答: Since $\arctan(1+x^2)-\tfrac{\pi}{4}=\arctan\tfrac{x^2}{2+x^2}\sim\tfrac{x^2}{2}$ and$$\tfrac{\pi}{4}-\arcsin\tfrac{\cos x}{\sqrt{2}}=\arcsin\tfrac{\sqrt{2-\cos^2x}-\cos x}{2}\sim\tfrac{\sin^2x}{\sqrt{2-\cos^2x}+\cos x}\sim\tfrac{x^2}{2},$$the limit is $\tfrac12+\tfrac12=1$.

Title: Evaluating $\lim_{x\to 0}\frac1{x^2}({\arctan(1+x^2)-\arcsin\frac{\cos x}{\sqrt{2}}})$

Body:
I've been struggling for a while with the following limit:
$$L = \lim_{x\to 0}\frac{\arctan(1+x^2) - \arcsin\left(\frac{\cos x}{\sqrt{2}}\right)}{x^2}$$

I know that $L = 1$, since it seems obvious from the graph and WolframAlpha confirms it, yet I simply can't find a way to theoretically evaluate it. I've tried all sorts of trig substitutions, two of which seem pretty intuitive:

$\cos x = \sin\alpha+\cos\alpha = \sqrt{2}\sin\left(\alpha + \frac{\pi}{4}\right)$ (suitable for the right part)

$1+x^2 = \tan\left(\theta + \frac{\pi}{4}\right)$ (suitable for the left part)

I thought of those because the limit kind of hints at them and both new variables also tend to $0$ and I feel I'm really close, but I couldn't finish anything and also started doubtingthis might not be the way to go. Any ideas are really welcome and greatly appreciated!

Edit: I forgot the mention, this limit is somehow supposed te be evaluable without using L'Hopital's rule, that's what all the fuzz is about! :D I also grew suspicious if it's even possible, but it's an interesting challenge nonetheless!


问:
how did they arrive to the rule of addition of vectors?
I'm an A2 igsce math student and I'm taking mechanics for the first time in math this session.

I wanted to ask about how they arrived to the rule of addition of vectors. How did they know that if we add the X's and Y's of two vectors they would get a third vector which has exactly the same direction and magnitude of the force that could replace these two vectors or forces.

I'm convinced that it's correct and can feel or see the direction a point will accelerate in if two certain forces are applied to it. And can feel how the resultant force tends to get closer to the bigger force and how if two equal forces are applied with the same angle the resultant force is going to be exactly between them.

So I'm sure who invented the vectors had the same feelings and visions too but how did he arrive at this simple method to get such fascinating and exact results, not only did he manage to get the direction but the magnitude of the resultant too!

答:
I think it is important to realize that reality is special. There is some structure to it, whether we can figure out what it is or not. And there is even radically different good approximations at different scales. At the human scale, objects in space seem to obey euclidean geometry. What does this mean? Well, for one thing, rigid motions (combinations of translations and rotations) preserve lengths and angles! Have you ever wondered if a solid object can be moved in some manner that its shape is changed? It just does not. Under any rigid motion $f$ in euclidean geometry, $|f(P)-f(Q)| = |P-Q|$ for every points $P,Q$, and $∠PQR = ∠f(P)f(Q)f(R)$ for every points $P,Q,R$. The fact that objects at the human scale seem to obey this tells us that euclidean geometry is a good approximation of the geometry of reality at the human scale (not too big or small). In fact, not just euclidean geometry by specifically the euclidean space $ℝ^3$ seems to be a good approximation for human-scale geometry of reality. And $ℝ^3$ is a vector space and obeys vector addition in the sense that you noted.

In other words, we would not be so interested in $ℝ^3$ and vector spaces in general if reality did not in the first place have structure that looks like $ℝ^3$ at a scale we can observe! Similarly, you could ask why people came up with the axiomatization of the naturals as a discrete ordered semi-ring plus induction, and the answer is again the same! Namely, counting seems to obey these properties!

As for the addition of forces specifically, it stems from the relation between force and acceleration in newtonian mechanics (which is at the human-scale). We have that a force $F$ on a mass $m$ induces acceleration $a$ given by $F = m·a$, and $a$ is the second derivative of position $x$. Now a position is a vector in $ℝ^3$ (according to our good approximation), so acceleration is also a vector in $ℝ^3$, and adding two forces is observed to induce the sum of the corresponding accelerations, and hence two forces applied on the same mass induce exactly the same acceleration as the vector sum of those forces applied on that mass. Take a while to think through this; it is indeed not merely an issue of discovery, but there is some underlying mathematical constraint given the observed relation between forces and positions via acceleration.

ContentLicense: CC BY-SA 4.0


问:
How to construct an LP problem that makes a (partial) theorem fail?
I am following a course on linear programming, and one of the exercises calls for an example, that may show that a theorem fails, if a assumption is omitted from the theorem.

The theorem is Theorem 5.5 from Vasek Chvatals book Linear Programming. The theorem states that:

If a linear programming problem has at least one non-degenerate basic optimal solution, then there is a positive $\varepsilon$ with the following property: If $|t_i| \leq \varepsilon$ for all $ i = 1, 2, ..., m$, then the problem

$$
\begin{align}
\mathrm{maximize } & \sum^{n}_{j=1}c_jx_j \\
\mathrm{subject\ to} & \sum^n_{j=1} a_{ij}x_j \leq b_i + t_i & (i = 1, 2, ..., m) \\
& x_j \geq 0 & (j=1, 2,...,n)
\end{align}
$$

has an optimal solution and its optimal value equals

$$
z^* + \sum^m_{i=1} y_i^* t_i
$$

with $z^*$ standing for the optimal value of the linear programming problem and with $y^*_1, y^*_2, ..., y^*_m $ standing for the optimal solution of its dual.

The hypothesis that should be omitted, is the one stating that the original LP problem has at least one non-degenerate basic optimal solution.

So far, I haven't been able to grok what would happen if the hypothesis is omitted and my search for alternate literature hasn't turned up much.

I assume, the solution is to construct a LP problem with a degenerate solution in the primal and then show that the conclusion doesn't hold. But how do I go about finding an LP problem that shows this?

Or are there smarter ways to solve this problem?

答: Let $m=n=1$, $a_{11}=1, b_1=-1, c_1=1$.  You will find that no small $t_1$ (including $t_1=0$) has any solution to the LP.</p>

ContentLicense: CC BY-SA 3.0


问:
Asymptote Angle rate of change for hyperbola
Assume there is a canonical hyperbola where the focus are $(c,0)$ and $(-c,0)$. The hyperbola curve intersects with the major axis at $(a,0)$ and $(-a,0)$. We can calculate $b^2 = c^2 - a^2$ so that the canonical hyperbola equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $.
From this hyperbola we have the asymptote lines $y=\pm\frac{b}{a}x $.

Let us say that we have two data: $a$ and $a+\delta$. Each has its own asymptotes and there will be an angle between asymptote from $a$ and $a+\delta$. How to derive the equation for this angle as the function of $a$? And how to derive the angle rate of change with respect to $a$?

The value of $a$ is $0<a<c$. I know from Geogebra simulation (the figure below) that with bigger $a$ I will have bigger angle rate of change. But near $c$ this is no longer applies. I don't know how to explain it in mathematical formula. Thank you in advance.

答:
For a given $a$ you have asymptotic lines
$$ y = \pm\frac ba x = \pm\frac{\sqrt{c^2 - a^2}}{a} x. $$

By symmetry, what happens to one line when you add $\delta$ to $a$
happens to the other line when you subtract $\delta$ from $a.$
So let's just consider one line; let's assume the $\pm$ sign is $+$.

The slope of the line is $b/a$ but the angle of the line (counterclockwise from the $x$ axis) is
$$ \theta(a) = \arctan\left(\frac ba\right)
 = \arctan\left(\frac{\sqrt{c^2 - a^2}}{a}\right). $$

This function is discontinuous at $a = 0$ (undefined at $a = 0$ exactly, and it jumps from an angle near $\frac\pi2$ (a right angle) to an angle near $-\frac\pi2$ as $a$ goes from a small positive value to a small negative value).
You could define the function differently for $a<0$ and add a definition of $\theta(0)$ to make a continuous function with output between $0$ and $\pi,$
but perhaps you would be satisfied just to consider $a > 0$ and not be concerned about what happens as we cross over from positive to negative $a$.

If you're looking for the difference in angle when you add a particular $\delta$ to $a,$ it's

$$ \left\lvert \arctan\left(\frac{\sqrt{c^2 - (a+\delta)^2}}{a+\delta}\right)
 - \arctan\left(\frac{\sqrt{c^2 - a^2}}{a}\right) \right\rvert. $$

Not a nice expression, though it might be possible to get a reasonable small-$\delta$ approximation with Taylor series.

If you mean the instantaneous rate of change of the angle when you refer to
"angle rate of change", then you're asking about

$$ \theta'(a) =
\frac{\mathrm d}{\mathrm da} \arctan\left(\frac{\sqrt{c^2 - a^2}}{a}\right).$$

For the line with slope $\frac ba$ this value is always negative,
that is, the line rotates clockwise as $a$ increases.
As your Geogebra graphs hint, the rate of change becomes more negative
(hence larger in magnitude) as $a$ gets closer to $c.$
The magnitude of the rate of change grows without bound as $a$ gets close to $c.$


问:
Long time behaviour of population that dies out with probability $0.2$, stays constant with probability $0.5$ and doubles with probability $0.3$
What is the expected long term behaviour of a population that dies out with probability $0.2$, stays the same with probability $0.5$ and doubles with probability $0.3$?

Of course,
$$
E[X_{n+1}] = 0.5X_n + 0.3\cdot2X_n = 1.1X_n
$$
and thus the expectated population size goes to infinity for $n\to\infty$.

But the hint given for the exercise was that this is is classic Markov chain problem that can be solved easily with a transition probability matrix. I don't really see how that could make the solution easier or did I get the question wrong?

答:
Assume the population size at time $0$ is $P_0$.

After 1 unit of time, the size will be either $0,P_0$ or $2P_0$.

After $N$ units of time, the population will be one of $0,P_0,2 P_0,...,2^N P_0$.

Take $N$ very large but fixed and assume those are the possible states and find the probability of being in each state at each time using the transition matrix.

It is possible to directly calculate the probabilities, but it is a good exercise to try to do it with the transition matrix and that is what you are expected to do from the hint.

The probability of never reaching $0$ is $0.8^N$, hence the probability of reaching $0$ is $1-0.8^N$.

The probability of having size $2^k P_0$ is ${{N}\choose{k}}0.5^{N-k} 0.3^k$.

The expected value can be calculated directly from there and it is equal to $1.1^N P_0$.

You can check for the case $N=1$ that the expected value is $0.5 P_0+0.3 (2) P_0=1.1 P_0$ and use induction for the general case.


问:
On a tempered distribution which comes from a locally integrable function
Let $ f\colon \mathbb{R}^n \to \mathbb{C} $ be a measurable function. We say that $ f $ is locally integrable if $ f|_K $ is integrable for all compact subsets $ K \subseteq \mathbb{R}^n $. A locally integrable function $ f $ yields a distribution $ T_f $, which is given by
$$
\langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x) \phi(x) \,dx
$$
for $ \phi \in \mathscr{D}(\mathbb{R}^n) $ (test functions).

Question. Suppose that $ T_f $ is tempered as a distribution. Then $ T_f $ is extended so that $ \langle T_f, \phi \rangle $ is defined for all $ \phi \in \mathscr{S}(\mathbb{R}^n) $ (rapidly decreasing functions).

Is $ f \phi $ integrable for all $ \phi \in \mathscr{S}(\mathbb{R}^n) $?

For $ \phi \in \mathscr{S}(\mathbb{R}^n) $ with $ f \phi $ integrable, is it true that $ \langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x) \phi(x) \,dx $?

答:
Is $ f\phi $ integrable for all $ \phi \in \mathscr{S}(\mathbb{R}^n) $?

Not necessarily. Since $ F(x) = \sin e^{x^2} $ is tempered as a distribution, its differentiation $ f(x) = 2x e^{x^2} \cos e^{x^2} $ is also tempered as a distribution. $ \phi(x) = e^{-x^2/2} $ is rapidly decreasing, but $ f\phi $ is not integrable.

For $ \phi \in \mathscr{S}(\mathbb{R}^n) $ with $ f \phi $ integrable, is it true that $ \langle T_f, \phi \rangle = \int_{\mathbb{R}^n} f(x)\phi(x) \,dx $?

Yes. To prove this, we take a sequence $ (\phi_k)_{k \in \mathbb{N}} $ in $ \mathscr{D}(\mathbb{R}^n) $ such that $ \lvert \phi_k \rvert \leq \lvert \phi \rvert $ and $ \phi_k \to \phi $ in $ \mathscr{S}(\mathbb{R}^n) $. Since $ T_f $ is tempered, $ \langle T_f, \phi_k \rangle \to \langle T_f, \phi \rangle $. On the other hand, since $ f\phi $ is integrable,
$$
\int_{\mathbb{R}^n} f(x)\phi_k(x) \,dx \to \int_{\mathbb{R}^n} f(x)\phi(x) \,dx
$$
by Lebesgue’s dominated convergence theorem. Therefore, we get
$$
\langle T_f, \phi \rangle
= \lim_{k \to \infty} \langle T_f, \phi_k \rangle
= \lim_{k \to \infty} \int_{\mathbb{R}^n} f(x)\phi_k(x) \,dx
= \int_{\mathbb{R}^n} f(x)\phi(x) \,dx.
$$


问:
Decomposing regular polygons into other regular polygons
I came across the german Wikipedia page for the Dodecagon and there it said (my translation) "The regular Hexagon and the regular Dodecagon are the only regular polygons that can be decomposed into smaller regular polygons".

https://de.wikipedia.org/wiki/Zw%C3%B6lfeck#Zerlegung_in_regelm%C3%A4%C3%9Fige_Polygone

I object to that as both the equilateral triangle and the square can be decomposed into four smaller copies of themselves.

