问:
Solving rank equations
Suppose we are given a (finite) matrix $A$ with entries $a_{ij} \in k[w_1, \ldots, w_n]$ which are polynomial expressions of $w_1, \ldots, w_n$ over some field $k$. For a fixed integer $m$, is there a way to determine all points $p=(w_1, \ldots, w_n) \in \mathbb A^n(k) = k^n$ such that
$$\operatorname{rank} A=m,$$ i.e. which $p\in k^n$ makes the rank of $A$ to be $m$?

答: The rank of a matrix is the size of its largest non-vanishing minor, so you can express your condition by polynomial equations in the matrix entries.</p>

Title: Solving rank equations


问:
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.

First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.

So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?

Can anyone help?

ContentLicense: CC BY-SA 4.0

Body:
Hint: We have that
$$f(x)+f(1-x)=\frac{1}{3^{x} + \sqrt{3}}+\frac{1}{3^{1-x} + \sqrt{3}}=\frac{1}{3^{x} + \sqrt{3}}+\frac{3^{x}/\sqrt{3}}{\sqrt{3} + 3^x}=\frac{1}{\sqrt{3}}$$


Body:
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.

First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.

So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?

Can anyone help?

Body:
Let $\sqrt3=a$

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

If $\dfrac{a^{2x-1}}{1+a^{2x-1}}=f(px+q)=\dfrac1{1+a^{2(px+q)-1}}$

$\implies 1-2x=2(px+q)-1$

$\implies p=-1,q=1$

$$\implies f(x)+f(1-x)=a$$

Can you take it from here?


问:
Formal Rules for Categorical Duality
EDIT: My question is not about what categorical duality is. I generally understand the idea, but my problem is that for every theorem, finding the dual theorem requires us to think about what it means, so that we can translate it into the opposite category by reversing the arrows. I am looking for a concrete set of "algebraic manipulations" that lets us take a statement to its dual, without understanding what anything means. In the example of negating a logical statement (in the final paragraph of the question), we can swap the symbols around without knowing what $P$ is, and more generally without understanding what things mean.

Duality is very important in category theory, but as far as I have seen, statements that "follow by duality" are often taken to be obvious, even though in reality there are lots of details to be checked. People seem to follow some informal procedure of replacing all instances of "left" with "right", "right" with "left", "limit" with "colimit" and so on.

This is reminiscent of the method for negating logical statements. For example, the statement $(\forall x, \exists y: P(x,y))$ has negation $(\exists x:\forall y, \neg P(x,y))$. We can obtain this negation by swapping $(\forall)$ with $(\exists)$, $(:)$ with $(,)$, and $(P)$ with $(\neg P)$. The set of rules for converting a statement of this form is well-known and easy to implement. It is also completely rigorous because we can actually prove in general that the rules apply.

Is there a similar set of rules for converting categorical statements into their duals?

答:
The idea of duality is that, whenever you have a purely categorical statement (or concept), you can look at what that statement means for a category $ C $ when applied to $ C^{\mathrm{op}} $: this gives you the dual statement or concept. Then, if you have a statement that is true in any category, then its dual statement is also necessarily true, since it is true for $ C^\mathrm{op} $. Finding out the dualized statement is then just a matter of properly writing down the statement in the dual category and interpreting it in the original one (which can often be tedious).

As an example, you can see how the definition of an initial object in $ C^{\mathrm{op}} $ coincides with that of a terminal object in $ C $! Thus, the dualized statement of "all initial items are uniquely isomorphic" is "all terminal objects are uniquely isomorphic". Same goes for limits/colimits, continuous/co-continuous functors, left/right-adjoints (note here that there are two categories involved, both of which you end up dualizing), left/right Kan extensions, kernels/cokernels, images/coimages in an abelian category, etc. In the end, you can apply duality to whatever statement you want, it is just a matter of practicing that conceptually simple but powerful tool.

I do think though that the lack of details is just typical of users of categories in general, and not just of duality, but that is another topic altogether.

Title: Formal Rules for Categorical Duality


问:
Prove the indication of the continuity of a function on its Jacobian
Suppose that a function $f: \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable on $\mathbb{R}^n$. Further, the Jacobian of $f$ at a point $x_0 \in \mathbb{R}^n$ is non-zero, that is,
\begin{equation}
\begin{vmatrix}
\frac{\partial f_1\left(x_{0}\right)}{\partial x_{1}} & \frac{\partial f_1\left(x_{0}\right)}{\partial x_{2}} & \dots & \frac{\partial f_1\left(x_{0}\right)}{\partial x_{n}} \\
\frac{\partial f_2\left(x_{0}\right)}{\partial x_{1}} & \frac{\partial f_2\left(x_{0}\right)}{\partial x_{2}} & \dots & \frac{\partial f_2\left(x_{0}\right)}{\partial x_{n}} \\
\vdots & \vdots & \ddots & \vdots\\
\frac{\partial f_n\left(x_{0}\right)}{\partial x_{1}} & \frac{\partial f_n\left(x_0\right)}{\partial x_{2}} & \dots & \frac{\partial f_n\left(x_0\right)}{\partial x_{n}}\\
\end{vmatrix}
\neq 0.
\end{equation}
How to prove the following proposition: there exists a neighborhood around $x_{0}$ where the Jacobian of $f$ is non-zero? That is, deduction of the following proposition is needed:
\begin{equation}
\exists \delta > 0, \forall x \in \mathbb{R}^{n},\lVert x-x_{0} \rVert < \delta \longrightarrow
\begin{vmatrix}
\frac{\partial f_{1}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{1}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{1}\left(x\right)}{\partial x_{n}} \\
\frac{\partial f_{2}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{2}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{2}\left(x\right)}{\partial x_{n}} \\
\vdots & \vdots & \ddots & \vdots\\
\frac{\partial f_{n}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{n}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{n}\left(x\right)}{\partial x_{n}}\\
\end{vmatrix}
\neq 0.
\end{equation}

Below is my own thinking and doubts about the proof.

Using the condition that $f$ is continuously differentiable, we have the following set of propositions:
\begin{equation}
\forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R}^{n}, \lVert x - x_{0} \rVert < \delta \longrightarrow \left\vert \frac{\partial f_{i}\left(x\right)}{\partial x_{j}} - \frac{\partial f_{i}\left(x_{0}\right)}{\partial x_{j}} \right\vert < \epsilon, i,j = 1,2,\dots,n.
\end{equation}
We should assume the following:
\begin{equation}
\forall \delta > 0, \exists x \in \mathbb{R}^{n},\lVert x-x_{0} \rVert < \delta \wedge
\begin{vmatrix}
\frac{\partial f_{1}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{1}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{1}\left(x\right)}{\partial x_{n}} \\
\frac{\partial f_{2}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{2}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{2}\left(x\right)}{\partial x_{n}} \\
\vdots & \vdots & \ddots & \vdots\\
\frac{\partial f_{n}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{n}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{n}\left(x\right)}{\partial x_{n}}\\
\end{vmatrix}
= 0,
\end{equation}
and draw a contradiction. However, I am not sure what trick to use to relate these partial derivatives to the Jacobian. Expanding the Jacobian using its definition in linear algebra looks really formidable. However, it seems that there is little chance to circumvent expanding it.

答:
You're going too deep into the details. If you stay at a higher level, you'll realize that the determinant of a square matrix is a polynomial function of the entries of the matrix, and is therefore a continuous function of its inputs. So you can see the determinant of the Jacobian as a composition of two functions: the first one from $\mathbb R^n$ to $\mathbb R^{n \times n}$ sends a vector to the Jacobian matrix, and this is a continuous function because each partial derivative is continuous by assumption; and the second one sends a matrix to its determinant, which is continuous because it is polynomial in the entries.

So if the determinant of the Jacobian is non-zero at a point, you can find a neighborhood of that point where the determinant remains say positive (if the value of the point was positive) or negative (in the other case).

Hope that helps,


问:
Theories question
The question: Let $L=${S}, S is a two-place relation. I need to show that there exists a theory - $T$, in the vocabulary L such that for each structure $M$ to interpret $L$:

$M\models T \leftrightarrow S^M$ is a partial order on the world and on the world there exist infinite maximum elements in a relation to $S^M$.

I have tried using the compatible theorem but without any success, and now I'm stuck.

答:
First we can knock out the axioms that $S$ is a partial order, which are standard:

$\psi_1\equiv\forall v(v S v)$                                                       (reflexivity)

$\psi_2\equiv\forall v,w([vSw\wedge wSv]\rightarrow v=w)$        (anti-symmetry)

$\psi_3\equiv\forall u,v,w([uSv\wedge vSw]\rightarrow uSw)$        (transitivity)

Now, to express that there are infinitely many maximal elements with respect to $S$, we can use a common trick. Recall that a theory $T$ can have infinitely many axioms. In particular, if we want to express "there are infinitely many elements satisfying property $P$", we can add infinitely many axioms to $T$, one for each $n\in\mathbb{N}$, each expressing "there are greater than $n$ elements satisfying property $P$", and this will suffice. So, consider the following $L$-sentence:

$\phi_n\equiv\exists v_1,\dots,v_n\big[\left(\bigwedge_{i<j}v_i\neq v_j\right)\wedge\left(\bigwedge_{i=1}^n \forall w[ v_i S w\rightarrow v_i=w]\right)\big]$

$\phi_n$ states precisely that there are $n$ distinct elements, each maximal with respect to $S$. So, if $T$ contains all $\phi_n$ for every $n\in\mathbb{N}$, then any model of $T$ must have infinitely many maximal elements with respect to $S$. In particular, letting $T=\{\psi_1,\psi_2,\psi_3\}\cup\{\phi_n\}_{n\in\mathbb{N}}$ gives the desired theory.

Title: Theories question


问:
Showing whether $\tau=\inf\{t:X_t=\sup_{0\leq t\leq T}X_t\}$ is a stopping time or not
As the title suggests, I am not sure how to show if $\tau=\inf\{t:X_t=\sup_{0\leq t\leq T}X_t\}$ is a stopping time or not, where $\{X_t\}$ is an adapted process. Intuitively, I see that it is not a stopping time from the finance context — if $X_t$ refers to stock price and we want to exercise an American call option, then if $\tau$ was a stopping time this would contradict the assertion that early exercise of an American call option is not optimal. However, this is as far I can go, as my intuition tells me this is not in $\mathcal{F}_t$ generated by $X_t$, but how should I start proving this rigorously, and a bigger concern is what about $\mathcal{F}_{t^+}$? Would be nice to see in particular for the geometric Brownian motion case too.

(NB: I am asking this in the Math SE instead of QF SE as I am searching for a more mathematical justification, but I am placing my American option argument in here for some context.)

答:
Assume, for the sake of contradiction, that $\tau$ is a stopping time.

I will assume $X$ Brownian motion so that it's clearer.

Since $0 \leq \tau \leq T$, we can you the optional stopping time theorem and say that $\mathbb{E}[X_\tau] = \mathbb{E}[X_0] = 0$.

However, it's clear that $X_\tau > 0$ almost surely, so we obtain a contradiction.


问:
How to solve ODE with cosine involved
I sought for this really long in the internet but didn't happen to find a general solution.
So here is the equation:

$y' = y + x\,\cos(2\,x)$

I know that for the particular solution $y_p$ following assumption is made:

$y_p = (a_0+a_1\,x)\,\sin(2\,x)+(b_0+b_1\,x)\,\cos(2\,x)$

so that

$\begin{align}
{y_p}' &= a_1\,\sin(2\,x)+(a_0+a_1\,x)\,2\,\cos(2\,x)+ b_1\,\cos(2\,x)-(b_0+b_1\,x)\,2\,\sin(2\,x)
 \\ &=y + x\,\cos(2\,x) = {y_p}'
\end{align}$

But I don't know at all how to determine $a_1, a_0, b_1, b_0$

I could also just solve $y = C\,e^{x} +e^{x}\,\int{\frac{\cos(2\,x)}{e^x}}\mathrm{d}x$ thanks to another answer here, but how to do it generally?

Title: How to solve ODE with cosine involved

Body:
Grouping by $sin(2x)$ gives the equation:

$$a_1 - 2b_0 = a_0$$

Grouping by $x sin(2x)$:

$$-2b_1 = a_1$$

Grouping by $cos(2x)$:

$$2a_0 + b_1 = b_0$$

Grouping by $x cos(2x)$:
$$2a_1 = b_1 + 1$$

Solve and you get $b_1 = -1/5$, $a_1 = 2/5$, $a_0 = 4/25$, $b_0 = 3/25$.


问:
How can I prove that $\sin (10^\circ), \sin(1^\circ), \sin(2^\circ), \sin(3^\circ), \tan(10^\circ)$ are irrational

Prove that $\sin (10^\circ)$, $\sin(1^\circ)$, $\sin(2^\circ)$, $\sin(3^\circ)$, and $\tan(10^\circ)$ are irrational.

