问:
$\operatorname{Spec} A$ as a colimit
I was reading the lecture notes on algebraic geometry and came across with the following definition:

Let $A$ be a commutative ring with unit, then $\operatorname{Spec} A$ is defined as the collection of all ring homomorphism $A\rightarrow K$, where $K$ is some field, and where we identify two maps $f:A\rightarrow K$ and $f':A\rightarrow K'$ if there exists a field homomorphism $\alpha:K\rightarrow K'$ such that $f'=\alpha \circ f$.

And then the author went on: this is a rather categorical definition, and in fact we could rephrase it as $$\operatorname{Spec} A=\operatorname*{colim}_{K\in \mathfrak{Field}} \operatorname{Hom}_{\mathfrak{Ring}}(A,K).$$
where $\mathfrak{Field}$ is the category of fields and $\mathfrak{Ring}$ is the category of rings.

I am trying to prove this but not sure if I am on the right track: Let $\mathfrak{C}$ be the subcategory of the coslice category of $\mathfrak{Ring}$ over $A$ where the objects are $f:A\rightarrow K$ and the morphisms are $\alpha: f\rightarrow f'$ such that $f'=\alpha \circ f$. But then I don't see how to proceed from there. Is there any literature about this definition of spectrum of a ring rather than the usual definition as the set of all prime ideals of $A$? Thank you.

答:
I'd like to briefly mention a slight set theoretic issue. The category of $\mathbf{Fields}$ is not small, so strictly speaking this colimit could "diverge" and no longer yield a set as an answer. In fact, the definition you gave in the second paragraph is a bit sketchy as well. The collection of all maps $A \longrightarrow K$ for all fields is definitely not a set, so modding out by an equivalence relation is not strictly in set theory. I won't dwell on these issues much, as they can be rectified (as below), and are beyond the scope of this question and of my understanding. But suffice it so say that this is a reason to use the usual definition of $\operatorname{Spec}$ as prime ideals.

I'll show this by directly proving that $\operatorname{Spec} A$ as you have it defined in the second paragraph satisfies the universal property of the below colimit. What is that universal property? Well it means that any cocone out of the indexing diagram is a map out of the colimit. More formally, let's fix some set $X$. Then a cocone from the indexing diagram to $X$ is a collection of commutative diagrams

for all maps $i: K \longrightarrow L$ of fields. To say that $\operatorname{Spec} A$ is a colimit of this diagram means that for any such cocone $(X, \{f_K\}_{K \in \mathbf{Fields}})$, we have the following commutative diagram, again for every such map $i: K \longrightarrow L$:

We can see therefore that it is, strictly speaking, incorrect to say that $\operatorname{Spec} A$ is the colimit of this diagram. Instead, we need to say that the cocone $(\operatorname{Spec} A, \{\alpha_K\}_{K \in \mathbf{Fields}})$ is  the colimit. In other words, that this cocone is initial among all cocones. In particular, we need our maps $\alpha_K: \operatorname{Hom}(A, K) \longrightarrow \operatorname{Spec} A$ for all fields $K$. Well per your second paragraph, the elements of $\operatorname{Spec} A$ are define as equivalence classes of such maps. So we take $\alpha_K(f) = [f]$, where the brackets denote the equivalence class of $f: A \longrightarrow K$ viewed in $\operatorname{Spec} A$.

Now we have to check that this is actually a cocone, i.e. that for $i: K \longrightarrow L$ that $\alpha_L \circ h_i = \alpha_K$. Indeed, take some $f: A \longrightarrow K$. The right hand side is $\alpha_K(f) = [f]$ as above. The left hand side is $\alpha_L(i \circ f) = [i \circ f] = [f]$ by the very definition of the equivalence relation you gave! I hope now that you can sort of see what @AlexKruckman in the comments meant. The construction you gave in the second paragraph fits the colimit like a glove, and that is precisely because it is the standard construction of a colimit of sets.

EDIT: For this next paragraph, see the comments for a discussion of an error. I assumed $f \sim i \circ f$ was the equivalence relation, but it merely generates it. The proof I wrote below can be adapted to this correct equivalence relation but I'll leave it as is. The key ideas remain the same.

Anyways, we now have two tasks. Construct a map $F: \operatorname{Spec} A \longrightarrow X$ that fits in the diagram and proving that it is unique. We need, of course, for any $f \in \operatorname{Hom}(A, K)$ to have $f_K(f) = F(\alpha_K(f))$. So we'll do the naive thing and literally define $F([f]) = \alpha_K(f)$, given a representative $f:A \longrightarrow K$. But why is this well defined? It comes down again to commutativity - this time of the cocone with vertex $X$. Indeed, suppose $[f] = [f']$ where $f \in \operatorname{Hom}(A, K)$ and $f' \in \operatorname{Hom}(A, L)$. Then by definition, there is a map $i: K \longrightarrow L$ such that $i \circ f = f'$. In other words, $h_i(f) = f'$. Now, recall that we assumed the outer triangle commuted, i.e. $f_L \circ h_i = f_K$. Hence, $f_L(h_i(f)) = f_K(f)$ so $f_L(f') = f_K(f)$. And by my definition, $F([f]) = f_K(f)$ and $F([f']) = f_L(f') = f_K(f)$. Thus, $F([f]) = F([f'])$ whenever $[f] = [f']$ and we have well definition. Furthermore, we need to show that $F$ actually commutes in the diagram. This again fits like a glove. $F(\alpha_K(f)) = F([f]) = f_K(f)$, so $F \circ \alpha_K = f_K$ for all fields $K$.

So far we have found that given any cocone to $X$, we have a map from the cocone of $\operatorname{Spec} A$ to the cocone of $X$. All we nee to do now is show uniqueness. Indeed, suppose we had some other $G: \operatorname{Spec} A \longrightarrow X$ that fits into that commutative diagram. Then $G \circ \alpha_K = f_K$ so $f_K(f) = G(\alpha_K(f)) = G([f])$. This is exactly how $F$ was defined, so $F = G$. Once again, this all came to us for free. The hard work was done in the definition you gave.

Now, let's return to the classical definition of $\operatorname{Spec} A$ as the collection of prime ideals. As @WhatsUp pointed out in the comments, the relation between the two is that every prime ideal $\mathfrak p$ of $A$ gives us a map $A \longrightarrow A/\mathfrak p \longrightarrow \operatorname{Frac}(A/\mathfrak p)$, meaning the quotient field of the domain $A/\mathfrak p$. This is often called the residue field of $\mathfrak p$ and is denoted $k(\mathfrak p)$. On the other hand, every map $f: A \longrightarrow K$, $K$ a field, will have $\ker(f)$ a prime ideal by the first isomorphism theorem. These very closely related. $\ker(A \longrightarrow k(\mathfrak p)) = \mathfrak p$ and given $f: A \longrightarrow K$, $k(\ker(f))$ has an embedding into $K$ via extending the inclusion $A/\ker(f) \longrightarrow K$ to the field of fractions. So in particular, this gives us inverse bijections between your $\operatorname{Spec} A$ and the usual prime spectrum. We take $\mathfrak p \mapsto [A \longrightarrow k(\mathfrak p)]$ and $[f] \mapsto \ker(f)$. This is well defined because a map between fields in necessarily an injection and therefore does not affect the kernel. In other words, we can restrict the index diagram from ranging over the entire category of $\mathbf{Fields}$ to ranging over the full subcategory whose objects are the residue fields $k(\mathfrak p)$. This also resolves the set theoretic issue. The colimit over this restricted diagram is the same as the colimit ranging over the entire large category. Since the restricted diagram is small, the set theoretic issues can go away. This idea of restricting to a subdiagram is related to the general notion of a final functor. I think that the subset of the category of open sets of a space consisting of basic open set is another such example, but to be completely honest I'm not too familiar with final functors so please take what I'm saying with a grain of salt.

Title: $\operatorname{Spec} A$ as a colimit


问:
Does the converse of uniform continuity -> Preservance of Cauchy sequences hold?

We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.

Now, I would like to ask the following question:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.

How do we go about doing it?

答: You have to prove is that if $(x_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ and $(y_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ converge to the same limit $x \in [0,1]$, then $(g(x_n))_{n \ge 0}$ and $(g(y_n))_{n \ge 0}$ have the same limit (you may define $z_{2n} = x_n$, $z_{2n+1} = y_n$, and use uniqueness of the limit). Then you can the deduce that $g$ is continuous on the compact $[0,1]$ (or can be extended to such a function), hence uniformly continuous.</p>

ContentLicense: CC BY-SA 3.0


问:
Zero, a mystery.

Is zero even?

Is zero a multiple of ANY number?

I did some reading, and I found the following:

for 1) I see a whole Wiki article saying that this is true, but I just can't understand why?

for 2) I think it is, but I don't have a precise definition, and the only thing I remember is that in the junior classes, I used to write that the smallest multiple of any number is the number itself and it was approved by my teachers...so a proof on 1) and 2) would be highly appreciated...
Thanks a Ton!

答: An integer $a$ is a multiple of an integer $b$ if there exists some third integer $c$ such that $ a = b \times c$. Hence since $0=m \times 0$, for any integer $m$, $0$ is a multiple of any number. In particular, a number is even if it is a multiple of $2$, so $0$ is even.</p>

Title: Zero, a mystery.


问:
Square root of a $3×3$ matrix
I was trying to find square root of this $3×3$ matrix
$$\begin{pmatrix}
12&8&31\\
7&14&28\\
9&35&6\\
\end{pmatrix}$$
I searched Wikipedia and some site for helpful information and also try to
$${ \begin{pmatrix}
a&b&c\\
d&e&f\\
g&h&i\\
\end{pmatrix} }^3 =
\begin{pmatrix}
12&8&31\\
7&14&28\\
9&35&6\\
\end{pmatrix}$$
Though I know that a similar question has been posted for a $2×2$ matrix, the workings are different so I'll need some help on this
Thanks

ContentLicense: CC BY-SA 4.0

Title: Square root of a $3×3$ matrix

Body:
And here is how to get that result with SageMath

sage: M = matrix(QQbar,3,3,[12,8,31,7,14,28,9,35,6])
sage: a,b=M.diagonalization()
sage: c=diagonal_matrix([x.sqrt() for x in a.diagonal()])
sage: b*c*~b
[ 2.994275474024927? + 0.151836493602799?*I  1.197978250858210? + 2.057588565728163?*I  3.000734901853716? - 2.183155335351262?*I]
[0.777514011271777? + 0.1253357713833934?*I  3.460596511446746? + 1.698468161085322?*I  2.737777768777000? - 1.802119184349875?*I]
[0.9410435413715063? - 0.201547233197786?*I  3.344874508620062? - 2.731235901474129?*I  2.789009650240784? + 2.897912794483287?*I]


问:
How to show that there is no positive rational number a such that $a^3$ = 2?
I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.

Can someone tell what exactly is going on here and how should I prove this?

答:
Your proof is on the right lines but there are no words in it. You need to explain what you are doing to your readers. Here is how I would prove this:

Assume, for the sake of contradiction, that there is a rational number $a$ such that $a^3=2$.

Since $a$ is rational, we can write it in the form $m/n$, where $m$ and $n$ are coprime integers.

This means that $$\frac{m^3}{n^3}=2 \, .$$

From this we see that $m^3=2n^3$, and so $m^3$ is even. That implies that $m$ is even, i.e. it can be written in the form $2p$, where $p$ is an integer. It follows that

$$\frac{8p^3}{n^3}=2$$
which implies $2n^3=8p^3$ and so $n^3=4p^3=2(2p^3)$. Since $n^3$ is even, $n$ must also be even. But earlier we said that $m$ and $n$ are coprime, which is a contradiction. Hence, our original assumption—that there is a positive rational number $a$ such that $a^3=2$—must be false, meaning that no rational number satisfies the desired property.


问:
Confused about a proof involving concave functions
This proof is from a book about linear programming; it is about interior point methods, but I think that might not be important for my question.

We have:

$$f_{\mu}(\mathbf{x}) = \mathbf{c}^T\mathbf{x} + \mu\sum_{i=1}^{m}\log(b_{i} - \mathbf{a}_{i}\cdot\mathbf{x}) $$

For vectors $\mathbf{c},\mathbf{x},\mathbf{a}_i$ for $i = 1,2,3,...,m$ and $\mu > 0$. The book claims that $f_{\mu}$ has a unique maximum inside the polytope defined by $A\mathbf{x} < \mathbf{b}$ where the $\mathbf{a_i}$'s are the rows (not the columns) of $A$; we also assume that this polytope is bounded. $A$ is an $m \times n$ real matrix of rank $m$; so the polytope is in $\mathbb{R}^n$.

I don't understand a line in the proof of the uniqueness claim. The books states that if there were two different maxima $\mathbf{x}$ and $\mathbf{y}$ then $f_{\mu}$ must be constant along the line segment $\mathbf{xy}$ because the polytope is convex so all of the line segment is inside the polytope and $f_{\mu}$ is concave.

The book then states that "since the logarithm is strictly concave, this can happen only if $A\mathbf{x} = A\mathbf{y}.$" This is the part I don't understand; why does the logarithm being concave imply that $A\mathbf{x} = A\mathbf{y}$? I don't have much experience with convexity.