I believe what the article wants to say is "...into regular polygons with fewer sides".

Is that a better way of phrasing it? Is the fact correct at all? Can someone give me a citation for it? How would one go about proving this?

Regards, Marian

答: An interesting additional example can be found in spherical geometry.  On a sphere, a properly sized regular pentagon can be divided into five equilateral triangles. Simply inscribe a regular icosahedron in the sphere and use the contact points to define the vertices of the pentagon and its component triangles.</p>


问:
What does nonlocal coupling mean in the context of oscillators?
I am trying to understand some papers on chimera type behaviour but I don't understand what nonlocal coupling means in the context "... in nonlocal coupling oscillator system". This is the paper: https://ui.adsabs.harvard.edu/abs/2014JPSJ...83c4002L/abstract

Any explanations would be very appreciated.

答:
According to the rather detailed history in Martens, Thutupalli, et al. (2013) "Chimera states in mechanical oscillator networks" the oscillators have a spatial arrangement. Local coupling would be interaction between close or direct neighbors, non-local coupling thus interaction with distant oscillators. Usually the non-local coupling is weaker than the local coupling, for instance falling with the distance.

The example first discussed is metronomes on horizontally moving platforms that are bound to some center location (here by being swings). Metronomes on the same platform are locally coupled. Connecting the platforms with (weak) springs then introduces non-local coupling.

Metronomes on a single isolated platform tend to synchronize, non-local coupling can destroy this synchronization among the metronomes on the same platform.


问:
Find the area of this pentagon

Let $BCDK$ be a convex quadrilateral with $BC=BK$ and $DC=DK$. $A$ and $E$ are points such that $AB=BC$, $DE=DC$ and such that $ABCDE$ is a convex pentagon. Point $K$ lies in the interior of pentagon $ABCDE$. If $\angle ABC=120^{\circ}$, $\angle CDE=60^{\circ}$ and $BD=2$, find the area of pentagon $ABCDE$.

I know that this can be solved using trigonometry.But i prefer to have a solution by euclidean geometry over that. I tried extending some lines and finding other trivia, but they didn't seem to work.

答:
Rotate $\triangle BCD$ $60$ degrees w.r.t point $D$ to $\triangle B'ED$. The brown area will be the desired area.

Connect $\overline{BB'}$. Since $\angle BDB'=60^{\circ}$ and $\overline{BD}=\overline{B'D}$, we can say that $\triangle BDB'$ is equilateral.

Assume $\overline{AE}$ and $\overline{BB'}$ intersect at $F$.

Let $\angle ABF=a$ and $\angle EB'F=b$.
By symmetry, $\angle FBK=b$ and $\angle KBD=\angle CBD=60^{\circ}-b$. Therefore we have $$a+b+(60^{\circ}-b)+(60^{\circ}-b)=120^{\circ}\implies a=b$$ So $$\triangle ABF\cong\triangle EB'F\quad(A.A.S.)$$ Therefore the answer will be the area of equilateral $\triangle BDB'$, which equals $\color{blue}{\sqrt3}$.

Title: Find the area of this pentagon


问:
Linearly independent set in $\mathbb R$ over $\mathbb Q$
The set $$\left\{\sqrt[2^n]{2},\quad n\in\mathbb N\right\}\subset\mathbb R$$ is linearly independent over $\mathbb Q$.

Is there an easy or elementary way to see this? I.e., why does the equation $\alpha_1 2^{2^{-i_1}}+\cdots +\alpha_N 2^{2^{-i_N}}=0$ not have a non-trivial solution in $\alpha_i\in\mathbb Q$ for any choice of $N\in\mathbb N$ and $i_1,\dots,i_N\in\mathbb N$?

答: More generally, for each $N \in \mathbb N$, the set $\left\{\sqrt[2^n]{2} : n=0,\dots,N\right\}$ is linearly independent because $X^{2^N}-2$ is irreducible over $\mathbb Q$ due to Eisenstein's criterion.


问:
Why does A(i,j) := ? (Symetric, positive definite matrices)
Hope you're having a great day.

Well I've been learning about Analytic Geometry in the mml book, but I've found myself stuck the question statement presented in the photo. Why does A(i,j) := ? I'm really confused.

The quote in the text book goes as follows:

"Consider an n-dimensional vector space V with an inner product h·, ·i :
V × V → R (see Definition 3.3) and an ordered basis B = (b1, . . . , bn) of
V . Recall from Section 2.6.1 that any vectors x, y ∈ V can be written as
linear combinations of the basis vectors so that $
\boldsymbol{x}=\sum_{i=1}^{n} \psi_{i} \boldsymbol{b}_{i} \in V$ and
$y=\sum_{j=1}^{n} \lambda_{j} b_{j} \in V$ for suitable ψi
, λj ∈ R. Due to the bilinearity of the
inner product, it holds for all x, y ∈ V that

$\langle\boldsymbol{x}, \boldsymbol{y}\rangle=\left\langle\sum_{i=1}^{n} \psi_{i} \boldsymbol{b}_{i}, \sum_{j=1}^{n} \lambda_{j} \boldsymbol{b}_{j}\right\rangle=\sum_{i=1}^{n} \sum_{j=1}^{n} \psi_{i}\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle \lambda_{j}=\hat{\boldsymbol{x}}^{\top} \boldsymbol{A} \hat{y}$

where $
A_{i j}:=\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle
$ and $
\hat{\boldsymbol{x}}, \hat{\boldsymbol{y}}
$ are the coordinates of x and y with respect
to the basis B."

Thank you in advance for your responses.

答:
It is just how matrix and vector multiplication works. If you set
$$
\hat x =\begin{pmatrix} \psi_1 \\ \vdots \\ \psi_n \end{pmatrix}
\;\;\;\mathrm{and} \;\;\;
\hat y =\begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{pmatrix}
$$
you will note that
$$
\sum\limits_{i=1}^n \sum\limits_{j=1}^n \psi_i A_{ij} \lambda_j = \hat x^T A \hat y\; .
$$
So you get
$$
\sum\limits_{i=1}^n \sum\limits_{j=1}^n \psi_i \,\langle \mathbf{b}_i , \mathbf{b}_j\rangle \,\lambda_j = \hat x^T A \hat y
$$
if you set
$$
A_{ij} = \langle \mathbf{b}_i , \mathbf{b}_j\rangle
$$

Title: Why does A(i,j) := <b(i),b(j)>? (Symetric, positive definite matrices)


问:
How to find $ f'(a)?$
I have some doubt in this answer

Here is the outline of the given answer

For $a\in \mathbb D,$ define the well known automorphism of $\mathbb D$

$$\varphi_a(z)=\frac{a-z}{1-\bar a z}.$$

Then $\varphi_a'(0) = |a|^2-1,$ $\varphi_a'(a) = 1/(|a|^2-1).$

Now take $f$ to be your function, so that $f(0)=a\ne 0.$ If we consider

$$\varphi_{f(a)}\circ f \circ \varphi_{a},$$

we have a map that takes $0$ to $0.$ By the Schwarz Lemma, the derivative of this function at $0$ has modulus $\le 1.$ Use the chain rule and $f(a)\ne 0$ to see this implies

$$|f'(a)| \le \frac{1-|f(a)|^2}{1-|a|^2} < \frac{1}{1-|a|^2}.$$

Since $|\varphi_a'(a)| = 1/(1-|a|^2),$ we have the desired solution.

My  doubt : How  to find $ f'(a)?$

My attempt :take $h(z)=\varphi_{f(a)}\circ f \circ \varphi_{a},$  . here h is satisfied schwarz  lemma  so $h(0)=0$ and  $h(z) \le |z|$

$h(z)=\varphi_{f(a)}\circ f \circ \varphi_{a}\le |z|\implies |f ( \varphi_{a} )|\le|\varphi_{f(a)}^{-1}(z) |$

therefore $$|f ( \varphi_{a} )| \le \frac{f(z)- f(a)}{ 1- f(a)f(z)}$$

After that im  not able to proceed further

答:
In that answer I didn't find $f'(a)$ exactly. But I showed $|f'(a)|<|\varphi_a'(a)|,$ and that was enough to answer the question there.

I'm not sure what you are doing in your attempt. I am not using $|\varphi_{f(a)}\circ f \circ \varphi_{a}(z)|\le |z|.$ I am using $|(\varphi_{f(a)}\circ f \circ \varphi_{a})'(0)| \le 1.$

Title: How to find $ f'(a)?$


问:
Prove that gradient descent is a contraction for strongly convex smooth functions
I am trying to prove that gradient descent on a $\alpha$-strongly convex, $\beta$-smooth function $f$ is a contractive operator. I am looking for a general proof which does not assume existence of the second derivative, that also obtains the optimal bound, which I believe is $\max\{\left|1 - \lambda\alpha\right|, \left|1 - \lambda\beta\right|\}$, where $\lambda$ is the step size (see here page 15 (page 13 using the pdf page counter)). This is my effort so far:

Gradient descent can be written in the form of the operator $T = (I - \lambda\nabla f)$.

$$
\|Tx - Ty \|^{2} = \|x - \lambda\nabla f(x) - y +\lambda \nabla f(y) \|^{2} \\
= \|(x- y) - \lambda(\nabla f(x) - \nabla f(y))\|^{2} \\
= \|x - y\|^{2} + \lambda^{2}\|\nabla f(x) - \nabla f(y)\|^{2} - 2\lambda\left<x - y, \nabla f(x) - \nabla f(y)\right>
$$

By using the co-coercivity property of the $(\beta - \alpha)$ smooth function $f(x) - \frac{\alpha}{2}\|x\|^{2}$. We get the bound:

$$
\left<x - y, \nabla f(x) - \nabla f(y)\right> \geq \frac{\alpha\beta}{\alpha + \beta}\|x - y\|^{2} + \frac{1}{\alpha + \beta}\|\nabla f(x) - \nabla f(y)\|^{2}
$$

Combining this inequality with the above we have:

$$
\|Tx - Ty \|^{2} \leq \|x - y\|^{2} + \lambda^{2}\|\nabla f(x) - \nabla f(y)\|^{2} - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\|x - y\|^{2} + \frac{1}{\alpha + \beta}\|\nabla f(x) - \nabla f(y)\|^{2}\right)
$$

Simplifying we have:
$$
\|Tx - Ty \|^{2} \leq \left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right)\|x - y\|^{2} + \left(\lambda^{2} - \frac{2\lambda}{\alpha + \beta}\right)\|\nabla f(x) - \nabla f(y)\|^{2}
$$

Note setting $\lambda = \frac{2}{\alpha + \beta}$ recovers the convergence rate for gradient descent here. How do I proceed from this point? One idea I have is to try and ignore the last term by ensuring

$$
\lambda^{2} - \frac{2\lambda}{\alpha + \beta} \leq 0 \implies \lambda \leq \frac{2}{\alpha + \beta}
$$

This gives the inequality:

$$
\|Tx - Ty \|^{2} \leq \left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right)\|x - y\|^{2}
$$

EDIT: I guess we also need to ensure that

$$
\left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right) \geq 0 \implies \lambda \leq \frac{\alpha + \beta}{2\alpha\beta}
$$

But then I am unsure on how to proceed from here. Perhaps there is a way I can use $\beta$-smoothness instead to bound the $\|\nabla f(x) - \nabla f(y)\|$ term?

答:
Okay, the proof is actually simple from this point. Choose the domain $\lambda \in (0, \frac{2}{\beta})$. This is chosen as at $\frac{2}{\beta}$, the map becomes non-expansive (this is easy to verify).

Recall the bound we had earlier:

$$
\|Tx - Ty \|^{2} \leq \left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right)\|x - y\|^{2} + \left(\lambda^{2} - \frac{2\lambda}{\alpha + \beta}\right)\|\nabla f(x) - \nabla f(y)\|^{2}
$$

When $\lambda \leq \frac{2}{\alpha +\beta}$, the second term on the right-hand side is nonpositive and thus we can replace $\|\nabla f(x) - \nabla f(y)\|$ with a lower bound. We will use the lower bound that follows trivially from $\alpha$-strong convexity:

$$
\|\nabla f(x) - \nabla f(y)\| \geq  \alpha \|x - y\|
$$

Plugging this in then gives:

$$
\|Tx - Ty \|^{2} \leq \left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right)\|x - y\|^{2} + \left(\lambda^{2}\alpha^{2} - \frac{2\alpha^{2}\lambda}{\alpha + \beta}\right)\|x - y\|^{2}
$$

Rearranging and simplifying then yields:

$$
\|Tx - Ty\|^{2} \leq (\lambda{2}\alpha^{2} - 2\lambda\alpha + 1)\|x-y\|^{2}
$$

Now factorising we have

$$
\|Tx - Ty\|^{2} \leq (1- \lambda\alpha)^{2}\|x-y\|^{2} \iff \|Tx - Ty\| \leq (1- \lambda\alpha)\|x - y\|
$$

Note that when $\lambda \leq \frac{2}{\alpha + \beta}$, that $1 - \lambda\alpha = \max\{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$:

$1 - \lambda\alpha \geq 1- \lambda\beta$ follows from $\beta > \alpha$

$1 - \lambda\alpha \geq \lambda\beta - 1 \iff  2 \geq \lambda(\alpha + \beta) \iff \frac{2}{\alpha + \beta} \geq \lambda$ which is true by assumption.

$1 - \lambda\alpha \geq \alpha\lambda - 1$ follows from the above fact and the fact that $\beta > \alpha$.

