问:
Proving that the product of two invertible matrices is also invertible, without determinants
I want to prove that if there are two matices A and B and A and B are both invertible, then the product A * B is also invertible.

This question is similar: Prove that the product of two invertible matrices also invertible but the answer uses determinants, which I have not learned about yet.

There is an answer provided in the thread above which does not use determinants.

Ans 2 is:

It is that $(AB)^{-1}=B^{-1}A^{-1}$, because $AB(AB)^{-1}=ABB^{-1}A^{-1}=1\!\!1$, but only for $n\times n$ matrices.

But I can't see how this shows that AB = matrix C such that C is invertible (such that there exists a matrix D where CD = DC = I, which is the definition of invertible given to me).

The other answer starts off by saying:

C is invertible iff for all y there is some x such that Cx = y.

I thought C is invertible iff there exists a matrix D such that CD = DC = I. How is it the case that for every y, there is some x such that Cx = y? Is y a matrix in this case? and is x a scalar?

答:
I don’t know if I understood your doubt. However, there is a very simple answer to the first question of your post. Suppose $A$ and $B$ are $n\times n $ matrices, both invertible. Then there are matrices $A^{-1}$ and $B^{-1}$ such that $A^{-1}A=AA^{-1}=I$ and $B^{-1}B=BB^{-1}=I$. We want to find a matrix that satisfies the same kind of relations for $AB$. Well, such matrix is $B^{-1}A^{-1}$. We prove this by direct verification;
$$B^{-1}A^{-1} AB=B^{-1}IB=B^{-1}B=I, $$
and, analogously,
$$
ABB^{-1}A^{-1}=I.$$

This concludes the proof.


问:
How many different strings of 5 lower case letters are there?
How many different strings of 5 lower case letters are there? Assume that letters may repeat and the string must contain at least one letter a?
I tried  this:
5×26^5=5×26 ×26 ×26 ×26=2684880</p>

答:
Break it into five sub-problems:  exactly 1 $A$, exactly 2 $A$s, ..., exactly 5 $A$s.

One $A$:  There are five places for the required $A$, and $25^4$ ways to fill the remaining four slots without the use of another $A$.

Two $A$s:  There are ${5 \choose 2}$ ways to place the two $A$s, and $25^3$ ways to fill the remaining $s$ slots with letters other than $A$.

...

Five $A$s:  There is ${5 \choose 5} = 1$ way to fill with all five $A$s.

Add these up.

ContentLicense: CC BY-SA 4.0


问:
Real Analysis - Closed and Open Sets
I have the next question from Rudin and want to know if it was answered correctly: Is every point of every closed set E $\subset$ $R^2$ a limit point of E?

I have answered as follows:

Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point. It is proved that all points in $R^2$ are limit points. Therefore $E^c$ must contain all limit points. But $E^c$ by definition is open ($E^c=R^2\setminus E$). This is a contradiction because $E^c$ has 2 limit points, namely $E=[0,1]$. Therefore E contains points which are not limit points.

答: The answer is no. Take a subspace of $\mathbb{R}^2$ consisting of a single point $P$. Then $\{P\}$ is closed, since its complement $\mathbb{R}^2\setminus\{P\}$ is open in the usual topology of $\mathbb{R}^2$. Now, $P$ is a point of $\{P\}$ but not a limit point for $\{P\}$ since you can't define in any way $P$ as the limit of a sequence of points of $\{P\}$ different from the constant sequence.</p>

ContentLicense: CC BY-SA 3.0

Title: Real Analysis - Closed and Open Sets


问:
Best book that covers all / most topics in logic
I am looking for a textbook or a series of textbooks that covers all or most of the topics in logic, that includes Aristotelian term logic, Stoic logic, Islamic logic, medieval logic, predicate calculus, all or most of non-classical logics, algebraic logic, model theory, proof theory, universal algebra, etc. In short, something that covers all historical and contemporary areas of logic in a most comprehensive way.</p>

答:
I don't think one can do much better than the 100+ page literature guide on introductory logic texts on logicmatters.net. The target audience here is not exclusively mathematicians, but also philosophy majors, so you'll not just be limited to the first- and second- order variants of mostly classical predicate logical logic that math faculties have by and large settled at 100 years ago.

In closing - while I know of topics such as Indian Jaina seven-valued logic - in view of the scope you want to cover in your question, I'm not sure if you will find a text that deeply discusses Stoic and Islamic logic on the one hand, and on the same time formal model theory.


答:
You set, to put it mildly, a rather high bar:

includes Aristotelian term logic, Stoic logic, Islamic logic, medieval logic, predicate calculus, all or most of non-classical logics, algebraic logic, model theory, proof theory, universal algebra, etc.

This is simply far too much material to be contained in a single book, or even a small set of books. The closest thing I can think of is the Handbook of the history of logic ... a series comprising eleven volumes, each with more than $600$ pages. And even then, many important topics are omitted. Meanwhile, even the "simple" topic of nonclassical propositional logics can lend itself to a ~$1500$-page treatment.

Title: Best book that covers all / most topics in logic

Body: I am looking for a textbook or a series of textbooks that covers all or most of the topics in logic, that includes Aristotelian term logic, Stoic logic, Islamic logic, medieval logic, predicate calculus, all or most of non-classical logics, algebraic logic, model theory, proof theory, universal algebra, etc. In short, something that covers all historical and contemporary areas of logic in a most comprehensive way.</p>


问:
Counterexample regarding basic properties of limits
Let $f, g$ be two functions defined on a deleted neighbourhood of
$a ∈ R$. If $g(x) \neq 0$ for every x on the deleted neighbourhood and $$\lim_{x\to a}{f(x)\over g(x)} = 1$$

then

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

I'm asked to prove or disprove this claim, along with others. My instinct in these type of questions is to start figuring out if there are any piecewise function that contradict the statements but I almost feel like I'm searching in the dark with no clear strategy.. is there an agreed upon process regarding searching for counter examples?

答:
Step One: Consider whether the problem can be simplified. In this case, we can eliminate the irrelevant distraction of $a$. Define $u=x-a$. Then instead of $\lim_{x \rightarrow a}f(x)$, we have $\lim_{u \rightarrow 0}f(u)$. Any proof or counterexample using the latter expression easily translates (no pun intended) to a solution of the former.

Step Two: State the negation. A statement of the form "if A then B" is false if there is a case where A is true but B is not. So we need a case where the first limit is 1, but the second is not zero.

Step Three: Enumerate the possibilities. There are four basic categories of limits: goes to zero, goes to a constant, goes to infinity, oscillates. The second two are both divergent on the real line, but the third is convergent on the extended real line. We're looking for a limit other than 0, so that leave a constant, infinity, or oscillates. The simplest is constant, and there isn't anything suggesting that one constant is better than another, so we can take it to be 1.

Step Four: Note the conditions/qualifications. The problem states that the functions need be defined only on a deleted neighborhood. Why would they include that detail? That suggests that we don't need to have them be defined for $u=0$. Why would they not be defined for $u=0$? Maybe they go to infinity there. What's the simplest function that goes to infinity? $\frac 1 u$. If one function is $\frac 1 u$, and the difference between that function and the other one is 1, then what is the other function? The simplest answer is $\frac 1 u+1$.

Step Five: Check whether you've accomplished the goal. It's generally easier to analyze a fraction when the more complicated expression is on top, so let's take $f(u) = \frac 1 u +1$ and $g(u) = \frac 1 u$. Now $\frac {f(u)}{g (u)} = u(\frac 1 u +1) = 1+u$. When we take the limit as $u$ goes to $0$, this indeed goes to $1$. And of course by construction the difference goes to $1$.

If you want to put this back in the original variable, you can substitute $u=x-a$ back in to get $f(x) =\frac 1 {x-a}+1, g(x)=\frac 1 {x-a}+1$, but I would expect that most instructors would find this last step trivial and the functions in term of $u$ to be sufficient.

If you combine Step One and Step Four, you might notice that if we take the substitution $u=\frac 1 {x-a}$, then the problem becomes to find two functions whose ration goes to $1$ when $u$ goes to infinity, but their difference doesn't go to $0$. If you're familiar with Big-O notation, that's asking for two functions in the same Big-O family whose difference doesn't go to zero, and $u+1$ and $u$ clearly satisfy that. When it comes to limits taking the independent variable to zero versus limits taking the independent variable to infinity, there are a lot of phenomena in one  that have an equivalent in the other, but one is easier to analyze than the other. If you're stuck with one of these types of limits, consider converting it to the other to see whether anything comes to mind.


问:
Verification that Chebyshev's Inequality Holds for $f(x) = \begin{cases} -2e^{-2x}+2e^{-x} &: 0 <x\\ 0 &: \text{else} \end{cases}$
I have the pdf $f(x) = \begin{cases}
-2e^{-2x}+2e^{-x} &: 0 <x\\
0 &: \text{else}
\end{cases}$

and I want to verify that Chebyshev's Inequality for holds for $k=2$.

After finding the mean and variance I have

$\mu = \frac{3}{2}$, $\sigma^2=\frac{5}{4} \implies \sigma=\sqrt{\frac{5}{4}}=1.118$,$1-\frac{1}{k^2}=1-\frac{1}{4}=\frac{3}{4}$ and then

$\mu \pm k\sigma$ which gives me $\frac{3}{2}-2\left(1.18\right)=-0.736$ and $\frac{3}{2}+2\left(1.118\right)=3.736$

So I have $P(-0.736 < X < 3.736)\geq 0.75$

but to verify this I need $\int^{3.736}_{-0.736}f(x)dx$ which does not seem correct.

Could anyone provide some feedback? Thanks.

答: You have to remember that the pdf is only defined for $0 < x$. So when checking that Chebyshev's Inequality holds, you only need to verify it for $\int_0^{3.736} f(x) dx$ which you can evaluate to $0.9528$ which is larger than $0.75$, so it does satisfy Chebyshev's Inequality.


问:
If $S$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that $S \circ S = S$, prove that $S$ can be:
We have that $S \circ S = S$, I want to prove that $S$ can be

$S = 0$ (The zero transformation) or

$S(x) = x$ (The Identity transformation) or

$S$ can have a basis $A$ such that $[S]_{A}^{A} = \begin{pmatrix}
1 && 0 \\
0 && 0
\end{pmatrix}$

I have that all of the three propositions are True, 3 is like to say that S can be a projection, in this case, over the $x$-axis, 2 can be done by see that $S(x) = S(S(x)) = x$, but what do you think? I try to find more about in the some books (I study for myself, not attending a school for now, but I love math, I try to solve the most, but some are really strange for me) thanks

答:
If $S$ is such that $S^2 = S$ then

$$
S^2 - S = 0
$$

and so $S$ is annihilated by the polynomial $x^2 - x = x(x - 1)$. Therefore, the minimal polynomial $\mu_S(x)$ of $S$ divides $x(x - 1)$. This leaves only three possibilities:

$\mu_S(x) = x$ which means that $S = 0$.

$\mu_S(x) = x - 1$ which means that $S = I$.

$\mu_S(x) = x(x - 1)$ which means that both $0$ and $1$ are the eigenvalues of $S$ and therefore we have

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

in the eigenbasis of $S$.


问:
The sum of Brownian motion increments
As I am studying about Brownian motion, as a preparation for my exam, I came across the following problem that I would like some help for:
Let $\{B(t), t ≥ 0\}$ be a standard Brownian motion and set $Y_n = \sum_{i=1}^{2^n} |B(\frac{i}{2^n}) - B(\frac{i-1}{2^n}) |$. Show that $E[Y_n] = 2^{n/2}\sqrt{2/\pi}$ and $\sum^∞_{k=1} P(Y_k < k) < \infty$.

Any help would be appreciated.

答:
Let $Z_1, Z_2, \dots$ be the sequence of i.i.d. standard normal variables. Then

$$ Y_n
\stackrel{d}= \frac{1}{2^{n/2}} \sum_{k=1}^{2^n} \left| Z_k \right|. $$

So by the Chebyshev's inequality,

$$ \mathbb{P}(\left| Y_n - \mathbb{E}[Y_n]\right| > \lambda) \leq \frac{\operatorname{Var}(Y_n)}{\lambda^2} = \frac{\operatorname{Var}(\left|Z_1\right|)}{\lambda^2}. $$

Using this, we estimate that

$$ \mathbb{P}(Y_n < n)
= \mathbb{P}(Y_n - \mathbb{E}[Y_n] < n - \mathbb{E}[Y_n])
\leq \frac{\operatorname{Var}(\left|Z_1\right|)}{(\mathbb{E}[Y_n] - n)^2}
\sim \frac{C}{2^n} $$

as $n\to\infty$, where $C = \frac{\pi}{2}\operatorname{Var}(\left|Z_1\right|)$ is an absolute constant. So $\sum_{n=1}^{\infty} \mathbb{P}(Y_n < n)$ converges by the limit comparison test.

