问:
Improper integral $\sin(x)/x $ converges absolutely, conditionally or diverges?
Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?

We have
$$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$

Integrating by parts
$$u=\frac{1}{x}$$
$$\text{d}u=-\frac{1}{x^2}\text{d}x$$
$$\text{d}v=\sin x\;\text{d}x$$
$$v=-\cos x$$
$$
\begin{aligned}
\int_1^{\infty} \frac{\sin x}{x} \text{d}x
& = \frac{-\cos x}{x} \Big|_1^{\infty}
    - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \\
& = \cos 1 - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x
\end{aligned}
$$

$\int_1^{\infty} \frac{\cos x}{x^2} \text{d}x$ converges absolutely
(using the Comparison Test For Improper Integrals):

$$
\int_1^{\infty} \frac{|\cos x|}{x^2} \text{d}x <
\int_1^{\infty} \frac{1}{x^2} \text{d}x
$$

So $\int_1^{\infty} \frac{\sin x}{x} \text{d}x$
converges.

Now I need to find out if
$\int_1^{\infty} |\frac{\sin x}{x}| \text{d}x$
converges or diverges.

答:
Let $N \in \Bbb N, N > 1$, we have:

\begin{align}
\int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\
&\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)}
\end{align}

The last sum diverges as $N \to \infty$, and so does the original integral.

Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.


问:
Derivative of trace-based linear scalar field
I'm just getting into matrix calculus, so this question might be really easy for you guys out there. I'm trying to understand some of the simpler derivations in The Matrix Cookbook. I've been looking at derivatives of the trace predominantly. The following formulas are given:

$$
\frac{\delta}{\delta X} {\rm Tr}(XA) = A^T
$$

$$
\frac{\delta}{\delta X} {\rm Tr}(AX^T) = A
$$

I can't completely reproduce these example though. When I try to take the derivative with a small example manually, I always get $A$.

\begin{align}
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
x_1 & x_2 \\
x_3 & x_4
\end{bmatrix}
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\right) &=
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1x_1 + a_3x_2 & a_2x_1 + a_4x_2 \\
a_1x_3 + a_3x_4 & a_2x_3 + a_4x_4
\end{bmatrix}\right)
\\
&=
{\rm Tr}\left(
\begin{bmatrix}
\begin{bmatrix}
a_1 & a_2 \\
0 & 0
\end{bmatrix} &
\begin{bmatrix}
a_3 & a_4 \\
0 & 0
\end{bmatrix} \\
\begin{bmatrix}
0 & 0\\
a_1 & a_2
\end{bmatrix} &
\begin{bmatrix}
0 & 0 \\
a_3 & a_4
\end{bmatrix}
\end{bmatrix}
\right)\\
&=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\end{align}

\begin{align}
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\begin{bmatrix}
x_1 & x_3 \\
x_2 & x_4
\end{bmatrix}
\right) &=
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1x_1 + a_2x_2 & a_1x_3 + a_2x_4 \\
a_3x_1 + a_4x_2 & a_3x_3 + a_4x_4
\end{bmatrix}\right)
\\
&=
{\rm Tr}\left(
\begin{bmatrix}
\begin{bmatrix}
a_1 & 0 \\
a_3 & 0
\end{bmatrix} &
\begin{bmatrix}
0 & a_2 \\
0 & a_4
\end{bmatrix} \\
\begin{bmatrix}
a_1 & 0\\
a_3 & 0
\end{bmatrix} &
\begin{bmatrix}
0 & a_2 \\
0 & a_4
\end{bmatrix}
\end{bmatrix}
\right)\\
&=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\end{align}

I'm sorry if the notation is wrong and thanks for your help.

答:
Solving the problem for a small case isn't helpful at all. In general, the basic idea is to expand the product $XA$ in terms of the elements $X_{ij}$ and $A_{ij}$ of the matrices $X$ and $A$ . Clearly, it'll turn out to be a sum in terms of $X_{ij}$ and $A_{ij}$ for $i,j = 1,2,\cdots , n$ .

Let $F(X) = \text{tr}(XA)$ . Now, you can partially differentiate the whole thing w.r.t. $X_{ij}$ , and eventually you'll obtain only $A_{ji}$ . Therefore, the Jacobian matrix corresponding to the linear transformation $DF$ has $A_{ji}$ in its $(i,j)^\text{th}$ position, and this holds for every pair $(i,j)$ . Thus, $DF_X = A^T$ .

Note : If you want a clearly written proof, you may refer to this link. Thank you.

ContentLicense: CC BY-SA 4.0


问:
About finding the closure of a set.
I want to learn how should I find the closure of a set $S$. It's a long that I was not worked with topological concepts, and I am afraid if I am mistaken about some easy things. Here I will list some of the facts which will help me to solve my problems and pursue my purposes.

Suppose that a topological space $X$ is given, and let $R, S \subseteq X$ be two sets.

Is this true? The closure of $S$ is the union of $S$ with all limit points of $S$. (I think it should be the definition of the closure, or at least it should be equivalent to the definition, any other equivalent definitions are welcome.)

If $S$ is dense in $R$, then $R \subseteq \overline S$.

$\overline S$ is the largest set, in which $S$ is dense. (i.e. if $\overline S \subsetneq R'$, then any $x\in R'\backslash \overline S$ has a positive distance from $S$, therefore is not a limit point)

The next two points, are not related to the closure, but I have some doubts.

A function $f: X \rightarrow Y$ between two topological spaces is called continuous, iff for every open $V \subseteq Y$, $f^{-1}(V)$ is open in $X$. (I think this was a standard definition, any other equivalent definition are welcome.)

A function $f: X \rightarrow Y$ between two topological spaces, is continuous, iff for every close $V \subseteq Y$, $f^{-1}(V)$ is close in $X$.

My intuition says that, the first $4$ points are true. Am I right? If I am mistaken about these facts, please tell me, and if it is possible please give me a counter-example. Also, I think the last point is also equivalent to the $4^{th}$ point, because if a set is closed iff and only if its complement is open. Having this in mind it seems the last two points are equivalent to each other as the definition of a continuous function.

Any equivalent definitions to the $1^{st}$ point and $4^{th}$ point are welcome. Especially those equivalent definitions, which are useful for computing the closure. Using the first three points I am able to find the closure of some simple sets in $p$-adic topology.

答:
Yes, a set is "closed"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set.  (The closure of a set is also the intersection of all closed sets containing it.)

2.Yes, that is pretty much the definition of "dense".

Yes, again that follows directly from the definition of "dense".

The spelling is "continuous", not "continues".  But, yes, that is a standard definition of "continuous".

Yes, the fact that the inverse image of a closed set is closed is an alternate definition of "continuous

Title: About finding the closure of a set.


问:
Is this derivation of $\frac{dy}{dx}$ correct?
Derivatives show us how fast something is changing at any point. For example; the gradient of the graph of $y = x^2$ at any point is twice the value of $x$ thereat. The process of finding the derivation of a gradient / slope of a function $y=f(x)$ or $y = x^2$ is as follow.

Pick any two points $A$ and $B$ close to each other on the curve of $y =x^2$. The coordinates of $A$ on the curve are $(x, y)$ or $(x, x^2)$. Add $Δx$ at $A$ as usual. When $x$ increases by $Δx$, then $y$ increases by $Δy$. The $x$ changes from $x$ to $(x +Δx)$ while $y$ changes from $y$ to $(y + Δy)$ or $f(x)$ to $(x+Δx)^2$. Thus the $x$ and $y$ coordinates of $B$ on the curve are $(x + Δx, y + Δy)$ or $([x+Δx, (x+Δx)^2].$ Now the instantaneous rate of change is given by

$$\frac{Δy}{Δx} = \frac{[(x + Δx)^2 – x^2]}{[x + Δx - x]}$$

$$\frac{Δy}{Δx} = \frac{[x^2 + Δx^2+2xΔx − x^2]}{ Δx}$$

$$\frac{Δy}{Δx} = \frac{[2x + Δx]}{ 1}$$

Reduce $Δx$ close to zero by taking limit ($Δx$ to $dx$ and $Δy$ to $dy$)

$$\frac{Δy}{Δx} = 2x + dx$$

$$\frac{Δy}{Δx} = 2x \tag1\label{eq1} $$
OR

$$dy = 2x.dx \tag2\label{eq2}$$

ABC is an infinitesimal triangle made by $dx, dy$, and hypotenuse or slope of tangent where point $A$ and $C$ are always on the curve. Length of $AB$ = Base = $dx$, Length of $BC$ = Perpendicular= $dy$ and Length of Hypotenuse = $AC$. $\angle CAB$ or $\angle BAC$ is the slope of a tangent

According to the aforementioned $\eqref{eq1}$ or $\eqref{eq2}$

•$\frac{dy}{dx}$ is directly proportional to $x$ or $\angle CAB$ is directly proportional to $x$.

• $dx$ is indirectly proportional to $x$ OR $x$ is inversely proportional to $dx$.

• $dy$ is directly proportional to $x.dx$ or $dx$

The length of $dx > dy$ when $\angle CAB < 45$ degrees The length of $dx = dy$ when $\angle CAB = 45$ degrees The length of $dx < dy$ when $\angle CAB > 45$ degrees.

The proportionality of both the $\angle CAB$ and $dy$ with $x$ are in contradiction with the proportionality of $x$ and $dx$ in the triangle $ABC$ after probing the equation of $\frac{dy}{dx} = 2x$ beyond its derivation on a graph of $y = x^2$. When $x$ increases; $dx$ decreases, $dy$ increases, and $\angle CAB$ increases. This means $AC$ also increases and ultimately SECANT when $x$ increases. Our goal is to bring $dx, dy$ and $AC$ to zero (not away from zero either positively or negatively - Point $C$ has to be on the curve) or secant to tangent by reducing them close to zero but here $dx$ heads toward zero but $dy$ and $AC$ increases when $x $ increases on axis mathematically.

Although the difference in the length of $dx$ and $dy$ can be noticeable clearly on the graph if we examine the triangle $ABC$ at two different points for a gradient ($\frac{dy}{dx}$), say when an $\angle BAC = 0.1$ degrees and 89.9 degrees on the curve but UNIT CIRCLE is the best example for observing the change in an $\angle CAB$ (say 0.1 and 89.9 degrees) of a triangle $ABC$ for $dy$ and $dx$ and the comparison of their lengths.

$RISE = dy = 2x$ and $RUN = dx = 1$ (always constant) in a GRADIENT of 1 in $2x$ which we obtained from the $\eqref{eq1}$ of $\frac{dy}{dx}=\frac{2x}{1}$ at any point on the curve when there is no difference between secant and tangent – No idea how do we get $\frac{dy}{dx} = 2x.dx$ but above said contradiction may be due to the introduction of another curve of $y =(x+dx)^2$ at a point where we seize $x$ or $y=x^2$  deliberately and introduce delta $x$ OR when function $y = f(x)$ changed to $y=f(x+Δx)^2$. The value of $x$ has reached to its maximum value instead of unlimited when a curve $y=x^2$ doesn’t continue anymore at a point where we introduce delta $x$ or $dx$ as $y=x^2$ and $y =(x+dx)^2$ are two different types of curve (two diffrent functions).

Further, integration is the reverse process of differentiation. Although delta $x$ or $dx$ is ignored during the process of derivation of $\frac{dy}{dx}$ because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, $dy$ is the small vertical change in $y$, therefore, we take the sum of all the small vertical lengths [dy(s)] not the whole slice or y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of $dy = 2xdx$ but it turns into function of $x^2$ or area under the graph – no idea how but summation of vertical lengths on a graph gives vertical length only not curve?

The derivation of the natural relationship of a gradient of 1 in $2x$ at any point with $y=x^2$ or $\frac{dy}{dx}=2x$ is still unbeknownst to illuminates - Anyone who is in agreement with all above unless satisfied by logic.

Below figure may help with the above.

Image

答:
The derivative of a univariate function $f(x)$ at the point $x$ is defined to be

$$f'(x):=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ when this limit exists. Full stop. This definition is non-negociable.

The geometric interpretation is that of the local slope of the graph of the function, and this is correct as well.


问:
What is minimum number of edges should be removed from $K_6$ to get planar graph?
What is minimum number of edges should be removed from $K_6$ to get planar graph?

It is easy to show it just with picture.. but is it possible to prove it analytically?

答: Hint: What is the maximum number of edges in an $n$-vertex planar graph? How many edges are in $K_6$? How do these numbers compare?


问:
Is $(\log n)^3$ the same as $\log^3 n$?
I saw the notation

Where I would have expected $(\log n)^3$.

Are both the same? Just a different style?

答: Yes, they are the same. Actually the notation $\log^3 n$ should mean the composition of $\log$ with itself three times, meaning $\log \log \log n$, to be coherent with the fact that some inverses are denoted as, for example, $\sin^{-1} x$, but for a conventions reasons it is accepted that $\log^3 n =(\log n)^3$.

Title: Is $(\log n)^3$ the same as $\log^3 n$?


问:
Probability with binomial model of 3 events
If one person will guess exactly $7$ correct answers, we have a binomial model with $k=7$ successes and $n=10$ trials with probability $p$. So for the first one, the probability will be $binopdf(7,10,0.8)=0.2013$, for the second is $0.057$ and for the third is $0.25$. Now from here, would it be fine to add them, to answer the exercise?

