问:
Prove this geometric inequality
Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$

My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have to proove $a+b+c\geq 3z$, i.e. $a+b\geq 2c$
Since $R(K,N;L,M)=R(B,C;D,E)$ so $R(K,N;L,M)= \frac{1}{2}:\frac{1}{2}$ thus $4ac=(a+b)(b+c)$.

I am not able to move further from here.

答:
Another way.

Let $KL=a$, $LM=x$ and $MN=b$.

Thus, since $$\frac{KM}{KN}:\frac{LM}{LN}=\frac{BE}{BC}:\frac{DE}{DC},$$ we obtain:
$$\frac{(a+x)(b+x)}{(a+b+x)x}=\frac{4}{3}$$ or
$$x^2+(a+b)x-3ab=0$$ or
$$x=\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ and we need to prove that
$$\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}+a+b\geq3\cdot\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ or
$$2(a+b)\geq\sqrt{a^2+14ab+b^2},$$
which is true by AM-GM:
$$\sqrt{a^2+14ab+b^2}=\sqrt{(a+b)^2+12ab}\leq\sqrt{(a+b)^2+12\left(\frac{a+b}{2}\right)^2}=2(a+b).$$

Title: Prove this geometric inequality


问:
$GL(n, \mathbb{C})$ is algebraically closed?
Let $GL(n,\mathbb{C})$ the group of non-singular matrices. Is it algebraically closed? For $GL(1,\mathbb{C})$ is it true; but if I take linear combinations of elements in $GL(n,\mathbb{C})$ with coefficients matrices in $GL(n,\mathbb{C})$ could I find solutions in $GL(n, \mathbb{C})$? We know that diagonalizable applications are dense: maybe it is useful. If $n=2$ can we use the isomorphis of $GL(2, \mathbb{C})$ with quatarnions?</p>

答: At the moment there is a problem with your question: $\text{GL}_n(\Bbb{C})$ is not a field. The notion of being algebraically closed only applies to fields. In fact $\text{GL}_n(\Bbb{C})$ is not even a ring because the ring axioms mean that we need it to be closed under addition. But then $diag(1,\ldots 1) + diag(-1,\ldots,-1) = 0$ so it is not a ring.

ContentLicense: CC BY-SA 3.0


问:
About joint probability divided by the product of the probabilities
Let $X$ and $Y$ be two events.

So $P(X)$ is the probability of $X$ happens, and $P(Y)$ is the probability of $Y$ happens.
So $P(X,Y)$ is probability of both $X$ and $Y$ happen.

So what is the meaning of the following function: $h(X,Y)=\frac{P(X,Y)}{P(X)P(Y)}?$

I know that when $h=1$, it means $X$ and $Y$ are independent. So what is the situation when

$h>1$ or $h<1$?

答: Mathematical Statistics is the branch of math that is area of specialization dealing with topics such as this. I can explain it to some point only here. But; it is my second major and leave it to those whose amin ateş of expertise is Applied Mathematics.</p>

ContentLicense: CC BY-SA 4.0


问:
Vector Notation
I just wanna ask, is it wrong if I write vector $v$ as $\bar{v}$ instead of $\textbf{v}$ or $\vec{v}$ in my handwriting? I ask this because someone told me that I was wrong. But, my professor also write vector $v$ as $\bar{v}$ and no one complains, because we know that our context is vector, not line segment or else. Can you guys share your opinion about this?

In my opinion, writing vector $v$ as $\bar{v}$ is just a shorthand notation for handwriting. Is it true?

答: I had a professor write their vectors as $\bar{v}$. Of course no one can say that it's wrong, it's just notation, what matters is that your audience understands what you write.

Title: Vector Notation


问:
Solving polynomials with compound angle formulae

Solve $64x^6-96x^4+36x^2-3=0$ (Hint: Consider expanding $\cos 6\theta$ in terms of $\cos \theta$)

To solve this exercise I can simply derive an identity using de-Moivre theorem then shuffle the equation around. It turns out that substituting $x=\cos \theta$ reduces the equation to $\cos 6\theta = \frac{1}{2}$.

However, I'm wondering in the general case, is there any way to see this without the monster of a hint? In other words, given a polynomial, is there something I could look for in the coefficients that would make me think that deriving a compound angle formula may be a good idea?

答:
It not always works in the general.

By the way, there are equations, which have nice real roots, but it not so easy to get these roots.

For example: $$x^5+x^4-12x^3-21x^2+x+5=0.$$
One of roots it's:

 $$2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)$$

In another hand, there are equations, which we can solve by the similar idea, which you showed, but without any hint.

For example:

Solve the following equation.
$$8x(2x^2-1)(8x^4-8x^2+1)=1.$$


问:
Find a continuous function such that$f:\mathbb{R}\to \mathbb{R}$, $f(0)=0$ and $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.
Could you give me a hint for this problem?

I have to find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and  $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.

I think this might be all the $f(x)=|x|$,  but can it be any other functions?

答:
$f(x) = x$ is a valid solution.

To see this is the only one, define $g(x) = f(x) - x$. It suffices to show that $g$ is constant. Then $f(2x) \geq x + f(x) \Leftrightarrow g(2x) \geq g(x)$, so $g(x) \geq g(\frac{x}{2^n})$ by induction. Similarly, and $f(3x) \leq  2x + f(x) \Leftrightarrow g(3x) \leq g(x)$ so $g(x) \leq g(\frac{x}{3^n})$ by induction. For any $a$ we have $$g(0) = \lim_{n \rightarrow \infty} g(\frac{a}{2^n}) \leq g(a) \leq \lim_{n \rightarrow \infty} g(\frac{a}{3^n}) = g(0)$$ so $g(a) = g(0)$ and thus $g$ is constant. Plug $f(x) = x + g(0)$ back in to conclude $g(0) = 0$.


问:
Prove that $UV^T$ is the nearest orthogonal matrix to $A$ given $||A||=\max_{x\neq 0}\dfrac{||Ax||}{||x||}=\sigma_1$

Prove that $UV^T$ is the nearest orthogonal matrix to $A$
ie., $||A-Q||\geq ||A-UV^T||$ for all orthogonal matrices $Q$
given the norm of the matrix is defined as $||A||=\max_{x\neq 0}\dfrac{||Ax||}{||x||}=\sigma_1$

The SVD of $A$ is $A=U\Sigma V^T=\sum_{i=1}^{r}\sigma_i u_i v_i^T$

$$
||A-Q||=||U\Sigma V^T-Q||=||U^T\Big(U\Sigma V^T-Q\Big)V||=||\Sigma-U^TQV||=||\Sigma-Q'||
$$

So we need to solve for $Q'$ when $||\Sigma-Q'||$ is minimized for the orthogonal matrix $Q'$.

How do I prove it ?

Note:

Similar question has been asked at Showing that matrix $Q=UV^T$ is the nearest orthogonal matrix to A., but it is for the Frobenius norm

Body:
\begin{aligned}
\|\Sigma-Q'\|
&=\max_{\|u\|=1}\|\Sigma u-Q'u\|\\
&\ge\max_{\|u\|=1}\big|\,\|\Sigma u\|-\|Q'u\|\,\big|\\
&=\max_{\|u\|=1}\big|\,\|\Sigma u\|-1\,\big|\\
&=\max(|\sigma_1-1|,|\sigma_n-1|)\\
&=\|\Sigma-I\|.
\end{aligned}


问:
Proof of Volume of sphere
I want a simple proof for the formula of volume of sphere. Does the proof/explanation without integration possible?</p>

答:
We have that the area of lateral surface $C$ of the trucated cone with basis circles of radii $r$ and $R$ is $$ A(C)= 2 \pi l \frac{r+R}{2} $$
with $l$ the meridian. If the cone is inscripted in the unit sphere $S^2$ we have that $A(C)=2 \pi d h$ with $h$ the hight of truncated cone and $d$ the distance from origin to $C$. If we cut $S^2$ with parallel planes with distance $h=\frac{1}{n}$ we can put $S^2$ in a lot of truncated. So is it easy to show that the sum of lateral surfaces $C_k$ of thes cones is
$$ \sum_k A(C_k)=\frac{2\pi}{n} \sum_{k=1}^{2n} d_n. $$ Now if $n \rightarrow + \infty$ we obtain the area of sphere. Using this principle we can calculate the volume os sphere. For all details you can look at http://www.proofwiki.org/wiki/Volume_of_Sphere/Proof_by_Archimedes.</p>

Title: Proof of Volume of sphere


问:
when removing a point from a closed set is still a closed set
What are the criteria for such that if you remove a point x, from a set A such that A{x} is still closed. We are working on the Real numbers here with standard topology.

The only solution I can think of is if a closed set is some kind of closed interval union with some isolated point, then removing that isolated point should leave the set closed. Also a singleton set is closed and if you remove the only element in the set, you get the empty set which is still closed. So this leads me to believe that if you remove a isolated point from a closed set, it is always still a closed set. Am I correct or are there counterexamples?

答: Correct:(when $X \in A$) $A\setminus\{x\}$ is closed when $A$ is closed iff $x$ is isolated in $A$. The proof is obvious (what if $x$ is not?)


问:
given a language, show L-terms are polynomials with integer coefficients
Let $L=\{0,1,+,-, \cdot \}$.

Show: for every $L$-term $t=t(x_1,x_2, \cdots, x_n)$, there exists a polynomial $P_t \in \mathbb{Z}[X_1, \cdots, X_n]$ such that for every communities unitary ring $\mathcal{R} $ viewed as an $L$-structure, we have $t^{\mathcal{R}}=P_t$. And conversely, every polynomial $P \in \mathbb{Z}[X_1, \cdots, X_n] $ is of the from $P_t$ for some L-term $t=t(x_1,x_2, \cdots, x_n)$.

For the first part of the question, we are given a $L$-term $t=t(x_1,x_2, \cdots, x_n)$. By definition of a $L$-term, we know

every variable is an $L$-term.

every constant symbol of $L$ is an $L$-term

every function in $L^F$ is an $L$-term.

But I'm not sure what exactly we are trying to show here. Is it clear that by 3), we are done?

And for the converse statement, I'm not sure what I'm supposed to show either.

Any hints or ideas would be greatly appreciated!

Thanks!

答:
I think it's worth taking a second to discuss what "terms" are in this context. From your question it looks like you might be slightly confused about this, and it's causing you to overcomplicate the rest of the question.

A term in a language $\mathcal{L}$ is (as you said) built up from variables and the constant/function symbols of the language.

I feel this is best given as a definition by examples, so let's take a second to look at some:

In $\mathcal{L}_\text{group} = (\cdot, {}^{-1}, e)$, we have terms like

$x \cdot y$

$x \cdot y^{-1}$

$(e \cdot y) \cdot (x \cdot z)^{-1}$

Notice terms coincide with a certain kind of function. Once we have a group $G$ in mind, then the term $x^{-1}y^{-1}xy$ defines a function $G^2 \to G$ where we simply substitute the actual group elements for the variables and simplify.

Now let's look at $\mathcal{L}_\text{ring} = (-, +, \times, 0, 1)$. Ideally, we should find that the terms are exactly the polynomials in with coefficients in $\mathbb{Z}$.

$x + 1$

$xy + (1 + 1)x - z$

$0x + (1+1+1)xx + (-(1+1)y) + z$

Do you see that these are all polynomials? If you play around with some terms, you should quickly get an intuitive understanding for why you always get a polynomial. Of course, intuition does not make a proof, so how can we formally show that the terms are always $\mathbb{Z}$ polynomials?

Since terms are defined recursively, we will prove things about them inductively. As you mentioned, terms are

variables $x,y,z,\ldots$

constant symbols $c$

a function symbol applied to preexisitng terms $f(t_1, \ldots, t_n)$

So let's try to show that each of these is a polynomial with integer cofficients!

the function $x$ is indeed a polynomial with coefficients in $\mathbb{Z}$, and indeed every variable is automatically a polynomial of degree 1.

The constant symbols are $0$ and $1$ both of which correspond to (degree 0) polynomials with coefficients in $\mathbb{Z}$

Lastly, if $t_1$ and $t_2$ are terms, we must show that $t_1 + t_2$, $-t_1$, and $t_1 \times t_2$ are all polynomials too. By induction, $t_1$ and $t_2$ are both polynomials with coefficients in $\mathbb{Z}$. Then negating, adding, and multiplying polynomials gives another polynomial.

So by induction on the structure of terms, we're done ^_^

What about the other direction, though? Why can every polynomial with coefficients in $\mathbb{Z}$ be written as a term?

Each polynomial looks like the sum of things of the form $a x_1^{n_1} x_2^{n_2} \ldots x_k^{n_k}$. Since we know we're allowed to take sums, it suffices to show that each of these is a term. But these all look like products of variables (which we can make) and some integer $a$.

So it suffices to show we can find a term for every integer $a$... Of course, we have access to a constant $1$ and an operation for negation. Do you see how to finish the proof from here?

I hope this helps ^_^


问:
Pigeon Hole Principle - 14 days, 10 physical exercises. How will you do PE on consecutive days at least once in the next fortnight.?
I was given this question. Maths is a weakness of mine and I'm really struggling to understand what this question is asking, and how to solve it. My notes indicate it's a pigeon hole principle but I'm unsure.

"You prepare a schedule of your physical exercises for the next fortnight (14 days). You don’t do physical exercise more than once a day. If you have 10 PE sessions planned, explain using the counting principles covered in class how this means you will do PE on consecutive days at least once in the next fortnight."