Title: The sum of Brownian motion increments


答: There are 3 mutually exclusive and exhaustive cases. One is, kernel is of dimension 2 in which case you get the zero operator i.e. S=0, second case,  when kernel is of dimension zero in which case you get the identity operator i.e. S=I and third case,  kernel is of dimension one in which case image is of dimension one and you get the matrix you have given in your third case.</p>

Title: If $S$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that $S \circ S = S$, prove that $S$ can be:

Body:
We have that $S \circ S = S$, I want to prove that $S$ can be

$S = 0$ (The zero transformation) or

$S(x) = x$ (The Identity transformation) or

$S$ can have a basis $A$ such that $[S]_{A}^{A} = \begin{pmatrix}
1 && 0 \\
0 && 0
\end{pmatrix}$

I have that all of the three propositions are True, 3 is like to say that S can be a projection, in this case, over the $x$-axis, 2 can be done by see that $S(x) = S(S(x)) = x$, but what do you think? I try to find more about in the some books (I study for myself, not attending a school for now, but I love math, I try to solve the most, but some are really strange for me) thanks


问:
Let $a$ and $b$ be integers. Show that if $a \equiv b \pmod{p}$ for every prime $p$, then $a = b$.
Let $a$ and $b$ be integers. Show that if $a \equiv b \pmod{p}$ for every prime $p$, then $a = b$.

My attempt:

Suppose that $p\mid (a-b)$ and $a-b\neq 0$. Then $p\leq|a-b| $.

Since the set of prime numbers is infinite we can take $p'$ such that $p\leq |a-b|<p'$. But $p'$ divides $a-b$ as well, thus $p'\leq|a-b|$, which is a contradiction. So $a-b=0$, that is, $a=b$. I am right?

答: If $a\neq b$, there exists a prime which does not divides $a-b$, take $p>|a-b|$, $a-b\neq 0$ mod $p$.


答: The hypothesis implies that $a-b$ is divisible by every prime. But any nonzero integer is only divisible by finitely many primes. Thus $a-b=0$ and hence $a=b$

Body:
Let $a$ and $b$ be integers. Show that if $a \equiv b \pmod{p}$ for every prime $p$, then $a = b$.

My attempt:

Suppose that $p\mid (a-b)$ and $a-b\neq 0$. Then $p\leq|a-b| $.

Since the set of prime numbers is infinite we can take $p'$ such that $p\leq |a-b|<p'$. But $p'$ divides $a-b$ as well, thus $p'\leq|a-b|$, which is a contradiction. So $a-b=0$, that is, $a=b$. I am right?


答:
No, your argument is not correct. I’ll comment on it in detail.

Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point.

While it is true that every point of $[0,1]$ is a limit point of $[0,1]$, this is not a matter of definition: there is nothing in the definition of closed set that requires every point of a closed set to be a limit point of that set. Moreover, $[0,1]$ isn’t a subset of $\Bbb R^2$. I suspect that you’re thinking of the set $[0,1]\times\{0\}$, the set of points on the $x$-axis between $0$ and $1$ inclusive.

It is proved that all points in $R^2$ are limit points.

Limit points of what? They’re all limit points of $\Bbb R^2$, but that has nothing to do with $E$, which isn’t even a subset of $\Bbb R^2$. And if your $E$ was really supposed to be $[0,1]\times\{0\}$, then it’s not true that all points of $\Bbb R^2$ are limit points of $E$: if they were, they’d have to be in $E$, since $E$ is closed, and a closed set contains all of its limit points.

Therefore $E^c$ must contain all limit points.

Again, all limit points of what? And what does $\Bbb R^2$ have to do with $E^c$? $E^c=(\leftarrow,0)\cup(1,\to)$, a subset of $\Bbb R$, not of $\Bbb R^2$.  For that matter, why are you bringing in $E^c$ at all? The question is whether every point of $E$ is a limit point of $E$, and $E^c$ isn’t going to help you to answer that question.

But $E^c$ by definition is open ($E^c=R^2\setminus E$).

See above: $E^c=\Bbb R\setminus E$, not $\Bbb R^2\setminus E$.

This is a contradiction because $E^c$ has 2 limit points,

No, every point of $(\leftarrow,0]\cup[1,\to)$ is a limit point of $E^c$; that’s far more than two limit points!

namely $E=[0,1]$.

No, the only points of $E$ that are limit points of $E^c$ are $0$ and $1$; the points of $E$ that are in $(0,1)$ are not limit points of $E^c$.

Therefore E contains points which are not limit points.

This doesn’t follow from anything that you’ve said, and in fact it contradicts your very first claim about $E$ up above. Moreover, it isn’t true, assuming that you’re talking about limit points of $E$: every point of $E$ is a limit point of $E$.

In order to answer the question, you must do one of two things: either prove that whenever $E$ is a closed subset of $\Bbb R^2$, then every point of $E$ is a limit point of $E$, or find an example of a closed subset $E$ of $\Bbb R^2$ containing a point that is not a limit point of that set $E$. You won’t be able to do the former, because it isn’t true. For the latter, just take any non-empty finite subset of $\Bbb R^2$, for instance $E=\{\langle 0,0\rangle\}$: $E$ is closed, and it has no limit points at all. In particular, $\langle 0,0\rangle$ is not a limit point of $E$, because it is not true that every open neighborhood of $\langle 0,0\rangle$ contains a point of $E$ different from $\langle 0,0\rangle$. If you want $E$ to be an infinite set, you can use $E=\big(\Bbb R\times\{0\}\big)\cup\{\langle 0,1\rangle\}$, for instance. This is a closed set, but $\langle 0,1\rangle$ is a point in $E$ that is not a limit point of $E$: if $\epsilon\le 1$, the open ball of radius $\epsilon$ centred at $\langle 0,1\rangle$ does not contain and point of $E$ besides the point $\langle 0,1\rangle$ itself. For a more extreme example, let $E=\{\langle m,n\rangle:m,n\in\Bbb Z\}$. You should try to prove that $E$ is an infinite closed set that has no limit points at all.

A closed set always contains all of its limit points. That is, suppose that $E$ is closed. If $p$ is a limit point of $E$, then $p\in E$. The point of this exercise is to show that the converse is not true: there can be a point $p\in E$ that is not a limit point of $E$.

Title: Real Analysis - Closed and Open Sets

Body:
I have the next question from Rudin and want to know if it was answered correctly: Is every point of every closed set E $\subset$ $R^2$ a limit point of E?

I have answered as follows:

Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point. It is proved that all points in $R^2$ are limit points. Therefore $E^c$ must contain all limit points. But $E^c$ by definition is open ($E^c=R^2\setminus E$). This is a contradiction because $E^c$ has 2 limit points, namely $E=[0,1]$. Therefore E contains points which are not limit points.


问:
Just out of curiousity. If we view a formal system as an algebra, what do 'prime' sentences look like?
$\varphi$ is prime iff $(\psi\wedge\eta\to \varphi)\to(\psi\to \varphi\vee \eta\to \varphi)$ is true for every $\psi,\eta$.

For instance, in language of set theory(assuming no set axioms), $(\exists x)(x=x)$ should be prime. For if $\psi\wedge\eta\to (\exists x)(x=x)$, the empty set is not a model of $\psi \wedge \eta$. Therefore empty set can't be model of $\psi,\eta$ both.

Another example. $\varphi=(\exists x)(x=\{x\})$ is not prime. $((\exists x)(x=\{\{x\}\})\wedge (\forall x)(x=\{\{x\}\}\to x=\{x\}))\to \varphi$. Denote the two sentences by $\psi,\eta$. It is obvious that $\eta\not\to \varphi$, since $\eta$ has an empty model. To show $\psi\not\to\varphi$, consider $M=\{0,1\}$ and let $\{0\}=1$ and $\{1\}=0$. Then $\{\{0\}\}=0$ in $M$ but no $x\in M$ satisfies $\{x\}=x$.

Is there some interesting facts about such algebras?

答:
The algebras you refer to in your question are called Lindenbaum-Tarski algebras. For any language $L$, the set of $L$-sentences up to logical equivalence naturally forms a Boolean algebra. More generally, we can fix an $L$-theory $T$ and a context of free variables $(x_1,\dots,x_n)$, and the set of $L$-formulas with free variables from $(x_1,\dots,x_n)$ up to equivalence modulo $T$ forms a Boolean algebra (when the context is empty, we get the case of sentences). These Boolean algebras are studied extensively in the field of model theory.

Your question is about prime elements of the algebra of sentences modulo logical equivalence. In the comments, halrankard2 suggested that if $\lnot \varphi$ is complete, then $\varphi$ is prime. This is true, and in fact this is a characterization of the prime sentences. [Caveat: I think you definition of "prime" should also include the condition that $\varphi$ is not a tautology, i.e. $\varphi$ generates a prime (and, in particular, proper) ideal in the Boolean algebra of sentences. If we do this, then we should take "complete" to mean "complete and consistent", i.e. $\varphi$ is complete if $\varphi$ is not a contradiction and for any sentence $\psi$, either $\varphi\rightarrow \psi$ or $\varphi\rightarrow \lnot \psi$. Alternatively, if you want to consider a tautology to be prime, then you can allow contradictions to be complete and you'll still get the equivalence.]

Here's a proof:

Suppose $\lnot \varphi$ is complete. Suppose $\psi\land \eta\rightarrow\varphi$. Then $\lnot \varphi\rightarrow \lnot (\psi \land \eta)$. Now since $\lnot \varphi$ is complete, $\lnot \varphi\rightarrow \psi$ or $\lnot \varphi\rightarrow \lnot \psi$. Similarly, $\lnot \varphi\rightarrow \eta$ or $\lnot \varphi\rightarrow \lnot \eta$. But if $\lnot \varphi\rightarrow \psi$ and $\lnot \varphi\rightarrow \eta$, then $\lnot \varphi\rightarrow (\psi\land \eta)$. Since $\lnot \varphi$ does not imply a contradiction, we conclude that $\lnot \varphi\rightarrow \lnot \psi$ or $\lnot \varphi\rightarrow \lnot \eta$. So $\psi\rightarrow \varphi$ or $\eta\rightarrow \varphi$. Hence $\varphi$ is prime.

Conversely, suppose $\varphi$ is prime. Since $\varphi$ is not a tautology, $\lnot \varphi$ is not a contradiction. Now for any sentence $\psi$, we have $\psi\land \lnot \psi \rightarrow \varphi$, so $\psi\rightarrow \varphi$ or $\lnot \psi \rightarrow \varphi$. Thus $\lnot \varphi\rightarrow \lnot \psi$ or $\lnot\varphi\rightarrow \psi$. So $\lnot \varphi$ is complete.

Similarly, fixing a theory $T$, a sentence $\varphi$ is prime in the algebra of sentences modulo $T$ if and only if $\lnot \varphi$ is complete modulo $T$, i.e., for every sentence $\varphi$, $T\vdash \lnot\varphi\rightarrow \psi$ or $T\vdash \lnot\varphi\rightarrow \lnot\psi$.

The presence of the negation in this characterization is a little bit annoying - it's often more convenient in logic to think about filters, rather than ideals, on Boolean algebras. We get that $\varphi$ is complete if and only if $\varphi$ generates a prime filter: If $\varphi\rightarrow \psi\lor \eta$, then $\varphi\rightarrow \psi$ or $\varphi\rightarrow \eta$.

And that last fact is really just a rephrasing of the statement that every prime filter on a Boolean algebra is a maximal filter (ultrafilter). Equivalently, every prime ideal in a Boolean ring is maximal.


问:
Prove $\sum_{n=1}^{+\infty}{\left| C_n \right|}<+\infty $
Assume there's a real function $f\in C^{1}[0,\pi]$. Let
$$
C_n=\int_0^{\pi}{f\left( x \right) \cos \left( nx \right) \text{d}x},\quad n=1,2,3,\cdots
$$
Prove that, $$\sum_{n=1}^{+\infty}{\left| C_n \right|}<+\infty $$

I don't have any ideas after a long time of thinking. Can anyone help?

答:
If I am not grossly mistaken, then I think it works as follows:

Expand $f$ over to the whole $I = [-\pi,\pi]$ by reflecting along the $y$ -  axis. This gives an even function $g\in C^1[-\pi,\pi].$ Then, $g'$ is an $L^2$ function with Fourier coefficients $c_n\sim O(n^{-1}).$ This means $g$ has Fourier coefficients that are of the order $O(n^{-2})$ and so it is absolutely summable.


问:
Normalizer in $S_5$
I'm considering the normalizer of the subgroup generated by the element $(12)(345)$. I know that this is exactly the elements of $S_5$ that commute with it and it's a subgroup of $S_5$, but not necessarily of the subgroup generated by $(12)(345)$ itself. I certainly have the identity permutation, $(12)(345)$ itself, and its inverse. However, I do not know if there are any others or an efficient algorithm to use. Is there a way to use the conjugacy class of $(12)(345)$?

答: Hint: Conjugation of permutations preserves cycle types.

Title: Normalizer in $S_5$


问:
$f(0)=0$ and $f'(x)=f(x)^2$
At first sight, this exercise seems to be already seen many times in this website but I could not find a Analysis-1 level proof of the following :
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.

My attempt is : Since $f'(x) \geq 0$ $\forall x$, we have that $\forall x \in \mathbb R^+ f(x) \ge 0$. By contradiction, suppose that $\exists a > 0$ such that $f(a)>0$. Since $f'=f^2\geq0$, we get that $f(x)>0$  for all $x \in (a, +\infty)$.

Let $g: (a,+\infty) \to \mathbb{R}$ be given by  $g(x)=\frac{1}{f(x)}$. As $g'(x)=-1$ we have $g(x)=-x+c$, $c \in \mathbb{R}$. So, $f(x)=\frac1{-x+c}, \forall x \in (a,+\infty)$. From $f(a)>0$ we get that $c>a$. But then $f$ isn't continuous at $x=c$, a contradiction.