答:
Now from here, would it be fine to add them, to answer the exercise?

Let's make the following modifications: the total questions are 2 and for each contestant the probability to guess correct answer is 0.5.  Find the probability that one contestant (any one) answers exactly 1 question correctly, while the other two give any other number of correct answers.

The probability is $\frac{1}{2}$ for any partecipant...do you think you can add

$$\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}=150\%$$

??

Using the probabilities you calculated, the answer is

$$0.2013(1-0.057)(1-0.25)+0.057(1-0.2013)(1-0.25)+0.25(1-0.2013)(1-0.057)$$


问:
A property of Lyndon words.
A word is a Lyndon word if it is strictly smaller than any of its proper right factors. Let $A$ be an alphabet and $L$ be the set of all Lyndon words over $A$. For a word $w \in L-A$, the pair $(l, m)$, $l, m \in L$, such that $w=lm$ and $m$ is of maximal length is called the standard factorization of $w$ and it is denoted by $\sigma(w) = (l,m)$.

In the book Combinatorics on words, on page 67, there is a proposition left as an excise:

Let $w \in L - A$, and $\sigma(w) = (l, m)$ be its standard factorization. Then for any $n \in L$ such that $w < n$, the pair $(w,n)$ is the standard factorization of $wn \in L$ if and only if $n \le m$.

The necessary condition is proved as follows. Suppose that $n>m$. Then $mn \in L$. The length of $mn$ is larger than the length of $n$. Therefore $(w,n)$ is not the standard factorization of $wn$.

How to prove the sufficient condition? Thank you very much.

答:
This is Lemma 1.6 in K. T. Chen, R. H. Fox & R. C. Lyndon. Free Differential Calculus, IV. The Quotient Groups of the Lower Central Series (1958). Annals of Mathematics 68(1), 86. Below I rewrite their proof with notation and terminology following the conventions of the book you mention. (Citations for posterity: D. Perrin, Factorizations of Free Monoids (chapter 5 in Combinatorics on words (1983). Cambridge Univ. Press, edited by M. Lothaire)).
(The terminology in the Chen–Fox–Lyndon paper is certainly nicer and simpler though, in my opinion at least.)

Suppose that $e$ is a proper right factor of $wn$ that is longer than $n$. We will show that $e\notin L$. Either (1) $e=fn$ where $f$ is a proper right factor of $m$, or (2) $e=mn$, or (3) $e=gmn$ where $g$ is a proper right factor of $l$. (You should now try proving it before looking at the solution.)

 (1) If $e=fn$, we have $m<f$ as $m\in L$. Thus $n\le m<f<fn$ and so $e\notin L$.

 (2) If $e=mn$, we have $n\le m<mn$, so $e\notin L$.

 (3) If $e=gmn$, we suppose for contradiction that $e\in L$, and that $g$ is the shortest proper right factor of $l$ for which $gmn\in L$. From cases (1) and (2) we see that $n$ must be the maximal proper right factor of $gmn$ that is Lyndon. As such, by Proposition 5.1.3, we see that $gm\in L$. But this contradicts the assumption that $m$ is the maximal proper right factor in $lm$ satisfying $m\in L$.

It follows that $n$ is the maximal proper right factor of $wn$; that is, $\sigma(wn)=(w,n)$.

Title: A property of Lyndon words.


问:
Alternate definition of Entropy
I am going through the book Computational Optimal Transport by Peyré and Cuturi.
In it (see Formula (4.1), P. 65/209) I came across the definition of the discrete entropy, which is not what I expected:
$$
H(P) = - \sum_{i,j}P_{i,j}(\log(P_{i,j})-1),
$$
with $P \in [0,1]^{n,m}$ being a coupling matrix, i.e. describing the transformation of discrete propability distributions.
What confuses me is the $-1$ in the sum and I would like to know, why it is included. Since  $\sum_{i,j} P_{i,j}=1$ it is shifting the result by $-1$.  Still it confuses me.

答: The obvious reason $-\sum_{i,\,j}P_{i,\,j}\ln P_{i,\,j}$ is a more common definition is that it's extensive, but we often don't care about that, so much as about relative entropies. In that case, one may as well add a constant $c$ to taste, giving $c-\sum_{i,\,j}P_{i,\,j}\ln P_{i,\,j}=-\sum_{i,\,j}P_{i,\,j}(\ln P_{i,\,j}-c)$. The choice $c=1$ has the unique attraction that the whole expression is just $-\sum_{i,\,j}\int_0^{P_{i,\,j}}\ln pdp$.

Title: Alternate definition of Entropy


问:
Properties of a an entire function: Solution verification
I would like to answer the following question

My solution:

(i) Since $f$ is entire it follows that $f$ is holomorphic and hence $f$ is analytic. Since $f$ is analytic on the whole $\mathbb{C}$ it follows that $f$ is analytic on the origin. Therefore we can write $f(z)= \sum a_n z^n $. If $f(z)/z \rightarrow 0$ as $|z| \rightarrow \infty$ then

$$\lim_{|z|\rightarrow \infty}\frac{\sum a_n z^n}{z}=\lim_{|z|\rightarrow \infty} \frac{a_0}{z} + a_1 + \sum_{n=2}^{\infty}a_n z^{n-1}=\lim_{|z|\rightarrow \infty}a_1 + \sum_{n=2}^{\infty}a_n z^{n-1}=0 $$

This must be true for all $z$ outside of the disc $D(0,N)$ (where $N$ is sufficiently large).  Since the function we are taking a limit of is analytic then by analytic continuation it follows that $a_1 + \sum_{n=2}^{\infty} a_nz^{n-1}=0$ for all $z$.  From this we get that
$$f(z)=a_0$$
and hence $f$ is constant.

(ii) Since $f$ is analytic it follows that if $f'(z)\neq 0$ then $f$ has an analytic isomorphism (by the inverse mapping theorem). The existence of the analytic isomorphism implies that $f$ is surjective and hence $f$ maps the whole of $\mathbb{C}$ to $\mathbb{C}$.  Now suppose that the condition in (ii) true. Since $f$ is subjective we can pick $z_0$ s.t. $f(z_0)=b$ and hence $|f(z_0)-f(z_0|=0$ which is a contradiction (could have also used the continuity of analytic functions here so that we have $|f(w=z_0+\delta)-f(z_0)|<\epsilon)$. Therefore the assumption that $f'(z)\neq 0$ must be false. Hence $f$ is constant.

My question: Is this correct?

答:
Part $1$ is fine, it is proved correctly.

For part $2$, we use Liouville's theorem on the function $\frac 1{f(z)-b}$, which is an entire function because the denominator is non-zero (and division, shifting etc. preserve entirety if valid), and also is a bounded function since $$\left|\frac 1{f(z)-b}\right|= \frac 1{|f(z)-b|} < \frac 1{\epsilon}$$

for all $z$. Hence $\frac 1{f(z)-b}$ is constant, therefore $f$ is constant.

The inverse mapping theorem tells you this : if $f'(z) \neq 0$ at a point , then there exist neighbourhoods $U,V \subset \mathbb C$ with $z \in U$ and $f(z) \in V$ such that $f: U \to V$ is analytic and a bijection.

The problem with what you want to say is two-fold. First of all, what if $f'(z)=0$ at some point? Note that if $f'$ is entirely $0$, only then is $f$ constant, but if $f$ is not constant it is still possible for $f'$ to have zeros. So at these points you can't create such a map.

Next, even if the function has non-zero derivative at all points, it still doesn't mean it is surjective. This is because, even if for each point you take a neighbourhood around it for which the isomorphism exists, the $V$s put together don't give you the entire of $\mathbb C$. Take the exponential function as an example of this phenomena.

Here is a nice result on what the image of an entire map must be.

Little Picard Theorem : Every nonconstant analytic function omits at most one complex number from its range. So at most one number cannot be the image of a non-constant entire map. The exponential map omits $0$, for example, but nothing else.


问:
Example of $f$, $g$ and $a$, such that $(f \circ g)'(a)$ exists, and $g'(a)$ exists, but $f'(g(a))$ does not.
Find an example of 2 functions $f$ and $g$ and a point $a \in
\mathbb{R}$, such that $(f \circ g)'(a)$ and $g'(a)$ exists, but $f'(g(a))$ does not exist, and also $f$ and $g$ must take on all values in $\mathbb{R}$.

Same problem as this one.

The solution I came up with was

$a = 0 ~~~~~~~~ f(x) =
\begin{cases}
x:  & x \geq0 \\
2x: & x \leq 0
\end{cases} \Bigr\} ~~~~~~~~ g(x) = x^3$

That's fine, but now I'm trying to find an example where $g'(a) \neq 0$, or alternatively prove $g'(a) \neq 0 \implies f'(g(a))$ exists.

I'm not even sure which conclusion is correct.

答:
If you assume in addition that $g'$ is continuous then $g'(a)\ne0$, $f\circ g$ differentiable imply $f$ differentiable.

Because now $g$ is a local homeoomorphism; if $h$ is a local inverse for $g$ near $a$ then $f=(f\circ g)\circ h$ is differentiable, being the composition of two differentiable functions.


问:
Contour integration and Green's function
I have a question which is related to changing an improper integral into a contour integral that 'looks' similar, but is not the same. The author (D.H. Griffel) says that it is 'right' to do it, but I don't understand why. It appears this is related to the standard thesis in the use of Green's functions. The author and I want to 'solve' the PDE
$$
u_{tt} - \nabla^2 u = f.
$$
The author rewrites the problem as
$$
(\partial_t ^2 - \nabla_x^2) w(\vec{x},t;\vec{y},s) = \delta(x-y) \delta(t-s).
$$
By using the Fourier transform he finds
$$
-(\omega^2 - |\vec{k}|^2) v = e^{i(\vec{k}\cdot \vec{y} - \omega s)}
$$
with $v$ being the Fourier transform of $w$. Now the question of contour integration arises. The author says that
$$
w(\vec{x},t;\vec{y},s) = \frac{1}{4 \pi^3 |\vec{y} - \vec{x}|} \int_{0}^\infty k \sin k |\vec{y} - \vec{x}| I(k)\,  \, dk
$$
with
$$
I(k) = \int_{- \infty}^\infty \frac{e^{i\omega(s-t)}}{k^2 - \omega^2} d\omega.
$$
The author can't evaluate this integral because it has poles at $\omega = \pm k$. Instead, the author evaluates different contour integral that avoids the poles and is closed in the upper- or lower half-plane.

But my 'gripe' is that these integrals are not the same integral as $I(k)$. The author says that each choice of integration contour corresponds to a different $w$. My question:

Why is it relevant to look at these alternative integral contours; after all, we're not interested in them. I can't see a reason why they are related to the integral we want to determine that would give us $w$. Is this anecdotal, or something rigorous?

How can the author prove that each choice of contour corresponds to different boundary conditions?

What do I mean by 'the integrals are not the same':

$I(k)$ is not a complex number for almost any $k$ (in the sense of lebesgue or riemann)

The integral of the integrand of $I$ over a contour avoiding the poles is a complex number (in the sense of riemann).

In the case so-called $i \varepsilon$-prescription is something mathematical, it would be great if you could provide a reference to the theorem, or just post the theorem and idea for the proof here.

答:
Different $i\epsilon$ prescriptions in fact correspond to different boundary conditions on the Green function.

Your function $w$ is defined by a second order partial differential equation which does not completely specify it until boundary conditions are given. The different contours give you different functions which satisfy the same differential equation but have different homogeneous parts - they differ by solutions to the homogeneous differential equation and these parts parameterise the possible boundary conditions on your equation.

As pointed out in the comments, the integral is not well posed due to the poles and requires either a regularisation or modification to its definition. This could be done in some cases by using the Principal Value but there are other options such as deforming the contour to avoid the poles (these coincide for certain choice of contour).


问:
A closed form for $\int_0^\pi \lvert \sin(m t) \cos(n t) \rvert \, \mathrm{d} t$
Motivated by this nice question I have been trying to compute the function $f: \mathbb{R}^+ \to \left[0,\frac{1}{2}\right]$ defined by

$$f(\alpha) = \lim_{x \to \infty} \frac{1}{x} \int \limits_0^x \lvert\sin(\alpha s) \cos(s)\rvert \, \mathrm{d} s \, .$$

Note that $f(\alpha) \leq \frac{1}{2}$ follows from the Cauchy-Schwarz inequality.

In the answers to the original question the equidistribution theorem is used to show that $f(3 - 2 \sqrt{2}) = \frac{4}{\pi^2}$ holds. This argument can be extended to every irrational number, so we have $f(\alpha) = \frac{4}{\pi^2} \approx 0.405285$ for any $\alpha \in \mathbb{R}^+ \setminus \mathbb{Q}^+$ .

