问:
Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$
where F is a differentiable function of  $x, y,z, t$ and $x, y, z$ differentiable functions of $t$, Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$

We define $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ as the position vector

I would like you to give me some suggestions. It has been difficult for me to know where to start

I did this but I don't know if it's okay

$\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$

$\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$

$\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$

答:
Use the chain rule to get, $$\dfrac{dF}{dt}=\dfrac{\partial F}{\partial t}+\dfrac{\partial F}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial F}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial F}{\partial z}\dfrac{dz}{dt}$$

Should be clear from there.


问:
Guidance with proof regarding divisibility and congruence.
If 11 ∤ n, prove that $n^{5}$ ≡ 1 mod 11 or $n^{5}$ ≡ -1 mod 11 .

I am taking my first proof writing class, and am unsure how to solve this problem. I tried using the definition of congruency to set $n^{5}$ ≡ ±1 mod 11 equivalent to $n^{5}$ - 1 = 11k, where k is any integer. Then I factored the left hand side, but I am unsure if this is leading me to the right direction. Any help, guidance, or answer would be very much so appreciated. Thank you

答:
Write $n = 11k + r$, where $k,r$ are integers with $1 \leq r \leq 10$. Note that we have
$$
(11 k + r)^5 = \binom {5}0 (11k)^5 + \cdots + \binom {5}4 (11k)r^4 + \binom 55 r^5\\
= 11 \cdot \left[\binom {5}0 11^4k^5 + \cdots + \binom {5}4 kr^4 \right] + r^5.
$$
That is, we have $n^5 = r^5$ modulo $11$.

With that, it suffices to show that we have $r^5 \equiv \pm 1 \pmod {11}$ for each $r \in \{1,2,\dots,10\}$.

If you prefer, you could consider the set $\{\pm 1, \pm 2, \dots, \pm 5\}$.

ContentLicense: CC BY-SA 4.0


问:
Liu's Algebraic Geometry Ex 2.12
The exercise to be shown is

Let $f:X \rightarrow Y$ be a morphism of ringed topological spaces. Let $V$ be an open subset of $Y$ containing $f(X)$. Show that there exists a unique morphism $g:X \rightarrow V$ whose composition with the open immersion $i:V \hookrightarrow Y$ is f.

My attempt so far:

Define the map $g$ on topological spaces to be $g(x) = f(x)$. Define the map $g^\#:\mathcal{O}_V \rightarrow g_*\mathcal{O}_X$ as given by for every $W \subset V$ open, $g^\#_W:\mathcal{O}_V(W) \rightarrow g_*\mathcal{O}_X(W) = \mathcal{O}_X(g^{-1}(W))$ by $s \mapsto f^\#(s)|_{g^{-1}(W)}$.

I'm not sure if that latter map is right.

Clearly we have that the composition of the maps of topological spaces is f. But I'm not sure about the map of sheaves. Clearly, we have that $(i \circ g)^\#_V:\mathcal{O}_Y(V) \rightarrow \mathcal{O}_X ((i \circ g)^{-1}(V)) = \mathcal{O}_X(f^{-1}(V))$ goes to the right place. But I'm confused about the induced map of sheaves. Will it is given by $s \mapsto g^\#(i^\#(s)) = g^\#(s) = f^\#_{f^{-1}(V)}(s)$? I'm fairly certain I'm misunderstanding something about these maps.

Any help understanding the sheaf map and the latter part of this exercise would be appreciated

答:
I'm not sure why you write $s \mapsto f^\#(s)|_{g^{-1}(W)}$.  The set $g^{-1}(W)$ is equal to the set $f^{-1}(W)$, so $f^\#(s)$ is already an element of $\mathcal{O}_X(g^{-1}(W))=\mathcal{O}_X(f^{-1}(W))$.  In other words, you can just define $g^\#_W$ to be literally the same thing as $f^\#_W$, since $\mathcal{O}_V(W)=\mathcal{O}_Y(W)$ and $\mathcal{O}_X(g^{-1}(W))=\mathcal{O}_X(f^{-1}(W))$.

Here's how you then verify that $(i\circ g)^\#=f^\#$.  For any open $W\subseteq Y$, $i^\#:\mathcal{O}_Y(W)\to\mathcal{O}_V(i^{-1}(W))=\mathcal{O}_V(V\cap W)=\mathcal{O}_Y(V\cap W)$ is just the restriction map.  So $(i\circ g)^\#_W:\mathcal{O}_Y(W)\to\mathcal{O}_X(g^{-1}(V\cap W))=\mathcal{O}_X(f^{-1}(V\cap W))$ is just the map that takes a section, restricts it to $V\cap W$, and then applies $g^\#_{V\cap W}=f^\#_{V\cap W}$.  Since $f^\#$ is compatible with restriction, this is the same as first applying $f^\#_W$ to get an element of $\mathcal{O}_X(f^{-1}(W))$ and then restricting it to $\mathcal{O}_X(f^{-1}(V\cap W))$.  But $f^{-1}(W)=f^{-1}(V\cap W)$, so this last restriction does nothing, so this says $(i\circ g)^\#_W=f^\#_W$.

Title: Liu's Algebraic Geometry Ex 2.12


问:
Dirac Delta Function "removing integral"
I am trying to proof the following  identity:
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
where $\delta$ represents the dirac delta function.
I found that
\begin{align}
  \int_{-\infty}^\infty  f(x) \delta(ax) \,\,dx = \frac{1}{|a|} \int_{-\infty}^\infty f(x)\delta(x)\,\,dx.
\end{align}
Consequently we get
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
by "removing" the integrals. Is this step actually valid?

答:
Shortly: yes, but talking about "removing the integrals" is an oversimplification.

A bit more complete answer: $\delta(x)$ is defined by
$$\int_{-\infty}^\infty f(x)\,\delta(x)\,dx=f(0)$$
(it must hold for any sufficiently regular $f(x)$).
You found that, for any sufficiently regular $f(x)$,
$$\int_{-\infty}^\infty f(x)\,\delta(ax)\,dx
 = \frac{1}{|a|}\int_{-\infty}^\infty f(x)\,\delta(x)\,dx
 = \frac{1}{|a|}f(0)\;,$$
so, after multiplying both sides by $|a|$, you obtain
$$\int_{-\infty}^\infty f(x)\,|a|\delta(ax)\,dx=f(0)\;.$$
Since it holds for any $f(x)$, it means (by definition) that
$$|a|\delta(ax)=\delta(x)\;.$$


问:
Check if polynomial can be rook polynomial
How to verify if a given polynomial is a rook polynomial? Let’s assume I have a chessboard 5x5 and those 5 polynomials:

$r(x) = 2 + 11x + 35x^2 + 50x^3 + 26x^4 + 5x^5$

$r(x) = 1 + 15x + 35x^2 + 50x^3 + 26x^4 + 5x^6$

$r(x) = 1 + 10x + 35x^2 + 50x^3 + 26x^4 + 5x^5$

$r(x) = 1 + 15x + 35x^2 + 50x^3 + 5x^5$

$r(x) = 1 + 26x + 35x^2 + 50x^3 + 26x^4 +5x^5$

I know the first one can not be rook polynomial because there is 2 instead of 1 on the first position.

Second is not valid because six rooks can't be placed on 5x5 board($x^6$).

4th is also not valid because if 4 rooks can not be placed on the board then it's not possible to place 5 rooks(missing $x^4$).

I do not have a clue how to verify if 3rd and 5th are valid or not, any ideas?

答:
According to wikipedia the coefficient of $x^5$ is the number of ways to place 5 non-attacking rooks on a 5x5 board, and this is not $5$.

Every row and columm must have exactly one rook in such configurations, so you have 5 spots available for the first columm, 4 spots for the second and so on. So the $x^5$ coefficient must be $5!$. For this reason, none of these polynomials could be rook polynomials.


问:
Is there a continuous, or smooth, generalization of the iterated logarithm $\log^* n$?
I was wondering about the properties of certain classes of functions related to algorithmic runtimes in this post.

So I know that $n!$ has a continuous/smooth generalization to the gamma function. Is there a continuous/smooth generalization of the iterated logarithm $\log_*n$?

答:
Let $f(x)$ be smooth s.t. $f(0) = 1$, $f(x) = 0$ if $|x| > 1$ - bump function, for example $f(x) = \exp(-\frac{1}{1 - x^2})$ if $|x| < 1$ and $f(x) = 0$ otherwise.

Then $g(x) = \sum\limits_{n=1}^\infty \log^*n \cdot f(x - n)$ (this series has at most one non-zero term for any given $x$, so convergence isn't an issue) is smooth extension of iterated logarithm: $g(n) = \log^* n$. It's quite useless, however.

You may also be interested in super logarithm - it has the same asymptotic, $\log^*n = \lceil \operatorname{slog}n\rceil$, and it has analytic extension. For details, check Uniqueness of Analytic Abel Functions in
Absence of a Real Fixed Point by Trappman, Kouznetsov and references from it.


答: Write the multivariate chain rule in the language of infinitesimals, viz.$$dF=\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz.$$Note there is nothing different about $t$ here as compared with $x,\,y,\,z$. But if $x,\,y,\,z$ are differentiable functions of $t$ so an $F(t,\,x,\,y,\,z)$ differentiable with respect to each of its four arguments simplifies to a differentiable function of $t$, division by $dt$ gives$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}.$$Finally, write the last three terms as a dot product:$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot\frac{d\vec{r}}{dt}.$$

Body:
where F is a differentiable function of  $x, y,z, t$ and $x, y, z$ differentiable functions of $t$, Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$

We define $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ as the position vector

I would like you to give me some suggestions. It has been difficult for me to know where to start

I did this but I don't know if it's okay

$\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$

$\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$

$\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$


问:
Counting the number of ways to break the symmetry of an N x N grid by placing one X and one O at a time
I have recently been working out the full game tree for tic-tac-toe, just for fun.  I am using the well known equivalence relation of rotations/reflections to simplify this tree in the standard way (which begins by noting that there are only 3 opening moves: edge, corner, and center).  I actually could not find an image of the full game tree for tic-tac-toe, so if anyone can provide a link I'd appreciate that.

This work motivates the following;

Let there be given an N x N grid; and let $m$ be a natural number.

Let $I$ be the set of all possible ways to place $m$ copies of the letter $X$ in the grid, and $m$ copies of the letter $O$ in the grid (with the restriction that we can only place one letter per space of the grid; in other words, just imagine playing a game of tic-tac-toe on an N x N board for an even number of moves).

Problem:

How many ways are there to completely break the rotational/reflection symmetry of an N x N grid by placing $m$ copies of $X$ and $m$ copies of $O$?

答: This is not the full game tree, but the complete optimal tree for tic-tac-toe.

ContentLicense: CC BY-SA 2.5

ContentLicense: CC BY-SA 3.0


问:
A question on countability of isolated points of a subset of R
The question is to prove that with respect to the euclidean metric on the Real numbers prove that if A is any subset of R, isoA is countable and hence deduce  that if A is uncountable the A' is uncountable.

I am still not that comfortable with proving things like countability so any answers or help is appreciated

thanks

答:
Let $A \subset \mathbb{R}$ and $\text{iso}(A)$ the set of isolated points of $A$. For any $x \in \text{iso}(A)$, there exist two rationals $a<b$ such that $x \in (a,b)$ and $(a,b) \cap A= \{x\}$. Let $\phi(x)=(a,b) \in \mathbb{Q}^2$.

Then $\phi : \text{iso}(A) \to \mathbb{Q}^2$ is injective and $\text{iso}(A)$ is at most countable.

For your second question, write $A= \text{iso}(A) \coprod A'$ and use the fact that the union of two countable sets is countable.

Remark: The axiom of choice is not needed to construct $\phi$.


问:
Prove $\sum_{k=1}^{\lfloor n/2\rfloor} (n+1-2k)\sin\frac{(2k-1)\pi}n= \frac n2\csc\frac\pi n $
Encountered the series summation below

$$\sum_{k=1}^{\lfloor n/2\rfloor} (n+1-2k)\sin\frac{(2k-1)\pi}n
$$

which is supposed to reduce to the close-form expression $\frac n2\csc\frac\pi n$. I have struggled though to see it quickly with familiar  algebraic manipulations. The proportional term $2k$ in the sequence is a bit annoying.

答:
I won't provide a complete answer since I think you're more than capable of working out the details, but the first thing to consider is writing $$\sin \frac{(2k-1)\pi}{n} = \sin \left( \frac{(2k-1-n)\pi}{n} + \pi \right) = -\sin \frac{(2k-1-n)\pi}{n}.$$

Next, consider the function $$f(z) = \sum_{k=1}^{\lfloor n/2 \rfloor} \cos (2k-1-n)z.$$  Then look at the derivative of this function with respect to $z$, evaluated at $z = \pi/n$.