Thanks in advance :)

答: Let $\ n\ $ be the number of PE lessons scheduled for days immediately followed by a day free of any PE lesson. These lessons plus the following lesson-free day must occupy a total of $\ 2n\ $ days, at most one of which (if it is the day following the last lesson) can lie outside the allowed $14$-day period.  This leaves at most $\ 15-2n\ $ days available to be occupied by the remaining $\ 10-n\ $ lessons. Thus, $\ 10-n\le15-2n\ $, giving $\ n\le5\ $.  Thus, there must be at least $\ 5\ $ lessons on days that are immediately followed by a day with another lesson.


答: Given that $A$ is closed, $A-\{x\}$ is closed iff x is an isolated point of $A$, as in, there is some open neighbourhood $u$ of $x$ so that $u \cap A = \{x\}$. To see this, if for every open neighbourhood $u$ of $x$ there is some $x \neq y \in A$ we can choouse $u_n = (x-\frac{1}{n}, x+\frac{1}{n}),  y_n \in u_n \cap (A-\{x\})$ and note that $y_n$ converges to $x$, so $A-\{x\}$ can't be closed.

Title: when removing a point from a closed set is still a closed set

Body:
What are the criteria for such that if you remove a point x, from a set A such that A{x} is still closed. We are working on the Real numbers here with standard topology.

The only solution I can think of is if a closed set is some kind of closed interval union with some isolated point, then removing that isolated point should leave the set closed. Also a singleton set is closed and if you remove the only element in the set, you get the empty set which is still closed. So this leads me to believe that if you remove a isolated point from a closed set, it is always still a closed set. Am I correct or are there counterexamples?


问:
Restriction caused by substitution
We often come across integrals where we use various substitutions to reduce our problem. A common one is the trigonometric substitution.

When we substitute $x=\sin \theta$,  aren't we restricting $x$'s values to $[-1,1]$?

Here's a quick example:

$$\int_2^3 xdx$$

In this case, $x$ is always in $[2,3]$. So is $x=\sin \theta$ a valid substitution here?

In general, I would like to say that a substitution in an indefinite integration is valid, if and only if the variable substituted has the same range as the new variable.

答:
You 'run' a definite integral over a set. In a way you can think of a ' definite integral' as a procedure which takes in a subset(S) of $\mathrm{R}$ and tells the area of a function underneath the curve and x-axis in that interval.

When you do a change of variables, since you are changing the function you are integrating, you must also change the set you are integrating over. If you have are integrating over a set $[a,b]$ and you do a substitution of the form below:

$$ x= g(t)$$

Then, in the new domain of $t$, your domain is transformed like so:

$$ [a,b] \to [ g^{-1} (a) , g^{-1} (b) ] $$

You can think that as we vary 't', we vary $g(t)$ and this, in turn, varies the output. And, so, yes we can only do a substitution if the $g(t)$ can span the original domain which we were integrating over.

Given the above, if you are doing a change of variables, first of all, you need to make sure there is a well-defined inverse function of the new function you are integrating such that you can change your bounds properly.

In the particular case, it seems we can't find $ \sin^{-1} (2)$ or $ \sin^{-1} (3)$ whilst constrained to the set of reals. And hence, that substitution is not feasible in this case.

Title: Restriction caused by substitution


问:
Convergence of an alternating infinite series
What does the following infinite series converge to ?

$\displaystyle \sum _{k=1}^{\infty } (-1)^k \left(\tan \left(\frac{\pi }{2}-\frac{1}{k}\right)-k\right)$

It seems that the given series converges to a specific numerical value. It is possible to verify this with Mathematica and Wolfran Alpha.

I'm wondering, is there a closed form analytical solution to this infinite series?

Thanks.

答:
$$S= \sum _{k=1}^{\infty } (-1)^k \left(\tan \left(\frac{\pi }{2}-\frac{1}{k}\right)-k\right)=\sum _{k=1}^{\infty } (-1)^k \left(\cot\left(\frac{1}{k}\right)-k\right)$$

Numerically the result is $0.253374555381592$ which is not recognized by inverse symbolic calculators.

If you compute the first $p$ terms
$$S=\sum _{k=1}^{p} (-1)^k \left(\cot\left(\frac{1}{k}\right)-k\right)+\sum _{k=p+1}^{\infty } (-1)^k \left(\cot\left(\frac{1}{k}\right)-k\right)$$ by Taylor, we have
$$a_k=\cot\left(\frac{1}{k}\right)-k=-\frac{1}{3 k}-\frac{1}{45 k^3}-\frac{2}{945
   k^5}+O\left(\frac{1}{k^7}\right)$$ The summation
$$S_p=\sum _{k=p+1}^{\infty } (-1)^k a_k$$ is explicit in terms of  the Hurwitz-Lerch transcendent function and the Hurwitz zeta function. Expanding again
$$S_p=(-1)^p \left(\frac{1}{6 p}-\frac{1}{12 p^2}+\frac{1}{90 p^3}+\frac{1}{40
   p^4}\right)+O\left(\frac{1}{p^5}\right)$$ Using $p=100$ gives
$$S \approx 0.253374555381544$$

Edit

Using the infinite series

$$\cot\left(\frac{1}{k}\right)-k=\sum_{n=0}^\infty (-1)^{n+1}\frac{ 2^{2 n+1}  \left(B_{2 n+2}+B_{2 n+2}(1)\right)}{(2 n+2)!\,\,k^{2 n+1}}$$ we could have
$$S_p=\frac{\log (2)}{3}+\sum_{n=1}^p (-1)^n\frac{2  \left(2^{2 n}-1\right) \zeta (2 n+1) \left(B_{2 (n+1)}+B_{2n+2}(1)\right)}{(2n+2)!}$$ which converges quite fast. $S_{50}$ gives $50$ exact decimal figures
$$S_{50}=0.25337455538159208879429837870426655264941867607205$$


问:
spivak calculus on manifold thm 1.11

I think the proof of thm 1.11 of spivak has little problem.
Please focus on the yellow highlight part. Actually M($a$,f,$\delta$) is only defined for  $a\in A$.
but in this proof, we don't have any guarantee  $y \in A$, actually, if  $y \notin A$ imply $y \in R^n-B$. so thm can be proven perfectly.

Please talk to me if there is any problem in my opinion.

答: $\{x: o_f(x) <\epsilon\}$ is an open subset of $A$ in the subspace topology (by the argument given in the book, restricting $y$ to points of $A$). Hence $\{x: o_f(x) \geq \epsilon\}$ is closed in $A$ which also implies it is closed in $\mathbb R$ because $A$ is closed in $\mathbb R$.

Title: spivak calculus on manifold thm 1.11


问:
Projective module of rank n and faithfully flat R-algebra
Suppose P is a projective R-module of rank n.

Show that there exists a finitely generated faithfully flat R-algebra A such that tensoring with P gives a free module of rank n.

Any hint is appreciated.

答:
$P$ is the image of some idempotent $e \in M_m(R)$ for some $m \ge n$. We want to find an fppf (this is very slightly stronger than what you asked for: finitely presented faithfully flat) extension $R \to A$ such that $\text{im}(e) \cong A^n$ is free when extended by scalars to $A$. This is equivalent to the condition that there exist linear maps $f : A^n \to A^m, g : A^m \to A^n$ such that $fg = e$ and $gf = \text{id}_{A^n}$.

Writing $f$ and $g$ as matrices, this is a system of $n^2 + m^2$ polynomial equations in $2mn$ variables. Now we can construct $A$ by formally adjoining to $R$ a solution to this system; that is, we take $A = R[f_{ij}, g_{ji}]/I$ where $1 \le i \le m, 1 \le j \le n$ and $I$ is the ideal generated by the conditions $fg = \text{id}_{A^m}, gf = e$. By construction $A$ is finitely presented and $P \otimes_R A$ is free.

It remains to check that $A$ is faithfully flat. To check that it's flat we can check that the localization $A \otimes_R R_P$ is flat for every prime ideal $P$ in $R$. Over a local ring $R_P$ every finitely generated projective module is free, so the system above always has solutions, or equivalently $\text{im}(e)$ is always already free. This lets us determine all possible values of $f, g$ over $R_P$ as follows: $GL_n(R_P)$ acts freely and transitively on the possible values of $f : R_P^n \to R_P^m$ (since $f$ must be an isomorphism onto $\text{im}(e)$), and given $f$ (and $e$), $g$ is uniquely determined: it satisfies

$$gfg = g = ge$$

so it must factor through $e$ and hence, as a map $\text{im}(e) \to A^n$, must be the inverse of $f : A^n \to \text{im}(e)$. Moreover this argument continues to apply over any $R_P$-algebra, since an extension of scalars of a free module remains free. It follows that $A \otimes_R R_P$ is isomorphic to the algebra of functions on the affine group scheme $GL_n$ over $R_P$, which is a localization of a polynomial algebra on $n^2$ variables over $R_P$ and hence flat.

Finally, to check that $A$ is faithfully flat given that it's flat, we can check that $A \otimes_R R/m$ is nonzero for every maximal ideal $m$ of $R$. But over a field $R/m$ it's again true as above that every finitely generated projective module is free, so again the system always has solutions.

There's probably a slightly cleaner way to say all this. Morally $A$ is the universal extension of $R$ which trivializes $P$, except that a priori the construction above might depend on the choice of $e \in M_n(R)$ presenting $P$ and I haven't checked that it doesn't.


问:
Cubic diophantine equation in two variables
I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.

I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now.  Any help would be appreciated.

答:
Do you have to find solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$?
The pair $(0,0)$ is a solution.

You can simplify further $x(y+1)^2=243y$.

We know that $y\mid x(y+1)^2$. But $\operatorname{gcd}(y,(y+1)^2)=1$. You can proof that simply.

So $y\mid x$. And we can express $x=yc$. Which would lead to a cubic equation in $y$, which you can solve, as $y$ now accours in every term, we even can factor this, which then leads to a quadratic equation. This can be solved easily with the known formulas/methods.


问:
Is an $\epsilon$-neighborhood same as the open interval $(a-\epsilon, a+\epsilon)$
While reading real analysis, I came across this definition:

Definition:($\epsilon$-neighbourhood).Given a real number $a\in\mathbb{R}$ and a positive number $\epsilon > 0$,
the set
$$V_{\epsilon}(a)=\{x\in\mathbb{R}:|x-a|<\epsilon\}$$
is called the $\epsilon$-neighbourhood of $a$.

From this definiton I noticed that $V_{\epsilon}(a)$ consists of all those points whose distance from $a$ is less than $\epsilon$. From this, I concluded that:

$$V_{\epsilon}(a)=(a-\epsilon,a+\epsilon)$$
is this alternate definition (or say, theorem) true?

答: Yes. This isn't a 'theorem', the definition of an open interval is already $(a,b) = \{ x \in \mathbb{R} | a<x<b\}$, so $(a-\epsilon, a+\epsilon) = \{ x \in \mathbb{R} | a-\epsilon < x < a+\epsilon\}$ and $ a-\epsilon < x < a+\epsilon $ is equivalent to saying $|a-x|<\epsilon$


问:
Can anyone please tell me how to say that the following graph is $\frac{1}{x^2 - 4}$?
Can anyone please tell me how to tell that  the following graph is $\frac{1}{x^2 - 4}$ ?

My Attempt : From the graph , we can see $\lim_{x \to 2+} = \infty = \lim_{x \to -2-} $ and $\lim_{x \to 2-} = -\infty = \lim_{x \to -2+} $ .

But I can not conclude from these that the graph will be $ \frac{1}{x^2 - 4}$.

Can anyone please tell me how to show that the graph will be $ \frac{1}{x^2 - 4}$.

答:
Roadmap:

0)$f$ is symmetric about $y-$axis (Why?)

Consider $f$ for $x \ge 0$, $x \not=2$.

1)Interval $[0,2);$ $f(0)=0;$

What happens for $x \rightarrow 2^{-}$?

4)Interval $(2,\infty);$

What happens for $x \rightarrow 2^{+}$?

5)Consider $x \rightarrow \infty.$

6)Reflect the graph about $y-$axis.


问:
Algorithm to solve n variable problem with range
During Software designing our team come across the following problem: we have linear equation with n variables with each variable have positive permissible range or allowed range.

In This Problem the constant a_i are the number of items person wants to be bought of type x_i which has minimum value of purchase l_i and maximum value of purchase r_i and now, buyer know about this ranges that why he smartly ask that is this possible within value b. So now we have to choose one solution if possible and give the invoice accordingly.

So in math form,

```<code> a1x1+a2x2+....+anxn = b
 known parameters:
 a1,a2,...,an and b
 unknown parameter x1,x2,....,xn
 but we know that
 x1 belongs to range [l1,r1]
 x2 belongs to range [l2,r2]
 .
 .
 .
 xn belongs to range [ln,rn]
 0<li<ri
 ai>0
 b>0
</code>```

and we need to find values of all xi in above equation(may be float or integer but must be range). I need to solve this problem as soon as possible.
Example :

```<code>  Let n be 5
  And equation is
  10*x_1+15*x_2+10*x_3+11*x_4+17*x_5=2365
   And allowed ranges are
   For
   X_1 [30,60]
   X_2 [20,40]
   X_3 [20,80]
   X_4 [50,120]
   X_5 [10,140]
  So one of the solution will be
  For
  X_1 =40
  X_2 =36
  X_3 =34
  X_4 =60
  X_5 =25
</code>```

Want to find only one solution.
I need solution in bigO(nlogn)
Or nearly linear as possible as.

答:
Consider
$$\Phi= \left(\sum_{k=1}^n a_k\,x_k-b\right)^2$$ and minimize $\Phi$ with $n$ bound constraints $l_k \leq x_k \leq r_k$.