For rational arguments we can take $m , n \in \mathbb{N}$ with $\gcd(m,n) =1$ . The change of variables $s = n t$ yields
$$ f \left(\frac{m}{n}\right) = \lim_{x \to \infty} \frac{1}{x} \int \limits_0^x \lvert\sin(m t) \cos(n t)\rvert \, \mathrm{d} t \, .$$
The integrand is periodic with period $\pi$, so
$$ f \left(\frac{m}{n}\right) = \lim_{x \to \infty} \frac{1}{x} \left[\bigg\lfloor \frac{x}{\pi} \bigg\rfloor \int \limits_0^\pi \lvert\sin(m t) \cos(n t)\rvert \, \mathrm{d} t + \mathcal{O} (1)\right] = \frac{1}{\pi} \int \limits_0^\pi \lvert\sin(m t) \cos(n t)\rvert \, \mathrm{d} t\, .$$

Now the idea is to split the interval of integration into subintervals on which the sign of the product is constant and then use
$$ \sin(m t) \cos(n t) = \frac{\sin[(m+n)t] + \sin[(m-n)t]}{2}$$
to find the integrals. The result is basically given by a finite sum of cosines evaluated at the zeroes of the integrand.

The first few results are
\begin{align}
f(1) &= \frac{1}{\pi} \approx 0.318301 \, , \\
f(2) &= \frac{4}{3 \pi} \approx 0.424413 \, , \\
f\left(\frac{1}{2}\right) &= \frac{2(2\sqrt{2}-1)}{3\pi} \approx 0.388004 \, .
\end{align}
Interestingly, most of the other values (especially those with large $m$ and $n$) seem to be very close to $\frac{4}{\pi^2}$ .

This method can be used (at least in principle) to compute $f$ at every rational argument, but it becomes increasingly complicated for larger values of $m$ and $n$ . Maybe I am overlooking a simple trick, maybe there is a better method. My question is:

How can we find a general expression for $f\left(\frac{m}{n}\right)$ with arbitrary coprime $m,n \in \mathbb{N}$ ?

Edit 1 July 2020

Thanks to Zacky's bounty and River Li's and asgeige's nice answers we have some closed-form results for special cases and a promising (but still somewhat complicated) conjecture for the general case. Using similar methods, I have also found
$$ f(m) = \frac{2}{\pi (m^2-1)} \left[m \csc \left(\frac{\pi}{2m}\right) \cos \left(\frac{\pi}{2m} 1_{2\mathbb{N}}(m)\right) - 1_{2 \mathbb{N}-1}(m)\right]$$
for $m \in \mathbb{N} \setminus \{1\}$ ($1_A$ is the indicator function of the set $A$), which implies $\lim_{m \to \infty} f(m) = \frac{4}{\pi^2}$. While a simple expression for general values of $m,n$ seems unlikely, it may be possible to show that $\lim_{n \to \infty} f \left(\frac{m}{n}\right) = \frac{4}{\pi^2}$ and $\lim_{m \to \infty} f \left(\frac{m}{n}\right) = \frac{4}{\pi^2}$ do indeed hold for fixed $m \in \mathbb{N}$ and $n \in \mathbb{N}$, respectively.

Edit 11 January 2021

Thanks to River Li's second answer we now have a proof of the conjectured limits. We can even use partial fractions and the pole expansions of $\csc$ and $\cot$ to simplify the remaining series and obtain the following general result (valid for $m, n \in \mathbb{N}$ coprime and not both equal to $1$):
$$ f\left(\frac{m}{n}\right) = \frac{g_m\left(\frac{\pi}{2m}\right) - g_m \left(\frac{\pi}{2n}\right)}{m^2-n^2} \, , \, g_m = \begin{cases} x \mapsto \frac{\csc(x)}{x} &, \, m \in 2 \mathbb{N} - 1 \\ x \mapsto \frac{\cot(x)}{x} &, \, m \in 2 \mathbb{N}\end{cases} \, . $$

Body:
Assume that $m\ge 1, n\ge 1$, $m\ne n$ and $\mathrm{gcd}(m, n) = 1$.

By the identity
$$\left|\sin x\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k x),$$
we have
$$|\sin mt | = \frac{2}{\pi}-\sum_{i = 1}^\infty \frac{4}{\pi(4i^2-1)}\cos(2i mt)$$
and
$$|\cos nt| = |\sin (\pi/2 - nt)|
 = \frac{2}{\pi}-\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}(-1)^j \cos(2jnt).$$
Then we have
\begin{align}
|\sin (mt) \cos (nt)| &= \frac{4}{\pi^2}
- \frac{2}{\pi}\sum_{i = 1}^\infty \frac{4}{\pi(4i^2-1)}\cos(2i mt)\\
&\quad - \frac{2}{\pi}\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}(-1)^j \cos(2jnt)\\
&\quad + \sum_{i = 1}^\infty \sum_{j = 1}^\infty \frac{16(-1)^j}{\pi^2(4i^2-1)(4j^2-1)}\cos(2i mt)\cos(2jnt).
\end{align}
Note that
$$\int_0^\pi \cos(2i mt)\cos(2jnt) \mathrm{d} t = \left\{\begin{array}{ll}
                                                           0 & im - jn \ne 0 \\[4pt]
                                                           \frac{\pi}{2} & im - jn = 0.
                                                         \end{array}
\right.$$
Then we have
\begin{align}
\frac{1}{\pi}\int_0^\pi |\sin (mt) \cos (nt)| \mathrm{d} t
&= \frac{4}{\pi^2} + \sum_{i = 1}^\infty \sum_{j = 1}^\infty \frac{8(-1)^j}{\pi^2 (4i^2-1)(4j^2-1)}\delta(i m - jn)\\
&= \frac{4}{\pi^2} + \sum_{k=1}^\infty  \frac{8(-1)^{km}}{\pi^2 (4k^2n^2-1)(4k^2m^2-1)}\\
&= \frac{4}{\pi^2} + \frac{1}{m^2n^2}\sum_{k=1}^\infty  \frac{8(-1)^{km}}{\pi^2 (4k^2-1/n^2)(4k^2-1/m^2)}.
\end{align}
Note that
$$\left|\sum_{k=1}^\infty  \frac{8(-1)^{km}}{\pi^2 (4k^2-1/n^2)(4k^2-1/m^2)}
\right| \le \sum_{k=1}^\infty  \frac{8}{\pi^2 (4k^2-1)(4k^2-1)} = \frac{\pi^2 - 8}{2\pi^2}.$$
Thus, we have
$$\lim_{\max(m, n) \to \infty} \frac{1}{\pi}\int_0^\pi |\sin (mt) \cos (nt)| \mathrm{d} t = \frac{4}{\pi^2}.$$


问:
Convergence of $\int_0^\infty \sin(t)/t^\gamma \mathrm{d}t$

For what values of $\gamma\geq 0$ does the improper integral $$\int_0^\infty \frac{\sin(t)}{t^\gamma} \mathrm{d}t$$ converge?

In order to avoid two "critical points" $0$ and $+\infty$ I've thought that it would be easier to test the convergence of the sum (is this coherent?):
$$
\int_0^1 \frac{\sin(t)}{t^\gamma}\mathrm{d}t +
\int_1^\infty \frac{\sin(t)}{t^\gamma}\mathrm{d}t.
$$
For the second integral, it converges if $\gamma > 1$ (comparision) and also converges if $0 <\gamma \leq 1$. I'm stuck on proving the last part and the fact that the first integral converges for $\gamma < 2$. Any help would be appreciated. Thanks in advance.

PD: I've checked the answers for this question but I would not like to solve this integral using $(n\pi,(n+1)\pi)$ intervals.

答:
Informally, $ \sin x \approx x $ near $ 0 $. Hence, $\displaystyle \int_0^1 \frac{\sin x}{x^\gamma} \ dx $ converges if $\displaystyle \int_0^1 \frac{1}{x^{\gamma - 1}} \ dx $ which occurs for $ \gamma < 2 $.

Likewise, for large values of $ x $, $ \sin x $ oscillates between negative and positive. As long as the numerator is increasing, the sum of these areas is an alternating series whose terms are decreasing monotonically to zero in absolute value and so it will converge if $ \gamma > 0 $ by the Alternating Series Test.

Hence, the interval of convergence is $ (0, 2) $.

ContentLicense: CC BY-SA 3.0


问:
Integral of a binomial-like function through Integration by parts
I wonder how to calculate the following through integration by parts:
\begin{align*}
\int_{0}^{1}\binom{n}{i}u^{i}(1-u)^{n-i}\, du
\end{align*}
$i=0,1,2,\cdots,n$. I suppose Gamma function might be useful here?

答:
Using $\beta$-integral:

$$I={n \choose k} \int_{0}^{1}x^k(1-x)^{n-k} dx={n \choose k} \frac{\Gamma(k+1) \Gamma(n-k+1)}{\Gamma(n+2)}={n \choose k}\frac{k! (n-k)!}{(n+1)!}=\frac{1}{n+1}$$


问:
determine $\iiint zdv$ using spherical coordinates
Let $D$ be the region enclosed by the surface $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$

Determine:

$$\iiint zdv$$

using Spherical coordinates.

I'm trying to solve the above but I'm unsure how to go about it.

My attempt has been:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} zdv$$

which I then integrate the first layer leaving me with:
$$\int_0^{2\pi}\int_0^{2} 2-x^2-y^2dv$$

Should I have subbed in the spherical coordinates right at the start and integrated it as:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}p\cos\phi p^2\sin\phi dp d\theta d\phi$$

Based on a users input, would I have to first rewrite $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$

In terms of spherical coordinates?

答:
The given region is that from a cone up to the sphere "capping" the cone.  $\rho$ goes from 0 to 2 (the radius of the sphere) and $\theta$ goes from 0 to $2\pi$ (the complete circle).  $\phi$ goes from 0 (vertical) down to the cone, $\pi/4$. Since $z=\rho cos(\phi)$ in spherical coordinates, the integral is
$\int_0^2\int_0^{2\pi}\int_0^{\pi/4} \rho cos(\phi) (\rho^2 sin(\phi)d\phi d\theta d\rho)$$= \int_0^2\int_0^{2\pi}\int_0^{\pi/4}\rho^3sin(\phi)cos(\phi)d\phi d\theta d\rho$.

That can easily be done as
$\left(\int_0^2\rho^3d\rho\right)\left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi/4} sin(\phi)cos(\phi) d\phi\right)$


问:
Simple algebra simplification question?
Hello everyone I have the following question.

I have the following fraction

$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$

But how would I reduce it I know I have to use the multiply the opposite numerator by denominator and I got.

$(x-1)^2(4)=4x^2-8x+4$

$1(x^2)=x^2$

So I got got $f(x)=-\frac{4x^2-8x+4+x^2}{x^2(x-1)^2}$

But this is incorrect.What am I doing wrong.

答:
Already answered, but I would like to point out the reduction:

$$f(x) = -{(3 x-2)(x-2)\over x^2 (x-1)^2}$$

Title: Simple algebra simplification question?


问:
Homeomorphism between $R^1 \times S^1 $ & $ R^2 -(0,0)$
My question is in the context of topological spaces. I can show that if $X_1$ & $X_2$ are homeomorphism and $Y_1$ & $Y_2$ are homeomorphism, then $X_1 \times Y_1$ & $X_2 \times Y_2$ are homeomorphism too.
Can I use this property to show homeomorphism between
$R^1 \times S^1 $ & $ R^2 -(0,0)$ ?

If not, can we do this using only a function like $f(x,t)=x^t$?

答:
Let's go from the punctured plane to the cylinder. We know that the positive reals $(0,\infty)$ are homeomorphic to the real line $\mathbb R$. Let $g$ be a homeomorphim from the positive reals to the reals.

We use polar coordinates in the punctured plane and define $f$ by $$f(r,\theta)=(1, \theta, g(r))$$

Notice the $1$ in the codomain. Technically speaking, we can remove the $1$ and give a two dimensional parameterization of the cylinder which is expected if it is homeomorphic to a subset of the plane.

I leave it to you to prove that this is a homeomorphism.


问:
Difference between $\equiv$ and $=$?
What is the difference between $\equiv$ and $=$?

My thought is, that, when $\equiv$ is used $=$ could have been used as well. The resulting expression would not be wrong, but just take on a slightly different meaning. But what exactly is the relation between those symbols?

To give a practical example, consider those:
$$
=:⟺∀:(x∈\iff∈)
$$
$$
=:=∀:(x∈\iff∈)
$$
$$
=:\equiv∀:(x∈\iff∈)
$$

Or those:
$$
5+7=12
$$
$$
5+7\equiv12
$$
$$
5+7=7+5
$$
$$
5+7\equiv7+5
$$

答: $\equiv$ means "equal for all values of the variables", while $=$ just means "equal" (perhaps for only some values of the variables). For example, compare $$\cos(\alpha+\beta)\equiv\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ with $$ \cos(\alpha+\beta)=\cos(\alpha-\beta).$$The first statement is true because the two sides are equal for all $\alpha$ and $\beta$, while the second can be true for certain values of $\alpha$ and $\beta$ (namely when one of them is a multiple of $\pi$).

Title: Difference between $\equiv$ and $=$?


问:
Sine Approximation of Bhaskara
An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)

$$\sin x \approx \frac{{16x\left( {\pi  - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi  - x} \right)}}$$

for $(0,\pi)$

Here's an image. Cyan for the sine and blue for the approximation.

¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations?

答:
That's a pretty neat visualization of how can you intiutively build, proove and understand the approximation.