My Attempt:

Let $x = 10^\circ$. Then

$$
\begin{align}
x &= 10^\circ\\
3x &= 30^\circ\\
\sin (3x) &= \sin (30^\circ)\\
3\sin (10^\circ)-4\sin^3(0^\circ) &= \frac{1}{2}
\end{align}
$$

Now let $y = \sin (10^\circ)$. Then

$$
\begin{align}
3y-4y^3 &= \frac{1}{2}\\
6y-8y^3 &= 1\\
\tag18y^3-6y+1 &= 0
\end{align}
$$

How can I calculate the roots of $(1)$?

答:
Here is a different approach for $\sin(1^\circ)=\sin\left(\frac{2\pi}{360}\right)$ (the other cases are similar).

The complex numbers $\zeta_{1,2}=e^{\pm\frac{2\pi}{360}i}$ are algebraic integers (are roots of $x^{360}-1$) and therefore $\zeta_1+\zeta_2=2\sin\left(\frac{2\pi}{360}\right)$ is algebraic integer. If $\sin\left(\frac{2\pi}{360}\right)$ is rational then $2\sin\left(\frac{2\pi}{360}\right)$ is rational and therefore integer (the only rational algebraic integers are the integers).

So $2\sin(1^\circ)=-2,-1,0,1$ or $2 \Rightarrow\Leftarrow$.

ContentLicense: CC BY-SA 3.0


问:
Spivak's treatement of Conic Sections
In his book Calculus, Spivak goes on to treat conic sections as follows ::

One of the simplest subsets of this three-dimensional space is the (infinite) cone illustrated in Figure 2; this cone may be produced by rotating a "generaling line," of slope $C$ say, around the third axis.

$\quad$

For any given first two coordinates $x$ and $y$, the point $(x,y,0)$ in the horizontal plane has distance $\sqrt{x^2+y^2}$ from the origin, and thus $$\tag1 (x,y,z)\text{ is in the cone if and only if }z=\pm C\sqrt{x^2+y^2}.$$ We can descend from these three-dimensional vistas to the more familiar two-dimnensional one by asking what happens when we intersect this cone with some plane $P$ (Figure 3).

$\quad$

If the plane is parallel to the horizontal plane, there's certainly no mystery--the intersection is just a circle. Otherwise, the plane $P$ intersects the horizontal plane in a straight line. We can make things a lot simpler for ourselves if we rotate everything so that this intersection line points straight out from the plane of the paper, while the first axis is in the usual position that we are familiar with. The plane $P$ is thus viewed "straight on," so that all we see (Figure 4) is its intersection $L$ with  the plane of the first and third axes; from this view-point the cone itself simply appears as two straight lines.

$\quad$

In the plane of the first and third axes, the line $L$ can be described as the collection of all points of the form $$(x,Mx+B),$$ where $M$ is the slope of $L$. For an arbitrary point $(x,y,z)$ it follows that $$\tag2 (x,y,z)\text{ is in the plane }P\text{ if and only if }z=Mx+B.$$ Combining $(1)$ and $(2)$, we see that $(x,y,z)$ is in the intersection of the cone and the plane if and only if $$\tag{$*$} Mx+B=\pm C\sqrt{x^2+y^2}.$$ Now we have to choose coordinate axes in the plane $P$. We can choose $L$ as the first axis, measuring distances from the intersection $Q$ with the horizontal plane (Figure 5); for the second axis we just choose the line through $Q$ parallel to our original second axis. If the first coordinate of a point in $P$ with respect to these axis is $x$, then the first coordinate of this point with respect to the original axes can be written in the form $$\alpha x+\beta$$ for some $\alpha$ and $\beta$. On the other hand, if the second coordinate of the point with respect to these axes is $y$, then $y$ is also the second coordinate with respect to the original axes.

$\quad$

Consequently, $(*)$ says that the point lies on the intersection of the plane and the cone if and only if $$M(\alpha x+\beta)+B=\pm C\sqrt{(\alpha x+\beta)^2+y^2}.$$ Although this looks fairly complicated, after squaring we can write this as $$\alpha^2C^2y^2+\alpha^2(M^2-A^2)x^2+Ex+F=0$$ for some $E$ and $F$ that we won't bother writing out. Dividing by $\alpha^2$ simplifies this to $$C^2y^2+(C^2-M^2)x^2+Gx+H=0.$$

My Question:

Why did he have to add the $\alpha x+\beta$ step? Why did he choose these particular coordinates? Is that so that he can translate the coordinates on the plane to those of the $xyz$ coordinate system? Any clarifications are welcome!

答: We're looking for the equation that gives us the set of points where the plane intersects the cone. If we didn't introduce $\alpha x + \beta$, we merely get the projection of the intersecting points onto the x-y plane, which would be the same shape but with distorted proportions. If we wanted to see how the intersection would look on the plane P, we have to change the input for $x$ so that a single unit change in the original $x$-axis corresponds to the same change on the "planar" axis, thus giving us an ellipse/hyperbola/circle with the correct dimensions.

Title: Spivak's treatement of Conic Sections


Title: Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.

Body:
$\sqrt{3}f(x) = \frac{1}{\sqrt{3^{2x-1}}+1}$ and $\frac{1}{\sqrt{3^{a}}+1}+\frac{1}{\sqrt{3^{-a}}+1}=1$.
You can pair like $(-11,11),(-9,9),(-7,7)\cdots(-1,1)$ then answer is $6$.


答:
You can start with the usual axioms for a partial order, and for each $n\in\Bbb N$ add the following formula:
$$\exists x_1,\dots,x_n: \bigwedge_{i< j}\lnot (x_i=x_j)\,\land\,\bigwedge_i\psi(x_i)$$
where $\psi(x)$ is a formula stating that $x$ is a maximal element.

Title: Theories question

Body:
The question: Let $L=${S}, S is a two-place relation. I need to show that there exists a theory - $T$, in the vocabulary L such that for each structure $M$ to interpret $L$:

$M\models T \leftrightarrow S^M$ is a partial order on the world and on the world there exist infinite maximum elements in a relation to $S^M$.

I have tried using the compatible theorem but without any success, and now I'm stuck.


答:
You're ever so slightly off. Notice the median of $(-5, -4, ..., 5, 6)$ is $\frac{-5+6}{2}=\frac 12$ which hints at trying $$f(\frac12+x)+f(\frac 12-x)\overbrace{=}^{y=x+\frac 12}f(y)+f(1-y)$$
We see that: $$\frac{1}{3^x+\sqrt 3}+\frac{1}{3^{1-x}+\sqrt 3}=\frac{3^x+3^{1-x}+2\cdot 3^\frac 12}{3^{x+\frac 12}+3^{\frac32-x}+2\cdot 3^1}=\frac{\alpha}{3^\frac 12 \cdot \alpha}=\frac{1}{\sqrt 3}$$


问:
Voiculescu theorem for non-separable $C^*$-algebras
In many refernce books, when the authors mention the Voiculescu theorem. They often state that the $C^*$-algebras are separable. See A proposition about Voiculescu's Theorem in C*-algebra

If the $C^*$-algebras are non-separable, does the theorem also hold?

答:
Already Glimm's Lemma (of which Voiculescu's Theorem is a generalization) fails for (some) non-separable C$^*$-algebras.

For instance consider $B(H)$, with $H=\ell^2(\mathbb N)$ and let $A$ be the diagonal masa (that is, the copy of $\ell^\infty(\mathbb N)$ acting as diagonal operators). It is well-known that the pure states of $A$ that are zero on $A\cap K(H)$ are in correspondence with the non-principal ultrafilers on $\mathbb N$. And there are $2^c$ of these. On the other hand, the number of sequences in $H$ is $c$. So by cardinality there are pure states which cannot be limits of pointwise evaluations on a sequence.


问:
How torsion arise in homotopy groups of spheres?
The example I have in mind of a torsion element in fundamental group is what happens on the projective plane: there is a cell $D^2$, but its boundary doubly covers a copy of $S^1$. The latter has then order two in the homotopy group. Let me call this a "wrapped 2-disk" sitting inside the projective plane.

In general, a torsion element $\gamma \in \pi_k(S^n) $ can be thought as a commutative diagrams

$$ \begin{array}{ccc}
\partial D^{k+1} & \rightarrow & D^{k+1} \\
\downarrow & & \downarrow \\
S^k & \rightarrow & S^n
\end{array}
$$

For the left vertical arrow having finite non zero index, and this is a wrapped $(k+1) $ disk. It is known that even for low indices torsion element exists; do you know an explicit representative? I suspect it could be symmetric enough.

Any geometric idea on how this "finite wrapping" between different dimensions should arise is welcomed.

答:
With the classical description of homotopy groups of spheres, the best answer is probably "It appears because there is no reason for it not to." If I recall correctly, every finite, simply-connected, and noncontractible CW complex has torsion in its homotopy groups - this is a result of Serre.

If you allow different descriptions of homotopy groups of spheres, then there can be more interesting explanations. In the comments Кряжев Арсений, alludes to framed manifolds. It is a result of Thom, I believe, that says the stable homotopy groups of spheres are isomorphic to the cobordism ring of framed manifolds. Here stable refers to a specific subcollection of the homotopy groups of spheres (or alternatively the colimit of the homotopy groups under the suspension homomorphism). Framed means manifolds equipped with a trivialization of their stable normal bundle, and cobordism means take these decorated manifolds up to the equivalence relation generated by $M \simeq N$ if there is a decorated manifold with boundary $W$ so that $\partial W = M \cup N$.

Then under this isomorphism, the suspension of the Hopf map is represented by the circle and instead of the canonical framing of its normal bundle in $\mathbb{R}^3$, we twist it once as we go along the circle. This can't be extended along a manifold with boundary $S^1$, so it gives a nontrivial element of the stable homotopy groups, but two copies of this bound a cylinder the framing can be extended along, so the element has order 2.

There is a another reasonable description I like of higher homotopy groups in the homotopy groups of spheres, and that is that it is required by the existence of cohomology operations. It's a classical result that the spaces with a single nontrivial homotopy group in degree n represent nth cohomology with coefficients in that homotopy group. By the Yoneda lemma, the cohomology of these spaces represents natural transformations between cohomology groups. There are many calculations of these natural transformations, so called cohomology operations, and they surely exist.

Since the cohomology of $S^n$ is nontrivial only in degree n, the existence of cohomology operations implies that this cannot represent nth singular cohomology, so it must have multiple nontrivial homotopy groups. Additionally, it is not hard to prove that if n is odd, $S^n$ has no higher homotopy groups with nontorsion in them. So we see in this case, that torsion is forced to exist as a result of cohomology operations existing.

Of course there are direct ways to show many nontorsion elements, and essentially all the computational ways will leave you with the feeling "This is just here by chance." And they really are, the homotopy groups of spheres are incredibly chaotic, and they are definitely not understood through the classical definition of maps of spheres that bound a disk. Topologists try to abstract away until we are very far from this original definition, using it only when necessary. If you have a geometric mind you will like Кряжев Арсений's description, and if you have a more algebraic mind you will like things like the Adams Spectral sequence. Both explain why torsion arises, and both are very far from the original definition.


问:
Solving $x^2 y''-2xy'=0$ using power series
Given the following ODE
$$x^2 y''-2xy'=0$$
I'm trying to solve it using power series. Since
$$y=\sum_{n=0}^{\infty} a_{n}x^{n}$$
then
$$x^{2}y''=\sum_{n=2}a_{n}n(n-1)x^{n}$$
and
$$2xy'=\sum_{n=1}^{\infty}2a_{n}nx^{n}=2a_{1}x+\sum_{n=2}^{\infty}2a_{n}nx^{n}$$
Therefore, in the equation, I get
$$-2a_{1}x+\sum_{n=2}^{\infty}(n^{2}-3n)a_{n}x^{n}=0$$
So
$$-2a_{1}=0\Rightarrow a_{1}=0$$
and
$$(n^{2}-3n)a_{n}=0,\quad\forall n\geq2$$
Since I'm looking for non-trivial solutions, $a_{n} \ne 0\Rightarrow n^{2}-3n=0$ which gives me $n=0$ or $n=3$.

I want to know if it's right and if so, what I'm supposed to do with these values of $n$.

答:
The $n=0$ does not come from the equation:
$$n^2-3n=0 \,\, n \ge 2$$.
This gives you only $n=3$.

The $n=0$ comes from the fact that you have no constraint on $a_0$. So that you have $a_1=0, a_0 \ne 0, \,$  $a_3 \ne 0$. And $a_n=0 $ for $n \ge 2$ and $n \ne 3$.

The solution is therefore:
$$y(x)=a_0x^0+a_3x^3=a_0+a_3x^3$$


问:
Calculate parameters of regression line
I'm learning statistics and trying to calculate the parameters of a regression line based on the data:

$(x,y)=(0,-1),(1,2),(2,9/2)$

Could you please tell me how would I do so? I'm not sure how to calculate the parameters based just on a table of values.

