问:
Prove that $\mathbb K[x] \times \mathbb K[y] \rightarrow \mathbb K[x,y] \,$ induces an isomorphism of $\mathbb K$-algebras.
I'm having some trouble with the following problem.

Let $x, y$ be two commuting variables, and let $\mathbb K$ a field. We define the following function $$\mathbb K[x] \times \mathbb K[y] \rightarrow \mathbb K[x,y], \, (P(x),Q(y))\mapsto P(x)Q(y).$$

I have two questions.

1.) I don't see why this function is bilinear.

2.) Why can we say that our function induces an  isomorphic $\mathbb K[x] \otimes \mathbb K[y] \rightarrow \mathbb K[x,y]?$

(For question 2., can we use the universal property of the tensor product?)

答:
Your question makes little sense. You also need to specify what kind of isomorphism. Of vector spaces over $\mathbb{K}$? As $\mathbb{K}$-algebras? As $\mathbb{Z}$-algebras (= rings)?

You can show that
$$\mathbb{K}[x] \otimes \mathbb{K}[y]\cong \mathbb{K}[x,y]$$ as $\mathbb{K}$-algebras.

I'll give a brief sketch. You can use the comments to ask for clarification if necessary.

Define the mapping
$$\mathbb{K}[x]  \times \mathbb{K}[y]\to \mathbb{K}[x,y]: (P(x), Q(y)) \mapsto P(x)Q(y)$$

This is a bilinear map by distributivity in $\mathbb{K}[x,y]$. The universal property of the tensor product gives a unique linear map
$$\mathbb{K}[x] \otimes \mathbb{K}[y] \to \mathbb{K}[x,y]: P(x) \otimes Q(y) \mapsto P(x)Q(y)$$

You have to show this map is a bijection and you will be done. Surjectivity ought to be clear, but you need to work a bit to show injectivity.

One possible approach (see the top comment for another approach): You can try to define an explicit inverse. For example, define
$$\mathbb{K}[x,y] \to \mathbb{K}[x] \otimes \mathbb{K}[y]$$
by $x \mapsto x \otimes 1$ and $y \mapsto 1 \otimes y$ and extend multiplicatively and linearly.


问:
Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$?
I'm trying to solve this equation:
$$2-x=-\sqrt{x}$$
Multiply by $(-1)$:
$$\sqrt{x}=x-2$$
power of $2$:
$$x=\left(x-2\right)^2$$
then:
$$x^2-5x+4=0$$
and that means:
$$x=1, x=4$$

But $x=1$ is not a correct solution to the original equation.

Why have I got it? I've never got a wrong solution to an equation before. What is so special here?

答: Overshoot, with subsequent back-tracking to the destination, is sometimes the easiest path, a common example being the fact that, due to tricky lane changes and / or confusing roadway signage, it could well be easier to overshoot the destination and then wend your way back to it.  The phenomenon of such overshoot is just a special case of the fact that a larger footprint is easier than a smaller footprint. (cf: ‘collateral damage’, and cf: “I would have written a shorter letter, but I didn’t have the time.” -- Blaise Pascal) An example of mathematically necessary overshoot is the fact that the vertical coordinate of the brachistochrone overshoots the vertical coordinate of the destination, and, even more famously, that ‘the shortest distance between two truths in the real domain often passes through the complex domain’ – the passing through the complex domain being a stratospheric instance of overshoot, two examples of which are the explanation of the radius of convergence of 1/(1 + x^2), and the proof of the fact that the product of two numbers, each representable as the sum of two squares, is itself representable as the sum of two squares. A very common example is the squaring of both sides of an equation (in order to remove a radical), which may introduce an ‘extraneous’ root. You then ‘pull back’ to the actual solution set by inspecting each root of the second equation individually.</p>

ContentLicense: CC BY-SA 4.0


问:
Confusion about limits, specifically the definition of a limit
$$\lim_{x\to0}f(x)$$
The most common way to read this is "limit of f(x) as x approaches 0". But the thing is, I find that there is a difference between these two limits for example
$$\lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1$$
$$\lim_{x\to0}\frac{x}{5} = 0$$
Looking at the first one, technically speaking, the value won't actually ever hit 1. But it's converging to it. But with the second one, it does become 0, even though it's correct to say "as x approaches 0, $\frac{x}{5}$ approaches 0", it's also correct to say "as x approaches 0, $\frac{x}{5}$ will eventually equal 0". The latter statement doesn't sound so correct when you try to apply it with the first limit.

This kind of confused me, and I was unable to find answers online, so for a while I made a deduction myself and kept in mind that "yes, $\frac{x^2-1}{x^2+1}$ won't actually ever equal 1, but it will be infinitesmally close to it, to the point that this infinitesmally small distance doesn't matter, thus concluding that from a larger scale it's equal to 1".

My question: is this interpretation correct?

Graphing the function of the first limit shows this even better

Edit: Sorry I edited the example because I realized the infinity limit is a bad a example!

答:
I guess the major issue you are having is that you are using an intuitive understanding of limit, instead of a rigorous definition. The rigorous definition for limit at a finite point $x_{0}$ in calculus is as follows:
$$\lim_{x\to x_{0}}f\left(x\right) = s \longleftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x, \lvert x-x_{0}\rvert < \delta \longrightarrow \lvert f\left(x\right) - s \rvert < \epsilon.$$
Similarly, the definition for limit at infinity is as follows:
$$\lim_{x\to\infty}f\left(x\right) = s \longleftrightarrow \forall \epsilon > 0, \exists y \in \mathbb{R}, \forall x, x > y \longrightarrow \lvert f\left(x\right) - s\rvert < \epsilon.$$

I am now using the definition to prove that $\lim_{x\to 0}\frac{x}{5} = 0$, and leave the rest to you. Given an arbitrary $\epsilon$, as long as $\lvert x-0\rvert < 5\epsilon$, we have
$$\lvert\frac{x}{5}-0\rvert = \frac{\lvert x\rvert}{5} < \frac{5\epsilon}{5} = \epsilon.$$
According to the definition, we take $\delta = 5\epsilon$ for each $\epsilon$, and we have $\lim_{x\to 0}\frac{x}{5} = 0$.

Intuitive understanding is good for introduction of a concept. However, when it comes to learning rigorous math, understanding and using formal definition is crucial. Sometimes, a function is not defined at a point, but limit at that point exists. For instance, we have

$$\lim_{x\to 0}\frac{\sin{x}}{x} = 1.$$ The function $\frac{\sin{x}}{x}$ is not defined at $x = 0$, but according to the definition, the limit exists at that point. Thus, you should follow strictly the formal definition of limit, instead of intuitive understanding. Math is extracted from intuition, but it is not intuition itself. It is a set of deduction rules and axioms that incorporate intuition while show something farther.


答:
Limits tell you about the behaviour of a function when you get close to a point, but not actually when you get there. For example,
$$
\lim_{x \to 1}\frac{x^2-1}{x-1}=2
$$
but this does not mean that $0/0=2$. This is one of many examples of where you can't just plug in the value being approached and expect to get the answer. When we write
$$
\lim_{x \to 1}\frac{x^2-1}{x-1}=2
$$
what we mean is that the function $f(x)=\frac{x^2-1}{x-1}$ will get arbitrarily close to $2$ as $x$ approaches $1$. Pick a tiny number—say $0.00001$. As $x$ gets closer and closer to $1$ (but not does ever reach it), $f(x)$ will get within that number of $2$; that is, it will get within the range $(1.99999,2.00001)$. Hopefully this is what you meant when you said 'infinitesimally close'. As you have noticed, sometimes
$$
\lim_{x \to a}f(x)=f(a) \, .
$$
If this happens, then the function is said to be continuous at $a$. In your example,
$$
\lim_{x \to 0}\frac{x}{5}=0 \, .
$$
The LHS evaluates to $0$ because as $x$ gets arbitrarily small, so does the value of $x/5$. It just so happens that plugging in $x=0$ also gets you the right answer. However, there are countless examples of functions that aren't continuous. A slightly contrived example is this
$$
f(x) =
\begin{cases}
x+10 &\text{if $x\neq0$} \\
147 &\text{if $x=0$}
\end{cases}
$$
The limit as $x$ approaches $0$ of $f(x)$ equals $10$, but $f(0)=147$. The function appears to 'jump' when we get to $x=0$. (Incidentally, this is called a 'jump discontinuity'.) Now, limits to infinity are a different beast altogether.
$$
\lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1
$$
has quite a different meaning to what we have been discussing so far. Here, $\infty$ is just a shorthand for saying that as $x$ gets larger and larger,
$$
\frac{x^2-1}{x^2+1}
$$
gets closer and closer to $1$. It doesn't make sense for $x$ to get 'arbitrarily close to infinity'—as you have probably been told countless times, infinity is not a number. Consider the sequence
$$
\{1,2,3,4,\ldots\} \, .
$$
However high you count, you don't ever get 'close' to infinity, but the values in the sequence do increase without bound. That's the sense in which $x$ 'approaches' infinity. And again,
$$
\frac{x^2-1}{x^2+1}
$$
gets closer and closer to $1$ as $x$ gets bigger and bigger, but does not ever reach it.

Title: Confusion about limits, specifically the definition of a limit

Body:
$$\lim_{x\to0}f(x)$$
The most common way to read this is "limit of f(x) as x approaches 0". But the thing is, I find that there is a difference between these two limits for example
$$\lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1$$
$$\lim_{x\to0}\frac{x}{5} = 0$$
Looking at the first one, technically speaking, the value won't actually ever hit 1. But it's converging to it. But with the second one, it does become 0, even though it's correct to say "as x approaches 0, $\frac{x}{5}$ approaches 0", it's also correct to say "as x approaches 0, $\frac{x}{5}$ will eventually equal 0". The latter statement doesn't sound so correct when you try to apply it with the first limit.

This kind of confused me, and I was unable to find answers online, so for a while I made a deduction myself and kept in mind that "yes, $\frac{x^2-1}{x^2+1}$ won't actually ever equal 1, but it will be infinitesmally close to it, to the point that this infinitesmally small distance doesn't matter, thus concluding that from a larger scale it's equal to 1".

My question: is this interpretation correct?

Graphing the function of the first limit shows this even better

Edit: Sorry I edited the example because I realized the infinity limit is a bad a example!


问:
For a circle-Apollonian-packing problem, finding the smallest circle such that all natural numbers may be packed.
In section $1$ of the paper "The sequence of radii of the apollonian packing" of David W. Boyd, the author said that the question for the smallest circle into which disks of radius $1/n,n=1,2,...$ can all be packed is posted by H.S.M. Coxeter in 1979, the paper named "Problem P.276", and Boyd found the answer is $3/2=1.5$ and published it in the paper named "Solution to problem P.276" in 1980.
I tried to find the paper, but I cannot. So I wonder whether anyone may suggest proof, a reference, or a source for the question. Thanks.

Something here we can know straight forward is that the radius of the smallest circle is $\ge \frac{\pi }{{\sqrt 6 }} \approx 1.28254983016$ (quite closed to the answer)  because the Riemann zeta function $\zeta (2) = \frac{{{\pi ^2}}}{6}$.

Integer circle packing in higher dimensions should be very relevant to the values of the zeta function. But I cannot find many papers towards to the relation.

答:
This is a very interesting problem and just for the fun, let's prove it ourselves.

First, it is impossible to pack all those circles inside a circle of radius $< \frac32$. If a circle of radius $R$ is large enough to hold the circle of radius $1$ and $\frac12$, then $2R$ will be larger than the length of the line segment passing through the centers of these two circles. Since the length of that line segment is at least $3 = 2 +1$. This forces $2R \ge 3 \implies R \ge \frac32$.

Next, start from an outer circle $\mathcal{C}_V$ of radius $\frac32$, we can place a circle $\mathcal{C}_U$ of radius $1$ and a circle $\mathcal{C}_0$ of radius $\frac12$ along a diameter of $\mathcal{C}_V$. If one work out the radii of circles in the Pappus chain associated with the pair $\mathcal{C}_U, \mathcal{C}_V$.
The radii are given by the formula

$$\frac{3}{6+ n^2}\quad\text{ for }\quad n \in \mathbb{Z}$$
The $n = 0$ corresponds to our circle $\mathcal{C}_0$. Aside from this, one
can find

two circles $\mathcal{C}_{\pm 1}$ with radius $\frac37$,

two circles $\mathcal{C}_{\pm 2}$ with radius $\frac3{10}$,

two circles $\mathcal{C}_{\pm 3}$ with radius $\frac15$,

two circles $\mathcal{C}_{\pm 4}$ with radius $\frac3{22}$.

Using the lemma at end of this answer, we can place all circles of radii $\frac19, \frac1{10}, \frac1{11}, \ldots$ inside $\mathcal{C}_1$. For the remaining $6$ circles, we can place

the circle with radius $\frac13$ into $\mathcal{C}_{-1}$.

the circles with radii $\frac14$ and $\frac15$ into $\mathcal{C}_{\pm 2}$.

the circles with radius $\frac16$ and $\frac17$ into $\mathcal{C}_{\pm 3}$.

the circle with radius $\frac18$ into $\mathcal{C}_{4}$.

In this manner, we can place all the circles inside a single circle of radius $\frac32$.

Lemma

It is possible to pack mutually disjoint open disks of radii $\frac19$, $\frac1{10},\ldots$ inside a circle of radius $\frac37$.

Proof of Lemma

Let $A = \frac37$, $\kappa = 3$ and $B = \frac{\kappa+1}{\kappa-1} = 2$.

For $n \ge 9$, let  $r_n = A - \frac{B}{n}$, $\rho_n = nr_n = An - B$ and $\delta_n = \sin^{-1}\frac{1}{\rho_n}$.

Define angles $\theta_n$ by $\theta_9 = 0$ and
$$\theta_{n} - \theta_{n-1} = \delta_n + \delta_{n-1}\quad\text{ for }n > 9\tag{*1}$$
Let $D_n$ be the open disc with radius $\frac1n$ centered at $(r_n\cos\theta_n,r_n\sin\theta_n)$. Notice for these $n$, $r_n > \frac1n$ and  disc $D_n$ subtend a cone $C_n$ of angle $2\delta_n$ at origin $O$. The relation $(*1)$ forces $$C_n \cap C_{n+1} \subset \{O\} \implies D_n \cap D_{n+1} = \emptyset$$
ie. $D_n$ will be disjoint from its neighbours $D_{n-1}$ and $D_{n+1}$.

