问:
How do I find the minimal polynomial of $\sqrt{q}$ with coefficients in $\mathbb{Q}(\sqrt{p})$?
I am trying to find $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q}]$ where $p,q$ are distinct primes in $\mathbb{N}$.

I know that the elements of $\mathbb{Q}(\sqrt{p})$ can be obtained by considering $F[\sqrt{p}]=\{f(\sqrt{p}): f \in \mathbb{Q}[x]\}$ and as the minimal polynomial of $\sqrt{p}$ is $x^2-p$ from the division theorem, we obtain that the form of the elements of $\mathbb{Q}(\sqrt{p})$ is $a+b\sqrt{p}$.

Now I need to do it again, so the elements of $\mathbb{Q}(\sqrt{p})(\sqrt{q})$ can be obtained by considering $G[\sqrt{q}]=\{f(\sqrt{q}): f \in \mathbb{Q}(\sqrt{p})[x]\}$ where $\mathbb{Q}(\sqrt{p})[x]$ are the polynomials with coefficients in $\mathbb{Q}(\sqrt{p})$, I think this should yield polynomials such as:

$$(a_{n} +b_{n}\sqrt{p})X^{n}+(a_{n-1} +b_{n-1}\sqrt{p})X^{n-1} + \dots$$

Now I am a bit confused: How do I find the minimal polynomial of $\sqrt{q}$ with coefficients in $\mathbb{Q}(\sqrt{p})$?

I tried to use this proposition:

From this, we know that $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q(\sqrt{p})}]$ has degree $1$ or $2$ but supposing that $[\mathbb{Q}(\sqrt{p},\sqrt{q}),\mathbb{Q(\sqrt{p})}]=1$ and obtaining a contradiction seemed harder. If that is true: It means that $\mathbb{Q}(\sqrt{p}, \sqrt{q})=\mathbb{Q(\sqrt{p})}$ which implies that $\sqrt{q} \in \mathbb{Q(\sqrt{p})}$ and hence:

$$\sqrt{q}=a+b{\sqrt{p}}$$

But I was not able to find a contradiction on the existence of the coefficients $a,b$.

答:
Your idea is good! Indeed, $\sqrt q\notin\mathbb Q(\sqrt p)$ and we can in fact derive a contradiction assuming that

$$\sqrt q=a+b\sqrt p$$

for some $a,b\in\mathbb Q$. First note that $a\ne0$ as writing $b=\frac xy$ and squaring both sides would give

$$q=b^2p\implies qy^2=x^2p$$

and hence a contradiction to unique factorization as the occuring factors $p,q$ on the LHS and the RHS differ. Now observe that

\begin{align*}
\sqrt q=a+b\sqrt p\implies(\sqrt q-b\sqrt p)^2=a^2\iff \sqrt{pq}=\frac{a^2-b^2p}{2b}
\end{align*}

given that $b\ne0$ (which is clear as $\sqrt p$ is irrational). Hence, if  $\sqrt q\notin\mathbb Q(\sqrt p)$ then $\sqrt{pq}$ is rational. But a variation of the classical argument for square roots of prime numbers shows that $\sqrt{pq}$ is irrational as well and hence the desired contradiction is found.

This generalizes nicely to finite sets of pairwise distinct primes, i.e. for

$$p_1,\dots,p_n\quad \text{we have}\quad [\mathbb Q(\sqrt{p_1},\dots,\sqrt{p_n}):\mathbb Q]=2^n$$

and the minimal polynomials at each step are just $X^2-p_i$. The proof follows the same argument as above using induction on the number of primes.

A more general note: if you have a field extenstion of the form $K(\alpha,\beta)$ and minimal polynomials $f=\min_K(\alpha)$, $g=\min_K(\beta)$ you can always first check whether or not, say, $g$ is irreducible in $K(\alpha)$ viewed as element of $K(\alpha)[X]$. If so, it will again be the minimal polynomial of $\beta$ in $K(\beta)$ and $[K(\alpha,\beta):K(\alpha)]=[K(\beta):K]$. This strategy might bypass a tedious calculation of minimal polynomials.


问:
An example of a module that is not flat.
My professor defined a flat module as follows:

$M$ is flat if $M \otimes_R(.)$ is exact.

Then she gave us this example for a module that is not flat:

For $n \in \mathbb Z, n \geq 2.$ the following sequence $0 \rightarrow \mathbb Z \xrightarrow{f} \mathbb Z \xrightarrow{g} \mathbb Z /n \mathbb Z \rightarrow 0$ with $f: 1 \mapsto n$ and $g$ surjective is exact but $$0 \rightarrow \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z  \xrightarrow{f \otimes 1} \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z \xrightarrow{g \otimes 1} \mathbb Z /n \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z \rightarrow 0$$ where we have $$t \otimes 1 \mapsto nt \otimes 1 \mapsto 0$$ and $nt \otimes 1 = 0$ is not exact. And she added, so $\mathbb Z /n \mathbb Z$ is not a flat $\mathbb Z-$module.

My question:

I know that $\mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z = \mathbb Z /n \mathbb Z$ and  $\mathbb Z /n \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z = \mathbb Z /n \mathbb Z.$ Also, I know that tensoring is right exact, we proved that before, so why the second sequence she gave is not left exact?

Any explanation to why my professor said that is appreciated.

答:
The crucial observation was already pointed out in the comments by the user Alex Wertheim. I will just add an answer spelling it out.

We first note that $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/n\mathbb Z\cong\mathbb Z/n\mathbb Z$ as usual. Along this identification the tensored map $f\otimes1$ becomes

$$\mathbb Z/n\mathbb Z\to\mathbb Z/n\mathbb Z,\,\overline x\mapsto\overline{nx}=\overline0$$

i.e. the zero map as $n\equiv0\mod n$. This map is clearly not injective for $n\ge2$ and hence the tensored sequence not left exact.

ContentLicense: CC BY-SA 4.0


问:
Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$
$\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$

So it is bounded by 1 and hence it is convergent.

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?

答:
Let $\displaystyle a_n = \frac{\log^2 n}{n^2}$.
Then
$$\lim_{n\to\infty} n\left(\frac{a_{n+1}}{a_n}-1\right) = -2 < -1.$$
The series converges by Raabe's test. </p>

ContentLicense: CC BY-SA 3.0


问:
Derivative involving inverse matrix
Let $\alpha_i\in\mathbb{R}$, $x_i\in\mathbb{R}^d$ for all $i\in[k]$, with $k \geq d$. I am looking for this derivative

$$
\frac{\partial}{\partial\alpha_i} \left(\sum_{i=1}^k\alpha_ix_ix_i^T\right)^{-1} = \frac{\partial}{\partial\alpha_i} \left(X\Lambda X^\top\right)^{-1},
$$

where we define $X: \text{col(X)}= \{x_i\}_{i\in[k]}$, $\Lambda = \text{diag}(\alpha)$, and we assume $\left(X\Lambda X^\top\right)$ is invertible.

答: Let's call $A=X\Lambda X^T$. Now, $$\frac{\partial}{\partial \alpha_i}A^{-1} = -A^{-1} \left( \frac{\partial}{\partial \alpha_i} A \right)A^{-1}$$ (see Derivative of the inverse of a matrix), provided $A^{-1}$ exists. Finally, $$\frac{\partial}{\partial \alpha_i} A= x_i x_i^T$$.

Title: Derivative involving inverse matrix


问:
Elementary proofs about $f(t)=\lim_{n\to\infty}(1+\frac tn)^n$
Please read carefully what I mean by elementary before marking the question as duplicate.

I'm starting calculus lessons in High School; that is: no series, no derivatives, no limits of functions, no continuity... I have just stated the axiom of supreme and defined sequence and limit of sequence. I have proved that a monotone, bounded sequence has limit. Just that.

Under so 'elementary' conditions, I'd like to prove:

For each real $t$, the sequence $\left(1+\dfrac tn\right)^n$ converges. (I have problems for $t<0$). Then I'll define $e^t$ as the limit of this sequence.

For every pair $t,s$, $e^{t+s}=e^te^s$.

Any ideas or bibliography?

答:
My favorite proof i found in Best and Penner's Calculus.

The idea is to use a single interval  Riemann sum for the area under the curve  $1/x $ between  $1$ and $1+t/n $.

We get that $t/(n+t)\le\int_1^{1+t/n}1/x\rm dx\le t/n\implies e^{t/(n+t)}\le1+t/n\le e^{t/n}\implies e^{nt/(n+t)}\le (1+t/n)^n\le e^t$, and the result follows by letting  $n\to\infty $.


问:
Euclidean Space and Open Sets
Let $x_1$ and $x_2$ be two distinct points in $R^n$. Prove that there exist disjoint open sets $U_1$ and $U_2$ such that $x_1$ is in $U_1$ and $x_2$ is in $U_2$.

When I think about open sets, the first thing that always comes up is the consideration of interior points in which I should prove that there exists a small neighbourhood for both $x_1$ and $x_2$ which are contained in the set $U_1$ and $U_2$. However, the main issue I struggled with was the fact of how to construct and come up with these sets. Pretty sure that if anyone suggests the sets themselves, proving that those points lie there would be easier.

答:
Let $R=\|x_1 - x_2\| >0$ be their distance.

Consider the open neighbourhoods $U_i = \{x  \in \Bbb R^n: \|x - x_i\| < \frac{R}{2}\}$, for $i=1,2$.

Title: Euclidean Space and Open Sets


问:
Why does the conductor ideal contain a power of the radical?
Let $X'$ be a singular algebraic curve with singularity $Q \in X'$ and normalization $X \to X'$. Suppose $\mathcal O_Q'$ is the local ring of $X'$ at $Q$ and $\mathcal O_Q$ is its integral closure in the total ring of fractions $Q(\mathcal O_Q')$. Let $\mathfrak c = \operatorname{Ann}_{\mathcal O_Q'}(\mathcal O_Q / \mathcal O_Q') $ be the conductor ideal, and $\mathfrak r \subset \mathcal O_Q$ the Jacobson radical.

We have $\delta = \dim \mathcal O_Q / \mathcal O_Q' < \infty$.

I have two questions:

Why exactly is $A = \mathcal O_Q' / \mathfrak c$ finite-dimensional?

We know that $M = \mathcal O_Q' / \mathcal O$ is a finite-dimensional $A$-modul with $\operatorname{Ann}_A(M) = 0$. Does that imply that $M$ contains a copy of $A$? Or can we somehow conclude that $A$ is artinian?

Why does $\mathfrak c$ contain a power of the radical ideal $\mathfrak r$, i.e. $\sqrt{\mathfrak c} = \mathfrak r$?

By 1. we know $d = \dim \mathcal O_Q / \mathfrak c < \infty$. Hence the ascending sequence
$$ \dim \mathcal O_Q / (\mathfrak r + \mathfrak c) \leq \dim \mathcal O_Q / (\mathfrak r^2 + \mathfrak c) \leq \dotsb$$
is bounded by $d$, and so there exists $n \in \mathbb N$ such that for all $k \geq n$
$$ \mathfrak r^k + \mathfrak c = \mathfrak r^n + \mathfrak c.$$
But does this already imply $\mathfrak r^n \subset \mathfrak c$?

Here is a little additional constraint: I would like to apply all of this to germs of analytic curves, so I would like to avoid thinking of $\mathcal O_Q'$ as the localization of a finitely generated $\Bbb C$-algebra.

[1] Serre, Algebraic groups and class fields

答:
For 1. we show that $\mathcal O_x' / \mathfrak c_x$ is an artinian ring (since $\mathcal O_x'$ has residue field $k$, this is sufficient as $\mathcal O_x'/\mathfrak c_x$ is then filtered by a sequence with pairwise quotients $\mathcal O_x' / \mathfrak m$). Being artinian is the same as being noetherian and having Krull dimension $0$. Since $\mathcal O_x'$ has Krull dimension $1$, we have to show that $\mathfrak c_x$ is not contained in any minimal prime $\mathfrak p \subset \mathcal O_x'$. But $(\mathcal O_x)_{\mathfrak p}$ is a field and hence normal, so $(\mathcal O_x')_{\mathfrak p} = (\mathcal O_x)_{\mathfrak p}$, which shows that $\mathfrak p$ is not contained in the support of $\mathfrak c_x = \operatorname{Ann}(\mathcal O_x' / \mathcal O_x)$.

The second question is an application of Nakayama's lemma:
Consider the projection $\pi: \mathcal O_Q \to \mathcal O_Q/\mathfrak c$. Then $\pi(\mathfrak r)$ is contained in the radical of $\mathcal O_Q / \mathfrak c$, and satisfies
$$ \pi(\mathfrak r) \cdot \pi(\mathfrak r)^n = \pi(\mathfrak r)^n.$$
Hence $\pi(\mathfrak r)^n = 0$ by Nakayama, which means $\mathfrak r^n \subset \mathfrak c$.


问:
Counter examples on uniform convergence
I found some false statements on the uniform convergence and I tried to compute a counter-example for each of them. I would really appreciate is someone could tell me if my counter examples hold, please. The statements are:

Let $f_n$ be a sequence of functioons such that $f_n\to f$ pointwise on a  set $E$
as $n\to\infty$

1) If each $f_n$ is continuous on $E$, then $f$ is continuous on $E$ as well.

2) If each $f_n$ is differentiable on $E$, then $f$ is differentiable on $E$ as well

3) Suppose here that $f_n\to f$ uniformly on $E$. If each $f_n$ is differentiable on $E$, then $f$ is differentiable on $E$ as well

Thank you in advance for help!

Counter-examples:

1) Consider $f_n(x)=x^n$ on  the interval $[0,1]$. Clearly, $f_n\to f$ pointwise on $[0,1]$ with $f(x)=0$ if $0\le x<1$ and $f(x)=1$ if $x=1$. But, $f$ is not continuous on $[0,1]$. So, the statement doesn't hold.