问:
How would I use Green's Theorem to evaluate this Line Integral?
If I have the closed loop $C$ (https://ibb.co/hK7xC5V) which is the union of $C_1,C_2,C_3$ and $C_4$ where
$$C_1=\sin(x)-2\pi,\text{where $x$ goes from 0 to $2\pi$}$$
$$C_2=-\sin(y)+2\pi,\text{where $y$ goes from $-2\pi$ to 0}$$
$$C_3=\sin(x),\text{where $x$ goes from $2\pi$ to 0}$$
$$C_4=-\sin(y),\text{where $y$ goes from 0 to $-2\pi$}$$
How would I evaluate $$\oint\limits_C\vec{F}\cdot d\vec{r}$$ using Green's Theorem?

(An example would also be appreciated with any field $\vec{F}$ which has nonzero $x$ and $y$ components)

答:
(1) Trivial case: If $\vec F$ is conservative, then $\oint_C\vec F\cdot\mathrm d\vec r=0$ for any path $C$.

(2) Simple, not-as-trivial case: If $\vec F(x,y)=\left(-\frac y2,\frac x2\right)$, then $\oint_C\vec F\cdot\mathrm d\vec r$ is simply the area of the region $D$ bounded by $C$, since

$$\frac{\partial\left(\frac x2\right)}{\partial x}-\frac{\partial\left(-\frac y2\right)}{\partial y}=\frac12+\frac12=1$$

$$\implies\oint_C\vec F\cdot\mathrm d\vec r=\iint_D\mathrm dA$$

Take $D'$ to be the region (shaded red below) to be

$$D'=\left\{(x,y)\mid0\le x\le2\pi\,\land\,\sin(x)-2\pi\le y\le\sin(x)\right\}$$

The area of blue-shaded regions is accounted for in $D'$ by virtue of symmetry, so that both $D$ and $D'$ share the same total area.

Then $\iint_D\mathrm dA$ is simply the area between two curves,

$$\iint_D\mathrm dA=\int_0^{2\pi}\int_{\sin(x)-2\pi}^{\sin(x)}\mathrm dy\,\mathrm dx=4\pi^2$$

(3.a) The more general case: Unfortunately, the symmetry argument wouldn't hold up for arbitrary vector field $\vec F$. In this case, I'm not sure there's any way to avoid splitting up $D$ into smaller, easy-to-integrate-over regions. Perhaps a clever change of coordinates can be employed, but nothing immediately comes to mind that would help in integrating over the entirety of $D$ in one fell swoop.

Consider partitioning $D$ along the lines $x=\frac\pi2$, $x=\frac{3\pi}2$, $y=\frac\pi2$, and $y=\frac{3\pi}2$. This splits up $D$ into $9$ smaller regions $D_1,D_2,\ldots,D_9$ whose boundaries are easy enough to deal with.

The red region highlighted here is the set

$$D_1=\left\{(x,y)\mid -\sin(y)\le x\le\frac\pi2\,\land\,-\frac\pi2\le y\le0\right\}\cup\left\{(x,y)\mid \sin^{-1}(y)\le x\le\frac\pi2\,\land\,0\le y\le1\right\}$$

and integrating over $D_1$ requires splitting up into two more integrals.

Suppose $\vec F(x,y)=(0,x^2)$ (just to make things relatively easy). Then this first region's contribution to the total line integral would be

$$\iint_{D_1}2x\,\mathrm dA=\int_{-\frac\pi2}^0\int_{-\sin(y)}^{\frac\pi2}x\,\mathrm dx\,\mathrm dy+\int_0^1\int_{-\sin^{-1}(y)}^{\frac\pi2}x\,\mathrm dx\,\mathrm dy=1-\frac\pi8+\frac{\pi^3}{16}$$

While manageable, this method forces one to painstakingly set up and compute $13$ total integrals.

(3.b) Better method for the general case: As Mark S. points out, you can indeed simplify the integral over $D$ by summing the five integrals over the square $[0,2\pi]\times[-2\pi,0]$ and the integrals over the "lobes" of $D$ that extend outside of the square (two of these lobes are highlighted in blue in the second case above), then subtract the four integrals over the lobes that dip into the square.

In terms of this sketch, you would integrate over the square and the lighter-shaded lobes, then subtract the integrals over the darker-shaded ones.


问:
Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid?
Suppose that we have a convex polyhedron $P$, such that the symmetry group of $P$ acts transitively on its vertices, edges, and faces (that is, it is isogonal, isotoxal, and isohedral). It then follows that $P$ is one of the Platonic solids, which have a number of equivalent definitions but usually includes some stipulation that the faces are regular. (Once you have regular faces plus these symmetry conditions, most other properties follow easily.)

This can be shown via a rather gross casework argument on the number of sides at each face. By face-transitivity, all faces are congruent (in fact, we only need that the polyhedron be monohedral for this). By edge-transitivity, said faces are equilateral polygons. Then we break into cases by the number of sides of the polygons:

If the faces are triangles, being equilateral implies being regular.

If the faces are non-regular equilateral quadrilaterals, they are rhombi. Then, at every vertex there must be an equal number of "skinny" and "wide" angles meeting, and there must be at least 3 faces meeting at a vertex. But then there are at least two skinny and two wide angles (since we assume the angles are different), so the angles at each vertex add up to at least 360 degrees, which is not possible in a convex polyhedron.

If the faces are non-regular equilateral pentagons, then they have angles which are not all the same; however, each vertex must have the same angles around it, so each vertex has a proportional fraction of the angles of the pentagons. But for any distribution other than the uniform one, the only way to have a proportional amount of every angle is to have some $n$ copies of every angle with multiplicity, since $5$ is prime. (E.g. if there are $2$ angles of one size and $3$ of another, we can't get a $2:3$ ratio without having at least $5$ angles at every vertex.) But the angles of a pentagon add to more than $360^\circ$, so we can't have them all at once.

If the faces are hexagons or higher, they cannot form a topological sphere by Euler's formula.

However, I don't like this proof very much; I walk away from it with no understanding of why this behavior occurs, and it certainly doesn't suggest any generalizations to other dimensions (where I believe the analogous statement - that transitivity on every degree of cell in a convex polytope implies regularity - is true, though I would appreciate a reference for this fact).

Is there a natural or more elegant proof of this statement?

One thing to note is that any proof must somehow use the fact that we are working on the sphere. In the case of planar tilings, the corresponding implication is false: being transitive on faces, vertices, and edges does not imply that one is looking at a tiling of regular polygons! As a counterexample, consider a rhombic tiling like this one:

As another aside, relaxing any one of these conditions allows for polyhedra whose faces are not congruent regular polygons: the rhombic dodecahedron is face-transitive and edge-transitive, an isosceles tetrahedron is face-transitive and vertex-transitive, and the cuboctahedron is edge-transitive and vertex-transitive.

答:
I shall use Euler's polyhedron formula. (It does generalise to a higher dimension!)  Let each face have $e$ edges and let $\rho$ edges meet at each vertex and then we have the standard result that $$\frac{1}{\rho}+ \frac{1}{e}>\frac{1}{2} \text { and therefore one of } e,\rho \text { is } 3.$$ If $e=3$ then we have triangular faces with edges of equal length and we are finished.

Therefore $\rho=3$. If the three angles at a vertex are not all equal the ratio of different sized angles is either $1:2$ or $1:1:1$. This ratio must be the same for the angles in a face and so $e$ is a multiple of $3$ and then $\frac{1}{\rho}+ \frac{1}{e}$ is too small.


问:
Number of points of the elliptic curve $y^2 + y = x^3$ over $F_q$
Given the elliptic curve $y^2 + y = x^3$ over $\mathbb{F}_q$ ($q=p^r$, where $p$ is prime), I want to prove that if $q \equiv 2 \bmod 3$, then the elliptic curve has $q + 1$ points.

My exercise says that $q$ can be even, but I thought this kind of curve assumed that $q = 2^r$. And I don't know how to apply the Legendre symbol to this case because the left hand side contains $+ y$.

答:
It is actually easier than you thought.

It is known that the multiplicative group $\Bbb F_q^\times$ is a cyclic group with $q - 1$ elements. If $q \equiv 2\mod 3$, then its order is prime to $3$.

This means that the map $z \mapsto z^3$ is an automorphism of $\Bbb F_q^\times$.

Therefore, for any $y \in \Bbb F_q$, there exists a unique $x$ such that $x^3 = y^2 + y$.

Together with the infinite point, we get $q + 1$ points in total.


问:
Given the orthonormal base of a given vector, find the coordinates of that vector on another base

In an orthonormal base, the coordinates of the vector $\vec{v}=(2,-5)$. Find the coordinates of $\vec{v}$ on the base $B=((1,-1),(0,-1))$

Attempt. Actually not 100% sure on what the intuition of a base in general is, I'm trying to understand it through exercises. Anyways, my attempt:

Any base $B$ is defined as $B=a\vec{x}+b\vec{y}$ where $a$ and $b$ are constants. If the base is orthonormal, it means the vectors that form that base form 90 degrees and have a length of 1. I related this base to the usual $\hat{i}$ and $\hat{j}$ one. On any case, I did: $$\vec{v}=a\hat{i}+b\hat{j}$$ and found $a=2$ and $b=-5$. The new base will have the same constants $a$ and $b$? Don't know at all if this method is valid but here it is: the new base is $B'=a(1,-1)+b(0,-1)$ and as we know $a=2$ and $v=-5$, $\vec{v}$ will be this? $$\vec{v}=2(1,-1)-5(0,-1)=(2,3)$$ which actually gives the right answer in the book but I'm afraid the procedure is bad or if I feel like I have some concepts wrong about bases.

答:
2 ways to do this.

One is to say $e_1 = (1,-1), e_2 = (0,-1)$  These are our basis vectors of $B.$

There exists some combination basis vectors such that $a e_1 + b e_2 = v = (2,-5)$

This provides a system of equations to solve.

$(a,-a) + (0,-b) = (a, -a-b) = (2,-5)\\
a = 2\\
-a-b = -5\\
a = 2, b = 3$

And, this is probably how your Prof. expects you do to solve this.

But, this is worth knowing.

$B = \begin{bmatrix} 1&0\\-1&-1 \end {bmatrix}$

$Bu$ will take a vector $u$ represented in the basis $B$ and translate it to the standard basis.

That means that $B^{-1}v$ will take a vector $v$ in the standard basis and represent it in terms of the basis $B$

The reason the method you use above gives the right answer is that in this case $B^{-1} = B$


问:
Does anyone know any resources for Quaternions for truly understanding them?
I've been studying Quaternions for a week, on my own. I've learned various facts about them but I still don't understand them. My goal is to understand rotation quaternions specifically. I don't want to just memorize the formulas for using rotation quaternions, I want to truly understand them.

It's the most difficult thing I've ever tried to learn. Does anyone know any good resources for understanding quaternions?

答:
I know that I am very late in this response, and that probably you found some paths to follow on quaternions, however, I thought that you might like to know these original resources on quaternions that I discovered when doing my own research to learn about quaternions myself:

An Elementary Treatise On Quaternions_Peter Guthrie Tait

Introduction to Quaternions_Philip Kelland, Peter Guthrie Tait

On Quaternions_William Rowan Hamilton

Elements of Quaternions_William Rowan Hamilton

Also, to grasp Quaternions intuitively, I recommend this book that I recently found: Visualizing Quaternions by Andrew Hanson.

Best of luck in your learning ventures!


答:
We assume $n>1$. Let the sum be $S_n$; observe that $\sum_{k=1}^{\lfloor n/2\rfloor}=\frac12\sum_{k=1}^n$ because 1) the term is invariant under $k\mapsto n+1-k$ and 2) if $n$ is odd, the term at $k=(n+1)/2\color{gray}{[=n+1-k]}$ is zero. Hence,
\begin{align*}
4S_n\sin\frac{\pi}{n}&=\sum_{k=1}^n(n+1-2k)\left(\cos\frac{2(k-1)\pi}{n}-\cos\frac{2k\pi}{n}\right)
\\&=\sum_{k=0}^{n-1}(n-1-2k)\cos\frac{2k\pi}{n}-\sum_{k=1}^n(n+1-2k)\cos\frac{2k\pi}{n}
\\&=\underbrace{(n-1)}_{k=0}-\underbrace{(1-n)}_{k=n}-2\underbrace{\sum_{k=1}^{n-1}\cos\frac{2k\pi}{n}}_{=-1}=2n.
\end{align*}

Title: Prove $\sum_{k=1}^{\lfloor n/2\rfloor} (n+1-2k)\sin\frac{(2k-1)\pi}n= \frac n2\csc\frac\pi n $

Body:
Encountered the series summation below

$$\sum_{k=1}^{\lfloor n/2\rfloor} (n+1-2k)\sin\frac{(2k-1)\pi}n
$$

which is supposed to reduce to the close-form expression $\frac n2\csc\frac\pi n$. I have struggled though to see it quickly with familiar  algebraic manipulations. The proportional term $2k$ in the sequence is a bit annoying.


问:
CLT infinite mean
I am trying to see if we can construct $(X_n)_{n\ge 1}$, a sequence of i.i.d random variables such that $E|X_n|=\infty$ while $n^{-1/2}(X_1+...+X_n)$ converges in distribution goes to a $N(0,1)$ random variable. I tried using
$$ n^{1/2}(1/n(X_1+..+X_n))$$ and using SLLN on the inside but then the limit would vanish. Is it even possible to construct such a sequence?