Let $\bar{C}_n$ be the closure of $C_n$. Consider finite union of these closed cones start at some $p \ge 11$. When enough cones are added to the union, it will cover the plane. Let $q$ be the smallest integer for this to happen:
$$\bigcup_{\ell=p}^q \bar{C}_\ell \supset \mathbb{R}^2
\quad\iff\quad \pi \le \sum_{\ell=p}^q \delta_\ell\tag{*2}$$
Notice $\delta_\ell = \sin^{-1}\frac{1}{\rho_\ell} = \sin^{-1}\frac{1}{A\ell - b}$ is a decreasing function in $\ell$. By bounding $\delta_\ell$ with corresponding integral over $[\ell,\ell+1]$, one find:

$$\pi \le \sum_{\ell=p}^q \delta_\ell \le \frac1A\int_{x_0}^{kx_0}\sin^{-1}\frac{1}{x} dx\tag{*3}
$$
where $x_0 = A(p-1)-B$ and $kx_0= Aq-B$.

Since $p \ge 11$, $x_0 \ge \frac{16}{7} \implies \frac1x \le \frac12$ for $x \in [x_0,kx_0]$.  For $y \in (0,\frac12)$, we have following bound for $\sin^{-1}y$,

$$\sin^{-1}y \le y + 4y^3\left(\frac{\pi}{3}-1\right)$$
Substitute $y$ by $\frac1x$ and plug these into RHS of $(*3)$, we obtain:

$$\pi A \le \log k + \frac{2}{x_0^2}\left(\frac{\pi}{3}-1\right)\left(1 - \frac1{k^2}\right)
$$
Treat RHS as a function of $k, p$ and call it $E(k,p)$. For fixed $k$, it is a decreasing function in $p$. If $\pi A > E(k,p )$ for a $p$, then $\pi A > E(k,p')$ for all $p' \ge p$.

In particular, we have $\pi A > E(k_0,11)$ for $k_0 = \frac{15}{4}$.
This means for any $p \ge 11$ and corresponding $q$ in $(*2)$, we have

$$\frac{Aq-B}{A(p-1)-B} = k > k_0 \implies q - 3p > \frac{9p-199}{12}$$
For $p \ge 21$, RHS of last expression $> -1$. Since $q - 3p$ is an integer, we find $q \ge \kappa p = 3p$ whenever $p \ge 21$.
This leads to

$$r_q - r_p = \frac{B}{p} - \frac{B}{q}
\ge \frac{B}{p}\left(1 - \frac{1}{\kappa}\right)
= \frac{1}{p}\left(1 + \frac{1}{\kappa}\right)
\ge \frac1{p} + \frac1q\\
$$
This implies $D_p \cap D_q = \emptyset$. Since $r_n$ is increasing while $\frac1n$ is decreasing, we have a similar inequality for all $n > q$. As a result, $D_p$ and $D_n$ will be disjoint for all $n \ge q$.

Based on this, we find

It is possible to pack mutually disjoint open discs of radii $\frac1{21}, \frac{1}{22}, \ldots$ inside a circle of radius
$\frac37$.

If we modify definition of $r_n$ to

$$r_n = \begin{cases}
\frac17, & n = 9\\
A - \frac{B}{n} - \frac23(\frac1n - \frac1{21}), & 9 < n < 21\\
A - \frac{B}{n}& n \ge 21
\end{cases}$$
and redefine $\rho_n, \delta_n, \theta_n, D_n$ using same formula as above. It is not hard to see for $n \ge 21$, the new location of $D_n$ can be obtained from the old location by an rotation independent of $n$. This means the new $D_n$ are also mutually disjoint for $n \ge 21$.

By brute force, one can verify $D_n \cap D_m = D_m \cap D_m'$ for any $n \le 21$ and $21 < m \ne m' \le 9$.

Combine all these, the lemma follows...

Sometimes a picture worth more than a thousand words.
The figure below illustrate what the configuration looks like.

the outer circle in red has radius $A = \frac37$.

the red disc is $D_9$.

the yellow discs are $D_{10}$ to $D_{20}$

the green discs are $D_{21}$ to $D_{200}$.

As one can see, aside from those that are adjacent to each other, the rest are clearly mutually disjoint.


问:
Simplifying fractions with fractions
I am trying to simplify

$$\frac{\frac{y}{x} - \frac{x}{y}}{\frac{1}{y} - \frac{1}{x}}$$

I make the top part into

$$\frac{y^2 - x ^2}{xy}$$

I know the bottom can be rewritten to just be multiplied into the top as its inverse.

$$\frac{y^2 - x ^2}{xy} * (y - x)$$

$$\frac{y - x }{xy} $$

This of course is wrong, but I do not know why, it appears to be correct to me. Nothing I did is mathematically wrong, it follows all the rules. What went wrong?

答:
No, you didn’t follow the rules: there is nothing in the rules that lets you transform

$$\frac1{\frac1y-\frac1x}$$

into $y-x$, which is what you’ve done when you go from

$$\frac{\frac{y^2-x^2}{xy}}{\frac1y-\frac1x}$$

to

$$\frac{y^2-x^2}{xy}\cdot(y-x)\;.$$

You can’t invert the denominator until you’ve transformed it into a single fraction:

$$\frac1y-\frac1x=\frac{x-y}{xy}\;.$$

Now you can invert and multiply:

$$\frac{\frac{y^2-x^2}{xy}}{\frac1y-\frac1x}=\frac{y^2-x^2}{xy}\cdot\frac{xy}{x-y}=\frac{y^2-x^2}{x-y}=\frac{(y-x)(y+x)}{x-y}=-(y+x)\;.$$

ContentLicense: CC BY-SA 3.0

Title: Simplifying fractions with fractions


问:
Is the resultant will be positive definite of not?
I have two square matrices $A=\begin{bmatrix}
a_{11}&a_{12}&0&0\\
a_{21}&a_{22}&0&0\\
0&0&a_{33}&a_{34}\\
0&0&a_{43}&a_{44}
\end{bmatrix}$ and $B=\begin{bmatrix}
b_{11}&0&0&0\\
0&b_{22}&0&0\\
0&0&b_{33}&0\\
0&0&0&b_{44}
\end{bmatrix},$which are in product form as shown in the following equation
\begin{equation}
A^{-1}BA
\end{equation}
where $B$ is a diagonal positive definite matrix. $A$ can be positive, negative or indefinite matrix.

If the main diagonal elements $B$ are equal (i.e., B is scalar positive definite), then obviously $A^{-1}BA=BI,$ which is a positive definite. But I am not sure what will be the result if $B$ is a diagonal with different elements on its main diagonal?

答:
Counterexample: Take
$$
A = \pmatrix{2&1&0&0\\1&2&0&0\\0&0&1&0\\0&0&0&1}, \quad
B = \pmatrix{10\\&1\\&&1\\&&&1}.
$$
We find that
$$
A^{-1}BA = \pmatrix{13&6&0&0\\-6&-2&0&0\\0&0&1&0\\0&0&0&1}
$$
has a negative diagonal entry and therefore cannot be positive definite.


问:
3D transform cylinder to make center line z axis
I have an array of points that represent the surface of a cylinder, I was able to calculate the center line of this cylinder but it is translated and rotated away from origin. I want to transform this cylinder to make the center line become the Z axis, I remember it was done by multiplying with a transform matrix but I don't remember how.

here is an example of the cylinder, I have the center line 2 points (start and end), bottom point is supposed to become the new (0,0,0) origin, I added some numbers for simplification.

What I want:

any help is much appreciated

答:
With the rotation matrix
$$R=\left(
\begin{array}{ccc}
 \frac{1}{6} \left(\sqrt{3}+3\right) & \frac{1}{6} \left(\sqrt{3}-3\right) & -\frac{1}{\sqrt{3}} \\
 \frac{1}{6} \left(\sqrt{3}-3\right) & \frac{1}{6} \left(\sqrt{3}+3\right) & -\frac{1}{\sqrt{3}} \\
 \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
\end{array}
\right)$$
the points $(2,2,2)$ and $(3,3,3)$ become $\left(0,0,2 \sqrt{3}\right)$ and $\left(0,0,3 \sqrt{3}\right)$.

Thus, given a point $P=(x,y,z)$ the transformation
$$P'=RP+(0,0,-2\sqrt 3)$$
should do the job.

Edit

The rotation matrix for a generic vector $(d x,d y,d z)=(x_2-x_1,y_2-y_1,z_2-z1)$  is the following. To simplify the notation I substituted $d=\sqrt{dx^2+dy^2+dz^2}$

$$
R=\left(
\begin{array}{ccc}
 \frac{\frac{dx^2 dz}{d}+dy^2}{dx^2+dy^2} & \frac{dx dy \left(\frac{dz}{d}-1\right)}{dx^2+dy^2} & -\frac{dx}{d} \\
 \frac{dx dy \left(\frac{dz}{d}-1\right)}{dx^2+dy^2} & \frac{\frac{dy^2 dz}{d}+dx^2}{dx^2+dy^2} & -\frac{dy}{d} \\
 \frac{dx}{d} & \frac{dy}{d} & \frac{dz}{d} \\
\end{array}
\right)
$$


问:
Evaluate $\lim\limits_{n \to \infty} nx_n.$
Suppose $$x_2 \in \left(0,\frac{\pi}{2}\right), x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n(n \ge 2).$$ Evaluate $\lim\limits_{n \to \infty} nx_n.$

Note that $$x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n\le \sin x_n\le x_n,$$ and $$|x_{n+1}|=\left|\left(1-\frac{1}{n}\right)\sin x_n\right|\le 1,$$ it follows that $\{x_n\}$ converges. Therefore, we can readily obtain $x_n \to 0$.

As for the limit wanted, we can consider apply Stolz theorem. But it's too complicated.

答:
Work in progress. We can conclude that $(nx_n)_{n\in\mathbb{N}}$ converges ...

Using induction we can show that $x_n \geq 0$ and, thus, $nx_n\geq0$.

Function $\sin{x}\geq0$, for $x\in\left[0,\frac{\pi}{2}\right]$. As a result $\sin{x_2}>0$ and
$$0\leq x_3=\frac{\sin{x_2}}{2}\leq 1<\frac{\pi}{2} \Rightarrow x_3\in\left[0,\frac{\pi}{2}\right]$$
And using the induction hypothesis for $x_n\in \left[0,\frac{\pi}{2}\right]$
$$0\leq x_{n+1}=\left(1-\frac{1}{n}\right)\sin{x_n}\leq 1<\frac{\pi}{2} \Rightarrow x_{n+1}\in\left[0,\frac{\pi}{2}\right]$$

From

$$|x_{n+1}|=
\left|1-\frac{1}{n}\right||\sin{x_n}|\leq
\left|1-\frac{1}{n}\right||x_n|\leq \\
\left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right||\sin{x_{n-1}}|\leq ... \\
\leq \left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right|\left|1-\frac{1}{n-2}\right|...\left|1-\frac{1}{2}\right||\sin{x_2}|=\frac{|\sin{x_2}|}{n}$$
Thus
$$|(n+1)x_{n+1}|\leq \left(1+\frac{1}{n}\right)|\sin{x_2}|$$
and $(nx_n)_{n\in\mathbb{N}}$ is bounded.

Now:

$$0\leq \frac{(n+1)x_{n+1}}{nx_{n}}=\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right) \frac{\sin{x_n}}{x_n} \leq 1-\frac{1}{n^2}<1$$

and $(nx_n)_{n\in\mathbb{N}}$ is descending.


答:
No, by definition, a limit never "hits" the target value of the argument. To evaluate a limit, you must not use the value of $f(x_0)$.

Intuitively, a limit is the value we should reach for the given value of the argument, by comparing to the values in the neighborhood. It is not the value we do reach by using the function value.

When the two values are defined and coincide (I mean $\lim_{x\to x_0}f(x)$ and $f(x_0)$), we say that the function is continuous. But they don't have to.


问:
Cl(A) if and only if each element of base on $X$ meets A.
Let $(X, \tau)$ topological space, $A \subseteq X$
Prove Cl(A) $ if and only if each element of base on $X$ meets $A$.

答:
I think $CI(A)$ is supposed to mean $Cl(A)$ or the closure of $A$, $\overline{A}$.

$\implies$ suppose $\overline{A} = X$. If there is a basic open set $U$ such that $U \cap A = \emptyset$, then for any $x \in U$, $x \not \in \overline{A} = X$, a contradiction.

$\impliedby$ Suppose every basic open set meets $A$. Then given any $x \in X$, then pick any open set $U$ containing $x$. Then there is a basic open $V \subseteq U$ containing $x$. Then by assumption $V \cap A \not = \emptyset$, which means that $U \cap A \not = \emptyset$, meaning $x \in \overline{A}$ and thus $X = \overline{A}$.


问:
Find polynomials : $ xP(x-1)=(x-11)P(x)$
Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ xP(x-1)=(x-11)P(x)$$

(Attempted work has been moved to answer)

答:
We have
$$\frac{P(x)}{P(x-1)} = \frac{Q(x)}{Q(x-m)}$$
Consider $S(x) =Q(x)Q(x-1) \cdots Q(x-m-1)$ then we have
$$\frac{P(x)}{P(x-1)} = \frac{S(x)}{S(x-1)}$$
or
$$\frac{P(x)}{S(x)} = \frac{P(x-1)}{S(x-1)}$$
and so $\frac{P(x)}{S(x)}$ is constant.


问:
Fast way to calculate Eigen of 2x2 matrix using a formula
I found this site: http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html

Which shows a very fast and simple way to get Eigen vectors for a 2x2 matrix. While harvard is quite respectable, I want to understand how this quick formula works and not take it on faith

Part 1 calculating the Eigen values is quite clear, they are using the characteristic polynomial to get the Eigen values.