2) Here we can consider the same $f_n(x)$ as in 1). Each $f_n$ is differentiable on $[0,1]$, but as $f$ is not continuous on [0,1], it is not differentiable on $[0,1]$. (Not really sure about this example)

3) Consider $f_n(x)=\sqrt{x+\frac{1}{n}}$ on $[0,1]$. Clearly, $f_n(x)\to f(x)=\sqrt{x}$ pointwise on $[0,1]$. We can prove now that $f_n\to f$ uniformly on $[0,1]$. $\forall \epsilon>0 \ \forall n\ge N \forall x \in [0,1]$ we have:

$|f_n(x)-f(x)|=|\sqrt{x+\frac{1}{n}}-\sqrt{x}|\underbrace{=\sqrt{x+\frac{1}{n}}-\sqrt{x}}_{\sqrt{x} \ increasing}=\frac{x+\frac{1}{n}-x}{\sqrt{x+\frac{1}{n}}+\sqrt{x}}=\frac{1}{n(\sqrt{x+\frac{1}{n}}+\sqrt{x})}\le \frac{1}{\sqrt{n}}$. Choosing $N=\frac{1}{\epsilon^2}$, we have that $|f_n(x)-f(x)|<\epsilon \ \forall \epsilon>0$. So, $f_n(x)\to f(x)$ uniformly on $[0,1]$. Each $f_n$ is differentiable on $[0,1]$, but $f(x)=\sqrt{x}$ is not differentiable on $[0,1]$ as the derivative on $x=0$ doesn't exist.

答:
Your counterexamples are great. Obviously, as 1) has a counterexample, 2) also as you mentioned.

You can have a look here for more counterexamples on the derivative of sequences of maps.

Title: Counter examples on uniform convergence


问:
Evaluating lambda expression call by value Beta reduction
$ + =  λmnab.m a((n a) b)$

I have to show that $2 + 3     \triangleright_\beta $ 5

what I understand from the lambda expression of + is that it takes

4 arguments m, n , a , b  But when I have to evaluate $2 + 3$, I only have $2$ arguments

How do I go about evaluating it.

答:
I guess, the numbers here are Church numerals:
\begin{align}
2 &= \lambda f.\lambda x. f (f x) \\
3 &= \lambda f.\lambda x. f (f (f x))
\end{align}
Applying the first argument (numeral 2) to addition, we get:
$$\begin{split}
& (\lambda m n a b.ma ((na)b))(\lambda fx. f (f x))&~~~~~~~~~\text{($\beta$-reduction)}\\
\to~ & \lambda n a b.((\lambda fx. f (f x))a) ((na)b)
\end{split}$$
Applying the second argument (numeral 3) to the above result, we get:
$$\begin{split}
& \big(\lambda n a b.((\lambda fx. f (f x))a) ((na)b)\big)\bigl(\lambda fx. f(f (f x))\bigr)&~~~~~~~~~\text{($\beta$-reduction)}\\
\to~ & \lambda a b.((\lambda fx. f (f x))a) (((\lambda fx. f(f (f x)))a)b)
\end{split}$$


问:
Which of these variables affect y more?
I have two equations:

1: $\displaystyle y=\frac{0.0029}{\sqrt{k}}$

2: $\displaystyle y=\frac{0.0060}{\sqrt{m}}$

Which variable has a greater effect on $y$? i.e., does doubling $m$ have a greater effect on $y$ than doubling $k$? How do I describe/prove that?

Thanks!

(I asked a similar question previously, here. However I feel that the square root nature of this question makes it an entirely different question, and so I've posted a new question.)

答:
From your first related question, you asked:

Of two variables, which affects $y$ more?
  I have two equations:

$\displaystyle y = \frac {0.0060}{k}$

$\displaystyle y= \frac{0.00016}{m}$

How do I determine which variable has a greater affect on $y$?

E.g., how do I determine if doubling $k$ as a greater effect on $y$ than doubling $m$ and so forth?

Note also that since  $\displaystyle y =\frac{0.0060}{k}\quad$ and $\quad\displaystyle y =\frac{0.00016}{m}$, then

$$\frac{0.0060}{k} = \frac{0.00016}{m} \implies \frac{k}{m} = \frac{0.0060}{0.00016} = 37.5$$

We call $37.5$ the proportionality constant (the value of the ratio $\displaystyle \frac km$). That is, $k$ and $m$ are directly proportional.  Another way to express this proportionality is that $k$ is a multiple of $m$: $k = 37.5m$.

(It would be good to take a look at Wikipedia-proportionality).

In either of these forms, it is easy to see that multiplying both $k$ and $m$ by the same factor does nothing to change the proportionality constant: and doing so for each of $k$ and $m$ results in the same effect on $y$.

E.g. if we look at $k = 37.5 m$, multiply both $m$ and $k$ by, say, $3$, we have:

$(3k) = 37.5(3m) \implies 3(k) = 3(37.5m)$ which simplifies back to our original equation: $k = 37.5m$.

In relation to $y$, $y$ and $k$, $y$ and $m$ are inversely proportional, hence:

$\displaystyle y = \frac{0.0060}{k} = \frac{0.00016}{m}$ then $\displaystyle \frac{0.0060}{3k} = \frac {1}{3} y = \frac{0.00016}{3m}$

Going back to the ratio of $k$ to $m$: $\displaystyle \frac{k}{m} = 37.5$  Now, we have seen that multiplying both $k$ and $m$ by the same factor (of any value) does not change the proportionality of $k$ to $m$, any any such change in the values of $k$ and $m$ impact the value of $y$ in exactly the same manner.

Similarly, with your post today:  as others have shown, multiplying both $m$ and $k$ by the same number has the same impact on $y$, regardless of the fact that you now have $y$ being expressed as functions of the square-root of $k$ and of $m$, respectively, and for similar reasons:

In this case, we have that $\displaystyle y = \frac{0.0029}{\sqrt{k}} = \frac{0.0060}{\sqrt{m}}$ and so we have the proportion of $\displaystyle \frac{\sqrt{k}}{\sqrt{m}}=\frac{0.0029}{0.0060} \approx 0.5$

Multiplying both $k$ and $m$ by $2$ is equivalent to multiplying both $\sqrt{k}$ and $\sqrt{m}$ by $\sqrt{2}$, thus cancelling out, leaving the proportionality constant unchanged.

However (I suspect this is what your intuition was with respect to your post:

IF (as in the first post, but choosing simpler numerators) $\displaystyle y = \frac{4}{k} = \frac{1}{m}$, then we have that $\displaystyle \frac{k}{m} = 4/1 = 4$, or equivalently, $k = 4m$.

NOW if you were to ask: what is the impact on $y$ if you were to take the square root of both $k$ and $m$, then, indeed, one of the two will have a greater impact on $y$ than the other, under a few conditions.

In this case, let $\displaystyle y_k = \frac{4}{\sqrt{k}}$ and $\displaystyle y_m = \frac{1}{\sqrt{m}}$, then squaring both sides of each equation gives us:
$$y_{k}^{2} = \frac{4^2}{k} = 4\left(\frac{4}{k}\right) = 4y$$
$$y_{m}^{2} =\frac{1^2}{m}= \frac{1}{m} = y$$

Hence we have that replacing $k$ from with the square root of $k$ results in a value of $y$ which is $4$ times greater than the value of $y$ when $m$ is replaced with the square root of $m$.

Similarly, for any exponent $p \neq 1$ and any equations of the form $\displaystyle y = \frac xk$ and $\displaystyle y = \frac ym, \;x \neq y$, replacing $k$ with $k^p$ and $m$ with $m^p$ (where $p \neq 0$) will result in differing impacts on the original value of $y$.  The magnitude of the difference in impact depends on the given values for the numerators, and the value of the exponent $p$.

Try playing around with this a bit!

Title: Which of these variables affect y more?


问:
Let $R$ be a ring, $S\subset R$ a subring and $I\subset R$ an ideal of $R$. Prove that every ideal of $S$ need not be of the form $S\cap I$.
Thoughts:
My idea to prove this is to find an explicit ring structure (in the context of the prompt) such that there exists some ideal $J$ of $S$ with $J\neq S\cap I$ for any ideal $I$ of $R$. I believe one way to do this would be to obtain some relatively small and easy to handle $R$ that has very few ideals, yet also has some proper non-trivial subring $S$ that has more ideals than $R$. That way, it would not be possible for the ideals of $S$ of the form $S\cap I$ to actually realize all ideals of $S$.

Attempt:
Consider the field $\mathbb{R}$, which has exactly the ideals $(0)$ and $(1)$. We then need to find a subring with more than two ideals. We have that $\mathbb{Z}\subset \mathbb{R}$ is a subring, and $\mathbb{Z}$ has infinitely many ideals of the form $n\mathbb{Z}=(n)$ for integers $n\geq 2$.

Does this seem reasonable?

答:
This is completely correct (as pointed out by the user Crostul already). You could have chosen $\mathbb Q$ instead of $\mathbb R$ which is a field as well but somewhat "closer" to the ring $\mathbb Z$ ($\mathbb Q$ is the quotient field of $\mathbb Z$ and so in some sense a natural extension). But this is just a matter of taste.

This argument generalizes to arbitrary fields $K$ with proper subrings $R$. Indeed, recall the standard fact that a (commutative unital) ring is a field iff its only ideals are trivial. Hence, if we have a proper subring, this ring has to have some non-trivial ideal $I$ which then is not the intersection of $R$ with the trivial ideals $(0)$ and $K$ of $K$ (as these intersections are just $(0)$ and $R$, respectively).

Claim. Let $R$ be a (commutative unital) ring. Then $R$ is a field iff its only ideals are $(0)$ and $R$.

Proof. Suppose $R$ is a field and take some ideal $I$. If $I\ne(0)$ we can find $r\ne0$ in $I$. As $R$ is a field there is some $s\in R$ such that $sr=1$. But then

$$x=x\cdot 1=x(sr)=(xs)r\in I$$

for all $x\in R$. Hence $I=R$. Conversely, suppose that $R$ is a ring with solely the ideals $(0)$ and $R$. Take $r\ne0$ in $R$ and consider the principal ideal $(r)$. As $r\ne0$ we have $(r)\ne(0)$ and hence by assumption $(r)=R$. But then $1\in(r)$ which is equivalent to $sr=1$ for some $s\in R$. This shows that every non-zero element of $R$ is invertible. $\square$

The underlying idea can be phrased as "for $r\in R$ we have $(r)=R$ iff $r$ is a unit". The proof of this proposition follows the same strategy as the one I gave above (see here for example).

So the contraposition of the claim reads as "a ring $R$ is not a field iff it has some non-trivial ideal $(0)<I<R$". Generally, one way such ideals arise is as $I=(r)$ for some non-unit $r$ (which exists due to $R$ not being a field). This should be clear from the given proof and the cited proposition about principal ideals genearted by units.


问:
Extremal of $J[y] = \int_{0}^{1}e^x \sqrt{1 + y'^2} dx$ where $y \in C^2[0,1]$
This is a problem from a mathematical contest. We have to find the form of the extremal for the variational problem  $J[y] = \int_{0}^{1}e^x \sqrt{1 + y'^2} dx$ where $y \in C^2[0,1]$.

The answer given is : $y(x) = \sec ^{-1}\left( \frac{x}{c_1} \right) + c_2$ where $|c_1|<1$ and $c_2$ is arbitrary.

Now my problem is that using the Euler Lagrange equation I have:

$$\frac{d}{dx} \left( \frac{e^x y'}{ \sqrt{1 +y'^2 } } \right) =0 $$

$$ \implies e^{2x}y'^2 = c_1^2 + c_1^2y'^2 $$

$$ \implies y' = \pm \frac{c_1}{\sqrt{e^{2x}-c_1^2}} $$

but this does not give the required form so I must be making some mistakes. Can someone please guide me?

Any help is highly appreciated.

答:
I think there must be some typing mistake, as my answer is $~y=\sec^{-1}\left(\dfrac{e^x}{c_1}\right)~+~c~,~~~|c_1|<1$

After your work,
$$y' = \pm \dfrac{c_1}{\sqrt{e^{2x}-c_1^2}}\implies y=\int \dfrac{c_1}{\sqrt{e^{2x}-c_1^2}}~dx~+~c~~~~\text{as} ~~y \in C^2[0,1]~.$$
Putting $~z=e^{2x}-c_1^2\implies dz=2e^{2x}dx~,$
$$y= \dfrac {c_1}{2}\int \dfrac{dz}{\sqrt z~(z+c_1^2)}~+~c$$Again substituting, $~t=\frac{\sqrt z}{c_1}\implies dt=\frac{1}{2c_1\sqrt z}dz\implies dz=2tc_1^2~dt~,$
$$y= \dfrac {1}{2}\int \dfrac{2tc_1^2~dt}{t~(t^2c_1^2+c_1^2)}~+~c= \int \dfrac{dt}{(t^2+1)}~+~c=\tan^{-1}(t)~+~c=\tan^{-1}\left(\dfrac{\sqrt{e^{2x}-c_1^2}}{c_1}\right)~+~c$$where $~c~$ is integrating constant.

Now we know that $~\tan^2\theta+1=\sec^2\theta~.$

Let $~\theta=\tan^{-1}\left(\dfrac{\sqrt{e^{2x}-c_1^2}}{c_1}\right)\implies \tan^2\theta=\dfrac{e^{2x}-c_1^2}{c_1^2}=\sec^2\theta-1\implies \sec^2\theta=1+\dfrac{e^{2x}-c_1^2}{c_1^2}=\dfrac{e^{2x}}{c_1^2}\implies \sec\theta=\dfrac{e^x}{|c_1|}\implies \theta=\sec^{-1}\left(\dfrac{e^x}{|c_1|}\right)\implies \tan^{-1}\left(\dfrac{\sqrt{e^{2x}-c_1^2}}{c_1}\right)=\sec^{-1}\left(\dfrac{e^x}{c_1}\right)~,~~~|c_1|<1$

Hence the extremal is $~y=\sec^{-1}\left(\dfrac{e^x}{c_1}\right)~+~c~,~~~|c_1|<1$


问:
Let $A, B$ be closed in $\mathbb{R}.$ Show that the product $A \times B$ is closed in $\mathbb{R}^2$.

