问:
How to find a continuous path between matrices?

I have matrix $A=\begin{pmatrix} 0&-1&0\\1&0&0\\0&0&1\\\end{pmatrix}$ and matrix $B=\begin{pmatrix} 1&0&0\\0&\frac{1}{2}&-\frac{\sqrt{3}}{2}\\0&\frac{\sqrt{3}}{2}&\frac{1}{2}\\\end{pmatrix}$ and I want to find a continuous path between them. How do I do this?

I know that matrices in $SO(n)$ should be path connected but how do you actually construct such a path?

I'm not sure if just taking $\gamma(t)=A(t)+B(1-t)$ is going to work.

答:
Correct me if I'm wrong, doesn't any element in $\text{SO}(n)$ have an exponential representation?

I.e for any $U \in \text{SO}$, there is a matrix $H$ such that:

$U= e^{H} $

If this is true, the connecting  path is apparent, isn't it?

Disclaimer: I'm not expert in algebra or geometry.


问:
prove that $(1 + x)^\frac{1}{b}$ is a formal power series
How to prove that $(1 + x)^\frac{1}{b}$ (where $b$ is an integer) can be expressed as a formal power series without using Binomial theorem?</p>

答:
We do it for $b>0$.  A similar argument will work for $b<0$.  Note that the argument is purely formal.

So we are trying to find $a_0$, $a_1$, $a_2$, and so on such that
$$1+x=(a_0+a_1x+a_2x^2+ \cdots +a_nx^n +\cdots)^b$$
where equality is meant in the formal sense.  It is clear that we want $a_0=1$ and $a_1=1/b$.

There is a natural proof by induction, if we choose the induction hypothesis correctly.

Let $n \ge 1$. Assume by way of induction hypothesis that there are numbers $a_0, a_1,  \dots, a_n$ such that
$$(a_0+a_1x+\cdots+a_nx^n)^b$$
formally agrees with $1+x$ in positions $0, 1,\dots,n$.  We will show that we can choose $a_{n+1}$ such that

$$(a_0+a_1x+\cdots+a_{n+1}x^{n+1})^b$$
formally agrees with $1+x$ in positions $0, 1,\dots,n+1$.

That is easy.  Let $c_{n+1}$ be the coefficient of $x^{n+1}$ in $(a_0+a_1x+ \cdots +a_nx^n)^b$.  Define $a_{n+1}$ by $ba_{n+1}=-c_{n+1}$.

ContentLicense: CC BY-SA 3.0


问:
How can I get a marginal PDF from a joint PDF (probability density function)?
Let X and Y be random variables with a joint probability density function (joint PDF) given by

$ f_{X,Y}(x,y) \quad=\quad  \begin{cases} \frac{c}{1+x^2+y^2} & \text{ if } x^2+y^2<1\,, \\ 0 & \text{ otherwise,} \end{cases} $

where the positive constant c is determined by the requirement that $ f_{X,Y} $ is a PDF.

What is the correct formula for the marginal PDF of X?

I think I have to start off by integrating $ \frac{c}{1+x^2+y^2} $ with respect to y. Which gives me

$ \int \frac{c}{1+x^2+y^2} dy = \frac{c*\arctan{\frac{y}{\sqrt{1+x^2}}}}{\sqrt{1+x^2}} $

But don't know how to continue now.

I'd also like to know if there's any software out there where I can compute this kind of symbolic stuff.

Thanks

答:
I think I finally managed to recall.

[1] I think you made a mistake, you should integrate w.r.t. $x$ not $y$.

That will give you the margin PDF of $Y$.

[2] Make this equation:

$$\int_{-1}^{1} \left(\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{c}{1+x^2+y^2} \,dx\right)\,dy = 1$$

Solving this equation should give you $c$.

[3] Also, the inner integral gives you the marginal distribution of $Y$.

$$f(y) = \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{c}{1+x^2+y^2} \,dx $$

So just solve this as an indefinite integral treating $y$ as constant (similar to what you did but reversed), but then compute the definite integral too (i.e. apply the boundaries). That will give you a function of just $y$, your marginal PDF.

ContentLicense: CC BY-SA 4.0


问:
Convergence of $\int_1^\infty\left[\ln\left(\frac{1}{1-x^{-M}}\right)\right]^pdx$
So I have this integral:
$$
\int_1^\infty\left[\ln\left(\frac{1}{1-x^{-M}}\right)\right]^pdx.
$$
The question is for what $M$ does the integral converge for a given $p > 0$.

So the problematic points are at $x=1$ and $x=\infty$, and I assumed that $M>0$ for every $p>0$.

My idea was to try and approximate the integrand in the limit with a polynomial and then procede from there to test for convergence.

So for $x\to\infty$ I got the following:

$$
\lim_{x\to\infty}\frac{\ln\left(\frac{1}{1-x^{-M}}\right)}{x^{\alpha}}\stackrel{\text{L'H}}{=}\lim_{x\to\infty}\frac{M\left(1-x^{-M}\right)x^{-M-1}}{\alpha x^{\alpha -1}}=\frac{M}{\alpha}\lim_{x\to\infty}\left(1-x^{-M}\right)x^{-\alpha-M}
$$

so for the limit to be non-zero, $\alpha$ must be equal to $-M$ and therefore:
$$
\lim_{x\to\infty}\ln\left(\frac{1}{1-x^{-M}}\right) = \lim_{x\to\infty}x^{-M}.
$$

So now when I substitute $\ln(...)$ with $x^{-M}$, i get:

$$
\lim_{x\to\infty}\left[\ln\left(\frac{1}{1-x^{-M}}\right)\right]^p=\lim_{x\to\infty}x^{-Mp}
$$

and in order for it to converge, $-Mp$ must be smaller than $-1$. And from there for $M$:
$$
-Mp < -1 \Rightarrow M>{1\over p}.
$$
But when I want to do the same for $x\to 1$, I get stuck:
$$
\lim_{x\to1}\frac{\ln\left(\frac{1}{1-x^{-M}}\right)}{(1-x)^{\alpha}}\stackrel{\text{L'H}}{=}\lim_{x\to\infty}\frac{M\left(1-x^{-M}\right)x^{-M-1}}{\alpha (1-x)^{\alpha -1}}=\;?
$$
So my questions are:

Is the first part for $x\to\infty$ correct and

How to solve the problem for $x\to1$

答:
Your first part is correct. I derived it differently, but I also came to the conclusion that $Mp > 1$.

To get the $x\to 1$ side, we can use a change of variables. Let $u = \log\left( \frac{1}{1-x^{-M}} \right)$. Then your integral becomes:

$$
\frac{1}{M}\int_0^\infty \frac{u^p e^{-u}}{\left(1-e^{-u}\right)^{1+\frac{1}{M}}}\,\mathrm{d}u
$$

The limit $x\to 1$ corresponds to $u\to\infty$ in this new integral. As $u\to\infty$ the denominator goes to $1$. More formally, for $u > \log(2)$, you have $\frac{1}{1-e^{-u}} < 2$ so you can upper-bound the tail of the integral by a constant times $\int_{\log{2}}^\infty u^p e^{-u}\,\mathrm{d}u$. This integral converges for all $p$, which shows the tail of the original integral converges for all $p$.


问:
Unsure about double integral bounds of integration in polar coordinates
I'm trying to convert the bounds of integration to polar coordinates but I'm stumped on one of the bounds.

$$\int_{x=0}^{6}\int_{y=\frac{1}{\sqrt{3}}x}^{\sqrt{8x-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$

The only thing that left me stumped was converting $y=\frac{1}{\sqrt{3}}x$ to polar.

Right now I have $\int_{\theta=0}^{\frac{\pi}{6}}\int_{?}^{8\cos{\theta}}r^2\,dr\,d\theta$

Where do I go from here?  It's nowhere in my notes and I'm having a tough time finding anything online about it. Thank you!

答:
The domain of integration is delimited by the line with equation $y=\frac1{\sqrt 3}$ and the part of the circle with centre $(4,0)$ and radius $4$ since
$$y=\sqrt{8x-x^2}\iff y^2+x^2-8x=0,\; y\ge 0\iff   (x-4)^2+y^2-16=0,\;y\ge 0.$$

On the other hand, by the law of sines, we have the relation
$$\frac r{\sin2\theta}=\frac4{\sin\theta}\quad\text{whence the polar equation of the circle}\quad r=\frac{4\sin2\theta}{\sin\theta}=8\cos\theta,$$
and the integral becomes
$$\int_{\tfrac\pi6}^\tfrac\pi2\int_0^{8\cos\theta}\mkern-12mu r^2\,\mathrm d r\,\mathrm d\theta.$$


Title: Arg of $(1-i\sqrt{3})^6$. Did I do it right?

Body:
Did I do it right?

$(1-i\sqrt{3})^6$

module = $\sqrt{1+3}=2$

$(1-i\sqrt{3})^6=2^6\cdot(\frac{1}{2}-\frac{\sqrt{3}}{2}i)^6$ $\implies$ Arg = $\frac{\pi}{3}$

Body:
$Re^{i\theta} =(1-i\sqrt{3})^6$

$\Rightarrow $$Re^{i\theta}=2^6(\frac{1} {2} - i\frac{\sqrt{3}}{2})^6$

$\Rightarrow $$Re^{i\theta}= 2^6 (e^{\frac{-i\pi}{3}}) ^{6}$

$\Rightarrow $$
Re^{i\theta}= 2^6e^{-i2\pi} $

$\Rightarrow$$ R=2^6$, $\theta=-2\pi+2k\pi$

$K=1$

So:

$R=64 $ and $ \theta=0$


问:
Solving recurrence relation from high-order derivative of reciprocal
How do I solve this recurrence relation?

$$
\left\{{\begin{aligned} u_0 &= \frac{1}{v_0} \\ u_n &= -\frac{1}{v_0}\sum_{i=0}^{n-1}{n\choose i}u_iv_{n-i} \end{aligned}}\right.
$$

I've got the first couple of terms figured out:

$$
\begin{aligned}
u_1&=\frac{-v_1}{v_0^2} \\
u_2&=\frac{2v_1^2-v_0v_2}{v_0^3} \\
u_3&=\frac{-6v_1^3+6v_0v_1v_2-v_0^2v_3}{v_0^4}\\
u_4&=\frac{24v_1^4-36v_0v_1^2v_2+6v_0^2v_2^2+8v_0^2v_1v_3-v_0^3v_4}{v_0^5}
\end{aligned}
$$

It looks like for $u_n$, the first term on the numerator is always $(n!)v_1^n$, and the rest has to do with interger partition of $n$. The sign of the coefficient is deterimined by the order of $v_0$.

This came from trying to express $\mathrm{d}^nu/{\mathrm{d}x^n}$ with derivatives of $v$ where $u(x)v(x)=1$.

答:
We have the following formal power series identity $$\left(\sum_{n\geqslant 0}u_n\frac{t^n}{n!}\right)\left(\sum_{n\geqslant 0}v_n\frac{t^n}{n!}\right)=1.$$

To get rid of signs, and powers of $v_0$ in denominators, let's put $u_n=a_n/v_0$ and $v_n=-b_n v_0$: $$\left\{\begin{aligned}a_0&=1,\\a_n&=\sum_{k=0}^{n-1}\binom{n}{k}a_k b_{n-k},\end{aligned}\right.\qquad\left(1+\sum_{n>0}a_n\frac{t^n}{n!}\right)\left(1-\sum_{n>0}b_n\frac{t^n}{n!}\right)=1.$$

Now, using operations on formal power series, we deduce
$$\sum_{n>0}a_n\frac{t^n}{n!}=\left(1-\sum_{n>0}b_n\frac{t^n}{n!}\right)^{-1}-1=\sum_{m>0}\left(\sum_{n>0}b_n\frac{t^n}{n!}\right)^m
\\=\sum_{m>0}\sum_{n_1,\ldots,n_m>0}\prod_{k=1}^m\frac{b_{n_k}t^{n_k}}{n_k!}=\sum_{m>0}\sum_{n_1,\ldots,n_m>0}t^{n_1+\cdots+n_m}\frac{b_{n_1}}{n_1!}\cdots\frac{b_{n_m}}{n_m!}.$$

The question asks us to group the terms by fixed powers of $t$, $b_1$, $b_2$, etc.; that is, given $n>0$ and $k_1,k_2,\ldots,k_n\geqslant 0$ such that $1k_1+2k_2+\cdots+nk_n=n$, we want the coefficient of $t^n b_1^{k_1}b_2^{k_2}\cdots b_n^{k_n}$ in the last expression above.

To get it, we count the number of tuples $(n_1,\ldots,n_m)$ that have $k_1$ $1$'s, $k_2$ $2$'s, ..., $k_n$ $n$'s (so we must have $m=k_1+k_2+\cdots+k_n$). This number is the well-known multinomial coefficient $\frac{(k_1+k_2+\cdots+k_n)!}{k_1!\,\cdot\,k_2!\,\cdots\,k_n!}$: $$a_n=n!\sum_{\substack{k_1,\,k_2,\,\ldots,\,k_n\,\geqslant\,0\\1k_1+2k_2+\cdots+nk_n=n}}\frac{(k_1+k_2+\cdots+k_n)!}{k_1!\,\cdot\,k_2!\,\cdots\,k_n!}\left(\frac{b_1}{1!}\right)^{k_1}\left(\frac{b_2}{2!}\right)^{k_2}\cdots\left(\frac{b_n}{n!}\right)^{k_n}.$$ To get back to $u_n$ and $v_n$, as stated at the beginning,  put $b_n=-v_n/v_0$ and take $u_n=a_n/v_0$.

One may recognize a big gun here.


问:
How do I scale one operand of a vector addition to get sum of a given magnitude?
Vector A + Vector B = Vector C

Vector A is known

The normal of Vector B is known

The magnitude of Vector C is known

I'm looking for the scalar coefficient x I can multiply B with so that A + xB = C where |C| is the known magnitude.

On the request of a replier, here's a diagram:

答:
You can also write $|A+xB|=c$ where $c$ is the known magnitude of $C$, which means that (by squaring both sides of the equation and using the inner produc $v\cdot v=|v|^2$)

$$|A|^2+x^2|B|^2+2(A\cdot B)x=c^2$$

This is a quadratic equation which gives you the solutions for $x$, which is $\frac{-2(A\cdot B)\pm\sqrt{4(A\cdot B)^2-4|B|^2(|A|^2-c^2)}}{2|B|^2}$, and there is no solution if $4(A\cdot B)^2-4|B|^2(|A|^2-c^2)<0$.