The full proof with video and explanation.

Title: Sine Approximation of Bhaskara


答:
Unfortunately there are a couple of different uses for $\equiv$, and it can be either stronger or weaker than $=$.

One common meaning is "is identically equal to". A typical case where you would see this is $f(x)\equiv g(x)$, and it is stressing that the functions $f$ and $g$ are equal, rather than just their values being equal for a specific $x$. It's basically the same as "for every $x$, $f(x)=g(x)$". We can also use $\equiv$ to distinguish a relation that always holds, such as $cos^2 x\equiv 1-\sin^2 x$, from an equation to be solved for $x$, such as $\cos x=1-\sin x$.

Another common use is in modular, or "clock", arithmetic. Here we say two integers are congruent modulo $m$ if they differ by a multiple of $m$. The connection to clocks is that two times of 4 o'clock are not necessarily the same time, but they have to be a multiple of 12 hours apart. We write $a\equiv b\pmod m$ for "$a$ is congruent to $b$ modulo $m$". Here $16\equiv 4\pmod{12}$, but of course $16\neq 4$.

Title: Difference between $\equiv$ and $=$?

Body:
What is the difference between $\equiv$ and $=$?

My thought is, that, when $\equiv$ is used $=$ could have been used as well. The resulting expression would not be wrong, but just take on a slightly different meaning. But what exactly is the relation between those symbols?

To give a practical example, consider those:
$$
=:⟺∀:(x∈\iff∈)
$$
$$
=:=∀:(x∈\iff∈)
$$
$$
=:\equiv∀:(x∈\iff∈)
$$

Or those:
$$
5+7=12
$$
$$
5+7\equiv12
$$
$$
5+7=7+5
$$
$$
5+7\equiv7+5
$$


问:
Fixed points of $ e^{x+1/x} = x$ in $\mathbb{R}$
This particular question was one of the options in a question in amock test i am solving of masters in mathematics.

Does $f(x) = e^{x+1/x}$ has unique fixed point in $\mathbb{R}$?

Attempt: the equation  can be simplified to $x+\frac{1}{x} = \log x$ and then $ x^2 +1 = x \log x $.

But I don't know the graph of $x \log x$ and hence can't solve the problem and don't know any other idea.

So, can you tell a elegant method to solve this question.

Thank you!!

答:
Hint :

as stated in the comments by @kccu $e^{x+\frac{1}{x}}$ has no fixed points (in $\mathbb{R}$, so maybe it's in $\mathbb{C}$ as stated in the answer of @Yves Daoust :P) so I assume it is $e^{\frac{x+1}{x}}$ .

Surely you have to search for it when $x>0$ , if you study the derivative of $$f(x) = e^{\frac{x+1}{x}}-x$$

what can you say about the number of fixed points? Now you need just to find a point where the function is positive and a point where the function is negative so that you can apply Bolzano's theorem to conclude.


问:
Uniqueness of powers mod n
I was doing this problem

Let n be a positive integer with $n ≥ 3$. Show that $n^{n^{n^{n}}}-n^{n^{n}}$ is divisible by $1989$.

The answer considered the expression mod $9$, $13$, and $17$. For mod $9$, when $3 \vert  n$ it is true; for $3\nmid n$, by Euler's theorem the answer took the exponents mod $6$ and said that $n^{n^{n}}\equiv n^n (\mod 6)$.

I was wondering why you can assume that the exponents are necessarily equal mod $6$. After all, couldn't there be a case where $n^{n^{n}}\not\equiv n^n (\mod 6)$ but $n^{n^{n^{n}}}\equiv n^{n^{n}} (\mod 9)$?

Also, is there a rule that dictates whether $x^n$ is unique mod $m$ for coprime integers $x, m$, x is taken mod $m$ and $n$ is taken mod $\phi(m)$? For example, $2^n$ mod $7$ isn't unique since $2^1 \equiv 2^4$ $(\mod 7)$ and $1\not\equiv4 (\mod 6)$.

答:
To answer the first part of the question,

either n is even or odd. If n is even, so are $n^{n^n}$ and $n^n$. If n is odd, both are odd. So they are congruent mod 2.

If $n \equiv 0 \mod 3$. Then both expressions are 0 mod 3.

If $n \equiv 1 \mod 3$. Then both expressions are 1 mod 3 because 1 to any power is 1.

If $n \equiv -1 \mod 3$. Then both expressions take -1 to an even power, or both to an odd power. So either both expressions are 1 mod 3, or both expressions are -1 mod 3.

So in all cases, the two expressions are congruent mod 3.

So using the chinese remainder theorem they are both congruent mod 6.

Title: Uniqueness of powers mod n


问:
Books to improve mathematical problem solving abilty
I don't want books like "How to solve it". Instead I am looking for books that actually have problems in them. And preferably solutions too. Basically books with problems that help you develop mathematical problem solving ability.

For my background, I have graduated from high school and know high school math fairly well. But lot of the problems at high school level are reiterative and can be done by memorizing stuff. I am looking for problems that challenge you and at the same time make you a better problem solver.

答:
The art and craft of problem solving by Paul Zeitz is a very good book and it meets your requirements. It has many head scratching problems which are easy to understand but hard to do.

The IMO compendium is also a good book, which includes the questions (and answers) from the previous IMO, as well as some additional questions. It also teaches some concepts.


答:
The $:$ is crucial here, indicating that what's on the left of $:\Longleftrightarrow$, $:=$ or $:\equiv$ is defined as what's on the right. These three two-character symbols all mean the same thing. But if you ask for a comparison of the meanings of the "naked" $\Longleftrightarrow$, $=$, $\equiv$, well, those are all different.

The first two are easy: $\Longleftrightarrow$ means iff, and $=$ means equals. But $\equiv$ can denote identities (making is stronger than $=$) or equivalence relations (which are weaker than $=$, and often denoted $\sim$, though congruence in particular is always represented with $\equiv$.)

Title: Difference between $\equiv$ and $=$?


问:
Find the ratio $\frac{R}{r}$ rounded to four decimals

A large circle of radius R is surrounded by 12 smaller circles of radius r. Find the ratio $\frac{R}{r}$ rounded to four decimals.

Hint: Draw an isosceles triangle with one vertex at the centre of the R-circle
and the other vertices at the centres of two adjacent r-circles.

I drew the triangle, and the two equal sides (let's call them A and B) have a length of $R + r$, and the different side (let's call it C) has a length of $2r$.

I understand that the angle AB is 30 degrees (I divided 360 by 12), and as it's an isosceles triangle the other two angles are equal. They both measure 75 degrees.

I don't understand how to calculate the ratio though.

答:
You can use the law of sines.
\begin{align}
\frac{2r}{\sin{30^\circ}}&=\frac{R+r}{\sin75^\circ}\\
2r\sin75^\circ&=(R+r)\sin30^\circ\\
2r\left(\frac{\sqrt2+\sqrt6}4\right)&=(R+r)\left(\frac12\right)\\
(\sqrt2+\sqrt6)r&=R+r\\
(\sqrt2+\sqrt6-1)r&=R\\
\frac{R}r&=\sqrt2+\sqrt6-1\approx2.8637
\end{align}


问:
Is there an operation over a finite set of integers (which size is $2^{128}$, or a power of two) that makes a non-abelian group?
I know that there is at least one operation on permutations that makes a non-abelian group: we can "multiply*" them, if we consider them as permutation matrices.
And naturals are directly mapped to permutations.
However, I need a non-abelian group which is not only finite, but also with a size which is a power of two due to the need of direct correspondence between them and hash digests (128bits).

Besides the non-abelian requirement, the results of the operation should be uniformly distributed along the $2^{128}$ numbers, just like the result of permutation matrix multiplication is.

P.S.: I suppose any power of two is welcome, since a pair of 64-bit  numbers, e.g., can be used to represent a 128-bit number; with the disadvantage of two separate operations - probably not ideal due to the uniformity requirement cited above.

-> "compose"?

答:
Here's one simple way with a few parameters to vary. Don't know how well they mesh with the hashes (you need to be more specific, if special constraints come from there).

Split that $128$-bit integer $x$ into six chunks. Five chunks of $24$ bits each and a single chunk with only $8$ bits. So write
$$
x=x_6\cdot 2^{120}+x_5\cdot2^{96}+x_4\cdot2^{72}+x_3\cdot2^{48}+x_2\cdot 2^{24}+x_1,
$$
where $x_1,x_2,x_3,x_4,x_5$ have $24$ bits each, treated as integers in the range $0\le x_i\le 2^{24}-1$, and $0\le x_6\le 255$ has $8$ bits.

Map $x$ into a $4\times 4$ matrix $M(x)$ like follows:
$$
M(x)=\left(\begin{array}{cccc}
1&2^{16}x_6&x_5&x_3\\
0&1&x_4&x_2\\
0&0&1&x_1\\
0&0&0&1\end{array}\right).
$$
We view all the entries as integers modulo $2^{24}$. That explains the extra factor $2^{16}$ of $x_6$. Without it $x_6$ would be determined only modulo $2^8$, so we tag $16$ zeros to the LSB end.

The operation, call it $\star$, is then the usual product of $4\times4$ matrices. In other words, we define $x\star y$ to be the $128$-bit chunk determined by
$$
M(x\star y)=M(x)M(y),
$$
where on the right hand side we perform the usual matrix multiplication with all arithmetic on the entries done modulo $2^{24}$. It is easy to check that the product $M(x)M(y)$ is of the same form for all pairs $x,y\in[0,2^{128}-1]$.

You get "more non-commutativity" by using larger matrices. The method is to split those
128 bits into equal size chunks and fill the upper triangular part with them. One chunk will necessarily be smaller than the others, and you should put that into the $(1,2)$ position, padded with an appropriate number of zeros. Other variations with more complicated zero padding schemes are also possible.

The real question may be whether this is non-abelian enough? Using large chunks leaves a lot of linearity that an attacker may be able to exploit. For example, in the above example if $x$ and $y$ both have the highest $56$ bits all zeros, then
$$
\left(\begin{array}{cccc}
1&0&0&x_3\\
0&1&0&x_2\\
0&0&1&x_1\\
0&0&0&1\end{array}\right)\left(\begin{array}{cccc}
1&0&0&y_3\\
0&1&0&y_2\\
0&0&1&y_1\\
0&0&0&1\end{array}\right)
=\left(\begin{array}{cccc}
1&0&0&x_3+y_3\\
0&1&0&x_2+y_2\\
0&0&1&x_1+y_1\\
0&0&0&1\end{array}\right).
$$
In other words, we only have three additions modulo $2^{24}$. May be such problems, if at all relevant, should go into a follow up question?

I'm not sure, but the comments underneath leave nagging fears. Here's what the operation comes to
$$
\left(
\begin{array}{cccc}
 1 & 2^{16} (x_6+y_6) & x_5+2^{16} x_6 y_4+y_5 &
   x_3+x_5 y_1+2^{16} x_6y_2+y_3 \\
 0 & 1 & x_4+y_4 & x_2+x_4 y_1+y_2 \\
 0 & 0 & 1 & x_1+y_1 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right).
$$
So if $x\star y=z$, then
$$
\begin{aligned}
z_6&\equiv x_6+y_6\pmod{2^8},\\
z_5&\equiv x_5+y_5+2^{16}x_6y_4\pmod{2^{24}},\\
z_4&\equiv x_4+y_4\pmod{2^{24}},\\
z_3&\equiv x_3+y_3+x_5y_1+2^{16}x_6y_2\pmod{2^{24}},\\
z_2&\equiv x_2+y_2+x_4y_1\pmod{2^{24}},\\
z_1&\equiv x_1+y_1\pmod{2^{24}}.
\end{aligned}
$$
You see that the various parts $z_j$ are only affected by a limited number of parts of the factors $x$ and $y$.


问:
Group variety $PGL(n)$
Sorry for my bad English.

In Hartshorne’s algebraic geometry III.example 10.7.2, projective space $\mathbb{P}_k^n$ is acted by $G＝$PGL($n)$.

But,I can’t understand why is $G$ group variety?

In general, when we are gave any group $H$, can we construct group variety $H$？

答: One can define the group variety $\operatorname{PGL}(n)$ as an open subvariety of $P M_n$; the Zariski open subset where the determinant is nonzero. The group law is given by polynomial equations, so that gives an algebraic group.

Title: Group variety $PGL(n)$


问:
Scope in $\forall x P(x) \rightarrow Q(x)$
For $ \exists x P(x) \wedge Q(\color{blue}x) $

The scope of the quantifier, is just the predicate $P(x)$, thus the occurrence of x in blue is free.

I am unsure exactly the scope of the quantifier in this formula:

$ \forall x P(x) \rightarrow Q(\color{red}x) $

Based off the previous formula, I think the occurrence of the x in red is free but I'm not sure.

Is the occurrence of $\color{red}{x}$ free or bound? Would this vary based off convention?

答:
In first-order logic, quantifiers bind more strongly than logical connectives; the precedence convention is:

quantification

negation

conjunction, disjunction

conditional, biconditional


答: $\equiv$ is used for $\text{identity}$ mostly, but $=$ is used for equation

Title: Difference between $\equiv$ and $=$?