Let $A, B$ be closed in $\mathbb{R}.$ Show that the product $A \times B$ is closed in $\mathbb{R}^2$.

Let $x=(x_A,x_B) \in (A\times B)^c$ such that $x_A \in A^c$ and $x_B \in B^c$. What I wanted to show was that since $A$ and $B$ are closed, then there is an open neighborhoods $B(x_A,r) \subset A^c$ and $B(x_B,r') \subset B^c$ for all $r,r' >0$. This would imply that $x \in B(x_A,r) \times B(x_B,r')$ so  $B(x_A,r) \times B(x_B,r')$ is an open neighborhood of $x$ and since $x \in (A\times B)^c$ we would have that $(A \times B)$ is closed since it's complement is open. This doesn't seem to be correct though since $(A\times B)^c \ne A^c \times B^c$ so I wouldn't neccessarily have that $x_A \in A^c$ and $x_B \in B^c$?

答:
Suppose $(x,y) \notin A \times B$. There are twe possibilities:

a) $x \notin A$

b) $y \notin B$.

Suppose  $x \notin A$. There exists $r >0$ such that $B(x,r) \subset A^{c}$. Verify that $B((x,y),r) \subset (A \times B)^{c}$.

Case b) is similar.


答:
You are correct that typically $(A\times B)^c\neq A^c\times B^c$. But $(x_A,x_B)\in (A\times B)^c$ implies that $x_A\in A^c$ or $x_B\in B^c$. WLOG assume that $x_A\in A^c$. Then $(x_A,x_B)\in A^c\times \mathbb{R}$ which is an open neighbourhood of $(x_A,x_B)$ that doesn't intersect $A\times B$. By the arbitrary choice of $(x_A,x_B)$ we conclude that $(A\times B)^c$ is open.

Note that this approach can be easily generalized to any topological space product $X\times Y$, not only $\mathbb{R}^2$.

Body:
Let $A, B$ be closed in $\mathbb{R}.$ Show that the product $A \times B$ is closed in $\mathbb{R}^2$.

Let $x=(x_A,x_B) \in (A\times B)^c$ such that $x_A \in A^c$ and $x_B \in B^c$. What I wanted to show was that since $A$ and $B$ are closed, then there is an open neighborhoods $B(x_A,r) \subset A^c$ and $B(x_B,r') \subset B^c$ for all $r,r' >0$. This would imply that $x \in B(x_A,r) \times B(x_B,r')$ so  $B(x_A,r) \times B(x_B,r')$ is an open neighborhood of $x$ and since $x \in (A\times B)^c$ we would have that $(A \times B)$ is closed since it's complement is open. This doesn't seem to be correct though since $(A\times B)^c \ne A^c \times B^c$ so I wouldn't neccessarily have that $x_A \in A^c$ and $x_B \in B^c$?


问:
Lie bracket of a vector field at a point $p \in M$.
Let $M$ be a manifold and let  $ X , Y $ two vector fields on $M$.  I know ths definition of the  Lie bracket between $X $ and $Y$ $[X,Y]: C^\infty(M)\rightarrow C^\infty(M)$ is an $\mathbb R$-linear map such that $[X,Y]f:=X(Yf)-Y(Xf)$.

However, if $p \in M $ I'm not sure what is the explicit expression of $[X,Y](p)$?

答:
Suppose $(x^1,\ldots,x^n)$ is a local coordinate patch and $\left(\partial_1,\ldots,\partial_n\right)$ are the associated tangent vector fields. Then locally, there exist functions $X^i$ and $Y^j$ such that:
\begin{align}
X &= \sum_{i=1}^nX^i\partial_i, & Y &=\sum_{j=1}^nY^j\partial_j.
\end{align}
If $f$ is a function, then:
\begin{align}
Y\cdot f = \mathrm{d}f(Y) = \sum_{i=1}^n\left(\frac{\partial f}{\partial x^i}\mathrm{d}x^i\right) Y = \sum_{i=1}^n Y^i\frac{\partial f}{\partial x^i}.
\end{align}
Thus:
\begin{align}
X\cdot(Y\cdot f) &= \sum_{j=1}^n \left(\frac{\partial}{\partial x^j}\left(Y\cdot f \right)\mathrm{d}x^j\right)X \\
 &= \sum_{j=1}^nX^j\left(\frac{\partial}{\partial x^j}\left(\sum_{i=1}^n Y^i \frac{\partial f}{\partial x^i} \right) \right) \\
&=\sum_{i,j=1}^nX^j \left(\frac{\partial Y^i}{\partial x^j}\frac{\partial f}{\partial x^i} + Y^i \frac{\partial^2 f}{\partial x^j\partial x^i} \right).
\end{align}
Therefore, it follows that:
\begin{align}
[X,Y]f &= X\cdot(Y\cdot f) - Y\cdot(X \cdot f)\\
&=\sum_{i,j=1}^n X^j\frac{\partial Y^i}{\partial x^j}\frac{\partial f}{\partial x^i} - Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i} + X^jY^i\frac{\partial^2 f}{\partial x^j\partial x^i} - Y^jX^i\frac{\partial^2 f}{\partial x^j\partial x^i}\\
&= \sum_{i=1}^n\left(\sum_{j=1}^n X^j\frac{\partial Y^i}{\partial x^j}- Y^j\frac{\partial X^i}{\partial x^j} \right) \frac{\partial f}{\partial x^i} + 0 \tag{$\star$}\\
&= \mathrm{d}f \left(\sum_{i=1}^n \left(\sum_{j=1}^n X^j\frac{\partial Y^i}{\partial x^j}- Y^j\frac{\partial X^i}{\partial x^j} \right) \partial_i \right),
\end{align}
where ($\star$) is a consequence of Swcharz's lemma, that is $\frac{\partial^2f}{\partial x^j \partial x^i} = \frac{\partial^2f}{\partial x^i\partial x^j}$. Hence, in local coordinates:
$$
[X,Y] = \sum_{i=1}^n\left(\sum_{j=1}^n X^j \frac{\partial Y^i}{\partial x^j} - Y^j\frac{\partial X ^i}{\partial x^j}\right) \partial_i.
$$
Note that thanks to Schwarz's lemma, it is canonically defined. Otherwise, as the second partial derivative of functions depends on the choosen chart, it would not have been a canonical expression.

To answer your comment: if $X(m)=0$, the above expression shows that
$$
[X,\xi](m) = - \sum_{i=1}^n\left(\sum_{j=1}^n \xi^j(m)\frac{X^i}{\partial x^j}(m)\right) \partial_i,
$$
which does not depend on the local behaviour of $\xi$ around $m$ but just on its value $\xi(m)$.


问:
How to tackle a statistics problem given a random variable X?
I'm having trouble getting an overall grasp of certain aspects of statistics, namely knowing when to use which formula/approach to solve a problem.

Let's consider a random variable X, after you determine whether the random variable is continuous or discrete, how would you know whether to know whether to use a binomial distribution, exponential distribution, Poisson distribution, moment generating function, etc.?

For example, I know that if I'm asked to model the time between random events I know I have an exponential distribution problem.

It's just that there are so many different mean, variance, and other equations, I got lost along the way on which equation to apply where. Some clarification would be great.

答: The first step is to thoroughly understand the scenario that is being described, and  what quantity the random variable represents.  Each of the common distributions has a typical paradigm for what it represents.  For example, the binomial distribution would represent the number of "successes" in a fixed number of independent trials.  On the other hand a hypergeometric distribution also represents the number of "successes", but where you're selecting a fixed number of objects selected without replacement from a given population.</p>


问:
Axioms of a coalgebra restated using Sweedler's notation
I'm struggling with understanding manipulation using Sweedler's notation at a very fundamental level. I don't understand the equivalence of the axioms of coalgebras in the standard notation
[Coproduct Map] $$\Delta:A\rightarrow A\otimes A$$ is a coassociative map;

$$(\Delta\otimes id_C)\circ\Delta=(id_C\otimes\Delta)\circ\Delta$$

in terms of Sweedler's notation;

Coassociativity $$\sum\Delta(a_{(1)})\otimes a_{(2)} =\sum a_{(1)}\otimes \Delta(a_{(2)})$$

I get upto being able to expand the left hand side of the above;
$$
\sum\sum(a_{(1)})_{(1)}\otimes (a_{(1)})_{(2)}\otimes(a_{(2)})
$$

but I don't know how to make it equivalent to $\sum\sum(a_{(1)})\otimes (a_{(2)})_{(1)}\otimes(a_{(2)})_{(2)}$. I'm assuming it would involve applying the coassociativity axiom.

For the counit axiom, I can't even scratch the surface and make a start.
$\eta:C\rightarrow K$ is the counit map that satisfies $(\eta\otimes id_C)\circ\Delta=(id_C\otimes \eta)\circ\Delta$ as
$$\sum\eta(c_1)c_2=\sum c_1\eta(c_2)=c, \forall c=\in C$$

Could someone help with a basic, step by step walk through of showing the equivalence of the two expressions?

答:
You say:

$\dots$ but I don't know how to make it equivalent to $\cdots$

You don't have to make it an equality, it is an equality by precisely the assumption that $\Delta$ is coassociative.

Let's step back a little. A comultiplication is a linear map $\Delta\colon C\to C\otimes C$ satisfying coassociativity, i.e.~$(\Delta\otimes 1)\circ \Delta=(1\otimes \Delta)\circ \Delta$. As every element of $C\otimes C$ can be written as a finite linear combination of elementary tensors (elements of the form $c\otimes c'$ with $c,c'\in C$), we can write $$\Delta(x)=\sum_{i=1}^nx_{1,i}\otimes x_{2,i}$$
with all $x_{j,i}\in C$. The above is not yet the Sweedler notation but is certainly correct notation. The annoying thing is that the above notation gets complicated really quickly when doing multiple operations. That's where the Sweedler notation comes in!

Instead of writing $$\Delta(x)=\sum_{i=1}^nx_{1,i}\otimes x_{2,i},$$
we surpress the index of the summation and obtain $$\Delta(x)=\sum x_{(1)}\otimes x_{(2)}.$$
By using these round squares, we remind ourselves that actually an index is omitted from the notation, but the labels $1$ and $2$ recall the order in which we 'blew up' the element $x$. Coassociativity precisely says that when doing repeated 'blow ups', the order doesn't matter.

So with this explanation, let's use this notation to understand the counit property $(\eta\otimes id_C)\circ\Delta=(id_C\otimes \eta)\circ\Delta$.

We find that $$\begin{aligned} (\eta\otimes id_C)\circ\Delta(x)&=(\eta\otimes id_C)(\Delta(x))\\ &= (\eta\otimes id_C)(\sum x_{(1)}\otimes x_{(2)}) \\ &= \sum \eta(x_{(1)})\otimes id_C(x_{(2)})\end{aligned}$$
Note that the above actually lives in $k\otimes C$ (where $k$ is the field we're working over). Clearly $k\otimes_k C\cong C$ by mapping $\lambda\otimes c$ to $\lambda c$. Hence we can identify $\sum \eta(x_{(1)})\otimes id_C(x_{(2)})$ with $\sum \eta(x_{(1)})x_{(2)}$. The other side can be computed similarly and yields the desired equality.


问:
Defining equivariant maps of bimodules
In this paper of Shulman, at page 4, he describes the (pseudo-)double category of rings and bimodules. This is supposed to be a double category with rings as objects, ring homomorphisms as tight 1-morphisms, bimodules as loose 1-morphisms and the two morphisms are "equivariant maps of bimodules". I do not understand the last bit (in scare quotes).

For instance, if we have $M$, a $R-S$ bimodule and $N$, a $R'-S'$ bimodule, what are the "equivariant" maps between $M$ and $N$? As far as I know, there are no relations/morphisms between $R,S,R',S'$.

Update: I am guessing that the data of an "equivariant map" $\psi: {_RM}_S \to {_{R'}M'}_{S'} $ is a triple $(\alpha:R \to R', \beta:S \to S', \psi:M \to M')$ where $\alpha, \beta$ are ring homomorphisms and $\psi$ is a map of bimodules with the induced action via the ring homomorphisms.
Is this right?

答:
Yes, your update is basically answering your question.

In a (pseudo) double category each cell has specific boundary arrows, two horizontal and two vertical.

It means that if $\psi$ is a cell, if its boundary is a bimodule $M:A-B$ on the top and a bimodule $M':A'-B'$ on the bottom, then its left side must be a ring homomorphism $\alpha:A\to A'$ and its right side $\beta:B\to B'$.

With these settings, the requirement for $\psi$ (beyond that it must be a homomorphism of the underlying Abelian groups) is simply:
$$\psi(a\cdot m\cdot b)=\alpha(a)\cdot \psi(m)\cdot\beta(b)$$
for any $a\in A,\ b\in B,\ m\in M$.

Title: Defining equivariant maps of bimodules


问:
Recurrent relation- integral
How to calculate $I_n=\int_{-n}^n \frac{y^n+y^3\cos{y}}{(|y|+1)^{n+2}} dy$? For $n=2k+1$ $I_n=0$, but what for $n=2k$?

I'm stuck with this problem, any help is welcome. Thanks in advance.

答: I solved a problem in two ways, by binomial theorem and by substitution $-\frac{1}{y+1}=t$. It is easy to check that $I_n=2\int_0^n (\frac{y}{1+y})\frac{dy}{(y+1)^2}=2\int_0^n (1-\frac{1}{y+1})^n d(\frac{-1}{y+1})$, since the other integral in $I_n$ is zero because $y^3$ is odd, and $\cos{y}, (|y|+1)$ are even functions.

Title: Recurrent relation- integral


答:
With sequences: let $((x_n,y_n))$ be a convergent sequence in $A \times B$ with limit $(x,y)$.

Then $x_n \to x$ and $y_n \to y$. Since $A$ and $B$ are closed , we have $x \in A$ and $y \in B$, thus $(x,y) \in A \times B$.

Title: Let $A, B$ be closed in $\mathbb{R}.$ Show that the product $A \times B$ is closed in $\mathbb{R}^2$.