Part 2, where they calculate the Eigen vectors is what I don't understand and have tried to prove but cannot.

I understand that that what matters with Eigen vectors is the ratio, not the value. For example, an Eigen value of 2, with vector 3, 4, I could have any other vector, example 6, 8, or 12, 16, etc... any scalar multiple.

In their example, given a matrix in the form a b c d, if b & c are zero, then the vectors are 1 0 and 0 1, which makes sense as you can scale these to any other size.

I don't understand the other two cases (when b=0, or c=0), or I presume the case when b & c are non-zero.

Can somebody offer an explanation or proof of this?

答:
Computing the eigenvalues comes down to finding the roots of $\lambda^2 -(a+d)\lambda + (ad-bc) = 0$. That part you know already.

So if the eigenvalues are $\lambda_1$ and $\lambda_2$, then assume $c\neq 0$ and then the claim is that the eigenvectors are $v_i = (\lambda_i-d,c)$. Then
$$Av = (a\lambda_i-ad + bc, c\lambda_i - cd + cd) = (a\lambda_i - (ad-bc),c\lambda_i),$$
and
$$\lambda_iv = (\lambda_i^2 - d\lambda_i,c\lambda_i).$$

The last coordinates are clearly equal, and we know that $\lambda_i^2 -(a+d)\lambda_i + (ad-bc) = 0$, which implies $\lambda_i^2 - d\lambda_i = a\lambda_i - (ad-bc)$, so the first coordinates are equal too.

If $c=0$ but $b\neq 0$, then the math works out the same way for $v_i=(b,\lambda_i-a)$.


问:
Prove that $Im(AA^*)=Im(A)$
Let A be an operator in a Unitary or Euclidean space.

How can I prove the following statement?
$$Im(AA^*)=Im(A)$$

答:
You have $\ker AA^*=\ker A^*$. Then, assuming finite dimension (otherwise you need closures)
$$
\operatorname{Im} A=\ker(A^*)^\perp=\ker (AA^*)^\perp=\operatorname{Im}AA^*.
$$

Title: Prove that $Im(AA^*)=Im(A)$


问:
Analyzing the Coefficient of a generating function and its asymptotic
The generating function
$\sum_{n\geq 0} D(n) x^n = \frac{1}{\sqrt{1-6x+x^2}}$ is the gf of the Delannoy number.

See last paragraph in https://en.wikipedia.org/wiki/Delannoy_number

In this link, they mentioned that the coefficient of this gf  is $D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}$.

My question is how can this coefficient be calculated/extracted from its gf.And how they conclude that it behaves asymptotically as

${\displaystyle D(n)={\frac {c\,\alpha ^{n}}{\sqrt {n}}}\,(1+O(n^{-1}))}$

where
$\alpha =3+2{\sqrt {2}}\approx 5.828$

and

$c=(4\pi (3{\sqrt {2}}-4))^{-1/2}\approx 0.5727$.

How to calculate/extract $D(n)$ from its gf and how to analyze it asymptotically then?
Is that possible to do that with a direct calculations?

Edit
I tried to find the asymptotic of the codfficidnt in the above gf this way:

The gf has two poles: $r_1=3+\sqrt{8}$, $r_2=3-\sqrt{8}$. So the smallest/dominant pole is $r_2=3-\sqrt{8}$.
We can write the gf as:
$A(x)=1/ \sqrt{r_2-x} * 1/ \sqrt{r_1-x}$
Then A(x) behaves as:

$1/ \sqrt {r_1-r_2} * 1/ \sqrt{r_2-x}=
1/ \sqrt {2 * 8^{1/4}} * \sum_{n\geq 0} {-1/2 \choose n} (-1)^n (r_1)^{n+1}x^n$
Here I used the expansion's formula  of $(1+x)^a$ for $a=-1/2$ and $x=-x/r_2$.
So the coefficient behaves asymptotically as:
$1/ \sqrt 2 * 1/ 8^{1/4} * (3+\sqrt 8)^{n+1/2} {-1/2 \choose n} (-1)^n$.

We can simplify it more, by using the asymptotic of ${-1/2 \choose n}$.

Is that approach fine?

答:
The formula you refer to
$$\frac{\left(3+2 \sqrt{2}\right)^n}{2 \sqrt{n\pi\left(3 \sqrt{2}-4\right)  } }$$ is an asymptotics.

Let $n=10^k$ and computing the logarithms of these monstrous numbers
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
 1 & 15.91875383 & 15.90706013 \\
 2 & 173.4147070 & 173.4135332 \\
 3 & 1758.735871 & 1758.735754 \\
 4 & 17622.30914 & 17622.30913 \\
 5 & 176268.4035 & 176268.4035
\end{array}
\right)$$ However, even for small values of $n$, it does not seems too bad
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
 1 & 3 & 3 \\
 2 & 14 & 13 \\
 3 & 65 & 63 \\
 4 & 330 & 321 \\
 5 & 1723 & 1683 \\
 6 & 9165 & 8989 \\
 7 & 49457 & 48639 \\
 8 & 269637 & 265729 \\
 9 & 1481680 & 1462563 \\
 10 & 8192698 & 8097453
\end{array}
\right)$$

In fact, you could find here a much better approximation for the central Delannoy numbers

$$\frac{\left(3+2 \sqrt{2}\right)^n}{2 \sqrt{n\pi\left(3 \sqrt{2}-4\right)  } }\times $$
$$\Bigg[1-\frac{23}{19 \left(8+3 \sqrt{2}\right) n}+\frac{2401}{1024 \left(113+72
   \sqrt{2}\right) n^2}+O\left(\frac{1}{n^3}\right) \Bigg]$$

For $n=10$, this would give $8112585$

Edit

In fact, there several definitions and in particular $P_n(3)$ (Legendre polynomials). I suggest that you have look at @Zurab Silagadze's anwer to this question (make $x=\cosh ^{-1}(3)$).

Another definition of $D_n$ is
$$D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}=\, _2F_1(-n,n+1;1;-1)=P_n(3)$$

Have a look here for more information.


问:
Why $\omega\left[\begin{pmatrix} 0 &Q\\ R &0 \end{pmatrix}\right]\leq\omega\left[\begin{pmatrix} P &Q\\ R &S \end{pmatrix}\right]$?
Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$. The numerical radius of an operator $T\in\mathcal{L}(E)$ is given by
$$\omega(T)=\sup_{\|x\|=1}|\langle Tx, x\rangle|.$$

Let $P,Q,R,S\in\mathcal{L}(E)$. I want to prove that
$\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right]\leq\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right].$
Here $\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix},\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\in \mathcal{L}(E\oplus E)$.

My attempt: One can remark that
$$
 \begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
P &Q\\
R &S
\end{pmatrix} + \frac{1}{2} \begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix},
$$
This implies that
$$
\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right] \leq \frac{1}{2} \omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right] + \frac{1}{2} \omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].
$$

But I'm facing difficulties to prove that
$$\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right]=\omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].$$
Let $T=\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}$. I want to find an unitary operator $U$ such that
$$U^*TU=\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}.$$
In this case we get the desired result since
$$\omega(U^*TU)=\omega(T).$$

答:
Note that $\omega(T) = \omega(-T)$. Thus it suffices to find a unitary operator such that satisfies
$$U^*TU=\begin{pmatrix}
P & -Q\\
-R &S
\end{pmatrix}.$$
But for this we can take
$$U=\begin{pmatrix}
I & 0\\
0 & -I
\end{pmatrix}.$$


问:
Solving $\left(\sin\left(\sqrt{x}+5\right)-\cos\left(\sqrt{x}+5\right)\right)e^\sqrt{x}=0$
I tried to solve the following problem; but I couldn't.

$$\left(\;\sin\left(\sqrt{x}+5\right)-\cos\left(\sqrt{x}+5\right)\;\right)e^\sqrt{x}=0$$

I placed it into Microsoft Math Solver and the following was the answer:

$$x=\frac{(-4\pi n_1+20-9\pi)^2}{16}, n_1\in Z$$

How do I solve it to reach the correct answer?

My solution:

$$\sin(\sqrt{x}+5)=\cos(\sqrt{x}+5)$$

$$\sqrt{x}+5=\pi n_1+\frac{\pi}{4}$$

And after solving it I obtained:

$$x=\frac{(4\pi n_1-20+\pi)^2}{16}$$

My solution lacks the $9$ that there's in the correct answer.

Am I doing something wrong?

答:
We need

$$-4m\pi+20-9\pi=\pm(4\pi n-20+\pi)$$

Considering the '-' sign, $$-4m\pi+20-9\pi=-(4\pi n-20+\pi)$$

$$\iff-4m-9=-4n-1 \iff4n=4m+8\iff n=m+2$$

Clearly, we don't need to consider the $+$ sign (why?)


问:
Show series converges
Show that the series

$$
\sum_{n=1}^{\infty} \frac{n!^n}{n^{n^2}}$$ converges.

I tried the ratio test but didn't come to a conclusion.
Wolfram says the series converges by comparison but I don't know what series to compare to.

答:
If you don't want to use Stirling the simpler is to go for AM-GM inequality followed by rough majoration

$$(n!)^{\frac 1n}\le \frac 1n\times\frac{n(n+1)}2=\frac{n+1}2\le \frac{2n}3$$ for $n$ large enough.

The term of your series is then to be compared to $(\frac 23)^{n^2}$

Title: Show series converges


问:
Multinomial Distribution Sufficient Statistic
I've read that for distributions that belong to the exponential family, there is a specific solution for finding their sufficient statistic.

First, I show that the multinomial belongs to the exponential family.

$p(x|\pi)=\frac{N!}{x_{1}!....x_{k}!}\pi_{1}^{x_{1}}\pi_{2}^{x_{2}}...\pi_{k}^{x_{k}}=A(x)e^{\sum_{k=1}^{K}x_{k}log(\pi_{k})},$

where $A(x)=\frac{N!}{x_{1}!....x_{k}!}$, hence it belongs to the exponential family.

However, I have to use further calculations in order to brint the form of $p(x|\pi)$ as

$p(x|\pi)=A(x)e^{C(\pi)T(x)+B(\pi)},$

where $C(\pi)$ and $B(\pi)$ are functions of $\pi_{1},\pi_{2},...,\pi_{k}$ and $T(x)$ corresdpons to the sufficient statistic of the Multinomial.

Thus, I would like some help on how to make that further calculation, or if there is a different way that the sufficient statistic of a Multinomial can be calculated.

答:
We suppose that $x_i$ is a number of trials in which outcome number $i$ appeared, $1 \le i \le $.  Hence $x_1 + x_2 + \ldots + x_N = n$ where $n$ is the size of the sample $X$.

Statistic $T(X)$ is sufficient iff $p(X| \pi) = h(X) g_{\pi}(T(X))$. Thus $(x_1, \ldots, x_N)$ is sufficient. Moreover, as $x_N = n - x_1 - \ldots - x_{N-1}$ then

$$T = (x_1, x_2, \ldots, x_{N-1})$$
is sufficient statistics.


答:
You might not be able to find the desired $U$.  E.g. if $P=S=I$, and $Q=R=0$, then
$$
  \pmatrix{P & Q \cr R & S}
  $$
is not conjugate to
$$
  \pmatrix{-P & Q \cr R & -S},
  $$
since they have different eigenvalues.  However, letting
$$
  V=\pmatrix{-I & 0 \cr 0 & I},
  $$
one has
$$
  V\pmatrix{P & Q \cr R & S}V^{-I} =   \pmatrix{P & -Q \cr -R & S} = -\pmatrix{-P & Q \cr R & -S}
  $$
so you get what you want because $\omega (-T)=\omega (T)$.

Title: Why $\omega\left[\begin{pmatrix} 0 &Q\\ R &0 \end{pmatrix}\right]\leq\omega\left[\begin{pmatrix} P &Q\\ R &S \end{pmatrix}\right]$?

Body:
Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$. The numerical radius of an operator $T\in\mathcal{L}(E)$ is given by
$$\omega(T)=\sup_{\|x\|=1}|\langle Tx, x\rangle|.$$

Let $P,Q,R,S\in\mathcal{L}(E)$. I want to prove that
$\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right]\leq\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right].$
Here $\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix},\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\in \mathcal{L}(E\oplus E)$.

My attempt: One can remark that
$$
 \begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
P &Q\\
R &S
\end{pmatrix} + \frac{1}{2} \begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix},
$$
This implies that
$$
\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right] \leq \frac{1}{2} \omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right] + \frac{1}{2} \omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].
$$

But I'm facing difficulties to prove that
$$\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right]=\omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].$$
Let $T=\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}$. I want to find an unitary operator $U$ such that
$$U^*TU=\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}.$$
In this case we get the desired result since
$$\omega(U^*TU)=\omega(T).$$


问:
Compactness of a subset of $\ell^2$
Let $K \subset \ell^2(\mathbb{N})$ be a set defined as follows:
$$ K := \left\{x = (x_1, x_2, \dots) \in \ell^2(\mathbb{N}) \,|\, |x_n| \le \frac{1}{n}\right\}.$$
Since $\ell^2(\mathbb{N})$ is a reflexive space, we know that any sequence of $K$, being bounded, admits a converging subsequence by the Eberlein-Smulian theorem. Moreover, since $K$ is closed, the weakly converging subsequence has a limit in $K$, and thus $K$ is weakly sequentially compact.

Since $(\ell^2(\mathbb{N}))^*$ is separable, we know that $\overline{B(0,1)}$ is metrizable for the weak topology and hence $K$ is not only weakly sequentially compact but compact for the weak topology.

Is it compact for the strong topology as well?

答:
Yes it is.
It is a standard (I mean "very common") argument in functional analysis, you can pick a sequence of points $(x_n)_n \in K^{\mathbb{N}}$ and proceed a Cantor's diagonal argument to extract a convergence subsequence. Then, it is just a matter of applying Lebesgue Dominated Convergence theorem to conclude convergece in $||.||_{\ell_2}$ as well.