答: No, it is not possible. For the CLT, we need to have almost two finite absolute moments and if the first absolute moment is infinite, so are all higher order moments (this follows from Hölder's inequality). Hence, the CLT does not hold.</p>

Title: CLT infinite mean


问:
Where does $k^{n-1}$ come from in this solution
Solution

I started this problem by plugging $\mathrm{det}(kA)$ into the first row cofactor expansion formula we're given, and I got stuck at approximately the second step. This solution manual pulled the $k$ out and put it to the power of $n-1$. Is this because you're using every row except the last/first one when you multiply $k$? If so, why?

答:
In (c), the proof of $\det(kA)=k^n\det(A)$ of an $n\times n$ matrix uses induction, so it is natural to step from  $\det(kA)=k^{n-1}\det(A)$ for an $(n-1)\times (n-1)$ matrix using the inductive hypothesis.

The remaining $k$ is a factor in the additional row.


问:
'Weird' Indexing In The MILP Literature
Fairly new to MILP. I posted this question as an issue on the Gurobi examples repo too, but it might be more appropriate to ask it here.

NOTE: Since I "need at least 10 reputation to post images" , the images won't render, but they are available on the github issue linked above and I have linked to the urls (sorry).

I was following this great example from Gurobi.

In the model formulation, we have:
[link to image of formulation]https://user-images.githubusercontent.com/19332617/108038384-70a65200-7043-11eb-9fbd-1eb9005240b6.png

I see that we have:
d∈D={n+1,n+2,...,n+m}: Index and set of depots (service centers), where m is the number of depots.
I am confused as to how there can be n+m depots.
I am also confused as to why the set starts from n+1 instead of 1. In the example there are two depots, not 9 i.e 2+7.

I have run into something similar when studying the literature (Ramkumar et al 2012, their formulation also has confusing indices for the number of depots:
link to image of Ramkumar formulation
)

Edit: Indeed, the paper referenced in the notebook also notes "(1,..,n+m) where the m depots are represented by n+1,...,n+m" but does not state their reasoning for doing so. (S. Salhi, A. Imran, N. A. Wassan. The multi-depot vehicle routing problem with heterogeneous vehicle fleet: Formulation and a variable neighborhood search implementation. Computers & Operations Research 52 (2014) 315-325.)

Edit 2: However, in the Supply Network I  example, we have something that makes more sense (to me).:
link to Supply Network formulation image

Here there is no (n+1...n+m) indexing.

Any enlightenment on this indexing and why it used would be greatly appreciated.

答: The idea is that there are $n+m$ nodes in total, with $n$ of one type (customer) and $m$ of the other type (depot).  The first $n$ are customers $\{1,\dots,n\}$, and the last $m$ are depots $\{n+1,\dots,n+m\}$.

Title: 'Weird' Indexing In The MILP Literature


问:
Tag condition for Riemann integrability
As part of the definition for a definite integral we say that the choice of tag in the partition should not affect the limit of the Riemann sum for that function to be considered Riemann integrable. What is an example of a function where the location of the tag changes the limit of the sum?</p>

答:
I will assume that you mean the choice of intermediate points associated with a given partition $P.$ Consider the function $\chi := \chi_{\mathbb{Q} \cap [0, 1]},$ the indicator function of the rationals in $[0, 1].$ Let us choose a partition $P := (x_0 = 0 < x_1 < \dots < x_{n - 1} < x_n = 1)$ of the interval $[0, 1]$ and choose two systems of intermediate points $\xi := (\xi_1, \dots, \xi_n),$ $\zeta := (\zeta_1, \dots, \zeta_n),$ where $\xi_j, \zeta_j \in [x_{j - 1}, x_j]$ for each $1 \leq j \leq n$ such that $\xi_j \in \mathbb{Q}$ and $\zeta_j \not\in \mathbb{Q}$ for all $1 \leq j \leq n.$ Then $\sum_{1 \leq j} \chi(\xi_j)(x_j - x_{j - 1}) = \sum_{1 \leq j \leq n} (x_j - x_{j - 1}) = 1$ and $\sum_{1 \leq j \leq n} \chi(\zeta_j) (x_j - x_{j - 1}) = 0,$ so this function is not Riemann integrable because it matters which system of intermediate points you choose to compute the Riemann sum.

The behaviour of this function under Riemann integrability is even nastier than this in fact. You can easily show that you may find some partitions and certain systems of intermediate points such that the associated Riemann sum converges to any number you like $\lambda \in [0, 1].$ However, this function is Lebesgue integrable and its integral is equal to $0.$ Hope this helps. :)

Title: Tag condition for Riemann integrability


问:
Determining instantaneous rates of changes for quartic functions
I have 5 quartic functions that were found with quartic regression. Each function models a country's relationship with rotavirus vaccination rates (x) against years (y, years are integers and not real values, ex. year 1 -> x=1, etc.)

f(x) = ax4 + bx3 + cx2 + dx + e
Domain and Range

Morocco
y = -0.220x⁴ + 5.44x³ - 47.3x² + 171x - 123
{x | 1≤x≤10} {y | 5.92 ≤y≤97}

Sudan
y = -0.199x⁴ + 5.27x³ - 50.1x² + 205x - 220
{x | 2≤x≤10} {y | 28.6≤y≤100}

Rwanda
y = -0.318x⁴ + 9.12x³ - 95.1x² + 426x - 592
{x | 3≤x≤10} {y | 50.6≤y≤98}

Ghana
y = -0.290x⁴ + 8.20x³ - 84.6x² + 377x - 518
{x | 3≤x≤10} {y | 49.5≤y≤ 92}

Botswana
y = -0.034x⁴ + 1.28x³ - 16.8x² + 90.2x - 78.3
{x | 3≤x≤10} {y | 72.9≤y≤83.7}

I'm now trying to find instantaneous rates of change for each country. I want to find instantaneous rates for each integer x value within a country's domain. so for Rwanda, it'd be instantaneous rates of change for x=3,4,5,6,7,8,9,10. I'm super new to calculus so I'm not sure how I'd go about finding the derivative of my quartic functions to calculate instantaneous rates of change - if someone could help me through this and possibly show a sample calculation for one instantaneous rate of change for one of my countries, that'd be really appreciated!!

PS: my goal afterwards is to graph each country’s instantaneous rates of change values (y) over time (x) and perform linear regression each. and then, whichever country (instantaneous rate of change linear graph) has the highest slope value = increased the most in vaccination rates over x many years. please let me know if you think this makes sense if you can!

答:
The rate of change over time t of a function is given by its rise over its run, so by $\frac {f(a+t)-f(a)}{t}$. Try to graph a function to see it graphically.

It's clear that by reducing $t$ more and more you get a rate of change over a smaller and smaller amount of time, so ideally as $t$ approaches $0$ (ax $x \to 0$) you'll get the instantaneous rate of change.

$f'(a)$ is none other than $lim_{t \to 0} \frac {f(a+t)-f(a)}{t}$.

Polynomials like yours are quite easy to differentiate; without going into a proof you should know that whenever $f(x)= b x^n$ then $f'(x)=b \;nx^{n-1}$; if $f(x)=c$ for some constant $c$ then $f'(x)=0$ and that $(f+g)'(a)=f'(a)+g'(a)$.

So in the first case we'll have $f'(x)=-0.220 *4 x^3+5.44*3 x^2+47.3*2x+171+0$.


问:
Show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y)=x^2+y$ is continuous in $\mathbb{R}^2$ by $\epsilon - \delta$ definition of continuity.
I need to show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x,y)=x^2+y$ is continuous in $\mathbb{R}^2$  (everywhere) by $\epsilon - \delta$ definition of continuity.

First, I write the definition.

$f$ continuous $\Leftrightarrow \forall a=(a_{1},a_{2}) \in \mathbb{R}^2 \ \ \forall \epsilon >0 \ \ \exists \delta>0 \ \ \forall x=(x_{1},x_{2})\in \mathbb{R}^2 \ \ (\sqrt{(x_{1}-a_{1})^2+(x_{2}-a_{2})^2}<\delta \Rightarrow |f(x)-f(a)|<\epsilon)$

Than, I rewrite that for the problem,

$\sqrt{(x_{1}-a_{1})^2+(x_{2}-a_{2})^2}<\delta \Rightarrow |x_{1}^2+x_{2}-a_{1}^2-a_{2}|<\epsilon$

After, I tried many things to solve this problem. Especially, I think the triangle inequality but i couldn't find the solution.

What should I do? How can I solve this problem?

Thanks a lot for your helps.

Body: Choose $\delta = \min\{1, \frac{\varepsilon}{2(1+2|a_1|)}\}$. Now, $|x_1| \leq |x_1-a_1| + |a_1| \leq 1 + |a_1|$. So, $$|x_1^2-a_1^2+x_2-a_2| \leq |x_1-a_1| \cdot |x_1+a_1| + |x_2-a_2| \leq |x_1-a_1| \cdot (|x_1|+|a_1|) + |x_2-a_2| \leq \delta \cdot (1+2|a_1|) + \delta \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2(1+2|a_1|)} < \varepsilon$$


问:
inclusion of sets -transitive?
show that if $A\subseteq B$ and $B\subseteq C$ then $A\subseteq C$

Can I do it with using injective functions?

$A\subseteq B$  means there exists an injective fcn $f:A\to B$

$B\subseteq C$  means there exists an injective fcn $g:B\to C$

then the composition  $g\circ f:A\to C$ is also an injective function then $A\subseteq C$

in each case all the functions are identity functions

答:
Caution: The existence of an injection between $A$ and $C$ doesn't necessarily imply $A \subseteq C$.

For example, consider set $A$, the set of all even integers, and set $B$, the set of all odd integers. Certainly, there exists an injection: $f: A \to C$, $\;f(x) = x + 1\;$ (which is not only injective, but surjective, as well). But clearly, $\;A \nsubseteq C$.

The converse is true: if $A \subseteq C$, then an injection $h: A \to C$ exists.

But we can easily prove the inclusion $A \subseteq C$ by "element chasing:"  a standard way to prove set inclusions, and/or  set equivalencies.

We have $A \subseteq B$ and $ B \subseteq C$. And we want to prove that this necessarily implies $A\subseteq C$.

$(1)$ Suppose $x \in A\quad $ (Assumption)

$(2)$ We know $A \subseteq B$ means $x \in A \implies x \in B.\;$ So given $(1)$, we have $x \in B$.

$(3)$ We know $B \subseteq C$ means $x \in B \implies x \in C$. So given $(2)$, we have $x \in C$.

$(4)\;\;x \in A \implies x \in C$. $\quad[(1) - (3)]$

Therefore, $A \subseteq C$.

Title: inclusion of sets -transitive?


问:
For the given functions $f$ and $g$, find $(g\circ f)(x)$.
$f(x)=4x^2+4x+6; g(x)=4x-4$

I have always had troubles with functions and need some help solving this one. I have been working on it for too long now and thought that maybe someone else's perspective would be better. I simply don't understand the path to solving it. If someone could walk me through the process that would be much appreciated.

Body: $$ (g\circ f)(x)=g(f(x))=4f(x)-4=4(4x^2+4x+6)-4=16x^2+16x+20$$


问:
find $\left \| {N}' \right \|$
Let $\gamma :I\rightarrow R^{3}$ be a parameterized curve by Arc Length with curvature $\left \| k \right \|=2$ and torsion $\left \|t  \right \|=3$ then find  $\left \| {N}' \right \|$

i tried using frenet formulas but i can only get that $ \left \| {N}' \right \|=\left \| -kT+tB \right \|\leq \left \| -kT \right \|+\left \| tB \right \|\leq 5$

by the fundamental theorem I know this is a propeller?(I don't know how it's called in English) but I don't know if it helps somewhere

答: You can do better than the triangle inequality (which you did wrong, by the way). Actually compute the length of that sum of vectors. What do you know about $T$ and $B$?

Title: find $\left \| {N}' \right \|$


答:
A more intuitive way of writing $(g \circ f)(x)$ is $g(f(x))$. This just means that you write the function $g$ with $f(x)$ instead of $x$ inside the parentheses. Since
$$
g(\text{something}) = 4 \cdot \text{something} - 4
$$
(which gives us $g(x) = 4x - 4$), just putting something = $f(x)$ gives

$$
(g \circ f)(x) = g(f(x)) = 4(f(x)) - 4 = 4(4x^{2} + 4x + 6) - 4 = \text{blah}
$$

Title: For the given functions $f$ and $g$, find $(g\circ f)(x)$.

Body:
$f(x)=4x^2+4x+6; g(x)=4x-4$

I have always had troubles with functions and need some help solving this one. I have been working on it for too long now and thought that maybe someone else's perspective would be better. I simply don't understand the path to solving it. If someone could walk me through the process that would be much appreciated.


问:
For any integers $a,b,c$, if $a+b+c=0$, then $a,b$, and $c$ are not all odd integers.
Prove or disprove the claim below from the definition of even and odd integers.

For any integers $a,b,c$, if $a+b+c=0$, then $a,b,$ and $c$ are not all odd integers

For this problem I set up the three variables into three different definitions, $A=(2k+1)$, $B=(2j+1)$, $C=(2l+1)$. Because the sum of integers is an integer we can do $k+j+l=t$, which when simplified would equal $2t+3=0$.