问:
Let $f_n\to f$ pointwise on a set $E$. If $|f_n(x)-f(x)|\le a_n \ \forall x\in E$ with $a_n\to 0$, then the convergence is uniform

Let $f_n$ and $f$ be a real valued functions defined on a set $E$. If $f_n\to f$ pointwise on $E$ and there is a real-valued sequence $a_n$ such that $a_n\to 0$ and $|f_n(x)-f(x)|\le a_n \ \forall x \in E$. Show that $f_n\to f$ uniformly. I would like to know if my proof holds, please. I also have a question on the statement in the end. Thank you in advance for help!

First, as $f_n\to f$ pointwise, then we have $\forall x \in E:\forall \epsilon>0 \ \exists N \ \forall n>N:$ $|f_n(x)-f(x)|<\epsilon$.

Then, as $a_n\to 0$ as $n\to\infty$, we have: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N':$ $|a_n|<\epsilon$

Moreover, $|f_n(x)-f(x)|\le a_n \ \forall x \in E$. So, $\forall \epsilon>0 \ \forall n\ge n_0:=\max(N,N') \ \forall x:$ $|f_n(x)-f(x)|\le|a_n|<\epsilon$. So, $f_n\to f$ uniformly.

I'm not sure that I have to use $\max(N,N')$. I think that using only $N'$ the inequality might work, as the pointwise convergence is needed here to ensure simply that the limit of $f_n$ exists and as $f_n$ is bounded by $a_n$ independently of $n$. Is it correct?

答:
You don't need the pointwise convergence hypothesis. The second hypothesis alone implies uniform convergence.

To see it, take $\epsilon \gt 0$. As $a_n \to 0$, you can find $N \in \mathbb N$ such that for $n \ge N$ you have $0 \le a_n \le \epsilon$. And then for $n \ge N$ you also have $\vert f_n(x) - f(x) \vert \le a_n \le \epsilon$ for all $x \in E$.


答:
Given fat matrix ${\rm V} \in \Bbb R^{d \times n}$, let

$${\rm F} ({\rm x}) := {\rm V} \,\mbox{diag} ({\rm x}) {\rm V}^\top, \qquad {\rm G} ({\rm x}) := \left( {\rm F} ({\rm x}) \right)^{-1}$$

Using Sherman-Morrison,

$$\begin{aligned} {\rm G} ({\rm x} + h \, {\rm e}_i) = \left( {\rm F} ({\rm x} + h \, {\rm e}_i) \right)^{-1} = \left( {\rm F} ({\rm x}) + h \, {\rm v}_i {\rm v}_i^\top \right)^{-1} &= {\rm G} ({\rm x}) - h \,\frac{{\rm G} ({\rm x}) \, {\rm v}_i {\rm v}_i^\top {\rm G} ({\rm x})}{1 + h \, {\rm v}_i^\top {\rm F} ({\rm x}) \,{\rm v}_i} \\ &= {\rm G} ({\rm x}) - h \,{\rm G} ({\rm x}) \, {\rm v}_i {\rm v}_i^\top {\rm G} ({\rm x}) + \mathcal O \left( h^2 \right)\end{aligned}$$

and, thus,

$$\partial_i {\rm G} ({\rm x}) = \color{blue}{- {\rm G} ({\rm x}) \, {\rm v}_i {\rm v}_i^\top {\rm G} ({\rm x})}$$

which is what Господин Лисица obtained.

matrices matrix-calculus derivatives

Title: Derivative involving inverse matrix

Body:
Let $\alpha_i\in\mathbb{R}$, $x_i\in\mathbb{R}^d$ for all $i\in[k]$, with $k \geq d$. I am looking for this derivative

$$
\frac{\partial}{\partial\alpha_i} \left(\sum_{i=1}^k\alpha_ix_ix_i^T\right)^{-1} = \frac{\partial}{\partial\alpha_i} \left(X\Lambda X^\top\right)^{-1},
$$

where we define $X: \text{col(X)}= \{x_i\}_{i\in[k]}$, $\Lambda = \text{diag}(\alpha)$, and we assume $\left(X\Lambda X^\top\right)$ is invertible.


答: Your proof is wrong because it assumes that $N$ is independent of $x$. The fact that $|f_n(x)-f(x)| \leq a_n <\epsilon$ for $n> N'$  ($N'$ as chosen by you) is enough to say that $f_n \to f $ uniformly.

Title: Let $f_n\to f$ pointwise on a set $E$. If $|f_n(x)-f(x)|\le a_n \ \forall x\in E$ with $a_n\to 0$, then the convergence is uniform

Body:
Let $f_n$ and $f$ be a real valued functions defined on a set $E$. If $f_n\to f$ pointwise on $E$ and there is a real-valued sequence $a_n$ such that $a_n\to 0$ and $|f_n(x)-f(x)|\le a_n \ \forall x \in E$. Show that $f_n\to f$ uniformly. I would like to know if my proof holds, please. I also have a question on the statement in the end. Thank you in advance for help!

First, as $f_n\to f$ pointwise, then we have $\forall x \in E:\forall \epsilon>0 \ \exists N \ \forall n>N:$ $|f_n(x)-f(x)|<\epsilon$.

Then, as $a_n\to 0$ as $n\to\infty$, we have: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N':$ $|a_n|<\epsilon$

Moreover, $|f_n(x)-f(x)|\le a_n \ \forall x \in E$. So, $\forall \epsilon>0 \ \forall n\ge n_0:=\max(N,N') \ \forall x:$ $|f_n(x)-f(x)|\le|a_n|<\epsilon$. So, $f_n\to f$ uniformly.

I'm not sure that I have to use $\max(N,N')$. I think that using only $N'$ the inequality might work, as the pointwise convergence is needed here to ensure simply that the limit of $f_n$ exists and as $f_n$ is bounded by $a_n$ independently of $n$. Is it correct?


问:
checking that a group is $A_n$ or $S_n.$
Suppose I have two (random) permutations of $[n].$ We know (Dixon, Babai) that they almost surely generate $A_n$ or $S_n$ (if both even, then $A_n,$ otherwise $S_n.$) But the question is: what is the best (theoretical or practical) algorithm to check that the generated subgroup really is one of those big groups?

答:
The algorithm described in Seress' book (and also in Section 4.2, page 81 of "Handbook of Computational Group Theory" by Holt, Eick and O'Brien) is one of the best examples of a Monte-Carlo algorithm.

It is very fast (almost linear time), and its output is either "yes, the group generated is definitely $A_n$ or $S_n$" or "the group is almost certainly not $A_n$ or $S_n$. It is extremely simple to describe and program, although of course it has to be justified theoretically.

It can be given an error probability $\epsilon$ as input, which means that the probability of a false negative is less than $\epsilon$.

It works as follows. Let $G = \langle \alpha,\beta \rangle$ be the input group. First check that $G$  is transitive (that's easy)  - if not, answer no.

Now choose a collection of $N(n,\epsilon)$ (pseudo)random elements $g \in G$, where $N(n,\epsilon)$ is a function to be defined below. (There are good ways of choosing random elements, which I won't go into here - the Product Replacement Algorithm is used in practice.) For each $g$ compute all powers  of $g$ that have prime order $p$. If one of these is a $p$-cycle with $n/2 < p < n-2$ then, by a theorem of Jordan, $G = A_n$ or $S_n$, and so we return the answer yes. If none of them have this property then return the answer probably no.

The probability estimate for the false negative is based on a result (Lemma 10.2.3 of Seress' book) that the proportion of elements of $A_n$ and $S_n$ that power to a $p$-cycle with $n/2 < p < n-2$ is at least $$\sum_{p\ {\rm prime},\ n/2<p<n-2}\frac{1}{p},$$
which is approximately $\log(2)/\log(n)$.

Choose $c(n)$ such that this proportion is at least $c(n)\log(2)/\log(n)$  (for example, we put $c(n)=0.34$ for $8 \le n \le 16$ and $c(n) = 0.57$ for $n>16$) and then let $d(n) := c(n)\log(2)/\log(n)$.
Then, setting $N(n,\epsilon) = -\log(\epsilon)/d(n)$ gives the required error probability.

To be precise, if the group really was $A_n$ or $S_n$, then the probability of choosing $N(n,\epsilon)$ random elements without finding one that powers to a $p$-cycle with $n/2<p<n-2$ is less than $\epsilon$.

Theoretically, the weakest part of the method is the difficulty of choosing genuinely random elements, but that is not a serious obstacle in practice, because Product Replacement works very well when the group is $A_n$ or $S_n$.

I should perhaps add that the standard Schreier-Sims algorithm for computing the order of $G$ runs in polynomial-time (and it has various refinements that make it run faster in practice) , so you can use that to verify a negative answer if you want to.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
Get one of these things:

Load it up in your favorite 3D modelling software and let it calculate the surface area for you:


问:
Finding laplace inverse of $\frac{1}{(p+2)^2(p^2+\omega^2)}$
I am finding the laplace inverse of the below :-
$$\frac{1}{(p+2)^2(p^2+\omega^2)}$$

I can apply convolution theorem by taking $$f(p)=\frac{1}{(p+2)^2}, g(p)=\frac{1}
{(p^2+\omega^2)}$$
and calculating the function which yield the above functions on laplace transform  i.e. essentially ,
$$F(t)=e^{-2t}t,G(t)=\frac{\sin{\omega}t}{\omega}$$
then solve by routine integration .

But this seems a little cumbersome . Is there a better way to solve the above ?

Thanks for reading !

答:
Hint:

Use  Partial Fraction Decomposition

$$\dfrac1{(p^2+w^2)(p+2)^2}=\dfrac a{p+2}+\dfrac b{(p+2)^2}+\dfrac{cp+d}{p^2+w^2}$$

Multiply both sides by $(p^2+w^2)(p+2)^2$

and  use the coefficients of different exponents of $p$ to find $a,b,c,d$

Finally use the  standard formulae


问:
Natural boundary conditions for a variational problem when $~f~$ is a function of higher order derivatives of $~y~$
I am not sure about the boundary conditions for a variational problem when $~f~$ is a function of higher order derivatives of $~y~.$

i.e. if we have: $~f(x,y,y',y'')~$ then what is the natural boundary condition in this case?

Any help is highly appreciated.

答:
When the functional is of the form $$I[y(x)] = \int_{x_1}^{x_2} F \left(x,y(x),y'(x)\right) \,\rm{d}x~,$$then the natural boundary conditions are given in my answer of the following question: What are natural boundary conditions in the calculus of variations?

If the functional involves higher derivatives of $~y(x)~,$ such as,
$$I[y(x)] = \int_{x_1}^{x_2} F \left(x,y(x),y'(x),y''(x)\right) \,\rm{d}x$$then the minimum (or maximum) of $~I[y(x)]~$ satisfies the following Euler-Lagrange equation,
$$\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)+\dfrac{d^2}{dx^2}\left(\dfrac{\partial F}{\partial y''}\right)=0$$provided that $~y(x)~$ is subjected to the following essential or natural boundary conditions,

\begin{array}{|c|c|}
\hline
\text{essential}& \text{natural}\\ \hline
y(x_1)=y_1 &\frac{\partial F}{\partial y'}-\frac{d}{dx}\left(\frac{\partial F}{\partial y''}\right)=0~~\text{at}~~x=x_1 \\ \hline
y'(x_1)=y_1' & \frac{\partial F}{\partial y''}=0 ~~\text{at}~~x=x_1 \\ \hline \hline
y(x_2)=y_2 & \frac{\partial F}{\partial y'}-\frac{d}{dx}\left(\frac{\partial F}{\partial y''}\right)=0~~\text{at}~~x=x_2  \\ \hline
y'(x_2)=y_2'& \frac{\partial F}{\partial y''}=0 ~~\text{at}~~x=x_2  \\ \hline
\end{array}


问:
How fast does this sum over primes converge?
Let $$A = \lim_{x\to \infty} \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k}$$

I want to show that $\displaystyle A - \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k} \in \mathcal{O}(\frac{1}{\sqrt{x}})$

Generally, I want to know how fast $\displaystyle \sum_{k=2}^\infty \sum_{p^k\le x} \frac{\ln(p)}{p^k}$ converges.

Some results:

It is possible to express $A$ as $\displaystyle \sum_{p \in \mathbb{P}} \frac{\ln(p)}{p(p-1)}$ and I was able to show that $\displaystyle A - \sum_{p \le x} \frac{\ln(p)}{p(p-1)} \in \mathcal{O}(\frac{1}{x})$ using Abel summation. I wasn't able to use the same technique on the double sum above. I noticed, that there is a connection to Chebyshev's functions by $\displaystyle \psi(x) - \vartheta(x) = \sum_{k=2}^\infty \sum_{p^k \le x} \ln(p)$

Also  $\displaystyle A = -\sum_{k=2}^\infty P'(k)$ where $P(s)$ denotes the Prime-Zeta function, but I don't think that this is useful for the stated problem.

I tried to approximate $A$ with the sum over all natural numbers instead of primes, and approximate that with an Integral.

I tried some other minor things but nothing seems to lead anywhere. If you got some idea, any help is appreciated.

答:
I'm using the Prime Number Theorem below. Let $\kappa(p,x):=\max\{1,\lfloor\ln x/\ln p\rfloor\}$; then $$R(x):=\sum_{k=2}^\infty\sum_{p^k>x}\frac{\ln p}{p^k}=\sum_{p\in\mathbb{P}}\sum_{k>\kappa(p,x)}\frac{\ln p}{p^k}=\sum_{p\in\mathbb{P}}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}.$$ Note that $\kappa(p,x)=1$ if and only if $p^2>x$; otherwise $\kappa(p,x)>\frac{\ln x}{\ln p}-1$ and $p^{\kappa(p,x)}>\frac{x}{p}$.