Title: Compactness of a subset of $\ell^2$


问:
Is it true that $f(x)=k ,k\in \mathbb{R}$ is a decreasing and increasing and constant function in the same time over it's domain of definition?
It is known that $f(x)=k,k\in \mathbb{R}$ is a constant function  ,one of my friend argued me that he has an example of functions which it is increasing and decreasing and constant in the same time over its domain of definition such that he gave me the following example $f(4)=5 $, I said him that is just a constant function and we can't say that is satisfies monotonicity namely increasing and decreasing and constant in the same time since that is needed  to be compared with other point , Any comment?

答:
It depends on the definitions actually. If for increasing and decreasing he means "weakly", then he is correct. Recall a function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called increasing(decreasing) if when $x<y$ then $f(x) \le f(y)$ ($f(x) \ge f(y)$). So a constant function is both increasing than decreasing. Of course it is not strictly increasing or strictly decreasing.

A same thing can happen with convexity. You may say that a linear function like $f(x)=x$ is both convex and concave, indeed it is linear.


问:
Trace of $p \wedge p$
Hello everyone here is a little problem I have some trouble with :

Let $V$ a vector space with $v_1,v_2,v_3$ as a basis and we define $p$ an endomorphism of $V$ with the following matrix :

$$A=\begin{pmatrix}
 a&b  &c \\
 d&e  &f \\
 g&h  &i
\end{pmatrix}$$

I want to prove that the trace of $p\wedge p$ is the sum of the three minors of order $2$ of $A$

A hint that a one of my friend gave me is to compute the values of the diagonal elements of the matrix $p\wedge p$ in the basis $v_1\wedge v_2,v_1\wedge v_3,v_2\wedge v_3$ of $V\wedge V$ but I don't know how to do if anyone can help compute one or two coefficients of this matrix so I can do it for the others it would be a lot appreciated , thanks in advance

答:
Let $\alpha_1,\alpha_2,\alpha_3$ denote the dual basis associated with $v_1,v_2,v_3$. Note that $\alpha_1 \wedge \alpha_2, \alpha_1 \wedge \alpha_3, \alpha_2 \wedge \alpha_3$ is the dual basis associated with $v_1 \wedge v_2, v_1 \wedge v_3, v_2 \wedge v_3$, where we have
$$
(\alpha \wedge \beta)(v \wedge w) = \det \pmatrix{\alpha(v) & \alpha(v)\\ \beta(v) & \beta(w)}.
$$
Note that for a space $V$ with basis $B = (v_1,\dots,v_n)$ and dual basis $\alpha_1,\dots,\alpha_n$, the $i$th diagonal element of the matrix of $\phi:V \to V$ with respect to $B$ is $\alpha_i(\phi(v_i))$. With that established, note that
$$
(\alpha_i \wedge \alpha_j)(p\wedge p)(v_i \wedge v_j) =
(\alpha_i \wedge \alpha_j)(pv_i \wedge pv_j) = \\
\det \pmatrix{\alpha_i \circ pv_i & \alpha_i \circ pv_j\\
\alpha_j \circ pv_i & \alpha_j \circ pv_j}.
$$

Proof of the determinant formula:

Let $m_{ij} = \alpha_i (pv_j)$. We have $pv_j = \sum_i m_{ij}v_i$. Thus, we have
$$
\begin{align}
(p \wedge p)(v_i \wedge v_j) &=
pv_i \wedge pv_j
\\ & =
\left(\sum_{k} m_{ki}v_k \right) \wedge
\left(\sum_{\ell} m_{\ell j}v_\ell \right)
\\ & =
\sum_{k,\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
\\ & =
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k>\ell} -m_{ki}m_{\ell j} (v_\ell \wedge v_k)
\\ & =
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k'<\ell'} -m_{\ell'i}m_{k' j} (v_{k'} \wedge v_{\ell'})
\\ & =
\sum_{k<\ell} (m_{ki}m_{\ell j} - m_{kj}m_{\ell i}) (v_k \wedge v_\ell)
=
\sum_{k<\ell} \det\pmatrix{m_{ki} & m_{\ell i}\\ m_{kj} & m_{\ell j}} \cdot (v_k \wedge v_\ell).
\end{align}
$$

Title: Trace of $p \wedge p$


问:
Solutions of $2^{1978}(1+2^{4a-2030})=y(y+1)$

I am given that $a,y$ positive integers and
$$2^{1978}(1+2^{4a-2030})=y(y+1).$$
I want to maximize $a$.

I guessed that $a$ is maximized when $2^{1978}=y$ and hence $4a-2030=1978\implies a =1002$. However, there is the other case of when a factor of $1+2^{4a-2030}$ is "contained" in $2^{1978}$, that is: $(2^{1978}x, \frac{1+2^{4a-2030}}{x}) = (y,y+1) \text{ or }=(y+1,y)$. I have proven for the $(y,y+1)$ case that $2^{4a-30} \ge 2^{1978}x^2$ which imply that there may be infinite solutions and so there is no maximum.

Body:
If we multiply given equation $$2^{1978}(1+2^{4a-2030})=y(y+1)$$

with $4$ and then add $1$ we get $$\boxed{2^{1980}+2^{4a-50}+1=(2y+1)^2}$$

Now suppose there exists $a>1002$, then $$1980<2a-24$$ so $$\color{red}{2^{1980}+2^{4a-50}+1} <(2^{2a-25}+1)^2$$
and since $$2^{4a-50}<\color{red}{2^{1980}+2^{4a-50}+1}$$

so $$2^{4a-50}<\color{red}{(2y+1)^2}<(2^{2a-25}+1)^2$$ which implies $$2^{2a-25}<2y+1<2^{2a-25}+1$$

and thus no solution. So $a\leq 1002$ and thus $a_{\max} = 1002$ (which clearly works).


问:
Find the form of $Y(t)$,$(D-2)^3(D+1)Y=3e^{2t}-te^{-t}$

$(D-2)^3(D+1)Y=3e^{2t}-te^{-t}$

(a) Show that $(D − 2)$ and $(D + 1)^2$
respectively are annihilators of the right side of the equation,
and that the combined operator $(D − 2)(D + 1)^2$ annihilates both terms on the right side of the
equation simultaneously.

(b) Find Y (t)

I've managed to solve (a) and I'm not sure how to proceed from here.

I think I need to use that $(D − 2)(D + 1)^2(3e^{2t}-te^{-t})\in Ker((D − 2)(D + 1)^2)$ and Y(t) will have the form of some quasi-polynomial?

答:
$\textbf{Hint}$: The "minimal" polynomial of annihilators is $(D-2)^4(D+1)^3Y = 0$ which means the solution will be of the form

$$Y(t) = Ae^{2t}+Bte^{2t}+Ct^2e^{2t}+Dt^3e^{2t}+Ee^{-t}+Fte^{-t}+Gt^2e^{-t}$$

This means the particular solution will require you to explicitly solve for $D,F,G$

$$Y_p(t) = Dt^3e^{2t}+Fte^{-t}+Gt^2e^{-t}$$

in the original equation. Can you take it from here?


问:
evolute of an astroid is another astroid
I want to prove that the evolute of the astroid, $(a\cos^3{t},a\sin^3{t})$, $t \in [0,2\pi)$, is another astroid turned by a $\frac{\pi}{4}$ angle.

I have tried a variable change from the parametric equation of $t=r-\frac{\pi}{4}$ and a $\frac{\pi}{4}$ angle turn but I can´t get an equation of the following form: $$(b\cos^3{r},b\sin^3{r})$$

答:
The rotated astroid, twice the given one, has parametric equations
$$\left\{\sqrt{2} a\cos ^3 t-\sqrt{2} a\sin ^3 t,\sqrt{2}a\sin ^3 t+\sqrt{2} a\cos ^3 t\right\}$$
Because the rotation matrix is
$$R(\pi/4)=\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$

In the picture below the graph for $a=1$


问:
how would $f(T)$ look like if ...
I know the result that if $T:V_F\to V_F$ is a linear operator then for any polynomial $f(x)\in F[x],~f(T)$ is a linear operator. Now my question is how would $f(T)$ look like if

$f(x)$ is the zero polynomial

$f(x)$ is a constant polynomial other than $0$

答:
Then $f(T)$ is the zero operator, i.e. it sends every vector to the zero vector.

If $f(x)$ is the constant polynomial $c$, then $f(T)$ is the operator $cI$, i.e. it sends a vector $\vec v$ to $c\vec v$.

Title: how would $f(T)$ look like if ...


问:
How to show mapping $w = \sin z$ is one to one?
If I consider the region in $xy$ plane
$$\frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \,\text{ and } \,y \geq 0$$

is mapped to upper half $v \geq 0$ of $w$ plane under $w = \sin z$. It can be easily seen that the line $x = c$ where $c$ is between $\frac{-\pi}{2} \leq c \leq \frac{\pi}{2}$ is mapped to a hyperbola and consequently it is easily observed that the above region is mapped to upper half plane in $uv$ plane. I wonder if this map is a one to one. What do you think? If yes how can I prove it ?

答:
$$w = \sin z = \frac{1}{2i} ( e^{iz} - e^{-iz})$$

$$2 i w e^{iz} = e^{2iz} -1$$

$$ e^{2iz} - 2iw e^{iz} -1 = 0$$

$$ e^{iz} = iw \mp \sqrt{1-w^2 }$$

$$z = -i \log \left[ iw \mp \sqrt{1-w^2} \right]$$

We need to see if both or just one of these lie in the half-strip ...

The function $w=\sin z$ is periodic with period $2\pi$.  Also, $$\sin (-z) = - \sin z$$ so we only need to show that the function is one-to-one in the half strip $0\le x\le \pi/2$ and $y\ge 0$.


问:
In search for a variety that would describe semigroups faithfully representable as right translations.
If $G$ is a group, we see that the bijection $y\mapsto y_r$, where $(x)a_r = xa$, functions written from the right, is a group isomorphism.

In this way, we have an interesting interpretation of groups as translations.

Inspired by this, let $S$ be any magma. Let's consider the same operation of a right translation on $S$. We can ask a question, when is this map $y\mapsto y_r$ a homomorphism.

Clearly, this is iff $(x)a_rb_r = (x)(ab)_r$ for any $x, a, b$, that is, $(xa)b = x(ab)$.

In this way, from a simple question about translations forming a homomorphism we naturally arrived at the definition of a semigroup.

We can ask ourselves another question, when is this map an isomorphism? This is an isomorphism iff $y\mapsto y_r$ is injective, and that is iff $(x)a_r = (x)b_r$, that is, $xa = xb$ implies $a = b$ for all $x, a, b$.

Such semigroups are called left weakly reductive, lwr semigroups for short.

However, there is a serious issue with this, lwr semigroups don't form a variety among structures with binary operation. Clearly, if $S$ is a semigroup which isn't a lwr semigroup, then $S^1$, where $S\mapsto S^1$ is the operation of adding an identity element, is a lwr semigroup. But then $S^1$ contains $S$ as its subsemigroup, so that a subsemigroup of a lwr semigroup doesn't have to be an lwr semigroup.

But there is a class of semigroups which does form a variety, maybe not as structures with binary operation, but with addition of a nullary element. Those are left monoids $(S, \cdot, e)$ where $\cdot$ is a binary operation, $e$ is nullary element, and there are two identities: $x(yz)\approx (xy)z$, $ex \approx x$.

As a class of semigroups which comes from forgetting the operation $e$ i. e. $(S, \cdot, e)\mapsto (S, \cdot)$, all left monoids are lwr semigroups.

My question is, if we could find some structure $(S, \cdot, \mathscr{F})$ where $\mathscr{F}$ denotes the set of all $n$-ary functions for $n\in\mathbb{N}_0$ other than $\cdot$ , the binary function, and identities, so that a set of algebras satisfying those identities forms a variety, and the map $(S, \cdot, \mathscr{F})\mapsto (S, \cdot)$ for any algebra from those variety turns it into an lwr semigroup, and so that the class of sets obtained this way is maximal, or if it's impossible.

For starters, if a variety which would cover a class of semigroups between left monoids and lwr semigroups.

Update:

I've been able to find a potential candidate for such structure. Consider $(S, \cdot, e)$ where $e:S\to S$ is an unary operation, with identities $e(x)x \approx x$ and $e(x)e(y)e(x)y \approx y$.

Such structures generalize monoids since if $f$ is the left identity, then $e(x) \equiv f$ gives us a structure of above type. Moreover, any semigroup which satisfies those relations is a lwr semigroup, since $xa = xb$ implies $a = e(a)b$ and $b = e(b)a$ so that $a = e(a)e(b)e(a)b = b$.

However, I don't have an example which isn't a left monoid.

Update 2:

I found that $e(x)x\approx x$, $e(x)e(y)e(x)y \approx y$ are equivalent to $e(x)^2 y \approx y$, $e(y)e(x)y \approx e(x)y$.

In particular, we can fix $x\in S$ and take $ e = e(x)^2$, so that $ey = y$ for all $y\in S$. This means $S$ is a left monoid.

Update 3:

Small lwr semigroups which aren't left monoids:

There is a unique lwr semigroup of order $3$ which isn't a left monoid. It's given by the matrix \begin{bmatrix} 1 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 3 \end{bmatrix}

By my calculations, there is $18$ semigroups of order $4$ which are lwr but not left monoids, with Id's $54, 67, 69, 69^t, 70, 70^t, 77^t, 88, 92, 98, 99^t, 100, 101^t, 102, 102^t, 103^t, 110^t, 112$ in GAP package Smallsemi, where $t$ means the multiplication table transposed (anti-isomorphism).

New structure:

Let's consider $(S, \cdot, e)$ where $e$ is a binary operation $e:S\times S\to S$, with identities $e(x, y) \approx e(y, x)$ and $e(x, y)x\approx x$.

You can think the binary operation as a choice of local left identities, which bind two elements $x, y$ together by demanding $e(x, y)$ to be both local left identity of $x$ and $y$.

If $S$ is a left monoid with left identity $f$, we can take $e(x, y) \equiv f$.

Any such structure is a lwr semigroup because $xa = xb$ for all $x$ implies $e(a, b)a = e(a, b)b$, and that means $a = b$.

Such structures are discussed here and here.

