问:
Residue field of embedded ring
Suppose I have an embedding $R\rightarrow S$ where $R$ is a subring of $S$. Given a prime ideal $\mathfrak{p}\subset S$ and its preimage $\mathfrak{q}\subset R$, is it true that $\kappa(\mathfrak{q})\subseteq\kappa(\mathfrak{p})$?

答:
As pointed out in the comments, $\subseteq$ is not quite the right definition, but you do get a canonical embedding of these fields.

Consider the composition $R \longrightarrow S \longrightarrow S/\mathfrak p$. What is its kernel? Well the preimage of 0 along the latter map is $\mathfrak p$ and the preimage of $\mathfrak p$ along the inclusion $R \longrightarrow S$ is $\mathfrak q$. Then by the first isomorphism theorem, we have an injective map $R/\mathfrak q \longrightarrow S/\mathfrak p$. You can think of this map as essentially being an "inclusion," but not literally. It sends the coset $a + \mathfrak q \mapsto a + \mathfrak p$.

Now, we can embed $S/\mathfrak p \longrightarrow qf(S/\mathfrak p)$. Then by universal property of the quotient field, the composite $R/\mathfrak q \longrightarrow S/\mathfrak p \longrightarrow qf(S/\mathfrak p)$ yields a map $qf(R/\mathfrak q) \longrightarrow qf(S/\mathfrak p)$. Explicitly, this map takes $\frac{a + \mathfrak q}{b + \mathfrak q} \mapsto \frac{a + \mathfrak p}{b + \mathfrak p}$. Of course, $\kappa(\mathfrak q) = qf(R/\mathfrak q)$ and $\kappa(\mathfrak p) = qf(S/\mathfrak p)$, so this is our desired map $\kappa(\mathfrak q) \longrightarrow \kappa(\mathfrak p)$. This is not a literal set theoretic inclusion, but it's a canonical embedding of fields. We do often abuse notation and say that $\kappa(\mathfrak q) \subseteq \kappa(\mathfrak p)$ when we have such an embedding.

Title: Residue field of embedded ring


问:
Compute $\int_{\gamma} f$
Let $\gamma_1 = S_1 + L - S_2 - L$ and $\gamma_2 = S_1 + L + S_2 - L$,
$$S_1(t) = e^{it} , t\in [0,2\pi]
$$
$$S_2(t) = 2e^{it} , t\in [0,2\pi]
$$
$$ L = [1,2]
$$
Let $f(z) = (\cos z)/z$. By writing cos z as a power series and considering $f(z) = (1/z)+g(z)$, compute $\int_{\gamma_1} f$ and $\int_{\gamma_2} f$.

Computing the integration seems long so I thought about using the Generalized Version of Cauchy Theorem. Since the winding numbers about the origin add up to 0, the integral is equaled to 0. Am I going about this the right way? (I have not learned the Residue Theorem yet)

答: Cauchy's Integral formula says $\frac 1 {2\pi i}\int_{\gamma} \frac {f{(z)}} {z-0} dz=f(0)$. Taking $f(z)=\cos z$ we get the value of the integral immediately as $2\pi i$.

ContentLicense: CC BY-SA 4.0

Title: Compute $\int_{\gamma} f$


问:
If $f$ is a contraction then it is continuous.
The proof might seem intuitive if just has one or more jump points which have a distance $d$ from each other.

But I am struggling, with the following problem:

If f is not continuous, such that for all intervals: $$[a,b], \text{with } a < b < c < d: f \text{ is not continuous in } [b,c]$$
where $a,d$ are constant and $b,c$ freely selectable.

Can it then be a contraction?

Edit:

$$\exists a,d: \forall b,c: (a < b < c  <d \implies f \text{ is not continuous in } [b,c])$$

I would prefer to use the definition of continuity using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition.

答:
Let $c: (M,d) \to  (M,d)$ with $d(c(x_1),c(x_2)) \leq \gamma d(x_1,x_2)$ with $\gamma \in [0,1)$ be a contraction.

If $c$ is continuous, then

$$\forall x_0 \in M: \forall \epsilon > 0: \exists \delta > 0: c(B(x_0, \delta)) \subset B(c(x_0),\epsilon)$$

Choose $\delta = \epsilon$

Then: $c(B(x_0, \delta)) = c(B(x_0, \epsilon)) \subset  B(c(x_0), \gamma\epsilon) \subset B(c(x_0),\epsilon)$

Therefore all contractions are continous.


问:
Delta function and $\sum_{t}\exp\{ i k t\} $
I am interested in evaluating the following, which appears in an integrand,
$$
\sum_{t=0}^{\infty}e^{ikt}
$$
where $k$ is real. Using the following relation
$$
\sum_{t=-\infty}^{\infty}e^{ikt} = 2\pi\delta(k),
$$
the real part of it is

\begin{eqnarray}
\Re\left[\sum_{t=0}^{\infty}e^{ikt}\right]&=&\frac{1}{2}+\frac{1}{2}\Re\left[\sum_{t=-\infty}^{\infty}e^{ik}\right]\\&=&\frac{1}{2}+\pi\delta(k).
\end{eqnarray}

Note the addition of $\frac{1}{2}$ term! This $\frac{1}{2}$ term does not agree with the following derivation using $\epsilon$ trick!

\begin{eqnarray}
\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\sum_{t=0}^{T}e^{ikt}e^{-\epsilon t}&=&\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\frac{e^{\epsilon}-e^{ik\left(T+1\right)}e^{-\epsilon T}}{e^{\epsilon}-e^{ik}}\\
&=&\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}
\end{eqnarray}

which becomes, when $k \rightarrow 0$,

\begin{eqnarray}
\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}&=&\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{\epsilon-ik}\\
&=&\lim_{k\rightarrow0}\lim_{\epsilon\rightarrow0_{+}}\frac{i}{k+i\epsilon}\\&=&\lim_{k\rightarrow0}\left[\pi\delta\left(k\right)+i\mathcal{P}\frac{1}{k}\right]
\end{eqnarray}

where Sokhotsky's formula is used.

Now when $k\nrightarrow0$,
$$
\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}=\frac{1}{1-e^{ik}}
$$

Therefore, for all $k$,
$$
\sum_{t=0}^{\infty}e^{ikt}=\pi\delta\left(k\right)+\mathcal{P}\frac{1}{1-e^{ik}}
$$

Note that the real part of it does not have $\frac{1}{2}$ addition term, instead has $\Re\mathcal{P}\frac{1}{1-e^{ik}}$.

What am I doing wrong?

Building on Svyatoslav's answer which in turn supported by vitamin d, the conclusion is the following:
$$
\sum_{t=0}^{\infty}e^{ikt} = \frac{1}{2} + \pi \delta(k) + \frac{i}{2} \mathcal{P} \cot (k/2)
$$

答:
The Dirichlet Kernel is a Fourier Series approximations to the Dirac delta. It relies on cancellation rather than decay away from $0$. The Fejér Kernel makes a better approximation to the Dirac delta since it is positive and decays to $0$ away from $0$.
$$
\sum_{k=-n}^ne^{ikx}=\underbrace{\frac{\sin\left(\frac{2n+1}2x\right)}{\sin\left(\frac12x\right)}}_{2\pi\times\text{Dirichlet Kernel}}
$$
If we multiply this by $e^{inx}$ we get
$$
\begin{align}
\sum_{k=0}^{2n}e^{ikx}
&=\frac{\sin\left(\frac{2n+1}2x\right)}{\sin\left(\frac12x\right)}\,e^{inx}\\[3pt]
&=\frac12\left(\vphantom{\frac{\frac12}{\frac12}}\!\right.\underbrace{\frac{\sin\left(\frac{4n+1}2x\right)}{\sin\left(\frac12x\right)}}_{2\pi\times\text{Dirichlet Kernel}}+1\left.\vphantom{\frac{\frac12}{\frac12}}\!\right)+\frac i2\cot\left(\frac12x\right)\left(\vphantom{\frac{\frac12}{\frac12}}\!\right.1-\underbrace{\frac{\cos\left(\frac{4n+1}2x\right)}{\cos\left(\frac12x\right)}}_{\substack{2\pi\times\text{Dirichlet Kernel}\\\text{rotated by $\pi$,}\\\text{which is killed}\\\text{by $\cot\left(\frac12x\right)$}}}\left.\vphantom{\frac{\frac12}{\frac12}}\!\right)
\end{align}
$$
So the real part tends to $\frac12(2\pi\delta+1)$ the imaginary part tends to $\frac12\cot\left(\frac12x\right)$.


问:
Relation between the topology definition and the real example
I'm new to topology and I find that its really difficult to relate some basic idea to a real example. Like the definition of topology below

The definition comes from the book "topology without tears".

Personally. I can understand each of the condition in the definition, however, when I also saw some real example, like the coffee cup and donut below.

How to explain these two things are the same according to the basic definition? Maybe its a wield question, but, i was stuck here for a long time.

答:
Welcome to MSE!

I had the exact same question when I first took a topology class. What you're experiencing is the difference between point set topology (which we need in order to formalize the notion of "closeness" without a notion of "distance") and algebraic topology (which is where we use this to classify "nice" spaces). For instance, the "rubber sheet" geometry you may have heard of lets you stretch and squish things because they only need to preserve which points are close to each other, without preserving any notion of "distance" between the points.

Open sets formalize this notion of "closeness", but describing exactly how is more or less the content of a topology class, so I won't be able to fit all that intuition into an MSE answer. You might think about how the open sets from a metric space encode the notion of closeness from the metric without encoding the specific distances.

If you want a good introduction to algebraic topology, I can't recommend Henle's "A Combinatorial Introduction to Topology" enough. It's set at a very approachable level, and will get you to the point where you can reason about donuts and coffee cups without worrying too much about the nitty gritty point-set topological details (which working algebraic topologists don't really worry about either, in my experience).

Edit:

I felt like I'd seen this before, and indeed I have. Here are some other questions on similar topics: (x, x, x)

I hope this helps ^_^


问:
Finding a number sequence as an upper bound
For the sequence $\frac{1}{n^2 (z-\frac{1}{n})}$ ($n=1,2,....$) on the left half plane, where $z$ is a complex number I would like to find a number sequence that bounds it. I have tried for days and for some reason I cannot find something that works. Thanks!

答:
Let $z=x+iy$ with $x <0$. Then $|\frac  1 {n^{2}(z-\frac 1 n)}|=\frac  1 {n^{2}} \frac {|x-\frac 1 n|+|y|} {(x-\frac  1 n)^{2}+y^{2}}$. Split this into two terms. The first term is bounded by  $\frac  1 {n^{2}} \frac 1 {|x-\frac  1 n|} \leq \frac1 n$. The second term is bounded by $\frac  1 {n^{2}} $ if $|y| \geq 1$ and it is bounded by $1$ if $|y| \leq 1$. Hence, the function is bounded by the constant $2$.

A better bound: Use the inequlity $2ab \leq a^{2}+b^{2}$ with $a=|x-\frac 1 n|$ and $ b=|y|$. You will see that the second term is bounded by $\frac 1 {2n}$. Hence, a bettre bound is $\frac 1n +\frac 1 {2n}=\frac 3 {2n}$.


问:
Simplification of the Traveling Salesman Problem
I'm not sure I can call it a simplification of TSP but here it is:

Suppose you're given the Traveling Salesman Problem but in addition, you're also given the shortest possible distance (not the route) for the whole journey. How much faster would it be to find the shortest possible route with this extra bit of information?

$\\$

Note:
I realize that the answer depends on what algorithm you choose. So you feel free to use your favourite algorithm to make this comparison.

答: In the worst case, it is as difficult as the original problem without the hint.  For example, suppose that all weights are $1$, and I tell you that the shortest tour has length $n$, where $n$ is the number of nodes.


问:
Simplifying Complex Propagation Constant of an Electromagnetic Plane Wave
I am attempting to study for an exam and encountered a problem involving an electromagnetic plane wave  with a provided electric field and propagation constant, but despite setting up the solution correctly I am unable to understand how to simplify the complex propagation constant with the provided information.

Question and Solution :

An electromagnetic plane wave with an angular frequency of $\omega=4.518 \times 10^{10} \mathrm{rad} / \mathrm{s}$ propagates in the $+z$ direction through a conducting medium. The medium is characterized by $\sigma=0.2 \mathrm{~S} / \mathrm{m}, \varepsilon=\varepsilon_{0}=8.8534 \times 10^{-12} \mathrm{~F} / \mathrm{m}$,and $\mu=\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$. The associated electric field is represented in phasor form as:
$$
\vec{E}=E_{x 0} e^{-\gamma z} \hat{x}
$$
where $\gamma$ is the propagation constant given by:
$$
\gamma=j \omega \sqrt{\mu_{0} \varepsilon_{0}} \sqrt{1-\frac{j \sigma}{\omega \varepsilon_{0}}}
$$
The value of $z$ (in meters) for which the magnitude of the electric field is $0.01 E_{x 0}$ is most nearly:

A. $\quad 0.00273$

B. $\quad 0.0289$

C. $\quad 0.0297$

D. $\quad 0.1259$

Using the equations in the item statement:
$$
\begin{array}{l}
\frac{1 \angle 90^{\circ}}{2.998 \times 10^{8}} \times 4.518 \times 10^{10} \sqrt{1-j 0.5}=159.3 \angle 76.72^{\circ}=36.59+j 155 \\
e^{-36.59 z}=0.01 \\
z=\frac{\ln (0.01)}{-36.59}=0.1259
\end{array}
$$

Essentially my issue comes with the first and second intermediary steps - I understand substituting the three terms with the provided values and simplifying the third term to $$(1-0.5j)^{0.5}$$, but what I do not understand is why the solution has a phasor in the first step instead of a complex j term, and why the $$2.998\cdot10^8$$ term is in the denominator of the first term.