I was given some hints, like this problem should be proved and not disproved, as well you can use the contrapositive to find a solution. Any help would be appreciated, thank you.

答: You already solved the problem. Since $2t+3=0$ has no solution when $t$ is an integer ($2t$ is even and 3 is odd), then the statement is proved.


问:
Bring close 3D point to another 3D point equivalence
Having 2 point in 3D field   - $\text{p1(x1,y1,z1) , p2(x2,y2,z2)}$ . </p>

<p>How could I generate the equivalence , $f$ which take parameter $t\in[0,1]$ and bring $p1$ closer to $p2$ as well as $t$ increase such that $f(1)=p2$ and $f(0)=p1$  ?   </p>

答: Use a convex combination: $f=(1-t) p_1 + t p_2$, $t\in [0,1]$.</p>


问:
Formula for alternating sum of odd powers of consecutive integers
I am trying to work through this identity on page 151 of Paulo Ribenboim's "13 Lectures on Fermat's Last Theorem"
$\sum_{j=1}^n (-1)^{j-1} j^{2k+1} = (-1)^{n+1}\{ \frac{n^{2k+1}}{2} + \binom{2k+1}{1}\frac{2^2-1}{2}B_{2}n^{2k} +\binom{2k+1}{3}\frac{2^4-1}{4}B_{4}n^{2k-2}+ \dots+ \binom{2k+1}{2r-1}\frac{2^{2r}-1}{2r}B_{2r}n^{2k+2-2r} + \dots +\frac{2^{2k+2}-1}{2k+2}B_{2k+2} \} +\frac{2^{2k+2}-1}{2k+2}B_{2k+2}$

I have attempted to use the formulae for Euler and Bernoulli polynomials found here: https://dlmf.nist.gov/24 and attempted to use a formula for alternating sum of powers of consecutive integers found here https://arxiv.org/pdf/math/0508233.pdf together with the formulae previously mentioned but I either have powers of $n$ to odd powers that I cant get rid of, or I find myself getting lost in notation and unable to simplify. Is this the correct method to solve this problem? Or should I be attacking it in a different way, if so how? Thanks in advance.

答:
Hmm, don't know whether this is something helpful, I'm a bit dense at the moment.

What I recognize is a difficult reformulation of the Faulhaber-polynomials using the Euler's $ \zeta() \to \eta() $  - conversion. I've not yet(?) checked all the documented conversion-coefficients... ;-) but I think that in general you've taken this correctly.

Here is how I've done the alternating sum of like powers using the inverse of the (Pascal-matrix + I), which lead implicitely to your complicated modified Faulhaber/Bernoulli-coefficicients: it seems much simpler and better to understand.

This uses initially
$$ H = (P +  I)^{-1} =
\small \begin{bmatrix}
 1/2 & . & . & . & . & . & . & . \\
 -1/4 & 1/2 & . & . & . & . & . & . \\
 0 & -1/2 & 1/2 & . & . & . & . & . \\
 1/8 & 0 & -3/4 & 1/2 & . & . & . & . \\
 0 & 1/2 & 0 & -1 & 1/2 & . & . & . \\
 -1/4 & 0 & 5/4 & 0 & -5/4 & 1/2 & . & . \\
 0 & -3/2 & 0 & 5/2 & 0 & -3/2 & 1/2 & . \\
 17/16 & 0 & -21/4 & 0 & 35/8 & 0 & -7/4 & 1/2
 \end{bmatrix} \\
$$
With this we introduce something I call "Vandermonde-vectors", simply a vector of consecutive powers of its argument, say $V(x)= [1,x,x^2,x^3,...,x^{n-1}]$ up to some chosen dimension $n$.

If we postmultiply $H \cdot V(1)$ to create polynomials from the row-coefficients, then we get in each row $r$ the infinite alternating sum  $\eta(-r,1) = 1^r - 2^r - 3^r + \cdots - \cdots $ or in general
$$H \cdot V(x) = E(x) \\ \small \text{ where } E=\operatorname{Column}[\eta(0,x),\eta(-1,x),\eta(-2,x),...,\eta(1-n,x)] $$
Of course, if we use two $V(x)$ -vectors we can compute the finite sums, let's say $$ \small \begin{array} {lll} H \cdot (V(1) + V(2)) = \operatorname{Col}[1,1,1,1...,1]\\
 H \cdot (V(1) - V(3)) = \operatorname{Col}[1-1,1-2,1-2^2,1-2^3,...,1-2^{n-1}] \\
 H \cdot (V(1) + V(4)) = \operatorname{Col}[1-1+1,1-2+3,1-2^2+3^2,1-2^3+3^3,...,1-2^{n-1}+3^{n-1}] \\ \end{array}$$
I've put the examples together:
$$ H \cdot \small{ \operatorname{Concat}[V(1), V(1)+V(2) , V(1)-V(3), V(1)+V(4),  V(1)-V(5)]} \\
=\small \begin{bmatrix}
 1/2 & 1 & 0 & 1 & 0 \\
 1/4 & 1 & -1 & 2 & -2 \\
 0 & 1 & -3 & 6 & -10 \\
 -1/8 & 1 & -7 & 20 & -44 \\
 0 & 1 & -15 & 66 & -190 \\
 1/4 & 1 & -31 & 212 & -812 \\
 0 & 1 & -63 & 666 & -3430 \\
 -17/16 & 1 & -127 & 2060 & -14324
 \end{bmatrix}
$$
If you need only the odd powers in the exponents, take just the results of odd row indexes.

This matrixmultiplication is in the spirit of the Faulhaber-polynomials only that we find different coefficients, not the Bernoulli-numbers/polynomials (which are hidden $\zeta()$-values at negative arguments) but numbers, which one could call $\eta$-numbers (perhaps they have already a name, don't know).

If you need only the solutions, this might be enough, if you need the derivation, it is not difficult, and I have it in some small treatize see treatize at web-index scroll down to "Summing of like powers - solved using the Pascalmatrix...," . The treatize not so much a compact mathematical article, but more-or-less the protocol of my naîve exploration of how I found out about this nice problems of alternating and non-alternating sums of like powers.

update: Upps - I shouldn't forget to remark that this is all explored, calculated and documented using the software Pari/GP which supplies amazing matrix-calculation functions. I seem to always forget to mention this option because it is so easy usable that it is for me like using a pocket-calculator, not worth a specific mention. But if you don't have it, Pari/GP can be downloaded free for personal use from the Pari-group at university Bordeaux.

Over the years I've come across some more learned articles on the Pascal-matrix, but one is perhaps the most interesting for your case:

*Sheng-Liang Yang and  Zhong-Kui Liu: "Explicit inverse of the Pascal Matrix plus One"

Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.

This paper presents a simple approach to invert the matrix $P_n + I_n$ by applying the Euler polynomials and Bernoulli numbers, where $P_n$ is the Pascal matrix.


问:
How does Ian Stewart's Complex Analysis Textbook compare to the more commonly used text?
I am trying to find a good complex analysis text for self study and I was wondering what people think of Ian Stewart's complex analysis text. I was trying to find some reviews on this forum but most people recommend other books and it seems like most people have not actually tried Stewart's text. Since I haven't really studied complex analysis, I can't tell exactly how good the text does at covering each topic. Has anyone tried self learning with this text?</p>

答:
Ian Stewart's book Complex Analysis is both, a gentle and solid introductory text into complex analysis.

There are two recurrent themes the author uses to motivate and guide the reader. On the one hand the historical development of complex numbers and complex analysis is nicely addressed, on the other hand he often uses familiar properties, facts and theorems from real analysis to show analogies and to also point out differences in complex analysis.

The last three chapters

Infinitesimals in Real and Complex Analysis

Homology Version of Cauchy's Theorem

The Road Goes Ever On ...

are a supplement to previous chapters and a view to other topics somewhat beyond the canon of an introductory text.

A reason that this book is not that often discussed might be that Ian Stewart is more widely known for his popular mathematical books. In fact the first book I've had the pleasure to read from him was The Problems of Mathematics nearly $30$ years ago.


问:
An integral representation of the Mittag-Leffler function
I met the following interesting identity of the Mittag-Leffler function:
$$E_\beta (-\lambda t^\beta)=\frac{\lambda}{\pi}\int_0^\infty e^{-t r}r^{\beta-1} \frac{\sin(\beta \pi)}{(r^\beta\cos(\beta \pi)+\lambda)^2+(r^\beta\sin(\beta \pi))^2}dr.$$

I cannot see how to relate this identity to the usual series expansion
$$E_\beta (-\lambda t^\beta)=\sum_{k=0}^\infty \frac{(-\lambda t^\beta)^k}{\Gamma(\beta k+1)},$$
or the countour integral form.

Feel free to attempt it through the contour integral way or through the probability interpretation.

Any help or reference is appreciated.

答:
I am going to assume $\lambda, \beta, t > 0$. Clearly, your equation cannot hold for $\beta \in \mathbb{N}$, since the right-hand side is zero in that case. In fact, it only holds for $\beta \in (0,1)$ (see equation (23) in this paper). However, we can compute the integral for any $\beta \in \mathbb{R}^+ \setminus \mathbb{N}$.

The substitution $r = \lambda^{1/\beta} u$ shows that it is equal to $f_\beta (\lambda^{1/\beta} t)$, where $f_\alpha \colon \mathbb{R}^+ \to \mathbb{R}$ is defined by
$$ f_\alpha (x) = \frac{\sin(\pi \alpha)}{\pi \alpha} \int \limits_0^\infty \frac{\mathrm{e}^{-x t^{1/\alpha}}}{1 + 2 \cos(\pi \alpha) t + t^2} \, \mathrm{d} t = \frac{\sin(\pi \alpha)}{\pi} \int \limits_0^\infty \frac{u^{\alpha - 1} \mathrm{e}^{-x u}}{1 + 2 \cos(\pi \alpha) u^{\alpha} + u^{2\alpha}} \, \mathrm{d} u $$
for $\alpha \in \mathbb{R}^+ \setminus \mathbb{N}$. We have (using the principal branch of the complex power function)
\begin{align}
f_\alpha (x) &= \frac{1}{2 \pi \mathrm{i}} \int \limits_0^\infty \frac{\mathrm{e}^{\mathrm{i} \pi \alpha} - \mathrm{e}^{-\mathrm{i} \pi \alpha}}{(u^\alpha + \mathrm{e}^{-\mathrm{i} \pi \alpha})(u^\alpha + \mathrm{e}^{\mathrm{i} \pi \alpha})} \, u^{\alpha - 1} \mathrm{e}^{-x u} \, \mathrm{d} u \\
&= \frac{1}{2 \pi \mathrm{i}} \int \limits_0^\infty \left(\frac{u^{\alpha - 1} \mathrm{e}^{-x u} }{u^\alpha + \mathrm{e}^{-\mathrm{i} \pi \alpha}} - \frac{u^{\alpha - 1} \mathrm{e}^{-x u} }{u^\alpha + \mathrm{e}^{\mathrm{i} \pi \alpha}} \right) \mathrm{d} u \\
&= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \!\! \int \limits_{1/R}^R \! \left(\! \frac{(u \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)})^{\alpha - 1} \mathrm{e}^{-x u} }{(u \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)})^\alpha + 1} \, \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)} - \frac{(u \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)})^{\alpha - 1} \mathrm{e}^{-x u}}{(u \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)})^\alpha + 1} \, \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)} \! \right) \mathrm{d} u \\
&= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \left(~\int \limits_{\Gamma_{R, \varepsilon}^+} \frac{z^{\alpha - 1} \mathrm{e}^{x \mathrm{e}^{\mathrm{i} \varepsilon} z} }{z^\alpha + 1} \, \mathrm{d} z + \int \limits_{\Gamma_{R, \varepsilon}^-} \frac{z^{\alpha - 1} \mathrm{e}^{x \mathrm{e}^{-\mathrm{i} \varepsilon} z}}{z^\alpha + 1} \, \mathrm{d} z\right) \\
&= \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \int \limits_{\Gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z \, .
\end{align}
Here $\Gamma_{R, \varepsilon}^+$ is a straight line from $\frac{1}{R} \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)}$ to $R \mathrm{e}^{\mathrm{i}(\pi - \varepsilon)}$, $\Gamma_{R, \varepsilon}^-$ is a straight line from $R \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)}$ to $\frac{1}{R} \mathrm{e}^{-\mathrm{i}(\pi - \varepsilon)}$ (depicted in the plots below), $\Gamma_{R,\varepsilon} = \Gamma_{R, \varepsilon}^+ \, \cup \, \Gamma_{R, \varepsilon}^-$ and $g_\alpha(x, z) = \frac{z^{\alpha - 1} \mathrm{e}^{x z}}{z^\alpha + 1}$. In the last step we have dropped the $\mathrm{e}^{\pm \mathrm{i} \varepsilon}$ factors in the exponentials. This does not change the value of the limit.