答:
Suppose $a_i \cdot x \neq a_i \cdot y$ for some $i$, then you get a contradiction that $x$ and $y$ are global minima:
\begin{align}
f_\mu(0.5x + 0.5y) &= c^T(0.5x+0.5y)+\mu \sum_{i=1}^m \log(b_i-a_i\cdot(0.5x+0.5y)) \\
&= c^T(0.5x+0.5y)+\mu \sum_{i=1}^m \log(0.5(b_i-a_i\cdot x) + 0.5(b_i - a_i\cdot y)) \\
&> 0.5c^Tx+0.5c^Ty+0.5\mu\sum_{i=1}^m \log(b_i-a_i\cdot x)+0.5\mu\sum_{i=1}^m \log(b_i-a_i\cdot y) \\
& = 0.5f_\mu(x) + 0.5 f_\mu(y) \\
& = \max_x f_\mu(x).
\end{align}
The strict inequality comes from the strict concavity of the logarithm: if $b_i-a_i\cdot x \neq b_i-a_i\cdot y$, then the midpoint has a lower function value.


问:
Inequality containing trigonometric functions with absolute values
While practicing trigonometry online I came across this one exercise which was too hard for me to solve:

$$ (\tan{4x} + \cot{4x}-\frac{5}{2})(\cos(2x) - |\sin{x}|) > 0 $$

In my humble opinion this is kind of overkill, because it exceeds even WolframAlpha's computation time and even then the solutions are uncommon values.

Frankly I don't even know where to start, but I'd start by noticing this. Let's say we have two expressions divided by brackets. Our condition will be true if either both of them are $>0$ or both of them are $<0$.

Now where I would have trouble is taking the intersection of all those sets of real numbers, so I was thinking that there must be an easier way to do this.

Can anyone help?

答:
Hints:

For the first term set $t=\tan(4x)$ and you get a quadratic in $t$ to solve.

Then solve $\tan(4x)=t$ for the $2$ roots found.

For the second one set $s=\sin(x)$ and rewrite $\cos(2x)$ in function of $s^2$ another quadratic (well two in fact because of absolute value) to solve.

Then solve $\sin(x)=s$ for the $4$ roots found.

Find the minimal interval for describing the solution and extend all solutions by periodicity


问:
Solve ODE $xy+(3xy^2+x^2)y'=-1$
How to solve $xy+(3xy^2+x^2)y'=-1$ ?

I have tried all the possible ways to solve an ODE but could not solve this one.

答: Beginning with @AleksasDomarkas's rearrangement and an integration factor Ansatz $\mu=x^ay^b$ viz.$$df=f_xdx+f_ydy=0,\,f_x=x^{a+1}y^{b+1}+x^ay^b,\,f_y=3x^{a+1}y^{b+2}+x^{a+2}y^b,$$for such a function $f$ to exist we require$$0=\frac{f_{xy}-f_{yx}}{x^ay^b}=(b+1)x+\frac{b}{y}-3(a+1)y^2-(a+2)x.$$Clearly $b-a-1=0,\,b=0,\,a+1=0$ has solution $a=-1,\,b=0$, so$$\mu=\frac1x,\,f_x=y+\frac1x,\,f_y=3y^2+x.$$In particular, $f:=xy+\ln|x|+y^3$ satisfies $df=0$. This $f$ is locally constant, allowed to have different values either side of $x=0$.

Title: Solve ODE $xy+(3xy^2+x^2)y'=-1$


问:
Ring of continuous functions being an integral domain implies all continuous functions are constant?
Given a (nonempty) topological space $X$, if the ring of continuous functions from $X$ to the real numbers $\mathbb{R}$ (henceforth denoted as $C(X,\mathbb{R})$) is an integral domain, must it contain only constant functions?

Given any continuous function $f$, define $Z(f)$ to be the set of zeros of $f$. Then, it is easily seen that $Z(fg)=Z(f) \cup Z(g)$, and in particular, $fg=0$ if and only if $Z(f) \cup Z(g)=X$. Also, since $\{0\}$ is closed in $\mathbb{R}$ and $f$ is continuous, $Z(f)$ is closed in $X$.

So, if $X$ is not the union of any two proper closed sets (i.e. $X$ is an irreducible space), then  $C(X,\mathbb{R})$ is an integral domain. But in fact, since $\mathbb{R}$ is Hausdorff, any continuous real-valued function on $X$ must be constant, and so $C(X,\mathbb{R})$ is actually isomorphic to $\mathbb{R}$.

But perhaps, there might be a non-irreducible space $X$ for which $C(X,\mathbb{R})$ is an integral domain, and it may or may not be isomorphic to $\mathbb{R}$.

答:
The answer to the question in the title is yes.

If $f$ is a non constant, continuous function, let $a$ and $b$ be distinct points in the image of $f$.

Also let $g$ and $h$ be continuous functions from $\mathbb R$ to itself, with disjoint supports, and such that $g(a)\neq 0$ and $h(b)\neq 0$.  Then the functions
$g(f(x))$ and $h(f(x))$ are nonzero, but their product vanishes.


答: Assume that $a=p/q$, where $p$ and $q$ are not both even (or make the stronger assumption that $\gcd(p,q)=1$), and that $a^3=2$.  Then $(p/q)^3=2$, or $p^3=2q^3$.  Since the right-hand side is even, the left-hand side must be as well; and since the cube of an odd number is odd, $p$ must be even.  So $p=2p'$ for some integer $p'$.  Now $8(p')^3=2q^3$, or $q^3=4(p')^3$, and hence $q$ must be even as well.  This contradicts the original assumption, completing the proof.

Title: How to show that there is no positive rational number a such that $a^3$ = 2?

Body:
I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.

Can someone tell what exactly is going on here and how should I prove this?


Title: Inequality containing trigonometric functions with absolute values

Body:
While practicing trigonometry online I came across this one exercise which was too hard for me to solve:

$$ (\tan{4x} + \cot{4x}-\frac{5}{2})(\cos(2x) - |\sin{x}|) > 0 $$

In my humble opinion this is kind of overkill, because it exceeds even WolframAlpha's computation time and even then the solutions are uncommon values.

Frankly I don't even know where to start, but I'd start by noticing this. Let's say we have two expressions divided by brackets. Our condition will be true if either both of them are $>0$ or both of them are $<0$.

Now where I would have trouble is taking the intersection of all those sets of real numbers, so I was thinking that there must be an easier way to do this.

Can anyone help?

Body:
HINT:

$(\tan{4x} + \cot{4x}-\frac{5}{2})(\cos(2x) - |\sin{x}|) > 0$

$(\tan{4x} + \frac{1}{\tan{4x}}-\frac{5}{2})$$(1-2|sinx|^2 - |\sin{x}|)$ > 0

A= $\tan{4x} + \frac{1}{\tan{4x}}-\frac{5}{2}$

B=$1-2|sinx|^2 - |\sin{x}|$

So either (A <0 and B<0) or (A>0, B>0) , we can continue from here...


问:
Regarding convex functions that attain infimum at infinity
Let function $f: \mathbb{R}^n \to \mathbb{R}$ be continuously differentiable, lower-bounded and convex. Let $\nabla f$ be the gradient of $f$, i.e., $\nabla f = \begin{bmatrix} \frac{\partial f}{\partial x_1} \dots \frac{\partial f}{\partial x_n} \end{bmatrix}$. If $x^1,x^2,\dots$ is a sequence in $\mathbb{R}^n$ such that

$$\lim_{n\to \infty} f(x^i) = \inf_{x \in \mathbb{R}^n} f(x)$$

then is the following true?

$$\lim_{n\to \infty} \left\| \nabla f(x^i) \right\| = 0$$

where $\|\cdot\|$ is the usual Euclidean norm. If this is not true, what additional assumptions on $f$ can be added so that this is true?

Thank you.

答:
Take $n=2$ and consider $f(t_1,t_2) = \max(e^{-t_1}, |t_2|)$.  Being the maximum of two convex functions, this is convex.  The points $x^i = (i, 2 e^{-i})$ have $f(x^i) = 2 e^{-i} \to 0 = \inf f$, but $|\nabla f(x^i)| = 1$.

This example is not continuously differentiable, but can be smoothed.


问:
Question about product of invertible matrix
Let $A$ be an $n\,\rm{x}\,n$ matrix. Prove that there exists a matrix $B$ of the same size and $B$ is invertible such that $BA$ is upper-triangular.

答:
Can you make $A$ upper triangular by performing elementary row operations?

Can you write an elementary row operation as left-multiplication by a simple matrix?

Can you write a sequence of elementary row operations as left-multiplication by a single matrix (which matrix is the product of a bunch of single operation matrices)?


Body:
Assume $ a=\frac pq $ with $gcd(p,q)=1$

$$p^3=2q^3\implies$$

$$(p-q)(p^2+pq+q^2)=q^3 \implies$$

$$q|( p^2+pq+q^2)\implies$$
$$q|p$$


问:
Prove that $\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$
I'm trying to prove that:

$$\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$$

I've tried to prove this by induction and directly, without luck. Any help would be appreciated.

答:
HINT: Use the identity

$$\binom{n}m\frac1{m+1}=\binom{n+1}{m+1}\frac1{n+1}\,.$$

I’ve taken it a step further in the spoiler block below.

 $$\begin{align*}\sum_{m=1}^n(-1)^{m+1}\binom{n}m\frac1{m+1}&=\frac1{n+1}\sum_{m=1}^n(-1)^{m+1}\binom{n+1}{m+1}\\&=\frac1{n+1}\sum_{k=2}^{n+1}(-1)^k\binom{n+1}k\end{align*}$$


问:
Source is smooth if target is smooth and fibers are smooth of constant dimension
This is Vakil 25.2 E b)

We are to show that if $\pi: X \to Y$ is a morphism of $k$-varieties with $\dim X = m > \dim Y = n$ and the scheme-theoretic fibers over closed points are smooth of dimension $m - n$, then if $Y$ is smooth, so is $X$.

We are allowed to assume Theorem 25.2.2, but I do not see how this gives us much. We need to essentially show that the sheaf of differentials $\Omega_{X/k}$ is locally free of rank $n$. We have the cotangent exact sequence

$$\pi^*\Omega_{Y/k} \to \Omega_{X/k} \to \Omega_{X/Y} \to 0$$

but it seems all that we know that the the first sheaf is locally free of rank $n$. How can we conclude?

答:
Let me show the slightly stronger assertion that $\pi$ is smooth. As you already touched upon in the comments, there are essentially two things to show here, neither of which is trivial (at least to me): 1) that $\pi$ is flat, and 2) that $\Omega_{X/Y}$ is locally free of rank $m-n$. I never understood how the conormal sequence would help us with that, especially with 1), so I'll give a different argument.

Part 1. We prove that $\pi$ is flat. Let's first show that $X$ is regular, which is actually easy. Pick a closed point $x\in X$. Then $\dim\mathcal O_{X,x}=m$ by assumption, and in order for $\mathcal O_{X,x}$ to be regular, we need to show that the maximal ideal $\mathfrak m_{X,x}$ can be generated by $m$ elements. The image of $X$ under $\pi$ is a closed point $y\in Y$ (see this question). We know that $\mathcal O_{Y,y}$ is regular of dimension $n$ by assumption on $Y$, and $\mathcal O_{X,x}/\mathfrak m_{Y,y}\mathcal O_{X,x}$ is regular of dimension $m-n$ by assumption on the fibres of $\pi$. Hence $\mathfrak m_{X,x}/\mathfrak m_{Y,y}\mathcal O_{X,x}$ can be generated by $m-n$ elements, whereas $\mathfrak m_{Y,y}$ can be generated by $n$ elements. This shows that $\mathfrak m_{X,x}$ can indeed be generated by $m$ elements, as desired.

Now that we know $X$ is regular, flatness of $\pi$ follows by a miracle: Since flatness of ring extensions can be checked locally at maximal ideals, it suffices to check flatness of $\pi$ at closed points. That is, we have to check that $\mathcal O_{X,x}$ is flat over $\mathcal O_{Y,y}$ for all $x$ and $y$ as above. This follows from the following lemma.

Lemma (miracle flatness theorem). Let $R\rightarrow S$ be a local morphism between noetherian local $(R,\mathfrak m_R)$ and $(S,\mathfrak m_S)$. If $R$ is regular, $S$ is Cohen-Macauly, and $\dim S=\dim R+\dim S/\mathfrak mS$, then $R\rightarrow S$ is flat.

Proof sketch. See [Stacks Project, Tag 00R4] for a proof in general. However, in our concrete situation $R=\mathcal O_{Y,y}$ and $S=\mathcal O_{X,x}$, we can give a simpler proof (or rather the same proof with less technicalities). Let $f_1,\dotsc,f_n$ be generators of $\mathfrak m_R$ (the $f_i$ are necessarily a minimal system of generators since $n=\dim R=\dim Y$). As seen above, in our concrete situation we can find $g_1,\dotsc,g_{m-n}\in\mathfrak m_S$ such that the $f_i$ and $g_j$ generate $\mathfrak m_S$ (and are thus again a minimal system of generators).

We do induction on $n=\dim R$. If $n=0$, then $R$ is a field and everything is clear. So assume $n>0$. Since we've chosen minimal systems of generators, $R/f_1R$ and $S/f_1S$ are regular again of dimensions $n-1$ and $m-1$, respectively. Hence the inductive hypothesis shows that $R/f_1R\rightarrow S/f_1S$ is flat. Hence so is $R\rightarrow S$ by the slicing criterion ([Vakil, Thm. 24.6.5]; note that the non-zero divisor hypothesis is satisfied since regular local rings are integral domains). $\square$

What to take away from this. There are some really cool flatness criteria: The local flatness criterion, the slicing criterion, the fibrewise criterion ([Stacks Project, Tag 00MP or [Vakil, Thm. 24.6.9]), and the miracle flatness theorem.