I think I showed that the function is constantly equal to $0$ on $(0,+\infty)$, but how do I show it on $(-\infty,0)$ ? Should I just do the same proof with the g function on $(-\infty,a)$ with $a < 0$?

答:
Let $M_{\epsilon} = \sup_{x\in [-\epsilon, \epsilon]} |f(x)|$. We have $M_{\epsilon} \to 0$ as $\epsilon \to 0$. Now, for any $x\in [-\epsilon, \epsilon]$ we have
$$|f(x)| = |f(x) - f(0)| \le \sup_{t\in [-\epsilon, \epsilon]}|f'(t)|\cdot|x|\le \epsilon M_{\epsilon}^2$$
so
$$M_{\epsilon} \le \epsilon \cdot M_{\epsilon}^2$$
Now, if $M_{\epsilon} > 0$ we would get from the above $M_{\epsilon} > \frac{1}{\epsilon}$. Since $M_{\epsilon} \to 0$ we have $M_{\epsilon}= 0$ for $\epsilon$ small enough.

Conclusion: the set $\{x \ | \ f(x)=0\}$ is not only closed, but also open, so it must be the whole $\mathbb{R}$.

Note that we can generalize it to
$$|f'(x)| \le \phi(f(x))$$ and $f(0)=0$
where $\phi(t)\le C\cdot |t|$ on a neighborhood of $0$. We only use the intermediate value theorem.

Title: $f(0)=0$ and $f'(x)=f(x)^2$


问:
Proof for an integral identity
Is it true that $\int_0^A dx \int_0^B dy f(x) f(y) = 2 \int_0^A dx \int_0^x dy f(x) f(y)$ ? If so, can this be proved?</p>

答:
Let's assume for simplicity that $f$ has an anti-derivative $F$. To simplify notation, write $F_t = F(t)$. Then the left-hand side yields

$$\int_0^A f(x) dx \times \int_0^B f(y) dy = (F_A-F_0)\cdot (F_B-F_0)$$

while the right-hand side becomes

$$\begin{split}
2 \int_0^A f(x)dx \int_0^x f(y) dy
 &= 2 \int_0^A f(x) (F_x - F_0)dx \\
 &= 2 \int_0^A f(x) F_x dx - 2F_0 (F_A - F_0) \\
 &= (F_A^2 - F_0^2) - 2F_0 (F_A - F_0) \\
 &= (F_A - F_0) \cdot (F_A + F_0 - 2F_0) \\
 &= (F_A - F_0)^2
\end{split}
$$

So these coincide if and only if $F_A = F_B$ or $F_A=F_0$.

EDIT A clarification comment. $\int F_x f(x) dx$ is integrated by subtitution with $u=F_x$ and then $du = f(x) dx$.

Title: Proof for an integral identity


答:
As others have already pointed out, this is really just a classification of the possible ranks of $S$ ("rank" here meaning the dimension of the image $S(\mathbb{R}^2)$, which can only equal $0,1$ or $2$). Here is a detailed argument that avoids using minimal polynomials:

If $S$ is rank $0$, then $S(\mathbb{R}^2)$ has dimension zero and is hence $\{0\}$, meaning $S$ is the $0$ transformation.

If $S$ is rank $2$, then it is invertible (it has trivial kernel by rank-nullity), so precomposing the equation $S^2 = S$ with $S^{-1}$ gives $S = I$, the identity.

Finally, suppose $S$ is rank $1$. The space $S(\mathbb{R}^2)$ is a subspace of $\mathbb{R}^2$, so take a basis vector $u$ of this subspace. Also note $S(u)=u$ (which can be seen by writing $u=S(x)$ and using $S^2=S$). Since the kernel of $S$ also has dimension $1$, we can pick a basis vector $v$ of this subspace as well. We claim $\{u,v\}$ is a basis of $\mathbb{R}^2$. By dimension it suffices to show $\{u,v\}$ is a linearly independent set. If $\lambda_1u+\lambda_2v=0$, then applying $S$ to this equation gives $\lambda_1u = 0$, so $\lambda_1=0$. But then $\lambda_2v = 0$, so $\lambda_2=0$ as well. This proves the claim.

Finally, put $A = \{u,v\}$. Since $S(u) = (1)u+(0)v$ and $S(v) = (0)u+(0)v$, it has the desired matrix representation
$$[S]_A^A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}.$$

As a final remark: the property $S^2=S$ is called idempotence, and this result generalizes to higher dimension. See, for example, this question.


问:
What is the probability to draw a black card, given draws without replacement?
If given a standard deck of 52 cards, what is the probability of drawing a black card on the 1st, 2nd, 3rd, 4th ... $n^{th}$ draw?

I understand that the first is $26/52$, but the second gets a bit more complicated because there is two scenarios:
 a) The first draw is black so the probability is $26/52$ * $25/51$

b) The first draw is non-black, so the probability is $26/52$ * $26/51$

I'm not sure what the formulaic expression of this is and how to consider both of these options in one calculation. I know it uses combinatorics, but I'm lost.

Thanks!

答:
Say you shuffle the cards randomly, and that you are picking from the top. Then you want the probability of the first n cards being all black. There are $n! {26 \choose n}$ possible permutation of black cards being at the top. Now, you may choose any permutation for the remaining cards, so there are $(52-n)!$ choices left. Hence, the probability of getting permutation that gives us our desired situation is

$$\frac{n! {26 \choose n} (52-n)!}{52!}$$


答:
If $\text{im}(S)$, the image of $S$, is $0$-dimensional then $S=0$.

If $\text{im}(S)$ is $2$-dimensional then any vector in $\mathbb{R}^2$ can be written as $S(x)$ for some $x\in \mathbb{R}^2$, so the relation $S(S(x))=I(S(x))$ implies that $S=I$.

Consider the final case where $\text{im}(S)$ is $1$-dimensional. We claim there is a non-zero vector which $S$ fixes. Indeed, take an $x\in \mathbb{R}^2$. If $S(x)=x$ we are done. If not, then the relation $S\circ S=S$ implies that $S$ fixes $y=x-S(x)$. Now take any vector in $\mathbb{R}^2$ which is linearly independent of this $S$-fixed vector and you have the basis for the final case.


问:
inside equilateral triangle $ABC$, $BD=CE$, $AE \cap CD = F$. Find maximum area of $BDFE$
inside equilateral triangle $ABC$, $BD=CE$, $AE \cap CD = F$. Find maximum area of $BDFE$

I could prove through extremely tedious computation that $S_{BDFE}$ maximizes when $D$ and $E$ are the midpoints of $AB,BC$, but I would like to see if there is a more elegant solution.

答:
Draw a third line from $B$ to cut $AC$ at $G$, so that $AG = BD = CE$.

By symmetry, the three lines $AE$, $BG$, $CD$ cut an equilateral triangle inside the triangle (with one vertex $F$), and leave an outer ring which has an area equals to three times that of $BDFE$.

To maximise the area of the outer ring and therefore $BDFE$, the area of the inner equilateral triangle should be minimised. This happens when the perpendicular distance from $AE$ to the common centre of the equilateral triangles is minimised to $0$.


答:
Just to give a different approach to the proof that $f(0)=0$ and $f'(x)=f(x)^2$ imply $f(x)=0$ for all $x\gt0$, suppose that were not the case. Then there would exist $a$ and $b$ with $0\le a\lt b\lt a+1$ such that $f(a)=0$ and $0\lt f(b)\lt1$.  (I.e., if $f$ has nonzero values, it'll have one close to a zero, in which case continuity of $f$ means the value will be small. And as the OP observed, $f'(x)=f(x)^2$ and $f(0)=0$ imply $f(x)\ge0$ for all $x\gt0$, so any nonzero values are positive.)

Now the Fundamental Theorem of Calculus, on the one hand, and integration by parts (with $u=f(x)^2$ and $v=x-a$), on the other hand, give

$$f(b)=f(b)-f(a)=\int_a^bf'(x)\,dx=\int_a^bf(x)^2\,dx=(x-a)f(x)^2\Big|_a^b-2\int_a^b(x-a)f(x)f'(x)\,dx\\=(b-a)f(b)^2-2\int_a^b(x-a)f(x)^3\,dx$$

and thus

$$2\int_a^b(x-a)f(x)^3\,dx=f(b)((b-a)f(b)-1)\lt f(b)(1\cdot1-1)=0$$

But the non-negativity of $f(x)$ for $x\gt0$ implies the integral $\int_a^b(x-a)f(x)^3\,dx$ cannot be negative. This contradiction proves that $f(x)=0$ for all $x\gt0$ (and as Marco observes in comments below the OP, symmetry shows that $f(x)=0$ for all $x\lt0$ as well).

Title: $f(0)=0$ and $f'(x)=f(x)^2$

Body:
At first sight, this exercise seems to be already seen many times in this website but I could not find a Analysis-1 level proof of the following :
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.

My attempt is : Since $f'(x) \geq 0$ $\forall x$, we have that $\forall x \in \mathbb R^+ f(x) \ge 0$. By contradiction, suppose that $\exists a > 0$ such that $f(a)>0$. Since $f'=f^2\geq0$, we get that $f(x)>0$  for all $x \in (a, +\infty)$.

Let $g: (a,+\infty) \to \mathbb{R}$ be given by  $g(x)=\frac{1}{f(x)}$. As $g'(x)=-1$ we have $g(x)=-x+c$, $c \in \mathbb{R}$. So, $f(x)=\frac1{-x+c}, \forall x \in (a,+\infty)$. From $f(a)>0$ we get that $c>a$. But then $f$ isn't continuous at $x=c$, a contradiction.

I think I showed that the function is constantly equal to $0$ on $(0,+\infty)$, but how do I show it on $(-\infty,0)$ ? Should I just do the same proof with the g function on $(-\infty,a)$ with $a < 0$?


问:
Series Summation: $\sum\limits_{k=1}^{N-1}\frac{1}{z-w_k}$ where $w_k=e^{\frac{2\pi i k}{N}}$
I have the series

$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum

$$\sum_{k=1}^{N-1}\frac{1}{z-w_k}   $$

答:
$$\frac{1}{1-w^{r}}+\frac{1}{1-w^{n-r}}=\frac{2-w^{r}-w^{n-r}}{2-w^{r}-w^{n-r}}=1$$ where $w=e^{i2\pi/n}$

If you continue cancelling in pairs the sum will turn out to be = $\dfrac{N-1}{2}$

In case you have a middle term ,the term = $\dfrac{1}{1-e^{i\pi/2}}=\dfrac{1}{2}$


问:
Properties of Adjunction Space
I'm having a difficult time fully grasping what is happening in an adjunction space. Let me start by writing out the definition:

Let X, Y be topological spaces. $A$ is a closed subspace of $Y$, $f:A\rightarrow X$ is a continuous map. Let ~ be the equivalence relation on the disjoint union $X\bigsqcup Y$ generated by $a$~$f(a) \forall a \in A$, and denote the resulting quotient space by
  \begin{equation} X\bigcup_f Y=(X\bigsqcup Y)/\sim \end{equation}
  This is called the adjunction space and is said to be formed by attaching Y to X along $f$.

Ok, so I am trying to prove a proposition:

Let $ X\bigcup_f Y$ be an adjunction space and let $q:X\bigsqcup Y \rightarrow X\bigcup_f Y$ be the associated quotient map. Then the restriction of $q$ to $X$ is a topological embedding whose image set $q(x)$ is a closed subspace of $X\bigcup_f Y$.

The proof is:

We begin by showing $q|_X$ is a closed map, suppose $B$ is a closed subset of $X$, to show that $q(B)$ is closed in the quotient space, we need to show $q^{-1}(q(B))$ is closed in $X\bigsqcup Y$, which is equivalent to showing that its intersections with $X$ and $Y$ are closed in $X$ and $Y$ resp.  (ALL OK SO FAR).
  From the form of the equivalent relation, it follows that
  \begin{equation}
q^{-1}(q(B))\bigcap X=B\quad (*)\end{equation}
   which is closed in $X$ by assumption, and $q^{-1}(q(B))\bigcap Y=f^{-1}(B)$ which is closed in $A$ by continuity of $f$ and thus is closed in $Y$ because $A$ is closed in $Y$.

(*) is what I don't understand. It is a really simple statement, right? But I think it's my lack of total understanding of that is actually happening in the adjunct space that makes me unable to see why $q^{-1}(q(B))\bigcap X=B$. Is $q$ injective? It seems like $q$ must be atleast injective for this statement to hold, but I don't know how to show this.

Thanks for the help!

答:
In my opinion, you may go back to the initial definition of the adjunction space and of the quotient map $q$.

The equivalence relation (based on a continuous function $f$ on $A$) defining the adjunction space is stating the basic fact that "It identifies each point $x \in X$ with all of the points (if any) in $f^{-1}(x) \subset A$"

Then, what does the quotient map $q$ do? You should bear in mind that the quotient map just maps all the points belonging to the same equivalence class to a single set(this set is the equivalence class). You can imagine this quotient map "glues" a $x$ and all points in $A$ which can map to $x$ (by $f$) together.

So, what does $q(B)$ do? It finds for all $x$'s in $B$ and "glues" those points which map to them, respectively.