Thanks

答:
Let´s say the regression function is $y=a+bx$ Then you have to minimize

$$S=\sum_{i=1}^3 (y_i-\alpha-\beta x_i)^2,$$

where $\alpha$ and $\beta$ are the estimators for the parameters $a$ and $b$. To obtain the minimum you differentiate $S$ w.r.t a and b respectively and set them equal to $0$.
$$\frac{\partial S}{\partial \alpha}=-2\cdot \sum_{i=1}^3 (y_i-\alpha-bx_i)=0$$

$$\frac{\partial S}{\partial \beta}=-2\cdot \sum_{i=1}^3 (y_i-\alpha-\beta x_i)\cdot x_i=0$$

The solutions of that two equations are

$$\beta=\frac{ \sum\limits_{i=1}^3 (x_i - \bar{x})(y_i - \bar{y}) }{ \sum\limits_{i=1}^3 (x_i - \bar{x})^2 } \ \ (I), \qquad\qquad \alpha=\overline y-\beta\overline x \ \ (II)$$

So in your case $$\beta=\frac{ (0-1)\cdot (-1-5.5/3)+(1-1)\cdot (2-5.5/3)+(2-1)\cdot (4.5-5.5/3) }{ (0-1)^2+(1-1)^2+(2-1)^2 }=2.75$$

With the help of $(II)$ it is straightforward to calculate the value of $\alpha$.

Title: Calculate parameters of regression line


问:
Reference for representation theory of classical groups over arbitrary fields
I am looking for a reference for the representation theory of (classical) algebraic groups, that would ideally be something like the second part of Jantzen's "Representations of Algebraic Groups", but with a treatment over arbitrary fields (especially: of non-split classical groups). It might very well be that there is no such book currently, in which case I would welcome a reference to some well-written articles that give a good idea of what is currently known, and the basic methods.</p>

答:
I think probably what you might want is in some articles of Tits. You can look in [1, §25.d] for an overview of these results with references.

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:
Computing localizations in $\mathbb{Z}/n\mathbb{Z}$
I am trying to find a way to easily compute localisations of the ring $\mathbb{Z}/n\mathbb{Z}$ ($n>1$). Is there any general result for this? I found here that when the multiplicative subset is $S=\{1,b,b^2,\dots\}$ and the prime factorization of $n$ is $n = \prod p_i^{l_i}$ then
$$S^{−1}(\mathbb{Z}/n\mathbb{Z}) \cong \prod_{p_i \nmid b} \mathbb{Z}/p_i^{l_i}\mathbb{Z}$$

Does this formula somehow generalize to an arbitrary multiplicative subset $S \subset \mathbb{Z}/n\mathbb{Z}$?

答:
By Chinese remainder theorem, we have a ring homomorphism
$$\Bbb Z/n\Bbb Z\xrightarrow\sim\prod_i\Bbb Z/p_i^{l_i}\Bbb Z$$
By exactness of localization, we get an isomorphism:
$$S^{-1}(\Bbb Z/n\Bbb Z)\xrightarrow\sim\prod_iS^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)$$
If $p_i$ divide some element in $S$, then $S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)=\{0\}$, otherwise $S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)\cong\Bbb Z/p_i^{l_i}\Bbb Z$.

If $T$ is the saturation of $S$, that's the set of all divisors of elements in $S$, then
$$S^{-1}(\Bbb Z/n\Bbb Z)\cong\prod_{p_i\notin T}S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)$$


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: For Field Theory, we have The Fundamental Theorem of Galois Theory. It is a basic tool not only for fields, but for algebraic number theory among other fields, and it establishes a beautiful connection between group theory and field theory.

Title: Landmarks of subjects of mathematics


问:
Probability regarding a contracting sequence of events
The aim of the exercice is to prove that if $A_1, A_2, ...$ is a contracting sequence of events, that is : $A_1 \supset A_2 \supset ...$ then, $P(A_1 \cap A_2 \cap ...) = lim_{n\to\infty} P(A_n)$

a Hint which has been given to us was to apply De Morgan's law in order to get to the solution, so here I started : $$\Omega \backslash (A_1 \cap A_2 \cap ...) = (\Omega \backslash A_1) \cup (\Omega \backslash A_2) \cup ...$$
Then, I've applied the definition of a probability measure :
$$ P(\Omega \backslash (A_1 \cap A_2 \cap ...)) = P(\Omega \backslash A_1) + P(\Omega \backslash A_2) +... = \sum_{i=1}^{n}P(\Omega \backslash A_i)$$

At this point I don't know what's missing to conclude to the equality with the limit, I understand intuitively that here, as our sequence is contracting, we'll "converge" to what we're looking for but I don't know what I may be allowed to introduce here to get to the proof or if it's explicit enough here to conclude.

答: For every $n$, $A_1\cap A_2...\cap A_n=A_n$ Therefore $P(A_1\cap A_2...\cap A_n)=P(A_n)$  The limits are then equal.


问:
Calculating flight distance of drone without losing line of sight due to obstacles
I hope everyone is off to a good star of the new year!

So I'm trying to figure out a way of calculate how far I'm able to fly my drone without losing visual sight of it when having an obstacle (a forest for example, but a house or hill or similar would also be applicable) between me and the drone. This would be helpful when planning the flight and choosing a take off site.

The problem:
Where I live we have lots of forests. There are clearings in the forests of different sizes however and I want to launch and control my drone from inside of these clearings. The problem is that I need to keep a clear visual line of sight (VLOS) to the drone, the reason for this is twofold: It's the law and the drone will lose its connection eventually if the obstacle is removing enough signal strength. Without knowing how far I'm able to fly the drone in VLOS while planning the mission I risk not being able to cover the planned area. So I need to be able to calculate the possible flight distance during planning and i figured simple trigonometry would be a logic way of doing that.

The solution:
Calculating the flight distance using simple trigonometry, using the Pythagorean theorem and Tangents, Sines, and Cosines.

How:
The illustration below shows the problem. The line i call VLOS need to be clear of any obstacle, in this case, trees. I want to know the maximum flight distance before the VLOS line hits a tree.

Assuming I know the height of the obstacle and the distance between it and where I will be standing with the remote, and also the altitude of the drone, I should be able to calculate the flying distance without losing VLOS.

As you see in this illustration I made (Flight distance illustration) if the drone goes any further the VLOS will hit the trees and thus making it impossible to see, and also weakening the signal. Assuming I'm already at maximum height (120 m) I'm unable to increase the altitude.

In this illustration (Flight distance illustration with values) I have changed the names of the lines to a,b,c etc. I have also made another triangle consisting of the distance between the RC and the bottom of the tree (b1), the height of the tree (a1) and the distance between the RC and the top of the tree. (c1).

Here are the values i will use:

a1= 36,7m

b1= 33m

c1= ?

a2= 120m

b2= ? (what I'm after)

c2= ?

v= ?

To start I would use the pythagorean theorem to get the length of c1 (distance between RC and treetop):

√(a1^2+b1^2)= c1

√36,7^2+33^2= 49,35m

The next step then would be to calculate the angle of v using arccos (degrees)

arccos(b1/c1)=v

arccos(33/49,35)=48,03°

Now the last step is to get the flight distance (b2) using Tangents

a2/Tan(v)=b2

120/Tan(48,03)=107m

So the maximum flight distance (while still being able to unhindered see the drone) is calculated to be 107m, however I have tried this exact scenario and the flight distance was instead 172m. Why is that?

The weaknesses that I have found to this method is:

My math. To put i simply I'm not very good at math and I could have easily made a mistake in my equations.

Using the wrong value for a1. This the easiest error to make here, since I don't have the exact height of the trees. Accuracy depending on method used for measuring the height. (in my case a tree height measuring application)

Using the wrong value for b1. Also easy to make. Depends on method used to measure. (in my case I measured in QGIS on satellite imagery)

Note: The drone doesn't change its altitude and it is relative to where the RC is, so the terrain under the drone doesn't affect the result of a2, it is constant)

If I change a1 and b1 to be -5m and +5m (a1=31,7 and b1=38) I still only get 148 m of flight distance. I find it rather unlikely that I would be that much off on both of these values.

While I was out doing this test I was also able to make a test where a1=26,7m and b1=94m. The result using the same calculation as above is 422 m  flight distance (or b2). The real result however was 843 m of flight distance!

So what is going on here? Why is the calculated flight distance much lower than the real flight distance?

Am I using the equations wrong?

I look forward to any input on this, thanks in advance!

UPDATE: I will do a third test using an obstacle with known height and distance since it all points to that there is something off with the values of a1 and b1. I will post the result as soon as this is done. Thanks for all the input so far!

Best regards
Patrik Forsberg

答:
The triangles are "similar" so the ratio "c1:a1" holds for other sides too. What you found with $a2/Tan(v)=b2$ is that the horizonal distance is $107.91$ and, if that is what you want, you are correct.

$$c1=\sqrt{a1^2+b1^2}=\sqrt{36.7^2+33^3}=49.35$$

$$r1=\frac{a2}{a1}=\frac{120}{36.7}\approx 3.27$$

$$b2=r1\cdot b1 \approx 3.27 \cdot 33 = 107.91$$
$$c2=r1\cdot c1 \approx 3.27 \cdot 49.35 = 161.3745$$

$$V=\arcsin{\frac{a1}{c1}}
=\arcsin{\frac{36.7}{49.35}}\approx\arcsin{(0.7436)} \approx 0.8385 \space rad \approx 48.04^\circ$$

With the change in a,b th new $c1$ length is hardly changed
$c1=\sqrt{31.7^2+38^2}\approx 49.48$

Assuming the max altitude is the same $(120)$, we have a  greater ratio:
$$r2=\frac{120}{31.7}\approx 3.79$$
thus $$b2=r2\cdot b1=3.79\cdot 38\approx 143.85$$

I can only guess that there is a difference between the a-value(s) and the actual height of the trees.


问:
$(,) = x^2$ How to determine if injective or surjective?
I just learned to determine wether $f(x)=x^2$ is injective or surjective but how can I solve this when there are two variables? I've been struggling to get my head around this.
I know that for a function to be injective the following condition must be met:
$()≠()$ implies $≠$, or the contrapositive.
Surjective:
$∀ ∈ , ∃ ∈  \text{ such that } f(x) = y$.

How could I scale these definitions to handle two variables?
$(,) = x^2$
mapping from $\Bbb R \times \Bbb R$ to $\Bbb R$

Edit: I want to thank you all for your time and help. It's greatly appreciated! I can clearly see through these problems now.

答:
Take a deep breath and do it literally.

If $f(x_1, y_1) = x_1^2 = M$ and $f(x_2, y_2) = x_2^2 = M$. so $x_1^2 = x_2^2$ does that mean $x_1 = x_2$ and $y_1 = y_2$?  If yes then it is injective (so explain why).  If no it is not injective (if not explain why not and/or give a counter example).

So does $x_1^2 = x_2^2$ mean $x_1 = x_2$ and $y_1 = y_2$?  the answer is of course not.  As the $y_k$ have nothing to do with anything they can be any thing.  A counter example if $f(2, 9) = 2^2 =4$ and $f(2, \frac {\log_7 \sqrt \pi}{e^{2.7}}) = 2^2 = 4 = f(2,9)$.  But $(2,9) \ne(2, \frac {\log_7 \sqrt \pi}{e^{2.7}})$ So it is not injective.

Not to mention $f(x,y) = x^2 = (-x)^2 = f(-x,tony\ the\ tiger)$ and $x\ne -x$ (if $x \ne 0$)

so not injective.

Surjective.

If $w\in \mathbb R$ then is there always as $(x,y)\in \mathbb R^2$ so that $f(x,y) = x^2 = w$.

If we try to solve for $x^2 = w$ we get that $x = \pm \sqrt w$ and $y =\text{any dang thing we want}$ will be a solution but does $(\pm \sqrt w,\text{any dang thing we want})$ always exist.  Well, $\text{any dang thing we want}$ always exists, but if $w < $ then $\pm \sqrt w$ does not.

So it is not surjective.  A counter example is $f(x,y) = -7$ has not solution as there are no $(x,y)$ where $x^2 = -7$ exist.


问:
Is $\int x\,dx \cdot x=\int x^2 \,dx$ or $\int x\,dx \cdot x=(\int x \, dx)\cdot x$?
Is $\int x\,dx \cdot x=\int x^2 \,dx$ or $\int x\, dx \cdot x=(\int x \,dx)\cdot x$?

Edit: My integrals looks like this:
\begin{align}
& \int_0^t e^{-\xi(t-t^\prime)} \, d t^\prime \int_0^t e^{-\xi(t-t^{\prime \prime})} \frac{2 C}{m^2} \delta(t^{\prime}-t^{\prime \prime}) \, dt^{\prime \prime} \\[6pt]
= {} &\int_0^t e^{-\xi(t-t^\prime)} \, d t^\prime\left(e^{-\xi(t-t^\prime)} \frac{2 C}{m^{2}} \Theta(\mathrm{t}-\mathrm{t}^\prime)\right)
\end{align}
And I am not sure if the term in the parenthesis on the last line should be included in the integral.