Part 2. We show that $\Omega_{X/Y}$ is locally free of rank $m-n$. For this we use the following lemma.

Lemma. Let $R$ be a reduced Jacobson ring and $M$ a finitely generated $R$-module. If there's an integer $d$ such that $\dim_{\kappa(\mathfrak m)}M/\mathfrak mM=d$ for all maximal ideals $\mathfrak m$, then $M$ is a locally free $R$-module of rank $d$.

Proof. Choose some maximal ideals $\mathfrak m$ and elements $m_1,\dotsc,m_d\in M$ whose images generate $M/\mathfrak mM$. By a version of Nakyama's lemma, there is an $f\in R\setminus \mathfrak m$ such that the $m_i$ generate $M_f$ over $R_f$. Replacing $M$ and $R$ by $M_f$ and $R_f$ (note that all maximal ideals of $R_f$ correspond to maximal ideals of $R$ by the Jacobson assumption), we may assume that the $m_i$ generate $M$. We claim that they do so freely. Indeed, if $\sum r_im_i=0$, then for every every maximal ideal $\mathfrak m'$ of $R$, the images of the $r_i$ in $\kappa(\mathfrak m')=R/\mathfrak m'$ must vanish, as $\dim_{\kappa(\mathfrak m')}M/\mathfrak m'M=d$. Hence the $r_i$ are contained in the intersection of all maximal ideals, which is $0$ since $R$ is reduced Jacobson. $\square$

To be able to apply the lemma, we must verify that $(\Omega_{X/Y})_x/\mathfrak m_{X,x}(\Omega_{X/Y})_x$ has dimension $m-n$ over the residue field $\kappa(x)$ for every closed point $x\in X$. But if $\pi(x)=y$ as above, then $(\Omega_{X/Y})_x/\mathfrak m_{Y,y}(\Omega_{X/Y})_x$ is free of rank $m-n$ over the local ring $\mathcal O_{X,x}/\mathfrak m_{Y,y}\mathcal O_{X,x}$ by assumption on the fibres of $\pi$. Hence the lemma applies and we conclude that $\Omega_{X/Y}$ is locally free of dimension $m-n$, as desired.

What to take away from this. Over a reduced ring $R$, a finitely generated module $M$ is already free if the vector spaces $M\otimes_R \kappa(\mathfrak p)$ have the same dimension for all $\mathfrak p\in\operatorname{Spec} R$ (essentially by the same argument as above). For the same to work with only the maximal ideals, we need the technical Jacobson condition.


问:
Prove that if $a$ is a local minimum of $f$ restricted to $[a,b]$, then $f'(a)\geqslant0$.
Let $f:(\alpha,\beta)\to\mathbb{R}$ be differentiable in $(\alpha,\beta)$
and let $a<b$ be such that $[a,b]\subset(\alpha,\beta)$. Prove that
if $a$ is a local minimum of $f$ restricted to $[a,b]$, then $f'(a)\geqslant0$.

Wouldn't $f'(a)\geqslant 0 $ anyways (if it is a local minimum)? I'm not sure if I understand the question correctly; could anyone clarify please?

答:
hint

$a $ local minimum of $ f $ at $[a,b]\implies$

$$\forall x\in (a,b]\; \;\;\frac{f(x)-f(a)}{x-a}\geqslant 0$$

$$\implies \lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}\geqslant 0$$
$$\implies f'(a)\geqslant0$$


问:
Showing $\int_{-1}^1\frac{m(2m-1)x^{2m-2}(1-x^{2m})+m^2x^{4m-2}}{(m^2x^{4m-2}+1-x^{2m})\sqrt{1-x^{2m}}}dx=\pi$, algebraically
Is there a nice algebraic way to solve the following geometrically motivated integral?

$$\int_{-1}^1\frac{m(2m-1)x^{2m-2}(1-x^{2m})+m^2x^{4m-2}}{(m^2x^{4m-2}+1-x^{2m})\sqrt{1-x^{2m}}}dx,$$ where $m$ is a positive integer.

In fact, this integral can be shown to be the integral of the curvature of the plane curve $x^{2m}+y^2=1, y\geq0$, which is the angle rotated by the tangent vector of the curve as it traverses along the curve. So this integral is $\pi$, but I would like to see some alternative, algebraic solutions.

答:
This is a supplement to @RiverLi's answer which makes @user10354138's statement that the substitution
\begin{align*}
\phi=\arctan\left(\frac{mx^{2m-1}}{\sqrt{1-x^{2m}}}\right)\tag{1}
\end{align*}
is obvious, somewhat plausible.

On the one hand we get from the definition of OPs plane curve
\begin{align*}
x^{2m}+y^2&=1, y\geq 0\\\\
\color{blue}{y}&\color{blue}{=\sqrt{1-x^{2m}}}
\end{align*}
On the other hand we have according to formula (5) the tangential angle $\phi$ of a curvature is
\begin{align*}
\tan\left(\phi\right)&=\frac{dy}{dx}\\
&=\frac{d}{dx}\sqrt{1-x^{2m}}\\
&\,\,\color{blue}{=-\frac{mx^{2m-1}}{\sqrt{1-x^{2m}}}}
\end{align*}
which makes the substitution (1) plausible.


问:
Computing limit of $(1+1/n)^{n^2}$
How can I compute the limit $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n^2}$? Of course $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n} = e$, and then $\left(1+\frac{1}{n}\right)^{n^2} = \left(\left(1+\frac{1}{n}\right)^n\right)^n$. Since the term inside converges to $e$, the whole thing is $e^n$, and can I conclude that the limit is infinity as $n\rightarrow\infty$? </p>

答:
This question is ancient but I thought I'd give another easy way to solve this that basically only requires that you know $(1+\frac{1}{n})^n \to e$.

By  $(1+\frac{1}{n})^n \to e>2$, there is an $N \in \mathbb{N}$ such that $n>N$ implies $(1+\frac{1}{n})^n > 2$. Then for $n> N$ we have $(1+\frac{1}{n})^{n^2} > 2^n$. For any large $M$, we thus have $(1+\frac{1}{n})^{n^2} > M$ for all $n> \max (N, \log_{2} M)$.

Title: Computing limit of $(1+1/n)^{n^2}$


答: Hint: Try to multiply by $n+1$ and note that ${n\choose m}\frac{n+1}{m+1}= {n+1\choose m+1}$. Then use Newton's binomial to expand $(1-1)^{n+1}$.

Title: Prove that $\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$

Body:
I'm trying to prove that:

$$\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$$

I've tried to prove this by induction and directly, without luck. Any help would be appreciated.


问:
Why does $\nabla\delta(t-\frac{r}{c}) = \frac{-\vec{e_r}}{c}\delta(t-\frac{r}{c})$?
I am trying to show that when you multiply $$G(t,\textbf{r}) = -\frac{1}{4\pi r}\delta(t-\frac{r}{c})  $$ by the d'Alembertian you get $$\square G = \delta(t)\delta(\textbf{r})$$ and immensely struggling, so any help would be much appreciated!

答:
I have been trying to answer your question since you posted it, this is my current state now. Not yet an answer! Maybe this helps you move forward while i struggle.

$$\nabla f(x-x')=\left(\partial_xf,\partial_yf,\partial_zf\right)=f'(x-x')$$

$$\square G(x-x')=\delta(x-x') $$

$$\phi=\delta(x-x_0,y-y_0)$$

This is the distribution sending $f$ to $f(x_0,y_0)$

$$\phi(f)=f(x_0,y_0)$$

rule:
https://math.stackexchange.com/a/1413826/738388

$$\langle f,\delta'_a\rangle=-\langle f',\delta_a\rangle=-f'(a)$$

$$(\delta')(f)=-\delta(f')=-f'(0)$$

$$\frac{d\phi}{dx}(f)=\frac{d}{dx}\delta(x-x_0,y-y_0)(f)=-\delta(x-x_0,y-y_0)(f')=-\frac{d}{dx}f(x_0,y_0))$$

https://math.stackexchange.com/a/445092/738388

$$\langle\nabla f,\cdot\rangle=df$$

$$\nabla\phi=\frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy$$

where $dx$ and $dy$ are differentials

$$\vec{\nabla}_x \delta^{(3)}(\vec{x}-\vec{x}')=-\vec{\nabla}_{x'} \delta^{(3)}(\vec{x}-\vec{x}').$$

where $\delta^{(3)}$ is the generalized 3 dimensional Dirac-delta.
Source https://www.physicsforums.com/threads/gradient-of-dirac-delta.880397/


问:
I don't understand how implication works in logic
The statement $(p \wedge q ) \Rightarrow (p \Rightarrow q)$ is a tautology, i.e., always true.

Now, imagine I am writing a proof and I want to show that $p \Rightarrow q$. I show that statement $p$ is always true, then I show that statement $q$ is always true, then I can end my proof and claim $p \Rightarrow q.$

This doesn't make sense to me. If I prove that $\sqrt2$ is irrational and then I prove that $7$ is odd, I don't understand how "$\sqrt2$ is irrational" $\Rightarrow$ "$7$ is odd".

Can someone help this disconnect I am having? Perhaps, you can provide information about what these logical statements actually mean and how they are used.

答:
The meaning of the conditional $\implies$ has a different meaning than in ordinary language.

$\sqrt{2}$ is irrational $\implies$ 7 is odd

Is a true statement precisely for the reasons you claimed. In ordinary language we use "if-then" or $P \implies Q$ statements almost exclusively in situations where there is some set of "distribution" over $P$; that is, $P$ can be true or false, and then we ask whether or not in every situation where $P$ is true does it also follow that $Q$ is true. There is also often a notion of "causality."

Again; crucially, in normal language when we say $P \implies Q$ there must be at least two possible scenarios: one in which $P$ could be true and one in which $P$ could be false. Furthermore, things get a little murkier here, but $P$ being true must be the "reason" that $Q$ is true.

The fact that this is the way the statement gets used in ordinary language is why the mathematical statement feels odd. Modal logic can more closely mirror the "natural language" usage. https://en.wikipedia.org/wiki/Modal_logic


答: This is one of the biggest issues with the conditional connective in propositional logic. This problem arises from the fact  that $p\rightarrow q\equiv \neg p\vee q$. All of the logical connectives in the propositional calculus produce a sentence whose truth value is only determined by the truth values of the propositions it connects. In general when we think or say $p$ implies $q$ we are also implies some form of causation which is not captured by the logical connective $\rightarrow$. This means that $\sqrt(2)\notin \mathbb{Q}\rightarrow 7\text{ is odd}$. is a true sentence even though there seems to be no real connection between the two things being true.

Title: I don't understand how implication works in logic

Body:
The statement $(p \wedge q ) \Rightarrow (p \Rightarrow q)$ is a tautology, i.e., always true.

Now, imagine I am writing a proof and I want to show that $p \Rightarrow q$. I show that statement $p$ is always true, then I show that statement $q$ is always true, then I can end my proof and claim $p \Rightarrow q.$

This doesn't make sense to me. If I prove that $\sqrt2$ is irrational and then I prove that $7$ is odd, I don't understand how "$\sqrt2$ is irrational" $\Rightarrow$ "$7$ is odd".

Can someone help this disconnect I am having? Perhaps, you can provide information about what these logical statements actually mean and how they are used.


问:
About quotient spaces of dual spaces
So when looking around online about dual spaces and quotient spaces I found that if $V$ is a Banach space, $Z \subseteq V^*$ is a closed subspace then there is a set $Z^T \subseteq V$ such that $V^* / Z \cong (V/Z^T )^*$. Could anybody perhaps explain to me what this set $Z^T$ exactly is? I could not easily find something about it online, so that is why I am asking if someone here maybe knows more about it.

答:
$Z^\top$ is the pre-annihilator of $Z$, i.e.,

$$Z^\top = \{x\in V\colon \langle f,x\rangle = 0\; (f\in Z)\}.$$

To see that indeed we have this identification in the case $Z$ is weak*-closed, note that we may define

Every functional $f\in (V/Z^\top)^*$ canonically defines $\Phi(f)\in V^* / Z$ by
$$\langle \Phi(f) + Z, v  \rangle = \langle f,v+Z^\perp\rangle \quad (v\in V, z\in Z^\top).$$

Title: About quotient spaces of dual spaces


问:
Zero, a mystery.

Is zero even?

Is zero a multiple of ANY number?

I did some reading, and I found the following:

for 1) I see a whole Wiki article saying that this is true, but I just can't understand why?

for 2) I think it is, but I don't have a precise definition, and the only thing I remember is that in the junior classes, I used to write that the smallest multiple of any number is the number itself and it was approved by my teachers...so a proof on 1) and 2) would be highly appreciated...
Thanks a Ton!

答:
$1$. It depends on your definition of even number. If you define the set of even numbers as
$$\mathbb{E} = \{x \in \mathbb{R}: x = 2n, \text{ where } n \in \mathbb{Z}^+\}$$
then $0$ is not even. However, if you define the set of even numbers as
$$\mathbb{E} = \{x \in \mathbb{R}: x = 2n, \text{ where } n \in \mathbb{Z}\}$$
then it is.