Thus we have proved that gradient descent is a $\max \{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$ contraction for all $\lambda \leq \frac{2}{\alpha+\beta}$. For $\lambda \geq \frac{2}{\alpha +\beta}$, note that the second term on the RHS of our inequality is nonnegative, and thus we can replace $\|\nabla f(x) - \nabla f(y)\|$ with an upper bound using $\beta$-smoothness:

$$
\|\nabla f(x) - \nabla f(y)\| \leq  \beta \|x - y\|
$$

Plugging this in yields:

$$
\|Tx - Ty \|^{2} \leq \left(1 - 2\lambda\left(\frac{\alpha\beta}{\alpha + \beta}\right)\right)\|x - y\|^{2} + \left(\lambda^{2}\beta^{2} - \frac{2\beta^{2}\lambda}{\alpha + \beta}\right)\|x - y\|^{2}
$$

Simplifying and factorising as before we find that:

$$
\|Tx - Ty\|^{2} \leq (\lambda\beta - 1)^{2}\|x-y\|^{2} \iff \|Tx - Ty\| \leq (\lambda\beta - 1)\|x - y\|
$$

Similarly to before, when $\lambda \geq \frac{2}{\alpha + \beta}$, we see that $\lambda\beta - 1 = \max \{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$. Thus we have shown that gradient descent is a $\max \{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$ contraction for all $\lambda$ such that $\frac{2}{\beta} > \lambda \geq \frac{2}{\alpha + \beta}$.

Combining both cases ($\lambda \geq \frac{2}{\alpha+\beta}$, $\lambda \leq \frac{2}{\alpha +\beta}$) we see that gradient descent is a  $\max \{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$ contraction for all $\lambda \in \left(0, \frac{2}{\beta}\right)$, i.e.

$$
\|Tx - Ty\| \leq \max\{|1 - \lambda\alpha|, |1 - \lambda\beta|\}\|x - y\|
$$

It now also immediately obvious why $\frac{2}{\beta}$ is an upper bound on the step size. Setting $\lambda \geq \frac{2}{\alpha + \beta}$ we have

$$
\lambda\beta - 1 = \max\{|1 - \lambda\alpha|, |1 - \lambda\beta|\}
$$

But

$$
\lambda\beta - 1 < 1 \iff \lambda < \frac{\beta}{2}
$$

So bounding $\lambda$ above by $\frac{2}{\beta}$ ensures $\max\{|1 - \lambda\alpha|, |1 - \lambda\beta|\}$ is less than 1.


问:
Find all values of $c$ such that $\{(1,c,0),(1,0,1),(0,1,-c)\}$ are linearly independent.
Find all values of $c$ such that $\{(1,c,0),(1,0,1),(0,1,-c)\}$ are linearly independent.

So, my strategy is to consider these vectors  as column vectors, and then form a matrix with them as the columns. Once I do that, I will employ Gauss-Jordan elimination to find an equivalent matrix except in reduced row echelon form.

I typed the matrix into matlab and asked it to find the reduced row echelon form, and it gave me the identity matrix. Thus, I believe that no matter what value I pick for $c$, these vectors will be linearly independent.

However, upon trying to employ the Gauss-Jordan elimination by hand, I have a $1-c^2$ in the $(3,3)$ position. Then I got to thinking, if $c \in \{ \pm 1\}$ then this would be zero, and I could not use it to eliminate the entry above it, and thus I would not be able to reduce this to the identity matrix.

Then I thought some more, and wondered if I would have employed different row operations get to reduced row echelon form, then maybe a different situation like this would have arisen... So now I am very confused and don't know how to do the problem!!

Help is appreciated, thank you!

答: The simplest way is to compute the determinant of the matrix$$A=\begin{bmatrix}1&1&0\\c&0&1\\0&1&-c\end{bmatrix},$$which is $c^2-1$ Therefore, the vectors are linearly independent if and only if $c\ne\pm1$. Or you can row reduce the matrix $A$. If you add to the second line the first one times $-c$, you get$$\begin{bmatrix}1&1&0\\0&-c&1\\0&1&-c\end{bmatrix}.$$Now, if you exchange the last two rows, you get$$\begin{bmatrix}1&1&0\\0&1&-c\\0&-c&1\end{bmatrix}.$$If now you subtract the second row from the first one and if you add to the third row the second one times $c$, you get$$\begin{bmatrix}1&0&c\\0&1&-c\\0&0&1-c^2\end{bmatrix}.$$This matrix is invertible if and only if $c^2\ne1$, and so those three vectors are linearly independent if and only if $c^2\ne1$.


问:
What does $\{0 , 1\}^+$ mean?
I'm trying to do an assignment in my theory of computation class, and its talking about some abstract mapping $\sigma$ as defined  $\Sigma^+ \mapsto \Sigma^*$. I am familar with what $\Sigma^*$ means, as has already been clarified in this question : What does $\{0, 1\}^*$ mean? , however i am very puzzled by what $\Sigma^+$ refers to. Could anyone help me? Here $\Sigma$ is a set $\{0 , 1\}$

答: $\Sigma^+$ is by definition $\Sigma\Sigma^*$, so in your case it would be all non-empty strings of 0s and 1s.

Title: What does $\{0 , 1\}^+$ mean?


问:
How to evaluate $ \int_{0}^{\pi/2} \arctan(\sin(x)) \,dx + \int_{0}^{\pi/4} \arcsin(\tan(x)) \,dx$?
Consider the integral $$ \int_{0}^{\pi/2} \arctan(\sin(x)) \,dx  + \int_{0}^{\pi/4} \arcsin(\tan(x))
\,dx$$

I tried using this substitution for the first integral
$$\arctan(\sin(x))=t \rightarrow x= \arcsin(\tan(t))$$
$$ dx= \frac{1}{\cos(t)*\sqrt{\cos(2t)}} dt$$ and I noticed that I obtained the form of the second integral, but I don't know how to use it.

答:
Proof without words:

(Ok, a very few words: the key observation is that $\arctan(\sin(x))$ and $\arcsin(\tan(x))$ are inverse functions of each other.)


问:
Asymptotics for $y'(t)=y(t-1)$
Let $y : \mathbb{R}\to\mathbb{R}$ be a differentiable function satisfying $y'(t)=y(t-1)$ for all $t$.

Is it possible to give an asymptotics for $y(t)$ as $t\to \infty$?

It is clear to me that there should be an asymptotics of the form $y(t) \sim C\frac{t^{\lfloor t \rfloor +1}}{(\lfloor t\rfloor +1)! }$, inducting on intervals of lenghts 1. How is it possible to prove it?

答:
With the help of the Laplace transform

$$
Y(s) = \frac{\int_{-1}^0 e^{-s t} y(t) \, dt+e^s y(0)}{e^s s-1}
$$

now considering $y(t)=0,\ \ -1\le t\le 0$ and $y(0)=1$

$$
Y(s) =  \frac{1}{s-e^{-s}}
$$

now taking the denominator $d(s) = s-e^{-s}$ and making $s=x + iy$ we have

$$
d(s) =x - e^{-x} \cos (y)+i \left(e^{-x} \sin (y)+y\right)
$$

so the denominator has infinite zeros but we pick one of them which is located at $x = W(1), y = 0$ where $W(\cdot)$ indicates the Lambert function. Here $W(1) = 0.567143$ so $y(t)$ is unstable because $Y(s)$ has at least one pole in the right complex plane hence it grows exponentially.

Title: Asymptotics for $y'(t)=y(t-1)$


问:
Prove that algebra is sigma-algebra
Let's assume that $\mathcal{A}$ is an algebra over $\mathit{X}$, and $\mathcal{A}$ is closed under countable intersections of descending set sequences; if $A_1 \supseteq A_2 \supseteq A_3 \supseteq ... $ and $A_i \in \mathcal{A}, i \in \mathbb{N}$, then $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A}$. Prove that $\mathcal{A}$ is $\sigma$-algebra on $X$.

My attempt:
Let $\forall n \in \mathbb{N} \; B_n \in \mathcal{A}$. Then define $A_n = B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i$. Then $A_n$ is descending family of sets, $A_n \in \mathcal{A}$ as this is algebra. We know that $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A} \leftrightarrow \bigcap\limits^{\infty}_{i=1}(B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i) \in \mathcal{A}$.
Now, I was thinking about using De Morgan's laws, etc., but I can't obtain the result and I don't know if I haven't messed up in the beginning.

答:
You want the unions of countable increasing chains of sets to be in the algebra, and you have decreasing intersections, which in a way, the complement. So given such an increasing chain, think of the chain of complements, show why it is decreasing, etc, consider the obtained intersection, and then look at the complement of the intersection.

In rigorous terms: if ${A_i}$ is the increasing chain, consider $A_i^c$ which is a decreasing chain, in the algebra, and then $\bigcap A_i^c$ is in the algebra. Finish by noting that $(\bigcap A_i^c)^c=\bigcup A_i$

Title: Prove that algebra is sigma-algebra


问:
when a closed map is locally affine?
Exercise 12.3 of the Gortz and Wedhorn book says that if $f:X \to Y$ is a closed map between two schemes and $y\in Y$ is a point such that $f^{-1}(y)$ is contained in some open affine subspace of $X$ then there is an open neighborhood $U$ of $y$ such that the morphism $f:f^{-1}(U)\to U$ is affine.

I can't solve this problem and my main problem is that why if you take an open affine neighborhood of $y$ then I'm not sure why there exist an open affine neighborhood of $f^{-1}(y)$ inside $f^{-1}(U)$. I can prove that this is equivalent to the problem.

this is problem three and we don't even have any finiteness condition so It shouldn't be hard but I don't have any idea. any help is appreciated.

答:
Let $V \subset X$ be an open affine subset containing $f^{-1}(y)$. Then $F=X\backslash V$ be a closed subset of $X$, so that $U_1=Y \backslash f(F)$ is an open subset of $Y$ containing $y$.

Let $U \subset U_1$ be an affine open subset containing $y$: then $f^{-1}(U)=X \times_Y U=V \times_Y U$.

Assume $Y$ is separated, then $V \times_Y U \rightarrow V\times_{\mathbb{Z}} U$ is a closed immersion to an affine scheme so that $V \times_Y U=f^{-1}(U)$ is affine.

Title: when a closed map is locally affine?


答:
You need to show $\mathcal{A}$ is closed under countable unions. We can first show it is closed under countable intersections, and if $A_n, n \in \omega$ are in $\mathcal{A}$ we can write $\bigcap_n A_n = \bigcap_n B_n$ where $B_n = \bigcap_{i=0}^n A_i$ forms a decreasing sequence of subsets of $\mathcal{A}$, and so $\bigcap_n A_n$ is in $\mathcal{A}$.

Now to get unions we apply de Morgan:

$$\bigcup_n A_n = \left( \bigcap_n A_n^\complement \right)^\complement$$

and we already know the algebra is closed under countable intersections...

Title: Prove that algebra is sigma-algebra

Body:
Let's assume that $\mathcal{A}$ is an algebra over $\mathit{X}$, and $\mathcal{A}$ is closed under countable intersections of descending set sequences; if $A_1 \supseteq A_2 \supseteq A_3 \supseteq ... $ and $A_i \in \mathcal{A}, i \in \mathbb{N}$, then $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A}$. Prove that $\mathcal{A}$ is $\sigma$-algebra on $X$.

My attempt:
Let $\forall n \in \mathbb{N} \; B_n \in \mathcal{A}$. Then define $A_n = B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i$. Then $A_n$ is descending family of sets, $A_n \in \mathcal{A}$ as this is algebra. We know that $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A} \leftrightarrow \bigcap\limits^{\infty}_{i=1}(B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i) \in \mathcal{A}$.
Now, I was thinking about using De Morgan's laws, etc., but I can't obtain the result and I don't know if I haven't messed up in the beginning.


问:
Every uniformly continuous function $\mathbb Q\to \mathbb Z$ is constant.
First of all, is continuity not enough?

Secondly, my attempt is since it is uniform continuous.

For $\epsilon=1/2$, for every ball on $\mathbb Z$

i.e. $B(n,\epsilon)\subset \mathbb Z$, with $B(n,\epsilon)\cap \mathbb Z=\{n\}$

on the inverse image $f^{-1}(B(n,\epsilon))$, $f$ is constant.

Since I have balls on $\mathbb Q$ with finite length, I can cover $\mathbb Q$ with infinitely many of them (centered at every rationals) and since every ball (interval) intersects and if two ball intersects their value under $f$ must be the same so I can conclude that $f$ is constant on whole $\mathbb Q$.

I believe the proof is correct (of course some details should be added), is there more elegant/clever way to prove it? Such as in the case for $f:\mathbb R\to \mathbb Z$ case using connectedness?

答:
If all you know about $f$ is that it's continuous, then $f$ may well fail to be constant. Take, for instance$$\begin{array}{rccc}f\colon&\Bbb Q&\longrightarrow&\Bbb Z\\&x&\mapsto&\begin{cases}0&\text{ if }x<\sqrt2\\1&\text{ otherwise.}\end{cases}\end{array}$$So, your proof cannot work, since it does not use uniform continuity.

Now, take an uniformly continuous function $f\colon\Bbb Q\longrightarrow\Bbb Z$. Suppose that there are numbers $a,b\in\Bbb Q$ such that $a<b$ and that $f(a)\ne f(b)$. Let $c=\sup\{x\in[a,b]\mid f(x)\ne f(b)\}$. There is a $\delta>0$ such that$$|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<1.$$The interval $\left(c-\frac\delta2,c+\frac\delta2\right)$ contains rational numbers $x$ such that $f(x)=f(b)$ and contains rational numbers $y$ such that $f(y)\ne f(b)$. But this is impossible, since $|x-y|<\delta$ and $\bigl|f(x)-f(y)\bigr|\geqslant1$.


问:
A Book Errata in a Derivative. I am missing something?
This part is drive me crazy. Or I am missing something in the derivative problem?

My answer:

I will appreciate any help, thank you

Body:
You are correct, differentiating $$y = \frac{z}{2(z+1)}$$ by the quotient rule give us:
$$y' = \frac{2(z+1)(1) - z(2)}{4(z+1)^2}=\frac{2z + 2 - 2z}{4(z+1)^2} = \frac{1}{2(z+1)^2}.
$$


问:
$\frac13n<\sqrt[n]{n!}<\frac12n$

Prove that for every $n\in N_{\ge6}$ the inequality $\frac{1}{3}n<\sqrt[n]{n!}<\frac{1}{2}n$ holds.