答:
I think it depends always on the context. The $dx$ part does not need to be at the end of the integral, for example, in quantum mechanics you see a lot of integrals written like:

$$\langle \psi| \hat V |\psi \rangle = \frac{1}{2} m \omega^2 |B|^2 \int_{-\infty}^{+\infty}dx x^4 e^{-2 \alpha x^2}$$

and in this case, the integrand is $x^4 e^{-2 \alpha x^2}$. I think that they do this in integrals with large integrands so it's more visible and to make stand out what variable you are integrating over. So if $\int x dx x$ is $\int x^2 dx$ or  $x\cdot \int x dx$ always depends on the context of what you are doing and of what you are working with.


问:
How to graph the equation: $y=\frac {x-2}{x+1}$?
the title says it all.
I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?

"Graph the equation $y=\frac {x-2}{x+1}$"

I know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.

Any help would be appreciated.
If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!

答:
Full Function investigation, check of asymptotic (infinite and non setting points), extreme points and inflection points.
nice tool to work with is desmos:
 $y=(x−2)/(x+1)$ - Click to view

In addition : Investigating Functions - Click to view

hope its helped you.


问:
Number of solutions of $x^2+y^2=1 \mod p^2$ with prime $p=3\mod 4$
I am asked to find the number of solutions of $x^2+y^2=1 \mod p^2$, with $p=3 \mod 4$ prime.
I thought a possible way was to apply Hensel's lemma, to shift the $\mod p$ solutions to $\mod p^2$ solutions, but I am not able to conclude my argument. Can someone help me?

答:
Suppose we have $ x^2 + y^2 = -1 \pmod{p}$.

We want to lift it to a solution, namely:

$$ -1 \equiv (x+k_y p)^2 + (y+k_yp)^2 \equiv x^2 + y^2 + 2(k_xx + k_yy)p \pmod{p^2}.$$

This happens when $k_x + k_y = A$ (for you to determine).

Hence, with every initial solution in mod $p$, there are $B$ (for you to determine) solutions in mod $p^2$.


问:
How to prove the concurrence of three lines?

Let quadrilateral $ABCD$ inscribe in circle whose center is $O$ and let $AC$ and $BD$ intersect at $P$,while $Q$ be a point on the straight line $OP$.The circumcenters of $\Delta QAB$, $\Delta QBC$, $\Delta QCD$, $\Delta QDA$ are $O_1$, $O_2$, $O_3$, $O_4$ respectively.Then how to prove the concurrence of the straight lines $O_1 O_3$, $O_2 O_4$, $OP$?($O$ is not coincide with $P$)

I'm sorry that I can't provide any useful ideas.(I have tried to violently calculate through trigonometric functions,but it only made me crazy)

And I'm very sorry for my possible wrong grammars and strange expressions because I'm still an English learner and my first language is very different from the English system.

I'd appreciate it if someone could share his ideas about the question.

答:
Here's an outline of my coordinate proof.

Lemma. For $\overline{RR'}$ a diameter of $\bigcirc O$ with radius $r$, and $\overline{AA}$ a chord through $P$ with $p:=|OP|$, and $\alpha :=\angle RR'A$ and $\alpha':=\angle R'RA'$,
$$\frac{r-p}{r+p}=\frac{(r-p)/|PA|}{(r+p)/|PA|}\quad\underbrace{\overbrace{\;\qquad=\qquad\;}^{\triangle PRA'\sim\triangle PAR'}}_{\triangle PR'A'\sim\triangle PAR}\quad\frac{|RA'|/|R'A|}{|R'A'|/|RA|}=\frac{|RA|/|R'A|}{|R'A'|/|RA'|}=\frac{\tan\alpha}{\tan\alpha'} \tag1$$

With this, and the Inscribed Angle Theorem, we can coordinatize chords $\overline{AA'}$ and $\overline{BB'}$ passing through $P$ as (abusing notation so that $\operatorname{cis}\theta=(\cos\theta,\sin\theta)$)
$$
A  = r \operatorname{cis} 2\alpha \qquad
A' = r \operatorname{cis}(2\alpha'-\pi) \qquad
B  = r \operatorname{cis} 2\beta \qquad
B' = r \operatorname{cis}(2\beta'-\pi)
\tag2$$
such that
$$\frac{\tan\alpha}{\tan\alpha'}=\frac{r-p}{r+p}=\frac{\tan\beta}{\tan\beta'}\tag3$$

With $P=(p,0)$ and $Q=(q,0)$, we find after a bit of symbol-crunching that the circumcenters of $\triangle QAB$ and $\triangle QA'B'$ are
$$\begin{align}
K\phantom{'} &:= \phantom{-}\frac{r^2 - q^2}{2((r - q) + \tan\alpha \tan\beta (r + q))}
\;\left( 1 - \tan\alpha \tan\beta, \tan\alpha + \tan\beta\right) \tag4 \\[8pt]
K' &:= - \frac{r^2 - q^2}{2 ((r + q) + \tan\alpha' \tan\beta' (r - q))}
\;\left( 1 - \tan\alpha' \tan\beta', \tan\alpha' + \tan\beta'\right) \tag5
\end{align}$$
Then, $\overleftrightarrow{KK'}$ meets $\overleftrightarrow{OP}$ (aka, the $x$-axis) at

$$K_\star := \left(\frac{p (r^2-q^2)}{2 (r^2-p q)}, 0\right) \tag6$$

This is independent of $\alpha$, $\beta$, $\alpha'$, and $\beta'$. That is, it's defined solely by $P$ and $Q$ along a diameter of $\bigcirc O$.

Consequently, not only does $K_\star$ also lie on the line through the circumcenters of $\triangle QA'B$ and $\triangle QAB'$ (thus completing the proof), but it lies on the line through the circumcenters of any pair of "opposite" triangles determined by $Q$ and two chords through $P$.

In particular, taking $\overline{AA'}\perp\overline{PQ}$, and $B=A$ and $B'=A'$, we have the case where the circumcircles are internally tangent to $\bigcirc O$. This gives us a construction of $K$ as the intersection of $\overline{OA}$ with the perpendicular bisector of $\overline{QA}$ (likewise for $K'$), and we see that $K_\star$ is the projection of this point onto $\overline{PQ}$:

Be that as it may ...

The independence of $K_\star$ seems like it's trying to tell me something (ideally, something that would have helped me avoid coordinatizing the problem), but I'm not sure what it is. I may need to come back to it with fresh eyes at a later date.


答:
Simplify dividing by $x$
$$x y''(x)-2 y'(x)=0$$

set $y'(x)=p(x)$, The equation becomes
$$x p'(x)-2p(x)=0$$
separate variables
$$\frac{dp}{p(x)}=\frac{2dx}{x}$$
integrate both sides
$$\log p(x)=2\log x+C$$
$$p(x)=k_1x^2\to y'(x)=k_1x^2$$
integrate again to get $y(x)$
$$y(x)=k_1\frac{x^3}{3}+k_2$$

Title: Solving $x^2 y''-2xy'=0$ using power series

Body:
Given the following ODE
$$x^2 y''-2xy'=0$$
I'm trying to solve it using power series. Since
$$y=\sum_{n=0}^{\infty} a_{n}x^{n}$$
then
$$x^{2}y''=\sum_{n=2}a_{n}n(n-1)x^{n}$$
and
$$2xy'=\sum_{n=1}^{\infty}2a_{n}nx^{n}=2a_{1}x+\sum_{n=2}^{\infty}2a_{n}nx^{n}$$
Therefore, in the equation, I get
$$-2a_{1}x+\sum_{n=2}^{\infty}(n^{2}-3n)a_{n}x^{n}=0$$
So
$$-2a_{1}=0\Rightarrow a_{1}=0$$
and
$$(n^{2}-3n)a_{n}=0,\quad\forall n\geq2$$
Since I'm looking for non-trivial solutions, $a_{n} \ne 0\Rightarrow n^{2}-3n=0$ which gives me $n=0$ or $n=3$.

I want to know if it's right and if so, what I'm supposed to do with these values of $n$.


问:
Are these subspaces of $\mathbb{C}^2$?
I want to know whether the following are subspaces of $\mathbb{C}^2$ over $\mathbb{C}$

$$S = \{(2a-1,4a-2)\mid a\in \mathbb{C}\}$$
$$U = \{(2a-1,4a-2)\mid a\in \mathbb{R}\}$$
$$T = \{(2a+3b,4a-c)\mid a,b,c\in \mathbb{C}\}$$
$$V = \{(a-1,a-b)\mid a,b\in \mathbb{C}\}$$

I got that S,T and V are subspaces but U is not due to scalar multiplication by $i$.

Am I correct?

答:
Indeed, $U$ is not a complex vector space, since it is stable with respect to multiplication by $i$. But you should provide a concrete example. For instance, $(1,2)\in U$, but $(i,2i)\notin U$.

All others are complex vector subspaces.

Title: Are these subspaces of $\mathbb{C}^2$?


问:
Show that $\mathbb{E}\left|\hat{f_n}-f \right| \leq \frac{2}{n^{1/3}}$ where $\hat{f_n}$ is a density estimator for $f$

Question

Suppose we have a continuous probability density $f : \mathbb{R} \to [0,\infty)$ such that $\text{sup}_{x \in \mathbb{R}}(\left|f(x)\right| + \left|f'(x)\right|) \leq 1. \;$ Define the density estimator:

$$\hat{f_n} = \frac{1}{n^{2/3}}\sum^n_{i=1}\mathbf{1}\{-1/2 \leq n^{1/3}(x-X_i) \leq 1/2\}, \: x \in \mathbb{R}$$

Show that, for every $x \in \mathbb{R} \:$ and every $n \in \mathbb{N},$
$$\mathbb{E}\left|\hat{f_n}(x)-f(x)\right| \leq \frac{2}{n^{1/3}}$$

Attempt

It's easy to show that $$\mathbb{E}\hat{f_n} = \frac{F(x+\frac{1}{2}n^{-1/3}) - F(x-\frac{1}{2}n^{-1/3})}{n^{-1/3}}$$
where $F$ is the CDF of $f$. From here, using the condition $\:\text{sup}_{x \in \mathbb{R}}(\left|f(x)\right| + \left|f'(x)\right|) \leq 1 \:$ and in particular, $\:\text{sup}_{x \in \mathbb{R}}\left|f'(x)\right|\leq 1\:$ we can show (geometrically) that,
$$\left|\mathbb{E}\hat{f_n}-f\right| \leq \frac{1}{4n^{1/3}}$$

However, I'm not sure how to get a bound on $\mathbb{E}\left|\hat{f_n}-f\right|$.

答:
General

The density estimator is a special case of a kernel density estimator (KDE) with rectangular kernel $$K(x)=\mathbf{1}\{-1/2\leq x\leq 1/2\}$$ and bandwidth $h=n^{-1/3}$. It would therefore be possible to use the general theory of KDEs and deduce the result.

I will not follow that approach but instead derive properties of this specific estimator directly and link the results to the general KDE theory if possible.

Distribution of $\hat{f_n}$

As (apart from a multiplicative constant) the estimator is effectively counting the number of times an observation of $X_i$ is falling into the interval $\left[x-\frac{1}{2}n^{-1/3}, x+\frac{1}{2}n^{-1/3}\right]$ when calculating $\hat{f_n}(x)$ we can immediately see that $$n^{2/3} \hat{f_n}\sim\operatorname{Binomial}(n,p)$$ with $p$ equal to $$F\left(x+\frac{1}{2}n^{-1/3}\right)-F\left(x-\frac{1}{2}n^{-1/3}\right).$$ This secretly assumes that all the $X_i$ are independent and identically distributed with CDF $F$. From this we can infer directly that $\mathbb{E}\hat{f_n}(x)=n^{1/3}p$ and $\operatorname{Var}(\hat{f_n}(x))=n^{-1/3}p(1-p)$.

Asymptotics

To get the asymptotic behaviour of the expectation and variance we have to analyse $p$ from above. Using a Taylor approximation at $x$ we see that $$F\left(x\pm\frac{1}{2}n^{-1/3}\right)=F(x)\pm f(x)\frac{1}{2}n^{-1/3}+f'(\xi_\pm)\frac{1}{8}n^{-2/3}$$ for some $\xi_\pm$ between $x$ and $x\pm\frac{1}{2}n^{-1/3}$. Knowing that $|f'|\leq 1$ we get $$\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\leq\frac{1}{4}n^{-1/3}\tag1.$$

Remark
Assuming that also $f''$ exists and is continuous we could get a stronger result. In this case the bias of $\hat{f_n}$ would tend to zero with the order $n^{-2/3}$. This is in line with the more general result that under mild assumptions on the kernel $h$ we have that $$\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|=\mathcal{O}(h^2)$$ for $n$ towards infinity.

