, where $\phi$
takes a coordinate vector relative to $\alpha$, and then gets transformed to be relative to the standard basis?

Am I just mixing the change of basis linear map, and a transformation
linear map together?

答:
If $\alpha = (v_1,\dots,v_n)$ is an ordered basis for $V$, the map $\phi_\alpha : \mathbb F^n \to V$ given by $$\phi_\alpha \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} = a_1v_1+\cdots+a_nv_n$$ is a bijection (for every element in the codomain there is a unique element in the domain that is mapped to it). Now, for $v \in V$, the column vector $x = \phi_\alpha^{-1}(v)$ (that is, $x$ is the unique vector in $\mathbb F^n$ such that $\phi_\alpha(x)=v$) is called the coordinate vector of $v$ with respect to $\alpha$.

Note also that if $\beta$ is the standard basis for $\mathbb F^n$, then the coordinate vector of $x \in \mathbb F^n$ with respect to $\beta$ is $x$ itself. So, the answer to your main question is: both, depending on what exactly you mean by "standard basis".


问:
Show that $N_1N_2 \cdots N_k \cong N_1\times N_2\times\cdots\times N_k$ .

Let $G$ be a group having finite normal subgroups $N_1,... , N_k$ , such that the gcd
of $|N_i |$ and $|N_j |$ is $1$ whenever $i\neq j$. Show that $N_1N_2 \cdots N_k \cong N_1\times N_2\times\cdots\times N_k$.

I think one way to solve this exercise is by proving two things: (1) $N_1N_2 \cdots N_k=G$  and (2) $(N_1\cdots N_s)\cap N_{s+1}=\{e\}$. Since in this way we can conclude that $G$ is the internal direct product of $N_1,... , N_k$ and therefore isomorphic to $N_1\times\cdots\times N_k$. The problem is, I don't know exactly how to start. Any hint?

Edition

For $(2)$, let $N=N_1N_2\cdots N_{k-1}$. We claim that $N\cap N_{k}=\{e\}$. Suppose otherwise, that $e\neq a\in N\cap N_{k} $, then $|a|$ must divide the order of both subgroups $N$ and $N_k$, but $(|N|,|N_k|)=1$; thus, $|a|=1$ and so $a=e$. For $(1)$, since $N\cap N_{k}=\{e\}$, we know that $|NN_k|=|N||N_k|\leq |G|$. Here is my problem, how can I prove that Indeed $|N||N_k|= |G|$?

I have not been able to advance much with the Grossmann's hint

答: Hint: Because $N_k$ is normal, the quotient map $N_1\cdots N_{k-1}N_k \to (N_1 \cdots N_{k-1}N_k)/N_k \cong N_1 \cdots N_{k-1}$ is a homomorphism.


问:
$K$ is weakly*-compact $\Rightarrow $ $K$ is norm-bounded
What I did was considering the collection of linear bounded functionals $\left \{ Jx ; x\in E \right \}$ where $J$ is the canonical injection from $E$ to its bidual . ($\left \langle Jx,f \right \rangle=\left \langle f,x \right \rangle$)

For all $x$ in $E$ , $Jx$ are continuous in the weak* topology (by its definition) and $K$ is weakly*-compact , then $Jx(K)$ is bounded in $\mathbb{R}$.

Then, $\underset{x\in E}{sup} \left | \left \langle Jx,f \right \rangle \right | < \infty $

(for all $f$ in $K$)

Then by uniform boundedness principle , $\underset{x\in E}{sup} \left \| Jx \right \|< \infty$

I'm stuck here , what I want to show  , is that $\left \| f \right \|< \infty$

(I know that $\left \| Jx \right \|=\left \| x \right \|$ , is it useful ? )

答:
It seems you are making a mistake in the first line with a formula:

If $K$ ist weak$^{\ast}$-compact and $Jx$ is weak$^{\ast}$-continuous for each $x \in E$, then for every $x \in E$ we have

$$ \sup_{f \in K} \vert \langle Jx, f \rangle \vert < \infty $$

The $\sup$ does not go over $x \in E$, but over $f \in K$.

Can you take it from here?


问:
Parametrizing a curve between surfaces with boundary values
I'm evaluating a line integral of the function $T= x^2 + 4xy + 2yz^3$ from $a = (0,0,0)$ to $b=(1,1,1)$ on the path $z = x^2$, and $y = x$ without using the fundamental theorem.

My question is how to factor in the boundaries of the integral when I parameterize the path in terms of $t$

So far I have:

Let $x = t$

so $r= \langle t,t,t^2\rangle$ and $dr = \langle 1,1,2t\rangle$

how do I factor in the boundaries $a=(0,0,0)$ and $b=(1,1,1)$ for my integral? After I have the boundaries, solving the line integral is not a problem

Thank you

Body:
From the formula here

$r=(t,t,t^2);\;r'(t)=(1,1,2t)\to |r'(t)|=\sqrt{1^2+1^2+(2t)^2}=\sqrt{2+4t^2}$

$$\int_r(x^2+4 x y+2 y z^3)\,ds=\int_0^1 \left(t^2+4t\cdot t+2t\cdot(t^2)^3\right)\sqrt{2+4t^2}\,dt=$$
$$=\int_0^1  \left(2 t^7+5 t^2\right) \sqrt{4 t^2+2}\, dt=$$
$$\small=\left[\frac{\sqrt{4 t^2+2} \left(1120 t^8+80 t^6-48 t^4+6300 t^3+32 t^2+1575 t-32\right)}{5040}-\frac{5}{16} \text{arcsinh}\left(\sqrt{2} t\right)\right]_0^1=$$
$$=\frac{1}{5040}\left(32 \sqrt{2}+9027 \sqrt{6}-1575 \text{arcsinh}\sqrt{2}\right)\approx 4.038$$


问:
Finding the singularity type at $z=0$ of $\frac{1}{\cos(\frac{1}{z})}$
I have the following homework problem:

What kind of singular point does the function
  $\frac{1}{\cos(\frac{1}{z})}$ have at $z=0$ ?

What I tried:

We note (visually) that $z_{0}$ is the same type of singularity for
both $f,f^{2}$ hence the type of singularity of $f(z)=\frac{1}{\cos(\frac{1}{z})}$
have at $z=0$ is the same type of singularity $f^{2}(z)=\frac{1}{\cos^{2}(\frac{1}{z})}$.
We recall
$$
\frac{1}{\cos^{2}(z)}=1+\tan^{2}(z)
$$

We also note that for any constant $z_{0}\in\mathbb{C}$ the singularity
of $f,f+z_{0}$ are the same, this can be proved by noting the Laurent
expansions of both functions are the same, up to an additive constant.

It remains to determine the singularity type at the origin of
$$
\tan(\frac{1}{z})
$$

This is where I'm stuck, we didn't study what the Taylor series of
$\tan(z)$.

I also know that type of singularity $$\cos(\frac{1}{z})$$ have, but
I don't know how to connect this with the singularity type of $$\frac{1}{\cos(\frac{1}{z})}$$

Can someone please hint me in the right direction ?

答:
Notice that neither of the limits $\lim_{z \to 0} \frac{1}{\cos(1/z)}$ and $\lim_{z \to 0} \cos(1/z)$ exist. Therefore the singularity is of essential type.

Edit: The conclusion is wrong since $0$ is an accumulation point of poles. (See Ted Shifrin's answer below.)


问:
Sequences satisfying an identity involving a discrete convolution
Let $(a_n)_{n=0}^\infty \cdot (b_n)_{n=0}^\infty = (c_n)_{n=0}^\infty$ where $c_n = \sum_{k=0}^n a_k b_{n-k}$.