Using the other three contours shown below we can rewrite the remaining integral as
$$ f_\alpha (x) = \frac{1}{2 \pi \mathrm{i}} \lim_{R \to \infty} \lim_{\varepsilon \to 0^+} \left(~\int \limits_{C_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z - \int \limits_{c_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z - \int \limits_{\gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z\right) . $$
The limit of the third integral vanishes, as
$$ \left \lvert ~ \int \limits_{\gamma_{R, \varepsilon}} g_\alpha(x, z) \, \mathrm{d} z \, \right \rvert \leq \frac{2 (\pi - \varepsilon) \mathrm{e}^{x/R}}{R^\alpha - 1} \stackrel{\varepsilon \to 0^+}{\longrightarrow} \frac{2 \pi \mathrm{e}^{x/R}}{R^\alpha - 1} \stackrel{R \to \infty}{\longrightarrow} 0 \, . $$
For small values of $\varepsilon$ the contour $c_{R,\varepsilon}$ encloses all poles of $g_\alpha(x,\cdot)$, so the second integral can be evaluated using the residue theorem. The limit of the first integral is just the usual contour integral representation of the Mittag-Leffler function. Therefore,
\begin{align}
f_\alpha (x) &= \operatorname{E}_\alpha (- x^\alpha) - \!\!\!\!\!\!\!\!\!\!\!\!\! \sum \limits_{z \in S^1 \setminus \{-1\} : \, z^\alpha = - 1} \!\!\!\!\!\!\!\!\!\!\!\!\! \operatorname{Res} (g_\alpha(x,\cdot), z) \\
&= \operatorname{E}_\alpha (- x^\alpha) - \frac{2}{\alpha} \sum \limits_{k=0}^{\left \lfloor \frac{\alpha-1}{2}\right \rfloor} \exp\left[x \cos \left(\frac{(2k+1) \pi}{\alpha}\right)\right] \cos\left[x \sin \left(\frac{(2k+1) \pi}{\alpha}\right)\right] \, .
\end{align}
For $\alpha \in (0,1)$ the additional sum vanishes ($g_\alpha(x,\cdot)$ does not have any poles enclosed by $c_{R,\varepsilon}$), so $f_\alpha (x) = \operatorname{E}_\alpha (-x^\alpha)$ holds.

The five contours used above for $R > 2$. The radii of the larger and smaller circles are $2$ and $\frac{1}{R}$, respectively.


问:
Prove that all natural numbers can be divisible by $9999$
I know this is clearly false, but I was given a "proof" by contradiction that seems to make a case. Why does it fail?

Begin by assuming that all natural numbers cannot be divisible by $9999$

Let $A = \{a \in \mathbb{N} | a$ is not divisble by $9999 \}$ represent all the counterexamples. We know that $A \neq \emptyset$ and must contain a least element of $b$.

Since $0$ is divisible by $9999$ we can conclude that $0 \not \in A$ and therefore $b \neq 0$. Now for an example $b - 9999$ where we know that $b - 9999 < b$. Therefore, we know that $b - 9999$ is not a valid counterexample and cannot be in $A$. Therefore, $b - 9999$ is divisible by $9999$ such that there's an integer $c$ where $b - 9999 = 9999c$.

We then get $b = 9999 + 9999c = 9999(1+c)$ and prove that $b$ is divisible by $9999$, contradicting $b \in A$.

答:
$\mathbb{N}$ is not closed under subtraction. For any natural number $n$ and any natural number $m \geq n$, subtracting another natural number $n-m$ gives a number that is not a natural number. So, in order for $b-9999$ to be in $A$ at all, as $A$ is restricted to only natural numbers, they have to first show that $b \geq 9999$.

Also, even if the rest of the proof was somehow correct, they negated their quantifier at the beginning incorrectly. The negation of "for all $a$, $p$ is true" is not "for no $a$, $p$ is is true", but "there exists at least one $a$ such that $p$ is false. So, in this proof, if you are trying to "prove" that all natural numbers are divisible by $9999$, then if using contradiction, you would assume that there exists at least one natural number such that it is not divisible by $9999$.


问:
Solve $x^2 \equiv -1 \pmod{41} $
Using Legendre's symbol, knowing that $41 \equiv 1 \pmod{4}$ we can solve said equation. Then, I looked at the residue classes and I just counted my way to the solution $x = 9$. Verifying, $9^2 = 81$ and $81 \equiv -1 \pmod{41} $. I was wondering if there was an actual way to solve the equation $$x^2+1 \equiv 0 \pmod{41}$$ using quadratic residue formulas or any other formula for that matter, instead of counting to get the solution.

答:
If $p$ is a prime, and $p\equiv 1\pmod{4}$, then we can express $p$ as the sum of two squares: $p=x^2+y^2$. Then you have $x^2\equiv -y^2\pmod{p}$, and since $\gcd(y,p)=1$, you can find $z$ such that $zy\equiv 1\pmod{p}$. Then $(xz)^2 = x^2z^2 \equiv -y^2z^2 = -1\pmod{p}$.

Here, $41=5^2+4^2$. Now, $4(10)=40\equiv -1\pmod{41}$, so $4(-10)\equiv 1\pmod{41}$. Thus, one solution is $5(-10)= -50\equiv 9\pmod{41}$.

Of course, part of this is cheating, since one way to establish that a prime $p$ can be written as a sum of two squares when $p\equiv 1\pmod{4}$ is to rely on the fact that the equation $x^2\equiv -1\pmod{p}$ has a solution.

Title: Solve $x^2 \equiv -1 \pmod{41} $


问:
Does $\sum_{k=1}^{\infty}k(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}})$ converge?
Does the sum:

$$\sum_{k=1}^{\infty}k(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}})$$
$$ p\in\mathbb{R}|0{\leq}p<1$$
converse, and if so, to what function?

答:
Let's simplify your expression to get Carl Najafi's expression :

\begin{align}
\sum_{k=1}^{\infty}k\left(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}}\right)&=\sum_{k=1}^{\infty}k\;p^{\frac{(k-1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}(k+1)\;p^{\frac{(k+1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}\;p^{\frac{k(k+1)}2}\\
\end{align}
Proving Carl's claim.

After that you'll simply have to use the definition of the second theta function
$$\theta_2(0,\sqrt{p})=2\sum_{k=0}^{\infty}\;\sqrt{p}^{(k+1/2)^2}$$
to get the Alpha result  (since $\sqrt{p}^{1/4}=\sqrt[8]{p}$) :
$$\frac {\theta_2(0,\sqrt{p})}{2\;\sqrt[8]{p}}\quad\text{for}\ 0<p<1$$


问:
Conditional Probability Distribution of Max$(X, Y)$ given $X$
Say $X$ and $Y$ are I.I.D uniform variables in $[0,1]$. Conditional Probability Distribution of Max$(X, Y)$ given $X$. I think it would be simply $P(X)$ when $X>=Y$ and $P(Y)$ the other way around. Is that so?

答:
Let $Z := \max(X, Y).$ I assume that you are actually interested in something like $\mathbb{P}(Z \geq t \vert X \geq s)$ for some $s, t \in [0, 1],$ but please do correct me if I'm wrong. Note that if $t \leq s,$ then obviously $\mathbb{P}(Z \geq t \vert X \geq s) = 1.$ If $t \geq s,$ then $\mathbb{P}(Z \geq t \vert X \geq s) = \frac{\mathbb{P}((Z \geq t) \cap (X \geq s))}{\mathbb{P}(X \geq s)} = \frac{\mathbb{P}(X \geq t, X \geq s) + \mathbb{P}(X < t, Y \geq t, X \geq s)}{1 - s} = \frac{t + (t - s)(1 - t)}{1 - s} = \frac{- t^2 + (2 + s)t - s}{1 - s}.$

Also note that the CDF of $Z$ can be computed in the following way ($t \in [0, 1]$ is fixed). $\mathbb{P}(Z \leq t) = \mathbb{P}(X \leq t, Y \leq t) = \mathbb{P}(X \leq t) \mathbb{P}(Y \leq t) = t^2.$

I hope this helps. :)


问:
Definition of the ring of weakly modular forms over $\mathbb{Z}_{(p)}$
I am an undergraduate in a small mathematics course focused on a single project. Right now we are in the preliminary stage, the part where our advisor has given us a high-level reference, referred us to sections, and asks we do our best to make left from right of it.

In doing this, I need some help. Early into Behrens' paper here, specifically on page 8, he defines the following:

$MF_*$ is the ring of weakly modular forms over $\mathbb{Z}_{(p)}$.
$$MF_* = \mathbb{Z}_{(p)}[c_4,c_6,\Delta^{\pm 1}]/(c_4^3-c_6^2 = 1728\Delta)$$
We must invert $\Delta$ because we are taking sections over the complement of the singular locus of the moduli space of generalized elliptic curves.

I do not understand where this definition of $MF_*$ comes from. This is what I would like clarification of. I would guess it is not a straightforward definition to motivate, so I would be fine with references that would help me answer my question (in particular, references that lead nothing to presumption.)

If it is useful, here is some additional context:

I am interested in the $p=3$ case.

I have very light background in algebraic geometry.

I have very light background in elliptic curves/modular forms (having only introduced these  to myself about two weeks ago.)

答:
This was going to be a comment but got way too long. I'm not familiar with weakly modular forms but there are a couple potential motivations that come to mind. Over $\mathbb{C}$ there is a ring of modular forms (of all weights), and this is generated as a $\mathbb{C}$-algebra by forms $E_4, E_6$ of weights 4 and 6 respectively. There is also a cusp form commonly called $\Delta$ of weight 12 that satisfies the relation $E_4^3 - E_6^2 = 1728 \Delta$, or something similar depending on your normalization. This can be found in any reasonable book about modular forms and doesn't require knowledge of algebraic geometry. I recommend Diamond and Shurman.

Another motivation behind the definition is the idea of a moduli space. You may be aware that modular forms can have many equivalent definitions. One is as sections of a sheaf over some moduli space (be that a scheme, stack, or something else) of elliptic curves. A detailed explanation of this can be found in Arithmetic Moduli by Katz and Mazur, but this may be impenetrable until you learn more algebraic geometry. The rough idea is that points of your space correspond to elliptic curves. Each elliptic curve $E$ has a discriminant $\Delta(E)$, and this discriminant is invertible since $E$ is smooth (by definition). I recommend taking a look at Silverman's book on elliptic curves if you're not comfortable with this.

We can then think of $\Delta$ as a function on our moduli space of elliptic curves, that takes a point $x$ (necessarily associated to some elliptic curve $E_x$) to the value $\Delta(E_x)$ in our base ring/field. If we call our sheaf $\omega$ and pretend its the structure sheaf (which it is locally) then sections of $\omega$ are just functions on our space. So $\Delta$ is an element of the ring of global sections on our moduli space. Similarly, we can define certain quantities $c_4, c_6$ for every elliptic curve, and view these as functions as well. I think the idea should be that every generalized elliptic curve $E$ (up to iso) is determined by the values $c_4(E), c_6(E), \Delta(E)$, with the caveat that we always know
$$
c_4(e)^3 - c_6(E)^2 = 1728 \Delta(E).
$$

A good example to keep in mind is that the (coarse) moduli space of elliptic curves over $\mathbb{C}$ is isomorphic to $\mathbb{A}^1_{\mathbb{C}}$ via the $j$-invariant; the moduli space is just $\operatorname{Spec}\mathbb{C}[j]$. In the paper you linked, they consider a moduli space of generalized elliptic curves (ref for this is the paper by Deligne-Rapoport, but it may not be accessible to you yet), so not all points will be smooth curves. Some points will be singular, and these correspond to where $\Delta$ vanishes. Inverting $\Delta$ corresponds to taking the complement of the singular locus, i.e., only considering the points corresponding to smooth curves. Finally, I would also check out the paper referenced as [8] in the link you gave (P. Deligne, Courbes elliptiques: formulaire d’après J. Tate), as this appears immediately after the definition you copied and will probably be quite helpful.


问:
Does this simple proof use the axiom of choice?
The axiom of choice lets us choose and single elements from an arbitrary amount of sets.  I'm wondering if we can choose a single element from a single set and build another set from it without using choice.  Do we need choice for this proof?  Say we want to show that for any two non-empty sets $A$ and $B$, there exists a function $f:A \rightarrow B$.

Proof: Because $B$ is non-empty, there exists at least one element in $B$.  Fix $x \in B$. Then define $f = \{(a, x): a \in A\}$.  Then $f:A \rightarrow B$.

Does this proof only use existential elimination and not choice?  Or can we only fix and an arbitrary $x \in B$ when using universal introduction to prove formulas like $\forall x \in B (\dots)$?

答:
The proof you've given is completely choice-free: it just boils down to existential elimination. This can be checked by writing out the fully-formal proof corresponding to that rigorous natural-language argument, and if you want you can even do this in such a way that a computer can verify the final result (although this takes a bit more work).

It's worth noting that an elaboration on this idea yields a stronger result:

$\mathsf{ZF}$ proves "Every finite collection of nonempty sets has a choice function."

This is not as trivial as it may seem, since in a model of $\mathsf{ZF}$ there may be "internally-finite" sets which are not in fact finite. Instead, we have to carefully check that the relevant induction argument can be run internally: reasoning in $\mathsf{ZF}$, what can you say about the least finite ordinal $n$ such that there is some $n$-sized collection of nonempty sets without a choice function?