Update 4:

There seems to be another class of semigroups which are weakly reductive! Those are semigroups induced by (small) categories. The construction can be found here in the answer of @J.-E.Pin.

If $(S, \cdot)$ is induced by category $\mathcal{C}$, and $x\cdot f = x\cdot g$ for all $x\in S$ and morphisms $f, g$ of $\mathcal{C}$, then taking $x = 1_X$ where $X$ is the domain of $f$, $f = 1_X\cdot g$. In particular, $1_X\cdot g = 1_X\circ g$ is defined so equal to $g$. If $x\cdot f = x\cdot 0 = 0$ for all $x\in S$ and morphism $f$ of $\mathcal{C}$, then as before $f = 0$. Contradiction.

This proves all semigroups induced by (small) categories are weakly reductive.

答:
You can describe the class of all lwrs this way, using one extra binary and one ternary operation. Denoting the two operations by $w(a,b)$ and $r(a,b,c)$, consider the variety defined by associativity of $\cdot$ and identities
$$r(a,b,w(a,b)\cdot a)=a,\\
r(a,b,w(a,b)\cdot b)=b.$$
Note that if those identities hold, then for any $a\neq b$ we must have $w(a,b)\cdot a\neq w(a,b)\cdot b$, which implies the semigroup $(S,\cdot)$ is lwr for any such algebra $(S,\cdot,w,r)$.

Conversely, if the semigroup $(S,\cdot)$ is lwr, then for any $a\neq b$ there is some $w(a,b)$ such that $w(a,b)\cdot a\neq w(a,b)\cdot b$. Use those values to define $w$, setting $w(a,a)$ arbitrarily. We can now define $r(a,b,c)$ in such a way that it equals $a$ if $c=w(a,b)\cdot a$, $b$ if $c=w(a,b)\cdot b$ and takes any value for all other $c$. Then $(S,\cdot,w,r)$ satisfies the identities above.


问:
Ratio of Dirac delta functions
My Professor tells me that

$\frac{\delta(x-a)}{\delta(x-a)} = 1$

and justifies it by considering the Dirac delta as a limit of normalized Gaussians whose width approaches zero. I don't quite buy this explanation because I've been told that this definition is only an approximation of the Dirac delta.

Is this proof fine? Or is there a more rigorous proof of this? Or is this not even true?

Thanks.

答:
The fact that you're calling these "delta functions" in the title seems to indicate that you're dealing with these things in a setting where rigour isn't emphasized. In physics, delta distributions are often treated like ordinary functions, more often than not with sensible results.

However, your question seems to be looking for more rigour. A rigorous treatment of what you call "delta functions" leads to distribution theory. Distributions can be regarded as generalizations of ordinary functions, but they don't share all the properties of ordinary functions.

In particular, as user8268 pointed out in the comments, multiplication of distributions works differently. The product of two general distributions (e.g. the product of two delta distributions) cannot be defined (at least not in a straightforward sense; see here). The product of a distribution with an ordinary function can be defined, but as user8268 pointed out, this is not invertible. Division is the inverse of multiplication, so if multiplication is not invertible, division makes no sense. If you applied your professor's division rule to user8268's example, you could cancel the delta distributions and get $1=1+x-a$, which is obviously wrong.

I'd suggest that you look at what your professor was trying to use this relationship for and see whether it can be justified rigorously without resorting to division of distributions. For instance, if you have two charge distributions and you know that they're both localized on a sphere of radius $R$, you can write them as $\rho_i(\theta,\phi)\delta(r-R)$; then if by some argument you find that they're the same, $\rho_1(\theta,\phi)\delta(r-R)=\rho_2(\theta,\phi)\delta(r-R)$, you can conclude that $\rho_1(\theta,\phi)=\rho_2(\theta,\phi)$ by other means than "dividing through" by the delta distribution.

Title: Ratio of Dirac delta functions


问:
Does category theory provide a rigorous definition of the cartesian product?
Let $X$ and $Y$ be sets. As far as I know, the defining property of the cartesian product
\begin{equation}
X\times Y=\left\{(x,y):x\in X\text{ and }y\in Y\right\}
\end{equation}
is the "universal property"
\begin{equation}
(x,y)=(x',y')\Leftrightarrow x=x'\text{ and }y=y'
\end{equation}
for all $x,x'\in X$ and $y,y'\in Y$. It is known that the definition $(x,y):=\{\{x\},\{x,y\}\}$ has the above property. But I guess this is only one of many possible definitions.

I have no background in category theory, but I got the impression that it is always needed to get the full picture when dealing with something that has a universal property (e.g. the tensor product), that's why I thought that it might help me get a more precise definition of the cartesian product.

If your answer contains terms specific to category theory, it would be much appreciated if you also gave some reference, so that I have a starting point for a rigorous introduction to the topic.

答:
The categorical version looks like this.  Let two sets $X,Y$ be given. A product of $X$ and $Y$ is  (1) a set $P$, (2) a map $\pi_X : P \to X$, (3) a map $\pi_Y : P\to Y$, such that:  For every set $Z$, and every pair of maps $f : Z \to X$, $g : Z \to Y$, there is a unique map $h : Z \to P$ such that $ f = \pi_X \circ h$ and $g = \pi_Y \circ h$.

One then proves the idea with ordered pairs does define such a product.  I that case $\pi_X(x,y) = x$, $\pi_Y(x,y) = y$, $h(x,y) = (f(x),g(x))$.


问:
Why are determinants computed the way they are?
I understand the formula on how to compute determinants and I understand how we get to the $(a\times d)-(c\times b)$.

My question is why do we multiply the way we multiply? Is there any reason or proof?

答:
By Gaussian elimination,

$$\begin{cases}ax+by=e,\\cx+dy=f\end{cases}$$

then

$$\begin{cases}ax+by=e,\\0x+\left(d-b\dfrac ca\right)y=f-e\dfrac ca\end{cases}.$$

The second equation can also be written

$$(ad-bc)y=(af-ec).$$

Does this ring a bell ?

More deeply, the expression of a determinant is the simplest algebraic expression that cancels out when the rows or columns of the matrix are linearly dependent (it is a multilinear antisymmetric polynomial), and it is no surprise that it appears in the resolution of linear systems of equations.


问:
Show that the isometries $\phi: M\rightarrow M$ of a surface $M$ form a group

Show that the isometries $\phi: M\rightarrow M$ of a surface $M$ form a group

Wikipedia states:

The set of bijective isometries from a metric space to itself forms a group with respect to function composition, called the isometry group.

I'm not sure how to show that.

答:
The isometry group of a metric space is the set of all bijective(thus have inverses) isometries from the metric space onto itself, with the function composition(associativity) as group operation. Its identity element is the identity function.

So we can see that the group laws are all satisfied.


问:
Proofs using modular arithmetic.
In my textbook, it says,

The sum of the cubes of any three consecutive integers is divisible by
$9$

This statement is equivalent to proving,

$(n^3 + (n+1)^3 + (n+2)^3 ) \mod 9 =0$  is true for $n=0,1, \ldots, 8$

Now I understand how the modulo function works but I just don't see how the latter part is the same as the former. Can anyone walk me through it?

Note that, I'm not really looking for the proof, what I am looking for is how the first and the second statement are related.

Follow up question which begs to be asked,

Given some expression of $a \in \mathbb{Z}$, say $P(a)$, are the following 2 statements equivalent?

$n$ divides $P(a), \, \forall a \in \mathbb{Z}$ and for some fixed $n \in \mathbb{N}$

$P(a) \mod n = 0, \forall a \in \{0,1, \ldots, n-1\}$

Can anyone explain me whether or not the two statements are equivalent? Thanks!

答:
Given some expression of $a \in \mathbb{Z}$, say $P(a)$, are the following 2 statements equivalent?

$n$ divides $P(a), \, \forall a \in \mathbb{Z}$ and for some fixed $n \in \mathbb{N}$

$P(a) \bmod n = 0, \forall a \in \{0,1, \ldots, n-1\}$

Yes, let's derive the equivalence step-by-step.

$$\begin{align}
\forall\, a\,\in\, \Bbb Z:&\ \ \ n\,\mid\, P(a)\\[.2em]
\iff \forall\, a\,\in\, \Bbb Z:&\ \ \ 0 = P(a)\bmod n\\[.2em]
\iff \forall\, a\,\in\, \Bbb Z:&\ \ \ 0 = P(\bar a)\bmod n,\,\ \bar a = a\bmod n, \ \rm by\ \color{#C00}{PCR}\\[.2em]
\iff \forall\, \bar a\in[n]\!:&\ \ \ 0 = P(\bar a)\bmod n,\,\ [n] = \{0,1,\ldots,n\!-\!1\}\\[.2em]
\iff \forall\, \bar a\in\Bbb Z_n\!:&\ \ \ 0 = P(\bar a)
\end{align}\qquad$$

where we employed the following Polynomial Congruence Rule $\, =\rm\color{#C00}{PCR}$

$\quad \bmod n\!:\ a\equiv \bar a\,\Rightarrow\, P(a)\equiv P(\bar a),\ $ for $\,P(x)\,$ any polynomial with integer coef's

Generally, given any expression composed of sums and products of integers, its value  $\bmod n\,$ is unchanged when we replace the arguments of the sums and products by congruent arguments.  In particular this is true if we replace them by their least nonnegative reps $\, a\mapsto a\bmod n$.

Above we could replace the standard residue system $\,[n] = 0,1,2,\ldots\,n\!-\!1\,$ by any complete system of residues $\cal R,\,$ i.e. any set of integers such that every integer is congruent to exactly one $\,r\in \cal R.\,$ Then verifying that every $\,r\in\cal R\,$ is a root is sufficient to guarantee that every integer is a root, since every integer is congruent to some normal rep $\,r\in\cal R,\,$ i.e. we have $\,n\equiv r\in {\cal R}\,\Rightarrow\, P(n)\equiv P(r)\equiv 0,\,$ by $\,\rm\color{#c00}{PCR}$

A "normal" rep in $\cal R$ can be view as "name' or "label" for the entire set of integers equivalent to it, i.e. $\,r\mapsto [r]_n = r + n\Bbb Z,\,$ the set of all integers $\equiv r\pmod{\!n}.\,$ Then the congruence sum and product rules imply that addition and multiplication is well-defined on these classes by the rules $[a]+[b] = [a+b],\,$ $[a]*[b] = [a*b].\,$ In turn this implies that the arithmetical (ring) laws of $\,\Bbb Z\,$ are inherited by the classes with these operations, so they form a new system of numbers (ring) $\,\Bbb Z_n\,$ sharing many of the same properties of the ring of integers - so we can reuse our well-honed integer arithmetic intuition in this new modular number system. In particular, viewed in the ring $\,\Bbb Z_n\,$ your statement simplifies to the final form in the string of equivalences, viz. $\,\forall\,\bar a \in \Bbb Z_n\!:\ P(\bar a) = 0,\,$ i.e. every element of $\,\Bbb Z_n\,$ is a root of $\,P\,$ (here $P$ is viewed as a polynomial over $\,\Bbb Z_n\,$ by mapping its integer coef's into $\,\Bbb Z_n\,$ via $\,p_i\mapsto \bar p_i = [p_i]_n)$.

Title: Proofs using modular arithmetic.


问:
Finding the inverse of $a^9$ in a group $G$ when given the order of $a$ is $18$.
The order of $a$ is $18$, and I'm trying to find the inverse of $a^9$.

I know that $o(a^9)=\frac{o(a)}{\gcd(o(a),9)}\frac{18}{\gcd(18,9)}=2$.

I have also seen that $o(a)=o(a^{-1})$ from the properties of element order in a group.

I have tried to find which k satisfies $o(a^k)=\frac{18}{\gcd(o(a),k)}=2$ and I have seen that the order will be $2$ is when $9*k$ and $k$ is coprime to $18$.

Do that mean that the smallest $k\neq9$ that satisfies $o(a^k)=2$ will be the inverse of $a^9$ which is $a^{27}$ ?

答:
Hint: For any element $g$ of any group $G$ with identity $e$, we have $$g^2=e\iff g=g^{-1}.$$

Proof: ($\Rightarrow$) Multiply on the left, say, by $g^{-1}$. ($\Leftarrow$) Multiply on the right, say, by $g$. $\square$

 But $e=a^{18}=(a^9)^2$, so $(a^9)^{-1}=a^9$.


问:
A field $\mathbb K$ is commutative iff for every matrix on $\mathbb K $ the rank of its column vector is equal to the one of its transpose
I have a little tricky question if anyone could help me see through it , it would be really appreciated :

We have $\mathbb K$ a field such that : $\alpha a \neq a\alpha$ I don't see why the rank of the column matrix :
$$\begin{pmatrix}
 a&\alpha a \\
 1&\alpha
\end{pmatrix}$$

is $=2$ , and that the rank of the column matrix (#)

$$\begin{pmatrix}
 a& 1 \\
 \alpha a&\alpha
\end{pmatrix}$$

is $=1$

(PS: How can we deduce from this that a field $\mathbb K$ is commutative iff for every matrix on $\mathbb K $ the rank of its column vector is equal to the one of its transpose)

I know that the column vectors are a vector space with the external multiplication ( right side  ) and to prove (#) I tried to write the second column as the first times a factor scalar $x$ I don't know what I should do next, maybe eliminate $x$ from the equations but isn't a contradiction? Any help would be welcomed thanks in advance.

答:
I assume that $K^2$ is viewed as a right-vector space over $K$.

Then, in the second matrix the second row is the first one times$ a$, so the rank is one (since the matrix is nonzero, it cannot be zero).

To prove that the rank of the first one is $2$, we need to prove that the columns are linearly independent. It is equivalent to show that the kernel of the matrix is zero (note that multiplying a matrix by a vector yields a right linear combination of the columns and vice versa).