Following on from a previous question, I am considering sequences $(a_n)_{n=0}^\infty$ of complex numbers which satisfy

$$(a_n z^n)_{n=0}^\infty \cdot (a_n w^n)_{n=0}^\infty = (a_n (z+w)^n)_{n=0}^\infty \tag{1}$$

for all $z,w \in \mathbb{C}$. I know that one family of such sequences is $a_n = \frac{c^n}{n!}$ for some constant $c \in \mathbb{C}$, since

$$\sum_{k=0}^n (a_k z^k) (a_{n-k} w^{n-k}) = \sum_{k=0}^n \frac{c^k}{k!} z^k \frac{c^{n-k}}{(n-k)!} w^{n-k} = \frac{c^n}{n!} \sum_{k=0}^n \binom{n}{k} z^k w^{n-k} = \frac{c^n}{n!} (z+w)^n$$ as required. In fact, it is easy to see that the requirement

$$\sum_{k=0}^n a_k a_{n-k} z^k w^{n-k} = a_n (z+w)^n = \sum_{k=0}^n a_n \binom{n}{k} z^k w^{n-k}$$

is equivalent to

$$a_n \binom{n}{k} = a_k a_{n-k} \tag{2}$$

for all $n,k \in \mathbb{N}$ with $0 \leq k \leq n$.

Are the sequences given by $a_n = \frac{c^n}{n!}$ the only sequences satisfying (2)?

答:
Yes, all such sequences are of the form $c^n/n!$, except for zero sequence $a_n=0$. Letting $k=1$ in $(1)$, you get
$$
a_n=\frac1{n}a_1a_{n-1},
$$
for all $n\ge 1$. You can then prove by induction that this implies
$$
a_n=\frac{a_1^{n}}{n!}\qquad \text{ for all }n\ge 0.
$$
For the base case, use $a_n\binom{n}k=a_ka_{n-k}$ with $n=k=0$ to get $a_0=a_0^2$, so $a_0$ is $0$ or $1$. If $a_0=0$, you get the zero sequence, and if $a_0=1$, the base case $a_0=(a_1)^0/0!$ is proven, and the inductive step is easy.


问:
Difference of sets and power set question
I am asked to state whether the following statements are TRUE or FALSE

$\emptyset$  is in $P(A)$.

$\emptyset$ is subset of A.

My answer is that both of these statements are true.

答:
Your reasoning is perfect. To expand a little bit:

$\varnothing \subseteq B$, so $\varnothing \in P(B)$, so $\varnothing \notin P(A) \setminus P(B)$.

$\varnothing \in A$, so $\{\varnothing\} \subseteq A$, so $\{\varnothing\} \in P(A)$. At the same time, $\varnothing \notin B$, so $\{\varnothing\} \nsubseteq B$, so $\{\varnothing\} \notin P(B)$. Thus, $\{\varnothing\} \in P(A) \setminus P(B)$.


问:
Two versions of Kleene's recursion theorem - what's the relationship between them?
Below are two versions of Kleene's recursion theorem. How are they related? Are they equivalent? If not, does one of them (which one?) imply the other?

Note that both $U(n,x)$ and $\phi_n(x)$ is the result of application of program number $n$ to input $x$.

Version 1:

If $H:N\to N$ is a total computable
function and $U$ is a Godel universal computable function (defined
here), then there is $n$ such that $U(n,x)=U(H(n),x)$ for all $x$ (i.e., there's a "program" that does the same thing before and after the application of $H$).

(The construction of the fixed point is given here.)

Version 2:

For every number $p$, there exists $q_0$ such that for all $x$, $$\phi_p(q_0,x)=\phi_{q_0}(x)$$

答:
The question of whether they're equivalent is a little imprecise since they're both true (whether one follows immediately from the other might depend on how the formalisms are developed), but I think both versions say essentially the same thing.

In version 2, we think of $p$ as encoding an interpreter that takes a program representation $q$ and runs it on input $x$ (using its own internal interpretation of $q$ as a program). The theorem states that for any such interpreter, there's a number $q_0$ that represents the same program in $p$'s "language" as it does in our "canonical language" of program numberings.

In version 1, given a Godel universal computable function $U$ (which we think of as a universal interpreter in the sense that for any interpreter $V$, we can computably translate programs from $V$'s language to $U$'s language), we can think of $H$ as the translation function for some arbitrary interpreter $V$. The theorem says that there's a number $n$ that represents the same program in $V$'s language that it does in the "universal" language of $U$.


问:
In a geometric sequence $u_1=125$ and $u_6 =1/25$
In a geometric sequence $u_1=125$ and $u_6 =1/25$.

a) Find the value of $r$ (the common ratio).

b) Find the largest value of $n$ for which $S_n < 156.22$.

c) Explain why there is no value of $n$ for which $S_n > 160$.

Can someone please help me figure this out?

Body:
hint

$$u_6=u_1(r)^{6-1}\implies$$
$$ \frac{1}{5^2}=5^3r^5 \implies$$
$$r^5=(\frac 15)^5$$
$$S_n=u_1+u_2+...+u_n=u_1\frac{r^n-1}{r-1}$$

$$\implies \lim_{n\to+\infty}S_n=\frac{625}{4}=156,25$$
$$\implies (\forall n\ge 1)\;\; S_n\le 156,25<160$$
because $ (S_n) $ is increasing.


问:
Initial Value Problem to find series solution

Consider the following initial value problem:
$$e^{-x}y'' + \ln(1 + x)y' - x^2y = 0$$
$$y(0) = 1$$
$$y'(0) = 2$$

(done) Show that $x = 0$ is an ordinary point in the differential equation.

Find the first $4$ non-zero terms of the series solution centered in $x = 0$;

Which is the radius of convergence of the series solution found in $(2)$?

I have solved number 1, but number 2 seems to be giving me way too much trouble. First, I can't find the series solutions. I feel like I'm missing some quick corners, as I may not need to spend one full hour trying to work out the math in this. Could anyone help me?

答:
This answer specifically addresses one way to work through part (2).

Assuming a solution of the form

$$y=\sum_{n\ge0}a_nx^n=a_0+a_1x+a_2x^2+\cdots$$

notice that $y(0)=a_0=1$, and similarly you have $y'(0)=a_1=2$.

Differentiate twice to find

$$y'=\sum_{n\ge0}na_nx^{n-1}=\sum_{n\ge1}na_nx^{n-1}=\sum_{n\ge0}(n+1)a_{n+1}x^n\\
y''=\sum_{n\ge1}n(n-1)a_nx^{n-2}=\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$$

Recall that

$$e^{-x}=\sum_{n\ge0}\frac{(-x)^n}{n!}=\sum_{n\ge0}\frac{(-1)^n}{n!}x^n\\
\ln(1+x)=\sum_{n\ge1}\frac{(-x)^n}n=\sum_{n\ge1}\frac{(-1)^n}nx^n$$

Now in the ODE replace everything you can with these series expansions:

$$e^{-x}y''+\ln(1+x)y'-x^2y=0$$

becomes

$$\left(\sum_{n\ge0}\frac{(-1)^n}{n!}x^n\right)\left(\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n\right)+\left(\sum_{n\ge1}\frac{(-1)^n}nx^n\right)\left(\sum_{n\ge0}(n+1)a_{n+1}x^n\right)-\sum_{n\ge0}a_nx^{n+2}=0$$

Simplifying this looks like a sordid business, but fortunately the exercises asks you for just first several non-zero terms of the series expansion of $y$. To that end, we only care about extracting what information we can about the first several $a_n$, so consider truncating the expansions up to, say, $4$ terms each:

$$\color{red}{\left(1-x+\frac{x^2}2-\frac{x^3}6\right)\left(2\cdot1a_2+3\cdot2a_3x+4\cdot3a_4x^2+5\cdot4a_5x^3\right)}\\
\quad\quad+\color{green}{\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4\right)\left(a_1+2a_2x+3a_3x^2+4a_4x^3\right)}\\
\quad\quad-\color{blue}{\left(a_0x^2+a_1x^3+a_2x^4+a_3x^5\right)}=0$$

Expand the left side and collect terms with the same power of $x$:

$$\begin{array}{c|rl}
x^0&\color{red}{2a_2}+\color{green}{0}-\color{blue}{0}&=0\\
x^1&\color{red}{(-2a_2+6a_3)}+\color{green}{a_1}-\color{blue}{0}&=0\\
x^2&\color{red}{(a_2-6a_3+12a_4)}+\color{green}{\left(-\frac{a_1}2+2a_2\right)}-\color{blue}{a_0}&=0\\
x^3&\color{red}{\left(-\frac{a_2}3+3a_3-12a_4+20a_5\right)}+\color{green}{\left(\frac{a_1}3-a_2+3a_3\right)}-\color{blue}{a_1}&=0
\end{array}$$