Hence $R(x)=S(x)+T(x)$, where $$S(x)=\sum_{p^2\leqslant x}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}\leqslant\frac1x\sum_{p^2\leqslant x}\frac{p\ln p}{p-1}\in\mathcal{O}(x^{-1/2})$$ because $\frac{p\ln p}{p-1}\leqslant 2\ln p\leqslant \ln x$ and $\#\{p : p^2\leqslant x\}\in\mathcal{O}\left(\frac{x^{1/2}}{\ln x}\right)$ by PNT, and $$T(x)=\sum_{p^2>x}\frac{\ln p}{p^{\kappa(p,x)}(p-1)}=\sum_{p^2>x}\frac{\ln p}{p(p-1)}\in\mathcal{O}(x^{-1/2})$$ is what you've shown already (moreover, this can't be enhanced).


问:
Conic Sections Checking my solution: My Foci seams to be bigger than my Vertices (I think there is something wrong with my solution)
Which one is my error? Pls help!

Body: Which one is my error? Pls help!

Body:
Note that

$$(2y-4)^2 \ne 2(y-2)^2$$

It should be

$$(2y-4)^2=4(y-2)^2$$


问:
If $s_{n}$ converges why does $\sqrt{s_{n}}$ converge only when $s_{n}$ is positive.
So I've been trying to figure out if it's true that if $s_{n}$ converges $\sqrt{s_{n}}$ converges. I've found proofs on here that this is true if $s_{n}>0$, but I can't find out why it isn't true if there are negative numbers. Why does the introduction of imaginary numbers break this rule? Can someone explain it, or provide me with an example of a series where $s_{n}$ converges and $\sqrt{s_{n}}$ doesn't.

Edit: I think it may have to do with the fact that the complex numbers aren't an ordered field, but I'm still not sure how to formalize this.

答:
The problem is that $\sqrt{s_n}$ is ambiguous if $s_n<0$. Which square root do you have in mind?

But that's not the only problem. Take the sequence $(s_n)_{n\in\Bbb N}$ defined by $s_n=-1$. For each $n\in\Bbb N$, let $t_n=(-1)^ni$. Then, for each $n\in\Bbb N$, $t_n$ is a square root of $s_n$. However, $(s_n)_{n\in\Bbb N}$ converges, whereas $(t_n)_{n\in\Bbb N}$ doesn't.


问:
Evaluate ${ \int_{0}^{ 1}{ \int_{0}^{1}{x\max(x^2,y)\,dy} \,dx} }$
${ \int_{0}^{ 1}{ \int_{0}^{1}{x\max(x^2,y)\,\text{d}y} \,\text{d}x} }$

How would you sketch the graph for this integral?

Body:
\begin{align}
=\int_0^1\left[\int_0^{x^2}x^3\,\text{d}y+\int_{x^2}^1xy\,\text{d}y\right]\text{d}x
\end{align}


问:
On Kunen's Set Theory, Definition II.6.17
Definition II.6.17 of Kunen's Set Theory, is as follows:

Fix $\Gamma : \mathbf{HF} \xrightarrow[\text{onto}]{1-1} \omega$, as in Lemma I.10.17. Working with $\mathcal{L} = \{\in\}$, list the set of all variables $\{v_r : r \in \omega\}$, where $r < s \to \Gamma(v_r) < \Gamma(v_s)$. Call a formula $\varphi$ good iff for some $n \geq 0$, the free variables of $\varphi$ are exactly $v_0,v_1,\dots,v_n$. List the set of all good formulas as $\{\varphi_i : i \in \omega\}$ with $i < j \to \Gamma(\varphi_i) < \Gamma(\varphi_j)$; let $\varphi_i$ have $n_i + 1$ free variables.

Whenever $A \neq \emptyset$, $i \in \omega$ and $\vec{b} \in A^{n_i}$, let $D(A,i,\vec{b}) = \{a \in A : A \models \varphi[b_0,\dots,b_{n_i - 1},a]$}. Then, for $S \in \mathcal{D}^+(A)$, let $i(S)$ be the least $i$ such that $S = D(A,i,\vec{b})$ for some $\vec{b} \in A^{n_i}$. Write $n(S)$ for $n_{i(S)}$.

I am unable to wrap my head around this definition.

Here, $v_r,v_s$ etc are variables. In that case, what does it mean to have $\Gamma(v_r) < \Gamma(v_s)$, since these variables can take up various values? E.g. $v_r = 0$ and $v_s = 1$, vs $v_r = 1$ and $v_s = 0$.

If $\varphi_i$ is a formula, how does one interpret $\Gamma(\varphi_i)$? In the case that we have $\varphi_i$ is "$x = y$", wouldn't this make no sense?

What is $\varphi$ in the definition of $D(A,i,\vec{b})$? Should it be $\varphi_i$ instead?

Thanks for any help in advance.

Remark. For completion sake, Lemma I.10,17 is as follows:

Define $E \subseteq \omega \times \omega$ by: $nEm \leftrightarrow 2 \nmid \lfloor{m2^{-n}}\rfloor$ (equivalently, there is a $1$ in place $n$ in the binary representation of $m$ (counting from the right;sp $43 = 101011_b$ has $1$ in places $0,1,3,5$)). Then $(R(\omega);\in) \cong (\omega;E)$, with the isomorphism $\Gamma$ defined recursively by $\Gamma(y) = \sum\{2^{\Gamma(x)} : x \in y\}$. For example:

$\Gamma(\emptyset) = \sum\emptyset = 0$

$\Gamma(1) = \sum\{\emptyset\} = 2^0 = 1$

$\Gamma(\{1\}) = 2^1 = 2$

$\Gamma(2) = 2^0 + 2^1 = 3$

$\Gamma(\{\{1\},0\}) = 2^2 + 2^0 = 5$

$\Gamma(\{\{\{1\},0\},2,1,0\}) = 2^5 + 2^3 + 2^1 + 2^0 = 43$

$\Gamma^{-1}$ is the Mostoski function $\mathrm{mos}_{\omega,E}$.

答:
I guess Kunen deems terms and formulas over the language $\{\in\}$ as hereditarily finite sets, but it is often convenient to regard them as natural numbers. Hence he needs to 'encode' formulas into natural numbers.

I am not sure that why Kunen imposed the conditions you asked, like $r<s\to \Gamma(v_r)<\Gamma(v_s)$, but you may agree that imposing these conditions makes the encoding simpler and more 'reasonable.' (Nobody would think the encoding of $v_{7}$ is larger than that of $v_{3900}$.) Or you may find the reason for this requirement in a proof of propositions later.

Since $\Gamma$ is already fixed, we have to view this requirement as for variables, not encoding functions. That is, we have to choose variables $v_i$ under the restriction $\Gamma(v_i)<\Gamma(v_j)$ for $i<j$. (One possible way to choose variables is choosing $v_i$ whose rank is larger than that of all $v_j$ for $j<i$.)

For your second question, since every formula is hereditarily finite, $\Gamma$ encodes other formulas well if we determine what $v_i$ are. For example, $\Gamma$ encodes $v_i=v_j$ with no problem.

In your example, however, you have to clarify what $x$ and $y$ are, in terms of $v_i$. (The natural choice would be $v_0$ and $v_1$.)

I agree that $\varphi$ should be $\varphi_i$ as commented. Then $D(A,i,\vec{b})$ means a definable subset of $A$ by $i$th formulas with parameters $\vec{b}$.


答:
You can rewrite by expressing the integrand as $x^3$ where $x^2>y$ and $xy$ where $y>x^2$.

$$I=\int_0^1\int_0^{x^2} x^3\,dy\,dx+\int_0^1\int_{x^2}^1 xy\,dy\,dx=\int_0^1x^5\,dx+\int_0^1 x\frac{1-x^4}2\,dx=\frac16+\frac14-\frac1{12}.$$

Title: Evaluate ${ \int_{0}^{ 1}{ \int_{0}^{1}{x\max(x^2,y)\,dy} \,dx} }$

Body:
${ \int_{0}^{ 1}{ \int_{0}^{1}{x\max(x^2,y)\,\text{d}y} \,\text{d}x} }$

How would you sketch the graph for this integral?


问:
Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$.
Let $x = \sqrt[3]{3} + \sqrt{5}$

Notice that $(x+y)^2 = x^2 + 2xy + y^2$

Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$

Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$

Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$

Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$

Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.

Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.

But I just don't how get it from both the ways I provide above.

Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?

Body: If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}


答:
Set of algebraic numbers form a field, so the sum of two algebraic numbers $\sqrt[3]{3} + \sqrt{5}$ is again an algebraic number.

To find the minimal polynomial, you want to continue in either of the two ways that you have started.

If it is the first way,
you have to take the suitable powers of $x^2-5$ and continue until getting a rational polynomial.

Title: Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$.

Body:
Let $x = \sqrt[3]{3} + \sqrt{5}$

Notice that $(x+y)^2 = x^2 + 2xy + y^2$

Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$

Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$

Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$

Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$

Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.

Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.

But I just don't how get it from both the ways I provide above.

Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?


问:
Continuous function in the complex plane
There are two define regions:
$K_1: z \in C: 1<|z-4i|<3$ and $K_2: z \in C: 5<|z-4i|<7$

I've to determine whether exists continuous $f: K_{1}\rightarrow R$ that $f(2i)=\frac{9\pi}{2}$ and $ z= |z|e^{f(z)}$ for all $z\in K_1$. If it's exists I've to find the function.

Similiraly, I've to determine whether exists continuous $f: K_{2}\rightarrow R$ that $f(-2i)=-\frac{9\pi}{2}$ and $ z= |z|e^{f(z)}$ for all $z\in K_2$. If it's exists I've to find the function.

I'm kind of lost in this question. I tried to use branch point and branch cuts, but I did not succeed in solving it.
Any ideas on how to approach this problem will help a lot!

答:
Denote $f(z) = u(z)+i v(z)$ where $u,v$ are real maps. The equality
$$ z= |z|e^{f(z)}$$ implies $e^{u(z)} = 1$ for all $z \neq 0$, and therefore $u \equiv 0$ as $u$ has to be continuous.

You then get the equation $$ z= |z|e^{i v(z)}$$ where $v$ is a real function. The question is then related to the complex logarithm.

As $K_1 \subseteq \mathbb C \setminus \mathbb R_{\le 0}$ and the equation $z = \vert z \vert e^{f(z)}$ holds if we assume $f(2i)=i\frac{9\pi}{2}$, we can find a continuous $f$ satisfying the requirements.

For $K_2$, it is not possible because you can find paths included in $K_2$ with non zero winding number around zero.


答:
You have two mistakes.

$(x+1)^2+(2y-4)^2 = 4$ should be

$\displaystyle \small \frac{(x+1)^2}{4}+\frac{(y-2)^2}{1} = 1$

$c^2 = a^2-b^2 = 3 \implies c = \pm \sqrt3$

You also write $c^2 = a^2+b^2$ but that is not correct.

Given the center, the foci will be $(-1 \pm \sqrt3, 2)$

Title: Conic Sections Checking my solution: My Foci seams to be bigger than my Vertices (I think there is something wrong with my solution)

Body: Which one is my error? Pls help!


问:
Contour integral of $\displaystyle\oint_{c} \frac{e^{ix}}{1+z^{2}} dz$
I want to find $\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx$

using

$$\oint_{c} \frac{e^{ix}}{1+z^2} \cdot d z$$

over the upper half of a large semicircle enclosing $z=i$  (not $\oint_{c} \frac{e^{iz}}{1+z^{2}} dz$)

For very large semicircle,  the above integral   reduces to $$\oint \frac{e^{i x}}{1+z^{2}} d z=\int_{-\infty}^{\infty} \frac{e^{i x}}{1+x^{2}}dx=\int_{-\infty}^{\infty} \frac{\cos x d x}{1+x^{2}}+i \int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x$$

So we have $$\oint \frac{e^{i x}}{1+z^{2}} d z=\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x+i\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x$$

Also by residue theorem $$\oint \frac{e^{i x}}{1+z^{2}} d z=2 \pi i b_{1}$$  and then

$$b_{1}=\lim _{z \rightarrow i} \frac{e^{i x}(z-i)}{1+z^{2}}=\frac{1}{2 i}$$ ( as $z$--> $i$ , the real part $x$--> $0$ )

Therefore we have$$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x+\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x=\frac{2 \pi i}{2 i}$$

And hence $$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x=\pi$$  which is incorrect.

Why am I getting wrong result. I suspect the above highlighted portion.
Can anyone please help me. Thank you so much

答:
Your error is that you did not change consequently the real variable $x$ to complex variable $z$. As a consequence your integrated function is wrong as seemingly $x$ is independent of $z$.

As a result you have wrongly computed the residue at $z=i$. The correct value is:
$$
\operatorname{res}(f(z),i)=\lim _{z \rightarrow i} \frac{e^{i \color{red}z}(z-i)}{1+z^{2}}=\frac{e^{-1}}{2 i}.
$$

UPDATE (after OP question was edited clarifying that symbol $x$ used in the integrated function is assumed to mean the real part of $z$)

You cannot integrate the function $$f(z)=\frac{e^{i\operatorname{Re}z}}{1+z^2}$$
with the help of the residue theorem because the function is holomorphic at no point of the complex plane, whereas the residue theorem requires it to be holomorphic at any point of the part of the complex plane surrounded by the contour except for a finite set of isolated points.

To avoid this you shall instead use the meromorphic function
$$
f(z)=\frac{e^{iz}}{1+z^2}
$$
as suggested above.


问:
Given a transient Markov chain with state space $S$, what is $\lim_{n \rightarrow \infty} \sum_{y \in S} \Pi^n (x, y)$?
Given an irreducible and transient Markov chain, we know that $\lim_{n \rightarrow \infty} \Pi^n (x, y) = 0$ for all pairs $x, y \in S$.

However, we also know that $\lim_{n \rightarrow \infty} \sum_{y \in S} \Pi^n (x, y) = 1$ too. How is this limit possible, since every term in the summation converges to $0$ a.s.?

答: In a finite state space irreducibale MC you cannot have transient states. If the state space is infinite there is no contradiction. For example,  $ \sum\limits_{k=1}^{n} \frac 1 n =1$ and each term tends to $0$. That is not a contradiction becuase you cannot take the limit inside the sum.


问:
Sum of non uniformly continuous and uniformly continuous is non uniformly continuous?
I have this question:

Prove/disprove: $f(x)=x\ln x+\sin(x^3)\frac1x$ is uniformly continuous in $(0,\infty)$.