So we have to solve $ ax+ \alpha  ay=0$ and $x+\alpha y=0$. Hence $x=-\alpha y$ and
$-a \alpha y + \alpha a y=(-a\alpha +\alpha a) y=0$. By assumption on $a$ ana $\alpha$, $-a\alpha +\alpha a $ is a nonzero element of $K$, hence invertible. Consequently, $y=0$ and $x=0$. QED

This proves the desired equivalence: if $K$ is commutative, then it is known that a matrix and its transpose have same rank. If $K$ is not commutative, then by definition yo ucan find $a,\alpha\in $ such that $a\alpha\neq \alpha a$. The matrix of your post is then an example of a matrix which has not the same rank as its transpose.


问:
If $M_t$ is a martingale, when $\frac{1}{M_t}$ be a sub-martingale?
I am wondering what are constrains for using the argument that "martingale + convex function -> sub-martingale".

The problem I have is:
if $M_t$ is within $]0, +\infty[$, then whether $(\frac{1}{M_t}, t\geq 0)$ a sub-martingale ?

Generally, let $f$ a convex function, then by Jensen's inequality, we have:
$$\forall t\geq 0, \mathbb{E}[f(M_t)] \geq f(\mathbb{E}[M_t])$$

Here for this problem, $f$ will be $f(x) = 1/x$ and since $M_t$ is positive, we have the convexity condition satisfied. And since $M_t$ cannot be $0$ or $+\infty$, we have $f(\mathbb{E}[M_t])$ well defined.

So am I right to write:
$$\mathbb{E}[\frac{1}{M_t}|\mathcal{F_s}] \geq \frac{1}{\mathbb{E}[M_t|\mathcal{F}_s]} = \frac{1}{M_s}, \quad 0\leq s \leq t$$

so that $(\frac{1}{M_t}, t\geq 0)$ is a sub-martingale ? Or I have overmitted some important points ?

Besides, If we discard the constrain that  $M_t$ is within $]0, +\infty[$, and take brownian motion $(B_t, t\geq 0)$ as an example, can I say that   $(\frac{1}{B_t}, t\geq 0)$ a sub-martingale ? If not, what's the nature of $(\frac{1}{B_t}, t\geq 0)$ ?

Thanks !

答: If you can live with the fact that $M^{-1}_t$ may not be integrable, so that $\Bbb E[M^{-1}_t\mid\mathcal F_s]$ has to be understood in a generalized sense, then Jensen can be applied and $M^{-1}_t$ is a  "submartingale".


问:
Laplace transform of delays Sinc function
I am trying to compute the Laplace transform of the following Sinc function,

$$\operatorname{sinc}(a(t-t_0)) = \frac{\sin(a(t-t_0))}{a(t-t_0)} $$

Where, $t_0$ is the delay and $a$ is constant value.

I know the Laplace transform for $\frac{\sin(at)}{at}$ without delay is $\frac{1}{a}\left(\frac{\pi}{2} - \operatorname{atan}\left(\frac{s}{a}\right)\right)$. I am concerned whether we can use the phase shift property (i.e. multiply by $\exp(-st_0)$) or not, as the function has non-zero values on the negative axis as well?

答:
Well, we have the following Laplace transform:

$$\mathscr{L}_x\left[\text{sinc}\left(\alpha\left(x-x_0\right)\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{sinc}\left(\alpha\left(x-x_0\right)\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$

It is not hard to show that (using Taylor series) $\forall\text{n}$:

$$\text{sinc}\left(\text{n}\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\text{n}^{2\text{k}}}{\left(2\text{k}+1\right)!}\tag2$$

So, using that we can rewrite $(1)$ as follows:

$$\mathscr{L}_x\left[\text{sinc}\left(\alpha\left(x-x_0\right)\right)\right]_{\left(\text{s}\right)}=\int_0^\infty\left\{\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(\alpha\left(x-x_0\right)\right)^{2\text{k}}}{\left(2\text{k}+1\right)!}\right\}\cdot\exp\left(-\text{s}x\right)\space\text{d}x=$$
$$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\alpha^{2\text{k}}}{\left(2\text{k}+1\right)!}\cdot\left\{\int_0^\infty\left(x-x_0\right)^{2\text{k}}\exp\left(-\text{s}x\right)\space\text{d}x\right\}\tag3$$

Now, let's take a look at a slightly different integral:

$$\mathcal{I}_\beta:=\int_0^\beta\left(x-x_0\right)^{2\text{k}}\exp\left(-\text{s}x\right)\space\text{d}x\tag4$$

First, let $\text{u}:=\left(x-x_0\right)^{2\text{k}+1}$, this leads to:

$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{2\text{k}+1}\int_{\left(-x_0\right)^{2\text{k}+1}}^{\left(\beta-x_0\right)^{2\text{k}+1}}\exp\left(-\text{s}\cdot\text{u}^\frac{1}{2\text{k}+1}\right)\space\text{du}\tag5$$

Now, let $\text{v}:=\text{us}^{2\text{k}+1}$, this leads to:

$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\int_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}\exp\left(-\text{v}^\frac{1}{2\text{k}+1}\right)\space\text{dv}\tag6$$

And this is a special function, called the Incomplete Gamma function:

$$\mathcal{I}_\beta=\frac{\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\cdot\left[-\left(2\text{k}+1\right)\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}=$$
$$\frac{-\left(2\text{k}+1\right)\exp\left(-\text{s}x_0\right)}{\left(2\text{k}+1\right)\text{s}^{2\text{k}+1}}\cdot\left[\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(-x_0\text{s}\right)^{2\text{k}+1}}^{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}=$$
$$\frac{\exp\left(-\text{s}x_0\right)}{\text{s}^{2\text{k}+1}}\cdot\left[\Gamma\left(2\text{k}+1,\text{v}^\frac{1}{2\text{k}+1}\right)\right]_{\left(\left(\beta-x_0\right)\text{s}\right)^{2\text{k}+1}}^{\left(-x_0\text{s}\right)^{2\text{k}+1}}\tag7$$


问:
Prove that {A ⇒ ¬C, ¬A ⇒ B} ⊢ C ⇒ B using only Modus Ponens, the typical theorem (A → ¬C) → (C → ¬A) and 3 axioms.
I have an exercise where I have to prove the given sentence
{A ⇒ ¬C, ¬A ⇒ B} ⊢ C ⇒ B using only Modus Ponens, the typical theorem (A → ¬C) → (C → ¬A)
and the following three axioms:

A→(B→A)

(A→(B→C))→((A→B)→(A→C))

(¬A→¬B)→((¬A→B)→A)

what I've done so far is

A ⇒ ¬C   Premise

(A → ¬C) → (C → ¬A)   from the Typical Theorem

(C → ¬A)  M.P 1,2

¬A ⇒ B  Premise
...

However I'm getting a bit confused on what substitutions to do given the 3 axioms in order proceed.

Any ideas anyone?
Thank you!

答:
You need an intermediate result (sometimes called Hypothetical Syllogism):

Lemma : $A \rightarrow B, B \rightarrow C \vdash A \rightarrow C$

(1) $A \rightarrow B$ --- assumed

(2) $B \rightarrow C$ --- assumed

(3) $\vdash (B \rightarrow C) \rightarrow (A \rightarrow (B \rightarrow C))$ --- Ax1

(4) $A \rightarrow (B \rightarrow C)$ --- from (2) and (3) by Modus Ponens

(5) $\vdash [A \rightarrow (B \rightarrow C)] \rightarrow [(A \rightarrow B) \rightarrow (A \rightarrow C)]$ --- Ax2

(6) $(A \rightarrow B) \rightarrow (A \rightarrow C)$ --- from (4) and (5) by Modus Ponens

(7) $A \rightarrow C$ --- from (1) and (6) by MP.

Now the main result:

(1) $A \to ¬C$ --- premise

(2) $¬A \to B$ --- premise

(3) $C \to \lnot A$ --- from (1) and the Typical Theorem by MP

(4) $C \to B$ --- from (3), (1) and Lemma.


问:
Elipse to a spheroid?
I'm having a problem trying to prove that an elipse, given by
$$ \xi = \left\{(x,y,z)\in \mathbb{R}^3 : y = 0, \frac{x^2}{a^2}+\frac{z^2}{c^2} = 1\right\}  $$
if we rotate to the z-axis is a spheroid given by
$$M = \left\{(x,y,z)\in \mathbb{R}^3 :\frac{x^2}{a^2} + \frac{y^2}{a^2}+\frac{z^2}{c^2} = 1 \right\}$$

答:
You can rename the initial $x$-axis as $r$. (Because now it is something like the (perpendicular) distance from the $z$-axis) Then you can use polar coordinates for the $xy$-plane.

$$ x = r \cos \theta \\
y = r \sin \theta$$

and perhaps a trigonometric identity to get the final answer that you seek.

Title: Elipse to a spheroid?


问:
Simplifying fractions with fractions
I am trying to simplify

$$\frac{\frac{y}{x} - \frac{x}{y}}{\frac{1}{y} - \frac{1}{x}}$$

I make the top part into

$$\frac{y^2 - x ^2}{xy}$$

I know the bottom can be rewritten to just be multiplied into the top as its inverse.

$$\frac{y^2 - x ^2}{xy} * (y - x)$$

$$\frac{y - x }{xy} $$

This of course is wrong, but I do not know why, it appears to be correct to me. Nothing I did is mathematically wrong, it follows all the rules. What went wrong?

Title: Simplifying fractions with fractions

Body: $$\frac{\frac{y^2-x^2}{xy}}{\frac{x-y}{xy}}=\frac{y^2-x^2}{x-y}=\frac{(y-x)(y+x)}{-(y-x)}=\frac{y+x}{-1}=-y-x$$


问:
Proving that $\forall n\in \mathbb N, 2^n |(2n)!$
I'm having some trouble proving the following statement:

$$\forall n\in \mathbb N, \ \ 2^n |(2n)!$$

The first thing that came to mind was mathematical induction but this is my first time using it with this kind of the question (with divisibility instead of an equality/inequality) and I'm not quite sure how to proceed. How can I prove this? Should I use the properties of the divisibility of two numbers and manipulate $2^n |(2n)!$ until I get $2^{n+1} |(2(n+1))!$ or should I use the fact that $\exists k \in \mathbb Z: (2n)! = k2^n $ and try to conclude that $\exists l\in \mathbb Z:\ (2(n+1)) = l 2^{n+1}!$ ?

答:
Induction works best.

For $n=1$, $2^1 =2$ divides $2!=2$.

Suppose $2^n$ divides $(2n)!$.
Then consider $2^{n+1}=2\cdot 2^n$. By induction hypthesis, $2^n$ divides $(2n)!$ and
$(2(n+1))! =(2n)! \cdot [((2n)!+1)\cdot ((2n)^!+2)\cdots (2(n+1))!]$. The latter partial product $[\ldots]$ contains obviously a factor which is divisible by $2$. I.e., $2^{n+1}$ divides $(2(n+1))!$ as claimed.


问:
Assumption in prooving the Inverse Function Theorem (in Spivak's "Calculus on manifolds")
My question follows up with an additional remark from Spivak's proof of Inverse Function Theorem. The problem I have is the statement which immediately follows the If the theorem is true for $λ^{−1}∘f$ , it is clearly true for f... statement (from the link I've posted), in which Spivak assumes "at the outset" that $λ$ is the identity function, i.e. $λ=I$, while $λ$ was clearly defined as $λ=Df(a)$.

How can he even assume this without loss of generality? He's basically limiting himself to functions $f$ such that $Df(a)=I$. Did I get this wrong?

答:
If $\mathcal J Df(a)\neq 0$ then the linear transformation $Df(a):=\lambda:\mathbb R^n\to \mathbb R^n$ is invertible in some neighborhood $U\ni a$. Note that $D\lambda(x)=\lambda$ since $\lambda$ is a linear transformation. The same is true of course, for $\lambda^{-1}.$

Now consider $g:=\lambda^{-1}\circ f.$ We have then by the chain rule,

$Dg(a)=D\lambda^{-1}(f(a))\circ Df(a)=\lambda^{-1}\circ Df(a)=I.$

If the theorem is true for $g$ then $g$ is invertible (in some neighborhood of $a$) and so $f$ is also invertible. Indeed, $g^{-1}=f^{-1}\circ\lambda\Rightarrow f^{-1}=g^{-1}\circ\lambda^{-1}.$

So we may as well assume that $Df(a)=I$ in the first place.


问:
What will happen if we change the order of variable in the double integral?
$$\int_{0}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)\,dx\,dy$$

If we change the order of variable $(x,y)$ to $(y,x)$ in the double integral, how will it behave? Will limits be changed due to changing the variable?

答:
Under the Fubini-Tonelli assumptions,

$$\int_0^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y)\,dxdy=\int_0^1\int_{-1}^11_A(x,y)f(x,y)\,dxdy=\int_{-1}^1\int_{0}^11_A(x,y)f(x,y)\,dydx$$

Where $A=\left\{(x,y)\in\Bbb [0,1]\times[-1,1]\,:\, -\sqrt{1-y^2}\le x\le\sqrt{1-y^2}\right\}$. Notice that $$\begin{cases}0\le y\le 1\\ -1\le x\le 1\\ -\sqrt{1-y^2}\le x\le\sqrt{1-y^2}\end{cases}\Leftrightarrow\begin{cases}0\le y\le \sqrt{1-x^2}\\ -1\le x\le 1\end{cases}$$

and therefore $A=\{(x,y)\in[0,1]\times[-1,1]\,:\, 0\le y\le\sqrt{1-x^2}\}$. Therefore, since $0\le \sqrt{1-x^2}\le 1$, $$\int_{-1}^1\int_0^11_A(x,y)f(x,y)\,dydx=\int_{-1}^1\int_0^{\sqrt{1-x^2}} f(x,y)\,dydx$$


问:
Cardinality of Real numbers in arbitrary small interval.
I have been learning about the cardinality of Real numbers and it got me thinking. Since the cardinality of real numbers is uncountable and is the same as the cardinality of real numbers in the interval $(0,1)$ or for that matter any given interval.

So my question is how small this interval can get and still retain the same cardinality. Does $[0, \epsilon)$ have some cardinality as real numbers as epsilon gets smaller and smaller. It is trivial to see that in the limiting case, the set contains only one element.

So the question is how small can epsilon get before the cardinality of the set $[0, \epsilon)$ gets smaller than that of Real numbers.