Now, with $a_0=1$ and $a_1=2$, compute the next few coefficients:

$$\begin{align}
2a_2=0\implies a_2&=0\\
a_1-2a_2-6a_3=0\implies a_3&=-\frac13\\
-a_0-\frac{a_1}2+3a_2-6a_3+12a_4=0\implies a_4&=0\\
-\frac{2a_1}3-\frac{4a_2}3+6a_3-12a_4+20a_5=0\implies a_5&=-\frac16
\end{align}$$

So we find

$$y(x)\approx1+2x+\color{lightgray}{0x^2}-\frac{x^3}3+\color{lightgray}{0x^4}-\frac{x^5}6$$


问:
Second order linear differential ODEs Reduction of order method

Background

1.4.1 Reduction of Order

If a solution $y_1(x)$ of the homogeneous Linear ODE (1.2) is known, then a second linearly independent solution $y_2(x)$ can be found by solving a first-order linear differential equation. A solution of (1.2) is sought in the form

$$y=u(x)y_1(x),\tag{1.4}$$

where $u(x)$ is a new unknown function. Inserting (1.4) into the homogeneous Linear ODE (1.2) and collecting terms with $u,u'$ and $u''$, we find

$$u''y_1+u'(2y'_1+py_1)+u[y''_1+py'_1+qy_1]=0.$$

The last term on the left-hand side is zero because $y_1(x)$ is a solution of (1.2), and we arrive at a first-order differential equation with respect to $U(x)=u'(x):$

$$y''+p(x)y'+q(x)y=0$$
$$y(x)=y_1(x)u(x)$$
$$u''y_1+u'(2y'_1+py_1)+u(y_1''+py_1'+qy_1)=0$$

The Question

Can someone please explain in layman's terms why $y_1(x)$ being a solution means that the last term on the left hand side is zero.

答:
You have a differential equation:
$$y''+py'+qy=0$$
Then $y_1$ is a solution of the DE this means:
$$y_1''+py_1'+qy_1=0$$
This is why $u(y_1''+py_1'+qy_1)=0$.


问:
Dense sequence definition
A subset $A$ of a space $X$ is called dense if every point $x$ in $X$ either belongs to $A$ or is a limit point of $A$ ; that is, the closure of $A$ constitutes the whole set $X$.

That the definition of a dense set, what about sequences? what is the meaning of dense sequence?

答: A sequence is dense if the set whose elements are the sequence elements, is dense according to your definition.</p>

Title: Dense sequence definition


问:
End - extension in the complete theory that isn't elementary extension
Could you prompt me,  example of complete theory,  that has, say,  two models, $\mathfrak{A}$ and $\mathfrak{B}$, that $\mathfrak{A} \le_{end} \mathfrak{B}$, but $\mathfrak{A} \npreceq \mathfrak{B}$?

For example,  for this reason I leave considering of arithmetic vector spaces,  with different dimensions,  say,  R^3 and R^7 - R^7 is elementary extension of R^3, so it isn't fit.

答:
Let $L = \{\leq\}$, and let $T$ be the complete theory of $N = (\mathbb{N}^*,\leq)$, the reverse order on the natural numbers. Let $M = ((\mathbb{N}\cup \{-1\})^*,\leq)$, so $M$ adds a new maximal element $-1$ on top of $N$.

Then $M$ is an end extension of $N$, and it is also a model of $T$ since it is isomorphic to $N$, but it is not an elementary extension: let $\varphi(x)$ be the formula $\forall y(y\leq x)$ which expresses that $x$ is a maximal element. Then $N\models \varphi(0)$ but $M\not\models \varphi(0)$.


答:
A sequence is not dense. It’s a common abuse of terminology that identifies the sequence, which is a function from the natural numbers to the real numbers (usually), with its image set, which is a subset of the codomain of the sequence.

A sequence $(x_n)_{n}$ is a function from the natural numbers to a set $X$ and, formally, can be written as $(x_n)_{n} \in X^{\mathbb{N}}$. The image of any function is a subset of the codomain, and so we write $\{x_n : n \in\mathbb{N}\}$. Hence, to be precise, we should say that the image set of the sequence is dense in $X$, that is $\overline{\{x_n : n \in\mathbb{N}\}}=X$

The same Rudin, Principles of Mathematical Analysis, definition 2.7: "By a sequence, we mean a function $f$ defined on the set $J$ of all positive integers".

Title: Dense sequence definition

Body:
A subset $A$ of a space $X$ is called dense if every point $x$ in $X$ either belongs to $A$ or is a limit point of $A$ ; that is, the closure of $A$ constitutes the whole set $X$.

That the definition of a dense set, what about sequences? what is the meaning of dense sequence?


问:
Least regularity of a surface for the curvature to be continuous
What is the minimal regularity required for a set $\Omega \subset \Bbb{R}^3 (\Bbb{R}^N)$ such that the mean curvature $H$ is a continuous function $H: \partial \Omega \to \Bbb{R}$?

I know that $C^2$ is enough, but can we have less regularity than that (for example $C^{1,1}$) and the curvature to still be continuous?

答: I would say that $C^{1,1}$ is not enough to guarantee continuity of the mean curvature. In $\mathbb{R}^2$, think of a stadium-like domain (the convex hull of two tangent balls having the same radius R): then $H=1/R$ on the curved parts, but $H=0$ on the straight lines.</p>


答: A sequence can be viewed as a set by forgetting the order of the elements. The sequence $(x_n)_{n\in\Bbb N}$ is dense if the set $\{x_n:n\in\Bbb N\}$ is dense.

Title: Dense sequence definition


问:
Differentiable at a certain point proof checking
Given $f(x)$ is differentiable at $x=0$ and $f(0)=0$ and $f'(0) > 0$

Prove that there exists an $\varepsilon > 0$ such that $f(x) > 0$ for all $0<x<\varepsilon$

My proof:

$$ f'(0) = \lim_{x \to 0} \frac{f(x) -f(0)}{x-0} = \lim_{x \to 0} \frac{f(x) - 0}{x} = \lim_{x \to 0} \frac{f(x)}{x} = f'(0) > 0$$

Thus:

$$ \frac{f(x)}{x} > 0 $$

So, assuming there is no such $\varepsilon>0$, it means our function is always negative ($f(x)<0$) otherwise we would have such $\varepsilon$ right?

But, if $x>0$ and $f(x)<0$ (because it is negative for all $x$'s) it means that $f(x)/x < 0$ contradicting the given information we got from the limit above...

I am really unsure of this proof, why does it seems very wrong? intuitively, I don't know where the red line is being drawn when talking about proving by contradiction..

Thank you!

答:
Yes, you correctly sense that the negation of "$f(x)$ is positive in some right-sided neighborhood of $0$" is not "$f(x)$ is negative in some right-sided neighborhood of $0$."

The correct negation is "every right-sided neighborhood of $0$ contains a point $a$ at which $f(x)$ is nonpositive." This negation would imply that there exists a strictly decreasing sequence of such points, $a_{n} > 0$, converging to $0$:
$$
{f(a_{n}) \over a_{n}} \leq 0 \mbox{ for all natural numbers }n.  \quad \mbox{(*)}
$$
Knowing that $f$ is differentiable at zero, the limit
$$
\lim_{n \rightarrow +\infty} {f(a_{n}) \over a_{n}}
$$
exists and equals $f'(0)$, which is positive.  But, by inequality (*), this limit must be nonpositive, which is the contradiction you were after.


问:
if $f(nt) \rightarrow 0$ as $n \rightarrow \infty$ then $f(x) \rightarrow 0$
Suppose $f$ is continuous, $f:[0,\infty)\rightarrow \mathbb{R}$ with the property that $f(nt)\rightarrow 0$ as $n \rightarrow \infty$ for all $t >0$.  We want to show that $f(x) \rightarrow 0$ as $x \rightarrow \infty$.

My attempt:

Fix $\varepsilon > 0$. We consider sets $A_n=\{t>0| m\geq n \implies |f(mt)|< \varepsilon\}$. Since $\bigcup_{n \in \mathbb{N}}A_n=(0,\infty)$ is nonmeager we have by Baire that $\overline{A_n}$ has a nonempty intertior for some $n$.  Thus WLOG we can take $[a,b] \in A_n$ with $n(b-a)\geq a$

Why is this allowed? I understand we get the interval, but why should I be guaranteed that inequality? Is this where I'm meant to use continuity?