My approach was to use induction. I started with the LHS and I noticed that the left inequality holds for every integer that is greater or equal to $1$. The base case is obvious. Now assume that $\frac{1}{3}k<\sqrt[k]{k!}$ holds for some $k > 1$, and let's consider the $n = k+1$ case.
So we want to prove that $\frac{1}{3}(k+1)<\sqrt[k+1]{(k+1)!}$ or $[\frac{1}{3}(k+1)]^{k+1}<(k+1)!$ or $(1+\frac{1}{k})^k<\frac{3^{k+1}k!}{k^n}$, now I wanted to use the fact that $2\le (1+\frac{1}{n})^n < 3$, so if I can prove $\frac{3^{k+1}k!}{k^n}<3$ I am done, but here the inequality goes in a wrong way. What am I missing? Any help appreciated.

答:
Your use of the letters $k$ and $n$ is a bit confusing. It seems to me that the chain of inequalities goes in the right way.

For the induction step, we have to show that
$$n^n<3^n n!\implies (n+1)^{n+1}<3^{n+1}(n+1)!$$
We have that
$$(n+1)^{n+1}=(n+1)n^n\left(1+\frac{1}{n}\right)^n<(n+1)3^nn!\cdot 3=3^{n+1}(n+1)!$$
because $\left(1+\frac{1}{n}\right)^n<3$ for $n>1$.
Similarly, $2^nn!<n^n$ implies
$$(n+1)^{n+1}=(n+1)n^n\left(1+\frac{1}{n}\right)^n>(n+1)2^nn!\cdot 2=2^{n+1}(n+1)!$$
because $\left(1+\frac{1}{n}\right)^n>2$ for $n>1$.

Title: $\frac13n<\sqrt[n]{n!}<\frac12n$


问:
Why can they cancel the denominator after finding the gradient of a secant?

In the image provided my textbook cancels out the denominator after finding the gradient of a secant without writing "undefined for values of $h = 0$". My question is that why can they cancel it out without writing that is undefined at $h = 0$. I know if you were finding the limit then you can cancel it out and the question does require finding the limit. But shouldn't the book write that it is undefined at $h = 0$ when finding the gradient and then leave it out when it is finding the limit?

Thanks

答: You're absolutely correct that generally if there is any ambiguity about whether or not $h = 0$ you should specify that the new expression is undefined at $h = 0$ after the cancellation.  However, I think here they are justified in omitting it.  Since they are asking for the secant line between $(2,8)$ and $(2+h, (2+h)^3)$, there is an underlying assumption that $h>0$ else this wouldn't be a secant line.


问:
Maximal volume for given surface area of an $n$-hedron
Is there a term for a polyhedron with $n$ faces (or, similarly, $n$ vertices) that maximises the enclosed volume for a given surface area (equivalently, minimises the surface area for a given volume)?

The equivalent in polygons produces the regular polygons, and can be considered a property of regular polygons. However, as we all know, there are only five "regular polyhedra" (not counting the star polyhedra), the platonic solids. I expect that these five solids maximise for either their respective numbers of faces or vertices (probably not both, as that would result in the dodecahedron and icosahedron having identical volume for the same surface area, which seems counterintuitive), but I'm interested in what the maximising shape would be for, for instance, a pentahedron.

答:
This will wind up similar to the Thomson problem of determining the minimum energy configuration of N electrons on the surface of a sphere.  </p>

<p>Those point configurations, which are available from the Wikipedia link, can at least provide a starting point. Put hulls around the points, then calculate surface area and volume.  Then perturb the points slightly and recalculate a million times, and see if a better solution pops up.  Repeat until perturbing finds a new local minimum.   My bet is that all the Thomson configurations will already be local minima. Thomson will be one of these 1. always optimal, 2. always nonoptimal, 3. optimal in certain cases.  </p>

Body:
This will wind up similar to the Thomson problem of determining the minimum energy configuration of N electrons on the surface of a sphere.

Those point configurations, which are available from the Wikipedia link, can at least provide a starting point. Put hulls around the points, then calculate surface area and volume.  Then perturb the points slightly and recalculate a million times, and see if a better solution pops up.  Repeat until perturbing finds a new local minimum.   My bet is that all the Thomson configurations will already be local minima. Thomson will be one of these 1. always optimal, 2. always nonoptimal, 3. optimal in certain cases.


问:
Inequality with a twice differentiable function
Let $f:\mathbb R \to \mathbb R$ be a twice differentiable function.

$f''(x)>0$ for every $x\in \mathbb R$.

Prove for every $x\in \mathbb R$:
$\frac{f(x+1)+f(x-1)}{2}>f(x)$

I was trying to use MVT but then I got stuck after not being able to get to the right side of the inequality.

答:
A function with $f''(x)>0$ is strictly convex and that implies the desired conclusion.

For a direct proof consider for a fixed $x \in \Bbb R$ the function
$$
 g(h) = f(x+h) + f(x-h) \, .
$$
Show that

$g''(h) > 0$ for all $h \in \Bbb R$.

$g'(h) > g'(0) = 0$ for $h > 0$.

$f(x+1)+f(x-1) = g(1) > g(0) = 2f(x)$.


问:
Valid inequality with chvatal gomory for problem with binary variables
I am not able to find a valid inequality with chvatal gomory for my problem that increases the value of the relaxed problem.

My problem:

Determine the optimal allocation of n workers to m machines in two instants (t ∈ {1, 2}) in such a way that: (1) each worker is assigned to a single machine, (2) each machine has at least one worker assigned, (3) one worker
cannot be assigned to the same machine in the two instants, (4) the sum of the costs
of assignment of workers to machines does not exceed a fixed value Uc and (5) the objective function is to minimized
the highest allocation cost between a worker and a machine.
enter image description here

答: Please use MathJax rather than linking to an image.  Your formulation is almost correct, but $z > c_{i,j}$ should instead be $z \ge c_{i,j} x_{i,j,t}$, which you can strengthen to $z \ge \sum_t c_{i,j} x_{i,j,t}$ because of $(3)$.


问:
Homomorphisms from $\mathbb{Z}[x] \to R$

Problem: Let $R$ be a ring with identity. Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R.$

I can't seem to find any approach for this. There are no generators of $\mathbb{Z}[x]$. Can we try something else? Any hint is highly appreciated.

Thank you.

答:
The evaluation map at $r \in \mathbb{R}$ from $\mathbb{Z}[x] \to \mathbb{R}$ is a ring homomorphism.

Thus for any $r \in \mathbb{R}$, there exists $p(x) \in \mathbb{Z}[x]$ such that $$p(x) \mapsto p(r)$$  is the ring homomorphism. Hence the number of ring homomorphisms is the cardinality of $\mathbb{R}$.


问:
Why does getting end coordinates using start coordinates, distance and angle seem to fall short?
While working on a problem today related to movement of a ball on a grid I wrote these formulas:

end_x = start_x+ distance * cos( deg2rad(angle))

end_y = start_y + distance * sin( deg2rad(angle))

When it is a simple angle, it works:
start_x=0, start_y=0 angle=0, distance=10 -> end_x = 10, end_y = 0.

But if I run it and change angle to 225 (southwest) it gives me:
end_x = -7.07, end_y = -7.07, which is just under 3 coordinates short.
Because if it's traveling 10 units directly southwest, surely it ends up at -10, -10/

Thanks in advance for any help

答:
You're program is fine. Moving a distance of $10$ in a SW direction is the same as moving $\sqrt 50$ South and $\sqrt 50$ West.

(You'll find it helps to draw a diagram and apply Pythagoras.)


问:
By what fractional part is exceeding?
I was reading the following puzzle:

Look here, George, said his cousin Reginald:  "by what fractional part
does four-fourths exceed three-fourths?"

"By one-fourth!" shouted everybody at once.

"Try another one," George suggested.

"With pleasure, when you have answered that one correctly," was
Reginald’s reply.

According to the solution the correct answer is $\frac{1}3$ because $3$ of anything if increased by $\frac{1}3$ become $4$.

I don't really get it though. I can think of having a set of $N$ objects and somehow consider that each object is $\frac{1}N$ to the set so to add one more object you consider it as equivalent to adding $\frac{1}N$ but that seems a very weird way to state this.

Can someone please explain what this means?

答:
The amount added is $\frac44-\frac34=\frac14$, so the amount by which $\frac44$ exceeds $\frac34$ is $\frac14$. However, the question does not ask for the amount of the excess: it asks what fraction of the original amount of $\frac34$ has been added to make the new amount of $\frac44$. Thus, what’s wanted is

$$\frac{\text{amount added}}{\text{original amount}}=\frac{\frac14}{\frac34}=\frac13\,.$$

Title: By what fractional part is exceeding?


问:
Calculate $\int_{\mathbb{R}} {e^{-x^2}\, d\lambda (x) } $ using Fubini's theorem

Calculate $\int_{\mathbb{R}} {e^{-x^2}\, d\lambda (x) } $ using Fubini's theorem.

A hint is added to the task, suggesting we first calculate: $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)}.$$
I calculated the integral using polar coordinates and using Tonelli's theorem:   $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)} = \pi $$ but I don't know how this hint relates to the integral of the prompt.

答: Just use the fact that\begin{align}\int_{\Bbb R^2}e^{-(x^2+y^2)}\,\mathrm dx\,\mathrm dy&=\int_{\Bbb R}\left(\int_{\Bbb R}e^{-x^2}e^{-y^2}\,\mathrm dy\right)\,\mathrm dx\\&=\left(\int_{\Bbb R}e^{-x^2}\,\mathrm dx\right)\left(\int_{\Bbb R}e^{-y^2}\,\mathrm dy\right).\end{align}


问:
Bianchi Identity in Chern's book
I'm reading Chern's book "complex manifolds without potential theory".

In chapter 5 he claims the following: Let $E\to M$ be a vector bundle, $D$ a connection, $\omega$ the corresponding connection form and $\Omega := D\omega = d\omega -\omega \land \omega$ the curvature form. Then he claims that taking exterior derivative for the equation $\Omega = d\omega - \omega\land \omega$ yields: $$D\Omega = d\Omega + \Omega \land \omega - \omega \land \Omega = 0 $$which is Bianchi identity.

I'm not familiar with these notations, and therefore struggling to understand this equation:

Shouldn't $D\Omega$ be just $d\Omega + \Omega \land \omega$, by the definition of exterior covariant derivative $D\left(\alpha \otimes s
 \right) = d\alpha \otimes s + \left( -1\right)^\left|\alpha\right|\alpha\land Ds$?

Why is the RHS equals zero?

答:
I'll answer in reverse order.

First, recall the definition of curvature: $\Omega = d\omega-\omega\wedge\omega$, so that
$$d\omega = \Omega + \omega\wedge\omega.$$
If you differentiate
$$\Omega - d\omega + \omega\wedge\omega = 0,$$
you get
$$0=d\Omega + (\Omega+\omega\wedge\omega)\wedge\omega - \omega\wedge(\Omega+\omega\wedge\omega) = d\Omega +\Omega\wedge\omega-\omega\wedge\Omega,$$
since the $\omega\wedge\omega\wedge\omega$ terms cancel.

Second, $\Omega$ is a $2$-form with values in $\text{Hom}(E,E)$, so you have to use the formula (5.28) to define the covariant derivative.

Title: Bianchi Identity in Chern's book


问:
Solve the limit $\lim_{x\to 0+} x\int_x^1 \frac{f(t)}{t}dt$
Solve the limit $\displaystyle \lim_{x\to 0+} ~ x\int_x^1 \frac{f(t)}{t}dt$.

$f$ is continuos on $[0,1]$.

I tried using mean value theorem (there exists $c\in[0,1] s.t ...) $to see where $\displaystyle\int_x^1 \frac{f(t)}{t}dt$ as $x\to 0+$ so I get the expression $\displaystyle\int_0^1 \frac{f(t)}{t}dt=f(c)\int_0^1 \frac{dt}{t}$ and then solve the improper integral, which diverges.

I'm not sure if I can apply that theorem, nor how to approach this initial limit.

答:
Just to give another approach, note that if $\int_0^1{f(t)\over t}dt$ converges, then the requested limit is clearly $0$, while if it tends to $\pm\infty$, then L'Hopital and the Fundamental Theorem of Calculus tell us that

$$\lim_{x\to0^+}x\int_x^1{f(t)\over t}\,dt=\lim_{x\to0^+}{\int_x^1{f(t)\over t}\,dt\over1/x}=\lim_{x\to0^+}{-f(x)/x\over-1/x^2}=\lim_{x\to0^+}xf(x)=0\cdot f(0)=0$$


问:
How to evaluate $ \int_{0}^{\pi/2} \arctan(\sin(x)) \,dx + \int_{0}^{\pi/4} \arcsin(\tan(x)) \,dx$?
Consider the integral $$ \int_{0}^{\pi/2} \arctan(\sin(x)) \,dx  + \int_{0}^{\pi/4} \arcsin(\tan(x))
\,dx$$

I tried using this substitution for the first integral
$$\arctan(\sin(x))=t \rightarrow x= \arcsin(\tan(t))$$
$$ dx= \frac{1}{\cos(t)*\sqrt{\cos(2t)}} dt$$ and I noticed that I obtained the form of the second integral, but I don't know how to use it.