$2$. For the second question again it depends on your definition of multiple of a number. If you define, the set of multiples of $x$ as
$$M_x = \{y \in \mathbb{R}: y = z \times x, \text{ where }z \in \mathbb{Z} \}$$
then $0$ is indeed a multiple of any number since $0 = 0 \times x$ for any $x$ and in fact, $0$ is the only number present in all the sets, i.e., $\cap_{x \in \mathbb{R}} M_x = \{0\}$, i.e., $0$ is the only number that is a multiple of all numbers.

Title: Zero, a mystery.


答: You seem to be missing a derivative. Since by the chain rule$$\partial_if(r)=\frac{1}{2r}\partial_i(r^2)f^\prime(r)=\frac{x_i}{r}f^\prime\implies\nabla f=\vec{e}_rf^\prime$$for radially symmetric $f$ (in this case $f$ is a distributional measure rather than a function), the final result follows from another application of the chain rule, which gives the $-\tfrac1c$ factor.

Title: Why does $\nabla\delta(t-\frac{r}{c}) = \frac{-\vec{e_r}}{c}\delta(t-\frac{r}{c})$?

Body: I am trying to show that when you multiply $$G(t,\textbf{r}) = -\frac{1}{4\pi r}\delta(t-\frac{r}{c})  $$ by the d'Alembertian you get $$\square G = \delta(t)\delta(\textbf{r})$$ and immensely struggling, so any help would be much appreciated!


问:
basic forces question about the reaction force
For a object like this, is the upward force applied of 2000N counted as part of the total reaction force to gravity? or is it separate and therefore the object will motion upwards.

答:
$R$ is not a reaction to gravity, it is a force that is necessary to keep the object on a surface. $R$ is a reaction to the force with which the object pushes on the surface. What it means that the sum of the forces in the vertical direction is zero. You can get this from second principle and superposition principle (sum of the forces equals to mass times acceleration, and you say that the acceleration in the vertical direction is zero). You have $R$ and $2000N$ acting up, and $mg=700\cdot 10N=7000N$ acting down. $$R+2000N=7000N\\R=5000N$$
The action-reaction pair that corresponds to $700g$ is the force (at the center of the Earth) with which the Earth is attracted to the object (also 700g).


答:
Note that logical implication (aka ‘material conditional’) does not translate directly into our intuitive understanding of implication. We often think of implication as synonymous with causality, whereas the material conditional exists irrespective of causality. Mathematics isn’t a science, and its truths are almost always absolute (relative to the chosen axioms), which makes mathematical truth rather than causation its object of study. You can try and shift your thinking in this direction. Try and think of statements like $a\implies b$ as “if $a$ is true, then $b$ also happens to be true” rather than “$a$ causes $b$”.

Note that the main instrument of mathematics is proof, and as used in proofs, the material implication’s purpose is not to formalise some causal connection between mathematical facts, but rather to verify that one of the statements is indeed a fact. Implications of the sort you mention, where the consequent is tautologically true, are of little interest to mathematicians, even though they are correct owing to the truth-functional definition of the material conditional.

To help you come to terms with this definition, remember that the only situation where $a\implies b$ fails to be true is when $a$ is true and $b$ is false. When doing proofs and saying that $b$ follows from $a$, what we really mean is that whenever $a$ is true, $b$ cannot possibly be false (hence is also true). The material conditional guarantees that. And it is that guarantee that is its sole purpose, and what makes making logical deductions in proofs at all possible.


问:
How to Factorise these expressions?
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

答:
I am going go try and answer my question. So you have to use the forms right:
$(a + b)^2 = a^2 + 2ab + b^2$

and
$(a − b)^2 = a^2 − 2ab + b^2)$

So for $9m^2−24mn+16n^2$ I would use the the first form right? So $(a + b)^2 = 9m^2 + 16n^2 + 24mn$

So $(3m^2 + 4n^2)^2$ as $3^2 = 9$ and  $4^2 = 16$

$(a + b)^2 = x^2 + 16 = 8x$

So $(x^2 + 16)^2$

It's $(x + 4)^2$ as $x$ x $x = x^2$ and $4$ x $4$ $= 16$ with the $^2$.

Title: How to Factorise these expressions?


问:
$(n!)^{1/n}$ behaves like $n$ as $n$ approaches infinity

Riley Hobson Mathematical Methods in Physics Page $831$

Since $(n!)^{1/n}$ behaves like $n$ as $n \rightarrow \infty$, we find $\lim(1/n!)^{1/n} =0$

Now I don't understand how does $(n!)^{1/n}$ behave like $n$ as $n \rightarrow \infty$.

I have seen the graph on Desmos and it seems like it becomes a straight line after $n=-1$. I don't understand this idea of it approaching $n$.

答: More precisely, $(n!)^{1/n}$ behave like $n/e$ where $e$ is Euler's constant. One way to see this is Stirling's approximation. This still implies that $1/[(n!)^{1/n}]$ approaches zero.


答: Stirling's formula: $n!\approx \sqrt{2\pi n}(\frac{n}{e})^n$  so $(n!)^{1/n}\approx \frac{n}{e}$.

Title: $(n!)^{1/n}$ behaves like $n$ as $n$ approaches infinity

Body:
Riley Hobson Mathematical Methods in Physics Page $831$

Since $(n!)^{1/n}$ behaves like $n$ as $n \rightarrow \infty$, we find $\lim(1/n!)^{1/n} =0$

Now I don't understand how does $(n!)^{1/n}$ behave like $n$ as $n \rightarrow \infty$.

I have seen the graph on Desmos and it seems like it becomes a straight line after $n=-1$. I don't understand this idea of it approaching $n$.


问:
Equilateral Triangles On Sides of a Parallelogram
On the sides of a parallelogram $ABCD$, construct similarly oriented equilateral triangles $ABC_1, BCA_1, CDD_1$, and $DAB_1$. Let $N,P,Q$ and $M$ be the centers of the triangles, respectively. Show that the quadrilateral $MNPQ$ happens to be a parallelogram.

I think this is a really interesting problem. I found it in the geometric transformations unit in my textbook. The given hint is to use the (central) symmetry with center the intersection of the diagonals of the given parallelogram ($ABCD$). I am not sure I see it though. Any help would be appreciated. Thank you in advance!

答:
Let $ R(X)$ be the $180^\circ$ rotation of the point $X$ about the center of parallelogram $ABCD$.

Show that

$R(A) = C, R(B) = D.$

$R(C_1) = D_1, R(A_1) = B_1$.   -> Note this isn't as symmetric as the rest because of the naming in the question. If we used cyclic notation like $ ABC_1, CDA_1$, then we would have $ R(C_1) = A_1$. In fact, $R(ABC_1) = R(CDA_1)$.

$R(N) = Q, R(P) = M$.

$NQ, PM$ bisect each other about the center,

These points form a parallelogram.


问:
Probability of a set having no elements removed when randomly removing elements from *m* sets
I have n elements uniformly distributed over m sets, if I were to randomly remove x elements, what’s the number of sets y that have no elements removed. Specifically, what’s the relationship between x and y.

It is safe to assume that n is much greater than m.

答:
This assumes that the $x$ elements are selected without replacement, uniformly from the pool of $n$ elements.  Alternative interpretations would be selection with replacement (which simplifies the analysis), or selection of a random element of a random set, which is initially simple but also gets complicated when $x$ is large enough to eliminate an entire set).

Fix a particular target set $S$ and we compute the probability that all of its elements survive each of the $x$ removals.  In the first removal, there are $n$ elements and $n/m$ in $S$, so the probability of hitting $S$ is $1/m$.  On the second removal, conditionally on $S$ being untouched, there are $n-1$ elements and $n/m$ in $S$, so the probability of hitting $S$ is $n/((n-1)m)$.

In general, the conditional probability of $S$ surviving the $k$th removal intact is $1 - n/((n-k+1)m)$, so the probability of surviving all removals is:

$$\prod_{k=1}^x \bigg( 1 - \frac{n}{m(n-k+1)}\bigg).$$

By linearity of expectation, the expected value of $y$ is just $m$ times the above product.  Note that the product becomes identically zero when $x > n-n/m$, which is expected.  For $x$ very small compared to $n/m$, we can approximate $n-k+1$ with $n$ to simplify this to $\approx m ( 1 - 1/m )^x \approx m \exp(-x/m)$.  This also holds for large $x$ in the case that we draw with replacement.


问:
Showing a functor that preserves coequalizers is faithful
While studying for an algebra exam I came across this problem:

Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and suppose that every pair of morphisms in $\mathcal{C}$ admits a coequalizer. Let $F : \mathcal{C} \rightarrow \mathcal{D}$ be a functor that preserves coequalizers. Suppose also that if $h$ is a morphism in $\mathcal{C}$ such that $F(h)$ is an isomorphism, then $h$ is an isomorphism. Show that $F$ is faithful.

I was thinking to prove this by contradiction, so I considered two morphisms $f : A \rightarrow B$ and $g : A \rightarrow B$ in $\mathcal{C}$ such that $F(f) = F(g)$ but $f \neq g$. I took $\pi : B \rightarrow Q$ to be the coequalizer of $f, g$ in $\mathcal{C}$; then since $F$ preserves coequalizers, we know that the coequalizer of $F(f), F(g)$ in $\mathcal{D}$ is $F(\pi) : F(B) \rightarrow F(Q)$. My hope was that I could then use the universal property to get a contradiction, but I didn't know what to contradict, and I also didn't see how to use the fact that $F(h)$ is an isomorphism implies $h$ is an isomorphism.

How should one proceed from here?

答:
No need to use contradiction. You start well: say $f,g\colon A\to B$ in $\mathscr{C}$ have $F(f)=F(g)$. Let $\pi\colon B\to Q$ be the coequalizer of $f$ and $g$.

Prove that the coequalizer of a map with itself (in any category) is the identity morphism on the image.

Conclude that $F(\pi)$ is an isomorphism.

Deduce that $\pi$ is an isomorphism.

Conclude that $\mathrm{id}_B$ is also a coequalizer of $f$ and $g$.

Use that to conclude that $f=g$.


问:
On the general definition and nature of preordered sets
I found the following statement in a book after the usual definition of preorder and preordered set and I am not sure about how I should interpret it.

"While a preordered set $(X, \succsim)$ is not a set, it is convenient to talk as if it is a set when referring to properties that apply only to $X$. For instance, by a 'finite preordered set', we understand a preordered set $(X, \succsim)$ with $|X|< \infty$."

So the question (probably trivial) is, in the end is the preordered set a set or not?

My interpretation is that here the only set is $X$, and even if we put a preorder on it, and we call the result a preordered set, still the only set around is $X$, even if we had some characteristics (i.e. the predorder) that describe the behavior of the elements of $X$ with more details.

Is this the right interpretation of this paragraph?

PS I was not sure about the title of this question. I hope it is ok.

答:
This is a small, but sometimes relevant, key issue. Are ordered pairs sets?

Recall that preordered sets are ordered pairs, $(X,R)$ such that $X$ is a [non-empty] set and $R$ is a preorder of $X$. If you treat ordered pairs as non-set objects, then preorders are not sets for that very reason.

But when we talk about the preordered set $(X,R)$, we are saying that $X$ is a set, and it is preordered by $R$. This is how you should interpret this paragraph. Often, however, it is very good to confuse between $(X,R)$ and $X$. It helps us read things better, as the example about finiteness proposes.


问:
Gradient in tensor form
I found a problem which had  $$\partial_i (A_i \vec{G})= (\vec{\nabla} .\vec{ A} )\vec{G}+ (\vec{A}.\nabla) \vec{G} $$ but my problem is what does   $$\partial_i (A_i \vec{B})$$ even mean? it doesn't look like $(\vec{\nabla} .\vec{ A} )\vec{G} $ neither does it looks like $\nabla (\vec{A}.\vec{B})$ So what is it in terms of $\nabla $.
Since the term  $$\partial_i (A_i \vec{B})$$ needs to be integrated so it would be easier to write it in  $\nabla$

Source: Problem 1.11 Modern Electromagantism by Zangwill

答:
Let $C$ be an ordinary vector, and dot it with the dyad of the $A$ and $G$ vectors $$C\cdot(AG) = (C\cdot A)G$$
Similarly, the equation in your question can be written in dyadic notation as $$\nabla\cdot(AG) \;=\; (\nabla\cdot A)G \color{red}{+ A\cdot(\nabla G)}$$
Note that dot product only involves the $A$ vector,
but differentiation occurs for both $A$ and $G$, and this gives rise to the $\rm\color{red}{second}$ term. The appearance of extra terms like this is what makes nabla notation tricky.

Also note that the dyadic product of two vectors is a second order tensor, i.e. $$M = AG \quad\iff\quad M_{ij} = A_iG_j$$

The quantity $(\nabla\cdot A)$ is the divergence of $A$ and is a scalar function.

The quantity $(\nabla\cdot M)$ is the divergence of $M$ and is a vector function.

The quantity $(\nabla G)$ is the gradient of $G;\,$ a ${\tt2}^{nd}$ order tensor function.

Title: Gradient in tensor form


问:
Inflection point and 3rd derivative f'''(x)
Given f(x), by wikipedia, the sufficient existence condition for x = c to be an inflection point is:

f″(c + ε) and f″(c − ε) to have opposite signs in the neighborhood of c.

Is that equal to:

f''(c) = 0, and f'''(c) != 0 ?