If you want to get rid of the constraints, define
$$x_k=l_k+\frac{r_k-l_k }{1+e^{-y_k} }$$

As @vvgiri  commented, you can face an infinite number of solutions (may be less if one or more bound constraints are active). For your example, the first I obtained is
$$\left\{34,\frac{85}{4},\frac{83}{4},85,\frac{2255}{68}\right\}$$


问:
Finding ratio of cevian lines
I am preparing for an exam and doing some pratice problems. So I'm having a difficult time with this problem. At first I thought the ratio was 2:1 and then I also thought I would be able to use the angle bisector theorem but then I realized that cevians could be altitudes, medians or angle bisectors. So theorems are not helpful with this problem.

Basically, I asked my teacher for help and he told me that the answer is

EF:FC is $10:7$

and DF:FA is $3:14$

I've been trying to find out how he got these ratios but i'm not getting anywhere. Any ideas?

答:
Use mass points.
We assign B a mass of 4. So A has a mass of 3 and C has a mass of 10. E has a mass of 7 and D has a mass of 14. So we have $$ \frac{AF}{FD}= \frac{14}{3}$$ and $$\frac{EF}{FC}= \frac{10}{7}$$.

Title: Finding ratio of cevian lines


问:
Is Lower Envelope of Compact and Convex Set always continuous?
From a picture in $\mathbb{R}^2$, say a square $[0, 1]\times [0, 1]$, the lower envelope of the square is $[0, 1]\times 0$ and can be seen as a continuous function over $[0, 1]$. I wonder if there exists a general theorem proving this intuition or there exist some examples showing things might go wrong. Thanks in advance.

答:
For a compact and connected set $S \subset \Bbb R^2$ the projection of $S$ onto the x-axis is a compact interval $[a, b] \subset \Bbb R$, and one can define the lower envelope of $S$ as the function
$$
f: [a, b] \to \Bbb R, f(x) = \min \{ y \mid (x, y) \in S \} \, .
$$

$f$ is not necessarily continuous, but in the case of a convex set $S$ it is. Here is a proof in three steps. (We can exclude the degenerate case by assuming that $a < b$).

Step 1. $f$ is lower
semi-continuous, i.e.
$$  \liminf_{x \to x_0} f(x) \ge f(x_0) $$ for all $x_0 \in [a, b]$.

Proof: If $x_0 \in [a, b]$ and $(x_n)$ is a sequence in $[a, b]\setminus \{ x_0 \}$ with $x_n \to x_0$ and $f(x_n) \to \liminf_{x \to x_0} f(x)$ then
$$
 \forall n: (x_n, f(x_n)) \in K
\implies (x_0, \liminf_{x \to x_0} f(x)) \in K
$$
and that implies $f(x_0) \le \liminf_{x \to x_0} f(x)$.

Step 2. If, in addition,  $S$ is convex, then $f$ is convex.

Proof: Consider $x, y \in [a, b]$ with $x < y$. Then $(x, f(y)) \in S$ and $(x, f(y)) \in S$, so that for $0 < \lambda < 1$
$$
 ((1-\lambda)x + \lambda y, (1-\lambda)f(x) + \lambda f(y)) \in S
$$
and that implies
$$
 f((1-\lambda)x + \lambda y) \le (1-\lambda)f(x) + \lambda f(y) \, .
$$

Remark: It is known this implies that $f$ is continuous on $(a, b)$, but that does not help at the boundary points.

Step 3. A convex function $f: [a, b] \to \Bbb R$ is upper semi-continuous, i.e.
$$  \limsup_{x \to x_0} f(x) \le f(x_0) $$ for all $x_0 \in [a, b]$.

Proof: Let $a \le x_0 < x_1 \le b$. Then
$$
 f(x) \le \frac{x_1-x}{x_1-x_0}f(x_0) + \frac{x-x_0}{x_1-x_0}f(x_1)
$$
for $x_0 < x < x_1$, and that implies
$$
 \limsup_{x \to x_0+} f(x) \le f(x_0)
$$
and
$$
 \limsup_{x \to x_1-} f(x) \le f(x_0) \, .
$$

Combining these three results, we see that if $S$ is compact and convex then its lower envelope function is both upper and lower semicontinuous, and therefore continuous.


答:
This is the plot of

$$\frac{x^2+10}{10(x^2-4)}.$$

It is virtually impossible to distinguish it from the given one.

Title: Can anyone please tell me how to say that the following graph is $\frac{1}{x^2 - 4}$?

Body:
Can anyone please tell me how to tell that  the following graph is $\frac{1}{x^2 - 4}$ ?

My Attempt : From the graph , we can see $\lim_{x \to 2+} = \infty = \lim_{x \to -2-} $ and $\lim_{x \to 2-} = -\infty = \lim_{x \to -2+} $ .

But I can not conclude from these that the graph will be $ \frac{1}{x^2 - 4}$.

Can anyone please tell me how to show that the graph will be $ \frac{1}{x^2 - 4}$.


问:
Basic question on stone-cech compactification
Let $\widetilde{X}$ be a compactification of a space $X$.

This implies that $\widetilde{X}$ is compact together with an embedding $e : X \rightarrow \widetilde{X}$ such that $e(X)$ is dense in $\widetilde{X}$.

Let $X' = e(X)$

Now $X$ is homeomorphic to $e(X)$ and we have that  $\widetilde{X}$ = $\bar X^{'} $  i.e closure of $X'$ is $\widetilde{X}$.

But I have seen that at various places, it has been written that $\bar X = \widetilde{X}$(by identifying $X$ with $X'$.

Now using this if $B((0,1])$ is stone-cech compactification of $(0,1]$ then $Cl((0,1])$ = $[0,1]$ should be equal to $B((0,1])$.

What is wrong in this?

I am definitely overlooking something. Please help!

答:
The homeomorphism class of the closure is not a topological invariant: specifically, it is not true that for any two embeddings $f:(A,\tau)\hookrightarrow (B,\sigma)$ and $g:(A,\tau)\hookrightarrow (C,\rho)$ there is a homeomorphism $u:\operatorname{cl}_B{f[A]}\to \operatorname{cl}_C{g[A]}$.

Even if you add the condition that $B$ and $C$ should be compact Hausdorff, this fact does not change.


问:
What does it mean for a subset $A$ of the universe $M$ of an $L$-structure $N$ to be closed under $t^N$ for all $L$-terms $t = t(x_1, ..., x_n)$?
Initially I thought it meant closed in the sense of functions i.e if we apply an $L$-term $t$ to some elements of $A$, then the result stays in $A$. But, well, I am not so sure anymore since $L$-terms can be constants, so it doesn't make sense to 'apply' a constant.

答:
Initially I thought it meant closed in the sense of functions i.e if we apply an $\mathcal L$-term $t$ to some elements of $A$, then the result stays in $A$.

Quite correct...

The domain $M$ of a structure $\mathfrak A$ is a collection of "objects" (e.g. numbers) and we interpret the language in such a way that formulas are statements about the objects and their properties.

Terms are "names": the individual constant $0$ in the language of arithmetic is the name for the number zero.

To say that a subset $A$ of $M$is "closed under" (the interpretation of term) $t$, for every $t$, means that:

according to the given interpretation, every term $t$ denotes an element of $A$, when its free variables are interpreted in $A$.

This is obvious for individual constants: either $t^{\mathfrak A} \in A$ or not.

But terms are built also from function symbols, like e.g. $+(x,y)$ in the language of arithmetic.

To say that $A$ is closed with respect to it means that every time we assign to variables $x,y$ values in $A$, the interpretation will assign to term $+(x,y)$ a value in $A$.

Some examples with $\mathfrak A = (\mathbb N, 0, s, +, \times)$.

Not all subsets $A$ of $\mathbb N$ are closed under $+^{\mathfrak M}$.

Consider e.g $A = \{ 1,2,3 \}$; we have that for $x=1$ and $y=3$, their sum is not in $A$.

The subset of $\mathbb N: \ A = \{ n \mid n \text { is Even } \}$, instead, is closed with respect to $+^{\mathfrak M}$.

Final example: no proper subset of $\mathbb N$ is closed under the successor function $s(x)$.


问:
Finding ramification points of a map to $\mathbb{P}^1$ given by a divisor
Let $X$ be a elliptic curve over $k$ ("curve" in the sense of Hartshorne Chapter IV). A closed point $P_0\in X$ gives rise to a base-point free linear system $|2P_0|$ of dimension 1, which in turn gives rise to a morphism $X\rightarrow \mathbb{P}^1$ of degree 2. This morphism is given by selecting a basis $f_0,f_1$ of $\Gamma(X,\mathcal{L}(2P_0))$ and setting $P\mapsto [f_0(P):f_1(P)]$. How do I spot the ramification points of this map? It seems obvious that $P_0$ should be a ramification point, but I don't quite see why. I believe answering this question will help me understand a more fundamental question: why is this map degree 2? We can think of $\Gamma(X,\mathcal{L}(2P_0))$ as rational functions on $X$ with a pole of order $\leq 2$ at $P_0$, so is there then some canonical choice of basis $f_0,f_1$ that makes the answers to the above questions clear?

Edit: I found this thread that seems to answer the question. I mostly understand the solution here, although the very last step is still giving me trouble. We choose $x\in \Gamma(X,\mathcal{L}(2P_o))$ is such that $\{1,x\}$ form a basis of this vector space. Since $x$ has a pole of order 2 at $P_0$ (and no other poles), we conclude that $x$ has exactly two zeros (by Riemann-Roch). We can say the same thing for $x-\alpha$ for any $\alpha\in \mathbb{C}$. The claim is that for generic $\alpha\in \mathbb{C}$, these two zeros are distinct. I don't yet understand this claim.

答:
(I mostly have the case $k = \Bbb C$ in mind, but it should work more or less for other field.)

Any elliptic curve is a smooth cubic curve $X \subset \Bbb P^2$ (by Riemann-Roch), and there is a group law such that $Q_1 +Q_2+Q_3 = 0$ if there is $L \subset \Bbb P^2$ such that $L \cap X = Q_1 + Q_2 + Q_3$ (as divisor). In particular, let $Q_0 = -P/2$. By definition, $Q_0$ is the point such that $L_{P_0}X \cap X = 2P_0 + Q_0$.

Lines through $Q_0$ are parametrised by $\Bbb P^1$, call this family $\{\ell_t\}_{t \in \Bbb P^1}$. Consider the morphism $X \to \Bbb P^1, P \mapsto t$ if $P \in \ell_t$. You can check that this is simply the morphism induced by $|2P_0|$. It follows that ramification points of the map are geometrically points $P$ such that $Q_0 \in T_{P}X$. Algebraically, there are element of order $2$ in the group law.

By Riemann-Hurwitz, you can check that there will be $4$ ramification points. It is a fun exercise to check it using this algebraic or geometric description of the ramification point.


问:
Proving $A \implies B$ or $C$
To show that a statement $A$ implies $B$ or $C$, is it sufficient to show that if $A$ and $ \neg B$ hold, then $C$ must hold?
It is clear in the case of a set where we have to show that $A$ is a subset of $B$ or $C$, showing the above must be sufficient.

答:
An impliciation $P\Rightarrow Q$ is equivalent to $\neg P \vee Q$. In your situation this means that $A\Rightarrow(B\vee C)$ is equivalent to
$$
\neg A \vee B\vee C.
$$
However, using de Morgan's law, you can also write this as
$$
\neg(A \wedge \neg B) \vee C
$$
which is then equivalent to
$$
(A \wedge \neg B) \Rightarrow C.
$$

Title: Proving $A \implies B$ or $C$


问:
Which pair of groups are isomorphic?

For a ring $R,$ let $R^*$ denote the set of units of $R.$ Are the pair of groups isomorphic?

(a) $\left (\Bbb C^*, \cdot \right ), \left (\left (\Bbb R^2 \right )^*,\cdot \right )$

(b) $\left (\Bbb R, + \right ),$ $\left (\Bbb C, + \right )$

I think (a) is false because $\Bbb C^*$ contains an element of order $3$ but $\left (\Bbb R^2 \right )^*$ doesn't contain any element of order $3.$ Now we know that two additive groups are isomorphic if and only if they are isomorphic as $\Bbb Q$-vector spaces. So (b) is true if and only if $\Bbb R$ and $\Bbb C$ are isomorphic as $\Bbb Q$-vector spaces. But I think I have read somewhere that two $\Bbb Q$-vector spaces are isomorphic if and only if both of them have the same cardinality (Though I am not quite sure about that). I believe that for any $\Bbb Q$ vector space $V$ we have $\dim_{\Bbb Q} V = \text {Card}\ (V)$ and two $\Bbb Q$-vector spaces are isomorphic if and only if they both have the same $\Bbb Q$-dimension. So if my argument were correct then (b) is true since $\text {Card}\ (\Bbb C) = \text {Card}\ (\Bbb R).$
Am I right?

Thanks in advance.

答:
(a) is correct.

Now we know that two additive groups are isomorphic if and only if they are isomorphic as $\mathbb{Q}$-vector spaces.

Only if they actually are $\mathbb{Q}$-spaces. Which is the case for $\mathbb{R}$ and $\mathbb{C}$, but not in general (e.g. any finite abelian group or $\mathbb{Z}$ is not a $\mathbb{Q}$-space). Also be warned that for "$\Rightarrow$" you have to show that every group homomorphism is actually $\mathbb{Q}$-linear map (which indeed holds).