Body:
Substitute $ x = \sin^{-1}(\tan t)$. Then

$$ \int_{0}^{\pi/2} \tan^{-1}(\sin x) dx = \int_{0}^{\pi/4} t \>d(\sin^{-1}\tan t)\overset{IBP} = \frac{\pi^2}8
- \int_{0}^{\pi/4} \sin^{-1}(\tan t) dt$$

Thus

$$ \int_{0}^{\pi/2} \tan^{-1}(\sin x) \,dx + \int_{0}^{\pi/4} \sin^{-1}(\tan x)dx = \frac{\pi^2}8
$$


答:
Consider
$$
\int_\Bbb{R\times R}e^{-(x^2+y^2)}d(x\times y)\stackrel{\rm Fubini}{=}\int_\Bbb{R}\int_\Bbb{R}e^{-(x^2+y^2)}d\lambda(x)\,d\lambda(y)=\int_\Bbb{R} \int_\Bbb{R}e^{-x^2}e^{-y^2}d\lambda(x)\,d\lambda(y)=\int_\Bbb{R}e^{-x^2}d\lambda(x) \int_\Bbb{R}e^{-y^2}d\lambda(y)=\left(\int_\Bbb{R}e^{-x^2}d\lambda(x)\right)^2.
$$
Thus:
$$
\Leftrightarrow \int_\Bbb{R}e^{-x^2}d\lambda(x)=\sqrt{\int_\Bbb{R\times R}e^{-(x^2+y^2)}d(x\times y)}=\sqrt{\pi}.
$$

Body:
Calculate $\int_{\mathbb{R}} {e^{-x^2}\, d\lambda (x) } $ using Fubini's theorem.

A hint is added to the task, suggesting we first calculate: $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)}.$$
I calculated the integral using polar coordinates and using Tonelli's theorem:   $$\int_{\mathbb{R}^2} {e^{-(x^2+y^2)}\,d(\lambda \times \lambda)} = \pi $$ but I don't know how this hint relates to the integral of the prompt.


问:
Consider a pair of random variables X, Y with constant joint density on the triangle with vertices at (0, 0), (3, 0), and (0, 3).
a. For 0 ≤ y ≤ 3, find the conditional density $f(x | y)$ of $X$, given $Y = y$.

Using area of $x$ and $y$ region $\cdot$ density $= 1$

I got area of triangle $\cdot$ density $= 1$ AKA $\cfrac{1}{2}\cdot3 \cdot 3 \cdot density = 1$ AKA $4.5 \cdot density = 1 $

so density $= f(x,y) = \cfrac{2}{9}$

Then $f(y) = \int_0^{3-y}\cfrac{2}{9}dx = \cfrac{2}{9}(3-y)$ for $0\leq y \leq 3$

so $f(x|y) = \cfrac{\cfrac{2}{9}}{\cfrac{2}{9}(3-y)} = \cfrac{1}{(3-y)}$ for $0\leq x, 0\leq y, x+y\leq3 $

b. Find the conditional probability that $X ≤ 1$, given $Y = 1$.

$P(X\leq1|Y=1) = \int_0^1\cfrac{1}{3-1}dx=\cfrac{1}{2}$

c. Find the conditional probability that $X ≤ 1$, given $Y ≤ 1$.

$P(X\leq1|Y\leq1) = ???$

I would appreciate it if someone could tell me if I have the bounds and calculations right for parts a) and b) and if someone could give me a hint on how to solve c) I don't know what to do if the given part $Y\leq1$ is an inequality and not $=$ to something, so any tips or help would be appreciated.

Thank You!

答:
It's all correct. By a simple drawing yo get that

$$\mathbb{P}(X\leq 1|Y\leq 1)=\frac{\mathbb{P}(X\leq 1;Y\leq 1)}{\mathbb{P}(Y\leq 1)}=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}$$

Purple area divided by the area of the red trapezoid

Observation:

The conditional density found in (a) is

$$f(x|y)=\frac{1}{3-y}$$

where $y \in[0;3)$ and $x \in [0;3-y]$

This means that $X$ conditioned to a value of a given $y$ is uniform in $[0;3-y]$ thus any probability can be immediately calcualted without integrals


问:
Is $\{t\mapsto \cos(at) \}_{a\in \mathbb{R}^+}$ linearly independent of $\mathcal{F}(\mathbb{R},\mathbb{R})$ over $\mathbb{R}$?
We consider the family  $F=\{t\mapsto \cos(at) \}_{a\in \mathbb{R}^+}$ of functions of space $\mathcal{F}(\mathbb{R},\mathbb{R})$.
Is $F$   linearly independent of $\mathcal{F}(\mathbb{R},\mathbb{R})$ over $\mathbb{R}$?

I know that $\{t\mapsto \cos(at) \}_{a\in \mathbb{R}^+_*}$ is linearly independent over $\mathbb{R}$. But I have no idea about $F$.
An idea please.

答:
I am spelling out the approach I mentioned in the comments.

Let $a_1, ..., a_n \in \mathbb R^+_*$ be distinct. Since you've already shown that $\{t \mapsto \cos(a_j t) \ | \ j = 1,..., n\}$ is linearly independent, suppose that $\lambda_0, ..., \lambda_n$ are real numbers such that

$$\lambda_0 \cdot 1 + \lambda_1 \cos(a_1 t) + ... + \lambda_n \cos(a_n t) = 0 \text{ for all } t \in \mathbb R.$$

In other words, the function $g \colon \mathbb R \to \mathbb R$ defined by

$$t \mapsto \lambda_0 \cdot 1 + \lambda_1 \cos(a_1 t) + ... + \lambda_n \cos(a_n t)$$

is the zero function. This function is differentiable twice with

$$g''(t) = -a_1^2\lambda_1 \cos(a_1t) - ... - a_n^2 \lambda_n \cos(a_n t) = 0.$$

As you've already shown, the condition that $g''(t) = 0$ implies that

$$-a_1^2 \lambda_1 = ... = -a_n^2 \lambda_n = 0,$$

hence $\lambda_1 = ... = \lambda_n = 0$. We obtain $\lambda_0 = 0$, too.


问:
Condition for an integral to be zero
For a bounded function $\operatorname{F}: \mathbb{R}_{\,\ge\ 0} \to \mathbb{R}$ ( not necessarily non-negative ), is it true that
$$
\int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad
\forall s > 0  \iff \operatorname{F} \equiv 0
$$
where $k \in \mathbb{N}$ is a positive constant $?$ Of course, one implication ($\leftarrow$) is true. What about the other one $?$.

答:
Write

$$f_s(x) = \frac{x^ks}{(s^2+x^2)^{(k+3)/2}}$$

Notice that any such function vanishes at infinity.  Let $A$ be the subalgebra of $\textbf{C}_0((0, \infty), \mathbb{R})$ generated by the family $f_s$.  All these functions are positive on $(0, \infty)$ so the family doesn't vanish at any point, and it clearly separates points.

Thus by the locally compact version of the Stone-Weierstrass Theorem, $A$ is dense in $\textbf{C}_0((0, \infty), \mathbb{R})$, and $F$ satisfies your integral equation with $f_s$ replaced by any element of $A$.  Thus as a functional, using the Riesz Representation Theorem (using the uniqueness part) you get that $F \equiv 0$ a.e.

Title: Condition for an integral to be zero


问:
Proving slope inequality when function is concave up
I'm trying to prove the following question:

Suppose that $f$ is concave up on an interval $I$ and $x_1, x_2, x_3$
and $x_4$ are numbers in $I$ with $x_1 < x_2 < x_3 < x_4$. Show that
the slope of the  secant line through the points at $x=x_1$ and
$x=x_3$ is less than the slope of the  secant line through the points
at $x=x_2$ and $x=x_4$.

Now using concavity theorem, I have been able to derive the following equations:

$\dfrac{f(x_3) - f(x_1)}{x_3 - x_1} < \dfrac{f(x_4) - f(x_3)}{x_4 - x_3}$

$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < \dfrac{f(x_4) - f(x_2)}{x_4 - x_2}$

Rewriting the above equations as:

$m_1 < \dfrac{f(x_4) - f(x_3)}{x_4 - x_3}$

$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < m_2$

My objective is to prove that $m_1 < m_2$

I also know that:

$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < \dfrac{f(x_3) - f(x_4)}{x_3 - x_4}$

Now i'm stuck at this point. I'm not exactly sure how to proceed so as
to prove $m_1 < m_2$.

答:
You can prove:

$$m_1 < m < m_2$$
where $m = \dfrac{f(x_3) - f(x_2)}{x_3-x_2}.$ The most elementary way to proceed would be to make use of the following simple fact: if $c,d > 0$ and $\dfrac ac < \dfrac bd,$ then

$$\dfrac{a}{c} < \dfrac{a+b}{c+d} < \dfrac{b}{d}.$$

You can prove the above easily by expanding and then use this for appropriately chosen $a,b,c,d$ to obtain both sides of the equality you want to prove.


问:
Wolfram Alpha: How to show functions with abs() and sgn() as a piecewise function?
I was trying to obtain this result:

```<code>integrate e^|x| dx
</code>```

The result is

$$ \int e^{|x|} dx = \frac12 e^{-x} \left( (e^x-1)^2 \operatorname{sgn}(x) -2e^x +e^{2x} - 1 \right) \color{gray}{\,+ \text{constant}} $$

which is fine but I would like to see the 2 part solution for $x>0$ and $x<0$, I know you can see that because just after before the final output I can see it but then it switches to the one line solution above.
Is there a way to see the other solution?

If it's not ok to ask about a tool used in mathematics I'm sorry. You can close it.

答:
You can use the following to get the two cases:

```<code>Assuming[t>0, integrate e^|x| dx from x=0 to t]
Assuming[t<0, integrate e^|x| dx from x=0 to t]
</code>```