问:
sum of binomial coefficients expansion to prove equation
I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$

I already expanded

$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?

答:
A variation based upon the  binomial theorem and the finite geometric series formula.

We consider the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We can write this way
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^n\binom{i}{2}}&=\sum_{i=0}^n[x^2](1+x)^i\tag{2}\\
&=[x^2]\sum_{i=0}^n(1+x)^i\tag{3}\\
&=[x^2]\frac{(1+x)^{n+1}-1}{(1+x)-1}\tag{4}\\
&=[x^3]\left((1+x)^{n+1}-1\right)\tag{5}\\
&\,\,\color{blue}{=\binom{n+1}{3}}\tag{6}
\end{align*}
and the claim follows.

Comment:

In (2) we apply the coefficient of operator according to (1). The lower limit of the index $i$ is set to $i=0$ which doesn't change anything, since $\binom{0}{2}=[x^2](1+x)^0=[x^2]1=0$.

In (3) we use the linearity of the coefficient of operator: $[x^p]A(x)+[x^p]B(x)=[x^p]\left(A(x)+B(x)\right)$.

In (4) we use the finite geometric summation formula.

In (5) we observe the denominator is $x$ and we use the rule $[x^p]\frac{1}{x}A(x)=[x^{p+1}]A(x)$.

In (6) we select the coefficient of $x^3$.


问:
Determine the number of elements of order 2 in AR

So i have completed parts a and b. For b i reduced R to smith normal form and ended up with diagonals 1,2,6. From this i have said that the structure of the group is $Z_2 \oplus Z_6 \oplus Z$. But i have no idea what so ever about part c.

答:
As stated in the comments, let us focus on $\,\Bbb Z_2\times\Bbb Z_6\,$ , which for simplicity (for me, at least) we'll better write multiplicatively as $\,C_2\times C_6=\langle a\rangle\times\langle b\rangle\,\;,\;\;a^2=b^6=1$

Suppose the first coordinate is $\,1\,$ , then the second one has to have order $\,2\,$ , and the only option is $\,(1,b^3)\,$ , so we go over elements with non-trivial first coordinate, and thus the second coordinate has to have order dividing two:

$$(a,1)\;,\;\;(a,b^3)$$

and that seems to be pretty much all there is: three involutions as any such one either has first coordinate trivial or not...


答:
For all $x\ne0$, $f(x)>0$ and there can be no fixed-point in the negatives. Then for $x>0$, $f(x)\ge 1+x+\dfrac1x>x$ and there can be no fixed point in the negatives.

Note that if $x$ can take values in $\mathbb C$,

$$\overline{e^{x+1/x}}=e^{\bar x+1/\bar x}$$and the conjugate of a fixed point is also a fixed point !

Title: Fixed points of $ e^{x+1/x} = x$ in $\mathbb{R}$

Body:
This particular question was one of the options in a question in amock test i am solving of masters in mathematics.

Does $f(x) = e^{x+1/x}$ has unique fixed point in $\mathbb{R}$?

Attempt: the equation  can be simplified to $x+\frac{1}{x} = \log x$ and then $ x^2 +1 = x \log x $.

But I don't know the graph of $x \log x$ and hence can't solve the problem and don't know any other idea.

So, can you tell a elegant method to solve this question.

Thank you!!


问:
Proof: $\arg\left(\bar z\right) = -\arg\left( z\right)$
I need to prove that $\arg\left(\bar z\right) = -\arg\left( z\right)$ and a few other properties of the complex numbers. I'm being able to prove the others with no problem, but I just don't see where to start this one from. I find it kind of obvious but don't know how to prove it.

We have $z=a+bi$ and $\bar z=a-bi$, $z$ being a complex number.

Therefore the arguments are $\arg(z)=\arctan\left(\frac{b}{a}\right)$ and $\arg(\bar z)=\arctan\left(\frac{-b}{a}\right)$

Any hint on how to approach this?

答: Let $z=r e^{0\theta}$ so $Arg(z)=\theta$, then $\bar z=r e^{-i\theta} \implies Arg(\bar z)=-\theta$. This proves that $Arg(\bar z)=-Arg(z)$


问:
Inverse Laplace transform of $e^{-\sqrt s}$
How could one possibly find the inverse Laplace transform of $e^{-\sqrt{s}}$ using a table of Laplace transforms?   </p>

答:
Here I just want to mention that the Laplace transform of one side Levy distribution is $\exp(-s^\alpha)$ with $0<\alpha<1$.

In your case, $\alpha=1/2$, you can find the exact form of one side levy distribution.

enter link description here


问:
What goes wrong in my computation of $\iiint _S 1 dz dy dx$ where $S=\{(x,y,z)\mid ax+by+cz=0, 1\leq x,y,z \leq N\}$?
I am trying to calculate this triple integral but I don't really know if it is correct. I don't have much experience with triple integrals. Here it is:

$\iiint _S 1 dz dy dx$ where $S$ is the set $\{(x,y,z)\mid ax+by+cz=0, 1\leq x,y,z \leq N\}$. Here $a$, $b$ and $c$ are fixed non-zero reals, $N$ is a fixed positive real. It seems that $z$ should run from $1$ to $-(ax+by)/c$ and we can let $x$ and $y$ run from $1$ to $N$, my reasoning being that if $x$ and $y$ are fixed, $z$ gets automatically fixed.

Then our integral transforms into $$\iiint _S 1 dz dy dx=\int_1^N \int_1^N \int_1^{-(ax+by)/c} 1 dz dy dx=\int_1^N \int_1^N -(ax+by+c)/c  \mathrm{d}x dy$$ which is equal to $$\int_1^N-(ax^2/2+bxy+cx)|^{N}_1 dy=\int_1^N \left(-(aN^2/2+bNy+cN)+a/2+by+c\right)dy=-aN^3/2-bN^3/2-cN^2+aN/2+bN^2/2+cN+aN^2+bN/2+cN-a/2-b/2-c.$$

I feel like i am making an error somewhere, but I am not sure. Thanks!

答:
The integration region of your problem isn't well-defined. My guess is that the equation $ax+by+cz=0$ should be an inequation, $ax+by+cz\leq0$ or $ax+by+cz\geq0$, because if the equation $ax+by+cz=0$ does define your set, you will actually have a surface and not a volume.

Anyway, it is always easier to solve multiple integrals by visualizing their respective integration regions. Doing so, it is possible to conclude that your question is definitely harder than it looks.

$1 \leq x,y,z,\leq N$ defines a cube of edge length $N-1$ and $ax+by+cz=0$ defines the plane $z=-(ax+by)/c$. From here, obvious questions arise: Does the plane cut the cube? If does, how exactly? You have to study the all the possibilities.

For now, your solution is only valid if $ 1 \leq z=-(ax+by)/c \leq N$ for all points $(x,y)$ in the square $[1,N] \, \text{x} \, [1,N]$ (and if the inequality $ax+by+cz\leq0$ takes the place of $ax+by+cz=0$).

It could be interesting to highlight that the upper bound value of the integral is known: since the integrand is $f(x,y,z)=1$, it is the volume of the cube, $(N-1)^3$.


问:
Prove the inequality $1/x^2 + x^2/4 \le 5/4$
Exercise for reference down below:

For any x you pick in the interval [1,2] you have to prove that the inequality in the title is true.

Would like some help in solving this or maybe an idea that I can work on, I don't know how & if I can solve it by using the common denominator 4*x^2.

in case you might spot a mistake in my attempt at solving, I am going to write it here:

$(4+x^4)/(4*x^2)$ <= $5/4$

$4+x^4 <= 5x^2$

$4+x^4$ is between [5,20] and $5*x^2$ is between [5,20]

After "step" 3 I didn't know how to continue as I thought it was a dead end.

答: As an idea:take $$f(x)=\frac{1}{x^2}+\frac{x^2}{4} , x\in[1,2]\\ $$show that range of that function is $y\leq \frac54$


答:
Hint: Multiply throughout by $x^2$, shift terms to one side, and factorize.

The inequality you want to show becomes

 $(x^2 - 4) (x^2 - 1) \leq 0$

which is obvious in the given domain.

Title: Prove the inequality $1/x^2 + x^2/4 \le 5/4$

Body:
Exercise for reference down below:

For any x you pick in the interval [1,2] you have to prove that the inequality in the title is true.

Would like some help in solving this or maybe an idea that I can work on, I don't know how & if I can solve it by using the common denominator 4*x^2.

in case you might spot a mistake in my attempt at solving, I am going to write it here:

$(4+x^4)/(4*x^2)$ <= $5/4$

$4+x^4 <= 5x^2$

$4+x^4$ is between [5,20] and $5*x^2$ is between [5,20]

After "step" 3 I didn't know how to continue as I thought it was a dead end.


答:
As long as the $\arg$ function is not precisely defined, this question cannot be answered. For example, if you define the argument to be an angle in range $(0,2\pi]$, it is clear that it is impossible for

$$\arg(a)=-\arg(b)$$ to hold.

Title: Proof: $\arg\left(\bar z\right) = -\arg\left( z\right)$

Body:
I need to prove that $\arg\left(\bar z\right) = -\arg\left( z\right)$ and a few other properties of the complex numbers. I'm being able to prove the others with no problem, but I just don't see where to start this one from. I find it kind of obvious but don't know how to prove it.

We have $z=a+bi$ and $\bar z=a-bi$, $z$ being a complex number.

Therefore the arguments are $\arg(z)=\arctan\left(\frac{b}{a}\right)$ and $\arg(\bar z)=\arctan\left(\frac{-b}{a}\right)$

Any hint on how to approach this?


问:
Minimization of sum of ratios objective with linear constraints
I am trying to solve a minimization mathematical model toward global
optimality. The objective function is the sum of some ratios. The numerator of
each fraction is quadratic, and the denominator is linear as follows

$ Min\ \sum_k\left(\frac{\sum_i\left(P_{ik}^{\left(1\right)}+P_{ik}^{\left(2\right)}\right).\sum_i\left({\tau{}}^{(1)}P_{ik}^{\left(1\right)}+{\tau{}}^{(2)}P_{ik}^{\left(2\right)}\right)\
\ \
}{2h_k\left(h_k-\sum_iP_{ik}^{\left(1\right)}-\sum_iP_{ik}^{\left(2\right)}\right)}\right)$

$\sum_kP_{ik}^{\left(1\right)}=r_i^{\left(1\right)},\forall{}i$

$\sum_kP_{ik}^{\left(2\right)}=r_i^{\left(2\right)},\ \forall{}i$

$\sum_iP_{ik}^{\left(1\right)}+\sum_iP_{ik}^{\left(2\right)}\leq{}h_k,\forall{}k$

$P_{ik}^{\left(1\right)},\ P_{ik}^{\left(2\right)}\geq{}0,\ \forall{},i,k$

All constraints are linear. My perception is that the objective function is not convex as it is the summation of some quasi-convex functions. The number of variables of each type is around 20. Therefore, an optimization software product like LINGO, MAPLE etc. can solve the model fairly quickly but the global optimality is not guaranteed.

Do you have any idea how we can convincingly solve the model to reach the optimal global solution or at least ensure our solutions are close to it?

答:
HINT

Sounds like each term of the sum is independent of the others since it uses different variables. So let's minimize term-by-term. Let
$$
f(x,y) = \frac{ax^2+(a+b)x y+b y^2}{2c_k(c_k-x-y)}
$$
Can you find the extrema of this 2D function using the usual techniques from multi-variable calculus? e.g., solve $\vec{\nabla} f = \vec{0}$?


问:
Is iteration process through equal sign possible?
Is iteration process through equal sign possible?

for an example: repeatedly pressing on the equal button of calculator

Repeatedly pressing on the equal button of calculator until full converging the Answer.

this is (Ahmad Reza Estakhr's) symbol of iteration of equal sign $\Re_{i=1}^{\infty}=$ which means repeatedly pressing on equal sign.

答:
yes, it is possible, for example M-set $ z_{n+1}=z_{n}^2+c $ in term of Estakhr's Iteration symbol is this like:

$$Ans^2+c\Re_{i=1}^{\infty}=Ans$$

Interestingly M-set under Estakhr's iteration equation of the complex quadratic polynomial $Ans^2+c\Re_{i=1}^{\infty}=Ans$ remains bounded, that which means M-set remains bounded (because of imaginary number) but it does not converge to a limit.


答:
Equality holds at the ends of the interval ($x=1$ and $x=2$),

and the function $x^{-2}+x^2/4$ is concave upward

because its second derivative is $6x^{-4}+1/2$ is always positive.


答:
The centers of circles and one of the point of tangency of two smaller circles determine a right triangle with $90,75,15$

$$\sin 15=\frac{r}{R+r}$$

$$\frac{R+r}{r}=\frac{1}{\sin 15}$$

$$\frac{R}{r}=\frac{1}{\sin 15}-1=\sqrt2+\sqrt6-1$$
where
$$\sin 15=\frac{\sqrt6-\sqrt2}{4}$$

Body:
A large circle of radius R is surrounded by 12 smaller circles of radius r. Find the ratio $\frac{R}{r}$ rounded to four decimals.

Hint: Draw an isosceles triangle with one vertex at the centre of the R-circle
and the other vertices at the centres of two adjacent r-circles.