But I think I have read somewhere that two $\mathbb{Q}$-vector spaces are isomorphic if and only if both of them have the same cardinality (Though I am not quite sure about that)

This is only true for infinite dimensional spaces over $\mathbb{Q}$. That's because if $V$ is a vector space over $F$ and $B\subseteq V$ its basis then $V$ can be represented as

$$\{f:B\to F\ |\ f(b)=0\text{ for all but finitely many }b\}$$

which for infinite $F$ has cardinality $|V|=|F|\cdot |B|=\max\{|F|, |B|\}$. Now if $|F|=\aleph_0$ (which is our case) and $V$ is of infinite dimension then this simplifies to $|V|=|B|$, i.e. $|V|=\dim V$.

And so, yes, once we have this the rest is simple, since $|\mathbb{R}|=|\mathbb{C}|$ and both are of infinite dimension over $\mathbb{Q}$ and thus they have to be isomorphic.

Title: Which pair of groups are isomorphic?


问:
Find number of normal subgroups in $F_3$ such that its factor is isomorphic to a given Abelian group
On the upcoming test I will be given a problem of type:

Find all normal subgroups $H$ in $F_n$ such that $F_n/H \cong G$.

Here $n$ is a small integer, likely 2 or 3, and $G$ is an Abelian group given as product of some cyclic groups.

Question: is there a more-or-less routine algorithm for such problems?

I do know an approach. I count two parameters: the number of epimorphisms $\#(F_n \twoheadrightarrow G)$ and the number of automorphisms $\#Aut(G)$. The answer then is $\frac{\#(F_n\twoheadrightarrow G)}{\#Aut(G)}$. This solution is based on the following facts: 1) every normal subgroup is a kernel of some homomorphism; 2) $F_n/Ker\,\phi \cong G$ if $\phi$ is an epimorphism; 3) $Ker\,\phi=Ker\,\psi$ iff $\psi = \alpha\circ\phi$ for some $\alpha \in Aut(G)$.

The problem is, very often I'm not sure how do I count one or both of these parameters. I could go through all homomorphisms and automorphisms manually, but the numbers in problem are usually too large.

Here are some cases to show how big numbers might be on the actual test. No need to solve them!

$n=3$, $G=\mathbb Z_{13}^{3}$

$n=3$, $G=\mathbb Z_{10}^3\times \mathbb Z$

$n=3$, $G=\mathbb Z_{70}\times \mathbb Z_{30} \times \mathbb Z_{15}$

$n=2$, $G=\mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_5$

$n=3$, $G=\mathbb Z_{55} \times \mathbb Z_{7} \times \mathbb Z_{77}$

$n=3$, $G=\mathbb Z_{75} \times \mathbb Z_{375} \times \mathbb Z_{125} \times \mathbb Z_{333}$

答:
Of the problems that you mention, I think that only the first two, which you can do in your head, are suitable for an exam. As I said in my comment, Example 4 is relatively easy, but even then there are $72$ distinct subgroups $H$, and it would not be reasonable to expect you to enumerate them by hand in a test.

Magma has a function for enumerating subgroups of a finite abelian group using the Hermite Normal Form for the generating sets of the subgroups, and I used this on Example 3. (This is essentially the second method proposed by Qiaochu Yuan in his answer.)

The group $G$ in Example 3 has exponent $210$, so all of the subgroups that we are looking for contain $[F_3,F_3]F_3^{210}$, and we are looking for those subgroups $N$ of the finite abelian group ${\mathbb Z}_{210}^3$ with quotient $G$.

Unfortunately the Magma function has no facility for looking for subgroups of a specific order (I wrote the code for the function, and I might consider adding that as an option), and so there was no alternative but to find all subgroups, and there are $332595$ of these.

Of these $5187$ have the required order and quotient group $G$.

In example 5 there are $235011$ subgroups $H$, and I am afraid Example 6, in which $G$ has exponent $41625 = 3^2\cdot 5^3\cdot 37$ is beyond the scope of the software!


问:
Integration with Kronecker delta
Suppose the functions $E(k)$ and $H(k')$ with units $[V]$ and $[A]$ and the unit of $k$ and $k'$ being $\left[\frac{1}{m}\right]$. Using the Kronecker delta function $\delta_{i,j}$ we can write
$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} E(k)H(k') \delta_{k,-k'} dk dk' = \int\limits_{-\infty}^{\infty} E(k)H(-k) dk$$

If now the units on both sides of the equation are compared, different results are obtained. The left side has unit $ \left[V A \frac{1}{m}\frac{1}{m} = \frac{VA}{m^2}\right]$ whereas the right side has $\left[ \frac{VA}{m}  \right]$. How can this be explained?

The Kronecker delta function can be defined as
$$\delta_{k,-k'} = \lim\limits_{X\to\infty} \frac{1}{X}\int\limits_{-X}^{X} e^{i(k+k')x}dx $$
and is thus dimensionless.

答: If $k$ has units $[ \frac{1}{m}]$ then the kroncker delta for k has units $[m]$. note that for any function f, $\int_{\mathbb{R}} f(x)\delta(x)dx = f(0)$ so if $dx$ has units of length and $f$ is, say, unitless, see that $\delta(x)$ must have units of $\frac{1}{length}$. Your intuition could be that under a change of units $x \to \alpha x$, we get $\delta(\alpha x) = \frac{1}{| \alpha |} \delta(x)$

Title: Integration with Kronecker delta


问:
If $\int|f|^p\ d\mu=K$, then $\int \lvert f \rvert^p\ \chi_E\ d\mu\le K\mu(E)$
Let $f\in L_p$, so

$$\lVert f \rVert_p <+\infty \Longrightarrow \int\vert f\rvert^p\ d\mu=K<\infty$$

For a demonstration of an exercise I assumed that,

$$\int \lvert f \rvert^p\ \chi_E\ d\mu\le\int K\chi_E\ d\mu=K\mu(E)$$

But a colleague questioned me about that statement, and I ended up not so sure about it anymore, because it’s an old proof and I don’t remember exactly what I thought when I wrote this. I was unable to demonstrate this statement or find a counterexample.

So, my question is whether the above implication is true or not.

答:
The inequality can be written as $$\frac{1}{\mu(E)}\int_E|f|\le\|f\|$$ (Put $|f|^p$ instead, if you wish.)

That the average of a function on a subset need not be less than the total function (or even the total average) is obvious:
Take $f(x)=\chi_{[0,1/2)}+2\chi_{[1/2,1)}$ and $E=[1/2,1)$,
$$2>\frac{2+1}{2}$$


问:
Relationship between density function

Could someone help me on this question. I'm not entirely sure how to relate the two different density functions to each other.

答:
First let's find $F_Y(y)$ with the definition

$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[e^X\leq y]=\mathbb{P}[X\leq log(y)]=F_X(log(y))$$

Now let's calculate its derivative to get the requested density

$$f_Y(y)=f_X(log(y))\cdot \frac{1}{y}\mathbb{1}_{(e;e^2)}(y)$$

...that is (a)

Title: Relationship between density function


问:
Why does the automorphism used to construct the group have to be non-inner?
I have a question on why a particular assumption is made that the automorphism used to construct a certain group be non-inner.

In [Herstein, Topics in Algebra, p. 69], a construction of a nonabelian group of order 21 is given as follows:  Let $G=C_7 = \langle a \rangle$, $\phi: a \mapsto a^2$ be an automorphism of $G$ of order 3, and consider all formal symbols $x^i a^j$ subject to conditions $x^3=a^7=e, x^{-1}ax=a^2$, and we get a nonabelian group of order 21.

The text continues that, more generally, if $G$ is a group, $T$ an automorphism of order $r$ of $G$ which is not an inner automorphism, pick a symbol $x$ and consider all elements $x^i g$, $i \in \mathbb{Z}, g \in G$, subject to $x^ig=x^{i'}g'$ iff $i \equiv i'(r), g =g'$, and $x^{-1} g^i x = gT^i$ for all $i$.  Then, we obtain a larger group $\{G,T\}$, with $G \trianglelefteq \{G,T\}, \{G,T\}/G \cong C_r$.  I think it is common to denote this group by the semidirect product $G \rtimes C_r$.  My question here is why is the assumption made that $T$ is ``not an inner automorphism.''

答:
Inner automorphisms are bijective homomorphisms $\theta_g:G\rightarrow G$ which have the form $$\theta_g(x)=g^{-1}xg$$
given a fixed $g\in G$.  If $G$ is abelian, then $xg=gx$ for every $x,g\in G$, which we can rewrite as as $g^{-1}xg=x$.  Thus every inner automorphism has the form $$\theta_g(x)=x\hspace{30pt}\text{for all }x\in G.$$
In other words, all inner automorphisms are the identity automorphism $\operatorname{id}_G$.  If one forms a semidirect product $G\rtimes_\phi H$ with the mapping $\phi(h)=\operatorname{id}$ for every $h\in H$, then $$G\rtimes_\phi H\cong G\times H.$$
In the context of semidirect products, this is a degenerate case.  To see a "non-trivial semidirect product," $\phi$ must map at least one $h$ to a non-identity automorphism of $G$, and thus when $G$ is abelian this automorphism cannot be inner.</p>


问:
Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$
Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$

I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:

$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$

I haven't been taught that as equal to $tan x$.

答:
Hint:

Use $$\cos2x=2\cos^2x-1=1-2\sin^2x\text{ and }\sin2x=2\sin x\cos x$$

Or use Weierstrass substitution

Or $$\dfrac{1+\sin2x}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{\cos^2x-\sin^2x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}\text { assuming }\cos x+\sin x\ne0$$

Apply Componendo et Dividendo


答:
\begin{align}
\frac{1-\cos 2x + \sin 2x}{1+\cos 2x + \sin 2x} &=\frac{1-(1-2\sin^2 x) + 2\sin x \cos x}{1+(2\cos^2 x -1) + 2\sin x \cos x} \\
&=\frac{2\sin^2x +2\sin x \cos x}{2\cos^2x + 2\sin x \cos x}
\end{align}

Can you simplify from here?

Title: Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$

Body:
Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$

I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:

$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$

I haven't been taught that as equal to $tan x$.


问:
Integration as inverse of differentiation
I have been reading "Mathematical Methods of Physics and Engineering" By K. F. Riley, M. P. Hobson and S. J. Bence. In chapter $2$  of this book (Preliminary Calculus) Under the sub-topic $2.2.2$ "Integration as the inverse of differentiation", I'm having trouble understanding the following proof:

$$ \text{Let } F(x)=\int_a^xf(u) \mathrm{d}u \ .\text{We then have:}$$
\begin{align}
 F(x+\Delta x)&=\int_a^{x+\Delta x}\ f(u) \mathrm{d}u\\[1ex]
              &=\int_a^x f(u)\ \mathrm{d}u\ + \ \int_x^{x+\Delta x} f(u)\ \mathrm{d}u\\[1ex]
 F(x+\Delta x)&= F(x)+\int_x^{x+\Delta x} f(u)\  \mathrm{d}u\\[1ex]
\implies\frac{F(x+\Delta x)-F(x)}{\Delta x}&=\frac{1}{\Delta x}\int_x^{x+\Delta x}f(u) \ \mathrm{d}u
\end{align}
Letting $\Delta x \to 0$ the LHS becomes $\frac{\mathrm{d}F}{\mathrm{d}x}$ whereas the RHS becomes $f(x)$. The latter conclusion follows because when $\Delta x$ is small the value of the integral on the RHS approximately becomes $f(x)\Delta x$, and in the limit $\Delta x \to 0$ no approximation is involved.

I didn't understand the last statement. How did we approximate the integral?

答: Since $f$ is supposed to be continuous the values of $f(y)$ for $x <y<x+\Delta x$ are all close to $f(x)$. So we can approximate $\int_x^{x+\Delta x} f(y)dy$ by $\int_x^{x+\Delta x} f(x)dy$ (in which the integrand is a  constant as far as integration w.r.t. $y$ is concerned) and the value of this integral $f(x)\Delta x$.


问:
Upper bounding convolution between compactly supported function and logarithm
Let $h:\mathbb{R}^2\rightarrow \mathbb{C}$ be a $C^\infty$ function with compact support which satisfies $\int h(z)\mathrm{d}x\, \mathrm{d}y = 0$. How can one show that
$$f(w) = \log \frac{1}{|\cdot|}\ast h(w) = \int_{\mathbb{R}^2}\log \frac{1}{|z|}h(w-z)\mathrm{d}x\, \mathrm{d}y$$
satisfies
$|f(w)|\leq \frac{C}{|w|}$ with some bound $C$.

答:
I came up with my own solution:

We have that

$$f(w) = \int_{\mathbb{R}^2}\log \frac{1}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y = \int_{\mathbb{R}^2}\log \frac{|w|}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y $$

by the specified condition. Now we may assume that $h$ has support in $B(0,R)$ and therefore for large enough $w$

$$|f(w)w|\leq \int_{B(0,R)}|w|\log \frac{|w|}{|w-z|}h(z)\mathrm{d}x\, \mathrm{d}y\leq \int_{B(0,R)}|w|\log \frac{|w|}{|w|-R}h(z)\mathrm{d}x\, \mathrm{d}y = |w|\log \frac{|w|}{|w|-R}\int_{B(0,R)}h(z)\mathrm{d}x\, \mathrm{d}y.$$

The term $x\log \frac{x}{x-R}$ has the finite limit $R$ as $x\rightarrow \infty$.