答:
Take $$f(x)=x^5$$

$(0,0) $ is an inflection point of the curve $ C_f $ but
$$\;\; f'''(0)=0$$


答:
inflection points occur at the points when the function switches concavity. Concave up corresponds to positive second derivate and concave down has a negative 2nd derivative. So at infection points, 2nd derivate must be $0$. So inflection points occur when the second derivate changes signs. The third derivative is the rate at which the second derivative is changing. So, when the third derivate is $0$,  the second derivative is constant and the rate of slope changing is fixed.

Hope this helps you figure it out!

Title: Inflection point and 3rd derivative f'''(x)

Body:
Given f(x), by wikipedia, the sufficient existence condition for x = c to be an inflection point is:

f″(c + ε) and f″(c − ε) to have opposite signs in the neighborhood of c.

Is that equal to:

f''(c) = 0, and f'''(c) != 0 ?


答:
In the above picture, we can use three pairs of congruent triangles to prove the conclusion.

1.For $\triangle MDQ$ $ and $ $\triangle PBN$,

$\alpha=\alpha'$, MD=BP,NB=DQ, $\triangle MDQ$ $\equiv$ $\triangle PBN$ (SAS), $\to$ MQ=NP (1)

2.For $\triangle MAN$ and  $\triangle PCQ$,

$\beta=\beta'$,MA=PC,AN=QC,$\triangle MAN$ $\equiv$ $\triangle PCQ$  (SAS), $\to$ MN=PQ (2)

3.For $\triangle MNP$ and  $\triangle MPQ$,
bb
MP=MP ,MQ=NP and MN=PQ  from (1)(2), $\triangle MNP$$\equiv$
$\triangle MPQ$ (SSS) $\to$

$\angle NMP=\angle QMP$, NP//MQ, with NP=MQ from (1),  MNPQ  is  a parallelogram.

Title: Equilateral Triangles On Sides of a Parallelogram

Body:
On the sides of a parallelogram $ABCD$, construct similarly oriented equilateral triangles $ABC_1, BCA_1, CDD_1$, and $DAB_1$. Let $N,P,Q$ and $M$ be the centers of the triangles, respectively. Show that the quadrilateral $MNPQ$ happens to be a parallelogram.

I think this is a really interesting problem. I found it in the geometric transformations unit in my textbook. The given hint is to use the (central) symmetry with center the intersection of the diagonals of the given parallelogram ($ABCD$). I am not sure I see it though. Any help would be appreciated. Thank you in advance!


问:
Smoothness of morphisms gives an exact cotangent sequence
This is Vakil 21.2 S, self-study.

We are to show that if $\pi: X \to Y$ and $\rho: Y \to Z$ are smooth morphisms of schemes, then the relative cotangent sequence

$$\pi^*\Omega_{Y/Z} \to \Omega_{X/Z} \to \Omega_{X/Y} \to 0$$

is also left-exact. By "smooth" we mean smooth of relative dimension $n$ for some $n$; see definition 12.6.2 in the linked notes; it's a local condition on the Jacobian.

We know the composition of smooth maps of relative dimension $m$ and $n$ is smooth of relative dimension $m+n$, so $\rho \circ \pi$ is smooth.  We also know that if $\pi : X \to Y$ is smooth of relative dimension $n$, then $\Omega_{X/Y}$ is locally free of rank $n$. This would give us that all our sheaves are locally free.

I do not follow the hint given in this exercise; I did not use a block upper triangular matrix in 12.6 D as the hint suggests I should have, so if there is a solution taking this hint in a different direction or opting for another route, that would be appreciated.

答:
Let me give another argument, which is hopefully more in tune with Vakil's definition of smoothness. The argument will use the fact that all sheaves in the conormal sequence are locally free, which Vakil only proves in Theorem 25.2.2 (i.e. after the exercise in question), but it seems that Exercise 21.2.S isn't referenced in between, so we don't get a circular argument.

The key is the following lemma.

Lemma ([EGA IV/1, Ch. 0 (19.1.12)]). Let $R$ be a ring, let $M$ be a finitely generated $R$-module, and let $P$ be a finitely generated projective $R$-module. Then for a $R$-module map $u\colon M\rightarrow P$, the following are equivalent.

$u$ is a locally split injection (i.e. there are elements $f_1,\dotsc,f_n\in R$ generating the unit ideal such that all $u_{f_i}\colon M_{f_i}\rightarrow
 P_{f_i}$ are split injections; in particular, $M$ is projective too).

For all prime ideals $\mathfrak p\in\operatorname{Spec} R$, the map $u\otimes\kappa(\mathfrak p)\colon M\otimes_R\kappa(\mathfrak p)\rightarrow P\otimes_R\kappa(\mathfrak p)$ is injective.

Proof sketch. 1. $\Rightarrow$ 2. is clear. For the converse, take a prime ideal $\mathfrak p$ and elements $m_1,\dotsc,m_r\in  M$ which form a basis for $M\otimes_R\kappa(\mathfrak p)$. Since $u\otimes\kappa(\mathfrak p)$ is injective, we find $p_1,\dotsc,p_s\in P$ such that the $u(m_i)$ and the $p_j$ form a basis of $P\otimes_R\kappa(\mathfrak p)$. By a version of Nakayama's lemma, we find an $f\in R\setminus \mathfrak p$ such that $M_f$ is generated by the $m_i$ and $P_f$ is free with basis given by the $u(m_i)$ and the $p_j$. Then a split $P_f\rightarrow M_f$ is given by sending $u(m_i)$ to $m_i$ and all $p_j$ to $0$. $\square$

To apply the lemma, we must show that $(\pi^*\Omega_{Y/Z})_{x}\otimes_{\mathcal O_{X,x}} \kappa(x)\rightarrow (\Omega_{X/Z})_x\otimes_{\mathcal O_{X,x}}\kappa(x)$ is injective for all $x\in X$. Since tensoring is right-exact, we certainly get an exact sequence $$(\pi^*\Omega_{Y/Z})_{x}\otimes_{\mathcal O_{X,x}} \kappa(x)\rightarrow (\Omega_{X/Z})_x\otimes_{\mathcal O_{X,x}}\kappa(x)\rightarrow (\Omega_{X/Y})_x\otimes_{\mathcal O_{X,x}}\kappa(x)\rightarrow 0\,.$$
But these are vector spaces over $\kappa(x)$ of dimensions $\dim_y(Y/Z)$ (the relative dimension of $Y$ over $Z$ at $y=\pi(x)$), $\dim_x(X/Z)$, and $\dim_x(X/Y)$ respectively. Since smooth morphisms are flat, [Stacks Project, Tag 02JS] ensures $$ \dim_y(Y/Z)+ \dim_x(X/Y) = \dim_x(X/Z)\,,$$
so the above sequence of $\kappa(x)$-vector spaces must in fact be a short exact sequence. In particular, the left arrow must be injective. This shows that the lemma is applicable.


问:
Monotone permutation of sequence
The following is the question:

What sequences have a permutation, which is monotone increasing or monotone decreasing?

So first of all I believe that if a sequence is convergent, then there definitely exists such a permutation. I haven't proved it yet, but I think it shouldn't be that hard.

Now there are divergent sequences which doesn't have such permutations ($0,1,0,1,0,1,\dots$ as an example), but the trivial divergent sequence $(1,2,3,\dots)$ clearly works.

So in general, is it true that it has such a permutation if and only if it has a limit (perhaps $\pm \infty$)?

答:
Clearly such a sequence must have a limit in the extended reals, but as was noted in the comments, that is not sufficient. Suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ converges to $x$, and suppose further that there are $k,\ell\in\Bbb N$ such that $x_k<x<x_\ell$. Then there is an $m\in\Bbb N$ such that $x_k<x_n<x_\ell$ whenever $n\ge m$; clearly, however, there is no permutation of $\sigma$ that puts all of the terms $x_n$ with $n\ge m$ between $x_k$ and $x_\ell$, so $\sigma$ has no monotone permutation. Thus, if $\sigma$ does have a monotone permutation, we must have either $x_n\le x$ for all $n\in\Bbb N$, or $x_n\ge x$ for all $n\in\Bbb N$. Now I’ll show that this additional condition is also sufficient.

Suppose that $\sigma$ converges to $x$, and $x_n\le x$ for all $n\in\Bbb N$; the other case is similar. Let

$$A_0=\{n\in\Bbb N:x_n\le x-1\}\,,$$

and for $k\in\Bbb Z^+$ let

$$A_k=\left\{n\in\Bbb N:\frac1n<x\le\frac1{n+1}\right\}\,.$$

Since $\sigma$ converges to $x$, $A_k$ is finite for each $k\in\Bbb N$, and it follows easily that $\sigma$ has a monotone increasing permutation.

Thus, $\sigma$ has a monotone permutation if and only if it converges to some $x$ in the extended reals, and either $x_n\le x$ for all $n\in\Bbb N$, or $x_n\ge x$ for all $n\in\Bbb N$.

Title: Monotone permutation of sequence


答:
Other people have already explained why this is true - but I thought I'd add a little more too.

Another version of Stirling's formula is the following:
$$
\ln(n!) \sim n\ln(n) - n
$$
what do I mean here by the ${\sim}$ symbol? It's the "asymptotic to" symbol. It means if you have two functions ${f(x),g(x)}$, we say that ${f\sim g}$ at infinity if
$$
\lim_{x\to \infty}\frac{f(x)}{g(x)}=1
$$
you can see how this essentially tells us ${f(x)}$ and ${g(x)}$ behave the same at infinity. If we exponentiate both sides, we get
$$
n! \sim \frac{n^n}{e^n}
$$
and so now,
$$
(n!)^{\frac{1}{n}}\sim \left(\frac{n^n}{e^n}\right)^{\frac{1}{n}} = \frac{n}{e}
$$
i.e.
$$
(n!)^{\frac{1}{n}}\sim \frac{n}{e}
$$
this tells you at infinity, ${(n!)^{\frac{1}{n}}}$ behaves like ${\frac{n}{e}}$. A straight line with gradient ${\frac{1}{e}}$.

Asymptotic relations are helpful as they allow you to deduce things like the above, and also help you in calculating limits. See my answers in the following posts to see how: Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero and Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$


问:
Proving that a sequence does not converge?
I cannot figure out how to prove that the sequence $x_n\:=\:\left(-1\right)^n\:+\:\frac{1}{n}$ does not converge. I tried a proof by contradiction but the sequence ends up being $\frac{1}{n}+1\:-\:\frac{1}{n}$ and I don't know where to go from there. I also tried showing that its subsequences don't converge but I did not know what to do with them

$x_{2n} = (-1)^2n + \frac{1}{2n}$

$x_{2n-1} = (-1)^2n-1 + \frac{1}{2n}-1$

答:
Hint: \begin{align*}
(-1)^{2n} + \frac{1}{2n} &= 1 + \frac{1}{2n}  \\
(-1)^{2n+1} + \frac{1}{2n} &= -1 + \frac{1}{2n}
\end{align*}
For large enough $n$, one of these is greater than $1/2$ and the other is less than $-1/2$...


问:
Integration with complex constants
So I saw a youtube video of the following integral, which can be solved by letting u=tanx, then you get

$\int \frac{1}{1+x^2} dx $ = ${tan}^{-1}x + C$

But it should also be solvable through partial fraction decomposition.

$\int \frac{1}{1+x^2}dx$ = $\int {\frac{A}{x-i} + \frac{B}{x+i}dx}$

$\frac{1}{1+x^2}$ = $\frac{A}{x-i}$ + $\frac{B}{x+i}$

$1$ = $A(x+i)$ + $B(x-i)$

$x=i$:  $1$ = $2iA$

$A$=$\frac 1{2i}$ = -$\frac {i}{2}$

$x=-i$: $1 = -2iB$

$B$ = $\frac{i}2$

$\int \frac{1}{1+x^2}dx$ = $\int -\frac{i}{2(x-i)}$ + $ \frac{i}{2(x+i)}dx$ = $-\frac{i}2$ln$\lvert {x-i}\rvert$ + $\frac{i}2$$ln\lvert {x+i}\rvert$ + C = $\frac{i}2ln\lvert {x+i}\rvert - ln\lvert {x-i}\rvert$ + C

This should be equal to ${tan}^{-1}x + C$ right?

${tan}^{-1}({1})$ =  $\frac\pi4$

$\frac{i}2(ln\lvert {1+i}\rvert - ln\lvert {1-i}\rvert)$ = $\frac{i}2(ln{\sqrt2}-ln{\sqrt2})$=O

So these are obviously not equal, then I thought why must i have absolute value sign in the logarithmic expressions, that rule I know only applies for integarting with real constants.

Thus,

$\int \frac{1}{1+x^2} dx = $$\frac{i}2(ln({x+i}) - ln({x-i}))$ + C = $\frac{i}2ln(\frac{({x+i})^2}{x^2+1}) + C$

Evaluated at x=1 gives us

$\frac{i}2ln(\frac{({1+i})^2}{1^2+1})$ = $\frac{i}2ln(i)$ = $\frac{i}2\cdot \frac{i\pi}{2}$ = $-\frac\pi4$

So, im quite close, but I must have missed something somewhere, and I cant find the mistake. But this triggers two other questions: How would I show that

$\frac{i}2ln(\frac{({x+i})^2}{x^2+1})$ = $\tan^{-1}x$ (or some other logarithmic expression if that expression is where the error lies)

and, why dont I need to take absolute value when integrating

$\int \frac{1}{x+a}dx$

with complex constants? Is the explanation as easy that I can evaluate all logarithmic expressions using complex numbers?