Geometrically, let $A=\vec{PQ}$, then you can draw a circle centered at $P$ with radius $c$ and a line passing by $Q$ in the direction of $B$. This line can intersect with the circle at $0,1$ or $2$ points and would give you the correct vector $xB$. (Or you can draw a circle centered at $Q$ and a line passing by $P$, too).


问:
Bicycle Question
In a survey, pupils were asked if they owned a bicycle.

Results:

3/8 of the pupils said ‘Yes’.

5/8 of the pupils said ‘No’.

46 more pupils said ‘No’ than said ‘Yes’.

Altogether, how many pupils were in the survey?

How do you work this out? I know should always find 1...

答:
Let there are  $8r$ pupils all together,

$\frac38\cdot8=3r$ pupils said 'yes' and $\frac58\cdot8=5r$ pupils said 'no'

$\implies (5r-3r)=46$

Can you take it from here?

Title: Bicycle Question


Body:
Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$

Case $S(1)$ is true:

$$2((1)+2)\leq((1)+2)^{2}$$

$$6\leq9$$

Case $S(n)$ is true for all $n=1,2,...$

$$2(n+2)\leq(n+2)^{2}(i)$$

Case $S\left(n+1\right)$

$$2(n+3)\leq(n+3)^{2}(ii)$$

From (i)

$$2(n+2+1-1)\leq(n+2+1-1)^{2}$$

$$2(n+3)-2(1)\leq(n+3)^{2}-2(n+3)+1$$
$$2(n+3)\leq(n+3)^{2}-2(n+3)+3$$
$$2(n+3)\leq(n+3)^{2}-(2n+3)$$
$$2(2n+3)+3\leq(n+3)^{2}$$

Thus $(i)$ is true for all $n=1,2,...$

My question: How can I get the same expression as (ii)?
I got to $2(2n+3)+3\leq(n+3)^{2}$ but it's clearly wrong

Body:
Recurrence :

$2(n+1)\leq(n+2)^2 $

For n=1

$4\leq 9$ that is correct

Suppose : $ 2(n+1)\leq(n+2)^2 $

Let's show $ 2(n+2)\leq(n+3)^2 $

$2(n+2)\leq(n+3)^2 $

$\Leftrightarrow $ $2(n+2)\leq ((n+2)+1)^2 $

$\Leftrightarrow $$2(n+2)\leq (n+2)^2 +2(n+2)+1$

$\Leftrightarrow $$0\leq(n+2)^2 +1$ that is correct

So:

$ 2(n+2)\leq(n+3)^2 $

Finally :

$ 2(n+1)\leq(n+2)^2 $


问:
Understanding singular complexes
I'm just getting started on singular homology and I've found that when we compute the singular $q$-simplices of a topological space X (which is the set of all linear mappings $\sigma: \Delta_n \to X$) we only use a few generators, but there'd be more. Let me explain myself:

Suppose we're computing the singular $q$-simplices of a line with vertices $v_1$ and $v_2$. We usually say that the singular 0-simplex is the free abelian group spanned by $\langle\sigma_1, \sigma_2\rangle$, where $\sigma_i: \Delta_0 \mapsto v_i$. However, when we compute the singular 1-simplex we say that it is the free abelian group spanned by $\langle \rho \rangle$, where $\rho: \Delta_2 \mapsto [v_1 v_2]$ (here $[v_1 v_2]$ denotes the line with edges $v_1$ and $v_2$). Wouldn't $\rho': \Delta_2 \mapsto v_1$ be another generator of the singular 1-simplex?

In general, I feel like I can choose infinitely many generators for every singular $q$-simplex.

I'm pretty sure I'm not understanding the point of the theory well enough, so please I'd really appreciate if anyone could answer me.

答:
You confuse a lot of things. A singular $n$-simplex of $X$ is a continuous map $$\sigma_\alpha\colon \Delta^n \to X$$

The singular $n$-simplices of $X$ are the basis elements of the free abelian (chain) group $C_n(X)$, i.e. they generate $C_n(X)$.

A singular $0$-simplex is not being generated by singular $0$-simplices, but if your space $X$ admits precisely two singular $0$-simplices, then the chaingroup $C_0(X)$ is generated by these two $0$-simplices.

Also, it's certainly not true that you can somehow choose infinitely many generators for a chaingroup, say $C_1(X)$. The free abelian group $C_n(X)$ is a free $\mathbb{Z}$-module and its elements are formal sums $\sum_i n_i\sigma_i$ with $\mathbb{Z}$-coefficients $n_i$. These elements are called $n$-chains.

In summary, $n$-chains of $C_n(X)$ are linear combinations of basis elements the singular $n$-simplices $\sigma_\alpha$ that generate $C_n(X)$.

In order to get a better intuitive picture i would highly suggest you to look up the idea of homology, it's very well illustrated in Hatcher on p.99.

Title: Understanding singular complexes


问:
Having trouble finding the fault in my logic for this coding question(it's heavily math related)
So I have been doing some coding questions to practice my coding and I ran into this problem: http://www.usaco.org/index.php?page=viewproblem2&cpid=989

The general gist of the problem is that there is a person named Bessie is racing a length of $K\ge1$ meters. She starts at $0$ meters per second at the $0$ second mark, and then every second after that, she can either increase her speed by $1$ meter per second, stay at the same speed, or decrease her speed by $1$ meters per second. Her speed can not drop below zero(so if Bessie is at $0$ meters per second, she can not decrease her speed.).

Bessie wants to finish the race in an integral amount of seconds, either finishing at the $K$ meter mark or past the $K$ meter mark. But Bessie also doesn't want to finish the race too quick. At the instant when she finishes $K$ meters, she wants to be traveling at a speed of no more than $X\ge1$ meters per second. Bessie wants to know the minimum speed required to finish $K$ meters, given $K$ and $X$.

The logic I use to solve this problem only works for the first 4 test cases, and I'm sure it's not because of an coding error. So my logic is as follows:

Before we do anything, we first have to test whether or not a speed of $X$ meters per second can be reached, as the following solution assumes that $X$ meters per second is reachable.

To do so, we first note that the quickest way to get to $X$ meters per second is to increase the speed by $1$ each second for $X$ seconds. We then note that if after increasing $X-1$ times, if the distance covered is $<K$ meters, then it is guaranteed that $X$ meters per second is reachable. But if after increasing $X-1$ times the distance covered is $\ge K$, then we know that $X$ meters per second is unobtainable.

To calculate the distance covered after $X-1$ increases, we can calculate the following sum:$$1+2+\cdots+(X-2)+(X-1)$$which can be represented as$$\frac{X(X-1)}2$$. We want to test whether or not$$\frac{X(X-1)}2\ge K$$. If this inequality is false, then go to the solution under the gray line. If this inequality is true, then we know $X$ meters per second is unobtainable, and therefore we need to calculate how many increases are required to surpass $K$ meters. We will call this amount $n$. To find the value of $n$ that will cause the distance to go over $K$ meters, we first need to find the formula of the distance covered after $n$ increases. That can be represented with the sum $$1+2+3+\cdots+n=\frac{n(n+1)}2$$. So then we set this sum to be $<K$, then use this inequality to maximize $n$:$$\frac{n(n+1)}2<K\\\frac{n^2+n}2<K\\n^2+n<2K\\(n+1/2)^2-1/4<2K\\n+1/2<\sqrt{2K+1/4}\\n<\frac{\sqrt{8K+1}-1}2$$So the value of $n$ would be:$$n=\left\lceil\frac{\sqrt{8K+1}-1}2\right\rceil$$(without the ceiling function we would be calculating the largest amount of increases that doesn't surpass $K$ meters, instead of actually passing $K$ meters)

First we want to find the maximum speed in which Bessie can go at. Let's say that this maximum speed is $m$ and the target speed(the speed we want to have at $K$ meters) be $X$ meters per second(as stated in the problem). We can find the maximum speed by allowing Bessie to increase its speed every second until it reaches $m$, then immediately start decreasing her speed until she hits $X$ meters per second. We then know that the total distance traveled after this increase and decrease is(which I will denote as $d$): $$d=\underbrace{1+2+3+\cdots+m}_{\text{increasing speed}}+\underbrace{(m-1)+(m-2)+\cdots+(X+1)+X}_{\text{decreasing speed}}$$. We can then find the formula for this sum to be: $$d=m^2-\frac{X(X-1)}2$$. This sum has to be $\le K$(or else we can't decrease enough in time), so we have the following inequality:$$m^2-\frac{X(X-1)}2\le K$$. $K$ and $X$ are already given as inputs in the problem, so we just have to isolate $m$. We get that: $$m\le \sqrt{K+\frac{X(X-1)}2}$$(positive square root). To get the highest $m$, we just need to take the floor of the RHS so:$$m=\left\lfloor\sqrt{K+\frac{X(X-1)}2}\right\rfloor$$. Then if $d$ is $<K$, we need to find out the remaining distance that we need to cover. That is easy to calculate: $K-d$. From this we can calculate how many seconds we need to stay at $m$ meters per second(if we stay at a speed $<m$, we can always stay at a higher speed to potentially reach $K$ meters quicker. Not too sure about this logic though). Each second we stay at $m$ meters per second adds an extra $m$ meters to our distance. So we need to divide $K-d$ by $m$ to see how many times we need to add $m$ to $d$ to reach $K$(I will denote this as $s$). So we get that we need to stay$$s=\left\lceil\frac{K-d}m\right\rceil$$seconds at $m$ meters per second to pass $K$ meters. Then we need to calculate the amount of seconds that passed for traveling $d$ meters using the method stated. To calculate this, we need to count how many terms we added together in the sum. So we need to find the length of this list:$$1,2,3\dots,m,(m-1),(m-2),\cdots,(X-1),X$$This can be calculated with the following formula: $$2m-X$$So finally we calculate $$2m-X+s$$ to obtain the final answer.

The problem is this only works for the first 4 test cases, and this strategy only works for certain values of $K$ and $X$ past test case 4, and is really close to the actual answer for other values(yes, I downloaded the test data), so I'm actually not too sure where I went wrong here. If you guys want the code I can also put it here, but this is more of a math problem, so I decided not to put the code for now.

答:
This is how I would approach:

First let us assume that $K \ge \frac{X(X-1)}{2}$; otherwise accelerate as quickly as you can to the finish.

ETA This is the problem with the formula in the OP; it is correct only if Bessi reaches a speed of $X$ or greater. She will not if the inequality $K < \frac{X(X-1)}{2} -X$ is satisfied. If $K \le \frac{X(X-1)}{2}$ then let $l$ be the smallest integer such that $\frac{l'-1}{l'} \ge K$ is satisfied, then that is the time needed to run, and the schedule is $1+2+ \ldots + (l'-1)+ l'$.

For each integer $l >X$, let $f(l)$ be the furthest one can travel in $l$ seconds so that one is traveling at speed no more than $X$ at the end.
Then for $l \ge X$:

$$f(l) = 1+2 + \ldots + m_l + (m_l-1) + \ldots + x$$

where $m_l \doteq \frac{l+X}{2}$ if $l+X$ is even and

$$f(l)= 1+2 + \ldots + m_l + m_l + (m_l-1)+ \ldots + x$$

where $m_l \doteq \left(\left \lfloor \frac{l+X}{2} \right \rfloor\right)$            if  $l+X$ is odd.

Also: if there is an $l'$ such that $f(l')=K$ then that is the time it needs to finish the race. Note that $l' > X$ so use the above to get a schedule that goes precisely $f(l')$ in $l'$. Otherwise, let $l'$ be the largest integer such that $f(l') < K$. Then $l' \ge X$, and also, the time to finish the race is at least $l'+1$. We now show that there is indeed a schedule that finishes in time $l'$, where the maximum speed is $X$:

If $l+X$ is odd let $d=K-f(l')$; then $1 \le d \le m_l$. If $d \le X$ go at paced $X$ at second $l'+1$; otherwise go at speed $d$ for another second.

If $l+X$ is even let $d=K-f(l')$; then $1 \le d \le m_l-1$. If $d \le X$ go at paced $X$ at second $l'+1$; otherwise go at speed $d$ for another second.


答:
If $A + xB  = C$,  then $xB = C-A$, so you can take the norm of both sides and get

$$ ||C - A|| = || xB ||= |x| \cdot ||B||,$$ which yields

$$ |x| = \frac{||C-A||}{||B||}.$$

So you just have to test the two possible values of $x$ (positive and negative) and see which one works.

HOWEVER, your equation doesn't always have a solution. You need to have $C-A$ and $B$ be vectors pointing in the identical direction. If they aren't then this equation will give you values of $x$, but neither one will satisfy your original equation. You could test that they point the same way (i.e. are "parallel" vectors) before you compute the values of $x$, but since you have to check both the positive and negative options anyways I'd just find my $x$'s, compute $A+ xB$, and see if either value works.

There is an interesting question of whether you could compute $x$ via a formula that doesn't require checking at the end, but I doesn't look like you care about that in you situation.

Title: How do I scale one operand of a vector addition to get sum of a given magnitude?

Body:
Vector A + Vector B = Vector C

Vector A is known

The normal of Vector B is known

The magnitude of Vector C is known

I'm looking for the scalar coefficient x I can multiply B with so that A + xB = C where |C| is the known magnitude.

On the request of a replier, here's a diagram:


问:
Translation equivariance of AV@R (average value at risk), proof
I am trying to prove that the average value at risk is translation equivariant:
$$AV@R_\alpha[Z+\tau] = AV@R_\alpha[Z] + \tau$$

where
$$AV@R_\alpha[Z] := \inf_{t\in \mathbb{R}} \{t+\alpha^{-1} \mathbb{E}[(Z-t)_+] \}.$$

I started by plugging in the definition, but I don't know how to get $\tau$ out of the expected value operator.

$$AV@R_\alpha[Z+\tau] = \inf_{t\in \mathbb{R}} \{t+\alpha^{-1} \mathbb{E}[(Z+\tau-t)_+] \}$$

Any help would be appreciated.

答: Note that $\inf_t \phi(t) = \inf_t \phi(t-\tau)$.