Now I'm supposed to show $|f(x)| < \varepsilon$ if $x \geq na$. We have that if $x \in [ka,kb]_{k\geq n}$, $|f(x)|\leq \varepsilon$ by the definition of $A_n$. I claim for any intervals $[ka,kb]_{k \geq n}$ we have $(k+1)a \leq kb$. If this is true then the intervals $[ka,kb]_{k\geq n}$ will cover $[na,\infty)$ And I will be done.  Clearly the statement is true for $k=n$ since $n(b-a)\geq a$. Suppose its true for $n+1,n+2...k-1$. Then:

$$(k+1)a=ak+a \leq (k-1)b + a = kb - (b-a)<kb$$

This completes the induction(does it?) and I'm done? I'm skeptical because I never (at least explicitly) use the continuity of $f$? Thanks in advance.

答:
The reason you can guarantee $n(b-a)\ge a$ is because the sets get larger as $n\to\infty$. If you choose $n$ and $[a,b]\subseteq A_n$ where $n(b-a)\not\ge a$, you could just choose a large $N>n$ so that $N(b-a)>a$ does hold, and you will still have $[a,b]\subset A_{N}$.

The rest of the proof is correct. You did use the continuity of $f$ when you said that $\bigcup_{n=1}^\infty A_n= (0,\infty)$. The reason that every $t$ is a part of some set $A_n$ is because the sequence $f(t),f(2t),f(3t)\dots$ converges to $0$, so there exists an $m$ such that $n\ge m$ implies $|f(nt)|<\varepsilon$.


问:
Proof that the Matrix of Orthogonal Projection can be calculated by $P = A (A^t A)^{-1} A^t$.
Prove that the Matrix of Orthogonal Projection $P$ in any subspace $W$, can be calculated by $P = A (A^t A)^{-1} A^t$ where ColA forms a basis for W.

Am absolutely lost in this one, please help.

答:
Okay. Let's take it one step at a time then.

First, let's prove that $P$ is a projection, that is to say $P=P^2$. Then, that $P$ is an orthogonal projection, i.e. that $P^t=P$. Lastly, let's prove that $P$ has the correct image.

So

\begin{align}
P^2 &=(A(A^tA)^{-1}A^t)^2=A(A^tA)^{-1}A^tA(A^tA)^{-1}A^t=A(A^tA)^{-1}A^t \\
P^t &= (A(A^tA)^{-1}A^t)^t=(A^t)^t((A^tA)^{-1})^tA^t=A((A^tA)^t)^{-1}A^t=A(A^tA)^{-1}A^t
\end{align}
Finally, we need to argue for the image of $P$. For this, you only need the fact that $A^t$ is surjective (which you must know in order to know that $A^tA$ is invertible).


问:
Solving system of equation using Macaulay2
As an algebraic curve, the Klein quartic can be viewed as a projective algebraic curve over the complex numbers $\mathbb{C}$, defined by the following quartic equation in homogeneous coordinates $[x:y:z]$ on $\mathbb{P}^2_{\mathbb{C}}$:

$$x^3 y + y^3 z + z^3 x = 0.$$

Now we want to find the eigenvectors of this curve, therefore consider the following matrix:

$$
 \begin{bmatrix}
 3x^2y+z^3 & x^3 + 3y^2z & y^3+3z^2  \\
 x & y & z  \\
 \end{bmatrix},
 $$
after computing the $2$-minors we got the following system:
\begin{equation*}
\left\{
\begin{alignedat}{3}
% R & L   &  R & L   &  R & L
3x^2y^2+z^3y & -{} &  (3y^2zx + x^4) &  = 0 \\
3z^2x^2+y^3x & -{} &  (3x^2yz + z^4)& = 0 \\
3y^2z^2+x^3z & -{} &  (3z^2xy + y^4)& = 0
\end{alignedat} \ .
\right.
\end{equation*}
I am new to Macaualy 2 and I am wondering if Macaulay2 can help me to solve the above system numerically with loading NumericalAlgebraicGeomety package, (I know that because of infinity many solution we need normalize the system on unit sphere).

Body:
I divided by $z^4$ all equations and set $$\frac{x}{z}=X;\;\frac{y}{z}=Y$$
so I got
$$
\begin{cases}
-X^4+3 X^2 Y^2-3 X Y^2+Y=0\\
-3 X^2 Y+3 X^2+X Y^3-1=0\\
X^3-3 X Y-Y^4+3 Y^2=0\\
\end{cases}
$$
solved by Mathematica I got these solutions
$$
\begin{array}{rr}
 X & Y\\
\hline
 1 & 1 \\
 0.307979 & 1.55496 \\
 0.643104 & 0.198062 \\
 5.04892 & 3.24698 \\
 -1.80194 & 0.801938 \\
 -0.445042 & -0.554958 \\
 1.24698 & -2.24698 \\
 -0.610856 & 0.106042 \\
 -0.173596 & -1.63705 \\
 0.667631 & 1.26654 \\
 0.789553 & 0.52713 \\
 1.89706 & 1.49783 \\
 9.4302 & -5.76459 \\
\end{array}
$$


问:
T1 space in general topology
Let $(X,\tau)$ be the topological space where $τ = \{A \subseteq X : p \in A\} \cup \{\varnothing\}$ (i.e., $\tau$ is the $p$-inclusion topology). Then show that $(X,\tau)$ is not a $T_1$-space.

I know that the definition

A space $X$ is called a $T_1$-space if for every $x,y \in X$ with $x \neq y$ there exists two open sets $U$ and $V$ such that $x \in U$, $y \notin U$ and $y \in V$, $x \notin V$.

So $p$ in $A$ and $A$ in $X$.

答:
The singleton $\{p\}$ is not closed, as $X\setminus \{p\}$ is not open: it does not contain $p$, nor is it empty (if $|X| \ge 2$, which is necessary anyway, so the original exercise was incomplete too).

If you're unfamiliar with the fact $X$ is $T_1$ iff for all $x \in X$, $\{x\}$ is closed, which I used above, you can pick $q \neq p$ by the assumption $|X| \ge 2$ again and note that there is no open set $U$ that contains $q$ but not $p$: the only open set that does not contain $p$ is $\emptyset$ and that is ruled out by $q \in U$..

Title: T1 space in general topology


问:
Get $x,y$ coord of a rectangle when rotation is $0$
I have this square div in the page. I need to get the $x,y$ coords (left/top) of the box based on rotation 0 even when the box is rotated at a different angle.
Is there a formula to calculate the the $x,y$ coords of a square with rotation angle of $0$.

Here is a demo - http://jsfiddle.net/blessenm/XrVTp/

You can see the rectangle is rotated to $22$ degree. And I get the value $'x: 208, y: 234'$.
Where $x$ is bottom left corner and $y$ is top left corner.

If the angle is at a rotation $0$. The value I get is $'x: 208, y: 234'$. This exactly what want irrespective of the rotation.

I know this question has a bit programming but I think strong math people might be able to solve it.

Btw, I get can get the following values as input.

The distance between the left most point in the rotated rect and the left of the container.

The distance between the top most point in the rotated rect and the top of the container.

The distance between the right most point in the rotated rect and the left of the container.

The distance between the bottom most point in the rotated rect and the top of the container.