问:
Equation of a circle passing through the intersection of 'two circle' or 'one circle and one line'.
In our mathmatics textbook , there is a question of finding out the Equation of a circle passing through the intersection of 'two circle' or 'one circle and one line'. They do it the following way.Lets say:

Equation of a circle            : $f(x,y) = 0$

Equation of another circle/line : $g(x,y) = 0$

Then the equation of a circle passing through the intersection of $f$ and $g$ is:

$f(x,y) + k.g(x,y) = 0 $ (Here k is any constant)

But I didn't really understood this last part because it didn't really have any explanation or proof along with it.What is the proof of this theorem?

In an another Analytic Geometry paper.I Saw this quote with no proof:

"If $S = 0$ and $S'= 0$ are equations of two curves, then $S + KS'= 0$ is the general equation of all curves, passing through all the points common to the two curves"

The above quote is clearly a theorem.But what does it intuitively mean , what is the proof of the theorem and Does it have anything to do with the original problem I had with circle? (It is okay if calculus is involved)

答:
First part with circles

Dropping of arguments $ (x,y)=0$ for brevity:

The coefficients of (x,y) in $f, g $ equations can be made unity by division to represent a circle.

$f-g $ is first degree equation of radical axis straight line
$f+ g, f+ k g $ are circles  which are concurrent with the radical axis.

The circles need not always intersect. They can be disjunct, with no real intersection..Four tangents are possible and they can be drawn  from the non-intersecting radical axis. These circles can be externally disjunct or internally disjunct. ( one circle enclosed inside another)

This can be proved in consideration of pole/polar valid for all conics in the plane, starting with zero constant term or zero power for curves passing through the origin.

These details are included in several middle level analytic geometry text books in India also.

Second part,  to find if such theorems can be found with any such general algebraic, polynomial curves.

Two cassinian ovals, their sum  and two circles.

We are not in a situation to conclude that such concurrency or equal co-tangentiality can exist in this family of five curves.


问:
Meaning of G(k) being dense in G for an algebraic group G/k
Let $G$ be an algebraic group over a field $k$ of characteristic 0. I have read that: if $G$ is connected, then $G(k)$ is dense in $G$ for the Zariski topology. I do not understand what kind of density this might be since $G(k)$ is abstractly defined as the set of morphisms $\text{Spec}(k) \to G$. If $G$ is an affine variety in $\mathbb{A}_k^n$ for instance, one might think of $G(k)$ as a subset of $k^n$. But how does this connect to the abstract definition of $G$ as a scheme or as a functor from $k$-algebras to groups? I do not see a direct relation or any mean to relate the group $G(k)$ (a set of points) to the abstract scheme (or functor) $G$, let alone saying that $G(k)$ is dense in $G$ for the Zariski topology. How is this topology even defined on $G(k)$ and how is $G(k)$ even a subset of $G$? Could someone please resolve this issue for me? Thank you in advance.

答:
$\newcommand{\Spec}{\mathrm{Spec}}$$\newcommand{\h}{\mathcal{O}}$$\newcommand{\mf}[1]{\mathfrak{#1}}$While hunter's answer is perfectly fine, I do think it's worth phrasing it in a way that is independent of embedding.

The point is the following:

Fact: Let $X$ be a scheme, $A$ ring, and $L$ a field. Suppose that $X$ and $L$ have the structure of a $\Spec(A)$-scheme. Then, there is
a natural equality (really bijection)
$$\mathrm{Hom}_{\Spec(A)}(\Spec(L),X)=\left\{(x,\varphi):\begin{aligned}(1)&\quad
 x\in X\\ (2)&\quad \varphi:k(x)\to L\text{ is an }\\ &\quad A\text{-algebra
 homomorphism}\end{aligned}\right\}$$

Here $k(x)$ is the residue field of $x$.

Proof: Let us use $p$ to denote the single point of $\Spec(L)$ (as a topological space it's one point). Suppose first that $f:\Spec(L)\to X$ is a morphism of schemes over $\Spec(A)$. Then, $x:=f(p)$ is a point of $X$. By the definition of a morphism of schemes one has a local ring map $f^\sharp_x:\h_{X,x}\to \h_{\Spec(L),p}=L$. But, since this map is local and $(0)$ is the maximal ideal of $L$ we see that $\ker(f_x^\sharp)=(f^\sharp_x)^{-1}((0))=\mf{m}_x$ (this last ideal being hte maximal ideal of $\h_{X,x}$). In particular, $f_x^\sharp$ factorizes uniquely through $\h_{X,x}/\mf{m}_x=k(x)$. Thus, from $f$ we get the pair $(x,f_x^\sharp)$. Conversely, if we have a pair $(x,\varphi)$ we get a morphism $\Spec(\varphi):\Spec(L)\to \Spec(k(x))$. But, note that we have obvious morphisms of $\Spec(A)$-schemes

$$\Spec(k(x))\to \Spec(\h_{X,x})\to X$$

where the first is the one coming from the quotient map $\h_{X,x}\to k(x)$ and the second is the tautological one (i.e. for any affine open $\Spec(B)$ of $X$ containing $x$ there is the natural map $\Spec(\h_{X,x})=\Spec(B_\mf{q})\to \Spec(B)\to X$ where $\mf{q}$ is the prime ideal of $B$ corresponding to $x$--this composition can be checked to be indpendent of the choice of $B$). We then get a morphism $\Spec(L)\to X$ as the composition

$$\Spec(L)\xrightarrow{\Spec(\varphi)}\Spec(k(x))\to\Spec(\h_{X,x})\to X$$

It's easy to check that these two associations are inverse to one another. $\blacksquare$

What's the upshot of this?

Corollary: Let $X$ be a scheme over $\Spec(k)$. Then, there is a natural equality (bijection)
$$X(k):=\mathrm{Hom}_{\Spec(k)}(\Spec(k),X)=\{x\in X:k(x)=k\}$$

Proof: Applying the last fact we see that $X(k)$ is in bijection with points $(x,\varphi)$ where $x$ is in $X$ and $\varphi:k(x)\to k$ is a $k$-morphism. But, note that there is a $k$-morphism $k(x)\to k$ if and only if $k(x)=k$ (more precisely the natural morphism $k\to k(x)$ is an isomorphism) in which case it's unique. Thus, we get our result! $\blacksquare$

In particular, for any scheme $X$ over $\Spec(k)$ we see that we can naturally view $X(k)$ as a subset of $X$. In particular, to say that $X(k)$ is dense in $X$ means that the subset $X(k)$ is dense in the topological space $X$ (the scheme $X$ considered with the Zariski topology).

The following is a good exercise:

Exercise: Let $X$ be an affine scheme which is of finite type over $\Spec(k)$. Choose a closed embedding $X\hookrightarrow \mathbb{A}^n_k=:\Spec(k[T_1,\ldots,T_n])$
over $\Spec(k)$. Then, $X(k)$ (as described above) agrees with $X\cap
 k^n$ where $k^n\subseteq \mathbb{A}^n_k$ is the usual embedding of sets
(i.e. $(x_1,\ldots,x_n)\in k^n$ corresponds to the maximal ideal
$(T_1-x_1,\ldots,T_n-x_n)$).

This shows the connection to hunter's answer.

Then, your result more accurately/generally is the following:

Theorem ([1, Theorem 17.93]): Let $k$ be an infinite perfect field. Let $G$ be a connected, affine, geometrically reduced, group
scheme of finite type over $\Spec(k)$. Then, $G(k)$ is dense in $G$.

In particular, given Cartier's theorem (i.e. [1, Theorem 3.23]) and the fact that a field of characteristic $0$ is automatically perfect and infinite, one deduces the following:

Corollary: Let $k$ be a field of characteristic $0$ and let $G$ be an affine group scheme of finite type over $\Spec(k)$. Then, $G(k)$
is dense in $G$.

References:

[1] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.


问:
Proving $\sinh{x}$ is strictly increasing over all the reals and $\cosh{x}$ is strictly decreasing on $(- \infty , 0]$
How would I be proving $\sinh{x}$ is strictly increasing and $\cosh{x}$ is strictly decreasing on $(- \infty , 0]$

I succeded in showing $\cosh{x}$ is strictly increasing on the interval $[0, \infty)$ but having trouble with the rest. The $\sinh$ one is really messing with me.

*I can not use derivatives. I know how to but am not allowed - for some reason.

答:
hint

For any real $ x$,

$$S(x)=\sinh(x)=\frac{e^x-e^{-x}}{2}$$

is the sum of two strictly increasing functions.


问:
Problem 17(v) Topology Without Tears (pg 96)

The Question

My Understanding(final edit)

(My proof will probably suck,but it is the best I can do)

We are required to prove that $R\setminus C \in \tau$  and  $R\setminus A\in\tau$ are Hausdorff but not regular.

Assumptions

1.Let X be the set,of all real numbers and
$ S=\{1/n:n \in N\}$

2.Define a set $C\subseteq\Bbb R$ to be closed if $C=A \cup T$. is $A$ is closed in the Euclidean topology on $\Bbb R$. $T\subset S$

If C is closed by Def 1.2.4 R\C is open. Similarly if A is closed
in Euclidean space A\C is open

But T$\subset S $. Then every point of S is an isolated point, since a sufficiently small interval about 1/n doesn’t contain 1/m for any integer m$\ne$ n, and S has no interior points. The set of boundary points ofS, S ∪ {0}. The point 0 $\notin$ S is the only accumulation point ofS, since every open interval about 0 contains 1/n for sufficiently large n.

 Source
This implies that $\bigcup_{n\in N} S\cup {0}=[0,1]$
Thus R\S=$(-\infty,0) \cup (1,\infty$)

Next l will show it is a topology

$\emptyset$,X $\in\tau$, since R\C and R\A are open
and byProp 1.2.2 (ii) & (iii) are satisfied cause both sets
are open Thus R\C,R\A $\in\tau$ Finally R\S by prop 1.2.2 (ii)
is satisfied.

Thus C=A $\cup T \in \tau$

Then we can find x$\in (-\infty,0)\subset$ R\S
and y$\in. (1,\infty)\subset $R\S so that
-$\infty$<x<0 and 1<y<$\infty$

.

Let U=(-$\infty$,0) V=(1,$\infty$) Then U $\cap V=\emptyset$
So $\tau $ on R is Hausdorff and since O$\notin S$ we can’t
separate T$\in\tau$ so it is not regular.

答:
You should perhaps start by proving that this definition of closed sets of $\mathcal{T}$ as sets of the form $A \cup S$, where $A$ is Euclidean closed and $S \subseteq T = \{\frac1n\mid n \in \Bbb N\}$ is a valid one, in that it obeys the axioms for closed sets for a topology.

Secondly note that if $O$ is Euclidean open, $\Bbb R\setminus O$ is Euclidean closed, so also closed in $\mathcal{T}$ (take $S=\emptyset$ in the definition) and hence $O$ is open in $\mathcal{T}$. So because we can separate any two points of $\Bbb R$ in the Euclidean topology, we can use these same open sets to separate them in $(\Bbb R, \mathcal T)$. So the newly defined space is still Hausdorff.

Moreover, $T$ is closed in $\mathcal T$ (take $A=\emptyset, S=T$ in the definition) and we can show that $0 \notin T$ cannot be separated from $T$ in $\mathcal T$ and this will show that $(\Bbb R, \mathcal T)$ is not regular.

Try to focus on this giving last proof, and forget about the (incomprehensible, at least for me) considerations in your post. The proof could go by contradiction, and should at least use that $0 \in \overline{T}$ in the Euclidean topology...


问:
Probably a really easy question...does the following ring exist?
This may be an easy couplet of questions but I have never seen a ring of this form ever brought up:

Does the following exist as a ring:

$R=\mathbb{Z}[x]/\{x^2=0; 7x = 0\}$

2.Does $R$ have the form $R=\{ax+b; b \in \mathbb{Z}; a \in \mathbb{F}_7\}$,
where $a'x+b'$ and $ax+b$ are distinct elements as long as either $a' \not = a$ or $b'\not = b$. Put another way, for example, $6x$ and $13x$ are the same element in $R$, but $6$ and $13$ are distinct elements in $R$.

If the answer to either is NO, then why not.

Thanks for looking at this!

答:
Yes, $R$ certainly exists as a ring.  If nothing else, $R=\Bbb Z[x]/(x^2, 7x)$.  The question is whether that ring is isomorphic to a more familiar ring.  The answer is that I don't think so.

It's not $\Bbb F_7 \oplus \Bbb Z$ as suggested in another answer (now deleted) because all products of two elements of the first component are $0$ and because multiplication isn't performed componentwise.  I guess you could define a trivial ring structure on the additive group $\Bbb Z/7 \Bbb Z$, give that a name, and define an appropriate product across components, but it seems easier just to work with the quotient ring.

All elements of $R$ can be presented in the form you suggest.  Just be careful to remember, when multiplying, that $x^2=0$.


问:
Calculating the mean square distance between two points on the sphere.