For example, if the function $f$ is in the simplest form: $ a_1 \rightarrow b_1 $ and $ a_2 \rightarrow b_1 $ , leaving all other points unspecified. Then the result $q(B)$ shuold be like $\{ \{a_1,a_2,b_1\},\{b_2\}, \{b_3\}, \cdots \}$

Then, what does $q^{-1}q(B)$ look like? With the same example above, it just "unglue" those glued points, the  $q^{-1}q(B)$ looks like $ \{a_1,a_2,b_1,b_2, b_3, \cdots \}$

It follows that $q^{-1}q(B) \cap X = B$. It is also obvious that $q^{-1}q(B) \cap Y = f^{-1}(B)$.

Many times you come across problems, just go back to the definitions. You will find the answer.

If you need more clarification, let me know.

Title: Properties of Adjunction Space


问:
Can the hypothesis in this theorem on commutative monoids be weakened?
Let $(M,*,1)$ be a commutative monoid. Define the binary relation $R$ on $M$, such that $xRy$ iff $(\exists z)(x*z=y)$. It is easy to show that, since $M$ is a commutative monoid, the relation $R$ is
both transitive and reflexive. I read in a paper the theorem that if additionally $M$ is cancellative and pure, (where pure means $1$ is the only invertible element), then $R$ is also antisymmetric. I am wondering if this hypothesis can be weakened. That is, if $M$ is merely cancellative, is $R$ antisymmetric? And also, if $M$ is merely pure, is $R$ antisymmetric? All of this is given the background assumption that $M$ is a commutative monoid.

答: After some tinkering around, I found the answer to my question. Pure is needed, because otherwise $x | 1$ and $1 | x$. As for cancellative, here is a counter-model I found where pure alone is not sufficient: Let $M = \{0,1,2\}$, let the identity of $M$ be $0$, and let $+$ be defined as $0+x=x+0=x$, $1+1=2+2=1$, and $1+2=2+1=2$. Then $(M,+,0)$ is a commutative pure monoid, where the divisor relation $R$ is not anti-symmetric, as both $1R2$ and $2R1$


问:
proof validation of dimension inequalities
The Porblem Is:

$A+B=\left \{ a+b : a\in A, b\in B \right \}$ and $C$ is a vectorspace and $A \subset C, B \subset C$.

Let $a_1,..,a_n$ be elements of $A$ and let $b_1,..,b_k$ be elements of $B$ be two families of vectors in $C$. Let $A=\operatorname{Span}(a_1,..,a_n)$ and $b=\operatorname{Span}(b_1,..,b_k)$

Now, Show that $\mathbf{dim}(A+B)\leq n+k$ and show that $\mathbf{dim}(A+B)\leq \mathbf{dim}(C)$

The proof i found:

The Questions

Now, I have 2 things that I would like explained/validated.

For the first proof; Why is the index for $\beta$ and $b$ now $i$ and the sum index is $i$ and not $t$?

For the second proof I just do not understand the "proof" from "Basis of $C$ is... [...]". It does not seem clear to me what the author is trying to say. Does someone have an idea or can further clarify?

答:
The first point is just a notational choice. The author wanted to make it clear that the two sets of indices for the $\alpha$'s and $\beta's$ were distinct (and thus ranged over a potentially different number of terms: $n$ and $k$ respectively in this case)

For the second part, the author is just using one of the definitions of a basis, i.e., "a linearly independent set with the largest possible number of elements". The cardinality of such a set is defined to be the dimension of $C$, so the dimension can be thought of as the maximum number of linearly independent vectors you can find. Since $(A+B)\subset C$ a basis $\{e_1,\dotsc, e_s\}$ of $A+B$ is a linearly independent set sitting inside $C$ with $s$ elements. Thus $s\leq \dim C$.


问:
$\frac{x}{1+x^2}$ is injective on $[1,\infty)$
Prove $\frac{x}{1+x^2}$ is injective on $[1,\infty)$. I'm having a huge brain-fart... I'm assuming $\frac{x}{1+x^2}=\frac{y}{1+y^2}$, yada yada, $\implies \frac{1}{x}+x=\frac{1}{y}+y$... but I can't get past this step!!!

Body:
if $1 \leq x < y,$  then $y-x > 0 $  and $xy - 1 > 0$ and
$$  \frac{x}{1+x^2} - \frac{y}{1 + y^2}  = \frac{(y-x)(xy-1)}{(1+x^2)(1+y^2)} >0$$


问:
$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$
Prove that for natural numbers $n_1 < n_2 < ... < n_k$:

$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$

I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2 < ... < n_k$ and induction would not recognise that fact.

What's funny is that I think I saw once that exercise on stack exchange but wasn't able to find it this time.

答:
I have a proof without induction. It's not as elegant as the induction proof but some of the results might be of interest to you. The proof is long but there's very little calculation.

Suppose we fix the sum of $n_j$'s and we want to minimize the sum of cubes. Because there are finite number of $k$-tuples $(n_1, \ldots, n_k)$, a global minima $(a_1, a_2, \ldots, a_k)$ exists, where $a:=a_1 < a_2 < \cdots < a_k$ and $\sum a_j = \sum n_j$. Notice that

$$\forall i \ne j, n_i^3 + n_j^3 = (n_i+n_j)\left[(n_i+n_j)^2-3n_i n_j\right]\tag1$$

This means if the sum of $n_i$ and $n_j$ is fixed, we can make them closer to each other, as long as the order of all $n_l$'s is preserved, to decrease the sum of cubes.  We now prove the following properties of $\{ a_j: 1 \le j \le k\}$.

Property 1: $a_j-a_{j-1} \le 2, \forall 1<j\le k$. If not, we change $(a_{j-1}, a_j)$ to $(a_{j-1}+1, a_j-1)$ and the sum of cubes gets smaller, a contradiction.

Property 2: There can be no more than one $j$ such that $a_j-a_{j-1}=2$. If not, suppose $a_j-a_{j-1}=a_k-a_{k-1}=2, j < k$, then we can replace $(a_{j-1}, a_k)$ with $(a_{j-1}+1, a_k-1)$ and the sum of cubes will decrease, another contradiction.

With the above properties, we know  $a_i$'s are mostly consecutive with at most one gap where the difference is $2$.

Case 1: All $a_i$'s are consecutive. Then
$$
\sum_{i=0}^{k-1} (a+i)^3 - \left( \sum_{i=0}^{k-1} (a+i) \right)^2 =\sum_1^{a+k-1} j^3 - \sum_1^{a-1} j^3 - \left(\sum_1^{a+k-1} j - \sum_1^{a-1} j\right)^2 \\
= \left(\sum_1^{a+k-1} j\right)^2 - \left(\sum_1^{a-1} j \right)^2 - \left(\sum_1^{a+k-1} j - \sum_1^{a-1} j\right)^2 \\
=\left(\sum_1^{a+k-1} j - \sum_1^{a-1} j\right) \left[ \sum_1^{a+k-1} j + \sum_1^{a-1} j  - \left( \sum_1^{a+k-1} j - \sum_1^{a-1} j \right) \right]\\
=  \left( \sum_a^{a+k-1} j  \right) \left( 2 \sum_1^{a-1} j \right) \ge 0
$$
and $"=" \iff a=1$.

Case 2: One of the $a_j-a_{j-1}$ is $2$. Then $a_k=a+k$. In this case we can insert the number $x=a_{j-1}+1=a_j-1$ between $a_j$ and $a_{j-1}$ so that we have $k+1$ consecutive numbers, and we need to prove that

$$
f(x) := \sum_{i=0}^k (a+i)^3 - x^3 - \left[ \sum_{i=0}^k (a+i) - x \right]^2 \ge 0 \tag2
$$
where $x \in \{ a+1, a+2, \cdots, a+k-1 \}$. Let's expand the domain of $f$ to $[0,a+k]$. Notice the following:

(i) $f''(x) = -6x-2 <0, \forall x\ge0$;

(ii) $f(0)\ge 0$ from Case #1;

(iii) $f(a+k) \ge 0$, also from Case #1.

Therefore $(2)$ holds and we have strict inequality.$\blacksquare$

One last thing: What does this minima sequence look like? We give a simple example when $k=5$.

If $\sum n_i$ is equal to sum of $5$ consecutive natural numbers, say $35$, then the minima is just $\{5,6,7,8,9\}$.

If $\sum n_i$ is greater than $35$ but less than the next sum of $k=5$ consecutive numbers $(40=6+7+8+9+10)$,  assuming it's equal to $38$, then we just shift the last $38-35=3$ numbers of the sequence $\{5,6,7,8,9\}$ one size bigger to $\{5,6,8,9,10\}$.


问:
How do you scale along a non canonical direction?
Silly question but I am not figuring it out on my own.

In 3D you can scale a point / shape / field along a canonical direction quite easily:

$p = (x_0, x_1, x_2)$ becomes $p = (ax_0, x_1, x_2)$ for example if we wanted to scale along the first axis. This map has the property that distances in any plane parallel to the plane $(x_1, x_2)$ remain unaffected.

Now given an arbitrary vector $N$ I want to induce the same transformation along the direction of $N$.

One possible solution, of course, is to find a rotation that transforms $N$ into one of the canonical directions, scale that direction, then invert the transformation. i.e. $RSR^{-1}$.

However this is inelegant, I am trying to do it in a way that involves only $N$ and requires no change of basis. i.e. I want to scale points along $N$ without changing the coordinate space, and perhaps better, in a coordinate free way.

答:
Given any vector $\vec{v}$, there is a unique way to decompose $\vec{v}$ into a component along $N$ and a component orthogonal to $N$, namely $\vec{v} = \operatorname{proj}_N(\vec{v}) + (\vec{v} - \operatorname{proj}_N(\vec{v}))$.  Moreover, there's a nice formula: $\operatorname{proj}_N(\vec{v}) = \left(\frac{N \cdot \vec{v}}{N \cdot N}\right)N$.

If I understand the question correctly, the thing you want to do is send $\vec{v}$ to $a\operatorname{proj}_N(\vec{v}) + (\vec{v} - \operatorname{proj}_N(\vec{v}))$.

In particular, in the case where $N = (1, 0, 0)$, and $\vec{v} = (x_0, x_1, x_2)$, we have
$$a\operatorname{proj}_N(\vec{v}) + (\vec{v} - \operatorname{proj}_N(\vec{v})) = a(x_0, 0, 0) + (x_0, x_1, x_2) - (x_0, 0, 0) = (ax_0, x_1, x_2)$$
This shows that the construction I described agrees with what you're saying should happen when you scale along one of the coordinate axes.


问:
Proving there is a function in a noncompact set $A$ that is bounded yet does not attain its maximum
I'm not very good at abstract proofs, quite awful at it really, so I'm struggling a lot with this seemingly basic question in my textbook.

"If $A$ is a noncompact subset of $R^n$, show that there is a bounded continuous real-valued function on $A$ that does not attain its maximum"

If a real set $A$ is compact then, by the Heine-Borel Theorem, it must be closed and bounded. We are dealing with a real set $A$ that is noncompact thus it must either be open or unbounded (or both). The question states that the function we wish to find is bounded, real, and continuous.

Am I right to assume that what we need to find is a function that is bounded yet open? An obvious example of this would be $y=\arctan(x)$, which exists in the set $(-\pi/2,\pi/2)$. All other sigmoidal curves are also open, bounded, and never achieve their maximum. How would I answer this question with abstract proofing instead of by direct example?

答:
Let $X$ be a metrizable compactification of $\mathbb R^n$ (e.g., an $n$-dimensional sphere or torus) and let $d$ be a compatible metric on $X$. Since $X$ is compact and $A$ is not, $A$ is not closed in $X$. Let $p$ be a limit point of $A$ in $X$ which does not belong to $A$. Then the function $x\mapsto-d(x,p)$ is a bounded continuous function on $X$ which does not attain a maximum on $A$.

If you don't want to use a compactification then I guess you have to consider two cases, $A$ is not closed or $A$ is not bounded. E.g., if $A$ is unbounded, use something like $x\mapsto\frac{-1}{1+d(x,0)}$.


问:
Confusion about the solution of a convex problem
I require the solution of the following problem:

$$ \hat{W} = \arg \min_{W} \frac{1}{2} {\left\| A W x - y \right\|}_{2}^{2} $$

Here, $A$ is a $10 \times 4$ sized matrix; $W$ is a $4 \times 32$ sized matrix and $x$ is a $32$ dimensional vector. $y$ is a $10$ dimensional vector.

The problem is convex; so I need to have at least one solution for $\hat{W}$.

I do the following; I calculate the derivative of $W$ and set it equal to zero:

$$\begin{align*}
\hat{W} = \arg \min_{W} \frac{1}{2} {\left\| A W x - y \right\|}_{2}^{2} & \Leftrightarrow \frac{\partial \frac{1}{2} {\left\| A \hat{W} x - y \right\|}_{2}^{2} }{\partial \hat{W}} = 0 \\
& \Leftrightarrow {A}^{T} \left( A \hat{W} x - y \right) {x}^{T} = 0 \\
& \Leftrightarrow {A}^{T} A \hat{W} x {x}^{T} = {A}^{T} y {x}^{T} \\
\end{align*}$$

At this stage; in order to solve for $\hat{W}$; I use the following equation, involving Kronecker Produts and vectorization; from $vec(AXB)=(B^T\otimes A)vec(X)$:

$$vec({A}^{T} A \hat{W} x {x}^{T}) = (xx^T \otimes A^TA) vec(\hat{W}) = vec(A^Tyx^T)$$.

Now, it becomes a linear system which can be solved for $\hat{W}$.