I drew the triangle, and the two equal sides (let's call them A and B) have a length of $R + r$, and the different side (let's call it C) has a length of $2r$.

I understand that the angle AB is 30 degrees (I divided 360 by 12), and as it's an isosceles triangle the other two angles are equal. They both measure 75 degrees.

I don't understand how to calculate the ratio though.


问:
When is it valid for me to just integrate a trig function?
I am having a problem identifying when I need to use some kind of integration technique or am I just over complicating things. Could someone please explain to me when I need to or not?

Normally, I would just integrate as normal when the problem is like this

$$\int \cos(x) dx = \sin(x) + c$$

But when things become like the following, I don't have rules to fall back on and just rely on memory and experience of solving problems.

$$\iiint \cos(x+y+z) dz dx dy$$

$$\int \sin(2x) dx$$

答:
Just apply the ordinary rules
and treat the variables that you are not
integrating with respect to as constants.

For example, the general forms are
$\int_u^v f(x+a)dx = \int_{u+a}^{v+a} f(y) dy$
(with $y = x-a$)
and $\int_u^v f(cx)dx = (1/c)\int_{cu}^{cv} f(y) dy$
(with $y = x/c$).

If the integrals are indefinite,
these become
$\int f(x+a)dx = \int f(y) dy$
and $\int f(cx)dx = (1/c)\int f(y) dy$.

Putting these in your examples,
$\int \cos(x+y+z) dz = \sin(x+y+z)$,
since the "$x+y$" is a just a constant here,
and
$\int \sin(2x) dx = (1/2)\int \sin(y) dy
\text{ (with } y = 2x)
= (1/2) (-\cos(y))
= (-1/2) \cos(2x)
$.

For the iterated integral, keep doing this
until you run out of integrals.


问:
Understanding Fibonacci Sequence with Rabbits?
Find a recurrence relation for the number of pairs of rabbits after n months if (1)
initially there is one pair of rabbits who were just born, and (2) every month each
pair of rabbits that are over one month old have a pair of offspring (a male and a
female)

I know that this is a fibonacci sequence where the recurrence relation is

$a_n=a_{n-1}+a_{n-2}$

but I'm a little confused on something.

In this situation,

$a_{n-1}$ represents the pairs of rabbits that are grown up/old/fertile correct?

and

$a_{n-2}$ represents the pairs of rabbits that are just born/young correct?

thus, you would sum the two together to find the total amount of pairs of rabbits correct?

I test this with

$a_0=1$ which represent a pair rabbits that were just born/are young

$a_1=1$ which represent those pair of rabbits growing up/ready for breeding

so from that I think that I understand what $a_{n-1}$ and $a_{n-2}$ represent but I want to make sure I do by checking with this site.

Do I have a good understanding of what $a_{n-1}$ and $a_{n-2}$ represent?

答:
For each $n, a_n$ is the total number of pairs of rabbits after $n$ months. So

$a_{n-1}$ is the total number of pairs one month ago. This includes both adults and newborns. It is how many rabbit pairs there are before the pregnant rabbits give birth for this month.

$a_{n-2}$ is the total number of pairs two months ago. By one month ago, they were all adults, and so all became pregnant at that time. So they are the ones giving birth now, each to one new pair. Therefore $a_{n-2}$ is also be the number of newborn pairs at this time.

Thus the number of adult rabbit pairs at this point in time is $a_{n-1}$ and the number of newborn rabbit pairs at this point of time is $a_{n-2}$. Since every rabbit pair is either adult or newborn, that is all of the rabbits.


问:
Solution to exercise 2.5.9 of Marker's Model theory
I'm trying to solve this exercise 2.5.9 of David Marker's model theory. The exercise is:

Take three L-structures: $\cal{M}_0 \subset\cal{M}_1 \subset\cal{M}_2$. If $\cal{M}_0 \prec \cal{M}_2$ and $\cal{M}_1 \prec \cal{M}_2$, then show that $\cal{M}_0 \prec \cal{M}_1$.

I know that this is the coherence property of abstract elementary classes. However, I have no idea how to prove the exercise. Does it use the compactness theorem?

答: Use the Tarski Vaught test. Suppose $\varphi(x,y)$ is a formula in the language, and $a \in M_0$ (I'm abusing notation wrt tuples and such). Then if there is $b \in M_1$ such that $M_1 \vDash \varphi(b, a) $, then since $M_1 \prec M_2$, $M_2 \vDash \varphi(b, a)$, and then by the Tarski vaught test between $M_0, M_2$, there is $b' \in M_0$ such that $M_2 \vDash \varphi(b', a)$, but then $b' \in M_1$ as well, so $M_1 \vDash \varphi(b', a)$. Thus the Tarski Vaught test is satisfied.


问:
Understanding transformations of the projective line

I am going through Stillwell's the Four Pillars of Geometry and having a hard time understanding this part of the text. I know that in $\mathbb{RP^2}$, the "points" are lines through $O$ in $\mathbb{R^3}$ and it's "lines" are planes through $O$ in $\mathbb{R^3}$. How can I visualize $\mathbb{RP^1}$? Does $\mathbb{RP^1}$ consist only of a single horizontal or vertical line?

For figure 7.2 they says that 'the single line that does not meet $y=1$, namely, the x-axis, naturally gets the label $\infty$. But why doesn't the x-axis meet line y=1? My understanding so far is that lines at parallel lines meet at infinity in projective geometry. So why not here?

For figure 7.3, I have no idea what is going on here. What are they trying to show? What does 'labels' mean for these lines? What plane are these lines in? $\mathbb{R P^1}$?  $\mathbb{R^2}?$

Any help would be appreciated, very confused about projective geometry in general and i'm still very much a beginner at all of this.

答:
The purpose of figures 7.2 and 7.3 is to establish the connection  $1D \leftrightarrow 2D$ by placing a "screen" at $y=1$.

What do I mean by "screen" ? It will be better understood on the equivalent correspondance with one more dimension (connection $2D \leftrightarrow 3D$) as shown on the right figure below (an excerpt of my lectures) where the eye is at the origin and "sees" the cube (or any other object as "projectively projected" on the screen positionned at height $z=1$. Please note that a "projective point" is completely identified with the corresponding (red) line issued from the origin.

The left hand side figure accounts for the connection between this screen plane, which is the plane of (projective) points, and the parallel plane at $z=0$ which is the plane of vectors. Moreover, it justifies barycentric expressions like $C=\tfrac23 A+\tfrac23 B$ by a 3D interpretation.


问:
Uniqueness of values of automorphisms on a primitive root
Let $L$ be a field and let $\zeta \in L$ be a primitive root of $n$-th degree. Let $\alpha$ and $\beta$ be automorphisms belonging to $Aut(L)$. Is it possible that $\alpha(\zeta) \ne \beta(\zeta)$? It strongly seems to me that the equality $\alpha(\zeta) = \beta(\zeta)$ must hold. Here is my reasoning behind this statement:

as $\alpha$ is automomorphism, it must be that $\alpha(1) = 1$. Since $\zeta$ is a primitive root of $n$-th degree, we have that $\zeta^n = 1$. Hence, we have that $\alpha(\zeta^n) = 1$. Similarly $\beta(\zeta^n) = 1$. So we have $$\alpha(\zeta^n) = \beta(\zeta^n)$$ Thus: $$(\alpha(\zeta^n))^{1-n} = (\beta(\zeta^n))^{1-n}$$ and finally we obtain: $$\alpha(\zeta) = \beta(\zeta)$$

答:
You've made an algebra error in the final step. In fact, there are many field automorphisms that act nontrivially on primitive roots of unity: the cyclotomic field $F = \mathbb{Q}(\zeta_n)$ is a Galois extension of degree $\varphi(n) = \lvert (\mathbb{Z}/n\mathbb{Z})^\times \rvert$ (where $\varphi$ is the Euler phi function), and has $\varphi(n)$ automorphisms. Given $d \in \mathbb{Z}$ with $\gcd(d, n) = 1$, the assignment $\zeta_n \mapsto \zeta_n^d$ extends to a field automorphism of $F$, and all automorphisms are of this form. (In particular, the only automorphism of $F$ that fixes the generator $\zeta_n$ is the identity automorphism. This is a general phenomenon: if an automorphism fixes the generators of a field extension, then it's the identity automorphism.)

More generally, one can show using Galois theory that the only algebraic numbers that are fixed by every field automorphism are rational numbers.


问:
How to get the conditional probability of $X|Y\sim Y N(0,\sigma_1^2)+(1-Y) N(0,\sigma_2^2)$?
If we have a random variable $Y\sim Ber(p)$ and
$$X|Y\sim Y N(0,\sigma_1^2)+(1-Y)N(0,\sigma_2^2)$$

How to get $f(x|y)$ proportion to the product of two normal distribution?

Is it just the sum of two normal distributions?
The expectation of $X|Y$ is $\mathbb{E}(X|Y)=0$. Do we have $ Var(X|Y)=Y^2\sigma_1^2+(1-Y)^2\sigma_2^2$

答:
Setting $Z_i\sim N(0;\sigma_i^2)$ and assuming independence between $Z_1$,$Z_2$ and $Y$,

your conditional density is a mixture of the two gaussian densities

$$f_{X|Y=y}(t)=yf_{Z_1}(t) +(1-y)f_{Z_2}(t)$$

The marginal distribution is:

$$f_X(x)=\frac{p}{\sigma_1\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma_1^2}}+\frac{1-p}{\sigma_2\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma_2^2}}$$


问:
A man sells ﬁrewood. To make standard portions, he uses a standard length of rope $l$ to surround a pack of logs

A man sells ﬁrewood. To make standard portions, he uses a standard length of rope $l$ to surround a pack of logs. One day, a customer asks him for a double portion of ﬁrewood. What length of rope should he use to measure this order? Assume the packs of logs are circular in shape.

This is a typical order:

$l$ is equal to $2\pi r$

This is the new order:

I understand that the perimeter in this case is:

The two sides A, plus the two half-circles, A = $2r$

$4r + 2\pi r$

However, the answer is $\sqrt{2}l$.

How can I get to this result?

答: I would think you need to find the circumference of a disk with twice the area, right? That is, find $$2\pi r_2$$ given that that $$\pi r_2^2 =2\cdot \pi r^2$$


答:
A double portion bundled with a circular cross section has area $2\pi r^2$ where $r$ is the radius of a standard portion.

A circle with twice the area $2\pi r^2$ has radius $R$ such that $\pi R^2=2\pi r^2$ which gives us that $R=\sqrt{2}r$.

A rope that ties the standard portion has length equal to the circumference of the circle and is equal to $2\pi r$. For the larger circle the rope has to be $$2\pi R=2\pi\sqrt{2}r=\sqrt{2}l$$

Body:
A man sells ﬁrewood. To make standard portions, he uses a standard length of rope $l$ to surround a pack of logs. One day, a customer asks him for a double portion of ﬁrewood. What length of rope should he use to measure this order? Assume the packs of logs are circular in shape.

This is a typical order:

$l$ is equal to $2\pi r$

This is the new order:

I understand that the perimeter in this case is:

The two sides A, plus the two half-circles, A = $2r$

$4r + 2\pi r$

However, the answer is $\sqrt{2}l$.

How can I get to this result?


问:
Alternate solution to indefinite integral
$\int{\frac{x}{(-10+7x-x^2)^{3/2}}}dx$

Is there any alternative way to solve this indefinite integral?(Other than trig substituition)

Body:
Let
$$I(x)=\int \frac{xdx}{(7x-10-x^2)^{3/2}}=\frac{ax+b}{(7x-10-x^2)^{1/2}}$$
D.w.r.t. $x$ gives
$$\frac{x}{(7x-10-x^2)^{3/2}}=\frac{(-20a-7b)/2+(7a/2+b)x}{(7x-10-x^2)^{3/2}}$$
So we get $a=14/9, b=-40/9$.


问:
Existence of isolated point.
Suppose: A is a countable closed subset of a complete metric space.

Prove: A contains at least one isolated point.

(the Baire category theorem brings me no further than Int(A) (the interior of A) is empty)

答: Consider $A$ as its own metric space now with the subspace topology. Notice that $A$ is complete so the Baire Category theorem applies. Now suppose that for every $x \in A$, $\{x\}$ is not open. In particular this means that $\{x\}$ is nowhere dense. But then since $A$ is countable, $A$ is a countable union of nowhere dense singletons, so $A$ is meager. But the Baire Category theorem applies on $A$, and by the Baire Category theorem, a complete metric space is not meager. So we have a contradiction.

Title: Existence of isolated point.


问:
Write $\mathbb Z_2 \times \mathbb Z_{12} \times \mathbb Z_{15}$ as direct product of two cyclic groups.
Consider the finite Abelian group $G = \mathbb Z_2 \times \mathbb Z_{12} \times \mathbb Z_{15}$. Find an isomorphic group a $G$ which is the direct product of only two cyclic groups. From this it can be deduced that $G$ can be generated by two elements. Give an example of such elements.

I suppose the order of $G$ is $360$, but I don't find two cyclic groups whose direct product makes $360$, I find: $360= 2^3 \times 5 \times 3^2$, but it's: $\mathbb Z _{24} \times \mathbb Z _{15}$, but only $15$ is cyclic...