For the variance we also use a shortened version of the Taylor approximation $$F\left(x\pm\frac{1}{2}n^{-1/3}\right)=F(x)\pm f(\zeta_\pm)\frac{1}{2}n^{-1/3}$$ for some $\zeta_\pm$ between $x$ and $x\pm\frac{1}{2}n^{-1/3}$. The variance can now be written as $n^{-1/3}p(1-p)=n^{-1/3}(p-p^2)$ with $$p-p^2=f(x)n^{-1/3}+(f'(\xi_+)-f'(\xi_-))\frac{1}{8}n^{-2/3} - (f(\zeta_+)+f(\zeta_-))^2\frac{1}{4}n^{-2/3}.$$ Again, using the given upper bounds for $f$ and $f'$ we can bound the variance from above by $$\operatorname{Var}(\hat{f_n}(x))\leq n^{-1/3}\left(n^{-1/3}+\frac{1}{4}n^{-2/3}\right)\leq\frac{5}{4}n^{-2/3}.\tag2$$

Remark
This result is similar to a result from the general theory of KDEs, which gives $$\operatorname{Var}(\hat{f_n}(x)) =\mathcal{O}((nh)^{-1})$$ for $n$ towards infinity.

Combining the results

To get the desired result we use a decomposition of the mean absolute error (MAE) similar to that of the mean squared error. We start with \begin{align}\mathbb{E}\left|\hat{f_n}(x)-f(x)\right|&\leq\mathbb{E}\left|\hat{f_n}(x)-\mathbb{E}\hat{f_n}(x)\right|+\mathbb{E}\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\\&\leq\left(\mathbb{E}\left|\hat{f_n}(x)-\mathbb{E}\hat{f_n}(x)\right|^2\right)^{1/2}+\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\\&=\operatorname{Var}\left(\hat{f_n}(x)\right)^{1/2}+\left|\mathbb{E}\hat{f_n}(x)-f(x)\right|\end{align} using the triangle and Hölder inequalities. Using bounds $(1)$ and $(2)$ we finally get $$\mathbb{E}\left|\hat{f_n}(x)-f(x)\right|\leq\left(\frac{\sqrt{5}}{2}+\frac{1}{4}\right)n^{-1/3}\leq 2n^{-1/3}$$ and have the result.


问:
Show that $\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$ converges for $a > 0$ in $[-a,a]$
Given:
$$\sum_{n=1}^{\infty} {(-1)^n \sin(\frac{x}{n})}$$

We need to show that it converges in $[-a,a]$ $\forall a >0$.

Now what I figured out:

This is a function series summation. we need first to know what $C_n$ actually is and then do the Radius test.

But how do I exactly get it in the general form ?

答:
What you should realise is that $\sum\limits_{n=1}^{\infty}(-1)^{n}\sin\left( \frac{x}{n} \right)$ converges iff $\sum\limits_{n=N+1}^{\infty}(-1)^{n}\sin\left( \frac{x}{n} \right)$ converges
where $N\in\mathbb{N}$ is such that $N \leq \vert a \vert$ and $N+1 > \vert a \vert$ as changing a finite amount of summands has no effect on the convergence. Now you should carefully argue that you may apply the Leibniz criterion. </p>


问:
Notation for Tensor Indexing
Given an $8\times 8\times 8 \times 8$ tensor $T$, what is the best terse notation for referring to the $8\times 8$ matrix formed by $T(2,3,i,j)$ for all $-1 < i < 8$ and $-1 < j < 8$. For example:

$$ T(2,3),\ T(2,3,0-7,0-7),\ T(2,3,*, *) $$

Is there a standard notation?

答:
I would use just regular matrix notation:

$$M \in M_{8\times8} : M_{ij} = T(2,3,i,j)$$

Or if you want you can also say:

$$M= \{T(2,3,i,j)\}_{i,j \in \{1,...,8\}}$$

I think that these two are the easiest ways to define the matrix.

Title: Notation for Tensor Indexing


问:
Fourier series: if $f$ is $C^\infty$ then $f(x) = s_n(x) + o(1/n^k) $ as $n\to \infty$.
Suppose $f$ is differentiable any number of times and is $2\pi$-periodical. Let $k\in \mathbb{N}$. Show that $$\lim_{n\to\infty}n^k|f(x)-s_n(x)|= 0$$
i.e that $f(x)=s_n(x)+o(1/n^k)$ as $n\to \infty$. The $s_n$ are the $n$th Fourier partial sums.

Intuitively I would like to show this using an argument along the following lines:
\begin{align*}
s_n(x) - f(x) &= \int_{-\pi}^{\pi}[f(x+y)-f(x)]D_n(y)dy\\
&= \int_{\delta\leq |y| \leq \pi}[f(x+y)-f(x)]D_n(y)dy+\int_{0\leq |y|\leq \delta}[f(x+y)-f(x)]D_n(y)dy\\
&\approx 0 +  \int_{0\leq |y|\leq \delta}o\Big(\dfrac{1}{n^k} \Big)D_n(y)dy
\end{align*}
as $n\to\infty$, for some $\delta>0$ small enough (since $f$ is differentiable). However, this seems hard to make rigorous, and having both sides depend on $n$ makes me hesitant to use this little-o notation. Some hints/solutions would be appreciated.

Body: $f=\sum_n c_n(f) e^{inx}$ and $(in)^{k+2} c_n(f)= c_n(f^{(k+2)})$ is bounded by $\|f^{(k+2)}\|_{L^1[0,2\pi]}$ thus $$|\sum_{|n|> N} c_n(f)e^{inx}|\le N^{-k-1}\|f^{(k+2)}\|_{L^1[0,2\pi]}$$


问:
Tesselation beeing used in a game board
I like to know what is the math behind the outlines of these boards

https://www.boardgamegeek.com/image/4383021/spirit-island

https://www.boardgamegeek.com/image/4880004/spirit-island

https://www.boardgamegeek.com/image/2706606/spirit-island

https://www.boardgamegeek.com/image/3692997/spirit-island

https://www.boardgamegeek.com/image/4013584/spirit-island

I wonder how to come up with these exact shape and how to express this in a formula.
The maker of the Boards used the term tessellation and mentioned that it was initial based on a hex map.
I like how they fit nicely and still give the impression of uniqueness.
Is there anything else interesting (from an mathematicians point of view) about these?

答:
The underlying shape for each board piece is a rhombus with angles of $60°$ and $120°$.

$2$ of these form a parallelogram, $3$ can be put together to form a regular hexagon, and $4$ can form a larger rhombus.

$3$ of the edges of each rhombus are replaced with identical rotationally symmetric curves. The rotational symmetry ensures these edges fit together in all layouts.

The fourth edge of the rhombus is replaced with a different curve, but always forms the water boundary so will only appear on the edge of the layout and doesn't need to match the other edges.

Title: Tesselation beeing used in a game board


问:
How come $\sum_{k=0}^\infty \frac{2}{k}$ and $\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2}$ equal the same thing?
$$\sum_{k=0}^\infty \frac{2}{k} \tag1$$

and

$$\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2} \tag2$$

Are equal to the same thing:

$$\begin{align}
\text{UNDEF} &+ 2 + 1 + 0.\overline{66} + 0.5 + 0.4 + 0.\overline{33} + 0.\overline{285714} + 0.25 \\ &+ 0.\overline{22} + 0.2 + 0.\overline{18} +  0.1\overline{66} +0.\overline{153846}+ 0.\overline{142857}+ \ldots
\end{align} \tag3$$

This is quite interesting. The sigma expression at the bottom is basically taking a value, and seeing how big it is in relation to the "additive factorial sum" of it the number below it. Now, additive factorial sum is not a proper term I think, but I used it to describe the sum you get when you put a number into a binomial coefficient like $\frac{n^2 + 2}{2}$. So if n is 3, the answer is 6, because $1+2+3 =6$. Now, $3+1 =4$ and I wondered, how does 4 stack up against 6, or how does n+1 stack up against $\frac{n^2 + 2}{2}$?

The second sigma expression answers this, and it happens to be the same as the fist one, weirdly. How come?

答:
$$(k+1)\div\frac{k^2+k}{2}=(k+1)\cdot \frac{2}{k^2+k}=\frac{2(k+1)}{k(k+1)}=\frac 2k$$

In other words, the summands are both the same, although both sums are divergent, so you can't say that the sums are equal.

Also, it should be summed from $1$ to $\infty$, as $\frac 2k$ and $(k+1)\div\frac{k^2+k}{2}$ are undefined at $k=0$. When doing infinite sums, it is okay for the overall sum to be $\infty$, but the summand should always be defined on the range given.


问:
A corollary of commutation theorem for tensor product of von Neumann algebras.
Please help me to understand the proof of the following corollary from the book 'Lectures on von Neumann algebras' by Stratila and Zsido.

Corollary: Let $\mathscr{M}_1$ and $\mathscr{M}_2$ be von Neumann algebras, $\mathscr{Z}_1,\,\mathscr{Z}_2$ their centers, respectively. Then the center of $\mathscr{M}_1\overline{\otimes}\mathscr{M}_2$ is $\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2$.

To prove the corollary, they have used the commutation theorem for tensor products (i.e. for two von Neumann algebras $\mathscr{M}_1$ and $\mathscr{M}_2$, we have $(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)'=\mathscr{M}_1'\overline{\otimes}\mathscr{M}_2'$). Here is my attempt to understand the proof.

Let $\mathscr{Z}$ be the center of $\mathscr{M}_1\overline{\otimes}\mathscr{M}_2$. Then $$\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2=(\mathscr{M}_1\cap\mathscr{M}_1')\overline{\otimes}(\mathscr{M}_2\cap\mathscr{M}_2')\subseteq (\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cap(\mathscr{M}_1'\overline{\otimes}\mathscr{M}_2')=(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cap(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)'=\mathscr{Z}.$$
Now conversely, we have to prove $\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2\supseteq\mathscr{Z}$, i.e. $(\mathscr{Z}_1\overline{\otimes}\mathscr{Z}_2)'\subseteq\mathscr{Z}'$, i.e. $\mathscr{Z}_1'\overline{\otimes}\mathscr{Z}_2'\subseteq\mathscr{Z}'$, i.e. $\mathfrak{A}(\mathscr{M}_1\cup\mathscr{M}_1')\overline{\otimes}\mathfrak{A}(\mathscr{M}_2\cup\mathscr{M}_2')\subseteq\mathfrak{A}((\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)\cup(\mathscr{M}_1\overline{\otimes}\mathscr{M}_2)')$, where $\mathfrak{A}(A)$ stands for the von Neumann algebra generated by the set $A$. But I   did not get the way the authors prove it. For the proof given in the book see here.

Thanks in advance for any help.

答:
In your attempt for the easy inclusion, I'm not comfortable enough with tensor products to assess your exchange of tensor and intersection. But that's not needed. Elements of $Z_1\otimes Z_2$ commute with every elementary tensor in $M_1\otimes M_2$, so commute with all elements of the algebraic tensor product, and then commute with limits of these. This shows that
$$
Z_1\otimes Z_2\subset Z.
$$

For the reverse inclusion, what the authors do is to get, from the definition of centre, that
$$\tag1
M_1\otimes M_2\subset Z'.
$$
Then, since every element of $M_1'\otimes M_2'$ commutes with every elements of $M_1\otimes M_2$,
$$\tag2
M_1'\otimes M_2'\subset (M_1\otimes M_2)'\subset Z',
$$
the last inclusion being simply $Z\subset M_1\otimes M_2$. Then,
$$\tag3
Z_1'\otimes Z_2'=(M_1\cap M_1')'\otimes (M_2\cap M_2')'
=(M_1'\cup M_1)''\otimes (M_2'\cup M_2)''\subset Z'.
$$
The last inclusion follows from $(1)$ and $(2)$, because if you have $a\otimes b$ with $a\in M_1'$, $b\in M_2$, then
$$
a\otimes b=(a\otimes 1)(1\otimes b)\in Z'
$$
since $a\otimes 1\in Z'$ by $(2)$ and $1\otimes b\in Z'$ by $(1)$. The other cases are treated similarly, to get $(M_1'\cup M_1)\otimes (M_2'\cup M_2)\subset Z'$. Taking sums and limits in the first coordinate we get $(M_1'\cup M_1)''\otimes (M_2'\cup M_2)\subset Z'$. And then taking sums and limits in the second coordinate, $(3)$ follows.

Finally, from $(3)$ one gets
$$
Z\subset (Z_1'\otimes Z_2')'=Z_1''\otimes Z_2''=Z_1\otimes Z_2.
$$


问:
Show that $E[ E[Z\lvert \mathcal{F}_{T}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]$
Let $T$ and $S$ be stopping times and $Z$ some integrable random variable. I already know that:

$\mathcal{F}_{T\land S}=\mathcal{F}_{T}\cap\mathcal{F}_{S}$ and $\{T>S\},\;\{T\leq S\},\; \{T<S\},\;\{T\geq S\} \in \mathcal{F}_{T\land S}$. And furthermore I have shown that:

$E[Z\lvert \mathcal{F}_{T}]=E[Z\lvert \mathcal{F}_{S\land T}]\; \; \mathbb P$-a.s. on $\{T\leq S\} \; (*)$

I want to show that:

$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]\; \; \mathbb P$-a.s.

I guess we have to use $(*)$ but I get stuck:

$E[ E[Z\lvert \mathcal{F}_{T}]\lvert \mathcal{F}_{S}]=E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T\leq S\}}\lvert \mathcal{F}_{S}]+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]= E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T\leq S\}}+E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]$

How can I go about showing that the second term

$$ E[ E[Z\lvert \mathcal{F}_{T}]1_{\{T> S\}}\lvert \mathcal{F}_{S}]=E[Z\lvert \mathcal{F}_{T\land S}]1_{\{T>S\}}$$ or am I going about this the wrong way?