But the thing is; $xx^T$ is a one rank matrix. And it is $\text{rank} \left(A \otimes B \right) = \text{rank}A\cdot \text{rank}B$, which means $(xx^T \otimes A^TA)$ has at most a rank of $rank (A^T A)=4$. Since in the linear system of $Ax=b$, $A$ is rank-deficient; $x$ can only have a solution either which meets the least squares criterion or it can have infinitely many exact solutions. If, let's say; we don't have an exact solution for $vec(\hat{W})$ then this means we don't have a solution which sets the derivative $\frac{\partial \frac{1}{2} {\left\| A \hat{W} x - y \right\|}_{2}^{2} }{\partial \hat{W}}$ to zero, which must not be possible, since the original problem is convex (it has to have a global minimum). I need some directions here: I am sure that I am mixing up something in this thought process but I can't point it out.

答:
You do not have a generic linear system, but a linear system where the right hand side depends on $A$ and $x$. The system always has at least one solution, as you concluded from convexity.

The kronecker product just creates copies of $A^TA$:
$$\begin{pmatrix} x_1 x_1 A^TA & \cdots & x_n x_1 A^TA \\ x_1 x_2 A^TA & \ddots \\ \vdots \\ x_1 x_n A^TA & & x_n x_n A^TA \end{pmatrix}\text{vec}(\hat{W}) = \begin{pmatrix}x_1A^Ty \\ \vdots \\ x_n A^T y\end{pmatrix}.$$
This is similar to a system $A^TAx = A^Tb$ for regular least squares, which always has a solution. The $x_i$ do not ruin the party: if all $x_i$ are $0$, any $\hat{W}$ solves this system. Otherwise, there is at least one $i$ for which $x_i \neq 0$, and the system above can be solved using only the column $i$ of $\hat{W}$ (setting the other columns to $0$).


问:
left/ right hand side limits
I want to calculate the left and right handside limit at $0$ for these functions:

$$\frac{2x}{\sqrt{x^2+6x^4}}$$

$$\frac{2x^2}{\sqrt{x^2+6x^4}}$$

So plotting the functions would make it very clear, why the one sided limits are different for the first one and the same for the second one.

However what I do not understand how to show that the left and right hand side are $-2$ resp. $+2$ for the first function.

So for $\lim\limits_{x\to 0^+} \frac{2x}{\sqrt{x^2+6x^4}}$ I know that the nominator is positive, and also the denominator of course. However how to show that the limit is $2$? Is this done with L'Hospital? I tried it with L'hospital and got a more complicated expression

and also for $\frac{2x^2}{\sqrt{x^2+6x^4}}$, since we have a square in the nominator, the left hand side and the right hand side will be the same. But what I do not understand is why the limit here becomes zero, and in the other one $+-2$? Is it because x is in a higher degree in the nominator? could someone maybe show me few steps?

答:
$$f(x) = \frac{2x}{\sqrt{x^2+6x^4}} = \frac{2x}{\sqrt{x^2}\sqrt{1+6x^2}},$$ since $x^2 \ge 0$ for all real $x$.  Then since $\sqrt{x^2} = |x|$, we get $$f(x) = \frac{2 g(x)}{\sqrt{1+6x^2}},$$ where we will take $$g(x) = \frac{|x|}{x} = \begin{cases}1, & x > 0 \\ \text{undefined}, & x = 0 \\ -1, & x < 0. \end{cases}$$  This makes the limiting behavior from either side readily apparent.

There is nothing magical about $g$.  The only thing to understand here is that when considering any particular one-sided limit, $g$ does not change value.  For instance, evaluation of $$\lim_{x \to 0^-} f(x)$$ has $x < 0$ for all such $x$, thus $g(x) = -1$ and $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{2}{\sqrt{1 + 6x^2}}.$$

Title: left/ right hand side limits


答: Yes, it's true by symmetry. The function $F(x,y)=f(x)f(y)$ is symmetric about the line $y=x$. The first integral is over the square, the second is over the bottom triangle. To give a formula, add $\int_0^1 dy\int_0^y dx$, which is clearly equal to the given integral on the RHS.</p>

Title: Proof for an integral identity

Body: Is it true that $\int_0^A dx \int_0^B dy f(x) f(y) = 2 \int_0^A dx \int_0^x dy f(x) f(y)$ ? If so, can this be proved?</p>


问:
Testing graph isomorphism with powers of the adjacency matrix
Suppose $A$ and $B$ are two square, symmetric, binary matrices with null diagonal representing the adjacency matrices of undirected graphs.  If the multisets of the column-sums of $A$ and $B$ differ, then they are definitely not isomorphic.  If the column-sum multisets are the same, then $A$ and $B$ might be isomorphic.  Now I am wondering how the effectiveness of this test is improved by running it on $A^s$ and $B^s$ with a large integer $s$.  Are there any two non-isomorphic graphs that "look" the same in this sense for all $s$ or even for an infinite number of values of $s$?</p>

答:
Testing graph isomorphism by multiset of column sums of adjacency with any power is the same as what first order Weisfeiler-Lehman (1-WL test known as vertex coloring) test does. So that approach is definitely not powerful than 1-WL test and you cannot distinguish not just isospectral graphs but also the graphs whose just maximum eigenvalues are the same.

1-WL test starts checking column-sum (actually not necessarily sum but any injective function) of Adjacency matrix. Let's think all vertices are the same, If $a_0=b_0={\bf1}$ is a vector of enough 1s, 1-WL test checks if multiset of  $Aa_0$ and  $Bb_0$ are equal or not. To do that generally, we can use the sort function, so we can check if $sort(Aa_0)=sort(Bb_0)$. If they are the same, then 1-WL relabel the result after concatenating of the current color of the vertex and the previous step ($a_1=[a_0,~Aa_0]$ and $b_1=[b_0,~Bb_0]$), relabel $a_1$ and $b_1$ and repeat the process. Relabbellin process can be seen as assigning a unique identifier for each unique row of the concatenated colors of both graphs ($[a_1;b_1]$).

In the second turn, it checks if $sort(Aa_1)=sort(Bb_1)$. If they are the same, again relabelling and again checking. If they are not equal in any step it concludes that the graphs are not isomorphic at all, or if the relabelling process cannot change the color of vertices ($a_{t+1}=a_{t}$) then the algorithm stops and supposed they are isomorphic (still they might not be isomorphic). So, what you describe is a simplified version of 1-WL test which performs $S$ number of iteration. Because if we enroll the update and change the relabelling process by just assign new color of vertex (neglect the old color of vertex by $a_1=Aa_0$), after $S$ iteration, it checks $sort(A^S{\bf1})=sort(B^S{\bf1})$ which is exactly what you asked!

This algorithm definitely not powerful and be able to just find very very easy cases. Check higher-order Weisfeiler-Lehman test procedures.


问:
Practical question about wine and pricing from a non-math expert
I work at a restaurant with a fancy wine list but I have noticed that there doesn't seem to be any method to the way the wines are priced.

Ideally you would want your expensive wines to have a much lower markup and your less expensive wines to have a steeper markup.  The list I inherited is all over the place and it seems both inelegant and bad for business.  I think there must be a way to figure out a function for a curve that I could apply to the bottles based on their cost that would determine price.

If I have a bottle that costs me only 10 dollars, it should be around a 300 percent markup (30 dollar menu price).  30 dollar bottles should be about 200 percent, and get lower and lower but never getting below 150 percent.  If the y-axis is the markup as a decimal, and the x-axis is the cost of the bottle, then the coordinates would be something like (10,3), (30,2), (400,1.5).  The last one is just because we have no bottles that exceed $400 cost per bottle. ( guess it could just be an asymptote but I don't know how that works.

答:
Here's a function that behaves pretty close to the way you're describing: $$m(x)=\frac{16.7}{x+0.83}+1.5$$ where $x$ is the cost of the bottle in dollars, and $m(x)$ is the markup of the bottle, expressed as a decimal. It has a horizontal asymptote at $m=1.5$, so you'll never get a markup lower than 150 percent.

For example, it gives a markup of 3.04 for a price of \$10, a markup of 2.04 for a price of \$30, and a markup of 1.54 for a price of \$400.


问:
Solution verification: Derivative of the infinite power tower $y(x) = x^{x^⋰}$
I was doing the problem $(x^{x^{⋰}})'$ and I would like someone to verify my solution:
\begin{align*}
&\left(y=x^{x^{^{⋰}}}\right)'\\
\implies & \;\left(y=x^{y}\right)'\\
\implies & \;\left(y^{\frac{1}{y}}=x\right)'\\
\implies & \;\left(\frac{\ln\left(y\right)}{y}=\ln\left(x\right)\right)'\\
\implies & \;\frac{y'-y'\ln\left(y\right)}{y^{2}}=\frac{1}{x}\\
\implies & \; y'-y'\ln\left(y\right)=\frac{y^{2}}{x}\\
\implies & \; y'=\frac{y^{2}}{\left(1-\ln\left(y\right)\right)x}\\
\implies & \; \left(x^{x^{⋰}}\right)'=\frac{\left(x^{x^{⋰}}\right)^{2}}{\left(1-\ln\left(x^{x^{⋰}}\right)\right)x}
\end{align*}

答:
May be easier would be to use Lambert function since$$ x^{x^⋰}= y \implies x^y = y\implies 1 = yx^{-y} = ye^{-y\log x}\implies
y = \frac{W(-\log x)}{(-\log x)}$$ Let $t=-\log(x)$

$$\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$$
$$\frac d{dt} \Big[\frac{W(t)}t\Big]=-\frac{W(t)^2}{t^2 (W(t)+1)}$$
$$\frac{dt}{dx}=-\frac 1x$$


问:
Is there only one logic in mathematics, or are there multiple variants of logic?
Is there only one logic in mathematics, or are there multiple variants of logic? I heard that some people believed in logical pluralism. I am wondering if mathematics provide the tools necessary for alternative logic to be explored and used in mathematical abstractions. What are some or all of these variants, and is there a root logic to all these variants or do wayward logics exist that are completely unrelated to the more conventional logic we know?</p>

答:
Welcome to MSE!

The answer to this question depends on how philosophical you want to be. I'll start with a formalist idea.

If by "logic" you mean "a formal method of proving things", then yes. Lots of people study lots of systems in which theorems can be proven. And it's very interesting (at least to me) what the differences in these systems are. What theorems can you prove in one but not another? What do proofs look like in the different systems? Which (if any) of these systems could be adapted to work as a "bottom floor" on which the rest of mathematics is built? After all, when we prove things as mathematicians, what system are we working in!?

Some common things that people think about include intuitionistic logic, which restricts the notion of "theorem" to only encompass those things which have a witness to their truth. Another popular one in the CS community is linear logic, which restricts the notion of "theorem" to only those things which (roughly) "use each assumption exactly once".

Of course, we can prove things about these proof systems. Some people call this "metamathematics", and it brings up an obvious question (which I hinted at before): In which system are we proving facts about these proof systems?

It is at this point that the question becomes philosophical in nature, and I'm not sure MSE is the best place for it.

It's useful to know that mathematics can be formalized inside some proof system, because it lets us use theorems from logic in order to prove things in other branches of math. However, I personally, don't really care what that system is, and I don't think it matters. Maybe one day I'll regret saying that, but in the moment it's how I feel.

Of course, many people disagree with me. We know that math can all be done internal to the theory of sets (ZFC, or others). So in order to formally encode all mathematics (and thus, potentially argue that all mathematicians are really working inside some logical system) we need exactly enough logic to talk about ZFC.... In fact we can go deeper. We need enough logic to describe a logic which can describe ZFC.

I'm not sure how deep the rabbit hole goes, but there are papers who are trying to build the weakest system possible (which minimizes the possibility of incompleteness happening) which can still, at some point, do set theory (and thus mathematics). Whether we're "really" working in one of these systems or not is unanswerable, of course. So it doesn't particularly interest me personally.

So, to summarize: Yes. There are lots of different logics which people study, and mathematics absolutely furnishes us with the tools to study and compare what is provable in the different systems. Asking for "some or all" of the variants in a MSE answer is a bit of a tall order. Lots and lots of ink has been spilled on this topic. The question of if a "root logic" exists is a very subtle one, and personally I think it's best left to the mathematical philosophers.

As one last aside, if you're interested in different logical systems, and want to see these ideas applied to mathematics that nonlogicians care about, you might be interested in reverse mathematics it's a relatively popular subject right now, and it's full of lots of interesting questions (and some answers!) in this vein.

Edit: I should clarify that the "logical systems" that are studied in reverse mathematics are different from the proof systems I was describing earlier. Rather than studying different ways of proving things, we fix a way of proving things (a classical one) and instead ask what theorems we can prove when we vary our axioms. These are "alternative logic"s in the sense that most of them are not strong enough to prove everything an analyst (say) is interested in. I wasn't sure if this was relevant, but since it's something I'm interested in I thought I would mention it anyways.

I hope this helps ^_^


答:
We can go further to even state the biconditional equality condition. The following is a quote of Proposition 3 from Sum of Cubes is Square of Sum by Barbeau and Seraj, with permission from me, one of the authors:

We prove by induction that:

For integers $a_k$, if $1 \leq a_1 < a_2 < \cdots < a_n$ then
$$a_1^3 + a_2^3 + \cdots + a_n^3 \ge (a_1 + a_2 + \cdots + a_n)^2,$$
with equality if and only if $a_k = k$ for $1 \le k \leq n$.