问:
Proof without solving equation that if $(1+z)^n = (1-z)^n$ then Re(z) = 0. Did I do it right?
So, I have to prove that if $(1+z)^n = (1-z)^n$ then $Re(z) = 0$ 'without actually solving the equation.' I see that I can easily transform the equation by interpreting $z$ as $a+bi$: $(1+ a + bi)^n = (1- a - bi)^n$. Then I see that the real parts on both sides must be equal, so

$$1+a = 1-a \Longrightarrow a =-a$$
$$ a = 0 \Longrightarrow Re(z) = 0 $$

And here comes my question - is that any good? I mean it seems too simple, I think I don't understand something.

答:
The question may be easily answered.

Since $(1+z)^n = (1-z)^n, ~~ |1+z|^n = |1 - z|^n.$

Therefore, the distance from $z$ to $1$ equals the distance from $z$ to $-1$.

Therefore, $z$ lies on the perpendicular bisector between $1$ and $-1$.

Therefore the real part of $z$ is equal $0.$


问:
Determine the $n$-th power of matrix
Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1  \\ }$$ May you help me with the answer since the book states :  $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n &  3\cdot4^n & 2+4^n  \\ }$$ while mine misses the ${1\over 6}$. So who is right?

$$$$Thanks in advance for any help you are able to provide.

答:
I get for the symbolic form depending on n and the symbolic eigenvalues $\alpha,\beta,\gamma$ (where we insert later the known values $\small \alpha=0,\beta=1,\gamma=4$)
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 \cdot \beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & -4\cdot\beta^n+3\cdot\alpha^n+\gamma^n & \\
 -2\cdot\beta^n+2\cdot\gamma^n & 3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n-3\cdot\alpha^n+\gamma^n \\
 -2\cdot\beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n+3\cdot\alpha^n+\gamma^n
 \end{array}
$$
and replacing with the actual values for $\small \alpha,\beta,\gamma$ we get
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & -4 +3\cdot 0^n+4^n & \\
 -2+2\cdot 4^n & 3\cdot 0^n+3\cdot 4^n & 2-3\cdot 0^n+4^n \\
 -2+2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & 2+3\cdot 0^n+4^n
 \end{array}
$$

For $n \in \mathbb N \text{ and } n \gt 0$ this reduces to
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & 3\cdot 4^n & -4+4^n & \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n
 \end{array}
$$
$ \qquad \qquad $ which matches the solution of the book.

For $n=0$ this gives with $0^0=1$
$$ A^0 = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2 & -3+3 & -4 +3+1 & \\
 -2+2 & 3+3 & 2-3+1 \\
 -2+2 & -3+3 & 2+3+1
 \end{array} = \frac16 \small \begin{array} {|r|r|r|r}
 6 & 0 & 0 & \\
 0 & 6 & 0 \\
 0 & 0 & 6
 \end{array} = I
$$
(So the textbook-solution should be commented with the required $0^0$-definition)