Reference to Problem 2.17 from Karatzas and Shreve p. 9 Brownian Motion and Stochastic Calculus

答:
What seems to be missing here is the following fact:

If $Y$ is $\mathcal{F}_{S}$-measurable, then $Y 1_{\{S < T\}\}}$ is
$\mathcal{F}_{S \wedge T}$-measurable.

(You already seem to have noticed that $Y 1_{\{S \leq T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable.)

Now notice that if $A \in \mathcal{F}_{S \wedge T}$, then $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ and, thus,
\begin{align*}
E[E[Z\mid\mathcal{F}_{S}] 1_{\{S < T\}} : A] &= E[Z : A \cap \{S < T\}] \\
&= E[E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}} : A].
\end{align*}
Thus, since $E[Z \mid \mathcal{F}_{S}]1_{\{S < T\}}$ is $\mathcal{F}_{S \wedge T}$-measurable by the observation above, the arbitrariness of $A$ implies that
\begin{equation*}
E[Z \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S < T\}}.
\end{equation*}

Finally, notice that if $A \in \mathcal{F}_{S}$, then our observation gives $A \cap \{S < T\} \in \mathcal{F}_{S \wedge T}$ so we can write
\begin{equation*}
E[E[Z\mid\mathcal{F}_{T}] : A \cap \{S  < T\}] = E[Z: A \cap \{S < T\}] = E[Z 1_{\{S < T\}} : A].
\end{equation*}
By the arbitrariness of $A$ and the inclusion $\{S < T\} \in \mathcal{F}_{S}$, this implies
\begin{equation*}
E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[E[Z \mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[Z \mid \mathcal{F}_{S}] 1_{\{S  <T\}}.
\end{equation*}

Combining the two identities obtained above, we find the equation we needed:
\begin{equation*}
E[E[Z\mid \mathcal{F}_{T}] 1_{\{S < T\}} \mid \mathcal{F}_{S}] = E[E[Z\mid \mathcal{F}_{T}] \mid \mathcal{F}_{S}] 1_{\{S < T\}} = E[Z \mid \mathcal{F}_{S \wedge T}] 1_{\{S  <T\}}.
\end{equation*}


问:
Question on compact subsets of finite dimensional normed spaces
I'm trying to prove this fact:

If $X$ is a finite dimensional normed space and $K$ a compact subset of $X$ that satisfies that if $x, y \in K$ then $\frac{x-y}{2} \in K$ then, given $a \in X \setminus K$ there exists an $\varepsilon >1$ such that $a \not \in\varepsilon K$

I know that the distance from $a$ to $K$ is greater than zero and is attained by some point $x \in K$, so I'm trying to prove that is enough to take $1 < \varepsilon < d(a,K)+1$ but I'm not able to reach a contradiction if I suppose that $a \in \varepsilon K$.

Any ideas?

答: This is true for all compact $K$ and, in fact, for $a\notin K$ the set $\{\varepsilon\in \Bbb R\,:\, a\notin \varepsilon K\}$ contains a neighbourhood of $1$ regardless of any other condition on $K$. For, if there were a sequence $\varepsilon_n\to 1$ such that $a\in\varepsilon_n K$ for all $n$, then there would be a sequence $x_n\in K$ such that $a=\varepsilon_nx_n$ for all $n$. By considering a convergent subsequence $x_{n_k}\to x\in K$, we obtain $a=\lim_{k\to\infty}\varepsilon_{n_k}x_{n_k}=x$.


问:
Show relation for integrals
Let $f \in C^{1}([a,b];\mathbb{R})$ and $|f'(x)-f'(y)| \le L |x-y|$
then we have $|\int_a^b f(x) dx -f(\frac{a+b}{2})(b-a)| \le L\frac{(b-a)^3}{4}$.

I have troubles to show this inequality. the problem is that i need to have a difference of derivatives of the function in order to use the Lipschitz condition.

答:
Denote $c=\frac{a+b}{2}$. By mean value theorem, for every $x\in[a,c]$(resp. $[c,b]$), there exists $t\in[x,c]$(resp. $[c,x]$), such that
$$f(x)-f(c)=f'(t)(x-c).$$
It follows that
$$|f(x)-f(c)-f'(c)(x-c)|=|(f'(t)-f'(c))(x-c)|\le L|(t-c)(x-c)|\le L(x-c)^2.$$
Integrating the inequality above over $[a,b]$, it follows that
$$\big|\int_a^bf(x)dx-f(\frac{a+b}{2})(b-a)\big|\le\frac{L(b-a)^3}{12}.$$</p>

Title: Show relation for integrals


答:
First sum is divergent (it's harmonic series), so you can't say about eqaulity in this case, only for finite sums $\sum_{k=1}^n \frac{2}{k}$ and $\sum_{k=1}^n(k+1) \div \frac{k^2 + k}{2}$.
Also, I suppose there was a mistake and $k$ starts from 1.

Title: How come $\sum_{k=0}^\infty \frac{2}{k}$ and $\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2}$ equal the same thing?

Body:
$$\sum_{k=0}^\infty \frac{2}{k} \tag1$$

and

$$\sum_{k=0}^\infty (k+1) \div \frac{k^2 + k}{2} \tag2$$

Are equal to the same thing:

$$\begin{align}
\text{UNDEF} &+ 2 + 1 + 0.\overline{66} + 0.5 + 0.4 + 0.\overline{33} + 0.\overline{285714} + 0.25 \\ &+ 0.\overline{22} + 0.2 + 0.\overline{18} +  0.1\overline{66} +0.\overline{153846}+ 0.\overline{142857}+ \ldots
\end{align} \tag3$$

This is quite interesting. The sigma expression at the bottom is basically taking a value, and seeing how big it is in relation to the "additive factorial sum" of it the number below it. Now, additive factorial sum is not a proper term I think, but I used it to describe the sum you get when you put a number into a binomial coefficient like $\frac{n^2 + 2}{2}$. So if n is 3, the answer is 6, because $1+2+3 =6$. Now, $3+1 =4$ and I wondered, how does 4 stack up against 6, or how does n+1 stack up against $\frac{n^2 + 2}{2}$?

The second sigma expression answers this, and it happens to be the same as the fist one, weirdly. How come?


答:
In this case you can manipulate $ (k+1) \div \frac{k^2 + k}{2}$ and get $2/k$ as Rhys Hughes did in his answer, but note that, even if that is not possible, two sequences don't need to be equal in order to have the same limit!

For example. $1/n$ and $\sin(n)/e^n$ are diferent for all $n$ but note that:

$$\lim \frac{1}{n} = \lim\frac {\sin(n)}{e^n} = 0$$


问:
Sheafification of representable presheaves
Recall that if $\mathscr{C}$ is a site (which, for me, means a category equipped with a Grothendieck pretopology—not sieves) that it’s not in general true that the representable presheaf $h_X$ associated to an object $X$ of $\mathscr{C}$ is a sheaf.

In particular, one can consider the composition of functors

$$\mathscr{C}\to \mathsf{PSh}(\mathscr{C})\xrightarrow{(-)^\#}\mathsf{Sh}(\mathscr{C})$$

(where $(-)^\#$ denotes sheafification) and think of the full image—the full subcategory of $\mathsf{Sh}(\mathscr{C})$ whose objects are sheaves isomorphic to $h_X^\#$ for an object $X$ of $\mathscr{C}$. Let me denote this category with the (probably poorly chosen) notation $\mathscr{C}^\#$.

My question is then the following

Question: What is an explicit description of $\mathscr{C}^\sharp$?

To give an idea of what type of description I’m looking for, I think that one can write that there is a bijection

$$\mathrm{Hom}(h_X^\#,h_Y^\#)=\varinjlim_{\substack{\{U_i\to X\}\\ \mathscr{C}\text{-coverings}}}\check{H}^0_\mathrm{loc}(\{U_i\to X\}, h_Y)$$

where

$$\check{H}^0_\mathrm{loc}(\{U_i\to X\}, h_Y):=\left\{(f_i)\in\prod_i \mathrm{Hom}_{\mathscr{C}}(U_i,Y): f_i\text{ and }f_j\text{ agree locally on overlaps for all }i,j\right\}$$

where $f_i$ and $f_j$ agreeing locally on overlaps means that there exists a cover $\{V_k\to U_i\times_X U_j\}$ such that $(f_i)\mid_{V_k}=(f_j)\mid_{V_k}$.

This is literally, I believe, just because one has natural bijections

$$\mathrm{Hom}(h_X^\#,h_Y^\#)=\mathrm{Hom}(h_X,h_Y^\sharp)=h_Y^\#(X)$$

the first from the definition of sheafification, and the second from Yoneda’s lemma. The claimed equality then follows from the explicit description of sheafification.

Refined question: With the above description of $\mathrm{Hom}(h_X^\#,h_Y^\#)$ (or a slightly modified, but still
explicit version)

What is the actual map $h_X^\sharp(Z)\to h_Y^\#(Z)$ for $Z$ an object of $\mathscr{C}$?

What is the composition  $$\mathrm{Hom}(h_X^\#,h_Y^\#)\times \mathrm{Hom}(h_Y^\#,h_Z^\#)\to \mathrm{Hom}(h_X^\#,h_Z^\#)$$ look
like?

NB: I’m mostly looking for a reference. Obviously you can answer this question by tracing through all the definitions, but I’d like to save myself the time/writing and just cite something. I’d be a little surprised if a reference didn’t exist.

Thanks!

答:
I don't have the reference you want, but here's the concrete description of composition you ask for.

An element of $h_Y^\sharp(X)$ is witnessed by some cover $\{U_i\to X\}$ and maps $U_i\to Y$ agreeing on overlaps. An element of $h_Z^\sharp(Y)$ is witnessed by a cover $\{V_j\to Y\}$ and maps $V_j\to Z$ agreeing on overlaps. The composition in $h_Z^\sharp(X)$ is then witnessed by the cover $\{U_i\times_Y V_j \to U_i \to X\}$ and the maps $U_i\times_Y V_j \to V_j \to Z$.


问:
Eigen vector corresponding to eigen value
I want to find the eigen vector of the matrix $(0,1;-2,-3)$
When I find and put the the eigen value to find the eigen vector ; I find  $y=-2x$

When I give $x=1$, $y = -2$, but if I give $x=-1$ , I find $y=2$
Which one is eigent vector $v(1,-2)$ or $v(-1,2)$

I hope I could explain what i want to ask.

答: Both of them are eigenvectors, because multiplying eigenvector with constant leaves it eigen: $A v = \lambda v$, and for $w = \mu \cdot v$ and non-zero constant $\mu$ we have $Aw = A(\mu \cdot v) = \mu \cdot Av = \mu \cdot \lambda v = \lambda \cdot(\mu \cdot v) = \lambda w$, so $w$ is also an eigenvector with eigenvalue $\lambda$.


问:
help calculating $\sum_{k=0}^n k!{n \choose k} ^2= \text{?}$
I am trying to find the number of relations on $[n] = \{1,\ldots,n\}$ s.t. for all $x,y,z\in R.xRz \wedge yRz\to x=y$. One idea I had was choosing $k$ elements from $n$ then choosing another $k$ elements from $n,$ there are $k!$ bijections between these two sets so we just have to sum over $k$ and here I got stuck, tried finding another approach but came up with nothing.

答:
Your approach is missing relations as commented above, you are considering just injective partial functions. To complete your argument You can go as follows: You pick a set of $[n]$ for the domain(say of size $k$) of the relation and you choose a different set of the codomain (say one of size $\ell$). Then you partition the codomain(into $k$ non empty blocks) and assign each block to an element in the domain (so you permute the blocks in $k!$ ways).

 I get
$$\sum _{k,\ell}\binom{n}{\ell}\binom{n}{k}{\ell \brace k}k!=\sum _{\ell= 0}^n\binom{n}{\ell}\sum _{k=0}^n\binom{n}{k}{\ell \brace k}k!=\sum _{\ell= 0}^n\binom{n}{\ell}n^{\ell}=(n+1)^n.$$
Which makes sense, your relation is the functional relation for the right coordinate and you are just adding a dummy image to see when is the element not in the domain.


问:
Illustrative Example of Riemannian Manifold
Hi I have been trying to understand how to see the gradient of a function $f:M\to \mathbb{R}$, where $M$ is a Riemannian manifold. Obviously to do this I need to have a firm understanding of then tangent space, etc.

I have done some reading, but haven't come across the 'quintessential example' of a Riemannian manifold. The examples here https://en.wikipedia.org/wiki/Riemannian_manifold#Examples I don't find very illustrative. I was expecting to find a sphere or some other shape as a subset of $\mathbb{R}^n$.

What examples of Riemannian manifolds and their tangent spaces do you find most enlightening?