This is clear for $n = 1$. Assume that the proposition holds for some natural $n$; we will prove it for $n+1$. Note that $a_k \ge k$ and that $a_{n+1} - k \ge a_{n+1-k}$ for all values of $k$. We begin with the fact that
$$(a_{n+1} - n)(a_{n+1} - n - 1) \ge 0$$
with equality if and only if $a_{n+1} = n+1$. Then expanding gives
\begin{align*}
 a_{n+1}^2 - a_{n+1} & \ge 2na_{n+1} - n(n+1) = 2 \left[ na_{n+1} - {{n(n+1)}\over 2} \right]\\
 & = 2 \left( \sum_{k=1}^n (a_{n+1} - k) \right) \ge 2 \sum_{k=1}^n a_{n+1-k} = 2 \sum_{k=1}^n a_k,
\end{align*}
with equality if and only if $a_k = k$ for all $k$. By the induction hypothesis, we have that
\begin{align*}
 \sum_{k=1}^{n+1} a_k^3 &=  a_{n+1}^3 + \sum_{k=1}^n a_k^3 \ge a_{n+1}^2 + 2a_{n+1}\left(\sum_{k=1}^n a_k \right) + \left(\sum_{k=1}^n a_k \right)^2\\
 & = \left(a_{n+1} + \sum_{k=1}^n a_k \right)^2 = \left( \sum_{k=1}^{n+1} a_k \right)^2,
\end{align*}
as desired.

Title: $n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$

Body:
Prove that for natural numbers $n_1 < n_2 < ... < n_k$:

$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$

I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2 < ... < n_k$ and induction would not recognise that fact.

What's funny is that I think I saw once that exercise on stack exchange but wasn't able to find it this time.


问:
A dense, uncountable subset of reals whose complement is also dense and uncountable in every interval
Does there exist a dense set of reals which is uncountable in every interval, whose complement is also dense and uncountable in every interval? So, for example, the union of positive rationals and negative irrationals would not work for my question.</p>

答: Using the axiom of choice, well order the reals in type $2^{\aleph_0}$ and go through and chose a real to belong in set $A$ and which has not already been chosen for $A^c$ and belongs in a given interval. Do this then for set $A^c$. Organize so that each interval with rational endpoints occurs uncountably many times.


答:
I vividly remember this exact question being asked before, but I can't seem to find it at the moment. So here goes:

There are in fact natural examples of such pairs of sets - in particular, the axiom of choice isn't necessary. For example, the set $N$ of normal numbers has full measure (= complement has measure zero) but is meager (= union of countably many nowhere dense sets); in particular, for every nontrivial interval $I$ both $N\cap I$ and $I\setminus N$ have size continuum (which is a bit stronger than what you ask for). Besides providing an answer to the problem, $N$ and its variants are interesting objects in their own right.

What if we don't want to use measure/category ideas? Well, we can still get an explicit and choice-free construction, albeit a bit less natural: let $A$ be the set of all reals whose decimal expansion uses only finitely many $7$s (say). Then both $A$ and $\mathbb{R}\setminus A$ have size-continuum intersection with every nontrivial interval. This is definitely easier to construct and verify, but it's less interesting as an object in its own right; offhand I don't know of any natural examples which do not require at least a little bit of analysis to verify.

Title: A dense, uncountable subset of reals whose complement is also dense and uncountable in every interval

Body: Does there exist a dense set of reals which is uncountable in every interval, whose complement is also dense and uncountable in every interval? So, for example, the union of positive rationals and negative irrationals would not work for my question.</p>


问:
Is it true that $\mathbb E[M_tX|\mathcal F_s]=M_s\mathbb E[X|\mathcal F_s]$?
Suppose we have a martingale $\{M_t\}_{t\in[0,T]}$ on the natural filtered probability space $(\Omega,\mathcal F,\{\mathcal F_t\}_{t\in[0,T]},\mathbb P)$. $X$ is any measurable random variable with respect to $\mathcal F$. Is it true that for any $0\le s\le t\le T$,

$$\mathbb E[M_tX|\mathcal F_s]=M_s\mathbb E[X|\mathcal F_s]\ ?$$

I guess it should be correct. It is apparently true when $X$ is $F_s$-measurable. But for a general $\mathcal F$-measurable $X$, is it still true?

答: This is not necessarly true. Take for instance $X=M_t$. Since $M_t^2$ is not necessarly a martingale, we cannot have $\mathbb{E}[M_t^2|F_s] = M_s\mathbb{E}[M_t|F_s] = M_s^2$.


问:
How to evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?
How can I evaluate  $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?</p>

答:
HINT:

As we are dealing with the limit in real numbers, $x>0\implies x\to+\infty$

Put $x=y^2$

$$\implies \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}$$

$$=\sqrt{y^2+y}-\sqrt{y^2-y}$$

$$=\frac{y^2+y-(y^2-y)}{\sqrt{y^2+y}+\sqrt{y^2-y}} \text{ (Rationalizing the numerator )}$$

$$=\frac2{\sqrt{1+\frac1y}+\sqrt{1-\frac1y}} (\text{Dividing the numerator & the denominator by } y)$$

Now, as $x\to\infty, y\to\infty$

Alternatively, put $x=\frac1{h^2}$

$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}$$

$$= \frac{\sqrt{1+h}-\sqrt{1-h}}h$$

$$= \frac{1+h-(1-h)}{(\sqrt{1+h}+\sqrt{1-h})h} \text{ (Rationalizing the numerator )}$$

$$=\frac2{\sqrt{1+h}+\sqrt{1-h}}\text{ if }h\ne0$$

As $x\to\infty, h\to0\implies h\ne0$


问:
Calculating azimuth and elevation angles from direction cosines
I've been scratching my head on this one for a while, I'm not very well versed in this subject area. Hopefully someone more mathematically gifted than I can assist in  explaining why this code is not working as intended. The expected outcome is to be able to input vector i, j, k, calculate the direction cosines, transform the cartesian components x, y, z to the spherical components r, theta, phi, and have it output the calculated elevation and azimuth. For all the code examples below, I am running it in Python with NumPy and have omitted other portions of the code for brevity.

The rundown

These lines take an input of three variables i, j, k . These variables are transformed into direction cosines via

```<code>magnitude  = np.sqrt(i**2 + j**2 + k**2)
x = i / magnitude
y = j / magnitude
z = k / magnitude
</code>```

These a transformed from cartesian coordinates to spherical via (In the program I’ve flipped theta to get the angles in the correct axis)

```<code>r = np.sqrt(x**2 + y**2 + z**2)
theta = np.arctan2(z, np.sqrt(x**2 + y**2))
phi = np.arctan2(y, x)
</code>```

Finally, theta and phi are converted from radians to degrees and output

```<code>elevation = np.degrees(theta)
azimuth = np.degrees(phi)
</code>```

A few observations

Azimuth output works entirely as expected.

```<code>IN: i=1.0, j=0.0, k=0.0
OUT: el=0.0, az=0.0

IN: i=0.7071067811865476, j=0.7071067811865475, k=0.0
OUT: el=0.0, az=44.99999999999999
</code>```