I used the diagonalization-procedure of Pari/GP for this and to get a textform for this answer:

```<code>M = [2,2,0;1,2,1;1,2,1]
tmpM=mateigen(M)
tmpW=tmpM^-1
tmpD=tmpW*M*tmpM
MPow = tmpM*matdiagonal(['a,'b,'c])*tmpW
\\ on text-level we have later to replace a by \alpha^n and so on
</code>```

Title: Determine the $n$-th power of matrix

Body:
I get for the symbolic form depending on n and the symbolic eigenvalues $\alpha,\beta,\gamma$ (where we insert later the known values $\small \alpha=0,\beta=1,\gamma=4$)
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 \cdot \beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & -4\cdot\beta^n+3\cdot\alpha^n+\gamma^n & \\
 -2\cdot\beta^n+2\cdot\gamma^n & 3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n-3\cdot\alpha^n+\gamma^n \\
 -2\cdot\beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n+3\cdot\alpha^n+\gamma^n
 \end{array}
$$
and replacing with the actual values for $\small \alpha,\beta,\gamma$ we get
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & -4 +3\cdot 0^n+4^n & \\
 -2+2\cdot 4^n & 3\cdot 0^n+3\cdot 4^n & 2-3\cdot 0^n+4^n \\
 -2+2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & 2+3\cdot 0^n+4^n
 \end{array}
$$

For $n \in \mathbb N \text{ and } n \gt 0$ this reduces to
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & 3\cdot 4^n & -4+4^n & \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n
 \end{array}
$$
$ \qquad \qquad $ which matches the solution of the book.

For $n=0$ this gives with $0^0=1$
$$ A^0 = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2 & -3+3 & -4 +3+1 & \\
 -2+2 & 3+3 & 2-3+1 \\
 -2+2 & -3+3 & 2+3+1
 \end{array} = \frac16 \small \begin{array} {|r|r|r|r}
 6 & 0 & 0 & \\
 0 & 6 & 0 \\
 0 & 0 & 6
 \end{array} = I
$$
(So the textbook-solution should be commented with the required $0^0$-definition)

I used the diagonalization-procedure of Pari/GP for this and to get a textform for this answer:

M = [2,2,0;1,2,1;1,2,1]
tmpM=mateigen(M)
tmpW=tmpM^-1
tmpD=tmpW*M*tmpM
MPow = tmpM*matdiagonal(['a,'b,'c])*tmpW
\\ on text-level we have later to replace a by \alpha^n and so on


问:
Is vanishing Gradient implies locally constant?
It is given that $f\in C^{\infty}(\mathbb R^2)$ and $f$ is constant along one direction . At some point $(a,b)\ne (0,0)$ , function vanishes as well as its gradient vanishes . Can we conclude that function is at least locally constant there ?

Any Help/ Hint will be greatly appreciated

答: $f(x,y)=x^2$ with $(a,b)=(0,0)$ satisfies your condition, but it doesn't seem to meet your expectations.


问:
Invertibility of second derivative operator

Let $H$ be the Sobolev space of twice $L^2$-differentiable functions, and define the operator $T:H\to L^2(\mathbb R/2\pi\mathbb Z)$ by
$$
 T=-\frac{d^2}{dx^2}.
 $$

Prove that if $\lambda\in\mathbb C$ is not the square of any integer, then the map $\lambda I-T$ is invertible.

In particular, prove that if also $\lambda<0$ then $\|(\lambda I-T)^{-1}\|\leq1/|\lambda|$.

I'm not sure how to attempt this problem. I began by writing the definitions I know:

$L^2(\mathbb R/2\pi\mathbb Z)$ is the space of square integrable, $2\pi$-periodic functions

$h_2\subset\ell^2(\mathbb Z)$ is the space of sequences $(x_n)_{n\in\mathbb Z}$ such that $\sum_{n\in\mathbb Z}(1+n^2)^2|x_n|^2<\infty$.

$F:L^2(\mathbb R/2\pi\mathbb Z)\to\ell^2(\mathbb Z)$ is the Fourier map defined by

$$
Ff=\left(\langle f(x),(2\pi)^{-1/2}e^{inx}\rangle_{L^2}\right)_{n\in\mathbb Z}=\left(\frac1{\sqrt{2\pi}}\int f(x)e^{-inx}dx\right)_{n\in\mathbb Z}
$$

$H\subset L^2(\mathbb R/2\pi\mathbb Z)$ is the function space defined by $H=\{f\in L^2(\mathbb R/2\pi\mathbb Z):Ff\in h_2\}$.

Help or hints on this question? Thanks!

答:
Let $Tf=if'$ on $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f : [0,2\pi]\rightarrow\mathbb{C}$ with $f'\in L^2[0,2\pi]$, and suppose $g\in L^2[0,2\pi]$ is in the domain of $\mathcal{D}(T^*)$. Then $h=T^*g$ is in $L^2[0,2\pi]$ and satisfies
$$
            \int_0^{2\pi}if'(\theta)\overline{g(\theta)}d\theta = \int_0^{2\pi}f(\theta)\overline{h(\theta)}d\theta,\;\;\; h\in\mathcal{D}(T).
$$
For example, let $f$ be $0$ on $[0,x-h]$, be $1$ on $[x,y]$, and be $0$ on $[y+k,2\pi]$, and be linear on $[x-h,x]$ and $[y,y+k]$ so that $f$ is continuous on $[0,2\pi]$. Then
$$
                 f'=\frac{1}{h}\chi_{[x-h,x]}-\frac{1}{k}\chi_{[y,y+k]}
$$
and
$$
         \langle if',g\rangle=\langle f,ig'\rangle
$$
Letting $h,k\downarrow 0$ gives
$$
    \lim_{h,k\downarrow 0} \left(  \frac{1}{h}\int_{x-h}^{x}g(t)-\frac{1}{k}\int_{y}^{y+k}g(t)dt \right)=-i\int_x^y f(t)dt, \\
            g(x)-g(y)=-i\int_x^y g(t)dt
$$
Therefore, $g$ is absolutely continuous with $g'=if$ a.e. on $[0,2\pi]$. So $T$ is self-adjoint on $\mathcal{D}(T)$.

The estimate that you want is a general consequence of having a positive self-adjoint operator $T$, meaning that $T^*=T$ and $\langle Tx,x\rangle \ge 0$ for all $x\in\mathcal{D}(T)$. For $\lambda < 0$, and $x\in\mathcal{D}(T)$,
$$
             \langle (T-\lambda I)x,x\rangle \ge -\lambda\langle x,x\rangle \\
          (-\lambda)\|x\|^2 \le \|(T-\lambda I)x\|\|x\| \\
             (-\lambda)\|x\| \le \|(T-\lambda I)x\|.
$$
Setting $x=(T-\lambda I)^{-1}y$ gives
$$
             \|(T-\lambda I)^{-1}y\| \le \frac{1}{(-\lambda)}\|y\|.
$$
Therefore, $\|(T-\lambda I)^{-1}\| \le 1/(-\lambda)$ for $\lambda\in\mathbb{R}$, $\lambda < 0$.


问:
Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?
This is a follow-up question of a previous one.

Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?

It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.

答:
By transfer, every element in $^*\mathbb Q$ is a quotient of two elements from $^*\mathbb Z$. By transfer, not every element in $^*\mathbb R$ is the quotient of two elements from $^*\mathbb Z$. So, the two enlargements are not equal. </p>

<p>What you are alluding to (I think) in the question is that both enlargements give rise the same completion of $\mathbb Q$, namely $\mathbb R$. </p>

Title: Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?

Body:
By transfer, every element in $^*\mathbb Q$ is a quotient of two elements from $^*\mathbb Z$. By transfer, not every element in $^*\mathbb R$ is the quotient of two elements from $^*\mathbb Z$. So, the two enlargements are not equal.

What you are alluding to (I think) in the question is that both enlargements give rise the same completion of $\mathbb Q$, namely $\mathbb R$.


问:
Extension of a self-diffeomorphism of unit ball
Let $B$ be the unit open ball in $\mathbb R^n$ and $\phi$ a self diffeomorphism of $B$. Can we find a self diffeomorphism $f$ of $\mathbb R^n$ such that
$$
f=\phi
$$
on $B$?

答: Certainly not, without making some further assumptions about $ f $.  Consider, for example $ f(x) = \left( 1 - \sqrt{1 - |x|^2} \right) \ x $.  Any extension cannot be differentiable on the boundary $ |x| = 1 $.


问:
prove that $\phi(a+b \sqrt{2})$=$ \begin{pmatrix}{} a & 2b \\ b & a \end{pmatrix}$ is an isomorphism
Prove that  $\phi:R \longrightarrow R^{\prime}$ where $R=\lbrace a+b \sqrt{2}:a,b\in \mathbb{Z}\rbrace$ and $R^{\prime}= \lbrace \begin{pmatrix}{}
a & 2b \\ b & a \end{pmatrix}:a,b\in \mathbb{Z} \rbrace$ given by $\phi(a+b \sqrt{2})$=$ \begin{pmatrix}{}
a & 2b \\ b & a \end{pmatrix}$ is a ring isomorphism between $R$ and $R^{\prime}$.
I tried to do it, and I concluded that $\phi$ is a group homomorphism with the $+$ operation, but it is not a group homomorphism with the product of two elements.

Note that the sum and product are the usual in $M_{2 \times 2}(\mathbb{Z})$, and the sum and the product in $R$ are the same as that in $\mathbb{Z}$.
Can someone tell me if I had an error while doing the comparison between $\phi((a+b\sqrt{2})(c+d\sqrt{2}))$ and $\phi(a+b\sqrt{2}) \phi(c+d\sqrt{2})$?

Body:
$$(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(bc+ad)\sqrt2$$

$$\pmatrix{a&2b\\b&a}\pmatrix{c&2d\\d&c}=\pmatrix{ac+2bd&2bc+2ad\\bc+ad&ac+2bd}$$


问:
Conditional Probability of $P(X|X^2)$ for uniform distribution.
Assume $X \sim U[-1,1]$ is a uniform distribution. We can tell intuitively that
$$P(X=k|X^2=x^2) =
  \begin{cases}
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=-x  \\
    1/2 \;\;\;\;\;\;\;\text{if} \;\; k=x
  \end{cases}$$

But I want to confirm this using regular conditional probability. However, when I try the definition as $P(X\in A|Y=y):=\lim\limits_{\{Y=y\}\in U}\frac{P(A\cap U)}{P(U)}$, the conditional probability is $0$ since for example if trying to compute $P(X=2|X^2=4):=\lim\limits_{\epsilon  \to 0}\frac{P(X=2 \;\cap\; X^2 \in [4-\epsilon,4+\epsilon])}{P(X^2 \in [4-\epsilon,4+\epsilon])}$, which is just $0$ because $X$ is continuous distribution. Maybe I got something wrong. Please help me compute this conditional probability formally.

答:
I'm not convinced something like $P(X=1|X^2=1)$ is well$-$defined if $X\sim \mathcal{U}(-2,2)$, and it's not because $P(X^2=1)=0$; it is possible (in fact, common) to condition on events with vanishing probabilities in higher$-$dimensional space. For example, if $(X,Y)\sim f_{XY}$ is a continuous random vector that's supported on $\mathbb{R}^2$, then $$P(X<0|Y=X^2)=\frac{\int_{-\infty}^{0}f(t,t^2)\sqrt{4t^2+1}dt}{\int_{-\infty}^{\infty}f(t,t^2)\sqrt{4t^2+1}dt}$$ even though $P(Y=X^2)=0$.

Here is why I don't believe something like $P(X=1|X^2=1)$ is well defined when $X \sim \mathcal{U}(-2,2)$. Notice how conditioning on $\{X^2=1\}$ is equivalent to conditioning on the event that $X$ belongs to the finite set $\{-1,1\}$. That being said, define intervals $I_1,I_2\subseteq (-2,2)$ by $I_1=(-1-\epsilon_1,-1+\epsilon_1)$ and $I_2=(1-\epsilon_2,1+\epsilon_2)$ where $\epsilon_1,\epsilon_2$ are very small positive real numbers. Notice $$P(X\in I_2|X\in I_1 \cup I_2)=\frac{\epsilon_2}{\epsilon_1 +\epsilon_2}$$ Now if $P(X=1|X^2=1)$ were well$-$defined, we would certainly have $$P(X=1|X^2=1)=\lim_{(\epsilon_1,\epsilon_2)\rightarrow (0^+,0^+)}\bigg(\frac{\epsilon_2}{\epsilon_1+\epsilon_2}\bigg)$$ But this limit doesn't exist.

My original intuition made me think that if $X\sim f_X$ is a continuous random variable supported on $\mathbb{R}$ and if $S\subseteq \mathbb{R}$ is a finite set, then for any $s\in S$ we have $$P(X=s|X\in S)=\frac{f_X(s)}{\sum_{x\in S}f_X{(x)}}$$ but I no longer believe this is the case based off my previous comments. Any thoughts?


问:
Question about Column Space and Row Space of a Matrix
If $rank(A)=1$ and $Null(A) \perp Col(A)$, then we have that $dim(Col(A))=1$ right? Then that also means that $dim(Row(A))=1$. So is it true that $$Col(A)=Row(A)$$ since they are both perpendicular to $Null(A)$ and they both have dimension of $1$?

答:
By the rank-nullity theorem, $\dim(\text{Col}(A))+\dim(\text{null}(A))=n$, so if $\dim(\text{Col}(A))=1$, then $\dim(\text{null}(A))=n-1$.

If you know what the orthogonal complement of a vector subspace $W$ is, which is denoted by $W^\perp$, then you probably know $\dim(W^\perp)+\dim(W)=n$, which in this case implies $\dim((\text{null}(A))^\perp)=1$. Since any vector subspace orthogonal to $\text{null}(A)$ must be a subspace of $(\text{null}(A))^\perp$, then $\text{Col}(A),\text{Row}(A)\subseteq(\text{null}(A))^\perp$. Since the three of them have dimension $1$, they are all equal.


问:
Is there a theory of extensible definitions?
We can define $+$ as a function $\mathbb{N}^2 \rightarrow \mathbb{N}$, and then prove:

Theorem 1. The range of $+$ is $\mathbb{N}$.

If we later wish to extend $+$ to a function $\mathbb{Z}^2 \rightarrow \mathbb{Z}$, then we have a problem; according to our new definition, the theorem we proved isn't true any more. To be rigorous, we must therefore define a new operation $ +' : \mathbb{Z}^2 \rightarrow \mathbb{Z}$. But this is clumsy. In practice, what we usually do is refrain from talking about $+$ in certain ways, and this allows us to extend the definition. So for instance, the following sentence is fine:

Theorem 2. For all $a,b \in \mathbb{N}$ it holds that $a \leq a+b$.

By writing sentences like Theorem 2 only, and refraining from sentences like Theorem 1, we're able to extend $+$ without invalidating our previous theorems, as opposed without having to introduce a new operation $+'$.

So what I'm looking for is a rigorous approach to the idea that, by pairing a definition with some additional information about how that definition is allowed to be used, it becomes extensible. And the more stringent we are about how a definition can be used, the more extensible it becomes.

Is there a theory of extensible definitions?

P.S. I don't know which tags are appropriate for this question. Please feel free to retag it.

答:
Perhaps what you're looking for is model theory. Model theory is a branch of logic which deals with structures defined in particular languages (which just consist of some symbols and rules of syntax). As an example, let's suppose your language $\cal{L}$ consists of the constant symbol "$0$", a two-place function symbol "+", and a binary predicate "$\leq$". Then on the structure $\mathcal{N} = \langle \mathbb{N},0^\mathbb{N},+^\mathbb{N},\leq^\mathbb{N} \rangle$, where "$s^\mathbb{N}$" denotes the standard interpretation of these symbols on $\mathbb{N}$, your (i) and (ii) hold in this structure, i.e. $\mathcal{N} \vDash \forall x\forall y(x \leq x + y)$ and $\text{ran}(+^\mathbb{N}) = \mathbb{N}$. Now define another structure $\mathcal{Z} = \langle \mathbb{Z},0^\mathbb{Z},+^\mathbb{Z},\leq^\mathbb{Z} \rangle$ in a similar way. In this case, it's not longer the case that $\mathcal{Z} \vDash \forall x\forall y(x \leq x + y)$, but what we have is $\mathcal{Z} \vDash \forall x,y \geq 0(x \leq x + y)$ (where "$\geq$" is an appropriate abbreviation), so (i) is still good. As for (ii), we then just note that, on these structures, $+^\mathbb{Z} \mid \mathcal{N} = +^\mathbb{N}$, where "$\mid$" is now denotes restriction to a substructure. So the range of $+^\mathbb{Z} \mid \mathcal{N}$ is $\mathbb{N}$, as before, even though the range of full $+^\mathbb{Z}$ is not $\mathbb{N}$.