答:
I'll write a short list, just for concreteness. Dealing with tangent spaces does not require a metric at all.

if $A$ is an affine space with translation vector space $V$, then $T_pA = V$ for all $p \in A$. An inner product in $V$ defines a Riemannian metric on $A$. This is a mild generalization of what people do in vector spaces, as we "forget the origin".

$\Bbb S^n =\{ x \in \Bbb R^{n+1} \mid \|x\|=1$, and we have $T_p(\Bbb S^n)= x^\perp$ for all $p \in \Bbb S^n$. This inherits a Riemannian metric from $\Bbb R^{n+1}$. I don't think I need to explain why spheres are relevant.

if $V$ is a vector space, then the collection ${\rm Gr}_k(V)$ of all $k$-dimensional vector subspaces of $V$ is a manifold and $T_W{\rm Gr}_k(V) \cong {\rm Hom}(W,V/W)$. For $k=1$, this is the projective space over $V$. It not as straightforward, but this also inherits a Riemannian metric from any inner product in $V$ (keyword: Fubini-Study metric). These are a nice source of examples, especially in the complex case (where we obtain compact homogeneous Kähler-Einstein manifolds).

if $V$ is a vector space with an indefinite scalar product, and $\mathcal{C} =\{x \in V \setminus \{0\}\mid \langle x,x\rangle=0$ is the lightcone of $V$, then $T_x\mathcal{C} = x^\perp$. If we define $\sim$ on $\mathcal{C}$ by saying that $x\sim y$ if $y=\lambda x$ with $\lambda \neq 0$, then the quotient $\mathbb{E}=\mathcal{C}/_{\sim}$ is also a manifold (it's the collection of all lightrays in $V$) and $T_L\mathbb{E} = x^\perp/\Bbb R x$, where $x$ is any vector with $L = \Bbb R x$. These do not inherit Riemannian or pseudo-Riemannian metrics from $V$. The scalar product restricted to $\mathcal{C}$ is degenerate, and what survives in the quotient $\mathbb{E}$ is just a conformal structure, which turns out to be conformally flat (to wit, $\mathbb{E}$ is diffeomorphic to a product of an anti-Euclidean sphere -- with negative-definite metric --- and an Euclidean sphere, whose dimensions are the positive and negative indices of the indefinite scalar product in $V$, both subtracted by $1$). This $\mathbb{E}$ is called "the Einstein manifold" (and is unrelated to the notion of Einstein manifolds, where the Ricci tensor is a multiple of the metric).

If $G$ is a Lie group, then $T_gG = {\rm d}(L_g)_e[\mathfrak{g}]$, where $\mathfrak{g}$ is the Lie algebra if $G$. In particular, you get things such as $$T_S{\rm SO}(n) = \{SA \in \Bbb R^{n\times n} \mid A \mbox{ is skew-symmetric}\},$$etc. Any inner product in $\mathfrak{g}$ gives rise to a left-invariant Riemannian metric on $G$, via left translations. In particular, metrics in ${\rm SO}(3)$ of the form $$\langle\!\langle V,W\rangle\!\rangle = \int_{\Bbb R^3} \langle V\xi,W\xi\rangle\,{\rm d}\mathfrak{m}(\xi),$$where $\mathfrak{m}$ is a finite measure not supported in any line through the origin, are used in the study of rigid bodies, in classical mechanics.


问:
the domain for $\dfrac{1}{x}\leq\dfrac{1}{2}$
What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$

according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$

however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?

答:
No, you are not missing anything, that rule only works for $A; B > 0$, so $\dfrac{1}{x} \le \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l} x < 0 \\ x \ge 2 \end{array} \right.$

Of course it's easy to see that for $x < 0$, we'll always have $\dfrac{1}{x} < 0 < \dfrac{1}{2}$.


问:
$f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$
Suppose that $f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$

$(a)$. Prove that $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b)$. Prove that $f$ is differentiable  on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ Prove that $f$ is differentiable at $1$, then $f^{\prime}(x)=\dfrac{f^{\prime}(1)}{x}$ for all $x\in (0,\infty)$.

$\underline {Attempt}$

$(a).$  If $f$ is continuous on $(0,\infty)$  then $f$ is also continuous at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is continuous at $1,$ We have  $$ \forall \space \varepsilon \space \exists \space \delta \space\text{such that} \space|x-1|<\delta \space \text{whenever} \space|f(x)-f(1)|=|f(x)|< \varepsilon $$

$\space$ $ \space $$\space$ $\space$ Now let $a \in (0,\infty)$ and if $|x-a|<\delta_1 $,

$$|f(x)-f(a)|=\left|f\left(\dfrac{x}{a}\right)\right| <\varepsilon$$

$\therefore$  $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b).$ If $f$ is differentiable $(0,\infty)$ then $f$ is also differentiable at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is differentiable  at $1$, We have
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$

$\space$ $ \space $$\space$ $\space$ set $h=\dfrac{t}{x}$ implies when $h \to 0$, $t \to 0$ so $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=\lim_{t\to 0 } \frac{f\left(1+\dfrac{t}{x}\right)}{\dfrac{t}{x}}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t} $$

$\therefore$  $f$ is differentiable  on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ If $f$ is differentiable at $1$,We have
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$
$\space$ $ \space $$\space$ $\space$ this implies,
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t}=xf^{\prime}(x) $$

I referred Functional equation $f(xy)=f(x)+f(y)$ and differentiability and Show a function for which $f(x + y) = f(x) + f(y) $ is continuous at zero if and only if it is continuous on $\mathbb R$ but I don't know exactly my attempt is correct or not if not give some advises.

Thank you!

答:
a) I would have used the characterisation of continuity with sequences.

If $f$ is continuous in $1$, then for $x\geqslant 0$, if $x_n\rightarrow x$, then $f(x_n) = f(x)+f(\frac{x_n}{x})\rightarrow f(x) + f(1) = f(x)$. So $f$ is continuous in $x$.

b) If $f$ is differentiable on $1$, then for $x_0\geqslant 0$ and $x\neq x_0$, $\frac{f(x)-f(x_0)}{x-x_0}=\frac{1}{x_0}\cdot \frac{f(\frac{x}{x_0})-f(1)}{\frac{x}{x_0}-1}\underset{x\rightarrow x_0}{\longrightarrow}\frac{1}{x_0}f'(1)$ so $f$ is differentiable on $x$.

c) If you derive the relation with $y$ as a variable, then you get $$-f'(y) = -\frac{x}{y^2} f'(\frac{x}{y})$$ so by replacing $y$ with $1$, you get $-xf'(x)=-f'(1)$, so $$f'(x) = \frac{f'(1)}{x}$$


问:
Implicit function theorem when the denominator is zero and L'Hospital rule.
Suppose we have a strictly increasing and continuous function $f(x)$ such that for some $\bar{x}: \exists\ f'(\bar{x}) = 0$ while the derivative exists and is positive in a small neighborhood of $\bar{x}$. Function $y(x)$ is given implicitly by $$af(x)+bf(y)=(a+b)f(\bar{x})$$ where $a, b$ are some positive constants. I am interested in $\frac{dy(x)}{dx} \bigg|_{x = \bar{x}}$.

From the Implicit function theorem we know that $$\frac{dy(x)}{dx} = -\frac{af'(x)}{bf'(y)}$$ as long as $f'(y) \neq 0$ which is not true at the point $x = \bar{x}$, since $y(\bar{x}) = \bar{x}$ and $f'(\bar{x})  =0$.

My question is: Suppose that the derivative above actually exists, moreover there exist finite, non-zero left and right second derivatives $f''_{+}(\bar{x})$ and $f''_{-}(\bar{x})$ which have different signs.

Would it be true to write: $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af'(x)}{bf'(y(x))}\ ?$$
And then to use L'Hospital rule to get $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af''_{+}(x)}{bf''_{-}(y(x)) y'(x)}$$ Concluding that $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = -\sqrt{\frac{a}{b}\bigg|\frac{f''_{+}(\bar{x})}{f''_{-}(\bar{x})}\bigg|}$$ where the minus sign follows from the fact that clearly $y(x)$ is decreasing since $f(x)$ is increasing and $a, b > 0$

If the conclusion above is incorrect which additional assumptions I need for it to work?

Would be very grateful for any help. Thanks in advance!

答:
This is not a case for using l'Hospital.  Instead use Taylor expansion on the two sides. Set $\lambda_+=f''_+(\bar{x})>0$
and $\lambda_-= f''_-(\bar{x})<0$.

For
$h>0$: $f(\bar{x}+h) -f(\bar{x})= \frac12 \lambda_+ h^2+o(h^2) = \frac12(\lambda_+ + \epsilon(h)) h^2$.

For $k<0$:
$f(\bar{x}+k) - f(\bar{x})= \frac12 (\lambda_-+\epsilon(k))k^2$.

Suppose now that $x=\bar{x}+h$ with $h>0$ (small) which implies that $y=y(x)=\bar{x}+k$ with $k=k(h)<0$. Then
$$ a (\lambda_+ +\epsilon(h)) h^2 + b (\lambda_- + \epsilon(k)) k^2 = 0,$$
which implies $y'_+(\bar{x}) = \lim_{h \rightarrow 0^+} k(h)/h = -\sqrt{-\frac{a\lambda_+}{b\lambda_-}}$
similarly  $y'_-(\bar{x}) = \lim_{h \rightarrow 0^-} k(h)/h = -\sqrt{-\frac{a\lambda_-}{b\lambda_+}}$. Note that the derivatives from the left and from the right may be different.


问:
Differentiating an integral general result
I saw a result a while ago which stated if $f(Y) = \int^{Y} g(t)dt$, then $f'(Y) = g(Y)$, where we differentiate with respect to Y. I'm unable to find any sources on this result (in particular, what it is called). I believe it is used in this question

What is the name of this result?

答:
This result is the first part of the fundamental theorem of calculus. Let $f$ be a function that is continuous on the interval $[a,b]$. Then, let $F:[a,b] \mapsto \mathbb{R}$ be the function defined by
$$
F(x)=\int_{a}^{x}f(t) \, dt
$$
The fundamental theorem of calculus states that $F$ is continuous on $[a,b]$, and
$$
F'(x)=f(x)
$$
for all $x$ in $(a,b)$. This result can be summarised as
$$
\frac{d}{dx}\int_{a}^{x}f(t) \, dt = f(x) \, .
$$
Integrating a function, followed by differentiating it, gets you back where you started. Note that integrating refers to finding the area under the curve. Really, that's what an integral is. The fact that functions of the form
$$
\int f(x) \, dx
$$
are often called 'indefinite integrals' is a result of the fundamental theorem of calculus, since
$$
\int_{a}^{x}f(t) \, dt = \int f(x) \, dx \, .
$$


问:
express $f$ in the interval $[0, 2]$
In a question of an exercise, I have to "express the following function in the interval $[0, 2]$ " :

$f\left(x\right)=\left|x-2\left\lfloor\frac{x+1}{2}\right\rfloor\right|$

My initial idea would be that I have to express the function without the absolute value...

Therefore, I tried working with inequalities, ie:
we have, if we define
$g(x)= x-2 \lfloor\frac{x+1}{2}\rfloor$

$0 \le  x \le  2$

and

$-x-1\le \:-2\lfloor \frac{x+1}{2}\rfloor < -x+1$

summing them:

$-x-1\le \:x-2\lfloor \frac{x+1}{2}\rfloor < 3-x$

$-x-1\le g(x) < 3-x$

however I don't get anything meaningful out of this.

By seeing the plot of $g$ ($f$ without the absolute value), it is both positive and negative in this interval.

What am I supposed to do? Did I understand the question wrong?

答: Since $x\in[0,2]$, $\frac{x+1}2\in\left[\frac12,\frac32\right]$. If $x\in[0,1)$, then $\frac{x+1}2\in\left[\frac12,1\right)$, and therefore $\left\lfloor\frac{x+1}2\right\rfloor=0$; otherwise,  $\left\lfloor\frac{x+1}2\right\rfloor=1$. So, $x-2\left\lfloor\frac{x+1}2\right\rfloor=x$ if $x\in[0,1)$ and $x-2\left\lfloor\frac{x+1}2\right\rfloor=x-2$ otherwise. Can you take it from here?

Title: express $f$ in the interval $[0, 2]$


答:
In $[0,1)$, $\lfloor \frac{x+1}{2} \rfloor =0$ and your function is $|x|=x$.

In $[1,2]$, the floor evaluates to $1$ and your function is $$|x-2| =2-x$$ as $x-2$ will be non-positive here.

Title: express $f$ in the interval $[0, 2]$

Body:
In a question of an exercise, I have to "express the following function in the interval $[0, 2]$ " :

$f\left(x\right)=\left|x-2\left\lfloor\frac{x+1}{2}\right\rfloor\right|$

My initial idea would be that I have to express the function without the absolute value...

Therefore, I tried working with inequalities, ie:
we have, if we define
$g(x)= x-2 \lfloor\frac{x+1}{2}\rfloor$

$0 \le  x \le  2$

and

$-x-1\le \:-2\lfloor \frac{x+1}{2}\rfloor < -x+1$

summing them:

$-x-1\le \:x-2\lfloor \frac{x+1}{2}\rfloor < 3-x$

$-x-1\le g(x) < 3-x$

however I don't get anything meaningful out of this.