Elevation output is very close when angle is near 0 or 180 degrees, however error is significant at higher values (this applies for moving away from 0 degrees and 180 degrees).

```<code>IN: i=1.0, j=0.0, k=0.17364817766693033
OUT: el=9.851076116583906, az: 0.0 (Expected 10 degrees)

IN: i=1.0, j=0.0, k=0.766044443118978
OUT: el=37.453719557105146, az: 0.0 (Expected 50 degrees)
</code>```

I'm assuming I must have made an error in the cartesian to spherical calculations and would greatly appreciate a fresh set of eyes!

答:
For the coordinates
\begin{align}
 i&=1.0,\\ j&=0.0,\\ k&=0.17364817766693033,
\end{align}

consider the triangle with vertices
$(0,0,0),$ $(1,0,0),$ $(1,0,0.17364817766693033)$.
This is a right triangle with right angle at $(1,0,0).$
The other two vertices are the origin and your chosen point $(i,j,k).$

In this triangle, the leg opposite the origin has length $0.17364817766693033$ and the leg adjacent to the origin has length $1,$ so the ratio of those two legs,
$0.17364817766693033/1 = 0.17364817766693033,$
is the tangent of the angle at the origin. That angle is
$\arctan(0.17364817766693033) \approx 0.17193371309896594$ (in radians),
which is approximately $9.8510761165839062$ degrees.
Your program is giving the correct output.

You may have been thinking that by setting $k$ to the sine of $10$ degrees you should get a $10$ degree elevation. But the sine is the ratio of the opposite side of the triangle to the hypotenuse, that is, the ratio of $k$ to the magnitude $\sqrt{i^2+j^2+k^2}$. If you want to use $k=0.17364817766693033$ to get an elevation of $10$ degrees you need to ensure that $\sqrt{i^2+j^2+k^2} = 1,$
not that $i = 1.$
You can do this by setting $i$ to the cosine of $10$ degrees,
or you can set $i$ to $1$ but set $k$ to the tangent of $10$ degrees.


问:
There is a way to know how many partitions has a number?
The title says it all, there is a way to know how many partitions has a number?

With "a way" i mean if there is any formula or polynomials to provide the number of ways to "part" a number.

The question arose up after i answer the query "What are the partitions of 5?"
I found seven, in fact is seven, but i would appreciate a double check

答:
There is a function called the partition function which allows you to calculate exactly how many integer partitions there are of a given integer by calculating the coefficients of the $n^\textrm{th}$ term of the series
$$(1+x+x^2+\cdots)(1+x^2+x^4 + \cdots)(1+x^3+x^6+\cdots)\cdots.$$ As you found, there are $7$ partitions of $5$ which are the following.

\begin{align}
5&=5 \\
5&=4+1 \\
5&=3+1+1 \\
5&=2+1+1+1 \\
5&=1+1+1+1+1 \\
5&=2+3 \\
5&=2+2+1
\end{align}

There isn't a known way to write these coefficients as a polynomial or anything, but we do know that it grows like $\frac{1}{4n\sqrt{3}}e^{\pi \sqrt{2n/3}}$ as $n \to \infty$. So we can figure out approximately how many integer partitions there are for large $n$.


问:
Evaluate the antiderivative at a particular point without integrating
If $g(x)$ is the antiderivative of $f(x)=\frac{e^x}{x^2}\:$   and $\:g(3)=0$, what is $g(6)$? $\:\:\:$If it were a function I could integrate, it'd be a snap - but not this one.
It feels like it should be relatively easy and using FTC 2. I'm sure I'm missing the obvious...
Can anyone help?
Thanks!!
Anna

答: We have $$g(x)=\int_a^x\frac{e^t}{t^2}\,\mathrm{d}t,$$ for some value of $a$. Since $g(3)=0$ it must be that $a=3$, and $$g(6)=\int_3^6\frac{e^t}{t^2}\,\mathrm{d}t$$  I don't see how one can do any better than this.


问:
Moment generating function for $\frac{1}{2\theta}e^{-|x|/\theta}$
Let $X$ be the random variable with pdf $\frac{1}{2\theta}e^{-|x|/\theta}$ with $\theta>0$.  I want to find its moment generating function.  I think I’ve found it but when I try to use it to find the second moment something must have gone wrong.  I calculate it as:

$$\int_{-\infty}^\infty e^{-tx}\frac{1}{2\theta}e^{-|x|/\theta} = \frac{1}{2\theta}\left( \int_{-\infty}^0 e^{-tx+x/\theta}dx+
\int_0^\infty e^{-tx-x/\theta}dx \right)$$

$$ = \frac{1}{2\theta}\left( \left[ \frac{ e^{ x(1/\theta-t) } }{1/\theta-t}\right]_{-\infty}^0 + \left[ \frac{ e^{ -x(1/\theta+t) } }{-(1/\theta+t)}\right]_0^\infty \right) $$

$$ = \frac{1}{2\theta}\left(\frac{1}{1/\theta-t}-\frac{1}{1/\theta+t}(0-1)\right) $$

$$ = \frac{1}{2}\left(\frac{1}{1-t\theta}+\frac{1}{1+t\theta}\right) $$

I justify that this all works out if $t$ is chosen in an interval about 0 no larger in radius than $1/\theta$.  Now Evaluating at $t=0$ tells me the expectation of $X$ is 1, sounds good.  But if I compute the derivative to find the second moment,

$$M_X’(t)=\frac{1}{2}\left(-\frac{1}{(1-t\theta)^2}(-\theta)-\frac{1}{1+t\theta)^2}(\theta)\right) $$

$$ = \frac{1}{2}\left(\frac{\theta}{(1-t\theta)^2}-\frac{\theta}{(1+t\theta)^2}\right)$$

Evaluated with $t=0$ this gives that $E[X^2]=0$ which cannot be correct.  Yet the MGF seemed valid in a domain around 0, and infinitely differentiable in that domain, no?

答: Although I posted a comment referring you to another answer, it is worth pointing out that $M_X'(0) = \operatorname{E}[X]$, not $\operatorname{E}[X^2]$.  The general formula is $$\operatorname{E}[X^k] = \left[ \frac{d^k M_X}{dt^k} \right]_{t=0}.$$  This means that we must always have $M_X(0) = 1$ whenever $M_X$ is defined in a neighborhood of $0$, since $X^0 = 1$.


问:
Puiseux expansion of logarithm of lower incomplete gamma function
I'm currently exploring different methods of evaluating the lower incomplete gamma function:
$$
\gamma(s,x)=\int_0^x{t^{s-1}}e^{-t}\;\mathrm{d}t.
$$
When I look at series representations on WolframAlpha, it returns a generalized Puiseux series expansion around $x=0$ (at least if $s$ is not too large). For example,
$$
\gamma(100,x)=\log\bigg(\frac{x^{100}}{100}\bigg) - \frac{100}{101}x + \frac{25}{520251}x^2 + \frac{550}{1804057051}x^3 + \frac{12367525}{5798643470693424}x^4 + O(x^5)
$$
This series is a complete mystery to me. How is such a thing derived? Is there a formula for generating this series for general $s$?

Body:
$$I=\int_0^x{t^{s-1}}e^{-t}\, dt=\Gamma (s)-\Gamma (s,x)$$ So, around $x=0$, we have
$$I=x^s \left(\frac{1}{s}-\frac{x}{s+1}+\frac{x^2}{2 (s+2)}-\frac{x^3}{6
   (s+3)}+\frac{x^4}{24 (s+4)}+O\left(x^5\right)\right)$$ that is to say
$$I=\frac {x^s}s\left(1-\frac{s x}{s+1}+\frac{s x^2}{2 (s+2)}-\frac{s x^3}{6 (s+3)}+\frac{s x^4}{24
   (s+4)}+O\left(x^5\right)  \right)$$

$$\log(I)=\log \left(\frac{x^s}{s}\right)+$$ $$\log \left(1-\frac{s x}{s+1}+\frac{s x^2}{2 (s+2)}-\frac{s x^3}{6 (s+3)}+\frac{s x^4}{24
   (s+4)}+O\left(x^5\right)\right)$$ Now, expand the logarithm to make
$$\log(I)=\log \left(\frac{x^s}{s}\right)+\sum_{n=1}^p a_n x^n$$
$$a_1=-\frac s{1+s}$$
$$a_2=\frac{s}{2 (s+1)^2 (s+2)}$$
$$a_3=\frac{(s-1) s}{3 (s+1)^3 (s+2) (s+3)}$$
$$a_4=\frac{s \left(s^3-s^2-6 s+2\right)}{4 (s+1)^4 (s+2)^2 (s+3) (s+4)}$$
and so on.


问:
MATH LOGIC :: proof that fourmula is derivable. Use Axioms
Without using completeness theorem, proof that fourmula  ( → ( → )) ↔ (¬ ∨ ¬ ∨ )  is derivable. Use Axioms
A→(B→A),
(A→(B→C))→((A→B)→(A→C)),
A∧B→A,
A∧B→B,
A→(B→(A∧B)),
A→A∨B,
B→A∨B,
(A→C)→((B→C)→(A∨B→C)),
¬A→(A→B),
(A→B)→((A→¬B)→¬A),
A∨¬A.</p>

答:
Start with the comparatively simple subproofs for $\def\to{{\rightarrow}}\def\lor{{\vee}}\rm B\to C\dashv\vdash \lnot B\lor C$.

You should see that $\rm B\to C\vdash \lnot B\lor C$ can be proven, via disjunction elimination and the law of excluded middle [ie axiom $a\lor\lnot a$], should we have proofs for $\rm B\to C\vdash B\to( \lnot B\lor C )$ and $\rm~\vdash\lnot B\to(\lnot B\lor C)$, so we just need to validate those.   Well, the second is axiomatic [$a\to(a\lor b)$], and and the first just takes a little work.

The other direction, $\rm\lnot B\lor C\vdash B\to C$ is almost trivial, since $\rm~\vdash\lnot B\to(B\to C)$ and $\rm~\vdash C\to(B\to C)$ are axiomatic [$\lnot a\to(a\to b)$ and $a\to (b\to a)$], thus we can again use disjunction elimination [which corresponds to axiom $(a\to c)\to((b\to c)\to((a\lor b)\to c)~$].

Then you can expand this on foundation to prove $\rm A\to(B\to C)\dashv\vdash \lnot A\lor\color{silver}(\lnot B\lor C\color{silver})$ .


答: Just consider the unit interval, and take the decimal expansion of the reals. Let A be the set of reals which, from some point on, all even decimal positions are zero. Then given an interval there will be a finite expansion so that all completions of the finite decimal to an infinite decimal lie in the given interval. There are unaccountably many in $A$, as the odd places can be chosen at will. $A^c$ clearly has the same property.


问:
Proving whether quotient ring is a field

Consider $R[x]/(x^6 + 5x + 5)$ where $R = \mathbb{Z}, \mathbb{Q}, \mathbb{C}$ in turn. Determine whether $R[x]/(x^6 + 5x + 5)$ is an integral domain and whether it is a field for each choice of $R$.

The only case I have figured out is when $R = \mathbb{Q}$. I proved a lemma stating that when $F$ is a field $F[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible. I proved by Eisenstein's criteria that $x^5 + 5x + 5$ is irreducible in $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$. Since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]/(x^6 + 5x + 5)$ is a field and therefore an integral domain.

I cannot use this same result for $\mathbb{Z}$ because $\mathbb{Z}$ is not a field. I can use the result for $\mathbb{C}[x]$, but I need to show that $\mathbb{C}[x]/(x^6 + 5x + 5)$ is not an integral domain, finding a zero divisor, but I don't have any intuition for how to do this.

Any help on the $\mathbb{Z}$ and $\mathbb{C}$ cases would be great.

答:
A quotient ring is a field (integral domain) iff the ideal we quotient by is maximal (prime).

For $\Bbb Z$, the ideal is clearly not maximal, as it is strictly contained in, for instance, $(x^6+5x+5, 5)$ (which also isn't maximal, but that's beside the point).

For $\Bbb C$, consider the fundamental theorem of algebra to show that $(x^6+5x+5)$ isn't maximal (or even prime).


问:
Using probability to determine dependent events
In my HS math class, I'm teaching Prob & Stats.  We are studying how the intersection of events can show dependence.  For example, we took a survey of students in the HS:  57% of those surveyed are in Extra-curricular Activities, 65% of those surveyed are on the Honor Roll, and 46% are both.  If these two events were independent, the the "both" (intersection) would be equal to their product: P(EC) * P(HR) = 37%.  Since the observed intersection is higher, these have a positive correlation.

Is this valid statistics?  I.e., is this kind of data collection and statistical analysis something that is done in research, in order to show events are independent or dependent?  I'd like to be able to tell my students that "This is how it is done in the 'Real World.'"  Can anyone give me examples of real-world situations where this is done?

答:
In the real world you would need to be a bit careful with just saying the size of the intersection is a bit large therefore there's a correlation.

You would need to work out what the probability that the intersection were higher than the one you observed if they were independent)
(That is if you assume each pupil is has a $35\%$ chance of being on both the honour roll and doing ECA (the null hypothesis that the events are independent)  what is the probability that 46% of your survey were both the honour roll and doing ECA.  This depends on the size of your survey. (a small survey might produce the result by chance)

There are three absolutely classic examples.  The first is the link between smoking and lung cancer.  which was shown in exactly the way you argued.   There's also Florence Nightingale who used similar ideas to demonstrate a link between various diseases and conditions in military hospitals during the Crimean war, another great one is John Snow who proved that cholera was a waterborne disease by showing that people who used a particular street pump were vastly more likely to get cholera.  Both these last two examples lead to huge improvements in sanitation around the world.


问:
Number Theoretic Transform (NTT) principal root of unity modulo a prime
I'm not sure why a certain Number Theoretic Transform (NTT) works. Suppose the following:

Let $n = 2^c$ with integer $c > 0$.

Let $k$ be the smallest integer such that $p = nk + 1$ is
prime.

Let $g$ be a generator of $Z_p^\star$, the multiplicative
group mod $p$.

One example has $n=4$, $k=1$, $p=5$, and $g=2$, since the powers of 3 (mod 5) are $3,4,2,1,3\ldots$.

Consider

$$
S_m = \sum_{j=0}^{n-1} g^{kmj} = \frac{g^{kmn} - 1}{g^{km} - 1} = \prod_{\ell = 0}^{c-1}\left( g^{2^\ell km} + 1\right)
$$

It is claimed in the book CLRS that $S_m$ is divisible by $p$ for $m = 1,2 \ldots n-1$, i.e. $g^k$ is a principal root of unity, and thus it can be used to construct a "unitary" DFT matrix for the NTT.

It's unclear to me how to prove this claim.

答:
For any $p$ prime and $k | p-1$

$g$ has order $p-1$ so $g^k$ has order $(p-1)/k$

For $m\not \equiv 0\bmod (p-1)/k$ then $$g^{km} S_m = S_m \implies (g^{km}-1)S_m=0\implies S_m=0$$


问:
Order of intersection between $A_7$ and $N_G(H)$
I have $G = S_7$ and $H = \langle (1\ 2\ 3\ 4\ 5\ 6\ 7) \rangle$, I'm finding the number of elements in $A_7\ \cap\ N_G(H)$.

I have $|N_G(H)| = 5040/120 = 42$ (120 is the number of conjugates of $H$ in $S_7$). And $A_7 = 2520$.

By the internal product formula, $|N_G(H) \ A_7| = \frac{35*2520}{|A_7\ \cap\ N_G(H)|}$, however I'm not sure how to go from there.

I would really appreciate any help.

答:
The sign homomorphism $S_n\to\mathbb{Z}_2$ can be restricted to any subgroup $N\subseteq S_n$. The kernel of the homomorphism is $N\cap A_n$ which by Lagrange's theorem either has index $1$ or $2$ in $N$. Indeed $N\cap A_n$ has index one iff $N\subseteq A_n$ iff $N$ has no odd elements, which can be rephrased as saying $N\cap A_n$ has index two iff $N$ contains an odd element. Applying this to $N=N_G(H)$, since $N$ contains the odd element $(27)(36)(45)$ (which conjugates $(1234567)$ to its inverse), $N\cap A_7$ must have index two in $N$.

For completeness, we know $H$ has $120$ conjugates because there are $7!/7=6!$-many $7$-cycles and the map from $7$-cycles to cyclic subgroups they generate is a $6$-to-$1$ map (since $H$ has $6$ cyclic generators) which means the number of such subgroups is $6!/6=5!=120$. Then we know $|G|/|N_G(H)|=120$ by orbit-stabilizer, so $|N|=7!/5!$. Note this means $N\cong\mathbb{Z}_7\rtimes\mathbb{Z}_7^{\times}$.