答: Since $\gcd(2,15)=1$, we have, by the Chinese Remainder Theorem, that $$\Bbb Z_2\times\Bbb Z_{15}\cong\Bbb Z_{30},$$ which is cyclic; hence a group isomorphic to $G$ would be $$\Bbb Z_{12}\times \Bbb Z_{30}.$$


问:
($a$ - algebraic element and $I_a$ - maximal ideal) => ($K[a]$ is isomorphic with $K[x]/I_a$)
The following is from A. Białynicki-Birula, "Algebra" (the translation is mine).

Let $K$ be a field. (...) Let $I_a$ denote ideal of these polynomials $f \in K[x]$ for which $f(a) = 0$.
 (...)

Theorem 1.4 Let $L$ be an extension of the field $K$ and let $a \in L$. Then the following conditions are equivalent:

(a) $a$ is an algebraic element over $K$,

(b) $K[a]$ is a field, that is $K[a] = K(a)$

(...)

Proof. (a) => (b). If $a$ is an algebraic element over $K$, then $I_a \neq (0)$. Since beside this $I_a$ is a prime ideal, and $K[x]$ is a PID, then $I_a$ is a maximal ideal. Hence the ring $K[a]$ is isomorphic with $K[x]/I_a$ , hence it is a field.

How can one show that "the ring $K[a]$ is isomorphic with $K[x]/I_a$"? I searched the book for various results related to maximal ideals but I wasn't able to use any of them to prove this fact.

答: Consider the evaluation morphism $K[x] \to K$ which sends a polynomial $f$ to $f(a)$. Its image is $K[a]$ and its kernel is $f\in K[x]$ with $f(a) = 0$, that is $I_a$. The first isomorphism theorem for rings then gives $K[a] \cong K[x]/I_a$.


问:
Why $0$ is not irreducible?
We took the following definition of irreducible:

If $R$ is a commutative ring, $r \in R$ is irreducible if $r = ab$ for $a,b \in R$ implies $a \in R^{*}$ or $b \in R^{*}.$

Then my professor said: "so $0$ is not irreducible" but I do not understand that as I know that $0 = 1.0$ and so we have $1$ is a unit and so $0$ can be irreducible. Could anyone explain that to me, please?

答:
It's worth emphasis that is not the definition of irreducible that is used by many ring theorists studying factorization theory and related matters in general commutative rings. As described in the paper excerpted below, common (integral) domain notions like "associate" and "irreducible" bifurcate into a few inequivalent notions in the presence of zero-divisors. Motivated by extending results on unique factorization to non-domains, this gives more insight for the most useful notion of irreducible and associate, including whether or not $0$ should be considered irreducible. It turns out that using such definitions yields that $0$ is irreducible iff the ring is a domain (for all notions of associate and irreducible). In particular this means that $p$ prime $\Rightarrow$ $p$ irreducible is true for all $p$ (including $p = 0)$. Below are the definitions in use.

Excerpted from: Factorization in Commutative Rings with Zero-divisors.

D.D. Anderson, Silvia Valdes-Leon.

Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480

Title: Why $0$ is not irreducible?


问:
Help with convergence of sequence
question is:
by choosing right $x_0$ prove that sequence that written below is convergent to ${\sqrt a}$
$$x_{n+1} = \frac {x_n(x_n^2-3a)}{3x_n^2+a}$$
where $n\ge0$ and $a>0$.

I tried this way:
$$x = \frac {x(x^2-3a)}{3x^2+a} \Rightarrow x(x^2-3a)=x(3x^2+a) \Rightarrow 2x(x^2+2a)=0 \, , \, since \, a>0 \quad  x=0$$
and $x =0$ is only solution

Thanks in advance for helping me find out how to solve this

答:
This (with the sign correction) is the Halley method applied to $f(x)=x^2-a$ or, in one way to construct that method, the Newton method to $f(x)=x^{3/2}-ax^{-1/2}$.

One can show relatively easily with the third degree binomial formula that
$$
\frac{x_{n+1}-\sqrt a}{x_{n+1}+\sqrt a}=\left(\frac{x_{n}-\sqrt a}{x_{n}+\sqrt a}\right)^3
$$
from where the third order convergence follows.

See also Convergence behaviour of an iteration of a square root using Halley's method.

Title: Help with convergence of sequence


问:
Modified probability sample
Consider a mathematical sample $X_1, X_2,...X_n$ where $X_1 \sim \mathbb{P}_\lambda, \lambda \in \Theta$. For $ r>0$ and let $\psi_r: \mathbb{R} \rightarrow\mathbb{R}$ with
$$
\psi(x)=
\begin{cases}
r, \,x<-r\\
x, \,|x|\leq r\\
r, \,x>r
\end{cases}
$$
Let $M_r$ be the set of zeros of the function
$$
\Psi_r:\mathbb{R} \rightarrow\mathbb{R}, y \mapsto \Psi_r(y):= \frac{1}{n} \sum_{k=1}^n \psi_r(X_k - y), \, y \in \mathbb{R}
$$
Define the random variable $\hat{m}_r = \inf M_r$

Now regarding the median - how do I show that $\hat{m}_r$ converges almost surely to a median of $X_1,...,X_n$ as $r \rightarrow 0$(as $r \rightarrow \infty$)?

I would very much appreciate your help!

答:
This is not a solution but some thoughts I have had, this is a real interesting problem by the way. First that $\psi_r(x) \geq 0$ for all $x$ and $\psi_i(x) = 0$ iff $x=0$. For each sample $\omega \in \Omega$ then $$\Psi_r(y,\omega) =\frac{1}{n} \sum_{k=1}^n\psi_r(X_k(\omega)-y)$$
By positivity of the $\psi_r$, we can only have $\Psi_r(y,\omega) = 0$ iff $\psi_r(X_k(\omega)-y) = 0$ for all $k$ which occurs iff $X_k(\omega)=y$ for all $k$. This shows that $$M_r(\omega)=\{y: X_k=y\,\,\,\text{for all $k$}\} $$
The second thing I noticed is that $|\Psi_r(y,\omega)|\leq \frac{r}{n}$ and so $\Psi_r\to 0$ uniformly. The third thing is that unless there is a typo, I am not sure that it is true that $\mathbb{P}(M_r =\emptyset)\not=0$, as a consequence we could have a set of positive measure where $\hat{m}_r$ is not defined. Not sure how helpful these are, but good luck!

Title: Modified probability sample


问:
Every type II$_1$ von Neumann algebra has a unique faithful trace
Here I am working with the definition that a von Neumann factor $M$ is a type-II factor if there is no non-zero, minimal projection but there are non-zero finite projections. It is a type-II$_1$ factor if it is a type-II factor and the identity is finite.

I've read that every type-II$_1$ factor $M$ has a unique, faithful trace. Namely there is a unique linear map $\omega: M \to [0 , \infty]$ so that:

It is a trace, i.e. $\omega \left( x y \right) = \omega \left( y x \right)$

It is faithful $\omega \left( x^{*}x \right)=0$ implies $x=0$.

However I haven't been able to find a proof of this statement anywhere. It would be appreciated if someone could provide a proof, reference or suggest how to approach the proof myself.

答:
This is done in several places, for instance in chapter 8 in Kadison-Ringrose; it's done in more generality, as they show that any finite von Neumann algebras has a unique central valued trace.

The "Murray-von Neumann way" is done in Sunder's An Invitation to von Neumann Algebras. Here is the (very rough) idea.

Show that two projections with the same central carrier are comparable. Conclude that in a factor all projections are comparable.

Prove an "Euclidean Algorithm" for projections: given projections $p,q\in M$, there exist projections $q_1,\ldots,q_m,r\in M$ such that $q_j\sim q$, $r\prec q$, and $p=r+\sum_jq_j$.

If $q|1$, that is if there exist $q_1,\ldots,q_m$ with $q_j\sim q$ and $\sum_jq_j=1$, define $\tau(q)=1/m$.

If $q$ admits a subprojection $q_1$ that divides both $1$ and $q$, define $\tau(q)=n/m$ where $m$ is the number of copies of $q_0$ in $1$, and $n$ the number of copies of $q_0$ in $q$.

For an arbitrary projection $p$, if $p=\sum_jq_j$ with each $q_j$ "rational" as above, define $\tau(p)=\sum_j\tau(q_j)$.

Note that the above decomposition of $p$ can always be done, as in a II$_1$ factor we have infinite divisibility of projections. With a bit of care, one can show that you can always "halve" a projction $p$, in the sense that there exist subprojections $p_1,p_2$, with $p_1p_2=0$, $p_1\sim p_2$, and $p_1+p_2=p$.

Now extend $\tau$ to linear combinations of pairwise orthogonal projections, and by taking limits to selfadjoint operators. And finally, by linearity again, to arbitrary elements.


问:
Completeness of the Fell topology
Recall that the Fell topology $\tau_F$ is a topology on the hyperspace $F(X)$ of closed subsets of a Hausdorff space (maybe you can define it in a more general context, but I am interested in $\mathbb{R}^n$). A prebase for the Fell topology is given by the sets of the form

$$\{ F\in F(X) : F\cap K =\emptyset  \} $$
$$\{ F\in F(X) : F\cap U \neq \emptyset  \}  $$

for some $K\subset X$ compact and some non-empty $U\subset X$ open.

I read in [1, Thm. 5.1.5] that $(F(\mathbb{R}^n),\tau_F)$ is compact metrizable, and becomes Polish if we restrict it to $(F(\mathbb{R}^n)\backslash \{\emptyset\},\tau_F)$. I assume that it is implicitly saying that $(F(\mathbb{R}^n),\tau_F)$ is not complete.

How can this be proved explicitly? I would guess that, since $(F(\mathbb{R}^n),\tau_F)$ is metric, there should be a Cauchy non-convergent sequence, but it is not clear to me what is the metric on $(F(\mathbb{R}^n),\tau_F)$.

[1]: Beer, Gerald, Topologies on closed and closed convex sets, Mathematics and its Applications (Dordrecht). 268. Dordrecht: Kluwer Academic Publishers. xi, 340 p. (1993). ZBL0792.54008.

答:
It's true that every compact metric space is complete. Since Beer proves in Theorem 5.1.5 that $F(\mathbb{R}^n)$ is compact and metrisable, it must be completely metrisable. I agree that Beer does not openly record this, but he does use it implicitly in the proof of the theorem.

btw, +1 for the well written question.

Title: Completeness of the Fell topology


问:
Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$
I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$.

I found a solution by myself 10 hours after I posted it, here it is:

$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$

then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.

Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get

$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$

where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function
$$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that
$$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$

But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!

Thanks to everyone who contributed.

答:
Actually the above is not quite complete, the missing piece is the proof that we can drop the contribution from the pole at $s=-1,$ which is $x/24.$ To verify this we have to show that
$$\int_{-i\infty}^{i\infty}
\frac{1}{\pi^{s+1}} \Gamma(s+1) (1-2^{-s})^2 \zeta(s+1)\zeta(s) ds = 0.$$
Now from the functional equation of the Riemann Zeta function we see that this integral is equal to
$$-\int_{-i\infty}^{i\infty}
\frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) ds$$
Actually doing the accounting we find that the kernel
$$ g(s) = \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s})
\zeta(s) $$ of this integral has the property that $g(s) = - g(-s)$ on the imaginary axis, so the integral is zero.

To see this consider what effect negation has on the individual terms.
$$\zeta(-s)\zeta(s) \to \zeta(s)\zeta(-s),$$
$$(2^s-1)(1-2^{-s}) \to (2^{-s}-1)(1-2^s) =  (2^s-1)(1-2^{-s}),$$
$$\sin(1/2 s\pi) \to \sin(1/2 (-s)\pi) = -\sin(1/2 s\pi).$$
The first two terms are even and the last one is odd, QED.

Note that we have taken advantage of the fact that $x=1$ ... for other values of $x$ this trick will not go through. Also relevant is that negation (rotation by 180 degrees about the origin) takes the imaginary axis to itself (this is not the case when we are integrating along some other line parallel to the imaginary axis in the right half plane).


问:
Weak convergence implies almost everywhere?
Suppose that $L^2(R^n;dx) \ni f_n\geq 0$ and $\int f_ng\to 0$ for all continuous function $g$ with compact  support. Then, does $f_n$ convergence to $0$ almost everywhere?

Intuitively, it convergence to $0$ by taking function $g$ s.t., $g=1$ around $x$ for almost every $x$. However I cannot prove.

答:
As the example by Ruy shows, you may not have pointwise convergence a.e. However, you can get a subsequence which converges a.e.:

Let $B_m$ denote the ball with radius $n$. Due to $f_n \ge 0$, we have
$$\|f_n\|_{L^1(B_m)} = \int_{B_m} f_n \, \mathrm d x = \int_{\mathbb R^d} \chi_{B_m} \, f_n \, \mathrm d x \to 0$$
as $n \to \infty$.
Note that $\chi_{B_m} \in L^2(\mathbb R^d)$ since $B_m$ has finite measure.
By weak convergence, $\|f_n\|_{L^2(B_m)}$ is bounded w.r.t. $n$. Using interpolation,
we get $\|f_n\|_{L^p(B_m)} \to 0$ as $n \to \infty$, where $p \in (1,2)$ is arbitrary (and fixed).
This implies the existence of a subsequence of $f_n$ which converges pointwise a.e. on $B_m$ for any fixed $m$.
A standard diagonal sequence argument yields a subsequence $f_{n_k}$ which converges pointwise a.e. on $B_m$ for all $m$.
From $\mathbb R^d = \bigcup_{m = 1}^\infty B_m$,
$f_{n_k}$ converges pointwise a.e. on $\mathbb R^d$.