答:
So the question is how small can epsilon get before

the cardinality of the set $[0, \epsilon)$ gets smaller than that of real numbers.

If $\epsilon \gt 0$, it can never get smaller than the cardinality of the real numbers.

That means if $\epsilon \gt 0$ the interval $[0,\epsilon)$ has the same cardinality as $[0,1)$, or as the real numbers.


问:
How to find this probability with bags of balls?
I have two bags with different color of balls in it. In bag 1 there are 1 white ball and 3 black balls and in bag 2 there are 1 white ball and 2 black balls. I wanna know what is the probability for me getting one black ball and one white ball if I choose one of the bag randomly and grab two balls out from it at once. Hence, what would be the probability for me taking out a black ball and a white ball from the bag 2 instead of bag 1?

my attempt: using binomial to test out the chance but I have gotten some really complicated answer. Would anyone be able to help me with it? Any help would be appreciated

答:
$$P(\text{one black, one white})=P(\text{one black, one white}|\text{Bag 1})\cdot P(\text{Bag 1}) + P(\text{one black, one white}|\text{Bag 2})\cdot P(\text{Bag 2}).$$

$$ P(\text{one black, one white})= \frac{\pmatrix{3\\1} }{\pmatrix{ 4\\2}}\cdot \frac{1}{2} + \frac{\pmatrix{2\\1} }{\pmatrix{ 3\\2}}\cdot \frac{1}{2}$$

$$ P(\text{one black, one white})=\frac{1}{4} + \frac{1}{3}=\frac{7}{12}.$$

$$ P(\text{one black, one white from Bag 2})=\frac{\pmatrix{2\\1} }{\pmatrix{ 3\\2}}\cdot \frac{1}{2}=\frac{1}{3}.$$


问:
High school contest question
Some work on it reveals the possibility of using gamma function. Is there any easy way to compute it?
$$\lim_{n\to\infty}\left(\frac{1}{n!} \int_0^e \log^n x \ dx\right)^n$$</p>

答:
I'm not totally sure of this, but Mathematica seemed to agree. Via integration by parts, $u=\log^{n}(x)\implies du = (n/x)\log^{n-1}(x)$ and $dv = dx\implies v = x$. Therefore,:
$$
\int_{0}^{e}\log^{n}(x)dx = e - n\int_{0}^{e}\log^{n-1}(x)
$$

By defining $a_{n} = \int_{0}^{e}\log^{n}(x)dx$, we see that $a_{n} = e - n a_{n-1}$. If $n$ is very large, the term iwth $e$ becomes unimportant, and effectively $a_{n} = -na_{n-1}$, which means that $a_{n} \sim (-1)^n n!$. Subbing this in, the limit $L$ becomes
$$
L\sim\lim_{n\to\infty}\left(\frac{(-1)^n n!}{n!}\right)^n = \lim_{n\to\infty}\left((-1)^n\right)^n = \lim_{n\to\infty}(-1)^{n^2}
$$

Which oscillates between $-1$ and $1$ for odd and even $n$, so the limit does not exist.

Title: High school contest question


问:
Identifying all but one face of $[0,1]^n$
Suppose $D=[0,1]^n, n \ge 2$. Then the quotient space $\frac{D}{\delta(D)\setminus [0,1]^{n-1}\times\{0\}}$ i.e. all but the bottom faces.

I could prove when $n=2$. In this case, we identify the three sides of the unit square. First, we take
$f: [0,1]^2\rightarrow S^1\times[0,1]$ by $f(x,y)=(e^x,y)$ If and after quotienting, we have $\frac{[0,1]^2}{\{(0,y),(1,y):y\in[0,1]\}}$ is homeo to $S^1\times [0,1]$. Further quotienting we get that the space we are looking for is homeo to reduced cone of $S^1$ which is actually homeo to the disc.

But how can we generalize it further? Are there any other ways to prove the same? I can easily visualize up to n=3 what might happen but can't produce an explicit map as in the case of n=2.

答:
The resulting space will always be homeomorphic to the $n$-dimensional disk $D = [0,1]^n$. I guess writing explicit formulas becomes a little annoying, but it is fairly clear what should happen if you divide the problem into two steps:

First convince yourself that the cone $\operatorname{cone}([0,1]^{n-1}) = D/[0,1]^{n-1} \times \{1\}$ of the $(n-1)$-disk is homeomorphic to the $n$-disk $D$.

Then convince yourself that there is a homeomorphism
\begin{aligned}
\phi: \frac{D}{\partial(D)\setminus [0,1]^{n-1}\times\{0\}} \xrightarrow{\;\;\;\cong \;\;\;} \frac{D}{[0,1]^{n-1} \times \{1\}}.
\end{aligned}

One could define $\phi$ for example by letting it be piecewise linear, determined by

letting $\phi(1/2,1/2,\dots,1/2,t) = (1/2,1/2,\dots,1/2,t)$ for all $t \in [0,1]$,

sending $\partial [0,1]^{n-1} \times \{0\}$ homeomorphically to $\partial [0,1]^{n-1} \times \{1\}$,

sending $\partial [\frac{1}{3},\frac{2}{3}]^{n-1} \times \{0\}$ homeomorphically to $\partial [0,1]^{n-1} \times \{0\}$,

sending $\partial [0,1]^{n-1} \times \{1\}$ homeomorphically to $\partial [\frac{1}{3},\frac{2}{3}]^{n-1} \times \{1\}$.


问:
What does it mean that multiplication and division have the same precedence?
I have read and heard in many places that multiplication and division are of equal precedence. So what is the value of:

$$
80 \div 10 \times 2
$$

Is it

$$(80 \div 10) \times 2 = 16$$

or

$$
80 \div (10 \times 2) = 4
$$

? The answer here seems to be that by convention we operate from left-to-right, so the first solution is correct. But I would have thought we should get the same answer either way if multiplication and division are of equal precedence. Isn't the whole point of operator precedence to resolve expressions that would otherwise have more than one solution?

For example, people also say that addition and subtraction are the same level of precedence, but no such ambiguity exists there:

$$
80 + 10 - 2 = (80 + 10) - 2 = 80 + (10 - 2) = 88
$$

So what is meant when people say that multiplication and division have the same precedence?

答:
For this discussion, operations have 2 properties of interest:

Precedence

Associativity

Both, yes, are used to disambiguate expressions.  (Note, e.g., that prefix and postfix expressions never need to rely on precedence, and never need parenthesis to force evaluation.)

Precedence applies when 2 operators are adjacent; the operator with the higher precedence is evaluated first.  $2 + 3 \times 5$ is always 17.  In such examples, associativity plays no role.  (When I say always, I mean, in math.  Every programming language I know will do it this way, but that doesn't make them authoritative.  Not all calculators yield this answer, though (infix) scientific calculators should.)

When 2 operators with the same precedence are adjacent, then associativity applies.  The four basic arithmetic operators associate left-to-right.  So, $8 - 3 - 2$ is always 3.

There are other operations which associate right-to-left.  Exponentiation.  $2^{3^2}$, which might be written as 2 ^ 3 ^ 2 in some languages, is always $2^9 = 512$.  We use parenthesis to force evaluation otherwise:  $\left(2^3\right)^2 = 8^2 = 64$.

While we're here, for clarity, 'cause I've seen too many CS professors get this wrong, associativity does not apply when evaluating, e.g., $ 3 \times 7 + 64 \div 16$  It does not matter (in arithmetic) whether the multiplication is performed first, or the division, just so they are both performed before the addition is applied (per the precedence rules).  Languages like C, Python and Java have no rules about which is evaluated first, either.

Someone commented above that math teachers should learn programming.  Languages are not authoritative; it is up to them to implement the rules of math correctly (or, not).  For example, while writing this, the Tex engine balked at $2^3^2$, insisted I put braces in to clarify, though math is quite clear on the subject.  OTOH, Python3 evaluates 2**3**2 just fine.

I would say it is important for programmers to be good at math, and know a bit about language theory.  Side-effects are not a concern in arithmetic, but must be considered in programming.


问:
What is the smallest $2l$ for which the most efficient sequence of numbers is possible?
So I'm searching for the a sequence of numbers such that when any $2$ are picked their sum will add up to any even number $2m \geq 2l$ where $2l$ is the smallest even number thier sum is equal to. Then, what is the smallest $2l$ for which the most efficient sequence is possible? To define the most efficient sequence is possible we use the following. Let,

$$S = a_1 + a_2 + a_3+\dots +a_n$$

where $a_1 < a_2 < \dots < a_n$ and are elements of our sequence. Then:

$$ (2 n -1) S = \sum_{i \leq j \leq N} (a_i + a_j)$$

Now we know the sum of these $a_i + a_j$ should yield an even number $ \geq 2 l$. Let us assume that each even number only occurs once. Then:

$$ (2 n -1) S = \sum_{r=l}^{a_n} 2r = a_n(a_n +1) - l(l+1)$$

Diving by $2n-1$:

$$ S  = \frac{a_n(a_n +1)}{2n-1} - \frac{l(l+1)}{2n-1}$$

Hence,

$$ a_n = \frac{a_n(a_n +1)}{2n-1} - \frac{l(l+1)}{2n-1} - \frac{a_{n-1}(a_{n-1} +1)}{2n-3} + \frac{l(l+1)}{2n-3} $$

When $l$ is odd it makes sense to assume $a_1 = 2l$ if $l$ is odd. But for even $l$ $a_1$ can also be $l$. However, all of this assumes the existence of the "the most efficient sequence" and thus we circle back to what is the smallest $2l$ for which the most efficient sequence is possible?

答:
This answer assumes that "the most efficient sequence" $\{a_n\}$ is a sequence satisfying

$a_1\lt a_2\lt \cdots\lt a_n$

$\{a_i+a_j\mid 1\le i\le j\le n\}=\bigg\{2a_1,2(a_1+1),\cdots, 2\bigg(a_1+\dfrac{n(n+1)}{2}-1\bigg)\bigg\}$

This answer proves that the most efficient sequence exists if and only if $n\le 2$.

Proof :

For $n=1$, the most efficient sequence exists.

For $n=2$, the most efficient sequence exists since $a_2=a_1+2$ works.

For $n=3$, suppose that the most efficient sequence exists.

Then, since $$a_1+a_1\lt a_1+a_2\lt a_2+a_2,$$
$$a_1+a_2\lt a_1+a_3\lt a_2+a_3\lt a_3+a_3,$$
$$a_3+a_3=2a_1+10\tag1$$we have to have
$$a_1+a_2=2a_1+2\implies a_2=a_1+2$$from which we get $a_2+a_2=2a_1+4$.

So, we have to have $$a_1+a_3=2a_1+6\implies a_3=a_1+6$$which contradicts $(1)$.

For $n\ge 4$, suppose that the most efficient sequence exists.

Then, since $$a_1+a_1\lt a_1+a_2\lt a_2+a_2,\qquad a_1+a_2\lt a_1+a_3\lt a_2+a_3\lt a_3+a_3,$$
$$a_1+a_3\lt a_1+a_4$$we have to have
$$a_1+a_2=2a_1+2\implies a_2=a_1+2$$from which we get $a_2+a_2=2a_1+4$.

So, we have to have $$a_1+a_3=2a_1+6\implies a_3=a_1+6$$from which we get $a_2+a_3=2a_1+8$ and $a_3+a_3=2a_1+12$.

So, we have to have$$a_1+a_4=2a_1+10\implies a_4=a_1+10$$implying $a_2+a_4=a_3+a_3$, which contradicts that $\{a_n\}$ is the most efficient sequence.$\quad\blacksquare$


问:
Why can't I use ratio in this ??
The Question:

Average salary of 60 employees is 12000 per month.

Number of executives is twice the no of non-executives.

If average salary of non-executives be 2/5th of the average salary of
executives then what is the average salary of non-executive employees
??

Answer is 6000.

I want to know why this is incorrect:

salaries in ratio =2:5

then  2x+5x=12000

x=12000/7

Please help me !!
And if anything is wrong let me know,

Thank you .

答:
you have 20 non executives with an average salary $S$ and 40 executives with average salary $2.5 S$

thus

$$\frac{20\times S+40\times 2.5\times S}{60}=12,000$$

$$S=\frac{12,000\times 60}{120}=6,000$$

Title: Why can't I use ratio in this ??


问:
On what page did Shannon axiomatize self-information?
I am trying to understand who first derived the formula I(x) = -log(p(x)) for self-information from  axioms. The wiki entry on information content and several other online sources all say it is in Shannon's well known article ``A mathematical theory of communication'' Bell Journal July 1948.  However, I have looked closely at this article and can only find an axiomatization of Shannon entropy. I cannot find an axiomatization of self-information. Can anyone please tell me the page number, equation number, or section number where this supposed axiomatization of self-information is hiding?    I have found references after Shannon 1948 eg Luce 1960 with the axiomatization of self-information I am seeking.

答:
Most of the time, people cited Shannon's work to appreciate his contributions, not necessarily the case that all results are derived by Shannon or presented in the same way as Shannon did. As you already see from Shannon's original paper, he considered uncertainty for a random variable (or information conveyed in a random variable) and directly obtained the entropy function.

As I know, the term "self-information" first appears in Prof. Fano's textbook (and later in Prof. Gallager's IT textbook). However, they all began with a more general quantity $i(E_1; E_2)$ called the "mutual information between events $E_1$ and $E_2$." The self-information $h(E)$ of an event $E$ is then defined as $h(E)=i(E;E)$. In fact, when taking expectation for $i(E;E)=h(E)$ with respect to all elementary events $E$ associated with a random variable $X$, we obtain this known formula $I(X;X)=H(X)$, where $I(\cdot)$ and $H(\cdot)$ are the usual mutual information and entropy, respectively.

In Prof. Fano's approach, you may find a different set of axioms to obtain a mathematical expression for the mutual information between two events. From that expression, an expression for the self-information is immediate.  However, I personally like to use the four axioms as shown on the Wiki page to define self-information since it is more straightforward. I hope the above helps.