问:
Is there a way this change of coordinates can yield positive results?
I have a system $x' = x - \cosh(x + y)^2 + 1, \ y' = \sinh(x + y)^2 - y$ and I noticed that I can take
$x' + y' = x - y.$ From here, I played around with the concept of changing non-linear coordinates by saying
$(t,x(t),y(t)) \rightarrow (t^*,z(t^*),1/z(t^*))$ because it gives me
$
\begin{align*}
z' - \frac{z'}{z^2} = z - 1/z
\end{align*}
$
which by some amazing coincidence simplifies into
$z' = z$ which has the trivial solution $z(t^*) = z_0e^{t^*}.$

What is the proper way to unchange these coordinates back into regular $t$-time to see if the system is solvable in some way? Or is there anything else that can be done to utilize this?

答:
If you set $u=x+y$ and $v=x-y$, then you get
\begin{align}
u'&=v\\
v'&=u-2\sinh^2u
\end{align}
This is the first order system for a second order equation
$$
u''-u+2\sinh^2u=0.
$$
This is a conservative system, that is, all solutions move inside the level curves of the energy function
$$
E=u'^2-(u+1)^2+\sinh(2u).
$$


问:
Being at an initial condition with zero initial velocity?
Suppose we're solving an autonomous DE of the form $\dot{x} = f(x)$ (where dot is derivative w.r.t. time) with some initial condition $x(0)=x_0$ (with $x_0 \in \mathbb{R}$), What significance does it have if we also know that $\dot{x}(0) = 0$? Say this system describes the movement of a particle, does that mean the particle will remain at $x_0$ for all time since it's at that point at initial time $t=0$ with zero initial velocity, so it can't move? If so, do we need any requirements for f? Perhaps Lipschitzness for the solution to be uniquely $x(t)=x_0$?

答:
Yes, if $f(x_0)=0$, then the IVP with $x(0)=x_0$ has the constant solution $x(t)=x_0$, as is easy to check.

You need the Lipschitz condition for uniqueness to exclude any other solution, like they occur for the classical example $\dot x=\sqrt{|x|}$, $x(0)=0$.


Body:
By Tangent half-angle formulas with $t=\tan x$

$\cos 2x=\frac{1-t^2}{1+t^2}$

$\sin 2x=\frac{2t}{1+t^2}$

we have

$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\frac{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}=\frac{1+t^2-1+t^2+2t}{1+t^2+1-t^2+2t}=\frac{2t+2t^2}{2+2t}=t$$


答:
$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \frac{1- (2\cos^2x-1)+2\sin x .\cos x}{1+ 2\cos^2x-1+2\sin x .\cos x}$$
$$ = \frac{2- 2\cos^2 x +2\sin x .\cos x}{ 2\cos^2 x +2\sin x .\cos x} = \frac{1- \cos^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x}$$
$$ = \frac{\sin^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x} = \frac{\sin x (\sin x + \cos x )}{ \cos x (\sin x + \cos x )} = \frac{\sin x }{ \cos x } = \tan x$$


问:
Calculate $\pi$ precisely using integrals?
This is probably a very stupid question, but I just learned about integrals so I was wondering what happens if we calculate the integral of $\sqrt{1 - x^2}$ from $-1$ to $1$.

We would get the surface of the semi-circle, which would equal to $\pi/2$.

Would it be possible to calculate $\pi$ this way?

答:
The answer is more no than yes.

There are plenty definite integrals giving an answer which is a function of $\pi$. In particular yours,

$$2\int_{-1}^1\sqrt{1-x^2}\,dx=\left.\left(x\sqrt{1-x^2}+\arcsin(x)\right)\right|_{-1}^1=\pi.$$

But this does not bring you any closer to the numerical value of $\pi$.

There are many ways to obtain a desired number of decimals of $\pi$, using finite approximations of various sequences, series or integrals (this is a broad topic). In particular, you can estimate the above area using the Newton-Cotes numerical method or similar, but this will be very slow and is not used in practice. You can also evaluate the antiderivative at the bounds (again using a numerical approximation such as a truncated Taylor series) but it is also dead slow in this particular setting.

ContentLicense: CC BY-SA 2.5


问:
Summation $\sum_{n=0}^{\infty} \frac{P_n(\cos \theta)}{n+1} = \log\frac{\sin(\theta/2) +1}{\sin(\theta/2)}$
$P_n$ is legendres polynomial of $n$ degree terms. I tried to use the formula summation of $C_n P_n(x) = f(x)$ but not able to find $f(x)$. Also I tried using generating functions but that too didn't help.

Body:
Let $\cos \theta =c$,
$$S=\sum_{n=0}^{\infty} \frac{P_n(c)}{n+1}= \sum_{n=0}^{\infty} \int_{0}^{1} P_n(c) t^n dt= \int_{0}^{1} \frac{dt}{\sqrt{1-2ct+t^2}}=\int_{0}^{1} \frac{dt}{\sqrt{(t
-c)^2+1-c^2}}$$
$$\implies S=ln[2((t-c)+\sqrt{(t-c)^2+s^2})|_{0}^{1}=\ln\frac{1-c+\sqrt{2(1-c)}}{(1-c)}=\ln \frac{1+\sin(\theta/2)}{\sin(\theta/2)}.$$


问:
Second moment and distribution of $Z=IX+(1-I)Y$ with $I\sim \text{Bernoulli}(p)$
Let $Z=IX+(1-I)Y$ with $I\sim \text{Bernoulli}(p)$ and $I,X,Y$ independent.

I want to determine the expected value of $Z^2$:

Using independence of $X,Y,I$ and linearity I arrive at:

$\begin{equation}E[Z^2]=\underbrace{E[I^2]}_{=p}E[X^2]+2E[X]E[Y]\underbrace{E[I(1-I)]}_{=p-p=0}+\underbrace{E[(1-I)^2]}_{=1-2p+p=1-p}E[Y^2]\end{equation}$

Is that correct?

Now assume $X,Y$ are i.i.d $\mathcal N(0,1)$.
How can I determine the distribution of $Z$?

I thought about using the mgf:

$m_Z(t)=E[e^{tIX}e^{tY}e^{-IY}]$ but I think this doesn't work out because $e^{tY}$ and $e^{-IY}$ aren't independent. So I need help with this task. Thanks!

答:
If $I \sim \operatorname{Bernoulli}(p)$, then $I^2 = I$ identically, and we immediately have $$Z^2 = I^2 X^2 + 2I(1-I)XY + (1-I)^2 Y^2 = IX^2 + (1-I)Y^2,$$ hence $$\operatorname{E}[Z^2] = p \operatorname{E}[X^2] + (1-p) \operatorname{E}[Y^2].$$

If $X, Y$ are IID standard normal, then conditioned on $I$, $$Z \mid I = IX + (1-I)Y \mid I \sim \operatorname{Normal}(\mu = 0, \sigma^2 = I^2 + (1-I)^2).$$  But since we already stated above that $I^2 = I$, this illustrates that the conditional distribution of $Z \mid I$ is standard normal irrespective of $I$.  So the unconditional distribution of $Z$ is also standard normal.

Another way to conceptualize this is to note that we may equivalently write $$Z \mid I = \begin{cases} X, & I = 1 \\ Y, & I = 0 \end{cases}$$ and since $X$ and $Y$ are IID standard normal, it doesn't matter what value $I$ takes on.  MGF calculations are unnecessary.

What $Z$ basically represents is "flip a coin that has probability $p$ of being heads.  If it's heads, then pick standard normal $X$ as the value of $Z$.  If it's tails, then pick standard normal $Y$ as the value of $Z$."  Since you're always picking from a standard normal random variable, the coin flip doesn't matter.


答:
Your first answer is correct but this method is faster:

The rv Z is the following

$$Z =
\begin{cases}
X,  & p \\
Y, & (1-p)
\end{cases}$$

Thus the second moment is simply

$$\mathbb{E}[Z^2]=p\mathbb{E}[X^2]+(1-p)\mathbb{E}[Y^2]$$

For the same reason, Z density is a mixture of two iid Standard gaussian

$$f_Z(z)=p\cdot \phi(z)+(1-p)\phi(z)=\phi(z)$$

....in other words

$$Z\sim N(0;1)$$

Title: Second moment and distribution of $Z=IX+(1-I)Y$ with $I\sim \text{Bernoulli}(p)$

Body:
Let $Z=IX+(1-I)Y$ with $I\sim \text{Bernoulli}(p)$ and $I,X,Y$ independent.

I want to determine the expected value of $Z^2$:

Using independence of $X,Y,I$ and linearity I arrive at:

$\begin{equation}E[Z^2]=\underbrace{E[I^2]}_{=p}E[X^2]+2E[X]E[Y]\underbrace{E[I(1-I)]}_{=p-p=0}+\underbrace{E[(1-I)^2]}_{=1-2p+p=1-p}E[Y^2]\end{equation}$

Is that correct?

Now assume $X,Y$ are i.i.d $\mathcal N(0,1)$.
How can I determine the distribution of $Z$?

I thought about using the mgf:

$m_Z(t)=E[e^{tIX}e^{tY}e^{-IY}]$ but I think this doesn't work out because $e^{tY}$ and $e^{-IY}$ aren't independent. So I need help with this task. Thanks!


问:
Domain whose boundry has non zero volume.
Can There be a domain in $\mathbb{R^n}$, for any $n$ such that some domain has non zero boundry volume? I.E. volume of boundry is non zero?

Motivation:

In some theorems, it is specified that volume of boundary is non zero. But I cannot think of domains where volume of boundry is non zero.

EDIT

If domain by definition is expected to be open subset of $\mathbb{R^n}$, then I would be looking for such open subsets.

Thank You.

答:
Let $ν \colon ℕ → [0,1]∩ℚ$ be a enumeration of the rationals there.
Then the set $M := \bigcup_{k∈ℕ} B_{r(k)}\big(ν(k)\big)$ with $r(k) = \tfrac{1}{3}·\big(\tfrac{1}{2}\big)^k$ has measure less than $\sum_{k=1}^∞ 2r(k) = \tfrac{1}{3}·1 < 1$ (as $B_{r(k)}$ has measure $2r(k)$), so it’s not all $[0,1]$.
But $M$ lies dense in $[0,1]$ since it contains all the rationals.
Therefore, the boundary of $M$ has to be $[0,1]\setminus{M}$, having nonzero measure, viz measure $\tfrac{2}{3}$.
Also $M$ is open because it’s the union of open balls.</p>


问:
Finding system of equations from critical points
We've gotten a function $f(x,y)=3x^3+3x^2y-y^3-15x$. For this function, it is told that the critical points of $f$ is $A = (1,1), B = (-1,-1),C = (\sqrt5,-\sqrt5), D = (-\sqrt5, \sqrt 5)$. These are the solutions to a system of two equations, and I can't seem to find any information on how to find these two equations? Note that it is not required to solve the equations, just find them.

Thanks in advance.

答:
The two equations are

$f_x(x,y)=0$ and $f_y(x,y)=0.$


问:
Why do we need to invoke the discrete-topology in the following definition of right-continuity?
In a book I read the following sentence:

"Let $V$ be a countable set, equipped with the discrete topology. A function $f : [0, \infty) \rightarrow V$ is right-continuous if $\lim_{s \downarrow t} f(s) = f(t)$.
Since $V$ is equipped with the discrete topology, we have that $f$ is right-continuous if and only if for every $t \geq 0$ there exists $\epsilon >0$ such that $f(s) = f(t)$ for every $s \in [t, t+\epsilon]$."

My question is: why does the if and only if condition stated in the second sentence follow from the fact that we have the discrete topology? What is then the definition of $\lim_{s \downarrow t} f(s) = f(t)$ in the discrete topology?

答: The singleton set $\{f(t)\}$ is an open set in the discrete topology. Hence the condition $f(s) \to f(t)$ as $s$ decreases to $t$ is equvalent to the condition $f(s)=f(t)$ whenever $t <s< t+\delta$ for some $\delta >0$.


问:
Definition of sigma-algebra of a continuous time stochastic process in a countable set
Let $X_{t} : t \geq 0$ be a continuous time stochastic process taking values in a countable set $I$, namely $X_t$ is a function defined in a measure space, for any $t \geq 0$.
How is the sigma-algebra of the stochastic process,
$$
\sigma \Big (  X_{t} : t \geq 0  \big )
$$
defined?

答:
It's the $\sigma -$algebra generated by cylinder.

Denote $C_{t_1,...,t_n}(B_1,...,B_n)=\{X_{t_1}\in B_1,...,X_{t_n}\in B_n\}$. Then

$$\sigma (X_t\mid t\geq 0)=\sigma \{C_{t_1,...,t_n}(B_1,...,B_n)\mid n\in\mathbb N, B_i\in \mathcal B(\mathbb R), 0\leq t_1<...<t_n\}.$$


问:
$\Bbb{R}$ with a given subbasis
Let us consider $\Bbb{R}$ with topology $\tau$ generated by subbasis consisting of $\Bbb{Q}$ and all open intervals $(a,b)$.

Let $U$ be an open set in $(\Bbb{R},\tau)$ such that $U \subset \Bbb{Q}$

Then how can I show that there exists an open set $V$ in $(\Bbb{R},usual)$ such that $U = V \cap \Bbb{Q}$?

If $U$ is empty or $\Bbb{Q}$, then we are done.

What about the rest of the cases?

答:
For every point in $U$, find its open basic neighborhood in $\mathbb R^\dagger$ (that's what I will call your topology for the moment) which is also contained in $U$ (and hence in $\mathbb Q$).

This neighborhood is a finite intersection of some subbasis sets of $\mathbb R^\dagger$.