Angle of rotation

答:
Note that the centre of the red square will not change when rotated. So we can find its coordinates first (by taking averages), then figure out the top left corner (based off the fact that the size of the red square is given beforehand; in this case, it was $256\times 256$). Here's a simple implementation:</p>

```<code>document.querySelectorAll('input')[0].addEventListener('click',function() {
var left = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().left),
    right = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().right),
    top = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().top),
    bottom = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().bottom),
    x_mid = (left+right)/2.0,
    y_mid = (top+bottom)/2.0,
    x = x_mid - 256/2, // Subtract by half the knob's width.
    y = y_mid - 256/2  // Subtract by half the knob's height.
alert('left: '+left+', right: '+right+', top: '+top+', bottom: '+bottom+'\n'
      +'centre: ('+x_mid+', '+y_mid+')\n'
      +'x: '+x+', y: '+y);

//Alerts x: 208, y: 234 when angle is 0. Need to get this value all the time irrespective of the rotation.
});
</code>```

Body:
Note that the centre of the red square will not change when rotated. So we can find its coordinates first (by taking averages), then figure out the top left corner (based off the fact that the size of the red square is given beforehand; in this case, it was $256\times 256$). Here's a simple implementation:

document.querySelectorAll('input')[0].addEventListener('click',function() {
var left = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().left),
    right = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().right),
    top = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().top),
    bottom = Math.round(document.querySelectorAll('#knob')[0].getBoundingClientRect().bottom),
    x_mid = (left+right)/2.0,
    y_mid = (top+bottom)/2.0,
    x = x_mid - 256/2, // Subtract by half the knob's width.
    y = y_mid - 256/2  // Subtract by half the knob's height.
alert('left: '+left+', right: '+right+', top: '+top+', bottom: '+bottom+'\n'
      +'centre: ('+x_mid+', '+y_mid+')\n'
      +'x: '+x+', y: '+y);

//Alerts x: 208, y: 234 when angle is 0. Need to get this value all the time irrespective of the rotation.
});


问:
Condition for a subset of an affine space to be affine
Let $E$ be an affine space attached to a $K$-vector space $T$ and $a \in V\subset E$. Consider the bijection $f:E\rightarrow T$ send $x$ to $x-a$, where $x=(x-a)+a$. Clearly $f(E)$ is a $K$-subspace of $T$.

Suppose for every family $(x_i)_{i\in I}$ of elements of $V$ and every
family $(\lambda_i)_{i\in I}$ such that $\sum_{i\in I}\lambda_i=1$, we
have $\sum_{i\in I}\lambda_i(x_i-a)+a\in V$. Then $f(V)$ is a
$K$-subspace of $T$.

Let $x,y\in V$ and $\lambda,\mu\in K$. We want to show that $\lambda f(x)+\mu f(y)\in f(V)$. There exists $z\in E$ such that $\lambda f(x)+\mu f(y)=f(z)$: i.e. $\lambda(x-a)+\mu(y-a)=z-a$. I want to show that $z\in V$, but I am not able to use the assumption since the sum of the coefficients is not necessarily $1$. Any suggestions?

答: Note that $a\in V$, so you can use $a$ as one of the $x_i$.  Since $a-a=0$, this means that the condition that the sum of the coefficients is $1$ doesn't do anything, since you can change the coefficient of $a-a$ arbitrarily.  In your situation, you can rewrite $z-a$ as $$\lambda(x-a)+\mu(y-a)+\nu(a-a)$$ for whatever value of $\nu$ you want, and in particular you can choose $\nu$ such that $\lambda+\mu+\nu=1$.


问:
Proving a and b are perfect squares if and only if $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares

Let $a$, $b$ be two positive integers. Prove that $a$ and $b$ are both perfect squares if and only if both $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ are perfect squares.

I believe the proof is based in prime factorization, as if $a$ is a perfect square, then $a = (p_1^{a_1}\cdot\ldots\cdot p_n^{a_n})^2$, and the same goes for the gcd and lcm when they are perfect squares. I'm just a little lost on how to connect the dots and prove both directions.

答:
\begin{align}
   \gcd(a,b) &= m^2 \tag{A} \\
   \operatorname{lcm}(a,b) &= n^2 \tag{B}
\end{align}

Let $p_i$ represent the $i^{\text{th}}$ prime number.

Then we can say

$a = \prod p_i^{\alpha_i}$ where all but a finite number of the $\alpha_i$ are non-zero.

$b = \prod p_i^{\beta_i}$ where all but a finite number of the $\beta_i$ are non-zero.

Then (A) implies $\min(\alpha_i, \beta_i)$ is even for all $i$ and
(B) implies $\max(\alpha_i, \beta_i)$ is even for all $i$

It follows that $\alpha_i$ and $\beta_i$ are both even for all $i$; that is to say, $a$ and $b$ are perfect squares.


问:
Prove that there is a subsequence $(x_{n_k})$ that is bounded above, where $(x_n)$ is a sequence of real numbers that does not diverge to infinity
Suppose that $(x_n)$ is a sequence of real numbers that does not diverge to infinity. Prove that there is a subsequence $(x_{n_k})$ that is bounded above. I did a proof but that proof showed that $(x_n)$ is bounded above, which is not necessarily true, so something was wrong with that proof.

答:
We have that $x_{n} \not\to \infty$, then  $\exists M>0$ such that for all $N>0$ there exists some $n>N$ such that $x_{n} \leq M$

So if we first take $N = 1$ we have that there exist some $n_{1}>1$ such that $x_{n_{1}} \leq M$

Then we take $N = n_{1}$, so there exist some $n_{2} > n_{1}$ such that $x_{n_{2}}\leq M$

Thus we can construct a subsequence $\{ x_{n_{k}} \}_{k\in \mathbb{N}}$ such that $x_{n_{k}} \leq M$ for all $k \in \mathbb{N}$


问:
Highest-weight module over $\mathfrak{sl}_2$
I'm reading the following example, which I have trouble understanding:

Let $\mathfrak{g}=\mathfrak{sl}_2$ and $\lambda=m\Lambda_1$. Consider the irreducible module $V$. Then the vectors in the representation are $\{v,fv,\cdots,f^mv\}$ with weights $\{\lambda,\lambda-\alpha_1,\cdots,\lambda-m\alpha_1\}$

It's not hard to show that the vectors have corresponding weights as described above. I'm assuming we think of $V$ as a highest-weight module with highest weight $\lambda$, but what are $\Lambda_1$ exactly? What importance does $\lambda=m\Lambda_1$ have upon determining the weights and corresponding vectors as described above?

My guess is that $\lambda=m\Lambda_1$ has something to do with $f^{m+1}v=0$ so the vectors and corresponding weights are exactly as stated above. However this is not clear to me, if it's the case.

I should mention that this example is thought of in the setting of Kac-Moody algebras.

答: No, $\Lambda_1$ is not a random element of $\mathfrak h^*$. It is a very specific element of $\mathfrak h^*$: it's the only linear form $\Lambda_1\colon\mathfrak h\longrightarrow\Bbb C$ such that$$\Lambda_1\left(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right)=1.$$


问:
What is a term-by-term explanation of Fourier Transform?
I've been self-studying Fourier Transform for a while ("vanilla" transform, not FFT, DFT or other types of that sort) but I cannot seem to understand how the mathematical expression reflects the idea. Most sources that claim to be "intuitive" seem to recite similar ideas, like wrapping the function around a circle or representing functions as movements in circular patterns, etc. I kind of understand that process, but I cannot understand how it is reflected in the equation. I cannot understand the phrase " $$(e^{-2\pi(i)\alpha x})$$ represents wrapping the input wave function $f(x)$ around the origin of the complex plane at some frequency $\alpha$ " which has been shown here, in particular. I know that at the heart of it lies the Euler's formula, which represents rotations with a given frequency.