Title: Integration with complex constants

Body:
$$e^{xi} = \cos{x}+i\sin{x}$$
$$e^{-xi} = \cos{x}-i\sin{x}$$
$$\sin{x} =  \frac{ e^{xi}-e^{-xi} }{2i}$$
$$ \cos{x} = \frac{ e^{xi}+e^{-xi} }{2}$$
$$\tan{x} = \frac{\sin{x}}{\cos{x}} =\frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
$$\tan{x} = \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
Say $ \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$
$$ \tan{x} = y , x = \arctan{y}$$
$$ \frac{1}{i}  \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$$
Say $x = i \log{z}$
$$ \frac{1}{i}  \frac{ z^{-1}-z }{z^{-1}+z} = y$$
$$ z^{-1}-z = i y (z^{-1}+z) $$
$$1-z^2 = i y (1+z^2)$$
$$i y z^2+z^2 = 1-i y$$
$$ z = \sqrt{ \frac{ 1-i y}{1+i y} }$$
$$\arctan{y} = i \log{  \sqrt{ \frac{ 1-i y}{1+i y} }}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ 1-i y}{1+i y }}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ (1-i y)^2}{(1+i y )(1-i y)}}$$
$$\arctan{y} = \frac{i}{2} \log{   \frac{ 1-2y i -y^2}{1+y^2 }}$$


问:
Decompose real positive-definite symmetric matrix
Can any $M\in \mathbb{R}^{n\times n}_{\text{sym}+}$ (real, positive-definite, symmetric matrix) be decomposed in $M=C^{T}.C$ with $C\in \text{GL}_n(R)$ and vice versa ($C\in \text{GL}_n(R)\implies C^{T}.C\in \mathbb{R}^{n\times n}_{\text{sym}+}$)?

This question is related to inner products on real n-dim vector spaces with associated matrices $M\in \mathbb{R}^{n\times n}_{\text{sym}+}$. Under a basis transformation with associated matrix $C\in \text{GL}_n(R)$, the Gram matrix $M$ transforms as follows
$$M_b=C^T\cdot M_e\cdot C$$
If we choose $M_e=\text{id}$ then all other possible Gram matrices $M_b$ can be constructed by changing the basis so that $M_b=C^T\cdot C$ (at least if the statement in the beginning is true). If this is true, is their a way to characterize a real inner product on the basis of its Gram matrix? Maybe the basis choice for which $M_e=\text{id}$? And if so, when is this choice Euclidean?

答:
I guess you're thinking of $M$ as the Gram matrix for a particular basis (which is just the matrix for the inner product as a symmetric bilinear form). There is an orthonormal basis on $\mathbb R^n$ with respect to this inner product, and so, yes, you can change basis to make $M$ turn into the identity matrix. (BTW, this works with any nondegenerate symmetric form, not necessarily positive-definite.)

But, to answer your more general question, for any $C\in GL_n(\mathbb R)$, the matrix $C^\top C$ is symmetric (just take its transpose) and positive-definite ($C^\top Cx\cdot x = Cx\cdot Cx  = \|Cx\|^2\ne 0$, since $C$ is nonsingular). Conversely, by the Spectral Theorem, there is an orthonormal basis for $\mathbb R^n$ consisting of eigenvectors of your given $M$. Moreover, since $M$ is positive definite, every eigenvalue is positive (see the calculation just above), and so $M=Q \Lambda Q^\top$ for some orthogonal matrix $Q$ and some diagonal matrix $\Lambda$ with positive entries. Now take $C=Q\sqrt\Lambda Q^\top$, where $\sqrt\Lambda$ denotes the diagonal matrix whose entries are the (positive) square roots of the entries of $\Lambda$. $C$ is a symmetric matrix with $C^2 = M$.


问:
What does it mean when Newton's method encounters a non-invertible matrix?
I'm trying to solve some systems of polynomials. I'm using Newton's method as described here. But my code breaks if my initial guess or any subsequent iteration lands on a point which is a solution for some but not all of the functions.

For example, say I want to find $(x,y,z)$ such that:
$$
f_1(x,y,z)=x^2-z \\
f_2(x,y,z)=y^2-z \\
f_3(x,y,z)=z
$$
are all zero. (The solution is $(0,0,0)$.) I believe the Jacobian and inverse are:
$$
\boldsymbol{J}=
\begin{pmatrix}
2x &  0 & -1 \\
 0 & 2y & -1 \\
 0 &  0 &  1 \\
\end{pmatrix}
,
\boldsymbol{J}^{-1}=
\begin{pmatrix}
\frac{1}{2x} &  0           & \frac{1}{2x} \\
0            & \frac{1}{2y} & \frac{1}{2x} \\
0            &  0           & 1 \\
\end{pmatrix}
$$
If we happen to guess any starting point on $x=0$ or $y=0$, then the corresponding column of $\boldsymbol{J}$ becomes zero, and non-invertible. Or, $\boldsymbol{J}^{-1}$ would contain $\frac{1}{0}$.

Am I Doing It Wrong™? How should the algorithm proceed in this case? Or should I be using a different algorithm?

I found this question which seems to say that $\boldsymbol{J}$ should always be invertible, which I don't understand.

答:
The Jacobian is invertible when there are several different directions you can travel which result in similar changes overall. Consider the one-dimensional problem of $f(x)=1-x^2$ starting from $x=0$. In such an instance, either direction is valid, so Newton's method cannot pick a direction to go.

Another situation is when the derivative is vanishing at a point even though the function is monotone, such as with $f(x)=1-x^3$.

There are a lot of different ways these situations can be tackled.

Line search let's you continue as long as you have a direction to go in. If Newton's method can't give you a direction, you can try choosing random ones.

Using a pseudoinverse may help. You could try estimating $\mathbf J^{-1}\mathbf f(\vec x)$ iteratively this way using something like the Gauss-Seidel method.

Move a small random amount when the Jacobian is not invertible, or perhaps if its determinant is sufficiently small. Intuitively, you can see from the above examples that this should help Newton's method "get away" from the troublesome point and be able to start progressing again.


问:
Identification of $\ell_\infty$ with $C(\beta\mathbb{N})$ and of $\ell_\infty^*$ with $C(\beta\mathbb{N})^*$
The Stone-Cech compactification of $\mathbb{N}$, denoted by $\beta\mathbb{N}$ has the property that every $x\in\ell_\infty(\mathbb{N})$ is identified with (extended uniquely) to $\beta x\in C(\beta\mathbb{N})$.

For example, $x=(x_n)_n$, with $x_n=1$, for exery $n\in\mathbb{N}$ I think has $\beta x=1$ (the constant function identically equal to $1$).

The dual of $C(\beta\mathbb{N})$, namely, $C(\beta\mathbb{N})^*$ is identified with the space of finite Borel measures on $\beta\mathbb{N}$ by the Riesz's Representation Theorem: - for every $F\in C(\beta\mathbb{N})^*$ there is a unique finite Borel measure on $\beta\mathbb{N}$ such that
$$
\forall f\in C(\beta\mathbb{N}),\ F(f)=\int_{\beta\mathbb{N}} f d\mu
$$
$F$ is denoted by $\mu$ in the sequel

My questions are:

If $\mu\in C(\beta\mathbb{N})^*$ and $x=(x_n)_n\in \ell_\infty$ can we find a closed formula for $\mu(\beta x)$?

For example, if $\delta^k\in\ell_\infty$ is given by $(\delta^k)_n=1$, if $n=k$; $(\delta^k)_n=0$, for $n\neq k$ can we say what $\beta \delta^k$ is and then what $\mu(\beta \delta^k)$ is?

When  $x=(x_n)_n$, with $x_n=1$, for exery $n\in\mathbb{N}$, $\beta x=1$, so $\mu(\beta x)=\mu(\beta\mathbb{N})$ is the (finite) measure of $\beta\mathbb{N}$.

Could you suggest a book where this subject is treated?

答:
In your special example, sure: $\beta \delta^k$ is the function on $\beta \mathbb{N}$ defined by $(\beta \delta^k)(y) = 1$ when $y=k \in \mathbb{N} \subset \beta \mathbb{N}$ and $0$ otherwise.  This is easily verified by showing that the function $f : \beta \mathbb{N} \to \mathbb{R}$ defined by this formula is continuous on $\beta \mathbb{N}$ (recall that $\mathbb{N}$ is open in $\beta \mathbb{N}$, hence $\{k\}$ is open as well) and that it agrees with $\delta^k$ on $\mathbb{N}$.

As such, $\mu(\beta \delta^k) = \mu(\{k\})$, the measure of the singleton set $\{k\}$.

For more general $x \in \ell^\infty$, you probably can't do this in closed form, precisely because you can't express $\mu$ in closed form except in very special cases.

Sorry, I don't know a reference for this off the top of my head.


问:
How should I refer to this random variable?
Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space and let $(\Gamma,\mathscr{G})$ be a measurable space. Then a random probability measure on $(\Gamma,\mathscr{G})$ can be defined as a function $M\colon \Omega \times \mathscr{G} \to [0,1]$ such that

For all $\omega \in \Omega$, the function $M_\omega\colon A \mapsto M(\omega,A)$ is a probability measure on $(\Gamma,\mathscr{G})$,

For all $A \in \mathscr{G}$, the function $M_A\colon \omega \mapsto M(\omega,A)$ is a random variable on $(\Omega,\mathscr{F},\mathbb{P})$.

This definition is based on chapter VI of Çınlar's Probability and Stochastics.

Let $X$ be a measurable function on $(\Gamma,\mathscr{G})$. Then $X$ is a random variable on each probability space $(\Gamma,\mathscr{G},M_\omega)$, and Fubini's theorem implies that the function
$$
\omega \mapsto \int_\Gamma  X \,\mathrm{d}M_\omega
$$
is a random variable on $(\Omega,\mathscr{F},\mathbb{P})$. Is there any particular name for such a random variable? From what I understand here, I would be tempted to call this random variable the "random expectation" of $X$ with respect to the random probability measure $M$. However, I've never seen the phrase "random expectation" anywhere, so I'm not sure if it is indeed a correct interpretation, or if another name is usually used.

答:
A variation is to realize everything on the same space (the sensible thing to do in probability: enrich your measured space as needed):

$$\begin{align}
Y & = \Omega \times \Gamma, \\
\mathcal{Y} & = \mathcal{F} \otimes \mathcal{G}, \\
\mu (B \times A) & = \int_B M_\omega (A) \ d\mathbb{P} (\omega).
\end{align}$$

Then $(Y, \mathcal{Y}, \mu)$ is a probabilized space. We can define $X = X(\gamma)$ on $Y$, but depending only on the second coordinate. Then

$$\int_\Gamma X(\gamma) \ d M_\omega (\gamma) = \mathbb{E} (X|\omega),$$

so you are looking at the conditional expectation of $X$ knowing $\omega$.


问:
Failure of a prop. regarding absolutely continuous measures

Proposition A: Let $\left(X, \mathcal{M}, \mu\right)$ be a measure space and $\nu$ a finite measure on the measurable space $\left(X, \mathcal{M}\right)$. Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if for each $\varepsilon>0$, there is a $\delta>0$ such that for any set $E\in\mathcal{M}$,
$$\text{if } \mu(E)<\delta,\text{ then } \nu(E)<\varepsilon.\tag{1}$$

My question is, why is the finiteness of $\nu$ needed? That is, I should be able to find a non-finite measure $\nu$ on a measurable space $\left(X, \mathcal{M}\right)$ with $\nu\ll\mu$ such that $(1)$ fails to hold, even if $\mu$ is finite on the measurable space $\left(X, \mathcal{M}\right)$.

My though was to consider $X:=\mathbb{N}$, $\mathcal{M}:=\mathscr{P}(\mathbb{N})$, and letting $\nu$ be the counting measure. As $\nu(X) = +\infty$, $\nu$ is non-finite on the measurable space $\left(X, \mathcal{M}\right)$. Then, let $\mu$ be another measure on the measurable space $\left(X, \mathcal{M}\right)$ for which $\nu\ll\mu$.

This is where I'm at right now. I'd expect a solution would follow by contradiction, but not sure.

答: Your thought is good; you just need to specify a measure $\mu$.  Try the "weighted counting measure" for which $\mu(\{n\}) = 2^{-n}$.  You can then easily check that $\nu \ll \mu$, but that as soon as $\epsilon < 1$, there can be no $\delta$ satisfying (1) for all $E$.


问:
Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$
In this post it is proved that, if $(a_n)$ is a sequence in $\mathbb C$:

$$
\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}a_n=k\implies\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k.
$$

Is the converse to this statement true? In other words, is there a sequence $(a_n)$ with

$$
\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k
$$

but $\lim_{x\to\infty}\frac1x\sum_{n\leq x}a_n$ converges to some other limit, or diverges? Thanks.

Body: What do you get with $$a_n = 10+ \Re( (n+1)^{1+i}-n^{1+i})$$


问:
Probability that Brownian motion makes a loop around the origin
Let $B$ be a complex Brownian motion started at $\epsilon > 0$. I am trying to prove that as $\epsilon \rightarrow 0$, the probability that $B$ makes a loop around the origin before hitting the unit circle tends to 1.

I tried looking at $\operatorname{arg} B_t$, and rescaling the problem, but neither of these seemed to get me anywhere. Does anyone have any suggestions?

答:
For each $\epsilon$, by scaling, this is the probability that a Brownian motion starting at $1$ makes a loop around the origin before hitting the circle of radius $1/\epsilon$. When $\epsilon\to0$, the hitting time of the circle of radius $1/\epsilon$ goes to infinity hence, in the limit, one is after the probability that a Brownian motion starting at $1$ makes a loop around the origin, ever.