The Question

What I want to know is how to calculate the mean square distance between a fixed point which is any random point in the sphere and rest of the points.

My Understanding

I'm clueless on how to calculate it. I'm a beginner. First I started from the straight line equation between the two points then equate it to radius of the sphere. then we can take another point. but this is not gonna help right? I think I should use integration here..

Why it is an useful quantity to calculate?

For example, in calculating magnetic moment of an electron around the nucleus the mean squared radius is used.

答:
Hint:

To compute a mean square distance of a point distribution from a given point $(x_0,y_0,z_0)$  you need to compute the integral:
$$
\langle r^2\rangle=\frac{\int_V[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]\rho(x,y,z)\,dV}{\int_V\rho(x,y,z)\,dV}.\tag1
$$
where $\rho(x,y,z)$ is the distribution density. In your example:
$$
\rho(x,y,z)\sim\delta(x^2+y^2+z^2-R^2)
$$
and
$$x_0=(0,0,R),
$$
where the origin of coordinates is assumed to be at the center of the sphere.

It remains only to substitute the expressions into (1) and evaluate the integral in spherical coordinates ($x=R\sin\theta\cos\phi,\,y=R\sin\theta\sin\phi, z=R\cos\theta)$.

Can you take it from here?


问:
$C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx$
Consider $C([0, 1])$, the linear space of continuous complex-valued functions
on the interval $[0, 1]$, with the norm

$$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$

I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.

Let $f_n \in C[0,1]$ be given by

$$f_n(x) := \begin{cases}
 0   & \text{if $0 \le x \le \frac1{2}$}\\
 n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\
 1              & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $}
\end{cases}$$

How to prove that $f_n$  is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$?

If I use basic definition then I have to prove that $\lVert f_n - f_m\rVert_1 < \epsilon$ $\forall n, m > N$. But I am finding it difficult to prove this.

Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.

Thanks

答:
Let $m\leq n$ both natural numbers, then
$$\|f_n-f_m\|_1 = \int_0^1 |f_n(x)-f_m(x)|\,\mathrm{d}x $$
$$ = \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n}}(n-m)\left(x-\frac{1}{2}\right)\,\mathrm{d}x + \int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-m\left(x-\frac{1}{2}\right)\right)\,\mathrm{d}x.$$

Now try to bound these integrals for $n,m\geq N$.


问:
Is the spectrum of a stalk a subscheme?
Given a scheme $X$ and a point $p \in X$, I know that $\text{Spec}(\mathcal{O}_{X,p})$ consists of the point $p$ together with all generic points of irreducible closed subsets containing $p$.

Can we thus view the spectrum of the stalk, $\text{Spec}(\mathcal{O}_{X,p})$, as an open, closed, or locally closed subscheme of $X$?

答: In general, $\operatorname{Spec} \mathcal{O}_{X,p}$ cannot be viewed as an open, closed, or locally closed subscheme of $X$. Consider $X=\Bbb A^1_k$ for a field $k$: then $\operatorname{Spec} \mathcal{O}_{X,p}$ is either the generic point, or a single closed point and the generic point topologized so that the only nontrivial open set consists of the generic point. Neither set is open nor closed in $X$, so $\operatorname{Spec} \mathcal{O}_{X,p}$ cannot be an open or closed subscheme of $X$. To see it is not a locally closed subscheme, we note that the closed points are dense in any locally closed subset of a scheme of finite type over a field, in contrast to the two spaces we found above.

Title: Is the spectrum of a stalk a subscheme?


答: Let $Z=\max\{X,Y\}$ with X, Y iid uniform on [0,1]. Then $f(\max\{X, Y\}|X=\epsilon)=f(\max\{\epsilon, Y\})$ has a mixed distribution. $P(Z=\epsilon|X=\epsilon)=\epsilon$. For $\epsilon <Z\le1$, $f(Z|X=\epsilon)=1$, and $f(Z|X=\epsilon)=0$ otherwise.

Body: Say $X$ and $Y$ are I.I.D uniform variables in $[0,1]$. Conditional Probability Distribution of Max$(X, Y)$ given $X$. I think it would be simply $P(X)$ when $X>=Y$ and $P(Y)$ the other way around. Is that so?


问:
About proving that the interior of the boundary of a closed set is empty
Using the notation of Int, Cl and $\partial$ for interior, closure and boundary respectively, I'm trying to prove that Int$(\partial A) = \emptyset $, with A a closed set. However, the material that I find on the internet in this regard uses as a definition of boundary "The set of points $x \in X$ such that every neighborhood of $x$ contains at least one point of $S$ and at least one point not of $S$", while I'm using $\partial S = \text{Cl}(S) - \text{Int}(S)$.

I would like some hint on how to start the proof from the point of view of the definition shown above. Thanks!

答:
It's quite easy to see that your definition of $\partial S = \text{Cl}(S) - \text{Int}(S)$ is equivalent to the one you saw: every neighbourhood of $x$ intersecting $S$ says $x \in \text{Cl}(S)$ and every neighbourhood of $x$ containing a point not in $S$ is the same as saying $x \notin \text{Int}(S)$.

Also consider $\partial A$ for $A$ closed, we write it as $ \text{Cl}(A) - \text{Int}(A) = A - \text{Int}(A)$, as $A$ is closed. So we subtracted from $A$ the maximal open subset of $A$ which means there cannot be any  non-empty open subsets inside $A - \text{Int}(A) (\subseteq A)$, because that would contradict the maximality of $\text{Int}(A)$. Hence $\partial A$ has empty interior.


问:
Is there a name for a semigroup whose idempotents form a subsemigroup?
For a semigroup $S,$ I will denote by $E(S)$ the set of all idempotents of $S$. For $X\subseteq S,$ let $X^2$ mean $\{xy\,|\,x,y\in X\}.$

Is there a name for the class of semigroups $S$ such that $$\left(E(S)\right)^2\subseteq E(S)?$$

To have an example, in every inverse semigroup, the idempotents form a subsemigroup. More generally, as rschwieb points out in a comment, any semigroup such that the idempotents commute with each other satisfies this condition.

I need a name to be able to search for information about such semigroups. So any contribution besides the name will be welcomed.

答:
In a very important paper, C.J. Ash, 'Finite semigroups with commuting idempotents', J. Austra. Math. Soc. 43 (1987), 81-90, Chris Ash proved that a finite semigroup has commuting idempotents if and if it divides (that is, is a homomorphic image of a subsemigroup) of an inverse semigroup.

There are examples of finite semigroups that have commuting idempotents, but do not embed in any inverse semigroup.

Birget, Margolis, Rhodes, `Semigroups whose idempotents form a subsemigroup' Bull. Austral. Math. Soc.  41 (1990) [161-184] prove that a finite semigroup has idempotents that form a semigroup if and only if it divides (but not necessarily embeds into) an orthodox finite semigroup, that is a regular semigroup with idempotents a subsemigroup. You can preserve the variety of bands in which the idempotents of the original semigroup reside.

These were all preliminaries for the so called Rhodes Type II conjecture which was also proved by Ash.

For details in book form, see Rhodes-Steinberg, "The q-theory of finite semigroups", Springer, 2009.
u


答:
Unsure if this is what you are asking for.

$x \in ~\text{interior of} ~S$ denotes that there is a neighborhood around $x$ that is completely contained in $S$.  This implies, per the definition that you have stated for the boundary of $S$, that $x \in ~\text{interior of} ~S$ implies that it is not the case that $x \in ~\text{the boundary of} ~S$.

Title: About proving that the interior of the boundary of a closed set is empty

Body:
Using the notation of Int, Cl and $\partial$ for interior, closure and boundary respectively, I'm trying to prove that Int$(\partial A) = \emptyset $, with A a closed set. However, the material that I find on the internet in this regard uses as a definition of boundary "The set of points $x \in X$ such that every neighborhood of $x$ contains at least one point of $S$ and at least one point not of $S$", while I'm using $\partial S = \text{Cl}(S) - \text{Int}(S)$.

I would like some hint on how to start the proof from the point of view of the definition shown above. Thanks!


问:
How do I show that $e^{x+1} > 2x + 2$ for all $x>0$?
Hello I would need help to solve the equation above.
So far I have tried letting $$f(x) = e^{x+1}- 2x -2$$

$$f'(x)= e^{x+1} -2$$

$$e^ {x+1} > 2$$

$$x> \ln 2-1$$

but I do not know how to continue.

答:
Using familiar properties of $e^x$, namely  $e\gt2$, $\int e^udu=e^u+C$, and $e^x\ge1$ for $x\ge0$, we have

$$e^{x+1}=e(e^x-1)+e\gt2\int_0^xe^u\,du+2\ge2\int_0^xdu+2=2x+2$$


问:
By Using the Binomial expansion as follow that the expression for the first three term is mathematically correct or not?
Let the expression is defined as $(1-\frac{1}{i})^{2i}$.

For example: Using the Binomial expansion as follow  $$(1-\frac{1}{i})^{2i}=\sum_{n=0}^{n}\frac{(2i)!}{n!(2i-n)!}(-1)^n(\frac{1}{i})^n$$

$$\approx1-\frac{1}{i}+O((-\frac{1}{i})^3)$$.

Please need your help that the above expression for the first three term is mathematically correct or not, where the big $O$ notation means that the first neglected or unknown term is of the order $(-\frac{1}{i})^3$.

答: The terms I get are the followings: $$\left(1-\dfrac{1}{i}\right)^{2i}=\sum\limits_{n=0}^{2i}\binom{2i}{n}(-1)^n\dfrac{1}{i^n}=\overbrace{\underbrace{\binom{2i}{0}(-1)^0\dfrac{1}{i^0}}_{1\cdot 1\cdot 1=1}}^{1^{st}}+\overbrace{\underbrace{\binom{2i}{1}(-1)^1\dfrac{1}{i^1}}_{2i\cdot (-1)\cdot 1/i=-2}}^{2^{nd}}+\overbrace{\underbrace{\binom{2i}{2}(-1)^2\dfrac{1}{i^2}}_{i(2i-1)\cdot 1\cdot 1/i^2=\frac{i(2i-1)}{i^2}}}^{3^{rd}}+\ldots$$ The second term does not seem to depend on $i$.


问:
When does $({\bf x}^\mathrm{T}{\bf y})(A{\bf z})=({\bf x}^\mathrm{T}A{\bf y}){\bf z}$?
Under what conditions is $({\bf x}^\mathrm{T}{\bf y})(A{\bf z})=({\bf x}^\mathrm{T}A{\bf y}){\bf z}$? I tried to see if there was any way to determine the conditions through index notation, but that just leaves me at $x_iy_iA_{jk}z_k$ with nowhere to proceed. Any help is greatly appreciated, thanks! For now I treat $A$ as a matrix and the remaining as vectors, with appropriate dimensions that allow for the products to be taken.

答: Setting $c_1 = x^T y$ and $c_2 = x^T Ay$, your equation becomes $c_1 Az = c_2 z$. Assuming that $c_1 \neq 0$, this means $Az = (c_2 / c_1)z$ and $c_2 / c_1$ is an eigenvalue. If $c_1 = 0$, then you get $c_2 z = 0$, so either $c_2 = 0$ or $z = 0$. Are these the kinds of conditions you are looking for?


答:
This condition can never be true:

Let $x,y,z$ be $n$- dimensional vectors.

For $(x^Ty)(Az)$ to be defined, $A$ would need to be an $1 \times n$ matrix.

For $(x^T A y)z$ to be defined, then $A$ would need to be an $n \times n$ matrix.

So, for the same $A$ you can't have that $(x^Ty)(Az) = (x^T A y)z$, because if one is defined, then the other is not

Title: When does $({\bf x}^\mathrm{T}{\bf y})(A{\bf z})=({\bf x}^\mathrm{T}A{\bf y}){\bf z}$?

Body: Under what conditions is $({\bf x}^\mathrm{T}{\bf y})(A{\bf z})=({\bf x}^\mathrm{T}A{\bf y}){\bf z}$? I tried to see if there was any way to determine the conditions through index notation, but that just leaves me at $x_iy_iA_{jk}z_k$ with nowhere to proceed. Any help is greatly appreciated, thanks! For now I treat $A$ as a matrix and the remaining as vectors, with appropriate dimensions that allow for the products to be taken.


问:
Permutation of Matrices on a Vector

We have the following two matrices:
$A=\left( \begin{matrix}1 & 1 \\ 0 & 1\end{matrix} \right)$ and $B=\left( \begin{matrix}1 & 0 \\ 1 & 1\end{matrix} \right).$ Take any finite product of them and multiply this by the vector $x=\left<1,1\right>$. What possible vectors $(x',y')$ can we get?

E.g. $ABx=\left<3,2 \right>$.

答:
Call $S$ the set of vectors $(p,q)$ such that $(p,q)$ can be obtained from an arbitrary product of $A$ and $B$, multiplied by $(1,1)$. For convenience, to avoid writing transpose vectors everywhere, I write $(p,q)$ for the column vector.

If $(p,q)$ is possible, $p$ and $q$ are coprime.