One of these sets is $\mathbb Q$, otherwise the said neighborhood would be an intersection of subbasis sets of usual $\mathbb R$, and hence contain some open interval of $\mathbb R$, and hence also contain some irrational point(s), which in fact it doesn't.

The intersection of the rest of the mentioned subbasis sets is an open set in usual $\mathbb R$...

...see where this is going?

Title: $\Bbb{R}$ with a given subbasis


问:
Specific family of functions?
Let $\Phi$ be the class of all fonctions  $\phi$ (let's say from $\mathbb{R}^+$ to $\mathbb{R}^+$) such that:

If there exist an increasing sequences (not necessarily strictly)  $\{t_n\}\subset \mathbb{R}^+$ that converges, then the sequence $\{\phi(t_n)\}$ converges also.

Is there any specific name for this class of functions?

It will be interesting to see some interesting examples/counterexamples.

See similar discussion here and here.

答:
$\phi \in \Phi$ if and only if

$f$ has left limits at all points,  i.e. $\lim_{s<t, s \to t} f(s) $ exists and

$\lim_{t\to \infty} f(t)$ exists.

[The limit in 1) need not be $f(t)$ so $f$ need not be left continuous].

All monotone bounded functions have these properties.

Title: Specific family of functions?


问:
What is the dimension of a vector space without a basis?
It's my understanding that without the axiom of choice, the vector space of $\mathbb{R}$ over $\mathbb{Q}$ lacks a set of basis vectors. Now it seems straight forward that with the axiom of choice that the following holds.
$$\text{dim }(\mathbb{R}) = \aleph_1$$
However, what happens if we assume the axiom of choice is false? Does the dimension simply become undefined? As we can't define any basis for the space, it seems like the question of dimension becomes meaningless; you can't have the cardinality of something that doesn't exist. Alternatively, do we have an alternate way of defining dimension for these cases?

答:
There is no good notion of dimension, in general, when the axiom of choice is omitted. Even if there is a basis, there may be different bases of different cardinality, so there need not be a "smallest" size.

Sometimes it's convenient to say that a space is "infinite dimensional" if it is not finite dimensional, but not necessarily be concrete about whether the dimension is this or that. And since a vector space without a basis is not finite dimensional, that means that any vector space without a basis is infinite dimensional.

But since the use of the notion of dimension in this sort of situation is so sparse, there's no commonly accepted convention, and any author would always caution on the side of simply setting up their own context.


问:
Elementary Inequality regarding Sum of Squares
\begin{align}
&\mbox{If}\quad
A = \sum_{i = 1}^{m}a_{i}\,x_{i}^{2}
\quad\mbox{and}\quad
B = \sum_{j = 1}^{n}b_{j}\,y_{j}^{2}
\\[2mm] &\
\mbox{where}\quad
x_{i}, y_{j}, a_{i}, b_{j} \geq 1, \quad x_{i} \neq y_{j}
\end{align}
Then, find necessary and sufficient conditions such that $A - B > 1$ when $m = n$ is odd, $\ A > B$.

My Attempt:

Because there are total $m = n\ $  subtractions $a_{i}\,x_{i}^{2} - b_{j}\,y_{j}^{2}$ and each
$a_{i}\,x_{i}^{2} - b_{j}y_{j}^{2} > 2$ ( minimum possible case is
$1 \cdot 2^{2} -  1\cdot 1^{1} = 3$ ),  we need to show  $\displaystyle\sum_{i = 1}^{n - 1}a_{i}\,x_{i}^{2} - b_{i}\,y_{i}^{2}$ has to be $\geq 0$.

so the sum of  $a_{i}\,x_{i}^{2} - b_{j}\,y_{j}^{2}$  is greater than $0$, thus $A - B > 1$.

Is there anything in the literature imply the proposition? In that case plz comment.

The post consider the non-trivial case, if any specification is required, plz let me know.

ADDED NOTE:

All $x_{i}, y_{j}, a_{i}, b_{j}$ are integers.

Is linear form of logarithm  relevant to this problem $?$.

答:
\begin{align}
&\mbox{If}\quad
A = \sum_{i = 1}^{n}a_{i}\,x_{i}^{2}
\quad\mbox{and}\quad
B = \sum_{j = 1}^{n}b_{j}\,y_{j}^{2}
\\[2mm]
\end{align}

with integer
$x_{i}, y_{j}, a_{i}, b_{j} \geq 1 , \quad x_{i} \neq y_{j} $, and $A > B$, we are to find conditions for which  $A-B > 1$.

As OP already posted, the  given condition $A > B$, together with the integer condition of the variables, implies that the difference $A - B$ cannot become arbitrarily small. So we start to investigate the smallest such difference under the given conditions.

This can be used to construct first of all an example where the assertion does not hold true.  Let $A = 1\cdot 3^2 = 9$ and $B = 2\cdot 2^2  = 8$, then $A>B$ and $A-B=1$ so the required $A-B>1$ fails. There are many more such examples possible for other integer numbers
$x_{i}, y_{j}, a_{i}, b_{j} \geq 1$. So, the only statement we can make in general is that, due to the integer condition and $A>B$, the minimum possible difference is  $A-B=1$.

From this it follows that it's difficult resp. unlikely to give general explicit conditions for making $A-B> 1$. Such conditions can  be  implicitely stated:   Choose integer numbers
$x_{i}, y_{j}, a_{i}, b_{j} \geq 1$ such that  $A-B\ge 2$. (Then $A-B > 1$ is obviously obtained.)


问:
Using Ratio test to determine convergence
In this series I need to use the Ratio test to determine whether the series diverges or converges and I'm not sure about my calculations
$$
\sum_{n=1}^\infty \left( \frac{\pi}{n}\right)^n * n!
$$
plug in n+1 * 1/n

$$
\left( \frac{\pi}{n+1}\right)^{n+1}   (n+1)!  * \left( \frac{n}{\pi}\right)^{n} \left( \frac{1}{n!}\right)
$$

$$\left( \frac{\pi}{n+1}\right)^{n+1}   (n+1)* \left( \frac{n}{\pi}\right)^{n}$$
$$\left( \frac{\pi}{n+1}\right)^{n}   \pi* \left( \frac{n}{\pi}\right)^{n}$$
$$\left( \frac{n}{n+1}\right)^{n}   \pi$$
$$\left( \frac{1}{2}\right)^{n}   \pi = 0$$
which would be 0 so it is convergent ??

答:
$$\left(\frac{n}{n+1} \right)^n =\frac{1}{\left(\frac{n+1}{n}\right)^n} =\frac{1}{\left(1+\frac{1}{n}\right)^n}\to \frac{1}{e}$$

So the limit of the ratio test should be $\dfrac{\pi}{e}>1$, so your series is divergent.


问:
If $X$ is a projective variety, then $O_X(n)$ has no global sections
I have a question about Example 7.1.2 in the link https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/alggeom-2002.pdf. In (ii), it is written that there are no global sections of $O_X(n)$ for $n<0$. Here $X$ is a projective variety over an algebraically closed field $k$. But why is this true? By definition, a section of $O_X(n)$ is of the form $f/g$ with $f,g$ homogeneous and $\deg(f)=\deg(g)+n$.

Edit: I see this when $X=\Bbb P^n$, because every nonconstant homogeneous polynomial should have a zero. But I can't see this in the general case.

答:
Let $i : X \to \Bbb P^r$ an embedding, then we can define $\mathcal O_X(n) = i^* \mathcal O(n)$. As you said, section of $\mathcal O_X(n)$ are quotients of homogeneous polynomials of appropriate degree, restricted to $X$.

In particular, any $f/g$ (as a function on $\Bbb P^r$) has poles on an hypersurface $Y \subset \Bbb P^r$.

If $X$ has positive dimension, then $Y \cap X \neq \emptyset$ and hence $f/g$ also has poles on $X$. It follows that $\mathcal O_X(n)$ has no global sections when $n < 0$, assuming $\dim X > 0$.


问:
When should we use conditional probability in a coin toss question given x number of tails have already been tossed?
I'm having a little trouble understanding when we would use conditional probability in the question below.

I was thinking since we are already given that there are at least 6 tails, we wouldn't need to consider that probability, and only calculate the probability of tossing exactly 1 tail (and hence 1 head) in 6th and 7th tosses, which would give 1/2 since the possible outcomes are TH, HT, TT, HH, and 2/4 outcomes have exactly one tail. Could someone please verify if that approach is right or if we'd need to use conditional probability in this case?

Original question:

Assume that the outcome of either heads or tails is equally likely in
coin tosses, and each coin toss event occurs independently. You toss
the coin exactly 8 times. Given that at least 6 of those tosses
resulted in tails, what is the probability that exactly 7 tosses were
tails?

答:
tossing the coin 8 times, the probability of any single realization is the same for any realization. This because $\mathbb{P}[H]=\mathbb{P}[T]=\frac{1}{2}$

Thus it can be wasted and we are interested only in the combinations.

Thus the answer is

$$\frac{\binom{8}{7}}{\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}=\frac{\binom{8}{1}}{\binom{8}{2}+\binom{8}{1}+\binom{8}{0}}=\frac{8}{28+8+1}=\frac{8}{37}$$


问:
Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
$\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$

Can we apply  Stolz-Cesaro?

$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}} {{\ln(n+1)-\ln(n)}}$ =
$\lim\limits_{n \to \infty}\frac{\sqrt n - \sqrt {n+1}}{\sqrt n \sqrt{n+1}\ln(1+\frac{1}{n})}$ = $\lim\limits_{n \to \infty}\frac{1-\sqrt{1+\frac{1}{n}}}{\sqrt{n+1}\ln(1+\frac{1}{n})}$

What can I do from here?

答:
The limit is $+\infty$.

You made an error in the first step. Applying Stolz-Cesaro, you should rather consider:

\begin{eqnarray*} \frac{\frac 1{\sqrt{n+1}}}{\ln{(n+1)}-\ln n}
& = & \frac 1{\sqrt{(n+1)}}\cdot \frac 1{\ln \left(1+\frac 1 n\right)}\\
& = & \frac n{\sqrt{(n+1)}}\cdot \underbrace{\frac{\frac 1n}{\ln \left(1+\frac 1 n\right)}}_{\stackrel{n \to \infty}{\longrightarrow}1} \\
& \stackrel{n \to \infty}{\longrightarrow} & +\infty
\end{eqnarray*}


问:
if $a,b,c$ are positive prove that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$

if $a,b,c$ are positive prove that $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$

I have tried to use the  p-q-r method.

let $p=a+b+c,q=ab+bc+ca,r=abc$

we have to prove $$p^2-3q+r+2-p\ge 0$$

as $p^2\ge 3q$ we have to prove $$r+2-p\ge 0$$.
but i don't think this is true. Also i know the following inequalities but i can't figure out how to use them.

$pq\ge 9r$

$2p^3+9r\ge 7pq$

$q^3\ge 27r^2$ etc. etc.

Please note that i am looking for a solution only using pqr.I would be grateful if anyone can explain their logic behind the proof as it is the first time i am using 'pqr' method.

答:
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that:
$$a^2+b^2+c^2+abc+2\geq a+b+c+ab+ac+bc$$ or $f(v^2)\geq0,$ where
$$f(v^2)=9u^2-6v^2+w^3+2-(3u+3v^2)=9u^2-9v^2+w^3+2-3u.$$
But $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $v^2$.

Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$-x^3+3ux^2+w^3=3v^2x.$$
Now, let $u$ and $w^3$ are constants and $v^2$ is changed. In this situation $a$, $b$ and $c$ are changed.

Since the last equation has three roots $a$, $b$ and $c$, we see that a graph of $g(x)=-x^3+3ux^2+w^3$ and the line $y=3v^2x$ have three common points and $v^2$ gets a maximal value for equality case of two numbers between $a$, $b$ and $c$.

Actually, if $a\leq b\leq c$ we can see that it happens for $b=c$.

Indeed, easy to draw a graph of $g$: $(0,w^3)$ it's a minimum point of $g$, $x_{max}=2u$ and the line $y=3v^2x$ goes across the origin.

If $v^2$ increases, so the line $y=3v^2x$ is  rotated and we can see, that $v^2$ may be maximal, if we want that our equation would have three positive roots, when $b=c$.

Id est, it's enough to prove our inequality for equality case of two variables.

Since our inequality is symmetric, it's enough to assume $b=c$ and we obtain a quadratic inequality of $a$ and the rest is smooth, which you can see here: show that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$


问:
Critical point by Taylor polynomial?
Use the second-degree Taylor polynomial to determine if the point $(1,1)$ is a critical point.

$f(x,y)=xy-x-y-2\cos\left(x-y\right)$ around the point $(x,y)=(1,1)$

$$f_{x}=y+2\sin \left(x-y\right)-1\\
f_{xx}=2\cos \left(x-y\right)\\
f_{xy}=-2\cos \left(x-y\right)+1\\
f_{y}=x-2\sin \left(x-y\right)-1\\
f_{yy}=2\cos(x-y)\\
\ \\
f(1,1)=-3\\
f_{x}(1,1)=0\\
f_{xx}(1,1)=2\\
f_{xy}(1,1)=-1\\
f_{y}(1,1)=0\\
f_{yy}(1,1)=2\\
\ \\
-3+0(x-1)+0(y-1)+\frac{2}{2!}(x-1)^2+\frac{-1}{1!1!}(x-1)(y-1)+\frac{2}{2!}(y-1)^2=x^2-xy-x+y^2-y-2
$$

If a insert $(x,y)=(1,1)$ in $x^2-xy-x+y^2-y-2$ i get $=-3$.

How do i use the taylor polynomial to determine if $(1,1)$ is a critical point?