Additionally, I do not understand why this mixed phasor + complex number is able to be represented in an equivalent step as a phasor in the next step, unless the provided solution is skipping a ton of information?

From there I understand the conversion to a complex number from a phasor, then setting up the desired field strength to the electric field strength with the calculated propagation constant, then simplifying for the magnitude of the distance from the field to obtain the desired strength - I am mainly just getting caught up in how the simplification is made as indicated in the solution on the first line.

答:
why the solution has a phasor in the first step instead of a complex j term

I think it's just easier to join an angle to a magnitude than multiply by $j$, maybe easier to calculate with. (For the unitiated reader: engineers and physicists often mark the imaginary unit with a $j$ because of $i$'s resemblance of the symbol of alternating current.)

and why the 2.998*10^8 term is in the denominator

$$\sqrt{\mu_0\varepsilon_0}=\sqrt{8.8534\cdot4\pi\cdot10^{-19}}\approx 3.3354925513\cdot10^{−9} \Rightarrow$$
$$\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}=\dfrac{1}{3.3354925513\cdot10^{−9}}=299805796.182\approx2.998\cdot10^8$$

Thus $$\sqrt{\mu_0\varepsilon_0}\approx\frac{1}{2.998\cdot10^{8}}$$

I do not understand why this mixed phasor + complex number is able to be represented in an equivalent step as a phasor in the next step

Both of them are numbers, you can check the physical quantities with dimensional analysis, but both things at the two side of the equals sign represent the same numerical value (apart from the approximations) they are just written differently. Yes, steps were skipped but you can change complex numbers into many forms.

Three main forms i know of, the algebraic form:
$$z=a+bi$$
The polar form:$$z=r\cdot e^{i\varphi}$$
where $\varphi$ is the degree of the complex number's vector, in electrical engineering subjects often marked  $$z=r\angle\varphi$$
and the trigonometric form: $$r(cos\varphi+i\sin\varphi)$$
They are described formally in this article.

How to switch between algebraic and phase form:
$$z=a+bi$$

$$r=\sqrt{a^2+b^2}\hspace{5mm}\varphi=\arctan\left(\frac{b}{a}\right)$$

The trigonometric form uses the same parameters as the phasor one.

Complex division because this is the one people usually struggle with:

$$z=\dfrac{a+bi}{c+di}=\dfrac{a+bi}{c+di}\cdot1=\dfrac{a+bi}{c+di}\cdot\dfrac{c-di}{c-di}=\dfrac{(a+bi)(c-di)}{(c+di)(c-di)}=\dfrac{ac+i(bc-ad)-bdi^2}{c^2-d^2i^2}=$$

$$z=\dfrac{ac+i(bc-ad)+bd}{c^2+d^2}=\color{blue}{\dfrac{ac+bd}{c^2+d^2}}+\color{red}{\dfrac{bc-ad}{c^2+d^2}}i$$


问:
When $x= \pm \sqrt{y-1}$, is $y$ a function of $x$? How to deal with the $\pm$?
I am solving equations to find if y is a function of x.  The equation I am solving for y is; $$x= \pm \sqrt{y-1}$$

I am wondering what to do with the $\pm$ sign?  I did take the one equation and made two out of it.  I think if I can do that then the rest of the answer is all algebraic.  Here are my two equations.

$x=\sqrt{y-1}$, and $x=- \sqrt{y-1}$

Then I square both sides of each equation;

$x^2=y-1$, and $x^2=-(y-1)$

Then distribute the negative in the second equation

$x^2=y-1$, and $x^2= -y+1$

Then I isolate the y on one side of the equation

$-y=-x^2-1$, and $y=-x^2+1$

Then the first equation there is -y.  I multiply that equation by -1 to finish getting y by itself.

$y=x^2+1$, and $y=-x^2+1$
Is this the right method to solve for $\pm$?  Can it be solved any further?  I thought it would be complete.

All the books says is that y is a function of x for this equation.  It doesn’t show how to solve.

答:
We start with

$x = \pm \sqrt{y-1} \implies x^2 = (\pm \sqrt{y-1})^2 = \lvert \sqrt{y-1} \rvert^2 = y-1$.

Now we have

$x^2 = y-1 \implies y = x^2 + 1$

In order for $y$ to be a function of $x$ it is necessary that for each input of $x$ we get exactly one output $y$. In other words, if we take $x_1$ we mustn't have $x_1^2+1 = y_1$ and $x_1^2+1 = y_2$ where $y_1 \neq y_2$.

Can you finish from here?


问:
Isometry fixing two points of a geodesic line
Let $H$ be a hyperbolic space, and let $\Gamma \subset H$ be a geodesic line, i.e., the image of an isometry from $\mathbb{R}$ to $H$. If $f$ is an isometry of $H$ that fixes two distinct points of $\Gamma$, is it true that $f(\Gamma) = \Gamma$?</p>

答:
The answer is yes. This is a simple application of the fact isometries send geodesics to geodesics, and every geodesic can be uniquely defined by two points in the union of the hyperbolic plane and its boundary circle.

If you've not seen either of these facts justified, feel free to ask and I'll try to spend the time explaining them.

To help clarify why every two points define a unique geodesic, I'll explain for the special case when both points $z$ and $w$ are in the hyperbolic plane $\mathbb{H}$. First, recall that geodesics in the hyperbolic upper half plane are either vertical Euclidean straight lines or semicircles with center on the real axis. Recall also that for every pair of points in $\mathbb{H}$, there exists at least one geodesic which passes through both points.

Suppose $z$ and $w$ share the same real coordinate and differ on the imaginary coordinate. Then, there are no semicircles with center on the real line which pass through both $z$ and $w$ and so it must be a vertical straight line which passes through them. It should be clear that such a line is unique.

Now, suppose that $z$ and $w$ differ on their real coordinate. There are a couple of ways that we could show that there is a unique geodesic which passes through both $z$ and $w$. One way would be to show that there is a unique $p\in\mathbb{R}\subset\partial\mathbb{H}$ such that $|z-p|=|w-p|$. This can be shown algebraically. You can also see this geometrically by drawing small circles around $z$ and $w$ of the same radius and then letting them grow continuously. It should be clear that there will be exactly one time when these two circles intersect on the real axis and they intersect at exactly one point.

The reason we want $|z-p|=|w-p|$ is because the point $p$ satisfies the necessary condition for the center of the semi circle passing through both $z$ and $w$. As $p$ is unique, and the length $|z-p|$ has to be the radius of such a semicircle, the geodesic passing through both $z$ and $w$ is uniquely determined by these two numbers.

The other way to see that a unique geodesic passes through $z$ and $w$ is to notice that there exists an isometry of the hyperbolic plane which maps both $z$ and $w$ to the imaginary axis. To describe this isometry, first take any geodesic $\Gamma_0$ which passes through both points (we know at least one geodesic passes through them) and call the end points $a$ and $b$ with $a<b$. Let $f$ be the isometry given by $$f(x)=\frac{x-b}{x-a}.$$ The isometry $f$ maps $b$ to $0$ and $a$ to $\infty$ and so must map $\Gamma_0$ to $\{x\:|\:\mbox{Re}(x)=0\}$ the imaginary axis. As $z,w\in\Gamma_0$, we also know that $f(z)$ and $f(w)$ are on the imaginary axis. By the previous discussion, $f(\Gamma_0)$ is the unique geodesic which passes through $f(z)$ and $f(w)$ but then $f^{-1}(f(\Gamma_0))$ must be the unique geodesic passing through $z$ and $w$ and so $\Gamma_0$ is unique.

From a quick search, you can find a proof of this as well in these lecture notes.

ContentLicense: CC BY-SA 3.0


答:
Be careful. $x=\pm y\implies x^2=y^2$ not $x^2=\pm y^2$. Indeed, $$y=1-x^2\implies x^2=1-y\implies x=\pm\sqrt{1-y}\neq\pm\sqrt{y-1}$$

However, $y=x^2+1$ is valid: $$y=x^2+1\implies x^2=y-1\implies x=\pm\sqrt{y-1}$$

Title: When $x= \pm \sqrt{y-1}$, is $y$ a function of $x$? How to deal with the $\pm$?

Body:
I am solving equations to find if y is a function of x.  The equation I am solving for y is; $$x= \pm \sqrt{y-1}$$

I am wondering what to do with the $\pm$ sign?  I did take the one equation and made two out of it.  I think if I can do that then the rest of the answer is all algebraic.  Here are my two equations.

$x=\sqrt{y-1}$, and $x=- \sqrt{y-1}$

Then I square both sides of each equation;

$x^2=y-1$, and $x^2=-(y-1)$

Then distribute the negative in the second equation

$x^2=y-1$, and $x^2= -y+1$

Then I isolate the y on one side of the equation

$-y=-x^2-1$, and $y=-x^2+1$

Then the first equation there is -y.  I multiply that equation by -1 to finish getting y by itself.

$y=x^2+1$, and $y=-x^2+1$
Is this the right method to solve for $\pm$?  Can it be solved any further?  I thought it would be complete.

All the books says is that y is a function of x for this equation.  It doesn’t show how to solve.


问:
How to get as product $\sin 2A-\sin 2B+\sin 2C$ in few steps and in a more orderly way?
The problem is as follows:

Using:

$A-B+C=360^\circ$

Find as product the expression from below:

$\sin 2A-\sin 2B+\sin 2C$

The alternatives given by my book are as follows:

$\begin{array}{ll}
1.&2\sin A\sin B\sin C\\
2.&4\cos A \cos B \cos C\\
3.&4\sin A \sin B \sin C\\
4.&-2\sin A \cos B \sin C\\
\end{array}$

Please take notice that in order to avoid any sort of duplication with this question and other similars. Please note that the signs are not the same.
And the angle condition is not the same.

I'm rephrasing my question as an individual situation based on my strategy on how to solve this problem provided that I landed in the right answer. Thus the sort of answer which I'm intending to get is how to solve or prove this without just many steps and more logically and less prone to get errors in algebraic manipulation.

What I've attempted to so, was a similar strategy as seen in the Cagnoli's identity.

Thus:

$A-B+c=2\pi$

Then:

$\sin (A-B+C)=\sin 2\pi=0$

$\sin (A-B)\cos C+\cos (A-B)\sin C=0$

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C + \sin A \sin B \sin C=0$

Then I could spot:

$-\sin A \sin B \sin C=\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

Now the thing is how to get what it is requested:

Then I thought:

On the right side of the equation I'm getting:

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

$\frac{1}{2}(\sin (A+B)+\sin (A-B))\cos C - \frac{1}{2}\cos C (\sin (A+B)-\sin (A-B))+\frac{1}{2}(\cos (A+B)+\cos (A-B))\sin C$

Multiplying by $2$ on the whole expression and replacing from:

$A-B=2\pi-C$

$(\sin (A+B)+\sin (2\pi-C))\cos C - \cos C (\sin (A+B)-\sin (2\pi-C))+(\cos (A+B)+\cos (2\pi-C))\sin C = -2\sin A \sin B \sin C$

$(\sin (A+B)-\sin (C))\cos C - \cos C (\sin (A+B)+\sin (C))+(\cos (A+B)+\cos (C))\sin C$

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B)-\frac{1}{2}\sin 2C +\sin C \cos (A+B) + \frac{1}{2}\sin 2C$

Simplifying:

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B) +\sin C \cos (A+B)$

Multiplying by $2$:

$2\cos C \sin (A+B)-\sin 2C- 2\cos C \sin (A+B) +2\sin C \cos (A+B)=-4\sin A \sin B \sin C$

Then product to sum:

$\sin (A+B+C)-\sin (C-A-B)-\sin 2C - \sin (A+B+C)+\sin (C-A-B)+\sin (A+B+C)+\sin (C-A-B)$

Luckily there are some terms which can be simplified:

$-\sin 2C +\sin (A+B+C)+\sin (C-A-B)$

Then I guess that what is left to replace is : $A-B+C=2\pi$

$-\sin 2C + \sin (2\pi+2B)+\sin (2\pi-2A)$

Then this becomes into:

$-\sin 2C + sin 2B -\sin 2A= -4\sin A \sin B \sin C$

Since it is mentioned that it requests to obtain the identity from the mentioned expression, this translates into:

Multiplying by $-1$:

$\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Therefore that correspond with the option 3. And I believe this must be the answer.
If my algebra is correct. I wonder, if it does exist a less painful way to get the same result without just relying in too many lines and perhaps. Did I overlooked something?. It would help me a lot if someone could help me with this. As my methodology seems too long and I had to do several trials to get this right.

答:
Solving by algebra and identities is absolutely the hardest possible way to proceed.