The conclusion then comes from a representation of the argument process of a plane Brownian motion... which you might know?


问:
Logic - Natural Language vs. Symbolic Form Confusion
Let $M(x, y)$ be “$x$ has sent $y$ an e-mail message” and $T(x, y)$ be “$x$ has telephoned $y$,” where the domain consists of all students in your class. Use quantifiers to express each of these statements. (Assume that all e-mail messages that were sent are received, which is not the way things often work.)

a) There is someone in your class who has either sent an e-mail message or telephoned everyone else in your class.

b) Every student in the class has either received an e-mail message or received a telephone call from another student in the class.

The solution manual answers are as follows:

a) $$\exists x \forall y (y \neq x \rightarrow (M(x, y) \vee T(x, y)))$$

This makes sense to me, because we don't know if the person has sent an e-mail to him/herslef, or telephoned him/herself, and that's not emphasized in part a, and that's why even if $x = y$ the statement would still be true.

b) Here $y$ is "another student": $$\forall x \exists y (x \neq y \wedge ((M(y, x) \vee T(y, x)))$$

Here the word "another" is taken to mean that these students are different, and so $x = y$ would make the statement false, but we don't consider "else" in part a to be as strong as "another" in part b. So what rules are exactly to be followed when translating natural language to symbolic form? Is there a logic behind all this?

答: The words "everyone else" in part a corresponds to $y \neq x$ in the proposed solution. As for your second point, there are no rigid rules about translating natural language into formal logic. Natural language is riddled with ambiguities. The value of translating natural language statements into formal logic is to expose the ambiguities. For example, in your example a, it is just about plausible that "everyone else" means everyone apart from you (the speaker): the given solution would not agree with that reading.


问:
Uniform Convergence of $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ [proof verification]
As the title suggests, I'm trying to show that $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ converges uniformly. I tried showing that $f_n(x)$ converges uniformly to the pointwise limit $\frac{1-2x}{2}$ but I couldn't get it to work out, so I've tried to show it's uniformly Cauchy like so:

Let $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$.
Let $m, n \in \mathbb{N}$ such that $m > n$. Consider
$$|f_n(x) - f_m(x)| = |\frac{(m-n)(2\cos(x) - \sin^2(x))}{(2m+\sin^2(x))(2m+\sin^2(x))}| \leq \frac{|m-n|(2)}{4mn} = \frac{|n - m|}{2mn} < \frac{n}{2n^2} < \frac{1}{2n} < \epsilon$$
So let $\epsilon > 0$.
Choose $N \in \mathbb{N}$ such that $N > \frac{1}{2\epsilon}$.
Then for all $m > n > N$,
$$|f_n(x) - f_m(x)| < \epsilon$$
So $f_n$ is uniformly Cauchy, so it must converge uniformly.

But after checking this result graphically I think that it's wrong, and I'm not really sure what else to do.
Any help would be greatly appreciated.

答:
The convergence is uniform for all $x \in \mathbb{R}$.

The pointwise limit is clearly $f(x) =(1-2x)/2$ and proceeding directly,

$$|f_n(x) - f(x) | = \left|\frac{\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}}{1 + \frac{\sin^2x}{2n}}- \frac{1-2x}{2} \right| \\ = \frac{\left|\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}- \frac{1-2x}{2}- \frac{\sin^2x}{2n}\frac{1}{2} + \frac{\sin^2 x}{2n} \frac{2x}{2}\right|}{1 + \frac{\sin^2x}{2n}} \\ \leqslant \left|\frac{\cos x}{2n}- \frac{\sin^2x}{4n} \right|\leqslant \frac{3}{4n}\underset{n \to \infty}\longrightarrow 0$$

Checking your approach, with $m >n$, we have

$$f_m(x) - f_n(x) \\=\frac{m(1-2x)2n + m(1-2x)\sin^2x + (\cos x - x\sin^2x)2n + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ -\frac{n(1-2x)2m + n(1-2x)\sin^2x + (\cos x - x\sin^2x)2m + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ =  \frac{(m-n)(1-2x)\sin^2 x +2(n-m)(\cos x - x \sin^2 x)}{(2m+\sin^2x)(2n+ \sin^2 x)}\\ =  \frac{(m-n)(\sin^2 x -2\cos x)}{(2m+\sin^2x)(2n+ \sin^2 x)},$$

and, thus,

$$|f_m(x) - f_n(x)| = \frac{(m-n)|\sin^2 x -2\cos x|}{(2m+\sin^2x)(2n+ \sin^2 x)}\leqslant \frac{3(m-n)}{4mn} = \frac{3}{4n} - \frac{3}{4m} < \frac{3}{4n}$$

For all $m > n > N(\epsilon)$ where $N(\epsilon) > \frac{4}{3\epsilon}$ we have $|f_m(x) - f_n(x)| < \epsilon$.


问:
Understanding the notion of the fundamental group
Can anyone explain what we mean by homotopy classes and for how can for example the fundamental group of the real number line be 0? And what exactly does the fundamental group consist of?</p>

答:
If you want a full, rigorous, Mathematical description of what is meant by the fundamental group you will need to watch some lectures or read some notes. But I guess an informal explanation would help with understanding intuitively what it all means.

Two continuous functions, ${f : X\to Y}$ and ${g:X\to Y}$ are said to be homotopic, written as ${f\simeq g}$, if there is some nice continuous way to "deform" $f$ into $g$. It turns out that the ${\simeq}$ symbol actually gives us an equivalence relation. That means that given some continuous function $f$, we can "slot it into" exactly one class of functions that are all homotopic to each other basically.

Now - a loop in a topological space is a nice continuous path that starts and ends at the same point. Loops are continuous functions, and so these have their own "homotopy classes". In other words - every loop can be fit nicely into exactly one class of loops, and all loops within their classes can be "deformed" nicely into one another.

With this in mind, the fundamental group is essentially the set of these "loop homotopy classes", where we define some nice group operation on these classes of loops (we do the "path product").

The reason the fundamental group of the real line, ${\pi_1(\mathbb{R})}$, is $0$ ($0$ in this context means "the group with just the identity element") is because any loop in ${\mathbb{R}}$ can be deformed continuously into any other loop in ${\mathbb{R}}$ (it's not difficult to see this). So there is only one "loop" class, and it must be the identity element (since groups need at the minimum the identity element).


问:
Find all roots of a quintic polynomial given 1?
Let's say I have solved for one of the roots $a$ of a quintic equation using Newton's method. I read somewhere that you can "simply" divide the equation by $x - a$ to get a quartic and solve from there.

How would I get this quartic equation from my quintic euqation (using a formula), and after doing this, how would I go about solving it numerically?

答:
Expanding on Gribouillis' answer, you don't need to factor to find more roots. The Durand-Kerner method allows for simultaneously finding all polynomial roots without this, and it allows you to include known roots if you wish as well. As it goes, it will also further polish the root you initially estimated, so if your known root isn't completely accurate then that is fine too.

Let $r_{i,n}$ be the $n$th estimate of the $i$th root for $i\in\{1,...,5\}$, with $r_{1,1}$ known and $r_{i,1}$ picked for $i\ne1$. One may then estimate

$$r_{i,n}=r_{i,n-1}-\frac{f(r_{i,n-1})}{P_{i,n}Q_{i,n}}$$

where

$$P_{i,n}=\prod_{j<i}(r_{i,n-1}-r_{j,n})=(r_{i,n-1}-r_{1,n})\cdots(r_{i,n-1}-r_{i-1,n})$$
$$Q_{i,n}=\prod_{j>i}(r_{i,n-1}-r_{j,n-1})=(r_{i,n-1}-r_{i+1,n-1})\cdots(r_{i,n-1}-r_{5,n-1})$$

or alternatively, one may use the simpler but slightly slower

$$r_{i,n}=r_{i,n-1}-\frac{f(r_{i,n-1})}{R_{i,n}}$$

where

$$R_{i,n}=\prod_{j\ne i}(r_{i,n-1}-r_{j,n-1})=(r_{i,n-1}-r_{1,n-1})\cdots(r_{i,n-1}-r_{5,n-1})$$

and the expanded RHS does not include the term $(r_{i,n-1}-r_{i,n-1})$.

The downside to this method is you will likely want to deal with complex arithmetic even if you only want real roots. There are also several similar algorithms on Wikipedia.


问:
Dirichlet Eta Function
Consider the set
$$U=\{s\in \mathbb{C}:\ \Re(s)>0\}\setminus \left\{1+(\ln(2))^{-1}2\pi ik:\ k\in \mathbb{Z}\right\}$$
Clearly $U$ is open and connected, and so $\zeta$ has a unique analytic continuation from
$$V=\{s\in \mathbb{C}:\ \Re(s)>1\}$$
to the set $U$.

Let $\{x\}=x-[x]$ be the fractional part of $x>0$. It is well-known that
$$\zeta(s)=\underbrace{\frac{s}{s-1}-s\int_1^\infty \frac{\{x\}}{x^{s+1}}dx}_{f(s)}\qquad \forall s\in V$$
$$\zeta(s)=\frac{1}{1-2^{1-s}}\underbrace{\sum_{n\in \mathbb{N}}\frac{(-1)^{n-1}}{n^s}}_{\eta(s)}\qquad \forall s\in V$$
It's easy to see that both $f$ and $\frac{1}{1-2^{1-s}}\eta$ are analytic on $U,V$. Hence, by uniqueness of analytic continuation to connected open sets, $f(s)=\frac{1}{1-2^{1-s}}\eta(s)$ for all $s\in U$. My question is, how do we prove this by direct computation?

My attempt:

Since the integral in $f$ is convergent, we have that
$$\begin{aligned}
 &\ s\int_0^\infty\frac{x-[x]}{x^{s+1}}dx \\
=&\ s\sum_{n\in \mathbb{N}}\int_n^{n+1}\frac{x-[x]}{x^{s+1}}dx \\
=&\ \sum_{n\in \mathbb{N}}s\int_n^{n+1}\frac{x-n}{x^{s+1}}dx \\
=&\ \sum_{n\in \mathbb{N}}\left[s\int_n^{n+1}\frac{1}{x^s}dx+\int_n^{n+1}n\left(\frac{-s}{x^{s+1}}\right)dx\right] \\
=&\ \sum_{n\in \mathbb{N}}\left[\left.\frac{s}{1-s}\frac{1}{x^{s-1}}\right|_{x=n}^{x=n+1}+\left.\frac{n}{x^s}\right|_{x=n}^{x=n+1}\right] \\
=&\ \sum_{n\in \mathbb{N}}\left[\frac{s}{1-s}\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)+\left(\frac{n}{(n+1)^s}-\frac{n}{n^s}\right)\right] \\
=&\ \sum_{n\in \mathbb{N}}\left[\frac{s}{1-s}\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)+\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{(n+1)^s}-\frac{1}{n^{s-1}}\right)\right] \\
=&\ \sum_{n\in \mathbb{N}}\left[\left(\frac{s}{1-s}+1\right)\left(\frac{1}{(n+1)^{s-1}}-\frac{1}{n^{s-1}}\right)-\frac{1}{(n+1)^s}\right] \\
=&\ \sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{(n+1)^s}\right] \\
=&\ 1+\sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{n^s}\right] \\
=&\ \frac{s}{s-1}-\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)-\frac{1}{n^s}\right]
\end{aligned}$$
Therefore,
$$f(s)=\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{n^s}-\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)\right]$$
Now it suffices to show that for all $s\in U$,
$$\frac{1}{s-1}+\sum_{n\in \mathbb{N}}\left[\frac{1}{n^s}-\frac{1}{s-1}\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}}\right)\right]=\frac{1}{1-2^{1-s}}\sum_{n\in \mathbb{N}}\frac{(-1)^{n-1}}{n^s}$$

But I haven't found a way to do it so far. Can anyone offer any hint or help? Thanks!

Title: Dirichlet Eta Function

Body:
The proof I can think to, for $\Re(s)\in (0,1)$: $$\eta(s)= \sum_{n\ge 1} (-1)^{n+1}s\int_n^\infty x^{-s-1}dx=s\int_1^\infty (\sum_{n\le x}(-1)^{n+1})x^{-s-1}dx$$

$$\frac{s}{s-1}-s\int_1^\infty \{x\}x^{-s-1}dx=-s\int_0^\infty \{x\}x^{-s-1}dx$$
$$2^{1-s} s\int_0^\infty \{x\}x^{-s-1}dx= s\int_0^\infty 2\{x/2\}x^{-s-1}dx$$

$$\sum_{n\le x}(-1)^{n+1}=2\{x/2\}-\{x\}$$


问:
A linear operator and Stone-Cech compactification
I am reading the following article

Jean-Pierre Gossez. “On the range of a coercive maximal monotone operator in a
nonreflexive Banach space”. In: Proc. Amer. Math. Soc. 35 (1972), pp. 88–92

in which the following linear operator $A:\ell_1(\mathbb{N})\to\ell_\infty(\mathbb{N})$ is defined by

$$
(Ax)_n:=-\sum_{k<n}x_{k}+\sum_{k>n}x_{k}=\sum_{k\in\mathbb{N}}x_{k}\alpha(k,n),\ x=(x_{k})_{k\in\mathbb{N}}\in\ell_{1},\ n\in\mathbb{N},
$$

where $\alpha(k,n)=\left\{ \begin{array}{lll}
-1, & \mathrm{if} & k<n\\
0, & \mathrm{if} & k=n\\
1, & \mathrm{if} & k>n
\end{array}\right.$
, $k,n\in\mathbb{N}$.