How does it imply "wrapping"? How does this relate to the rest of the formula? I think my confusion is caused by notation, rather than the concept itself, but please feel free to correct me.

答: If it comes to visualization $3$Blue$1$Brown is one of the best. Here he talks about the Fourier Transform and explains the idea of wrapping intuitively (Minute: 2:55). Also this video may give you a new perspective too.


问:
How to pass minimum count of apples so that every one around the table has same amount of apples?
With three ($A_1,A_2 and A_3)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=4 $ apples,to ensure each one has equal amount of 3 apples,  A3 just  gives A1 one apple A3 $\to$ A1 =1. In this case, the minimum count of ample exchanged is 1.

With four ($A_1,A_2,A_3 and A_4)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=6,a_3=9 $ apples, to ensure each one has equal amount of 5 apples,  A1 $\to$  A2 =0, A2 $\to$ A3 =-2,  A3 $\to$ A4 ==-1, A4 $\to$ A1 =3. In this case, the minimum count of apples exchanged is 0+2+1+3=6.

Question : With n people $A_i$(i from 1 to n) having  $a_i$ apples initially, what  the minimum count of apples exchanged needed so that each will have  same amount of apples in the end? (please note the question is about the minimum count of apples exchanged, not the minimum count of apple exchanges: we are counting the apples, not the exchanges.

Clarification:   The exchange can only happen between 2 neighbors. For any person  $A_i$, you can  give away $x_i$ to $A_{i+1}$ and take in $x_{i-1}$ from $ A_{i-1}$. $x_{i-1}$ and $x_{i+1}$ can be any integers (including 0).

答:
First assume it is a linear table where $1$ can pass to $2$, $n$ can pass to $n-1$ but everyone else can pass to the person to their left or right.

Use induction strategy.

Let $n$ be the number of people at the table and assume it is known how many apple passes it takes to make everybody have the same number of apples for all numbers of people less than $n$.

$k=\frac{1}{n}\sum_{i=m+1}^n a_i$ is the number of apples you want everyone to eventually end up with.

If the first person has $k$ apples, leave that person out and solve the known problem of re-distributing among the remaining $n-1$ people.
If the first person has more than $k$ apples, they will have to pass the excess apples to person $2$ and now the problem has be reduced to a known answer with $n-1$ people remaining.

What happens if the first person has less than $k$ apples?
Find the smallest $j$ such that $\sum_{i=1}^j a_i \ge k$.

Have those people $2$ through $j-1$ pass all their apples to person $1$ and have person $j$ pass as much as needed so that person $1$ will end up having exactly $k$.

Then, solve the problem of making the remaining $n-1$ people each have $k$ apples.

By induction, this strategy always does it in the minimum number of apples passes possible.

Now, solve the circular table problem.

It seems like if you try each linear strategy treating persons $1$ through $n$ as at the head of the line of the linear table and minimize that over all of them, you will get the minimum for the circular table too. That assumes the mimium for the circular table has at least one pair of people sitting next to each other that don't have to pass apples to each other. At any rate, you have an upper bound for the minimum.

I would be interested if anyone can find an initial configuration such that the minimum for the circular table method is lower than the minimum of all possible linear solutions putting persons $1$ through $n$ as the first seat of the linear table.


答:
Essentially, at the heart of the result is that $B = (A^\top A)^{-1} A^\top$ is the Moore-Penrose inverse of $A$ whenever $(A^\top A)^{-1}$ exists (which is when the columns of $A$ are linearly independent). It's OK if you don't know this term, because it's defined by $4$ simple properties:

$ABA = A$

$BAB = B$

$(AB)^* = AB$

$(BA)^* = BA$

As it turns out, for every matrix $A$, there is exactly one matrix $B$ satisfying all four conditions above, and that matrix is the Moore-Penrose inverse of $A$. If $A$ is invertible, then $B = A^{-1}$. We don't need to prove any of this, but if we have a $B$ that satisfies the above four conditions (which we do: check them using our formula for $B$!), then $P = AB$ is the projection matrix onto the columnspace of $A$. So, we should be able to prove this using only the above four properties.

First, note that $P$ is a projection matrix, as
$$P^2 = A\color{red}{BAB} = A\color{red}B = P,$$
using the second property. Next, $P$ is self-adjoint, which is precisely the third property, making $P$ an orthogonal projection onto its columnspace. The only thing we now need to prove is that $\operatorname{colspace}(P) = \operatorname{colspace}(A)$.

Recall that one way to represent the columnspace of a matrix is
$$\operatorname{colspace}(M) = \{Mx : x \text{ is a column vector of appropriate size}\}.$$
So, the columnspace of $P$ is made up of vectors of the form $Px = A(Bx)$, which are elements of $\operatorname{colspace}(A)$ by the above characterisation.

On the other hand, if we take arbitrary $Ax \in \operatorname{colspace}(A)$, then we can write
$$Ax = ABAx = P(Ax) \in \operatorname{colspace}(P),$$
using the first property. So, our column spaces for $A$ and $P$ coincide, hence $P$ is the orthogonal projection matrix onto $\operatorname{colspace}(A)$.

Title: Proof that the Matrix of Orthogonal Projection can be calculated by $P = A (A^t A)^{-1} A^t$.

Body:
Prove that the Matrix of Orthogonal Projection $P$ in any subspace $W$, can be calculated by $P = A (A^t A)^{-1} A^t$ where ColA forms a basis for W.

Am absolutely lost in this one, please help.


问:
Maxima and Minima of Sin(x)/x
I am trying to calculate the maximum and minimum points (between $-3\pi$ and $3\pi$) of $$f(x)=\frac{\sin(x)}{x}$$

I have found the derivative of the function and let it equal to zero.

$$f'(x)=\frac{x\cos(x) - \sin(x)}{x^2}$$

$$f'(x)=0$$
$$\frac{x\cos(x) - \sin(x)}{x^2}=0$$
$$x\cos(x) - \sin(x)=0$$
$$x\cos(x)=\sin(x)$$
$$x=\tan(x)$$

I am unaware as to how to find $x$. I assume that once I find $x$, I can use a sign diagram or second derivative test to determine the minimum and maximum values. Any help would be highly appreciated.

答:
Without proof, the $m^{th}$ extremum of $y(x) =\dfrac{\sin(x)}{x}$ is given by the root of the following function:
$$ \sqrt{1-y_m^2} +y_m \sin^{-1}(y_m)-(-1)^{m}\left(m+\dfrac{1}{2}\right)\pi y_m=0$$

It can be simplied to the following:

$$ \sin^{-1}(y_m)+\cot(\sin^{-1}(y_m))=(-1)^m \left(m+\dfrac{1}{2}\right)\pi$$

It is an exact expression and you can find all the extrema of $y(x) =\dfrac{\sin(x)}{x}$ using it. It can be easily verified that $y_0=1$ (which is the $0^{th}$ extremum) is a root of this equation. However, for $m\ge1$ you cannot solve this equation analytically. But no worries, if you would look at the graph of $y(x) =\dfrac{\sin(x)}{x}$ you would notice that it decays pretty fast owing to the $\dfrac{1}{x}$ factor and $y_m\ll1$ should not be a bad approximation. Under this approximatiom, we will have the following $O(y_m^3)$ approximations:

$$\sin^{-1}(y_m)\approx y_m+O(y_m^3)$$
$$\sqrt{1-y_m^2}\approx 1-\dfrac{1}{2}y_m^2+O(y_m^3)$$

Plugging these into the exact equation, we have:

$$ \dfrac{1}{2}y_m^2-(-1)^{m}\left(m+\dfrac{1}{2}\right)\pi y_m+1=0$$

We get a quadratic equation which we can exactly solve. So, on solving it we have found an expression that gives approximate extrema values for $y(x) =\dfrac{\sin(x)}{x}$.

$$\boxed{y_m\approx\pi\left(m+\dfrac{1}{2}\right)(-1)^m-(-1)^m\sqrt{\pi^2\left(m+\dfrac{1}{2}\right)^2-2}}$$