问:
Complex number quadratic with conjugation
I have the following equation:
$$
3z^2+z^*-1=0
$$
z is a complex number and $z^*$ is its conjugate. I'm stuck solving it. There's what I did so far:
$$
z = a+bi
$$
$$
3a^2+6abi-3b^2+a+bi=1
$$
Assuming $Im(z) = 0$ I'm getting following equations:
$$
\begin{cases}
3a^2-3b^2+a=1\\
6abi-bi=0
\end{cases}
$$
$$
bi(6a-1)=0=>b=0 \lor a=-\frac{1}{6}
$$

Since $b=0$, I'm getting
$$
3a^2+a=1
$$
which will have 2 solutions. I have no idea what should I do next since I will be getting 3 different $a$'s, but wolfram shows me only 2 solutions.
$$
z = -\frac{1}{6} - \frac{\sqrt{13}}{6}
$$
$$
z = \frac{\sqrt{13}}{6} -\frac{1}{6}
$$

答:
Well, we are trying to solve:

$$3\text{z}^2+\overline{\text{z}}-1=0\tag1$$

As you did, let's set $\text{z}:=\alpha+\beta i$ where $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}$. So we get:

$$3\left(\alpha+\beta i\right)^2+\alpha-\beta i-1=0\tag2$$

Expanding $\left(\alpha+\beta i\right)^2$, gives:

$$\left(\alpha+\beta i\right)^2=\alpha^2-\beta^2+2\alpha\beta i\tag3$$

So, we get:

$$3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1=0\tag4$$

Finding the real and imaginary parts, we can set:

$$
\begin{cases}
\Re\left(3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1\right)=0\\
\\
\Im\left(3\left(\alpha^2-\beta^2+2\alpha\beta i\right)+\alpha-\beta i-1\right)=0
\end{cases}\tag5
$$

This gives:

$$
\begin{cases}
3\alpha^2-3\beta^2+\alpha-1=0\\
\\
6\alpha\beta-\beta=0
\end{cases}\tag6
$$

Solving the second equation for $\beta$ gives:

$$\beta=0\tag7$$

So:

$$3\alpha^2-3\cdot0^2+\alpha-1=0\space\Longleftrightarrow\space\alpha=\frac{\pm\sqrt{13}-1}{6}\tag8$$

As pointed out by @Gae. S., try to prove or disprove that $\alpha=\frac{1}{6}$ is or isn't a solution to your problem.


答:
If $a=-\dfrac16$, then $3a^2-3b^2+a=\dfrac1{12}-3b^2-\dfrac16=1\iff 3b^2=\dfrac1{12}-\dfrac16-1$,

which has no real solutions for $b$, so there really are only the two solutions from when $b=0$.

Title: Complex number quadratic with conjugation

Body:
I have the following equation:
$$
3z^2+z^*-1=0
$$
z is a complex number and $z^*$ is its conjugate. I'm stuck solving it. There's what I did so far:
$$
z = a+bi
$$
$$
3a^2+6abi-3b^2+a+bi=1
$$
Assuming $Im(z) = 0$ I'm getting following equations:
$$
\begin{cases}
3a^2-3b^2+a=1\\
6abi-bi=0
\end{cases}
$$
$$
bi(6a-1)=0=>b=0 \lor a=-\frac{1}{6}
$$

Since $b=0$, I'm getting
$$
3a^2+a=1
$$
which will have 2 solutions. I have no idea what should I do next since I will be getting 3 different $a$'s, but wolfram shows me only 2 solutions.
$$
z = -\frac{1}{6} - \frac{\sqrt{13}}{6}
$$
$$
z = \frac{\sqrt{13}}{6} -\frac{1}{6}
$$


问:
Computation of $\int_0^{\pi} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta$

Show that
  $$\begin{align*} \forall x \in [-1,1]: \int_0^{\pi} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= c_n \tag{1} \\ \int_0^{\pi} \frac{\sin^{n+2} \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= a_n \cdot x^2+b_n \tag{2} \end{align*}$$ where $a_n, b_n, c_n$ are constants (which do not depend on $x \in [-1,1]$), $n \geq 3$.

I tried several approaches (differentiation to show that the derivative of $(1)$ is equal to $0$, Weierstraß substitution, ...), but always got stuck. For example, one can show that the integrand in $(1)$ equals

$$\begin{align*} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} &= \left( \sqrt{ \left( \frac{x-\cos \theta}{\sin \theta} \right)^2+1} \right)^{-n} \\ &= \left( \sqrt{4 \left( \frac{\sin \frac{\theta+\varrho}{2} \cdot \sin \frac{\theta-\varrho}{2}}{\sin \theta}  \right)^2+1} \right)^{-n} \end{align*}$$ where $x=\cos \varrho$. I hoped to get some kind of symmetrization out of it, but (as far as I can see) it doesn't work.

Any ideas? (The aim is to find a rather quick or direct proof - a lengthy one is already known, using a recursive/inductive approach.)

Thanks!

答:
The Gegenbauer polynomials were built to solve problems like this.
Suppose $m\in\mathbb{N}$.
We have
$$\begin{eqnarray*}
\int_0^{\pi} d\theta \,
    \frac{\sin^{n+2m} \theta}{(1-2x \cos \theta+x^2)^{n/2}}
&=& \int_0^\pi  d\theta \, \sin^{n+2m} \theta
    \sum_{k=0}^\infty C_k^{(n/2)}(\cos\theta) x^k \\
&=& \int_{-1}^1  d u \, (1-u^2)^{n/2-1/2} (1-u^2)^m
    \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k
    \hspace{5ex} (u=\cos\theta) \\
&=& \int_{-1}^1  d u \, (1-u^2)^{n/2-1/2}
    \sum_{l=0}^m \beta_{2l} C_{2l}^{(n/2)}(u)
    \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k \\
&=& \sum_{l=0}^m \sum_{k=0}^\infty  \beta_{2l} x^k
    \underbrace{\int_{-1}^1  d u \, (1-u^2)^{n/2-1/2}
    C_{2l}^{(n/2)}(u) C_k^{(n/2)}(u)}_{\gamma_{2l}^{(n/2)} \delta_{2l,k}} \\
&=& \sum_{l=0}^m \beta_{2l} \gamma_{2l}^{(n/2)} x^{2l}.
\end{eqnarray*}$$
This proves the claim for any $m\in\mathbb{N}$.
The restriction $n\ge 3$ is too loose.
The result holds for any $n$ such that $n>-1$.

For a given $m$ it is a straightforward exercise to find the $\beta$s and $\gamma$s.
For example, using the fact that $1=C_0^{(n/2)}(u)$ and the normalization relation for the Gegenbauer polynomials we find
$$\int_0^{\pi} d\theta \, \frac{\sin^{n} \theta}{(1-2x \cos \theta+x^2)^{n/2}}
= \frac{\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}.$$

Some details:

$1/(1-2x \cos \theta+x^2)^{n/2}$
is the generating function for the Gegenbauer polynomials.

The polynomials have the parity property.

The polynomials provide an orthogonal basis for functions on $[-1,1]$ with weight function $(1-u^2)^{n/2-1/2}$.


问:
Diameters of state spaces

The authors mention that if $M$ is not a factor, the diameter of $S_0(M)/Int(M)$ is equal to 2.

When $M$ is not a factor, it is easy to check that the $d(,)\leq 2$, how to check that $d(,)\geq2$?

答:
If $M$ is not a factor then $M\cap M'\neq \mathbb{C}I$. Since the linear span of unitaries is norm-dense in any von Neumann algebra, this means there exists a unitary $u\in M\cap M'$ which is not a scalar multiple of the identity. Consider the self adjoint unitary $v=\frac{u+u^*}{2}\in M\cap M'$. Its spectrum, $\sigma(v)$, is contained in $\{-1,1\}$. We claim that in fact $\sigma(v)=\{-1,1\}$.

Indeed, if $\sigma(v)=\{-1\}$, then $\sigma(v+I)=\{0\}$, hence, since $v+I$ is self adjoint, we get $v+I=0$, i.e. $v=-I$. For all unit vectors $x$ we then obtain
$$-1=\langle -x, x\rangle=\langle vx,x\rangle=\frac{1}{2}\langle ux,x \rangle+\frac{1}{2}\langle x,ux \rangle.$$

Since $|\langle ux,x \rangle|, |\langle x,ux \rangle|\leq 1$ this is only possible if $\langle ux,x\rangle=\langle x, ux\rangle=-1$ for all unit vectors $x$ which implies that $u=-I$, which is a contradiction. Hence $\sigma(v)\neq\{-1\}$ and by a similar argument, $\sigma(v)\neq\{1\}$.

Then $\sigma(v)=\{-1,1\}$ and it is a standard fact that this set is contained in $\overline{\{\langle vx,x\rangle: \|x\|=1\}}$. Choose unit vectors $y$ and $z$ such that $\langle vy,y\rangle$ and $\langle vz,z\rangle$ approximate $1$ and $-1$, respectively. Consider the normal states $\omega=\langle \cdot y,y\rangle $ and $\phi=\langle \cdot z,z\rangle$ on $M$.

For any unitaries $u_1, u_2 \in M$ we have,
$$\|\omega \circ \text{Ad}(u_1)-\phi\circ \text{Ad}(u_2)\|\geq |\omega(u_1vu_1^*)-\phi(u_2vu_2^*)|=|\omega(v)-\phi(v)|.$$

(The last equality is due to the fact that $v$ commutes with $M$). By choosing $y$ and $z$ appropriately this last quantity can be made as close to $2$ as we wish. We conclude that $\text{diam}(S_0(M)/\text{Int}(M))\geq 2.$

Title: Diameters of state spaces


问:
Uniform bound for law of large numbers
Let $(X_i)_i$ be a sequence of iid real valued random variables with finite variance. Is it true that given a bounded measurable function $f:\mathbb R^2\to \mathbb R$, then almost surely and uniformly in $k\in\{1,...,n\},$

$$\frac 1 n \sum_{i=1}^n f(X_i,X_{i+k}) \to \mathbb E(f(X_1,X_{2}))\,\,?$$

If I am not requiring the uniformity in $k$, then this result should simply follow from the ergodic law of large numbers. How could I get this uniformity?

答:
Here we only need boundedness of $f$ (no assumption on the variance of $X_1$); using Baum-Katz results for martingales with identically distributed increments, it is likely that we only need $\lvert f(X_1,X_2)\rvert^p$ to be integrable for some $p$.

We have to prove that
$$
\max \limits_{1 \leqslant k \leqslant n} \left\lvert \frac{1}{n}\sum \limits_{i=1}^n f(X_i, X_{i+k}) - \mathbb{E}(f(X_1,X_2))\right\vert\to 0
$$
almost surely. To this aim, we will introduce a martingale by defining first the $\sigma$-algebras $\mathcal F_j:=\sigma(X_i,i\leqslant j)$ and let $d_{k,i}:=f(X_i,X_{i+k})-\mathbb E\left[f(X_i,X_{i+k})\mid\mathcal F_{i+k-1}\right]$. We have to show that
$$\max \limits_{1 \leqslant k \leqslant n} \left\lvert \frac{1}{n}\sum \limits_{i=1}^n d_{k,i}\right\vert\to 0\mbox{ a.s. and }\max \limits_{1 \leqslant k \leqslant n} \left\lvert \frac{1}{n}\sum \limits_{i=1}^n \mathbb E\left[f(X_i,X_{i+k})\mid\mathcal F_{i+k-1}\right]-\mathbb E\left[f(X_i,X_{i+k}) \right]\right\vert\to 0\mbox{ a.s.}.$$
For the first part, it suffices to show the almost sure convergence to $0$ of
$$
Y_N:=\max \limits_{1 \leqslant k \leqslant 2^N}\max_{1\leqslant n\leqslant 2^N} \left\lvert \frac{1}{2^N}\sum \limits_{i=1}^n d_{k,i}\right\vert.
$$
By the Borel-Cantelli lemma, it suffices to show that for each positive $\varepsilon$, the series
$\sum_N \mathbb P\left(Y_N>\varepsilon\right)$ converges. To do so, we first use a union bound
$$
\mathbb P\left(Y_N>\varepsilon\right)\leqslant \sum_{k=1}^{2^N}\mathbb P\left(\ \max_{1\leqslant n\leqslant 2^N} \left\lvert \frac{1}{2^N}\sum \limits_{i=1}^n d_{k,i}\right\vert\gt\varepsilon\right).
$$
Since for each $k$, $(d_{k,i})_{i\geqslant 1}$ is a bounded martingale difference sequence, we get by Azuma-Hoeffding's inequality (the version with maximum, or the classical version combined with Doob's inequality) that
$$
\mathbb P\left(\ \max_{1\leqslant n\leqslant 2^N} \left\lvert \frac{1}{2^N}\sum \limits_{i=1}^n d_{k,i}\right\vert\gt\varepsilon\right)\leqslant c\exp\left(-2^N C\right),
$$
where $c$ and $C$ are independent of $k$ and $N$.

We now have to treat $\max \limits_{1 \leqslant k \leqslant n} \left\lvert \frac{1}{n}\sum \limits_{i=1}^n \mathbb E\left[f(X_i,X_{i+k})\mid\mathcal F_{i+k-1}\right]-\mathbb E\left[f(X_i,X_{i+k}) \right]\right\vert$. Observe that by independence, $\mathbb E\left[f(X_i,X_{i+k})\mid\mathcal F_{i+k-1}\right]=\mathbb E\left[f(X_i,X_{i+k})\mid X_i\right]=\mathbb E\left[f(X_i,X_{0})\mid X_i\right]=g(X_i)$, where $g(x)=\mathbb E\left[f(x,X_{0})\right]$ hence
$$\max \limits_{1 \leqslant k \leqslant n} \left\lvert \frac{1}{n}\sum \limits_{i=1}^n \mathbb E\left[f(X_i,X_{i+k})\mid\mathcal F_{i+k-1}\right]-\mathbb E\left[f(X_i,X_{i+k}) \right]\right\vert=  \left\lvert \frac{1}{n}\sum \limits_{i=1}^n g(X_i ) -\mathbb E\left[g(X_i) \right]\right\vert$$
and we can use the strong law of large numbers.

Title: Uniform bound for law of large numbers