In addition to being a good tool for talking about all these sorts of things, model theory does have a number of results regarding "definability" in a model, i.e. when a model can define certain objects/properties and when it can't, and also what things follow from such definable objects/properties. There are also a number of "preservation" results, stating when certain sentences (usually of a certain syntactic form) remain true in extensions/substructures of the original structure you're looking at. So for instance, $\mathcal{Z}$ doesn't satisfy the sentence $\forall x \forall y (x \leq x + y)$ (since $y$ could be negative); model theory let's you prove something more general, viz. $\forall_1$ sentences (which have the syntactic form of a string of $\forall$s followed by an atomic formula) aren't necessarily preserved in extensions of structures, though they're preserved in substructures. For a nice introduction to the material, see Hodges's A Shorter Model Theory or Marker's Model Theory: An Introduction (note, though, these assume you're already familiar with the basics of mathematical logic).


问:
If there are two positive numbers then prove if there are two numbers $q$ and $r$ such that $a=bq+r$
Question:

Prove:

$a,b\in \Bbb Z^+\, \Rightarrow \exists q,r\, \ni a=bq+r\, :\, -\frac  b2 \lt r \leq \frac b2$

My attempt:

Let $a$ to be an odd number and $b$ to be an even number. Then we can write $b=2r,\, r\in \Bbb N$ and $a=2x+1,x\in \Bbb N\cup \{0\}$. Since we know $\text{Odd $\times$ Odd $=$ Odd}$, we can write $a=2x+1=r(2q+1)$ and $r$ must be an odd number. Then we get $a=2rq+r = bq+r$.

I'm stuck at here. This is the best proof i can think, but i can't show how is  $-\frac
b2 \lt r \leq \frac b2$. Can you help me please? I have no idea to show the second part. Thanks in advance

答:
Don't by distracted by even oddness.

This is just a restatement of the division theorem that there is unique $q,r$ so that $a=bq + r$ and $0 \le r < b$.

Case 1:  $r \le \frac b2$.  Then $a = bq + r$ and $-\frac b2 < 0 \le r \le \frac b2$.

Case 2:  $\frac b2 < r < b$.  (Note: $\frac b2$ can be a non-integer)  The we have:

$a  = bq + r = b(q+1) + (r -b)$.  But $\frac b2 < r < b$ so $-\frac b2 < r-b < 0 < \frac b2$.  So if we replace $q$ with $q' = q+1$ and $r$ with $r' = r -b$ then we have $a = bq' + r'$ and $-\frac b2 < r' < 0 < \frac b2$.

That's it.

......

Of course this does assume you have already proven the division algorithm.  If not.... you can prove it from scratch be instead of shooting for a positive remainder, shoot for a remainder between $-\frac b2$ and $\frac b2$.

Do you need to see how to do that?

=====

I don't know what class you are taking or to what degree I should rely on induction of well ordered principal of natural numbers is.

But first note:  $b > 0$ so $0*b < 1*b < 2*b < 3*b < 4*b <.......$ is an infinite chain of inequalities that is unbounded.  $a > 0$ and $a$ is a finite value so $a$ has to fit in there somewhere so there is some $m*b$ where $m*b \le a < (m+1)b$. ($m$ could be zero if $a < b$.)

(That's actually too informal for a basic number theory class but is the most intuitive and obvious explaination.  To say the same thing formally would be to let $A= \{n\in \mathbb N| nb > a\}$.  We know $A$ is non-empty as $ab \ge a$ so $(a+1)b > a$ and $a+1 \in A$.  So by well-ordered principal $A$ has a least element $L$. Let $m = L-1$.  Then $m \not \in L$ so $mb \not > a$ so $mb \le a$.  And $m+1 = L\in A$ so $(m+1)a > a$.)

(Alternatively, if $B = \{w\in \mathbb Z| wb \le a\}$ we know $B$ is not empty because $0 \in B$ and that if $k \ge a+1$ then $kb=(a+1)a > a$ so $k \not \in B$ so $B$ is bounded above.  So $\sup B$ exists.  If $\sup B$ is not in $B$ we get a contradiction: Between $\sup B -\frac 12$ and $\sup B$ there must be an integer $w$ in $B$. But $w < \sup B$ so there must be another element $w'$ in $B$ between $w$ and $\sup B$. so $\sup B-\frac 12 \le w < w'< \sup B$ but we can't have two integers withing a half unit of each other.  So $\sup B\in B$ so $m=\sup B$ is the integer we want.)

Let $d= a- mb$.  Then $0 \le d < b$.

If $0\le d \le \frac b2$ then let $q=m$ and let $r= d$.  Then $a=qb +r; -\frac b2 < r \le \frac b2$.

If $b> d > \frac b2$ then let $q = m+1$ and $r = d-b$ os $0> r >-\frac b2$.  Then $a=qb + r$ and $-\frac b2 < r \le \frac b2$.


问:
Reference request: Where does this proposition about ellipses appear in the literature?
In this question I conjectured that a simple proposition about ellipses holds. In the accepted answer, "Chrystomath" proves it.

Is this in some published refereed source?

quote:

Consider a closed bounded set with non-empty interior in the plane. Suppose it is strictly convex, i.e. every point between two of its points is one of its interior points. This entails that a line that intersecting its boundary but not its interior intersects it at only one point. Call such a line a tangent line. It follows that for every tangent line, there is  exactly one other tangent line parallel to it. Suppose that for every line parallel to those two and between them, the midpoint of the intersection of that line with our closed bounded convex set is on the line connecting the two points of tangency.
Does it follow that our closed bounded set is the convex hull of an ellipse?

答:
Posted and answered on mathoverflow.net

Yes, this characterization is a theorem proven by Blaschke in "Kreis
und Kugel" (1916). The theorem has a higher dimensional version,
characterizing ellipsoids as the unique strongly convex bodies with
the property that all centroids of codimension 1 sections that are
parallel to a fixed plane lie on a line. See the paper
"Characterizations of ellipsoids by section-centroid location" by
Meyer and Reisner.


问:
All smooth functions have Taylor expansion?
The theories are:

$f(x)$ has Taylor expansion equals the remainder of Taylor polynomial converge to $0$.

a smooth function such as $f(x) = e^{-1/x^2}$, $x>0$, $f(x) = 0$, $x \leqslant 0$ do not has Taylor expansion near $0$.

However, I think about these propositions and find:

smooth - exist Taylor polynomial - select Peano remainder = $o(x^n$) - the remainder must converge to $0$ - smooth function must have Taylor expansion near $0$

I will be appreciated if someone can point out the error in this infer.

答:
Assume $f(x)$ as a smooth function and consider its $Maclaurin formula$,

then for every n:
$f(x) = \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k + o(x^n)$

select a sufficiently small neighborhood of zero, then:
$\lim\limits_{n\to+\infty}R_{n}(x) = \lim\limits_{n\to+\infty}(f(x) - \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k)=\lim\limits_{n\to+\infty}o(x^n)=0$

then smooth function $f(x)=\lim\limits_{n\to+\infty}\sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k=\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$.

then smooth function $f(x)$ has Taylor expansion.

the above conclusion is wrong, because $e^{-1/x^2}$ is smooth and do not have Taylor expansion. However, what's the error during my infer?


问:
Show equivalence (combinations)
Show that for integers $1 \leq r < n$ we have

${n\choose r-1} = {n+2\choose r+1} - 2\times{n+1\choose r+1} + {n\choose r+1}$

I have expanded the RHS thusly, but I'm not sure where to go from here algebraically:

$\frac{(n+2)!}{(r+1)!(n-r+1)!} - \frac{2(n+1)!}{(r+1)!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!}$

Suggestions?

答:
$\frac{(n+2)!}{(r+1)!(n-r+1)!} - \frac{2(n+1)!}{(r+1)!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!}$

put them under a common denominator of $(r+1)!(n-r+1)!$

$=\frac{(n+2)!-2(n-r+1)(n+1)! + (n-r+1)(n-r)n!}{(r+1)!(n-r+1)!}$

$=\frac{(n+2)(n+1)n!-2(n-r+1)(n+1)n! + (n-r+1)(n-r)n!}{(r+1)!(n-r+1)!}$

So factor the $n!$

$=\frac{n!((n+2)(n+1)-2(n-r+1)(n+1)+(n-r+1)(n-r))}{(r+1)!(n-r+1)!}$

$=\frac{n!(n^2+3n+2-2(n^2+n-rn-r+n+1)+n^2-2rn+r^2+n-r)}{(r+1)!(n-r+1)!}$

$=\frac{n!(r^2+r)}{(r+1)!(n-r+1)!}$

$=\frac{n!(r+1)r}{(r+1)!(n-r+1)!}$

$=\frac{n!}{(r-1)!(n-r+1)!}$

$=$ $n \choose r-1$

But a better way to solve the problem would be to use the recurrence relation for binomial coefficients:

${n \choose k}={n-1 \choose k}+{n-1 \choose k-1}$

So we have

${n \choose k}-{n-1 \choose k}={n-1 \choose k-1}$

We can apply this to the rhs of your initial equation:

${n+2\choose r+1} - 2\times{n+1\choose r+1} + {n\choose r+1}$

$={n+2\choose r+1} - {n+1\choose r+1} -{n+1\choose r+1} + {n\choose r+1}$

applying our recurrence relation to the first two terms we get

$={n+1\choose r} -{n+1\choose r+1} + {n\choose r+1}$

$={n+1\choose r} -({n+1\choose r+1} - {n\choose r+1})$

applying the recurrence relation to the 2nd two terms we get:

$={n+1\choose r} -{n\choose r}$

applying the recurrence relation again we get:

$={n\choose r-1}$

Title: Show equivalence (combinations)


答:
I highly recommend the beautiful combinatorial approaches given in the comments. However, just to finish out the algebraic approach you've started. First, judiciously expand the first few terms of the factorials $(n+2)!$ and $(r+1)!$ and "un-expand" some of the other factorials:

$\frac{(n+2)!}{(r+1)!(n-r+1)!} -2 \frac{(n+1)!}{(r+1)!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!}=$

$\frac{(n+2)(n+1)n!}{(r+1)(r)(r-1)!(n-r+1)!} - 2\frac{(n+1)n!(n-r+1)}{(r+1)(r)(r-1)!(n-r+1)!} + \frac{n!(n-r+1)(n-r-1)}{(r+1)(r)(r-1)!(n-r+1)!}=$

$\left(\frac{(n+2)(n+1)}{(r+1)(r)} - 2\frac{(n+1)(n-r+1)}{(r+1)(r)} + \frac{(n-r+1)(n-r-1)}{(r+1)(r)}\right)\frac{n!}{(r-1)!(n-r+1)!}= $

$\left(\frac{(n+2)(n+1)}{(r+1)(r)} - 2\frac{(n+1)(n-r+1)}{(r+1)(r)} + \frac{(n-r+1)(n-r)}{(r+1)(r)}\right)\binom{n}{r-1}$

What's left to show then is that

$$\left(\frac{(n+2)(n+1)}{(r+1)(r)} - 2\frac{(n+1)(n-r+1)}{(r+1)(r)} + \frac{(n-r+1)(n-r)}{(r+1)(r)}\right)=1.$$

Title: Show equivalence (combinations)

Body:
Show that for integers $1 \leq r < n$ we have

${n\choose r-1} = {n+2\choose r+1} - 2\times{n+1\choose r+1} + {n\choose r+1}$

I have expanded the RHS thusly, but I'm not sure where to go from here algebraically:

$\frac{(n+2)!}{(r+1)!(n-r+1)!} - \frac{2(n+1)!}{(r+1)!(n-r)!} + \frac{n!}{(r+1)!(n-r-1)!}$

Suggestions?


问:
Upper bound for the total curvature of a shortest path in the boundary of a convex polyhedron in $\mathbb{R}^3$.
Problem : Consider finitely many points in
$\mathbb{R}^3$. The boundary of the convex hull is $\Sigma$. When
$f_i$ is a face and $u_i$ is unit outnormal to $f_i$, then assume
that $$(-u_1)\cdot u_i >\eta>0 \ \ast$$ for all $i>1$ where $\cdot $ is an
inner product. When $P$ is a shortest path in $\Sigma -f_1$,

then
the total curvature of $P$ is smaller than $\frac{\pi}{\eta}$.

I need another proof, because the following well-known proof is not easy.

Proof : If $P$ contains line segments $[z_iz],\ [z z_{i+1}]$ in faces
$f_i, \ f_{i+1}$ where $z\in f_i\bigcap f_{i+1}$, then we define $$ x_i =\frac{z-z_i}{|z-z_i|},\
x_{i+1} = \frac{z_{i+1}-z}{|z_{i+1} -z|}$$

Similarly, we have any line segments $[\overline{z}_iz],\ [z
\overline{z}_{i+1}]$ in faces $f_i, \ f_{i+1}$ s.t. they are
orthogonal to $f_i\bigcap f_{i+1}$. Similarly we have
$\overline{x}_i,\ \overline{x}_{i+1}$. Then there is $$ x_i-x_{i+1}
= C_i (\overline{x}_i-\overline{x}_{i+1} )=\lambda_i( u_i + u_{i+1})
$$ where $C_i,\ \lambda_i>0$ : Here the second inequality is followed from drawing directly. If we set $x_i=A \overline{x}_i +B u_i\times \overline{x_i}$, then note that $x_{i+1}=A \overline{x}_{i+1} +B u_{i+1}\times \overline{x}_{i+1}$. Hence the first inequality is followed.

Hence $ -u_1\cdot( x_i-x_{i+1} ) \geq \lambda_i (2\eta ) $ Hence \begin{align*}
 \angle \
(x_i,x_{i+1}) &\leq \frac{\pi}{2}|x_i-x_{i+1}| \\
&=\frac{\pi}{2} |\lambda_i (u_i+u_{i+1}) |
 \\&\leq \pi \lambda_i
 \\&\leq \pi\frac{-u_1\cdot (
 x_i-x_{i+1}) }{2\eta } \end{align*}

Hence $\sum_{i=1}^{n-1} \angle(x_i,x_{i+1}) \leq \frac{\pi}{2\eta} |u_1\cdot (x_1-x_n)| \leq \frac{\pi}{\eta}$.

答:
Consider the case where $\Sigma$ is smooth Riemannian surface in
$\mathbb{R}^3$ of positive Gaussian curvature, homeomorphic to a
sphere. Define a region $D =\bigg\{ x\in \Sigma \bigg| N(x)\cdot v >\eta \bigg\}$
for some $|v|=1$ and $\eta>0$ where $N$ is unit out normal to
$\Sigma$.

Assume that $c:[0,l]\rightarrow D$ is a $\Sigma$-minimizing geodesic of unit
speed in $D$. Hence we have $|w(t)|=1,\ v\perp w(t)$ s.t. $$
N(t)=a(t)w(t) + b(t) v,\ b\geq \eta ,\ a^2+b^2=1$$

If $k$ is a curvature for $c$, then $c''(t)=-k(t)N(t),\ k>0$. Hence
\begin{align*} -c''\cdot v&=k b \geq k\eta \\
\int_0^l\ k(t)dt &\leq \int_0^l\ \frac{-c''\cdot v}{\eta} \ dt \\&=
-\frac{1}{\eta} c'(t)\cdot v \bigg|_0^l \\&\leq \frac{2}{\eta }
\end{align*}


问:
The larger the commutator subgroup is, the "less abelian" the group is
On the page of commutator subgroup Wikipedia says that "the larger the commutator subgroup is, the "less abelian" the group is."

I know that for every group $G$ and $ N\trianglelefteq G$ the quotient $G/N$ is Abelian if and only if $[G,G] \le N$.

But what does "the larger the commutator subgroup is, the "less abelian" the group is" mean?

答: It's a somewhat vague comment, which nevertheless makes some sense.  Since the commutator is, as you noticed, the minimum you need to "mod out by" in order to get an abelian group, its size does indeed correspond to how far the group is from being abelian.</p>


问:
Permutations with repetition beginning and/or ending with members of a subset
For the sake of this example, take the random letters OOOAEESTMLHHXX. How would I go about finding all of the permutations, accounting for repetition, that start and end with members of a subset, in this case any of the vowels?

I'm aware of the formula for permutations with repetition n!/(n1!n2!...) where n1/n2... are the counts of each item, but I can't figure out how to split up the cases to not include unique permutations multiple times.

I originally wanted to do cases for each vowel where it started at the beginning (O___) or end (___O), add them, then subtract away the cases where the string begins AND ends with the particular vowel (since O___O would be included in each of the added cases, we take one of them out). The problem I see here is that for future vowels, I'm repeating cases like A___O, which I could be counting again but already included when calculating for O.

I also thought about calculating every case that begins and ends with consonants (which I still think is probably the way to go), but I can't see any way other than manually pairing every case of consonants for the front and back of the word, then using the repetition formula, which seems like SO MANY calculations.

Is there something easier that I'm missing?
Appreciated highly.

答:
The key to this problem is to combine elegance and brute force.

The string is "OOOAEESTMLHHXX".

There is one constraint: the first and last letters must each be