答:
I don't know how to use the second degree Taylor polynomial in a sensible way to do this, but if you must, here it goes:

The 2nd degree Taylor polynomial of a scalar function expanded at $a$ is
$$T^2 f(x;a)=f(a)+\nabla f(a)\cdot (x-a)+\frac{1}{2}(x-a)^T\mathrm H f(a)(x-a),$$
where $\nabla f$ is the gradient and $\mathrm Hf$ is the Hessian. Now $f$ has a critical point at $a$ if $\nabla f(a)=0$, in other words, if the second degree Taylor polynomial reduces to
$$f(a)+(x-a)^T\mathrm Hf(a)(x-a).$$
But that's just a detour around simply calculating $\nabla f$, which is the actually important condition, which you have to do anyway to find the Taylor polynomial.

What you can do with the 2nd degree Taylor polynomial, though, is to determine the type of critical point: if the Hessian is positive [negative] definite, then we have a minimum [maximum], and if it is indefinite, then we have a saddle point. If it is semi-definite, then the second degree Taylor polynomial can't be used to find the type of critical point.

Title: Critical point by Taylor polynomial?


问:
Prove A is symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$
Prove A is a symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$. (for all $x,y \in \mathbb{R}^n$)

Going from the assumption that it is symmetric to the two conditions is fairly straightforward.

However, going the other way, I am stuck at proving $A^T = A$ using the second and condition, being stuck at $X^T (A-A^T)y=0$.

Note T is for transpose!

答:
Since you already accepted an answer I can show a more abstract method.

Consider $B_x=x^TA-(Ax)^T$, which is a row and so defines a linear map

$$\mathbb{R}^n\to\mathbb{R}$$

that, under your hypothesis, is the zero map. This means that $B_x=0$ for all $x$, because the rank of $B_x$ is zero. Now

$$B_x=x^TA-(Ax)^T=x^TA-x^TA^T=x^T(A-A^T)$$

so

$$B_x^T=(A-A^T)^Tx=(A^T-A)x$$

is the zero matrix for all $x$. Hence the linear map $\mathbb{R}^n\to\mathbb{R}^n$ associated to $C=A^T-A$ (the usual $x\mapsto Cx$) is the zero map and therefore $C$ has rank zero, so $C=0$.


答:
The inequality $r+2-p \geqslant 0$ does not true, example $a = 2, b = 1, c = \frac 12.$

We need to prove
$$p^2-p-3q+r+2 \geqslant 0,$$
equivalent to
$$(p^2+2r+1-4q)+(p^2-2p-2q+3) \geqslant 0.$$
Because
$$p^2-2p-2q+3 = (a-1)^2+(b-1)^2+(c-1)^2 \geqslant 0.$$
It's remain to prove that
$$p^2+2r+1 \geqslant 4q.$$
Using the AM-GM inequality, we have
$$2r+1 \geqslant 3\sqrt[3]{r^2} \geqslant \frac{9r}{p}.$$
Therefore, we will show that
$$p^2+ \frac{9r}{p}\geqslant 4q.$$
Which is Schur inequality.

Note. See here (page 4, Vietnamese).

Title: if $a,b,c$ are positive prove that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$

Body:
if $a,b,c$ are positive prove that $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$

I have tried to use the  p-q-r method.

let $p=a+b+c,q=ab+bc+ca,r=abc$

we have to prove $$p^2-3q+r+2-p\ge 0$$

as $p^2\ge 3q$ we have to prove $$r+2-p\ge 0$$.
but i don't think this is true. Also i know the following inequalities but i can't figure out how to use them.

$pq\ge 9r$

$2p^3+9r\ge 7pq$

$q^3\ge 27r^2$ etc. etc.

Please note that i am looking for a solution only using pqr.I would be grateful if anyone can explain their logic behind the proof as it is the first time i am using 'pqr' method.


问:
CNF for modilisation of a word
Imagine that we are interested in problem of words. A word is a sequence of letters from a $\Sigma$ alphabet. For encoding a word in SAT we are using variable like $x_{i,a}$ which means that in position $i$ we find the letter $a$.

So now my problem is this: Find a CNF formula which expresses that the variables $x_{i,a}$, with $i$ ranging from 1 to $n$ (the size of the word), encode a word of size $n$. That is for any position $i$, there is exactly one letter of $a \in \Sigma$ that appears in that position.

So far I came up with this but I am not sure if it is true:

$$\bigwedge_{i=1}^n \; \bigwedge_{a \in \Sigma \\ a' \in \Sigma} \neg x_{i,a} \lor \neg x_{i,a'}$$
So this means that we can't have in the same time two letters in position $i$.

答:
Long comment

Quite...

Let start approaching the problem from a different point of view: to say that position $i$ is filled with symbol $a$ can be written in predicate logic as follows:

$\forall s \in \Sigma (\text {In}(i,a) \to \lnot \text {In}(i, s))$,

which is equivalent to: $\forall s \in \Sigma (\lnot \text {In}(i,a) \lor \lnot \text {In}(i, s))$.

Assuming a finite alphabet of symbols $\Sigma$, we may replace the universal quantifier with a finite conjunction.

But we have to take care of the case: $s_k=a$, for some $k$: when "scanning" the symbols of $\Sigma$, at a certain point we will find $a$ itself.

In that case we have that the conjunct: $(\lnot \text {In}(i,a) \lor \lnot \text {In}(i, s_k))$ is false.

Title: CNF for modilisation of a word


答:
The case you are modelling is that you have tossed 6 tails in a row and now you will toss two more times, and yes then the probability of 7 tails is 1/2.

But the case mentioned in the problem and rightly modelled by @tommik is
that someone has tossed 8 coins already (and you cannot see the outcome) and tells you that at least 6 are tails, and then asks you what is the probability that there are exactly 7 tails. Which is modelled as follows:

$$P(\#T=7 | \#T>6) = \frac{P((\#T =7) \& (\#T > 6) )}{P(\#T>6)} = \frac{P(\#T = 7)}{P(\#T>6)}$$

Yeah in retrospect I see that there is slight problem in the way the question is framed, because if you are tossing the coin then it seems to mean that you have seen 6 tails in a row. But I am quite sure what it wants to say is that given such 8 tosses take place and you only have the information that at least 6 are tails then what is the probability of 7 tails.

Title: When should we use conditional probability in a coin toss question given x number of tails have already been tossed?

Body:
I'm having a little trouble understanding when we would use conditional probability in the question below.

I was thinking since we are already given that there are at least 6 tails, we wouldn't need to consider that probability, and only calculate the probability of tossing exactly 1 tail (and hence 1 head) in 6th and 7th tosses, which would give 1/2 since the possible outcomes are TH, HT, TT, HH, and 2/4 outcomes have exactly one tail. Could someone please verify if that approach is right or if we'd need to use conditional probability in this case?

Original question:

Assume that the outcome of either heads or tails is equally likely in
coin tosses, and each coin toss event occurs independently. You toss
the coin exactly 8 times. Given that at least 6 of those tosses
resulted in tails, what is the probability that exactly 7 tosses were
tails?


问:
Submodules of $k^n$ as $T_n(k)$-module
Let $A = T_n(k)$ the set of upper triangular matrices be an algebra and let $k^n$  be a $A$-module given by left multiplication, I need find all submodules of $k^n$.

I know that if $N$ is an A-submodule of $k^n$ then $\frac{k^n}{N}$ has structure of $A$-module and the projection $\pi: k^n \to \frac{k^n}{N} $ should be $A$-linear.

My problem is that it is based on that idea, I don't know how to do to see specifically which are the submodules.
Could you give me some help? Or if you can give me another idea of how to find the submodules I would also appreciate it.

Thanks

答:
Let $e_1, \dots, e_n$ be the canonical basis of $k^n$. I assume that $T_n(k)$ is given by the set of all triangular matrices (not strictly triangular).

Let $W$ be a sub-$T_n(k)$-module of $k^n$. Let $I$ the smallest index such that $a_ie_i + \dots + a_ne_n \in W$ ($a_i \neq 0$). Then, acting by a diagonal matrix we get $e_i + w' \in W$. Using $E_{i,j}e_i = e_j$ if $j > i$ we get $w' \in W$ and hence $e_i \in W$. From this, it is easy to conclude that $W = \langle e_i, \dots, e_n \rangle $. Hence each $T_n(k)$-module is on this form.

Title: Submodules of $k^n$ as $T_n(k)$-module


问:
Finding expected value through a different RV

I am currently studying probability and statistics at second year university level and have came across this question. Previously I was not sure on how to approach this. I initially thought I needed to work out the ImY but now know that this is the right approach. Having now looked over the replies and course notes again, I realise that using the concept 'Expected value of g(x)' was the obvious approach and proceeding on with integrating the multiple of the two function with the boundaries 0 and +infinity. I apologise for not using the correct maths coding as I'm not quite sure how to use it yet.

答:
Your random variable is a negative exponential

$$X\sim Exp(4)$$

the requested expectation is easy to calculate with the definition

$$\mathbb{E}[g(X)]=\int_{-\infty}^{+\infty}g(x)f(x)dx$$

That in your case is

$$\mathbb{E}[e^{-5X}]=\int_0^{+\infty}e^{-5x}4e^{-4x}dx$$

$$\mathbb{E}[e^{-5X}]=\frac{4}{9}\underbrace{\int_0^{+\infty}9e^{-9x}dx}_{=1}=\frac{4}{9}$$


问:
$\mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}$ are not equivalent as abelian group models.
I am facing a problem about showing the proposition above, where the language $\mathscr{L}$ is the language of abelian groups. I am thinking that $\mathbb{Z}$ has a generator $1$ and therefore $\mathbb{Z}\vDash \exists x \forall y (\exists n, y=nx)$, while the sentence is false in model $\mathbb{Z}^2$. However, $n$ is not in the language and I am confused if this idea make sense.

Thank you all!

答:
The sentence involving a generator is not first-order, as you suspect. In fact, there is no first-order way to capture the property of being cyclic. More precisely, cyclic groups do not form an elementary class.

So, here is a hint: Consider remainders mod $n$ in your groups.


问:
Limit of the modulus of a complex number
Given $\alpha\left(\lambda\right)=\left(1-\lambda\right)+\lambda\alpha$ where $\lambda\in\mathbb{R}$ and $\alpha$ is a complex number, consider the function $f_{\lambda}=\|\alpha\left(\lambda\right)\|^{\frac{1}{\lambda}}$. Obviously $\|\alpha\left(\lambda\right)\|$ is the modulus of $\alpha\left(\lambda\right)$. The limit of $f_{\lambda}$ for $\lambda$ going to zero is given by $e^{Re\left(\alpha\right)-1}$ where $Re\left(\alpha\right)$ is the real part of $\alpha$.
How do you derive this result?

答:
If you write $\alpha=a+bi$ and expand the module, you'll get that
$$f(\lambda)=\left( 1 + 2 (-1 + a) \lambda + (1 - 2 a + a^2 + b^2) \lambda^2\right)^{1/2\lambda} = \left[\left( 1 + 2 (-1 + a) \lambda + (1 - 2 a + a^2 + b^2) \lambda^2\right)^{\frac{1}{ 2 (-1 + a) \lambda + (1 - 2 a + a^2 + b^2) \lambda^2}} \right]^{\frac{ 2 (-1 + a) \lambda + (1 - 2 a + a^2 + b^2) \lambda^2}{2\lambda}}
$$

The part in Brackets goes to $e$ as $\lambda$ goes to $0$.

The exponent $\dfrac{2 (-1 + a) \lambda + (1 - 2 a + a^2 + b^2) \lambda^2}{2\lambda}$ clearly goes to $-1+a$.

So you're done.


问:
Prove Lower bound of a set ${x^3 : x<10}$
I just started studying Calculus and I'm confused on how to prove Lower bounds.
I understood how to prove upper bounds but I just don't know how to proceed.

let $B={({x^3 : x<10 , x∈R})}$

The upper bound is obviously $m=1000$ and all the numbers that are bigger than that.

Lower bound - I assumed that exists lower bound $y∈R$ such that every $b∈B$  : $b≥
y$.

and Then I'm stuck , I don't know how to proceed with the proof to show that no such $y$ exists.

答:
Let $b$ be the lower bound (certainly a negative number), i.e. $\forall x<10:x^3\ge b$.

But if we take $x=2\sqrt[3]b$, we do have $x<10$, but $x^3=8b<b$, a contradiction.


问:
Basic book about mathematical proofs
I'm looking for an understandable book about mathematical proofs. </p>

<p>As an engineer/computer scientist major I'm used to do higher math on a daily base, but whenever I'm asked in an exercise to "show" something I get the "where should I start? and what exactly do they want me to do?" syndrome. </p>

答: Read Pólya's How to Solve It. See also these slides.</p>

Title: Basic book about mathematical proofs


问:
Definition of sigma-algebra of a continuous time stochastic process in a countable set
Let $X_{t} : t \geq 0$ be a continuous time stochastic process taking values in a countable set $I$, namely $X_t$ is a function defined in a measure space, for any $t \geq 0$.
How is the sigma-algebra of the stochastic process,
$$
\sigma \Big (  X_{t} : t \geq 0  \big )
$$
defined?

答:
Let $(X_t)_{t\geq 0}$ be a (real-valued) stochastic process on some measurable space $(\Omega,\mathcal{F})$, i.e., for all $t\geq 0$ $X_t:\Omega\to\mathbb{R}$ is a random variable.