I know that $\sin(x^3)\frac1x$ is uniformly continuous and $x\ln x$ is non uniformly continuous. I tried proving that the sum of uniformly continuous function and non uniformly continuous function is non uniformly continuous, but is this even true?

答:
The sum of two uniform continuous maps $f,g$ is uniformly continuous.

If $f$ is uniformly continuous and $g$ is not, $f+g$ can't be uniformly continuous as otherwise $g = (f+g) - f$ would be uniformly continuous.


问:
Proving Laplace expansion using exterior algebra
Let $A = (a_{ij})$ be an $n\times n$ matrix with entries in a ring $R$, $M$ a free $R$-module of rank $n$ with an ordered basis $(e_i)_{i \leq n}$ and $\phi\colon M\to M$ is an endomorphism which $A$ represents (such that $\phi(e_j) = \sum_{i = 1}^n a_{ij}e_i$ fop all $j$). Then $\bigwedge^n M$ has rank $1$ with a basis $\{e_1\wedge ... \wedge e_n\}$ and there is a unique scalar $r \in R$ such that $$\left(\bigwedge^n \phi\right)(e_1\wedge ... \wedge e_n) = \phi(e_1)\wedge ... \wedge \phi(e_n) = r(e_1\wedge ... \wedge e_n).$$ This scalar is precisely the determinant $\det(A)$ of $A$. It can be easily shown that the value of $\det(A)$ doesn't depend on the choice of $M$. I take this as the definition of a determinant. From this other definitions can be deduced as theorems, including the formula
$$\det(A) = \sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)a_{\sigma(1)1}...a_{\sigma(n)n}.$$

I understand that there is a shorter proof of the Laplace expansion of a determinant using the definition of a determinant in question. However, a book I consulted has a serious gap in the proof. I will state what I want to prove:

Let $A = (a_{ij})$ be an $n\times n$ matrix with entries in a commutative ring $R$. Denote by $A_{ij}$ the $(n-1)\times(n-1)$ matrix obtained from $A$ by deleting the $i$-th row and the $j$-th column of $A$. Then, for all $i$,
  $$\det(A) = \sum_{j = 1}^n (-1)^{i + j} a_{ij}\det(A_{ij}).$$

答:
I would like to present here an alternative categorification.

Step 1: defining the general adjugation process.
We start by taking any free $R$-module $M$ of finite rank $n$, and observing that the natural map

$$\bigwedge^kM\otimes_R\bigwedge^{n-k}M\xrightarrow{\wedge}\bigwedge^nM$$

is a perfect pairing. We thus have a canonical isomorphism

$$\bigwedge^kM\xrightarrow{\phi}\text{Hom}\left(\bigwedge^{n-k}M,\bigwedge^nM\right).$$

So for any $\alpha\in\bigwedge^kM$ we have $\phi(\alpha)=\phi_\alpha$ is the morphism taking $\beta\in\bigwedge^{n-k}M$ and sending it to $\alpha\wedge\beta$:

$$\phi_\alpha(\beta)=\alpha\wedge\beta.$$

This enables us to define adjugate morphisms, i.e. for any endomorphism $f:\bigwedge^{n-k}M\longrightarrow\bigwedge^{n-k}M$ one has a morphism $f^\dagger:\bigwedge^kM\longrightarrow\bigwedge^kM$ verifying

$$\forall\alpha\in\bigwedge^kM,\beta\in\bigwedge^{n-k}M,\quad
\alpha\wedge f(\beta)=f^\dagger(\alpha)\wedge\beta.$$

Step 2: fundamental lemma concerning general adjugates. We prove the following for any endomorphism $f:M\longrightarrow M$ and any $k=0,\ldots,n$

$$\left(\bigwedge^{n-k}f\right)^\dagger\circ\bigwedge^kf=
\det(f)\cdot\text{Id}_{\wedge^kM}.$$

Indeed, write $F$ for the above composition and take any $\alpha\in\bigwedge^kM,\beta\in\bigwedge^{n-k}M$:

$$F(\alpha)\wedge\beta=\left(\bigwedge^kf\right)(\alpha)\wedge
\left(\bigwedge^{n-k}f\right)(\beta)=
\left(\bigwedge^nf\right)(\alpha\wedge\beta)=\det(f)\alpha\wedge\beta=
\left(\det(f)\alpha\right)\wedge\beta,$$

thus, considering this equality is made for arbitrary $\beta\in\bigwedge^{n-k}M$, we have $F(\alpha)=\det(f)\alpha$.

Step 3: coming back down to the $k=1$ case and decategorification.
The classical adjugate is defined for the special case $k=1$, where we identify $M=\bigwedge^1M$:

$$f^{ad}:=\left(\bigwedge^{n-1}f\right)^\dagger.$$

We now wish to show that going from $f$ to $f^{ad}$ corresponds at the matrix level, to going from a matrix $A$ to its adjugate. For this consider $M=R^n$ with canonical basis $e_1,\ldots,e_n$ and take $A$ the matrix of $f$ in this basis:

$$f(e_j)=\sum_{i=1}^nA_{i,j}e_i.$$

By a classic result on exterior products, we know that if we set

$$\epsilon_k=e_1\wedge\ldots\wedge e_{k-1}\wedge\widehat{e_k}
\wedge e_{k+1}\wedge\ldots\wedge e_n,$$

then $\epsilon_1,\ldots,\epsilon_n$ forms a basis for $\bigwedge^{n-1}M$. Remark then that

$$e_k\wedge\epsilon_l=\left\{\begin{array}{ll}
(-1)^{k-1}e_1\wedge\ldots\wedge e_n & \text{if }k=l\\
0 & \text{otherwise}
\end{array}\right.$$

We now wish to show how does the map $\left(\bigwedge^{n-1}f\right)^\dagger$ act on $V$. For this it suffices to show what happens on each base vector $e_k$, that we wedge with $\epsilon_l$:

\begin{align*}
\left(\bigwedge^{n-1}f\right)^\dagger(e_k)\wedge\epsilon_l&=e_k\wedge
\left(\bigwedge^{n-1}f\right)(\epsilon_l)\\
&=e_k\wedge f(e_1)\wedge\ldots\wedge f(e_{l-1})\wedge f(e_{l+1})
\wedge\ldots\wedge f(e_n)\\
&=(-1)^{k-1}\det(A^{(k,l)})e_1\wedge\ldots\wedge e_n.
\end{align*}

where $A^{(k,l)}$ is the matrix obtained from $A$ by eliminating the $k$-th row and $l$-th column (this is just Maxime's second commutative square). So we clearly have

$$\left(\bigwedge^{n-1}f\right)^\dagger(e_k)=\sum_{l=1}^n(-1)^{k+l}\det(A^{(k,l)})e_l$$

thus showing that the matrix of $f^{ad}$ in the canonical basis is $A^{ad}={}^t\left((-1)^{k+l}\det(A^{(k,l)})\right)_{1\leqslant k,l\leqslant n}$, i.e. the classical adjugate matrix of $A$. By step two, we get that $A^{ad}A=\det(A)I_n$. This shows the classic Laplace expansion, by considering the computation of every $(i,i)$ coefficient in the product.

Remark: Suppose $\det(A)\in R^\times$, then we wish to show that $A$ is invertible.
An application of Nakayama's lemma shows that if an endomorphism of fintely generated modules is surjective, then it is injective, and thus an isomorphism, so $A^{ad}$ is invertible considering $A^{ad}A=\det(A)I_n$ and since $\det(A)I_n$ is invertible. But then $\det(A)^{-1}A$ is a right inverse and by a simple computation, we can show that it is also a left inverse to $A^{ad}$. Thus we have also shown that $A$ is invertible.

Edit: In fact, Nakayama isn't necessary to conclude. Since $\det(A)\in R^\times$, then $\det(A^{ad})\det(A)=\det(A)^n$, so $\det(A^{ad})=\det(A)^{n-1}$ is also invertible, so the identity $(A^{ad})^{ad}A^{ad}=\det(A^{ad})I_n$ shows that $A^{ad}$ also has a left inverse. A simple computation shows that the two inverses are identical.

Any comments and errata are welcome. I hope this helps.


问:
Find the limit of function
I do not know how to approach this problem. Is there something related to integrating?

The function $f$ is defined on $[0,1]$, continuous, and positive. Find the following limit:
$$\lim_{n\to \infty} \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}$$

答:
Take log on both sides,

Let $L_n= \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}\implies \ln L_n=\frac {\sum_{r=1}^n\ln f(\frac rn)}{n}$, which corresponds to Riemann sum for the integral $\int_0^1 \ln f(x) dx$.

$\lim(\ln L_n)=\int_0^1 \ln f(x) dx\implies L_n\to \exp (\int_0^1 \ln f(x) dx)$

In this case, you may convert to Riemann sum like this also: Replace $1/n$ by $dx$, put $r/n =x$. When $r=1, x=1/n\to 0$ as $n\to \infty$ (This gives lower limit for the above integral) and when $r=n, x=n/n\to 1 $ as $n\to \infty$ (for the upper limit).

Title: Find the limit of function


答:
You have a product and powers. These are nasty; we like sums, sums are nice.
So take the logarithm to convert products into sums:
$$
\log \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}
= \frac{1}{n}\sum_{k=1}^n \log f\!\left(\frac{k}{n}\right) \tag{1}
$$
Now, this looks like a Riemann sum, applied to the continuous, well-defined (can you see why?) function $g = \log \circ f$. In particular,

$$
\lim_{n\to \infty}
\frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right) =
\int_0^1 g(x)dx \tag{2}
$$
Now, since $\exp$ is continuous, we have
$$
\lim_{n\to \infty} e^{\frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right)}
= e^{\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n g\!\left(\frac{k}{n}\right)}
= e^{\int_0^1 g(x)dx} \tag{3}
$$
i.e., your limit is
$$
\lim_{n\to \infty} \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}
= \boxed{e^{\int_0^1 \log(f(x))dx}}
$$

Sanity check: let's verify on a couple examples!

$f$ is constant: if $f(x)=a>0$ for all $x$, then the limit is easily shown to be $\sqrt[n]{a^n}=a$, and $e^{\int_0^1 \log(a) dx} = e^{\log a} = a$. This checks out.

$f(x)=x+1$ (this is continuous, positive, etc.). Then
$$
\sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}
= \sqrt[n]{ \frac{n+1}{n}\cdot\frac{n+2}{n}\cdots \frac{2n}{n} }
= \sqrt[n]{ \frac{(2n)!}{n!\cdot n^n} }
$$
and,, e.g., using Sterling's inequality, you can check the limit is $4e^{-1}$.

On the other hand,
$e^{\int_0^1 \log(x+1) dx} = e^{\log 4-1} = 4e^{-1}$. This  again checks out.

Title: Find the limit of function

Body:
I do not know how to approach this problem. Is there something related to integrating?

The function $f$ is defined on $[0,1]$, continuous, and positive. Find the following limit:
$$\lim_{n\to \infty} \sqrt[n]{f \left( \frac{1}{n} \right) \cdot f\left(\frac{2}{n}\right)\cdot f\left(\frac{3}{n}\right)\cdots f\left(\frac{n}{n}\right)}$$


问:
minimal polynomial of $7^{th}$ root of unity

Let $\alpha=\zeta_7+\zeta_7^2+\zeta_7^4$. What's the minimal polynomial of $\zeta_7$ over $\mathbb{Q}(\alpha).$

I have managed to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and $[\mathbb{Q}(\zeta):\mathbb{Q}]=6$. By tower rule, the minimal polynomial has degree 3. As long as we find a polynomial with degree 3 such that $\zeta$ is a root, then we are done. But I got stuck in finding such polynomial. Any hint is appreciated.

答: Put $\beta=\zeta^{-1}+\zeta^{-2}+\zeta^{-4}$ and note $\beta=-1-\alpha\in\mathbb{Q}(\alpha)$. Now form the cubic with roots $\zeta, \zeta^2,\zeta^4$. The sum of the roots is $\alpha$, the sum of the roots two at a time is $\beta$, the product of the roots is 1.


问:
Definition of multivariable function classes doubt
I was given this example by my teacher:

$f(x)=\begin{cases}
      0 & (x,y) = 0 \\
      \frac{x^2y^2}{x^4 + y^2} & (x,y)\neq (0,0)\
   \end{cases}
$

He proved diferentiablity for all $\mathbb{R}^2$.

Then he went to prove wether the function was class $C^1$ or not. He used $y=x^2$ to prove it wasn't. And that's what put me in doubt.

Doesn't diferentiability in $\mathbb{R}^2$ imply it's partial derivatives are continuous, therefore $C^1$. Am I missing something in this example?

答: No, differentiability does not imply that the partial derivatives are continuous (it's the other way around). Take, for instance$$f(x,y)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$


问:
$\lim_{t\to\infty}a\sqrt{\log \log t}+bt$
What is the reason for

$\lim_{t\to\infty}a\sqrt{\log \log t}+bt=\lim_{t\to \infty}bt\quad\quad  a,b\in \mathbb R$

How can I show this? de L'Hospital?

答: Assuming $b \neq 0$ you can argue as follows: L'Hopital's Rule shows that $\frac {\log \log t} {t^{2}} \to 0$. Now write $a\sqrt {\log \log t}+bt$ as $bt[1+ \frac  a b\sqrt {\frac {\log \log t} {t^{2}}}]$.


问:
Can the notions of "path" and "connectedness" be generalized to discrete spaces?
A path in a topological space X is defined as a continuous function f from the closed interval [0,1] into X. A connected space X is defined as a space that cannot be represented as the union of two or more disjoint non-empty open subsets.

Neither of these definitions can be satisfied for discrete spaces. But what I'm wondering is whether there's a natural (and non-disjunctive) generalization of either of these notions that applies to both discrete and non-discrete spaces, and that generates results such as the following: in ℤ, the sequence (1, 2, 3, 4) is a "path" but the sequence (1, 3, 4, 2) is not, and the subspace {1, 2, 3, 4} is "connected" but the subspace {1, 2, 5, 6} is not.