either "O", "A", or "E".  Although the string could (for example)
begin and end with "O", since there is only 1 "A", "A" can not
appear at both ends of the string.

This means that there are the letter groups "OOO", "EE", "HH", and "XX".

There are also the single letters "A", "S", "T", "M", and "L".

Therefore, the string can vary in length from 2 through 14, inclusive.

As I see it, you are going to need 6 variables.

Let $L_O \equiv ~$ the number of "O" letters used.

Let $L_E\equiv ~$ the number of "E" letters used.

Let $L_H \equiv ~$ the number of "H" letters used.

Let $L_X \equiv ~$ the number of "X" letters used.

Let $L_Z \equiv ~$ the number of letters used from the group "A", "S", "T", "M", and "L"

Let $T \equiv ~$ the total number of letters used in the string.

The preliminary portion of this analysis will resemble
the following problem.

When you study this problem you will notice a few points.

There are two ways to attack the # of solutions -
[Stars and Bars + Inclusion - Exclusion] or generating functions.

The # of solutions that you generate will actually be a function of $T$, which
represents the variable length of the string (i.e. from 2 through 14 inclusive).

All this analysis is doing is identifying all of the possible non-negative
integer solutions for each of the variables $L_O, L_E, L_H, L_X, L_Z.$

At each possible value for $T$, you must focus on each of the individual
solutions separately.

You then have to compute a formula that represents the number of possible
strings, as a function of the 6 variables.

Note that you have to be very careful to prune away any of the solutions
that do not have at least two vowels.  Since the # of vowels are actually
reflected in the interaction between $L_O, L_E,~$ and whether $a$ makes an
appearance, you have a couple of options here.

You could split "a" off from the other group and make it its own variable.

Or you could decline to split "a" off, and simply ignore any strings that
don't begin and end with a vowel.

Believe it or not, unless someone has a more elegant approach, this is where
the fun starts.  For a given set of values for the variables,
you have to enumerate the # of possible distinct strings that can be formed.

I am going to give you a starter enumeration.  However, you will have to
polish the formula re the first and last letters need to begin with a vowel.

You can think of the # of strings as a fraction which equals

$$\frac{T!}{D\text{(enominaor)}}.$$

Here,

$$D = (L_O!) \times (L_E!) \times (L_H!) \times (L_X!).$$

The reason for this is because (for example) if the "O" appears in a specific
group of $L_O$ slots, you have to compensate for it being counted $(L_O!)$ times.
This approach is the easiest way to adjust for the over-counting.

I emphasize: I am leaving the constraint that both ends of the string
are vowels to you.

Obviously, you are going to have to program your computer very carefully.

There is an obvious alternative.  Since it would be ridiculous to try to
count the number of solutions (without using a computer) using a whole group of
elegant formulas, you have two alternatives.

Come up with a more elegant approach than I have suggested.

Abandon all elegance and simply use brute force to have the computer
spin through all of the various possible strings, seeing which ones satisfy
the constraints.  Here, the # of possible strings would be strictly less than
or equal to the following:

$$\sum_{k=2}^{14} ~\binom{14}{k} \times k!$$.

Addendum

Keep in mind that I am totally ignorant of both generating functions and exponential generating functions.  I suspect that a more elegant approach is available through exponential generating functions.  However, I could be mistaken.

You would have to first study generating functions and then
study exponential generating functions to make this work.  Further, you would somehow have to include the constraint that the string begins and ends with a vowel.


问:
Geometric Nakayama's Lemma
This is Vakil 13.7 E, self-study.

We are to show that if $X$ is a scheme and $\mathcal F$ is a finite type quasicoherent sheaf on $X$, then if  $p \in U \subset X$ is an open neighborhood of $p$ and $a_1, ... , a_n \in \mathcal F(U)$ have images generating the fiber $\mathcal F_p \otimes \kappa(p)$, then there must be an affine open neighborhood $p \in \operatorname{Spec} A \subset U$ such that the $a_i$ each restricted to $\operatorname{Spec} A$ generate $\mathcal F(\operatorname{Spec}A)$ as an $A$-module, and for each $q \in \operatorname{Spec} A$, the (images of) the $a_i$ generate $\mathcal F_q$ as an $\mathcal O_{X, q}$-module.

Here is my attempt, but something feels off about it:

If we assume $U$ is already an affine open $\operatorname{Spec}A$, then we know $\mathcal F$ is locally a finite type $A$-module $M$ on $U$. Then the fiber at $p$ is isomorphic to $M_p/pM_p$. Since being a finite type $A$-module is a local property, $M_p$ is a finite type $A_p$-module. Since $p$ is a prime ideal in $A$, by version 8 of Nakayama's Lemma from the Stacks Project's tag 07RC, $M_p$ is generated by the images of the $a_i$. Since $p$ was arbitrary, again by the localness of being finite type, $M$ is finitely generated by the $a_i$.

Using localness once more, $M_q$ is generated by the images of the $a_i$ for any $q \in \operatorname{Spec}A$.

Something about assuming $U$ was affine feels off, almost like I did not quite show what was asked. Also, we did not show that finite type was local in the notes thus far, only that $M$ is finite type over $A$ if and only if $M_{f_i}$ is finite type over $A_{f_i}$, where the $f_i$ generate $A$. I am not sure this allows me to conclude the same about localizing at a prime. It also feels like I used localness "too much."

答:
You have already reduced to the affine case (and there is nothing wrong with your reduction) so I will write things purely in terms of rings and modules.

We are in a situation where we have a ring $A$, a finitely-generated $A$-module $M$, and elements $a_1,\dots,a_n\in M$ whose images generate $M_p$ as an $A_p$-module. Our goal is to find some $g\in A\smallsetminus p$ for which the images of $a_i$ generate $M_g$ as an $A_g$-module, because then $\operatorname{Spec}(A_g)$ is the affine open neighborhood of $p$ you are looking for.

Now you should use the fact that $M$ is finitely generated over $A$ and write down a generating set, say $x_1,\dots,x_m$, and notice that the $x_i$ will generate $M_g$ over $A_g$ for any $g$ we choose. Now we know that the $a_i$ generate $M_p$ over $A_p$, so for each $i$ we can write
$$x_i=\sum_j \big(\frac{b_{ij}}{s_{ij}}\big)a_j\:\:\:\:\:\:\:\:\text{in $M_p$}$$
for some elements $b_{ij}/s_{ij}\in A_p$, so $s_{ij}\in A_p\smallsetminus p$. I claim we should take $g:=\prod_{i,j}s_{ij}$; since the $x_i$ already generate $M_g$ over $A_g$, you just need to verify that each of these can be generated with the $a_i$ in $M_g$, and this is easy to see by our choice of $g$.

Edit: the comments below are correct, one needs a tweak to the argument above that I'm too lazy to type out right now.

Title: Geometric Nakayama's Lemma


问:
Tensor Product of a Vector Space & the corresponding Field extension
Let $K$ be a field and $V$ a vector space of finite dimension $n$ over $K$  also suppose $K\subset{F}$ and let ${V_F}={F}\otimes_K V$ the tensor product over $K$ of $V$ and ${F}$.

I have proved that $V_F$ forms a vector space over $F$ naturally but I need to prove that $\dim _F  V_F \leq \dim_K V_F$. I  also have to find the example when the strict inequality holds.

Kindly help!!

Any hints will be appreciated.

thanks & regards

答:
$$\dim V_F(F) \times \dim (F(K)) = \dim V_F(K)$$

See I have thought of like extension of fields. Is it helpful?


问:
Number of surjections from 8-element set to 6-element set
My attempt:

Surjections from 8-element set to 6-element can be divided into 2 cases.

Three of the elements from 8-element set are mapped to single element in 6-element set.

Four of the elements from 8-element set, which are divided into groups of two, are mapped to two elements in 6-element set.

For the first case, we have ${8}\choose{3}$ $\cdot$ $6!$ possible surjections.

The second case has ${8}\choose{2}$ $\cdot$ ${6}\choose{2}$ $\cdot$ $6!$ possible surjections.

In total, there are ${8}\choose{3}$ $\cdot$ $6!$ + ${8}\choose{2}$ $\cdot$ ${6}\choose{2}$ $\cdot$ $6!$
$= 342720$ possible surjections.

Is this a correct approach to solve the problem?

Any advice would be appreciated.

Thank you!

答:
Almost perfect.  You simply have to apply the $(1/2)$ scalar to the second case to compensate for over-counting.

For example:
If you are sending $\{a,b,c,d,e,f,g,h\}$ to $\{1,2,3,4,5,6\}$

your present formula counts twice $(a,b) \to 1, ~~(c,d) \to 2$.

Your approach is very elegant in construing double or triple letters as a single unit. The common trap here is if two of your units each have a size greater than 1, and each are the same size,
as in your second case, then you get over-counting.


问:
Calculate the joint PDF of the following random variables.
Suppose $\theta$ and $R$ are independent random variables, with $\theta$ being Uniform$(−\pi/2,\pi/2)$ and $R$ having the probability density function given by:

$$f_R(r)=\frac{2r}{(1+r^2)^2} \:\text{for } r>0, \quad f_R(r)=0 \:\text{otherwise}.$$

Determine the joint PDF for $X=R\cos\theta$ and $Y=R\sin\theta$.

It is simple enough to calculate the cumulative distribution functions of $\theta$ and $R$, having

$$F_\theta(x)=\frac{1}{\pi}x+\frac{1}{2} \:\text{ on }\: (-\pi/2,\pi/2),\quad \text{and } \:F_r(x)=\frac{x^2}{x^2+1}\:\text{ on }\:(0,\infty),$$

with my aim being eventually to calculate $F_{XY}(x,y)$ from which, by taking partial derivatives, I would be able to obtain $f_{XY}(x,y).$

However, $F_{XY}(x,y)=\mathbb{P}[X=R\cos\theta\le x,Y=R\sin\theta\le y],$ and no matter how much I try to interpret this expression, I can't seem to find a way to bring it into terms of $F_r$ and $F_\theta$ as would lead (I imagine) to the solution.

Is this the correct way to tackle this sort of problem? If so, any hints?

答:
You only want the probability density function.   Thus you do not need to know what the cummulative density function is, just how to differentiate it, w.r.t. $x,y$.

Using $~\arctan:\Bbb R\mapsto (-\pi/2.. \pi/2)$

$$\begin{align}f_{\small X,Y}(x,y)&=\dfrac{\mathrm d^2 F_{\small X,Y}(x,y)}{\mathrm d x~\mathrm dy}\\[1ex]&=\dfrac{\mathrm d^2~F_{\small R,\Theta}(\surd (x^2+y^2),\arctan(x/y))}{\mathrm d x~\mathrm d y}\\[1ex]&=\left\lVert\dfrac{\partial\langle\surd(x^2+y^2),\arctan(x/y) \rangle}{\partial\langle x,y\rangle}\right\rVert \left.\dfrac{\mathrm d^2 F_{_{R,\Theta}(r,\theta)}}{\mathrm d r\,\mathrm d \theta}\right\vert_{r=\surd(x^2+y^2)\\\theta=\arctan(x/y)}\\[1ex]&=\left\lVert\dfrac{\partial\langle\surd(x^2+y^2),\arctan(x/y) \rangle}{\partial\langle x,y\rangle}\right\rVert f_{\small R,\Theta}(\surd(x^2+y^2),\arctan(x/y))\tag{1}\\[1ex]&=\begin{Vmatrix}\dfrac{\partial \surd(x^2+y^2)}{\partial x}& \dfrac{\partial\surd(x^2+y^2)}{\partial y}\\ \dfrac{\partial \arctan(x/y)}{\partial x}&\dfrac{\partial\arctan(x/y)}{\partial y}\end{Vmatrix}\,f_{\small R}(\surd(x^2+y^2))\,f_{\small\Theta}(\arctan(x/y))\\[1ex]&~~\vdots\end{align}$$

(1): This is known as the Jacobian Transformation. When $U=g(X,Y), V=h(X,Y)$ then: $$f_{\small X,Y}(x,y)=\left\lVert\dfrac{\partial \langle g(x,y), h(x,y)\rangle}{\partial\langle x,y\rangle}\right\rVert\,f_{\small U,V}(g(x,y),h(x,y))$$


问:
Is there any finite deductive system for propositional logic which only uses unary rules?
I'm not sure if this has ever been proven/disproven, but, assuming the usual grammar of propositional logic, is there any deductive system which derives exactly the tautologies of classical logic while only using finitely many unary rules and axiom schemes? This would of course be equivalent to proving a similar statement for usual kinds of intutionistic, minimal or
even subminimal logic.

By a unary rule, I mean two propositional formulas (premise and conclusion) built from variables and the usual connectives.

A concept of substitution needs to be implicitly assumed,
as well as syntactic rules.
The restriction to unary rules explicitly forbids the use of binary rules like modus ponens or usual conjunction introduction, as their premises consist both of an implication and its antecedent.

I don't think such a system exists, simply because I cannot think of any real "pattern" in true implications which would apply in such a case. Has any similar statement been proven or is there
some reasonably difficult approach to such questions?

答:
There exists a deductive system which derives exactly the tautologies of classical propositional logic while using finitely many at-most-unary rules and axiom schemata.

We will restrict our attention to classical propositional logic given by the two connectives $\neg, \rightarrow$, where other connectives are defined as abbreviations, as customary in Hilbert-style calculi (besides, a nigh-identical strategy would work even if we gave the other connectives explicitly). We abbreviate $\neg (A \rightarrow \neg B)$ as $A \wedge B$. For parenthesis-management we write $\wedge$ and $\rightarrow$ as right-associative, so that $A \wedge B \wedge C$ denotes $A \wedge (B \wedge C)$, while $A \rightarrow B \rightarrow C$ denotes $A \rightarrow (B \rightarrow C)$.

Consider the deductive system (called "our system" from here on) that has the following (nullary and unary) rules of inference.

Axiom rules

We call a formula a logical axiom if it occurs as a substitution instance of one of the following: $P \rightarrow (Q \rightarrow P), (P \rightarrow Q \rightarrow R) \rightarrow (P \rightarrow Q) \rightarrow P \rightarrow R, (\neg Q \rightarrow \neg P) \rightarrow P \rightarrow Q$. Let $\varphi$ denote a logical axiom. We admit the following inference rules:

Infer $\varphi$.

From $C$ infer $\varphi \wedge C$.

From $C$ infer $C \wedge \varphi \wedge \varphi$.

Modus ponens rules

From $C \wedge D \wedge (A \wedge (A \rightarrow B) \wedge E)$ infer $C \wedge D \wedge (A \wedge (A \rightarrow B) \wedge B \wedge E)$.

From $C \wedge D \wedge ((A \rightarrow B) \wedge A \wedge E)$ infer $C \wedge D \wedge ((A \rightarrow B) \wedge A \wedge B \wedge E)$.

Shunting rules

From $(A \wedge C) \wedge D \wedge E$ infer $C \wedge (A \wedge D) \wedge E$.