You can also convert this single equation into two equations, one for the maxima and one for the minima just by setting $m$ as even or odd integer.

Now, checking for how accurate our formula is. The exact and the approximate formulas are solved for their roots. The extrema $y_0$, as expected under our assumption of $y_m\ll1$, is the least accurate.

$$\begin{array}{|c|c|c|c|}  \hline
     &exact         &approx        &error\%           \\  \hline
y_0  &+1.000000     &+0.887129     &11.2871\%         \\  \hline
y_1  &-0.217234     &-0.217213     &0.00963963\%       \\  \hline
y_2  &0.128375      &0.128373      &0.0011467\%        \\  \hline
y_3  &-0.0913252    &-0.0913249    &0.000291779\%       \\  \hline
y_4  &+0.0709135    &0.0709134     &0.000105792\%       \\  \hline
y_5  &-0.0579718    &-0.0579718    &0.0000471872\%       \\  \hline
y_6  &+0.0490296    &0.0490296     &0.0000241245\%       \\  \hline
y_7  &-0.0424796    &-0.0424796    &0.0000135875\%        \\  \hline
\end{array}$$

Title: Maxima and Minima of Sin(x)/x


问:
Convergence/divergence of improper integrals
I have a task like that and I'm hopeless with that:

Find the function so that $$J_0=\int_0^\infty f(x)\,{\rm d}x $$ converges and the integrals $$J_1=\int_0^\infty f(x)\sin x \,{\rm d}x,\quad J_2=\int_0^\infty f(x)\sin (x^2)\, {\rm d}x$$ diverge.

Does anyone have a clue for searched function? Thank you so much.

答:
Obviously $J_0$ should converge only conditional. The idea is to choose $f$ with the same sign as $\sin x$ or $\sin(x^2)$, so after multiplication it will become positive.

Let $g(x) = \frac{\sin x}{x}$ and $h(x) = \frac{\sin\left(x^2\right)}{x}$. It's easy to check that $\int_1^\infty g(x)\,\rm dx$ and $\int_1^\infty h(x)\,\rm dx$ converge, and $\int_1^\infty g(x)\sin x\,\rm dx$ and $\int_1^\infty h(x)\sin\left(x^2\right)\,\rm dx$ diverge.

Now, we have three possibilities.

If $\int_1^\infty g(x)\sin\left(x^2\right)\,\rm dx$ diverges, take $f(x) = g(x)$.

If $\int_1^\infty h(x)\sin\left(x\right)\,\rm dx$ diverges, take $f(x) = h(x)$.

If they both converge, take $f(x) = g(x) + h(x)$.


问:
Can similar convex heptagons tile the plane?
It is a fairly straightforward matter to apply Euler's formula $V-E+F=2$ for planar graphs to see that congruent convex heptagons cannot tile the plane. The graph associated to the heptagons within distance $R$ of the origin, with edges between heptagons that overlap along an edge, has degree $\ge7$ at every "interior" vertex but (for sufficiently large $R$) insufficiently many exterior vertices to balance things out. This can be extended to any family of convex heptagons whose area and perimeter are bounded - see e.g. this thread on the /r/mathriddles subreddit for a more detailed proof.

I am curious whether it is possible to tile the plane with convex heptagons which are similar to each other; I suspect it is impossible, but the above sort of argument seems to fail, because we cannot assume boundedness.

Note that there do exist tilings of the plane by heptagons if we relax these boundedness constraints; for instance, one can take a tiling of the hyperbolic plane by heptagons in the disc model and apply a map $(r,\theta)\mapsto(\frac{r}{1-r},\theta)$ to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).

Is it possible to tile the plane with similar convex heptagons? If the answer is "no", I am interested in the case of finitely many heptagon shapes, which I suspect is also impossible.

Edit: Edward H.'s answer made me realize that there are some additional tiling conditions that seem relevant when allowing for arbitrarily-scaled pieces.

Say that a tiling is lower-bounded if there is some $\epsilon>0$ such that every tile contains a ball of radius $\epsilon$.

Say that a tiling is locally finite if the boundary of every tile is covered by finitely many other tiles.

Say that a tiling is traversable if you can reach any tile from any other tile by crossing over shared edges.

Say that a tiling is complete if every point in the plane is in the closure of a tile (as opposed to being in the closure of the union of all tiles).

I'm most interested in the case of tilings that are locally finite, traversable, and complete, but I'm also curious to hear about possibility/impossibility results for any of the above restrictions. (I'm also interested in any implications between them - I think that being lower-bounded implies the other three conditions, but I don't have a proof yet.)

答: Here's a construction that tiles a half-plane with two similar heptagons (four if discounting reflections). Two half-planes tile a plane. I'm having trouble coming up with a tiling using one similar heptagon, but at least the finite case is settled :)


问:
Using the chain rule to differentiate composite functions (exponential of an exponential)
How would I differentiate the following two functions using the chain rule.

$t=(e^{2y})^{e^{y^2}+1}$

$t=e^{{(y^3-\ln(y)+1)}^5}$.

Body:
$$t(y)=(e^{2y})^{e^{y^2}+1}=e^{(2y)(e^{y^2}+1)}=e^{f(y)}=g(f(y))\\
t'(y)=(g(f(y)))'=f'(y)g'(f(y))$$
since we have defined $g(x)$ as $e^x$, we get $g'(x)=e^x$, thus
$$\boxed{g'(f(y))=e^{f(y)}=e^{(2y)(e^{y^2}+1)}}$$
now, we need to differentiate $f(y)=(2y)(e^{y^2}+1)=g(y)h(y)$, with $g(y)=2y$ and $h(y)=e^{y^2}+1$.
$$f'(y)=g'(y)h(y)+g(y)h'(y)$$
calculating $g'(y)$ is easy; $\boxed{g'(y)=2}$.

as for $h'(y)$, then you could further look at $h(y)=e^{y^2}+1$ as a composite function $h(y)=e^{z(y)}+1=w(z(y))$, with $z(y)=y^2$ and $w(x)=e^x+1$.
$$h'(y)=(w(z(y)))'=z'(y)w'(z(y))$$
easily, we get $z'(y)=2y$ and $w'(x)=e^x$, thus $w'(z(y))=e^{z(y)}=e^{y^2}$.
$$\boxed{h'(y)=2ye^{y^2}}$$
let's put everything together:
$$\boxed{f'(y)=2(e^{y^2}+1)+2y(2ye^{y^2})}$$
and, finally:
$$t'(y)=f'(y)g'(f(y))=\left[2(e^{y^2}+1)+2y(2ye^{y^2})\right]e^{(2y)(e^{y^2}+1)}$$
the rationale for the second function is the same.


问:
On the centres of the dihedral groups
In an proof that I recently read, the following 'fact' is used, where $D_{2n}$ denotes the dihedral group of order $2n$:

If $n$ is even, then $D_{2n} \cong C_2 \times D_n$.

The (short) given justification is that the centre $Z(D_{2n}) \cong C_2$, whenever $n$ is even, and is trivial provided that $n$ is odd.

However, here is a result that a friend of mine found in the litterature which contradicts the previous one.

Suppose that 4 divides $n$. Then $D_{2n}$ is not isomorphic to $C_2 \times D_n$.

Proof: Suppose otherwise. We know that $Z(D_{2n}) \cong C_2$. By assumption, $D_{2n} \cong C_2 \times D_n$, therefore $Z(D_{2n}) \cong Z(C_2 \times D_{n})$. Since in a direct product $A \times B$, the groups $A$ and $B$ commute, we obtain that $Z(C_2 \times D_{n}) = Z(C_2) \times Z(D_n) \cong C_2 \times C_2$, a contradiction. QED.

Now comes my first question : Is the above proof correct?

Second question : Given a finite group $G$ whose centre is not trivial, does it always exist a group $H$ such that $G \cong Z(G) \times H$?

If so, then given $n$ a multiple of 4, the dihedral group $D_{2n}$ could be written as a direct product $C_2 \times H$. Here comes my third question : What should be $H$ since we know that it cannot have nontrivial centre (and, in particular, since $H$ cannot be $D_n$)?

答:
A comment about the assertion (first question).

"when n is even then $D_{2n}\cong C_2 \times D_{n}$".

This looks to be not true: consider a special class of dihedral groups namely $2-$groups, i.e. dihedral groups of order $2^n$, with $n\geq 3$. These are non-abelian $2-$ groups. As you said, their centers are non-trivial (should be; it is true for any p-group); the center of the dihedral group $D_{2^n}$ of order $2^n$ is $C_2$.