By seeing the plot of $g$ ($f$ without the absolute value), it is both positive and negative in this interval.

What am I supposed to do? Did I understand the question wrong?


问:
Is Hahn-Kolmogorov theorem a direct result of Carathéodory's extension theorem?
Both theorems assume the same condition and conclusion, except that

Hahn-Kolmogorov  theorem extends a premeasure on a field of subsets to a measure on a sigma algebra generated by the field of subsets

Carathéodory's extension theorem extends a premeasure on a ring of subsets to a measure on a sigma algebra generated by the ring of subsets.

Since a field of subsets is also a ring of subsets, is Hahn-Kolmogorov  theorem a direct result of Carathéodory's extension theorem?

Thanks!

答:
Yes. As you said, the various versions of extension theorems all extend a premeasure on some family of subsets $\mathcal{F}$ of $\Omega$ to a measure on $\sigma(\mathcal{F})$, and differ in what conditions $\mathcal{F}$ satisfies.

The most general form of Carathéodory's extension theorem that I'm
aware of requires $\mathcal{F}$ to be a semi-ring, i.e., (1). $\emptyset \in \mathcal{F}$, (2). $A, B \in \mathcal{F} \implies A \cap B \in \mathcal{F}$, and (3). $A, B \in \mathcal{F} \implies \exists n, \exists C_{1,2,...,n} \in \mathcal{F}, A - B = \dot\bigcup_i C_i$.  See, e.g., the textbook by Schilling, "Measures, Integrals and Martingales".

The Wikipedia version of Carathéodory's extension theorem requires $\mathcal{F}$ to be a ring, i.e., (1). $\emptyset \in \mathcal{F}$, (2). $A, B \in \mathcal{F} \implies A \cup B \in \mathcal{F}$, and (3). $A, B \in \mathcal{F} \implies A - B \in \mathcal{F}$.

The Hahn-Kolmogorov theorem requires $\mathcal{F}$ to be a field (also called an algebra), i.e., (1). $\emptyset \in \mathcal{F}$, (2). $A, B \in \mathcal{F} \implies A \cup B \in \mathcal{F}$, and (3). $A \in \mathcal{F} \implies A^c \in \mathcal{F}$.  This seems to be the version used in the Ash and Dade textbook.

It's easy to show that a ring is closed under intersection (write $A \cap B = A - (A - B))$, and condition (3) for a semi-ring trivially holds or a ring, so a ring is a semi-ring. Also, it's clear that a field is a ring (that happens to also contain $\Omega$).

Therefore, version 1 ("Carathéodory's extension theorem") implies version 2 implies version 3 ("Hahn-Kolmogorov theorem").


问:
What does this notation mean? ( $X=\inf\{ x : U \le F(x) \}$ )
What does ( $X= \inf\{ x : U \le F(x) \}$ ) mean?

I know the concept of inf, found here

I found the notation in the title at the end of the chapter on the Inverse Transform Sampling Method where $F(x)$ is the CDF cumulative density function of X and $U$ a random number generate with Uniform distribution $[0,1]$

答:
$U$ is a number and $F$ is a function of one variable. When you put a number $x$ into that function, you get a number $F(x).$ Whether $U\le{}$that number depends of course on what $F(x)$ is, and that depends on what $x$ is. Thus you may put in one number $x$ and find that $U\le F(x)$ and another value of $x$ and find that $U\not\le F(x).$ This distinguishes some values of $x$ from others. Thereby it defines a set of certain values of $x.$ The notation in the question refers to the infimum, i.e. the greatest lower bound, of that set.

For example, suppose $F(x) = 1 - e^{-x}$ for $x\ge0$ and $U= 0.75.$ Then
\begin{align}
& \{x : U \le F(x)\} = \{x : x\ge \ln 4\} = [\ln4,+\infty)
\end{align}
and that set has an infimum.


问:
How to graph the equation: $y=\frac {x-2}{x+1}$?
the title says it all.
I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?

"Graph the equation $y=\frac {x-2}{x+1}$"

I know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.

Any help would be appreciated.
If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!

答:
When you want draw the graph of the function, you need to check a couple of things:

Check if there are any points in the domain where the function is undefined, you know there's something happening there. For instance here you saw that when $x=-1$ it is undefined.

Then, $y$ can be equal to $0$, there's no problem with that, and you should find all the $x$ such that $y=0$. In this case, when $x=2, y=0$. Thus you know that the function goes through the point $(2,0)$.

Now you have all the points where there's something special happening, you can check where the function is positive, and where it is negative. You need to look at all the intervals between the special points. In your case : $(-\infty,-1),\,(-1,2),\,(2,\infty)$. Take $(-\infty,-1)$ for example. On this interval, you know that the numerator is always negative and the denominator is always negative as well, so the value for $y\in(-\infty,-1)$ will always be positive. Do that for all the intervals.

Finally, you need to check some limits to see what happens at those special points and also when you tend to infinity (in your case only $-1$ is a special point because you already now that the value of your function when $x=2$ is $0$). So you should check :

$$\lim_{x\rightarrow\infty}f(x),\; \lim_{x\rightarrow-\infty}f(x),\; \lim_{x\rightarrow-1^+}f(x),\; \lim_{x\rightarrow-1^-}f(x),\;$$

That's it, you have everything you need to graph your function! (If you want to go into further details you can also check the derivative for extremas and the second derivative for inflexion points. This means that you have to compute the derivatives and check for which values of $x$ they are equal to $0$).


问:
Basic integration techniques question $\int\frac{e^{3x}}{e^x+1}dx$ (work is shown)
Integration problem with work shown

I am trying to figure out how I have created the $-\frac{3}{2}$ in the problem solution. I have been searching worlfram and symbolab and cant seem to find the step I messed up on. Any pointers are greatly appreciated. Thank you for the help.

答:
To compute $$I=\int \frac{e^{2x}}{e^x+1}e^xdx$$

put $ t=e^x$

then

$$I=\int \frac{t^2-1+1}{t+1}dt$$
$$=\int (t-1+\frac{1}{t+1})dt $$

$$=\frac{t^2}{2}-t+\ln(|t+1|)+K$$

replace $ t $ by $ e^x$.

You have taken $ K =C-\frac 32$.


答:
Every relationship from $[n]$ to $[n]$ is just a subset of $[n]\times [n]$, so consider the $n\times n$ grid, what you have calculated is placing $k$ non-threating rooks on the grid. Your answer is equal to the sum of coefficients of the "rook polynomial" and does not have a closed form.

However, what this problem is asking is to place $k$ rooks on the $n\times n$ grid while they threat each other only on one axis, for example just vertically.

To count the answer over all $k$s, for each column of the grid, one can pick the position of the corresponding rook with a number between $0$ to $n$, where $0$ is leaving the rook outside of the grid.

This leads to $(n+1)^n$ possible outcomes.

Title: help calculating $\sum_{k=0}^n k!{n \choose k} ^2= \text{?}$

Body: I am trying to find the number of relations on $[n] = \{1,\ldots,n\}$ s.t. for all $x,y,z\in R.xRz \wedge yRz\to x=y$. One idea I had was choosing $k$ elements from $n$ then choosing another $k$ elements from $n,$ there are $k!$ bijections between these two sets so we just have to sum over $k$ and here I got stuck, tried finding another approach but came up with nothing.


问:
Proof Verification of Exercise 2.9 Baby Rudin: Prove that the complement of $E^{o}$ is the closure of the complement of $E$.
I am stuck on part d of this exercise:

Let $E^{o}$ denote the set of all interior points of a set. Prove that the complement of $E^{o}$ is the closure of the complement of $E$.

I solved it using double inclusion:

Let $p$ $\in (E^{o})^{c}$. Since $p$ is not an interior point of $E$, every neighborhood of $p$ contains a point not in $E$ (or else we have a neighborhood of $p$ that is contained in $E$ and p is an interior point) so $p$ is a limit point of $E^{c}$ so $p$ $\in$ $\overline{E^{c}}$. Conversely, let $q$ $\in$ $\overline{E^{c}}$. Then $q$ belongs to $E^{c}$ and is not an interior point of E, or $q$ is a limit point of $E^{c}$ and no neighborhood of $q$ is contained in $E$, so $q$ is not an interior point of $E$ and so $q \in (E^{o})^{c})$

Is this proof correct? If not, where exactly does it break down? Thank you for your time.

答: The only thing that I would add to that is a justification why is it that, if $q$ is a limit point of $E^\complement$, then no neighborhood of $q$ is contained in $E$. That's easy, of course: since $q$ is a limit point of $E^\complement$, every neighborhood of $q$ intersects $E^\complement$, and therefore no neighborhood of $q$ is contained in $E$.


问:
On Sharp Poincaré-Friedrichs inequality
I am concerned with the sharp Poincaré-Friedrichs inequality which I can find proof.

Assume $f: [-a,a]\to \Bbb R$ ($a>0$) is $C^1$ such that $f(-a)=f(a)=0$. Then I want to show that

$$\int_{-a}^a|f(t)|^2d t\leq a^2\int_{-a}^a|f'(t)|^2d t.$$

When I use the fundamental theorem of calculus, I get the constant factor $2a^2$ instead of $a^2$.

$$\begin{split}
f(t)= \int_t^a f'(s)((a-t) ds \implies |f(t)|^2 \leq (a-t)\int_t^a |f'(s)|^2  ds.
\end{split}$$

Integrating both sides over $(-a,a)$ yileds
$$\begin{split}
\int_{-a}^a|f(t)|^2 dt& \leq \int_{-a}^a(a-t)\int_t^a |f'(s)|^2  dsd t\\
&\leq \int_{-a}^a(a-t)\int_{-a}^a |f'(s)|^2  dsd t\\
&= 2a^2 \int_{-a}^a |f'(s)|^2  ds.
\end{split}$$

How can I optimize the above inequality with constant factor $a^2$?

More generally, as stated
[https://en.wikipedia.org/wiki/Friedrichs%27s_inequality] assume $f\in C^k(\Bbb R)$ $k\in \Bbb N$ such that $supp \,\,f\subset (-a,a)$ then how can I prove that

$$\int_{-a}^a|f(t)|^2d t\leq a^{2k}\int_{-a}^a|f^{(k)}(t)|^2 d t.$$

答:
For the second part of your question, if $f\in C^k$, for $k>1$, then $f'\in C^{k-1}$ and you can apply the first result to $\int_{-a}^a\lvert f'\rvert^2dt$.
By induction then, $$\lVert f \rVert_{L^2}^2 \leq a^{2n}\lVert f^{(n)}\rVert_{L^2}^2 \text{, for all } n\leq k.$$

Concerning the first part, I can't find a way to tighten the control, but I think your first inequality is wrong, namely $f(t)\neq \int_t^a f'(s)(a-t)ds$. Instead, use Cauchy-Schwarz inequality on
$$\lvert f(t)\rvert \leq \int_{-a}^t \vert f'(s)\rvert ds$$
and the fact that $(t+a)^{1/2} \leq (2a)^{1/2}$ to get
$$\lVert f \rVert_{L^2} \leq 4a^2\lVert f'\rVert_{L^2}.$$

Title: On Sharp Poincaré-Friedrichs inequality


答:
Let $K=C-\frac{3}{2}$. Then, you have found the same solution as the one in the book, since both K and C are just arbitrary constants. What you have shown is that the derivative of any function of the form
$$
f(x)=\frac{1}{2}e^{2x}-e^x-\frac{3}{2}-\ln(e^x-1) + \text{some arbitrary constant}
$$
is
$$
\frac{e^{3x}}{e^x+1} \, .
$$
It doesn't matter how you label that constant. Sometimes different notations can 'conflict'—your '$C$' is different to the '$C$' in the book because you have added $-3/2$. However, this fact is immaterial. Try not to let notation get in the way of the actual mathematics.

Title: Basic integration techniques question $\int\frac{e^{3x}}{e^x+1}dx$ (work is shown)

Body:
Integration problem with work shown

I am trying to figure out how I have created the $-\frac{3}{2}$ in the problem solution. I have been searching worlfram and symbolab and cant seem to find the step I messed up on. Any pointers are greatly appreciated. Thank you for the help.


答:
If you get
$$
\Big(\text{some function of } x \Big) - \frac 3 2 + \text{constant}
$$
and someone else gets
$$
\Big(\text{some function of } x \Big) + \text{constant}
$$
then you both got the same thing, but the other person is expressing it more efficiently.

For example, if your "constant" is $14.5$ and that other person's "constant" is $13$ than you both have the same function. In either case, if the "constant" is any number at all, then you have an antiderivative of the function you started with, and in either case, there exist no other antiderivatives than those that can be expressed by the expression given as the answer.


问:
Is the empty set a superset of itself?
I know the empty set is a subset of itself, so I'm assuming that it's also the case that it is a superset of itself?

∅ ⊇ ∅

答: Yes. In the same way that every set is a superset of itself, but not a proper superset.

Title: Is the empty set a superset of itself?