答:
Just use the definition: let $a$ be a tuple from $M_0$, and let $\varphi$ be a formula. We want to show $\mathcal{M_0}\models \varphi(a)$ iff $\mathcal{M_1}\models\varphi(a)$

Since $\mathcal{M_0}\prec \mathcal{M_2}$, and $a\in M_0$, $\mathcal{M_0}\models \varphi(a)$ iff $\mathcal{M_2}\models\varphi(a)$. Since $\mathcal{M_1}\prec \mathcal{M_2}$, and $a\in M_1$, $\mathcal{M_2}\models \varphi(a)$ iff $\mathcal{M_1}\models\varphi(a)$. And we're done!

Title: Solution to exercise 2.5.9 of Marker's Model theory

Body:
I'm trying to solve this exercise 2.5.9 of David Marker's model theory. The exercise is:

Take three L-structures: $\cal{M}_0 \subset\cal{M}_1 \subset\cal{M}_2$. If $\cal{M}_0 \prec \cal{M}_2$ and $\cal{M}_1 \prec \cal{M}_2$, then show that $\cal{M}_0 \prec \cal{M}_1$.

I know that this is the coherence property of abstract elementary classes. However, I have no idea how to prove the exercise. Does it use the compactness theorem?


答:
Seeing the shape of the denominator, we can investigate the following two derivatives:

$$\left(\dfrac1{\sqrt{-10+7x-x^2}}\right)'=-\frac12\frac{7-2x}{(-10+7x-x^2)^{3/2}}$$

and

$$\left(\dfrac x{\sqrt{-10+7x-x^2}}\right)'=\frac{-10+7x-x^2}{(-10+7x-x^2)^{3/2}}-\frac12\frac{x(7-2x)}{(-10+7x-x^2)^{3/2}}
\\=\frac{-10+\dfrac72x}{(-10+7x-x^2)^{3/2}}.$$

Now you can form a linear combination to obtain the numerator $x$.

Title: Alternate solution to indefinite integral

Body:
$\int{\frac{x}{(-10+7x-x^2)^{3/2}}}dx$

Is there any alternative way to solve this indefinite integral?(Other than trig substituition)


问:
Use of Mean Value Theorem
Question

Let $f$ be continuous on $[0, 1]$ and differentiable on $(0, 1)$ such that $f(0) = 1$ and $f(1) = 0\ $.

(i) Show that there exists $c$ in $(0, 1)$ such that $f(c) = kc\ $, where $k$ is a positive constant.

(ii) Show that there exist $a$ and $b$ in $(0, 1)$ such that $cf'(a) + (1 - c)f'(b) = -1\ $.

My working

I believe I am able to show (i).

Consider $g(x) = f(x) - kx\ $, where $k$ is a positive constant.

$\implies g(0) = 1 > 0$ and $g(1) = -k < 0\ $.

By IVT, $\exists\ c \in (0, 1)$ such that $g(c) = 0\ $.

$\implies f(c) = kc\ $, where $k$ is a positive constant.

However, I am not sure how to go about (ii), although I have a hunch that it requires MVT and may be linked to (i). Any help/suggestions on how to work (ii) will be greatly appreciated :)

答:
I have solved (ii)!

By MVT, $\exists\ a \in (0, c)$ and $b \in (c, 1)$ such that $f'(a) = \frac {f(c) - f(0)} {c - 0} = \frac {kc - 1} {c}$ and $f'(b) = \frac {f(1) - f(c)} {1 - c} = \frac {-kc} {1 - c}\ $.

$\implies cf'(a) + (1 - c)f'(b) = -1\ $.

I am so happy :-)

Title: Use of Mean Value Theorem


问:
Finding parametric equations for the tangent line at a point on a curve

Find parametric equations for the tangent line at the point $(\cos(-\frac{4 \pi}{6}), \sin(-\frac{4 \pi}{6}), -\frac{4 \pi}{6}))$ on the curve $x = \cos(t), y = \sin(t), z=t$

I understand that in order to find the solution, I need to use partial derivatives. However, the method in my textbook works for simpler problems -- I seem to be making a calculation error when I try to apply the method to this problem.

Can anyone suggest how to approach this problem?

I found a very similar problem and solution here, but the solution by the person who answered is hard for me to follow. Unfortunately, I get stuck at the line where he subtracts $\frac{\pi}{6}$ from $\pi$ within the trigonometric functions.

Here is the "simple" method that I was originally using.

Any sincere help would be appreciated. Thank you.

答:
the correct answer for $x$ and $y$ are:
$$
\cos(-4\pi/6) + t(-\sin(-4\pi/6))\\
\sin(-4\pi/6) + t(\cos(-4\pi/6))
$$

this is because the point that lies on the parametric lines are defined by inputting the $t$ value into the original parametric equation

ContentLicense: CC BY-SA 2.5


问:
Loss function for regression
On Bishop, it is explained that the decision stage for a regression problem consists in finding a properly $y(\textbf{x})$ associated to the target $t$. To do this we must consider a Loss function $L(t,y(\textbf{x}))$ and what we do next is minimizing its expected value

$$\mathbb{E}[L] = \iint L(y(\textbf{x}),t)p(\textbf{x},t)\,d\textbf{x}\,dt$$

On this why do we considerthe joint distribution $p(\textbf{x},t)$? What I was thinking is that $t$ should not be a random variable since it is a deterministic value..

Also choosing the loss $L(y(\textbf{x},t)) = \{y(\textbf{x})-t\}^2$ how do we exactly obtain that

$$\frac{\partial\mathbb{E}[L]}{\partial y(\textbf{x})} = 2 \int \{y(\textbf{x})-t\}p(\textbf{x},t) \, dt $$ ??

Thanks!

EDIT:

Ok based on what I've read about variational calculus the functional derivative can be defined as (if $S$ is a functional)

$$\frac{d}{d\varepsilon}S[y(x)+\varepsilon g(x)]_{\varepsilon=0}$$

where $g$ is any function satisfying eventual boundary conditions. I then tried to apply this definition to the above loss getting

\begin{align}
\frac{d}{d\varepsilon}L[(y(\textbf{x})+\varepsilon g(\textbf{x}), t)]_{\varepsilon = 0} &= \iint \frac{d}{d\varepsilon} \{(y(\textbf{x})+\varepsilon g(\textbf{x}))-t)\}^2p(\textbf{x},t) \, d\textbf{x}dt \\
&= \iint 2\{y(\textbf{x})-t\}g(\textbf{x})p(\textbf{x},t)\,d\textbf{x}dt
\end{align}

but this of course is not equal to what it should to be.. can you please help me through these steps?

答:
The value of $t$ for a given ${\bf x}$ is not assumed to be fixed. For instance, see page 28 where for a given $x$, $t$ is assumed to be Gaussian with certain parameters depending on $x$. The book notes in Section 1.2.3 that it takes a Bayesian approach, where probabilities provide information about uncertainty. So, for instance, in the digit classification problem, $p(C_k|{\bf x})$ is interpreted as the level of certainty that the image ${\bf x}$ should be classified as belonging to the digit class $C_k$. Certainly there is a "correct" classification based on the intentions of the person who drew the digit, but as outside observers we only have degrees of certainty/uncertainty about classification (for instance, how certain are you in your own classification of these images?).

For your second question, there is a note next to the equation to see Appendix D which explains calculus of variations and functional derivatives. You left out the $p({\bf x},t)$ term - perhaps that's where your confusion lies?

Title: Loss function for regression


问:
How to compute $\int_0^x\dfrac{e^\frac{t^2}{2}}{\sqrt{x-t}}~dt$
Hi I am working through this integral

$$\int_0^x\dfrac{e^\frac{t^2}{2}}{\sqrt{x-t}}~dt$$

But I tried doing it by parts and nothing, it seems it is solved with erf, but I don't know the method. Any help you could give, or orientation to solve this integral ?. I think one possible solution is to approximate $e^\frac{t^2}{2}$ by Taylor but will be and approximate the integral.

$$e^\frac{t^2}{2}=1+\frac{1}{2}t^2+\frac{1}{8}t^4+\frac{1}{48}t^6+\frac{1}{384}t^8+......$$


问:
Question regarding the notation of $\equiv$ and $\iff$
What difference does the notation in those two terms make?
$$
=:⟺∀:(∈\iff∈)
$$
$$
=:⟺∀:(∈\equiv∈)
$$

答:
One way of reading it is as a sigle formula, meaning: two sets are equal iff "condition".

In this case, it is an inconsistency to use two different symbols for the same concept: the bi-conditional.

Another reading is to consider it as an "abbreviation", meaning: we write $X=Y$ exactly when "condition" holds.

In this case, there is no benefit from translating the leftmost "iff" with a symbol. The abbreviation is not a formula of the object language but a statement in the meta-language, and it is not necessary to "formalize" it.


问:
On the dimension of tangents spaces of smooth manifolds
Let $M \subset \mathbb{R}^k$ be a smooth $m$-dimensional manifold and fix a point $p$ on $M$.

Let $U_0 \subset M$ be an $M$-open set with $p \in U_0$ and let $\phi_0 : U_0 → \Omega_0$
be a diffeomorphism onto an open subset $Ω_0 ⊂ \mathbb{R}^m$.

Define $x_0 := \phi_0(p)$ and let $\psi_0 := \phi_0^{-1}: \Omega_0 \rightarrow U_0$ be the inverse map. Then:

$$T_pM = (\text{im  }d\psi_0(x_0): \mathbb{}  \mathbb{R}^m \rightarrow \mathbb{R}^k) $$

An inclusion argument is trivial to make, i.e. for a small enough $t$ and $\xi \in \mathbb{R}^m$, such that $x_0 + t \xi \in \Omega_0$, we can define a curve $\gamma: (-\epsilon, \epsilon) \rightarrow M$, such that $\gamma(t) := \psi_0(x_0 + t\xi)$. Naturally, we get:

$$\gamma(0) = p, \text{ } \dot{\gamma}(0) = \frac{d}{dt} \bigg|_{t=0} \psi_0(x_0 + t\xi) = d\psi_0(x_0) \xi$$

This shows that $d\psi_0\xi \in T_pM$, which is all fine. The author then goes to say that since $\dim \text{im  }d\psi_0(x_0) = m$, the problem is solved.

I'm missing the justification behind this $\dim \text{im  }d\psi_0(x_0) = m$. Indeed, given the smoothness of $\psi_0 := \phi_0^{-1}$ I assume some argument can be made regarding the rank/surjectivity of the Jacobian $d\psi(x_0)$ via some form of the inverse function theorem, but I just can't get my finger on it.

Otherwise, regarding $d\psi(x_0)$ as the best linear approximation to $\psi$ around $x_0$, one may write:

$$\psi(x_0 + t \xi) = \psi(x_0) + td\psi(x_0)\xi + o(\Vert \xi \Vert)$$

But given $m \neq k$ I find it non-trivial to argue for the surjectivity of $d\psi(x_0)$.
A solution would be welcome.

The text I'm studying is Introduction to Differential Geometry, Robin and Salamon (18 April 2020 version), page 26.

答:
We follow J. Milnor, Topology from the differentiable viewpoint, page 5, almost verbatim.

Since $\phi_0\colon U_0\to\Omega_0$ is a smooth mapping, we can choose an open set $W$ containing $p$ and a smooth map $F\colon W\to\mathbf{R}^m$ that coincides with $\phi_0$ on $W\cap U_0$. Letting $V=\phi_0(W\cap U_0)$ and writing $\iota\colon V\to\mathbf{R}^m$ for the inclusion map, we have $\iota=F\circ\psi_0$ and therefore $\text{id}_{\mathbf{R}^m}=dF_p\circ d(\psi_0)_{x_0}$. It follows that $d(\psi_0)_{x_0}$ has rank $m$, and hence its image $T_pM$ has dimension $m$.


问:
Why are addition and multiplication commutative, but not exponentiation?
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation
where

$H_0(a,b) = b+1$   (successor op)

$H_1(a,b) = a+b$   (addition op)

$H_2(a,b) = ab $   (multiplication op)

$H_3(a,b) = a^b$   (exponentiation op)

$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )

etc.

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?

Thanks.

答:
I totally agree that this "breaking of symmetry" happens and that it is because we do not understand addition and multiplication completely. There are many types of rings, also e.g. power-associative rings, etc. but that's not the point. What exactly makes it possible to so easily interchange the "operation" with the "quantity" under the operator for addition?

For your suggestion:

I'd like to see an answer that either: defines a "new" exponentiation (by generalizing differently) which is both associative and commutative

I thought for ordinal numbers exponentiation is associative, it isn't. But there might be mathematical constructs, Jacobsthal's exponentiation(?), for which it is the case.