Instead, pick values of $A$, $B$, and $C$ that satisfy the relation $A - B + C = 360^\circ$ and then see which of the potential answers are eliminated.  (It can take a few rounds of choices to become familiar with how certain choices make certain terms be zero or $\pm 1$, and I haven't shown these exploratory steps.  The choices I've made below were my first choices and come from much prior practice.)

For instance, take $A = 360^\circ$, and $B = C = 0^\circ$.  We find $\sin 2A - \sin 2B + \sin 2C = 0$.  Any product with a $\sin A$, a $\sin B$, or a $\sin C$ is also zero, so we have probably eliminated potential answer 2.  We compute $4 \cos A \cos B \cos C = 4 \cdot 1 \cdot 1 \cdot 1 \neq 0$, so we have eliminated potential answer 2.

All the other answers have $\sin A$ instead of $\cos A$, so it's time to pick values for $A$ that have $\sin A \neq 0$.

Now try $A = C = 90^\circ$, $B = -180^\circ$.  Again, $\sin 2A - \sin 2B + \sin 2C = 0$.  Here, potential answer 4 gives $-2 \sin A \cos B \sin C = -2 \cdot 1 \cdot -1 \cdot 1 \neq 0$, so we have eliminated potential answer 4.

The surviving potential answers differ only by a factor of $2$, so we want all three of $\sin A$, $\sin B$, and $\sin C$ different from 0.

Take $A = 270^\circ$, $B = 30^\circ$ and $C = 120^\circ$.  We find $\sin 2A - \sin 2B + \sin 2C = 0 - \sqrt{3}/2 -\sqrt{3}/2 = -\sqrt{3}$.  Then, $\sin A \sin B \sin C = -1 \cdot 1/2 \cdot \sqrt{3}/2 = -\sqrt{3}/4$, eliminating potential answer 1.

So we know that potential answer 3 is the correct answer.


问:
Question about Hochschild extension
An extension of a ring $C'$ by an ideal $N$ can be expressed by an exact sequence $0\to N\to C\to C'\to 0$. Such an extension is called a Hochschild extension if $C$ is isomorphic to $C'\oplus N$ as additive group, while the multiplication is given by $(a,x)(b,y)=(ab,ay+bx+f(a,b))$ $(a,b\in C';x,y\in N )$, where the map $f:C'\times C'\to N$ is symmetric and bilinear satisfying the cocycle condition:

$af(b,c)-f(ab,c)+f(a,bc)-f(a,b)c=0$

Then, it's said that this extension is trivial(the exact sequence is trivial) iff there exists a function $g:C'\to N$, s.t. $f(a,b)=ag(b)-g(ab)+g(a)b$.

I don't know why this follows? Hope someone could help. Thanks!

答: The sequence splits precisely if such $g$ exists. The splitting is given by $a \mapsto (a, -g(a))$. To check it is a homomorphism do: $$ (a, -g(a))(b, -g(b)) = (ab, -ag(a) - g(b)b + f(a,b)) = (ab, -g(ab))$$

Title: Question about Hochschild extension


问:
Equivalent sets for naturals
Prove that the sets $\{0, 1\}^\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ are equivalent.

My friend gave me the following proof:

Proof.

It suffices to show that $\{0, 1\}^\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ have the same cardinality so we need to find a function. In particular, we need

$$\{f:\mathbb{N} \rightarrow \{0, 1\}\} \longleftrightarrow \{f:\mathbb{N} \rightarrow \mathbb{N}\}$$

Consider $f: \mathbb{N} \rightarrow \mathbb{N}$.

Define $\phi(f) = (1 ... 1 0 1 ... 1 0 ...) = ((1 ... 1) 0 (1 ... 1) 0 ...)$, where the $i$th grouping of $1$s can be treated as $f(i)$.

We have $\phi:\{f:\mathbb{N} \rightarrow \mathbb{N}\} \rightarrow \{0, 1\}^{\mathbb{N} \times \mathbb{N}}$, with $\{0, 1\}^{\mathbb{N} \times \mathbb{N}}$ having the same cardinality as $\{0, 1\}^\mathbb{N}$.

Then we also have $\phi(f) = 1_{\{(n, f(n)) | n \in \mathbb{N}\}}$.

End.

He barely explained this and did not explain any further. I have been trying to make sense of it but cannot figure out what is happening or how to formalize it. Can somebody write it down as a clear, formal proof and/or explain it? Any assistance is appreciated.

答:
$\{0,1\}^{\mathbb N}= \{$ countable sequences of $0$s and $1$s.$\}$.

$\mathbb N^{\mathbb N}= \{$ countable sequences of natural numbers$\}$.

Let $s\in \mathbb N^{\mathbb N} = \{a_i\} = \{a_1, a_2, a_3.....\}$.

We can map $s\to s' \in \{0,1\}^{\mathbb N}$ by replacing every $a_i$  with a string of $a_i$ $1$s followed be a zero.

It's a simple concept. If the sequence $s= \{3,1,5,0, 7, 2,.... \}$ then $s'$ would be $s' = \{\underbrace{1,1,1}_3,0,\underbrace{1}_1,0,\underbrace{1,1,1,1,1}_5,0,\underbrace{}_00,\underbrace{1,1,1,1,1,1,1}_7,0, \underbrace{1,1}_2,0,.....\}$

Defining this formally and recursively can by done but.... we can see what is happening and doing it formally will be tedious.

Your friend actually did this in the opposite direction.  He took something like $t \in \{0,1\}^{\mathbb N}$ as mapped something like $t = \{0,1,1,1,0,0,1,0,1,1,0,0,....\}= \{\underbrace{0}_0,\underbrace{1,1,1,0}_3,\underbrace{0}_0,\underbrace{1,0}_1,\underbrace{1,1,0}_2,\underbrace{0}_0,....\}\to \{0,3,0,1,2,0,....\}=t'\in \mathbb N^{\mathbb N}$

That function migh actually be easier to define formally but I thought it wasn't as easy to describe.

....

Anyhooo.... we can prove this is bijective by... saying "Aw, come on, isn't it obvious... any sequence of ones and zeros can be 'blocked' as a set of ones separated by a zero so that string and the following zero is mapped to be then natural number that is number of ones, and we can do that for each block, so it's gotta be surjective.  And if a string has a specific number of ones it can only be mapped from that number.  So it's gotta be injective."

That the proves their cardinalities are equal.

....

I must confess I found the wording confusing.  I that he was view $\{0,1\}^{\mathbb N}$ and $\mathbb N^{\mathbb N}$ by their alternative and equivalent definition of $A^B = \{$functions $B\to A\}$.

This is equivalent ans $s\in A^{\mathbb N}$ where $s= \{a_1, a_2, a_3, ....\}$ is equivaly to a function $\sigma:\mathbb N \to A$ via $\sigma(n) = a_n$.

I got very confused reading his work.

Title: Equivalent sets for naturals


问:
Visually stunning math concepts which are easy to explain
Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

答:
The surface obtained by spinning a cube on two diametrically opposite corners:

$\hskip{4.5cm}$

All the surfaces are ruled surfaces. The top and bottom are simply conical caps. The curved part in the middle is part of a hyperboloid of one sheet. It can be obtained by taking a cylinder of radius and height $\sqrt{\frac23}$ times the edge length of the cube and giving one of the ends a $60^{\large\circ}$ twist.


答:
Let's do some transformation first to turn this into a symmetric problem. Set  A=a, B=-b, C= c

then $a+b+c=360^\circ$

$sin (a+b)=-sinc$

$\cos (a+b)=cosc$

$sin2a+sin2b+sin2c$

$= 2sin(a+b)*cos(a-b)+sin2c$
$= -2sin c* cos(a-b)+2sinc*cosc$

$=2sinc* (cosc-cos(a-b))$
$= 2sinc*(cos(a+b)-cos(a-b))$
$=2sinc*(-2sina*sinb)$
$=-4sina*sinb*sinc$

Replace a,b,c with A,B,C, we have $\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Title: How to get as product $\sin 2A-\sin 2B+\sin 2C$ in few steps and in a more orderly way?

Body:
The problem is as follows:

Using:

$A-B+C=360^\circ$

Find as product the expression from below:

$\sin 2A-\sin 2B+\sin 2C$

The alternatives given by my book are as follows:

$\begin{array}{ll}
1.&2\sin A\sin B\sin C\\
2.&4\cos A \cos B \cos C\\
3.&4\sin A \sin B \sin C\\
4.&-2\sin A \cos B \sin C\\
\end{array}$

Please take notice that in order to avoid any sort of duplication with this question and other similars. Please note that the signs are not the same.
And the angle condition is not the same.

I'm rephrasing my question as an individual situation based on my strategy on how to solve this problem provided that I landed in the right answer. Thus the sort of answer which I'm intending to get is how to solve or prove this without just many steps and more logically and less prone to get errors in algebraic manipulation.

What I've attempted to so, was a similar strategy as seen in the Cagnoli's identity.

Thus:

$A-B+c=2\pi$

Then:

$\sin (A-B+C)=\sin 2\pi=0$

$\sin (A-B)\cos C+\cos (A-B)\sin C=0$

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C + \sin A \sin B \sin C=0$

Then I could spot:

$-\sin A \sin B \sin C=\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

Now the thing is how to get what it is requested:

Then I thought:

On the right side of the equation I'm getting:

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

$\frac{1}{2}(\sin (A+B)+\sin (A-B))\cos C - \frac{1}{2}\cos C (\sin (A+B)-\sin (A-B))+\frac{1}{2}(\cos (A+B)+\cos (A-B))\sin C$

Multiplying by $2$ on the whole expression and replacing from:

$A-B=2\pi-C$

$(\sin (A+B)+\sin (2\pi-C))\cos C - \cos C (\sin (A+B)-\sin (2\pi-C))+(\cos (A+B)+\cos (2\pi-C))\sin C = -2\sin A \sin B \sin C$

$(\sin (A+B)-\sin (C))\cos C - \cos C (\sin (A+B)+\sin (C))+(\cos (A+B)+\cos (C))\sin C$

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B)-\frac{1}{2}\sin 2C +\sin C \cos (A+B) + \frac{1}{2}\sin 2C$

Simplifying:

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B) +\sin C \cos (A+B)$

Multiplying by $2$:

$2\cos C \sin (A+B)-\sin 2C- 2\cos C \sin (A+B) +2\sin C \cos (A+B)=-4\sin A \sin B \sin C$

Then product to sum:

$\sin (A+B+C)-\sin (C-A-B)-\sin 2C - \sin (A+B+C)+\sin (C-A-B)+\sin (A+B+C)+\sin (C-A-B)$

Luckily there are some terms which can be simplified:

$-\sin 2C +\sin (A+B+C)+\sin (C-A-B)$

Then I guess that what is left to replace is : $A-B+C=2\pi$

$-\sin 2C + \sin (2\pi+2B)+\sin (2\pi-2A)$

Then this becomes into:

$-\sin 2C + sin 2B -\sin 2A= -4\sin A \sin B \sin C$

Since it is mentioned that it requests to obtain the identity from the mentioned expression, this translates into:

Multiplying by $-1$:

$\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Therefore that correspond with the option 3. And I believe this must be the answer.
If my algebra is correct. I wonder, if it does exist a less painful way to get the same result without just relying in too many lines and perhaps. Did I overlooked something?. It would help me a lot if someone could help me with this. As my methodology seems too long and I had to do several trials to get this right.


问:
Norm and metrics
Let $E=C^1[a,b]$ and $\|\cdot\|_0$, $\|\cdot\|_1$ norms defined as $$\|f\|_0=\max_{x\in [a,b]} |f(x)|$$ and $$\|f\|_1=\max_{x\in [a,b]} |f(x)|+\max_{x\in [a,b]} |f'(x)|.$$
For $r>0$, consider the open ball with center in the origin defined by: $$B_r^0(0)=\{f \in E;\|f\|_0<r\}$$ and $$B_r^1(0)=\{f \in E;\|f\|_1<r\}.$$

Prove that $B_r^1(0) \subset B_r^0(0)$.

Prove that there is no $\epsilon>0$ such that $B_\epsilon^0 (0) \subset B_r^1(0)$.

Can I have some hints? This does not look very intuitive and I'm having some trouble to prove it.

答:
To get 1. let $f \in B_r^1(0)$, i.e. $\lVert f \rVert_0 + \lVert f' \rVert_0 < r$. Now we conclude as norms are non-negative:
$$
\lVert f \rVert_0 \leq \lVert f \rVert_0 + \lVert f' \rVert_0 < r
$$
So $f \in B_r^0(0)$.

To get 2. let us assume that such $\varepsilon$ exists. Then set $f_\delta: [a, b] \rightarrow \mathbb{R}$,
$$
f_\delta(x) := \frac{\varepsilon}{\sqrt{b-a +\varepsilon}} \cdot \sqrt{x-a+\delta}
$$
where $\delta \in (0, \varepsilon)$ is arbitrary. Clearly, $f_\delta \in C^1[a, b]$. Hence, for all $x \in [a, b]$:
$$
\lvert f_\delta(x) \rvert = \frac{\varepsilon}{\sqrt{b-a+\varepsilon}} \cdot \sqrt{x-a+\delta} \leq \frac{\varepsilon}{\sqrt{b-a+\varepsilon}} \cdot \sqrt{b-a+\varepsilon} = \varepsilon
$$
So $f_\delta(x) \in B_\varepsilon^0(0)$. Then, $f \in B_r^1(0)$ by assumption. So, for all $x \in [a, b]$:
$$
r > \lVert f_\delta(x) \rVert_0 + \lVert f_\delta'(x) \rVert_0 \geq \lVert f_\delta'(x) \rVert_0 = \max_{x \in [a, b]}\frac{\varepsilon}{2\sqrt{b-a +\varepsilon}} \cdot \frac{1}{\sqrt{x-a+\delta}} = \frac{\varepsilon}{2\sqrt{b-a +\varepsilon}} \cdot \frac{1}{\delta} \overset{\delta \downarrow 0}{\longrightarrow} \infty
$$
This is a contradiction.