I can see how one might construct analogues of the notions of path and connectedness that apply only to discrete spaces. But what I'm having trouble with is finding a generalization that subsumes the original notions, while also being satisfiable for discrete spaces in the manner above. I should mention also that I'd be happy to find a generalization that applies only to metric spaces.

Thanks for your help, and apologies for any unclarity in the question.

答:
Any form of connectedness for topological spaces must by definition be a topological invariant, i.e. must be invariant under homeomorphism. More generally, if non-homeomorphic spaces can be distinguished by some property $\mathcal P$, then $\mathcal P$ must refer to other structural components than the topologies of spaces.

In your question you implicitly refer to the natural ordering on $\mathbb Z$. This has nothing to do with topology.

Thus, there is no concept of connectedness for discrete spaces. In fact, discrete spaces are uniquley determined by their underlying sets and there is no set-theoretic property other than the cardinality allowing to distinguish between sets. Therefore you must consider sets endowed with an additional structure not related to topology.

By the way, in your examples you do not consider sets , but sequences in ordered sets and it seems that you call such a sequence a path if its elements are arranged in ascending order. This is okay as a definition, but it has nothing to do with topology.


问:
Matrix sums along rows and columns
I have two vectors $x$ and $y$ of length $m$ and $n$ respectively.

I'm trying to determine whether a matrix $N$ of dimension $(m \times n)$ exists, such that the sum along its i'th row is equal to $x_i$ and the sum along every column $j$ (possibly with elements weighted by $\omega_{ij}$) is equal to or greater than $y_j$

Can anyone give me some pointers?

答:
The matrix may not exist, for instance if $\mathbf{x} = (0,0)$ and $\mathbf{y} = (1,1)$.

To try to find a solution in the general case (with exact equality to the $y_j$), I would go about it as follows. We want to find an $m \times n$ matrix $N$ such that
$$\begin{matrix}
n\left\{ \vphantom{\begin{pmatrix}
\mathbf{1}_m & \mathbf{0}_m & \cdots & \cdots & \cdots & \mathbf{0}_m \\
\mathbf{0}_m & \mathbf{1}_m & \mathbf{0}_m & \cdots & \cdots & \mathbf{0}_m \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\mathbf{0}_m & \cdots & \cdots & \cdots & \mathbf{0}_m & \mathbf{1}_m \end{pmatrix}} \right. \\
m\left\{ \vphantom{\begin{pmatrix}
1 & \mathbf{0}_{m-1} & 1 & \mathbf{0}_{m-1} & \cdots & \cdots\\
0 & 1 & \mathbf{0}_{m-1} & 1 & \mathbf{0}_{m-1} & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\mathbf{0}_{m-1} & 1 & \cdots & \cdots & \mathbf{0}_{m-1} & 1
\end{pmatrix}} \right.
\end{matrix}
\begin{pmatrix}
\mathbf{1}_m & \mathbf{0}_m & \cdots & \cdots & \cdots & \mathbf{0}_m \\
\mathbf{0}_m & \mathbf{1}_m & \mathbf{0}_m & \cdots & \cdots & \mathbf{0}_m \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\mathbf{0}_m & \cdots & \cdots & \cdots & \mathbf{0}_m & \mathbf{1}_m \\
1 & \mathbf{0}_{m-1} & 1 & \mathbf{0}_{m-1} & \cdots & \cdots\\
0 & 1 & \mathbf{0}_{m-1} & 1 & \mathbf{0}_{m-1} & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\mathbf{0}_{m-1} & 1 & \cdots & \cdots & \mathbf{0}_{m-1} & 1
\end{pmatrix} \text{vec}(N)
= \begin{pmatrix} \mathbf{x} \\ \mathbf{y}
\end{pmatrix}$$
where $\text{vec}(N)$ is the vectorization of $N$ obtained by stacking its columns together, and $\mathbf{0}_k$ and $\mathbf{1}_k$ are the row vectors of zeros and ones of size $k$ respectively. The left-hand matrix is known, of size $(n+m, nm)$, and so is the right-hand vector of size $(n+m)$, so it remains only to find a solution to this problem now written in standard form $Ax = b$ with $A$ and $b$ known, where the unknown $x$ stands for $\text{vec}(N)$. Use any solver (eg Matlab, NumPy, SymPy) to find a solution (if it exists).

Title: Matrix sums along rows and columns


问:
Laurent polynomial ring which is a PID
Let $R$ be a commutative ring with identity. We know that a polynomial ring $R[x]$ is a PID if and only if $R$ is a field. (You may check the proof here.) The ring of formal series $R[[x]]$ satisfies a similar property with a similar proof. Meanwhile, when it comes to the Laurent polynomial ring, it is true that if $R$ is a field, then $R[x,x^{-1}]$ is a PID. (The proof can be found here.)

Is it true that $R[x,x^{-1}]$ is a PID if and only if $R$ is a field? (In other words, does the converse hold?)

Yes, unfortunately we can't use the factor ring $R[x,x^{-1}]/\langle x\rangle$ since $x$ is a unit in the Laurent polynomial ring. Does the converse still hold in this case? Thank you very much.

答:
Assuming that $R$ is Noetherian, the Krull dimension is given by
$$
\dim (R[x,x^{-1}])=1+\dim (R).
$$
Assume that $R[x,x^{-1}]$ is a PID. Then its Krull dimension is $1$, so that it follows $\dim(R)=0$. If $R$ is a commutative ring with $1$, then $R$ is a field.

Reference:

Dimension of a quotient ring

Edit: If $R$ is not Noetherian, then the Krull dimension can be bigger and we have (also for $R[x,x^{-1}]$ instead of $R[x]$)
$$
\dim R+1\le \dim R[x] \le 2\dim R+1.
$$
See the papers by Seidenberg. If $1+\dim (R)\le \dim (R[x,x^{-1}])=1$, then again $\dim (R)=0$.

Title: Laurent polynomial ring which is a PID


问:
Category of abelian groups has enough injectives
This is Weibel's exercise 2.3.2, posted on MSE before here but without answer (and not following the hint given in the book)..

For $A$ an abelian group, $I(A)$ is the product of copies of $\mathbb{Q}/\mathbb{Z}$ indexed by $\text{Hom}_{\mathbf{Ab}}(A, \mathbb{Q}/\mathbb{Z})$.
I know that $I(A)$ is injective, and I wish to show first that there is a canonical map $A \to I(A)$.

Following the hint, a map $f: a\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ for which $f(a) \neq 0$ can be sending $a$ to the coset $[1/n]$ for an integer $n > 1$, and the kernel is $a n \mathbb{Z}$.
I can still construct a map $a \mathbb{Z} \to I(A)$ given by $a \mapsto f(a) \times f(a) \times \dots$, and so there is also a map $A \to I(A)$ lifting this map.
However the map $a \mathbb{Z} \to I(A)$ has nonzero kernel, so it is not clear that the lifted map is an injection.

On the other hand, I could also send $a \in a\mathbb{Z}$ to $f_1(a) \times f_2(a) \times \dots$ where the $f_i \in \text{Hom}_{\mathbf{Ab}}(A, \mathbb{Q}/\mathbb{Z})$ index $I(A)$. But again, since $f_i$ sends $a$ to some coset $[1/n_i]$, the kernel is $an_i\mathbb{Z}$, so the kernel of this map would look like $a n_1 \mathbb{Z} \times an_2 \mathbb{Z} \times \dots$, which also isn't correct - this is where I am stuck, I am not sure what else can be tried.

答:
Attempted solution which I think is correct:

The map $\coprod_{a \in A} a\mathbb{Z} \to A$ is an injection, and I can define also a map $\phi: \coprod_{a \in A} a\mathbb{Z} \to I(A)$ by sending $a$ to $f_1(a) \times f_2 (a) \times \dots$, where the $f_i$ run over all elements in $\text{Hom}(A, \mathbb{Q}/\mathbb{Z})$ as I wrote in my question (this map is actually canonical too).
Since we found a map $f: a\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ for which $f(a) \neq 0$, then the kernel of $\phi$ is zero, and so the lifted map $A \to I(A)$ is also an injection.

I think this is correct, but I am not 100% sure, so I won't accept this as an answer until I can confirm its correctness.


问:
Does the sequence sequence of bounded linear transformations have a supremum?
This is an argument I used in one of the questions that I had in my Functional Analysis Exam.
I have a sequence of transformations $T_n$ from a Banach Space $X$ to another Banach space $Y$ where $T_n \in B(X,Y)$. I used the argument that since, $$T_n \in B(X,Y) \rightarrow \exists M_n>0 \hspace{0.15cm} s.t \hspace{0.15cm}||T_n(x)||_Y \leq M_n||x||_X$$
And since, we know that all these transformations are bounded, implies the set $S = \{M_n : \forall n\}$ is also bounded. And hence, it has a supremum $M$. Hence, $\forall n \hspace{0.15cm}||T_n(x)||_y\leq M||x||_X \hspace{0.15cm} \forall x\in X$.

Can anyone explain what is the flaw in my reasoning?

答: What makes $(M_n)$ bounded? Take $T_n=nI$ (and $M_n=n$) to see where you are wrong.


问:
An exercise about compactness

Let $X$ be the set of all bounded sequences of complex numbers. That is, $\left\{x_{n}\right\} \in X$ $\iff$ sup $\left\{\left|x_{n}\right|: n \geq 1\right\}<\infty .$ If $x=\left\{x_{n}\right\}$ and $y=\left\{y_{n}\right\},$ define $d(x, y)=\sup \left\{\left|x_{n}-y_{n}\right|: n \geq 1\right\} .$ Show that for each $x$ in $X$ and $\epsilon>0, \bar{B}(x ; \epsilon)$
is not totally bounded although it is complete. (Hint: you might have an easier time of it if you first show that you can assume $x=(0,0, \ldots$.)

I don't have any way to start.

答:
Hint: First consider the case $x = (0, 0, \ldots)$. Define a sequence $(y_n)_{n \in \mathbb{N}}$ in $\bar{B}(x; \varepsilon)$ by

$$y_n(m) = \begin{cases}
\varepsilon, & \text{if m = n} \\
0, & \text{otherwise}.
\end{cases}$$

Now consider $d(y_n, y_m)$ for $n \neq m$ and conclude that $(y_n)_{n \in \mathbb{N}}$ cannot have a convergent subsequence.

Title: An exercise about compactness


答:
Following you're analysis of the indices, the Galois group of the extension $\mathbb{Q}(\zeta_7)\supseteq \mathbb{Q}(\alpha)$ is a (the) subgroup of order $3$ of $\text{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong (\mathbb{Z}/7\mathbb{Z})^\times\cong\mathbb{Z}/6\mathbb{Z}$. This subgroup is the cyclic subgroup generated by the element $$
\sigma\in\text{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q}), \,\,\,\sigma(\zeta_7)=\zeta_7^2
$$
A great trick which often helps in finding polynomials (or really any type of object in the context of galois/representation theory) which are invariant under the action of some group of transformations is taking the 'average'.
More explicitly, if we're interested in finding a polynomial in $\mathbb{Q}(\alpha)[x]=(\mathbb{Q}(\zeta_7))^{\langle \sigma \rangle}[x]$ which has $\zeta_7$ as a root, we can construct it by taking the product of the orbit of $(x-\zeta_7)$ under $\langle \sigma \rangle$: $$
p(x)=(x-\zeta_7)(x-\sigma(\zeta_7))(x-\sigma^2(\zeta_7))=(x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4)
$$ Because of the construction, clearly $p(x)$ is invariant under the action of $\langle \sigma \rangle$ (thought of as an automorphism of the ring $\mathbb{Q}(\zeta_7)[x]$ by acting on coefficients) and hence the coefficients of the polynomial $p(x)$ must all lie in $(\mathbb{Q}(\zeta_7))^{\langle \sigma\rangle}=\mathbb{Q}(\alpha)$.

At this point, I think it's quite satisfying to verify this last statement directly:$$
p(x)=(x-\zeta_7)(x-\zeta_7^2)(x-\zeta_7^4)= $$ $$=x^3-(\zeta_7+\zeta_7^2+\zeta_7^4)x^2+(\zeta_7^3+\zeta_7^5+\zeta_7^6)x-1=x^3-\alpha x^2 + \frac{\alpha^2-\alpha}{2}x-1.
$$ And low and behold the coefficients in fact are in $\mathbb{Q}(\alpha)$.

Hope this helps :)

POST: bare in mind that this is all possible because $\mathbb{Q}(\zeta_7)\supseteq \mathbb{Q}$ is Galois.

Title: minimal polynomial of $7^{th}$ root of unity

Body:
Let $\alpha=\zeta_7+\zeta_7^2+\zeta_7^4$. What's the minimal polynomial of $\zeta_7$ over $\mathbb{Q}(\alpha).$

I have managed to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and $[\mathbb{Q}(\zeta):\mathbb{Q}]=6$. By tower rule, the minimal polynomial has degree 3. As long as we find a polynomial with degree 3 such that $\zeta$ is a root, then we are done. But I got stuck in finding such polynomial. Any hint is appreciated.


问:
Lower bound for the cardinality of a filter
I know that an infinite sigma algebra has to be uncountable and I was wondering if a similar statement holds for filters. I think that I can say that given an infinite set $X$, the principal filter generated by $x$ must be uncountable.

I think that given a finite set $F\subseteq X$ where $\vert X\vert \geq \aleph_0$, the principal filter $\mathcal{F}_F$ generated by $F$ must be uncountable, because we can construct an injection from $2^{X\setminus F}$ to $\mathcal{F}_F$.

I also think that going by this logic, if $\mathcal{F}$ is a set filter over $X$ such that $A\in \mathcal{F}$ and $\vert X\setminus A\vert\geq \kappa$, then $\vert \mathcal{F}\vert \geq 2^\kappa$.

I have the following questions:

$(i)$ Do these arguments seem right?

$(ii)$ Can we also conclude that if $\mathcal{F}$ is an infinite filter, then is $\mathcal{F}$ necessarily uncountable?