From $(A \wedge C) \wedge D \wedge E$ infer $C \wedge D \wedge (A \wedge E)$.

From $C \wedge (A \wedge D) \wedge E$ infer $(A \wedge C) \wedge D \wedge E$.

From $C \wedge D \wedge (A \wedge E)$ infer $(A \wedge C) \wedge D \wedge E$.

Conjunction elimination

From $A \wedge B$ infer $A$.

Our system clearly satisfies soundness for propositional classical logic. It also satisfies completeness: we prove this by reducing the completeness of our system to that of Hilbert's proof calculus.

Lemma. Given a derivation of length $n$,

{1) $Q_1$

(2) $Q_2$

(3) $\dots$

(n) $Q_n$

in the Hilbert calculus, we can find a derivation of $Q_n \wedge \dots \wedge Q_2 \wedge Q_1$ in our system.

Proof. By induction on the length of the Hilbert calculus derivation $\delta$. If the derivation has length 1, then $Q_1$ is a substitution instance of an axiom $\varphi$, so we can use the first axiom rule of our system to prove $Q_1$. From here on assume that the derivation has length $n+1$. By induction hypothesis, our system has a derivation of $Q_n \wedge \dots \wedge Q_1$. We have two cases to consider.

Case 1: The last rule of the derivation $\delta$ is an axiom rule of the Hilbert system. In this case $Q_{n+1}$ is a substitution instance of an axiom, and from $Q_n \wedge \dots \wedge Q_1$ we can infer $Q_{n+1} \wedge Q_n \wedge \dots \wedge Q_1$ using the second axiom rule of our system.

Case 2: The last rule of the derivation $\delta$ is a modus ponens rule of the Hilbert system, inferring $Q_{n+1}$ from $Q_k$ and $Q_\ell$ (w.lo.g. assume $k > \ell > 1$). Take your favorite axiom $\varphi$, then argue in our system as follows:

Have $Q_n \wedge \dots \wedge Q_1$ by induction hypothesis.

Infer $(Q_n \wedge \dots \wedge Q_1) \wedge \varphi \wedge \varphi$ using the third axiom rule.

Infer $(Q_k \wedge \dots \wedge Q_1) \wedge (Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge \varphi$ by using the first shunting rule repeatedly.

Infer $(Q_{k-1} \wedge \dots \wedge Q_1) \wedge (Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_k \wedge \varphi)$ by using the second shunting rule.

Infer $(Q_\ell \wedge \dots \wedge Q_1) \wedge (Q_{\ell + 1} \wedge \dots \wedge Q_{k-1} \wedge Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_k \wedge \varphi)$ by using the first shunting rule repeatedly.

Infer $(Q_{\ell-1} \wedge \dots \wedge Q_1) \wedge (Q_{\ell + 1} \wedge \dots \wedge Q_{k-1} \wedge Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_\ell \wedge Q_k \wedge \varphi)$ using the second shunting rule.

Infer $(Q_{\ell-1} \wedge \dots \wedge Q_1) \wedge (Q_{\ell + 1} \wedge \dots \wedge Q_{k-1} \wedge Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_\ell \wedge Q_k \wedge Q_{n+1} \wedge \varphi)$ using the relevant modus ponens rule.

Infer $(Q_{\ell} \wedge \dots \wedge Q_1) \wedge (Q_{\ell + 1} \wedge \dots \wedge Q_{k-1} \wedge Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_k \wedge Q_{n+1} \wedge \varphi)$ using the fourth shunting rule.

Infer $(Q_{k-1} \wedge \dots \wedge Q_1) \wedge (Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_k \wedge Q_{n+1} \wedge \varphi)$ using the third shunting rule repeatedly.

Infer $(Q_{k} \wedge \dots \wedge Q_1) \wedge (Q_{k+1} \wedge \dots \wedge Q_n \wedge \varphi) \wedge (Q_{n+1} \wedge \varphi)$ using the fourth shunting rule.

Infer $(Q_{n} \wedge \dots \wedge Q_1) \wedge \varphi \wedge (Q_{n+1} \wedge \varphi)$ using the third shunting rule repeatedly.

Infer $(Q_{n+1} \wedge \dots \wedge Q_1) \wedge \varphi \wedge \varphi$ using the fourth shunting rule.

Infer $Q_{n+1} \wedge \dots \wedge Q_1$ using conjunction elimination.

Qed.

As a corollary, we get completeness for our system.

Proof. Take a classical tautology $P$. By completeness for the Hilbert calculus, we can find a derivation $\delta$ of $P$ in the Hilbert calculus. By our previous lemma, we can find a derivation of $P \wedge Q_n \wedge \dots \wedge Q_1$ for some $n \in \mathbb{N}$ in our system. Using conjunction elimination, we can infer $P$ in our system. Qed.


问:
Game on $[0;1]$ about repeating decimal
Two players play the following game. Before the beginning $"0."$ is written on the board. The first player writes any (finite) sequence of digits. The second one then writes only one digit. Then the players take turns. The first player wins if the number on the board is of repeating decimal type. The question is, which of players has a winning strategy.

My guess is that the first player can win despite the turns of the second player. I tried to combine facts that for any $m$ set of numbers $0.1a_1a_2\dots a_m$ is open, and that repeating decimals are dense in $[0;1]$, but I can't explicitly show the winning strategy.

Any help would be appreciated. Thanks!

答: The second player will always have a winning strategy if the "target" set ($\mathbb{Q}$ in this case) is countable: If that set is $\{x_n\}_{n\in\mathbb{N}}$ then the second player only has to choose, on the $n^{th}$ round, a digit that makes the decimal expansion of the number constructed up to that point different from the decimal expansion of $x_n$. To avoid ambiguities, the digit can be limited to just two values, say 1 and 2 (so there's no issue with non unique decimal expansions)

Title: Game on $[0;1]$ about repeating decimal


问:
Vanishing criterion for sections of module on a product
This queston can be regarded as a variant of this one:

Does a section that vanishes at every point vanish?

Let $X,Y$ be varieties, $M$ be an $\mathcal{O}_X$-Module and
$$\pi:X\times Y\rightarrow X$$
be the projection. Let $s$ be a local section of $M$, which vanishes along each "horizontal line":

$$\forall y: i_y^* s=0$$

where denotes the map $i_y:x\mapsto (x,y)$. Does it follow, that $s=0$?

答:
If you take $X$ to be a point, then this reduces to your previous question: you are just taking fibres at each point $y \in Y$, and so the answer is "no" for the same reason.

Just to see that this isn't just being pedantic, let me explain how to bootstrap the answer
to the previous question to give examples in this case which truly have a non-trivial product
structure.  For this, suppose that $Y = \mathbb A^1$ and that $X$ is the variety attached to
the ring $A$, so that $X \times Y$ corresponds to the ring $A[t]$.  Let $M$ be the module
$A[t]/t^2,$ and let $s$ be the section $t \bmod t^2$.  Then $s$ will vanish along each "horizontal line", but is non-zero.  (The case $X = $ a point gives the counterexample to
your previous question.)

Again, if $M$ is torsion free and finitely generated then you should be okay.

ContentLicense: CC BY-SA 2.5


问:
Distribution of the maximum of noisy random variables
Let $X_1, \ldots, X_N$ be $N$ hidden random iid variables, all with the same standard distribution, let's say uniform $\mathcal{U}(0, 1)$ or Gaussians $\mathcal{N}(0, 1)$ (probably easiest).
I observe $N$ corresponding 'noisy' variables $Z_n = X_n + \mathcal{N}(0, \sigma^2)$.
I know how to derive the distribution of $X^* = \max({X_1, \ldots, X_N})$ (and correspondingly $Z^*$ if the $X$'s are Gaussians).
What I would like to know is how to compute the distribution (or at least the expectation) of $X_{\mathrm{argmax}_n(Z_n)}$.

Intuitively if $\sigma^2$ is small my observed variables will closely follow the hidden ones, and the distribution will be close to $X^*$, while if it is big, they will be dominated by the noise, and the distribution will be the original one of the $X$'s.

答:
Let's say $Z_n = X_n + Y_n$ with $X_n \sim N(0,1)$, $Y_n \sim N(0,\sigma^2)$,
and all $X_n$ and $Y_n$ independent.  Let $W = X_{{\rm argmax}_n Z_n} = \sum_n X_n \prod_{j \ne n} I_{Z_n > Z_j}$.  Thus $E[W] = N E[X_1 \prod_{j > 1} I_{Z_1 > Z_j}$.
Now $Z_n \sim N(0,1+\sigma^2)$. Moreover $X_1$ and $I_{Z_1 > Z_j}$ are conditionally independent given $Z_1$, so
$E[W] = N E[E[X_1 | Z_1] \prod_{j > 1} E[I_{Z_1 > Z_j}|Z_1]]
= N E[Z_1 \Phi(Z_1/\sqrt{1+\sigma^2})^{N-1}$.  This is $N \sqrt{1+\sigma^2} E[Z \Phi(Z)^{N-1}]$ where $Z \sim N(0,1)$.</p>


问:
Show that $\mathbb{Z}[\sqrt{3}]$ is dense in $\mathbb{R}.$
Here is the question I want to answer:

Show that $\mathbb{Z}[\sqrt{3}]$ is dense in $\mathbb{R}.$

I got this hint:

Hint: Find $u \in \mathbb{Z}[\sqrt{3}]^*$ with $u > 1$ and consider $lim_{n\rightarrow \infty} u^{-n}.$

My question is:

How can we find this $\zeta$? how can I prove that it is a unit? and why we are taking it from the set of units?should this unit be greater than 1 or less than 1?

答:
You don't really need a unit $\zeta$, just an element $\zeta$ such that $0 < \zeta < 1$.  You can start with any element $r$ in your ring that is not integral. Then $\zeta =\{r\} = r - [r]$ is in the interval $(0,1)$, and can be used in your nice argument.

$\bf{Added:}$  For example, consider $r=\sqrt{3}$. We have $[r]=1$.  Take $\zeta= \sqrt{3}-1$, in $(0,1)$, but not a unit ( since its norm $(\sqrt{3}-1)(\sqrt{3}+1)=2>1$).


问:
Which family of sets is a $\sigma$-algebra?
Which family of sets is a $\sigma$-algebra?

a) Subsets $X$ of $\mathbb {N}$, for $X$: $X$ is finite or $\mathbb {N} \setminus X$ is finite.

b) At most countable subsets of $\mathbb {Q}$.

c) At most countable subsets of $[-1,1]$.

答:
a) Each $\{2n\}$ is finite, but $\bigcup_{n=1}^\infty\{2n\}$ is the set of even numbers, which neither is finite nor is $\Bbb N\setminus(\bigcup_{n=1}^\infty\{2n\})$ the set of odd numbers.

b) Every subset of $\Bbb Q$ is at most countable, since $\Bbb Q$ is countable itself. Then we actually have the power set $\mathcal P(\Bbb Q)$, which is always a $\sigma$-algebra.

c) $\Bbb Q\cap[-1,1]$ is countable but $[-1,1]\setminus(\Bbb Q\cap[-1,1])$ is the set of irrationals in $[-1,1]$, which is uncountable.


问:
Surface with non-zero mean curvature means orientable
Let $M$ be a surface in $\Bbb R^3$ with non-zero mean curvature for every point. How could I  show that this implies that $M$ is orientable? By our definition, orientable means that an unitary, normal vector can be defined continuously for every point in the surface.

This is a homework question, so just hints would be appreciated :)

答: Think through all the bits and pieces required to define mean curvature.


问:
Indefinite symmetric quadratic form with inequality constraint
I have a scalar quantity $\Phi =u^TAu$ that I am tracking as part of a greater optimization problem.
$A=I-(1+\mu^2)nn^T$, where $u$ is a 3 x 1 vector, $I$ is the 3 x 3 identity matrix, $\mu$ is a non-negative scalar, and $n$ is a normalized 3 x 1 vector ($n^Tn=1$). The eigenvalues of $A$ are $(1,1,-\mu^2)$. Therefore, $A$ is an indefinite, symmetric matrix. Everything is real-valued.

I wish to constrain my choice of $u$ such that $\Phi\leq0$. Because $A$ is indefinite, this condition is dependent on $u$. How can I enforce this constraint? I was hoping there was a way to define a matrix $X$ such that $u=Xv$, to transform this into an unconstrained optimization problem (similar to a subspace method) but I will appreciate any advice. Thank you!

答:
If you project $u$ onto the eigenspace of $A$ corresponding to eigenvalue $1$, the feasible region is a circle whose radius depends on the remaining component of $u$. So, you can either solve a problem with a quadratic constraint, or do a polar transformation.

An eigenvector of $A$ corresponding to eigenvalue $-\mu^2$ is $n$. You can write $u = Xw$ where $X$ is $3 \times 3$ whose first column is $n$ and the other two columns an orthogonal basis for the space orthogonal to $n$, and $w$ is $3 \times 1$.

The condition $u^TAu \leq 0$ is equivalent with $w_2^2 + w_3^2 \leq w_1^2 \mu^2$. Writing $w_1 = r \cos \theta$ and $w_2 = r \sin \theta$, this becomes $0 \leq r \leq w_1 \mu$.

So now we have three options:

use $u^T A u \leq 0$ as a constraint

use $w_2^2 + w_3^2 \leq w_1^2 \mu^2$ and $u = Xw$ as constraints

reformulate the entire problem with polar coordinates and add the linear constraint $0 \leq r \leq w_1 \mu$

The constraint $w_2^2 + w_3^2 \leq w_1^2 \mu^2$ is convex (second order cone) only if $w_1 \geq 0$ or if $w_1 \leq 0$, but not for $w_1 \in \mathbb{R}$. So, if you can solve two constrained convex optimization problems (one for $w_1 \geq 0$, one for $w_1 \leq 0$), option 2 probably has the best numerical performance.


问:
Orthogonal Linear Transformation of Subspace
given that U is an orthogonal n by n matrix, and linear transformation T map from $R^n$ to $R^n$ be $T(x) = Ux$. Let W be a subspace of $R^n$ such that $T(W)  \subseteq   W$. Prove $T(W) = W$ and $T(W^\perp) = W^\perp$.
Here is my attempt:

For the first question, I constructed an orthogonal basis U for W, and showed that $T(U)$ is orthogonal and linearly independent with dimension dim(W).

Any hint will be appreciated. Thanks in advance.

答:
If $w\in T(W^\bot)$ then there is some $v\in W^\bot$ such that $T(v)=Uv=w$. Take any $u\in W$.

Orthogonal matrices preserve inner product, just note that $\langle a,b\rangle=b^\mathsf t a=b^\mathsf tU^{-1}Ua=b^{\mathsf t}U^{\mathsf t}Ua=(Ub)^{\mathsf t}(Ua)=\langle Ua,Ub\rangle$,

and $U^{-1}u\in W$, since

 $T^{-1}(T(W))=W$ because of $T$ being bijective ($U$ is invertible),

so we have that

$0=\langle U^{-1}u,v\rangle=\langle U(U^{-1}u),Uv\rangle=\langle u,w\rangle$,

so $w\in W^\bot$, and therefore $T(W^\bot)\subseteq W^\bot$.

You can do something similar for the other inclusion, and then you get the equality.


问:
Free module is a right $R$-module with scaling operation.
Let $R$ be a ring and ruppose $F$ is a set of all functions from $S$ to $R$. Then, every element $u\in S$ can be viewed as a function from $S$ to $R$:
$$u(t)=\begin{cases}
1 \hbox{ if } t=u\\
0 \hbox{ if } t\neq u
\end{cases}$$
So $S$ is a subset of $F$.

Note that $F$ is an abelian group with $$[f+g](t):=f(t)+g(t)$$ for all $f,g\in F$, and $t\in S$.

I'm trying to prove that $F$ is a right $R$-module with scaling operation $$[f\cdot r](t):=f(t)r$$ for all $f\in F$, $r\in R$ and $t\in S$, but I'm having trouble how to show this.

答:
To show the required claim we have to check that for any $f,g\in F$, $r,s\in R$, and $t\in S$ we have

1)) $[(f+g)\cdot r](t)=([f\cdot r]+ [g\cdot r])(t)$

2)) $[f\cdot (r+s)](t)=([f\cdot r]+ [f\cdot s])(t)$.

For (1) we have

$[(f+g)\cdot r](t)=([f+g](t))r=(f(t)+g(t))r=f(t)r+g(t)r=[f\cdot r](t)+[g\cdot r](t)=([f\cdot r]+ [g\cdot r])(t)$

For (2) we have

$[f\cdot (r+s)](t)=f(t)(r+s)=f(t)r+f(t)s=[f\cdot r](t)+ [f\cdot s](t)=([f\cdot r]+ [f\cdot s])(t)$.


问:
Substituting in values in a cubic equation
I am stuck on the following problem. I also mean to say greet you with a Hello! but for some reason it won't save it haha.

This problem was part of my homework, which is now past due and I have received partial credit for. The answer was released and I am trying to understand it.

I am completely stuck on the "Solving the equation $1+ \frac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 $ for $a$" part. No matter how I break it down, I can't get that answer. Also, it seems like the $a=-6$ as the coefficient of the $x^2$ term has nothing to do with this part of the problem. Can someone explain please?

答:
It is confusing for them to use $a$ in the hint, since it coincides with the coefficient of the $x^2$ term, and has nothing to do with the purpose of the hint.

Now the hint was

\begin{align}1+ \frac{2\sqrt{21}}{9} &= (3a+a\sqrt{21})^3 \\&=a^3(3+\sqrt{21})^3
\\&= a^3(3^3+3\cdot3^2\sqrt{21}+3\cdot3\sqrt{21}^2+\sqrt{21}^3)
\\&=a^3(27+27\sqrt{21}+189+21\sqrt{21})
\\&=a^3(216+48\sqrt{21})
\\\leadsto a^3=\frac 19\frac{9+2\sqrt{21}}{216+48\sqrt{21}}&=\frac19\frac1{24}=\frac1{216}\end{align} (Factor the denominator and the numerator appears so a cancellation takes place)

which gives $a=\dfrac16$. Hence $1+ \dfrac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 = \left(\dfrac12+\dfrac16\sqrt{21}\right)^3$.

This hint is only used to simplify the root

$$u=\sqrt[3]{1+ \frac{2\sqrt{21}}{9}}$$


问:
Variation of inverse trigonometric function with intervals
How can I prove this property

$2\sin^{-1}x =  \begin{cases}\sin^{-1}(2x \sqrt{1-x^2})&\text{for }-\frac1{\sqrt2}<x<\frac1{\sqrt2} \\
\pi- \sin^{-1}(2x\sqrt{1-x^2})&\text{for }\frac1{\sqrt2}<x<1 \\
-\pi-\sin^{-1}(2x\sqrt{1-x^2})&\text{for }-1<x<-\frac1{\sqrt2}\end{cases}$

The first one is quite easy but how to proceed with other two

答:
For the second case,

$\begin{align}\frac1{\sqrt2}<x<1 &\Rightarrow \frac\pi4<\sin^{-1}x<\frac\pi2 \\&\Rightarrow \frac\pi2<2\sin^{-1}x < \pi \\&\Rightarrow 0<\pi-2\sin^{-1}x <\frac{\pi}{2}\end{align}$.

Now let $\theta =\pi-2\sin^{-1}x $ and it lies in the range $(0,\frac\pi2)$. This is done to make the angle lie in the principal range $[-\frac\pi2,\frac\pi2]$ of the arcsine function.

$\begin{align}&\Rightarrow x = \sin\frac{\pi-\theta}{2 } = \cos\frac\theta2\\
&\Rightarrow 2x\sqrt{1-x^2} = 2\cos\frac\theta2\sin\frac\theta2 = \sin\theta \\
&\Rightarrow  2x\sqrt{1-x^2}= \sin(\pi-2\sin^{-1}x) \color{blue}{\text{ and the angle lies in the principal range of arcsine }}\\
&\Rightarrow\boxed{2\sin^{-1}x = \pi-\sin^{-1}(2x\sqrt{1-x^2})}\end{align}$

Similarly do with the third case, making the angle lie within the required range.


问:
Diagonalizable iff $V$ is the direct sum of eigenspace
Here comes the fundamental definition of diagonalizable : $A=PDP^{-1}$
And we can prove that $A$ is diagonalizable iff $A$ has $n$ linear independent eigenvectors.
But here is another theorem saying that A is diagonalizable iff $V$ decomposes as a direct sum of eigenspaces, i.e. $V=V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_k}$ for $dim V=n$, but I am not sure whether the k here equals to n, because the basis of each eigenspace $V_{\lambda_i}$ may contain more than one element, so it seems that $k\leq n$.
Or is it that for the diagonalizable cases, we have $n$ distinct eigenvalues, so the cardinality of basis of each eigenspace is exactly one? Could anyone help me? Thank you

答:
If n distinct eigenvalues are there then diagonalizability follows.
But matrices with repeated eigenvalues can also be diaganozibale.

If $V$ decomposes into direct sum of eigenspaces, then take any basis for each eigen space and then their union. This union will have exactly n elements and will consist of eigen vectors, proving diagonalizability. We can see that dimension of individual eigenspaces do not matter. The main thing all these eigen spaces together GENERATE the whole of V; in general they may not.


问:
ordinal spaces and order topology
What is the difference between general order topology and ordinal space topology?
I'm reading the first answer in Is every locally compact Hausdorff space paracompact?, and I don't understand why $E = [0,\omega) \times \{ \omega_1 \}$ and $F = \{ \omega \} \times [0,\omega_1)$ are closed. Every basis element containing $ \omega \times \omega_1 \in X - E  $ must be the form of $(a,\omega] \times (c, \omega_1]$ in the order topology and since there must be an element of $[0,\omega)$ in $(a,\omega]$, $(a,\omega] \times (c, \omega_1]$ must intersect $E$. Then why is $ X - E$ open?

答:
$X\setminus E=[0,\omega)\times[0,\omega_1]$; this is the product of an open set in $[0,\omega]$ and an open set in $[0,\omega_1]$, so it’s open in $[0,\omega]\times[0,\omega_1]$ and therefore open in the subspace $X$. $E$ is therefore the complement of an open subset of $X$, so $E$ is closed in $X$. The proof that $F$ is closed in $X$ is similar.

The point $\langle\omega,\omega_1\rangle$ is altogether irrelevant, since it’s not in $X$.

Title: ordinal spaces and order topology


问:
How to solve $\frac{1}{\kappa}\tanh^{-1}(\kappa t)+bt=c$ for $t$
I want to solve this equation for $t$:
$$
\frac{1}{\kappa}\tanh^{-1}(\kappa t)+bt=c,
$$
given that $\kappa$ is a constant, and $b$ and $c$ are functions not in terms of $t$.

I tried to convert it into the form of natural logarithms and transform it with the power of $e$ like below:
$$
e^{bt}+\kappa te^{bt}+\kappa te^c=e^c
$$

However, $t$ lies on both power of $e$ and the linear position, and I have been stuck at extracting $t$ for hours.

How can I get the solution to the equation? Is there any clever way to extract the solution?

答:
$$\frac{1}{\kappa}\tanh^{-1}(\kappa t)+bt=c\tag 1$$ is an highly transcendental equation and then you will require some numerical method for obtainin the solution.

In fact, there is an analytical solution. Rewrite $(1)$ as
$$\kappa t=\tanh (\kappa  (c-b t)) \tag 2$$
Let $x=\kappa  (c-b t)$ to make
$$\frac{c \kappa -x}{b}=\tanh(x)= \frac {e^{2x}-1 } {e^{2x}+1 } \tag 3$$ which leads to
$$e^{-2x}=-\frac{x+(b-c \kappa)}{x-(b+c \kappa)} \tag 4$$ which has solutions in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).

This being said, it is not very useful from a practical point of view.

Just think about numerical methods.

Edit

Equation $(2)$ can rewrite and look for the zero of function
$$f(x)=x+b\tanh(x)-c \kappa\qquad \text{where} \qquad x=\kappa  (c-b t)$$

Assuming $b>0$ and $c \kappa >0$, we face an increasing function since
$$f'(x)=1+b \,\text{sech}^2(x)\quad > 0 \quad \forall x$$ Starting Newton method with $x_0=0$ will give as an estimate
$$x_1=\frac{c \kappa }{b+1}$$

Trying with $b=\pi$ and $c \kappa=12.345$, Newton iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0 \\
 1 & 2.9807374 \\
 2 & 9.0248926 \\
 3 & 9.2034074
\end{array}
\right)$$

Doing the same with $b=\frac 1\pi$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0 \\
 1 & 9.3642626 \\
 2 & 12.026690
\end{array}
\right)$$


问:
prove , if $ \frac {dy}{dx} = 0 $ then $ y = c$ for some constant $c$ .
we all know that if $ y = c $ for some constant $c$ then , $ \frac{dy}{dx} = 0 $

but how can we prove the other way ? , i mean , how can we

prove that , if $ \frac {dy}{dx} = 0 $ then $ y = c$ for some constant $c$ .

that is not a homework . it's just a question i want to know its answer !

i hope that you take my question seriously :)

so , can anyone give us the simplest prove ?!!

thanx for all

答: If $f'=0$, then for every $x,y$ we have $\displaystyle f(y)=f(x)+ \int_x^yf'(t)dt=f(x)$. Therefore, $f$ is constant.</p>


问:
Problem : $(1+z)^n=(1-z)^n$ $\Rightarrow $ $Re(z) =0$
I am trying, but I found $im(z) =0$

I want to know where is my mistake

My attempt :

$(1+z)^n=(1-z)^n$

We put $z =e^{i\theta}$

$(1+e^{i\theta})^n=(1-e^{i\theta})^n$

$\Leftrightarrow $$(-e^{-i\pi} +e^{i\theta})^n=(-e^{-i\pi} - e^{i\theta})^n$

$\Leftrightarrow $$(e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}-e^{\frac{-i(\theta+\pi) }{2}}) ) ^n=(-((e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}+e^{\frac{-i(\theta+\pi) }{2}}) ) ^n$

$\Leftrightarrow $$(e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}-e^{\frac{-i(\theta+\pi) }{2}}) ) ^n=(e^{-\pi} ((e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}+e^{\frac{-i(\theta+\pi) }{2}}) ) ^n$

$\Leftrightarrow $ $ (sin (\frac{\theta +\pi}{2}) e^{i\frac{\theta}{2}}) ^n=(cos(\frac{\theta+\pi}{2})(e^{i\frac{\theta-\pi}{2} - \pi})^n$

$\Leftrightarrow $$-sin(\frac {\theta}{2}) =cos(\frac {\theta}{2}) $ and $e^{i\frac{\theta}{2}}=e^{i\frac{\theta-\pi}{2} - \pi}$

$e^{i\frac{\theta}{2}}=e^{i\frac{\theta-\pi}{2} - \pi}$$\Rightarrow $$cos(\frac {\theta}{2})  +isin(\frac {\theta}{2}) =cos(\frac{\theta-\pi}{2} - \pi) +isin(\frac{\theta-\pi}{2} - \pi)  $

$\Rightarrow $ $cos(\frac {\theta}{2})  +isin(\frac {\theta}{2}) =-cos(\frac {\theta-\pi}{2})  - isin(\frac {\theta-\pi}{2})$

$\Rightarrow $ $cos(\frac {\theta}{2})  +isin(\frac {\theta}{2})=-sin(\frac {\theta}{2}) +icos(\frac {\theta}{2})$

We know that:$-sin(\frac {\theta}{2}) =cos(\frac {\theta}{2}) $

So $2 isin(\frac{\theta}{2}))= 0$

$\Rightarrow $ $im(z) =0$

答:
\begin{gather*}
\left( 1+e^{i\theta }\right)^{n} =\left( 1-e^{i\theta }\right)^{n}\\
\frac{1+e^{i\theta }}{1-e^{i\theta }} =e^{i\frac{2k\pi }{n}}\\
e^{i\theta } =\frac{e^{i\frac{2k\pi }{n}} -1}{e^{i\frac{2k\pi }{n}} +1} =itan\left(\frac{k\pi }{n}\right)\\
\end{gather*}
Hence, z is purely imaginary.
Hope you find this method simpler and hope this answers your question.


问:
How to show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$
Given $f(x) < g(x)$ and $f(x), g(x), t \in[0,1]$, how can I show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$?

It is easy to show $f(x) \leq (1-t)f(x)+tg(x)$. Indeed, $(1-t)f(x)+tg(x) = f(x)-tf(x)+tg(x) = f(x)+t(g(x)-f(x))$. Given that $f(x) \leq g(x)$ it follows $f(x) \leq (1-t)f(x) +g(x).$

However, I am having difficulties to show $(1-t)f(x)+tg(x) \leq g(x)$. One argument would be to show at $t=0$ we have $(1-t)f(x) \leq g(x)$ and $t=1$ we will get $g(x) \leq g(x)$. But I am not sure if this is good enough since we need to check for $t \in (0,1)$.

I would like an answer that does not depend on the specific choices of $t$. When I showed $f(x) \leq (1-t)f(x)+tg(x)$, the argument did not depend on the choice of $t$ and some sort of squeeze theorem.

答:
$(1-t)f(x)+tg(x)\leq g(x) \iff (1-t)f(x) \leq (1-t)g(x) \iff f(x) \leq g(x).$

The $\iff$ is not quite since $(1-t)$ could possibly be $0$. It just outlines the idea.


问:
How to find a continuous path between matrices?

I have matrix $A=\begin{pmatrix} 0&-1&0\\1&0&0\\0&0&1\\\end{pmatrix}$ and matrix $B=\begin{pmatrix} 1&0&0\\0&\frac{1}{2}&-\frac{\sqrt{3}}{2}\\0&\frac{\sqrt{3}}{2}&\frac{1}{2}\\\end{pmatrix}$ and I want to find a continuous path between them. How do I do this?

I know that matrices in $SO(n)$ should be path connected but how do you actually construct such a path?

I'm not sure if just taking $\gamma(t)=A(t)+B(1-t)$ is going to work.

答:
You can check that any two dimensional rotation matrix can be written as $$R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} = e^{\theta J} \quad \text{where} \quad J = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}.$$

Using this fact, we have $$R_{\pi/2} = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = e^{(\pi/2)J} \quad \text{and} \quad R_{\pi/3} = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = e^{(\pi/3)J}.$$

Block-diagonal matrices $M = \begin{bmatrix}M_1 & 0 \\ 0 & M_2\end{bmatrix}$ satisfy $e^M = \begin{bmatrix}e^{M_1} &  \\  & e^{M_2}\end{bmatrix}$.

Therefore, $$A = \begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} = e^{C} \quad \text{where} \quad C = \begin{bmatrix} \tfrac{\pi}{2}J & \\ & 0 \end{bmatrix} = \dfrac{\pi}{2}\begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$ and $$B = \begin{bmatrix}1 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2} \\ 0 & \tfrac{\sqrt{3}}{2} & \tfrac{1}{2}\end{bmatrix} = e^{D} \quad \text{where} \quad D = \begin{bmatrix} 0 & \\ & \tfrac{\pi}{3}J \end{bmatrix} = \dfrac{\pi}{3}\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{bmatrix}.$$

Finally, you can check that for any anti-symmetric real matrix $H$, $e^H$ is in $SO(n)$. Can you use these facts to connect $A = e^C$ to $B = e^D$ by a continuous path in $SO(3)$? You have the right idea to use something like $t \cdot (\text{start point}) + (1-t) \cdot (\text{end point})$, but you need to tie in the matrix exponential somehow to guarantee the path stays in $SO(3)$.


答:
Alternatively, observe that $$(1+z)^n=(1-z)^n\implies |1+z|=|1-z|$$
Now, we square the equation and use the fact that $|a|^2 = a\overline{a}$ for $a\in\mathbb{C}$.
$$|1+z|^2=|1-z|^2\iff (1+z)(1+\overline{z})=(1-z)(1-\overline{z})\iff z+\overline{z} = 0=2\:\text{Re}(z)$$
The problem is solved.

Title: Problem : $(1+z)^n=(1-z)^n$ $\Rightarrow $ $Re(z) =0$

Body:
I am trying, but I found $im(z) =0$

I want to know where is my mistake

My attempt :

$(1+z)^n=(1-z)^n$

We put $z =e^{i\theta}$

$(1+e^{i\theta})^n=(1-e^{i\theta})^n$

$\Leftrightarrow $$(-e^{-i\pi} +e^{i\theta})^n=(-e^{-i\pi} - e^{i\theta})^n$

$\Leftrightarrow $$(e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}-e^{\frac{-i(\theta+\pi) }{2}}) ) ^n=(-((e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}+e^{\frac{-i(\theta+\pi) }{2}}) ) ^n$

$\Leftrightarrow $$(e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}-e^{\frac{-i(\theta+\pi) }{2}}) ) ^n=(e^{-\pi} ((e^{\frac{i\theta-\pi}{2}}(e^{\frac{i\theta+\pi}{2}}+e^{\frac{-i(\theta+\pi) }{2}}) ) ^n$

$\Leftrightarrow $ $ (sin (\frac{\theta +\pi}{2}) e^{i\frac{\theta}{2}}) ^n=(cos(\frac{\theta+\pi}{2})(e^{i\frac{\theta-\pi}{2} - \pi})^n$

$\Leftrightarrow $$-sin(\frac {\theta}{2}) =cos(\frac {\theta}{2}) $ and $e^{i\frac{\theta}{2}}=e^{i\frac{\theta-\pi}{2} - \pi}$

$e^{i\frac{\theta}{2}}=e^{i\frac{\theta-\pi}{2} - \pi}$$\Rightarrow $$cos(\frac {\theta}{2})  +isin(\frac {\theta}{2}) =cos(\frac{\theta-\pi}{2} - \pi) +isin(\frac{\theta-\pi}{2} - \pi)  $

$\Rightarrow $ $cos(\frac {\theta}{2})  +isin(\frac {\theta}{2}) =-cos(\frac {\theta-\pi}{2})  - isin(\frac {\theta-\pi}{2})$

$\Rightarrow $ $cos(\frac {\theta}{2})  +isin(\frac {\theta}{2})=-sin(\frac {\theta}{2}) +icos(\frac {\theta}{2})$

We know that:$-sin(\frac {\theta}{2}) =cos(\frac {\theta}{2}) $

So $2 isin(\frac{\theta}{2}))= 0$

$\Rightarrow $ $im(z) =0$