Or, if you don't care about choosing the constant of integration such that the two pieces fit together, simply:

```<code>Assuming[x>0, integrate e^|x| dx]
Assuming[x<0, integrate e^|x| dx]
</code>```

(although then it's so easy that I don't see why you would need Wolfram Alpha).

ContentLicense: CC BY-SA 2.5

Body:
You can use the following to get the two cases:

Assuming[t>0, integrate e^|x| dx from x=0 to t]
Assuming[t<0, integrate e^|x| dx from x=0 to t]

Or, if you don't care about choosing the constant of integration such that the two pieces fit together, simply:

Assuming[x>0, integrate e^|x| dx]
Assuming[x<0, integrate e^|x| dx]

(although then it's so easy that I don't see why you would need Wolfram Alpha).


问:
Riemann integral question
Suppose that $ f : [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $[a,b]$ and $g:[a,b] \rightarrow \mathbb{R}$ differs from $f$ at only one point $x_0 \in [a,b]$, that is, $g(x)=f(x)$ for $x \neq x_0$ and $g(x_0) \neq f(x_0)$. Show that $g$ is Riemann integrable on $[a,b]$.

I'm having a little trouble, my thing was that maybe find a partition and look at how it behaves in the partition containing $x_0$

Appreciate any help

答:
Here is an unnecessarily slick answer:

There is a famous Lebesgue criterion for Riemann integrability of a function $f: [a,b] \rightarrow \mathbb{R}$ (my colleague Roy Smith informs me that it can actually be found already in the work of Riemann!): it is necessary and sufficient that $f$ be bounded and that its set of discontinuities have (Lebesgue!) measure zero.

Given this: it is an easy exercise to show that modifying a function by changing its values at any finite set $S$ does not change its boundedness/unboundedness, and similarly could only create or destroy discontinuities at $x$ for $x \in S$.  So the Lebesgue criterion applies here.   (Beware: changing a function at a countable set of values only can change the continuity at every point: I leave it to the reader to supply the canonical example of this.)

Of course one can -- and should -- also show this directly from the definition of Riemann integrability.

Title: Riemann integral question


问:
Ideal of finite intersection of algebraic sets
In general if $X_1$ and $X_2$ are two algebraic sets on $k^n$ with $k$ a field of characteristic zero, we have that $I( X_1 \cap X_2 ) = \sqrt{ I(X_1) + I(X_2) }.$
Is posible in general compute $I(X_1 \cap X_2 \cap \dots \cap X_n)$ in terms of $I(X_1), \dots , I(x_n)$?.

Thanks in advance.

答:
Remember that for any ring $R$ and ideal $J\subseteq R$,
$$\sqrt{J}=\bigcap_{\large\substack{\text{prime ideals}\\ P\supseteq J}} \!\!\!\!\!\!\!P.$$
Also, remember that for ideals $J_1,J_2,J_3\subseteq R$, we have
$$J_1\supseteq J_2+J_3\iff J_1\supseteq J_2\quad\text{and}\quad J_1\supseteq J_3.$$
Use this to show that for any ideals $J_1,J_2$ of $R$,
$$\sqrt{J_1+\sqrt{J_2}}=\sqrt{J_1+J_2}$$
so that you can use induction to prove what you want.</p>


问:
A question related to Primitive 2n - roots of unity
Consider the following question asked in an assignment worksheet which I am solving by myself.

If n is an odd integer such that K contains a primitive nth root of unity and $ char
K \neq  2 $, then K also contains a primitive 2n th root of unity. Here K is a field.

Let x be primitive n th root of unity. Then x can be written as $ x= y^{d/n}$ , (d,n)=1. n=2m+1 for some integer m. What I need to prove that $y^{d'/2n}$ , where (d', 2n)=1. I have proved that $(-y)^{d/n} \neq -1$   for all the  d lying in {1,..., n-1} but unable to prove that $(-y)^{2n-1 /2n}\neq 1$ which will complete my proof as $(-x)^{2n}=1$ as $x^{n}=1$. I tried to prove that -x is a primitive root of unity.

So, I request your help.

答: As you noted we only need to prove that $(-x)^k\ne1$ for $0<k<2n$. If $(-x)^k=1$ for such a $k$ then $x^{2k}=1$, hence $n\mid 2k$ as $x$ is a primitive $n$-th root. As $n$ is odd we get $n\mid k$ and therefore $n=k$. But then $1=(-x)^k=(-1)^nx^n=-1\ne1$ as are not in characteristic $2$ and $n$ is odd. Hence $(-x)^k\ne1$ for $0<k<2n$, so it follows that $-k$ is a primitive $2n$-th root of unity.


问:
3d quasilinear PDE
I want to solve the following:

$$ (I) \ u_t+e^ue_x=0  \ \ \ \ ; x>0,t>0 \\
(II) \ u(0,x)=2   \ \ \ \ ; x>0 \\
(III) \ u(t,0)=a>0  \ \ \ \ ; t>0$$
I just can't get a solution which satiesfies all of the requirements.

Obviously, a constant function $2$ solves (I) and (II) and a constant function $a$ solves $(I)$ and $(III)$.

Using method of characteristics, I found an implicit solution $u(x,t)= C(x-e^u t)$ but this also doesn't work for $(II)$ and $(III)$.

I am also wondering: is it even possible for a continous function to satisfie both boundary conditions?

答:
$u$ is constant along characteristic curves.

The characteristic curves of (II) with $u=2$ are $x-e^2t=x_0$ for $x_0>0$. They cross the line $x=0$ at $t=-e^{-2}x_0<0$, so never in the quadrant under consideration.

The curves of (III) with $u=a$ are $x-e^at=-e^at_0$ for $t_0>0$. They cross the line $t=0$ at $x=-e^at_0<0$, thus also not in the quadrant.

If $a<2$ the lines from (II) and (III) cross, so that there is a shock front on the line $x=\frac12(e^2+e^a)t$. For $a\ge2$ one gets a dispersion wave filling the remaining sector, if $2\le x/t\le a$ then this point has value $u(x,t)=\ln(x/t)$.

Title: 3d quasilinear PDE


问:
Calculate the Directional Derivative of $f (x, y, z) = x z + \sin (x z + y)$ at $[−4,−2,−2]$ in the direction of the vector $[3, −3, −2]$
I am trying to calculate the directional derivative of the function

$$f (x, y, z) = x z + \sin (x z + y)$$

at the point $[−4, −2, −2]$ in the direction given by the vector $[3, −3, −2]$.

I have done several calculations but the answer seems to be wrong. Can anyone spot where I have done some error or if I have missed any step in the solution? Here are my calculations:

Thanks in advance!

答: It looks like you just have a simple error in the last line: $$-6\cos(6) - 3\cos(6) + 8\cos(6) = -\cos(6)$$ instead of  $$-3\cos(6).$$


问:
Symbols: If but not only if
I was searching for a Latex symbol that indicates $A \Rightarrow B$ and $A \not\Leftarrow B$ ($B$ if not only if $A$, $B$ ifnf $A$). I thought of using $A \Leftrightarrow B$ with the left arrow tick < crossed out. Since I did not find such a symbol:

Is there a Latex symbol for this?

How common or understandable is this symbol?

If it isn't common: How easily is it confused with the symbol $\not\Leftrightarrow$?

Update:

I need it for a sequence $A$ ifnf $B$ ifnf $C$ ifnf $D$, which I find more understandable than $A \Leftarrow B \Leftarrow C \Leftarrow D$  and $A \not\Rightarrow B \not\Rightarrow C \not\Rightarrow D$.

Of course I will prove both directions.

答:
You need to be careful with this connective (in classical logic) because it doesn't let you play fast and loose with the scope of free variables the way that $\to$ and $\leftrightarrow$ do.

I think using words in your expression is probably the best way to go, since it doesn't require additional context to understand.

$$ \varphi \;\;\text{if but not only if}\;\; \psi $$

If you need a compact notation for a chain of these things, I would recommend using plain $<$, but also call out what you are doing explicitly and give a definition of $<$. Comparison chaining should be intuitive to readers without additional comment.

$$ \varphi_1 < \varphi_2 < \varphi_3 $$
$$ \text{where $\xi$ < $\psi$ if and only if
the values of $\xi$ are a proper subset of the values of $\psi$ } $$

Also, if you can collect the things you are talking about into a class or set, then you can use the notation $\subsetneq$ or proper subset or proper subclass to denote the relationship you are interested in.

Title: Symbols: If but not only if


问:
There are pawns at each of two ending cells of 1 × 20 strip
There are pawns at each of two ending cells of 1 × 20 strip. In a single step a player can shift one of the pawns towards the second one by one or two cells. It is forbidden to jump over a pawn. The player who cannot make a step loses the game. Who has a winning strategy - the player who starts the game or his opponent?</p>

答:
The game is equivalent to a simple takeaway game. The game starts with a pile of $18$ stones. Each player in turn must take either one or two stones from the pile; the player who takes the last stone wins, since the other player cannot make a move.

Observe that if there are $3$ stones left, the player whose turn it is must lose: no matter what that player does, the other player can always take the last stone. Thus, if you can leave your opponent $3$ stones, you’ll win. If you leave your opponent $4$ or $5$ stones, however, your opponent can leave you $3$ stones and win. What if you leave your opponent $6$ stones? $9$ stones?


答: In general we have $\sqrt{I+\sqrt{J}}=\sqrt{I+J}$, since both ideals are the smallest radical ideals containing $I$ and $J$. It follows by induction that $I(X_1 \cap \dotsc \cap X_n)=\sqrt{I(X_1) + \dotsc + I(X_n)}$.</p>

Title: Ideal of finite intersection of algebraic sets

Body:
In general if $X_1$ and $X_2$ are two algebraic sets on $k^n$ with $k$ a field of characteristic zero, we have that $I( X_1 \cap X_2 ) = \sqrt{ I(X_1) + I(X_2) }.$
Is posible in general compute $I(X_1 \cap X_2 \cap \dots \cap X_n)$ in terms of $I(X_1), \dots , I(x_n)$?.

Thanks in advance.


问:
What maths concepts should you know to model infectious diseases?
I'm a high school student and for my math research paper, I've to model and predict spread of an infectious disease (e.g. covid) in a country of my choice. Which math concepts should I know to do this activity? I've heard that I should learn differential equations, but is there anything else that I must know?

Could you also suggest some good books to learn how to do this?

Thank you,

答:
Yes, differential equations are at core of it and for finding the parameters, non-linear optimization is a way to go. This makes it easy to jump to a software tool (e.g. Pyomo for Python) and actually solve it.
Actually, here you can find a repo where the SIR-model is used and solved.

When you are talking about designing an optimal strategy on how to deal with pandemics, markov decision processes and dynamic programming comes in handy.

Some good mathematical references:

A biologist's guide to mathematical modeling in ecology and evolution, Otto S.P., Troy D., 2007.

Optimal control of a birth and death epidemic process, Lefevre C., 1981.

Lectures on mathematical modelling of biological systems, Basting G., 2018.


问:
Two variable definite integral
Q) Let $f:R^+\rightarrow R$ be differentiable function with $f(1)=3$ and satisfying
$$\int_{1}^{xy}f(t)dt=y\int_{1}^{x}f(t)dt+x\int_{1}^{y}f(t)dt; \forall x,y\in R^+$$,
then $f(e)=?$

My Attempt:I tried Leibnitz Rule which states that
$$\frac{d}{dx}\int_{\beta(x)}^{\alpha(x)}f(t)dt=f(\alpha(x))\frac{d\alpha(x)}{dx}-f(\beta(x))\frac{d\beta(x)}{dx}$$
but nothing useful seem to come out.

So, please help and $\mathcal THANKS\, IN\, ADVANCE!$

NOTE: ANSWER IS '6'

答:
Differentiating with respect to $x$ gives us $$f(xy)y=yf(x)+\int_1^yf(t)dt$$
Now we differentiate again, this time with respect to $y$, to get: $$f(xy)+f'(xy)xy=f(x)+f(y)$$
Now plug in $x=1$:
\begin{align*}
f(y)+f'(y)y&=3+f(y)\\
\implies f'(y)&=\frac{3}{y}\\
\implies f(y)&=3\log y+C
\end{align*}
As $f(1)=3$ we see $C=3$ and hence $f(e)=3+3=6$.

Title: Two variable definite integral


问:
Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$
Hopefully I am reading the correct line from LMFDB.

Let $a$ be the algebraic number solving $a^4 - a^3  - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3  - x^2 + x + 1)$.

Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.

Body:
Note that
$$
(a^2-a^3)^6-1=(a^6 - 2a^5 + a^4 - a^3 + a^2 + 1)(a^4 - a^3 - a^2 + a + 1)(a^3 - a^2 + 1)(a^3 - a^2 - 1)(a^2 - a + 1),
$$
so that the claim of the title follows.


问:
Upper bound on the difference between two elements of an eigenvector
Let $W$ be the non-negative, symmetric adjacency/affinity matrix for some connected graph. If $W_{ij}$ is large, then vertex $i$ and vertex $j$ have a heavily weighted edge between them. If $W_{ij} = 0$, then no edge connects vertex $i$ to vertex $j$.

Now $L = \mathrm{diag}(W\mathbf{1})-W$ is the (unnormalized) graph Laplacian. Let $v$ be the Fiedler vector of $L$, that is, a unit eigenvector corresponding to the second smallest eigenvalue of $L$. As $W_{ij}$ increases, all else equal, $|v_i - v_j|$ tends to decrease---at least this is the idea behind spectral clustering. What is an upper bound on $|v_i - v_j|$, given quantities that don't require computing $v$, like $W_{ij}$ and $\|W\|$?

Any suggestions or thoughts would be greatly appreciated.

答:
I've seen peopl use Davis and Kahan (1970), "The Rotation of Eigenvectors by a Perturbation". It's sometimes a bit tough going, but incredibly useful for problems like this.

More info would also be useful. Is $W$ stochastic? Are there underlying latent classes that control the distribution of $W_{ij}$, e.g., with $E(W_{ij})$ larger if $i$ and $j$ possess the same class? If so, then I suggest first considering $E(W_{ij})$ with the latent classes given. Reorder the rows and columns according to latent class. You'll end up with a block matrix with entries constant on each block. It's not that difficult to then compute eigenvalues and eigenvectors of this block matrix. You'll find the eigenvector entries for the same latent class are the same. Now perturb the matrix from $E(W)$ back to the original $W$ and use Davis and Kahan to bound differences in the eigenvector entries.


问:
Show that continuous functions on $\mathbb R$ are Borel-measurable