问:
Solving a Polynomial Function Given 3 Zeros, 1 Having a Square Root
I have not been taught how to do this and I'm stumped on a problem. I found similar questions but I'd like to be sure of the correct answer.
The problem reads: Give a polynomial function that has the zeros $0$, $1$, and $3-\sqrt{5}$.

答: The simplest solution is $$x(x-1)(x-(3-\sqrt{5}))$$ but this will have non rational coefficients. If these are the only roots then you are done. If you want rational coefficients then take $$x(x-1)(x-(3-\sqrt{5}))(x-(3+\sqrt{5}))$$


问:
Bipartite Version of Schur's Lemma
Let $V$ be a finite-dimensional vector space, let $\pi$ be a representation of some group $G$ (I'm really interested in $G=SO(n)$) on $V$, and suppose that a linear operator $S$ on $V\otimes V$ commutes with all $U\otimes U$, where $U=\pi(g)$ for some $g\in G$. I think it should be the case that $S$ is a linear combination of the identity and the linear operator $\text{SWAP}:x\otimes y\mapsto y\otimes x$. Is this true? I'm really not sure where to start.

Edit: As Qiaochu points out, $\pi$ should of course be an irrep.

答:
This is true for $G = GL(V)$ acting on $V$ (and the ground field doesn't hae characteristic $2$), and this is a special case of Schur-Weyl duality. If we're working over $\mathbb{C}$ we can also take $G = U(n)$ acting on $\mathbb{C}^n$. There's no reason it should be true in general; note that you didn't make any kind of irreducibility assumption on $V$, so $G$ could be acting trivially.

What we can say in general is that $V \otimes V$ always decomposes under the action of the flip map as a direct sum

$$V \otimes V \cong \text{Sym}^2(V) \oplus \text{Alt}^2(V)$$

of the symmetric and the antisymmetric tensors, where the symmetric tensors are the $+1$ eigenspace and the antisymmetric tensors are the $-1$ eigenspace. (This is why the ground field can't have characteristic $2$.) If both of these are irreducible representations of $G$ then Schur's lemma applied to this decomposition implies the desired result (and the converse holds if the action of $G$ is completely reducible, e.g. if $G$ is finite or compact). This requires in particular that $V$ is an irreducible representation of $G$ but this condition is only necessary, not sufficient.

The desired result is not true for $G = O(n)$ acting on $\mathbb{R}^n$ (which is irreducible). The issue is that, by definition, $G$ leaves $\sum_{i=1}^n x_i^2$ is invariant, which spans a trivial subrepresentation of $\text{Sym}^2(\mathbb{R}^n)$. So the symmetric square is reducible. The exterior square is also reducible when $n = 4$; in this case $\text{Alt}^2(\mathbb{R}^4)$ is famously known to break up into two further irreducible representations due to the action of the Hodge star operator. $\text{Alt}^2(\mathbb{R}^n)$ is an irreducible representation of $O(n)$ for $n = 2, 3$ and $n \ge 5$, though.

Title: Bipartite Version of Schur's Lemma


问:
Other than integrating Dirichlet function on $[0,1]$, what are practical usage of lebesgue measure?
My question is motivated an extremely easy calculus question:

Find the area enclosed by $y=x$ and $y=x^{2}, 0 \leq x \leq 1$.

By simply computing $$\int^1_0x-x^2 dx=\frac{1}{2}x^2-\frac{1}{3}x^3 \big|^1_0=\frac{1}{6},$$ To me it is just a Riemann integration thing and I don't know how to connect this simple question with concepts such as product measure which I have learned.

For example, we have introduced the concept of the section as:

(a) Let $X$ denote a function on $\Omega \times \Omega^{\prime} .$ For each $\omega$ in $\Omega,$ the function $X_{\omega}(\cdot)$ on $\Omega^{\prime}$ defined by $X_{\omega}\left(\omega^{\prime}\right) \equiv X\left(\omega, \omega^{\prime}\right),$ for each $\omega^{\prime}$ in $\Omega^{\prime},$ is called an $\omega$ -section of $X(\cdot, \cdot) .$ An $\omega^{\prime}-$section of $X(\cdot, \cdot)$ is defined analogously.

(b) Let $C$ be a subset of $\Omega \times \Omega^{\prime} .$ For each $\omega$ in $\Omega,$ the set $C_{\omega}=\left\{\omega^{\prime}:\left(\omega, \omega^{\prime}\right) \in C\right\}$ is called the $\omega$ -section of $C$. An $\omega^{\prime}-$section of $C$ is defined analogously.

Also, if I think of the area enclosed by these two functions as a product measure, I want to apply the following theorem to solve the problem:

Let $(\Omega, \mathcal{A}, \mu)$ and $\left(\Omega^{\prime}, \mathcal{A}^{\prime}, \nu\right)$ denote finite measure spaces. Let $C \in \mathcal{A} \times \mathcal{A}^{\prime}$,
then for any $C \in \mathcal{A} \times \mathcal{A}^{\prime}$
$C_{\omega^{\prime}} \in \mathcal{A} $ and $ C_{\omega} \in \mathcal{A}^{\prime},$ for any $\omega$ and $\omega^{\prime}$
$\mu\left(C_{\omega^{\prime}}\right)$ is $ \mathcal{A}^{\prime}$ -measurable and $\nu\left(C_{\omega}\right)$ is $\mathcal{A}$ -measurable
$$
\begin{aligned}
\phi(C) & \equiv \int_{\Omega^{\prime}} \mu\left(C_{\omega^{\prime}}\right) d \nu\left(\omega^{\prime}\right) \\
&=\int_{\Omega} \nu\left(C_{\omega}\right) d \mu(\omega)
\end{aligned}
$$

It seems to me by applying the definition of a section, $\mu(C_x)=x-x^2$, $\mu(C_y)=\sqrt{y}-y$, and $$\phi(C)=\int_{[0,1]}x-x^2 dx=\int_{[0,1]}\sqrt{y}-y dy=1/6$$

The form $$\int_{\Omega^{\prime}} \mu\left(C_{\omega^{\prime}}\right) d \nu\left(\omega^{\prime}\right)$$ look identical to how I would find the area using Riemmen integration.

I am very puzzled because I feel whenever there's a concrete example (when I am expecting to compute an integration), I am always using some techniques that were taught in high school. This is just a naive example provided by our teacher where we are guided to

Use the ordinary calculus to evaluate the area of $C$.

Describe the sections $C_{x}$ and $C_{y} .$ Evaluate $\mu\left(C_{y}\right)$ and $\nu\left(C_{x}\right)$ by the Lebesgue measures $\mu$ and $\nu$

Use $\mu\left(C_{y}\right)$ and $\nu\left(C_{x}\right)$ to calculate the area of $C$.

I don't know what exactly I was expecting, I think I am expecting some different forms ended up producing the same output. However, I feel that I gained no insights from this example and have no idea if any steps I have done later would be invalid without the knowledge of the product measure, the section, and the Lebesgue measure. I was told Lebesgue measure is generally more flexible when solving integration problem but to me, it is more like the other way around, even in probability and statistics, if we really want to find the expectation of a random variable, solving $E(X)=\int XF(x)$ is usually done by $E(x)=\int xf(x)dx$. Again, I am using Riemann. I am wondering could someone provide me with an example where the computation will be simpler using a non-Riemann method? I have been puzzled by this confusion for a very long time.

答:
Let me put it this way : Lebesgue measure is not intended to make explicit integral computations easier.

From a practical point of view, any Riemann integrable function is Lebesgue integrable, and in this case Lebesgue integration coincides with the Riemann one (but the converse is not true). From a calculus point of view, most of the integrals we compute in practice are done using Riemann integration techniques. The key role of Lebesgue measure is to provide a powerful, general and theoretical tool for measure theory.

Indeed, in such context, we no longer define the equality between two functions as $f(x) = g(x) \forall x$ but rather $\int_E f ~d \mu = \int_E g ~d \mu$ for all $E$ measurable. An element is no longer defined strictly by its expression but rather by what we can measure from it. Lebesgue measure is the right tool to deal with such equality condition for real functions.

This concept of equality might appear unnecessarily complicated at first, but it is the right way to define it when measure comes into play as we do not want to consider different two elements that will yield the same result for every possible measure evaluation we can make. As you said in the title of the question, $f(x)$ or $f(x)+\mathbf{1}_{\mathbb{Q}}$ are equal from the Lebesgue measure viewpoint.


问:
Looking for example of an order homomorphism that doesn't preserve joins.
I know that not every order homomorphism preserves joins. But, I can't think of an example!

Both minimal examples and 'natural' examples welcome.

答:
Let $S$ be the poset

and let $T$ be the poset

Then the map $\phi:S\to T$ defined by
$$\phi(a)=y,\quad \phi(b)=z,\quad\phi(c)=w$$
is an order homomorphism, but
$$\phi(a\vee b)=\phi(c)=w\neq x=\phi(a)\vee\phi(b).$$


问:
Find polynomials $r(x)$ and $s(x)$ such that $r(x)(x^2+x+1)+s(x)(x^3-x-1)=1$.
I realize this has something to do with the Euclidean algorithm over $\mathbb{Q}[x]$, and I know how to show that the $\gcd(x^2+x+1, x^3-x-1)=1$, but I have no idea where to even begin looking for polynomials satisfying the Euclidean algorithm.

Is it largely just trial and error? Or is there some methodic process that I can go about solving for these polynomials...

Body:
Let $p(x) = x^{2}+x+1$ and $q(x) = x^{3}-x-1$.

Then if we divide, we have $q(x) = (x-1)\cdot p(x) - x$.

Then we divide $p(x)$ with $-x$, we have $p(x) = (-x-1)\cdot (-x) + 1$.

Thus

$$-x = q(x) - (x-1)\cdot p(x),$$
i.e.,
$$1 = p(x) - (-x-1)\cdot(-x).$$

Then $$\begin{align}1 &= p(x) - (-x-1)\cdot (q(x)-(x-1)\cdot p(x))\\
&= (x+1)\cdot q(x) + ((-x-1)\cdot (x-1)+1)\cdot p(x).\end{align}$$

Just take $s(x) = x+1$, and $r(x) = -x^{2}+2$.


问:
Find binary vector furthest away from set of binary vectors?
How can I find the binary vector furthest away from a set of binary vectors?</p>

答:
For $i\in\{1,\dots,n\}$ and $j\in\{1,\dots,d\}$, let $x_{i,j}\in\{0,1\}$ be the given value  of vector $i$, coordinate $j$.  For $j\in\{1,\dots,d\}$, let binary decision variable $y_j$ be coordinate $j$ of the desired vector.  You want to maximize
$$\sum_{i=1}^n \sum_{j=1}^d (x_{i,j} - y_j)^2 = \sum_{i=1}^n \sum_{j=1}^d (x^2_{i,j} - 2 x_{i,j} y_j + y_j^2) = \sum_{i=1}^n \sum_{j=1}^d (x_{i,j} - 2 x_{i,j} y_j + y_j).$$
Equivalently, maximize
$$\sum_{j=1}^d y_j \sum_{i=1}^n (1-2x_{i,j}) = \sum_{j=1}^d y_j \left(n-2\sum_{i=1}^n x_{i,j}\right).$$
So take $y_j=1$ if $\sum_{i=1}^n x_{i,j} < n/2$ and $y_j=0$ otherwise.


问:
Evaluation map infinite dimensional vector space
In a finite-dimensional vector space, we have an isomorphism ev: $ V \to V^{**}$, where ev(v)$: V^* \to F$ and $f \mapsto f(v)$. But what exactly happens in the infinite-dimensional case? Why is ev$(v)$ not surjective?

答: If you follow the argument you see that there is a dual basis in $V^*$. So if $v_1, ...,v_n$ is a basis  of $V$, then $f_i$ where $f_i(v_j)=\delta_{ij}$ is a dual basis. But in the case of infinite dimensions this is not a basis, because there are many functions which take nonzero values on infinitly many of the basis $\{v_i\}$ of $V$. In any case they dont even have the same cardinality, $|V^*|=2^{|V|}$.


问:
Proving that $\int_{0}^{1} \frac{dx}{1+x^a}>\frac{a}{1+a}, a>0$
Binding an integral from one or both sides is always interesting even if the integral is doable.
For instance, $$a>0, x \in (0,1) \implies  1>1-x^{2a} \implies \frac{1}{1+x^a} > 1-x^a.$$
Hence $$\int_{0}^{1} \frac{dx}{1+x^a}> \int_{0}^{1} (1-x^a)~ dx=\frac{a}{1+a}, a>0 ~~~(1)$$

The question is: How else (1) can be proved?

答: By cauchy schwarz $$\int_0^1 dx\le \sqrt{\int_0^1 (1+x^a)dx \int_0^1 \frac{1}{1+x^a}dx}$$ $$\Rightarrow\int_0^1\frac{1}{1+x^a} \ge \frac{a+1}{a+2}>\frac{a}{a+1}$$


问:
Prove that the number of common tangents to two circles exterior to one another is 4
Given two circles outside each other, the maximum number of common tangents to the two circles is 4 according to Wikipedia. How can this be proven?

That's a picture above showing what I'm talking about. How can we prove that no other common tangents can be drawn aside those four?

答:
This approach is by geometric construction.

Given circles exterior to one another with centers $A$, $B$:
join $AB$, cutting the circles at $D$, $E$, and extend $AB$ to $C$ such that$$\frac{AC}{BC}=\frac{AD}{BE}$$From $C$ draw tangent $CF$ and join $BF$. Draw $AG\parallel BF$.

It is clear that $AF$ extended makes$$\triangle CBF\sim\triangle CAG$$so that $\angle AGC$ is right, and $CFG$ is a common tangent to both circles.

Next, take $H$ on $DE$ such that$$\frac{HD}{HE}=\frac{AD}{BE}$$Draw tangent $HJ$ and join $BJ$. Make $AK\parallel BJ$. Again it is clear that $JH$ extended makes$$\triangle HJB\sim\triangle HKA$$so that $\angle HKA$ is right and $JHK$ is tangent to both circles.

Since each of these two tangents has its symmetrical counterpart, we then have four common tangents.  Further, except in the special case of equal radii, a common tangent must intersect $AB$ or its extension. And since $C$ on $AB$ outside the circles, and $H$ on $DE$ between the circles, are unique points, then only four common tangents are possible.


问:
Quotient module of module generated by integral basis
Let $K/\mathbb{Q}$ be a number field of degree $n$ and $\mathcal O_K$ be its ring of integers. Let $\{v_i \}_{1\leq i\leq n}\subseteq \mathcal{O}_K$ be a basis of $K$. Let $M\subseteq \mathcal{O}_K$ be the abelian group generated by $\{v_i \}$ over $\mathbb{Z}$. Let $\{w_i \}$ be the dual basis with respect to $\text{Tr}_{K/\mathbb{Q}}$, and $M'$ generated by $\{w_i \}$ over $\mathbb{Z}$. We then have
$$M'=\{x\in K|\text{Tr}_{K/\mathbb{Q}}(xy)\in\mathbb{Z},\forall y\in M \}   $$
I want to show that $M'/M$ is finite with cardinality |Disc$(v_1,\dots,v_n)$|. Could anyone please help?

答:
As $M \subseteq M'$ and they are both finitely generated submodules of $K$ from a well-known identity we have that $\text{disc}(M) = [M':M]^2 \cdot \text{disc}(M')$. If you are not familiar with this identity you can find a proof of it under Proposition $2.12$ in Neukirch's "Algebraic Number Theory". In fact, I'm pretty sure most of algebraic number theory textbooks include it. So from this we have that:

$$[M':M]^2 = \left| \frac{\text{disc}(M)}{\text{disc}(M')} \right|$$

As $\text{disc}(M) = |\text{disc}(v_1,\dots,v_n)|$, it remains to prove that $\text{disc}(M')^{-1} = |\text{disc}(v_1,\dots,v_n)|^{-1}$. This follows almost immediately from the duality of the basis. Indeed

$$1 = (\det I)^2 = \det (\text{Tr}(v_iw_j))^2 = \det\left(\sum_k (\sigma_kv_i)(\sigma_kw_j)\right)^2 \\ = \det(\sigma_kv_i)^2 \cdot \det(\sigma_kw_j)^2 = \text{disc}(M) \cdot \text{disc}(M') $$