Title: Norm and metrics


问:
Matrix multiplication efficiency
I have a multiplication $A * B * A^T$, where $B$ is a covariance matrix (i.e. symmetric, square) and $A$ is an arbitrary matrix with compatible dimensions. Given the symmetry of $B$ and the fact that I multiply by $A$ and $A^T$, it really feels like there must be a more efficient way to compute the result than two multiplications, but I haven't been able to find anything. Does anyone have any insight on the problem, or why there isn't any better way of doing it?

Thanks!

答:
Since $B$ is symmetric, $ABA^T$ is also symmetric, and so you only need to determine the entries above the diagonal.  Thus, instead of doing two matrix multiplications, you can make do with one and a half.

However, without there being anything else special about $A$ or $B$, I am skeptical that any speedup you get will be more than a constant factor.  In particular, it is very unlikely that computing $ABA^T$ can be done faster than computing $AB$.  Algorithmically, this means that your won't change the order of magnitude of the run time of whatever you are using this for, and so unless you are doing a ton of matrix multiplications, looking for a clever optimization won't likely net you great gains.

However, on a related note, something that will speed up the multiplication, at least when you are dealing with large matrices, is Strassen's algorithm, which is an algorithm for multiplying $2\times 2$ matrices with 7 multiplications instead of 8, which can recursively be used to multiply large matrices together quickly ($O(n^{\log_2 7+o(1)})$ instead of $O(n^3)$).

Title: Matrix multiplication efficiency


问:
In a competition of $45$ people, how many received medals in exactly two categories?

Background

I've been learning the basics of set theory and this is a reference to Business Mathematics and Statistics
by Asim Kumar Manna, section 2.56:

In a competition, a school awarded medals in different categories. $36$
medals in dance, $12$ medals in dramatics and $18$ medals in music. If
these medals went to a total of $45$ people and only $4$ people got
medals in all the three categories, how many received medals in
exactly two of these categories?

My attempt

Let $A$ = set of persons who got medals in dance.

$B$ = set of persons who got medals in dramatics.

$C$ = set of persons who got medals in music.

Given,

$n(A) = 36$

$n(B) = 12$

$n(C) = 18$

$n(A \cup B \cup C) = 45$

$n(A \cap B \cap C) = 4$

We know that number of elements belonging to exactly two of the three sets $A, B, C$

$= n(A \cap B) + n(B \cap C) + n(A \cap C) - 3n(A \cap B \cap C)$

$= n(A \cap B) + n(B \cap C) + n(A \cap C) - 3 \times 4 \dots(i)$

$=n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

Therefore

$n(A \cap B) + n(B \cap C) + n(A \cap C) = n(A) + n(B) + n(C) + n(A \cap B \cap C) - n(A \cup B \cup C)$

From $(i)$ required number

$= n(A) + n(B) + n(C) + n(A\cap B \cap C) - n(A \cup B \cup C) - 12$

$= 36 + 12 + 18 + 4 - 45 - 12$

$= \boxed{13}$

13 medals received. Is there a better way go about this? Is this process succinct?

答:
In words

There were $36+12+18 =66$ medals

$45$ teams won at least one

leaving $66-45=21$ extra medals

of which $4\times(3-1)=8$ were extra medals for teams winning three matches

leaving $21-8=13$ extra medals for teams winning two matches

so there were $13$ teams winning exactly two matches

As equations, if $m_1$ teams won one medal, $m_2$ teams won two medals and $m_3$ teams won three medals then

$m_1+m_2+m_3 = 45$

$m_1+2m_2+3m_3=36+12+18$

$m_3=4$

which are three linear equations in three unknowns, with the unique solution $m_1=28, m_2=13, m_3=4$ and the question asks for $m_2$


问:
Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $
I have the conjecture for the integral

$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$

where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.

I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral

$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$

is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.

Edit:

Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$

which may not reduce the difficulty, albeit appearing compact.

答:
Rewriting

Use the trigonometric identity $\csc^2(x) = \cot^2+1$. Rewrite the integral as
$$\int\limits_{\alpha}^{\beta}\csc^2(x)\left(-\frac{\log\cot x}{\cot^2+1}\right)\,\mathrm{d}x$$
After substituting $u=\cot x$ we get the integral
$$-\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u^2+1}\,\mathrm{d}u=-\left[\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u+i}\,\mathrm{d}u-\frac{i}{2}\int\limits_{\cot(\alpha)}^{\cot(\beta)}\frac{\log u}{u-i}\,\mathrm{d}u\right].$$

First integral

Sub $v=u+i$. You should get
$$\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log v-i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{\log iv+1}{v}\,\mathrm{d}v+\log(-i)\int\limits_{\cot(\alpha)+i}^{\cot(\beta)+i}\frac{1}{v}\,\mathrm{d}v.$$
Integral 1

Sub $w=-iv$. This will get us to
$$-\int\limits_{1-i\cot(\alpha)}^{1-i\cot(\beta)}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(1-i\cot(\beta))-\operatorname{Li}_2(1-i\cot(\alpha))\Big]$$

Integral 2

Note that $\log(-i)=-i\pi/2$. We get:
$$-i\pi/2\Bigg[\log(\cot(\beta)+i)-\log(\cot(\alpha)+i)\Bigg].$$

Second Integral

Sub $v=u-i$. You should get
$$\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log v+i}{v}\,\mathrm{d}v = \int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{\log 1-iv}{v}\,\mathrm{d}v+\log(i)\int\limits_{\cot(\alpha)-i}^{\cot(\beta)-i}\frac{1}{v}\,\mathrm{d}v.$$
Integral 1

Sub $w=iv$. This will get us to
$$-\int\limits_{i\cot(\alpha)-1}^{i\cot(\beta)-1}-\frac{\log 1-w}{w}\,\mathrm{d}w = -\Big[\operatorname{Li}_2(i\cot(\beta)-1)-\operatorname{Li}_2(i\cot(\alpha)-1)\Big]$$

Integral 2

Note that $\log(i)=i\pi/2$. We get:
$$i\pi/2\Bigg[\log(\cot(\beta)-i)-\log(\cot(\alpha)-i)\Bigg].$$

Values for $\alpha$ and $\beta$

For the first orignal integral we have

$$-\frac{{\pi}\left(\log\left(\tan^2\left(\frac{7{\pi}}{36}\right)+1\right)-\log\left(\tan^2\left(\frac{5{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{7{\pi}}{36}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{7{\pi}}{36}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{5{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{5{\pi}}{36}\right)\right)\right)}{4}$$

For the second orignal integral we have

$$-\dfrac{{\pi}\left(\log\left(\tan^2\left(\frac{{\pi}}{12}\right)+1\right)-\log\left(\tan^2\left(\frac{{\pi}}{36}\right)+1\right)\right)+{i}\left(2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{12}\right)+1\right)-2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{12}\right)\right)-2\operatorname{Li}_2\left({i}\tan\left(\frac{{\pi}}{36}\right)+1\right)+2\operatorname{Li}_2\left(1-{i}\tan\left(\frac{{\pi}}{36}\right)\right)\right)}{4}$$

A few last words

I added these values numerically and I got the right result. The integral is solved but it still needs some long (algebraic) operations to prove the end result. But the Polylogarithm $\operatorname{Li}_2$ looks promising. Maybe I will add the final derivation later on, but first I have to take a break of this monster challenge.
Also, if someone finds a faster way and all of this is useless I would be twice as happy. Nonetheless I hope it has helped you at least a bit.


问:
Expressing $\sqrt{3 − \sqrt{5}}$ in the form $\frac{\sqrt{a}−\sqrt{b}}{c}$
Express $\sqrt{3 − \sqrt{5}}$ in the form $\frac{\sqrt{a}−\sqrt{b}}{c}$ for some integers $a,b,c$.

I'm not quite sure how to start this, can anyone give me a little hint or two

答:
I remember I solved this kind of problem when I was a pupil, a lot of nostalgia :D.

You have

\begin{align}
\sqrt{3-\sqrt{5}} &= \frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}} \\
&= \frac{\sqrt{1-2\sqrt{5} +5}}{\sqrt{2}} \\
&= \frac{\sqrt{(\sqrt{5}-1)^2}}{\sqrt{2}} \\
&= \frac{\sqrt{5}-1}{\sqrt{2}} \\
&= \frac{\sqrt{10}-\sqrt{2}}{2} \\
\end{align}


Title: Expressing $\sqrt{3 − \sqrt{5}}$ in the form $\frac{\sqrt{a}−\sqrt{b}}{c}$

Body:
Express $\sqrt{3 − \sqrt{5}}$ in the form $\frac{\sqrt{a}−\sqrt{b}}{c}$ for some integers $a,b,c$.

I'm not quite sure how to start this, can anyone give me a little hint or two

Body: $\dfrac{3-\sqrt{5}}{2}= \dfrac{6-2\sqrt{5}}{4}= \dfrac{(1-\sqrt{5})^2}{4}\implies \sqrt{\dfrac{3-\sqrt{5}}{2}}=\dfrac{\sqrt{5}-1}{2}\implies \dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5}-1}{2}\implies \sqrt{3-\sqrt{5}}=\sqrt{2}\cdot \dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{10}-\sqrt{2}}{2}$.


答:
Your set analysis seems solid. There's a minor shortcut available since we know the total winning population is being overcounted by the category medal counts due to those who won multiple medals.

Given total medals of $36+12+18=66$, we know we could also express this total medal count as  $M_1 + 2M_2 + 3M_3 = 66$, where $M_i$ is the number of people who won $i$ medals.

Then we also know that the number of medal winners $M_1 + M_2 + M_3 = 45$,

So subtracting, $M_2+2M_3 = 21$ and of course $M_3 = 4$ giving the consistent result $M_2 = 13$

Title: In a competition of $45$ people, how many received medals in exactly two categories?

Body:
Background

I've been learning the basics of set theory and this is a reference to Business Mathematics and Statistics
by Asim Kumar Manna, section 2.56:

In a competition, a school awarded medals in different categories. $36$
medals in dance, $12$ medals in dramatics and $18$ medals in music. If
these medals went to a total of $45$ people and only $4$ people got
medals in all the three categories, how many received medals in
exactly two of these categories?

My attempt

Let $A$ = set of persons who got medals in dance.

$B$ = set of persons who got medals in dramatics.

$C$ = set of persons who got medals in music.

Given,

$n(A) = 36$

$n(B) = 12$

$n(C) = 18$

$n(A \cup B \cup C) = 45$

$n(A \cap B \cap C) = 4$

We know that number of elements belonging to exactly two of the three sets $A, B, C$

$= n(A \cap B) + n(B \cap C) + n(A \cap C) - 3n(A \cap B \cap C)$

$= n(A \cap B) + n(B \cap C) + n(A \cap C) - 3 \times 4 \dots(i)$

$=n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

Therefore

$n(A \cap B) + n(B \cap C) + n(A \cap C) = n(A) + n(B) + n(C) + n(A \cap B \cap C) - n(A \cup B \cup C)$

From $(i)$ required number

$= n(A) + n(B) + n(C) + n(A\cap B \cap C) - n(A \cup B \cup C) - 12$

$= 36 + 12 + 18 + 4 - 45 - 12$

$= \boxed{13}$

13 medals received. Is there a better way go about this? Is this process succinct?


问:
Find a unique $2\pi$-periodic solution for a parametrical differential equation of the second order.
Abstract

I was given this question in class, and I am quite confused whether the proper solution for it even exists. The content of the task is as follows:

Given  differential  equation  $y''+2ay'+b^{2}y = \cos(x)$  $[*]$ with $0 \leqslant a < b < 1$.  Find  the  unique  periodic solution of this equation, whose period is $2\pi$. For what value of $a$ the amplitude of this solution takes the greatest value?

I have found a similar question on MathExchange here, but it didn't seem to be helpful, as I have already found the general solution for given $a$ and $b$.

General solution for $a$ and $b$

By given $0 \leqslant a < b < 1$, thus, for homogeneous case of $[*]$ characteristic polynomial, $\lambda^{2}+2a\lambda+b^{2}=0, D_{\lambda} = 4a^{2} - 4b^{2}<0$, hence $$y_{h} = C_{1}e^{x(-a+i\sqrt{b^{2}-a^{2}})} + C_{2}e^{x(-a-i\sqrt{b^{2}-a^{2}})}.$$
Thus: (partial solution derivation is omitted)$$y(x) = C_{1}e^{x(-a+i\sqrt{b^{2}-a^{2}})} + C_{2}e^{x(-a-i\sqrt{b^{2}-a^{2}})} +\small\frac{(b^{2}-1)\cos(x)+2a\sin(x)}{4a^{2}+b^{4}-2b^{2}+1}.$$
But then, $y(x)$ is $2\pi$-periodic only if either $(a,b) = (0,0)$ or $(0,1)$, as otherwise $e^{-ax}$ will undermine periodicity, however such cases are restricted by the given conditions on $a$ and $b$.