Proof: we start initially with $(1,1)$, and $1$ is coprime to $1$. For any $(p,q)\in S$, we can write $(p,q)=M_nM_{n-1}\cdots M_1 (1,1)$ with $M_i=A$ or $M_i=B$. Now, by induction, if $(u,v)$ are coprime, then $M_i (u,v)=(u',v')$, and either:

$u'=u+v$, $v'=v$ and $(u',v')$ are coprime,

$u'=u$, $v'=u+v$ and again $(u',v')$ are coprime.

And since the initial values $(1,1)$ are coprime, so are $(p,q)$, by induction.

Therefore, for all $(p,q)\in S$, $p$ and $q$ are coprime.

If $p$ and $q$ are coprime, then $(p,q)$ is possible.

To see why, we will build a decreasing sequence $(u_i,v_i)$ such that if $(u_i,v_i)$ is possible, then so is $(p,q)$. If we can prove the sequence ends with $(1,1)$, we are done.

Let's start from a given $(p,q)$ with coprime $p,q$.

If $(u-v,v)$ is possible, then $A(u-v,v)=(u,v)$ is possible. Likewise, if $(u,v-u)$ is possible, $B(u,v-u)=(u,v)$ is possible.

We start the sequence with $u_0=p, v_0=q$. At each step $i$, $u_i$ and $v_i$ are positive and coprime (hence different, unles they are both equal to $1$).

If $u_i=v_i$, then $u_i=v_i=1$ (they are coprime) and we are done: we can rewrite the sequence as a product $M_0M_{1}\cdots M_{n-1}(1,1)$.

Otherwise, either $u_i>v_i$ or $u_i<v_i$.

If $u_i>v_i$, let $(u_{i+1},v_{i+1})=(u_i-v_i,v_i)$. And if $u_i<v_i$, let $(u_{i+1},v_{i+1})=(u_i,v_i-u_i)$. Thus, either way, $(u_{i},v_{i})=A(u_{i+1},v_{i+1})$ or $(u_{i},v_{i})=B(u_{i+1},v_{i+1})$. That is, $(u_{i},v_{i})=M_i(u_{i+1},v_{i+1})$ with $M_i=A$ or $M_i=B$.

Since $u_i$ and $v_i$ are positive and coprime, then so are $u_{i+1}$ and $v_{i+1}$, and $0<u_{i+1}+v_{i+1}<u_i+v_i$. Since we have a strictly decreasing sequence $s_i=u_i+v_i$ of positive numbers, it must stop at some point, and it can only stop if $u_n=v_n=1$ for some $n$.

Then, $(p,q)=(u_0,v_0)=M_0M_1\cdots M_{n-1}(1,1)$.

Therefore, if $p$ and $q$ are positive and coprime, then $(p,q)\in S$.

All in all, $S$ is the set of pairs of positive coprime integers.

Title: Permutation of Matrices on a Vector


答:
(Assuming $A$ is $n\times n$ and $\bf x,y,z$ are $n\times 1$ for some natural $n$.)

Note that if we instead write
$$
(A{\bf z})({\bf x}^\mathrm{T}{\bf y})={\bf z}({\bf x}^\mathrm{T}A{\bf y})
$$
then all multiplications involved are actually standard matrix multiplications. And dropping the parentheses (which are no longer needed) we see that it is not actually $A$ and $\bf y$ that commute, but $A$ and ${\bf zx}^T$.

More generally, some elementary manipulations of the above equation tells us that we have equality iff $\bf y$ is in the kernel of the commutator
$$
[A,{\bf zx}^T]=A{\bf z x}^T-{\bf zx}^TA
$$


问:
Lower bound for Strongly convex and Lipschitz gradient function
I was reading this paper but could not understand one of it's condition. It says a function $f(x)$ is twice differentiable and strongly convex with parameter $m$ and Lipschitz continuous gradient with parameter $M$ and $M\geq m$. Then:
$$(x-y)^T(\nabla f(x)-\nabla f(y) \geq \frac{mM}{m+M} \lVert x-y \rVert^2+\frac{1}{m+M}\lVert \nabla f(x)-\nabla f(y) \rVert ^2$$

I know that if $f$ is strongly convex and Lipshcitz continuous gradient, then :
$$(x-y)^T(\nabla f(x)-\nabla f(y)) \geq m \rVert x-y\lVert ^2$$
and
$$(x-y)^T(\nabla f(x)-\nabla f(y)) \geq \frac{1}{M} \lVert \nabla f(x)-\nabla f(y) \rVert  ^2 $$. I tried multiply each by $M$ and $m$ and adding up but that didn't quite work.

答:
I emailed the author of the paper, here is the reply I got: "It is a non trivial result which is often used in convex optimization. I guess you can find it in Nesterov's book
Introductory Lectures on Convex Optimization: A Basic Course. Theorem 2.1.12"</p>


问:
For what class of functions $f(x)$ is it true that $F\left(s^*\right)=F(s)^*$ where $F(s)$ is the Mellin transform of $f(x)$?
The Mellin transform of $f(x)$ is defined in formula (1) below.

(1) $\quad F(s)=\int\limits_0^\infty f(x)\,x^{s-1}\,dx$

Question: For what class of functions $f(x)$ is it true that $F\left(s^*\right)=F(s)^*$ where $F(s)$ is the Mellin transform of $f(x)$ and $^*$ is the complex conjugate?

I know of examples of functions $f(x)$ where $F\left(s^*\right)\ne F(s)^*$ but in these examples $f(x)$ has an imaginary component as well as a real component for $x>0$. I'm wondering whether $f(x)\in\mathbb{R}\ \forall\ x>0\implies F\left(s^*\right)=F(s)^*$ and if not how to define the class of functions $f(x)$ for which $F\left(s^*\right)=F(s)^*$ and what are some counter examples where $f(x)\in\mathbb{R}\ \forall\ x>0$ and $F\left(s^*\right)\ne F(s)^*$.

In this question I'm assuming the Mellin transform $F(s)$ of $f(x)$ exists, so I'm not asking about conditions for the existence of the Mellin transform in general.

答:
$\overline {F(\bar s)}=\overline {\int\limits_0^\infty {f(x)x^{\bar s-1}}dx} =\int\limits_0^\infty \overline {f(x)x^{\bar s-1}}dx=\int\limits_0^\infty \overline {f(x)}x^{s-1}dx=M(\bar f, s)$

Hence $M(f, \bar s)=\overline {M(\bar f, s)}$, so if $f$ is real then yes $M(f, \bar s)=\overline {M(f, s)}$

In particular if $f$ is not real, there is no easy relation between $M(f, \bar s), {M(f, s)}$ since both $\bar f, x^{s}$ have imaginary parts and $f \ne \bar f$

(note $f({\bar x})=\overline {f(x)}$ means $f$ is real on $(0, \infty)$ so the property is true)


问:
$\epsilon , \delta$-proof and choosing correct $\delta$
Let $$f(x)=\sqrt{1+x}$$ Show that if $x \rightarrow 0$ a limit value does exist. Furtheremore, find the limit value and explain the choice of $\delta$ w.r.t $\epsilon$ when the definitions of a limit should be shown.

I need help with $\epsilon - \delta$-proofs and existence. I thought that showing a limit must exist was the whole point of $\epsilon - \delta$-proofs and that it would also specify the value by giving us a $\delta$ thats equal to some expression with $\epsilon$.

Would someone please walk me through how such a problem would be tackled? I.e how would you first show that a limit does indeed exist, formally. Then, find what it is and explain the choice of $\delta$ in the most simple way.

Thanks!

答:
It's intuitive to see that, if exists, the limit must be $1$. Let's show that. Let $\epsilon\in\mathbb{R}^+$. Then: $$|\sqrt{1
+x}-1|=\left|\dfrac{(\sqrt{1+x}-1)\cdot(\sqrt{1+x}+1)}{\sqrt{1+x}+1}\right|=\left|\dfrac{x}{\sqrt{1+x}+1}\right|$$

Obviously, $1\leq|\sqrt{1+x}+1|=\sqrt{1+x}+1$ for all $x\in\mathbb{R}$, because of $\sqrt{}$ being a non negative function. Then, $$\left|\dfrac{1}{\sqrt{1+x}+1}\right|\leq 1$$

From this two inequalities we can conclude that $|\sqrt{1+x}-1|\leq \epsilon$ whenever $|x|<\epsilon$. So it suffices to choose $\delta=\epsilon$.


问:
Let $\pi$ be a stationary distribution of a Markov Chain. Show that if $\pi (x) >0$ and x leads to y, then $\pi (y) >0$.
Let $\pi$ be a stationary distribution of a Markov Chain. Show that if $\pi (x) >0$ and x leads to y, then $\pi (y) >0$.

I understand that since x leads to y, $P^{n}(x,y)>0$ for some $n < \infty$. By definition, the stationary distribution satisfies that $\pi(y) = \sum_{i} \pi(i)P^{n}(i,y)$.
I think I can break this up as $\pi(y) = \pi (x) P^{n}(x,y) +\sum_{i \neq x} \pi(i)P^{n}(i,y)$.
I understand the first part of this sum is greater than 0 by assumptions and definitions stated earlier, but how can I prove or how do I know the second part of this sum is greater than 0?

Please let me know if you know! Thanks!

答: $\pi(y) = \sum_{i} \pi(i)P^{n}(i,y) \geq \pi (x)P^{n} (x,y)$ since $\pi(x)P^{n}(x,y)$ is one of the terms in the sum and all the terms are non-negative. Hence, $\pi(y)>0$.


答:
Often times, $\epsilon-\delta$ proofs are written to show that a given element is in-fact what the limit is. In order to show that a sequence is convergent without a-priori knowing the limit, one has to use other lines of reasoning. Say, if you are working in the reals, you can perhaps talk about the sequence being Cauchy, or bounded and monotone in a particular direction. In other general spaces, there are more sophisticated arguments that will help you determine if a sequence converges somewhere or not, and some arguments won't necessarily even tell you what the sequence converges to. Even in $\mathbb{R}$ we have methods that tell us $\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k^3}$ converges (to what is called Aperi's constant), but we have no closed form for the actual number it coverges to, and thus $\epsilon-\delta$ is of little use in this case.

So the best approach to concrete problems such as the one you have (where the function is explicitly given to you) is to hypothesize a certain limit; and then prove using $\epsilon-\delta$ that said hypothesis holds. In this case, from elementary calculus, you are aware that the limit you seek is going to be equal to 1. Now, we use $\epsilon-\delta$ to show that it is indeed 1.

For an actual walk through on how the $\epsilon - \delta$ proof should go, I recommend looking through @R.V.N's answer. My answer would be more or less a repetition of theirs anyway haha

Title: $\epsilon , \delta$-proof and choosing correct $\delta$

Body:
Let $$f(x)=\sqrt{1+x}$$ Show that if $x \rightarrow 0$ a limit value does exist. Furtheremore, find the limit value and explain the choice of $\delta$ w.r.t $\epsilon$ when the definitions of a limit should be shown.

I need help with $\epsilon - \delta$-proofs and existence. I thought that showing a limit must exist was the whole point of $\epsilon - \delta$-proofs and that it would also specify the value by giving us a $\delta$ thats equal to some expression with $\epsilon$.

Would someone please walk me through how such a problem would be tackled? I.e how would you first show that a limit does indeed exist, formally. Then, find what it is and explain the choice of $\delta$ in the most simple way.

Thanks!


问:
How to find the number of lattice paths
How to find the number of increasing lattice paths from $(0,0)$ to $(n,n)$ that never cross but may touch the main diagonal (for example the line joining $(0,0)$ with $(n,n)$).
I've found a similar question here (I did not manage to find it again) but I guess this case is a little different (because the final answer is $2c_n$, however in the question I saw it was $c_n$).
Thanks for clarifying.

答:
JMoravitz's comment answers this.  Turning that into an answer for completeness:

The Catalan numbers count various things including paths which stay on or below the diagonal (i.e. do not go above it): a previous question here asked for a proof.

The number of paths which stay on or above the diagonal is clearly the same number, and for $n \ge 1$ a path must go above or below the diagonal at the start so the combined number of paths which do not cross the diagonal is double the Catalan numbers.

OEIS A068875 gives the values for small $n$.

Title: How to find the number of lattice paths


问:
Show that a differential is null at a maximum
I know it is a silly question, but I have a hard time proving that the differential at a critical point (say a maximum) is null.

I try to prove it using Taylor expansion around the optimum, but I can't figure out how to rearrange this expression:

$$ f(M+h)=f(M)+d_{M}f(h) + o(\lvert\lvert h \rvert\rvert)$$

答:
If $f\colon [a,b]\to\Bbb R$ has a local maximum at $c\in(a,b)$ (i.e., for all $h$ with $|h|$sufficiently small, we have $f(c+h)\le f(c)$) and $f'(x)$ exists, then $f'(c)=0$.

Indeed, if we had $f'(c)>0$, then for sufficiently small positive $h$, we'd have $\frac{f(c+h)-f(c)}{h}>\frac 12f'(c)>0$; and if $f'(c)<0$, then for sufficiently (absolutely) small negative $h$, $\frac{f(c+h)-f(c)}{h}<\frac 12f'(c)<0$. Both cases lead to $f(c+h)-f(c)>0$, contradiction.