Here are my thoughts. We want to show that $f^{-1}$ maps Borel sets to Borel sets.

Let $f:\mathbb R\to\mathbb R$ be a continuous function and $\mathcal B(\mathbb R)$ a Borel $\sigma$-algebra. Let $B\in\mathcal B(\mathbb R)$ and $B$ open. Then $f^{-1}(B)$ is open as well, that is, there exists neighborhoods around every point of $f^{-1}(B)$ contained within. Since these neighborhoods are members of $\mathcal B(\mathbb R)$, their union are also members of $\mathcal B(\mathbb R)$. Hence $f^{-1}(B)$ is a member of $\mathcal B(\mathbb R)$.

Here's where I'm stuck. I don't know how to show countably many such members can have their unions taken to give me $f^{-1}(B)$. I also don't know what to do with the closed sets in $\mathcal B(\mathbb R)$, especially singletons and those only generatable via complements.

Hints?

答:
One thought that helped me understand the statement, and may be useful is:

Let $f:\mathbb R\to\mathbb R$ be a continuous function and $\mathcal B(\mathbb R)$ a Borel $\sigma$-algebra. The ($b$,+$\infty$) is an open set (It can be written as $\cup^{\infty}_{n=1}$B($x_0$,$\epsilon$$_x$)) therefore ($b$,+$\infty$)$\in$ $\mathcal B(\mathbb R)$. Then [-$\infty$,$b$] = $\mathbb R$$\setminus$($b$,+$\infty$) = ($b$,+$\infty$)$^c$$\in$ $\mathcal B(\mathbb R)$ (because it is a $\sigma$-algebra).

We already know that $f$=continuous and [-$\infty$,$b$] = closed $\in$ $\mathcal B(\mathbb R)$ $\Rightarrow$ $f$$^{-1}$([-$\infty$,$b$]) = closed $\in$ $\mathcal B(\mathbb R)$ and since* [$f$ $\leq$ $b$] = $f$$^{-1}$([-$\infty$,$b$])$\in$ $\mathcal B(\mathbb R)$ then the "$f$", by definition, is Borel-measurable.

PS1:* [$f$ $\leq$ $b$] = $f$$^{-1}$([-$\infty$,$b$]) = {$x$$\in$$\mathbb R$: $f(x)$ $\leq$ $b$}.


问:
Permutations of race positions

Recently, I got this math question:

Alice, Bob, Caroline, and Derek are competing on a 400 m run.

a) How many different ways are there for them to arrive if participants can have joint places?

b) Is there a way to generalize this mathematically?

I started on a) then got the number 75 after laboring for hours, but I still don't know how to start on b).

I made a table of different possibilities if $n$ is the number of participants and $P(n)$ is the function to get the number of arrangements of the participants:

$n$
$P(n)$

0
1

1
1

2
3

3
13

4
75

but I still can't seem to generalise this (even though I keep getting super close).
Could you please help me?  I've been thinking about this for hours, but I can't get it...

答:
Suppose there are $n$ racers. We consider the following cases:

If there is one winner, there are $\binom{n}{1}$ ways to choose the winner and $P(n-1)$ ways to order the rest.

If there are two (tied) winners, there are $\binom{n}{2}$ ways to choose the two winners, and $P(n-2)$ ways to order the rest

$\vdots$

If there are $n$ (tied) winners, there are $\binom{n}{n}$ ways to choose the $n$ winners, and $P(0)$ ways to order the rest.

So we get the recurrence relation
$$P(n)=\sum_{k=1} ^n\binom{n}{k}P(n-k).$$
Starting with $P(0)=1$, you can use this formula to get all the values for $P(n)$. For example, for $P(5)$ (the smallest value that you haven't yet computed), we have

$$\begin{align*}
P(5)&=\binom{5}{1}\times75+\binom{5}{2}\times13+\binom{5}{3}\times3\\
&\qquad+\binom{5}{4}\times1+\binom{5}{5}\times1\\
&=541.
\end{align*}$$

The sequence $P(n)$ being described is called the ordered Bell numbers.

Title: Permutations of race positions


答:
Say that you have $n$ contestants, and they finish in $k$ groups. There are $n\brace k$ ways to partition the $n$ contestants into $k$ non-empty sets, and the $k$ sets can be permuted in $k!$ different orders, so there are $k!{n\brace k}$ possible ways for the contestants to finish in $k$ groups. Here $n\brace k$ is a Stirling number of the second kind. Thus,

$$P(n)=\sum_{k=0}^nk!{n\brace k}\,.$$

(I include $k=0$ to cover the case $n=0$: ${0\brace 0}=1$, and ${n\brace 0}=0$ for $n\ge 1$.)

These are the so-called Fubini numbers, OEIS A000670; you’ll find many references at the link.

Title: Permutations of race positions

Body:
Recently, I got this math question:

Alice, Bob, Caroline, and Derek are competing on a 400 m run.

a) How many different ways are there for them to arrive if participants can have joint places?

b) Is there a way to generalize this mathematically?

I started on a) then got the number 75 after laboring for hours, but I still don't know how to start on b).

I made a table of different possibilities if $n$ is the number of participants and $P(n)$ is the function to get the number of arrangements of the participants:

$n$
$P(n)$

0
1

1
1

2
3

3
13

4
75

but I still can't seem to generalise this (even though I keep getting super close).
Could you please help me?  I've been thinking about this for hours, but I can't get it...


问:
$|G| = pqr^2$, $G$ not simple
If $G$ is a group with cardinality $pqr^2$ and $p,q,r$ different primes $\ge 2$, and given $1+np$ doesn't divide $qr^2$ for any $n \in \mathbb{Z}^+$,

How can you show $G$ is not simple?

I feel like you would use Sylow's Theorem but i'm not sure how to apply it to this scenario.

答:
Let $n_p$ be the number of Sylow $p$-subgroups. By the Sylow theorems we have \begin{align*}
n_p&\equiv 1\pmod p\\
n_p&\mid qr^2
\end{align*}
By the first line we can write $n_p=1+np$ for some $n\in\Bbb N_0$.  By assumption $1+np\nmid qr^2$ if $n>0$, hence $n=0$, i.e. $G$ has a unique Sylow $p$-subgroup and is therefore not simple.

Title: $|G| = pqr^2$, $G$ not simple


问:
Q: Let $X$ be a continous random variable s.t. $X\sim U(-1,3)$. Find $F_Y$, for $Y=X^4$.
Q: Let $X$ be a continous random variable s.t. $X\sim U(-1,3)$. Find $F_Y$, for $Y=X^4$.

My attempts so far:

$F_Y(t)=P(Y\le t)=P(X^4\le t)=P(-\sqrt[4]{t}\le X\le\sqrt[4]{t})=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.

Since $X\sim U(-1,3)$
then $F_X(t)=\begin{cases}0&,  t<-1 \\ \frac{t+1}{4}&,  -1\le t\le 3 \\ 1&, 3<t &\end{cases}$

What I don't understand is how find the right way to split $F_Y(t)=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.

答:
To go on first observe that $Y$ support is $y \in [0;3^4]$

Do a drawing of the transformation function

and observe that where $y \in [0;1]$ the transformation is not monotonic and thus

$$F_Y(y)=F_X[\sqrt[4]{y}]-F_X[-\sqrt[4]{y}]=\frac{\sqrt[4]{y}}{2}$$

In the other part of the support, the transformation is monotonic thus

$$F_Y(y)=F_X[\sqrt[4]{y}]=\frac{\sqrt[4]{y}+1}{4}$$

Concluding...

$$  F_Y(y) =
\begin{cases}
0,  & \text{if $y<0$ } \\
\frac{\sqrt[4]{y}}{2},  & \text{if $0 \leq y<1$ } \\
\frac{\sqrt[4]{y}+1}{4},  & \text{if $1 \leq y<3^4$ } \\
1,  & \text{if $ y\geq 3^4$ }
\end{cases}$$


问:
Trace of matrices is equal to zero or one
I want to prove that if matrices $E_{i,j}$ in $\mathbb{K}^{m,n}$ are matrices with all zeros except for one 1 on the $i,j$ index, then $\text{tr}((E_{i,j})^TE_{k,l})$ equals $1$ if $E_{i,j}=E_{k,l}$ and 0 in other cases.
I have an intuition that this is true but I don't know how to formally prove it.

答: Given a matrix $A \in {\rm Mat}(n\times m, \Bbb R)$, let $\widetilde{A} \in \Bbb R^{nm}$ be the vector whose entries are the entries of $A$, listed row after row. You can show (using the definitions of trace and matrix multiplication) that ${\rm tr}(A^TB)$ equals the inner product of $\widetilde{A}$ and $\widetilde{B}$ in $\Bbb R^{nm}$. Very obviously the matrices $E_{ij}$ correspond to the standard basis of $\Bbb R^{nm}$, and the standard basis is orthonormal.


问:
Finding gradient of $\det(\mathbf{\Sigma})^{\frac{-1}{2}}$ with respect to $\mathbf{\Sigma}$ (a Covariance matrix)
In the book Mathematics for Machine Learning (chapter on Gaussian Mixture Models), I got stuck trying to figure out how they came up with

$$\frac{\partial \det(\mathbf{\Sigma})^{\frac{-1}{2}}}{\partial \mathbf{\Sigma}} = \frac{-1}{2}\det(\mathbf{\Sigma})^\frac{-1}{2} \mathbf{\Sigma}^{-1}$$

where $\mathbf{\Sigma}$ is the covariance matrix. The book mentions that the following identity was used:
$$\frac{\partial \det(f(\mathbf{X}))}{\partial \mathbf{X}} = \det(f(\mathbf{X}))Trace\left( f(\mathbf{X})^{-1} \frac{\partial f(\mathbf{X})}{\partial \mathbf{X}} \right)$$

But the question is, how did they use it to arrive at the first equation ? I can't seem to figure it out. Been stuck for quite a while.

Thanks!

答: Your question has two parts: an apparent matrix/scalar mismatch in the general result, and how to choose and substitute a suitable $f$. To address the first part, I'll assume your $\operatorname{Tr}$ shouldn't be in the general result; after all, you've already edited it out of the special case of interest. So onto the second part: if we take $f(y)=y^{-1/2}$ so $f^\prime(y)=-\tfrac12y^{-3/2}$, the putative identity gives$$\partial_\Sigma(\det \Sigma)^{-1/2}=\det(\Sigma^{-1/2})\operatorname{Tr}(\Sigma^{1/2}\cdot-\tfrac12\Sigma^{-3/2})=-\tfrac12(\det\Sigma)^{-1/2}\Sigma^{-1}.$$


答:
Let's step back for a minute and think about what happens to $Y$ for various values of $X$.

When $-1 \le X \le 1$, then $0 \le Y \le 1$.  So when we pick some $y$ between $0$ and $1$, $$\Pr[Y \le y] = \Pr[-y^{1/4} \le X \le y^{1/4}].$$  But when $1 < y \le 3^4$, instead of writing $-y^{1/4} \le X \le y^{1/4}$, the left-hand side inequality stops at $-1$ because that's where the support of $X$ ends.  And when $y > 81$, then both left and right sides of the inequality becomes $-1 \le X \le 3$.  So in summary we have $$\Pr[Y \le y] = \begin{cases} 0, & y < 0 \\
\Pr[-y^{1/4} \le X \le y^{1/4}], & 0 \le y \le 1 \\
\Pr[-1 \le X \le y^{1/4}], & 1 < y \le 81 \\
1, & y > 81. \end{cases}$$  Now there are no further restrictions and we can proceed with the evaluation of each case.

Body:
Q: Let $X$ be a continous random variable s.t. $X\sim U(-1,3)$. Find $F_Y$, for $Y=X^4$.

My attempts so far:

$F_Y(t)=P(Y\le t)=P(X^4\le t)=P(-\sqrt[4]{t}\le X\le\sqrt[4]{t})=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.

Since $X\sim U(-1,3)$
then $F_X(t)=\begin{cases}0&,  t<-1 \\ \frac{t+1}{4}&,  -1\le t\le 3 \\ 1&, 3<t &\end{cases}$

What I don't understand is how find the right way to split $F_Y(t)=F_X(\sqrt[4]{t})-F_X(-\sqrt[4]{t})$.


问:
Improper integral convergence with parameter
Hello I need to examine the convergence of this followintg improper integral:

$\int_{0}^{\infty} \frac{1-e^{-a^{2} x^{2}}}{x^{2}}$

If I understand it correctly, I have to examine it in points where it is not continuous which is 0 and infinity. In infinity I can compare it with: $\frac{1}{x^{2}}$ and that is convergent for all values of parameter "a".

In my materials it is written that it has $\lim _{x \rightarrow 0}$ = -a^2 but I do not understand why this is sufficient? It would help me greatly if someone could explain to me what are other options for examining covergence other than comparisons.

Thank you very much.

答:
hint

Let $$f(x)=\frac{1-e^{-a^2x^2}}{x^2}$$

$ f $ is continuous and locally integrable at $ (0,+\infty)$.

near 0

$$\lim_{x\to 0^+}f(x)=a^2 \implies$$
$$\int_0^1f(x)dx \;\;converges$$
near infinity

$$\lim_{x\to+\infty} x^2f(x)=\color{red}{1}\implies$$

with $ \epsilon=\frac 12 $ and for $ x $ great enough,
$$\color{red}{1}-\epsilon<x^2f(x)<\color{red}{1}+\epsilon=\frac 32$$

or

$$0<f(x)<\frac{3}{2x^2}$$
thus,
$$\int_1^{+\infty}f(x)dx\;\; converges$$

we conclude that
$$\int_0^{+\infty}f(x)dx\;\;\text{ is convergent}$$


问:
complex operator
I have asked this question before. I deleted it as no one answered it and I received negative feedback. Though I didn't understand what is going on now I think I do. I just want to verify my solution and also ask about complex structure on $T_U,u$. This is my question again.

I started learning things from Claire Voisin. I am confused about how the author arrived at 1.3 from 1.1 (partial Solution below). I am also confused about the natural complex structure on , can someone explain that part as well?

We know that $\frac{\partial}{\partial z} = a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y}$.

Therefore

$\frac{\partial x}{\partial z} = a$.

We also know that $a = dx = \frac{1}{2}(dz + d\bar{z})$. Similarly we know that $b = dy = \frac{1}{2i}(dz - d\bar{z})$.

Therefore $\frac{\partial}{\partial z} = \frac{1}{2}(dz + d\bar{z}) - \frac{i}{2}(dz - d\bar{z}) = \frac{1}{2} (dz - i dz) + \frac{1}{2}(d\bar{z} + id\bar{z})$.

I think I am almost there though little bit confused.

答:
$\frac{\partial}{\partial z} = \frac{1}{2}(dz + d\bar{z}) - \frac{i}{2}(dz - d\bar{z})$ is nonsense (LHS is a differential operator, RHS is a differential form and they are dual to one another but not the same type of objects) and denotes a lack of understanding of basic differential calculus.

For the derivation of 1.3 from 1.1, one notices that 1.1 implies $dz=dx+idy, d\bar z=dx-idy$ so using that by definition we must have $\frac{\partial}{\partial z}$ is the dual of $dz$ and all the rest of similar duality results between the other $3$ variables $x,y, \bar z$, we need to invert the matrix of change of variables from $dx,dy$ to $dz, d\bar z$ and get 1.3

Or if matrix computation is mysterious, just do it by hand with $\frac{\partial}{\partial z}=a\frac{\partial}{\partial x}+bi\frac{\partial}{\partial y}$ using $<\frac{\partial}{\partial z}, dz>=1, <\frac{\partial}{\partial z}, d\bar z>=0$ and get $a-b=1, a+b=0$ giving $a=1/2, b=-1/2$ as expected; same for the other one, or just use conjugation noticing that $\overline {\frac{\partial}{\partial z}}=\frac{\partial}{\partial \bar z}$ while $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$ are real operators so conjugate invariant and only the $i$ changes sign

Title: complex operator


问:
On the definition of cohomological dimension