At the moment I have no clue, whether the task has incorrect conditions on $a$ and $b$,or I am just blind...

Thank you in advance for the help.

答:
To bring this problem to a conclusion, note that the general solution $$y=C_1 y_1+C_2 y_2+y_p$$ represents the full two-dimensional space of solutions of the inhomogeneous second-order linear ODE $$y''+2ay'+b^2 y=\cos x.$$
Every choice of $C_1,C_2$ corresponds to a valid solution of the ODE for some initial condition. In particular, $y_p$ is itself a solution. So if the particular solution happens to be periodic, then this is the desired example.

A side question is how one finds such a periodic particular solution. In the present case, the method of undetermined coefficients suffices: we guess a solution of the form $y_p = A \cos t+B\sin t$ and then determine what $A,B$ will work. (If the forcing function were more complicated, then we could instead use variation of parameters or the Fourier transform method.)

Lastly, this solution is unique: If two periodic solutions $y_{p1},y_{p2}$ existed, then $y_{p1}-y_{p2}$ would be a $2\pi$-periodic solution to the homogeneous ODE. But the only such solution is $y=0$. So once a periodic particular solution is found, then this must be the unique periodic solution.


问:
$H$ is a subgroup of $G$ and every coset of $H$ in $G$ is a subgroup of $G$.Then which of the following is true?
$H$ is a subgroup of $G$ and every coset of $H$ in $G$ is a subgroup of $G$.Then which of the following is true?

(A) $H=${$e$}

(B) $H=G$

(C) $G$ must have prime order.

(D) $H$ must have prime order.

答:
Hint:-

(B) is true. every subgroup of a group must contain the identity and any two distinct cosets are disjoint.


问:
Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$
Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$

I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that

$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$

Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and  consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.

Update:

I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try  find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.

Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.

Body:
Here is a solution with elementary math.

Let's denote
$$S_n = \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$$

You have

$$2S_n = 1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}$$

Then

\begin{align}
S_n = 2S_n - S_n &= 1+\left(\frac{3}{2}-\frac{1}{2} \right)+\ldots+\left(\frac{2n-1}{2^{n-1}} - \frac{2n-3}{2^{n-1}} \right) - \frac{2n-1}{2^{n}}  \\
&= 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\ldots+ \frac{1}{2^{n-2}}  \right) - \frac{2n-1}{2^{n}}  \\
&= 1+2\left(1- \frac{1}{2^{n-1}}  \right) - \frac{2n-1}{2^{n}}  \\
&= 3-\frac{1}{2^{n-2}} - \frac{2n-1}{2^{n}}  \\
\end{align}

Hence,
$$\lim_{n\to\infty} S_n=3$$


问:
Find the values of $a$ and $c$ to make $f(x)$ continuous at $x = 4$
Let $$f(x)=\begin{cases} 2cx, \quad x<4\\ x+a, \quad x=4\\ x^{2}-6, \quad x>4\end{cases}$$

I have to solve for $c$ and $a$. So far I got $5/4$ for $c$ and $4$ for $x$. Not sure how to go to $a$.

答:
In order for a function $f$ to be continuous at $x = a$, we must have the following:

$ \lim_{x \to a} {f(x)} $ exists

$f$ is defined at $x = a$

$ \lim_{x \to a} {f(x)} = f(a) $

Given

$$ f(x) = \begin{cases}
2cx, & x<4 \\
x+a, & x=4 \\
x^2-6, & x>4
\end{cases} , $$

the first part you've solved (for $c$) is correct. This comes from the limits from the left and right needing to be equal in order for the limit to exist (condition 1 above):

$$ \lim_{x \to 4^-} {f(x)} = \lim_{x \to 4^-} {f(x)} $$
$$ \Rightarrow \lim_{x \to 4^-} {2cx} = \lim_{x \to 4^-} {x^2-6} $$
$$ \Rightarrow 8c = 10 $$
$$ \Rightarrow c = \frac{5}{4} . $$

We also need $f$ to be defined at $x = 4$, which it is: $f(4) = 4+a$. This satisfies condition 2 above.

The final condition is that the limit is equal to the function at $x = 4$:

$$ \lim_{x \to 4} {f(x)} = f(4) $$
$$ \Rightarrow 2c*4 = 4+a $$
$$ \Rightarrow 2*\frac{5}{4}*4 = 4+a $$
$$ \Rightarrow 10 = 4 + a $$
$$ \Rightarrow a = 6 . $$

So, we need $c = \frac{5}{4}$ in order for the limit to exist at $x = 4$, and we need $a = 6$ in order for the limit to equal the function at $x = 4$.


问:
Proving $\sum\limits_{n=1}^\infty\frac{|\mu(n)|}{n^s}=\frac{\zeta(s)}{\zeta(2s)}.$
I am trying to show that
$$D(|\mu|,s)=\sum_{n=1}^\infty\frac{|\mu(n)|}{n^s}=\frac{\zeta(s)}{\zeta(2s)}.$$
By a previous exercise I know that $D(\lambda,s)=\zeta(2s)/\zeta(s)$, where $\lambda$ is Liouville's function $\lambda(n)=(-1)^{\Omega(n)}$. But I don't see how to use this fact (I am not sure I should, but it looks like it would be useful).</p>

答:
This is how I convinced myself of this fact.
Here's an argument which utilizes the Euler product. So I am going to take as a given:

$$\prod _{p} 1-p^{-s}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s }=\frac{1}{\zeta(s)}, \hspace{1 cm}  \prod _{p} 1+p^{-s}=\sum_{n=1}^\infty \frac{|\mu(n)|}{n^s } $$

Then $$\frac{\zeta(s)}{\zeta(2s)}= \frac{\prod_p 1-p^{-2s}}{\prod_p 1-p^{-s}}=\prod_{p} \frac{1-p^{-2s}}{1-p^{-s}}=\prod_p \frac{(1-p^{-s})(1+p^{-s})}{1-p^{-s}}=\prod_{p}1+p^{-s}=\sum_{n=1}^\infty \frac{|\mu(n)|}{n^s}$$


问:
Good bound for $\sum_{p\mid n}\frac{1}{p}$
I am trying to get a good asymptotic bound for $$\sum_{\substack{p\mid n\\ \text{$p$ prime}}}\frac{1}{p}.$$ One way I can think of is $$\sum_{p\mid n}\frac1p\leqslant \sum_{p\leqslant n}\frac{1}{p} \ll \log\log n,$$ by  Mertens' theorem.
Can we do better?

As a follow-up question, if you see a sum like this, is there a way to intuitively guess how large it could be?  Thanks!

答:
Similar to how Gronwall deduced

$$
\limsup_{n\to\infty}{\sigma(n)\over e^\gamma n\log\log n}=1
$$

We consider defining $a_n$ such that

$$
\prod_{p\le p_{a_n}}p\le n\le\prod_{p\le p_{a_n+1}}p
$$

Taking logarithms gives

$$
{\vartheta(p_{a_n})\over p_{a_n}}\le{\log n\over p_{a_n}}\le{p_{a_n+1}\over p_m}\cdot{\vartheta(p_{a_n+1})\over p_{a_n+1}}
$$

Now, due to prime number theorem we have $p_n\sim p_{n+1}$ and $\vartheta(x)\sim x$, so that

$$
\log n\sim p_{a_n}\tag1
$$

With these priliminaries, we can begin working on the problem

$$
\sum_{p|n}\frac1p\le\sum_{p\le p_{a_n}}\frac1p=\log\log p_{a_n}+\mathcal O(1)
$$

By (1), we also see that

$$
\sum_{p\le p_{a_n}}\frac1p\sim\log\log\log n
$$

To see how tight this bound is, we set

$$
n_k=(p_1p_2\cdots p_k)^{\lfloor\log p_k\rfloor}
$$

so that

$$
\log n_k=\lfloor\log p_k\rfloor\vartheta(p_k)\sim p_k\log p_k
$$

and

$$
\log\log n_k\sim\log p_k
$$

Plugging these into the original formula, we get

$$
\sum_{p|n}\frac1p=\sum_{p\le p_k}\frac1p\sim\log\log p_k\sim\log\log\log n
$$

Therefore, $\log\log\log n$ is not only an upper bound for $\sum_{p|n}\frac1p$, but also a maximal order for $\sum_{p|n}\frac1p$:

$$
\limsup_{n\to\infty}{1\over\log\log\log n}\sum_{p|n}\frac1p=1
$$


问:
Write output of linear dynamical system in terms of adjugate matrix
I'm trying to understand a statement given by a book ("Analysis and Design of Autonomous Microwave Circuits", Almudena Suarez). It defines a linear dynamical system in the Laplace domain as $sX=AX+GU$ or equivalently $(sI-A)X=G(s)U(s)$ where $X$ is a vector of system states, $A$ is the system matrix, $U$ is a vector with inputs to the system, $G$ is a matrix that combines the inputs in some manner, and $s$ is the Laplace variable.

With zero input to the system ($U=0$), we can write $(sI-A)X=0$ from which we obtain the characteristic polynomial of the system, $P(s)=det(sI-A)$.

The textbook then states that some output of the system can be written as $$Y(s)=B(sI-A)^+$$ where the "+" denotes the transpose of the cofactor matrix and $B$ is some matrix. This is where I get confused. My understanding is that the transpose of the cofactor matrix is called the adjugate, but I'm not sure what significance the adjugate has in this context.

The book goes further to state in the case of a single-input, single-output system (ie $U$ and $Y$ are both scalar), then we can write the transfer function of this system as
$$H(s)=\frac{Y(s)}{U(s)}=\frac{B(sI-A)^+G}{P(s)}$$

I'm having trouble understanding how the author goes from the definitions of $U$ and $Y$ to obtain the transfer function.

答:
Thanks to @Semiclassical for providing a source that led to an answer.

Essentially, we can write the following:
$$sX=AX+GU$$
$$Y=BX$$

Then we can solve for $X$ in the first equation:
$X=(sI-A)^{-1}GU$

and substitute into the second equation: $Y=B(sI-A)^{-1}GU$

Finally, we write the transfer function as
$$H=\frac{Y}{U}=B(sI-A)^{-1}G=\frac{B(sI-A)^+G}{det(sI-A)}=\frac{B(sI-A)^+G}{P(s)}$$

Regarding the invertibility of $(sI-A)$, we look at whether $P(s)$ is zero or not. Since $P(s)$ is the characteristic polynomial of $A$, it is clearly only zero when $s$ is an eigenvalue of $A$, or in this case a pole of the system. So this behavior is expected.


答:
1)First you need to proof that this sequence is a absolute convergent sequence.
Hence the convergence properties wont be affected by the finite basic algebraic manipulation ,such as multiplying by a real number ,or exchanging the position of elements of the S(n).

let S=lim S(n) (n⇒+∞)
1/2 S=1/2 limS(n) (n⇒+∞)

S（n）-1/2S（n）=1/2+1/2+1/2²+...1/2ⁿ⁻¹-（2n+1）/2ⁿ⁺¹
let n⇒+∞ ，THEN $1/2S=1/2+1-0=3/2$
THUS $S=3$

Title: Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$

Body:
Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$

I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that

$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$

Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and  consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.

Update:

I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try  find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.

Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.


答: In spite of being already given the final answer, think of the following: if we have a coset $\,gH\neq H\,$ , then it can not be that $\,1\in gH\,$ (why?) , thus...

Body:
$H$ is a subgroup of $G$ and every coset of $H$ in $G$ is a subgroup of $G$.Then which of the following is true?

(A) $H=${$e$}

(B) $H=G$

(C) $G$ must have prime order.

(D) $H$ must have prime order.


问:
Problem on Lagrange multipliers
This problem has two parts:
$a)$ Let $k>0$, find the minimum of the function $f(x,y)=x+y$ over the set S=$\{(x,y) \in \mathbb R^2_{> 0}:xy=k\}$. $b)$ Prove that for every $(x,y) \in \mathbb R^2_{> 0}$ the inequality $\frac{x+y} {2}\geq \sqrt{xy}$ holds.$$$$
 I want to find the minimum of $f$ restricted to the level set $g(x,y)=k$. The function $f$ and $g$ are differentiable and $\nabla g(x,y)=(y,x)\neq0$ for every $(x,y) \in \mathbb R^2_{> 0}$, then I can apply lagrange multipliers to find the critical points of f restricted to that region. By the equation $\nabla f(x,y)=\lambda\nabla g(xy)$ I get that $(1,1)=\lambda (y,x)$, which means $x=y$. By the conditions of the problem, it is required that the points of the form $(x,x)$ live in the level set. So, the only critical point is $(\sqrt{k},\sqrt{k})$. $$$$My question is, how can I prove that at $(\sqrt{k},\sqrt{k})$, $f|_S$ attains a minimum?$$$$For part $b)$, let's assume I could prove that $f(\sqrt{k},\sqrt{k})$ is a minimum of $f|_S$.Then, for all $(x,y) \in S$ it immediately follows that $x+y\ge \sqrt{k}+\sqrt{k}=2\sqrt{k}$, so dividing both sides by $2$, I get $\frac{x+y} {2}\ge\sqrt{k}=\sqrt{xy}$. I don't see why this inequality would also hold for $(x,y) \in \mathbb R^2_{>0}\setminus S$</p>

答:
You already had proven that the point $ \ (\sqrt{k} \ ,  \sqrt{k}) \ $ is at the minimum when you made your gradient calculation.  You found that this point lies on the "level surface" $ \ x \ + \ y \ = \ 2\sqrt{k} \ $ , which represents a plane through the point with gradient $ \ \nabla f \ = \ \langle 1 \ , \ 1 \rangle \ $ there (really, everywhere on the plane).