答:
Your first argument is correct. But in order to prove that every infinite filter is uncountable, using the first argument, you'd have to prove that in every infinite filter there is a co-infinite set.

Unfortunately, for your argument, this isn't the case. Easily, $\{A\subseteq X\mid X\setminus A\text{ is finite}\}$ is a filter. And indeed, if $X$ is countable, it is a countable filter.


问:
Seifert-van-Kampen and free product with amalgamation
I would like to apply Seifert van Kampen to a simple example taken from Wikipedia: I have $X = S^2$ and $A = S^2 - n$, where $n$ is the north pole and $B = S^2 - s$, where $s$ is the south pole.

According to my understanding, which might be wrong, Seifert van Kampen tells me that $\pi_1(X) = \pi_1(A) *_{\pi_1(A \cap B)} \pi_1(B)$, where the right hand side is the free product with amalgamation.

$A \cap B$, the sphere minus the two points has a fundamental group isomorphic to $\mathbb{Z}$.

The free product with amalgamation of two groups $G, H$ is $ G * H / N$, where $N$ is the smallest normal subgroup in $G * H$ (according to the Wikipedia entry about free product with amalgamation).

Note : I am aware that in the sphere example both $G$ and $H$ are trivial and so quotienting them with anything will be trivial again. This question is not about actually computing the fundamental group of $S^2$!

In the sphere example, this means I have to find the smallest normal subgroup of $\mathbb{Z}$.

Question 1: Is my understanding of Seifert van Kampen correct?

Question 2: What is the smallest normal subgroup of $\mathbb{Z}$?

As for question 2, what do you think about $\lim\limits_{k \rightarrow \infty} k\mathbb{Z}$?

Thanks for your help.

答:
The only thing I'd like to add is that in general, if $i:A\hookrightarrow X$ is the inclusion, then $i_\#:\pi_1(A)\rightarrow\pi_1(X)$, defined by $i_\#:[\alpha]\mapsto[i\circ\alpha]$, usually isn't an inclusion.

For example, for the inclusion $i:\mathbb{S}^1 \hookrightarrow\mathbb{B}^2$, the homomorphism $i_\#$ isn't an inclusion, since $\pi_1(\mathbb{S}^1)\cong\mathbb{Z}$ and $\pi_1(\mathbb{B}^2)\cong\{1\}$.

Even if $i$ maps $[\alpha]$ to $[\alpha]$, the first equivalency class of $\alpha$ is computed via homotopies in $A$, while the second via homotopies in $X$! The codomain of the loop is vital. However, if $A$ is a retract of $X$ (i.e. $A\!\subseteq\!X$ and $\exists$ continuous $r:X\rightarrow A$ with $r|_A=id_A$), then $i_\#:\pi_1(A)\rightarrow\pi_1(X)$ is an inclusion.

Hence in the formulation of Seifert van Kampen theorem, when one has the inclusions $i:X_1\!\cap\!X_2\hookrightarrow X_1$, $j:X_1\!\cap\!X_2\hookrightarrow X_2$, the theorem states (under the hypotheses) that
$$\pi_1(X)\cong\pi_1(X_1)\ast\pi_1(X_2)/\langle\!\langle i_\#([\alpha])j_\#([\alpha])^{-1};\;[\alpha]\!\in\!\pi_1(X_1\!\!\cap\!\!X_2)\rangle\!\rangle$$
and NOT that
$$\pi_1(X)\cong\pi_1(X_1)\ast\pi_1(X_2)/\langle\!\langle [\alpha][\alpha]^{-1};\;[\alpha]\!\in\!\pi_1(X_1\!\!\cap\!\!X_2)\rangle\!\rangle.$$


问:
Existence of random variables $D,E$ such that $g(D,E)=A$, where $A$ and $g(B,C)$ are identically distributed
Questions:

For random variables with $A,B,C$ and some function (probably Borel-measurable and bounded/integrable) $g$, if $A$ and $g(B,C)$ have the same distribution, then does there necessarily exist random variables $D$ and $E$ such that $A=g(D,E)$ (at least almost surely)?

If this doesn't work in general, then are there conditions where such $D$ and $E$ will exist?

Context:

Sum of iid Bernoulli is binomial. But also conversely, for every binomial random variable $X$, what I understand is that there exists iid Bernoulli's whose sum equals $X$. Like, not just equal in distribution but really at least almost surely equal. Also, I don't think this is trivial because it's not like binomial distribution's definition necessarily depends on having a definition of bernoulli distribution. Well at least I recall in undergrad that we learned Bernoulli after Binomial.

This question (This random variable $Z$ seems to have the same distribution as $\min{X,Y}$ as $|X-Y|$, for $X,Y \sim \operatorname{Unif}(0,1)$.) and that question (Why does $\min(X,Y)$ and $|X-Y|$ have the same distribution when $X,Y\sim U(0,1)$?).

For $R=\min\{X,Y\}$ and $Q=|X-Y|$, I was able to find random variables $I,J,G,H$ such that $R=|I-J|$ and $Q=\min\{G,H\}$. However, neither $(I,J)$ nor $(G,H)$ necessarily have the same relationship as $(X,Y)$ (which is that $X$ and $Y$ are iid). Also, each of $I,J,G,H$ isn't necessarily Unif(0,1). I was thinking to ask later on if there were such $(I,J)$ or $(G,H)$, but I wanted to settle the existence of $D,E$ 1st.

答:
The answer to your first question is no, at least not in general. The answer to the second question might be a bit more subtle. Of course a necessary and sufficient condition would be that $D,E$ exists and $A=g(D,E)$, but that isn't really useful.

Let me argue why the answer to the first question is no. Consider the probability space $\Omega = \{0,1,2\}$ with the $\sigma$-algebra $\mathcal{P}(\Omega)$ and the probability measure given by
$$\mathbb{P}(\{x\}) = \begin{cases} \frac14 & ,x=0 \\ \frac12 & ,x=1 \\ \frac14 &,x=2 \end{cases}.$$
Furthermore consider the random variable $X:\Omega \rightarrow \mathbb{R}$ defined as $X(\omega)=\omega$. Clearly $\mathbb{P}(X=x)=\mathbb{P}(\{x\})$ for $x=0,1,2$ and therefore $X\sim \operatorname{Binomial}(2,\frac12)$. Now, since our probability space $\Omega$ is quite small, there isn't many ways to construct $\operatorname{Bernoulli}(\frac12)$ variables, in fact there are only 2 ways:
$$Y(\omega)=\begin{cases} 1 &,\omega =0,2 \\ 0 &,\omega =1 \end{cases} \quad \text{and} \quad Z(\omega)=\begin{cases} 1 &,\omega=1 \\ 0 &,\omega=0,2\end{cases}.$$
Now first of all $Y$ and $Z$ are not independent, since $Z=1-Y$, and furthermore $Y+Z = 1 \neq X$. Hence it is clear that $X$ cannot be written as a sum of $i.i.d.$ $\operatorname{Bernoulli}(\frac12)$ variables.


问:
Formulas for lcm of three integers
I know that both of the following formulas equal to $\text{lcm}(a,b,c)$, but how can we show that

$\dfrac{abc \cdot \gcd(a,b,c)}{\gcd(a,b) \cdot \gcd(a,c) \cdot  \gcd(b,c)} = \dfrac{abc }{\gcd(ab,ac,bc)}$ ?

答:
Hint:

WLOG the highest exponents of prime $p$ that divides $a,b,c$ be $A,B,C$

WLOG $A\ge B\ge C\ge0$

So, the  highest exponents of prime $p$ that divides the numerator of the LHS will be $$A+B+C+C$$

For the denominator of the left hand side  the highest exponent $=B+C+C$

For the denominator of the right hand side the highest exponent  $$=B+C$$

The above will hold true for any prime that divides $abc$

Title: Formulas for lcm of three integers


问:
nonlinear partial-differential-equations
I have a problem with  partial differential equation

$$u_xu_y=xy$$
where $$u(0,y)=-y.$$

Therefore $x_0(s)=0$,$y_0(s)=s$,$u_0(s)=-s.$
I wrote my equation as $F(x,y,u,p,q)=pq-xy=0$ and defined the characteristic equations:
$$\frac{dx}{dt}=q$$
$$\frac{dy}{dt}=p$$
$$\frac{du}{dt}=2pq$$
$$\frac{dp}{dt}=y$$
$$\frac{dq}{dt}=x.$$

I know that
$F(x_0(s),y_0(s),u_0(s),p_0(s),q_0(s))=0$ and $p_0(s)x_0'(s)+q_0(s)y_0'(s)=u_0'(s)$,
then
$p_0(s)q_0(s)=0$ and $q_0(s)=-1$, therefore $p_0(s)=0.$ Then $p(t,s)=0$ and $q(t,s)=-1$ and then I solved characteristic equations: $x(t,s)=-t, y(t,s)=s, u(t,s)=-s.$ But u(x,y)=-y is not a solution, and I don't know what I did wrong.

答:
I solved it and now I have:
$$x(t,s)=-\frac{1}{2}(e^t-e^{-t})$$
$$q(t,s)=-\frac{1}{2}(e^t+e^{-t})$$
$$y(t,s)=\frac{s}{2}(e^t+e^{-t})$$
$$p(t,s)=\frac{s}{2}(e^t-e^{-t})$$
and $u(t,s)=-\frac{s}{4}(e^{2t}+e^{-2t})-\frac{s}{2}.$

I checked the initial conditions and I think that it is a good solution, but I saw that

$$u(x,y)=xy-\frac{s}{2}$$ and I have a problem with $s$, because the function should depend only on $x$ and $y$.


问:
A basic question on the derivative of a continuous function
Is the following condition necessary for the existence of derivative of a continuous function at point $x$: </p>

<p>$$\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^-}\frac{f(x)-f(x+h)}{h}$$</p>

答: If you change the denominator of the second limit to $-h$ then yes but as it stands that's actually not true in general (think of the signs of the numerator and denominator)</p>


答:
You can solve this limit also avoiding de l'Hospital. In particular you have only to use the hierarchy of infinities:
$$\lim_{t\to\infty}a\sqrt{\log{(\log{t})}}+bt=\lim_{t\to\infty}t\Big(\frac{a[\log({\log{t})}]^{1/2}}{t}+b\Big)$$
Now since the logarithm goes to infinity slower than $t$, or in general than $t^k$ (with $k$ positive) so it remains:
$$\lim_{t\to\infty}bt$$

Title: $\lim_{t\to\infty}a\sqrt{\log \log t}+bt$

Body:
What is the reason for

$\lim_{t\to\infty}a\sqrt{\log \log t}+bt=\lim_{t\to \infty}bt\quad\quad  a,b\in \mathbb R$

How can I show this? de L'Hospital?


问:
Antiderivative of a polynomial is always a polynomial?
At first it seems like a simple enough question, but I couldn't quickly find any proofs on this. I would assume that they are, and moreover that the degree is raised by $1$. This is just due to the reverse power rule. Consider what happens if we take the antiderivative of $$a_n x^n + \cdots+ a_1x+a_0,$$

where $a_i$ are real coefficients. The degree would be raised and we would get another zero-degree term in the form of the constant of integration $c$. Is this a total triviality or are there hidden complications?

Edit : $x\in \Bbb{R}$

答:
The antiderivative of a polynomial
$$\int a_nx^n+ \cdots + a_1x+a_0 dx=\frac{a_n}{n+1}x^{n+1}+\cdots+\frac{a_1}{2}x^2+a_0x+C$$
I am assuming that $a_i$ are real numbers. We have $n\geq 0$ is an integer, thus $i+1>0$ for all $0\leq i\leq n$. Thus $\frac{a_i}{i+1}$ is a real number. Therefore, result is a polynomial. (One would need an additional condition to solve for $C$).

Edit: The argument above does not necessarily hold for a disconnected domain. Consider the following example:

$$f:[0,1]\cup [1,2] \rightarrow \mathbb{R} \hspace{25pt} \text{ defined by }f(x)=3x^2+1$$
Then, a possible antiderivative is:
$$F(x)=\left\{\begin{array}{cc} x^3+x & x \in [0,1] \\ x^3+x+2 & x \in [1,2] \end{array}\right.$$
which is not a polynomial.


问:
Splitting of square diagram in chain complexes
Let us consider a square in chain complexes over a field $k$  \begin{array}{ccc}A & \xrightarrow{f} & B \\ \downarrow{g} & & \downarrow{h} \\ C & \xrightarrow{k} & D\end{array}

If A is chain equivalent to a direct sum of chain complexes $\bigoplus_i A_i$ such that each of the summands $A_i$ is concentrated in exactly one degree but any two of $A_i$ s are in different degrees, then is it true that the square is equivalent (in the diagram category) to a direct sum of the squares with top left corner is $A_i$?

Any help will be appreciated.

答:
I don’t think so. Colimits in functor categories are computed objectwise, so we would need not only $A$ but also $B,C,D$ to be direct sums with the same indices and the morphisms would be restricted.

What I think your question is motivated by is the following. Consider the square
$$\tag{$\star$}\begin{array}{ccc}
\bigoplus\limits_{i\in I} A_i & \rightarrow &B\\
\downarrow && \downarrow\\
C &\rightarrow &D
\end{array}$$
By precomposition with the inclusion $A_j \hookrightarrow \bigoplus_i A_i$ we obtain a commutative square of the form
$$\tag{$\star(j)$}\begin{array}{ccc}
A_j& \rightarrow &B\\
\downarrow && \downarrow\\
C &\rightarrow &D
\end{array}$$
for every $j$.

Conversely, if we have a square of the form $\star(j)$ for every $j$, denote the top and left morphisms by $b_j: A_j \rightarrow B$ and $c_j:A_j \rightarrow C$ respectively. By the universal property of the direct sum, they determine unique morphisms $b:\bigoplus_i A_i \rightarrow B$ and $c: \bigoplus_i A_i \rightarrow C$. Moreover uniqueness of the universal property implies that these morphisms yield a commutative square of the form $\star$.

So there is a bijective correspondence between squares of the form $\star$ and a collection of squares of the form $\star(j)$, but it is not just direct sums in the functor category.