Also, the adjoint of $A$  is considered $A^*:\ell_{\infty}^{*}\to\ell_{\infty}$ which is given by $\langle y,A^{*}\mu\rangle=\langle\mu,Ay\rangle$, for $y\in\ell_{1}$, $\mu\in\ell_{\infty}^{*}$, where $\langle\cdot,\cdot\rangle$ is the natural duality formed by a Banach space $X$ with its topological dual $X^*$, $\langle x,x^*\rangle:=x^*(x)$.

The author says "It is easy to verify that"
$$
\langle\mu,-A^{*}\mu\rangle=(\mu(\beta\mathbb{N}\smallsetminus\mathbb{N}))^{2}.\hspace{3cm}(\star)
$$
Here $\beta\mathbb{N}$ is the Stone-Cech compactification of $\mathbb{N}$, $\ell_{\infty}^{*}$ is identified with $M(\beta\mathbb{N})$ the space of finite Borel measures on $\beta\mathbb{N}$, and $\mu(\beta\mathbb{N}\smallsetminus\mathbb{N})$ denotes the measure of the set $\beta\mathbb{N}\smallsetminus\mathbb{N}$. I am mainly interested in the proof of ($\star$).

I tried getting ($\star$) by finding the expression of $A^*$ as follows.
Every $x\in\ell_\infty$ is uniquely extended to (identified with) $\beta x\in C(\beta\mathbb{N})$ while $C(\beta\mathbb{N})^*$ is identified with $\ell^*_\infty$ via Riesz's Representation Theorem as follows: every $\mu\in C(\beta\mathbb{N})^*$ is a finite Borel measure (on $\beta\mathbb{N}$) such that $\mu(f)=\int f d\mu$, for $f\in C(\beta\mathbb{N})$.

I wonder whether for $\mu\in \ell^*_\infty$, $x\in\ell_\infty$, $\mu(x)=\sum_{n\in\mathbb{N}}x_n \mu(\{n\})$, where $\mu(\{n\}$ denoted the measure of the set $\{n\}$?

Any other way to approach ($\star$)?

答:
As for your second question

I wonder whether for $\mu\in \ell^*_\infty$, $x\in\ell_\infty$, $\mu(x)=\sum_{n\in\mathbb{N}}x_n \mu(\{n\})$, where $\mu(\{n\}$ denoted the measure of the set $\{n\}$?

The answer is no, not every measure on $\beta \mathbb N$ is discrete. For example, take a continuous surjection $f\colon \beta \mathbb N\to [0,1]$. Then $\mu(A) = m(f[A])$ ($A$ is a Borel subset of $\beta \mathbb N$), where $m$ is the Lebesgue measure, is an atomless measure so it cannot have the form you specified.

Compact spaces on which every measure is discrete (in particular, of the form you specified) are precisely the scattered spaces. (This was noted by W. Rudin in the 1950s.)

Every measure $\mu$ however naturally decomposes into

$$\langle \mu ,f \rangle = \sum_{n=1}^\infty f_n \mu(\{n\}) + \int_{\beta N\setminus N} (\beta f)\,{\rm} d\mu\quad (f\in \ell_\infty)$$

so the first step is to show that the former summand always vanishes for $\langle \mu, -A^*\mu\rangle$.


问:
How do you determine which plot has smaller or larger residuals from a box plot?
Which plot has the smallest residuals?

答: There is no formula for determining this, you just have to make a judgement based on what we can infer about the data from the box plots. A set of points with higher variance will have larger residuals than a set of points with lower variance.</p>

Body: Which plot has the smallest residuals?


问:
Economic Elasticity: where elasticity-equation come from?
I know the equation for economic elasticity is:

$$\varepsilon = \frac{\%\,\Delta Y}{\%\,\Delta X}\frac{X}{Y} = \frac{\partial Y(X)}{\partial X}\frac{X}{Y} = \frac{\partial \log(Y)}{\partial \log(X)}$$

In fact, this is the generalization for any sensitive-analysis or elasticity for a given function; in this case $Y(X)$.

But, where does come it from? I mean, why is it the equation and no -for example- just the partial derivate of the given function?

Thanks in advance!

答: The elasticity gives you the percentage change of the dependent variable with respect to the percentage change of the independent variable. Elasticity by definition is dimensionless and I believe this is part of the motivation behind using this over the derivative like you mentioned. </p>


问:
Proof: accumulation point is limit of a sequence of distinct points in metric space
Let $(X, d)$ be a metric space. Let $E \subseteq X$ be a subset of $X$.
Let $p \in X$ be a accumulation point of $E$. So by definition for every neighbourhood $V \subseteq X$ of $p$ this intersection $(V \setminus \{p \} ) \cap E$ is not empty. I know how to show, that there is a sequence $(b_k)$, where for all $k \in \mathbb{N}: b_k \in E$ and $b_k \to p$. (by axiom of choice). My textbook states, that there also exists a sequence $(a_k)$, where for all $k \in \mathbb{N}: a_k \in E$ and $a_k \to p$ and $$\textbf{ for all } n,m \in \mathbb{N} \text{ with } n \neq m \textbf{  also  } a_n \neq a_m .$$
So $(a_n)$ should be a sequence of distinct members of $E$.

My idea for a proof is to use $(b_k)$ and remove all members of the sequence that appear twice. Because $(b_k)$ is never constant, there will remain infinitely many points in the new sequence. Since it is a subsequence of a convergent sequence it will also have the limit $p$. Is this a good idea and how does one remove maybe countable infinite many different points from a set?

答: Let $n_1=1$. Now, let $n_2$ be the smallest natural $n$ such that $d(b_n,p)<d(b_{n_1},p)$. Then, let $n_3$ be the smallest natural $n$ such that $d(b_n,p)<d(b_{n_2},p)$, and so on. With this definition, $\lim_{k\to\infty}b_{n_k}=p$ and $k\ne l\implies b_{n_k}\ne b_{n_l}$.


问:
Mumford's regularity theorem
I am reading the section on Castelnuovo-Mumford regularity from Lazarsfeld's Positivity in algebraic geometry.

Theorem 1.8.3 in the book reads as follows:

Let $F$ be an $m$-regular sheaf on $P^n$. Then for every $k \geq 0$

(i). $F(m+k)$ is generated by its global sections.

(ii). The natural maps
$$H^0(P^n, F(m))\otimes H^0(P^n,O_{P^n}(k))\rightarrow H^0(P^n,F(m+k))$$
are surjective.

(iii). $F$ is $(m + k)$-regular.

The proof claims (i) is a consequence of (ii): it says that, since for $l\gg 0$, $F(m+l)$ is globally generated, the surjectivities in (ii) imply that $F(m)$ itself must be globally generated. I am not able to understand this statement.

Choose $l\gg 0$ such that $F(m+l)$ itself is globally generated. Now consider the morphism
$$H^0(F(m))\otimes O_{P^n}\rightarrow F(m)\,.$$
We need to prove that this is surjective.

Tensoring by $O(l)$, we get
$$H^0(F(m))\otimes O_{P^n}(l)\rightarrow F(m+l)\,.$$

A priori we do not know the injectiveness or surjectiveness of the above morphism. But if we take global sections, by (ii), the global sections morphism is surjective.

Does this mean $H^0(F(m))\otimes O_{P^n}(l)\rightarrow F(m+l)$ is surjective? If so, I can tensor back by $O(-l)$ and get my required result.

Thanks in advance.

答:
Lemma: If $\mathcal{F}_1\to\cdots\to\mathcal{F}_n$ is an exact sequence of coherent sheaves on $\Bbb P^n$, then there exists an $l_0>0$ so that for all $l\geq l_0$, the sequence of global sections $\mathcal{F}_1(l)(\Bbb P^n)\to\cdots\to\mathcal{F}_n(l)(\Bbb P^n)$ is exact.

Proof: Saying the initial sequence is exact means we have short exact sequences $0\to\mathcal{G}_i\to\mathcal{F}_i\to\mathcal{G}_{i+1}\to 0$ for all $i$, where $\mathcal{G}_i=\ker(\mathcal{F}_i\to\mathcal{F}_{i+1})=\operatorname{im}(\mathcal{F}_{i-1}\to\mathcal{F}_i)$ are coherent. By a result of Serre, there exists an $l_i>0$ so that for all $l\geq l_i$ we have $H^1(\mathcal{G}_i(l))=0$, or that the sequence of global sections $0\to\mathcal{G}_i(l)(\Bbb P^n)\to\mathcal{F}_i(l)(\Bbb P^n)\to\mathcal{G}_{i+1}(l)(\Bbb P^n)\to 0$ is exact. Letting $l_0=\max(l_i)$ and stitching together the exact sequences on global sections, we see the result. $\blacksquare$

Now let $\mathcal{K}$ and $\mathcal{R}$ be the kernel and cokernel of $H^0(\mathcal{F}(m))\otimes\mathcal{O}_{\Bbb P^n} \to \mathcal{F}(m)$, respectively. Consider the exact sequence $$0\to \mathcal{K}(l)\to H^0(\mathcal{F}(m))\otimes\mathcal{O}_{\Bbb P^n}(l) \to \mathcal{F}(m+l)\to \mathcal{R}(l)\to 0.$$
As $\mathcal{K}$ and $\mathcal{R}$ are coherent, there exists an $l_0$ so that for all $l\geq l_0$ we have that the sequence of global sections is exact and the middle map is surjective. Therefore $\mathcal{R}(l)$ has vanishing global sections for all large enough $l$, which means $\mathcal{R}=0$. Therefore the map $H^0(\mathcal{F}(m))\otimes\mathcal{O}_{\Bbb P^n} \to \mathcal{F}(m)$ is surjective and $\mathcal{F}(m)$ is globally generated.

Title: Mumford's regularity theorem


问:
degree of field extension / splitting field
i was wondering if anyone could give me a quick explanation about a problem concerning the degree of a field extension $Z\mid \mathbb{Q}$.  The problem is as follows :

Let $Z\subset \mathbb{C}$ be the splitting field of $f(X)=X^7-3$. Also let $\alpha = 3^{\frac{1}{7}}$ and $\zeta=e^{\frac{2\pi i}{7}}$. Calculate $[Z : \mathbb{Q}]$.

I know the Polynomial $f$ to be irreducible because of Eisenstein (and also about the product rule for the degree of extension fields), but every time the field extension is slightly more complicated than simply adjoining the square root of a prime to $\mathbb{Q}$, i get confused because i cant simply say $[Z : \mathbb{Q}]=7$, because that would leave out the roots that have non zero imaginary parts(?). What exactly should my thought process be in these types of exercises?

答:
Hint:

The roots of your polynomial are $\;\alpha,\,\alpha\zeta,\ldots,\alpha\zeta^6\;$, yet observe that in fact $\;Z=\Bbb Q(\alpha,\,\zeta)\;$, and thus for its degree you can try

$$|Z:\Bbb Q|=\left|\Bbb Q(\alpha,\zeta):\Bbb Q(\zeta)\right|\,\left|\Bbb Q(\zeta):\Bbb Q\right|$$

Can you get it from here? Try to check stuff about irreducible polynomials and etc.


问:
Prove that for $n\ge 1$, $1^3+2^3+\dotsb+n^3> \frac{n^4}{4}$.

Prove that for $n\ge 1$, $1^3+2^3+\dotsb+n^3> \frac{n^4}{4}$.

I'm having a hard time to prove it. I would appreciate if you could give me some help.

答: Hint: $1^3+2^3+\cdots n^3 = \dfrac{n^2(n+1)^2}{4}$. Perhaps you want to show this by induction first before proceeding to show your inequality.


答:
hint

If you use induction, you will have to prove that

$$\frac{n^4}{4}+(n+1)^3> \frac{(n+1)^4}{4}$$

or

$$4(n+1)^3>(n+1)^4-n^4$$

Body:
Prove that for $n\ge 1$, $1^3+2^3+\dotsb+n^3> \frac{n^4}{4}$.

I'm having a hard time to prove it. I would appreciate if you could give me some help.


问:
Question about product of invertible matrix
Let $A$ be an $n\,\rm{x}\,n$ matrix. Prove that there exists a matrix $B$ of the same size and $B$ is invertible such that $BA$ is upper-triangular.

答:
We prove this by induction on $n$. For $n = 1$, the statement is obvious.

Let $X$ be the first column of $A$. If $X \ne 0$, there is some invertible $B_1$ such that $B_1 X = (1, 0, \dots, 0)^t$. If $X = 0$, we let $B_1$ be an arbitrary $n \times n$ invertible matrix.

Now write
$$
B_1 A =
\begin{pmatrix}
 * & * \\
0 & A_1
\end{pmatrix},
$$
in block form, where $A_1$ is an $(n-1) \times (n - 1)$ matrix.

By the induction hypothesis, there is some invertible $B_2$ such that $B_2 A_1$ is upper triangular.

Now let
$$
B =
\begin{pmatrix}
1 & 0 \\
0 & B_2
\end{pmatrix}
B_1.
$$
Then $B$ satisfies the required conditions.


Title: Prove that for $n\ge 1$, $1^3+2^3+\dotsb+n^3> \frac{n^4}{4}$.

Body: Hint: $$\frac{n^4}4 = \int_{0}^n t^3 \; dt$$