For a line with slope $ \ m \ $ in the $ \ xy-$ plane, its direction vector is   $ \ \langle 1 \ , \ m \rangle $ ; the associated unit vector is u  =  $ \ \frac{1}{\sqrt{1 + m^2}} \ \langle 1 \ , \ m \rangle \ $ .  Consequently, the directional derivative at  $ \ (\sqrt{k} \ ,  \sqrt{k}) \ $ is

$$ D_{\mathbf{u}} \  f \ \ = \ \ \nabla f \ \centerdot \ \mathbf{u} \ \ = \ \ \langle 1 \ , \ 1 \rangle \ \centerdot \ \frac{1}{\sqrt{1 + m^2}} \ \langle 1 \ , \ m \rangle \ \ = \ \ \frac{1 \ + \ m}{\sqrt{1 + m^2}} \ \ , $$

which is positive for $ \  1 \ + \ m \ > \ 0 \ \ \Rightarrow \ \ m \ > \ -1 \ \ . $ Since $ \ S \ $ is restricted to the first quadrant, this condition is certainly met by any ray within this quadrant extending from $ \ (\sqrt{k} \ ,  \sqrt{k}) \  . $  So the directional derivative of $ \ f(x,y) \ $ in the level surface at this point is always positive, which tells us that the point marks the minimum for $ \ x \ + \ y \ $ on  $ \ xy \ = \ k \ $  .

[Note that we are abandoned by a "Hessian matrix" calculation, since the plane $ \ x \ + \ y \ = \ 2\sqrt{k} \ $ has no critical points; the second partial derivatives tell us nothing of use.]

Title: Problem on Lagrange multipliers


问:
How to get as product $\sin 2A-\sin 2B+\sin 2C$ in few steps and in a more orderly way?
The problem is as follows:

Using:

$A-B+C=360^\circ$

Find as product the expression from below:

$\sin 2A-\sin 2B+\sin 2C$

The alternatives given by my book are as follows:

$\begin{array}{ll}
1.&2\sin A\sin B\sin C\\
2.&4\cos A \cos B \cos C\\
3.&4\sin A \sin B \sin C\\
4.&-2\sin A \cos B \sin C\\
\end{array}$

Please take notice that in order to avoid any sort of duplication with this question and other similars. Please note that the signs are not the same.
And the angle condition is not the same.

I'm rephrasing my question as an individual situation based on my strategy on how to solve this problem provided that I landed in the right answer. Thus the sort of answer which I'm intending to get is how to solve or prove this without just many steps and more logically and less prone to get errors in algebraic manipulation.

What I've attempted to so, was a similar strategy as seen in the Cagnoli's identity.

Thus:

$A-B+c=2\pi$

Then:

$\sin (A-B+C)=\sin 2\pi=0$

$\sin (A-B)\cos C+\cos (A-B)\sin C=0$

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C + \sin A \sin B \sin C=0$

Then I could spot:

$-\sin A \sin B \sin C=\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

Now the thing is how to get what it is requested:

Then I thought:

On the right side of the equation I'm getting:

$\sin A\cos B\cos C-\cos A\sin B\cos C + \cos A \cos B \sin C$

$\frac{1}{2}(\sin (A+B)+\sin (A-B))\cos C - \frac{1}{2}\cos C (\sin (A+B)-\sin (A-B))+\frac{1}{2}(\cos (A+B)+\cos (A-B))\sin C$

Multiplying by $2$ on the whole expression and replacing from:

$A-B=2\pi-C$

$(\sin (A+B)+\sin (2\pi-C))\cos C - \cos C (\sin (A+B)-\sin (2\pi-C))+(\cos (A+B)+\cos (2\pi-C))\sin C = -2\sin A \sin B \sin C$

$(\sin (A+B)-\sin (C))\cos C - \cos C (\sin (A+B)+\sin (C))+(\cos (A+B)+\cos (C))\sin C$

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B)-\frac{1}{2}\sin 2C +\sin C \cos (A+B) + \frac{1}{2}\sin 2C$

Simplifying:

$\cos C \sin (A+B)-\frac{1}{2}\sin 2C- \cos C \sin (A+B) +\sin C \cos (A+B)$

Multiplying by $2$:

$2\cos C \sin (A+B)-\sin 2C- 2\cos C \sin (A+B) +2\sin C \cos (A+B)=-4\sin A \sin B \sin C$

Then product to sum:

$\sin (A+B+C)-\sin (C-A-B)-\sin 2C - \sin (A+B+C)+\sin (C-A-B)+\sin (A+B+C)+\sin (C-A-B)$

Luckily there are some terms which can be simplified:

$-\sin 2C +\sin (A+B+C)+\sin (C-A-B)$

Then I guess that what is left to replace is : $A-B+C=2\pi$

$-\sin 2C + \sin (2\pi+2B)+\sin (2\pi-2A)$

Then this becomes into:

$-\sin 2C + sin 2B -\sin 2A= -4\sin A \sin B \sin C$

Since it is mentioned that it requests to obtain the identity from the mentioned expression, this translates into:

Multiplying by $-1$:

$\sin 2A-\sin 2B + \sin 2C = 4 \sin A \sin B \sin C$

Therefore that correspond with the option 3. And I believe this must be the answer.
If my algebra is correct. I wonder, if it does exist a less painful way to get the same result without just relying in too many lines and perhaps. Did I overlooked something?. It would help me a lot if someone could help me with this. As my methodology seems too long and I had to do several trials to get this right.

答:
Note

\begin{align}
& \sin 2A-\sin 2B+\sin 2C\\
= & 2\sin A \cos A -2\sin(B-C) \cos(B+C)\\
= &2\sin A \sin(2\pi +B- C) -2\sin (A-2\pi ) \cos(B+ C )\\
= & 2\sin A\sin (B-C) -2\sin A \cos(B+ C )\\
 = & 2\sin A[ \cos ( B-C)  -\cos(B+C)] \\
 = & 4\sin A \sin B\sin C \\
\end{align}


问:
A supposedly obvious equation in Hörmander's The Analysis of Linear Partial Differential Operators I
To quote page 8.

Another important example of a $C^1$ map is for two Banach spaces $U$
and $V$ the map $f$ taking an invertible element $T \in L(U,V)$ (T is
a continuous linear map between U  and V), to its inverse $T^{-1}$. If
$S \in L(U,V)$ we have $(T+S)T^{-1}=id_v+ST^{-1}$ so if   $\|S\| \|T^{-1}\| \lt 1$ a right inverse of $T+S$ is given by

$T^{-1}(id_v+ST^{-1})^{-1}=\sum_{0}^{\infty}T^{-1}(-ST^{-1})^k$

The infinite series in the last equation I do not understand, even though it's supposed to be obvious.

答: If $\|A\|<1,$ then the series $$\sum\limits_{n=0}^\infty A^n =(I-A)^{-1},$$ with the convergence being in operator norm. This is called a Neumann series. In your situation, $A=-ST^{-1}.$ Then, $$\|A\|=\|-ST^{-1}\|\leq \|S\|\|T^{-1}\|<1\implies (I-ST^{-1})^{-1}=\sum\limits_{n=0}^\infty (-ST^{-1})^n.$$ Thus, $$T^{-1}(I+ST^{-1})^{-1}=\sum\limits_{n=0}^\infty T^{-1}(-ST^{-1})^n=\sum\limits_{n=0}^\infty(-1)^nS^n(T^{-1})^{n+1}.$$


问:
How to prove this moment generating function's property?
The moment generating function of a random variable $X$ is defined, for all $\theta \in \mathbb{R}$, by: $M_{X}(\theta) := E(e^{\theta X})$. We denote $\Lambda_{X}(\theta) = \log(M_{X}(\theta))$.

From Cramer-Chernoff theorem, we know that: $\Lambda_{X}^{*}(t) = \sup(\theta t - \Lambda_{X}(\theta))$ for $\theta > 0 $

How to prove that, if $X_1,\ldots,X_n$ are random variables and if $X=X_1+\cdots+X_n$ then $\Lambda_{X}^{*}(t) = n \Lambda_{X_1}^{*}(\frac{t}{n})$ for all $t \in \mathbb{R}$?

How to prove that? I didn't figure out any formula or property, which would lead to this.

答:
As @Semiclassical pointed out in a comment, you probably want to assume that the random variables are independent. Moreover, I will also assume that they are identically distributed.

Now, let $Y = X_1 + \ldots + X_n$, and note that by the independence assumption we have that the moment generating function for $Y$ is given by $M_Y(\theta) = M_{X_1}(\theta) \ldots M_{X_n}(\theta)$. This means that
$$\Lambda_{Y}(\theta) = \log(M_{X_1}(\theta)) + \ldots + \log(M_{X_n}(\theta)) = n \log(M_{X_1}(\theta)),$$
where the last equality follows by the assumption that the variables are identically distributed.

Next, the transformation $f^*(t) := \sup(x t - f(x))$ is known as the convex conjugate (or Fenchel conjugate) of the function $f(x)$. A direct computation (which is also a well-known result for convex conjugates) gives that
$$
(a f)^*(t) := \sup_x(x \ t - af(x)) = \sup_x(a (x \ t/a - f(x))) = a \sup_x(x \ t/a - f(x)) = af^*(t/a)
$$
Applying this result gives you your sought expression, namely that
$$
\Lambda_{Y}^*(t) = (n \Lambda_{X_1}(t))^* = n \Lambda_{X_1}^*(t/n).
$$


问:
Show basic open sets in $\{0,1\}^{A}$ are both open and closed
Consider the topological space $\{0,1\}^A$. The set of functions from an arbitrary index set $A$ to $\{0,1\}$. Then a subbasis for the product topology is $S=\{U(a,\delta)\mid a \in A,\delta \in\{0,1\}\}$ where $U(a,0)=\pi_a^{-1}(0),U(a,1)=\pi_a^{-1}(1)$. Show that every basic open set is both open and closed.

Attempt: A basic open set is of the form $$\pi_{a_1}^{-1}(b_1) \cap \dots \cap \pi_{a_n}^{-1}(b_n)$$

where $b_i \in \{0,1\},a_i \in A$. Then $$\{0,1\}^A-(\pi_{a_1}^{-1}(b_1) \cap \dots \cap \pi_{a_n}^{-1}(b_n))=\pi_{a_1}^{-1}(c_1) \cap \dots \cap \pi_{a_n}^{-1}(c_n)$$ where $c_i \neq b_i$. Another basic open set. So basic open sets are both open and closed.

答: Assuming that $\{0,1\}$ is endowed with the discrete topology, then every subbasic open set of the product space $\{0,1\}^{A}$ is a preimage under a continuous function of a clopen subset of $\{0,1\}$. Then, every basic open set is a finite intersection of clopen subsets.


问:
Prove this point is on the circumcircle

I have a triangle $DCE$ with $DC>DE$. The exterior angle bisector of $\angle CDE$ meets the perpendicular bisector of $CE$ at $H$. How do I prove that $H$ always lies on the circumcircle of $CDE$?

答: The center of circumcircle ($O$) is on perpendicular bisector of $CE$. It's sufficient to prove that $OD = OH$. Let $G$ the point on $CE$ such that $DG$ is perpendicular to $CE$. Let $K$ the point on CE such that $HK$ is perpendicular to $CE$. It's clear that $DG$ is parallel to $HK$. We note that $\angle{CDO} = \angle{EDG}$ because $\angle{EDG} = \frac{\pi}{2}-\angle{DEC} = \frac{\pi}{2}-\frac{\angle{COD}}{2} = \frac{1}{2}(\pi-\angle{COD}) = \angle{CDO}$, the last equality is due $\triangle{CDO}$ is isosceles. Let $x = \angle{EDG}$, we see that $\angle{OHD} = x + \angle{EDG}$ (because $DG$ is parallel to $HK$) and $\angle{HDO} = x+\angle{CDO} = x+\angle{EDG} = \angle{OHD}$ and this equality implies that $OD = OH$ and we conclude.

Title: Prove this point is on the circumcircle


问:
Show that two finite two generator groups are isomorphic
I was faced by a question that I can't solve. Any help would be great! Let $A$ and $B$ be groups with the following properties:

\begin{cases}
    |A| = 9 \cdot 3=27\\
A = \left \\
a^{9} = e \\
b^{3} = e \\
ba = a^{4}b
\end{cases}

\begin{cases}
    |B| = 9 \cdot3=27\\
B = \left<\alpha,\beta\right> \\
\alpha^{9} = e \\
\beta^{3} = e \\
\beta\alpha = \alpha^{7}\beta
\end{cases}
I need to prove that $A \cong B$.

I really can't think of any isomorphism (or how to use some theorem) to solve this. Thanks in advance!