But by Sylow-theory, it should intersect with every normal subgroup of $D_{2^n}$ non-trivially, hence should be contained in that normal subgroup(as center has order 2); in particular the center must be contained in every maximal subgroup of $D_{2^n}$, and hence in particular, it should be contained in subgroups isomorphic to $D_{2^{n-1}}$.

Therefore, $D_{2^n}$ can not be (internal) semidirect product of $C_2$ by $D_{2^{n-1}}$.; hence in particular it can be direct product of $C_2$ by $D_{2^{n-1}}$.

[Another simple argument can be given: if for $n$ even, $D_{2n}\cong C_2 \times D_{n}$, then in particular we should have (by induction)

$D_{2^n} \cong C_2 \times D_{2^{n-1}} \cong C_2 \times (C_2 \times D_{2^{n-2}})\cong \cdots \cong C_2 \times C_2 \times \cdots \times D_4$, which is elementary abelian $2$-group, contradiction.

Already very simple answers given above to second and third question, so nothing to add more.]

Title: On the centres of the dihedral groups


问:
help with $\lim_{(x,y) \to (0,0)} {\ln(xy + 1) \over x}$
Compute:

$\lim_{(x,y) \to (0,0)} f(x,y)$ if $f(x,\ y)={\ln(xy + 1) \over x}$

According to Wolfram the limit is 0.

I'm trying to verify it by approaching the origin from different paths.

When approaching the origin from $x$:

$\lim_{(x,y=0) \to (0)} f(x,0) = {\ln(1) \over x}={0\over x}= 0$

However, I don't know how to proceed when approaching the origin from $y$. In this case, $x=0$ and the denominator is zero.

$\lim_{(x=0,y) \to (0)} f(0,y) = {\ln(1) \over 0}={0\over 0}$

I guess I should try get rid of that zero in the denominator, but I don't see how as if I set $x=0$ all the variables disappear.

答:
The expression
$${\log(1+xy)\over x}$$
is not defined in a full neighborhood of $(0,0)$. Therefore the limit doesn't make sense without further comment. One way out could be the following: The function
$$g(z):=\cases{{\log(1+z)\over z}\quad&$(z\ne 0)$ \cr 1&$(z=0)$\cr}$$
is analytic in a neighborhood of $z=0$ and has a Taylor expansion $g(z)=1-{z\over2}+{z^2\over3}-\ldots$ at $0$. More important, we have
$$\log(1+z)= z\>g(z)\qquad\bigl(|z|<1\bigr)\ .$$
It follows that for $|x|<1$, $\>|y|<1$ we can write
$$\log(1+xy)=xy\>g(xy)$$
and therefore
$${\log(1+xy)\over x}=y\> g(xy)\qquad(x\ne0)\ .$$
As the right side of the last equation is continuous in a neighborhood of $(0,0)$ the expression in question can be extended continuously into this full neighborhood, and the limit when $(x,y)\to(0,0)$ is obviously $=0$.</p>


问:
Showing $\cos(P − Q) = \frac{33}{85}$ for obtuse $P$ and reflex $Q$ such that $\sin P = \frac{8}{17}$ and $\tan Q = \frac43$,

$\sin P = \frac{8}{17}$ and $\tan Q = \frac43$. If $P$ is obtuse and $Q$ is reflex, show clearly that $\cos(P − Q) = \frac{33}{85}$.

My working:

$$\cos P \times \cos Q + \sin P \times \sin Q$$
$$\left(-\frac{15}{17} \times -\frac35\right) + \left(\frac{8}{17} \times -\frac45\right) = \frac{13}{85}$$
which is not $33/85$. Where have I gone wrong?

Body:
$\cos(p-q) =\cos(p)×\cos(q)+\sin(p)×\sin(q)$

$\Rightarrow$$\cos(p-q)= \cos(q) (\cos(p) +\sin(p) ×\tan(q)) $

$\Rightarrow$$\cos(p-q) =\cos(q) (\cos(p) +\frac{8}{17}×\frac{4}{3})$

We know $\sin(p) =\frac{8}{17}$

$\Rightarrow$$1-\sin(p)^2=1-(\frac{8}{17})^2$

$\Rightarrow$$ \cos(p) =\frac{15}{17}$ or $\cos(p) =-\frac{15}{17}$

And we know $\tan(q) =\frac{4}{3}$$\Rightarrow$$ 1+\tan(q)^2=1+(\frac{4}{3})^2$$\Rightarrow$$ \cos(q) =\frac{3}{5} $ or $ \cos(q) =-\frac{3}{5}$

$\cos(p-q) =\cos(q) (\cos(p) +\frac{8}{17}×\frac{4}{3})>0$$\Rightarrow $$ \cos(q) >0 $ and $\cos(p) >\frac{-32}{51}$
So In this Cas $\cos(q) =\frac{3}{5}$ and $ \cos (p) =\frac{15}{17}$

So $\cos(p-q) =\frac{3}{5} (\frac{15}{17}+\frac{8}{17}×\frac{4}{3})=\frac{77}{85}$

In other Case $\cos(q) <0 $ and $\cos(p) <\frac{-32}{51}$ so $\cos(q) =-\frac{3}{5}$
and  $\cos(p) =\frac{-15}{17}$

So $\cos(p-q) =\frac{-3}{5} (\frac{-15}{17}+\frac{8}{17}×\frac{4}{3})=\frac{13}{85}$


问:
How to find the remaining percentage of lemon juice?
The problem is as follows:

Assume that you have three mugs. Each mug has a capacity of
$1\textrm{liter}$. The first mug is full and it has $60\%$ of orange
juice and $40\%$ of water. The second mug which is also full is $80\%$
of orange juice and $20\%$ of lemon juice. Finally the third mug is
empty. Now, you want to fill the third mug using the contents of the
other two mugs mentioned. The condition is that such mug must have the
double amount of orange juice that of water. Using this information.
What percentage of lemon juice will remain in the third mug?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{2 percent}\\
2.&\textrm{4 percent}\\
3.&\textrm{5 percent}\\
4.&\textrm{6 percent}\\
\end{array}$

I'm confused exactly how should I approach this question. Is there more than meets the eye here?.

What I tried to do is this:

Assuming $x$ is the volume to be taken from the first mug. Then my equation becomes into:

$$\left(\frac{6}{10}\right)\cdot x + \left(\frac{8}{10}\right)(1-x)=\left(\frac{2}{3}\right)\cdot 1$$

The two thirds comes that the requested concentration of the orange juice is two times that of water. (Or at least that's how I'm understanding it).

Hence

$$2:1\equiv\frac{2}{2+1}$$

Therefore solving all of this yields $x=\frac{2}{3}$, which I believe should the volume of orange juice to be taken from the first mug.

Now, because $1-x=1-\frac{2}{3}=\frac{1}{3}$

Is taken out from the second mug. Then I conclude that one third of liter has in its content one third of the $\frac{2}{10}$ of lemon juice.

Thus:

$$\frac{1}{3}\times\frac{2}{10}=\frac{2}{30}=\frac{1}{15}$$

which in terms of percentage would be 6.67 % but this answer does not appear in any of the choices. What went wrong?. Can someone help me here to spot exactly which part did I derailed from the intended interpretation?.

It would help a lot that an answer could explain this to me so I can be adequately guided.

答:
If $x$ if the percentage of volume you remove from the first mug, the equation you should be solving is (because you want your total volume of orange juice in the third mug [LHS] to be equal to twice the volume of water in the third mug [RHS])
$$\frac{6}{10}x + \frac{8}{10}(1-x) = 2 \ \frac{4}{10}x$$
which yields $x= \frac{4}{5}$ and consequently the percentage of lemon juice is given by
$$\frac{1}{5} \ \frac{2}{10} = \frac{1}{25} = 4\%.$$


问:
Modular Arithmetic/Congruence
Hello I was wondering if anyone could help better my understanding of finding remainders using congruence and modular arithmetics as I cannot wrap my head around it.

The question I have been presented is to find the remainder when $4444^{4444}$ is divided by 9.

Now I have started by working out $4444$ (mod 9)

$4444\equiv16\equiv-2 \pmod 9$

but for the remaining part I am unsure how to work out.

I have been told through the answers that it should be done by the following:

$4444^{4444} \equiv-2^{4444}\equiv 2^{3*1481+1} \equiv -2 \equiv 7$

I follow the first step as I am aware from a Theorem that;

$a \equiv b \pmod m$ implies $a^{k} \equiv b^{k}\pmod m$ for any integer $k\geq 0$

But I don't understand why we split up the power and how to integer 2 flips between positive and negative.

Thanks in advance for any help

答:
This is what you showed.

$4444^{4444} \equiv-2^{4444}\equiv 2^{3*1481+1} \equiv -2 \equiv 7$

The reason for $2^3$ is that $2^2 \equiv 8 \equiv -1 \pmod 9$

So
\begin{align}
   2^{3\times1481+1} \\
      &\equiv (2^3)^{1481} \cdot 2^1  \pmod 9\\
      &\equiv (-1)^{1481} \cdot 2  \pmod 9\\
      &\equiv (-1)\cdot 2  \pmod 9\\
      &\equiv -2  \pmod 9\\
      &\equiv 7 \pmod 9
\end{align}

Title: Modular Arithmetic/Congruence