Let $G$ be a group and $R$ a commutative unital ring. We define the $R$-cohomological dimension of $G$ to be
$$cd_R(G) := \sup \{ n : H^n(G, M) \neq 0 \text{ for some } R[G]\text{-module } M \}.$$
I have read somewhere that this may be equivalently defined as
$$cd_R'(G) := \sup \{ n : H^n(G, R[G]) \neq 0 \},$$
but in a recent discussion on a previous post it was pointed out to me that this is only true if $cd_R(G) < \infty$. Is this the case? What are some counterexamples? Could you point me to a reference?

I would like to understand this in general but I am especially interested in a specific case: $G$ is finitely presented, $R$ is finite, and $cd_R'(G) \leq 1$. Can I deduce in this case that $cd_R(G) = cd_R'(G)$, or even just that $cd_R(G) \leq 1$ as well?

答:
One class of groups that have $\operatorname{cd}_R(G)=\infty$ but $\operatorname{cd'}_R(G)<\infty$ are groups with a finite index subgroup $H$ such that $\operatorname{cd}_R(G)<\infty$, and with an element $x$ of finite order whose order is not invertible in $R$. Such groups necessarily have torsion. I don't know whether there are other examples that are torsion free, but there may be examples that are well known to experts.

For example, this includes (taking $H$ to be the trivial subgroup) all finite groups $G$ with $|G|$ not invertible in $R$. But also many infinite groups.

The references in the following proof are to Brown's Cohomology of Groups. Actually I state results for general $R$ that are only stated for the case $R=\mathbb{Z}$ in Brown, but the generalizations are straightforward.

First, a group with $\operatorname{cd}_R(G)<\infty$ has no elements $x$ with finite order not invertible in $R$ [Corollary (2.5). In the case $R=\mathbb{Z}$ this just says it is torsion free, and relies on the fact that $H^n(C,\mathbb{Z})$ is nonzero for infinitely many $n$, where $C=\langle x\rangle$ is the cyclic subgroup generated by $x$. But if $|C|$ is not invertible in $R$ then $H^n(C,R)$ is nonzero for infinitely many $n$, and the proof goes through.]

Next, by Shapiro's Lemma [Proposition (6.2)],
$$H^n(G,\operatorname{coind}^G_H R[H])\cong H^n(H, R[H]),$$
which is zero for all but finitely many $n$, and since $|G:H|<\infty$, $\operatorname{coind}^G_H R[H]\cong R[G]$, so $\operatorname{cd'}_R(G)<\infty$.

One positive result is [Proposition (2.3)]:
$$\operatorname{cd}_R(G)=\sup\left\{n\mid H^n(G,F)\neq0\text{ for some free $R[G]$ module $F$}\right\}$$
if $\operatorname{cd}_R(G)<\infty$,
but I don't see why this free module could necessarily be taken to be $R[G]$. Maybe there are examples where $F$ necessarily has infinite rank that are well known to experts.


问:
Centroid of the Region bounded by the functions: $y = x, x = \frac{64}{y^2}$, and $y = 8$.
Can someone help and teach me how to solve this problem?

Find the centroid of the region bounded by the graphs of $$\begin{align}
&y = x\\
&x = \frac{64}{y^2}\\
&y = 8
\end{align}$$

I'm having trouble solving this problem, please help. Thank you!

I currently don't know how to solve this problem, if someone would show a step-by-step solution would be very much appreciated!

答:
HINT

We can find the $x$ and $y$ coordianates of the centroid separately. There are several formulae that we can use:
$$\sum m_i x_i=\bar x\sum m_i$$
$$\sum m_i y_i=\bar y\sum m_i$$
$$\bar x=\frac{\int_a^bxy ~dx}{\int_a^b y ~dx}$$
$$\bar y=\frac{\int_a^b\frac{1}{2}y^2dx}{\int_a^b y ~dx}$$
If that doesn't help, or if you have any questions, I will be happy to elaborate.


问:
Is it possible to make integers a field?
Is it possible to define addition and/or multiplication on the set of

a) natural numbers (including $0$: $0,1,2,3,...$)

b) integers $(..., -2, -1, 0, 1, 2, ...)$

in such way that they will become fields?

Thanks in advance.

答:
Let $X$ be any countably infinite set (such as $\mathbb{Z}$ or $\mathbb{N}$). Choose any bijection $f:X\to\mathbb{Q}$.  Define operations $\oplus,\otimes$ on $X$ by
$$a\oplus b=f^{-1}(f(a)+f(b)),\qquad a\otimes b=f^{-1}(f(a)\cdot f(b))$$
where $+$ and $\cdot$ are the operations on $\mathbb{Q}$. Then these operations make $X$ into a field, indeed one that is isomorphic to $\mathbb{Q}$.</p>


问:
Prove that $A \subseteq B$ and $B$ is an ideal of $R$.
Let $R$ be a commutative ring with unity, let $A$ be an ideal of $R$ and let $a \in R$. Define $B = \{b \in R ~|~ ab \in A \}$.

Prove that $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$.

$\textbf{My Attempt:}$

Prove of $A \subseteq B$:

Since, $A$ is an ideal of $R$.

Then, by definition of ideal, $A$ is a subring $R$ and $A \subseteq R$.

Also, by definition of ideal, $A$ is an ideal of $R$, if for all $r \in R$ and for all $x \in A$ both $rx$ and $xr$ are in $A$.

Since, $R$ is a commutative ring with unity. Then, $A$ is a commutative ring with unity as well.

So, $rx = xr$ is in $A$.

Since, $A \subseteq R$ then by def of subset, for all $x \in A$ $\implies$ $x \in R$.

So, for all $r \in R$ and for all $x \in R$ such that $rx = xr$ is in $A$.

Since, we defined $B = \{ b \in R ~|~ ab \in A \}$. So, $rx = xr$ is also in $B$.

So, this proves that elements that in $A$ is also in $B$. Which means that $A \subseteq B$.

Prove of $B$ is an ideal of $R$:
Let $y \in B$ and since we let $a \in R$

Since, we define $B = \{ b \in R ~|~ ab \in A \}$. So, $y \in R$ and $ay \in A$.

Since, $A \subseteq R$. So, $ay \in R$.

So, we have that $B \subseteq R$.

Then, we need to apply the ideal test.

ideal test case (1):

Let $u,v \in B$. WTP: $u - v \in B$.

Since, $u, v \in B$. Then, $u, v \in R$ and $au \in A$ and $av \in A$.

Since, $R$ is a commutative ring with unity.

Then, by def of commutative ring, $u - v \in R$ as well.

Since, $A$ is an ideal of $R$. Then, by definition of ideal, $A$ is a subring $R$.

Which means $A$ is a commutative ring with unity as well.

Since, $au \in A$ and $av \in A$.

So, $au - av = a(u - v)$ which $a(u - v) \in A$.

Since, we prove that if $u, v \in B$ both $u - v \in R$ and $a(u - v) \in A$.

So, $u - v \in B$ which means ideal test case (1) passed.

ideal test case (2):

Let $\beta \in B$ and $\gamma \in R$. WTP: both $\gamma\beta$ and $\beta\gamma$ are in $B$.

Since, $\beta \in B$. Then, $\beta \in R$ and $a\beta \in A$.

Since, both $R$ and $A$ are commutative ring with unity. So, $\gamma \beta = \beta \gamma \in R$.

Since, $A$ is an ideal of $R$ and $\gamma \in R$ and $a\beta \in A$.

So, $a\beta\gamma = a\gamma\beta \in A$.

So, we proved that both $\gamma\beta$ and $\beta\gamma$ are in $B$. Which means ideal test case (2) passed.

Therefore, we proved that $B$ is an ideal of $R$.

$\textbf{Questions:}$

I don't feel my prove of both $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$ are quite right. Are there any other ways to prove better than mine ? Or are there anyting I can fixes for my prove to make it better ?

答:
Your proves look quite complicated. I would do something like that.

As $R$ is supposed to be commutative, it is enough for proving that $I \subseteq R$ is an ideal to prove that $I$ is a left ideal.

Take $x \in A$. As $A$ is a (left) ideal, we have $ax \in A$. Which implies $x \in B$ by definition of $B$. This proves that $A \subseteq B$.

Let's now prove that $B$ is a left ideal of $R$.

$b,b^\prime \in B$ mean that $ab, ab^\prime \in A$ and therefore
$$ab + ab^\prime = a(b + b^\prime) \in A$$ as $A$ is an ideal. And bydefinition of $B$, $b+b^\prime$ belongs to $B$.

Also if $b \in B$ and $x \in R$, we have $ab \in A$ (by definition of $B$) and $rb \in B$ as

$$a(rb)= r(ab) \in A$$ because $R$ is supposed to be commutative and $A$ to be an ideal.

This concludes the proves of the requested questions.


问:
Proving that $\min\limits_{x \neq 0} \frac{\|Ax\|_2}{\|x\|_2}=\sigma_n$
For $A \in \mathbb R^{m \times n}, m \geq n, x \in \mathbb R^{n \times 1}$, how to prove $\min\limits_{x \neq 0} \frac{\|Ax\|_2}{\|x\|_2}=\sigma_n$? As I'm new to SVD, can anyone help me in solving this problem?</p>

答:
The most direct way to prove it, without any "educated guesses", is to solve the constrained optimization problem via Lagrange Multipliers. First, we rewrite the problem in standard form:

$$ \min_x \underbrace{\|Ax\|_2}_{f(x)} \qquad \text{s.t.}\qquad \underbrace{\|x\|_2-1}_{g(x)}=0$$

To make our life easier, we will consider the squared version $f(x)=\|Ax\|^2_2$, $g(x) = \|x\|_2^2-1$, which is equivalent (though you can also work with the original equation). Then the Lagrangian is:

$$\mathcal L(x, \lambda) = \|Ax\|_2^2 - \lambda (\|x\|_2^2-1)$$

And the first order necessary conditions imply

$$0\overset{!}{=}\frac{\partial}{\partial x} \mathcal L(x, \lambda) = 2A^T Ax-2\lambda x \iff A^T A x= \lambda x$$

So the optimum is obtained for an eigenvector corresponding to an eigenvalue of $A^TA$. I think you can finish from here.


Title: Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$

Body:
Hopefully I am reading the correct line from LMFDB.

Let $a$ be the algebraic number solving $a^4 - a^3  - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3  - x^2 + x + 1)$.

Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.

Body:
Note

\begin{align}
\frac{(-a^3 + a^2)^6 - 1 }{(-a^3 + a^2)^3 - 1  }&
=(-a^3 + a^2)^3 +1 = (-a^3 + a^2 +1 )[  (a^3 -a^2)^2 + (a^3 - a^2)+ 1]\tag1\\
\end{align}
where
\begin{align}
&  (a^3 -a^2)^2 +(a^3 - a^2)+ 1 \\
= & (a^3 -a^2+a-a)^2 +(a^3 - a^2+a -a)+ 1 \\
=& (a^3 -a^2+a)^2 -2a (a^3 -a^2+a) +(a^3 - a^2+a ) +a^2-a+ 1 \\
=& (a^2 -a+1)(a^4-a^3-a^2+a+1)=0
\end{align}
Plug into (1) to obtain
$$(-a^3 + a^2)^6 =1$$


答:
It’s correct, but it’s much wordier than necessary. For instance, the proof that $A\subseteq B$ could be reduced to this:

Let $x\in A$. $A$ is an ideal, so $ax\in A$, and that by definition means that $x\in B$. Thus, $A\subseteq B$.

In fact it can even be reduced to this:

$aA\subseteq RA=A$, so $A\subseteq B$.

Here $RA=\{rx:r\in R\text{ and }x\in A\}$. $RA\subseteq A$ because $A$ is an ideal, and $RA=A$ because $R$ is a ring with unity.

The proof that $B$ is an ideal can also be drastically shortened.

Let $b_0,b_1\in B$. Then $a(b_0-b_1)=ab_0-ab_1$, where $ab_0,ab_1\in A$. $A$ is a subring of $R$, so $ab_0-ab_1\in A$, and therefore $b_0-b_1\in B$, i.e., $B$ is closed under subtraction.

Now let $b\in B$ and $r\in R$. Then $a(br)=(ab)r$, where $ab\in A$, and $A$ is an ideal of $R$, so $(ab)r\in A$ as well. Thus, $br\in B$, and since $R$ is commutative, $rb$ is also in $B$, and we’ve shown that $B$ is an ideal of $R$.

Indeed, if I were writing it up for myself, I’d shorten it even further:

If $b_0,b_1\in B$, then $a(b_0-b_1)=ab_0-ab_1\in A-A=A$, so $b_0-b_1\in B$. If $b\in B$ and $r\in R$, then $a(br)=(ab)r\in Ar\subseteq A$, so $rb=br\in B$. Thus, $B$ is an ideal.

When you’re just learning, it’s generally better to err on the side of giving more detail than necessary, but it’s also a good idea to work on cutting out parts that really can safely be left to the reader: including too much detail makes it harder to read. With practice one learns to strike an appropriate balance for a given audience.

Title: Prove that $A \subseteq B$ and $B$ is an ideal of $R$.

Body:
Let $R$ be a commutative ring with unity, let $A$ be an ideal of $R$ and let $a \in R$. Define $B = \{b \in R ~|~ ab \in A \}$.

Prove that $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$.

$\textbf{My Attempt:}$

Prove of $A \subseteq B$:

Since, $A$ is an ideal of $R$.

Then, by definition of ideal, $A$ is a subring $R$ and $A \subseteq R$.

Also, by definition of ideal, $A$ is an ideal of $R$, if for all $r \in R$ and for all $x \in A$ both $rx$ and $xr$ are in $A$.

Since, $R$ is a commutative ring with unity. Then, $A$ is a commutative ring with unity as well.

So, $rx = xr$ is in $A$.

Since, $A \subseteq R$ then by def of subset, for all $x \in A$ $\implies$ $x \in R$.

So, for all $r \in R$ and for all $x \in R$ such that $rx = xr$ is in $A$.

Since, we defined $B = \{ b \in R ~|~ ab \in A \}$. So, $rx = xr$ is also in $B$.

So, this proves that elements that in $A$ is also in $B$. Which means that $A \subseteq B$.

Prove of $B$ is an ideal of $R$:
Let $y \in B$ and since we let $a \in R$

Since, we define $B = \{ b \in R ~|~ ab \in A \}$. So, $y \in R$ and $ay \in A$.

Since, $A \subseteq R$. So, $ay \in R$.

So, we have that $B \subseteq R$.

Then, we need to apply the ideal test.

ideal test case (1):

Let $u,v \in B$. WTP: $u - v \in B$.

Since, $u, v \in B$. Then, $u, v \in R$ and $au \in A$ and $av \in A$.

Since, $R$ is a commutative ring with unity.

Then, by def of commutative ring, $u - v \in R$ as well.

Since, $A$ is an ideal of $R$. Then, by definition of ideal, $A$ is a subring $R$.

Which means $A$ is a commutative ring with unity as well.

Since, $au \in A$ and $av \in A$.

So, $au - av = a(u - v)$ which $a(u - v) \in A$.

Since, we prove that if $u, v \in B$ both $u - v \in R$ and $a(u - v) \in A$.

So, $u - v \in B$ which means ideal test case (1) passed.

ideal test case (2):

Let $\beta \in B$ and $\gamma \in R$. WTP: both $\gamma\beta$ and $\beta\gamma$ are in $B$.

Since, $\beta \in B$. Then, $\beta \in R$ and $a\beta \in A$.

Since, both $R$ and $A$ are commutative ring with unity. So, $\gamma \beta = \beta \gamma \in R$.

Since, $A$ is an ideal of $R$ and $\gamma \in R$ and $a\beta \in A$.

So, $a\beta\gamma = a\gamma\beta \in A$.

So, we proved that both $\gamma\beta$ and $\beta\gamma$ are in $B$. Which means ideal test case (2) passed.

Therefore, we proved that $B$ is an ideal of $R$.

$\textbf{Questions:}$

I don't feel my prove of both $A \subseteq B$ (where $A$ does not have to be the proper subset of $B$) and $B$ is an ideal of $R$ are quite right. Are there any other ways to prove better than mine ? Or are there anyting I can fixes for my prove to make it better ?