答:
Thanks to Jeremy Rickard for his comments on my previous, incorrect, answer.

First, it's easy to prove by induction on $n$ that $ba^n=a^{4n}b$.  It's then easy to prove by induction on $k$ that $b^ka^n=a^{{4^k}n}b^k.$  Similarly, $\beta^k\alpha^n=\alpha^{{7^k}n}\beta^k.$  Thus, we observe that $b^2a=a^{16}b^2=a^7b^2$.  This observation motivates the following effort.

Define $\varphi:A \to B$ by $\varphi(a)=\alpha, \varphi(b)=\beta^2$.  It's easy to confirm that $\varphi(a^9)=\varphi(b)^3=1_B$, that $\varphi(a)$ in fact has order $9$, and that $\varphi(b)$ in fact has order $3$.  Also:

$$\varphi(ba)=\varphi(a^4b)=\varphi(a)^4\varphi(b)=\alpha^4\beta^2=\alpha^{49}\beta^2=\beta^2\alpha=\varphi(b)\varphi(a).$$

Thus, $\varphi$ is well-defined and a homomorphism, so $\varphi(A)$ is a subgroup of $B$, which meanns its order is a power of $3$ no greater than $27$.  But $\alpha \in \varphi(A) \Rightarrow \vert \varphi(A) \vert \geq 9$ and $\beta^2 \in \varphi(A) \setminus \langle \alpha \rangle$, so in fact $\vert \varphi(A) \vert \gt 9$.  It follows that $\varphi(A)=B$ and since $\vert A \vert = \vert B \vert, \varphi:A \to B$ is $1$-$1$, and hence is an isomorphism.


问:
Lagrange Multiplier Problem - Distance from point With a Circle Constraint
Given a point $P:(x_0,y_0)$ in $\mathbb{R}^2$, and a constraint function
$$x^2+y^2=R^2$$where $R$ is the radius of the circle. The distance from $P$ to any point on the circle is to be minimized using the method of Lagrange Multiplier. The distance $d$ is given by
$$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}.$$
Then the Lagrange function becomes
$$\mathcal{L}(x,y,\lambda)=d(x,y)+\lambda\ (x^2+y^2-R^2)$$
with the optimality conditions $\partial_x\mathcal{L}=0$, $\partial_y\mathcal{L}=0$ and $\partial_\lambda\mathcal{L}=0$. These conditions yield:
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(x-x_0)+2\ \lambda\ x=0$$
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(y-y_0)+2\ \lambda\ y=0$$
$$x^2+y^2-R^2=0$$
How should I continue from this point?</p>

答:
Accepting @lab's hint, the Lagrange function can be rewritten as
$$\mathcal{L}(x,y,\lambda)=d^2(x,y)+\lambda\ (x^2+y^2-R^2)=(x-x_0)^2+(y-y_0)^2+\lambda\ (x^2+y^2-R^2)$$
Then the optimality conditions would become
$$\partial_x\mathcal{L}=2(x-x_0)+2x\lambda=0,\ \partial_y\mathcal{L}=2(y-y_0)+2y\lambda=0$$
$$\partial_\lambda\mathcal{L}=x^2+y^2-R^2=0$$
The first two conditions yield
$$x=\cfrac{x_0}{1+\lambda},\ y=\cfrac{y_0}{1+\lambda}\tag{1}$$
These can be substituted in the third condition to obtain
$$\cfrac{x_0^2+y_0^2}{(1+\lambda)^2}=R^2$$
Solving for $\lambda$ yields
$$\lambda=\left[\cfrac{x_0^2+y_0^2}{R^2}\right]^{\frac{1}{2}}-1$$
$\lambda$ can be substituted in Equation 1 to obtain
$$(x,y)=\left(x_0 \left[\cfrac{x_0^2+y_0^2}{R^2}\right]^{\frac{-1}{2}},\, y_0 \left[\cfrac{x_0^2+y_0^2}{R^2}\right]^{\frac{-1}{2}}\right)$$
This point represents the point closest to $P$ and the minimum distance can be found by using the original function
$$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}.$$</p>


问:
Given a circle, is there a relationship between the cycloid it generates and its circumscribed square?
This question is motivated by an answer I provided to a question here on the arc length of a cycloid.

I noticed that the ratio of the circumference of the generating circle (which is also the horizontal distanced traveled through one cycle) to the arc length of the cycloid is

$$ \frac{2\pi r}{8r} = \frac{\pi}{4} . $$

If we take that generating circle and circumscribe a square around it, the ratio of the areas of the circle to the square is the same:

$$ \frac{\pi r^2}{(2r)^2} = \frac{\pi}{4} . $$

Given a circle, is there some sort of deeper relationship between the cycloid it generates and its circumscribed square? Or is this just a coincidence? I don't have much experience with roulettes, so I'm at a bit of a loss here - would appreciate any insight anybody has on whether there's some deeper geometric relationship here.

答:
Yes there is a relationship, based on a couple hidden features of cycloidal loci.  And there is also a similar relationship between the four-cusped hypocycloid and a regular hexagon inscribed in the same circle, derivable from thecsame concepts.

Making a clean sweep

You already know that when a circle is rolled along a fixed straight line, a fixed point on the circle generates a cycloid loop whose arc length is four times the diameter of the circle.  Since the center of the circle also slides through a length equal to (of course) $\pi$ times the diameter, the arc length of the cycloid loop may also be rendered as $4/\pi$ time the length traversed by the center.

Now suppose the straight line is replaced by a circular arc to make an epicycloidal loop (if the fixed arc is externally tangent to the roller) or a hypocycloidal loop (if internally tangent).  The calculus is more complicated now, but we find that the arc length of the epicycloidal/hypocycloidal loop is still $4/\pi$ times the arc length traversed by the center of the roller.

To explain this geometrically, consider the picture below in which a cycloid is generated from point $A$ through $A'$ to $A''$.  The roller sweeps through an area, which is equivalent to a rectangle whose dimensions are the length of the center's path and the diameter of the roller.  Replacing the fixed line with a circular arc leaves this relationship unchanged; half the area is stretched but the other half is squeezed to balance it.  (We assume that in the hypocloidal case the fixed arc has large enough radius so that its center is not in the interior of the roller.  This assumption will be obeyed in what is to follow.)

We imagine this area as being filled with crisscrossing cycloidal loops whose points on the roller are distributed over the circumference of the same.  Thus the arc length of a loop should be proportional to the swept area divided by the circumference.  This quotient contains only the length of the centerline path.

Through this proportionality we may relate the length of the cycloid loop to various epicycloid or hypocycloid loops.  For instance, suppose that the rolling circle from the cycloid generation is now held fixed and we make a new roller, internally tangent with half the radius.  The fixed point now moves from $B$ through $B'$ to $B''$ where it re-contacts the larger circle.

In this case the center of the roller has moved only half its circumference and the roller itself is only half the circumference of that used for the cycloid.  Therefore, the length of this path should be one-fourth the length of the cycloid loop.  If the $B$ to $B''$ path corresponds to one side of a circumscribed square, then the cycloid loop will match the whole square.  But, can we prove the former claim?  And is there something to the hypocycloid loop in my second picture looking like a straight line?

Spiro-mania!

In my younger days I played with a drawing toy in which you place a pen inside a wheel (the pen being a fixed point in the wheel) and roll it around a wheel, ring or rack that is fixed to the paper, thus producing various patterns.  If you choose a wheel with $48$ teeth and roll it in a ring interior having $96$ teeth (a $1:2$ ratio), you get just an oval, which becomes longer and narrower if you place the pen near the outer edge of the wheel.  Evidently if the pen could be placed at the very edge of the wheel (physically impossible because the gear-teeth are there), this oval would be elongated and flattened to a full diameter of the ring.  Which would match a side of a circumscribed square ... .

To prove this mathematically, consider the figure below for the case where the roller is moved up to 90° from the initial contact.

Because of the $1:2$ circumference ratio the roller has the center $O$ of the fixed circle on its boundary (not in the interior, see above), and also the arc on the roller from $C$ to $B'$ has twice as much angular measure as the arc from $B$ to $C$.  Now, central $\angle BOC$ has the same measure as arc $BC$, and inscribed $\angle B'OC$ has half the angular measure as arc $B'C$ -- so the two angles are congruent, and this indeed forces $B'$ to lie on the radius $\overline{OB}$.

For a greater rolling angle, first move the roller $180°$ to recontact the locus at $B''$, then roll back through the supplementary acute angle and apply the above reasoning replacing $B$ with $B''$.

Fitting a square peg around a round hole...

So our hypocycloidal loop, with a $1:2$ ratio of diameters or circum-ferences, indeed matches a diameter of the circle and thus one side of a circumscribed square, and measures one-fourth the length of the corresponding cycloid.  Therefore the cycloid loop measures the full perimeter of the square.

...and a hexagonal peg inside

At the beginning of this geo-metrical trip, I mentioned that there is a similar relationship between a four-cusped hypocycloid and a regular hexagon inscribed in the same circle.  I present that here in the form of a "homework" problem.

A four-cusped hypocycloid is generated as the fixed-point locus obtained when a roller of diameter $r/2$ is rolled inside a fixed circle of diameter $2r$.  Using the fact that the hypocloid can be rendered with the parametric equations $x=r\cos^3\theta,y=r\sin^3\theta$, apply standard calculus techniques to find that the length of the complete hypocycloid matches the perimeter of an inscribed regular hexagon.

Compare the arc length traversed by the center of the roller over all four hypocycloidal loops to the corresponding arc length associated with diameter $\overline {BB''}$ above.  Conclude that the length of the hypocycloid matches three diameters of the circle, which can be matched by obvious geometric means with the inscribed regular hexagon.


问:
Simplification of the Traveling Salesman Problem
I'm not sure I can call it a simplification of TSP but here it is:

Suppose you're given the Traveling Salesman Problem but in addition, you're also given the shortest possible distance (not the route) for the whole journey. How much faster would it be to find the shortest possible route with this extra bit of information?

$\\$

Note:
I realize that the answer depends on what algorithm you choose. So you feel free to use your favourite algorithm to make this comparison.

答:
Suppose that your original algorithm had complexity $O(N)$. Like, the brute force method requires $N=n!$ with $n$ the number of destinations. Then depending on at which of your checks you find the minimum, it could be the 1st, 2nd, third check, ..., all the way to the last one. On average, you will need $\frac{1+2+...+n!}{n!}=\frac{n!(n!+1)}{2n!}=\frac{n!+1}{2}$. So I say that the complexity in the brute force method is still on the order of $O(n!)$.