The sigma algebra generated by the stochastic process $(X_t)_{t\geq 0}$ is the smallest sigma algebra such that $X_t$ is measurable for all $t\geq 0$, i.e.,
$$\sigma\left(X_{t}\colon t\geq 0\right)=\left\{A\in\mathcal{F}\mid \exists t\geq 0 \;\exists B ∈ \mathcal{B}(\mathbb{R}):\; A=X^{-1}_t(B)\right\},$$
where $\mathcal{B}(\mathbb{R})$ denote the Borel sets over the real line.


问:
Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$
Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:

I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?

答:
$$f(x)=\dfrac{1+\sin x+\cos x}{\sin x\cos x}$$

Clearly $f(x)$ is undefined for $2x=n\pi$ where $n$ is any integer

$$\dfrac{f(x)}2=\dfrac{1+\sin x+\cos x}{(\sin x+\cos x)^2-1}=\dfrac1{\sin x+\cos x-1} $$

Now use $-\sqrt2\le\sin x+\cos x\le\sqrt2$ to find

$$\implies-\sqrt2-1\le\sin x+\cos x-1\le\sqrt2-1$$

If $\sin x+\cos x-1\ge0, f(x)\ge2+2\sqrt2$

and if $\sin x+\cos x-1\le0,f(x)\le2-2\sqrt2$


问:
Every cofinal real closed subfield is dense?
Consider a real closed subfield $F_0$ of a real closed field $F_1$ that is cofinal.
Here, "$F_0$ is cofinal in $F_1$" means "For each $x \in F_1$, there exists $y \in F_1$ such that $x < y$". Then, is $F_0$ order dense in $F_1$?

答: It need not be dense. For instance, consider the real-closed field $F_0$ of Puiseux series with real algebraic coefficients, and the real-closed field $F_1$ of Puiseux series with real coefficients. The natural inclusion $F_0\subset F_1$ is cofinal, but not dense since for instance we have $(\pi - \varepsilon,\pi+\varepsilon) \cap F_0 = \varnothing$ for every positive infinitesimal $\varepsilon \in F_1$.


问:
Matrix representation of universal enveloping $U(sl(2,\mathbb{C}))$ generators $ e^k,f^k$
Wiki describes the Lie algebra of universal enveloping algebra $U(sl(2,\mathbb{C}))$ with generators $$e^k,f^k,h$$
but Wiki says $e^k,f^k$ are not the matrix power of $sl(2,\mathbb{C}) $   generators $$e ,f ,h.$$

Question; What are the matrix representation of $U(sl(2,\mathbb{C}))$   generators $ e^k,f^k$?

答:
This is a legitimate confusion and I think the downvote is not justified.

You should not think of $e^k$ as $e \cdot \dots \cdot e \in M_2(\Bbb C)$, but more as formal objects that acts like $e$ composed $k$ times on any representation of $sl_2$.  The matrix product $e \cdot e = 0$. It represents how $e^2 \in U(sl_2)$ acts on $\Bbb C^2$, but it doesn't mean it that $e^2$ will act by zero on other representation !

This is because for a representation $V$, $e^k$ will acts as $\rho(e)^k$ (matrix product) when $\rho : sl_2 \to End(V)$ is a representation.

For example, the matrix form of $e^2$ acting on $Sym^2(\Bbb C^2)$ is $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix}$.

More generally, to answer your question "what are the matrix form of $e^k$ ? " you need to pick a representation of $\rho : sl_2 \to End(V)$ and compute $\rho(e)^k$ . You can work out quite easily the general pattern, remembering that finite-dimensional irreducible representations of $\mathfrak {sl}_2$ are given by $Sym^n(\Bbb C^2)$.


问:
Deficiency index of symmetric operator is locally constant
Let $A: D_A \rightarrow H$ be a symmetric linear operator defined on $D_A \subseteq H$. Define the deficiency index of $A$ at $z \in \mathbb{C} \setminus \mathbb{R}$ to be $\dim( \ker (A^* - \bar{z}))$. Then I want to show that the deficiency index is constant on the upper half-plane and constant on the lower half-plane.

The assertion is given in Lax, section 33.2, Theorem 3 (i), but he does not prove it.

If $A$ is also closed, then I am able to show that if the deficiency index is $0$ then it is so on the whole half plane, which is a consequence of the fact that the resolvent set is open. However, I don't know about the general case.

答: The proof is in Reed/Simon “Methods of Modern Mathematicsl Physics”, Vol. 2, Section X:  Theorem X.1</p>


问:
Handling more than three indices/super indices, tensorial calculus
I need to code an equation such as the following one :

$$ \frac{\partial u^j}{\partial q^i} = \frac{\partial \mathrm A^j_{pl}}{\partial q^i}\dot q^p \dot q^l + \frac{\partial \mathrm B^{jl}}{\partial q^i} \frac{\partial \mathrm X}{\partial q^l} $$

Where $\mathbf u$ and $\mathbf q$ are $n\times 1$ size vectors, $\mathbf A$ and $\mathbf B$ are $n\times n$ size matrices and $\mathrm X$ is just a $1\times 1$ variable.

Now, I don't really feel at ease when it comes to handling more than three indices/superindices but I do understand that I have to obtain a vector of the same dimensions as the $\mathbf u$ vector in the equation...

So I guess I'll have to sum over the $i$ index, or the $j$ super index. Can anybody tell me which proposition is likely to be the right one?

First proposition

$$ \sum\limits_{i=1}^{n}\left(\frac{\partial u^j}{\partial q^i} = \frac{\partial \mathrm A^j_{pl}}{\partial q^i}\dot q^p \dot q^l + \frac{\partial \mathrm B^{jl}}{\partial q^i} \frac{\partial \mathrm X}{\partial q^l}\right) $$

Second proposition

$$ \sum\limits_{j=1}^{n}\left(\frac{\partial u^j}{\partial q^i} = \frac{\partial \mathrm A^j_{pl}}{\partial q^i}\dot q^p \dot q^l + \frac{\partial \mathrm B^{jl}}{\partial q^i} \frac{\partial \mathrm X}{\partial q^l}\right) $$

答:
Well, as it turns out, I was chasing the wrong thing all along. In fact, the equation

$$ \frac{\partial u^j}{\partial q^i} = \frac{\partial \mathrm A^j_{pl}}{\partial q^i}\dot q^p \dot q^l + \frac{\partial \mathrm B^{jl}}{\partial q^i} \frac{\partial \mathrm X}{\partial q^l} $$

is to be interpreted as a $n\times n$ sized matrix over the $i$ and $j$ indices...


问:
Solution verification: 2 simple problems involving Conditional Probability and Hypergeometric Distribution
Q1) An experiment is carried out to investigate how well 100 nano-particles disperse uniformly, when introduced into a container of 900 other particles of similar size and shape.
After the nano-particles are added at the top of the container, the researchers wait 10
hours before randomly selecting a sample of 20 particles without replacement from the
bottom of the container.

a) What is the probability that the sample will contain 10 nano-particles?

b) What assumption did you need to make to answer part (a)?

Q2 Player A who enters a golf tournament is not certain whether player B will enter.
Player A has probability 1=6 of winning the tournament if player B enters and probability
3=4 of winning if player B does not enter the tournament. The probability that player B
enters is 1=3. On Monday morning you read that player A won the tournament. What
is the probability that player B entered the tournament?

My Work:

Q1 (a): $P(\text{Getting 10 nano particles})= \frac{\binom{100}{10} \binom{900}{10}}{\binom{1000}{20}}$

Q1 (b) The assumption that we made is that all subsets of size 20 i.e $\binom{1000}{20} $ are equally likely to be chosen.

For Q2)   We have

$P(A_w|B)= \frac{P(A_w \cap B)}{P(B)}=\frac{1}{6}$

$P(A_w|B^c)= \frac{P(A_w \cap B^c)}{P(B^c)}= \frac{3}{4}$

$P(B)= \frac{1}{3}$

We need to find $ P(B | A_w)$.

$ P(B | A_w)= \frac{P(B\cap A_w)}{P(A_w)}$

$P(B \cap A_w) = \frac{1}{6} \text{ x } \frac{1}{3}= \frac{1}{18}$

$P(A_w)= P(A_w\cap B) + P(A_w \cap B^c) = \frac{1}{6}\text{ x } \frac{1}{3} + \frac{3}{4}\text{ x } \frac{2}{3} = \frac{5}{9}$

Then $P(B | A_w)= \frac{\frac{1}{18}}{\frac{5}{9}}= \frac{1}{10}$  $\text { (Final answer)}$

答:
(1) The probability of getting exactly 10 of the particles newly introduced is quite small: 4.659696e-06 $\approx$ 0.000004660.

```<code>dhyper(10, 100,900,  20)
[1] 4.659696e-06
choose(100,10)*choose(900,10)/choose(1000,20)
[1] 4.659696e-06
</code>```

It is unclear why one would expect to get 10 new particles out of 20, if the new particles are randomly dispersed. Recapturing about 2 in 20 (at the bottom) seems most likely. Here is a plot of the relevant hypergeometric distribution, for the number of new particles sampled.

```<code>x = 0:20; pdf=dhyper(x, 100,900, 20)
sum(pdf)
[1] 1
plot(x, pdf, type="h", lwd=2)
abline(h=0, col="green2")
</code>```

Getting 5 or more of the new particles out of 20 would
lead to rejection (5% level) of the null hypothesis that
particles are randomly mixed. Getting 10 or more would be extremely strong evidence against random mixing (P-value
5.121643e-06.)

```<code>qhyper(.95, 100, 900, 20)
[1] 4
phyper(4, 100, 900, 20)
[1] 0.9585121
1 - phyper(9, 100, 900, 20)
[1] 5.121643e-06
</code>```

Note: In R, dhyper, phyper, and qhyper designate
hypergeometric PDF, CDF, and inverse CDF (quantile function), respectively.

Adddendum per question in comment: In mathematical terms, your answer to 1(b) is OK. In practical terms, you are assuming that the new particles have become randomly mised wit the others.

If the P-value is smaller than 5% then you reject the null hypothesis at the 5% level.

When you use printed normal tables you are doing a CDF lookup or an inverse CDF lookup.  If you look for z-score 1.96 in the margins of the table and find
probability corresponding to 0.9750 in the body of the table, that is a CDF look-up. If you scan the body of the table for the nearest probability to 0.0750 and use that to find z-score 1.96 in the margins, that is an inverse CDF look-up. Om R. it would be:

```<code>pnorm(1.96)
[1] 0.9750021
qnorm(.975)
[1] 1.959964
</code>```

Body:
(1) The probability of getting exactly 10 of the particles newly introduced is quite small: 4.659696e-06 $\approx$ 0.000004660.

dhyper(10, 100,900,  20)
[1] 4.659696e-06
choose(100,10)*choose(900,10)/choose(1000,20)
[1] 4.659696e-06

It is unclear why one would expect to get 10 new particles out of 20, if the new particles are randomly dispersed. Recapturing about 2 in 20 (at the bottom) seems most likely. Here is a plot of the relevant hypergeometric distribution, for the number of new particles sampled.

x = 0:20; pdf=dhyper(x, 100,900, 20)
sum(pdf)
[1] 1
plot(x, pdf, type="h", lwd=2)
abline(h=0, col="green2")

Getting 5 or more of the new particles out of 20 would
lead to rejection (5% level) of the null hypothesis that
particles are randomly mixed. Getting 10 or more would be extremely strong evidence against random mixing (P-value
5.121643e-06.)

qhyper(.95, 100, 900, 20)
[1] 4
phyper(4, 100, 900, 20)
[1] 0.9585121
1 - phyper(9, 100, 900, 20)
[1] 5.121643e-06

Note: In R, dhyper, phyper, and qhyper designate
hypergeometric PDF, CDF, and inverse CDF (quantile function), respectively.

Adddendum per question in comment: In mathematical terms, your answer to 1(b) is OK. In practical terms, you are assuming that the new particles have become randomly mised wit the others.

If the P-value is smaller than 5% then you reject the null hypothesis at the 5% level.

When you use printed normal tables you are doing a CDF lookup or an inverse CDF lookup.  If you look for z-score 1.96 in the margins of the table and find
probability corresponding to 0.9750 in the body of the table, that is a CDF look-up. If you scan the body of the table for the nearest probability to 0.0750 and use that to find z-score 1.96 in the margins, that is an inverse CDF look-up. Om R. it would be:

pnorm(1.96)
[1] 0.9750021
qnorm(.975)
[1] 1.959964


问:
An inverse problem in PDE cours
Consider the family of characteristic projections given by
$x(\tau,s)=s\cos(\tau) , Y(\tau,s) = \epsilon s \sin(\tau)$

How do i use the characterist ODE's to show that the quadratic for $Q= x^2+\frac{y^2}{\epsilon^2}$ is conserved.

Currently I have eliminated $\tau$ to a single equation that determin the shape of the curves (ellipse).

I guess knowing how to start would be greatly appreciated.

答:
After time I worked it out. Will leave answer here for the sake of completeness:
It was a lot simpler than I thought:

$Q=x^2+\frac{y^2}{\epsilon^2} = s^2\cos^2(\tau)+ \frac{\epsilon^2 s^2 \sin^2(\tau)}{\epsilon^2}$

$ Q= s^2\cos^2(\tau)+s^2 \sin^2(\tau)$

$\frac{dQ}{d\tau}(s^2\cos^2(\tau)+s^2 \sin^2(\tau)) = -2\sin(\tau)\cos(tau)+2\sin(\tau)cos(\tau)$

Hence $\frac{dQ}{d\tau}=0$  as required.

Title: An inverse problem in PDE cours