If you're using something like branch and bound algorithm, let's use an example where you have 5 cities to visit total and you start with the first one and must return back to the first one. Let the distance between cities be given by the matrix

```<code>A="
5 4 3 2 11
4 6 10 3 5
3 10 3 2 3
2 3 2 8 3
11 5 3 3 11"
</code>```

Without knowledge of the shortest path, you need 54 cycles through the branch and bound algorithm.

With knowledge of the shortest path, which is length 16, the algorithm for this particular problem finds the best route after 13 cycles. In general, you should be able to stop as soon as you come upon the length of the route matching the shortest route. You might also be able to make some improvements in "pruning" branches.

Body:
Suppose that your original algorithm had complexity $O(N)$. Like, the brute force method requires $N=n!$ with $n$ the number of destinations. Then depending on at which of your checks you find the minimum, it could be the 1st, 2nd, third check, ..., all the way to the last one. On average, you will need $\frac{1+2+...+n!}{n!}=\frac{n!(n!+1)}{2n!}=\frac{n!+1}{2}$. So I say that the complexity in the brute force method is still on the order of $O(n!)$.

If you're using something like branch and bound algorithm, let's use an example where you have 5 cities to visit total and you start with the first one and must return back to the first one. Let the distance between cities be given by the matrix

A="
5 4 3 2 11
4 6 10 3 5
3 10 3 2 3
2 3 2 8 3
11 5 3 3 11"

Without knowledge of the shortest path, you need 54 cycles through the branch and bound algorithm.

With knowledge of the shortest path, which is length 16, the algorithm for this particular problem finds the best route after 13 cycles. In general, you should be able to stop as soon as you come upon the length of the route matching the shortest route. You might also be able to make some improvements in "pruning" branches.


问:
Is how I evaluated the following product correct $\prod_{n=1}^{20}(1+\frac{2n+1}{n^2})$?
The product is $$\prod_{n=1}^{20}\left(1+\frac{2n+1}{n^2}\right)$$. I can rewrite this product as $$\prod_{n=1}^{20}\left(\frac{n^2+2n+1}{n^2}\right)$$ which can be further simplified to $$\prod_{n=1}^{20}\frac{(n+1)^2}{n^2}$$. Here is the odd step I am going to do. I am going to rewrite this product as $$(\prod_{n=1}^{20}\frac{(n+1)}{n})^2$$. Now, I evaluate the product in the base $$\prod_{n=1}^{20}\frac{(n+1)}{n}$$. Writing out all the terms, I get $\frac{\color{red}{2}}{1} * \frac {\color{blue}{3}}{\color{red}{2}} * \frac {\color{red}{4}}{\color{blue}{3}} ... \frac{\color{blue}{19}}{\color{red}{18}} * \frac{\color{red}{20}}{\color{blue}{19}} * \frac {{21}}{\color{red}{20}}$. Terms of same color near each other cancel, black terms do not, I hope my use of blue and red helps anyone with colorblindness. Thus, we see that only $21 \over 1$ remains, making this product $21$. Now, we square it to get the original product of $441$. So, was it valid to write the product of a square as the square of a product?

答: Absolutely.  You are using an expanded form of $(ab)^2=a^2b^2$, which relies on the fact that multiplication is commutative.  Your telescoping works without pulling the square out, too.  All the terms in the product are squared compared to what you did and the squares cancel nicely.


问:
Wrapping a Rubber Band around a Cube
Imagine a cube wrapped with string so that each of the six sides has two linked "L's" on it, like this image (or some reflection or rotation thereof).

You can get a lot of different knots and links depending on the particular choices made.

Are any of them the unknot or the unlink?

In other words, can any of them be achieved by wrapping real rubber bands around a real box?

答:
Here's the geometric net of a cube:

I represented the cube as a graph $g$ of size $6$ vertices, each vertex of degree four (links to its neighbors).

Then I searched for an Eulerian path through the graph $g$ with the conditions that no three consecutive sides were collinear.  (That is, there must be an L turn at each vertex.)  Here is one such Eulerian path:

$\{a \to b \to f \to a \to d \to f \to e \to d \to c \to e \to b \to c \to a \}$

There are then $2^5$ distinguishable choices for "over-under."  Some will be trivial, others (ummm.... "knot"!).

Title: Wrapping a Rubber Band around a Cube


问:
adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ is homeomorphic to $\mathbf{S}^{2n-1}$
Consider the adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ by gluing $\mathbf{S}^{n-1}\times \mathbf{S}^{n-1}$ together, is there a good way to prove that $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}\cong\mathbf{S}^{2n-1}$? When $n=1$, I can check manually that it is true (attach four lines together by glueing their end points, constructing a square that is homeomorphic to $\mathbf{S}^{1}$), but I don't know how to prove it in general.
Thanks!

答:
$S^{n-1} \times D^{n} \cup D^{n} \times S^{n-1}$ is the boundary of $D^n \times D^n$

(by the formula $\partial(M \times N) = \partial M \times N \cup M \times \partial N$)

And since $D^{n} \times D^n$ is homeomorphic to $D^{2n}$ its boundary is $S^{2n-1}$.

Thus we conclude that $S^{n-1} \times D^{n} \cup D^{n} \times S^{n-1} \approx S^{2n-1}$


问:
Irreducible representations of direct product of compact groups
I am trying to prove that for $G$ and $H$ two compact groups, it's true that any irreducible $G \times H$-representation $U$ is isomorphic to a tensor product of an irreducible $G$-representation $V$ and an irreducible $H$-representation $W$, ie. that $U \simeq V \otimes W$.

After thinking a lot about this, I found the following argument in some online lecture notes (Lemma 22.6 in https://math.berkeley.edu/~teleman/math/RepThry.pdf):

"From the properties of the tensor product of matrices, it follows
that the character of $V \otimes W$ at the element $g \times h$ is
$\chi_{V \otimes W} (g \times h) = \chi_V (g) \cdot \chi_W (h)$. Now, a conjugacy class in $G \times H$ is a Cartesian product of conjugacy classes in $G$ and $H$, and character theory ensures that the $\chi_V$ and $\chi_W$ form a Hilbert space basis of the $L^2$ class functions on the two groups. It follows that the $\chi_V (g) \cdot \chi_W (h)$ form a Hilbert space basis of the class functions on $G\times H$, so this is a complete list of irreducible characters. $\blacksquare$"

I understand the claim about the conjugacy classes of the direct product and everything else, except how the conclusion that $\chi_V \cdot \chi_W$ are a full set of irreducible characters for the product group follows from this.

I just don't get why a class function $f : G \times H \rightarrow \mathbb{C}$ can be written as a linear combination of the various $\chi_V \cdot \chi_W$, as $V$ and $W$ vary through all irreps of $G$ and $H$. I tried to convince myself of this by looking at, say,
\begin{align*}
f_G : G \times \lbrace e \rbrace &\longrightarrow \mathbb{C}, \text{ and}\\
f_H : \lbrace e \rbrace \times H &\longrightarrow \mathbb{C},
\end{align*}
and use that since $f_G$ and $f_H$ are class functions, they can each be written as a linear combination of $\lbrace \chi_V \rbrace_V$ and $\lbrace \chi_W \rbrace_W$ respectively. But I can't seem to connect this fact to the conclusion I'm trying to reach.

Any tips on why the specific argument works or how the result might be proved in a different way are most welcome!

答: The basic idea is that $L^2(G)\otimes L^2(H) \cong L^2(G\times H)$ by the canonical isomorphism $A\otimes B \mapsto A(g)B(h)$. Then the statement follows.


问:
How to prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$ diverges?

Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.

Please provide me a hint or a full solution.

答:
$\sum_{n=2^m}^{2^{m+1}-1}\frac{u_n}{n}
>\sum_{n=2^m}^{2^{m+1}-1}\frac{u_{2^{m+1}}}{2^{m+1}}
= 2^m \frac{u_{2^{m+1}}}{2^{m+1}}
=u_{2^{m+1}}/2
$.

Therefore
$\sum_{n=1}^{2^m-1} \frac{u_n}{n}
> \frac1{2}\sum_{k=1}^m u_{2^k}
$
and, if the latter sum diverges, so does the former.

Note that an almost identical argument will show that
if $(a_m)_{m=1}^{\infty}$
is an increasing sequence of positive integers
such that there is a $c < 1$ such that
$\frac{a_{m}}{a_{m+1}} \le c$ for all $m$,
and $u_n \ge u_{n+1}$ for all $n$,
then $\sum_{m=1}^{\infty} u_{a_m}$ diverges
implies that
$\sum_{n=1}^{\infty} \frac{u_n}{n}$ diverges.

Here is the proof by copy and paste:

$\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}
>\sum_{n=a_m}^{a_{m+1}-1}\frac{u_{a_{m+1}}}{a_{m+1}}
= (a_{m+1}-a_m) \frac{u_{a_{m+1}}}{a_{m+1}}
=(1-\frac{a_{m}}{a_{m+1}})u_{a_{m+1}}
=(1-c)u_{a_{m+1}}
$.
Therefore
$\sum_{n=1}^{a_m-1} \frac{u_n}{n}
> (1-c)\sum_{k=1}^m u_{a_k}
$
and, if the latter sum diverges, so does the former.

The fact that $1-\frac{a_{m}}{a_{m+1}}$
occurs leads to the requirement on $a_m$.

We can also go the other way
and see what is needed.

$\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}
<\sum_{n=a_m}^{a_{m+1}-1}\frac{u_{a_{m}}}{a_{m}}
= (a_{m+1}-a_m) \frac{u_{a_{m}}}{a_m}
=(\frac{a_{m+1}}{a_m}-1)u_{a_{m}}
$.

From this we see that if
$\frac{a_{m+1}}{a_m}-1 < d$
for some real $d$ and all $m$,
then
$u_{a_{m}} > \frac1{d}\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}$,
so that
$\sum \frac{u_n}{n}$ diverges
implies that
$\sum u_{a_m}$ diverges.

Combining these two,
we get this:

If $(a_m)_{m=1}^{\infty}$
is an increasing sequence of positive integers
such that there exists positive reals $c$ and $d$ such that
$c \le \frac{a_{m+1}}{a_{m}}-1 \le d$ for all $m$,
and $u_n \ge u_{n+1}$ for all $n$,
then $\sum_{m=1}^{\infty} u_{a_m}$ converges
if and only if
$\sum_{n=1}^{\infty} \frac{u_n}{n}$ converges.


问:
Discrete math - Show that an argument is valid using premises
I've seen things like this where we are asked to validate some argument in the form below for instance. $$p\rightarrow q$$
$$p \lor r$$ $$r$$

Now, I get why logic is important. Any cursory glance at some proofs can testify to its relevance. But I really have a hard time seeing how it benefits a math major at all when taught in this way. We have to look at this and then argue using things like modus ponens, modus tollens, disjunctive simplification and a bunch of tedious things and we literally never think of it again no matter how many proofs we write or how many problems we tackle. Why are we ever bothering with this?

Is this something like what Lockhart's Lament would decry as sort of sterile and joyless that ultimately ends up being kinda ugly and boring? I don't want to bash any subject here and I hope that's not what it seems like.

答:
My perspective: I use mathematical logic as the primary organizing force in all the mathematics I do in any subject. Of course, this is as it should be, since mathematical arguments in any subject are mediated through logical relationships — as you point out in your post.

I think the problem you point out is right: logic isn’t taught (in most schools) to be useful. It’s still often couched in philosophical terminology from antiquity. This is indeed a shame, because complicated logical arguments are very difficult to carry out in natural language, but are effortless to manipulate using mathematics.

The style of logic I learned was a calculational, algebraic approach championed by the Dutch mathematicians and computing scientists Edsger Dijkstra, Wim Feijen, and Netty van Gasteren. I’d be happy to give you more information if you’re interested — it’s not clear from your post whether you’re a student or teacher or where your question/rant is coming from.


问:
Compute $\mathbb{E}[X]$ and $Var[X]$ for $\lambda=1$
Let X be a discrete random variable such that ($\lambda\geq0)$
$$P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}\quad\text{for}\;k\in\{0,1,...\} $$

Check that $\displaystyle{\sum^\infty_{k=0}P(X=k)=1}$

Compute $\mathbb{E}[X]$ for $\lambda=1$

Compute Var$[X]$ for $\lambda=1$

My attempt:

For the first part, we can use the fact that the taylor expansion of $e^x$ is $e^x=\displaystyle{\sum_{k=1}^\infty\frac{x^k}{k!}}\;\forall\;x\in\mathbb{R}$.
Then, $$\displaystyle{\sum^\infty_{k=0}P(X=k)=\sum^\infty_{k=0}\frac{\lambda^k}{k!}e^{-\lambda}=e^\lambda\cdot e^{-\lambda}=e^0=1}$$

For the second part, since we know that this is a valid p.d.f (from the first part), we can use the formula $\mathbb{E}[X]=\displaystyle{\sum_{k\in X}kp(k)}$. Then,
$$\mathbb{E}[X]=\sum k\cdot\frac{1^k}{k!}e^{-1}=\sum \frac{1^k}{(k-1)!} e^{-1}=1$$

This is where I am stuck. I don't know if I am finding the expectation correctly because the result seems strange.

Once I am able to find this, I can find the third part by using the formula $\text{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ to find the variance.

I also have a question about finding $\mathbb{E}[X^2]$. Is it $\displaystyle{\sum_{k\in X}(kp(k))^2}$ or $\displaystyle{\sum_{k\in X} k^2p(k^2)}$?

答:
You have $\displaystyle \sum_{x=0}^\infty \frac{\lambda^x e^{-\lambda}}{x!} = 1.$ Then
$$
\operatorname E(X) = \sum_{x=0}^\infty x \Pr(X=x) = \sum_{x=0}^\infty x\cdot \frac{\lambda^x e^{-\lambda}}{x!}
$$
Now notice that in the line above

I am carefully distinguishing between two different things, denoted capital $X$ and lower-case $x$ (and I partially cleaned up the notation in the question);

When $x\ge1$ we have $x\cdot\dfrac 1 {x!} = \dfrac 1 {(x-1)!},$ but when $x=0$ that cannot be done. And need not be done, since that term is $0.$

Since the $x=0$ term of the last sum is $0$ we can drop it and get $\displaystyle\sum_{x=1}^\infty x\cdot\frac{\lambda^x e^{-\lambda}}{x!}$ (starting at $x=1,$ not at $x=0$). And then we can do the cancelation noted above and get $\displaystyle \sum_{x=1}^\infty \frac{\lambda^x e^{-\lambda}}{(x-1)!}.$ And this is equal to
$$
\lambda\sum_{x=1}^\infty \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!}.
$$
If we let $y=x-1$ then we notice that as $x$ goes from $1$ to $\infty$ then $y$ goes from $0$ to $\infty,$ and the sum becomes
$$
\lambda\sum_{x=1}^\infty \frac{\lambda^y e^{-\lambda}}{y!}
$$
and that sum is equal to $1,$ so the expected value is $\lambda.$

Maybe the simplest way to find $\operatorname E(X^2),$ and thus to get $\operatorname{var}(X) = \operatorname E(X^2) - \big(\operatorname E(X) \big)^2,$ is to find $\operatorname E(X(X-1)) = \operatorname E(X^2) - \operatorname E(X),$ as follows:
$$
\operatorname E(X(X-1)) = \sum_{x=0}^\infty x(x-1)\cdot\frac{\lambda^x e^{-\lambda}}{x!}.
$$
Now notice that the first two terms, with $x=0$ and $x=1,$ are $0.$ And so as above we get
$$
\lambda^2 \sum_{x=2}^\infty \frac{\lambda^{x-2} e^{-\lambda}}{(x-2)!} = \lambda^2 \sum_{y=0}^\infty \frac{\lambda^y e^{-\lambda}}{y!}
$$
where this time $y=x-2.$ Again the sum is $1$ and we have $\operatorname E(X(X-1)) = \lambda^2.$ Therefore $\operatorname E(X^2) = \lambda^2+\lambda,$ and so $\operatorname{var}(X) = \lambda.$